From 7c8f340dad36e16737374299f541160f7d508dee Mon Sep 17 00:00:00 2001 From: Sean Fitzpatrick Date: Mon, 29 Jun 2026 13:32:42 -0600 Subject: [PATCH 1/5] Update devcontainer.json Need new version with updated asymptote --- .devcontainer/devcontainer.json | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/.devcontainer/devcontainer.json b/.devcontainer/devcontainer.json index cb728e469..4fd49c2b2 100644 --- a/.devcontainer/devcontainer.json +++ b/.devcontainer/devcontainer.json @@ -13,9 +13,9 @@ // /////////////////////////////////////////////////////////////// { - "image": "pretextbook/pretext-full:1.10", // uses latest image from https://hub.docker.com/r/PreTeXtBook/pretext-full/tags + "image": "pretextbook/pretext-full:1.11", // uses latest image from https://hub.docker.com/r/PreTeXtBook/pretext-full/tags // If you don't need sagemath, you can use a smaller base image. Comment out the line above and uncomment the line below to use a smaller image. - // "image": "pretextbook/pretext:1.10", + // "image": "pretextbook/pretext:1.11", "features": {"ghcr.io/devcontainers/features/github-cli": {}}, // The pretext-full image above includes pretext, prefigure, and enough parts of latex and sagemath for most cases. Here we install additional dependencies. From e9c58a7636515d0dab9722cfe0019ef508bccb37 Mon Sep 17 00:00:00 2001 From: sean-fitzpatrick Date: Mon, 6 Jul 2026 14:49:35 -0600 Subject: [PATCH 2/5] fix some schema errors --- apex-assembled.xml | 210304 +++++++++++++++++++++++++++++++ apex-validation.txt | 30735 +++++ ptx/sec_conic_sections.ptx | 8 +- ptx/sec_deriv_intro.ptx | 11 +- ptx/sec_graph_sketch.ptx | 2 +- ptx/sec_hyperbolic.ptx | 2 +- ptx/sec_limit_intro.ptx | 6 +- ptx/sec_line_int_intro.ptx | 2 +- ptx/sec_multi_limit.ptx | 4 +- ptx/sec_optimization.ptx | 6 +- ptx/sec_polar.ptx | 2 +- ptx/sec_space_coord.ptx | 2 +- ptx/sec_stokes_divergence.ptx | 4 +- ptx/sec_substitution.ptx | 2 +- ptx/sec_taylor_poly.ptx | 17 +- ptx/sec_vector_fields.ptx | 2 +- 16 files changed, 241080 insertions(+), 29 deletions(-) create mode 100644 apex-assembled.xml create mode 100644 apex-validation.txt diff --git a/apex-assembled.xml b/apex-assembled.xml new file mode 100644 index 000000000..e040583ba --- /dev/null +++ b/apex-assembled.xml @@ -0,0 +1,210304 @@ + + + + + APEX + Exercise + Part + Discussion Questions + + + APEX + + + + A traditional calculus textbook with many exercises and few proofs, + covering calculus from limits to vector calculus. + + + + + + \newcommand\blank[2]{\,\colorbox{gray}{$\phantom{\rule{#1pt}{#2pt}}$}\,} + \newcommand{\highlight}[1]{{\color{blue}{#1}}} + \newcommand{\ds}{\displaystyle} + \newcommand{\fp}{f\hskip.75pt '} + \newcommand{\fpp}{f\hskip.75pt ''} + + % Leibniz notation + % Usage: \lz{y}{x} + \newcommand{\lz}[2]{\frac{d#1}{d#2}} + % + % higher Leibniz notation + % Usage: \lzn{n}{y}{x} + \newcommand{\lzn}[3]{\frac{d^{#1}#2}{d#3^{#1}}} + % + % Leibniz operator + % Usage: \lzo{x} + \newcommand{\lzo}[1]{\frac{d}{d#1}} + % + % Leibniz operator on .... + % Usage: \lzoo{x}{y} + \newcommand{\lzoo}[2]{{\frac{d}{d#1}}{\left(#2\right)}} + % + % higher Leibniz operator + % Usage: \lzon{n}{x}{y} + \newcommand{\lzon}[2]{\frac{d^{#1}}{d#2^{#1}}} + % + % Leibniz operator at .... + % Usage: \lzoa{y}{x}{a} + \newcommand{\lzoa}[3]{\left.{\frac{d#1}{d#2}}\right|_{#3}} + % + % partial Leibniz notation + % Usage: \plz{y}{x} + \newcommand{\plz}[2]{\frac{\partial#1}{\partial#2}} + % + % partial Leibniz operator at .... + % Usage: \plzoa{y}{x}{a} + \newcommand{\plzoa}[3]{\left.{\frac{\partial#1}{\partial#2}}\right|_{#3}} + + %Limit at infinity for sequences + %Usage: \inflim e^{-n} or \inflim[m] e^{-m} + \newcommand{\inflim}[1][n]{\lim\limits_{#1 \to \infty}} + + %Infinite Series with index starting n=1 + %Usage: \infser \frac{1}{n} or \infser[0] 0.5^n + \newcommand{\primeskip}{\hskip.75pt } + \newcommand{\infser}[1][1]{\sum_{n=#1}^\infty} + \newcommand{\Fp}{F\hskip.75pt '} + \newcommand{\Fpp}{F\hskip.75pt ''} + \newcommand{\yp}{y\hskip.75pt '} + \newcommand{\gp}{g\hskip.75pt '} + \newcommand{\dx}{\Delta x} + \newcommand{\dy}{\Delta y} + \newcommand{\ddz}{\Delta z} + \newcommand{\thet}{\theta} + \newcommand{\norm}[1]{\left\lVert#1\right\rVert} + \newcommand{\vnorm}[1]{\left\lVert\vec #1\right\rVert} + \newcommand{\snorm}[1]{\left|\left|\,#1\,\right|\right|} + \newcommand{\la}{\left\langle} + \newcommand{\ra}{\right\rangle} + \newcommand{\dotp}[2]{\vec #1 \cdot \vec #2} + \newcommand{\proj}[2]{\text{proj}_{\,\vec #2}{\,\vec #1}} + \newcommand{\crossp}[2]{\vec #1 \times \vec #2} + \newcommand{\veci}{\vec i} + \newcommand{\vecj}{\vec j} + \newcommand{\veck}{\vec k} + \newcommand{\vecu}{\vec u} + \newcommand{\vecv}{\vec v} + \newcommand{\vecw}{\vec w} + \newcommand{\vecx}{\vec x} + \newcommand{\vecy}{\vec y} + \newcommand{\vrp}{\vec r\hskip0.75pt '} + \newcommand{\vrpp}{\vec r\hskip0.75pt ''} + \newcommand{\vsp}{\vec s\hskip0.75pt '} + \newcommand{\vrt}{\vec r(t)} + \newcommand{\vst}{\vec s(t)} + \newcommand{\vvt}{\vec v(t)} + \newcommand{\vat}{\vec a(t)} + \newcommand{\px}{\partial x} + \newcommand{\py}{\partial y} + \newcommand{\pz}{\partial z} + \newcommand{\pf}{\partial f} + \newcommand{\unittangent}{\vec{{}T}} + \newcommand{\unitnormal}{\vec{N}} + \newcommand{\unittangentprime}{\vec{{}T}\hskip0.75pt '} + \newcommand{\R}{mathbb{R}} + \newcommand{\mathN}{\mathbb{N}} + \newcommand{\surfaceS}{\mathcal{S}} + \newcommand{\zerooverzero}{\displaystyle \raisebox{8pt}{\text{``\ }}\frac{0}{0}\raisebox{8pt}{\textit{ ''}}} + \newcommand{\abs}[1]{\left\lvert #1\right\rvert} + \newcommand{\sech}{\operatorname{sech}} + \newcommand{\csch}{\operatorname{csch}} + \newcommand{\curl}{\operatorname{curl}} + \newcommand{\divv}{\operatorname{div}} + \newcommand{\Hess}{\operatorname{Hess}} + + + + \usepackage[dvipsnames]{xcolor} + \usepackage{pgfplots} + \usepackage{tikz-cd} + \usetikzlibrary{positioning,matrix,arrows,arrows.meta} + + \usetikzlibrary{shapes,decorations,shadows,fadings,plotmarks} + \usepgfplotslibrary{fillbetween,decorations.softclip,polar} + + \usepackage{tkz-euclide} + %\usetkzobj{all} + %%% uncomment above line if on TeXLive 2017 or earlier %%% + + \definecolor{magenta}{rgb}{0.79216,0.12156,0.48236} + \colorlet{firstcolor}{blue} + \colorlet{secondcolor}{red} + \colorlet{thirdcolor}{magenta} + \colorlet{treestump}{brown} + \colorlet{treetop}{green!50!black} + \colorlet{yellowcolfill}{yellow} + \pgfplotsset{firstcurvestyle/.style={color=firstcolor,mark=none,thick,-,solid,fill=firstcolor!50,fill=none}} + \pgfplotsset{secondcurvestyle/.style={color=secondcolor,mark=none,thick,-,dashdotted,fill=secondcolor!50,fill=none}} + \pgfplotsset{thirdcurvestyle/.style={color=thirdcolor,mark=none,thick,-,dashdotdotted,fill=thirdcolor!50,fill=none}} + \pgfplotsset{areastyle/.style={fill,opacity=0.5,draw=none}} + \pgfplotsset{tangentline/.style={color=black,mark=none,thick,{Kite}-{Kite},solid}} + \pgfplotsset{tangentlineseg/.style={color=black,mark=none,thick,-,solid}} + \pgfplotsset{lineseg/.style={color=black,mark=none,solid}} + \pgfplotsset{normalline/.style={color=black,mark=none,thick,{Kite}-{Kite},dashed}} + \pgfplotsset{normallineseg/.style={color=black,mark=none,thick,-,dashed}} + \pgfplotsset{secantline/.style={color=black,mark=none,thin,{Kite}-{Kite},dashed}} + \pgfplotsset{secantlineseg/.style={color=black,mark=none,thin,-,dashed}} + \pgfplotsset{asymptote/.style={color=black,mark=none,thick,{Kite}-{Kite},dashed}} + \pgfplotsset{guideline/.style={color=black,mark=none,-}} + \pgfplotsset{symmetryline/.style={color=black,mark=none,-,dashed}} + \pgfplotsset{openinterval/.style={color=black,mark=none,ultra thick,shorten <=-2.4pt,shorten >=-2.4pt,{Parenthesis}-{Parenthesis}}} + \pgfplotsset{closedinterval/.style={color=black,mark=none,ultra thick,shorten <=-2.4pt,shorten >=-2.4pt,{Bracket}-{Bracket}}} + + \pgfplotscreateplotcyclelist{curvestylelist}{% + firstcurvestyle\\% + secondcurvestyle\\% + thirdcurvestyle\\% + } + + %To disable arrows, edit these and the above lines accordingly + \pgfplotsset{leftarrow/.style={{Kite}-}} + \pgfplotsset{rightarrow/.style={-{Kite}}} + + % Redefine these to be empty to turn off axis discontinuities + \pgfplotsset{xdiscontinuity/.style={axis x discontinuity=parallel}} + \pgfplotsset{ydiscontinuity/.style={axis y discontinuity=crunch}} + + \pgfplotsset{hollowdot/.style={color=firstcolor,fill=white,only marks,mark=*}} + \pgfplotsset{soliddot/.style={color=firstcolor,fill=firstcolor,only marks,mark=*}} + \pgfplotsset{squaremark/.style={color=black,fill=black,only marks,mark size=6pt,mark=square*}} + + \pgfplotsset{open/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle[open]}-{Circle[open]}}} + \pgfplotsset{openclosed/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle[open]}-{Circle}}} + \pgfplotsset{closed/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle}-{Circle}}} + \pgfplotsset{closedopen/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle}-{Circle[open]}}} + \pgfplotsset{infiniteopen/.style={shorten >=-2.4pt,{Kite}-{Circle[open]}}} + \pgfplotsset{openinfinite/.style={shorten <=-2.4pt,{Circle[open]}-{Kite}}} + \pgfplotsset{infiniteclosed/.style={shorten >=-2.4pt,{Kite}-{Circle}}} + \pgfplotsset{closedinfinite/.style={shorten <=-2.4pt,{Circle}-{Kite}}} + \pgfplotsset{infinite/.style={{Kite}-{Kite}}} + \pgfplotsset{infiniteleft/.style={{Kite}-}} + \pgfplotsset{infiniteright/.style={-{Kite}}} + \pgfplotsset{openleft/.style={shorten <=-2.4pt,{Circle[open]}-}} + \pgfplotsset{openright/.style={shorten >=-2.4pt,-{Circle[open]}}} + \pgfplotsset{closedleft/.style={shorten <=-2.4pt,{Circle}-}} + \pgfplotsset{closedright/.style={shorten >=-2.4pt,-{Circle}}} + + \pgfplotsset{every axis/.append style = { + cycle list name = curvestylelist, + %tick label style = {font = \scriptsize}, + axis x line = middle, + axis y line = middle, + xlabel = {$x$}, + ylabel = {$y$}, + %x label style = {font = \scriptsize}, + %y label style = {font = \scriptsize}, + minor x tick num = 1, + minor y tick num = 1, + %width = {ifthenelse(.9\linewidth>240pt,240pt,.9\linewidth)}, + name=myplot, + }} + + \pgfplotsset{numberline/.style = { + xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + axis y line=none, + every axis x label/.style={at={(current axis.right of origin)},anchor=west}, + y=15pt, + axis lines=middle, + enlarge x limits, + xlabel={}, + grid=none, + clip=false, + axis background/.style={}, + }} + %%temporary fix to get circles working right in TexLive 2019 + + \makeatletter + \def\pgfplots@install@path@replacements{% + \ifpgfplots@path@replace@ellipse + \let\tikz@do@circle=\pgfplots@path@@tikz@do@circle + \let\tikz@do@ellipse=\pgfplots@path@@tikz@do@circle + \expandafter\def\expandafter\pgfinterruptpicture\expandafter{\pgfinterruptpicture + \let\tikz@do@circle=\pgfplots@path@@tikz@do@circle@orig + \let\tikz@do@ellipse=\pgfplots@path@@tikz@do@ellipse@orig + }% + \fi + }% + \let\pgfplots@path@@tikz@do@circle@orig=\tikz@do@circle + \let\pgfplots@path@@tikz@do@ellipse@orig=\tikz@do@ellipse + + \let\pgfplots@path@@tikz@do@circle@oldandbroken=\pgfplots@path@@tikz@do@circle + \def\pgfplots@path@@tikz@do@circle#1{\pgfplots@path@@tikz@do@circle@oldandbroken{#1}{#1}} + \def\pgfplots@path@@tikz@do@ellipse#1#2{\pgfplots@path@@tikz@do@circle@oldandbroken{#1}{#2}} + \makeatother + + + + + + + import graph3; + bool incolor; + incolor = true; + pen apexmeshpen=rgb(0,0,.7); + pen blackmeshpen=rgb(0,0,0); + pen surfacepen=rgb(.6,.6,1)+opacity(.7); + pen surfacepen2=rgb(1,.6,.6)+opacity(1); + material simplesurfacepen=emissive(rgb(.6,.6,1)+opacity(0.7)); + material simplesurfacepen2=emissive(rgb(1,.6,.6)+opacity(0.7)); + material simplesurfacepen3=emissive(rgb(.5,.9,.5)+opacity(0.7)); + pen bluepen=blue; + pen bluemeshpen=rgb(0,0,.5); + pen bluecurvepen=rgb(.1,.1,.7); + pen dotblue=rgb(.6,.6,1); + pen redpen=red; + pen redmeshpen=rgb(.7,0,0); + pen redmeshpen2=rgb(.5,0,0); + pen redcurvepen=rgb(.9,0,0); + pen greenmeshpen=rgb(0,.5,0); + pen greencurvepen=rgb(0,.7,0); + pen curvepen=.4mm+bluepen; + pen curvepen2=.4mm+redpen; + pen darksurfacepen=rgb(.2,.2,1)+opacity(.7); + if(settings.outformat == "html") currentlight.background=opacity(0.0); + + + + + Key Idea + + + + APEX Calculus + + + + + + Gregory Hartman, Ph.D. + Department of Applied Mathematics + Virginia Military Institute + + + + Sean Fitzpatrick, Ph.D. + Department of Mathematics and Computer Science + University of Lethbridge + + + + Alex Jordan, Ph.D. + Department of Mathematics + Portland Community College + + + + Carly Vollet, M.S. + Department of Mathematics + Portland Community College + + + + + + Contributors to the 4th Edition + Jennifer Bowen, Troy Siemers, Brian Heinold, Dimplekumar Chalishajar + + + 5 + + + apexcalculus.com + + + + 2021 + Gregory Hartman + Creative Commons BY NC + + Licensed to the public under Creative Commons Attribution-Noncommercial 4.0 International Public License + + + + + + + + + + + + + Thanks +

+ There are many people who deserve recognition for the important role they have played in the development of this text. + First, I thank Michelle for her support and encouragement, + even as this project from work occupied my time and attention at home. + Many thanks to Troy Siemers, whose most important contributions extend far beyond the sections he wrote or the 227 figures he coded in Asymptote for 3D interaction. + He provided incredible support, advice and encouragement for which I am very grateful. + My thanks to Brian Heinold and Dimplekumar Chalishajar for their contributions and to Jennifer Bowen for reading through so much material and providing great feedback early on. + Thanks to Troy, Lee Dewald, Dan Joseph, Meagan Herald, Bill Lowe, John David, + Vonda Walsh, Geoff Cox, Jessica Libertini and other faculty of VMI who have given me numerous suggestions and corrections based on their experience with teaching from the text. + (Special thanks to Troy, Lee and Dan for their patience in teaching Calc III while I was still writing the Calc III material.) + Thanks to Randy Cone for encouraging his tutors of VMI's Open Math Lab to read through the text and check the solutions, + and thanks to the tutors for spending their time doing so. + A very special thanks to Kristi Brown and Paul Janiczek who took this opportunity far above and beyond what I expected, + meticulously checking every solution and carefully reading every example. + Their comments have been extraordinarily helpful. I am also thankful for the support provided by Wane Schneiter, + who as my Dean provided me with extra time to work on this project. + I am blessed to have so many people give of their time to make this book better. +

+ + + + + A Note on Using this Text +

+ Thank you for reading this short preface. + Allow us to share a few key points about the text so that you may better understand what you will find beyond this page. +

+ +

+ This text comprises a threevolume series on Calculus. + The first part covers material taught in many Calc 1 courses: + limits, derivatives, and the basics of integration, found in Chapters 1 through 6.1. + The second text covers material often taught in Calc 2: integration and its applications, + including an introduction to differential equations, + along with an introduction to sequences, series and Taylor Polynomials, + found in Chapters 5 through 8. + The third text covers topics common in Calc 3 or multivariable calc: + parametric equations, polar coordinates, vector-valued functions, and functions of more than one variable, + found in Chapters 10 through 15. + All three are available separately for free at apexcalculus.com, + and HTML versions of the book can be found at opentext.uleth.ca. +

+ +

+ These three texts are intended to work together and make one cohesive text, + APEX Calculus, which can also be downloaded from the website. +

+ +

+ Printing the entire text as one volume makes for a large, heavy, cumbersome book. + One can certainly only print the pages they currently need, but some prefer to have a nice, + bound copy of the text. Therefore this text has been split into these three manageable parts, + each of which can be purchased for about $15 at Amazon.com. +

+ + + + For Students: How to Read this Text +

+ Mathematics textbooks have a reputation for being hard to read. + Highlevel mathematical writing often seeks to say much with few words, + and this style often seeps into texts of lowerlevel topics. + This book was written with the goal of being easier to read than many other calculus textbooks, + without becoming too verbose. +

+ +

+ Each chapter and section starts with an introduction of the coming material, + hopefully setting the stage for why you should care, + and ends with a look ahead to see how the justlearned material helps address future problems. +

+ +

+

    +
  • Please read the text +

    + It is written to explain the concepts of Calculus. + There are numerous examples to demonstrate the meaning of definitions, + the truth of theorems, and the application of mathematical techniques. + When you encounter a sentence you don't understand, read it again. + If it still doesn't make sense, read on anyway, + as sometimes confusing sentences are explained by later sentences. +

    +
    • +
  • You don't have to read every equation +

    + The examples generally show all the steps needed to solve a problem. + Sometimes reading through each step is helpful; sometimes it is confusing. + When the steps are illustrating a new technique, + one probably should follow each step closely to learn the new technique. + When the steps are showing the mathematics needed to find a number to be used later, + one can usually skip ahead and see how that number is being used, + instead of getting bogged down in reading how the number was found. +

    +
    • +
  • Most proofs have been omitted +

    + In mathematics, proving something is always true is extremely important, + and entails much more than testing to see if it works twice. + However, students often are confused by the details of a proof, + or become concerned that they should have been able to construct this proof on their own. + To alleviate this potential problem, we do not include the proofs to most theorems in the text. + The interested reader is highly encouraged to find proofs online or from their instructor. + In most cases, one is very capable of understanding what a theorem means + and how to apply it without knowing fully why it is true. +

    +
    • +
+

+ + + + Interactive, 3D Graphics +

+ Versions 3.0 and 4.0 of the textbook include interactive, + 3D graphics in the PDF version. Nearly all graphs of objects in space can be rotated, + shifted, and zoomed in/out so the reader can better understand the object illustrated. + However, the only pdf viewers that support these 3D graphics are Adobe Reader Acrobat + (and only the versions for PC/Mac/Unix/Linux computers, not tablets or smartphones). +

+ +

+ The latest version of the book, which is authored in , + is available in HTML. + In HTML, the 3D graphics are rendered using WebGL, + and should work in any modern web browser. +

+ +

+ Interactive graphics are no longer supported within the PDF, + but clicking on any 3D graphic within the PDF + will take you directly to the interactive version on the web. +

+ + + + APEX <ndash/> Affordable Print and Electronic teXts +

+ APEX is a consortium of authors who collaborate to produce high quality, low cost textbooks. + The current textbookwriting paradigm is facing a potential revolution as desktop publishing and electronic formats increase in popularity. + However, writing a good textbook is no easy task, as the time requirements alone are substantial. + It takes countless hours of work to produce text, write examples and exercises, edit and publish. + Through collaboration, however, the cost to any individual can be lessened, + allowing us to create texts that we freely distribute electronically and sell in printed form for an incredibly low cost. + Having said that, nothing is entirely free; someone always bears some cost. + This text cost the authors of this book their time, and that was not enough. + APEX Calculus would not exist had not the Virginia Military Institute, + through a generous JacksonHope grant, + given the lead author significant time away from teaching so he could focus on this text. +

+ +

+ Each text is available as a free .pdf, protected by a Creative Commons Attribution - Noncommercial 4.0 copyright. + That means you can give the .pdf to anyone you like, print it in any form you like, + and even edit the original content and redistribute it. + If you do the latter, you must clearly reference this work and you cannot sell your edited work for money. +

+ +

+ We encourage others to adapt this work to fit their own needs. + One might add sections that are missing or remove sections that your students won't need. + The source files can be found at github.com/APEXCalculus. +

+ +

+ You can learn more at www.vmi.edu/APEX. +

+ + + + First <pretext/> Edition (Version 5.0) +

+ Key changes from Version 4.0 to 5.0: +

    +
  • +

    + The underlying source code has been completely rewritten, + to use the language, + instead of the original \LaTeX. +

    +
    • +
  • +

    + Using allows us to produce the books in multiple formats, + including HTML, which is both more accessible and more interactive than the original PDF. + HTML versions of the book can be found at opentext.uleth.ca. +

    +
    • +
  • +

    + The appendix on differential equations from the Calculus for Quarters + version of the book has been included as Chapter 8, just after applications of integration. + Chapters 8 14 are now numbered 9 15 as a result. +

    +
    • +
  • +

    + In the HTML version of the book, + many of the exercises are now interactive, and powered by WeBWorK. +

    +
    • +
+

+ +

+ Key changes from Version 3.0 to 4.0: +

    +
  • +

    + Numerous typographical and small mathematical corrections (again, thanks to all my close readers!). +

    +
    • +
  • +

    + Large mathematical corrections and adjustments. + There were a number of places in Version 3.0 where a definition/theorem was not correct as stated. + See www.apexcalculus.com for more information. +

    +
    • +
  • +

    + More useful numbering of Examples, Theorems, . + Definition 11.4.2 refers to the second definition of Chapter 11, Section 4. +

    +
    • +
  • +

    + The addition of Section 13.7: Triple Integration with Cylindrical and Spherical Coordinates +

    +
    • +
  • +

    + The addition of Chapter 14: Vector Analysis. +

    +
    • +
+

+ + + + A Brief History of Calculus +

+ Calculus means a method of calculation or reasoning. + When one computes the sales tax on a purchase, + one employs a simple calculus. + When one finds the area of a polygonal shape by breaking it up into a set of triangles, + one is using another calculus. + Proving a theorem in geometry employs yet another calculus. +

+ +

+ Despite the wonderful advances in mathematics that had taken place into the first half of the 17th century, + mathematicians and scientists were keenly aware of what they could not do. + (This is true even today.) In particular, + two important concepts eluded mastery by the great thinkers of that time: + area and rates of change. +

+ +

+ Area seems innocuous enough; + areas of circles, rectangles, parallelograms, + etc., are standard topics of study for students today just as they were then. + However, the areas of arbitrary + shapes could not be computed, + even if the boundary of the shape could be described exactly. +

+ +

+ Rates of change were also important. + When an object moves at a constant rate of change, + then \text{distance} = \text{rate}\times\text{time}. + But what if the rate is not constantcan distance still be computed? + Or, if distance is known, can we discover the rate of change? +

+ +

+ It turns out that these two concepts were related. + Two mathematicians, Sir Isaac Newton and Gottfried Leibniz, + are credited with independently formulating a system of computing that solved the above problems and showed how they were connected. + Their system of reasoning was a calculus. + However, as the power and importance of their discovery took hold, + it became known to many as the calculus. + Today, we generally shorten this to discuss calculus. +

+ + + + + Limits + +

+ The foundation of the calculus + is the limit. + It is a tool to describe a particular behavior of a function. + This chapter begins our study of the limit by approximating its value graphically and numerically. + After a formal definition of the limit, + properties are established that make + finding limits tractable. + Once the limit is understood, + then the problems of area and rates of change can be approached. +

+ + + + +
+ An Introduction To Limits +

+ We begin our study of limits + by considering examples that demonstrate key concepts that will be explained as we progress. +

+ + + +

+ Consider the function y = \frac{\sin(x) }{x}. + When x is near the value 1, what value (if any) + is y near? +

+ +

+ While our question is not precisely formed + (what constitutes near the value 1?), + the answer does not seem difficult to find. + One might think first to look at a graph of this function to approximate the appropriate + y values. + Consider , + where y = \frac{\sin(x) }{x} is graphed. + For values of x near 1, it seems that y takes on values near 0.85. + In fact, when x=1, then y=\frac{\sin(1) }{1} \approx 0.84, + so it makes sense that when x is near + 1, y will be near 0.84. +

+ + +
+ \sin(x)/x + + +

+ Graph of \sin(x)/x, shown for x between + -7 and 7, and y between 0 and + 1. The x intercepts are at + x=-2\pi, -\pi, \pi, and 2\pi, and a y intercept is at y = 1. + The graph has a downward curve for -\pi \lt x \lt \pi and an + upward curve for -2\pi \lt x \lt -\pi, and + \pi \lt x \lt 2\pi. The graph is undefined for x = 0. +

+ + + Graph of sin(x)/x, shows values for the domain -7 <=x <=7, undefined at x = 0. + + + \begin{tikzpicture}[ + declare function = {func(\x) = (\x != 0) * (sin(x*180/pi)/x) + (\x == 0) * (1);} + ] + \begin{axis}[ + ymin=-0.25, + ymax=1.2, + xmin=-7, + xmax=7, + ] + \addplot+[infinite,domain=-7:7,samples=50] {func(x)}; + \addplot[hollowdot] coordinates {(0,1)}; + \end{axis} + \end{tikzpicture} + + +
+ +
+ \sin(x)/x near x=1 + + + +

+ Graph of \sin(x)/x zoomed in on values where x is near 1. + This view of the graph shows x from 0.5 to 1.5. + The graph has only a slight downward curve. It shows that for x = 1, + \sin(x)/x is approximately 0.84 +

+ + + Graph of sin(x)/x zoomed in on values where x is near 1. Shows that for x = 1, sin(x)/x is approx. 0.84. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=.3, + xmax=1.6, + ymin=.4, + ymax=1.1, + xtick={0.5,1,1.5}, + ytick={0.6,0.8,1}, + xdiscontinuity, + ydiscontinuity, + grid=both, + ] + \addplot+[infinite,domain=0.5:1.5] {sin(x*180/pi)/x}; + \end{axis} + \end{tikzpicture} + + + +
+ + +

+ Consider this same function again at a different value for x. + When x is near 0, what value + (if any) + is y near? + By considering , + one can see that it seems that y takes on values near 1. + But what happens when x=0? + We have + + y \rightarrow \frac{\sin(0) }{0} \rightarrow {\genfrac{}{}{0pt}{0}{\text{“}}{}}\frac{0}{0}{\genfrac{}{}{0pt}{0}{\text{”}}{}} + . + The expression 0/0 has no value; it is + + limitindeterminate form + indeterminate form + + indeterminate. + Such an expression gives no information about what is going on with the function nearby. + We cannot find out how y behaves near x=0 + for this function simply by letting x=0. +

+ +
+ \sin(x)/x near x=0 + + + +

+ Graph of \sin(x)/x zoomed in on values where x is near 0. + The image shows the portion of the graph where x is from -1 to 1. + The graph has a downward curve and is symmetric about x=0. + The height of the graph approaches y = 1 when x is near 0. + A hollow dot at the point (0,1) shows that the function is undefined when x = 0; + that is, f(0) = undefined. +

+ + + Graph of sin(x)/x zoomed in on values where x is near 0. Shows that when x = 0 + sin(x)/x is undefined. + + + \begin{tikzpicture}[ + declare function = {func(\x) = (\x != 0) * (sin(x*180/pi)/x) + (\x == 0) * (1);} + ] + \begin{axis}[ + xmin=-1.1, + xmax=1.2, + ymin=.75, + ymax=1.05, + ytick={.8,.9,1}, + extra y ticks={.1,.3,...,.9}, + extra y tick labels={}, + ] + \addplot+[infinite,domain=-1:1] {func(x)}; + \addplot[hollowdot] coordinates {(0,1)}; + \end{axis} + \end{tikzpicture} + + + +
+ + + +

+ Finding a limit entails understanding how a function behaves near a particular value of + x. + Before continuing, it will be useful to establish some notation. + Let y=f(x); that is, + let y be a function of x for some function f. + The expression the limit of y as x approaches + 1 describes a number, often referred to as L, + that y nears as x nears 1. + We write all this as + + \lim_{x\to 1} y = \lim_{x\to 1} f(x) = L + . + This is not a complete definition (that will come in the next section); + this is a pseudo-definition that will allow us to explore the idea of a limit. + limitpseudo-definition +

+ +

+ Above, where f(x) = \sin(x)/x, we approximated + + \lim_{x\to 1} \frac{\sin(x) }{x} \approx 0.84 \quad \text{ and } \quad \lim_{x\to 0}\frac{\sin(x) }{x} \approx 1 + . + (We approximated these limits, + hence used the \approx symbol, + since we are working with the pseudo-definition of a limit, + not the actual definition.) +

+ + + +

+ Once we have the true definition of a limit, + we will find limits analytically; + that is, determining exact values using a variety of mathematical tools. + For now, we will approximate + limits both graphically and numerically. + Graphing a function can provide a good approximation, + though often not very precise. + Numerical methods can provide a more accurate approximation. + We have already approximated limits graphically, + so we now turn our attention to numerical approximations. +

+ +

+ Consider again \lim_{x\to 1}\frac{\sin(x)}{x}. + To approximate this limit numerically, + we can create a table of x and f(x) values where x is near 1. + This is done in . +

+ +

+ Notice that for values of x near 1, + we have \sin(x)/x near 0.841. + The x=1 row is included, + but we stress the fact that when considering limits, + we are not concerned with the value of the function at that particular x value; + we are only concerned with the values of the function when x is near 1. +

+ +
+ Values of \sin(x)/x with x near 1 + + + x + \sin(x)/x + + + 0.9 + 0.870363 + + + 0.99 + 0.844471 + + + 0.999 + 0.841772 + + + 1 + 0.841471 + + + 1.001 + 0.841170 + + + 1.01 + 0.838447 + + + 1.1 + 0.810189 + + +
+ +

+ Now approximate \lim_{x\to 0} \frac{\sin(x)}{x} numerically. + We already approximated the value of this limit as 1 graphically in + . + + shows the value of \sin(x)/x for values of x near 0. + Ten places after the decimal point are shown to highlight how close to 1 the value of + \sin(x)/x gets as x takes on values very near 0. + We include the x=0 row but again stress that we are not concerned with the value of our function at x=0, + only on the behavior of the function near 0. +

+ +
+ Values of \sin(x)/x with x near 0 + + + x + \sin(x)/x + + + -0.1 + 0.9983341665 + + + -0.01 + 0.9999833334 + + + -0.001 + 0.9999998333 + + + 0 + not defined + + + 0.001 + 0.9999998333 + + + 0.01 + 0.9999833334 + + + 0.1 + 0.9983341665 + + +
+ +

+ This numerical method gives confidence to say that 1 is a good approximation of + \lim_{x\to 0} \frac{\sin(x)}{x}; that is, + + \lim_{x\to 0} \frac{\sin(x)}{x} \approx 1 + . +

+ +

+ Later we will be able to prove that the limit is + exactly 1. +

+ + + +

+ We now consider several examples that allow us explore different aspects of the limit concept. +

+ + + Approximating the value of a limit + +

+ Use graphical and numerical methods to approximate + + \lim_{x\to 3} \frac{x^2-x-6}{6x^2-19x+3} + . +

+ + +

+ To graphically approximate the limit, graph + + y = \frac{x^2-x-6}{6x^2-19x+3} + + on a small interval that contains 3. + To numerically approximate the limit, + create a table of values where the x values are near 3. + This is done in + and , respectively. +

+ + +
+ Graphically approximating a limit in + + + +

+ Graph of f(x)=\frac{x^2 - x - 6}{6x^2 - 19x + 3}, + zoomed on values near x = 3, and showing the portion of the graph + for x from 2.5 to 3.5. +

+

+ There is a slight upward curve to the graph. + The graph suggests that the limit of the function + as x approaches 3 is 0.294. The graph also + shows that the function is undefined for x = 3. +

+ + + Graph of the function for this example, which shows that when x = 3, f(x) is undefined, + but near 0.294. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=2.3, + xmax=3.7, + ymin=0.2, + ymax=.35, + xdiscontinuity, + ydiscontinuity, + extra y tick labels={}, + grid=both, + ] + \addplot+[infinite,domain=2.5:3.45] {(x^2 - x - 6)/(6*x^2 - 19*x + 3)}; + \addplot [hollowdot] coordinates{(3,0.294)}; + \end{axis} + \end{tikzpicture} + + + +
+ +
+ Numerically approximating a limit in + + + x + \frac{x^2-x-6}{6x^2-19x+3} + + + 2.9 + 0.29878 + + + 2.99 + 0.294569 + + + 2.999 + 0.294163 + + + 3 + not defined + + + 3.001 + 0.294073 + + + 3.01 + 0.293669 + + + 3.1 + 0.289773 + + +
+ + +

+ The graph shows that when x is near 3, + the value of y is very near 0.3. + By considering values of x near 3, + we see that y=0.294 is a better approximation. + The graph and the table imply that + + \lim_{x\to 3} \frac{x^2-x-6}{6x^2-19x+3} \approx 0.294 + . +

+ + + + +

+ This example may bring up a few questions about approximating limits + (and the nature of limits themselves). +

    +
  1. +

    + If a graph does not produce as good an approximation as a table, + why bother with it? +

    +
    2. +
  2. +

    + How many values of x in a table are + enough? In the previous example, + could we have just used x=3.001 and found a fine approximation? +

    +
    3. +
+

+ +

+ Graphs are useful since they give a visual understanding concerning the behavior of a function. + Sometimes a function may act erratically + near certain x values which is hard to discern numerically but very plain graphically + (see ). + Since graphing utilities are very accessible, + it makes sense to make proper use of them. +

+ +

+ Since tables and graphs are used only to + approximate the value of a limit, + there is not a firm answer to how many data points are enough. + Include enough so that a trend is clear, and use values + (when possible) + both less than and greater than the value in question. + In , + we used both values less than and greater than 3. + Had we used just x=3.001, + we might have been tempted to conclude that the limit had a value of 0.3. + While this is not far off, we could do better. + Using values on both sides of 3 + helps us identify trends. +

+ + + Approximating the value of a limit + +

+ Graphically and numerically approximate the limit of + f(x) as x approaches 0, where + + f(x) = \begin{cases}x+1 \amp x\lt 0 \\ -x^2+1 \amp x \gt 0\end{cases} + . +

+ + +

+ Again we graph f(x) and create a table of its values near + x=0 to approximate the limit. + Note that this is a piecewise defined function, + so it behaves differently on either side of 0. + shows a graph of f(x), + and on either side of 0 it seems the y values approach 1. + Note that f(0) is not actually defined, + as indicated in the graph with the open circle. +

+ + +
+ Graphically approximating a limit in + + + +

+ Graph of the piecewise-defined function in . + For values of x \lt 0 the graph is straight + with a slope of 1 and for values of x \gt 0 the graph + curves downward. A hollow dot at the point (0,1) shows that at x = 0, + f(x) is undefined. + However, both parts of the graph, for x\lt 0 and for x\gt 0, + get close to the point (0,1) as x gets close to 0. +

+ + + Graph of the piecewise function, shows that at x = 0, y is undefined, + but is near 1. + + + \begin{tikzpicture}[ + declare function = {func(\x) = (\x < 0) * (x+1) + (\x > 0) * (-x^2+1);} + ] + \begin{axis}[ + xmin=-1.1, + xmax=1.2, + ymin=-.1, + ymax=1.1, + ] + \addplot+[infinite,domain=-1:1] {func(x)}; + \addplot[hollowdot] coordinates {(0,1)}; + \end{axis} + \end{tikzpicture} + + + +
+ +
+ Numerically approximating a limit in + + + x + f(x) + + + -0.1 + 0.9 + + + -0.01 + 0.99 + + + -0.001 + 0.999 + + + 0.001 + 0.999999 + + + 0.01 + 0.9999 + + + 0.1 + 0.99 + + +
+ + +

+ + shows values of f(x) for values of x near 0. + It is clear that as x takes on values very near 0, + f(x) takes on values very near 1. + It turns out that if we let x=0 for either piece + of f(x), 1 is returned; + this is significant and we'll return to this idea later. +

+ +

+ The graph and table allow us to say that \lim_{x\to 0}f(x) \approx 1; + in fact, we are probably very sure it equals 1. +

+ + + + + + Identifying When Limits Do Not Exist +

+ A function may not have a limit for all values of x. + That is, we cannot write that \lim_{x\to c}f(x)=L + (where L is some real number) + for all values of c, + for there may not be a number that f(x) is approaching. + There are three common ways in which a limit may fail to exist. + + limitdoes not exist +

    +
  1. +

    + The function f(x) may approach different values on either side of c. +

    +
    2. + +
  2. +

    + The function may grow without upper or lower bound as x approaches c. +

    +
    3. + +
  3. +

    + The function may oscillate as x approaches c + without approaching a specific value. +

    +
    4. +
+ We'll explore each of these in turn. +

+ + + + + Different Values Approached From Left and Right + +

+ Explore why \lim_{x\to 1} f(x) does not exist, where + + f(x) = \begin{cases}x^2-2x+3 \amp x\leq 1 \\ x \amp x \gt 1\end{cases} + . +

+ + +

+ A graph of f(x) around x=1 and a table are given in + Figures and + , respectively. + It is clear that as x approaches 1, + f(x) does not seem to approach a single number. + Instead, it seems as though f(x) approaches two different numbers. + When considering values of x less than 1 + (approaching 1 from the left), + it seems that f(x) is approaching 2; + when considering values of x greater than 1 + (approaching 1 from the right), + it seems that f(x) is approaching 1. + Recognizing this behavior is important; + we'll study this in greater depth later. + Right now, it suffices to say that the limit does not exist since + f(x) is approaching two different + values as x approaches 1. +

+ + +
+ Observing no limit as x\to 1 in + + + +

+ Graph of piecewise function in . + For values of x \leq 1 the graph + has a upward curve, and the graph ends at the point (1,2), + illustrating the fact that f(1)=2. +

+

+ For values of x \gt 1 the graph is a straight line with a positive slope. + Moving left to right, the line begins at the point (1,1), + at which there is a hollow dot, indicating that to the right of x=1, + the value of f(x) approaches 1. +

+

+ The most important feature of the graph is that it shows how f(x) + approaches two different values as x approaches 1, + depending on whether x\lt 1 or x\gt 1. +

+ + + Graph of the piecewise function, which shows that as x approaches 1 there + is no limit. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.1, + xmax=2.1, + ymin=-.1, + ymax=3.3, + ] + \addplot[firstcurvestyle,leftarrow, domain=0:1] {x^2 - 2*x + 3}; + \addplot[firstcurvestyle,rightarrow, domain=1:2] {x}; + \addplot[soliddot] coordinates {(1,2)}; + \addplot[hollowdot] coordinates {(1,1)}; + \end{axis} + \end{tikzpicture} + + + +
+ +
+ Values of f(x) near x=1 in + + + x + f(x) + + + 0.9 + 2.01 + + + 0.99 + 2.0001 + + + 0.999 + 2.000001 + + + 1.001 + 1.001 + + + 1.01 + 1.01 + + + 1.1 + 1.1 + + +
+ + + + + + + + The Function Grows Without Bound + +

+ Explore why \lim_{x\to 1} \frac{1}{(x-1)^2} does not exist. +

+ + +

+ A graph and table of f(x) = \frac{1}{(x-1)^2} are given in + + and , respectively. + Both show that as x approaches 1, + f(x) grows larger and larger. +

+ + +
+ Observing no limit as x\to 1 in + + + +

+ Graph of the function for . + The graph hows a horizontal + asymptote at y = 0 and a vertical asymptote at x = 1. + Because of the vertical asymptote at x = 1 the function + has no limit as x approaches 1. +

+ + + Graph of the function f(x), showing that as x approaches 1, there is a vertical asymptote. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.1, + xmax=2.1, + ymin=-1, + ymax=110 + ] + \addplot[firstcurvestyle,infinite,domain=0:0.9] {1/(x - 1)^2}; + \addplot[firstcurvestyle,infinite,domain=1.1:2] {1/(x - 1)^2}; + \addplot[asymptote,rightarrow] coordinates {(1,1) (1,100)}; + \end{axis} + \end{tikzpicture} + + + +
+ +
+ Values of f(x) near x=1 in + + + x + f(x) + + + 0.9 + 100. + + + 0.99 + 10000. + + + 0.999 + 1.\times 10^6 + + + 1.001 + 1.\times 10^6 + + + 1.01 + 10000. + + + 1.1 + 100. + + +
+ + +

+ We can deduce this on our own, + without the aid of the graph and table. + If x is near 1, then (x-1)^2 is very small, and: + + \frac{1}{\text{ very small number } } = \text{ very large number } + . +

+ +

+ Since f(x) is not approaching a single number, we conclude that + \lim_{x\to 1}\frac{1}{(x-1)^2} does not exist. +

+ + + + + The Function Oscillates + +

+ Explore why \lim_{x\to 0}\sin(1/x) does not exist. +

+ + +

+ Two graphs of f(x) = \sin(1/x) are given in . + + shows f(x) on the interval [-1,1]; + notice how f(x) seems to oscillate near x=0. + One might think that despite the oscillation, + as x approaches 0, + f(x) approaches 0. + However, zooms in on \sin(1/x), + on the interval [-0.1,0.1]. + Here the oscillation is even more pronounced. + Finally, in , + we see \sin(1/x) evaluated for values of x near 0. + As x approaches 0, + f(x) does not appear to approach any value. +

+ +
+ Observing that f(x)=\sin(1/x) has no limit as x\to0 in + +
+ + + + +

+ The graph of f(x)=sin(1/x) is shown, for x values between -1 and 1. + Like any sinusoidal graph, the curve oscillates back and forth between y=1 and y=-1. + However, as x gets close to 0, the argument of the sine function increases rapidly, + causing the distance between successive peaks to get smaller and smaller as the graph nears the y axis. + As x gets close to zero, the oscillations get so close together that it is no longer possible to distinguish them, + and the curve appears to become a solid, vertical strip. +

+ + + Graph of the function sin(1/x), showing oscillations that become so rapid near the origin that they blur together. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1.1, + xmax=1.1, + ymin=-1.1, + ymax=1.2, + ] + \addplot[firstcurvestyle,leftarrow, + domain=-1.018591636:-0.318309886, % 1/(15/48 pi) to 1/(48/48 pi) + samples=34, % sample every pi/48 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=-0.318309886:-0.106103295, % 1/pi to 1/(3 pi) + samples=25, % sample every pi/12 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=-0.106103295:-0.063661977, % 1/(3 pi) to 1/(5 pi) + samples=49, % sample every pi/24 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=-0.063661977:-0.045472841, % 1/(5 pi) to 1/(7 pi) + samples=97, % sample every pi/48 through phase + ] {sin(1/x * 180 / pi)}; + \draw [color=firstcolor,fill=firstcolor] (axis cs:-0.045472841,-1) rectangle (axis cs:0.045472841,1); + %\addplot[firstcurvestyle,-,smooth, + % domain=-0.045472841:-0.0106103295, % 1/(7 pi) to 1/(30 pi) + % samples=93, % sample every pi/2 through phase + %] {sin(1/x * 180 / pi)}; + %\addplot[firstcurvestyle,-,smooth, + % domain=0.0106103295:0.045472841, % 1/(30 pi) to 1/(7 pi) + % samples=93, % sample every pi/2 through phase + %] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=0.045472841:0.063661977, % 1/(7 pi) to 1/(5 pi) + samples=97, % sample every pi/48 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=0.063661977:0.106103295, % 1/(5 pi) to 1/(3 pi) + samples=49, % sample every pi/24 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=0.106103295:0.318309886, % 1/(3 pi) to 1/pi + samples=25, % sample every pi/12 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,rightarrow, + domain=0.318309886:1.018591636, % 1/(48/48 pi) to 1/(15/48 pi) + samples=34, % sample every pi/48 through phase + ] {sin(1/x * 180 / pi)}; + \end{axis} + \end{tikzpicture} + + + +
+ +
+ + + + +

+ Another graph of f(x)=\sin(1/x) is shown, + this time zoomed in to show only the x interval from -0.1 to 0.1. + The features of the graph are the similar to what is visible over the larger interval: + further from the origin, we see the graph oscillating (rapidly) between y=1 and y=-1. + Near the orgin, the oscillations become so rapid that we can no longer tell them apart. + What we conclude from the graph is that on any interval containing x=0, + f(x)=\sin(1/x) takes on every y value between -1 and 1. + (In fact, f(x) attains every value infinitely many times!) +

+ + + Graph of the same function, sin(1/x), shown with a smaller x interval, -0.1 to 0.1. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.11, + xmax=.11, + ymin=-1.1, + ymax=1.1 + ] + \addplot[firstcurvestyle,leftarrow, + domain=-0.100518911:-0.063661977, % 1/(152/48 pi) to 1/(240/48 pi) + samples=89, % sample every pi/48 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=-0.063661977:-0.045472841, % 1/(5 pi) to 1/(7 pi) + samples=25, % sample every pi/12 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=-0.045472841:-0.035367765, % 1/(7 pi) to 1/(9 pi) + samples=49, % sample every pi/24 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=-0.035367765:-0.024485376, % 1/(9 pi) to 1/(13 pi) + samples=145, % sample every pi/36 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=-0.024485376:-0.012732395, % 1/(13 pi) to 1/(25 pi) + samples=73, % sample every pi/6 through phase + ] {sin(1/x * 180 / pi)}; + \draw [color=firstcolor,fill=firstcolor] (axis cs:-0.012732395,-1) rectangle (axis cs:0.012732395,1); + %\addplot[firstcurvestyle,-,smooth, + % domain=-0.0031830988618379:-0.0010610329539459, % 1/(100 pi) to 1/(300 pi) + % samples=401, % sample every pi/2 through phase + %] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,leftarrow, + domain=0.100518911:0.063661977, % 1/(152/48 pi) to 1/(240/48 pi) + samples=89, % sample every pi/48 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=0.063661977:0.045472841, % 1/(5 pi) to 1/(7 pi) + samples=25, % sample every pi/12 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=0.045472841:0.035367765, % 1/(7 pi) to 1/(9 pi) + samples=49, % sample every pi/24 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=0.035367765:0.024485376, % 1/(9 pi) to 1/(13 pi) + samples=145, % sample every pi/36 through phase + ] {sin(1/x * 180 / pi)}; + \addplot[firstcurvestyle,-,smooth, + domain=0.024485376:0.012732395, % 1/(13 pi) to 1/(25 pi) + samples=73, % sample every pi/6 through phase + ] {sin(1/x * 180 / pi)}; + %\addplot[firstcurvestyle,-,smooth, + % domain=0.0031830988618379:0.0010610329539459, % 1/(100 pi) to 1/(300 pi) + % samples=401, % sample every pi/2 through phase + %] {sin(1/x * 180 / pi)}; + \end{axis} + \end{tikzpicture} + + + +
+ +
+ +
+ Observing that f(x)=\sin(1/x) has no limit as x\to0 in + + + x + \sin(1/x) + + + 0.1 + -0.544021 + + + 0.01 + -0.506366 + + + 0.001 + 0.82688 + + + 0.0001 + -0.305614 + + + 1.\times 10^{-5} + 0.0357488 + + + 1.\times 10^{-6} + -0.349994 + + + 1.\times 10^{-7} + 0.420548 + + +
+ +

+ It can be shown that in reality, + as x approaches 0, \sin(1/x) takes on all values between + -1 and 1 infinitely many times! + Because of this oscillation, + \lim_{x\to 0}\sin(1/x) does not exist. +

+ + + + + + + + + Limits of Difference Quotients +

+ We have approximated limits of functions as x approached a particular number. + We will consider another important kind of limit after explaining a few key ideas. + + limitdifference quotient + +

+ + + +

+ Let f(x) represent the position function, in feet, + of some particle that is moving in a straight line, + where x is measured in seconds. + Let's say that when x=1, + the particle is at position 10 ft., and when x=5, + the particle is at 20 ft. + Another way of expressing this is to say +

+ +
+

+ f(1)=10 and f(5) = 20. +

+
+ +

+ Since the particle traveled 10 feet in 4 seconds, + we can say the particle's + + average velocity + + velocityaverage velocity + + average velocity was 2.5 ft/s. + We write this calculation using a + quotient of differences, or, a + + difference quotient + + difference quotient: + + \frac{f(5) - f(1)}{5-1} \frac{\text{ ft}}{\text{s}} = \frac{10 \text{ ft}}{4\text{ s}} = 2.5 \text{ ft/s } + . +

+ +

+ This difference quotient can be thought of as the familiar + rise over run used to compute the slopes of lines. + In fact, that is essentially what we are doing: + given two points on the graph of f, + we are finding the slope of the secant line + through those two points. + See . +

+ +
+ Interpreting a difference quotient as the slope of a secant line + + + +

+ The image shows the graph of a function, along with a line that intersects the graph at two points. + The graph has the shape of a parabola that opens downward, + and is displayed over the region 0\leq x\leq 6, + with a y range from 0 to 25. + There are two points plotted on the graph at coordinates + (1, 10) and (5, 20), and the line through these points is an example of a secant line. +

+ + + A downward curved graph, with marked points at (1,10) and (5,20), and a + line intercepting both points. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1, + xmax=6.5, + ymin=-1, + ymax=25, + ] + \addplot+[infinite,domain=0:6] {-1.5*x^2+11.5*x}; + \addplot+[infinite,domain=0:6] {2.5*(x-1)+10}; + \addplot[soliddot] coordinates {(1,10) (5,20)}; + \end{axis} + \end{tikzpicture} + + + +
+ +

+ Now consider finding the average speed on another time interval. + We again start at x=1, + but consider the position of the particle h seconds later. + That is, consider the positions of the particle when x=1 + and when x=1+h. The difference quotient (excluding units) + is now + + \frac{f(1+h)-f(1)}{(1+h)-1} = \frac{f(1+h)-f(1)}{h} + . +

+ +

+ Let f(x) = -1.5x^2+11.5x; + note that f(1)=10 and + f(5) = 20, as in our discussion. + We can compute this difference quotient for all values of h + (even negative values!) + except h=0, for then we get 0/0, + the indeterminate form introduced earlier. + For all values h\neq 0, + the difference quotient computes the average velocity of the particle over an interval of time of length h starting at x=1. +

+ +

+ For small values of h, + , values of h close to 0, + we get average velocities over very short time periods and compute secant lines over small intervals. + See . + This leads us to wonder what the limit of the difference quotient is as + h approaches 0. + That is, + + \lim_{h\to 0} \frac{f(1+h)-f(1)}{h} = \text{ ? } + +

+ + + +
+ Secant lines of f(x) at x=1 and x=1+h, for shrinking values of h (, h\rightarrow 0) + +
+ h=2 + + + +

+ Graph of the function from , with the points on the graph (1,10) + and (3,21) marked. A secant line is drawn through these points; it has a steeper slope than in + . Here the value of h is 2. +

+ + + Graph of the same function as the previous figure, with the points (1,10) + and (3,21). The secant line has a steeper slope, equal to 5.5. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1, + xmax=6.5, + ymin=-1, + ymax=25, + ] + \addplot+[infinite,domain=0:6] {-1.5*x^2+11.5*x}; + \addplot+[infinite,domain=0:3.5454] {5.5*(x-1)+10}; + \addplot[soliddot] coordinates {(1,10) (3,21)}; + \end{axis} + \end{tikzpicture} + + + +
+ +
+ h=1 + + + +

+ Graph of the function from , but + with the points (1,10) and (2,17) on the graph marked. + These points correspond to a value of h=1, + and the secant line through these points has a steeper slope than in . +

+ + + Graph of the same function as the previous figure, with the points (1,10) + and (2,17) marked. The secant line has a steeper slope equal to 7. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1, + xmax=6.5, + ymin=-1, + ymax=25, + ] + \addplot+[infinite,domain=0:6] {-1.5*x^2+11.5*x}; + \addplot+[infinite,domain=0:3] {7*(x-1)+10}; + \addplot[soliddot] coordinates {(1,10) (2,17)}; + \end{axis} + \end{tikzpicture} + + + +
+ +
+ h=0.5 + + + +

+ Graph of the function from , but + with the points (1,10) and (1.5,13.875) on the graph marked, corresponding to the value h=0.5. + The secant line through these points again has a steeper slope than in the previous figures. +

+ + + Graph of the same function, with the points (1,10) and + (1.5,13.875). The secant line has a steeper slope equal to 7.75. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1, + xmax=6.5, + ymin=-1, + ymax=25, + ] + \addplot+[infinite,domain=0:6] {-1.5*x^2+11.5*x}; + \addplot+[infinite,domain=0:2.80645] {7.75*(x-1)+10}; + \addplot[soliddot] coordinates {(1,10) (1.5,13.875)}; + \end{axis} + \end{tikzpicture} + + + +
+ +
+ +

+ As we do not yet have a true definition of a limit nor an exact method for computing it, + we settle for approximating the value. + While we could graph the difference quotient + (where the x-axis would represent h values and the + y-axis would represent values of the difference quotient) + we settle for making a table. + See . + The table gives us reason to assume the value of the limit is about 8.5. +

+ +
+ The difference quotient evaluated at values of h near 0 + + + h + \frac{f(1+h)-f(1)}{h} + + + -0.5 + 9.25 + + + -0.1 + 8.65 + + + -0.01 + 8.515 + + + 0.01 + 8.485 + + + 0.1 + 8.35 + + + 0.5 + 7.75 + + +
+ + +

+ Proper understanding of limits is key to understanding calculus. + With limits, + we can accomplish seemingly impossible mathematical things, + like adding up an infinite number of numbers + (and not get infinity) + and finding the slope of a line between two points, + where the two points are actually the same point. + These are not just mathematical curiosities; + they allow us to link position, + velocity and acceleration together, + connect cross-sectional areas to volume, + find the work done by a variable force, and much more. +

+ + + +

+ In the next section we give the formal definition of the limit and begin our study of finding limits analytically. + In the following exercises, + we continue our introduction and approximate the value of limits. +

+ + + + + + + Terms and Concepts + + + + +

+ In your own words, what does it mean to + find the limit of f(x) as x approaches 3? +

+ + + + + + + + + + +

+ An expression of the form \frac{0}{0} is called . + +

+ + + + + indeterminate + + + undefined + +

+ Yes, but there is a more specific answer. +

+ + + + + + + + + + + +

+ + The limit of f(x) as x approaches 5 is f(5). +

+ + +

+ The limit considers values of x near to 5, + but not equal to 5. +

+ +

+ The case where the limit is equal to f(5) has a special name; + we will consider this in detail in . +

+ + + + + + + +

+ Describe three situations where \lim\limits_{x\to c}f(x) does not exist. +

+ + + + +

+ The function may approach different values from the left and right, + the function may grow without bound, + or the function might oscillate. +

+ + + + + + + +

+ In your own words, what is a difference quotient? +

+ + + + + + + + + +

+ When x is near 0, + \dfrac{\sin(x)}{x} is near the value . +

+ + + + + + + + + + + +

+ That is a reasonable guess, but no. +

+ + + + + +

+ Try values of x close to 0, such as 0.0001. + A calculator reveals that \dfrac{\sin 0.0001}{0.0001}\approx0.999999998\ldots. + This is near 1. +

+ + + + + + + Problems + + +

+ Approximate the limit numerically and graphically. +

+ + + + + + do { + $a = list_random(-1,1); + $b = non_zero_random(-5,5,1); + $c = non_zero_random(-5,5,1); + if ($envir{problemSeed} == 1){ + ($a,$b,$c) = (1,3,-5); + }; + $f = Formula("x^2 + $b x + $c")->reduce; + $vx = $b/2; + $vy = $f->eval(x => $vx); + $l = $f->eval(x => $a); + } until($l != 0); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + + +

+ \lim\limits_{x\to }\left(\right) +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph shows a parabola opening upward, + with vertex at (,), and domain [-2,2]. + When x is close to , the value of f(x) appears to be close to . +

+ + A parabola opening upward, with vertex at (,), + also passing through (,) + + \begin{tikzpicture} + \begin{axis}[grid=both] + \addplot+[infinite,domain=-2:2] {$f}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , + that is close to . + So \lim_{x\to }\left(\right)=. +

+ + + + + + + + do { + $a = random(-1,1,1); + $b = non_zero_random(-5,5,1); + $c = non_zero_random(-5,5,1); + $d = non_zero_random(-5,5,1); + if ($envir{problemSeed} == 1){ + ($a,$b,$c,$d) = (0,-3,1,-5); + }; + $f = Formula("x^3 + $b x^2 + $c x + $d")->reduce; + $l = $f->eval(x => $a); + } until ($l != 0); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + + +

+ \lim\limits_{x\to }\left(\right) +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph of . + When x is close to , + the value of f(x) appears to be close to ; + however, this is difficult to tell at the scale used for the graph. +

+ + The graph of the function for this exercise. When x is close to , y appears to be close to + + \begin{tikzpicture} + \begin{axis}[grid=both] + \addplot+[infinite,domain=-2:2] {$f}; + % Be sure to include some area above and below x-axis + \addplot[mark=none] coordinates {(0,5) (0,-5)}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , + that is close to . + So \lim_{x\to }\left(\right)=. +

+ + + + + + + + $a = 0; + ($b,$c) = random_subset(2,-5..-1,1..5); + if ($envir{problemSeed} == 1){ + ($b,$c) = (-1,-3); + }; + ($left,$right) = num_sort(0,$c); + # for graphing + $xmin = $left - 2; + $xmax = $right + 2; + @d = ($left - 0.02, $left + 0.02, $right - 0.02, $right + 0.02); + $f = Formula("(x - $b)/(x^2 - $c x)")->reduce; + $l = Compute("DNE"); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + + +

+ \lim\limits_{x\to }\left(\right) +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The image shows the graph of a function with vertical asymptotes at x=0 + and at x=. The x axis is also a horizontal asymptote. +

+

+ On one side of x=, values of f(x) appear large and positive. + On the other side, the values of f(x) are large and negative. +

+ + Graph of the function for this problem, showing a vertical asymptote at the limit point + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + xmin=$xmin, + xmax=$xmax, + ymin=-25, + ymax=25, + ] + \addplot[firstcurvestyle,infinite, domain=$xmin:$d[0]] {$f}; + \addplot[firstcurvestyle,infinite, domain=$d[1]:$d[2]] {$f}; + \addplot[firstcurvestyle,infinite, domain=$d[3]:$xmax] {$f}; + \addplot[asymptote] coordinates {($left,-25) ($left,25)}; + \addplot[asymptote] coordinates {($right,-25) ($right,25)}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , + that grows without bound. + So \lim_{x\to }\left(\right) does not exist (DNE). +

+ + + + + + + + ($a,$b,$c) = random_subset(3,-5..-1,1..5); + if ($envir{problemSeed} == 1){ + ($a,$b,$c) = (3,-1,1); + }; + $f = Formula("(x^2 - ($a + $b)x + $a*$b)/(x^2 - ($a + $c)x + $a*$c)")->reduce; + ($left,$right) = num_sort($a,$c); + # for graphing + @d = ($c - 0.5, $c + 0.5); + Context("Fraction"); + $l = Fraction(($a - $b)/($a - $c)); + $lreal= ($a - $b)/($a - $c); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + $xmin = -abs($a) - 2; + $xmax = abs($a) + 2; + $xmid1 = $c - 0.01; + $xmid2 = $c + 0.01; + $ymin = -ceil(abs(Real($l))) - 2; + $ymax = ceil(abs(Real($l))) + 2; + + +

+ \lim\limits_{x\to }\left(\right) +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph for this problem has a vertical asymptote at x=, + and a hole in the graph at x=. +

+ +

+ When x is close to , the value of f(x) is close to . +

+ + A graph with a hole at and a vertical asymptote at + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + xmin=$xmin, + xmax=$xmax, + ymin=$ymin, + ymax=$ymax, + ] + \addplot[firstcurvestyle,infinite, domain=$xmin:$d[0]] {$f}; + \addplot[firstcurvestyle,infinite, domain=$d[1]:$xmax] {$f}; + \addplot[asymptote] coordinates {($c,$ymin) ($c,$ymax)}; + \addplot[hollowdot] coordinates {($a,$lreal)}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , + that is close to . + So \lim_{x\to }\left(\right)=. +

+ + + + + + + + ($a,$c) = random_subset(2,-5..-1,1..5); + $b = list_random(-9..-6,6..9); + if ($envir{problemSeed} == 1){ + ($a,$b,$c) = (-1,-7,-5); + }; + $f = Formula("(x^2 - ($a + $b)x + $a*$b)/(x^2 - ($a + $c)x + $a*$c)")->reduce; + ($left,$right) = num_sort($a,$c); + # for graphing + @d = ($c - 0.1, $c + 0.1); + Context("Fraction"); + $l = Fraction(($a - $b)/($a - $c)); + $lreal = ($a - $b)/($a - $c); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + $xmin = $left - 2; + $xmax = $right + 2; + $xmid1 = $c - 0.01; + $xmid2 = $c + 0.01; + $ymin = -ceil(abs(Real($l))) - 2; + $ymax = ceil(abs(Real($l))) + 2; + + +

+ \lim\limits_{x\to }\left(\right) +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph for this problem has a vertical asymptote at x=, + and a hole in the graph at x=. +

+ +

+ When x is close to , the value of f(x) is close to . +

+ + A graph with a hole at and a vertical asymptote at + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + xmin=$xmin, + xmax=$xmax, + ymin=$ymin, + ymax=$ymax, + ] + \addplot[firstcurvestyle,infinite, domain=$xmin:$d[0]] {$f}; + \addplot[firstcurvestyle,infinite, domain=$d[1]:$xmax] {$f}; + \addplot[asymptote] coordinates {($c,$ymin) ($c,$ymax)}; + \addplot[hollowdot] coordinates {($a,$lreal)}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , + that is close to . + So \lim_{x\to }\left(\right)=. +

+ + + + + + + + ($a,$b,$c) = random_subset(3,-9..-1,1..9); + if ($envir{problemSeed} == 1){ + ($a,$b,$c) = (2,-5,-2); + }; + $f = Formula("(x^2-($c+$b)x+$a*$b)/(x^2-(2*$a)x+($a)^2)")->reduce; + $l = (($a-$c)*($a-$b) > 0) ? OneOf("DNE","inf") : OneOf("DNE","-inf"); + # for graphing + @d = ($a - 0.1, $a + 0.1); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + $xmin = -abs($a) - 2; + $xmax = abs($a) + 2; + $xmid1 = $a - 0.01; + $xmid2 = $a + 0.01; + $ymin = -100; + $ymax = 100; + + +

+ \lim\limits_{x\to }\left(\right) +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph for this problem has a vertical asymptote at x=. +

+ +

+ When x is close to , the value of f(x) increases without bound. +

+ + A graph with a vertical asymptote at + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + xmin=$xmin, + xmax=$xmax, + ymin=$ymin, + ymax=$ymax, + ] + \addplot[firstcurvestyle,infinite, domain=$xmin:$d[0], samples=40] {$f}; + \addplot[firstcurvestyle,infinite, domain=$d[1]:$xmax, samples=40] {$f}; + \addplot[asymptote] coordinates {($a,$ymin) ($a,$ymax)}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , + that grows without bound. + So \lim_{x\to }\left(\right) does not exist (DNE). +

+ + + + + + + + $a = non_zero_random(-5,5,1); + $b = non_zero_random(-5,5,1); + $c = list_random(-4..-1,2..4); + $diff = non_zero_random(-3,3,1); + $d = $a + $b - ($c*$a) - $diff; + if ($envir{problemSeed} == 1){ + ($a,$b,$c,$d) = (2,2,3,-5); + }; + $f1 = Formula("x + $b")->reduce; + $f2 = Formula("$c*x + $d")->reduce; + Context("PiecewiseFunction"); + Context()->flags->set(tolType=>"absolute",tolerance => 0.00001); + $f = PiecewiseFunction("x <= $a" => "$f1", "x > $a" => "$f2"); + $l = Compute("DNE"); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + $xmin = -abs($a) - 2; + $xmax = abs($a) + 2; + + +

+ \lim\limits_{x\to }f(x), where f(x)= +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph shows a piecewise-defined function. + At x=, the y value jumps, + with values of y on one side of + different from those on the other side. +

+ + A piecewise-linear graph with a jump at x= + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + xmin=$xmin, + xmax=$xmax, + ] + \addplot[firstcurvestyle, domain=$xmin:$a, infiniteopen] {$f1}; + \addplot[firstcurvestyle, domain=$a:$xmax, closedinfinite] {$f2}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , + that f(x) approaches different values from the left and right. + So \lim_{x\to }f(x) does not exist. +

+ + + + + + + + do { + $a = non_zero_random(-3,3,1); + $b = random(-4,-1,1); + $c = non_zero_random(-5,5,1); + $d = non_zero_random(-3,3,1); + if ($envir{problemSeed} == 1){ + ($a,$b,$c,$d) = (3,-1,1,2); + }; + $e = $a**2 + $b*$a + $c - $d*$a; + $f1 = Formula("x^2+$b*x+$c")->reduce; + $f2 = Formula("$d*x+$e")->reduce; + Context("PiecewiseFunction"); + Context()->flags->set(tolType=>"absolute",tolerance => 0.00001); + $f = PiecewiseFunction("x <= $a" => "$f1", "x > $a" => "$f2"); + $l = $f1->eval(x => $a); + } until ($l!=0); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + $xmin = -4; + $xmax = 4; + + +

+ \lim\limits_{x\to }f(x), where f(x)= +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph of a piecewise-defined function. + To the left of x=, + the graph is a parabola. To the right, the graph is a line. +

+ +

+ The two parts of the graph appear to have the same y value when x=. +

+ + A piecewise-defined graph. One part is a parabola, and the other, a line. The two parts meet when x= + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + xmin=$xmin, + xmax=$xmax, + ] + \addplot[firstcurvestyle, domain=$xmin:$a, infiniteleft] {$f1}; + \addplot[firstcurvestyle, domain=$a:$xmax, infiniteright] {$f2}; + % Be sure to include some area above and below x-axis + \addplot[mark=none] coordinates {(0,5) (0,-5)}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , + that f(x) approaches . + So \lim_{x\to }f(x)=. +

+ + + + + + + + $a=0; + $b = non_zero_random(-5,5,1); + if ($envir{problemSeed} == 1){ + $b = 3; + }; + $f1 = Formula("cos(x)")->reduce; + $f2 = Formula("x^2+$b*x+1")->reduce; + Context("PiecewiseFunction"); + Context()->flags->set(tolType=>"absolute",tolerance => 0.00001); + $f = PiecewiseFunction("x <= $a" => "$f1", "x > $a" => "$f2"); + $l = $f1->eval(x => $a); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + $xmin = -3; + $xmax = 3; + + +

+ \lim\limits_{x\to }f(x), where f(x)= +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph of a piecewise-defined function. + To the left of x=, + the graph is a cosine function. To the right, the graph is a parabola. +

+ +

+ The two parts of the graph appear to have the same y value when x=. +

+ + A piecewise-defined graph. One part is a parabola, and the other, a cosine wave. The two parts meet when x= + + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + xmin=$xmin, + xmax=$xmax, + ] + \addplot[firstcurvestyle, domain=$xmin:$a, infiniteleft] {cos(x*180/pi)}; + \addplot[firstcurvestyle, domain=$a:$xmax, infiniteright] {$f2}; + % Be sure to include some area above and below x-axis + \addplot[mark=none] coordinates {(0,2) (0,-2)}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , + that f(x) approaches . + So \lim_{x\to }f(x)=. +

+ + + + + + + + Context()->flags->set(reduceConstants => 0); + $a = list_random(Formula("pi/2"),Formula("pi/3"),Formula("pi/6")); + if ($envir{problemSeed} == 1){ + $a = Formula("pi/2") + }; + $f1=Formula("sin(x)")->reduce; + $f2=Formula("cos(x)")->reduce; + Context("PiecewiseFunction"); + Context()->flags->set(tolType=>"absolute",tolerance => 0.00001); + $areal = $a->eval(x => 0); + $f = PiecewiseFunction("x <= $areal" => "$f1", "x > $areal" => "$f2"); + $l = Compute("DNE"); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_->eval(x => 0))} (@tabin); + $xmin = -1; + $xmax = 3; + + +

+ \lim\limits_{x\to }f(x), + where f(x)=\begin{cases}\sin(x)\amp x\leq \\\cos(x)\amp x\gt \end{cases} +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph of a piecewise-defined function. + The two parts are portions of sine and cosine graphs, + and they do not approach the same y values near x=. +

+ + A piecewise-defined graph consisting of two sine waves that approach different values near x= + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + xmin=$xmin, + xmax=$xmax, + ] + \addplot[firstcurvestyle, domain=$xmin:$a, infiniteopen] {sin(x*180/pi)}; + \addplot[firstcurvestyle, domain=$a:$xmax, closedinfinite] {cos(x*180/pi)}; + % Be sure to include some area above and below x-axis + \addplot[mark=none] coordinates {(0,2) (0,-2)}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , + that f(x) approaches different values from the left and right. + So \lim_{x\to }f(x) does not exist. +

+ + + + + + + + $f = Formula("abs(x)^x"); + $a = 0; + $l = 1; + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + + +

+ \lim\limits_{x\to 0}\lvert x\rvert^x +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph shows a wave-like curve that approaches the point (0,1) + when x is close to zero, for both positive and negative values of x. + However, a hollow dot at the point (0,1) indicates that the function is undefined at this point. +

+ + A wave-like graph with a hole at the point (0,1). + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + ymin=-1, + ymax=2, + ] + \addplot[firstcurvestyle,domain=-1.5:-0.005,infiniteleft] {(-x)^x}; + \addplot[only marks,mark=o,color=firstcolor] coordinates { (0,1) }; + \addplot[firstcurvestyle,domain=0.005:1.5,infiniteright] {x^x}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to 0, that \lvert x \rvert^x is close to 1. + So \lim_{x\to 0}\lvert x\rvert^x=1. +

+ + + + + + + + $f = Formula("e^(-e^(1/x))"); + $a = 0; + $l = Compute("DNE"); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{$f->eval(x => $_)} (@tabin); + + +

+ \lim\limits_{x\to0}e^{-e^{1/x}} +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph of a piecewise-defined function. + For negative values of x, + the graph climbs slowly, from values near 0.5 to values near 1 as x approaches 0. + For positive values of x, the graph is close to 0 when x is close to 0, + and then climbs slowly as x increases. +

+ + The graph of a piecewise-defined function that jumps from 1 to 0 at x=0. + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + ymin=-2, + ymax=2, + ] + \addplot[firstcurvestyle, domain=-2:-0.001,infiniteleft] {e^(-e^(1/x))}; + \addplot[only marks,mark=o,color=firstcolor] coordinates { (0,1) }; + \addplot[firstcurvestyle, domain=0.2:2, infiniteright] {e^(-e^(1/x))}; + \addplot[only marks,mark=o,color=firstcolor] coordinates { (0,0) }; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to 0, that e^{-e^{1/x}} approaches different values from the left and right. + So \lim_{x\to 0}e^{-e^{1/x}} does not exist. +

+ + + + + + + + %fact = (0 => 1, 1 => 1, 2 => 2, 3 => 6, 4 => 24, 5 => 120, 6 => 720); + sub f { + my $x = shift; + return $fact{int(abs($x))}; + } + $a = list_random(-5..-2,2..5); + $l = (abs($a) <= 1) ? Real(1) : Compute("DNE"); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{f($_)} (@tabin); + $ymax = f(abs($a)+0.5)+10; + $ymin = max(-10, -1*f(abs($a)+0.5)); + + +

+ \lim\limits_{x\to}\big\lfloor\lvert x\rvert\big\rfloor !, + where \lvert x\rvert is the absolute value of x, + \lfloor x\rfloor is the floor of x (the greatest integer less than or equal to x), + and x! is x factorial. +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + \big\lfloor\lvert x\rvert\big\rfloor ! + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph is piecewise-defined, and consists of a collection of horizontal line segments. + For negative values of x, the segments include the right endpoint, but not the left endpoint. + For positive value of x, the segments include the left endpoint, but not the left endpoint. +

+ +

+ The y values for the segments are small when \abs{x} is small, + and then get larger for larger values of \abs{x}. +

+ +

+ Near x=, the y value is different just to the left and right of . +

+ + A piecewise-defined graph consisting of many short horizontal line segments + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + xmin=$xmin, + xmax=$xmax, + ymin=$ymin, + ymax=$ymax, + ] + \addplot[firstcurvestyle, open] coordinates {(-2,1) (2,1)}; + \addplot[firstcurvestyle, closedopen] coordinates {(2,2) (3,2)}; + \addplot[firstcurvestyle, closedopen] coordinates {(3,6) (4,6)}; + \addplot[firstcurvestyle, closedopen] coordinates {(4,24) (5,24)}; + \addplot[firstcurvestyle, closedopen] coordinates {(5,120) (6,120)}; + \addplot[firstcurvestyle, closedopen] coordinates {(-2,2) (-3,2)}; + \addplot[firstcurvestyle, closedopen] coordinates {(-3,6) (-4,6)}; + \addplot[firstcurvestyle, closedopen] coordinates {(-4,24) (-5,24)}; + \addplot[firstcurvestyle, closedopen] coordinates {(-5,120) (-6,120)}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , that \big\lfloor\lvert x\rvert\big\rfloor ! approaches different values from the left and right. So + + \lim_{x\to}\big\lfloor\lvert x\rvert\big\rfloor !\text{ does not exist (DNE).} + +

+ + + + + + + + %fact = (0 => 1, 1 => 1, 2 => 2, 3 => 6, 4 => 24, 5 => 120, 6 => 720); + sub f { + my $x = shift; + return $fact{int(abs($x))}; + } + $a = random(-1,1,1); + $l = (abs($a) <= 1) ? Real(1) : Compute("DNE"); + @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); + @tabout = map{f($_)} (@tabin); + + +

+ \lim\limits_{x\to}\big\lfloor\lvert x\rvert\big\rfloor !, + where \lvert x\rvert is the absolute value of x, + \lfloor x\rfloor is the floor of x (the greatest integer less than or equal to x), + and x! is x factorial. +

+ + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+ + +

+ For a numerical approximation, make a table: +

+ + + x + \big\lfloor\lvert x\rvert\big\rfloor ! + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ For a graphical approximation: +

+ + +

+ The graph is piecewise-defined, and consists of several horizontal line segments. + For negative values of x, the segments include the right endpoint, but not the left endpoint. + For positive value of x, the segments include the left endpoint, but not the left endpoint. +

+ +

+ The y values for the segments are small when \abs{x} is small, + and then get larger for larger values of \abs{x}. +

+ +

+ Near x=, the y appears to be equal to 1. +

+ + A piecewise-defined graph consisting of several short horizontal line segments + + \begin{tikzpicture} + \begin{axis}[ + grid=both, + xmin=$xmin, + xmax=$xmax, + ymin=-1, + ymax=3, + ] + \addplot[firstcurvestyle, open] coordinates {(-2,1) (2,1)}; + \addplot[firstcurvestyle, closedopen] coordinates {(2,2) (3,2)}; + \addplot[firstcurvestyle, closedopen] coordinates {(3,6) (4,6)}; + \addplot[firstcurvestyle, closedopen] coordinates {(4,24) (5,24)}; + \addplot[firstcurvestyle, closedopen] coordinates {(5,120) (6,120)}; + \addplot[firstcurvestyle, closedopen] coordinates {(-2,2) (-3,2)}; + \addplot[firstcurvestyle, closedopen] coordinates {(-3,6) (-4,6)}; + \addplot[firstcurvestyle, closedopen] coordinates {(-4,24) (-5,24)}; + \addplot[firstcurvestyle, closedopen] coordinates {(-5,120) (-6,120)}; + \end{axis} + \end{tikzpicture} + + +

+ It appears that when x is close to , that \big\lfloor\lvert x\rvert\big\rfloor ! is always equal to 1. So + + \lim_{x\to}\big\lfloor\lvert x\rvert\big\rfloor !=1 + +

+ + + + + + + +

+ Approximate the limit of the difference quotient, + \lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h}, + using h = \pm 0.1, \pm 0.01. +

+ + + + + #$m = non_zero_random(-9,9,1); + #do { + # $b = non_zero_random(-9,9,1); + #} until (abs($m) != abs($b)); + #$a = random(1,5,1); + #if ($envir{problemSeed} == 1){ + $a=3; + $m=-7; + $b=2; + #}; + $f = Formula("$m*x + $b")->reduce; + Context()->variables->add(h => 'Real'); + $dq = ($f->substitute(x => Formula("$a + h")) - ($f->eval(x => $a)))/Formula("h"); + @tabin = (Real(-0.1), Real(-0.01), Real(0.01), Real(0.1)); + @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); + @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); + $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ + my ($correct,$student,$self)=@_; + my @cor = @{$correct}; + my @stu = @{$student}; + my @return = (1,1,1,1,1,1,1,1); + for my $i (0,2,4,6) { + do { + $return[$i+1] = 0; + $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); + } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); + do { + $return[$i] = 0; + $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); + } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); + }; + do { + $return[2] = 0; + $self->setMessage(3,'You already used this h-value.'); + } unless ($stu[2] != $stu[0] or $stu[2] eq ""); + do { + $return[4] = 0; + $self->setMessage(5,'You already used this h-value.'); + } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); + do { + $return[6] = 0; + $self->setMessage(7,'You already used this h-value.'); + } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); + return [@return]; + }); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

+ f(x)=, + a= +

+ + + + + h + \frac{f(a+h)-f(a)}{h} + + + + + + + + + + + + + + + + + + + + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+

+ +

+ + + + + + + h + \frac{f(a+h)-f(a)}{h} + + + -0.1 + + + + -0.01 + + + + 0.01 + + + + 0.1 + + + +

+ \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. +

+ + + + + + + + #$m = non_zero_random(-9,9,1); + #$b = random(0.01,0.09,0.01) + random(0.1,0.9,0.1) + random(-9,9,1); + #$a = random(-5,5,1); + #if ($envir{problemSeed} == 1){ + $a = -1; + $m = 9; + $b = 0.06; + #}; + $f = Formula("$m*x + $b")->reduce; + Context()->variables->add(h => 'Real'); + $dq = ($f->substitute(x => Formula("$a + h")) - ($f->eval(x => $a)))/Formula("h"); + @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); + @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); + @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); + $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ + my ($correct,$student,$self)=@_; + my @cor = @{$correct}; + my @stu = @{$student}; + my @return = (1,1,1,1,1,1,1,1); + for my $i (0,2,4,6) { + do { + $return[$i+1] = 0; + $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); + } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); + do { + $return[$i] = 0; + $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); + } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); + }; + do { + $return[2] = 0; + $self->setMessage(3,'You already used this h-value.'); + } unless ($stu[2] != $stu[0] or $stu[2] eq ""); + do { + $return[4] = 0; + $self->setMessage(5,'You already used this h-value.'); + } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); + do { + $return[6] = 0; + $self->setMessage(7,'You already used this h-value.'); + } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); + return [@return]; + }); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

+ f(x)=, + a= +

+ + + + + h + \frac{f(a+h)-f(a)}{h} + + + + + + + + + + + + + + + + + + + + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+

+ +

+ + + + + + + h + \frac{f(a+h)-f(a)}{h} + + + -0.1 + + + + -0.01 + + + + 0.01 + + + + 0.1 + + + +

+ It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. +

+ + + + + + + + #do { + # $b = non_zero_random(-5,5,1); + # $c = non_zero_random(-9,9,1); + # $a = random(-5,5,1); + # if ($envir{problemSeed} == 1){ + $a = 1; + $b = 3; + $c = -7; + # }; + $f = Formula("x^2+$b*x+$c")->reduce; + $m = $f->D('x')->eval(x => $a); + #} until ($m!=0); + Context()->variables->add(h => 'Real'); + $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a)))/Formula("h"); + @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); + @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); + @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); + $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ + my ($correct,$student,$self)=@_; + my @cor = @{$correct}; + my @stu = @{$student}; + my @return = (1,1,1,1,1,1,1,1); + for my $i (0,2,4,6) { + do { + $return[$i+1] = 0; + $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); + } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); + do { + $return[$i] = 0; + $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); + } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); + }; + do { + $return[2] = 0; + $self->setMessage(3,'You already used this h-value.'); + } unless ($stu[2] != $stu[0] or $stu[2] eq ""); + do { + $return[4] = 0; + $self->setMessage(5,'You already used this h-value.'); + } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); + do { + $return[6] = 0; + $self->setMessage(7,'You already used this h-value.'); + } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); + return [@return]; + }); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

+ f(x)=, + a= +

+ + + + + h + \frac{f(a+h)-f(a)}{h} + + + + + + + + + + + + + + + + + + + + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+

+ +

+ + + + + + + h + \frac{f(a+h)-f(a)}{h} + + + -0.1 + + + + -0.01 + + + + 0.01 + + + + 0.1 + + + +

+ It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. +

+ + + + + + + + #do { + # $b = non_zero_random(-5,5,1); + # $a = random(-5,5,1); + #} until ($a + $b!= 0); + #if ($envir{problemSeed} == 1){ + $a = 2; + $b = 1; + #}; + $f = Formula("1/(x+$b)")->reduce; + $m = $f->D('x')->eval(x => $a)->with(tolerance => .01); + Context()->variables->add(h => 'Real'); + $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a)))/Formula("h"); + @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); + @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); + @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); + $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ + my ($correct,$student,$self)=@_; + my @cor = @{$correct}; + my @stu = @{$student}; + my @return = (1,1,1,1,1,1,1,1); + for my $i (0,2,4,6) { + do { + $return[$i+1] = 0; + $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); + } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); + do { + $return[$i] = 0; + $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); + } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); + }; + do { + $return[2] = 0; + $self->setMessage(3,'You already used this h-value.'); + } unless ($stu[2] != $stu[0] or $stu[2] eq ""); + do { + $return[4] = 0; + $self->setMessage(5,'You already used this h-value.'); + } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); + do { + $return[6] = 0; + $self->setMessage(7,'You already used this h-value.'); + } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); + return [@return]; + }); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

+ f(x)=, + a= +

+ + + + + h + \frac{f(a+h)-f(a)}{h} + + + + + + + + + + + + + + + + + + + + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+

+ +

+ + + + + + + h + \frac{f(a+h)-f(a)}{h} + + + -0.1 + + + + -0.01 + + + + 0.01 + + + + 0.1 + + + +

+ It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. +

+ + + + + + + + #do { + # do { + # $d = non_zero_random(-5,5,1); + # $b = non_zero_random(-5,5,1); + # $c = non_zero_random(-5,5,1); + # } until (gcd($d,gcd($b,$c))==1); + # $a = random(-5,5,1); + # if ($envir{problemSeed} == 1){ + $a = -3; + $d = -4; + $b = 5; + $c = -1; + # }; + $f = Formula("$d*x^2+$b*x+$c")->reduce; + $m = $f->D('x')->eval(x => $a); + #} until ($m!=0); + Context()->variables->add(h => 'Real'); + $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a)))/Formula("h"); + @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); + @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); + @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); + $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ + my ($correct,$student,$self)=@_; + my @cor = @{$correct}; + my @stu = @{$student}; + my @return = (1,1,1,1,1,1,1,1); + for my $i (0,2,4,6) { + do { + $return[$i+1] = 0; + $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); + } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); + do { + $return[$i] = 0; + $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); + } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); + }; + do { + $return[2] = 0; + $self->setMessage(3,'You already used this h-value.'); + } unless ($stu[2] != $stu[0] or $stu[2] eq ""); + do { + $return[4] = 0; + $self->setMessage(5,'You already used this h-value.'); + } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); + do { + $return[6] = 0; + $self->setMessage(7,'You already used this h-value.'); + } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); + return [@return]; + }); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

+ f(x)=, + a= +

+ + + + + h + \frac{f(a+h)-f(a)}{h} + + + + + + + + + + + + + + + + + + + + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+

+ +

+ + + + + + + h + \frac{f(a+h)-f(a)}{h} + + + -0.1 + + + + -0.01 + + + + 0.01 + + + + 0.1 + + + +

+ It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. +

+ + + + + + + + #$a = random(2,9,1); + #if ($envir{problemSeed} == 1){ + $a = 5; + #}; + $f = Formula("ln(x)"); + $m = $f->D('x')->eval(x => $a)->with(tolerance => .01); + Context()->variables->add(h => 'Real'); + $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a)))/Formula("h"); + @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); + @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); + @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); + $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ + my ($correct,$student,$self)=@_; + my @cor = @{$correct}; + my @stu = @{$student}; + my @return = (1,1,1,1,1,1,1,1); + for my $i (0,2,4,6) { + do { + $return[$i+1] = 0; + $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); + } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); + do { + $return[$i] = 0; + $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); + } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); + }; + do { + $return[2] = 0; + $self->setMessage(3,'You already used this h-value.'); + } unless ($stu[2] != $stu[0] or $stu[2] eq ""); + do { + $return[4] = 0; + $self->setMessage(5,'You already used this h-value.'); + } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); + do { + $return[6] = 0; + $self->setMessage(7,'You already used this h-value.'); + } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); + return [@return]; + }); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

+ f(x)=, + a= +

+ + + + + h + \frac{f(a+h)-f(a)}{h} + + + + + + + + + + + + + + + + + + + + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+

+ +

+ + + + + + + h + \frac{f(a+h)-f(a)}{h} + + + -0.1 + + + + -0.01 + + + + 0.01 + + + + 0.1 + + + +

+ It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. +

+ + + + + + + + Context()->flags->set(reduceConstants => 0); + #$a = list_random(Formula("pi/3"),Formula("2pi/3"),Formula("pi"),Formula("4pi/3"),Formula("5pi/3"),); + #if ($envir{problemSeed} == 1){ + $a = Formula("pi"); + #}; + $f = Formula("sin(x)"); + $m = $f->D('x')->eval(x => $a->eval(x => 0))->with(tolerance => .01); + Context()->variables->add(h => 'Real'); + $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a->eval(x => 0))))/Formula("h"); + @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); + @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); + @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); + $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ + my ($correct,$student,$self)=@_; + my @cor = @{$correct}; + my @stu = @{$student}; + my @return = (1,1,1,1,1,1,1,1); + for my $i (0,2,4,6) { + do { + $return[$i+1] = 0; + $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); + } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); + do { + $return[$i] = 0; + $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); + } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); + }; + do { + $return[2] = 0; + $self->setMessage(3,'You already used this h-value.'); + } unless ($stu[2] != $stu[0] or $stu[2] eq ""); + do { + $return[4] = 0; + $self->setMessage(5,'You already used this h-value.'); + } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); + do { + $return[6] = 0; + $self->setMessage(7,'You already used this h-value.'); + } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); + return [@return]; + }); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

+ f(x)=, + a= +

+ + + + + h + \frac{f(a+h)-f(a)}{h} + + + + + + + + + + + + + + + + + + + + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+

+ +

+ + + + + + + h + \frac{f(a+h)-f(a)}{h} + + + -0.1 + + + + -0.01 + + + + 0.01 + + + + 0.1 + + + +

+ It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. +

+ + + + + + + + Context()->flags->set(reduceConstants => 0); + #$a = list_random(Formula("pi/3"),Formula("2pi/3"),Formula("pi"),Formula("4pi/3"),Formula("5pi/3"),); + #if ($envir{problemSeed} == 1){ + $a = Formula("pi"); + #}; + $f = Formula("cos(x)"); + $m = $f->D('x')->eval(x => $a->eval(x => 0))->with(tolerance => .01); + Context()->variables->add(h => 'Real'); + $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a->eval(x => 0))))/Formula("h"); + @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); + @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); + @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); + $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ + my ($correct,$student,$self)=@_; + my @cor = @{$correct}; + my @stu = @{$student}; + my @return = (1,1,1,1,1,1,1,1); + for my $i (0,2,4,6) { + do { + $return[$i+1] = 0; + $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); + } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); + do { + $return[$i] = 0; + $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); + } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); + }; + do { + $return[2] = 0; + $self->setMessage(3,'You already used this h-value.'); + } unless ($stu[2] != $stu[0] or $stu[2] eq ""); + do { + $return[4] = 0; + $self->setMessage(5,'You already used this h-value.'); + } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); + do { + $return[6] = 0; + $self->setMessage(7,'You already used this h-value.'); + } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); + return [@return]; + }); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

+ f(x)=, + a= +

+ + + + + h + \frac{f(a+h)-f(a)}{h} + + + + + + + + + + + + + + + + + + + + If the limit does not exist, you may type DNE. + If you need to enter \infty, you may type inf. + +

+ +

+

+ +

+ + + + + + + h + \frac{f(a+h)-f(a)}{h} + + + -0.1 + + + + -0.01 + + + + 0.01 + + + + 0.1 + + + +

+ It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. +

+ + + + + + +
+
+ Epsilon-Delta Definition of a Limit +

+ This section introduces the formal definition of a limit. + Many refer to this as the epsilon-delta definition, + referring to the letters \varepsilon and \delta + of the Greek alphabet. +

+ + + +

+ Before we give the actual definition, + let's consider a few informal ways of describing a limit. + Given a function y=f(x) and an x-value, + c, we say that the limit of the function f, + as x approaches c, is a value L if: +

+
  • + Tends +

    + y tends to L + as x tends to c. +

    +
    • + +
  • + Approaches +

    + y approaches L + as x approaches c. +

    +
    • + +
  • + Near +

    + y is near L + whenever x is near c. +

    +
    • +
    +

    + +

    + The problem with these definitions is that the words + tends, approach, + and especially near are not exact. + In what way does the variable x tend to, or approach, + c? + How near do x and y have to be to c and L, + respectively? +

    + +

    + The definition we describe in this section comes from formalizing + Near. + A quick restatement gets us closer to what we want: +

    +
  • + Tolerance Levels +

    + If x is within a certain + tolerance level of c, + then the corresponding value y=f(x) is within a certain + tolerance level of L. +

    +
    • +
    +

    + +

    + The traditional notation for the x-tolerance is the lowercase Greek letter delta, + or \delta, + and the y-tolerance is denoted by lowercase epsilon, + or \varepsilon. + One more rephrasing of Tolerance Levels + nearly gets us to the actual definition: +

    +
  • + Named Tolerance Levels +

    + If x is within \delta units of c, + then the corresponding value of y is within + \varepsilon units of L. +

    +
    • +
    +

    + +

    + We can write x is within \delta units of c + mathematically as + + \abs{x-c} \lt \delta + , + which is equivalent to + + c-\delta \lt x \lt c+\delta + . +

    + + + +

    + Letting the symbol \implies + represent the word implies, + we can rewrite Named Tolerance Levels as + + \abs{x - c} \lt \delta \implies \abs{y - L} \lt \varepsilon + + or + + c - \delta \lt x \lt c + \delta \implies L - \varepsilon \lt y \lt L + \varepsilon + . +

    + +

    + The point is that \delta and \varepsilon, + being tolerances, can be any positive + (but typically small) + values satisfying this implication. + Finally, we have the formal definition of the limit with the notation seen in the previous section. +

    + + + + + The Limit of a Function <m>f</m> at a point + +

    + Let I be an open interval containing c, + and let f be a function defined on I, + except possibly at c. + The statement that the limit of f(x), + as x approaches c, + is L is denoted by + + \lim_{x\to c} f(x) = L + , + and means that given any \varepsilon \gt 0, + there exists \delta \gt 0 such that for all x in I, + where x \neq c, + if \abs{x - c} \lt \delta, + then \abs{f(x) - L} \lt \varepsilon. + + limitdefinition + +

    +     +     + + + + + + +

    + Mathematicians often enjoy writing ideas without using any words. + Here is the wordless definition of the limit: + + \lim_{x\to c} f(x) = L + \iff + \forall \, \varepsilon \gt 0, \exists \, \delta \gt 0 \text{ s.t. }0\lt \abs{x - c} \lt \delta \implies \abs{f(x) - L} \lt \varepsilon + . +

    + +

    + Note the order in which \varepsilon and \delta are given. + In the definition, the y-tolerance + \varepsilon is given first + and then the limit will exist if + we can find an x-tolerance \delta that works. +

    + +

    + An example will help us understand this definition. + Note that the explanation is long, + but it will take one through all steps necessary to understand the ideas. +

    + + + Evaluating a limit using the definition + +

    + Show that \lim\limits_{x\to 4} \sqrt{x} = 2. +

    +     + +

    + Before we use the formal definition, + let's try some numerical tolerances. + What if the y tolerance is 0.5, + or in other words \varepsilon =0.5? + How close to 4 does x have to be so that y + is within 0.5 units of 2? + That is, 1.5 \lt y \lt 2.5? + In this case, we can proceed as follows: + + 1.5 \amp\lt y \lt 2.5 + 1.5 \amp\lt \sqrt{x} \lt 2.5 \amp\amp \text{(Let } y=\sqrt{x} \text{)} + 1.5^2 \amp\lt x \lt 2.5^2\amp\amp \text{(Square the inequality)} + 2.25 \amp\lt x \lt 6.25 + 2.25-4 \amp\lt x-4 \lt 6.25-4\amp\amp \text{(Subtract }4\text{ from both sides)} + -1.75 \amp\lt x-4 \lt 2.25 + +

    + +

    + So, what is the desired x tolerance? + Remember, we want to find a \delta so that + \abs{x-4} is smaller than \delta. + Since 1.75\lt2.25, + then if we require \abs{x-4}\lt 1.75, then we have + + \abs{x-4}\lt1.75 + \implies-1.75\lt x-4\lt1.75\lt2.25 + + Therefore we can have \delta \leq 1.75. + See . +

    + +
    + Illustrating the \varepsilon-\delta process. + With \varepsilon=0.5, we pick any \delta \lt 1.75 + + +
    + + + +

    + Graph of the function \sqrt{x}. + There are three points on the graph, the first is at x = 2.25, + f(2.25) = 1.5, giving the point (2.25, 1.5). For the + second point we have x = 4, making f(4) = 2, which gives + the point (4, 2). The third point found at x = 6.25 + which gives f(6.25) = 2.5, giving the point (6.25, 2.5). +

    +

    + Taking f(2.25) and adding \varepsilon we get 2, which is + f(4) = 2 and taking f(6.25) and subtracting \varepsilon + we again get 2. This shows the range given by \varepsilon=0.5. +

    +

    + There are two vertical lines near the y axis, at y = 2. One which goes + down and the other goes up. The length of the lines represents + \varepsilon = 0.5 and are used to show that the points (2.25, 1.5) + and (6.25, 2.5) are within \varepsilon of the point + (4, 2). +

    +     + + Graph of square root x with 3 points, shows that two points are within epsilon + of the other point. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.1, + xmax=7.1, + ymin=-.1, + ymax=3.1, + xtick={2,4,6}, + ytick={1,2}, + ] + \addplot+[infiniteright,domain=0:2.64575] ({x^2},{x}); + \addplot[soliddot] coordinates {(4,2)}; + \addplot[guideline] coordinates {(0,1.5) (2.25,1.5)}; + \addplot[guideline] coordinates {(0,2.5) (6.25,2.5)}; + \addplot[guideline,|->] coordinates {(0.2,2) (0.2,2.5)} node [below right]{$\varepsilon=0.5$} ; + \addplot[guideline,|->] coordinates {(0.2,2) (0.2,1.5)} node [above right]{$\varepsilon=0.5$} ; + \addplot[soliddot] coordinates {(2.25,1.5)}; + \addplot[soliddot] coordinates {(6.25,2.5)}; + \addplot[mark=none] coordinates {(0,2.5)} node [above right]{Choose $\varepsilon>0$. Then \ldots} ; + \end{axis} + \end{tikzpicture} + + + +
    + +
    + + + +

    + Graph of the equation sqrt{x}. + There are three points on the graph, the first is at x = 2.25, + f(2.25) = 1.5, giving the point (2.25, 1.5). For the + second point we have x = 4, making f(4) = 2, which + gives the point (4, 2). The third point found at x = 6.25 + which gives f(6.25) = 2.5, giving the point (6.25, 2.5). +

    +

    + This graph has the same vertical lines as + There are also horizontal lines near the x axis, showing the distance from + 2.25 to 4 is 1.75 and the distance from 6.25 to + 4 is 2.25. +

    +     + + Graph of square root x with 3 points, shows that two points are within epsilon + of the other point. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.1, + xmax=7.1, + ymin=-.1, + ymax=3.1, + xtick={2,4,6}, + ytick={1,2}, + ] + \addplot+[infiniteright,domain=0:2.64575] ({x^2},{x}); + \addplot[soliddot] coordinates {(4,2)}; + \addplot[guideline] coordinates {(0,1.5) (2.25,1.5)}; + \addplot[guideline] coordinates {(0,2.5) (6.25,2.5)}; + \addplot[guideline] coordinates {(2.25,0) (2.25,1.5)}; + \addplot[guideline] coordinates {(6.25,0) (6.25,2.5)}; + \addplot[guideline,|->] coordinates {(0.2,2) (0.2,2.5)} node [below right]{$\varepsilon=0.5$} ; + \addplot[guideline,|->] coordinates {(0.2,2) (0.2,1.5)} node [above right]{$\varepsilon=0.5$} ; + \addplot[guideline,|->] coordinates {(4,0.2) (2.25,0.2) } node [above right]{\parbox{3em}{width\\ 1.75}} ; + \addplot[guideline,|->] coordinates {(4,0.2) (6.25,0.2)} node [above left]{\parbox{3em}{\raggedleft width\\ 2.25}} ; + \addplot[soliddot] coordinates {(2.25,1.5)}; + \addplot[soliddot] coordinates {(6.25,2.5)}; + \addplot[mark=none] coordinates {(2.25,1.5)} node [below right]{\parbox{9em}{\ldots choose $\delta$ smaller\\ than each of these:}}; + \end{axis} + \end{tikzpicture} + + +
    + +     +
    + +

    + Given the y tolerance \varepsilon =0.5, + we have found an x tolerance, \delta \lt 1.75, + such that whenever x is within \delta units of 4, + then y is within \varepsilon units of 2. + That's what we were trying to find. +

    + +

    + Let's try another value of \varepsilon. +

    + +

    + What if the y tolerance is 0.01, + \varepsilon =0.01? + How close to 4 does x have to be in order for + y to be within 0.01 units of 2? + (In other words for 1.99 \lt y \lt 2.01?) + Again, we just square these values to get 1.99^2 \lt x \lt 2.01^2, or + + 3.9601 \amp\lt x \lt 4.0401 + -0.0399\amp\lt x-4 \lt 0.0401 + +

    + +

    + What is the desired x tolerance? + In this case we must have \delta \lt 0.0399, + which is the minimum distance from 4 of the two bounds given above. +

    + +

    + What we have so far: if \varepsilon =0.5, + then \delta \lt 1.75 leads to f(x) being less than + \varepsilon from f(4) and if \varepsilon = 0.01, + then \delta \lt 0.0399 being less than \varepsilon from f(4). + A pattern is not easy to see, + so we switch to general \varepsilon try to determine an adequate \delta symbolically. + We start by assuming y=\sqrt{x} is within \varepsilon units of 2: + + \amp\abs{y - 2} \lt \varepsilon + -\varepsilon \amp\lt y - 2 \lt \varepsilon + -\varepsilon \amp\lt \sqrt{x} - 2 \lt \varepsilon \amp\amp (y=\sqrt{x}) + 2 - \varepsilon \amp\lt \sqrt{x} \lt 2+ \varepsilon \amp\amp \text{(Add 2)} + (2 - \varepsilon)^2 \amp\lt x \lt (2+ \varepsilon) ^2 \amp\amp \text{(Square all)} + 4 - 4\varepsilon + \varepsilon^2 \amp\lt x \lt 4 + 4\varepsilon + \varepsilon^2 \amp\amp \text{(Expand)} + -4\varepsilon + \varepsilon^2 \amp\lt x-4 \lt 4\varepsilon + \varepsilon^2\amp\amp \text{(Subtract 4)} + +

    + +

    + The desired form in the last step is + 4-\textit{something} \lt x \lt 4 + \textit{something}. + Since we want this last interval to describe an x tolerance around 4, + we have that either \delta \lt 4\varepsilon - \varepsilon^2 or + \delta \lt 4\varepsilon + \varepsilon^2, whichever is smaller: + + \delta \lt \min\{4\varepsilon - \varepsilon^2, 4\varepsilon + \varepsilon^2\} + . +

    + +

    + Since \varepsilon \gt 0, + we have 4\varepsilon - \varepsilon^2 \lt 4\varepsilon + \varepsilon^2, + the minimum is \delta \leq 4\varepsilon - \varepsilon^2. + That's the formula: given an \varepsilon, + set \delta \leq 4\varepsilon-\varepsilon^2. +

    + +

    + We can check this for our previous values. + If \varepsilon=0.5, the formula gives + \delta \lt 4(0.5) - (0.5)^2 = 1.75 and when \varepsilon=0.01, + the formula gives \delta \lt 4(0.01) - (0.01)^2 = 0.0399. +

    + +

    + So given any \varepsilon \gt 0, + set \delta \lt 4\varepsilon - \varepsilon^2. + Then if \abs{x-4}\lt \delta + (and x\neq 4), + then \abs{f(x) - 2} \lt \varepsilon, + satisfying the definition of the limit. + We have shown formally + (and finally!) + that \lim_{x\to 4} \sqrt{x} = 2. +

    +     + +     + +

    + The previous example was a little long in that we sampled a few specific cases of + \varepsilon before handling the general case. + Normally this is not done. + The previous example is also a bit unsatisfying in that \sqrt{4}=2; + why work so hard to prove something so obvious? + Many \varepsilon-\delta proofs are long and difficult to do. + In this section, + we will focus on examples where the answer is, frankly, + obvious, because the non-obvious examples are even harder. + In the next section we will learn some theorems that allow us to evaluate limits + analytically, that is, + without using the \varepsilon-\delta definition. +

    + + + + + Evaluating a limit using the definition + +

    + Show that \lim\limits_{x\to 2} x^2 = 4. +

    +     + +

    + Let's do this example symbolically from the start. + Let \varepsilon \gt 0 be given; + we want \abs{y-4} \lt \varepsilon, + , \abs{x^2-4} \lt \varepsilon. + How do we find \delta such that when \abs{x-2} \lt \delta, + we are guaranteed that \abs{x^2-4}\lt \varepsilon? +

    + +

    + This is a bit trickier than the previous example, + but let's start by noticing that \abs{x^2-4} = \abs{x-2}\cdot\abs{x+2}. + Consider: + + \abs{x^2-4} \lt \varepsilon \implies \abs{x-2}\cdot\abs{x+2} \lt \varepsilon \implies \abs{x-2} \lt \frac{\varepsilon}{\abs{x+2}} + . +

    + +

    + Could we not set \delta = \frac{\varepsilon}{\abs{x+2}}? +

    + +

    + We are close to an answer, + but the catch is that \delta must be a constant value + (so it can't depend on x). + There is a way to work around this, + but we do have to make an assumption. + Remember that \varepsilon is supposed to be a small number, + which implies that \delta will also be a small value. + In particular, we can (probably) assume that \delta \lt 1. + If this is true, + then \abs{x-2} \lt \delta would imply that \abs{x-2} \lt 1, + giving 1 \lt x \lt 3. +

    + +

    + Now, back to the fraction \frac{\varepsilon}{\abs{x+2}}. + If 1\lt x\lt 3, then 3\lt x+2\lt 5 + (add 2 to all terms in the inequality). + Taking reciprocals, we have + + \frac15 \lt \frac1{\abs{x+2}} \lt \frac 13 + , + which implies + + \frac15 \lt \frac1{\abs{x+2}} + , + which implies + + \frac{\varepsilon}{5}\lt \frac{\varepsilon}{\abs{x+2}} + . +

    + +

    + This suggests that we set \delta \lt \frac{\varepsilon}{5}. + To see why, let consider what follows when we assume \abs{x-2}\lt \delta: + + \abs{x-2}\amp\lt\delta + \abs{x-2}\amp\lt\frac{\varepsilon}{5}\amp\amp\text{(Our choice of } \delta\text{)} + \abs{x-2}\cdot\abs{x+2}\amp\lt\abs{x+2}\cdot\frac{\varepsilon}{5}\amp\amp\text{(Multiply by }\abs{x+2}\text{)} + \abs{x^2-4}\amp\lt\abs{x+2}\cdot\frac{\varepsilon}{5}\amp\amp\text{(Simplify left side)} + \abs{x^2-4}\amp\lt\abs{x+2}\cdot\frac{\varepsilon}{\abs{x+2}}\amp\amp\text{(}Inequality, \delta \lt 1\text{)} + \abs{x^2-4}\amp\lt\varepsilon + +

    + +

    + We have arrived at \abs{x^2-4}\lt \varepsilon as desired. + Note again, in order to make this happen we needed \delta + to first be less than 1. That is a safe assumption; + we want \varepsilon to be arbitrarily small, + forcing \delta to also be small. +

    + +

    + We have also picked \delta to be smaller than necessary. + We could get by with a slightly larger \delta, + as shown in . + The outer lines show the boundaries defined by our choice of \varepsilon. + The inner lines show the boundaries defined by setting \delta = \varepsilon/5. + Note how these dotted lines are within the dashed lines. + That is perfectly fine; + by choosing x within the dotted lines we are guaranteed that + f(x) will be within \varepsilon of 4. +

    + +
    + Choosing \delta = \varepsilon/5 in + + + +

    + Graph of x^2 zoomed in where x = 2. There are three points on + the graph: +

      +
    1. (1.73, 3), since when x = 1.73 we get f(1.73) = 3
      2. +
    2. (2, 4), because when x = 2, f(2) = 4
      3. +
    3. (2.24, 5) when x = 2.24 we get f(2.24) = 5
      4. +
    +

    +

    + There is a vertial line showing the point (1.73, 3) is \varepsilon + from the point (2, 4), because 3 + \varepsilon = 4. Another vertical + line is used to show the point (2.24, 5) is \varepsilon from the + point 2, 4, since 5 - \varepsilon = 4. +

    +

    + There are also vertical lines at x = 2.24, x = 1.73, + x = 2.2, and x = 1.8. This creates two sets of vertical lines, + the outer set is at x = 1.73 and x = 2.24. The inner set + is at x = 1.8 and x = 2.2. The out set of lines mark where + \delta = 1, while the inner set marks + \delta = \epsilon / 5. +

    +

    + The inner set of vertical lines shows that the points at (1.73, 3) + and 2.24, 5 are not within \delta of the point 2, 4. +

    +

    + There are horizontal lines that are between the inner vertical lines + marking 2 \pm (\varepsilon / 5). +

    +     + + Graph of x squared, has 3 points. Shows that two points are within epsilon + and delta of the third point. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=0, + xmax=4, + ymin=2, + ymax=6, + xtick={2}, + ytick={4}, + ydiscontinuity + ] + \addplot+[infinite,domain=0:2.5] {x^2}; + \addplot[soliddot] coordinates {(2,4)}; + \addplot[soliddot] coordinates {(1.73,3)}; + \addplot[soliddot] coordinates {(2.24,5)}; + \addplot[guideline] coordinates {(0,3) (1.73,3)}; + \addplot[guideline] coordinates {(0,5) (2.24,5)}; + \addplot[guideline,dashed] coordinates {(2.24,0) (2.24,5)}; + \addplot[guideline,dashed] coordinates {(1.73,0) (1.73,3)}; + \addplot[guideline,dotted] coordinates {(2.2,0) (2.2,4.84)}; + \addplot[guideline,dotted] coordinates {(1.8,0) (1.8,3.24)}; + \addplot[guideline,|->] coordinates {(0.2,4) (0.2,5)} node [below right]{$\varepsilon$}; + \addplot[guideline,|->] coordinates {(0.2,4) (0.2,3)} node [above right]{$\varepsilon$}; + \addplot[guideline,<-|] coordinates {(2.2,2.2) (2,2.2)}; + \addplot[guideline,<-|] coordinates {(1.8,2.2) (2,2.2)}; + \node[pin={[pin distance=2em, pin edge={inner sep=0pt}]60:{$\delta=\varepsilon/5$}}] at (axis cs:2.1,2.2) {}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + In summary, given \varepsilon \gt 0, + set \delta=\varepsilon/5. + Then \abs{x-2} \lt \delta implies + \abs{x^2 - 4}\lt\varepsilon (\abs{y-4}\lt\varepsilon) as desired. + This shows that \lim_{x\to 2} x^2 = 4. + + gives a visualization of this; + by restricting x to values within \delta = \varepsilon/5 of 2, + we see that f(x) is within \varepsilon of 4. +

    +     + +     + +

    + Make note of the general pattern exhibited in these last two examples. + In some sense, each starts out backwards. + That is, while we want to + +

      +
    1. +

      + start with \abs{x-c}\lt \delta and conclude that +

      +
      2. + +
    2. +

      + \abs{f(x)-L}\lt \varepsilon, +

      +
      3. +
    +

    +

    + we actually start by doing what is essentially some + scratch-work first: + +

      +
    1. +

      + assume \abs{f(x)-L}\lt \varepsilon, + then perform some algebraic manipulations to give an inequality of the form +

      +
      2. + +
    2. +

      + \abs{x-c}\lt something. +

      +
      3. +
    +

    +

    + When we have properly done this, + the something on the greater than + side of the inequality becomes our \delta. + We can refer to this as the scratch-work phase of our proof. + Once we have \delta, + we can formally start the actual proof with + \abs{x-c}\lt \delta and use algebraic manipulations to conclude that + \abs{f(x)-L}\lt \varepsilon, + usually by using the same steps of our + scratch-work in reverse order. +

    + +

    + We highlight this process in the following example. +

    + + + Evaluating a limit using the definition + +

    + Prove that \lim\limits_{x\to 1}(x^3-2x) = -1. +

    +     + +

    + We start our scratch-work by considering \abs{f(x) - (-1)} \lt \varepsilon: + + \abs{f(x)-(-1)} \amp \lt \varepsilon + \abs{x^3-2x + 1}\amp \lt \varepsilon \amp\amp \text{(Now factor)} + \abs{(x-1)(x^2+x-1)}\amp \lt \varepsilon + \abs{x-1}\amp\lt\frac{\varepsilon}{\abs{x^2+x-1}} + . +

    + +

    + We are at the phase of saying that \abs{x-1}\lt something, + where \textit{something}=\varepsilon/\abs{x^2+x-1}. + We want to turn that something into \delta. +

    + +

    + Since x is approaching 1, + we are safe to assume that x is between 0 and 2. + But we need to be careful! The quadratic x^2+x-1 has roots \frac{-1\pm\sqrt{5}}{2}, + and \frac{-1+\sqrt{5}}{2}\approx 0.62 is in the interval [0,2]! + We want to divide by x^2+x-1, so to avoid any division by zero issues, + we choose a smaller interval containing 1: 0.75\lt x\lt 1.25. + So + + \frac34 \amp \lt x\lt \frac54 + \frac{9}{16}\amp \lt x^2\lt \frac{25}{16}\amp\amp \text{(Squared each term.)} + Adding the above two inequalities, we get: + \frac{21}{16}\amp \lt x^2+x\lt \frac{45}{16} + \frac{5}{16}\amp \lt x^2+x-1\lt \frac{29}{16}\amp\amp \text{(Subtracted 1 from each part.)} + +

    + +

    + In Inequality, + we wanted \abs{x-1}\lt \varepsilon/\abs{x^2+x-1}. + The above shows that given any x in (0.75,1.25), + we know that + + \abs{x^2+x-1} \amp \lt \frac{29}{16} \amp\amp \text{which implies that} + \frac{16}{29} \amp \lt \frac{1}{\abs{x^2+x-1}} \amp\amp \text{which implies that} + \frac{16\varepsilon}{29} \amp \lt \frac{\varepsilon}{\abs{x^2+x-1}}\text{.}\amp\amp + +

    + +

    + So we set \delta \lt 16\epsilon/29, on the interval (0.75, 1.25). + This ends our scratch-work, and we begin the formal proof + (which also helps us understand why this was a good choice of \delta). +

    + +

    + Given \varepsilon\gt 0, let \delta \lt \min\{1/4, 16\varepsilon/29\}. + We want to show that when \abs{x-1}\lt \delta, + then \abs{(x^3-2x)-(-1)}\lt \varepsilon. + We start with \abs{x-1}\lt \delta: + + \abs{x-1} \amp \lt \delta + \abs{x-1} \amp \lt \frac{16\varepsilon}{29} + \abs{x-1} \amp \lt \frac{\varepsilon}{\abs{x^2+x-1}} \amp\amp \text{(}Inequality, x\text{ near 1)} + \abs{x-1}\cdot \abs{x^2+x-1} \amp \lt \varepsilon + \abs{x^3-2x+1} \amp \lt \varepsilon + \abs{(x^3-2x)-(-1)} \amp \lt \varepsilon + , + which is what we wanted to show. + (Note that the condition x near 1 is satisfied by the fact that \delta\lt 1/4, + so that 3/4\lt x\lt 5/4.) + Thus \lim_{x\to 1}(x^3-2x) = -1. +

    +     + +     + + + + + +

    + We illustrate evaluating limits once more. +

    + + + Evaluating a limit using the definition + +

    + Prove that \lim\limits_{x\to 0} e^x = 1. +

    +     + +

    + Symbolically, we want to take the inequality + \abs{e^x - 1} \lt \varepsilon and unravel it to the form \abs{x-0} \lt \delta. + Here is our scratch-work: + + \amp\abs{e^x - 1}\lt \varepsilon + -\varepsilon \amp\lt e^x - 1 \lt \varepsilon\amp\amp \text{(Definition of absolute value)} + 1-\varepsilon \amp\lt e^x \lt 1+\varepsilon \amp\amp \text{(Add 1)} + \ln(1-\varepsilon) \amp\lt x \lt \ln(1+\varepsilon) \amp\amp \text{(Take natural logs)} + +

    + +

    + Making the safe assumption that + \varepsilon\lt 1 ensures the last inequality is valid (, so that \ln(1-\varepsilon) is defined). + We can then set \delta to be the minimum of + \abs{\ln(1-\varepsilon)} and \ln(1+\varepsilon); , + + \delta = \min\{\abs{\ln(1-\varepsilon)}, \ln(1+\varepsilon)\} = \ln(1+\varepsilon)\text{.} \quad \text{(See marginal note.)} + +

    + +

    + Now, we work through the actual the proof: + + \amp\abs{x - 0}\lt \delta + -\delta \amp \lt x \lt \delta \amp\amp \text{(Definition of absolute value)} + -\ln(1+\varepsilon) \amp \lt x \lt \ln(1+\varepsilon) + \ln(1-\varepsilon) \amp \lt x \lt \ln(1+\varepsilon) \amp \amp \text{(since \(\ln(1-\varepsilon) \lt -\ln(1+\varepsilon)\)). } + The above line is true by our choice of \delta and by the fact that since \abs{\ln(1-\varepsilon)} \gt \ln(1+\varepsilon) and \ln(1-\varepsilon)\lt 0, we know \ln(1-\varepsilon) \lt -\ln(1+\varepsilon ). + 1-\varepsilon \amp \lt e^x \lt 1+\varepsilon \amp \amp \text{(Exponentiate)} + -\varepsilon \amp \lt e^x - 1 \lt \varepsilon \amp\amp \text{(Subtract 1)} + +

    + +

    + In summary, given \varepsilon \gt 0, + let \delta = \ln(1+\varepsilon). + Then \abs{x - 0} \lt \delta implies \abs{e^x - 1}\lt \varepsilon as desired. + We have shown that \lim_{x\to 0} e^x = 1. +

    +     +     + +

    + We note that we could actually show that + \lim_{x\to c} e^x = e^c for any constant c. + We do this by factoring out e^c from both sides, + leaving us to show \lim_{x\to c} e^{x-c} = 1 instead. + By using the substitution u=x-c, + this reduces to showing \lim_{u\to 0} e^u = 1 which we just did in the last example. + As an added benefit, this shows that in fact the function + f(x)=e^x is continuous + at all values of x, + an important concept we will define in . +

    + +

    + This formal definition of the limit is not an easy concept grasp. + Our examples are actually easy examples, + using simple functions like polynomials, + square roots and exponentials. + It is very difficult to prove, + using the techniques given above, + that \lim_{x\to 0}\frac{\sin(x)}{x} = 1, + as we approximated in . +

    +

    + There is hope. + + shows how one can evaluate complicated limits using certain basic limits as building blocks. + While limits are an incredibly important part of calculus + (and hence much of higher mathematics), + rarely are limits evaluated using the definition. + Rather, the techniques of are employed. +

    + + + + + + + + Terms and Concepts + + + + +

    + What is wrong with the following + definition of a limit? +

    + + +
    +

    + The limit of f(x), as x approaches a, + is K means that given any \delta \gt 0 there exists + \varepsilon \gt 0 such that whenever \abs{f(x)-K}\lt\varepsilon, + we have \abs{x-a}\lt \delta. +

    +
    + +     + + + +

    + \varepsilon should be given first, + and the restriction \abs{x-a}\lt \delta implies + \abs{f(x)-K}\lt\varepsilon, not the other way around. +

    +     + +     + + + + + +

    + Which is given first in establishing a limit? +

    + +     + + + +

    + x-tolerance +

    +     +     + + +

    + y-tolerance +

    +     +     +     + +     + + + + + +

    + + \varepsilon must always be positive. +

    +     + +     + + + + + +

    + + \delta must always be positive. +

    +     + +     +     + + + Problems + + + +

    + Prove the given limit using an \varepsilon-\delta proof. +

    +     + + + + +

    + \lim\limits_{x\to 4} (2x+5) = 13 +

    + +     + + + +

    + Let \varepsilon \gt 0 be given. + We wish to find \delta \gt 0 such that when |x-4|\lt \delta, + \abs{f(x)-13}\lt \epsilon. +

    +

    + Consider \abs{f(x)-13}\lt \varepsilon: + + \abs{f(x) -13} \lt \varepsilon + \abs{(2x+5)-13}\lt \varepsilon + \abs{2x-8} \lt \varepsilon + 2\abs{x-4} \lt \varepsilon + -\varepsilon/2 \lt x-4 \lt \varepsilon/2 + . +

    +

    + This implies we can let \delta =\varepsilon/2. + Then: + + \abs{x-4}\lt \delta + -\delta \lt x-4 \lt \delta + -\varepsilon/2 \lt x-4 \lt \varepsilon/2 + -\varepsilon \lt 2x-8 \lt \varepsilon + -\varepsilon \lt (2x+5)-13 \lt \varepsilon + \abs{(2x+5) - 13} \lt \epsilon + , + which is what we wanted to prove. +

    +     + +     + + + + +

    + \lim\limits_{x\to 5}(3-x)=-2 +

    + +     + + + +

    + Let \varepsilon \gt 0 be given. + We wish to find \delta \gt 0 such that when \abs{x-5}\lt\delta, + \abs{f(x)-(-2)}\lt\varepsilon. +

    +

    + First, some preliminary investigation to find a suitable \delta. + Consider \abs{f(x)-(-2)}\lt\varepsilon: + + \abs{f(x)+2}\amp\lt\varepsilon + \abs{(3-x)+2}\amp\lt\varepsilon + \abs{5-x}\amp\lt\varepsilon + \abs{x-5}\amp\lt\varepsilon + + Since we want to start with \abs{x-5}\lt\delta, + this suggests we let \delta=\varepsilon. +

    +

    + Now we can apply the definition. + + \abs{x-5}\amp\lt\delta + \abs{x-5}\amp\lt\varepsilon + -\varepsilon\lt x-5\amp\lt\varepsilon + -\varepsilon\lt (x-3)-2\amp\lt\varepsilon + \varepsilon\gt(-x+3)-(-2)\gt\amp-\varepsilon + \abs{(3-x)-(-2)}\amp\lt\varepsilon + . + In other words, + \abs{x-5}\lt\delta implies \abs{(3-x)-(-2)}\lt\varepsilon. + This is what we needed to prove. +

    +     + +     + + + + +

    + \lim\limits_{x\to 3}\left(x^2-3\right)=6 +

    + +     + + + +

    + Let \varepsilon \gt 0 be given. + We wish to find \delta \gt 0 such that when \abs{x-3}\lt\delta, + \abs{f(x)-6}\lt\varepsilon. +

    +

    + First, some preliminary investigation to find a suitable \delta. + Consider \abs{f(x)-6}\lt\varepsilon: + + \abs{\left(x^2-3\right)-6}\amp\lt\varepsilon + \abs{x^2-9}\amp\lt\varepsilon + \abs{x-3}\cdot\abs{x+3}\amp\lt\varepsilon + \abs{x-3}\amp\lt\frac{\varepsilon}{\abs{x+3}} + + Since x is near 3, + we can safely assume that, for instance, 2\lt x\lt 4. + Thus + + 2+3\lt x+3\lt4+3 + 5\lt x+3\lt 7 + \frac{1}{7}\lt\frac{1}{x+3}\lt\frac{1}{5} + \frac{\varepsilon}{7}\lt\frac{\varepsilon}{x+3}\lt\frac{\varepsilon}{5} + + Since we need to begin the actual proof with \abs{x-3}\lt\delta, + this suggests that we take \delta=\frac{\varepsilon}{7}. +

    +

    + Now we can apply the definition. + + \abs{x-3}\amp\lt\delta + \abs{x-3}\amp\lt\frac{\varepsilon}{7} + \abs{x-3}\amp\lt\frac{\varepsilon}{\abs{x+3}} + \abs{x-3}\cdot\abs{x+3}\amp\lt\varepsilon + \abs{x^2-9}\amp\lt\varepsilon + \abs{\left(x^2-3\right)-6}\amp\lt\varepsilon + + In other words, + \abs{x-3}\lt\delta implies \abs{\left(x^2-3\right)-6}\lt\varepsilon. + This is what we needed to prove. +

    +     + +     + + + + +

    + \lim\limits_{x\to 4}\left(x^2+x-5\right)=15 +

    + +     + + + +

    + Let \varepsilon \gt 0 be given. + We wish to find \delta \gt 0 such that when \abs{x-4}\lt\delta, + \abs{f(x)-15}\lt\varepsilon. +

    +

    + First, some preliminary investigation to find a suitable \delta. + Consider \abs{f(x)-15}\lt\varepsilon: + + \abs{\left(x^2+x-5\right)-15}\amp\lt\varepsilon + \abs{x^2+x-20}\amp\lt\varepsilon + \abs{x-4}\cdot\abs{x+5}\amp\lt\varepsilon + \abs{x-4}\amp\lt\frac{\varepsilon}{\abs{x+5}} + + Since x is near 4, + we can safely assume that, for instance, 3\lt x\lt 5. + Thus + + 3+5\amp\lt x+5\lt5+5 + 8\amp\lt x+5\lt 10 + \frac{1}{10}\amp\lt\frac{1}{x+5}\lt\frac{1}{8} + \frac{\varepsilon}{10}\amp\lt\frac{\varepsilon}{x+5}\lt\frac{\varepsilon}{8} + + Since we need to begin the actual proof with \abs{x-4}\lt\delta, + this suggests that we take \delta=\frac{\varepsilon}{10}. +

    +

    + Now we can apply the definition. + + \abs{x-4}\amp\lt\delta + \abs{x-4}\amp\lt\frac{\varepsilon}{10} + \abs{x-4}\amp\lt\frac{\varepsilon}{\abs{x+5}} + \abs{x-4}\cdot\abs{x+5}\amp\lt\varepsilon + \abs{x^2+x-20}\amp\lt\varepsilon + \abs{\left(x^2+x-5\right)-15}\amp\lt\varepsilon + + In other words, + \abs{x-4}\lt\delta implies \abs{\left(x^2+x-5\right)-15}\lt\varepsilon. + This is what we needed to prove. +

    +     + +     + + + + +

    + \lim\limits_{x\to 1} \left(2x^2+3x+1\right) = 6 +

    + +     + + + +

    + Let \varepsilon \gt 0 be given. + We wish to find \delta \gt 0 such that when \abs{x-1}\lt \delta, + \abs{f(x)-6}\lt \varepsilon. +

    +

    + First, some prelimary investigation to find a suitable \delta. + Consider \abs{f(x)-6}\lt \varepsilon, + keeping in mind we want to make a statement about \abs{x-1}: + + \abs{f(x) -6} \lt \varepsilon + \abs{(2x^2+3x+1)-6}\lt \varepsilon + \abs{2x^2+3x-5} \lt \varepsilon + \abs{2x+5}\cdot\abs{x-1} \lt \varepsilon + \abs{x-1} \lt \varepsilon/\abs{2x+5} + +

    +

    + Since x is near 1, we can safely assume that, + for instance, 0\lt x\lt 2. + Thus + + 0+5\lt 2x+5\lt 4+5 + 5 \lt 2x+5 \lt 9 + \frac{1}{9} \lt \frac{1}{2x+5} \lt \frac{1}{5} + \frac{\varepsilon}{9} \lt \frac{\varepsilon}{2x+5} \lt \frac{\varepsilon}{5} + +

    +

    + Let \delta =\frac{\varepsilon}{9}. + Then: + + \abs{x-1}\lt \delta + \abs{x-1} \lt \frac{\varepsilon}{9} + \abs{x-1} \lt \frac{\varepsilon}{2x+5} + \abs{x-1}\cdot\abs{2x+5} \lt \frac{\epsilon}{2x+5}\cdot\abs{2x+5} + +

    +

    + Assuming x is near 1, 2x+5 is positive and we can drop the absolute value signs on the right. + + \abs{x-1}\cdot\abs{2x+5} \lt \frac{\varepsilon}{2x+5}\cdot(2x+5) + \abs{2x^2+3x-5} \lt \varepsilon + \abs{(2x^2+3x+1) -6} \lt \varepsilon + , + which is what we wanted to prove. +

    +     + +     + + + + +

    + \lim\limits_{x\to 2}\left(x^3-1\right)=7 +

    + +     + + + +

    + Let \varepsilon \gt 0 be given. + We wish to find \delta \gt 0 such that when \abs{x-2}\lt\delta, + \abs{f(x)-7}\lt\varepsilon. +

    +

    + First, some preliminary investigation to find a suitable \delta. + Consider \abs{f(x)-7}\lt\varepsilon: + + \abs{\left(x^3-1\right)-7}\amp\lt\varepsilon + \abs{x^3-8}\amp\lt\varepsilon + \abs{x-2}\cdot\abs{x^2+2x+4}\amp\lt\varepsilon + \abs{x-2}\amp\lt\frac{\varepsilon}{\abs{x^2+2x+4}} + + Since x is near 2, + we can safely assume that, for instance, 1\lt x\lt 3. + Thus + + 1^2+2(1)+4\amp\lt x^2+2x+4\lt3^2+2(3)+4 + 7\amp\lt x^2+2x+4\lt 19 + \frac{1}{19}\amp\lt\frac{1}{x^2+2x+4}\lt\frac{1}{7} + \frac{\varepsilon}{19}\amp\lt\frac{\varepsilon}{x^2+2x+4}\lt\frac{\varepsilon}{7} + + Since we need to begin the actual proof with \abs{x-2}\lt\delta, + this suggests that we take \delta=\frac{\varepsilon}{19}. +

    +

    + Now we can apply the definition. + + \abs{x-2}\amp\lt\delta + \abs{x-2}\amp\lt\frac{\varepsilon}{19} + \abs{x-2}\amp\lt\frac{\varepsilon}{\abs{x^2+2x+4}} + \abs{x-2}\cdot\abs{x^2+2x+4}\amp\lt\varepsilon + \abs{x^3-8}\amp\lt\varepsilon + \abs{\left(x^3-1\right)-7}\amp\lt\varepsilon + + In other words, + \abs{x-2}\lt\delta implies \abs{\left(x^3-1\right)-7}\lt\varepsilon. + This is what we needed to prove. +

    +     + +     + + + + +

    + \lim\limits_{x\to 2}5=5 +

    + +     + + + +

    + Let \varepsilon \gt 0 be given. + We wish to find \delta \gt 0 such that when \abs{x-2}\lt\delta, + \abs{f(x)-5}\lt\varepsilon. +

    +

    + First, some preliminary investigation to find a suitable \delta. + Consider \abs{f(x)-5}\lt\varepsilon: + + \abs{5-5}\amp\lt\varepsilon + \abs{0}\amp\lt\varepsilon + + Well, this is just plain true no matter what \varepsilon is + (as long as its positive). + So it really doesn't matter what \delta is. +

    +

    + Now we can apply the definition. + + \abs{x-2}\amp\lt\delta + \abs{0}\amp\lt\varepsilon + \abs{5-5}\amp\lt\varepsilon + + In other words, + \abs{x-2}\lt\delta (vacuously) implies \abs{5-5}\lt\varepsilon. + This is what we needed to prove. +

    +     + +     + + + + +

    + \lim\limits_{x\to 0}\left(e^{2x}-1\right)=0 +

    + +     + + + +

    + Let \varepsilon \gt 0 be given. + We wish to find \delta \gt 0 such that when \abs{x-0}\lt\delta, + \abs{f(x)-0}\lt\varepsilon. +

    +

    + First, some preliminary investigation to find a suitable \delta. + Consider \abs{f(x)-0}\lt\varepsilon: + + \abs{\left(e^{2x}-1\right)-0}\amp\lt\varepsilon + \abs{e^{2x}-1}\amp\lt\varepsilon + -\varepsilon\amp\lt e^{2x}-1\lt\varepsilon + 1-\varepsilon\amp\lt e^{2x}\lt1+\varepsilon + \ln(1-\varepsilon)\amp\lt 2x\lt\ln(1+\varepsilon) + \frac{1}{2}\ln(1-\varepsilon)\amp\lt x\lt\frac{1}{2}\ln(1+\varepsilon) + \frac{1}{2}\ln(1-\varepsilon)\amp\lt x-0\lt\frac{1}{2}\ln(1+\varepsilon) + + Since we will need to start with \abs{x-0}\lt\delta, + this suggests we take \delta=\min\left\{\frac{1}{2}\abs{\ln(1-\varepsilon)},\frac{1}{2}\ln(1+\varepsilon)\right\}=\frac{1}{2}\ln(1+\varepsilon). +

    +

    + Now we can apply the definition. + + \abs{x-0}\amp\lt\delta + \abs{x}\amp\lt\frac{1}{2}\ln(1+\varepsilon)\lt\frac{1}{2}\abs{\ln(1-\varepsilon)} + \frac{1}{2}\ln(1-\varepsilon)\amp\lt x\lt\frac{1}{2}\ln(1+\varepsilon) + \ln(1-\varepsilon)\amp\lt 2x\lt\ln(1+\varepsilon) + 1-\varepsilon\amp\lt e^{2x}\lt1+\varepsilon + -\varepsilon\amp\lt e^{2x}-1\lt\varepsilon + \abs{e^{2x}-1}\amp\lt\varepsilon + \abs{\left(e^{2x}-1\right)-0}\amp\lt\varepsilon + + In other words, + \abs{x-0}\lt\delta implies \abs{\left(e^{2x}-1\right)-0}\lt\varepsilon. + This is what we needed to prove. +

    +     + +     + + + + +

    + \lim\limits_{x\to 1} \frac1x = 1 +

    + +     + + + +

    + Let \epsilon \gt 0 be given. + We wish to find \delta \gt 0 such that when |x-1|\lt \delta, + |f(x)-1|\lt \epsilon. +

    +

    + Consider |f(x)-1|\lt \epsilon, + keeping in mind we want to make a statement about |x-1|: + + |f(x) -1 | \lt \epsilon + |1/x-1 |\lt \epsilon + | (1-x)/x | \lt \epsilon + | x-1 |/|x| \lt \epsilon + | x-1 | \lt \epsilon\cdot|x| + +

    +

    + Since x is near 1, we can safely assume that, + for instance, 1/2\lt x\lt 3/2. + Thus + \epsilon/2 \lt \epsilon\cdot x. +

    +

    + Let \delta =\frac{\epsilon}{2}. + Then: + + \abs{x-1}\lt \delta + \abs{x-1} \lt \frac{\epsilon}{2} + \abs{x-1} \lt \epsilon\cdot x + \abs{x-1}/x \lt \epsilon + +

    +

    + Assuming x is near 1, x is positive and we can bring it into the absolute value signs on the left. + + \abs{(x-1)/x} \lt \epsilon + \abs{1-1/x} \lt \epsilon + \abs{(1/x) -1} \lt \epsilon + , + which is what we wanted to prove. +

    +     + +     + + + + +

    + \lim\limits_{x\to 0} \sin(x)= 0 +

    + +     + + + +

    + Use the fact that \abs{\sin(x)} \leq \abs{x}, + with equality only when x=0. +

    +     + +

    + Let \varepsilon \gt 0 be given. + We wish to find \delta \gt 0 such that when \abs{x-0}\lt\delta, + \abs{f(x)-0}\lt\varepsilon. +

    +

    + First, some preliminary investigation to find a suitable \delta. + Consider \abs{f(x)-0}\lt\varepsilon: + + \abs{\sin(x)-0}\amp\lt\varepsilon + \abs{\sin(x)}\amp\lt\varepsilon + + Using the hint that \abs{\sin(x)} \leq \abs{x}, + then if \abs{x}\lt0, it is automatic that \abs{\sin(x)}\lt0. + This suggests we take \delta=\varepsilon. +

    +

    + Now we can apply the definition. + + \abs{x-0}\amp\lt\delta + \abs{x}\amp\lt\varepsilon + \abs{\sin(x)}\amp\lt\varepsilon + \abs{\sin(x)-0}\amp\lt\varepsilon + + In other words, + \abs{x-0}\lt\delta implies \abs{\sin(x)-0}\lt\varepsilon. + This is what we needed to prove. +

    +     + +     +     +     +     +
    +
    + Finding Limits Analytically + +

    + In + we explored the concept of the limit without a strict definition, + meaning we could only make approximations. + In the previous section we gave the definition of the limit and demonstrated how to use it to verify our approximations were correct. + Thus far, our method of finding a limit is +

      +
    1. + make a really good approximation either graphically or numerically, and +
      2. +
    2. + verify our approximation is correct using a + \varepsilon-\delta proof. +
      3. +
    +

    + +

    + Recognizing that \varepsilon-\delta proofs are cumbersome, + this section gives a series of theorems which allow us to find limits much more quickly and intuitively. +

    + +

    + Suppose that \lim_{x\to 2} f(x)=2 and \lim_{x\to 2} g(x) = 3. + What is \lim_{x\to 2}(f(x)+g(x))? + Intuition tells us that the limit should be 5, + as we expect limits to behave in a nice way. + The following theorem states that already established limits do behave nicely. +

    + + + Basic Limit Properties + +

    + Let b, c, L and K be real numbers, + let n be a positive integer, + and let f and g be functions defined on an open interval I containing c with the following limits: + + limitproperties + + + \lim_{x\to c}f(x)\amp = L\amp \lim_{x\to c} g(x)\amp = K + . +

    + +

    + The following limits hold. +

    +
  • + Constants +

    + \lim\limits_{x\to c} b = b +

    +
    • +
  • + Identity +

    + \lim\limits_{x\to c} x = c +

    +
    • +
  • + Sums/Differences +

    + \lim\limits_{x\to c}(f(x)\pm g(x)) = L\pm K +

    +
    • +
  • + Scalar Multiples +

    + \lim\limits_{x\to c}(b\cdot f(x)) = bL +

    +
    • +
  • + Products +

    + \lim\limits_{x\to c} (f(x)\cdot g(x)) = LK +

    +
    • +
  • + Quotients +

    + \lim\limits_{x\to c} (f(x)/g(x)) = L/K, when K\neq 0 +

    +
    • +
  • + Powers +

    + \lim\limits_{x\to c} f(x)^n = L^n +

    +
    • +
  • + Roots +

    + \lim\limits_{x\to c} \sqrt[n]{f(x)} = \sqrt[n]{L} +

    + +

    + (If n is even then require f(x)\geq 0 on I.) +

    +
    • +
  • + Compositions +

    + Adjust the limit requirements to + + \lim_{x\to c}f(x)\amp=L\amp\lim_{x\to L}g(x)\amp=K\amp g(L)\amp=K + . + Then \lim\limits_{x\to c}g(f(x)) = K. +

    +
    • +
    +

    +     +     + + + + + + +

    + We apply the theorem to an example. +

    + + + Using basic limit properties + +

    + Let + + \lim_{x\to 2} f(x)\amp=2\amp\lim_{x\to 2} g(x)\amp= 3\amp p(x)\amp = 3x^2-5x+7 + . + Find the following limits: +

      +
    1. \lim\limits_{x\to 2}(f(x) + g(x))
      2. +
    2. \lim\limits_{x\to 2}(5f(x) + g(x)^2)
      3. +
    3. +

      + \lim\limits_{x\to 2}p(x) +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + Using the property, + we know that + + \lim_{x\to 2}(f(x) + g(x)) \amp = \lim_{x\to 2}f(x) + \lim_{x\to 2}g(x) + \amp = 2+3 =5 + . +

      +
      2. +
    2. +

      + Using the , + , + and properties, + we find that + + \lim_{x\to 2}(5f(x) + g(x)^2) \amp = \lim_{x\to 2}(5f(x))+\lim_{x\to 2}(g(x)^2) + \amp = 5\lim_{x\to 2}f(x) + \mathopen{}\left(\lim_{x\to 2}g(x)\right)^2\mathclose{} + \amp = 5\cdot 2 + 3^2 = 19 + . +

      +
      3. +
    3. +

      + Here we combine the , + , + and properties. + We show quite a few steps, but in general these can be omitted: + + \lim_{x\to 2} p(x) \amp = \lim_{x\to 2}\left(3x^2-5x+7\right) + \amp = \lim_{x\to 2}\mathopen{}\left(3x^2\right)\mathclose{}-\lim_{x\to 2}(5x)+\lim_{x\to 2}7 + \amp = 3\bigl(\lim_{x\to 2}x\bigr)^2-5\lim_{x\to 2}(x) +7 + \amp = 3\cdot 2^2 - 5\cdot 2+7 + \amp = 9 + . +

      +
      4. +
    +

    +     + +     + +

    + Part + of the previous example demonstrates how the limit of a quadratic polynomial + can be determined using the properties of + . + Not only that, recognize that + + \lim_{x\to 2} p(x) = 9 = p(2); + + , the limit at 2 could have been found just by plugging 2 + into the function. This holds true for all polynomials, + and also for rational functions + (which are quotients of polynomials), + as stated in the following theorem. +

    + + + + Limits of Polynomial and Rational Functions + +

    + Let p(x) and q(x) be polynomials and c a real number. + Then: + + limitof polynomial functions + limitof rational functions + continuityof polynomial functions + continuityof rational functions + polynomial functioncontinuity of + rational functioncontinuity of + +

      +
    1. +

      + \lim\limits_{x\to c} p(x) = p(c) +

      +
      2. +
    2. +

      + \lim\limits_{x\to c} \frac{p(x)}{q(x)} = \frac{p(c)}{q(c)}, + when q(c) \neq 0. +

      +
      3. +
    +

    +     +     + + + + + + + Finding a limit of a rational function + +

    + Using , find + + \lim_{x\to -1} \frac{3x^2-5x+1}{x^4-x^2+3} + . +

    +     + +

    + Using , we can quickly state that + + \lim_{x\to -1}\frac{3x^2-5x+1}{x^4-x^2+3} \amp = \frac{3(-1)^2-5(-1)+1}{(-1)^4-(-1)^2+3} + \amp = \frac{9}{3} =3 + . +

    +     +     + +

    + It was likely frustrating in + to do a lot of work with \varepsilon and \delta to prove that + + \lim_{x\to 2} x^2 = 4 + + as it seemed fairly obvious. + The previous theorems state that many functions behave in such an + obvious fashion, + as demonstrated by the rational function in + . +

    + +

    + Polynomial and rational functions are not the only functions to behave in such a predictable way. + The following theorem gives a list of functions whose behavior is particularly + nice in terms of limits. + In , + we will give a formal name to these functions that behave nicely. +

    + + + + + + Limits of Common Functions + +

    + Let c be a real number in the domain of the given function and let + n be a positive integer. + The following limits hold: + + limitof trigonometric functions + limitof exponential functions + limitof logarithmic functions + continuityof trigonometric functions + continuityof exponential functions + continuityof logarithmic functions + trigonometric functioncontinuity of + exponential functioncontinuity of + logarithmic functioncontinuity of + +

      +
    1. \lim\limits_{x\to c}\sin(x) = \sin(c)
      2. + +
    2. \lim_{x\to c}\cos(x) = \cos(c)
      3. + +
    3. \lim_{x\to c}\tan(x)= \tan(c)
      4. + +
    4. \lim_{x\to c}\csc(x) = \csc(c)
      5. + +
    5. \lim_{x\to c}\sec(x) = \sec(c)
      6. + +
    6. \lim_{x\to c}\cot(x) = \cot(c)
      7. + +
    7. \lim_{x\to c}a^x = a^c, if a \gt 0
      8. + +
    8. \lim_{x\to c}\ln(x) = \ln(c)
      9. + +
    9. \lim_{x\to c}\sqrt[n]{x} = \sqrt[n]{c}
      10. +
    +

    +     +     + + + Evaluating limits analytically + +

    + Evaluate the following limits. +

      +
    1. \lim\limits_{x\to \pi} \cos(x)
      2. +
    2. \lim\limits_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right)
      3. +
    3. \lim\limits_{x\to \pi/2}(\cos(x)\sin(x))
      4. +
    4. \lim\limits_{x\to 1} e^{\ln(x)}
      5. +
    5. \lim\limits_{x\to 0} \dfrac{\sin(x)}{x}
      6. +
    +

    +     + +

    +

      +
    1. +

      + This is a straightforward application of + : + \lim\limits_{x\to \pi} \cos(x) = \cos(\pi) = -1. +

      +
      2. +
    2. +

      + We can approach this in at least two ways. + First, by directly applying + , we have: + + \lim_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right) = \sec^2(3)-\tan^2(3) + . + Using the Pythagorean Theorem, + this last expression is 1; therefore + + \lim_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right) = 1 + . +

      +

      + We can also use the Pythagorean Theorem from the start. + + \lim_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right) = \lim_{x\to 3} 1 = 1 + , + using the rule. + Either way, we find the limit is 1. +

      +
      3. +
    3. +

      + Applying the rule and + gives + + \lim\limits_{x\to \pi/2} \cos(x)\sin(x) = \cos(\pi/2)\sin(\pi/2) = 0\cdot 1 = 0 + . +

      +
      4. +
    4. +

      + Again, we can approach this in two ways. + First, we can use the exponential/logarithmic identity that + e^{\ln(x)} = x and evaluate + \lim\limits_{x\to 1} e^{\ln(x)} = \lim\limits_{x\to 1} x = 1. +

      + +

      + We can also use the rule. + Using , + we have \lim\limits_{x\to 1}\ln(x) = \ln(1) = 0 and + \lim_{x\to 0} e^x= e^0=1, + satisfying the conditions of the + rule. + Applying this rule, + + \lim_{x\to 1} e^{\ln(x)} = e^{\lim_{x\to 1} \ln(x)}=e^{\ln(1)} = e^0 = 1 + . + Both approaches are valid, giving the same result. +

      +
      5. + +
    5. +

      + We encountered this limit in . + Applying our theorems, we attempt to find the limit as + + \lim_{x\to 0}\frac{\sin(x)}{x}\rightarrow \frac{\sin(0) }{0} + , + which is of the form \frac{0}{0}. + This, of course, + violates a condition of the + rule, + as the limit of the denominator is not allowed to be 0. + Therefore, we are still unable to evaluate this limit with tools we currently have at hand. +

      +
      6. +
    +

    +     +     + +

    + Based on what we've done so far, this section could have been titled + Using Known Limits to Find Unknown Limits. + By knowing certain limits of functions, + we can find limits involving sums, products, + powers, , of these functions. + We further the development of such comparative tools with the Squeeze Theorem, + a clever and intuitive way to find the value of some limits. +

    + +

    + Before stating this theorem formally, + suppose we have functions f, + g, and h where g always takes on values between f and h; + that is, for all x in an interval, + + f(x) \leq g(x) \leq h(x) + . +

    + +

    + If f and h have the same limit at c, + and g is always squeezed between them, + then g must have the same limit as well. + That is what the Squeeze Theorem states. + This is illustrated in . +

    + + + Squeeze Theorem + +

    + Let f, g, and h be functions on an open interval I + containing c such that for all x in I, + + f(x)\leq g(x) \leq h(x) + . + + limitSqueeze Theorem + Squeeze Theorem + + If + + \lim_{x\to c} f(x) = L = \lim_{x\to c} h(x) + , + then + + \lim_{x\to c} g(x) = L + . +

    +     +     + +
    + An illustration of the Squeeze Theorem + + +

    + An illustration of the squeeze theorem. There are graphs of three functions shown, labeled h(x), + g(x), and f(x). On the yaxis, there is a marker at y = 4, + labeled L and on the xaxis there is a marker at x = 5, labeled + c. +

    +

    + For all values of x \leq c f(x) \leq L and h(x) \geq L. + For all values of x f(x) \leq g(x) \leq h(x), that is, the graph + of the function g(x) lies between the graphs of the functions g(x) + and f(x). +

    +

    + The image shows that where x = c, f(x) and h(x) converge + on y = L = 4. Because f(x) \leq g(x) \leq h(x), we can extrapolate + that \lim\limits_{x\to \c} g(x) = L too. +

    +     + + Graph that illustrates the squeeze theorem using functions h, g, and f. + + + \begin{tikzpicture} + \begin{axis}[ + xtick={5}, + ytick={4}, + xticklabel={$c$}, + yticklabel={$L$} + ] + \addplot+[smooth] coordinates {(1,3) (2,5) (3,5) (4,4.3) (5,4) (6,3.7) (7,4.5) (8,5) (9,6)} node[pos=0.3, above] {$g$}; + \addplot+[smooth] coordinates {(1,1) (2,1.5) (3,2.5) (4,3.9) (5,4) (6,3.5) (7,4) (8,4.1) (9,5.3)} node[pos=0.25, above left] {$f$}; + \addplot+[smooth] coordinates {(1,6.5) (2,6.9) (3,6.5) (4,5) (5,4) (6,4.2) (7,5) (8,5.3) (9,7)} node[pos=0.25, above right] {$h$}; + \end{axis} + \end{tikzpicture} + + +
    + + + +

    + It can take some work to figure out appropriate functions by which to + squeeze a given function. + However, that is generally the only place where work is necessary; + the theorem makes the evaluating the limit part very simple. +

    + + + + + +

    + We use the + in the following example to finally prove that + \lim\limits_{x\to 0} \frac{\sin(x)}{x} = 1. +

    + + + Using the Squeeze Theorem + +

    + Use the to show that + + \lim_{x\to 0} \frac{\sin(x)}{x} = 1 + . +

    +     + +

    + We begin by considering the unit circle. + Each point on the unit circle has coordinates + (\cos(\theta),\sin(\theta)) for some angle \theta as shown in + . + Using similar triangles, + we can extend the line from the origin through the point to the point + (1,\tan(\theta)), as shown. + (Here we are assuming that 0\leq \theta \leq \pi/2. + Later we will show that we can also consider \theta \leq 0.) +

    + +
    + The unit circle and related triangles + + + +

    + Illustration of a unit circle, with a right and acute triangle in + quadrant one. The right triangle purtudes out of the unit circle, + while the acute triangle stays within the bounds of the circle. + The acute triangle is also contained within the right triangle. + There are horizontal lines and vertical lines inside the circle, + these lines act only as guides. + The horizontal and vertical lines intersect at (0, 0). + The horizontal line starts at (-1, 0) and ends at (1, 0), + the vertical line starts at (0, 1) and ends at (0, -1). +

    +

    + The right triangle is made up of the points (0, 0), + (1, \tan(\theta)), and (1, 0). The acute Triangle is made + of the points (0, 0), (\cos(\theta), \sin(\theta)), and + (1, 0). The angle at point (0, 0) is labeled as \theta +

    +     + + Illustration of a unit circle, with a right and acute triangle in + the first quadrant. + + + \begin{tikzpicture} + \begin{axis}[ + compat=1.5.1, + clip=false, + ticks=none, + xmin=-1.1, xmax=1.1, + ymin=-1.1, ymax=1.1, + xlabel={}, + ylabel={}, + axis equal image + ] + \draw (axis cs:0,0) circle [radius=1]; + \draw (axis cs:0,0) node[shift={(22.5:25)}]{$\theta$} -- (axis cs:1,0.839) node[above]{$(1,\tan(\theta) )$} -- (axis cs:1,0) node[below right]{$(1,0)$} -- (axis cs:0.766,0.643) node [left]{$(\cos(\theta) ,\sin(\theta))$}; + \addplot[soliddot] coordinates {(1,0.839) (1,0) (0.766,0.643)}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + + shows three regions have been constructed in the first quadrant, + two triangles and a sector of a circle, + which are also drawn below. + The area of the large triangle is \frac{1}{2}\tan(\theta); + the area of the sector is \theta/2; + the area of the triangle contained inside the sector is + \frac{1}{2}\sin(\theta). + It is then clear from that + + \frac{\tan(\theta)}{2}\geq\frac{\theta}{2}\geq\frac{\sin(\theta)}{2} + . + (You may need to recall that the area of a sector of a circle is + \frac{1}{2}r^2 \theta with \theta measured in radians.) +

    + +
    + Bounding the sector between two triangles + + +
    + + + +

    + A right triangle with two dotted lines inside it. One is curved and + represents a sector of a circle inside the triangle. + The sector of a circle has an area of \frac{1}{2}r^2 \theta. + The other dotted line represents an acute triangle housed within the + larger right triangle. +

    +

    + The right triangle has the side lengths 1 for its base and \tan(\theta) + for the vertical line. Where these lines connect is the right angle. There is + an angle \theta that is shared by both triangles and the circle sector. +

    +     + + Right triangle housing an acute triangle and sector of a circle. + + + \begin{tikzpicture}[x=100pt,y=100pt,scale=0.9] + \fill [draw=black,thick,fill=firstcolor!20] (0,0) node [shift={(22.5:0.2)}] {$\theta$} -- (1,.839) -- node [pos=.5,below,rotate=90] {$\tan(\theta)$} (1,0) -- cycle; + \draw (.5,0) node [below] {$1$}; + \draw [black,dashed,thick] (1,0) arc (0:40:1); + \draw [black,dashed,thick] (1,0) -- (.766,.643); + \end{tikzpicture} + + +
    + + +
    + + + +

    + Sector of a circle from with dotted lines + creating the right triangle from , there is + also a dotted line showing the same acute triangle from + . The acute triangle is contained inside + the circle sector, while the right triangle purtrudes out of it. +

    +

    + There is an angle \theta that is shared by both triangles and + the circle sector. They also share the length 1 for the horizontal line. +

    +     + + Sector of a circle from previous image with dotted lines + creating the right and acute triangles. + + + \begin{tikzpicture}[x=100pt,y=100pt] + \fill [draw=black,thick,fill=firstcolor!20] (0,0) node [shift={(22.5:0.2)}] {$\theta$} -- (1,0) arc(0:40:1) -- cycle; + \draw (.5,0) node [below] {$1$}; + \draw [black,dashed,thick] (1,0) arc (0:40:1); + \draw [black,dashed,thick] (1,0) -- (.766,.643) -- (1,.839) -- node [pos=.5,below,rotate=90,opacity=0] {$\tan(\theta)$} cycle; + \end{tikzpicture} + + +
    + + +
    + + + +

    + Acute triangle from with dotted lines + representing the circle sector from and + right triangle from . There is an additional + dotted line that creates a second right triangle within the acute triangle. + Both the sector of a circle and right triangle fall outside the bounds of the + acute triangle. +

    +

    + The second right triangle a height of \sin(\theta). As with + and the horizontal + line of the triangle is 1. +

    +     + + Acute triangle from previous image with dotted lines creating the + two right triangle an circle sector. + + + \begin{tikzpicture}[x=100pt,y=100pt] + \fill [draw=black,thick,fill=firstcolor!20] (0,0) node [shift={(22.5:0.2)}] {$\theta$} -- (1,0) --(.766,.643) -- cycle; + \draw [dashed,thick] (.766,0) -- node [pos=.4,above,rotate=90] {$\sin(\theta)$}(.766,.643); + \draw (.766,5pt) -- ++(5pt,0) -- ++(0,-5pt); + \draw (.5,0) node [below] {$1$}; + \draw [black,dashed,thick] (1,0) arc(0:40:1) -- (1,.839) -- node [pos=.5,below,rotate=90,opacity=0] {$\tan(\theta)$} cycle; + \end{tikzpicture} + + +
    + +     +
    + +

    + Multiply all terms by \frac{2}{\sin(\theta)}, giving + + \frac{1}{\cos(\theta)} \geq \frac{\theta}{\sin(\theta)} \geq 1 + . +

    + +

    + Taking reciprocals reverses the inequalities, giving + + \cos(\theta) \leq \frac{\sin(\theta)}{\theta} \leq 1 + . + (These inequalities hold for all values of \theta near 0, + even negative values, + since \cos(-\theta) = \cos(\theta) and + \sin(-\theta) = -\sin(\theta).) +

    + +

    + Now take limits. + + \lim_{\theta\to 0} \cos(\theta)\amp \leq \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} \leq \lim_{\theta\to 0} 1 + \cos(0) \amp\leq \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} \leq 1 + 1\amp \leq \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} \leq 1 + +

    + +

    + Clearly this means that + \lim\limits_{\theta\to 0} \frac{\sin(\theta)}{\theta}=1. +

    +     + +     + + + + + +

    + Two notes about the are worth mentioning. + First, one might be discouraged by this application, + thinking I would never + have come up with that on my own. + This is too hard! Don't be discouraged; + within this text we will guide you in your use of the . + As one gains mathematical maturity, + clever proofs like this are easier and easier to create. +

    + +

    + Second, this limit tells us more than just that as x approaches 0, + \sin(x)/x approaches 1. + Both x and \sin(x) are approaching 0, + but the ratio of x and \sin(x) approaches 1, + meaning that they are approaching 0 in essentially the same way. + Another way of viewing this is: + for small x, the functions y=x and + y=\sin(x) are essentially indistinguishable. +

    + + + +

    + We include this special limit, + along with three others, in the following theorem. +

    + + + Special Limits + +

    +

      +
    1. \lim\limits_{x\to 0} \dfrac{\sin(x)}{x} = 1
      2. + +
    2. \lim\limits_{x\to 0} \dfrac{\cos(x)-1}{x} = 0
      3. + +
    3. \lim\limits_{x\to 0} (1+x)^{1/x} = e
      4. + +
    4. \lim\limits_{x\to 0} \dfrac{e^x-1}{x} = 1
      5. +
    +

    +     +     + +

    + A short word on how to interpret the latter three limits. + We know that as x goes to 0, + \cos(x) goes to 1. + So, in the second limit, + both the numerator and denominator are approaching 0. + However, since the limit is 0, + we can interpret this as saying that + \cos(x) is approaching 1 faster than x is approaching 0. +

    + +

    + In the third limit, + inside the parentheses we have an expression that is approaching 1 + (though never equaling 1), + and we know that 1 raised to any power is still 1. + At the same time, the power is growing toward infinity. + What happens to a number near 1 raised to a very large power? + In this particular case, + the result approaches Euler's number, + e, approximately 2.718. +

    + +

    + In the fourth limit, we see that as x\to 0, + e^x approaches 1 just as fast as x\to 0, + resulting in a limit of 1. +

    + + + +

    + The special limits stated in + are called indeterminate forms; + + limitindeterminate form + + in this case they are of the form 0/0, + except the third limit, which is of a different form. + You'll learn techniques to find these limits exactly using calculus in + . +

    + +

    + Our final theorem for this section will be motivated by the following example. +

    + + + Using algebra to evaluate a limit + +

    + Evaluate the following limit: + + \lim_{x\to 1}\frac{x^2-1}{x-1} + . +

    +     + +

    + We begin by attempting to apply + and substituting 1 for x in the quotient. + This gives: + + \lim_{x\to 1}\frac{x^2-1}{x-1} = \frac{1^2-1}{1-1} + + which is of the form \frac{0}{0}, an indeterminate form. + We cannot apply the theorem. +

    + +

    + By graphing the function, + as in , + we see that the function seems to be linear, + implying that the limit should be easy to evaluate. + Recognize that the numerator of our quotient can be factored: + + \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} + . + The function is not defined when x=1, + but for all other x, + + \frac{x^2-1}{x-1}\amp = \frac{(x-1)(x+1)}{x-1} + \amp = \frac{\cancel{(x-1)}(x+1)}{\cancel{(x-1)} } + \amp = x+1, \quad \text{if } x \neq 1 + +

    + +
    + Graphing f in to understand a limit + + + +

    + Graph of the function (x^2-1)/(x-1), showing the region with + y from 0 to 3 and x from 0 + to 2. The graph is a straight line, with slope 1 and a y intercept at y = 1, + except for a hole at the point (1,2), since the function is undefined at x = 1. +

    +     + + Graph of the polynomial x squared minus 1 divided by the polynomial x minus 1. + It is the same as the line y=x+1, except that it is undefined at x = 1. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.1, + xmax=2.2, + ymin=-.1, + ymax=3.2, + ] + \addplot+[] coordinates {(0,1) (2,3)}; + \addplot[hollowdot] coordinates {(1,2)}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + Clearly \lim\limits_{x\to 1}(x+1) = 2. + Recall that when considering limits, + we are not concerned with the value of the function at 1, + only the value the function approaches as x approaches 1. + Since (x^2-1)/(x-1) and x+1 are the same at all points except at x=1, + they both approach the same value as x approaches 1. + Therefore we can conclude that + + \lim_{x \to 1} \frac{x^2-1}{x-1} \amp =\lim_{x \to 1} (x+1) + \amp =2 + +

    +     +     + +

    + The key to + is that the functions y=(x^2-1)/(x-1) and y=x+1 are identical except at x=1. + Since limits describe a value the function is approaching, + not the value the function actually attains, + the limits of the two functions are always equal. +

    + + + Limits of Functions Equal At All But One Point + +

    + Let g(x) = f(x) for all x in an open interval, + except possibly at c, + and let \lim\limits_{x\to c} g(x) = L for some real number L. + Then + + \lim_{x\to c}f(x) = L + . +

    +     +     + +

    + The Fundamental Theorem of Algebra tells us that when dealing with a rational function of the form + g(x)/f(x) and directly evaluating the limit + \lim\limits_{x\to c} \frac{g(x)}{f(x)} returns 0/0, + then (x-c) is a factor of both g(x) and f(x). + One can then use algebra to factor this binomial out, + cancel, then apply . + We demonstrate this once more. +

    + + + Evaluating a limit using <xref ref="thm_limit_allbut1"/> + +

    + Evaluate + + \lim\limits_{x\to 3} \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15} + . +

    +     + +

    + We attempt to apply + by substituting 3 for x. + This returns the familiar indeterminate form of 0/0. + Since the numerator and denominator are each polynomials, + we know that (x-3) is factor of each. + Using whatever method is most comfortable to you, + factor out (x-3) from each + (using polynomial division, + synthetic division, a computer algebra system, etc.). + We find that + + \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15} = \frac{(x-3)\left(x^2+x-2\right)}{(x-3)\left(2 x^2+9 x-5\right)} + . + We can cancel the (x-3) factors as long as x\neq 3. + Using we conclude: + + \lim_{x\to 3} \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15} \amp = \lim_{x\to 3}\frac{(x-3)\left(x^2+x-2\right)}{(x-3)\left(2 x^2+9 x-5\right)} + \amp =\lim_{x\to 3} \frac{x^2+x-2}{2 x^2+9 x-5} + \amp = \frac{10}{40} + \amp = \frac{1}{4} + . +

    +     + +     + + + Evaluating a Limit with a Hole + +

    + Evaluate + + \lim\limits_{x\to 9} \frac{\sqrt{x}-3}{x-9} + . +

    +     + +

    + We begin by trying to apply the limit rule, + but the denominator evaluates to zero. + In fact, this limit is of the indeterminate form 0/0. + We will do some algebra to resolve the indeterminate form. + In this case, + we multiply the numerator and denominator by the conjugate of the numerator. + + \frac{\sqrt{x}-3}{x-9} \amp = \frac{\sqrt{x}-3}{x-9} \cdot \frac{\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)} + \amp= \frac{x-9}{(x-9)(\sqrt{x}+3)} + + We can cancel the (x-9) factors as long as x\neq 9. + Using we conclude: + + \lim_{x\to 9} \frac{\sqrt{x}-3}{x-9} \amp =\lim_{x\to 9} \frac{x-9}{(x-9)\left(\sqrt{x}+3\right)} + \amp = \lim_{x\to 9 }\frac{1}{\sqrt{x}+3} + \amp = \frac{1}{\lim_{x\to 9}\sqrt{x}+\lim_{x \to 9}3} + \amp = \frac{1}{\sqrt{\lim_{x\to 9}x}+3} + \amp = \frac{1}{\sqrt{3+3}} + \amp = \frac{1}{6} + . +

    +     + +     + + + +

    + We end this section by revisiting a limit first seen in + , + a limit of a difference quotient. + Let f(x) = -1.5x^2+11.5x; + we approximated the limit + \lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h}\approx 8.5. + We formally evaluate this limit in the following example. +

    + + + Evaluating the limit of a difference quotient + +

    + Let f(x) = -1.5x^2+11.5x; + find \lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h}. +

    +     + +

    + Since f is a polynomial, + our first attempt should be to employ + and substitute 0 for h. + However, we see that this gives us 0/0. + Knowing that we have a rational function hints that some algebra will help. + Consider the following steps: + + \lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\amp = \lim_{h\to 0}\frac{-1.5(1+h)^2 + 11.5(1+h) - \left(-1.5(1)^2+11.5(1)\right)}{h} + \amp =\lim_{h\to 0}\frac{-1.5(1+2h+h^2) + 11.5+11.5h - 10}{h} + \amp =\lim_{h\to 0}\frac{-1.5h^2 +8.5h}{h} + \amp =\lim_{h\to 0}\frac{h(-1.5h+8.5)}h + \amp =\lim_{h\to 0}(-1.5h+8.5) \quad (\text{using } , \text{ as } h\neq 0 ) + \amp =8.5 \quad (\text{using } ) + + This matches our previous approximation. +

    +     +     + +

    + This section contains several valuable tools for evaluating limits. + One of the main results of this section is + ; + it states that many functions that we use regularly behave in a very nice, + predictable way. + In + we give a name to this nice behavior; + we label such functions as continuous. + Defining that term will require us to look again at what a limit is and what causes limits to not exist. +

    + + + + + Terms and Concepts + + + + +

    + Explain in your own words, + without using \varepsilon-\delta formality, + why \lim\limits_{x\to c}b=b. +

    + +     + + +     + + + + +

    + Explain in your own words, + without using \varepsilon-\delta formality, + why \lim\limits_{x\to c}x=c. +

    + +     + + +     + + + + +

    + What does the text mean when it says that certain functions' + behavior is nice in terms of limits? + What, in particular, is nice? +

    + +     + + +     + + + +

    + Sketch a graph that visually demonstrates the Squeeze Theorem. +

    +     + +     + + + + +

    + You are given the following information: + + \lim_{x\to 1}f(x)\amp=0\amp\lim_{x\to 1}g(x)\amp=0\amp\lim_{x\to 1}\frac{f(x)}{g(x)}\amp=2 + + What can be said about the relative sizes of f(x) and g(x) as x approaches 1? +

    + +     + + +

    + As x is near 1, + both f and g are near 0, + but f is approximately twice the size of g. (That is, + f(x)\approx2g(x).) +

    +     + +     + + + + + +

    + Consider the following logarithmic limit: +

    +     + + +

    + + \lim\limits_{x\to 1}\ln(x) = 0. +

    +     + +

    + Look for a list of limits of common functions in this section. + Then, review the known values for the logarithm. +

    +     +     + + +

    + Use a theorem to defend your answer to the first part. +

    +     + + + + + +

    + True. We know that limits involving \ln(x) + can be evaluated by direct substitution; therefore, + \lim_{x\to 1}\ln x = \ln 1 =0. +

    +     +     + +     +     + + + Problems + + + +

    + Use the following information to evaluate the given limit, when possible. + + \lim\limits_{x\to9}f(x)\amp=6\amp\lim\limits_{x\to6}f(x)\amp=9\amp f(9)\amp=6 + \lim\limits_{x\to9}g(x)\amp=3\amp\lim\limits_{x\to6}g(x)\amp=3\amp g(6)\amp=3 + +

    +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 9; + $q = 6; + $r = 3; + %Lf = ($p => $q, $q => $p); + %Lg = ($p => $r, $q => $r); + %f = ($p => $q); + %g = ($q => $r); + $a = $p; + if (exists $Lf{$a} and exists $Lg{$a}) + {$L=$Lf{$a}+$Lg{$a};} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }(f(x)+g(x)) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 9; + $q = 6; + $r = 3; + %Lf = ($p => $q, $q => $p); + %Lg = ($p => $r, $q => $r); + %f = ($p => $q); + %g = ($q => $r); + $a = $p; + if (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} != 0)) + {$L=3*$Lf{$a}/$Lg{$a};} + elsif (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} == 0) and ($Lf{$a} != 0)) + {$L=Compute('does not exist');} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }\left(\frac{3f(x)}{g(x)}\right) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     + +

    + Using the constant and quotient rules, + + \lim\limits_{x\to }\left(\frac{3f(x)}{g(x)}\right)\amp = + \frac{\lim\limits_{x\to }(3f(x))}{\lim\limits_{x\to }g(x)} + \amp = \frac{3\lim\limits_{x\to }f(x)}{\lim\limits_{x\to }g(x)} + \amp \frac{3(6)}{3} = 6 + . +

    +     +     +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 9; + $q = 6; + $r = 3; + %Lf = ($p => $q, $q => $p); + %Lg = ($p => $r, $q => $r); + %f = ($p => $q); + %g = ($q => $r); + $a = $p; + if (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} != 0)) + {$L=($Lf{$a}-2*$Lg{$a})/$Lg{$a};} + elsif (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} == 0) and ($Lf{$a}-2*$Lg{$a} != 0)) + {$L=Compute('does not exist');} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }\left(\frac{f(x)-2g(x)}{g(x)}\right) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 9; + $q = 6; + $r = 3; + %Lf = ($p => $q, $q => $p); + %Lg = ($p => $r, $q => $r); + %f = ($p => $q); + %g = ($q => $r); + $a = $q; + if (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} != 3)) + {$L=$Lf{$a}/(3-$Lg{$a});} + elsif (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} == 3) and ($Lf{$a} != 0)) + {$L=Compute('does not exist');} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }\left(\frac{f(x)}{3-g(x)}\right) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 9; + $q = 6; + $r = 3; + %Lf = ($p => $q, $q => $p); + %Lg = ($p => $r, $q => $r); + %f = ($p => $q); + %g = ($q => $r); + $a = $p; + if (exists $Lf{$a} and exists $Lg{$Lf{$a}} and ($g{$Lf{$a}} == $Lg{$Lf{$a}})) + {$L=$Lg{$Lf{$a}};} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }g(f(x)) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 9; + $q = 6; + $r = 3; + %Lf = ($p => $q, $q => $p); + %Lg = ($p => $r, $q => $r); + %f = ($p => $q); + %g = ($q => $r); + $a = $q; + if (exists $Lg{$a} and exists $Lf{$Lg{$a}} and ($f{$Lg{$a}} == $Lf{$Lg{$a}})) + {$L=$Lf{$Lg{$a}};} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }f(g(x)) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 9; + $q = 6; + $r = 3; + %Lf = ($p => $q, $q => $p); + %Lg = ($p => $r, $q => $r); + %f = ($p => $q); + %g = ($q => $r); + $a = $q; + if (exists $Lf{$a} and exists $Lf{$Lf{$a}} and exists $Lg{$Lf{$Lf{$a}}} and ($f{$Lf{$a}} == $Lf{$Lf{$a}}) and ($g{$Lf{$Lf{$a}}} == $Lg{$Lf{$Lf{$a}}})) + {$L=$Lg{$Lf{$Lf{$a}}};} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }g(f(f(x))) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 9; + $q = 6; + $r = 3; + %Lf = ($p => $q, $q => $p); + %Lg = ($p => $r, $q => $r); + %f = ($p => $q); + %g = ($q => $r); + $a = $q; + if (exists $Lf{$a} and exists $Lg{$a}) + {$L=$Lf{$a}*$Lg{$a} - ($Lf{$a})**2 + ($Lg{$a})**2;} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }\left(f(x)g(x)-f(x)^2+g(x)^2\right) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     +     + + + +

    + Use the following information to evaluate the given limit, when possible. + If it is not possible to determine the limit, state why not. + + \lim_{x\to1}f(x)\amp=2\amp\lim_{x\to10}f(x)\amp=1\amp f(1)\amp=1/5 + \lim_{x\to1}g(x)\amp=0\amp\lim_{x\to10}g(x)\amp=\pi\amp g(10)\amp=\pi + +

    +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 1; + $q = 10; + $r = 2; + $s = 0; + $t = Compute("pi"); + Context("Fraction"); + $u = Fraction("1/5"); + %Lf = ($p => $r, $q => $p); + %Lg = ($p => $s, $q => $t); + %f = ($p => $u); + %g = ($q => $t); + $a = $p; + if (exists $Lf{$a} and exists $Lg{$a}) + {$L=$Lf{$a}*$Lg{$a}} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }\left(f(x)g(x)\right) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 1; + $q = 10; + $r = 2; + $s = 0; + $t = Compute("pi"); + Context("Fraction"); + $u = Fraction("1/5"); + %Lf = ($p => $r, $q => $p); + %Lg = ($p => $s, $q => $t); + %f = ($p => $u); + %g = ($q => $t); + $a = $q; + if (exists $Lg{$a}) + {$L=Compute("cos($Lg{$a})")} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }\cos(g(x)) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 1; + $q = 10; + $r = 2; + $s = 0; + $t = Compute("pi"); + Context("Fraction"); + $u = Fraction("1/5"); + %Lf = ($p => $r, $q => $p); + %Lg = ($p => $s, $q => $t); + %f = ($p => $u); + %g = ($q => $t); + $a = $p; + if (exists $Lf{$a} and exists $Lg{5*$Lf{$a}} and ($Lg{5*$Lf{$a}} == $g{5*$Lf{$a}})) + {$L=$Lg{5*$Lf{$a}}} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }g(5f(x)) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('not possible to know'=>{}); + Context()->strings->add('NPK'=>{alias=>'not possible to know'}); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $p = 1; + $q = 10; + $r = 2; + $s = 0; + $t = Compute("pi"); + Context("Fraction"); + $u = Fraction("1/5"); + %Lf = ($p => $r, $q => $p); + %Lg = ($p => $s, $q => $t); + %f = ($p => $u); + %g = ($q => $t); + $a = $p; + if (exists $Lf{$a} and exists $Lg{$a} and ($Lf{$a} > 0)) + {$L=($Lf{$a})**($Lg{$a});} + else + {$L=Compute('not possible to know');} + $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; + + +

    + \lim\limits_{x\to }5^{g(x)} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. + +

    + +

    +

    + +

    +     +     +     + +     + + + +

    + Evaluate the given limit. +

    +     + + + + + $a=random(2,8,1); + $b=non_zero_random(-5,5,1); + $c=non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=3;$b=-3;$c=7}; + $f=Formula("x^2+$b x+$c")->reduce; + $L=$f->eval(x=>$a); + + +

    + \lim\limits_{x\to }\left(\right) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     + +

    + + \lim\limits_{x\to }\left(\right)\amp=\lim\limits_{x\to }x^2+\lim\limits_{x\to }x+\lim\limits_{x\to } + \amp=\left(\lim\limits_{x\to }x\right)^2+\lim\limits_{x\to }x+\lim\limits_{x\to } + \amp=\left(\right)^2+ \left(\right)+ + \amp= + +

    +     +     +     + + + + + ($b,$c) = random_subset(2,-9..-1,1..9); + $n=random(3,9,1); + if($envir{problemSeed}==1){$b=3;$c=5;$n=7}; + $f=Formula("((x-$b)/(x-$c))^($n)")->reduce; + $L=$f->substitute(x=>Formula("pi")); + + +

    + \lim\limits_{x\to\pi} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     + +

    + + \lim\limits_{x\to\pi}\amp=\left(\lim\limits_{x\to\pi}\frac{x- }{x- }\right)^{} + \amp=\left(\frac{\lim\limits_{x\to\pi}(x- )}{\lim\limits_{x\to\pi}(x- )}\right)^{} + \amp=\left(\frac{\lim\limits_{x\to\pi}x-\lim\limits_{x\to\pi}}{\lim\limits_{x\to\pi}x-\lim\limits_{x\to\pi}}\right)^{} + \amp= + +

    +     +     +     + + + + + $d=list_random(3,4,6); + if($envir{problemSeed}==1){$d=4;}; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $a=Formula("pi/$d"); + %f=(3=>Formula("sqrt(3)/4"),4=>Formula("1/2"),6=>Formula("sqrt(3)/4")); + $L=$f{$d}; + %p=(3=>Formula("1/2"),4=>Formula("sqrt(2)/2"),6=>Formula("sqrt(3)/2")); + $l=$p{$d}; + $r=$p{$d**2/2-5.5*$d+18}; + + +

    + \lim\limits_{x\to }\cos(x)\sin(x) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     + +

    + + \lim_{x\to }\cos(x)\sin(x)\amp=\lim_{x\to }\cos(x)\cdot\lim_{x\to }\sin(x) + \amp=\cos\mathopen{}\left(\right)\mathclose{}\cdot\sin\mathopen{}\left(\right)\mathclose{} + \amp=\cdot + \amp= + +

    +     +     +     + + + + + ($a,$d) = random_subset(2,-6..-1,1..6); + $b=non_zero_random(-5,5,1); + $c=non_zero_random(-5,5,1); + if($envir{problemSeed}==1){$a=1;$b=2;$c=2;$d=4}; + $f=Formula("$b*x-$c")->reduce; + $g=Formula("x-$d")->reduce; + $h=$f/$g; + Context("Fraction"); + $L=Fraction($b*$a-$c,$a-$d); + + +

    + \lim\limits_{x\to} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     + +

    + + \lim_{x\to} \amp = \frac{-}{-} + \amp = + +

    +     +     +     + + + + + $L=Compute("DNE"); + + +

    + \lim\limits_{x\to0}\ln(x) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     + +

    + This limit does not exist for at least two reasons. + For one reason, + \ln(x) can be arbitrarily large negative if you allow x to be positive and close enough to 0. + Also, since \ln(x) is undefined for negative x and + \lim\limits_{x\to0}\ln(x) is an expression that depends on outputs of \ln using inputs + both slightly smaller and + slightly larger than 0, + we cannot hope to give meaning to \lim\limits_{x\to0}\ln(x). +

    +     +     +     + + + + + ($a,$b) = random_subset(2,2,3,4); + $c=$a**2-random(1,2,1); + if($envir{problemSeed}==1){$a=3;$b=4;$c=8;}; + $f=Formula("($b)^(x^3-$c x)"); + $L=$f->eval(x=>$a); + + +

    + \lim\limits_{x\to } +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     + +

    + + \lim\limits_{x\to }\amp=^{\lim\limits_{x\to }\left(x^3- x\right)} + \amp=^{\lim\limits_{x\to }x^3-\lim\limits_{x\to }x} + \amp=^{\left(\lim\limits_{x\to }x\right)^3- \lim\limits_{x\to }x} + \amp=^{^3- \cdot } + \amp= + +

    +     +     +     + + + + + $d=list_random(3,4,6); + if($envir{problemSeed}==1){$d=6;}; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $a=Formula("pi/$d"); + %f=(3=>Formula("2sqrt(3)/3"),4=>Formula("sqrt(2)"),6=>Formula("2")); + $L=$f{$d}; + + +

    + \lim\limits_{x\to }\csc(x) +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     + +

    + + \lim_{x\to }\csc(x)\amp=\csc\mathopen{}\left(\right)\mathclose{} + \amp= + +

    +     +     +     + + + + + Context()->flags->set(reduceConstantFunctions=>0); + $a=random(1,9,1); + if($envir{problemSeed}==1){$a=1;}; + $f=Formula("ln($a+x)"); + $L=($a==1)?0:Formula("ln($a)"); + + +

    + \lim\limits_{x\to0} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + #ensure rational function doesn't reduce + do{$a=non_zero_random(-5,5,1); + $b=non_zero_random(-5,5,1); + $c=random(2,5,1); + do{$d=non_zero_random(-5,5,1);}until($d!=-$a*$c); + $e=non_zero_random(-5,5,1); + }until((($b*$c+$e)/(-$a*$c-$d))**2+$a*(($b*$c+$e)/(-$a*$c-$d))+$b); + if($envir{problemSeed}==1){$a=3;$b=5;$c=5;$d=2;$e=2}; + $f=Formula("x^2+$a x+$b")->reduce; + $g=Formula("$c x^2-$d x-$e")->reduce; + $r=$f/$g; + $L=$r->substitute(x=>Formula("pi")); + + +

    + \lim\limits_{x\to\pi} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     + +

    + + \lim\limits_{x\to\pi}\amp=\frac{\lim\limits_{x\to\pi}\left(\right)}{\lim\limits_{x\to\pi}\left(\right)} + \amp=\frac{\lim\limits_{x\to\pi}x^2+\lim\limits_{x\to\pi}(x)+\lim\limits_{x\to\pi}()}{\lim\limits_{x\to\pi}(x^2)+\lim\limits_{x\to\pi}(- x)+\lim\limits_{x\to\pi}- } + \amp=\frac{\lim\limits_{x\to\pi}x^2+ \lim\limits_{x\to\pi}x+\lim\limits_{x\to\pi}()}{\lim\limits_{x\to\pi}x^2- \lim\limits_{x\to\pi}x-\lim\limits_{x\to\pi}} + \amp= + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $a=non_zero_random(-5,5,1); + $b=non_zero_random(-5,5,1); + $c=non_zero_random(-5,5,1); + do{$d=non_zero_random(-5,5,1);}until($a*$c!=-$b*$d); + if($envir{problemSeed}==1){$a=3;$b=1;$c=1;$d=-1;}; + $f=Formula("$a x+$b")->reduce; + $g=Formula("$c-$d x")->reduce; + $r=$f/$g; + $L=$r->substitute(x=>Formula("pi")); + + +

    + \lim\limits_{x\to\pi} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + ($a,$b,$c) = random_subset(3,-9..-1,1..9); + if($envir{problemSeed}==1){$a=6;$b=-2;$c=7;}; + $f=Formula("x^2-($a+$b)x+$a*$b")->reduce; + $g=Formula("x^2-($a+$c)x+$a*$c")->reduce; + $r=$f/$g; + Context("Fraction"); + $L=Fraction($a-$b,$a-$c); + + +

    + \lim\limits_{x\to } +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + ($b,$c) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$b=-2;$c=2;}; + $f=Formula("x^2-($b)x")->reduce; + $g=Formula("x^2-($c)x")->reduce; + $r=$f/$g; + Context("Fraction"); + $L=Fraction($b,$c); + + +

    + \lim\limits_{x\to0} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + ($a,$b,$c) = random_subset(3,-9..-1,1..9); + if($envir{problemSeed}==1){$a=2;$b=-8;$c=1;}; + $f=Formula("x^2-($a+$b)x+$a*$b")->reduce; + $g=Formula("x^2-($a+$c)x+$a*$c")->reduce; + $r=$f/$g; + Context("Fraction"); + $L=Fraction($a-$b,$a-$c); + + +

    + \lim\limits_{x\to } +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + ($a,$b,$c) = random_subset(3,-9..-1,1..9); + if($envir{problemSeed}==1){$a=2;$b=8;$c=-1;}; + $f=Formula("x^2-($a+$b)x+$a*$b")->reduce; + $g=Formula("x^2-($a+$c)x+$a*$c")->reduce; + $r=$f/$g; + Context("Fraction"); + $L=Fraction($a-$b,$a-$c); + + +

    + \lim\limits_{x\to } +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + ($a,$b,$c) = random_subset(3,-9..-1,1..9); + if($envir{problemSeed}==1){$a=-2;$b=7;$c=-8;}; + $f=Formula("x^2-($a+$b)x+$a*$b")->reduce; + $g=Formula("x^2-($a+$c)x+$a*$c")->reduce; + $r=$f/$g; + Context("Fraction"); + $L=Fraction($a-$b,$a-$c); + + +

    + \lim\limits_{x\to } +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + ($a,$b,$c) = random_subset(3,-9..-1,1..9); + if($envir{problemSeed}==1){$a=-1;$b=-8;$c=7;}; + $f=Formula("x^2-($a+$b)x+$a*$b")->reduce; + $g=Formula("x^2-($a+$c)x+$a*$c")->reduce; + $r=$f/$g; + Context("Fraction"); + $L=Fraction($a-$b,$a-$c); + + +

    + \lim\limits_{x\to } +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     +     + + + +

    + Use the Squeeze Theorem to evaluate the limit. +

    +     + + + + $showwork = '[@ explanation_box(message=>"Show your work.") @]*'; + + +

    + \lim\limits_{x\to0}\left(x\sin\mathopen{}\left(\frac{1}{x}\right)\mathclose{}\right) +

    +

    + +

    +

    + +

    +     +     +     + + + + + $showwork = '[@ explanation_box(message=>"Show your work.") @]*'; + + +

    + \lim\limits_{x\to0}\left(\sin(x)\cos\mathopen{}\left(\frac{1}{x^2}\right)\mathclose{}\right) +

    +

    + +

    +

    + +

    +     +     +     + + + + + $showwork = '[@ explanation_box(message=>"Show your work.") @]*'; + + +

    + \lim\limits_{x\to1} f(x), + where 3x-2\leq f(x) \leq x^3 +

    +

    + +

    +

    + +

    +     +     +     + + + + + $showwork = '[@ explanation_box(message=>"Show your work.") @]*'; + + +

    + \lim\limits_{x\to3} f(x), + where 6x-9\leq f(x) \leq x^2 +

    +

    + +

    +

    + +

    +     +     +     +     + + + +

    + The following exercises challenge your understanding of limits but can be evaluated using the knowledge gained in . +

    +     + + + + + $L=random(2,9,1); + if($envir{problemSeed}==1){$L=3;}; + $f=Formula("sin($L x)/x"); + + +

    + \lim\limits_{x\to0} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + ($a,$b) = random_subset(2,2..9); + if($envir{problemSeed}==1){$a=5;$b=8}; + $f=Formula("sin($a x)/($b x)"); + Context("Fraction"); + $L=Fraction($a,$b); + + +

    + \lim\limits_{x\to0} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + $L=1; + + +

    + \lim\limits_{x\to0}\frac{\ln(1+x)}{x} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0); + $L=Formula("pi/180"); + + +

    + \lim\limits_{x\to0}\frac{\sin(x)}{x}, + where x is measured in degrees, not radians. +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     +     + + + + +

    + Let f(x)=0 and g(x)=\frac{x}{x}. +

    +     + + + +

    + Explain why \lim\limits_{x\to2}f(x)=0. +

    + +     + + + +

    + Apply Part 1 of . +

    +     +     + + + +

    + Explain why \lim\limits_{x\to0}g(x)=1. +

    + +     + + + +

    + Apply ; + g(x)=\frac{x}{x} is the same as g(x)=1 everywhere except at x=0. + Thus \lim\limits_{x\to0}g(x)=\lim_{x\to0}1=1. +

    +     +     + + + +

    + Explain why \lim\limits_{x\to2} g(f(x)) does not exist. +

    + +     + + + +

    + The function f always gives output 0, + so g(f(x)) is never defined as g is not defined for an input of 0. + Therefore the limit does not exist. +

    +     +     + + + +

    + Explain why the previous statement does not violate the Composition Rule of . +

    + +     + + + +

    + The Composition Rule requires that + \lim\limits_{x\to0}g(x) be equal to g(0). + They are not equal, + so the conditions of the Composition Rule are not satisfied, + and hence the rule is not violated. +

    +     +     + +     +     +     +
    +
    + One-Sided Limits + +

    + We introduced the concept of a limit gently, + approximating their values graphically and numerically. + Next came the rigorous definition of the limit, + along with an admittedly tedious method for evaluating them. + gave us tools + (which we call theorems) + that allow us to compute limits with greater ease. + Chief among the results were the facts that polynomials and + rational, trigonometric, exponential and logarithmic functions + (and their sums, products, etc.) + all behave nicely. + In this section we rigorously define what we mean by nicely. +

    +

    + In + we saw three ways in which limits of functions can fail to exist: +

      +
    1. +

      + The function approaches different values from the left + and right. +

      +
      2. +
    2. +

      + The function grows without bound. +

      +
      3. +
    3. +

      + The function oscillates. +

      +
      4. +
    +

    + +

    + In this section we explore in depth the concepts behind + by introducing the one-sided limit. + We begin with formal definitions that are very similar to the + definition of the limit given in , + but the notation is slightly different and x\neq c + is replaced with either x\lt c or x \gt c. +

    +

    + There is a slightly different definition for a left-hand limit, + than for a right-hand limit, + but both have a lot in common with . +

    + + + One Sided Limits: Left- and Right-Hand Limits + +

    +

    +
  • + Left-Hand Limit + limitone-sided + limitleft-handed + +

    + Let f be a function defined on (a,c) + for some a\lt c and let L be a real number. + The statement that the limit of f(x), + as x approaches c from the left, + is L, (alternatively, + that the left-hand limit of f at + c is L) is denoted by + + \lim_{x\to c^-} f(x) = L + , + and means that for any \varepsilon \gt 0, + there exists \delta \gt 0 such that for all + x\in (a,c), if \abs{x - c} \lt \delta, + then \abs{f(x) - L} \lt \varepsilon. +

    +
    • + +
  • + Right-Hand Limit + limitone-sided + limitright-handed + +

    + Let f be a function defined on (c,b) + for some b \gt c and let L be a real number. + The statement that the limit of f(x), + as x approaches c from the right, + is L, (alternatively, + that the right-hand limit of f at + c is L) is denoted by + + \lim_{x\to c^+} f(x) = L + , + and means that for any \varepsilon \gt 0, + there exists \delta \gt 0 such that for all + x\in (c,b), if \abs{x - c} \lt \delta, + then \abs{f(x) - L} \lt \varepsilon. +

    +
    • +
    +

    +     +     + + + + + +

    + Practically speaking, when evaluating a left-hand limit, + we consider only values of x + to the left of c, + , where x\lt c. + The admittedly imperfect notation x\to c^- is used to + imply that we look at values of x to the left of c. + The notation has nothing to do with positive or negative values + of either x or c. + It's more like you are adding very small negative values to + c to get values for x. + A similar statement holds for evaluating right-hand limits; + there we consider only values of x to the right of c, + , x \gt c. + We can use the theorems from previous sections to help us evaluate these limits; + we just restrict our view to one side of c. +

    + + + + + + + +

    + We practice evaluating left- and right-hand + limits through a series of examples. +

    + + + Evaluating one-sided limits + +

    + Let f(x) = \begin{cases} + x \amp 0\leq x\leq 1 \\ + 3-x \amp 1\lt x\lt 2 + \end{cases}, + as shown in . + Find each of the following: + +

      +
    1. + \lim_{x\to 1^-} f(x) +
      2. +
    2. + \lim_{x\to 1^+} f(x) +
      3. +
    3. + \lim_{x\to 1} f(x) +
      4. +
    4. + f(1) +
      5. +
    5. + \lim_{x\to 0^+} f(x) +
      6. +
    6. + f(0) +
      7. +
    7. + \lim_{x\to 2^-} f(x) +
      8. +
    8. + f(2) +
      9. +
    +

    + +
    + A graph of f in + + + +

    + Graph of the piecewise function + f(x) = \begin{cases} x \amp 0\leq x\leq 1 \\ 3-x \amp 1\lt x\lt 2 \end{cases}. + There are two line segments: for 0\leq x\leq 1 we have a line segment + with positive slope, and for 1\lt x\lt 2 we have a line segment with negative slope. +

    +

    + The line segment with a positive slope starts at the point (0, 0) + and ends at (1, 1). The line segment with a negative slope starts + at (1, 2) and ends at (2, 1). +

    +

    + The start and end points of the line segment with a positive slope + are solid dots, indicating that those points are part of the graph. + The start and end points of the line segment + with a negative slope are hollow dots. + This tells that although the second line segment gets arbitrarily close + to the points (1,2) and (2,1), these points are not part of the graph. +

    + +

    + Since f(x) is close to 1 when x is close to 1, but x\lt 1, + while f(x) is close to 2 when x is close to 1, but x\gt 1, + we can conclude that the left and right hand limits are different. +

    +     + + Graph of a piecewise function that has different left and right hand limits when x=1. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.4, + xmax=2.4, + ymin=-.4, + ymax=2.4, + ] + \addplot[firstcurvestyle] coordinates {(0,0) (1,1)}; + \addplot[firstcurvestyle] coordinates {(1,2) (2,1)}; + \addplot[soliddot] coordinates {(0,0) (1,1)}; + \addplot[hollowdot] coordinates {(1,2) (2,1)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + For these problems, + the visual aid of the graph is likely more effective in + evaluating the limits than using f itself. + Therefore we will refer often to the graph. +

    + +

    +

      +
    1. +

      + As x goes to 1 from the left, + we see that f(x) is approaching the value of 1. +

      +

      + Therefore \lim\limits_{x\to 1^-} f(x) =1. +

      + +
      2. +
    2. +

      + As x goes to 1 from the right, + we see that f(x) is approaching the value of 2. + Recall that it does not matter that there is an + open circle there; + we are evaluating a limit, not the value of the function. +

      +

      + Therefore \lim\limits_{x\to 1^+} f(x)=2. +

      +
      3. +
    3. +

      + The limit of f as x approaches + 1 does not exist, as discussed in + . + The function does not approach one particular value, + but two different values from the left and the right. +

      +
      4. +
    4. +

      + Using the definition, and by looking at the graph, + we see that f(1) = 1. +

      +
      5. +
    5. +

      + As x goes to 0 from the right, + we see that f(x) is approaching 0. + Therefore \lim_{x\to 0^+} f(x)=0. + Note we cannot consider a left-hand limit at 0 + as f is not defined for values of x\lt 0. +

      +
      6. +
    6. +

      + Using the definition and the graph, f(0) = 0. +

      +
      7. +
    7. +

      + As x goes to 2 from the left, + we see that f(x) is approaching the value of 1. +

      +

      + Therefore \lim\limits_{x\to 2^-} f(x)=1. +

      +
      8. +
    8. +

      + The graph and the definition of the function show + that f(2) is not defined. +

      +
      9. +
    +

    +     + +     + +

    + Note how the left- and right-hand limits were different at x=1. + This, of course, causes the limit to not exist. + The following theorem states what is fairly intuitive: + the limit exists precisely when the left- and right-hand limits are equal. +

    + + + Limits and One-Sided Limits + +

    + Let f be a function defined on an open interval I + containing c, except possibly at c. + + limitdoes not exist + + Then + + \lim_{x\to c}f(x) = L + + if, and only if, + + \lim_{x\to c^-}f(x) = L \text{ and } \lim_{x\to c^+}f(x) = L + . +

    +     +     + +

    + The phrase if, and only if + means the two statements are equivalent: + they are either both true or both false. + If the limit equals L, + then the left and right hand limits both equal L. + If the limit is not equal to L, + then at least one of the left and right-hand limits is not equal to + L (it may not even exist). +

    + + + +

    + One thing to consider in Examples + is that the value of the function may/may not be equal to the value(s) + of its left/right-hand limits, even when these limits agree. +

    + + + Evaluating limits of a piecewise-defined function + +

    + Let f(x) =\begin{cases} + 2-x \amp 0\lt x\lt 1 \\ + (x-2)^2 \amp 1\lt x\lt 2 + \end{cases} + . + + Evaluate the following: +

      +
    1. + \lim_{x\to 1^-} f(x) +
      2. +
    2. + \lim_{x\to 1^+} f(x) +
      3. +
    3. + \lim_{x\to 1} f(x) +
      4. +
    4. + f(1) +
      5. +
    5. + \lim_{x\to 0^+} f(x) +
      6. +
    6. + f(0) +
      7. +
    7. + \lim_{x\to 2^-} f(x) +
      8. +
    8. + f(2) +
      9. +
    +

    +     + +

    + In this example, we evaluate each expression using just the definition of f, + without using a graph as we did in the previous example. +

    + +

    +

      +
    1. +

      + As x approaches 1 from the left, + we consider a limit where all x-values are less than 1. + This means we use the 2-x piece of the piecewise-defined function f. + As the x-values near 1, 2-x approaches 1; + that is, f(x) approaches 1. +

      +

      + Therefore \lim\limits_{x\to 1^-} f(x)=1. +

      +

      + A concise mathematical presentation of the above argument could be written as follows: + + \lim_{x\to 1^-}f(x) \amp = \lim_{x\to 1^-}(2-x) \quad (f(x)=x-2 \text{ for } 0\lt x\lt 1) + \amp = 2-1 = 1 \quad (\text{ properties of limits }) + + + +

      +
      2. +
    2. +

      + As x approaches 1 from the right, + we consider a limit where all x-values are greater than 1. + This means we use the (x-2)^2 piece of f. + As the x-values near 1, (x-2)^2 approaches 1; + that is, we see that again f(x) approaches 1. +

      +

      + Therefore \lim\limits_{x\to 1+} f(x)=1. +

      +

      + Once again, we can present our work computationally as follows: + + \lim_{x\to 1^+}f(x) \amp = \lim_{x\to 1^+}(x-2)^2 \quad (f(x)=(x-2)^2 \text{ for } 1\lt x\lt 2) + \amp = (1-2)^2=1 \quad (\text{ properties of limits }) + + +

      +
      3. +
    3. +

      + The limit of f as x approaches 1 exists and is 1, + as f approaches 1 from both the right and left. +

      +

      + Therefore \lim\limits_{x\to 1} f(x)=1. +

      +
      4. +
    4. +

      + Neither piece of f is defined for the x-value of 1; in other words, + 1 is not in the domain of f. Therefore f(1) is not defined. + +

      +
      5. +
    5. +

      + As x approaches 0 from the right, we consider a limit where all x-values are greater than 0. + This means we use the 2-x piece of f. As the x-values near 0, 2-x approaches 2; + that is, f(x) approaches 2. +

      +

      + So \lim\limits_{x\to 0^+} f(x)=2. + +

      +
      6. +
    6. +

      + f(0) is not defined as 0 is not in the domain of f. +

      +
      7. +
    7. +

      + As x approaches 2 from the left, we consider a limit where all x-values are less than 2. + This means we use the (x-2)^2 piece of f. As the x-values near 2, (x-2)^2 nears 0; + that is, f(x) approaches 0. +

      +

      + So \lim\limits_{x\to 2^-} f(x)=0. + +

      +
      8. +
    8. +

      + f(2) is not defined as 2 is not in the domain of f. +

      +
      9. +
    +

    + +

    + We can confirm our analytic result by consulting the graph of f shown in . + Note the open circles on the graph at x=0, 1 and 2, where f is not defined. +

    +
    + A graph of f from + + + +

    + Graph of f from . + The graph consists of two parts. + The first part is a line that starts at the point (0, 2) and ends at + (1, 1). The second part is a curve that starts at (1, 1) and ends + at (2, 0). + The points (0,2), (1,1), and (2,0) are all marked with hollow dots, + indicating that although the graph gets close to these points, they are not part of the graph. +

    +

    + The function is undefined for x = 1, + but the graph shows that f(x) approaches the same value (namely, 1) + from both the left and the right, allowing us to conclude that \lim_{x\to 1}f(x) exists, + and is equal to 1. +

    +     + + Graph of a piecewise-defined function. It is undefined when x=1, but has a limit at this point. + + + \begin{tikzpicture}[declare function = {func(\x) = (\x < 1) * (2 - x) + (\x > = 1) * ((x - 2)^2);}] + \begin{axis}[ + xmin=-.4, + xmax=2.4, + ymin=-.4, + ymax=2.4 + ] + \addplot+[domain=0:2] {func(x)}; + \addplot[hollowdot] coordinates {(0,2) (1,1) (2,0)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + + + Evaluating limits of a piecewise-defined function + +

    + Let f(x) =\begin{cases} (x-1)^2 \amp 0\leq x\leq 2, x\neq 1\\ 1 \amp x=1 + \end{cases} as shown in . + Evaluate the following: +

      +
    1. + \lim_{x\to 1^-} f(x) +
      2. +
    2. + \lim_{x\to 1^+} f(x) +
      3. +
    3. + \lim_{x\to 1} f(x) +
      4. +
    4. + f(1) +
      5. +
    +

    + +
    + Graphing f in + + + +

    + The graph of f(x) in . + The graph is a parabola, opening upward, plotted from + (0, 1) to (2, 1), with its vertex at (1,0). + However, there is a hole in the graph at the vertex, indicated by a hollow dot. + The function is still defined at x = 1, + because there is a solid dot at (1, 1). This shows that + f(1) = 1, but the limit of f as x approaches 1 + is 0. +

    +     + + Graph of f(x), showing an upward curved line and a solid dot at (1, 1). + Line of the equation is undefined at x = 1. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.4, + xmax=2.4, + ymin=-.4, + ymax=1.4 + ] + \addplot+[domain=0:2] {(x-1)^2}; + \addplot[soliddot] coordinates {(0,1) (1,1) (2,1)}; + \addplot[hollowdot] coordinates {(1,0)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + It is clear by looking at the graph that both the left- and right-hand limits of f, + as x approaches 1, are 0. + Thus it is also clear that the limit is 0; + , \lim_{x\to 1} f(x) = 0. + It is also clearly stated that f(1) = 1. +

    +     + +     + + + Evaluating limits of a piecewise-defined function + +

    + Let f(x) = \begin{cases} x^2 \amp 0\leq x\leq 1 \\ 2-x \amp 1\lt x\leq 2 + \end{cases} as shown in . + Evaluate the following: +

      +
    1. + \lim_{x\to 1^-} f(x) +
      2. +
    2. + \lim_{x\to 1^+} f(x) +
      3. +
    3. + \lim_{x\to 1} f(x) +
      4. +
    4. + f(1) +
      5. +
    +

    + +
    + Graphing f in + + + +

    + Graph of a piecewise-defined function, on the interval [0,2]. + For x values between 0 + and 1 (inclusive) the graph is an upward curved parabola. + For x values 1 to 2 (inclusive) the + graph is a straight line with a negative slope. +

    + +

    + The parabola and the line meet at the point (1,1). +

    +

    + There are three solid dots plotted on the graph at the points, + (0, 0), (1, 1), and (2, 0), to indicate where each part of the graph begins and ends. +

    +     + + Graph of the piecewise function for the above example. + + + \begin{tikzpicture}[declare function = {func(\x) = (\x < 1) * (x^2) + (\x > 1) * (2-x);}] + \begin{axis}[ + xmin=-.4, + xmax=2.4, + ymin=-.4, + ymax=1.4 + ] + \addplot+[domain=0:2] {func(x)}; + \addplot[soliddot] coordinates {(0,0) (1,1) (2,0)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + It is clear from the definition of the function and its graph that all of the following are equal: + + \lim_{x\to 1^-} f(x) = \lim_{x\to 1^+} f(x) =\lim_{x\to 1} f(x) =f(1) = 1 + . +

    +     + +     + +

    + In Examples + we were asked to find both \lim_{x\to 1}f(x) and f(1). + Consider the following table: +

    + + + + + \lim\limits_{x\to 1}f(x) + f(1) + + + + does not exist + 1 + + + + 1 + not defined + + + + 0 + 1 + + + + 1 + 1 + + + +

    + Only in + do both the function and the limit exist and agree. + This seems nice; in fact, + it seems normal. This is in fact an important situation which we explore in + entitled Continuity. In short, + a continuous function + is one in which when a function approaches a value as x\to c (, when \lim_{x\to c} f(x) = L), + it actually attains that value at c. + Such functions behave nicely as they are very predictable. +

    + + + + + + Terms and Concepts + + + + +

    + What are the three ways in which a limit may fail to exist? +

    + +     + + + +

    + The function approaches different values from the left and right; + the function grows without bound; + the function oscillates. +

    +     + +     + + + + + +

    + + If \lim\limits_{x\to 1^-}f(x)=5, + then \lim\limits_{x\to 1}f(x)=5. +

    +     + +

    + We aren't told the value of the right-hand limit, + and the right-hand limit might not equal 5. +

    +     + +     + + + + + +

    + + If \lim\limits_{x\to 1^-}f(x)=5, + then \lim\limits_{x\to 1^+}f(x)=5. +

    +     + +

    + Just because the left-hand limit equals some number, + that doesn't mean the right-hand limit exists or equals the same number. +

    +     + +     + + + + + +

    + + If \lim\limits_{x\to 1}f(x)=5, + then \lim\limits_{x\to 1^-}f(x)=5. +

    +     + +

    + When the limit exists, both the left-hand and right-hand limits also exist + and have the same value. +

    +     + +     + +     + + + Problems + + + +

    + Evaluate each expression using the given graph of f. +

    +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + @x=num_sort((0,random_subset(2,1..5))); + @y=random_subset(3,1..5); + $z=list_random(1..$y[1]-1,$y[1]+1..5); + $b=list_random($x[0],$x[2]); + if($envir{problemSeed}==1){@x=(0,1,2);@y=(1,2,0);$z=1;$b=0;}; + $a=$x[1]; + $f=Formula("($y[1]-$y[0])/($x[1]-$x[0])*(x-$x[0])+$y[0]"); + $g=Formula("($y[1]-$y[2])/($x[1]-$x[2])**2*(x-$x[2])^2+$y[2]"); + $xmin=min(@x)-1;$xmax=max(0,@x)+1; + $ymin=min(0,@y,$z)-1;$ymax=max(@y,$z)+1; + @L=($y[1],$y[1],$y[1],$z); + ($L[4],$L[5])=($b==0)?(Compute("DNE"),$y[0]):($y[2],Compute("DNE")); + + + + +

    + Graph that shows the domain to + . For x values less than + the graph is a straight line and for + x values where \lt x \leq + the graph is slightly curved. When x equals + the graph is defined by the point + (, ). +

    +

    + The straight line is defined by the points + (, ) and + (, ). + The curved line is defined by the points + (, ) and + (, ). +

    +     + + Graph for exercise problem 5. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = $ymin, + ymax = $ymax, + ] + \addplot[firstcurvestyle, domain=$x[0]:$x[1]] {$f}; + \addplot[firstcurvestyle, domain=$x[1]:$x[2]] {$g}; + \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[1],$z) ($x[2],$y[2])}; + \addplot[hollowdot] coordinates {($x[1],$y[1])}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + @x=num_sort((0,random_subset(2,1..5))); + @y=random_subset(2,1..5); + $y[2]=random(0,5,1); + $z=list_random(1..$y[1]-1,$y[1]+1..5); + $b=list_random($x[0],$x[2]); + $a=$x[1]; + $f=Formula("($y[1]-$y[0])/($x[1]-$x[0])*(x-$x[0])+$y[0]"); + $g=Formula("($z-$y[2])/($x[1]-$x[2])*(x-$x[2])+$y[2]"); + $xmin=min(0,@x)-1;$xmax=max(0,@x)+1;$ymin=min(0,@y,$z)-1;$ymax=max(0,@y,$z)+1; + @L=($y[1],$z,Compute("DNE"),$z); + ($L[4],$L[5])=($b==0)?(Compute("DNE"),$y[0]):($y[2],Compute("DNE")); + + + + +

    + Graph that shows the domain to . + For x values in the interval \leq x \lt + the graph is a line segment from (,) (a solid dot) + to (, ) (a hollow dot). + At x equals there is a jump in the y value. + For \leq x\leq the graph is the line segment + from (, ) to (,) + (both endpoints are solid dots). +

    +     + + Graph for exercise problem 6. See long description for details. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = $ymin, + ymax = $ymax, + ] + \addplot[firstcurvestyle, domain=$x[0]:$x[1]] {$f}; + \addplot[firstcurvestyle, domain=$x[1]:$x[2]] {$g}; + \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[1],$z) ($x[2],$y[2])}; + \addplot[hollowdot] coordinates {($x[1],$y[1])}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + @x=num_sort((0,random_subset(2,1..5))); + $y[0]=random(0,5,1); + $y[1]=Compute("inf"); + $y[2]=random(0,5,1); + $b=list_random($x[0],$x[2]); + if($envir{problemSeed}==1){@x=(0,1,2);@y=(0,Compute("inf"),0);$b=2;}; + $a=$x[1]; + $c=($b==$x[0])?$x[2]:$x[0]; + $f=Formula("sec((x-$x[0])/($x[1]-$x[0])*90)-1+$y[0]"); + $g=Formula("sec((x-$x[2])/($x[1]-$x[2])*90)-1+$y[2]"); + $xmin=min(0,@x)-1;$xmax=max(0,@x)+1;$ymin=-1;$ymax=max(0,$y[0],$y[2])+4; + $xmidlow = asec($ymax + 1 - $y[0])/90*180/pi * ($x[1]-$x[0]) + $x[0]; + $xmidhigh = asec($ymax + 1 - $y[2])/90*180/pi * ($x[1]-$x[2]) + $x[2]; + @L=(OneOf(Compute("DNE"),Compute("INF")),OneOf(Compute("DNE"),Compute("INF")),OneOf(Compute("DNE"),Compute("INF")),Compute("DNE")); + ($L[4],$L[5])=($b==0)?(Compute("DNE"),Compute("DNE")):($y[2],$y[0]); + + + + +

    + Graph that shows the domain to . There is a + vertical asymptote at . The graph on either side has + an upward curve. The line on the left of the asymptote starts at + (, ) and has a positive slope, + as x gets closer to the slope of the line + gets steeper. The line to the right of the asymptote starts at + (, ) and has a negative slope, + as x gets closer to the slope of the line + gets steeper. +

    +     + + Graph for exercise problem 7. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = $ymin, + ymax = $ymax, + ] + \addplot[firstcurvestyle,infiniteright, domain=$x[0]:$xmidlow] {$f}; + \addplot[firstcurvestyle, infiniteleft, domain=$xmidhigh:$x[2]] {$g}; + \addplot[asymptote, -] coordinates {($x[1],$ymin) ($x[1],$ymax)}; + \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[2],$y[2])}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + @x=num_sort((0,random_subset(2,1..5))); + # actual y values the function takes + @y=random_subset(3,1..5); + # y values for limit points (w from left, z from right) + ($w,$z)=random_subset(2,1..$y[1]-1,$y[1]+1..5); + $a=$x[1]; + $f=Formula("($w-$y[0])/($x[1]-$x[0])*(x-$x[0])+$y[0]"); + $g=Formula("($y[2]-$z)/($x[2]-$x[1])^2*(x-$x[1])^2+$z"); + $xmin=min(0,@x)-1;$xmax=max(0,@x)+1;$ymin=min(0,@y,$z,$w)-1;$ymax=max(0,@y,$z,$w)+1; + @L=($w,$z,Compute("DNE"),$y[1]); + + + + +

    + Graph that shows the domain to . + The graph is piecewise, with two parts. + The first part is a line from a solid dot at + (, ) to a hollow dot at + (, ). + The second part is a parabolic arc from a hollow dot at (, ) + to a solid dot at (, ). + There is also a point marked by a solid dot at (, ). +

    +     + + Graph for exercise problem 8. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = $ymin, + ymax = $ymax, + ] + \addplot[firstcurvestyle, domain=$x[0]:$x[1]] {$f}; + \addplot[firstcurvestyle, domain=$x[1]:$x[2]] {$g}; + \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[1],$y[1]) ($x[2],$y[2])}; + \addplot[hollowdot] coordinates {($x[1],$w) ($x[1],$z)}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + @x=num_sort((0,random_subset(2,1..5))); + @y=random_subset(2,1..5); + $y[2]=$y[0]; + if($envir{problemSeed}==1){@x=(0,1,2);@y=(0,2,0);}; + $a=$x[1]; + $f=Formula("($y[0]-$y[1])/($x[0]-$x[1])^2*(x-$x[1])^2+$y[1]"); + $g=Formula("($y[2]-$y[1])/($x[2]-$x[1])*(x-$x[1])+$y[1]"); + $xmin=min(0,@x)-1;$xmax=max(0,@x)+1;$ymin=min(0,@y,0)-1;$ymax=max(0,@y,0)+1; + @L=($y[1],$y[1],$y[1],$y[1]); + + + + +

    + Graph that shows the domain to . The graph + is defined on the interval \leq x \leq + . At (, ) + there is a change in the graphs curve, it goes from a curved line + to a straight line. +

    +     + + Graph for exercise problem 9. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = $ymin, + ymax = $ymax, + ] + \addplot[firstcurvestyle, domain=$x[0]:$x[1]] {$f}; + \addplot[firstcurvestyle, domain=$x[1]:$x[2]] {$g}; + \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[2],$y[2])}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $x[0]=random(-6,-2,1); + $x[1]=$x[0]+random(3,5,1); + $x[2]=$x[1]+random(3,5,1); + ($y[1],$y[2],$y[3])=random_subset(3,-6..6); + $y[0]=list_random(-6..$y[1]-1,$y[1]+1..6); + $y[4]=list_random(-6..$y[3]-1,$y[3]+1..6); + if($envir{problemSeed}==1){@x=(-4,0,4);@y=(-4,4,0,-4,4);}; + $a=$x[1]; + $f=Formula("cos((x-$x[1])/($x[0]-$x[1])*180)*($y[1]-$y[0])/2+($y[0]+$y[1])/2"); + $g=Formula("cos((x-$x[2])/($x[1]-$x[2])*180)*($y[4]-$y[3])/2+($y[3]+$y[4])/2"); + $xmin=min(0,@x)-1;$xmax=max(0,@x)+1;$ymin=min(0,@y)-1;$ymax=max(0,@y)+1; + @L=($y[1],$y[3],Compute("DNE"),$y[2]); + + + + +

    + Graph that shows the domain to . There are two + lines, one on the interval \leq x \lt + .The second line is on the interval + \lt x \leq . +

    +

    + The first line is defined by the solid point (, ) + and hollow point (, ). + The second line is defined by the hollow point (, ) + and solid point (, ). + For x = 0 the graph is defined by the point + (, ). +

    +     + + Graph for exercise problem 10. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = $ymin, + ymax = $ymax, + ] + \addplot[firstcurvestyle, domain=$x[0]:$x[1]] {cos((x-$x[1])/($x[0]-$x[1])*180)*($y[1]-$y[0])/2+($y[0]+$y[1])/2}; + \addplot[firstcurvestyle, domain=$x[1]:$x[2]] {$g}; + \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[1],$y[2]) ($x[2],$y[4])}; + \addplot[hollowdot] coordinates {($x[1],$y[1]) ($x[1],$y[3])}; + \end{axis} + \end{tikzpicture} + + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + @L=(2,2,2,0,2,2,2,Compute("DNE")); + + + + +

    + Graph that shows the domain -4 to + 4. + There are four lines that make up this graph. The first + going from left to right is defined by the solid point + (-4, 0) and hollow point (-2. 2), and has + a positive slope. The second line is defined by the points + (-2, 2), 0, 0 with a negative slope. The + third line by (0, 0), 2, 2 with a positive slope. + And the last line is defined by the hollow point (2, 2) + and solid point (4, 0), with a negative slope. +

    +

    + For x = -2, F(x) = 0 +

    +     + + Graph for exercise problem 11. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = -5, + xmax = 5, + ymin = -4, + ymax = 4, + ] + \addplot[firstcurvestyle, domain=-4:-2] {x+4}; + \addplot[firstcurvestyle, domain=-2:0] {-x}; + \addplot[firstcurvestyle, domain=0:2] {x}; + \addplot[firstcurvestyle, domain=2:4] {4-x}; + \addplot[soliddot] coordinates {(-4,0) (-2,0) (4,0)}; + \addplot[hollowdot] coordinates {(-2,2) (2,2)}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to -2^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to -2^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to -2} f(x) +

    +

    + +

    +     +     + + + +

    + f(-2) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to 2^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to 2^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to 2} f(x) +

    +

    + +

    +     +     + + + +

    + f(2) +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + Context()->variables->are(a=>'Real'); + @L=(Formula("a-1"),Formula("a"),Compute("DNE"),Formula("a")); + + + + +

    + The graph is + made up of eight separate lines that do not connect to each + other. Each of the lines is parallel with the x axis, defined + by a solid dot on its left and a hollow dot on its right, and + has a length of 1 unit. The first line is inside quadrant 3 + of the graph and the last is in quadrant 1, there are no + lines in quadrants 2 or 4. Each of the other six lines is between + the first and last line moving along a straight diagonal path. +

    +

    + The following are the points that define each line: +

      +
    1. solid dot at (-4, -4) and hollow dot at (-3, -4)
      2. +
    2. solid dot at (-3, -3) and hollow dot at (-2, -3)
      3. +
    3. solid dot at (-2, -2) and hollow dot at (-1, -2)
      4. +
    4. solid dot at (-1, -1) and hollow dot at (0, -1)
      5. +
    5. solid dot at (0, 0) and hollow dot at (1, 0)
      6. +
    6. solid dot at (1, 1) and hollow dot at (2, 1)
      7. +
    7. solid dot at (2, 2) and hollow dot at (3, 2)
      8. +
    8. solid dot at (3, 3) and hollow dot at (4, 3)
      9. +
    +

    +     + + Graph for exercise problem 12. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = -5, + xmax = 5, + ymin = -5, + ymax = 5, + ] + \foreach \i in {-4,...,3} { + \addplot[firstcurvestyle, domain=\i:\i+1] {\i}; + \addplot[soliddot] coordinates {(\i,\i)}; + \addplot[hollowdot] coordinates {(\i+1,\i)}; + + } + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + + +

    + Let a be an integer with -3\leq a\leq3. +

    +     + + + +

    + \lim\limits_{x\to a^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to a^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to a} f(x) +

    +

    + +

    +     +     + + + +

    + f(a) +

    +

    + +

    +     +     +     +     +     + + + +

    + Evaluate the given limits of the piecewise defined function. +

    +     + + + + + $a=random(-2,4,1); + $b=non_zero_random(-5,5,1); + do{$c=non_zero_random(-5,5,1);}until($a+$b!=$a**2+$c); + if($envir{problemSeed}==1){$a=1;$b=1;$c=-5}; + @f=(Formula("x+$b")->reduce,Formula("x^2+$c")->reduce); + Context("PiecewiseFunction"); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $g=PiecewiseFunction("x <= $a" => "$f[0]", "x > $a" => "$f[1]"); + @L=($f[0]->eval(x=>$a),$f[1]->eval(x=>$a)); + $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); + $L[3]=$g->eval(x=>$a); + + +

    + f(x) = +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     +     +     + + + + + $a=random(-2,4,1); + $b=non_zero_random(-2,2,1); + ($c,$d)=random_subset(2,-5..-1,1..5); + if($envir{problemSeed}==1){$a=0;$b=2;$c=5;$d=-1;}; + @f=(Formula("$b x^2 + $c x+$d")->reduce,Formula("sin(x-$a)")->reduce); + Context("PiecewiseFunction"); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $g=PiecewiseFunction("x < $a" => "$f[0]", "x >= $a" => "$f[1]"); + @L=($f[0]->eval(x=>$a),$f[1]->eval(x=>$a)); + $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); + $L[3]=$g->eval(x=>$a); + + +

    + f(x) = +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     +     +     + + + + + ($a,$b)=num_sort(random_subset(2,-5..-1,1..5)); + if($envir{problemSeed}==1){$a=-1;$b=1;}; + @f=(Formula("x^2 + ($a+1) x+(($a)^3+1-($a)^2-($a+1)*$a)")->reduce,Formula("x^3+1")->reduce,Formula("x^2 + ($b-1) x+(($b)^3+1-($b)^2-($b-1)*$b)")->reduce); + Context("PiecewiseFunction"); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $g=PiecewiseFunction("x < $a" => "$f[0]", "$a <= x <= $b" => "$f[1]", "x > $b" => $f[2]); + @L=($f[0]->eval(x=>$a),$f[1]->eval(x=>$a)); + $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); + $L[3]=$g->eval(x=>$a); + ($L[4],$L[5])=($f[1]->eval(x=>$b),$f[2]->eval(x=>$b)); + $L[6]=($L[4]==$L[5])?$L[4]:Compute("DNE"); + $L[7]=$g->eval(x=>$b); + + +

    + f(x) = +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + @f=(Formula("cos(x)"),Formula("sin(x)")); + @L=(-1,0,Compute("DNE"),0); + + +

    + f(x) = \begin{cases}\amp x\lt\pi\\ \amp x\geq\pi\end{cases} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to\pi^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to\pi^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to\pi} f(x) +

    +

    + +

    +     +     + + + +

    + f(\pi) +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + Context()->variables->add(a=>'Real'); + @L=(Formula("1-cos^2(a)"),Formula("sin^2(a)"),OneOf(Formula("1-cos^2(a)"),Formula("sin^2(a)")),Formula("sin^2(a)")); + + +

    + f(x) = \begin{cases}1-\cos^2(x)\amp x<a\\\sin^2(x)\amp x\geq a\end{cases} + where a is a real number. +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to a^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to a^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to a} f(x) +

    +

    + +

    +     +     + + + +

    + f(a) +

    +

    + +

    +     +     +     +     + + + + + $a=random(-3,3,1); + @y= random_subset(3,-2..2); + if($envir{problemSeed}==1){$a=1;@y=(1,0,-1);}; + @f=(Formula("x+$y[0]")->reduce,Formula("x+$y[1]")->reduce,Formula("x+$y[2]")->reduce); + Context("PiecewiseFunction"); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $g=PiecewiseFunction("x < $a" => "$f[0]", "x = $a" => "$f[1]", "x > $a" => $f[2]); + @L=($f[0]->eval(x=>$a),$f[2]->eval(x=>$a)); + $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); + $L[3]=$g->eval(x=>$a); + + +

    + f(x) = +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     +     +     + + + + + $a=random(-3,3,1); + @y=random_subset(2,-5..5); + if($envir{problemSeed}==1){$a=2;@y=(4,3);}; + @f=(Formula("x^2+($y[0]-2*$a)x+($y[0]-($a)^2-($y[0]-2*$a)*$a)")->reduce->reduce,Formula("x+($y[1]-$a)")->reduce,Formula("-x^2+$a*x+$y[0]")->reduce); + Context("PiecewiseFunction"); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + $g=PiecewiseFunction("x < $a" => "$f[0]", "x = $a" => "$f[1]", "x > $a" => $f[2]); + @L=($f[0]->eval(x=>$a),$f[2]->eval(x=>$a)); + $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); + $L[3]=$g->eval(x=>$a); + + +

    + f(x) = +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + f() +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + Context()->variables->add(c=>'Real'); + $L=Formula("c"); + + +

    + f(x) = \begin{cases}a(x-b)^2+c\amp x\lt b\\a(x-b)+c\amp x\geq b\end{cases} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to b^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to b^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to b} f(x) +

    +

    + +

    +     +     + + + +

    + f(b) +

    +

    + +

    +     +     +     +     + + + + + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + @L=(-1,1,Compute("DNE"),0); + + +

    + f(x) = \begin{cases}\frac{\abs{x}}{x}\amp x\neq0\\0\amp x=0\end{cases} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to 0^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to 0^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to 0} f(x) +

    +

    + +

    +     +     + + + +

    + f(0) +

    +

    + +

    +     +     +     +     +     +     +     +
    +
    + Continuity + +

    + As we have studied limits, + we have gained the intuition that limits measure + where a function is heading. That is, + if \lim\limits_{x\to 1} f(x) = 3, + then as x is close to 1, + f(x) is close to 3. + We have seen, though, + that this is not necessarily a good indicator of what f(1) actually is. + This can be problematic; + functions can tend to one value but attain another. + This section focuses on functions that + do not exhibit such behavior. +

    + + + Continuous Function + +

    + Let f be a function whose domain contains an open interval I. + + continuous function + functioncontinuous + +

      +
    1. +

      + f is continuous at a point c in I + if \lim\limits_{x\to c}f(x) = f(c). + + continuousat a point + +

      +
      2. +
    2. +

      + f is continuous on the open interval I + if f is continuous at c for all values of + c in I. + + continuouson an interval + + If f is continuous on (-\infty,\infty), + we say f is continuous everywhere + (or everywhere continuous). + + continuouseverywhere + everywhere continuous + +

      +
      3. +
    +

    +     +     + +

    + Note that in , + a function f can only be continuous at a point c if c is in the domain of f. +

    + + + +

    + A useful way to establish whether or not a function f + is continuous at c is to verify the following three things: + +

      +
    1. +

      + \lim\limits_{x\to c} f(x) exists, +

      +
      2. +
    2. +

      + f(c) is defined, and +

      +
      3. +
    3. +

      + \lim\limits_{x\to c} f(x) = f(c). +

      +
      4. +
    +

    + + + + + Finding intervals of continuity + +

    + Let f be defined as shown in . + Give the interval(s) on which f is continuous. +

    + +
    + A graph of f in + + + +

    + The graph of a piecewise-defined function is shown, for x from 0 to 3. + For 0 \leq x \lt 1 + the graph looks like a parabola opening downward. + This part of the graph approaches, but does not reach, the point (1,1). + There is a hollow dot at (1,1), indicating that f(1) is undefined. + For 1 \lt x \leq 2 the graph + is a horizontal line segment, with y=1. + For 2 \leq x \leq 3 the graph again has the appearance of a downward-facing parabola + that begins at (2,1) and ends at (3,1). +

    +     + + Graph of a function with a discontinuity when x=1. Although the limit at 1 exists, f(1) is undefined. + + + \begin{tikzpicture}[declare function = {func(\x) = (\x >= 0)*(\x <= 1)*(-(\x-1/4)^2+1/16+1.5) + (\x > 1)*(\x <= 2) + (\x > 2)*(\x <= 3)*((2-\x)*(\x-3)+1);}] + \begin{axis}[ + xmin=-.4, + xmax=3.4, + ymin=-.4, + ymax=1.7 + ] + \addplot+[domain=0:3,samples=48] {func(x)}; + \addplot[soliddot] coordinates {(0,1.5) (2,1) (3,1)}; + \addplot[hollowdot] coordinates {(1,1)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + We proceed by examining the three criteria for continuity. +

      +
    1. +

      + The limits \lim\limits_{x\to c} f(x) exists for all + c between 0 and 3. +

      +
      2. +
    2. +

      + f(c) is defined for all c between 0 and 3, + except for c=1. + We know immediately that f cannot be continuous at x=1. +

      +
      3. +
    3. +

      + The limit \lim\limits_{x\to c} f(x) = f(c) for all + c between 0 and 3, + except, of course, for c=1. +

      +
      4. +
    +

    + +

    + We conclude that f is continuous at every point of the interval + (0,3) except at x=1. + Therefore f is continuous on (0,1) and (1,3). +

    + +     +     + + + Finding intervals of continuity + +

    + The floor function, + + floor function + functionfloor + + f(x) = \lfloor x \rfloor, + returns the largest integer smaller than, or equal to, + the input x. (For example, + f(\pi) = \lfloor \pi \rfloor = 3.) The graph of f in + + demonstrates why this is often called a step function. +

    + +

    + Give the intervals on which f is continuous. +

    + +
    + A graph of the step function in + + + +

    + Shows the graph of the greatest integer function, for x from -2 to 3. There are five + horizontal line segments in a staircase configuration, ascending from left to right. Each segment is + one unit in length and includes its left endpoint, but the right endpoint of each segment is not included. + The first segment is from (-2, -2) to (-1, -2), + the second from (-1, -1) to (0, -1), the third from (0, 0) to (1, 0), + the fourth from (1, 1) to (2, 1), and the fifth from (2, 2) to (3, 2). +

    +     + + Graph of a step function whose value at x is the greatest integer less than or equal to x. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-2.4, + xmax=3.4, + ymin=-2.4, + ymax=2.4 + ] + \addplot[firstcurvestyle,-] coordinates {(-2,-2) (-1,-2)}; + \addplot[firstcurvestyle,-] coordinates {(-1,-1) (0,-1)}; + \addplot[firstcurvestyle,-] coordinates {(0,0) (1,0)}; + \addplot[firstcurvestyle,-] coordinates {(1,1) (2,1)}; + \addplot[firstcurvestyle,-] coordinates {(2,2) (3,2)}; + \addplot[soliddot] coordinates {(-2,-2) (-1,-1) (0,0) (1,1) (2,2)}; + \addplot[hollowdot] coordinates {(-1,-2) (0,-1) (1,0) (2,1) (3,2)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + We examine the three criteria for continuity. +

      +
    1. +

      + The limits \lim\limits_{x\to c} f(x) + do not exist at the jumps from one step to the next, + which occur at all integer values of c. + Therefore the limits exist for all c except when + c is an integer. +

      +
      2. +
    2. +

      + The function is defined for all values of c. +

      +
      3. +
    3. +

      + The limit \lim\limits_{x\to c} f(x) = f(c) for all values of + c where the limit exist, + since each step consists of just a line. +

      +
      4. +
    +

    + +

    + We conclude that f is continuous everywhere except at integer values of c. + So the intervals on which f is continuous are + + \ldots, (-2,-1), (-1,0), (0,1), (1,2), \ldots + . + We could also say that f is continuous on all intervals of the form + (n,n+1) where n is an integer. +

    +     +     + + + +

    + Our definition of continuity on an interval specifies the interval is an open interval. + We can extend the definition of continuity to closed intervals of the form [a,b] + by considering the appropriate one-sided limits at the endpoints. +

    + + + + + Continuity on Closed Intervals + +

    + Let f be defined on the closed interval + [a,b] for some real numbers a\lt b. +

    + +

    + We say f is continuous on the closed interval + [a,b] if: +

      +
    1. +

      + f is continuous on (a,b), +

      +
      2. +
    2. +

      + \lim\limits_{x\to a^+} f(x) = f(a) and +

      +
      3. +
    3. +

      + \lim\limits_{x\to b^-} f(x) = f(b). +

      +
      4. +
    +

    +     +     + +

    + We can make the appropriate adjustments to talk about continuity on half-open intervals such as + [a,b) or (a,b] if necessary. +

    + +

    + If the domain of f includes values less than a, + we say that Item in + indicates that f is continuous from the right at a. + But if f is undefined for x\lt a, + we can say that f is continuous at a without ambiguity. +

    + +

    + Similarly, Item indcates that f + is continuous from the left at b, + and if f is not defined for x\gt b, + we can simply say that f is continuous at b. +

    + +

    + For example, it makes sense to say that the function f(x)=\sqrt{1-x^2} + is continuous at 1 and -1, + while the floor function in is continuous from the left at 1 and -1, + but is not continuous at these points. +

    + +

    + Using this new definition, + we can adjust our answer in + by stating that f is continuous on [0,1) and (1,3], + as mentioned in that example. + We can also revisit + and state that the floor function is continuous on the following half-open intervals + + \ldots, [-2,-1), [-1,0), [0,1), [1,2), \ldots + . +

    + + + +

    + This can tempt us to conclude that f is continuous everywhere; + after all, if f is continuous on [0,1) and [1,2), + isn't f also continuous on [0,2)? + Of course, the answer is no, + and the graph of the floor function immediately confirms this. +

    +

    + Continuous functions are important as they behave in a predictable fashion: + functions attain the value they approach. + Because continuity is so important, + most of the functions you have likely seen in the past are continuous on their domains. + This is demonstrated in the following example where we examine the intervals of continuity of a variety of common functions. +

    + + + Determining intervals on which a function is continuous + +

    + For each of the following functions, + give the domain of the function and the interval(s) on which it is continuous. +

      +
    1. +

      + f(x) = 1/x +

      +
      2. +
    2. +

      + f(x) = \sin(x) +

      +
      3. +
    3. +

      + f(x) = \sqrt{x} +

      +
      4. +
    4. +

      + f(x) = \sqrt{1-x^2} +

      +
      5. +
    5. +

      + f(x) = \abs{x} +

      +
      6. +
    +

    +     + +

    + We examine each in turn. +

      +
    1. +

      + The domain of f(x) = 1/x is (-\infty,0) \cup (0,\infty). + As it is a rational function, + we apply + to recognize that f is continuous on all of its domain. +

      +
      2. +
    2. +

      + The domain of f(x) = \sin(x) is all real numbers, + or (-\infty,\infty). + Applying + shows that \sin(x) is continuous everywhere. +

      +
      3. +
    3. +

      + The domain of f(x) = \sqrt{x} is [0,\infty). + Applying + shows that f(x) = \sqrt{x} is continuous on its domain of + [0,\infty). +

      +
      4. +
    4. +

      + The domain of f(x) = \sqrt{1-x^2} is [-1,1]. + Applying Theorems + and + shows that f is continuous on all of its domain, + [-1,1]. +

      +
      5. +
    5. +

      + The domain of f(x) = \abs{x} is (-\infty,\infty). + We can define the absolute value function as + + f(x) = \begin{cases}-x \amp x\lt 0 \\ x \amp x\geq 0\end{cases} + . + Each piece of this piecewise defined function is continuous on all of its domain, + giving that f is continuous on + (-\infty,0) and [0,\infty). + We cannot assume this implies that f is continuous on (-\infty,\infty); + we need to check that \lim\limits_{x\to 0}f(x) = f(0), + as x=0 is the point where f transitions from one piece + of its definition to the other. + It is easy to verify that this is indeed true, + hence we conclude that f(x) = \abs{x} is continuous everywhere. +

      +
      6. +
    +

    +     + +     + +

    + Continuity is inherently tied to the properties of limits. + Because of this, + the properties of limits found in Theorems + and apply to continuity as well. + Further, now knowing the definition of continuity we can re-read + + as giving a list of functions that are continuous on their domains. + The following theorem states how continuous functions can be combined to form other continuous functions, + followed by a theorem which formally lists functions that we know are continuous on their domains. +

    + + + Properties of Continuous Functions + +

    + Let f and g be continuous functions on an interval I, + let c be a real number and let n be a positive integer. + The following functions are continuous on I. +

    +
  • + Sums/Difference +

    + f\pm g +

    +
    • +
  • + Constant Multiple +

    + c\cdot f +

    +
    • +
  • + Product +

    + f\cdot g +

    +
    • +
  • + Quotient +

    + f/g (as long as g\neq 0 on I) +

    +
    • +
  • + Power +

    + f^n +

    +
    • +
  • + Root +

    + \sqrt[n]{f} (If n is even then require f(x)\geq 0 on I.) +

    +
    • +
  • + Compositions +

    + Adjust the definitions of f and g to: Let f be continuous on I, + where the range of f on I is J, + and let g be continuous on J. + Then g\circ f, , g(f(x)), + is continuous on I. +

    +
    • +
    +

    +     +     + + + + + + + + + Continuous Functions + +

    + Let n be a positive integer. + The following functions are continuous on their domains. + + continuous functionproperties + +

      +
    1. +

      + f(x) = \sin(x) +

      +
      2. +
    2. +

      + f(x) = \tan(x) +

      +
      3. +
    3. +

      + f(x) = \sec(x) +

      +
      4. +
    4. +

      + f(x) = \ln(x) +

      +
      5. +
    5. +

      + f(x) = a^x (a\gt 0) +

      +
      6. +
    6. +

      + f(x) = \cos(x) +

      +
      7. +
    7. +

      + f(x) = \cot(x) +

      +
      8. +
    8. +

      + f(x) = \csc(x) +

      +
      9. +
    9. +

      + f(x) = \sqrt[n]{x}, where n is a positive integer. +

      +
      10. +
    +

    +     +     + +

    + Note also that we can interpret as telling us that polynomial + and rational functions are also continuous on their domains. +

    + + + + + Checking continuity at a given point + +

    + Let f(x) = \begin{cases}x^2+2x+1,\amp \text{ if } x\lt 0\\2, \amp \text{ if } x=0\\5x+\cos(x^3), \amp \text{ if } x\gt 0\end{cases}. + For each point below, determine whether or not f is continuous at that point, and explain why. +

      +
    1. x=0
      2. +
    2. x=-2
      3. +
    3. x=2
      4. +
    +

    +     + +

    +

      +
    1. +

      + At x=0, we must rely on . + If x\lt 0, then f(x) = x^2+2x+1; therefore, + + \lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(x^2+2x+1)=1 + . + Similarly, for x\gt 0, f(x)=5x+\cos(x^3), so + + \lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}(5x+\cos(x^3)) = 0+\cos(0)=1 + . + Therefore, \lim_{x\to 1}f(x)=1 exists. +

      + +

      + However, f is not continuous at 0, since + + f(0)=2 \neq 1 = \lim_{x\to 0}f(x) + . +

      +
      2. + +
    2. +

      + If x is near -2, then we can assume that x\lt 0, + and therefore f(x)=x^2+2x+1. + This is a polynomial function, and by , + every polynomial function is continuous on its domain. + Therefore, f is continuous at -2. +

      +
      3. + +
    3. +

      + If x is near 2, then we can assume x\gt 0. + Since 2x and x^3 are polynomial, + we know that these functions are continuous, by . + We know that \cos(x) is continuous by , + and by , the composition of continuous functions is continuous, + so \cos(x^3) is continuous. + Also by , the sum of continuous functions is continuous. + Therefore, for any x\gt 0, f(x)=2x+\cos(x^3) is continuous, and in particular, + f is continuous at x=2. +

      +
      4. +
    +

    +     +     + + + + + + +

    + We apply these theorems in the following Example. +

    + + + Determining intervals on which a function is continuous + +

    + State the interval(s) on which each of the following functions is continuous. +

      +
    1. +

      + f(x) = \sqrt{x-1} + \sqrt{5-x} +

      +
      2. +
    2. +

      + f(x) = x\sin(x) +

      +
      3. +
    3. +

      + f(x) = \tan(x) +

      +
      4. +
    4. +

      + f(x) = \sqrt{\ln(x)} +

      +
      5. +
    +

    +     + +

    + We examine each in turn, + applying Theorems + and as appropriate. +

    + +
    + A graph of f(x)=\sqrt{x-1}+\sqrt{5-x} + + + +

    + Shows the graph of f(x)=\sqrt{x-1}+\sqrt{5-x} on its domain [1,5]. + The graph looks somewhat like the top of a slice of bread, + rising from the point (1,2) to it highest point near x=3, + and then falling to the point (5,2). +

    +     + + Graph of function that is the sum of two square root functions. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.4, + xmax=5.4, + ymin=-.4, + ymax=3.2, + ] + \addplot+[domain=0:180] ({3+2*cos(x)},{sqrt(2+2*cos(x))+sqrt(2-2*cos(x))}); + \addplot[soliddot] coordinates {(1,2) (5,2)}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    +

      +
    1. +

      + The square root terms are continuous on the intervals + [1,\infty) and (-\infty,5], respectively. + As f is continuous only where each term is continuous, + f is continuous on [1,5], + the intersection of these two intervals. + A graph of f is given in + . +

      +
      2. +
    2. +

      + The functions y=x and + y=\sin(x) are each continuous everywhere, + hence their product is, too. +

      +
      3. +
    3. +

      + + states that f(x) = \tan(x) is continuous on its domain. + Its domain includes all real numbers except odd multiples of \pi/2. + Thus the intervals on which f(x) = \tan(x) is continuous are + + \ldots \left(-\frac{3\pi}{2},-\frac{\pi}2\right),\ \left(-\frac{\pi}2,\frac{\pi}2\right),\ \left(\frac{\pi}2,\frac{3\pi}2\right),\ldots, + . +

      +
      4. +
    4. +

      + Here, f(x) is the composition g(h(x)), where + g(x) = \sqrt{x} and h(x)=\ln(x). + The domain of g is [0,\infty), while + the range of h is (-\infty,\infty). + If we restrict the domain to + [1,\infty), then the output from h(x)=\ln(x) is restricted to + [0,\infty), on which g(x) = \sqrt{x} is defined. + Thus the domain of f(x) = \sqrt{\ln(x)} is [1,\infty). +

      +
      5. +
    +

    +     + +     + + + + Classification of discontinuities +

    + We now know what it means for a function to be continuous, + so of course we can easily say what it means for a function to be discontinuous; + namely, not continuous. + However, to better understand continuity, + it is worth our time to discuss the different ways in which a function can fail to be discontinuous. + By definition, a function f is continuous at a point + a in its domain if \lim\limits_{x\to a}f(x) = f(a). + If this equality fails to hold, then f is not continuous. + We note, however, that there are a number of different things that can go wrong with this equality. +

    + + + +

    +

      +
    1. +

      + \lim\limits_{x\to a}f(x)=L exists, + but L\neq f(a), or f(a) is undefined. + Such a discontinuity is called a + removable discontinuity + discontinuityremovable. +

      + +

      + A removable discontinuity can be pictured as a + hole in the graph of f. + The term removable refers to the fact that + by simply redefining f(a) to equal L + (that is, changing the value of f at a single + point), we can create a new function that is + continuous at x=a, and agrees with f + at all x\neq a. +

      +
      2. + +
    2. +

      + \lim\limits_{x\to a^+}f(x) = L and + \lim\limits_{x\to a^-}f(x)=M exist, + but L\neq M. + In this case the left and right hand limits both exist, + but since they are not equal, the limit of f as + x\to a does not exist. + Such a discontinuity is called a + jump discontinuity. + discontinuityjump +

      + +

      + The phrase jump discontinuity is meant to + represent the fact that visually, + the graph of f jumps from one value + to another as we cross the value x=a. +

      +
      3. + +
    3. +

      + The function f is unbounded + near x=a. This means that the value of + f becomes arbitrarily large (or large and + negative) as x approaches a. + Such a discontinuity is called an + infinite discontinuity. + discontinuityinfinite +

      + +

      + Infinite discontinuities are most easily understood + in terms of infinite limits, which are + discussed in . +

      +
      4. +
    +

    + +
    + Illustrating three common types of discontinuity + +
    + The graph of a function with a removable discontinuity at x=2 + + +

    + A portion of the graph of a function is shown, for x from 0 to 4. + The graph has the shape of a parabola opening downward, + but at x=2 there is a hole in the graph, + and instead the point (2,1) (which is not on the graph) is plotted. + The graph of this function illustrates a removable discontinuity because + \lim_{x\to 2}f(x) exists, but does not equal f(2). +

    +     + + Graph showing a removable discontinuity: a hole in the graph when x=2 shows that the limit and function values disagree. + + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.1, + ymax=3.5, + xmin=-0.1, + xmax=4.3, + axis equal + ] + \addplot+[leftarrow,domain=0:1.98]{-0.35*x^2+x+2}; + \addplot [firstcurvestyle,rightarrow,domain=2.02:4]{-0.35*x^2+x+2}; + \addplot[soliddot] coordinates{(2,1)}; + \addplot[hollowdot] coordinates{(2,2.6)}; + \end{axis} + \end{tikzpicture} + + +
    + +
    + The graph of a function with a jump discontinuity at x=2 + + +

    + The graph of a function is shown for x from 0 to 4. + As x approaches 2 from the left, + the graph of f approaches a point that is not part of the graph, + as indicated by a hollow dot. + As x approaches 2 from the right, + the graph of f approaches a point that is part of the graph, + as indicated by a solid dot. + The point marked by the solid dot lies below the point marked by the hollow dot, + illustrating that the left and right hand limits are different as x\to 2. +

    + +

    + On the interval 0 \lt x \leq 2 the graph is curved downward and on the + interval 2 \leq x \leq 4 the graph is a straight line with a + positive slope. +

    +     + + Graph of a function with a jump discontinuity at x = 2. + + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.1, + ymax=3.5, + xmin=-0.1, + xmax=4.3, + axis equal + ] + \addplot+[leftarrow,domain=0:1.98]{-0.35*x^2+x+2}; + \addplot [firstcurvestyle,rightarrow,domain=2.02:4]{x-1}; + \addplot[soliddot] coordinates{(2,1)}; + \addplot[hollowdot] coordinates{(2,2.6)}; + \end{axis} + \end{tikzpicture} + + +
    + +
    + The graph of a function with an infinite discontinuity at x=2 + + +

    + The graph of a function is shown for x from 0 to 4. + There is a + vertical dotted line at x = 2 illustrating a vertical asymptote. + As x approaches 2 from either side, + the graph of f extends upward along the asymptote, + indicating that the value of f(x) is increasing without bound. +

    +     + + Graph of a function with an infinite discontinuity at x = 2, + illustrated by a vertical asymtote. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.1, + xmax=4.3, + ymin=-1, + ymax=110, + ] + \addplot[firstcurvestyle,domain=0:1.9] {1/(x-2)^2}; + \addplot[firstcurvestyle,domain=2.1:4] {1/(x-2)^2}; + \addplot[asymptote,rightarrow] coordinates {(2,1) (2,100)}; + \end{axis} + \end{tikzpicture} + + +
    +     +
    + + +     + + + Consequences of continuity +

    + A common way of thinking of a continuous function is that + its graph can be sketched without lifting your pencil. That is, + its graph forms a continuous curve, + without holes, breaks or jumps. + This pseudo-definition glosses over some of the finer points of continuity. + There are some very strange continuous functions that one would be hard pressed to actually sketch by hand. +

    + +

    + However, this intuitive notion of continuity does help us understand another important concept as follows. + Suppose f is defined on [1,2], and f(1) = -10 and f(2) = 5. + If f is continuous on [1,2] (, its graph can be sketched as a continuous curve + from (1,-10) to (2,5)) then we know intuitively that somewhere on the interval [1,2] + f must be equal to -9, and -8, and -7,-6,\ldots,0,1/2, etc. + In short, f takes on all intermediate + values between -10 and 5. + It may take on more values; + f may actually equal 6 at some time, for instance, + but we are guaranteed all values between -10 and 5. This concept is illustrated in . +

    + +
    + Illustration of the Intermediate Value Theorem: the output 3 is in between -10 and 5, + and therefore any continuous function on [1,2] with f(1) = -10 and f(2) = 5 + will achieve the output 3 somewhere in [1,2] + + +

    + The image shows the graph of three functions + defined on the interval 1 \leq x \leq 2. + One function is plotted in blue, using a solid line style. + Two other functions are plotted in red, with a dash-dot line style. + All three functions are continuous, and satisfy f(1)=-10, and f(2)=5. +

    + +

    + There are other lines marking certain values on the graphs. + A horizontal line at y=-10 indicates that all graphs start at the point (1,-10). + Another horizontal line at y=5 indicates that all graphs end at the point (2,5). + A third horizontal line indicates the value y=3, which is a value between -10 and 5. + There are also three shorter vertical lines, marking an x value on each of the three graphs where f(x)=3. +

    +     + + Graph illustrating the Intermediate Value Theorem. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=0.5, + xmax=2.5, + ymin=-15, + ymax=10, + minor ytick={-15,-14,...,10} + ] + \addplot+[smooth] coordinates {(1,-10) (1.5,-7) (1.8,3) (2,5)}; + \addplot+[smooth] coordinates {(1,-10) (1.2,6) (1.5,3) (2,5)}; + \addplot+[smooth] coordinates {(1,-10) (1.1,-12) (1.2,-5) (1.3,-7) (1.4,0) (1.5,-3) (1.6,3) (1.7,0) (1.8,5) (1.9,3) (2,5)}; + \addplot[soliddot] coordinates {(1,-10) (2,5)}; + \addplot[guideline] coordinates {(1,-10) (0.5,-10)}; + \addplot[guideline] coordinates {(2,5) (0.5,5)}; + \addplot[guideline] coordinates {(1.8,0) (1.8,3) (0.5,3)}; + \addplot[guideline] coordinates {(1.5,0) (1.5,3) (0.5,3)}; + \addplot[guideline] coordinates {(1.6,0) (1.6,3) (0.5,3)}; + \end{axis} + \end{tikzpicture} + + +
    + +

    + While this notion seems intuitive, + it is not trivial to prove and its importance is profound. + Therefore the concept is stated in the form of a theorem. +

    + + + Intermediate Value Theorem + +

    + Let f be a continuous function on [a,b] and, + without loss of generality, let f(a) \lt f(b). + Then for every value y, + where f(a) \lt y \lt f(b), + there is at least one value c in (a,b) such that f(c) = y. + + Intermediate Value Theorem + theoremIntermediate Value + +

    +     +     + + + + +

    + One important application of the is root finding. + Given a function f, + we are often interested in finding values of x where f(x) = 0. + These roots may be very difficult to find exactly. + Good approximations can be found through successive applications of this theorem. + Suppose through direct computation we find that + f(a) \lt 0 and f(b)\gt 0, where a\lt b. + The states that there is at least one c in (a,b) such that f(c) = 0. + The theorem does not give us any clue as to where to find such a value in the interval (a,b), + just that at least one such value exists. +

    + +

    + There is a technique that produces a good approximation of c. + Let d be the midpoint of the interval [a,b], + with f(a) \lt 0 and + f(b) \gt 0 and consider f(d). + There are three possibilities: +

      +
    1. +

      + f(d) = 0: We got lucky and stumbled on the actual value. + We stop as we found a root. +

      +
      2. +
    2. +

      + f(d) \lt 0: Then we know there is a root of f on the interval [d,b] we have halved the size of our interval, + hence are closer to a good approximation of the root. +

      +
      3. +
    3. +

      + f(d) \gt 0: Then we know there is a root of f on the interval [a,d] again,we have halved the size of our interval, + hence are closer to a good approximation of the root. +

      +
      4. +
    +

    + +

    + Successively applying this technique is called the + Bisection Method + + Bisection Method + + of root finding. + We continue until the interval is sufficiently small. + We demonstrate this in the following example. +

    + + + Using the Bisection Method + +

    + Approximate the root of f(x) = x-\cos(x), + accurate to three places after the decimal. +

    +     + +

    + Consider the graph of f(x) = x-\cos(x), + shown in . + It is clear that the graph crosses the x-axis somewhere near x=0.8. + To start the Bisection Method, + pick an interval that contains 0.8. + We choose [0.7,0.9]. + Note that all we care about are signs of f(x), + not their actual value, so this is all we display. +

    + +

    +

    +
  • + Iteration 1: +

    + f(0.7) \lt 0, f(0.9) \gt 0, and f(0.8) \gt 0. + So replace 0.9 with 0.8 and repeat. +

    +
    • +
  • + Iteration 2: +

    + f(0.7)\lt 0, f(0.8) \gt 0, + and at the midpoint, 0.75, + we have f(0.75) \gt 0. + So replace 0.8 with 0.75 and repeat. + Note that we don't need to continue to check the endpoints, + just the midpoint. + Thus we put the rest of the iterations in . +

    +
    • +
    +

    + +
    + Graphing a root of f(x) = x-\cos(x) + + + +

    + Graph of the function f(x)=x-\cos(x) on [0,1]. + The graph has an upward curve and intersects the x axis around + x = 0.8. There are two vertical guide lines, one at + x = 0.7, the other at x = 0.9. The guide lines + mark the interval in which the intercept occurs. +

    +     + + Graph of f(x) = x - cos(x), showing the x intercept between + values x = 0.7 and x = 0.9. + + + \begin{tikzpicture} + \begin{axis}[ + minor xtick={0,0.1,...,1}, + xmin=-.05, + xmax=1.07, + ymin=-1.1, + ymax=.7 + ] + \addplot+[domain=0:1] {x - cos(deg(x))}; + \addplot[guideline] coordinates {(0.7,0) (0.7,-0.0648)}; + \addplot[guideline] coordinates {(0.9,0) (0.9,0.278)}; + \end{axis} + \end{tikzpicture} + + + +
    + + + Iterations of the Bisection Method of Root Finding + + + Iteration # + Interval + Midpoint Sign + + + + + + + + 1 + [\highlight{0.7},0.9] + f(0.8) \gt 0 + + + 2 + [\highlight{0.7},0.8] + f(0.75) \gt 0 + + + 3 + [\highlight{0.7},\highlight{0.75}] + f(0.725)\lt 0 + + + 4 + [0.725,\highlight{0.75}] + f(0.7375)\lt 0 + + + 5 + [\highlight{0.7375},\highlight{0.75}] + f(0.7438)\gt + + + 6 + [\highlight{0.7375},0.7438] + f(0.7407)\gt 0 + + + 7 + [\highlight{0.7375},0.7407] + f(0.7391)\gt 0 + + + 8 + [\highlight{0.7375},\highlight{0.7391}] + f(0.7383)\lt 0 + + + 9 + [0.7383,\highlight{0.7391}] + f(0.7387)\lt 0 + + + 10 + [0.7387,\highlight{0.7391}] + f(0.7389)\lt 0 + + + 11 + [0.7389,\highlight{0.7391}] + f(0.7390)\lt 0 + + + 12 + [0.7390,\highlight{0.7391}] + + + + + + + +
    + +

    + Notice that in the 12th iteration we have the endpoints of the interval each starting with 0.739. + Thus we have narrowed the zero down to an accuracy of the first three places after the decimal. + Using a computer, we have + + f(0.7390) = -0.00014, f(0.7391) = 0.000024 + . +

    + +

    + Either endpoint of the interval gives a good approximation of where f is 0. + The states that the actual zero is still within this interval. + While we do not know its exact value, + we know it starts with 0.739. +

    + +

    + This type of exercise is rarely done by hand. + Rather, it is simple to program a computer to run such an algorithm and stop when the endpoints differ by a preset small amount. + One of the authors did write such a program and found the zero of f to be + 0.7390851332, accurate to 10 places after the decimal. + While it took a few minutes to write the program, + it took less than a thousandth of a second for the program to run the necessary 35 iterations. + In less than 8 hundredths of a second, + the zero was calculated to 100 decimal places + (with less than 200 iterations). +

    +     + +     + + + +

    + It is a simple matter to extend the Bisection Method to solve problems similar to Find x, + where f(x) = 0. For instance, + we can find x, where f(x) = 1. + It actually works very well to define a new function g where g(x) = f(x) - 1. + Then use the Bisection Method to solve g(x)=0. +

    + +

    + Similarly, given two functions f and g, + we can use the Bisection Method to solve f(x) = g(x). + Once again, create a new function h where + h(x) = f(x)-g(x) and solve h(x) = 0. +

    + +

    + In + another equation solving method will be introduced, called Newton's Method. + In many cases, Newton's Method is much faster. + It relies on more advanced mathematics, though, + so we will wait before introducing it. +

    + +

    + This section formally defined what it means to be a continuous function. + Most functions that we deal with are continuous, + so often it feels odd to have to formally define this concept. + Regardless, it is important, and forms the basis of the next chapter. +

    +     + + + + + + + Terms and Concepts + + + + +

    + In your own words, describe what it means for a function to be continuous. +

    + +     + + +     + + + + +

    + In your own words, describe what the Intermediate Value Theorem states. +

    + +     + + +     + + + + +

    + What is a root of a function? +

    + +     + + + +

    + A root of a function f is a value c such that f(c)=0. +

    +     + +     + + + + +

    + Given functions f and g on an interval I, + how can the Bisection Method be used to find a value c where f(c) = g(c)? +

    + +     + + + +

    + Consider the function h(x) = g(x) - f(x), + and use the Bisection Method to find a root of h. +

    +     + +     + + + + + +

    + + If f is defined on an open interval containing c, + and \lim\limits_{x\to c}f(x) exists, + then f is continuous at c. +

    +     + +

    + There is more to the definition of continuity than existence of the limit! + What else needs to be true? +

    +     + +     + + + + + +

    + + If f is defined on an open interval containing c, and + f is continuous at c, + then \lim\limits_{x\to c}f(x) exists. +

    +     + +

    + The definition of continuity requires that \lim_{x\to c}f(x)=f(c). + This is not possible if the limit does not exist! +

    +     + +     + + + + + +

    + + If f is defined on an open interval containing c, + and f is continuous at c, + then \lim\limits_{x\to c^+}f(x) = f(c). +

    +     + +

    + If f is continuous at c, + then the limit of f at c must exist. + If the limit exists, then both one-sided limits must also exist. +

    +     + +     + + + + + +

    + + If f is continuous on [a,b], + then \lim\limits_{x\to a^-}f(x) = f(a). +

    +     + +

    + The left-hand limit at a involves points outside the interval [a,b], + and continuity on [a,b] only tells us about points inside [a,b]. + In particular, it is the right-hand limit at a that must equal f(a). +

    +     + +     + + + + + +

    + + If f is continuous on [0,1) and [1,2), + then f is continuous on [0,2). +

    +     + +

    + Review the discussion following . +

    +     + +     + + + + + +

    + + The sum of continuous functions is also continuous. +

    +     + +

    + See . +

    +     + +     +     + + + Problems + + + +

    + Use the graph to determine if the function is continuous at the given point. +

    +     + + + + parserPopUp.pl + + + $a=1; + @b=(1,2,1,2,0.5,0); + $left=Formula("(($b[1]-$b[0])/(1-0))*(x-0)+$b[0]"); + $right=Formula("(x-1)*(x-1.5)*$b[5]/((2-1)*(2-1.5))+(x-1)*(x-2)*$b[4]/((1.5-1)*(1.5-2))+(x-1.5)*(x-2)*$b[3]/((1-1.5)*(1-2))"); + $answer=DropDown(['Yes.','No.'],1,showInStatic=>0); + $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; + + +

    + Is f in the graph below continuous at ? +

    + + +

    + The graph of a piecewise-defined function defined on [0,2], consisting of two parts. + The first part is a straight line with intercept (0,1) and a positive + slope. It ends at (1,2), but this point is not part of the graph. + The second part begins at (1,2), but does not include this point, and curves downward, ending at (2,0). + There is also a solid dot at (1,1), indicating that f(1)=1. +

    +     + + The graph used for the current exercise. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-0.5, + xmax=2.5, + ymin=-0.5, + ymax=2.5 + ] + \addplot[firstcurvestyle, domain=0:1] {$left}; + \addplot[firstcurvestyle, domain=1:2] {$right}; + \addplot[soliddot] coordinates {(0,$b[0]) (1, $b[2]) (2,$b[5])}; + \addplot[hollowdot] coordinates {(1,$b[1]) (1,$b[3])}; + \end{axis} + \end{tikzpicture} + + +

    + +

    +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $a=1; + @b=(0,1,2,2,1,0); + $left=Formula("(($b[1]-$b[0])/(1-0))*(x-0)+$b[0]"); + $right=Formula("(($b[5]-$b[3])/(2-1))*(x-1)+$b[3]"); + $answer=DropDown(['Yes.','No.'],1,showInStatic=>0); + $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; + + +

    + Is f in the graph below continuous at ? +

    + + +

    + Graph of a piecewise-defined function with the domain [0,2]. The graph has + two straight lines, the first is defined by the solid point (0 ,0) + and the hollo point (1, 1), and has a positive slope. + The second line is defined by the solid points (1, 2) + and (2, 2), and has a negative slope. +

    +     + + The graph used for the current exercise. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-0.5, + xmax=2.5, + ymin=-0.5, + ymax=2.5 + ] + \addplot[firstcurvestyle, domain=0:1] {$left}; + \addplot[firstcurvestyle, domain=1:2] {$right}; + \addplot[hollowdot] coordinates {(1,$b[1]) (1,$b[3])}; + \addplot[soliddot] coordinates {(0,$b[0]) (1, $b[2]) (2,$b[5])}; + \end{axis} + \end{tikzpicture} + + +

    + +

    +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $a=1; + @b=(0,Compute("INF"),Compute("DNE"),Compute("INF"),1,0); + $left=Formula("sec(x*90)-1"); + $right=Formula("-sec(x*90)-1"); + $lowmidx = acos(1/3.5)*2/pi; + $highmidx = 2 - acos(1/3.5)*2/pi; + $answer=DropDown(['Yes.','No.'],1,showInStatic=>0); + $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; + + +

    + Is f in the graph below continuous at ? +

    + + +

    + The image shows the graph of a function that is defined on the interval [0,2], + except at x=, where it has a vertical asymptote. + On either side of the vertical asymptote, as x gets close to , + the y value increases without bound. +

    + +     + + The graph used for the current exercise. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-0.5, + xmax=2.5, + ymin=-0.5, + ymax=2.5 + ] + \addplot[firstcurvestyle, infiniteright, domain=0:$lowmidx] {$left}; + \addplot[firstcurvestyle, infiniteleft, domain=$highmidx:2] {$right}; + \addplot[asymptote, -] coordinates {(1,-0.5) (1,2.5)}; + \addplot[soliddot] coordinates {(0,$b[0]) (2,$b[5])}; + \end{axis} + \end{tikzpicture} + + +

    + +

    +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $a=0; + ($b[1],$b[2],$b[3],$b[4]) = random_subset(4,0,0.5,1,1.5,2); + for my $i(0,5) {$b[$i]=random(0,2,0.5)}; + $left=Formula("(($b[1]-$b[0])/(1-0))*(x-0)+$b[0]"); + $right=Formula("(x-1)*(x-1.5)*$b[5]/((2-1)*(2-1.5))+(x-1)*(x-2)*$b[4]/((1.5-1)*(1.5-2))+(x-1.5)*(x-2)*$b[3]/((1-1.5)*(1-2))"); + $answer=DropDown(['Yes.','No.'],0,showInStatic=>0); + $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; + + +

    + Is f in the graph below continuous at ? +

    + + +

    + Graph of a piecewise-defined function with domain [0,2]. The graph has + two lines; the first is straight and defined by the solid + point (0, ) and hollow point + (1, ). The second line + is defined by the hollow point (1, + ) and solid point + (2, ). When + x = 1 f(x) = +

    +     + + The graph used for the current exercise. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-0.5, + xmax=2.5, + ymin=-0.5, + ymax=2.5 + ] + \addplot[firstcurvestyle, domain=0:1] {$left}; + \addplot[firstcurvestyle, domain=1:2] {$right}; + \addplot[hollowdot] coordinates {(1,$b[1]) (1,$b[3])}; + \addplot[soliddot] coordinates {(0,$b[0]) (1,$b[2]) (2,$b[5])}; + \end{axis} + \end{tikzpicture} + + +

    + +

    +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $a=1; + for my$i(0..5){$b[$i]=random(0,2,0.5)}; + $b[2] = $b[1]; + $b[3] = $b[1]; + $right=Formula("(($b[1]-$b[0])/(1-2))*(x-2)+$b[0]"); + $left=Formula("(x-0.5)*(x-1)*$b[5]/((0-0.5)*(0-1))+(x-0)*(x-1)*$b[4]/((0.5-0)*(0.5-1))+(x-0)*(x-0.5)*$b[3]/((1-0)*(1-0.5))"); + $answer=DropDown(['Yes.','No.'],0,showInStatic=>0); + $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; + + +

    + Is f in the graph below continuous at ? +

    + + +

    + The graph + changes at x = 1 and is defined by the solid dots + (0, ) and + (2, ). + At the interval 0 \leq x \lt 1 the graph is curved + and on the interval 1 \lt x \leq 2 the line is straight. +

    +     + + The graph used for the current exercise. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-0.5, + xmax=2.5, + ymin=-0.5, + ymax=2.5 + ] + \addplot[firstcurvestyle, domain=0:1] {$left}; + \addplot[firstcurvestyle, domain=1:2] {$right}; + \addplot[soliddot] coordinates {(0,$b[5]) (2,$b[0])}; + \end{axis} + \end{tikzpicture} + + +

    + +

    +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $a=4; + ($b[1],$b[3]) = random_subset(2,-4..4); + for my $i (0,4) {$b[$i]=random(-4,4,1)}; + $answer=DropDown(['Yes.','No.'],0,showInStatic=>0); + $ymin=min($b[0],$b[1]-$b[0],$b[3],$b[4]-$b[3])-1; + $ymax=max($b[0],$b[1]-$b[0],$b[3],$b[4]-$b[3])+1; + $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; + + +

    + Is f in the graph below continuous at ? +

    + + +

    + The graph is made up of two curves. The first is defined + by the solid points (-4, ) and + (0, ). + The second is defined by the hollow point + (0, ) and + solid point (4, ). +

    +     + + The graph used for the current exercise. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-5, + xmax=5, + ymin=$ymin, + ymax=$ymax + ] + \addplot[firstcurvestyle, domain=-4:0] {($b[0] + $b[1] + sin((x+2)*45) * ($b[1]-$b[0]))/2}; + \addplot[firstcurvestyle, domain=0:4] {($b[3] + $b[4] - sin((x+2)*45) * ($b[4]-$b[3]))/2}; + \addplot[hollowdot] coordinates {(0,$b[3])}; + \addplot[soliddot] coordinates {(-4,$b[0]) (0,$b[1]) (4,$b[4])}; + \end{axis} + \end{tikzpicture} + + +

    + +

    +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + @b=(0,2,0,2,0,0,0,2,Compute("DNE"),2,0); + for my$i(0..3){$f[$i]=Formula("($b[3*$i+1]-$b[3*$i])/2*(x-(2*$i-4))+$b[3*$i]");} + @choices=('Yes.','No.'); + for my$i(0..2){if($b[3*$i+2]==Compute("DNE")){$correct[$i]=1;} + elsif($f[$i]->eval(x=>(2*$i-2))!=$f[$i+1]->eval(x=>(2*$i-2))){$correct[$i]=1;} + elsif($f[$i]->eval(x=>(2*$i-2))!=$b[3*$i+2]){$correct[$i]=1;} + else{$correct[$i]=0;}; + $answer[$i]=DropDown(~~@choices,$correct[$i],showInStatic=>0); + }; + $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; + + +

    + Is f in the graph below continuous at -2, + 0, and 2? +

    + + +

    + There are four lines that make up this graph. The first + going from left to right is defined by the solid point + (-4, 0) and hollow point (-2. 2), and has + a positive slope. The second line is defined by the points + (-2, 2), 0, 0 with a negative slope. The + third line by (0, 0), 2, 2 with a positive slope. + And the last line is defined by the hollow point (2, 2) + and solid point (4, 0), with a negative slope. +

    +

    + For x = -2, f(x) = 0, so there is a solid point at + (-2, 0). +

    +     + + The graph used for the current exercise. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = -5, + xmax = 5, + ymin = -4, + ymax = 4, + ] + \addplot[firstcurvestyle, domain=-4:-2] {x+4}; + \addplot[firstcurvestyle, domain=-2:0] {-x}; + \addplot[firstcurvestyle, domain=0:2] {x}; + \addplot[firstcurvestyle, domain=2:4] {4-x}; + \addplot[soliddot] coordinates {(-4,0) (-2,0) (4,0)}; + \addplot[hollowdot] coordinates {(-2,2) (2,2)}; + \end{axis} + \end{tikzpicture} + + +

    + At -2: +

    +

    + +

    +

    + At 0: +

    +

    + +

    +

    + At 2: +

    +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + Context()->flags->set(reduceConstants=>0); + $a=Formula("3pi/2"); + $g=Formula("1+sin(x*180/pi)"); + $answer=DropDown(['Yes.','No.'],0,showInStatic=>0); + $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; + + +

    + Is f in the graph below continuous at ? +

    + + +

    + The graph has a downward curve on the + interval 0 \leq x \leq \pi and an upward curve on + the interval \pi \leq x \leq 2\pi. The peak of the + graph is at x = \frac{\pi}{2} and the lowest point + is at x = \frac{3\pi}{2}. There are also + no breaks in the graph. +

    +     + + The graph used for the current exercise. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = -pi/2, + xmax = 2.5*pi, + ymin = -0.5, + ymax = 2.5, + xtick={ + -2*pi, -(3*pi)/2, -pi, -pi/2, + pi/2, pi, (3*pi)/2, 2*pi + }, + xticklabels={ + \(-2\pi\), \(-\frac{3\pi}{2}\), \(-\pi\), \(-\frac{\pi}{2}\), + \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\), \(2\pi\) + } + ] + \addplot[firstcurvestyle, domain=0:2*pi, smooth] {$g}; + \end{axis} + \end{tikzpicture} + + +

    + +

    +

    + +

    +     +     +     +     + + + +

    + Determine if f is continuous at the indicated values. +

    +     + + + + + parserPopUp.pl + + + Context("PiecewiseFunction"); + @choices=('Yes.', + 'No.' + ); + @answer=(DropDown(~~@choices,0,showInStatic=>0),DropDown(~~@choices,0,showInStatic=>0)); + $showwork = '[@ explanation_box(message => "If not, explain why not.") @]*'; + + +

    + f(x)=\begin{cases}1\amp x=0\\\frac{\sin(x)}{x}\amp x\neq0\end{cases} +

    +     + + + +

    + Is f is continuous at 0? +

    +

    + +

    +

    + +

    +     +     + + + +

    + Is f is continuous at \pi? +

    +

    + +

    +

    + +

    +     +     +     +     + + + + + parserPopUp.pl + + + Context("PiecewiseFunction"); + #for my$i(0..3){$a[$i]=non_zero_random(-2,2,1)}; + #if($envir{problemSeed}==1){@a=(1,-1,1,-2)}; + @a=(1,-1,1,-2); + $left=Formula("$a[0]x^3+$a[1]x^2")->reduce; + $right=Formula("$a[2]x+$a[3]")->reduce; + $f=PiecewiseFunction("x<1" => "$left","x>=1"=>"$right"); + @choices=('Yes.', + 'No.' + ); + $correct[1]=($left->eval(x=>1)==$right->eval(x=>1))?0:1; + @answer=(DropDown(~~@choices,0,showInStatic=>0),DropDown(~~@choices,$correct[1],showInStatic=>0)); + $showwork = '[@ explanation_box(message => "If not, explain why not.") @]*'; + + +

    + f(x)= +

    +     + + + +

    + Is f is continuous at 0? +

    +

    + +

    +

    + +

    +     +     + + + +

    + Is f is continuous at 1? +

    +

    + +

    +

    + +

    +     +     +     +     + + + + + parserPopUp.pl + + + Context("PiecewiseFunction"); + #if($envir{problemSeed}==1){@a=(-1,-4,-2,10)}; + @a=(-1,-4,-2,10); + $f=PiecewiseFunction("x!=$a[0]" => Formula("(x^2-($a[0]+$a[1])x+($a[0]*$a[1]))/(x^2-($a[0]+$a[2])x+($a[0]*$a[2]))")->reduce,"x=$a[0]"=>Real(($a[0]-$a[1])/($a[0]-$a[2]))); + @choices=('Yes.', + 'No.' + ); + @answer=(DropDown(~~@choices,0,showInStatic=>0),DropDown(~~@choices,0,showInStatic=>0)); + $showwork = '[@ explanation_box(message => "If not, explain why not.") @]*'; + + +

    + f(x)= +

    +     + + + +

    + Is f is continuous at ? +

    +

    + +

    +

    + +

    +     +     + + + +

    + Is f is continuous at ? +

    +

    + +

    +

    + +

    +     +     +     +     + + + + + parserPopUp.pl + + + Context("PiecewiseFunction"); + #if($envir{problemSeed}==1){@a=(0,8,-8,3,5,0)}; + @a=(0,8,-8,3,5,0); + $f=PiecewiseFunction("x!=$a[1]" => Formula("(x^2-($a[1]+$a[2])x+($a[1]*$a[2]))/(x^2-($a[1]+$a[3])x+($a[1]*$a[3]))")->reduce,"x=$a[1]"=>$a[4]); + @choices=('Yes.', + 'No.' + ); + $correct[1]=($a[4]==($a[1]-$a[2])/($a[1]-$a[3]))?0:1; + @answer=(DropDown(~~@choices,0,showInStatic=>0),DropDown(~~@choices,$correct[1],showInStatic=>0)); + $showwork = '[@ explanation_box(message => "If not, explain why not.") @]*'; + + +

    + f(x)= +

    +     + + + +

    + Is f is continuous at ? +

    +

    + +

    +

    + +

    +     +     + + + +

    + Is f is continuous at ? +

    +

    + +

    +

    + +

    +     +     +     +     +     + + + +

    + Give the intervals on which the function is continuous. +

    +     + + + + + Context("Interval"); + $a=random(-9,9,1); + $b=non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=-3;$b=9;}; + $f=Formula("x^2+$a x+$b")->reduce; + $interval=List(Interval("(-inf,inf)")); + + +

    + f(x)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     + +

    + Since f is a polynomial function, + it is continuous on (-\infty,\infty). +

    +     +     +     + + + + + Context("Interval"); + $a=random(1,5,1); + if($envir{problemSeed}==1){$a=2;}; + $f=Formula("sqrt(x^2-$a**2)")->reduce; + $interval=List("(-inf,-$a], [$a,inf)"); + + +

    + f(x)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     + +

    + The domain of f is , + and since f is a composition of continuous functions on that domain, + it is continuous on . +

    +     +     +     + + + + + Context("Interval"); + $a=random(1,5,1); + if($envir{problemSeed}==1){$a=2;}; + $f=Formula("sqrt($a**2-x^2)")->reduce; + $interval=List("[-$a,$a]"); + + +

    + f(x)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     + +

    + The domain of f is , + and since f is a composition of continuous functions on that domain, + it is continuous on . +

    +     +     +     + + + + + Context("Interval"); + $a=random(1,5,1); + $b=random(1,5,1); + if($envir{problemSeed}==1){$a=1;$b=1}; + $f1=Formula("sqrt($a-x)"); + $f2=Formula("sqrt(x+$b)"); + $f=$f1+$f2; + $interval=List(Interval("[-$b,$a]")); + + +

    + f(x)=+ +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     + +

    + The domain of is (-\infty,], + and the domain of is [-,\infty). + So the domain of f is . + And since f is the sum of two compositions of continuous functions, + it is continuous on . +

    +     +     +     + + + + + Context("Interval"); + $a=random(2,5,1); + $b=random(2,9,1); + if($envir{problemSeed}==1){$a=5;$b=6}; + Context()->variables->are(t=>'Real'); + $f=Formula("sqrt($a t^2-$a*$b)")->reduce; + $interval=List("(-inf,-sqrt($b)], [sqrt($b),inf)"); + + +

    + f(t)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     +     +     + + + + + Context("Interval"); + $a=random(1,9,1); + if($envir{problemSeed}==1){$a=1}; + Context()->variables->are(t=>'Real'); + $f=Formula("1/sqrt($a**2-t^2)")->reduce; + $interval=List(Interval("(-$a,$a)")); + + +

    + g(t)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     +     +     + + + + + Context("Interval"); + $a=random(1,9,1); + $b=random(1,9,1); + if($envir{problemSeed}==1){$a=1;$b=1;}; + Context()->variables->are(t=>'Real'); + $f=Formula("1/($a+$b*t^2)")->reduce; + $interval=Compute("(-inf,inf)"); + + +

    + g(t)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     +     +     + + + + + Context("Interval"); + $b=list_random(2..9,Formula("e"),Formula("pi")); + if($envir{problemSeed}==1){$b=Formula("e");}; + $f=Formula("$b^x")->reduce; + $interval=List(Interval("(-inf,inf)")); + + +

    + f(x)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     +     +     + + + + + Context("Interval"); + Context()->variables->are(s=>'Real'); + $b=list_random(2..9,Formula("e"),Formula("pi")); + if($envir{problemSeed}==1){$b=Formula("e");}; + $f=($b==Formula("e"))?'\ln(s)':"\log_{$b}(s)"; + $interval=List(Interval("(0,inf)")); + + +

    + g(s)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     +     +     + + + + + Context("Interval"); + Context()->variables->are(t=>'Real'); + $f=list_random(Formula("cos(t)"),Formula("sin(t)")); + if($envir{problemSeed}==1){$f=Formula("cos(t)");}; + $interval=List(Interval("(-inf,inf)")); + + +

    + h(t)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     +     +     + + + + + Context("Interval"); + $a=random(1,9,1); + if($envir{problemSeed}==1){$a=1}; + Context()->variables->are(k=>'Real'); + $f=Formula("sqrt($a-e^k)"); + $interval=($a==1)?List(Interval("(-inf,0]")):List(Interval("(-inf,ln($a)]")); + + +

    + f(k)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     +     +     + + + + + Context("Interval"); + $a=list_random('sin','cos'); + $b=list_random(2,3,'e'); + $n=random(2,5,1); + if($envir{problemSeed}==1){$a='sin';$b='e';$n=2;}; + $f=Formula("$a($b^x+x^($n))"); + $interval=List(Interval("(-inf,inf)")); + + +

    + f(x)= +

    + + Use interval notation. If the answer is a list of more than one interval, use commas to separate them. + +

    + +

    +     +     +     +     + + + +

    + Test your understanding of the Intermediate Value Theorem. +

    +     + + + + +

    + Let f be continuous on [1,5] where + f(1) = -2 and f(5) = -10. + Does a value 1\lt c\lt 5 exist such that f(c) = -9? + Why/why not? +

    + +     + + + +

    + Yes, by the Intermediate Value Theorem. +

    +     + +     + + + + +

    + Let g be continuous on [-3,7] where + g(0) = 0 and g(2) = 25. + Does a value -3\lt c\lt 7 exist such that g(c) = 15? + Why/why not? +

    + +     + + + +

    + Yes, by the Intermediate Value Theorem. + In fact, we can be more specific and state such a value c + exists in (0,2), not just in (-3,7). +

    +     + +     + + + + +

    + Let f be continuous on [-1,1] where + f(-1) = -10 and f(1) = 10. + Does a value -1\lt c\lt 1 exist such that f(c) = 11? + Why/why not? +

    + +     + + + +

    + We cannot say; + the Intermediate Value Theorem only applies to function values between -10 and 10; + as 11 is outside this range, we do not know. +

    +     + +     + + + + +

    + Let h be a function on [-1,1] where + h(-1) = -10 and h(1) = 10. + Does a value -1\lt c\lt 1 exist such that h(c) = 0? + Why/why not? +

    + +     + + + +

    + We cannot say; + the Intermediate Value Theorem only applies to continuous functions. + As we do know know if h is continuous, we cannot say. +

    +     + +     +     + + + +

    + Use the Bisection Method to approximate, + accurate to two decimal places, + the value of the root of the given function in the given interval. +

    +     + + + + + parserPopUp.pl + + + Context("Interval"); + Context()->flags->set(tolType=>'absolute',tolerance=>0.00000005); + #do{$b=non_zero_random(-9,9,1);$c=non_zero_random(-9,9,1);}until($b**2>4*$c and int(sqrt(abs($b**2-4*$c)))!=sqrt(abs($b**2-4*$c))); + #if($envir{problemSeed}==1){$b=2;$c=-4;}; + $b=2;$c=-4; + $f=Formula("x^2+$b x+$c")->reduce; + $left=floor((-$b+sqrt($b**2-4*$c))/2); + $right=$left+1; + $mid=($left+$right)/2; + ($l[0],$r[0])=($f->eval(x=>$mid)>0)?($left,$mid):($mid,$right); + $in[0]=Compute("[$l[0],$r[0]]"); + for my$i(1..7){$m[$i-1]=($l[$i-1]+$r[$i-1])/2; + ($l[$i],$r[$i])=($f->eval(x=>$m[$i-1])>0)?($l[$i-1],$m[$i-1]):($m[$i-1],$r[$i-1]); + $popup[$i-1]=($f->eval(x=>$m[$i-1])>0)?DropDown(['+','-'],'0',showInStatic=>0):DropDown(['+','-'],'1',showInStatic=>0); + $in[$i]=Compute("[$l[$i],$r[$i]]"); + }; + $z=Real(($l[7]+$r[7])/2)->with(tolType=>'absolute',tolerance=>0.01); + #$showwork = '[@ explanation_box(message => "Show the steps you used applying the Bisection Method.") @]*'; + + +

    + f(x)= on the interval +

    + + If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. + + + + Iteration + Interval + Midpoint Sign + + + 1 + + + + + 2 + + + + + 3 + + + + + 4 + + + + + 5 + + + + + 6 + + + + + 7 + + + + + 8 + + + + + + Enter the solution to f(x)=0, accurate to two decimal places. + +

    + +

    + +     +     +     + + + + + parserPopUp.pl + + + Context("Interval"); + Context()->flags->set(tolType=>'absolute',tolerance=>0.00000005,reduceConstants=>0); + #$b=random(2,6,1); + #if($envir{problemSeed}==1){$b=2;}; + $b=2; + $f=Compute("sin(x)-1/$b"); + $root=asin(1/$b); + ($l[0],$r[0])=(floor(20*$root)/20,floor(20*$root)/20+0.05); + $in[0]=Compute("[$l[0],$r[0]]"); + for my$i(1..7){$m[$i-1]=($l[$i-1]+$r[$i-1])/2; + ($l[$i],$r[$i])=($f->eval(x=>$m[$i-1])>0)?($l[$i-1],$m[$i-1]):($m[$i-1],$r[$i-1]); + $popup[$i-1]=($f->eval(x=>$m[$i-1])>0)?DropDown(['+','-'],'0',showInStatic=>0):DropDown(['+','-'],'1',showInStatic=>0); + $in[$i]=Compute("[$l[$i],$r[$i]]"); + }; + $z=Real(($l[7]+$r[7])/2)->with(tolType=>'absolute',tolerance=>0.01); + #$showwork = '[@ explanation_box(message => "Show the steps you used applying the Bisection Method.") @]*'; + + +

    + f(x)= on the interval +

    + + If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. + + + + Iteration + Interval + Midpoint Sign + + + 1 + + + + + 2 + + + + + 3 + + + + + 4 + + + + + 5 + + + + + 6 + + + + + 7 + + + + + 8 + + + + + + Enter the solution to f(x)=0, accurate to two decimal places. + +

    + +

    + +     +     +     + + + + + parserPopUp.pl + + + Context("Interval"); + Context()->flags->set(tolType=>'absolute',tolerance=>0.00000005); + #$b=random(1.5,3,0.1); + #if($envir{problemSeed}==1){$b=2;}; + $b=2; + $f=Compute("e^x-$b"); + $root=ln($b); + ($l[0],$r[0])=(floor(20*$root)/20,floor(20*$root)/20+0.05); + $in[0]=Compute("[$l[0],$r[0]]"); + for my$i(1..7){$m[$i-1]=($l[$i-1]+$r[$i-1])/2; + ($l[$i],$r[$i])=($f->eval(x=>$m[$i-1])>0)?($l[$i-1],$m[$i-1]):($m[$i-1],$r[$i-1]); + $popup[$i-1]=($f->eval(x=>$m[$i-1])>0)?DropDown(['+','-'],'0',showInStatic=>0):DropDown(['+','-'],'1',showInStatic=>0); + $in[$i]=Compute("[$l[$i],$r[$i]]"); + }; + $z=Real(($l[7]+$r[7])/2)->with(tolType=>'absolute',tolerance=>0.01); + #$showwork = '[@ explanation_box(message => "Show the steps you used applying the Bisection Method.") @]*'; + + +

    + f(x)= on the interval +

    + + If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. + + + + Iteration + Interval + Midpoint Sign + + + 1 + + + + + 2 + + + + + 3 + + + + + 4 + + + + + 5 + + + + + 6 + + + + + 7 + + + + + 8 + + + + + + Enter the solution to f(x)=0, accurate to two decimal places. + +

    + +

    + +     +     +     + + + + + parserPopUp.pl + + + Context("Interval"); + Context()->flags->set(tolType=>'absolute',tolerance=>0.00000005); + #$b=random(1.5,3,0.1); + #if($envir{problemSeed}==1){$b=2;}; + $b=2; + $f=Compute("cos(x)-sin(x)"); + $root=pi/4; + ($l[0],$r[0])=(floor(10*$root)/10,floor(10*$root)/10+0.1); + $in[0]=Compute("[$l[0],$r[0]]"); + for my$i(1..7){$m[$i-1]=($l[$i-1]+$r[$i-1])/2; + ($l[$i],$r[$i])=($f->eval(x=>$m[$i-1])<0)?($l[$i-1],$m[$i-1]):($m[$i-1],$r[$i-1]); + $popup[$i-1]=($f->eval(x=>$m[$i-1])>0)?DropDown(['+','-'],'0',showInStatic=>0):DropDown(['+','-'],'1',showInStatic=>0); + $in[$i]=Compute("[$l[$i],$r[$i]]"); + }; + $z=Real(($l[7]+$r[7])/2)->with(tolType=>'absolute',tolerance=>0.01); + #$showwork = '[@ explanation_box(message => "Show the steps you used applying the Bisection Method.") @]*'; + + +

    + f(x)= on the interval +

    + + If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. + + + + Iteration + Interval + Midpoint Sign + + + 1 + + + + + 2 + + + + + 3 + + + + + 4 + + + + + 5 + + + + + 6 + + + + + 7 + + + + + 8 + + + + + + Enter the solution to f(x)=0, accurate to two decimal places. + +

    + +

    + +     +     +     +     +     +     +
    +
    + Limits Involving Infinity + +

    + In + we stated that in the equation \lim_{x\to c}f(x) = L, + both c and L were numbers. + In this section we relax that definition a bit by considering situations when it makes sense to let c and/or L be infinity. +

    + +

    + As a motivating example, consider f(x) = 1/x^2, + as shown in . + Note how, as x approaches 0, f(x) grows very, + very large in fact, it grows without bound. + It seems appropriate, and descriptive, to state that + + \lim_{x\to 0} \frac1{x^2}=\infty + . +

    + +

    + Also note that as x gets very large, + f(x) gets very, very small. + We could represent this concept with notation such as + + \lim_{x\to \infty} \frac1{x^2}=0 + . +

    + +
    + Graphing f(x) = 1/x^2 for values of x near 0 + + + +

    + Graph of f(x)=1/x^2 for x between -1 and 1. There is a vertical + asymptote at x = 0 and a horizontal asymptote at y = 0. + For x values near the left and right edges of the image, + the y value is close to 0. + For x values near 0, the graph extends to the top of the image (and presumably beyond), + suggesting that y approaches \infty. +

    +     + + Graph of 1 over x squared. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1.1, + xmax=1.1, + ymin=-.1, + ymax=110, + ] + \addplot[firstcurvestyle,infinite,domain=-1:-0.1] {1/x^2}; + \addplot[firstcurvestyle,infinite,domain=0.1:1] {1/x^2}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + We explore both types of use of \infty in turn. +

    +     + + + Infinite limits + + + Limit of Infinity, <m>\infty</m> + +

    + Let I be an open interval containing c, + and let f be a function defined on I, + except possibly at c. +

      +
    • +

      + The limit of f(x), + as x approaches c, + is infinity, denoted by + + \lim_{x\rightarrow c} f(x) = \infty + , + if given any N \gt 0, + there exists \delta \gt 0 such that for all x in I, + where x\neq c, + if \abs{x - c} \lt \delta, then f(x) \gt N. +

      +
      • +
    • +

      + The limit of f(x), + as x approaches c, + is negative infinity, denoted by + + \lim_{x\rightarrow c} f(x) = -\infty + , + if given any N \lt 0, + there exists \delta \gt 0 such that for all x in I, + where x\neq c, + if \abs{x - c} \lt \delta, then f(x) \lt N. +

      +
      • +
    + limitof infinity + +

    +     +     + + + + + + +

    + The first definition is similar to the + \varepsilon-\delta definition in + from . + In that definition, given any (small) value \varepsilon, + if we let x get close enough to c + (within \delta units of c) + then f(x) is guaranteed to be within \varepsilon of L. + Here, given any (large) value N, + if we let x get close enough to c + (within \delta units of c), + then f(x) will be at least as large as N. + In other words, if we get close enough to c, + then we can make f(x) as large as we want. +

    + +

    + It is important to note that by saying + \lim_{x\to c}f(x) = \infty we are implicitly stating that + the limit of f(x), + as x approaches c, + does not exist. A limit only exists when f(x) approaches an actual numeric value. + We use the concept of limits that approach infinity because it is helpful and descriptive. + It is one specific way + in which a limit can fail to exist. +

    + +

    + We define one-sided limits that approach infinity in a similar way. +

    + + + One-Sided Limits of Infinity + +

    +

      +
    • +

      + Let f be a function defined on (a,c) for some a\lt c. + We say the limit of f(x), + as x approaches c from the left, is infinity, or, + the left-hand limit of f at c is infinity, + denoted by + + \lim_{x\rightarrow c^-} f(x) = \infty + , + if given any N \gt 0, + there exists \delta \gt 0 such that for all + a\lt x\lt c, + if \abs{x - c} \lt \delta, then f(x) \gt N. +

      +
      • +
    • +

      + Let f be a function defined on (c,b) for some b \gt c. + We say the limit of f(x), + as x approaches c from the right, is infinity, or, + the right-hand limit of f at c is infinity, + denoted by + + \lim_{x\rightarrow c^+} f(x) = \infty + , + if given any N \gt 0, + there exists \delta \gt 0 such that for all + c\lt x\lt b, + if \abs{x - c} \lt \delta, then f(x) \gt N. +

      +
      • +
    • +

      + The term left- (or, + right-) hand limit of f at c is negative infinity + is defined in a manner similar to . +

      +
      • +
    +

    +     +     + + + Evaluating limits involving infinity + +

    + Find \lim\limits_{x\to 1}\frac1{(x-1)^2} as shown in . +

    + +
    + Observing infinite limit as x\to 1 in + + + +

    + Graph of f(x)=\frac{1}{(x-1)^2} for x between 0 and 1. + There is a vertical + asymptote at x = 1 and a horizontal asymptote at y = 0. + As x gets near 1 from either side of the vertical asymptote, + y approaches \infty. + For x values near the left and right edges of the image, + the value of y approaches 0. +

    +     + + Graph of 1 over (x-1) squared. Has a vertical asymptote at x = 1. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.1, + xmax=2.1, + ymin=-1, + ymax=110, + ] + \addplot[firstcurvestyle,infinite,domain=0:0.9] {1/(x-1)^2}; + \addplot[firstcurvestyle,infinite,domain=1.1:2] {1/(x-1)^2}; + \addplot[asymptote,rightarrow] coordinates {(1,1) (1,100)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + In + of , + by inspecting values of x close to 1 we concluded that this limit does not exist. + That is, it cannot equal any real number. + But the limit could be infinite. + And in fact, + we see that the function does appear to be growing larger and larger, + as f(0.99)=10^4, f(0.999)=10^6, f(0.9999)=10^8. + A similar thing happens on the other side of 1. + From the graph and the numeric information, + we could state \lim_{x \to 1}1/(x-1)^2=\infty. + We can prove this by using +

    + +

    + In general, let a large value N be given. + Let \delta=1/\sqrt{N}. + If x is within \delta of 1, + , if \abs{x-1}\lt 1/\sqrt{N}, then: + + \abs{x-1} \amp \lt \frac{1}{\sqrt{N}} + (x-1)^2 \amp \lt \frac{1}{N} + \frac{1}{(x-1)^2} \amp \gt N + , + which is what we wanted to show. + So we may say \lim_{x\to 1}1/{(x-1)^2}=\infty. +

    +     + +     + + + Evaluating limits involving infinity + +

    + Find \lim\limits_{x\to 0}\frac1x, + as shown in . +

    + +
    + Evaluating \lim\limits_{x\to 0}\frac1x in + + + +

    + Graph of y=1/x, for x between -1 and 1. + There is a vertical asymptote at x = 0 and a horizontal asymptote at + y = 0. As x approaches 0 from the left, y + approaches -\infty and from the right, y approaches + \infty. As y approaches 0 from the bottom, x + approaches -\infty and from the top, x approaches + \infty. The graph conists of two parts; one in quadrant one and the other in + quadrant three. +

    +     + + Graph of 1 over x. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1.1, + xmax=1.1, + ymin=-55, + ymax=55, + ] + \addplot[firstcurvestyle,domain=-1:-0.125,leftarrow] {1/x}; + \addplot[firstcurvestyle,domain=-8:-50,rightarrow] ({1/x},{x}); + \addplot[firstcurvestyle,domain=1:0.125,leftarrow] {1/x}; + \addplot[firstcurvestyle,domain=8:50,rightarrow] ({1/x},{x}); + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + It is easy to see that the function grows without bound near 0, + but it does so in different ways on different sides of 0. + Since its behavior is not consistent, + we cannot say that \lim_{x\to 0}\frac{1}{x}=\infty. + Instead, we will say \lim_{x\to 0}\frac{1}{x} does not exist. + However, we can make a statement about one-sided limits. + We can state that \lim_{x\to 0^+}\frac1x=\infty and \lim_{x\to 0^-}\frac1x=-\infty. +

    +     + +     + + +     + + + Vertical asymptotes +

    + The graphs in the two previous examples demonstrate that if a function f has a limit (or, + left- or right-hand limit) of infinity at x=c, + then the graph of f looks similar to a vertical line near x=c. + This observation leads to a definition. +

    + + + Vertical Asymptote + +

    + Let I be an interval that either contains c or has c as an endpoint, + and let f be a function defined on I, + except possibly at c. +

    + +

    + If the limit of f(x) as x approaches c from either the left or right + (or both) + is \infty or -\infty, + then the line x=c is a vertical asymptote of f. + + asymptotevertical + +

    +     +     + + + + + Finding vertical asymptotes + +

    + Find the vertical asymptotes of f(x)=\frac{3x}{x^2-4}. +

    +     + +

    + Vertical asymptotes occur where the function grows without bound; + this can occur at values of c where the denominator is 0. + When x is near c, the denominator is small, + which in turn can make the function take on large values. + In the case of the given function, + the denominator is 0 at x=\pm 2. + Substituting in values of x close to 2 and -2 seems to indicate that the function tends toward \infty or -\infty at those points. + We can graphically confirm this by looking at . + Thus the vertical asymptotes are at x=\pm2. +

    + +
    + Graphing f(x) = \frac{3x}{x^2-4} + + + +

    + Graph of f(x)=\frac{3x}{x^2-4}. There are + two vertical asymptotes, one at x = -2 and the other at + x = 2. For x values less than -2, f(x) is + less than 0 and the graph is curved downward. For x + values greater than 2, f(x) is greater than 0 + and the graph is curved upward. For the interval -2 \lt x \lt 0 + the graph is curved upward, at x = 0 the graph changes and + starts to curve downward for the interval 0 \lt x \lt 2. +

    +

    + Because of the asymptote at x = -2, as x gets near + -2 from the left f(x) approaches -\infty. + But coming from the right f(x) approaches \infty. + Because of the other asymptote at x = 2, as x get near + 2 from the left f(x) approaches -\infty. But + coming from the right f(x) approaches \infty. +

    +     + + Graph of a rational function with two vertical asymptotes. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-6.1, + xmax=6.1, + ymin=-16, + ymax=16, + ] + \addplot[firstcurvestyle,domain=-6:-2.5,leftarrow] {3*x/(x^2-4)}; + \addplot[firstcurvestyle,domain=-3.333:-15.366,rightarrow] ({(3+sqrt(9+16*x^2))/(2*x)},{x}); + \addplot[firstcurvestyle,domain=14.615:2.571,leftarrow] ({(3-sqrt(9+16*x^2))/(2*x)},{x}); + \addplot[firstcurvestyle,domain=-1.5:1.5,-] {3*x/(x^2-4)}; + \addplot[firstcurvestyle,domain=-2.571:-14.615,rightarrow] ({(3-sqrt(9+16*x^2))/(2*x)},{x}); + \addplot[firstcurvestyle,domain=3.333:15.366,rightarrow] ({(3+sqrt(9+16*x^2))/(2*x)},{x}); + \addplot[firstcurvestyle,domain=2.5:6,rightarrow] {3*x/(x^2-4)}; + \addplot[asymptote] coordinates {(-2,-16) (-2,16)}; + \addplot[asymptote] coordinates {(2,-16) (2,16)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + +

    + When a rational function has a vertical asymptote at x=c, + we can conclude that the denominator is 0 at x=c. + However, just because the denominator is 0 at a certain point does not mean there is a vertical asymptote there. + For instance, + f(x)=(x^2-1)/(x-1) does not have a vertical asymptote at x=1, + as shown in . + While the denominator does get small near x=1, + the numerator gets small too, + matching the denominator step for step. + In fact, factoring the numerator, we get + + f(x)=\frac{(x-1)(x+1)}{x-1} + . +

    + +

    + Canceling the common term, we get that f(x)=x+1 for x\not=1. + So there is clearly no asymptote; + rather, a hole exists in the graph at x=1. +

    + +
    + Graphically showing that f(x) = \frac{x^2-1}{x-1} does not have an asymptote at x=1 + + + +

    + The graph is a single straight line with a positive slope. + At x = 1 there is a hollow dot indicating a removable discontinuity. + The exact position of the discontinuity is (1, 2). +

    +     + + Graph of a rational function. A zero in the denominator is a hole, not a vertical asymptote. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1.2, + xmax=2.2, + ymin=-.2, + ymax=3.2, + ] + \addplot+[infinite,domain=-1:2] {x+1}; + \addplot[hollowdot] coordinates {(1,2)}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + The above example may seem a little contrived. + Another example demonstrating this important concept is f(x)= (\sin(x) )/x. + We have considered this function several times in the previous sections. + We found that \lim_{x\to0}\frac{\sin(x) }{x}=1; + , there is no vertical asymptote. + No simple algebraic cancellation makes this fact obvious; + we used the in to prove this. +

    + +

    + If the denominator is 0 at a certain point but the numerator is not, + then there will usually be a vertical asymptote at that point. + On the other hand, + if the numerator and denominator are both zero at that point, + then there may or may not be a vertical asymptote at that point. + This case where the numerator and denominator are both zero returns us to an important topic. +

    +     + + + Indeterminate Forms +

    + limitindeterminate form + indeterminate form +

    + +

    + We have seen how the limits \lim_{x\to 0}\frac{\sin(x) }{x} and + \lim_{x\to1}\frac{x^2-1}{x-1} each return the indeterminate form 0/0 when we blindly plug in x=0 and x=1, + respectively. + However, 0/0 is not a valid arithmetical expression. + It gives no indication that the respective limits are 1 and 2. +

    + +

    + With a little cleverness, + one can come up with 0/0 expressions which have a limit of \infty, 0, or any other real number. + That is why this expression is called + indeterminate. +

    + +

    + A key concept to understand is that such limits do not really return 0/0. + Rather, keep in mind that we are taking limits. + What is really happening is that the numerator is shrinking to 0 while the denominator is also shrinking to 0. + The respective rates at which they do this are very important and determine the actual value of the limit. +

    + +

    + An indeterminate form indicates that one needs to do more work in order to compute the limit. + That work may be algebraic + (such as factoring and canceling), + it may involve using trigonometric identities or logarithm rules, + or it may require a tool such as the . + In + we will learn yet another technique called L'Hospital's Rule that provides another way to handle indeterminate forms. +

    + +

    + Some other common indeterminate forms are + \infty-\infty, \infty\cdot 0, + \infty/\infty, 0^0, + \infty^0 and 1^{\infty}. + Again, keep in mind that these are the blind + results of directly substituting c into the expression, + and each, + in and of itself, has no meaning. + The expression \infty-\infty does not really mean + subtract infinity from infinity. Rather, + it means One quantity is subtracted from the other, + but both are growing without bound. What is the result? + It is possible to get every value between -\infty and \infty. +

    + +

    + Note that 1/0 and \infty/0 are not indeterminate forms, + though they are not exactly valid mathematical expressions, either. + In each, the function is growing without bound, + indicating that the limit will be \infty, -\infty, + or simply not exist if the left- and right-hand limits do not match. +

    +     + + + Limits at Infinity and Horizontal Asymptotes +

    + At the beginning of this section we briefly considered what happens to + f(x) = 1/x^2 as x grew very large. + Graphically, it concerns the behavior of the function to the + far right of the graph. + We make this notion more explicit in the following definition. +

    + + + + + Limits at Infinity and Horizontal Asymptotes + +

    + Let L be a real number. +

      +
    1. +

      + Let f be a function defined on + (a,\infty) for some number a. + The limit of f at infinity is L, + denoted \lim_{x\to\infty} f(x)=L, + if for every \epsilon \gt 0 there exists M \gt a such that if x \gt M, + then \abs{f(x)-L}\lt \epsilon. + + limitat infinity + asymptotehorizontal + +

      +
      2. +
    2. +

      + Let f be a function defined on + (-\infty,b) for some number b. + The limit of f at negative infinity is L, + denoted \lim_{x\to-\infty} f(x)=L, + if for every \epsilon \gt 0 there exists M\lt b such that if x \lt M, + then \abs{f(x)-L}\lt \epsilon. +

      +
      3. +
    3. +

      + If \lim_{x\rightarrow\infty} f(x)=L or \lim_{x\rightarrow-\infty} f(x)=L, + we say the line y=L is a + horizontal asymptote of f. +

      +
      4. +
    +

    +     +     + +

    + We can also define limits such as + \lim_{x\to\infty}f(x)=\infty by combining this definition with + . +

    + + + Approximating horizontal asymptotes + +

    + Approximate the horizontal asymptote(s) of f(x)=\frac{x^2}{x^2+4}. +

    +     + +

    + We will approximate the horizontal asymptotes by approximating the limits + \lim_{x\to-\infty} \frac{x^2}{x^2+4} and + \lim_{x\to\infty} \frac{x^2}{x^2+4}. + (A rational function can have at most + one horizontal asymptote. + So we could get away with only taking x \to \infty). +

    + +

    + shows a sketch of f, + and the table in + gives values of f(x) for large magnitude values of x. + It seems reasonable to conclude from both of these sources that f has a horizontal asymptote at y=1. +

    + +
    + Using a graph and a table to approximate a horizontal asymptote in + +
    + + +

    + Graph of f(x)=\frac{x^2}{x^2+4} showing x values from -20 to 20. + There is a horizontal asymptote at y = 1. + As x approaches -\infty and \infty, f(x) gets near 1, + but never equals 1. The graph lies between y=0 and y=1, + and drops to the point (0,0) + as x approaches 0 from either direction. +

    +     + + Graph of a rational function showing a horizontal asymptote at y = 1. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-21, + xmax=21, + ymin=-.2, + ymax=1.1, + ] + \addplot+[infinite,domain=-84:84,samples=177] ({2*tan(x)},{-(cos(x))^2+1}); + \addplot[asymptote] coordinates {(-21,1) (21,1)}; + \end{axis} + \end{tikzpicture} + + + +
    + +
    + + + x + f(x) + + + 10 + 0.9615 + + + 100 + 0.9996 + + + 10000 + 0.999996 + + + -10 + 0.9615 + + + -100 + 0.9996 + + + -10000 + 0.999996 + + +
    +     +
    + +

    + Later, we will show how to determine this analytically. +

    +     +     + + + + + +

    + Horizontal asymptotes can take on a variety of forms. + + shows that f(x) = x/(x^2+1) has a horizontal asymptote of y=0, + where 0 is approached from both above and below. +

    + +

    + + shows that f(x) =x/\sqrt{x^2+1} has two horizontal asymptotes; + one at y=1 and the other at y=-1. +

    + +

    + + shows that f(x) = \sin(x)/x has even more interesting behavior than at just x=0; + as x approaches \pm\infty, + f(x) approaches 0, + but oscillates as it does this. +

    + + +
    + Considering different types of horizontal asymptotes + +
    + + + + +

    + As x approaches -\infty and \infty, f(x) gets near 0. + The graph dips to a minimum value just to the left of the y axis, + then crosses the x axis at (0,0), rising to a maximum value just to the right of the x axis, + before falling again toward the horizontal asymptote y=0. +

    +     + + Graph of a rational function with x axis as horizontal asymptote. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-21, + xmax=21, + ymin=-1.1, + ymax=1.1, + ] + \addplot+[infinite,domain=-87:87,samples=101] ({tan(x)},{sin(x)*cos(x)}); + \end{axis} + \end{tikzpicture} + + + +
    + +
    + + + + +

    + The graph of f(x)=\frac{x}{\sqrt{x^2+1}}, which has two horizontal asymptotes, + one at y = -1 (representing the limit as x\to -\infty) + and the other at y = 1 (representing the limit as x\to\infty). +

    +     + + Graph of a function with two horizontal asymptotes. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-21, + xmax=21, + ymin=-1.1, + ymax=1.1, + ] + \addplot+[infinite,domain=-87:87,samples=101] ({tan(x)},{sin(x)}); + \end{axis} + \end{tikzpicture} + + + +
    + +
    + + + + +

    + The graph of f(x)=\sin(x)/x, + which oscillates around the x axis. + As the absolute value of x gets bigger, the oscillations get smaller. + The limit as x\to \pm \infty is zero, so f(x) has a horizontal asymptote that it crosses an infinite number of times. +

    +     + + Graph of sin(x) over x, showing that a function can cross a horizontal asymptote infinitely many times. + + + \begin{tikzpicture}[declare function = {func(\x) = (\x != 0) * (sin(x*180/pi)/x) + (\x == 0) * (1);}] + \begin{axis}[ + xmin=-21, + xmax=21, + ymin=-.3, + ymax=1.1 + ] + \addplot+[infinite,domain=-20:20,samples=101] {func(x)}; + \end{axis} + \end{tikzpicture} + + + +
    +     +
    + +

    + We can analytically evaluate limits at infinity for rational functions once we understand \lim_{x\to\infty}\frac{1}{x}. + As x gets larger and larger, + 1/x gets smaller and smaller, approaching 0. + We can, in fact, + make 1/x as small as we want by choosing a large enough value of x. + Given \varepsilon, + we can make 1/x\lt \varepsilon by choosing x \gt 1/\varepsilon. + Thus we have \lim_{x\to\infty} 1/x=0. +

    + +

    + It is now not much of a jump to conclude the following: + + \lim_{x\to\infty}\frac1{x^n}\amp=0\amp\lim_{x\to-\infty}\frac1{x^n}\amp=0 + . +

    + + + + +

    + Now suppose we need to compute the following limit: + + \lim_{x\to\infty}\frac{x^3+2x+1}{4x^3-2x^2+9} + . +

    + +

    + A good way of approaching this is to divide through the numerator and denominator by x^3 + (hence multiplying by 1), + which is the largest power of x to appear in the denominator. + Doing this, we get + + \lim_{x\to\infty}\frac{x^3+2x+1}{4x^3-2x^2+9} \amp = + \lim_{x\to\infty}\frac{1/x^3}{1/x^3}\cdot\frac{x^3+2x+1}{4x^3-2x^2+9} + \amp =\lim_{x\to\infty}\frac{x^3/x^3+2x/x^3+1/x^3}{4x^3/x^3-2x^2/x^3+9/x^3} + \amp = \lim_{x\to\infty}\frac{1+2/x^2+1/x^3}{4-2/x+9/x^3} + . +

    + +

    + Then using the rules for limits + (which also hold for limits at infinity), + as well as the fact about limits of 1/x^n, + we see that the limit becomes + + \frac{1+0+0}{4-0+0}=\frac14 + . +

    + +

    + This procedure works for any rational function. + In fact, it gives us the following theorem. +

    + + + Limits of Rational Functions at Infinity + +

    + Let f(x) be a rational function of the following form: + + f(x)=\frac{a_nx^n + a_{n-1}x^{n-1}+\dots + a_1x + a_0}{b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0} + , + where m,n are positive integers and where any of the coefficients may be 0 except for a_n and b_m. + Then: +

      +
    1. +

      + If n=m, then + + \lim_{x\to\infty} f(x) = \lim_{x\to-\infty} f(x) = \frac{a_n}{b_m} + . +

      +
      2. +
    2. +

      + If n\lt m, then + + \lim_{x\to\infty} f(x) = \lim_{x\to-\infty} f(x) = 0 + . +

      +
      3. +
    3. +

      + If n \gt m, + then \lim_{x\to\infty} f(x) and \lim_{x\to-\infty} f(x) are both infinite. +

      +
      4. +
    +

    +     +     + +

    + We can see why this is true. + If the highest power of x is the same in both the numerator and denominator ( n=m), + we will be in a situation like the example above, + where we will divide by x^n and in the limit all the terms will approach 0 except for + a_nx^n/x^n and b_mx^m/x^n. + Since n=m, this will leave us with the limit a_n/b_m. + If n\lt m, then after dividing through by x^m, + all the terms in the numerator will approach 0 in the limit, + leaving us with 0/b_m or 0. + If n \gt m, and we try dividing through by x^m, + we end up with the denominator tending to b_m while the numerator tends to \infty. +

    + +

    + Intuitively, as x gets very large, + all the terms in the numerator are small in comparison to a_nx^n, + and likewise all the terms in the denominator are small compared to b_mx^m. + If n=m, looking only at these two important terms, + we have (a_nx^n)/(b_mx^m). + This reduces to a_n/b_m. + If n\lt m, the function behaves like a_n/(b_mx^{m-n}), + which tends toward 0. + If n \gt m, the function behaves like a_nx^{n-m}/b_m, + which will tend to either \infty or -\infty depending on the values of n, + m, + a_n, + b_m and whether you are looking for + \lim_{x\to\infty} f(x) or \lim_{x\to-\infty} f(x). +

    + + + Finding a limit of a rational function + +

    + Confirm analytically that y=1 is the horizontal asymptote of f(x) = \frac{x^2}{x^2+4}, + as approximated in . +

    +     + +

    + Before using , + let's use the technique of evaluating limits at infinity of rational functions that led to that theorem. + The largest power of x in f is 2, + so divide the numerator and denominator of f by x^2, + then take limits. + + \lim_{x\to\infty}\frac{x^2}{x^2+4} \amp = \lim_{x\to\infty}\frac{x^2/x^2}{x^2/x^2+4/x^2} + \amp =\lim_{x\to\infty}\frac{1}{1+4/x^2} + \amp =\frac{1}{1+0} + \amp = 1 + . +

    + +

    + We can also use directly; + in this case n=m so the limit is the ratio of the leading coefficients of the numerator and denominator, + , 1/1 = 1. +

    +     + +     + + + Finding limits of rational functions + +

    + Use + to evaluate each of the following limits. +

      +
    1. \lim_{x\to-\infty}\dfrac{x^2+2x-1}{x^3+1}
      2. +
    2. \lim_{x\to\infty}\dfrac{x^2+2x-1}{1-x-3x^2}
      3. +
    3. \lim_{x\to\infty}\dfrac{x^2-1}{3-x}
      4. +
    +

    +     + +

    +

      +
    1. +

      + The highest power of x is in the denominator. + Therefore, the limit is 0; + see . +

      +
      2. +
    2. +

      + The highest power of x is x^2, + which occurs in both the numerator and denominator. + The limit is therefore the ratio of the coefficients of x^2, + which is -1/3. + See . +

      +
      3. +
    3. +

      + The highest power of x is in the numerator so the limit will be \infty or -\infty. + To see which, + consider only the dominant terms from the numerator and denominator, + which are x^2 and -x. + The expression in the limit will behave like + x^2/(-x) = -x for large values of x. + Therefore, the limit is -\infty. + See . +

      +
      4. +
    +

    + +
    + Visualizing the functions in + +
    + + + + +

    + Graph of \dfrac{x^2+2x-1}{x^3+1} for x\lt 0. + The graph shows that + as x approaches -\infty, the limit of f(x) + is 0. +

    +     + + Graph that illustrates the highest power of x is in the denominator. Therefore, the limit is 0. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-41, + xmax=2, + ymin=-.6, + ymax=.6, + ] + \addplot[firstcurvestyle,domain=-40:-8,leftarrow] {(x^2+2*x-1)/(x^3+1)}; + \addplot[firstcurvestyle,domain=-8:-2,rightarrow] {(x^2+2*x-1)/(x^3+1)}; + \end{axis} + \end{tikzpicture} + + + +
    + +
    + + + + +

    + The graph of f(x)=\frac{x^2+2x-1}{1-x-3x^2} is shown for x\gt 0. + The graph has a horizontal asymptote at y = -\frac{1}{3}, + which it approaches from below. + The graph illustrates that the limit of f(x) as x\to\infty is -\frac{1}{3}. + The coefficient of the term in the numerator of f(x) with the highest + power of x is 1, + while the coefficient of the term in the dennominator with the + highest power of x is -3. + The ratio of these two coefficients gives the limit as x\to\pm \infty + when the highest power of x is the same in both the numerator and the denominator. +

    +     + + Graph that shows the limit is the ratio of the coefficients of the highest power of x. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1, + xmax=41, + ymin=-.55, + ymax=.55, + ] + \addplot[firstcurvestyle,domain=2:8,leftarrow] {(x^2+2*x-1)/(1-x-3*x^2)}; + \addplot[firstcurvestyle,domain=8:40,rightarrow] {(x^2+2*x-1)/(1-x-3*x^2)}; + \addplot[asymptote] coordinates {(-1,-0.33333) (41,-0.33333)}; + \end{axis} + \end{tikzpicture} + + + +
    + +
    + + + + +

    + The graph of f(x)=\frac{x^2-1}{3-x} is shown for x\gt 3. + Near x=3 the graph appears to be heading down a vertical asymptote, + suggesting that \lim_{x\to 3^+}f(x)=-\infty. + The graph then rises to a peak, before beginning to descend again. + Beyond x=10, the graph appears almost straight, + and continues downward at a slope close to -1, + showing that there is no horizontal asymptote in this case. +

    +

    + The graph shows that the limit of f(x) will be determined + by dominant terms from the numerator and denominator, which are + x^2 and -x. Since \frac{x^2}{-x} = -x for large values + of x, the graph of f(x) behaves approximately the same as that of y=-x. +

    +     + + Graph that shows the behaviour of a function can depend on the dominant terms of the numerator and denominator. + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-5, + xmax=41, + ymin=-50, + ymax=5 + ] + \addplot[firstcurvestyle,domain=3.5:8,leftarrow] {(x^2-1)/(3-x)}; + \addplot[firstcurvestyle,domain=8:40,rightarrow] {(x^2-1)/(3-x)}; + \end{axis} + \end{tikzpicture} + + + +
    +     +
    +     +     + + + +

    + With care, we can quickly evaluate limits at infinity for a large number of functions by considering the + long run behavior using dominant terms of f(x). + For instance, + consider again \lim_{x\to\pm\infty}\frac{x}{\sqrt{x^2+1}}, + graphed in . + The dominant terms are x in the numerator and \sqrt{x^2} in the denominator. + When x is very large, x^2+1 \approx x^2. + Thus + + \sqrt{x^2+1}\amp\approx \sqrt{x^2} = \abs{x} \amp \frac{x}{\sqrt{x^2+1}} \amp\approx \frac{x}{\abs{x}} + . +

    + +

    + This expression is 1 when x is positive and -1 when x is negative. + Hence we get asymptotes of y=1 and y=-1, + respectively. + We will show this more formally in the next example. +

    + + + Finding a limit using dominant terms + +

    + Confirm analytically that y=1 and y=-1 are the horizontal asymptote of \lim_{x\to\pm\infty}\frac{x}{\sqrt{x^2+1}}, + as graphed in . +

    +     + +

    + The dominating term of f in the denominator is + \sqrt{x^2}=\abs{x} so divide the numerator and denominator of f by \sqrt{x^2}, + then take limits. + + \lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}} + \amp = \lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}}\cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}} + \amp = \lim_{x\to\infty} \frac{\frac{x}{\abs{x}}}{\sqrt{\frac{x^2+1}{x^2}}} + \amp = \lim_{x\to\infty} \frac{1}{\sqrt{1+\frac{1}{x^2}}} \text{ for } x\gt 0 + \amp =\frac{1}{\sqrt{1+0}} + \amp = 1 + . +

    + +

    + As x \to -\infty, + the only thing that changes is the value of \frac{x}{\abs{x}}. + For x \lt 0, we have \frac{x}{\abs{x}}=-1, + making \lim_{x\to-\infty} \frac{x}{\sqrt{x^2+1}}=-1. + Therefore, the horizontal asymptotes are y=1 and y=-1. +

    +     +     + + + +     + + + + + + + Terms and Concepts + + + + + + +

    + + If \lim\limits_{x\to 5} f(x) = \infty, + then we are implicitly stating that the limit exists. +

    +     + +

    + When we say that a limit exists, we mean that the limit has a finite, numerical value. + An infinite limit describes a particular way in which a limit can fail to exist. +

    +     + +     + + + + + +

    + + If \lim\limits_{x\to 5} f(x) = 5, + then we are implicitly stating that the limit exists. +

    +     + +

    + In this case, the limit is a finite, numerical value; namely, 5. + This means that the limit exists. +

    +     + +     + + + + + +

    + + If \lim\limits_{x\to 1^-} f(x) = -\infty, + then \lim\limits_{x\to 1^+} f(x) = \infty. +

    +     + +

    + The sign of a function does not necessarily change at every vertical asymptote. + Consider, for example, f(x)=1/x versus f(x)=1/x^2. + We always need to carefully determine the sign of f(x) on both sides of each + vertical asymptote. +

    + +

    + It could also happen that the limit from one side is infinite, + while the limit from the other side is finite, or fails to exist in some other way. + (Can you think of an example?) +

    +     + +     + + + + + +

    + + If \lim\limits_{x\to 5} f(x) = \infty, + then f has a vertical asymptote at x=5. +

    +     + +

    + According to , + a function f has a vertical asymptote at c + if f approaches infinity from either side of c. +

    +     + +     + + + + + +

    + + \infty/0 is not an indeterminate form. +

    +     + +

    + Both the numerator approaching infinity + and the denominator approaching zero indicate an infinite limit, + so this is not indeterminate. + We may not be able to tell immediately if the result will be +\infty or -\infty, + but this is true of most infinite limits. +

    +     + +     + + + + +

    + List five indeterminate forms. +

    + +     + + +     + + + + +

    + Construct a function with a vertical asymptote at x=5 and a horizontal asymptote at y=5. +

    + +     + + +     + + + + +

    + Let \lim\limits_{x\to 7} f(x) = \infty. + Explain how we know that f is or is not continuous at x=7. +

    + +     + + + +

    + The limit of f as x approaches 7 does not exist, + hence f is not continuous. + (Note: f could be defined at 7!) +

    +     + +     +     + + + Problems + + + +

    + Evaluate the given limits using the graph of the function. +

    +     + + + + + $a=non_zero_random(-3,3,1); + $b=1; + $n=random(2,5,1); + $f=Formula("$b/(x-$a)^($n)")->reduce; + $L[0]=($f->eval(x=>$a-1)>0)?Compute("inf"):Compute("-inf"); + $L[1]=($f->eval(x=>$a+1)>0)?Compute("inf"):Compute("-inf"); + for my$i(0,1){$ev[$i]=$L[$i]->cmp()->withPostFilter(AnswerHints(Compute("DNE")=> "Technically correct, but this question is asking for more specific information."));}; + $xmin=min(-1,$a-2);$xmax=max(1,$a+2); + $e=1/50**(1/$n); + $below=$a-$e; + $above=$a+$e; + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + f(x) = has the graph: +

    + + +

    + Graph with the domain to . There + is a vertical asymptote at x = , + and f(x) grows without bound on both sides of the asymptote. + To the left of x = , f(x) is negative, + and to the right of the asymptote, f(x) is positive. +

    +     + + Graph of a reciprocal power function with a vertical asymptote at x =. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = -50, + ymax = 50, + ] + \addplot[firstcurvestyle, infiniteright, domain=$xmin:$below, samples=101] {$b/(x-$a)^($n)}; + \addplot[firstcurvestyle, infiniteleft, domain=$above:$xmax, samples=101] {$b/(x-$a)^($n)}; + \addplot[asymptote,-] coordinates {($a,-50) ($a,50)}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     +     +     + + + + + $a0 = non_zero_random(-2,3,1); + $b = random(1,2,1); + $a1 = $a0+$b; + @a = ($a0,$a1); + $f=Formula("1/((x-$a[0])(x-$a[1])^(2))")->reduce; + $L[0]=($f->eval(x=>$a[0]-0.1)>0)?Compute("inf"):Compute("-inf"); + $L[1]=($f->eval(x=>$a[0]+0.1)>0)?Compute("inf"):Compute("-inf"); + $L[3]=($f->eval(x=>$a[1]-0.1)>0)?Compute("inf"):Compute("-inf"); + $L[4]=($f->eval(x=>$a[1]+0.1)>0)?Compute("inf"):Compute("-inf"); + for my$i(2,5){$L[$i]=($L[$i-2]==$L[$i-1])?$L[$i-2]:Compute("DNE");}; + for my$i(0..5){$ev[$i]=($L[$i]==Compute("DNE"))?$L[$i]->cmp():$L[$i]->cmp()->withPostFilter(AnswerHints(Compute("DNE")=> "Technically correct, but this question is asking for more specific information."));}; + $xmin=min(0,@a)-1;$xmax=max(@a)+1; + $d = abs($a[1]-$a[0]); + @x = num_sort( + (sqrt($d*(25*$d**3+4))-5*$d**2)/(20*$d) + $a[0], + -(sqrt($d*(25*$d**3+4))-5*$d**2)/(20*$d) + $a[0], + sqrt(1/(50*$d)) + $a[1], + -sqrt(1/(50*$d)) + $a[1] + ); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + f(x) = has the graph: +

    + + +

    + There are + two vertical asymptotes, one at x = + and the other at x = . On the interval + \lt x \lt the + graph curves and as x gets near + and , f(x) + approaches \infty. For the interval + -\infty \lt x \lt as x gets near + , f(x) approaches -\infty. Lastly, + for the interval \lt x \lt \infty as x + gets near , f(x) approaches \infty. +

    +     + + Graph for problem 10. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = -50, + ymax = 50, + ] + \addplot[firstcurvestyle, infiniteright, domain=$xmin:$x[0], samples=101] {1/((x-$a[0])*(x-$a[1])^(2))}; + \addplot[firstcurvestyle, infinite, domain=$x[1]:$x[2], samples=101] {1/((x-$a[0])*(x-$a[1])^(2))}; + \addplot[firstcurvestyle, infiniteleft, domain=$x[3]:$xmax, samples=101] {1/((x-$a[0])*(x-$a[1])^(2))}; + \addplot[asymptote,-] coordinates {($a[0],-50) ($a[0],50)}; + \addplot[asymptote,-] coordinates {($a[1],-50) ($a[1],50)}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^-} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to ^+} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to } f(x) +

    +

    + +

    +     +     +     +     + + + + + $a=random(1,5,1); + $b=random(-1,1,2); + $f=Formula("$a/(e^($b*x)+1)")->reduce; + $L[0]=($b>0)?$a:0; + $L[1]=($b>0)?0:$a; + $L[2]=$f->eval(x=>0); + $L[3]=$f->eval(x=>0); + $xmin=-11;$xmax=11; + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + f(x) = has the graph: +

    + + +

    + There are two horizontal asymptotes, one at y = 0 and the other at y = . + For some versions of this question, the graph f(x)= begins with y + near when x is negative, and then descends toward 0 as x increases. +

    + +

    + For other versions of this question, f(x) will be close to 0 when x is negative, + and then will rise toward as x increases. +

    +     + + Graph for problem 11. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = -1, + ymax = $a+1, + ] + \addplot[firstcurvestyle, domain=$xmin:$xmax, smooth] {$a/(e^($b*x)+1)}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to-\infty} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to\infty} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to 0^{-}} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to 0^{+}} f(x) +

    +

    + +

    +     +     +     +     + + + + + $a=random(1,5,1); + $n=random(1,4,2); + $f=Formula("x^($n)*sin($a pi x)")->reduce; + $L[0]=Compute("DNE"); + $L[1]=Compute("DNE"); + $L[2]=0; + $L[3]=0; + $xmax = int(11/(2/$a))*(2/$a); + $xmin = -$xmax; + $samples = 1 + int(11/(2/$a))*80; + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + f(x) = has the graph: +

    + + +

    + Graph of f(x)=. + The graph is oscillating, as x approaches -\infty and \infty + the amplitude gets larger, but as x gets near 0 the amplitude + shrinks. +

    +     + + Oscillating graph with high amplitude at infinity and -infinty, but very small amplitude near 0. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ] + \addplot[firstcurvestyle, domain=$xmin:$xmax, samples=$samples] {x^($n)*sin(deg($a*pi*x))}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to-\infty} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to\infty} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to 0^{-}} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to 0^{+}} f(x) +

    +

    + +

    +     +     +     +     + + + + + $a=random(1,5,1); + $trigfun = list_random('sin','cos'); + $f=Formula("$trigfun($a x)")->reduce; + $L[0]=Compute("DNE"); + $L[1]=Compute("DNE"); + $xmax = int(14/(2*pi/$a))*(2*pi/$a); + $xmin = -$xmax; + $samples = 1 + int(14/(2*pi/$a))*80; + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + f(x) = has the graph: +

    + + +

    + The graph + is oscillating, with a peak of y = and minimum of y = -. + The graph oscillates continually between these values over the domain of the function. +

    +     + + Oscillating graph with a peak of y = 1 and minimum of y = -1. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = -1.2, + ymax = 1.2, + ] + \addplot[firstcurvestyle, domain=$xmin:$xmax, samples=$samples] {$trigfun(deg($a*x))}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to-\infty} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to\infty} f(x) +

    +

    + +

    +     +     +     +     + + + + + $b=random(1.5,2.5,0.1); + $c=non_zero_random(-12,12,1); + $n=random(-1,1,2); + $f=Formula("($b)^($n x)+$c")->reduce; + $L[0]=($n>0)?$c:Compute("inf"); + $L[1]=($n>0)?Compute("inf"):$c; + $xmin=-11;$xmax=11; + $start = $xmin; + $stop = log(20 - $c)/log($b) / $n; + if ($n < 1) { + ($start,$stop) = ($stop,$xmax); + } + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + f(x) = has the graph: +

    + + +

    + Graph with the domain to . The + line y = is a horizontal asymptote, but only in one direction. +

    + +

    + In some versions of the question, the function grows without bound as x\to \infty, + and approaches the horizontal asymptote as x\to -\infty. + In other version fo the question, these roles are reversed. +

    +     + + Graph for problem 14. + + + \begin{tikzpicture} + \begin{axis}[ + xmin = $xmin, + xmax = $xmax, + ymin = -20, + ymax = 20, + ] + \addplot[firstcurvestyle, domain=$start:$stop, smooth] {$f}; + \end{axis} + \end{tikzpicture} + + + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to-\infty} f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to\infty} f(x) +

    +

    + +

    +     +     +     +     +     + + + +

    + Numerically approximate the limits. +

    +     + + + + + @b = random_subset(4,-9..9); + if($envir{problemSeed}==1){@b=(3,1,-1,-2);}; + $a=$b[0]; + $f=Formula("(x^2-($b[1]+$b[2])*x+$b[1]*$b[2])/(x^2-($a+$b[3])*x+$a*$b[3])")->reduce; + @x=($a-0.1,$a-0.01,$a-0.001,$a+0.1,$a+0.01,$a+0.001); + @y=map{$f->eval(x=>$_)}(@x); + $L[0]=($y[2]>0)?Compute("inf"):Compute("-inf"); + $L[1]=($y[5]>0)?Compute("inf"):Compute("-inf"); + $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); + $limit=($L[2]==Compute("DNE"))?"does not exist":"is `".$L[2]->TeX."`"; + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + f(x)= +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to^{-}}f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to^{+}}f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to}f(x) +

    +

    + +

    +     +     + + +

    + Tables will vary. +

    +

    +

      +
    1. + + + x + f(x) + + + + + + + + + + + + + + +

      + It seems \lim\limits_{x\to^{-}}f(x)=. +

      +
      2. +
    2. + + + x + f(x) + + + + + + + + + + + + + + +

      + It seems \lim\limits_{x\to^{+}}f(x)=. +

      +
      3. +
    3. +

      + It seems \lim\limits_{x\to3}f(x) . +

      +
      4. +
    +

    +     +     +     + + + + + @b = random_subset(4,-9..9); + if($envir{problemSeed}==1){@b=(3,4,-9,-1);}; + $a=$b[0]; + $f=Formula("(x^2-($b[1]+$b[2])*x+$b[1]*$b[2])/(x^3-(2*$a+$b[3])*x^2+(2*$a*$b[3]+($a)^2)*x-($a)^2*$b[3])")->reduce; + @x=($a-0.1,$a-0.01,$a-0.001,$a+0.1,$a+0.01,$a+0.001); + @y=map{$f->eval(x=>$_)}(@x); + $L[0]=($y[2]>0)?Compute("inf"):Compute("-inf"); + $L[1]=($y[5]>0)?Compute("inf"):Compute("-inf"); + $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); + $limit=($L[2]==Compute("DNE"))?"does not exist":"is `".$L[2]->TeX."`"; + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + f(x)= +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to^{-}}f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to^{+}}f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to}f(x) +

    +

    + +

    +     +     + +

    + Tables will vary. +

    +

    +

      +
    1. + + + x + f(x) + + + + + + + + + + + + + + +

      + It seems \lim\limits_{x\to^{-}}f(x)=. +

      +
      2. +
    2. + + + x + f(x) + + + + + + + + + + + + + + +

      + It seems \lim\limits_{x\to^{+}}f(x)=. +

      +
      3. +
    3. +

      + It seems \lim\limits_{x\to3}f(x) . +

      +
      4. +
    +

    +     +     +     + + + + + @b = random_subset(4,-9..9); + if($envir{problemSeed}==1){@b=(3,5,6,-2);}; + $a=$b[0]; + $f=Formula("(x^2-($b[1]+$b[2])*x+$b[1]*$b[2])/(x^3-(2*$a+$b[3])*x^2+(2*$a*$b[3]+($a)^2)*x-($a)^2*$b[3])")->reduce; + @x=($a-0.1,$a-0.01,$a-0.001,$a+0.1,$a+0.01,$a+0.001); + @y=map{$f->eval(x=>$_)}(@x); + $L[0]=($y[2]>0)?Compute("inf"):Compute("-inf"); + $L[1]=($y[5]>0)?Compute("inf"):Compute("-inf"); + $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); + $limit=($L[2]==Compute("DNE"))?"does not exist":"is `".$L[2]->TeX."`"; + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + f(x)= +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to^{-}}f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to^{+}}f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to}f(x) +

    +

    + +

    +     +     + +

    + Tables will vary. +

    +

    +

      +
    1. + + + x + f(x) + + + + + + + + + + + + + + +

      + It seems \lim\limits_{x\to^{-}}f(x)=. +

      +
      2. +
    2. + + + x + f(x) + + + + + + + + + + + + + + +

      + It seems \lim\limits_{x\to^{+}}f(x)=. +

      +
      3. +
    3. +

      + It seems \lim\limits_{x\to3}f(x) . +

      +
      4. +
    +

    +     +     +     + + + + + @b = random_subset(3,-9..9); + if($envir{problemSeed}==1){@b=(3,6,-2);}; + $a=$b[0]; + $f=Formula("(x^2-($a+$b[1])*x+$a*$b[1])/(x^2-($a+$b[2])*x+$a*$b[2])")->reduce; + @x=($a-0.1,$a-0.01,$a-0.001,$a+0.1,$a+0.01,$a+0.001); + @y=map{$f->eval(x=>$_)}(@x); + $L[0]=Real(($a-$b[1])/($a-$b[2])); + $L[1]=Real(($a-$b[1])/($a-$b[2])); + $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); + $limit=($L[2]==Compute("DNE"))?"does not exist":"is `".$L[2]->TeX."`"; + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + f(x)= +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +     + + + +

    + \lim\limits_{x\to^{-}}f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to^{+}}f(x) +

    +

    + +

    +     +     + + + +

    + \lim\limits_{x\to}f(x) +

    +

    + +

    +     +     + +

    + Tables will vary. +

    +

    +

      +
    1. + + + x + f(x) + + + + + + + + + + + + + + +

      + It seems \lim\limits_{x\to^{-}}f(x)=. +

      +
      2. +
    2. + + + x + f(x) + + + + + + + + + + + + + + +

      + It seems \lim\limits_{x\to^{+}}f(x)=. +

      +
      3. +
    3. +

      + It seems \lim\limits_{x\to3}f(x) . +

      +
      4. +
    +

    +     +     +     +     + + + +

    + Identify the horizontal and vertical asymptotes, + if any, of the given function. +

    +     + + + + + @b = random_subset(3,-9..-1,1..9); + do {@c=(random(2,5,1),non_zero_random(-9,9,1))} until ($b[1] != $c[1]/$c[0] and $b[2] != $c[1]/$c[0]); + if($envir{problemSeed}==1){@c=(2,-2);@b=(2,4,-5)}; + $f=Formula("($c[0] x^2-($c[1]+$b[0]*$c[0])*x+$b[0]*$c[1])/(x^2-($b[1]+$b[2])*x+$b[1]*$b[2])")->reduce; + parser::Assignment->Allow; + Context()->variables->are(x=>'Real',y=>'Real'); + Context()->flags->set(reduceConstants=>0); + $asymptotes=Formula("y=$c[0],x=$b[1],x=$b[2]"); + + +

    + f(x)= +

    + + Submit your answer as a list, using commas. + Each asymptote is line of the form x=... or y=.... + If there are no such asymptotes, + you may type NONE. + +

    + +

    +     + +

    + There are horizontal/vertical asymptotes at . +

    +     +     +     + + + + + @b = random_subset(3,-9..-1,1..9); + @c=(list_random(-5..-2,2..5),non_zero_random(-9,9,1)); + if($envir{problemSeed}==1){@c=(-3,6);@b=(-1,3,5)}; + Context()->noreduce('(-x)+y','(-x)-y'); + $f=Formula("($c[0] x^2-($c[1]+$b[0]*$c[0])*x+$b[0]*$c[1])/($b[2]*x^2-$b[2]*($b[0]+$b[1])*x+$b[0]*$b[1]*$b[2])")->reduce; + parser::Assignment->Allow; + Context()->variables->are(x=>'Real',y=>'Real'); + Context()->flags->set(reduceConstants=>0); + $asymptotes=Compute("y=$c[0]/$b[2],x=$b[1]"); + + +

    + f(x)= +

    + + Submit your answer as a list, using commas. + Each asymptote is line of the form x=... or y=.... + If there are no such asymptotes, + you may type NONE. + +

    + +

    +     + +

    + There are horizontal/vertical asymptotes at . +

    +     +     +     + + + + + @b = random_subset(3,-9..-1,1..9); + @c=(random(1,5,1),non_zero_random(-9,9,1)); + if($envir{problemSeed}==1){@c=(1,-4);@b=(3,-1,7)}; + Context()->noreduce('(-x)+y','(-x)-y'); + $f=Formula("($c[0] x^2-($c[1]+$b[0]*$c[0])*x+$b[0]*$c[1])/($b[2]*x^3-$b[2]*($b[0]+$b[1])*x^2+$b[0]*$b[1]*$b[2]*x)")->reduce; + parser::Assignment->Allow; + Context()->variables->are(x=>'Real',y=>'Real'); + Context()->flags->set(reduceConstants=>0); + $asymptotestring="y=0,x=0,x=$b[1]"; + $asymptotes=Compute("$asymptotestring"); + + +

    + f(x)= +

    + + Submit your answer as a list, using commas. + Each asymptote is line of the form x=... or y=.... + If there are no such asymptotes, + you may type NONE. + +

    + +

    +     + +

    + There are horizontal/vertical asymptotes at . +

    +     +     +     + + + + + @b = random_subset(3,-9..-1,1..9); + @c=(random(1,5,1),non_zero_random(-9,9,1)); + if($envir{problemSeed}==1){@c=(1,3);@b=(-3,1,9)}; + Context()->noreduce('(-x)+y','(-x)-y'); + $f=Formula("($c[0] x^2-($c[1]+$b[0]*$c[0])*x+$b[0]*$c[1])/($b[2]*x-$b[2]*$b[1])")->reduce; + parser::Assignment->Allow; + Context()->variables->are(x=>'Real',y=>'Real'); + Context()->flags->set(reduceConstants=>0); + $asymptotestring="x=$b[1]"; + $asymptotes=Compute("$asymptotestring"); + + +

    + f(x)= +

    + + Submit your answer as a list, using commas. + Each asymptote is line of the form x=... or y=.... + If there are no such asymptotes, + you may type NONE. + +

    + +

    +     + +

    + There are horizontal/vertical asymptotes at . +

    +     +     +     + + + + + @b = random_subset(3,-9..-1,1..9); + @c=(random(1,5,1),non_zero_random(-9,9,1)); + if($envir{problemSeed}==1){@c=(1,3);@b=(-3,1,9)}; + Context()->noreduce('(-x)+y','(-x)-y'); + $f=Formula("($c[0] x^2-($c[1]+$b[0]*$c[0])*x+$b[0]*$c[1])/($b[2]*x-$b[2]*$b[0])")->reduce; + parser::Assignment->Allow; + Context()->variables->are(x=>'Real',y=>'Real'); + Context()->flags->set(reduceConstants=>0); + $asymptotes=Compute("NONE"); + + +

    + f(x)= +

    + + Submit your answer as a list, using commas. + Each asymptote is line of the form x=... or y=.... + If there are no such asymptotes, + you may type NONE. + +

    + +

    +     + +

    + There are horizontal/vertical asymptotes at . +

    +     +     +     + + + + + @b = random_subset(2,-9..-1,1..9); + $b[2]=random(-2,2,1); + $b[3]=random(1,5,1); + @c=(random(1,5,1),non_zero_random(-5,-1,1)); + if($envir{problemSeed}==1){@c=(1,-1);@b=(1,-1,0,1)}; + Context()->noreduce('(-x)+y','(-x)-y'); + $f=Formula("($c[0] x^2-($b[0]+$b[1])*$c[0]*x+$b[0]*$b[1]*$c[0])/($c[1] x^2-2*$b[2]*$c[1]*x+(($b[2])^2+$b[3])*$c[1])")->reduce; + parser::Assignment->Allow; + Context()->variables->are(x=>'Real',y=>'Real'); + Context()->flags->set(reduceConstants=>0); + $asymptotes=Compute("y=$c[0]/$c[1]"); + + +

    + f(x)= +

    + + Submit your answer as a list, using commas. + Each asymptote is line of the form x=... or y=.... + If there are no such asymptotes, + you may type NONE. + +

    + +

    +     + +

    + There are horizontal/vertical asymptotes at . +

    +     +     +     +     + + + +

    + Evaluate the given limit. +

    +     + + + + + $a=list_random(Compute("inf"),Compute("-inf")); + @b = random_subset(3,-9..-1,1..9); + @c = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$a=Compute("inf");@b=(2,0,1);@c=(1,-5)}; + $f=Formula("(x^3+$b[0]*x^2+$b[1]*x+$b[2])/($c[0]*x+$c[1])")->reduce; + $L=($c[0]>0)?Compute("inf"):Compute("-inf"); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + \lim\limits_{x\to} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + $a=list_random(Compute("inf"),Compute("-inf")); + @b = random_subset(3,-9..-1,1..9); + @c = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$a=Compute("inf");@b=(2,0,1);@c=(-1,5)}; + $f=Formula("(x^3+$b[0]*x^2+$b[1]*x+$b[2])/($c[0]*x+$c[1])")->reduce; + $L=($c[0]>0)?Compute("inf"):Compute("-inf"); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + \lim\limits_{x\to} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + $a=list_random(Compute("inf"),Compute("-inf")); + @b = random_subset(3,-9..-1,1..9); + @c = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$a=Compute("-inf");@b=(2,0,1);@c=(1,-5)}; + $f=Formula("(x^3+$b[0]*x^2+$b[1]*x+$b[2])/($c[0]*x^2+$c[1])")->reduce; + $L=($c[0]>0 and $a==Compute("inf") or $c[0]<0 and $a==Compute("-inf"))?Compute("inf"):Compute("-inf"); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + \lim\limits_{x\to} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     + + + + + $a=list_random(Compute("inf"),Compute("-inf")); + @b = random_subset(3,-9..-1,1..9); + @c = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$a=Compute("-inf");@b=(2,0,1);@c=(-1,5)}; + $f=Formula("(x^3+$b[0]*x^2+$b[1]*x+$b[2])/($c[0]*x^2+$c[1])")->reduce; + $L=($c[0]>0 and $a==Compute("inf") or $c[0]<0 and $a==Compute("-inf"))?Compute("inf"):Compute("-inf"); + Context()->strings->add('does not exist'=>{alias=>'DNE'}); + + +

    + \lim\limits_{x\to} +

    + + If you need to enter \infty, you may type infinity, or just INF. + If the limit does not exist, you may type does not exist, or just DNE. + +

    + +

    +     +     +     +     +     +     +
    + + + Chapter Summary +

    + In this chapter we: +

      +
    • +

      + defined the limit, +

      +
      • +
    • +

      + found accessible ways to approximate their values numerically and graphically, +

      +
      • +
    • +

      + developed a not-so-easy method of proving the value of a limit (\varepsilon-\delta proofs), +

      +
      • +
    • +

      + explored when limits do not exist, +

      +
      • +
    • +

      + defined continuity and explored properties of continuous functions, and +

      +
      • +
    • +

      + considered limits that involved infinity. +

      +
      • +
    +

    + +

    + Why? + Mathematics is famous for building on itself and calculus proves to be no exception. + In the next chapter we will be interested in + dividing by 0. That is, + we will want to divide a quantity by a smaller and smaller number and see what value the quotient approaches. + In other words, we will want to find a limit. + These limits will enable us to, among other things, + determine exactly how fast something is moving when we are only given position information. +

    + +

    + Later, we will want to add up an infinite list of numbers. + We will do so by first adding up a finite list of numbers, + then take a limit as the number of things we are adding approaches infinity. + Surprisingly, this sum often is finite; + that is, we can add up an infinite list of numbers and get, + for instance, + 42. +

    + +

    + These are just two quick examples of why we are interested in limits. + Many students dislike this topic when they are first introduced to it, + but over time an appreciation is often formed based on the scope of its applicability. +

    +     + + +     + + + Derivatives + +

    + introduced the most fundamental of calculus topics: + the limit. + This chapter introduces the second most fundamental of calculus topics: + the derivative. + Limits describe where a function is going; + derivatives describe how fast the function is going. +

    +     + +
    + Instantaneous Rates of Change: The Derivative + + Introduction + + + +

    + A common amusement park ride lifts riders to a height then allows them to freefall a certain distance before safely stopping them. + Suppose such a ride drops riders from a height of 150 feet. + Students of physics may recall that the height + (in feet) + of the riders, t seconds after freefall + (and ignoring air resistance, etc.) + can be accurately modeled by f(t) = -16t^2+150. +

    + +

    + Using this formula, it is easy to verify that, without intervention, + the riders will hit the ground when f(t)=0 so at t=2.5\sqrt{1.5} \approx 3.06 seconds. + Suppose the designers of the ride decide to begin slowing the riders' fall after 2 seconds (corresponding to a height of f(2)= + 86). + How fast will the riders be traveling at that time? +

    + +

    + We have been given a position function, + but what we want to compute is a velocity at a specific point in time, + , we want an instantaneous velocity. + We do not currently know how to calculate this. +

    + +

    + However, we do know from common experience how to calculate an + average velocity. + (If we travel 60 miles in 2 hours, + we know we had an average velocity of + 30.) + We looked at this concept in + when we introduced the difference quotient. + We have + + \frac{\text{change in distance}}{\text{change in time}} = \frac{\text{“rise”}}{\text{“run”}} = \text{average velocity.} + +

    + +

    + We can approximate the instantaneous velocity at t=2 by considering the average velocity over some time period containing t=2. + If we make the time interval small, + we will get a good approximation. + (This fact is commonly used. + For instance, + high speed cameras are used to track fast moving objects. + Distances are measured over a fixed number of frames to generate an accurate approximation of the velocity.) +

    + +

    + Consider the interval from t=2 to t=3 + (just before the riders hit the ground). + On that interval, the average velocity is + + \frac{f(3)-f(2)}{3-2} = \frac{6-86}{1} =-80\,\text{ft/s} + , + where the minus sign indicates that the riders are moving down. + By narrowing the interval we consider, + we will likely get a better approximation of the instantaneous velocity. + On [2,2.5] we have + + \frac{f(2.5)-f(2)}{2.5-2} = \frac{50-86}{0.5} =-72\,\text{ft/s} + . +

    + + + +

    + We can do this for smaller and smaller intervals of time. + For instance, over a time span of one tenth of a second, + , on [2,2.1], we have + + \frac{f(2.1)-f(2)}{2.1-2} = \frac{79.44-86}{0.1} =-65.6\,\text{ft/s} + . +

    + +

    + Over a time span of one hundredth of a second, + on [2,2.01], the average velocity is + + \frac{f(2.01)-f(2)}{2.01-2} = \frac{85.3584-86}{0.01} =-64.16\,\text{ft/s} + . +

    + +

    + What we are really computing is the average velocity on the interval [2,2+h] for small values of h. + That is, we are computing + + \frac{f(2+h) - f(2)}{h} + + where h is small. +

    + +

    + We really want to use h=0, but this, of course, + returns the familiar 0/0 indeterminate form. + So we employ a limit, + as we did in . +

    + +

    + We can approximate the value of this limit numerically with small values of h as seen in . + It looks as though the velocity is approaching + -64. +

    + +
    + Approximating the instantaneous velocity with average velocities over a small time period h + + + h + Average Velocity () + + + 1 + \phantom{Average}{-80} + + + 0.5 + \phantom{Average}{-72} + + + 0.1 + \phantom{Average}{-65.6} + + + 0.01 + \phantom{Average}{-64.16} + + + 0.001 + \phantom{Average}{-64.016} + + +
    + +

    + Computing the limit directly gives + + \lim_{h\to 0} \frac{f(2+h)-f(2)}{h} \amp = \lim_{h\to 0}\frac{-16(2+h)^2+150 - (-16(2)^2+150)}{h} + \amp = \lim_{h\to 0}\frac{-16(4+4h+h^2)+150 - 86}{h} + \amp = \lim_{h\to 0}\frac{-64-64h-16h^2+64}{h} + \amp = \lim_{h\to 0}\frac{-64h-16h^2}{h} + \amp = \lim_{h\to 0}(-64 -16h) + \amp =-64 + . +

    + +

    + Graphically, + we can view the average velocities we computed numerically as the slopes of secant lines on the graph of f going through the points + (2,f(2)) and (2+h,f(2+h)). + In Figures, + the secant line corresponding to h=1 is shown in three contexts. + shows a zoomed out + version of f with its secant line. + In , + we zoom in around the points of intersection between f and the secant line. + Notice how well this secant line approximates f between those two points it is a common practice to approximate functions with straight lines. +

    + + + + + +
    + The function f(t) and its secant line corresponding to t=2 and t=3 + + + + A downward-curving graph of function f(t) with its secant line corresponding to t=2 and t=3. + + +

    + The graph starts at the origin and is drawn on the first quadrant. + The horizontal t axis is drawn between 0 to 3, + and the y axis is drawn between 0 to 150. + The graph has a decreasing slope, gently declining from points (0, 150) to (3,0). + A secant line is drawn on the curve from t=2 and t=3. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.5, + xmax=3.49, + ymin=-20, + ymax=180, + xlabel={$t$} + ] + \addplot+[infinite,domain=0:3.06] {-16*x^2+150}; + \addplot+[secantline,domain=1.5:3.2] {-80*(x-2)+86}; + \addplot[soliddot] coordinates {(2,86) (3,6)}; + \end{axis} + \end{tikzpicture} + + + +
    + +
    + The function f(t) and a secant line corresponding to t=2 and t=3, zoomed in near t=2 + + + + The previous graph of function f(t) that is zoomed in near t=2, with its secant line from t=2 to t=3. + + +

    + The graph appears to start at the point (1.8,0) and is drawn on the first quadrant. + The horizontal t axis is drawn between 1.8 to 3.4, and the y axis + is drawn between 0 to 120. + The curve appears to be coinciding with the secant line. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=1.8, + xmax=3.4, + ymin=0, + ymax=129, + xlabel={$t$}, + xdiscontinuity + ] + \addplot+[infinite,domain=0:3.06] {-16*x^2+150}; + \addplot+[secantline,domain=1.8:3.4] {-80*(x-2)+86}; + \addplot[soliddot] coordinates{(2,86) (3,6)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + + +
    + The function f(t) with the same secant line, zoomed in further + + + + The graph of function f(t) that is further zoomed in near t=2. + + +

    + The graph appears to start at the point (1.4,0) and is drawn on the first quadrant. + The horizontal t axis is drawn between 1.4 to 2.6, and the y axis + is drawn between 0 to 120. + The curve appears to be coinciding with the secant line that is drawn on the function at t=2. + The secant line is above the curve on the left of the point of intersection and is below the curve to its right. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=1.4, + xmax=2.6, + ymin=0, + ymax=130, + xlabel={$t$}, + xdiscontinuity + ] + \addplot+[infinite,domain=0:3.06] {-16*x^2+150}; + \addplot+[secantline,domain=1.5:2.6] {-80*(x-2)+86}; + \addplot[soliddot] coordinates{(2,86)}; + \end{axis} + \end{tikzpicture} + + + +
    + +
    + The function f(t) with its tangent line at t=2 + + + + The graph of function f(t) that is highly zoomed in near t=2. + + +

    + The graph appears to start at the point (1.4,0) and is drawn on the first quadrant. + The horizontal t axis is drawn between 1.4 to 2.6, and the y axis + is drawn between 0 to 120. + The curve appears to be coinciding with the tangent line at t=2. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=1.4, + xmax=2.6, + ymin=0, + ymax=140, + xlabel={$t$}, + xdiscontinuity + ] + \addplot+[infinite,domain=0:3.06] {-16*x^2+150}; + \addplot+[tangentline,domain=1.4:2.6] {-64*(x-2)+86}; + \addplot[soliddot] coordinates{(2,86)}; + \end{axis} + \end{tikzpicture} + + +
    +     +     + +

    + As h\to 0, these secant lines approach the + tangent line, + a line that goes through the point + (2,f(2)) with the special slope of -64. + In + and , + we zoom in around the point (2,86). + We see the secant line, which approximates f well, + but not as well the tangent line shown in . +

    + +

    + We have just introduced a number of important concepts that we will flesh out more within this section. + First, we formally define two of them. +

    + + + Derivative at a Point + +

    + Let f be a continuous function on an open interval I and let c be + in I. + + derivativeat a point + + The derivative of f at c, denoted \fp(c), is + + \lim_{h\to 0}\frac{f(c+h)-f(c)}{h} + , + provided the limit exists. + If the limit exists, we say that + f is differentiable at c; + if the limit does not exist, + then f is not differentiable at c. + If f is differentiable at every point in I, + then f is differentiable on I. + + differentiable + +

    +     +     + + + + + Tangent Line + +

    + Let f be continuous on an open interval I and differentiable at c, + for some c in I. + The line with equation \ell(x) = \fp(c)(x-c)+f(c) is the tangent line + to the graph of f at c; + that is, it is the line through + (c,f(c)) whose slope is the derivative of f at c. + + tangent line + derivativetangent line + +

    +     +     + +

    + Some examples will help us understand these definitions. +

    + + + Finding derivatives and tangent lines + +

    + Let f(x) = 3x^2+5x-7. + Find: +

      +
    1. +

      + \fp(1) +

      +
      2. +
    2. +

      + The equation of the tangent line to the graph of f at x=1. +

      +
      3. +
    3. +

      + \fp(3) +

      +
      4. +
    4. +

      + The equation of the tangent line to the graph f at x=3. +

      +
      5. +
    +

    +     + +

    +

      +
    1. +

      + We compute this directly using . + + \fp(1)\amp = \lim_{h\to 0} \frac{f(1+h)-f(1)}{h} + \amp = \lim_{h\to 0} \frac{3(1+h)^2+5(1+h)-7 - (3(1)^2+5(1)-7)}{h} + \amp = \lim_{h\to 0} \frac{3(1+2h+h^2)+5+5h-7 - 1}{h} + \amp = \lim_{h\to 0} \frac{3+6h+3h^2+5+5h-8}{h} + \amp = \lim_{h\to 0} \frac{3h^2+11h}{h} + \amp = \lim_{h\to 0} (3h+11) + \amp = 11 + . +

      +
      2. +
    2. +

      + The tangent line at x=1 has slope \fp(1) and goes through the point + (1,f(1)) = (1,1). + Thus the tangent line has equation, + in point-slope form, y = 11(x-1) + 1. + In slope-intercept form we have y = 11x-10. +

      +
      3. +
    3. +

      + Again, using the definition, + + \fp(3)\amp = \lim_{h\to 0} \frac{f(3+h)-f(3)}{h} + \amp = \lim_{h\to 0} \frac{3(3+h)^2+5(3+h)-7 - (3(3)^2+5(3)-7)}{h} + \amp = \lim_{h\to 0} \frac{3(9+6h+h^2)+15+3h-7 - 35}{h} + \amp = \lim_{h\to 0} \frac{27+18h+3h^2+15+3h-42}{h} + \amp = \lim_{h\to 0} \frac{3h^2+23h}{h} + \amp = \lim_{h\to 0} 3h+23 + \amp = 23 + . +

      +
      4. +
    4. +

      + The tangent line at x=3 has slope 23 and goes through the point + (3,f(3)) = (3,35). + Thus the tangent line has equation y=23(x-3)+35 = 23x-34. +

      +
      5. +
    +

    + +

    + A graph of f is given in + along with the tangent lines at x=1 and x=3. +

    + +
    + A graph of f(x) = 3x^2+5x-7 and its tangent lines at x=1 and x=3 + + + + A graph of function 3x^2+5x-7 and its tangent line at x =1 and x=3. + + +

    + The graph starts at origin (0,0) the x axis ends at 4 + while the y axis ends at 60. + The curve starts in the third quadrant and moves into the first quadrant with an + x intercept at 1, the curve reaches point (4,60) and continues further. + Two tangents are drawn on the curve at x=1 and x=3. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1, + xmax=4.1, + ymin=-11, + ymax=66 + ] + \addplot+[infinite,domain=-1:4]{3*x^2+5*x-7}; + \addplot+[tangentline,domain=0:4] {11*(x-1)+1}; + \addplot+[tangentline,domain=1.8:3.8] {23*(x-3)+35}; + \addplot[<->,soliddot] coordinates{(1,1) (3,35)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + + + + + + + + +

    + Another important line that can be created using information from the derivative is the + normal line. + It is perpendicular to the tangent line, + hence its slope is the negative-reciprocal of the tangent line's slope. +

    + + + Normal Line + +

    + Let f be continuous on an open interval I and differentiable at c, + for some c in I. + The normal line to the graph of f at c is the line with equation + + n(x) =\frac{-1}{\fp(c)}(x-c)+f(c) + , + when \fp(c)\neq 0. (When \fp(c)=0, + the normal line is the vertical line through \left(c,f(c)\right); + that is, x=c.) + + derivativenormal line + normal line + +

    +     +     + + + Finding equations of normal lines + +

    + Let f(x) = 3x^2+5x-7, as in . + Find the equations of the normal lines to the graph of f at x=1 and + x=3. +

    +     + +

    + In , + we found that \fp(1)=11. + Hence at x=1, the normal line will have slope -1/11. + An equation for the normal line is + + n(x) = \frac{-1}{11}(x-1)+1 + . +

    + +

    + The normal line is plotted with y=f(x) in . + Note how the line looks perpendicular to f. (A key word here is + looks. Mathematically, + we say that the normal line is + perpendicular to f at x=1 as the slope of the normal line is the negative-reciprocal of the slope of the tangent line. + However, normal lines may not always + look perpendicular. +

    + +
    + A graph of f(x)=3x^2+5x-7, along with its normal line at x=1 + + + + A graph of function 3x^2+5x-7 along with its normal line at x=1. + + +

    + The graph starts at the origin. The x axis ends at 4 in the diagram while the y axis ends at 3. + The function is a straight line that rises steeply from points (0.9,0) to point (1.1, 3.2). + The normal is drawn on the line at point (1,1). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.1, + ymax=3.5, + xmin=-0.1, + xmax=4.3, + axis equal + ] + \addplot+[infinite,domain=0.9:1.2]{3*x^2+5*x-7}; + \addplot[normalline,domain=0.1:2.5] {(-1/11)*(x-1)+1}; + \addplot[soliddot] coordinates{(1,1)}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + The aspect ratio of the picture of the graph plays a big role in this. + When using graphing software, + there is usually an option called Zoom Square that keeps the aspect ratio + 1:1 +

    + +

    + We also found that \fp(3) = 23, + so the normal line to the graph of f at x=3 will have slope -1/23. + An equation for the normal line is + + n(x) = \frac{-1}{23}(x-3)+35 + . +

    +     +     + +

    + Linear functions are easy to work with; + many functions that arise in the course of solving real problems are not easy to work with. + A common practice in mathematical problem solving is to approximate difficult functions + with not-so-difficult functions. + Lines are a common choice. + It turns out that at any given point on the graph of a differentiable function f, + the best linear approximation to f is its tangent line. + That is one reason we'll spend considerable time finding tangent lines to functions. +

    + +

    + One type of function that does not benefit from a tangent line approximation is a line; + it is rather simple to recognize that the tangent line to a line is the line itself. + We look at this in the following example. +

    + + + Finding the derivative of a linear function + +

    + Consider f(x) = 3x+5. + Find the equation of the tangent line to f at x=1 and x=7. +

    +     + +

    + We find the slope of the tangent line by using + Definition. + + \fp(1) \amp = \lim_{h\to 0}\frac{f(1+h)-f(1)}{h} + \amp = \lim_{h\to 0} \frac{3(1+h)+5 - (3+5)}{h} + \amp = \lim_{h\to 0} \frac{3h}{h} + \amp = \lim_{h\to 0} 3 + \amp = 3 + . +

    + +

    + We just found that \fp(1) = 3. + That is, we found the instantaneous rate of change + of f(x) = 3x+5 is 3. + This is not surprising; lines are characterized by being the + only functions with a + constant rate of change. + That rate of change is called the + slope of the line. + Since their rates of change are constant, + their instantaneous rates of change are always the same; + they are all the slope. +

    + +

    + So given a line f(x) = ax+b, + the derivative at any point x will be a; + that is, \fp(x) = a. +

    + +

    + It is now easy to see that the tangent line to the graph of f at x=1 + is just f, + with the same being true at x=7. +

    +     +     + +

    + We often desire to find the tangent line to the graph of a function without knowing + the actual derivative of the function. + While we will eventually be able to find derivatives of many common functions, + the algebra and limit calculations on some functions are complex. + Until we develop further techniques, + the best we may be able to do is approximate the tangent line. + We demonstrate this in the next example. +

    + + + Numerical approximation of the tangent line + +

    + Approximate the equation of the tangent line to the graph of f(x)=\sin(x) at x=0. +

    +     + +

    + In order to find the equation of the tangent line, + we need a slope and a point. + The point is given to us: (0,\sin(0)) = (0,0). + To compute the slope, we need the derivative. + This is where we will make an approximation. + Recall that + + \fp(0) \approx \frac{\sin(0+h)- \sin(0) }{h} + + for a small value of h. + We choose (somewhat arbitrarily) to let h=0.1. + Thus + + \fp(0) \approx \frac{\sin(0.1)-\sin(0) }{0.1} \approx 0.9983 + . +

    + +

    + Thus our approximation of the equation of the tangent line is + y = 0.9983(x-0) +0 = 0.9983x; + it is graphed in . + The graph seems to imply the approximation is rather good. +

    + + + +
    + f(x) = \sin(x) graphed with an approximation to its tangent line at x=0 + + + + A graph of function sin(x) with an approximation to its tangent line at x=0. + + +

    + The y axis of the graph is drawn from -1 to 1 + and the x axis is drawn from -\pi and \pi. + The sine graph has an x intercept at -\pi then enters the + third quadrant and forms an upward facing parabola + with its vertex at point (-\pi/2, -1). + It passes throgh the origin then enters the first quadrant forming a downward + facing parabola with vertex at (\pi/2, 1). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.1, + ymax=1.1, + xmin=-3.5, + xmax=3.5, + xtick={-3.14,-1.57,1.57,3.14}, + xticklabels={$-\pi$,$-\frac{\pi}2$,$\frac{\pi}2$,$\pi$} + ] + \addplot+[infinite,domain=-3.5:3.5,samples=101]{sin(deg(x))}; + \addplot+[tangentline,domain=-1:1] {sin(deg(1))*x}; + \addplot [soliddot] coordinates{(0,0)}; + \end{axis} + \end{tikzpicture} + + + +
    + +     +     + +

    + Recall from + that \lim_{x\to 0}\frac{\sin(x)}x =1, + meaning for values of x near 0, \sin(x) \approx x. + Since the slope of the line y=x is 1 at x=0, + it should seem reasonable that the slope of f(x)=\sin(x) + is near 1 at x=0. + In fact, since we approximated + the value of the slope to be 0.9983, + we might guess the actual value is 1. + We'll come back to this later. +

    + +

    + Consider again . + To find the derivative of f at x=1, + we needed to evaluate a limit. + To find the derivative of f at x=3, + we needed to again evaluate a limit. + We have this process: + + + \begin{array}{c}\text{input specific}\\\text{number }c\end{array}\longrightarrow + \begin{array}{|c|}\hline\text{do something}\\\text{to }f\text{ and }c\\\hline\end{array} + \longrightarrow\begin{array}{c}\text{return}\\\text{number }f'(c)\end{array} + +

    + +

    + This process describes a function; + given one input + (the value of c), + we return exactly one output (the value of \fp(c)). + The do something box is where the tedious work + (taking limits) + of this function occurs. +

    + +

    + Instead of applying this function repeatedly for different values of c, + let us apply it just once to the variable x. + We then take a limit just once. + The process now looks like: + + + \begin{array}{c}\text{input}\\\text{variable }x\end{array}\longrightarrow + \begin{array}{|c|}\hline\text{do something}\\\text{to }f\text{ and }x\\\hline\end{array} + \longrightarrow\begin{array}{c}\text{return}\\\text{function }f'(x)\end{array} + +

    + +

    + The output is the derivative function, \fp(x). + The \fp(x) function will take a number c as input and return the + derivative of f at c. + This calls for a definition. +

    + + + Derivative Function + +

    + Let f be a differentiable function on an open interval I. + The function + + \fp(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} + + is the derivative of f. + + derivativeas a function + derivativenotation + +

    + +

    + Let y = f(x). + The following notations all represent the derivative of f: + + \fp(x)\ =\ \yp\ =\ \frac{dy}{dx}\ =\ \frac{df}{dx}\ =\ \frac{d}{dx}(f)\ =\ + \frac{d}{dx}(y) + . +

    +     +     + + + + + +

    + Important: The notation \frac{dy}{dx} is one symbol; + it is not the fraction dy/dx. + The notation, + while somewhat confusing at first, was chosen with care. + A fraction-looking symbol was chosen because the derivative has many fraction-like properties. + Among other places, + we see these properties at work when we talk about the units of the derivative, + when we discuss the Chain Rule, + and when we learn about integration + (topics that appear in later sections and chapters). +

    + + + +

    + Examples will help us understand this definition. +

    + + + + + Finding the derivative of a function + +

    + Let f(x) = 3x^2+5x-7 as in . + Find \fp(x). +

    +     + +

    + We apply . + + \fp(x) \amp = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} + \amp = \lim_{h\to 0} \frac{3(x+h)^2+5(x+h)-7-(3x^2+5x-7)}{h} + \amp = \lim_{h\to 0} \frac{3h^2 +6xh+5h}{h} + \amp = \lim_{h\to 0} (3h+6x+5) + \amp = 6x+5 + +

    + +

    + So \fp(x) = 6x+5. + Recall earlier we found that + \fp(1) = 11 and \fp(3) = 23. + Note our new computation of \fp(x) affirms these facts. +

    +     +     + + + Finding the derivative of a function + +

    + Let f(x) = \frac{1}{x+1}. + Find \fp(x). +

    +     + +

    + We apply . + + \fp(x) \amp = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} + \amp = \lim_{h\to 0} \frac{\frac{1}{x+h+1}-\frac{1}{x+1}}{h} + Now find common denominator then subtract; pull 1/h out front to facilitate reading. + \amp = \lim_{h\to 0} \frac{1}{h}\cdot\left(\frac{x+1}{(x+1)(x+h+1)} - + \frac{x+h+1}{(x+1)(x+h+1)}\right) + Now simplify algebraically. + \amp = \lim_{h\to 0} \frac 1h\cdot\left(\frac{x+1-(x+h+1)}{(x+1)(x+h+1)}\right) + \amp = \lim_{h\to 0} \frac1h\cdot\left(\frac{-h}{(x+1)(x+h+1)}\right) + Finally, apply the limit. + \amp = \lim_{h\to 0} \frac{-1}{(x+1)(x+h+1)} + \amp = \frac{-1}{(x+1)(x+1)} + \amp = \frac{-1}{(x+1)^2} + . +

    + +

    + So \fp(x) = \frac{-1}{(x+1)^2}. + To practice using our notation, we could also state + + \frac{d}{dx}\left(\frac{1}{x+1}\right) = \frac{-1}{(x+1)^2} + . +

    +     + +     + + + + + Finding the derivative of a function + +

    + Find the derivative of f(x) = \sin(x). +

    +     + +

    + Before applying , + note that once this is found, + we can find the actual tangent line to f(x) = \sin(x) at x=0, + whereas we settled for an approximation in . + + \fp(x) \amp = \lim_{h\to 0} \frac{\sin(x+h)-\sin(x)}{h} \amp \amp + \text{Derivative definition} + \amp = \lim_{h\to 0} \frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h} \amp \amp + \text{Angle addition identity} + \amp = \lim_{h\to 0} \frac{\sin(x)(\cos(h)-1) + \cos(x)\sin(h)}{h} \amp \amp + \text{Regrouped and factored} + \amp = \lim_{h\to 0} \left(\frac{\sin(x)(\cos(h)-1)}{h} + \frac{\cos(x)\sin(h)}{h} + \right) \amp \amp \text{Split into two fractions} + \amp = \lim_{h\to 0} \sin(x) \cdot \lim_{h\to 0}\frac{\cos(h)-1}{h} + \amp\quad{}+\lim_{h\to 0}\cos(x)\cdot \lim_{h\to 0}\frac{\sin(h)}{h} \amp \amp + \text{Product/sum limit rules} + \amp = \sin(x)\cdot 0 + \cos(x) \cdot 1 \amp \amp \text{Applied } + + \amp = \cos(x). \amp \amp + +

    + +

    + We have found that when f(x) = \sin(x), \fp(x) = \cos(x). + This should be somewhat amazing; + the result of a tedious limit process on the sine function is a nice function. + Then again, perhaps this is not entirely surprising. + The sine function is periodic it repeats itself on regular intervals. + Therefore its rate of change also repeats itself on the same regular intervals. + We should have known the derivative would be periodic; + we now know exactly which periodic function it is. +

    + +

    + Thinking back to , + we can find the slope of the tangent line to + f(x)=\sin(x) at x=0 using our derivative. + We approximated the slope as 0.9983; + we now know the slope is exactly \cos(0) =1. +

    + + + + + +     + +     + + + Finding the derivative of a piecewise defined function + +

    + Find the derivative of the absolute value function, + + f(x) = \abs{x} = \begin{cases} -x \amp x\lt 0 \\ x \amp x\geq 0\end{cases} + . +

    + +

    + See . +

    + +
    + The absolute value function f(x) = \abs{x}. Notice how the slope of + the lines (and hence the tangent lines) abruptly changes at x=0. + + + + The graph of function of absolute value of x, the function forms a v shape with it's vertex at origin. + + +

    + The y axis of the graph is drawn from -0.2 to 1 + and the x axis from -1 to 1. + The function is a straight line, in the second quadrant it appears to pass through point + (-1,1) and decreases until it meets the origin. + From the origin, it steeply increases after it enters the first quadrant + and appears to pass through point (1,1). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.2, + ymax=1.1, + xmin=-1.1, + xmax=1.1, + ] + \addplot+[infinite,domain=-1:1]{abs(x)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + We need to evaluate \lim_{h\to0}\frac{f(x+h)-f(x)}{h}. + As f is piecewise-defined, + we need to consider separately the limits when x\lt 0 and when x \gt 0. +

    + +

    + When x\lt 0: + + \frac{d}{dx}(-x) \amp = \lim_{h\to 0}\frac{-(x+h) - (-x)}{h} + \amp = \lim_{h\to 0}\frac{-h}{h} + \amp = \lim_{h\to 0}-1 + \amp = -1 + . +

    + +

    + When x \gt 0, + a similar computation shows that \frac{d}{dx}(x) = 1. +

    + +

    + We need to also find the derivative at x=0. + By the definition of the derivative at a point, we have + + \fp(0) = \lim_{h\to0}\frac{f(0+h)-f(0)}{h} + . +

    + +

    + Since x=0 is the point where our function's definition switches from + one piece to the other, we need to consider left and right-hand limits. + Consider the following, + where we compute the left and right hand limits side by side. +

    + + +

    + + \amp\lim_{h\to0^-}\frac{f(0+h)-f(0)}{h} + \amp= \lim_{h\to0^-}\frac{-h-0}{h} + \amp=\lim_{h\to0^-}-1 + \amp =-1 + +

    + +

    + + \amp \lim_{h\to0^+}\frac{f(0+h)-f(0)}{h} + \amp =\lim_{h\to0^+}\frac{h-0}{h} + \amp =\lim_{h\to0^+}1 + \amp =1 + +

    +     + +

    + The last lines of each column tell the story: + the left and right hand limits are not equal. + Therefore the limit does not exist at 0, + and f is not differentiable at 0. + So we have + + \fp(x) = \begin{cases} -1 \amp x\lt 0 \\ 1 \amp x \gt 0\end{cases} + . +

    + +

    + At x=0, \fp(x) does not exist; + there is a jump discontinuity at 0; + see . + So f(x) = \abs{x} is differentiable everywhere except at 0. +

    + +
    + A graph of the derivative of f(x) = \abs{x} + + + + The graph of derivative of function of absolute value of x. + + +

    + The y axis and the x axis are drawn from -1 to 1. + In the first quadrant the derivative of the function starts from the open point (0,1) + and continues as a function y=1 that moves towards right parallel to the x axis at + y=1 for all values of x, but value for x=0 does not exist. + In the third quadrant the derivative of the function starts from the open point (0,-1) + and continues as a function y=-1 that moves towards left parallel to the x axis at + y=-1 for all values of x, but value for x=0 does not exist. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.4, + ymax=1.4, + xmin=-1.1, + xmax=1.1 + ] + \addplot [firstcurvestyle,leftarrow,domain=-1:0]{-1}; + \addplot [firstcurvestyle,rightarrow,domain=0:1]{1}; + \addplot[hollowdot] coordinates{(0,-1) (0,1)}; + \end{axis} + \end{tikzpicture} + + + +
    +     +     + +

    + The point of non-differentiability came where the piecewise defined function + switched from one piece to the other. + Our next example shows that this does not always cause trouble. +

    + + + Finding the derivative of a piecewise defined function + +

    + Find the derivative of f(x), + where + + f(x) = \begin{cases}\sin(x) \amp x\leq \pi/2 \\ 1 \amp x \gt \pi/2\end{cases} + . + See . +

    + +
    + A graph of f(x) as defined in + + + + Graph of function sin(x) until pi/2 and then y = 1. + + +

    + The y axis is from 0 to 1.5 and the x axis from 0 + to \pi/2 the curve starts at 0 and increases until point (\pi/2,1) + and then moves horizontal to the x axis at y=1. +

    +     + + \begin{tikzpicture}[declare function = {func(\x) = (\x < 1.5708) * (sin(x*180/pi)) + (\x >= 1.5708) * (1);}] + \begin{axis}[ + ymin=-.4, + ymax=1.4, + xmin=-.1, + xmax=2.1, + xtick={1.57}, + xticklabels={$\frac{\pi}2$} + ] + \addplot+[infinite,domain=-.1:2,samples=101]{func(x)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + Using , + we know that when x\lt \pi/2, \fp(x) = \cos(x). + It is easy to verify that when x \gt \pi/2, + \fp(x) = 0; consider: + + \lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to0}\frac{1-1}{h} = \lim_{h\to0}0 =0 + . +

    + +

    + So far we have + + \fp(x) = \begin{cases}\cos(x) \amp x\lt \pi/2\\ 0 \amp x\gt\pi/2\end{cases} + . +

    + +

    + We still need to find \fp(\pi/2). + Notice at x=\pi/2 that both pieces of \fp are 0, + meaning we can state that \fp(\pi/2)=0. +

    + +

    + Being more rigorous, + we can again evaluate the difference quotient limit at x=\pi/2, + utilizing again left- and right-hand limits. + We will begin with the left-hand limit: +

    + +

    + + \amp \lim_{h\to0^-}\frac{f(\pi/2+h)-f(\pi/2)}{h} + \amp =\lim_{h\to0^-}\frac{\sin(\pi/2+h)-\sin(\pi/2)}{h} + \amp =\lim_{h\to0^-}{ \frac{\sin(\frac{\pi}{2})\cos(h)+\sin(h)\cos(\frac{\pi}{2})-\sin(\frac{\pi}{2})}{h}} + \amp =\lim_{h\to0^-}\frac{1\cdot\cos(h)+\sin(h)\cdot 0-1}{h} + \amp =\lim_{h\to0^-}\frac{\cos(h)-1}{h} \cdot \lim_{h\to0^-}\frac{\sin(h)}{h} + + \amp=1\cdot 0 + \amp =0 + . + Notice we used to finally evaluate the limit. +

    + +

    + Now we will find the right-hand limit: + + \amp \lim_{h\to0^+}\frac{f(\pi/2+h)-f(\pi/2)}{h} + \amp =\lim_{h\to0^+}\frac{1-1}{h} + \amp =\lim_{h\to0^+}\frac{0}{h} + \amp =0 + . +

    + +

    + Since both the left and right hand limits are 0 at x=\pi/2, + the limit exists and \fp(\pi/2) exists + (and is 0). + Therefore we can fully write \fp as + + \fp(x) = \begin{cases}\cos(x) \amp x\leq\pi/2\\ 0 \amp x \gt \pi/2\end{cases} + . +

    + +

    + See + for a graph of this derivative function. +

    + +
    + A graph of \fp(x) in . + + + + + Graph of derivative of function for this example. + + +

    + Graph of the derivative function, the derivative exists at \pi/2 + and follows a curve of \cos(x) for x\leq\pi/2 and 0 after x=\pi. + There is a sharp bend at \pi/2, hence it is not differentiable at that point, + since for x=\pi there exists f(x)=0 function is continuous. +

    +     + + \begin{tikzpicture}[declare function = {func(\x) = (\x < 1.5708) * (cos(x*180/pi)) + (\x >= 1.5708) * (0);}] + \begin{axis}[ + ymin=-.4, + ymax=1.4, + xmin=-.1, + xmax=2.1, + xtick={1.57}, + xticklabels={$\frac{\pi}2$}, + ] + \addplot+[infinite,domain=-.1:2,samples=101]{func(x)}; + \end{axis} + \end{tikzpicture} + + + +
    +     +     + + + + + +

    + Recall we pseudo-defined a continuous function as one in which we could sketch its graph without lifting our pencil. + We can give a pseudo-definition for differentiability as well: + it is a continuous function that does not have any sharp corners + or a vertical tangent line. + One such sharp corner is shown in . + Even though the function f in is piecewise-defined, + the transition is smooth + hence it is differentiable. + Note how in the graph of f in + it is difficult to tell when f switches from one piece to the other; + there is no corner. +

    +     + + + Differentiability on Closed Intervals +

    + When we defined the derivative at a point in , + we specified that the interval I over which a function f was defined needed to be an open interval. + Open intervals are required so that we can take a limit at any point c in I, + meaning we want to approach c from both the left and right. +

    + +

    + differentiableon a closed interval +

    + +

    + Recall we also required open intervals in + when we defined what it meant for a function to be continuous. + Later, we used one-sided limits to extend continuity to closed intervals. + We now extend differentiability to closed intervals by again considering one-sided limits. +

    + +

    + Our motivation is three-fold. + First, we consider common sense. + In + we found that when f(x) = 3x^2+5x-7, \fp(x) = 6x+5, + and this derivative is defined for all real numbers, + hence f is differentiable everywhere. + It seems appropriate to also conclude that f is differentiable on closed intervals, + like [0,1], as well. + After all, \fp(x) is defined at both x=0 and x=1. +

    + +

    + Secondly, consider f(x) = \sqrt{x}. + The domain of f is [0,\infty). + Is f differentiable on its domain specifically, + is f differentiable at 0? (We'll consider this in the next example.) +

    + +

    + Finally, in later sections, + having the derivative defined on closed intervals will prove useful. + One such place is + where the derivative plays a role in measuring the length of a curve. +

    + +

    + After a formal definition of differentiability on a closed interval, + we explore the concept in an example. +

    + + + Differentiability on a Closed Interval + +

    + Let f be continuous on [a,b] and differentiable on (a,b). + If the one-sided limits + + \lim_{h\to0^+}\frac{f(a+h)-f(a)}{h} \amp \amp \amp \lim_{h\to0^-}\frac{f(b+h)-f(b)}{h} + + exist, then we say f is differentiable on [a,b]. +

    +     +     + +

    + For all the functions f in this text, + we can determine differentiability on [a,b] by considering the limits + \lim_{x\to a^+}\fp(x) and \lim_{x\to b^-}\fp(x). + This is often easier to evaluate than the limit of the difference quotient. +

    + + + Differentiability at an endpoint + +

    + Consider f(x) = \sqrt{x}=x^{1/2} and g(x) = \sqrt{x^3}=x^{3/2}. + The domain of each function is [0,\infty). + It can be shown that each is differentiable on (0,\infty); + determine the differentiability of each at x=0. +

    +     + +

    + We start by considering f and take the right-hand limit of the difference quotient: + + \lim_{h\to0^+}\frac{f(a+h)-f(a)}{h} \amp = \lim_{h\to0^+}\frac{\sqrt{0+h}-\sqrt{0}}h + \amp = \lim_{h\to0^+}\frac{\sqrt{h}}h + \amp = \lim_{h\to0^+}\frac{1}{h^{1/2}}\ =\ \infty + . +

    + +

    + The one-sided limit of the difference quotient does not exist at x=0 for f; + therefore f is differentiable on + (0,\infty) and not differentiable on [0,\infty). +

    + +

    + We state (without proof) that \fp(x) = 1/\big(2\sqrt{x}\big). + Note that \lim_{x\to 0^+}\fp(x) = \infty; + this limit was easier to evaluate than the limit of the difference quotient, + though it required us to already know the derivative of f. +

    + +

    + Now consider g: + + \lim_{h\to0^+}\frac{g(a+h)-g(a)}{h} \amp = \lim_{h\to0^+}\frac{\sqrt{(0+h)^3}-\sqrt{0}}h + \amp = \lim_{h\to0^+}\frac{h^{3/2}}h + \amp = \lim_{h\to0^+}h^{1/2}\ =\ 0 + . +

    + +

    + As the one-sided limit exists at x=0, + we conclude g is differentiable on its domain of [0,\infty). +

    + +

    + We state (without proof) that \gp(x) = 3\sqrt{x}/2. + Note that \lim_{x\to 0^+}\gp(x) = 0; + again, this limit is easier to evluate than the limit of the difference quotient. +

    + +
    + A graph of y=x^{1/2} and y=x^{3/2} in + + + Two curves, one of function of square root of x and the other of x to the power 3/2. + + +

    + The graph has two curves that form a leaf shape. + The y axis is drawn from 0 to 1 and the x axis is + drawn from -0.2 to 1.2. + The curve on top is of function x^(1/2) and its a downward facing curve. + The one below is of function x^(3/2) and it is an upward facing curve. + The two graphs intersect at points (0,0) and (1,1). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.2, + ymax=1.2, + xmin=-.1, + xmax=1.1, + axis equal + ] + \addplot+[domain=0:1]({\x^2},{\x^3}); + \addplot+[solid,domain=0:1]({\x^2},{\x}); + \draw (axis cs:.25,.7) node {$y=x^{1/2}$}; + \draw (axis cs:.7,.35) node {$y=x^{3/2}$}; + \end{axis} + \end{tikzpicture} + + +
    + +

    + We state (without proof) that \gp(x) = 3\sqrt{x}/2. + Note that \lim_{x\to0^+}\gp(x) = 0; + again, this limit is easier to evaluate than the limit of the difference quotient. +

    + +

    + The two functions are graphed in . + Note how f(x) = \sqrt{x} seems to go vertical + as x approaches 0, implying the slopes of its tangent lines are growing toward infinity. + Also note how the slopes of the tangent lines to + g(x)= \sqrt{x^3} approach 0 as x approaches 0. +

    +     +     + +

    + Most calculus textbooks omit this topic and simply avoid specific cases where it could be applied. + We choose in this text to not make use of the topic unless it is needed. + Many theorems in later sections require a function f to be differentiable on an + open interval I; + we could remove the word open + and just use \ldots on an interval I, + but choose to not do so in keeping with the current mathematical tradition. + Our first use of differentiability on closed intervals comes in , + where we measure the lengths of curves. +

    + +

    + This section defined the derivative; + in some sense, it answers the question of What + is the derivative? + The next section addresses the question + What does the derivative mean? +

    +     + + + + Terms and Concepts + + + + +

    + + Let f be a position function. + The average rate of change on [a,b] is the slope of the line through the points (a, f(a)) and (b,f(b)). +

    +     + +

    + Average velocity is the change in position, divided by the change in time. +

    +     + +     + + + + +

    + + The definition of the derivative of a function at a point involves taking a limit. +

    +     + +

    + A limit is necessary, since the difference quotient is undefined when h=0. +

    +     + +     + + + + +

    + In your own words, + explain the difference between the average rate of change and instantaneous rate of change. +

    + +     + + + +     + + + + +

    + In your own words, + explain the difference between Definitions + and . +

    + +     + + + +     + + + + +

    + Let y=f(x). + Give three different notations equivalent to f'(x). +

    + +     + + + +     + + + + +

    + If two lines are perpendicular, + what is true of their slopes? +

    + +     + + + +

    + The two lines have negative-reciprocal slopes. + That is, if m_1 and m_2 are the slopes, + then m_2 = -\frac{1}{m_1}. + Equivalently, the slopes of perpendicular lines satisfy m_1m_2=-1. +

    +     + +     +     + + + Problems + + + + +

    + Use the definition of the derivative to compute the derivative of the given function. +

    +     + + + + + $d = Formula('0'); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

    + f(x) = 6 +

    +

    + +

    +

    + +

    +     +     +     + + + + + $d = Formula('2'); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

    + f(x) = 2x +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->are(t=>'Real'); + $d = Formula('-3'); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

    + f(t) = 4-3t +

    +

    + +

    +

    + +

    +     +     +     + + + + + $d = Formula('2x'); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

    + g(x) = x^2 +

    +

    + +

    +

    + +

    +     +     +     + + + + + $d = Formula('3x^2'); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

    + h(x) = x^3 +

    +

    + +

    +

    + +

    +     +     +     + + + + + $d = Formula('6x-1'); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

    + f(x) = 3x^2-x+4 +

    +

    + +

    +

    + +

    +     +     +     + + + + + $d = Formula('-1/x^2'); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

    + r(x) = \frac{1}{x} +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->are(s=>'Real'); + $d = Formula('-1/(s-2)^2'); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

    + r(s) = \frac{1}{s-2} +

    +

    + +

    +

    + +

    +     +     +     +     + + + +

    + A function and an x-value are given. + (Note: these functions are the same as those given in Exercises.) + Give the equations of the tangent line and the normal line at that x-value. +

    +     + + + + + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $tangentline=Formula("y=6"); + $normalline=Formula("x=-2"); + + +

    + f(x)=6 at x=-2 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $tangentline=Compute("y=2x"); + $normalline=Compute("y=-(1/2)(x-3)+6"); + + +

    + f(x)=2x at x=3 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $tangentline=Compute("y=-3x+4"); + $normalline=Compute("y=(1/3)(x-7)-17"); + + +

    + f(x)=4-3x at x=7 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $tangentline=Compute("y=4(x-2)+4"); + $normalline=Compute("y=-(1/4)(x-2)+4"); + + +

    + g(x)=x^2 at x=2 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $tangentline=Compute("y=48(x-4)+64"); + $normalline=Compute("y=-1/48(x-4)+64"); + + +

    + h(x)=x^3 at x=4 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $tangentline=Compute("y=-7(x+1)+8"); + $normalline=Compute("y=(1/7)(x+1)+8"); + + +

    + f(x)=3x^2-x+4 at x=-1 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $tangentline=Compute("y=-(1/4)(x+2)-1/2"); + $normalline=Compute("y=4(x+2)-1/2"); + + +

    + r(x)=\frac{1}{x} at x=-2 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $tangentline=Compute("y=-(x-3)+1"); + $normalline=Compute("y=(x-3)+1"); + + +

    + r(x)=\frac{1}{x-2} at x=3 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     +     + + + +

    + A function f and an x-value a are given. + Approximate the equation of the tangent line to the graph of f at x=a by numerically approximating f'(a), + using h=0.1. +

    +     + + + + + $b=non_zero_random(-5,5,1); + $c=non_zero_random(-9,9,1); + $a=non_zero_random(-5,5,1); + if($envir{problemSeed}==1){$a=3;$b=2;$c=1;}; + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $f=Formula("x^2+$a x + $b")->reduce; + $fa=$f->eval(x=>$a); + $h=0.1; + $m=($f->eval(x=>$a+$h) - $fa)/$h; + $tangentline=Compute("y=$m(x-$a)+$fa"); + + +

    + f(x) = and a= +

    +

    + +

    +     +     +     + + + + + $b=non_zero_random(-10,10,1); + $c=non_zero_random(-9,9,1); + $a=non_zero_random(-9,9,1); + do {$a=non_zero_random(-9,9,1)} until ($a+$c != 0); + if($envir{problemSeed}==1){$a=10;$b=1;$c=9;}; + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $f=Formula("$b/(x+$c)")->reduce; + $fa=$f->eval(x=>$a); + $h=0.1; + $m=($f->eval(x=>$a+$h) - $fa)/$h; + $tangentline=Compute("y=$m(x-$a)+$fa"); + + +

    + f(x) = and a= +

    +

    + +

    +     +     +     + + + + + $a=non_zero_random(-5,5,1); + if($envir{problemSeed}==1){$a=2;}; + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $f=Formula("e^x")->reduce; + $fa=$f->eval(x=>$a); + $h=0.1; + $m=($f->eval(x=>$a+$h) - $fa)/$h; + $tangentline=Compute("y=$m(x-$a)+$fa"); + + +

    + f(x) = and a= +

    +

    + +

    +     +     +     + + + + + Context("ImplicitPlane"); + Context()->variables->are(x=>'Real',y=>'Real'); + $tangentline=Compute("y=-0.04996x+1"); + + +

    + f(x) = \cos(x) and a=0 +

    +

    + +

    +     +     +     +     + + + + + @approx=( + Compute("-2")->with(tolType=>'absolute',tolerance=>0.4), + Compute("0")->with(tolType=>'absolute',tolerance=>0.1), + Compute("4")->with(tolType=>'absolute',tolerance=>2) + ); + $approx_list=List($approx[0],$approx[1],$approx[2]); + $derivative=Formula("2x"); + $slopes=List(-2,0,4); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

    + The graph of f(x)=x^2-1 is shown. +

    + + + Graph of function x squared minus 1, curve is U-shaped. + + +

    + The x axis is between -2 and 2 and the y axis + is drawn between -1 and 3. + The graph is U-shaped reaching a minimum at (0,-1) and moving + further upwards after crossing point (-1,0) to the left and (1,0) to the right. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ytick={-1,1,2,3}, + ymin=-1.3,ymax=3.5,% + xmin=-2.1,xmax=2.1,% + grid=major + ] + + \addplot [firstcurvestyle,infinite,domain=-2.1:2.1] {x^2-1}; + + \end{axis} + \end{tikzpicture} + + + +     + + +

    + Use the graph to approximate the slope of the tangent line to f at (-1,0), (0,-1), and (2,3). +

    + + + Enter your answers as a comma-separated list of values. + + +

    + +

    +     +     + + + +

    + Using the definition of the derivative, find \fp(x). +

    +

    + +

    +

    + +

    +     +     + + +

    + Use the derivative to find the slope of the tangent line at the points (-1,0), (0,-1) and (2,3). +

    + + Enter your answers as a comma-separated list of values. + +

    + +

    +     +     +     +     + + + + + @approx=( + Compute("-1")->with(tolType=>'absolute',tolerance=>0.2), + Compute("-1/4")->with(tolType=>'absolute',tolerance=>0.1) + ); + $approx_list=List($approx[0],$approx[1]); + $derivative=Formula("-1/(x+1)^2"); + $slopes=List(-1,'-1/4'); + $showwork = '[@ explanation_box(message => "Show your work.") @]*'; + + +

    + The graph of f(x)=\frac{1}{x+1} is shown. +

    + + + Graph of function f(x) = 1 divided by (x+1). + + +

    + Graph of function f(x) = \frac{1}{x+1}. + The x axis is drawn from -1 + to 3 and the y axis from 0 to 5. + The curve begins in the second quadrant and steeply declines + from infinty to point (0,1) then it enters the first quadrant declining gently + and seems to come very close to the x axis on the right. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ytick={1,2,3,4,5}, + ymin=-.5,ymax=5.5,% + xmin=-1.1,xmax=3.1,% + grid=major + ] + + \addplot [firstcurvestyle,infinite,domain=-.9:3.1,samples=50] {1/(x+1)}; + + \end{axis} + \end{tikzpicture} + + +     + + + +

    + Use the graph to approximate the slope of the tangent line to f at (0,1) and (1,0.5). +

    + + Enter your answers as a comma-separated list of values. + +

    + +

    +     +     + + + +

    + Using the definition of the derivative, find \fp(x). +

    +

    + +

    +

    + +

    +     +     + + + +

    + Use the derivative to find the slope of the tangent line at the points (0,1) and (1,0.5). +

    + + Enter your answers as a comma-separated list of values. + +

    + +

    +     +     +     +     + + + + +

    + A graph of a function f(x) is given. + Using the graph, sketch \fp(x). +

    +     + + + + + + Graph of function f(x) that is a straight line. + + +

    + The y axis if drawn from -1 to 3 + and the x axis from -2 to 5. + The graph is a decreasing straight line primarily in + the first quadrant that passes through 2 on the y axis and 4 on the x axis. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-2.1, + xmax=5.1, + ymin=-1.5, + ymax=3.5 + ] + \addplot+[infinite,domain=-2.1:5.1] {-.5*x+2}; + \end{axis} + \end{tikzpicture} + + +     +     + + + + + + Graph of function is U shaped primarily in the second and third quadrant. + + +

    + The y axis is from -3 to 3 and the x axis -6 and 2. + The U-shaped curve has a minimun at point (-2,-3) + it intersects the x axis at -4.5 on the left and 0.5 on the right. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-7.1, + xmax=3.1, + ymin=-3.5, + ymax=3.5 + ] + \addplot+[infinite,domain=-5.6:1.6,samples=40] {.5*(x+2)^2-3}; + \end{axis} + \end{tikzpicture} + + +     +     + + + + + + Graph of curve that crosses the origin with a mixima at x=-1 and a minima at x=1. + + +

    + The x axis is drawn from -3 to 3 and the y axis from -5 to 5. + The graph has a maxima near point (-1,3) and a minima near point (1,-3). + It crosses the origin then passes the maxima then crosses x axis at -2 + and decreases in the third quadrant and continues to infinity on the left. + On the right side it decreases from the origin passes the minima and goes to infinity + after crossing the x axis at 2 in the first quadrant. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-3.1, + xmax=3.1, + ymin=-5.5, + ymax=5.5 + ] + \addplot+[infinite,domain=-2.45:2.45,samples=40] {(x-2)*x*(x+2)}; + \end{axis} + \end{tikzpicture} + + +     +     + + + + + + Graph of function that has maxima at even multiple of pi and minima at odd multiple of pi. + + +

    + The y axis is drawn from -1 to 1 and the x axis is drawn from + -2* \pi to 2* \pi. + At x=0 the function has a maximum of 1 on the left of the y axis the + function decreases from 1 and reaches a minima of -1 at x=-\pi after crossing -\p/2. + Then it increases to cross the x axis at x=-3*\pi/2 and increases further + to reach a maximum of 1 in the second quadrant. +

    +

    + Similarly, on the right of the y axis the function decreases from 1 and reaches + a minima of -1 at x=\pi after crossing \p/2. + Then it increases to cross the x axis at x=3*\pi/2 and increases further to reach + a maximum of 1 in the first quadrant. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-6.9, + xmax=6.9, + ymin=-1.1, + ymax=1.1, + xtick={-6.28,-3.14,3.14,6.28}, + xticklabels={$-2\pi$, $-\pi$, $\pi$, $2\pi$} + ] + \addplot+[infinite,domain=-6.8:6.8,samples=101] {cos(deg(x))}; + \end{axis} + \end{tikzpicture} + + +     +     +     + + + +

    + Use the graph of the function to answer the following questions. +

      +
    1. +

      + Where is g(x)\gt 0? +

      +
      2. +
    2. +

      + Where is g(x)\lt 0? +

      +
      3. +
    3. +

      + Where is g(x) = 0? +

      +
      4. +
    4. +

      + Where is \gp(x) \lt 0? +

      +
      5. +
    5. +

      + Where is \gp(x) \gt 0? +

      +
      6. +
    6. +

      + Where is \gp(x) = 0? +

      +
      7. +
    +

    +     + + + + Context("Interval"); + $pos=Compute("(-2,0)U(2,inf)"); + $neg=Compute("(-inf,-2)U(0,2)"); + $zer=Compute("{-2,0,2}"); + $dec=Compute("(-1,1)"); + $inc=Compute("(-inf,-1)U(1,inf)"); + $fla=Compute("{-1,1}"); + + + + + Graph of curve that crosses the origin with a mixima at x=-1 nd a minima at x=1. + + +

    + The x axis is drawn from -3 to 3 and the y axis from -5 to 5. + The graph has a maxima near point (-1,3) and a minima near point (1,-3). + It crosses the origin passes the maxima, crosses x axis at -2 + and decreases in the third quadrant and continues to infinity on the left. + On the right side it decreases from the origin passes the minima and increases + to infinity after crossing the x axis at 2 in the first quadrant. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xtick={-2,-1,1,2}, + ymin=-5.6,ymax=5.6,% + xmin=-3,xmax=3,% + ] + + \addplot [firstcurvestyle,infinite,domain=-2.5:2.5,samples=40] {(x-2)*x*(x+2)}; + + \end{axis} + \end{tikzpicture} + + + + Answer using interval notation or set notation, as appropriate. If you need to write \infty, you may type inf or infinity. + If you need the union symbol, \cup, you may type the capital letter U. + + + Enter the set on which g(x) \gt 0 using interval notation. + +

    + +

    + + Enter the set on which g(x) \lt 0 using interval notation. + +

    + +

    + + Enter the set of points where g(x) = 0 using the syntax {a,b}. + +

    + +

    + + Enter the set on which g^\prime(x) \lt 0 using interval notation. + +

    + +

    + + Enter the set on which g^\prime(x) \gt 0 using interval notation. + +

    + +

    + + Enter the set of points where g^\prime(x) = 0 using the syntax {a,b}. + +

    + +

    +     +     +     + + + + + Context("Interval"); + $pos=Compute("(-2,2)"); + $neg=Compute("(-inf,-2)U(2,inf)"); + $zer=Compute("{-2,2}"); + $dec=Compute("(-1,0)U(1,inf)"); + $inc=Compute("(-inf,-1)U(0,1)"); + $fla=Compute("{-1,0,1}"); + + + + + An M-shaped graph with two maxima. + + +

    + The y axis is drawn from -2 to 5 and the x axis + is drawn from -2 and 2. + At x=0 y has a value of 4, to the left the function increases to reach a maximum + of 4.5 in the second quadrant + and decreses and continues into the third quadrant after crossing x axis at -2. + Similarly, on the right side, the function increases to reach a maximum of 4.5 in the + first quadrant and decreses and continues into the fourth quadrant after crossing x axis at 2. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xtick={-2,-1,1,2}, + ymin=-3,ymax=6,% + xmin=-2.8,xmax=2.8,% + ] + + \addplot [firstcurvestyle,infinite,domain=-2.2:2.2,samples=50] {(-2)*(x^4/4-x^2/2)+4}; + + \end{axis} + \end{tikzpicture} + + + + Answer using interval notation or set notation, as appropriate. If you need to write \infty, you may type inf or infinity. + If you need the union symbol, \cup, you may type the capital letter U. + + + Answer using interval notation or set notation, as appropriate. If you need to write \infty, you may type inf or infinity. + If you need the union symbol, \cup, you may type the capital letter U. + + + Enter the set on which g(x) \gt 0 using interval notation. + +

    + +

    + + Enter the set on which g(x) \lt 0 using interval notation. + +

    + +

    + + Enter the set of points where g(x) = 0 using the syntax {a,b}. + +

    + +

    + + Enter the set on which g^\prime(x) \lt 0 using interval notation. + +

    + +

    + + Enter the set on which g^\prime(x) \gt 0 using interval notation. + +

    + +

    + + Enter the set of points where g^\prime(x) = 0 using the syntax {a,b}. + +

    + +

    +     +     +     +     + + + +

    + A function f(x) is given, + along with its domain and derivative. + Determine if f(x) is differentiable on its domain. +

    +     + + + + +

    + f(x) = \sqrt{x^5(1-x)}, + domain is [0,1], + \fp(x) = \frac{(5-6x)x^{3/2}}{2\sqrt{1-x}} +

    + +     + +

    + \lim_{h\to0^+}\frac{f(0+h)-f(0)}h =0; + note also that \lim_{x\to0^+}\fp(x) = 0. + So f is differentiable at x=0. +

    +

    + \lim_{h\to0^-}\frac{f(1+h)-f(1)}h =-\infty; + note also that \lim_{x\to1^-}\fp(x) = -\infty. + So f is not differentiable at x=1. +

    +

    + f is differentiable on [0,1), not its entire domain. +

    +     + +     + + + + +

    + f(x) = \cos\left(\sqrt{x}\right), + domain is [0,\infty), + \fp(x) = -\frac{\sin\left(\sqrt{x}\right)}{2\sqrt{x}} +

    + +     + +

    + The limit of the difference quotient is difficult to evaluate. + Using , + we can determine \lim_{x\to0^+}\fp(x) = -1/2. +

    +

    + Since \fp is defined on (0,\infty), + we conclude f is differentiable on [0,\infty). +

    +     + +     +     +     +     +
    +
    + Interpretations of the Derivative + +

    + + defined the derivative of a function and gave examples of how to compute it using its definition (, using limits). + The section also started with a brief motivation for this definition, that is, + finding the instantaneous velocity of a falling object given its position function. + + will give us more accessible tools for computing the derivative; + tools that are easier to use than repeated use of limits. +

    + +

    + This section falls in between the + What is the definition of the derivative? + and How do I compute the derivative? sections. + Here we are concerned with What does the derivative mean?, + or perhaps, + when read with the right emphasis, + What is the derivative? + We offer two interconnected interpretations of the derivative, + hopefully explaining why we care about it and why it is worthy of study. + derivativeinterpretation +

    +     + + + Interpretation of the Derivative as Instantaneous Rate of Change +

    + + started with an example of using the position of an object + (in this case, a falling amusement park rider) + to find the object's velocity. + This type of example is often used when introducing the derivative because we tend to readily recognize that velocity is the + instantaneous rate of change in position. In general, + if f is a function of x, + then \fp(x) measures the instantaneous rate of change of f with respect to x. + Put another way, the derivative answers + When x changes, + at what rate does f change? + Thinking back to the amusement park ride, + we asked When time changed, + at what rate did the height change? + and found the answer to be By -64 feet per second. +

    + +

    + Now imagine driving a car and looking at the speedometer, + which reads + 60 + . Five minutes later, + you wonder how far you have traveled. + Certainly, lots of things could have happened in those 5 minutes; + you could have intentionally sped up significantly, + you might have come to a complete stop, you might have slowed to + + 20 + + as you passed through construction. + But suppose that you know, as the driver, + none of these things happened. + You know you maintained a fairly consistent speed over those 5 minutes. + What is a good approximation of the distance traveled? +

    + +

    + One could argue the only good approximation, + given the information provided, + would be based on \text{distance} = \text{rate}\times\text{time.} In this case, + we assume a constant rate of + + 60 + + with a time of 5 minutes or 5/60 of an hour. + Hence we would approximate the distance traveled as 5 miles. +

    + +

    + Referring back to the falling amusement park ride, + knowing that at t=2 the velocity was -64 ft/s, we could reasonably approximate that 1 second later the riders' height would have dropped by about 64 feet. + Knowing that the riders were accelerating + as they fell would inform us that this is an + under-approximation. + If all we knew was that f(2) = 86 and \fp(2) = -64, + we'd know that we'd have to stop the riders quickly otherwise they would hit the ground. +

    + +

    + In both of these cases, + we are using the instantaneous rate of change to predict future values of the output. +

    + + + +     + + + Units of the Derivative +

    + It is useful to recognize the units + of the derivative function. + If y is a function of x, + , y=f(x) for some function f, + and y is measured in feet and x in seconds, + then the units of \yp = \fp are + feet per second, + commonly written as ft/s. In general, + if y is measured in units P and x is measured in units Q, + then \yp will be measured in units + P per Q, + or P/Q. Here we see the fraction-like behavior of the derivative in the notation: + the units of \frac{dy}{dx}are \frac{\text{units of }y}{\text{units of }x}. +

    + + + The meaning of the derivative: World Population + +

    + Let P(t) represent the world population t minutes after 12:00 a.m., January 1, 2012. + It is fairly accurate to say that P(0) = 7{,}028{,}734{,}178 + (). + It is also fairly accurate to state that P'(0) = 156; + that is, at midnight on January 1, 2012, + the population of the world was growing by about 156 + people per minute + (note the units). + Twenty days later + (or 28{,}800 minutes later) + we could reasonably assume the population grew by about 28{,}800\cdot156 = 4{,}492{,}800 people. +

    +     +     + + + The meaning of the derivative: Manufacturing + +

    + The term widget is an economic term for a generic unit of manufacturing output. + Suppose a company produces widgets and knows that the market supports a price of \$10 per widget. + Let P(n) give the profit, in dollars, + earned by manufacturing and selling n widgets. + The company likely cannot make a (positive) profit making just one widget; + the start-up costs will likely exceed \$10. + Mathematically, we would write this as P(1) \lt 0. +

    + +

    + What do P(1000) = 500 and P'(1000)=0.25 mean? + Approximate P(1100). +

    +     + +

    + The equation P(1000)=500 means that selling 1000 widgets returns a profit of \$500. + We interpret P'(1000) = 0.25 as meaning that when we are selling 1000 widgets, + the profit is increasing at rate of \$0.25 per widget + (the units are dollars per widget.) + Since we have no other information to use, + our best approximation for P(1100) is: + + P(1100) \amp \approx P(1000) + P'(1000)\times100 + \amp= \$500 + (100\text{ widgets })\cdot \$0.25/\text{widget} + \amp= \$525 + . +

    + +

    + We approximate that selling 1100 widgets returns a profit of \$525. +

    +     +     + + + + + +

    + The previous examples made use of an important approximation tool that we first used in our previous driving a car at + + 60 + example at the beginning of this section. + Five minutes after looking at the speedometer, + our best approximation for distance traveled assumed the rate of change was constant. + In Examples + and + we made similar approximations. + We were given rate of change information which we used to approximate total change. + Notationally, we would say that + + f(c+h) \approx f(c) + \fp(c)\cdot h + . +

    + +

    + This approximation is best when h is small. + Small is a relative term; + when dealing with the world population, + h=22\text{ days} = 28{,}800\text{ minutes} is small in comparison to years. + When manufacturing widgets, + 100 widgets is small when one plans to manufacture thousands. +

    +     + + + The Derivative and Motion +

    + One of the most fundamental applications of the derivative is the study of motion. + Let s(t) be a position function, + where t is time and s(t) is distance. + For instance, + s could measure the height of a projectile or the distance an object has traveled. +

    + +

    + Let s(t) measure the distance traveled, + in feet, of an object after t seconds of travel. + Then s'(t) has units feet per second, + and s'(t) measures the instantaneous rate of distance change with respect to time + it measures velocity. + velocity +

    + +

    + Now consider v(t), a velocity function. + That is, at time t, v(t) gives the velocity of an object. + The derivative of v, v'(t), + gives the instantaneous rate of velocity change with respect to time + acceleration. + (We often think of acceleration in terms of cars: + a car may go from 0 to 60 in 4.8 seconds. + This is an average acceleration, + a measurement of how quickly the velocity changed.) + If velocity is measured in feet per second, + and time is measured in seconds, + then the units of acceleration (, the units of v'(t)) are + feet per second per second, + or (ft/s)/s. + We often shorten this to feet per second squared, or + + + , but this tends to obscure the meaning of the units. + + acceleration +

    + +

    + Perhaps the most well known acceleration is that of gravity. + In this text, + we use g=32\,\text{ft}/\text{s}^2 or g=9.8\,\text{m}/\text{s}^2. + What do these numbers mean? +

    + +

    + A constant acceleration of + 32\,\frac{\text{ft}/\text{s}}{\text{s}} means that the velocity changes by 32\,\text{ft}/\text{s} each second. + For instance, + let v(t) measure the velocity of a ball thrown straight up into the air, + where v has units ft/s and t is measured in seconds. + The ball will have a positive velocity while traveling upwards and a negative velocity while falling down. + The acceleration is thus -32\,\text{ft}/\text{s}^2. + If v(1) = 20\,\text{ft}/\text{s}, + then 1 second later, + the velocity will have decreased by 32\,\text{ft}/\text{s}; + that is, v(2) = -12\,\text{ft/s}. + We can continue: v(3) = -44\,\text{ft/s}. + Working backward, + we can also figure that v(0) = 52\,\text{ft}/\text{s}. +

    + +

    + These ideas are so important we write them out as a Key Idea. +

    + + + The Derivative and Motion +

    +

      +
    1. +

      + Let s(t) be the position function of an object. + Then s'(t)=v(t) is the velocity function of the object. +

      +
      2. +
    2. +

      + Let v(t) be the velocity function of an object. + Then v'(t)=a(t) is the acceleration function of the object. +

      +
      3. +
    + derivativemotion + derivativevelocity + derivativeacceleration +

    +     +     + + + Interpretation of the Derivative as the Slope of the Tangent Line +

    + We now consider the second interpretation of the derivative given in this section. + This interpretation is not independent from the first by any means; + many of the same concepts will be stressed, + just from a slightly different perspective. +

    + +

    + Given a function y=f(x), + the difference quotient \frac{f(c+h)-f(c)}{h} gives a change in y values divided by a change in x values; + , it is a measure of the rise over run, or slope, + of the secant line that goes through two points on the graph of f: + (c, f(c)) and (c+h,f(c+h)). + As h shrinks to 0, + these two points come close together; + in the limit we find \fp(c), + the slope of a special line called the tangent line that intersects f only once near x=c. +

    + +

    + Lines have a constant rate of change, their slope. + Nonlinear functions do not have a constant rate of change, + but we can measure their instantaneous rate of change + at a given x value c by computing \fp(c). + We can get an idea of how f is behaving by looking at the slopes of its tangent lines. + We explore this idea in the following example. +

    + + + Understanding the derivative: the rate of change + +

    + Consider f(x) = x^2 as shown in . + It is clear that at x=3 the function is growing faster than at x=1, + as it is steeper at x=3. + How much faster is it growing at 3 compared to 1? +

    + +
    + A graph of f(x)=x^2 + + + A graph of f(x)=x^2 + +

    + The curve is parabolic with the focus being at the origin. + As the x values get bigger and bigger the curve gers further from the y axis + on both sides as both the ends approach infinity. Both the rate at which + it is getting further gets slower as the x values increase. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1, + xmax=4.1, + ymin=-.4, + ymax=17, + xtick={-1,0,...,4}, + minor xtick={-1,0,...,4}, + ytick={0,5,...,15}, + minor ytick={0,1,...,17}, + ] + \addplot+[infinite,domain=-.8:4] {x^2}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + We can answer this exactly + (and quickly) + after , + where we learn to quickly compute derivatives. + For now, we will answer graphically, + by considering the slopes of the respective tangent lines. +

    + +
    + A graph of f(x)=x^2 and tangent lines at x=1 and x=3 + + + a graph of f(x)=x^2 + +

    + The curve is parabolic with the focus being at the origin. As the x values get bigger and bigger the curve gets + further from the y axis on both sides as both the ends approach infinity. The rate at which it is getting further + gets slower as the x values increase. The tangent lines are drawn at x=1 and x=3. +

    + +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1, + xmax=4.1, + ymin=-.4, + ymax=17, + xtick={-1,0,...,4}, + minor xtick={-1,0,...,4}, + ytick={0,5,...,15}, + minor ytick={0,1,...,17}, + grid=major, + ] + \addplot+[infinite,domain=-.8:4] {x^2}; + \addplot[tangentline,domain=0.4:3.5]{2*(x-1)+1}; + \addplot[tangentline,domain=1.6:4]{6*(x-3)+9}; + \addplot[soliddot] coordinates{(1,1) (3,9)}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + With practice, + one can fairly effectively sketch tangent lines to a curve at a particular point. + In , + we have sketched the tangent lines to f at x=1 and x=3, + along with a grid to help us measure the slopes of these lines. + At x=1, the slope is 2; + at x=3, the slope is 6. + Thus we can say not only is f growing faster at x=3 than at x=1, + it is growing three times as fast. +

    +     +     + + + Understanding the graph of the derivative + +

    + Consider the graph of f(x) and its derivative, + \fp(x), in . + Use these graphs to find the slopes of the tangent lines to the graph of f at x=1, + x=2, and x=3. +

    + +
    + Graphs of f and \fp in + + + Graphs of a function and its derivative. + +

    + The curve of f(x) intersects y axis at y=4 and slopes downwards. After intersecting the x axis at around x=0.5. + It continues sloping downwards in the negative y direction until about y=-2 and then starts sloping upwards again. + This time it intersects the x axis at x=2 and then continue upwards until y=5 then slope downwards again. + The graph of \fp is a downward parabolic curve that intersects f(x) at (2.5, 4) in the positive y direction and + y=-2 at negative y direction. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + minor y tick num=4, + minor x tick num=0, + xtick={1,2,3}, + ymin=-8.9, + ymax=5.9, + xmin=-.1, + xmax=3.49, + clip=false + ] + \addplot+[infinite,domain=0:3.5] {-x^3+7*x^2-12*x+4} node[above left] {$f(x)$}; + \addplot+[infinite,domain=0.5:3.3] {-12+14*x-3*x^2} node[pos=0.5, above left] {$f'(x)$}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +

    + To find the appropriate slopes of tangent lines to the graph of f, + we need to look at the corresponding values of \fp. +

      +
    • +

      + The slope of the tangent line to f at x=1 is \fp(1); + this looks to be about -1. +

      +
      • +
    • +

      + The slope of the tangent line to f at x=2 is \fp(2); + this looks to be about 4. +

      +
      • +
    • +

      + The slope of the tangent line to f at x=3 is \fp(3); + this looks to be about 3. +

      +
      • +
    +

    + +

    + Using these slopes, + tangent line segments to f are sketched in . + Included on the graph of \fp in this figure are points where x=1, + x=2 and x=3 to help better visualize the y value of \fp at those points. +

    + +
    + Graphs of f and \fp in + + + Graphs of function f and its derivative + +

    + The curve of f(x) starts at at y=4 and slopes downwards intersecting the x axis at around x=0.5. + It continues sloping downwards in the negative y direction until about y=-2 and then starts sloping upwards again, + this time intersecting the x axis at x=2 and then continues upwards until y=5 then slope downwards again. + The graph of \fp is a downward parabolic curve that intersects f(x) at (2.5, 4) in the positive y direction + and y=-2 at negative y direction. The tangent lines of f are drawn at (1, 2), (2, 0), (3, 4). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + minor y tick num=4, + minor x tick num=0, + xtick={1,2,3}, + ymin=-8.9, + ymax=5.9, + xmin=-.1, + xmax=3.49, + clip=false + ] + \addplot+[infinite,domain=0:3.5] {-x^3+7*x^2-12*x+4} node[above left] {$f(x)$}; + \addplot+[infinite,domain=0.5:3.3] {-12+14*x-3*x^2} node[pos=0.5, above left] {$f'(x)$}; + \addplot [tangentlineseg,domain=0.5:1.5] {-1*(x-1)-2}; + \addplot [tangentlineseg,domain=1.5:2.5] {4*(x-2)}; + \addplot [tangentlineseg,domain=2.5:3.5] {3*(x-3)+4}; + \addplot[soliddot] coordinates{(1,-2) (2,0) (3,4)}; + \addplot[soliddot] coordinates{(1,-1) (2,4) (3,3)}; + \end{axis} + \end{tikzpicture} + + + +
    +     +     + + + Approximation with the derivative + +

    + Consider again the graph of f(x) and its derivative \fp(x) in . + Use the tangent line to f at x=3 to approximate the value of f(3.1). +

    +     + +

    + + shows the graph of f along with its tangent line, + zoomed in at x=3. + Notice that near x=3, + the tangent line makes an excellent approximation of f. + Since lines are easy to deal with, + often it works well to approximate a function with its tangent line. + (This is especially true when you don't actually know much about the function at hand, + as we don't in this example.) +

    + +

    + While the tangent line to f was drawn in , + it was not explicitly computed. + Recall that the tangent line to f at x=c is y = \fp(c)(x-c)+f(c). + While f is not explicitly given, + by the graph it looks like f(3) = 4. + Recalling that \fp(3) = 3, + we can compute the tangent line to be approximately y = 3(x-3)+4. + It is often useful to leave the tangent line in point-slope form. +

    + +
    + Zooming in on f and its tangent line at x=3 for the function given in Examples and + + + Graph of a function f that is tracing the path of a projectile and its tangent line drawn at x=3. + +

    + The graph of f starts close to (0, 3). It curves upwards tracing the path of a projectile. Up to (3.1, 4.2) in the first quadrant it moves straight + upwards to the right. Then the rate at which it was moving upwards becomes slower and the graph makes a slower movement where it is still moving upwards but we can see + that the movement also has a downward component which pulling it downwards. The tangent line is drawn at (3, 4). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + minor y tick num=4, + ymin=1.9, + ymax=4.9, + xmin=2.7, + xmax=3.3, + xdiscontinuity, + ydiscontinuity, + grid=major + ] + \addplot+[infinite,domain=2.7:3.3] {-x^3+7*x^2-12*x+4} node[pos=0.85,below] {$f(x)$}; + \addplot[tangentlineseg,domain=2.7:3.3,thick] {3*(x-3)+4}; + \addplot[soliddot] coordinates{(3,4)}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + To use the tangent line to approximate f(3.1), + we simply evaluate y at 3.1 instead of f. + + f(3.1) \amp \approx y(3.1) + \amp= 3(3.1-3)+4 + \amp= 0.1\cdot3+4 + \amp = 4.3 + . +

    + +

    + We approximate f(3.1) \approx 4.3. +

    +     +     + +

    + To demonstrate the accuracy of the tangent line approximation, + we now state that in , + f(x) = -x^3+7x^2-12x+4. + We can evaluate f(3.1) = 4.279. + Had we known f all along, + certainly we could have just made this computation. + In reality, we often only know two things: +

      +
    1. +

      + what f(c) is, for some value of c, and +

      +
      2. +
    2. +

      + what \fp(c) is. +

      +
      3. +
    +

    + +

    + For instance, + we can easily observe the location of an object and its instantaneous velocity at a particular point in time. + We do not have a function f + for the location, just an observation. + This is enough to create an approximating function for f. +

    + +

    + This last example has a direct connection to our approximation method explained above after . + We stated there that + + f(c+h) \approx f(c)+\fp(c)\cdot h + . +

    + +

    + If we know f(c) and \fp(c) for some value x=c, + then computing the tangent line at (c,f(c)) is easy: + y(x) = \fp(c)(x-c)+f(c). + In , + we used the tangent line to approximate a value of f. + Let's use the tangent line at x=c to approximate a value of f near x=c; + , compute y(c+h) to approximate f(c+h), + assuming again that h is small. Note: + + y(c+h) \amp = \fp(c)\left((c+h)-c\right)+f(c) + \amp = \fp(c)\cdot h + f(c) + . +

    + +

    + This is the exact same approximation method used above! + Not only does it make intuitive sense, + as explained above, it makes analytical sense, + as this approximation method is simply using a tangent line to approximate a function's value. +

    + +

    + The importance of understanding the derivative cannot be understated. + When f is a function of x, + \fp(x) measures the instantaneous rate of change of f with respect to x and gives the slope of the tangent line to f at x. +

    +     + + + + Terms and Concepts + + + + +

    + What is the instantaneous rate of change of position called? +

    +

    + +

    +     + + + + + + + + +

    + Your answer includes the correct word but has extra text. + Unless stated otherwise, velocity is assumed to be instantaneous. +

    +     +     +     +     + + +     + + + + +

    + Given a function y=f(x), + in your own words describe how to find the units of \fp(x). +

    + +     + + + +     + + + + +

    + What functions have a constant rate of change? +

    +

    + + +

    +     + + + + + linear|linear functions? + + + + + +

    + Your answer includes the correct word but has extra text. +

    +     +     +     +     + +     +     + + Problems + + + + + $a=random(1,9,1); + $fa=random(10,20,1); + $fpa=non_zero_random(-4,4,1); + if($envir{problemSeed}==1){$a=5;$fa=10;$fpa=2;}; + $b=$a+1; + $ans=$fa+$fpa*($b-$a); + + +

    + Given f()= and \fp() = , approximate f(). +

    +

    + +

    +     +     +     + + + + + $a=random(50,150,10); + $fa=non_zero_random(-99,99,1); + $fpa=non_zero_random(-9,9,1); + $b=$a+random(5,15,5); + if($envir{problemSeed}==1){$a=100;$fa=-67;$fpa=5;$b=10;}; + $ans=$fa+$fpa*($b-$a); + + +

    + Given P()= and P'() = , approximate P(). +

    +

    + +

    +     +     +     + + + + + $a=random(10,95,5); + $fa=random(101,199,1); + $fpa=non_zero_random(-20,20,1); + $b=$a-random(5,15,5); + if($envir{problemSeed}==1){$a=25;$fa=187;$fpa=17;$b=20;}; + $ans=$fa+$fpa*($b-$a); + + +

    + Given z()= and z'() = , + approximate z(). +

    +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $popup = DropDown(['f(10.1)','f(11)','f(20)'],0,showInStatic=>0); + $showwork = '[@ explanation_box(message => "Explain your reasoning.") @]*'; + + +

    + Knowing f(10)=25 and + \fp(10) = 5 and the methods described in this section, + which approximation is likely to be most accurate? +

    +

    + +

    +     + +

    + f(10.1) is likely most accurate, + as accuracy is lost the farther from x=10 we go. +

    +     +     +     + + + + + $a=random(1,9,1); + $fa=random(11,99,1); + $fpa=non_zero_random(-10,10,1); + if($envir{problemSeed}==1){$a=7;$fa=26;$fpa=-4;}; + $b=$a+1; + $fb=$fa+$fpa*($b-$a); + + +

    + Given f()= and f() = , + approximate \fp(). +

    +

    + +

    +     +     +     + + + + + $a=random(-5,5,1); + $fa=random(11,99,1); + $fpa=non_zero_random(-10,10,1); + $b=$a+random(2,6,1); + if($envir{problemSeed}==1){$a=0;$fa=17;$fpa=6;$b=2;}; + $fb=$fa+$fpa*($b-$a); + + +

    + Given H()= and H() = , + approximate H'(). +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('decibels per customer'=>{}); + Context()->strings->add( + 'decibel per customer' => {alias=>'decibels per customer'}, + 'db per customer' => {alias=>'decibels per customer'}, + 'decibels / customer' => {alias=>'decibels per customer'}, + 'decibel / customer' => {alias=>'decibels per customer'}, + 'db / customer' => {alias=>'decibels per customer'}, + 'decibels/customer' => {alias=>'decibels per customer'}, + 'decibel/customer' => {alias=>'decibels per customer'}, + 'db/customer' => {alias=>'decibels per customer'}, + ); + $ans=Compute("decibels per customer"); + + +

    + Let V(x) measure the volume, in decibels, + measured inside a restaurant with x customers. + What are the units of V'(x)? +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('foot per second squared'=>{}); + Context()->strings->add( + 'foot per square second' => {alias=>'foot per second squared'}, + 'foot per second per second' => {alias=>'foot per second squared'}, + 'ft per second squared' => {alias=>'foot per second squared'}, + 'ft per square second' => {alias=>'foot per second squared'}, + 'ft per second per second' => {alias=>'foot per second squared'}, + 'foot per s^2' => {alias=>'foot per second squared'}, + 'foot per s per s' => {alias=>'foot per second squared'}, + 'ft per s^2' => {alias=>'foot per second squared'}, + 'ft per s per s' => {alias=>'foot per second squared'}, + 'foot / second squared' => {alias=>'foot per second squared'}, + 'foot / square second' => {alias=>'foot per second squared'}, + 'foot / second / second' => {alias=>'foot per second squared'}, + 'ft / second squared' => {alias=>'foot per second squared'}, + 'ft / square second' => {alias=>'foot per second squared'}, + 'ft / second / second' => {alias=>'foot per second squared'}, + 'foot / s^2' => {alias=>'foot per second squared'}, + 'foot / s / s' => {alias=>'foot per second squared'}, + 'ft / s^2' => {alias=>'foot per second squared'}, + 'ft / s / s' => {alias=>'foot per second squared'}, + 'foot/second squared' => {alias=>'foot per second squared'}, + 'foot/square second' => {alias=>'foot per second squared'}, + 'foot/second/second' => {alias=>'foot per second squared'}, + 'ft/second squared' => {alias=>'foot per second squared'}, + 'ft/square second' => {alias=>'foot per second squared'}, + 'ft/second/second' => {alias=>'foot per second squared'}, + 'foot/s^2' => {alias=>'foot per second squared'}, + 'foot/s/s' => {alias=>'foot per second squared'}, + 'ft/s^2' => {alias=>'foot per second squared'}, + 'ft/s/s' => {alias=>'foot per second squared'}, + 'feet per square second' => {alias=>'foot per second squared'}, + 'feet per second per second' => {alias=>'foot per second squared'}, + 'feet per s^2' => {alias=>'foot per second squared'}, + 'feet per s per s' => {alias=>'foot per second squared'}, + 'feet / second squared' => {alias=>'foot per second squared'}, + 'feet / square second' => {alias=>'foot per second squared'}, + 'feet / second / second' => {alias=>'foot per second squared'}, + 'feet / s^2' => {alias=>'foot per second squared'}, + 'feet / s / s' => {alias=>'foot per second squared'}, + 'feet/second squared' => {alias=>'foot per second squared'}, + 'feet/square second' => {alias=>'foot per second squared'}, + 'feet/second/second' => {alias=>'foot per second squared'}, + 'feet/s^2' => {alias=>'foot per second squared'}, + 'feet/s/s' => {alias=>'foot per second squared'}, + ); + $ans=Compute("foot per second squared"); + + +

    + Let v(t) measure the velocity, + in ft/s, of a car moving in a straight line t seconds after starting. + What are the units of v'(t)? +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('foot per hour'=>{}); + Context()->strings->add( + 'foot per hr' => {alias=>'foot per hour'}, + 'foot per h' => {alias=>'foot per hour'}, + 'foot / hour' => {alias=>'foot per hour'}, + 'foot / hr' => {alias=>'foot per hour'}, + 'foot / h' => {alias=>'foot per hour'}, + 'foot/hour' => {alias=>'foot per hour'}, + 'foot/hr' => {alias=>'foot per hour'}, + 'foot/h' => {alias=>'foot per hour'}, + 'feet per hour' => {alias=>'foot per hour'}, + 'feet per hr' => {alias=>'foot per hour'}, + 'feet per h' => {alias=>'foot per hour'}, + 'feet / hour' => {alias=>'foot per hour'}, + 'feet / hr' => {alias=>'foot per hour'}, + 'feet / h' => {alias=>'foot per hour'}, + 'feet/hour' => {alias=>'foot per hour'}, + 'feet/hr' => {alias=>'foot per hour'}, + 'feet/h' => {alias=>'foot per hour'}, + 'ft per hour' => {alias=>'foot per hour'}, + 'ft per hr' => {alias=>'foot per hour'}, + 'ft per h' => {alias=>'foot per hour'}, + 'ft / hour' => {alias=>'foot per hour'}, + 'ft / hr' => {alias=>'foot per hour'}, + 'ft / h' => {alias=>'foot per hour'}, + 'ft/hour' => {alias=>'foot per hour'}, + 'ft/hr' => {alias=>'foot per hour'}, + 'ft/h' => {alias=>'foot per hour'}, + ); + $ans=Compute("foot per hour"); + + +

    + The height H, in feet, + of a river is recorded t hours after midnight, April 1. + What are the units of H'(t)? +

    +

    + +

    +     +     +     + + + + +

    + P is the profit, in thousands of dollars, + of producing and selling c cars. +

    +     + + +

    + What are the units of P'(c)? +

    + +     + + + +

    + Thousands of dollars per car. +

    +     +     + + +

    + What is likely true of P(0)? +

    + +     + + + +

    + It is likely that P(0)\lt 0. + That is, negative profit for not producing any cars. +

    +     +     + +     + + + + +

    + T is the temperature in degrees Fahrenheit, + h hours after midnight on July 4 in Sidney, NE. +

    +     + + +

    + What are the units of T'(h)? +

    + +     + + + +

    + Degrees Fahrenheit per hour. +

    +     +     + + +

    + Is T'(8) likely greater than or less than 0? + Why? +

    + +     + + + +

    + It is likely that T'(8) \gt 0 since at 8 in the morning, + the temperature is likely rising. +

    +     +     + + +

    + Is T(8) likely greater than or less than 0? + Why? +

    + +     + + + +

    + It is very likely that T(8) \gt 0, + as at 8 in the morning on July 4, we would expect the temperature to be well above 0. +

    +     +     + +     + + + +

    + Graphs of functions f and g are given. + Identify which function is the derivative of the other. +

    +     + + + + + parserPopUp.pl + + + $h=non_zero_random(-2,2,1); + $k=non_zero_random(-2,2,1); + $a=non_zero_random(-1,1,0.05); + if($envir{problemSeed}==1){$a=0.25;$h=1;$k=2;}; + $f=Formula("$a*(x-$h)^2+$k"); + $fp=$f->D('x'); + if ($a>0) { + $fleft = $h - sqrt(5/$a - $k/$a); + $fright = $h + sqrt(5/$a - $k/$a); + @fpos = ('center','top'); + $fpleft = $h-5/2/$a; + $fpright = $h+5/2/$a; + @fppos = ('left','top'); + } else { + $fleft = $h - sqrt(-5/$a - $k/$a); + $fright = $h + sqrt(-5/$a - $k/$a); + @fpos = ('center','bottom'); + $fpleft = $h+5/2/$a; + $fpright = $h-5/2/$a; + @fppos = ('left','bottom'); + }; + $fleft=max($fleft,-4); + $fright=min($fright,4); + $fpleft=max($fpleft,-4); + $fpright=min($fpright,4); + $radio=DropDown([ + 'f is the derivative of g.', + 'g is the derivative of f.' + ],0,showInStatic=>0); + + + + f(x) is a line with slope 2 and x intercept . g(x) is a parabola with vertex at (,) + +

    + The graph of f is a line with slope 2a and the x intercept h. + The value of the random variable a decides what the graphs look like. + Note that the graph of f is always a line and the graph of g is always a hyperbola. + The description below is of one of the randomly generated graphs with h=-2, + a=-1 and k=1. If h=-2 and a=-1, we have a line that intersects + the negative x axis at x=-2 and the negative y axis at y=-4. + The graph of g is a hyperbola. The vertex is at (h, k). + With h=-2, a=-1 and k=1 we have the graph of g, + a downwards parabola with its vertex at (-2, 1). One of its arms intersect + the negative x axis at x=-1 and the other at x=-1. + The graph of f intersects the graph of g twice, once in the second quadrant + and once in the fourth quadrant. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-5.1,ymax=5.1,% + xmin=-4.3,xmax=4.3,% + ] + + \addplot [firstcurvestyle,domain=$fpleft:$fpright] {2*$a*(x-$h)}; + \addplot [secondcurvestyle,domain=$fleft:$fright] {$a*(x-$h)^2+$k}; + \addlegendentry{\(f(x)\)}; + \addlegendentry{\(g(x)\)}; + + \end{axis} + \end{tikzpicture} + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + ($r,$s,$t) = num_sort(random_subset(3,-3..3)); + $a=non_zero_random(-1,1,0.05); + if($envir{problemSeed}==1){$a=1;$r=0;$s=1;$t=2;}; + $f=Formula("$a*(x-$r)*(x-$s)*(x-$t)"); + $fp=$f->D('x'); + $xmin=$r-2; + $xmax=$t+2; + @cp=($xmin,$xmax); + @cv; + for $i (@cp) { + push(@cv, $f->eval(x=>$i)); + }; + $ymin=min(-5,map{Round($_)-1}(@cv)); + $ymax=max(5,map{Round($_)+1}(@cv)); + $infl=($r+$s+$t)/3; + $inflval=$fp->eval(x=>$infl); + $radio=DropDown([ + 'f is the derivative of g.', + 'g is the derivative of f.' + ],1,showInStatic=>0); + + + + f is a parabola and g is a cubic graph that has three x intercepts. + +

    + The graph of g is a parabola and the graph of f + is a graph of a cubic function that intersects the x axis three times. Note that the graph of f is always + a cubic graph and the graph of g is always a hyperbola. + The description below describes one of the randomized images for the excercise. The vertex of g is at + (-0.667,-0.367) in the third qudrant just below the negative x + and slightly to the left of the negative y axis. + From there the right arm travels upwards to the right, intersects the positive x + axis and continues travelling upwards while also moving to its right. The graph of + g is the graph of a cubic function that travels upwards from the bottom of the fourth quadrant, + intersecting the negative x axis at x=-3 it travels upwards + for a shortwhile then slopes down and again intersects the x axis at x=-2. + Next it travels downwards to the right. After crossing the negative y axis + at y=-0.9 and then it slopes upwards again. It intersects the positive x + axis for the last time at x=3 and continues upwards. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=$ymin,ymax=$ymax,% + xmin=$xmin,xmax=$xmax,% + ] + + \addplot [firstcurvestyle,domain=$xmin:$xmax,samples=101] {$a*(x-$r)*(x-$s)*(x-$t)}; + \addplot [secondcurvestyle,domain=$xmin:$xmax] {$a*(3*x^2-2*($r+$s+$t)*x+$r*$s+$r*$t+$s*$t)}; + \addlegendentry{\(f(x)\)}; + \addlegendentry{\(g(x)\)}; + + \end{axis} + \end{tikzpicture} + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $radio=DropDown([ + 'f is the derivative of g.', + 'g is the derivative of f.' + ],1,showInStatic=>0); + + + + f is a hyperbola with curves in the first and the third quadrant and g is a hyperbola with curves in the third and the fourth quadrant. + +

    + The graph of f is a hyperbola with curves in the first and the third quadrant. The curves look like infinite bows with the branches extending in + positive and negative x and y directions. The curve in the first quadrant has its vertex at (1, 1). The two branches of the curve extend + towards positive x and y axis. They continue moving in their respective directions while they get closer to the x and the y axis + without ever merging with them. The curve in the third qudrant has its vertex at (-1, -1) and the two branches of this one move towards the negative x + and y axis in the exact manner as the curve in the first quadrant creating a mirror image of the first curve. The graph of g also consists of two + curves, one in the third and one in the fourth quadrant that are mirror images of each other. The curve in the fourth quadrant is shaped like an inverted L. The + focus of the curve is located close to (1, -0.5). From there, one of branches move upwards to the right until it merges with the positive x axis and + continues along with it. The other branch moves downwards while moving slightly to its left and then contines vertically downwards parallel to the negative y + axis. The curve in the third qudrant is a mirror image of the previous curve. It follows the part of f in the third quadrant closely intersecting f + at (-1, -1). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-5.5,ymax=5.5,% + xmin=-5.5,xmax=5.5,% + ] + + \addplot [firstcurvestyle,domain=-5.5:-.182,samples=40] {1/x}; + \addplot [secondcurvestyle,domain=-5.5:-.426,samples=40] {-1/x^2}; + \addplot [firstcurvestyle,domain=.182:5.5,samples=40] {1/x}; + \addplot [secondcurvestyle,domain=.426:5.5,samples=40] {-1/x^2}; + \addlegendentry{\(f(x)\)}; + \addlegendentry{\(g(x)\)}; + + \end{axis} + \end{tikzpicture} + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + parserPopUp.pl + + + $h=random(-2,2,1); + $p=random(2,4,1); + if($envir{problemSeed}==1){$h=1;$p=2;}; + $f=Formula("cos((pi/$p)*(x-$h))"); + $fp=$f->D('x'); + $ymin=-max(int(pi/$p),1)-1; + $ymax=max(int(pi/$p),1)+1; + $radio=DropDown([ + 'f is the derivative of g.', + 'g is the derivative of f.' + ],1,showInStatic=>0); + + + + f is making a curve similar to a sine wave while g looks similar to a cosine wave both intersecting the x axis multiple times as they oscillate. + +

    + The graph of f looks like a sine wave. Starting at (0, -4) it slopes downward in the + negative y direction until (-2, -1) then starts moving upwards towards the origin. It continues + upwards through the origin until it reaches (2, 1), then starts sloping down again and eventually + intersects the x axis at (4, 0). g looks similar to a cosine curve. Starting close to (-4, -0.75) + it slopes upwards intersecting f and then intersecting the x axis at (-2, 0). It keeps moving + up and intersects the y axis close to (0, 0.75) which is where the top of the bell is. Then it starts + moving down to the right, crosses the x axis at (2, 0) then slopes down. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=$ymin,ymax=$ymax,% + xmin=-4,xmax=4,% + ] + + \addplot [firstcurvestyle,domain=-4:4,samples=101] {cos(deg((3.14/$p)*(x-$h)))}; + \addplot [secondcurvestyle,domain=-4:4,samples=40] {-(3.14/$p)*sin(deg((3.14/$p)*(x-$h)))}; + \addlegendentry{\(f(x)\)}; + \addlegendentry{\(g(x)\)}; + + \end{axis} + \end{tikzpicture} + + +

    + +

    +     +     +     +     +     +     +
    +
    + Basic Differentiation Rules +

    + The derivative is a powerful tool but is admittedly awkward given its reliance on limits. + Fortunately, one thing mathematicians are good at is + abstraction. For instance, + instead of continually finding derivatives at a point, + we abstracted and found the derivative function. +

    + +

    + Let's practice abstraction on linear functions, y=mx+b. + What is \yp? + Without limits, + recognize that linear functions are characterized by being functions with a constant rate of change + (the slope). + The derivative, \yp, gives the instantaneous rate of change; + with a linear function, this is constant, m. + Thus \yp=m. +

    + +

    + Let's abstract once more. + Let's find the derivative of the general quadratic function, + f(x) = ax^2+bx+c. + Using the definition of the derivative, we have: + + \fp(x) \amp = \lim_{h\to 0}\frac{a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h} + \amp = \lim_{h\to 0}\frac{ax^2+2ahx+ah^2+bx+bh+c-ax^2-bx-c}{h} + \amp = \lim_{h\to 0} \frac{ah^2+2ahx+bh}{h} + \amp = \lim_{h\to 0} ah+2ax+b + \amp = 2ax+b + . +

    + +

    + So if y = 6x^2+11x-13, + we can immediately compute \yp = 12x+11. +

    + +

    + In this section + (and in some sections to follow) + we will learn some of what mathematicians have already discovered about the derivatives of certain functions and how derivatives interact with arithmetic operations. + We start with a theorem. +

    + + + Derivatives of Common Functions + +

    +

      +
    1. + Constant Rule + + derivativeConstant Rule + +

      + \lzoo{x}{c} = 0, where c is a constant. +

      +
      2. + +
    2. + Power Rule + + derivativePower Rule + Power Ruledifferentiation + +

      + \lzoo{x}{x^n}= nx^{n-1}, + where n is an integer, n \gt 0. +

      +
      3. + +
    3. + +

      + \lzoo{x}{\sin(x)} = \cos(x) +

      +
      4. + +
    4. +

      + \lzoo{x}{\cos(x)} = {-\sin(x)} +

      +
      5. + +
    5. +

      + \lzoo{x}{e^x} = e^x +

      +
      6. + +
    6. +

      + \lzoo{x}{\ln(x)} = \frac{1}{x}, for x \gt 0. +

      +
      7. +
    + + derivativebasic rules + +

    +     +     + + + + + +

    + This theorem starts by stating an intuitive fact: + constant functions have zero rate of change as they are constant. + Therefore their derivative is 0 + (they change at the rate of 0). + The theorem then states some fairly amazing things. + The states that the derivatives of Power Functions + (of the form y=x^n) + are very straightforward: + multiply by the power, then subtract 1 from the power. + We see something incredible about the function y=e^x: + it is its own derivative. + We also see a new connection between the sine and cosine functions. +

    + +

    + One special case of the is when n=1, + , when f(x) = x. + What is \fp(x)? + According to the , + + \fp(x) = \lzoo{x}{x} = \lzoo{x}{x^1} = 1\cdot x^0 = 1 + . +

    + +

    + In words, we are asking At what rate does f change with respect to x? + Since f is x, + we are asking At what rate does x change with respect to x? The answer is: + 1. + They change at the same rate. + We can also interpret the derivative as the slope of the tangent line to the function at a point (c,f(c)). + Since f(x)=x is a linear function with constant slope 1, + we can say that the derivative of f(x)=x is \fp(x)=1. +

    + + + + + + +

    + Let's practice using this theorem. +

    + + + Using common derivative rules to find, and use, derivatives + +

    + Let f(x)=x^3. +

      +
    1. +

      + Find \fp(x). +

      +
      2. +
    2. +

      + Find the equation of the line tangent to the graph of f at x=-1. +

      +
      3. +
    3. +

      + Use the tangent line to approximate (-1.1)^3. +

      +
      4. +
    4. +

      + Sketch f, + \fp and the tangent line from on the same axis. +

      +
      5. +
    +

    +     + +

    +

      +
    1. +

      + The states that if f(x) = x^3, + then \fp(x) = 3x^2. +

      +
      2. +
    2. +

      + To find the equation of the line tangent to the graph of f at x=-1, + we need a point and the slope. + The point is (-1,f(-1)) = (-1, -1). + The slope is \fp(-1)= 3. + Thus the tangent line has equation y = 3(x-(-1))+(-1) = 3x+2. +

      +
      3. +
    3. +

      + We can use the tangent line to approximate + (-1.1)^3 since -1.1 is close to -1. + We have + + (-1.1)^3 \approx 3(-1.1)+2 = -1.3 + . + We can easily find the actual value: + (-1.1)^3 = -1.331. +

      +
      4. +
    4. +

      + See . +

      +
      5. +
    +

    + +
    + A graph of f(x) = x^3, along with its derivative \fp(x) = 3x^2 and its tangent line at x=-1 + + + Graph of function x^3, its derivative and a tangent line drawn at point (-1,-1). + + +

    + The y axis is drawn between -4 and 4 and the x axis + is drawn between -2 and 2. + The graph of function f(x)=x^3 is a curve that appears to conincide the + x axis from x=-0.5 to x=0.5, + then it appears to increase to infinity from x=0.5 in the first quadrant, + and curves downward from to infinity from x=-0.5 in the third quadrant. +

    +

    + The derivative \fp(x) is a parabola opening upwards. + It intersects the function f(x) at the origin and increases to infinity on + either side of the positive y axis. +

    +

    + A tangent line l(x)is drawn on the function at point (-1,-1). + It has a positive slope and and intersects the x axis at approximately + x=-0.6 and the y axis at y=2. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-2.1, + xmax=2.1, + ymin=-5.1, + ymax=5.1, + clip=false + ] + \addplot+[infinite,domain=-1.5:1.5] {x^3} node[below right] {$f(x)$}; + \addplot+[infinite,domain=-1.2:1.2] {3*x^2} node[right] {$f'(x)$}; + \addplot[tangentline,domain=-2:1] {3*(x+1)-1} node[above left] {$\ell(x)$}; + \addplot[soliddot] coordinates {(-1,-1) (-1,3)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + +

    + gives useful information, + but we will need much more. + For instance, using the theorem, + we can easily find the derivative of y=x^3, + but it does not tell how to compute the derivative of y=2x^3, + y=x^3+\sin(x) nor y=x^3\sin(x). + The following theorem helps with the first two of these examples + (the third is answered in the next section). +

    + + + + + Properties of the Derivative + +

    + Let f and g be differentiable on an open interval I and let c be a real number. + Then: +

      +
    1. + Sum/Difference Rule +

      + + \lzoo{x}{f(x) \pm g(x)} = \lzoo{x}{f(x)} \pm \lzoo{x}{g(x)} = \fp(x)\pm \gp(x) + + + derivativeSum/Difference Rule + Sum/Difference Ruleof derivatives + +

      +
      2. + +
    2. + Constant Multiple Rule +

      + + \lzoo{x}{c\cdot f(x)} = c\cdot\lzoo{x}{f(x)} = c\cdot\fp(x) + . + + derivativeConstant Multiple Rule + Constant Multiple Ruleof derivatives + +

      +
      3. +
    +

    +     +     + + + + + + + + + + + +

    + + allows us to find the derivatives of a wide variety of functions. + It can be used in conjunction with the to find the derivatives of any polynomial. + Recall in that we found, + using the limit definition, the derivative of f(x) = 3x^2+5x-7. + We can now find its derivative without expressly using limits: + + \lzoo{x}{3x^2+5x-7} \amp = 3\lzoo{x}{x^2} + 5\lzoo{x}{x} - \lzoo{x}{7} + \amp = 3\cdot 2x+5\cdot 1- 0 + \amp = 6x+5 + . +

    + +

    + We were a bit pedantic here, showing every step. + Normally we would do all the arithmetic and steps in our head and readily find \lzoo{x}{3x^2+5x+7}= 6x+5. +

    + + + Using the tangent line to approximate a function value + +

    + Let f(x) = \sin(x) + 2x+1. + Approximate f(3) using an appropriate tangent line. +

    +     + +

    + This problem is intentionally ambiguous; + we are to approximate using an + appropriate tangent line. + How good of an approximation are we seeking? + What does appropriate mean? +

    + +

    + In the real world, people solving problems deal with these issues all time. + One must make a judgment using whatever seems reasonable. + In this example, the actual answer is f(3) = \sin(3) + 7, + where the real problem spot is \sin(3). + What is \sin(3)? +

    + +

    + Since 3 is close to \pi, + we can assume \sin(3) \approx \sin(\pi) = 0. + Thus one guess is f(3) \approx 7. + Can we do better? + Let's use a tangent line as instructed and examine the results; + it seems best to find the tangent line at x=\pi. +

    + +

    + Using + we find \fp(x) = \cos(x) + 2. + The slope of the tangent line is thus \fp(\pi) = \cos(\pi) + 2 =1. + Also, f(\pi) = 2\pi+1 \approx 7.28. + So the tangent line to the graph of f at x=\pi is y=1(x-\pi)+ 2\pi+1 =x+\pi+1 \approx x+4.14. + Evaluated at x=3, our tangent line gives y=3+4.14 = 7.14. + Using the tangent line, + our final approximation is that f(3) \approx 7.14. +

    + +

    + Using a calculator, + we get an answer accurate to four places after the decimal: + f(3) = 7.1411. + Our initial guess was 7; + our tangent line approximation was more accurate, at 7.14. +

    + +

    + The point is not Here's a cool way to do some math without a calculator. Sure, + that might be handy sometime, + but your phone could probably give you the answer. + Rather, the point is to say that tangent lines are a good way of approximating, + and many scientists, + engineers and mathematicians often face problems too hard to solve directly. + So they approximate. +

    + +

    + The graphs in + shows the graph of the function f(x) along with the tangent line constructed at x=\pi. + The graph in + shows the same tangent line and function. + Once zoomed in, + you can barely distinguish the tangent line from the function. + This indicates that the tangent line is a good a approximation of the function so long as we are near the point of tangency. +

    + + +
    + A graph of f(x) = \sin(x)+2x+1 along with its tangent line approximation at x=\pi + + + Graph of the function for this example and its tangent at x=pi. + + +

    + The y axis is drawn from -1 to 10 and x axis is drawn from -1 to 5. + The graph of f(x)is drawn in the first quadrant, the curve passes the + y axis at y=1, it increases until point (5,10). + It has a slight undulation between x=2 to x=4. +

    +

    + A tangent line is drawn on the curve at approximately point (3,7) or x=\pi. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1.1, + xmax=5.1, + ymin=-1, + ymax=10, + ] + \addplot+[infinite,domain=-.5:5] {sin(deg(x))+2*x+1}; + \addplot[tangentline,domain=1:4.5] {x+4.14}; + \addplot[soliddot] coordinates {(3,7.1411) (3.14,7.28)}; + \draw (axis cs:0.5,1.2) node { $f(x)$}; + \draw (axis cs:0.5,5) node { $l(x)$}; + \end{axis} + \end{tikzpicture} + + + +
    + +
    + A graph of f(x) = \sin(x)+2x+1 along with its tangent line approximation at x=\pi, zoomed in + + + Highly zoomed in view of the previous graph and its tangent line, near x=pi. + + +

    + The y axis is drawn from 6.8 to 7.4 and x axis + is drawn from 2.6 to 3.4. + The graph of f(x) is drawn in the first quadrant, the curve and + its tangent line are aligned except near shifted origin around point (2.6, 6.8) + where they are slightly apart. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=6.7, + ymax=7.4, + xmin=2.5, + xmax=3.5, + xdiscontinuity, + ydiscontinuity + ] + \addplot+[infinite,domain=2.6:3.25] {sin(deg(x))+2*x+1}; + \addplot[tangentline,domain=2.6:3.25] {x+4.14}; + \addplot[soliddot] coordinates {(3,7.1411) (3.14,7.28)}; + %\draw (axis cs:0.5,1.2) node { $f(x)$}; + %\draw (axis cs:0.5,5) node { $l(x)$}; + \end{axis} + \end{tikzpicture} + + + +
    +     +     + +     + + + Higher Order Derivatives + +

    + The derivative of a function f is itself a function, + therefore we can take its derivative. + The following definition gives a name to this concept and introduces its notation. +

    + + + Higher Order Derivatives + +

    + Let y=f(x) be a differentiable function on I. + The following are defined, provided the corresponding limits exist. + + derivativehigher order + derivativenotation + +

      +
    1. +

      + The second derivative of f is: + + \fp'(x) = \lzoo{x}{\fp(x)} = \lzoo{x}{\lz{y}{x}} = \lzn{2}{y}{x}=\yp' + . +

      +
      2. +
    2. +

      + The third derivative of f is: + + \fp''(x) = \lzoo{x}{\fp'(x)} = \lzoo{x}{\lzn{2}{y}{x}} = \lzn{3}{y}{x}=\yp'' + . +

      +
      3. +
    3. +

      + The nth derivative of f is: + + f^{(n)}(x) = \lzoo{x}{f^{(n-1)}(x)} = \lzoo{x}{\lzn{n-1}{y}{x}} = \lzn{n}{y}{x}=y^{(n)} + . +

      +
      4. +
    +

    +     +     + + + + + +

    + In general, when finding the fourth derivative and on, + we resort to the f\,^{(4)}(x) notation, not \fp'''(x); + after a while, too many ticks is confusing. +

    + +

    + Let's practice using this new concept. +

    + + + Finding higher order derivatives + +

    + Find the first four derivatives of the following functions: +

      +
    1. f(x) = 4x^2
      2. +
    2. f(x) = \sin(x)
      3. +
    3. f(x) = 5e^x
      4. +
    +

    +     + +

    +

      +
    1. +

      + Using the Power and Constant Multiple Rules, + we have: \fp(x) = 8x. + Continuing on, we have + + \fp'(x)\amp=\lzoo{x}{8x}=8\amp\fp''(x)\amp=0\amp f^{(4)}(x)\amp=0 + . + Notice how all successive derivatives will also be 0. +

      +
      2. +
    2. +

      + We employ repeatedly. + + \fp(x)\amp=\cos(x)\amp \fp''(x)\amp= -\cos(x) + \fp'(x)\amp= -\sin(x)\amp f^{(4)}(x)\amp = \sin(x) + + Note how we have come right back to f(x) again. (Can you quickly figure what f^{(23)}(x) is?) +

      +
      3. +
    3. +

      + Employing + and the Constant Multiple Rule, we can see that + + \fp(x) = \fp'(x) = \fp''(x) = f^{(4)}(x) = 5e^x + . +

      +
      4. +
    +

    +     + +     +     + + + Interpreting Higher Order Derivatives +

    + What do higher order derivatives mean? + What is the practical interpretation? + + derivativehigher orderinterpretation +

    + +

    + Our first answer is a bit wordy, + but is technically correct and beneficial to understand. + That is, +

    + +
    +

    + The second derivative of a function f is the rate of change of the rate of change of f. +

    +
    + +

    + One way to grasp this concept is to let f describe a position function. + Then, as stated in , + \fp describes the rate of position change: velocity. + We now consider \fp', which describes the rate of velocity change. + Sports car enthusiasts talk of how fast a car can go from 0 to + + 60 + ; they are bragging about the + acceleration of the car. +

    + +

    + We started this chapter with amusement park riders free-falling with position function f(t) = -16t^2+150. + It is easy to compute \fp(t)=-32t ft/s and \fp'(t) = -32 (ft/s)/s. + We may recognize this latter constant; + it is the acceleration due to gravity. + In keeping with the unit notation introduced in the previous section, + we say the units are feet per second per second. + This is usually shortened to feet per second squared, + written as ft/s^2. +

    + +

    + It can be difficult to consider the meaning of the third, + and higher order, derivatives. + The third derivative is the rate of change of the rate of change of the rate of change of f. + That is essentially meaningless to the uninitiated. + In the context of our position/velocity/acceleration example, + the third derivative is the rate of change of acceleration, + commonly referred to as jerk. +

    + +

    + Make no mistake: + higher order derivatives have great importance even if their practical interpretations are hard + (or impossible) + to understand. + The mathematical topic of series + makes extensive use of higher order derivatives. +

    +     + + + + Terms and Concepts + + + + +

    + What is the name of the rule which states that \lzoo{x}{x^n} = nx^{n-1}, + where n \gt 0 is an integer? +

    +

    + + +

    +     + + + + + power|power rule|the power rule + + + + + +     + + + + + $ans=Formula("1/x"); + + +

    + What is \lzoo{x}{\ln(x)}? +

    +

    + +

    +     +     +     + + + + + $ans=Compute("e^x"); + $ev=$ans->cmp(checker=>sub{ + my($correct,$student,$self )=@_; + return 0 unless ($student->D('x') == $student); + return 1; + }); + + +

    + Give an example of a function f(x) where \fp(x) = f(x). +

    + + Enter the formula for the function without the f(x)=. + +

    + +

    +     + +

    + One answer is f(x) = 10e^x. + Any constant multiple of e^x will do. +

    +     +     +     + + + + + $ans=Formula("10"); + $ev=$ans->cmp(checker=>sub{my($correct,$student,$self)=@_; + return 0 unless ($student->D('x') == Formula("0")); + return 1; + }); + + +

    + Give an example of a function f(x) where \fp(x) = 0. +

    + + Enter the formula for the function without the f(x)=. + +

    + +

    +     + +

    + One answer is f(x) = 10. + Any constant function will do. +

    +     +     +     + + + + +

    + The derivative rules introduced in + explain how to compute the derivative of which of the following functions? +

    + +     + + + +

    + f(x)=3x^2-x+17 +

    +     +     + + +

    + f(x)=5\ln(x) +

    +     +     + + +

    + f(x)=\frac{3}{x^2} +

    +     +     + + +

    + f(x) = \sin(x)\cos(x) +

    +     +     + + +

    + f(x)=e^{x^2} +

    +     +     + + +

    + f(x)=\sqrt{x} +

    +     +     +     + +     + + + + +

    + Explain in your own words how to find the third derivative of a function f(x). +

    + +     + + + +     + + + + + $ans=Compute("17x-205"); + $ev=$ans->cmp(checker=>sub{my($correct,$student,$self )=@_; + return 0 if ($student->D('x') == Formula("0")); + return 0 unless ($student->D('x')->D('x') == Formula("0")); + return 1; + }); + + +

    + Give an example of a function where + \fp(x) \neq 0 and \fpp(x) = 0. +

    + + Enter the formula for the function without the f(x)=. + +

    + +

    +     + +

    + One answer is f(x) = 17x-205. + Any linear function with nonzero slope will do. +

    +     +     +     + + + + +

    + Explain in your own words what the second derivative means. +

    + +     + + + +     + + + + + Context()->strings->add('a velocity function'=>{}); + Context()->strings->add( + 'velocity'=>{alias=>'a velocity function'}, + 'velocity function'=>{alias=>'a velocity function'}, + ); + $velocity = Compute("a velocity function"); + Context()->strings->add('an acceleration function'=>{}); + Context()->strings->add( + 'acceleration'=>{alias=>'an acceleration function'}, + 'acceleration function'=>{alias=>'an acceleration function'}, + ); + $acceleration = Compute("an acceleration function"); + + +

    + If f(x) describes a position function, + then \fp(x) describes what kind of function? + What kind of function is \fpp(x)? +

    + + What kind of function is \fp(x)? + +

    + +

    + + What kind of function is \fpp(x)? + +

    + +

    +     + +

    + \fp(x) is a velocity function, + and \fpp(x) is acceleration. +

    +     +     +     + + + + + Context()->strings->add('pound per foot squared'=>{}); + Context()->strings->add( + 'pound per square foot' => {alias=>'pound per foot squared'}, + 'pound per foot per foot' => {alias=>'pound per foot squared'}, + 'lb per foot squared' => {alias=>'pound per foot squared'}, + 'lb per square foot' => {alias=>'pound per foot squared'}, + 'lb per foot per foot' => {alias=>'pound per foot squared'}, + 'pound per ft^2' => {alias=>'pound per foot squared'}, + 'pound per ft per ft' => {alias=>'pound per foot squared'}, + 'lb per ft^2' => {alias=>'pound per foot squared'}, + 'lb per ft per ft' => {alias=>'pound per foot squared'}, + 'pound / foot squared' => {alias=>'pound per foot squared'}, + 'pound / square foot' => {alias=>'pound per foot squared'}, + 'pound / foot / foot' => {alias=>'pound per foot squared'}, + 'lb / foot squared' => {alias=>'pound per foot squared'}, + 'lb / square foot' => {alias=>'pound per foot squared'}, + 'lb / foot / foot' => {alias=>'pound per foot squared'}, + 'pound / ft^2' => {alias=>'pound per foot squared'}, + 'pound / ft / ft' => {alias=>'pound per foot squared'}, + 'lb / ft^2' => {alias=>'pound per foot squared'}, + 'lb / ft / ft' => {alias=>'pound per foot squared'}, + 'pound/foot squared' => {alias=>'pound per foot squared'}, + 'pound/square foot' => {alias=>'pound per foot squared'}, + 'pound/foot/foot' => {alias=>'pound per foot squared'}, + 'lb/foot squared' => {alias=>'pound per foot squared'}, + 'lb/square foot' => {alias=>'pound per foot squared'}, + 'lb/foot/foot' => {alias=>'pound per foot squared'}, + 'pound/ft^2' => {alias=>'pound per foot squared'}, + 'pound/ft/ft' => {alias=>'pound per foot squared'}, + 'lb/ft^2' => {alias=>'pound per foot squared'}, + 'lb/ft/ft' => {alias=>'pound per foot squared'}, + 'pounds per square foot' => {alias=>'pound per foot squared'}, + 'pounds per foot per foot' => {alias=>'pound per foot squared'}, + 'pounds per ft^2' => {alias=>'pound per foot squared'}, + 'pounds per ft per ft' => {alias=>'pound per foot squared'}, + 'pounds / foot squared' => {alias=>'pound per foot squared'}, + 'pounds / square foot' => {alias=>'pound per foot squared'}, + 'pounds / foot / foot' => {alias=>'pound per foot squared'}, + 'pounds / ft^2' => {alias=>'pound per foot squared'}, + 'pounds / ft / ft' => {alias=>'pound per foot squared'}, + 'pounds/foot squared' => {alias=>'pound per foot squared'}, + 'pounds/square foot' => {alias=>'pound per foot squared'}, + 'pounds/foot/foot' => {alias=>'pound per foot squared'}, + 'pounds/ft^2' => {alias=>'pound per foot squared'}, + 'pounds/ft/ft' => {alias=>'pound per foot squared'}, + 'lbs per foot squared' => {alias=>'pound per foot squared'}, + 'lbs per square foot' => {alias=>'pound per foot squared'}, + 'lbs per foot per foot' => {alias=>'pound per foot squared'}, + 'lbs per ft^2' => {alias=>'pound per foot squared'}, + 'lbs per ft per ft' => {alias=>'pound per foot squared'}, + 'lbs / foot squared' => {alias=>'pound per foot squared'}, + 'lbs / square foot' => {alias=>'pound per foot squared'}, + 'lbs / foot / foot' => {alias=>'pound per foot squared'}, + 'lbs / ft^2' => {alias=>'pound per foot squared'}, + 'lbs / ft / ft' => {alias=>'pound per foot squared'}, + 'lbs/foot squared' => {alias=>'pound per foot squared'}, + 'lbs/square foot' => {alias=>'pound per foot squared'}, + 'lbs/foot/foot' => {alias=>'pound per foot squared'}, + 'lbs/ft^2' => {alias=>'pound per foot squared'}, + 'lbs/ft/ft' => {alias=>'pound per foot squared'}, + ); + $ans=Compute("pound per foot squared"); + + +

    + Let f(x) be a function measured in pounds (lb), + where x is measured in feet + (ft). + What are the units of \fpp(x)? +

    +

    + +

    +     +     +     +     + + Problems + + + +

    + Compute the derivative of the given function. +

    +     + + + + + $a=list_random(-9..-2,2..9); + $b=list_random(-9..-2,2..9); + $c=list_random(-9..-2,2..9); + if($envir{problemSeed}==1){$a=7;$b=-5;$c=7;}; + $f=Formula("$a x^2+$b x+$c")->reduce; + $fp=Formula("2*$a x + $b")->reduce; + + +

    + f(x) = +

    +

    + +

    +     +     +     + + + + + $a=list_random(-29..-2,2..29); + $b=list_random(-29..-2,2..29); + $c=list_random(-29..-2,2..29); + $d=list_random(-29..-2,2..29); + if($envir{problemSeed}==1){$a=14;$b=7;$c=11;$d=-9;}; + $f=Formula("$a x^3+$b x^2+$c x+ $d")->reduce; + $fp=Formula("3*$a x^2+2*$b x+$c")->reduce; + + +

    + g(x) = +

    +

    + +

    +     +     +     + + + + + $a=list_random(-9..-2,2..9); + $b=list_random(-8,-7,-5,-4,-2,2,4,5,6,8); + $c=list_random(-9..-2,2..9); + $d=list_random(-9..-2,2..9); + if($envir{problemSeed}==1){$a=9;$b=-8;$c=3;$d=-8;}; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + $f=Formula("$a t^5 + 1/$b t^3 + $c t + $d")->reduce; + $fp=Formula("5*$a t^4 + 3/$b t^2 + $c")->reduce; + + +

    + m(t) = +

    +

    + +

    +     +     +     + + + + + $a=list_random(-19..-2,2..19); + $b=list_random(-19..-2,2..19); + if($envir{problemSeed}==1){$a=9;$b=10;}; + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $f=Formula("$a sin(theta) + $b cos(theta)")->reduce; + $fp=Formula("$a cos(theta) - $b sin(theta)")->reduce; + + +

    + f(\theta) = +

    + + To enter \theta, type theta. + +

    + +

    +     +     +     + + + + + $a=list_random(-9..-2,2..9); + if($envir{problemSeed}==1){$a=6;}; + Context()->variables->are(r=>'Real'); + $f=Formula("$a e^r")->reduce; + $fp=Formula("$a e^r")->reduce; + + +

    + f(r) = +

    +

    + +

    +     +     +     + + + + + $a=non_zero_random(-19,19,1); + $n=random(2,5,1); + $b=non_zero_random(-9,9,1); + $c=non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=10;$n=4;$b=-1,$c=7;}; + Context()->variables->are(t=>'Real'); + $f=Formula("$a t^$n+$b cos(t) +$c sin(t)")->reduce; + $fp=Formula("$a*$n t^($n-1) -$b sin(t) + $c cos(t)")->reduce; + + +

    + g(t) = +

    +

    + +

    +     +     +     + + + + + $a=non_zero_random(-9,9,1); + $b=non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=2;$b=-1;}; + $f=Formula("$a ln(x) + $b x")->reduce; + $fp=Formula("$a/x + $b")->reduce; + + +

    + f(x) = +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->variables->are(s=>'Real'); + $f=Formula("1/4s^4+1/3s^3+1/2s^2+s+1")->reduce; + $fp=Formula("s^3+s^2+s+1")->reduce; + + +

    + p(s) = +

    +

    + +

    +     +     +     + + + + + $a=non_zero_random(-1,1,2); + $b=non_zero_random(-1,1,2); + $c=non_zero_random(-1,1,2); + if($envir{problemSeed}==1){$a=1;$b=-1;$c=-1;}; + Context()->variables->are(t=>'Real'); + $f=Formula("$a e^t +$b sin(t) + $c cos(t)")->reduce; + $fp=Formula("$a e^t +$b cos(t) - $c sin(t)")->reduce; + + +

    + h(t) = +

    +

    + +

    +     +     +     + + + + + $a=non_zero_random(2,9,1); + $b=non_zero_random(2,9,2); + if($envir{problemSeed}==1){$a=5;$b=3;}; + $f=Formula("ln($a x^$b)")->reduce; + $fp=Formula("$b/x")->reduce; + + +

    + f(x) = +

    +

    + +

    +     +     +     + + + + + $a=random(2,20,1); + $b=random(2,9,1); + $c=list_random('sin','cos'); + $d=list_random(2,3,4); + if($envir{problemSeed}==1){$a=17;$b=2;$c='sin';$d=2;}; + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstantFunctions=>0); + Context()->flags->set(reduceConstants=>0); + $f=Formula("ln($a)+e^$b+$c(pi/$d)"); + $fp=Formula("0"); + + +

    + f(t) = +

    +

    + +

    +     +     +     + + + + + $a=random(1,9,1); + $b=random(1,9,1); + if($envir{problemSeed}==1){$a=1;$b=3;}; + Context()->variables->are(t=>'Real'); + $f=Formula("($a+$b t)^2")->reduce; + $fp=Formula("2*($b)^2 t+2*$a*$b")->reduce; + + +

    + g(t) = +

    +

    + +

    +     +     +     + + + + + $a=random(1,4,1); + $b=non_zero_random(-4,4,1); + if($envir{problemSeed}==1){$a=2;$b=-5;}; + $f=Formula("($a x+$b)^3")->reduce; + $fp=Formula("3*($a)^3 x^2 + 6*($a)^2*$b x+3*$a*($b)^2")->reduce; + + +

    + g(x) = +

    +

    + +

    +     +     +     + + + + + $a=random(1,4,1); + $b=non_zero_random(-4,4,1); + if($envir{problemSeed}==1){$a=1;$b=-1;}; + $f=Formula("($a+$b x)^3")->reduce; + $fp=Formula("3*($b)^3 x^2 + 6*($b)^2*$a x+3*$b*($a)^2")->reduce; + + +

    + f(x) = +

    +

    + +

    +     +     +     + + + + + $a=random(1,9,1); + $b=non_zero_random(1,9,1); + if($envir{problemSeed}==1){$a=2;$b=-3;}; + $f=Formula("($a+$b x)^2")->reduce; + $fp=Formula("2*($b)^2 x+2*$a*$b")->reduce; + + +

    + f(x) = +

    +

    + +

    +     +     +     +     + + + + +

    + A property of logarithms is that \log_a(x) = \frac{\log_b(x)}{\log_b(a)}, + for all bases a,b \gt 0,\neq 1. +

    +     + + +

    + Rewrite this identity when b=e, + , using \log_e(x) =\ln(x), with a=10. +

    +     +     + + +

    + Use part (a) to find the derivative of y=\log_{10}(x). +

    +     +     + + +

    + Find the derivative of y=\log_a(x) for any a\gt0,\neq1. +

    + + +     +     + +     + + + +

    + Compute the first four derivatives of the given function. +

    +     + + + + + $n=random(5,10,1); + if($envir{problemSeed}==1){$n=6;}; + $f=Formula("x^$n")->reduce; + $d[0] = $f->D('x')->reduce; + for my $i (1..3) { + $d[$i] = $d[$i-1]->D('x')->reduce; + } + + +

    + f(x) = +

    + + Enter the first derivative. + +

    + +

    + + Enter the second derivative. + +

    + +

    + + Enter the third derivative. + +

    + +

    + + Enter the fourth derivative. + +

    + +

    +     +     +     + + + + + $a=non_zero_random(-9,9,1); + $b=list_random('sin','cos'); + if($envir{problemSeed}==1){$a=2;}; + $f=Formula("$a $b(x)")->reduce; + $d[0] = $f->D('x')->reduce; + for my $i (1..3) { + $d[$i] = $d[$i-1]->D('x')->reduce; + } + + +

    + g(x) = +

    + + Enter the first derivative. + +

    + +

    + + Enter the second derivative. + +

    + +

    + + Enter the third derivative. + +

    + +

    + + Enter the fourth derivative. + +

    + +

    +     +     +     + + + + + $a=non_zero_random(-4,4,1); + $b=random(-4,4,1); + $c=random(-1,1,2); + if($envir{problemSeed}==1){$a=1;$b=0;$c=-1;}; + Context()->variables->are(t=>'Real'); + $f=Formula("$a t^2 + $b t + $c e^t")->reduce; + $d[0] = $f->D('t')->reduce; + for my $i (1..3) { + $d[$i] = $d[$i-1]->D('t')->reduce; + } + + +

    + h(t) = +

    + + Enter the first derivative. + +

    + +

    + + Enter the second derivative. + +

    + +

    + + Enter the third derivative. + +

    + +

    + + Enter the fourth derivative. + +

    + +

    +     +     +     + + + + + ($a,$b) = random_subset(2,2..9); + $c=random(-1,1,2); + $d=random(-1,1,2); + if($envir{problemSeed}==1){$a=4;$b=3;$c=1;$d=-1;}; + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $f=Formula("$c theta^$a + $d theta^$b")->reduce; + $d[0] = $f->D('theta')->reduce; + for my $i (1..3) { + $d[$i] = $d[$i-1]->D('theta')->reduce; + } + + +

    + p(\theta) = +

    + + Enter the first derivative. + To enter \theta, type theta. + +

    + +

    + + Enter the second derivative. + To enter \theta, type theta. + +

    + +

    + + Enter the third derivative. + To enter \theta, type theta. + +

    + +

    + + Enter the fourth derivative. + To enter \theta, type theta. + +

    + +

    +     +     +     + + + + + $a=random(-1,1,2); + $b=random(-1,1,2); + if($envir{problemSeed}==1){$a=1;$b=-1;}; + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $f=Formula("$a sin(theta)+$b cos(theta)")->reduce; + $d[0] = $f->D('theta')->reduce; + for my $i (1..3) { + $d[$i] = $d[$i-1]->D('theta')->reduce; + } + + +

    + f(\theta) = +

    + + Enter the first derivative. + To enter \theta, type theta. + +

    + +

    + + Enter the second derivative. + To enter \theta, type theta. + +

    + +

    + + Enter the third derivative. + To enter \theta, type theta. + +

    + +

    + + Enter the fourth derivative. + To enter \theta, type theta. + +

    + +

    +     +     +     + + + + + $a=random(1,2000,1); + if($envir{problemSeed}==1){$a=1100}; + $f=Formula("$a"); + $d[0] = $f->D('x')->reduce; + for my $i (1..3) { + $d[$i] = $d[$i-1]->D('x')->reduce; + } + + +

    + f(x) = +

    + + Enter the first derivative. + +

    + +

    + + Enter the second derivative. + +

    + +

    + + Enter the third derivative. + +

    + +

    + + Enter the fourth derivative. + +

    + +

    +     +     +     +     + + + +

    + Find the equations of the tangent and normal lines to the graph of the function at the given point. +

    +     + + + + + $a=non_zero_random(-3,3,1); + do {$b=non_zero_random(-9,9,1)} until ($b != -3*$a**2); + if($envir{problemSeed}==1){$a=1;$b=-1}; + Context("Fraction"); + $f=Formula("x^3+$b x")->reduce; + $k=$f->eval(x=>$a); + $fp=$f->D('x'); + $m=$fp->eval(x=>$a); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=$m(x-$a)+$k"); + $n=Formula("y=-1/$m(x-$a)+$k"); + + +

    + f(x)= at x= +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + $a=random(-3,3,1); + $b=non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=0;$b=3;}; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $f=Formula("e^t+$b")->reduce; + $k=$f->substitute(t=>$a); + $fp=$f->D('t'); + $m=$fp->substitute(t=>$a); + Context()->variables->are(y=>'Real',t=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=$m(t-$a)+$k"); + $n=Formula("y=-1/$m(t-$a)+$k"); + + +

    + f(t)= at t= +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0); + Context()->variables->are(y=>'Real',x=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=x-1"); + $n=Formula("y=-(x-1)"); + + +

    + g(x)=\ln(x) at x=1 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + $a=list_random(2,3,4,6); + $b=non_zero_random(-9,9,1); + $c=list_random('sin','cos'); + if($envir{problemSeed}==1){$a=2;$b=4,$c='sin';}; + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $f=Formula("$b $c(x)")->reduce; + %trig = ( + 'sin'=>{ + 2=>Formula("1"), + 3=>Formula("sqrt(3)/2"), + 4=>Formula("sqrt(2)/2"), + 6=>Formula("1/2"), + }, + 'cos'=>{ + 2=>Formula("0"), + 3=>Formula("1/2"), + 4=>Formula("sqrt(2)/2"), + 6=>Formula("sqrt(3)/2"), + }, + ); + %trigrecip = ( + 'sin'=>{ + 2=>Formula("1"), + 3=>Formula("2*sqrt(3)/3"), + 4=>Formula("sqrt(2)"), + 6=>Formula("2"), + }, + 'cos'=>{ + 2=>Formula("DNE"), + 3=>Formula("2"), + 4=>Formula("2*sqrt(2)"), + 6=>Formula("2*sqrt(3)/3"), + }, + ); + %trigd = ( + 'sin'=>'cos', + 'cos'=>'sin', %intentional + ); + $k=Formula("$b")*$trig{$c}{$a}; + $fp=$f->D('x'); + $m=Formula("$b")*$trig{$trigd{$c}}{$a}; + $m = Formula("-$b")->reduce*$trig{$trigd{$c}}{$a} if ($c eq 'cos'); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=$m(x-pi/$a)+$k"); + $t=Formula("y=$k") if ($m == 0); + if (($c eq 'sin') and ($a==2)) + {$n= Formula("x=pi/$a");} + else { + $rm = Formula("1/$m"); + $n= Formula("y=-$rm(x-pi/$a)+$k"); + }; + + +

    + f(x)= at x=\pi/ +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + $a=list_random(2,3,4,6); + $b=non_zero_random(-9,9,1); + $c=list_random('sin','cos'); + if($envir{problemSeed}==1){$a=4;$b=-2,$c='cos';}; + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $f=Formula("$b $c(x)")->reduce; + %trig = ( + 'sin'=>{ + 2=>Formula("1"), + 3=>Formula("sqrt(3)/2"), + 4=>Formula("sqrt(2)/2"), + 6=>Formula("1/2"), + }, + 'cos'=>{ + 2=>Formula("0"), + 3=>Formula("1/2"), + 4=>Formula("sqrt(2)/2"), + 6=>Formula("sqrt(3)/2"), + }, + ); + %trigrecip = ( + 'sin'=>{ + 2=>Formula("1"), + 3=>Formula("2 sqrt(3)/3"), + 4=>Formula("sqrt(2)"), + 6=>Formula("2"), + }, + 'cos'=>{ + 2=>Formula("DNE"), + 3=>Formula("2"), + 4=>Formula("2 sqrt(2)"), + 6=>Formula("2 sqrt(3)/3"), + }, + ); + %trigd = ( + 'sin'=>'cos', + 'cos'=>'sin', %intentional + ); + $k=Formula("$b")*$trig{$c}{$a}; + $fp=$f->D('x'); + $m=Formula("$b")*$trig{$trigd{$c}}{$a}; + $m = Formula("-$b")->reduce*$trig{$trigd{$c}}{$a} if ($c eq 'cos'); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=$m(x-pi/$a)+$k"); + $t=Formula("y=$k") if ($m == 0); + if (($c eq 'sin') and ($a==2)) + {$n= Formula("x=pi/$a");} + else { + $rm = Formula("1/$b")*$trigrecip{$trigd{$c}}{$a}; + $rm = Formula("1/-$b")->reduce*$trigrecip{$trigd{$c}}{$a} if ($c eq 'cos'); + $n= Formula("y=-$rm(x-pi/$a)+$k"); + $t=Formula("y=$k") if ($m == 0); + }; + + +

    + f(x)= at x=\pi/ +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + $a=random(-9,9,1); + $b=non_zero_random(-9,9,1); + $c=non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=5;$b=2;$c=3}; + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $f=Formula("$b x + $c")->reduce; + $k=$f->eval(x=>$a); + $t=Formula("y=$f"); + $n=Formula("y=-1/$b(x-$a)+$k"); + + +

    + f(x)= at x= +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     +     +     +     +
    +
    + The Product and Quotient Rules +

    + showed that, + in some ways, derivatives behave nicely. + The and established that the derivative of f(x) = 5x^2+\sin(x) was not complicated. + We neglected computing the derivative of things like + g(x) = 5x^2\sin(x) and h(x) = \frac{5x^2}{\sin(x) } on purpose; + their derivatives are not as straightforward. + (If you had to guess what their respective derivatives are, + you would probably guess wrong.) + For these, we need the Product and Quotient Rules, respectively, + which are defined in this section. + We begin with the Product Rule. +

    + + + + + + Product Rule + +

    + Let f and g be differentiable functions on an open interval I. + Power Ruledifferentiation + derivativeProduct Rule + Then fg is a differentiable function on I, and + + \lzoo{x}{f(x)g(x)} = \fp(x)g(x) + f(x)g'(x) + . +

    +     +     + + +

    + \lzoo{x}{f(x)g(x)}\neq \fp(x)g'(x)! + While this would be simpler than the , it is wrong. +

    +     + +

    + We practice using this new rule in an example, + followed by an example that demonstrates why this theorem is true. +

    + + + Using the Product Rule + +

    + Use the Product Rule to compute the derivative of y=5x^2\sin(x). + Evaluate the derivative at x=\pi/2. +

    +     + +

    + To make our use of the explicit, + let's set f(x) = 5x^2 and g(x) = \sin(x). + We easily compute/recall that + \fp(x) = 10x and g'(x) = \cos(x). + Employing the rule, we have + + \lzoo{x}{5x^2\sin(x)} \amp= \lzoo{x}{5x^2}\sin(x) + 5x^2\lzoo{x}{\sin(x)} + \amp= 10x\sin(x) + 5x^2\cos(x) + . +

    + +

    + At x=\pi/2, we have + + \yp(\pi/2) = 10\cdot\frac{\pi}2 \sin\left(\frac{\pi}{2}\right) + 5\left(\frac{\pi}{2}\right)^2\cos\left(\frac{\pi}2\right) = 5\pi + . +

    + +

    + We graph y and its tangent line at x=\pi/2, + which has a slope of 5\pi, + in . + While this does not prove + that the Product Rule is the correct way to handle derivatives of products, + it helps validate its truth. +

    + +
    + A graph of y = 5x^2\sin(x) and its tangent line at x=\pi/2 + + + The graph of the function used in this example, along with its tangent line at pi/2 + +

    + The graph of f(x)=5x^2\sin(x) is shown, beginning at (0,0), + and ending at (\pi,0). + From (0,0) the graph moves upward with a slope that initially increases, + and then becomes steady around the point on the graph where x=\pi/2. + The slope then decreases until approximately x=3\pi/4, where the graph reaches its peak. + It then descends steeply toward the point (\pi,0). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-0.1, + xmax=3.5, + ymin=-1.1, + ymax=21, + xtick={1.57,3.14}, + xticklabels={$\frac{\pi}2$,$\pi$}, + ] + \addplot+[infinite,domain=0:3.14,samples=101] {5*(x^2)*sin(x*180/3.14)}; + \addplot [tangentline,domain=0.75:2] {15.7*(x-1.57)+12.34}; + \addplot[soliddot] coordinates{(1.57,12.34)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + + + +

    + We now investigate why the is true. +

    + + + Proof of Product Rule + + +

    + We can use the definition of the derivative to prove . +

    + +

    + By the limit definition, we have + + \lzoo{x}{f(x)g(x)} =\lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} + . +

    + +

    + We now do something a bit unexpected; + add 0 to the numerator + (so that nothing is changed) + in the form of {}-f(x)g(x+h)+f(x)g(x+h), + then do some regrouping as shown. +

    + + + +

    + + \amp\lzoo{x}{f(x)g(x)} = \lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} + \amp \text{ (now add 0 to the numerator) } + \amp = \lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h} + \amp \text{ (regroup) } + \amp = \lim_{h\to0} \frac{\left[f(x+h)g(x+h)-f(x)g(x+h)\right]+\left[f(x)g(x+h)-f(x)g(x)\right]}{h} + \amp \text{ (split fraction)} + \amp = \lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x+h)}{h}+\lim_{h\to0}\frac{f(x)g(x+h)-f(x)g(x)}{h} + \amp \text{ (factor) } + \amp = \lim_{h\to0} \left(\frac{f(x+h)-f(x)}{h}g(x+h)\right)+\lim_{h\to0}\left(f(x)\frac{g(x+h)-g(x)}{h}\right) + \amp \text{ (apply limit properties) } + \amp = \lim_{h\to0} \frac{f(x+h)-f(x)}{h}\cdot \lim_{h\to0}g(x+h)+f(x)\cdot\lim_{h\to0}\frac{g(x+h)-g(x)}{h} + \amp \text{ (apply limits) } + \amp =\fp(x)g(x) + f(x)g'(x) + \amp \text{ (by definition of the derivative)} + . + We have proven the product rule as desired. (In the last step, + we also relied on the fact that since g is differentiable, + it is also continuous, which guarantees that \lim_{h\to0}g(x+h)=g(x).) +

    +     + +

    + It is often true that we can recognize that a theorem is true through its proof yet somehow doubt its applicability to real problems. + In the following example, + we compute the derivative of a product of functions in two ways to verify that the is indeed right. +

    + + + Exploring alternate derivative methods + +

    + Let y = (x^2+3x+1)(2x^2-3x+1). + Find \yp two ways: first, + by expanding the given product and then taking the derivative, + and second, + by applying the . + Verify that both methods give the same answer. +

    +     + +

    + We first expand the expression for y; + a little algebra shows that y = 2x^4+3x^3-6x^2+1. + It is easy to compute \yp: + + \yp = 8x^3+9x^2-12x + . +

    + +

    + Instead, let's apply the to the original factored form: + + \yp \amp = \lzoo{x}{x^2+3x+1}(2x^2-3x+1)+(x^2+3x+1)\lzoo{x}{2x^2-3x+1} + \amp = (2x+3)(2x^2-3x+1)+(x^2+3x+1)(4x-3) + \amp = \left(4x^3-7x+3\right)+\left(4x^3+9x^2-5x-3\right) + \amp = 8x^3+9x^2-12x + . +

    + +

    + The uninformed usually assume that + the derivative of the product is the product of the derivatives. + Thus we are tempted to say that \yp = (2x+3)(4x-3) = 8x^2+6x-9. + Obviously this is not correct. +

    +     + +     + + + Using the Product Rule with a product of three functions + +

    + Let y = x^3\ln(x) \cos(x). + Find \yp. +

    +     + +

    + We have a product of three functions while the only specifies how to handle a product of two functions. + Our method of handling this problem is to simply group the latter two functions together, + and consider y = x^3\cdot\left[\ln(x) \cos(x) \right]. + Following the , we have + + \yp \amp = \lzoo{x}{x^3}\ln(x) \cos(x) + (x^3)\lzoo{x}{\ln(x) \cos(x)} + To evaluate \lzoo{x}{\ln(x) \cos(x)}, we apply the Product Rule again: + \yp \amp = 3x^2\left[\ln(x) \cos(x) \right] + (x^3)\left[\frac1x\cos(x) + \ln(x) (-\sin(x) )\right] + \amp = 3x^2\ln(x) \cos(x) + x^3\frac{1}{x}\cos(x) + x^3\ln(x) (-\sin(x) ) + . +

    + +

    + Recognize the pattern in our answer above: + when applying the to a product of three functions, + there are three terms added together in the final derivative. + Each term contains only one derivative of one of the original functions, + and each function's derivative shows up in only one term. + It is straightforward to extend this pattern to finding the derivative of a product of four or more functions. +

    + +

    + Ultimately though, we would simplify our final computation to: + + \yp=3x^2\ln(x)\cos(x) + x^2\cos(x) + -x^3\ln(x)\sin(x) + + If you check this answer with a CAS, + it may factor and give the answer: + + \yp=-x^2\left[x\ln(x)\sin(x) + \cos(x) + 3\ln(x)\cos(x)\right] + +

    +     + +     + + + + +

    + We consider one more example before discussing another derivative rule. +

    + + + Using the Product Rule + +

    + Find the derivatives of the following functions. +

      +
    1. f(x) = x\ln(x)
      2. +
    2. g(x) = x\ln(x) - x
      3. +
    +

    +     + +

    + Recalling that the derivative of \ln(x) is 1/x, + we use the to find our answers. +

    + +

    +

      +
    1. +

      + Applying the : + + \lzoo{x}{x\ln(x)}\amp= 1\cdot \ln(x) + x\cdot 1/x + \amp = \ln(x) + 1 + . +

      +
      2. +
    2. +

      + Using the result from above, we compute + + \lzoo{x}{x\ln(x) -x}\amp= \ln(x) + 1 - 1 + \amp= \ln(x) + . +

      +
      3. +
    +

    + +

    + This seems significant; + if the natural log function \ln(x) is an important function + (it is), + it seems worthwhile to know a function whose derivative is \ln(x). + We have found one. + (We leave it to the reader to find another; + a correct answer will be very similar to this one.) +

    +     +     + +

    + We have learned how to compute the derivatives of sums, + differences, and products of functions. + We now learn how to find the derivative of a quotient of functions. +

    + + + Quotient Rule + +

    + Let f and g be differentiable functions defined on an open interval I, + where g(x) \neq 0 on I. + derivativeQuotient Rule + Quotient Rule + Then f/g is differentiable on I, and + + \lzoo{x}{\frac{f(x)}{g(x)}} = \frac{g(x)\fp(x) - f(x)g'(x)}{g(x)^2} + . +

    +     +     + + + +

    + The is not hard to use, + although it might be a bit tricky to remember. + A useful mnemonic works as follows. + Consider a fraction's numerator and denominator as HI + and LO, respectively. + Then + + \lzoo{x}{\frac{\text{ HI } }{\text{ LO } }} = \frac{\text{ LO}\cdot\text{dHI}-\text{HI}\cdot\text{dLO }}{\text{ LOLO } } + , + read low dee high minus high dee low, + over low low. Said fast, + that phrase can roll off the tongue, + making it easy to memorize. + The dee high and dee low + parts refer to the derivatives of the numerator and denominator, + respectively. +

    + +

    + Let's practice using the Quotient Rule. +

    + + + Using the Quotient Rule + +

    + Let f(x) = \frac{5x^2}{\sin(x) }. + Find \fp(x). +

    +     + +

    + Directly applying the gives: + + \lzoo{x}{\frac{5x^2}{\sin(x) }} \amp = \frac{\sin(x) \cdot \lzoo{x}{5x^2} - 5x^2\cdot \lzoo{x}{\sin(x)} }{\sin^2(x) } + \amp = \frac{10x\sin(x) - 5x^2\cos(x) }{\sin^2(x) } + . +

    +     + +     + + + +

    + The allows us to fill in holes in our understanding of derivatives of the common trigonometric functions. + We start with finding the derivative of the tangent function. +

    + + + Using the Quotient Rule to find <m>\lzoo{x}{\tan(x)}</m> + +

    + Find the derivative of y=\tan(x). +

    +     + +

    + At first, one might feel unequipped to answer this question. + But recall that \tan(x) = \sin(x) /\cos(x), + so we can apply the . + + \lzoo{x}{\tan(x)}\amp = \lzoo{x}{\frac{\sin(x) }{\cos(x) }} + \amp = \frac{\cos(x) \lzoo{x}{\sin(x)} - \sin(x)\lzoo{x}{\cos(x)}}{\cos^2(x) } + \amp = \frac{\cos(x) \cos(x) - \sin(x) (-\sin(x) )}{\cos^2(x) } + \amp = \frac{\cos^2(x) +\sin^2(x) }{\cos^2(x) } + \amp = \frac{1}{\cos^2(x) } + \amp = \sec^2(x) + . +

    + +

    + This is a beautiful result. + To confirm its truth, + we can find the equation of the tangent line to y=\tan(x) at x=\pi/4. + The slope is \sec^2(\pi/4) = 2; + y=\tan(x), along with its tangent line, + is graphed in . +

    + +
    + A graph of y=\tan(x) along with its tangent line at x=\pi/4 + + + A graph of the tangent function between its vertical asymptotes at -pi/2 and pi/2, along with a tangent line at pi/4. + +

    + A graph of y=\tan(x) between the vertical asymptotes x=-\pi/2 and x=\pi/2. + As x approaches -\pi/2 from the right, \tan(x) approaches -\infty. + Moving right from the x=-\pi/2 asymptote, the graph initialy climbs steeply, + and then levels off, continuing to climb with a much smaller slope from -\pi/4 to \pi/4, + and passing through the origin. Past x=\pi/4, the graph again climbs steeply toward the vertical asymptote at x=\pi/2. +

    + +

    + A tangent line to the graph at the point (\pi/4,1) is also shown. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1.65, + xmax=1.7, + ymin=-11, + ymax=11, + xtick={-1.57,-.785,.785,1.57}, + xticklabels={$-\frac{\pi}{2}$,$-\frac{\pi}{4}$,$\frac{\pi}{4}$,$\frac{\pi}{2}$}, + ] + \addplot+[infinite,samples=101,domain=-11:11] ({atan(x)*pi/180},{x}); + \addplot [tangentline,domain=-1.0:1.5] {2*(x-pi/4)+1}; + \addplot [soliddot] coordinates{(0.785,1)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + +

    + We include this result in the following theorem about the derivatives of the trigonometric functions. + Recall we found the derivative of + y=\sin(x) in + and stated the derivative of the cosine function in . + The derivatives of the cotangent, + cosecant and secant functions can all be computed directly using + and the . +

    + + + Derivatives of Trigonometric Functions + derivativetrigonometric functions + +

    +

      +
    1. \lzoo{x}{\sin(x)} = \cos(x)
      2. +
    2. \lzoo{x}{\cos(x)} = -\sin(x)
      3. +
    3. \lzoo{x}{\tan(x)} = \sec^2(x)
      4. +
    4. \lzoo{x}{\cot(x)} = -\csc^2(x)
      5. +
    5. \lzoo{x}{\sec(x)} = \sec(x) \tan(x)
      6. +
    6. \lzoo{x}{\csc(x)} = -\csc(x) \cot(x)
      7. +
    +

    +     +     + +

    + To remember the above, + it may be helpful to keep in mind that the derivatives of the trigonometric functions that start with c + have a minus sign in them. +

    + + + Exploring alternate derivative methods + +

    + In + the derivative of f(x) = \frac{5x^2}{\sin(x) } was found using the . + Rewriting f as f(x) = 5x^2\csc(x), + find \fp using + and verify the two answers are the same. +

    +     + +

    + We found in + that \fp(x) = \frac{10x\sin(x) - 5x^2\cos(x) }{\sin^2(x) }. + We now find \fp using the , considering f as f(x) = 5x^2\csc(x). + + \fp(x) \amp = \lzoo{x}{5x^2\csc(x)} + \amp = 5x^2\lzoo{x}{\csc(x)} + \lzoo{x}{5x^2}\csc(x) + \amp = 5x^2\left(-\csc(x) \cot(x) \right) + 10x\csc(x) \amp\amp\text{ (now rewrite trig functions) } + \amp = 5x^2\cdot \frac{-1}{\sin(x) }\cdot \frac{\cos(x) }{\sin(x) } + \frac{10x}{\sin(x) } + \amp = \frac{-5x^2\cos(x) }{\sin^2(x) }+\frac{10x}{\sin(x) } \amp\amp\text{ (get common denominator) } + \amp = \frac{10x\sin(x) - 5x^2\cos(x) }{\sin^2(x) } + . +

    + +

    + Finding \fp using either method returned the same result. + At first, the answers looked different, + but some algebra verified they are the same. + In general, there is not one final form that we seek; + the immediate result from the Product Rule is fine. + Work to simplify your results into a form that is most readable and useful to you. +

    +     +     + +

    + The gives other useful results, + as shown in the next example. +

    + + + Using the Quotient Rule to expand the Power Rule + +

    + Find the derivatives of the following functions. +

      +
    1. +

      + f(x) = \dfrac{1}{x} +

      +
      2. +
    2. +

      + f(x)= \dfrac{1}{x^n}, where n \gt 0 is an integer. +

      +
      3. +
    +

    +     + +

    + We employ the . +

      +
    1. +

      + + \fp(x)\amp = \frac{x\cdot 0 - 1\cdot 1}{x^2} + \amp = -\frac{1}{x^2} + +

      +
      2. +
    2. +

      + + \fp(x)\amp = \frac{x^n\cdot 0 - 1\cdot nx^{n-1}}{(x^n)^2} + \amp = -\frac{nx^{n-1}}{x^{2n}} + \amp = -\frac{n}{x^{n+1}} + . +

      +
      3. +
    +

    +     + +     + +

    + The derivative of y=\frac{1}{x^n} turned out to be rather nice. + It gets better. + Consider: + + \lzoo{x}{\frac{1}{x^n}} \amp = \lzoo{x}{x^{-n}} \amp\amp\text{ (apply result from } \text{)} + \amp = -\frac{n}{x^{n+1}} \amp\amp \text{ (rewrite algebraically) } + \amp = -nx^{-(n+1)}\amp + \amp = -nx^{-n-1}\amp + . +

    + +

    + This is reminiscent of the : + multiply by the power, then subtract 1 from the power. + We now add to our previous Power Rule, + which had the restriction of n \gt 0. +

    + + + Power Rule with Integer Exponents + +

    + Let f(x) = x^n, where n\neq 0 is an integer. + derivativePower Rule + Power Ruledifferentiation + Then + + \fp(x) = n\cdot x^{n-1} + . +

    +     +     + +

    + Taking the derivative of many functions is relatively straightforward. + It is clear + (with practice) + what rules apply and in what order they should be applied. + Other functions present multiple paths; + different rules may be applied depending on how the function is treated. + One of the beautiful things about calculus is that there is not the right way; + each path, when applied correctly, + leads to the same result, the derivative. + We demonstrate this concept in an example. +

    + + + Exploring alternate derivative methods + +

    + Let f(x) = \frac{x^2-3x+1}{x}. + Find \fp(x) in each of the following ways: +

      +
    1. +

      + By applying the , +

      +
      2. +
    2. +

      + by viewing f as f(x) = \left(x^2-3x+1\right)\cdot x^{-1} and applying the and , and +

      +
      3. +
    3. +

      + by simplifying first through division. +

      +
      4. +
    + Verify that all three methods give the same result. +

    +     + +

    +

      +
    1. +

      + Applying the gives: + + \fp(x) \amp =\frac{x\cdot\lzoo{x}{x^2-3x+1}-\left(x^2-3x+1\right)\lzoo{x}{x}}{x^2} + \amp = \frac{x\cdot(2x-3)-\left(x^2-3x+1\right)\cdot 1}{x^2} + \amp = \frac{x^2-1}{x^2} + \amp = 1-\frac{1}{x^2} + . +

      +
      2. +
    2. +

      + By rewriting f, + we can apply the and as follows: + + \fp(x) \amp = \left(x^2-3x+1\right)\lzoo{x}{x^{-1}} + \lzoo{x}{x^2-3x+1} x^{-1} + \amp = \left(x^2-3x+1\right)\cdot (-1)x^{-2} + (2x-3)\cdot x^{-1} + \amp = -\frac{x^2-3x+1}{x^2}+\frac{2x-3}{x} + \amp = -\frac{x^2-3x+1}{x^2}+\frac{2x^2-3x}{x^2} + \amp = \frac{x^2-1}{x^2} = 1-\frac{1}{x^2} + , + the same result as above. +

      +
      3. +
    3. +

      + As x\neq 0, we can divide through by x first, + giving f(x) = x-3+x^{-1}. + Now apply the . + + \fp(x) = 1-\frac{1}{x^2} + , + the same result as before. +

      +
      4. +
    +

    +     + +     + +

    + + demonstrates three methods of finding \fp. + One is hard pressed to argue for a best method + as all three gave the same result without too much difficulty, + although it is clear that using the required more steps. + Ultimately, the important principle to take away from this is: + reduce the answer to a form that seems + simple and easy to interpret. + In that example, + we saw different expressions for \fp, including: + + \amp 1-\frac{1}{x^2} + \amp \frac{x\cdot(2x-3)-\left(x^2-3x+1\right)\cdot 1}{x^2} + \amp \left(x^2-3x+1\right)\cdot (-1)x^{-2} + (2x-3)\cdot x^{-1} + . +

    + +

    + They are equal; they are all correct; + only the first is simple. + Work to make answers simple. +

    + +

    + In the next section we continue to learn rules that allow us to more easily compute derivatives than using the limit definition directly. + We have to memorize the derivatives of a certain set of functions, + such as the derivative of \sin(x) is \cos(x). + The , + , + , + and show us how to find the derivatives of certain combinations of these functions. + The next section shows how to find the derivatives when we compose + these functions together. +

    + + + + Terms and Concepts + + + + +

    + + The Product Rule states that \lzoo{x}{x^2\sin(x)}= 2x\cos(x). +

    +     + +

    + The product rule results in something more complicated than the product of the derivatives. + Double-check the statement of . +

    +     + +     + + + + +

    + + The Quotient Rule states that \lzoo{x}{\frac{x^2}{\sin(x) }} = \frac{\cos(x) }{2x}. +

    +     + +

    + The quotient rule results in something more complicated than the quotient of the derivatives. + Double-check the statement of . +

    +     + +     + + + + +

    + + The derivatives of the trigonometric functions that start with c + have minus signs in them. +

    +     + +

    + We could even go so far as to say this is true of the trig functions starting with co! +

    +     + +     + + + + +

    + What derivative rule is used to extend the Power Rule to include negative integer exponents? +

    +

    + + +

    +     + + + + + quotient|quotient rule|the quotient rule + + + + + +     + + + + +

    + + Regardless of the function, + there is always exactly one right way of computing its derivative. +

    +     + +

    + Sometimes you have options! +

    +     + +     + + + + +

    + In your own words, + explain what it means to make your answers clear. +

    + +     + + + +     +     + + + Problems + + + +

    +

      +
    1. + Use the Product Rule to differentiate the function. +
      2. +
    2. + Manipulate the function algebraically and differentiate without using the Product Rule. +
      3. +
    3. + Show that the two derivatives are equivalent. +
      4. +
    +

    +     + + + + + $x = 'x'; + Context()->variables->are($x=>'Real'); + $f = Formula("$x($x^2+3$x)"); + $d = $f->D($x); + + +

    + f() = +

    +

    + +

    +     +     +     + + + + + $x = 'x'; + Context()->variables->are($x=>'Real'); + $f = Formula("2$x^2(5$x^3)"); + $d = $f->D($x); + + +

    + f() = +

    +

    + +

    +     +     +     + + + + + $x = 's'; + Context()->variables->are($x=>'Real'); + $f = Formula("(2$x-1)($x+4)"); + $d = $f->D($x); + + +

    + f() = +

    +

    + +

    +     +     +     + + + + + $x = 'x'; + Context()->variables->are($x=>'Real'); + $f = Formula("($x^2+5)(3-$x^3)"); + $d = $f->D($x); + + +

    + f() = +

    +

    + +

    +     +     +     +     + + + +

    +

      +
    1. + Use the Quotient Rule to differentiate the function. +
      2. +
    2. + Manipulate the function algebraically and differentiate without using the Quotient Rule. +
      3. +
    3. + Show that the two derivatives are equivalent. +
      4. +
    +

    +     + + + + + $x = 'x'; + Context()->variables->are($x=>'Real'); + $f = Formula("($x^2+3)/$x"); + $d = $f->D($x); + + +

    + f(x) = +

    +

    + +

    +     +     +     + + + + + $x = 'x'; + Context()->variables->are($x=>'Real'); + $f = Formula("($x^3-2$x^2)/(2$x^2)"); + $d = $f->D($x); + + +

    + f(x) = +

    +

    + +

    +     +     +     + + + + + $x = 's'; + Context()->variables->are($x=>'Real'); + $f = Formula("3/(4$x^3)"); + $d = $f->D($x); + + +

    + f(x) = +

    +

    + +

    +     +     +     + + + + + $x = 't'; + Context()->variables->are($x=>'Real'); + $f = Formula("($x^2-1)/($x+1)"); + $d = $f->D($x); + + +

    + f(x) = +

    +

    + +

    +     +     +     +     + + + +

    + Compute the derivative of the given function. +

    +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$fname='f';$x='x';$trig='sin';}; + Context()->variables->are($x => 'Real'); + $f = Formula("$x $trig($x)"); + $df = $f->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig = list_random('sin','cos'); + $n = random(2,9,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$trig='cos';$n=2;}; + Context()->variables->are($x => 'Real'); + $f = Formula("$x^$n $trig($x)"); + $df = $f->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + if($envir{problemSeed}==1){$fname='f';$x='x';}; + Context()->variables->are($x => 'Real'); + $f = Formula("e^$x ln($x)"); + $df = $f->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig = list_random('sec','csc','tan','cot'); + $n = random(2,9,1); + $a = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$fname='f';$x='t';$trig='csc';$n=2;$a=-4;}; + Context()->variables->are($x => 'Real'); + $f = Formula("1/$x^$n ($trig($x)+$a)")->reduce; + $df = $f->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + ($a,$b) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$fname='g';$x='x';$a=7;$b=-5;}; + Context()->variables->are($x => 'Real'); + $f = Formula("($x+$a)/($x+$b)")->reduce; + $df = $f->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig = list_random('sin','cos'); + $m = random(2,9,1); + $n = random(2,9,1); + $a = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$fname='g';$x='t';$trig='cos';$m=2;$n=5;$a=-2;}; + Context()->variables->are($x => 'Real'); + $f = Formula("$x^$n/($trig($x)+$a $x^$m)")->reduce; + $df = $f->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig = list_random('cot','sec','csc'); + Context()->variables->are($x => 'Real'); + $pm = list_random('+','-'); + if($envir{problemSeed}==1){$fname='h';$x='x';$trig='cot';$pm='-';}; + Context()->variables->are($x => 'Real'); + $formula = Formula("$trig($x) $pm e^$x"); + $df = $formula->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig = list_random('tan','cot','sec','csc'); + if($envir{problemSeed}==1){$fname='f';$x='x';$trig='tan';}; + Context()->variables->are($x => 'Real'); + $formula = Formula("$trig($x) ln($x)"); + $df = $formula->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $a = random(3,9,1); + $b = non_zero_random(-9,9,1); + $c = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$fname='h';$x='t';$a=7;$b=6;$c=-2;}; + Context()->variables->are($x => 'Real'); + $formula = Formula("$a $x^2+$b$x+$c")->reduce; + $df = $formula->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $a = list_random(-9..-2,2..9); + $n = random(2,5,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$a=2;$n=3;}; + Context()->variables->are($x => 'Real'); + $formula = Formula("($x^($n+1)+$a $x^($n))/($x+$a)")->reduce; + $df = Formula("$n $x^($n-1)")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + ($a,$b,$c) = random_subset(3,3..9); + if($envir{problemSeed}==1){$fname='f';$x='x';$a=3;$b=8;$c=7;}; + Context()->variables->are($x => 'Real'); + $formula = Formula("($a $x^2 + $b $x + $c) e^($x)")->reduce; + $df = Formula("($a $x^2 + ($b+2*$a) $x + ($c+$b)) e^($x)")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + ($a,$b) = random_subset(2,3,5,7,9); + $pm = list_random('+','-'); + if($envir{problemSeed}==1){$fname='g';$x='t';$a=5;$b=3;$pm='-';}; + Context()->variables->are($x => 'Real'); + $formula = Formula("($x^$a $pm $x^$b) / e^($x)")->reduce; + $df = Formula("(-$x^$a + $a $x^($a-1) - $pm $x^$b $pm $b $x^($b-1)) / e^($x)")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + ($a,$d) = random_subset(2,1..30); + ($b,$c,$e) = random_subset(3,-30..-1,1..30); + if($envir{problemSeed}==1){$fname='f';$x='x';$a=16;$b=24;$c=3;$d=7;$e=-1;}; + Context()->variables->are($x => 'Real'); + $formula = Formula("($a $x^3+$b $x^2 + $c $x) (($d $x + $e)/($a $x^3+$b $x^2 + $c $x))")->reduce; + $df = Formula("$d")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $n = random(4,6,1); + $trig = list_random('cos','sin','sec','csc','tan','cot'); + if($envir{problemSeed}==1){$fname='f';$x='t';$n=5;$trig='sec';}; + Context()->variables->are($x => 'Real'); + $formula = Formula("$x^$n($trig($x) + e^$x)"); + $df = $formula->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $a = random(2,9,1); + ($triga,$trigb) = random_subset(2,'cos','sin','sec','csc','tan','cot'); + if($envir{problemSeed}==1){$fname='f';$x='x';$a=3;$triga='sin';$trigb='cos';}; + Context()->variables->are($x => 'Real'); + $formula = Formula("$triga($x)/($trigb($x) + $a)"); + $df = $formula->D("$x")->reduce; + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $x = Formula('theta'); + $n = random(2,6,1); + $trig = list_random('cos','sin','sec','csc','tan','cot'); + if($envir{problemSeed}==1){$fname='f';$n=3;$trig='sin';}; + $formula = Formula("theta^$n $trig(theta) + $trig(theta)/theta^$n"); + $df = $formula->D("$x"); + + +

    + () = +

    + + To enter \theta, type theta. + +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + ($triga,$trigb) = random_subset(2,'cos','sin','sec','csc','tan','cot'); + if($envir{problemSeed}==1){$fname='f';$x='x';$triga='cos';$trigb='tan';}; + Context()->variables->are($x => 'Real'); + $formula = Formula("$triga($x)/$x + $x/$trigb($x)"); + $df = $formula->D("$x"); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $a = random(2,5,1); + $b = list_random(3,4,6); + $c = random(1,5,1); + $trig = list_random('\cos','\sin','\sec','\csc','\tan','\cot'); + if($envir{problemSeed}==1){$fname='g';$x='x';$a=2;$b=4;$c=1;$trig='sin';}; + Context()->variables->are($x => 'Real'); + $formula = "e^$a($trig(" . '\pi' . "/$b) - $c)"; + $df = Formula("0"); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + ($a,$b) = random_subset(2,2..9); + $trig = list_random("sin($x)*cos($x)","sec($x)*csc($x)"); + $pm = list_random('+','-'); + if($envir{problemSeed}==1){$fname='g';$x='t';$a=4;$b=3;$trig="sin($x)*cos($x)";$pm='-';}; + Context()->variables->are($x => 'Real'); + $formula = Formula("$a $x^$b e^$x $pm $trig"); + $df = $formula->D("$x"); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $n = random(2,3,1); + ($a,$b) = random_subset(2,-9..-1,1..9); + ($triga,$trigb) = random_subset(2,'sin','cos'); + if($envir{problemSeed}==1){$fname='h';$x='t';$n=2;$a=3;$b=2;$triga='sin';$trigb='cos';}; + Context()->variables->are($x => 'Real'); + $formula = Formula("($x^$n $triga($x)+$a)/($x^$n $trigb($x)+$b)")->reduce; + $df = $formula->D("$x"); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $n = random(2,5,1); + $tran = list_random("e^$x","ln($x)"); + $trig = list_random('cos','sin','sec','csc','tan','cot'); + if($envir{problemSeed}==1){$fname='f';$x='x';$n=2;$tran="e^$x";$trig='tan';}; + Context()->variables->are($x => 'Real'); + $formula = Formula("$x^$n $tran $trig($x)")->reduce; + $df = $formula->D("$x"); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $a = random(2,9,1); + $trig = list_random("sin($x) sec($x)","cos($x) csc($x)","cos($x) tan($x)","sin($x) cot($x)"); + if($envir{problemSeed}==1){$fname='g';$x='x';$a=2;$trig="sin($x) sec($x)";}; + Context()->variables->are($x => 'Real'); + $formula = Formula("$a $x $trig")->reduce; + $df = $formula->D("$x"); + + +

    + () = +

    +

    + +

    +     +     +     +     + + + +

    + Find the equations of the tangent and normal lines to the graph of g at the indicated point. +

    +     + + + + + $n = random(2,3,1); + $a = list_random(-9..-1,1..9); + if($envir{problemSeed}==1){$n=2;$a=2}; + Context("Fraction"); + $formula = Formula("e^x(x^$n+$a)")->reduce; + $x0=0; + $y0=Fraction($formula->eval(x=>$x0)); + $df=$formula->D('x'); + $m=Fraction($df->eval(x=>$x0)); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=$m x+$y0")->reduce; + if ($m == 0) { + $n=Formula("x=$x0"); + } + else { + $mn = Fraction(-1/$m); + $n=Formula("y=$mn x+$y0")->reduce; + } + + +

    + g(x) = at \left(,\right) +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0); + $trig = list_random('sin','cos','tan','cot','sec','csc'); + if ($trig eq 'sin' or $trig eq 'csc') { + $x0 = list_random(Formula("pi/6"),Formula("pi/4"),Formula("pi/3"),Formula("pi/2"),Formula("2pi/3"),Formula("3pi/4"),Formula("5pi/6"),Formula("7pi/6"),Formula("5pi/4"),Formula("4pi/3"),Formula("3pi/2"),Formula("5pi/3"),Formula("7pi/4"),Formula("11pi/6")); + } else { + $x0 = list_random(Formula("pi/6"),Formula("pi/4"),Formula("pi/3"),Formula("2pi/3"),Formula("3pi/4"),Formula("5pi/6"),Formula("7pi/6"),Formula("5pi/4"),Formula("4pi/3"),Formula("5pi/3"),Formula("7pi/4"),Formula("11pi/6")); + } + if($envir{problemSeed}==1){$trig='sin';$a=Formula("3pi/2")}; + $formula = Formula("x $trig(x)"); + $y0=$formula->eval(x=>($x0->eval(x=>0)))->value; + $sign = ($y0 > 0) ? '' : '-'; + Context("Fraction"); + $frac = Fraction(($y0/pi)**2); + ($num,$den)=$frac->value; + if ($num == 1 or $num == 4 or $num == 9 or $num == 16) {$N = sqrt($num)} else {$N = "sqrt($num)"}; + if ($den == 1 or $den == 4 or $den == 9 or $den == 16) {$D = sqrt($den)} else {$D = "sqrt($den)"}; + $y0 = Formula("($sign$N pi)/$D"); + $df=$formula->D('x'); + $m=$df->eval(x=>($x0->eval(x=>0))); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=$m (x-$x0)+$y0")->reduce; + if ($m == 0) { + $n=Formula("x=$x0"); + } + else { + $mn = Fraction(-1/$m); + $n=Formula("y=$mn (x-$x0)+$y0")->reduce; + } + + +

    + g(x) = at \left(,\right) +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + $a = list_random(-9..-1,1..9); + $x0=$a+list_random(-2,-1,1,2); + if($envir{problemSeed}==1){$a=1;$x0=2;}; + Context("Fraction"); + $formula = Formula("x^2/(x-$a)"); + $y0=Fraction($formula->eval(x=>$x0)); + $df=$formula->D('x'); + $m=Fraction($df->eval(x=>$x0)); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=$m (x-$x0)+$y0")->reduce; + if ($m == 0) { + $n=Formula("x=$x0"); + } + else { + $mn = Fraction(-1/$m); + $n=Formula("y=$mn (x-$x0)+$y0")->reduce; + } + + +

    + g(x) = at \left(,\right) +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + ($a,$b) = random_subset(2,-9..-1,1..9); + $trig = list_random('cos','sin'); + if($envir{problemSeed}==1){$a=-8;$b=1;$trig='cos'}; + Context("Fraction"); + $formula = Formula("($trig(x)+$a x)/(x+$b)")->reduce; + $x0=0; + $y0=Fraction($formula->eval(x=>$x0)); + $df=$formula->D('x'); + $m=Fraction($df->eval(x=>$x0)); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=$m (x-$x0)+$y0")->reduce; + if ($m == 0) { + $n=Formula("x=$x0"); + } + else { + $mn = Fraction(-1/$m); + $n=Formula("y=$mn (x-$x0)+$y0")->reduce; + } + + +

    + g(x) = at \left(,\right) +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     +     + + + +

    + Find the x-values where the graph of the function has a horizontal tangent line. +

    +     + + + + + Context("Fraction"); + Context()->noreduce('(-x)+y','(-x)-y'); + $a = non_zero_random(-9,9,1); + ($b,$c) = random_subset(2,-30..-1,1..30); + if($envir{problemSeed}==1){$a=6;$b=-18;$c=-24;}; + $formula = Formula("$a x^2 + $b x + $c")->reduce; + @x = (Fraction(-$b,2*$a)); + $answer = List(@x); + + +

    + f(x) = +

    + + Use commas to separate more than one x-value. + If there are no such x-values, + answer with NONE. + +

    + +

    +     +     +     + + + + + $points = List("0"); + + +

    + f(x) = x\sin(x) on [-1,1] +

    + + Use commas to separate more than one x-value. + If there are no such x-values, + answer with NONE. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->noreduce('(-x)+y','(-x)-y'); + ($a,$b,$c) = random_subset(3,-3..-1,1..3); + if($envir{problemSeed}==1){$a=1;$b=1;$c=1;}; + $formula = Formula("($a x)/($b x + $c)")->reduce; + $points = Compute("NONE"); + + +

    + f(x) = +

    + + Use commas to separate more than one x-value. + If there are no such x-values, + answer with NONE. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->noreduce('(-x)+y','(-x)-y'); + ($a,$b,$c) = random_subset(3,-3..-1,1..3); + if($envir{problemSeed}==1){$a=1;$b=1;$c=1;}; + $formula = Formula("($a x^2)/($b x + $c)")->reduce; + @x = (Fraction(0),Fraction(-(2*$c)/$b)); + $points = List(@x); + + +

    + f(x) = +

    + + Use commas to separate more than one x-value. + If there are no such x-values, + answer with NONE. + +

    + +

    +     +     +     +     + + + +

    + Find the requested higher order derivative. +

    +     + + + + + $answer=Formula("2cos(x)-x*sin(x)"); + + +

    + \fpp(x), where f(x) = x\sin(x) +

    +

    + +

    +     +     +     + + + + + $answer=Formula("-4cos(x)+x*sin(x)"); + + +

    + f^{(4)}(x), where f(x) = x\sin(x) +

    +

    + +

    +     +     +     + + + + + $trig = list_random('csc','cot','sec','tan'); + if($envir{problemSeed}==1){$trig = 'csc'}; + $formula = Formula("$trig(x)"); + $answer=$formula->D('x')->D('x'); + + +

    + f''(x), where f(x) = +

    +

    + +

    +     +     +     + + + + + ($a,$b,$c,$d) = random_subset(4,-9..-1,1..9); + $n = random(7,9,1); + if($envir{problemSeed}==1){$a=-5;$b=2;$c=1;$d=-7;$n=8;}; + $formula = Formula("(x^3+$a x+$b)(x^2+$c x+$d)")->reduce; + $answer=Formula("0"); + + +

    + f^{()}(x), where f(x) = +

    +

    + +

    +     +     +     +     +     + +     +
    +
    + The Chain Rule + +

    + We have covered almost all of the derivative rules that deal with combinations of two + (or more) + functions. + The operations of addition, subtraction, multiplication + (including by a constant) + and division led to the , + the , + the , + the and the . + To complete the list of differentiation rules, + we look at the last way two + (or more) + functions can be combined: + the process of composition ( one function + inside another). +

    + + + +

    + One example of a composition of functions is f(x) = \cos(x^2). + We currently do not know how to compute this derivative. + If forced to guess, one might guess \fp(x) = -\sin(2x), + where we recognize -\sin(x) as the derivative of \cos(x) and 2x as the derivative of x^2. + However, this is not the case; + \fp(x)\neq -\sin(2x). + One way to see this is to examine the graph of + y=\cos\mathopen{}\left(x^2\right)\mathclose{} in + and its tangent line at x=\pi/2. + Clearly the slope of the tangent line there is nonzero, + but -2\sin(2\cdot\pi/2)=0. + So it can't be correct to say that \yp=-\sin(2x). +

    + +
    + A graph of y=\cos(x^2) and a tangent line at \pi/2 + + + A cosine wave with increasing frequency with a negatively sloped tangent line. + + +

    + A cosine wave with increasing frequency. + The distance between the peaks decreases as x increases. + There is a point drawn on the curve at x= \frac{\pi}{2}, roughly at y=-0.8. + A tangent line to the point is drawn, sloping downward. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ymin=-1.5,ymax=1.5, + xmin=-.2,xmax=3.5, + clip=false, + ] + \addplot+[infinite,domain=-0.2:3.5,samples=101] {cos(deg(x^2))}; + \addplot[tangentline,domain=1:1.93] {-sin(deg((pi/2)^2))*2*pi/2*(x-pi/2)+cos(deg((pi/2)^2))}; + \addplot[soliddot] coordinates {(1.571,-0.781)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +

    + In + we'll see the correct way to compute the derivative of \sin\mathopen{}\left(x^2\right)\mathclose{}, + which employs the new rule this section introduces, the Chain Rule. +

    + +

    + Before we define this new rule, + recall the notation for composition of functions. + We write (f \circ g)(x) or f(g(x)), + read as f of g of x, + to denote composing f with g. + In shorthand, we simply write + f \circ g or f(g) and read it as + f of g. + Before giving the corresponding differentiation rule, + we note that the rule extends to multiple compositions like + f(g(h(x))) or f(g(h(j(x)))), etc. +

    + +

    + To motivate the rule, + let's look at three derivatives we can already compute. +

    + + + Exploring similar derivatives + +

    + Find the derivatives of F_1(x) = (1-x)^2, + F_2(x) = (1-x)^3, and F_3(x) = (1-x)^4. + (We'll see later why we are using subscripts for different functions and an uppercase F.) +

    +     + +

    + In order to use the rules we already have, + we must first expand each function as + + F_1(x)\amp = 1 - 2x + x^2 + F_2(x)\amp = 1 - 3x + 3x^2 - x^3 + F_3(x)\amp = 1 - 4x + 6x^2 - 4x^3 + x^4 + + It is not hard to see that: + + F_1'(x)\amp = -2 + 2x + F_2'(x)\amp = -3 + 6x - 3x^2 + F_3'(x)\amp = -4 + 12x - 12x^2 + 4x^3 + + An interesting fact is that these can be rewritten as: + + F_1'(x)\amp = -2(1-x) + F_2'(x)\amp = -3(1-x)^2 + F_3'(x)\amp = -4(1-x)^3 + +

    + +

    + A pattern might jump out at you; + note how the we end up multiplying by the old power and the new power is reduced by 1. + We also always multiply by (-1). +

    + +

    + Recognize that each of these functions is a composition, + letting g(x) = 1-x: + + F_1(x)\amp = f_1(g(x)),\amp\amp \text{ where } f_1(x) = x^2, + F_2(x)\amp = f_2(g(x)),\amp\amp \text{ where } f_2(x) = x^3, + F_3(x)\amp = f_3(g(x)),\amp\amp \text{ where } f_3(x) = x^4 + . +

    + +

    + We'll come back to this example after giving the formal statements of the Chain Rule; + for now, we are just illustrating a pattern. +

    +     +     + + + + + Chain Rule + +

    + Let g be a differentiable function on an interval I, + let the range of g be a subset of the interval J, + and let f be a differentiable function on J. + + derivativeChain Rule + Chain Rule + Then y=f(g(x)) is a differentiable function on I, and + + \yp = \fp(g(x))\cdot \gp(x) + . +

    +     +     + + + +

    + Here is the Chain Rule in words: +

    + +
    +

    + The derivative of the outside function, + evaluated at the inside function, + multiplied by the derivative of the inside function. +

    +
    + + + + + + + + + +

    + To help understand the Chain Rule, + we return to . +

    + + + Using the Chain Rule + +

    + Use the Chain Rule to find the derivatives of the functions F_1(x), + F_2(x), and F_3(x), + as given in . +

    +     + +

    + + ended with the recognition that each of the given functions was actually a composition of functions. + To avoid confusion, we ignore most of the subscripts here. +

    + +

    +

      +
    • + <m>F_1(x) = (1-x)^2</m> +

      + We found that + + y=(1-x)^2 = f(g(x)) + , + where f(x) = x^2 and g(x) = 1-x. + To find \yp, + we apply the . + We need to note that \fp(x)=2x and \gp(x)=-1. +

      + +

      + Part of the uses \fp(g(x)). + This means substitute g(x) for x in the equation for \fp(x). + That is, \fp(x) = 2(1-x). + Finishing out the we have + + \yp \amp = \fp(g(x))\cdot \gp(x) + \amp = 2(1-x)\cdot (-1) + \amp = -2(1-x) + \amp = 2x-2 + . +

      +
      • + +
    • + <m>F_2(x) = (1-x)^3</m> +

      + Let y = (1-x)^3 = f(g(x)), + where f(x) = x^3 and g(x) = (1-x). + We have \fp(x) = 3x^2, + so \fp(g(x)) = 3(1-x)^2. + The then states + + \yp \amp = \fp(g(x))\cdot \gp(x) + \amp = 3(1-x)^2\cdot (-1) + \amp = -3(1-x)^2 + . +

      +
      • + +
    • + <m>F_3(x) = (1-x)^4</m> +

      + Finally, when y = (1-x)^4, + we have f(x)= x^4 and g(x) = (1-x). + Thus \fp(x) = 4x^3 and \fp(g(x)) = 4(1-x)^3. + Thus + + \yp \amp = \fp(g(x))\cdot \gp(x) + \amp = 4(1-x)^3\cdot (-1) + \amp = -4(1-x)^3 + . +

      +
      • +
    +

    +     +     + +

    + + demonstrated a particular pattern: + when f(x)=x^n, then \yp =n\cdot (g(x))^{n-1}\cdot \gp(x). + This is called the Generalized Power Rule. +

    + + + Generalized Power Rule + +

    + Let g(x) be a differentiable function and let n\neq 0 be an integer. + + derivativeGeneralized Power Rule + Generalized Power Rule + Then + + \lzoo{x}{g(x)^n} = n\cdot\left(g(x)\right)^{n-1}\cdot \gp(x) + . +

    +     +     + +

    + This allows us to quickly find the derivative of functions like y = (3x^2-5x+7+\sin(x) )^{20}. + While it may look intimidating, + the Generalized Power Rule states that + + \yp = 20(3x^2-5x+7+\sin(x) )^{19}\cdot (6x-5+\cos(x) ) + . +

    + +

    + Treat the derivative-taking process step-by-step. + In the example just given, + first multiply by 20, then rewrite the inside of the parentheses, + raising it all to the 19th power. + Then think about the derivative of the expression inside the parentheses, + and multiply by that. +

    + +

    + We now consider more examples that employ the . +

    + + + Using the Chain Rule + +

    + Find the derivatives of the following functions: + +

      +
    1. y = \sin(2x).
      2. + +
    2. y= \ln(4x^3-2x^2).
      3. + +
    3. y = e^{-x^2}.
      4. +
    +

    +     + +

    +

      +
    1. +

      + Consider y = \sin(2x). + Recognize that this is a composition of functions, + where f(x) = \sin(x) and g(x) = 2x. + Thus + + \yp \amp = \fp(g(x))\cdot \gp(x) + \amp = \cos(2x)\cdot \lzoo{x}{2x} + \amp = \cos(2x)\cdot 2 + \amp = 2\cos(2x) + . +

      +
      2. + +
    2. +

      + Recognize that y = \ln\mathopen{}\left(4x^3-2x^2\right)\mathclose{} is the composition of + f(x) = \ln(x) and g(x) = 4x^3-2x^2. + Also, recall that + + \lzoo{x}{\ln(x)} = \frac{1}{x} + . + This leads us to: + + \yp \amp = \frac{1}{4x^3-2x^2} \cdot \lzoo{x}{4x^3-2x^2} + \amp = \frac{1}{4x^3-2x^2} \cdot \left(12x^2-4x\right) + \amp = \frac{12x^2-4x}{4x^3-2x^2} + \amp = \frac{4x(3x-1)}{2x(2x^2-x)} + \amp = \frac{2(3x-1)}{2x^2-x} + . + Note that \ln\mathopen{}\left(4x^3-2x^2\right)\mathclose{}=\ln\mathopen{}\left(4x^2(x-1/2)\right)\mathclose{} was only defined for x \gt 1/2, + so the result of \yp=\frac{2(3x-1)}{2x^2-x} is only valid for x \gt 1/2 as well. +

      +
      3. + +
    3. +

      + Recognize that y = e^{-x^2} is the composition of + f(x) = e^x and g(x) = -x^2. + Remembering that \fp(x) = e^x, we have + + \yp \amp = e^{-x^2}\cdot \lzoo{x}{-x^2} + \amp = e^{-x^2}\cdot (-2x) + \amp = -2xe^{-x^2} + . +

      +
      4. +
    +

    +     + +     + + + Using the Chain Rule to find a tangent line + +

    + Let f(x) = \cos(x^2). + Find the equation of the line tangent to the graph of f at x=1. +

    +     + +

    + The tangent line goes through the point + (1,f(1)) \approx (1,0.54) with slope \fp(1). + To find \fp, + we need the . +

    + +

    + \fp(x) = -\sin(x^2) \cdot(2x) = -2x\sin(x^2). + Evaluated at x=1, we have \fp(1) = -2\sin(1) \approx -1.68. + Thus the equation of the tangent line is approximated by + + y \approx -1.68(x-1)+0.54 + . +

    + +

    + The tangent line is sketched along with f in . +

    + +
    + f(x) = \cos(x^2) sketched along with its tangent line at x=1 + + + + A wave with increasing frequency with a negatively sloped tangent line at x=1. + + +

    + A cosine wave with increasing frequency as the graph moves further from the y-axis. + To the left, there is one peak and valley before the graph moves to the y-axis, where the graph becomes gradually horizontal. + To the right, the graph has another valley and peak, symmetrical to the left side. + At the point x=1, there is a tangent line drawn. This line has a clear negative slope. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-3.1,xmax=3.1, + ymin=-1.1,ymax=1.1,] + \addplot+[infinite,domain=-3:3,samples=200] ({x},{cos(deg(x^2))});% {}; + \addplot [tangentline,domain=.7:1.3] {-1.68*(x-1)+0.54}; + \addplot [soliddot] coordinates{(1,0.54)}; + \end{axis} + \end{tikzpicture} + + + + +
    +     +     + +

    + The is used often in taking derivatives. + Because of this, + one can become familiar with the basic process and learn patterns that facilitate finding derivatives quickly. + For instance, + + \lzoo{x}{\ln(\text{anything})} = \frac{1}{\text{anything}}\cdot\lzoo{x}{\text{anything}} = \frac{\lzoo{x}{\text{anything}}}{\text{anything}} + . +

    + +

    + A concrete example of this is + + \lzoo{x}{\ln(3x^{15}-\cos(x) +e^x)} = \frac{45x^{14}+\sin(x) +e^x}{3x^{15}-\cos(x) +e^x} + . +

    + +

    + While the derivative may look intimidating at first, look for the pattern. + The denominator is the same as what was inside the natural log function; + the numerator is simply its derivative. +

    + +

    + This pattern recognition process can be applied to lots of functions. + In general, instead of writing anything, + we use u as a generic function of x. + We then say + + \lzoo{x}{\ln(u)} = \frac{u'}{u} + . +

    + +

    + The following is a short list of how the can be quickly applied to familiar functions. + +

      +
    1. \lzoo{x}{u^n} = n\cdot u^{n-1}\cdot u'.
      2. + +
    2. \lzoo{x}{e^u} = e^u \cdot u'.
      3. + +
    3. \lzoo{x}{\sin(u)} = \cos(u) \cdot u'.
      4. + +
    4. \lzoo{x}{\cos(u)} = -\sin(u)\cdot u'.
      5. + +
    5. \lzoo{x}{\tan(u)} = \sec^2(u) \cdot u'.
      6. +
    +

    + +

    + Of course, the can be applied in conjunction with any of the other rules we have already learned. + We practice this next. +

    + + + Using the Product, Quotient and Chain Rules + +

    + Find the derivatives of the following functions. + +

      +
    1. f(x) = x^5 \sin(2x^3).
      2. + +
    2. f(x) = \dfrac{5x^3}{e^{-x^2}}.
      3. +
    +

    +     + +

    +

      +
    1. +

      + We must use the and . + Do not think that you must be able to see + the whole answer immediately; + rather, just proceed step-by-step. + + \fp(x) \amp = x^5 \cdot \lzoo{x}{\sin\mathopen{}\left(2x^3\right)\mathclose{}} + \sin\mathopen{}\left(2x^3\right)\mathclose{}\cdot \lzoo{x}{x^5} + \amp = x^5\left(\cos\mathopen{}\left(2x^3\right)\mathclose{}\cdot \lzoo{x}{2x^3} \right) + 5x^4\left(\sin\mathopen{}\left(2x^3\right)\mathclose{} \right) + \amp = x^5\big(6x^2\cos\left(2x^3\right) \big) + 5x^4\big(\sin\left(2x^3\right) \big) + \amp = 6x^7\cos\mathopen{}\left(2x^3\right)\mathclose{}+5x^4\sin\mathopen{}\left(2x^3\right)\mathclose{} + . +

      +
      2. + +
    2. +

      + We must employ the along with the . + Again, proceed step-by-step. + + \fp(x) \amp= \frac{e^{-x^2}\cdot \lzoo{x}{5x^3} - 5x^3\cdot \lzoo{x}{e^{-x^2}}}{\left(e^{-x^2}\right)^2} + \amp= \frac{e^{-x^2}\cdot 15x^2 - 5x^3\cdot e^{-x^2}\cdot \lzoo{x}{-x^2}}{\left(e^{-x^2}\right)^2} + \amp= \frac{e^{-x^2}\left(15x^2\right) - 5x^3\left((-2x)e^{-x^2}\right)}{\left(e^{-x^2}\right)^2} + \amp =\frac{e^{-x^2}\left(10x^4+15x^2\right)}{e^{-2x^2}} + \amp = e^{x^2}\left(10x^4+15x^2\right) + . +

      +
      3. +
    +

    +     + +     + +

    + A key to correctly working these problems is to break the problem down into smaller, + more manageable pieces. + For instance, + when using the and together, + just consider the first part of the at first: + f(x)\gp(x). + Just rewrite f(x), then find \gp(x). + Then move on to the \fp(x)g(x) part. + Don't attempt to figure out both parts at once. +

    + +

    + Likewise, using the , + approach the numerator in two steps and handle the denominator after completing that. + Only simplify afterward. +

    + +

    + We can also employ the itself several times, + as shown in the next example. +

    + + + Using the Chain Rule multiple times + +

    + Find the derivative of y = \tan^5(6x^3-7x). +

    +     + +

    + Recognize that we have the + g(x)=\tan\mathopen{}\left(6x^3-7x\right)\mathclose{} function inside + the f(x)=x^5 function; + that is, we have y = \left(\tan\mathopen{}\left(6x^3-7x\right)\mathclose{}\right)^5. + We begin using the ; + in this first step, we do not fully compute the derivative. + Rather, we are approaching this step-by-step. + + \yp = 5\left(\tan\mathopen{}\left(6x^3-7x\right)\mathclose{}\right)^4\cdot \gp(x) + . +

    + +

    + We now find \gp(x). + We again need the ; + + \gp(x) \amp = \sec^2\mathopen{}\left(6x^3-7x\right)\mathclose{}\cdot \lzoo{x}{6x^3-7x}. + \amp = \sec^2\mathopen{}\left(6x^3-7x\right)\mathclose{}\cdot\left(18x^2-7\right) + . +

    + +

    + Combine this with what we found above to give + + \yp \amp = 5\left(\tan\mathopen{}\left(6x^3-7x\right)\mathclose{}\right)^4\cdot\sec^2\mathopen{}\left(6x^3-7x\right)\mathclose{}\cdot\left(18x^2-7\right) + \amp = \left(90x^2-35\right)\sec^2\mathopen{}\left(6x^3-7x\right)\mathclose{}\tan^4\mathopen{}\left(6x^3-7x\right)\mathclose{} + . +

    + +

    + This function is frankly a ridiculous function, + possessing no real practical value. + It is very difficult to graph, + as the tangent function has many vertical asymptotes and 6x^3-7x grows so very fast. + The important thing to learn from this is that the derivative can be found. + In fact, it is not hard; + one can take several simple steps and should be careful to keep track of how to apply each of these steps. +

    +     + +     + +

    + It is a traditional mathematical exercise to find the derivatives of arbitrarily complicated functions just to demonstrate that it can be done. + Just break everything down into smaller pieces. +

    + + + Using the Product, Quotient and Chain Rules + +

    + Find the derivative of f(x) = \dfrac{x\cos(x^{-2})-\sin^2(e^{4x})}{\ln(x^2+5x^4)}. +

    +     + +

    + This function likely has no practical use outside of demonstrating derivative skills. + The answer is given below without simplification. + It employs the , + the , + and the three times. + + + f'(x)=\frac{\begin{matrix}\ln\left(x^2+5x^4\right)\cdot\bigg[\Big(x\cdot\left(-\sin\left(x^{-2}\right)\right)\cdot\left(-2x^{-3}\right)+1\cdot \cos\left(x^{-2}\right)\Big)\hspace{60pt}\\ + \hspace{72pt}-2\sin\left(e^{4x}\right)\cdot\cos\left(e^{4x}\right)\cdot\left(4e^{4x}\right)\bigg]\\ + \hspace{120pt}-\left(x\cos\left(x^{-2}\right)-\sin^2\left(e^{4x}\right)\right)\cdot\frac{2x+20x^3}{x^2+5x^4}\end{matrix}}{\left(\ln\left(x^2+5x^4\right)\right)^2} + . +

    + +

    + The reader is highly encouraged to look at each term and recognize why it is there. (, the is used; + in the numerator, identify the LOdHI term, + etc.) This example demonstrates that derivatives can be computed systematically, + no matter how arbitrarily complicated the function is. +

    +     +     + +

    + The also has theoretic value. + That is, it can be used to find the derivatives of functions that we have not yet learned as we do in the following example. +

    + + + The Chain Rule and exponential functions + +

    + Use the Chain Rule to find the derivative of y= 2^x. +

    +     + +

    + We only know how to find the derivative of one exponential function, + y = e^x. + We can accomplish our goal by rewriting 2 in terms of e. + Recalling that e^x and \ln(x) are inverse functions, + we can write + 2=e^{\ln 2} and so + + y=2^x = \left(e^{\ln 2}\right)^x = e^{x(\ln(2))} + , + using the power to a power property of exponents. +

    + +

    + The function is now the composition y=f(g(x)), + with f(x) = e^x and g(x) = x(\ln(2)). + Since \fp(x) = e^x and + \gp(x) = \ln(2), the gives + + \yp = e^{x (\ln(2))} \cdot \ln 2 + . +

    + +

    + Recall that the e^{x(\ln(2))} term on the right hand side is just 2^x, + our original function. + Thus, the derivative contains the original function itself. + We have + + \yp = y \cdot \ln(2) = 2^x\cdot \ln(2) + . +

    + +

    + We can extend this process to use any base a, + where a \gt 0 and a\neq 1. + All we need to do is replace each 2 + in our work with a. The Chain Rule, + coupled with the derivative rule of e^x, + allows us to find the derivatives of all exponential functions. +

    +     +     + +

    + The comment at the end of previous example is important and is restated formally as a theorem. +

    + + + Derivatives of Exponential Functions + +

    + Let f(x)=a^x, for a \gt 0, a\neq 1. + + derivativeexponential functions + + Then f is differentiable for all real numbers (, differentiable everywhere) and + + \fp(x) = \ln(a) \cdot a^x + . +

    +     +     + + + + + Alternate Chain Rule Notation +

    + It is instructive to understand what the + looks like using + \lz{y}{x} + notation instead of \yp notation. + Suppose that y=f(u) is a function of u, + where u=g(x) is a function of x, + as stated in . + Then, through the composition f \circ g, + we can think of y as a function of x, as y=f(g(x)). + Thus the derivative of y with respect to x makes sense; + we can talk about \lz{y}{x}. + This leads to an interesting progression of notation: +Chain RulenotationderivativeChain Rule + + \yp \amp = \fp(g(x))\cdot \gp(x) + \lz{y}{x} \amp = \yp(u) \cdot u'(x)\amp\amp \text{ since } y=f(u) \text{ and }u=g(x) + \lz{y}{x} \amp = \lz{y}{u} \cdot \lz{u}{x}\amp\amp \text{(using “fractional notation” for the derivative)} + +

    + +

    + Here the fractional aspect of the derivative notation stands out. + On the right hand side, it seems as though the du + terms cancel out, leaving + + \frac{dy}{dx} = \frac{dy}{dx} + . +

    + +

    + It is important to realize that we + are not canceling these terms; + the derivative notation of \lz{y}{u} is one symbol. + It is equally important to realize that this notation was chosen precisely because of this behavior. + It makes applying the easy with multiple variables. + For instance, + + \lz{y}{t} = \lz{y}{\bigcirc} \cdot \lz{\bigcirc}{\triangle} \cdot \lz{\triangle}{t} + . + where \bigcirc and \triangle are any variables you'd like to use. +

    + +

    + One of the most common ways of visualizing + the is to consider a set of gears, + as shown in . + The gears have 36, 18, + and 6 teeth, respectively. + That means for every revolution of the x gear, + the u gear revolves twice. + That is, the rate at which the u gear makes a revolution is twice as fast as the rate at which the x gear makes a revolution. +

    + + +
    + A series of gears to demonstrate the Chain Rule. Note how \lz{y}{x} = \lz{y}{u}\cdot\lz{u}{x} + + + + 3 gears of various sizes demonstrating the chain rule. + + + Three gears, connected in the order x,u,y. + x is the largest gear, having 36 teeth. It is rotating counter-clockwise. + u is connected to x, and it has 18 teeth. To the left of the connection is \frac{du}{dx} = 2. + y is connected to u, and it has 6 teeth. Below the connection is \frac{dy}{du}=3. + To the right of the gears is the expression \frac{dy}{dx} = 6. + + + + \begin{tikzpicture}[>=latex,scale=0.9] + + \begin{scope}[shift={(0,-200pt)}] + \begin{scope} + \foreach \x in {0,1,2,...,35} + {% + \draw [rotate around={{\x*10}:(0,0)}] (60pt,0)--(65pt,0) arc (0:{4.}:65pt); + \draw [rotate around={{\x*10+4.}:(0,0)}] (65pt,0) -- (60pt,0) arc (0:6:60pt); + } + \draw [->] (40pt,0) arc (0:170:40pt); + \draw (0,0) node {$x$}; + \end{scope} + + \begin{scope}[shift={(4.5pt,-99pt)}] + \foreach \x in {0,1,2,...,17} + {% + \draw [rotate around={{\x*20}:(0,0)}] (30pt,0)--(35pt,0) arc (0:{9}:35pt); + \draw [rotate around={{\x*20+9}:(0,0)}] (35pt,0) -- (30pt,0) arc (0:11:30pt); + } + \draw [->] (0,25pt) arc (90:-80:25pt); + \draw (0,0) node {$u$}; + \draw (45pt,-30pt) node {\small $\ds \frac{dy}{du} = 3$}; + \draw (-50pt,30pt) node {\small $\ds \frac{du}{dx} = 2$}; + \draw (60pt,40pt) node {\small $\ds \frac{dy}{dx} = 6$}; + \end{scope} + + \begin{scope}[shift={(53.5pt,-100pt)}] + \foreach \x in {0,1,2,...,5} + {% + \draw [rotate around={{\x*60}:(0,0)}] (10pt,0)--(15pt,0) arc (0:{29}:15pt); + \draw [rotate around={{\x*60+29}:(0,0)}] (15pt,0) -- (10pt,0) arc (0:31:10pt); + } + \draw [->] (0,-20pt) arc (-90:70:20pt); + \draw (0,0) node {$y$}; + \end{scope} + \end{scope} + \end{tikzpicture} + + + + +
    + +

    + Using the terminology of calculus, + the rate of u-change, with respect to x, + is \lz{u}{x} = 2. +

    + +

    + Likewise, every revolution of u causes 3 revolutions of y: + \lz{y}{u} = 3. + How does y change with respect to x? + For each revolution of x, + y revolves 6 times; that is, + + \frac{dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx} = 2\cdot 3 = 6 + . +

    + +

    + We can then extend the with more variables by adding more gears to the picture. +

    + +

    + It is difficult to overstate the importance of the . + So often the functions that we deal with are compositions of two or more functions, + requiring us to use this rule to compute derivatives. + It is also often used in real life when actual functions are unknown. + Through measurement, we can calculate (or, + approximate) \lz{y}{u} and \lz{u}{x}. + With our knowledge of the , + we can find \lz{y}{x}. +

    + +

    + In , + we use the to justify another differentiation technique. + There are many curves that we can draw in the plane that fail the + vertical line test. For instance, + consider x^2+y^2=1, which describes the unit circle. + We may still be interested in finding slopes of tangent lines to the circle at various points. + + shows how we can find \lz{y}{x} without first + solving for y. + While we can in this instance, + in many other instances solving for y is impossible. + In these situations, + implicit differentiation is indispensable. +

    +     + + + + Terms and Concepts + + + + +

    + + The Chain Rule describes how to evaluate the derivative of a composition of functions. +

    +     + +

    + This is precisely what it is for! +

    +     + +     + + + + +

    + + The Generalized Power Rule states that \lzoo{x}{g(x)^n} = n\left(g(x)\right)^{n-1}. +

    +     + +

    + Don't forget to multiply by the derivative of the inside function! +

    +     + +     + + + + +

    + + \lzoo{x}{\ln\mathopen{}\left(x^2\right)\mathclose{}} = \frac{1}{x^2}. +

    +     + +

    + Don't forget to multiply by the derivative of the inside function! + Note also that in this case, we can use the property \ln\mathopen{}\left(x^2\right)\mathclose{}=2\ln(x) + to compute the derivative without the chain rule. +

    +     + +     + + + + +

    + + \lzoo{x}{3^x}\approx 1.1\cdot 3^x. +

    +     + +

    + Note that \ln(3)\approx 1.1. +

    +     + +     + + + + +

    + + \lz{x}{y} = \lz{x}{t}\cdot\lz{t}{y}. +

    +     + +

    + This is one way of stating the chain rule using Leibniz notation. +

    +     + +     + + + + +

    + + Taking the derivative of f(x) = x^2\sin(5x) requires the use of both the Product and Chain Rules. +

    +     + +

    + The chain rule is needed for the argument 5x of the sine function, + and the product rule is needed since we also multiply by x^2. +

    +     + +     +     + + + Problems + + + +

    + Compute the derivative of the given function. +

    +     + + + + + + $fp=Formula("10(4x^3-x)^9(12x^2-1)"); + + +

    + f(x) = \left(4x^3-x\right)^{10} +

    +

    + +

    +     +     +     + + + + + + Context()->variables->are(t=>'Real'); + $fp=Formula("15(3t-2)^4"); + + +

    + f(t) = (3t-2)^{5} +

    +

    + +

    +     +     +     + + + + + + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $fp=Formula("3(sin(theta)+cos(theta))^2(cos(theta)-sin(theta))"); + + +

    + g(\theta)=(\sin(\theta)+\cos(\theta))^3 +

    + + For \theta, type theta. + +

    + +

    +     +     +     + + + + + + Context()->variables->are(t=>'Real'); + $fp=Formula("(6t+1)e^(3t^2+t-1)"); + + +

    + h(t) = e^{3t^2+t-1} +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $a = random(2,5,1); + $b = random(3,8,1); + $pm = list_random('+','-'); + if($envir{problemSeed}==1){$fname='f';$x='x';$a=2;$b=3;$pm='+'}; + Context()->variables->are($x=>'Real'); + $f=Formula("(ln($x)$pm $x^$a)^$b"); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $a = random(2,9,1); + $n = random(2,5,1); + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$a=2;$n=3;$b=3}; + Context()->variables->are($x=>'Real'); + $f=Formula("$a^($x^$n+$b $x)")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $pm = list_random('+','-'); + $n = random(3,6,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$pm='+';$n=4;}; + Context()->variables->are($x=>'Real'); + $f=Formula("($x$pm 1/$x)^$n"); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig= list_random('sin','cos'); + $a = random(2,9,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$trig='cos';$a=3;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$trig($a $x)"); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig= list_random('tan','cot'); + $a = random(2,9,1); + if($envir{problemSeed}==1){$fname='g';$x='x';$trig='tan';$a=5;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$trig($a $x)"); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $trig= list_random('sin','cos','tan','cot'); + $a = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$fname='h';$trig='tan';$a=4;}; + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $f=Formula("$trig(theta^2+$a)")->reduce; + $fp=$f->D('theta'); + + +

    + (\theta) = +

    + + For \theta, type theta. + +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig= list_random('sin','cos','tan','cot'); + $n = random(2,9,1); + $m = random(1,4,1); + if($envir{problemSeed}==1){$fname='g';$x='t';$trig='sin';$n=2;$m=1;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$trig($x^$n+1/$x^$m)")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig= list_random('sin','cos'); + $n = random(3,9,1); + $a = random(2,9,1); + if($envir{problemSeed}==1){$fname='h';$x='t';$trig='sin';$n=4;$a=2;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$trig^$n($a $x)")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig= list_random('sin','cos'); + $n = random(3,9,1); + ($a,$b) = random_subset(2,-9..-2,2..9); + if($envir{problemSeed}==1){$fname='p';$x='t';$trig='sin';$n=3;$a=3;$b=1;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$trig^$n($x^2+$a $x + $b)")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $trig= list_random('sin','cos','sec','csc'); + if($envir{problemSeed}==1){$fname='f';$x='x';$trig='cos';}; + Context()->variables->are($x=>'Real'); + $f=Formula("ln($trig($x))")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $n = random(2,9,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$n=2;}; + Context()->variables->are($x=>'Real'); + $f=Formula("ln($x^$n)")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $n = random(2,9,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$n=2;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$n ln($x)")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $b = random(2,9,1); + if($envir{problemSeed}==1){$fname='g';$x='r';$b=4;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$b^$x")->reduce; + Context()->flags->set(reduceConstantFunctions=>0); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + $b = random(2,9,1); + $trig= list_random('sin','cos','sec','csc','tan','cot'); + if($envir{problemSeed}==1){$fname='g';$x='t';$b=5;$trig='cos'}; + Context()->variables->are($x=>'Real'); + $f=Formula("$b^($trig($x))")->reduce; + Context()->flags->set(reduceConstantFunctions=>0); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p'); + $x = list_random('q','r','t','x','y','z'); + ($a,$b) = random_subset(2,2..20); + if($envir{problemSeed}==1){$fname='g';$x='t';$b=15;$a=2;}; + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstants=>0); + $f=Formula("$b^$a"); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + do {($a,$b) = random_subset(2,2..9);} until (gcd($a,$b) == 1); + if($envir{problemSeed}==1){$fname='m';$x='w';$a=3;$b=2;}; + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstantFunctions=>0); + $f=Formula("$a^$x/$b^$x"); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + do {($a,$b) = random_subset(2,2..9);} until (gcd($a,$b) == 1); + if($envir{problemSeed}==1){$fname='h';$x='t';$a=2;$b=3;}; + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstantFunctions=>0); + $f=Formula("($a^$x+$b)/($b^$x+$a)"); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + do {($a,$b) = random_subset(2,2..9);} until (gcd($a,$b) == 1); + $c = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$fname='m';$x='w';$a=3;$b=2;$c=1;}; + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstantFunctions=>0); + $f=Formula("($a^$x+$c)/($b^$x)")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + do {($a,$b) = random_subset(2,2..9);} until (gcd($a,$b) == 1); + $pm = list_random('+','-'); + if($envir{problemSeed}==1){$fname='f';$x='x';$a=3;$b=2;$pm='+';}; + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstantFunctions=>0); + $f=Formula("($a^($x^2)$pm$x)/($b^($x^2))"); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $n = random(2,3,1); + $trig = list_random('sin','cos','tan','cot'); + $a = random(2,9,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$n=2;$trig='sin';$a=5;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$x^$n $trig($a $x)"); + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + ($m,$n,$p) = random_subset(3,2..6); + ($a,$b,$c) = random_subset(3,1..9); + if($envir{problemSeed}==1){$fname='f';$x='x';$m=5;$n=4;$p=3;$a=1;$b=3;$c=2;}; + Context()->variables->are($x=>'Real'); + $f=Formula("($x^2+$a $x)^$m($b $x^$n+$c $x)^$p")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + ($triga,$trigb) = random_subset(2,'sin','cos'); + ($m,$n) = random_subset(2,0,2); + ($a,$b,$c,$d) = random_subset(4,1..9); + $b=$b*random(-1,1,2); + $d=$d*random(-1,1,2); + if($envir{problemSeed}==1){$fname='g';$x='t';$m=2;$n=0;$a=1;$b=3;$c=5;$d=-7;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$triga($a $x^$m + $b $x)$trigb($c $x + $d $x^$n)")->reduce->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + ($triga,$trigb) = random_subset(2,'sin','cos'); + ($a,$b,$c,$d) = random_subset(4,1..9); + $b=$b*random(-1,1,2); + $d=$d*random(-1,1,2); + if($envir{problemSeed}==1){$fname='f';$x='x';$a=3;$b=4;$c=5;$d=-2;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$triga($a $x + $b)$trigb($c + $d $x)")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $trig = list_random('sin','cos'); + $a = random(2,9,1); + $n = random(2,3,1); + if($envir{problemSeed}==1){$fname='g';$x='t';$a=5;$n=2;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$trig(1/$x)e^($a $x^$n)")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $trig = list_random('sin','cos'); + ($a,$b,$c,$d) = random_subset(4,1..9); + $b=$b*random(-1,1,2); + $d=$d*random(-1,1,2); + $n = random(2,4,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$a=4;$b=1;$c=5;$d=-9;$n=3;}; + Context()->variables->are($x=>'Real'); + $f=Formula("$trig($a $x + $b)/($c $x + $d)^$n")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $trig = list_random('sin','cos','tan','cot'); + ($a,$b,$c) = random_subset(3,1..9); + $n = random(2,4,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$a=4;$b=1;$c=5;$n=2;}; + Context()->variables->are($x=>'Real'); + $f=Formula("($a $x + $b)^$n/$trig($c $x)")->reduce; + $fp=$f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     +     + + + +

    + Find the equations of tangent and normal lines to the graph of the function at the given point. + Note: the functions here are the same as in Exercises. + +

    +     + + + + + + Context("Numeric")->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=0"); + $n=Formula("x=0"); + + +

    + f(x) = \left(4x^3-x\right)^{10} at x=0 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + + Context("Numeric")->variables->add(y=>'Real'); + parser::Assignment->Allow; + Context()->flags->set(reduceConstants=>0); + $t=Formula("y=15(x-1)+1"); + $n=Formula("y=-1/15(x-1)+1"); + + +

    + f(x) = (3x-2)^5 at x=1 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + + Context("Numeric")->variables->add(y=>'Real'); + parser::Assignment->Allow; + Context()->flags->set(reduceConstants=>0); + $t=Formula("y=-3(x - pi/2)+1"); + $n=Formula("y=1/3(x - pi/2)+1"); + + +

    + g(x) = (\sin(x)+\cos(x))^3 at x=\pi/2. +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     + + + + + + Context("Numeric")->variables->add(y=>'Real'); + parser::Assignment->Allow; + Context()->flags->set(reduceConstants=>0); + $t=Formula("y=-5e(x+1)+e"); + $n=Formula("y=1/(5e)(x+1)+e"); + + +

    + h(x) = e^{3x^2+x-1} at x=-1 +

    + + Enter the equation of the tangent line here. + +

    + +

    + + Enter the equation of the normal line here. + +

    + +

    +     +     +     +     + + + + + Context()->variables->add(k=>'Real'); + $f = Formula("ln(k x)"); + $d = Formula("1/x"); + $showwork = '[@ explanation_box(message => "Show the work for both parts.") @]*'; + + +

    + Compute \lzoo{x}{\ln(kx)} two ways. + First by using the Chain Rule. + Second, by using the logarithm rule \ln(ab) = \ln(a) + \ln(b) and then taking the derivative. +

    +

    + +

    +

    + +

    +     + +

    + In both cases the derivative is the same: 1/x. +

    +     +     +     + + + + + Context()->variables->add(k=>'Real'); + $f = Formula("ln(x^k)"); + $d = Formula("k/x"); + $showwork = '[@ explanation_box(message => "Show the work for both parts.") @]*'; + + +

    + Compute \lzoo{x}{\ln\mathopen{}\left(x^k\right)\mathclose{}} two ways. + First by using the Chain Rule. + Second, by using the logarithm rule \ln\mathopen{}\left(a^p\right)\mathclose{} = p\ln(a) + (for positive a) and then taking the derivative. +

    +

    + +

    +

    + +

    +     + +

    + In both cases the derivative is the same: k/x. +

    +     +     +     +     + +     +
    +
    + Implicit Differentiation + +

    + In the previous sections we learned to find the derivative, + \lz{y}{x}, or \yp, + when y is given explicitly + as a function of x. + That is, if we know y=f(x) for some function f, + we can find \yp. + For example, given y=3x^2-7, + we can easily find \yp=6x. + (Here we explicitly state how y depends on x. + Knowing x, we can directly find y.) +

    + + + +

    + Sometimes the relationship between y and x is not explicit; + rather, it is implicit. + For instance, we might know that x^2-y=4. + This equality defines a relationship between x and y; + if we know x, we could figure out y. + Can we still find \yp? + In this case, sure; we solve for y to get y=x^2-4 + (hence we now know y explicitly) + and then differentiate to get \yp=2x. +

    + +

    + Sometimes the implicit + relationship between x and y is complicated. + Suppose we are given \sin(y)+y^3=6-x^3. + A graph of this implicit relationship is given in . + In this case there is absolutely no way to solve for y in terms of elementary functions. + The surprising thing is, however, + that we can still find \yp via a process known as + implicit differentiation. + + implicit differentiation + derivativeimplicit + +

    + +
    + A graph of the implicit relationship \sin(y)+y^3=6-x^3 + + + + A curve beginning in the second quadrant, forming a gentle corner in the first quadrant, and decreasing into the fourth quadrant. + + +

    + The curve begins in the second quadrant. + From the left, the curve decreases as x increases. + The curve slowly flattens out, almost becoming horizontal as the curve crosses the y-axis near the point (0,1.8). + When x comes close to 0.75, the curve begins decreasing in the shape of a gentle corner. + The curve continues decreasing, becoming steepest around the point (0,1.8), at which it also crosses into the fourth quadrant. + When x is close to 2, the curve begins to decrease more gently, at around the same rate as the beginning of the curve. +

    +     + + + \begin{tikzpicture}[declare function = {cbrt(\x) = (\x < 0) * -(-\x)^(1/3) + (\x >= 0) * (\x)^(1/3);}] + \begin{axis}[,xmin=-3.2,xmax=3.2, + ymin=-3.2,ymax=3.2] + \addplot[firstcurvestyle,variable=\t,domain=-2.7:1.5,leftarrow] ({cbrt(6-sin(deg(t))-t^3)},{t}); + \addplot[firstcurvestyle,variable=\t,domain=1.5:1.8] ({cbrt(6-sin(deg(t))-t^3)},{t}); + \addplot[firstcurvestyle,variable=\t,domain=1.8:3,rightarrow] ({cbrt(6-sin(deg(t))-t^3)},{t}); + \end{axis} + \end{tikzpicture} + + + + +
    +     + + + The method of implicit differentiation +

    + Implicit differentiation is a technique based on the that is used to find a derivative when the relationship between the variables is given implicitly rather than explicitly + (solved for one variable in terms of the other). +

    + +

    + We begin by reviewing the Chain Rule. + Let f and g be functions of x. + Then + + \lzoo{x}{f(g(x))} = \fp(g(x))\cdot g'(x) + . +

    + +

    + Suppose now that y=g(x). + We can rewrite the above as + + \lzoo{x}{f(y)}=\fp(y)\cdot \yp, \quad \text{ or } \quad \lzoo{x}{f(y)}=\fp(y)\cdot\lz{y}{x} + . +

    + +

    + These equations look strange; + the key concept to learn here is that we can find \yp even if we don't exactly know how y and x relate. +

    + +

    + We demonstrate this process in the following example. +

    + + + Using Implicit Differentiation + +

    + Find \yp given that \sin(y) + y^3=6-x^3. +

    +     + +

    + We start by taking the derivative of both sides + (thus maintaining the equality.) + We have: + + \lzoo{x}{\sin(y) + y^3}=\lzoo{x}{6-x^3} + . +

    + +

    + The right hand side is easy; + it returns -3x^2. +

    + +

    + The left hand side requires more consideration. + We take the derivative term-by-term. + Using the technique derived from Equation above, + we can see that + + \lzoo{x}{\sin(y)} = \cos(y) \cdot \yp + . +

    + +

    + We apply the same process to the y^3 term. + + \lzoo{x}{y^3} = \lzoo{(y)^3} = 3(y)^2\cdot \yp + . +

    + +

    + Putting this together with the right hand side, we have + + \cos(y)\yp+3y^2\yp = -3x^2 + . +

    + +

    + Now solve for \yp. + It's important to treat \yp as an algebraically independent variable from y and x. + + \cos(y)\yp+3y^2\yp\amp = -3x^2 + \left(\cos(y) +3y^2\right)\yp \amp =-3x^2 + \yp\amp =\frac{-3x^2}{\cos(y) +3y^2} + +

    + +

    + This equation for \yp probably seems unusual for it contains both x and y terms. + How is it to be used? + We'll address that next. +

    +     + +     + + + + + +

    + Implicit functions are generally harder to deal with than explicit functions. + With an explicit function, given an x value, + we have an explicit formula for computing the corresponding y value. + With an implicit function, + one often has to find x and y values + at the same time + that satisfy the equation. + It is much easier to demonstrate that a given point satisfies the equation than to actually find such a point. +

    + +

    + For instance, we can affirm easily that the point + \left(\sqrt[3]{6},0\right) lies on the graph of the implicit function \sin(y) + y^3=6-x^3. + Plugging in 0 for y, + we see the left hand side is 0. + Setting x=\sqrt[3]6, we see the right hand side is also 0; + the equation is satisfied. + The following example finds the equation of the tangent line to this function at this point. +

    + + + Using implicit differentiation to find a tangent line + +

    + Find the equation of the line tangent to the curve of the implicitly defined function + \sin(y) + y^3=6-x^3 at the point \left(\sqrt[3]6,0\right). +

    +     + +

    + In we found that + + \yp = \frac{-3x^2}{\cos(y) +3y^2} + . +

    + +

    + We find the slope of the tangent line at the point \left(\sqrt[3]6,0\right) by substituting + \sqrt[3]6 for x and 0 for y. + Thus at the point \left(\sqrt[3]6,0\right), we have the slope as + + \yp = \frac{-3\left(\sqrt[3]{6}\right)^2}{\cos(0) + 3\cdot0^2} = \frac{-3\sqrt[3]{36}}{1} \approx -9.91 + . +

    + +

    + Therefore the equation of the tangent line to the implicitly defined function + \sin(y) + y^3=6-x^3 at the point \left(\sqrt[3]{6},0\right) is + + y = -3\sqrt[3]{36}\left(x-\sqrt[3]{6}\right)+0 \approx -9.91x+18 + . +

    + +

    + The curve and this tangent line are shown in . +

    + +
    + The function \sin(y) +y^3 = 6-x^3 and its tangent line at the point (\sqrt[3]{6},0) + + + + A decreasing curve with a negative tangent line through the negative x-axis + + +

    + The same curve as , but with a tangent line drawn at + x=\sqrt[3]{6}. The tangent line is pointing sharply downward. +

    +     + + + \begin{tikzpicture}[declare function = {cbrt(\x) = (\x < 0) * -(-\x)^(1/3) + (\x >= 0) * (\x)^(1/3);}] + \begin{axis}[,xmin=-3.2,xmax=3.2, + ymin=-3.2,ymax=3.2] + \addplot[firstcurvestyle,variable=\t,domain=-2.7:1.5,leftarrow] ({cbrt(6-sin(deg(t))-t^3)},{t}); + \addplot[firstcurvestyle,variable=\t,domain=1.5:1.8] ({cbrt(6-sin(deg(t))-t^3)},{t}); + \addplot[firstcurvestyle,variable=\t,domain=1.8:3,rightarrow] ({cbrt(6-sin(deg(t))-t^3)},{t}); + \addplot [tangentline,domain=1.617:2.017] {-9.9057*x+18}; + \end{axis} + \end{tikzpicture} + + + + +
    +     + +     + +

    + This suggests a general method for implicit differentiation. + For the steps below assume y is a function of x. +

    + +

    +

      +
    1. +

      + Take the derivative of each term in the equation. + Treat the x terms like normal. + When taking the derivatives of y terms, + the usual rules apply except that, + because of the , + we need to multiply each term by \yp. +

      +
      2. + +
    2. +

      + Get all the \yp terms on one side of the equal sign and put the remaining terms on the other side. +

      +
      3. + +
    3. +

      + Factor out \yp; solve for \yp by dividing. +

      +
      4. +
    +

    + +

    + (Practical Note: when working by hand, + it may be beneficial to use the symbol + \frac{dy}{dx} instead of \yp, + as the latter can be easily confused for y or y^1.) +

    + + + Using Implicit Differentiation + +

    + Given the implicitly defined function y^3+x^2y^4=1+2x, + find \yp. +

    +     + +

    + We will take the implicit derivatives term by term. + The derivative of y^3 is 3y^2\yp. +

    + +

    + The second term, x^2y^4, is a little tricky. + It requires the as it is the product of two functions of x: + x^2 and y^4. + Its derivative is x^2(4y^3\yp) + 2xy^4. + The first part of this expression requires a \yp because we are taking the derivative of a y term. + The second part does not require it because we are taking the derivative of x^2. +

    + +

    + The derivative of the right hand side is easily found to be 2. + In all, we get: + + 3y^2\yp + 4x^2y^3\yp + 2xy^4 = 2 + . +

    + +

    + Move terms around so that the left side consists only of the \yp terms and the right side consists of all the other terms: + + 3y^2\yp + 4x^2y^3\yp = 2-2xy^4 + . +

    + +

    + Factor out \yp from the left side and solve to get + + \yp = \frac{2-2xy^4}{3y^2+4x^2y^3} + . +

    + +

    + To confirm the validity of our work, + let's find the equation of a tangent line to this function at a point. + It is easy to confirm that the point (0,1) lies on the graph of this function. + At this point, \yp = 2/3. + So the equation of the tangent line is y = 2/3(x-0)+1. + The function and its tangent line are graphed in . +

    + +
    + A graph of the implicitly defined function y^3+x^2y^4=1+2x along with its tangent line at the point (0,1) + + + + A curve with two distinct segments and a tangent line with a positive slope + + + Two curves are drawn in the xy-plane. + The left curve stretches upwards from the left side of the y axis, curving slightly to the left. + As y approaches -2, the curve begins to widen to the left, creating a bump in the curve. + As the curve crosses the x axis, the curve moves towards the right, no longer increasing and becoming more horizontal as x increases. + At the point (0,1), a tangent line is drawn, with a moderate positive slope. + This point corresponds to the corner at which the curve begins to become horizontal. + At this point, the curve passes the vertical line test, but does not at most other points on the graph. + The second curve begins to the right of the y-axis, as a line stretching upwards from the bottom of the y-axis. + As x approaches 1, the curve also begins to become horizontal as x increases. + The entire second curve lies in the fourth quadrant. + + + + \begin{tikzpicture} + \begin{axis}[xmin=-1.5,xmax=10.5, + ymin=-10.5,ymax=2.49,] + + \addplot[firstcurvestyle,infinite,smooth] coordinates {(-0.324,-9.34) (-0.336,-8.66) (-0.351,-7.95) (-0.371,-7.13) (-0.396,-6.3) (-0.414,-5.68) (-0.451,-4.91) (-0.472,-4.48)(-0.497,-4.02) (-0.525,-3.57) (-0.56,-3.15) (-0.61,-2.63)(-0.664,-2.18) (-0.727,-1.71) (-0.762,-1.29) (-0.668,-0.805)(-0.513,-0.29) (-0.453,0.439) (-0.308,0.71) (0.0816,1.05)(0.452,1.15) (0.893,1.13) (1.33,1.08) (1.79,1.02) (2.24,0.975)(2.82,0.925) (3.4,0.881) (4.11,0.844) (4.91,0.781) (5.77,0.766)(6.76,0.74) (7.59,0.722) (8.26,0.706) (8.93,0.689) (9.69,0.671)}; + + \addplot[firstcurvestyle,infinite,smooth] coordinates {(9.6,-0.677) (9.13,-0.696) (8.66,-0.697) (8.21,-0.707) (7.72,-0.722)(7.3,-0.738) (6.79,-0.754) (5.98,-0.778) (5.1,-0.81) (4.34,-0.858)(3.84,-0.893) (3.4,-0.927) (2.77,-0.99) (2.27,-1.07) (1.82,-1.17)(1.38,-1.34) (0.97,-1.7) (0.779,-2.1) (0.678,-2.5) (0.593,-3.)(0.534,-3.66) (0.483,-4.38) (0.444,-5.04) (0.415,-5.71) (0.393,-6.38)(0.375,-7.05) (0.362,-7.59) (0.343,-8.38) (0.334,-8.9) (0.322,-9.33)(0.317,-9.83) }; + + \addplot [tangentline,domain=-1:2] {2/3*(x)+1}; + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Notice how our curve looks much different than for functions we have seen. + For one, it fails the vertical line test, + and so the complete curve is not truly representing y as a function of x. + But when we indicate we are interested in the derivative at (0,1), + we are indicating that we want the function defined by the small portion of the curve that passes through (0,1), + and that small portion does pass the vertical line test. + Such functions are important in many areas of mathematics, + so developing tools to deal with them is also important. +

    +     + +     + + + Using Implicit Differentiation + +

    + Given the implicitly defined function \sin\mathopen{}\left(x^2y^2\right)\mathclose{}+y^3=x+y, + find \yp. +

    +     + +

    + Differentiating term by term, + we find the most difficulty in the first term. + It requires both the and . + + \lzoo{x}{\sin\mathopen{}\left(x^2y^2\right)\mathclose{}} \amp = \cos\mathopen{}\left(x^2y^2\right)\mathclose{}\cdot\lzoo{x}{x^2y^2} + \amp = \cos\mathopen{}\left(x^2y^2\right)\mathclose{}\cdot\left(x^2(2y\yp)+2xy^2\right) + \amp = 2\left(x^2y\yp+xy^2\right)\cos\mathopen{}\left(x^2y^2\right)\mathclose{} + . +

    + +

    + We leave the derivatives of the other terms to the reader. + After taking the derivatives of both sides, we have + + 2\left(x^2y\yp+xy^2\right)\cos\mathopen{}\left(x^2y^2\right)\mathclose{} + 3y^2\yp = 1 + \yp + . +

    + +

    + We now have to be careful to properly solve for \yp, + particularly because of the product on the left. + It is best to multiply out the product. + Doing this, we get + + 2x^2y\cos\mathopen{}\left(x^2y^2\right)\mathclose{}\yp + 2xy^2\cos\mathopen{}\left(x^2y^2\right)\mathclose{} + 3y^2\yp = 1 + \yp + . +

    + +

    + From here we can safely move around terms to get the following: + + 2x^2y\cos\mathopen{}\left(x^2y^2\right)\mathclose{}\yp + 3y^2\yp - \yp = 1 - 2xy^2\cos\mathopen{}\left(x^2y^2\right)\mathclose{} + . +

    + +

    + Then we can solve for \yp to get + + \yp = \frac{1 - 2xy^2\cos\mathopen{}\left(x^2y^2\right)\mathclose{}}{2x^2y\cos\mathopen{}\left(x^2y^2\right)\mathclose{}+3y^2-1} + . +

    + +

    + A graph of this implicit function is given in . +

    + +
    + A graph of the implicitly defined curve \sin\mathopen{}\left(x^2y^2\right)\mathclose{}+y^3=x+y + + + + A curve beginning in the third quadrant passing through the points (0,-1), (0,0), (0,1). + + +

    + The curve begins in the third quadrant. + From there, the curve bends slightly back and increases, crossing above itself. + The curve extends to the right, increasing almost linearly as it crosses the y-axis at y = -1 into the fourth quadrant. + The curve continues to increase as such until it reaches a point close to (\frac{1}{2},-\frac{3}{4}. + The curve then bends back, increasing towards the top left linearly. + It then crosses the origin and passes into the second quadrant. + The curve quickly bends towards the right, crossing the y-axis at y = 1 into the first quadrant. + From there, the curve continues towards the right while slightly increasing. + The curves rises sharply at x = 1.5, before decreasing again. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-2.05,xmax=2.05, + ymin=-1.95,ymax=1.95,] + \addplot+[smooth,infinite] coordinates {(-2.03,-1.513)(-1.988,-1.536)(-1.972,-1.543)(-1.879,-1.586)(-1.786,-1.621)(-1.68,-1.572)(-1.707,-1.464)(-1.749,-1.357)(-1.739,-1.261)(-1.621,-1.264)(-1.5,-1.299)(-1.348,-1.357)(-1.179,-1.428)(-1.074,-1.467)(-0.9643,-1.494)(-0.8524,-1.495)(-0.6789,-1.429)(-0.5,-1.303)(-0.3182,-1.179)(-0.1408,-1.073)(0,-1.)(0.1397,-0.9317)(0.2397,-0.8826)(0.3496,-0.8214)(0.4457,-0.7329)(0.4619,-0.6071)(0.4189,-0.5)(0.276,-0.2954)(0.07143,-0.07178)(-0.07143,0.07183)(-0.2684,0.3031)(-0.3426,0.4855)(-0.3214,0.6721)(-0.2591,0.7857)(-0.1429,0.9072)(0,1.)(0.1009,1.042)(0.2857,1.083)(0.5178,1.089)(0.7793,1.065)(0.9789,1.05)(1.179,1.08)(1.248,1.145)(1.275,1.275)(1.279,1.393)(1.3,1.514)(1.393,1.582)(1.5,1.558)(1.607,1.514)(1.712,1.467)(1.804,1.426)(1.856,1.405)}; + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + It is easy to verify that the points (0,0), + (0,1) and (0,-1) all lie on the graph. + We can find the slopes of the tangent lines at each of these points using our formula for \yp. + +

      +
    • At (0,0), the slope is -1.
      • + +
    • At (0,1), the slope is 1/2.
      • + +
    • At (0,-1), the slope is also 1/2.
      • +
    + + The tangent lines have been added to the graph of the function in . +

    + +
    + A graph of the implicitly defined curve \sin\mathopen{}\left(x^2y^2\right)\mathclose{}+y^3=x+y and certain tangent lines + + + + A curve beginning in the third quadrant passing through the points (0,-1), (0,0), (0,1), with tangent lines at those points. + + +

    + The graph in , with tagent lines drawn at (0,-1), (0,0), and (0,1). + The tangent line at (0,-1) has a positive slope less than 1. + The tangent line at (0,0) has a negative slope, close to -1. + The tangent line at (0,1) has a positive slope, less than 1. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-2.05,xmax=2.05, + ymin=-1.95,ymax=1.95,] + \addplot+[infinite,smooth] coordinates {(-2.03,-1.513)(-1.988,-1.536)(-1.972,-1.543)(-1.879,-1.586)(-1.786,-1.621)(-1.68,-1.572)(-1.707,-1.464)(-1.749,-1.357)(-1.739,-1.261)(-1.621,-1.264)(-1.5,-1.299)(-1.348,-1.357)(-1.179,-1.428)(-1.074,-1.467)(-0.9643,-1.494)(-0.8524,-1.495)(-0.6789,-1.429)(-0.5,-1.303)(-0.3182,-1.179)(-0.1408,-1.073)(0,-1.)(0.1397,-0.9317)(0.2397,-0.8826)(0.3496,-0.8214)(0.4457,-0.7329)(0.4619,-0.6071)(0.4189,-0.5)(0.276,-0.2954)(0.07143,-0.07178)(-0.07143,0.07183)(-0.2684,0.3031)(-0.3426,0.4855)(-0.3214,0.6721)(-0.2591,0.7857)(-0.1429,0.9072)(0,1.)(0.1009,1.042)(0.2857,1.083)(0.5178,1.089)(0.7793,1.065)(0.9789,1.05)(1.179,1.08)(1.248,1.145)(1.275,1.275)(1.279,1.393)(1.3,1.514)(1.393,1.582)(1.5,1.558)(1.607,1.514)(1.712,1.467)(1.804,1.426)(1.856,1.405)}; + + \addplot [tangentline,domain=-.5:.5] {-x}; + \addplot [tangentline,domain=-.5:.5] {0.5*x+1}; + \addplot [tangentline,domain=-.5:.5] {0.5*(x-0)-1}; + \end{axis} + \end{tikzpicture} + + + + +
    +     + +     + +

    + Quite a few famous curves have equations that are given implicitly. + We can use implicit differentiation to find the slope at various points on those curves. + We investigate two such curves in the next examples. +

    + + + Finding slopes of tangent lines to a circle + +

    + Find the slope of the tangent line to the circle + x^2+y^2=1 at the point \left(1/2, \sqrt{3}/2\right). +

    +     + +

    + Taking derivatives, we get 2x+2y\yp=0. + Solving for \yp gives: + + \yp = \frac{-x}{y} + . +

    + +

    + This is a clever formula. + Recall that the slope of the line through the origin and the point (x,y) on the circle will be y/x. + We have found that the slope of the tangent line to the circle at that point is the opposite reciprocal of y/x, namely, + -x/y. + Hence these two lines are always perpendicular. +

    + +

    + At the point \left(1/2, \sqrt{3}/2\right), + we have the tangent line's slope as + + \yp = \frac{-1/2}{\sqrt{3}/2} = \frac{-1}{\sqrt{3}} \approx -0.577 + . +

    + +

    + A graph of the circle and its tangent line at + \left(1/2,\sqrt{3}/2\right) is given in , + along with a thin dashed line from the origin that is perpendicular to the tangent line. + (It turns out that all normal lines to a circle pass through the center of the circle.) +

    + +
    + The unit circle with its tangent line at (1/2,\sqrt{3}/2) + + + + A circle of radius 1 centered at the origin with a tangent line drawn at a point in the first quadrant. + + +

    + A circle of radius 1 centered at the origin. + A dashed line extends from the origin to a tangent line at the point (\frac{1}{2},\frac{\sqrt{3}}{2}). + At that point a tangent line is drawn with a slight negative slope. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-1.2,xmax=1.2, + ymin=-1.2,ymax=1.2, + axis equal] + \addplot+[domain=0:pi,samples=200] ({cos(deg(2*x))},{sin(deg(2*x))}); + \addplot [tangentline,domain=.1:.9] {-.577*(x-.5)+.866}; + \addplot [soliddot] coordinates {(0.5,.866)} node[above right] {$\left(1/2,\sqrt{3}/2\right)$}; + \addplot [normallineseg,domain=0:0.5]{1/0.577*x}; + \end{axis} + \end{tikzpicture} + + + + +
    +     +     + +

    + This section has shown how to find the derivatives of implicitly defined functions, + whose graphs include a wide variety of interesting and unusual shapes. + Implicit differentiation can also be used to further our understanding of + regular differentiation. +

    + +

    + One hole in our current understanding of derivatives is this: + what is the derivative of the square root function? + That is, + + \lzoo{x}{\sqrt{x}} = \lzoo{x}{x^{1/2}} = \text{?} + +

    + +

    + We allude to a possible solution, + as we can write the square root function as a power function with a rational (or, + fractional) power. + We are then tempted to apply the and obtain + + \lzoo{x}{x^{1/2}} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} + . +

    + +

    + The trouble with this is that the was initially defined only for positive integer powers, + n \gt 0. + While we did not justify this at the time, + generally the is proved using something called the Binomial Theorem, + which deals only with positive integers. + The allowed us to extend the to negative integer powers. + Implicit Differentiation allows us to extend the to rational powers, + as shown below. +

    + +

    + Let y = x^{m/n}, + where m and n are integers with no common factors + (so m=2 and n=5 is fine, + but m=2 and n=4 is not). + We can rewrite this explicit function implicitly as y^n = x^m. + Now apply implicit differentiation. + + y \amp = x^{m/n} + y^n \amp = x^m + \lzoo{x}{y^n} \amp = \lzoo{x}{x^m} + n\cdot y^{n-1}\cdot \yp \amp = m\cdot x^{m-1} + \yp\amp = \frac{m}{n} \frac{x^{m-1}}{y^{n-1}}\amp\amp\text{(now substitute }x^{m/n}\text{ for }y\text{)} + \amp = \frac{m}{n} \frac{x^{m-1}}{(x^{m/n})^{n-1}}\amp\amp\text{(apply lots of algebra)} + \amp = \frac{m}n x^{(m-n)/n} + \amp = \frac{m}n x^{m/n -1} + . +

    + +

    + The above derivation is the key to the proof extending the to rational powers. + Using limits, + we can extend this once more to include all + powers, including irrational + (even transcendental!) + powers, giving the following theorem. +

    + + + Power Rule for Differentiation + +

    + Let f(x) = x^n, where n\neq 0 is a real number. + Then f is differentiable on its domain, + except possibly at x=0, + and \fp(x) = n\cdot x^{n-1}. + derivativePower Rule + Power Ruledifferentiation +

    +     +     + +

    + This theorem allows us to say the derivative of x^\pi is \pi x^{\pi -1}. +

    + +

    + We now apply this final version of the in the next example, + the second investigation of a famous curve. +

    + + + + + Using the Power Rule + +

    + Find the slope of x^{2/3}+y^{2/3}=8 at the point (8,8). +

    +     + +

    + This is a particularly interesting curve called an astroid. + It is the shape traced out by a point on the edge of a circle that is rolling around inside of a larger circle, + as shown in . +

    + +
    + An astroid, traced out by a point on the smaller circle as it rolls inside the larger circle + + + + A four pointed star with rounded edges, surrounded by a dashed circle. + + +

    + A dashed circle of radius 20 entirely contains the curve. + In each quadrant curves connect the points on the x and y axis which also lie on the circle. + This gives the overall curve the appearence of a diamond with sides curved towards the inside. + In the third quadrant a smaller circle is drawn which touches both the outer circle and the curve. + The point on the circle touching the curve is highlighted blue. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-24,xmax=24, + ymin=-24,ymax=24, + axis equal] + \addplot+[domain=0:pi,samples=200] ({22.63*cos(deg(2*x))^3},{22.63*sin(deg(2*x))^3}); + \addplot+[domain=0:pi,samples=200] ({22.63*cos(deg(2*x))},{22.63*sin(deg(2*x))}); + \addplot+[domain=0:pi,samples=101] ({5.66*cos(deg(2*x))-12},{5.66*sin(deg(2*x))-12}); + \addplot [soliddot] coordinates {(-8,-8)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +

    + To find the slope of the astroid at the point (8,8), + we take the derivative implicitly. + + \frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}\yp\amp =0 + \frac{2}{3}y^{-1/3}\yp \amp = -\frac{2}{3}x^{-1/3} + \yp\amp =-\frac{x^{-1/3}}{y^{-1/3}} + \yp\amp =-\frac{y^{1/3}}{x^{1/3}} = -\sqrt[3]{\frac{y}{x}} + . +

    + +

    + Plugging in x=8 and y=8, + we get a slope of -1. + The astroid, with its tangent line at (8,8), + is shown in . +

    + +
    + An astroid with a tangent line + + + + A previously described astroid with a tangent line in the first quadrant. + + +

    + The curve sketched in with a tangent line at (8,8). + It has a slope of -1. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-24,xmax=24, + ymin=-24,ymax=24, + axis equal] + \addplot+[domain=0:pi,samples=200] ({22.63*cos(deg(2*x))^3},{22.63*sin(deg(2*x))^3}); + \addplot [tangentline,domain=1:15] {-1*(x-8)+8}; + \addplot [soliddot] coordinates {(8,8)} node[above right] {$(8,8)$}; + \end{axis} + \end{tikzpicture} + + + + +
    +     +     +     + + + Implicit Differentiation and the Second Derivative +

    + We can use implicit differentiation to find higher order derivatives. + In theory, this is simple: + first find \lz{y}{x}, then take its derivative with respect to x. + In practice, + it is not hard, but it often requires a bit of algebra. + We demonstrate this in an example. +

    + + + Finding the second derivative + +

    + Given x^2+y^2=1, find \lzn{2}{y}{x} = \yp'. +

    +     + +

    + We found that \yp = \lz{y}{x} = -x/y in . + To find \yp', we apply implicit differentiation to \yp. + + \yp' \amp = \lzoo{x}{\yp} + \amp = \lzoo{x}{-\frac{x}{y}}\amp\amp\text{(Now use the Quotient Rule.)} + \amp = -\frac{y\cdot1 - x(\yp)}{y^2}\amp\amp\text{replace }\yp\text{ with }-x/y\text{:} + \amp = -\frac{y-x(-x/y)}{y^2} + \amp = -\frac{y+x^2/y}{y^2} + . +

    + +

    + While this is not a particularly simple expression, it is usable. + We can see that \yp' \gt 0 when y\lt 0 and \yp'\lt 0 when y \gt 0. + In , + we will see how this relates to the shape of the graph. +

    + +

    + Also, if we remember that we are only considering points on the curve x^2+y^2=1, + then we know that x^2=1-y^2. + So we can replace the x^2 in the expression for \yp' to get + + \yp'=-\frac{y+\left(1-y^2\right)/y}{y^2}=-\frac{1}{y^3} + + which is a simpler expression. + Recognizing when simplifications like this are possible is not always easy. +

    +     + +     +     + + + Logarithmic Differentiation +

    + Consider the function y=x^x; + it is graphed in . + It is well-defined for x \gt 0 and we might be interested in finding equations of lines tangent and normal to its graph. + How do we take its derivative? + + logarithmic differentiation + derivativelogarithmic +

    + + + +
    + A plot of y=x^x + + + + An exponential curve with a discontinuity at (0,1). It decreases slightly before increasing. + + +

    + The curve is entirely contained within the first quadrant. + At the point (0,1) there is a discontinuity. + The curve begins decreasing, reaching a minimum at around (0.3,0.7). + After that point, the curve increases exponentially. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-0.1,xmax=2.2, + ymin=-.1,ymax=4.2] + \addplot+[domain=0.001:2,rightarrow,samples=50] {x^x}; + \addplot[hollowdot] coordinates{(0,1)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +

    + The function is not a power function: + it has a power of x, not a constant. + It is not an exponential function either: + it has a base of x, not a constant. +

    + +

    + A differentiation technique known as + logarithmic differentiation becomes useful here. + The basic principle is this: + take the natural log of both sides of an equation y=f(x), + then use implicit differentiation to find \yp. + We demonstrate this in the following example. +

    + + + Using Logarithmic Differentiation + +

    + Given y=x^x, use logarithmic differentiation to find \yp. +

    +     + +

    + As suggested above, + we start by taking the natural log of both sides then applying implicit differentiation. + + y \amp = x^x\amp + \ln(y) \amp = \ln(x^x)\amp\amp\text{(apply logarithm rule)} + \ln(y) \amp = x\ln(x) \amp\amp\text{(now use implicit differentiation)} + \lzoo{x}{\ln(y)} \amp = \lzoo{x}{x\ln(x)} + \frac{\yp}{y} \amp = \ln(x) + x\cdot\frac{1}{x}\amp + \frac{\yp}{y} \amp = \ln(x) + 1\amp + \yp \amp = y\left(\ln(x) +1\right) \amp\amp\text{(substitute }y=x^x\text{)} + \yp \amp = x^x\left(\ln(x) +1\right)\amp + . +

    + +

    + To test our answer, + let's use it to find the equation of the tangent line at x=1.5. + The point on the graph our tangent line must pass through is \left(1.5, 1.5^{1.5}\right) \approx (1.5, 1.837). + Using the equation for \yp, we find the slope as + + \yp = 1.5^{1.5}\left(\ln(1.5) +1\right) \approx 1.837(1.405) \approx 2.582 + . +

    + +

    + Thus the equation of the tangent line is (approximately) y \approx 2.582(x-1.5)+1.837. + + graphs y=x^x along with this tangent line. +

    + +
    + A graph of y=x^x and its tangent line at x=1.5 + + + + The previously described exponential curve with a tangent line in the increasing part of the curve. + + +

    + The graph shown in , with a tangent line at + (1.5,1.5^{1.5}). It has a slope of around 2.6. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-0.1,xmax=2.2, + ymin=-.1,ymax=4.2] + \addplot+[domain=0.001:2,rightarrow,samples=50] {x^x}; + \addplot[hollowdot] coordinates{(0,1)}; + \addplot[tangentline,domain=1:2] {2.582 * (x - 1.5) + 1.837}; + \addplot[soliddot] coordinates{(1.5,1.837)} node[below right] {$\left(1.5,1.5^{1.5}\right)$}; + \end{axis} + \end{tikzpicture} + + + + +
    +     + +     + + + + + +

    + Implicit differentiation proves to be useful as it allows us to find the instantaneous rates of change of a variety of functions. + In particular, + it extended the to rational exponents, + which we then extended to all real numbers. + In , + implicit differentiation will be used to find the derivatives of + inverse functions, + such as y=\sin^{-1}(x). +

    +     + + + + Terms and Concepts + + + + +

    + In your own words, + explain the difference between implicit functions and explicit functions. +

    + +     + + + +     + + + + +

    + Implicit differentiation is based on what other differentiation rule? +

    +

    + + +

    +     + + + + + chain|chain rule|the chain rule + + + + + +     + + + + +

    + + Implicit differentiation can be used to find the derivative of y=\sqrt{x}. +

    +     + +

    + The function \sqrt{x} is one of two solutions to the equation y^2=x. +

    +     + +     + + + + +

    + + Implicit differentiation can be used to find the derivative of y=x^{3/4}. +

    +     + +

    + x^{3/4} is one of two (real) solutions to y^4=x^3. +

    +     + +     +     + + + Problems + + +

    + Compute the derivative of the given function. +

    +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $pm = list_random('+','-'); + if($envir{problemSeed}==1){$fname='f';$x='x';$pm='+'}; + Context()->variables->are($x=>'Real'); + Context()->variables->set($x=>{limits=>[0,4]}); + $f = Formula("sqrt($x) $pm 1/sqrt($x)")->reduce; + $fp = $f->D($x); + + +

    + ()= +

    + + To enter \sqrt{x}, type sqrt(x). + +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $pm = list_random('+','-'); + $n = random(3,6,1); + if($envir{problemSeed}==1){$fname='f';$x='x';$pm='+';$n=3;}; + Context("Fraction"); + parser::Root->Enable; + Context()->variables->are($x=>'Real'); + Context()->variables->set($x=>{limits=>[0,4]}); + $f = Formula("root($n,$x) $pm $x^(($n-1)/$n)"); + $fp = $f->D($x); + + +

    + ()= +

    + + To enter \sqrt{x}, type sqrt(x). + To enter \sqrt[n]{x}, type root(n,x). + +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $pm = list_random('+','-'); + $a = random(1,4,1)**2; + $n = random(2,4,1); + if($envir{problemSeed}==1){$fname='f';$x='t';$pm='-';$a=1;$n=2;}; + Context()->variables->are($x=>'Real'); + Context()->variables->set($x=>{limits=>[-$a**(1/$n),$a**(1/$n)]}); + $f = Formula("sqrt($a $pm $x^$n)"); + $fp = $f->D($x); + + +

    + ()= +

    + + To enter \sqrt{x}, type sqrt(x). + +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $trig = list_random('sin','cos','tan'); + if($envir{problemSeed}==1){$fname='g';$x='t';$trig='sin';}; + Context()->variables->are($x=>'Real'); + Context()->variables->set($x=>{limits=>[0,pi/2]}); + $f = Formula("sqrt($x)$trig($x)"); + $fp = $f->D($x); + + +

    + ()= +

    + + To enter \sqrt{x}, type sqrt(x). + +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $n = random(1,3,1) + random(0.1,0.9,0.1); + $trig = list_random('sin','cos','tan'); + if($envir{problemSeed}==1){$fname='h';$x='x';$n=1.5;}; + Context()->variables->are($x=>'Real'); + Context()->variables->set($x=>{limits=>[0,4]}); + $f = Formula("$x^$n"); + $fp = $f->D($x); + + +

    + ()= +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $n = random(1,3,1) + random(0.1,0.9,0.1); + $b = list_random(Formula("pi"),Formula("e")); + if($envir{problemSeed}==1){$fname='f';$x='x';$n=1.9;$b=Formula("pi")}; + Context()->variables->are($x=>'Real'); + Context()->variables->set($x=>{limits=>[0,4]}); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $f = Formula("$x^$b + $x^$n + $b^$n"); + $fp = $f->D($x); + + +

    + ()= +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $a = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$fname='g';$x='x';$a=7;}; + Context()->variables->are($x=>'Real'); + Context()->variables->set($x=>{limits=>[0,4]}); + $f = Formula("($x+$a)/sqrt($x)"); + $fp = $f->D($x); + + +

    + ()= +

    + + To enter \sqrt{x}, type sqrt(x). + +

    + +

    +     +     +     + + + + + Context("Fraction"); + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $n = random(3,9,1); + $trig = list_random('sin','cos','tan','sec','cot','csc'); + if($envir{problemSeed}==1){$fname='f';$x='t';$n=5;$trig='sec';}; + Context()->variables->are($x=>'Real'); + Context()->variables->set($x=>{limits=>[0,4]}); + parser::Root->Enable; + $f = Formula("root($n,$x)($trig($x)+e^$x)"); + $fp = $f->D($x); + + +

    + ()= +

    + + To enter \sqrt{x}, type sqrt(x). + To enter \sqrt[n]{x}, type root(n,x). + +

    + +

    +     +     +     +     + + + +

    + Find \lz{y}{x} using implicit differentiation. +

    +     + + + + Context()->variables->add(y=>'Real'); + $dydx=Formula("-4x^3/(2y+1)"); + @y=(random(-2,2,0.2),random(-2,2,0.2),random(-2,2,0.2),random(-2,2,0.2),random(-2,2,0.2)); + @x=map{(7-($_)-($_)**2)**(1/4)*(-1)**(random(-1,1,2))}(@y); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + x^4+y^2+y=7 +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $dydx=Formula("-y^(3/5)/x^(3/5)"); + @y=(random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01)); + @x=map{(1-($_)**(2/5))**(5/2)}(@y); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + x^{2/5}+y^{2/5}=1 +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $dydx=Formula("sin(x)sec(y)"); + @y=(random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01)); + @x=map{acos(1-sin($_))}(@y); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + \cos(x)+\sin(y)=1 +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $dydx=Formula("y/x"); + @x=(random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1)); + @y=map{$_/10}(@x); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + \dfrac{x}{y}=10 +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $dydx=Formula("y/x"); + $a = random(5,15,1); + if($envir{problemSeed}==1){$a=10;}; + #$a=10; + @y=(random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1)); + @x=map{$_/$a}(@y); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + \dfrac{y}{x}= +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + Context()->flags->set(reduceConstantFunctions=>0); + $m=list_random(2,3,5); + $b=random(2,9,1); + $c=random(4,9,1); + if($envir{problemSeed}==1){$m=2;$b=2;$c=5;}; + #$m=2;$b=2;$c=5; + $dydx=Formula("-(e^x x(x+$m) $b^(-y))/(ln($b))"); + @x=(random(-4,1,0.1),random(-4,1,0.1),random(-4,1,0.1),random(-4,1,0.1),random(-4,1,0.1)); + @y=map{ln($c-($_)**$m*exp($_))/ln($b)}(@x); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + x^e^x+^y= +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $m = random(2,9,1); + $trig = list_random('sin','cos','tan','cot'); + $c = random(1,9,1); + if($envir{problemSeed}==1){$m=2;$trig='tan';$c=50;}; + #$m=2;$trig='tan';$c=50; + if ($trig eq 'sin') {$dydx=Formula("-$m tan(y)/x");}; + if ($trig eq 'cos') {$dydx=Formula("$m cot(y)/x");}; + if ($trig eq 'tan') {$dydx=Formula("-$m sin(y) cos(y)/x");}; + if ($trig eq 'cot') {$dydx=Formula("$m sin(y) cos(y)/x");}; + # need to avoid x==0 + @x=(random(3,7,0.2),random(3,7,0.2),random(3,7,0.2),random(3,7,0.2),random(3,7,0.2)); + if ($trig eq 'sin') {@y=map{asin($c/($_)**$m)}(@x);}; + if ($trig eq 'cos') {@y=map{acos($c/($_)**$m)}(@x);}; + if ($trig eq 'tan') {@y=map{atan($c/($_)**$m)}(@x);}; + if ($trig eq 'cot') {@y=map{pi/2 - atan($c/($_)**$m)}(@x);}; + $left = Formula("x^$m $trig(y)"); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + = +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + parser::Root->Enable; + Context()->variables->add(y=>'Real'); + ($a,$b) = random_subset(2,2..5); + $c = random(1,9,1); + ($m,$n,$p) = random_subset(3,2..5); + if($envir{problemSeed}==1){$a=3;$b=2;$c=2;$m=2;$n=3;$p=4;}; + #$a=3;$b=2;$c=2;$m=2;$n=3;$p=4; + $w = ($c**(1/$p)/$a)**(1/$m); + $left = Formula("($a x^$m + $b y^$n)^$p"); + $right = Formula("$c"); + $frac = Fraction(-$a*$m,$b*$n); + $dydx=Formula("$frac x^($m-1) / y^($n-1)")->reduce; + do { + @x=(random(-$w,$w,0.01),random(-$w,$w,0.01),random(-$w,$w,0.01),random(-$w,$w,0.01),random(-$w,$w,0.01)); + if ($n % 2 == 1) { + @y=map{root($n,(root($p,$c) - $a*($_)**$m)/$b)}(@x); + } else { + @y=map{random(-1,1,2)*root($n,(root($p,$c) - $a*($_)**$m)/$b)}(@x); + } + } until (!grep( /^0$/, @y )); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + = +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + parser::Root->Enable; + Context()->variables->add(y=>'Real'); + $a = non_zero_random(-9,9,1); + $b = random(100,200,5); + $pm = list_random('+','-'); + $m = random(2,4,1); + $n = random(2,4,1); + if($envir{problemSeed}==1){$a=2;$b=200;$pm='-';$m=2;$n=2;}; + #$a=2;$b=200;$pm='-';$m=2;$n=2; + $left = Formula("(y^$m+$a y $pm x)^$n"); + $right = Formula("$b"); + $dydx=Formula("-($pm 1/($m y^($m-1)+$a))")->reduce->reduce; + @y=(random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2)); + @x=map{(root($n,$b) - $_**$m - $_) * (($pm eq '+') ? 1 : -1);}(@y); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + = +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->variables->add(y=>'Real'); + ($m,$n) = random_subset(2,1..2); + $c = random(10,20,1); + if($envir{problemSeed}==1){$m=2;$n=1;$c=17}; + #$m=2;$n=1;$c=17; + $left = Formula("(x^$m + y^$n)/(x^$n + y^$m)")->reduce; + $right = Formula("$c"); + $dydx = Formula("(($n-$m) x^2 + $n x^($n-1) y^$n - $m x^($m-1) y^$m)/($n x^$n y^($n-1) + ($n-$m) y^2 - $m x^$m y^($m-1))")->reduce; + @y=(random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2)); + if ($m == 2) { + @x=map{($c+(-1)**(random(-1,1,2))*sqrt($c**2 - 4*($_)*(1 - $c*($_))))/2}(@y); + } else { + @x=map{(1+(-1)**(random(-1,1,2))*sqrt(1 - 4*$c*($_)*($c - ($_))))/(2*$c)}(@y); + } + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + = +

    +

    + +

    +

    + +

    +     + +

    + If one takes the derivative of the equation, + as shown, using the Quotient Rule, + one finds \frac{dy}{dx}=\frac{x^2+2 x y^2-y}{2 x^2 y-x+y^2}. +

    +

    + If one first clears the denominator and writes + x^2+y=17(x+y^2) then takes the derivative of both sides, + one finds \frac{dy}{dx}= \frac{2x-17}{34y-1}. +

    +

    + These expressions, by themselves, are not equal. + However, for values of x and y that satisfy the original equation + (i.e, for x and y such that \frac{x^2+y}{x+y^2}=17), + these expressions are equal. +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + ($triga,$trigb) = random_subset(2,'sin','cos'); + if($envir{problemSeed}==1){$triga='sin';$trigb='cos';}; + #$triga='sin';$trigb='cos'; + $left = Formula("($triga(x)+y)/($trigb(y)+x)"); + $right = Formula("1"); + if ($triga eq 'sin') { + $dydx = Formula("(1-cos(x))/(sin(y)+1)")->reduce; + $dydx->{test_points}=[[-3.1,-3.83041],[-0.41,0.732269],[0.59,0.759097],[2.6,1.82908],[5.3,6.93017]]; + } else { + $dydx = Formula("(sin(x)+1)/(1-cos(y))")->reduce; + $dydx->{test_points}=[[0.831958,1],[1.329583,2],[-0.0952339,-2],[-3.70455,-3],[0.7390851,0]]; + } + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + = +

    +

    + +

    +

    + +

    +     + +

    + If one takes the derivative of the equation, + as shown, using the Quotient Rule, + one finds \frac{dy}{dx} = \frac{-\cos (x) (x+\cos (y))+\sin (x)+y}{\sin (y) (\sin (x)+y)+x+\cos (y)}. +

    +

    + If one first clears the denominator and writes + \sin(x)+y = \cos(y)+x then takes the derivative of both sides, + one finds \frac{dy}{dx} = \frac{1-\cos(x)}{1+\sin(y)}. +

    +

    + These expressions, by themselves, are not equal. + However, for values of x and y that satisfy the original equation + (i.e, for x and y such that \frac{\sin(x)+y}{\cos(y)+x}=1), + these expressions are equal. +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $dydx=Formula("-x/y"); + @t=(random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01)); + @x=map{exp(exp(1)/2)*cos($_)}(@t); + @y=map{exp(exp(1)/2)*sin($_)}(@t); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + \ln\mathopen{}\left(x^2+y^2\right)\mathclose{}=e +

    +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $dydx=Formula("-(2x+y)/(2y+x)"); + @t=(random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01)); + @u=map{sqrt(2/3*exp(1))*cos($_)}(@t); + @v=map{sqrt(2*exp(1))*sin($_)}(@t); + @x=map{($u[$_]-$v[$_])/sqrt(2)}(0..4); + @y=map{($u[$_]+$v[$_])/sqrt(2)}(0..4); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $dydx->{test_points}=~~@xy; + $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; + + +

    + \ln\mathopen{}\left(x^2+xy+y^2\right)\mathclose{} = 1 +

    +

    + +

    +

    + +

    +     +     +     +     + + + +

    + Show that \lz{y}{x} is the same for each of the following implicitly defined functions. +

    +     + + + +

    + xy=1 +

    +     +     + + +

    + x^2y^2=1 +

    +     +     + + +

    + \sin(xy) = 1 +

    +     +     + + +

    + \ln(xy) =1 +

    +     +     +     + + + + +

    + Find the equation of the tangent line to the graph of the implicitly defined function at the indicated points. + As a visual aid, the function is graphed. +

    +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + @eq=(Formula("y=0"),Formula("y=-1.859(x-0.1)+0.2811")); + + +

    + On the curve x^{2/5}+y^{2/5} = 1. +

    + + + An astroid of radius 1 with a point drawn at (0.1,0.281). + + +

    + An astroid of radius 1. + A point is drawn on the curve, at (0.1,0.281). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.1,ymax=1.1,% + xmin=-1.1,xmax=1.1,% + ] + + \addplot [firstcurvestyle,domain=0:360,samples=101] ({(cos(x))^5},{(sin(x))^5}); + \filldraw [] (axis cs:.1,.28) node [above right] {\((0.1,0.281)\)} circle (1pt); + \end{axis} + \end{tikzpicture} + + +     + + +

    + At (1,0). +

    +

    + +

    +     +     + + +

    + At (0.1,0.2811) (which does not exactly + lie on the curve, but is very close). +

    +

    + +

    +     +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + @eq=(Formula("x=1"),Formula("y=-3sqrt(3)/8(x-sqrt(0.6))+sqrt(0.8)"),Formula("y=1")); + + +

    + On the curve x^{4}+y^{4} = 1. +

    + + + A square with rounded corners and edges with a point in the first quadrant. + + + A curve that lies in all 4 quadrants. + It has the appearance of a square with rounded sides and corners. + A point is drawn at (\sqrt{0.6},\sqrt{0.8}). + + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.1,ymax=1.1,% + xmin=-1.1,xmax=1.1,% + ] + + \addplot [firstcurvestyle] coordinates {(1.,0) (0.98901,0.45597) (0.9558,0.63776) (0.89945,0.76667) (0.818,0.86206) (0.70711,0.9306) (0.55589,0.97522) (0.32331,0.99726) (-0.32331,0.99726) (-0.55589,0.97522) (-0.70711,0.9306) (-0.818,0.86206) (-0.89945,0.76667) (-0.9558,0.63776) (-0.98901,0.45597) (-1.,0) (-1.,0) (-0.98901,-0.45597) (-0.9558,-0.63776) (-0.89945,-0.76667) (-0.818,-0.86206)(-0.70711,-0.9306) (-0.55589,-0.97522) (-0.32331,-0.99726)(0.32331,-0.99726) (0.55589,-0.97522) (0.70711,-0.9306)(0.818,-0.86206) (0.89945,-0.76667) (0.9558,-0.63776)(0.98901,-0.45597) (1.,0) }; + \filldraw [] (axis cs:.775,.894) node [below left] {\((\sqrt{0.6},\sqrt{0.8})\)} circle (1pt); + + \end{axis} + \end{tikzpicture} + + +     + + +

    + At (1,0). +

    +

    + +

    +     +     + + +

    + At \left(\sqrt{0.6},\sqrt{0.8}\right). +

    +

    + +

    +     +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + @eq=(Formula("y=4"),Formula("y=3/108^(1/4)(x-2)-108^(1/4)")); + + +

    + On the curve (x^2+y^2-4)^3 = 108y^2. +

    + + + A combination of 2 circles, with cusps at (-2,0) and (2,0) with a tangent line in the fourth quadrant. + + +

    + A curve which seems like two circles combined. + Beginning at the top of the curve, at the point (0,4), the curve decreases towards the right. + It continues to decreases, bending towards the y-axis as the curve nears the x-axis. + The curve passes through the x-axis at a point close to (2.1,0), forming a cusp at that point. + In the fourth quadrant, the curve bends outwards slightly, before curving wide towards the y-axis. + The curve passes through the y-axis at the point (0,-4). + The left side of the curve is symmetrical to the right, again curving inwards and forming a cusp at (-2,0). + From there the curve bends out, before curving in and passing through the point (0,4). + A point is drawn in the lower left of the curve at (2,-\sqrt[4]{108}). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-4.5,ymax=4.5,% + xmin=-4.5,xmax=4.5,% + ] + + \addplot [firstcurvestyle] coordinates {(-0.22282,-3.9915) (-0.57143,-3.9441) (-1.,-3.8244) (-1.2959,-3.7041)(-1.5714,-3.551) (-1.8453,-3.3571) (-2.0412,-3.1841) (-2.2149,-3.)(-2.3706,-2.7991) (-2.4727,-2.6429) (-2.5832,-2.4403)(-2.7019,-2.1552) (-2.783,-1.8571) (-2.8275,-1.4582)(-2.8002,-1.0859) (-2.7143,-0.74808) (-2.5958,-0.5)(-2.4708,-0.31491) (-2.3571,-0.19279) (-2.2143,-0.081938)(-2.0659,-0.005515) (-2.2078,0.077938) (-2.4036,0.23922) + (-2.5608,0.4392) (-2.6877,0.68771) (-2.786,1.) (-2.8262,1.3571)(-2.8065,1.7143) (-2.7075,2.136) (-2.5114,2.5714) (-2.2471,2.9614)(-2.0396,3.1825) (-1.6513,3.5) (-1.2857,3.7088) (-1.0714,3.8007)(-0.78571,3.8938) (-0.52476,3.9533) (-0.28571,3.9855) + (-0.070563,3.9991) (0.27289,3.9872) (0.64411,3.9298) (1.082,3.7963)(1.4286,3.6354) (1.7143,3.4554) (1.9675,3.2532) (2.1511,3.0714)(2.3284,2.8571) (2.4288,2.7143) (2.5351,2.5351) (2.6283,2.3426)(2.7079,2.1365) (2.7939,1.7939) (2.8255,1.3969) (2.8017,1.0874)(2.7481,0.85714) (2.6429,0.58296) (2.552,0.42857) (2.438,0.27631) + (2.3214,0.16103) (2.1786,0.06028) (2.1143,-0.028574)(2.2857,-0.13046) (2.4439,-0.28571) (2.5958,-0.5) (2.7143,-0.74808)(2.8002,-1.0859) (2.827,-1.4286) (2.7984,-1.773) (2.7079,-2.1365)(2.6052,-2.3948) (2.4292,-2.7137) (2.2453,-2.9596) (2.,-3.2231)(1.7143,-3.4554) (1.4682,-3.611) (1.1839,-3.7553) (0.86968,-3.8697) + (0.57143,-3.9441) (0.33765,-3.9805) (0.072336,-3.9991) (-0.22282,-3.9915)}; + + \filldraw [] (axis cs:2,-3.22) node [shift={(15pt,-10pt)}] {\((2,-\sqrt[4]{108})\)} circle (1pt); + + \end{axis} + \end{tikzpicture} + + +     + + +

    + At (0,4). +

    +

    + +

    +     +     + + +

    + At \left(2,-\sqrt[4]{108}\right). +

    +

    + +

    +     +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + @eq=(Formula("y=-x+1"),Formula("y=3sqrt(3)/4")); + + +

    + On the curve (x^2+y^2+x)^2 = x^2+y^2. +

    + + + An oval with a cusp on the right side. + + + An oval with a cusp on the right side. + A majority of the curve lies to the left of the y-axis. + From the top of the curve, the curve decreases towards the right. + It enters the first quadrant through the point (0,1). + As the curve nears the x-axis, it bends back toward the y-axis, forming a cusp at the origin. + The curve then bends outwards into the fourth quadrant. + The curve continues downwards and to the left, passing the y-axis through the point (0,-1). + In the third quadrant the curve bends upwards, passing vertically into the second quadrant through the point (-2,0). + The curve bends upwards and to the right, once again meeting the top of the curve. + The point at the top of the curve is drawn at (-\frac{3}{4},\frac{3\sqrt{3}}{4}). + + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.5,ymax=1.5,% + xmin=-2.5,xmax=0.5,% + ] + + \addplot [firstcurvestyle,samples=150] coordinates {(-2.,0) (-1.9547,-0.34466) (-1.8227,-0.66341) (-1.616,-0.93301)(-1.3529,-1.1352) (-1.056,-1.2584) (-0.75,-1.299) (-0.459,-1.2611)(-0.2038,-1.1558) (0,-1.) (0.14349,-0.8138) (0.22504,-0.6183)(0.25,-0.43301) (0.22961,-0.27364) (0.17922,-0.15038)(0.11603,-0.066987) (0.05667,-0.020626) (0.014961,-0.0026381) (0,0) + (0.014961,0.0026381) (0.05667,0.020626) (0.11603,0.066987)(0.17922,0.15038) (0.22961,0.27364) (0.25,0.43301) (0.22504,0.6183)(0.14349,0.8138) (0,1.) (-0.2038,1.1558) (-0.459,1.2611)(-0.75,1.299) (-1.056,1.2584) (-1.3529,1.1352) (-1.616,0.93301)(-1.8227,0.66341) (-1.9547,0.34466) (-2.,0)}; + + \filldraw [] (axis cs:-.75,1.3) node [shift={(0pt,-12pt)}] {\(\left(-\frac{3}{4},\frac{3\sqrt{3}}{4}\right)\)} circle (1pt); + + \end{axis} + \end{tikzpicture} + + +     + + +

    + At (0,1). +

    +

    + +

    +     +     + + +

    + At \left(-\frac{3}{4}, \frac{3\sqrt{3}}{4}\right). +

    +

    + +

    +     +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + @eq=(Formula("y=-1/sqrt(3)(x-7/2)+(6+3sqrt(3))/2"),Formula("y=sqrt(3)(x-(4+3sqrt(3)))/2+3/2")); + + +

    + On the curve (x-2)^2+(y-3)^2=9. +

    + + + A circle of radius 3 centered at (2,3). Two points are drawn on the circle. + + +

    + A circle of radius 3 centered at (2,3). + Two points are drawn, one at (3.5,\frac{6+3\sqrt{3}}{2}), + and the other at (\frac{4+3\sqrt{3}}{2},1.5). + The first point is in the upper right side of the circle. + The second point is in the lower right side of the circle. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.5,ymax=6.5,% + xmin=-1.5,xmax=6.5,% + ] + + \addplot [firstcurvestyle,samples=101,domain=0:360] ({(2+3*cos(x))},{(3+3*sin(x))}); + \filldraw [] (axis cs:4.6,1.5) node [left] {\(\left(\frac{4+3\sqrt{3}}{2},1.5\right)\)} circle (1pt); + \filldraw [] (axis cs:3.5,5.6) node [below left] {\(\left(3.5,\frac{6+3\sqrt{3}}{2}\right)\)} circle (1pt); + + \end{axis} + \end{tikzpicture} + + +     + + +

    + At \left(\frac{7}{2},\frac{6+3\sqrt{3}}{2}\right). +

    +

    + +

    +     +     + + +

    + At \left(\frac{4+3\sqrt{3}}{2},\frac{3}{2}\right). +

    +

    + +

    +     +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + @eq=(Formula("y=1"),Formula("y=-2/sqrt(5)(x+1)+1/2(-1+sqrt(5))"),Formula("y=2/sqrt(5)(x+1)+1/2(-1-sqrt(5))")); + + +

    + On the curve x^2+y^3+2xy=0. +

    + + + A curve that begins in the third quadrant, forms a loop in the third quadrant, and decreases in the fourth quadrant. + + +

    + The curve begins in the third quadrant. + The curve increases to the right, passing through the origin almost vertically. + From there, the curve forms a loop in the third quadrant, then decreasing and passing through the origin again. + The curve passes through the origin into the fourth qudrant, where the curve gently decreases. + Three points are drawn on the curve. + The first point is in the third quadrant, at the point (-1,\frac{-1-\sqrt{5}}{2}). + The second point is in the second quadrant, at the point (-1,1). + The third point also lies in the second quadrant, at the point (-1,\frac{-1+\sqrt{5}}{2}). + The second point lies on the top of the loop, while the third point lies on the bottom of the loop. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-2.5,ymax=2.5,% + xmin=-2.5,xmax=2.5,% + ] + \addplot [firstcurvestyle,samples=101,domain=45:157.5] ({((-sin(2*x)-(cos(x))^2)*cos(x)/((sin(x))^3))},{((-sin(2*x)-(cos(x))^2)/((sin(x))^2))}); + \filldraw [] (axis cs:-1,1) node [above] {\((-1,1)\)} circle (1pt); + \filldraw [thin,->,>=latex] (axis cs:1.25,-1.5) node [fill=white] {\(\left(-1,\frac{-1-\sqrt{5}}2\right)\)} -- (axis cs: -.9,-1.62); + \filldraw [] (axis cs: -1,-1.62) circle (1pt); + \filldraw [->,thin,>=latex] (axis cs:1.25,0.5) node [ fill=white] {\(\left(-1,\frac{-1+\sqrt{5}}2\right)\)} -- (axis cs: -.9,0.62); + \filldraw [] (axis cs:-1,0.62) circle (1pt); + + \end{axis} + \end{tikzpicture} + + +     + + +

    + At (-1,1). +

    +

    + +

    +     +     + + +

    + At \left(-1,\frac12(-1+\sqrt{5})\right). +

    +

    + +

    +     +     + + +

    + At \left(-1,\frac12(-1-\sqrt{5})\right). +

    +

    + +

    +     +     +     +     +     + + + +

    + An implicitly defined function is given. + Find \lzn{2}{y}{x}. + Note: these are the same functions used in Exercises through . +

    +     + + + + Context()->variables->add(y=>'Real'); + $ddyddx=Formula("-((2y+1)(12x^2)-4x^3(2*-(4x^3)/(2y+1)))/(2y+1)^2"); + @y=(random(-2,2,0.1),random(-2,2,0.1),random(-2,2,0.1),random(-2,2,0.1),random(-2,2,0.1)); + @x=map{(7-($_)-($_)**2)**(1/4)*(-1)**(random(-1,1,2))}(@y); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $ddyddx->{test_points}=~~@xy; + + +

    + x^4+y^2+y=7 +

    +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $ddyddx=Formula("-(x^(3/5)*3/5y^(-2/5)(-y^(3/5)/x^(3/5))-y^(3/5)*3/5x^(-2/5))/x^(6/5)"); + @y=(random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01)); + @x=map{(1-($_)**(2/5))**(5/2)}(@y); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $ddyddx->{test_points}=~~@xy; + + +

    + x^{2/5}+y^{2/5}=1 +

    +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $ddyddx=Formula("sin^2(x)sec^2(y)tan(y)+cos(x)sec(y)"); + @y=(random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01)); + @x=map{acos(1-sin($_))}(@y); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $ddyddx->{test_points}=~~@xy; + + +

    + \cos(x)+\sin(y)=1 +

    +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $ddyddx=Formula("0"); + @x=(random(-4,4,0.1),random(-4,4,0.1),random(-4,4,0.1),random(-4,4,0.1),random(-4,4,0.1)); + @y=map{$_/10}(@x); + @xy=map{[$x[$_],$y[$_]]}(0..4); + $ddyddx->{test_points}=~~@xy; + + +

    + \dfrac{x}{y}=10 +

    +

    + +

    +     +     +     +     + + + +

    + Use logarithmic differentiation to find \lz{y}{x}, + then find the equation of the tangent line at the indicated x-value. +

    +     + + + + Context()->variables->set(x=>{limits=>[-0.9,4]}); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $dydx=Formula("(1+x)^(1/x)(1/(x(x+1))-ln(1+x)/x^2)"); + Context("Numeric"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=(1-2ln(2))(x-1)+2"); + $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; + + +

    + y=(1+x)^{1/x} at x=1 +

    + + Enter \lz{y}{x} here. + +

    + +

    + + Enter the equation of the tangent line here. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->set(x=>{limits=>[0.01,4]}); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $dydx=Formula("(2x)^(x^2)(2x ln(2x)+x)"); + Context("Numeric"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=(2+4ln(2))(x-1)+2"); + $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; + + +

    + y=(2x)^{x^2} at x=1 +

    + + Enter \lz{y}{x} here. + +

    + +

    + + Enter the equation of the tangent line here. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->set(x=>{limits=>[0.01,4]}); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $dydx=Formula("x^x/(x+1)(ln(x)+1-1/(x+1))"); + Context("Numeric"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=1/4*(x-1)+1/2"); + $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; + + +

    + y=\dfrac{x^{x}}{x+1} at x=1 +

    + + Enter \lz{y}{x} here. + +

    + +

    + + Enter the equation of the tangent line here. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->variables->set(x=>{limits=>[0.01,4]}); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $dydx=Formula("x^(sin(x)+2)(cos(x)*ln(x)+(sin(x)+2)/x)"); + Context("Numeric"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=(3pi^2)/4*(x-pi/2)+(pi/2)^3"); + $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; + + +

    + y=x^{\sin(x)+2} at x=\pi/2 +

    + + Enter \lz{y}{x} here. + +

    + +

    + + Enter the equation of the tangent line here. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $dydx=Formula("(x+1)/(x+2)(1/(x+1)-1/(x+2))"); + Context("Numeric"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=1/9(x-1)+2/3"); + $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; + + +

    + y=\dfrac{x+1}{x+2} at x=1 +

    + + Enter \lz{y}{x} here. + +

    + +

    + + Enter the equation of the tangent line here. + +

    + +

    +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $dydx=Formula("((x+1)(x+2))/((x+3)(x+4))(1/(x+1)+1/(x+2)-1/(x+3)-1/(x+4))"); + Context("Numeric"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $t=Formula("y=11/72x+1/6"); + $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; + + +

    + y=\dfrac{(x+1)(x+2)}{(x+3)(x+4)} at x=0 +

    + + Enter \lz{y}{x} here. + +

    + +

    + + Enter the equation of the tangent line here. + +

    + +

    +

    + +

    +     +     +     +     +     +     +
    +
    + Derivatives of Inverse Functions + + + +

    + Recall that a function y=f(x) is said to be one-to-one + if it passes the horizontal line test; + that is, for two different x values x_1 and x_2, + we do not have f\mathopen{}\left(x_1\right)\mathclose{}=f\mathopen{}\left(x_2\right)\mathclose{}. + In some cases the domain of f must be restricted so that it is one-to-one. + For instance, consider f(x)=x^2. + Clearly, f(-1)= f(1), + so f is not one-to-one on its regular domain, + but by restricting f to + (0,\infty), f is one-to-one. + + derivativeinverse function +

    + +

    + Now recall that one-to-one functions have inverses. + That is, if f is one-to-one, + it has an inverse function, denoted by f^{-1}, + such that if f(a)=b, then f^{-1}(b) = a. + The domain of f^{-1} is the range of f, and vice-versa. + For ease of notation, we set + g=f^{-1} and treat g as a function of x. +

    + +

    + Since f(a)=b implies g(b)=a, + when we compose f and g we get a nice result: + + f\big(g(b)\big) = f(a) = b + . +

    + +

    + In general, f\big(g(x)\big) = x and g\big(f(x)\big) = x. + This gives us a convenient way to check if two functions are inverses of each other: + compose them and if the result is x + (on the appropriate domains), + then they are inverses. +

    + + + +

    + When the point (a,b) lies on the graph of f, + the point (b,a) lies on the graph of g. + This leads us to discover that the graph of g is the reflection of f across the line y=x. + In + we see a function graphed along with its inverse. + See how the point (1,1.5) lies on one graph, + whereas (1.5,1) lies on the other. + Because of this relationship, + whatever we know about f can quickly be transferred into knowledge about g. +

    + +
    + A function f along with its inverse f^{-1}. (Note how it does not matter which function we refer to as f; the other is f^{-1}.) + + + A function f along with its inverse. Since they are inverses of each other It doesn't matter which one we call f. + +

    + The graph of f starts at (-1,-0.5) below the negative x axis then slopes upwards while moving to the right. + A little after passing throught he point (-0.5,0.375) it slopes further and faster to the right and starts moving + horizontally parallel to the x axis. Then it starts to move upagain continuuing through (1, 1.5). The graph of g is a + similar curve starting below the x axis close to the y axis in the third quadrant it moves to the right while moving up + and then after passing through (0.375,-0.5) it moves upwards vertically for a short while and then starts moving to + the right while also moving upwards passing through (1.5,1). +

    +     + + + \begin{tikzpicture} + \begin{axis}[axis equal image, + ymin=-1.3,ymax=2, + xmin=-1.3,xmax=2, + clip=false] + \addplot+[infinite,domain=-1:1.14] ({x},{x^3+.5}); + \addplot+[infinite,domain=-1:1.14] ({x^3+.5},{x}); + \addplot [symmetryline] coordinates {(-1.3,-1.3) (1.7,1.7)}; + \addplot [soliddot] coordinates{(-.5,.375)} node[above left] {$(-0.5,0.375)$}; + \addplot [soliddot] coordinates{(.375,-.5)} node[below right] {$(0.375,-0.5)$}; + \addplot [soliddot] coordinates{(1,1.5)} node[above left] {$(1,1.5)$}; + \addplot [soliddot] coordinates{(1.5,1)} node[below right] {$(1.5,1)$}; + \end{axis} + \end{tikzpicture} + + + + +
    + +

    + For example, consider + where the tangent line to f at the point (1,1.5) is drawn. + That line has slope 3. + Through reflection across y=x, + we can see that the tangent line to g at the point (1.5,1) has slope 1/3. + Their slopes are reciprocals. + This should make sense since reflecting a line + (such as a tangent line) + across the line y=x switches the x and y values. + Also consider the point (0,0.5) on the graph of f, + where the tangent line is horizontal. + At the point (0.5,0) on g, + the tangent line is vertical. +

    + +

    + More generally, + consider the tangent line to f at the point (a,b). + That line has slope \fp(a). + Through reflection across y=x, + we can extend our above observation to say that the tangent line to g at the point (b,a) should have slope 1/\fp(a). + This then tells us that \gp(b)=1/\fp(a). +

    + +
    + Corresponding tangent lines drawn to f and f^{-1} + + + Corresponding tangent lines drawn to f and its inverse + +

    + The graph of f starts close to (-1,-0.5) below the negative x axis and slopes upwards and to the right. + After passing through (-0.5,0.375) it starts moving to the right. After intersecting the y axis at + (0,0.5) it continues to move horizontally for a short while then starts moving upwards to the right + and continues upwards through the point (1,1.5). The tangent line of f is drawn at (1,1.5), + it is a straight line that slightly touches the the curve at (1,1.5) and keeps moving straight. g starts + close to the negative y axis and starts moving to the right while sloping upwards. Once it moves through the point + (0.375,-0.5) then starts moving vertically upward and eventually starts moving again upwards and to the right. + The tangent line of the curve is drawn at (1.5,1). +

    +     + + + \begin{tikzpicture} + \begin{axis}[axis equal image, + ymin=-1.3,ymax=2, + xmin=-1.3,xmax=2, + clip=false, + %grid=both, + xtick={-1,1}, + ytick={-1,1}, + minor xtick={-1.25,-1,...,2}, + minor ytick={-1.25,-1,...,2},] + \addplot+[infinite,domain=-1:1.14] ({x},{x^3+.5}); + \addplot+[infinite,domain=-1:1.14] ({x^3+.5},{x}); + \addplot [symmetryline] coordinates {(-1.3,-1.3) (1.7,1.7)}; + \addplot [tangentlineseg,domain=0.75:1.25]{3*(x-1)+1.5}; + \addplot [tangentlineseg,domain=0.75:2]{1/3*(x-1.5)+1}; + \addplot [soliddot] coordinates{(-.5,.375)} node[above left] {$(-0.5,0.375)$}; + \addplot [soliddot] coordinates{(.375,-.5)} node[below right] {$(0.375,-0.5)$}; + \addplot [soliddot] coordinates{(1,1.5)} node[above left] {$(1,1.5)$}; + \addplot [soliddot] coordinates{(1.5,1)} node[below right] {$(1.5,1)$}; + \end{axis} + \end{tikzpicture} + + + + +
    + +

    + The information from these two graphs is summarized in below: +

    + + + + <tabular> + <row bottom="medium"> + <cell>Information about <m>f</m></cell> + <cell>Information about <m>g=f^{-1}</m></cell> + </row> + <row> + <cell><m>(1,1.5)</m> lies on <m>f</m></cell> + <cell><m>(1.5,1)</m> lies on <m>g</m></cell> + </row> + <row> + <cell> + <!-- <p> --> + Slope of tangent line to <m>f</m> at <m>x=1</m> is <m>3</m> + <!-- </p> --> + </cell> + <cell> + <!-- <p> --> + Slope of tangent line to <m>g</m> at <m>x=1.5</m> is <m>1/3</m> + <!-- </p> --> + </cell> + </row> + <row> + <cell><m>\fp(1) = 3</m></cell> + <cell><m>\gp(1.5) = 1/3</m></cell> + </row> + </tabular> + + </table> + + + + <p> + We have discovered a relationship between <m>\fp</m> and <m>\gp</m> in a mostly graphical way. + We can realize this relationship analytically as well. + Let <m>y = g(x)</m>, where again <m>g = f^{-1}</m>. + We want to find <m>\yp</m>. + Since <m>y = g(x)</m>, we know that <m>f(y) = x</m>. + Using the <xref ref="thm_chain_rule" text="title"/> and Implicit Differentiation, + take the derivative of both sides of this last equality. + <md> + <mrow>\lzoo{x}{f(y)} \amp = \lzoo{x}{x}</mrow> + <mrow>\fp(y)\cdot \yp \amp = 1</mrow> + <mrow>\yp \amp = \frac{1}{\fp(y)}</mrow> + <mrow>\yp \amp = \frac{1}{\fp(g(x))}</mrow> + </md>. + </p> + + <p> + This leads us to the following theorem. + </p> + + <theorem xml:id="thm_deriv_inverse_functions"> + <title>Derivatives of Inverse Functions + +

    + Let f be differentiable and one-to-one on an open interval I, + where \fp(x) \neq 0 for all x in I, + let J be the range of f on I, + let g be the inverse function of f, + and let f(a) = b for some a in I. + Then g is a differentiable function on J, + and in particular, + +

      +
    1. + \left(f^{-1}\right)'(b)=\gp(b) = \dfrac{1}{\fp(a)} +
      2. + +
    2. + \left(f^{-1}\right)'(x)=\gp(x) = \dfrac{1}{\fp(g(x))} +
      3. +
    +

    +     + + + + +

    + The results of are not trivial; + the notation may seem confusing at first. + Careful consideration, along with examples, + should earn understanding. +

    + + + +

    + In the next example we apply to the arcsine function. +

    + + + + + + + + Finding the derivative of an inverse trigonometric function + +

    + Let y = \arcsin(x) = \sin^{-1}(x). + Find \yp using . +

    +     + + +

    + Adopting our previously defined notation, + let g(x) = \arcsin(x) and f(x) = \sin(x). + Thus \fp(x) = \cos(x). + Applying the theorem, we have + + \gp(x) \amp = \frac{1}{\fp(g(x))} + \amp = \frac{1}{\cos(\arcsin(x))} + . +

    + +

    + This last expression is not immediately illuminating. + Drawing a figure will help, + as shown in . + Recall that the sine function can be viewed as taking in an angle and returning a ratio of sides of a right triangle, + specifically, + the ratio opposite over hypotenuse. + This means that the arcsine function takes as input a ratio of sides and returns an angle. + The equation y=\arcsin(x) can be rewritten as y=\arcsin(x/1); + that is, consider a right triangle where the hypotenuse has length 1 and the side opposite of the angle with measure y has length x. + This means the final side has length \sqrt{1-x^2}, + using the Pythagorean Theorem. +

    + +
    +
    + + + The right triangle defined by the equation sin(y)=x/1 + +

    + The right angle triangle is defined by y=\sin^{-1}(x/1). + The length of the base is \sqrt{1-x^2} and the length of the perpendicular is x. The length of the + hypotenuse is 1. The angle between the base and the hypotenuse is y. +

    +     + + + \begin{tikzpicture}[scale=1.06] + \coordinate (O) at (0,0); + \coordinate (A) at (5,0); + \coordinate (B) at (5,3.75); + \draw (O)--(A)--(B)--cycle; + + \tkzLabelSegment[below=0pt](O,A){$\sqrt{1-x^2}$} + \tkzLabelSegment[above left=0pt](O,B){$1$} + \tkzLabelSegment[right=0pt](A,B){$x$} + + \tkzMarkAngle[fill= orange, + size=1, + opacity=.4](A,O,B) + \tkzLabelAngle[pos = 0.75](A,O,B){$y$} + \end{tikzpicture} + + + + + + +

    + Therefore + + \cos\mathopen{}\left(\sin^{-1}(x)\right)\mathclose{} \amp= \cos(y) + \amp = \frac{\sqrt{1-x^2}}{1} + \amp = \sqrt{1-x^2} + , + resulting in + + \lzoo{x}{\arcsin(x)} = \frac{1}{\sqrt{1-x^2}} + . +

    + + + + +

    + Remember that the input x of the arcsine function is a ratio of a side of a right triangle to its hypotenuse; + the absolute value of this ratio will never be greater than 1. + Therefore the inside of the square root will never be negative. +

    + +

    + In order to make y=\sin(x) one-to-one, + we restrict its domain to [-\pi/2,\pi/2]; + on this domain, the range is [-1,1]. + Therefore the domain of y=\arcsin(x) is [-1,1] and the range is [-\pi/2,\pi/2]. + When x=\pm 1, + note how the derivative of the arcsine function is undefined; + this corresponds to the fact that as x\to \pm1, + the tangent lines to arcsine approach vertical lines with undefined slopes. +

    + +
    +
    + + + + The graph of sin(x) and arcsin(x) with their corresponding tangent line + +

    + The graph of \sin(x) starts at (-\pi/2,-1). It first curves upwards to the right + and then moves towards the origin making a slope. Once the slope passes through the origin it continues + upwards to the right in the first quadrant. Once it reaches the point (\pi/2,1), the graph starts + to move downwards again. The tangent line of the graph is drawn at (\pi/3,\sqrt{3/2}). +

    +     + + + \begin{tikzpicture} + \begin{axis}[ymin=-1.7,ymax=1.7, + xmin=-1.7,xmax=1.7, + xtick={-1.57,-.79,.79,1.57}, + xticklabels={$-\frac{\pi}{2}$,$-\frac{\pi}{4}$,$\frac{\pi}{4}$,$\frac{\pi}{2}$}, + axis equal] + + \addplot+[domain=-1.57:1.57] (x,{sin(deg(x))}) node [pos=0.5,above,sloped]{$y=\sin(x)$}; + \addplot [tangentline,domain=.42:1.7] {.5*(x-pi/3)+sqrt(3)/2}; + \addplot [soliddot] coordinates {(1.06,.866)} node [above left] {$\left(\frac{\pi}{3},\frac{\sqrt{3}}{2}\right)$}; + \end{axis} + \end{tikzpicture} + + + + + + Graphs of sin(x) and arcsin (x) with corresponding tangent line + +

    + Starting at the third quadrant at (-1,-\pi/2) the graph of \arcsin(x) + begins with a nearly vertical slope. The slope decreases to 1 as the graph passes through the origin. + The graph then moves toward its end at (1, \pi/2) with increasing slope. The tangent line is drawn at + (\sqrt{3/2},\pi/3). The tangent line of \sin^{-1}(x) is steeper than the tangent line of \sin(x). +

    +     + + + \begin{tikzpicture} + \begin{axis}[ymin=-1.7,ymax=1.7, + xmin=-1.7,xmax=1.7, + ytick={-1.57,-.79,.79,1.57}, + yticklabels={$-\frac{\pi}{2}$,$-\frac{\pi}{4}$,$\frac{\pi}{4}$,$\frac{\pi}{2}$}, + axis equal] + \addplot+[domain=-1.57:1.57] ({sin(deg(x))},x) node [pos=0.5,above,sloped]{$y=\sin^{-1}(x)$}; + \addplot [tangentline,domain=.42:1.7] ({.5*(x-pi/3)+sqrt(3)/2},{x}); + \addplot [soliddot] coordinates {(0.866,1.06)} node [below right] {$\left(\frac{\sqrt{3}}{2},\frac{\pi}{3}\right)$}; + \end{axis} + \end{tikzpicture} + + + +     + + +

    + In + we see f(x) = \sin(x) and + f^{-1}(x) = \sin^{-1}(x) graphed on their respective domains. + The line tangent to \sin(x) at the point + \left(\pi/3, \sqrt{3}/2\right) has slope \cos(\pi)/3 = 1/2. + The slope of the corresponding point on \sin^{-1}(x), + the point \left(\sqrt{3}/2,\pi/3\right), is + + \frac{1}{\sqrt{1-\left(\sqrt{3}/2\right)^2}} \amp = \frac{1}{\sqrt{1-3/4}} + \amp = \frac{1}{\sqrt{1/4}} + \amp = \frac{1}{1/2}=2 + , + verifying yet again that at corresponding points, + a function and its inverse have reciprocal slopes. +

    + +

    + Using similar techniques, + we can find the derivatives of all the inverse trigonometric functions. + In + we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible. +

    + +
    A right triangle defined by y=\sin^{-1}(x/1) with the length of the third leg found using the Pythagorean TheoremGraphs of \sin(x) and \sin^{-1}(x) along with corresponding tangent lines
    + Domains and ranges of the trigonometric and inverse trigonometric functions + + + Function + Domain + Range + + + \sin(x) + [-\pi/2, \pi/2] + [-1,1] + + + \sin^{-1}(x) + [-1,1] + [-\pi/2, \pi/2] + + + \cos(x) + [0,\pi] + [-1,1] + + + \cos^{-1}(x) + [-1,1] + [0,\pi] + + + \tan(x) + (-\pi/2,\pi/2) + (-\infty,\infty) + + + \tan^{-1}(x) + (-\infty,\infty) + (-\pi/2,\pi/2) + + + \csc(x) + [-\pi/2,0)\cup (0, \pi/2] + (-\infty,-1]\cup [1,\infty) + + + \csc^{-1}(x) + (-\infty,-1]\cup [1,\infty) + [-\pi/2,0)\cup (0, \pi/2] + + + \sec(x) + [0,\pi/2)\cup (\pi/2,\pi] + (-\infty,-1]\cup [1,\infty) + + + \sec^{-1}(x) + (-\infty,-1]\cup [1,\infty) + [0,\pi/2)\cup (\pi/2,\pi] + + + \cot(x) + (0,\pi) + (-\infty,\infty) + + + \cot^{-1}(x) + (-\infty,\infty) + (0,\pi) + + + +
    + + + Derivatives of Inverse Trigonometric Functions + +

    + The inverse trigonometric functions are differentiable on all open sets contained in their domains + (as listed in ) + and their derivatives are as follows: + +

      +
    1. \lzoo{x}{\sin^{-1}(x)} = \frac{1}{\sqrt{1-x^2}}
      2. + +
    2. \lzoo{x}{\cos^{-1}(x)} = -\frac{1}{\sqrt{1-x^2}}
      3. + +
    3. \lzoo{x}{\tan^{-1}(x)} = \frac{1}{1+x^2}
      4. + +
    4. \lzoo{x}{\csc^{-1}(x)} = -\frac{1}{\abs{x}\sqrt{x^2-1}}
      5. + +
    5. \lzoo{x}{\sec^{-1}(x)} = \frac{1}{\abs{x}\sqrt{x^2-1}}
      6. + +
    6. \lzoo{x}{\cot^{-1}(x)} = -\frac{1}{1+x^2}
      7. +
    + + derivativeinverse trig. + +

    +     +     + +

    + Note how each derivative is the negative of the derivative of its co function. + Because of this, derivatives of + \sin^{-1}(x), \tan^{-1}(x), + and \sec^{-1}(x) are used almost exclusively throughout this text. +

    + + + + +

    + In , + we stated without proof or explanation that \lzoo{x}{\ln(x)}=\frac{1}{x}. + We can justify that now using , + as shown in the example. +

    + + + Finding the derivative of <m>y=\ln(x)</m> + +

    + Use + to compute \lzoo{x}{\ln(x)}. +

    +     + +

    + View y= \ln(x) as the inverse of y = e^x. + Therefore, using our standard notation, + let f(x) = e^x and g(x) = \ln(x). + We wish to find \gp(x). + gives: + + \gp(x) \amp = \frac{1}{\fp(g(x))} + \amp = \frac{1}{e^{\ln(x) }} + \amp = \frac{1}{x} + . +

    +     +     + +

    + In this chapter we have defined the derivative, + given rules to facilitate its computation, + and given the derivatives of a number of standard functions. + We restate the most important of these in the following theorem, + intended to be a reference for further work. +

    + +

    + + + Glossary of Derivatives of Elementary Functions + +

    + Let f and g be differentiable functions, + and let a, + c and n be real numbers, + a \gt 0, n\neq 0. +

    + +

    +

      +
    1. \lzoo{x}{c} = 0
      2. + +
    2. \lzoo{x}{x} = 1
      3. + +
    3. \lzoo{x}{x^n} = nx^{n-1}
      4. + +
    4. \lzoo{x}{f(x)\pm g(x)} = f'(x) \pm \gp(x)
      5. + +
    5. \lzoo{x}{c\cdot f(x)} = c\cdot f'(x)
      6. + +
    6. \lzoo{x}{f(x)\cdot g(x)} = f'(x)\cdot g(x)+f(x)\cdot \gp(x)
      7. + +
    7. \lzoo{x}{f(g(x))} = f'(g(x)) \cdot \gp(x)
      8. + +
    8. \lzoo{x}{\frac{f(x)}{g(x)}} = \frac{f'(x)\cdot g(x)-f(x) \cdot \gp(x)}{(g(x))^2}
      9. + +
    9. \lzoo{x}{e^x} = e^x
      10. + +
    10. \lzoo{x}{\ln(x)} = \frac{1}{x}
      11. + +
    11. \lzoo{x}{a^x} = \ln(a)\cdot a^x
      12. + +
    12. \lzoo{x}{\log_a x} = \frac{1}{\ln(a)}\cdot\frac{1}{x}
      13. + +
    13. \lzoo{x}{\sin(x)} = \cos(x)
      14. + +
    14. \lzoo{x}{\cos(x)} = -\sin(x)
      15. + +
    15. \lzoo{x}{\tan(x)} = \sec^2(x)
      16. + +
    16. \lzoo{x}{\csc(x)} = -\csc(x)\cot(x)
      17. + +
    17. \lzoo{x}{\sec(x)} = \sec(x)\tan(x)
      18. + +
    18. \lzoo{x}{\cot(x)} = -\csc^2(x)
      19. + +
    19. \lzoo{x}{\sin^{-1}(x)} = \frac{1}{\sqrt{1-x^2}}
      20. + +
    20. \lzoo{x}{\cos^{-1}(x)} = -\frac{1}{\sqrt{1-x^2}}
      21. + +
    21. \lzoo{x}{\tan^{-1}(x)} = \frac{1}{1+x^2}
      22. + +
    22. \lzoo{x}{\csc^{-1}(x)} = -\frac{1}{\abs{x}\sqrt{x^2-1}}
      23. + +
    23. \lzoo{x}{\sec^{-1}(x)} = \frac{1}{\abs{x}\sqrt{x^2-1}}
      24. + +
    24. \lzoo{x}{\cot^{-1}(x)} = -\frac{1}{1+x^2}
      25. +
    +

    + + + + + + Terms and Concepts + + + + +

    + + Every function has an inverse. +

    +     + +

    + A necessary condition is that the function is one-to-one. +

    +     + +     + + + + +

    + In your own words explain what it means for a function to be one-to-one. +

    + +     + + + +     + + + + +

    + If (1,10) lies on the graph of y=f(x), + what can be said about the graph of y=f^{-1}(x)? +

    + +     + + + +

    + The point (10,1) lies on the graph of y=f^{-1}(x) + (assuming f is invertible). +

    +     + +     + + + + +

    + If (1,10) lies on the graph of y=f(x) and \fp(1) = 5, + what can be said about y=f^{-1}(x)? +

    + +     + + + +

    + The point (10,1) lies on the graph of y=f^{-1}(x) + (assuming f is invertible) + and (f^{-1})'(10) = 1/5. +

    +     + +     +     + + Problems + + +

    + Verify that the given functions are inverses. +

    +     + + + + +

    + f(x) = 2x+6 and + g(x) = \frac{1}{2}x-3 +

    + +     + + + +

    + Compose f(g(x)) and g(f(x)) to confirm that each equals x. +

    +     + +     + + + + +

    + f(x) = x^2+6x+11, x\geq 3 and + g(x) = \sqrt{x-2}-3, x\geq 2 +

    + +     + + + +

    + Compose f(g(x)) and g(f(x)) to confirm that each equals x. +

    +     + +     + + + + +

    + f(x) = \frac{3}{x-5}, x\neq 5 and + g(x) = \frac{3+5x}{x}, x\neq 0 +

    + +     + + + +

    + Compose f(g(x)) and g(f(x)) to confirm that each equals x. +

    +     + +     + + + + +

    + f(x) = \frac{x+1}{x-1}, x\neq 1 and + g(x) = f(x) +

    + +     + + + +

    + Compose f(g(x)) and g(f(x)) to confirm that each equals x. +

    +     + +     +     + + + +

    + An invertible function f(x) is given along with a point that lies on its graph. + Using , + evaluate \left(f^{-1}\right)'(x) at the indicated value. +

    +     + + + + + Context("Fraction"); + ($m,$b,$x0) = random_subset(3,2..10); + if($envir{problemSeed}==1){$m=5; $b=10; $x0=2;}; + $f = Formula("$m x + $b"); + $y0 = $f->eval(x=>$x0); + $point = Point($x0,$y0); + $fip = Fraction(1,$m); + + +

    + The point is on the graph of f(x) = . + Find \left(f^{-1}\right)'(). +

    +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)-y', '(-x)+y'); + ($h,$c,$x0) = random_subset(3, -9 .. -1, 1 .. 9); + if($envir{problemSeed}==1){$h=1; $c=4; $x0=3;}; + $f = Formula("x^2 - 2*$h x + $c")->reduce; + $y0 = $f->eval(x => $x0); + $point = Point($x0,$y0); + $fp = $f->D('x'); + $fpx0 = $fp->eval(x => $x0); + $fipy0 = Fraction(1/$fpx0); + $sign = ($x0 > $h) ? '\geq' : '\leq'; + + +

    + The point is on the graph of f(x) = , x . + Find \left(f^{-1}\right)'(). +

    +

    + +

    +     +     +     + + + + + Context("Fraction")->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $n = random(2,4,1); + $d = list_random(6*$n,4*$n,3*$n); + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$n=2; $d=6; $trig='sin';}; + $x0 = Formula("pi/$d"); + $f = Formula("$trig($n x)"); + ($num2,$den2) = Fraction($f->eval(x=>$x0->eval(x=>0))**2)->value; + if ($num2 == 3) { + $y0 = Formula("sqrt(3)/2"); + } elsif ($den2 == 2) { + $y0 = Formula("sqrt(2)/2"); + } else { + $y0 = Formula("1/2"); + }; + $fp = $f->D('x'); + $fpx0 = $fp->eval(x=>$x0->eval(x=>0)); + $fipy0 = 1/$fpx0; + if ($fipy0 == 1/sqrt(3)) {$fipy0 = Formula("sqrt(3)/3");} + elsif ($fipy0 == 2/(3*sqrt(3))) {$fipy0 = Formula("2 sqrt(3)/3");} + elsif ($fipy0 == 1/(2*sqrt(3))) {$fipy0 = Formula("sqrt(3)/6");} + elsif ($fipy0 == 1/sqrt(2)) {$fipy0 = Formula("sqrt(2)/2");} + elsif ($fipy0 == sqrt(2)/3) {$fipy0 = Formula("sqrt(2)/3");} + elsif ($fipy0 == 1/(2*sqrt(2))) {$fipy0 = Formula("sqrt(2)/4");} + elsif ($fipy0 == 1) {$fipy0 = Formula("1");} + elsif ($fipy0 == 2/3) {$fipy0 = Formula("2/3");} + elsif ($fipy0 == 1/2) {$fipy0 = Formula("1/2");}; + $d2 = 2*$n; + if ($trig eq 'sin') { + $low = Formula("-pi/$d2"); + $high = Formula("pi/$d2"); + } else { + $low = Formula("0"); + $high = Formula("pi/$n"); + } + + +

    + The point \left(,\right) is on the graph of f(x) = , + \leq x\leq . + Find \left(f^{-1}\right)'\left(\right). +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + ($h,$k,$c) = random_subset(3,-9..-1,1..9); + $x0 = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$h=2; $k=1; $c=-2; $x0=1;}; + $f = Formula("x^3 - (3*$h) x^2 + 3*(($h)^2+$k) x + $c")->reduce; + $y0 = $f->eval(x=>$x0); + $point = Point($x0,$y0); + $fp = $f->D('x'); + $fpx0 = $fp->eval(x=>$x0); + $fipy0 = Fraction(1/$fpx0); + + +

    + The point is on the graph of f(x) = . + Find \left(f^{-1}\right)'(). +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + $n = random(2,5,1); + $x0 = random(1,2,1); + if($envir{problemSeed}==1){$n=2; $x0=1;}; + $f = Formula("1/(1+x^$n)")->reduce; + $y0 = Fraction($f->eval(x=>$x0)); + $point = Point($x0,$y0); + $fp = $f->D('x'); + $fpx0 = $fp->eval(x=>$x0); + $fipy0 = Fraction(1/$fpx0); + + +

    + The point is on the graph of f(x) = , x\geq 0. + Find \left(f^{-1}\right)'\left(\right). +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + ($a,$k) = random_subset(2,2..9); + if($envir{problemSeed}==1){$a=6; $k=3;}; + $x0 = 0; + $f = Formula("$a e^($k x)"); + $y0 = $f->eval(x=>$x0); + $point = Point($x0,$y0); + $fp = $f->D('x'); + $fpx0 = $fp->eval(x=>$x0); + $fipy0 = Fraction(1/$fpx0); + + +

    + The point is on the graph of f(x) = . + Find \left(f^{-1}\right)'(). +

    +

    + +

    +     +     +     +     + + + +

    + Compute the derivative of the given function. +

    +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $k = random(2,9,1); + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$fname='h'; $x='t'; $k = 2; $trig = 'sin'}; + Context("Fraction"); + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->set($x=>{limits=>[-1/$k,1/$k]}); + $f = Formula("$trig^-1($k $x)"); + $fp = $f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $k = random(2,9,1); + $trig = list_random('sec','csc'); + if($envir{problemSeed}==1){$fname='f'; $x='t'; $k = 2; $trig = 'sec'}; + Context("Fraction"); + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->set($x=>{limits=>[1/$k,10/$k]}); + $f = Formula("$trig^-1($k $x)"); + $fp = $f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $k = random(2,9,1); + $trig = list_random('tan','cot'); + if($envir{problemSeed}==1){$fname='g'; $x='x'; $k = 2; $trig = 'tan'}; + Context("Fraction"); + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $f = Formula("$trig^-1($k $x)"); + $fp = $f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$fname='f'; $x='x'; $trig = 'sin'}; + Context("Fraction"); + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->set($x=>{limits=>[-1,1]}); + $f = Formula("$x $trig^-1($x)"); + $fp = $f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + ($triga,$trigb) = random_subset(2,'sin','cos','tan'); + if($envir{problemSeed}==1){$fname='g'; $x='t'; $triga = 'sin'; $trigb ='cos';}; + Context("Fraction"); + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->set($x=>{limits=>[-1,1]}); + $f = Formula("$triga($x) $trigb^-1($x)"); + $fp = $f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->set(t=>{limits=>[1,5]}); + $fp=Formula("e^t/t+ln(t)e^t"); + + +

    + f(t) = \ln(t) e^t +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + ($triga,$trigb) = random_subset(2,'sin','cos','tan'); + if($envir{problemSeed}==1){$fname='h'; $x='x'; $triga = 'sin'; $trigb ='cos';}; + Context("Fraction"); + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->set($x=>{limits=>[-1,1]}); + $f = Formula("$triga^-1($x)/$trigb^-1($x)"); + $fp = $f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $trig = list_random('tan','cot'); + $n = random(2,4,1); + if($envir{problemSeed}==1){$fname='g'; $x='x'; $trig = 'tan'; $n = 2}; + Context("Fraction"); + parser::Root->Enable; + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->set($x=>{limits=>[0,pi/2]}); + $f = Formula("$trig(root($n,$x))"); + $f = Formula("$trig(sqrt($x))") if ($n == 2); + $fp = $f->D($x); + + +

    + () = +

    + + To enter \sqrt{x}, type sqrt(x). + To enter \sqrt[n]{x}, type root(n,x). + +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $trig = list_random('sec','csc'); + $n = random(1,4,1); + if($envir{problemSeed}==1){$fname='f'; $x='x'; $trig = 'sec'; $n = 1}; + Context("Fraction"); + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->set($x=>{limits=>[0,pi/2]}); + $f = Formula("$trig(1/$x^$n)")->reduce; + $fp = $f->D($x); + + +

    + () = +

    +

    + +

    +     +     +     + + + + + $fname = list_random('f','g','h','j','k','p','m'); + $x = list_random('q','r','t','x','y','z','w'); + $trig = list_random('sin','cos','tan','cot','sec','csc'); + if($envir{problemSeed}==1){$fname='f'; $x='x'; $trig = 'sin'}; + Context()->variables->are($x=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->set($x=>{limits=>[-1,1]}); + $f = Formula("$trig($trig^-1($x))"); + $fp = Formula("1"); + + +

    + () = +

    +

    + +

    +     + +

    + () = x, so \fp(x) = 1. +

    +     +     +     +     + + + +

    + Compute the derivative of the given function in two ways: + +

      +
    1. +

      + By simplifying first, then taking the derivative, and +

      +
      2. +
    2. +

      + by using the Chain Rule first then simplifying. +

      +
      3. +
    + Verify that the two answers are the same. +

    +     + + + + +

    + f(x)=\sin(\sin^{-1}(x)) +

    + +     + + + +

    +

      +
    1. +

      + f(x) = x, so \fp(x) = 1. +

      +
      2. +
    2. +

      + \fp(x) = \cos(\sin^{-1}(x) )\frac{1}{\sqrt{1-x^2}} = 1. +

      +
      3. +
    +

    +     + +     + + + + +

    + f(x)=\tan^{-1}(\tan(x)) +

    + +     + + + +

    +

      +
    1. +

      + Recall that the inverse tangent function is defined to have range (-\pi/2, \pi/2). + If -\pi/2 \lt x\lt \pi/2, then f(x) = x. + If x is in the interval (-\pi/2+k\pi,\pi/2+k\pi) for some integer k, + then f(x)=x-k\pi. On any interval where f(x) is defined, + f(x)=x+C for some constant C, so \fp(x) = 1. +

      +
      2. +
    2. +

      + \fp(x) = \frac{1}{1+\tan^2(x) }\sec^2(x) = 1. +

      +
      3. +
    +

    +     + +     + + + + +

    + f(x)=\sin(\cos^{-1}(x)) +

    + +     + + + +

    +

      +
    1. +

      + f(x) = \sqrt{1-x^2}, so \fp(x) = \frac{-x}{\sqrt{1-x^2}}. +

      +
      2. +
    2. +

      + \fp(x) = \cos(\cos^{-1}(x) ) \frac{1}{\sqrt{1-x^2}} =\frac{-x}{\sqrt{1-x^2}}. +

      +
      3. +
    +

    +     + +     + + + + +

    + f(x)=\sin(2\sin^{-1}(x)) +

    + +     + + + +

    +

      +
    1. +

      + f(x) = 2\sin(\sin^{-1}(x))\cos(\sin^{-1}(x))=2x\sqrt{1-x^2}, + so \fp(x) = 2\sqrt{1-x^2} -\frac{2x^2}{\sqrt{1-x^2}}. +

      +
      2. +
    2. +

      + \fp(x) = \cos(2\sin^{-1}(x) ) \frac{2}{\sqrt{1-x^2}}, + but to compare, we need to simplify \cos(2\sin^{-1}(x)). + Using \cos(2\theta)=1-2\sin^2(\theta), + we get \fp(x) = (1-2x^2)\frac{2}{\sqrt{1-x^2}}. +

      +

      + This is the same as the first answer, since + + 2\sqrt{1-x^2} -\frac{2x^2}{\sqrt{1-x^2}} \amp = \frac{2(1-x^2)-2x^2}{\sqrt{1-x^2}} + \amp =\frac{2(1-2x^2)}{\sqrt{1-x^2}} + . +

      +
      3. +
    +

    +     + +     +     + + + +

    + Find the equation of the line tangent to the graph of f at the indicated x value. +

    +     + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $x0 = list_random('1/2','sqrt(2)/2','sqrt(3)/2'); + $sign = list_random('-',''); + if($envir{problemSeed}==1){$x0='sqrt(2)/2';$sign=''}; + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $f = Formula("sin^-1(x)"); + if ($x0 eq '1/2') { + $m = Formula("2sqrt(3)/3"); + $y0 = Formula("pi/6"); + } + elsif ($x0 eq 'sqrt(2)/2') { + $m = Formula("sqrt(2)"); + $y0 = Formula("pi/4"); + } + elsif ($x0 eq 'sqrt(3)/2') { + $m = Formula("2"); + $y0 = Formula("pi/3"); + }; + $x0 = Formula("$sign $x0"); + $y0 = Formula("$sign $y0"); + $t=Formula("y=$m(x-$x0)+$y0"); + + +

    + f(x)=\sin^{-1}(x) at x= +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $x0 = list_random('1/4','sqrt(2)/4','sqrt(3)/4'); + $sign = list_random('-',''); + if($envir{problemSeed}==1){$x0='sqrt(3)/4';$sign=''}; + Context()->variables->add(y=>'Real'); + parser::Assignment->Allow; + $f = Formula("cos^-1(2x)"); + if ($x0 eq '1/4') { + $m = Formula("-4sqrt(3)/3"); + if ($sign eq '') {$y0 = Formula("pi/3")} + else {$y0 = Formula("2pi/3")}; + } + elsif ($x0 eq 'sqrt(2)/4') { + $m = Formula("-2sqrt(2)"); + if ($sign eq '') {$y0 = Formula("pi/4")} + else {$y0 = Formula("3pi/4")}; + } + elsif ($x0 eq 'sqrt(3)/4') { + $m = Formula("-4"); + if ($sign eq '') {$y0 = Formula("pi/6")} + else {$y0 = Formula("5pi/6")}; + }; + $x0 = Formula("$sign $x0"); + $t=Formula("y=$m(x-$x0)+$y0"); + + +

    + f(x)=\cos^{-1}(2x) at x= +

    +

    + +

    +     +     +     +     +     + +     +
    + + +     + + + The Graphical Behavior of Functions + +

    + Our study of limits led to continuous functions, + a certain class of functions that behave in a particularly nice way. + Limits then gave us an even nicer class of functions, + functions that are differentiable. +

    + +

    + This chapter explores many of the ways we can take advantage of the information that continuous and differentiable functions provide. +

    +     + +
    + Extreme Values +

    + Given any quantity described by a function, + we are often interested in the largest and/or smallest values that quantity attains. + For instance, if a function describes the speed of an object, + it seems reasonable to want to know the fastest/slowest the object traveled. + If a function describes the value of a stock, + we might want to know the highest/lowest values the stock attained over the past year. + We call such values extreme values. +

    + + + Extreme Values + +

    + Let f be defined on an interval I containing c. + + extreme values + absolute minimum + minimumabsolute + absolute maximum + maximumabsolute + +

      +
    1. +

      + f(c) is the minimum (also, + absolute minimum) of f on I if + f(c) \leq f(x) for all x in I. +

      +
      2. + +
    2. +

      + f(c) is the maximum (also, + absolute maximum) of f on I if + f(c) \geq f(x) for all x in I. +

      +
      3. +
    +

    + +

    + The maximum and minimum values are the + extreme values, + or extrema, of f on I. + extremaabsolute +

    +     +     + + + + + +

    + Consider . + The function displayed in has a maximum, + but no minimum, + as the interval over which the function is defined is open. + In , the function has a minimum, + but no maximum; + there is a discontinuity in the natural + place for the maximum to occur. + Finally, the function shown in has both a maximum and a minimum; + note that the function is continuous and the interval on which it is defined is closed. +

    + + + +
    + Graphs of functions with and without extreme values + +
    + + + + A parabolic graph, opening downward, from -2 to 2, not including the endpoints + +

    + The image shows the graph of a function with the appearance of a parabola that opens downward, + with its vertex at (0,5). + The domain for the function is (-2,2), so the graph approaches, but does not reach, + the points (-2,1) and (2,1). These points are shown with hollow dots, + to illustrate the fact that they are not part of the graph. +

    + +

    + The maximum value f(0)=5 is reached, but there is no minimum value, + since the graph approaches, but does not reach, its end points. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ ymin=-.9,ymax=5.5,xmin=-2.2,xmax=2.4,] + \addplot+[open,domain=-2:2] {-x^2+5}; + \addplot[openinterval] coordinates {(-2,0) (2,0)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + A parabolic graph, opening downward, from -2 to 2. The vertex at (0,5) is missing. + +

    + The image shows the graph of a function with the appearance of a parabola that opens downward, + with its vertex at (0,5). + The domain for the function is [-2,2], this time, + the points (-2,1) and (2,1) at the ends of the graph are included. +

    + +

    + However, there is a hollow dot at (0,5), indicating a hole at the vertex. + A solid dot at (0,3) suggests that this is the graph of a function with f(0)=3. +

    + +

    + This graph illustrates a function with a minimum value f(-2)=f(2)=1, + but there is no maximum value, since the parabola approaches, but does not reach, its vertex. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ ymin=-.9,ymax=5.5,xmin=-2.2,xmax=2.4,] + \addplot+[domain=-2:2,closed] {-x^2+5}; + \addplot[soliddot] coordinates {(0,3)}; + \addplot[closedinterval] coordinates {(-2,0) (2,0)}; + \addplot[hollowdot] coordinates {(0,5)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + The graph of a parabola opening downward from a vertex at (0,5), for x from -2 to 2 + +

    + This graph shows a parabola on the interval [-2,2], + opening downward from its vertex at (0,5). +

    + +

    + There are no holes in the graph, and the endpoints are included. +

    + +

    + The graph illustrates a maximum value f(0)=5 and a minimum value of 1, + which is attained at both x=2 and x=-2. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ ymin=-.9,ymax=5.5,xmin=-2.2,xmax=2.4,] + \addplot+[domain=-2:2,closed] {-x^2+5}; + \addplot[closedinterval] coordinates {(-2,0) (2,0)}; + \end{axis} + \end{tikzpicture} + + + + +
    +     +
    + +

    + It is possible for discontinuous functions defined on an open interval to have both a maximum and minimum value, + but we have just seen examples where they did not. + On the other hand, + continuous functions on a closed interval always + have a maximum and minimum value. +

    + + + The Extreme Value Theorem + +

    + Let f be a continuous function defined on a closed interval I=[a,b]. + Then f has both a maximum and minimum value on I. + + Extreme Value Theorem + +

    +     +     + + + + + +

    + This theorem states that f has extreme values, + but it does not offer any advice about how/where to find these values. + The process can seem to be fairly easy, + as the next example illustrates. + After the example, + we will draw on lessons learned to form a more general and powerful method for finding extreme values. +

    + + + Approximating extreme values + +

    + Consider f(x) = 2x^3-9x^2 on I=[-1,5], + as graphed in . + Approximate the extreme values of f. +

    + +
    + A graph of f(x) = 2x^3-9x^2 as in + + + + The graph of a cubic function on the interval [-1,5], with maximum and minimum values + +

    + The image shows the graph of f(x)=2x^3-9x^2 on the interval [-1,5]. + Beginning at the point (-1,-11), the graph rises to a peak at (0,0), + before falling to its minimum value at (3,-27). + The graph then climbs to the endpoint (5,25), where it reaches its maximum value. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-2,xmax=5.7, + ymin=-33,ymax=31,] + \addplot+[domain=-1:5, samples=40] {2*x^3-9*x^2}; + \addplot[soliddot] coordinates{(-1,-11)} node[below] {$(-1,-11)$}; + \addplot[soliddot] coordinates{(0,0)} node[above right] {$(0,0)$}; + \addplot[soliddot] coordinates{(3,-27)} node[below] {$(3,-27)$}; + \addplot[soliddot] coordinates{(5,25)} node[above] {$(5,25)$}; + \end{axis} + \end{tikzpicture} + + + + +
    + +     + +

    + The graph is drawn in such a way to draw attention to certain points. + It certainly seems that the smallest y-value is -27, + found when x=3. + It also seems that the largest y-value is 25, + found at the endpoint of I, x=5. + We use the word seems, + for by the graph alone we cannot be sure the smallest value is not less than -27. + Since the problem asks for an approximation, + we approximate the extreme values to be 25 and -27. +

    +     +     + +

    + Notice how the minimum value came at + the bottom of a hill, + and the maximum value came at an endpoint. + Also note that while 0 is not an extreme value, + it would be if we narrowed our interval to [-1,4]. + The idea that the point (0,0) is the location of an extreme value for some interval is important, + leading us to a definition of a + relative maximum. + In short, a relative max + is a y-value that's the largest y-value nearby. +

    + + + + + + + Relative Minimum and Relative Maximum + +

    + Let f be defined on an interval I containing c. + +

      +
    1. +

      + If there is a \delta \gt 0 such that + f(c) \leq f(x) for all x in I where \abs{x-c}\lt \delta, + then f(c) is a relative minimum of f. + We also say that f has a relative minimum at (c,f(c)). + minimumrelative/local + extremarelative +

      +
      2. + +
    2. +

      + If there is a \delta \gt 0 such that + f(c) \geq f(x) for all x in I where \abs{x-c}\lt \delta, + then f(c) is a relative maximum of f. + We also say that f has a relative maximum at (c,f(c)). + maximumrelative/local + extremarelative +

      +
      3. +
    +

    + +

    + The relative maximum and minimum values comprise the + relative extrema of f. +

    +     +     + +

    + We briefly practice using these definitions. +

    + + + Approximating relative extrema + +

    + Consider f(x) = (3x^4-4x^3-12x^2+5)/5, + as shown in . + Approximate the relative extrema of f. + At each of these points, evaluate \fp. +

    + +
    + A graph of f(x) = (3x^4-4x^3-12x^2+5)/5 as in + + + The graph of a degree 4 function with several critical points. + +

    + The graph of f(x)=\frac15(3x^4-4x^3-12x^2+5) is shown, for x between -2 and 3. + Arrows at either end of the graph indicate that f(x) will approach \infty in either direction beyond the interval shown. + The graph shows three relative extrema: there are minima when x=-1 and x=2, and a maximum at x=0. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-2.2,xmax=3.2, + ymin=-6.5,ymax=6.9,] + \addplot+[infinite,domain=-1.8:3, samples=50] {(3*x^4-4*x^3-12*x^2+5)/5}; + \end{axis} + \end{tikzpicture} + + + + +
    +     + +

    + We still do not have the tools to exactly find the relative extrema, + but the graph does allow us to make reasonable approximations. + It seems f has relative minima at x=-1 and x=2, + with values of f(-1)=0 and f(2) = -5.4. + It also seems that f has a relative maximum at the point (0,1). +

    + +

    + We approximate the relative minima to be 0 and -5.4; + we approximate the relative maximum to be 1. +

    + +

    + It is straightforward to evaluate + \fp(x) =\frac{1}{5}\left(12x^3-12x^2-24x\right) at x=0, 1 and 2. + In each case, \fp(x) = 0. +

    +     +     + + + Approximating relative extrema + +

    + Approximate the relative extrema of f(x) = (x-1)^{2/3}+2, + shown in . + At each of these points, evaluate \fp. +

    + +
    + A graph of f(x) = (x-1)^{2/3}+2 as in + + + A graph in the shape of a curved V, with a sharp point at (1,2). + +

    + The graph of f(x)=(x-1)^{2/3}+2 resembles the outstretched wings of a bird. + It is in the shape of a wide, curved V, with a vertex at (1,2). +

    + +

    + This graph illustrates the case of a local minimum that occurs at a point of non-differentiability. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-.6,xmax=2.6, + ymin=-.5,ymax=3.5] + \addplot+[rightarrow,domain=2:3.2] ({(x-2)^1.5+1},{x}); + \addplot[firstcurvestyle,rightarrow,domain=2:3.2] ({-(x-2)^1.5+1},{x}); + \end{axis} + \end{tikzpicture} + + + + +
    +     + +

    + The figure implies that f does not have any relative maxima, + but has a relative minimum at (1,2). + In fact, the graph suggests that not only is this point a relative minimum, + y=f(1)=2 is the minimum value of the function. +

    + +

    + We compute \fp(x) = \frac23(x-1)^{-1/3}. + When x=1, \fp is undefined. +

    +     +     + + + +

    + What can we learn from the previous two examples? + We were able to visually approximate relative extrema, + and at each such point, + the derivative was either 0 or it was not defined. + This observation holds for all functions, + leading to a definition and a theorem. +

    + + + Critical Numbers and Critical Points + +

    + Let f be defined at c. + The value c is a critical number + (or critical value) + of f if \fp(c)=0 or \fp(c) is not defined. + + critical number + critical point + +

    + +

    + If c is a critical number of f, + then the point (c,f(c)) is a + critical point of f. +

    +     +     + + + + + Relative Extrema and Critical Points + +

    + Let a function f be defined on an open interval I containing c, + and let f have a relative extremum at the point (c,f(c)). + Then c is a critical number of f. +

    +     +     + + + + + +

    + Be careful to understand that this theorem states + Relative extrema on open intervals occur at critical points. + It does not say All critical numbers produce relative extrema. For instance, + consider f(x) = x^3. + Since \fp(x) = 3x^2, + it is straightforward to determine that x=0 is a critical number of f. + However, f has no relative extrema, + as illustrated in . +

    + + +
    + A graph of f(x)=x^3 which has a critical value of x=0, but no relative extrema + + + A graph of the cubic function, which has a horizontal tangent at the origin, but no maximum or minimum + +

    + The graph of f(x)=x^3 rises steadily as x increases. + It briefly levels off at (0,0), where there is a horizontal tangent. + The horizontal tangent corresponds to a critical point, but it is not a relative maximum nor a relative minimum. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-1.1,xmax=1.1, + ymin=-1.1,ymax=1.1] + \addplot+[infinite,domain=-1:1] {x^3}; + \end{axis} + \end{tikzpicture} + + + + +
    + +

    + + states that a continuous function on a closed interval will have both an absolute maximum and an absolute minimum. + Common sense tells us extrema occur either at the endpoints or somewhere in between. + It is easy to check for extrema at endpoints, + but there are infinitely many points to check that are in between. + tells us we need only check at the critical points that are in between the endpoints. + We combine these concepts to offer a strategy for finding extrema. +

    + + + Finding Extrema on a Closed Interval +

    + Let f be a continuous function defined on a closed interval [a,b]. + To find the maximum and minimum values of f on [a,b]: + extremafinding + +

      +
    1. +

      + Evaluate f at the endpoints a and b of the interval. +

      +
      2. + +
    2. +

      + Find the critical numbers of f in [a,b]. +

      +
      3. + +
    3. +

      + Evaluate f at each critical number. +

      +
      4. + +
    4. +

      + The absolute maximum of f is the largest of these values, + and the absolute minimum of f is the least of these values. +

      +
      5. +
    +

    +     + +

    + We practice these ideas in the next examples. +

    + + + Finding extreme values + +

    + Find the extreme values of f(x) = 2x^3+3x^2-12x on [0,3], + graphed in . +

    + +
    + A graph of f(x) = 2x^3+3x^2-12x on [0,3] as in + + + A graph of a cubic function on the interval [0,3]. + +

    + The image shows the graph of f(x)=2x^3+3x^2-12x on the interval [0,3]. + There appears to be a relative minimum near x=1, and no other critical points. + The absolute maximum value of the function appears to occur at the right endpoint of the interval. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-.4,xmax=3.3, + ymin=-7.5,ymax=45.5] + \addplot+[closed,domain=0:3] {2*x^3+3*x^2-12*x}; + \end{axis} + \end{tikzpicture} + + + + +
    +     + +

    + We follow the steps outlined in . + We first evaluate f at the endpoints: + + f(0)\amp=0\amp f(3)\amp=45 + . +

    + +

    + Next, we find the critical values of f on [0,3]. + \fp(x) = 6x^2+6x-12 = 6(x+2)(x-1); + therefore the critical values of f are x=-2 and x=1. + Since x=-2 does not lie in the interval [0,3], + we ignore it. + Evaluating f at the only critical number in our interval gives: + f(1) = -7. +

    + +

    + + gives f evaluated at the important + x values in [0,3]. + We can easily see the maximum and minimum values of f: + the maximum value is 45 and the minimum value is -7. +

    + +
    + Finding the extreme values of f(x)= 2x^3+3x^2-12x in + + + x + f(x) + + + 0 + 0 + + + 1 + -7 + + + 3 + 45 + + +
    + +     + +     + +

    + Note that all this was done without the aid of a graph; + this work followed an analytic algorithm and did not depend on any visualization. + + shows f and we can confirm our answer, + but it is important to understand that these answers can be found without graphical assistance. +

    + +

    + We practice again. +

    + + + Finding extreme values + +

    + Find the maximum and minimum values of f on [-4,2], where + + f(x) = \begin{cases}(x-1)^2 \amp x\leq 0 \\ x+1 \amp x \gt 0\end{cases} + . + +

    +     + +

    + Here f is piecewise-defined, + but we can still apply + as it is continuous on [-4,2] + (one should check to verify that \lim\limits_{x\to 0}f(x) =f(0)). +

    + +

    + Evaluating f at the endpoints gives: + + f(-4)\amp=25\amp f(2)\amp=3 + . +

    + +

    + We now find the critical numbers of f. + We have to define \fp in a piecewise manner; it is + + \fp(x) =\begin{cases}2(x-1) \amp x \lt 0 \\ 1 \amp x \gt 0\end{cases} + . +

    + +

    + Note that while f is defined for all of [-4,2], + \fp is not, + as the derivative of f does not exist when x=0. + (From the left, the derivative approaches -2; + from the right the derivative is 1.) + Thus one critical number of f is x=0. +

    + +

    + We now set \fp(x) = 0. + When x \gt 0, \fp(x) is never 0. + When x\lt 0, + \fp(x) is also never 0, so we find no critical values from setting \fp(x)=0. +

    + +

    + So we have three important x-values to consider: + x= -4, 2 and 0. + Evaluating f at each gives, respectively, 25, + 3 and 1, shown in . + + Thus the absolute minimum of f is 1, + the absolute maximum of f is 25. + Our answer is confirmed by the graph of f in . +

    + + + +
    + Finding the extreme values of a piecewise-defined function in + + + x + f(x) + + + -4 + 25 + + + 0 + 1 + + + 2 + 3 + + +
    + +
    + A graph of f(x) on [-4,2] as in + + + The graph of a piecewise-defined function on the interval [-4,2] + +

    + The image shows the graph of a piecewise-defined function on the interval [-4,2]. + For -4\leq x\leq 0, the graph appears to be part of a parabola, + descending from (-4,25) to (0,1). + For 0\leq x\leq 2, the graph is a straight line with positive slope, from (0,1) to (2,3). +

    + +

    + The point (0,1) is a relative minimum, and a point of non-differentiability. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-4.5,xmax=2.5, + ymin=-.9,ymax=25.5] + \addplot+[domain=-4:0] {(x-1)^2}; + \addplot[firstcurvestyle,domain=0:2] {x+1}; + \addplot[soliddot] coordinates {(-4,25) (2,3)}; + \end{axis} + \end{tikzpicture} + + + + +
    +     +     + +     + + + Finding extreme values + +

    + Find the extrema of f(x) = \cos\mathopen{}\left(x^2\right)\mathclose{} on [-2,2]. +

    +     + +

    + We again use . + Evaluating f at the endpoints of the interval gives: + f(-2) = f(2) = \cos(4) \approx -0.6536. + We now find the critical values of f. +

    + +

    + Applying the , + we find \fp(x) = -2x\sin\mathopen{}\left(x^2\right)\mathclose{}. + Set \fp(x) = 0 and solve for x to find the critical values of f. +

    + +

    + We have \fp(x) = 0 when x = 0 and when \sin\mathopen{}\left(x^2\right)\mathclose{}. + In general, \sin(t) = 0 when + t = \ldots -2\pi, -\pi, 0, \pi, \ldots Thus \sin\mathopen{}\left(x^2\right)\mathclose{} = 0 when + x^2 = 0, \pi, 2\pi, \ldots (x^2 is always nonnegative so we ignore -\pi, + etc.) So \sin\mathopen{}\left(x^2\right)\mathclose{}=0 when x= 0, \pm \sqrt{\pi}, \pm\sqrt{2\pi}, \ldots. + The only values to fall in the given interval of [-2,2] are 0 and \pm\sqrt{\pi}, + where \sqrt{\pi} \approx 1.77. +

    + +

    + We again construct a table of important values in . + In this example we have five values to consider: + x= 0, \pm 2, \pm\sqrt{\pi}. + From the table it is clear that the maximum value of f on [-2,2] is 1; + the minimum value is -1. + The graph in confirms our results. +

    + + + +
    + Finding the extrema of f(x)= \cos\mathopen{}\left(x^2\right)\mathclose{} in + + + x + f(x) + + + -2 + -0.65 + + + -\sqrt{\pi} + -1 + + + 0 + 1 + + + \sqrt{\pi} + -1 + + + 2 + -0.65 + + +
    + +
    + A graph of f(x)=\cos\mathopen{}\left(x^2\right)\mathclose{} on [-2,2] as in + + + A graph that resembles a flattened cosine wave for small values of x + +

    + The graph of f(x)=\cos(x^2) resembles a cosine wave, except that it is much flatter when x is near 0. + From an apparent relative maximum at (0,1), the graph descends in both directions toward relative minimum values close to the endpoints, + at x=2 and x=-2. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-2.2,xmax=2.2, + ymin=-1.1,ymax=1.1] + \addplot+[closed,samples=40,smooth, domain=-2:2] {cos(deg(x^2))}; + \end{axis} + \end{tikzpicture} + + + + +
    +     +     + +     + +

    + We consider one more example. +

    + + + Finding extreme values + +

    + Find the extreme values of f(x) = \sqrt{1-x^2}. +

    +     + +

    + A closed interval is not given, + so we find the extreme values of f on its domain. + f is defined whenever 1-x^2\geq 0; + thus the domain of f is [-1,1]. + Evaluating f at either endpoint returns 0. +

    + + +
    + A graph of f(x)=\sqrt{1-x^2} on [-1,1] as in + + + A semi-circle equivalent to the top half of the unit circle + +

    + The graph y=\sqrt{1-x^2} is the half of the unit circle x^2+y^2=1, where y\geq 0. + It has absolute minimum values at its endpoints, which are (-1,0) and (1,0). + It reaches its (absolute and relative) maximum at the top of the circle, which is the point (0,1). +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-1.2,xmax=1.2, + ymin=-1.2,ymax=1.2, + axis equal] + \addplot+[closed,domain=0:180] ({cos(x)},{sin(x)}); + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + Finding the extrema of the half-circle in + + + x + f(x) + + + -1 + 0 + + + 0 + 1 + + + 1 + 0 + + +
    + +     + +

    + Using the , we find \fp(x) = -x\big/\sqrt{1-x^2}. + The critical points of f are found when + \fp(x) = 0 or when \fp is undefined. + It is straightforward to find that \fp(x) = 0 when x=0, + + and \fp is undefined when x=\pm 1, + the endpoints of the interval + (which are in the domain of f.) + The table of important values is given in . + The maximum value is 1, + and the minimum value is 0. +

    +     +     + + + + + +

    + We have seen that continuous functions on closed intervals always have a maximum and minimum value, + and we have also developed a technique to find these values. + In , + we further our study of the information we can glean from nice + functions with the Mean Value Theorem. + On a closed interval, we can find the + average rate of change of a function + (as we did at the beginning of ). + We will see that differentiable functions always have a point at which their instantaneous + rate of change is same as the average rate of change. + This is surprisingly useful, as we'll see. +

    + + + + Terms and Concepts + + + + +

    + Describe what an extreme value + of a function is in your own words. +

    + +     + + + +     + + + +

    + Sketch the graph of a function f on (-1,1) that has both a maximum and minimum value. +

    +     + + + +     + + + + +

    + Describe the difference between absolute and relative maxima in your own words. +

    + +     + + + +     + + + +

    + Sketch the graph of a function f where f has a relative maximum at x=1 and \fp(1) is undefined. +

    +     + + + +     + + + + +

    + + If c is a critical value of a function f, + then f has either a relative maximum or relative minimum at x=c. +

    +     + +     + + + + + $zero = Real(0); + Context()->strings->add('undefined'=>{}); + Context()->strings->add('not defined'=>{alias=>'undefined'}); + $undefined = String("undefined"); + + +

    + Fill in the blanks: + The critical points of a function f are found where \fp(x) is equal to + or where \fp(x) is . +

    +     +     +     + +     + + + Problems + + + +

    + Identify each of the marked points as being an absolute maximum or minimum, + a relative maximum or minimum, or none of the above. +

    +     + + + + + for my $A ('A'..'G') {Context()->strings->add($A=>{})}; + $absmax = List("B"); + $absmin = List("None"); + $relmax = List("B", "G"); + $relmin = List("C","F"); + + + + A graph with several possible relative maximum and minimum values, some of which are omitted + +

    + The image shows a graph with labeled points A,B,C,D,E,F,G, going from left to right. + Each point is to be considered as potentially a relative or absolute maximum or minimum. +

    + +

    + The point A is an endpoint of the graph, but it is a hollow dot, indicating that it is not part of the graph. + Points on the graph near A are lower than anywhere else on the graph. +

    + +

    + The graph goes up until it reachs the point B, a solid dot, before going back down to the point C. + The y value at B is higher than at any other point on the graph. + At C the graph turns sharply: there is a cusp. To the right of C the graph goes up again, + until the point D. The point D is a hollow dot, indicating that it is not part of the graph. +

    + +

    + From D the graph goes down until it reaches E, which is a solid dot. + To the right of E, the graph continues to go down, + until it reaches F. + The point F is also a solid dot. To the right of F the graph goes back up, until ending at the point G. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-2.5,ymax=3.5,% + xmin=-.5,xmax=6.5,% + ] + + \draw [firstcolor] (axis cs:.1,-2) parabola [bend at end] (axis cs:1,3) parabola (axis cs:2,1) parabola [bend at end] (axis cs:3,2) parabola (axis cs:4,1) parabola [bend at end] (axis cs:5,-1) parabola (axis cs:6,2.5); + + \draw [firstcolor,fill=white] (axis cs: .1,-2) node [black,right] {\(A\)} circle (2.4pt) + (axis cs:3,2) node [above,black] {\(D\)} circle (2.4pt); + + \draw [firstcolor,fill=firstcolor] (axis cs: 1,3) node [above,black] {\(B\)} circle (2.4pt) + (axis cs:2,1) node [below,black] {\(C\)} circle (2.4pt) + (axis cs:4,1) node [right,black] {\(E\)} circle (2.4pt) + (axis cs:5,-1) node [below,black] {\(F\)} circle (2.4pt) + (axis cs:6,2.5) node [left,black] {\(G\)} circle (2.4pt); + + \end{axis} + \end{tikzpicture} + + + + Enter the absolute maxima points here, using commas to separate them if there is more than one. If there are none, enter NONE. + +

    + +

    + + Enter the absolute minima points here, using commas to separate them if there is more than one. If there are none, enter NONE. + +

    + +

    + + Enter the relative maxima points here, using commas to separate them if there is more than one. If there are none, enter NONE. + +

    + +

    + + Enter the relative minima points here, using commas to separate them if there is more than one. If there are none, enter NONE. + +

    + +

    +     +     +     + + + + + for my $A ('A'..'E') {Context()->strings->add($A=>{})}; + $absmax = List("C"); + $absmin = List("A"); + $relmax = List("C"); + $relmin = List("A","E"); + + + + A graph on the interval [1,5], with 5 marked points. + +

    + The image shows a graph on the intervl [1,5], with marked points A,B,C,D,E. + All five points are part of the graph. + The point A is the left endpoint of the graph, and its y value is below any other point on the graph. + The graph goes up toward the point B. + At B there appears to be a point of non-differentiability, + but to the right of B the graph continues to go up toward the point C. +

    + +

    + At C there is another cusp. To the right of C, the graph begins to go down, toward the point D. + There is a cusp at D as well, but to the right of D, the graph again goes down, until it reaches the endpoint at E. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-2.5,ymax=2.5,% + xmin=-.5,xmax=5.5,% + ] + + \draw [firstcolor] (axis cs:1,-2) parabola (axis cs:2,1) + -- (axis cs:3,2) + parabola [bend at end] (axis cs:4,1) + -- (axis cs:5,-1); + + \draw [firstcolor,fill=firstcolor] (axis cs: 1,-2) node [above,black] {\(A\)} circle (2.4pt) + (axis cs:2,1) node [left,black] {\(B\)} circle (2.4pt) + (axis cs:3,2) node [above,black] {\(C\)} circle (2.4pt) + (axis cs:4,1) node [right,black] {\(D\)} circle (2.4pt) + (axis cs:5,-1) node [below,black] {\(E\)} circle (2.4pt); + + \end{axis} + \end{tikzpicture} + + + + Enter the absolute maxima points here, using commas to separate them if there is more than one. If there are none, enter NONE. + +

    + +

    + + Enter the absolute minima points here, using commas to separate them if there is more than one. If there are none, enter NONE. + +

    + +

    + + Enter the relative maxima points here, using commas to separate them if there is more than one. If there are none, enter NONE. + +

    + +

    + + Enter the relative minima points here, using commas to separate them if there is more than one. If there are none, enter NONE. + +

    + +

    +     +     +     +     + + + +

    + Evaluate \fp(x) at the points indicated in the graph. +

    +     + + + + + $f=Formula("2/(x^2+1)"); + $zero=Real(0); + $dne = String("DNE"); + + +

    + f(x) = +

    + + A bell-shaped curve with a relative maximum at the marked point (0,2). + +

    + The graph of f(x)=\frac{2}{x^2+1} is bell-shaped, with the peak of the bell at the point (0,2). + The point (0,2) appears to be a relative maximum, and it is also the indicated point, + meaning that the goal of the question is to compute f'(0). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.5,ymax=2.5,% + xmin=-5.5,xmax=5.5,% + ] + + \addplot [firstcurvestyle,samples=101] {2/((x^2+1))}; + + \draw [firstcolor,fill=firstcolor] (axis cs: 0,2) node [above right,black] {\((0,2)\)} circle (2.4pt); + + \end{axis} + \end{tikzpicture} + + + + Enter f^{\prime}(0). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    +     +     +     + + + + + $f=Formula("x^2*sqrt(6-x^2)"); + $zero=Real(0); + $dne = String("DNE"); + + +

    + f(x) = +

    + + A graph consisting of a pair of arches, resembling a curved letter M. The points (0,0) and (2,4*sqrt(2)) are marked. + +

    + The graph of f(x)=x^2\sqrt{6-x^2} resembles a curved letter M, + with a pair of arches, one on either side of the y axis. +

    + +

    + The marked points are (0,0), which appears to be at a relative minimum between the two arches, + and (2,4\sqrt{2}), which appears to be at a relative maximum at the top of the right arch. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.9,ymax=6.5,% + xmin=-3.5,xmax=3.5,% + ] + + \addplot [firstcurvestyle,domain=-2.449489:2.449489,samples=104] {x^2*sqrt(6-x^2)}; + + \draw [firstcolor,fill=firstcolor] (axis cs: 0,0) node [below right,black] {\((0,0)\)} circle (2.4pt); + \draw [firstcolor,fill=firstcolor] (axis cs: 2,5.66) node [left,black] {\((2,4\sqrt{2})\)} circle (2.4pt); + \addplot[soliddot] coordinates {(2.449489,0) (-2.449489,0)}; + + \end{axis} + \end{tikzpicture} + + + + Enter f^{\prime}(0). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    + + Enter f^{\prime}(2). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    +     +     +     + + + + + $f=Formula("sin(x)"); + $zero=Real(0); + $dne = String("DNE"); + + +

    + f(x) = +

    + + Graph of a sine wave from 0 to 2 pi. The marked points are (pi/2, 1) and (3 pi/2,-1) + +

    + The graph of f(x)=\sin(x) from 0 to \pi. + There are two points marked on the graph: (\pi/2,1), at which f(x) appears to have a relative maximum, + and (3\pi/2,-1), at which f(x) appears to have a relative minimum. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.5,ymax=1.5,% + xmin=-.5,xmax=6.5,% + ] + + \addplot [firstcurvestyle,domain=0:6.28] {sin(deg(x))}; + + \draw [firstcolor,fill=firstcolor] (axis cs: 1.57,1) node [above,black] {\((\pi/2,1)\)} circle (2.4pt); + \draw [firstcolor,fill=firstcolor] (axis cs: 4.71,-1) node [below,black] {\((3\pi/2,-1)\)} circle (2.4pt); + \addplot[soliddot] coordinates {(0,0) (6.28,0)}; + + \end{axis} + \end{tikzpicture} + + + + Enter f^{\prime}(\pi/2). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    + + Enter f^{\prime}(3\pi/2). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    +     +     +     + + + + + $f=Formula("x^2*sqrt(4-x)"); + $zero=Real(0); + $dne = String("DNE"); + + +

    + f(x) = +

    + + A curved graph with a relative minimum at (0,0), and a relative maximum when x=16/5 + +

    + The graph of f(x)=x^2\sqrt{4-x} is plotted for x between -2 and 4. + There are three marked points: a relative minimum at (0,0), + a relative maximum at \left(\frac{16}{5},\frac{512}{25\sqrt{5}}\right), + and the point (4,0), which is an endpoint, since the domain of f(x) is (-\infty, 4]. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.8,ymax=11,% + xmin=-2.5,xmax=4.5,% + ] + + \addplot [firstcurvestyle,infiniteleft,domain=-2:4,samples=101] {x^2*sqrt(4-x)}; + + \draw [firstcolor,fill=firstcolor] (axis cs: 0,0) node [below right,black] {\((0,0)\)} circle (2.4pt); + \draw [firstcolor,fill=firstcolor] (axis cs: 3.2,9.16) node [above,black] {\(\left(\frac{16}{5},\frac{512}{25\sqrt{5}}\right)\)} circle (2.4pt); + \draw [firstcolor,fill=firstcolor] (axis cs: 4,0) node [above left,black] {\((4,0)\)} circle (2.4pt); + + \end{axis} + \end{tikzpicture} + + + + Enter f^{\prime}(0). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    + + Enter f^{\prime}(16/5). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    + + Enter f^{\prime}(4). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0); + $f=Formula("1+(x-2)^(2/3)/x"); + $zero=Real(0); + $dne = String("DNE"); + + +

    + f(x) = +

    + + A curve that descends steeply to a cusp at (2,1), and that is relatively flat for x greater than 1. + +

    + The graph of f(x)=1+\frac{(x-2)^{2/3}}{x} has a vertical asymptote at x=0. + It descends toward a cusp at (2,1), which is a relative minimum, and one of the marked points. + For x\gt 1 the graph rises slowly toward the point \left(6,1+\frac{\sqrt[3]{2}}{3}\right), which is also marked. + The graph is quite flat near this point, and it is hard to tell whether or not it could be a relative maximum. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.5,ymax=6.5,% + xmin=-1,xmax=11,% + ] + + \addplot [firstcurvestyle,infinite,domain=.2:11,samples=101] {(((x-2)^2)^(1/3))/x+1}; + + \draw [firstcolor,fill=firstcolor] (axis cs: 2,1) node [below,black] {\((2,1)\)} circle (2.4pt); + \draw [firstcolor,fill=firstcolor] (axis cs: 6,1.42) node [above,black] {\(\left(6,1+\frac{\sqrt[3]{2}}{3}\right)\)} circle (2.4pt); + + \end{axis} + \end{tikzpicture} + + + + + Enter f^{\prime}(2). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    + + Enter f^{\prime}(6). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0); + parser::Root->Enable; + $f=Formula("root(3,x^4-2x^2+1)"); + $zero=Real(0); + $dne = String("DNE"); + + +

    + f(x) = +

    + + A graph in the shape of a curved letter W. There are two cusps, both relative minima, at marked points (-1,0) and (1,0). + +

    + The graph of f(x)=\sqrt[3]{x^4-2x^2+1} has two cusps, at the points (-1,0) and (1,0), + which are the indicated points for this problem. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xtick={-2,-1,1,2}, + ymin=-.5,ymax=3.5,% + xmin=-2.5,xmax=2.5,% + ] + + \addplot [firstcurvestyle,infiniteleft,domain=-2:-1,samples=25] {(x^4-2*x^2+1)^(1/3)}; + \addplot [firstcurvestyle,domain=-1:1,samples=50] {(x^4-2*x^2+1)^(1/3)}; + \addplot [firstcurvestyle,infiniteright,domain=1:2,samples=25] {(x^4-2*x^2+1)^(1/3)}; + + \draw [firstcolor,fill=firstcolor] (axis cs: 1,0) node [above right,black] {\((1,0)\)} circle (2.4pt); + \draw [firstcolor,fill=firstcolor] (axis cs: -1,0) node [above left,black] {\((-1,0)\)} circle (2.4pt); + + \end{axis} + \end{tikzpicture} + + + + + Enter f^{\prime}(-1). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    + + Enter f^{\prime}(1). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    +     +     +     + + + + + $zero=Real(0); + $dne = String("DNE"); + + +

    + f(x) = \begin{cases}x^2,\amp x\leq0\\x^5,\amp x\gt0\end{cases} +

    + + The graph of a piecewise-defined function with a relative minimum at (0,0). + +

    + The piecewise-defined function for this problem is equal to x^2 when x is negative, + and x^5 when x is positive. +

    + +

    + The indicated point is (0,0), where there is a relative minimum. + Since the slope of the tangent line approaches 0 as x\to 0 for both x^2 and x^5, + it appears as though the graph will have a horizontal tangent line at the origin. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.5,ymax=1.1,% + xmin=-1.1,xmax=1.1,% + ] + + \addplot [firstcurvestyle,infiniteleft,domain=-1:0] {x^2}; + \addplot [firstcurvestyle,infiniteright,domain=0:1] {x^5}; + + \draw [firstcolor,fill=firstcolor] (axis cs: 0,0) node [below right,black] {\((0,0)\)} circle (2.4pt); + + \end{axis} + \end{tikzpicture} + + + + Enter f^{\prime}(0). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    +     +     +     + + + + + $zero=Real(0); + $dne = String("DNE"); + + +

    + f(x) = \begin{cases}x^2,\amp x\leq0\\x,\amp x\gt0\end{cases} +

    + + The graph of a piecewise-defined function with a relative minimum at the origin. + +

    + For this piecewise-defined function, there is a relative minimum at (0,0), which is the indicated point. + For x\gt 0, f(x)=x, so the slope of the tangent line for positive values of x close to 0 is 1. + For x\lt 0, f(x)=x^2, and the slope of the tangent line approaches 0 as x\to 0^-. + This suggests that f'(0) will be undefined. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-.5,ymax=1.1,% + xmin=-1.1,xmax=1.1,% + ] + + \addplot [firstcurvestyle,infiniteleft,domain=-1:0] {x^2}; + \addplot [firstcurvestyle,infiniteright,domain=0:1] {x}; + + \draw [firstcolor,fill=firstcolor] (axis cs: 0,0) node [below right,black] {\((0,0)\)} circle (2.4pt); + + \end{axis} + \end{tikzpicture} + + + + Enter f^{\prime}(0). If the derivative does not exist at the indicated point, enter DNE for does not exist. + +

    + +

    +     +     +     +     + + + +

    + Find the extreme values of the function on the given interval. + +

    +     + + + + ($b,$c) = random_subset(2,-9..-1,1..9); + ($l,$h) = random_subset(2,1..4); + if($envir{problemSeed}==1){$b=1;$c=4;$l=1;$h=2}; + $low = int(-$b/2) - $l; + $high = int(-$b/2) + $h; + $f = Formula("x^2+$b x+$c")->reduce; + Context("Fraction-NoDecimals"); + $max = Fraction(max($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>-$b/2))); + $min = Fraction(min($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>-$b/2))); + + +

    + f(x) = on [,] +

    + + Enter the maximimum value here, or enter NONE if there isn't one. + +

    + +

    + + Enter the minimimum value here, or enter NONE if there isn't one. + +

    + +

    +     +     +     + + + + + Context("Fraction-NoDecimals"); + $r = random(-9,-1,1); + $s = random(1,9,1); + $d = non_zero_random(-9,9,1); + $high = $s+random(1,2,1); + if($envir{problemSeed}==1){$r=-2;$s=5;$d=3;$high=6}; + $low = Real(0); + $b = Fraction(-3*($r+$s)/2); + $c = 3*$r*$s; + $f = Formula("x^3+$b x^2+$c x+$d")->reduce; + $max = Fraction(max($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>$s))); + $min = Fraction(min($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>$s))); + + +

    + f(x) = on [,] +

    + + Enter the maximimum value here, or enter NONE if there isn't one. + +

    + +

    + + Enter the minimimum value here, or enter NONE if there isn't one. + +

    + +

    +     +     +     + + + + + Context("Fraction-NoDecimals"); + $c = random(2,4,1); + $trig = list_random('sin','cos'); + ($lowd,$highd) = random_subset(2,3,4,6); + if ($trig eq 'sin') { + if (list_random(0,1)) {$low = Formula("pi/$lowd"); $high = Formula("(($highd-1)pi)/$highd")} + else {$low = Formula("(($lowd+1)pi)/$lowd"); $high = Formula("((2*$highd-1)pi)/$highd")} + } else { + if (list_random(0,1)) {$low = Formula("-pi/$lowd"); $high = Formula("pi/$highd")} + else {$low = Formula("(($lowd-1)pi)/$lowd"); $high = Formula("(($highd+1)pi)/$highd")} + } + if($envir{problemSeed}==1){$c=3;$trig='sin';$low=Formula("pi/4");$high=Formula("(2pi)/3")}; + $f = Formula("$c $trig(x)"); + $max = max($f->eval(x=>$low->eval(x=>0)),$f->eval(x=>$high->eval(x=>0)),$c * $f->eval(x=>$low->eval(x=>0)) / abs($f->eval(x=>$low->eval(x=>0)))); + $min = min($f->eval(x=>$low->eval(x=>0)),$f->eval(x=>$high->eval(x=>0)),$c * $f->eval(x=>$low->eval(x=>0)) / abs($f->eval(x=>$low->eval(x=>0)))); + + +

    + f(x) = on \left[,\right] +

    + + Enter the maximimum value here, or enter NONE if there isn't one. + +

    + +

    + + Enter the minimimum value here, or enter NONE if there isn't one. + +

    + +

    +     +     +     + + + + + Context("Fraction-NoDecimals"); + $n = list_random(2,4,6); + $r = random(1,4,1); + if($envir{problemSeed}==1){$n=2;$r=2;}; + $low = -$r; + $high = $r; + $f = Formula("x^$n sqrt($r^2-x^2)"); + $a = -sqrt($n*$r**2/($n+1)); + $b = sqrt($n*$r**2/($n+1)); + $max = max($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>$a),$f->eval(x=>$b),$f->eval(x=>0)); + $min = min($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>$a),$f->eval(x=>$b),$f->eval(x=>0)); + + +

    + f(x) = on [,] +

    + + Enter the maximimum value here, or enter NONE if there isn't one. + +

    + +

    + + Enter the minimimum value here, or enter NONE if there isn't one. + +

    + +

    +     +     +     + + + + + Context("Fraction-NoDecimals"); + $a = list_random(2,3,5,6,7); + $low = random(1,int(sqrt($a)),1); + $high = random(int(sqrt($a))+1,int(sqrt($a))+6); + if($envir{problemSeed}==1){$a=3;$low=1;$high=5;}; + $f = Formula("x+$a/x"); + $max = Fraction(max($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>sqrt($a)))); + $min = min($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>sqrt($a))); + + +

    + f(x) = on [,] +

    + + Enter the maximimum value here, or enter NONE if there isn't one. + +

    + +

    + + Enter the minimimum value here, or enter NONE if there isn't one. + +

    + +

    +     +     +     + + + + + Context("Fraction-NoDecimals"); + $n = random(2,4,2); + $a = random(1,9,1); + $low = random(-9,-1,1); + $high = random(1,9,1); + if($envir{problemSeed}==1){$n=2;$a=5;$low=-3;$high=5;}; + $f = Formula("x^$n/(x^$n+$a)"); + $max = Fraction(max($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>0))); + $min = Fraction(min($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>0))); + + +

    + f(x) = on [,] +

    + + Enter the maximimum value here, or enter NONE if there isn't one. + +

    + +

    + + Enter the minimimum value here, or enter NONE if there isn't one. + +

    + +

    +     +     +     + + + + + Context("Numeric"); + $trig='cos'; + $low = Formula("0"); + $high = Formula("pi"); + $f = Formula("e^x $trig(x)"); + $a = Formula("pi/4"); + $max = Compute("e^($a)/sqrt(2)"); + $min = Compute("-e^(pi)"); + + +

    + f(x) = on [,] +

    + + Enter the maximimum value here, or enter NONE if there isn't one. + +

    + +

    + + Enter the minimimum value here, or enter NONE if there isn't one. + +

    + +

    +     +     +     + + + + + Context("Numeric"); + $trig='sin'; + $low = Formula("0"); + $high = Formula("pi"); + $f = Formula("e^x $trig(x)"); + $a = Formula("3pi/4"); + $max = Compute("e^($a)/sqrt(2)"); + $min = Compute("0"); + + +

    + f(x) = on [,] +

    + + Enter the maximimum value here, or enter NONE if there isn't one. + +

    + +

    + + Enter the minimimum value here, or enter NONE if there isn't one. + +

    + +

    +     +     +     + + + + + Context("Numeric"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $n = random(1,4,1); + $high = random(3,9,1); + if($envir{problemSeed}==1){$n=1;$high=4;}; + $low = Compute("1"); + $a = exp(1/$n); + $f = Formula("ln(x)/x^$n")->reduce; + $max = Compute("1/($n e)"); + $min = Real(0); + + +

    + f(x) = on [,] +

    + + Enter the maximimum value here, or enter NONE if there isn't one. + +

    + +

    + + Enter the minimimum value here, or enter NONE if there isn't one. + +

    + +

    +     +     +     + + + + + Context("Fraction-NoDecimals"); + $d = random(2,5,1); + do {$n = random(1,2*$d-1,1)} until (gcd($n,$d) == 1); + $m = random(int($n/$d)+1,int($n/$d)+5,1); + $high = random(2,4,1); + if($envir{problemSeed}==1){$d=3;$n=2;$m=1;$high=2;}; + $low = Compute("0"); + $md = $m*$d; + $mdn = $md - $n; + $f = Formula("x^($n/$d) - x^$m")->reduce; + $max = max($f->substitute(x=>$low),$f->substitute(x=>$high),Formula("($mdn/$md) ($n/$md)^($n/$mdn)")); + $min = min($f->substitute(x=>$low),$f->substitute(x=>$high),Formula("($mdn/$md) ($n/$md)^($n/$mdn)")); + + +

    + f(x) = on [,] +

    + + Enter the maximimum value here, or enter NONE if there isn't one. + +

    + +

    + + Enter the minimimum value here, or enter NONE if there isn't one. + +

    + +

    +     +     +     +     +     +     +
    +
    + The Mean Value Theorem +

    + We motivate this section with the following question: Suppose you leave your house and drive to your friend's house in a city 100 miles away, + completing the trip in two hours. + At any point during the trip do you necessarily have to be going 50 miles per hour? +

    + +

    + In answering this question, + it is clear that the average + speed for the entire trip is + + 50 + + ( 100 miles in 2 hours), + but the question is whether or not your + instantaneous speed is ever exactly + + 50 + . + More simply, does your speedometer ever read exactly + + 50 + ? The answer, + under some very reasonable assumptions, is yes. +

    + + + +

    + Let's now see why this situation is in a calculus text by translating it into mathematical symbols. +

    + +

    + First assume that the function y = f(t) gives the distance + (in miles) + traveled from your home at time t + (in hours) + where 0\le t\le 2. + In particular, this gives f(0)=0 and f(2)=100. + The slope of the secant line connecting the starting and ending points + (0,f(0)) and (2,f(2)) is therefore + + \frac{\Delta f}{\Delta t} \amp = \frac{f(2)-f(0)}{2-0} + \amp = \frac{100-0}{2} + \amp = 50 \text{ mph} + . +

    + +

    + The slope at any point on the graph itself is given by the derivative \fp(t). + So, since the answer to the question above is yes, + this means that at some time during the trip, + the derivative takes on the value of + + 50 + . + Symbolically, + + \fp(c) = \frac{f(2)-f(0)}{2-0} = 50 + + for some time 0\le c \le 2. +

    + +

    + How about more generally? + Given any function y=f(x) and a range + a\le x\le b does the value of the derivative at some point between a and b have to match the slope of the secant line connecting the points (a,f(a)) and (b,f(b))? + Or equivalently, does the equation + \fp(c) = \frac{f(b)-f(a)}{b-a} have to hold for some a \lt c \lt b? +

    + +

    + Let's look at two functions in an example. +

    + + + Comparing average and instantaneous rates of change + +

    + Consider functions + + f_1(x)\amp=\frac{1}{x^2}\amp f_2(x)\amp= \abs{x} + + with a=-1 and b=1 as shown in . + Both functions have a value of 1 at a and b. + Therefore the slope of the secant line connecting the end points is 0 in each case. + But if you look at the plots of each, + you can see that there are no points on either graph where the tangent lines have slope zero. + Therefore we have found that there is no c in [-1,1] such that + + \fp(c) = \frac{f(1)-f(-1)}{1-(-1)} = 0 + . +

    + +
    + Graphs of two misbehaving functions + +
    + A graph of f_1(x) = 1/x^2 + + + + A graph of a Mean Value Theorem counterexample function. + + +

    + The graph illustrates the function f_1(x) = \frac{1}{x^2}, showcasing a scenario where the Mean Value Theorem is not applicable. + As x approaches zero, the function demonstrates a vertical asymptote, highlighting its discontinuity and the tendency of the values to become infinite. +

    +

    + Within the interval [-1, 1], although the function's value is the same at both endpoints, there is no point where the function's derivative, which represents the slope of the tangent, is zero. + This absence violates the Mean Value Theorem's requirement for the function to be both continuous on the closed interval and differentiable on the open interval, specifically at x = 0. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-1.95,xmax=1.95, + ymin=-.6,ymax=3.5,] + \addplot+[infinite,domain=-1.8:-0.58]{1/x^2}; + \addplot[firstcurvestyle,infinite,domain=.58:1.8]{1/x^2}; + \addplot[tangentlineseg, domain=-1:1]{1}; + \addplot[soliddot] coordinates {(-1,1) (1,1)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + A graph of f_2(x) = \abs{x} + + + + A graph of the absolute value function as a Mean Value Theorem counterexample. + + +

    + Displayed is the function f_2(x) = \abs{x}, chosen to highlight a case where the Mean Value Theorem cannot be applied. + The graph forms a V shape, characteristic of the absolute value function, and is continuous over the interval [-1, 1]. +

    +

    + At the endpoints of the interval, the function attains the same value, yielding a secant line with a slope of zero. + However, due to the sharp corner at the origin x = 0, the function is not differentiable at this point. + This lack of differentiability means that there does not exist a point in the interval where the slope of the tangent line is equal to the slope of the secant line, as required by the Mean Value Theorem. + Thus, the function f_2(x) serves as an example of when the Mean Value Theorem's conditions are not fulfilled. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-1.95,xmax=1.95, + ymin=-.6,ymax=3.5,] + \addplot+[infinite,domain=-1.8:1.8]{abs(x)}; + \addplot[tangentlineseg, domain=-1:1]{1}; + \addplot[soliddot] coordinates {(-1,1) (1,1)}; + \end{axis} + \end{tikzpicture} + + + + +
    +     +
    +     +     + +

    + So what went wrong? + It may not be surprising to find that the discontinuity of f_1 and the corner of f_2 play a role. + If our functions had been continuous and differentiable, + would we have been able to find that special value c? + This is our motivation for the following theorem. +

    + + + The Mean Value Theorem of Differentiation + +

    + Let y=f(x) be a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b). + There exists a value c, a \lt c \lt b, + such that Mean Value Theoremof differentiationderivativeMean Value Theorem + + \fp(c) = \frac{f(b)-f(a)}{b-a} + . +

    + +

    + That is, there is a value c in (a,b) where the instantaneous rate of change of f at c is equal to the average rate of change of f on [a,b]. +

    +     +     + +

    + Note that the reasons that the functions in + fail are indeed that f_1 has a discontinuity on the interval [-1,1] and f_2 is not differentiable at the origin. +

    + + + + +

    + We will give a proof of the Mean Value Theorem below. + To do so, we use a fact, called Rolle's Theorem, stated here. +

    + + + Rolle's Theorem + +

    + Let f be continuous on [a,b] and differentiable on (a,b), + where f(a) = f(b). + There is some c in (a,b) such that \fp(c) = 0. + Rolle's Theorem +

    +     +     + + + +

    + Consider + where the graph of a function f is given, where f(a) = f(b). + It should make intuitive sense that if f is differentiable + (and hence, continuous) + that there would be a value c in (a,b) where \fp(c)=0; + that is, there would be a relative maximum or minimum of f in (a,b). + Rolle's Theorem guarantees at least one; there may be more. +

    + +
    + A graph of f(x) = x^3-5x^2+3x+5, where f(a) = f(b). Note the existence of c, where a\lt c\lt b, where \fp(c)=0. + + + + A visual representation of Rolle's Theorem with a corresponding graph. + + +

    + The image showcases a graph of the polynomial function f(x) = x^3 - 5x^2 + 3x + 5, which is used to illustrate Rolle's Theorem. The function is plotted to show the existence of at least one point c in the open interval (a, b) where the derivative \fp(c) is zero, indicated by the horizontal tangent at the peak of the curve. + The points labeled a and b on the x-axis denote the interval on which the function's endpoints have equal values, satisfying the precondition for Rolle's Theorem. +

    +

    + Rolle's Theorem is stated, asserting that for a function continuous on the closed interval [a, b] and differentiable on the open interval (a, b), where the function values at the endpoints are equal (f(a) = f(b)), there exists at least one point c within (a, b) where the derivative f'(c) is zero. This theorem is a specific case of the Mean Value Theorem. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-1.5,xmax=2.5, + ymin=-5.2,ymax=8, + extra x ticks={-.39,1.22,.333}, + extra x tick labels={$a$,$b$,$c$},] + \addplot+[infinite,domain=-1:2.2, samples=40] {x^3-5*x^2+3*x+5}; + \addplot [secantlineseg,domain=-1:2] {3}; + \addplot [soliddot] coordinates {(-.39,3) (1.22,3) (.3333,5.48)}; + \addplot [tangentlineseg,domain=-.75:1.5] {5.48}; + \end{axis} + \end{tikzpicture} + + + + +
    + + + + + +

    + Rolle's Theorem is really just a special case of the Mean Value Theorem. + If f(a) = f(b), then the average + rate of change on (a,b) is 0, + and the theorem guarantees some c where \fp(c)=0. + We will prove Rolle's Theorem, + then use it to prove the Mean Value Theorem. +

    + + + Proof of Rolle's Theorem +

    + Let f be differentiable on (a,b) where f(a)=f(b). + We consider two cases. +

    + Case 1 +

    + Consider the case when f is constant on [a,b]; + that is, f(x) = f(a) = f(b) for all x in [a,b]. + Then \fp(x) = 0 for all x in [a,b], + showing there is at least one value c in (a,b) where \fp(c)=0. +

    +     + Case 2 +

    + Now assume that f is not constant on [a,b]. + The Extreme Value Theorem guarantees that f has a maximal and minimal value on [a,b], + found either at the endpoints or at a critical value in (a,b). + Since f(a)=f(b) and f is not constant, + it is clear that the maximum and minimum cannot both + be found at the endpoints. + Assume, without loss of generality, + that the maximum of f is not found at the endpoints. + Therefore there is a c in (a,b) such that f(c) is the maximum value of f. + By , + c must be a critical number of f; + since f is differentiable, + we have that \fp(c) = 0, + completing the proof of the theorem. +

    +     +     + +

    + We can now prove the Mean Value Theorem. +

    + + + Proof of the Mean Value Theorem + + +

    + Define the function + + g(x) = f(x) - \frac{f(b)-f(a)}{b-a}x + . +

    + +

    + We know g is differentiable on (a,b) and continuous on [a,b] since f is. + We can show g(a)=g(b) (it is actually easier to show g(b)-g(a)=0, + which suffices). + We can then apply Rolle's theorem to guarantee the existence of + c in (a,b) such that \gp(c) = 0. + But note that + + 0= \gp(c) = \fp(c) - \frac{f(b)-f(a)}{b-a} + ; + hence + + \fp(c) = \frac{f(b)-f(a)}{b-a} + , + which is what we sought to prove. +

    +     + +

    + Going back to the very beginning of the section, + we see that the only assumption we would need about our distance function f(t) is that it be continuous and differentiable for t from 0 to 2 hours + (both reasonable assumptions). + By the Mean Value Theorem, + we are guaranteed a time during the trip where our instantaneous speed is + + 50 + . + This fact is used in practice. + Some law enforcement agencies monitor traffic speeds while in aircraft. + They do not measure speed with radar, + but rather by timing individual cars as they pass over lines painted on the highway whose distances apart are known. + The officer is able to measure the average + speed of a car between the painted lines; + if that average speed is greater than the posted speed limit, + the officer is assured that the driver exceeded the speed limit at some time. +

    + +

    + Note that + is an existence theorem. + It states that a special value c exists, + but it does not give any indication about how to find it. + It turns out that when we need the Mean Value Theorem, + existence is all we need. +

    + + + Using the Mean Value Theorem + +

    + Consider f(x) = x^3+5x+5 on [-3,3]. + Find c in [-3,3] that satisfies . +

    +     + +

    + The average rate of change of f on [-3,3] is: + + \frac{f(3)-f(-3)}{3-(-3)} \amp =\frac{47-(-37)}{6} + \amp =\frac{84}{6} + \amp = 14 + . +

    + +

    + We want to find c such that \fp(c) = 14. + We find \fp(x) = 3x^2+5. + We set this equal to 14 and solve for x. + + \fp(x) \amp = 14 + 3x^2 +5 \amp = 14 + x^2 \amp = 3 + x \amp = \pm \sqrt{3} \approx \pm 1.732 + +

    + +

    + We have found two values c in [-3,3] where the instantaneous rate of change is equal to the average rate of change; + the Mean Value Theorem guaranteed at least one. + In , + f is graphed with a line representing the average rate of change; + the lines tangent to f at x=\pm \sqrt{3} are also given. + Note how these lines are parallel (, have the same slope) to the secant line. +

    + +
    + Demonstrating the Mean Value Theorem in + + + + Graphical illustration of the Mean Value Theorem in action. + + +

    + This graph depicts the function f(x) = x^3 + 5x + 5 over the interval [-3, 3]. + The function is shown as a solid blue curve, and there is a black dashed line that represents the average rate of change of the function over the given interval. +

    +

    + Two red dashed lines are drawn on the graph, indicating the tangent lines to the curve at the points where the function's derivative equals the average rate of change. + These points, marked on the x-axis, are where the slope of the tangent is the same as the slope of the secant line (the black dashed line), which is the essence of the Mean Value Theorem. + The exact points on the x-axis where these tangents touch the curve correspond to the solutions where f'(x) = 3x^2 + 5 is set equal to 14, the average rate of change. +

    +     + + + \begin{tikzpicture} + \begin{axis}[xmin=-3.2,xmax=3.2, + ymin=-45,ymax=51,] + \addplot+[domain=-3:3] {x^3+5*x+5}; + \addplot [secantlineseg, domain=-3:3]{14*(x-3)+47}; + \addplot [soliddot] coordinates {(3,47) (-3,-37) (-1.732,-8.86) (1.732, 18.86)}; + \addplot [tangentlineseg,domain=-2.8:-.1] {14*(x+1.732)-8.86}; + \addplot [tangentlineseg,domain=-.1:3] {14*(x-1.732)+18.86}; + \end{axis} + \end{tikzpicture} + + + + +
    +     + +     + + + +

    + While has practical use + (for instance, + the speed monitoring application mentioned before), + it is mostly used to advance other theory. + We will use it in the next section to relate the shape of a graph to its derivative. +

    + +

    + Before ending this section, we give two important consequences of the Mean + Value Theorem. Each of these consequences has important applications to + mathematical theory, and can be easily understood in the context of the + position and velocity of objects in motion. +

    + +

    + First, we recall that the derivative of any constant function is zero. + Is the converse true? That is, are constant functions the only ones whose derivative is zero? + The Mean Value Theorem says yes. + This officially + establishes our intuition about objects in (or, actually, not in) + motion: if the velocity of an object is 0, then the object's position is + unchanged; it is constant. + Second, if two functions f and g have the same derivative, + what does this tell us about f and g? + The Mean Value Theorem implies that these functions must only differ by a constant; + that is, f(x)=g(x)+C, for some constant C. +

    + +

    + This has an application to motion that is not intuitive to some. Suppose + two objects start moving while 5 apart, + and always move with the same velocity. + Then the two objects will always be 5 apart. + (If two pennies are dropped from the 30th and 31st stories of a tall building at + the same time, they will always be 1 story apart as they fall.) +

    + + + Consequences of the Mean Value Theorem + +

    + Let f, g, and h be differentiable + (and therefore continuous) functions on an in terval I. +

      +
    1. +

      + If \fp(x)=0 for all x in the interval I, + then f is a constant function on I. +

      +
      2. +
    2. +

      + If \gp(x)=h'(x) for all x in I, + then there is a constant C such that g(x)=h(x)+C + for all x in I. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + Choose any two points a and b in the interval I. + By the Mean Value Theorem, + we must have + + \fp(c) = \frac{f(b)-f(a)}{b-a} + + for some c between a and b. + But \fp(c)=0, so f(b)-f(a)=0, or f(a)=f(b). + Since a and b were any two points, + this tells us that f must have the same value at every point; + that is, f must be constant. +

      +
      2. +
    2. +

      + Suppose \gp(x)=h'(x) for each point x in I, + and consider the function f(x)=g(x)-h(x). + By the difference rule for derivatives, we have + + f'(x) = \gp(x)-h'(x)=0 + , + since \gp(x)=h'(x). +

      + +

      + By the previous result, this means that f(x) is a constant function. + That is, f(x)=C for each x in I, giving us + g(x)-h(x)=C, or g(x)=h(x)+C. +

      +
      3. +
    +

    +     +     + + + + + + + + + + + + + + Terms and Concepts + + + + +

    + Explain in your own words what the Mean Value Theorem states. +

    + +     + + + +     + + + + +

    + Explain in your own words what Rolle's Theorem states. +

    + +     + + + +     +     + + + Problems + + +

    + A function f(x) and interval [a,b] are given. + Check if can be applied to f on [a,b]; + if so, find c in (a,b) such that \fp(c)=0. +

    +     + + + + Context("Interval"); + Context()->strings->add('does not apply'=>{}); + $c=Compute("(-1,1)"); + $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; + + +

    + f(x) = 6 on [-1,1] +

    +

    + +

    + + If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. + You may either list individual numbers separated by commas, or use interval notation. + If Rolle's Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Interval"); + Context()->strings->add('does not apply'=>{}); + $c=Compute("does not apply"); + $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; + + +

    + f(x) = 6x on [-1,1] +

    +

    + +

    + + If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. + You may either list individual numbers separated by commas, or use interval notation. + If Rolle's Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->strings->add('does not apply'=>{}); + $c=Fraction("-1/2"); + $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; + + +

    + f(x) = x^2+x-6 on [-3,2] +

    +

    + +

    + + If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. + You may either list individual numbers separated by commas, or use interval notation. + If Rolle's Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->strings->add('does not apply'=>{}); + $c=Fraction("-1/2"); + $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; + + +

    + f(x) = x^2+x-2 on [-3,2] +

    +

    + +

    + + If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. + You may either list individual numbers separated by commas, or use interval notation. + If Rolle's Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Interval"); + Context()->strings->add('does not apply'=>{}); + $c=Compute("does not apply"); + $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; + + +

    + f(x) = x^2+x on [-2,2] +

    +

    + +

    + + If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. + You may either list individual numbers separated by commas, or use interval notation. + If Rolle's Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Interval"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Formula("pi/2"); + $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; + + +

    + f(x) = \sin(x) on [\pi/6,5\pi/6] +

    +

    + +

    + + If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. + You may either list individual numbers separated by commas, or use interval notation. + If Rolle's Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Interval"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Compute("does not apply"); + $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; + + +

    + f(x) = \cos(x) on [0,\pi] +

    +

    + +

    + + If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. + You may either list individual numbers separated by commas, or use interval notation. + If Rolle's Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Interval"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Compute("does not apply"); + $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; + + +

    + f(x) = \frac{1}{x^2-2x+1} on [0,2] +

    +

    + +

    + + If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. + You may either list individual numbers separated by commas, or use interval notation. + If Rolle's Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     +     + + + +

    + A function f(x) and interval [a,b] are given. + Check if can be applied to f on [a,b]; + if so, find c in (a,b) guaranteed by the Mean Value Theorem. +

    +     + + + + Context("Interval"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Formula("0"); + $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; + + +

    + f(x) = x^2+3x-1 on [-2,2] +

    +

    + +

    + + If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. + You may either list individual numbers separated by commas, or use interval notation. + If the Mean Value Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Fraction("5/2"); + $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; + + +

    + f(x) = 5x^2-6x+8 on [0,5] +

    +

    + +

    + + If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. + You may either list individual numbers separated by commas, or use interval notation. + If the Mean Value Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Formula("3sqrt(2)/2"); + $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; + + +

    + f(x) = \sqrt{9-x^2} on [0,3] +

    +

    + +

    + + If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. + You may either list individual numbers separated by commas, or use interval notation. + If the Mean Value Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Fraction("19/4"); + $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; + + +

    + f(x) = \sqrt{25-x} on [0,9] +

    +

    + +

    + + If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. + You may either list individual numbers separated by commas, or use interval notation. + If the Mean Value Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Compute("does not apply"); + $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; + + +

    + f(x) = \frac{x^2-9}{x^2-1} on [0,2] +

    +

    + +

    + + If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. + You may either list individual numbers separated by commas, or use interval notation. + If the Mean Value Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Formula("4/ln(5)"); + $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; + + +

    + f(x) = \ln(x) on [1,5] +

    +

    + +

    + + If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. + You may either list individual numbers separated by commas, or use interval notation. + If the Mean Value Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Compute("-asec(2/sqrt(pi)), asec(2/sqrt(pi))"); + $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; + + +

    + f(x) = \tan(x) on [-\pi/4,\pi/4] +

    +

    + +

    + + If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. + You may either list individual numbers separated by commas, or use interval notation. + If the Mean Value Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Fraction("-2/3"); + $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; + + +

    + f(x) = x^3-2x^2+x+1 on [-2,2] +

    +

    + +

    + + If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. + You may either list individual numbers separated by commas, or use interval notation. + If the Mean Value Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Compute("5+7sqrt(7)/6, 5-7sqrt(7)/6"); + $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; + + +

    + f(x) = 2x^3-5x^2+6x+1 on [-5,2] +

    +

    + +

    + + If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. + You may either list individual numbers separated by commas, or use interval notation. + If the Mean Value Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->strings->add('does not apply'=>{}); + $c=Compute("sqrt(pi^2-4)/pi, -sqrt(pi^2-4)/pi"); + $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; + + +

    + f(x) = \sin^{-1}(x) on [-1,1] +

    +

    + +

    + + If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. + You may either list individual numbers separated by commas, or use interval notation. + If the Mean Value Theorem does not apply, enter does not apply. + +

    + +

    +     +     +     +     +     +     +
    +
    + Increasing and Decreasing Functions +

    + Our study of nice functions f in this chapter has so far focused on individual points: + points where f is maximal/minimal, + points where \fp(x) = 0 or \fp does not exist, + and points c where \fp(c) is the average rate of change of f on some interval. +

    + +

    + In this section we begin to study how functions behave + between special points; + we begin studying in more detail the shape of their graphs. +

    + +

    + We start with an intuitive concept. + Given the graph in , + where would you say the function is increasing? + Decreasing? + Even though we have not defined these terms mathematically, + one likely answered that f is increasing when x \gt 1 and decreasing when x\lt 1. + We formally define these terms here. +

    + +
    + A graph of a function f used to illustrate the concepts of increasing and decreasing + + + The graph of a function with a relative minimum at x=1 + +

    + The image shows the graph of a function that has a relative minimum at x=1. + At the left edge, the graph is shown with y values close to 4 when x\lt 0. + The y values then get smaller until the minimum is reached at x=1. + Continuing to the right of x=1, the y values start to get bigger again, + reaching values above y=4 when x\gt 2. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.9, + xmax=3.2, + ymin=-.5, + ymax=5.1, + ] + \addplot+[domain=-.5:2.5] {(x-1)^2+2}; + \end{axis} + \end{tikzpicture} + + + +
    + + + + Increasing and Decreasing Functions + +

    + Let f be a function defined on an interval I. + increasing function + decreasing function + + + +

      +
    1. +

      + f is increasing + on I if for every a\lt b in I, f(a) \lt f(b). +

      +
      2. +
    2. +

      + f is decreasing + on I if for every a\lt b in I, f(a) \gt f(b). +

      +
      3. +
    +

    +     +     + + + +

    + Informally, a function is increasing if as x gets larger (, looking left to right) f(x) gets larger. +

    + + + + + + +

    + Our interest lies in finding intervals in the domain of f on which f is either increasing or decreasing. + Such information should seem useful. + For instance, if f describes the speed of an object, + we might want to know when the speed was increasing or decreasing (, when the object was accelerating vs. decelerating). + If f describes the population of a city, + we should be interested in when the population is growing or declining. +

    + +

    + To find such intervals, we again consider secant lines. + Let f be an increasing, + differentiable function on an open interval I, + such as the one shown in , + and let a\lt b be given in I. + The secant line on the graph of f from x=a to x=b is drawn; + it has a slope of (f(b)-f(a))/(b-a). +

    + +

    + But note, since b \gt a and f is increasing, + f(b) \gt f(a). + And these facts imply b-a \gt 0 and f(b)-f(a) \gt 0. + Therefore: +

    + +
    + Examining the secant line of an increasing function + +

    + + \amp\frac{f(b)-f(a)}{b-a} \gt 0 + \implies\amp \text{slope of the secant line} \gt 0 + \implies\amp \text{Average rate of change of }f + \amp\text{ on }[a,b]\text{ is } \gt 0 + . +

    + + + + The graph of a function is shown, along with the secant line between two points on the graph. + +

    + The graph of some function is shown, for x and y between 0 and 2. + Two points (a,f(a)) and (b,f(b)) are marked on the graph, + with b\gt a and f(b)\gt f(a). +

    + +

    + A secant line (dashed) is drawn between these points. + The slope of the secant line is positive, indicating a positive average rate of change for the function on the interval [a,b]. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.2, + xmax=2.2, + ymin=-.5, + ymax=2.2, + extra x ticks={.218,1.67}, + extra x tick labels={$a$,$b$} + ] + \addplot+[domain=-80:-5] ({2*cos(x)-.3},{2*sin(x)+2.3}); + \addplot [secantlineseg] coordinates { ({2*cos(-75)-.3},{2*sin(-75)+2.3}) ({2*cos(-10)-.3},{2*sin(-10)+2.3})}; + \addplot [soliddot] coordinates {(.218,.368)} node [below]{$(a,f(a))$}; + \addplot [soliddot] coordinates {(1.67,1.95)} node [above left]{$(b,f(b))$}; + \end{axis} + \end{tikzpicture} + + + +     +
    + +

    + We have shown mathematically what may have already been obvious: + when f is increasing, its secant lines will have a positive slope. + Now recall that the Mean Value Theorem + guarantees that there is a number c, + where a\lt c\lt b, such that + + \fp(c) = \frac{f(b)-f(a)}{b-a} \gt 0 + . +

    + +

    + By considering all such secant lines in I, + we strongly imply that \fp(x) \gt 0 on I. + A similar statement can be made for decreasing functions. +

    + +

    + Our above logic can be summarized as + If f is increasing, + then \fp is probably positive. + + below turns this around by stating + If \fp is positive, + then f is increasing. + This leads us to a method for finding when functions are increasing and decreasing. +

    + + + Test For Increasing/Decreasing Functions + +

    + Let f be a continuous function on [a,b] and differentiable on (a,b). +

    + +

    +

      +
    1. +

      + If \fp(c) \gt 0 for all c in (a,b), + then f is increasing on [a,b]. +

      +
      2. +
    2. +

      + If \fp(c) \lt 0 for all c in (a,b), + then f is decreasing on [a,b]. +

      +
      3. +
    3. +

      + If \fp(c) =0 for all c in (a,b), + then f is constant on [a,b]. +

      +
      4. +
    +

    + +

    + The conclusions of and also hold if + \fp(c) = 0 for a finite number of nonadjacent values of c in I. +

    +     +     + + + + + +

    + Let f be differentiable on an interval I and let a and b be in I where + \fp(a) \gt 0 and \fp(b)\lt 0. + If \fp is continuous on [a,b], + it follows from the Intermediate Value Theorem that there must be some value + c between a and b where \fp(c) = 0. + (It turns out that this is still true even if \fp is not continuous on [a,b].) + This leads us to the following method for finding intervals on which a function is increasing or decreasing. +

    + + + Finding Intervals on Which <m>f</m> is Increasing or Decreasing +

    + Let f be a continuous function on an interval I. + To find intervals on which f is increasing and decreasing: + increasing functionfinding intervals + decreasing functionfinding intervals +

    + +

    +

      +
    1. +

      + If not stated, find the domain of f, D. Begin a number line that only includes D. +

      +
      2. +
    2. +

      + Find the critical values of f. + That is, find all c in the domain of f where + \fp(c) = 0 or \fp is not defined. + (Note: Any values of c not + in the domain of f where \fp(c) is undefined should already be marked on your number line from Step). +

      +
      3. +
    3. +

      + Use the critical values to divide D into subintervals. +

      +
      4. +
    4. +

      + Pick any point p in each subinterval, + and find the sign of \fp(p). +

      + +

      +

        +
      1. +

        + If \fp(p) \gt 0, then f is increasing on that subinterval. +

        +
        2. +
      2. +

        + If \fp(p)\lt 0, then f is decreasing on that subinterval. +

        +
        3. +
      +

      +
      5. +
    +

    + +

    + Note that although allows us to use determine that a function is increasing or decreasing on a closed interval, + it is conventional to state the intervals of increase and decrease as open intervals. + We will follow this convention in the examples that follow, + but it is also acceptable to answer using closed intervals. +

    + +

    + In particular, one should note the following: +

      +
    • +

      + If \fp(x)\gt 0 on (a,b) and on (b,c), with \fp(b)=0, + then we should say that f is increasing on (a,c) (or on [a,c]) + the zero of the derivative should be included. +

      +
      • +
    • +

      + If \fp(x)\gt 0 on (a,b) and on (b,c), + but f(b) is undefined (or f is discontinuous at b), + then we should not include the point b in our interval. + Instead, we say that f is increasing on (a,b) and (b,c), + or on [a,b) and (b,c]. +

      +
      • +
    +

    +     + + + +

    + We demonstrate using this process in the following example. +

    + + + Finding intervals of increasing/decreasing + +

    + Let f(x) = x^3+x^2-x+1. + Find intervals on which f is increasing or decreasing. +

    +     + +

    + Since an interval was not specified for us to consider, using , + the domain of f is \mathbb{R} or (-\infty,\infty). + Next, we find the critical values of f. + We have \fp(x) = 3x^2+2x-1 = (3x-1)(x+1), + so \fp(x) = 0 when x=-1 and when x=1/3. + \fp is never undefined. +

    + +

    + We thus break the domain + (in this case the (-\infty, \infty)) + into three subintervals based on the two critical values we just found: + (-\infty,-1), (-1,1/3) and (1/3,\infty). + This is shown in . +

    + +
    + Number line for f in + + + Number line showing the critical points of f + +

    + A number line is shown with two marked points. + The points are labeled with the values -1 and 1/3, + which are the critical points of the function f(x)=x^3+x^2-x+1. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + numberline, + xmin=-2, + xmax=1, + extra x ticks={-1, 0.3333}, + extra x tick labels={$-1$,$1/3$}, + ] + \addplot[guideline] coordinates {(-1,0) (-1,2)}; + \addplot[guideline] coordinates {(0.3333,0) (0.3333,2)}; + \end{axis} + \end{tikzpicture} + + +
    + +

    + We now pick a value p in each subinterval and find the sign of \fp(p). + All we care about is the sign, + so we do not actually have to fully compute \fp(p); + pick nice values that make this simple. +

    + +

    +

    +
  • + Subinterval 1: <m>(-\infty,-1)</m> +

    + We (arbitrarily) pick p=-2. + We can compute \fp(-2) directly: + \fp(-2) = 3(-2)^2+2(-2)-1=7\gt 0. + We conclude that f is increasing on (-\infty,-1). +

    + +

    + Note we can arrive at the same conclusion without computation. + For instance, we could choose p=-100. + The first term in \fp(-100), + , 3(-100)^2 is clearly positive and very large. + The other terms are small in comparison, + so we know \fp(-100)\gt 0. + All we need is the sign. +

    +
    • +
  • + Subinterval 2: <m>(-1,1/3)</m> +

    + We pick p=0 since that value seems easy to deal with. + \fp(0) = -1\lt 0. + We conclude f is decreasing on (-1,1/3). +

    +
    • +
  • + Subinterval 3: <m>(1/3,\infty)</m> +

    + Pick an arbitrarily large value for p\gt 1/3 and note that \fp(p) =3p^2+2p-1 \gt 0. + We conclude that f is increasing on (1/3,\infty). +

    +
    • +
    +

    + +

    + summarizes our work. +

    + +
    + Completed number line for f in + + + + +

    + A number line is shown with two marked points. + The points are labeled with the values -1 and 1/3, + which are the critical points of the function f(x)=x^3+x^2-x+1. +

    + +

    + The marked points divide the number line into three intervals. + Above each interval, there is text indicating the sign of f', + and whether the function f is increasing or decreasing. +

    + +

    + The number line indicates the following: +

      +
    • +

      + For x\lt -1, \fp(x)\gt 0, + and f is increasing. +

      +
      • + +
    • +

      + For -1\lt x\lt 1/3, \fp(x)\lt 0, + and f is decreasing. +

      +
      • + +
    • +

      + For x\gt 1/3, \fp(x)\gt 0, + and f is increasing. +

      +
      • +
    +

    +     + + \begin{tikzpicture} + \begin{axis}[ + numberline, + xmin=-2, + xmax=1, + extra x ticks={-1, 0.3333}, + extra x tick labels={$-1$,$1/3$}, + ] + \addplot[guideline] coordinates {(-1,0) (-1,2)}; + \addplot[guideline] coordinates {(0.3333,0) (0.3333,2)}; + \addplot[mark=none] coordinates {(-1.6666,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; + \addplot[mark=none] coordinates {(-0.3333,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }}; + \addplot[mark=none] coordinates {(1,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + We can verify our calculations by considering , + where f is graphed. + The graph also presents \fp; + note how \fp\gt 0 when f is increasing and + \fp\lt 0 when f is decreasing. +

    + +
    + A graph of f(x) in , showing where f is increasing and decreasing + + + The graphs of a function and its derivative. + +

    + In the image, we see the graph of f(x)=x^3+x^2-x+1 (solid, in blue), + along with the graph of its derivative f'(x)=3x^2+2x-1 (dash-dotted, in red). +

    + +

    + In the regions where f'(x) is positive (above the x axis), f(x) is increasing. + Where f'(x) is negative (below the x axis), f(x) is decreasing. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-2.2, + xmax=2.2, + ymin=-3, + ymax=11, + extra x ticks={.33}, + extra x tick labels={$1/3$}, + ] + \addplot+[domain=-2:2] {x^3+x^2-x+1} node[pos=0.6, right]{$f(x)$}; + \addplot+[domain=-1.8:1.5] {3*x^2+2*x-1} node[pos=0.1, above right] {$\fp(x)$}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + +

    + One is justified in wondering why so much work is done when the graph seems to make the intervals very clear. + We give three reasons why the above work is worthwhile. +

    + +

    + First, the points at which f switches from increasing to decreasing are not precisely known given a graph. + The graph shows us something significant happens near x=-1 and x=0.3, + but we cannot determine exactly where from the graph. +

    + +

    + One could argue that just finding critical values is important; + once we know the significant points are x=-1 and x=1/3, + the graph shows the increasing/decreasing traits just fine. + That is true. + However, the technique prescribed here helps reinforce the relationship between increasing/decreasing and the sign of \fp. + Once mastery of this concept + (and several others) + is obtained, + one finds that either (a) just the critical points are computed and the graph shows all else that is desired, + or (b) a graph is never produced, + because determining increasing/decreasing using \fp is straightforward and the graph is unnecessary. + So our second reason why the above work is worthwhile is this: + once mastery of a subject is gained, + one has options for finding needed information. + We are working to develop mastery. +

    + +

    + Finally, our third reason: + many problems we face in the real world are very complex. + Solutions are tractable only through the use of computers to do many calculations for us. + Computers do not solve problems + on their own, however; + they need to be taught (, programmed) to do the right things. + It would be beneficial to give a function to a computer and have it return maximum and minimum values, + intervals on which the function is increasing and decreasing, + the locations of relative maxima, etc. + The work that we are doing here is easily programmable. + It is hard to teach a computer to + look at the graph and see if it is going up or down. + It is easy to teach a computer to + determine if a number is greater than or less than 0. +

    + +

    + In + we learned the definition of relative maxima and minima and found that they occur at critical points. + We are now learning that functions can switch from increasing to decreasing + (and vice-versa) + at critical points. + This new understanding of increasing and decreasing creates a great method of determining whether a critical point corresponds to a maximum, minimum, + or neither. + Imagine a function increasing until a critical point at x=c, + after which it decreases. + A quick sketch helps confirm that f(c) must be a relative maximum. + A similar statement can be made for relative minimums. + We formalize this concept in a theorem. +

    + + + First Derivative Test + +

    + Let f be continuous on an interval I, + and differentiable on I, + except possibly at c, where c is a critical number in I. + + First Derivative Test + derivativeFirst Deriv. Test + extremaand First Deriv. Test + maximumand First Deriv. Test + minimumand First Deriv. Test + +

      +
    1. +

      + If the sign of \fp switches from positive to negative at c, + then f(c) is a relative maximum of f. +

      +
      2. +
    2. +

      + If the sign of \fp switches from negative to positive at c, + then f(c) is a relative minimum of f. +

      +
      3. +
    3. +

      + If \fp is positive (or, + negative) before and after c, + then f(c) is not a relative extrema of f. +

      +
      4. +
    +

    +     +     + + + + + + Importance of Continuity +

    + The continuity of f when using the first derivative test is very important. + Without continuity, almost anything can happen at a critical number. + For example, + we can construct a piecewise function where the sign of \fp switches to positive to negative at c and f(c) is + not a local maximum. + This is shown in . +

    + +
    + A discontinuous function where \fp changes sign at 1, but f(1) is not a local maximum + + The graph of a piecewise-linear function with a discontinuity at x=1. + +

    + The image is the graph of a piecewise-linear function. + For x\lt 1, the graph is a straight line with positive slope, + ending at the point (1,2), where there is a hollow dot, indicating that this point is not part of the graph. + Beginning at the point (1,1) (a solid dot, which is included), + we have another line, with negative slope, for x\geq 1. +

    + +

    + The graph gives an example of a function whose derivative changes sign at x=1. + But since f is not continuous at 1, there is neither a maximum nor a minimum at that point. + The first derivative test can only be applied to continuous functions. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1, + xmax=3, + ymin=-3, + ymax=4, + ] + \addplot+[leftarrow,domain=-.9:1] {x+1}; + \addplot+[firstcurvestyle,rightarrow,domain=1:2.5] {-1*(x-1)+1}; + \addplot[soliddot] coordinates{(1,1)}; + \addplot[hollowdot] coordinates{(1,2)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + + + Using the First Derivative Test + +

    + Find the intervals on which f is increasing and decreasing, + and use the First Derivative Test + to determine the relative extrema of f, where + + f(x) = \frac{x^2+3}{x-1} + . +

    +     + +

    + We start by noting the domain of f: + (-\infty,1)\cup(1,\infty). +

    + +

    + Since f is not defined at x=1 + (it has a vertical asymptote), + the increasing/decreasing nature of f could switch at this value. + We know that \fp(1) will be undefined since f is discontinuous at 1. + We do not formally consider x=1 to be a critical value of f, + but we will use 1 to subdivide the real number line. +

    + +

    + Using the , we find + + \fp(x) = \frac{x^2-2x-3}{(x-1)^2} + . +

    + +

    + We need to find the critical values of f; + we want to know when \fp(x)=0 and when \fp is not defined. + That latter is straightforward: + when the denominator of \fp(x) is 0, \fp is undefined. + That occurs when x=1, + which we've already recognized as an important value, + but not a critical number. +

    + +

    + \fp(x)=0 when the numerator of \fp(x) is 0. + That occurs when x^2-2x-3 = (x-3)(x+1) = 0; + , when x=-1,3. +

    + +

    + We have found that f has two critical numbers, x=-1,3, + and at x=1 something important might also happen. + These three numbers divide the real number line into four subintervals: + + (-\infty,-1), (-1, 1), (1,3), \text{ and } (3,\infty) + . +

    + +

    + Pick a number p from each subinterval and test the sign of \fp at p to determine whether f is increasing or decreasing on that interval. + Again, we do well to avoid complicated computations; + notice that the denominator of \fp is always + positive so we can ignore it during our work. +

    + +

    +

    +
  • + Interval 1: <m>(-\infty,-1)</m> +

    + Choosing a very small number (, a negative number with a large magnitude) p returns + p^2-2p-3 in the numerator of \fp; + that will be positive. + Hence f is increasing on (-\infty,-1). +

    +
    • +
  • + Interval 2: <m>(-1,1)</m> +

    + Choosing 0 seems simple: + \fp(0)=-3\lt 0. + We conclude f is decreasing on (-1,1). +

    +
    • +
  • + Interval 3: <m>(1,3)</m> +

    + Choosing 2 seems simple: + \fp(2) = -3\lt 0. + Again, f is decreasing. +

    +
    • +
  • + Interval 4: <m>(3,\infty)</m> +

    + Choosing an very large number p from this subinterval will give a positive numerator and + (of course) + a positive denominator. + So f is increasing on (3,\infty). +

    +
    • +
    +

    + +

    + In summary, f is increasing on the intervals (-\infty,-1) and + (3,\infty) and is decreasing on the intervals (-1,1) and (1,3). + Since at x=-1, + the sign of \fp switched from positive to negative, + + states that f(-1) is a relative maximum of f. + At x=3, + the sign of \fp switched from negative to positive, + meaning f(3) is a relative minimum. + At x=1, f is not defined, + so there is no relative extremum at x=1. + As previously stated, x=1 is a vertical asymptote of f. +

    + +
    + Number line for f in + + + + Number line for this example, showing critical points of f and intervals of increase and decrease + +

    + On a number line, three points are marked. + The first point is labeled below with -1, and above with rel max. + This indicates that f has a relative maximum at the critical point x=-1. +

    + +

    + The next point is labeled below with 1 and above with VA, + indicating that f has a vertical asymptote at x=1. +

    + +

    + The last point is labeled below with 3 and above with rel min, + indicating that f has a relative minimum at the critical point x=3. +

    + +

    + In between these points there is text indicating the sign of \fp(x), + and whether f is increasing or decreasing, as follows: +

      +
    • +

      + For x\lt -1, f'\gt 0 and f is increasing +

      +
      • + +
    • +

      + For -1\lt x\lt 1, \fp\lt 0 and f is decreasing +

      +
      • + +
    • +

      + For 1\lt x\lt 3, \fp\lt 0 and f is decreasing +

      +
      • + +
    • +

      + For x\gt 3, \fp\gt 0 and f is increasing +

      +
      • +
    +

    +     + + \begin{tikzpicture} + \begin{axis}[ + numberline, + xmin=-3, + xmax=4.9, + extra x ticks={-1, 1, 3}, + extra x tick labels={$-1$,$1$, $3$}, + ] + \addplot[guideline] coordinates {(-1,0) (-1,2)}; + \addplot[guideline] coordinates {(1,0) (1,2)}; + \addplot[guideline] coordinates {(3,0) (3,2)}; + \addplot[mark=none] coordinates {(-2,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; + \addplot[mark=none] coordinates {(0,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }}; + \addplot[mark=none] coordinates {(2,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }}; + \addplot[mark=none] coordinates {(4,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; + \addplot[mark=none] coordinates {(-1,2)} node[above] {\parbox{3em}{\centering rel\\max}}; + \addplot[mark=none] coordinates {(1,2)} node[above] {\parbox{3em}{\centering VA}}; + \addplot[mark=none] coordinates {(3,2)} node[above] {\parbox{3em}{\centering rel\\min}}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + This is summarized in the number line shown in . + Also, shows a graph of f, + confirming our calculations. + This figure also shows \fp, + again demonstrating that f is increasing when \fp\gt 0 and decreasing when \fp\lt 0. +

    + +
    + A graph of f(x) in , showing where f is increasing and decreasing + + + The graphs of a function and its derivative for the current example. + +

    + The graphs of f(x)=x^{8/3}-4x^{2/3} (solid, in blue) and its derivative (dash-dotted, in red) are shown. + The graph of f resembles a large letter W. + It has two relative minima, at x=1 and x=-1, both with horizontal tangents, + and a relative maximum at x=0, where there is a cusp. +

    + +

    + The graph of f'(x) is also shown. The y axis is a vertical asymptote for f'(x), + so the graph appears as two pieces, one on either side of the axis. + We can see that the x intercepts for f'(x) correspond to the two relative minima of f(x). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-4.2, + xmax=4.2, + ymin=-21, + ymax=21, + ] + \addplot[firstcurvestyle,domain=-4:.8,samples=40] {(x^2+3)/(x-1)}; + \addplot[firstcurvestyle,domain=1.25:4] {(x^2+3)/(x-1)} node[pos=0.2,right]{$f(x)$}; + \addplot[secondcurvestyle,domain=-4:.5,samples=40] {(x^2-2*x-3)/((x-1)^2)} node[pos=0.05,above]{$\fp(x)$}; + \addplot[secondcurvestyle,domain=1.5:4] {(x^2-2*x-3)/((x-1)^2)} ; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + +

    + One is often tempted to think that functions always alternate + increasing, decreasing, increasing, + decreasing, around critical values. + Our previous example demonstrated that this is not always the case. + While x=1 was not technically a critical value, + it was an important value we needed to consider. + We found that f was decreasing on + both sides of x=1. +

    + +

    + We examine one more example. +

    + + + Using the First Derivative Test + +

    + Find the intervals on which + f(x) = x^{8/3}-4x^{2/3} is increasing and decreasing and identify the relative extrema. +

    +     + +

    + The domain of f is \mathbb{R} + (you can take the odd root of both positive and negative nubmers). + Next, we take the first derivative. + Since we know we want to solve \fp(x) = 0, + we will do some algebra after taking the derivative. + + f(x) \amp = x^{\frac{8}{3}}-4x^{\frac{2}{3}} + \fp(x) \amp = \frac{8}{3} x^{\frac{5}{3}} - \frac{8}{3}x^{-\frac{1}{3}} + \amp = \frac{8}{3}x^{-\frac{1}{3}}\left(x^{\frac{6}{3}}-1\right) + \amp =\frac{8}{3}x^{-\frac{1}{3}}\left(x^2-1\right) + \amp =\frac{8}{3}x^{-\frac{1}{3}}(x-1)(x+1) + . +

    + +

    + This derivation of \fp shows that + \fp(x) = 0 when x=\pm 1 and \fpis not defined when x=0. + Thus we have three critical values, + breaking the number line into four subintervals as shown in . +

    + +
    + Number line for f in + + + + Number line showing critical points and intervals of increase and decrease + +

    + On a number line, three points are marked. + The first point is labeled below with -1, and above with rel min, + indicating that f has a relative minimum at the critical point x=-1. +

    + +

    + The next point is labeled below with 0 and above with rel max, + indicating that f has a relative maximum at the critical point x=0. +

    + +

    + The last point is labeled below with 1 and above with rel min, + indicating that f has a relative minimum at the critical point x=1. +

    + +

    + In between these points there is text indicating the sign of \fp(x), + and whether f is increasing or decreasing, as follows: +

      +
    • +

      + For x\lt -1, f'\tt 0 and f is decreasing +

      +
      • + +
    • +

      + For -1\lt x\lt 0, \fp\gt 0 and f is increasing +

      +
      • + +
    • +

      + For 0\lt x\lt 1, \fp\lt 0 and f is decreasing +

      +
      • + +
    • +

      + For x\gt 1, \fp\gt 0 and f is increasing +

      +
      • +
    +

    +     + + \begin{tikzpicture} + \begin{axis}[ + numberline, + xmin=-2, + xmax=2, + extra x ticks={-1, 0, 1}, + extra x tick labels={$-1$,$0$, $1$}, + ] + \addplot[guideline] coordinates {(-1,0) (-1,2)}; + \addplot[guideline] coordinates {(0,0) (0,2)}; + \addplot[guideline] coordinates {(1,0) (1,2)}; + \addplot[mark=none] coordinates {(-1.5,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }}; + \addplot[mark=none] coordinates {(-0.5,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; + \addplot[mark=none] coordinates {(0.5,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }}; + \addplot[mark=none] coordinates {(1.5,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; + \addplot[mark=none] coordinates {(-1,2)} node[above] {\parbox{3em}{\centering rel\\min}}; + \addplot[mark=none] coordinates {(0,2)} node[above] {\parbox{3em}{\centering rel\\max}}; + \addplot[mark=none] coordinates {(1,2)} node[above] {\parbox{3em}{\centering rel\\min}}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    +

    +
  • + Interval 1: <m>(\infty,-1)</m> +

    + We choose p=-2; we can easily verify that \fp(-2)\lt 0. + So f is decreasing on (-\infty,-1). +

    +
    • +
  • + Interval 2: <m>(-1,0)</m> +

    + Choose p=-1/2. + Once more we practice finding the sign of \fp(p) without computing an actual value. + We have \fp(p) = (8/3)p^{-1/3}(p-1)(p+1); + find the sign of each of the three terms at the chosen value of p. + + \fp(p) = \frac{8}{3} \cdot \underbrace{p^{-\frac{1}{3}}}_{\lt 0}\cdot \underbrace{(p-1)}_{\lt 0}\underbrace{(p+1)}_{\gt 0} + . +

    + +

    + We have a negative negative positive + giving a positive number; + f is increasing on (-1,0). +

    +
    • +
  • + Interval 3: <m>(0,1)</m> +

    + We do a similar sign analysis as before, + using p in (0,1). + + \fp(p) = \frac{8}{3} \cdot \underbrace{p^{-\frac{1}{3}}}_{\gt 0}\cdot \underbrace{(p-1)}_{\lt 0}\underbrace{(p+1)}_{\gt 0} + . +

    + +

    + We have two positive factors and one negative factor; + \fp(p)\lt 0 and so f is decreasing on (0,1). +

    +
    • +
  • + Interval 4: <m>(1,\infty)</m> +

    + Similar work to that done for the other three intervals shows that + \fp(x)\gt 0 on (1,\infty), + so f is increasing on this interval. +

    +
    • +
    +

    + +

    + We conclude by stating that f is increasing on the intervals (-1,0) and + (1,\infty) and decreasing on the intervals (-\infty,-1) and (0,1). + The sign of \fpchanges from negative to positive around x=-1 and x=1, + meaning by + that f(-1) and f(1) are relative minima of f. + As the sign of \fp changes from positive to negative at x=0, + we have a relative maximum at f(0). + + shows a graph of f, confirming our result. + We also graph \fp, + highlighting once more that f is increasing when \fp\gt 0 and is decreasing when \fp\lt 0. +

    + +
    + A graph of f(x) in , showing where f is increasing and decreasing + + + + + \begin{tikzpicture} + \begin{axis}[ + xmin=-3.2, + xmax=3.2, + ymin=-5, + ymax=11, + ] + \addplot+[rightarrow,domain=0.0001:3,samples=50] {x^(8/3)-4*x^(2/3)}; + \addplot[firstcurvestyle,leftarrow,domain=-3:-0.001,samples=50] {(-x)^(8/3)-4*(-x)^(2/3)} node[pos=0.1, above right]{$f(x)$}; + \addplot[secondcurvestyle,domain=0.2:2,samples=50] {(8*(x^2 - 1))/(3 *x^(1/3))} node[pos=0.7,left] {$\fp(x)$}; + \addplot[secondcurvestyle,domain=-2:-0.2,samples=50] {(8*((-x)^2 - 1))/(3 *(-x)^(1/3))}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + +

    + We have seen how the first derivative of a function helps determine when the graph of a function is going up + or down. In the next section, + we will see how the second derivative helps determine how the graph of a function curves. +

    + + + + Terms and Concepts + + + + +

    + In your own words describe what it means for a function to be increasing. +

    + +     + + + +     + + + + +

    + What does a decreasing function look like? +

    + +     + + + +

    + Answers will vary, but should indicate that + the graph slopes downward from left to right. +

    +     + +     + + + +

    + Sketch a graph of a function on [0,2] that is increasing, + where it is increasing quickly + near x=0 and increasing slowly near x=2. +

    +     + + + +

    + Answers will vary; + graphs should be steeper near x=0 than near x=2. +

    +     +     + + + + +

    + Give an example of a function describing a situation where it is bad + to be increasing and good to be decreasing. +

    + +     + + + +     + + + + +

    + + Functions always switch from increasing to decreasing, + or decreasing to increasing, at critical points. +

    +     + +

    + False; for instance, + y=x^3 is always increasing though it has a critical point at x=0. +

    +     + +     + + + + +

    + A function f has derivative \fp(x) = (\sin(x) +2)e^{x^2+1}, + where \fp(x) \gt 1 for all x. + Is f increasing, decreasing, + or can we not tell from the given information? Why or why not? +

    + +     + + + +

    + The function is increasing. + Since \fp(x) \gt 1, we know in + particular that \fp(x) \gt 0, + so f is increasing by + . +

    +     + +     +     + + + Problems + + + +

    + A function f(x) is given. + Graph f and \fp on the same axes + (using technology is permitted) + and verify . +

    +     + + + +

    + f(x) = 2x+3 +

    +     +     + + + +

    + f(x) = x^2-3x+5 +

    +     +     + + + +

    + f(x) = \cos(x) +

    +     +     + + + +

    + f(x) = \tan(x) +

    +     +     + + + +

    + f(x) = x^3-5x^2+7x-1 +

    +     +     + + + +

    + f(x) = 2x^3-x^2+x-1 +

    +     +     + + + +

    + f(x) = x^4-5x^2+4 +

    +     +     + + + +

    + f(x) = \frac{1}{x^2+1} +

    +     +     +     + + + +

    + A function f(x) is given. +

      +
    1. +

      + Give the domain of f. +

      +
      2. +
    2. +

      + Find the critical numbers of f. +

      +
      3. +
    3. +

      + Find the intervals on which f is increasing. +

      +
      4. +
    4. +

      + Find the intervals on which f is decreasing. +

      +
      5. +
    5. +

      + Use the First Derivative Test to determine which critical numbers are a relative maximum. +

      +
      6. +
    6. +

      + Use the First Derivative Test to determine which critical numbers are a relative minimum. +

      +
      7. +
    +

    +     + + + + + Context("Fraction"); + ($r,$s) = random_subset(2,-9..9); + $m = ($r+$s)/2; + $crit=List("$m"); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $f = Formula("x^2-($r+$s)x+$r*$s")->reduce; + $domain = Compute("(-inf,inf)"); + $inc = List(Interval("[$m,inf)")); + $dec = List(Interval("(-inf,$m]")); + $max = List("none"); + $min = List("$m"); + + +

    + f(x)= +

    + + Give the domain of f using interval notation. + Use U for union if the domain consists of more than one interval. + +

    + +

    + + Enter the critical numbers of f, + separating with commas if needed. + If there are no critical numbers, + enter none. + +

    + +

    + + List the maximal intervals where f is increasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is decreasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the critical numbers at which there is a relative maximum. + If there are none, enter none. + +

    + +

    + + List the critical numbers at which there is a relative minimum. + If there are none, enter none. + +

    + +

    +     + + +     +     + + + + + Context("Fraction"); + ($a,$b) = random_subset(2,-9..-1,1..9); + $m = Fraction(-2*$a,3); + ($m,$n) = num_sort(0,$m); + $crit=List($m,$n); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $f = Formula("x^3 + $a x^2 + $b")->reduce; + $domain = Compute("(-inf,inf)"); + $inc = List(Interval("(-inf,$m]"),Interval("[$n,inf)")); + $dec = List(Interval("[$m,$n]")); + $max = List("$m"); + $min = List("$n"); + + +

    + f(x)= +

    + + Give the domain of f using interval notation. + Use U for union if the domain consists of more than one interval. + +

    + +

    + + Enter the critical numbers of f, + separating with commas if needed. + If there are no critical numbers, + enter none. + +

    + +

    + + List the maximal intervals where f is increasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is decreasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the critical numbers at which there is a relative maximum. + If there are none, enter none. + +

    + +

    + + List the critical numbers at which there is a relative minimum. + If there are none, enter none. + +

    + +

    +     + + +     +     + + + + + Context("Fraction"); + #roots of derivative + do { + $s = Fraction(list_random(1,2,4,5,7,8)*list_random(-1,1), list_random(3,6,9)); # make b div by 3 + ($a,$b) = $s->value; + if ($a % 2 == 0) {$t = Fraction(non_zero_random(-8,8,2),random(1,9,2));} + elsif ($b % 2 == 0) {$t = Fraction(random(-9,9,2),random(2,8,2));} + else {$t = Fraction(random(-9,9,2),random(1,9,2));} + } until ($s != $t); + ($c,$d) = $t->value; + # f'(x) = (bx - a)(dx - c) = bd x^2 - (bc+ad) x + ac + ($m,$n) = num_sort($s,$t); + $crit=List($m,$n); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $A = $b*$d/3; + $B = -($b*$c+$a*$d)/2; + $C = $a*$c; + $D = non_zero_random(-9,9,1); + $f = Formula("$A x^3 + $B x^2 + $C x + $D")->reduce; + $domain = Compute("(-inf,inf)"); + $inc = $inc = List(Interval("(-inf,$m]"),Interval("[$n,inf)")); + $dec = List(Interval("[$m,$n]")); + $max = List($m); + $min = List($n); + + +

    + f(x)= +

    + + Give the domain of f using interval notation. + Use U for union if the domain consists of more than one interval. + +

    + +

    + + Enter the critical numbers of f, + separating with commas if needed. + If there are no critical numbers, + enter none. + +

    + +

    + + List the maximal intervals where f is increasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is decreasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the critical numbers at which there is a relative maximum. + If there are none, enter none. + +

    + +

    + + List the critical numbers at which there is a relative minimum. + If there are none, enter none. + +

    + +

    +     + + +     +     + + + + + $r = non_zero_random(-5,5,1); + $crit=List($r); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $A = 1; + $B = -3*$r; + $C = 3*($r)**2; + $D = -($r)**3; + $f = Formula("$A x^3 + $B x^2 + $C x + $D")->reduce; + $domain = Compute("(-inf,inf)"); + $inc = List(Interval("(-inf,inf)")); + $dec = List(Compute("None")); + $max = List(Compute("None")); + $min = List(Compute("None")); + + +

    + f(x)= +

    + + Give the domain of f using interval notation. + Use U for union if the domain consists of more than one interval. + +

    + +

    + + Enter the critical numbers of f, + separating with commas if needed. + If there are no critical numbers, + enter none. + +

    + +

    + + List the maximal intervals where f is increasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is decreasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the critical numbers at which there is a relative maximum. + If there are none, enter none. + +

    + +

    + + List the critical numbers at which there is a relative minimum. + If there are none, enter none. + +

    + +

    +     + + +     +     + + + + + $r = non_zero_random(-5,5,1); + $crit=List($r); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $A = 1; + $B = -2*$r; + $C = ($r)**2+random(1,9,1); + $f = Formula("1/($A x^2 + $B x + $C)")->reduce; + $domain = Compute("(-inf,inf)"); + $inc = List(Interval("(-inf,$r]")); + $dec = List(Interval("[$r,inf)")); + $max = List(Compute($r)); + $min = List(Compute("None")); + + +

    + f(x)= +

    + + Give the domain of f using interval notation. + Use U for union if the domain consists of more than one interval. + +

    + +

    + + Enter the critical numbers of f, + separating with commas if needed. + If there are no critical numbers, + enter none. + +

    + +

    + + List the maximal intervals where f is increasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is decreasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the critical numbers at which there is a relative maximum. + If there are none, enter none. + +

    + +

    + + List the critical numbers at which there is a relative minimum. + If there are none, enter none. + +

    + +

    +     + + +     +     + + + + + ($r,$t) = random_subset(2,1..9); + $R = ($r)**2; + $T = ($t)**2; + $crit=List(0); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $f = Formula("(x^2-$R)/(x^2-$T)")->reduce; + $domain = Compute("(-inf,-$t)U(-$t,$t)U($t,inf)"); + $inc = List(Interval("[0,$t)"),Interval("($t,inf)")); + $dec = List(Interval("(-inf,-$t)"),Interval("(-$t,0]")); + $max = List(Compute("None")); + $min = List(0); + if ($r < $t) { + ($inc,$dec) = ($dec,$inc); + ($max,$min) = ($min,$max); + } + + +

    + f(x)= +

    + + Give the domain of f using interval notation. + Use U for union if the domain consists of more than one interval. + +

    + +

    + + Enter the critical numbers of f, + separating with commas if needed. + If there are no critical numbers, + enter none. + +

    + +

    + + List the maximal intervals where f is increasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is decreasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the critical numbers at which there is a relative maximum. + If there are none, enter none. + +

    + +

    + + List the critical numbers at which there is a relative minimum. + If there are none, enter none. + +

    + +

    +     + + +     +     + + + + + ($r,$s) = num_sort(random_subset(2,-9..-1,1..9)); + $rs = $r*$s; + $crit = ($rs > 0) ? List(Compute(-sqrt($rs)),-Compute(-sqrt($rs))) : List(Compute("None")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $f = Formula("x/(x^2-($r+$s)x+$rs)")->reduce; + $domain = Compute("(-inf,$r)U($r,$s)U($s,inf)"); + if ($r > 0) { + $inc = List(Interval("[-sqrt($rs),$r)"),Interval("($r,sqrt($rs)]")); + $dec = List(Interval("(-inf,-sqrt($rs)]"),Interval("[sqrt($rs),$s)"),Interval("($s,inf)")); + $max = List(Compute("sqrt($rs)")); + $min = List(Compute("-sqrt($rs)")); + } + elsif ($s > 0) { + $inc = List(Compute("none")); + $dec = List(Interval("(-inf,$r)"),Interval("($r,$s)"),Interval("($s,inf)")); + $max = List(Compute("none")); + $min = List(Compute("none")); + } + else { + $inc = List(Interval("[-sqrt($rs),$s)"),Interval("($s,sqrt($rs)]")); + $dec = List(Interval("(-inf,$r)"),Interval("($r,-sqrt($rs)]"),Interval("[sqrt($rs),inf)")); + $max = List(Compute("sqrt($rs)")); + $min = List(Compute("-sqrt($rs)")); + } + + +

    + f(x)= +

    + + Give the domain of f using interval notation. + Use U for union if the domain consists of more than one interval. + +

    + +

    + + Enter the critical numbers of f, + separating with commas if needed. + If there are no critical numbers, + enter none. + +

    + +

    + + List the maximal intervals where f is increasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is decreasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the critical numbers at which there is a relative maximum. + If there are none, enter none. + +

    + +

    + + List the critical numbers at which there is a relative minimum. + If there are none, enter none. + +

    + +

    +     + + +     +     + + + + + $r = non_zero_random(-9,9,1); + $crit = List($r,3*$r); + Context("Interval")->flags->set(reduceConstants=>0); + Context()->flags->set(ignoreEndpointTypes => 1); + $f = Formula("(x-$r)^(2/3)/x"); + $domain = Compute("(-inf,0)U(0,inf)"); + if ($r > 0) { + $inc = List(Interval("[$r,3*$r]")); + $dec = List(Interval("(-inf,0)"),Interval("(0,$r]"),Interval("[3*$r,inf)")); + $max = List(3*$r); + $min = List($r); + } else { + $inc = List(Interval("[3*$r,$r]")); + $dec = List(Interval("(-inf,3*$r]"),Interval("[$r,0)"),Interval("(0,inf)")); + $max = List($r); + $min = List(3*$r); + } + + +

    + f(x)= +

    + + Give the domain of f using interval notation. + Use U for union if the domain consists of more than one interval. + +

    + +

    + + Enter the critical numbers of f, + separating with commas if needed. + If there are no critical numbers, + enter none. + +

    + +

    + + List the maximal intervals where f is increasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is decreasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the critical numbers at which there is a relative maximum. + If there are none, enter none. + +

    + +

    + + List the critical numbers at which there is a relative minimum. + If there are none, enter none. + +

    + +

    +     + + +     +     + + + + + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $domain = Compute("(-pi,pi)"); + $crit = List("-3pi/4,-pi/4,pi/4,3pi/4"); + $inc = List(Interval("(-pi,-3pi/4)"),Interval("(-pi/4,pi/4)"),Interval("(3pi/4,pi)")); + $dec = List(Interval("(-3pi/4,-pi/4)"),Interval("(pi/4,3pi/4)")); + $max = List("-3pi/4,pi/4"); + $min = List("-pi/4,3pi/4"); + + +

    + f(x)=\sin(x)\cos(x) on (-\pi,\pi) +

    + + Give the domain of f using interval notation. + Use U for union if the domain consists of more than one interval. + +

    + +

    + + Enter the critical numbers of f, + separating with commas if needed. + If there are no critical numbers, + enter none. + +

    + +

    + + List the maximal intervals where f is increasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is decreasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the critical numbers at which there is a relative maximum. + If there are none, enter none. + +

    + +

    + + List the critical numbers at which there is a relative minimum. + If there are none, enter none. + +

    + +

    +     + + +     +     + + + + + $n = random(3,9,1); + if ($n == 3) { + $b = non_zero_random(-9,9,1); + } + elsif ($n == 4) { + $b = non_zero_random(-5,5,1); + } + elsif ($n == 5) { + $b = non_zero_random(-3,3,1); + } + elsif ($n >= 6 and $n <= 8) { + $b = non_zero_random(-2,2,1); + } + else { + $b = non_zero_random(-1,1,1); + } + $a = ($b)**($n-1); + Context("Interval")->flags->set(reduceConstants=>0); + Context()->flags->set(ignoreEndpointTypes => 1); + $domain = Compute("(-inf,inf)"); + $f = Formula("x^$n - $a*$n x")->reduce; + if ($n % 2 == 0) { + $crit = List($b); + $inc = List(Interval("[$b,inf)")); + $dec = List(Interval("(-inf,$b]")); + $max = List(Compute("none")); + $min = List($b); + } else { + $crit = List($b,-$b); + $inc = List(Interval("(-inf,-abs($b)]"),Interval("[abs($b),inf)")); + $dec = List(Interval("[-abs($b),abs($b)]")); + $max = List(-abs($b)); + $min = List(abs($b)); + } + + +

    + f(x)= +

    + + Give the domain of f using interval notation. + Use U for union if the domain consists of more than one interval. + +

    + +

    + + Enter the critical numbers of f, + separating with commas if needed. + If there are no critical numbers, + enter none. + +

    + +

    + + List the maximal intervals where f is increasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is decreasing, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the critical numbers at which there is a relative maximum. + If there are none, enter none. + +

    + +

    + + List the critical numbers at which there is a relative minimum. + If there are none, enter none. + +

    + +

    +     + + +     +     +     +     +     +
    +
    + Concavity and the Second Derivative + + + +

    + Our study of nice functions continues. + The previous section showed how the first derivative of a function, + \fp, can relay important information about f. + We now apply the same technique to \fp itself, + and learn what this tells us about f. +

    + +

    + The key to studying \fp is to consider its derivative, + namely \fp', which is the second derivative of f. + When \fp'\gt 0, \fp is increasing. + When \fp'\lt 0, \fp is decreasing. + \fp has relative maxima and minima where \fp'=0 or is undefined. +

    + +

    + This section explores how knowing information about \fpp gives information about f. +

    +     + + + Concavity +

    + We begin with a definition, then explore its meaning. +

    + + + Concave Up and Concave Down + +

    + Let f be continuous on an interval I. + The graph of f is concave up + on I if for any a\lt b in I, + + f\left(\frac{a+b}{2}\right) \lt \frac{f(a)+f(b)}{2} + . + The graph of f is concave down + on I if for any a\lt b in I, + + f\left(\frac{a+b}{2}\right) \gt \frac{f(a)+f(b)}{2} + . + concavity + concave up + concave down +

    +     +     + + + + + +

    + Geometrically, the condition in Equation + states that a graph is concave up if the midpoint of the secant line from (a,f(a)) + to (b,f(b)) (and hence, the secant line itself) is above the graph y=f(x). + Similarly, Equation states that the secant line lies below the graph. +

    +

    + In order for equality to hold instead of Equation or Equation, + the function would have to be of the form f(x)=mx+c, + in which case the graph is a straight line. + Straight lines are considered to have no concavity. +

    + +
    + Illustrating the nature of concave up and concave down + +
    + A graph that is concave up. Notice how the secant line lies above the graph. + + + A graph of a function with concave up curvature. + + +

    + The image shows a graph where the curve is shaped upwards, resembling a bowl. + There is a secant line from x=-1 to x=2 that lies above the curve. + The points on the curve are all below this secant line, implying that the slope of the function is increasing as you move from left to right. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1.5,xmax=2.5, + ymin=-0.5,ymax=7, + ] + \addplot+[infinite,domain=-1.4:2.4] {x^2}; + \addplot+[secantline,-,domain=-1:2] {x+2}; + \addplot[soliddot] coordinates {(-1,1) (2,4)}; + \end{axis} + \end{tikzpicture} + + +
    +
    + A graph that is concave down. Notice how the secant line lies below the graph. + + + A graph representing a function with a concave down shape. + + +

    + This image illustrates a graph with a curve that opens downwards, similar to an inverted bowl, which characterizes a concave down function. + There's a secant line from x=-1 to x=2 that lies below the curve. + The points on the curve are all above this secant line, implying that the function's slope is decreasing as you move from left to right. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1.5,xmax=2.5, + ymin=-0.5,ymax=7, + ] + \addplot+[infinite,domain=-1.4:2.3] {6-x^2}; + \addplot+[secantline,-,domain=-1:2] {-x+4}; + \addplot[soliddot] coordinates {(-1,5) (2,2)}; + \end{axis} + \end{tikzpicture} + + +
    +     +
    + + + +

    + Consider a function f such that f is continuous on [a,b] and differentiable on (a,b). + Note that \frac{a+b}{2} is the midpoint of the interval [a,b]. + By the , + there must be a point c_1 in \left[a,\frac{a+b}{2}\right] such that + + \fp(c_1) = \frac{f\left(\frac{a+b}{2}\right)-f(a)}{\frac{a+b}{2}-a}=\frac{2}{b-a}\left(f\left(\frac{a+b}{2}\right)-f(a)\right) + . + Similarly, there must be a point c_2 in \left[\frac{a+b}{2},b\right] such that + + \fp(c_2) = \frac{f(b)-f\left(\frac{a+b}{2}\right)}{b-\frac{a+b}{2}} = \frac{2}{b-a}\left(f(b)-f\left(\frac{a+b}{2}\right)\right) + . + But then we have + + \fp(c_2)-\fp(c_1) \amp = \frac{2}{b-a}\left(f(b)-f\left(\frac{a+b}{2}\right)-f\left(\frac{a+b}{2}\right)+f(a)\right) + \amp = \frac{4}{b-a}\left(\frac{f(a)+f(b)}{2}-f\left(\frac{a+b}{2}\right)\right) + . +

    + +

    + Now, let us suppose that \fp(x) is an increasing function on (a,b). + In that case, \fp(c_2)-\fp(c_1)\gt 0, and since b-a\gt 0, + this implies that + + \frac{f(a)+f(b)}{2}-f\left(\frac{a+b}{2}\right)\gt 0 + , + which, by means that the graph of f is concave up. +

    + +

    + Similarly, if \fp(x) is a decreasing function on (a,b), + then the graph of f will be concave down. + Using , we arrive at the following theorem. +

    + + + +

    + Let f be a continuous function on [a,b] and differentiable on (a,b). +

    + +

    +

      +
    1. +

      + If \fpp(c) \gt 0 for all c in (a,b), + then f is concave up on [a,b]. +

      +
      2. +
    2. +

      + If \fpp(c) \lt 0 for all c in (a,b), + then f is concave down on [a,b]. +

      +
      3. +
    3. +

      + If \fpp(c) =0 for all c in (a,b), + then f is linear on [a,b]. +

      +
      4. +
    +

    +     +     + + + +

    + The graph of a function f is concave up + when \fp is increasing. + That means as one looks at a concave up graph from left to right, + the slopes of the tangent lines will be increasing. + Consider , + where a concave up graph is shown along with some tangent lines. + Notice how the tangent line on the left is steep, + downward, corresponding to a lesser + (large negative) + value of \fp. + On the right, the tangent line is steep, + upward, corresponding to a greater + (large positive) + value of \fp. +

    + +
    + A function f with a concave up graph. Notice how the slopes of the tangent lines, when looking from left to right, are increasing. (The slope values pictured are -12, -6, 6 and 12). + + + + A graph of a function with concave up curvature and tangent lines. + + +

    + The image displays a graph of a concave up function f with tangent lines drawn at various points along the curve. + The function is shown as a continuous curve that opens upwards, like a parabola. + Tangent lines at points with x-values of (-2, -1, 1, 2) are depicted, illustrating how the slope of these lines increases as we move from left to right. + On the left side of the graph, the tangent lines slope downwards, indicating negative slopes, while on the right side, the tangent lines slope upwards, indicating positive slopes. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-3.5, + xmax=3.5, + ymin=-1, + ymax=35, + ] + \addplot+[infinite,domain=-3:3] {3*x^2+5}; + \addplot [tangentlineseg,domain =-3:-1] {-12*(x+2)+17}; + \addplot [tangentlineseg,domain =1:3] {12*(x-2)+17} ; + \addplot [tangentlineseg,domain =-2.5:.5] {-6*(x+1)+8}; + \addplot [tangentlineseg,domain =-.5:2.5] {6*(x-1)+8} ; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + If a function is decreasing and concave up, + then its rate of decrease is slowing; + it is leveling off. You can see this in the left side of . + If the function is increasing and concave up, + then the rate of increase is increasing. + The function is increasing at a faster and faster rate. + You can see this in the right side of . +

    + +

    + Now consider a function which is concave down. + We essentially repeat the above paragraphs with slight variation. +

    + +

    + The graph of a function f is concave down + when \fp is decreasing. + That means as one looks at a concave down graph from left to right, + the slopes of the tangent lines will be decreasing. + Consider , + where a concave down graph is shown along with some tangent lines. + Notice how the tangent line on the left is steep, + upward, corresponding to a greater + (large positive) + value of \fp. + On the right, the tangent line is steep, + downward, corresponding to a lesser + (large negative) + value of \fp. +

    + +
    + A function f with a concave down graph. Notice how the slopes of the tangent lines, when looking from left to right, are decreasing. + + + + A graph of a function with concave down curvature and tangent lines. + + +

    + The image shows a graph of a concave down function f with tangent lines drawn at various points along the curve. + The function forms a continuous curve that opens downwards. + Tangent lines at points with x-values of (-2, 1, 1, 2) are shown, demonstrating how the slope of these lines decreases as we move from left to right across the graph. + On the left side of the graph, the tangent lines slope upwards, indicating positive slopes, while on the right side, the tangent lines slope downwards, indicating negative slopes. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-3.5, + xmax=3.5, + ymin=-1, + ymax=35, + ] + \addplot+[infinite,domain=-3:3] {-3*x^2+31}; + \addplot [tangentlineseg, domain =-3:-1] {12*(x+2)+19}; + \addplot [tangentlineseg, domain =1:3] {-12*(x-2)+19} ; + \addplot [tangentlineseg, domain =-2.5:-.2] {6*(x+1)+28}; + \addplot [tangentlineseg, domain =.2:2.5] {-6*(x-1)+28} ; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + If a function is increasing and concave down, + then its rate of increase is slowing; + it is leveling off. If the function is decreasing and concave down, + then the rate of decrease is decreasing. + The function is decreasing at a faster and faster rate. +

    + +

    + Our definition of concave up and concave down is given in terms of when the first derivative is increasing or decreasing. + We can apply the results of the previous section to find intervals on which a graph is concave up or down. + That is, we recognize that \fp is increasing when \fpp \gt 0, etc. +

    + + + Test for Concavity + +

    + Let f be twice differentiable on an interval I. + The graph of f is concave up if \fpp \gt 0 on I, + and is concave down if \fpp\lt 0 on I. + concavitytest for + concavityinflection point +

    +     +     + + +
    + Demonstrating the four ways that concavity interacts with increasing/decreasing, along with the relationships with the first and second derivatives + + +
    + \fp\gt 0, f increasing; \fpp\lt0, f is concave down + + + A graph illustrating a function that is increasing and concave down. + + +

    + Diagram a shows a portion of a curve where the function f is increasing as the first derivative \fp\gt 0 suggests. + Despite the increase, the curve is concave down, indicated by the second derivative \fpp\lt0, meaning the rate of increase is slowing down. +

    +     + + \begin{tikzpicture} + \draw [thick] (0,0) arc (180:90:2); + \end{tikzpicture} + + +
    + +
    + \fp\lt0, f decreasing; \fpp\lt0, f is concave down + + + A graph illustrating a function that is decreasing and concave down. + + +

    + Diagram b depicts a curve that is decreasing, with the first derivative \fp\lt0, and is also concave down, as shown by the second derivative \fpp\lt0. + This represents a function f that is decreasing at an increasing rate. +

    +     + + \begin{tikzpicture} + \draw [thick] (0,0) arc (90:0:2); + \end{tikzpicture} + + +
    +     + + +
    + \fp\lt0, f decreasing; \fpp\gt 0, f is concave up + + + A graph illustrating a function that is decreasing and concave up. + + +

    + In diagram c, the curve represents a function f that is decreasing since \fp\lt0, but the concavity is up, which is indicated by the second derivative \fpp\gt 0. + This means the function is decreasing but the rate of decrease is slowing. +

    +     + + \begin{tikzpicture} + \draw [thick] (0,0) arc (180:270:2); + \end{tikzpicture} + + +
    + +
    + \fp\gt 0, f increasing; \fpp\gt 0, f is concave up + + + A graph illustrating a function that is increasing and concave up. + + +

    + Diagram d displays a curve that is increasing as the first derivative \fp\gt 0 indicates, and it is concave up as the second derivative \fpp\gt 0 confirms. + This represents a function f that is increasing at an increasing rate. +

    +     + + \begin{tikzpicture} + \draw [thick] (0,0) arc (0:-90:2); + \end{tikzpicture} + + +
    +     +     +
    + +

    + If knowing where a graph is concave up/down is important, + it makes sense that the places where the graph changes from one to the other is also important. + This leads us to a definition. +

    + + + Point of Inflection + +

    + A point of inflection + is a point on the graph of f at which the concavity of f changes. + point of inflection + inflection point +

    +     +     + +

    + + shows a graph of a function with inflection points labeled. +

    + +
    + A graph of a function with its inflection points marked. The intervals where concave up/down are also indicated. + + + + A graph depicting a function with labeled inflection points and concavity indications. + + +

    + The image shows the graph of a function featuring distinct inflection points at x=2 and x=3. + The section of the function bounded by x=0 and x=2 is concave up with \fpp\gt 0, while the section between x=2 and x=3 is concave down with \fpp\lt 0. + The section bouned by x=3 and x=4 is concave up with \fpp\gt 0. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.5, + xmax=4.5, + ymin=-1, + ymax=16, + ] + \addplot+[infinite,domain=.7:4.2, samples=40] {2*(x-1)*(x-2)*(x-3)*(x-4)+5}; + \addplot[soliddot] coordinates {(1.85,4.39) (3.15, 4.39)}; + \addplot[guideline] coordinates {(1.85,-1) (1.85,16)} node[left, anchor=north east] {\parbox{4em}{\flushright$\fpp>0$, $f$ is concave up}}; + \addplot[guideline] coordinates {(3.15,-1) (3.15,16)} node[right, anchor=north west] {\parbox{4em}{\flushleft$\fpp>0$, $f$ is concave up}}; + \draw (axis cs:2.5,16) node[anchor=north] {\parbox{4em}{\centering$\fpp<0$, $f$ is concave down}}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + If the concavity of f changes at a point (c,f(c)), + then \fp is changing from increasing to decreasing (or, + decreasing to increasing) at x=c. + That means that the sign of \fppis changing from positive to negative (or, + negative to positive) at x=c. + A sign change may occur when \fpp=0 or \fpp is undefined. + This leads to the following theorem. +

    + + + Points of Inflection + +

    + If (c,f(c)) is a point of inflection on the graph of f, + then either \fpp(c)=0 or \fpp is not defined at c. +

    +     +     + +

    + We have identified the concepts of concavity and points of inflection. + It is now time to practice using these concepts; + given a function, + we should be able to find its points of inflection and identify intervals on which it is concave up or down. + We do so in the following examples. +

    + + + Finding intervals of concave up/down, inflection points + +

    + Let f(x)=x^3-3x+1. + Find the inflection points of f and the intervals on which it is concave up/down. +

    +     + +

    + We start by finding \fp(x)=3x^2-3 and \fpp(x)=6x. + To find the inflection points, + we use + and find where \fpp(x)=0 or where \fpp is undefined. + We find \fpp is always defined, + and is 0 only when x=0. + So the point (0,f(0))=(0,1) is the only possible point of inflection. +

    + +

    + This possible inflection point divides the real line into two intervals, + (-\infty,0) and (0,\infty). + We use a process similar to the one used in the previous section to determine increasing/decreasing. + Pick any c\lt 0; + \fpp(c)\lt 0 so f is concave down on (-\infty,0). + Pick any c\gt 0; + \fpp(c)\gt 0 so f is concave up on (0,\infty). + Since the concavity changes at x=0, + the point (0,1) is an inflection point. +

    + +

    + The number line in + illustrates the process of determining concavity; + + shows a graph of f and \fpp, confirming our results. + Notice how f is concave down precisely when + \fpp(x)\lt 0 and concave up when \fpp(x)\gt 0. +

    + +
    + A number line determining the concavity of f in + + + + + A number line indicating the concavity of a function based on the second derivative. + + +

    + The image illustrates a number line used to determine the concavity of a function f. + The number line is divided at zero, with the left side showing \fpp\lt 0, indicating that the function is concave down in this region. + Conversely, the right side of the number line where \fpp\gt 0 indicates that the function is concave up. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + numberline, + xmin=-0.9, + xmax=0.9, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + ] + \addplot[guideline] coordinates {(0,0) (0,2)}; + \addplot[mark=none] coordinates {(-0.5,1)} node {\parbox{8em}{\centering \small $\fpp<0$\\$f$ is concave down}}; + \addplot[mark=none] coordinates {(0.5,1)} node {\parbox{8em}{\centering \small $\fpp>0$\\$f$ is concave up}}; + \end{axis} + \end{tikzpicture} + + + +
    + +
    + A graph of f(x) used in + + + + A graph showing the a function and its second derivative. + + +

    + The graph shows the function f(x) and its second derivative \fpp. + The function f(x) exhibits both concave up and concave down behavior within different intervals. + The second derivative \fpp crosses the x-axis at points where the concavity of f(x) changes, indicating the locations of possible inflection points. + Above the x-axis, where \fpp is positive, the function f(x) is concave up. Below the x-axis, where \fpp is negative, the function f(x) is concave down. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-2.2, + xmax=2.2, + ymin=-3.5, + ymax=3.5, + ] + \addplot+[infinite,domain=-2.1:2,samples=60] {x^3-x*3+1} node[below right, pos=0.2] {$f(x)$} ; + \addplot+[infinite,domain=-.5:.5] {6*x} node[below right]{$\fpp(x)$}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + + + + + Finding intervals of concave up/down, inflection points + +

    + Let f(x)=x/(x^2-1). + Find the inflection points of f and the intervals on which it is concave up/down. +

    +     + +

    + We need to find \fpand \fpp. + Using the and simplifying, we find + + \fp(x)\amp=\frac{-(1+x^2)}{(x^2-1)^2}\amp\fpp(x)\amp= \frac{2x(x^2+3)}{(x^2-1)^3} + . +

    + +

    + To find the possible points of inflection, + we seek to find where \fpp(x)=0 and where \fpp is not defined. + Solving \fpp(x)=0 reduces to solving 2x(x^2+3)=0; + we find x=0. + We find that \fpp is not defined when x=\pm 1, + for then the denominator of \fpp is 0. + We also note that f itself is not defined at x=\pm1, + having a domain of (-\infty,-1)\cup(-1,1)\cup(1,\infty). + Since the domain of f is the union of three intervals, + it makes sense that the concavity of f could switch across intervals. + We technically cannot say that f has a point of inflection at x=\pm1 as they are not part of the domain, + but we must still consider these x-values to be important and will include them in our number line. +

    + +

    + The important x-values at which concavity might switch are x=-1, + x=0 and x=1, + which split the number line into four intervals as shown in . + We determine the concavity on each. + Keep in mind that all we are concerned with is the sign + of \fpp on the interval. +

    + +

    +

    +
  • + Interval 1: <m>(-\infty,-1)</m> +

    + Select a number c in this interval with a large magnitude + (for instance, c=-100). + The denominator of \fp'(x) will be positive. + In the numerator, + the \left(c^2+3\right) factor will be positive and the 2c factor will be negative. + Thus the numerator is negative and \fpp(c) is negative. + We conclude f is concave down on (-\infty,-1). +

    +
    • + +
  • + Interval 2: <m>(-1,0)</m> +

    + For any number c in this interval, + the factor 2c in the numerator will be negative, + the factor \left(c^2+3\right) in the numerator will be positive, + and the factor \left(c^2-1\right)^3 in the denominator will be negative. + Thus \fpp(c)\gt 0 and f is concave up on this interval. +

    +
    • + +
  • + Interval 3: <m>(0,1)</m> +

    + Any number c in this interval will be positive and small. + Thus the numerator is positive while the denominator is negative. + Thus \fpp(c)\lt 0 and f is concave down on this interval. +

    +
    • + +
  • + Interval 4: <m>(1,\infty)</m> +

    + Choose a large value for c. + It is evident that \fpp(c)\gt 0, + so we conclude that f is concave up on (1,\infty). +

    +
    • +
    +

    + +
    + Number line for f in + + + + + A number line indicating intervals of concavity for a function. + + +

    + The image shows a number line that is used to determine the concavity of the function f. + The line is divided into four intervals by the points -1, 0 and 1, + each of which is an inflection point. +

    + +

    + The intervals are marked as follows: +

      +
    • +

      + For x\lt -1, \fpp\lt 0, and f is concave down. +

      +
      • + +
    • +

      + For -1\lt x\lt 0, \fpp\gt 0, and f is concave up. +

      +
      • + +
    • +

      + For 0\lt x\lt 1, \fpp\lt 0, and f is concave down. +

      +
      • + +
    • +

      + For x\gt 1, \fpp\gt 0, and f is concave up. +

      +
      • +
    +

    +     + + \begin{tikzpicture} + \begin{axis}[ + numberline, + xmin=-2, + xmax=2, + extra x ticks={-1, 1, 0}, + extra x tick labels={$-1$,$1$, $0$}, + ] + \addplot[guideline] coordinates {(-1,0) (-1,2)}; + \addplot[guideline] coordinates {(0,0) (0,2)}; + \addplot[guideline] coordinates {(1,0) (1,2)}; + \addplot[mark=none] coordinates {(-1.5,1.5)} node {\parbox{3em}{\centering \small $\fpp\lt0$\\$f$ conc down}}; + \addplot[mark=none] coordinates {(-0.5,1.5)} node {\parbox{3em}{\centering \small $\fpp>0$\\$f$ conc up}}; + \addplot[mark=none] coordinates {(0.5,1.5)} node {\parbox{3em}{\centering \small $\fpp\lt0$\\$f$ conc down}}; + \addplot[mark=none] coordinates {(1.5,1.5)} node {\parbox{3em}{\centering \small $\fpp>0$\\$f$ conc up}}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + We conclude that f is concave up on (-1,0) and + (1,\infty) and concave down on (-\infty,-1) and (0,1). + There is only one point of inflection, + (0,0), as f is not defined at x=\pm 1. + Our work is confirmed by the graph of f in . + Notice how f is concave up whenever \fppis positive, + and concave down when \fppis negative. + The inflection in f occurs where \fpp changes sign. +

    + +
    + A graph of f(x) and \fpp(x) in + + + + A graph comparing a function and its second derivative. + + +

    + The graph displays two curves on the same set of axes: one for the function f(x) in blue and another for its second derivative \fpp(x) in red. + The function f(x) shows a vertical asymptote, where the curve approaches but never reaches the line x=0. + The red curve of the second derivative \fpp(x) also approaches this vertical asymptote. + Both curves demonstrate that as x gets closer to zero from either side, the values of f(x) and \fpp(x) increase or decrease without bound, indicating a point of inflection at x=0 for the function f(x). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-3.2, + xmax=3.2, + ymin=-11, + ymax=11, + ] + \addplot[firstcurvestyle, infinite,samples=25, domain=-3.1:-1.05] {x/(x^2-1)}; + \addplot[firstcurvestyle, infinite,samples=25, domain=-.95:.95] {x/(x^2-1)} node[left, pos=0.1]{$f(x)$}; + \addplot[firstcurvestyle, infinite,samples=25, domain=1.05:3.1] {x/(x^2-1)}; + \addplot[secondcurvestyle,infinite, samples=25, domain=-3.1:-1.5] {(2*x*(x^2+3))/((x^2-1)^3)}; + \addplot[secondcurvestyle,infinite, samples=25, domain=-.5:.5] {(2*x*(x^2+3))/((x^2-1)^3)}; + \addplot[secondcurvestyle, infinite,domain=1.5:3.1] {(2*x*(x^2+3))/((x^2-1)^3)} node[right, pos=0.1] {$\fpp(x)$}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + +

    + Recall that relative maxima and minima of f are found at critical points of f; + that is, they are found when + \fp(x)=0 or when \fp is undefined. + Likewise, the relative maxima and minima of \fp are found when + \fpp(x)=0 or when \fpp is undefined; + note that these are the inflection points of f. +

    + +

    + What does a relative maximum of \fp mean? + The derivative measures the rate of change of f; + maximizing \fp means finding where f is increasing the most where f has the steepest tangent line. + A similar statement can be made for minimizing \fp; it corresponds to where f has the steepest negatively-sloped tangent line. +

    + +

    + We utilize this concept in the next example. +

    + + + Understanding inflection points + +

    + The sales of a certain product over a three-year span are modeled by S(t)= t^4-8t^2+20, + where t is the time in years, + shown in . +

    + +
    + A graph of S(t) in , modeling the sale of a product over time + + + + A graph representing the sales model of a product over time. + + +

    + The figure illustrates a graph of the function S(t), which models the sales of a product over a three-year period. + The sales are modeled by the equation S(t)= t^4-8t^2+20, where t is the time in years. + The graph shows an initial decline in sales, reaching a minimum before rising again. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.2, + xmax=3.2, + ymin=-1, + ymax=21, + xlabel={$t$}, + ] + \addplot+[rightarrow, domain=0:2.8, samples=40] {x^4-8*x^2+20} node[above right, pos=0.3]{$S(t)$}; + \end{axis} + \end{tikzpicture} + + + +
    + +

    + Over the first two years, sales are decreasing. + Find the point at which sales are decreasing at their greatest rate. +

    +     + +

    + We want to maximize the rate of decrease, + which is to say, we want to find where S' has a minimum. + To do this, we find where S'' is 0 and S'' changes from negative to positive. + We find S'(t)=4t^3-16t and S''(t)=12t^2-16. + Setting S''(t)=0 and solving, + we get t=\sqrt{4/3}\approx 1.16 + (we ignore the negative solution for t since it does not lie in the domain of our function S). +

    + +

    + Since S''(1)=-4\lt 0 and S''(2)=32\gt 0, + we can say S'(\sqrt{4/3}) is a local minimum of S'. + This is both the inflection point and the point of maximum decrease. + This is the point at which things first start looking up for the company. + After the inflection point, sales are still decreasing, + but not decreasing quite as quickly as they had been. +

    + +

    + A graph of S(t) and S'(t) is given in . + When S'(t)\lt 0, sales are decreasing; + note how at t\approx 1.16, + S'(t) is minimized. + That is, sales are decreasing at the fastest rate at t\approx 1.16. + On the interval of (1.16,2), + S is decreasing but concave up, + so the decline in sales is leveling off. +

    + +
    + A graph of S(t) in , along with S'(t) + + + + A graph showing both the sales function and its rate of change over time. + + +

    + The graph depicts two curves: the blue curve represents the sales function S(t), and the red curve represents its rate of change, S'(t), over a time period t. + The sales function curve shows a parabolic shape, indicating a period of declining sales followed by an upturn. + The rate of change curve S'(t) crosses the t-axis at its lowest point, signifying the moment when the sales rate is changing from decreasing to increasing. + The point where S'(t) is most negative corresponds to the fastest rate of decrease in sales, which is the primary point of interest in this example. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-.2, + xmax=3.2, + ymin=-16, + ymax=21, + xlabel={$t$}, + ] + \addplot+[rightarrow,domain=0:2.8, samples=40] {x^4-8*x^2+20} node[above right, pos=0.3]{$S(t)$}; + \addplot+[domain=0:3, samples=40] {4*x^3-16*x} node[below right, pos=0.2]{$S'(t)$}; + \addplot[soliddot] coordinates {(1.16,11.04) (1.16,-12.32)}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + +

    + Not every critical point corresponds to a relative extrema; + f(x)=x^3 has a critical point at (0,0) but no relative maximum or minimum. + Likewise, just because \fpp(x)=0 we cannot conclude concavity changes at that point. + We were careful before to use terminology + possible point of inflection + since we needed to check to see if the concavity changed. + The canonical example of \fpp(x)=0 without + concavity changing is f(x)=x^4. + At x=0, + \fpp(x)=0 but f is always concave up, + as shown in . +

    + +
    + A graph of f(x) = x^4. Clearly f is always concave up, despite the fact that \fpp(x) = 0 when x=0. In this example, the possible point of inflection (0,0) is not a point of inflection. + + + + A graph depicting a U-shaped curve representing a function raised to the fourth power. + + +

    + The graph of f(x) = x^4 illustrates a U-shaped curve that is concave up across its domain. Despite the second derivative \fpp(x) equalling zero when x = 0, indicating a potential inflection point, the function remains concave up on both sides of the y-axis. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-1.1, + xmax=1.1, + ymin=-.1, + ymax=1.1, + ] + \addplot+[infinite,domain=-1:1] {x^4}; + \end{axis} + \end{tikzpicture} + + + +
    +     + + + The Second Derivative Test +

    + The first derivative of a function gave us a test to find if a critical value corresponded to a relative maximum, minimum, + or neither. + The second derivative gives us another way to test if a critical point is a local maximum or minimum. + The following theorem officially states something that is intuitive: + if a critical value occurs in a region where a function f is concave up, + then that critical value must correspond to a relative minimum of f, etc. + See for a visualization of this. +

    + + + The Second Derivative Test + +

    + Let c be a critical value of f where \fpp(c) is defined. + derivativeSecond Deriv. + Test + Second Derivative Test + extremaand Second Deriv. Test + maximumand Second Deriv. Test + minimumand First Deriv. Test + +

      +
    1. +

      + If \fpp(c)\gt 0, + then f has a local minimum at (c,f(c)). +

      +
      2. + +
    2. +

      + If \fpp(c)\lt 0, + then f has a local maximum at (c,f(c)). +

      +
      3. +
    +

    +     +     + +
    + Demonstrating the fact that relative maxima occur when the graph is concave down and relative minima occur when the graph is concave up + + + + A graph depicting points of maximum and minimum on a curve with changing concavity. + + +

    + The figure showcases a curve that illustrates the concepts of relative maxima and minima in relation to concavity. + The curve rises upwards, dips downward and then rises upward again, forming a valley-like shape. + At the peak of the first curve, where the shape is bending downwards indicating concave down, a point is marked at (-1, 10) signifying a relative maximum. + Similarly, at the lowest point in the valley at (1,-10), where the curve starts to bend upwards denoted as concave up, a point signifies a relative minimum. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-2.1, + xmax=2.1, + ymin=-11, + ymax=11, + ] + \addplot+[infinite,domain=-2:2, samples=50] {5*x^3-15*x}; + \addplot[soliddot] coordinates {(-1,10) (1,-10)}; + \addplot[mark=none] coordinates {(-1,7.5)} node {\parbox{6em}{\centering concave down\\$\implies$ rel max}}; + \addplot[mark=none] coordinates {(1,-7.5)} node {\parbox{6em}{\centering concave up\\$\implies$ rel min}}; + \end{axis} + \end{tikzpicture} + + + +
    + + + +

    + The Second Derivative Test relates to the in the following way. + If \fpp(c)\gt 0, + then the graph is concave up at a critical point c and \fp itself is growing. + Since \fp(c)=0 and \fp is growing at c, + then it must go from negative to positive at c. + This means the function goes from decreasing to increasing, + indicating a local minimum at c. +

    + + + Using the Second Derivative Test + +

    + Let f(x)=100/x + x. + Find the critical points of f and use the to label them as relative maxima or minima. +

    +     + +

    + We find \fp(x)=-100/x^2+1 and \fpp(x) = 200/x^3. + We set \fp(x)=0 and solve for x to find the critical values + (note that \fp is not defined at x=0, + but neither is f so this is not a critical value.) + We find the critical values are x=\pm 10. + We now evaluate the second derivative at these critical numbers. + Evaluating \fpp(10)=0.1\gt 0, + so there is a local minimum at x=10. + Evaluating \fpp(-10)=-0.1\lt 0, + determining a relative maximum at x=-10. + These results are confirmed in . +

    + +
    + A graph of f(x) in . The second derivative is evaluated at each critical point. When the graph is concave up, the critical point represents a local minimum; when the graph is concave down, the critical point represents a local maximum. + + + + A graph illustrating the evaluation of the second derivative at critical points. + + +

    + This graph depicts a function f(x) with two marked critical points where the second derivative \fpp(x) has been evaluated. + On the left, a point at x = -10 is marked with \fpp(-10)\lt 0, indicating that the graph is concave down at this point, which typically corresponds to a local maximum. + On the right, a point at x = 10 is marked with \fpp(10)\gt 0, signifying that the graph is concave up at this point, suggesting the presence of a local minimum. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xmin=-21, + xmax=21, + ymin=-51, + ymax=51, + ] + \addplot+[infinite,domain=-20:-2.2] {100/x+x}; + \addplot[firstcurvestyle,domain=2.2:20] {100/x+x}; + \addplot[soliddot] coordinates{(10,20)} node [below] {$\fpp(10)>0$}; + \addplot[soliddot] coordinates{(-10,-20)} node [above] {$\fpp(-10)\lt0$}; + \end{axis} + \end{tikzpicture} + + + +
    +     + +     + +

    + The second derivative test can only be used on a function that is twice differentiable at c. + For functions that are not twice differentiable at c, + you will need to use the . + If you've already determined the sign diagram for \fp, + the is usually + easier to apply, and it applies in cases when + does not. +

    + +

    + We have been learning how the first and second derivatives of a function relate information about the graph of that function. + We have found intervals of increasing and decreasing, + intervals where the graph is concave up and down, + along with the locations of relative extrema and inflection points. + In + we saw how limits explained asymptotic behavior. + In the next section we combine all of this information to produce accurate sketches of functions. +

    +     + + + + Terms and Concepts + + + +

    + Sketch a graph of a function f(x) that is concave up on (0,1) and is concave down on (1,2). +

    +     + + + +     + + + +

    + Sketch a graph of a function f(x) that is: +

    + +

    +

      +
    • + increasing, concave up on (0,1), +
      • + +
    • + increasing, concave down on (1,2), +
      • + +
    • + decreasing, concave down on (2,3), and +
      • + +
    • + increasing, concave down on (3,4). +
      • +
    +

    +     + + + +     + + + +

    + Is is possible for a function to be increasing and concave down on + (0,\infty) with a horizontal asymptote of y=1? + If so, give a sketch of such a function. +

    +     + + + +     + + + +

    + Is is possible for a function to be increasing and concave up on + (0,\infty) with a horizontal asymptote of y=1? + If so, give a sketch of such a function. +

    +     + + + +     +     + + + Problems + + + + +

    + A function f(x) is given. + Graph f and \fpp on the same axes + (using technology is permitted) + and verify . +

    +     + + + +

    + f(x) = -7x+3 +

    +     +     + + + +

    + f(x) = -4x^2+3x-8 +

    +     +     + + + +

    + f(x) = 4x^2+3x-8 +

    +     +     + + + +

    + f(x) = x^3-3x^2+x-1 +

    +     +     + + + +

    + f(x) = -x^3+x^2-2x+5 +

    +     +     + + + + + +

    + f(x) = \sin(x) +

    +     +     + + + +

    + f(x) = \tan(x) +

    +     +     + + + +

    + f(x) = \dfrac{1}{x^2+1} +

    +     +     + + + +

    + f(x) = \frac{1}{x} +

    +     +     + + + +

    + f(x) = \frac{1}{x^2} +

    +     +     +     + + + +

    + A function f(x) is given. +

      +
    1. +

      + Find the possible points of inflection of f. +

      +
      2. +
    2. +

      + Find the intervals on which the graph of f is concave up. +

      +
      3. +
    3. +

      + Find the intervals on which the graph of f is concave down. +

      +
      4. +
    +

    +     + + + + + $r = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$r=1;}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("x^2 - 2*$r x + ($r)^2")->reduce; + $inflecs = List(Compute("None")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,inf)")); + $dn = List(Compute("None")); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + ($a,$b) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$a=-5; $b=7}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("-x^2 + $a x + $b")->reduce; + $inflecs = List(Compute("None")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Compute("None")); + $dn = List(Interval("(-inf,inf)")); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + $s = random(1,9,1); + $c = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$s=1; $c=1;}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("x^3 - $s x + $c")->reduce; + $inflecs = List(0); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("[0,inf)")); + $dn = List(Interval("(-inf,0]")); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + $a = non_zero_random(-9,9,1); + $b = non_zero_random(-9,9,1); + $c = ($a > 0) ? random(int(($b)**2/(3*$a))+1,int(($b)**2/(3*$a))+9,1) : random(int(($b)**2/(3*$a))-9,int(($b)**2/(3*$a))-1,1); + $d = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=2; $b=-3; $c=9; $d=5}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("$a x^3 + $b x^2 + $c x + $d")->reduce; + Context("Fraction"); + $i = Fraction(-2*$b,6*$a); + $inflecs = List($i); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("[$i,inf)")); + $dn = List(Interval("(-inf,$i]")); + ($up,$dn) = ($dn,$up) if ($a < 0); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + $s = non_zero_random(-9,9,1); + $k = ($s % 2 == 0) ? non_zero_random(2,8,2) : non_zero_random(1,9,2); + $d = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$s=-1; $k=-1; $d=3;}; + $r = ($s + $k)/-2; + $t = $k*$s; + Context("Fraction"); + $b = Fraction(-($r+2*$s),3); + ($bn,$bd) = $b->value; + $c = -$r*(($s)**2+$t); + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("x^4/4 + $bn x^3/$bd + $c x + $d")->reduce; + $i = Fraction(3*$s-$k,3); + ($i,$j) = num_sort($i,0); + $inflecs = List($i,$j); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); + $dn = List(Interval("[$i,$j]")); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + ($r,$s,$t) = random_subset(3,-5..-1,1..5); + $k = list_random(-12,-8,-4,4,8,12); + if ($k % 3 != 0) { + do {$t = non_zero_random(-($r+$s) % 3 - 6, -($r+$s) % 3 + 6, 3)} until ($t != $r and $t != $s); + } + ($r,$s,$t) = num_sort($r,$s,$t); + $e = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$r=-1; $s=1; $t=2; $k=12;}; + $a = $k/4; + $b = $k*($r+$s+$t)/3; + $c = $k*($s*$t+$r*$s+$r*$t)/2; + $d = $k*$r*$s*$t; + Context("Fraction"); + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; + $disc = 9*($b)**2 - 24*$a*$c; + ($discsquare,$discnonsquare) = (1,$disc); + for my $p (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251) { + while ($discnonsquare % ($p**2) == 0) { + ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); + } + } + $G = gcd(gcd(3*$b,sqrt($discsquare)),12*$a); + $A = -3*$b/$G; + $B = sqrt($discsquare)/$G; + $C = 12*$a/$G; + if ($discnonsquare == 1) { + $i = Fraction($A+$B,$C); + $j = Fraction($A-$B,$C); + } else { + $i = Formula("($A+$B sqrt($discnonsquare))/$C"); + $j = Formula("($A-$B sqrt($discnonsquare))/$C"); + } + ($i,$j) = num_sort($i,$j); + $inflecs = List($i,$j); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); + $dn = List(Interval("[$i,$j]")); + ($up,$dn) = ($dn,$up) if ($a < 0); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + $r = non_zero_random(-4,4,1); + if($envir{problemSeed}==1){$r=1;}; + $a = 1; + $b = -4*$r; + $c = 6*($r**2); + $d = -4*($r**3); + $e = ($r)**4; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; + $inflecs = List($r); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,inf)")); + $dn = List(Compute("none")); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("sec(x)"); + $inflecs = List(Compute("none")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-pi/2,pi/2)")); + $dn = List(Interval("(-3pi/2,-pi/2)"),Interval("(pi/2,3pi/2)")); + + +

    + f(x)= on (-3\pi/2,3\pi/2) +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + $r = random(-9,9,1); + $s = random(1,9,1); + if($envir{problemSeed}==1){$r=0; $s = 1;}; + $b = -2*$r; + $c = ($r**2) + $s; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("1/(x^2 + $b x + $c)")->reduce; + $disc = 3*$s; + ($discsquare,$discnonsquare) = (1,$disc); + for my $p (2,3,5) { + while ($discnonsquare % ($p**2) == 0) { + ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); + } + } + $G = gcd(gcd(3*$3,sqrt($discsquare)),3); + $A = 3*$r/$G; + $B = sqrt($discsquare)/$G; + $C = 3/$G; + Context("Fraction"); + if ($discnonsquare == 1) { + $i = Fraction($A+$B,$C); + $j = Fraction($A-$B,$C); + } else { + $i = Formula("($A+$B sqrt($discnonsquare))/$C"); + $j = Formula("($A-$B sqrt($discnonsquare))/$C"); + } + ($i,$j) = num_sort($i,$j); + $inflecs = List($i,$j); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); + $dn = List(Interval("[$i,$j]")); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + ($r,$s) = num_sort(random_subset(2,-9..-1,1..9)); + if($envir{problemSeed}==1){$r=-1; $s = 1;}; + $b = -$r - $s; + $c = $r*$s; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("1/(x^2 + $b x + $c)")->reduce; + $inflecs = List(Compute("none")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$r)"),Interval("($s,inf)")); + $dn = List(Interval("($r,$s)")); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("sin(x) + cos(x)"); + $inflecs = List(Compute("-pi/4"),Compute("3pi/4")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-pi,-pi/4]"),Interval("[3pi/4,pi)")); + $dn = List(Interval("[-pi/4,3pi/4]")); + + +

    + f(x)= on (-\pi,\pi) +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("x^2e^x"); + $inflecs = List(Compute("-2+sqrt(2)"),Compute("-2-sqrt(2)")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,-2-sqrt(2)]"),Interval("[-2+sqrt(2),inf)")); + $dn = List(Interval("[-2-sqrt(2),-2+sqrt(2)]")); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("x^2ln(x)"); + $inflecs = List(Compute("1/e^(3/2)")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("[1/e^(3/2),inf)")); + $dn = List(Interval("(0,1/e^(3/2)]")); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("e^(-x^2)"); + $inflecs = List(Compute("1/sqrt(2)"),Compute("-1/sqrt(2)")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,-1/sqrt(2)]"),Interval("[1/sqrt(2),inf)")); + $dn = List(Interval("[-1/sqrt(2),1/sqrt(2)]")); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the maximal intervals where f is concave up, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    + + List the maximal intervals where f is concave down, + using interval notation, + and separating with commas if needed. + If there are no such intervals, + enter none. + +

    + +

    +     + + +     +     +     + + + +

    + A function f(x) is given. + Find the critical points of f and use the Second Derivative Test, + when possible, + to determine the relative extrema. + (Note: these are the same functions as in .) +

    +     + + + + + $r = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$r=1;}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("x^2 - 2*$r x + ($r)^2")->reduce; + $inflecs = List(Compute("None")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,inf)")); + $dn = List(Compute("None")); + Context("Fraction"); + $crit = List($r); + $max = List(Compute("none")); + $min = List($r); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + ($a,$b) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$a=-5; $b=7}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("-x^2 + $a x + $b")->reduce; + $inflecs = List(Compute("None")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Compute("None")); + $dn = List(Interval("(-inf,inf)")); + Context("Fraction"); + $m = Fraction($a,2); + $crit = List($m); + $max = List($m); + $min = List(Compute("none")); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $s = random(1,9,1); + $c = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$s=1; $c=1;}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("x^3 - $s x + $c")->reduce; + $inflecs = List(0); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("[0,inf)")); + $dn = List(Interval("(-inf,0]")); + Context("Fraction"); + $rad = 3*$s; + ($radsquare,$radnonsquare) = (1,$rad); + for my $p (2,3,5) { + while ($radnonsquare % ($p**2) == 0) { + ($radsquare,$radnonsquare) = ($radsquare * $p**2, $radnonsquare / $p**2); + } + } + $G = gcd(sqrt($radsquare),3); + $B = sqrt($radsquare)/$G; + $C = 3/$G; + if ($discnonsquare == 1) { + $m = Fraction($B,$C); + $n = Fraction(-$B,$C); + } else { + $m = Formula("($B sqrt($radnonsquare))/$C"); + $n = Formula("(-$B sqrt($radnonsquare))/$C"); + } + ($m,$n) = num_sort($m,$n); + $crit = List($m,$n); + $max = List($m); + $min = List($n); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $a = non_zero_random(-9,9,1); + $b = non_zero_random(-9,9,1); + $c = ($a > 0) ? random(int(($b)**2/(3*$a))+1,int(($b)**2/(3*$a))+9,1) : random(int(($b)**2/(3*$a))-9,int(($b)**2/(3*$a))-1,1); + $d = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=2; $b=-3; $c=9; $d=5}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("$a x^3 + $b x^2 + $c x + $d")->reduce; + Context("Fraction"); + $i = Fraction(-2*$b,6*$a); + $inflecs = List($i); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("[$i,inf)")); + $dn = List(Interval("(-inf,$i]")); + ($up,$dn) = ($dn,$up) if ($a < 0); + $crit = List(Compute("none")); + $max = List(Compute("none")); + $min = List(Compute("none")); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $s = non_zero_random(-9,9,1); + $k = ($s % 2 == 0) ? non_zero_random(2,8,2) : non_zero_random(1,9,2); + $d = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$s=-1; $k=-1; $d=3;}; + $r = ($s + $k)/-2; + $t = $k*$s; + Context("Fraction"); + $b = Fraction(-($r+2*$s),3); + ($bn,$bd) = $b->value; + $c = -$r*(($s)**2+$t); + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("x^4/4 + $bn x^3/$bd + $c x + $d")->reduce; + $i = Fraction(3*$s-$k,3); + ($i,$j) = num_sort($i,0); + $inflecs = List($i,$j); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); + $dn = List(Interval("[$i,$j]")); + $crit = List($r); + $max = ($i <= $r and $r <= $j) ? List($r) : List(Compute("none")); + $min = ($i <= $r and $r <= $j) ? List(Compute("none")) : List($r); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + ($r,$s,$t) = random_subset(3,-5..-1,1..5); + $k = list_random(-12,-8,-4,4,8,12); + if ($k % 3 != 0) { + do {$t = non_zero_random(-($r+$s) % 3 - 6, -($r+$s) % 3 + 6, 3)} until ($t != $r and $t != $s); + } + ($r,$s,$t) = num_sort($r,$s,$t); + $e = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$r=-1; $s=1; $t=2; $k=12;}; + $a = $k/4; + $b = $k*($r+$s+$t)/3; + $c = $k*($s*$t+$r*$s+$r*$t)/2; + $d = $k*$r*$s*$t; + Context("Fraction"); + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; + $disc = 9*($b)**2 - 24*$a*$c; + ($discsquare,$discnonsquare) = (1,$disc); + for my $p (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251) { + while ($discnonsquare % ($p**2) == 0) { + ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); + } + } + $G = gcd(gcd(3*$b,sqrt($discsquare)),12*$a); + $A = -3*$b/$G; + $B = sqrt($discsquare)/$G; + $C = 12*$a/$G; + if ($discnonsquare == 1) { + $i = Fraction($A+$B,$C); + $j = Fraction($A-$B,$C); + } else { + $i = Formula("($A+$B sqrt($discnonsquare))/$C"); + $j = Formula("($A-$B sqrt($discnonsquare))/$C"); + } + ($i,$j) = num_sort($i,$j); + $inflecs = List($i,$j); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); + $dn = List(Interval("[$i,$j]")); + ($up,$dn) = ($dn,$up) if ($a < 0); + $crit = List($r,$s,$t); + $max = ($a > 0) ? List($s) : List($r,$t); + $min = ($a > 0) ? List($r,$t): List($s); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $r = non_zero_random(-4,4,1); + if($envir{problemSeed}==1){$r=1;}; + $a = 1; + $b = -4*$r; + $c = 6*($r**2); + $d = -4*($r**3); + $e = ($r)**4; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; + $inflecs = List($r); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,inf)")); + $dn = List(Compute("none")); + $crit = List($r); + $max = List(Compute("none")); + $min = List(Compute("none")); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("sec(x)"); + $inflecs = List(Compute("none")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-pi/2,pi/2)")); + $dn = List(Interval("(-3pi/2,-pi/2)"),Interval("(pi/2,3pi/2)")); + $crit = List(Compute("-pi"),0,Compute("pi")); + $max = List(Compute("-pi"),Compute("pi")); + $min = List(0); + + +

    + f(x)= on (-3\pi/2,3\pi/2) +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $r = random(-9,9,1); + $s = random(1,9,1); + if($envir{problemSeed}==1){$r=0; $s = 1;}; + $b = -2*$r; + $c = ($r**2) + $s; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("1/(x^2 + $b x + $c)")->reduce; + $disc = 3*$s; + ($discsquare,$discnonsquare) = (1,$disc); + for my $p (2,3,5) { + while ($discnonsquare % ($p**2) == 0) { + ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); + } + } + $G = gcd(gcd(3*$3,sqrt($discsquare)),3); + $A = 3*$r/$G; + $B = sqrt($discsquare)/$G; + $C = 3/$G; + Context("Fraction"); + if ($discnonsquare == 1) { + $i = Fraction($A+$B,$C); + $j = Fraction($A-$B,$C); + } else { + $i = Formula("($A+$B sqrt($discnonsquare))/$C"); + $j = Formula("($A-$B sqrt($discnonsquare))/$C"); + } + ($i,$j) = num_sort($i,$j); + $inflecs = List($i,$j); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); + $dn = List(Interval("[$i,$j]")); + $crit = List($r); + $max = List($r); + $min = List("none"); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + ($r,$s) = num_sort(random_subset(2,-9..-1,1..9)); + if($envir{problemSeed}==1){$r=-1; $s = 1;}; + $b = -$r - $s; + $c = $r*$s; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("1/(x^2 + $b x + $c)")->reduce; + $inflecs = List(Compute("none")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$r)"),Interval("($s,inf)")); + $dn = List(Interval("($r,$s)")); + Context("Fraction"); + $m = Fraction($r+$s,2); + $crit = List($m); + $max = List($m); + $min = List("none"); + + +

    + f(x)= +

    + + Enter the possible points of inflection, + separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimum. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("sin(x) + cos(x)"); + $inflecs = List(Compute("-pi/4"),Compute("3pi/4")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-pi,-pi/4]"),Interval("[3pi/4,pi)")); + $dn = List(Interval("[-pi/4,3pi/4]")); + $crit = List(Compute("-3pi/4"),Compute("pi/4")); + $max = List(Compute("pi/4")); + $min = List(Compute("-3pi/4")); + + +

    + f(x)= on (-\pi,\pi) +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("x^2e^x"); + $inflecs = List(Compute("-2+sqrt(2)"),Compute("-2-sqrt(2)")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,-2-sqrt(2)]"),Interval("[-2+sqrt(2),inf)")); + $dn = List(Interval("[-2-sqrt(2),-2+sqrt(2)]")); + $crit = List(-2,0); + $max = List(-2); + $min = List(0); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("x^2ln(x)"); + $inflecs = List(Compute("1/e^(3/2)")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("[1/e^(3/2),inf)")); + $dn = List(Interval("(0,1/e^(3/2)]")); + $crit = List(Compute("e^(-1/2)")); + $max = List(Compute("none")); + $min = List(Compute("1/sqrt(e)")); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("e^(-x^2)"); + $inflecs = List(Compute("1/sqrt(2)"),Compute("-1/sqrt(2)")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,-1/sqrt(2)]"),Interval("[1/sqrt(2),inf)")); + $dn = List(Interval("[-1/sqrt(2),1/sqrt(2)]")); + $crit = List(0); + $max = List(0); + $min = List(Compute("none")); + + +

    + f(x)= +

    + + Enter the critical points, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative maximum. + If there are none, + enter none. + +

    + +

    + + List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. + If there are none, + enter none. + +

    + +

    +     + + +     +     +     + + + +

    + A function f(x) is given. + Find the x values where \fp(x) has a relative maximum or minimum. + (Note: these are the same functions as in .) +

    +     + + + + + $r = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$r=1;}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("x^2 - 2*$r x + ($r)^2")->reduce; + $inflecs = List(Compute("None")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,inf)")); + $dn = List(Compute("None")); + Context("Fraction"); + $crit = List($r); + $max = List(Compute("none")); + $min = List($r); + $fpmax = List(Compute("none")); + $fpmin = List(Compute("none")); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + ($a,$b) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$a=-5; $b=7}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("-x^2 + $a x + $b")->reduce; + $inflecs = List(Compute("None")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Compute("None")); + $dn = List(Interval("(-inf,inf)")); + Context("Fraction"); + $m = Fraction($a,2); + $crit = List($m); + $max = List($m); + $min = List(Compute("none")); + $fpmax = List(Compute("none")); + $fpmin = List(Compute("none")); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $s = random(1,9,1); + $c = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$s=1; $c=1;}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("x^3 - $s x + $c")->reduce; + $inflecs = List(0); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("[0,inf)")); + $dn = List(Interval("(-inf,0]")); + Context("Fraction"); + $rad = 3*$s; + ($radsquare,$radnonsquare) = (1,$rad); + for my $p (2,3,5) { + while ($radnonsquare % ($p**2) == 0) { + ($radsquare,$radnonsquare) = ($radsquare * $p**2, $radnonsquare / $p**2); + } + } + $G = gcd(sqrt($radsquare),3); + $B = sqrt($radsquare)/$G; + $C = 3/$G; + if ($discnonsquare == 1) { + $m = Fraction($B,$C); + $n = Fraction(-$B,$C); + } else { + $m = Formula("($B sqrt($radnonsquare))/$C"); + $n = Formula("(-$B sqrt($radnonsquare))/$C"); + } + ($m,$n) = num_sort($m,$n); + $crit = List($m,$n); + $max = List($m); + $min = List($n); + $fpmax = List(Compute("none")); + $fpmin = List(0); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $a = non_zero_random(-9,9,1); + $b = non_zero_random(-9,9,1); + $c = ($a > 0) ? random(int(($b)**2/(3*$a))+1,int(($b)**2/(3*$a))+9,1) : random(int(($b)**2/(3*$a))-9,int(($b)**2/(3*$a))-1,1); + $d = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=2; $b=-3; $c=9; $d=5}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("$a x^3 + $b x^2 + $c x + $d")->reduce; + Context("Fraction"); + $i = Fraction(-2*$b,6*$a); + $inflecs = List($i); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("[$i,inf)")); + $dn = List(Interval("(-inf,$i]")); + ($up,$dn) = ($dn,$up) if ($a < 0); + $crit = List(Compute("none")); + $max = List(Compute("none")); + $min = List(Compute("none")); + $fpmax = List(Compute("none")); + $fpmin = List($i); + ($fpmax,$fpmin) = ($fpmin,$fpmax) if ($a < 0); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $s = non_zero_random(-9,9,1); + $k = ($s % 2 == 0) ? non_zero_random(2,8,2) : non_zero_random(1,9,2); + $d = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$s=-1; $k=-1; $d=3;}; + $r = ($s + $k)/-2; + $t = $k*$s; + Context("Fraction"); + $b = Fraction(-($r+2*$s),3); + ($bn,$bd) = $b->value; + $c = -$r*(($s)**2+$t); + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("x^4/4 + $bn x^3/$bd + $c x + $d")->reduce; + $i = Fraction(3*$s-$k,3); + ($i,$j) = num_sort($i,0); + $inflecs = List($i,$j); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); + $dn = List(Interval("[$i,$j]")); + $crit = List($r); + $max = ($i <= $r and $r <= $j) ? List($r) : List(Compute("none")); + $min = ($i <= $r and $r <= $j) ? List(Compute("none")) : List($r); + $fpmax = List($i); + $fpmin = List($j); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + ($r,$s,$t) = random_subset(3,-5..-1,1..5); + $k = list_random(-12,-8,-4,4,8,12); + if ($k % 3 != 0) { + do {$t = non_zero_random(-($r+$s) % 3 - 6, -($r+$s) % 3 + 6, 3)} until ($t != $r and $t != $s); + } + ($r,$s,$t) = num_sort($r,$s,$t); + $e = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$r=-1; $s=1; $t=2; $k=12;}; + $a = $k/4; + $b = $k*($r+$s+$t)/3; + $c = $k*($s*$t+$r*$s+$r*$t)/2; + $d = $k*$r*$s*$t; + Context("Fraction"); + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; + $disc = 9*($b)**2 - 24*$a*$c; + ($discsquare,$discnonsquare) = (1,$disc); + for my $p (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251) { + while ($discnonsquare % ($p**2) == 0) { + ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); + } + } + $G = gcd(gcd(3*$b,sqrt($discsquare)),12*$a); + $A = -3*$b/$G; + $B = sqrt($discsquare)/$G; + $C = 12*$a/$G; + if ($discnonsquare == 1) { + $i = Fraction($A+$B,$C); + $j = Fraction($A-$B,$C); + } else { + $i = Formula("($A+$B sqrt($discnonsquare))/$C"); + $j = Formula("($A-$B sqrt($discnonsquare))/$C"); + } + ($i,$j) = num_sort($i,$j); + $inflecs = List($i,$j); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); + $dn = List(Interval("[$i,$j]")); + ($up,$dn) = ($dn,$up) if ($a < 0); + $crit = List($r,$s,$t); + $max = ($a > 0) ? List($s) : List($r,$t); + $min = ($a > 0) ? List($r,$t): List($s); + $fpmax = List($i); + $fpmin = List($j); + ($fpmax,$fpmin) = ($fpmin,$fpmax) if ($a < 0); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $r = non_zero_random(-4,4,1); + if($envir{problemSeed}==1){$r=1;}; + $a = 1; + $b = -4*$r; + $c = 6*($r**2); + $d = -4*($r**3); + $e = ($r)**4; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; + $inflecs = List($r); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,inf)")); + $dn = List(Compute("none")); + $crit = List($r); + $max = List(Compute("none")); + $min = List(Compute("none")); + $fpmax = List(Compute("none")); + $fpmin = List(Compute("none")); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("sec(x)"); + $inflecs = List(Compute("none")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-pi/2,pi/2)")); + $dn = List(Interval("(-3pi/2,-pi/2)"),Interval("(pi/2,3pi/2)")); + $crit = List(Compute("-pi"),0,Compute("pi")); + $max = List(Compute("-pi"),Compute("pi")); + $min = List(0); + $fpmax = List(Compute("none")); + $fpmin = List(Compute("none")); + + +

    + f(x)= on (-3\pi/2,3\pi/2) +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $r = random(-9,9,1); + $s = random(1,9,1); + if($envir{problemSeed}==1){$r=0; $s = 1;}; + $b = -2*$r; + $c = ($r**2) + $s; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("1/(x^2 + $b x + $c)")->reduce; + $disc = 3*$s; + ($discsquare,$discnonsquare) = (1,$disc); + for my $p (2,3,5) { + while ($discnonsquare % ($p**2) == 0) { + ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); + } + } + $G = gcd(gcd(3*$3,sqrt($discsquare)),3); + $A = 3*$r/$G; + $B = sqrt($discsquare)/$G; + $C = 3/$G; + Context("Fraction"); + if ($discnonsquare == 1) { + $i = Fraction($A+$B,$C); + $j = Fraction($A-$B,$C); + } else { + $i = Formula("($A+$B sqrt($discnonsquare))/$C"); + $j = Formula("($A-$B sqrt($discnonsquare))/$C"); + } + ($i,$j) = num_sort($i,$j); + $inflecs = List($i,$j); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); + $dn = List(Interval("[$i,$j]")); + $crit = List($r); + $max = List($r); + $min = List("none"); + $fpmax = List($i); + $fpmin = List($j); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + ($r,$s) = num_sort(random_subset(2,-9..-1,1..9)); + if($envir{problemSeed}==1){$r=-1; $s = 1;}; + $b = -$r - $s; + $c = $r*$s; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("1/(x^2 + $b x + $c)")->reduce; + $inflecs = List(Compute("none")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,$r)"),Interval("($s,inf)")); + $dn = List(Interval("($r,$s)")); + Context("Fraction"); + $m = Fraction($r+$s,2); + $crit = List($m); + $max = List($m); + $min = List("none"); + $fpmax = List(Compute("none")); + $fpmin = List(Compute("none")); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("sin(x) + cos(x)"); + $inflecs = List(Compute("-pi/4"),Compute("3pi/4")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-pi,-pi/4]"),Interval("[3pi/4,pi)")); + $dn = List(Interval("[-pi/4,3pi/4]")); + $crit = List(Compute("-3pi/4"),Compute("pi/4")); + $max = List(Compute("pi/4")); + $min = List(Compute("-3pi/4")); + $fpmax = List(Compute("-pi/4")); + $fpmin = List(Compute("3pi/4")); + + +

    + f(x)= on (-\pi,\pi) +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("x^2e^x"); + $inflecs = List(Compute("-2+sqrt(2)"),Compute("-2-sqrt(2)")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,-2-sqrt(2)]"),Interval("[-2+sqrt(2),inf)")); + $dn = List(Interval("[-2-sqrt(2),-2+sqrt(2)]")); + $crit = List(-2,0); + $max = List(-2); + $min = List(0); + $fpmax = List(Compute("-2-sqrt(2)")); + $fpmin = List(Compute("-2+sqrt(2)")); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("x^2ln(x)"); + $inflecs = List(Compute("1/e^(3/2)")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("[1/e^(3/2),inf)")); + $dn = List(Interval("(0,1/e^(3/2)]")); + $crit = List(Compute("e^(-1/2)")); + $max = List(Compute("none")); + $min = List(Compute("1/sqrt(e)")); + $fpmax = List(Compute("none")); + $fpmin = List(Compute("1/e^(3/2)")); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     + + + + + $f = Formula("e^(-x^2)"); + $inflecs = List(Compute("1/sqrt(2)"),Compute("-1/sqrt(2)")); + Context("Interval"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up = List(Interval("(-inf,-1/sqrt(2)]"),Interval("[1/sqrt(2),inf)")); + $dn = List(Interval("[-1/sqrt(2),1/sqrt(2)]")); + $crit = List(0); + $max = List(0); + $min = List(Compute("none")); + $fpmax = List(Compute("-1/sqrt(2)")); + $fpmin = List(Compute("1/sqrt(2)")); + + +

    + f(x)= +

    + + Enter the points where \fp(x) has a relative maximum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    + + Enter the points where \fp(x) has a relative minimum, separating with commas if needed. + If there are none, + enter none. + +

    + +

    +     + + +     +     +     +     +     +
    +
    + Curve Sketching +

    + We have been learning how we can understand the behavior of a function based on its first and second derivatives. + While we have been treating the properties of a function separately + (increasing and decreasing, concave up and concave down, etc.), + we combine them here to produce an accurate graph of the function without plotting lots of extraneous points. +

    + +

    + Why bother? + Graphing utilities are very accessible, + whether on a computer, a hand-held calculator, or a smartphone. + These resources are usually very fast and accurate. + We will see that our method is not particularly fast it will require time + (but it is not hard). + So again: why bother? +

    + +

    + We are attempting to understand the behavior of a function f based on the information given by its derivatives. + While all of a function's derivatives relay information about it, + it turns out that most of the behavior we care about is explained by \fpand \fpp. + Understanding the interactions between the graph of f and \fpand \fppis important. + To gain this understanding, + one might argue that all that is needed is to look at lots of graphs. + This is true to a point, + but is somewhat similar to stating that one understands how an engine works after looking only at pictures. + It is true that the basic ideas will be conveyed, + but hands-on access increases understanding. +

    + + + +

    + + summarizes what we have learned so far that is applicable to sketching graphs of functions and gives a framework for putting that information together. + It is followed by several examples. +

    + + + Curve Sketching +

    + To produce an accurate sketch a given function f, + consider the following steps. + curve sketching +

      +
    1. + +

      + Find the domain of f. + Generally, we assume that the domain is the entire real line then find restrictions, + such as where a denominator is 0 or where negatives appear under the radical. +

      +
      2. + +
    2. +

      + Find the critical values of f. +

      +
      3. + +
    3. +

      + Find the possible points of inflection of f. +

      +
      4. + +
    4. +

      + Find the location of any vertical asymptotes of f + (usually done in conjunction with ). +

      +
      5. + +
    5. +

      + Consider the limits \lim\limits_{x\to-\infty}f(x) and + \lim\limits_{x\to\infty}f(x) to determine the end behavior of the function. +

      +
      6. + +
    6. +

      + Create a number line that includes all critical points, + possible points of inflection, + and locations of vertical asymptotes. + For each interval created, + determine whether f is increasing or decreasing, + concave up or down. +

      +
      7. + +
    7. +

      + Evaluate f at each critical point and possible point of inflection. + Plot these points on a set of axes. + Connect these points with curves exhibiting the proper concavity. + Sketch asymptotes and x and y intercepts where applicable. +

      +
      8. +
    +

    +     + + + + + Curve sketching + +

    + Use + to sketch f(x) = 3x^3-10x^2+7x+5. +

    +     + +

    + We follow the steps outlined in . + +

      +
    1. +

      + The domain of f is the entire real line; + there are no values x for which f(x) is not defined. +

      +
      2. + +
    2. +

      + Find the critical values of f. + We compute \fp(x) = 9x^2-20x+7. + Use the Quadratic Formula to find the roots of \fp: + + x \amp= \frac{20\pm \sqrt{(-20)^2-4(9)(7)}}{2(9)} + \amp = \frac19\left(10\pm\sqrt{37}\right) + x \amp \approx 0.435, 1.787 + . +

      +
      3. + +
    3. +

      + Find the possible points of inflection of f. + Compute \fpp(x) = 18x-20. + We have + + \fpp(x) \amp = 0 + 18x-20 \amp =0 + x \amp = 10/9 + \amp \approx 1.111 + . +

      +
      4. + +
    4. +

      + There are no vertical asymptotes. +

      +
      5. + +
    5. +

      + We determine the end behavior using limits as x approaches \pm \infty . + + \lim_{x\to -\infty} f(x) = -\infty \qquad \lim_{x\to \infty}f(x) = \infty + . + We do not have any horizontal asymptotes. +

      +
      6. + +
    6. + +

      + We place the values x=(10\pm\sqrt{37})/9 and x=10/9 on a number line, + as shown in . + We mark each subinterval as increasing or decreasing, concave up or down, + using the techniques used in Sections. +

      +
      7. + +
    7. +

      + Evaluate f at each critical number and possible inflection point. + + f(0.435)\amp\approx6.400\amp f(1.111)\amp\approx4.547\amp f(1.787)\amp\approx 2.695 + +

      + +

      + We plot the appropriate points on axes as shown in + and connect the points with straight lines + (to show increasing/decreasig behavior). + In + we adjust these lines to demonstrate the proper concavity. + In + we show a graph of f drawn with a computer program, + verifying the accuracy of our sketch. +

      +
      8. +
    +

    + +
    + Number line for f in + + + Number line showing intervals where f is increasing/decreasing and concave up/down + +

    + A number line is shown, on which three points are marked. + The first point is marked as \frac19(10-\sqrt{37}), or approximately 0.435; + it is a critical point of f, and a relative maximum. +

    + +

    + The second point is marked as \frac{10}{9}, or approximately 1.111; + it is an inflection point of f. +

    + +

    + The last point is marked as \frac19(10+\sqrt{37}), or approximately 1.787; + it is a critical point of f, and a relative minimum. +

    + +

    + These points divide the number line into four intervals. + Above each interval, the signs of both \fp and \fpp are given, + along with whether f is increasing or decreasing, and concave up or down. +

    + +

    + This information is as follows: +

      +
    • +

      + For x\lt 0.435, \fp\gt 0, so f is increasing, + and \fpp\lt 0, so f is concave down. +

      +
      • + +
    • +

      + For 0.435\lt x\lt 1.111, \fp\lt 0, so f is decreasing, + and \fpp\lt 0, so f is concave down. +

      +
      • + +
    • +

      + For 1.111\lt x\lt 1.787, \fp\lt 0, so f is decreasing, + and \fpp\gt 0, so f is concave up. +

      +
      • + +
    • +

      + For x\gt 1.787, \ft\gt 0, so f is increasing, + and \fpp\gt 0, so f is concave up. +

      +
      • +
    +

    +     + + \begin{tikzpicture} + \begin{axis}[numberline, + xmin=0,xmax=2.222, + xtick={10}, + minor xtick={{}}, + extra x ticks={0.435, 1.111, 1.787}, + extra x tick labels={\parbox{6em}{\centering \scriptsize $\frac{1}{9}\left(10-\sqrt{37}\right)$ $\approx0.435$}, + \parbox{2em}{\scriptsize $\frac{10}{9}\approx1.111$}, \parbox{6em}{\centering \scriptsize $\frac{1}{9}\left(10+\sqrt{37}\right)$ $\approx1.787$}},] + \addplot[guideline] coordinates {(0.435,0) (0.435,2)}; + \addplot[guideline] coordinates {(1.111,0) (1.111,2)}; + \addplot[guideline] coordinates {(1.787,0) (1.787,2)}; + \addplot[mark=none] coordinates {(0.18,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\gt 0$,\\ $f$ incr\\$\fpp\lt 0$,\\ $f$ c.\ down }}; + \addplot[mark=none] coordinates {(0.77,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\lt 0$,\\ $f$ decr\\$\fpp\lt 0$,\\ $f$ c.\ down }}; + \addplot[mark=none] coordinates {(1.45,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\lt 0$,\\ $f$ decr\\$\fpp\gt 0$,\\ $f$ c.\ up }}; + \addplot[mark=none] coordinates {(2.1,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\gt 0$,\\ $f$ incr\\$\fpp\gt 0$,\\ $f$ c.\ up }}; + \end{axis} + \end{tikzpicture} + + +
    + +
    + Sketching f in + +
    + + + + A hand-drawn graph based on the critical points from the number line. + +

    + The graph is hand-drawn with plotted significant points from the number line. + It connects these points with straight lines to give a general impression of the graph's shape. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ymin=-5.5,ymax=10.9,xmin=-1.2,xmax=3.2,] + \addplot[firstcurvestyle,leftarrow,domain=-1:0.435] {(5-6.4)/(0-0.435)*(x-0.435)+6.4}; + \addplot[firstcurvestyle] coordinates {(0.435,6.4) (1.79,2.69)}; + \addplot[firstcurvestyle,rightarrow,domain=1.79:2.7] {(5-6.4)/(0-0.435)*(x-1.79)+2.69}; + \addplot[soliddot] coordinates{(0.435,6.4) (1.11,4.55) (1.79,2.69)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + Adjusted graph for proper concavity. + +

    + The image shows an adjusted graph of the piecewise linear function. + The function is now a smooth, continuous curve that crosses the y-axis at y=5. + This adjustment demonstrates the proper concavity of the function. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ymin=-5.5,ymax=10.9,xmin=-1.2,xmax=3.2,] + \addplot[firstcurvestyle,leftarrow,domain=-0.8:0.435] {(x-0.435)^2/(0-0.435)^2*(5-6.4) + 6.4}; + \addplot[firstcurvestyle,domain=0.435:1.79] {(6.4-2.69)*6/(1.79-0.435)^3*((x-0.435)*(x-1.79)^2/2 - (x-1.79)^3/6)+2.69}; + \addplot[firstcurvestyle,rightarrow,domain=1.79:2.8] {-(x-1.79)^2/(0-0.435)^2*(5-6.4) + 2.69}; + \addplot[soliddot] coordinates{(0.435,6.4) (1.11,4.55) (1.79,2.69)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + Computer-generated graph for accuracy verification. + +

    + The image shows a computer-generated graph of the function. The graph verifies the accuracy of our sketch: + there is very little difference between this graph and the previous graph, + showing that accounting for concavity helps to ensure an accurate sketch. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ymin=-5.5,ymax=10.9,xmin=-1.2,xmax=3.2,] + \addplot [firstcurvestyle,infinite,domain=-.6:2.8] {3*x^3-10*x^2+7*x+5}; + \addplot[soliddot] coordinates{(0.435,6.4) (1.11,4.55) (1.79,2.69)}; + \end{axis} + \end{tikzpicture} + + + + + +
    +     +
    +     + +     + + + Curve sketching + +

    + Sketch \ds f(x) = \frac{x^2-x-2}{x^2-x-6}. +

    +     + +

    + We again follow the steps outlined in . +

    + +

    +

      +
    1. +

      + In determining the domain, + we assume it is all real numbers and look for restrictions. + We find that at x=-2 and x=3, + f(x) is not defined. + So the domain of f is D = \{ x \mid x\neq -2,3\}. +

      +
      2. + +
    2. +

      + To find the critical values of f, + we first find \fp(x). + Using the , we find + + \fp(x) = \frac{-8x+4}{(x^2+x-6)^2} = \frac{-8x+4}{(x-3)^2(x+2)^2} + . + We get \fp(x) = 0 when x = 1/2, + and \fp is undefined when x=-2,3. + Since \fpis undefined only when f is also undefined, + these are not critical values. + The only critical value is x=1/2. +

      +
      3. + +
    3. +

      + To find the possible points of inflection, we find \fpp(x), + again employing the : + + \fpp(x) = \frac{24x^2-24x+56}{(x-3)^3(x+2)^3} + . + We find that \fpp(x) is never 0 + (setting the numerator equal to 0 and solving for x, + we find the only roots to this quadratic are not real numbers) + and \fppis undefined when x=-2,3. + Thus concavity will possibly only change at x=-2 and x=3 + (which are not in the domain of f, + so these won't be inflection points). +

      +
      4. + +
    4. +

      + The vertical asymptotes of f are at x=-2 and x=3, + the places where f is undefined. +

      +
      5. + +
    5. +

      + There is a horizontal asymptote of y=1, + as \lim\limits_{x\to -\infty}f(x) = 1 and \lim\limits_{x\to\infty}f(x) =1. +

      +
      6. + +
    6. +

      + We place the values x=1/2, + x=-2 and x=3 on a number line as shown in . + We mark in each interval whether f is increasing or decreasing, + concave up or down. + We see that f has a relative maximum at x=1/2; + concavity changes only at the vertical asymptotes. +

      +
      7. + +
    7. +

      + Evaluate f at each critical number. + + f(0)\amp=1/3\amp f(1/2)\amp=9/25 + +

      + +

      + In , + we plot the points from the number line on a set of axes and connect the points with straight lines to get a general idea of what the function looks like + (these lines effectively only convey increasing/decreasing information). + In , + we adjust the graph with the appropriate concavity. + We also show f crossing the x-axis at x=-1 and x=2 and crossing the y-axis at y=1/3. + Finally, + shows a computer generated graph of f, + which verifies the accuracy of our sketch. +

      +
      8. +
    +

    + +
    + Number line for f in + + Number line showing intervals where f is increasing/decreasing and concave up/down + +

    + On a number line there are three marked points: -2, \frac12, and 3. + At x=-2 and x=3, the graph of f has a vertical asymptote. + The point x=\frac12 is a critical point of f, and a relative maximum. +

    + +

    + These three points divide the number line into four intervals. + Above each interval, the following information is indicated: +

      +
    • +

      + For x\lt -2, \fp\gt 0, so f is increasing, + and \fpp\gt 0, so f is concave up. +

      +
      • + +
    • +

      + For -2\lt x\lt \frac12, \fp\gt 0, so f is increasing, + and \fpp\lt 0, so f is concave down. +

      +
      • + +
    • +

      + For \frac12\lt x\lt 3, \fp\lt 0, so f is decreasing, + and \fpp\lt 0, so f is concave down. +

      +
      • + +
    • +

      + For x\gt 3, \fp\lt 0, so f is decreasing, + and \fpp\gt 0, so f is concave up. +

      +
      • +
    +

    +     + + \begin{tikzpicture} + \begin{axis}[numberline, + xmin=0,xmax=2.222, + minor xtick={{}}, + xtick={10}, + extra x ticks={0.435, 1.111, 1.787}, + extra x tick labels={\scriptsize $-2$, \scriptsize $\frac{1}{2}$, \scriptsize $3$},] + \addplot[guideline] coordinates {(0.435,0) (0.435,2)}; + \addplot[guideline] coordinates {(1.111,0) (1.111,2)}; + \addplot[guideline] coordinates {(1.787,0) (1.787,2)}; + \addplot[mark=none] coordinates {(0.18,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\gt 0$,\\ $f$ incr\\$\fpp\gt 0$,\\ $f$ c.\ up }}; + \addplot[mark=none] coordinates {(0.77,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\gt 0$,\\ $f$ incr\\$\fpp\lt 0$,\\ $f$ c.\ down }}; + \addplot[mark=none] coordinates {(1.45,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\lt 0$,\\ $f$ decr\\$\fpp\lt 0$,\\ $f$ c.\ down }}; + \addplot[mark=none] coordinates {(2.1,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\lt 0$,\\ $f$ incr\\$\fpp\lt 0$,\\ $f$ c.\ down }}; + \end{axis} + \end{tikzpicture} + + +
    + +
    + Sketching f in + +
    + + + + A linear piecewise graph showing changes in concavity. + +

    + The figure shows a linear piecewise graph on a Cartesian plane with the x-axis ranging from -5 to 5, + and the y-axis from -5 to 5. + The graph consists of two line segments forming a V shape, with the tip intersecting the x-axis at x = 0.5. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ymin=-5.2,ymax=5.2,xmin=-5.1,xmax=5.1,] + \addplot[firstcurvestyle,domain=-4.5:-2.5] {(x+3)+3}; + \addplot[firstcurvestyle,leftarrow,domain=-1.7:0.5] {(0-0.36)/(-1-0.5)*(x-0.5)+0.36}; + \addplot[firstcurvestyle,rightarrow,domain=0.5:2.7] {(0-0.36)/(2-0.5)*(x-0.5)+0.36}; + \addplot[firstcurvestyle,domain=3.5:5] {-(x-4)+3}; + \addplot[asymptote] coordinates {(-2,-5) (-2,5)}; + \addplot[asymptote] coordinates {(3,-5) (3,5)}; + \addplot[asymptote] {1}; + \addplot[soliddot] coordinates{(0.5,0.36)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + A concave adjusted graph. + +

    + The image displays an adjusted version of the piecewise graph, now with concavity. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ymin=-5.2,ymax=5.2,xmin=-5.1,xmax=5.1,] + \addplot[firstcurvestyle,domain=-5.1:-2.3] {-1/(x+2)+1}; + \addplot[firstcurvestyle,domain=-1.8:2.8] {(x^2-x-2)/(x^2-x-6)}; + \addplot[firstcurvestyle,domain=3.3:5.1] {1/(x-3)+1}; + \addplot[asymptote] coordinates {(-2,-5) (-2,5)}; + \addplot[asymptote] coordinates {(3,-5) (3,5)}; + \addplot[asymptote] {1}; + \addplot[soliddot] coordinates{(0.5,0.36) (-1,0) (2,0)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + Computer generated graph for accuracy verification. + +

    + The image shows a computer-generated graph of the function. The graph verifies the accuracy of our sketch: + there is very little difference between this graph and the previous graph, + showing that accounting for concavity helps to ensure an accurate sketch. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ymin=-5.5,ymax=5.5,xmin=-5.2,xmax=5.2,] + \addplot [firstcurvestyle,infinite,domain=-5:-2.1] {((x-2)*(x+1))/((x-3)*(x+2))}; + \addplot [firstcurvestyle,infinite,domain=-1.9:2.9] {((x-2)*(x+1))/((x-3)*(x+2))}; + \addplot [firstcurvestyle,infinite,domain=3.1:5] {((x-2)*(x+1))/((x-3)*(x+2))}; + \addplot [asymptote] coordinates {(-2,5.5) (-2,-5.5)}; + \addplot [asymptote] coordinates {(3,5.5) (3,-5.5)}; + \addplot [asymptote] coordinates {(-5.5,1) (5.5,1)}; + \addplot [soliddot] coordinates{(0.5,.36)}; + \end{axis} + \end{tikzpicture} + + + + + + +
    +     +
    +     + +     + + + Curve sketching + +

    + Sketch f(x) = \dfrac{5(x-2)(x+1)}{x^2+2x+4}. +

    +     + +

    + We again follow . +

    + +

    +

      +
    1. +

      + We assume that the domain of f is all real numbers and consider restrictions. + The only restrictions could come when the denominator is 0, + but this never occurs because the denominator is a quadratic polynomial with no real roots. + Therefore the domain of f is all real numbers, \mathbb{R}. +

      +
      2. + +
    2. +

      + We find the critical values of f by setting + \fp(x)=0 and solving for x. + We find + + \fp(x) \amp = \frac{15x(x+4)}{(x^2+2x+4)^2} + 0 \amp =\frac{15x(x+4)}{(x^2+2x+4)^2} + x \amp =-4,0 + . + Since the denominator of \fp is just the square of the denominator of f, + there are no values of x for which \fp is undefined. +

      +
      3. + +
    3. +

      + We find the possible points of inflection by solving \fpp(x) = 0 for x (again, + there are no values of x for which \fpp is undefined.) We find + + \fpp(x) = -\frac{30x^3+180x^2-240}{(x^2+2x+4)^3} + . + The cubic in the numerator does not factor very nicely. + We instead approximate the roots + (using a CAS) + at x= -5.759, x=-1.305 and x=1.064. +

      +
      4. + +
    4. +

      + There are no vertical asymptotes as the denominator never equals zero. +

      +
      5. + +
    5. +

      + We have a horizontal asymptote of y=5, + as \lim\limits_{x\to-\infty}f(x) = \lim\limits_{x\to\infty}f(x) = 5. +

      +
      6. + +
    6. +

      + We place the critical points and possible points on a number line as shown in + and mark each interval as increasing/decreasing, + concave up/down appropriately. +

      +
      7. + +
    7. +

      + Evaluate f at each critical number, + possible inflection point. + + f(-5.759)\amp\approx7.200\amp f(-4)\amp=7.5 + f(-1.305)\amp\approx1.630\amp f(0)\amp=2.5 + f(1.064)\amp\approx-1.331 + +

      + +

      + In + we plot the significant points from the number line as well as the x- and y-intercepts, + and connect the points with straight lines to get a general impression about the graph + (this graph only includes increasing/decreasing information). + In , + we add concavity, drawing the function so that it is smooth + (since f is differentiable everywhere, + there should be no kinks or corners). + + shows a computer generated graph of f, affirming our results. +

      +
      8. +
    +

    + +
    + Number line for f in Example + + Number line showing intervals where f is increasing/decreasing and concave up/down + +

    + A number line is shown, on which five points are marked. + The points are labeled, from left to right, as -5.579, + -4, -1.305, 0, and 1.064. + The points x=-4 and x=0 are the critical points of f. +

    + +

    + The three decimal values are the possible inflection points of f, + which had to be approximated using software. + These points divide the number line into six intervals; + above each interval, the following information is indicated: +

    + +

    +

      +
    • +

      + For x\lt -5.579, \fp\gt 0, so f is increasing, + and \fpp \gt 0, so f is concave up. +

      +
      • + +
    • +

      + For -5.579\lt x\lt -4, \fp\gt 0, so f is increasing, + and \fpp \lt 0, so f is concave down. +

      +
      • + +
    • +

      + For -4\lt x\lt -1.305, \fp\lt 0, so f is decreasing, + and \fpp \lt 0, so f is concave down. +

      +
      • + +
    • +

      + For -1.305\lt x\lt 0, \fp\lt 0, so f is decreasing, + and \fpp \gt 0, so f is concave up. +

      +
      • + +
    • +

      + For 0\lt x\lt 1.064, \fp\gt 0, so f is increasing, + and \fpp \gt 0, so f is concave up. +

      +
      • + +
    • +

      + For x\gt 1.064, \fp\gt 0, so f is increasing, + and \fpp \lt 0, so f is concave down. +

      +
      • +
    +

    +     + + \begin{tikzpicture} + \begin{axis}[axis y line=none, + x=18pt, + y=20pt, + xtick={10}, + xlabel={}, + ymax=2.5, + xmin=-7,xmax=6, + minor xtick={{}}, + extra x ticks={-5,-2.8,-0.6,1.3,3.2}, + extra x tick labels={\scriptsize $-5.579$,\scriptsize $-4$,\scriptsize $-1.305$,\scriptsize $0$,\scriptsize $1.064$},] + \addplot[guideline] coordinates {(-5,0) (-5,2)}; + \addplot[guideline] coordinates {(-2.8,0) (-2.8,2)}; + \addplot[guideline] coordinates {(-0.6,0) (-0.6,2)}; + \addplot[guideline] coordinates {(1.3,0) (1.3,2)}; + \addplot[guideline] coordinates {(3.2,0) (3.2,2)}; + \addplot[mark=none] coordinates {(-6,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\gt0$,\\ $f$ incr\\$\fpp\gt0$,\\ $f$ c.\ up }}; + \addplot[mark=none] coordinates {(-3.9,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\gt0$,\\ $f$ incr\\$\fpp\lt0$,\\ $f$ c.\ down }}; + \addplot[mark=none] coordinates {(-1.7,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\lt0$,\\ $f$ decr\\$\fpp\lt0$,\\ $f$ c.\ down }}; + \addplot[mark=none] coordinates {(0.4,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\lt0$,\\ $f$ decr\\$\fpp\gt0$,\\ $f$ c.\ up }}; + \addplot[mark=none] coordinates {(2.3,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\gt0$,\\ $f$ incr\\$\fpp\gt0$,\\ $f$ c.\ up }}; + \addplot[mark=none] coordinates {(4.4,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\gt0$,\\ $f$ incr\\$\fpp\lt0$,\\ $f$ c.\ down }}; + \end{axis} + \end{tikzpicture} + + +
    + + +
    + Sketching f in + +
    + + + + A hand-drawn graph based on the critical points from the number line. + +

    + The graph is hand-drawn with plotted significant points from the number line. + It connects these points with straight lines to give a general impression of the graph's shape. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ ymin=-3.1,ymax=8.1,xmin=-7.5,xmax=6.5,] + \addplot[firstcurvestyle,leftarrow,domain=-7:-4] {(7.2-7.5)/(-5.579+4)*(x+4)+7.5}; + \addplot[firstcurvestyle,domain=-4:-1.305] {(1.630-7.5)/(-1.305+4)*(x+4)+7.5}; + \addplot[firstcurvestyle,domain=-1.305:-1] {(0-1.630)/(-1+1.305)*(x+1)}; + \addplot[firstcurvestyle,domain=-1:0] {(-2.5-0)/(0+1)*(x+1)}; + \addplot[firstcurvestyle,domain=0:1.064] {(-1.331+2.5)/(1.064-0)*(x-1.064)-1.331}; + \addplot[firstcurvestyle,rightarrow,domain=1.064:5] {(0+1.331)/(2-1.064)*(x-2)}; + \addplot[asymptote,domain=-7.5:6.5] {5}; + \addplot[soliddot] coordinates{(0,-2.5) (-1,0) (2,0) (-5.579,7.2) (-1.305,1.630) (1.064,-1.331) (-4,7.5)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + A hand-drawn graph with concavity. + +

    + The graph is hand-drawn with plotted significant points from the number line. + It connects these points with smooth curves to give a general impression of the graph's shape. +

    +     + + + + \begin{tikzpicture} + \begin{axis}[ ymin=-3.1,ymax=8.1,xmin=-7.5,xmax=6.5,] + \addplot[firstcurvestyle,leftarrow,domain=-7.5:-4] {7.5-0.3/(exp(-1/2/(-5.579+4)^2*(-5.579+4)^2) - 1)*(-1+exp(-1/2/(-5.579+4)^2*(x+4)^2))}; + \addplot[firstcurvestyle,domain=-4:-1.305] {(1.630-7.5)/(-1.305+4)^2*(x+4)^2+7.5}; + \addplot[firstcurvestyle,domain=-1.305:-1] {(0-1.630)/(-1+1.305)*(x+1)+0}; + \addplot[firstcurvestyle,domain=-1:0] {2.5/(-1)^2*x^2-2.5}; + \addplot[firstcurvestyle,domain=0:1.064] {(-1.331+2.5)/(1.064)^2*x^2-2.5}; + \addplot[firstcurvestyle,rightarrow,domain=1.064:6] {4*(-1.331+2.5)/1.064 * sqrt(1.064+((1.331*1.064/4/(-1.331+2.5)+1.064)^2 - 1.064*2) / (1.064+2 - 2*(1.331*1.064/4/(-1.331+2.5)+1.064)))*sqrt(x+((1.331*1.064/4/(-1.331+2.5)+1.064)^2 - 1.064*2) / (1.064+2 - 2*(1.331*1.064/4/(-1.331+2.5)+1.064)))-(4*(-1.331+2.5)/1.064 * sqrt(1.064+((1.331*1.064/4/(-1.331+2.5)+1.064)^2 - 1.064*2) / (1.064+2 - 2*(1.331*1.064/4/(-1.331+2.5)+1.064))))*sqrt(2+((1.331*1.064/4/(-1.331+2.5)+1.064)^2 - 1.064*2) / (1.064+2 - 2*(1.331*1.064/4/(-1.331+2.5)+1.064)))}; + \addplot[asymptote,domain=-7.5:6.5] {5}; + \addplot[soliddot] coordinates{(0,-2.5) (-1,0) (2,0) (-5.579,7.2) (-1.305,1.630) (1.064,-1.331) (-4,7.5)}; + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + + A computer generated graph of the function. + + +

    + The graph is a computer generated graph of the function, + showing the same critical points as the hand-drawn graph. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ymin=-3.1,ymax=8.1,xmin=-7.5,xmax=6.5,] + \addplot [asymptote] coordinates {(-7.5,5) (6.5,5)}; + \addplot [firstcurvestyle,infinite,domain=-7.5:6.5] {(5*(x-2)*(x+1))/(x^2+2*x+4)}; + \addplot [soliddot] coordinates {(-5.75,7.2) (-4., 7.5) (-1.31, 1.63) (-1,0) (0,-2.5) (1.064, -1.33) (2,0)}; + \end{axis} + \end{tikzpicture} + + + + + + +
    +     +
    +     +     + + + + + + + + + +

    + In each of our examples, + we found a few significant points on the graph of f that corresponded to changes in increasing/decreasing or concavity. + We connected these points with straight lines, + then adjusted for concavity, + and finished by showing a very accurate, computer generated graph. +

    + + + +

    + Why are computer graphics so good? + It is not because computers are + smarter than we are. + Rather, it is largely because computers are much faster at computing than we are. + In general, computers graph functions much like most students do when first learning to draw graphs: + they plot equally spaced points, + then connect the dots using lines. + By using lots of points, + the connecting lines are short and the graph looks smooth. +

    + +

    + This does a fine job of graphing in most cases + (in fact, this is the method used for many graphs in this text). + However, in regions where the graph is very curvy, + this can generate noticeable sharp edges on the graph unless a large number of points are used. + High quality computer algebra systems, + such as Mathematica and Sage, + use special algorithms to plot lots of points only where the graph is curvy. +

    + +

    + In , + two graph of y=\sin(x) is given, + generated by Sage and Mathematica. + The small points represent each of the places where each CAS sampled the function. + Notice how at the bends + of \sin(x), lots of points are used; + where \sin(x) is relatively straight, fewer points are used. + (In the Mathematica plot, + many points are also used at the endpoints to ensure the + end behavior is accurate.) +

    + +
    + CAS plots of y=\sin(x) illustrating the sample points + +
    + Sage output + + A graph of the sine function generated by Sage. + +

    + The plot features a solid blue curve representing the sine wave. + The x-axis is labeled from 0 to 6, + and the y-axis ranges from -1 to 1. + Sample points are marked along the curve with blue dots, + indicating the places where the function was sampled. +

    +     + + + + + +
    + +
    + Mathematica output + + Mathematica plot of the sine graph + +

    + The same plot of the sine function, showing the location of the sample points. + This version comes from the software Mathematica. + In addition to more frequent sampling in areas with more curvature, + Mathematica also uses extra sample points at the ends of the domain. +

    +     + +
    +     +
    + +

    + How does Sage know where the graph is curvy? + Calculus. + When we study curvature in a later chapter, + we will see how the first and second derivatives of a function work together to provide a measurement of curviness. + Sage employs algorithms to determine regions of high curvature + and plots extra points there. +

    + +

    + Again, the goal of this section is not + How to graph a function when there is no computer to help. Rather, + the goal is Understand that the shape of the graph of a function is largely determined by understanding the behavior of the function at a few key places. + In , + we were able to accurately sketch a complicated graph using only five points and knowledge of asymptotes! +

    + +

    + There are many applications of our understanding of derivatives beyond curve sketching. + The next chapter explores some of these applications, + demonstrating just a few kinds of problems that can be solved with a basic knowledge of differentiation. +

    + + + + Terms and Concepts + + + + + +

    + Why is sketching curves by hand beneficial even though technology is ubiquitous? +

    + + +     + + + +     + + + + + + + + +

    + When sketching graphs of functions, + it is useful to find the critical points. + +

    +     + +     + + + + + +

    + When sketching graphs of functions, + it is useful to find the possible points of inflection. + +

    +     + +     + + + + + +

    + When sketching graphs of functions, + it is useful to find the horizontal and vertical asymptotes. + +

    +     + +     + +     + + + Problems + + + +

    + Practice using + by applying the principles to the given functions with familiar graphs. +

    +     + + + +

    + Use + to sketch a graph of f(x) = 2x+4 +

    +     + +

    + A good sketch will include the x and y intercepts and draw the appropriate line. +

    +     +     + + + +

    + Use + to sketch a graph of f(x) = -x^2+1 +

    +     + +

    + A good sketch will include the x and y intercepts.. +

    +     +     + + + +

    + Use + to sketch a graph of f(x) = \sin(x) +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of f(x) = e^x +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = \frac{1}{x} +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = \frac{1}{x^2} +

    +     + +

    + Use technology to verify sketch. +

    +     +     + +     + + + +

    + Sketch a graph of the given function using . + Show all work; check your answer with technology. +

    +     + + + +

    + Use + to sketch a graph of \ds f(x) = x^3-2x^2+4x+1 +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = -x^3+5x^2-3x+2 +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = x^3+3x^2+3x+1 +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = x^3-x^2-x+1 +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = (x-2)\ln(x-2) +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = (x-2)^2\ln(x-2) +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = \frac{x^2-4}{x^2} +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = \frac{x^2-4x+3}{x^2-6x+8} +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = \frac{x^2-2x+1}{x^2-6x+8} +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = x\sqrt{x+1} +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = x^2e^x +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = \sin(x) \cos(x) on [-\pi,\pi] +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = (x-3)^{2/3} + 2 +

    +     + +

    + Use technology to verify sketch. +

    +     +     + + + +

    + Use + to sketch a graph of \ds f(x) = \frac{(x-1)^{2/3}}{x} +

    +     + +

    + Use technology to verify sketch. +

    +     +     + +     + + + +

    + A function with the parameters a and b are given. + Describe the critical points and possible points of inflection of f in terms of a and b. +

    +     + + + + +

    + \ds f(x) = \frac{a}{x^2+b^2} +

    + +

    +

      +
    1. +

      + Find the critical points of f. +

      + +
      2. + +
    2. +

      + Find the inflection points of f. +

      + +
      3. +
    +

    +     + +

    + Critical point: x=0 + Points of inflection: \pm b/\sqrt{3} +

    +     + +     + + + + +

    + \ds f(x) = \sin(ax+b) +

    + +

    +

      +
    1. +

      + Find the critical points of f. +

      + +
      2. + +
    2. +

      + Find the inflection points of f. +

      + +
      3. +
    +

    +     + +

    + Critical points: x=\frac{n\pi/2-b}{a}, + where n is an odd integer + Points of inflection: (n\pi-b)/a, + where n is an integer. +

    +     + +     + + + + +

    + \ds f(x) = (x-a)(x-b) +

    + +

    +

      +
    1. +

      + Find the critical points of f. +

      + +
      2. + +
    2. +

      + Find the inflection points of f. +

      + +
      3. +
    +

    +     + +

    + Critical point: x=(a+b)/2 + Points of inflection: none +

    +     + +     + + + + +

    + Given x^2+y^2=1, use implicit differentiation to find + \frac{dy}{dx} and \frac{d^2y}{dx^2}. + Use this information to justify the sketch of the unit circle. +

    +     + +

    + \frac{dy}{dx} = -x/y, + so the function is increasing in second and fourth quadrants, + decreasing in the first and third quadrants. +

    + +

    + \frac{d^2y}{dx^2} = -1/y - x^2/y^3, + which is positive when y\lt 0 and is negative when y\gt0. + Hence the function is concave down in the first and second quadrants and concave up in the third and fourth quadrants. +

    +     + +     + +     +     +     +
    + + +     + + + Applications of the Derivative + +

    + In , + we learned how the first and second derivatives of a function influence its graph. + In this chapter we explore other applications of the derivative. +

    +     + +
    + Newton's Method +

    + Solving equations is one of the most important things we do in mathematics, + yet we are surprisingly limited in what we can solve analytically. + For instance, equations as simple as x^5+x+1=0 or + \cos(x) =x cannot be solved by algebraic methods in terms of familiar functions. + Fortunately, + there are methods that can give us approximate + solutions to equations like these. + These methods can usually give an approximation correct to as many decimal places as we like. + In + we learned about the Bisection Method. + This section focuses on another technique + (which generally works faster), + called Newton's Method. +

    + +

    + Newton's Method is built around tangent lines. + The main idea is that if x is sufficiently close to a root of f(x), + then the tangent line to the graph at + (x,f(x)) will cross the x-axis at a point closer to the root than x. +

    + +

    + We start Newton's Method with an initial guess about roughly where the root is. + Call this x_0. + (See .) + Draw the tangent line to the graph at + (x_0,f(x_0)) and see where it meets the x-axis. + Call this point x_1. + Then repeat the process draw the tangent line to the graph at + (x_1, f(x_1)) and see where it meets the x-axis. + (See .) + Call this point x_2. + Repeat the process again to get x_3, x_4, etc. + This sequence of points will often converge rather quickly to a root of f. +

    + +
    + Demonstrating the geometric concept behind Newton's Method + +
    + + + + A tangent line is drawn to the graph at (x_0,f(x_0)) and and meets on the x-axis at point x_1. + + +

    + This graph depicts the iterations of Newton's method for finding the root of a function. + The y axis represents function values (1,0.5,0,-0.5,-1), + while the x axis represents iterations (x_0, x_1). + The graph shows the initial guess x_0 and the subsequent approximation x_1. +

    +

    + A tangent line is drawn from the point (x_0,f(x_0)) to intersect the x axis at x_1, + providing a refined approximation of the function’s root. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xtick=\empty,% + extra x ticks={.45,2.52}, + extra x tick labels={$x_0$,$x_1$}, + ymin=-1.1,ymax=1.1, + xmin=-.3,xmax=2.7 + ] + \addplot+ [infinite,domain=-.3:2.7] {cos(deg(x))}; + \addplot [tangentlineseg,domain=.45:2.52] {-0.434*(x-.45)+.9}; + \addplot [guideline] coordinates {(0.45,0) (0.45,0.9)}; + \addplot [soliddot] coordinates {(0.45,0.9)}; + \end{axis} + \end{tikzpicture} + + +
    + +
    + + + A tangent line is drawn to the graph at (x_1,f(x_1)) and and meets on the x-axis at point x_2. + +

    + The second graph has values (1, 0.5, 0, -0.5, -1) on the y axis,while the x axis represents iterations (x_0, x_2, x_1). + A tangent line is drawn from the point (x_1) to intersect the x axis and the new point is called x_2. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xtick=\empty, + extra x ticks={.5,2.52,1.12}, + extra x tick labels={$x_0$,$x_1$,$x_2$}, + ymin=-1.1,ymax=1.1,% + xmin=-.3,xmax=2.7% + ] + \addplot+ [infinite,domain=-.3:2.7] {cos(deg(x))}; + \addplot [tangentlineseg,domain=1.12:2.52] {-0.58*(x-2.52)-.813}; + \addplot [guideline] coordinates {(2.52,0) (2.52,-0.813)}; + \addplot [soliddot] coordinates {(2.52,-.813)}; + \end{axis} + \end{tikzpicture} + + +
    + +
    + + + A tangent line is drawn to the graph at (x_2,f(x_2)) and and meets on the x-axis at point x_3. + +

    + On the third graph, the y axis has values (1,0.5, 0, -0.5, -1), + while the x axis represents iterations (x_0,x_1,x_2,x_3). + A tangent line is drawn from the point (x_2) to intersect the x-axis and the new point is called x_3. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xtick=\empty,% + extra x ticks={.5,2.52,1.12,1.6}, + extra x tick labels={$x_0$,$x_1$,$x_2$,$x_3$}, + ymin=-1.1,ymax=1.1,% + xmin=-.3,xmax=2.7% + ] + \addplot+ [infinite,domain=-.3:2.7] {cos(deg(x))}; + \addplot [tangentlineseg,domain=1.12:1.6] {-.9*(x-1.12)+.432}; + \addplot [guideline] coordinates {(1.12,0) (1.12,0.432)}; + \addplot [soliddot] coordinates {(1.12,.432)}; + \end{axis} + \end{tikzpicture} + + +
    +     +
    + + + +

    + We can use this geometric + process to create an algebraic process. + Let's look at how we found x_1. + We started with the tangent line to the graph at (x_0,f(x_0)). + The slope of this tangent line is + \fp(x_0) and the equation of the line is + + y=\fp(x_0)(x-x_0)+f(x_0) + . +

    + +

    + This line crosses the x-axis when y=0, + and the x-value where it crosses is what we called x_1. + So let y=0 and replace x with x_1, + giving the equation: + + 0 = \fp(x_0)(x_1-x_0)+f(x_0) + . +

    + +

    + Now solve for x_1: + + x_1=x_0-\frac{f(x_0)}{\fp(x_0)} + . +

    + +

    + Since we repeat the same geometric process to find x_2 from x_1, we have + + x_2=x_1-\frac{f(x_1)}{\fp(x_1)} + . +

    + +

    + In general, given an approximation x_n, + we can find the next approximation, + x_{n+1} as follows: + + x_{n+1} = x_{n} - \frac{f(x_{n})}{\fp(x_{n})} + . +

    + +

    + We summarize this process as follows. +

    + + + Newton's Method +

    + Let f be a differentiable function on an interval I with a root in I. + To approximate the value of the root, + accurate to d decimal places: + Newton's Method +

    + +

    +

      +
    1. +

      + Choose a value x_0 as an initial approximation of the root. + (This is often done by looking at a graph of f.) +

      +
      2. + +
    2. +

      + Create successive approximations iteratively; + given an approximation x_n, + compute the next approximation x_{n+1} as + + x_{n+1} = x_n - \frac{f(x_n)}{\fp(x_n)} + . +

      +
      3. + +
    3. +

      + Stop the iterations when successive approximations do not differ in the first d places after the decimal point. +

      +
      4. +
    +

    +     + + + +

    + Let's practice Newton's Method with a concrete example. +

    + + + Using Newton's Method + +

    + Approximate the real root of x^3-x^2-1=0, + accurate to the first three places after the decimal, + using Newton's Method and an initial approximation of x_0=1. +

    +     + +

    + To begin, we compute \fp(x)=3x^2-2x. + Then we apply the Newton's Method algorithm, + outlined in . +

      +
    • +

      + \begin{aligned} + x_1\amp=1-\frac{f(1)}{\fp(1)}\\ + \amp=1-\frac{1^3-1^2-1}{3\cdot 1^2-2\cdot 1}\\ + \amp =2\end{aligned} + +

      +
      • +
    • +

      + \begin{aligned} + x_2\amp =2-\frac{f(2)}{\fp(2)}\\ + \amp =2-\frac{2^3-2^2-1}{3\cdot 2^2-2\cdot 2}\\ + \amp=1.625\end{aligned} + +

      +
      • +
    • +

      + \begin{aligned} + x_3\amp =1.625-\frac{f(1.625)}{\fp(1.625)}\\ + \amp = 1.625-\frac{1.625^3-1.625^2-1}{3\cdot 1.625^2-2\cdot 1.625}\\ + \amp \approx 1.48579\end{aligned} + +

      +
      • +
    • +

      + \begin{aligned} + x_4 \amp =1.48579-\frac{f(1.48579)}{\fp(1.48579)}\\ + \amp \approx 1.46596\end{aligned} + +

      +
      • +
    • +

      + \begin{aligned} + x_5 \amp = 1.46596 - \frac{f(1.46596)}{\fp(1.46596)}\\ + \amp \approx 1.46557\end{aligned} + +

      +
      • +
    +

    + +

    + We performed five iterations of Newton's Method to find a root accurate to the first three places after the decimal; + our final approximation is 1.465. + The exact value of the root, to six decimal places, + is 1.465571; It turns out that our x_5 is accurate to more than just three decimal places. +

    + +

    + A graph of f(x) is given in . + We can see from the graph that our initial approximation of x_0=1 was not particularly accurate; + a closer guess would have been x_0=1.5. + Our choice was based on ease of initial calculation, + and shows that Newton's Method can be robust enough that we do not have to make a very accurate initial approximation. +

    + +
    + A graph of f(x) = x^3-x^2-1 in + + + A cubic graph features a curve with two turning points. + + +

    + The y axis represents function values (0.5, 0, -0.5, -1,-1.5), while the x axis represents values of (0,0.5,1,1.5). + The graph intersects the y axis at -1 and 1.5 on x axis and forms a curve. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.5,ymax=.6,% + xmin=-0.1,xmax=1.6% + ] + \addplot+ [infinite,domain=-.05:1.58] {x^3-x^2-1}; + \end{axis} + \end{tikzpicture} + + +
    +     +     + +

    + We can automate this process on a calculator that has an ANS key that returns the result of the previous calculation. + Start by pressing 1 and then Enter. (We have just entered our initial guess, + x_0=1.) Now compute + + \texttt{ ANS } - \frac{f(\texttt{ ANS } )}{\fp(\texttt{ ANS } )} + + by entering the following and repeatedly press the Enter key. + + ANS-(ANS^3-ANS^2-1)/(3*ANS^2-2*ANS) + +

    + +

    + Each time we press the Enter key, + we are finding the successive approximations, x_1, + x_2, + , and each one is getting closer to the root. + In fact, once we get past around x_7 or so, + the approximations don't appear to be changing. + They actually are changing, + but the change is far enough to the right of the decimal point that it doesn't show up on the calculator's display. + When this happens, + we can be pretty confident that we have found an accurate approximation. +

    + +

    + Using a calculator in this manner makes the calculations simple; + many iterations can be computed very quickly. +

    + + + Using Newton's Method to find where functions intersect + +

    + Use Newton's Method to approximate a solution to \cos(x) = x, + accurate to five places after the decimal. +

    +     + +

    + Newton's Method provides a method of solving f(x) = 0; + it is not (directly) a method for solving equations like f(x) = g(x). + However, this is not a problem; + we can rewrite the latter equation as + f(x) - g(x)=0 and then use Newton's Method. +

    + +

    + So we rewrite \cos(x) =x as \cos(x)-x=0. + Written this way, we are finding a root of f(x)=\cos(x) -x. + We compute \fp(x)=-\sin(x) - 1. + Next we need a starting value, x_0. + Consider , + where f(x) = \cos(x) -x is graphed. + It seems that x_0=0.75 is pretty close to the root, + so we will use that as our x_0. (The figure also shows the graphs of y=\cos(x) and y=x. + Note how they intersect at the same x value as when f(x) = 0.) +

    + +
    + A graph of f(x)=\cos(x) -x used to find an initial approximation of its root + + + A graph of f(x) = cos x - x used to find an initial approximation of its root. + + +

    + The y axis and x axis both have values of (-1,0.5,0,0.5,1). + A broken line with equation y=x crosses the origin. + A second broken line depicts a downward parabola of equation y=\cos(x), the parabola intersects the first line y=x. + A solid line from the point where y=1 on the y axis to approximately x=0.75 on the x axis demonstrates an approach to estimating the root of the equation f(x) = \cos(x) - x. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.1,ymax=1.1,% + xmin=-1.1,xmax=1.1% + ] + \addplot+ [infinite,domain=-.1:1] {cos(deg(x))-x}; + \addplot+ [infinite,domain=-1:1] {cos(deg(x))}; + \addplot+ [infinite,domain=-1:1] {x}; + \addplot [guideline] coordinates {(0.739,0) (0.739,0.739)}; + \end{axis} + \end{tikzpicture} + + +
    + +

    + We now compute x_1, x_2, etc. + The formula for x_1 is + + x_1 \amp= 0.75 - \frac{\cos(0.75)-0.75}{-\sin(0.75)-1} + \amp \approx 0.7391111388 + . +

    + +

    + Apply Newton's Method again to find x_2: + + x_2 \amp= 0.7391111388 - \frac{\cos(0.7391111388)-0.7391111388}{-\sin(0.7391111388)-1} + \amp \approx 0.7390851334 + . +

    + +

    + We can continue this way, but it is really best to automate this process. + On a calculator with an ANS key, + we would start by entering 0.75, + then Enter, + inputting our initial approximation. + We then enter: + + ANS - (cos(ANS)-ANS)/(-sin(ANS)-1) + +

    + +

    + Repeatedly pressing the Enter key gives successive approximations. + We quickly find: + + x_3 \amp = 0.7390851332 + x_4 \amp = 0.7390851332 + . +

    + +

    + Our approximations x_2 and x_3 did not differ for at least the first five places after the decimal, + so we could have stopped. + However, using our calculator in the manner described is easy, + so finding x_4 was not hard. + It is interesting to see how we found an approximation, + accurate to as many decimal places as our calculator displays, + in just four iterations. +

    +     +     + + + +

    + If you know how to program, + you can translate the following pseudocode into your favorite language to perform the computation in this problem. +

    + + 
    +  x = 0.75
+  while true
+      oldx = x
+      x = x - (cos(x)-x)/(-sin(x)-1)
+      print x
+      if abs(x-oldx) < 0.0000000001
+          break
+
    + +

    + This code calculates x_1, x_2, + etc., storing each result in the variable x. + The previous approximation is stored in the variable oldx. + We continue looping until the difference between two successive approximations, abs(x-oldx), + is less than some small tolerance, + in this case, 0.0000000001. +

    + + + Convergence of Newton's Method +

    + What should one use for the initial guess, x_0? + Generally, the closer to the actual root the initial guess is, + the better. + However, some initial guesses should be avoided. + For instance, consider + where we sought the root to f(x) = x^3-x^2-1. + Choosing x_0=0 would have been a particularly poor choice. + Consider , + where f(x) is graphed along with its tangent line at x=0. + Since \fp(0)=0, + the tangent line is horizontal and does not intersect the x-axis. + Graphically, we see that Newton's Method fails. +

    + +
    + A graph of f(x) = x^3-x^2-1, showing why an initial approximation of x_0=0 with Newton's Method fails + + + A cubic graph features an "S" shaped curve with two turning points. + + +

    + The y axis represents function values (0.5, 0, -0.5, -1,-1.5), while the x axis represents values of (-0.5,0,0.5,1,1.5). + The graph intersects the y axis at -1 and 1.5 on the x axis, and forms an S shaped curve. +

    +

    + A horizontal tangent line is drawn at -1 on the y axis and 0 on the x axis +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.5,ymax=.6, + xmin=-0.6,xmax=1.6 + ] + + \addplot+ [infinite,domain=-.5:1.55] {-1}; + \addplot [tangentline,domain=-.5:1.55] {x^3-x^2-1}; + + \end{axis} + \end{tikzpicture} + + + +
    + +

    + We can also see analytically that it fails. + Since + + x_1 = 0 -\frac{f(0)}{\fp(0)} + + and \fp(0)=0, we see that x_1 is not well defined. +

    + +

    + This problem can also occur if, for instance, + it turns out that \fp(x_5)=0. + Adjusting the initial approximation x_0 by a very small amount will likely fix the problem. +

    + +

    + It is also possible for Newton's Method to not converge while each successive approximation is well defined. + Consider f(x) = x^{1/3}, + as shown in . + It is clear that the root is x=0, + but let's approximate this with x_0=0.1. + + shows graphically the calculation of x_1; + notice how it is farther from the root than x_0. + + and + show the calculation of x_2 and x_3, + which are even farther away; + our successive approximations are getting worse. + (It turns out that in this particular example, + each successive approximation is twice as far from the true answer as the previous approximation.) +

    + +
    + Newton's Method fails to find a root of f(x) = x^{1/3}, regardless of the choice of x_0. + +
    + + + + The graph illustrates the cube root function with initial guess x_0 as 0.1 + + +

    + The y axis represents function values (-1,0,1), while the x axis represents values of (-1,0,1). + The graph illustrates the cube root function, f(x) = x^{1/3}. It exhibits a curve that passes through the origin (0,0) + and extends into the positive and negative regions of the x axis. +

    +

    + It starts with an initial guess x_0=0.1. A tangent line is drawn at this point and intersects the x axis at the next + approximation, bringing it further from the root x=0. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + xtick={-1,1}, + extra x ticks={.1,-.2}, + extra x tick labels={$x_0$,$x_1$}, + ymin=-1.6,ymax=1.6,% + xmin=-1.6,xmax=1.6% + ] + + \addplot+ [infinite,domain=-1.16:1.16] ({x^3},{x}); + \addplot [tangentlineseg,domain = -.2:.1] {1.547*(x-.1)+.464}; + \addplot [guideline] coordinates {(0.1,0) (0.1,0.464)}; + \addplot [soliddot] coordinates {(.1,.464)}; + + \end{axis} + \end{tikzpicture} + + + + +
    +
    + + + + The graph illustrates the cube root function with point x_1 + + +

    + The graph illustrates the cube root function, f(x) = x^{1/3}. It exhibits a curve that passes through the origin (0,0) + and extends into the positive and negative regions of the x axis. +

    +

    + Here x_1 is twice as much as the initial guess bringing it more further from the root. + A tangent line is drawn at this point and intersects the x axis at the next + approximation, x_2. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + xtick={-1,1}, + extra x ticks={.1,-.2,.4}, + extra x tick labels={$x_0$, $x_1$,$x_2$}, + ymin=-1.6,ymax=1.6,% + xmin=-1.6,xmax=1.6% + ] + + \addplot+ [infinite,domain=-1.16:1.16]({x^3},{x}); + \addplot [tangentlineseg,domain = -.2:.4] {.975*(x+.2)-.585}; + \addplot [guideline] coordinates {(-0.2,0) (-0.2,-0.585)}; + \addplot [soliddot] coordinates {(-.2,-.585)}; + + \end{axis} + \end{tikzpicture} + + + +
    + +
    + + + + The graph illustrates the cube root function with point x_2 + + +

    + The graph illustrates the cube root function, f(x) = x^{1/3}. It exhibits a curve that passes through the origin (0,0) + and extends into the positive and negative regions of the x axis. +

    +

    + Here x_2 is twice as much as the x_1bringing it even more further from the root. + A tangent line is drawn at this point and intersects the x axis at the next approximation,x_3. +

    + +

    + Each successive approximation is twice as far from the true answer as the previous approximation. + This shows that Newton's Method doesn't work every time. +

    + +     + + + \begin{tikzpicture} + \begin{axis}[ + xtick={-1,1}, + extra x ticks={.1,-.2,.4,-.8}, + extra x tick labels={$x_0$,$x_1$,$x_2$,$x_3$}, + ymin=-1.6,ymax=1.6,% + xmin=-1.6,xmax=1.6% + ] + + \addplot+ [infinite,domain=-1.16:1.16] ({x^3},{x}); + \addplot [tangentlineseg,domain = -.8:.4] {.614*(x-.4)+.737}; + \addplot [guideline] coordinates {(0.4,0) (0.4,0.737)}; + \addplot [soliddot] coordinates {(.4,.737)}; + + \end{axis} + \end{tikzpicture} + + + +
    +     +
    + +

    + There is no fix to this problem; Newton's Method simply will not work and another method must be used. (In this case the particular reason Newton's Method fails is that the tangent line is vertical at the root). +

    + + + +

    + While Newton's Method does not always work, + it does work most of the time, + and it is generally very fast. + Once the approximations get close to the root, Newton's Method can as much as double the number of correct decimal places with each successive approximation. + A course in Numerical Analysis will introduce the reader to more iterative root finding methods, + as well as give greater detail about the strengths and weaknesses of Newton's Method. +

    +     + + + + Terms and Concepts + + + + +

    + + Given a function f(x), Newton's Method produces an exact solution to f(x) = 0. +

    +     + +     + + + + +

    + + In order to get a solution to f(x)=0 accurate to d places after the decimal, + at least d+1 iterations of Newton's Method must be used. +

    +     + +     +     + + + + Problems + + + +

    + The roots of the function f(x) are known or are easily found. + Use five iterations of Newton's Method with the given initial approximation + to approximate the root. + Compare it to the known value of the root. +

    +     + + + + + $f=Compute('cos(x)'); + Context("LimitedNumeric"); + $x[0]=1.5; + $df=$f->D('x'); + for my $i (1..5) { + $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); + }; + $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; + + +

    + f(x) = , x_0= +

    + + Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. + +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +     +     +     + + + + + $f=Compute('sin(x)'); + Context("LimitedNumeric"); + $x[0]=1; + $df=$f->D('x'); + for my $i (1..5) { + $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); + }; + $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; + + +

    + f(x) = , x_0= +

    + + Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. + +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +     +     +     + + + + + $f=Compute('x^2+x-2'); + Context("LimitedNumeric"); + $x[0]=0; + $df=$f->D('x'); + for my $i (1..5) { + $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); + }; + $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; + + +

    + f(x) = , x_0= +

    + + Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. + +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +     +     +     + + + + + $f=Compute('x^2-2'); + Context("LimitedNumeric"); + $x[0]=1.5; + $df=$f->D('x'); + for my $i (1..5) { + $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); + }; + $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; + + +

    + f(x) = , x_0= +

    + + Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. + +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +     +     +     + + + + + $f=Compute('ln(x)'); + Context("LimitedNumeric"); + $x[0]=2; + $df=$f->D('x'); + for my $i (1..5) { + $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); + }; + $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; + + +

    + f(x) = , x_0= +

    + + Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. + +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +     +     +     + + + + + $f=Compute('x^3-x^2+x-1'); + Context("LimitedNumeric"); + $x[0]=2; + $df=$f->D('x'); + for my $i (1..5) { + $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); + }; + $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; + + +

    + f(x) = , x_0= +

    + + Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. + +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +

    + +

    +     +     +     +     + + + + +

    + Use Newton's Method to approximate all roots of the given function accurate to three places after the decimal. + If an interval is given, find only the roots that lie within that interval. + Use technology to obtain good initial approximations. +

    +     + + + + + $f = Compute('x^3+5x^2-x-1'); + Context("FiniteSolutionSets"); + Context()->flags->remove("NumberCheck"); + Context()->flags->set(preferSetNotation=>0); + Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); + $L = Formula("{-5.15632517465866,-0.369102386184855,0.525427560843517}"); + $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; + + +

    + f(x)= +

    + + Use commas to separate roots. + +

    + +

    +

    + +

    +     +     +     + + + + + $f = Compute('x^4+2x^3-7x^2-x+5'); + Context("FiniteSolutionSets"); + Context()->flags->remove("NumberCheck"); + Context()->flags->set(preferSetNotation=>0); + Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); + $L = Formula("{-3.71447874438776,-0.856722678160357,1,1.57120142254812}"); + $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; + + +

    + f(x)= +

    + + Use commas to separate roots. + +

    + +

    +

    + +

    +     +     +     + + + + + $f = Compute('x^17-2x^13-10x^8+10'); + Context("FiniteSolutionSets"); + Context()->flags->remove("NumberCheck"); + Context()->flags->set(preferSetNotation=>0); + Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); + $L = Formula("{-1.0134032173648748818979878,0.98831169481331833285038046,1.3934095863189322104023657}"); + $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; + + +

    + f(x)= on (-2,2) +

    + + Use commas to separate roots. + +

    + +

    +

    + +

    +     +     +     + + + + + $f = Compute('x^2cos(x) + (x-1)sin(x)'); + Context("FiniteSolutionSets"); + Context()->flags->remove("NumberCheck"); + Context()->flags->set(preferSetNotation=>0); + Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); + $L = Formula("{-2.16477423608072,0,0.524501487448536,1.81327843509828}"); + $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; + + +

    + f(x)= on (-3,3) +

    + + Use commas to separate roots. + +

    + +

    +

    + +

    +     +     +     +     + + + + +

    + Use Newton's Method to approximate when the given functions are equal, + accurate to 3 places after the decimal. + Use technology to obtain good initial approximations. +

    +     + + + + + + $f = Compute('x^2'); + $g = Compute('cos(x)'); + Context("FiniteSolutionSets"); + Context()->flags->remove("NumberCheck"); + Context()->flags->set(preferSetNotation=>0); + Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); + $L = Formula("{-0.82413231230252242296,0.82413231230252242296}"); + $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; + + +

    + f(x)=, g(x)= +

    + + Use commas to separate solutions. + +

    + +

    +

    + +

    +     +     +     + + + + + + $f = Compute('x^2-1'); + $g = Compute('sin(x)'); + Context("FiniteSolutionSets"); + Context()->flags->remove("NumberCheck"); + Context()->flags->set(preferSetNotation=>0); + Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); + $L = Formula("{-0.63673265080528201089,1.4096240040025962492}"); + $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; + + +

    + f(x)=, g(x)= +

    + + Use commas to separate solutions. + +

    + +

    +

    + +

    +     +     +     + + + + + + $f = Compute('e^(x^2)'); + $g = Compute('cos(x)'); + Context("FiniteSolutionSets"); + Context()->flags->remove("NumberCheck"); + Context()->flags->set(preferSetNotation=>0); + Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); + $L = Formula("{0}"); + $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; + + +

    + f(x)=, g(x)= +

    + + Use commas to separate solutions. + +

    + +

    +

    + +

    +     +     +     + + + + + + $f = Compute('x'); + $g = Compute('tan(x)'); + Context("FiniteSolutionSets"); + Context()->flags->remove("NumberCheck"); + Context()->flags->set(preferSetNotation=>0); + Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); + $L = Formula("{-4.49340945790906,0,4.49340945790906}"); + $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; + + +

    + f(x)=, g(x)= on [-6,6] +

    + + Use commas to separate solutions. + +

    + +

    +

    + +

    +     +     +     +     + + + + +

    + Why does Newton's Method fail in finding a root of f(x) = x^3-3x^2+x+3 when x_0=1? +

    + +     + + + +

    + The approximations alternate between x=1 and x=2. +

    +     + +     + + + + +

    + Why does Newton's Method fail in finding a root of f(x) = -17x^4+130x^3-301x^2+156x+156 when x_0=1? +

    + +     + + + +

    + The approximations alternate between x=1, + x=2 and x=3. +

    +     + +     +     +     +
    +
    + Related Rates + + + +

    + When two quantities are related by an equation, + knowing the value of one quantity can determine the value of the other. + For instance, + the circumference and radius of a circle are related by C=2\pi r; + knowing that C is + + 6\pi + + determines the radius must be + + 3 + . +

    + +

    + But what if both variables are changing with time? + If we know how two variables are related and we know how one of them changes with time, + can we find how the other variable changes with time? +

    + +

    + The topic of related rates + allows us to answer this question: + knowing the rate at which one quantity is changing can determine the + rate at which another changes. + + related rates + +

    + + + +

    + We demonstrate the concepts of related rates through examples. +

    + + + Understanding related rates + +

    + The radius of a circle is growing at a rate of + + 5 + . + At what rate is the circumference growing? +

    +     + +

    + The circumference and radius of a circle are related by C = 2\pi r. + We are given information about how the length of r changes with respect to time; + that is, we are told \lz{r}{t} is + + 5 + . + We want to know how the length of C changes with respect to time, + , we want to know \lz{C}{t}. +

    + +

    + Implicitly differentiate both sides of + C = 2\pi r with respect to t: + + C \amp = 2\pi r + \lzoo{t}{C} \amp = \lzoo{t}{2\pi r} + \lz{C}{t} \amp =2\pi \lz{r}{t} + . +

    + +

    + As we know \lz{r}{t} is + + 5 + , we know + + \lz{C}{t} = 2\pi 5 = 10\pi \approx 31.4\text{ in/hr } + . +

    + + + + + +     + +     + +

    + In related rates problems, + we will be presented with an application problem that involves two or more variables and one or more rate. + It is the job of the reader to construct the appropriate model that can be used to answer the posed question. + + outlines the basic steps for solving a related rates problem. +

    + + + Related Rates + Related Rates + +

    +

      +
    1. +

      + Understand the problem. + Clearly identify the quantity whose rate of change you need to determine. + Make sketch, if helpful. +

      +
      2. + +
    2. +

      + Identify other quantities relevant to the context of the problem + and create an equation that relates them to the quantity identified in Step. + If values for certain quantities are already known, do not substitute these values into your equation yet. + These instantaneous values will be used in Step. +

      +
      3. + +
    3. +

      + Implicitly differentiate both sides of the equation found in Step + with respect to t. +

      +
      4. + +
    4. +

      + Substitute in the known values of rates and known instantaneous values of the variables. +

      +
      5. + +
    5. +

      + Solve for the unknown rate identified in Step. +

      +
      6. + +
    6. +

      + Write a full sentence conclusion. +

      +
      7. +
    +

    +     + +

    + Consider another, similar example. +

    + + + Finding related rates + +

    + Water streams out of a faucet at a rate of + + 2 + + onto a flat surface at a constant rate, + forming a circular puddle that is + + 1/8 + + deep. +

    + +

    +

      +
    1. + +

      + At what rate is the area of the puddle growing? +

      +
      2. + +
    2. + +

      + At what rate is the radius of the circle growing? +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + We can answer this question two ways: + using common sense or related rates. + The common sense method states that the volume of the puddle is + growing by + + 2 + , where + + \text{volume of puddle}=\text{area of circle}\times\text{depth} + . + Since the depth is constant at + + 1/8 + , the area must be growing by + + 16 + + since 16\cdot \frac{1}{8}=2. + This approach reveals the underlying related rates principle. +

      + +

      + Now let's solve the problem using + . + Based on the problem description, + the quantities that change with time are the volume of water + (the volume of the puddle), + the area of the circular puddle and the radius of the circle. + We don't need a diagram for this problem. + The important variables for this part of the problem are the + volume and area. +

      + +

      + Let V and A represent the Volume and Area of the puddle. + We know V= A\times \frac{1}{8}. + Take the derivative of both sides with respect to t, + employing implicit differentiation. + + V \amp = \frac{1}{8}A + \lzoo{t}{V}\amp = \lzoo{t}{\frac{1}{8}A} + \lz{V}{t} \amp = \frac{1}{8}\lz{A}{t} + + We know the change in volume, \lz{V}{t} = 2, + so we substitute this value into our related rates equation: + 2 = \frac{1}{8}\lz{A}{t}, + and hence \lz{A}{t} = 16. + Thus the area is growing by + + 16 + . +

      +
      2. + +
    2. +

      + We already identified the quantities that are changing in + Part. + The variables of interest in this problem are the radius and the volume. + We need an equation that relates the volume of the circle to the radius. + Since the puddle is a right circular cylinder, + we will use a known volume formula, + V=\pi r^2 h where V is the volume of the puddle (in + + + , r is the radius + (in inches) + and h is the height ( depth) of the puddle in inches. + (Notice that this formula is equivalent to + V=\text{area} \times \text{depth}.) + We know that the height (depth) is a constant 1/8 inch. + Since this quantity does not change in the problem, + we can safely substitute this value now. +

      + +

      + Implicitly derive both sides of + V=\pi r^2 \frac{1}{8} with respect to t: + + V \amp = \frac{1}{8} \pi r^2 + \frac{d}{dt}\big(V\big) \amp = \frac{d}{dt}\left(\frac{1}{8}\pi r^2\right) + \lz{V}{t} \amp = \frac{1}{8} 2\pi r\lz{r}{t} + \lz{V}{t} \amp = \frac{1}{4} \pi r\lz{r}{t} + + We know that \lz{V}{t} is + + 2 + . + So we have: + + 2 = \frac14 \pi r\lz{r}{t} + + Solving for \lz{r}{t}, we have + + \lz{r}{t} = \frac{8}{\pi r} + . + Note how our answer is not a number, + but rather a function of r. + In other words, + the rate at which the radius is growing depends on how big + the circle already is. + If the circle is very large, adding + + 2 + + of water will not make the circle much bigger at all. + If the circle is dime-sized, + adding the same amount of water will make a radical change in the + radius of the circle. +

      + +

      + In some ways, our problem was (intentionally) ill-posed. + We need to specify a current (instantaneous) value of the radius + in order to know a rate of change. + When the puddle has a radius of + + 10 + , the radius is growing at a rate of + + \lz{r}{t} = \frac{8}{10\pi} = \frac{4}{5\pi} \approx 0.25\text{ in/s } + . +

      +
      3. +
    +

    +     + +     + + + + + + + Studying related rates + +

    + Radar guns measure the rate of distance change between the gun and the + object it is measuring. + For instance, a reading of + 55 + means the object is moving away from the gun at a rate + of 55 miles per hour, + whereas a measurement of + -25 + would mean that the object is approaching the gun at a + rate of 25 miles per hour. +

    + +

    + If the radar gun is moving (say, + attached to a police car) then radar readouts are only immediately + understandable if the gun and the object are moving along the same line. + If a police officer is traveling + + 60 + + and gets a readout of + + 15 + , he knows that the car ahead of him is moving away at a rate + of 15 miles an hour, + meaning the car is traveling + + 75 + . (This straight-line principle is one reason officers park + on the side of the highway and try to shoot straight back down the road. + It gives the most accurate reading.) +

    + +

    + Suppose an officer is driving due north at + + 30 + + and sees a car moving due east, + as shown in . + Using his radar gun, he measures a reading of + + 20 + . + By using landmarks, + he believes both he and the other car are about 1/2 mile from the + intersection of their two roads. +

    + +
    + A sketch of a police car (at bottom) attempting to measure the + speed of a car (at right) in + + + A passenger car is shown moving east, while a police car is moving north. Their roads are set to intersect. + + +

    + The passenger car is on an eastward road, while the police car is on a northward road, and these roads are perpendicular to each other. The police car is 1/2 units south of the intersection point, and the passenger car is 1/2 units east of the same intersection. +

    +

    + A straight line, labeled as C, connects the current positions of both the police car and the passenger car. +

    +     + + + + \begin{tikzpicture}[>=latex,scale=1.32] + \draw(0,0) -- (3,0) node [pos=.5,above] { $B=1/2$} -- (0,-3) node [pos=.4,below] { $C$}-- (0,0) node [pos=.5,rotate=90,shift={(.1,.2)}] { $A=1/2$}; + + \draw [thick] (0,-3.5) -- (0,.5) node [above] { $N$}; + \draw [thick] (-.5,0) -- (3.5,0) node [right] { $E$}; + + \filldraw [fill=firstcolor,draw=secondcolor] (0,-3) circle (2pt) node [right] { Officer}; + \filldraw [fill=black,draw=black] (3,0) circle (2pt) node [below ] { \quad Car}; + + \draw[->,thick] (-.2,-3.3) -- (-.2,-2.7); + \draw[->,thick] (2.7,.2) -- (3.3,.2); + + \end{tikzpicture} + + + +
    + +

    + If the speed limit on the other road is + + 55 + , is the other driver speeding? +

    +     + +

    + The important quantities that are changing are: + the distance of the officer to the intersection, + the distance of the car to the intersection, + and the distance of the officer to the car. + (There are other quantities that are changing as well such as the + angles and area of the triangle, + but these are not important to this problem.) +

    + +

    + Using the diagram in , + let's label what we know about the situation. + As both the police officer and other driver are 1/2 mile from + the intersection, + we have A = 1/2, B = 1/2, + and through the Pythagorean Theorem, + C = 1/\sqrt{2}\approx 0.707. + These values are instantaneous + values for our variables, + so we won't use them until the end of the problem. + Instead, we will use the variables A, B, + and C. +

    + +

    + We need an equation that relates A, B, and C. + The Pythagorean Theorem is a good choice: + A^2+B^2 = C^2. + Differentiate both sides with respect to t: + + A^2 + B^2 \amp = C^2 + \lzoo{t}{A^2+B^2} \amp = \lzoo{t}{C^2} + 2A\lz{A}{t} + 2B\lz{B}{t} \amp = 2C\lz{C}{t} + +

    + +

    + We know the police officer is traveling at + + 30 + ; that is, \lz{A}{t} = -30. + The reason this rate of change is negative is that A is getting + smaller; + the distance between the officer and the intersection is shrinking. + The radar measurement is \lz{C}{t} = 20. + We want to find \lz{B}{t}. +

    + + + +

    + We have values for everything except \lz{B}{t}. + Solving for this we have: + + \lz{B}{t} = \frac{C\lz{C}{t}- A\lz{A}{t}}{B} + . + Now we substitue in our known rates and instantaneous values of our + variables: + + \lz{B}{t} \amp \approx \frac{0.707(20)- 0.5(-30)}{(0.5)} + \amp= 58.28 \text{ mph } + . +

    + +

    + The other driver appears to be speeding slightly. +

    +     + +     + + + Studying related rates + +

    + A camera is placed on a tripod + + 10 + + from the side of a road. + The camera is to turn to track a car that is to drive by at + + 100 + + for a promotional video. + The video's planners want to know what kind of motor the tripod should + be equipped with in order to properly track the car as it passes by. + shows the proposed setup. +

    + +
    + Tracking a speeding car (at left) with a rotating camera + + + A tripod with a rotating camera is shown 10ft from the side of the road, tracking a speeding car. + + +

    + A car is shown to the left of the image, driving at a 100 mph east, + with a tripod placed 10 from the side of the road, set to track the car. +

    +

    + A right triangle is drawn, with the hypotenuse connecting the car to the tripod, and the other two sides parallel to the road and perpendicular to the road. + The side parallel to the road has length x, and the angle between the hypotenuse and the side perpendicular to the road is labeled as \theta. The side perpendicular to the road has a length of 10. +

    +     + + + \begin{tikzpicture}[>=latex,scale=1.32] + + \draw (-3,0) -- (0,0) -- (0,-2) node [shift={(-5pt,10pt)}] {\small $\theta$} node [right,pos=.5] { 10ft} -- cycle; + \draw [<->, thick] (-3.5,0) -- node [below,pos=.5] { $x$} (.5,0); + \draw [fill=firstcolor,draw=secondcolor] (-3,0) circle (2pt); + \draw [fill=black] (0,-2) circle (2pt); + \draw [->] (-3,.2) -- (-2,.2) node [above,pos=.5] { 100mph}; + + \end{tikzpicture} + + + +
    + +

    + How fast must the camera be able to turn to track the car? +

    +     + +

    + The quantities that changing are x and \theta as drawn on + . (The hypotenuse of the triangle is + also changing, but this isn't important to the problem). + We seek information about how fast the camera is to turn; + therefore, we need an equation that will relate an angle \theta + to the position of the camera and the speed and position of the car. +

    + +

    + + suggests we use a trigonometric equation. + Letting x represent the distance the car is from the point on the + road directly in front of the camera, we have + + \tan(\theta) = \frac{x}{10} + . +

    + +

    + Now take the derivative of both sides of Equation + using implicit differentiation: + + \tan(\theta) \amp = \frac{x}{10} + \lzoo{t}{\tan(\theta)} \amp = \lzoo{t}{\frac{x}{10}} + \sec^2(\theta)\lz{\theta}{t} \amp = \frac{1}{10}\lz{x}{t} + + Now we solve for \lz{\theta}{t}: + + \lz{\theta}{t} = \frac{\cos^2(\theta)}{10}\lz{x}{t} + +

    + +

    + As the car is moving at + + 100 + , we have that \lz{x}{t} is + + -100 + + (as in the last example, + since x is getting smaller as the car travels, + \lz{x}{t} is negative). + We need to convert the measurements so they use the same units + (we chose + + + ); + rewrite + + -100 + + in terms of + + + : + + \lz{x}{t} \amp = -100\frac{\text{mi} }{\text{ hr } } + \amp = -100\frac{\text{mi} }{\text{ hr } }\cdot5280\frac{\text{ ft } }{\text{mi} }\cdot\frac{1}{3600}\frac{\text{ hr } }{\text{s} } + \amp =-146.\overline{6}\text{ ft/s } + . +

    + +

    + We want to know the fastest the camera has to turn. + Common sense tells us this is when the car is directly in front of the + camera (, when \theta = 0). + Our mathematics bears this out. + In Equation + we see this is when \cos^2(\theta) is largest; + this is when \cos(\theta) = 1, + or when \theta = 0. + We also know that we should get an answer that is in + + + . + Since \cos(\theta) is a dimensionless measure, + it won't contribute to the units. + However, radians are also dimensionless. + This means we can write + (or erase) + the word radian without any unit consequences. (The same is not + true of degrees always convert degress to radians). +

    + +

    + With \lz{x}{t} approximately + + -146.7 + , we have + + \lz{\theta}{t} \amp \approx -\frac{1}{10\text{ ft } }146.67\text{ ft/s } + \amp = -14.667\text{ radians/s } + +

    + +

    + We find that \lz{\theta}{t} is negative; + this matches our diagram in + for \theta is getting smaller as the car approaches the camera. +

    + +

    + What is the practical meaning of + + -14.667 + ? + Recall that 1 circular revolution goes through 2\pi + radians, thus + + 14.667 + + means 14.667/(2\pi)\approx 2.33 revolutions per second. + The negative sign indicates the camera is rotating in a clockwise fashion. +

    +     + +     + + + + + +

    + We introduced the derivative as a function that gives the slopes of tangent + lines of functions. + This chapter emphasizes using the derivative in other ways. + Newton's Method uses the derivative to approximate roots of functions; + this section stresses the rate of change + aspect of the derivative to find a relationship between the rates of change + of two related quantities. +

    + +

    + In the next section we use Extreme Value concepts to + optimize quantities. +

    + + + + Terms and Concepts + + + + +

    + + Implicit differentiation is often used when solving + related rates type problems. +

    +     + +     + + + + +

    + + A study of related rates is part of the standard police officer training. +

    +     + +     +     + + + Problems + + + + + ($rate,$depth) = random_subset(2,4..10); + $a = random(1,4,1); + if($envir{problemSeed}==1){$rate=5; $depth=10; $a=1;}; + Context()->flags->set(reduceConstants=>0); + $rateU = NumberWithUnits("$rate cm^3/s"); + $depthU = NumberWithUnits("$depth mm"); + @r=($a,$a*10,$a*100); + @rU = map{NumberWithUnits("$_ cm")}(@r); + Context("Fraction"); + @f = map{Fraction(5*$rate,$_*$depth)}(@r); + @num = map{($_->value)[0]}(@f); + @den = map{($_->value)[1]}(@f); + @sU = map{NumberWithUnits("$num[$_]/($den[$_] pi) cm/s")}(0,1,2); + + +

    + Water flows onto a flat surface at a rate of + + forming a circular puddle deep. + How fast is the radius growing when the radius is: +

    +     + + + +

    + +

    +

    + +

    +     +     + + + +

    + +

    +

    + +

    +     +     + + + +

    + +

    +

    + +

    +     +     +     +     + + + + + $rate = random(4,12,1); + $a = random(1,4,1); + if($envir{problemSeed}==1){$rate=10; $a=1;}; + Context()->flags->set(reduceConstants=>0); + $rateU = NumberWithUnits("$rate cm^3/s"); + @r=($a,$a*10,$a*100); + @rU = map{NumberWithUnits("$_ cm")}(@r); + Context("Fraction"); + @f = map{Fraction($rate,4*$_**2)}(@r); + @num = map{($_->value)[0]}(@f); + @den = map{($_->value)[1]}(@f); + @sU = map{NumberWithUnits("$num[$_]/($den[$_] pi) cm/s")}(0,1,2); + + +

    + A spherical balloon is inflated with air flowing at a rate of + . + How fast is the radius of the balloon increasing when the radius is: +

    +     + + + +

    + +

    +

    + +

    +     +     + + + +

    + +

    +

    + +

    +     +     + + + +

    + +

    +

    + +

    +     +     +     +     + + + + + Context("Fraction"); + $f = list_random(Fraction(1,2),Fraction(3,4),Fraction(2,3),Fraction(3,5),Fraction(4,5)); + $speed = random(45,55,5); + $rate = random(70,90,5); + if($envir{problemSeed}==1){$f=Fraction(1,2);$speed=50;$rate=80;}; + Context("Numeric"); + Context()->flags->set(reduceConstants=>0); + $a = Compute("$rate*sqrt(2)-$speed"); + $a = NumberWithUnits("$a mi/h"); + + +

    + Consider the traffic situation introduced in . + How fast is the other car traveling if the officer and the other car are each + mile from the intersection, the other car is traveling due west, + the officer is traveling north at \,\text{mph}, + and the radar reading is -\,\text{mph}? +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + $f = list_random(Fraction(1,2),Fraction(3,4),Fraction(2,3),Fraction(3,5),Fraction(4,5)); + $speed = random(45,55,5); + $rate = random(70,90,5); + if($envir{problemSeed}==1){$f=Fraction(1,2);$speed=50;$rate=80;}; + Context("Numeric"); + Context()->flags->set(reduceConstants=>0); + $f2 = 1+$f**2; + $f3 = $speed*$f; + $a = Compute("$rate*sqrt($f2)-$f3"); + $a = NumberWithUnits("$a mi/h"); + $f4 = $rate/$f; + $f5 = $speed/$f; + $b = Compute("$f4 sqrt($f2)-$f5"); + $b = NumberWithUnits("$b mi/h"); + + +

    + Consider the traffic situation introduced in . + Calculate how fast the other car is traveling in each of the following situations. +

    +     + + + +

    + The officer is traveling due north at \,\text{mph} + and is mile from the intersection, + while the other car is 1 mile from the intersection + traveling west and the radar reading is -\,\text{mph}? +

    +

    + +

    +     +     + + + +

    + The officer is traveling due north at \,\text{mph} + and is 1 mile from the intersection, + while the other car is mile from the + intersection traveling west and the radar reading is -\,\text{mph}? +

    +

    + +

    +     +     +     +     + + + + + Context()->flags->set(reduceConstants=>0); + $spd = random(300,800,10); + $elev = random(5000,15000,100); + $a = Compute("($elev*$spd*5280)/(5280^2+$elev^2)"); + $aU = NumberWithUnits("$a rad/hr"); + $b = Compute("($elev*$spd*5280)/(1056^2+$elev^2)"); + $bU = NumberWithUnits("$b rad/hr"); + $c = Compute("$spd*5280/$elev"); + $cU = NumberWithUnits("$c rad/hr"); + + +

    + An F-22 aircraft is flying at \,\text{mph} + with an elevation of \,\text{ft} on a + straight-line path that will take it directly over an + anti-aircraft gun. +

    + + + An airplane is shown on the right, over horizontal ground with a gun on the ground to the left. + + +

    + To the left of a horizontal surface, a square marks an anti-aircraft gun. + Above the ground, on the right of the image, is an airplane flying to the left. + The horizontal distance along the ground from the gun to the airplane position is marked with the variable x. + The elevation of the plane is shown as 6600. +

    +

    + There is also a dashed line from the gun to the plane. + The angle that this line makes with the ground is marked with the variable \theta. +

    +     + + \begin{tikzpicture}[>=latex,scale=1.3] + \begin{scope} + \clip (0,0) rectangle (120pt,50pt); + \draw [inner color=firstcolor,draw = white] (-10,-25pt) rectangle (180pt,70pt); + \end{scope} + \draw [top color=treestump, bottom color=white,draw=white] (0,0) rectangle (120pt,-15pt); + \draw (0,0) -- (120pt,0); + \draw [dashed] (12.5pt,0pt) node [xshift=22pt,yshift=4pt] { \(\theta\)} -- (105pt,35pt); + \draw [ultra thick] (12.5pt,0) -- (18pt,4pt); + \filldraw [fill=white] (10pt,-2.5pt) rectangle (15pt,2.5pt); + \draw [<->] (15pt,-10pt) -- (110pt,-10pt) node [below,pos=.5] { \(x\)}; + \draw [<->] (120pt,0pt) -- (120pt,35pt) node [right, pos=.5] { $elev ft}; + \begin{scope}[shift={(105pt,35pt)}] + \draw [fill=black] (0,0) rectangle (10pt,2pt); + \draw [very thick] (3pt,1pt) -- (6pt,-1.5pt) -- (5pt,0pt); + \draw [very thick] (9pt,2pt) -- (9pt,4pt) -- (7pt,2pt); + \end{scope} + \end{tikzpicture} + + +

    + How fast (in radians per second) must the gun be able to turn to accurately track the + aircraft when the plane is: +

    +     + + + +

    + 1 mile away? +

    +

    + +

    + + Use rad/s for radians per second. + +     +     + + + +

    + 1/5 mile away? +

    +

    + +

    + + Use rad/s for radians per second. + +     +     + + + +

    + Directly overhead? +

    +

    + +

    + + Use rad/s for radians per second. + +     +     + +

    + There are 5280 feet in one mile. +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0); + $speed = random(400,600,50); + $speedU = NumberWithUnits("$speed mi/h"); + @a = (random(1000,2000,100), random(100,500,50), 0); + if($envir{problemSeed}==1){$a[0]=1000,$a[1]=100;}; + @aU = map{NumberWithUnits("$_ ft")}(@a); + @b = map{100*$speed*5280/3600/(100**2 + $_**2)}(@a); + @bU = map{NumberWithUnits("$_ rad/s")}(@b); + + +

    + An F-22 aircraft is flying at + + with an elevation of 100\,\text{ft} + on a straight-line path that will take it directly over an + anti-aircraft gun as in + (note the lower elevation here). +

    +

    + How fast (in radians per second) must the gun be able to turn to accurately track the + aircraft when the plane is: +

    +     + + + +

    + away? +

    + + Use rad/s for radians per second. + +

    + +

    +     +     + + + +

    + away? +

    + + Use rad/s for radians per second. + +

    + +

    +     +     + + + +

    + Directly overhead? +

    + + Use rad/s for radians per second. + +

    + +

    +     +     +     +     + + + + + + $f = Compute("x/sqrt(24^2-x^2)"); + $a = Compute($f->eval(x=>1)); + $a = NumberWithUnits("$a ft/s"); + $b = Compute($f->eval(x=>10)); + $b = NumberWithUnits("$b ft/s"); + $c = Compute($f->eval(x=>23)); + $c = NumberWithUnits("$c ft/s"); + $d = Compute("inf"); + + +

    + A 24 ladder + is leaning against a house while the base is pulled away at a + constant rate of + + 1 + . +

    + + + A ladder is shown leaning against a house, with its base being pulled away. + + +

    + A 24 ladder is positioned against the side of a house. The base of the ladder is being pulled away from the house at a rate of 1 ft/s. The exact distance of the ladder's base from the house varies based on different scenarios provided. +

    +     + + \begin{tikzpicture}[>=latex,xscale=4,yscale=3] + \draw [top color=treestump, bottom color=treestump!50!white,draw=black] (0,0) rectangle (20pt,30pt); + \draw [fill=black] (0,30pt) -- (10pt,38pt) -- (20pt,30pt) -- cycle; + \draw [fill=white] (7pt,15pt) rectangle (13pt,25pt); + \draw (10pt,15pt) -- (10pt,25pt) (7pt,20pt) -- (13pt,20pt); + \draw [thick] (20pt,25pt) -- node [pos=.5,rotate=-67,yshift=5pt] { 24 ft} (30pt,0); + \draw [thick] (0,0) -- (35pt,0); + \draw [->,>=latex] (34pt,2pt) -- node [above right,pos=.5] { 1 ft/s} (50pt,2pt); + \end{tikzpicture} + + +

    + At what rate is the top of the ladder sliding down the side of + the house when the base is: +

    +     + + + +

    + 1 foot from the house? +

    +

    + +

    + + Use ft/s for feet per second. + +     +     + + + +

    + 10 feet from the house? +

    +

    + +

    + + Use ft/s for feet per second. + +     +     + + + +

    + 23 feet from the house? +

    + +

    + +

    + + Use ft/s for feet per second. + +     +     + + + +

    + 24 feet from the house? +

    +

    + +

    + + Use ft/s for feet per second. + +     +     +     +     + + + + + + $f = Compute("30*sqrt(10^2+x^2)/x"); + $a = Compute($f->eval(x=>50)); + $a = NumberWithUnits("$a ft/min"); + $b = Compute($f->eval(x=>15)); + $b = NumberWithUnits("$b ft/min"); + $c = Compute($f->eval(x=>1)); + $c = NumberWithUnits("$c ft/min"); + + +

    + A boat is being pulled into a dock at a constant rate of + + 30 + + by a winch located 10 + above the deck of the boat. +

    + + + A boat is being pulled towards a dock by an overhead winch. + + +

    + A boat is approaching a dock, being pulled by a winch located 10 above the deck of the boat. The rope from the winch to the boat forms a hypotenuse, creating a right-angle triangle with the water's surface and the vertical winch post. The boat is being pulled at a constant rate of 30 ft/min towards the dock. +

    +     + + \begin{tikzpicture}[>=latex,scale=1.8] + \draw [top color=firstcolor!60!white, bottom color=firstcolor!10!white,draw=firstcolor] (-5pt,5pt) rectangle (100pt,-5pt); + \draw [thick,fill=treestump] (-10pt,20pt) rectangle (8pt,22pt); + \draw [top color=treestump!30!black, bottom color=treestump!70!white,draw=black] (0,0) rectangle (5pt,30pt); + \draw [fill=white] (50pt,0pt) -- (90pt,0pt) -- (90pt, 15pt) -- (45pt,15pt) -- cycle; + \draw [fill=black] (65pt,15pt) rectangle (67pt,20pt); + \draw [fill=yellowcolfill] (55pt,20pt) -- (80pt,20pt) -- (65pt,40pt) -- cycle; + \draw [fill=black] (8pt,35pt) circle (2pt); + \draw (45pt,15pt) [dashed] -- (8pt,15pt) -- (8pt,35pt) node [pos=.4,xshift=12pt] { 10 ft}; + \draw [-,thick] (8pt,35pt) -- (45pt,15pt); + \end{tikzpicture} + + +

    + At what rate is the boat approaching the dock when the boat is: +

    +     + + + +

    + 50 feet out? +

    +

    + +

    + + Use ft/min for feet per minute. + +     +     + + + +

    + 15 feet out? +

    +

    + +

    + + Use ft/min for feet per minute. + +     +     + + + +

    + 1 foot from the dock? +

    +

    + +

    + + Use ft/min for feet per minute. + +     +     + + + +

    + What happens when the length of rope pulling in the boat is less than 10 feet long? +

    + +

    + +

    +     +     +     +     + + + + + contextUnits.pl + + + # make $depth and $diameter two distinct numbers + ($depth, $diameter) = random_subset(2, 10 .. 30); + # the rate water is entering the tank + $rate = random(8, 12); + # the depths that water will be at when we ask how fast the water is rising + # the first depth will just be 1, and the other two should be no more than $depth + @a = (1, num_sort(random_subset(2, 2 .. $depth-1))); + Context("Units")->withUnitsFor('length', 'time'); + $depthU = Compute("$depth ft"); + $diameterU = Compute("$diameter ft"); + $rateU = Compute("$rate ft^3/s"); + # push the depths through the function that produces the rates + @b = map{$rate * 4 / pi * $depth**2 / $diameter**2 / $_**2} (@a); + @bU = map{Compute("$_ ft/s")} (@b); + $T = pi/12 * $diameter**2 * $depth / $rate; + $TU = Compute("$T s"); + + +

    + An inverted cylindrical cone, + deep and + across + at the top, is being filled with water at a rate of + . + At what rate is the water rising in the tank when the depth of + the water is: +

    +     + + + +

    + foot? +

    +

    + +

    +     +     + + + +

    + feet? +

    +

    + +

    +     +     + + + +

    + feet? +

    +

    + +

    +     +     + + + +

    + How long will the tank take to fill when starting at empty? +

    +

    + +

    +     +     +     +     + + + + + + $f = Compute("2x/sqrt(x^2+30^2)"); + $a = Compute($f->eval(x=>10)); + $a = NumberWithUnits("$a ft/s"); + $b = Compute($f->eval(x=>40)); + $b = NumberWithUnits("$b ft/s"); + $c = Compute("sqrt(60^2-30^2)"); + $c = NumberWithUnits("$c ft"); + + +

    + A rope, attached to a weight, + goes up through a pulley at the ceiling and back down to a worker. + The man holds the rope at the same height as the connection + point between rope and weight. +

    + + + A worker pulls a rope attached to a weight through a ceiling pulley. + + +

    + The diagram illustrates a weight connected to a 60 long rope, with the other end looped through a pulley affixed to the ceiling, 30 above the weight's resting position. Adjacent to the weight, a worker is depicted holding the rope, specifically at its attachment point to the weight. As the worker moves away at 2 ft/s, pulling the rope with him, the weight begins its ascent. +

    +     + + \begin{tikzpicture}[>=latex,x=1pt,y=1pt,scale=2.4] + \draw [thick](0,40) -- (55,40); + \draw [thick] (0,0) -- (55,0); + \draw (2,0) -- (12,0) -- (9,5) -- (5,5)--cycle; + \draw (7,5) -- (7,35) -- (33,5); + \draw (7,35) -- (7,40); + \draw [fill=gray] (7,35) circle (2pt); + \draw (35,10) circle (2pt); + \draw (35,8) -- (35,4) -- (33,0) (35,4) -- (37,0) (33,6) -- (37,6); + \draw (-4,5) -- (-1,5) (-2.5,5) -- (-2.5,35) node [pos=.5,draw,fill=white,draw=white,rotate=90] { 30 ft} + (-4,35) -- (-1,35); + \draw [thick,->](40,5) --(55,5) node[pos=.5,above] { 2 ft/s}; + \end{tikzpicture} + + +

    + Suppose the man stands directly next to the weight + (, a total rope length of 60 feet) and begins to + walk away at a rate of + + 2 + . + How fast is the weight rising when the man has walked: +

    +     + + + +

    + 10 feet? +

    +

    + +

    + + Use ft/s for feet per second. + +     +     + + + +

    + 40 feet? +

    +

    + +

    + + Use ft/s for feet per second. + +     +     + + + +

    + How far must the man walk to raise the weight all the way to + the pulley? +

    + +

    + +

    + + Use ft for feet. + +     +     +     +     + + + + + # Height is fixed at 30 ft + # Need an initial distance that makes a Pythagorean triple with 30 + $b = random(40,72,224); + $speed = random(2,4,1); + if($envir{problemSeed}==1){$b=40;$speed=2;}; + $bU = NumberWithUnits("$b ft"); + $speedU = NumberWithUnits("$speed ft/s"); + $c = sqrt(30**2 + $b**2); + $length = 30 + $c; + $lengthU = NumberWithUnits("$length ft"); + $newb = $b+10; + $newc = sqrt(30**2 + $newb**2); + $newrate = $newb/$newc * $speed; + $newrateU = NumberWithUnits("$newrate ft/s"); + $newnewb = $b+30; + $newnewc = sqrt(30**2 + $newnewb**2); + $newnewrate = $newnewb/$newnewc * $speed; + $newnewrateU = NumberWithUnits("$newnewrate ft/s"); + $far = sqrt($length**2 - 30**2); + $farU = NumberWithUnits("$far ft"); + + +

    + Consider the situation described in + . + Suppose the man starts from the weight and begins + to walk away at a rate of . +

    +     + + + +

    + How long is the rope? +

    +

    + +

    +     +     + + + +

    + How fast is the weight rising after the man has walked + 10 feet? +

    +

    + +

    +     +     + + + +

    + How fast is the weight rising after the man has walked + 30 feet? +

    +

    + +

    +     +     + + + +

    + How far must the man walk to raise the weight all the way + to the pulley? +

    +

    + +

    +     +     +     +     + + + + + $distance = random(80,120,5); + $height = random(5,6,1); + $rate = random(3,8,1); + if($envir{problemSeed}==1){$distance=100;$height=5;}; + $heightU = NumberWithUnits("$height ft"); + $elev = $distance + $height; + $elevU = NumberWithUnits("$elev ft"); + $speed = 2*$distance*$rate*pi/180; + $speedU = NumberWithUnits("$speed ft/min"); + + +

    + A hot air balloon lifts off from ground rising vertically. + From feet away, a + tall woman + tracks the path of the balloon. + When her sightline with the balloon makes a + 45^\circ angle with the horizontal, she notes the angle + is increasing at about ^\circ per minute. +

    +     + + + +

    + What is the elevation of the balloon? +

    +

    + +

    +     +     + + + +

    + How fast is it rising? +

    +

    + +

    +     +     +     +     + + + + + $rate = random(3,8,1); + $nd = list_random([2,3],[4,5],[4,7],[5,6],[5,7],[5,8]); + $height = random(20,40,5); + if($envir{problemSeed}==1){$rate=5;$nd=[2,3];$height=30;}; + $rateU = NumberWithUnits("$rate ft^3/s"); + Context("Fraction"); + $frac = Fraction($nd->[0],$nd->[1]); + Context("Numeric"); + $speed = 4*$frac**2/pi/$height**2*$rate; + $speedU = NumberWithUnits("$speed ft/s"); + + +

    + A company that produces landscaping materials is dumping sand + into a conical pile. + The sand is being poured at a rate of + . + The physical properties of the sand, + in conjunction with gravity, + ensure that the cone's height is roughly the length + of the diameter of the circular base. +

    +

    + How fast is the cone rising when it has a height of feet? +

    +

    + +

    +     +     +     +     +     +
    +
    + Optimization + + + + +

    + In + we learned about extreme values the largest and smallest values a + function attains on an interval. + We motivated our interest in such values by discussing how it made sense to + want to know the highest/lowest values of a stock, + or the fastest/slowest an object was moving. + In this section we apply the concepts of extreme values to solve + word problems, + , problems stated in terms of situations that require us to create the + appropriate mathematical framework in which to solve the problem. + optimization +

    + +

    + We start with a classic example which is followed by a discussion of the + topic of optimization. +

    + + + Optimization: perimeter and area + +

    + A man has 100 feet of fencing, + a large yard, and a small dog. + He wants to create a rectangular enclosure for his dog with the fencing + that provides the maximal area. + What dimensions provide the maximal area? +

    +     + +

    + One can likely guess the correct answer that is great. + We will proceed to show how calculus can provide this answer in a context + that proves this answer is correct. +

    + +

    + It helps to make a sketch of the situation. + Our enclosure is sketched twice in , + either with treetop grass and nice fence boards or as a simple rectangle. + Either way, drawing a rectangle forces us to realize that we need to + know the dimensions of this rectangle so we can create an area function + after all, + we are trying to maximize the area. +

    + +
    + A sketch of the enclosure in . + + + + + A rectangular enclosure with green grass and a fence with x and y dimensions. + + +

    + A 3-D rectangular enclosure is drawn with green grass and nice fence boards on the x and x axis. +

    +     + + + + \begin{tikzpicture}[>=latex,scale=1.2] + + \shadedraw [top color = treetop,bottom color = treetop!50!black] (0,0) -- (33.6pt,35.28pt) -- (95.2pt,35.28pt) -- (61.6pt,0) -- cycle; + + \begin{scope}[cm={1,1.05,0,1,(0,0)}] + \foreach \x in {0,...,7} + { + \shadedraw [xscale=.3,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++ (0,20pt) -- ++ (4pt,4pt) -- ++ (6pt,0pt) -- ++ (4pt,-4pt) -- ++ (0pt,-20pt) -- cycle; + } + \end{scope} + + \begin{scope}[shift={(33.6pt,35.28pt)}] + \foreach \x in {0,...,10} + { + \shadedraw [xscale=.4,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++ (0,20pt) -- ++ (4pt,4pt) -- ++ (6pt,0pt) -- ++ (4pt,-4pt) -- ++ (0pt,-20pt) -- cycle; + } + \end{scope} + + \foreach \x in {0,...,10} + { + \shadedraw [xscale=.4,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++ (0,20pt) -- ++ (4pt,4pt) -- ++ (6pt,0pt) -- ++ (4pt,-4pt) -- ++ (0pt,-20pt) -- cycle; + } + + \begin{scope}[cm={1,1.05,0,1,(61.6pt,0)}] + \foreach \x in {0,...,7} + { + \shadedraw [xscale=.3,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++ (0,20pt) -- ++ (4pt,4pt) -- ++ (6pt,0pt) -- ++ (4pt,-4pt) -- ++ (0pt,-20pt) -- cycle; + } + \end{scope} + + \draw [<->] (0,-5pt) -- (61.6pt,-5pt) node [below,pos=.5] {$x$}; + \draw [<->] (100.2pt,35.28pt) -- (66.6pt,0pt) node [right,pos=.5] {$y$}; + + \end{tikzpicture} + + + + + + + A rectangle shape is drawn with sides labeled x and y. + + +

    + The 2-D plane rectangle has sides labeled x and y. The dimensions are both unknown. +

    +     + + + \begin{tikzpicture}[>=latex,scale=0.66] + \draw (0,0) -- (8,0) node [pos=.5,below] {$x$} -- (8,5) node [pos=.5,right] {$y$} -- (0,5) -- cycle; + \end{tikzpicture} + + + + +     +
    +

    + We let x and y denote the lengths of the sides of the + rectangle. Clearly, + + \text{ Area } =xy + . +

    + +

    + We do not yet know how to handle functions with two variables; + we need to reduce this down to a single variable. + We know more about the situation: + the man has 100 feet of fencing. + By knowing the perimeter of the rectangle must be 100, + we can create another equation: + + \text{ Perimeter } = 100 = 2x+2y + . +

    + +

    + We now have two equations and two unknowns. + In the latter equation, we solve for y: + + y = 50-x + . +

    + +

    + Now substitute this expression for y in the area equation: + + \text{ Area } = A(x) = x(50-x) + . +

    + +

    + Note we now have an equation of one variable; + we can truly call the Area a function of x. +

    + +

    + This function only makes sense when 0\leq x \leq 50, + otherwise we get negative values of area. + So we find the extreme values of A(x) on the interval + [0,50] using . +

    + +

    + To find the critical points, + we take the derivative of A(x) and set it equal to 0, + then solve for x. + + A(x) \amp = x(50-x) + \amp = 50x-x^2 + A'(x) \amp = 50-2x + +

    + +

    + We solve 50-2x=0 to find x=25; + this is the only critical point. + We evaluate A(x) at the endpoints of our interval and at this + critical point to find the extreme values; + in this case, all we care about is the maximum. +

    + +

    + Clearly A(0)=0 and A(50)=0, + whereas A(25) = 625 + + . + This is the maximum. + Since we earlier found y = 50-x, + we find that y is also 25. + Thus the dimensions of the rectangular enclosure with perimeter of + + 100 + . with maximum area is a square, with sides of length + + 25 + . +

    +     +     + +

    + This example is very simplistic and a bit contrived. + (After all, most people create a design then buy fencing to meet their needs, + and not buy fencing and plan later.) + But it models well the necessary process: + create equations that describe a situation, + reduce an equation to a single variable, + then find the needed extreme value. +

    + +

    + In real life, problems are much more complex. + The equations are often not + reducible to a single variable + (hence multi-variable calculus is needed) + and the equations themselves may be difficult to form. + Understanding the principles here will provide a good foundation for the + mathematics you will likely encounter later. +

    + +

    + We outline here the basic process of solving these optimization problems. +

    + + + Solving Optimization Problems +

    +

      +
    1. +

      + Understand the problem. + Clearly identify what quantity is to be maximized or minimized. + Make a sketch if helpful. +

      +
      2. + +
    2. +

      + Create equations relevant to the context of the problem, + using the information given. One of these should describe the + quantity to be optimized. + We'll call this the fundamental equation. +

      +
      3. + +
    3. +

      + If the fundamental equation defines the quantity to be optimized as + a function of more than one variable, + reduce it to a single variable function using substitutions derived + from the other equations. We call these other equations constraint equations. +

      +
      4. + +
    4. +

      + Identify the domain of this function, + keeping in mind the context of the problem. +

      +
      5. + +
    5. +

      + Find the extreme values of this function on the determined domain. +

      +
      6. + +
    6. +

      + Identify the values of all relevant quantities of the problem and write a full sentence conclusion. +

      +
      7. +
    +

    +     + +

    + We will use in a variety of + examples. +

    + + + + + Optimization: perimeter and area + +

    + Here is another classic calculus problem: A woman has a 100 feet of + fencing, a small dog, and a large yard that contains a stream + (that is mostly straight). + She wants to create a rectangular enclosure with maximal area that uses + the stream as one side. + (Apparently her dog won't swim away.) + What dimensions provide the maximal area? +

    +     + +

    + We will follow the steps outlined by + . +

    + +

    +

      +
    1. +

      + We are maximizing area. + A sketch of the region will help; + + gives two sketches of the proposed enclosed area. + A key feature of the sketches is to acknowledge that one side is + not fenced. +

      + +
      + A sketch of the enclosure in + + + + A rectangular fenced yard with one side not fenced. + + +

      + A rectangular fenced yard with one side open for a stream. + The dimensions are labeled x and y and are not provided. +

      +       + + + \begin{tikzpicture}[>=latex,scale=1.2] + + \shadedraw [top color = treetop,bottom color = treetop!50!black] (0,0) -- (33.6pt,35.28pt) -- (95.2pt,35.28pt) -- (61.6pt,0) -- cycle; + + \begin{scope}[cm={1,1.05,0,1,(0,0)}] + \foreach \x in {0,...,7} + { + \shadedraw [xscale=.3,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++(0,20pt) -- ++(4pt,4pt) -- ++(6pt,0pt) -- ++(4pt,-4pt)-- ++(0pt,-20pt) -- cycle; + } + \end{scope} + + \begin{scope}[shift={(33.6pt,35.28pt)}] + \foreach \x in {0,...,10} + { + \shadedraw [xscale=.4,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++(0,20pt) -- ++(4pt,4pt) -- ++(6pt,0pt) -- ++(4pt,-4pt)-- ++(0pt,-20pt) -- cycle; + } + \end{scope} + + \begin{scope}[cm={1,1.05,0,1,(61.6pt,0)}] + \foreach \x in {0,...,7} + { + \shadedraw [xscale=.3,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++(0,20pt) -- ++(4pt,4pt) -- ++(6pt,0pt) -- ++(4pt,-4pt)-- ++(0pt,-20pt) -- cycle; + } + \end{scope} + + \shadedraw [top color=firstcolor, bottom color=firstcolor!50!white] (-10pt,0) sin (30pt,2pt) cos (70pt,0pt) -- (70pt,-10pt) sin (35pt,-8pt) cos (-10pt,-10pt) --cycle; + + \draw [<->] (0,-15pt) -- (61.6pt,-15pt) node [below,pos=.5] {$x$}; + \draw [<->] (100.2pt,35.28pt) -- (66.6pt,0pt) node [right,pos=.5] {$y$}; + + \end{tikzpicture} + + + + + + + A rectangle shape is drawn below also not closed on one side. + + +

      + A 2-D sketch is drawn below showing a rectangle shape with one side opened and the sides labeled x and y. + The dimensions are unknown. +

      +       + + + \begin{tikzpicture}[scale=0.66] + + \draw (0,0) -- (0,5) -- (8,5) node [pos=.5,below] {$x$} -- (8,0) node [pos=.5,right] {$y$}; + \draw [firstcolor,very thick] (0,0) sin (4,.2) cos (8,0); + + \end{tikzpicture} + + + + +       +
      + +
      2. + +
    2. +

      + We want to maximize the area; + as in the example before, + + \text{ Area } = xy + . + This is our fundamental equation. + This defines area as a function of two variables, + so we need another equation to reduce it to one variable. +

      + +

      + We again appeal to the perimeter; + here the perimeter is + + \text{ Perimeter } = 100 = x+2y + . + The perimeter is our constraint equation. + Note how this is a different equation for perimeter than in + , + since one of the sides does not need to be fenced. +

      +
      3. + +
    3. +

      + We now reduce the fundamental equation to a single variable using + our constraint equation. + In the perimeter equation, solve for y: y = 50 - x/2. + We can now write Area as + + \text{ Area } = A(x) \amp= x(50-x/2) + \amp = 50x - \frac12x^2 + . + Area is now defined as a function of one variable. +

      +
      4. + +
    4. +

      + We want the area to be non-negative. + Since A(x) = x(50-x/2), + we want x\geq 0 and 50-x/2\geq 0. + The latter inequality implies that x\leq100, + so 0\leq x\leq 100. +

      +
      5. + +
    5. +

      + We now find the extreme values. + At the endpoints, the minimum is found, + giving an area of 0. +

      + +

      + Find the critical points. + We have A'(x) = 50-x; + setting this equal to 0 and solving for x returns x=50. + This gives an area of + + A(50) = 50(25) = 1250 + . +

      +
      6. + +
    6. +

      + We earlier set y = 50-x/2; thus y = 25. + Thus our rectangle will have two sides of length 25 and one + side of length 50, + with a total area of + + 1250 + . +

      +
      7. +
    +

    +     + +     + +

    + Keep in mind as we do these problems that we are practicing a process; + that is, we are learning to turn a situation into a system of equations. + These equations allow us to write a certain quantity as a function of one + variable, which we then optimize. +

    + + + + + + Optimization: minimizing cost + +

    + A power line needs to be run from a power station located on the beach + to an offshore facility. + + shows the distances between the power station to the facility. +

    + +

    + It costs \$50/\text{ ft } to run a + power line along the land, and + \$130/\text{ ft } to run a power + line under water. + How much of the power line should be run along the land to minimize the + overall cost? What is the minimal cost? +

    + +
    + Running a power line from the power station to an offshore + facility with minimal cost in + + + a power line is ran from the power station to an offshore facility. + + +

    + A power line is run from a power station to an offshore facility 5000 underwater. + It is also run 1000 along the offshore facility forming a right angle triangle with the under water powerline. +

    +

    + The hypotenuse of the triangle shows the distance between the station and offshore facility. +

    +     + + + \begin{tikzpicture}[>=latex] + + \begin{scope} + \clip (0,0) rectangle (140pt,50pt); + \draw [inner color=firstcolor,draw = white] (-10,-25pt) rectangle (180pt,70pt); + + \end{scope} + \draw [top color=treestump, bottom color=white,draw=white] (0,0) rectangle (140pt,-15pt); + \draw (0,0) -- (150pt,0); + \filldraw [fill=white] (10pt,-5pt) rectangle (20pt,5pt); + %\filldraw [fill=white] (130pt,35pt) circle (7pt); + + \draw [<->] (15pt,-10pt) -- (130pt,-10pt) node [below,pos=.5] { 5000 ft}; + \draw [<->] (140pt,0pt) -- (140pt,35pt) node [right, pos=.5] { 1000 ft}; + + \draw [ultra thick] (20pt,0) -- (70pt,0); + \filldraw (70pt,0) circle (2pt); + \draw [dashed,thick] (70pt,0) -- (130pt,35pt); + \filldraw [fill=white] (130pt,35pt) circle (7pt); + + \end{tikzpicture} + + + +
    +     + +

    + We will follow the strategy of + implicitly, + without specifically numbering steps. +

    + +

    + There are two immediate solutions that we could consider, + each of which we will reject through + common sense. First, + we could minimize the distance by directly connecting the two + locations with a straight line. + However, this requires that all the wire be laid underwater, + the most costly option. + Second, we could minimize the underwater length by running a wire all + + 5000 + + along the beach, directly across from the offshore facility. + This has the undesired effect of having the longest distance of all, + probably ensuring a non-minimal cost. +

    + +

    + The optimal solution likely has the line being run along the ground for + a while, then underwater, as the figure implies. + We need to label our unknown distances the distance run along + the ground and the distance run underwater. + Recognizing that the underwater distance can be measured as the + hypotenuse of a right triangle, + we choose to label the distances as shown in + . +

    + +
    + Labeling unknown distances in + + + A power line is ran from the power station to an offshore facility with the distance run along + the ground and the distance run underwater unknown. + + +

    + A power line is run from a power station to an offshore facility 5000 underwater. Here, the distance run along + the ground and the distance run underwater is unknown and thereby labeled as x. + It is also run 1000 along the offshore facility forming a right angle triangle with the power line. +

    +

    + The horizontal line of the image is labelled 5000-x which is the distance of the power line laid along land,and + x from the middle to the pont of the offshore facility which is the distance the power line is not laid. + The hypotenuse of the right angle triangle measures underwater distance with the equation \sqrt{x^2-1000^2}. +

    +     + + + \begin{tikzpicture}[>=latex] + + \begin{scope} + \clip (0,0) rectangle (140pt,50pt); + \draw [inner color=firstcolor,draw = white] (-10,-25pt) rectangle (180pt,70pt); + + \end{scope} + \draw [top color=treestump, bottom color=white,draw=white] (0,0) rectangle (140pt,-15pt); + \draw (0,0) -- (150pt,0); + \filldraw [fill=white] (10pt,-5pt) rectangle (20pt,5pt); + %\filldraw [fill=white] (130pt,35pt) circle (7pt); + + \draw [<->] (15pt,-10pt) -- (70pt,-10pt) node [below,pos=.5] { $5000 -x$}; + \draw [<->] (70pt,-10pt) -- (130pt,-10pt) node [below,pos=.5] { $x$}; + \draw [<->] (140pt,0pt) -- (140pt,35pt) node [right, pos=.5] { 1000 ft}; + + \draw [ultra thick] (20pt,0) -- (70pt,0); + \filldraw (70pt,0) circle (2pt); + \draw [dashed,thick] (70pt,0) -- (130pt,35pt) ; + \draw (100pt,23pt) node [rotate=30] { $\sqrt{x^2+1000^2}$}; + \filldraw [fill=white] (130pt,35pt) circle (7pt); + + \end{tikzpicture} + + + +
    + +

    + By choosing x as we did + (instead of letting x be the distance along the land), + we make the expression under the square root simple. + We now create the cost function. + + \text{Cost} \amp ={}\amp\amp \text{land cost} \amp\amp +\text{water cost} + \amp\amp\amp \$50\times \text{land distance} \amp\amp +\$130 \times \text{water distance} + \amp\amp\amp 50(5000-x) \amp\amp +130\sqrt{x^2+1000^2} + . +

    + +

    + So we have c(x) = 50(5000-x)+ 130\sqrt{x^2+1000^2}. + This function only makes sense on the interval [0,5000]. + While we are fairly certain the endpoints will not give a minimal cost, + we still evaluate c(x) at each to verify. + + c(0) \amp = 380{,}000\amp c(5000) \amp \approx 662{,}873 + . + (Notice that if x=0, the line is run the full + + 5000 + + along land and a full + + 1000 + + under water. + If x=5000, the line is run the maximum distance underwater.) +

    + +

    + We now find the critical values of c(x). + We compute c'(x) as + + c'(x) = -50+\frac{130x}{\sqrt{x^2+1000^2}} + . +

    + +

    + Recognize that this is never undefined. + Setting c'(x)=0 and solving for x, we have: + + -50+\frac{130x}{\sqrt{x^2+1000^2}} \amp = 0 + \frac{130x}{\sqrt{x^2+1000^2}} \amp = 50 + \frac{130^2x^2}{x^2+1000^2} \amp = 50^2 + 130^2x^2 \amp = 50^2(x^2+1000^2) + 130^2x^2-50^2x^2 \amp = 50^2\cdot1000^2 + (130^2-50^2)x^2 \amp = 50,000^2 + x^2 \amp = \frac{50,000^2}{130^2-50^2} + x \amp = \frac{50,000}{\sqrt{130^2-50^2}} + x \amp = \frac{50,000}{120} =\frac{1250}3\approx 416.67 + . +

    + +

    + Evaluating c(x) at x=416.67 gives a minimal cost of about + \$370{,}000. + The distance the power line is laid along land is + 5000-416.67 = 4583.33 ft., and the underwater distance is + \sqrt{416.67^2+1000^2} \approx 1083 ft. +

    +     + +     + + + +

    + In the exercises you will see a variety of situations that require you to + combine problem-solving skills with calculus. + Focus on the process; + learn how to form equations from situations that can be manipulated into + what you need. + Eschew memorizing how to do this kind of problem + as opposed to that kind of problem. + Learning a process will benefit one far longer than memorizing a specific + technique. +

    + + + + + + +

    + + introduces our final application of the derivative: + differentials. + Given y=f(x), + they offer a method of approximating the change in y after x + changes by a small amount. +

    + + + + Terms and Concepts + + + + +

    + + An optimization problem + is essentially an extreme values + problem in a story problem setting. +

    +     + +     + + + + +

    + + This section teaches one to find the extreme values of a + function that has more than one variable. +

    +     + +     +     + + + Problems + + + + + $sum = random(50,200,2); + if($envir{problemSeed}==1){$sum=100;}; + $maxprod = ($sum/2)**2; + + +

    + Find the maximum product of two numbers + (not necessarily integers) + that have a sum of . +

    + + If there is no maximum product, enter DNE. + +

    + +

    +     +     +     + + + + + $prod = random(400,600,10); + if($envir{problemSeed}==1){$prod=500;}; + $minsum = Compute("2 sqrt($prod)"); + + +

    + Find the minimum sum of two positive numbers + whose product is . +

    + + If there is no minimum sum, enter DNE. + +

    + +

    +     +     +     + + + + + $prod = random(400,600,10); + if($envir{problemSeed}==1){$prod=500;}; + $maxsum = Compute("DNE"); + + +

    + Find the maximum sum of two positive numbers + whose product is . +

    + + If there is no maximum sum, enter DNE. + +

    + +

    +     + +

    + There is no maximum sum; + the fundamental equation has only 1 critical value that + corresponds to a minimum. +

    +     +     +     + + + + + $prod = random(400,600,10); + $upper = random(200,500,10); + if($envir{problemSeed}==1){$prod=500;$upper=300;}; + Context("Fraction-NoDecimals"); + $maxsum = Fraction("$upper + $prod/$upper"); + + +

    + Find the maximum sum of two numbers, + each of which is less than or equal to , + whose product is . +

    + + If there is no maximum sum, enter DNE. + +

    + +

    +     +     +     + + + + + $hyp = random(1,9,1); + if($envir{problemSeed}==1){$hyp=1;}; + $maxarea=Real($hyp**2/4); + + +

    + Find the maximal area of a right triangle with hypotenuse of + length . +

    +

    + +

    +     +     +     + + + + + contextUnits.pl + + + Context(context::Fraction::extending(Context('Units')->withUnitsFor('length'))); + + $total = random(800, 1500, 100); + if ($envir{problemSeed} == 1) { $total = 1000; } + + $length = Fraction($total, 6); + $width = Fraction($total, 8); + $l = FormulaWithUnits($length, 'ft'); + $w = FormulaWithUnits($width, 'ft'); + $multians = MultiAnswer($l, $w)->with( + singleResult => 1, + checker => sub { + my ($correct, $student, $self) = @_; + my ($stu1, $stu2) = @{$student}; + my ($cor1, $cor2) = @{$correct}; + if (($cor1 == $stu1 && $cor2 == $stu2) + || ($cor1 == $stu2 && $cor2 == $stu1)) + { + return 1; + } else { + return 0; + } + } + ); + + +

    + A rancher has feet of fencing in which to construct + adjacent, equally sized rectangular pens, as shown below. + What dimensions should these pens have to maximize the enclosed area? +

    + + + A rectangle divided into two parts. + + + A large rectangle divided into two equal rectangles. + + + \begin{tikzpicture} + \draw [thick] (0,0) rectangle (3,2); + \draw [thick] (1.5,0) -- (1.5,2); + \end{tikzpicture} + + + + Enter the two dimensions, one in each blank, for one of the two pens here. Use ft for feet. + +

    + +

    +

    + +

    +     +     +     + + + + + $V = NumberWithUnits("355 cm^3"); + parser::Root->Enable; + $r = Compute("root(3,355/(2pi))"); + $h = Compute("root(3,1420/pi)"); + $rU = NumberWithUnits("$r cm"); + $hU = NumberWithUnits("$h cm"); + $showwork = '[@ explanation_box(message => "Discuss whether or not your calculation suggests that a real world soda can is designed to minimize the materials cost.") @]*'; + + +

    + A standard soda can is roughly cylindrical and holds + of liquid. + What dimensions should the cylinder have to minimize the material + needed to produce the can? + Based on your dimensions, + determine whether or not the standard can is produced to + minimize the material costs. +

    + + Enter its radius here. + +

    + +

    + + Enter its height here. + +

    + +

    +

    + +

    +     +     +     + + + + + $V = NumberWithUnits("206 in^3"); + parser::Root->Enable; + $r = Compute("root(3,206/(2pi))"); + $h = Compute("root(3,824/pi)"); + $rU = NumberWithUnits("$r in"); + $hU = NumberWithUnits("$h in"); + $showwork = '[@ explanation_box(message => "Discuss whether or not your calculation suggests that a #10 can is designed to minimize the materials cost.") @]*'; + + +

    + Find the dimensions of a cylindrical can with a volume of + that minimizes the surface area. +

    + + Enter its radius here. + +

    + +

    + + Enter its height here. + +

    + +

    +

    + The #10 can is a standard sized can used by the restaurant + industry that holds about + with a diameter of + 6\,\frac{3}{16}\,\text{in} + and height of + 7\,\text{in}. + Does it seem these dimensions where chosen with minimization in mind? +

    +

    + +

    +     +     +     + + + + + $V = NumberWithUnits("355 cm^3"); + parser::Root->Enable; + $r = Compute("root(3,355/(4pi))"); + $h = Compute("root(3,5680/pi)"); + $rU = NumberWithUnits("$r cm"); + $hU = NumberWithUnits("$h cm"); + $showwork = '[@ explanation_box(message => "Discuss whether or not your calculation suggests that a real world soda can is designed to minimize the materials cost.") @]*'; + + +

    + A standard soda can is roughly cylindrical and holds + of liquid. + A real-world soda can has material on the top and bottom that is thicker + than the material around the side. + Assume that the top/bottom material is twice as thick as the material around the side. + What dimensions should the cylinder have to minimize the material + needed to produce the can? + Based on your dimensions and the assumption about material thickness, + determine whether or not the standard can is produced to + minimize the material costs. +

    + + Enter its radius here. + +

    + +

    + + Enter its height here. + +

    + +

    +

    + +

    +     +     +     + + + + + $V = NumberWithUnits("36*18**2 in^3"); + + +

    + The United States Postal Service charges more for boxes whose + combined length and girth exceeds 108 inches. + (The length of a package is the length of its longest side; + the girth is the perimeter of the cross section, i.e., 2w+2h). +

    +

    + What is the maximum volume of a package with a square cross + section (w=h) that does not exceed the + 108 inch standard? +

    +

    + +

    +     +     +     + + + + + $D = random(10,20,1); + if($envir{problemSeed}==1){$D=12;}; + $w = Compute("$D/sqrt(3)"); + $h = Compute("sqrt($D^2 - $w^2)"); + $wU = NumberWithUnits("$w in"); + $hU = NumberWithUnits("$h in"); + $multians = MultiAnswer($wU, $hU)->with( + singleResult => 1, + checker => sub { + my ( $correct, $student, $self ) = @_; + my ( $stu1, $stu2 ) = @{$student}; + my ( $cor1, $cor2 ) = @{$correct}; + if ( ($cor1 == $stu1 && $cor2 == $stu2) || + ($cor1 == $stu2 && $cor2 == $stu1) ) { + return 1; + } else { + return 0; + } + } + ); + + +

    + The strength S of a wooden beam is directly proportional + to its cross sectional width w and the square of its + height h. + that is, S = kwh^2 for some constant k. +

    + + + A circle with a rectangle divided diagonally inscribed inside it. + + +

    + A circular log with diameter of 12 has rectangle inscribed inside. The rectangle is divided diagonally forming + two equal right-angle triangles. The line that divides it is the diameter. +

    +

    + The shorter side of the rectangle is labeled w which serves as the cross sectional width and the + longer side is labelled h which is the height. +

    +     + + \begin{tikzpicture}[scale=1.8] + \draw [thick] (0,0) circle (1cm); + \draw [thick] (-.6,.8) -- node [pos=.5,right] { \(12\)} (.6,-.8); + \draw [thick] (-.6,-.8) rectangle (.6,.8); + \draw (.7,0) node { \(h\)} (0,.7) node { \(w\)}; + \end{tikzpicture} + + +

    + Given a circular log with diameter of inches, + what sized beam can be cut from the log with maximum strength? +

    + + Enter the cross-sectional dimensions here. + +

    + +

    +

    + +

    +     +     +     + + + + + do { + ($x,$y) = random_subset(2,2..6); + $ycost=random(30000,50000,5000); + $xcost=$ycost + random(30000,50000,5000); + } until ($y - $ycost*$x/sqrt($xcost**2 - $ycost**2) > 0); + if($envir{problemSeed}==1){$x=2;$y=5;$xcost=80000;$ycost=50000}; + Context("Currency")->flags->set(trimTrailingZeros=>1,tolerance=>1); + $xC = Currency($xcost); + $yC = Currency($ycost); + $g = Compute("50000"); + $s = Compute("80000"); + $a = Compute("5"); + $b = Compute("2"); + $z = max(0,$y - $ycost*$x/sqrt($xcost**2 - $ycost**2)); + $C = Currency($z*$ycost + $xcost*sqrt($x**2+($y - $z)**2)); + Context("Numeric"); + $zU = NumberWithUnits("$z mi"); + + +

    + A power line is to be run to an offshore facility in the manner + described in . + The offshore facility is miles at sea and + miles along the shoreline from the power plant. + It costs per mile to lay a power line underground and + to run the line underwater. +

    +

    + How much of the power line should be run underground? + What is the minimum overall cost? +

    + + Enter the length of line under ground here. + +

    + +

    + + Enter the minimum overall cost here. + +

    + +

    +     +     +     + + + + + do { + ($x,$y) = random_subset(2,2..6); + $ycost=random(30000,50000,5000); + $xcost=$ycost + random(30000,50000,5000); + } until ($y - $ycost*$x/sqrt($xcost**2 - $ycost**2) < 0); + if($envir{problemSeed}==1){$x=5;$y=2;$xcost=80000;$ycost=50000}; + Context("Currency")->flags->set(trimTrailingZeros=>1,tolerance=>1); + $xC = Currency($xcost); + $yC = Currency($ycost); + $g = Compute("50000"); + $s = Compute("80000"); + $a = Compute("5"); + $b = Compute("2"); + $z = max(0,$y - $ycost*$x/sqrt($xcost**2 - $ycost**2)); + $C = Currency($z*$ycost + $xcost*sqrt($x**2+($y - $z)**2)); + Context("Numeric"); + $zU = NumberWithUnits("$z mi"); + + +

    + A power line is to be run to an offshore facility in the manner + described in . + The offshore facility is miles at sea and + miles along the shoreline from the power plant. + It costs per mile to lay a power line underground and + to run the line underwater. +

    +

    + How much of the power line should be run underground? + What is the minimum overall cost? +

    + + Enter the length of line under ground here. + +

    + +

    + + Enter the minimum overall cost here. + +

    + +

    +     +     +     + + + + + ($x,$y) = random_subset(2,10..40); + $run = random(18,25,1); + $swim = random(1.2,2.4,0.1); + if($envir{problemSeed}==1){$x=15;$y=20;$run=22;$swim=1.5;}; + $runU = NumberWithUnits("$run ft/s"); + $swimU = NumberWithUnits("$swim ft/s"); + $z = max(0,$y - $swim*$x/sqrt($run**2 - $swim**2)); + $zU = NumberWithUnits("$z ft"); + + +

    + A woman throws a stick into a lake for her dog to fetch; + the stick is feet down the shore line + and feet into the water from there. + The dog may jump directly into the water and swim, + or run along the shore line to get closer to the stick before + swimming. The dog runs about + and swims about . +

    +

    + How far along the shore should the dog run to minimize the + time it takes to get to the stick? (Hint: the figure from + can be useful.) +

    +

    + +

    +     +     +     + + + + + ($x,$y) = random_subset(2,10..40); + $run = random(18,25,1); + $swim = random(1.2,2.4,0.1); + if($envir{problemSeed}==1){$x=30;$y=15;$run=22;$swim=1.5;}; + $runU = NumberWithUnits("$run ft/s"); + $swimU = NumberWithUnits("$swim ft/s"); + $z = max(0,$y - $swim*$x/sqrt($run**2 - $swim**2)); + $zU = NumberWithUnits("$z ft"); + + +

    + A woman throws a stick into a lake for her dog to fetch; + the stick is feet down the shore line + and feet into the water from there. + The dog may jump directly into the water and swim, + or run along the shore line to get closer to the stick before + swimming. The dog runs about + and swims about . +

    +

    + How far along the shore should the dog run to minimize the + time it takes to get to the stick? + (Google calculus dog to learn more about a dog's ability to minimize times.) +

    +

    + +

    +     +     +     + + + + + $w = Compute("sqrt(2)"); + $multians = MultiAnswer($w,$w)->with( + singleResult => 1, + checker => sub { + my ( $correct, $student, $self ) = @_; + my ( $stu1, $stu2 ) = @{$student}; + my ( $cor1, $cor2 ) = @{$correct}; + if ( ($cor1 == $stu1 && $cor2 == $stu2) || + ($cor1 == $stu2 && $cor2 == $stu1) ) { + return 1; + } else { + return 0; + } + } + ); + + +

    + What are the dimensions of the rectangle with largest area that + can be drawn inside the unit circle? +

    +

    + +

    +

    + +

    +     +     +     +     +     +
    +
    + Differentials + + + +

    + In + we explored the meaning and use of the derivative. + This section starts by revisiting some of those ideas. +

    + +

    + Recall that the derivative of a function f can be used to find the slopes of lines tangent to the graph of f. + At x=c, + the tangent line to the graph of f has equation + + y = \fp(c)(x-c)+f(c) + . +

    + +

    + The tangent line can be used to find good approximations of f(x) for values of x near c. +

    + +

    + For instance, we can approximate + \sin(1.1) using the tangent line to the graph of + f(x)=\sin(x) at x=\pi/3 \approx 1.05. + Recall that \sin(\pi/3) = \sqrt{3}/2 \approx 0.866, + and \fp(\pi/3)=\cos(\pi/3) = 1/2. + Thus the tangent line to f(x) = \sin(x) at x=\pi/3 is: + + \ell(x) = \frac12(x-\pi/3)+0.866 + . +

    + +
    + Graphing f(x) = \sin(x) and its tangent line at x=\pi/3 in order to estimate \sin(1.1) + +
    + + + + A graph showcasing the sine function and a tangent line at a point. + + +

    + The graph portrays f(x) = \sin(x) in its typical wave-like form, beginning from x=0. A tangent line intersects the curve at x=\pi/3, giving us an insight into the function's behavior around this point. A small rectangular region on the graph suggests a section of interest, likely to be examined more closely in a subsequent depiction. The overall visualization offers a comprehensive understanding of the sine function's progression and how the tangent line serves as a useful approximation around x=\pi/3. +

    +     + + + + \begin{tikzpicture} + \begin{axis}[ + xtick=\empty, + extra x ticks={1.047}, + extra x tick labels={$\frac{\pi}{3}$}, + ytick = {0,0.5,1}, + extra y ticks={0.866}, + extra y tick labels={$\frac{\sqrt{3}}{2}$}, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=2.36, + ] + \addplot+[domain=0:2.1, rightarrow] {sin(deg(x))}; + \addplot [tangentline, domain=-.1:1.5] {.5*(x-1.047197551)+0.866025}; + \addplot [soliddot] coordinates {(1.047,0.866)} node [below right] {\small $(\pi/3,\sqrt{3}/3)$}; + \draw (axis cs:1.04,.86) rectangle (axis cs:1.122,.9); + \end{axis} + \end{tikzpicture} + + + +
    + +
    + + + + A close-up view of the sine function and its tangent line around a specific point. + + +

    + Zooming in on the small rectangle from the prior graph, + we get a magnified view around the point x=\pi/3 for f(x) = \sin(x) and its tangent line. + This zoomed-in perspective demonstrates the precision with which the tangent line estimates \sin(1.1) at x=1.1. + We can see how points on the tangent line lie quite close to the graph of f(x), + illustrating the fact that the tangent line gives a good approximation to the original graph, + near the point where it meets the graph. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + xtick=\empty,% + extra x ticks={1.047,1.1}, + extra x tick labels={$\frac{\pi}{3}$,$1.1$}, + extra y ticks={0.866}, + extra y tick labels={$\frac{\sqrt{3}}{2}$}, + ytick={0.86,0.87,0.88,0.89}, + %extra y ticks={0.866}, + %extra y tick labels={$\frac{\sqrt{3}}{2}$}, + xdiscontinuity, + ydiscontinuity, + ymin=.86,ymax=.9,% + xmin=1.04,xmax=1.122, + ] + \addplot+[infinite,domain=1.04:1.12] {sin(deg(x))}; + \addplot [tangentline,domain=1.04:1.115] {.5*(x-1.0471)+0.866025}; + \addplot [soliddot] coordinates{(1.047,0.866)} node [below right] {\small $\left(\pi/3,\sqrt{3}/3\right)$}; + \addplot [soliddot,guideline] coordinates {(1.1,.8925)} node [pin={180:{$\ell(1.1)\approx \sin(1.1)$}}] {}; + \addplot [soliddot] coordinates {(1.1,.8912)} node [pin={0:{$\sin(1.1)$}}] {}; + + \end{axis} + \end{tikzpicture} + + + +
    +     +
    + +

    + In , + we see a graph of f(x) = \sin(x) graphed along with its tangent line at x=\pi/3. + The small rectangle shows the region that is displayed in . + In this figure, we see how we are approximating + \sin(1.1) with the tangent line, evaluated at 1.1. + Together, the two figures show how close these values are. +

    + +

    + Using this line to approximate \sin(1.1), we have: + + \ell(1.1) \amp = \frac{1}{2}(1.1-\pi/3)+0.866 + \amp = \frac12(0.053)+0.866 = 0.8925 + . +

    + +

    + (We leave it to the reader to see how good of an approximation this is.) +

    + + + + + + + + +

    + We now generalize this concept. + Given f(x) and an x-value c, + the tangent line is y=\ell(x), where \ell(x) = \fp(c)(x-c)+f(c). + Clearly, f(c) = \ell(c). + Let \dx be a small number, + representing a small change in the x-value. + We assert that: + + f(c+\dx) \approx \ell(c+\dx) + , + since the tangent line to a function approximates well the values of that function near x=c. + In fact, the tangent line is the graph of the linear function that best approximates the value of + f(x) for x near c. + Because of its value in applications, we give it a name. +

    + + + +

    + Let f be differentiable on an open inverval I containing c. + The function \ell(x) = \fp(c)(x-c)+f(c) is called the linearization, + or linear approximation of f at c. + + linearization + approximationtangent line + approximationlinear +

    +     +     + +

    + As the x-value changes from c to c+\dx, + the y-value of f changes from f(c) to f(c+\dx). + We call this change of y-value \dy. + That is: + + \dy = f(c+\dx)-f(c) + . +

    + +

    + Replacing f(c+\dx) with its tangent line approximation, we have + + \dy \amp \approx \ell(c+\dx) - f(c) + \amp = \fp(c)\big((c+\dx)-c\big)+f(c) - f(c) + \amp =\fp(c)\dx + . +

    + +

    + This final equation is important; + it becomes the basis of and . + In short, it says that when the x-value changes from c to c+\Delta x, + the y value of a function f changes by about \fp(c)\Delta x. +

    + + + +

    + We introduce two new variables, + dx and dy in the context of a formal definition. +

    + + + Differentials of <m>x</m> and <m>y</m> + +

    + Let y=f(x) be differentiable. + The differential of x, + denoted dx, is any nonzero real number + (usually taken to be a small number). + + differential + + derivativedifferential + + The differential of y, denoted dy, is + + dy = \fp(x)dx + . +

    +     +     + + + + +

    + We can solve for \fp(x) in the above equation: + \fp(x) = dy/dx. + This states that the derivative of f with respect to x is the differential of y divided by the differential of x; + this is not the alternate notation for the derivative, + \lz{y}{x}. + This latter notation was chosen because of the fraction-like qualities of the derivative, + but again, + it is one symbol and not a fraction. +

    + + + +

    + It is helpful to organize our new concepts and notations in one place. +

    + + + + + Differential Notation +

    + Let y = f(x) be a differentiable function. + + differentialnotation + +

    + +

    +

      +
    1. +

      + Let \dx represent a small, + nonzero change in x value. +

      +
      2. + +
    2. +

      + Let dx represent a small, + nonzero change in x value (, \dx = dx). +

      +
      3. + +
    3. +

      + Let \dy be the change in y value as x changes by \dx; hence + + \dy = f(x+\dx)-f(x) + . +

      +
      4. + +
    4. +

      + Let dy = \fp(x)dx which, + by Equation, + is an approximation of the change in y-value as x changes by \dx; + dy \approx \dy. +

      +
      5. +
    +

    +     + + + +

    + What is the value of differentials? + Like many mathematical concepts, + differentials provide both practical and theoretical benefits. + We explore both here. +

    + + + Finding and using differentials + +

    + Consider f(x) = x^2. + Knowing f(3) = 9, approximate f(3.1). +

    +     + +

    + The x-value is changing from x=3 to x=3.1; + therefore, we see that dx=0.1. + If we know how much the y-value changes from f(3) to f(3.1) (, if we know \dy), + we will know exactly what f(3.1) is (since we already know f(3)). + We can approximate \dy with dy. + + \dy \amp \approx dy + \amp = \fp(3)dx + \amp = 2\cdot 3\cdot 0.1 = 0.6 + . +

    + +

    + We expect the y-value to change by about 0.6, + so we approximate f(3.1) \approx 9.6. +

    + +

    + We leave it to the reader to verify this, + but the preceding discussion links the differential to the tangent line of f(x) at x=3. + One can verify that the tangent line, + evaluated at x=3.1, also gives y=9.6. +

    +     + +     + + + +

    + Of course, it is easy to compute the actual answer (by hand or with a calculator): + 3.1^2 = 9.61. (Before we get too cynical and say + Then why bother?, + note our approximation is really good!) +

    + +

    + So why bother? +

    + +

    + In most real life situations, + we do not know the function that describes a particular behavior. + Instead, we can only take measurements of how things change measurements of the derivative. +

    + +

    + Imagine water flowing down a winding channel. + It is easy to measure the speed and direction (, the + velocity) of water at any location. + It is very hard to create a function that describes the overall flow, + hence it is hard to predict where a floating object placed at the beginning of the channel will end up. + However, we can approximate + the path of an object using differentials. + Over small intervals, + the path taken by a floating object is essentially linear. + Differentials allow us to approximate the true path by piecing together lots of short, + linear paths. + This technique is called Euler's Method, + studied in introductory Differential Equations courses. +

    + + + +

    + We use differentials once more to approximate the value of a function. + Even though calculators are very accessible, + it is neat to see how these techniques can sometimes be used to easily compute something that looks rather hard. +

    + + + Using differentials to approximate a function value + +

    + Approximate \sqrt{4.5}. +

    +     + +

    + We expect \sqrt{4.5} \approx 2, yet we can do better. + Let f(x) = \sqrt{x}, and let c=4. + Thus f(4) = 2. + We can compute \fp(x) = 1/(2\sqrt{x}), so \fp(4) = 1/4. +

    + +

    + We approximate the difference between f(4.5) and f(4) using differentials, + with dx = 0.5: + + f(4.5)-f(4) = \dy \approx dy = \fp(4)\cdot dx = \frac14 \cdot \frac12 = \frac18 = 0.125 + . +

    + +

    + The approximate change in f from x=4 to x=4.5 is 0.125, + so we approximate \sqrt{4.5} \approx 2.125. +

    +     + +     + +

    + Differentials are important when we discuss integration. + When we study that topic, we will use notation such as + + \int f(x)\,dx + + quite often. + While we don't discuss here what all of that notation means, + note the existence of the differential dx. + Proper handling of integrals + comes with proper handling of differentials. +

    + +

    + In light of that, we practice finding differentials in general. +

    + + + Finding differentials + +

    + In each of the following, find the differential dy. +

    + +

    +

      +

    1. y = \sin(x)

      2. + +

    2. y = e^x\left(x^2+2\right)

      3. + +

    3. y = \sqrt{x^2+3x-1}

      4. +
    +

    +     + +

    +

      +
    1. +

      + y = \sin(x): As f(x) = \sin(x), \fp(x) = \cos(x). + Thus + + dy = \cos(x)dx + . +

      +
      2. + +
    2. +

      + y = e^x\left(x^2+2\right): Let f(x) = e^x\left(x^2+2\right). + We need \fp(x), + requiring the . +

      + +

      + We have \fp(x) = e^x\left(x^2+2\right) + 2xe^x, so + + dy = \left(e^x\left(x^2+2\right) + 2xe^x\right)dx + . +

      +
      3. + +
    3. +

      + y = \sqrt{x^2+3x-1}: Let f(x) = \sqrt{x^2+3x-1}; + we need \fp(x), + requiring the . +

      + +

      + We have \fp(x) = \frac{1}{2}\left(x^2+3x-1\right)^{-\frac{1}{2}}(2x+3) = \frac{2x+3}{2\sqrt{x^2+3x-1}}. + Thus + + dy = \frac{(2x+3)dx}{2\sqrt{x^2+3x-1}} + . +

      +
      4. +
    +

    +     +     + +

    + Finding the differential dy of y=f(x) is really no harder than finding the derivative of f; + we just multiply \fp(x) by dx. + It is important to remember that we are not simply adding the symbol + dx at the end. +

    + +

    + We have seen a practical use of differentials as they offer a good method of making certain approximations. + Another use is error propagation. + Suppose a length is measured to be x, + although the actual value is x+\dx + (where \dx is the error, which we hope is small). + This measurement of x may be used to compute some other value; + we can think of this latter value as f(x) for some function f. + As the true length is x+\dx, + one really should have computed f(x+\dx). + The difference between f(x) and + f(x+\dx) is the propagated error. +

    + +

    + How close are f(x) and f(x+\dx)? + This is a difference in y values: + + f(x+\dx)-f(x) = \dy \approx dy + . +

    + +

    + We can approximate the propagated error using differentials. +

    + + + Using differentials to approximate propagated error + +

    + A steel ball bearing is to be manufactured with a diameter of + + 2 + . + The manufacturing process has a tolerance of \pm 0.1 + + + in the diameter. + Given that the density of steel is about + + 7.85 + , estimate the propagated error in the mass of the ball bearing. +

    +     + + +

    + The mass of a ball bearing is found using the equation + mass = volume density. + In this situation the mass function is a product of the radius of the ball bearing, + hence it is m = 7.85\frac43\pi r^3. + The differential of the mass is + + dm = 31.4\pi r^2 dr + . +

    + +

    + The radius is to be + + 1 + ; the manufacturing tolerance in the radius is \pm 0.05 + + , or \pm 0.005 + + . + The propagated error is approximately: + + \Delta m \amp \approx dm + \amp = 31.4\pi (1)^2 (\pm 0.005) + \amp = \pm 0.493\text{g} + +

    + +

    + Is this error significant? + It certainly depends on the application, + but we can get an idea by computing the + relative error. + The ratio between amount of error to the total mass is + + \frac{dm}{m} \amp = \pm \frac{0.493}{7.85\frac43\pi} + \amp =\pm \frac{0.493}{32.88} + \amp =\pm 0.015 + , + or \pm 1.5\%. +

    + +

    + We leave it to the reader to confirm this, + but if the diameter of the ball was supposed to be + + 10 + , the same manufacturing tolerance would give a propagated error in mass of \pm12.33 + + , which corresponds to a + percent error of \pm0.188\%. + While the amount of error is much greater (12.33 \gt 0.493), + the percent error is much lower. +

    +     + +     + + + + Terms and Concepts + + + + +

    + Given a differentiable function y=f(x), + we are generally free to choose a value for dx, + which then determines the value of dy. +

    +     +     + + + + + +

    + + The symbols dx and \dx + represent the same concept. +

    +     + +     + + + + +

    + + The symbols dy and \dy + represent the same concept. +

    +     + +     + + + + +

    + + Differentials are important in the study of integration. +

    +     + +     + + + + +

    + How are differentials and tangent lines related? +

    + +     + + + +     + + + + +

    + + In real life, + differentials are used to approximate function values + when the function itself is not known. +

    +     + +     +     + + + Problems + + + +

    + Use differentials to approximate the given value by hand. +

    +     + + + + + $x0 = random(2,9,1); + $dx = random(0.03,0.09,0.01); + if($envir{problemSeed}==1){$x0=2;$dx=0.05}; + Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); + $f = Compute("x^2"); + $df = $f->D('x'); + $x1 = $x0 + $dx; + $dy = $dx * $df->eval(x=>$x0); + $y = $f->eval(x=>$x0); + $r = $y+$dy; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $given = $f->substitute(x=>Formula("$x1")); + + +

    + +

    +

    + +

    +     + +

    + Use y = ; + dy = \cdot dx with x= + and dx = . + Thus dy = ; knowing + =, + we have \approx . +

    +     +     +     + + + + $x0 = random(2,9,1); + $dx = -random(0.03,0.09,0.01); + if($envir{problemSeed}==1){$x0=6;$dx=-0.07}; + Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); + $f = Compute("x^2"); + $df = $f->D('x'); + $x1 = $x0 + $dx; + $dy = $dx * $df->eval(x=>$x0); + $y = $f->eval(x=>$x0); + $r = $y+$dy; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $given = $f->substitute(x=>Formula("$x1")); + + +

    + +

    +

    + +

    +     + +

    + Use y = ; + dy = \cdot dx with x= + and dx = . + Thus dy = ; knowing + =, + we have \approx . +

    +     +     +     + + + + + $x0 = random(2,9,1); + $dx = random(0.1,0.4,0.1); + if($envir{problemSeed}==1){$x0=5;$dx=0.1}; + Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); + $f = Compute("x^3"); + $df = $f->D('x'); + $x1 = $x0 + $dx; + $dy = $dx * $df->eval(x=>$x0); + $y = $f->eval(x=>$x0); + $r = $y+$dy; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $given = $f->substitute(x=>Formula("$x1")); + + +

    + +

    +

    + +

    +     + +

    + Use y = ; + dy = \cdot dx with x= + and dx = . + Thus dy = ; knowing + =, + we have \approx . +

    +     +     +     + + + + $x0 = random(2,9,1); + $dx = -random(0.1,0.4,0.1); + if($envir{problemSeed}==1){$x0=7;$dx=-0.2}; + Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); + $f = Compute("x^3"); + $df = $f->D('x'); + $x1 = $x0 + $dx; + $dy = $dx * $df->eval(x=>$x0); + $y = $f->eval(x=>$x0); + $r = $y+$dy; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $given = $f->substitute(x=>Formula("$x1")); + + +

    + +

    +

    + +

    +     + +

    + Use y = ; + dy = \cdot dx with x= + and dx = . + Thus dy = ; knowing + =, + we have \approx . +

    +     +     +     + + + + + $x0 = list_random(9,16,25,36,49); + $dx = random(0.3,0.9,0.1); + if($envir{problemSeed}==1){$x0=16;$dx=0.5}; + Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); + $f = Compute("sqrt(x)"); + $df = $f->D('x'); + $x1 = $x0 + $dx; + $dy = $dx * $df->eval(x=>$x0); + $y = $f->eval(x=>$x0); + $r = $y+$dy; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $given = $f->substitute(x=>Formula("$x1")); + + +

    + +

    +

    + +

    +     + +

    + Use y = ; + dy = \cdot dx with x= + and dx = . + Thus dy = ; knowing + =, + we have \approx . +

    +     +     +     + + + + $x0 = list_random(9,16,25,36,49); + $dx = -random(0.3,1.5,0.1); + if($envir{problemSeed}==1){$x0=25;$dx=-1}; + Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); + $f = Compute("sqrt(x)"); + $df = $f->D('x'); + $x1 = $x0 + $dx; + $dy = $dx * $df->eval(x=>$x0); + $y = $f->eval(x=>$x0); + $r = $y+$dy; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $given = $f->substitute(x=>Formula("$x1")); + + +

    + +

    +

    + +

    +     + +

    + Use y = ; + dy = \cdot dx with x= + and dx = . + Thus dy = ; knowing + =, + we have \approx . +

    +     +     +     + + + + + $x0 = list_random(8,27,64,125,216); + $dx = -random(0.3,1.5,0.1); + if($envir{problemSeed}==1){$x0=64;$dx=-1}; + Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); + parser::Root->Enable; + $f = Compute("root(3,x)"); + $df = $f->D('x'); + $x1 = $x0 + $dx; + $dy = $dx * $df->eval(x=>$x0); + $y = $f->eval(x=>$x0); + $r = $y+$dy; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $given = $f->substitute(x=>Formula("$x1")); + + +

    + +

    +

    + +

    +     + +

    + Use y = ; + dy = \cdot dx with x= + and dx = . + Thus dy = ; knowing + =, + we have \approx . +

    +     +     +     + + + + $x0 = list_random(8,27,64,125,216); + $dx = random(0.3,1.5,0.1); + if($envir{problemSeed}==1){$x0=8;$dx=0.5}; + Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); + parser::Root->Enable; + $f = Compute("root(3,x)"); + $df = $f->D('x'); + $x1 = $x0 + $dx; + $dy = $dx * $df->eval(x=>$x0); + $y = $f->eval(x=>$x0); + $r = $y+$dy; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $given = $f->substitute(x=>Formula("$x1")); + + +

    + +

    +

    + +

    +     + +

    + Use y = ; + dy = \cdot dx with x= + and dx = . + Thus dy = ; knowing + =, + we have \approx . +

    +     +     +     + + + + + $x0 = Compute("pi"); + $dx = 3-$x0; + Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); + $f = Compute("sin(x)"); + $df = $f->D('x'); + $x1 = 3; + $dy = $dx * $df->eval(x=>$x0); + $y = $f->eval(x=>$x0); + $r = $y+$dy; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $given = $f->substitute(x=>Formula("$x1")); + + +

    + +

    +

    + +

    +     + +

    + Use y = ; + dy = \cdot dx with x= + and dx = . + Thus dy = ; knowing + =, + we have \approx . +

    +     +     +     + + + + + $x0 = 0; + $dx = random(0.1,0.5,0.1); + if($envir{problemSeed}==1){$dx=0.1}; + Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); + parser::Root->Enable; + $f = Compute("e^x"); + $df = $f->D('x'); + $x1 = $x0 + $dx; + $dy = $dx * $df->eval(x=>$x0); + $y = $f->eval(x=>$x0); + $r = $y+$dy; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $given = $f->substitute(x=>Formula("$x1")); + + +

    + +

    +

    + +

    +     + +

    + Use y = ; + dy = \cdot dx with x= + and dx = . + Thus dy = ; knowing + =, + we have \approx . +

    +     +     +     +     + + + + +

    + Compute the differential dy. +

    +     + + + + $b = non_zero_random(-9,9,1); + $c = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b=3;$c=-5;}; + Context()->variables->add(dx=>"Real"); + $f = Compute("x^2+$b x + $c")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y= +

    +

    + +

    +     +     +     + + + + + ($m,$n) = random_subset(2,3..9); + $c = list_random(1,-1); + if($envir{problemSeed}==1){$m=7;$n=5;$c=-1}; + Context()->variables->add(dx=>"Real"); + $f = Compute("x^$m+$c x^$n")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y= +

    +

    + +

    +     +     +     + + + + + ($m,$n) = random_subset(2,1..9); + if($envir{problemSeed}==1){$m=4;$n=2;}; + Context()->variables->add(dx=>"Real"); + $f = Compute("1/($m x^$n)")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y=\displaystyle +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $n = random(2,3,1); + if($envir{problemSeed}==1){$a=2;$n=2;}; + Context()->variables->add(dx=>"Real"); + $f = Compute("($a x + sin(x))^$n")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y= +

    +

    + +

    +     +     +     + + + + + ($m,$n) = random_subset(2,2..9); + if($envir{problemSeed}==1){$m=2;$n=3;}; + Context()->variables->add(dx=>"Real"); + $f = Compute("x^$m + e^($n x)")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y= +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $b = random(2,9,1); + if($envir{problemSeed}==1){$a=4;$b=4;}; + Context()->variables->add(dx=>"Real"); + $f = Compute("$a/x^$b")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y=\displaystyle +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $b = random(1,9,1); + if($envir{problemSeed}==1){$a=2;$b=1;}; + Context()->variables->add(dx=>"Real"); + $f = Compute("($a x)/(tan(x) + $b)")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y=\displaystyle +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + if($envir{problemSeed}==1){$a=5;}; + Context()->variables->add(dx=>"Real"); + $f = Compute("ln($a x)")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y= +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','cos','tan','sec','csc','cot'); + if($envir{problemSeed}==1){$trig='sin';}; + Context()->variables->add(dx=>"Real"); + $f = Compute("e^x $trig(x)")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y= +

    +

    + +

    +     +     +     + + + + + ($trig1,$trig2) = random_subset(2,'sin','cos','tan','sec','csc','cot'); + if($envir{problemSeed}==1){$trig1='cos';$trig2='sin';}; + Context()->variables->add(dx=>"Real"); + $f = Compute("$trig1($trig2(x))")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y= +

    +

    + +

    +     +     +     + + + + + ($a,$b) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$a=1;$b=2;}; + Context()->variables->add(dx=>"Real"); + $f = Compute("(x+$a)/(x+$b)")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y=\displaystyle +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + if($envir{problemSeed}==1){$a=3;}; + Context()->variables->add(dx=>"Real"); + $f = Compute("$a^x ln(x)")->reduce; + $df = $f->D('x')->reduce; + $A = Formula("$df dx"); + + +

    + y= +

    +

    + +

    +     +     +     + + + + + $fstring = list_random('x ln(x) - x','x arctan(x) - (1/2) ln(1+x^2)'); + if($envir{problemSeed}==1){$fstring='x ln(x) - x';}; + Context()->variables->add(dx=>"Real"); + $f = Compute("$fstring")->reduce; + $df = ($fstring eq 'x ln(x) - x') ? Compute("ln(x)") : Compute("arctan(x)"); + $A = Formula("$df dx"); + + +

    + y= +

    +

    + +

    +     +     +     + + + + + $fstring = list_random('ln(sec(x))','ln(sin(x))',); + if($envir{problemSeed}==1){$fstring='ln(sec(x))';}; + Context()->variables->add(dx=>"Real"); + $f = Compute("$fstring")->reduce; + $df = ($fstring eq 'ln(sec(x))') ? Compute("tan(x)") : Compute("cot(x)"); + $A = Formula("$df dx"); + + +

    + y= +

    +

    + +

    +     +     +     +     + + + + + + $diameter = random(1,5,1); + $dx = random(1,$diameter,1); + if($envir{problemSeed}==1){$diamter=1;$dx=1;}; + $diameterU = NumberWithUnits("$diameter cm"); + $dxU = NumberWithUnits("$dx mm"); + $f = Compute("1/6*pi*x^3"); + $df = $f->D('x'); + $err = $df->eval(x=>$diameter) * $dx/10; + $errU = NumberWithUnits("$err cm^3"); + + +

    + A set of plastic spheres are to be made with a diameter of . + If the manufacturing process is accurate to , + what is the propagated error in volume of the spheres? +

    +

    + +

    +     +     +     + + + + + $dx = random(1,4,1); + $a = random(2,4,1); + $b = random(5,8,1); + if($envir{problemSeed}==1){$dx=2;$a=2;$b=5;}; + $f = Compute("16*x^2"); + $df = $f->D('x'); + $err_a = $df->eval(x=>$a) * $dx/10; + $err_aU = NumberWithUnits("$err_a ft"); + $err_b = $df->eval(x=>$b) * $dx/10; + $err_bU = NumberWithUnits("$err_b ft"); + + +

    + The distance, in feet, + a stone drops in t seconds is given by d(t) = 16t^2. + The depth of a hole is to be approximated by dropping a rock and listening for it to hit the bottom. + What is the propagated error if the time measurement is accurate to + /10 of a second and the measured time is: +

    +     + + +

    + seconds? +

    +

    + +

    +     +     + + +

    + seconds? +

    +

    + +

    +     +     +     +     + + + + + $diameter = random(12,20,1); + $dx = list_random(2,4,8,16); + if($envir{problemSeed}==1){$diameter=15;$dx=4}; + $f = Compute("pi/4*x^2"); + $df = $f->D('x'); + $err = $df->eval(x=>$diameter) * 1/$dx; + $errU = NumberWithUnits("$err in^2"); + + +

    + What is the propagated error in the measurement of the cross sectional area + of a circular log if the diameter is measured at '', + accurate to 1/''? +

    +

    + +

    +     +     +     + + + + + $h = random(8,10,1); + $lft = random(10,20,1); + $lin = random(1,11); + $dx = list-random(2,4,8,16); + if($envir{problemSeed}==1){$h=8;$lft=10;$lin=7;$dx=2}; + $f = Compute("$h*x"); + $df = $f->D('x'); + $x = $lft + $lin/12; + $err = $df->eval(x=>$x) * 1/$dx; + $errU = NumberWithUnits("$err ft^2"); + + +

    + A wall is to be painted that is ' high and is measured + to be ',\,'' long. + Find the propagated error in the measurement of the wall's surface area + if the measurement is accurate to 1/''. +

    +

    + +

    +     +     +     + + + +

    + The following exercises explore some issues related to surveying in which + distances are approximated using other measured distances and measured angles. + (Hint: Convert all angles to radians before computing.) +

    +     + + + + + $theta = 85.2; + $d = 25; + $f = Compute("$d*tan(x)"); + $l = $f->eval(x=>$theta*pi/180); + $lU = NumberWithUnits("$l ft"); + $df = $f->D('x'); + $err = $df->eval(x=>$theta*pi/180) * pi/180; + $errU = NumberWithUnits("$err ft"); + Context("Percent"); + $err_pct = Percent("$err/$l"); + + +

    + The length L of a long wall is to be approximated. + The angle \theta, as shown in the diagram + (not to scale), + is measured at a distance of 25 feet from the wall, and found to be ^\circ, + accurate to 1^\circ. + Assume that the triangle formed is a right triangle. +

    + + + A right-angled triangle representing an approximation challenge for a wall's length. + + +

    + The diagram presents a right-angled triangle utilized to navigate an approximation problem concerning the length, l, of a long wall. + The angle labeled as \theta is opposite the wall, + and the adjacent side is marked as 25. + The opposite side, representing the wall's length, is denoted by the variable l. + The triangle's hypotenuse is not specified in the given representation. +

    +     + + + \begin{tikzpicture}[scale=2.1] + \draw [ultra thick] (1,-1) -- node [pos=.5,right] {\(l=\)?}(1,1); + \draw [dashed] (1,1) -- (-1,-1) node [xshift=10pt,yshift=5pt] {\(\theta\)} -- node [pos=.5,below] {\(25'\)} (1,-1); + \end{tikzpicture} + + +     + + +

    + What is the measured length L of the wall? +

    +

    + +

    +     +     + + +

    + What is the propagated error? +

    +

    + +

    +     +     + + +

    + What is the percent error? +

    +

    + +

    +     +     +     +     + + + + + $theta = 71.5; + $d = 100; + $f = Compute("$d*tan(x)"); + $l = $f->eval(x=>$theta*pi/180); + $lU = NumberWithUnits("$l ft"); + $df = $f->D('x'); + $err = $df->eval(x=>$theta*pi/180) * pi/180; + $errU = NumberWithUnits("$err ft"); + Context("Percent"); + $err_pct = Percent("$err/$l"); + + +

    + The length L of a long wall is to be approximated. + The angle \theta, as shown in the diagram + (not to scale), + is measured at a distance of 100 feet from the wall, and found to be ^\circ, + accurate to 1^\circ. + Assume that the triangle formed is a right triangle. +

    + + + A right-angled triangle illustrating an approximation problem related to a long wall. + + +

    + The diagram showcases a right-angled triangle, + which aids in determining an approximation for the length, l, of a long wall. + The angle labeled as \theta is opposite the wall, + and the adjacent side is marked as 25. + The opposite side, representing the wall's length, is denoted by the variable l. + The triangle's hypotenuse is not specified in the given representation. +

    +     + + \begin{tikzpicture}[scale=2.1] + \draw [ultra thick] (1,-1) -- node [pos=.5,right] {\(l=\)?}(1,1); + \draw [dashed] (1,1) -- (-1,-1) node [xshift=10pt,yshift=5pt] {\(\theta\)} -- node [pos=.5,below] {\(100'\)} (1,-1); + \end{tikzpicture} + + +     + + +

    + What is the measured length L of the wall? +

    +

    + +

    +     +     + + +

    + What is the propagated error? +

    +

    + +

    +     +     + + +

    + What is the percent error? +

    +

    + +

    +     +     +     +     + + + + + $theta = 143; + $d = 50; + $f = Compute("2*$d*tan(x/2)"); + $l = $f->eval(x=>$theta*pi/180); + $lU = NumberWithUnits("$l ft"); + $df = $f->D('x'); + $err = $df->eval(x=>$theta*pi/180) * pi/180; + $errU = NumberWithUnits("$err ft"); + Context("Percent"); + $err_pct = Percent("$err/$l"); + + +

    + The length L of a long wall is to be calculated + by measuring the angle \theta shown in the diagram + (not to scale) at a distance of 50 feet from the wall. + Assume the formed triangle is an isosceles triangle. + The measured angle is ^\circ, + accurate to 1^\circ. +

    + + + An isosceles triangle used for calculating a wall's length, l. + + +

    + The diagram illustrates an isosceles triangle to facilitate a calculation related to determining the length, l, of a long wall. + The triangle's height, perpendicular to the base representing the wall, + is given as 50. The length of the wall is denoted as l. + The angle opposite the wall is labeled as \theta. +

    +     + + + \begin{tikzpicture}[scale=2.1] + \draw [ultra thick] (1,-1) -- node [pos=.5,right] {\(l=\)?}(1,1); + \draw [dashed] (1,1) -- (-1,0) node [xshift=10pt] {\(\theta\)} -- (1,-1); + \draw (-.5,0) -- node [pos=.5,draw=white,fill=white] {\(50'\)} (1,0); + \end{tikzpicture} + + +     + + +

    + What is the measured length L of the wall? +

    +

    + +

    +     +     + + +

    + What is the propagated error? +

    +

    + +

    +     +     + + +

    + What is the percent error? +

    +

    + +

    +     +     +     +     + + + + + + $theta = 143; + $d = 50; + $f = Compute("2*x*tan( ($theta*pi/180) /2)"); + $l = $f->eval(x=>$d); + $lU = NumberWithUnits("$l ft"); + $df = $f->D('x'); + $err = $df->eval(x=>$d) * 0.5; + $errU = NumberWithUnits("$err ft"); + Context("Percent"); + $err_pct = Percent("$err/$l"); + + +

    + Consider the setup in . + This time, assume the angle measurement of + 143^\circ is exact but the measured 50' from the wall is accurate to 6''. +

    + + + An isosceles triangle used for calculating a wall's length, l. + + +

    + The diagram illustrates an isosceles triangle to facilitate a calculation related to determining the length, l, of a long wall. + The triangle's height, perpendicular to the base representing the wall, + is given as 50. The length of the wall is denoted as l. + The angle opposite the wall is labeled as \theta. +

    +     + + \begin{tikzpicture} + \draw [ultra thick] (1,-1) -- node [pos=.5,right] {\scriptsize \(l=\)?}(1,1); + \draw [dashed] (1,1) -- (-1,0) node [xshift=10pt] {\scriptsize \(\theta\)} -- (1,-1); + \draw (-.5,0) -- node [pos=.5,draw=white,fill=white] {\scriptsize \(50'\)} (1,0); + \end{tikzpicture} + + +

    + What is the approximate percent error? +

    +

    + +

    +     +     +     + + + + + + parserPopUp.pl + + + $buttons = DropDown( + [ + 'Right triangle at 25 feet', + 'Right triangle at 100 feet', + 'Isosceles triangle at 50 feet', + ],2,showInStatic=>0); + + +

    + The length of the walls in + + are essentially the same. + Which setup gives the most accurate result? +

    +

    + +

    +     +     +     +     +     +     +
    + + +

    + We first learned of the derivative in the context of instantaneous rates of change and slopes of tangent lines. + We furthered our understanding of the power of the derivative by studying how it relates to the graph of a function + (leading to ideas of increasing/decreasing and concavity). + This chapter has put the derivative to yet more uses: +

    + +

    +

      +
    • +

      + Equation solving (Newton's Method), +

      +
      • + +
    • +

      + Related Rates (furthering our use of the derivative to find instantaneous rates of change), +

      +
      • + +
    • +

      + Optimization + (applied extreme values), + and +

      +
      • + +
    • +

      + Differentials + (useful for various approximations and for something called integration). +

      +
      • + + + + + + + +
    +

    + +

    + In the next chapters, we will consider the reverse + problem to computing the derivative: + given a function f, + can we find a function whose derivative is f? + Being able to do so opens up an incredible world of mathematics and applications. +

    +     + +     + + + Integration + +

    + We have spent considerable time considering the derivatives of a function and their applications. + In the following chapters, + we are going to starting thinking in + the other direction. That is, + given a function f(x), + we are going to consider functions F(x) such that F'(x) = f(x). + There are numerous reasons this will prove to be useful: + these functions will help us compute area, + volume, mass, force, pressure, work, and much more. +

    +     + +
    + Antiderivatives and Indefinite Integration + + + +

    + Given a function y=f(x), + a differential equation + is an equation that incorporates y, x, + and the derivatives of y. + For instance, a simple differential equation is: + + \yp = 2x + . +

    +

    + Solving a differential equation amounts to finding a function y + that satisfies the given equation. + Take a moment and consider that equation; + can you find a function y such that \yp = 2x? +

    +

    + Can you find another? +

    +

    + And yet another? +

    +

    + Hopefully you were able to come up with at least one solution: + y = x^2. + Finding another may have seemed impossible until one realizes that + a function like y=x^2+1 also has a derivative of 2x. + Once that discovery is made, finding + yet another is not difficult; + the function y = x^2 + 123{,}456{,}789 also has a derivative of + 2x. + The differential equation \yp = 2x has many solutions. + This leads us to some definitions. +

    + + + Antiderivatives and Indefinite Integrals + +

    + Let a function f(x) be given. + An antiderivative of f(x) is a function F(x) + such that \Fp(x) = f(x). + antiderivative + indefinite integral + integrationindefinite +

    +

    + The set of all antiderivatives of f(x) is the + indefinite integral of f, denoted by + + \int f(x) \,dx + . +

    +     +     + + + + + + +

    + Make a note about our definition: + we refer to an antiderivative of f, + as opposed to the antiderivative of f, + since there is always an infinite number of them. + We often use upper-case letters to denote antiderivatives. +

    +

    + When f is continuous, + knowing one antiderivative of f allows us to find infinitely more, + simply by adding a constant. + Not only does this give us more antiderivatives, + it gives us all of them. +

    + + + Antiderivative Forms + +

    + Let F(x) and G(x) be antiderivatives of a continuous + function f(x) on an interval I. + Then there exists a constant C such that, on I, + + G(x) = F(x) + C + . +

    +     +     + +

    + Given a continuous function f defined on an interval I and + one of its antiderivatives F, + we know all antiderivatives of f on I have the form + F(x) + C for some constant C. + Using , we can say that + + \int f(x) \,dx = F(x) + C + . +

    +

    + Note that we are abusing notation somewhat: + when we write F(x)+C on the right-hand side, + we really mean the set of all such functions, + for each real number value of C. + Let's analyze this indefinite integral notation. + integrationnotation +

    + +
    + Antiderivative notation + + + + + A labeled formula for the indefinite integral/ + +

    + The equation \int f(x) \cdot dx = F(x) + C, with labels for each part. + \int is labeled as the integral symbol. + f(x) is labeled as the integrand function. + dx is labeled as the differential of x. + F(x) is labeled as any "antiderivative of f." + C is labeled as the constant of integration. +

    +     + + + \begin{tikzpicture}[>=latex, node distance=6pt,scale=0.9] + + \node[] (integral) + {$\displaystyle\int$}; + \node[above left=of integral, anchor=south east] (l1) {Integral symbol}; + \draw[->] (l1) edge [out=-90, in=180] (integral); + \node[right=of integral] (integrand) + {$\displaystyle f(x)$}; + \node[below=of integrand, anchor=north east] (l2) {Integrand function}; + \draw[->] (l2) edge [out=0, in=-90] (integrand); + \node[right=of integrand] (dot) {$\cdot$}; + \node[right=of dot] (differential) + {$\displaystyle dx$}; + \node[above=of differential, anchor=south west] (l3) {Differential of $x$}; + \draw[->] (l3) edge [out=-90, in=90] (differential); + \node[right=of differential] (equals) + {$\displaystyle =$}; + \node[right=of equals] (antiderivative) + {$\displaystyle F(x)$}; + \node[below=of antiderivative, anchor=north west] (l4) {Any antiderivative of $f$}; + \draw[->] (l4) edge [out=180, in=-90] (antiderivative); + \node[right=of antiderivative] (plus) + {$\displaystyle {+}$}; + \node[right=of plus] (constant) + {$\displaystyle C$}; + \node[above right=of constant, anchor=south west] (l5) {Constant of integration}; + \draw[->] (l5) edge [out=-90, in=0] (constant); + + \end{tikzpicture} + + + + +
    + +

    + + shows the typical notation of the indefinite integral. + The integration symbol, \int, + is in reality an elongated S, + representing take the sum. + We will later see how sums + and antiderivatives are related. +

    +

    + The function we want to find an antiderivative of is called the + integrand. + It contains the differential of the variable we are integrating with respect to. + The \int symbol and the differential dx are not bookends + with a function sandwiched in between; + rather, the symbol \int means + find all antiderivatives of what follows, + and the function f(x) and dx are multiplied together; + the dx does not just sit there. +

    +

    + Another way of looking at the notation is that it tells us that + f(x)\,dx is the differential of F(x): dF(x) = f(x)\,dx, + confirming that F'(x)=f(x), as required of an antiderivative. + The integral symbol can then be viewed as an instruction to undo + the differential and recover the antiderivative F(x). +

    +

    + Another important aspect of the dx is that it tells us which + variable we're taking the antiderivative with respect to, + much like how \lzo{x} would mean to take the derivative with + respect to x, + while \lzo{t} would be the derivative with respect to t. +

    +

    + Let's practice using this notation. +

    + + + Evaluating indefinite integrals + +

    + Evaluate \int \sin(x) \,dx. +

    +     + +

    + We are asked to find all functions F(x) such that + \Fp(x) = \sin(x). + Some thought will lead us to one solution: + F(x) = -\cos(x), because \lzoo{x}{-\cos(x)} = \sin(x). +

    +

    + The indefinite integral of \sin(x) is thus -\cos(x), + plus a constant of integration. + So: + + \int \sin(x) \,dx = -\cos(x) + C + . +

    +     + +     + +

    + A commonly asked question is What happened to the dx? + The unenlightened response is Don't worry about it. + It just goes away. A full understanding includes the following. +

    +

    + This process of antidifferentiation + is really solving a differential question. + The integral + + \int \sin(x) \,dx + + presents us with a differential, dy = \sin(x) \, dx. + It is asking: What is y? + We found lots of solutions, + all of the form y = -\cos(x) +C. +

    +

    + Letting dy = \sin(x)\,dx, rewrite + + \int \sin(x) \,dx \text{ as } \int\,dy + . +

    +

    + This is asking: + What functions have a differential of the form dy? + The answer is Functions of the form y+C, + where C is a constant. What is y? + We have lots of choices, all differing by a constant; + the simplest choice is y = -\cos(x). +

    +

    + Understanding all of this is more important later as we try to find + antiderivatives of more complicated functions. + In this section, + we will simply explore the rules of indefinite integration, + and one can succeed for now with answering + What happened to the dx? + with It went away. +

    +

    + Let's practice once more before stating integration rules. +

    + + + Evaluating indefinite integrals + +

    + Evaluate \int\left(3x^2 + 4x+5\right)\,dx. +

    +     + +

    + We seek a function F(x) whose derivative is 3x^2+4x+5. + When taking derivatives, + we can consider functions term-by-term, + so we can likely do that here. +

    +

    + What functions have a derivative of 3x^2? + Some thought will lead us to a cubic, + specifically x^3+C_1, where C_1 is a constant. +

    +

    + What functions have a derivative of 4x? + Here the x term is raised to the first power, + so we likely seek a quadratic. + Some thought should lead us to 2x^2+C_2, + where C_2 is a constant. +

    +

    + Finally, what functions have a derivative of 5? + Functions of the form 5x+C_3, + where C_3 is a constant. +

    +

    + Our answer appears to be + + \int\left(3x^2+4x+5\right)\,dx = x^3+C_1+2x^2+C_2+5x+C_3 + . +

    +

    + We do not need three separate constants of integration; + combine them as one constant, + giving the final answer of + + \int\left(3x^2+4x+5\right)\,dx = x^3+2x^2+5x+C + . +

    +

    + It is easy to verify our answer; + take the derivative of x^3+2x^2+5x+C and see we indeed get + 3x^2+4x+5. +

    +     + +     + +

    + This final step of verifying our answer + is important both practically and theoretically. + In general, taking derivatives is easier than finding antiderivatives so + checking our work is easy and vital as we learn. +

    +

    + We also see that taking the derivative of our answer returns the function + in the integrand. + Thus we can say that: + + \lzoo{x}{\int f(x)\,dx} = f(x) + . +

    +

    + Differentiation undoes the work done by antidifferentiation. +

    +

    + + gave a list of the derivatives of common functions we had learned at that point. + We restate part of that list here to stress the relationship between + derivatives and antiderivatives. + This list will also be useful as a glossary of common antiderivatives + as we learn. +

    + + + Derivatives and Antiderivatives + +

    + Here are the Common Differentiation Rules and their Common Indefinite + Integral Rule counterparts. +

    +

    + + \amp\lzoo{x}{cf(x)}=c\cdot\fp(x)\amp\amp\int c\cdot f(x)\,dx= c\cdot \int f(x)\,dx\quad(c\neq 0) + \amp\lzoo{x}{f(x)\pm g(x)} =\fp(x)\pm \gp(x)\amp\amp \int \big(f(x)\pm g(x)\big)\,dx =\int f(x)\,dx\pm \int g(x)\,dx + \amp\lzoo{x}{C} = 0\amp\amp \int 0\,dx = C + \amp\lzoo{x}{x} = 1\amp\amp \int 1\,dx = \int dx = x+C + \amp\lzoo{x}{x^n} = n\cdot x^{n-1}\amp\amp \int x^n\,dx =\frac{1}{n+1}x^{n+1}+ C\quad(n\neq -1) + \amp\lzoo{x}{\sin(x)} = \cos(x)\amp\amp \int \cos(x) \,dx = \sin(x) +C + \amp\lzoo{x}{\cos(x)} = -\sin(x)\amp\amp \int \sin(x) \,dx = -\cos(x) +C + \amp\lzoo{x}{\tan(x)} = \sec^2(x)\amp\amp \int \sec^2(x) \,dx = \tan(x) +C + \amp\lzoo{x}{\csc(x)} = -\csc(x) \cot(x)\amp\amp \int \csc(x) \cot(x) \,dx = -\csc(x) +C + \amp\lzoo{x}{\sec(x)} = \sec(x) \tan(x)\amp\amp \int \sec(x) \tan(x) \,dx = \sec(x) +C + \amp\lzoo{x}{\cot(x)} = -\csc^2(x)\amp\amp \int \csc^2(x) \,dx = -\cot(x) +C + \amp\lzoo{x}{e^ x} = e^x\amp\amp \int e^x\,dx = e^x+C + \amp\lzoo{x}{a^x} = \ln(a) \cdot a^x\amp\amp \int a^x\,dx = \frac{1}{\ln(a) }\cdot a^x+C + \amp\lzoo{x}{\ln(x)} = \frac1 x,\ x \gt 0\amp\amp \int \frac{1}x\,dx = \ln\abs{x}+C + +

    +     +     + + + +

    + We highlight a few important points from . +

    +

    +

      +
    • +

      + + \int c\cdot f(x)\,dx = c\cdot \int f(x)\,dx + +

      +

      + This is the Constant Multiple Rule: + + Constant Multiple Ruleof integration + + we can temporarily ignore constants when finding antiderivatives, + just as we did when computing derivatives (, + \lzoo{x}{3x^2} is just as easy to compute as + \lzoo{x}{x^2}). An example: + + \int 5\cos(x) \,dx = 5\cdot\int \cos(x) \,dx = 5\cdot (\sin(x) +C) = 5\sin(x) + C + . + In the last step we can consider the constant as also being + multiplied by 5, but + 5 times a constant is still a constant, + so we just write C . +

      +
      • + +
    • +

      + + \int \big(f(x)\pm g(x)\big)\,dx =\int f(x)\,dx\pm \int g(x)\,dx + +

      +

      + This is the Sum/Difference Rule: + + integrationSum/Difference Rule + + Sum/Difference Ruleof integration + + we can split integrals apart when the integrand contains terms + that are added/subtracted, + as we did in . + So: + + \int(3x^2+4x+5)\, dx \amp = \int 3x^2\, dx + \int 4x\,dx + \int 5\,dx + \amp = 3\int x^2\, dx + 4\int x\,dx + \int 5 \,dx + \amp = 3\cdot \frac13x^3 + 4\cdot \frac12x^2+5x+C + \amp = x^3+2x^2+5x+C + + In practice we generally do not write out all these steps, + but we demonstrate them here for completeness. +

      +
      • + +
    • +

      + + \int x^n\,dx =\frac{1}{n+1}x^{n+1}+ C\quad(n\neq -1) + +

      +

      + This is the Power Rule of indefinite integration. + + integrationPower Rule + + Power Ruleintegration + + There are two important things to keep in mind: + +

        +
      1. +

        + Notice the restriction that n\neq -1. + This is important: \int \frac{1}{x}\,dx \neq + \frac{1}{0}x^0+C; + rather, see the last rule from the list. +

        +
        2. + +
      2. +

        + We are presenting antidifferentiation as the + inverse operation of differentiation. + Here is a useful quote to remember: +

        + +
        +

        + Inverse operations do the opposite things in the opposite + order. +

        +
        + +

        + When taking a derivative using the Power Rule, + we first multiply by the power, + then second subtract 1 from the power. + To find the antiderivative, do the opposite things in the + opposite order: + first add 1 to the power, + then second divide by the power. +

        +
        3. +
      +

      +
      • + +
    • +

      + + \int \frac{1}x\,dx = \ln\abs{x}+C + +

      +

      + Note that this rule uses the absolute value of x. + The exercises will work the reader through why this is the case; + for now, know the absolute value is important and cannot be ignored. +

      +
      • +
    +

    + + + + Initial Value Problems + +

    + In + we saw that the derivative of a position function gave a velocity function, + and the derivative of a velocity function describes acceleration. + initial value problem + We can now go the other way: + the antiderivative of an acceleration function gives a velocity function, + . While there is just one derivative of a given function, + there are infinitely many antiderivatives. + Therefore we cannot ask What is the + velocity of an object whose acceleration is + + -32 + ?, since there is more than one answer. +

    + + + +

    + We can find the answer if we provide more information with the + question, as done in the following example. + Often the additional information comes in the form of an initial value, + a value of the function that one knows beforehand. +

    + + + Solving initial value problems + +

    + The acceleration due to gravity of a falling object is + + -32 + . + At time t=3, a falling object had a velocity of + + -10 + . + Find the equation of the object's velocity. +

    +     + +

    + We want to know a velocity function, v(t). + We know two things: + +

      +
    • +

      + The acceleration, , v'(t)= -32, and +

      +
      • + +
    • +

      + the velocity at a specific time, , v(3) = -10. +

      +
      • +
    +

    +

    + Using the first piece of information, + we know that v(t) is an antiderivative of v'(t)=-32. + So we begin by finding the indefinite integral of -32: + + \int (-32)\,dt = -32t+C=v(t) + . +

    +

    + Now we use the fact that v(3)=-10 to find C: + + v(t) \amp = -32t+C + v(3) \amp = -10 + -32(3)+C \amp = -10 + C \amp = 86 + +

    +

    + Thus v(t)= -32t+86. + We can use this equation to understand the motion of the object: + when t=0, the object had a velocity of v(0) = 86 + + . + Since the velocity is positive, + the object was moving upward. +

    +

    + When did the object begin moving down? + Immediately after v(t) = 0: + + -32t+86 = 0 \implies t = \frac{43}{16} \approx 2.69\text{s} + . +

    +

    + Recognize that we are able to determine quite a bit about the path of + the object knowing just its acceleration and its velocity at a single + point in time. +

    +     + +     + + + Solving initial value problems + +

    + Find f(t), given that \fpp(t) = \cos(t), + \fp(0) = 3 and f(0) = 5. +

    +     + +

    + We start by finding \fp(t), + which is an antiderivative of \fpp(t): + + \int \fpp(t)\,dt \amp = \int \cos(t) \,dt + \amp = \sin(t) + C + \amp = \fp(t) + . +

    +

    + So \fp(t) = \sin(t) +C for the correct value of C. + We are given that \fp(0) = 3, so: + + \sin(0) +C \amp = 3 + C \amp = 3 + . +

    +

    + Using the initial value, we have found \fp(t) = \sin(t) + 3. + We now find f(t) by integrating again. + We will use a different integration constant since we have already + defined C to equal 3 above. + + f(t)=\int \fp(t) \,dt = \int (\sin(t) +3)\,dt = -\cos(t) + 3t + D + . +

    +

    + We are given that f(0) = 5, so + + -\cos(0) + 3(0) + D \amp = 5 + -1 + C \amp = 5 + C \amp = 6 + +

    +

    + Thus f(t) = -\cos(t) + 3t + 6. +

    +     + +     + + + +

    + This section introduced antiderivatives and the indefinite integral. + We found they are needed when finding a function given information about + its derivative(s). For instance, + we found a velocity function given an acceleration function. +

    +

    + In the next section, + we will see how position and velocity are unexpectedly related by the + areas of certain regions on a graph of the velocity function. + Then, in , + we will see how areas and antiderivatives are closely tied together. + This connection is incredibly important, + as indicated by the name of the theorem that describes it: + The Fundamental Theorem of Calculus. +

    +     + + + + Terms and Concepts + + + + +

    + Define the term antiderivative in your own words. +

    + +     + + + +     + + + + +

    + Is it more accurate to refer to the + antiderivative of f(x) or an + antiderivative of f(x)? +

    + +     + + + +

    + An antiderivative +

    +     +     + + +

    + The antiderivative +

    +     +     + +     + +     + + + + +

    + Use your own words to define the indefinite integral of f(x). +

    + +     + + + +     + + + + +

    + Fill in the blanks: + Inverse operations do the + things in the order. +

    +     + + + + opposite|reverse + + + + + opposite|reverse + + + + +     + + + + +

    + What is an initial value problem? +

    + +     + + + +     + + + + +

    + The derivative of a position function is a/an function. +

    +     + + + + + + + + +     + + + + +

    + An antiderivative of an acceleration function is a/an + function. +

    +     + + + + + + + + +     + + + + + parserFunction("F" => "exp(x)*sin(x)/sqrt(abs(x))"); + parserFunction("G" => "ln(abs(x))*cos(x)*sqrt(abs(x))"); + $answer = Formula('F(x)+G(x)'); + $evaluator = $answer->cmp(upToConstant=>1); + + +

    + If F(x) is an antiderivative of f(x), + and G(x) is an antiderivative of g(x), + give an antiderivative of f(x)+g(x). +

    +

    + +

    +     +     +     +     + + + Problems + + +

    + Evaluate the indefinite integral. + Don't forget your constant of integration! +

    +     + + + + + $a = random(2,9,1); + $n = random(2,9,1); + if($envir{problemSeed}==1){$a=3;$n=3}; + $f = Formula("$a x^$n"); + Context("Fraction"); + $b = Fraction($a,$n+1); + $F = FormulaUpToConstant("$b x^($n+1)"); + + +

    + \int \, dx +

    +

    + +

    +     +     +     + + + + + $n = random(2,9,1); + if($envir{problemSeed}==1){$n=9}; + $f = Formula("x^$n"); + Context("Fraction"); + $F = FormulaUpToConstant("1/($n+1) x^($n+1)"); + + +

    + \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(5,15,1); + $n = random(2,9,1); + $c = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=10;$n=2;$c=-2}; + $f = Formula("$a x^$n + $c")->reduce; + Context("Fraction"); + $b = Fraction($a,$n+1); + $F = FormulaUpToConstant("$b x^($n+1) + $c x")->reduce; + + +

    + \int \left(\right)\, dx +

    +

    + +

    +     +     +     + + + + + Context()->variables->are(t=>"Real"); + $F = FormulaUpToConstant("t"); + + +

    + \int dt +

    +

    + +

    +     +     +     + + + + + Context()->variables->are(s=>"Real"); + $F = FormulaUpToConstant("s"); + + +

    + \int 1\, ds +

    +

    + +

    +     +     +     + + + + + ($a,$n) = random_subset(2,2..9); + if($envir{problemSeed}==1){$a=3;$n=2}; + Context("Fraction"); + Context()->variables->are(t=>"Real"); + $f = Formula("1/($a t^$n)")->reduce; + $F = FormulaUpToConstant("-1/($a*($n-1) t^($n-1))")->reduce; + + +

    + \int \, dt +

    +

    + +

    +     +     +     + + + + + ($a,$n) = random_subset(2,2..9); + if($envir{problemSeed}==1){$a=3;$n=2}; + Context("Fraction"); + Context()->variables->are(t=>"Real"); + $f = Formula("$a/(t^$n)")->reduce; + ($num,$den) = Fraction($a,$n-1)->value; + $F = FormulaUpToConstant("-$num/($den t^($n-1))")->reduce; + + +

    + \int \, dt +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("2*sqrt(x)"); + + +

    + \int \frac{1}{\sqrt{x}}\, dx +

    +

    + +

    +     +     +     + + + + + $F = list_random('tan','cot','sec','csc'); + if($envir{problemSeed}==1){$F='tan';}; + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $F = Formula("$F(theta)"); + $f = $F->D('theta')->reduce; + $F = FormulaUpToConstant("$F"); + + +

    + \int \, d\theta +

    + + To enter \theta, type theta. + +

    + +

    +     +     +     + + + + + $F = list_random('-sin','-cos'); + if($envir{problemSeed}==1){$F='-cos';}; + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $F = Formula("$F(theta)"); + $f = $F->D('theta')->reduce; + $F = FormulaUpToConstant("$F"); + + +

    + \int \, d\theta +

    + + To enter \theta, type theta. + +

    + +

    +     +     +     + + + + + ($G,$H) = ('sec','csc','tan','cot'); + $c = list_random(-1,1); + if($envir{problemSeed}==1){$G='sec';$H=csc;$c=-1}; + $F = Formula("$G(x) + $c*$H(x)")->reduce; + $f = $F->D('x')->reduce; + $F = FormulaUpToConstant("$F"); + + +

    + \int \left(\right)\, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + if($envir{problemSeed}==1){$a=5;}; + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $F = Formula("$a e^(theta)"); + $f = $F->D('theta')->reduce; + $F = FormulaUpToConstant("$F"); + + +

    + \int \, d\theta +

    + + To enter \theta, type theta. + +

    + +

    +     +     +     + + + + + $b = random(2,9,1); + if($envir{problemSeed}==1){$b=3;}; + Context()->variables->are(t=>'Real'); + $f = Formula("$b^t"); + Context()->flags->set(reduceConstantFunctions=>0); + $F = FormulaUpToConstant("$b^t/ln($b)"); + + +

    + \int \, dt +

    +

    + +

    +     +     +     + + + + + ($b,$c) = random_subset(2,2..9); + if($envir{problemSeed}==1){$b=5;$c=2}; + Context()->variables->are(t=>'Real'); + $f = Formula("$b^t/$c"); + Context()->flags->set(reduceConstantFunctions=>0); + $F = FormulaUpToConstant("$b^t/($c ln($b))"); + + +

    + \int \, dt +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=2;$b=3}; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + $f = Formula("($a t + $b)^2")->reduce; + $A = Fraction($a**3,3*$a); + $B = Fraction(3*$a**2*$b,3*$a); + $C = Fraction(3*$a*$b**2,3*$a); + $D = Fraction($b**3,3*$a); + $F = FormulaUpToConstant("$A t^3 + $B t^2 + $C t + $D")->reduce; + + +

    + \int \, dt +

    +

    + +

    +     +     +     + + + + + $m = random(2,5,1); + ($a,$b) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$m=2;$a=3;$b=-2}; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + $n = $m+1; + $f = Formula("(t^$m+$a)(t^$n+$b t)")->reduce; + ($num2,$den2) = Fraction($b/($m+2) + $a/($n+1))->value; + ($num3,$den3) = Fraction($a*$b,2)->value; + $F = FormulaUpToConstant("t^($m+$n+1)/($m+$n+1) + $num2 t^($n+1)/$den2 + $num3 t^2/$den3")->reduce; + + +

    + \int \, dt +

    +

    + +

    +     +     +     + + + + + ($m,$n) = random_subset(2,2..9); + if($envir{problemSeed}==1){$m=2;$n=3}; + $f = Formula("x^$m x^$n")->reduce; + $F = FormulaUpToConstant("x^($m+$n+1)/($m+$n+1)")->reduce; + + +

    + \int \, dx +

    +

    + +

    +     +     +     + + + + + ($b,$p) = random_subset(2,'pi','e','sqrt(2)'); + if($envir{problemSeed}==1){$b='e';$p='pi'}; + Context()->flags->set(reduceConstants=>0); + $f = Formula("$b^$p")->reduce; + $F = FormulaUpToConstant("$b^$p x")->reduce; + + +

    + \int \, dx +

    +

    + +

    +     +     +     + + + + + $c = list_random('a'..'d','k','m','n','p'..'s'); + if($envir{problemSeed}==1){$c='a'}; + Context()->flags->set(reduceConstants=>0); + Context()->constants->add($c=>Real(exp(pi/sqrt(2)+sin(1)))); + $f = Formula("$c")->reduce; + $F = FormulaUpToConstant("$c x")->reduce; + + +

    + \int \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Consider the two integrals, \ds \int s^n\,ds and + \ds \int s^n\,dn. +

    +     + + +

    + What is the difference between these two indefinite integrals? +

    +     + + + +

    + The indefinite integrals have different differentials, + , different variables of integration. +

    +     +     + + +

    + Evaluate \ds \int s^n\,ds. +

    +     + + + +

    + \frac{s^{n+1}}{n+1}+C, n\neq -1 +

    +     +     + + +

    + Evaluate \ds \int s^n\,dn. +

    +     + + + +

    + \frac{s^{n}}{\ln s}+C, s>0 +

    +     +     +     + + + + +

    + This problem investigates why + states that \ds \int \frac1x\,dx = \ln\abs{x} + C. +

    +     + + +

    + What is the domain of y = \ln(x)? +

    +     +     + + +

    + Find \lzoo{x}{\ln(x) }. +

    +     +     + + +

    + What is the domain of y = \ln(-x)? +

    +     +     + + +

    + Find \lzoo{x}{\ln(-x)}. +

    +     +     + + +

    + You should find that 1/x has two types of antiderivatives, + depending on whether x \gt 0 or x\lt 0. + In one expression, give a formula for + \ds \int \frac{1}{x}\, dx that takes these different domains into account, + and explain your answer. +

    + + +     +     + +     + + + +

    + Find the function determined by the given initial value problem. +

    +     + + + + + $trig = list_random('-sin','-cos'); + $f0 = list_random(-9..-1,2..9); + if($envir{problemSeed}==1){$trig='-cos';$f0=2;}; + $F = Formula("$trig(x)"); + $f = $F->D('x'); + $F0 = $F->eval(x=>0); + $c = $f0 - $F0; + $F = Formula($F + $c)->reduce; + + +

    + \fp(x) = and f(0)= +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $b = random(2,9,1); + if($envir{problemSeed}==1){$a=5;$b=5;}; + $f0 = $a+$b; + $f = Formula("$a e^x"); + $F = Formula("$a e^x + $b"); + + +

    + \fp(x) = and f(0)= +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $b = non_zero_random(-9,9,1); + $m = random(1,2,1); + $x0 = non_zero_random(-2,2,1); + $f0 = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=4;$b=-3;$m=2;$x0=-1;$f0=9}; + $f = Formula("$a x^3 + $b x^$m")->reduce; + Context("Fraction"); + ($num1,$den1) = Fraction($a,4)->value; + ($num2,$den2) = Fraction($b,$m+1)->value; + $F = Formula("$num1 x^4/$den1 + $num2 x^($m+1)/$den2"); + $c = Fraction($f0 - $F->eval(x=>$x0)); + $F = Formula("$num1 x^4/$den1 + $num2 x^($m+1)/$den2 + $c")->reduce; + + +

    + \fp(x) = and f()= +

    +

    + +

    +     +     +     + + + + + $trig = list_random(['tan','pi/4'],['sec','pi/3'],['cot','pi/4'],['csc','pi/6']); + $f0 = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$trig=['tan','pi/4'];$f0=5;}; + $trf = $trig->[0]; + $x0 = $trig->[1]; + Context()->flags->set(reduceConstants=>0); + $x0 = Formula("$x0"); + $F = Formula("$trf(x)"); + $f = $F->D('x'); + $c = $f0 - $F->eval(x=>Real("$x0")); + $F = Formula("$F + $c"); + + +

    + \fp(x) = and f\mathopen{}\left(\right)\mathclose{}= +

    +

    + +

    +     +     +     + + + + + $b = random(2,9,1); + $f0 = random(1,9,1); + if($envir{problemSeed}==1){$b=7;$f0=1;}; + $f = Formula("$b^x"); + Context()->flags->set(reduceConstantFunctions=>0); + $F = Formula("$b^x/ln($b) - $b^2/ln($b) + $f0"); + + +

    + \fp(x) = and f(2) = +

    +

    + +

    +     +     +     + + + + + ($fpp,$fp0,$f0) = random_subset(3,2..9); + if($envir{problemSeed}==1){$fpp=5;$fp0=7;$f0=3;}; + Context("Fraction"); + $a = Fraction($fpp,2); + $F = Formula("$a x^2 + $fp0 x + $f0"); + + +

    + \fpp(x)= and \fp(0)=, f(0)= +

    +

    + +

    +     +     +     + + + + + ($fpp,$fp0,$f0) = random_subset(3,-10..-2,2..10); + if($envir{problemSeed}==1){$fpp=7;$fp0=-1;$f0=10;}; + Context("Fraction"); + $a = Fraction($fpp,6); + $a2 = Fraction($fpp,2); + $a3 = Fraction($fpp,3); + $b = Fraction($fp0 - $a2); + $c = Fraction($f0 + $a3 - $fp0); + $F = Formula("$a x^3 + $b x + $c"); + + +

    + \fpp(x)=x and \fp(1)=, f(1)= +

    +

    + +

    +     +     +     + + + + + $fpp = random(2,9,1); + ($fp0,$f0) = random_subset(2,-10..-2,2..10); + if($envir{problemSeed}==1){$fpp=5;$fp0=3;$f0=5;}; + $f = Formula("$fpp e^x"); + $F = Formula("$f + ($fp0-$fpp)x + ($f0-$fpp)")->reduce; + + +

    + \fpp(x)= and \fp(0)=, f(0)= +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','cos'); + $x0 = list_random(0,'pi/2','pi'); + ($fp0,$f0) = random_subset(2,2..9); + if($envir{problemSeed}==1){$trig='sin';$x0='pi';$fp0=2;$f0=4;}; + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $x0 = Formula("$x0"); + $f = Formula("$trig(theta)"); + $F = Formula("-$trig(theta)"); + $m = $fp0 - $F->D('theta')->eval(theta=>Real("$x0")); + $F = Formula("-$trig(theta) + $m theta + ($f0 + $trig($x0) - $m $x0 ) ")->reduce; + + +

    + \fpp(\theta)= and \fp\left(\right)=, f\left(\right)= +

    + + To enter \theta, type theta. + +

    + +

    +     +     +     + + + + + $a = random(10,30,1); + $m = random(2,4,1); + $b = random(2,9,1); + $trig = list_random('sin','cos'); + $c = list_random(1,-1); + ($fp0,$f0) = random_subset(2,-9..9); + if($envir{problemSeed}==1){$a=24;$m=2;$b=2;$c=-1;$trig='cos';$fp0=5;$f0=0;}; + Context("Fraction"); + Context()->flags->set(reduceConstants=>0); + ($num,$den) = Fraction($a,($m+1)*($m+2))->value; + $f = Formula("$a x^$m + $b^x + $c $trig(x)")->reduce; + $F = Formula("$num x^($m+2)/$den + $b^x/(ln($b))^2 - $c $trig(x)")->reduce; + $m = $fp0 - $F->D('x')->eval(x=>0); + $F = Formula("$F + $m x + ($f0 + $c*$trig(0)- 1/(ln($b))^2)")->reduce; + + +

    + \fpp(x)= and \fp(0)=, f(0)= +

    +

    + +

    +     +     +     + + + + + $x0 = non_zero_random(-5,5,1); + ($fp0,$f0) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$x0=1;$fp0=5;$f0=0;}; + $F = Formula("$fp0 x + ($f0 - $fp0*$x0)")->reduce; + + +

    + \fpp(x)=0 and \fp()=, f()= +

    +

    + +

    +     +     +     +     +     +     +
    +
    + The Definite Integral + + + + +

    + We start with an easy problem. + An object travels in a straight line at a constant velocity of + 5 + for 10 seconds. + How far away from its starting point is the object? +

    + +

    + We approach this problem with the familiar + \text{Distance } = \text{ Rate } \times \text{ Time} equation. + In this case, the distance traveled is + 5 + + 10 + = 50 feet. +

    + +

    + It is interesting to note that this solution of 50 feet can be represented + graphically. + Consider , + where the constant velocity of + 5 + is graphed on the axes. + Shading the area under the line from t=0 to t=10 gives a + rectangle with an area of 50 square units; + when one considers the units of the axes, + we can say this area represents + 50. +

    + + + +
    + The area under a constant velocity function corresponds to distance traveled + + + Graph of a linear function, the area under the curve is rectangular in shape. + + +

    + The y axis is drawn from 0 to 5 represents velocity in feets per second + and the x axis is drawn from 0 to 10 represents time in second. + The function y=5 for all values of x until 10, the function + is a straight line parallel to the x axis. + The area under the line to the x axis is shaded, and is drawn between x=0 and x=10. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + x label style={at={(axis description cs:0.85,0.05)},anchor=north}, + y label style={at={(axis description cs:0,.85)},rotate=90,anchor=south}, + xlabel={$t$ (s)}, + ylabel={$v$ (ft/s)}, + ytick={5}, + ymin=-.9,ymax=5.9, + xmin=-.9,xmax=10.9 + ] + \addplot [firstcurvestyle,areastyle,domain=0:10] {5} \closedcycle; + \addplot [firstcurvestyle,domain=0:10] {5}; + \end{axis} + \end{tikzpicture} + + +
    + +

    + Now consider a slightly harder situation (and not particularly realistic): + an object travels in a straight line with a constant velocity of + 5 + for 10 seconds, + then instantly reverses course at a rate of + 2 + for 4 seconds. + (Since the object is traveling in the opposite direction when reversing course, + we say the velocity is a constant + -2.) + How far away from the starting point is the object what is its + displacement? +

    + +

    + Here we use \text{Distance } = \text{ Rate}_1\, \times \text{ Time}_1\, + \text{ Rate}_2\, \times \text{ Time}_2, which is + + \text{ Distance } = 5\cdot10 + (-2)\cdot 4 = 42\text{ ft. } + +

    + +

    + Hence the object is 42 feet from its starting location. +

    + + + + + +

    + We can again depict this situation graphically. + In + we have the velocities graphed as straight lines on [0,10] and + [10,14], + respectively. + The displacement of the object is +

    + +
    +

    Area above the t-axis -Area below the t-axis,

    +
    + +

    + which is easy to calculate as 50-8=42 feet. +

    + +
    + The total displacement is the area above the t-axis minus the area below the t-axis + + + The graph has two areas shaded that are rectangular in shape. + + +

    + The y axis is drawn from 0 to 5 represents velocity in feets per second + and the x axis is drawn from 0 to 14 represents time in seconds. + There are two areas in the graph both rectangular in shape, + the bigger one lies in the first quadrant and the smaller one lies in the fourth quadrant. + The function y=5 is a straight line parallel to the x axis and creates the bigger area under graph, + and is drawn between x=0 and x=10. + The function y=-2, is a straight line parallel to the x axis and creates the smaller area, + and is drawn between x=10 and x = 14. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + x label style={at={(axis description cs:0.85,0.33)},anchor=south}, + y label style={at={(axis description cs:0.05,.75)},rotate=90,anchor=south}, + xlabel={$t$ (s)}, + ylabel={$v$ (ft/s)}, + ytick={-2,5}, + minor x tick num=4, + ymin=-2.9,ymax=5.9, + xmin=-.9,xmax=15.9 + ] + \addplot [firstcurvestyle,areastyle,domain=0:10] {5} \closedcycle; + \addplot [firstcurvestyle,domain=0:10] {5}; + \addplot [secondcurvestyle,areastyle,domain=10:14] {-2} \closedcycle; + \addplot [secondcurvestyle,domain=10:14] {-2}; + \end{axis} + \end{tikzpicture} + + +
    + +

    + Now consider a more difficult problem. +

    + + + Finding position using velocity + +

    + The velocity of an object moving straight up/down under the acceleration + of gravity is given as v(t) = -32t+48, + where time t is given in seconds and velocity is in + . + When t=0, the object had a height of + 0. +

    + +

    +

      +
    1. +

      + What was the initial velocity of the object? +

      +
      2. + +
    2. +

      + What was the maximum height of the object? +

      +
      3. + +
    3. +

      + What was the height of the object at time t=2? +

      +
      4. +
    +

    +     + +

    + It is straightforward to find the initial velocity; at time t=0, + + v(0) \amp =-32\cdot 0+48 + \amp = 48 + + The initial velocity was + 48. +

    + +

    + To answer questions about the height of the object, + we need to find the object's position function s(t). + This is an initial value problem, + which we studied in the previous section. + We are told the initial height is 0, , s(0) = 0. + We know s'(t) = v(t) = -32t+48. + To find s, we find the indefinite integral of v(t): + + s(t) \amp =\int v(t)\, dt + \amp = \int (-32t+48)\, dt + \amp = -16t^2+48t+C + . +

    + +

    + Since s(0) = 0, + we conclude that C=0 and s(t) = -16t^2+48t. +

    + +

    + To find the maximum height of the object, + we need to find the maximum of s. + Recalling our work finding extreme values, + we find the critical points of s by setting its derivative + (the velocity function) + equal to 0 and solving for t: + + 0 \amp = -32t+48 + t \amp =48/32 + \amp = 1.5\text{ s } + . +

    + +

    + (Notice how we ended up just finding when the velocity was 0ft/s!) + The first derivative test shows this is a maximum, + so the maximum height of the object is found at + + s(1.5) = -16(1.5)^2+48(1.5)=36\text{ ft } + . +

    + +

    + The height at time t=2 is now straightforward to compute: + + s(2) \amp =-16(2)^2+48(2) + \amp = 32 + . + The height is 32 + after 2 seconds. +

    + +

    + While we have answered all three questions + (using derivatives and antiderivatives), + let's look at them again graphically, + using the concepts of area that we explored earlier. +

    + +

    + + shows a graph of v(t) on axes from t=0 to t=3. + It is again straightforward to find v(0). + How can we use the graph to find the maximum height of the object? +

    + +
    + A graph of v(t)=-32t+48; the shaded areas help determine displacement + + + The graph of the function is a straight line that intersects the x axis. There are two triangular areas. + + +

    + The y axis is drawn from -40 to 40 and the x axis is drawn from 0 to 3. + The line starts from point (0, 48) and crosses the x axis at x=1.5, + the area under the curve forms a right angled triangle in the first quadrant with the positive x and y axes. + After crossing the x axis at x=1.5 the line goes down to point (3,-48). + Another right angle is formed on the x axis but in the fourth quadrant with the x axis and line x=3. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + ymin=-55,ymax=55, + xmin=-.5,xmax=3.2, + x label style={at={(axis description cs:0.85,0.3)},anchor=south}, + y label style={at={(axis description cs:0,.85)},rotate=90,anchor=south}, + xlabel={$t$ (s)}, + ylabel={$v$ (ft/s)}, + ] + \addplot [firstcurvestyle,areastyle,domain=0:3] {-32*x+48} \closedcycle; + \addplot [firstcurvestyle,domain=0:3] {-32*x+48}; + \end{axis} + \end{tikzpicture} + + +
    + +

    + Recall how in our previous work that the displacement of the object + (in this case, its height) + was found as the area under the velocity curve, as shaded in the figure. + Moreover, the area between the curve and the t-axis that is + below the t-axis counted as negative area. + That is, it represents the object coming back toward its starting position. + So to find the maximum distance from the starting point the + maximum height we find the area under the velocity line that + is above the t-axis, + , from t=0 to t=1.5. + This region is a triangle; its area is + + \text{ Area } \amp = \frac12\text{ Base } \times \text{ Height } + \amp =\frac12\times 1.5\text{ s } \times 48\text{ ft/s } + \amp = 36\text{ ft } + + which matches our previous calculation of the maximum height. +

    + +

    + Finally, to find the height of the object at time t=2 we + calculate the total signed area + (where some area is negative) + under the velocity function from t=0 to t=2. + This signed area is equal to s(2), + the displacement (, signed distance) from the starting position at + t=0 to the position at time t=2. + That is, +

    + +

    + Displacement = Area above the t-axis - Area below + t-axis. +

    + +

    + The regions are triangles, and we find + + \text{ Displacement } \amp = \frac12(1.5\text{s} )(48\text{ ft/s } ) - \frac12(0.5\text{s} )(16\text{ ft/s } ) + \amp = 32\text{ ft } + . +

    + +

    + This also matches our previous calculation of the height at t=2. +

    + +

    + Notice how we answered each question in this example in two ways. + Our first method was to manipulate equations using our understanding of + antiderivatives and derivatives. + Our second method was geometric: + we answered questions looking at a graph and finding the areas of + certain regions of this graph. +

    +     + +     + +

    + The above example does not prove + a relationship between area under a velocity function and displacement, + but it does imply a relationship exists. + + will fully establish fact that the area under a velocity function is + displacement. +

    + +

    + Given a graph of a function y=f(x), + we will find that there is great use in computing the area between the + curve y=f(x) and the x-axis. + Because of this, we need to define some terms. +

    + + + The Definite Integral, Total Signed Area + +

    + Let y=f(x) be defined on a closed interval [a,b]. + The total signed area from x=a to x=b under + f is: + integrationdefinite + definite integral + signed area + total signed area + integrationnotation + integrationarea +

    + +

    + (area under y=f(x) and above the x-axis on [a,b]) - + (area above y=f(x) and under the x-axis on [a,b]). +

    + +

    + The definite integral of f on [a,b] + is the total signed area of f on [a,b], denoted + + \int_a^b f(x)\, dx + , + where a and b are the + bounds of integration. +

    +     +     + + + + +

    + By our definition, the definite integral gives the + signed area under f. + We usually drop the word signed + when talking about the definite integral, + and simply say the definite integral gives + the area under f or, more commonly, + the area under the curve. +

    + +

    + The previous section introduced the indefinite integral, + which related to antiderivatives. + We have now defined the definite integral, + which relates to areas under a function. + The two are very much related, + as we'll see when we learn the Fundamental Theorem of Calculus in + . + Recall that earlier we said that the + \int symbol was an elongated S + that represented finding a sum. + In the context of the definite integral, + this notation makes a bit more sense, + as we are adding up areas under the function f. +

    + +

    + We practice using this notation. +

    + + + Evaluating definite integrals + +

    + Consider the function f given in . +

    + +
    + A graph of f(x) in + + + The graph of f(x) that shows the areas it forms with the x axis. + + +

    + The y axis is drawn between -1 to 1 and the x axis is drawn between 0 to 5. + There is a triangle in the first quadrant drawn on the x axis + with its base from x=0 to x=3 the peak of the triangle is at point (1,1). + The area in the first quadrant is the shaded portion inside the triangle. +

    +

    + The second area is a right angle triangle and is drawn on the x axis between x=3 and x=5. + The triangle lies in the fourth quadrant with its peak at point (5,-1) and the function forming the hypotenuse. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + ymin=-1.1,ymax=1.1, + xmin=-.5,xmax=5.5 + ] + \addplot [firstcurvestyle,areastyle] coordinates {(0,0) (1,1) (5,-1)} \closedcycle; + \addplot [firstcurvestyle,-] coordinates {(0,0) (1,1) (5,-1)}; + \end{axis} + \end{tikzpicture} + + +
    + +

    + Find: +

    + +

    +

      +
    1. \int_0^3 f(x)\, dx
      2. + +
    2. \int_3^5 f(x)\, dx
      3. + +
    3. \int_0^5 f(x)\, dx
      4. + +
    4. \int_0^3 5f(x)\, dx
      5. + +
    5. \int_1^1 f(x) \, dx
      6. +
    +

    +     + +

    +

      +
    1. +

      + \int_0^3 f(x)\, dx is the area under f on the + interval [0,3]. + This region is a triangle, + so the area is \int_0^3 f(x)\, dx=\frac12(3)(1) = 1.5. +

      +
      2. + +
    2. +

      + \int_3^5 f(x)\, dx represents the area of the triangle + found under the x-axis on [3,5]. + The area is \frac12(2)(1) = 1; + since it is found under + the x-axis, + this is negative area. Therefore \int_3^5 f(x)\, dx = -1. +

      +
      3. + +
    3. +

      + \int_0^5f(x)\, dx is the total signed area under f + on [0,5]. + This is 1.5 + (-1) = 0.5. +

      +
      4. + +
    4. +

      + \int_0^35f(x)\, dx is the area under 5f on [0,3]. + This is sketched in . + Again, the region is a triangle, + with height 5 times that of the height of the original triangle. + Thus the area is \int_0^35f(x)\, dx = \frac12(15)(1) = 7.5. +

      +
      5. + +
    5. +

      + \int_1^1f(x)\, dx is the area under f on the + interval [1,1]. + This describes a line segment, + not a region; it has no width. + Therefore the area is 0. +

      +
      6. +
    +

    +
    + + A graph of 5f in . + (Yes, it looks just like the graph of f in , just with a different y-scale.) + + + + The graph of f(x) that shows the areas it forms with the x axis. + + +

    + The y axis is drawn between -5 to 5 and the x axis is drawn between 0 to 5. + There is a triangle in the first quadrant drawn on the x axis with + its base from x=0 to x=3 the peak of the triangle is at point (1,5). + The area in the first quadrant is the shaded portion inside the triangle. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + ymin=-5.5,ymax=5.5, + xmin=-.5,xmax=5.5 + ] + \addplot [firstcurvestyle,areastyle] coordinates {(0,0) (1,5) (5,-5)} \closedcycle; + \addplot [firstcurvestyle,-] coordinates {(0,0) (1,5) (5,-5)}; + \end{axis} + \end{tikzpicture} + + +
    +     + +     + +

    + This example illustrates some of the properties of the definite integral, + given here. +

    + + + Properties of the Definite Integral + +

    + Let f and g be defined on a closed interval I that + contains the values a, + b and c, and let k be a constant. + The following hold: + integrationdefinite!properties + definite integralproperties +

    + +

    +

      +
    1. \int_a^a f(x)\, dx = 0
      2. + +
    2. \int_a^b f(x)\, dx + \int_b^c f(x)\, dx = \int_a^cf(x)\, dx
      3. + +
    3. \int_a^bf(x)\, dx = -\int_b^a f(x)\, dx
      4. + +
    4. \int_a^b\big(f(x)\pm g(x)\big)\, dx = \int_a^bf(x)\, dx \pm \int_a^bg(x)\, dx
      5. + +
    5. \int_a^bk\cdot f(x)\, dx = k\cdot\int_a^bf(x)\, dx
      6. +
    +

    +     +     + + + + + + +

    + We give a brief justification of here. +

    + +

    +

    +
  • + 1. +

    + As demonstrated in , + there is no area under the curve + when the region has no width; + hence this definite integral is 0. +

    +
    • + +
  • 2. +

    + This states that total area is the sum of the areas of subregions. + It is easily considered when we let a\lt b\lt c. + We can break the interval [a,c] into two subintervals, + [a,b] and [b,c]. + The total area over [a,c] is the area over [a,b] plus + the area over [b,c]. + It is important to note that this still holds true even if + a\lt b\lt c is not true. + We discuss this in the next point. +

    +
    • + +
  • 3. +

    + This property can be viewed a merely a convention to make other + properties work well. + (Later we will see how this property has a justification all its own, + not necessarily in support of other properties.) + Suppose b\lt a\lt c. + The discussion from the previous point clearly justifies + + \int_b^a f(x)\, dx + \int_a^c f(x)\, dx = \int_b^c f(x)\, dx + . + However, we still claim that, as originally stated, + + \int_a^b f(x)\, dx + \int_b^c f(x)\, dx = \int_a^c f(x)\, dx + . + How do Equations and + relate? + Start with Equation: + + \int_b^a f(x)\, dx + \int_a^c f(x)\, dx \amp = \int_b^c f(x)\, dx + \int_a^c f(x)\, dx \amp = -\int_b^a f(x)\, dx + \int_b^c f(x)\, dx + + Property (3) justifies changing the sign and switching the + bounds of integration on the \ds -\int_b^a f(x)\, dx term; + when this is done, Equations and + are equivalent. + The conclusion is this: + by adopting the convention of Property (3), Property (2) holds no + matter the order of a, + b and c. + Again, in the next section we will see another justification for this + property. +

    +
    • + +
  • + 4,5. +

    + Each of these may be non-intuitive. + Property (5) states that when one scales a function by, + for instance, 7, the area of the enclosed region also is scaled by a + factor of 7. + Both Properties (4) and (5) can be proved using geometry. + The details are not complicated but are not discussed here. +

    +
    • +
    +

    + + + Evaluating definite integrals using <xref ref="thm_defintprop"/> + +

    + Consider the graph of a function f(x) shown in + . +

    + +
    + A graph of a function in + + + The graph of function in this example. + + +

    + The x and the y axes are uncalibrated, there are three positions a, + b and c in order on the x axis where the inflection changes. + The curve in the first quadrant is smaller, the area under the curve forms a bell jar shape + it extends from x=a to x=b. + The other curve is twice in height as the first and lies in the fourth quadrant. + It is also bell jar shaped, but is inverted on the positive x axis and extends between x=b to x=c. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={.5,1.5,3}, + extra x tick labels={$a$,$b$,$\ c$}, + ytick=\empty, + ymin=-1.1,ymax=1.1, + xmin=-.2,xmax=3.5 + ] + \addplot [firstcurvestyle,areastyle,domain=0.5:3] {(x-.5)*(x-1.5)*(x-3)} \closedcycle; + \addplot [firstcurvestyle,domain=0.5:3,samples=40] {(x-.5)*(x-1.5)*(x-3)}; + \end{axis} + \end{tikzpicture} + + +
    + +

    + Answer the following: +

    + +

    +

      +
    1. +

      + Which value is greater: + \ds \int_a^b f(x)\, dx or \ds \int_b^c f(x)\, dx? +

      +
      2. + +
    2. +

      + Is \ds \int_a^c f(x)\, dx greater or less than 0? +

      +
      3. + +
    3. +

      + Which value is greater: + \ds \int_a^b f(x)\, dx or \ds \int_c^b f(x)\, dx? +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + \int_a^b f(x)\, dx has a positive value + (since the area is above the x-axis) + whereas \int_b^c f(x)\, dx has a negative value. + Hence \int_a^b f(x)\, dx is bigger. +

      +
      2. + +
    2. +

      + \int_a^c f(x)\, dx is the total signed area under f + between x=a and x=c. + Since the region below the x-axis looks to be larger than + the region above, + we conclude that the definite integral has a value less than 0. +

      +
      3. + +
    3. +

      + Note how the second integral has the bounds reversed. + Therefore \int_c^b f(x)\, dx=-\int_b^c f(x)\, dx represents + a positive number, + greater than the area described by the first definite integral. + Hence \int_c^b f(x)\, dx is greater. +

      +
      4. +
    +

    +     +     + +

    + The area definition of the definite integral allows us to use geometry to + compute the definite integral of some simple functions. +

    + + + Evaluating definite integrals using geometry + +

    + Evaluate the following definite integrals: + + 1. \, \int_{-2}^5 (2x-4)\, dx \qquad 2.\, \int_{-3}^3 \sqrt{9-x^2}\, dx + . +

    +     + +

    +

      +
    1. +

      + It is useful to sketch the function in the integrand, + as shown in . + We see we need to compute the areas of two regions, + which we have labeled R_1 and R_2. + Both are triangles, so the area computation is straightforward: + + R_1: \frac12(4)(8) = 16 \qquad R_2: \frac12(3)6 = 9 + . + Region R_1 lies under the x-axis, + hence it is counted as negative area + (we can think of the triangle's height as being -8), + so + + \int_{-2}^5(2x-4)\, dx = -16+9 = -7 + . +

      +
      2. + +
    2. +

      + Recognize that the integrand of this definite integral describes + a half circle, + as sketched in , + with radius 3. + Thus the area is: + + \int_{-3}^3 \sqrt{9-x^2}\, dx = \frac12\pi r^2 = \frac 92\pi + . +

      +
      3. +
    +

    + + +
    + f(x) = 2x-4 + + + Graph of function of a positive sloped straight line along with the areas it forms on the x axis. + + +

    + The y axis is drawn from -10 to 10 and the + x axis is drawn from -3 to 5. + The function is a straight line that starts from point (-2,-8) and ends at point (5,6). + The function intersects the x axis at x=2. +

    +

    + The function, below the x axis creates a right angled triangle + with the x axis and line x=-2 from y=0 to y=-8, + with its base from x=-2 to x=2. This area is named R1. +

    +

    + The function above the x axis forms a right angled triangle with the x axis and line x=5 + from y=0 to y=6, with its base from x=2 and x=5. This area is named R2. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + ymin=-11,ymax=11, + xmin=-3.5,xmax=5.5 + ] + \addplot [firstcurvestyle,areastyle,domain=-2:5] {2*x-4} \closedcycle; + \addplot [firstcurvestyle,domain=-2:5] {2*x-4}; + \addplot [soliddot] coordinates {(-2,-8)} node[right] {\small $(-2,-8)$}; + \addplot [soliddot] coordinates {(5,6)} node[left] {\small $(5,6)$}; + \draw (axis cs:-1,-2) node { $R_1$}; + \draw (axis cs:4.2,2) node { $R_2$}; + \end{axis} + \end{tikzpicture} + + +
    +
    + f(x) = \sqrt{9-x^2} + + + The graph area under the curve is a semicircle with radius 3 on the x axis with centre at origin. + + + The graph shows the area under the curve that is a semicircle with radius 3 on the x axis + with centre at origin. It lies on the first and the second quadrant. + + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + xtick={-3,3}, + extra y ticks={1,2,3,4}, + extra y tick labels=\empty, + ytick={5}, + ymin=-.5,ymax=5.5, + xmin=-3.5,xmax=3.5] + \addplot [firstcurvestyle,areastyle,domain=0:180] ({3*cos (x)},{3*sin(x)}) \closedcycle; + \addplot [firstcurvestyle,domain=0:180,samples=40] ({3*cos (x)},{3*sin(x)}); + \end{axis} + \end{tikzpicture} + + +
    +     +     + +     + + + Understanding motion given velocity + +

    + Consider the graph of a velocity function of an object moving in a + straight line, + given in , + where the numbers in the given regions gives the area of that region. + Assume that the definite integral of a velocity function gives displacement. + Find the maximum speed of the object and its maximum displacement from + its starting position. +

    + +
    + A graph of a velocity in + + + The graph of velocity for this example. + + +

    + The y axis is drawn from -10 to 15 and + it represents velocity in feets per second, the x axis is uncalibrated and represents time in seconds. + There are three points on the x axis marked a, b and c in that order. +

    +

    + In the fourth quadrant, on the x axis from 0 to a, a parabola is drawn with its peak at + y=-10. The number 11 is written inside the shaded portion. +

    +

    + From a to b there is a second parabola in the first quadrant with its peak at y= 15, + the number 38 is written inside it. +

    +

    + From b to c the third parabola is located in the fourth quadrant with its peak at x=-10, + the number 11 is written inside it, indicating same size as the first parabola. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + x label style={at={(axis description cs:0.85,0.35)},anchor=south}, + y label style={at={(axis description cs:-.05,.85)},rotate=90,anchor=south}, + xlabel={$t$ (s)}, + ylabel={$v$ (ft/s)}, + xtick=\empty, + extra x ticks={2,6,8}, + extra x tick labels={$\ a$,$b\ $,$\ c$}, + ytick={-5,5,10,15}, + ymin=-9,ymax=17, + xmin=-.5,xmax=8.5 + ] + \addplot [firstcurvestyle,areastyle,domain=0:8,samples=40] {15/64*x*(x-2)*(x-6)*(x-8)} \closedcycle; + \addplot [firstcurvestyle,domain=0:8,samples=70] {15/64*x*(x-2)*(x-6)*(x-8)}; + \draw (axis cs:.9,-4) node { $11$}; + \draw (axis cs:7.1,-4) node { $11$}; + \draw (axis cs:4,8) node { $38$}; + \end{axis} + \end{tikzpicture} + + +
    +     + +

    + Since the graph gives velocity, + finding the maximum speed is simple: + it looks to be 15ft/s. +

    + +

    + At time t=0, the displacement is 0; + the object is at its starting position. + At time t=a, the object has moved backward 11 feet. + Between times t=a and t=b, + the object moves forward 38 feet, + bringing it into a position 27 feet forward of its starting position. + From t=b to t=c the object is moving backwards again, + hence its maximum displacement is 27 feet from its starting position. +

    +     + +     + +

    + In our examples, + we have either found the areas of regions that have nice geometric shapes + (such as rectangles, triangles and circles) + or the areas were given to us. + Consider , + where a region below y=x^2 is shaded. + What is its area? + The function y=x^2 is relatively simple, + yet the shape it defines has an area that is not simple to find geometrically. +

    + +
    + What is the area below y=x^2 on [0,3]? The region is not a usual geometric shape. + + + Graph of function described above between x=0 and x=3. + + +

    + The y axis is drawn from 0 to 10 and the x axis is drawn from 0 to 3. + The curve is drawn in the first quadrant, it starts at the origin and + increases gently until x=1 and steeply from x=1 to x=3 and ends at point (3,9). +

    +

    + The area under the curve roughly looks like a right angled triangle. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + xtick={1,2,3}, + ymin=-1,ymax=11, + xmin=-1,xmax=3.5 + ] + \addplot [firstcurvestyle,areastyle,domain=0:3] {x^2} \closedcycle; + \addplot [firstcurvestyle,domain=0:3] {x^2}; + \end{axis} + \end{tikzpicture} + + +
    + +

    + In + we will explore how to find the areas of such regions. +

    + + + + Terms and Concepts + + + + +

    + What is total signed area? +

    + +     + + + +     + + + + +

    + What is displacement? +

    + +     + + + +     + + + + +

    + What is \ds \int_3^3 \sin(x) \, dx? +

    +

    + +

    +     + + + + + + + +

    + Notice that the upper and lower bounds of the integral are equal. +

    +     +     +     +     + +     + + + +

    + Give a single definite integral that has the same value as + I = \int_0^1(2x+3)\, dx + \int_1^2 (2x+3)\, dx. +

    +     + + + +

    + \int 0^2 (2x+3)\, dx +

    +     +     +     + + + Problems + + +

    + A graph of a function f(x) is given. + Using the geometry of the graph, + evaluate the definite integrals. +

    +     + + + + + @a = (3,4,3,0,-4,9); + + + + + + The graph of the function is a decreasing straight line drawn in the first and fourth quadrant. + + +

    + The y axis is drawn from -4 to 4 and the x axis is drawn from 0 to 4. + The function y=-2x+4 is a line that decreases from point (0,4) to point (4,-4). + The function is linear hence the graph is a straight line between the two points. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + ymin=-4.5, + ymax=4.5, + xmin=-.9, + xmax=4.5 + ] + \addplot+ [domain=0:4] {-2*x+4}; + \draw (axis cs:2,2.5) node { \(y = -2x+4\)}; + \end{axis} + \end{tikzpicture} + + +     + + + +

    + \int_0^1 (-2x+4)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^2 (-2x+4)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^3 (-2x+4)\, dx +

    +

    + +

    +     +     + + +

    + \int_1^3 (-2x+4)\, dx +

    +

    + +

    +     +     + + +

    + \int_2^4 (-2x+4)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^1 (-6x+12)\, d +

    +

    + +

    +     +     +     +     + + + + + @a = (-4,-5,-3,1,-2,10); + + + + + + Graph of function f(x). + + +

    + The y axis is drawn from -2 to 2 and the x axis is drawn from 0 to 5. + The graph of function is drawn in the fourth and the first quadrants. In the fourth quadrant, + the function is a straight line y=-2, from x=0 to x=2. + Then the function increases steeply and meets the x axis at x=3. + After intersecting the x axis at 3, + the function assumes a gentler slope and increases until point (5,2). +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + xtick={1,2,3,4,5}, + ymin=-2.2, + ymax=2.2, + xmin=-.9, + xmax=5.5 + ] + \addplot+ [domain=0:2] {-2}; + \addplot [firstcurvestyle,domain=2:3] {2*x-6}; + \addplot [firstcurvestyle,domain=3:5] {x-3}; + \draw (axis cs:3.2,-1.5) node { \(y = f(x)\)}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + \int_0^2 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^3 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^5 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_2^5 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_5^3 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^3 -2f(x)\, dx +

    +

    + +

    +     +     +     +     + + + + + @a = (4,2,4,2,1,2); + + + + + + The graph has two triangles from 0 to 2 and from 2 to 4, + the first one is twice the height of the second. + + +

    + The x axis is drawn from 0 to 4 and the y axis + is drawn from 0 to 4. + The function increases from the origin to peak at (1,4) + then it decreases sharply to point (2,0). + Then it increases again, gently to the second peak at (3,2) + after which it decreases to point (4,0). +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + xtick={1,2,3,4,5}, + ymin=-.5, + ymax=4.5, + xmin=-.9, + xmax=4.5 + ] + \addplot+ [domain=0:1] {4*x}; + \addplot [firstcurvestyle,domain=1:2] {-4*(x-2)}; + \addplot [firstcurvestyle,domain=2:3] {2*(x-2)}; + \addplot [firstcurvestyle,domain=3:4] {-2*(x-4)}; + \draw (axis cs:3,2.5) node { \(y = f(x)\)}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + \int_0^2 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_2^4 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_2^4 2f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^1 4x\, dx +

    +

    + +

    +     +     + + +

    + \int_2^3 (2x-4)\, dx +

    +

    + +

    +     +     + + +

    + \int_2^3 (4x-8)\, dx +

    +

    + +

    +     +     +     +     + + + + + Context("Fraction"); + @a = (Fraction(-1/2),Fraction(0),Fraction(3/2),Fraction(3/2),Fraction(9/2),Fraction(15/2)); + + + + + + The graph of the function is a straight line with a positive slope. + + +

    + The y axis is drawn from -1 to 3 and the x axis + is drawn from 0 to 4. + The graph of function y=x-1 is a straight line that + increases from point (0,-1) to point (4,3), + and it crosses the x axis at x=1. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + xtick={1,2,3,4,5}, + ytick={-1,1,2,3}, + ymin=-1.5, + ymax=3.5, + xmin=-.9, + xmax=4.5 + ] + \addplot+ [domain=0:4] {x-1}; + \draw (axis cs:2,2.5) node { \(y = x-1\)}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + \int_0^1 (x-1)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^2 (x-1)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^3 (x-1)\, dx +

    +

    + +

    +     +     + + +

    + \int_2^3 (x-1)\, dx +

    +

    + +

    +     +     + + +

    + \int_1^4 (x-1)\, dx +

    +

    + +

    +     +     + + +

    + \int_1^4 \big((x-1)+1\big)\, dx +

    +

    + +

    +     +     +     +     + + + + + @a = (Formula("pi"),Formula("pi"),Formula("2 pi"),Formula("10 pi")); + + + + + + Graph of function that is a semicircle on the x axis. + + +

    + The y axis is drawn from 0 to 3 and the x axis is + drawn from 0 to 4. + The graph of function f(x) = \sqrt{4-(x-2)^2} + is a semi circle drawn on the x axis with centre at point + (2,0) and radius of 2. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + xtick={1,2,3,4}, + ytick={1,2,3}, + ymin=-.5, + ymax=3.5, + xmin=-.9, + xmax=4.5 + ] + \addplot+ [samples=40,domain=0:3.14159] ({2*cos(deg(x))+2}, {2*sin(deg(x))}); + \draw (axis cs:3,2.4) node { \(f(x) = \sqrt{4-(x-2)^2}\)}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + \int_0^2 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_2^4 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^4 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^4 5f(x)\, dx +

    +

    + +

    +     +     +     +     + + + + + Context()->variables->add(a=>'Real',b=>'Real'); + @a = (15,12,0,Formula("3(b-a)")); + + + + + + The graph of a function is a straight line parallel to the x axis. + + +

    + The y axis is drawn from 0 to 4 and + the x axis is drawn from 0 to 10. + The graph of function f(x) = 3 is a straight + line parallel to the x axis and goes from x=0 to x=10. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + ytick={1,2,3}, + ymin=-.5, + ymax=5, + xmin=-0.5, + xmax=10.5 + ] + \addplot+[domain=0:10] ({x}, {3}); + \draw (axis cs:5,3.4) node[] { \(f(x) = 3\)}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + \int_0^5 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_3^7 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^0 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \ds \int_a^b f(x)\, dx, where + 0\leq a\leq b\leq 10 +

    +

    + +

    +     +     +     +     +     + + + +

    + A graph of a function f(x) is given; + the numbers inside the shaded regions give the area of that region. + Evaluate the definite integrals using this area information. +

    +     + + + + + @a = (-59,-48,-27,-33); + + + + + Graph of function f(x) for this exercise. + + +

    + The y axis is drawn from -100 to 50 and the + x axis is drawn from 0 to 3. The function + forms three parabolic shapes. The areas inside the parabolas + until the x axis are shaded. The biggest one lies in the + fourth quadrant and the other two in the first. +

    +

    + Between x=0 to x=1 the highest parabola is formed + with peak at y=-100, it is labeled with number 59. +

    +

    + Between x=1 to x=2 the smallest parabola is formed + with the peak at nearly y=20, it is labeled with number 11. +

    +

    + Between x=2 to x=3 the third parabola is formed with + the peak at nearly y=45, it is labeled with number 21. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + xtick={1,2,3}, + ymin=-105, + ymax=55, + xmin=-.9, + xmax=3.5 + ] + \addplot [firstcurvestyle,areastyle,domain=0:3,samples=40] {60*x*(1-x)*(x-2)*(x-2)*(x-3)} \closedcycle; + \addplot [firstcurvestyle,domain=0:3,samples=90] {60*x*(1-x)*(x-2)*(x-2)*(x-3)}; + \draw (axis cs:1.5,30) node { \(y=f(x)\)}; + \draw (axis cs:.45,-10) node { \(59\)}; + \draw (axis cs:1.4,10) node { \(11\)}; + \draw (axis cs:2.6,10) node { \(21\)}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + \int_0^1 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^2 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^3 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_1^2 -3f(x)\, dx +

    +

    + +

    +     +     +     +     + + + + + @a = (Formula("4/pi"),Formula("-4/pi"),Formula("0"),Formula("2/pi")); + + + + + + Graph of sine function with angle as multiple of pi/2. + + +

    + The y axis is drawn from -1 to 1 and + the x axis is drawn from 0 to 4. The function + crosses the x axis at 0, 2 and 4. +

    +

    + There is a downward facing parabola between x=0 and x=2 with + peak at point (1,1) and it lies in the first quadrant. + The shaded portion inside this downward facing parabola until the x axis is labeled 4/\pi. + From x=2 and x=4 the secong prabola is present + in the fourth quadrant. + The shaded portion inside this parabola until the x axis is labeled 4/\pi. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ymin=-1.5, + ymax=1.5, + xmin=-.9, + xmax=4.5 + ] + \addplot [firstcurvestyle,areastyle,domain=0:4] {sin(deg(3.14159*x/2))} \closedcycle; + \addplot [firstcurvestyle,domain=0:4,samples=50] {sin(deg(3.14159*x/2))}; + \draw (axis cs:3,0.75) node { \(f(x)=\sin(\pi x/2)\)}; + \draw (axis cs:1,.6) node { \(4/\pi\)}; + \draw (axis cs:3,-.6) node { \(4/\pi\)}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + \int_0^2 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_2^4 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^4 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^1 f(x)\, dx +

    +

    + +

    +     +     +     +     + + + + + @a = (4,4,-4,-2); + + + + + + Graph of function for this exercise that represents a parabola. + + +

    + The y axis is drawn from -5 to 10 and the x axis + is drawn from -2 to 2. + The function f(x) = 3x^2 -3 is parabolic, with vertex at point (0,-3). + It has x intercepts at x=-1 and x=1. +

    +

    + There are three distinct shaded regions, from x=-1 to x=-2 and from + x=1 to x=2 the area under the arms of the parabola to the x + axis are shaded and both are labeled with number 4. +

    +

    + From x=-1 to x=1, the area from above the parabola to the x axis + is shaded and labeled as -4. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + xtick={-2,-1,1,2}, + ymin=-5.5, + ymax=11, + xmin=-2.5, + xmax=2.5 + ] + \addplot [firstcurvestyle,areastyle,domain=-2:2] {3*x^2-3} \closedcycle; + \addplot [firstcurvestyle,domain=-2:2,samples=40] {3*x^2-3}; + \draw (axis cs:1.5,10) node { \(f(x)=3x^2-3\)}; + \draw (axis cs:-1.5,1) node { \(4\)}; + \draw (axis cs:1.5,1) node { \(4\)}; + \draw (axis cs:.3,-1) node { \(-4\)}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + \int_{-2}^{-1} f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_1^2 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_{-1}^1 f(x)\, dx +

    +

    + +

    +     +     + + +

    + \int_0^1 f(x)\, dx +

    +

    + +

    +     +     +     +     + + + + + Context("Fraction"); + @a = (Fraction("40/3"),Fraction("26/3"),Fraction("8/3"),Fraction("38/3")); + + + + + + Graph of function is a curve with a positive slope. + + +

    + The y axis is drawn from 0 to + 4 and the x axis is drawn from 0 to 2. + The function f(x)=x^2 is drawn in the first quadrant, + it starts at the origin and gently curves up to point (1,1) + then steeply to point (2,4). The area under the curve is + shaded from x=0 to x=2. It forms a roughly right-angled + triangle shape. +

    +

    + From x=0 to x=1, the number 1/3 is written + inside the shaded region, and from x=1 to x=2, + the number 7/3 is written. +

    + +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + xtick={1,2}, + ymin=-.1, + ymax=4.2, + xmin=-.2, + xmax=2.1 + ] + \addplot [firstcurvestyle,areastyle,domain=0:2] {x^2} \closedcycle; + \addplot [firstcurvestyle,domain=0:2] {x^2}; + \draw [thick,firstcolor,dashed] (axis cs:1,0) -- (axis cs:1,1); + \draw (axis cs:1,3) node { \(f(x)=x^2\)}; + \draw (axis cs:.7,.2) node { \(1/3\)}; + \draw (axis cs:1.5,.2) node { \(7/3\)}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + \int_{0}^{2} 5x^2\, dx +

    +

    + +

    +     +     + + +

    + \int_0^2 (x^2+3)\, dx +

    +

    + +

    +     +     + + +

    + \int_{1}^3 (x-1)^2\, dx +

    +

    + +

    +     +     + + +

    + \int_2^4 \big((x-2)^2+5\big)\, dx +

    +

    + +

    +     +     +     +     +     + + + +

    + A graph is given of the velocity function of an object moving in a straight line. + Answer the questions based on the graph. +

    +     + + + + + $mv = NumberWithUnits("2 ft/s"); + $md = NumberWithUnits("2 ft"); + $td = NumberWithUnits("1.5 ft"); + + + + + + Graph of function for this exercise. It is a straight line with a negative slope. + + +

    + The y axis is drawn from -1 to 3 and represents velocity in + feets per second and the x axis is drawn from 0 to 3 and + it represents time in seconds. + The graph of the function is an inclined line. It starts from point (0,2) + then decreases in the first quadrant, crosses the x axis at x=2 + and continues decreasing in the fourth quadrant. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + xtick={1,2,3}, + ymin=-1.1, + ymax=3, + xmin=-.2, + xmax=3.1, + xlabel={\(t\) (s)}, + ylabel={\(v\) (ft/s)} + ] + \addplot+ [domain=0:3] {-x+2}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + What is the object's maximum velocity? +

    +

    + +

    +     +     + + +

    + What is the object's maximum displacement? +

    +

    + +

    +     +     + + +

    + What is the object's total displacement on [0,3]? +

    +

    + +

    +     +     +     +     + + + + + $mv = NumberWithUnits("3 ft/s"); + $md = NumberWithUnits("9.5 ft"); + $td = NumberWithUnits("9.5 ft"); + + + + + + The graph of function for this example. + + +

    + The y axis is drawn from 0 to 4 and represents velocity + in feets per second and the x axis is drawn from 0 to 5 + and it represents time in seconds. + The graph is a straight line y=3 from x=0 to x=1 then + it decreases to point (2,0). The function then increases to (3,2) + then it becomes a straight line y=2. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + xtick={1,2,3,4,5}, + ymin=-.5, + ymax=4, + xmin=-.2, + xmax=5.3, + xlabel={\(t\) (s)}, + ylabel={\(v\) (ft/s)} + ] + \addplot+ [domain=0:1] {3}; + \addplot [firstcurvestyle,domain=1:2] {-3*(x-2)}; + \addplot [firstcurvestyle,domain=2:3] {2*(x-2)}; + \addplot [firstcurvestyle,domain=3:5] {2}; + \end{axis} + \end{tikzpicture} + + +     + + +

    + What is the object's maximum velocity? +

    +

    + +

    +     +     + + +

    + What is the object's maximum displacement? +

    +

    + +

    +     +     + + +

    + What is the object's total displacement on [0,5]? +

    +

    + +

    +     +     +     +     +     + + + + + $mv = NumberWithUnits("64 ft/s"); + $md = NumberWithUnits("64 ft"); + $md_t = NumberWithUnits("2 s"); + $h0_t = FormulaWithUnits("2+sqrt(7) s"); + + +

    + An object is thrown straight up with a velocity, + in ft/s, given by v(t) = -32t+64, + where t is in seconds, from a height of 48 feet. +

    +     + + +

    + What is the object's maximum velocity? +

    +

    + +

    +     +     + + +

    + What is the object's maximum displacement? +

    +

    + +

    +     +     + + +

    + When does the maximum displacement occur? +

    +

    + +

    +     +     + + +

    + When will the object reach a height of 0? (Hint: + find when the displacement is -48ft.) +

    +

    + +

    +     +     +     +     + + + + + $v0 = NumberWithUnits("96 ft/s"); + $d = NumberWithUnits("6 s"); + $h0_2 = NumberWithUnits("6 s"); + $hmax = NumberWithUnits("-16*3^2+96*3+64 ft"); + + +

    + An object is thrown straight up with a velocity, + in ft/s, given by v(t) = -32t+96, + where t is in seconds, from a height of 64 feet. +

    +     + + +

    + What is the object's initial velocity? +

    +

    + +

    +     +     + + +

    + When is the object's displacement 0? +

    +

    + +

    +     +     + + +

    + How long does it take for the object to return to its initial height? +

    +

    + +

    +     +     + + +

    + What is the maximum height the object reaches? +

    +

    + +

    +     +     +     +     + + + +

    + The values of several definite integrals are given as follows: + \int_0^2f(x)\,dx = 5 \quad \int_0^3f(x)\,dx = 7 \quad \int_0^2g(x)\,dx = -3 \quad \int_2^3g(x)\,dx = 5 + Use these values and properties of definite integrals to evaluate the indicated definite integral. +

    +     + + + + +

    + \int_0^2 \big(f(x)+g(x)\big) \, dx +

    +

    + +

    +     +     +     + + + + +

    + \int_0^3 \big(f(x)-g(x)\big) \, dx +

    +

    + +

    +     +     +     + + + + +

    + \int_2^3 \big(3f(x)+2g(x)\big) \, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->variables->set(a=>"Real",b=>"Real"); + parser::Assignment->Allow; + $a = Formula("a = -2/7*b"); + + +

    + Find a formula for a in terms of b such that + \ds \int_0^3 \big(af(x)+bg(x)\big) \, dx=0. +

    +

    + +

    +     +     +     +     + + + +

    + The values of several definite integrals are given as follows: + \int_0^3s(t)\,dt = 10 \quad \int_3^5s(t)\,dt = 8 \quad \int_3^5r(t)\,dt = -1 \quad \int_0^5r(t)\,dt = 11 + Use these values and properties of definite integrals to evaluate the indicated definite integral. +

    +     + + + + +

    + \int_0^3 \big(s(t) + r(t)\big)\, dt +

    +

    + +

    +     +     +     + + + + +

    + \int_5^0 \big(s(t) - r(t)\big)\, dt +

    +

    + +

    +     +     +     + + + + +

    + \int_3^3 \big(\pi s(t) - 7r(t)\big)\, dt +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->variables->set(a=>"Real",b=>"Real"); + parser::Assignment->Allow; + $a = Formula("a = -18/11*b"); + + +

    + Find a formula for a in terms of b such that + \ds \int_0^5 \big(ar(t)+bs(t)\big) \, dt=0. +

    +

    + +

    +     +     +     +     +     +     +
    +
    + Riemann Sums + + + +

    + In the previous section we defined the definite integral of a function on [a,b] to be the signed area between the curve and the x-axis. + Some areas were simple to compute; + we ended the section with a region whose area was not simple to compute. + In this section we develop a technique to find such areas. + Riemann Sum +

    + +

    + A fundamental calculus technique is to first answer a given problem with an approximation, + then refine that approximation to make it better, + then use limits in the refining process to find the exact answer. + That is what we will do here. +

    +     + + + Approximating area with rectangles + +

    + Consider the region given in , + which is the area under y=4x-x^2 on [0,4]. + What is the signed area of this region , what is \int_0^4(4x-x^2)\, dx? +

    + +
    + A graph of f(x) = 4x-x^2. What is the area of the shaded region? + + + A shaded downward opening parabola in the first quadrant. + +

    + A parabola with the equation y = 4x -x^2. + It is a downward opening parabola lying in the first quadrant. + It has a maximum of (2,4), and crosses the x-axis at x=0 and x = 4. + The area between the curve and the x-axis is shaded on the interval [0,4]. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4}, + ymin=-1,ymax=5, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; + \addplot [firstcurvestyle,domain=-0.2:4.2,samples=45] {4*x-x^2}; + + \end{axis} + \end{tikzpicture} + + + + +
    + +

    + We start by approximating. + We can surround the region with a rectangle with height and width of 4 and find the area is approximately 16 square units. + This is obviously an over-approximation; + we are including area in the rectangle that is not under the parabola. +

    + +
    + Approximating area under a curve with one rectangle + + The previously described parabola with a large rectangle containing the shaded area. + +

    + The curve shown in , with a rectangle entirely containing the shaded area. + The rectangle has a height and width of 4. + On the left and right sides of the parabola, area is included in the rectangle which is not shaded. + The rectangle clearly contains a greater area than is under the parabola. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4}, + ymin=-1.2,ymax=5, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2} \closedcycle; + \addplot [firstcurvestyle,domain=-.2:4.2,samples=40] {4*x-x^2} node [pos=1, left] { $y=4x-x^2$}; + + \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:4,4); + + \end{axis} + \end{tikzpicture} + + + + +
    + +

    + We have an approximation of the area, using one rectangle. + How can we refine our approximation to make it better? + The key to this section is this answer: + use more rectangles. +

    + +

    + Let's use four rectangles with an equal width of 1. + This partitions the interval [0,4] into 4 subintervals, + [0,1], [1,2], [2,3] and [3,4]. + On each subinterval we will draw a rectangle. +

    + +

    + There are three common ways to determine the height of these rectangles: + the Left Hand Rule, + the Right Hand Rule, + and the Midpoint Rule. + The Left Hand Rule says to evaluate the function at the left-hand endpoint of the subinterval and make the rectangle that height. + In , + the rectangle drawn on the interval [2,3] has height determined by the Left Hand Rule; + it has a height of f(2). (The rectangle is labeled LHR.) + Left Hand Rule + Right Hand Rule + Midpoint Rule +

    + +
    + Approximating \int_0^4(4x-x^2)\, dx using rectangles. The heights of the rectangles are determined using different rules. + + + The area of a downward opening parabola being approximated by 4 rectangles. + +

    + The parabola shown in . The shaded are is being approximated by 4 rectangles of width 1. + The left most rectangle has a height of 3 and is labeled "RHR," for "Right Hand Rule." + The height of the rectangle is equal to the height of the parabola om the rightmost side, where x = 1. + The second rectangle has a height of 3.75, and is labeled "MPR," for "Midpoint Rule." + The height of the rectangle is equal to the height of the parabola at the center of the rectangle, where x=1.5. + The third rectangle had a height of 4, and is labeled "LHR," for "Left Hand Rule." + The height of the rectangle is given by the height of the parabola on the leftmost side, where x = 2. + The rightmost rectangle has a height of 2.6, and is labeled "other". + The height is given by the height of the parabola at a point slightly to the right of the center of the rectangle. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4}, + ymin=-1,ymax=5, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; + \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; + + \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:1,3); + \draw [thick,secondcolor] (axis cs:1,0) rectangle (axis cs:2,3.75); + \draw [thick,secondcolor] (axis cs:2,0) rectangle (axis cs:3,4); + \draw [thick,secondcolor] (axis cs:3,0) rectangle (axis cs:4,1.666); + + \draw [fill=black] (axis cs:1,3) circle (1pt); + \draw [fill=black] (axis cs:1.5,3.75) circle (1pt); + \draw [fill=black] (axis cs:2,4) circle (1pt); + \draw [fill=black] (axis cs:3.53,1.666) circle (1pt); + + \draw (axis cs:.5,.3) node { RHR}; + \draw (axis cs:1.5,.3) node { MPR}; + \draw (axis cs:2.5,.3) node { LHR}; + \draw (axis cs:3.5,.3) node { other}; + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + The Right Hand Rule says the opposite: + on each subinterval, + evaluate the function at the right endpoint and make the rectangle that height. + In the figure, + the rectangle drawn on [0,1] is drawn using f(1) as its height; + this rectangle is labeled RHR.. +

    + +

    + The Midpoint Rule says that on each subinterval, + evaluate the function at the midpoint and make the rectangle that height. + The rectangle drawn on [1,2] was made using the Midpoint Rule, + with a height of f(1.5). + That rectangle is labeled MPR. +

    + +

    + These are the three most common rules for determining the heights of approximating rectangles, + but one is not forced to use one of these three methods. + The rectangle on [3,4] has a height of approximately f(3.53), + very close to the Midpoint Rule. + It was chosen so that the area of the rectangle is exactly + the area of the region under f on [3,4]. + (Later you'll be able to figure how to do this, too.) +

    + +

    + The following example will approximate the value of \int_0^4 (4x-x^2)\, dx using these rules. +

    + + + Using the Left Hand, Right Hand and Midpoint Rules + +

    + Approximate the value of \int_0^4 (4x-x^2)\, dx using the Left Hand Rule, + the Right Hand Rule, + and the Midpoint Rule, using 4 equally spaced subintervals. +

    +     + +

    + We break the interval [0,4] into four subintervals as before. + In + we see 4 rectangles drawn on + f(x) = 4x-x^2 using the Left Hand Rule. + (The areas of the rectangles are given in each figure.) +

    + +

    + Note how in the first subinterval, + [0,1], the rectangle has height f(0)=0. + We add up the areas of each rectangle (height width) for our Left Hand Rule approximation: + + \amp f(0)\cdot 1 + f(1)\cdot 1+ f(2)\cdot 1+f(3)\cdot 1 + =\amp 0+3+4+3 = 10 + . +

    + +

    + + shows 4 rectangles drawn under f using the Right Hand Rule; + note how the [3,4] subinterval has a rectangle of height 0. +

    + +

    + In this example, + these rectangles seem to be the mirror image of those found in . + This is because of the symmetry of our shaded region. + Our approximation gives the same answer as before, + though calculated a different way: + + \amp f(1)\cdot 1 + f(2)\cdot 1+ f(3)\cdot 1+f(4)\cdot 1 + \amp =3+4+3+0= 10 + . +

    + +

    + + shows 4 rectangles drawn under f using the Midpoint Rule. +

    + +

    + This gives an approximation of \int_0^4(4x-x^2)\, dx as: + + f(0.5)\cdot 1 + f(1.5)\cdot 1+ f(2.5)\cdot 1 \amp +f(3.5)\cdot 1 + \amp =1.75+3.75+3.75+1.75 = 11 + . +

    + +

    + Our three methods provide two approximations of \int_0^4(4x-x^2)\, dx: 10 and 11. +

    +
    + Approximating \int_0^4(4x-x^2)\, dx in + +
    + using the Left Hand Rule + + + A parabola approximated using the Left Hand Rule. + +

    + The parabola in approximated using 4 strips of width 1. + The height of each strip is fiven by the left hand rule. + The first strip has a height of 0, and an area of 0. + The second strip has a height of 3, and an area of 3. + The third strip has a height of 4, and an area of 4. + The fourth strip has a height of 3, and an area of 3. + Each rectangle touches the curve in the top left corner. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4}, + ymin=-1,ymax=5, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; + \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; + + \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:1,0); + \draw [thick,secondcolor] (axis cs:1,0) rectangle (axis cs:2,3); + \draw [thick,secondcolor] (axis cs:2,0) rectangle (axis cs:3,4); + \draw [thick,secondcolor] (axis cs:3,0) rectangle (axis cs:4,3); + + \draw [fill=black] (axis cs:0,0) circle (1pt); + \draw [fill=black] (axis cs:1,3) circle (1pt); + \draw [fill=black] (axis cs:2,4) circle (1pt); + \draw [fill=black] (axis cs:3,3) circle (1pt); + + \draw (axis cs:.5,.3) node {$0$}; + \draw (axis cs:1.5,1.5) node {$3$}; + \draw (axis cs:2.5,2) node {$4$}; + \draw (axis cs:3.5,1.5) node {$3$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + using the Right Hand Rule + + + A parabola approximated using the Right Hand Rule. + +

    + The parabola in approximated using 4 strips of width 1. + The height of each strip is fiven by the right hand rule. + The first strip has height of 3, and an area of 3. + The second strip has height of 4, and an area of 4. + The third strip has height of 3, and an area of 3. + The fourth strip has height of 0, and an area of 0. + Each rectangle touches the curve in the top right corner. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4}, + ymin=-1,ymax=5, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; + \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; + + \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:1,3); + \draw [thick,secondcolor] (axis cs:1,0) rectangle (axis cs:2,4); + \draw [thick,secondcolor] (axis cs:2,0) rectangle (axis cs:3,3); + \draw [thick,secondcolor] (axis cs:3,0) rectangle (axis cs:4,0); + + \draw [fill=black] (axis cs:4,0) circle (1pt); + \draw [fill=black] (axis cs:1,3) circle (1pt); + \draw [fill=black] (axis cs:2,4) circle (1pt); + \draw [fill=black] (axis cs:3,3) circle (1pt); + + \draw (axis cs:.5,1.5) node {$3$}; + \draw (axis cs:1.5,2) node {$4$}; + \draw (axis cs:2.5,1.5) node {$3$}; + \draw (axis cs:3.5,.3) node {$0$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + using the Midpoint Rule + + + A parabola approximated using the Midpoint Rule + +

    + The parabola in approximated using 4 strips of width 1. + The height of each strip is fiven by the Midpoint Rule. + The first strip has a height and area of 1.75. + The second strip has a height and area of 3.75. + The third strip has a height and area of 3.75. + The fourth strip has a height and area of 1.75. + Each rectangle touches the curve in the top center of the rectangle. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4}, + ymin=-1,ymax=5, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; + \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; + + \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:1,1.75); + \draw [thick,secondcolor] (axis cs:1,0) rectangle (axis cs:2,3.75); + \draw [thick,secondcolor] (axis cs:2,0) rectangle (axis cs:3,3.75); + \draw [thick,secondcolor] (axis cs:3,0) rectangle (axis cs:4,1.75); + + \draw [fill=black] (axis cs:.5,1.75) circle (1pt); + \draw [fill=black] (axis cs:1.5,3.75) circle (1pt); + \draw [fill=black] (axis cs:2.5,3.75) circle (1pt); + \draw [fill=black] (axis cs:3.5,1.75) circle (1pt); + + \draw (axis cs:.5,.875) node {\small $1.75$}; + \draw (axis cs:1.5,1.875) node {\small $3.75$}; + \draw (axis cs:2.5,1.875) node {\small $3.75$}; + \draw (axis cs:3.5,.875) node {\small $1.75$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + +     +
    +     + +     + + +     + + + Summation Notation +

    + It is hard to tell at this moment which is a better approximation: 10 or 11? + We can continue to refine our approximation by using more rectangles. + The notation can become unwieldy, + though, as we add up longer and longer lists of numbers. + We introduce summation notation + to ameliorate this problem. + summationnotation +

    + + + + +

    + Suppose we wish to add up a list of numbers a_1, + a_2, + a_3, , a_9. + Instead of writing + + a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8+a_9 + , + we use summation notation and write \sum_{i=1}^9 a_i. + + The upper case sigma, + Sigma represents the term sum. The index (counter) of summation in this example is i; any symbol can be used. + By convention, the index takes on only the integer values between + (and including) + the lower and upper bounds. + To the right of \Sigma, + the expression a_i is called the summand. + It tells us what we are summing. + This is summarized in Equation. + + \sum_{\underbrace{i=1}_{i \text{-index of summation}}}^{\overbrace{9}^\text{upper bound}} \underbrace{a_i}_\text{summand} + +

    + +

    + Let's practice using this notation. +

    + + + Using summation notation + +

    + Let the numbers \{a_i\} be defined as + a_i = 2i-1 for integers i, where i\geq 1. + So a_1 = 1, a_2 = 3, + a_3 = 5, etc. (The output is the positive odd integers). + Evaluate the following summations: + +

      +
    1. \sum_{i=1}^6 a_i
      2. + +
    2. \sum_{i=3}^7 (3a_i-4)
      3. + +
    3. \sum_{i=1}^4 (a_i)^2
      4. +
    +

    +     + +

    +

      +
    1. +

      + + \sum_{i=1}^6 a_i \amp = a_1+a_2+a_3+a_4+a_5+a_6 + \amp = 1+3+5+7+9+11 + \amp = 36 + . +

      +
      2. + +
    2. +

      + Note the starting value is different than 1: + + \sum_{i=3}^7 (3a_i-4) \amp = (3a_3-4)+(3a_4-4)+(3a_5-4)+(3a_6-4)+(3a_7-4) + \amp = 11+17+23+29+35 + \amp = 115 + . +

      +
      3. + +
    3. +

      + + \sum_{i=1}^4 (a_i)^2 \amp = (a_1)^2+(a_2)^2+(a_3)^2+(a_4)^2 + \amp = 1^2+3^2+5^2+7^2 + \amp = 84 + . +

      +
      4. +
    +

    +     + +     + +

    + It might seem odd to stress a new, + concise way of writing summations only to write each term out as we add them up. + It is. + The following theorem gives some of the properties of summations that allow us to work with them without writing individual terms. + Examples will follow. +

    + + + Properties of Summations + +

    + summationproperties + +

      +
    1. +

      + \ds \sum_{i=1}^n c = c\cdot n, + where c is a constant. +

      +
      2. + +
    2. \sum_{i=m}^n (a_i\pm b_i) = \sum_{i=m}^n a_i \pm \sum_{i=m}^n b_i
      3. + +
    3. \sum_{i=m}^n c\cdot a_i = c\cdot\sum_{i=m}^n a_i
      4. + +
    4. \sum_{i=m}^j a_i + \sum_{i=j+1}^n a_i = \sum_{i=m}^n a_i
      5. + +
    5. \sum_{i=1}^n i = \frac{n(n+1)}2
      6. + +
    6. \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}6
      7. + +
    7. \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}2\right)^2
      8. +
    +

    +     +     + + + + + + Evaluating summations using <xref ref="thm_summation"/> + +

    + Revisit and, + using , evaluate + + \sum_{i=1}^6 a_i = \sum_{i=1}^6 (2i-1) + . +

    +     + +

    + + \sum_{i=1}^6 (2i-1) \amp = \sum_{i=1}^6 2i - \sum_{i=1}^6 (1) + \amp = \left(2\sum_{i=1}^6 i \right)- 6 + \amp = 2\frac{6(6+1)}{2} - 6 + \amp = 42-6 = 36 + +

    +

    + We obtained the same answer without writing out all six terms. + When dealing with small sizes of n, + it may be faster to write the terms out by hand. + However, + is incredibly important when dealing with large sums as we'll soon see. +

    +     +     +     + + + Riemann Sums +

    + Consider again \int_0^4(4x-x^2)\, dx. + We will approximate this definite integral using 16 equally spaced subintervals and the Right Hand Rule in . + Before doing so, it will pay to do some careful preparation. + Riemann Sum +

    + +
    + Dividing [0,4] into 16 equally spaced subintervals + + A number line between 0 and 4 divided into 16 sections + +

    + A number line ranging between 0 and 4. + The space between each number is divided into 4 equal sections, giving a total of 16 equal sections. + x_0 is written under 0. + x_4 is written under 1. + x_8 is written under 2. + x_12 is written under 3. + x_16 is written under 4. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-.5,0) -- (4.5,0); + + \foreach \x in {0,.25,...,4} { + \draw (\x,-.1) -- (\x,.1); + } + \draw (0,-.25) node {\scriptsize $0$}; + \draw (1,-.25) node {\scriptsize $1$}; + \draw (2,-.25) node {\scriptsize $2$}; + \draw (3,-.25) node {\scriptsize $3$}; + \draw (4,-.25) node {\scriptsize $4$}; + + \draw (0,-.55) node {\scriptsize $x_0$}; + \draw (1,-.55) node {\scriptsize $x_4$}; + \draw (2,-.55) node {\scriptsize $x_8$}; + \draw (3,-.55) node {\scriptsize $x_{12}$}; + \draw (4,-.55) node {\scriptsize $x_{16}$}; + + \end{tikzpicture} + + + +
    + +

    + + shows a number line of [0,4] divided, + or partitioned, into 16 equally spaced subintervals. + We denote 0 as x_0; + we have marked the values of x_4, x_8, + x_{12} and x_{16}. + We could mark them all, but the figure would get crowded. + While it is easy to figure that x_{9} = 2.25, in general, + we want a method of determining the value of x_i without consulting the figure. + Consider: +

    + + + A labeled equation for caclulating values at subintervals + +

    + The equation x_i = x_0 + i\Delta x, with labels for each component. + x_0 is labeled as the starting value. + i is labeled as the numer of subintervals between x_0 and x_i. + \Delta x is labeled as the subinterval size. +

    +     + + + \begin{tikzpicture} + \draw (1.5,1) node {$\displaystyle x_i = x_0 + i\Delta x$}; + + \draw [firstcolor] (1,0) node [text width=32pt,align=center] (a) { \centering starting\\[-5pt] value}; + \draw [firstcolor,->] (a) -- (.95,.75); + + \draw [firstcolor] (2,2.5) node [text width=120pt,align=center] (b) { \centering number of subintervals\\[-5pt] between $x_0$ and $x_i$}; + \draw [firstcolor,->] (b) -- (1.95,1.25); + + \draw [firstcolor] (3,0) node [text width=60pt,align=center] (c) { \centering subinterval\\[-5pt] size}; + \draw [firstcolor,->] (c) -- (2.75,.75); + + \end{tikzpicture} + + + + +

    + So x_{9} = x_0 + 9(4/16) = 2.25. +

    + +

    + If we had partitioned [0,4] into 100 equally spaced subintervals, + each subinterval would have length \Delta x=4/100 = 0.04. + We could compute x_{31} as + + x_{31} = x_0 + 31(4/100) = 1.24 + . +

    + +

    + (That was far faster than creating a sketch first.) +

    + +

    + Given any subdivision of [0,4], + the first subinterval is [x_0,x_1]; + the second is [x_1,x_2]; + the ith subinterval is [x_{i-1},x_{i}]. +

    + +

    + When using the Left Hand Rule, + the height of the ith rectangle will be f(x_{i-1}). +

    + +

    + When using the Right Hand Rule, + the height of the ith rectangle will be f(x_{i}). +

    + +

    + When using the Midpoint Rule, + the height of the ith rectangle will be \ds f\left(\frac{x_{i-1}+x_{i}}2\right). +

    + +

    + Thus approximating \int_0^4(4x-x^2)\, dx with 16 equally spaced subintervals can be expressed as follows, + where \Delta x = 4/16 = 1/4: +

    + +

    +

    +
  • + Left Hand Rule +

    + \ds \sum_{i=1}^{16} f(x_{i-1})\Delta x +

    +
    • + +
  • + Right Hand Rule +

    + \ds \sum_{i=1}^{16} f(x_{i})\Delta x +

    +
    • + +
  • + Midpoint Rule +

    + \ds \sum_{i=1}^{16} f\left(\frac{x_{i-1}+x_{i}}2\right)\Delta x +

    +
    • +
    + Left Hand Rule + Right Hand Rule + Midpoint Rule +

    + +

    + We use these formulas in the next two examples. + The following example lets us practice using the Right Hand Rule and the summation formulas introduced in . +

    + + + Approximating definite integrals using sums + +

    + Approximate \int_0^4(4x-x^2)\, dx using the Right Hand Rule and summation formulas with 16 and 1000 equally spaced intervals. +

    +     + +

    + Using the formula derived before, + using 16 equally spaced intervals and the Right Hand Rule, + we can approximate the definite integral as + + \sum_{i=1}^{16}f(x_{i})\Delta x + . +

    + +

    + We have \Delta x = 4/16 = 0.25. + Since x_i = 0+i\Delta x, we have + + x_{i} = 0 + i\Delta x = i\Delta x + . +

    + +

    + Using the summation formulas, consider: + + \int_0^4 (4x-x^2)\, dx \amp \approx \sum_{i=1}^{16} f(x_{i})\Delta x + \amp = \sum_{i=1}^{16} f(i\Delta x) \Delta x + \amp = \sum_{i=1}^{16} \big(4i\Delta x - (i\Delta x)^2\big)\Delta x + \amp = \sum_{i=1}^{16} (4i\Delta x^2 - i^2\Delta x^3) + \amp = (4\Delta x^2)\sum_{i=1}^{16} i - \Delta x^3 \sum_{i=1}^{16} i^2 + \amp = (4\Delta x^2)\frac{16\cdot 17}{2} - \Delta x^3 \frac{16(17)(33)}6 + \amp = 4\cdot 0.25^2\cdot 136-0.25^3\cdot 1496 + \amp =10.625 + +

    + +

    + We were able to sum up the areas of 16 rectangles with very little computation. + In + the function and the 16 rectangles are graphed. + While some rectangles over-approximate the area, + other under-approximate the area + (by about the same amount). + Thus our approximate area of 10.625 is likely a fairly good approximation. +

    + +

    + Notice Equation; + by replacing 16 by 1,000 + (and appropriately changing the value of \Delta x), + we can use that equation to sum up 1000 rectangles! +

    + +
    + Approximating \int_0^4(4x-x^2)\, dx with the Right Hand Rule and 16 evenly spaced subintervals + + + A parabola approximated using 16 equal subintervals. + +

    + The parabola in approximated using 16 strips of equal width. + The height of each strips are given by the Right Hand Rule. + On the left side the area of the rectangles is slightly greater than the area of the parabola. + On the right side the area of the rectangles is slightly less than the area of the parabola. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4}, + ymin=-1,ymax=5, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; + \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; + + \foreach \x / \y in {0.25 / 0.9375, 0.5 / 1.75, 0.75 / 2.4375, 1. / 3., 1.25 / 3.4375, + 1.5 / 3.75, 1.75 / 3.9375, 2. / 4., 2.25 / 3.9375, 2.5 / 3.75, + 2.75 /3.4375, 3. / 3., 3.25 / 2.4375, 3.5 / 1.75, 3.75 / 0.9375, 4. / 0.} + {\addplot [thick,secondcolor] coordinates {(\x-.25,0) (\x-.25,\y) (\x,\y) (\x,0) (\x-.25,0)}; + } + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + We do so here, + skipping from the original summand to the equivalent of Equation to save space. + Note that \Delta x = 4/1000 = 0.004. + + \int_0^4 (4x-x^2)\, dx \amp \approx \sum_{i=1}^{1000} f(x_{i})\Delta x + \amp = (4\Delta x^2)\sum_{i=1}^{1000} i - \Delta x^3 \sum_{i=1}^{1000} i^2 + \amp = (4\Delta x^2)\frac{1000\cdot 1001}{2} - \Delta x^3 \frac{1000(1001)(2001)}6 + \amp =10.666656 + +

    + +

    + Using many, many rectangles, we have a likely good approximation of + \int_0^4 (4x-x^2)\dx. + That is, + + \int_0^4(4x-x^2)\, dx \approx 10.666656 + . +

    +     + +     + +

    + Before the above example, + we stated what the summations for the Left Hand, Right Hand and Midpoint Rules looked like. + Each had the same basic structure, which was: +

    + +

    +

      +
    1. +

      + each rectangle has the same width, + which we referred to as \Delta x, and +

      +
      2. + +
    2. +

      + each rectangle's height is determined by evaluating f at a particular point in each subinterval. + For instance, + the Left Hand Rule states that each rectangle's height is determined by evaluating f at the left hand endpoint of the subinterval the rectangle lives on. +

      +
      3. +
    +

    + +

    + One could partition an interval [a,b] with subintervals that do not have the same size. + We refer to the length of the + ith subinterval as \Delta x_i. + Also, one could determine each rectangle's height by evaluating f at any + point c_i in the ith subinterval. + Thus the height of the ith subinterval would be f(c_i), + and the area of the ith rectangle would be f(c_i)\Delta x_i. + These ideas are formally defined below. +

    + + + Partition + +

    + A partition \Delta x of a closed interval [a,b] is a set of numbers x_0, + x_1, + \ldots x_{n} where + + a=x_0 \lt x_1 \lt \ldots \lt x_{n-1} \lt x_{n}=b + . +

    + +

    + The length of the ith subinterval, + [x_{i-1},x_{i}], is \Delta x_i = x_{i}-x_{i-1}. + If [a,b] is partitioned into subintervals of equal length, + we let \Delta x represent the length of each subinterval. +

    + +

    + The size of the partition, + denoted \norm{\Delta x}, + is the length of the largest subinterval of the partition. + partition + partitionsize of +

    +     +     + + + + +

    + Summations of rectangles with area + f(c_i)\Delta x_i are named after mathematician Georg Friedrich Bernhard Riemann, + as given in the following definition. +

    + + + Riemann Sum + +

    + Let f be defined on a closed interval [a,b], + let \Delta x be a partition of [a,b] + as given in , + Riemann Sum + and let c_i denote any value in the ith subinterval. +

    + +

    + The sum + + \sum_{i=1}^n f(c_i)\Delta x_i + + is a Riemann sum of f on [a,b]. +

    +     +     + + + + +

    + + shows the approximating rectangles of a Riemann sum of \int_0^4(4x-x^2)\, dx. + While the rectangles in this example do not approximate well the shaded area, + they demonstrate that the subinterval widths may vary and the heights of the rectangles can be determined without following a particular rule. +

    + +
    + An example of a general Riemann sum to approximate \int_0^4(4x-x^2)\, dx + + + A parabola approximated by 3 rectangles of different width + +

    + The parabola in approximated using 3 rectangles of different width. + The height of each rectangle is not given by any particular rule. + The left most strip has a width of 2, and a height of around 0.95. This height is given by a point at around x=0.3. + The center strip has a width of 2.5. It has a height of around 2.8, given by a point at around x=3.2. + The right strip has a width of 0.5. It has a height of around 0.8, given by a point around x=3.8. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4}, + ymin=-1,ymax=5, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; + \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; + + \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:2,.9375); + \draw [thick,secondcolor] (axis cs:2,0) rectangle (axis cs:3.5,2.79); + \draw [thick,secondcolor] (axis cs:3.5,0) rectangle (axis cs:4,.76); + \draw [fill=black] (axis cs:.25,.9375) circle (1pt); + \draw [fill=black] (axis cs:3.1,2.79) circle (1pt); + \draw [fill=black] (axis cs:3.8,.76) circle (1pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Usually Riemann sums are calculated using one of the three methods we have introduced. + The uniformity of construction makes computations easier. + Before working another example, + let's summarize some of what we have learned in a convenient way. +

    + + + Riemann Sum Concepts +

    + Consider \ds \int_a^b f(x) \, dx \approx \sum_{i=1}^n f(c_i)\Delta x_i. +

    + +

    +

      +
    1. +

      + When the n subintervals have equal length, + \ds \Delta x_i = \Delta x = \frac{b-a}n. +

      +
      2. + +
    2. +

      + The ith term of an equally spaced partition is x_i = a + i\Delta x. + (Thus x_0=a and x_{n} = b.) +

      +
      3. + +
    3. +

      + The Left Hand Rule summation is: + \ds \sum_{i=1}^n f(x_{i-1})\Delta x. +

      +
      4. + +
    4. +

      + The Right Hand Rule summation is: + \ds \sum_{i=1}^n f(x_{i})\Delta x. +

      +
      5. + +
    5. +

      + The Midpoint Rule summation is: + \ds \sum_{i=1}^n f\left(\frac{x_{i-1}+x_{i}}{2}\right)\Delta x. +

      +
      6. +
    +

    +     + +

    + Let's do another example. +

    + + + Approximating definite integrals with sums + +

    + Approximate \int_{-2}^3 (5x+2)\, dx using the Midpoint Rule and 10 equally spaced intervals. +

    +     + +

    + Following , we have + + \Delta x = \frac{3 - (-2)}{10} = 1/2 \text{ and } x_i = (-2) + (1/2)(i) = i/2-2 + . +

    + +

    + As we are using the Midpoint Rule, + we will also need x_{i-1} and \ds \frac{x_{i-1}+x_{i}}2. + Since x_i = i/2-2,x_{i-1} = (i-1)/2 - 2 = i/2 -5/2. + This gives + + \frac{x_{i-1}+x_{i}}2 = \frac{(i/2-5/2) + (i/2-2)}{2} = \frac{i-9/2}{2} = i/2 - 9/4 + . +

    + +

    + We now construct the Riemann sum and compute its value using summation formulas. + + \int_{-2}^3 (5x+2)\, dx \amp \approx \sum_{i=1}^{10} f\left(\frac{x_{i-1}+x_{i}}{2}\right)\Delta x + \amp = \sum_{i=1}^{10} f(i/2 - 9/4)\Delta x + \amp = \sum_{i=1}^{10} \big(5(i/2-9/4) + 2\big)\Delta x + \amp = \Delta x\sum_{i=1}^{10}\left[\left(\frac{5}{2}\right)i - \frac{37}{4}\right] + \amp = \Delta x\left(\frac{5}2\sum_{i=1}^{10} (i) - \sum_{i=1}^{10}\left(\frac{37}{4}\right)\right) + \amp = \frac12\left(\frac52\cdot\frac{10(11)}{2} - 10\cdot\frac{37}4\right) + \amp = \frac{45}2 = 22.5 + +

    + +
    + Approximating \int_{-2}^3 (5x+2)\, dx using the Midpoint Rule and 10 evenly spaced subintervals in + + + The area under a line approximated with 10 even subintervals. + +

    + The graph 5x + 2, plotted from x = -2 to x = 3. + The line begins in the third quadrant and crosses the x-axis at x = -\frac{2}{5}. + The line crosses the y-axis at y=2. + The area between the line and the x-axis is shaded. + 10 rectangles of equal width are approximating the shaded area. + The height of each rectangle is given by the Midpoint Rule. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-2,-1,1,2,3}, + ytick={10,17,-8}, + ymin=-9,ymax=19, + xmin=-2.5,xmax=3.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=-2:3] {5*x+2} \closedcycle; + \addplot [firstcurvestyle,domain=-2:3] {5*x+2}; + + \foreach \x / \y in {-1.75 / -6.75, -1.25 / -4.25, -0.75 / -1.75, -0.25 / 0.75, + 0.25 /3.25, 0.75 / 5.75, 1.25 / 8.25, 1.75 / 10.75, 2.25 / 13.25, + 2.75 /15.75} + {\addplot [thick,secondcolor] coordinates {(\x-.25,0) (\x-.25,\y) (\x+.25,\y) (\x+.25,0) (\x-.25,0)}; + } + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Note the graph of f(x) = 5x+2 in . + The regions whose area is computed by the definite integral are triangles, + meaning we can find the exact answer without summation techniques. + We find that the exact answer is indeed 22.5. + One of the strengths of the Midpoint Rule is that often each rectangle includes area that should not be counted, + but misses other area that should. + When the partition size is small, + these two amounts are about equal and these errors almost + cancel each other out. In this example, + since our function is a line, + these errors are exactly equal and they do cancel each other out, + giving us the exact answer. +

    + +

    + Note too that when the function is negative, + the rectangles have a negative height. + When we compute the area of the rectangle, + we use f(c_i)\Delta x; + when f is negative, the area is counted as negative. +

    +     + +     + +

    + Notice in the previous example that while we used 10 equally spaced intervals, + the number 10 didn't play a big role in the calculations until the very end. + Mathematicians love to abstract ideas; + let's approximate the area of another region using n subintervals, + where we do not specify a value of n until the very end. +

    + + + Approximating definite integrals with a formula, using sums + +

    + Revisit \int_0^4(4x-x^2)\, dx yet again. + Approximate this definite integral using the Right Hand Rule with n equally spaced subintervals. +

    +     + +

    + Using , + we know \Delta x = \frac{4-0}{n} = 4/n. + We also find x_i = 0 + i\Delta x = 4i/n. +

    + +

    + We construct the Right Hand Rule Riemann sum as follows. + Be sure to follow each step carefully. + If you get stuck, + and do not understand how one line proceeds to the next, + you may skip to the result and consider how this result is used. + You should come back, though, + and work through each step for full understanding. + + \int_0^4(4x-x^2)\, dx \amp \approx \sum_{i=1}^n f(x_{i})\Delta x + \amp = \sum_{i=1}^n f\left(\frac{4i}{n}\right) \Delta x + \amp = \sum_{i=1}^n \left[4\frac{4i}n-\left(\frac{4i}n\right)^2\right]\Delta x + \amp = \sum_{i=1}^n \left(\frac{16\Delta x}{n}\right)i - \sum_{i=1}^n \left(\frac{16\Delta x}{n^2}\right)i^2 + \amp = \left(\frac{16\Delta x}{n}\right)\sum_{i=1}^n i - \left(\frac{16\Delta x}{n^2}\right)\sum_{i=1}^n i^2 + \amp = \left(\frac{16\Delta x}{n}\right)\cdot \frac{n(n+1)}{2} - \left(\frac{16\Delta x}{n^2}\right)\frac{n(n+1)(2n+1)}{6} + \amp =\frac{32(n+1)}{n} - \frac{32(n+1)(2n+1)}{3n^2} (\text{ recall } \Delta x = 4/n) + \amp = \frac{32}{3}\left(1-\frac{1}{n^2}\right) \text{ (after simplifying) } + +

    + +

    + The result is an amazing, easy to use formula. + To approximate the definite integral with 10 equally spaced subintervals and the Right Hand Rule, + set n=10 and compute + + \int_0^4 (4x-x^2)\, dx \approx \frac{32}{3}\left(1-\frac{1}{10^2}\right) = 10.56 + . +

    + +

    + Recall how earlier we approximated the definite integral with 4 subintervals; + with n=4, the formula gives 10, our answer as before. +

    + +

    + It is now easy to approximate the integral with 1,000,000 subintervals! + Hand-held calculators will round off the answer a bit prematurely giving an answer of 10.66666667. + (The actual answer is 10.666666666656.) +

    + +

    + We now take an important leap. + Up to this point, + our mathematics has been limited to geometry and algebra + (finding areas and manipulating expressions). + Now we apply calculus. + For any finite n, we know that + + \int_0^4 (4x-x^2)\, dx \approx \frac{32}{3}\left(1-\frac{1}{n^2}\right) + . +

    + +

    + Both common sense and high-level mathematics tell us that as n gets large, + the approximation gets better. + In fact, if we take the limit + as n\rightarrow \infty, + we get the exact area described by \int_0^4 (4x-x^2)\, dx. + That is, + + \int_0^4 (4x-x^2)\, dx \amp = \lim_{n\rightarrow \infty} \frac{32}{3}\left(1-\frac{1}{n^2}\right) + \amp = \frac{32}{3}\left(1-0\right) + \amp = \frac{32}{3} = 10.\overline{6} + +

    + +

    + This is a fantastic result. + By considering n equally-spaced subintervals, + we obtained a formula for an approximation of the definite integral that involved our variable n. + As n grows large without bound the error shrinks to zero and we obtain the exact area. +

    +     + +     + +

    + This section started with a fundamental calculus technique: + make an approximation, refine the approximation to make it better, + then use limits in the refining process to get an exact answer. + That is precisely what we just did. +

    + +

    + Let's practice this again. +

    + + + Approximating definite integrals with a formula, using sums + +

    + Find a formula that approximates + \int_{-1}^5 x^3\, dx using the Right Hand Rule and n equally spaced subintervals, + then take the limit as n\to\infty to find the exact area. +

    +     + +

    + Following , + we have \Delta x = \frac{5-(-1)}{n} = 6/n. + We have x_i = (-1) + i\Delta x, + which is the right endpoint of the ith subinterval. +

    + +

    + The Riemann sum corresponding to the Right Hand Rule is (followed by simplifications): + + \int_{-1}^5 x^3\, dx \amp \approx \sum_{i=1}^n f(x_{i})\Delta x + \amp = \sum_{i=1}^n f(-1+i\Delta x)\Delta x + \amp = \sum_{i=1}^n (-1+i\Delta x)^3\Delta x + \amp = \sum_{i=1}^n \big((i\Delta x)^3 -3(i\Delta x)^2 + 3i\Delta x -1\big)\Delta x \text{ (now distribute \(\Delta x\)) } + \amp = \sum_{i=1}^n \big(i^3\Delta x^4 - 3i^2\Delta x^3 + 3i\Delta x^2 -\Delta x\big) \text{ (now split up summation) } + \amp = \Delta x^4 \sum_{i=1}^ni^3 -3\Delta x^3 \sum_{i=1}^n i^2+ 3\Delta x^2 \sum_{i=1}^n i - \sum_{i=1}^n \Delta x + \amp = \Delta x^4 \left(\frac{n(n+1)}{2}\right)^2 -3\Delta x^3 \frac{n(n+1)(2n+1)}{6}+ 3\Delta x^2 \frac{n(n+1)}{2} - n\Delta x + (use \Delta x = 6/n) + \amp = \frac{1296}{n^4}\cdot\frac{n^2(n+1)^2}{4} - 3\frac{216}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} + 3\frac{36}{n^2}\frac{n(n+1)}2 -6 + (now do a sizable amount of algebra to simplify) + \amp =156 + \frac{378}n + \frac{216}{n^2} + +

    + +

    + Once again, we have found a compact formula for approximating the definite integral with n equally spaced subintervals and the Right Hand Rule. + Using 10 subintervals, we have an approximation of 195.96 + (these rectangles are shown in ). + Using n=100 gives an approximation of 159.802. +

    + +
    + Approximating \int_{-1}^5 x^3\, dx using the Right Hand Rule and 10 evenly spaced subintervals + + + A cubic graph approximated using 10 subintervals and the Right Hand Rule. + +

    + The curve begins at x=-1, at which point the curve is close to the x-axis. + The curve increases to the right, ending at x=5, at which point the curve has a value of 125. + The area under the curve is approsimated by 10 rectangles of equal width. + The height of each rectangle is given by the Right Hand Rule. + The left of each rectangle includes some unshaded area, so each rectangle overestimates the area under the curve. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,1,2,3,4,5}, + ymin=-2,ymax=130, + xmin=-1.5,xmax=5.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=-1:5] {x^3} \closedcycle; + \addplot [firstcurvestyle,domain=-1:5] {x^3}; + + \foreach \x / \y in {-0.4 / -0.064, 0.2 / 0.008, 0.8 / 0.512, 1.4 / 2.744, 2. / 8., 2.6 /17.576, 3.2 / 32.768, 3.8 / 54.872, 4.4 / 85.184, 5. / 125.} + {\addplot [thick,secondcolor] coordinates {(\x-.6,0) (\x-.6,\y) (\x,\y) (\x,0) (\x-.6,0)}; + } + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Now find the exact answer using a limit: + + \int_{-1}^5 x^3\, dx = \lim_{n\to\infty} \left(156 + \frac{378}n + \frac{216}{n^2}\right) = 156 + . +

    +     + +     +     + + + Limits of Riemann Sums +

    + We have used limits to evaluate given definite integrals. + Will this always work? + It can be shown, + given not-very-restrictive conditions, + that yes, it will always work this is the content of + below. +

    + +

    + The previous two examples demonstrated how an expression such as + + \sum_{i=1}^n f(x_{i})\Delta x + + can be rewritten as an expression explicitly involving n, + such as 32/3(1-1/n^2). +

    + +

    + Viewed in this manner, + we can think of the summation as a function of n. + An n value is given + (where n is a positive integer), + and the sum of areas of n equally spaced rectangles is returned, + using the Left Hand, Right Hand, or Midpoint Rules. +

    + +

    + Given a definite integral \int_a^b f(x)\, dx, let: +

    + +

    +

      +
    • +

      + \ds S_L(n) = \sum_{i=1}^n f(x_{i-1})\Delta x, + the sum of equally spaced rectangles formed using the Left Hand Rule, +

      +
      • + +
    • +

      + \ds S_R(n) = \sum_{i=1}^n f(x_{i})\Delta x, + the sum of equally spaced rectangles formed using the Right Hand Rule, and +

      +
      • + +
    • +

      + \ds S_M(n) = \sum_{i=1}^n f\left(\frac{x_{i-1}+x_{i}}{2}\right)\Delta x, + the sum of equally spaced rectangles formed using the Midpoint Rule. +

      +
      • +
    +

    + +

    + Recall the definition of a limit as n\to\infty: + \lim\limits_{n\to\infty}S_L(n) = K if, given any \varepsilon \gt 0, + there exists N \gt 0 such that + + \abs{S_L(n)-K} \lt \varepsilon \text{ when } n\geq N + . +

    + +

    + The following theorem states that we can use any of our three rules to find the exact value of a definite integral \int_a^b f(x)\, dx. + It also goes two steps further. + The theorem states that the height of each rectangle doesn't have to be determined following a specific rule, + but could be f(c_i), + where c_i is any point in the ith subinterval, + as discussed before Riemann Sums were defined in . +

    + +

    + The theorem goes on to state that the rectangles do not need to be of the same width. + Using the notation of , + let \Delta x_i denote the length of the + ith subinterval in a partition of [a,b] and let + \norm{\Delta x} represent the length of the largest subinterval in the partition: + that is, \norm{\Delta x} is the largest of all the \Delta x_i. + If \norm{\Delta x} is small, + then [a,b] must be partitioned into many subintervals, + since all subintervals must have small lengths. + Taking the limit as \norm{\Delta x} goes to zero + implies that the number n of subintervals in the partition is growing to infinity, + as the largest subinterval length is becoming arbitrarily small. + We then interpret the expression + + \lim_{\norm{\Delta x}\to 0}\sum_{i=1}^nf(c_i)\Delta x_i + + as the limit of the sum of the areas of rectangles, + where the width of each rectangle can be different but getting small, + and the height of each rectangle is not necessarily determined by a particular rule. + The theorem states that this Riemann Sum also gives the value of the definite integral of f over [a,b]. +

    + + + Definite Integrals and the Limit of Riemann Sums + +

    + Let f be continuous on the closed interval [a,b] and let S_L(n), + S_R(n), + S_M(n), \Delta x, + \Delta x_i and c_i be defined as before. + Then: + + Riemann Sumand definite integral + + integrationdefinite!Riemann Sums + +

    + +

    +

      +
    1. \begin{aligned}\lim_{n\to\infty} S_L(n) \amp = \lim_{n\to\infty} S_R(n)\\ + \amp = \lim_{n\to\infty} S_M(n)\\ + \amp = \lim_{n\to\infty}\sum_{i=1}^n f(c_i)\Delta x\end{aligned}
      2. + +
    2. \ds \lim_{n\to\infty}\sum_{i=1}^n f(c_i)\Delta x = \int_a^b f(x)\, dx
      3. + +
    3. \lim_{\norm{\Delta x}\to 0} \sum_{i=1}^n f(c_i)\Delta x_i = \int_a^b f(x)\, dx
      4. +
    +

    +     +     + + + + + + +

    + We summarize what we have learned over the past few sections here. +

    + +

    +

      +
    • +

      + Knowing the area under the curve can be useful. + One common example: the area under a velocity curve is displacement. +

      +
      • + +
    • +

      + We have defined the definite integral, \int_a^b f(x)\, dx, + to be the signed area under f on the interval [a,b]. +

      +
      • + +
    • +

      + While we can approximate a definite integral many ways, + we have focused on using rectangles whose heights can be determined using the Left Hand Rule, + the Right Hand Rule and the Midpoint Rule. +

      +
      • + +
    • +

      + Sums of rectangles of this type are called Riemann sums. +

      +
      • + +
    • +

      + The exact value of the definite integral can be computed using the limit of a Riemann sum. + We generally use one of the above methods as it makes the algebra simpler. +

      +
      • +
    +

    + +

    + We first learned of derivatives through limits then learned rules that made the process simpler. + We know of a way to evaluate a definite integral using limits; + in the next section we will see how the Fundamental Theorem of Calculus makes the process simpler. + The key feature of this theorem is its connection between the indefinite integral and the definite integral. +

    +     + + + + Terms and Concepts + + + + +

    + A fundamental calculus technique is to use + to refine approximations to get an exact answer. +

    +     + + + + limits? + + + + +     + + + + + $lb = random(2,9,1); + $ub = random(12,20,1); + $m = random(30,60,1); + $b = non_zero_random(-300,300,1); + if($envir{problemSeed}==1){$lb=7;$ub=14;$m=48;$b=-201;}; + Context()->variables->are(i=>'Real'); + $f = Formula("$m i + $b")->reduce; + + +

    + What is the upper bound in the summation + \sum\limits_{i=}^{} ()? +

    +

    + +

    +     +     +     + + + + + Context()->strings->add('rectangles'=>{}); + Context()->strings->add('rectangle'=>{alias=>'rectangles'}); + + +

    + This section approximates definite integrals using what geometric shape? +

    +

    + +

    +     +     +     + + + + +

    + + A sum using the Right Hand Rule is an example of a Riemann Sum. +

    +     + +     +     + + + Problems + + +

    + Write out each term of the summation and compute the sum. +

    +     + + + + + $lb = random(2,5,1); + $ub = $lb + random(2,3,1); + if($envir{problemSeed}==1){$lb=2;$ub=4;}; + @i = ($lb..$ub); + @fi = map{$_**2}(@i); + $string = join('+',@fi); + $sum = Real($string); + Context("Form"); + Context()->operators->set( + '+' => {class => 'bizarro::BOP::add', isCommand => 1}, + '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, + ); + $terms = Formula("$string"); + + +

    + \sum\limits_{i=}^{} i^2 +

    + + Enter the sum of terms here. For example, 1+2+3. + +

    + +

    + + Enter the actual sum here. + +

    + +

    +     +     +     + + + + + $lb = random(-3,-1,1); + $ub = random(1,3,1); + $m = random(2,5,1); + $b = non_zero_random(-3,3,1); + if($envir{problemSeed}==1){$lb=-1;$ub=3;$m=4;$b=-2;}; + Context()->variables->are(i=>'Real'); + $f = Formula("$m i + $b")->reduce; + @i = ($lb..$ub); + @fi = map{$m*$_ + $b}(@i); + $string = join('+',@fi); + $sum = Real($string); + Context("Form"); + Context()->operators->set( + '+' => {class => 'bizarro::BOP::add', isCommand => 1}, + '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, + ); + $terms = Formula("$string"); + + +

    + \sum\limits_{i=}^{} \left(\right) +

    + + Enter the sum of terms here. For example, 1+2+3. + +

    + +

    + + Enter the actual sum here. + +

    + +

    +     +     +     + + + + + $lb = random(-3,-1,1); + $ub = random(1,3,1); + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$lb=-2;$ub=2;$trig='sin';}; + Context()->variables->are(i=>'Real'); + $f = Formula("sin(pi i /2)")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + Context("Form"); + Context()->operators->set( + '+' => {class => 'bizarro::BOP::add', isCommand => 1}, + '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, + ); + $terms = Formula("$string"); + + +

    + \sum\limits_{i=}^{} +

    + + Enter the sum of terms here. For example, 1+2+3. + +

    + +

    + + Enter the actual sum here. + +

    + +

    +     +     +     + + + + + $lb = 1; + ($ub,$c) = random_subset(2,5..10); + if($envir{problemSeed}==1){$ub=10;$c=5;}; + Context()->variables->are(i=>'Real'); + $f = Formula("$c")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + Context("Form"); + Context()->operators->set( + '+' => {class => 'bizarro::BOP::add', isCommand => 1}, + '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, + ); + $terms = Formula("$string"); + + +

    + \sum\limits_{i=}^{} +

    + + Enter the sum of terms here. For example, 1+2+3. + +

    + +

    + + Enter the actual sum here. + +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(4,6,1); + if($envir{problemSeed}==1){$ub=5;}; + Context("Fraction"); + Context()->variables->are(i=>'Real'); + $f = Formula("1/i")->reduce; + @i = ($lb..$ub); + @fi = map{Fraction($f->eval(i=>$_))}(@i); + $string = join('+',@fi); + $sum = Fraction($string); + Context("Form"); + Context()->operators->set( + '+' => {class => 'bizarro::BOP::add', isCommand => 1}, + '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, + ); + $terms = Formula("$string"); + + +

    + \sum\limits_{i=}^{} +

    + + Enter the sum of terms here. For example, 1+2+3. + +

    + +

    + + Enter the actual sum here. + +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(4,8,1); + if($envir{problemSeed}==1){$ub=6;}; + Context()->variables->are(i=>'Real'); + $f = Formula("(-1)^i i")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + Context("Form"); + Context()->operators->set( + '+' => {class => 'bizarro::BOP::add', isCommand => 1}, + '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, + ); + $terms = Formula("$string"); + + +

    + \sum\limits_{i=}^{} +

    + + Enter the sum of terms here. For example, 1+2+3. + +

    + +

    + + Enter the actual sum here. + +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(3,5,1); + if($envir{problemSeed}==1){$ub=4;}; + Context()->variables->are(i=>'Real'); + $f = Formula("1/i - 1/(i+1)")->reduce; + @i = ($lb..$ub); + Context("Fraction"); + @fi = map{Fraction($f->eval(i=>$_))}(@i); + $string = join('+',@fi); + $sum = Fraction($string); + Context("Form"); + Context()->operators->set( + '+' => {class => 'bizarro::BOP::add', isCommand => 1}, + '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, + ); + $terms = Formula("$string"); + + +

    + \sum\limits_{i=}^{} \left(\right) +

    + + Enter the sum of terms here. For example, 1+2+3. + +

    + +

    + + Enter the actual sum here. + +

    + +

    +     +     +     + + + + + $lb = random(0,1,1); + $ub = random(4,6,1); + if($envir{problemSeed}==1){$lb=0;$ub=5;}; + Context()->variables->are(i=>'Real'); + $f = Formula("(-1)^i cos(pi i)")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + Context("Form"); + Context()->operators->set( + '+' => {class => 'bizarro::BOP::add', isCommand => 1}, + '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, + ); + $terms = Formula("$string"); + + +

    + \sum\limits_{i=}^{} +

    + + Enter the sum of terms here. For example, 1+2+3. + +

    + +

    + + Enter the actual sum here. + +

    + +

    +     +     +     +     + + + +

    + Write the sum in summation notation. +

    +     + + + + + $lb = 1; + $ub = random(4,6,1); + $m = random(2,9,1); + if($envir{problemSeed}==1){$ub=5;$m=3;}; + Context()->variables->are(i=>'Real'); + $f = Formula("$m i")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + Context()->flags->set(reduceConstants=>0); + $terms = Formula("$string"); + $multians = MultiAnswer($lb, $ub, $f)->with( + singleResult => 1, + checker => sub { + my ( $correct, $student, $self ) = @_; + my ( $stulb, $stuub, $stuf ) = @{$student}; + my ( $corlb, $corub, $corf ) = @{$correct}; + if ($stuub - $stulb != $corub - $corlb) {Value::Error("Your sum does not have the correct number of terms");}; + @stui = ($stulb..$stuub); + @stufi = map{$stuf->eval(i=>$_)}(@stui); + @cori = ($corlb..$corub); + @corfi = map{$corf->eval(i=>$_)}(@cori); + $score = 1; + for $j (0..$#corfi) { + if ($stufi[$j] != $corfi[$j]) { + $score = 0; + last; + } + } + return $score; + } + ); + + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the start value here. + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the end value here. + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the summand expression here. + +

    + +

    +     +     +     + + + + + $lb = 0; + $ub = random(5,9,1); + $b = non_zero_random(-2,2,1); + if($envir{problemSeed}==1){$ub=8;$b=-1;}; + Context()->variables->are(i=>'Real'); + $f = Formula("i^2 + $b")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + Context()->flags->set(reduceConstants=>0); + $terms = Formula("$string"); + $multians = MultiAnswer($lb, $ub, $f)->with( + singleResult => 1, + checker => sub { + my ( $correct, $student, $self ) = @_; + my ( $stulb, $stuub, $stuf ) = @{$student}; + my ( $corlb, $corub, $corf ) = @{$correct}; + if ($stuub - $stulb != $corub - $corlb) {Value::Error("Your sum does not have the correct number of terms");}; + @stui = ($stulb..$stuub); + @stufi = map{$stuf->eval(i=>$_)}(@stui); + @cori = ($corlb..$corub); + @corfi = map{$corf->eval(i=>$_)}(@cori); + $score = 1; + for $j (0..$#corfi) { + if ($stufi[$j] != $corfi[$j]) { + $score = 0; + last; + } + } + return $score; + } + ); + + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the start value here. + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the end value here. + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the summand expression here. + +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(4,6,1); + $b = random(1,5,1); + if($envir{problemSeed}==1){$ub=4;$b=1;}; + Context("Fraction"); + Context()->flags->set(reduceFractions=>0); + Context()->variables->are(i=>'Real'); + $f = Formula("i/(i+$b)")->reduce; + @i = ($lb..$ub); + @fi = map{Fraction($_,$_+$b)}(@i); + $string = join('+',@fi); + Context()->flags->set(reduceConstants=>0); + $terms = Formula("$string"); + $multians = MultiAnswer($lb, $ub, $f)->with( + singleResult => 1, + checker => sub { + my ( $correct, $student, $self ) = @_; + my ( $stulb, $stuub, $stuf ) = @{$student}; + my ( $corlb, $corub, $corf ) = @{$correct}; + if ($stuub - $stulb != $corub - $corlb) {Value::Error("Your sum does not have the correct number of terms");}; + @stui = ($stulb..$stuub); + @stufi = map{$stuf->eval(i=>$_)}(@stui); + @cori = ($corlb..$corub); + @corfi = map{$corf->eval(i=>$_)}(@cori); + $score = 1; + for $j (0..$#corfi) { + if ($stufi[$j] != $corfi[$j]) { + $score = 0; + last; + } + } + return $score; + } + ); + + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the start value here. + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the end value here. + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the summand expression here. + +

    + +

    +     +     +     + + + + + $lb = list_random(0,1); + $ub = $lb + 4; + if($envir{problemSeed}==1){$lb=0;}; + Context()->variables->are(i=>'Real'); + $f = ($lb == 0) ? Formula("(-1)^i") : Formula("(-1)^(i+1)"); + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_) * Formula("e^$_")}(@i); + $f = ($lb == 0) ? Formula("(-e)^i") : Formula("-(-e)^i"); + $string = join('+',@fi); + Context()->flags->set(reduceConstants=>0); + $terms = Formula("$string")->reduce; + $multians = MultiAnswer($lb, $ub, $f)->with( + singleResult => 1, + checker => sub { + my ( $correct, $student, $self ) = @_; + my ( $stulb, $stuub, $stuf ) = @{$student}; + my ( $corlb, $corub, $corf ) = @{$correct}; + if ($stuub - $stulb != $corub - $corlb) {Value::Error("Your sum does not have the correct number of terms");}; + @stui = ($stulb..$stuub); + @stufi = map{$stuf->eval(i=>$_)}(@stui); + @cori = ($corlb..$corub); + @corfi = map{$corf->eval(i=>$_)}(@cori); + $score = 1; + for $j (0..$#corfi) { + if ($stufi[$j] != $corfi[$j]) { + $score = 0; + last; + } + } + return $score; + } + ); + + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the start value here. + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the end value here. + +

    + +

    + + If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the summand expression here. + +

    + +

    +     +     +     +     + + + +

    + Evaluate the summation using . +

    +     + + + + + $lb = 1; + ($ub,$c) = random_subset(2,5..10); + if($envir{problemSeed}==1){$ub=10;$c=5;}; + Context()->variables->are(i=>'Real'); + $f = Formula("$c")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + + +

    + \sum\limits_{i=}^{} +

    +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(20,30,1); + if($envir{problemSeed}==1){$ub=25;}; + Context()->variables->are(i=>'Real'); + $f = Formula("i")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + + +

    + \sum\limits_{i=}^{} +

    +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(8,12,1); + $a = random(2,6,1); + $b = non_zero_random(-4,4,1); + if($envir{problemSeed}==1){$ub=10;$a=3;$b=-2;}; + Context()->variables->are(i=>'Real'); + $f = Formula("$a i^2 + $b i")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + + +

    + \sum\limits_{i=}^{} \left(\right) +

    +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(12,20,1); + $a = random(2,6,1); + $b = non_zero_random(-10,10,1); + if($envir{problemSeed}==1){$ub=15;$a=2;$b=-10;}; + Context()->variables->are(i=>'Real'); + $f = Formula("$a i^3 + $b")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + + +

    + \sum\limits_{i=}^{} \left(\right) +

    +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(8,12,1); + ($a,$b,$c,$d) = random_subset(4,-12..-1,1..12); + if($envir{problemSeed}==1){$ub=10;$a=-4;$b=10;$c=-7;$d=11;}; + Context()->variables->are(i=>'Real'); + Context()->noreduce('(-x)+y','(-x)-y'); + $f = Formula("$a i^3 + $b i^2 + $c i + $d")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + + +

    + \sum\limits_{i=}^{} \left(\right) +

    +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(8,12,1); + ($b,$c,$d) = random_subset(3,-9..-1,1..9); + if($envir{problemSeed}==1){$ub=10;$b=-3;$c=2;$d=7;}; + Context()->variables->are(i=>'Real'); + Context()->noreduce('(-x)+y','(-x)-y'); + $f = Formula("i^3 + $b i^2 + $c i + $d")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + + +

    + \sum\limits_{i=}^{} \left(\right) +

    +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(80,120,5); + if($envir{problemSeed}==1){$ub=100;}; + Context()->variables->are(i=>'Real'); + $f = Formula("i")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + $penultimate = $ub - 1; + + +

    + 1+2+3+\cdots++ +

    +

    + +

    +     +     +     + + + + + $lb = 1; + $ub = random(16,30,1); + if($envir{problemSeed}==1){$ub=20;}; + Context()->variables->are(i=>'Real'); + $f = Formula("i^2")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + $last = $ub**2; + $penultimate = ($ub - 1)**2; + + +

    + 1+4+9+\cdots++ +

    +

    + +

    +     +     +     +     + + + +

    + states + \sum\limits_{i=1}^na_i = \sum\limits_{i=1}^k a_i + \sum\limits_{i=k+1}^n a_i, + so + \sum\limits_{i=k+1}^na_i = \sum\limits_{i=1}^n a_i - \sum\limits_{i=1}^k a_i. + Use this fact, + along with other parts of , + to evaluate the summation. +

    +     + + + + + $lb = random(7,13,1); + $ub = $lb + random(8,12,1); + if($envir{problemSeed}==1){$lb=11;$ub=20;}; + Context()->variables->are(i=>'Real'); + $f = Formula("i")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + + +

    + \sum\limits_{i=}^{} +

    +

    + +

    +     +     +     + + + + + $lb = random(14,18,1); + $ub = $lb + random(8,12,1); + if($envir{problemSeed}==1){$lb=16;$ub=25;}; + Context()->variables->are(i=>'Real'); + $f = Formula("i**3")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + + +

    + \sum\limits_{i=}^{} +

    +

    + +

    +     +     +     + + + + + $lb = random(5,8,1); + $ub = $lb + random(5,8,1); + $c = random(3,8,1); + if($envir{problemSeed}==1){$lb=7;$ub=12;$c=4;}; + Context()->variables->are(i=>'Real'); + $f = Formula("$c")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + + +

    + \sum\limits_{i=}^{} +

    +

    + +

    +     +     +     + + + + + $lb = random(5,8,1); + $ub = $lb + random(5,8,1); + $c = random(3,8,1); + if($envir{problemSeed}==1){$lb=5;$ub=10;$c=4;}; + Context()->variables->are(i=>'Real'); + $f = Formula("$c i**3")->reduce; + @i = ($lb..$ub); + @fi = map{$f->eval(i=>$_)}(@i); + $string = join('+',@fi); + $sum = Real($string); + + +

    + \sum\limits_{i=}^{} +

    +

    + +

    +     +     +     +     + + + +

    + A definite integral \ds \int_a^b f(x) \, dx is given. +

      +
    1. +

      + Graph f(x) on [a,b]. +

      +
      2. +
    2. +

      + Add to the sketch rectangles using the provided rule. +

      +
      3. +
    3. +

      + Approximate \ds \int_a^b f(x) \, dx by summing the areas of the rectangles. +

      +
      4. +
    +

    +     + + + +

    + \ds \int_{-3}^3 x^2\, dx, + with 6 rectangles using the Left Hand Rule. +

    +     +     + + + +

    + \ds \int_{0}^2 (5-x^2)\, dx, + with 4 rectangles using the Midpoint Rule. +

    +     +     + + + +

    + \ds \int_{0}^\pi \sin(x) \, dx, + with 6 rectangles using the Right Hand Rule. +

    +     +     + + + +

    + \ds \int_{0}^3 2^x\, dx, + with 5 rectangles using the Left Hand Rule. +

    +     +     + + + +

    + \ds \int_{1}^2 \ln(x) \, dx, + with 3 rectangles using the Midpoint Rule. +

    +     +     + + + +

    + \ds \int_{1}^9 \frac1x\, dx, + with 4 rectangles using the Right Hand Rule. +

    +     +     +     + + + +

    + A definite integral is given below. + As demonstrated in Examples + and , do the following: +

      +
    1. +

      + Find a formula to approximate the definite integral + using n subintervals and the provided rule. +

      +
      2. +
    2. +

      + Evaluate the formula using n=10, + 100, and 1000. +

      +
      3. +
    3. +

      + Find the limit of the formula, as n\to \infty, + to find the exact value of the definite integral. +

      +
      4. +
    +

    +     + + + + + $rule = list_random('Left Hand Rule','Right Hand Rule'); + if($envir{problemSeed}==1){$rule = 'Right Hand Rule';}; + Context()->variables->are(n=>"Real"); + if ($rule eq 'Left Hand Rule') { + $sum = Formula("(n-1)^2/(4n^2)"); + } + elsif ($rule eq 'Right Hand Rule') { + $sum = Formula("(n+1)^2/(4n^2)"); + }; + $sum10 = $sum->eval(n=>10); + $sum100 = $sum->eval(n=>100); + $sum1000 = $sum->eval(n=>1000); + Context("Fraction"); + $exact = Fraction(1,4); + + +

    + \ds \int_{0}^1 x^3\, dx, using the . +

    + + Enter the formula in terms of n here. + +

    + +

    + + Evaluate the formula using n=10 here. + +

    + +

    + + Evaluate the formula using n=100 here. + +

    + +

    + + Evaluate the formula using n=1000 here. + +

    + +

    + + Enter the limit here. + +

    + +

    +     +     +     + + + + + $rule = list_random('Left Hand Rule','Right Hand Rule'); + $lb = random(-2,-1,1); + $ub = random(1,2,1); + $c = random(2,5,1); + if($envir{problemSeed}==1){$rule = 'Left Hand Rule';$lb=-1;$ub=1;$c=3;}; + Context("Fraction"); + Context()->variables->are(n=>"Real"); + $A = Fraction($c*($ub - $lb)*(6*$lb*$ub + 2*($ub - $lb)**2),6); + $B = Fraction($c*($ub - $lb)*(6*$lb*($ub - $lb) + 3*($ub - $lb)**2),6); + if ($rule eq 'Left Hand Rule') {$B = -$B;}; + $C = Fraction($c*($ub - $lb)**3,6); + ($Bn,$Bd)=$B->value; + ($Cn,$Cd)=$C->value; + $sum = Formula("$A + $Bn/($Bd n) + $Cn/($Cd n^2)"); + $sum10 = $sum->eval(n=>10); + $sum100 = $sum->eval(n=>100); + $sum1000 = $sum->eval(n=>1000); + $exact = $A; + + +

    + \ds \int_{}^{} x^2\, dx, using the . +

    + + Enter the formula in terms of n here. + +

    + +

    + + Evaluate the formula using n=10 here. + +

    + +

    + + Evaluate the formula using n=100 here. + +

    + +

    + + Evaluate the formula using n=1000 here. + +

    + +

    + + Enter the limit here. + +

    + +

    +     +     +     + + + + + $lb = random(-2,-1,1); + $ub = random(1,4,1); + $m = random(2,5,1); + $b = non_zero_random(-2,2,1); + if($envir{problemSeed}==1){$lb=-1;$ub=3;$m=3;$b=-1;}; + $f = Formula("$m x + $b")->reduce; + Context("Fraction"); + Context()->variables->are(n=>"Real"); + $sum = Fraction(($ub-$lb)*($m*($ub+$lb)+2*$b),2); + $sum = Formula("$sum"); + $sum10 = $sum->eval(n=>10); + $sum100 = $sum->eval(n=>100); + $sum1000 = $sum->eval(n=>1000); + $exact = Fraction("$sum"); + + +

    + \ds \int_{}^{} \left(\right)\, dx, using the Midpoint Rule. +

    + + Enter the formula in terms of n here. + +

    + +

    + + Evaluate the formula using n=10 here. + +

    + +

    + + Evaluate the formula using n=100 here. + +

    + +

    + + Evaluate the formula using n=1000 here. + +

    + +

    + + Enter the limit here. + +

    + +

    +     +     +     + + + + + $rule = list_random('Left Hand Rule','Right Hand Rule'); + $lb = random(1,2,1); + $ub = $lb + random(2,4,1); + $m = random(2,5,1); + $b = non_zero_random(-4,4,1); + if($envir{problemSeed}==1){$rule='Left Hand Rule';$lb=1;$ub=4;$m=2;$b=-3;}; + $f = Formula("$m x^2 + $b")->reduce; + Context("Fraction"); + $A = Fraction(($ub - $lb)*($m*$ub**2 + $m*$ub*$lb + $m*$lb**2 + 3*$b),3); + $B = Fraction($m*($ub - $lb)**2*($ub + $lb),2); + if ($rule eq 'Left Hand Rule') {$B = -$B;}; + $C = Fraction($m*($ub - $lb)**3,6); + ($Bn,$Bd)=$B->value; + ($Cn,$Cd)=$C->value; + Context()->variables->are(n=>"Real"); + $sum = Formula("$A + $Bn/($Bd n) + $Cn/($Cd n^2)"); + $sum10 = $sum->eval(n=>10); + $sum100 = $sum->eval(n=>100); + $sum1000 = $sum->eval(n=>1000); + $exact = $A; + + +

    + \ds \int_{}^{} \left(\right)\, dx, using the . +

    + + Enter the formula in terms of n here. + +

    + +

    + + Evaluate the formula using n=10 here. + +

    + +

    + + Evaluate the formula using n=100 here. + +

    + +

    + + Evaluate the formula using n=1000 here. + +

    + +

    + + Enter the limit here. + +

    + +

    +     +     +     + + + + + $rule = list_random('Left Hand Rule','Right Hand Rule'); + $lb = random(-12,-7,1); + $ub = -$lb; + $b = non_zero_random(3,8,1); + if($envir{problemSeed}==1){$rule='Right Hand Rule';$lb=-10;$ub=10;$b=5;}; + $f = Formula("$b - x")->reduce; + Context()->variables->are(n=>"Real"); + $sum = Formula("2*$ub*$b + (2*$ub**2)/n")->reduce; + if ($rule eq 'Left Hand Rule') { + $sum = Formula("2*$ub*$b - (2*$ub**2)/n")->reduce; + }; + $sum10 = $sum->eval(n=>10); + $sum100 = $sum->eval(n=>100); + $sum1000 = $sum->eval(n=>1000); + $exact = Real(2*$ub*$b); + + +

    + \ds \int_{}^{} \left(\right)\, dx, using the . +

    + + Enter the formula in terms of n here. + +

    + +

    + + Evaluate the formula using n=10 here. + +

    + +

    + + Evaluate the formula using n=100 here. + +

    + +

    + + Evaluate the formula using n=1000 here. + +

    + +

    + + Enter the limit here. + +

    + +

    +     +     +     + + + + + $rule = list_random('Left Hand Rule','Right Hand Rule'); + if($envir{problemSeed}==1){$rule='Right Hand Rule';}; + $f = Formula("x^3 - x^2")->reduce; + Context("Fraction"); + Context()->variables->are(n=>"Real"); + $sum = Formula("-1/12+1/(12n^2)"); + $sum10 = $sum->eval(n=>10); + $sum100 = $sum->eval(n=>100); + $sum1000 = $sum->eval(n=>1000); + $exact = Fraction(-1,12); + + +

    + \ds \int_{0}^{1} \left(\right)\, dx, using the . +

    + + Enter the formula in terms of n here. + +

    + +

    + + Evaluate the formula using n=10 here. + +

    + +

    + + Evaluate the formula using n=100 here. + +

    + +

    + + Evaluate the formula using n=1000 here. + +

    + +

    + + Enter the limit here. + +

    + +

    +     +     +     +     +     +     +
    +
    + The Fundamental Theorem of Calculus + + + + +

    + Let f(t) be a continuous function defined on [a,b]. + The definite integral \int_a^b f(x)\, dx is the + area under f on [a,b]. + We can turn this concept into a function by letting the upper + (or lower) + bound vary. +

    + +

    + Let F(x) = \int_a^x f(t)\, dt. + It computes the area under f on [a,x] as illustrated in . + We can study this function using our knowledge of the definite integral. + For instance, F(a)=0 since \int_a^af(t)\, dt=0. +

    + +
    + The area of the shaded region is F(x) = \int_a^x f(t)\, dt + + + A concave-down graph in the first quadrant, with the area underneath shaded from a to x. + + +

    + The t and the y axes are uncalibrated. + There are three distinct positions on the t axis, + a, x and b in the order from left to right. + The function f(t) is a curve facing downward, + the areas under the curve between a and x + are shaded. b lies to the right of x + outside of the shaded portion. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={1,2.5,3}, + extra x tick labels={$a$,$x$,$b$}, + ytick=\empty, + xlabel=$t$, + ymin=-.5,ymax=2, + xmin=-.5,xmax=3.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:2.5] {.5*sin(deg(x))+1} \closedcycle; + \addplot [firstcurvestyle,domain=0:3.25,samples=30] {.5*sin(deg(x))+1}; + + \end{axis} + \end{tikzpicture} + + + +
    + + + Exploring the <q>Area so far</q> function + +

    + Consider f(t)=2t pictured in Figure + and its associated area so far function, + F(x)=\int_1^x 2t\, dt. + Using the graph of f and geometry, + find an explicit formula for F. +

    + +
    + The area of the shaded region is F(x) = \int_1^x 2t\, dt + + + The graph y=2t, and below it, the trapezoidal region from t=1 to t=x is shaded. + + +

    + The y axis is drawn from -2 to 8 while the + t axis is drawn without calibration but it has + two distinct points 1 and x. + The graph is a straight line passing through the origin and + has a positive slope. The shaded portion is drawn under the + graph between 1 and x. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={1,2.5}, + extra x tick labels={$1$,$x$}, + %ytick=\empty, + xlabel=$t$, + ymin=-2,ymax=8, + xmin=-1,xmax=3.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:2.5] {2*x} \closedcycle; + \addplot [firstcurvestyle,domain=-1:3.25] {2*x}; + + \end{axis} + + \end{tikzpicture} + + + +
    +     + +

    + We can see from that for x \geq 1, + the area under the curve can be found by subtracting the area of two triangles. + The larger triangle will have a base of x and a height of f(x)=2x, + while the smaller triangle will have a base of 1 and a height of 2. + Therefore, the area under the curve for + x \geq 1 is given by A(x)=\frac12 (x)(2x)-\frac12 (1)(2)=x^2-1. +

    + +
    + The area of the shaded region is F(x) = \int_1^x 2t\, dt + + + The previous trapezoidal region, with heights 2 and 2x shown. + + +

    + Graph of function same as the one used for the previous example. +

    +

    + The y axis is drawn from -2 to 8 while the + t axis is drawn without calibration but it has two + distinct positions 1 and x. The graph is a straight + line passing through the origin and has a positive slope. + The shaded portion is drawn under the graph between 1 and x. +

    +

    + The distance from the origin to point x on the t + axis is marked x. The distance from origin to 1 + is labeled 1. The left boundary of the shaded region the + y value is labeled 2 and on the right boundaries + the y value is labeled 2x. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={1,2.5}, + extra x tick labels={$1$,$x$}, + xlabel=$t$, + ymin=-2,ymax=8, + xmin=-1,xmax=3.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:2.5] {2*x} \closedcycle; + \addplot [firstcurvestyle,domain=-1:3.25] {2*x}; + + \addplot [lineseg,domain=0:2]({1},{x}) node [pos=0.5, right] {$2$}; + \addplot [lineseg,domain=0:1]({x},{0}) node [pos=0.5, above] {$1$}; + \addplot [lineseg,<->, domain=0:2.5] ({x},{-1}) node [pos=0.5, below] {$x$}; + \addplot [lineseg,domain=0:5]({2.5},{x}) node [pos=0.5, right] {$2x$}; + + \end{axis} + + \end{tikzpicture} + + + +
    + +

    + Note that this same formula holds for x\lt 1. + If x \lt 1, + then F(x) = \int_1^x 2t\, dt=-\int_x^1 2t\, dt. + The areas to the left of x=1 will have opposite signs + (since they areas are accumulated + before x=1). + For example, when x=0, + F(0) = -\int_0^1 2t\, dt=-\frac12 (1)(2)=-1. + This is the same value we get from evaluating x^2-1 for x=0. + Also notice that F(-1)=\int_1^{-1} 2t \, dt=-\int_{-1}^1 2t\, dt. + This integral is clearly 0 since the areas over [-1,0] and [0,1] will sum to zero. + Again, this is the same answer obtained by evaluating x^2-1 for x=-1. +

    + +

    + Therefore, we can reasonably say that F(x)=x^2-1. + A plot of both f(x)=2x and + F(x)=x^2-1 are given in Figure . + You should notice a familiar relationship between these two functions. + This relationship is formally stated in . +

    + +
    + Graphs of f(x)=2x and F(x)=x^2-1 + + + Graph of f(x) = 2*x and F(x)= x^2-1. + + +

    + The y axis is drawn from -2 to 8 and + the x axis is drawn from -1 to 3. + The graphs of two functions are shown. + The first is a line through the origin with slope 2, + and the second is a parabola that opens upward, with its vertex at (0,-1). + Both functions intersect at approximately x=-0.4 + and x=2.4, it is also between these points that + the line is above the curve beyond which the curve is + above the line. +

    +

    + The straight line that passes through the origin and has + a positive slope. It lies in the first and the third + quadrant. +

    +

    + The curve appears to start from x intercept + 1. It has a y intercept at -1 and + the second x intercept at 1. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-2,ymax=8, + xmin=-1,xmax=3.5 + ] + + \addplot+ [domain=-1:3.25] {2*x}; + \addplot+ [domain=-1:3.25] {x^2-1}; + + \end{axis} + + \end{tikzpicture} + + + +
    +     + +     +     + + + Fundamental Theorem of Calculus, Parts 1 and 2 +

    + As hinted, + we can apply calculus ideas to F(x); + in particular, we can compute its derivative. + In , + F(x)=x^2-1, so F'(x)=2x=f(x). + While this may seem like an innocuous thing to do, + it has far-reaching implications, + as demonstrated by the fact that the result is given as an important theorem. +

    + + + The Fundamental Theorem of Calculus, Part 1 + +

    + Let f be continuous on [a,b] and let F(x) = \int_a^x f(t)\, dt. + Then F is continuous on [a,b], differentiable on (a,b), and + + Fundamental Theorem of Calculus + + integrationFun. Thm. of Calc. + + + \Fp(x)=f(x) + . + In other words: + + \lzoo{x}{\int_a^x f(t)\, dt}=f(x) + . +

    +     +     + + + + +

    + Initially this seems simple, as demonstrated in the following example. +

    + + + Using the Fundamental Theorem of Calculus, Part 1 + +

    + Let \ds F(x) = \int_{-5}^x (t^2+\sin(t) )\, dt. + What is \Fp(x)? +

    +     + +

    + Using the Fundamental Theorem of Calculus, + we have \Fp(x) = x^2+\sin(x). + That is, the derivative of the area so far function, + is simply the integrand replacing x with t. +

    + +

    + This simple example reveals something incredible: + F(x) is an antiderivative of x^2+\sin(x)! + Therefore, F(x) = \frac13x^3-\cos(x) +C for some value of C. (We can find C, + but generally we do not care. + We know that F(-5)=0, which allows us to compute C. + In this case, C=\cos(-5)+\frac{125}3.) +

    +     + +     + + + + + + + + +

    + What we have done in + was more than finding a complicated way of computing an antiderivative. + Consider a function f defined on an open interval containing a, + b and c. + Suppose we want to compute \int_a^b f(t)\, dt. + First, let + + F(x) = \int_c^x f(t)\, dt + . + Using the properties of the definite integral found in , we know + + \int_a^b f(t)\, dt \amp = \int_a^c f(t)\, dt + \int_c^b f(t)\, dt + \amp = -\int_c^a f(t)\, dt + \int_c^b f(t)\, dt + + Using Equation, + let x=a in the first integral and x=b in the second integral so that + \int_c^a f(t)\, dt =F(a) and \int_c^b f(t)\, dt =F(b). + Therefore: + + \int_a^b f(t)\, dt \amp =-F(a) + F(b) + \amp = F(b) - F(a) + . +

    + +

    + We now see how indefinite integrals and definite integrals are related: + we can evaluate a definite integral using antiderivatives! + In fact, this is exactly what we noticed in . + The area so far function was indeed an anti-derivative of the integrand. + This is the second part of the Fundamental Theorem of Calculus. +

    + + + Fundamental Theorem of Calculus, Part 2 + +

    + Let f be continuous on [a,b] and let F be any + antiderivative of f. + Then + + Fundamental Theorem of Calculus + + integrationFun. Thm. of Calc. + + + \int_a^b f(x)\, dx = F(b) - F(a) + . +

    +     +     + + + + + + + + + + Using the Fundamental Theorem of Calculus, Part 2 + +

    + We spent a great deal of time in the previous section studying \int_0^4(4x-x^2)\, dx. + Using the Fundamental Theorem of Calculus, + evaluate this definite integral. +

    +     + +

    + We need an antiderivative of f(x)=4x-x^2. + All antiderivatives of f have the form F(x) = 2x^2-\frac13x^3+C; + for simplicity, choose C=0. +

    + +

    + The Fundamental Theorem of Calculus states + + \int_0^4(4x-x^2)\, dx \amp= F(4)-F(0) + \amp = \big(2(4)^2-\frac134^3\big)-\big(0-0\big) + \amp = 32-\frac{64}3 = 32/3 + . +

    + +

    + This is the same answer we obtained using limits in the previous section, + just with much less work. +

    +     +     + +

    + Notation: + + integrationnotation + + A special notation is often used in the process of evaluating definite integrals using the Fundamental Theorem of Calculus. + Instead of explicitly writing F(b)-F(a), + the notation F(x)\Big|_a^b is used. + Thus the solution to would be written as: + + \int_0^4(4x-x^2)\, dx \amp = \left.\left(2x^2-\frac13x^3\right)\right|_0^4 + \amp = \big(2(4)^2-\frac134^3\big)-\big(0-0\big) = 32/3 + . +

    + +

    + The Constant C: Any + antiderivative F(x) can be chosen when using the Fundamental Theorem of Calculus to evaluate a definite integral, + meaning any value of C can be picked. + The constant always cancels out of the expression when evaluating F(b)-F(a), + so it does not matter what value is picked. + This being the case, we might as well let C=0. +

    + + + Using the Fundamental Theorem of Calculus, Part 2 + +

    + Evaluate the following definite integrals. + +

      +
    1. \int_{-2}^2 x^3\, dx
      2. + +
    2. \int_0^\pi \sin(x) \, dx
      3. + +
    3. \int_0^5 e^t\, dt
      4. + +
    4. \int_4^9 \sqrt{u}\, du
      5. + +
    5. \int_1^5 2\, dx
      6. +
    +

    +     + +

    +

      +
    1. +

      + + \int_{-2}^2 x^3\, dx \amp = \left.\frac14x^4\right|_{-2}^2 \amp + \amp = \left(\frac142^4\right) - \left(\frac14(-2)^4\right) + \amp = 0 + . +

      +
      2. + +
    2. +

      + + \int_0^\pi \sin(x) \, dx \amp = -\cos(x) \Big|_0^\pi + \amp = -\cos(\pi) - \big(-\cos(0) \big) + \amp = 1+1=2 + . + (This is interesting; it says that the area under one + hump of a sine curve is 2.) +

      +
      3. + +
    3. +

      + + \int_0^5e^t\, dt \amp = e^t\Big|_0^5 + \amp = e^5 - e^0 + \amp= e^5-1 \approx 147.41 + . +

      +
      4. + +
    4. +

      + + \int_4^9 \sqrt{u}\, du \amp= \int_4^9 u^\frac12\, du + \amp = \frac23u^\frac32\Big|_4^9 + \amp = \frac23\left(9^\frac32-4^\frac32\right) + \amp = \frac23\big(27-8\big) =\frac{38}3 + . +

      +
      5. + +
    5. +

      + + \int_1^5 2\, dx \amp = 2x\Big|_1^5 + \amp = 2(5)-2 + \amp =2(5-1)=8 + . + This integral is interesting; + the integrand is a constant function, + hence we are finding the area of a rectangle with width (5-1)=4 and height 2. + Notice how the evaluation of the definite integral led to 2(4)=8. + + In general, if c is a constant, + then \int_a^b c\, dx = c(b-a). +

      +
      6. +
    +

    +     + +     +     + + + Understanding Motion with the Fundamental Theorem of Calculus +

    + We established, + starting with , + that the derivative of a position function is a velocity function, + and the derivative of a velocity function is an acceleration function. + Now consider definite integrals of velocity and acceleration functions. + Specifically, if v(t) is a velocity function, + what does \ds \int_a^b v(t)\, dt mean? + + integrationdisplacement + + displacement + +

    + +

    + The Fundamental Theorem of Calculus states that + + \int_a^b v(t)\, dt = V(b) - V(a) + , + where V(t) is any antiderivative of v(t). + Since v(t) is a velocity function, + V(t) must be a position function, + and V(b) - V(a) measures a change in position, or displacement. +

    + + + Finding displacement and distance + +

    + A ball is thrown straight up with velocity given by + v(t) = -32t+20ft/s, where t is measured in seconds. + Find, and interpret, + \int_0^1 v(t)\, dt and \int_0^1 \abs{v(t)}\, dt. +

    +     + +

    + Using the Fundamental Theorem of Calculus, we have + + \int_0^1 v(t)\, dt \amp = \int_0^1 (-32t+20)\, dt + \amp = \left(-16t^2 + 20t\right)\Big|_0^1 + \amp = 4 + . +

    + +

    + Thus if a ball is thrown straight up into the air with velocity v(t) = -32t+20, + the height of the ball, 1 second later, + will be 4 feet above the initial height. +

    + +

    + Note that the ball has traveled much farther. + It has gone up to its peak and is falling down, + but the difference between its height at t=0 and t=1 is 4ft. +

    + +

    + If we wish to find the total distance traveled, + we must evaluate \int_0^1 \abs{v(t)}\, dt + (noting that negative velocities will reduce the diplacement, + but we want distance, not displacement). + In this case, + we know that the velocity changes sign once when v(t)=0,so t=20/32=5/8 seconds. + The velocity is positive over [0,5/8] and negative over [5/8,1]. + Therefore + + \int_0^1 \abs{v(t)}\, dt \amp = \int_0^{5/8}v(t)\,dt + \int_{5/8}^1 -v(t)\,dt + \amp =\int_0^{5/8}(-32t+20)\, dt-\int_{5/8}^1 (-32t+20)\, dt + \amp =\left(-16t^2 + 20t\right)\Big|_0^{5/8}-\left(-16t^2 + 20t\right)\Big|_0^1 + \amp = \frac{25}4-\left(-\frac94\right)=9 + . + So the total distance traveled over [0,1] is \int_0^1 \abs{-32t+20}\, dt=9 \text{ feet }. +

    + +

    + As we can see in , + the positive area between v(t) and the t-axis, + A_1=25/4, while the negative area, A_2=-9/4. + When we add these two areas, we get the displacement of 4 ft. + But when we add the absolute value of both of these areas + (as in ), + we get the total distance of 9 ft. +

    + + +
    + The area between v(t) and the t-axis can be used to represent displacement + + + Graph showing area between v(t) and the t axis can be used to represent displacement. + + +

    + The y axis is drawn from -10 to 20 + and the t axis is drawn from 0 to 1. + The function is a straight line with a negative slope. + The function has an y intercept of 20 and + an t intercept of 5/8. +

    +

    + The line forms two shaded regions with the t axis. + Between 0 to 5/8 on the t axis the region + under the graph is shaded and labeled A1. + The second area is between 5/8 and 1 that lies + in the fourth quadrant and the area is labeled A2. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={0.625,1}, + extra x tick labels={$5/8$,$1$}, + xlabel=$t$, + ymin=-15,ymax=25, + xmin=-1,xmax=1.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1] {-32*x+20} \closedcycle; + \addplot [firstcurvestyle,domain=-.5:1.5] {-32*x+20}; + + \node at (axis cs:.3125,5) { $A_1$}; + \node at (axis cs:.8125,-3) { $A_2$}; + + \end{axis} + + \end{tikzpicture} + + + +
    + +
    + The area between \abs{v(t)} and the t-axis can be used to represent distance + + + Graph showing area between absolute value of v(t) and the t axis can be used to represent displacement. + + +

    + The y axis is drawn from -10 to 20 and the + t axis is drawn from 0 to 1. + The function is a straight line with a negative slope. + The function has an y intercept of 20 and an + t intercept of 5/8. +

    +

    + The line forms two shaded regions with the t axis. + Between 0 to 5/8 on the t axis the region + under the graph is shaded and labeled A1. + Between 5/8 and 1 lies the second shaded region + labeled A3. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={0.625,1}, + extra x tick labels={$5/8$,$1$}, + xlabel=$t$, + ymin=-15,ymax=25, + xmin=-1,xmax=1.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:5/8] {-32*x+20} \closedcycle; + \addplot [firstcurvestyle,areastyle,domain=5/8:1] {32*x-20} \closedcycle; + + \addplot [firstcurvestyle,domain=-.5:5/8] {-32*x+20}; + \addplot [firstcurvestyle,domain=5/8:1.5] {32*x-20}; + + \node at (axis cs:.3125,5) { $A_1$}; + \node at (axis cs:.8125,3) { $A_3$}; + + \end{axis} + + \end{tikzpicture} + + + +
    +     +     +     + +

    + Integrating a rate of change function gives total change. + Velocity is the rate of position change; + integrating velocity gives the total change of position, + , displacement. +

    + +

    + Integrating a speed function gives a similar, though different, result. + Speed is also the rate of position change, + but does not account for direction. + That is, the speed an object is the absolute value of its velocity. + This is what we saw in + when we evaluated \int_0^1 \abs{v(t)}\, dt. + So integrating a speed function gives total change of position, + without the possibility of negative position change. + Hence the integral of a speed function gives + distance traveled. +

    + +

    + As acceleration is the rate of velocity change, + integrating an acceleration function gives total change in velocity. + We do not have a simple term for this analogous to displacement. + If a(t) = 5miles/h^2 and t is measured in hours, then + + \int_0^3 a(t)\, dt = 15 + + means the velocity has increased by 15m/h from t=0 to t=3. +

    +     + + + The Fundamental Theorem of Calculus and the Chain Rule +

    + Part 1 of the Fundamental Theorem of Calculus (FTC) states that given + + F(x) = \int_a^x f(t)\, dt + , + we have \Fp(x) = f(x). Using other notation, + + \lzoo{x}{F(x)}=\lzoo{x}{\int_a^x f(t)\, dt} = f(x) + . + While we have just practiced evaluating definite integrals, + sometimes finding antiderivatives is impossible and we need to rely on other techniques to approximate the value of a definite integral. + Functions written as F(x) = \int_a^x f(t)\, dt are useful in such situations. +

    + +

    + It may be of further use to compose such a function with another. + As an example, we may compose F(x) with g(x) to get + + F\big(g(x)\big) = \int_a^{g(x)} f(t)\, dt + . +

    + +

    + What is the derivative of such a function? + Fundamental Theorem of Calculusand Chain Rule + The Chain Rule can be employed to state + + \frac{d}{dx}\Big(F\big(g(x)\big)\Big) = \Fp\big(g(x)\big)\gp(x) = f\big(g(x)\big)\gp(x) + . +

    + + + +

    + An example will help us understand this. +

    + + + The FTC, Part 1, and the Chain Rule + +

    + Find the derivative of \ds F(x) = \int_2^{x^2} \ln(t) \, dt. +

    +     + +

    + We can view F(x) as being the function + G(x) = \int_2^x \ln(t) \, dt composed with g(x) = x^2; + that is, F(x) = G\big(g(x)\big). + The Fundamental Theorem of Calculus states that G'(x) = \ln(x). + The Chain Rule gives us + + F'(x) \amp = G'\big(g(x)\big) \gp(x) + \amp = \ln(g(x)) \gp(x) + \amp = \ln(x^2) 2x + \amp =2x\ln(x^2) + +

    + +

    + Normally, the steps defining G(x) and g(x) are skipped. +

    +     +     + + + +

    + Let's practice this once more. +

    + + + + + The FTC, Part 1, and the Chain Rule + +

    + Find the derivative of \ds F(x) = \int_{\cos(x) }^5 t^3\, dt. +

    +     + +

    + Note that \ds F(x) = -\int_5^{\cos(x) } t^3\, dt. + Viewed this way, the derivative of F is straightforward: + + F'(x) \amp= -\cos^3(x)\left(-\sin(x)\right) + \amp= \cos^3(x)\sin(x) + . +

    +     + +     +     + + + Area Between Curves + + +

    + Consider continuous functions f(x) and g(x) defined on [a,b], + where f(x) \geq g(x) for all x in [a,b], + as demonstrated in . + What is the area of the shaded region bounded by the two curves over [a,b]? + integrationarea between curves +

    + +
    + Finding the area bounded by two functions on an interval by subtracting the area under g from the area under f + +
    + + + + Graph showing area bounded by two functions f(x) and g(x) between x=a and x=b. + + +

    + The x and the y axes are uncalibrated but there + are two positions a and b marked on the x axis. + There are two functions f(x) and g(x) graphed. + The graph shows the area that f(x) forms on g(x) + between positions a and b. + The two functions intersect at one point. +

    +

    + The function f(x) is above g(x), + it first curves up and reaches a maxima after which it dips + down for a minima and then continues to increase. + The function g(x) has a dip between a and + b then it increases. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={.5,3}, + extra x tick labels={$a$,$b$}, + ytick=\empty, + ymin=-.5,ymax=2, + xmin=-.5,xmax=3.5 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.25:3.25,samples=40] {.25*sin(deg(2*x))+1.25} node [shift={(5pt,7pt)},black] { $f(x)$}; + \addplot [name path=B,firstcurvestyle,domain=-.25:3.25] {.25*(x-2)^2+.2} node [shift={(5pt,7pt)},black] { $g(x)$}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=.5:3}]; + + \end{axis} + + \end{tikzpicture} + + + +
    + +
    + + + + Graph showing area bounded by two functions f(x) and g(x) between x=a and x=b. + + +

    + Graph same as previous with additional details. + The x and the y axes are uncalibrated but there + are two positions a and b marked on the x axis. + There are two functions f(x) and g(x) graphed. + The graph shows the area that f(x) forms on g(x) + between positions a and b. + The two functions intersect at one point. +

    +

    + The function f(x) is above g(x), it first curves + up and reaches a maxima after which it dips down for a minima + and then continues to increase.The function g(x) has a + dip between a and b then it increases. + The graph shows areas under both curves as different shaded regions. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={.5,3}, + extra x tick labels={$a$,$b$}, + ytick=\empty, + ymin=-.5,ymax=2, + xmin=-.5,xmax=3.5 + ] + + \addplot [firstcurvestyle,areastyle,area style,domain=.5:3] {.25*sin(deg(2*x))+1.25} \closedcycle; + \addplot [secondcurvestyle,areastyle,domain=.5:3] {.25*(x-2)^2+.2} \closedcycle; + + \addplot [firstcurvestyle,domain=-.25:3.25,samples=40] {.25*sin(deg(2*x))+1.25} node [shift={(5pt,7pt)},black] { $f(x)$}; + \addplot [secondcurvestyle,domain=-.25:3.25] {.25*(x-2)^2+.2}node [shift={(5pt,7pt)},black] { $g(x)$}; + + \end{axis} + + \end{tikzpicture} + + + +
    +     +
    + +

    + The area can be found by recognizing that this area is + the area under f - the area under g. + Using mathematical notation, the area is + + \int_a^b f(x)\, dx - \int_a^b g(x)\, dx + . +

    + +

    + Properties of the definite integral allow us to simplify this expression to + + \int_a^b\big(f(x) - g(x)\big)\, dx + . +

    + + + Area Between Curves + +

    + Let f(x) and g(x) be continuous functions defined on [a,b] where + f(x)\geq g(x) for all x in [a,b]. + The area of the region bounded by the curves y=f(x), + y=g(x) and the lines x=a and x=b is + + \int_a^b \big(f(x)-g(x)\big)\, dx + . +

    +     +     + + + Finding area between curves + +

    + Find the area of the region enclosed by y=x^2+x-5 and y=3x-2. +

    +     + +

    + It will help to sketch these two functions, + as done in . +

    + +
    + Sketching the region enclosed by y=x^2+x-5 and y=3x-2 in + + + Graph of region enclosed by y=x^2+x-5 and y=3*x-2. + + +

    + The y axis is drawn from -5 to 15 and + the x axis is drawn from -2 to 4. + The first function x^2 +x-5is a curve that has a + y intercept at -5 and the x intercept + near 1.8. The second function 3*x-2 is a + straight line. The two lines intersect at points (-1,-5) + and (3,7). The area bounded by the two functions is shaded. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-2,-1,1,2,3,4}, + ytick={5,10,15}, + ymin=-10,ymax=16, + xmin=-2.5,xmax=4.5 + ] + + \addplot+ [name path=F,domain={-2:4},samples=40] {x^2+x-5} node [pos=1, left] { $y=x^2+x-5$}; + \addplot+ [name path=G,domain={-2:4}] {3*x-2} node [pos=0.7, above left] { $y=3x-2$}; + \addplot [gray!50] fill between [of=F and G, soft clip={domain=-1:3}]; + + \end{axis} + + \end{tikzpicture} + + + +
    + +

    + The region whose area we seek is completely bounded by these two functions; + they seem to intersect at x=-1 and x=3. + To check, set x^2+x-5=3x-2 and solve for x: + + x^2+x-5 \amp = 3x-2 + (x^2+x-5) - (3x-2) \amp = 0 + x^2-2x-3 \amp = 0 + (x-3)(x+1) \amp = 0 + x\amp =-1,\,3 + . +

    + +

    + Following , + the area is + + \int_{-1}^3\big(3x-2 -(x^2+x-5)\big)\, dx \amp = \int_{-1}^3 (-x^2+2x+3)\, dx + \amp =\left.\left(-\frac13x^3+x^2+3x\right)\right|_{-1}^3 + \amp =-\frac13(27)+9+9-\left(\frac13+1-3\right) + \amp = 10\frac23 = 10.\overline{6} + +

    +     + +     + + + + + +     + + + The Mean Value Theorem and Average Value +
    + A graph of a function f to introduce the Mean Value Theorem + + + A graph of a function to introduce the Mean Value Theorem. + + +

    + The y axis is uncalibrated and the x axis is + drawn from 0 to 4. + The curve first decreases from 0 to 3 then + it increases until x=4. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick=\empty, + ymin=-.5,ymax=1.75, + xmin=-.5,xmax=4.5 + ] + + \addplot+ [domain=-.5:4.5,samples=40] {.3*cos(deg(x))+1}; + + \end{axis} + + \end{tikzpicture} + + + +
    + +

    + Consider the graph of a function f in + and the area defined by \int_1^4 f(x)\, dx. + Three rectangles are drawn in ; + in , + the height of the rectangle is greater than f on [1,4], + hence the area of this rectangle is is greater than \int_1^4 f(x)\, dx. +

    + +

    + In , + the height of the rectangle is smaller than f on [1,4], + hence the area of this rectangle is less than \int_1^4 f(x)\, dx. +

    + +

    + Finally, in + the height of the rectangle is such that the area of the rectangle is exactly + that of \int_1^4 f(x)\, dx. + Since rectangles that are too big, + as in (a), and rectangles that are + too little, as in (b), + give areas greater/lesser than \int_1^4 f(x)\, dx, + it makes sense that there is a rectangle, + whose top intersects f(x) somewhere on [1,4], + whose area is exactly that of the definite integral. +

    + +
    + Differently sized rectangles give upper and lower bounds on \int_1^4 f(x)\, dx; the last rectangle matches the area exactly + +
    + + + + Graph of function with rectangle to show upper bound on function. + + +

    + The y axis is uncalibrated and the x axis is drawn + from 0 to 4. The curve first decreases from 0 + to 3 then it increases until x=4. + The area under the curve between x=1 to x=4 is shaded. + A rectangular boundary is drawn on the x axis with height + above the curve. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick=\empty, + ymin=-.5,ymax=1.75, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:4] {.3*cos(deg(x))+1} \closedcycle; + \addplot [firstcurvestyle,domain=-.5:4.5,samples=40] {.3*cos(deg(x))+1}; + + \draw [thick,secondcolor] (axis cs: 1,0) rectangle (axis cs:4,1.35); + + \end{axis} + + \end{tikzpicture} + + + +
    + +
    + + + + Graph of function with rectangle to show lower bound on function. + + +

    + The y axis is uncalibrated and the x axis is drawn + from 0 to 4. The curve first decreases from 0 + to 3 then it increases until x=4. + The area under the curve between x=1 to x=4 is shaded. + A rectangular boundary is drawn on the x axis with height + a little below the curve. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick=\empty, + ymin=-.5,ymax=1.75, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:4] {.3*cos(deg(x))+1} \closedcycle; + \addplot [firstcurvestyle,domain=-.5:4.5,samples=40] {.3*cos(deg(x))+1}; + + \draw [thick,secondcolor] (axis cs: 1,0) rectangle (axis cs:4,.65); + + \end{axis} + + \end{tikzpicture} + + + +
    + +
    + + + + Graph of function with rectangle that matches area of function exactly. + + +

    + The y axis is uncalibrated and the x axis is drawn from + 0 to 4. The curve first decreases from 0 to 3 + then it increases until x=4. + The area under the curve between x=1 to x=4 is shaded. + A rectangular boundary is drawn on the x axis such that from + 1 to 2 the curve is above the rectangle and from 2 + to 4 the curve is below the rectangle. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick=\empty, + ymin=-.5,ymax=1.75, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:4] {.3*cos(deg(x))+1} \closedcycle; + \addplot [firstcurvestyle,domain=-.5:4.5,samples=50] {.3*cos(deg(x))+1}; + + \draw [thick,secondcolor] (axis cs: 1,0) rectangle (axis cs:4,.84); + + \end{axis} + + \end{tikzpicture} + + + +
    +     +
    + +

    + We state this idea formally in a theorem. +

    + + + The Mean Value Theorem of Integration + +

    + Let f be continuous on [a,b]. + Mean Value Theoremof integration + integrationMean Value Theorem + There exists a value c in [a,b] such that + + \int_a^bf(x)\, dx = f(c)(b-a) + . +

    +     +     + + + + +

    + This is an existential statement; + c exists, but we do not provide a method of finding it. + + is directly connected to the Mean Value Theorem of Differentiation, given + as ; + we leave it to the reader to see how. +

    + + + +

    + We demonstrate the principles involved in this version of the Mean Value Theorem in the following example. +

    + + + Using the Mean Value Theorem + +

    + Consider \int_0^\pi \sin(x) \, dx. + Find a value c guaranteed by the Mean Value Theorem. +

    +     + +

    + We first need to evaluate \int_0^\pi \sin(x) \, dx. + (This was previously done in .) + + \int_0^\pi\sin(x) \, dx = -\cos(x) \Big|_0^\pi = 2 + . +

    + +

    + Thus we seek a value c in [0,\pi] such that \pi\sin(c) =2. + + \pi\sin(c) = 2\,\,\Rightarrow\,\,\sin(c) = 2/\pi\,\,\Rightarrow\,\,c = \arcsin(2/\pi) \approx 0.69 + . +

    + +
    + A graph of y=\sin(x) on [0,\pi] and the rectangle guaranteed by the Mean Value Theorem + + + Graph of function y=sin(x) and the rectangle guaranteed by the Mean Value Theorem. + + +

    + The y axis has two positions marked sin(0.69) and + 1, the x axis has c, 1, 2 and + \pi marked in order. The function is an inverted parabola + of height 1 and has x intercepts at x=0 and + x=\pi. The area under the parabola until the x axis + is shaded. + A rectangular boundary is drawn with height sin(0.69) and + width \pi +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2}, + extra x ticks={.69,3.14}, + extra x tick labels={$c$,$\pi$}, + ytick={1}, + extra y ticks={.637}, + extra y tick labels={$\sin(0.69) $}, + ymin=-.25,ymax=1.25, + xmin=-.5,xmax=3.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:3.14] {sin(deg(x))}; + \addplot [firstcurvestyle,domain=-.5:3.5,samples=40] {sin(deg(x))}; + + \draw [thick,secondcolor] (axis cs: 0,0) rectangle (axis cs:3.14,.637); + + \end{axis} + + \end{tikzpicture} + + + +
    + +

    + In + \sin(x) is sketched along with a rectangle with height \sin(0.69). + The area of the rectangle is the same as the area under \sin(x) on [0,\pi]. +

    +     +     + +

    + We now turn our attention to a related topic average value. + Let f be a function on [a,b] with c such that f(c)(b-a) = \int_a^bf(x)\, dx. + Consider \int_a^b\big(f(x)-f(c)\big)\, dx: + + \int_a^b\big(f(x)-f(c)\big)\, dx \amp = \int_a^b f(x) - \int_a^b f(c)\, dx + \amp = f(c)(b-a) - f(c)(b-a) + \amp = 0 + . +

    + +

    + When f(x) is shifted by -f(c), + the amount of area under f above the x-axis on [a,b] is the same as the amount of area below the x-axis above f; + see for an illustration of this. + In this sense, + we can say that f(c) is the average value + of f on [a,b]. +

    + + +
    + On the left, a graph of y=f(x) and the rectangle guaranteed by the Mean Value Theorem. On the right, y=f(x) is shifted down by f(c); the resulting area under the curve is 0 + + + + A graph of y = f(x) and the rectangle guaranteed by the Mean Value Theorem. + + +

    + The x and the y axes are uncalibrated. + The y axis is labeled as f(x). + The x axis has three distinct positions marked a, + c and b in the order. +

    +

    + The function represents a parabola and the graph shows + the area under the parabola. a and b are the + two ends for the area. c is to the left of the vertex + of the parabola almost at half the distance from a. +

    +

    + The rectangle guaranteed by the Mean Value Theorem lies above + the vertex of the parabola. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={1,4,1.63}, + extra x tick labels={$a$,$b$,$c$}, + ytick=\empty, + extra y ticks={1.75}, + extra y tick labels={$f(c)$}, + ymin=-1,ymax=3.5, + xmin=-.5,xmax=4.25 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:4] {(x-2.5)^2+1} \closedcycle; + \addplot [firstcurvestyle,domain=-.5:4.25,samples=40] {(x-2.5)^2+1}; + + \draw [thick,secondcolor] (axis cs: 1,0) rectangle (axis cs:4,1.75); + \draw (axis cs:3.35,3) node { $y=f(x)$}; + \end{axis} + + \end{tikzpicture} + + + + + + + Graph of f(x) is shifted down by f(c), the resulting “area under the curve” is 0. + + +

    + The x and the y axes are uncalibrated. + The y axis is labeled as f(x). + The x axis has three distinct positions marked + a, c and b in the order. +

    +

    + The function represents a parabola and the graph shows + the area under the parabola. a and b are the + two ends for the area. c is to the left of the + vertex of the parabola almost at half the distance from + a. +

    +

    + Sine the function is shifted by f(c), there are three + parts to the area, two are positive and are above the + x axis, from a to c and from a little + before b to b. From a little before b + to b the area is drawn in the fourth quadrant from + above the parabola on the x axis. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={1,4,1.63}, + extra x tick labels={$a$,$b$,$c$}, + ytick=\empty, + extra y ticks={1.75}, + extra y tick labels={$f(c)$}, + ymin=-1,ymax=3.5, + xmin=-.5,xmax=4.25 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:4] {(x-2.5)^2-.75} \closedcycle; + \addplot [firstcurvestyle,domain=.75:4.25,samples=40] {(x-2.5)^2-.75}; + + \draw (axis cs:3.2,2.5) node { $y=f(x)-f(c)$}; + \end{axis} + + \end{tikzpicture} + + + +     +
    + +

    + The value f(c) is the average value in another sense. + First, recognize that the Mean Value Theorem can be rewritten as + + f(c) = \frac{1}{b-a}\int_a^b f(x)\, dx + , + for some value of c in [a,b]. + + + Replacing the integral with the limit of a Riemann sum (as in ): + + f(c) \amp = \frac{1}{b-a}\int_a^b f(x)\, dx \amp + \amp = \frac{1}{b-a} \lim\limits_{n \to \infty}\sum_{i=1}^n f(c_i)\,\Delta x \amp \text{Using } + \amp =\frac{1}{b-a} \lim\limits_{n \to \infty}\sum_{i=1}^n f(c_i)\,\frac{b-a}{n}\amp \Delta x =\frac{b-a}{n} + \amp =\lim\limits_{n \to \infty}\sum_{i=1}^n f(c_i)\frac{1}{n} \amp \text{Cancelling the common factor of } b-a + . +

    + +

    + Examining this last line closely, + the expression \sum_{i=1}^n f(c_i)\frac{1}{n} represents adding up n sample values of f(x)and then dividing by n. + This is exactly what we do when we calculate the average of a set of n numbers. + Now when we consider taking the limit as n goes to \infty, + \lim\limits_{n \to \infty}\sum_{i=1}^n f(c_i)\frac{1}{n}, + we are adding up all of the function's output values over [a,b] and dividing by the + number of numbers. + In a sense, we are adding up an infinite number of output values and then dividing by the number of terms we summed + (which is again infinite). +

    + +

    + This leads us to a definition. +

    + + + The Average Value of <m>f</m> on <m>[a,b]</m> + +

    + Let f be continuous on [a,b]. + The average value of f on [a,b] is f(c), + where c is a value in [a,b] guaranteed by the Mean Value Theorem. + integrationaverage value + average value of function + , + + \text{ Average Value of \(f\) on \([a,b]\) } = \frac{1}{b-a}\int_a^b f(x)\, dx + . +

    +     +     + + + +

    + An application of this definition is given in the following example. +

    + + + Finding the average value of a function + +

    + An object moves back and forth along a straight line with a velocity given by v(t) = (t-1)^2 on [0,3], + where t is measured in seconds and v(t) is measured in ft/s. +

    + +

    + What is the average velocity of the object? +

    +     + +

    + By our definition, the average velocity is: + + \frac{1}{3-0}\int_0^3 (t-1)^2\, dt \amp =\frac13 \int_0^3 \big(t^2-2t+1\big)\, dt + \amp = \left.\frac13\left(\frac13t^3-t^2+t\right)\right|_0^3 + \amp =\frac13\left[\left(\frac13(3)^3-(3)^2+(3)\right)-\left(\frac13(0)^3-(0)^2+(0)\right)\right] + \amp = 1\text{ ft/s } + . +

    +     +     + +

    + We can understand the above example through a simpler situation. + Suppose you drove 100 miles in 2 hours. + What was your average speed? + The answer is simple: displacement/time = 100 miles/2 hours = 50 mph. +

    + +

    + What was the displacement of the object in ? + We calculate this by integrating its velocity function: + \int_0^3 (t-1)^2\, dt = 3 ft. + Its final position was 3 feet from its initial position after 3 seconds: + its average velocity was 1 ft/s. +

    + +

    + This section has laid the groundwork for a lot of great mathematics to follow. + The most important lesson is this: + definite integrals can be evaluated using antiderivatives. + Since + established that definite integrals are the limit of Riemann sums, + we can later create Riemann sums to approximate values other than + area under the curve, + convert the sums to definite integrals, + then evaluate these using the . + This will allow us to compute the work done by a variable force, + the volume of certain solids, + the arc length of curves, and more. +

    + +

    + The downside is this: generally speaking, + computing antiderivatives is much more difficult than computing derivatives. + + is devoted to techniques of finding antiderivatives so that a wide variety of definite integrals can be evaluated. + Before that, + + explores techniques of approximating the value of definite integrals beyond using the Left Hand, Right Hand and Midpoint Rules. + These techniques are invaluable when antiderivatives cannot be computed, + or when the actual function f is unknown and all we know is the value of f at certain x-values. +

    +     + + + + Terms and Concepts + + + + +

    + How are definite and indefinite integrals related? +

    + +     + + + +     + + + + +

    + What constant of integration is most commonly used when evaluating definite integrals? +

    +

    + +

    +     +     +     + + + + +

    + + If f is a continuous function, + then \ds F(x) = \int_a^x f(t)\, dt is also a continuous function. +

    +     + +     + + + + +

    + The definite integral can be used to find + the area under a curve. + Give two other uses for definite integrals. +

    + +     + + + +     +     + + + Problems + + +

    + Evaluate the definite integral. +

    +     + + + + + $b = list_random(-1,1); + $c = non_zero_random(-9,9,1); + $low = random(1,2,1); + $high = $low + random(1,3,1); + if($envir{problemSeed}==1){$b=-1;$c=1;$low=1;$high=3}; + Context()->noreduce('(-x)-y','(-x)+y'); + $F = Formula("x^3 + $b x^2 + $c x")->reduce; + $f = $F->D('x')->reduce; + $r = $F->eval(x=>$high) - $F->eval(x=>$low); + + +

    + \ds\int_{}^{} \left(\right)\, dx +

    +

    + +

    +     +     +     + + + + + $c = non_zero_random(-9,9,1); + $low = 0; + $high = random(2,6,1); + if($envir{problemSeed}==1){$c=-1;$high=4;}; + Context("Fraction"); + Context()->noreduce('(-x)-y','(-x)+y'); + $F = Formula("1/3(x + $c)^3")->reduce; + $f = Formula("(x + $c)^2")->reduce; + $r = Fraction($F->eval(x=>$high) - $F->eval(x=>$low)); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random(-1,1); + $low = random(-4,-1,1); + ($m,$n) = random_subset(2,3,5,7); + if($envir{problemSeed}==1){$b=-1;$low=-1;}; + $high = -$low; + Context("Fraction"); + Context()->noreduce('(-x)-y','(-x)+y'); + $F = Formula("x^($m+1)/($m+1) + $b x^($n+1)/($n+1)")->reduce; + $f = Formula("x^($m) + $b x^($n)")->reduce; + $r = $F->eval(x=>$high) - $F->eval(x=>$low); + + +

    + \ds\int_{}^{} \left(\right)\, dx +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','cos'); + $lh = list_random(['0','pi/2'],['0','pi'],['pi/2','pi']); + if($envir{problemSeed}==1){$trig='cos';$lh=['pi/2','pi'];}; + $low = Formula($lh->[0]); + $high = Formula($lh->[1]); + $f = Formula("$trig(x)"); + $F = -$f->D('x')->reduce; + $r = $F->eval(x=>Real("$high")) - $F->eval(x=>Real("$low")); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $trig = list_random('tan','sec'); + $lh = list_random(['0','pi/4'],['0','pi/3'],['pi/4','pi/3']); + if($envir{problemSeed}==1){$trig='tan';$lh=['0','pi/4'];}; + $low = Formula($lh->[0]); + $high = Formula($lh->[1]); + $F = Formula("$trig(x)"); + $f = $F->D('x')->reduce; + $r = $F->eval(x=>Real("$high")) - $F->eval(x=>Real("$low")); + Context()->flags->set(reduceConstantFunctions=>0); + if ($r == 1) {$r = 1;} + elsif ($r == sqrt(3)) {$r = Formula("sqrt(3)");} + elsif ($r == sqrt(3) - 1) {$r = Formula("sqrt(3) - 1");} + elsif ($r == sqrt(2) - 1) {$r = Formula("sqrt(2) - 1");} + elsif ($r == 2 - sqrt(2)) {$r = Formula("2 - sqrt(2)");}; + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $n = random(1,9,1); + if($envir{problemSeed}==1){$n=1;}; + $low = 1; + $high = Formula("e^$n")->reduce; + $F = Formula("ln(x)")->reduce; + $f = Formula("1/x")->reduce; + $r = $n; + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $b = random(2,9,1); + $low = random(-3,-1,1); + $high = random(1,3,1); + if($envir{problemSeed}==1){$b=5;$low=-1;$high=1;}; + $f = Formula("$b^x")->reduce; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $r = Formula("($b**$high - 1/$b**(-$low))/ln($b)"); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + ($low,$high) = num_sort(random_subset(2,-4..-1)); + ($a,$b) = random_subset(2,-9..-1,1..9); + $m = random(2,5,1); + if($envir{problemSeed}==1){$low=-2;$high=-1;$a=4,$b=-2;$m=3;}; + $f = Formula("$a + $b x^$m")->reduce; + Context("Fraction"); + $r = Fraction(($a*$high + $b/($m+1)*$high**($m+1)) - ($a*$low + $b/($m+1)*$low**($m+1))); + + +

    + \ds\int_{}^{} \left(\right)\, dx +

    +

    + +

    +     +     +     + + + + + ($a,$b) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$a=2;$b=-2;}; + $f = Formula("$a cos(x) + $b sin(x)")->reduce; + $r = 2*$b; + + +

    + \ds\int_{0}^{\pi} \left(\right)\, dx +

    +

    + +

    +     +     +     + + + + + ($low,$high) = num_sort(random_subset(2,1..4)); + if($envir{problemSeed}==1){$low=1;$high=4;}; + $f = Formula("e^x")->reduce; + $r = Formula("e^$high - e^$low"); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + ($low,$high) = num_sort(random_subset(2,0,1,4,9,16,25)); + if($envir{problemSeed}==1){$low=0;$high=4;}; + Context()->variables->are(t=>'Real'); + $f = Formula("sqrt(t)")->reduce; + Context("Fraction"); + $r = Fraction(2/3 * ($high**(3/2) - $low**(3/2))); + + +

    + \ds\int_{}^{} \, dt +

    +

    + +

    +     +     +     + + + + + ($low,$high) = num_sort(random_subset(2,1,4,9,16,25)); + if($envir{problemSeed}==1){$low=9;$high=25;}; + Context()->variables->are(t=>'Real'); + $f = Formula("1/sqrt(t)")->reduce; + Context("Fraction"); + $r = Fraction(2 * ($high**(1/2) - $low**(1/2))); + + +

    + \ds\int_{}^{} \, dt +

    +

    + +

    +     +     +     + + + + + $n = random(3,5,1); + ($low,$high) = num_sort(random_subset(2,0**$n,1**$n,2**$n,3**$n,4**$n)); + if($envir{problemSeed}==1){$n=3;$low=1;$high=8;}; + parser::Root->Enable; + $f = Formula("root($n,x)")->reduce; + Context("Fraction"); + $r = Fraction($high**(1/$n + 1)/(1/$n + 1) - $low**(1/$n + 1)/(1/$n + 1)); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $low = 1; + $high = random(2,9,1); + if($envir{problemSeed}==1){$high=2;}; + $f = Formula("1/x")->reduce; + Context()->flags->set(reduceConstantFunctions=>0); + $r = Formula("ln($high)"); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $low = 1; + $high = random(2,9,1); + if($envir{problemSeed}==1){$high=2;}; + $f = Formula("1/x^2")->reduce; + Context("Fraction"); + $r = Fraction(1/$low - 1/$high); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $n = random(3,6,1); + $low = 1; + $high = random(2,9,1); + if($envir{problemSeed}==1){$n=3;$high=2;}; + $f = Formula("1/x^$n")->reduce; + Context("Fraction"); + $r = Fraction($high**(-$n+1)/(-$n+1) - $low**(-$n+1)/(-$n+1)); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = 1; + $f = Formula("x")->reduce; + Context("Fraction"); + $r = Fraction(1/2); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = 1; + $f = Formula("x^2")->reduce; + Context("Fraction"); + $r = Fraction(1/3); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = 1; + $f = Formula("x^3")->reduce; + Context("Fraction"); + $r = Fraction(1/4); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $n = random(80,120,1); + if($envir{problemSeed}==1){$n=100;}; + $low = 0; + $high = 1; + $f = Formula("x^$n")->reduce; + Context("Fraction"); + $r = Fraction(1/($n+1)); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $low = random(-9,-2,1); + if($envir{problemSeed}==1){$low=-4;}; + $high = -$low; + $r = Real(2*$high); + + +

    + \ds\int_{}^{} dx +

    +

    + +

    +     +     +     + + + + + ($low,$high) = num_sort(random_subset(2,-9..-1)); + $c = random(2,9,1); + if($envir{problemSeed}==1){$low=-10;$high=-5;$c=3;}; + $f = Formula("$c")->reduce; + $r = Real($c*($high - $low)); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $low = random(-9,-2,1); + if($envir{problemSeed}==1){$low=-2;}; + $high = -$low; + $f = Formula("0")->reduce; + $r = Real(0); + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $trig = list_random('-cot','-csc'); + $lh = list_random(['pi/6','pi/4'],['pi/6','pi/3'],['pi/6','pi/2'],['pi/4','pi/3'],['pi/4','pi/2'],['pi/3','pi/2']); + if($envir{problemSeed}==1){$trig='-csc';$lh=['pi/6','pi/3'];}; + $low = Formula($lh->[0]); + $high = Formula($lh->[1]); + $F = Formula("$trig(x)"); + $f = $F->D('x')->reduce; + $r = $F->eval(x=>Real("$high")) - $F->eval(x=>Real("$low")); + Context()->flags->set(reduceConstantFunctions=>0); + if ($r == 1) {$r = 1;} + elsif ($r == sqrt(3) - 1) {$r = Formula("sqrt(3) - 1");} + elsif ($r == sqrt(3) - 1/sqrt(3)) {$r = Formula("sqrt(3) - 1/sqrt(3)");} + elsif ($r == sqrt(3)) {$r = Formula("sqrt(3)");} + elsif ($r == 1 - 1/sqrt(3)) {$r = Formula("1 - 1/sqrt(3)");} + elsif ($r == 1/sqrt(3)) {$r = Formula("1/sqrt(3)");} + elsif ($r == 2 - sqrt(2)) {$r = Formula("2 - sqrt(2)");} + elsif ($r == 2 - 2/sqrt(3)) {$r = Formula("2 - 2/sqrt(3)");} + elsif ($r == sqrt(2) - 2/sqrt(3)) {$r = Formula("sqrt(2) - 2/sqrt(3)");} + elsif ($r == sqrt(2) - 1) {$r = Formula("sqrt(2) - 1");} + elsif ($r == 2/sqrt(3) - 1) {$r = Formula("2/sqrt(3) - 1");} + + +

    + \ds\int_{}^{} \, dx +

    +

    + +

    +     +     +     +     + + + + + + +

    + Explain why \ds\int_{-1}^1 x^n\, dx=0 + when n is a positive, odd integer. +

    + + + +     + + +     + + + +

    + Explain why \ds\int_{-1}^1 x^n\, dx = 2\int_{0}^1 x^n\, dx when n is a positive, + even integer. +

    + + +     + + +     + +     + + + + +

    + Explain why \ds\int_{a}^{a+2\pi} \sin(t)\, dt = 0 for all values of a. +

    + +     + + + +     + + + +

    + Find all values c such that \int_a^b f(x)\, dx = f(c)(b-a), + as guaranteed by the . +

    +     + + + + + $low = 0; + $high = random(2,4,1); + if($envir{problemSeed}==1){$high=2;}; + $f = Formula("x^2"); + $D = $high - $low; + $F = Formula("x^3/3"); + $f_inv = Formula("sqrt(x)"); + $f_c = ($F->eval(x=>$high) - $F->eval(x=>$low))/$D; + $c = List($f_inv->eval(x=>$f_c)); + + +

    + \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $low = random(-9,-1,1); + if($envir{problemSeed}==1){$low=-2;}; + $high = -$low; + $f = Formula("x^2"); + $D = $high - $low; + $F = Formula("x^3/3"); + $f_inv = Formula("sqrt(x)"); + $f_c = ($F->eval(x=>$high) - $F->eval(x=>$low))/$D; + $c = $f_inv->eval(x=>$f_c); + $c = List(-$c,$c); + + +

    + \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = 1; + $f = Formula("e^x"); + $D = $high - $low; + $F = Formula("e^x"); + $f_inv = Formula("ln(x)"); + $f_c = ($F->eval(x=>$high) - $F->eval(x=>$low))/$D; + $c = $f_inv->eval(x=>$f_c); + $c = List($c); + + +

    + \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = list_random(1,4,9,16,25,36,49); + if($envir{problemSeed}==1){$high=16;}; + $f = Formula("sqrt(x)"); + $D = $high - $low; + $F = Formula("2/3 x^(3/2)"); + $f_inv = Formula("x^2"); + $f_c = ($F->eval(x=>$high) - $F->eval(x=>$low))/$D; + $c = $f_inv->eval(x=>$f_c); + $c = List($c); + + +

    + \int_{}^{} \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Find the average value of the function on the given interval. +

    +     + + + + + $trig = list_random('sin','cos'); + $lh = list_random(['0','pi/2'],['0','pi'],['pi/2','pi']); + if($envir{problemSeed}==1){$trig='sin';$lh=['0','pi/2'];}; + $low = Formula($lh->[0]); + $high = Formula($lh->[1]); + $D = $high - $low; + $f = Formula("$trig(x)"); + $F = -$f->D('x')->reduce; + $A = ($F->eval(x=>Real("$high")) - $F->eval(x=>Real("$low")))/$D; + $num = Real($A*pi); + $A = Formula("$num/pi"); + + +

    + f(x)= on \left[,\right] +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','cos'); + $lh = list_random(['0','pi/2'],['0','pi'],['pi/2','pi']); + if($envir{problemSeed}==1){$trig='sin';$lh=['0','pi'];}; + $low = Formula($lh->[0]); + $high = Formula($lh->[1]); + $D = $high - $low; + $f = Formula("$trig(x)"); + $F = -$f->D('x')->reduce; + $A = ($F->eval(x=>Real("$high")) - $F->eval(x=>Real("$low")))/$D; + $num = Real($A*pi); + $A = Formula("$num/pi"); + + +

    + y= on [,] +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = random(1,9,1); + if($envir{problemSeed}==1){$high=4;}; + $D = $high - $low; + $f = Formula("x"); + $F = Formula("x^2/2"); + Context("Fraction"); + $A = Fraction(($F->eval(x=>Real($high)) - $F->eval(x=>Real($low)))/$D); + + +

    + y= on [,] +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = random(1,9,1); + if($envir{problemSeed}==1){$high=4;}; + $D = $high - $low; + $f = Formula("x^2"); + $F = Formula("x^3/3"); + Context("Fraction"); + $A = Fraction(($F->eval(x=>Real($high)) - $F->eval(x=>Real($low)))/$D); + + +

    + y= on [,] +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = random(1,9,1); + if($envir{problemSeed}==1){$high=4;}; + $D = $high - $low; + $f = Formula("x^3"); + $F = Formula("x^4/4"); + Context("Fraction"); + $A = Fraction(($F->eval(x=>Real($high)) - $F->eval(x=>Real($low)))/$D); + + +

    + y= on [,] +

    +

    + +

    +     +     +     + + + + + $low = 1; + $m = random(1,9,1); + if($envir{problemSeed}==1){$m=1;}; + $high = Formula("e^$m")->reduce; + $D = $high - $low; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + $f = Formula("1/t"); + $F = Formula("ln(t)"); + $A = Formula("$m/(e^$m - 1)"); + + +

    + y=\ds on \left[,\right] +

    +

    + +

    +     +     +     +     + + + +

    + A velocity function is given for an object moving along a straight line. + Find the displacement of the object over the given time interval. +

    +     + + + + + $low = 0; + $high = random(4,8,1); + $v0 = random(10,30,2); + if($envir{problemSeed}==1){$high=5;$v0=20}; + $a = -32; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + $f = FormulaWithUnits("$a t + $v0 ft/s"); + $F = Formula("$a/2 t^2 + $v0 t"); + $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft'); + + +

    + v(t)= on [,] +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = random(7,15,1); + $v0 = random(100,300,20); + if($envir{problemSeed}==1){$high=10;$v0=200}; + $a = -32; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + $f = FormulaWithUnits("$a t + $v0 ft/s"); + $F = Formula("$a/2 t^2 + $v0 t"); + $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft'); + + +

    + v(t)= on [,] +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = random(2,9,1); + $v0 = random(5,20,1); + if($envir{problemSeed}==1){$high=3;$v0=10}; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + $f = FormulaWithUnits("$v0 ft/s"); + $F = Formula("$v0 t"); + $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft'); + + +

    + v(t)= on [,] +

    +

    + +

    +     +     +     + + + + + $low = random(-3,-1,1); + $high = random(1,3,1); + $b = random(2,9,1); + if($envir{problemSeed}==1){$low=-1;$high=1;$b=2;}; + Context()->variables->are(t=>'Real'); + $f = FormulaWithUnits("$b^t mph"); + $F = Formula("$b^t/ln($b)"); + $D = FormulaWithUnits($F->eval(t=>Real($high)) - $F->eval(t=>Real($low)), 'mi'); + + +

    + v(t)= on [,] +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','cos'); + $lh = list_random(['0','pi/2'],['0','pi'],['0','3*pi/2'],['pi/2','pi'],['pi/2','3*pi/2'],['pi','3*pi/2']); + if($envir{problemSeed}==1){$trig='cos';$lh=['0','3*pi/2'];}; + $low = Formula($lh->[0]); + $high = Formula($lh->[1]); + Context()->variables->are(t=>'Real'); + $f = FormulaWithUnits("$trig(t) ft/s"); + $F = Formula("-$trig(t)")->D('t')->reduce; + $D = FormulaWithUnits($F->eval(t=>Real("$high")) - $F->eval(t=>Real("$low")), 'ft'); + + +

    + v(t)= on \left[,\right] +

    +

    + +

    +     +     +     + + + + + $low = 0; + $n = random(3,5,1); + $high = list_random(2**$n,3**$n,4**$n); + if($envir{problemSeed}==1){$high=1;$n=4;$high=16;}; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + parser::Root->Enable; + $f = FormulaWithUnits("root($n,t) ft/s"); + $F = Formula("t^(1/$n+1)/(1/$n+1)"); + $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft'); + + +

    + v(t)= on [,] +

    +

    + +

    +     +     +     +     + + + +

    + An acceleration function of an object moving along a straight line is given. + Find the change of the object's velocity over the given time interval. +

    +     + + + + + $low = 0; + $high = random(1,9,1); + if($envir{problemSeed}==1){$high=2;}; + $a = -32; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + $f = FormulaWithUnits("$a ft/s^2"); + $F = Formula("$a t"); + $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft/s'); + + +

    + a(t)= on [,] +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = random(1,9,1); + $a = random(5,20,1); + if($envir{problemSeed}==1){$high=5;$a=10;}; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + $f = FormulaWithUnits("$a ft/s^2"); + $F = Formula("$a t"); + $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft/s'); + + +

    + a(t)= on [,] +

    +

    + +

    +     +     +     + + + + + $low = 0; + $high = random(1,9,1); + if($envir{problemSeed}==1){$high=2;}; + Context("Fraction"); + Context()->variables->are(t=>'Real'); + $f = FormulaWithUnits("t ft/s^2"); + $F = Formula("t^2/2"); + $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft/s'); + + +

    + a(t)= on [,] +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','cos'); + $lh = list_random(['0','pi/2'],['0','pi'],['0','3*pi/2'],['pi/2','pi'],['pi/2','3*pi/2'],['pi','3*pi/2']); + if($envir{problemSeed}==1){$trig='cos';$lh=['0','pi/2'];}; + $low = Formula($lh->[0]); + $high = Formula($lh->[1]); + Context()->variables->are(t=>'Real'); + $f = FormulaWithUnits("$trig(t) ft/s^2"); + $F = Formula("-$trig(t)")->D('t')->reduce; + $D = FormulaWithUnits($F->eval(t=>Real("$high")) - $F->eval(t=>Real("$low")), 'ft/s'); + + +

    + a(t)= on \left[,\right] +

    +

    + +

    +     +     +     +     + + + +

    + Sketch the given relations and find the area of the enclosed region. +

    +     + + + +

    + y=2x, y=5x, and x= 3 +

    +     +     + + + +

    + y=-x+1, y=3x+6, + x=2 and x= -1 +

    +     +     + + + +

    + y=x^2-2x+5, y=5x-5 +

    +     +     + + + +

    + y=2x^2+2x-5, y=x^2+3x+7, +

    +     +     +     + + + +

    + Find F'(x). +

    +     + + + + + $m = random(3,5,1); + $b = non_zero_random(-9,9,1); + $low = random(1,9,1); + if($envir{problemSeed}==1){$m=3;$b=1;$low=2;}; + Context()->variables->add(t=>'Real'); + $high = Formula("x^$m + $b x")->reduce; + $f = Formula("1/t"); + $D = $high->D('x')->reduce / $high; + + +

    + F(x) = \ds\int_{}^{} \, dt +

    +

    + +

    +     +     +     + + + + + $m = random(2,5,1); + $high = random(-9,9,1); + if($envir{problemSeed}==1){$m=3;$high=0;}; + Context()->variables->add(t=>'Real'); + $low = Formula("x^$m"); + $f = Formula("t^$m"); + $D = Formula("-$m x^($m**2+$m-1)")->reduce; + + +

    + F(x) = \ds\int_{}^{} \, dt +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + $m = random(2,4,1); + if($envir{problemSeed}==1){$b=1;$m=2}; + Context()->variables->add(t=>'Real'); + $low = Formula("x"); + $high = Formula("x^$m"); + $f = Formula("t+$b")->reduce; + $D = Formula("$m x^($m-1)(x^$m+$b) - (x+$b)")->reduce; + + +

    + F(x) = \ds\int_{}^{} ()\, dt +

    +

    + +

    +     +     +     + + + + + ($low,$high,$f) = random_subset(3,'e^','ln','sin','cos','sqrt'); + if($envir{problemSeed}==1){$low='ln';$high='e^';$f='sin';}; + Context()->variables->add(t=>'Real'); + # interval where all these functions have positive output + Context()->variables->set(x=>{limits=>[1,pi/2]}); + $low = Formula("$low(x)"); + $high = Formula("$high(x)"); + $f = Formula("$f(t)"); + $D = $high->D('x') * $f->substitute(t=>$high) - $low->D('x') * $f->substitute(t=>$low); + $D = $D->reduce; + + +

    + F(x) = \ds\int_{}^{} \, dt +

    +

    + +

    +     +     +     + + + + $b = non_zero_random(-9,9,1); + $m = random(2,4,1); + Context()->variables->add(t=>'Real'); + $low = random(1,9,1); + $high = Formula("x^$m"); + $f = Formula("sin($b t^2)")->reduce; + $D = Formula("$m x^($m-1)sin($b x^(2*$m))")->reduce; + + +

    + F(x) = \ds\int_{}^{} ()\, dt +

    +

    + +

    +     +     +     + + + + $b = non_zero_random(-9,9,1); + ($low,$high) = random_subset(2,'e^','ln','sin','cos','sqrt'); + Context()->variables->add(t=>'Real'); + $low = Formula("$low(x)"); + $high = Formula("$high(x)"); + $f = Formula("sqrt(t^4+$b t^2)")->reduce; + $D = $high->D('x') * $f->substitute(t=>$high) - $low->D('x') * $f->substitute(t=>$low); + $D = $D->reduce; + + +

    + F(x) = \ds\int_{}^{} ()\, dt +

    +

    + +

    +     +     +     +     +     +     +
    +
    + Numerical Integration + +

    + The Fundamental Theorem of Calculus gives a concrete technique for finding the exact value of a definite integral. + That technique is based on computing antiderivatives. + Despite the power of this theorem, + there are still situations where we must approximate + the value of the definite integral instead of finding its exact value. + The first situation we explore is where we cannot + compute the antiderivative of the integrand. + The second case is when we actually do not know the function in the integrand, + but only its value when evaluated at certain points. + integrationnumerical + numerical integration +

    + +

    + An elementary function + elementary function + is any function that is a combination of polynomial, + nth root, rational, + exponential, logarithmic and trigonometric functions. + We can compute the derivative of any elementary function, + but there are many elementary functions of which we cannot compute an antiderivative. + For example, + the following functions do not have antiderivatives that we can express with elementary functions: +

      +
    • e^{x^2},
      • +
    • \sin(x^3),
      • +
    • \frac{\sin(x)}{x}.
      • +
    +

    + +

    + The simplest way to refer to the antiderivatives of + e^{-x^2} is to simply write \int e^{-x^2}\, dx. +

    + +

    + This section outlines three common methods of approximating the value of definite integrals. + We describe each as a systematic method of approximating area under a curve. + By approximating this area accurately, + we find an accurate approximation of the corresponding definite integral. +

    + +

    + We will apply the methods we learn in this section to the following definite integrals: +

      +
    • \int_0^1 e^{-x^2} \, dx,
      • +
    • \int_{-\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(x^3) \, dx,
      • +
    • \int_{0.5}^{4\pi} \frac{\sin(x)}{x} \, dx,
      • +
    + as pictured in . +

    + +
    + Graphically representing three definite integrals that cannot be evaluated using antiderivatives + + +
    + + + + Integral of e^{-x^2} from 0 to 1 + +

    + This is a graph of f(x)=e^{-x^2}.starting at (0, 1) the line slopes down towards the x axis. The line goes from (0, 1) to close + to (1, 0.75). We consider the area between (0, 1), (1, 0.75),(1, 0) and (0, 0). + +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.2,ymax=1.2, + xmin=-.2,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1] {e^-x^2} \closedcycle; + \addplot [firstcurvestyle,domain=0:1] {e^-x^2}; + + \draw (axis cs:.75,1) node { $y=e^{-x^2}$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + + +
    + + + + Graph of the function sin(x^3) + +

    + The graph of \sin(x^3). Starting at (-0.75, -0.5) the graph moves upwards to the right. It merges to the x axis close to (-0.25, 0) and + continues along the x axis. It goes through and continues along the positive x axis for a short while then rises up until it reaches (1.25, 1) + then sharply descends back towards the x axis. It intersects the y axis close to (1.5, 0) and continues downwards. We consider the area above the part of the + curve that is in the fourth quadrant and below the negative x axis. Then below the curve that rises up from the positive x axis and comes down. And the area above the + part of the curve that continues downwards after intersecting the x axis. + +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.6,ymax=1.2, + xmin=-1,xmax=1.7 + ] + + \addplot [firstcurvestyle,areastyle,domain=-.7854:1.5708,samples=40] {sin(deg(x^3))} \closedcycle; + \addplot [firstcurvestyle,domain=-.7854:1.5708,samples=80] {sin(deg(x^3))}; + + \draw (axis cs:.6,.9) node { $y=\sin(x^3)$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + + +
    + + + + The graph of y=sin(x)/x and its integral from .5 to 4 pi + +

    + We consider the area under the graph of \sin(x)/x starting close to (0.5, 0.9) to (12.5, 0). The structure of the graph is similar to that + of a wave. The distance between the highest point of the graph in the positive y axis and the lowest point in the negative y axis shows a + decreasing trend as it continues in the positive x direction but the wavelength increases in the positive x direction.The graph starts at (0.5, 0.9) + close to the x axis and curves downwards. It crosses the x axis close to (3, 0) while travelling downwards. It continues downwards for + a short amount and the curves upwards and this time crosses the x axis at (6.25, 0). From (6.25, 0) it travels another wavelength crossing + the x axis once close to (10,0) while traveling upwards and at (12.5, 0) where the graph ends. We consider the area under the graph when + it is above the x axis and above it when it is below the x axis. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={5,10,15}, + ymin=-.3,ymax=1.1, + xmin=-.9,xmax=15.9 + ] + + \addplot [firstcurvestyle,areastyle,domain=0.5:12.57] {sin(deg(x))/x} \closedcycle; + \addplot [firstcurvestyle,domain=0.5:12.57,samples=60] {sin(deg(x))/x}; + + \draw (axis cs:10,.7) node { $\displaystyle y=\frac{\sin(x) }{x}$}; + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    +     + + + The Left and Right Hand Rule Methods +

    + In + we addressed the problem of evaluating definite integrals by approximating the area under the curve using rectangles. + We revisit those ideas here before introducing other methods of approximating definite integrals. + numerical integrationLeft/Right Hand Rule + Right Hand Rule + Left Hand Rule + integrationnumerical!Left/Right Hand Rule +

    + +

    + We start with a review of notation. + Let f be a continuous function on the interval [a,b]. + We wish to approximate \ds \int_a^b f(x)\, dx. + We partition [a,b] into n equally spaced subintervals, + each of length \ds\dx = \frac{b-a}{n}. + The endpoints of these subintervals are labeled as + + x_0=a,\,x_1 = a+\dx,\,x_2 = a+ 2\dx,\,\ldots,\,x_i = a+i\dx,\,\ldots,\,x_{n} = b + . +

    + +

    + + states that to use the Left Hand Rule we use the summation + \ds \sum_{i=1}^n f(x_{i-1})\dx and to use the Right Hand Rule we use \ds \sum_{i=1}^n f(x_{i})\dx. + We review the use of these rules in the context of examples. +

    + + + Approximating definite integrals with rectangles + +

    + Approximate \ds \int_0^1e^{-x^2}\, dx using the Left and Right Hand Rules with 5 equally spaced subintervals. +

    +     + +

    + We begin by partitioning the interval [0,1] into 5 equally spaced intervals. + We have \dx = \frac{1-0}5 = 1/5=0.2, so + + x_0 = 0,\,x_1 = 0.2,\,x_2 = 0.4,\,x_3 = 0.6,\,x_4 = 0.8,\,\text{ and } \,x_5 = 1 + . +

    + +

    + Using the Left Hand Rule, we have: + + \sum_{i=1}^n f(x_{i-1})\dx \amp = \big(f(x_0)+f(x_1) + f(x_2) + f(x_3) + f(x_4)\big)\dx + \amp = \big(f(0) + f(0.2) + f(0.4) + f(0.6) + f(0.8)\big)\dx + \amp \approx (1+0.9608 +0.8521 + 0.6977 + 0.5273)(0.2) + \amp \approx 0.8076 + . +

    + +

    + Using the Right Hand Rule, we have: + + \sum_{i=1}^n f(x_{i})\dx \amp = \big(f(x_1) + f(x_2) + f(x_3) + f(x_4)+f(x_5)\big)\dx + \amp = \big(f(0.2) + f(0.4) + f(0.6) + f(0.8)+f(1)\big)\dx + \amp \approx (0.9608 +0.8521 + 0.6977 + 0.5273 + 0.3678)(0.2) + \amp \approx 0.6812 + . +

    + +
    + Approximating \int_0^1e^{-x^2}\, dx in + + + +
    + Using the Left Hand Rule + + Using the Left Hand rule + +

    The graph of e^{-x^2} starts at (0, 1) and slowly slopes downwards to the right as it slowly gets closer to the positive x axis. + To use the left hand rule we divide the area between [0, 1] five rectangles of equal widhth and decreasing height. As the curve moves closer to the x axis, + the heights of the rectangles gets shorter even thought the widths stay the same. Because the curve is downwards sloping to the right, the top right corners + of the rectangles contain a little region which is above the curve. As the curve moves downwards and the height of the rectangles become shorter, the area + that is contained by the top right corners of the rectangles and the top of the curve gets larger. We calculate the sum of the areas of these rectangles. +

    + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={.2,.4,.6,.8,1}, + ymin=-.2,ymax=1.2, + xmin=-.2,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1] {e^-x^2} \closedcycle; + \addplot [firstcurvestyle,domain=0:1] {e^-x^2}; + + \foreach \x / \y in {0 / 1, 0.2 / 0.96, 0.4 / .85, 0.6 / .7, .8 / .53} + {\addplot [thick,secondcolor] coordinates {(\x+.2,0) (\x+.2,\y) (\x,\y) (\x,0) (\x+.2,0)}; + } + + \draw (axis cs:.75,1) node { $y=e^{-x^2}$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + + +
    + Using the Right Hand Rule + + Using the Right Hand rule + +

    For this image we consider the same downwards sloping curve of e^{-x^2}. We divide the area between [0,1] into five + rectangles of equal width and decreasing height. For the right hand rule the rectangles are drawn in such a way that the top right corners of the rectangles + touch the bottom of the curve. There is a small area between the top left of the triangles and the bottom of the curve. This area gets larger as the + curve moves downwards and the lengths of the rectangles decrease. For the left hand rule we add up the areas of these rectangles. +

    + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={.2,.4,.6,.8,1}, + ymin=-.2,ymax=1.2, + xmin=-.2,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1] {e^-x^2} \closedcycle; + \addplot [firstcurvestyle,domain=0:1] {e^-x^2}; + + \foreach \x / \y in {0.2 / 0.96, 0.4 / .85, 0.6 / .7, .8 / .53, 1/ .37} + {\addplot [thick,secondcolor] coordinates {(\x-.2,0) (\x-.2,\y) (\x,\y) (\x,0) (\x-.2,0)}; + } + + \draw (axis cs:.75,1) node { $y=e^{-x^2}$}; + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    + +

    + + shows the rectangles used in each method to approximate the definite integral. + These graphs show that in this particular case, + the Left Hand Rule is an over approximation and the Right Hand Rule is an under approximation. + To get a better approximation, + we could use more rectangles, + as we did in . + We could also average the Left and Right Hand Rule results together, giving + + \frac{0.8076 + 0.6812}{2} = 0.7444 + . +

    + +

    + The actual answer, accurate to 4 places after the decimal, + is 0.7468, showing our average is a good approximation. +

    +     +     + + + Approximating definite integrals with rectangles + +

    + Approximate \ds\int_{-\frac{\pi}4}^{\frac{\pi}2} \sin(x^3)\, dx using the Left and Right Hand Rules with 10 equally spaced subintervals. +

    +     + +

    + We begin by finding \dx: + + \frac{b-a}{n} = \frac{\pi/2 - (-\pi/4)}{10} = \frac{3\pi}{40}\approx 0.2356 + . +

    + +

    + It is useful to write out the endpoints of the subintervals in a table; + in , + we give the exact values of the endpoints, + their decimal approximations, + and decimal approximations of + \sin(x^3) evaluated at these points. +

    + +
    + Values used to approximate \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in + + + x_i + Exact + Approx. + \sin(x_i^3) + + + + + + + + + x_0 + -\pi/4 + -0.7854 + -0.4657 + + + x_1 + -7 \pi/40 + -0.5498 + -0.1654 + + + x_2 + -{\pi }/{10} + -0.3142 + -0.0310 + + + x_3 + -{\pi }/{40} + -0.0785 + -0.0005 + + + x_4 + {\pi }/{20} + 0.1571 + 0.0039 + + + x_5 + {\pi }/{8} + 0.3927 + 0.0605 + + + x_6 + {\pi }/{5} + 0.6283 + 0.2455 + + + x_7 + {11 \pi }/{40} + 0.8639 + 0.6011 + + + x_8 + {7 \pi }/{20} + 1.0996 + 0.9710 + + + x_{9} + {17 \pi }/{40} + 1.3352 + 0.6899 + + + x_{10} + {\pi }/{2} + 1.5708 + -0.6700 + + +
    + +

    + Once this table is created, + it is straightforward to approximate the definite integral using the Left and Right Hand Rules. + (Note: the table itself is easy to create, + especially with a standard spreadsheet program on a computer. + The last two columns are all that are needed.) + The Left Hand Rule sums the first 10 values of + \sin(x_i^3) and multiplies the sum by \dx; + the Right Hand Rule sums the last 10 values of + \sin(x_i^3) and multiplies by \dx. + Therefore we have: +

    + +

    + Left Hand Rule: + \ds \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx \approx (1.9093)(0.2356) \approx 0.4498. +

    + +

    + Right Hand Rule: + \ds \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx \approx (1.705)(0.2356) \approx 0.4017. +

    + +

    + Average of the Left and Right Hand Rules: 0.4258. +

    + +
    + Approximating \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in + + + +
    + + + + Using the Left Hand rule + +

    We calculate \int_{-\frac{\pi}4}^{\frac{\pi}/2}\sin(x^3)\, dx using the left hand rule. The graph of \sin(x^3) starts at (-0.75, -0.5) + in the fourth quadrant and curves upwards to the right. It merges with the negative x axis close to (-0.25, 0) and continues along the x axis + until (0.25, 0) and then slopes upwards to the right. The graph continues upwards until (1.25, 1) in the first quadrant and then descends sharply. It + crosses the positive x axis close to (1.5, 0) and continues downwards. We divide the areas under the curve into boxes of equal width but varying + height. The far left side of the graph is divided into three boxes of decreasing length that are below the negative x axis. THe rectangle farthest to the + left is the tallest and the other two get get shorter as the graph moves upwards towards the negative x axis. The region roughly from (-0.25, 0) to + (0.25, 0) where the curve moves along the x axis there are no boxes since the are of that part is zero. Then the part of the graph in the first + quadrant is divided into five rectangles. As the graph rises up, the heights of the rectangles increase. The first one being the shortest and the fourth one + being the tallest. The fifth rectangle is shorter than the fourth one but taller than the first three. We add up the areas of the rectangles to approximate the + integral.

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.7,ymax=1.2, + xmin=-1,xmax=1.7 + ] + + \addplot [firstcurvestyle,areastyle,domain=-.7854:1.5708,samples=40] {sin(deg(x^3))} \closedcycle; + \addplot [firstcurvestyle,domain=-.7854:1.5708,samples=80] {sin(deg(x^3))}; + + \foreach \x / \y in {-0.785/ -0.466, -0.550/-0.165, -0.314/ -0.0310, + -0.0785/-0.000484, 0.157/ 0.00388, 0.393/ 0.0605, + 0.628/ 0.246, 0.864/ 0.601, 1.10/ 0.971, 1.34/ 0.690} + {\addplot [thick,secondcolor] coordinates {(\x+.2356,0) (\x+.2356,\y) (\x,\y) (\x,0) (\x+.2356,0)}; + } + + \draw (axis cs:.6,.9) node { $y=\sin(x^3)$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + + +
    + + + + Using the Right Hand rule + +

    We calculate \int_{-\frac{\pi}4}^{\frac{\pi}/2}\sin(x^3)\, dx using the Right hand rule. The far left part of the graph is divided into two + rectangles of decreasing height. The region from (-0.25, 0) to (0.25, 0) on the x axis does not contribute to the area since the curve + here continues along the axis and therefore the area is zero. The part in the first quadrant is divided into five adjacent rectangles of increasing height. + The upper left side of the first four rectangles contain some area outside of the curve on the left. As the graph moves up, the heights of the rectangles increase, + the fourth one being the tallest. The fifth rectangle is below the graph with its upper right corner touching the inside of the graph. The sixth and the last rectangle + is below the positive x axis. It captures the area adjacent to the part of the graph that is below the positive x axis.

     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.7,ymax=1.2, + xmin=-1,xmax=1.7 + ] + + \addplot [firstcurvestyle,areastyle,domain=-.7854:1.5708,samples=40] {sin(deg(x^3))} \closedcycle; + \addplot [firstcurvestyle,domain=-.7854:1.5708,samples=80] {sin(deg(x^3))}; + + \foreach \x / \y in {-0.550/-0.165, -0.314/ -0.0310, + -0.0785/-0.000484, 0.157/ 0.00388, 0.393/ 0.0605, + 0.628/ 0.246, 0.864/ 0.601, 1.10/ 0.971, 1.34/ 0.690, 1.57/ -0.670} + {\addplot [thick,secondcolor] coordinates {(\x-.2356,0) (\x-.2356,\y) (\x,\y) (\x,0) (\x-.2356,0)}; + } + + \draw (axis cs:.6,1.1) node { $y=\sin(x^3)$}; + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    + +

    + The actual answer, accurate to 4 places after the decimal, + is 0.4609. + Our approximations were once again fairly good. + The rectangles used in each approximation are shown in . + It is clear from the graphs that using more rectangles + (and hence, narrower rectangles) + should result in a more accurate approximation. +

    +     +     +     + + + The Trapezoidal Rule +

    + In + we approximated the value of + \ds \int_0^1 e^{-x^2}\, dx with 5 rectangles of equal width. + + shows the rectangles used in the Left and Right Hand Rules. + These graphs clearly show that rectangles do not match the shape of the graph all that well, + and that accurate approximations will only come by using lots of rectangles. + Trapezoidal Rule + numerical integrationTrapezoidal Rule + integrationnumerical!Trapezoidal Rule +

    + +

    + Instead of using rectangles to approximate the area, + we can instead use trapezoids. + In , + we show the region under f(x) = e^{-x^2} on [0,1] approximated + with 5 trapezoids of equal width; + the top corners of each trapezoid lies on the graph of f(x). + It is clear from this figure that these trapezoids more accurately approximate + the area under f and hence should give a better approximation of + \int_0^1 e^{-x^2}\, dx. + (In fact, these trapezoids seem to give a great + approximation of the area!) +

    + +
    + Approximating \int_0^1 e^{-x^2}\, dx using 5 trapezoids of equal widths + + + Approximating the integral of e to the power of negative x squared from 0 to 1 using five trapezoids + +

    The graph of e^{-x^2} starts from (0,1) and slopes downwards to the right in the positive x direction. We consider the are under the curve from + (0,1) to (1, 0.25) in the first quadrant. The area is approximated with five adjacent trapezoids of equal width but decreasing heights. The upper arms + of the trapezoids are downwar sloping and they follow the graph of e^{-x^2} very closely which allows for more precise approximation. Both of the upper corners + of the trapezoids lie on the curve. The first trapezoid is the tallest one and the fifth one is the shortest. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={.2,.4,.6,.8,1}, + ymin=-.2,ymax=1.2, + xmin=-.2,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1] {e^-x^2} \closedcycle; + \addplot [firstcurvestyle,domain=0:1] {e^-x^2}; + + \foreach \x / \y / \z in {0/1 /0.96 , 0.2/0.96 / .85, 0.4 /.85 / .7, .6 / .7/ .53, .8 / .53 / .37} + {\addplot [thick,secondcolor] coordinates {(\x,0) (\x,\y) (\x+.2,\z) (\x+.2,0) (\x,0)}; + } + + \draw (axis cs:.75,1) node { $y=e^{-x^2}$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + The formula for the area of a trapezoid is given in . + We approximate \int_0^1 e^{-x^2}\, dx with these trapezoids in the following example. +

    + +
    + The area of a trapezoid + + + The area of a trapezoid. + +

    This is a picture of a trapezoid. The height is h. The left arm of the trapezoid is shorter than the right arm. The arm connecting the left and the right + arm is thus upward sloping. The left arm is named a and the right arm is named b. +

    + + + \begin{tikzpicture}[scale=1.3] + + \draw (0,0) -- node [pos=.5,left] {\small $a$} (0,1) -- (1,1.5) -- node [pos=.5,right] {\small $b$} (1,0) -- node [pos=.5,below] {\small $h$} (0,0); + + \draw (0,.1) -- (.1,.1) -- (.1,0); + \draw (.9,0) -- (.9,.1) -- (1,.1); + + \draw (3,.75) node {Area = $\frac{a+b}2h$}; + + \end{tikzpicture} + + + + +
    + + + Approximating definite integrals using trapezoids + +

    + Use 5 trapezoids of equal width to approximate \ds \int_0^1e^{-x^2}\, dx. +

    +     + +

    + To compute the areas of the 5 trapezoids in , + it will again be useful to create a table of values as shown in . +

    + +
    + A table of values of e^{-x^2} + + + x_i + e^{-x_i^2} + + + + + + + 0 + 1 + + + 0.2 + 0.9608 + + + 0.4 + 0.8521 + + + 0.6 + 0.6977 + + + 0.8 + 0.5273 + + + 1 + 0.3679 + + +
    + +

    + The leftmost trapezoid has legs of length 1 and 0.9607 and a height of 0.2. + Thus, by our formula, the area of the leftmost trapezoid is: + + \frac{1+0.9608}{2}(0.2) = 0.1961 + . +

    + +

    + Moving right, + the next trapezoid has legs of length 0.9607 and 0.8521 and a height of 0.2. + Thus its area is: + + \frac{0.9608+0.8521}2(0.2) = 0.1813 + . +

    + +

    + The sum of the areas of all 5 trapezoids is: + + \frac{1+0.9608}{2}(0.2) + \frac{0.9608+0.8521}2(0.2)+\frac{0.8521+0.6977}2(0.2)\amp + + \frac{0.6977+0.5273}2(0.2)+\frac{0.5273+0.3679}2(0.2)\amp = 0.7444 + . +

    + +

    + We approximate \int_0^1 e^{-x^2}\, dx \approx 0.7444. +

    +     +     + +

    + There are many things to observe in this example. + Note how each term in the final summation was multiplied by both 1/2 and by \dx = 0.2. + We can factor these coefficients out, + leaving a more concise summation as: + + \frac12(0.2)\amp\Big[(1+0.9608) + (0.9608+0.8521) + (0.8521+0.6977) + \amp\quad + ( 0.6977+ 0.5273) +(0.5273 + 0.3679)\Big] + . +

    + +

    + Now notice that all numbers except for the first and the last are added twice. + Therefore we can write the summation even more concisely as + + \frac{0.2}{2}\Big[1 + 2(0.9608+0.8521+0.6977+0.5273) + 0.3679\Big] + . +

    + +

    + This is the heart of the Trapezoidal Rule, + wherein a definite integral + \int_a^b f(x) \, dx is approximated by using trapezoids of equal widths + to approximate the corresponding area under f. + Using n equally spaced subintervals with endpoints x_0, + x_1, + \ldots, x_{n}, + we again have \ds \dx = \frac{b-a}n. + Thus: + + \int_a^b f(x)\, dx \amp \approx \sum_{i=1}^n \frac{f(x_{i-1})+f(x_{i})}2\dx + \amp = \frac{\dx}2 \sum_{i=1}^n \big(f(x_{i-1})+f(x_{i})\big) + \amp = \frac{\dx}2\Big[f(x_0)+ \left(2\sum_{i=1}^{n-1} f(x_i)\right) + + f(x_{n})\Big] + . +

    + + + Using the Trapezoidal Rule + +

    + Revisit + and approximate \ds\int_{-\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(x^3)\, dx using the Trapezoidal Rule and 10 equally spaced subintervals. +

    +     + +

    + We refer back to + for the table of values of \sin(x^3). + Recall that \dx = 3\pi/40 \approx 0.236. + Thus we have: + + \amp \int_{-\frac{\pi}4}^{\frac{\pi}2} \sin(x^3)\, dx + \amp \approx \frac{0.236}{2}\Big[-0.4657 + 2\Big(-0.1654+(-0.031)+\ldots+ + 0.68999\Big)+(-0.67)\Big] + \amp = 0.4258 + . + The actual answer, accurate to 4 decimal places is 0.4609. + So the Trapezoidal Rule with 10 subintervals is an under-approximation + by about 0.0351. +

    +     +     + +

    + Notice how quickly the Trapezoidal Rule can be implemented once the + table of values is created. + This is true for all the methods explored in this section; + the real work is creating a table of x_i and f(x_i) values. + Once this is completed, + approximating the definite integral is not difficult. + Again, using technology is wise. + Spreadsheets can make quick work of these computations and make + using lots of subintervals easy. +

    + +

    + Also notice the approximations the Trapezoidal Rule gives. + It is the average of the approximations given by the Left and Right Hand Rules! + This effectively renders the Left and Right Hand Rules obsolete. + They are useful when first learning about definite integrals, + but if a real approximation is needed, + one is generally better off using the Trapezoidal Rule instead of either + the Left or Right Hand Rule. + However, there are two other methods that are also generally more accurate + than the Left or Right Hand Rule. +

    +     + + + The Midpoint Rule +

    + Another method that can be more accurate than the Trapezoidal Rule is the Midpoint Rule: + + S_M(n)\amp =\sum_{i=1}^n f\left(\frac{x_{i-1}+x_{i}}{2}\right)\Delta x + \amp = \sum_{i=1}^n f\left(\overline{x_i}\right)\Delta x + \amp \text{ where } \overline{x_i} \text{ is the midpoint of each subinterval,} + \amp \overline{x_i}=a+\dx\left(i-\frac12\right) + +

    + + + Using the Midpoint Rule + +

    + Use the Midpoint Rule with n=5 to approximate \ds \int_0^1e^{-x^2}\, dx. +

    +     + +

    + We cannot use the table in + that we used for the Trapezoidal, Right and Left Hand Rules when using the Midpoint Rule. + The Trapezoidal rule averages the outputs + of the function to obtain a more accurate estimate of the definite integral. + The Midpoint Rule averages the inputs + of each subinterval to create a rectangle with height f\left(\frac{x_{i-1}+x_{i}}{2}\right). + Generally f\left(\frac{x_{i-1}+x_{i}}{2}\right)\neq \frac{f(x_{i-1})+f(x_{i})}{2}. +

    + +

    + So we will create a new table of values as shown in . + We have \dx=(1-0)/5=0.2. + The midpoint of the first subinteval is at + 0+0.2(1/2)=0.1 and each successive midpoint is 0.2 from the last. +

    + +
    + A table of values of e^{-x^2} + + + x_i + e^{-x_i^2} + + + + + + + 0.1 + 0.9900 + + + 0.3 + 0.9139 + + + 0.5 + 0.7788 + + + 0.7 + 0.6126 + + + 0.9 + 0.4449 + + +
    + +

    + So we have + + \int_0^1 e^{-x^2}\, dx \amp \approx 0.2(0.99+0.9139+0.7788+0.6126+0.4449) + \amp \approx 0.7480 + +

    + +

    + We approximate \ds \int_0^1 e^{-x^2}\, dx \approx 0.7480. +

    +     +     + + + Using the Midpoint Rule + +

    + Revisit + and approximate \ds\int_{-\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(x^3)\, dx using the Midpoint Rule and 10 equally spaced subintervals. +

    +     + +

    + Again, a table will be useful. + Recall that \dx = 3\pi/40 \approx 0.2356. + The midpoint of the first subinterval is \overline{x_1}=a+\dx/2=-\pi/4+3\pi/40(1/2)=-17\pi/80 + (notice that \overline{x_1} is half of a subinterval width to the right of a). + Each successive midpoint is + \dx=3 \pi/40=6\pi/80 to the right of the last. + So we have: +

    + +
    + Values used to approximate \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in + + + \overline{x_i} + Exact + Approx. + \sin(x_i^3) + + + + + + + + + \overline{x_1} + -\frac{17\pi}{80} + -0.6676 + -0.2932 + + + \overline{x_2} + -\frac{11 \pi}{80} + -0.4320 + -0.0805 + + + \overline{x_3} + -\frac{5\pi }{80} + -0.1963 + -0.0076 + + + \overline{x_4} + \frac{\pi }{80} + -0.0393 + 0.0001 + + + \overline{x_5} + \frac{7\pi }{80} + 0.2749 + 0.0208 + + + \overline{x_6} + \frac{13\pi }{80} + 0.5105 + 0.1327 + + + \overline{x_7} + \frac{19\pi}{80} + 0.7461 + 0.4035 + + + \overline{x_8} + \frac{25 \pi}{80} + 0.9817 + 0.8112 + + + \overline{x_9} + \frac{31 \pi}{80} + 1.2174 + 0.9729 + + + \overline{x_{10}} + \frac{37 \pi}{80} + 1.4530 + 0.0740 + + + + + + +
    + +

    + Thus we have: + + \amp \int_{-\frac{\pi}4}^{\frac{\pi}2} \sin(x^3)\, dx + \amp \approx 0.2356\Big[-0.2932+(-0.0805)+(-0.0076)+\dots+0.9729+0.0740\Big] + \amp=0.2356\cdot 2.0339 + \amp \approx 0.4792 + . + The actual answer, accurate to 4 decimal places is 0.4609. + So the Midpoint Rule with 10 subintervals is an overrapproximation by about 0.0183 . + Notice that this error is about half of the error in using the Trapezoidal Rule. +

    +     +     + + +

    + In many cases, + the Midpoint Rule will more accurate than the Trapezoidal Rule. + You may wonder though, + how can we improve on the Trapezoidal and Midpoint Rules, + apart from using more and more subintervals? + The answer is clear once we look back and consider what we have + really done so far. + The Left Hand Rule, Right Hand Rule and Midpoint Rules are not really + about using rectangles to approximate area. + Instead, they approximate a function f with constant functions on small + subintervals and then compute the definite integral of these constant functions. + The Trapezoidal Rule is really approximating a function f with a linear + function on a small subinterval, + then computing the definite integral of this linear function. + In all of these cases the definite integrals are easy to compute in geometric terms. +

    + +

    + So we have a progression: + we start by approximating f with a constant function and then with a linear function. + What is next? + A quadratic function. + By approximating the curve of a function with lots of parabolas, + we generally get an even better approximation of the definite integral. + We call this process Simpson's Rule, + named after Thomas Simpson (1710-1761), + even though others had used this rule as much as 100 years prior. +

    +     + + + Simpson's Rule +

    + Given one point, + we can create a constant function that goes through that point. + Given two points, + we can create a linear function that goes through those points. + Given three points, + we can create a quadratic function that goes through those three points + (given that no two have the same x-value). + Simpson's Rule + numerical integrationSimpson's Rule + integrationnumerical!Simpson's Rule +

    + +

    + Consider three points (x_0,y_0), + (x_1,y_1) and (x_2,y_2) whose x-values are equally spaced and x_0\lt x_1\lt x_2. + Let f be the quadratic function that goes through these three points. + It is not hard to show that + + \int_{x_0}^{x_2} f(x)\, dx = \frac{x_2-x_0}{6}\big(y_0+4y_1+y_2\big) + . +

    + + + + + +

    + Consider . + A function f goes through the 3 points shown and the parabola g that also goes through those points is graphed with a dashed line. + Using our equation from above, we know exactly that + + \int_1^3 g(x) \, dx = \frac{3-1}{6}\big(3+4(1)+2\big)= 3 + . +

    + +

    + Since g is a good approximation for f on [1,3], + we can state that + + \int_1^3 f(x)\, dx \approx 3 + . +

    + +
    + A graph of a function f and a parabola that approximates it well on [1,3] + + + A parabola approximating the function f on [1, 3]. + +

    The graph of the function f starts close to (1, 4) in the first quadrant and then slopes downwards to the right towards the x axis. Once + it reaches the point (2, 1), the graph starts sloping upwards to the right and continues upwards. It ends close to (3, 2). The shape of the graph + resembles the shape of a parabola. We approximate the graph of f using a parabola whose path follows f very closely. The parabola intersects f + three times, once at (1, 3) then at (2, 1) and at (3, 2). +

    + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3}, + ytick={1,2,3}, + ymin=-.2,ymax=4, + xmin=.5,xmax=3.5 + ] + + \addplot+ [domain=.9:3.1,samples=30] {11-12*x+4.5*x^2-.5*x^3}; + \addplot+ [domain=.9:3.1] {8-6.5*x+1.5*x^2}; + + \filldraw (axis cs:1,3) circle (1pt); + \filldraw (axis cs:2,1) circle (1pt); + \filldraw (axis cs:3,2) circle (1pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Notice how the interval [1,3] was split into two subintervals as we needed 3 points. + Because of this, whenever we use Simpson's Rule, + we need to break the interval into an even number of subintervals. +

    + +

    + In general, to approximate + \ds \int_a^b f(x)\, dx using Simpson's Rule, + subdivide [a,b] into n subintervals, + where n is even and each subinterval has width \dx = (b-a)/n. + We approximate f with n/2 parabolic curves, + using Equation to compute the area under these parabolas. + Adding up these areas gives the formula: + + \int_a^b f(x) \, dx \approx \frac{\dx}3\Big[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+\ldots+2f(x_{n-2})+4f(x_{n-1})+f(x_{n})\Big] + . +

    + +

    + Note how the coefficients of the terms in the summation have the pattern 1, 4, 2, 4, 2, 4, \ldots, 2, 4, 1. +

    + +

    + + illustrates how the area calculated by Simpson's Rule approximates + \int_0^5 f(x)\, dx for the function f(x)=\sin(\pi x). + In this case, 8 subintervals were used, + resulting in 4 quadratic curves + (dashed lines) + being fitted to each pair of subintervals. + The actual answer + (accurate to 4 decimal places) + is about 10.6366, while Simpson's rule gives 10.7294. + Of course more subintervals would result in better accuracy. + However 8 intervals were chosen specifically so that you could see how the parabolas compare to the original function. + With larger values of n, + it becomes difficult to distinguish the function and its quadratic approximations on each subinterval. +

    + +
    + An illustration of Simpson's rule on f(x)=\sin(\pi x)+2 over [0,5] using 8 subintervals, resulting in 4 quadratic approximations + + Approximating f(x)=sin(pi x)+2 using simpson's rule. + +

    The graph of f(x)=\sin(\pi x)+2 is shaped like wave that oscillates between y=3 and y=1 as it moves in the positive x direction. We consider + the area under the curve from x=0 to x=5. THe area is subdivided into eight intervals and each interval is approximated by quadratic curves. The first + quadratic curve is a downward parabola with its vertex at (3,0.5) which approximates the first two subintervals. The second quadratic curve is also a parabola + whose vertex is located at (1.5,1). This parabola approximates the second two subintervals between x=1.25 and x=2.5. The next two subintervals are + are approximated by the left half of a parabola whose vertex is at (3.75,1.25). The area under the last two subintervals are approximated by a downwards parabola + whose vertex is at (3,4.5). +

    + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.2,ymax=3.3, + xmin=-1,xmax=5.8, + ] + + \addplot [firstcurvestyle,name path=F,domain={-0.9:5.2},samples=120] {sin(deg(3.14159*x))+2} node [pos=.87, above] { $y=\sin(\pi x)+2$}; + \addplot [name path=G,domain={0:5}]{0}; + \addplot [color=firstcolor!50] fill between [of=F and G, soft clip={domain=0:5}]; + + \addplot [secondcurvestyle,name path=H,domain=0:1.25] {-3.27*x^2+3.52*x+2}; + \addplot [color=gray!30] fill between [of=H and G, soft clip={domain=0:1.25}]; + + \addplot [secondcurvestyle,name path=I,domain=1.25:2.5] {1.35*x^2-3.71*x+3.82}; + \addplot [color=gray!30] fill between [of=I and G, soft clip={domain=1.25:2.5}]; + + \addplot [secondcurvestyle,name path=J,domain=2.5:3.75] {1.35*x^2-9.83*x+19.11}; + \addplot [color=gray!30] fill between [of=J and G, soft clip={domain=2.5:3.75}]; + + \addplot [secondcurvestyle,name path=K,domain=3.75:5] {-3.27*x^2+29.18*x-62.145}; + \addplot [color=gray!30] fill between [of=K and G, soft clip={domain=3.75:5}]; + + \addplot [-,color=gray!70,dashed,thin,domain=0:2.924] ({.625},{x}); + \addplot [-,color=gray,dashed,thin,domain=0:1.293] ({1.25},{x}); + \addplot [-,color=gray!70,dashed,thin,domain=0:1.617] ({1.875},{x}); + \addplot [-,color=gray,dashed,thin,domain=0:3] ({2.5},{x}); + \addplot [-,color=gray!70,dashed,thin,domain=0:1.617] ({3.125},{x}); + \addplot [-,color=gray,dashed,thin,domain=0:1.293] ({3.75},{x}); + \addplot [-,color=gray!70,dashed,thin,domain=0:2.924] ({4.375},{x}); + \addplot [-,color=gray,dashed,thin,domain=0:3] ({5},{x}); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Let's demonstrate Simpson's Rule with a concrete example. +

    + + + Using Simpson's Rule + +

    + Approximate \ds\int_0^1 e^{-x^2}\, dx using Simpson's Rule and 4 equally spaced subintervals. +

    +     + +

    + We begin by making a table of values as we have in the past, + as shown in . +

    + +
    + A table of values to approximate \int_0^1e^{-x^2}\, dx, along with a graph of the function + + +
    + + + x_i + e^{-x_i^2} + + + + + + + 0 + 1 + + + 0.25 + 0.939 + + + 0.5 + 0.779 + + + 0.75 + 0.570 + + + 1 + 0.368 + + + + + +
    + +
    + + + + + Approximating the integral of e^{-x^2} using Simpson's rule + +

    The graph of e^{-x^2} starts from (0, 1) and slopes downwards to the right in the positive x direction. We consider the are under the curve from + (0, 1) to (1, 0.25) in the first quadrant. The area is subdivided into four equally spaced intervals. The top of the intervals are parts of parabolas that + track the curve of e^{-x^2}. These approximating curves are almost identical to the parts of the graph they are tracking thus resulting in a much more accurate + approximation. +

    + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={.25,.5,.75,1}, + ymin=-.2,ymax=1.2, + xmin=-.2,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1] {e^-x^2} \closedcycle; + \addplot [firstcurvestyle,domain=0:1] {e^-x^2}; + + \addplot [thick,secondcolor,domain=0:.5] {1-0.042*x-0.8*x^2}; + \addplot [thick,secondcolor,domain=.5:1] {1.218-0.907*x+0.0569*x^2}; + \addplot [thick,secondcolor] coordinates {(0,1) (0,0) (.5,0) (.5,.779)}; + \addplot [thick,secondcolor] coordinates {(.5,0) (1,0) (1,.368)}; + + \draw (axis cs:.75,1) node { $y=e^{-x^2}$}; + + \filldraw (axis cs:0,1) circle (1pt) + (axis cs:.25,.939) circle (1pt) + (axis cs:.5,.779) circle (1pt) + (axis cs:.75,.57) circle (1pt) + (axis cs:1,.368) circle (1pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    + +

    + Simpson's Rule states that + + \int_0^1e^{-x^2}\, dx \approx \frac{0.25}{3}\Big[1+4(0.939)+2(0.779)+4(0.570) + 0.368\Big] = 0.7468\overline{3} + . +

    + +

    + Recall in + we stated that the correct answer, + accurate to 4 places after the decimal, was 0.7468. + Our approximation with Simpson's Rule, with 4 subintervals, + is better than our approximation with the Trapezoidal Rule using 5! +

    + +

    + + shows f(x) = e^{-x^2} along with its approximating parabolas, + demonstrating how good our approximation is. + The approximating curves are nearly indistinguishable from the actual function. +

    +     +     + + + Using Simpson's Rule + +

    + Approximate \ds\int_{-\frac{\pi}4}^{\frac{\pi}2} \sin(x^3)\, dx using Simpson's Rule and 10 equally spaced intervals. +

    +     + +

    + + shows the table of values that we used in the past for this problem, + shown here again for convenience. + Again, \dx = (\pi/2+\pi/4)/10 \approx 0.236. +

    + +
    + Values used to approximate \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in + + + x_i + \sin(x_i^3) + + + -0.7854 + -0.4657 + + + -0.5498 + -0.1654 + + + -0.3142 + -0.0310 + + + -0.0785 + -0.0005 + + + 0.1571 + 0.0039 + + + 0.3927 + 0.0605 + + + 0.6283 + 0.2455 + + + 0.8639 + 0.6011 + + + 1.0996 + 0.9710 + + + 1.3352 + 0.6899 + + + 1.5708 + -0.6700 + + +
    + +

    + Simpson's Rule states that + + \int_{-\frac{\pi}4}^{\frac{\pi}2} \sin(x^3)\, dx \amp \approx \frac{0.2356}3\Big[(-0.4657)+4(-0.1654)+2(-0.0310) + \ldots + \amp \ldots + 2(0.9710) + 4(0.6899) + (-0.6700)\big] + \amp \approx 0.4701 + +

    + +
    + Approximating \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in with Simpson's Rule and 10 equally spaced intervals + + + Approximating the integral of sin(x^3) using Simpson's rule + +

    The area under the graph of \sin(x^3) is subdivided into ten equally spaced subintervals. The leftmost part of the graph that is below the negative + x axis is approximated by a straight line and the upper left part of a downward parabola on its right. The part of the graph on the first quadrant is + subdivided into six subintervals. These are approximated by parabolas that are almost identical to the parts of the graphs they are approximating. The last part + of the graph that is below the positive x axis, is approximated with the descending part of a downward parabola and a straight line. +

    + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.7,ymax=1.2, + xmin=-1,xmax=1.7 + ] + + \addplot [firstcurvestyle,areastyle,domain=-.7854:1.5708,samples=40] {sin(deg(x^3))} \closedcycle; + \addplot [firstcurvestyle,domain=-.7854:1.5708,samples=80] {sin(deg(x^3))}; + + \addplot [thick,secondcolor,domain=-.785:-.314] {-.11-.721*x-1.49*x^2}; + \addplot [thick,secondcolor] coordinates {(-.785,-.466) (-.785,0) (-.314,0) (-.314,-.03)}; + + \addplot [thick,secondcolor,domain=-.314:.157] {.004+.037*x-.236*x^2}; + \addplot [thick,secondcolor] coordinates {(-.314,0) (.157,0) (.157,.004)}; + + \addplot [thick,secondcolor,domain=.157:.628] {.037-.395*x+1.156*x^2}; + \addplot [thick,secondcolor] coordinates {(.157,0) (.628,0) (.628,.246)}; + + \addplot [thick,secondcolor,domain=.628:1.1] {-.632+1.32*x+.13*x^2}; + \addplot [thick,secondcolor] coordinates {(.628,0) (1.1,0) (1.1,.971)}; + + \addplot [thick,secondcolor,domain=1.1:1.57] {-11.98+22.46*x-9.72*x^2}; + \addplot [thick,secondcolor] coordinates {(1.1,0) (1.57,0) (1.57,-.67)}; + + \filldraw (axis cs:-.785,-.466 ) circle (1pt) + (axis cs:-.55,-.165 ) circle (1pt) + (axis cs:-.314,-.031 ) circle (1pt) + (axis cs:-.0785,0 ) circle (1pt) + (axis cs:.157,.004 ) circle (1pt) + (axis cs:.393,.061 ) circle (1pt) + (axis cs:.628,.246 ) circle (1pt) + (axis cs:.864,.601 ) circle (1pt) + (axis cs:1.1,.971 ) circle (1pt) + (axis cs:1.34,.69 ) circle (1pt) + (axis cs:1.57,-.67 ) circle (1pt); + + \draw (axis cs:.6,.9) node { $y=\sin(x^3)$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Recall that the actual value, + accurate to 3 decimal places, is 0.4609. + Our approximation is within one 1/100th of the correct value. + The graph in + shows how closely the parabolas match the shape of the graph. +

    +     +     +     + + + Summary and Error Analysis +

    + We summarize the key concepts of this section thus far in the following Key Idea. +

    + + + Numerical Integration +

    + Let f be a continuous function on [a,b], + let n be a positive integer, + and let \ds\dx = \frac{b-a}{n}. + integrationnumerical!Left/Right Hand Rule + integrationnumerical!Trapezoidal Rule + integrationnumerical!Simpson's Rule + Left/Right Hand Rule + Trapezoidal Rule + Simpson's Rule + numerical integrationLeft/Right Hand Rule + numerical integrationTrapezoidal Rule + numerical integrationSimpson's Rule +

    + +

    + Set x_0=a, x_1 = a+\dx, + \ldots, x_i = a+i\dx, x_{n}=b. +

    + +

    + Consider \ds\int_a^b f(x)\, dx. +

    + +

    + Left Hand Rule: + \ds\int_a^b f(x)\, dx \approx \dx\big[f(x_0) + f(x_1) + \ldots + f(x_{n-1})\big]. +

    + +

    + Right Hand Rule: + \ds\int_a^b f(x)\, dx \approx \dx\big[f(x_1) + f(x_2) + \ldots + f(x_{n})\big]. +

    + +

    + Trapezoidal Rule: + \ds\int_a^b f(x)\, dx \approx \frac{\dx}2\big[f(x_0) + 2f(x_1) + 2f(x_2) +\ldots + 2f(x_{n-1})+ f(x_{n})\big]. +

    + +

    + Midpoint Rule: + \ds \int_a^b f(x)\, dx \approx \sum_{i=1}^n f\left(\frac{x_{i-1}+x_{i}}{2}\right)\Delta x. +

    + +

    + Simpson's Rule: + \ds\int_a^b f(x)\, dx \approx \frac{\dx}3\big[f(x_0) + 4f(x_1) + 2f(x_2) +\ldots + 4f(x_{n-1})+ f(x_{n})\big] \text{ for } n \text{ even}. +

    +     + +

    + In our examples, + we approximated the value of a definite integral using a given method then compared it to the right answer. + This should have raised several questions in the reader's mind, such as: +

    + +

    +

      +
    1. +

      + How was the right answer computed? +

      +
      2. + +
    2. +

      + If the right answer can be found, + what is the point of approximating? +

      +
      3. + +
    3. +

      + If there is value to approximating, + how are we supposed to know if the approximation is any good? +

      +
      4. +
    +

    + +

    + These are good questions, and their answers are educational. + In the examples, the right answer was never computed. + Rather, an approximation accurate to a certain number of places after the decimal was given. + In , + we do not know the exact answer, + but we know it starts with 0.7468. + These more accurate approximations were computed using numerical integration but with more precision (, more subintervals and the help of a computer). +

    + +

    + Since the exact answer cannot be found, + approximation still has its place. + How are we to tell if the approximation is any good? +

    + +

    + Trial and error provides one way. + Using technology, make an approximation with, + say, 10, 100, and 200 subintervals. + This likely will not take much time at all, + and a trend should emerge. + If a trend does not emerge, try using yet more subintervals. + Keep in mind that trial and error is never foolproof; + you might stumble upon a problem in which a trend will not emerge. +

    + +

    + A second method is to use Error Analysis. + While the details are beyond the scope of this text, + there are some formulas that give bounds + for how good your approximation will be. + For instance, + the formula might state that the approximation is within 0.1 of the correct answer. + If the approximation is 1.58, + then one knows that the correct answer is between 1.48 and 1.68. + By using lots of subintervals, + one can get an approximation as accurate as one likes. + states what these bounds are. +

    + + + Error Bounds in the Trapezoidal Rule and Simpson's Rule + +

    + integrationnumerical!Trapezoidal Rule + integrationnumerical!Simpson's Rule + Trapezoidal Ruleerror bounds + Simpson's Ruleerror bounds + numerical integrationTrapezoidal Rule!error bounds + numerical integrationSimpson's Rule!error bounds +

      +
    1. +

      + Let E_T and E_Mbe the error in approximating + \ds \int_a^b f(x)\, dx using the Trapezoidal and Midpoint Rules respectively, with n subintervals. + + If f has a continuous second derivative on [a,b] and K is any upper bound of + \abs{\fpp(x)} on [a,b], then + + \ds E_T \leq \frac{(b-a)^3}{12n^2}K + . + and + + \ds E_M \leq \frac{(b-a)^3}{24n^2}K + . +

      +
      2. + +
    2. +

      + Let E_S be the error in approximating + \ds \int_a^b f(x)\, dx using Simpson's Rule with n subintervals.. + + If f has a continuous 4th derivative on [a,b] and K is any upper bound of + \abs{f^{(4)}(x)} on [a,b], then + + E_S \leq \frac{(b-a)^5}{180n^4}K + . +

      +
      3. +
    +

    +     +     + +

    + There are some key things to note about this theorem. + +

      +
    1. +

      + The larger the interval, the larger the error. + This should make sense intuitively. +

      +
      2. + +
    2. +

      + The error shrinks as more subintervals are used (, as n gets larger). +

      +
      3. + +
    3. +

      + The maximum error in the Midpoint Rule is half of the maximum error in the Trapezoidal Rule. + (Usually the errors in these two rules have opposite signs as well, + that is one will be an under approximation and the other will be an over approximation). +

      +
      4. + +
    4. +

      + The error in Simpson's Rule has a term relating to the 4th derivative of f. + Consider a cubic polynomial: + its 4th derivative is 0. + Therefore, the error in approximating the definite integral of a cubic polynomial with Simpson's Rule is 0 Simpson's Rule computes the exact answer! +

      +
      5. +
    +

    + +

    + We revisit Examples + and + and compute the error bounds using in the following example. +

    + + + Computing error bounds + +

    + Find the error bounds when approximating + \ds \int_0^1 e^{-x^2}\, dx using the Trapezoidal and Midpoint Rules and 5 subintervals, + and using Simpson's Rule with 4 subintervals. +

    +     + +

    + Trapezoidal and Midpoints Rules with n=5: +

    + +

    + We start by computing the 2nd derivative of f(x) = e^{-x^2}: + + \fpp(x) = e^{-x^2}(4x^2-2) + . +

    + +

    + + shows a graph of \fpp(x) on [0,1]. + It is clear that the largest value of \fpp, in absolute value, is 2. +

    + +
    + Graphing \fpp(x) in to help establish error bounds + + + Shows the double derivative of f. + +

    The graph of \fpp(x) starts at (0, -2) and curves upwards to the left towards the positive x axis. It intersects the x axis at + (0.7, 0). Then it moves futher to the right in the positive x direction while moving upwards and ends at (1, 0.5). +

    + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-2.1,ymax=.7, + xmin=-.1,xmax=1.1 + ] + + \addplot+ [domain=0:1,samples=30] {(e^-x^2)*(4*x^2-2)}; + + \draw (axis cs:.8,-1.5) node { $y=e^{-x^2}(4x^2-2)$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Thus we let K=2 and apply the error formula from . + + E_T \leq \frac{(1-0)^3}{12\cdot 5^2}\cdot 2 = 0.00\overline{6} + . + Since the maximum error in the Midpoint rule is half the error in the Trapezoidal Rule, + we can say: + E_M \leq 0.00\overline{3} +

    + + + +

    + Our error estimation formula states that our approximation of 0.7444 found in + is within 0.0067 of the correct answer. + Hence we know that the actual value is within + [0.7444-0.0067, 0.7444+0.0067]=[0.7377,0.7511]. So: + + 0.7377 \leq \int_0^1e^{-x^2}\, dx \leq 0.7511 + + But we can do better than this with the Midpoint Rule since its error is at most half of the error of the Trapezoidal Rule. + Our error estimate formula state that our approximate of 0.7480 found in + is within 0.0034 of the correct answer. + Hence Hence we know that the actual value is within [0.7480-0.0034, 0.7480+0.0033]=[0.7447,0.7513]. +

    + +

    + We had earlier stated the actual answer, + correct to 4 decimal places, to be 0.7468, + affirming the validity of . +

    + +

    + Simpson's Rule with n=4: +

    + +

    + We start by computing the 4th derivative of f(x) = e^{-x^2}: + + f^{(4)}(x) = e^{-x^2}(16x^4-48x^2+12) + . +

    + +

    + + shows a graph of f^{(4)}(x) on [0,1]. + It is clear that the largest value of f^{(4)}, in absolute value, is 12. + Thus we let K=12 and apply the error formula from . + + E_s =\leq \frac{(1-0)^5}{180\cdot 4^4}\cdot 12 = 0.00026 + . +

    + +
    + Graphing f^{(4)}(x) in to help establish error bounds + + + Graphing the fourth derivative of f to help establish error bounds. + +

    The graph of f^{(4)}(x) starts from (12,0) in the positive y-axis and curves downwards to the right towards the positive x axis. + It intersects the x axis close to x=0.5 and continues moving downwards while also moving to the right in the positive x direction. It stops at the point + (1, -8). +

    + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + minor y tick num=4, + ymin=-8,ymax=13.5, + xmin=-.1,xmax=1.1 + ] + + \addplot+ [domain=0:1,samples=30] {(e^-x^2)*(16*x^4-48*x^2+12)}; + + \draw (axis cs:.7,11) node { $y=e^{-x^2}(16x^4-48x^2+12)$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Our error estimation formula states that our approximation of + 0.7468\overline{3} found in + is within 0.00026 of the correct answer, + hence we know that the correct answer is in the interval [0.74683-0.00026 , 0.74683 + 0.00026]=[0.74657,0.74709]. So: + + 0.74657 \leq \int_0^1e^{-x^2}\, dx \leq 0.74709 + . +

    + +

    + Once again we affirm the validity of + since the answer to 4 decimal places is actually 0.7468. +

    +     +     + +

    + At the beginning of this section we mentioned two main situations where numerical integration was desirable. + We have considered the case where an antiderivative of the integrand cannot be computed. + We now investigate the situation where the integrand is not known. + This is, in fact, + the most widely used application of Numerical Integration methods. + Most of the time we observe behavior but do not know the + function that describes it. + We instead collect data about the behavior and make approximations based on this data. + We demonstrate this in an example. +

    + + + Approximating distance traveled + +

    + One of the authors drove his daughter home from school while she recorded their speed every 30 seconds. + The data is given in . + Approximate the distance they traveled. +

    +
    + Speed data collected at 30 second intervals for + + + TimeSpeed + + + (min)() + + + 0 + 0 + + + 1 + 25 + + + 2 + 22 + + + 3 + 19 + + + 4 + 39 + + + 5 + 0 + + + 6 + 43 + + + 7 + 59 + + + 8 + 54 + + + 9 + 51 + + + 10 + 43 + + + 11 + 35 + + + 12 + 40 + + + 13 + 43 + + + 14 + 30 + + + 15 + 0 + + + 16 + 0 + + + 17 + 28 + + + 18 + 40 + + + 19 + 42 + + + 20 + 40 + + + 21 + 39 + + + 22 + 40 + + + 23 + 23 + + + 24 + 0 + + +
    +     + +

    + Recall that by integrating a speed function we get distance traveled. + We have information about v(t); + we will use Simpson's Rule to approximate \ds \int_a^b v(t)\, dt. +

    + +

    + The most difficult aspect of this problem is converting the given data into the form we need it to be in. + The speed is measured in miles per hour, + whereas the time is measured in minutes. +

    + +

    + We need to compute \dx = (b-a)/n. + With 25 data points collected, + there are n=24 subintervals. + What are a and b? + Since we start at time t=0, we have a=0. + The final recorded time was t=12 minutes, which is 1/5 of an hour. + Thus we have + + \dx = \frac{b-a}{n} = \frac{1/5-0}{24} = \frac1{120}; \frac{\dx}{3} = \frac{1}{360} + . +

    + +

    + Thus the distance traveled is approximately: + + \int_0^{0.2}v(t)\, dt \amp \approx \frac{1}{360}\Big[f(x_0)+4f(x_1) + 2f(x_2) + \cdots + 4f(x_{n-1})+f(x_{n})\Big] + \amp = \frac{1}{360}\Big[0+4\cdot25+2\cdot 22 + \cdots + 2\cdot40+4\cdot 23 + 0\Big] + \amp \approx 6.2167 \,\text{ miles. } + +

    + +

    + We approximate the author drove 6.2 miles. + (Because we are sure the reader wants to know, + the author's odometer recorded the distance as about 6.05 miles.) +

    +     +     + +     + + + + Terms and Concepts + + + + + +

    + Simpson's Rule is a method of approximating antiderivatives. + +

    +     + +     + + + + + +

    + What are the two basic situations where approximating the value of a definite integral is necessary? +

    + + +     + + + +     + + + + + +

    + Why are the Left and Right Hand Rules rarely used? +

    +     + + + +

    + They are superseded by the Trapezoidal Rule; + it takes an equal amount of work and is generally more accurate. +

    +     + +     + + + +

    + Simpson's Rule is based on approximating portions of a function with what type of function? +

    +

    + +

    +     + + + + quadratic|quadratic function|parabola|parabolic + + + +     +     + + Problems + + + +

    + Approximate the definite integral with the Trapezoidal Rule and Simpson's Rule, + with n=4. Then find the exact value. +

    +     + + + + + $a = -1; + $b = 1; + $n = 4; + $f = Formula("x^2"); + $F = Formula("x^3/3"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + $exact = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + For the integral \int_{-1}^1 x^2\, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     + + + +

    + Find the exact value. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = 10; + $n = 4; + $f = Formula("5*x"); + $F = Formula("5*x^2/2"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + $exact = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + For the integral \int_{0}^{10} 5x\, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     + + + +

    + Find the exact value. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = pi; + $n = 4; + $f = Formula("sin(x)"); + $F = Formula("-cos(x)"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + $exact = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + For the integral \int_{0}^{\pi} \sin(x) \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     + + + +

    + Find the exact value. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = 4; + $n = 4; + $f = Formula("sqrt(x)"); + $F = Formula("2/3*x^(3/2)"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + $exact = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + For the integral \int_{0}^{4} \sqrt x\, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     + + + +

    + Find the exact value. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = 3; + $n = 4; + $f = Formula("x^3+2x^2-5x+7"); + $F = Formula("x^4/4+2x^3/3-5x^2/2+7x"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + $exact = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + For the integral \int_{0}^{3} (x^3+2x^2-5x+7)\, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     + + + +

    + Find the exact value. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = 1; + $n = 4; + $f = Formula("x^4"); + $F = Formula("x^5/5"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + $exact = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + For the integral \int_{0}^{1} x^4\, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     + + + +

    + Find the exact value. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = 2*pi; + $n = 4; + $f = Formula("cos(x)"); + $F = Formula("sin(x)"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + $exact = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + For the integral \int_{0}^{2\pi} \cos(x) \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     + + + +

    + Find the exact value. +

    + +

    + +

    +     +     +     +     + + + + + $a = -3; + $b = 3; + $n = 4; + $f = Formula("sqrt(9-x^2)"); + $F = Formula("1/2*(sqrt(9-x^2)*x + 9 *asin(x/3))"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + $exact = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + For the integral \int_{-3}^{3} \sqrt{9-x^2} \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     + + + +

    + Find the exact value. +

    + +

    + +

    +     +     +     +     + +     + + + +

    + Approximate the definite integral with the Trapezoidal Rule and Simpson's Rule, + with n=6. +

    +     + + + + + $a = 0; + $b = 1; + $n = 6; + $f = Formula("cos(x^2)"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + + + +

    + For the integral \int_{0}^{1} \cos\big(x^2\big) \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     +     +     + + + + + $a = -1; + $b = 1; + $n = 6; + $f = Formula("e^(x^2)"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + + + +

    + For the integral \int_{-1}^{1} e^{x^2} \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = 5; + $n = 6; + $f = Formula("sqrt(x^2+1)"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + + + +

    + For the integral \int_{0}^{5} \sqrt{x^2+1} \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = pi; + $n = 6; + $f = Formula("x*sin(x)"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + + + +

    + For the integral \int_{0}^{\pi} x\sin(x) \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = pi/2; + $n = 6; + $f = Formula("sqrt(cos(x))"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + + + +

    + For the integral \int_{0}^{\pi/2} \sqrt{\cos(x) } \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     +     +     + + + + + $a = 1; + $b = 4; + $n = 6; + $f = Formula("ln(x)"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + + + +

    + For the integral \int_{1}^{4} \ln(x) \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     +     +     + + + + + $a = -1; + $b = 1; + $n = 6; + $f = Formula("1/(sin(x)+2)"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + + + +

    + For the integral \int_{-1}^{1} \frac{1}{\sin(x) +2} \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = 6; + $n = 6; + $f = Formula("1/(sin(x)+2)"); + $delta_x = ($b - $a)/$n; + @l = (); + $trap = 0; + $simp = 0; + foreach $i (0..$n-1) { + push @l, $i*$delta_x + $a; + } + $flip = 1; + foreach $i (@l) { + $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); + if ($flip == 1) { + $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); + $flip = 0; + } else { + $flip = 1; + } + } + $trap = $trap*$delta_x/2; + $simp = $simp*$delta_x/3; + + + +

    + For the integral \int_{0}^{6} \frac{1}{\sin(x) +2} \, dx: +

    +     + + + +

    + Approximate using the trapezoidal rule. +

    + +

    + +

    +     +     + + + +

    + Approximate using Simpson's rule. +

    + +

    + +

    +     +     +     +     + +     + + + +

    + Find n such that the error in approximating the given definite integral is less than 0.0001 + when using the Trapezoidal Rule and Simpson's Rule. +

    +     + + + + + $a = 0; + $b = pi; + $err = 0.0001; + $Mt = 1; + $Ms = 1; + $nt = sqrt(($b-$a)**3*$Mt/(12*$err)); + $nt = int($nt + 0.5); + $ns = (($b-$a)**5*$Ms/(180*$err))**(1./4); + $ns = int($ns + .99999); + if ($ns%2==1) { + $ns++; + } + + + +

    + For the integral \int_{0}^{\pi} \sin(x) \, dx: +

    +     + + + +

    + Using the trapezoid rule. +

    + +

    + +

    +     + +

    + n=161 (using \max\big(\fpp(x)\big)=1) +

    +     +     + + + +

    + Using Simpson's rule. +

    + +

    + +

    +     + +

    + n=12 (using \max\big(f^{(4)}(x)\big)=1) +

    +     +     +     +     + + + + + $a = 1; + $b = 4; + $err = 0.0001; + $Mt = 0.75; + $Ms = 105/16; + $nt = sqrt(($b-$a)**3*$Mt/(12*$err)); + $nt = int($nt + 0.99999); + $ns = (($b-$a)**5*$Ms/(180*$err))**(1./4); + $ns = int($ns + .99999); + if ($ns%2==1) { + $ns++; + } + + + +

    + For the integral \int_{1}^{4} \frac{1}{\sqrt x} \, dx: +

    +     + + + +

    + Using the trapezoid rule. +

    + +

    + +

    +     +     + + + +

    + Using Simpson's rule. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = pi; + $err = 0.0001; + $Mt = 38.1865; + $Ms = 804.2; + $nt = sqrt(($b-$a)**3*$Mt/(12*$err)); + $nt = int($nt + 0.99999); + $ns = (($b-$a)**5*$Ms/(180*$err))**(1./4); + $ns = int($ns + .99999); + if ($ns%2==1) { + $ns++; + } + + + +

    + For the integral \int_{0}^{\pi} \cos\big(x^2\big) \, dx: +

    +     + + + +

    + Using the trapezoid rule. +

    + +

    + +

    +     +     + + + +

    + Using Simpson's rule. +

    + +

    + +

    +     +     +     +     + + + + + $a = 0; + $b = 5; + $err = 0.0001; + $Mt = 300; + $Ms = 24; + $nt = sqrt(($b-$a)**3*$Mt/(12*$err)); + $nt = int($nt + 0.99999); + $ns = (($b-$a)**5*$Ms/(180*$err))**(1./4); + $ns = int($ns + .99999); + if ($ns%2==1) { + $ns++; + } + + + +

    + For the integral \int_{0}^{5} x^4 \, dx: +

    +     + + + +

    + Using the trapezoid rule. +

    + +

    + +

    +     +     + + + +

    + Using Simpson's rule. +

    + +

    + +

    +     +     +     +     + +     + + + +

    + A region is given. + Find the area of the region using Simpson's Rule: +

    + +

    +

      +
    1. +

      + where the measurements are in centimeters, + taken in 1 cm increments, and +

      +
      2. + +
    2. +

      + where the measurements are in hundreds of feet, + taken in 100 ft increments. +

      +
      3. +
    +

    +     + + + + + $a = 0; + $b = 6; + $n = 6; + $delta_x = ($b - $a)/$n; + @y = (0, 4.7, 6.3, 6.9, 6.6, 5.1, 0); + @coeffs = (1, 4, 2, 4, 2, 4, 1); + $area = 0; + $size = scalar @y; + foreach $i (0..$size) { + $area += $coeffs[$i] * $y[$i]; + } + $area = $area*$delta_x/3; + $area_ft = $area * 100 * 100; + $area = NumberWithUnits("$area cm^2"); + $area_ft = NumberWithUnits("$area_ft ft^2"); + + + + + + Finding area using Simpson's rule. + +

    We have an area subdivided into six subintervals. The heights of the subintervals are 4.7,6.3,6.9,6.6 and 5.1. +

    + + + \begin{tikzpicture}[xscale=1.19,yscale=.7] + + \draw [firstcolor,thick,fill=firstcolor!30,smooth] plot coordinates { + (0,2.)(0.06639,2.527)(0.2407,2.839)(0.4857,3.009)(0.7641,3.112)(1.039, + 3.221)(1.294,3.367)(1.532,3.53)(1.759,3.689)(1.98,3.823)(2.2,3.915)(2. + 42,3.967)(2.64,3.992)(2.86,4.)(3.08,4.)(3.305,3.987)(3.543,3.93)(3. + 808,3.797)(4.107,3.558)(4.45,3.185)(4.815,2.703)(5.17,2.152)(5.483,1. + 577)(5.723,1.018)(5.872,0.5022)(5.938,0.02083)(5.93,-0.4363)(5.861,-0. + 88)(5.741,-1.319)(5.58,-1.742)(5.389,-2.128)(5.179,-2.456)(4.96,-2. + 705)(4.737,-2.865)(4.502,-2.953)(4.243,-2.991)(3.95,-3.)(3.614,-2.999) + (3.244,-2.984)(2.86,-2.935)(2.484,-2.829)(2.137,-2.646)(1.832,-2.378)( + 1.558,-2.06)(1.299,-1.734)(1.039,-1.443)(0.7641,-1.222)(0.4857,-0. + 9779)(0.2407,-0.5015)(0.06639,0.4201)(0,2.)}; + + \draw (1,3.22) -- (1,-1.443) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.7}; + + \draw (2,3.823) -- (2,-2.5) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.3}; + + \draw (3,4) -- (3,-2.95) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.9}; + + \draw (4,3.62) -- (4,-3) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.6}; + + \draw (5,2.4) -- (5,-2.7) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 5.1}; + + \end{tikzpicture} + + + + + + Enter your answer using the metric measurements. + +

    + +

    + + + Enter your answer using the imperial measurements. + +

    + +

    + +     +     +     + + + + + $a = 0; + $b = 6; + $n = 6; + $delta_x = ($b - $a)/$n; + @y = (0, 3.6, 3.6, 4.5, 6.6, 5.6, 0); + @coeffs = (1, 4, 2, 4, 2, 4, 1); + $area = 0; + $size = scalar @y; + foreach $i (0..$size) { + $area += $coeffs[$i] * $y[$i]; + } + $area = $area*$delta_x/3; + $area_ft = $area * 100 * 100; + $area = NumberWithUnits("$area cm^2"); + $area_ft = NumberWithUnits("$area_ft ft^2"); + + + + + + Finding area using Simpson's rule. + +

    We have an area subdivided into six subintervals. The heights of the subintervals are 3.6,3.6,4.5,6.6 and 5.6. +

    + + + \begin{tikzpicture}[xscale=1.19,yscale=.7] + + \draw [firstcolor,thick,fill=firstcolor!30,smooth] plot coordinates { + (0,1.)(0.05533,1.541)(0.2027,1.965)(0.414,2.278)(0.6613,2.483)(0.9167, + 2.583)(1.157,2.588)(1.382,2.521)(1.595,2.409)(1.799,2.282)(2.,2.167)( + 2.2,2.089)(2.4,2.068)(2.6,2.119)(2.8,2.257)(3.,2.5)(3.201,2.847)(3. + 411,3.233)(3.636,3.58)(3.885,3.807)(4.167,3.833)(4.483,3.604)(4.815,3. + 159)(5.139,2.561)(5.431,1.876)(5.667,1.167)(5.828,0.488)(5.917,-0. + 1427)(5.943,-0.7173)(5.912,-1.228)(5.833,-1.667)(5.715,-2.028)(5.564,- + 2.317)(5.389,-2.543)(5.199,-2.712)(5.,-2.833)(4.799,-2.912)(4.589,-2. + 943)(4.364,-2.917)(4.115,-2.828)(3.833,-2.667)(3.513,-2.432)(3.153,-2. + 149)(2.753,-1.851)(2.313,-1.568)(1.833,-1.333)(1.323,-1.155)(0.828,-0. + 944)(0.4053,-0.5893)(0.1107,0.02133)(0,1)}; + + \draw (1,2.6) -- (1,-1.) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 3.6}; + + \draw (2,2.167) -- (2,-1.43) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 3.6}; + + \draw (3,2.5) -- (3,-2) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.5}; + + \draw (4,3.8) -- (4,-2.75) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.6}; + + \draw (5,2.8) -- (5,-2.83) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 5.6}; + + \end{tikzpicture} + + + + + + Enter your answer using the metric measurements. + +

    + +

    + + + Enter your answer using the imperial measurements. + +

    + +

    + +     +     +     + +     +     +     +
    + + +

    + We started this chapter learning about antiderivatives and indefinite integrals. + We then seemed to change focus by looking at areas between the graph of a function and the x-axis. + We defined these areas as the definite integral of the function, + using a notation very similar to the notation of the indefinite integral. + The Fundamental Theorem of Calculus tied these two seemingly separate concepts together: + we can find areas under a curve, + , we can evaluate a definite integral, using antiderivatives. +

    + +

    + We ended the chapter by noting that antiderivatives are sometimes more than difficult to find: + they are impossible. + Therefore we developed numerical techniques that gave us good approximations of definite integrals. +

    + +

    + We used the definite integral to compute areas, + and also to compute displacements and distances traveled. + There is far more we can do than that. + In + we'll see more applications of the definite integral. + Before that, in + we'll learn advanced techniques of integration, + analogous to learning rules like the Product, Quotient and Chain Rules of differentiation. +

    +     + + +     + + + Techniques of Antidifferentiation + +

    + The previous chapter introduced the antiderivative and connected it to signed areas under a curve through the Fundamental Theorem of Calculus. + The next chapter explores more applications of definite integrals than just area. + As evaluating definite integrals will become important, + we will want to find antiderivatives of a variety of functions. +

    + +

    + This chapter is devoted to exploring techniques of antidifferentiation. + While not every function has an antiderivative in terms of elementary functions (a concept introduced in the section on Numerical Integration), + we can still find antiderivatives of a wide variety of functions. +

    +     + +
    + Substitution + + + + +

    + We motivate this section with an example. + Let f(x) = (x^2+3x-5)^{10}. + We can compute \fp(x) using the Chain Rule. + It is: + + \fp(x) \amp = 10(x^2+3x-5)^9\cdot(2x+3) + \amp = (20x+30)(x^2+3x-5)^9 + . +

    + +

    + Now consider this: What is \int (20x+30)(x^2+3x-5)^9\, dx? + We have the answer in front of us; + + \int (20x+30)(x^2+3x-5)^9\, dx = (x^2+3x-5)^{10}+C + . +

    + +

    + How would we have evaluated this indefinite integral without starting with f(x) as we did? +

    + +

    + This section explores integration by substitution. + It allows us to undo the Chain Rule. + Substitution allows us to evaluate the above integral without knowing the original function first. +

    + +

    + The underlying principle is to rewrite a complicated + integral of the form \int f(x)\, dx as a not-so-complicated integral \int h(u)\, du. + We'll formally establish later how this is done. + First, consider again our introductory indefinite integral, + \int (20x+30)(x^2+3x-5)^9\, dx. + Arguably the most complicated + part of the integrand is (x^2+3x-5)^9. + We wish to make this simpler; + we do so through a substitution. + Let u=x^2+3x-5. + Thus + + (x^2+3x-5)^9 = u^9 + . +

    + +

    + We have established u as a function of x, + so now consider the differential of u: + + du = (2x+3)dx + . +

    + +

    + Keep in mind that (2x+3) and dx are multiplied; + the dx is not just sitting there. +

    + +

    + Return to the original integral and do some substitutions through algebra: + + \int (20x+30)(x^2+3x-5)^9\, dx \amp = \int 10(2x+3)(x^2+3x-5)^9\, dx + \amp =\int 10(\underbrace{x^2+3x-5}_u)^9\underbrace{(2x+3)\, dx}_{du} + \amp =\int 10u^9\, du + \amp = u^{10} + C \quad \text{(replace \(u\) with \(x^2+3x-5\)) } + \amp = (x^2+3x-5)^{10} +C + +

    + +

    + One might well look at this and think I + (sort of) + followed how that worked, + but I could never come up with that on my own, + but the process is learnable. + This section contains numerous examples through which the reader will gain understanding and mathematical maturity enabling them to regard substitution as a natural tool when evaluating integrals. +

    + +

    + We stated before that integration by substitution + undoes the Chain Rule. + Specifically, + let F(x) and g(x) be differentiable functions and consider the derivative of their composition: + + \frac{d}{dx}\Big(F\big(g(x)\big)\Big) = \Fp(g(x))\gp(x) + . +

    + +

    + Thus + + \int \Fp(g(x))\gp(x)\, dx = F(g(x)) + C + . +

    +     + + + + Substitution for indefinite integrals + +

    + Integration by substitution works by recognizing the inside + function g(x) and replacing it with a variable. + By setting u=g(x), we can rewrite the derivative as + + \frac{d}{dx}\Big(F\big(u\big)\Big) = \Fp(u)u' + . +

    + +

    + Since du = \gp(x)dx, we can rewrite the above integral as + + \int \Fp(g(x))\gp(x)\, dx = \int \Fp(u) du = F(u)+C = F(g(x))+ C + . +

    + +

    + This concept is important so we restate it in the context of a theorem. +

    + + + Integration by Substitution + +

    + Let F and g be differentiable functions, + where the range of g is an interval I contained in the domain of F. + Then integrationby substitution + + \int \Fp(g(x))\gp(x)\, dx = F(g(x)) + C + . +

    + +

    + If u = g(x), then du = \gp(x)dx and + + \int \Fp(g(x))\gp(x)\, dx = \int \Fp(u)\, du = F(u)+C = F(g(x))+C + . +

    +     +     + + + +

    + The point of substitution is to make the integration step easy. + Indeed, the step \int \Fp(u)\, du = F(u) + C looks easy, + as the antiderivative of the derivative of F is just F, + plus a constant. + The work involved is making the proper substitution. + There is not a step-by-step process that one can memorize; + rather, experience will be one's guide. + To gain experience, we now embark on many examples. +

    + + + Integrating by substitution + +

    + Evaluate \ds \int x\sin(x^2+5)\, dx. +

    +     + +

    + Knowing that substitution is related to the Chain Rule, + we choose to let u be the inside + function of \sin(x^2+5). + (This is not always a good choice, + but it is often the best place to start.) +

    + +

    + Let u = x^2+5, hence du = 2x\,dx. + The integrand has an x\,dx term, + but not a 2x\,dx term. + (Recall that multiplication is commutative, + so the x does not physically have to be next to dx for there to be an x\,dx term.) + We can divide both sides of the du expression by 2: + + du = 2x\,dx \quad \Rightarrow \quad \frac12du = x\,dx + . +

    + +

    + We can now substitute. + + \int x\sin(x^2+5)\, dx \amp = \int \sin(\underbrace{x^2+5}_u) \underbrace{x\, dx}_{\frac12du} + \amp = \int \frac12\sin(u) \, du + \amp = -\frac12\cos(u) + C \quad \text{ (now replace \(u\) with \(x^2+5\)) } + \amp =-\frac12\cos(x^2+5) + C + . +

    + +

    + Thus \int x\sin(x^2+5)\, dx = -\frac12\cos(x^2+5)+C. + We can check our work by evaluating the derivative of the right hand side. +

    +     +     + + + Integrating by substitution + +

    + Evaluate \ds \int \cos(5x)\, dx. +

    +     + +

    + Again let u replace the inside function. + Letting u = 5x, we have du = 5\, dx. + Since our integrand does not have a 5\, dx term, + we can divide the previous equation by 5 to obtain \frac15du = dx. + We can now substitute. + + \int \cos(5x)\, dx \amp = \int \cos(\underbrace{5x}_u) \underbrace{dx}_{\frac15du} + \amp = \int \frac15\cos(u) \, du + \amp = \frac15\sin(u) + C + \amp = \frac15\sin(5x)+C + . +

    + +

    + We can again check our work through differentiation. +

    +     +     + +

    + The previous example exhibited a common, + and simple, type of substitution. + The inside function was a linear function + (in this case, y = 5x). + When the inside function is linear, + the resulting integration is very predictable, outlined here. +

    + + + Substitution With A Linear Function +

    + Consider \int \Fp(ax+b)\, dx, + where a\neq 0 and b are constants. + Letting u = ax+b gives + du = a\cdot dx, leading to the result + + \int \Fp(ax+b)\, dx = \frac{1}{a}F(ax+b) + C + . +

    +     + +

    + Thus \int \sin(7x-4)\, dx = -\frac17\cos(7x-4)+C. + Our next example can use , + but we will only employ it after going through all of the steps. +

    + + + Integrating by substituting a linear function + +

    + Evaluate \ds \int \frac{7}{-3x+1}\, dx. +

    +     + +

    + View the integrand as the composition of functions f(g(x)), + where f(x) = 7/x and g(x) = -3x+1. + Employing our understanding of substitution, + we let u = -3x+1, the inside function. + Thus du = -3\, dx. + The integrand lacks a -3; + hence divide the previous equation by -3 to obtain -du/3 = dx. + We can now evaluate the integral through substitution. + + \int \frac{7}{-3x+1}\, dx \amp = \int \frac{7}{u}\frac{du}{-3} + \amp = \frac{-7}3\int \frac{du}{u} + \amp = \frac{-7}3\ln\abs{u} + C + \amp =-\frac73\ln\abs{-3x+1} + C + . +

    + +

    + Using is faster, + recognizing that u is linear and a = -3. + One may want to continue writing out all the steps until they are comfortable with this particular shortcut. +

    +     +     + + + + + +

    + Not all integrals that benefit from substitution have a clear + inside function. + Several of the following examples will demonstrate ways in which this occurs. +

    + + + Integrating by substitution + +

    + Evaluate \ds \int \sin(x) \cos(x) \, dx. +

    +     + +

    + There is not a composition of functions here to exploit; + rather, just a product of functions. + Do not be afraid to experiment; + when given an integral to evaluate, + it is often beneficial to think + If I let u be this, + then du must be that + and see if this helps simplify the integral at all. +

    + +

    + In this example, let's set u = \sin(x). + Then du = \cos(x) \, dx, + which we have as part of the integrand! + The substitution becomes very straightforward: + + \int \sin(x) \cos(x) \, dx \amp = \int u\, du + \amp = \frac12u^2+ C + \amp = \frac12\sin^2(x) + C + . +

    + +

    + One would do well to ask What would happen if we let u = \cos(x)? + The result is just as easy to find, yet looks very different. + The challenge to the reader is to evaluate the integral letting + u = \cos(x) and discover why the answer is the same, + yet looks different. +

    +     + +     + +

    + Our examples so far have required + basic substitution. + The next example demonstrates how substitutions can be made that often strike the new learner as being nonstandard. +

    + + + Integrating by substitution + +

    + Evaluate \ds\int x\sqrt{x+3}\, dx. +

    +     + +

    + Recognizing the composition of functions, set u = x+3. + Then du = dx, + giving what seems initially to be a simple substitution. + But at this stage, we have: + + \int x\sqrt{x+3}\, dx = \int x\sqrt{u}\, du + . +

    + +

    + We cannot evaluate an integral that has both an x and an u in it. + We need to convert the x to an expression involving just u. +

    + +

    + Since we set u = x+3, we can also state that u-3 = x. + Thus we can replace x in the integrand with u-3. + It will also be helpful to rewrite \sqrt{u} as u^\frac12. + + \int x\sqrt{x+3} \, dx \amp = \int (u-3)u^\frac12\, du + \amp = \int \big(u^\frac32 - 3u^\frac12\big) \, du + \amp = \frac25u^\frac52 - 2u^\frac32 + C + \amp = \frac25(x+3)^\frac52 - 2(x+3)^\frac32 + C + . +

    + +

    + Checking your work is always a good idea. + In this particular case, + some algebra will be needed to make one's answer match the integrand in the original problem. +

    +     +     + + + Integrating by substitution + +

    + Evaluate \ds \int \frac{1}{x\ln(x) }\, dx. +

    +     + +

    + This is another example where there does not seem to be an obvious composition of functions. + The line of thinking used in is useful here: + choose something for u and consider what this implies du must be. + If u can be chosen such that du also appears in the integrand, + then we have chosen well. +

    + +

    + Choosing u = 1/x makes du = -1/x^2\, dx; + that does not seem helpful. + However, setting u = \ln(x) makes du = 1/x\, dx, + which is part of the integrand. + Thus: + + \int \frac1{x\ln(x)}\, dx \amp = \int \frac{1}{\underbrace{\ln(x)}_{u}}\underbrace{\frac1x\, dx}_{du} + \amp = \int \frac1u\, du + \amp = \ln\abs{u} + C + \amp = \ln\abs{ \ln(x) } + C + . +

    + +

    + The final answer is interesting; + the natural log of the natural log. + Take the derivative to confirm this answer is indeed correct. +

    +     +     + + + +     + + + Integrals Involving Trigonometric Functions +

    + + delves deeper into integrals of a variety of trigonometric functions; + here we use substitution to establish a foundation that we will build upon. +

    + +

    + The next three examples will help fill in some missing pieces of our antiderivative knowledge. + We know the antiderivatives of the sine and cosine functions; + what about the other standard functions tangent, + cotangent, secant and cosecant? + We discover these next. +

    + + + Integrating by substitution: the antiderivative of <m>\tan(x)</m> + +

    + Evaluate \int \tan(x) \, dx. +

    +     + +

    + The previous paragraph established that we did not know the antiderivatives of tangent, + hence we must assume that we have learned something in this section that can help us evaluate this indefinite integral. +

    + +

    + Rewrite \tan(x) as \sin(x) /\cos(x). + While the presence of a composition of functions may not be immediately obvious, + recognize that \cos(x) is + inside the 1/x function. + Therefore, we see if setting + u = \cos(x) returns usable results. + We have that du = -\sin(x) \, dx, + hence -du = \sin(x) \, dx. + We can integrate: + + \int \tan(x) \, dx \amp = \int \frac{\sin(x) }{\cos(x) }\, dx + \amp = \int \frac1{\underbrace{\cos(x) }_u}\underbrace{\sin(x) \, dx}_{-du} + \amp = \int \frac {-1}u \, du + \amp = -\ln\abs{u} + C + \amp = -\ln\abs{\cos(x) } + C + . +

    + +

    + Some texts prefer to bring the -1 inside the logarithm as a power of \cos(x), as in: + + -\ln\abs{\cos(x) } + C \amp = \ln\abs{(\cos(x) )^{-1}} + C + \amp = \ln\abs{ \frac{1}{\cos(x) }} + C + \amp = \ln\abs{\sec(x) } + C + . +

    + +

    + Thus the result they give is \int \tan(x) \, dx = \ln\abs{\sec(x) } + C. + These two answers are equivalent. +

    +     + +     + + + Integrating by substitution: the antiderivative of <m>\sec(x)</m> + +

    + Evaluate \int \sec(x) \, dx. +

    +     + +

    + This example employs a wonderful trick: + multiply the integrand by 1 + so that we see how to integrate more clearly. + In this case, we write 1 as + + 1 = \frac{\sec(x) + \tan(x) }{\sec(x) + \tan(x) } + . +

    + +

    + This may seem like it came out of left field, but it works beautifully. + Consider: + + \int \sec(x) \, dx \amp = \int \sec(x) \cdot \frac{\sec(x) + \tan(x) }{\sec(x) + \tan(x) }\, dx + \amp = \int \frac{\sec^2(x) + \sec(x) \tan(x) }{\sec(x) + \tan(x) }\, dx. + Now let u = \sec(x) +\tan(x); this means du = (\sec(x) \tan(x) + \sec^2(x) )\, dx, which is our numerator. Thus: + \amp = \int \frac{du}{u} + \amp = \ln\abs{u} + C + \amp = \ln\abs{\sec(x) +\tan(x) } + C + . +

    +     + +     + + + + +

    + We can use similar techniques to those used in Examples + and + to find antiderivatives of \cot(x) and \csc(x) + (which the reader can explore in the exercises.) + We summarize our results here. +

    + + + Antiderivatives of Trigonometric Functions + + +

    + integrationof trig. functions +

    + + + + +

    +

      +
    1. \int \sin(x) \, dx = -\cos(x) +C,
      2. + +
    2. \int \cos(x) \, dx = \sin(x) + C,
      3. + +
    3. \int \tan(x) \, dx = -\ln\abs{\cos(x) }+C,
      4. + +
    4. \int \csc(x) \, dx = -\ln\abs{\csc(x) +\cot(x) } +C,
      5. + +
    5. \int \sec(x) \, dx = \ln\abs{\sec(x) +\tan(x) } + C,
      6. + +
    6. \int \cot(x) \, dx = \ln\abs{\sin(x) }+C,
      7. +
    +

    +     +     + +

    + We explore one more common trigonometric integral. +

    + + + Integration by substitution: powers of <m>\cos(x)</m> and <m>\sin(x)</m> + +

    + Evaluate \int \cos^2(x) \, dx. +

    +     + +

    + We have a composition of functions as \cos^2(x) = \big(\cos(x) \big)^2. + However, setting u = \cos(x) means du = -\sin(x) \, dx, + which we do not have in the integral. + Another technique is needed. +

    + + + +

    + The process we'll employ is to use a Power Reducing formula for \cos^2(x), + which states + + \cos^2(x) = \frac{1+\cos(2x)}{2} + . +

    + +

    + The right hand side of this equation is not difficult to integrate. + We have: + + \int \cos^2(x) \, dx \amp = \int \frac{1+\cos(2x)}2\, dx + \amp = \int \left( \frac12 + \frac12\cos(2x)\right)\, dx + \amp = \frac12x + \frac12\frac{\sin(2x)}{2} + C + \amp = \frac12x + \frac{\sin(2x)}4 + C + , + where we used for the antiderivative of \cos(2x). +

    + +

    + We'll make significant use of this power-reducing technique in future sections. +

    +     +     + + +     + + + Simplifying the Integrand +

    + It is common to be reluctant to manipulate the integrand of an integral; + at first, our grasp of integration is tenuous and one may think that working with the integrand will improperly change the results. + Integration by substitution works using a different logic: + as long as equality is maintained, + the integrand can be manipulated so that its + form is easier to deal with. + The next two examples demonstrate common ways in which using algebra first makes the integration easier to perform. +

    + + + Integration by substitution: simplifying first + +

    + Evaluate \ds\int \frac{x^3+4x^2+8x+5}{x^2+2x+1}\, dx. +

    +     + +

    + One may try to start by setting u equal to either the numerator or denominator; + in each instance, the result is not workable. +

    + +

    + When dealing with rational functions (, quotients made up of polynomial functions), + it is an almost universal rule that everything works better when the degree of the numerator is less than the degree of the denominator. + Hence we use polynomial division. +

    + +

    + We skip the specifics of the steps, + but note that when x^2+2x+1 is divided into x^3+4x^2+8x+5, + it goes in x+2 times with a remainder of 3x+3. + Thus + + \frac{x^3+4x^2+8x+5}{x^2+2x+1} = x+2 + \frac{3x+3}{x^2+2x+1} + . +

    + +

    + Integrating x+2 is simple. + The fraction can be integrated by setting u = x^2+2x+1, + giving du = (2x+2)\, dx. + This is very similar to the numerator. + Note that du/2 = (x+1)\, dx and then consider the following: + + \int \frac{x^3+4x^2+8x+5}{x^2+2x+1}\, dx \amp = \int \left(x+2 + \frac{3x+3}{x^2+2x+1}\right)\, dx + \amp = \int (x+2)\, dx + \int \frac{3(x+1)}{x^2+2x+1}\, dx + \amp = \frac12x^2+2x+C_1 + \int \frac{3}{u}\frac{du}{2} + \amp = \frac12x^2+2x+C_1 + \frac32\ln\abs{u} + C_2 + \amp = \frac12x^2+2x+\frac32\ln\abs{x^2+2x+1} + C + . +

    + +

    + In some ways, we lucked out in that after dividing, + substitution was able to be done. + In later sections we'll develop techniques for handling rational functions where substitution is not directly feasible. +

    +     + +     + + + Integration by alternate methods + +

    + Evaluate \ds\int \frac{x^2+2x+3}{\sqrt{x}}\, dx with, + and without, substitution. +

    +     + +

    + We already know how to integrate this particular example. + Rewrite \sqrt{x} as + x^\frac12 and simplify the fraction: + + \frac{x^2+2x+3}{x^{1/2}} = x^\frac32 + 2x^\frac12 + 3x^{-\frac12} + . +

    + +

    + We can now integrate using the Power Rule: + + \int \frac{x^2+2x+3}{x^{1/2}}\, dx \amp = \int\left(x^\frac32 + 2x^\frac12 + 3x^{-\frac12}\right)\, dx + \amp = \frac25x^\frac52 + \frac43x^\frac32 + 6x^\frac12 + C + +

    + +

    + This is a perfectly fine approach. + We demonstrate how this can also be solved using substitution as its implementation is rather clever. +

    + +

    + Let u = \sqrt{x} = x^\frac12; therefore + + du = \frac{1}{2\sqrt{x}}\, dx \quad \Rightarrow \quad 2du = \frac{1}{\sqrt{x}}\, dx + . +

    + +

    + This gives us \ds \int \frac{x^2+2x+3}{\sqrt{x}}\, dx = \int (x^2+2x+3)\cdot2\, du. + What are we to do with the other x terms? + Since u = x^\frac12, u^2 = x, etc. + We can then replace x^2 and x with appropriate powers of u. + We thus have + + \int \frac{x^2+2x+3}{\sqrt{x}}\, dx \amp = \int (x^2+2x+3)\cdot2\, du + \amp = \int 2(u^4 + 2u^2 + 3)\, du + \amp = \frac25u^5 + \frac43u^3 + 6u + C + \amp = \frac25x^\frac52 + \frac43x^\frac32 + 6x^\frac12+C + , + which is obviously the same answer we obtained before. + In this situation, + substitution is arguably more work than our other method. + The fantastic thing is that it works. + It demonstrates how flexible integration is. +

    +     +     +     + + + Substitution and Inverse Trigonometric Functions +

    + When studying derivatives of inverse functions, we learned that + + \frac{d}{dx}\big(\tan^{-1}(x) \big) = \frac{1}{1+x^2} + . +

    + +

    + Applying the Chain Rule to this is not difficult; for instance, + + \frac{d}{dx}\big(\tan^{-1}(5x) \big) = \frac{5}{1+25x^2} + . +

    + +

    + We now explore how Substitution can be used to undo + certain derivatives that are the result of the Chain Rule applied to Inverse Trigonometric functions. + We begin with an example. +

    + + + Integrating by substitution: inverse trigonometric functions + +

    + Evaluate \ds \int \frac{1}{25+x^2}\, dx. +

    +     + +

    + The integrand looks similar to the derivative of the arctangent function. + Note: + + \frac{1}{25+x^2} \amp = \frac{1}{25\left(1+\frac{x^2}{25}\right)} + \amp = \frac{1}{25(1+\left(\frac{x}{5}\right)^2)} + \amp = \frac{1}{25}\frac{1}{1+\left(\frac{x}{5}\right)^2} + . +

    + +

    + Thus + + \int\frac{1}{25+x^2}\, dx = \frac{1}{25}\int \frac{1}{1+\left(\frac{x}{5}\right)^2}\, dx + . +

    + +

    + This can be integrated using Substitution. + Set u = x/5, hence du = dx/5 or dx=5\, du. + Thus + + \int\frac{1}{25+x^2}\, dx \amp = \frac{1}{25}\int \frac{1}{1+\left(\frac{x}{5}\right)^2}\, dx + \amp = \frac15\int \frac{1}{1+u^2}\, du + \amp = \frac15\tan^{-1}(u) + C + \amp = \frac15\tan^{-1}\left(\frac x5\right)+C + +

    +     + +     + +

    + + demonstrates a general technique that can be applied to other integrands that result in inverse trigonometric functions. + The results are summarized here. +

    + + + Integrals Involving Inverse Trigonometric Functions + +

    + Let a \gt 0. +

    + +

    +

      +
    1. \int \frac{1}{a^2+x^2}\, dx = \frac1a\tan^{-1}\left(\frac{x}{a}\right) + C
      2. + +
    2. \int \frac{1}{\sqrt{a^2-x^2}}\, dx = \sin^{-1}\left(\frac{x}{a}\right)+C
      3. + +
    3. \int \frac{1}{x\sqrt{x^2-a^2}}\, dx = \frac1a\sec^{-1}\left(\frac{\abs{x}}{a}\right)+C
      4. +
    +

    +     +     + +

    + Let's practice using . +

    + + + Integrating by substitution: inverse trigonometric functions + +

    + Evaluate the given indefinite integrals: +

      +
    1. \int \frac{1}{9+x^2}\, dx
      2. + +
    2. \int \frac{1}{\sqrt{5-x^2}}\, dx
      3. + +
    3. \int \frac{1}{x\sqrt{x^2-\frac{1}{100}}}\, dx
      4. +
    +

    +     + +

    + Each can be answered using a straightforward application of . + +

      +
    1. +

      + \ds \int \frac{1}{9+x^2}\, dx = \frac13\tan^{-1}\left(\frac x3\right) + C, + as a = 3. +

      +
      2. + +
    2. +

      + \ds \int \frac{1}{\sqrt{5-x^2}} = \sin^{-1}\left(\frac{x}{\sqrt{5}}\right)+C, + as a = \sqrt{5}. +

      +
      3. + +
    3. +

      + \ds \int \frac{1}{x\sqrt{x^2-\frac{1}{100}}}\, dx = 10\sec^{-1}(10x) + C, + as a = \frac1{10}. +

      +
      4. +
    +

    +     +     + +

    + Most applications of are not as straightforward. + The next examples show some common integrals that can still be approached with this theorem. +

    + + + Integrating by substitution: completing the square + +

    + Evaluate \ds \int\frac{1}{x^2-4x+13}\, dx. +

    +     + +

    + Initially, this integral seems to have nothing in common with the integrals in . + As it lacks a square root, + it almost certainly is not related to arcsine or arcsecant. + It is, however, related to the arctangent function. +

    + +

    + We see this by completing the square in the denominator. + We give a brief reminder of the process here. +

    + +

    + Start with a quadratic with a leading coefficient of 1. + It will have the form of x^2 + bx + c. + Take 1/2 of b, square it, + and add/subtract it back into the expression. + , + + x^2+bx+ c \amp = \underbrace{x^2 + bx + \frac{b^2}4}_{(x+b/2)^2} - \frac{b^2}4 + c + \amp = \left(x+\frac b2\right)^2 + c-\frac{b^2}4 + +

    + +

    + In our example, + we take half of -4 and square it, getting 4. + We add/subtract it into the denominator as follows: + + \frac{1}{x^2-4x+13} \amp = \frac{1}{\underbrace{x^2-4x+4}_{(x-2)^2}-4+13} + \amp =\frac{1}{(x-2)^2 + 9} + +

    + +

    + We can now integrate this using the arctangent rule. + Technically, we need to substitute first with u=x-2, + but we can employ instead. + Thus we have + + \int \frac{1}{x^2-4x+13}\, dx \amp = \int \frac{1}{(x-2)^2+9}\, dx + \amp = \frac13\tan^{-1}\left(\frac{x-2}{3}\right)+C + . +

    +     + +     + + + Integrals requiring multiple methods + +

    + Evaluate \ds \int \frac{4-x}{\sqrt{16-x^2}}\, dx. +

    +     + +

    + This integral requires two different methods to evaluate it. + We get to those methods by splitting up the integral into two terms: + + \int \frac{4-x}{\sqrt{16-x^2}}\, dx = \int \frac{4}{\sqrt{16-x^2}}\, dx - \int \frac{x}{\sqrt{16-x^2}}\, dx + . +

    + +

    + We handle each separately. + The first integral is handled using a straightforward application of : +

    + +

    + \ds \int \frac{4}{\sqrt{16-x^2}}\, dx = 4\sin^{-1}\left(\frac{x}{4}\right) + C. +

    + +

    + The second integral is handled by substitution, with u = 16-x^2. + \ds \int\frac{x}{\sqrt{16-x^2}}\, dx: Set u = 16-x^2, + so du = -2x\, dx and x\, dx = -du/2. + We have + + \int\frac{x}{\sqrt{16-x^2}}\, dx \amp = \int\frac{-du/2}{\sqrt{u}} + \amp = -\frac12\int \frac{1}{\sqrt{u}}\, du + \amp = - \sqrt{u} + C + \amp = -\sqrt{16-x^2} + C + . +

    + +

    + Combining these together, we have + + \int \frac{4-x}{\sqrt{16-x^2}}\, dx = 4\sin^{-1}\left(\frac{x}{4}\right) + \sqrt{16-x^2}+C + . + As with all definite integrals, + you can check your work by differentiation. +

    +     + +     +     + + + Substitution and Definite Integration + + + +

    + This section has focused on evaluating indefinite integrals as we are learning a new technique for finding antiderivatives. + However, much of the time integration is used in the context of a definite integral. + Definite integrals that require substitution can be calculated using the following workflow: +

    + +

    +

      +
    1. +

      + Start with a definite integral + \ds \int_a^b f(x)\, dx that requires substitution. +

      +
      2. + +
    2. +

      + Ignore the bounds; use substitution to evaluate + \ds \int f(x)\, dx and find an antiderivative F(x). +

      +
      3. + +
    3. +

      + Evaluate F(x) at the bounds; + that is, evaluate F(x)\Big|_a^b = F(b) - F(a). +

      +
      4. +
    +

    + +

    + This workflow works fine, + but substitution offers an alternative that is powerful and amazing + (and a little time saving). +

    + +

    + At its heart, + (using the notation of ) + substitution converts integrals of the form + \int \Fp(g(x))\gp(x)\, dx into an integral of the form + \int \Fp(u)\, du with the substitution of u = g(x). + The following theorem states how the bounds of a definite integral can be changed as the substitution is performed. +

    + + + Substitution with Definite Integrals + +

    + Let F and g be differentiable functions, + where the range of g is an interval I that is contained in the domain of F and u=g(x). + Then + + integrationdefinite!and substitution + + definite integraland substitution + + + \int_a^b \Fp\big(g(x)\big)\gp(x)\, dx = \int_{g(a)}^{g(b)} \Fp(u)\, du + . +

    +     +     + + + +

    + In effect, + states that once you convert to integrating with respect to u, + you do not need to switch back to evaluating with respect to x. + A few examples will help one understand. +

    + + + Definite integrals and substitution: changing the bounds + +

    + Evaluate \ds\int_0^2 \cos(3x-1)\, dx using . +

    +     + +

    + Observing the composition of functions, + let u=3x-1, hence du = 3\, dx. + As 3\, dx does not appear in the integrand, + divide the latter equation by 3 to get du/3 = dx. +

    + +

    + By setting u = 3x-1, we are implicitly stating that g(x) = 3x-1. + + states that the new lower bound is g(0) = -1; + the new upper bound is g(2) = 5. + We now evaluate the definite integral: + + \int_0^2 \cos(3x-1) \, dx \amp = \int_{-1}^5 \cos(u) \frac{du}{3} + \amp = \frac{1}{3} \sin(u) \Big|_{-1}^5 + \amp = \frac{1}{3}\big(\sin(5) - \sin(-1)\big) + \amp \approx -0.039 + . +

    + +

    + Notice how once we converted the integral to be in terms of u, + we never went back to using x. +

    + +
    + Graphing the areas defined by the definite integrals of + +
    + + + + + Graph of function cos(3x-1) with integrals shown as shaded portions under the curve between 0 and 2. + + +

    + The y axis is drawn from -1 to 1 and the x axis is drawn from + -1 to 5. + The function y=\cos(3x-1) has several maxima of 1 and minima of -1. + The function has several complete waves. At x intercept -0.9 the first wave + starts it forms a y intercept of 0.5 after which it reaches the maxima, it + decreases after and crosses the x axis at 0.9 then it decreases further to + reach the minima then it gets a positive slope and meets the x axis again at 2. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,1,2,3,4,5}, + ymin=-1.1,ymax=1.1, + xmin=-1.5,xmax=5.5] + + \addplot [firstcurvestyle,areastyle,domain=0:2] {cos(deg(3*x-1))} \closedcycle; + \addplot [firstcurvestyle,domain=-1.5:5.5,samples=140] {cos(deg(3*x-1))} node [pos=0.54, above right]{ $y=\cos(3x-1)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + Graph of a cosine wave with amplitude 1/3. + + +

    + The y axis is drawn from -1 to 1 + and the u axis is drawn from -1 to 5. + The area that the curve forms with the u axis is + shaded between -1 and 5. The function has + two u intercepts at u = 1.5 and u = 4.6. There is a parabolic curve above the u axis the major part of which is shaded from -1.4 to 1.5, there is a second shaded inverted parabolic curve of the same size as the first below the u axis, from 1.5 to 4.6. A third parabola is shaded from 4.6 to 5. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,1,2,3,4,5}, + ymin=-1.1,ymax=1.1, + xmin=-1.5,xmax=5.5, + xlabel={$u$}, + ] + + \addplot [firstcurvestyle,areastyle,domain=-1:5] {cos(deg(x))/3} \closedcycle; + \addplot [firstcurvestyle,domain=-1.5:5.5,samples=40] {cos(deg(x))/3} node [pos=0.45, above right] { $y=\frac13\cos(u)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    + +

    + The graphs in tell more of the story. + In + the area defined by the original integrand is shaded, + whereas in + the area defined by the new integrand is shaded. + In this particular situation, + the areas look very similar; + the new region is shorter + but wider, giving the same area. +

    +     +     + + + Definite integrals and substitution: changing the bounds + +

    + Evaluate \ds \int_0^{\pi/2} \sin(x) \cos(x) \, dx using . +

    +     + +

    + We saw the corresponding indefinite integral in . + In that example we set u = \sin(x) but stated that we could have let u = \cos(x). + For variety, we do the latter here. +

    + +

    + Let u = g(x) = \cos(x), + giving du = -\sin(x) \, dx and hence \sin(x) \, dx = -du. + The new upper bound is g(\pi/2) = 0; + the new lower bound is g(0) = 1. + Note how the lower bound is actually larger than the upper bound now. + We have + + \int_0^{\pi/2} \sin(x) \cos(x) \, dx \amp = \int_1^0 -u\, du \quad \text{(switch bounds and change sign) } + \amp = \int_0^1 u\, du + \amp = \frac12u^2\Big|_0^1= 1/2 + . +

    + +

    + In + we have again graphed the two regions defined by our definite integrals. + Unlike the previous example, they bear no resemblance to each other. + However, + guarantees that they have the same area. +

    + +
    + Graphing the areas defined by the definite integrals of + +
    + + + + + Graph of function showing area under the curve as the definite integrals. + + +

    + The y axis is drawn from -0.5 to 1 and the x + axis is drawn from a little before 0 to pi/2. + The function y=\sin(x)*\cos(x) has a parabolic shape shown in the + graph; it is drawn in the first quadrant. The function has two x + intercepts at x=0 and x = \pi/2. The area under the curve + between the x intercepts until the x axis is shaded. The + curve continues from the x intercepts under the x axis. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={1.57}, + extra x tick labels={$\frac{\pi}{2}$}, + ymin=-.6,ymax=1.1, + xmin=-.5,xmax=2 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1.57] {cos(deg(x))*sin(deg(x)} \closedcycle; + \addplot [firstcurvestyle,domain=-.5:2,samples=40] {cos(deg(x))*sin(deg(x)} node [pos=0.5, above right] { $y=\sin(x) \cos(x)$}; + + \end{axis} + \end{tikzpicture} + + + + + +
    + +
    + + + + + Graph of function y=u, showing shaded portion from 0 to 1. + + +

    + The y axis is drawn from -0.5 to 1 and the + u axis is drawn from 0 to \pi/2. The graph + of function y=u is a straight line with a positive slope + and passes through the origin. The area under the line is shaded + from 0 to 1 indicating the limit for the integral. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1}, + extra x ticks={1.57}, + extra x tick labels={$\frac{\pi}{2}$},ymin=-.6,ymax=1.1, + xmin=-.5,xmax=2, + xlabel={$u$} + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1] {x} \closedcycle; + \addplot [firstcurvestyle,domain=-.5:1.1] {x} node [pos=0.9, right] { $y=u$}; + + \end{axis} + \end{tikzpicture} + + + + +
    +     +
    +     + +     + +

    + Integration by substitution is a powerful and useful integration technique. + The next section introduces another technique, + called Integration by Parts. + As substitution undoes the Chain Rule, + integration by parts undoes the Product Rule. + Together, these two techniques provide a strong foundation on which most other integration techniques are based. +

    +     + + + + Terms and Concepts + + + +

    + Substitution undoes what derivative rule? +

    +

    + + +

    +     + + + + + chain|chain rule|the chain rule + + + + + +     + + + + +

    + + One can use algebra to rewrite the integrand of an integral to make it easier to evaluate. +

    +     + +     +     + + + Problems + + +

    + Evaluate the indefinite integral to develop an understanding of Substitution. +

    +     + + + + + $m = random(3,4,1); + $b = non_zero_random(-9,9,1); + $n = random(5,9,1); + if($envir{problemSeed}==1){$m=3;$b=-5;$n=7;}; + Context("Fraction"); + $u = Formula("x^$m+$b")->reduce; + $du = $u->D('x'); + $f = $du * Formula("($u)^$n"); + $F = FormulaUpToConstant("1/($n+1)*($u)^($n+1)"); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($m,$b) = random_subset(2,-9..-1,1..9); + $n = random(3,9,1); + if($envir{problemSeed}==1){$m=-5;$b=7;$n=3;}; + Context("Fraction"); + $u = Formula("x^2+$m x+$b")->reduce; + $du = $u->D('x'); + $f = $du * Formula("($u)^$n"); + $F = FormulaUpToConstant("1/($n+1)*($u)^($n+1)"); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + $n = random(3,9,1); + if($envir{problemSeed}==1){$b=1;$n=8;}; + Context("Fraction"); + $u = Formula("x^2+$b")->reduce; + $f = Formula("x ($u)^$n"); + $F = FormulaUpToConstant("1/(2*($n+1))*($u)^($n+1)"); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($a,$b,$c) = random_subset(3,-9..-1,1..9); + $n = random(3,9,1); + $s = random(2,4,1); + if($envir{problemSeed}==1){$a=3;$b=7;$c=-1;$n=5;$s=2;}; + Context("Fraction"); + $u = Formula("$a x^2 + $b x + $c")->reduce; + $f = Formula("(2*$a*$s x + $b*$s) ($u)^$n")->reduce; + $frac = Fraction($s,$n+1); + $F = FormulaUpToConstant("$frac*($u)^($n+1)"); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($m,$b) = random_subset(2,2..9); + if($envir{problemSeed}==1){$m=2;$b=7;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [-$b/$m-2,-$b/$m+2]}); + $f = Formula("1/($m x + $b)")->reduce; + $F = FormulaUpToConstant("1/$m ln(abs($m x + $b))")->reduce; + $F->{test_at} = [[-$b/$m-2],[-$b/$m+2]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($m,$b) = random_subset(2,2..9); + if($envir{problemSeed}==1){$m=2;$b=3;}; + Context("Fraction"); + $f = Formula("1/sqrt($m x + $b)")->reduce; + $frac = Fraction(2,$m); + $F = FormulaUpToConstant("$frac sqrt($m x + $b)")->reduce; + $F->{limits} = [-$b/$m,-$b/$m+4]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b=3;}; + Context("Fraction"); + $f = Formula("x/sqrt(x + $b)")->reduce; + $F = FormulaUpToConstant("2/3 (x - 2*$b) sqrt(x + $b)")->reduce; + $F->{limits} = [-$b,-$b+4]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + $m = random(3,5,1); + if($envir{problemSeed}==1){$m=3;$b=-1;}; + Context("Fraction"); + $f = Formula("(x^$m+$b x)/sqrt(x)")->reduce; + $frac = Fraction(2*$b,3); + $F = FormulaUpToConstant("x^(3/2) (2/(2*$m+1) x^($m-1) + $frac )")->reduce; + $F->{limits} = [0,4]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("2*e^sqrt(x)"); + $F->{limits} = [0,4]; + + +

    + \ds \int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + $m = random(4,8,1); + if($envir{problemSeed}==1){$m=5;$b=1;}; + Context("Fraction"); + $f = Formula("x^($m-1)/sqrt(x^$m+$b)")->reduce; + $frac = Fraction(2,$m); + $F = FormulaUpToConstant("$frac sqrt(x^$m+$b)")->reduce; + if ($b < 0) { + $absb = abs($b); + $F->{limits} = [$absb**(1/$m)+1,$absb**(1/$m)+4]; + } + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + $m = random(1,3,1); + if($envir{problemSeed}==1){$m=1;$b=1;}; + Context("Fraction"); + $f = Formula("(1/x^$m + $b)/x^($m+1)")->reduce; + $frac = Fraction(2,$m); + $F = FormulaUpToConstant("-1/(2*$m) (1/x^$m + $b)^2")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("ln(x)^2/2"); + + +

    + \ds \int \frac{\ln(x)}{x} dx +

    +

    + +

    +     +     +     +     + + + +

    + Use Substitution to evaluate the indefinite integral involving trigonometric functions. +

    +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=2;}; + Context("Fraction"); + $f = Formula("sin^$m(x) cos(x)")->reduce; + $F = FormulaUpToConstant("sin^($m+1)(x)/($m+1)")->reduce; + + +

    + \ds \int \sin^{}(x) \cos(x) \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=3;}; + Context("Fraction"); + $f = Formula("cos^$m(x) sin(x)")->reduce; + $F = FormulaUpToConstant("-cos^($m+1)(x)/($m+1)")->reduce; + + +

    + \ds \int \cos^{}(x) \sin(x) \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(-9,-1,1); + $b = random(1,9,1); + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$m=-6;$b=3;$trig='cos'}; + Context("Fraction"); + $f = Formula("$trig($b+$m x)")->reduce; + $trigAD =Formula("-$trig(x)")->D('x'); + $F = $trigAD->substitute(x=>Formula("$b+$m x")); + $F = FormulaUpToConstant("$F/$m")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(-9,-1,1); + $b = random(1,9,1); + if($envir{problemSeed}==1){$m=-1;$b=4;}; + Context("Fraction"); + $g = Formula("$b + $m x")->reduce; + $f = Formula("sec^2($b + $m x)")->reduce; + $F = FormulaUpToConstant("tan($b + $m x)/$m")->reduce; + + +

    + \ds \int \sec^2() \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=2;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [0,2*pi/$m]}); + $f = Formula("sec($m x)")->reduce; + $F = FormulaUpToConstant("1/$m ln(abs(sec($m x) + tan($m x)))")->reduce; + $F->{test_at} = [[pi/4/$m],[3*pi/4/$m],[5*pi/4/$m],[7*pi/4/$m]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=2;}; + Context("Fraction"); + $f = Formula("tan^$m(x) sec^2(x)")->reduce; + $F = FormulaUpToConstant("1/($m+1) tan^($m+1)(x)")->reduce; + + +

    + \ds \int \tan^{}(x)\sec^2(x) \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$m=2;$trig='cos'}; + Context("Fraction"); + $f = Formula("x^($m-1) $trig(x^$m)")->reduce; + $trigAD = Formula("-$trig(x)")->D('x'); + $F = $trigAD->substitute(x=>Formula("x^$m")); + $F = FormulaUpToConstant("1/($m) $F")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("tan(x)-x"); + + +

    + \ds \int \tan^2(x) dx +

    +

    + +

    +     +     +     + + + + + Context()->variables->set(x => {limits => [0,2*pi]}); + $F = FormulaUpToConstant("ln(abs(sin(x)))"); + $F->{test_at} = [[pi/4],[3*pi/4],[5*pi/4],[7*pi/4]]; + + +

    + \ds \int \cot(x) \, dx +

    +

    + Do not just refer to for the answer; + justify it through Substitution. +

    +

    + +

    +     +     +     + + + + + Context()->variables->set(x => {limits => [0,2*pi]}); + $F = FormulaUpToConstant("-ln(abs(csc(x) + cot(x)))"); + $F->{test_at} = [[pi/4],[3*pi/4],[5*pi/4],[7*pi/4]]; + + +

    + \ds \int \csc(x) \, dx +

    +

    + Do not just refer to for the answer; + justify it through Substitution. +

    +

    + +

    +     +     +     +     + + + +

    + Use Substitution to evaluate the indefinite integral involving exponential functions. +

    +     + + + + + $m = random(2,9,1); + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$m=3;$b=-1}; + Context("Fraction"); + $f = Formula("e^($m x+$b)")->reduce; + $F = FormulaUpToConstant("1/$m e^($m x+$b)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=3;}; + Context("Fraction"); + $f = Formula("e^(x^$m)x^($m-1)")->reduce; + $F = FormulaUpToConstant("1/$m e^(x^$m)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b=-1}; + Context("Fraction"); + Context()->variables->set(x => {limits => [-$b-2,-$b+2]}); + $f = Formula("e^(x^2 + 2*$b x + ($b)^2) (x + $b)")->reduce; + $F = FormulaUpToConstant("1/2 e^((x+$b)^2)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b=1}; + Context("Fraction"); + $f = Formula("(e^x+$b)/e^x")->reduce; + $F = FormulaUpToConstant("x - $b e^(-x)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(1,9,1); + if($envir{problemSeed}==1){$b=1}; + Context("Fraction"); + $f = Formula("e^x/(e^x+$b)")->reduce; + $F = FormulaUpToConstant("ln(e^x+$b)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $c = list_random(-1,1); + $m = random(2,5,1); + if($envir{problemSeed}==1){$c=-1;$m=2;}; + Context("Fraction"); + $f = Formula("(e^x + $c e^(-x))/e^($m x)")->reduce; + $F = FormulaUpToConstant("1/(1-$m) e^((1-$m)x) + $c/(-1-$m) e^((-1-$m)x)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = random(2,9,1); + if($envir{problemSeed}==1){$b=3;}; + Context("Fraction"); + $f = Formula("$b^($b x)")->reduce; + Context()->flags->set(reduceConstantFunctions=>0); + $F = FormulaUpToConstant("$b^($b x)/($b ln($b))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($b,$c) = random_subset(2,2..9); + if($envir{problemSeed}==1){$b=4;$c=2;}; + Context("Fraction"); + $f = Formula("$b^($c x)")->reduce; + Context()->flags->set(reduceConstantFunctions=>0); + $F = FormulaUpToConstant("$b^($c x)/($c ln($b))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Use Substitution to evaluate the indefinite integral involving logarithmic functions. +

    +     + + + + + + $F = FormulaUpToConstant("ln(x)^2/2"); + + +

    + \ds \int \frac{\ln(x)}{x} dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=2;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [0.1,10]}); + $f = Formula("(ln(x))^$m/x")->reduce; + $F = FormulaUpToConstant("(ln(x))^($m+1)/($m+1)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(3,9,2); + if($envir{problemSeed}==1){$m=3;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [0.1,10]}); + $f = Formula("ln(x^$m)/x")->reduce; + $frac = Fraction($m,2); + $F = FormulaUpToConstant("$frac (ln(x))^2")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,8,2); + if($envir{problemSeed}==1){$m=2;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [0.1,10]}); + $f = Formula("1/(x ln(x^$m))")->reduce; + $F = FormulaUpToConstant("1/$m ln(abs(ln(x^$m)))")->reduce; + $F->{test_at} = [[0.5],[2]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Use Substitution to evaluate the indefinite integral involving rational functions. +

    +     + + + + + ($a,$b) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$a=3;$b=1;}; + Context("Fraction"); + $f = Formula("(x^2 + $a x + $b)/x")->reduce; + $F = FormulaUpToConstant("x^2/2 + $a x + $b ln(abs(x))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + $f = Formula("(x^3 + x^2 + x + 1)/x")->reduce; + $F = FormulaUpToConstant("x^3/3 + x^2/2 + x + ln(abs(x))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $c = list_random(-1,1); + $b = non_zero_random(-9,9,1); + if ($b == $c) {$b = list_random(-9..-2,2..9);} + if($envir{problemSeed}==1){$c=1;$b=-1}; + Context("Fraction"); + Context()->variables->set(x => {limits => [-$c-2,-$c+2]}); + $f = Formula("(x^3 + $b)/(x + $c)")->reduce; + $frac = Fraction(3*$c,2); + $F = FormulaUpToConstant("1/3 (x+$c)^3 + $frac (x+$c)^2 + 3*($c)^2 (x+$c) + (($c)^3 + $b) ln(abs(x+$c))")->reduce; + $F->{test_at} = [[-$c-2],[-$c+2]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + do {($a,$b,$c) = random_subset(3,-9..-1,1..9);} until ((-$c)^2 + $a*(-$c) + $b != 0); + if($envir{problemSeed}==1){$a=2;$b=-5;$c=-3;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [-$c-2,-$c+2]}); + $f = Formula("(x^2 + $a x + $b)/(x + $c)")->reduce; + $F = FormulaUpToConstant("(x+$c)^2/2 + ($a-2*$c)(x+$c) + (($c)^2 - $a*$c + $b)ln(abs(x+$c))")->reduce; + $F->{test_at} = [[-$c-2],[-$c+2]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + do {($a,$b,$c,$d) = random_subset(4,-9..-1,1..9);} until ($a*(-$d)^2 + $b*(-$d) + $c != 0); + if($envir{problemSeed}==1){$a=3;$b=-5;$c=7;$d=1;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [-$d-2,-$d+2]}); + $f = Formula("($a x^2 + $b x + $c)/(x + $d)")->reduce; + $frac = Fraction($a,2); + $F = FormulaUpToConstant("$frac (x+$d)^2 + ($b - 2*$a*$d)(x+$d) + ($a*($d)^2 - $b*$d + $c)ln(abs(x+$d))")->reduce; + $F->{test_at} = [[-$d-2],[-$d+2]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($a,$b,$c) = random_subset(3,-9,-6,-3,0,3,6,9); + if($envir{problemSeed}==1){$a=3;$b=3;$c=0;}; + Context("Fraction"); + $denom = Formula("x^3 + $a x^2 + $b x + $c")->reduce; + $A = 2*$a/3; + $B = $b/3; + $numer = Formula("x^2 + $A x + $B")->reduce; + $f = Formula("($numer)/($denom)")->reduce; + $F = FormulaUpToConstant("1/3 ln(abs($denom))")->reduce; + $bound = abs($c)+abs($b)+abs($a); # be sure to test where $denom is pos and neg + $F->{test_at} = [[-1],[1],[-$bound-1],[$bound+1]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Use Substitution to evaluate the indefinite integral involving inverse trigonometric functions. +

    +     + + + + + $a = list_random(2,3,5,6,7,11,13,14,15,17); + if($envir{problemSeed}==1){$a=7;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$a/(x^2+$a)")->reduce; + $F = FormulaUpToConstant("sqrt($a) atan(x/sqrt($a))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + if($envir{problemSeed}==1){$a=3;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [-$a,$a]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$a/sqrt($a^2-x^2)")->reduce; + $F = FormulaUpToConstant("$a asin(x/$a)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = list_random(2,3,5,6,7,10,11,13,14,15); + $b = random(2,20,1); + if($envir{problemSeed}==1){$a=5;$b=14;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [-sqrt($a),sqrt($a)]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$b/sqrt($a-x^2)")->reduce; + $F = FormulaUpToConstant("$b asin(x/sqrt($a))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($a,$b) = random_subset(2,2..9); + if($envir{problemSeed}==1){$a=3;$b=2;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [$a,$a+10]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$b/(x sqrt(x^2-$a^2))")->reduce; + $frac = Fraction($b,$a); + $F = FormulaUpToConstant("$frac asec(abs(x)/$a)")->reduce; + $F->{test_at} = [[-$a-1],[-$a-2],[-$a-3],[-$a-4]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($a,$b) = random_subset(2,2..9); + if($envir{problemSeed}==1){$a=4;$b=5;}; + Context("Fraction"); + Context()->variables->set(x => {limits => [$a,$a+10]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("($b x)/sqrt(x^6 - ($a)^2 x^4)")->reduce; + $frac = Fraction($b,$a); + $F = FormulaUpToConstant("$frac asec(abs(x)/$a)")->reduce; + $F->{test_at} = [[-$a-1],[-$a-2],[-$a-3],[-$a-4]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + Context()->variables->set(x => {limits => [-1,1]}); + $F = FormulaUpToConstant("1/2*asin(x^2)"); + + +

    + \ds \int \frac{x}{\sqrt{1-x^4}} dx +

    +

    + +

    +     +     +     + + + + + $a = non_zero_random(-9,9,1); + $b = list_random(2,3,5,6,7,10,11,13,14,15); + if($envir{problemSeed}==1){$a=-1;$b=7;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("1/(x^2 + (2*$a) x + (($a)^2+$b))")->reduce; + $F = FormulaUpToConstant("1/sqrt($b) atan((x+$a)/$b)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = non_zero_random(-9,9,1); + ($b,$c) = random_subset(2,2..9); + if($envir{problemSeed}==1){$a=-3;$b=4;$c=2}; + Context("Fraction"); + Context()->noreduce('(-x)-y','(-x)+y'); + Context()->variables->set(x => {limits => [-$a-$b,-$a+$b]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$c/sqrt(-x^2 - (2*$a) x + (($b)^2-($a)^2))")->reduce; + $F = FormulaUpToConstant("$c asin((x+$a)/$b)")->reduce; + $F->{limits} = [-$b-$a,$b-$a]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = non_zero_random(-9,9,1); + ($b,$c) = random_subset(2,2..9); + if($envir{problemSeed}==1){$a=-4;$b=5;$c=3}; + Context("Fraction"); + Context()->noreduce('(-x)-y','(-x)+y'); + Context()->variables->set(x => {limits => [-$a-$b,-$a+$b]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$c/sqrt(-x^2 - (2*$a) x + (($b)^2-($a)^2))")->reduce; + $F = FormulaUpToConstant("$c asin((x+$a)/$b)")->reduce; + $F->{limits} = [-$b-$a,$b-$a]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = non_zero_random(-9,9,1); + $b = random(2,9,1); + if($envir{problemSeed}==1){$a=3;$b=5}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$b/(x^2+(2*$a)x+(($a)^2+($b)^2))")->reduce; + $F = FormulaUpToConstant("atan((x+$a)/$b)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Evaluate the indefinite integral. +

    +     + + + + + $m = random(3,9,1); + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$m=3;$b=3}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("x^($m-1)/(x^$m + $b)^2")->reduce; + $F = FormulaUpToConstant("-1/($m(x^$m + $b))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(1,9,1); + ($b,$c) = random_subset(2,-9..-1,1..9); + ($n,$m) = num_sort(random_subset(2,2..6)); + $p = random(4,9,1); + if($envir{problemSeed}==1){$a=5;$b=5;$c=2;$m=3;$n=2;$p=8;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $u = Formula("$a x^$m + $b x^$n + $c")->reduce; + $g = gcd($a*$m,$b*$n); + $A = $a*$m/$g; + $B = $b*$n/$g; + $f = Formula("($A x^($m-1) + $B x^($n-1))($u)^$p")->reduce; + $F = FormulaUpToConstant("1/(($p+1)*$g) ($u)^($p+1)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(1,9,1); + $c = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$a=1;$c=-1;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("x/sqrt($a + $c x^2)")->reduce; + $frac = Fraction(1,$c); + $F = FormulaUpToConstant("$frac sqrt($a + $c x^2)")->reduce; + if ($c < 0) { + $F->{limits} = [-sqrt(-$a/$c),sqrt(-$a/$c)]; + }; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(3,9,1); + $c = non_zero_random(-9,9,1); + $trig = list_random('tan','-cot'); + if($envir{problemSeed}==1){$m=3;$c=1;$trig='-cot'}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $dt = ($trig eq 'tan') ? '\sec^2' : '\csc^2'; + $h = Formula("x^$m+$c")->reduce; + $g = Formula("$trig(x)")->D('x')->reduce->substitute(x=>$h)->reduce; + $f = Formula("x^($m-1) $g")->reduce; + $F = FormulaUpToConstant("$trig(x^$m+$c)")->reduce; + + +

    + \ds \int x^{} \left(\right) \, dx +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','-cos'); + if($envir{problemSeed}==1){$trig='-cos'}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $g = Formula("$trig(x)")->D('x')->reduce; + $f = ($trig eq 'sin') ? Formula("cos(x) sqrt(sin(x))") : Formula("sin(x) sqrt(cos(x))"); + $F = FormulaUpToConstant("2/3*$trig(x)^(3/2)")->reduce; + $F->{limits} = [0,pi/2]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + $b = non_zero_random(-9,9,1); + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$trig='sin';$m=5;$b=1;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$trig($m x + $b)")->reduce; + $F = Formula("-$trig(x)")->D('x')->reduce->substitute(x=>Formula("$m x + $b")); + $F = FormulaUpToConstant("1/$m $F")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b=-5}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + Context()->variables->set(x => {limits => [-$b-2,-$b+2]}); + $f = Formula("1/(x+$b)")->reduce; + $F = FormulaUpToConstant("ln(abs(x+$b))")->reduce; + $F->{test_at} = [[-$b-2],[-$b+2]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($a,$b,$c) = random_subset(3,1..9); + if($envir{problemSeed}==1){$a=7;$b=3;$c=2;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + Context()->variables->set(x => {limits => [-$c/$b-2,-$c/$b+2]}); + $f = Formula("$a/($b x+$c)")->reduce; + $frac = Fraction($a,$b); + $F = FormulaUpToConstant("$frac ln(abs($b x+$c))")->reduce; + $F->{test_at} = [[-$c/$b-2],[-$c/$b+2]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + do {($b,$c) = random_subset(2,-9..-1,1..9)} until (($b)**2 != 4*$c); + $m = random(2,4,1); + $k = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b=3;$c=5;$m=3;$k=-5;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + if (($b)**2 > 4*$c) { + ($min, $max) = num_sort((-$b - sqrt(($b)**2 - 4*$c))/2,(-$b + sqrt(($b)**2 - 4*$c))/2); + $D = sqrt(($b)**2 - 4*$c); + Context()->variables->set(x => {limits => [$min-$D,$max+$D]}); + }; + $denom = Formula("x^2 + $b x + $c")->reduce; + $numer = Formula("$m x^3 + ($m*$b + $k)x^2 + ($m*$c + $b*$k + 2) x + ($c*$k + $b)")->reduce; + $f = Formula("($numer)/($denom)"); + $frac = Fraction($m,2); + $F = FormulaUpToConstant("$frac x^2 + $k x + ln(abs($denom))")->reduce; + if (($b)**2 > 4*$c) { + $F->{test_at} = [[$min-$D],[$max+$D]]; + }; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + do {($b,$c) = random_subset(2,-9..-1,1..9)} until (($b)**2 != 4*$c); + if($envir{problemSeed}==1){$b=7;$c=3;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + if (($b)**2 > 4*$c) { + ($min, $max) = num_sort((-$b - sqrt(($b)**2 - 4*$c))/2,(-$b + sqrt(($b)**2 - 4*$c))/2); + $D = sqrt(($b)**2 - 4*$c); + Context()->variables->set(x => {limits => [$min-$D,$max+$D]}); + }; + $denom = Formula("x^2 + $b x + $c")->reduce; + $numer = Formula("2x + $b")->reduce; + $f = Formula("($numer)/($denom)"); + $F = FormulaUpToConstant("ln(abs($denom))")->reduce; + if (($b)**2 > 4*$c) { + $F->{test_at} = [[$min-$D],[$max+$D]]; + }; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + do {($a,$b,$c) = random_subset(3,-9..-1,1..9)} until (($b)**2 != 4*$a*$c); + $k = random(2,5,1); + if($envir{problemSeed}==1){$a=3;$b=9;$c=7;$k=3;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + if (($b)**2 > 4*$a*$c) { + ($min, $max) = num_sort((-$b - sqrt(($b)**2 - 4*$a*$c))/2,(-$b + sqrt(($b)**2 - 4*$a*$c))/2); + $D = sqrt(($b)**2 - 4*$a*$c); + Context()->variables->set(x => {limits => [$min-$D,$max+$D]}); + }; + $denom = Formula("$a x^2 + $b x + $c")->reduce; + $g = gcd(2*$a,$b); + $A = 2*$a/$g; + $B = $b/$g; + $numer = Formula("$g*$k ($A x + $B)")->reduce; + $f = Formula("($numer)/($denom)"); + $F = FormulaUpToConstant("$k ln(abs($denom))")->reduce; + if (($b)**2 > 4*$c) { + $F->{test_at} = [[$min-$D],[$max+$D]]; + }; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + do {($b,$c) = random_subset(2,-9..-1,1..9)} until (($b)**2 != 4*$c); + $k = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b=-7;$c=1;$k=7;}; + $m = -1; + Context("Fraction"); + Context()->noreduce('(-x)-y','(-x)+y'); + Context()->flags->set(reduceConstantFunctions=>0); + if (($b)**2 > 4*$c) { + ($min, $max) = num_sort((-$b - sqrt(($b)**2 - 4*$c))/2,(-$b + sqrt(($b)**2 - 4*$c))/2); + $D = sqrt(($b)**2 - 4*$c); + Context()->variables->set(x => {limits => [$min-$D,$max+$D]}); + }; + $denom = Formula("x^2 + $b x + $c")->reduce; + $numer = Formula("$m x^3 + ($m*$b + $k)x^2 + ($m*$c + $b*$k + 2) x + ($c*$k + $b)")->reduce; + $f = Formula("($numer)/($denom)"); + $frac = Fraction($m,2); + $F = FormulaUpToConstant("$frac x^2 + $k x + ln(abs($denom))")->reduce; + if (($b)**2 > 4*$c) { + $F->{test_at} = [[$min-$D],[$max+$D]]; + }; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + if($envir{problemSeed}==1){$a=9;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("x/(x^4+$a^2)")->reduce; + $F = FormulaUpToConstant("1/(2*$a) atan(x^2/$a)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + if($envir{problemSeed}==1){$a=2;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$a/($a^2x^2+1)")->reduce; + $F = FormulaUpToConstant("atan($a x)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + if($envir{problemSeed}==1){$a=2;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + Context()->variables->set(x => {limits => [1/$a,1/$a+4]}); + $f = Formula("1/(x sqrt($a^2x^2-1))")->reduce; + $F = FormulaUpToConstant("asec(abs($a x))")->reduce; + $F->{test_at} = [[-1/$a-1],[-1/$a-2],[-1/$a-3],[-1/$a-4]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($a,$b) = random_subset(2,2..9); + if($envir{problemSeed}==1){$a=4;$b=3;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + Context()->variables->set(x => {limits => [-$a/$b,$a/$b]}); + $f = Formula("1/sqrt($a^2 - $b^2 x^2)")->reduce; + $F = FormulaUpToConstant("1/$b asin($b x/$a)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = non_zero_random(-9,9,1); + $b = random(2,9,1); + $c = non_zero_random(-9,9,1); + $k = random(2,9,1); + if($envir{problemSeed}==1){$a=-1;$b=3;$c=1;$k = 3;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $denom = Formula("x^2 + 2*$a x + (($a)^2 + $b^2)")->reduce; + $numer = Formula("$k x + ($k*$a + $c)")->reduce; + $f = Formula("($numer)/($denom)")->reduce; + $frac = Fraction($k,2); + $frac2 = Fraction($c,$b); + $F = FormulaUpToConstant("$frac ln(abs($denom)) + $frac2 atan((x+$a)/$b)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = non_zero_random(-9,9,1); + $b = random(2,9,1); + $c = non_zero_random(1,19,1); + $k = random(2,9,1); + if($envir{problemSeed}==1){$a=6;$b=5;$c=19;$k=-2;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $denom = Formula("x^2 + 2*$a x + (($a)^2 + $b^2)")->reduce; + $numer = Formula("$k x + ($k*$a + $c)")->reduce; + $f = Formula("($numer)/($denom)")->reduce; + $frac = Fraction($k,2); + $frac2 = Fraction($c,$b); + $F = FormulaUpToConstant("$frac2 atan((x+$a)/$b) + $frac ln(abs($denom))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = non_zero_random(-9,9,1); + $b = list_random(2,3,5,6,7,10,11,13,14,15); + $c = non_zero_random(1,50,1); + $k = random(2,20,1); + if($envir{problemSeed}==1){$a=-5;$b=7;$c=41;$k=15;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $denom = Formula("x^2 + 2*$a x + (($a)^2 + $b)")->reduce; + $numer = Formula("x^2 + (2*$a+$k) x + (($a)^2 + $b + $k*$a + $c)")->reduce; + $f = Formula("($numer)/($denom)")->reduce; + $frac = Fraction($k,2); + $F = FormulaUpToConstant("x+$c/sqrt($b) atan((x+$a)/sqrt($b)) + $frac ln(abs($denom))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = random(2,9,1); + if($envir{problemSeed}==1){$b=3;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("x^3/(x^2 + $b^2)")->reduce; + $frac = Fraction($b**2,2); + $F = FormulaUpToConstant("x^2/2 - $frac ln(abs(x^2+$b^2))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = non_zero_random(-9,9,1); + $b = list_random(2,3,5,6,7,10,11,13,14,15); + $c = non_zero_random(1,50,1); + $k = random(2,20,1); + if($envir{problemSeed}==1){$a=2;$b=5;$c=24;$k=6;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $denom = Formula("x^2 + 2*$a x + (($a)^2 + $b)")->reduce; + $numer = Formula("x^3 + ($b - 3*($a)^2 + $k)x + ((($a)^2+$b) (-2*$a) + $a*$k + $c)")->reduce; + $f = Formula("($numer)/($denom)")->reduce; + $frac = Fraction($k,2); + $F = FormulaUpToConstant("1/2 x^2 - (2*$a) x + $frac ln(abs($denom)) + $c/sqrt($b) atan((x+$a)/sqrt($b))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("-atan(cos(x))"); + + +

    + \ds \int \frac{\sin(x)}{\cos^2(x)+1} dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("atan(sin(x))"); + + +

    + \ds \int \frac{\cos(x)}{\sin^2(x)+1} dx +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$trig='sin'}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $num = ($trig eq 'cos') ? 'sin' : 'cos'; + $numtex = '\'.$num; + $trigtex = '\'.$trig; + $f = Formula("$num(x)/(1 - $trig^2(x))")->reduce; + $F = FormulaUpToConstant("ln(abs(sec(x) + tan(x)))")->reduce; + if ($trig eq 'cos') { + $F = FormulaUpToConstant("-ln(abs(csc(x) + cot(x)))")->reduce; + }; + $F->{test_at} = [[pi/4],[3*pi/4],[5*pi/4],[7*pi/4]]; + + +

    + \ds \int \frac{(x)}{1-^2(x)} \, dx +

    +

    + +

    +     +     +     + + + + + $h = non_zero_random(-9,9,1); + $k = random(-9,-1,1); + $c = random(3,9,1); + if($envir{problemSeed}==1){$h=1;$k=-7;$c=3;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + Context()->variables->set(x => {limits => [$h-sqrt(-$k),$h+sqrt(-$k)]}); + $numer = Formula("$c x - $c*$h")->reduce; + $quad = Formula("x^2 - (2*$h) x + (($h)^2+$k)")->reduce; + $f = Formula("($numer)/(sqrt($quad))")->reduce; + $F = FormulaUpToConstant("$c sqrt($quad)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $h = non_zero_random(-9,9,1); + $k = random(-9,-1,1); + if($envir{problemSeed}==1){$h=3;$k=-1;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + Context()->variables->set(x => {limits => [$h-sqrt(-$k),$h+sqrt(-$k)]}); + $numer = Formula("x - $h")->reduce; + $quad = Formula("x^2 - (2*$h) x + (($h)^2+$k)")->reduce; + $f = Formula("($numer)/(sqrt($quad))")->reduce; + Context()->variables->set(x => {limits => [sqrt(-$k)+$h,sqrt(-$k)+$h+4]}); + $F = FormulaUpToConstant("sqrt($quad)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Evaluate the definite integral. +

    +     + + + + + $b = non_zero_random(-9,9,1); + ($lb,$ub) = num_sort(random_subset(2,-$b-9..-$b-1)); + if($envir{problemSeed}==1){$b=-5;$lb=1;$ub=3}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("1/(x+$b)")->reduce; + $frac = Fraction($ub+$b,$lb+$b); + $A = Formula("ln($frac)"); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + ($lb,$ub) = num_sort(random_subset(2,0,1,4,9,16,25,36,49,64,81)); + $lb = $lb - $b; + $ub = $ub - $b; + if($envir{problemSeed}==1){$b=-2;$lb=2;$ub=6}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("x sqrt(x+$b)")->reduce; + $fracu = Fraction(2*(($ub+$b)/5 - $b/3)); + $fracl = Fraction(2*(($lb+$b)/5 - $b/3)); + $A = Fraction($fracu*($ub + $b)**(3/2) - $fracl*($lb + $b)**(3/2)); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','cos'); + $lb = list_random('-pi','-pi/2'); + $ub = list_random('pi','pi/2'); + $n = random(2,5,1); + if($envir{problemSeed}==1){$trig='sin';$lb='-pi/2';$ub='pi/2';$n=2;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("sin^$n(x) cos(x)"); + $ftex = '\sin^' . $n . '(x)\cos(x)'; + $A = Fraction(1/($n+1)*sin(Real($ub))**($n+1) - 1/($n+1)*sin(Real($lb))**($n+1)); + if ($trig eq 'cos') { + $f = Formula("cos^$n(x) sin(x)"); + $ftex = '\cos^' . $n . '(x)\sin(x)'; + $A = Fraction(-1/($n+1)*cos(Real($ub))**($n+1) + 1/($n+1)*cos(Real($lb))**($n+1)); + } + $lb = Formula("$lb"); + $ub = Formula("$ub"); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $n = random(3,9,1); + if($envir{problemSeed}==1){$n=4;}; + $lb = 0; + $ub = 1; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("2x (1 - x^2)^$n"); + $A = Fraction(-1/($n+1)*(1 - $ub**2)**($n+1) + 1/($n+1)*(1 - $lb**2)**($n+1)); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + ($lb,$ub) = num_sort(random_subset(2,-$b-2..-$b+2)); + if($envir{problemSeed}==1){$b=1;$lb=-2;$ub=-1}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("(x+$b) e^(x^2 + 2*$b x + ($b)^2)")->reduce; + $A = Formula("1/2 (e^(($ub + $b)^2) - e^(($lb + $b)^2))")->reduce; + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $A = Formula("pi/2")->reduce; + + +

    + \ds \int_{-1}^{1} \frac{1}{1+x^2} \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b=-3;}; + $lb = -$b-1; + $ub = -$b+1; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("1/(x^2 + 2*$b x + (($b)^2+1))")->reduce; + $A = Formula("pi/2")->reduce; + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + ($lbindex,$ubindex) = num_sort(random_subset(2,0..8)); + if($envir{problemSeed}==1){$lbindex=5;$ubindex=7;}; + ($lb,$ub) = (-2,'-sqrt(3)','-sqrt(2)',-1,0,1,'sqrt(2)','sqrt(3)',2)[$lbindex,$ubindex]; + Context()->flags->set(reduceConstantFunctions=>0); + $lb = Formula("$lb"); + $ub = Formula("$ub"); + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("1/sqrt(4 - x^2)")->reduce; + $A = Real("arcsin($ub/2) - arcsin($lb/2)"); + $frac = Fraction($A/pi); + $A = Formula("$frac pi")->reduce; + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     +     +     +     +
    +
    + Integration by Parts +

    + Here's a simple integral that we can't yet evaluate: + + \int x\cos(x) \,dx + . +

    + +

    + It's a simple matter to take the derivative of the integrand using the Product Rule, + but there is no Product Rule for integrals. + However, this section introduces + Integration by Parts, + a method of integration that is based on the Product Rule for derivatives. + It will enable us to evaluate this integral. +

    + + + +

    + The Product Rule says that if u and v are functions of x, + then (uv)' = u'v + uv'. + For simplicity, + we've written u for u(x) and v for v(x). + Suppose we integrate both sides with respect to x. + This gives + + \int (uv)'\,dx = \int (u'v+uv')\,dx + . +

    + +

    + By the Fundamental Theorem of Calculus, + the left side integrates to uv. + The right side can be broken up into two integrals, and we have + + uv = \int u'v\,dx + \int uv'\,dx + . +

    + +

    + Solving for the second integral we have + + \int uv'\,dx = uv - \int u'v\,dx + . +

    + +

    + Using differential notation, + we can write du = u'(x)dx and + dv=v'(x)dx and the expression above can be written as follows: + + \int u\,dv = uv - \int v\,du + . +

    + +

    + This is the Integration by Parts formula. + For reference purposes, we state this in a theorem. +

    + + + Integration by Parts + +

    + Let u and v be differentiable functions of + x on an interval I containing a and b. + Then + + \int u\, dv = uv - \int v\, du + , + and integrationby parts + + \int_{x=a}^{x=b} u\, dv = uv\Big|_a^b - \int_{x=a}^{x=b}v\, du + . +

    +     +     + + + + + +

    + Let's try an example to understand our new technique. +

    + + + Integrating using Integration by Parts + +

    + Evaluate \ds\int x\cos(x) \, dx. +

    +     + +

    + The key to Integration by Parts is to identify part of the integrand as + u and part as dv. + Regular practice will help one make good identifications, + and later we will introduce some principles that help. + For now, let u=x and dv=\cos(x) \, dx. +

    + +

    + It is generally useful to make a small table of these values as done below. + Right now we only know u and dv as shown on the left of + ; + on the right we fill in the rest of what we need. + If u = x, then du = dx. + Since dv = \cos(x)\, dx, + v is an antiderivative of \cos(x). + We choose v = \sin(x). +

    + +
    + Setting up Integration by Parts + +

    + + u\amp = x \amp v\amp =\mathord{?} + du\amp = \mathord{?} \amp dv\amp =\cos(x) \, dx + +

    + +

    + \implies +

    + +

    + + u\amp = x \amp v\amp =\sin(x) + du\amp = dx \amp dv\amp =\cos(x) \, dx + +

    +     +
    + +

    + Now substitute all of this into the Integration by Parts formula, giving + + \int x\cos(x)\,dx = x\sin(x) - \int \sin(x) \,dx + . +

    + +

    + We can then integrate \sin(x) to get + -\cos(x) + C and overall our answer is + + \int x\cos(x)\, dx = x\sin(x) + \cos(x) + C + . +

    + +

    + Note how the antiderivative contains a product, x\sin(x). + This product is what makes Integration by Parts necessary. +

    + +

    + We can check our work by taking the derivative: + + \lzoo{x}{x\sin(x) + \cos(x) + C} \amp =x\cos(x)+\sin(x)-\sin(x)+0 + \amp = x\cos(x) + . +

    +     + +     + +

    + You may wonder what would have happened in + if we had chosen our u and dv differently. + If we had chosen u=\cos(x) and + dv=x \, dx then du=-\sin(x)\, dx and v=x^2/2. + Our second integral is not simpler than the first; + we would have + + \int x\cos(x)\,dx=\cos(x)\frac{x^2}{2}-\int \frac{x^2}{2}\left(-\sin(x)\right)\,dx + . + The only way to approach this second integral would be yet another integration by parts. +

    + +

    + + demonstrates how Integration by Parts works in general. + We try to identify u and dv in the integral we are given, + and the key is that we usually want to choose u and dv + so that du is simpler than u and v + is hopefully not too much more complicated than dv. + This will mean that the integral on the right side of the Integration by Parts formula, + \int v\,du will be simpler to integrate than the original integral \int u\,dv. +

    + +

    + In the example above, we chose u=x and dv=\cos(x)\,dx. + Then du=dx was simpler than u and + v=\sin(x) is no more complicated than dv. + Therefore, instead of integrating x\cos(x) \,dx, + we could integrate \sin(x)\,dx, which we knew how to do. +

    + +

    + A useful mnemonic for helping to determine u is LIATE, where +

    + +

    + L = Logarithmic, I = Inverse Trig., A = + Algebraic (polynomials, roots, + power functions), T = Trigonometric, + and E = Exponential. +

    + +

    + If the integrand contains both a logarithmic and an algebraic term, + in general letting u be the logarithmic term works best, + as indicated by L coming before A in LIATE. +

    + + + +

    + We now consider another example. +

    + + + Integrating using Integration by Parts + +

    + Evaluate \displaystyle \int x e^x\,dx. +

    +     + +

    + The integrand contains an Algebraic term (x) and an + Exponential term + (e^x). + Our mnemonic suggests letting u be the algebraic term, + so we choose u=x and dv=e^x\,dx. + Then du=dx and v=e^x as indicated by the tables below. +

    + +
    + Setting up Integration by Parts + +

    + + u\amp = x \amp v\amp =\mathord{?} + du\amp = \mathord{?} \amp dv\amp =e^x \, dx + +

    + +

    + \implies +

    + +

    + + u\amp = x \amp v\amp =e^x + du\amp = dx \amp dv\amp =e^x \, dx + +

    +     +
    + +

    + We see du is simpler than u, + while there is no change in going from dv to v. + This is good. + The Integration by Parts formula gives + + \int x e^x\,dx = xe^x - \int e^x\,dx + . +

    + +

    + The integral on the right is simple; our final answer is + + \int xe^x\, dx = xe^x - e^x + C + . +

    + +

    + Note again how the antiderivatives contain a product term. +

    +     +     + + + Integrating using Integration by Parts + +

    + Evaluate \displaystyle \int x^2\cos(x) \,dx. +

    +     + +

    + The mnemonic suggests letting u=x^2 instead of the trigonometric function, + hence dv=\cos(x)\,dx. + Then du=2x\,dx and v=\sin(x) as shown below. +

    + +
    + Setting up Integration by Parts + +

    + + u\amp = x^2 \amp v\amp =\mathord{?} + du\amp = \mathord{?} \amp dv\amp =\cos(x) \, dx + +

    + +

    + \implies +

    + +

    + + u\amp = x^2 \amp v\amp =\sin(x) + du\amp = 2x\, dx \amp dv\amp =\cos(x) \, dx + +

    +     +
    + +

    + The Integration by Parts formula gives + + \int x^2\cos(x)\,dx = x^2\sin(x) - \int 2x\sin(x)\,dx + . +

    + +

    + At this point, + the integral on the right is indeed simpler than the one we started with, + but to evaluate it, we need to do Integration by Parts again. + Here we choose r=2x and + ds=\sin(x) and fill in the rest below. + (We are choosing new names since we have already used u and v. + Our integration by parts formula is now \int r\,ds=rs-\int s\,dr.) +

    + +
    + Setting up Integration by Parts (again) + +

    + + u\amp = 2x \amp v\amp =\text{?} + du\amp = \text{?} \amp dv\amp =\sin(x)\, dx + +

    + +

    + \Rightarrow +

    + +

    + + u\amp = 2x \amp v\amp =-\cos(x) + du\amp = 2\, dx \amp dv\amp =\sin(x)\, dx + +

    +     +
    + +

    + + \int x^2\cos(x)\,dx = x^2\sin(x) - \left(-2x\cos(x) - \int -2\cos(x)\,dx\right) + . +

    + +

    + The integral all the way on the right is now something we can evaluate. + It evaluates to -2\sin(x). + Then going through and simplifying, + being careful to keep all the signs straight, our answer is + + \int x^2\cos(x)\, dx = x^2\sin(x) + 2x\cos(x) - 2\sin(x) + C + . +

    +     + +     + + + Integrating using Integration by Parts + +

    + Evaluate \displaystyle \int e^x\cos(x) \,dx. +

    +     + +

    + This is a classic problem. + Our mnemonic suggests letting u + be the trigonometric function instead of the exponential. + In this particular example, + one can let u be either \cos(x) or e^x; + to demonstrate that we do not have to follow LIATE, we choose + u=e^x and hence dv = \cos(x)\,dx. + Then du=e^x\,dx and v=\sin(x) as shown below. +

    + +
    + Setting up Integration by Parts + +

    + + u\amp = e^x \amp v\amp =\mathord{?} + du\amp = \mathord{?} \amp dv\amp =\cos(x) \, dx + +

    + +

    + \implies +

    + +

    + + u\amp = e^x \amp v\amp =\sin(x) + du\amp = e^x \amp dv\amp =\cos(x) \, dx + +

    +     +
    + +

    + Notice that du is no simpler than u, + going against our general rule + (but bear with us). + The Integration by Parts formula yields + + \int e^x\cos(x)\, dx = e^x\sin(x) - \int e^x\sin(x)\,dx + . +

    + +

    + The integral on the right is not much different than the one we started with, + so it seems like we have gotten nowhere. + Let's keep working and apply Integration by Parts to the new integral, + using u=e^x and dv = \sin(x)\,dx. + This leads us to the following: +

    + +
    + Setting up Integration by Parts (again) + +

    + + r\amp = e^x \amp s\amp =\mathord{?} + dr\amp = \mathord{?} \amp ds\amp =\sin(x) \, dx + +

    + +

    + \implies +

    + +

    + + r\amp = e^x \amp s\amp =-\cos(x) + dr\amp = e^x\, dx \amp ds\amp =\sin(x) \, dx + +

    +     +
    + +

    + The Integration by Parts formula then gives: + + \int e^x\cos(x)\,dx \amp = e^x\sin(x) - \left(-e^x\cos(x) - \int -e^x\cos(x)\,dx\right) + \amp = e^x\sin(x)+ e^x\cos(x) - \int e^x\cos(x)\, dx + . +

    + +

    + It seems we are back right where we started, + as the right hand side contains \int e^x\cos(x)\,dx. + But this is actually a good thing. +

    + +

    + Add \ds\int e^x\cos(x)\, dx to both sides. + This gives + + 2\int e^x\cos(x) \, dx \amp = e^x\sin(x) + e^x\cos(x) + Now divide both sides by 2 and then add the integration constant: + \int e^x\cos(x) \, dx \amp = \frac{1}{2}\big(e^x\sin(x) + e^x\cos(x) \big)+C + . +

    + +

    + Simplifying a little, our answer is thus + + \int e^x\cos(x)\, dx = \frac12e^x\left(\sin(x) + \cos(x)\right)+C + . +

    +     + +     + + + Integrating using Integration by Parts: antiderivative of <m>\ln(x)</m> + +

    + Evaluate \int \ln(x)\,dx. +

    +     + +

    + One may have noticed that we have rules for integrating the familiar trigonometric functions and e^x, + but we have not yet given a rule for integrating \ln(x). + That is because \ln(x) can't easily be integrated with any of the rules we have learned up to this point. + But we can find its antiderivative by a clever application of Integration by Parts. + Set u=\ln(x) and dv=dx. + This is a good, + sneaky trick to learn as it can help in other situations. + This determines du=(1/x)\,dx and v=x as shown below. +

    + +
    + Setting up Integration by Parts + +

    + + u\amp = \ln(x) \amp v\amp =\mathord{?} + du\amp = \mathord{?} \amp dv\amp =1 \, dx + +

    + +

    + \implies +

    + +

    + + u\amp = \ln(x) \amp v\amp =x + du\amp = 1/x\, dx \amp dv\amp =1 \, dx + +

    +     +
    + +

    + Putting this all together in the Integration by Parts formula, + things work out very nicely: + + \int \ln(x)\,dx = x\ln(x) - \int x\,\frac1x\,dx + . +

    + +

    + The new integral simplifies to \int 1\,dx, + which is about as simple as things get. + Its integral is x+C and our answer is + + \int \ln(x) \, dx = x\ln(x) - x + C + . +

    +     + +     + + + Integrating using Int. by Parts: antiderivative of <m>\arctan x</m> + +

    + Evaluate \displaystyle \int \arctan x \,dx. +

    +     + +

    + The same sneaky trick we used above works here. + Let u=\arctan x and dv=dx. + Then du=1/(1+x^2)\,dx and v=x. + The Integration by Parts formula gives + + \int \arctan x \,dx = x\arctan x - \int \frac x{1+x^2}\,dx + . +

    + +

    + The integral on the right can be solved by substitution. + Taking w=1+x^2, we get dw=2x\,dx. + The integral then becomes + + \int \arctan x \,dx = x\arctan x - \frac12\int \frac 1{w}\,dw + . +

    + +

    + The integral on the right evaluates to \ln\abs{w}+C, + which becomes \ln(1+x^2)+C + (we can drop the absolute values as 1+x^2 is always positive). + Therefore, the answer is + + \int \arctan x\, dx = x\arctan x - \frac12\ln(1+x^2) + C + . +

    +     + +     + + + + + Substitution Before Integration +

    + When taking derivatives, it was common to employ multiple rules + (such as using both the Quotient and the Chain Rules). + It should then come as no surprise that some integrals are best evaluated by combining integration techniques. + In particular, here we illustrate making an unusual + substitution first before using Integration by Parts. +

    + + + Integration by Parts after substitution + +

    + Evaluate \ds \int \cos(\ln(x))\, dx. +

    +     + +

    + The integrand contains a composition of functions, + leading us to think Substitution would be beneficial. + Letting u=\ln(x), we have du = 1/x\, dx. + This seems problematic, as we do not have a 1/x in the integrand. + But consider: + + du = \frac 1x\, dx \quad \Rightarrow \quad x\cdot du = dx + . +

    + +

    + Since u = \ln(x), + we can use inverse functions and conclude that x = e^u. + Therefore we have that + + dx \amp = x\cdot du + \amp = e^u\, du + . +

    + +

    + We can thus replace \ln(x) with u and dx with e^u\, du. + Thus we rewrite our integral as + + \int \cos(\ln(x))\, dx = \int e^u\cos u \, du + . +

    + +

    + We evaluated this integral on the right in . (This integral can also be found in a table of integrals). + Using the result there, we have: + + \int \cos(\ln(x) )\, dx \amp = \int e^u\cos(u) \, du + \amp = \frac12e^u\big(\sin(u) + \cos(u) \big) + C + \amp = \frac12e^{\ln(x) } \big(\sin(\ln(x) ) + \cos(\ln(x) )\big)+C + \amp = \frac12x \big(\sin(\ln(x) ) + \cos(\ln(x) )\big)+C + . +

    +     + +     +     + + + Definite Integrals and Integration By Parts +

    + So far we have focused only on evaluating indefinite integrals. + Of course, we can use Integration by Parts to evaluate definite integrals as well, + as states. + We do so in the next example. +

    + + + Definite integration using Integration by Parts + +

    + Evaluate \displaystyle \int_1^2 x^2 \ln(x) \,dx. +

    +     + +

    + Our mnemonic suggests letting u=\ln(x), hence dv =x^2\,dx. + We then get du = (1/x)\,dx and v=x^3/3 as shown below. +

    + +
    + Setting up Integration by Parts + +

    + + u\amp = \ln(x) \amp v\amp =\text{?} + du\amp = \text{?} \amp dv\amp =x^2\, dx + +

    + +

    + \Rightarrow +

    + +

    + + u\amp = \ln(x)\amp v\amp =x^3/3 + du\amp = 1/x\, dx \amp dv\amp =x^2\, dx + +

    +     +
    + +

    + The Integration by Parts formula then gives + + \int_1^2 x^2 \ln(x)\,dx \amp = \left.\frac{x^3}3\ln(x)\right|_1^2 - \int_1^2 \frac{x^3}{3}\,\frac 1x\,dx + \amp = \left.\frac{x^3}3\ln(x)\right|_1^2 - \int_1^2 \frac{x^2}{3}\,dx + \amp = \left.\frac{x^3}3\ln(x)\right|_1^2 - \left.\frac{x^3}{9}\right|_1^2 + \amp = \left.\left(\frac{x^3}3\ln(x) - \frac{x^3}{9}\right)\right|_1^2 + \amp = \left(\frac83\ln(2) - \frac89\right)-\left(\frac13\ln(1) - \frac19\right) + \amp = \frac83\ln(2) - \frac79 + \amp \approx 1.07 + . +

    +     + +     + +

    + In general, Integration by Parts is useful for integrating certain products of functions, + like \int x e^x\,dx or \int x^3\sin(x)\,dx. + It is also useful for integrals involving logarithms and inverse trigonometric functions. +

    + +

    + As stated before, + integration is generally more difficult than derivation. + We are developing tools for handling a large array of integrals, + and experience will tell us when one tool is preferable/necessary over another. + For instance, + consider the three similar-looking integrals + + \int xe^x\,dx, \qquad \int x e^{x^2}\,dx \qquad \text{ and } \qquad \int xe^{x^3}\,dx + . +

    + +

    + While the first is calculated easily with Integration by Parts, + the second is best approached with Substitution. + Taking things one step further, + the third integral has no answer in terms of elementary functions, + so none of the methods we learn in calculus will get us the exact answer. +

    + +

    + Integration by Parts is a very useful method, + second only to Substitution. + In the following sections of this chapter, + we continue to learn other integration techniques. + + focuses on handling integrals containing trigonometric functions. +

    +     + + + + Terms and Concepts + + + + +

    + + Integration by Parts is useful in evaluating integrands that contain products of functions. +

    +     + +     + + + + +

    + + Integration by Parts can be thought of as the + opposite of the Chain Rule. +

    +     + +     + + + + +

    + For what is LIATE useful? +

    + +     + + + +

    + Determining which functions in the integrand to set equal to u + and which to set equal to dv. +

    +     + +     + + + + +

    + + If the integral that results from Integration by Parts appears to also need Integration by Parts, + then a mistake was made in the original choice of u. +

    +     + +

    + False; it is not uncommon to need to use Integration by Parts several times to fully evaluate an integral. +

    +     + +     +     + + + Problems + + + +

    + Evaluate the given indefinite integral. +

    +     + + + + + $F = FormulaUpToConstant("sin(x)-x*cos(x)"); + + +

    + \ds \int x\sin(x) \, dx +

    +

    + +

    +     + +

    + \sin(x) - x\cos(x) +C +

    +     +     +     + + + + + $F = FormulaUpToConstant("-e^(-x)*(x+1)"); + + +

    + \ds \int xe^{-x}\, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("-x^2*cos(x)+2x*sin(x)+2*cos(x)"); + + +

    + \ds \int x^2\sin(x) \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("-x^3*cos(x)+3x^2*sin(x)+6x*cos(x)-6*sin(x)"); + + +

    + \ds \int x^3\sin(x) \, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + $F = FormulaUpToConstant("1/2*e^(x^2)"); + + +

    + \ds \int xe^{x^2}\, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("e^x*(x^3-3x^2+6x-6)"); + + +

    + \ds \int x^3e^{x}\, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + $F = FormulaUpToConstant("-1/2*x*e^(-2x)-e^(-2x)/4"); + + +

    + \ds \int xe^{-2x}\, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + $F = FormulaUpToConstant("1/2*e^x*(sin(x)-cos(x))"); + + +

    + \ds \int e^x\sin(x) \, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + $F = FormulaUpToConstant("1/5*e^(2x)*(sin(x)+2*cos(x))"); + + +

    + \ds \int e^{2x}\cos(x) \, dx +

    +

    + +

    +     +     +     + + + + + ($m,$n) = random_subset(2,2..9); + if($envir{problemSeed}==1){$m=2;$n=3}; + $g = gcd($m,$n); + $M = $m/$g; + $N = $n/$g; + $f = Formula("e^($m x) sin($n x)"); + Context("Fraction"); + $frac = Fraction($g,$m**2+$n**2); + $F = FormulaUpToConstant("$frac e^($m x)*($M sin($n x) - $N cos($n x))"); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=5;}; + $f = Formula("e^($m x) cos($m x)"); + Context("Fraction"); + $frac = Fraction(1,2*$m); + $F = FormulaUpToConstant("$frac e^($m x) (sin($m x) + cos($m x))"); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("1/2*sin(x)^2"); + + +

    + \ds \int \sin(x) \cos(x) \, dx +

    +

    + +

    +     +     +     + + + + + $f = Formula("sin^-1(x)"); + $F = FormulaUpToConstant("sqrt(1-x^2)+x*asin(x)"); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m = 2;}; + $f = Formula("tan^-1($m x)"); + $F = FormulaUpToConstant("x*tan^-1($m x)-1/(2*$m)*ln($m^2x^2+1)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("1/2*x^2*atan(x)-x/2+1/2*atan(x)"); + + +

    + \ds \int x\tan^{-1}(x) \, dx +

    +

    + +

    +     +     +     + + + + + $f = Formula("cos^-1(x)"); + $F = FormulaUpToConstant("-sqrt(1-x^2)+x*acos(x)"); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + Context()->variables->set(x=>{limits=>[0,4]}); + $F = FormulaUpToConstant("1/2*x^2*ln(x)-x^2/4"); + + +

    + \ds \int x\ln(x) \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b = -2;}; + $f = Formula("(x+$b)ln(x)")->reduce; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[0,4]}); + $F = FormulaUpToConstant("1/2*x^2*ln(x)-x^2/4 + $b x ln(x) - $b x")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b = 1;}; + $f = Formula("x ln(x-$b)")->reduce; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[$b,$b+4]}); + $frac = Fraction(($b)**2,2); + $F = FormulaUpToConstant("1/2 x^2 ln(x-$b) - 1/4 (x-$b)^2 - $b x - $frac ln(x-$b)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("1/2*x^2*ln(x^2) - x^2/2"); + + +

    + \ds \int x\ln(x^2)\, dx +

    +

    + +

    +     +     +     + + + + + Context()->variables->set(x => {limits => [0,4]}); + $F = FormulaUpToConstant("1/3*x^3*ln(x) - x^3/9"); + + +

    + \ds \int x^2\ln(x) \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("2x+x*(ln(x))^2-2x*ln(x)"); + $F->{limits} = [1,5]; + + +

    + \ds \int \left(\ln(x) \right)^2\, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b = 1;}; + $f = Formula("(ln(x+$b))^2")->reduce; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[-$b,-$b+4]}); + $F = FormulaUpToConstant("2(x+$b)+(x+$b)*ln(x+$b)^2-2*(x+$b)*ln(x+$b)")->reduce; + $F->{limits} = [-$b+1,-$b+5]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("x*tan(x)+ln(abs(cos(x)))"); + $F->{test_at} = [[pi/4],[3*pi/4],[5*pi/4],[7*pi/4]]; + + +

    + \ds \int x\sec^2(x) \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("ln(abs(sin(x)))-x*cot(x)"); + $F->{test_at} = [[pi/4],[3*pi/4],[5*pi/4],[7*pi/4]]; + + +

    + \ds \int x\csc^2(x) \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$b = -2;}; + $f = Formula("x sqrt(x+$b)")->reduce; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[-$b,-$b+4]}); + $frac = Fraction(2*$b,3); + $F = FormulaUpToConstant("(2/5 (x+$b)^2 - $frac (x+$b))sqrt(x+$b)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = non_zero_random(2,9,1); + if($envir{problemSeed}==1){$b = 2;}; + $f = Formula("x sqrt(x^2-$b)")->reduce; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[sqrt($b),sqrt($b)+4]}); + $F = FormulaUpToConstant("1/3(x^2-$b)^(3/2)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("sec(x)"); + + +

    + \ds \int \sec(x) \tan(x) \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("x*sec(x)-ln(abs(sec(x)+tan(x)))"); + + +

    + \ds \int x\sec(x) \tan(x) \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("-x*csc(x)-ln(abs(csc(x)+cot(x)))"); + + +

    + \ds \int x\csc(x) \cot(x) \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Evaluate the indefinite integral after first making a substitution. +

    +     + + + + + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$trig = 'sin';}; + $f = Formula("$trig(ln(x))"); + Context()->variables->set(x=>{limits=>[0,4]}); + $F = FormulaUpToConstant("x/2*(sin(ln(x)) - cos(ln(x)))"); + $F = FormulaUpToConstant("x/2*(sin(ln(x)) + cos(ln(x)))") if ($trig eq 'cos'); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$trig = 'cos';}; + $f = Formula("e^(2x) $trig(e^x)"); + $F = FormulaUpToConstant("cos(e^x)+(e^x)*sin(e^x)"); + $F = FormulaUpToConstant("sin(e^x)-(e^x)*cos(e^x)") if ($trig eq 'sin'); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $trig = list_random('sin','cos'); + if($envir{problemSeed}==1){$trig = 'sin';}; + $f = Formula("$trig(sqrt(x))"); + Context()->variables->set(x=>{limits=>[0,4]}); + $F = FormulaUpToConstant("2*sin(sqrt(x)) - 2*sqrt(x)*cos(sqrt(x))"); + $F = FormulaUpToConstant("2*cos(sqrt(x)) + 2*sqrt(x)*sin(sqrt(x))") if ($trig eq 'cos'); + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("x ln(sqrt(x)) - x/2"); + Context()->variables->set(x=>{limits=>[0,4]}); + + +

    + \ds \int \ln(\sqrt{x})\, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("2*sqrt(x)*e^sqrt(x)-2*e^sqrt(x)"); + Context()->variables->set(x=>{limits=>[0,4]}); + + +

    + \ds \int e^{\sqrt{x}}\, dx +

    +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("x^2/2"); + Context()->variables->set(x=>{limits=>[0,4]}); + + +

    + \ds \int e^{\ln(x) }\, dx +

    +

    + +

    +     +     +     +     + + + +

    + Evaluate the definite integral. + Note: the corresponding indefinite integral appears in + Exercises. +

    +     + + + + + $b = list_random('\pi/2','\pi','3\pi/2','2\pi'); + if($envir{problemSeed}==1){$b = '\pi';}; + $F = Formula("sin(x)-x*cos(x)"); + %r = ('\pi/2'=>'1','\pi'=>'pi','3\pi/2'=>'-1','2\pi'=>'-2pi'); + $r = Compute($r{$b}); + + +

    + \ds \int_0^{} x\sin(x) \, dx +

    +

    + +

    +     +     +     + + + + + ($a,$b) = num_sort(random_subset(2,-2,-1,1,2)); + if($envir{problemSeed}==1){$a=-1;$b=1;}; + $f = Formula("x e^(-x)"); + $F = Formula("-e^(-x)*(x+1)"); + $r = Formula("-($b+1)e^(-$b)+($a+1)e^(-$a)")->reduce; + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random('\pi/6','\pi/4','\pi/3','\pi/2'); + if($envir{problemSeed}==1){$b='\pi/4';}; + $a = '-'.$b; + $f = Formula("x^2 sin(x)"); + $F = Formula("-x^2*cos(x)+2x*sin(x)+2*cos(x)"); + $r = Formula("0")->reduce; + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random('\pi/6','\pi/4','\pi/3','\pi/2'); + if($envir{problemSeed}==1){$b='\pi/2';}; + $a = '-'.$b; + $f = Formula("x^3 sin(x)"); + $F = Formula("-x^3*cos(x)+3x^2*sin(x)+6x*cos(x)-6*sin(x)"); + $r = Formula("-pi^3/(32sqrt(2))+3pi^2/(8sqrt(2))+3pi/sqrt(2)-12/sqrt(2)")->reduce; + $r = Formula("-pi^3sqrt(3)/216+pi^2/12+pi sqrt(3)-6")->reduce if ($b eq '\pi/6'); + $r = Formula("-pi^3/27+pi^2sqrt(3)/3+2pi-6*sqrt(3)")->reduce if ($b eq '\pi/3'); + $r = Formula("3pi^2/2-12")->reduce if ($b eq '\pi/2'); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $c = random(2,9,1); + if($envir{problemSeed}==1){$c=2;}; + $a = 0; + Context()->flags->set(reduceConstantFunctions=>0); + $b = Formula("sqrt(ln(2))"); + $f = Formula("x e^(x^2)"); + $F = Formula("1/2*e^(x^2)"); + Context("Fraction"); + $r = Fraction($c/2-1/2); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $a = 0; + $b = 1; + $F = Formula("x^3*e^x-3*x^2*e^x+6*x*e^x-6*e^x"); + $r = $F->eval(x=>$b) - $F->eval(x=>$a); + + +

    + \ds \int_0^1 x^3e^{x}\, dx +

    +

    + +

    +     +     +     + + + + + ($a,$b) = num_sort(random_subset(2,1..4)); + if($envir{problemSeed}==1){$a=1;$b=2}; + $f = Formula("x e^(-2x)"); + $F = Formula("(-x/2-1/4) e^(-2x)"); + Context("Fraction"); + $fracb = Fraction(-$b/2-1/4); + $fraca = Fraction(-$a/2-1/4); + $b2 = 2*$b; + $a2 = 2*$a; + $r = Formula("$fracb e^(-$b2) - $fraca e^(-$a2)"); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random('\pi/2','\pi','3\pi/2','2\pi'); + if($envir{problemSeed}==1){$b='\pi'}; + $a=0; + $f = Formula("e^x sin(x)"); + $F = Formula("1/2*e^x*(sin(x)-cos(x))"); + $r = Formula("1/2*e^(pi/2) + 1/2"); + $r = Formula("1/2*e^pi + 1/2") if ($b eq '\pi'); + $r = Formula("-1/2*e^(3pi/2) + 1/2") if ($b eq '3\pi/2'); + $r = Formula("-1/2*e^(2pi) + 1/2") if ($b eq '2\pi'); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random('\pi/2','\pi','3\pi/2','2\pi'); + if($envir{problemSeed}==1){$b='\pi/2'}; + $a='-'.$b; + $f = Formula("e^(2x) cos(x)"); + $F = Formula("1/5*e^(2x)*(sin(x)+2*cos(x))"); + $r = Formula("1/5 (e^(pi) + e^(-pi))"); + $r = Formula("2/5 (-e^(2pi) + e^(-2pi))") if ($b eq '\pi'); + $r = Formula("1/5 (-e^(3pi) - e^(-3pi))") if ($b eq '3\pi/2'); + $r = Formula("2/5 (e^(4pi) - e^(-4pi))") if ($b eq '2\pi'); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     +     +     +     +
    +
    + Trigonometric Integrals + +

    + Functions involving trigonometric functions are useful as they are good at describing periodic behavior. + This section describes several techniques for finding antiderivatives of certain combinations of trigonometric functions. +

    +     + + + Integrals of the form <m>\int \sin^m(x) \cos^n(x) \, dx</m> +

    + In learning the technique of Substitution, + we saw the integral \int \sin(x) \cos(x) \, dx in . + The integration was not difficult, + and one could easily evaluate the indefinite integral by letting + u=\sin(x) or by letting u = \cos(x). + This integral is easy since the power of both sine and cosine is 1. +

    + +

    + We generalize this integral and consider integrals of the form + \int \sin^m(x) \cos^n(x) \, dx, + where m,n are nonnegative integers. + Our strategy for evaluating these integrals is to use the identity + \cos^2(x) +\sin^2(x) =1 to convert high powers of one trigonometric function into the other, + leaving a single sine or cosine term in the integrand. + Let's see an example of how this technique works. +

    + + + Integrating powers of sine and cosine + +

    + Evaluate \ds\int\sin^3(x) \cos(x) \, dx. +

    +     + +

    + We have used substitution on problems similar to this problem in + . + If we let u=\sin(x), + then du=\cos(x)\,dx, and + + \int \sin^3(x)\cos(x)\,dx = \int u^3 \,du = \frac{u^4}{4}+C = \frac14 \sin^4(x)+C + . + But what if, for some reason, we wanted to let u=\cos(x) instead? + Unfortunately, + we have \sin^3(x) as part of our integrand, not just \sin(x). + The solution to this problem is to replace some of our powers of sine + (two of them to be exact) + with expressions that involve cosine. + We will use the Pythagorean Identity \sin^2(x)=1-\cos^2(x). + + \int\sin^3(x) \cos(x) \, dx \amp =\int\sin(x)\cdot \sin^2(x) \cos(x) \, dx + \amp =\int\sin(x)\left(1-\cos^2(x)\right) \cos(x) \, dx + . +

    + +

    + Now we let u=\cos(x) so that -du=\sin(x)\,dx. + + \int\sin^3(x) \cos(x) \, dx \amp =\int\sin(x)\left(1-\cos^2(x)\right) \cos(x) \, dx + \amp =\int -\left(1-u^2\right)u\,du + \amp =\int -\left(u-u^3\right)\,du + \amp =-\frac{u^2}{2}+\frac{u^4}{4}+C + \amp =-\frac{\cos^2(x)}{2}+\frac{\cos^4(x)}{4}+C + . +

    + +

    + This looks like a very different answer, so you might wonder if we went wrong somewhere. + But in fact, the two answers are equivalent, in the sense that they differ by a constant! + (So the +C is different in each case, if you like.) + Notice that + + \frac14\sin^4(x) \amp = \frac14(1-\cos^2(x))^2 + \amp = \frac14-\frac12\cos^2(x)+\frac14\cos^4(x) + , + so the difference between the two answers is the constant \frac14 . +

    +     + +     + +

    + We summarize the general technique in the following Key Idea. +

    + + + Integrals Involving Powers of Sine and Cosine +

    + Consider \ds \int \sin^m(x) \cos^n(x) \, dx, + where m,n are nonnegative integers. + integrationof trig. powers +

    + +

    +

      +
    1. +

      + If m is odd, then m=2k+1 for some integer k. + Rewrite + + \sin^m(x) \amp = \sin^{2k+1}(x) + \amp = \sin^{2k}(x) \sin(x) + \amp = (\sin^2(x) )^k\sin(x) + \amp = (1-\cos^2(x) )^k\sin(x) + . + Then + + \int \sin^m(x) \cos^n(x) \, dx \amp = \int (1-\cos^2(x) )^k\sin(x) \cos^n(x) \, dx + \amp = -\int (1-u^2)^ku^n\,du + , + where u = \cos(x) and du = -\sin(x) \, dx. +

      +
      2. + +
    2. +

      + If n is odd, + then using substitutions similar to that outlined above + (replacing all of the even powers of cosine using a Pythagorean identity) + we have: + + \int \sin^m(x) \cos^n(x) \, dx = \int u^m(1-u^2)^k\,du + , + where u = \sin(x) and du = \cos(x) \, dx. +

      +
      3. + +
    3. +

      + If both m and n are even, + use the power-reducing identities: + + \cos^2(x) = \frac{1+\cos(2x)}{2} \text{ and } \sin^2(x) = \frac{1-\cos(2x)}2 + + to reduce the degree of the integrand. + Expand the result and apply the principles of this Key Idea again. +

      +
      4. +
    +

    +     + +

    + We practice applying in the next examples. +

    + + + Integrating powers of sine and cosine + +

    + Evaluate \ds\int\sin^5(x) \cos^8(x) \, dx. +

    +     + +

    + The power of the sine term is odd, + so we rewrite \sin^5(x) as + + \sin^5(x) \amp = \sin^4(x) \sin(x) + \amp = (\sin^2(x) )^2\sin(x) + \amp = (1-\cos^2(x) )^2\sin(x) + . +

    + +

    + Our integral is now \ds \int (1-\cos^2(x) )^2\cos^8(x) \sin(x) \, dx. + Let u = \cos(x), hence du = -\sin(x) \, dx. + Making the substitution and expanding the integrand gives + + \int (1-\cos^2)^2\cos^8(x) \sin(x) \, dx \amp = -\int (1-u^2)^2u^8\,du + \amp = -\int \big(1-2u^2+u^4\big)u^8\,du + \amp = -\int \big(u^8-2u^{10}+u^{12}\big)\,du + . +

    + +

    + This final integral is not difficult to evaluate, giving + + -\int \big(u^8-2u^{10}+u^{12}\big)\, du \amp = -\frac19u^9 + \frac2{11}u^{11} - \frac1{13}u^{13} + C + \amp =-\frac19\cos^9(x) + \frac2{11}\cos^{11}(x) - \frac1{13}\cos^{13}(x) + C + . +

    +     + +     + + + Integrating powers of sine and cosine + +

    + Evaluate \ds \int\sin^5(x) \cos^9(x) \, dx. +

    +     + +

    + The powers of both the sine and cosine terms are odd, + therefore we can apply the techniques of + to either power. + We choose to work with the power of the cosine term since the previous example used the sine term's power. +

    + +

    + We rewrite \cos^9(x) as + + \cos^9(x) \amp = \cos^8(x) \cos(x) + \amp = \left(\cos^2(x)\right)^4\cos(x) + \amp = \left(1-\sin^2(x)\right)^4\cos(x) + . +

    + +

    + We rewrite the integral as + + \int\sin^5(x) \cos^9(x) \, dx = \int\sin^5(x) \left(1-\sin^2(x)\right)^4\cos(x) \, dx + . +

    + +

    + Now substitute and integrate, + using u = \sin(x) and du = \cos(x) \, dx. + Expand the binomial using algebra. +

    + +

    + + \amp \int u^5(1-u^2)^4\,du + \amp = \int u^5(1-4u^2+6u^4-4u^6+u^8)\,du + \amp = \int\big(u^5-4u^7+6u^9-4u^{11}+u^{13}\big)\,du + \amp = \frac16u^6-\frac12u^8+\frac35u^{10}-\frac13u^{12}+\frac{1}{14}u^{14}+C + \amp = \frac16\sin^6(x) -\frac12\sin^8(x) +\frac35\sin^{10}(x)-\frac13\sin^{12}(x) +\frac{1}{14}\sin^{14}(x) +C + . +

    +     +     + +

    + Technology Note: The work we are doing here can be a bit tedious, + but the skills developed + (problem solving, algebraic manipulation, etc.) + are important. + Nowadays problems of this sort are often solved using a computer algebra system. + The powerful program Mathematica integrates + \int \sin^5(x) \cos^9(x) \, dx as + + f(x) =\amp -\frac{45 \cos(2 x)}{16384}-\frac{5 \cos(4 x)}{8192}+\frac{19 \cos(6x)}{49152} + \amp+\frac{\cos(8 x)}{4096}-\frac{\cos(10 x)}{81920}-\frac{\cos(12x)}{24576}-\frac{\cos(14 x)}{114688} + , + which clearly has a different form than our answer in + , which is + + g(x)=\frac16\sin^6(x) -\frac12\sin^8(x) +\frac35\sin^{10}(x) -\frac13\sin^{12}(x) +\frac{1}{14}\sin^{14}(x) + . +

    + +

    + + shows a graph of f and g; + they are clearly not equal, but they differ + only by a constant. + That is g(x) = f(x) + C for some constant C. + So we have two different antiderivatives of the same function, + meaning both answers are correct. +

    + +
    + A plot of f(x) and g(x) from and the Technology Note + + + + Graph of f(x) and g(x) that are different only by a constant. + + +

    + The y axis is drawn from -0.002 and 0.004 and the x axis + is drawn from 0 to 3. The graph has two functions drawn. Both of them + are plateau shaped. +

    +

    + The first function g(x) is drawn on the x axis, the function starts at the + origin and rises up at x=0.25 and keeps rising steeply until x=1.75 then + it runs parallel to the x axis and starts declining at x=2.25, it declines + sharply till x=2.75 then it merged with the x axis. +

    +

    + The second function f(x) is the same as the first but the line on which it is placed + is x=0.0025 instead of x=0. +

    +     + + + \begin{tikzpicture} + \begin{axis}[% + axis on top,% + ytick={-.002,.002,.004}, + yticklabels={$-0.002$,$0.002$,$0.004$}, + ymin=-.003,ymax=0.005,% + xmin=-.1,xmax=3.15,% + scaled ticks=false + ] + + \addplot [secondcurvestyle] coordinates {(0,-0.0027879) (0.15708,-0.0027856) (0.31416,-0.0026798) (0.47124,-0.0020322) (0.62832,-0.00060997) (0.7854,0.00089518)(0.94248,0.0017233) (1.0996,0.0019485) (1.2566,0.0019734) (1.4137,0.0019741) (1.5708,0.0019741) (1.7279,0.0019741) (1.885,0.0019734) (2.042,0.0019485) (2.1991,0.0017233) (2.3562,0.00089518) (2.5133,-0.00060997) (2.6704,-0.0020322) (2.8274,-0.0026798) (2.9845,-0.0027856) (3.1416,-0.0027879)}; + + \draw (axis cs:2.6,0.004) node { $g(x)$}; + + \addplot [firstcurvestyle] coordinates {(0,0) (0.15708,0) (0.31416,0.00010807) (0.47124,0.0007557) (0.62832,0.0021779) (0.7854,0.003683) (0.94248,0.0045111) + (1.0996,0.0047364) (1.2566,0.0047612) (1.4137,0.0047619) + (1.5708,0.0047619) (1.7279,0.0047619) (1.885,0.0047612) (2.042,0.0047364) (2.1991,0.0045111) (2.3562,0.003683) (2.5133,0.0021779) (2.6704,0.0007557) (2.8274,0.00010807) (2.9845,0) (3.1416,0)}; + + \draw (axis cs:2.4,-0.002) node { $f(x)$}; + \end{axis} + + \end{tikzpicture} + + + + +
    + + + Integrating powers of sine and cosine + +

    + Evaluate \ds\int\cos^4(x) \sin^2(x) \, dx. +

    +     + +

    + The powers of sine and cosine are both even, + so we employ the power-reducing formulas and algebra as follows. + + \amp \int \cos^4(x) \sin^2(x) \, dx = \int\left(\frac{1+\cos(2x)}{2}\right)^2\left(\frac{1-\cos(2x)}2\right)\, dx + \amp = \int\frac{1+2\cos(2x)+\cos^2(2x)}4\cdot\frac{1-\cos(2x)}2\, dx + \amp = \int \frac18\big(1+\cos(2x)+\cos^2(2x)-\cos^3(2x)\big)\, dx + \amp = \frac18 \left(\underbrace{\int 1 \, dx}_{a}+\underbrace{\int \cos(2x)\, dx}_{b}-\underbrace{\int \cos^2(2x)\, dx}_{c}-\underbrace{\int \cos^3(2x)\, dx}_{d}\right) + +

    + +

    + The first integral labeled a is easy to integrate. + The \cos(2x) term is also easy to integrate, + especially with . + The \cos^2(2x) term is another trigonometric integral with an even power, + requiring the power-reducing formula again. + The \cos^3(2x) term is a cosine function with an odd power, + requiring a substitution as done before. + We integrate each in turn below. + + \underbrace{\int\cos(2x)\, dx}_{b} \amp = \frac12\sin(2x)+C + \underbrace{\int\cos^2(2x)\, dx}_{c} \amp = \int \frac{1+\cos(4x)}2\, dx + \amp = \frac12\big(x+\frac14\sin(4x)\big)+C + . +

    + +

    + Finally, we rewrite \cos^3(2x) as + + \cos^3(2x) \amp = \cos^2(2x)\cos(2x) + \amp = \big(1-\sin^2(2x)\big)\cos(2x) + . +

    + +

    + Letting u=\sin(2x), we have du = 2\cos(2x)\, dx, hence + + \underbrace{\int \cos^3(2x)\, dx}_{d} \amp = \int\big(1-\sin^2(2x)\big)\cos(2x)\, dx + \amp = \int \frac12(1-u^2)\,du + \amp = \frac12\Big(u-\frac13u^3\Big)+C + \amp = \frac12\Big(\sin(2x)-\frac13\sin^3(2x)\Big)+C + . +

    + +

    + Putting all the pieces together, we have + + \amp \int \cos^4(x) \sin^2(x) \, dx + \amp =\int \frac18\big(1+\cos(2x)-\cos^2(2x)-\cos^3(2x)\big)\, dx + \amp = \frac18\Big[x+\frac12\sin(2x)-\frac12\big(x+\frac14\sin(4x)\big)-\frac12\Big(\sin(2x)-\frac13\sin^3(2x)\Big)\Big]+C + \amp =\frac18\Big[\frac12x-\frac18\sin(4x)+\frac16\sin^3(2x)\Big]+C + . +

    +     + +     + +

    + The process above was a bit long and tedious, + but being able to work a problem such as this from start to finish is important. +

    + + + + +     + + + Integrals of the form <m>\int\sin(mx)\sin(nx)\, dx</m>, <m>\int \cos(mx)\cos(nx)\, dx</m>, and <m>\int \sin(mx)\cos(nx)\, dx</m> +

    + Functions that contain products of sines and cosines of differing periods are important in many applications including the analysis of sound waves. + Integrals of the form + + \int\sin(mx)\sin(nx)\, dx, \int \cos(mx)\cos(nx)\, dx \text{ and } \int \sin(mx)\cos(nx)\, dx + + are best approached by first applying the Product to Sum Formulas found in the back cover of this text, namely + + \sin(mx)\sin(nx) \amp = \frac12\Big[\cos\big((m-n)x\big)-\cos\big((m+n)x\big)\Big] + \cos(mx)\cos(nx) \amp = \frac12\Big[\cos\big((m-n)x\big)+\cos\big((m+n)x\big)\Big] + \sin(mx)\cos(nx) \amp = \frac12\Big[\sin\big((m-n)x\big)+\sin\big((m+n)x\big)\Big] + . +

    + + + Integrating products of <m>\sin(mx)</m> and <m>\cos(nx)</m> + +

    + Evaluate \ds\int\sin(5x)\cos(2x)\, dx. +

    +     + +

    + The application of the formula and subsequent integration are straightforward: + + \int\sin(5x)\cos(2x)\, dx \amp = \int \frac12\Big[\sin((5-2)x)+\sin((5+2)x)\Big]\, dx + \amp= \int \frac12\Big[\sin(3x)+\sin(7x)\Big]\, dx + \amp = -\frac16\cos(3x) - \frac1{14}\cos(7x) + C + +

    +     + +     +     + + + Integrals of the form <m>\int\tan^m(x) \sec^n(x) \, dx</m> +

    + When evaluating integrals of the form \int \sin^m(x) \cos^n(x) \, dx, + the Pythagorean Theorem allowed us to convert even powers of sine into even powers of cosine, + and vise-versa. + If, for instance, the power of sine was odd, + we pulled out one \sin(x) and converted the remaining even power of \sin(x) into a function using powers of \cos(x), + leading to an easy substitution. +

    + +

    + The same basic strategy applies to integrals of the form \int \tan^m(x) \sec^n(x) \, dx, + albeit a bit more nuanced. + The following three facts will prove useful: +

    + +

    +

      +
    • \frac{d}{dx}(\tan(x) ) = \sec^2(x),
      • + +
    • \frac{d}{dx}(\sec(x) ) = \sec(x) \tan(x),
      • + +
    • +

      + 1+\tan^2(x) = \sec^2(x) + (the Pythagorean Theorem). +

      +
      • +
    +

    + +

    + If the integrand can be manipulated to separate a + \sec^2(x) term with the remaining secant power even, + or if a \sec(x) \tan(x) term can be separated with the remaining \tan(x) power even, + the Pythagorean Theorem can be employed, + leading to a simple substitution. + This strategy is outlined in the following Key Idea. +

    + + + Integrals Involving Powers of Tangent and Secant +

    + Consider \ds\int\tan^m(x) \sec^n(x) \, dx, + where m,n are nonnegative integers. + integrationof trig. powers + +

    + +

    +

      +
    1. + +

      + If n is even, then n=2k for some integer k. + Rewrite \sec^n(x) as + + \sec^n(x) \amp = \sec^{2k}(x) + \amp = \sec^{2k-2}(x) \sec^2(x) + \amp = (1+\tan^2(x) )^{k-1}\sec^2(x) + . + Then + + \int\tan^m(x) \sec^n(x) \, dx \amp =\int\tan^m(x) (1+\tan^2(x) )^{k-1}\sec^2(x) \, dx + \amp =\int u^m(1+u^2)^{k-1}\,du + , + where u = \tan(x) and du = \sec^2(x) \, dx. +

      +
      2. + +
    2. + +

      + If m is odd, then m=2k+1 for some integer k. + Rewrite \tan^m(x) \sec^n(x) as + + \tan^m(x) \sec^n(x) \amp = \tan^{2k+1}(x) \sec^n(x) + \amp = \tan^{2k}(x) \sec^{n-1}(x) \sec(x) \tan(x) + \amp = (\sec^2(x) -1)^k\sec^{n-1}(x) \sec(x) \tan(x) + . + Then + + \int\tan^m(x) \sec^n(x) \, dx \amp =\int(\sec^2(x) -1)^k\sec^{n-1}(x) \sec(x) \tan(x) \, dx + \amp = \int(u^2-1)^ku^{n-1}\,du + , + where u = \sec(x) and du = \sec(x) \tan(x) \, dx. +

      +
      3. + +
    3. + +

      + If n is odd and m is even, + then m=2k for some integer k. + Convert \tan^m(x) to (\sec^2(x) -1)^k. + Expand the new integrand and use Integration By Parts, + with dv = \sec^2(x) \, dx. +

      +
      4. + +
    4. + +

      + If m is even and n=0, + rewrite \tan^m(x) as + + \tan^m(x) \amp = \tan^{m-2}(x) \tan^2(x) + \amp = \tan^{m-2}(x) (\sec^2(x) -1) + \amp = \tan^{m-2}\sec^2(x) -\tan^{m-2}(x) + . + So + + \int\tan^m(x) \, dx = \underbrace{\int\tan^{m-2}\sec^2(x) \, dx}_{\text{ apply rule 1 } } - \underbrace{\int\tan^{m-2}(x) \, dx}_{\text{ apply rule 4 again } } + . +

      +
      5. +
    +

    +     + +

    + The techniques described in Item + and Item + of + are relatively straightforward, + but the techniques in Item + and Item can be rather tedious. + A few examples will help with these methods. +

    + + + Integrating powers of tangent and secant + +

    + Evaluate \ds\int \tan^2(x) \sec^6(x) \, dx. +

    +     + +

    + Since the power of secant is even, + we use Rulefrom + and pull out a \sec^2(x) in the integrand. + We convert the remaining powers of secant into powers of tangent. + + \int \tan^2(x) \sec^6(x) \, dx \amp = \int\tan^2(x) \sec^4(x) \sec^2(x) \, dx + \amp = \int \tan^2(x) \big(1+\tan^2(x) \big)^2\sec^2(x) \, dx + Now substitute, with u=\tan(x), with du = \sec^2(x) \, dx. + \amp =\int u^2\big(1+u^2\big)^2\,du + We leave the integration and subsequent substitution to the reader. The final answer is + \amp =\frac13\tan^3(x) +\frac25\tan^5(x) +\frac17\tan^7(x) +C + . +

    +     + +     + + + + + + + Integrating powers of tangent and secant + +

    + Evaluate \ds\int \sec^3(x) \, dx. +

    +     + +

    + We apply Rule + from + as the power of secant is odd and the power of tangent is even (0 is an even number). + We use Integration by Parts; + the rule suggests letting dv = \sec^2(x) \, dx, + meaning that u = \sec(x). +

    + + +
    + Setting up Integration by Parts + +

    + + u\amp = \sec(x) \amp v\amp =\mathord{?} + du\amp = \mathord{?} \amp dv\amp =\sec^2(x) \, dx + +

    + +

    + \implies +

    + +

    + + u\amp = \sec(x) \amp v\amp =\tan(x) + du\amp = \sec(x) \tan(x) \, dx \amp dv\amp =\sec^2(x) \, dx + +

    +     +
    + +

    + Employing Integration by Parts, we have + + \int \sec^3(x) \, dx \amp = \int \underbrace{\sec(x) }_u\cdot\underbrace{\sec^2(x) \, dx}_{dv} + \amp = \sec(x) \tan(x) - \int \sec(x) \tan^2(x) \, dx. + This new integral also requires applying Rule + of : + \int \sec^3(x)\, dx\amp = \sec(x) \tan(x) - \int \sec(x) \big(\sec^2(x) -1\big)\, dx + \amp = \sec(x) \tan(x) - \int \sec^3(x) \, dx + \int \sec(x) \, dx + \amp = \sec(x) \tan(x) -\int \sec^3(x) \, dx + \ln\abs{\sec(x) +\tan(x) } + In previous applications of Integration by Parts, we have seen where the original integral has reappeared in our work. We resolve this by adding \int \sec^3(x) \, dx to both sides, giving: + 2\int \sec^3(x) \, dx \amp = \sec(x) \tan(x) + \ln\abs{\sec(x) +\tan(x) } + \int \sec^3(x) \, dx \amp = \frac12\Big(\sec(x) \tan(x) + \ln\abs{\sec(x) +\tan(x) }\Big)+C + +

    +     + +     + + + + + +

    + We give one more example. +

    + + + + Integrating powers of tangent and secant + +

    + Evaluate \ds\int\tan^6(x) \, dx. +

    +     + +

    + We employ Rule of + . + + \int \tan^6(x) \, dx \amp = \int \tan^4(x) \tan^2(x) \, dx + \amp = \int\tan^4(x) \big(\sec^2(x) -1\big)\, dx + \amp = \int\tan^4(x) \sec^2(x) \, dx - \int\tan^4(x) \, dx + Integrate the first integral with substitution, u=\tan(x); integrate the second by employing rule + Rule again. + \amp = \frac15\tan^5(x) -\int\tan^2(x) \tan^2(x) \, dx + \amp = \frac15\tan^5(x) -\int\tan^2(x) \big(\sec^2(x) -1\big)\, dx + \amp = \frac15\tan^5(x) -\underbrace{\int\tan^2(x) \sec^2(x) \, dx}_{a} + \underbrace{\int\tan^2(x) \, dx}_b + Again, use substitution (u=\tan(x)) for the first integral (a) and + Rule for the second (b). + \amp = \frac15\tan^5(x) -\frac13\tan^3(x) +\int\big(\sec^2(x) -1\big)\, dx + \int \tan^6(x)\, dx\amp = \frac15\tan^5(x) -\frac13\tan^3(x) +\tan(x) - x+C + . +

    +     + +     + + + +

    + These latter examples were admittedly long, + with repeated applications of the same rule. + Try to not be overwhelmed by the length of the problem, + but rather admire how robust this solution method is. + A trigonometric function of a high power can be systematically reduced to trigonometric functions of lower powers until all antiderivatives can be computed. +

    + +

    + + introduces an integration technique known as Trigonometric Substitution, + a clever combination of Substitution and the Pythagorean Theorem. +

    +     + + + + Terms and Concepts + + + + +

    + + \ds \int \sin^2(x) \cos^2(x) \, dx cannot be evaluated using the techniques described in this section + since both powers of \sin(x) and \cos(x) are even. +

    +     + +     + + + + +

    + + \ds \int \sin^3(x) \cos^3(x) \, dx cannot be evaluated using the techniques described in this section + since both powers of \sin(x) and \cos(x) are odd. +

    +     + +     + + + + +

    + + This section addresses how to evaluate indefinite integrals such as + \ds \int \sin^5(x) \tan^3(x) \, dx. +

    +     + +     + + + + +

    + + Sometimes computer programs evaluate integrals involving trigonometric functions differently + than one would using the techniques of this section. + When this is the case, + the techniques of this section have failed and one should only trust the answer given by the computer. +

    +     + +     +     + + + Problems + + +

    + Evaluate the indefinite integral. +

    +     + + + + + + $F = FormulaUpToConstant("-1/5*cos(x)^5"); + + +

    + \ds \int \sin(x) \cos^4(x) \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("1/4*sin(x)^4"); + + +

    + \ds \int \sin^3(x) \cos(x) \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=2;}; + Context("Fraction"); + $f = Formula("sin^3(x) cos^$m(x)"); + $F = FormulaUpToConstant("1/($m+3)*cos(x)^($m+3)-1/($m+1)*cos(x)^($m+1)"); + + +

    + \ds \int \sin^3(x)\cos^{}(x) \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + $n = 3; + if (list_random(-1,1)) { + ($m,$n) = ($n,$m); + } + if($envir{problemSeed}==1){$m=3;$n=3;}; + Context("Fraction"); + $f = Formula("sin^$m(x) cos^$n(x)"); + $F = FormulaUpToConstant("1/($m+1)*sin(x)^($m+1)-1/($m+3)*sin(x)^($m+3)"); + if ($n != 3) { + $F = FormulaUpToConstant("1/($n+3)*cos(x)^($n+3)-1/($n+1)*cos(x)^($n+1)"); + } + + +

    + \ds \int \sin^{}(x)\cos^{}(x) \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=6;}; + Context("Fraction"); + $f = Formula("sin^$m(x) cos^5(x)"); + $F = FormulaUpToConstant("1/($m+5)*sin(x)^($m+5)-2/($m+3)*sin(x)^($m+3)+1/($m+1)*sin(x)^($m+1)"); + + +

    + \ds \int \sin^{}(x)\cos^{5}(x) \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("-1/9*sin(x)^9+3/7*sin(x)^7-3/5*sin(x)^5+1/3*sin(x)^3"); + + +

    + \ds \int \sin^2(x) \cos^7(x) \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("x/8-1/32*sin(4x)"); + + +

    + \ds \int \sin^2(x) \cos^2(x) \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("1/2*(-1/8*cos(8x)-1/2*cos(2x))"); + + +

    + \ds \int \sin(5x)\cos(3x)\, dx +

    +

    + +

    +     +     +     + + + + + ($m,$n) = random_subset(2,2..9); + if (list_random(-1,1)) {$m = 1;} else {$n = 1;}; + if($envir{problemSeed}==1){$m=1;$n=2}; + Context("Fraction"); + $f = Formula("sin($m x) cos($n x)")->reduce; + $a = abs($m-$n); + $b = abs($m+$n); + $A = Fraction(-1,2*$a); + $B = Fraction(-1,2*$b); + $F = FormulaUpToConstant("$A cos($a x) + $B cos($b x)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($m,$n) = random_subset(2,2..9); + if($envir{problemSeed}==1){$m=3;$n=7}; + Context("Fraction"); + $f = Formula("sin($m x) sin($n x)")->reduce; + $a = abs($m-$n); + $b = abs($m+$n); + $A = Fraction(1,2*$a) * (abs($a)/$a); + $B = Fraction(1,2*$b) * (abs($b)/$b); + $F = FormulaUpToConstant("$A sin($a x) - $B sin($b x)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($m,$n) = random_subset(2,2..9); + if (list_random(-1,1)) {$m = 1;} else {$n = 1;}; + if($envir{problemSeed}==1){$m=1;$n=2}; + Context("Fraction"); + $f = Formula("sin($m pi x) sin($n pi x)")->reduce; + $a = abs($m-$n); + $b = abs($m+$n); + $A = Fraction(1,2*$a) * (abs($a)/$a); + $B = Fraction(1,2*$b) * (abs($b)/$b); + ($Anum,$Aden) = $A->value; + ($Bnum,$Bden) = $B->value; + $A = Formula("$Anum/($Aden pi)"); + $B = Formula("$Bnum/($Bden pi)"); + $F = FormulaUpToConstant("$A sin($a pi x) - $B sin($b pi x)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("1/2*(sin(x)+1/3*sin(3x))"); + + +

    + \ds \int \cos(x)\cos(2x)\, dx +

    +

    + +

    +     +     +     + + + + + ($m,$n) = random_subset(2,'pi/6','pi/4','pi/3','pi/2','pi'); + if($envir{problemSeed}==1){$m='pi/2';$n='pi'}; + Context("Fraction"); + $f = Formula("cos($m x) cos($n x)")->reduce; + $a = Fraction(abs((Real($m)-Real($n))/pi)); + $b = Fraction(abs((Real($m)+Real($n))/pi)); + $A = Fraction(1/(2*$a)); + $B = Fraction(1/(2*$b)); + ($anum,$aden) = $a->value; + ($bnum,$bden) = $b->value; + $a = Formula("$anum pi/$aden"); + $b = Formula("$bnum pi/$bden"); + ($Anum,$Aden) = $A->value; + ($Bnum,$Bden) = $B->value; + $A = Formula("$Anum/($Aden pi)"); + $B = Formula("$Bnum/($Bden pi)"); + $F = FormulaUpToConstant("$A cos($a pi x) + $B cos($b pi x)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("tan(x)^5/5"); + + +

    + \ds \int \tan^4(x) \sec^2(x) \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("tan(x)^5/5+tan(x)^3/3"); + + +

    + \ds \int \tan^2(x) \sec^4(x) \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=3;}; + Context("Fraction"); + $f = Formula("tan^$m(x) sec^4(x)"); + $F = FormulaUpToConstant("1/($m+3)*tan(x)^($m+3)+1/($m+1)*tan(x)^($m+1)"); + + +

    + \ds \int \tan^{}(x)\sec^{4}(x) \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=3;}; + Context("Fraction"); + $f = Formula("tan^$m(x) sec^2(x)"); + $F = FormulaUpToConstant("1/($m+1)*tan(x)^($m+1)"); + + +

    + \ds \int \tan^{}(x)\sec^{2}(x) \, dx +

    +

    + +

    +     +     +     + + + + + $n = random(2,9,1); + if($envir{problemSeed}==1){$n=3;}; + Context("Fraction"); + $f = Formula("tan^3(x) sec^$n(x)"); + $F = FormulaUpToConstant("1/($n+2)*sec(x)^($n+2)-1/($n)*sec(x)^($n)"); + + +

    + \ds \int \tan^{3}(x)\sec^{}(x) \, dx +

    +

    + +

    +     +     +     + + + + + $n = random(2,9,1); + if($envir{problemSeed}==1){$n=5;}; + Context("Fraction"); + Context()->flags->set(limits=>[-1,1]); + $f = Formula("tan^5(x) sec^$n(x)"); + $F = FormulaUpToConstant("1/($n+4)*sec(x)^($n+4)-2/($n+2)*sec(x)^($n+2) + 1/$n sec(x)^($n)"); + + +

    + \ds \int \tan^{5}(x)\sec^{}(x) \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("tan(x)^3/3-tan(x)+x"); + + +

    + \ds \int \tan^4(x) \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("1/4*tan(x)*sec(x)^3+3/8*(sec(x)*tan(x)+ln(abs(sec(x) + tan(x))))"); + + +

    + \ds \int \sec^5(x) \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("1/2*(sec(x)*tan(x)-ln(abs(sec(x)+tan(x))))"); + + +

    + \ds \int \tan^2(x) \sec(x) \, dx +

    +

    + +

    +     +     +     + + + + + + $F = FormulaUpToConstant("1/4*tan(x)*sec(x)^3-1/8*(sec(x)*tan(x)+ln(abs(sec(x)+tan(x))))"); + + +

    + \ds \int \tan^2(x) \sec^3(x) \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Evaluate the definite integral. + Note: the corresponding indefinite integrals appear in + Exercises. +

    +     + + + + + $b = list_random(pi/2,pi,3*pi/2,2*pi); + if($envir{problemSeed}==1){$b=pi}; + $a=0; + Context("Fraction"); + $F = Formula("-1/5*cos(x)^5"); + $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); + $a = Fraction($a/pi); + $b = Fraction($b/pi); + ($anum,$aden) = $a->value; + ($bnum,$bden) = $b->value; + $a = Formula("($anum pi)/$aden")->reduce; + $b = Formula("($bnum pi)/$bden")->reduce; + + +

    + \ds \int_{}^{} \sin(x) \cos^4(x) \, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random(pi/2,pi,3*pi/2,2*pi); + if($envir{problemSeed}==1){$b=pi}; + $a = -$b; + Context("Fraction"); + $F = Formula("1/4*sin(x)^4"); + $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); + $a = Fraction($a/pi); + $b = Fraction($b/pi); + ($anum,$aden) = $a->value; + ($bnum,$bden) = $b->value; + $a = Formula("($anum pi)/$aden")->reduce; + $b = Formula("($bnum pi)/$bden")->reduce; + + +

    + \ds \int_{}^{} \sin^3(x) \cos(x) \, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random(pi/2,pi,3*pi/2,2*pi); + if($envir{problemSeed}==1){$b=pi/2}; + $a = -$b; + Context("Fraction"); + $F = Formula("-1/9*sin(x)^9+3/7*sin(x)^7-3/5*sin(x)^5+1/3*sin(x)^3"); + $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); + $a = Fraction($a/pi); + $b = Fraction($b/pi); + ($anum,$aden) = $a->value; + ($bnum,$bden) = $b->value; + $a = Formula("($anum pi)/$aden")->reduce; + $b = Formula("($bnum pi)/$bden")->reduce; + + +

    + \ds \int_{}^{} \sin^2(x) \cos^7(x) \, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random(pi/4,pi/2,3*pi/4,pi,3*pi/2,2*pi); + if($envir{problemSeed}==1){$b=pi/2}; + $a = 0; + Context("Fraction"); + $F = Formula("1/2*(-1/8*cos(8x)-1/2*cos(2x))"); + $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); + $a = Fraction($a/pi); + $b = Fraction($b/pi); + ($anum,$aden) = $a->value; + ($bnum,$bden) = $b->value; + $a = Formula("($anum pi)/$aden")->reduce; + $b = Formula("($bnum pi)/$bden")->reduce; + + +

    + \ds \int_{}^{} \sin(5 x)\cos(3x)\, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random(pi/2,pi,3*pi/2,2*pi); + if($envir{problemSeed}==1){$b=pi/2}; + $a = -$b; + Context("Fraction"); + $F = Formula("1/2*(sin(x)+1/3*sin(3x))"); + $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); + $a = Fraction($a/pi); + $b = Fraction($b/pi); + ($anum,$aden) = $a->value; + ($bnum,$bden) = $b->value; + $a = Formula("($anum pi)/$aden")->reduce; + $b = Formula("($bnum pi)/$bden")->reduce; + + +

    + \ds \int_{}^{} \cos(x)\cos(2x)\, dx +

    +

    + +

    +     +     +     + + + + + $a = list_random(0,-pi/4); + if($envir{problemSeed}==1){$a=0}; + $b = pi/4; + Context("Fraction"); + $F = Formula("tan(x)^5/5"); + $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); + $a = Fraction($a/pi); + $b = Fraction($b/pi); + ($anum,$aden) = $a->value; + ($bnum,$bden) = $b->value; + $a = Formula("($anum pi)/$aden")->reduce; + $b = Formula("($bnum pi)/$bden")->reduce; + + +

    + \ds \int_{}^{} \tan^4(x) \sec^2(x) \, dx +

    +

    + +

    +     +     +     + + + + + $a = list_random(0,-pi/4); + if($envir{problemSeed}==1){$a=-pi/4}; + $b = pi/4; + Context("Fraction"); + $F = Formula("tan(x)^5/5+tan(x)^3/3"); + $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); + $a = Fraction($a/pi); + $b = Fraction($b/pi); + ($anum,$aden) = $a->value; + ($bnum,$bden) = $b->value; + $a = Formula("($anum pi)/$aden")->reduce; + $b = Formula("($bnum pi)/$bden")->reduce; + + +

    + \ds \int_{-\pi/4}^{\pi/4} \tan^2(x) \sec^4(x) \, dx +

    +

    + +

    +     +     +     +     +     +     +
    +
    + Trigonometric Substitution +

    + In + we defined the definite integral as the + signed area under the curve. + In that section we had not yet learned the Fundamental Theorem of Calculus, + so we only evaluated special definite integrals which described nice, + geometric shapes. + For instance, we were able to evaluate + + \int_{-3}^3\sqrt{9-x^2}\, dx = \frac{9\pi}{2} + + as we recognized that f(x) = \sqrt{9-x^2} described the upper half of a circle with radius 3. +

    + +

    + We have since learned a number of integration techniques, + including Substitution and Integration by Parts, + yet we are still unable to evaluate the above integral without resorting to a geometric interpretation. + This section introduces Trigonometric Substitution, + a method of integration that fills this gap in our integration skill. + This technique works on the same principle as Substitution as found in , + though it can feel backward. + In , we set u=f(x), + for some function f, and replaced f(x) with u. + In this section, we will set x=f(\theta), + where f is a trigonometric function, + then replace x with f(\theta). +

    + + + +

    + We start by demonstrating this method in evaluating the integral in Equation. + After the example, + we will generalize the method and give more examples. +

    + + + Using Trigonometric Substitution + +

    + Evaluate \ds \int_{-3}^3\sqrt{9-x^2}\, dx. +

    +     + +

    + We begin by noting that 9\left(\sin^2(\theta) + \cos^2(\theta)\right) = 9, + and hence 9\cos^2(\theta) = 9-9\sin^2(\theta). + If we let x=3\sin(\theta), + then 9-x^2 = 9-9\sin^2(\theta) = 9\cos^2(\theta). +

    + +

    + Setting x=3\sin(\theta) gives dx = 3\cos(\theta) \, d\theta. + We are almost ready to substitute. + We also wish to change our bounds of integration. + The bound x=-3 corresponds to + \theta = -\pi/2 (for when \theta = -\pi/2, + x=3\sin(\theta) = -3). + Likewise, the bound of x=3 is replaced by the bound \theta = \pi/2. + Thus + + \int_{-3}^3\sqrt{9-x^2}\, dx \amp = \int_{-\pi/2}^{\pi/2} \sqrt{9-9\sin^2(\theta) }\,(3\cos(\theta) )\, d\theta + \amp = \int_{-\pi/2}^{\pi/2} 3\sqrt{9\cos^2(\theta) } \cos(\theta) \, d\theta + \amp =\int_{-\pi/2}^{\pi/2} 3\abs{3\cos(\theta) } \cos(\theta) \, d\theta + . + On [-\pi/2,\pi/2], \cos(\theta) is always positive, so we can drop the absolute value bars, then employ a power-reducing formula: + + \int_{-3}^3\sqrt{9-x^2}\, dx \amp = \int_{-\pi/2}^{\pi/2} 9\cos^2(\theta) \, d\theta + \amp = \int_{-\pi/2}^{\pi/2} \frac{9}{2}\big(1+\cos(2\theta)\big)\, d\theta + \amp = \left.\frac92 \big(\theta +\frac12\sin(2\theta)\big)\right|_{-\pi/2}^{\pi/2} + \amp = \frac92\pi + . +

    + +

    + This matches our answer from before. +

    +     + +     + +

    + We now describe in detail Trigonometric Substitution. + This method excels when dealing with integrands that contain \sqrt{a^2-x^2}, + \sqrt{x^2-a^2} and \sqrt{x^2+a^2}. + The following Key Idea outlines the procedure for each case, + followed by more examples. + Each right triangle acts as a reference to help us understand the relationships between x and \theta. +

    + + + + Trigonometric Substitution +

    +

      +
    1. + Integrands containing <m>\sqrt{a^2-x^2}</m> + + +

      + integrationtrig. subst. + + Let x=a\sin(\theta), dx = a\cos(\theta) \, d\theta. + Thus \theta = \sin^{-1}(x/a), + for -\pi/2\leq \theta\leq \pi/2. + On this interval, \cos(\theta) \geq 0, + so \sqrt{a^2-x^2} = a\cos(\theta). +

      + +
      + + + + + Diagram showing trigonometric substitution with integrands containing square root of a^2-x^2. + + +

      + The diagram is of a right angled triangle. The perpendicular is marked x + the base as \sqrt{a^2 -x^2}. The hypotenuse is marked a. The angle + opposite to the perpendicular is marked as \theta. +

      +       + + + \begin{tikzpicture}[scale=1.5] + + \draw [very thick] (0,0) -- node [below,pos=.5] { $\sqrt{a^2-x^2}$} (3,0) -- node [right,pos=.5] { $x$} (3,2) -- node [pos=.5,above] { $a$} (0,0); + \draw [thick] (2.7,0) -- (2.7,.3) -- (3,.3); + \draw (.75,.25) node {$\theta$}; + + \end{tikzpicture} + + + + +
      +       +
      2. + +
    2. + Integrands containing <m>\sqrt{x^2+a^2}</m> + + +

      + Let x=a\tan(\theta), dx = a\sec^2(\theta) \, d\theta. + Thus \theta = \tan^{-1}(x/a), + for -\pi/2 \lt \theta \lt \pi/2. + On this interval, \sec(\theta) \gt 0, + so \sqrt{x^2+a^2} = a\sec(\theta). +

      + +
      + + + + + Diagram showing trigonometric substitution with integrands containing square root of x^2+a^2. + + +

      + The diagram is of a right angled triangle. The perpendicular is marked x + the base as a. The hypotenuse is marked \sqrt{x^2+a^2}. The angle + opposite to the perpendicular is marked as \theta. +

      +       + + + \begin{tikzpicture}[scale=1.5] + + \draw [very thick] (0,0) -- node [below,pos=.5] { $a$} (3,0) -- node [right,pos=.5] { $x$} (3,2) -- node [pos=.5,above,sloped] { $\sqrt{x^2+a^2}$} (0,0); + \draw [thick] (2.7,0) -- (2.7,.3) -- (3,.3); + \draw (.75,.25) node {$\theta$}; + + \end{tikzpicture} + + + + +
      +       +
      3. + +
    3. + Integrands containing <m>\sqrt{x^2-a^2}</m> + + +

      + Let x=a\sec(\theta), + dx = a\sec(\theta) \tan(\theta) \, d\theta. + Thus \theta = \sec^{-1}(x/a). + If x/a\geq 1, then 0\leq\theta\lt \pi/2; + if x/a \leq -1, then \pi/2\lt \theta\leq \pi. + We restrict our work to where x\geq a, + so x/a\geq 1, and 0\leq\theta\lt \pi/2. + On this interval, \tan(\theta) \geq 0, + so \sqrt{x^2-a^2} = a\tan(\theta). +

      + +
      + + + + + Diagram showing trigonometric substitution with integrands containing square root of x^2-a^2. + + +

      + The diagram is of a right angled triangle. The perpendicular is marked \sqrt{x^2-a^2} + the base as a. The hypotenuse is marked x. The angle opposite to the perpendicular + is marked as \theta. +

      +       + + + \begin{tikzpicture}[scale=1.5] + + \draw [very thick] (0,0) -- node [below,pos=.5] { $a$} (3,0) -- node [right,xshift=3mm,pos=.9,rotate=-90] { $\sqrt{x^2-a^2}$} (3,2) -- node [pos=.5,above] { $x$} (0,0); + \draw [thick] (2.7,0) -- (2.7,.3) -- (3,.3); + \draw (.75,.25) node {$\theta$}; + + \end{tikzpicture} + + + + +
      +       +
      4. +
    +

    +     + + + + + + + Using Trigonometric Substitution + +

    + Evaluate \ds \int \frac{1}{\sqrt{5+x^2}}\, dx. +

    +     + +

    + Using Item in , + we recognize a=\sqrt{5} and set x= \sqrt{5}\tan(\theta). + This makes dx = \sqrt{5}\sec^2(\theta) \, d\theta. + We will use the fact that \sqrt{5+x^2} = \sqrt{5+5\tan^2(\theta) } = \sqrt{5\sec^2(\theta) } = \sqrt{5}\sec(\theta). + Substituting, we have: + + \int \frac{1}{\sqrt{5+x^2}}\, dx \amp = \int \frac{1}{\sqrt{5+5\tan^2(\theta) }}\sqrt{5}\sec^2(\theta) \, d\theta + \amp = \int \frac{\sqrt{5}\sec^2(\theta) }{\sqrt{5}\sec(\theta) } \, d\theta + \amp = \int \sec(\theta) \, d\theta + \amp = \ln\abs{\sec(\theta) +\tan(\theta) }+C + . +

    + +

    + While the integration steps are over, we are not yet done. + The original problem was stated in terms of x, + whereas our answer is given in terms of \theta. + We must convert back to x. +

    + +

    + The reference triangle given in helps. + With x=\sqrt{5}\tan(\theta), we have + + \tan(\theta) = \frac x{\sqrt{5}} \text{ and } \sec(\theta) = \frac{\sqrt{x^2+5}}{\sqrt{5}} + . +

    + +

    + This gives + + \int \frac{1}{\sqrt{5+x^2}}\, dx \amp = \ln\abs{\sec(\theta) +\tan(\theta) }+C + \amp = \ln\abs{\frac{\sqrt{x^2+5}}{\sqrt{5}}+ \frac x{\sqrt{5}}}+C + . +

    + +

    + We can leave this answer as is, + or we can use a logarithmic identity to simplify it. + Note: + + \ln\abs{\frac{\sqrt{x^2+5}}{\sqrt{5}}+ \frac x{\sqrt{5}}}+C \amp = \ln\abs{\frac{1}{\sqrt{5}}\big(\sqrt{x^2+5}+ x\big)}+C + \amp = \ln\abs{\frac{1}{\sqrt{5}}} + \ln\abs{\sqrt{x^2+5}+ x}+C + \amp = \ln\abs{\sqrt{x^2+5}+ x}+C + , + where the \ln\big(1/\sqrt{5}\big) term is absorbed into the constant C. + (In + we will learn another way of approaching this problem.) +

    +     + +     + + + Using Trigonometric Substitution + +

    + Evaluate \ds \int \sqrt{4x^2-1}\, dx. +

    +     + +

    + We start by rewriting the integrand so that it looks like + \sqrt{x^2-a^2} for some value of a: + + \sqrt{4x^2-1} \amp = \sqrt{4\left(x^2-\frac14\right)} + \amp = 2\sqrt{x^2-\left(\frac12\right)^2} + . +

    + +

    + So we have a=1/2, + and following Part of , + we set x= \frac12\sec(\theta), + and hence dx = \frac12\sec(\theta) \tan(\theta) \, d\theta. + We now rewrite the integral with these substitutions: + + \int \sqrt{4x^2-1}\, dx \amp = \int 2\sqrt{x^2-\left(\frac12\right)^2}\, dx + \amp = \int 2\sqrt{\frac14\sec^2(\theta) - \frac14}\left(\frac12\sec(\theta) \tan(\theta) \right)\, d\theta + \amp =\int \sqrt{\frac14(\sec^2(\theta) -1)}\Big(\sec(\theta) \tan(\theta) \Big)\, d\theta + \amp =\int\sqrt{\frac14\tan^2(\theta) }\Big(\sec(\theta) \tan(\theta) \Big)\, d\theta + \amp =\int \frac12\tan^2(\theta) \sec(\theta) \, d\theta + \amp =\frac12\int \Big(\sec^2(\theta) -1\Big)\sec(\theta) \, d\theta + \amp =\frac12\int \big(\sec^3(\theta) - \sec(\theta) \big)\, d\theta + . +

    + +

    + We integrated \sec^3(\theta) in , + finding its antiderivatives to be + + \int \sec^3(\theta) \, d\theta = \frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C + . +

    + +

    + Thus + + \amp \int \sqrt{4x^2-1}\, dx =\frac12\int \big(\sec^3(\theta) - \sec(\theta) \big)\, d\theta + \amp = \frac12\left(\frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big) -\ln\abs{\sec(\theta) + \tan(\theta) }\right) + C + \amp = \frac14\left(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\right)+C + . +

    + +

    + We are not yet done. + Our original integral is given in terms of x, + whereas our final answer, as given, + is in terms of \theta. + We need to rewrite our answer in terms of x. + With a=1/2, and x=\frac12\sec(\theta), + the reference triangle in shows that + + \tan(\theta) = \sqrt{x^2-1/4}\Big/(1/2) = 2\sqrt{x^2-1/4} \text{ and } \sec(\theta) = 2x + . +

    + +

    + Thus + + \amp \frac14\Big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C + \amp\quad = \frac14\Big(2x\cdot 2\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C + \amp\quad = \frac14\Big(4x\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C + . +

    + +

    + The final answer is given in the last line above, repeated here: + + \int \sqrt{4x^2-1}\, dx = \frac14\Big(4x\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C + . +

    +     + +     + + + Using Trigonometric Substitution + +

    + Evaluate \ds \int \frac{\sqrt{4-x^2}}{x^2}\, dx. +

    +     + +

    + We use Part of with a=2, + x=2\sin(\theta), + dx = 2\cos(\theta) and hence \sqrt{4-x^2} = 2\cos(\theta). + This gives + + \int \frac{\sqrt{4-x^2}}{x^2}\, dx \amp = \int \frac{2\cos(\theta) }{4\sin^2(\theta) }(2\cos(\theta) )\, d\theta + \amp = \int \cot^2(\theta) \, d\theta + \amp = \int (\csc^2(\theta) -1)\, d\theta + \amp = -\cot(\theta) -\theta + C + . +

    + +

    + We need to rewrite our answer in terms of x. + Using the reference triangle found in , + we have \cot(\theta) = \sqrt{4-x^2}/x and \theta = \sin^{-1}(x/2). + Thus + + \int \frac{\sqrt{4-x^2}}{x^2}\, dx = -\frac{\sqrt{4-x^2}}x-\sin^{-1}\left(\frac x2\right) + C + . +

    +     + +     + +

    + Trigonometric Substitution can be applied in many situations, + even those not of the form \sqrt{a^2-x^2}, + \sqrt{x^2-a^2} or \sqrt{x^2+a^2}. + In the following example, + we apply it to an integral we already know how to handle. +

    + + + Using Trigonometric Substitution + +

    + Evaluate \ds \int\frac1{x^2+1}\, dx. +

    +     + +

    + We know the answer already as \tan^{-1}(x) +C. + We apply Trigonometric Substitution here to show that we get the same answer without inherently relying on knowledge of the derivative of the arctangent function. +

    + +

    + Using Part of , + let x=\tan(\theta), + dx=\sec^2(\theta) \, d\theta and note that x^2+1 = \tan^2(\theta) +1 = \sec^2(\theta). + Thus + + \int \frac1{x^2+1}\, dx \amp = \int \frac{1}{\sec^2(\theta) }\sec^2(\theta) \, d\theta + \amp = \int 1\, d\theta + \amp = \theta + C + . +

    + +

    + Since x=\tan(\theta), \theta = \tan^{-1}(x), + and we conclude that \ds \int\frac1{x^2+1}\, dx = \tan^{-1}(x) +C. +

    +     +     + +

    + The next example is similar to the previous one in that it does not involve a square-root. + It shows how several techniques and identities can be combined to obtain a solution. +

    + + + Using Trigonometric Substitution + +

    + Evaluate \ds\int\frac1{(x^2+6x+10)^2}\, dx. +

    +     + +

    + We start by completing the square, + then make the substitution u=x+3, + followed by the trigonometric substitution of u=\tan(\theta): + + \int \frac1{(x^2+6x+10)^2}\, dx \amp =\int \frac1{\big((x+3)^2+1\big)^2}\, dx = \int \frac1{(u^2+1)^2}\,du. + Now make the substitution u=\tan(\theta), du=\sec^2(\theta) \, d\theta: + \amp = \int \frac1{(\tan^2(\theta) +1)^2}\sec^2(\theta) \, d\theta + \amp = \int\frac 1{(\sec^2(\theta) )^2}\sec^2(\theta) \, d\theta + \amp = \int \cos^2(\theta) \, d\theta. + Applying a power reducing formula, we have + \amp = \int \left(\frac12 +\frac12\cos(2\theta)\right)\, d\theta + \amp = \frac12\theta + \frac14\sin(2\theta) + C + . +

    + +

    + We need to return to the variable x. + As u=\tan(\theta), \theta = \tan^{-1}(u). + Using the identity \sin(2\theta) = 2\sin(\theta) \cos(\theta) and using the reference triangle found in + , we have + + \frac14\sin(2\theta) = \frac12\frac u{\sqrt{u^2+1}}\cdot\frac 1{\sqrt{u^2+1}} = \frac12\frac u{u^2+1} + . +

    + +

    + Finally, we return to x with the substitution u=x+3. + We start with the expression in Equation: + + \frac12\theta + \frac14\sin(2\theta) + C \amp = \frac12\tan^{-1}(u) + \frac12\frac{u}{u^2+1}+C + \amp = \frac12\tan^{-1}(x+3) + \frac{x+3}{2(x^2+6x+10)}+C + . +

    + +

    + Stating our final result in one line, + + \int\frac1{(x^2+6x+10)^2}\, dx=\frac12\tan^{-1}(x+3) + \frac{x+3}{2(x^2+6x+10)}+C + . +

    +     + +     + + + +

    + Our last example returns us to definite integrals, + as seen in our first example. + Given a definite integral that can be evaluated using Trigonometric Substitution, + we could first evaluate the corresponding indefinite integral + (by changing from an integral in terms of x to one in terms of \theta, + then converting back to x) + and then evaluate using the original bounds. + It is much more straightforward, though, + to change the bounds as we substitute. +

    + + + Definite integration and Trigonometric Substitution + +

    + Evaluate \ds\int_0^5\frac{x^2}{\sqrt{x^2+25}}\, dx. +

    +     + +

    + Using Part of , + we set x=5\tan(\theta), + dx = 5\sec^2(\theta) \, d\theta, + and note that \sqrt{x^2+25} = 5\sec(\theta). + As we substitute, we can also change the bounds of integration. +

    + +

    + The lower bound of the original integral is x=0. + As x=5\tan(\theta), + we solve for \theta and find \theta = \tan^{-1}(x/5). + Thus the new lower bound is \theta = \tan^{-1}(0) = 0. + The original upper bound is x=5, + thus the new upper bound is \theta = \tan^{-1}(5/5) = \pi/4. +

    + +

    + Thus we have + + \int_0^5\frac{x^2}{\sqrt{x^2+25}}\, dx \amp = \int_0^{\pi/4} \frac{25\tan^2(\theta) }{5\sec(\theta) }5\sec^2(\theta) \, d\theta + \amp = 25\int_0^{\pi/4} \tan^2(\theta) \sec(\theta) \, d\theta + . +

    + +

    + We encountered this indefinite integral in where we found + + \int \tan^2(\theta) \sec(\theta) \, d\theta = \frac12\big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\big) + . +

    + +

    + So + + 25\int_0^{\pi/4} \tan^2(\theta) \sec(\theta) \, d\theta \amp = \left.\frac{25}2\big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\big)\right|_0^{\pi/4} + \amp = \frac{25}2\big(\sqrt2-\ln(\sqrt2+1)\big) + \amp \approx 6.661 + . +

    +     + +     + +

    + The following equalities are very useful when evaluating integrals using Trigonometric Substitution. +

    + + + Useful Equalities with Trigonometric Substitution +

    +

      +
    1. +

      + \sin(2\theta) = 2\sin(\theta) \cos(\theta) +

      +
      2. + +
    2. +

      + \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) -1 = 1-2\sin^2(\theta) +

      +
      3. + +
    3. +

      + \ds \int \sec^3(\theta) \, d\theta = \frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C +

      +
      4. + +
    4. +

      + \ds \int \cos^2(\theta) \, d\theta = \int \frac12\big(1+\cos(2\theta)\big)\, d\theta = \frac12\big(\theta+\sin(\theta) \cos(\theta) \big)+C. +

      +
      5. +
    +

    +     + +

    + The next section introduces Partial Fraction Decomposition, + which is an algebraic technique that turns complicated + fractions into sums of simpler fractions, + making integration easier. +

    + + + + Terms and Concepts + + + + +

    + Trigonometric Substitution works on the same principles as Integration by Substitution, + though it can feel . +

    +     + + + + backwards? + + + + +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=5;}; + Context()->variables->add(theta=>['Real',TeX=>'\theta']); + $integrand = Formula("sqrt($m^2-x^2)"); + $answer = OneOf(Formula("$m*sin(theta)"),Formula("$m*cos(theta)")); + + +

    + If one uses Trigonometric Substitution on an integrand containing , + then one should set x equal to what? +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=9;}; + Context()->variables->add(theta=>['Real',TeX=>'\theta']); + # Redefine tan() so the following equtaion is not an identity + package my::Function::numeric; + our @ISA = ('Parser::Function::numeric'); + sub tan { + shift; my $x = shift; + return CORE::sin($x+2)/CORE::cos($x+2); + } + package main; + Context()->functions->set(tan => {class => 'my::Function::numeric'}); + $I = ImplicitEquation("tan(theta)^2+1=sec(theta)^2"); + $S = Formula("$m sec(theta)^2"); + + +

    + Consider the Pythagorean Identity \sin^2(\theta) +\cos^2(\theta) = 1. +

    +

    +

      +
    1. +

      + What identity is obtained when both sides are divided by \cos^2(\theta)? +

      +

      + +

      +
      2. +
    2. +

      + Use the new identity to simplify 9\tan^2(\theta) + 9. +

      +

      + +

      +
      3. +
    +

    +     +     +     + + + + +

    + Why does Part of + state that \sqrt{a^2-x^2} = a\cos(\theta), + and not \abs{a\cos(\theta) }? +

    + +     + + + +

    + Because we are considering a>0 and x=a\sin(\theta), + which means \theta = \sin^{-1}(x/a). + The arcsine function has a domain of -\pi/2\leq \theta \leq \pi/2; + on this domain, \cos(\theta) \geq 0, + so a\cos(\theta) is always non-negative, + allowing us to drop the absolute value signs. +

    +     + +     +     + + + Problems + + +

    + Apply Trigonometric Substitution to evaluate the indefinite integral. +

    +     + + + + + Context("Fraction"); + $F = FormulaUpToConstant("1/2 (x sqrt(x^2 + 1) + ln(sqrt(x^2 + 1) + x))"); + + +

    + \ds \int \sqrt{x^2+1}\, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + $F = FormulaUpToConstant("x/2 sqrt(x^2 + 4) + 2 ln(sqrt(x^2 + 4)/2 + x/2)"); + + +

    + \ds \int \sqrt{x^2+4}\, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->variables->set(x=>{limits=>[-1,1]}); + $F = FormulaUpToConstant("1/2 asin(x) + x/2 sqrt(1 - x^2)"); + + +

    + \ds \int \sqrt{1-x^2}\, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->variables->set(x=>{limits=>[-3,3]}); + $F = FormulaUpToConstant("9/2 asin(x/3) + x/2 sqrt(9 - x^2)"); + + +

    + \ds \int \sqrt{9-x^2}\, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->variables->set(x=>{limits=>[1,5]}); + $F = FormulaUpToConstant("1/2 x sqrt(x^2 - 1) - 1/2 ln(abs(x+sqrt(x^2 - 1)))"); + $F->{test_at} = [-2,-3,-4]; + + +

    + \ds \int \sqrt{x^2-1}\, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->variables->set(x=>{limits=>[4,9]}); + $F = FormulaUpToConstant("1/2 x sqrt(x^2 - 16) - 8 ln(abs(x/4 + sqrt(x^2 - 16)/4))"); + $F->{test_at} = [-5,-6,-7]; + + +

    + \ds \int \sqrt{x^2-16}\, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=2;}; + Context("Fraction"); + $f = Formula("sqrt($m^2 x^2+1)"); + $F = FormulaUpToConstant("x/2 sqrt($m^2 x^2+1)+1/(2*$m)*ln($m x + sqrt($m^2 x^2+1))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=3;}; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[-1/$m,1/$m]}); + $f = Formula("sqrt(1 - $m^2 x^2)"); + $F = FormulaUpToConstant("x/2 sqrt(1 - $m^2 x^2) + 1/(2*$m) asin($m x)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=4;}; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[1/$m,1/$m+4]}); + $f = Formula("sqrt($m^2 x^2 - 1)"); + $F = FormulaUpToConstant("x/2 sqrt($m^2 x^2 - 1) - 1/(2*$m) ln(abs($m x + sqrt($m^2 x^2 - 1)))")->reduce; + $F->{test_at} = [-1/$m-1,-1/$m-2,-1/$m-3]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $b = list_random(2,3,5,6,7,10,11,13,14,15); + if($envir{problemSeed}==1){$a=8;$b=2}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$a/sqrt(x^2 + $b)"); + $F = FormulaUpToConstant("$a ln(x/sqrt($b) + sqrt(x^2/$b + 1))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $b = list_random(2,3,5,6,7,10,11,13,14,15); + if($envir{problemSeed}==1){$a=3;$b=7}; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[-sqrt($b),sqrt($b)]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$a/sqrt($b - x^2)"); + $F = FormulaUpToConstant("$a asin(x/sqrt($b))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $b = list_random(2,3,5,6,7,10,11,13,14,15); + if($envir{problemSeed}==1){$a=5;$b=6}; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[sqrt($b),sqrt($b)+4]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("$a/sqrt(x^2-$b)"); + $F = FormulaUpToConstant("$a ln(abs(x/sqrt($b) + sqrt(x^2/$b - 1)))")->reduce; + $F->{test_at} = [-sqrt($b)-1,-sqrt($b)-2,-sqrt($b)-3]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Evaluate the indefinite integral. + Trigonometric Substitution may not be required. +

    +     + + + + + $b = list_random(2,3,5,6,7,10,11,13,14,15); + if($envir{problemSeed}==1){$b=11}; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[sqrt($b),sqrt($b)+4]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("sqrt(x^2-$b)/x"); + $F = FormulaUpToConstant("sqrt(x^2 - $b) - sqrt($b) asec(x/sqrt($b))")->reduce; + $F->{test_at} = [-sqrt($b)-1,-sqrt($b)-2,-sqrt($b)-3]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + $F = FormulaUpToConstant("1/2 atan(x) + x/[2(x^2+1)]"); + + +

    + \ds \int \frac {1}{(x^2+1)^2}\, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random(2,3,5,6,7,10,11,13,14,15); + if($envir{problemSeed}==1){$b=3}; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[sqrt($b),sqrt($b)+4]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("x/sqrt(x^2-$b)"); + $F = FormulaUpToConstant("sqrt(x^2 - $b)")->reduce; + $F->{test_at} = [-sqrt($b)-1,-sqrt($b)-2,-sqrt($b)-3]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->variables->set(x=>{limits=>[-1,1]}); + $F = FormulaUpToConstant("1/8 asin(x) + x/8 sqrt(1-x^2) (2x^2 - 1)"); + + +

    + \ds \int x^2\sqrt{1-x^2}\, dx +

    +

    + +

    +     +     +     + + + + + $m = random(2,9,1); + if($envir{problemSeed}==1){$m=3}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("x/(x^2 + $m^2)^(3/2)"); + $F = FormulaUpToConstant("-1/sqrt(x^2 + $m^2)")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $b = list_random(2,3,5,6,7,10,11,13,14,15); + if($envir{problemSeed}==1){$a=5;$b=10;}; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[sqrt($b),sqrt($b)+4]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("($a x^2)/sqrt(x^2 - $b)"); + $frac = Fraction($a,2); + ($num,$den) = $frac->value; + $frac2 = Fraction($a*$b,2); + $F = FormulaUpToConstant("($num x)/$den sqrt(x^2 - $b) + $frac2 ln(abs(x/sqrt($b) + sqrt(x^2/$b - 1)))")->reduce; + $F->{test_at} = [-sqrt($b)-1,-sqrt($b)-2,-sqrt($b)-3]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $h = non_zero_random(-9,9,1); + $m = random(1,9,1); + if($envir{problemSeed}==1){$h=-2;$m=3;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("1/(x^2 - (2*$h) x + (($h)^2+$m^2))^2")->reduce; + $frac = Fraction(1,2*$m**2); + $frac2 = Fraction(1,2*$m**3); + $F = FormulaUpToConstant("$frac (x-$h)/(x^2 - (2*$h) x + (($h)^2+$m^2)) + $frac2 atan((x-$h)/$m) ")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + Context("Fraction"); + Context()->variables->set(x=>{limits=>[-1,1]}); + $F = FormulaUpToConstant("x/sqrt(1-x^2)-asin(x)"); + + +

    + \ds \int x^2(1-x^2)^{-3/2}\, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + $b = list_random(2,3,5,6,7,10,11,13,14,15); + if($envir{problemSeed}==1){$a=7;$b=5;}; + Context("Fraction"); + Context()->variables->set(x=>{limits=>[-sqrt($b),sqrt($b)]}); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("sqrt($b - x^2)/($a x^2)"); + $F = FormulaUpToConstant("-sqrt($b - x^2)/($a x) - 1/$a asin(x/sqrt($b))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $b = list_random(2,3,5,6,7,10,11,13,14,15); + if($envir{problemSeed}==1){$b=3;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("x^2/sqrt(x^2 + $b)"); + $frac = Fraction($b,2); + $F = FormulaUpToConstant("x/2 sqrt(x^2 + $b) - $frac ln(x/sqrt(3) + sqrt(x^2/$b + 1))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Evaluate the definite integral by making the proper trigonometric substitution and + changing the bounds of integration. + (Note: the corresponding indefinite integrals appeared previously in the exercises.) +

    +     + + + + + $answer = Compute("pi/2"); + + +

    + \ds \int_{-1}^1 \sqrt{1-x^2}\, dx +

    +

    + +

    +     +     +     + + + + + $b = random(5,10,1); + if($envir{problemSeed}==1){$b=8;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $arg = $b**2 - 16; + ($sq,$nsq) = (1,$arg); + while ($nsq % 4 == 0) { + $sq *= 2; + $nsq /= 4; + } + while ($nsq % 9 == 0) { + $sq *= 3; + $nsq /= 9; + } + $frac = Fraction($b*$sq,2); + $frac2 = Fraction($b,4); + $frac3 = Fraction($sq,4); + $answer = Formula("$frac sqrt($nsq) - 8 ln(abs($frac2 + $frac3 sqrt($nsq)))"); + + +

    + \ds \int_{4}^{} \sqrt{x^2-16}\, dx +

    +

    + +

    +     +     +     + + + + + $b = random(1,10,1); + if($envir{problemSeed}==1){$b=2;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $arg = $b**2 + 4; + ($sq,$nsq) = (1,$arg); + while ($nsq % 4 == 0) { + $sq *= 2; + $nsq /= 4; + } + while ($nsq % 9 == 0) { + $sq *= 3; + $nsq /= 9; + } + $frac = Fraction($b*$sq,2); + $frac2 = Fraction($b,2); + $frac3 = Fraction($sq,2); + $answer = Formula("$frac sqrt($nsq) + 2 ln($frac2 + $frac3 sqrt($nsq))"); + + +

    + \ds \int_{0}^{} \sqrt{x^2+4}\, dx +

    +

    + +

    +     +     +     + + + + + $b = random(1,10,1); + if($envir{problemSeed}==1){$b=1;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $frac = Fraction($b,$b**2 + 1); + $answer = Formula("atan($b) + $frac"); + + +

    + \ds \int_{-}^{} \frac1{(x^2+1)^2}\, dx +

    +

    + +

    +     +     +     + + + + + $b = random(1,2,1); + if($envir{problemSeed}==1){$b=1;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $arg = 9 - $b**2; + ($sq,$nsq) = (1,$arg); + while ($nsq % 4 == 0) { + $sq *= 2; + $nsq /= 4; + } + while ($nsq % 9 == 0) { + $sq *= 3; + $nsq /= 9; + } + $frac = Fraction($b,3); + $frac2 = Fraction($b*$sq,1); + $answer = Formula("9 asin($frac) + $frac2 sqrt($nsq)"); + + +

    + \ds \int_{-}^{} \sqrt{9-x^2}\, dx +

    +

    + +

    +     +     +     + + + + + $answer = Formula("pi/8"); + + +

    + \ds \int_{-1}^1 x^2\sqrt{1-x^2}\, dx +

    +

    + +

    +     +     +     +     +     +     +
    +
    + Partial Fraction Decomposition +

    + In this section we investigate the antiderivatives of rational functions. + Recall that rational functions are functions of the form f(x)= \frac{p(x)}{q(x)}, + where p(x) and q(x) are polynomials and q(x)\neq 0. + Such functions arise in many contexts, + one of which is the solving of certain fundamental differential equations. +

    + + + +

    + We begin with an example that demonstrates the motivation behind this section. + Consider the integral \ds\int \frac{1}{x^2-1}\, dx. + We do not have a simple formula for this + (if the denominator were x^2+1, + we would recognize the antiderivative as being the arctangent function). + It can be solved using Trigonometric Substitution, + but note how the integral is easy to evaluate once we realize: + + \frac{1}{x^2-1} = \frac{1/2}{x-1} - \frac{1/2}{x+1} + . +

    + +

    + Thus + + \int\frac{1}{x^2-1}\, dx \amp = \int\frac{1/2}{x-1}\, dx - \int\frac{1/2}{x+1}\, dx + \amp = \frac12\ln\abs{x-1} - \frac12\ln\abs{x+1} + C + . +

    + +

    + This section teaches how to decompose + + \frac{1}{x^2-1} \text{ into } \frac{1/2}{x-1}-\frac{1/2}{x+1} + . +

    + +

    + We start with a rational function f(x)=\frac{p(x)}{q(x)}, + where p and q do not have any common factors and the degree of p is less than the degree of q. + It can be shown that any polynomial, and hence q, + can be factored into a product of linear and irreducible quadratic terms. + The following Key Idea states how to decompose a rational function into a sum of rational functions whose denominators are all of lower degree than q. +

    + + + Partial Fraction Decomposition +

    + Let \ds \frac{p(x)}{q(x)} be a rational function, + where the degree of p is less than the degree of q. + integrationpartial fraction decomp. + +

    + +

    +

      +
    1. +

      + Linear Terms: Let (x-a) divide q(x), + where (x-a)^n is the highest power of (x-a) that divides q(x). + Then the decomposition of \frac{p(x)}{q(x)} will contain the sum + + \frac{A_1}{(x-a)} + \frac{A_2}{(x-a)^2} + \cdots +\frac{A_n}{(x-a)^n} + . +

      +
      2. + +
    2. +

      + Quadratic Terms: Let + x^2+bx+c be an irreducible quadratic that divides q(x), + where (x^2+bx+c)^n is the highest power of + x^2+bx+c that divides q(x). + Then the decomposition of \frac{p(x)}{q(x)} will contain the sum + + \frac{B_1x+C_1}{x^2+bx+c}+\frac{B_2x+C_2}{(x^2+bx+c)^2}+\cdots+\frac{B_nx+C_n}{(x^2+bx+c)^n} + . +

      +
      3. +
    +

    + + + +

    + To find the coefficients A_i, + B_i and C_i: +

    + +

    +

      +
    1. +

      + Multiply all fractions by q(x), + clearing the denominators. + Collect like terms. +

      +
      2. + +
    2. +

      + Equate the resulting coefficients of the powers of x and solve the resulting system of linear equations. +

      +
      3. +
    +

    +     + + +

    + The following examples will demonstrate how to put this Key Idea into practice. + + stresses the decomposition aspect of the Key Idea. +

    + + + Decomposing into partial fractions + +

    + Decompose \ds f(x)=\frac{1}{(x+5)(x-2)^3(x^2+x+2)(x^2+x+7)^2} without solving for the resulting coefficients. +

    +     + +

    + The denominator is already factored, + as both x^2+x+2 and x^2+x+7 cannot be factored further. + We need to decompose f(x) properly. + Since (x+5) is a linear term that divides the denominator, + there will be a + + \frac{A}{x+5} + + term in the decomposition. +

    + +

    + As (x-2)^3 divides the denominator, + we will have the following terms in the decomposition: + + \frac{B}{x-2}, \frac{C}{(x-2)^2} \text{ and } \frac{D}{(x-2)^3} + . +

    + +

    + The x^2+x+2 term in the denominator results in a \ds\frac{Ex+F}{x^2+x+2} term. +

    + +

    + Finally, the (x^2+x+7)^2 term results in the terms + + \frac{Gx+H}{x^2+x+7} \text{ and } \frac{Ix+J}{(x^2+x+7)^2} + . +

    + +

    + All together, we have + + \amp\frac{1}{(x+5)(x-2)^3(x^2+x+2)(x^2+x+7)^2} + \amp \quad\quad = \frac{A}{x+5} + \frac{B}{x-2}+ \frac{C}{(x-2)^2}+\frac{D}{(x-2)^3} + \amp \quad\quad\quad\quad + \frac{Ex+F}{x^2+x+2}+\frac{Gx+H}{x^2+x+7}+\frac{Ix+J}{(x^2+x+7)^2} + +

    + +

    + Solving for the coefficients A, + B \ldots J would be a bit tedious but not hard. +

    +     + +     + + + Decomposing into partial fractions + +

    + Perform the partial fraction decomposition of \ds \frac{1}{x^2-1}. +

    +     + +

    + The denominator factors into two linear terms: + x^2-1 = (x-1)(x+1). + Thus + + \frac{1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1} + . +

    + +

    + To solve for A and B, + first multiply through by x^2-1 = (x-1)(x+1): + + 1 \amp = \frac{A(x-1)(x+1)}{x-1}+\frac{B(x-1)(x+1)}{x+1} + \amp = A(x+1) + B(x-1) + \amp = Ax+A + Bx-B + + \amp = (A+B)x + (A-B) + , + by collecting like terms. +

    + +

    + The next step is key. + Note the equality we have: + + 1 = (A+B)x+(A-B) + . +

    + +

    + For clarity's sake, rewrite the left hand side as + + 0x+1 = (A+B)x+(A-B) + . +

    + +

    + On the left, the coefficient of the x term is 0; + on the right, it is (A+B). + Since both sides are equal, we must have that 0=A+B. +

    + +

    + Likewise, on the left, we have a constant term of 1; + on the right, the constant term is (A-B). + Therefore we have 1=A-B. +

    + +

    + We have two linear equations with two unknowns. + This one is easy to solve by hand, leading to + + A+B \amp = 0 + A-B \amp = 1 + + If we add these two equations, + we get 2A=1 \Rightarrow A=1/2. + Substitution into the first equation gives B=-1/2. +

    + +

    + Thus + + \frac{1}{x^2-1} = \frac{1/2}{x-1}-\frac{1/2}{x+1} + . +

    +     + +     + + + +

    + There is another method for finding the partial fraction decomposition called the + Heaviside method, + named after Oliver Heaviside. + We show a variation of this process using the same example as in . +

    + + + Decomposing into partial fractions using the Heaviside method + +

    + Perform the partial fraction decomposition of \ds \frac{1}{x^2-1}. +

    +     + +

    + As we saw in , + + \frac{1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1} + . +

    + +

    + To solve for A and B using the Heaviside method, + we will build to a common denominator: + + \frac{1}{x^2-1} \amp = \frac{A(x+1)}{(x-1)(x+1)}+\frac{B(x-1)}{(x+1)(x-1)} + \amp = \frac{A(x+1)+B(x-1)}{(x-1)(x+1)} + +

    + +

    + Now since the denomiators match, + we will only consider the numerator equation (essentially if we multiply both sides of the equation by (x-1)(x+1), + we will clear the denomiators): + + 1=A(x+1)+B(x-1) + + Now we substitute in convenient values of x. + When x=1, we get 1=2A \Rightarrow A=1/2. + When x=-1, we get 1=-2B \Rightarrow B=-1/2. +

    + +

    + You may note that x=1 and x=-1 were not in the domain of the original fraction. + However, + + \frac{1}{x^2-1}=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)} + + is an identity, meaning it is true for all values of x, + even those for which the equation is undefined. + We could have chosen any values of x to substitute. + Whenever possible, + we choose values of x that will make one of the factors zero. + In this way, we can avoid solving a system of equations. +

    + +

    + Thus as in , we get + + \frac{1}{x^2-1} = \frac{1/2}{x-1}-\frac{1/2}{x+1} + . +

    +     + +     + +

    + For the remaining examples, we will use a combination of systems of equations and the Heaviside method to get partial fraction decompositions. +

    + + + Integrating using partial fractions + +

    + Use partial fraction decomposition to integrate \ds\int\frac{1}{(x-1)(x+2)^2}\, dx. +

    +     + +

    + We decompose the integrand as follows, + as described by : + + \frac{1}{(x-1)(x+2)^2} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} + . +

    + +

    + To solve for A, B and C, + we multiply both sides by (x-1)(x+2)^2: + + 1 = A(x+2)^2 + B(x-1)(x+2) + C(x-1) + +

    + +

    + Now we collect like terms: + + 1 \amp = A(x+2)^2 + B(x-1)(x+2) + C(x-1) + \amp = Ax^2+4Ax+4A + Bx^2 + Bx-2B + Cx-C + \amp = (A+B)x^2 + (4A+B+C)x + (4A-2B-C) + +

    + + + +

    + We have + + 0x^2+0x+ 1 = (A+B)x^2 + (4A+B+C)x + (4A-2B-C) + + leading to the equations + + A+B = 0, 4A+B+C = 0 \text{ and } 4A-2B-C = 1 + . +

    + +

    + These three equations of three unknowns lead to a unique solution: + + A = 1/9, B = -1/9 \text{ and } C = -1/3 + . +

    + +

    + Thus + + \int\frac{1}{(x-1)(x+2)^2}\, dx = \int \frac{1/9}{x-1}\, dx + \int \frac{-1/9}{x+2}\, dx + \int \frac{-1/3}{(x+2)^2}\, dx + . +

    + +

    + Each can be integrated with a simple substitution with u=x-1 or u=x+2 + (or by directly applying + as the denominators are linear functions). + The end result is + + \int\frac{1}{(x-1)(x+2)^2}\, dx = \frac19\ln\abs{x-1} -\frac19\ln\abs{x+2} +\frac1{3(x+2)}+C + . +

    +     + +     + + + + + + + Integrating using partial fractions + +

    + Use partial fraction decomposition to integrate \ds \int \frac{x^3}{(x-5)(x+3)}\, dx. +

    +     + +

    + + presumes that the degree of the numerator is less than the degree of the denominator. + Since this is not the case here, + we begin by using polynomial division to reduce the degree of the numerator. + We omit the steps, but encourage the reader to verify that + + \frac{x^3}{(x-5)(x+3)} = x+2+\frac{19x+30}{(x-5)(x+3)} + . +

    + +

    + Using , + we can rewrite the new rational function as: + + \frac{19x+30}{(x-5)(x+3)} = \frac{A}{x-5} + \frac{B}{x+3} + + for appropriate values of A and B. + Clearing denominators, we have +

    + + +

    + + 19x+30 \amp = A(x+3) + B(x-5) + \amp = (A+B)x + (3A-5B). + This implies that: + 19\amp = A+B + 30\amp = 3A-5B. + Solving this system of linear equations gives + 125/8 \amp =A + 27/8 \amp =B + . +

    +

    + We can now integrate. + + \int \frac{x^3}{(x-5)(x+3)}\, dx \amp = \int\left(x+2+\frac{125/8}{x-5}+\frac{27/8}{x+3}\right)\, dx + \amp = \frac{x^2}2 + 2x + \frac{125}{8}\ln\abs{x-5} + \frac{27}8\ln\abs{x+3} + C + . +

    +     + +     + + + Integrating using partial fractions + +

    + Use partial fraction decomposition to evaluate \ds \int\frac{7x^2+31x+54}{(x+1)(x^2+6x+11)}\, dx. +

    +     + +

    + The degree of the numerator is less than the degree of the denominator so we begin by applying . + We have: + + + \frac{7x^2+31x+54}{(x+1)(x^2+6x+11)} \amp = \frac{A}{x+1} + \frac{Bx+C}{x^2+6x+11}. + Now clear the denominators. + 7x^2+31x+54 \amp = A(x^2+6x+11) + (Bx+C)(x+1) + + Now, letting x=-1 we have 30 = 6A \Rightarrow A=5. + When x=0, 54 = 11A+C. + But we know that A=5, + so 54 =55+C \Rightarrow C=-1 Finally, we choose x=1 + (with A=5, C=-1) + we have 92=90+(B-1)(2)\Rightarrow B=2. +

    + +

    + + Thus + + \int\frac{7x^2+31x+54}{(x+1)(x^2+6x+11)}\, dx = \int\left(\frac{5}{x+1} + \frac{2x-1}{x^2+6x+11}\right)\, dx + . +

    + +

    + The first term of this new integrand is easy to evaluate; + it leads to a 5\ln\abs{x+1} term. + The second term is not hard, + but takes several steps and uses substitution techniques. +

    + +

    + The integrand \ds \frac{2x-1}{x^2+6x+11} has a quadratic in the denominator and a linear term in the numerator. + This leads us to try substitution. + Let u = x^2+6x+11, so du = (2x+6)\, dx. + The numerator is 2x-1, not 2x+6, + but we can get a 2x+6 term in the numerator by adding 0 in the form of 7-7. + + \frac{2x-1}{x^2+6x+11} \amp = \frac{2x-1+7-7}{x^2+6x+11} + \amp = \frac{2x+6}{x^2+6x+11} - \frac{7}{x^2+6x+11} + . +

    + +

    + We can now integrate the first term with substitution, + leading to a \ln\abs{x^2+6x+11} term. + The final term can be integrated using arctangent. (We can tell there is no further factoring for this quadratic since the denominator has no real solutions). + First, complete the square in the denominator: + + \frac{7}{x^2+6x+11} = \frac{7}{(x+3)^2+2} + . +

    + +

    + An antiderivative of the latter term can be found using and substitution: + + \int \frac{7}{x^2+6x+11}\, dx = \frac{7}{\sqrt{2}}\tan^{-1}\left(\frac{x+3}{\sqrt{2}}\right)+C + . +

    + +

    + Let's start at the beginning and put all of the steps together. + + \amp \int\frac{7x^2+31x+54}{(x+1)(x^2+6x+11)}\, dx + \amp = \int\left(\frac{5}{x+1} + \frac{2x-1}{x^2+6x+11}\right)\, dx + \amp = \int\frac{5}{x+1}\, dx + \int\frac{2x+6}{x^2+6x+11}\, dx -\int\frac{7}{(x+3)^2+2}\, dx + \amp = 5\ln\abs{x+1}+ \ln\abs{x^2+6x+11} -\frac{7}{\sqrt{2}}\tan^{-1}\left(\frac{x+3}{\sqrt{2}}\right)+C + . +

    + +

    + As with many other problems in calculus, + it is important to remember that one is not expected to see + the final answer immediately after seeing the problem. + Rather, given the initial problem, + we break it down into smaller problems that are easier to solve. + The final answer is a combination of the answers of the smaller problems. +

    +     + +     + + + +

    + Partial Fraction Decomposition is an important tool when dealing with rational functions. + Note that at its heart, it is a technique of algebra, + not calculus, as we are rewriting a fraction in a new form. + Regardless, it is very useful in the realm of calculus as it lets us evaluate a certain set of + complicated integrals. +

    + +

    + introduces new functions, + called the Hyperbolic Functions. + They will allow us to make substitutions similar to those found when studying Trigonometric Substitution, + allowing us to approach even more integration problems. +

    + + + + Terms and Concepts + + + +

    + Partial Fraction Decomposition is a method of rewriting functions. +

    +     + + + + + rational + + + + + +     + + + + +

    + + It is sometimes necessary to use polynomial division before using Partial Fraction Decomposition. +

    +     + +     + + +

    + Decompose without solving for the coefficients, + as done in . +

    +     + + + + + $r = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$r=-3}; + Context("Form")->variables->add(s=>"Real",t=>"Real",A=>"Real",B=>"Real",C=>"Real"); + Context()->reductions->set("(-x)+y" => 1); + $f = Formula("1/(x^2 + $r x)")->reduce; + $template = Formula("s/x + t/(x+$r)")->reduce, + @answer = (); + for my $s (A..C) { + for my $t (A..C) { + if ($s ne $t) { + for my $ss ("","-") { + for my $st ("","-") { + push(@answer,$template->substitute(s=>Formula("$ss $s"),t=>Formula("$st $t"))->reduce); + } + } + } + } + } + $evaluator = $answer[0]->cmp( + bypass_equivalence_test => 1, + checker=> sub { + my ($correct,$student,$ansHash) = @_; + for my $ans (@answer) { + if ($ans->cmp->evaluate($student)->{score} == 1) {return 1}; + } + Value->Error("Your answer should use paramaters A, B, and/or C.") unless ($student->usesOneOf(A..C)); + return 0; + }); + + +

    + \ds +

    + + Use A, B, and C as needed. + Do not actually solve for these coefficients. + +

    + +

    +     +     +     + + + + + $r = random(1,9,1); + ($a,$b) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$r=3;$a=-1;$b=7}; + Context("Form")->variables->add(s=>"Real",t=>"Real",A=>"Real",B=>"Real",C=>"Real"); + Context()->reductions->set("(-x)+y" => 1); + $f = Formula("($a x+$b)/(x^2 - $r^2)")->reduce; + $template = Formula("s/(x-$r) + t/(x+$r)")->reduce, + @answer = (); + for my $s (A..C) { + for my $t (A..C) { + if ($s ne $t) { + for my $ss ("","-") { + for my $st ("","-") { + push(@answer,$template->substitute(s=>Formula("$ss $s"),t=>Formula("$st $t"))->reduce); + } + } + } + } + } + $evaluator = $answer[0]->cmp( + bypass_equivalence_test => 1, + checker=> sub { + my ($correct,$student,$ansHash) = @_; + for my $ans (@answer) { + if ($ans->cmp->evaluate($student)->{score} == 1) {return 1}; + } + Value->Error("Your answer should use paramaters A, B, and/or C.") unless ($student->usesOneOf(A..C)); + return 0; + }); + + +

    + \ds +

    + + Use A, B, and C as needed. + Do not actually solve for these coefficients. + +

    + +

    +     +     +     + + + + + $r = list_random(2,3,5,6,7,10,11,13,14,15); + $b = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$r=7;$b=-3}; + Context("Form")->variables->add(s=>"Real",t=>"Real",A=>"Real",B=>"Real",C=>"Real"); + Context()->reductions->set("(-x)+y" => 1); + $f = Formula("(x+$b)/(x^2 - $r)")->reduce; + $template = Formula("s/(x-sqrt($r)) + t/(x+sqrt($r))"); + @answer = (); + for my $s (A..C) { + for my $t (A..C) { + if ($s ne $t) { + for my $ss ("","-") { + for my $st ("","-") { + push(@answer,$template->substitute(s=>Formula("$ss $s"),t=>Formula("$st $t"))); + } + } + } + } + } + $evaluator = $answer[0]->cmp( + bypass_equivalence_test => 1, + checker=> sub { + my ($correct,$student,$ansHash) = @_; + for my $ans (@answer) { + if ($ans->cmp->evaluate($student)->{score} == 1) {return 1}; + } + Value->Error("Your answer should use paramaters A, B, and/or C.") unless ($student->usesOneOf(A..C)); + return 0; + }); + + +

    + \ds +

    + + Use A, B, and C as needed. + Do not actually solve for these coefficients. + +

    + +

    +     +     +     + + + + + $r = random(2,9,1); + ($a,$b) = random_subset(2,1..9); + if($envir{problemSeed}==1){$r=7;$a=2;$b=5}; + Context("Form")->variables->add(r=>"Real",s=>"Real",t=>"Real",A=>"Real",B=>"Real",C=>"Real"); + Context()->reductions->set("(-x)+y" => 1); + $f = Formula("($a x + $b)/(x^3 + $r x)")->reduce; + $template = Formula("r/x + (s x + t)/(x^2 + $r)")->reduce, + @answer = (); + for my $r (A..C) { + for my $s (A..C) { + if ($r ne $s) { + $t = 'C' if ($r.$s eq 'AB' or $r.$s eq 'BA'); + $t = 'B' if ($r.$s eq 'AC' or $r.$s eq 'CA'); + $t = 'A' if ($r.$s eq 'BC' or $r.$s eq 'CB'); + for my $sr ("","-") { + for my $ss ("","-") { + for my $st ("","-") { + push(@answer,$template->substitute(r=>Formula("$sr $r"), s=>Formula("$ss $s"),t=>Formula("$st $t"))->reduce); + } + } + } + } + } + } + $evaluator = $answer[0]->cmp( + bypass_equivalence_test => 1, + checker=> sub { + my ($correct,$student,$ansHash) = @_; + for my $ans (@answer) { + if ($ans->cmp->evaluate($student)->{score} == 1) {return 1}; + } + Value->Error("Your answer should use paramaters A, B, and/or C.") unless ($student->usesOneOf(A..C)); + return 0; + }); + + +

    + \ds +

    + + Use A, B, and C as needed. + Do not actually solve for these coefficients. + +

    + +

    +     +     +     +     +     + + + Problems + + +

    + Evaluate the indefinite integral. +

    +     + + + + + ($r,$s) = num_sort(random_subset(2,-9..-1,1..9)); + ($A,$B) = random_subset(2,1..9); + if($envir{problemSeed}==1){$r=-5;$s=2;$A=4;$B=3;}; + $f = Formula("(($A+$B)x-($A*$s+$B*$r))/(x^2 - ($r+$s)x + $r*$s)")->reduce; + $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(abs(x-$s))")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $r = non_zero_random(-9,9,1); + ($A,$B) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$r=-1;$A=9;$B=-2;}; + $s = 0; + $f = Formula("(($A+$B)x-($A*$s+$B*$r))/(x^2 - ($r+$s)x + $r*$s)")->reduce; + $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(abs(x-$s))")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $r = random(-9,-1,1); + $c = random(2,5,1); + $A = non_zero_random(-9,9,1); + if($envir{problemSeed}==1){$r=-2;$c=3;$A=1;}; + $B = -$A; + $s = -$r; + Context()->noreduce('(-x)/y'); + $f = Formula("($A($r-$s))/($c x^2 - $c*($r+$s)x + $c*$r*$s)")->reduce; + Context("Fraction"); + $Afrac = Fraction($A,$c); + $Bfrac = Fraction($B,$c); + $F = FormulaUpToConstant("$Afrac ln(abs(x-$r)) + $Bfrac ln(abs(x-$s))")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $r = non_zero_random(-9,9,1); + $m = list_random(-9..-2,2..9); + if($envir{problemSeed}==1){$r=-1;$m=3;}; + $f = Formula("((2*$m)x+(1-$m*$r))/($m x^2 + (1 - $r*$m)x - $r)")->reduce; + Context("Fraction"); + $F = FormulaUpToConstant("ln(abs(x-$r)) + ln(abs($m x + 1))")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $r = non_zero_random(-9,9,1); + ($A,$B) = random_subset(2,1..9); + if($envir{problemSeed}==1){$r=-5;$A=1;$B=2;}; + $f = Formula("($A x + ($B-$A*$r))/(x-$r)^2")->reduce; + Context("Fraction"); + $F = FormulaUpToConstant("$A ln(abs(x-$r)) - $B/(x - $r)")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $r = non_zero_random(-9,9,1); + ($A,$B) = random_subset(2,1..9); + if($envir{problemSeed}==1){$r=-8;$A=-3;$B=4;}; + Context()->noreduce('(-x)-y'); + $f = Formula("($A x + ($B-$A*$r))/(x-$r)^2")->reduce; + Context("Fraction"); + $F = FormulaUpToConstant("$A ln(abs(x-$r)) - $B/(x - $r)")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $r = non_zero_random(-9,9,1); + ($A,$B,$C) = random_subset(3,1..9); + if($envir{problemSeed}==1){$r=-1;$A=7;$B=2;$C=5}; + $f = Formula("(($A+$B)x^2 + (-2*$A*$r-$B*$r-$C)x + $A*($r)^2)/(x (x-$r)^2)")->reduce; + Context("Fraction"); + $F = FormulaUpToConstant("$A ln(abs(x)) + $B ln(abs(x-$r)) + $C/(x-$r)")->reduce; + $F->{test_at} = [[-0.25],[0.25],[$r-0.25],[$r+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($r,$s) = random_subset(2,-9..-1,1..9); + ($m,$n) = random_subset(2,-9..-2,2..9); + ($A,$B,$C) = random_subset(3,-9..-1,1..9); + if($envir{problemSeed}==1){$r=1;$s=-3;$m=-2;$n=3;$A=5;$B=2;$C=-1}; + Context()->noreduce('(-x)-y','(-x)+y'); + $f = Formula("(($A+$B*$m+$C*$m)x^2 + ($A*$m*(-$r-$s)+$B*($n-$m*$s)+$C*($n-$r*$m))x + ($A*$m*$r*$s-$B*$n*$s-$C*$r*$n))/((x-$r)(x-$s)($n + $m x))")->reduce; + Context("Fraction"); + $F = FormulaUpToConstant("$A ln(abs($m*x+$n)) + $B ln(abs(x-$r)) + $C ln(abs(x-$s))")->reduce; + $F->{test_at} = [[$s-0.1],[$s+0.1],[$r-0.1],[$r+0.1],[-$n/$m+0.1],[-$n/$m-0.1]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($m,$n,$p) = random_subset(3,2..9); + ($A,$B,$C) = random_subset(3,1..3); + $A = $A*random(-1,1,2); + $B = $B*random(-1,1,2); + $C = $C*random(-1,1,2); + $pm1 = list_random(-1,1); + do {$r = non_zero_random(-3,3,1);} until (($pm1 - $B*$r) % $A == 0 and ($pm1 - $B*$r) != 0); + $s = ($pm1 - $B*$r)/$A; + $t = -$C*$r*$s/$pm1; + if($envir{problemSeed}==1){$m=5;$n=3;$p=7;$r=-1;$s=-1;$t=3;$A=-1;$B=2;$C=3}; + $f = Formula("(($A*$n*$p+$B*$m*$p+$C*$m*$n)x^2 + ($A*($n*$t+$p*$s)+$B*($m*$t+$p*$r)+$C*($m*$s+$n*$r))x + ($A*$s*$t+$B*$r*$t+$C*$r*$s))/(($p x+$t)($m x+$r)($n x+$s))")->reduce; + Context("Fraction"); + $a =Fraction($A,$m); + $b =Fraction($B,$n); + $c =Fraction($C,$p); + $F = FormulaUpToConstant("$a ln(abs($m x+$r)) + $b ln(abs($n x+$s)) + $c/($p x+$t)")->reduce; + $F->{test_at} = [[-$r/$m-0.1],[-$r/$m+0.1],[-$s/$n-0.1],[-$s/$n+0.1],[-$t/$p-0.1],[-$t/$p+0.1]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($r,$s) = random_subset(2,-9..-1,1..9); + ($A,$B) = random_subset(2,-3..-1,1..3); + if($envir{problemSeed}==1){$r=1;$s=-2;$A=1;$B=-1;}; + $f = Formula("(x^2+($A+$B-$r-$s)x+($r*$s-$A*$s-$B*$r))/(x^2-($r+$s)x+$r*$s)")->reduce; + Context("Fraction"); + $F = FormulaUpToConstant("x + $A ln(abs(x-$r)) + $B ln(abs(x-$s))")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + ($r,$s) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$r=-4;$s=5;}; + $f = Formula("x^3/(x^2-($r+$s)x+$r*$s)")->reduce; + Context("Fraction"); + $A = Fraction($r*($r**2+$s**2+$r*$s) - ($r+$s)*$r*$s, $r-$s); + $B = Fraction(-$s*($r**2+$s**2+$r*$s) + ($r+$s)*$r*$s, $r-$s); + $F = FormulaUpToConstant("1/2 x^2 + ($r+$s) x + $A ln(abs(x-$r)) + $B ln(abs(x-$s))")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $h = non_zero_random(-4,4,1); + $k = random(1,5,1); + $c = random(2,4,1); + if($envir{problemSeed}==1){$h=1;$k=2;$c=2;}; + $f = Formula("($c x^2+(-2*$h*$c) x+$c*(($h)^2+$k))/(x^2+(-2*$h) x+(($h)^2+$k))")->reduce; + Context("Fraction"); + $F = FormulaUpToConstant("$c x")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $h = non_zero_random(-4,4,1); + $k = list_random(2,3,5); + if($envir{problemSeed}==1){$h=-1;$k=2;}; + $f = Formula("1/(x^3+(-2*$h) x^2+(($h)^2+$k)x)")->reduce; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $A = Fraction(1,($h)**2+$k); + $B = -$A/2; + $C = 2*$h*($A+$B); + $F = FormulaUpToConstant("$A ln(abs(x)) + $B ln(x^2+(-2*$h) x+(($h)^2+$k)) + $C/sqrt($k) atan((x-$h)/sqrt($k))")->reduce; + $F->{test_at} = [[-0.25],[0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $h = non_zero_random(-4,4,1); + $k = list_random(2,3,5,6); + ($m,$b) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$h=-2;$k=6;$m=-3;$b=-5;}; + $f = Formula("(x^2+(-2*$h+$m) x+(($h)^2+$k+$b))/(x^2+(-2*$h) x+(($h)^2+$k))")->reduce; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $A = Fraction($m,2); + $B = $b + 2*$A*$h; + $F = FormulaUpToConstant("x + $A ln(x^2+(-2*$h) x+(($h)^2+$k)) + $B/sqrt($k) atan((x-$h)/sqrt($k))")->reduce; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $a = random(2,9,1); + ($b,$c) = random_subset(2,-9..-1,1..9); + do {$r = non_zero_random(-9,9,1);} until ($a*($r)**2 + $b*$r + $c != 0); + $B = non_zero_random(-5,5,1); + if($envir{problemSeed}==1){$a=3;$b=5;$c=-1;$r=-1;$B=2;}; + $f = Formula("($a*(2+$B)x^2 + ($b-2*$a*$r+$B*$b)x + ($B*$c-$b*$r))/((x - $r)($a x^2 + $b x + $c))")->reduce; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $F = FormulaUpToConstant("ln(abs($a x^2 + $b x + $c)) + $B ln(abs(x-$r))")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25],[-$b/(2*$a)],[6]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $h = non_zero_random(-4,4,1); + $r = non_zero_random(-9,9,1); + $A = random(2,5,1); + $B = random(2,5,1); + $C = non_zero_random(-5,5,1); + if($envir{problemSeed}==1){$h=-3;$r=3;$A=2;$B=2;$C=-4;}; + $f = Formula("(($A+2*$B)x^2 + (-2*$A*$h - 2*$B*$r + $C - 2*$B*$h)x + ($A*($h)**2 + $A - $r*$C + 2*$r*$B*$h))/((x-$r)(x^2+(-2*$h) x+(($h)^2+1)))")->reduce; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(x^2+(-2*$h) x+(($h)^2+1)) + $C atan(x-$h)")->reduce; + $F->{test_at} = [[-0.25],[0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $k = list_random(4,9,16,25); + $r = non_zero_random(-9,9,1); + $a = random(2,9,1); + ($b,$c) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$k=9;$r=-1;$a=2;$b=1;$c=1;}; + $sqrtk = sqrt($k); + Context("Fraction"); + $C = Fraction(($c*$r - $k*$a*$r - $b*$k)/-$sqrtk/(($r)**2+$k)); + $B = ($b*$k - $C*$k*$sqrtk)/(-2*$r*$k); + $A = $a - 2*$B; + $f = Formula("(($A+2*$B)x^2 + (-2*$B*$r + $C*$sqrtk)x + ($A*$k - $r*$C*$sqrtk))/((x-$r)(x^2+$k))")->reduce; + Context()->flags->set(reduceConstantFunctions=>0); + $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(x^2+$k) + $C atan(x/$sqrtk)")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $h = non_zero_random(-4,4,1); + $k = list_random(4,9,16,25,36); + $r = non_zero_random(-9,9,1); + $A = non_zero_random(-5,5,1); + $C = non_zero_random(-5,5,1); + if($envir{problemSeed}==1){$h=-1;$k=16;$r=7;$A=-2;$C=1;}; + $sqrtk = sqrt($k); + Context("Fraction"); + $B = Fraction(1-$A,2); + $C = Fraction($C,2); + $f = Formula("(($A+2*$B)x^2 + (-2*$A*$h - 2*$B*$r + $C*$sqrtk - 2*$B*$h)x + ($A*($h)**2 + $A*$k - $r*$C*$sqrtk + 2*$r*$B*$h))/((x-$r)(x^2+(-2*$h) x+(($h)^2+$k)))")->reduce; + Context()->flags->set(reduceConstantFunctions=>0); + $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(x^2+(-2*$h) x+(($h)^2+$k)) + $C atan((x-$h)/$sqrtk)")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $h = non_zero_random(-4,4,1); + $k = list_random(2,3,5,6,7,10,11,13,14,15); + $r = non_zero_random(-9,9,1); + $A = non_zero_random(-5,5,1); + $B = non_zero_random(-5,5,1); + $C = non_zero_random(-5,5,1); + if($envir{problemSeed}==1){$h=1;$k=10;$r=9;$A=3;$B=3;$C=6;}; + Context("Fraction"); + $f = Formula("(($A+2*$B)x^2 + (-2*$A*$h - 2*$B*$r + $C - 2*$B*$h)x + ($A*($h)**2 + $A*$k - $r*$C + 2*$r*$B*$h))/((x-$r)(x^2+(-2*$h) x+(($h)^2+$k)))")->reduce; + Context()->flags->set(reduceConstantFunctions=>0); + $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(x^2+(-2*$h) x+(($h)^2+$k)) + $C/sqrt($k) atan((x-$h)/sqrt($k))")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     + + + + + $h = non_zero_random(-9,9,1); + $k = list_random(2,3,5); + $r = non_zero_random(-9,9,1); + $A = non_zero_random(-5,5,1); + $B = non_zero_random(-5,5,2); + $C = non_zero_random(-15,15,1); + if($envir{problemSeed}==1){$h=-5;$k=2;$r=-2;$A=5;$B=1;$C=-12;}; + Context("Fraction"); + $B = Fraction($B,2); + $f = Formula("(($A+2*$B)x^2 + (-2*$A*$h - 2*$B*$r + $C - 2*$B*$h)x + ($A*($h)**2 + $A*$k - $r*$C + 2*$r*$B*$h))/((x-$r)(x^2+(-2*$h) x+(($h)^2+$k)))")->reduce; + Context()->flags->set(reduceConstantFunctions=>0); + $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(x^2+(-2*$h) x+(($h)^2+$k)) + $C/sqrt($k) atan((x-$h)/sqrt($k))")->reduce; + $F->{test_at} = [[$r-0.25],[$r+0.25]]; + + +

    + \ds \int \, dx +

    +

    + +

    +     +     +     +     + + + +

    + Evaluate the definite integral. +

    +     + + + + + $a = 1; + $b = 2; + ($r,$s) = num_sort(random_subset(2,-9..-1,3..9)); + ($A,$B) = random_subset(2,1..9); + if($envir{problemSeed}==1){$r=-3;$s=-2;$A=3;$B=5;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("(($A+$B)x-($A*$s+$B*$r))/((x-$r)(x-$s))")->reduce; + $arg = Fraction(abs($b-$r)**$A * abs($b-$s)**$B * abs($a-$r)**(-$A) * abs($a-$s)**(-$B)); + $answer = Formula("ln($arg)"); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $a = 0; + $b = random(2,9,1); + $r = random(-9,-1,1); + ($m,$n) = random_subset(2,2..5); + ($A,$B) = random_subset(2,-9..-1,1..9); + if($envir{problemSeed}==1){$b=5;$r=-4;$m=3;$n=2;$A=5;$B=-1;}; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $f = Formula("(($A*$m+$B)x+($A*$n-$B*$r))/(($m x + $n)(x-$r))")->reduce; + $F = Formula("$A ln(abs(x-$r)) + $B/$m ln(abs($m x+$n))")->reduce; + $answer = $F->eval(x=>$b) - $F->eval(x=>$a); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + $a = -1; + $b = 1; + $h = non_zero_random(-4,4,1); + $r = list_random(-10..-2,2..10); + $A = random(2,5,1); + $B = random(2,5,1); + $C = non_zero_random(-5,5,1); + if($envir{problemSeed}==1){$h=-2;$r=10;}; + $f = Formula("(x^2 + (1-2*$h)x + (($h)^2+1-$r))/((x-$r)(x^2+(-2*$h) x+(($h)^2+1)))")->reduce; + Context("Fraction"); + Context()->flags->set(reduceConstantFunctions=>0); + $F = Formula("ln(abs(x-$r)) + atan(x-$h)")->reduce; + $frac=Fraction($b-$r,$a-$r); + $Fb = Formula("atan($b-$h)"); + if ($b-$h == 0) {$Fb = 0;}; + if ($b-$h == 1) {$Fb = Formula("pi/4");}; + if ($b-$h == -1) {$Fb = Formula("-pi/4");}; + $Fa = Formula("atan($a-$h)"); + if ($a-$h == 0) {$Fa = 0;}; + if ($a-$h == 1) {$Fa = Formula("pi/4");}; + if ($a-$h == -1) {$Fa = Formula("-pi/4");}; + $Fba = $Fb - $Fa; + $Fba = $Fb if ($Fa == 0); + $Fba = -$Fa if ($Fb == 0); + $answer = Formula("ln($frac) + $Fba"); + + +

    + \ds \int_{}^{} \, dx +

    +

    + +

    +     +     +     + + + + + + $a = 0; + $b = 1; + $F = Formula("-(2x+1)/(2*(x+1)^2)"); + Context("Fraction"); + $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); + + +

    + \ds \int_{0}^1 \frac{x}{(x+1)(x^2+2x+1)}\, dx +

    +

    + +

    +     +     +     +     +     +     +
    +
    + Hyperbolic Functions + +

    + The hyperbolic functions + are a set of functions that have many applications to mathematics, physics, + and engineering. + Among many other applications, + they are used to describe the formation of satellite rings around planets, + to describe the shape of a rope hanging from two points, + and have application to the theory of special relativity. + This section defines the hyperbolic functions and describes many of their properties, + especially their usefulness to calculus. +

    + +

    + These functions are sometimes referred to as the + hyperbolic trigonometric functions as there are many, + many connections between them and the standard trigonometric functions. + + demonstrates one such connection. + Just as cosine and sine are used to define points on the circle defined by x^2+y^2=1, + the functions hyperbolic cosine and + hyperbolic sine + are used to define points on the hyperbola x^2-y^2=1. +

    + + + +
    + Using trigonometric functions to define points on a circle and hyperbolic functions to define points on a hyperbola. The area of the shaded regions are included in them. + + +
    + + + + + Graph showing cosine and sine function used to define points on a circle. + + +

    + The y and the x axes are drawn from -1 to + 1. The function x^2+y^2=1 represents a circle of + radius 1 and center at origin. +

    +

    + A sector in the circle is shaded, it is present in the first + quadrant and is drawn with one side on the x axis. + It has an angle \theta/2 drawn from the x axis + and is marked inside the sector. The points between which the + sector is drawn on the circumference are (1,0) and + (\cos(\theta), \sin(\theta)). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis equal image, + axis on top, + extra x tick labels={$a$,$b$}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.4 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:45] ({cos(x)},{sin(x)}) -- (axis cs:0,0) -- cycle; + \addplot [firstcurvestyle,domain=0:360,samples=80] ({cos(x)},{sin(x)}); + + \filldraw (axis cs:.707,.707) circle (1pt) node [shift={(20pt,12pt)}] { ($\cos(\theta) $,$\sin(\theta) $)}; + + \draw (axis cs:.6,.25) node { $\ds\frac{\theta}{2}$}; + \draw (axis cs:-.75,1) node { $x^2+y^2=1$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + Graph showing hyperbolic cosine and hyperbolic sine function used to define points on a hyperbola. + + +

    + The y and the x axes are both drawn from -2 to 2. + The graph of function x^2 -y^2=1 has two x intercepts at x=1 + and x=-1. The function represents a hyperbola and has two conic sections + facing opposite to each other, opening along the positive and negative x + axis with vertices at the x intercepts. An angle of \theta/2 is + marked at the origin starting from the x axis, it is drawn from point + (1,0) to (\cos(\theta), \sin(\theta)) on the conic section to the + right of the y axis. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis equal image, + axis on top, + ymin=-3.1,ymax=3.1, + xmin=-3.1,xmax=3.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1.6] ({cosh(x)},{sinh(x)}) -- (axis cs:0,0) -- cycle; + + \addplot [firstcurvestyle,domain=-2:2,samples=40] ({cosh(x)},{sinh(x)}); + \addplot [firstcurvestyle,domain=-2:2,samples=40] ({-cosh(x)},{sinh(x)}); + + \filldraw (axis cs:2.577,2.376) circle (1pt) node [left] { ($\cosh(\theta) $,$\sinh(\theta) $)}; + + \draw (axis cs:.73,.32) node { $\ds\frac{\theta}{2}$}; + \draw (axis cs:-1.75,2.75) node { $x^2-y^2=1$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    +     + + + The Hyperbolic Functions and their Properties +

    + We begin with their definition. +

    + + + Hyperbolic Functions + +

    +

      +
    1. +

      + \ds \cosh(x) = \frac{e^x+e^{-x}}2 + + hyperbolic functiondefinition + +

      +
      2. + +
    2. \sinh(x) = \frac{e^x-e^{-x}}2
      3. + +
    3. \tanh(x) = \frac{\sinh(x) }{\cosh(x) }
      4. + +
    4. \sech(x) = \frac{1}{\cosh(x) }
      5. + +
    5. \csch(x) = \frac{1}{\sinh(x) }
      6. + +
    6. \coth(x) = \frac{\cosh(x) }{\sinh(x) }
      7. +
    +

    +     +     + + +

    + These hyperbolic functions are graphed in + and . +

    + +

    + In the graph of \cosh(x) in , + the graphs of e^x/2 and + e^{-x}/2 are included with dashed lines. + In the graph of \sinh(x) in , + the graphs of e^x/2 and + - + e^{-x}/2 are included with dashed lines. + As x gets large, + \cosh(x) and \sinh(x) each act like e^x/2; + when x is a large negative number, + \cosh(x) acts like e^{-x}/2 whereas + \sinh(x) acts like -e^{-x}/2. +

    + +
    + Graphs of \sinh(x) and \cosh(x) + + + +
    + + + + + Graph of hyperbolic cosine function. + + +

    + The y axis is drawn from -10 to 10 and the x + axis is drawn from -3 to 3. The function f(x)=\cosh(x) + is shown as a U shaped curve that opens upwards along the positive y + axis, it is symmetrical about the y axis. The graphs of e^{x}/2 + and e^{-x}/2 are also included. +

    +

    + For large values of x the function f(x)=\cosh(x) is approximately equal + to e^{x}/2. From left to right, the function e^{x}/2 appears to + coincide with the x axis in the fourth quadrant, it has a positive slope, + it is some distance apart from f(x) then it rises to coincide with f(x) + after x=1. +

    +

    + For large negative values of x, f(x) becomes equal to the function + e^{-x}/2. From right to left, the function e^{-x}/2 appears to start + from the first quadrant and enters the second quadrant with a positive slope. It + coincides with f(x) after approximately x= -1. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-11,ymax=11, + xmin=-3.5,xmax=3.5, + scaled ticks=false + ] + + \addplot+ [domain=-3.1:3.1] {cosh(x)} node [pos=0.9, above left] { $f(x)=\cosh(x)$}; + \addplot+ [domain=-3:3] {e^x/2} node [pos=0, above right] { $e^x/2$}; + \addplot [secondcurvestyle,domain=-3:3] {e^(-x)/2} node [pos=0.9, above right] { $e^{-x}/2$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + + +
    + + + + + Graph of hyperbolic sine function. + + +

    + The y axis is drawn from -10 to 10 and the x axis is drawn + from -3 to 3. The function f(x) = \sinh(x), e^{x}/2 + and e^-{x}/2 is also shown in the graph. +

    +

    + From left to right, the function e^{x}/2 starts in the second quadrant + and gets close to the x axis, it gains the positive slope into the first + quadrant. The function f(x) in the first quadrant, starts at the origin and + rises with a positive slope after a bend.It is separated by a small distance from + e^{x}/2, after approximately x=1 the function e^{x}/2 coincides + with f(x). +

    +

    + From right to left, the function e^{-x}/2 starts in the fourth quadrant and + coincides with the x axis, it moves downward after entering the third quadrant. + The function f(x) in the third quadrant, starts at the origin and moves downwards + from right to left. It is separated by a small distance from e^{-x}/2 after + approximately x=-1 the function e^{-x}/2 coincides with f(x). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-11,ymax=11, + xmin=-3.5,xmax=3.5, + scaled ticks=false + ] + + \addplot+ [domain=-3.1:3.1] {0.5*(e^x-e^(-x))} node [pos=0.9, above left] { $f(x)=\sinh(x)$}; + \addplot+ [domain=-3:3] {0.5*(e^x)} node [pos=0, above right] { $e^x/2$}; + \addplot+ [secondcurvestyle,domain=-3:3] {-0.5*(e^(-x)} node [pos=0.9, above right] { $-e^{-x}/2$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    + + + +

    + In Figure , + notice the domains of \tanh(x) and + \sech(x) are (-\infty,\infty), + whereas both \coth(x) and + \csch(x) have vertical asymptotes at x=0. + Also note the ranges of these functions, + especially \tanh(x): + as x\to\infty, both \sinh(x) and + \cosh(x) approach e^{-x}/2, + hence \tanh(x) approaches 1. +

    + +
    + Graphs of \tanh(x), \coth(x), \csch(x) and \cosh(x) + + +
    + + + + + Graph of hyperbolic tangent and hyperbolic cotangent functions. + + +

    + The y axis is drawn from -2 to 2 and the x axis + is drawn from -3 to 3. The functions \tanh(x) and + \coth(x) are shown. There are two lines drawn at y=-1 and y=1. +

    +

    + The \tanh(x) function is drawn in the third and the first quadrant. + In the first quadrant the function starts at the origin and gets a positive + slope then after x=2 it becomes parallel to the x axis at + y=1. In the third quadrant the function starts at the origin and + decreases until x=-2 after which it becomes parallel to the x + axis as y=-1. +

    +

    + The \coth(x) function is drawn in the first and the third quadrants. + It is hyperbolic in shape with the two parts being symmetrical about the + axis y=-x. It has a horizontal asymptote at x=0. This function + coincides with the \tanh(x) curve after x=2 and extends along + the positive x axis and x=-2 and extends further along the + negative x axis. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-3.5,ymax=3.5, + xmin=-3.5,xmax=3.5, + scaled ticks=false + ] + + \addplot+ [domain=-3:3] {(e^x-e^(-x))/(e^x+e^(-x))} node [pos=0.60,pin={120:$ \tanh(x)$},inner sep=0pt] {}; + + \addplot+ [solid,domain=-3:-.33] {(e^x+e^(-x))/(e^x-e^(-x))}; + \addplot [secondcurvestyle,solid,domain=.33:3,samples=30] {(e^x+e^(-x))/(e^x-e^(-x))} node [pos=0.5,pin={45:$ \coth(x)$},inner sep=0pt] {}; + \addplot+ [lineseg,dotted,domain=-3:3] {1}; + \addplot+ [lineseg,dotted,domain=-3:3] {-1}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + + +
    + + + + + Graph of hyperbolic secant and cosecant functions. + + +

    + The y axis is drawn from -2 to 2 and the x + axis is drawn from -3 to 3. The functions \sech(x) + and \csch(x) are shown. +

    +

    + The \sech(x) is drawn in the second and the first quadrant. + From point (1,0) the function slowly decreases moving left to + right, almost touching the x axis at x=3. It is symmetrical + about the x axis and in the third quadrant it decreases from + (1,0), moving from right to left, and almost touches the x axis at + x=-3. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-3.5,ymax=3.5, + xmin=-3.5,xmax=3.5, + scaled ticks=false + ] + + \addplot+ [domain=-3:3] {(2/(e^x+e^(-x))} node [pos=0.50,pin={120:$ \sech(x)$},inner sep=0pt] {}; + \addplot+ [solid,domain=-3:-.33] {2/(e^x-e^(-x))}; + \addplot+ [secondcurvestyle,solid,domain=.33:3,samples=30] {2/(e^x-e^(-x))} node [pos=0.5,pin={45:$ \csch(x)$},inner sep=0pt] {}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +     +
    + +

    + The following example explores some of the properties of these functions that bear remarkable resemblance to the properties of their trigonometric counterparts. +

    + + + Exploring properties of hyperbolic functions + +

    + Use + to rewrite the following expressions. +

    + +

    +

      +
    1. \cosh^2(x) -\sinh^2(x)
      2. + +
    2. \tanh^2(x) +\sech^2(x)
      3. + +
    3. 2\cosh(x) \sinh(x)
      4. + +
    4. \frac{d}{dx}\big(\cosh(x) \big)
      5. + +
    5. \frac{d}{dx}\big(\sinh(x) \big)
      6. + +
    6. \frac{d}{dx}\big(\tanh(x) \big)
      7. +
    +

    +     + +

    +

      +
    1. + +

      + By + + \cosh^2(x) -\sinh^2(x) \amp = \left(\frac{e^x+e^{-x}}2\right)^2 -\left(\frac{e^x-e^{-x}}2\right)^2 + \amp = \frac{e^{2x}+2e^xe^{-x} + e^{-2x}}4 - \frac{e^{2x}-2e^xe^{-x} + e^{-2x}}4 + \amp = \frac44=1 + . + So \cosh^2(x) -\sinh^2(x) =1. +

      +
      2. + +
    2. +

      + Again, use + + \tanh^2(x) +\sech^2(x) \amp =\frac{\sinh^2(x) }{\cosh^2(x) } + \frac{1}{\cosh^2(x) } + \amp= \frac{\sinh^2(x) +1}{\cosh^2(x) }\qquad \text{ Now use identity from } Part + \amp = \frac{\cosh^2(x) }{\cosh^2(x) } = 1 + . + So \tanh^2(x) +\sech^2(x) =1. +

      +
      3. + +
    3. +

      + Again, use + + 2\cosh(x) \sinh(x) \amp = 2\left(\frac{e^x+e^{-x}}2\right)\left(\frac{e^x-e^{-x}}2\right) + \amp = 2 \cdot\frac{e^{2x} - e^{-2x}}4 + \amp = \frac{e^{2x} - e^{-2x}}2 = \sinh(2x) + . + Thus 2\cosh(x) \sinh(x) = \sinh(2x). +

      +
      4. + +
    4. +

      + Again, use + + \frac{d}{dx}\big(\cosh(x) \big) \amp = \frac{d}{dx}\left(\frac{e^x+e^{-x}}2\right) + \amp = \frac{e^x-e^{-x}}2 + \amp = \sinh(x) + + So \frac{d}{dx}\big(\cosh(x) \big) = \sinh(x). +

      +
      5. + +
    5. +

      + Apply derivatives to : + + \frac{d}{dx}\big(\sinh(x) \big) \amp = \frac{d}{dx}\left(\frac{e^x-e^{-x}}2\right) + \amp = \frac{e^x+e^{-x}}2 + \amp = \cosh(x) + . + So \frac{d}{dx}\big(\sinh(x) \big) = \cosh(x). +

      +
      6. + +
    6. +

      + Apply derivatives to : + + \frac{d}{dx}\big(\tanh(x) \big) \amp = \frac{d}{dx}\left(\frac{\sinh(x) }{\cosh(x) }\right) + \amp = \frac{\cosh(x) \cosh(x) - \sinh(x) \sinh(x) }{\cosh^2(x) } + \amp = \frac{1}{\cosh^2(x) } + \amp = \sech^2(x) + . + So \frac{d}{dx}\big(\tanh(x) \big) = \sech^2(x). +

      +
      7. +
    +

    +     + +     + +

    + The following Key Idea summarizes many of the important identities relating to hyperbolic functions. + Each can be verified by referring back to . +

    + + + Useful Hyperbolic Function Properties + + + + + Basic Identities +
      +
    1. +

      + \cosh^2(x) -\sinh^2(x) =1 + + hyperbolic functionidentities + + hyperbolic functionderivatives + + hyperbolic functionintegrals + + derivativehyperbolic funct. + + integrationhyperbolic funct. + +

      +
      2. + +
    2. \tanh^2(x) +\sech^2(x) =1
      3. + +
    3. \coth^2(x) -\csch^2(x) = 1
      4. + +
    4. \cosh(2x) =\cosh^2(x) +\sinh^2(x)
      5. + +
    5. \sinh(2x) = 2\sinh(x) \cosh(x)
      6. + +
    6. \cosh^2(x) = \frac{\cosh(2x) +1}{2}
      7. + +
    7. \sinh^2(x) =\frac{\cosh(2x) -1}{2}
      8. +
    +     + + + Derivatives +
      +
    1. \frac{d}{dx}\big(\cosh(x) \big) = \sinh(x)
      2. + +
    2. \frac{d}{dx}\big(\sinh(x) \big) = \cosh(x)
      3. + +
    3. \frac{d}{dx}\big(\tanh(x) \big) = \sech^2(x)
      4. + +
    4. \frac{d}{dx}\big(\sech(x) \big) = -\sech(x) \tanh(x)
      5. + +
    5. \frac{d}{dx}\big(\csch(x) \big) = -\csch(x) \coth(x)
      6. + +
    6. \frac{d}{dx}\big(\coth(x) \big) = -\csch^2(x)
      7. +
    +     + + + Integrals +
      +
    1. \int \cosh(x) \, dx = \sinh(x) +C
      2. + +
    2. \int \sinh(x) \, dx = \cosh(x) +C
      3. + +
    3. \int \tanh(x) \, dx = \ln(\cosh(x) ) +C
      4. + +
    4. \int \coth(x) \, dx = \ln\abs{\sinh(x) \,}+C
      5. +
    +     + +     + + + +

    + We practice using . +

    + + + Derivatives and integrals of hyperbolic functions + +

    + Evaluate the following derivatives and integrals. +

    + +

    +

      +
    1. \frac{d}{dx}\big(\cosh(2x) \big)
      2. + +
    2. \int \sech^2(7t-3)\,dt
      3. + +
    3. \int_0^{\ln(2) } \cosh(x) \, dx
      4. +
    +

    +     + +

    +

      +
    1. +

      + Using the Chain Rule directly, + we have \frac{d}{dx} \big(\cosh(2x) \big) = 2\sinh(2x). + + Just to demonstrate that it works, + let's also use the Basic Identity found in : + \cosh(2x) = \cosh^2(x) +\sinh^2(x). + + \frac{d}{dx}\big(\cosh(2x) \big) \amp = \frac{d}{dx}\big(\cosh^2(x) +\sinh^2(x) \big) + \amp = 2\cosh(x) \sinh(x) + 2\sinh(x) \cosh(x) + \amp = 4\cosh(x) \sinh(x) + . + Using another Basic Identity, + we can see that 4\cosh(x) \sinh(x) = 2\sinh(2x). + We get the same answer either way. +

      +
      2. + +
    2. +

      + We employ substitution, with u = 7t-3 and du = 7dt. + Applying Key Ideas + and we have: + + \int \sech^2(7t-3)\,dt = \frac17\tanh(7t-3) + C + . +

      +
      3. + +
    3. +

      + + \int_0^{\ln(2) } \cosh(x) \, dx \amp = \sinh(x) \Big|_0^{\ln(2) } + \amp = \sinh(\ln(2) ) - \sinh(0) + \amp = \sinh(\ln(2) ) + . + We can simplify this last expression as + \sinh(x) is based on exponentials: + + \sinh(\ln(2) ) \amp = \frac{e^{\ln(2) }-e^{-\ln(2) }}2 + \amp = \frac{2-1/2}{2} + \amp = \frac34 + . +

      +
      4. +
    +

    +     + +     +     + + + Inverse Hyperbolic Functions +

    + Just as the inverse trigonometric functions are useful in certain applications, + the inverse hyperbolic functions are useful with others. + + shows restriction on the domain of + \cosh(x) to make the function one-to-one and the resulting domain and range of its inverse function. + Since \sinh(x) is already one-to-one, + no domain restriction is needed as shown in . + Since \sech(x) is not one to one, + it also needs a restricted domain in order to be invertible. + + shows the graph of \sech^{-1}(x). + You should carefully compare the graph of this function to the graph given in + to see how this inverse was constructed. + The rest of the hyperbolic functions area already one-to-one and need no domain restrictions. + Their graphs are also shown in . + + hyperbolic functioninverse + +

    + +

    + Because the hyperbolic functions are defined in terms of exponential functions, + their inverses can be expressed in terms of logarithms as shown in . + It is often more convenient to refer to + \sinh^{-1}(x) than to \ln\big(x+\sqrt{x^2+1}\big), + especially when one is working on theory and does not need to compute actual values. + On the other hand, when computations are needed, + technology is often helpful but many hand-held calculators lack a convenient + \sinh^{-1}(x) button. + (Often it can be accessed under a menu system, but not conveniently.) + In such a situation, the logarithmic representation is useful. + The reader is not encouraged to memorize these, + but rather know they exist and know how to use them when needed. +

    + + + + + Domains and ranges of the hyperbolic and inverse hyperbolic functions + + + + Function + Domain + Range + Function + Domain + Range + + + \cosh(x) + [0,\infty) + [1,\infty) + \cosh^{-1}(x) + [1,\infty) + [0,\infty) + + + \sinh(x) + (-\infty,\infty) + (-\infty,\infty) + \sinh^{-1}(x) + (-\infty,\infty) + (-\infty,\infty) + + + \tanh(x) + (-\infty,\infty) + (-1,1) + \tanh^{-1}(x) + (-1,1) + (-\infty,\infty) + + + \sech(x) + [0,\infty) + (0,1] + \sech^{-1}(x) + (0,1] + [0,\infty) + + + \csch(x) + (-\infty,0) \cup (0,\infty) + (-\infty,0) \cup (0,\infty) + \csch^{-1}(x) + (-\infty,0) \cup (0,\infty) + (-\infty,0) \cup (0,\infty) + + + \coth(x) + (-\infty,0) \cup (0,\infty) + (-\infty,-1) \cup (1,\infty) + \coth^{-1}(x) + (-\infty,-1) \cup (1,\infty) + (-\infty,0) \cup (0,\infty) + + + +
    + +
    + Graphs of the hyperbolic functions (with restricted domains) and their inverses + + + +
    + + + + + Graph of hyperbolic cosine function and its inverse. + + +

    + The y and the x axes are drawn from 0 to 10. + The functions y=\cosh(x) and y=\cosh^{-1}(x) are shown. + They are symmetrical about the axis y=x. +

    +

    + From left to right, the function y=\cosh(x) starts at point + (0,1) then slowly rises from (1,1) then it rises up + steeply and it appears to run almost parallel to the y axis. + The function \cosh^{-1}(x) starts at point (1,0) and + curves up steeply until (2,1) then it rises very slowly and + appears to almost run parallel to the x axis. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.9,ymax=10.9, + xmin=-.9,xmax=10.9, + scaled ticks=false + ] + + \addplot+ [domain=0:3] {cosh(x)}; + \addplot+ [domain=0:3] (cosh(x),x); + + \addplot [dashed,domain=-.5:10] {x}; + + \draw (axis cs:8,4) node { $y=\cosh^{-1}(x) $}; + \draw (axis cs:5.6,10) node { $y=\cosh(x) $}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + + +
    + + + + + Graph of hyperbolic sine function and its inverse. + + +

    + The y and the x axes are drawn from -10 to + 10. The functions y=\sinh(x) and y=\sinh^{-1}(x). + The axis y=x is shown. +

    +

    + From left to right, the \sinh(x) function starts in the third + quadrant and it rises steeply, very closely to the y axis. It + crosses the origin along the y=x line, has a dip then increases + very steeply and closely to the y axis in the first quadrant. +

    +

    + From left to right, in the third quadrant, the \sinh{-1}(x) + function runs very closely to the x axis, it crosses the origin + along the y=x line and bends to move very closely to the x + axis in the first quadrant. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-10.9,ymax=10.9, + xmin=-10.9,xmax=10.9, + scaled ticks=false + ] + + \addplot+ [domain=-3:3] {sinh(x)}; + \addplot+ [domain=-3:3] (sinh(x),x); + + \addplot [dashed,domain=-10:10] {x}; + + \draw (axis cs:-6,7) node { $y=\sinh(x) $}; + \draw (axis cs:6,-5) node { $y=\sinh^{-1}(x) $}; + \draw[->,>=stealth] (axis cs:-3,7) -- (axis cs:2,7); + \draw[->,>=stealth] (axis cs:7,-3) -- (axis cs:7,2); + + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + + + +
    + + + + + Graph of inverse of hyperbolic tangent and inverse of hyperbolic cotangent functions. + + +

    + The y and the x axes are drawn from -3 to 3. + There are two functions drawn, \coth^{-1}(x) and \tanh{-1}(x) + along with two dashed lines x=-1 and x=1. +

    +

    + The \tanh{-1}(x) function is drawn in the third and the first quadrant. + From left to right, in the third quadrant the function is aligned with the line + x=-1 at around y=-2 it diverges to the right side of the line, it + crosses the origin then bends and merges with the line x=1 from its left + in the first quadrant. +

    +

    + The coth^{-1}(x) is also drawn in the third and the first quadrant. From + left to right, in the third quadrant, the function appears to be parallel to the + x axis; it diverges and bends down to join the line x=-1. In the + first quadrant, from left to right the function is along the line x=1, it + decreases and diverges from the line, there is a bend after x=2 after which + it becomes parallel to the x axis. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ytick={-2,2}, + ymin=-3.2,ymax=3.2, + xmin=-3.2,xmax=3.2, + scaled ticks=false + ] + + \addplot+ [domain=-3:3] (tanh(x),x); + \addplot+ [solid,smooth] coordinates {(1.005,2.997)(1.01,2.65)(1.02,2.31)(1.03,2.11)(1.04,1.97)(1.05,1.86)(1.06,1.77)(1.07,1.69)(1.08,1.63)(1.09,1.57)(1.1,1.52)(1.2,1.2)(1.3,1.02)(1.4,0.896)(1.5,0.805)(1.6,0.733)(1.7,0.675)(1.8,0.626)(1.9,0.585)(2.,0.549)(2.1,0.518)(2.2,0.49)(2.3,0.466)(2.4,0.444)(2.5,0.424)(2.6,0.405)(2.7,0.389)(2.8,0.374)(2.9,0.36)(3.,0.347)}; + \addplot [secondcurvestyle,solid,smooth] coordinates {(-3.,-0.347)(-2.9,-0.36)(-2.8,-0.374)(-2.7,-0.389)(-2.6,-0.405)(-2.5,-0.424)(-2.4,-0.444)(-2.3,-0.466)(-2.2,-0.49)(-2.1,-0.518)(-2.,-0.549)(-1.9,-0.585)(-1.8,-0.626)(-1.7,-0.675)(-1.6,-0.733)(-1.5,-0.805)(-1.4,-0.896)(-1.3,-1.02)(-1.2,-1.2)(-1.1,-1.52)(-1.09,-1.57)(-1.08,-1.63)(-1.07,-1.69)(-1.06,-1.77)(-1.05,-1.86)(-1.04,-1.97)(-1.03,-2.11)(-1.02,-2.31)(-1.01,-2.65)(-1.005,-2.997)}; + + \draw (axis cs:2.2,1.5) node {\tiny $y=\coth^{-1}(x) $}; + \draw (axis cs:2.2,-1.5) node {\tiny $y=\tanh^{-1}(x) $}; + \draw [->,>=stealth] (axis cs:1.8,-1.3) -- (axis cs:.2,-.2); + + \addplot [asymptote] coordinates {(-1,-5) (-1,5)}; + \addplot [asymptote] coordinates {(1,-5) (1,5)}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + Graph of inverse of hyperbolic cosecant and inverse of hyperbolic secant functions. + + +

    + The sec{-1}(x) is drawn only in the first quadrant.From left to right, + it starts very close to the y axis without touching it, at around + y=3. It moves away from the y axis while declining then gets + a small bend before making an x intercept at x=1. +

    +

    + The \csch{-1}(x) is drawn in the third and the first quadrant. In the + third quadrant from left to right the function appears to be parallel to the + x axis. It bends toward the negative y axis and comes very close + to it at y=-3. In the first quadrant from left to right, the function + appears to start very close to the y axis coinciding with the sech{-1}(x) + function. It has a negative slope and it moves down, gets a bend and runs parallel + to the x axis. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ytick={-3,-2,-1,1,2,3}, + ymin=-3.5,ymax=3.5, + xmin=-3.5,xmax=3.5, + scaled ticks=false + ] + + \addplot+ [smooth] coordinates {(0.1,2.993)(0.2,2.292)(0.3,1.874)(0.4,1.567)(0.5,1.317) + (0.55,1.205)(0.6,1.099)(0.65,0.9961)(0.7,0.8956)(0.75,0.7954)(0.8,0.6931)(0.85,0.5857)(0.9,0.4671)(0.95,0.323)(1.,0)}; + + \addplot+ [solid,smooth] coordinates {(-3.,-0.3275)(-2.9,-0.3383)(-2.8,-0.35)(-2.7,-0.3624)(-2.6,-0.3757)(-2.5,-0.39)(-2.4,-0.4055)(-2.3,-0.4221)(-2.2,-0.4402)(-2.1,-0.4598)(-2.,-0.4812)(-1.9,-0.5046)(-1.8,-0.5303)(-1.7,-0.5587)(-1.6,-0.5901)(-1.5,-0.6251)(-1.4,-0.6643)(-1.3,-0.7085)(-1.2,-0.7585)(-1.1,-0.8156)(-1.,-0.8814)(-0.9,-0.9578)(-0.8,-1.048)(-0.7,-1.154)(-0.6,-1.284)(-0.5,-1.444)(-0.4,-1.647)(-0.3,-1.919)(-0.2,-2.312)(-0.1,-2.998)}; + \addplot+ [secondcurvestyle,solid,smooth] coordinates {(0.1,2.998)(0.2,2.312)(0.3,1.919)(0.4,1.647)(0.5,1.444)(0.6,1.284)(0.7,1.154)(0.8,1.048)(0.9,0.9578)(1.,0.8814)(1.1,0.8156)(1.2,0.7585)(1.3,0.7085)(1.4,0.6643)(1.5,0.6251)(1.6,0.5901)(1.7,0.5587)(1.8,0.5303)(1.9,0.5046)(2.,0.4812)(2.1,0.4598)(2.2,0.4402)(2.3,0.4221)(2.4,0.4055)(2.5,0.39)(2.6,0.3757)(2.7,0.3624)(2.8,0.35)(2.9,0.3383)(3.,0.3275)}; + + \draw (axis cs:-1.2,.5) node {\tiny $y=\sech^{-1}(x) $}; + \draw (axis cs:-2.2,-1.6) node {\tiny $y=\csch^{-1}(x) $}; + \draw [->,>=latex] (axis cs:-.2,.5) -- (axis cs:0.7,.5); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +     +     +
    + + + Logarithmic definitions of Inverse Hyperbolic Functions +

    +

      +
    1. +

      + \ds\cosh^{-1}(x) =\ln\big(x+\sqrt{x^2-1}\big);\, x\geq1 + + hyperbolic functioninverse!logarithmic def. + +

      +
      2. + +
    2. \tanh^{-1}(x) = \frac12\ln\left(\frac{1+x}{1-x}\right);\, \abs{x}\lt 1
      3. + +
    3. \sech^{-1}(x) = \ln\left(\frac{1+\sqrt{1-x^2}}x\right);\, 0\lt x\leq1
      4. + +
    4. \sinh^{-1}(x) = \ln\big(x+\sqrt{x^2+1}\big)
      5. + +
    5. \coth^{-1}(x) = \frac12\ln\left(\frac{x+1}{x-1}\right);\, \abs{x} \gt 1
      6. + +
    6. \csch^{-1}(x) = \ln\left(\frac1x+\frac{\sqrt{1+x^2}}{\abs{x}}\right);\, x\neq0
      7. +
    +

    +     + +

    + The following Key Ideas give the derivatives and integrals relating to the inverse hyperbolic functions. + In , + both the inverse hyperbolic and logarithmic function representations of the antiderivative are given, + based on . + Again, these latter functions are often more useful than the former. + Note how inverse hyperbolic functions can be used to solve integrals we used Trigonometric Substitution to solve in . +

    + + + + + Derivatives Involving Inverse Hyperbolic Functions +

    +

      +
    1. +

      + \ds\frac{d}{dx}\big(\cosh^{-1}(x) \big) = \frac{1}{\sqrt{x^2-1}};\\ x \gt 1 + + derivativeinverse hyper. + + hyperbolic functioninverse!derivative + +

      +
      2. + +
    2. \frac{d}{dx}\big(\sinh^{-1}(x) \big) = \frac{1}{\sqrt{x^2+1}}
      3. + +
    3. \frac{d}{dx}\big(\tanh^{-1}(x) \big) = \frac{1}{1-x^2};\\ \abs{x}\lt 1
      4. + +
    4. \frac{d}{dx}\big(\sech^{-1}(x) \big) = \frac{-1}{x\sqrt{1-x^2}};\\ 0\lt x\lt 1
      5. + +
    5. \frac{d}{dx}\big(\csch^{-1}(x) \big) = \frac{-1}{\abs{x}\sqrt{1+x^2}};\\ x\neq0
      6. + +
    6. \frac{d}{dx}\big(\coth^{-1}(x) \big) = \frac{1}{1-x^2};\\ \abs{x} \gt 1
      7. +
    +

    +     + + + + + Integrals Involving Inverse Hyperbolic Functions +

    + Assume a\gt 0. +

      +
    1. +

      + + \int \frac{1}{\sqrt{x^2-a^2}}\, dx \amp = \ln\abs{x+\sqrt{x^2-a^2}}+C + (\text{for } 0\lt x\lt a)\, \amp = \cosh^{-1}\left(\frac xa\right)+C + +

      +
      2. + +
    2. +

      + + \int \frac{1}{\sqrt{x^2+a^2}}\, dx \amp =\ln\abs{x+\sqrt{x^2+a^2}}+C + \amp = \sinh^{-1}\left(\frac xa\right)+C + +

      +
      3. + +
    3. +

      + + \int \frac{1}{a^2-x^2}\, dx \amp = \frac1{2a}\ln\abs{\frac{a+x}{a-x}}+C + \amp = \begin{cases} + \frac1a\tanh^{-1}\left(\frac xa\right)+C \amp x^2\lt a^2\\ + \frac1a\coth^{-1}\left(\frac xa\right)+C \amp a^2\lt x^2 + \end{cases} + +

      +
      4. + +
    4. +

      + + \int \frac{1}{x\sqrt{a^2-x^2}}\, dx \amp =\frac1a \ln\left(\frac{x}{a+\sqrt{a^2-x^2}}\right)+C + (\text{for } 0\lt x\lt a)\, \amp = -\frac1a\sech^{-1}\left(\frac xa\right)+C + +

      +
      5. + +
    5. +

      + + \int \frac{1}{x\sqrt{x^2+a^2}}\, dx \amp = \frac1a \ln\abs{\frac{x}{a+\sqrt{a^2+x^2}}}+C + \amp = -\frac1a\csch^{-1}\abs{\frac xa} + C + +

      +
      6. +
    + integrationinverse hyperbolic + + hyperbolic functioninverse!integration + +

    +     + + + + + +

    + We practice using the derivative and integral formulas in the following example. +

    + + + Derivatives and integrals involving inverse hyperbolic functions + +

    + Evaluate the following. +

    + +

    +

      +
    1. \frac{d}{dx}\left[\cosh^{-1}\left(\frac{3x-2}{5}\right)\right]
      2. + +
    2. \int\frac{1}{x^2-1}\, dx
      3. + +
    3. \int \frac{1}{\sqrt{9x^2+10}}\, dx
      4. +
    +

    +     + +

    +

      +
    1. +

      + Applying with the Chain Rule gives: + + \frac{d}{dx}\left[\cosh^{-1}\left(\frac{3x-2}5\right)\right] = \frac{1}{\sqrt{\left(\frac{3x-2}5\right)^2-1}}\cdot\frac35 + . +

      +
      2. + +
    2. +

      + Multiplying the numerator and denominator by (-1) gives: + \ds \int \frac{1}{x^2-1}\, dx = \int \frac{-1}{1-x^2}\, dx. + The second integral can be solved with a direct application of item #3 from , + with a=1. + Thus + + \int \frac{1}{x^2-1}\, dx \amp = -\int \frac{1}{1-x^2}\, dx + \amp = \left\{\begin{array}{ccc} -\tanh^{-1}\left(x\right)+C \amp \amp x^2\lt 1 \\ \\ + -\coth^{-1}\left(x\right)+C \amp \amp 1\lt x^2 + \end{array} \right. + \amp =-\frac12\ln\abs{\frac{x+1}{x-1}}+C + \amp =\frac12\ln\abs{\frac{x-1}{x+1}}+C + . + We should note that this exact problem was solved at the beginning of . + In that example the answer was given as \frac12\ln\abs{x-1}-\frac12\ln\abs{x+1}+C. + Note that this is equivalent to the answer given in Equation, + as \ln(a/b) = \ln(a) - \ln(b). +

      +
      3. + +
    3. +

      + This requires a substitution, + then item #2 of can be applied. + + Let u = 3x, hence du = 3dx. + We have + + \int \frac{1}{\sqrt{9x^2+10}}\, dx \amp = \frac13\int\frac{1}{\sqrt{u^2+10}}\,du. + Note a^2=10, hence a = \sqrt{10}. Now apply the integral rule. + \amp = \frac13 \sinh^{-1}\left(\frac{3x}{\sqrt{10}}\right) + C + \amp = \frac13 \ln\abs{3x+\sqrt{9x^2+10}}+C + . +

      +
      4. +
    +

    +     + +     + +

    + This section covers a lot of ground. + New functions were introduced, + along with some of their fundamental identities, + their derivatives and antiderivatives, their inverses, + and the derivatives and antiderivatives of these inverses. + Four Key Ideas were presented, + each including quite a bit of information. +

    + +

    + Do not view this section as containing a source of information to be memorized, + but rather as a reference for future problem solving. + + contains perhaps the most useful information. + Know the integration forms it helps evaluate and understand how to use the inverse hyperbolic answer and the logarithmic answer. +

    + +

    + The next section takes a brief break from demonstrating new integration techniques. + It instead demonstrates a technique of evaluating limits that return indeterminate forms. + This technique will be useful in , + where limits will arise in the evaluation of certain definite integrals. +

    +     + + + + Terms and Concepts + + + + + +

    + In , + the equation \ds \int \tanh(x) \, dx = \ln(\cosh(x) )+C is given. + Why is \ln\abs{\cosh(x) } + not used , why are absolute values not necessary? +

    + + +     + + + +

    + Because \cosh(x) is always positive. +

    +     + +     + + + + + +

    + The hyperbolic functions are used to define points on the right hand portion of the hyperbola x^2-y^2=1, + as shown in . + How can we use the hyperbolic functions to define points on the left hand portion of the hyperbola? +

    + + +     + + + +

    + The points on the left hand side can be defined as (-\cosh(x) , \sinh(x) ). +

    +     + +     +     + + Problems + + + +

    + Verify the given identity using , + as done in . +

    +     + + + + + +

    + Verify the identity \coth^2(x) -\csch^2(x) =1 using the definitions of the hyperbolic functions. +

    + + +     + + + +

    + \begin{aligned}\coth^2(x) -\csch^2(x) \amp = \left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right)^2 - \left(\frac{2}{e^x-e^{-x}}\right)^2 \\ + \amp = \frac{(e^{2x} + 2 + e^{-2x}) - (4)}{e^{2x} - 2 + e^{-2x}}\\ + \amp = \frac{e^{2x} - 2 + e^{-2x}}{e^{2x} - 2 + e^{-2x}}\\ + \amp = 1 + \end{aligned} +

    +     + +     + + + + + +

    + Verify the identity \cosh(2x) = \cosh^2(x) +\sinh^2(x) using the definitions of the hyperbolic functions. +

    + + +     + + + +

    + \begin{aligned}\cosh^2(x) +\sinh^2(x) \amp = \left(\frac{e^x+e^{-x}}{2}\right)^2 + \left(\frac{e^x-e^{-x}}{2}\right)^2 \\ + \amp = \frac{e^{2x} + 2 + e^{-2x}}{4} + \frac{e^{2x} - 2 + e^{-2x}}{4}\\ + \amp = \frac{2e^{2x} + 2e^{-2x}}{4}\\ + \amp = \frac{e^{2x} + e^{-2x}}{2} \\ + \amp = \cosh(2x) . + \end{aligned} +

    +     + +     + + + + + +

    + Verify the identity \ds\cosh^2(x) = \frac{\cosh(2x) +1}{2} using the definitions of the hyperbolic functions. +

    + + +     + + + +

    + \begin{aligned}\cosh^2(x) \amp = \left(\frac{e^x+e^{-x}}{2}\right)^2 \\ + \amp = \frac{e^{2x} + 2 + e^{-2x}}{4} \\ + \amp = \frac12\frac{(e^{2x} + e^{-2x})+2}{2}\\ + \amp = \frac12\left(\frac{e^{2x} + e^{-2x}}{2}+1\right)\\ + \amp = \frac{\cosh(2x) +1}{2}. + \end{aligned} +

    +     + +     + + + + + +

    + Verify the identity \ds\sinh^2(x) = \frac{\cosh(2x) -1}{2} using the definitions of the hyperbolic functions. +

    + + +     + + + +

    + \begin{aligned}\sinh^2(x) \amp = \left(\frac{e^x-e^{-x}}{2}\right)^2 \\ + \amp = \frac{e^{2x} - 2 + e^{-2x}}{4} \\ + \amp = \frac12\frac{(e^{2x} + e^{-2x})-2}{2}\\ + \amp = \frac12\left(\frac{e^{2x} + e^{-2x}}{2}-1\right)\\ + \amp = \frac{\cosh(2x) -1}{2}. + \end{aligned} +

    +     + +     + + + + + +

    + Verify the identity \ds\frac{d}{dx}\left[\sech(x) \right] = -\sech(x) \tanh(x) using the definitions of the hyperbolic functions. +

    + + +     + + + +

    + \begin{aligned}\frac{d}{dx}\left[\sech(x) \right] \amp = \frac{d}{dx}\left[\frac{2}{e^x+e^{-x}}\right] \\ + \amp = \frac{-2(e^x-e^{-x})}{(e^x+e^{-x})^2} \\ + \amp = -\frac{2(e^x-e^{-x})}{(e^x+e^{-x})(e^x+e^{-x})} \\ + \amp = -\frac{2}{e^x+e^{-x}}\cdot \frac{e^x-e^{-x}}{e^x+e^{-x}}\\ + \amp = -\sech(x) \tanh(x) \end{aligned} +

    +     + +     + + + + + +

    + Verify the identity \ds\frac{d}{dx}\left[\coth(x) \right] = -\csch^2(x) using the definitions of the hyperbolic functions. +

    + + +     + + + +

    + \begin{aligned}\frac{d}{dx}\left[\coth(x) \right] \amp = \frac{d}{dx}\left[\frac{e^x+e^{-x}}{e^x-e^{-x}}\right] \\ + \amp = \frac{(e^x-e^{-x})(e^x-e^{-x})-(e^x+e^{-x})(e^x+e^{-x})}{(e^x-e^{-x})^2}\\ + \amp = \frac{e^{2x}+e^{-2x} - 2 - (e^{2x}+e^{-2x}+2)}{(e^x-e^{-x})^2}\\ + \amp = -\frac{4}{(e^x-e^{-x})^2}\\ + \amp =-\csch^2(x) \end{aligned} +

    +     + +     + + + + + +

    + Verify the identity \ds \int \tanh(x) \, dx = \ln(\cosh(x) ) + C using the definitions of the hyperbolic functions. +

    + + +     + + + +

    + \ds \int \tanh(x) \, dx = \int \frac{\sinh(x) }{\cosh(x) }\, dx +

    + +

    + Let u = \cosh(x); du = (\sinh(x) ) dx +

    + +

    + \begin{aligned}\amp = \int \frac{1}{u}\,du \\ + \amp = \ln\abs{u} + C \\ + \amp = \ln(\cosh(x) ) + C. + \end{aligned} +

    +     + +     + + + + + +

    + Verify the identity \ds \int \coth(x) \, dx = \ln\abs{\sinh(x) } + C using the definitions of the hyperbolic functions. +

    + + +     + + + +

    + \begin{aligned}\int \coth(x) \, dx \amp = \int \frac{\cosh(x) }{\sinh(x) }\, dx\\ + \text{ Let \(u = \sinh(x) \); \(du = (\cosh(x) ) dx\) } \\ + \amp = \int \frac{1}{u}\,du \\ + \amp = \ln\abs{u} + C \\ + \amp = \ln\abs{\sinh(x) } + C. + \end{aligned} +

    +     + +     + +     + + + +

    + Find the derivative of the given function. +

    +     + + + + + $f = Formula("sinh(2x)"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = \sinh(2x). +

    + +

    + +

    +     +     +     + + + + + $f = Formula("(cosh(x))^2"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = \cosh ^2x. +

    + +

    + +

    +     + +

    + Taking the derivative of (\cosh x)^2 directly, + one gets 2\cosh x\sinh x; + using the identity \cosh^2x=\frac12(\cosh2x+1) first, + one gets \sinh 2x; + by known hyperbolic identities, + these are equal. +

    +     +     +     + + + + + + $f = Formula("tanh(x^2)"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = \tanh(x^2). +

    + +

    + +

    +     +     +     + + + + + $f = Formula("ln(sinh(x))"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = \ln(\sinh(x) ). +

    + +

    + +

    +     +     +     + + + + + $f = Formula("sinh(x)*cosh(x)"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = \sinh(x) \cosh(x). +

    + +

    + +

    +     +     +     + + + + + $f = Formula("x*sinh(x)-cosh(x)"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = x\sinh(x) -\cosh(x). +

    + +

    + +

    +     +     +     + + + + + $f = Formula("asech(x^2)"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = \sech^{-1}(x^2). +

    + +

    + +

    +     +     +     + + + + + $f = Formula("asinh(3x)"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = \sinh^{-1}(3x). +

    + +

    + +

    +     +     +     + + + + + $f = Formula("acosh(2x^2)"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = \cosh^{-1}(2x^2). +

    + +

    + +

    +     +     +     + + + + + $f = Formula("atanh(x+5)"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = \tanh^{-1}(x+5). +

    + +

    + +

    +     +     +     + + + + + $f = Formula("atanh(cos(x))"); + $df = $f->D('x'); + + + +

    + Find the derivative of f(x) = \tanh^{-1}(\cos(x) ). +

    + +

    + +

    +     +     +     + + + + + $f = Formula("acosh(sec(x))"); + $df = $f->D('x'); + + + + +

    + Find the derivative of f(x) = \cosh^{-1}(\sec(x) ). +

    + +

    + +

    +     +     +     + +     + + + +

    + Find the equation of the line tangent to the function at the given x-value. +

    +     + + + + + $x0 = 0; + $f = Formula("sinh(x)"); + $df = $f->D('x'); + $m = $df->eval(x=>$x0); + $y0 = $f->eval(x=>$x0); + $line = Formula("$m*(x - $x0) + $y0"); + + + +

    + Find the equation of the tangent line to y=f(x) at x=0, where f(x) = \sinh(x). +

    + +

    + y = +

    +     + +

    + y=x +

    + +     +     +     + + + + + $x0 = ln(2); + $f = Formula("cosh(x)"); + $df = $f->D('x'); + $m = $df->eval(x=>$x0); + $y0 = $f->eval(x=>$x0); + $line = Formula("$m*(x - $x0) + $y0"); + + + +

    + Find the equation of the tangent line to y=f(x) at x=\ln(2), where f(x) = \cosh(x). +

    + +

    + y = +

    +     + +

    + y=3/4(x-\ln(2) )+5/4 +

    + +     +     +     + + + + + $x0 = -ln(3); + $f = Formula("tanh(x)"); + $df = $f->D('x'); + $m = $df->eval(x=>$x0); + $y0 = $f->eval(x=>$x0); + $line = Formula("$m*(x - $x0) + $y0"); + + + + +

    + Find the equation of the tangent line to y=f(x) at x=-\ln(3), where f(x) = \tanh(x). +

    + +

    + y = +

    +     + +

    + y=\frac9{25}(x+\ln(3))-\frac45 +

    + +     +     +     + + + + + + $x0 = ln(3); + $f = Formula("sech(x)^2"); + $df = $f->D('x'); + $m = $df->eval(x=>$x0); + $y0 = $f->eval(x=>$x0); + $line = Formula("$m*(x - $x0) + $y0"); + + + +

    + Find the equation of the tangent line to y=f(x) at x=\ln(3), where f(x) = \sech^2(x). +

    + +

    + y = +

    +     + +

    + y=-72/125(x-\ln(3) )+9/25 +

    + +     +     +     + + + + + $x0 = 0; + $f = Formula("asinh(x)"); + $df = $f->D('x'); + $m = $df->eval(x=>$x0); + $y0 = $f->eval(x=>$x0); + $line = Formula("$m*(x - $x0) + $y0"); + + + +

    + Find the equation of the tangent line to y=f(x) at x=0, where f(x) = \sinh^{-1}(x). +

    + +

    + y = +

    +     + +

    + y=x +

    + +     +     +     + + + + + $x0 = sqrt(2); + $f = Formula("acosh(x)"); + $df = $f->D('x'); + $m = $df->eval(x=>$x0); + $y0 = $f->eval(x=>$x0); + $line = Formula("$m*(x - $x0) + $y0"); + + + +

    + Find the equation of the tangent line to y=f(x) at x=\sqrt 2, where f(x) = \cosh^{-1}(x). +

    + +

    + y = +

    +     + +

    + y=(x-\sqrt{2})+\cosh^{-1}(\sqrt{2}) \approx (x-1.414)+0.881 +

    + +     +     +     + +     + + + +

    + Evaluate the given indefinite integral. +

    +     + + + + + $F = FormulaUpToConstant("1/2*ln(cosh(2x))"); + + + +

    + Evaluate the indefinite integral \ds \int \tanh(2x)\, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("1/3*sinh(3x-7)"); + + + +

    + Evaluate the indefinite integral \ds \int \cosh(3x-7)\, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("1/2*sinh(x)^2"); + + + +

    + Evaluate the indefinite integral \ds \int \sinh(x) \cosh(x) \, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("x*sinh(x)-cosh(x)"); + + + +

    + Evaluate the indefinite integral \ds \int x\cosh(x) \, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("x*cosh(x)-sinh(x)"); + + + +

    + Evaluate the indefinite integral \ds \int x\sinh(x) \, dx. +

    + +

    + +

    +     +     +     + + + +

    + Evaluate the indefinite integral \ds\int \frac1{\sqrt{x^2+1}}\, dx. +

    +     + +

    + \sinh^{-1} x +C=\ln\big(x+\sqrt{x^2+1}\big)+C +

    +     +     + + + +

    + Evaluate the indefinite integral \ds\int \frac1{\sqrt{x^2-9}}\, dx. +

    +     + +

    + \cosh^{-1} x/3 +C=\ln\big(x+\sqrt{x^2-9}\big)+C +

    +     +     + + + + + + $F = FormulaUpToConstant("1/2*ln(abs(x+1))-1/2*ln(abs(x-1))"); + + + +

    + Evaluate the indefinite integral \ds \int \frac{1}{9-x^2}\, dx. +

    + +

    + +

    +     + +

    + \left\{\begin{array}{ccc} \frac13\tanh^{-1}\left(\frac x3\right)+C \amp \amp x^2\lt 9 \\ \\ + \frac13\coth^{-1}\left(\frac x3\right)+C \amp \amp 9\lt x^2 + \end{array} \right. = \frac12\ln\abs{x+1} - \frac12\ln\abs{x-1}+C +

    + +     +     +     + + + + + $F = FormulaUpToConstant("acosh(x^2/2)"); + + + +

    + Evaluate the indefinite integral \ds \int \frac{2x}{\sqrt{x^4-4}}\, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("2/3*asinh(x^(3/2))"); + + + +

    + Evaluate the indefinite integral \ds \int \frac{\sqrt{x}}{\sqrt{1+x^3}}\, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("-1/16*atan(x/2)+1/32*ln(abs(x-2))-1/32*ln(abs(x+2))"); + + + +

    + Evaluate the indefinite integral \ds \int \frac{1}{x^4-16}\, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("ln(x)-ln(abs(x+1))"); + + + +

    + Evaluate the indefinite integral \ds \int \frac{1}{x^2+x}\, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("atan(e^x)"); + + + +

    + Evaluate the indefinite integral \ds \int \frac{e^x}{e^{2x}+1}\, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("x*asinh(x)-sqrt(x^2+1)"); + + + +

    + Evaluate the indefinite integral \ds \int \sinh^{-1}(x) \, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("x*atanh(x)+1/2*ln(abs(x^2-1))"); + + + +

    + Evaluate the indefinite integral \ds \int \tanh^{-1}(x) \, dx. +

    + +

    + +

    +     +     +     + + + + + $F = FormulaUpToConstant("atan(sinh(x))"); + + + +

    + Evaluate the indefinite integral \ds \int \sech(x) \, dx. +

    + +

    + (Hint: mutiply by \frac{\cosh(x) }{\cosh(x) }; + set u = \sinh(x).) +

    + +

    + +

    +     +     +     + +     + + + +

    + Evaluate the given definite integral. +

    +     + + + + + $a = -1; + $b = 1; + $F = Formula("cosh(x)"); + $r = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + Evaluate the definite integral \ds \int_{-1}^1 \sinh(x) \, dx. +

    + +

    + +

    + +     +     +     + + + + + $a = -ln(2); + $b = ln(2); + $F = Formula("sinh(x)"); + $r = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + Evaluate the definite integral \ds \int_{-\ln(2) }^{\ln(2) } \cosh(x) \, dx. +

    + +

    + +

    + +     +     +     + + + + + $a = 0; + $b = 1; + $F = Formula("tanh(x)"); + $r = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + Evaluate the definite integral \ds \int_{0}^{1} \sech^{2}(x) \, dx. +

    + +

    + +

    + +     +     +     + + + + + $a = 0; + $b = 2; + $F = Formula("asinh(x)"); + $r = $F->eval(x=>$b) - $F->eval(x=>$a); + + + +

    + Evaluate the definite integral \ds \int_{0}^{2}\frac1{\sqrt{x^2+1}}\, dx. +

    + +

    + +

    +     + +     +     +     +     +     +
    +
    + L'Hospital's Rule + +

    + While this chapter is devoted to learning techniques of integration, + this section is not about integration. + Rather, it is concerned with a technique of evaluating certain limits that will be useful in the following section, + where integration is once more discussed. +

    + +

    + Our treatment of limits exposed us to the notion of + 0/0, an indeterminate form. + If \lim\limits_{x\to c}f(x)=0 and \lim\limits_{x\to c} g(x) =0, + we do not conclude that \lim\limits_{x\to c} f(x)/g(x) is 0/0; + rather, we use 0/0 as notation to describe the fact that both the numerator and denominator approach 0. + The expression 0/0 has no numeric value; + other work must be done to evaluate the limit. +

    + +

    + Other indeterminate forms exist; they are: + \infty/\infty, 0\cdot\infty, + \infty-\infty, 0^0, + 1^\infty and \infty^0. + Just as 0/0 does not mean divide 0 by 0, + the expression \infty/\infty + does not mean divide infinity by infinity. Instead, + it means a quantity is growing without bound and is being divided by another quantity that is growing without bound. + We cannot determine from such a statement what value, + if any, results in the limit. + Likewise, 0\cdot \infty + does not mean multiply zero by infinity. Instead, + it means one quantity is shrinking to zero, + and is being multiplied by a quantity that is growing without bound. + We cannot determine from such a description what the result of such a limit will be. +

    + + + +

    + This section introduces l'Hospital's Rule, + a method of resolving limits that produce the indeterminate forms 0/0 and \infty/\infty. + We'll also show how algebraic manipulation can be used to convert other indeterminate expressions into one of these two forms so that our new rule can be applied. +

    +     + + + L'Hospital's Rule with Indeterminate Forms <m>0/0</m> and <m>\infty/\infty</m> + + L'Hospital's Rule, Part 1 + +

    + Let \lim\limits_{x\to c}f(x) = 0 and \lim\limits_{x\to c}g(x)=0, + where f and g are differentiable functions on an open interval I containing c, + and \gp(x)\neq 0 on I except possibly at c. + If limitl'Hospital's Rule + l'Hospital's Rulezero over zero + + \lim_{x\to c} \frac{\fp(x)}{\gp(x)}=L + , + then + + \lim_{x\to c} \frac{f(x)}{g(x)}=L + , + where L is a real number, or L=\pm \infty. + The result applies to one-sided limits as well. +

    +     +     + + + +

    + We demonstrate the use of l'Hospital's Rule in the following examples; + we will often use LHR as an abbreviation of + l'Hospital's Rule. +

    + + + Using l'Hospital's Rule + +

    + Evaluate the following limits, + using l'Hospital's Rule as needed. +

    + +

    +

      +
    1. + \lim\limits_{x\to0}\frac{\sin(x) }x +
      2. + +
    2. + \lim\limits_{x\to 1}\frac{\sqrt{x+3}-2}{1-x} +
      3. + +
    3. + \lim\limits_{x\to0}\frac{x^2}{1-\cos(x) } +
      4. + +
    4. + \lim\limits_{x\to 2}\frac{x^2+x-6}{x^2-3x+2} +
      5. +
    +

    +

    + + +

    +

      +
    1. +

      + We proved this limit is 1 in using the Squeeze Theorem. + Here we use l'Hospital's Rule to show its power. + + \lim_{x\to0}\frac{\sin(x) }x \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to0} \frac{\cos(x) }{1}=1 + . +

      +
      2. + +
    2. + + \lim\limits_{x\to 1}\frac{\sqrt{x+3}-2}{1-x} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x \to 1} \frac{\frac12(x+3)^{-1/2}}{-1} =-\frac 14 + . +
      3. + +
    3. +

      + \lim\limits_{x\to 0}\frac{x^2}{1-\cos(x) } \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to 0} \frac{2x}{\sin(x) }. + + This latter limit also evaluates to the 0/0 indeterminate form. + To evaluate it, we apply l'Hospital's Rule again. + + \lim\limits_{x\to 0} \frac{2x}{\sin(x) } \stackrel{\,\text{ by LHR } \,}{=} \frac{2}{\cos(x) } = 2. + + Thus \lim\limits_{x\to0}\frac{x^2}{1-\cos(x) }=2. +

      +
      4. + +
    4. +

      + We already know how to evaluate this limit; + first factor the numerator and denominator. + We then have: + + \lim_{x\to 2}\frac{x^2+x-6}{x^2-3x+2} = \lim_{x\to 2}\frac{(x-2)(x+3)}{(x-2)(x-1)} = \lim_{x\to 2}\frac{x+3}{x-1} = 5 + . + We now show how to solve this using l'Hospital's Rule. + + \lim_{x\to 2}\frac{x^2+x-6}{x^2-3x+2}\stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to 2}\frac{2x+1}{2x-3} = 5 + . +

      +
      5. +
    +

    + + +     + +

    + Note that at each step where l'Hospital's Rule was applied, + it was needed: + the initial limit returned the indeterminate form of 0/0. + If the initial limit returns, + for example, 1/2, then l'Hospital's Rule does not apply. +

    + +

    + The following theorem extends our initial version of l'Hospital's Rule in two ways. + It allows the technique to be applied to the indeterminate form + \infty/\infty and to limits where x approaches \pm\infty. +

    + + + L'Hospital's Rule, Part 2 + +

    +

      +
    1. +

      + Let \lim\limits_{x\to a}f(x) = \pm\infty and \lim\limits_{x\to a}g(x)=\pm \infty, + where f and g are differentiable on an open interval I containing a. + If + + limitl'Hospital's Rule + l'Hospital's Ruleinfinity over infinity + + + \lim_{x\to a}\frac{\fp(x)}{\gp(x)}=L + , + then + + \lim_{x\to a}\frac{f(x)}{g(x)}=L + , + where L is a real number, or L=\pm\infty. + The result applies to one-sided limits as well. +

      +
      2. + +
    2. +

      + Let f and g be differentiable functions on the open interval + (a,\infty) for some value a, + where \gp(x)\neq 0 on (a,\infty) and + \lim\limits_{x\to\infty} f(x)/g(x) returns either 0/0 or \infty/\infty. + If + + \lim_{x\to \infty}\frac{\fp(x)}{\gp(x)}=L + , + then + + \lim_{x\to \infty}\frac{f(x)}{g(x)}=L + , + where L is a real number, or L=\pm \infty. + A similar statement can be made for limits where x approaches -\infty. +

      +
      3. +
    +

    +     +     + + + Using l'Hospital's Rule with limits involving <m>\infty</m> + +

    + Evaluate the following limits. +

    + +

    + \ds 1.\,\lim_{x\to\infty} \frac{3x^2-100x+2}{4x^2+5x-1000} \qquad\qquad 2. \,\lim_{x\to \infty}\frac{e^x}{x^3}. +

    +     + +

    +

      +
    1. +

      + We can evaluate this limit already using ; + the answer is 3/4. + We apply l'Hospital's Rule to demonstrate its applicability. + + \lim_{x\to\infty} \frac{3x^2-100x+2}{4x^2+5x-1000}\stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{6x-100}{8x+5} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac68 = \frac34 + . +

      +
      2. + +
    2. +

      + + \lim\limits_{x\to \infty}\frac{e^x}{x^3} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{e^x}{3x^2} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{e^x}{6x} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{e^x}{6} = \infty + . + + Recall that this means that the limit does not exist; + as x approaches \infty, + the expression e^x/x^3 grows without bound. + We can infer from this that e^x grows + faster than x^3; + as x gets large, e^x is far larger than x^3. + (This has important implications in computing when considering efficiency of algorithms.) +

      +
      3. +
    +

    +     + +     + + +     + + + Indeterminate Forms <m>0\cdot\infty</m> and <m>\infty-\infty</m> +

    + L'Hospital's Rule can only be applied to ratios of functions. + When faced with an indeterminate form such as + 0\cdot\infty or \infty-\infty, + we can sometimes apply algebra to rewrite the limit so that l'Hospital's Rule can be applied. + We demonstrate the general idea in the next example. + limitindeterminate form + indeterminate form +

    + + + Applying l'Hospital's Rule to other indeterminate forms + +

    + Evaluate the following limits. +

    + +

    +

      +
    1. \lim\limits_{x\to0^+} x\cdot e^{1/x}
      2. + +
    2. \lim\limits_{x\to0^-} x\cdot e^{1/x}
      3. + +
    3. \lim\limits_{x\to\infty} \ln(x+1)-\ln(x)
      4. + +
    4. \lim\limits_{x\to\infty} x^2-e^x
      5. +
    +

    +     + +

    +

      +
    1. +

      + As x\rightarrow 0^+, + x\rightarrow 0 and e^{1/x}\rightarrow \infty. + Thus we have the indeterminate form 0\cdot\infty. + We rewrite the expression x\cdot e^{1/x} as \ds\frac{e^{1/x}}{1/x}; + now, as x\rightarrow 0^+, + we get the indeterminate form + \infty/\infty to which l'Hospital's Rule can be applied. + + \lim_{x\to0^+} x\cdot e^{1/x} = \lim_{x\to 0^+} \frac{e^{1/x}}{1/x} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to 0^+}\frac{(-1/x^2)e^{1/x}}{-1/x^2} =\lim_{x\to 0^+}e^{1/x} =\infty + . + Interpretation: e^{1/x} grows faster + than x shrinks to zero, + meaning their product grows without bound. +

      +
      2. + +
    2. +

      + As x\rightarrow 0^-, + x\rightarrow 0 and e^{1/x}\rightarrow e^{-\infty}\rightarrow 0. + The the limit evaluates to + 0\cdot 0 which is not an indeterminate form. + We conclude then that + + \lim_{x\to 0^-}x\cdot e^{1/x} = 0 + . +

      +
      3. + +
    3. +

      + This limit initially evaluates to the indeterminate form \infty-\infty. + By applying a logarithmic rule, + we can rewrite the limit as + + \lim_{x\to\infty} \ln(x+1)-\ln(x) = \lim_{x\to \infty} \ln\left(\frac{x+1}x\right) + . + As x\rightarrow \infty, + the argument of the \ln term approaches \infty/\infty, + to which we can apply l'Hospital's Rule. + + \lim_{x\to\infty} \frac{x+1}x \stackrel{\,\text{ by LHR } \,}{=} \frac11=1 + . + Since x\rightarrow \infty implies \ds\frac{x+1}x\rightarrow 1, + it follows that + + x\rightarrow \infty \text{ implies } \ln\left(\frac{x+1}x\right)\rightarrow \ln(1) =0 + . + Thus + + \lim_{x\to\infty} \ln(x+1)-\ln(x) = \lim_{x\to \infty} \ln\left(\frac{x+1}x\right)=0 + . + Interpretation: + since this limit evaluates to 0, it means that for large x, + there is essentially no difference between \ln(x+1) and \ln(x); + their difference is essentially 0. +

      +
      4. + +
    4. +

      + The limit \lim\limits_{x\to\infty} x^2-e^x initially returns the indeterminate form \infty-\infty. + We can rewrite the expression by factoring out x^2; + \ds x^2-e^x = x^2\left(1-\frac{e^x}{x^2}\right). + We need to evaluate how e^x/x^2 behaves as x\rightarrow \infty: + + \lim_{x\to\infty}\frac{e^x}{x^2} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{e^x}{2x} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{e^x}{2} = \infty + . + Thus \lim_{x\to\infty}x^2(1-e^x/x^2) evaluates to \infty\cdot(-\infty), + which is not an indeterminate form; + rather, \infty\cdot(-\infty) evaluates to -\infty. + We conclude that + \lim\limits_{x\to\infty} x^2-e^x = -\infty. + + Interpretation: as x gets large, + the difference between x^2 and e^x grows very large. +

      +
      5. +
    +

    +     + +     +     + + + Indeterminate Forms<nbsp/> <m>0^0</m>, <m>1^\infty</m> and <m>\infty^0</m> +

    + When faced with an indeterminate form that involves a power, + it often helps to employ the natural logarithmic function. + The following Key Idea expresses the concept, + which is followed by an example that demonstrates its use. +

    + + + Evaluating Limits Involving Indeterminate Forms <m>0^0</m>, <m>1^\infty</m> and <m>\infty^0</m> +

    + If \lim\limits_{x\to c} \ln\big(f(x)\big) = L, then + \lim\limits_{x\to c} f(x) = \lim_{x\to c} e^{\ln(f(x))} = e^L. + + limitindeterminate form + indeterminate form +

    +     + + + Using l'Hospital's Rule with indeterminate forms involving exponents + +

    + Evaluate the following limits. +

    + +

    +

      +
    1. \lim_{x\to\infty} \left(1+\frac1x\right)^x
      2. +
    2. \lim_{x\to0^+} x^x
      3. +
    +

    +     + +

    +

      +
    1. +

      + This is equivalent to a special limit given in ; + these limits have important applications within mathematics and finance. + Note that the exponent approaches \infty while the base approaches 1, leading to the indeterminate form 1^\infty. + Let f(x) = (1+1/x)^x; + the problem asks to evaluate \lim\limits_{x\to\infty}f(x). + Let's first evaluate \lim\limits_{x\to\infty}\ln\big(f(x)\big). + + \lim_{x\to\infty}\ln\big(f(x)\big) \amp = \lim_{x\to\infty} \ln\left(1+\frac1x\right)^x + \amp = \lim_{x\to\infty} x\ln\left(1+\frac1x\right) + \amp = \lim_{x\to\infty} \frac{\ln\left(1+\frac1x\right)}{1/x} + This produces the indeterminate form 0/0, so we apply l'Hospital's Rule. + \amp = \lim_{x\to\infty} \frac{\frac{1}{1+1/x}\cdot(-1/x^2)}{(-1/x^2)} + \amp = \lim_{x\to\infty}\frac{1}{1+1/x} + \amp = 1 + . + Thus \lim\limits_{x\to\infty} \ln\big(f(x)\big) = 1. + We return to the original limit and apply . + + \lim_{x\to\infty}\left(1+\frac1x\right)^x = \lim_{x\to\infty} f(x) = \lim_{x\to\infty}e^{\ln(f(x))} = e^1 = e + . +

      +
      2. + +
    2. +

      + This limit leads to the indeterminate form 0^0. + Let f(x) = x^x and consider first \lim\limits_{x\to0^+} \ln\big(f(x)\big). + + \lim_{x\to0^+} \ln\big(f(x)\big) \amp = \lim_{x\to0^+} \ln\left(x^x\right) + \amp = \lim_{x\to0^+} x\ln(x) + \amp = \lim_{x\to0^+} \frac{\ln(x) }{1/x}. + This produces the indeterminate form -\infty/\infty so we apply l'Hospital's Rule. + \amp = \lim_{x\to0^+} \frac{1/x}{-1/x^2} + \amp = \lim_{x\to0^+} -x + \amp = 0 + . + Thus \lim\limits_{x\to0^+} \ln\big(f(x)\big) =0. + We return to the original limit and apply . + + \lim_{x\to0^+} x^x = \lim_{x\to0^+} f(x) = \lim_{x\to0^+} e^{\ln(f(x))} = e^0 = 1 + . + This result is supported by the graph of + f(x)=x^x given in . +

      +
      + A graph of f(x)=x^x supporting the fact that as x\to 0^+, f(x)\to 1 + + + + Graph of function x^x, used in this the example. + + +

      + The y axis is drawn from 0 to 4 and the + x axis is drawn from 0 to 2. The + function f(x) = x^x is drawn as a curve opening + towards the positive y axis with arrows towards the + ends. The function is drawn from point (0,1) from + where it dips gently then rises up slowly. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ytick={1,2,3,4}, + ymin=-.4,ymax=4.5, + xmin=-.1,xmax=2.2, + scaled ticks=false + ] + + \addplot+ [infinite,domain=0:2,samples=40] {x^x}; + + \draw (axis cs:1,1) node [below right] { $f(x)=x^x$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      3. +
    +

    +     + +     + + + + + +

    + Our brief revisit of limits will be rewarded in the next section where we consider + improper integration. So far, + we have only considered definite integrals where the bounds are finite numbers, + such as \ds \int_0^1 f(x)\, dx. + Improper integration considers integrals where one, or both, + of the bounds are infinity. + Such integrals have many uses and applications, + in addition to generating ideas that are enlightening. +

    +     + + + + Terms and Concepts + + + + +

    + List the different indeterminate forms described in this section. +

    + + +     + + + +

    + 0/0, \infty/\infty, 0\cdot\infty,\infty-\infty,0^0,1^\infty,\infty^0 +

    + +     + +     + + + + +

    + L'Hospital's Rule provides a faster method of computing derivatives. + +

    +     + +     + + + + +

    + L'Hospital's Rule states that + \ds \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\fp(x)}{\gp(x)}. + +

    +     + +     + + + + +

    + Explain what the indeterminate form + 1^\infty means. +

    + + +     + + + +

    + The base of an expression is approaching 1 while its power is growing without bound. +

    + +     + +     + + + + +

    + Fill in the blanks: The Quotient Rule is applied to + \ds \frac{f(x)}{g(x)} when taking ; + l'Hospital's Rule is applied when taking certain . +

    +     + + + + a derivative|derivatives? + + + + +

    + Your answer includes the correct word but has extra text. +

    +     +     +     + + + limits? + + + + +

    + Your answer includes the correct word but has extra text. +

    +     +     +     +     + +     + + + + +

    + Create (but do not evaluate!) a limit that returns + \infty^0. +

    + +     + + + +     + + + + +

    + Create a function f(x) such that + \lim\limits_{x\to 1}f(x) returns 0^0. +

    + + +     + + + +     + + + + +

    + Create a function f(x) such that \ds \lim_{x\to \infty}f(x) returns + 0\cdot\infty. +

    + + + +     + + + +     +     + + Problems + + + +

    + Evaluate the given limit using l'Hospital's rule. +

    +     + + + + + + $xlim = 1; + $fu = Formula("x^2+x-2"); + $fl = Formula("x-1"); + $dfu = $fu->D('x'); + $dfl = $fl->D('x'); + if ($dfl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to 1} \frac{ }{} +

    + +

    + +

    +     +     +     + + + + + + $xlim = 2; + $fu = Formula("x^2+x-6"); + $fl = Formula("x^2-7x+10"); + $dfu = $fu->D('x'); + $dfl = $fl->D('x'); + if ($dfl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to 2} \frac{ }{} +

    + +

    + +

    + +     +     +     + + + + + + $xlim = pi; + $fu = Formula("sin(x)"); + $fl = Formula("x-pi"); + $dfu = $fu->D('x'); + $dfl = $fl->D('x'); + if ($dfl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to \pi} \frac{ }{} +

    + +

    + +

    + +     +     +     + + + + + + $xlim = pi/4; + $fu = Formula("sin(x)-cos(x)"); + $fl = Formula("cos(2x)"); + $dfu = $fu->D('x'); + $dfl = $fl->D('x'); + if ($dfl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to \pi/4} \frac{ }{} +

    + +

    + +

    + +     +     +     + + + + + + $xlim = 0; + $fu = Formula("sin(5x)"); + $fl = Formula("x"); + $dfu = $fu->D('x'); + $dfl = $fl->D('x'); + if ($dfl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to 0} \frac{ }{} +

    + +

    + +

    + +     +     +     + + + + + + $xlim = 0; + $fu = Formula("sin(2x)"); + $fl = Formula("x+2"); + $l = $fu->eval(x=>$xlim) / $fl->eval(x=>$xlim); + + +

    + \lim\limits_{x\to 0} \frac{ }{} +

    + +

    + +

    + +     +     +     + + + + + + $xlim = 0; + $fu = Formula("sin(2x)"); + $fl = Formula("sin(3x)"); + $dfu = $fu->D('x'); + $dfl = $fl->D('x'); + if ($dfl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to 0} \frac{ }{} +

    + +

    + +

    + +     +     +     + + + + + + + Context()->variables->add(a=>"Real",b=>"Real"); + $xlim = 0; + $fu = Formula("sin(a*x)"); + $fl = Formula("sin(b*x)"); + $dfu = $fu->D('x'); + $dfl = $fl->D('x'); + if ($dfl->substitute(x=>$xlim)==Formula("0")) { + $l = Compute("inf"); + } + else { + $l = $dfu->substitute(x=>$xlim) / $dfl->substitute(x=>$xlim); + } + + +

    + \lim\limits_{x\to 0} \frac{ }{} +

    + +

    + +

    +     +     +     + + + + + + $xlim = 0; + $fu = Formula("e^x-1"); + $fl = Formula("x^2"); + $dfu = $fu->D('x'); + $dfl = $fl->D('x'); + if ($dfl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to 0^+} \frac{ }{} +

    + +

    + +

    + +     +     +     + + + + + + $xlim = 0; + $fu = Formula("e^x-x-1"); + $fl = Formula("x^2"); + $d2fu = $fu->D('x')->D('x'); + $d2fl = $fl->D('x')->D('x'); + if ($d2fl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $d2fu->eval(x=>$xlim) / $d2fl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to 0^+} \frac{ }{} +

    + +

    + +

    + +     +     +     + + + + + + $xlim = 0; + $fu = Formula("x-sin(x)"); + $fl = Formula("x^3-x^2"); + $d2fu = $fu->D('x')->D('x'); + $d2fl = $fl->D('x')->D('x'); + if ($d2fl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $d2fu->eval(x=>$xlim) / $d2fl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to 0^+} \frac{ }{} +

    + +

    + +

    + +     +     +     + + + + + + + $l = Compute("0"); + $fu = Formula("x^4"); + $fl = Formula("e^x"); + + +

    + \lim\limits_{x\to \infty} \frac{}{} +

    + +

    + +

    +     +     +     + + + + + + $l = Compute("0"); + $fu = Formula("sqrt(x)"); + $fl = Formula("e^x"); + + +

    + \lim\limits_{x\to \infty} \frac{}{} +

    + +

    + +

    +     +     +     + + + + + + $l = Compute("infinity"); + $fu = Formula("e^x"); + $fl = Formula("x^2"); + + +

    + \lim\limits_{x\to\infty} \frac{}{} +

    +     +     +     + + + + + + $fu = Formula("e^x"); + $fl = Formula("sqrt(x)"); + $l = Compute("infinity"); + + +

    + \lim\limits_{x\to \infty} \frac{}{} +

    + +

    + +

    +     +     +     + + + + + + $fu = Formula("e^x"); + $fl = Formula("2^x"); + $l = Compute("infinity"); + + +

    + \lim\limits_{x\to \infty} \frac{}{} +

    + +

    + +

    +     +     +     + + + + + + $fu = Formula("e^x"); + $fl = Formula("3^x"); + $l = Compute("0"); + + +

    + \lim\limits_{x\to \infty} \frac{}{} +

    + +

    + +

    +     +     +     + + + + + + $xlim = 3; + $fu = Formula("x^3-5x^2+3x+9"); + $fl = Formula("x^3-7x^2+15x-9"); + $d2fu = $fu->D('x')->D('x'); + $d2fl = $fl->D('x')->D('x'); + if ($d2fl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $d2fu->eval(x=>$xlim) / $d2fl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to 3} \frac{}{} +

    + +

    + +

    + +     +     +     + + + + + + $xlim = -2; + $fu = Formula("x^3+4x^2+4x"); + $fl = Formula("x^3+7x^2+16x+12"); + $d2fu = $fu->D('x')->D('x'); + $d2fl = $fl->D('x')->D('x'); + if ($d2fl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $d2fu->eval(x=>$xlim) / $d2fl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to -2} \frac{}{} +

    + +

    + +

    + +     +     +     + + + + + + $fu = Formula("ln(x)"); + $fl = Formula("x"); + $l = Compute("0"); + + +

    + \lim\limits_{x\to \infty} \frac{}{} +

    + +

    + +

    + +     +     +     + + + + + + $fu = Formula("ln(x^2)"); + $fl = Formula("x"); + $l = Compute("0"); + + + + +

    + \lim\limits_{x\to \infty} \frac{}{} +

    + +

    + +

    + +     +     +     + + + + + + $fu = Formula("ln(x)^2"); + $fl = Formula("x"); + $l = Compute("0"); + + + + +

    + \lim\limits_{x\to \infty} \frac{}{} +

    + +

    + +

    + +     +     +     + + + + + + $f1 = Formula("x"); + $f2 = Formula("ln(x)"); + $l = Compute("0"); + + +

    + \lim\limits_{x\to 0^+} \cdot +

    + +

    + +

    +     +     +     + + + + + + $f1 = Formula("sqrt(x)"); + $f2 = Formula("ln(x)"); + $l = Compute("0"); + + +

    + \lim\limits_{x\to 0^+} \cdot +

    + +

    + +

    + +     +     +     + + + + + + + $f1 = Formula("x"); + $f2 = Formula("e^(1/x)"); + $l = Compute("inf"); + + +

    + \lim\limits_{x\to 0^+} \cdot +

    + +

    + +

    +     +     +     + + + + + + $f1 = Formula("x^3"); + $f2 = Formula("x^2"); + $l = Compute("inf"); + + +

    + \lim\limits_{x\to \infty} - +

    + +

    + +

    + +     +     +     + + + + + + $f1 = Formula("sqrt(x)"); + $f2 = Formula("ln(x)"); + $l = Compute("inf"); + + +

    + \lim\limits_{x\to \infty} - +

    + +

    + +

    +     +     +     + + + + + + $f1 = Formula("x"); + $f2 = Formula("e^x"); + $l = Compute("0"); + + +

    + \lim\limits_{x\to -\infty} \cdot +

    + +

    + +

    +     +     +     + + + + + + $f1 = Formula("1/x^2"); + $f2 = Formula("e^(-1/x)"); + $l = Compute("0"); + + +

    + \lim\limits_{x\to 0^+} \cdot +

    + +

    + +

    +     +     +     + + + + + + $fb = Formula("(1+x)"); + $fp = Formula("1/x"); + $l = Compute("e"); + + +

    + \lim\limits_{x\to 0^+} ()^ +

    + +

    + +

    +     +     +     + + + + + + $fb = Formula("2x"); + $fp = Formula("x"); + $l = Compute("1"); + + +

    + \lim\limits_{x\to 0^+} ()^{} +

    + +

    + +

    +     +     +     + + + + + + $fb = Formula("2/x"); + $fp = Formula("x"); + $l = Compute("1"); + + +

    + \lim\limits_{x\to 0^+} ()^{} +

    + +

    + +

    +     +     +     + + + + + + $fb = Formula("sin(x)"); + $fp = Formula("x"); + $l = Compute("1"); + + +

    + \lim\limits_{x\to 0^+} ()^{} + Hint: use the Squeeze Theorem. +

    + +

    + +

    +     +     +     + + + + + + $fb = Formula("1-x"); + $fp = Formula("1-x"); + $l = Compute("1"); + + +

    + \lim\limits_{x\to 1^-} ()^{} +

    + +

    + +

    +     +     +     + + + + + + $fb = Formula("x"); + $fp = Formula("1/x"); + $l = Compute("1"); + + +

    + \lim\limits_{x\to \infty} ()^{} +

    + +

    + +

    +     +     +     + + + + + + $fb = Formula("1/x"); + $fp = Formula("x"); + $l = Compute("0"); + + +

    + \lim\limits_{x\to \infty} ()^{} +

    + +

    + +

    +     +     +     + + + + + + $fb = Formula("ln(x)"); + $fp = Formula("1-x"); + $l = Compute("1"); + + +

    + \lim\limits_{x\to 1^+} ()^{} +

    + +

    + +

    +     +     +     + + + + + + $fb = Formula("1+x"); + $fp = Formula("1/x"); + $l = Compute("1"); + + +

    + \lim\limits_{x\to \infty} ()^{} +

    + +

    + +

    +     +     +     + + + + + + $fb = Formula("1+x^2"); + $fp = Formula("1/x"); + $l = Compute("1"); + + +

    + \lim\limits_{x\to \infty} ()^{} +

    + +

    + +

    +     +     +     + + + + + + $f1 = Formula("tan(x)"); + $f2 = Formula("cos(x)"); + $l = Compute("1"); + + +

    + \lim\limits_{x\to \pi/2} +

    + +

    + +

    +     +     +     + + + + + + $f1 = Formula("tan(x)"); + $f2 = Formula("sin(2x)"); + $l = Compute("2"); + + +

    + \lim\limits_{x\to \pi/2} +

    + +

    + +

    +     +     +     + + + + + + $f1 = Formula("1/ln(x)"); + $f2 = Formula("1/(x-1)"); + $l = Compute("1/2"); + + +

    + \lim\limits_{x\to 1^+} - +

    + +

    + +

    +     +     +     + + + + + + $f1 = Formula("5/(x^2-9)"); + $f2 = Formula("x/(x-3)"); + $l = Compute("-inf"); + + +

    + \lim\limits_{x\to 3^+} - +

    + +

    + +

    +     +     +     + + + + + + $f1 = Formula("x"); + $f2 = Formula("tan(1/x)"); + $l = Compute("1"); + + +

    + \lim\limits_{x\to \infty} +

    + +

    + +

    +     +     +     + + + + + + $fu = Formula("ln(x)^3"); + $fl = Formula("x"); + $l = Compute("0"); + + +

    + \lim\limits_{x\to \infty} \frac{}{} +

    + +

    + +

    +     +     +     + + + + + + $xlim = 1; + $fu = Formula("x^2+x-2"); + $fl = Formula("ln(x)"); + $dfu = $fu->D('x'); + $dfl = $fl->D('x'); + if ($dfl->eval(x=>$xlim)==0) { + $l = Compute("inf"); + } + else { + $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); + } + + +

    + \lim\limits_{x\to 1} \frac{}{ } +

    + +

    + +

    + +     +     +     + +     +     +     +
    +
    + Improper Integration + +

    + We begin this section by considering the following definite integrals: +

    + +

    +

      +
    • \int_0^{100}\frac1{1+x^2}\, dx \approx 1.5608
      • + +
    • \int_0^{1000}\frac1{1+x^2}\, dx \approx 1.5698
      • + +
    • \int_0^{10,000}\frac1{1+x^2}\, dx \approx 1.5707
      • +
    +

    + +

    + Notice how the integrand is 1/(1+x^2) in each integral + (which is sketched in ). + As the upper bound gets larger, + one would expect the area under the curve would also grow. + While the definite integrals do increase in value as the upper bound grows, + they are not increasing by much. + In fact, consider: + + \int_0^b \frac{1}{1+x^2}\, dx = \tan^{-1}(x) \Big|_0^b = \tan^{-1}(b) -\tan^{-1}(0) = \tan^{-1}(b) + . +

    + +

    + As b\rightarrow \infty, + \tan^{-1}(b) \rightarrow \pi/2. + Therefore it seems that as the upper bound b grows, + the value of the definite integral + \ds \int_0^b\frac{1}{1+x^2}\, dx approaches \pi/2\approx 1.5708. + This should strike the reader as being a bit amazing: + even though the curve extends to infinity, + it has a finite amount of area underneath it. +

    + +
    + Graphing \ds f(x)=\frac{1}{1+x^2} + + + + Graph of function 1/(1+x^2) showing finite area under curves that extends to infinity. + + +

    + The y axis is drawn from 0 to 1 and the x axis is drawn + from 0 to 10. + The function f(x)=\frac{1}{1+x^2} starts at point (0,1), then decreases + sharply, has a dip at (2,0.2) then decreases gently and almost coincides with + the x axis after x=6. The area under the curve is shaded. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-1,xmax=11 + ] + + \addplot [firstcurvestyle,areastyle,domain=-0:1.5] {1/(1+x^2)} \closedcycle; + \addplot [firstcurvestyle,areastyle,domain=1.5:10.5] {1/(1+x^2)} \closedcycle; + + \addplot [firstcurvestyle,domain=0:1.5] {1/(1+x^2)}; + \addplot [firstcurvestyle,domain=1.5:10.5,samples=40] {1/(1+x^2)}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + + + +

    + When we defined the definite integral + \ds\int_a^b f(x)\, dx, we made two stipulations: +

    + +

    +

      +
    1. +

      + The interval over which we integrated, + [a,b], was a finite interval, and +

      +
      2. + +
    2. +

      + The function f(x) was continuous on [a,b] + (ensuring that the range of f was finite). +

      +
      3. +
    +

    + +

    + In this section we consider integrals where one or both of the above conditions do not hold. + Such integrals are called improper integrals. +

    +     + + + Improper Integrals with Infinite Bounds + + + Improper Integrals with Infinite Bounds; Converge, Diverge + +

    +

      +
    1. +

      + Let f be a continuous function on [a,\infty). + Define + + integrationimproper + improper integration + convergenceof improper int. + divergenceof improper int. + + + \int_a^\infty f(x)\, dx \text{ to be } \lim_{b\to\infty}\int_a^b f(x)\, dx + . +

      +
      2. + +
    2. +

      + Let f be a continuous function on (-\infty,b]. + Define + + \int_{-\infty}^b f(x)\, dx \text{ to be } \lim_{a\to-\infty}\int_a^b f(x)\, dx + . +

      +
      3. + +
    3. +

      + Let f be a continuous function on (-\infty,\infty). + Let c be any real number; define + + \int_{-\infty}^\infty f(x)\, dx \text{ to be } \lim_{a\to-\infty}\int_a^c f(x)\, dx\,+\,\lim_{b\to\infty}\int_c^b f(x)\, dx + . +

      +
      4. +
    +

    + +

    + An improper integral is said to converge + if its corresponding limit exists; + otherwise, it diverges. + The improper integral in part 3 converges if and only if both of its limits exist. +

    +     +     + + + Evaluating improper integrals + +

    + Evaluate the following improper integrals. +

    + +

    +

      +
    1. \int_1^\infty \frac1{x^2}\, dx
      2. + +
    2. \int_1^\infty \frac1x\, dx
      3. + +
    3. \int_{-\infty}^0 e^x\, dx
      4. + +
    4. \int_{-\infty}^\infty \frac1{1+x^2}\, dx
      5. +
    +

    +     + +

    +

      +
    1. +

      + + \int_1^\infty \frac{1}{x^2}\, dx = \lim_{b\to\infty} \int_1^b\frac1{x^2}\, dx \amp = \lim_{b\to\infty} \frac{-1}{x}\Big|_1^b + \amp = \lim_{b\to\infty} \frac{-1}{b} + 1 + \amp = 1 + . + + A graph of the area defined by this integral is given in . +

      + +
      + A graph of f(x) = \frac{1}{x^2} in + + + + Graph of the area under the curve 1/x^2 from 1 to infinity. + + +

      + The y axis is drawn from 0 to 1 and + the x axis is drawn from 0 to 10. The + function f(x) = \frac{1}{x^2} starts from point + (1,1), the curve drops sharply then bends before + x=5 after which it runs very close to the x + axis. The area under the curve is shaded from x=1 to + x=10. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,5,10}, + ymin=-.1,ymax=1.1, + xmin=-1,xmax=11 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:10.5] {1/(x^2)} \closedcycle; + \addplot [firstcurvestyle,domain=1:10.5,samples=50] {1/(x^2)}; + + \draw (axis cs:5,.75) node { $\ds f(x)=\frac{1}{x^2}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      2. + +
    2. +

      + + \int_1^\infty \frac1x\, dx \amp = \lim_{b\to\infty}\int_1^b\frac1x\, dx + \amp = \lim_{b\to\infty} \ln\abs{x}\Big|_1^b + \amp = \lim_{b\to\infty} \ln(b) + \amp = \infty + . + + The limit does not exist, + hence the improper integral \ds\int_1^\infty\frac1x\, dx diverges. + Compare the graphs in Figures + and ; + notice how the graph of f(x) = 1/x is noticeably larger. + This difference is enough to cause the improper integral to diverge. +

      + +
      + A graph of f(x) = \frac{1}{x} in + + + + Graph of area of the curve 1/x from 1 to 10. + + +

      + The y axis is drawn from 0 to 1 and the + x axis is drawn from 0 to 10. The function + f(x) = \frac{1}{x} starts from point (1,1), the + curve drops sharply then bends before x=5 after which it + runs almost parallel to the x axis about line y=0.2. + The area under the curve is shaded from x=1 to x=10. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,5,10}, + ymin=-.1,ymax=1.1, + xmin=-1,xmax=11 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:10.5] {1/x} \closedcycle; + \addplot [firstcurvestyle,domain=1:10.5,samples=40] {1/x}; + + \draw (axis cs:5,.75) node { $\ds f(x)=\frac{1}{x}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      3. + +
    3. +

      + + \int_{-\infty}^0 e^x \, dx \amp = \lim_{a\to-\infty} \int_a^0e^x\, dx + \amp = \lim_{a\to-\infty} e^x\Big|_a^0 + \amp = \lim_{a\to-\infty} e^0-e^a + \amp = 1 + . + + A graph of the area defined by this integral is given in . +

      + +
      + A graph of f(x) = e^x in + + + + Graph of area under the curve f(x)=e^x from -10 to 0. + + +

      + The graph of function f(x)= e^x is drawn in the + second quadrant.The y axis is drawn from 0 + to 1 and the x axis is drawn from -10 + to 0. From left to right the function appears to + coincide with the x axis, at about x=5 it gets + a positive slope and rises sharply until it reaches point (0,1). +

      +       + + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,-5,-10}, + ytick={1}, + ymin=-.1,ymax=1.1, + xmin=-11,xmax=1 + ] + + \addplot [firstcurvestyle,areastyle,domain=-10.5:0] {e^x} \closedcycle; + \addplot [firstcurvestyle,domain=-10.5:0,samples=50] {e^x}; + + \draw (axis cs:-5,.75) node { $\ds f(x)=e^x$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      4. + +
    4. +

      + We will need to break this into two improper integrals and choose a value of c as in part 3 of . + Any value of c is fine; + we choose c=0. + + \int_{-\infty}^\infty \frac1{1+x^2}\, dx \amp = \lim_{a\to-\infty} \int_a^0\frac{1}{1+x^2}\, dx + \lim_{b\to\infty} \int_0^b\frac{1}{1+x^2}\, dx + \amp = \lim_{a\to-\infty} \tan^{-1}(x) \Big|_a^0 + \lim_{b\to\infty} \tan^{-1}(x) \Big|_0^b + \amp = \lim_{a\to-\infty} \left(\tan^{-1}(0) -\tan^{-1}(a) \right) + \amp \quad\quad + \lim_{b\to\infty} \left(\tan^{-1}(b) -\tan^{-1}(0) \right) + \amp = \left(0-\frac{-\pi}2\right) + \left(\frac{\pi}2-0\right). + Each limit exists, hence the original integral converges and has value: + \amp = \pi + . + A graph of the area defined by this integral is given in . +

      + +
      + A graph of f(x) = \frac{1}{1+x^2} in + + + + Graph of function 1/(1+x^2). + + +

      + The y axis is drawn from 0 to 1 and the x + axis is drawn from -10 to 10.The function f(x)= \frac{1}{1+x^2} + is symmetrical about the y axis. From left to right the function + moves very closely to the x axis after x=-5 it bends and sharply + increases to point (0,1), from this point it decreases sharply and then + gets a dip then after x=5 runs very closely to the x axis. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ytick={1}, + ymin=-.1,ymax=1.1, + xmin=-11,xmax=11 + ] + + \addplot [firstcurvestyle,areastyle,domain=-10.5:-.1] {1/(1+x^2)} \closedcycle; + \addplot [firstcurvestyle,areastyle,domain=-.1:.1] {1/(1+x^2)} \closedcycle; + \addplot [firstcurvestyle,areastyle,domain=.1:10.5] {1/(1+x^2)} \closedcycle; + + \addplot [firstcurvestyle,domain=-10.5:-1,samples=40] {1/(1+x^2)}; + \addplot [firstcurvestyle,domain=-1:1,samples=30] {1/(1+x^2)}; + \addplot [firstcurvestyle,domain=1:10.5,samples=40] {1/(1+x^2)}; + + \draw (axis cs:6,.75) node { $\ds f(x)=\frac{1}{1+x^2}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      5. +
    +

    +     + +     + +

    + introduced L'Hospital's Rule, + a method of evaluating limits that return indeterminate forms. + It is not uncommon for the limits resulting from improper integrals to need this rule as demonstrated next. +

    + + + Improper integration and L'Hospital's Rule + +

    + Evaluate the improper integral \ds \int_1^\infty \frac{\ln(x) }{x^2}\, dx. +

    +     + +

    + This integral will require the use of Integration by Parts. + Let u = \ln(x) and dv = 1/x^2\, dx. + Then +

    + +
    + A graph of f(x) = \frac{\ln(x) }{x^2} in + + + + Graph of lx(x)/x^2. + + +

    + The y axis is drawn from 0 to 0.4 and the x + axis is drawn from 0 to 10. From left to right, the function + f(x) = \frac{\ln(x) }{x^2} starts at point (1,0) it rises sharply, + reaches a height of 0.2 near x= 1.5 then it decreases gently until + x=10 but stays a bit above the x axis. The area under the curve + until x=10 is shaded. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,5,10}, + ymin=-.1,ymax=.5, + xmin=-1,xmax=11 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:2] {ln(x)/x^2} \closedcycle; + \addplot [firstcurvestyle,areastyle,domain=2:10.5] {ln(x)/x^2} \closedcycle; + + \addplot [firstcurvestyle,domain=1:2] {ln(x)/x^2}; + \addplot [firstcurvestyle,domain=2:10.5] {ln(x)/x^2}; + + \draw (axis cs:5,.3) node { $\ds f(x)=\frac{\ln(x) }{x^2}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +

    + + \int_1^\infty\frac{\ln(x) }{x^2}\, dx \amp = \lim_{b\to\infty}\int_1^b\frac{\ln(x) }{x^2}\, dx + \amp = \lim_{b\to\infty}\left(-\frac{\ln(x) }{x}\Big|_1^b +\int_1^b \frac{1}{x^2} \, dx \right) + \amp = \lim_{b\to\infty} \left.\left(-\frac{\ln(x) }{x} -\frac1x\right)\right|_1^b + \amp = \lim_{b\to\infty} \left(-\frac{\ln(b) }{b}-\frac1b - \left(-\ln(1) -1\right)\right). + The 1/b and \ln(1) terms go to 0, leaving \lim\limits_{b\to\infty} -\frac{\ln(b) }b + 1. We need to evaluate \lim\limits_{b\to\infty} \frac{\ln(b) }{b} with l'Hospital's Rule. We have: + \lim_{b\to\infty}\frac{\ln(b) }b \amp \stackrel{\,\text{ by LHR } \,}{=} \lim_{b\to\infty} \frac{1/b}{1} + \amp = 0. + Thus the improper integral evaluates as: + \int_1^\infty\frac{\ln(x) }{x^2}\, dx \amp = 1 + . +

    +     + +     +     + + + Improper Integrals with Infinite Range + +

    + We have just considered definite integrals where the interval of integration was infinite. + We now consider another type of improper integration, + where the range of the integrand is infinite. +

    + + + Improper Integration with Infinite Range + +

    + Let f(x) be a continuous function on [a,b] except at c, + a\leq c\leq b, + where x=c is a vertical asymptote of f. + Define + + integrationimproper + improper integration + + + \int_a^b f(x)\, dx = \lim_{t\to c^-}\int_a^t f(x)\, dx + \lim_{t\to c^+}\int_t^b f(x)\, dx + . +

    +     +     + + + Improper integration of functions with infinite range + +

    + Evaluate the following improper integrals: +

    + +

    +

      +
    1. \int_0^1\frac1{\sqrt{x}}\, dx
      2. +
    2. \int_{-1}^1\frac{1}{x^2}\, dx
      3. +
    +

    +     + +

    +

      +
    1. +

      + A graph of f(x) = 1/\sqrt{x} is given in . + Notice that f has a vertical asymptote at x=0; + in some sense, + we are trying to compute the area of a region that has no top. + Could this have a finite value? + + \int_0^1 \frac{1}{\sqrt{x}}\, dx \amp = \lim_{a\to0^+}\int_a^1 \frac1{\sqrt{x}}\, dx + \amp = \lim_{a\to0^+} 2\sqrt{x}\Big|_a^1 + \amp = \lim_{a\to0^+} 2\left(\sqrt{1}-\sqrt{a}\right) + \amp = 2 + . + It turns out that the region does have a finite area even though it has no upper bound + (strange things can occur in mathematics when considering the infinite). +

      + + + + +
      + A graph of f(x)=\frac{1}{\sqrt{x}} in + + + + Graph of function 1 over square root of x. + + +

      + The y axis is drawn from 0 to 10 and the x + axis is drawn from 0 to 2. The function f(x)=\frac{1}{\sqrt{x}} + starts at (0,10) then it decreases sharply while being very close to the + y axis at about y=4 it starts diverging away, it moves away from + the y axis and appears to move almost parallel to the x axis until x=1. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=11, + xmin=-.1,xmax=2.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0.01:.2] {1/sqrt(x)} \closedcycle; + \addplot [firstcurvestyle,areastyle,domain=.2:1] {1/sqrt(x)} \closedcycle; + + \addplot [firstcurvestyle,domain=0.01:.2] {1/sqrt(x)}; + \addplot [firstcurvestyle,domain=.2:1] {1/sqrt(x)}; + + \draw (axis cs:.5,5) node { $\ds f(x)=\frac{1}{\sqrt{x}}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      2. + +
    2. +

      + The function f(x) = 1/x^2 has a vertical asymptote at x=0, + as shown in , + so this integral is an improper integral. + Let's eschew using limits for a moment and proceed without recognizing the improper nature of the integral. + This leads to: + + \int_{-1}^1\frac1{x^2}\, dx \amp = -\frac1x\Big|_{-1}^1 + \amp = -1 - (1) + \amp =-2. \,(!) + +

      + +
      + A graph of f(x)=\frac{1}{x^2} in + + + + Graph of function f(x) = 1/x^2. + + +

      + The y axis is drawn from -1 to 1 and the x axis is drawn + from 0 to 20. The graph of function f(x)=\frac{1}{x^2} is drawn + in the first and the second quadrants. From left to right, in the second quadrant the + function starts at x=-1 a little above the x axis then it curves up sharply + and runs parallel to the y axis, at x=-0.4. In the first quadrant, from + left to right the function runs parallel to the y axis and declines sharply, it + then bends and moves along the x axis until x=1. Between x=-1 and + x=1 and from y=0 to y=20 the area within the curves is shaded. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=21, + xmin=-1.1,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0.1:1,samples=30] {1/x^2} \closedcycle; + \addplot [firstcurvestyle,domain=0.1:1,samples=50] {1/x^2}; + \addplot [firstcurvestyle,areastyle,domain=-0.1:0.1] {22} \closedcycle; + \addplot [firstcurvestyle,areastyle,domain=-1:-0.1,samples=30] {1/x^2} \closedcycle; + \addplot [firstcurvestyle,domain=-1:-0.1,samples=50] {1/x^2}; + + \draw (axis cs:.7,7) node { $\ds f(x)=\frac{1}{x^2}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      + +

      + Clearly the area in question is above the x-axis, + yet the area is supposedly negative! + Why does our answer not match our intuition? + To answer this, + evaluate the integral using . + + \int_{-1}^1\frac1{x^2}\, dx \amp = \lim_{t\to0^-}\int_{-1}^t \frac1{x^2}\, dx + \lim_{t\to0^+}\int_t^1\frac1{x^2}\, dx + \amp = \lim_{t\to0^-}-\frac1x\Big|_{-1}^t + \lim_{t\to0^+}-\frac1x\Big|_t^1 + \amp = \lim_{t\to0^-}-\frac1t-1 + \lim_{t\to0^+} -1+\frac1t + \amp \Rightarrow \Big(\infty-1\Big)\,+ \,\Big(- 1+\infty\Big) + . + Neither limit converges hence the original improper integral diverges. + The nonsensical answer we obtained by ignoring the improper nature of the integral is just that: + nonsensical. +

      +
      3. +
    +

    +     + +     +     + + + Understanding Convergence and Divergence +

    + Oftentimes we are interested in knowing simply whether or not an improper integral converges, + and not necessarily the value of a convergent integral. + We provide here several tools that help determine the convergence or divergence of improper integrals without integrating. +

    + + + +

    + Our first tool is to understand the behavior of functions of the form \ds \frac1{x^p}. +

    + + + Improper integration of <m>1/x^p</m> + +

    + Determine the values of p for which \ds \int_1^\infty \frac1{x^p}\, dx converges. +

    +     + +

    + We begin by integrating and then evaluating the limit. + + \int_1^\infty \frac1{x^p}\, dx \amp = \lim_{b\to\infty}\int_1^b\frac1{x^p}\, dx + \amp = \lim_{b\to\infty}\int_1^b x^{-p}\, dx \qquad \text{ (assume \(p\neq 1\)) } + \amp = \lim_{b\to\infty} \frac{1}{-p+1}x^{-p+1}\Big|_1^b + \amp = \lim_{b\to\infty} \frac{1}{1-p}\big(b^{1-p}-1^{1-p}\big) + . +

    + +

    + When does this limit converge , when is this limit + not \infty? + This limit converges precisely when the power of b is less than 0: + when 1-p\lt 0 \Rightarrow 1\lt p. +

    + +
    + Plotting functions of the form 1/x^p in + + + + Graph of two functions 1/x^p and 1/x^q with p <1 <q. + + +

    + The y and the x axes are uncalibrated except that 1 + is marked in the middle of the x axis. There are two functions + f(x) = 1/x^p and 1/x^q shown in the graph. The two functions + intersect when x=1. Before x=1 both functions have negative + slopes and decline sharply along the y axis then gently as they approach + x=1, p<q hence 1/x^p graph is below 1/x^q. After + x=1, 1/x^p is above 1/x^q but they are only slightly apart. + There is a dashed line between the two functions. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1}, + ytick=\empty, + ymin=-.1,ymax=6, + xmin=-.1,xmax=2.1 + ] + + \addplot+ [smooth] coordinates {(0.1,31.62)(0.15,17.21)(0.2,11.18)(0.25,8.)(0.3,6.086)(0.35,4.829)(0.4,3.953)(0.45,3.313)(0.5,2.828)(0.55,2.452)(0.6,2.152)(0.65,1.908)(0.7,1.707)(0.75,1.54)(0.8,1.398)(0.85,1.276)(0.9,1.171)(0.95,1.08)(1.,1.)(1.05,0.9294)(1.1,0.8668)(1.15,0.8109)(1.2,0.7607)(1.25,0.7155)(1.3,0.6747)(1.35,0.6375)(1.4,0.6037)(1.45,0.5727)(1.5,0.5443)(1.55,0.5182)(1.6,0.4941)(1.65,0.4718)(1.7,0.4512)(1.75,0.432)(1.8,0.4141)(1.85,0.3974)(1.9,0.3818)(1.95,0.3672)(2.,0.3536)}; + \addplot+ [solid,smooth] coordinates {(.05,4.47)(0.1,3.162)(0.15,2.582)(0.2,2.236)(0.25,2.)(0.3,1.826)(0.35,1.69)(0.4,1.581)(0.45,1.491)(0.5,1.414)(0.55,1.348)(0.6,1.291)(0.65,1.24)(0.7,1.195)(0.75,1.155)(0.8,1.118)(0.85,1.085)(0.9,1.054)(0.95,1.026)(1.,1.)(1.05,0.9759)(1.1,0.9535)(1.15,0.9325)(1.2,0.9129)(1.25,0.8944)(1.3,0.8771)(1.35,0.8607)(1.4,0.8452)(1.45,0.8305)(1.5,0.8165)(1.55,0.8032)(1.6,0.7906)(1.65,0.7785)(1.7,0.767)(1.75,0.7559)(1.8,0.7454)(1.85,0.7352)(1.9,0.7255)(1.95,0.7161)(2.,0.7071)}; + \addplot [lineseg,dashed,thick,smooth] coordinates {(0.1,10.)(0.15,6.667)(0.2,5.)(0.25,4.)(0.3,3.333)(0.35,2.857)(0.4,2.5)(0.45,2.222)(0.5,2.)(0.55,1.818)(0.6,1.667)(0.65,1.538)(0.7,1.429)(0.75,1.333)(0.8,1.25)(0.85,1.176)(0.9,1.111)(0.95,1.053)(1.,1.)(1.05,0.9524)(1.1,0.9091)(1.15,0.8696)(1.2,0.8333)(1.25,0.8)(1.3,0.7692)(1.35,0.7407)(1.4,0.7143)(1.45,0.6897)(1.5,0.6667)(1.55,0.6452)(1.6,0.625)(1.65,0.6061)(1.7,0.5882)(1.75,0.5714)(1.8,0.5556)(1.85,0.5405)(1.9,0.5263)(1.95,0.5128)(2.,0.5)}; + + \draw (axis cs:.7,5) node { $\ds f(x)=\frac{1}{x^q}$}; + \draw (axis cs:.35,.8) node { $\ds f(x)=\frac{1}{x^p}$}; + \draw (axis cs:1.5,3) node { $p<1<q$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Our analysis shows that if p \gt 1, + then \ds\int_1^\infty \frac1{x^p}\, dx converges. + When p\lt 1 the improper integral diverges; + we showed in + that when p=1 the integral also diverges. +

    + +

    + + graphs y=1/x with a dashed line, + along with graphs of y=1/x^p, p\lt 1, + and y=1/x^q, q \gt 1. + Somehow the dashed line forms a dividing line between convergence and divergence. +

    +     + +     + +

    + The result of + provides an important tool in determining the convergence of other integrals. + A similar result is proved in the exercises about improper integrals of the form \ds \int_0^1\frac1{x^p}\, dx. + These results are summarized in the following Key Idea. +

    + + + Convergence of Improper Integrals involving <m>1/x^p</m> +

    +

      +
    1. +

      + The improper integral \ds \int_1^\infty\frac1{x^p}\, dx converges when p\gt 1 and diverges when p\leq 1. + + convergenceof improper int. + divergenceof improper int. +

      +
      2. + +
    2. +

      + The improper integral \ds \int_0^1\frac1{x^p}\, dx converges when p\lt 1 and diverges when p\geq 1. +

      +
      3. +
    +

    +     + +

    + A basic technique in determining convergence of improper integrals is to compare an integrand whose convergence is unknown to an integrand whose convergence is known. + We often use integrands of the form 1/x^p to compare to as their convergence on certain intervals is known. + This is described in the following theorem. +

    + + + + + Direct Comparison Test for Improper Integrals + +

    + Let f and g be continuous on [a,\infty) where + 0\leq f(x)\leq g(x) for all x in [a,\infty). +

    + +

    +

      +
    1. +

      + If \ds \int_a^\infty g(x)\, dx converges, + then \ds \int_a^\infty f(x)\, dx converges. + integrationimproper + convergenceDirect Comparison Test!for integration + divergenceDirect Comparison Test!for integration + Direct Comparison Testfor integration + convergenceof improper int. + divergenceof improper int. +

      +
      2. + +
    2. +

      + If \ds \int_a^\infty f(x)\, dx diverges, + then \ds \int_a^\infty g(x)\, dx diverges. +

      +
      3. +
    +

    +     +     + + + Determining convergence of improper integrals + +

    + Determine the convergence of the following improper integrals. +

    + +

    +

      +
    1. \int_1^\infty e^{-x^2}\, dx
      2. +
    2. \int_3^\infty \frac{1}{\sqrt{x^2-x}}\, dx
      3. +
    +

    +     + +

    +

      +
    1. +

      + The function f(x) = e^{-x^2} does not have an antiderivative expressible in terms of elementary functions, + so we cannot integrate directly. + It is comparable to g(x)=1/x^2, + and as demonstrated in , + e^{-x^2} \lt 1/x^2 on [1,\infty). + We know from + that \ds \int_1^\infty \frac{1}{x^2}\, dx converges, + hence \ds\int_1^\infty e^{-x^2}\, dx also converges. +

      + +
      + Graphs of f(x) = e^{-x^2} and f(x)= 1/x^2 in + + + + Graph of two functions e to the power negative x^2 and 1/x^2. + + +

      + The y axis is drawn from 0 to 1 and the x axis + is drawn from 0 to 4. There are two functions drawn + f(x) = e^{-x^2} and f(x)= 1/x^2. The function e appears + to start from (1,0.4) then decreases and merges with the x axis + after x=2. The function 1/x^2 is above the function e^{-x^2}, + it starts from point (1,1) and curves down and appears to be parallel to the + x axis after x=3. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=4.1 + ] + + \addplot+ [domain=1:4,samples=40] {e^-x^2}; + \addplot+ [solid,domain=1:4,samples=40] {1/x^2}; + + \draw (axis cs:.7,.45) node { $\ds f(x)=e^{-x^2}$}; + \draw (axis cs:1.9,.8) node { $\ds f(x)=\frac{1}{x^2}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      2. + +
    2. +

      + Note that for large values of x, + \ds \frac{1}{\sqrt{x^2-x}} \approx \frac{1}{\sqrt{x^2}} =\frac{1}{x}. + We know from + and the subsequent note that \ds \int_3^\infty \frac1x\, dx diverges, + so we seek to compare the original integrand to 1/x. + + It is easy to see that when x \gt 0, + we have x = \sqrt{x^2} \gt \sqrt{x^2-x}. + Taking reciprocals reverses the inequality, giving + + \frac1x \lt \frac1{\sqrt{x^2-x}} + . + Using , + we conclude that since \ds\int_3^\infty\frac1x\, dx diverges, + \ds\int_3^\infty\frac1{\sqrt{x^2-x}}\, dx diverges as well. + illustrates this. +

      + +
      + Graphs of f(x) = 1/\sqrt{x^2-x} and f(x)= 1/x in + + + + Graph of the two functions used in this example. + + +

      + The y axis is drawn from 0 to 0.4 and the x axis is + drawn from 0 to 6. Two functions f(x)=1/x and + f(x) = 1/\sqrt{x^2-x} are drawn. At x=3, the function 1/x + starts at y=3.33 then curves down with a negative slope. The function + f(x) = 1/\sqrt{x^2-x} at x=3 starts at y=0.41 and also + curves down with a negative slope. The function 1/x is below function + f(x) = 1/\sqrt{x^2-x} they are comparatively more apart at x=3 + than at x=6. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=.5, + xmin=-.1,xmax=6.2 + ] + + \addplot+ [domain=3:6] {1/sqrt((x^2)-x)}; + \addplot+ [solid,domain=3:6] {1/x}; + + \draw (axis cs:4,.45) node { $\ds f(x)=\frac{1}{\sqrt{x^2-x}}$}; + \draw (axis cs:2.5,.2) node { $\ds f(x)=\frac{1}{x}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      3. +
    +

    +     + +     + +

    + Being able to compare unknown + integrals to known + integrals is very useful in determining convergence. + However, some of our examples were a little + too nice. For instance, + it was convenient that \ds \frac{1}x \lt \frac{1}{\sqrt{x^2-x}}, + but what if the -x + were replaced with a +2x+5? + That is, what can we say about the convergence of \ds \int_3^\infty\frac{1}{\sqrt{x^2+2x+5}}\, dx? + We have \ds \frac{1}{x} \gt \frac1{\sqrt{x^2+2x+5}}, + so we cannot use . +

    + +

    + In cases like this + (and many more) + it is useful to employ the following theorem. +

    + + + Limit Comparison Test for Improper Integrals + +

    + Let f and g be continuous functions on + [a,\infty) where f(x) \gt 0 and g(x) \gt 0 for all x. + If + + \lim_{x\to\infty} \frac{f(x)}{g(x)} = L,\qquad 0\lt L\lt \infty + , + then + + \int_a^\infty f(x)\, dx \text{ and } \int_a^\infty g(x)\, dx + + either both converge or both diverge. + integrationimproper + convergenceLimit Comparison Test!for integration + divergenceLimit Comparison Test!for integration + Limit Comparison Testfor integration + convergenceof improper int. + divergenceof improper int. +

    +     +     + + + Determining convergence of improper integrals + +

    + Determine the convergence of \ds \int_3^{\infty} \frac{1}{\sqrt{x^2+2x+5}}\, dx. +

    +     + +

    + As x gets large, + the denominator of the integrand will begin to behave much like y=x. + So we compare \ds\frac{1}{\sqrt{x^2+2x+5}}to \ds\frac1xwith the Limit Comparison Test: + + \lim_{x\to\infty} \frac{1/\sqrt{x^2+2x+5}}{1/x} = \lim_{x\to\infty}\frac{x}{\sqrt{x^2+2x+5}} + . +

    + +

    + The immediate evaluation of this limit returns + \infty/\infty, an indeterminate form. + Using L'Hospital's Rule seems appropriate, + but in this situation, it does not lead to useful results. + (We encourage the reader to employ L'Hospital's Rule at least once to verify this.) +

    + +

    + The trouble is the square root function. + To get rid of it, + we employ the following fact: If \lim\limits_{x\to c} f(x) = L, + then \lim\limits_{x\to c} f(x)^2 = L^2. + (This is true when either c or L is \infty.) + So we consider now the limit + + \lim_{x\to\infty} \frac{x^2}{x^2+2x+5} + . +

    + +

    + This converges to 1, meaning the original limit also converged to 1. + As x gets very large, the function + \ds\frac{1}{\sqrt{x^2+2x+5}}looks very much like \ds\frac1x. + Since we know that \ds\int_3^{\infty} \frac1x\, dxdiverges, + by the Limit Comparison Test we know that \ds\int_3^\infty\frac{1}{\sqrt{x^2+2x+5}}\, dxalso diverges. + + graphs f(x)=1/\sqrt{x^2+2x+5} and f(x)=1/x, + illustrating that as x gets large, + the functions become indistinguishable. +

    + +
    + Graphing f(x)=\frac{1}{\sqrt{x^2+2x+5}} and f(x)=\frac1x in + + + + Graph of the two functions used in this example. + + +

    + The y axis is drawn from -0.1 to 0.3 and the x axis + is drawn from 0 to 20. Two functions are drawn f(x)= 1/x + and f(x)=\frac{1}{\sqrt{x^2+2x+5}}. The function f(x)=\frac{1}{\sqrt{x^2+2x+5}} + is drawn below 1/x. Both functions have negative slopes, they are apart by a small + distance around x=4, after x=10 they come very close and almost coincide. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=.35, + xmin=-.1,xmax=21 + ] + + \addplot+ [domain=3:20,samples=40] {1/sqrt(x^2+2*x+5)}; + \addplot+ [solid,domain=3:20,samples=50] {1/x}; + + \draw (axis cs:4.5,.08) node { $\ds f(x)=\frac{1}{\sqrt{x^2+2x+5}}$}; + \draw (axis cs:6,.3) node { $\ds f(x)=\frac{1}{x}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +     + + + +

    + Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. + There are versions that apply to improper integrals with an infinite range, + but as they are a bit wordy and a little more difficult to employ, + they are omitted from this text. +

    + + + +

    + This chapter has explored many integration techniques. + We learned Substitution, + which undoes the Chain Rule of differentiation, + as well as Integration by Parts, + which undoes the Product Rule. + We learned specialized techniques for handling trigonometric functions and introduced the hyperbolic functions, + which are closely related to the trigonometric functions. + All techniques effectively have this goal in common: + rewrite the integrand in a new way so that the integration step is easier to see and implement. +

    + +

    + As stated before, integration is, in general, hard. + It is easy to write a function whose antiderivative is impossible to write in terms of elementary functions, + and even when a function does have an antiderivative expressible by elementary functions, + it may be really hard to discover what it is. + The powerful computer algebra system + Mathematica has approximately 1,000 pages of code dedicated to integration. +

    + +

    + Do not let this difficulty discourage you. + There is great value in learning integration techniques, + as they allow one to manipulate an integral in ways that can illuminate a concept for greater understanding. + There is also great value in understanding the need for good numerical techniques: + the Trapezoidal and Simpson's Rules are just the beginning of powerful techniques for approximating the value of integration. +

    + +

    + The next chapter stresses the uses of integration. + We generally do not find antiderivatives for antiderivative's sake, + but rather because they provide the solution to some type of problem. + The following chapter introduces us to a number of different problems whose solution is provided by integration. +

    +     + + + + Terms and Concepts + + + + +

    + The definite integral was defined with what two stipulations? +

    + + +     + + + +

    + The interval of integration is finite, + and the integrand is continuous on that interval. +

    + +     + +     + + + + +

    + If \lim\limits_{b\to \infty} \int_0^b f(x)\, dx exists, + then the integral \ds \int_0^\infty f(x)\, dx is said to . +

    +     + + + + + + + + +

    + converge +

    +     + +     + + + + +

    + If \ds \int_1^\infty f(x)\, dx=10, + and 0\leq g(x)\leq f(x) for all x, + then we know that \ds \int_1^\infty g(x)\, dx . +

    +     + + + + converges|<10|< 10 + + + + +

    + converges; could also state \lt 10. +

    +     + +     + + + +

    + For what values of p will \ds \int_1^\infty \frac1{x^p}\, dx converge? +

    +     + + + +

    + p\lt 1 +

    +     +     + + +

    + p\leq 1 +

    +     +     + + +

    + p\gt 1 +

    +     +     + + +

    + p\geq 1 +

    +     +     +     +     + + + +

    + For what values of p will \ds \int_{10}^\infty \frac1{x^p}\, dx converge? +

    +     + + + +

    + p\lt 1 +

    +     +     + + +

    + p\leq 1 +

    +     +     + + +

    + p\gt 1 +

    +     +     + + +

    + p\geq 1 +

    +     +     +     +     + + + +

    + For what values of p will \ds \int_{0}^1 \frac1{x^p}\, dx converge? +

    +     + + + +

    + p\lt 1 +

    +     +     + + +

    + p\leq 1 +

    +     +     + + +

    + p\gt 1 +

    +     +     + + +

    + p\geq 1 +

    +     +     +     +     + +     + + Problems + + + +

    + Evaluate the given improper integral. +

    +     + + + + + $f = Formula("e^(5-2x)"); + $l = Compute("e^5/2"); + + +

    + \ds \int_0^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("1/x^3"); + $l = Compute("1/2"); + + +

    + \ds \int_1^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("x^-4"); + $l = Compute("1/3"); + + +

    + \ds \int_1^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("1/(x^2+9)"); + $l = Compute("pi/3"); + + +

    + \ds \int_{-\infty}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("2^x"); + $l = Compute("1/ln(2)"); + + +

    + \ds \int_{-\infty}^0 \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("(1/2)^x"); + $l = Compute("inf"); + + +

    + \ds \int_{-\infty}^0 \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("x/(x^2+1)"); + $l = Compute("inf"); + + +

    + \ds \int_{-\infty}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("x/(x^2-4)"); + $l = Compute("inf"); + + +

    + \ds \int_{3}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("1/(x-1)^2"); + $l = Compute("1"); + + +

    + \ds \int_{2}^\infty\, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("1/(x-1)^2"); + $l = Compute("inf"); + + +

    + \ds \int_{1}^2\, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("1/(x-1)"); + $l = Compute("inf"); + + +

    + \ds \int_{2}^\infty\, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("1/(x-1)"); + $l = Compute("inf"); + + +

    + \ds \int_{1}^2\, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("1/x"); + $l = Compute("inf"); + + +

    + \ds \int_{-1}^1 \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("1/(x-2)"); + $l = Compute("inf"); + + +

    + \ds \int_{1}^3\, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("sec(x)^2"); + $l = Compute("inf"); + + +

    + \ds \int_{0}^\pi \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("1/sqrt(abs(x))"); + $l = Compute("2+2*sqrt(2)"); + + +

    + \ds \int_{-2}^1 \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("x*e^-x"); + $l = Compute("1"); + + +

    + \ds \int_{0}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("x*e^(-x^2)"); + $l = Compute("1/2"); + + +

    + \ds \int_{0}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("x*e^(-x^2)"); + $l = Compute("0"); + + +

    + \ds \int_{-\infty}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("1/(e^x+e^-x)"); + $l = Compute("pi/2"); + + +

    + \ds \int_{-\infty}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("x*ln(x)"); + $l = Compute("-1/4"); + + +

    + \ds \int_{0}^1 \, dx +

    + +

    + +

    +     +     +     + + + + + + $f = Formula("x^2*ln(x)"); + $l = Compute("-1/9"); + + +

    + \ds \int_{0}^1 \, dx +

    + +

    + +

    +     +     +     + + + + + + $f = Formula("ln(x)/x"); + $l = Compute("inf"); + + +

    + \ds \int_{1}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("ln(x)"); + $l = Compute("-1"); + + +

    + \ds \int_{0}^1 \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("ln(x)/x^2"); + $l = Compute("1"); + + +

    + \ds \int_{1}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("ln(x)/sqrt(x)"); + $l = Compute("inf"); + + +

    + \ds \int_{1}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("e^-x*sin(x)"); + $l = Compute("1/2"); + + +

    + \ds \int_{0}^\infty \, dx +

    + +

    + +

    +     +     +     + + + + + $f = Formula("e^-x*cos(x)"); + $l = Compute("1/2"); + + +

    + \ds \int_{0}^\infty \, dx +

    + +

    + +

    +     +     +     + +     + + + +

    + Use the Direct Comparison Test or the Limit Comparison Test to determine whether the given definite integral converges or diverges. + Clearly state what test is being used and what function the integrand is being compared to. +

    +     + + + + + parserPopUp.pl + + + $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); + $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); + $converges = DropDown(['converges','diverges'],0,showInStatic=>0); + $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); + $comp_f = Formula("1/x"); + + +

    + \ds \int_{10}^\infty \frac{3}{\sqrt{3x^2+2x-5}} \, dx +

    + + + Select the test to be used. + + +

    + +

    + + + Determine whether the integral converges or diverges. + + +

    + +

    + + + Enter the function used for comparison. + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); + $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); + $converges = DropDown(['converges','diverges'],0,showInStatic=>0); + $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); + $comp_f = Formula("1/x^(3/2)"); + + +

    + \ds \int_{2}^\infty \frac{4}{\sqrt{7x^3-x}} \, dx +

    + + + Select the test to be used. + + +

    + +

    + + + Determine whether the integral converges or diverges. + + +

    + +

    + + + Enter the function used for comparison. + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); + $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); + $converges = DropDown(['converges','diverges'],0,showInStatic=>0); + $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); + $comp_f = Formula("1/x"); + + +

    + \ds \int_{0}^\infty \frac{\sqrt{x+3}}{\sqrt{x^3-x^2+x+1}} \, dx +

    + + + Select the test to be used. + + +

    + +

    + + + Determine whether the integral converges or diverges. + + +

    + +

    + + + Enter the function used for comparison. + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); + $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); + $converges = DropDown(['converges','diverges'],0,showInStatic=>0); + $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); + $comp_f = Formula("x*e^-x"); + + +

    + \ds \int_{1}^\infty e^{-x}\ln(x) \, dx +

    + + + Select the test to be used. + + +

    + +

    + + + Determine whether the integral converges or diverges. + + +

    + +

    + + + Enter the function used for comparison. + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); + $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); + $converges = DropDown(['converges','diverges'],0,showInStatic=>0); + $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); + $comp_f = Formula("e^-x"); + + +

    + \ds \int_{5}^\infty e^{-x^2+3x+1} \, dx +

    + + + Select the test to be used. + + +

    + +

    + + + Determine whether the integral converges or diverges. + + +

    + +

    + + + Enter the function used for comparison. + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); + $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); + $converges = DropDown(['converges','diverges'],0,showInStatic=>0); + $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); + $comp_f = Formula("x*e^-x"); + + +

    + \ds \int_{0}^\infty \frac{\sqrt{x}}{e^x} \, dx +

    + + + Select the test to be used. + + +

    + +

    + + + Determine whether the integral converges or diverges. + + +

    + +

    + + + Enter the function used for comparison. + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); + $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); + $converges = DropDown(['converges','diverges'],0,showInStatic=>0); + $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); + $comp_f = Formula("1/(x^2-1)"); + + +

    + \ds \int_{2}^\infty \frac{1}{x^2+\sin(x) } \, dx +

    + + + Select the test to be used. + + +

    + +

    + + + Determine whether the integral converges or diverges. + + +

    + +

    + + + Enter the function used for comparison. + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); + $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); + $converges = DropDown(['converges','diverges'],0,showInStatic=>0); + $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); + $comp_f = Formula("x/(x^2+1)"); + + + +

    + \ds \int_{0}^\infty \frac{x}{x^2+\cos(x) } \, dx +

    + + + Select the test to be used. + + +

    + +

    + + + Determine whether the integral converges or diverges. + + +

    + +

    + + + Enter the function used for comparison. + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); + $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); + $converges = DropDown(['converges','diverges'],0,showInStatic=>0); + $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); + $comp_f = Formula("1/e^x"); + + +

    + \ds \int_{0}^\infty \frac{1}{x+e^x} \, dx +

    + + + Select the test to be used. + + +

    + +

    + + + Determine whether the integral converges or diverges. + + +

    + +

    + + + Enter the function used for comparison. + + +

    + +

    +     +     +     + + + + + parserPopUp.pl + + + $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); + $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); + $converges = DropDown(['converges','diverges'],0,showInStatic=>0); + $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); + $comp_f = Formula("1/e^x"); + + +

    + \ds \int_{0}^\infty \frac{1}{e^x-x} \, dx +

    + + + Select the test to be used. + + +

    + +

    + + + Determine whether the integral converges or diverges. + + +

    + +

    + + + Enter the function used for comparison. + + +

    + +

    +     +     +     + +     +     +     +
    +     + + + Applications of Integration + +

    + We begin this chapter with a reminder of a few key concepts from . + Let f be a continuous function on [a,b] which is partitioned into n equally spaced subintervals as + + a=x_0 \lt x_1 \lt \cdots \lt x_n\lt x_{n}=b + . +

    + +

    + Let \dx=(b-a)/n denote the length of the subintervals, + and let c_i be any x-value in the ith subinterval. + states that the sum + + \sum_{i=1}^n f(c_i)\dx + + is a Riemann Sum. Riemann Sums are often used to approximate some quantity + (area, volume, work, pressure, etc.). + The approximation becomes + exact by taking the limit + + \lim_{n\to\infty} \sum_{i=1}^n f(c_i)\dx + . +

    + +

    + + connects limits of Riemann Sums to definite integrals: + + \lim_{n\to\infty} \sum_{i=1}^n f(c_i)\dx = \int_a^b f(x)\, dx + . +

    + +

    + Finally, the Fundamental Theorem of Calculus states how definite integrals can be evaluated using antiderivatives. +

    + +

    + This chapter employs the following technique to a variety of applications. + Suppose the value Q of a quantity is to be calculated. + We first approximate the value of Q using a Riemann Sum, + then find the exact value via a definite integral. + We spell out this technique in the following Key Idea. +

    + + + Application of Definite Integrals Strategy +

    + Let a quantity be given whose value Q is to be computed. + integrationgeneral application technique +

      +
    1. +

      + Divide the quantity into n smaller + subquantities of value Q_i. +

      +
      2. + +
    2. +

      + Identify a variable x and function f(x) such that each subquantity can be approximated with the product f(c_i)\dx, + where \dx represents a small change in x. + Thus Q_i \approx f(c_i)\dx. + A sample approximation f(c_i)\dx of Q_i is called a + differential element. +

      +
      3. + +
    3. +

      + Recognize that \ds Q= \sum_{i=1}^n Q_i \approx \sum_{i=1}^n f(c_i)\dx, + which is a Riemann Sum. +

      +
      4. + +
    4. +

      + Taking the appropriate limit gives \ds Q = \int_a^b f(x)\, dx +

      +
      5. +
    +

    +     + +

    + This Key Idea will make more sense after we have had a chance to use it several times. + We begin with Area Between Curves, + which we addressed briefly in . +

    +     + +
    + Area Between Curves +

    + We are often interested in knowing the area of a region. + Forget momentarily that we addressed this already in + and approach it instead using the technique described in . +

    + + + +

    + Let Q be the area of a region bounded by continuous functions f and g. + If we break the region into many subregions, + we have an obvious equation: +

    + +

    + Total Area = sum of the areas of the subregions. +

    + +

    + The issue to address next is how to systematically break a region into subregions. + A graph will help. + Consider + where a region between two curves is shaded. + While there are many ways to break this into subregions, + one particularly efficient way is to + slice it vertically, + as shown in , + into n equally spaced slices. +

    + +

    + We now approximate the area of a slice. + Again, we have many options, but using a rectangle seems simplest. + Picking any x-value c_i in the ith slice, + we set the height of the rectangle to be f(c_i)-g(c_i), + the difference of the corresponding y-values. + The width of the rectangle is a small difference in x-values, + which we represent with \dx. + + shows sample points c_i chosen in each subinterval and appropriate rectangles drawn. + (Each of these rectangles represents a differential element.) + Each slice has an area approximately equal to \big(f(c_i)-g(c_i)\big)\dx; + hence, the total area is approximately the Riemann Sum + + Q = \sum_{i=1}^n \big(f(c_i)-g(c_i)\big)\dx + . +

    + +

    + Taking the limit as n\to \infty gives the exact area as \int_a^b \big(f(x)-g(x)\big)\, dx. +

    + +
    + Subdividing a region into vertical slices and approximating the areas with rectangles + +
    + + + + +

    + Graph of two functions f(x) and g(x) lying on the xy plane. + The x-axis contains two marked points, a and b, which are both plotted on the positive x-axis. + The functions f(x) and g(x) cross at the y-axis before diverging. + The function f(x) lies above the function g(x) and resembles a sine wave. + The function g(x) is a quadratic function which lies below f(x) on the interval (a,b). + The area lying between the marked points a and b and below f(x) and above g(x) is shaded, which gives the exact area of this region. +

    +     + Graph of two curves f and g with a shaded region between two points on the x-axis and the two curves. + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={.5,3}, + extra x tick labels={$a$,$b$}, + ytick=\empty, + ymin=-.2,ymax=2, + xmin=-.5,xmax=3.5 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.25:3.25,samples=40] {.25*sin(deg(2*x))+1.25} node [shift={(5pt,7pt)} ,black] { $f(x)$}; + \addplot [name path=B,firstcurvestyle,domain=-.25:3.25] {.25*(x-2)^2+.2}node [shift={(5pt,7pt)} ,black] { $g(x)$}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0.5:3}]; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + +

    + Graph of the same two functions f(x) and g(x) as in the previous image. + Now, the area lying between the marked points a and b and below f(x) and above g(x) is subdivided into 10 exact vertical slices, whose areas added together give the exact area of this region. +

    +     + Graph of the shaded region between the points a and b and the two curves is subdivided into 10 vertical slices. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={.5,3}, + extra x tick labels={$a$,$b$}, + ytick=\empty, + ymin=-.2,ymax=2, + xmin=-.5,xmax=3.5 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.25:3.25,samples=40] {.25*sin(deg(2*x))+1.25} node [shift={(5pt,7pt)} ,black] { $f(x)$}; + \addplot [name path=B,firstcurvestyle,domain=-.25:3.25] {.25*(x-2)^2+.2}node [shift={(5pt,7pt)} ,black] { $g(x)$}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0.5:3}]; + + \draw [thick] (axis cs:0.5,0.7625)--(axis cs:0.5,1.46) (axis cs:1.,0.45)--(axis cs:1.,1.477) (axis cs:1.5,0.2625)--(axis cs:1.5,1.285) (axis cs:2.,0.2)--(axis cs:2.,1.061) (axis cs:2.5,0.2625)--(axis cs:2.5,1.01) (axis cs:3.,1.18)--(axis cs:3.,0.45) (axis cs:0.75,1.499)--(axis cs:0.75,0.5906) (axis cs:1.25,0.3406)--(axis cs:1.25,1.4) (axis cs:1.75,1.162)--(axis cs:1.75,0.2156) (axis cs:2.25,0.2156)--(axis cs:2.25,1.006) (axis cs:2.75,1.074)--(axis cs:2.75,0.3406); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + +

    + Graph of the same two functions f(x) and g(x) as in the previous two images. + Now, the area lying between the marked points a and b and below f(x) and above g(x) is subdivided into 10 approximate vertical slices. + These slices showcase the slight inaccuracy of approximation using Riemann sums for small n , while the first image shows us that taking the limit as n\to \infty gives the exact area of this region between the two functions. +

    +     + Graph of the shaded region between the two curves is approximated using 10 rectangles. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={.5,3}, + extra x tick labels={$a$,$b$}, + ytick=\empty, + ymin=-.2,ymax=2, + xmin=-.5,xmax=3.5 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.25:3.25,samples=40] {.25*sin(deg(2*x))+1.25} node [shift={(5pt,7pt)} ,black] { $f(x)$}; + \addplot [name path=B,firstcurvestyle,domain=-.25:3.25] {.25*(x-2)^2+.2}node [shift={(5pt,7pt)} ,black] { $g(x)$}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0.5:3}]; + + \draw [thick] (axis cs:0.5,0.6225) rectangle (axis cs:0.75,1.5) (axis cs:1.,1.487) rectangle (axis cs:0.75,0.4756) (axis cs:1.,0.4025) rectangle (axis cs:1.25,1.452) (axis cs:1.5,1.334) rectangle (axis cs:1.25,0.29) (axis cs:1.5,0.24) rectangle (axis cs:1.75,1.235) (axis cs:2.,1.097) rectangle (axis cs:1.75,0.2025) (axis cs:2.,0.21) rectangle (axis cs:2.25,1.012) (axis cs:2.5,1.002) rectangle (axis cs:2.25,0.2225) (axis cs:2.5,0.29) rectangle (axis cs:2.75,1.029)(axis cs:3.,1.134) rectangle (axis cs:2.75,0.4025); + + \filldraw (axis cs:.7,.6225) circle (1.3pt) + (axis cs:0.95,0.4756) circle (1.3pt) + (axis cs:1.1,0.4025) circle (1.3pt) + (axis cs:1.4,0.29) circle (1.3pt) + (axis cs:1.6,0.24) circle (1.3pt) + (axis cs:1.9,0.2025) circle (1.3pt) + (axis cs:2.2,0.21) circle (1.3pt) + (axis cs:2.3,0.2225) circle (1.3pt) + (axis cs:2.6,0.29) circle (1.3pt) + (axis cs:2.9,0.4025) circle (1.3pt) + (axis cs:0.7,1.496) circle (1.3pt) + (axis cs:0.95,1.487) circle (1.3pt) + (axis cs:1.1,1.452) circle (1.3pt) + (axis cs:1.4,1.334) circle (1.3pt) + (axis cs:1.6,1.235) circle (1.3pt) + (axis cs:1.9,1.097) circle (1.3pt) + (axis cs:2.2,1.012) circle (1.3pt) + (axis cs:2.3,1.002) circle (1.3pt) + (axis cs:2.6,1.029) circle (1.3pt) + (axis cs:2.9,1.134) circle (1.3pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + + + Area Between Curves (restatement of <xref ref="thm_areabtwncurves"/>) + +

    + Let f(x) and g(x) be continuous functions defined on [a,b] where + f(x)\geq g(x) for all x in [a,b]. + The area of the region bounded by the curves y=f(x), + y=g(x) and the lines x=a and x=b is + integrationarea between curves + + \int_a^b \big(f(x)-g(x)\big)\, dx + . +

    +     +     + + + + + Finding area enclosed by curves + +

    + Find the area of the region bounded by f(x) = \sin(x) +2, + g(x) = \frac12\cos(2x)-1, + x=0 and x=4\pi, + as shown in . +

    + +
    + Graphing an enclosed region in + + + +

    + Graph showing the area of the region bounded by f(x) = \sin(x) +2, g(x) = \frac12\cos(2x)-1, x=0 and x=4\pi. + The function f(x) is drawn starting from the y-axis and ends at the point x=4 \pi . + The function g(x) is also drawn starting from the y-axis and ending at the point x=4 \pi . + For the duration of the region between x=0 and x=4 \pi , the curve f(x) = \sin(x) +2 lies above the curve g(x) = \frac12\cos(2x)-1. +

    +     + Graph of the shaded region between two points on the x-axis and the functions f and g. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick = {2,4,6,8,10}, + extra x ticks={12.57}, + extra x tick labels={$4\pi$}, + ymin=-2,ymax=3.5, + xmin=-.5,xmax=13.5 + ] + + \addplot [name path=A,firstcurvestyle,domain=0:12.57,samples=70] {sin(deg(x))+2} node [shift={(-15pt,7pt)},black] { $f(x)$}; + \addplot [name path=B,firstcurvestyle,domain=0:12.57,samples=90] {.5*cos(deg(2*x))-1.2} node [shift={(-20pt,-16pt)},black] { $g(x)$}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B]; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + The graph verifies that the upper boundary of the region is given by f and the lower bound is given by g. + Therefore the area of the region is the value of the integral + + \int_0^{4\pi} \big(f(x)- g(x)\big)\, dx \amp = \int_0^{4\pi} \Big(\sin(x) +2 - \big(\frac12\cos(2x)-1\big)\Big)\, dx + \amp = -\cos(x) -\frac14\sin(2x)+3x\Big|_0^{4\pi} + \amp = 12\pi \approx 37.7\,\text{units}^2 + . +

    +     + +     + + + Finding total area enclosed by curves + +

    + Find the total area of the region enclosed by the functions f(x) = -2x+5 and + g(x) = x^3-7x^2+12x-3 as shown in . +

    + +
    + Graphing a region enclosed by two functions in + + + +

    + Graph showing the area of the region bounded by f(x) = -2x+5, g(x) = x^3 -7x^2 +12x -3. + The line given by f(x) is drawn starting approximately at x=1 and ending at the point x=5 . + The cubic function given by g(x) is drawn starting from the y-axis as it comes up, before intersecting and going above the line f(x) at the point (1,3) . + The function g(x) then meets f(x) at the point (2,1) and continues below f(x) until once again intercepting at the point (4,-3) . + After this point, the function g(x) continues above the x-axis while the line f(x) slopes downwards never to meet again. +

    +

    + The area encosed by the curves f(x) and g(x) contains two parts. + The first part begins when g(x) rises above f(x) at x=1 and ends at x=2 , where g(x) falls below f(x) . + The second part begins at the point x=2 where g(x) is below f(x) and ends at x=4 , where g(x) rises above f(x) . + The first region is bounded below by f(x) , above by g(x) , x=1 and x=2 . + The second region is bounded above by f(x) , below by g(x) , x=2 and x=4 . + The two regions are shaded to showcase the total area enclosed by the two functions. +

    +     + Graph of the shaded region enclosed by the functions f and g. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ymin=-5,ymax=3.5, + xmin=-.5,xmax=4.7 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.2:4.4,samples=60] {x^3-7*x^2+12*x-3}; + \addplot [name path=B,firstcurvestyle,domain=0:4.4] {5-2*x}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=1:4}]; + + \end{axis} + + \end{tikzpicture} + + + + + +
    +     + +

    + A quick calculation shows that f=g at x=1, 2 and 4. + One can proceed thoughtlessly by computing \ds \int_1^4\big(f(x)-g(x)\big)\, dx, + but this ignores the fact that on [1,2], g(x) \gt f(x). + (In fact, the thoughtless integration returns -9/4, + hardly the expected value of an area.) + Thus we compute the total area by breaking the interval [1,4] into two subintervals, + [1,2] and [2,4] and using the proper integrand in each. + + \text{Total Area} \amp = \int_1^2 \big(g(x)-f(x)\big)\, dx + \int_2^4\big(f(x)-g(x)\big)\, dx + \amp = \int_1^2 \big(x^3-7x^2+14x-8\big) \, dx + \int_2^4\big(-x^3+7x^2-14x+8\big)\, dx + \amp = 5/12 + 8/3 + \amp = 37/12 = 3.083\,\text{units}^2 + . +

    +     + +     + +

    + The previous example makes note that we are expecting area to be positive. + When first learning about the definite integral, + we interpreted it as signed area under the curve, + allowing for negative area. That doesn't apply here; + area is to be positive. +

    + +

    + The previous example also demonstrates that we often have to break a given region into subregions before applying . + The following example shows another situation where this is applicable, + along with an alternate view of applying the Theorem. +

    + + + Finding area: integrating with respect to <m>y</m> + +

    + Find the area of the region enclosed by the functions y=\sqrt{x}+2, + y=-(x-1)^2+3 and y=2, + as shown in . +

    +
    + Graphing a region for + + + +

    + Graph showing the area of the region bounded by y=\sqrt{x}+2, y=-(x-1)^2+3 and y=2. + The curve given by y=\sqrt{x}+2 is drawn starting at the y-axis and ending at the point (1,3) . + The curve given by y=-(x-1)^2+3 is drawn starting from the end of the previous curve, at the point (1,3) . + This curve then slopes downwards before intersecting the horizontal line y=2 at the point (2,2) . + Both curves lie entirely above the line y=2. + Additionally, the curve y=\sqrt{x}+2 lies to the left of y=-(x-1)^2+3 for the entirety of the enclosed region. +

    +     + Graph of the shaded region enclosed by the two functions and the line y=2. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2}, + ymin=-.1,ymax=3.5, + xmin=-.1,xmax=2.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(0,2.)(0.02,2.141)(0.04,2.2)(0.06,2.245)(0.08,2.283)(0.1,2.316)(0.12,2.346)(0.14,2.374)(0.16,2.4)(0.18,2.424)(0.2,2.447)(0.22,2.469)(0.24,2.49)(0.26,2.51)(0.28,2.529)(0.3,2.548)(0.32,2.566)(0.34,2.583)(0.36,2.6)(0.38,2.616)(0.4,2.632)(0.42,2.648)(0.44,2.663)(0.46,2.678)(0.48,2.693)(0.5,2.707)(0.52,2.721)(0.54,2.735)(0.56,2.748)(0.58,2.762)(0.6,2.775)(0.62,2.787)(0.64,2.8)(0.66,2.812)(0.68,2.825)(0.7,2.837)(0.72,2.849)(0.74,2.86)(0.76,2.872)(0.78,2.883)(0.8,2.894)(0.82,2.906)(0.84,2.917)(0.86,2.927)(0.88,2.938)(0.9,2.949)(0.92,2.959)(0.94,2.97)(0.96,2.98)(0.98,2.99)(1.,3.)(1.04,2.998)(1.08,2.994)(1.12,2.986)(1.16,2.974)(1.2,2.96)(1.24,2.942)(1.28,2.922)(1.32,2.898)(1.36,2.87)(1.4,2.84)(1.44,2.806)(1.48,2.77)(1.52,2.73)(1.56,2.686)(1.6,2.64)(1.64,2.59)(1.68,2.538)(1.72,2.482)(1.76,2.422)(1.8,2.36)(1.84,2.294)(1.88,2.226)(1.92,2.154)(1.96,2.078)(2.,2.)(0,2)}; + + \addplot [firstcurvestyle,domain=0:.1] {sqrt(x) + 2}; + \addplot [firstcurvestyle,domain=.1:1] {sqrt(x) + 2}; + \addplot [firstcurvestyle,domain=1:2] {-(x-1)^2+3}; + \addplot [firstcurvestyle,domain=0:2] {2}; + + \draw (axis cs:.5,3.1) node { $y=\sqrt{x}+2$} (axis cs:1.8,3.1) node { $y=-(x-1)^2+3$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + We give two approaches to this problem. + In the first approach, we notice that the region's top + is defined by two different curves. + On [0,1], the top function is y=\sqrt{x}+2; + on [1,2], the top function is y=-(x-1)^2+3. +

    + +

    + Thus we compute the area as the sum of two integrals: + + \text{Total Area} \amp = \int_0^1 \Big(\big(\sqrt{x}+2\big)-2\Big)\, dx + \int_1^2 \Big(\big(-(x-1)^2+3\big)-2\Big)\, dx + \amp = 2/3 + 2/3 + \amp =4/3 + . +

    + +

    + The second approach is clever and very useful in certain situations. + We are used to viewing curves as functions of x; + we input an x-value and a y-value is returned. + Some curves can also be described as functions of y: + input a y-value and an x-value is returned. + We can rewrite the equations describing the boundary by solving for x: + + y=\sqrt{x}+2 \Rightarrow x=(y-2)^2 + + + y=-(x-1)^2+3 \Rightarrow x=\sqrt{3-y}+1 + . +

    + +
    + The region used in with boundaries relabeled as functions of y + + + + Graph of the shaded region bounded by y=\sqrt{x}+2, y=-(x-1)^2+3 and y=2 with a red horizontal rectangle slice showing integration with respect to y. + The rectangle starts near the beginning of the graph of y=\sqrt{x}+2 at a y-level of about 2.2 and spans across to meet the graph of y=-(x-1)^2+3. + + Graph of the shaded region with boundaries relabeled as functions of y. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2}, + ymin=-.1,ymax=3.5, + xmin=-.1,xmax=2.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(0,2.)(0.02,2.141)(0.04,2.2)(0.06,2.245)(0.08,2.283)(0.1,2.316)(0.12,2.346)(0.14,2.374)(0.16,2.4)(0.18,2.424)(0.2,2.447)(0.22,2.469)(0.24,2.49)(0.26,2.51)(0.28,2.529)(0.3,2.548)(0.32,2.566)(0.34,2.583)(0.36,2.6)(0.38,2.616)(0.4,2.632)(0.42,2.648)(0.44,2.663)(0.46,2.678)(0.48,2.693)(0.5,2.707)(0.52,2.721)(0.54,2.735)(0.56,2.748)(0.58,2.762)(0.6,2.775)(0.62,2.787)(0.64,2.8)(0.66,2.812)(0.68,2.825)(0.7,2.837)(0.72,2.849)(0.74,2.86)(0.76,2.872)(0.78,2.883)(0.8,2.894)(0.82,2.906)(0.84,2.917)(0.86,2.927)(0.88,2.938)(0.9,2.949)(0.92,2.959)(0.94,2.97)(0.96,2.98)(0.98,2.99)(1.,3.)(1.04,2.998)(1.08,2.994)(1.12,2.986)(1.16,2.974)(1.2,2.96)(1.24,2.942)(1.28,2.922)(1.32,2.898)(1.36,2.87)(1.4,2.84)(1.44,2.806)(1.48,2.77)(1.52,2.73)(1.56,2.686)(1.6,2.64)(1.64,2.59)(1.68,2.538)(1.72,2.482)(1.76,2.422)(1.8,2.36)(1.84,2.294)(1.88,2.226)(1.92,2.154)(1.96,2.078)(2.,2.)(0,2)}; + + \addplot [firstcurvestyle,domain=0:.1] {sqrt(x) + 2}; + \addplot [firstcurvestyle,domain=.1:1] {sqrt(x) + 2}; + \addplot [firstcurvestyle,domain=1:2] {-(x-1)^2+3}; + \addplot [firstcurvestyle,domain=0:2] {2}; + + \addplot [secondcurvestyle,solid,fill] coordinates {(0.0625,2.2)(0.0625,2.3)(1.867,2.3)(1.867,2.2)(0.0625,2.2)}; + + \draw (axis cs:.5,3.1) node { $x=(y-2)^2$} (axis cs:1.8,3.2) node { $x=\sqrt{3-y}+1$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + + shows the region with the boundaries relabeled. + A differential element, a horizontal rectangle, is also pictured. + The width of the rectangle is a small change in y: + \Delta y. + The height of the rectangle is a difference in x-values. + The top x-value is the largest value, , the rightmost. + The bottom x-value is the smaller, , the leftmost. + Therefore the height of the rectangle is + + \big(\sqrt{3-y}+1\big) - (y-2)^2 + . +

    + +

    + The area is found by integrating the above function with respect to y with the appropriate bounds. + We determine these by considering the y-values the region occupies. + It is bounded below by y=2, + and bounded above by y=3. + That is, both the top and bottom + functions exist on the y interval [2,3]. + Thus + + \text{Total Area} \amp = \int_2^3 \big(\sqrt{3-y}+1 - (y-2)^2\big)\, dy + \amp = \Big(-\frac23(3-y)^{3/2}+y-\frac13(y-2)^3\Big)\Big|_2^3 + \amp = 4/3 + . +

    +     + +     + + + +

    + This calculus-based technique of finding area can be useful even with shapes that we normally think of as easy. + + computes the area of a triangle. + While the formula \frac12\times\,\text{base}\, \times\,\text{height} is well known, + in arbitrary triangles it can be nontrivial to compute the height. + Calculus makes the problem simple. +

    + + + Finding the area of a triangle + +

    + Compute the area of the regions bounded by the lines +

    + +

    + y=x+1, y=-2x+7 and y=-\frac12x+\frac52, + as shown in . +

    + +
    + Graphing a triangular region in + + + +

    + Graph showing the enclosed area of the triangular region bounded by y=x+1, y=-2x+7 and y=-\frac12x+\frac52. + The corners of the triangle are the points (1,2) , (2,3) and (3,1) . + The line y=x+1 lies above the line y=-\frac12x+\frac52 between x=1 and x=2. + The line y=-2x+7 also lies above the line y=-\frac12x+\frac52 between x=2 and x=3. + For integrating in terms of y, the line y=x+1 lies to the left of y=-2x+7 between y=2 and y=3. + The line y=-\frac12x+\frac52 lies to the left of y=-2x+7 between y=1 and y=2. +

    +     + Graph of the triangle having corners on the points (1,2), (2,3) and (3,1). + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3}, + ymin=-.1,ymax=3.5, + xmin=-.1,xmax=3.9 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(1,2) (2,3) (3,1) (1,2)}; + + \addplot [firstcurvestyle,domain=1:2] {x+1} node [shift={(-50pt,-20pt)},black] { $y=x+1$}; + \addplot [firstcurvestyle,domain=2:3] {-2*x+7} node [shift={(0pt,60pt)},black] { $y=-2x+7$}; + \addplot [firstcurvestyle,domain=1:3] {-.5*x+(5/2)} node [shift={(-50pt,0pt)},black] { $y=-\frac12x+\frac52$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + Recognize that there are two top + functions to this region, + causing us to use two definite integrals. + + \text{Total Area} \amp = \int_1^2\big((x+1)-(-\frac12x+\frac52)\big)\, dx + \amp \quad + \int_2^3\big((-2x+7)-(-\frac12x+\frac52)\big)\, dx + \amp = 3/4+3/4 + \amp =3/2 + . +

    + +

    + We can also approach this by converting each function into a function of y. + This also requires 2 integrals, + so there isn't really any advantage to doing so. + We do it here for demonstration purposes. +

    + +

    + The top function is always x=\frac{7-y}2 while there are two + bottom functions. + Being mindful of the proper integration bounds, we have + + \text{Total Area} \amp = \int_1^2\big(\frac{7-y}2 - (5-2y)\big)\, dy + \int_2^3\big(\frac{7-y}2-(y-1)\big)\, dy + \amp = 3/4 + 3/4 + \amp = 3/2 + . +

    + +

    + Of course, the final answer is the same. + (It is interesting to note that the area of all 4 subregions used is 3/4. + This is coincidental.) +

    +     + +     + +

    + While we have focused on producing exact answers, + we are also able to make approximations using the principle of . + The integrand in the theorem is a distance (top minus bottom); + integrating this distance function gives an area. + By taking discrete measurements of distance, + we can approximate an area using numerical integration techniques developed in . + The following example demonstrates this. +

    + + + Numerically approximating area + +

    + To approximate the area of a lake, + shown in , + the length of the lake is measured at 200-foot increments, as shown in . + The lengths are given in hundreds of feet. + Approximate the area of the lake. +

    + +
    + (a) A sketch of a lake, and (b) the lake with length measurements + +
    + + + + +

    + A sketch of a lake. This image contains no included measurements and no coordinate plots. +

    +     + A sketch of a lake. + + + \begin{tikzpicture}[scale=.6] + + \draw [firstcolor,thick,fill=firstcolor!15,smooth] plot coordinates {(0,3.2)(0.5,3.417)(1.,3.637)(1.5,4.041)(2.,4.505)(2.5,5.)(3.,5.495)(3.5,5.959)(4.,6.363)(4.5,6.683)(5.,6.898)(5.5,6.995)(6.,6.968)(6.5,6.819)(7.,6.556)(7.5,6.197)(8.,5.763)(8.5,5.282)(9.,4.784)(9.5,4.298)(10.,3.857)(10.5,3.486)(11.,3.21)(11.5,2.8)(11.5,2)(11.,1.545)(10.5,1.314)(10.,1.102)(9.5,0.9154)(9.,0.7585)(8.5,0.6361)(8.,0.5514)(7.5,0.5069)(7.,0.5038)(6.5,0.5421)(6.,0.6208)(5.5,0.7378)(5.,0.8897)(4.5,1.072)(4.,1.281)(3.5,1.509)(3.,1.751)(2.5,2.)(2.,2.249)(1.5,2.491)(1.,2.719)(0.5,2.9102)(0,3.15)(0,3.2)}; + + \end{tikzpicture} + + + + +
    + +
    + + + + +

    + A graph of the previous sketch of a lake with vertically strenching length measurements. + The curve begins at the y-axis at the point (0,3). + From this point, the curve slopes upwards until it reaches a peak near the point (6,7). + After reaching a peak, the curve slopes downwards towards the point (12,3). + From the point (0,3), the downward portion of the curve reaches its minimum near (7,1), after which it goes upwards to meet at the point (12,3). +

    +

    + There are 5 vertical length measurements given which occur at all even x starting at x=2. + The first vertical length measurement is 2.25. + The second is 5.08. + The third is 6.35. + The fourth is 5.21. + The fifth is 2.76. + There is then an increment of length 2 on the x-axis between the last measurement occuring at x=10m, and the edge of the lake occuring at x=12. +

    +     + A graph of the lake with five vertical length measurements. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4,5,6,7,8,9,10,11,12}, + ytick={1,2,3,4,5,6,7,8}, + ymin=-.1,ymax=8.5, + xmin=-.1,xmax=12.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(0,3.2)(0.5,3.417)(1.,3.637)(1.5,4.041)(2.,4.505)(2.5,5.)(3.,5.495)(3.5,5.959)(4.,6.363)(4.5,6.683)(5.,6.898)(5.5,6.995)(6.,6.968)(6.5,6.819)(7.,6.556)(7.5,6.197)(8.,5.763)(8.5,5.282)(9.,4.784)(9.5,4.298)(10.,3.857)(10.5,3.486)(11.,3.21)(11.5,2.8)(11.55,2)(11.,1.545)(10.5,1.314)(10.,1.102)(9.5,0.9154)(9.,0.7585)(8.5,0.6361)(8.,0.5514)(7.5,0.5069)(7.,0.5038)(6.5,0.5421)(6.,0.6208)(5.5,0.7378)(5.,0.8897)(4.5,1.072)(4.,1.281)(3.5,1.509)(3.,1.751)(2.5,2.)(2.,2.249)(1.5,2.491)(1.,2.719)(0.5,2.9102)(0,3.15)(0,3.2)}; + \addplot [firstcurvestyle,smooth] coordinates {(0,3.2)(0.5,3.417)(1.,3.637)(1.5,4.041)(2.,4.505)(2.5,5.)(3.,5.495)(3.5,5.959)(4.,6.363)(4.5,6.683)(5.,6.898)(5.5,6.995)(6.,6.968)(6.5,6.819)(7.,6.556)(7.5,6.197)(8.,5.763)(8.5,5.282)(9.,4.784)(9.5,4.298)(10.,3.857)(10.5,3.486)(11.,3.21)(11.5,2.8)(11.5,2)(11.,1.545)(10.5,1.314)(10.,1.102)(9.5,0.9154)(9.,0.7585)(8.5,0.6361)(8.,0.5514)(7.5,0.5069)(7.,0.5038)(6.5,0.5421)(6.,0.6208)(5.5,0.7378)(5.,0.8897)(4.5,1.072)(4.,1.281)(3.5,1.509)(3.,1.751)(2.5,2.)(2.,2.249)(1.5,2.491)(1.,2.719)(0.5,2.9102)(0,3.15)(0,3.2)}; + + \draw (axis cs:2,4.5) -- (axis cs:2,2.249) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 2.25} + (axis cs:4,6.36) -- (axis cs:4,1.28) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 5.08} + (axis cs:6,6.97) -- (axis cs:6,0.62) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.35} + (axis cs:8,5.76) -- (axis cs:8,.55) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 5.21} + (axis cs:10,3.86) -- (axis cs:10,1.1) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 2.76}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    +     + +

    + The measurements of length can be viewed as measuring + top minus bottom of two functions. + The exact answer is found by integrating \ds \int_0^{12} \big(f(x)-g(x)\big)\, dx, + but of course we don't know the functions f and g. + Our discrete measurements instead allow us to approximate. +

    + +

    + We have the following data points: + + (0,0),\,(2,2.25),\,(4,5.08),\,(6,6.35),\,(8,5.21),\,(10,2.76),\,(12,0) + . +

    + +

    + We also have that \dx=\frac{b-a}{n} = 2, so Simpson's Rule gives + + \text{Area} \amp \approx \frac{2}{3}\Big(1\cdot0+4\cdot2.25+2\cdot5.08+4\cdot6.35+2\cdot5.21+4\cdot2.76+1\cdot0\Big) + \amp = 44.01\overline{3} \,\text{units}^2 + . +

    + +

    + Since the measurements are in hundreds of feet, + square units are given by (100)^2 = 10,000, + giving a total area of 440,133. + (Since we are approximating, + we'd likely say the area was about 440,000, + which is a little more than 10 acres.) +

    +     +     + +

    + In the next section we apply our applications of integration techniques to finding the volumes of certain solids. +

    + + + + Terms and Concepts + + + +

    + The area between curves is always positive. + +

    +     + +     + + + + +

    + Calculus can be used to find the area of basic geometric shapes. + +

    +     + +     + + + + +

    + In your own words, + describe how to find the total area enclosed by y=f(x) and y=g(x). +

    + + +     + + + +     + + + + +

    + Describe a situation where it is advantageous to find an area enclosed by curves through integration with respect to y instead of x. +

    + + +     + + + +

    + Answers may vary; + one common answer is when the region has two or more + top or bottom + functions when viewing the region with respect to x, + but has only 1 top function and 1 bottom + function when viewed with respect to y. + The former area requires multiple integrals to compute, + whereas the latter area requires one. +

    +     + +     +     + + Problems + + + +

    + Find the area of the shaded region in the given graph. +

    +     + + + + + + $x0 = 0; + $x1 = Compute("2*pi"); + $f = Formula("x/2+3"); + $g = Formula("1/2*cos(x)+1"); + $F = Formula("x^2/4+3x"); + $G = Formula("1/2*sin(x)+x"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + +

    + Between y=\frac12 x +3 and y=\frac12\cos(x)+1, + for 0\leq x\leq 2\pi. +

    + +

    + +

    + + + + Graph of the region enclosed by the functions y=\frac12 x +3 and y=\frac12\cos(x)+1 between x = 0 and x = 2\pi. + The function y=\frac12 x +3 lies above the function y=\frac12\cos(x)+1 for the entirety of the region between x = 0 and x = 2\pi. + + Graph of the region between two functions between x=0 and x=2pi. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={3.14,6.28}, + extra x tick labels={\(\pi\),\(2\pi\)}, + ymin=-.2,ymax=6.5, + xmin=-1,xmax=7 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.9:6.5,samples=40] {.5*cos(deg(x))+1} node [shift={(-50pt,7pt)} ,black] { \(y=\frac12\cos(x) +1\)}; + \addplot [name path=B,firstcurvestyle,domain=-.9:6.5] {.5*x+3}node [shift={(-50pt,-5pt)} ,black] { \(y=\frac12x+3\)}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0:2*pi}]; + + \end{axis} + + \end{tikzpicture} + + + + + + + +     + +

    + 4\pi+\pi^2\approx 22.436 +

    +     +     +     + + + + + $x0 = Compute("-1"); + $x1 = Compute("1"); + $f = Formula("-3x^3+2x+2"); + $g = Formula("x^2+x-1"); + $F = Formula("-x^4/3+3x^2/2+2x"); + $G = Formula("x^3/3+x^2/2-x"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + +

    + Between y=-3x^3+3x+2 and y=x^2+x-1, + for -1\leq x\leq 1. +

    + +

    + +

    + + + + Graph of the region enclosed by the functions y=-3x^3+3x+2 and y=x^2+x-1 between x = -1 and x = 1. + The function y=-3x^3+3x+2 lies above the function y=x^2+x-1 for the entirety of the region between x = -1 and x = 1. + + Graph of the region enclosed by the two functions between x=0 and x=2pi. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,1}, + ytick={-1,1,2,3}, + ymin=-1.5,ymax=3.5, + xmin=-1.1,xmax=1.1 + ] + + \addplot [name path=A,firstcurvestyle,domain=-1.05:1.05] {x^2+x-1} node [shift={(-40pt,-65pt)},black] { \(y=x^2+x-1\)}; + \addplot [name path=B,firstcurvestyle,domain=-1.05:1.05,samples=60] {-3*x^3+3*x+2} node [shift={(-140pt,30pt)},black] { \(y=-3x^3+3x+2\)}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=-1:1}]; + + \end{axis} + + \end{tikzpicture} + + + + + + + +     + +

    + 10/3 +

    +     +     +     + + + + + $x0 = Compute("0"); + $x1 = Compute("pi"); + $f = Formula("2"); + $g = Formula("1"); + $F = Formula("2x"); + $G = Formula("x"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + +

    + Between y=1 and y=2, for 0\leq x\leq \pi. +

    + +

    + +

    + + + + Graph of the region enclosed by horizontal lines y=0 and y=1 between x = 0 and x = \pi. + + Graph of the rectangular region between y=1, y=2, and x=0, x=pi. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={3.14,1.57}, + extra x tick labels={\(\pi\),\(\pi/2\)}, + ytick={-1,1,2,3}, + ymin=-.5,ymax=2.4, + xmin=-.1,xmax=3.5 + ] + + \addplot [name path=A,firstcurvestyle,-] {1} node [shift={(-100pt,-10pt)},black] { \(y=1\)}; + \addplot [name path=B,firstcurvestyle,-] {2} node [shift={(-100pt,10pt)},black] { \(y=2\)}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0:pi}]; + + \end{axis} + + \end{tikzpicture} + + + + + + + +     + +

    + \pi +

    +     +     +     + + + + + $x0 = Compute("0"); + $x1 = Compute("pi"); + $f = Formula("sin(x)+1"); + $g = Formula("sin(x)"); + $F = Formula("-cos(x)+x"); + $G = Formula("-cos(x)"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + +

    + Between y=\sin(x)+1 and y=\sin(x), + for 0\leq x\leq \pi. +

    + +

    + +

    + + + + Graph of the region enclosed by the functions y=\sin(x)+1 and y=\sin(x) between x = 0 and x = \pi. + The function y=\sin(x)+1 lies above the function y=\sin(x) for the entirety of the region between x = 0 and x = \pi. + + Graph of the region between the two sine graphs between x=0 and x=pi. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={3.14,1.57}, + extra x tick labels={\(\pi\),\(\pi/2\)}, + ytick={-1,1,2,3}, + ymin=-.5,ymax=2.2, + xmin=-.1,xmax=3.5 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.1:3.5,samples=40] {sin(deg(x))} node [shift={(-80pt,45pt)} ,black] { \(y=\sin(x) \)}; + \addplot [name path=B,firstcurvestyle,domain=-.1:3.5,samples=40] {sin(deg(x))+1} node [shift={(-45pt,85pt)} ,black] { \(y=\sin(x) +1\)}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0:pi}]; + + \end{axis} + + \end{tikzpicture} + + + + + + + +     + +

    + \pi +

    +     +     +     + + + + + $x0 = Compute("0"); + $x1 = Compute("pi/4"); + $f = Formula("sec(x)^2"); + $g = Formula("sin(4x)"); + $F = Formula("tan(x)"); + $G = Formula("-cos(4x)/4"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + + +

    + Between y=\sin(4x) and y=\sec^2(x), + for 0\leq x\leq \pi/4. +

    + +

    + +

    + + + + Graph of the region enclosed by the functions y=\sin(4x) and y=\sec^2(x) between x = 0 and x = \pi/4. + The function y=\sec^2(x) lies above the function y=\sin(4x) for the entirety of the region between x = 0 and x = \pi/4. + + Graph of the region enclosed by the two functions between x=0 and x=pi/4. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={0.785,0.392}, + extra x tick labels={\(\pi/4\),\(\pi/8\)}, + ytick={-1,1,2,3}, + ymin=-.5,ymax=2.2, + xmin=-.1,xmax=1 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.1:.9,samples=60] {sin(deg(4*x))} node [shift={(-65pt,40pt)} ,black] { \(y=\sin(4x) \)}; + \addplot [name path=B,firstcurvestyle,domain=-.1:.9,samples=60] {sec(deg(x))^2} node [shift={(-60pt,-40pt)} ,black] { \(y=\sec^2(x)\)}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0:pi/4}]; + + \end{axis} + + \end{tikzpicture} + + + + + + + +     + +

    + 1/2 +

    +     +     +     + + + + + $x0 = Compute("pi/4"); + $x1 = Compute("5*pi/4"); + $f = Formula("sin(x)"); + $g = Formula("cos(x)"); + $F = Formula("-cos(x)"); + $G = Formula("sin(x)"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + +

    + Between y=\sin(x) and y=\cos(x), + for \pi/4\leq x\leq 5\pi/4. +

    + +

    + +

    + + + + Graph of the region enclosed by the functions y=\sin(x) and y=\cos(x) between x = \pi/4 and x = 5\pi/4. + The two functions intersect at x = \pi/4 and x = 5\pi/4. + The function y=\sec^2(x) lies above the function y=\sin(4x) for the entirety of the region between x = \pi/4 and x = 5\pi/4. + + Graph of the region enclosed by the two functions between x=pi/4 and x=5 pi/4. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={.785,1.57,2.36,3.14,3.92}, + extra x tick labels={\(\pi/4\),\(\pi/2\),\(3\pi/4\),\(\pi\),\(5\pi/4\)}, + ymin=-1.1,ymax=1.1, + xmin=-.1,xmax=4.1 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.1:4.1,samples=40] {sin(deg(x))} node [shift={(-30pt,90pt)},black] { \(y=\sin(x) \)}; + \addplot [name path=B,firstcurvestyle,domain=-.1:4.1,samples=40] {cos(deg(x))} node [shift={(-125pt,-5pt)},black] { \(y=\cos(x) \)}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=pi/4:5*pi/4}]; + + \end{axis} + + \end{tikzpicture} + + + + + + + +     + +

    + 2\sqrt{2} +

    +     +     +     + + + + + $x0 = Compute("0"); + $x1 = Compute("1"); + $f = Formula("4^x"); + $g = Formula("2^x"); + $F = Formula("4^x/ln(4)"); + $G = Formula("2^x/ln(2)"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + +

    + Between y=2^x and y=4^x, + for 0\leq x\leq 1. +

    + +

    + +

    + + + + Graph of the region enclosed by the functions y=2^x and y=4^x between x = 0 and x = 1. + The two functions intersect at x = 0 before y=4^x overtakes y=2^x. + The function y=4^x lies above the function y=2^x for the entirety of the region between 0 and 1. + + Graph of the region enclosed by the two functions between x=0 and x=1. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=4.1, + xmin=-.1,xmax=1.1 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.1:1.1] {2^x} node [shift={(-25pt,-20pt)},black] { \(y=2^x\)}; + \addplot [name path=B,firstcurvestyle,domain=-.1:1.1] {4^x} node [shift={(-45pt,-30pt)},black] { \(y=4^x\)}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0:1}]; + + \end{axis} + + \end{tikzpicture} + + + + + + + +     + +

    + 1/\ln(4) +

    +     +     +     + + + +

    + Bounded by the curves y=\sqrt{x}+1, y=\sqrt{2-x}+1, + and y=1. +

    + + Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1 and the horizontal line y=1. + The curve given by y=\sqrt{x}+1 is drawn starting at the y-axis and ending at the point (1,2) . + The curve given by y=\sqrt{2-x}+1 is drawn starting from the end of the previous curve, at the point (1,2) . + This curve then falls downwards before intersecting the horizontal line y=1 at the point (2,1) . + Both curves lie entirely above the horizontal line y=1. + The curve y=\sqrt{x}+1 also lies to the left of y=\sqrt{2-x}+1 throughout the enclosed region. + + Graph of the region enclosed by the two functions and the line y=1. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=2.5, + xmin=-.1,xmax=2.25 + ] + + \addplot [name path=A,firstcurvestyle,domain=0:1,samples=40] {1}; + \addplot [name path=B,firstcurvestyle,domain=0:1,samples=40] ({x^2},{x+1}) node [pos=.85,above left,black] { \(y=\sqrt{x}+1\)}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B]; + + \addplot [name path=C,firstcurvestyle,domain=1:2,samples=2] {1}; + \addplot [name path=D,firstcurvestyle,domain=0:1,samples=40] ({-x^2+2},{x+1}) node [pos=.9,above right,black] { \(y=\sqrt{2-x}+1\)}; + + \addplot [firstcurvestyle,areastyle] fill between [of=C and D]; + + \end{axis} + + \end{tikzpicture} + + + +     + +

    + 4/3 +

    +     +     + +     + + + +

    + Find the total area enclosed by the functions f and g. +

    +     + + + + + $x0 = Compute("-2"); + $x1 = Compute("1"); + $f = Formula("x^2+4x-1"); + $g = Formula("2x^2+5x-3"); + $F = Formula("x^3/3+2x^2-x"); + $G = Formula("2x^3/3+5x^2/2-3x"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + +

    + f(x) = 2x^2+5x-3, g(x) = x^2+4x-1 +

    + +

    + +

    + +     +     +     + + + + + $x0 = Compute("-1"); + $x1 = Compute("1"); + $f = Formula("-3x+3"); + $g = Formula("x^2-3x+2"); + $F = Formula("3x^2/2+3x"); + $G = Formula("x^3/3-3x^2+2x"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + +

    + f(x) = x^2-3x+2, g(x) = -3x+3 +

    + +

    + +

    + +     +     +     + + + + + $x0 = Compute("0"); + $x1 = Compute("pi/2"); + $f = Formula("sin(x)"); + $g = Formula("2x/pi"); + $F = Formula("-cos(x)"); + $G = Formula("x^2/pi"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + $area = 2*$area; + + +

    + f(x) = \sin(x), g(x) = 2x/\pi +

    + +

    + +

    + +     +     +     + + + + + $x0 = Compute("-1"); + $x1 = Compute("1"); + $x2 = Compute("3"); + + $f = Formula("x^3-4x^2+x-1"); + $g = Formula("-x^2+2x-4"); + $F = Formula("x^4/4-4x^3/3+x^2/2-x"); + $G = Formula("-x^3/3+x^2-4x"); + $D1 = Formula("$F-$G"); + $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); + $D2 = Formula("-$D1"); + $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); + $area = $area1 + $area2; + + +

    + f(x) = x^3-4x^2+x-1, g(x) = -x^2+2x-4 +

    + +

    + +

    + +     +     +     + + + + + $x0 = Compute("0"); + $x1 = Compute("1"); + $f = Formula("sqrt(x)"); + $g = Formula("x"); + $F = Formula("2/3*x^(3/2)"); + $G = Formula("x^2/2"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + +

    + f(x) = x, g(x) = \sqrt{x} +

    + +

    + +

    + +     +     +     + + + + + $x0 = Compute("-1"); + $x1 = Compute("1"); + $x2 = Compute("2"); + $f = Formula("3x^2+x+3"); + $g = Formula("-x^3+5x^2+2x+1"); + $F = Formula("x^3+x^2/2+3x"); + $G = Formula("-x^4/4+5x^3/3+x^2+x"); + $D1 = Formula("$F-$G"); + $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); + $D2 = Formula("-$D1"); + $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); + $area = $area1 + $area2; + + +

    + f(x) = -x^3+5x^2+2x+1, g(x) = 3x^2+x+3 +

    + +

    + +

    + +     +     +     + +     + + + +

    + The functions f(x) = \cos (x) and + g(x) = \sin(x) intersect infinitely many times, + forming an infinite number of repeated, enclosed regions. + Find the areas of these regions. +

    +     + +

    + All enclosed regions have the same area, + with regions being the reflection of adjacent regions. + One region is formed on [\pi/4,5\pi/4], + with area 2\sqrt{2}. +

    +     +     + + + + + + $x0 = Compute("pi/6"); + $x1 = Compute("5*pi/6"); + $x2 = Compute("3*pi/2"); + + $f = Formula("sin(x)"); + $g = Formula("cos(2x)"); + $F = Formula("-cos(x)"); + $G = Formula("sin(2x)/2"); + $D1 = Formula("$F-$G"); + $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); + $D2 = Formula("-$D1"); + $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); + $area = $area1 + $area2; + + +

    + The functions f(x) = \cos(2x) and + g(x) = \sin(x) intersect infinitely many times, + forming an infinite number of repeated, enclosed regions. + Find the areas of these regions. +

    + +

    + +

    + +     + +

    + On regions such as [\pi/6,5\pi/6], + the area is 3\sqrt{3}/2. + On regions such as [-\pi/2,\pi/6], + the area is 3\sqrt{3}/4. +

    +     +     +     + + + +

    + Find the area of the enclosed region in two ways: +

    + +

    +

      +
    1. +

      + by treating the boundaries as functions of x, and +

      +
      2. + +
    2. +

      + by treating the boundaries as functions of y. +

      +
      3. +
    +

    +     + + + + + $x0 = Compute("0"); + $x1 = Compute("1"); + $x2 = Compute("3"); + + $f1 = Formula("x^2+1"); + $f2 = Formula("1/4*(x-3)^2+1"); + $g = Formula("1"); + $F1 = Formula("x^3/3+x"); + $F2 = Formula("1/12*(x-3)^3+x"); + $G = Formula("x"); + $D1 = Formula("$F1-$G"); + $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); + $D2 = Formula("$F2-$G"); + $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); + $area = $area1 + $area2; + + +

    + Bounded by y=x^2+1, y=\frac14(x-3)^2+1, and y=1. +

    + + + + Graph of the region enclosed by the functions y=x^2+1, y=\frac14(x-3)^2+1 and the horizontal line y=1. + The curve given by y=x^2+1 is drawn starting at the y-axis and ending at the point (1,2) . + The curve given by y=\frac14(x-3)^2+1 is drawn starting from the end of the previous curve at the point (1,2) . + This curve then falls downwards before intersecting the horizontal line y=1 at the point (3,1) . + Both curves lie above the line y=1 for the entirety of the enclosed region. + The curve y=\frac14(x-3)^2+1 also lies to the right of the curve y=x^2+1 throughout the enclosed region. + + Graph of the region enclosed by the two functions and the line y=1. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=2.2, + xmin=-.5,xmax=3.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(0,1.)(0.1,1.01)(0.2,1.04)(0.3,1.09)(0.4,1.16)(0.5,1.25)(0.6,1.36)(0.7,1.49)(0.8,1.64)(0.9,1.81)(1.,2.)(1.1,1.903)(1.2,1.81)(1.3,1.723)(1.4,1.64)(1.5,1.563)(1.6,1.49)(1.7,1.423)(1.8,1.36)(1.9,1.303)(2.,1.25)(2.1,1.203)(2.2,1.16)(2.3,1.123)(2.4,1.09)(2.5,1.063)(2.6,1.04)(2.7,1.023)(2.8,1.01)(2.9,1.003)(3.,1.)(0,1)}; + + \addplot [firstcurvestyle,domain=1:3] {.25*(x-3)^2+1} node [shift={(-20pt,40pt)},black] { \(y=\frac14(x-3)^2+1\)}; + \addplot [firstcurvestyle,domain=0:1] {x^2+1} node [shift={(-40pt,-15pt)},black] { \(y=x^2+1\)}; + \addplot [firstcurvestyle,domain=0:3] {1} node [pos=.5,below,black] { \(y=1\)} (axis cs:3,1); + + \end{axis} + + \end{tikzpicture} + + + + + +

    + +

    + +     + +

    + 1 +

    +     +     +     + + + + + $x0 = Compute("0"); + $x1 = Compute("1"); + $x2 = Compute("2"); + + $f1 = Formula("sqrt(x)"); + $f2 = Formula("-2x+3"); + $g = Formula("-1/2*x"); + $F1 = Formula("2/3*x^(3/2)"); + $F2 = Formula("-x^2+3x"); + $G = Formula("-x^2/4"); + $D1 = Formula("$F1-$G"); + $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); + $D2 = Formula("$F2-$G"); + $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); + $area = $area1 + $area2; + + +

    + Bounded by y=\sqrt{x}, y=-2x+3, and y=-\frac12 x. +

    + + + + Graph of the region enclosed by the functions y=\sqrt{x}, y=-2x+3 and y=-\frac12 x. + The curve given by y=\sqrt{x} is drawn starting at the origin, from which it goes upwards and ends at the point (1,1) . + The curve given by y=-2x+3 is drawn starting from the end of the previous curve at the point (1,1) . + This curve then goes downwards before ending at the point (2,-1) . + The curve given by y=-\frac12 x is drawn starting from the origin, from which it goes downwards until it meets the previous curve at the point (2,-1) . + The curve y=\sqrt{x} graphed between x=0 and x=1 and the line y=-2x+3 graphed between x=1 and x=2 lie entirely above the line y=-\frac12 x. + The curve y=\sqrt{x} between y=0 and y=1 and the line y=-\frac12 x between y=-1 and y=0 lie to the left of the curve y=-2x+3. + + Graph of the region enclosed by the three functions. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-1.1,ymax=1.1, + xmin=-.5,xmax=2.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates { (0,0)(0.02,0.1414)(0.04,0.2)(0.06,0.2449)(0.08,0.2828)(0.1,0.3162)(0.2,0.4472)(0.3,0.5477)(0.4,0.6325)(0.5,0.7071)(0.6,0.7746)(0.7,0.8367)(0.8,0.8944)(0.9,0.9487)(1.,1.)(2,-1)(0,0)}; + + \addplot [firstcurvestyle,domain=0:1,samples=60] {sqrt(x)} node [shift={(-40pt,-10pt)} ,black] { \(y=\sqrt{x}\)}; + \addplot [firstcurvestyle,domain=0:2] {-.5*x} node [pos=.3,shift={(5pt,-25pt)},black] { \(y=-\frac12x\)}; + \addplot [firstcurvestyle,domain=1:2] {-2*x+3} node [pos=.5,shift={(25pt,25pt)},black] { \(y=-2x+3\)}; + + \end{axis} + + \end{tikzpicture} + + + + + +

    + +

    + +     + +

    + 5/3 +

    +     +     +     + + + + + $x0 = Compute("-1"); + $x1 = Compute("2"); + $f = Formula("x+2"); + $g = Formula("x^2"); + $F = Formula("x^2/2+2x"); + $G = Formula("x^3/3"); + $D = Formula("$F-$G"); + $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); + + +

    + Between the curves y=x+2 and y=x^2. +

    + + + + Graph of the region enclosed by the functions y=x+2 and y=x^2. + The line y=x+2 is drawn starting at the point (-1,1), from which it goes upwards and ends at the point (2,4) . + The parabola given by y=x^2 is also drawn starting from the point (-1,1), from which it slopes downwards until the point (0,0) . + The parabola then goes upwards before meeting the line at the point (2,4) . + The line y=x+2 graphed between x=-1 and x=2 lies entirely above the parabola y=x^2 graphed on the same bounds, only interesecting at the end points. + The line y=\sqrt{x} lies to the left of the parabola y=x^2 between y=1 and y=4. + Between y=0 and y=1, the enclosed region spans from the left side of y=x^2 to the right side of the y=x^2. + + Graph of the region enclosed by the line and parabola. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=4.5, + xmin=-1.5,xmax=2.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates { (-1.,1.)(-0.9,0.81)(-0.8,0.64)(-0.7,0.49)(-0.6,0.36)(-0.5,0.25)(-0.4,0.16)(-0.3,0.09)(-0.2,0.04)(-0.1,0.01)(0,0)(0.1,0.01)(0.2,0.04)(0.3,0.09)(0.4,0.16)(0.5,0.25)(0.6,0.36)(0.7,0.49)(0.8,0.64)(0.9,0.81)(1.,1.)(1.1,1.21)(1.2,1.44)(1.3,1.69)(1.4,1.96)(1.5,2.25)(1.6,2.56)(1.7,2.89)(1.8,3.24)(1.9,3.61)(2.,4.)(-1,1)}; + + \addplot [firstcurvestyle,domain=-1:2] {x^2} node [shift={(0pt,-60pt)} ,black] { \(y=x^2\)}; + \addplot [firstcurvestyle,domain=-1:2] {x+2} node [pos=.5,shift={(10pt,35pt)},black] { \(y=x+2\)} (axis cs:-1,1); + + \end{axis} + + \end{tikzpicture} + + + + + +

    + +

    + +     + +

    + 9/2 +

    +     +     +     + + + + + $x0 = Compute("0"); + $x1 = Compute("1/2"); + $x2 = Compute("2"); + + $f1 = Formula("sqrt(2*x)"); + $f2 = Formula("-2*(x-1)"); + $g = Formula("-sqrt(2*x)"); + $F1 = Formula("sqrt(2)*2/3*x^(3/2)"); + $F2 = Formula("-(x-1)^2"); + $G = Formula("-sqrt(2)*2/3*x^(3/2)"); + $D1 = Formula("$F1-$G"); + $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); + $D2 = Formula("$F2-$G"); + $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); + $area = $area1 + $area2; + + +

    + Between the curves x=-\frac12 y+1 and x=\frac12 y^2. +

    + + + + Graph of the region enclosed by the functions x=-\frac12 y+1 and x=\frac12 y^2. + The line y=x+2 is drawn starting at the point (0.5,1), from which it goes downwards and ends at the point (2,-2) . + The sideways parabola given by x=\frac12 y^2 is also drawn starting from the point (0.5,1), from which it goes backwards until the point (0,0). + The parabola then heads to the right before meeting the line at the point (2,-2) . + The line x=-\frac12 y+1 graphed between x=-0.5 and x=2 lies entirely above the parabola x=\frac12 y^2 between the same bounds, only interesecting at the end points. + However, between x=0 and x=0.5 , both the top and bottom of the enclosed region are made up by the parabola. + The sideways parabola x=\frac12 y^2 lies to the left of the line x=-\frac12 y+1 for the entirety of the region between y=-2 and y=1 . + + Graph of the region enclosed by the line and parabola. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-2.5,ymax=1.5, + xmin=-.5,xmax=2.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(2.,-2.)(1.805,-1.9)(1.62,-1.8)(1.445,-1.7)(1.28,-1.6)(1.125,-1.5)(0.98,-1.4)(0.845,-1.3)(0.72,-1.2)(0.605,-1.1)(0.5,-1.)(0.405,-0.9)(0.32,-0.8)(0.245,-0.7)(0.18,-0.6)(0.125,-0.5)(0.08,-0.4)(0.045,-0.3)(0.02,-0.2)(0.005,-0.1)(0,0)(0.005,0.1)(0.02,0.2)(0.045,0.3)(0.08,0.4)(0.125,0.5)(0.18,0.6)(0.245,0.7)(0.32,0.8)(0.405,0.9)(0.5,1.)(2,-2)}; + + \addplot [firstcurvestyle,domain=-2:1] ({.5*x^2},x) node [shift={(5pt,-110pt)},black] { \(x=\frac12y^2\)}; + \addplot [firstcurvestyle,domain=-2:1] ({-.5*x+1},x) node [pos=.85,shift={(40pt,0pt)},black] { \(x=-\frac12y+1\)}; + + \end{axis} + + \end{tikzpicture} + + + + + +

    + +

    + +     + +

    + 9/4 +

    +     +     +     + + + + + $x0 = Compute("0"); + $x1 = Compute("1/2"); + $x2 = Compute("1"); + + $f1 = Formula("x^(1/3)"); + $f2 = Formula("sqrt(x-1/2)"); + $g = Formula("0"); + $F1 = Formula("3/4*x^(4/3)"); + $F2 = Formula("2/3*(x-1/2)^(3/2)"); + $G = Formula("0"); + $D1 = Formula("$F1-$G"); + $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); + $D2 = Formula("$F1-$F2"); + $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); + $area = $area1 + $area2; + + +

    + Bounded by y=x^{1/3}, y=\sqrt{x-1/2}, + y=0, and x=1. +

    + + + + Graph of the region enclosed by the functions y=x^{1/3}, y=\sqrt{x-1/2} and the lines y=0, x=1. + The curve y=x^{1/3} is drawn starting at the origin, from which it goes upwards and ends at the point (1,1) . + The curve y=\sqrt{x-1/2} is drawn starting from the point (0.5,0), from which it goes upwards until ending at the point (1,sqrt{\frac12}). + The curve y=x^{1/3} graphed between x=0 and x=0.5 lies above x-axis. + Between x=0.5 and x=1 , the curve y=x^{1/3} lies above the curve y=\sqrt{x-1/2}. + The curve y=x^{1/3} lies to the left of the curve y=\sqrt{x-1/2} for the entirety of the length of the curve y=\sqrt{x-1/2}. + After this point, the curve y=x^{1/3} lies to the left of the boundary line occuring at x=1. + + Graph of the region enclosed by the two functions and y=0 and x=1. + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle] coordinates { (0,0)(0.01,0.2154)(0.02,0.2714)(0.03,0.3107)(0.04,0.342)(0.05,0.3684)(0.06,0.3915)(0.07,0.4121)(0.08,0.4309)(0.09,0.4481)(0.1,0.4642)(0.12,0.4932)(0.14,0.5192)(0.16,0.5429)(0.18,0.5646)(0.2,0.5848)(0.22,0.6037)(0.24,0.6214)(0.26,0.6383)(0.28,0.6542)(0.3,0.6694)(0.4,0.7368)(0.5,0.7937)(0.6,0.8434)(0.7,0.8879)(0.8,0.9283)(0.9,0.9655)(1.,1.)(1.,0.7071)(0.9,0.6325)(0.8,0.5477)(0.7,0.4472)(0.6,0.3162)(0.59,0.3)(0.58,0.2828)(0.57,0.2646)(0.56,0.2449)(0.55,0.2236)(0.54,0.2)(0.53,0.1732)(0.52,0.1414)(0.51,0.1)(0.5,0)(0,0)}; + + \addplot [firstcurvestyle,domain=0:1] (x^3,x) node [shift={(-40pt,0pt)},black] { \(y=x^{1/3}\)}; + \addplot [firstcurvestyle,domain=0:.7071] ({(x)^2+.5},x) node [shift={(-20pt,-50pt)},black] { \(y=\sqrt{x-1/2}\)}; + + \end{axis} + + \end{tikzpicture} + + + + + +

    + +

    + +     + +

    + \frac{1}{12}(9-2\sqrt{2})\approx 0.514 +

    +     +     +     + + + +

    + Bounded by the curves y=\sqrt{x}+1, y=\sqrt{2-x}+1, + and y=1. +

    + + Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1 and the line y=1. + The curve y=\sqrt{x}+1 is drawn starting at the point (0,1), from which it goes upwards and ends at the point (1,2) . + The curve y=\sqrt{2-x}+1 is drawn starting from the end of the previous curve, at the point (1,2), from which it goes downward until ending at the point (2,1). + Both of these curves lie entirely above the horizontal line y=1. + The curve y=\sqrt{x}+1 lies to the left of the curve y=\sqrt{2-x}+1 for the entirety of the enclosed region. + + Graph of the region enclosed by the two functions and the line y=1. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=2.5, + xmin=-.1,xmax=2.25 + ] + + \addplot [name path=A,firstcurvestyle,domain=0:1,samples=40] {1}; + \addplot [name path=B,firstcurvestyle,domain=0:1,samples=40] ({x^2},{x+1}) node [pos=.85,above left,black] { $y=\sqrt{x}+1$}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B]; + + \addplot [name path=C,firstcurvestyle,domain=1:2,samples=2] {1}; + \addplot [name path=D,firstcurvestyle,domain=0:1,samples=40] ({-x^2+2},{x+1}) node [pos=.9,above right,black] { $y=\sqrt{2-x}+1$}; + + \addplot [firstcurvestyle,areastyle] fill between [of=C and D]; + + \end{axis} + + \end{tikzpicture} + + + +     + +

    + 4/3 +

    +     +     + +     + + + +

    + Find the area of the triangle formed by the given three points. +

    +     + + + + + @p1 = (1,1); + @p2 = (2,3); + @p3 = (3,3); + $area = 1/2*($p1[1]*($p2[0]-$p3[0])+$p2[1]*($p3[0]-$p1[0]) + $p3[1]*($p1[0] - $p2[0])); + $area = Formula(abs($area)); + $s1 = join(", ", @p1); + $s2 = join(", ", @p2); + $s3 = join(", ", @p3); + + +

    + (),(), and () +

    + +

    + +

    + +     +     +     + + + + + @p1 = (-1,1); + @p2 = (1,3); + @p3 = (2,-1); + $area = 1/2*($p1[1]*($p2[0]-$p3[0])+$p2[1]*($p3[0]-$p1[0]) + $p3[1]*($p1[0] - $p2[0])); + $area = Formula(abs($area)); + $s1 = join(", ", @p1); + $s2 = join(", ", @p2); + $s3 = join(", ", @p3); + + +

    + (),(), and () +

    + +

    + +

    + +     +     +     + + + + + @p1 = (1,1); + @p2 = (3,3); + @p3 = (0,4); + $area = 1/2*($p1[1]*($p2[0]-$p3[0])+$p2[1]*($p3[0]-$p1[0]) + $p3[1]*($p1[0] - $p2[0])); + $area = Formula(abs($area)); + $s1 = join(", ", @p1); + $s2 = join(", ", @p2); + $s3 = join(", ", @p3); + + +

    + (),(), and () +

    + +

    + +

    + +     +     +     + + + + + @p1 = (0,0); + @p2 = (2,5); + @p3 = (5,2); + $area = 1/2*($p1[1]*($p2[0]-$p3[0])+$p2[1]*($p3[0]-$p1[0]) + $p3[1]*($p1[0] - $p2[0])); + $area = Formula(abs($area)); + $s1 = join(", ", @p1); + $s2 = join(", ", @p2); + $s3 = join(", ", @p3); + + +

    + (),(), and () +

    + +

    + +

    + +     +     +     + +     + + + + + + $a = 0; + $b = 6; + $n = 5; + $delta_x = ($b - $a)/$n; + @y = (0, 4.9, 5.2, 7.3, 4.5, 0); + @coeffs = (1, 2, 2, 2, 2, 1); + $area = 0; + $size = scalar @y; + foreach $i (0..$size) { + $area += $coeffs[$i] * $y[$i]; + } + $area = $area*$delta_x/2; + $area_ft = $area * 100 * 100; + $area = NumberWithUnits("$area cm^2"); + $area_ft = NumberWithUnits("$area_ft ft^2"); + + +

    + Use the Trapezoidal Rule to approximate the area of the pictured lake whose lengths, + in hundreds of feet, + are measured in 100-foot increments. +

    + + + + +

    + A sketch of a lake with four vertically strenching length measurements measured with lengths in hundreds of feet. + Each length measurement is given in 100-foot increments from the previous measurement, starting 100 feet from the leftmost point of the lake. + The first vertical length measurement is 4.9. + The second is 5.2. + The third is 7.3. + The fourth is 4.5. + There is then a 100-foot distance from the last measurement to the end of the lake. +

    +     + A sketch of a lake with four vertical length measurements. + + + \begin{tikzpicture}[xscal=1.19,yscale=.6] + + \draw [firstcolor,thick,fill=firstcolor!15,smooth] plot coordinates {(0,1.)(0.1013,1.899)(0.3567,2.785)(0.6934,3.42)(1.039,3.569)(1.355,3.224)(1.646,2.705)(1.925,2.355)(2.2,2.473)(2.475,2.968)(2.75,3.529)(3.025,3.844)(3.3,3.709)(3.575,3.249)(3.85,2.651)(4.125,2.098)(4.389,1.661)(4.624,1.285)(4.807,0.9071)(4.921,0.4743)(4.958,0)(4.921,-0.4743)(4.807,-0.9071)(4.624,-1.279)(4.389,-1.625)(4.125,-1.986)(3.85,-2.401)(3.575,-2.843)(3.3,-3.233)(3.025,-3.487)(2.75,-3.536)(2.475,-3.397)(2.2,-3.111)(1.925,-2.718)(1.646,-2.263)(1.355,-1.792)(1.039,-1.352)(0.6934,-0.9844)(0.3567,-0.6744)(0.1013,-0.3653)(0,0)(0,1)}; + + \draw (1,3.56) -- (1,-1.35) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.9}; + \draw (2,2.37) -- (2,-2.8) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 5.2}; + \draw (3,3.8) -- (3,-3.5) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 7.3}; + \draw (4,2.35) -- (4,-2.15) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.5}; + + \end{tikzpicture} + + + + + +

    + +

    + +     +     +     + + + + + + $a = 0; + $b = 6; + $n = 6; + $delta_x = ($b - $a)/$n; + @y = (0, 4.25, 6.6, 7.7, 6.45, 4.9, 0); + @coeffs = (1, 4, 2, 4, 2, 4, 1); + $area = 0; + $size = scalar @y; + foreach $i (0..$size) { + $area += $coeffs[$i] * $y[$i]; + } + $area = $area*$delta_x/3; + $area_ft = $area * 100 * 200; + $area = NumberWithUnits("$area cm^2"); + $area_ft = NumberWithUnits("$area_ft ft^2"); + + +

    + Use Simpson's Rule to approximate the area of the pictured lake whose lengths, + in hundreds of feet, + are measured in 200-foot increments. +

    + + + + + +

    + A sketch of a lake with five vertically strenching length measurements with lengths in hundreds of feet. + Each length measurement is given in 200-foot increments from the previous measurement, starting 200 feet from the leftmost point of the lake. + The first vertical length measurement is 4.25. + The second is 6.6. + The third is 7.7. + The fourth is 6.45. + The fifth is 4.9. + There is then a 200-foot distance from the last measurement to the end of the lake. +

    +     + A sketch of a lake with five vertical length measurements. + + + \begin{tikzpicture}[xscale=1.19,yscale=.6] + + \draw [firstcolor,thick,fill=firstcolor!15,smooth] plot coordinates {(0,2.)(0.1013,2.717)(0.3567,3.238)(0.6934,3.594)(1.039,3.818)(1.355,3.948)(1.646,4.035)(1.925,4.132)(2.2,4.28)(2.475,4.428)(2.75,4.471)(3.025,4.307)(3.305,3.88)(3.607,3.287)(3.952,2.652)(4.361,2.098)(4.815,1.661)(5.253,1.285)(5.614,0.9071)(5.843,0.4705)(5.938,-0.04167)(5.919,-0.6295)(5.807,-1.293)(5.624,-2.009)(5.389,-2.703)(5.125,-3.29)(4.849,-3.692)(4.562,-3.892)(4.243,-3.929)(3.871,-3.846)(3.432,-3.674)(2.956,-3.409)(2.484,-3.028)(2.057,-2.511)(1.692,-1.869)(1.363,-1.168)(1.039,-0.4819)(0.6934,0.1248)(0.3567,0.679)(0.1013,1.273)(0,2.)}; + + \draw (1,3.8) -- (1,-.45) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.25}; + \draw (2,4.15) -- (2,-2.45) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.6}; + \draw (3,4.3) -- (3,-3.4) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 7.7}; + \draw (4,2.6) -- (4,-3.85) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.45}; + \draw (5,1.45) -- (5,-3.45) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.9}; + + \end{tikzpicture} + + + + + +

    + +

    + +     +     +     +     +     +
    +
    + Volume by Cross-Sectional Area; Disk and Washer Methods +

    + The volume of a general right cylinder, + as shown in , is +

    + +
    +

    + Area of the base height. +

    +
    + +
    + The volume of a general right cylinder + + + + +

    + An image of a general right cylinder. + The area of the base of the cylinder is given to be A. + The base of the cylinder resemembles a parallelogram with curved edges, with each side slightly bowing in at the half way point between two corners. + The height of the cylinder is h. + The base of the general cylinder is identical to the top of the general cylinder, with the two faces being parallel. + These faces also coincide on top of eachother, which leads to right angles when the top and base are connected to form the general right cylinder. + The volume of the general right cylinder is V=A\cdot h. +

    +     + An image of a general right cylider with a base area of A and heigh h. + + + + + + //ASY file for figcross1.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(3.4,5.4,.9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={1,2,3}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,2.5); + pair ybounds=(-0.25,2.5); + pair zbounds=(-0.25,1.5); + + //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple f(pair t) { + return (t.y*(cos(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))+.5*sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))),t.y*(sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))),0); + } + surface s=surface(f,(0,0),(2*pi,1),16,2,Spline); + pen p=apexmeshpen; + draw(s,darksurfacepen,meshpen=p); + + triple f(pair t) { + return (cos(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))+.5*sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x)),sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x)),t.y); + } + surface s=surface(f,(0,0),(2*pi,1),16,2,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (t.y*(cos(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))+.5*sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))),t.y*(sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))),1); + } + surface s=surface(f,(0,0),(2*pi,1),16,2,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return (cos(t)*(1.5+.5*cos(t)*sin(3*t))+.5*sin(t)*(1.5+.5*cos(t)*sin(3*t)),sin(t)*(1.5+.5*cos(t)*sin(3*t)),1);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (cos(t)*(1.5+.5*cos(t)*sin(3*t))+.5*sin(t)*(1.5+.5*cos(t)*sin(3*t)),sin(t)*(1.5+.5*cos(t)*sin(3*t)),0);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + label("base area $= A$",(2,2,-.1)); + label("Volume $=A\cdot h$",(2,2,-.4)); + label("$h$",(-1.9,0,.5)); + + + + +
    + +

    + We can use this fact as the building block in finding volumes of a variety of shapes. +

    + + + +

    + Given an arbitrary solid, + we can approximate its volume by cutting it into n thin slices. + When the slices are thin, + each slice can be approximated well by a general right cylinder. + Thus the volume of each slice is approximately its cross-sectional area thickness. + (These slices are the differential elements.) +

    + +

    + By orienting a solid along the x-axis, + we can let A(x_i) represent the cross-sectional area of the ith slice, + and let \dx_i represent the thickness of this slice + (the thickness is a small change in x). + The total volume of the solid is approximately: + + \text{Volume} \amp \approx \sum_{i=1}^n \Big[\text{Area} \,\times\,\text{thickness} \Big] + \amp = \sum_{i=1}^n A(x_i)\dx_i + . +

    + +

    + Recognize that this is a Riemann Sum. + By taking a limit + (as the thickness of the slices goes to 0) + we can find the volume exactly. +

    + + + Volume By Cross-Sectional Area + +

    + The volume V of a solid, + oriented along the x-axis with cross-sectional area A(x) from x=a to x=b, is + integrationvolume!cross-sectional area + + V = \int_a^b A(x)\, dx + . +

    +     +     + + + Finding the volume of a solid + +

    + Find the volume of a pyramid with a square base of side length 10 + and a height of 5. +

    +     + +

    + There are many ways to orient + the pyramid along the x-axis; + gives one such way, + with the pointed top of the pyramid at the origin and the x-axis going through the center of the base. +

    + +
    + Orienting a pyramid along the x-axis in + + + + +

    + Three-dimensional plot of a pyramid with a square base of side length 10 and height 5. + The peak of the pyramid is placed on the origin, from which it expands towards the positive x-axis. + The square base of the pyramid is centered at the point (5,0). + Additionally, the plot contains an arbitrarily chosen point labeled x, lying on the x-axis. + Slicing the pyramid parallel to the base at this point x results in side lengths of the square slice to be 2x. +

    +     + A three-dimensional graph of a pyramid with a square base of side length 10 and height 5. + + + + + //ASY file for fig13_06_ex_143D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((8.1,15.5,23.4),(0,1,0),(0,0,0),1,(-0.13,0.0046)); + defaultrender.merge=true; + + //(-0.01,0.061,-0.029) + + // setup and draw the axes + real[] myxchoice={5}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,8); + pair ybounds=(-8,8); + pair zbounds=(-8,8); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + path3 p=(5,5,5)--(5,-5,5)--(5,-5,-5)--(5,5,-5); + draw(surface(p -- cycle), simplesurfacepen); + draw(p--cycle,bluepen+.3mm); + + path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); + draw(surface(p1 -- cycle), simplesurfacepen); + draw(p1--cycle,bluepen+.3mm); + + path3 p2=(0,0,0)--(5,5,5)--(5,5,-5); + draw(surface(p2 -- cycle), simplesurfacepen); + draw(p2--cycle,bluepen+.3mm); + + path3 p3=(0,0,0)--(5,5,-5)--(5,-5,-5); + draw(surface(p3 -- cycle), simplesurfacepen); + draw(p3--cycle,bluepen+.3mm); + + path3 p4=(0,0,0)--(5,-5,5)--(5,-5,-5); + draw(surface(p4 -- cycle), simplesurfacepen); + draw(p4--cycle,bluepen+.3mm); + + draw((3,-3,3) -- (3,-3,-3) -- (3,3,-3)-- (3,3,3)--cycle,black+.5mm); + + dotfactor=3; + dot((3,0,0)); + label("$x$",(3,0,0),S); + + label("$2x$",(3,0,3),W); + label("$2x$",(3,3,0),N); + label("$10$",(5,-5,0),S); + label("$10$",(5,0,-5),E); + + + + +
    + +

    + Each cross section of the pyramid is a square; + this is a sample differential element. + To determine its area A(x), + we need to determine the side lengths of the square. +

    + +

    + When x=5, the square has side length 10; + when x=0, the square has side length 0. + Since the edges of the pyramid are lines, + it is easy to figure that each cross-sectional square has side length 2x, + giving A(x) = (2x)^2=4x^2. +

    + +

    + If one were to cut a slice out of the pyramid at x=3, + as shown in , + one would have a shape with square bottom and top with sloped sides. + If the slice were thin, + both the bottom and top squares would have sides lengths of about 6, and thus the cross-sectional area of the bottom and top would be about + 36. + Letting \Delta x_i represent the thickness of the slice, + the volume of this slice would then be about 36\Delta x_i . +

    + +
    + Cutting a slice in the pyramid in at x=3 + + + + +

    + Three-dimensional plot of the same pyramid with a square base of side length 10 and height 5 as in the previous image. + This time, the plot contains a thin slice of the pyramid at x=3 which is parallel to the base of the pyramid. + The side lengths of the resulting square slice are 6, giving the square a surface area of 36 with an arbitrarily small thickness \Delta x. +

    +     + A three-dimensional plot of the same pyramid showing a small slice being used to approximate its volume. + + + + + //ASY file for fig13_06_ex_143D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((8.1,15.5,23.4),(0,1,0),(0,0,0),1,(-0.13,0.0046)); + defaultrender.merge=true; + + //(-0.01,0.061,-0.029) + + // setup and draw the axes + real[] myxchoice={3,5}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,8); + pair ybounds=(-8,8); + pair zbounds=(-8,8); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple f1(real t) { + return (t,t,t); + } + triple f2(real t) { + return (t,-t,t); + } + triple f3(real t) { + return (t,t,-t); + } + triple f4(real t) { + return (t,-t,-t); + } + //path3 mypath1=graph(f1,2.8,3.2,operator ..); + //draw(mypath1,bluepen+.3mm); + + real x1=2.8; + real x2=3.2; + + path3 p1=f1(x1)--f1(x2)--f2(x2)--f2(x1); + draw(surface(p1 -- cycle), simplesurfacepen); + draw(p1--cycle,bluepen+.3mm); + + path3 p2=f1(x1)--f1(x2)--f3(x2)--f3(x1); + draw(surface(p2 -- cycle), simplesurfacepen); + draw(p2--cycle,bluepen+.3mm); + + path3 p3=f4(x1)--f4(x2)--f3(x2)--f3(x1); + draw(surface(p3 -- cycle), simplesurfacepen); + draw(p3--cycle,bluepen+.3mm); + + path3 p4=f4(x1)--f4(x2)--f2(x2)--f2(x1); + draw(surface(p4 -- cycle), simplesurfacepen); + draw(p4--cycle,bluepen+.3mm); + + path3 p5=f1(x1)--f3(x1)--f4(x1)--f2(x1); + draw(surface(p5 -- cycle), simplesurfacepen); + draw(p5--cycle,bluepen+.3mm); + + path3 p6=f1(x2)--f3(x2)--f4(x2)--f2(x2); + draw(surface(p6 -- cycle), simplesurfacepen); + draw(p6--cycle,bluepen+.3mm); + + label("$\Delta x$",f3(3),N); + + path3 p=(5,5,5)--(5,-5,5)--(5,-5,-5)--(5,5,-5); + draw(p--cycle,bluepen+.2mm+dashed); + + path3 p1=(0,0,0)--(5,5,5); + draw(p1,bluepen+.2mm+dashed); + + path3 p2=(0,0,0)--(5,-5,-5); + draw(p2,bluepen+.2mm+dashed); + + path3 p3=(0,0,0)--(5,5,-5); + draw(p3,bluepen+.2mm+dashed); + + path3 p4=(0,0,0)--(5,-5,5); + draw(p4,bluepen+.2mm+dashed); + + //draw((3,-3,3) -- (3,-3,-3) -- (3,3,-3)-- (3,3,3)--cycle,black+.5mm); + + //dotfactor=3; + //dot((3,0,0)); + //label("$x$",(3,0,0),S); + + //label("$2x$",(3,0,3),W); + //label("$2x$",(3,3,0),N); + //label("$10$",(5,-5,0),S); + //label("$10$",(5,0,-5),E); + + + + +
    + +

    + Cutting the pyramid into n slices divides the total volume into n equally-spaced smaller pieces, + each with volume (2x_i)^2\Delta x, + where x_i is the approximate location of the slice along the x-axis and + \Delta x represents the thickness of each slice. + One can approximate total volume of the pyramid by summing up the volumes of these slices: + + \text{Approximate volume} = \sum_{i=1}^n (2x_i)^2\Delta x + . +

    + +

    + Taking the limit as n\to\infty gives the actual volume of the pyramid; + recoginizing this sum as a Riemann Sum allows us to find the exact answer using a definite integral, + matching the definite integral given by . +

    + +

    + We have + + V \amp = \lim_{n\to\infty} \sum_{i=1}^n (2x_i)^2\Delta x + \amp = \int_0^5 4x^2\, dx + \amp = \frac43x^3\Big|_0^5 + \amp =\frac{500}{3}\,\text{in}^3 \approx 166.67\,\text{in}^3 + . +

    + +

    + We can check our work by consulting the general equation for the volume of a pyramid + (a pyramid is a special type of cone see in the back matter under + Volume of A General Cone): +

    + +

    + \frac13\times \,\text{area of base}\, \times \,\text{height}. +

    + +

    + Certainly, using this formula from geometry is faster than our new method, + but the calculus-based method can be applied to much more than just cones. +

    +     + +     + +

    + An important special case of + is when the solid is a solid of revolution, that is, + when the solid is formed by rotating a shape around an axis. +

    + +

    + Start with a function y=f(x) from x=a to x=b. + Revolving this curve about a horizontal axis creates a three-dimensional solid whose cross sections are disks + (thin circles). + Let R(x) represent the radius of the cross-sectional disk at x; + the area of this disk is \pi R(x)^2. + Applying gives the Disk Method. +

    + + + The Disk Method +

    + Let a solid be formed by revolving the curve y=f(x) from x=a to x=b around a horizontal axis, + and let R(x) be the radius of the cross-sectional disk at x. + The volume of the solid is + integrationvolume!Disk Method + Disk Method + + V = \pi \int_a^b R(x)^2\, dx + . +

    +     + + + + + Finding volume using the Disk Method + +

    + Find the volume of the solid formed by revolving the curve y=1/x, + from x=1 to x=2, around the x-axis. +

    +     + +

    + A sketch can help us understand this problem. + In , the curve y=1/x is sketched along with the differential element a disk at x with radius R(x)=1/x. + In the whole solid is pictured, along with the differential element. +

    + +

    + The volume of the differential element shown in is approximately \pi R(x_i)^2\Delta x, + where R(x_i) is the radius of the disk shown and + \Delta x is the thickness of that slice. + The radius R(x_i) is the distance from the x-axis to the curve, + hence R(x_i) = 1/x_i. +

    + +
    + Sketching a solid in + +
    + + + + + +

    + Three-dimensional plot the curve y=1/x between x=1 and x=2 lying in the xy plane. + The volume of the inside of the rotated curve is then approximated using thin circular disks. + Taking an arbitrary circular disk, it would have a radius which whose length is equal to the distance between the x-axis and the curve y=1/x. + The image depicts a circular disk centered at some arbitrary value of x between x=1 and x=2, with the radius given by the function R(x)=1/x. +

    +     + A three-dimensional plot of the curve from the example showing a thin vertical slice being used to approximate its volume. + + + + + //ASY file for figdisk1.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((18,7.7,21.1),(0,1,0),(0,0,0),1,(-0.088,.0039)); + defaultrender.merge=true; + + + //(-0.01,0.061,-0.029) + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={.5,1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-0.1,2.2); + pair ybounds=(-1.2,1.2); + pair zbounds=(-1.2,1.2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + path3 p=(3/2,2/3,0)..(3/2,0,2/3)..(3/2,-2/3,0)..(3/2,0,-2/3); + draw(surface(p .. cycle), simplesurfacepen2); + draw(p..cycle,redpen+.4mm); + + triple g(real t) {return (t,1/t,0);} + path3 mypath=graph(g,1,2,operator ..); + draw(mypath,bluepen+linewidth(2)); + + draw((1.5,0,0)--(1.5,2/3,0),black+.3mm); + + triple pt=(1.3,Sin(-20),Cos(-20)); + + draw(pt--(1.48,.3,.05),linewidth(.75),Arrow3); + label("$R(x)=1/x$",pt,S); + + label("$y=1/x$",(1,1,0),N); + + //triple f(pair t) {return (t.x,1/t.x*cos(t.y),1/t.x*sin(t.y));} + //surface s=surface(f,(1,0),(2,2*pi),5,16,Spline); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + //path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); + //draw(surface(p1 -- cycle), simplesurfacepen); + //draw(p1--cycle,bluepen+.3mm); + + + + +
    + +
    + + + + + +

    + Three-dimensional plot the curve y=1/x between x=1 and x=2 lying in the xy plane. + The curve is then rotated in a full circle about the x-axis. + The solid is then bounded on both sides by the planes x=1/x and x=2/x which close off the solid with cirles of radius 1 and \frac12 respectively. + The volume of the inside of the curve is then approximated using thin circular disks which lie completely inside the solid as described in the previous image. +

    +     + A three-dimensional plot of the solid which comes from rotating the curve from the example. + + + + + + //ASY file for figdisk1b.asy in Chapter 7 + + size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((18,7.7,21.1),(0,1,0),(0,0,0),1,(-0.088,.0039)); + defaultrender.merge=true; + + //(-0.01,0.061,-0.029) + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={.5,1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-0.1,2.2); + pair ybounds=(-1.2,1.2); + pair zbounds=(-1.2,1.2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + path3 p2=(1,1,0)..(1,0,1)..(1,-1,0)..(1,0,-1); + draw(surface(p2 .. cycle), simplesurfacepen); + draw(p2..cycle,bluepen+.4mm); + + triple f(pair t) {return (t.x,1/t.x*cos(t.y),1/t.x*sin(t.y));} + surface s=surface(f,(1,0),(2,2*pi),5,16,Spline); + pen p=apexmeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + path3 p1=(2,.5,0)..(2,0,.5)..(2,-.5,0)..(2,0,-.5); + draw(surface(p1 .. cycle), simplesurfacepen); + draw(p1..cycle,bluepen+.4mm); + + path3 p=(3/2,2/3,0)..(3/2,0,2/3)..(3/2,-2/3,0)..(3/2,0,-2/3); + draw(surface(p .. cycle), simplesurfacepen2); + draw(p..cycle,redpen+.4mm); + + triple g(real t) {return (t,1/t,0);} + path3 mypath=graph(g,1.01,1.99,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); + //draw(surface(p1 -- cycle), simplesurfacepen); + //draw(p1--cycle,bluepen+.3mm); + + + + +
    +     + +
    + +

    + Slicing the solid into n equally-spaced slices, + we can approximate the total volume by adding up the approximate volume of each slice: + + \text{Approximate volume} = \sum_{i=1}^n \pi \left(\frac1{x_i}\right)^2\Delta x + . +

    + +

    + Taking the limit of the above sum as + n\to\infty gives the actual volume; + recognizing this sum as a Riemann sum allows us to evaluate the limit with a definite integral, + which matches the formula given in : + + V \amp = \lim_{n\to\infty}\sum_{i=1}^n \pi \left(\frac1{x_i}\right)^2\Delta x + \amp = \pi\int_1^2 \left(\frac1x\right)^2\, dx + \amp = \pi\int_1^2 \frac1{x^2}\, dx + + + \amp = \pi\left[-\frac1x\right]\Big|_1^2 + \amp = \pi \left[-\frac12 - \left(-1\right)\right] + \amp = \frac{\pi}{2}\,\text{units}^3 + . +

    +     + +     + + + + + + + + + +

    + While + is given in terms of functions of x, + the principle involved can be applied to functions of y when the axis of rotation is vertical, + not horizontal. + We demonstrate this in the next example. +

    + + + Finding volume using the Disk Method + +

    + Find the volume of the solid formed by revolving the curve y=1/x, + from x=1 to x=2, about the y-axis. +

    +     + +

    + Since the axis of rotation is vertical, + we need to convert the function into a function of y and convert the x-bounds to y-bounds. + Since y=1/x defines the curve, + we rewrite it as x=1/y. + The bound x=1 corresponds to the y-bound y=1, + and the bound x=2 corresponds to the y-bound y=1/2. +

    + +

    + Thus we are rotating the curve x=1/y, + from y=1/2 to y=1 about the y-axis to form a solid. + The curve and sample differential element are sketched in , + with a full sketch of the solid in . +

    + +
    + Sketching a solid in + +
    + + + + + +

    + Three-dimensional plot the curve y=1/x between x=1 and x=2 lying in the xy plane. + The volume of the inside of the rotated curve is approximated thin horizontal circular disks. + Taking an arbitrary circular disk, it would have a radius which whose length is equal to the distance between the y-axis and the curve x=1/y. + The image depicts a circular disk centered at some arbitrary value of y between y=\frac12 and y=1, with the radius given by the function R(y)=1/y. +

    +     + A three-dimensional plot of the curve from the example showing a thin horizontal slice being used to approximate its volume. + + + + + + //ASY file for figdisk1a.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((24.2,2.9,24.5),(0,1,0),(0,0,0),.95,(0.011,0.0033)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,-1,1,2}; + real[] myychoice={1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-2.3,2.3); + pair ybounds=(-.2,1.2); + pair zbounds=(-2.3,2.3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); + draw(surface(p .. cycle), simplesurfacepen2); + draw(p..cycle,redpen+.4mm); + + triple g(real t) {return (t,1/t,0);} + path3 mypath=graph(g,1,2,operator ..); + draw(mypath,bluepen+linewidth(2)); + + draw((0,2/3,0)--(1.5,2/3,0),black+.3mm); + + label("$x=1/y$",(1,1,0),N); + + triple pt=(1,.2,0); + draw(pt--(.75,.65,0),linewidth(.75),Arrow3); + label("$R(y)=1/y$",pt,S); + + triple f(pair t) {return (1/t.x*cos(t.y),t.x,1/t.x*sin(t.y));} + surface s=surface(f,(.5,0),(1,2*pi),5,16,Spline); + pen p=apexmeshpen+.1mm; + draw(s,invisible); + + //triple f(pair t) {return (t.x,1/t.x*cos(t.y),1/t.x*sin(t.y));} + //surface s=surface(f,(1,0),(2,2*pi),5,16,Spline); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + //path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); + //draw(surface(p1 -- cycle), simplesurfacepen); + //draw(p1--cycle,bluepen+.3mm); + + + + +
    + +
    + + + + + +

    + Three-dimensional plot the curve y=1/x from x=1 and x=2 lying in the xy plane. + The curve is then rotated in a full circle about the y-axis. + The solid is then bounded on both the top and bottom by the planes y=1 and y=\frac12. + The plane y=1 closes off the top of the solid with a circle of radius 1, and the plane y=\frac12 closes off the bottom of the solid with a circle of radius 2. + The volume of the inside of the curve is then approximated using thin circular disks which are contained completely inside the solid as described in the previous image. +

    +     + A three-dimensional plot of the solid which comes from rotating the curve from the example. + + + + + + //ASY file for figdisk2a.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((24.2,2.9,24.5),(0,1,0),(0,0,0),.95,(0.011,0.0033)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,-1,1,2}; + real[] myychoice={1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-2.3,2.3); + pair ybounds=(-.2,1.2); + pair zbounds=(-2.3,2.3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple g(real t) {return (t,1/t,0);} + path3 mypath=graph(g,1.005,1.99,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple f(pair t) {return (1/t.x*cos(t.y),t.x,1/t.x*sin(t.y));} + surface s=surface(f,(.5,0),(1,2*pi),5,16,Spline); + pen p=apexmeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + path3 p1=(1,1,0)..(0,1,1)..(-1,1,0)..(0,1,-1); + draw(surface(p1 .. cycle), simplesurfacepen); + draw(p1..cycle,bluepen+.4mm); + + path3 p2=(2,.5,0)..(0,.5,2)..(-2,.5,0)..(0,.5,-2); + draw(surface(p2 .. cycle), simplesurfacepen); + draw(p2..cycle,bluepen+.4mm); + + path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); + draw(surface(p .. cycle), simplesurfacepen2); + draw(p..cycle,redpen+.4mm); + + //triple f(pair t) {return (t.x,1/t.x*cos(t.y),1/t.x*sin(t.y));} + //surface s=surface(f,(1,0),(2,2*pi),5,16,Spline); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + //path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); + //draw(surface(p1 -- cycle), simplesurfacepen); + //draw(p1--cycle,bluepen+.3mm); + + + + +
    +     + +
    + +

    + We integrate to find the volume: + + V \amp = \pi\int_{1/2}^1 \frac{1}{y^2}\,dy + \amp = -\frac{\pi}y\Big|_{1/2}^1 + \amp = \pi\,\text{units}^3 + . +

    +     + +     + + + +

    + We can also compute the volume of solids of revolution that have a hole in the center. + The general principle is simple: + compute the volume of the solid irrespective of the hole, + then subtract the volume of the hole. + If the outside radius of the solid is R(x) and the inside radius + (defining the hole) + is r(x), then the volume is + + V = \pi\int_a^b R(x)^2 \,dx - \pi\int_a^b r(x)^2\,dx = \pi\int_a^b \left(R(x)^2-r(x)^2\right)\,dx + . +

    + +
    + Establishing the Washer Method; see also + +
    + + + + + +

    + Graph of two curves lying in the xy plane. + One of the curves is blue and resembles the peak of a wave. + The other curve is red, and resembles the trough of a wave. + Both curves are plotted between the points a and b on the x-axis. + For the interval [a,b] the blue curve lies entirely above the red curve. + Once the region between the blue and red curve is rotated about the y-axis, the blue curve which has a radius given by the function R(x) will be on the outside of the solid. + The red curve will be on the inside of the solid, which will have a radius given by the function r(x). + Both the functions R(x) and r(x) give the distance of the curve from the axis of rotation, which in this case is some arbitrarily chosen line paralell and above the x-axis. + The space on the inside of the red curve will then be completely empty once the region is rotated about the chosen axis of rotation. +

    +     + Graph of two curves which will be used to showcase the Washer Method. + + + + + //ASY file for figwasher_idea_a_3D.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.2,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-3.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*(-.5*(x-2)^2+3-((x-2)^2+1))+(x-2)^2+1}); + + //path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); + //draw(surface(p .. cycle), simplesurfacepen2); + //draw(p..cycle,redpen+.4mm); + + triple g1(real t) {return (t,((t-2)^2+1),0);} + path3 p1=graph(g1,1,3,operator ..); + draw(p1,redpen+.4mm); + + triple g2(real t) {return (t,(-.5*(t-2)^2+3),0);} + path3 p2=graph(g2,3,1,operator ..); + draw(p2,bluepen+.4mm); + + path3 p3=p1--p2; + draw(surface(p3 -- cycle), simplesurfacepen); + + draw((-.2,.5,0)--(3.5,.5,0),dashed+.2mm); + + draw((1,0,0)--(1,-.1,0)); + draw((3,0,0)--(3,-.1,0)); + label("$a$",(1,-.1,0),S); + label("$b$",(3,-.1,0),S); + + //\draw (axis cs:1.2,0,2.68) -- node [pos=.6,left] { $R(x)$} (axis cs:1.2,0,.5); + //\draw (axis cs: 2.8,0,1.64) -- node [pos=.5,right] { $r(x)$} (axis cs: 2.8,0,.5); + + triple dot1=(1.2,2.68,0); + triple dot2=(1.2,.5,0); + triple dot3=(2.8,1.64,0); + triple dot4=(2.8,.5,0); + + draw(dot1--dot2,black+.2mm); + draw(dot3--dot4,black+.2mm); + + dotfactor=2; + dot(dot1); + dot(dot2); + dot(dot3); + dot(dot4); + + label("$R(x)$",dot2+.3*(dot1-dot2),W); + label("$r(x)$",dot4+.4*(dot3-dot4),E); + + + + +
    + +
    + + + + + +

    + Three dimensional graph of the region between the blue and red curves on the interval [a,b] rotated about some line parallel and above x-axis. + The outside of the solid is coloured blue, which resembles the blue curve which has a radius given by the function R(x) being rotated about the axis of rotation. + The inside of the solid is coloured red, which resembles the red curve which has a radius by the function r(x) being rotated about the axis of rotation. + The space on the inside of the red region is completely empty, which makes us subtract the outside radius R(x) from the inside radius r(x) to create washers which will approximate the volume of the solid. +

    +     + Three-dimensional plot of solid coming from rotating the region between the two curves. + + + + + //ASY file for figwasher_idea_a_3D.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.2,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-3.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*(-.5*(x-2)^2+3-((x-2)^2+1))+(x-2)^2+1}); + + //path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); + //draw(surface(p .. cycle), simplesurfacepen2); + //draw(p..cycle,redpen+.4mm); + + pen p=apexmeshpen+.1mm; + + triple g1(real t) {return (t,((t-2)^2+1),0);} + path3 p1=graph(g1,1,3,operator ..); + draw(p1,redpen+.5mm); + + triple g2(real t) {return (t,(-.5*(t-2)^2+3),0);} + path3 p2=graph(g2,3,1,operator ..); + draw(p2,bluepen+.5mm); + + triple f2(pair t) {return (1,cos(t.x)*t.y+.5,sin(t.x)*t.y);} + surface s2=surface(f2,(0,1.5),(2*pi,2),16,1,Spline); + draw(s2,simplesurfacepen,meshpen=p); + + triple f3(pair t) {return (3,cos(t.x)*t.y+.5,sin(t.x)*t.y);} + surface s3=surface(f3,(0,1.5),(2*pi,2),16,1,Spline); + draw(s3,simplesurfacepen,meshpen=p); + + triple g3(real t) {return (1,1.5*cos(t)+.5,1.5*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,redpen+.3mm); + + triple g3(real t) {return (3,1.5*cos(t)+.5,1.5*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,redpen+.3mm); + + triple g3(real t) {return (1,2*cos(t)+.5,2*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.3mm); + + triple g3(real t) {return (3,2*cos(t)+.5,2*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.3mm); + + //path3 p3=p1--p2; + //draw(surface(p3 -- cycle), simplesurfacepen); + + draw((-.2,.5,0)--(3.5,.5,0),dashed+.2mm); + + triple f(pair t) {return (t.x,(-.5*(t.x-2)^2+3-.5)*cos(t.y)+.5,(-.5*(t.x-2)^2+3-.5)*sin(t.y));} + surface s=surface(f,(1,0),(3,2*pi),5,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + triple f1(pair t) {return (t.x,((t.x-2)^2+1-.5)*cos(t.y)+.5,((t.x-2)^2+1-.5)*sin(t.y));} + surface s1=surface(f1,(1,0),(3,2*pi),5,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s1,simplesurfacepen2,meshpen=p); + + //draw(p..cycle,redpen+.4mm); + + //label("$x=1/y$",(1,1,0),N); + + //triple pt=(1,.2,0); + //draw(pt--(.75,.65,0),linewidth(.75),Arrow3); + //label("$R(y)=1/y$",pt,S); + + //triple f(pair t) {return (1/t.x*cos(t.y),t.x,1/t.x*sin(t.y));} + //surface s=surface(f,(.5,0),(1,2*pi),5,16,Spline); + //pen p=apexmeshpen+.1mm; + //draw(s,invisible); + + //triple f(pair t) {return (t.x,1/t.x*cos(t.y),1/t.x*sin(t.y));} + //surface s=surface(f,(1,0),(2,2*pi),5,16,Spline); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + //path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); + //draw(surface(p1 -- cycle), simplesurfacepen); + //draw(p1--cycle,bluepen+.3mm); + + + + +
    +     + +
    + +

    + One can generate a solid of revolution with a hole in the middle by revolving a region about an axis. + Consider , + where a region is sketched along with a dashed, + horizontal axis of rotation. + By rotating the region about the axis, + a solid is formed as sketched in . + The outside of the solid has radius R(x), + whereas the inside has radius r(x). + Each cross section of this solid will be a washer (a disk with a hole in the center) as sketched in . This leads us to the Washer Method. +

    + +
    + Establishing the Washer Method; see also + + + + +

    + Three dimensional graph of the region between the blue and red curves on the interval [a,b]. + The graph also cotains a washer plotted on an arbitrary value of x in the interval [a,b], which will lie entirely inside the area between the blue and red curve. + The outside of the washer has a radius of R(x), while the inside of the washer has a radius of r(x). + In other words, the washer is a circle of radius R(x) with a circular hole of radius r(x) in the center. + Taking these washers to be arbitrarily thin, we can calculate the volume of the solid which comes from rotating the region. +

    +     + Three-dimensional plot of solid coming from rotating the region between the two curves. + + + + + //ASY file for figwasher_idea_a_3D.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.2,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-3.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*(-.5*(x-2)^2+3-((x-2)^2+1))+(x-2)^2+1}); + + //path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); + //draw(surface(p .. cycle), simplesurfacepen2); + //draw(p..cycle,redpen+.4mm); + + triple g1(real t) {return (t,((t-2)^2+1),0);} + path3 p1=graph(g1,1,3,operator ..); + draw(p1,redpen+.4mm); + + triple g2(real t) {return (t,(-.5*(t-2)^2+3),0);} + path3 p2=graph(g2,3,1,operator ..); + draw(p2,bluepen+.4mm); + + path3 p3=p1--p2; + draw(surface(p3 -- cycle), simplesurfacepen); + + draw((-.2,.5,0)--(3.5,.5,0),dashed+.2mm); + + draw((1,0,0)--(1,-.1,0)); + draw((3,0,0)--(3,-.1,0)); + label("$a$",(1,-.1,0),S); + label("$b$",(3,-.1,0),S); + + //\draw (axis cs:1.2,0,2.68) -- node [pos=.6,left] { $R(x)$} (axis cs:1.2,0,.5); + //\draw (axis cs: 2.8,0,1.64) -- node [pos=.5,right] { $r(x)$} (axis cs: 2.8,0,.5); + + triple dot1=(1.2,2.68,0); + triple dot2=(1.2,.5,0); + triple dot3=(2.8,1.64,0); + triple dot4=(2.8,.5,0); + + draw(dot1--dot2,black+.2mm); + draw(dot3--dot4,black+.2mm); + + dotfactor=2; + dot(dot1); + dot(dot2); + dot(dot3); + dot(dot4); + + label("$R(x)$",dot2+.3*(dot1-dot2),W); + label("$r(x)$",dot4+.4*(dot3-dot4),E); + + triple f2(pair t) {return (2,cos(t.x)*t.y+.5,sin(t.x)*t.y);} + surface s2=surface(f2,(0,.5),(2*pi,2.5),16,1,Spline); + draw(s2,simplesurfacepen2); + + triple g3(real t) {return (2,.5*cos(t)+.5,.5*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,redpen+.4mm); + + triple g3(real t) {return (2,2.5*cos(t)+.5,2.5*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.4mm); + + draw((2,1,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); + + + + +
    + + + The Washer Method +

    + Let a region bounded by y=f(x), y=g(x), + x=a and x=b be rotated about a horizontal axis that does not intersect the region, + forming a solid. + Each cross section at x will be a washer with outside radius R(x) and inside radius r(x). + The volume of the solid is + integrationvolume!Washer Method + Washer Method + + V = \pi\int_a^b \Big(R(x)^2-r(x)^2\Big)\, dx + . +

    +     + +

    + Even though we introduced it first, + the Disk Method is just a special case of the Washer Method with an inside radius of r(x)=0. +

    + + + Finding volume with the Washer Method + +

    + Find the volume of the solid formed by rotating the region bounded by + y=x^2-2x+2 and y=2x-1 about the x-axis. +

    +     + +

    + A sketch of the region will help, + as given in . + Rotating about the x-axis will produce cross sections in the shape of washers, + as shown in ; + the complete solid is shown in . + The outside radius of this washer is R(x) = 2x-1; + the inside radius is r(x) = x^2-2x+2. + As the region is bounded from x=1 to x=3, + we integrate as follows to compute the volume. + + V \amp = \pi\int_1^3 \Big((2x-1)^2-(x^2-2x+2)^2\Big)\, dx + \amp = \pi\int_1^3 \big(-x^4+4x^3-4x^2+4x-3\big)\, dx + \amp = \pi\Big[-\frac{1}{5}x^5+x^4-\frac43x^3+2x^2-3x\Big]\Big|_1^3 + \amp =\frac{104}{15}\pi \approx 21.78\,\text{units}^3 + . +

    + +
    + Sketching the differential element and solid in + +
    + + + + + +

    + Graph of the region enclosed by the curves y=x^2-2x+2 and y=2x-1. + The curves which create an enclosed region both start at the point (1,1) and end at the point (3,5). + The line y=2x-1 lies entirely above the curve y=x^2-2x+2 for the entirety of the enclosed region. + The radius of the outside of the washer is then given as distance between the x-axis and the line y=2x-1, and is given by the function R(x)=2x-1. + The inside radius of the washer is given as the distance between the x-axis and the inside curve y=x^2-2x+2, and is given by the function r(x)=x^2-2x+2. +

    +     + Graph of the region bounded by the two curves lying in the xy plane. + + + + + //ASY file for figwash1_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3}; + real[] myychoice={-5,5}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.2,3.3); + pair ybounds=(-5.6,5.6); + pair zbounds=(-5.6,5.6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); + + triple g1(real t) {return (t,(2*t-1),0);} + path3 p1=graph(g1,1,3,operator ..); + draw(p1,bluepen+.4mm); + + triple g2(real t) {return (t,(t^2-2*t+2),0);} + path3 p2=graph(g2,3,1,operator ..); + draw(p2,redpen+.4mm); + + path3 p3=p1--p2; + draw(surface(p3 -- cycle), simplesurfacepen); + + + + +
    + +
    + + + + + +

    + Graph of the region enclosed by the curves y=x^2-2x+2 and y=2x-1 with a washer centered at x=2. + The washer lies parallel to the y-axis, and has an outside radius of R(2)=3 and an inside radius of r(2)=2. + In the original graph of the functions, R(2)=3 is the distance between the line y=2x-1 and the x-axis, and r(2)=2 is the distance between the curve y=x^2-2x+2 and the x-axis. + The washer will lie entirely in the space enclosed by rotating the region between the two curves. +

    +     + Graph of the region bounded by the two curves with an arbitrary washer used to calculate the volume of the solid. + + + + + //ASY file for figwash1_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3}; + real[] myychoice={-5,5}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.2,3.3); + pair ybounds=(-5.6,5.6); + pair zbounds=(-5.6,5.6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); + + triple g1(real t) {return (t,(2*t-1),0);} + path3 p1=graph(g1,1,3,operator ..); + draw(p1,bluepen+.4mm); + + triple g2(real t) {return (t,(t^2-2*t+2),0);} + path3 p2=graph(g2,3,1,operator ..); + draw(p2,redpen+.4mm); + + path3 p3=p1--p2; + draw(surface(p3 -- cycle), simplesurfacepen); + + triple f2(pair t) {return (2,cos(t.x)*t.y,sin(t.x)*t.y);} + surface s2=surface(f2,(0,2),(2*pi,3),16,1,Spline); + draw(s2,simplesurfacepen2); + + triple g3(real t) {return (2,2*cos(t),2*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,redpen+.4mm); + + triple g3(real t) {return (2,3*cos(t),3*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.4mm); + + draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); + + + + +
    + +
    + + + + + +

    + Three-dimensional graph of the space enclosed by rotating the curves y=x^2-2x+2 and y=2x-1 about the x-axis. + The outside line y=2x-1 forms the outer border of the solid once rotated about the x-axis, whose surface is coloured in blue. + The inside curve y=x^2-2x+2 forms the inner border of the solid once rotated about the x-axis, whose surface is coloured in red. + The washers having outside radius R(x)=2x-1 and inner radius r(x)=x^2-2x+2 will lie entirely in the space enclosed by rotating the region between the two curves. +

    +     + Three-dimensional graph of the solid formed by rotating the region bounded by the two curves about the x axis. + + + + + //ASY file for figwash1_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3}; + real[] myychoice={-5,5}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.2,3.3); + pair ybounds=(-5.6,5.6); + pair zbounds=(-5.6,5.6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); + + pen p=apexmeshpen+.1mm; + pen q=redmeshpen+.1mm; + + triple g1(real t) {return (t,(2*t-1),0);} + path3 p1=graph(g1,1,3,operator ..); + draw(p1,bluepen+.4mm); + + triple g2(real t) {return (t,(t^2-2*t+2),0);} + path3 p2=graph(g2,3,1,operator ..); + draw(p2,redpen+.4mm); + + path3 p3=p1--p2; + draw(surface(p3 -- cycle), simplesurfacepen); + + triple f2(pair t) {return (t.x,cos(t.y)*(2*t.x-1),sin(t.y)*(2*t.x-1));} + surface s2=surface(f2,(1,0),(3,2*pi),4,16,Spline); + draw(s2,simplesurfacepen,meshpen=p); + + triple f2(pair t) {return (t.x,cos(t.y)*(t.x^2-2*t.x+2),sin(t.y)*(t.x^2-2*t.x+2));} + surface s2=surface(f2,(1,0),(3,2*pi),4,16,Spline); + draw(s2,simplesurfacepen2,meshpen=q); + + triple g3(real t) {return (1,cos(t),sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,black+.4mm); + + triple g3(real t) {return (3,5*cos(t),5*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,black+.4mm); + + + + +
    +     + +
    +     + +     + + + +

    + When rotating about a vertical axis, + the outside and inside radius functions must be functions of y. +

    + + + Finding volume with the Washer Method + +

    + Find the volume of the solid formed by rotating the triangular region with vertices at (1,1), + (2,1) and (2,3) about the y-axis. +

    +     + +

    + The triangular region is sketched in ; + the differential element is sketched in and the full solid is drawn in . + They help us establish the outside and inside radii. + Since the axis of rotation is vertical, + each radius is a function of y. +

    + +

    + The outside radius R(y) is formed by the line connecting (2,1) and (2,3); + it is a constant function, + as regardless of the y-value the distance from the line to the axis of rotation is 2. + Thus R(y)=2. +

    + +
    + Sketching the solid in + +
    + + + + + +

    + Graph of the triangular region with vertices at (1,1), (2,1) and (2,3). + As we are rotating about the y-axis, the outer radius is the outermost edge of the triangle, which is a horizontal line at x=2, giving us a constant R(y)=2. + The inner radius is formed by the line between (1,1) and (2,3), which is the line y=2x-1. + This gives us an inner radius r(y)=\frac12(y+1). + The triangular region which we will rotate about the y-axis then consists of the area to the right of the line y=2x-1 and to the left of the horizontal line x=2 for 1 \leq y \leq 3. +

    +     + Graph of the triangular region showing the inner and outer radius functions. + + + + + //ASY file for figwasher_idea_a_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0.029,-0.015)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,-2,1,2}; + real[] myychoice={1,2,3}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-2.2,2.2); + pair ybounds=(-.2,3.2); + pair zbounds=(-2.2,2.2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (axis cs:1,0,1) -- (axis cs:2,0,1)--(axis cs:2,0,3) -- cycle; + + //path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); + //draw(surface(p .. cycle), simplesurfacepen2); + //draw(p..cycle,redpen+.4mm); + + path3 p1=(1,1,0)--(2,1,0)--(2,3,0); + draw(surface(p1 -- cycle), simplesurfacepen); + draw(p1--cycle,bluepen+.4mm); + + //\draw (axis cs: 0,0,1.2) -- (axis cs:2,0,1.2); + //\draw (axis cs: .8,0,.5) node { $R(y)=2$}; + //\filldraw (axis cs:0,0,1.2) circle (1pt) + // (axis cs:2,0,1.2) circle (1pt) + // (axis cs:0,0,2.6) circle (1pt) + // (axis cs:1.8,0,2.6) circle (1pt); + //\draw (axis cs: 0,0,2.6) -- (axis cs:1.8,0,2.6); + //\draw (axis cs: 1.5,0,3.2) node { $r(y)=\frac12(y+1)$}; + + triple dot1=(0,1.2,0); + triple dot2=(2,1.2,0); + triple dot3=(0,2.6,0); + triple dot4=(1.8,2.6,0); + + draw(dot1--dot2,black+.2mm); + draw(dot3--dot4,black+.2mm); + + dotfactor=2; + dot(dot1); + dot(dot2); + dot(dot3); + dot(dot4); + + label("$R(y)$",dot2+.7*(dot1-dot2),S); + label("$r(y)$",dot4+.7*(dot3-dot4),N); + + + +
    + + + +
    + + + +

    + Graph of the triangular region with vertices at (1,1), (2,1) and (2,3) with a washer centered at y=2. + The washer lies parallel to the x-axis, and has an outside radius of R(2)=2 and an inside radius of r(2)=\frac32. + In the original graph of the functions, R(2)=2 is the distance between the rightmost edge of the triangular region and the y-axis. + Similarly, r(2)=\frac32 is the distance betweeen the leftmost edge of the triangular region given by the line y=2x-1, and the y-axis + The washer will lie entirely in the space enclosed by rotating the region between the two curves as described in the following image. +

    +     + Graph of the triangular region with an arbitrary washer which will be used to calculate the volume of the solid. + + + + + //ASY file for figwasher_idea_a_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0.029,-0.015)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,-2,1,2}; + real[] myychoice={1,2,3}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-2.2,2.2); + pair ybounds=(-.2,3.2); + pair zbounds=(-2.2,2.2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (axis cs:1,0,1) -- (axis cs:2,0,1)--(axis cs:2,0,3) -- cycle; + + //path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); + //draw(surface(p .. cycle), simplesurfacepen2); + //draw(p..cycle,redpen+.4mm); + + path3 p1=(1,1,0)--(2,1,0)--(2,3,0); + draw(surface(p1 -- cycle), simplesurfacepen); + draw(p1--cycle,bluepen+.4mm); + + //\draw (axis cs: 0,0,1.2) -- (axis cs:2,0,1.2); + //\draw (axis cs: .8,0,.5) node { $R(y)=2$}; + //\filldraw (axis cs:0,0,1.2) circle (1pt) + // (axis cs:2,0,1.2) circle (1pt) + // (axis cs:0,0,2.6) circle (1pt) + // (axis cs:1.8,0,2.6) circle (1pt); + //\draw (axis cs: 0,0,2.6) -- (axis cs:1.8,0,2.6); + //\draw (axis cs: 1.5,0,3.2) node { $r(y)=\frac12(y+1)$}; + + triple g1(real t) {return (1.5*cos(t),2,1.5*sin(t));} + path3 p1=graph(g1,0,2*pi,operator ..); + draw(p1,redpen+.4mm); + + triple g1(real t) {return (2*cos(t),2,2*sin(t));} + path3 p1=graph(g1,0,2*pi,operator ..); + draw(p1,redpen+.4mm); + + triple f2(pair t) {return (cos(t.x)*t.y,2,sin(t.x)*t.y);} + surface s2=surface(f2,(0,1.5),(2*pi,2),16,1,Spline); + draw(s2,simplesurfacepen2); + + draw((1.5,2,0)--(2,2,0),purple+linetype(new real[] {4,4})+.3mm); + + + +
    + + + +
    + + + +

    + Three-dimensional graph of the space enclosed by rotating the triangular region about the y-axis. + The outside line x=2 forms the outer border of the solid once rotated about the y-axis, whose surface is coloured in blue. + The inside curve y=2x-1, or written in terms of y as x=\frac12(y+1) forms the inner border of the solid once rotated about the y-axis, whose surface is coloured in red. + The washers having outside radius R(y)=2 and inner radius r(y)=\frac12(y+1) will lie entirely in the space enclosed by rotating the region between the two curves and will allow us to calculate the volume of this solid. +

    +     + Graph of the triangular region with an arbitrary washer which will be used to calculate the volume of the solid. + + + + + //ASY file for figwasher_idea_a_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0.029,-0.015)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,-2,1,2}; + real[] myychoice={1,2,3}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-2.2,2.2); + pair ybounds=(-.2,3.2); + pair zbounds=(-2.2,2.2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (axis cs:1,0,1) -- (axis cs:2,0,1)--(axis cs:2,0,3) -- cycle; + + pen p=apexmeshpen+.1mm; + pen q=redmeshpen+.1mm; + + path3 p1=(1,1,0)--(2,1,0)--(2,3,0); + draw(surface(p1 -- cycle), simplesurfacepen); + draw(p1--cycle,bluepen+.4mm); + + //\draw (axis cs: 0,0,1.2) -- (axis cs:2,0,1.2); + //\draw (axis cs: .8,0,.5) node { $R(y)=2$}; + //\filldraw (axis cs:0,0,1.2) circle (1pt) + // (axis cs:2,0,1.2) circle (1pt) + // (axis cs:0,0,2.6) circle (1pt) + // (axis cs:1.8,0,2.6) circle (1pt); + //\draw (axis cs: 0,0,2.6) -- (axis cs:1.8,0,2.6); + //\draw (axis cs: 1.5,0,3.2) node { $r(y)=\frac12(y+1)$}; + + triple g1(real t) {return (2*cos(t),3,2*sin(t));} + path3 p1=graph(g1,0,2*pi,operator ..); + draw(p1,bluepen+.4mm); + + triple g1(real t) {return (2*cos(t),1,2*sin(t));} + path3 p1=graph(g1,0,2*pi,operator ..); + draw(p1,bluepen+.4mm); + + triple g1(real t) {return (1*cos(t),1,1*sin(t));} + path3 p1=graph(g1,0,2*pi,operator ..); + draw(p1,redpen+.4mm); + + triple f2(pair t) {return (cos(t.x)*t.y,1,sin(t.x)*t.y);} + surface s2=surface(f2,(0,1),(2*pi,1),16,1,Spline); + draw(s2,simplesurfacepen); + + triple f2(pair t) {return (cos(t.x)*(2),t.y,sin(t.x)*2);} + surface s2=surface(f2,(0,1),(2*pi,3),16,2,Spline); + draw(s2,simplesurfacepen,meshpen=p); + + triple f2(pair t) {return (cos(t.x)*(t.y/2+1/2),t.y,sin(t.x)*(t.y/2+1/2));} + surface s2=surface(f2,(0,1),(2*pi,3),16,2,Spline); + draw(s2,simplesurfacepen2,meshpen=q); + + + +
    + +     +
    + +

    + The inside radius is formed by the line connecting (1,1) and (2,3). + The equation of this line is y=2x-1, + but we need to refer to it as a function of y. + Solving for x gives r(y) = \frac12(y+1). +

    + +

    + We integrate over the y-bounds of y=1 to y=3. + Thus the volume is + + V \amp = \pi\int_1^3\Big(2^2 - \big(\frac12(y+1)\big)^2\Big)\, dy + \amp = \pi\int_1^3\Big(-\frac14y^2-\frac12y+\frac{15}4\Big)\, dy + \amp = \pi\Big[-\frac1{12}y^3-\frac14y^2+\frac{15}4y\Big]\Big|_1^3 + \amp = \frac{10}3\pi \approx 10.47\,\text{units}^3 + . +

    +     + +     + +

    + This section introduced a new application of the definite integral. + Our default view of the definite integral is that it gives + the area under the curve. However, + we can establish definite integrals that represent other quantities; + in this section, we computed volume. +

    + +

    + The ultimate goal of this section is not to compute volumes of solids. + That can be useful, + but what is more useful is the understanding of this basic principle of integral calculus, + outlined in : + to find the exact value of some quantity, +

    + +

    +

      +
    • +

      + we start with an approximation (in this section, + slice the solid and approximate the volume of each slice), +

      +
      • + +
    • +

      + then make the approximation better by refining our original approximation (, use more slices), +

      +
      • + +
    • +

      + then use limits to establish a definite integral which gives the exact value. +

      +
      • +
    +

    + +

    + We practice this principle in the next section where we find volumes by slicing solids in a different way. +

    + + + + Terms and Concepts + + + + +

    + A solid of revolution is formed by revolving a shape around an axis. +

    +     + +     + + + + +

    + In your own words, + explain how the Disk and Washer Methods are related. +

    +     + + + +     + + + + +

    + Explain the how the units of volume are found in the integral of : + if A(x) has units of \text{in}^2, + how does \int A(x)\,dx have units of \text{in}^3? +

    +     + + + +

    + Recall that dx does not just sit there; + it is multiplied by A(x) and represents the thickness of a small slice of the solid. + Therefore dx has units of in, + giving A(x)\,dx the units of in^3. +

    +     + +     +     + + Problems + + + +

    + Use the Disk/Washer Method to find the volume of the solid of revolution + formed by revolving the given region about the x-axis. +

    +     + + + + +

    + The region between y=3-x^2 and the x axis: +

    + + + +

    + Graph of the region bounded by the curve y=3-x^2 and the x-axis. + The curve y=3-x^2 begins at x-axis at the point (-\sqrt{3},0). + From this point the curve rises until reaching the y-axis at the point (0,3). + After reaching the y-axis, the curve slopes down until reaching the x-axis at the point (\sqrt{3},0). + The region then contains the entire area below this curve, and above the x-axis. +

    +     + Graph of the region bounded by the curve and the x axis. + + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=3.5, + xmin=-2.1,xmax=2.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=-1.73:1.73] {3-x^2}; + \addplot [firstcurvestyle,domain=-1.73:1.73,samples=50] {3-x^2} node [shift={(-15pt,90pt)} ,black] { $y=3-x^2$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + 48\pi\sqrt{3}/5 units^3 +

    +     + +     + + + + +

    + The region between y=5x and the x axis, + for 1\leq x\leq 2: +

    + + + + +

    + Graph of the region between y=5x and the x-axis, for 1\leq x\leq 2. + The line y=5x begins at the point (1,5) from which it linearly increases until reaching the rightmost bound which is at the point (2,10). + The region then contains the entire area below the line y=5x and above the x-axis for 1\leq x\leq 2. +

    +     + Graph of the region bounded by the the line y=5x and the x-axis for x between 1 and 2. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=11, + xmin=-.1,xmax=2.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:2] {5*x} \closedcycle; + \addplot [firstcurvestyle,domain=0:2.1] {5*x} node [shift={(-30pt,-5pt)},black] { $y=5x$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + 175\pi/3 units^3 +

    +     + +     + + + + +

    + The region between y=\cos(x) and the x axis, + for 0\leq x\leq \pi/2: +

    + + + + +

    + Graph of the region between y=\cos(x) and the x-axis, for 0\leq x\leq \pi/2. + The curve y=\cos(x) begins at the point (0,1) from which it slopes downwards until ending after reaching the x-axis at the point (\pi/2,0). + The region then contains the entire area below the curve y=\cos(x) and above the x-axis for 0\leq x\leq \pi/2. +

    +     + Graph of the region bounded by the the cosine function and the two coordinate axes. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.7 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1.57] {cos(deg(x))} \closedcycle; + \addplot [firstcurvestyle,domain=0:1.57] {cos(deg(x))} node [shift={(-20pt,65pt)},black] { $y=\cos(x) $}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \pi^2/4 units^3 +

    +     + +     + + + + +

    + The region between the curves y=x and y=\sqrt{x}: +

    + + + + +

    + Graph of the region between the curves y=x and y=\sqrt{x}. + The curves y=x and y=\sqrt{x} both begin at the origin. + From this point the curve y=\sqrt{x} rises above the line y=x until reaching the point (1,1), where the curve once again intersects the line. + The region then contains the area below the curve y=\sqrt{x} and above the line y=x. +

    +     + Graph of the region bounded by the two curves. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1,samples=60] {sqrt(x)}; + + \addplot [firstcurvestyle,domain=0:.05] {sqrt(x)}; + \addplot [firstcurvestyle,domain=.05:1,samples=30] {sqrt(x)} node [shift={(-55pt,-5pt)} ,black] { $y=\sqrt{x}$}; + + \addplot [firstcurvestyle,domain=0:1] {x} node [shift={(-30pt,-45pt)},black] { $y=x$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \pi/6 units^3 +

    +     + +     + +     + + + +

    + Use the Disk/Washer Method to find the volume of the solid of revolution + formed by revolving the given region about the y-axis. +

    +     + + + + +

    + The region bounded by the curve y=3-x^2, + the x axis, and the y axis: +

    + + + + +

    + Graph of the region bounded by the curve y=3-x^2, the x-axis, and the y-axis. + Note that this region can reference both the regions to the left or right of the y-axis, but we will consider the region to the right of the y-axis. + The curve y=3-x^2 begins at the point (0,3) from which it slopes down until reaching the x-axis. + The region then contains the area to the right of the y-axis, and to the left of the curve y=3-x^2 for y values between 0 and 3. +

    +     + Graph of the region lying in the first quadrant bounded by the curve and the two coordinate axes. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=3.5, + xmin=-2.1,xmax=2.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1.73] {3-x^2} \closedcycle; + \addplot [firstcurvestyle,domain=-1.73:1.73,samples=50] {3-x^2} node [shift={(-15pt,90pt)},black] { $y=3-x^2$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + 9\pi/2 units^3 +

    +     + +     + + + + +

    + The region between y=5x and the y axis, + for 5\leq y\leq 10: +

    + + + + +

    + Graph of the region between y=5x and the y axis, for 5\leq y\leq 10. + The line y=5x begins at the point (1,5), from which it linearly increases until ending upper bound for y at the point (2,10). + The region then contains the entire area to the right of x=0 and to the left of the line y=5x for y between 5 and 10. +

    +     + Graph of the region bounded by the line and the y axis for y values between 5 and 10. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=11, + xmin=-.1,xmax=2.1 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(1,5) (2,10) (0,10) (0,5) (1,5)}; + \addplot [firstcurvestyle,domain=0:2.1] {5*x} node [shift={(-20pt,-40pt)},black] { $y=5x$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + 35\pi/3 units^3 +

    +     + +     + + + + +

    + The region between y=\cos(x) and the x axis, + for 0\leq x\leq \pi/2: +

    + +

    + (Hint: Integration By Parts will be necessary, twice. + First let u = \arccos^2x, + then let u=\arccos x.) +

    + + + +

    + Graph of the region between y=\cos(x) and the x axis, for 0\leq x\leq \pi/2. + The curve y=\cos(x) begins at the point (0,1) from which it slopes downwards until ending after reaching the x-axis at the point (\pi/2,0). + The region then contains the entire area to the right of x=0 and to the left of of the curve y=\cos(x) for y values between 0 and 1. +

    +     + Graph of the region bounded by the cosine curve and the coordinate axes. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.7 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1.57] {cos(deg(x))} \closedcycle; + \addplot [firstcurvestyle,domain=0:1.57] {cos(deg(x))} node [shift={(-20pt,65pt)},black] { $y=\cos(x) $}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \pi^2-2\pi units^3 +

    +     + +     + + + + +

    + The region between the curves y=x and y=\sqrt{x}: +

    + + + + +

    + Graph of the region between the curves y=x and y=\sqrt{x}. + The curves y=x and y=\sqrt{x} both begin at the origin. + From this point the curve y=\sqrt{x} rises above the line y=x until reaching the point (1,1), where the curve once again intersects the line. + The region consists of the area to the right of the curve y=\sqrt{x} and the to the left line y=x for y values between 0 and 1. +

    +     + Graph of the region bounded by the two curves. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1,samples=60] {sqrt(x)}; + + \addplot [firstcurvestyle,domain=0:.05] {sqrt(x)}; + \addplot [firstcurvestyle,domain=.05:1,samples=30] {sqrt(x)} node [shift={(-60pt,-10pt)} ,black] { $y=\sqrt{x}$}; + + \addplot [firstcurvestyle,domain=0:1] {x} node [shift={(-30pt,-45pt)},black] { $y=x$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + 2\pi/15 units^3 +

    +     + +     + +     + + + +

    + Use the Disk/Washer Method to find the volume of the solid of revolution + formed by rotating the given region about each of the given axes. +

    +     + + + + +

    + Region bounded by: y=\sqrt{x}, + y=0 and x=1. +

    +     + + + +

    + Rotate about the x axis. +

    +     + +

    + \pi/2 +

    +     +     + + + +

    + Rotate about y=1. +

    +     + +

    + 5\pi/6 +

    +     +     + + + +

    + Rotate about the y axis. +

    +     + +

    + 4\pi/5 +

    +     +     + + + +

    + Rotate about x=1. +

    +     + +

    + 8\pi/15 +

    +     +     + +     + + + + +

    + Region bounded by: y=4-x^2 and y=0. +

    +     + + + +

    + Rotate about the x axis. +

    +     + +

    + 512\pi/15 +

    +     +     + + + +

    + Rotate about y=4. +

    +     + +

    + 256\pi/5 +

    +     +     + + + +

    + Rotate about y=-1. +

    +     + +

    + 832\pi/15 +

    +     +     + + + +

    + Rotate about x=2. +

    +     + +

    + 128\pi/3 +

    +     +     + +     + + + + +

    + The triangle with vertices (1,1), + (1,2) and (2,1). +

    +     + + + +

    + Roate about the x axis. +

    +     + +

    + 4\pi/3 +

    +     +     + + + +

    + Roate about y=2. +

    +     + +

    + 2\pi/3 +

    +     +     + + + +

    + Rotate about the y axis. +

    +     + +

    + 4\pi/3 +

    +     +     + + + +

    + Rotate about x=1. +

    +     + +

    + \pi/3 +

    +     +     + +     + + + + +

    + Region bounded by y=x^2-2x+2 and y=2x-1. +

    +     + + + +

    + Rotate about the x axis. +

    +     + +

    + 104\pi/15 +

    +     +     + + + +

    + Rotate about y=1. +

    +     + +

    + 64\pi/15 +

    +     +     + + + +

    + Rotate about y=5. +

    +     + +

    + 32\pi/5 +

    +     +     + +     + + + + +

    + Region bounded by y=1/\sqrt{x^2+1}, x=-1, + x=1 and the x-axis. +

    +     + + + +

    + Rotate about the x axis. +

    +     + +

    + \pi^2/2 +

    +     +     + + + +

    + Rotate about y=1. +

    +     + +

    + \pi^2/2-4\pi\sinh^{-1}(1) +

    +     +     + + + +

    + Rotate about y=-1. +

    +     + +

    + \pi^2/2+4\pi\sinh^{-1}(1) +

    +     +     + +     + + + + +

    + Region bounded by y=2x, + y=x and x=2. +

    +     + + + +

    + Rotate about the x axis. +

    +     + +

    + 8\pi +

    +     +     + + + +

    + Rotate about y=4. +

    +     + +

    + 8\pi +

    +     +     + + + +

    + Rotate about the y axis. +

    +     + +

    + 16\pi/3 +

    +     +     + + + +

    + Rotate about x=2. +

    +     + +

    + 8\pi/3 +

    +     +     + +     + +     + + + +

    + Orient the given solid along the x-axis such that a cross-sectional area function A(x) can be obtained, + then apply + to find the volume of the solid. +

    +     + + + + +

    + A right circular cone with height of 10 and base radius of 5. +

    + + + + +

    + Image of a right circular cone with height of 10 and base radius of 5. + The base of the cone is a circle of radius 5. + From the center of the circle to the peak of the cone we measure the height to be 10. +

    +     + An image of a right circular cone with height of 10 and base radius of 5. + + + \begin{tikzpicture}[scale=.8] + + \begin{scope}[xscale=2] + + \draw [thick] (-1,0) arc (180:360:1); + \draw [thick,dashed] (1,0) arc (0:180:1); + + \end{scope} + + \draw [fill=black] (0,0) circle (1pt) -- node [pos=.5,above] {5} (2,0); + \draw (0,0) -- node [pos=.5,rotate=90,above] {10} (0,4); + \draw [thick] (-2,0) -- (0,4)-- (2,0); + + \end{tikzpicture} + + + + + +     + +

    + 250\pi/3 +

    +     + +

    + Placing the tip of the cone at the origin such that the x-axis runs through the center of the circular base, + we have A(x)=\pi x^2/4. + Thus the volume is 250\pi/3 units^3. +

    +     + +     + + + + +

    + A skew right circular cone with height of 10 and base radius of 5. (Hint: + all cross-sections are circles.) +

    + + + + +

    + Image of a skew right circular cone with height of 10 and base radius of 5. + The base of the cone is a circle of radius 5. + From the rightmost edge of the circle to the peak of the cone we measure the height to be 10. +

    +     + An image of a skew right circular cone with height of 10 and base radius of 5. + + + \begin{tikzpicture}[scale=.8] + + \begin{scope}[xscale=2] + + \draw [thick] (-1,0) arc (180:360:1); + \draw [thick,dashed] (1,0) arc (0:180:1); + \draw [dashed] (.4,1.7) circle (.6); + + \end{scope} + + \draw [fill=black] (0,0) circle (1pt) -- node [pos=.5,above] { 5} (2,0); + \draw [thick] (-2,0.2) -- (2,4)-- node [pos=.5,rotate=-90,above] { 10}(2,0); + + \end{tikzpicture} + + + + + +     + +

    + 250\pi/3 +

    +     + +

    + The cross-sections of this cone are the same as the cone in . + Thus they have the same volume of 250\pi/3 units^3. +

    +     + +     + + + + +

    + A right triangular cone with height of 10 and whose base is a right, + isosceles triangle with side length 4. +

    + + + +

    + Image of a right triangular cone with height of 10 and whose base is a right isosceles triangle with side length 4. + The base of the cone is a right isosceles triangle having two side lengths of 4. + The two sides of length 4 are connected by a right angle. + Additionally, the distance from the right angle occuring at the connection of the two sides of length 4 from the base of the triangle to the peak of the cone is measured to be 10. +

    +     + Image of a right triangular cone whose base is a right isosceles triangle. + + + \begin{tikzpicture}[scale=1.58] + + \draw [thick](-1,0) -- (1,0) -- (0,3)--cycle; + \draw [thick,dashed] (-1,0) -- node [pos=.5,above] {4} (0,.5) -- node [pos=.5,above] {4} (1,0) + (0,.5) -- node [pos=.3,above,rotate=90] {10} (0,3); + \draw (-.2,.4) -- (0,.3) -- (.2,.4); + + \end{tikzpicture} + + + +     + +

    + 80/3 +

    +     + +

    + Orient the cone such that the tip is at the origin and the x-axis is perpendicular to the base. + The cross-sections of this cone are right, + isosceles triangles with side length 2x/5; + thus the cross-sectional areas are A(x) = 2x^2/25, + giving a volume of 80/3 units^3. +

    +     + +     + + + + +

    + A solid with length 10 with a rectangular base and triangular top, + wherein one end is a square with side length 5 and the other end is a triangle with base and height of 5. +

    + + + +

    + Image of a solid with length 10 with a rectangular base and triangular top, wherein one end is a square with side length 5 and the other end is a triangle with base and height of 5.. + The rectangular base measures a length of 10 and a depth of 5. + The square of side length 5 connects to the rectangle on the left side of the solid. + The triangle with a base length and height of 5 connects to the rectangle on the right side of the solid. + The square and triangle are then connected by two slanted trapezoidal regions on the front and back of the solid. + On top of the solid is a triangle whose base of length 5 is connected to the square, having a height of 10. +

    +     + Image solid with a rectangular base, and a square and triangular sides which are connected by two trapezoids and a triangle. + + + \begin{tikzpicture}[x={(1,0)},z={(0,1)},y={(.5,.87)},scale=.35] + + \draw [thick] (0,0,0) -- node[pos=.5,below] {10} (10,0,0) -- (10,2.5,5) -- (10,5,0) -- node [pos=.5,below right] { 5} (10,0,0) + (0,0,0) -- node [pos=.5,left] {5} (0,0,5) -- (10,2.5,5) -- (0,5,5) -- node [pos=.5,left] { 5} (0,0,5); + + \draw [thick,dashed] (0,0,0) -- (0,5,0) -- (0,5,5) + (0,5,0) -- (10,5,0); + + \end{tikzpicture} + + + +     + +

    + 187.5 +

    +     + +

    + Orient the solid so that the x-axis is parallel to long side of the base. + All cross-sections are trapezoids + (at the far left, the trapezoid is a square; + at the far right, + the trapezoid has a top length of 0, making it a triangle). + The area of the trapezoid at x is A(x) = 1/2(-1/2x+5+5)(5) = -5/4x+25. + The volume is 187.5 units^3. +

    +     + +     + +     +     +     +
    +
    + The Shell Method +

    + Often a given problem can be solved in more than one way. + A particular method may be chosen out of convenience, + personal preference, or perhaps necessity. + Ultimately, it is good to have options. +

    + +

    + The previous section introduced the Disk and Washer Methods, + which computed the volume of solids of revolution by integrating the cross-sectional area of the solid. + This section develops another method of computing volume, + the Shell Method. Instead of slicing the solid perpendicular to the axis of rotation creating cross-sections, + we now slice it parallel to the axis of rotation, creating shells. +

    + + + +

    + Consider , + where the region shown in + is rotated around the y-axis forming the solid shown in + . + A small slice of the region is drawn in , + parallel to the axis of rotation. + When the region is rotated, + this thin slice forms a cylindrical shell, + as pictured in . + The previous section approximated a solid with lots of thin disks (or washers); + we now approximate a solid with many thin cylindrical shells. +

    + +
    + Introducing the Shell Method + +
    + + + + + +

    + Graph of the function y=\frac{1}{1+x^2} . + The curve begins at the point (0,1). + The curve then decreases until it meets the horizontal line x=1 where it stops at the point (1,\frac12) . + The entire area enclosed by y=\frac{1}{1+x^2} , x=0 , x=1 and y=0 is shaded. + In other words, the region consists of the area lying below the curve and above the x-axis + A vertical red rectangular slice with a base having an approximate length of 0.5 coming up from approximately the point (0.5,0) reaches up to meet the curve. + This rectangle when rotated about the y-axis will form a cylindrical shell in the following figures. +

    +     + A graph of a function drawn in the first quadrant in the xy plane. + + + + + //ASY file for figwasher_shell_intro_b_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0,0)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-1.2,1.2); + pair ybounds=(-.2,1.2); + pair zbounds=(-1.2,1.2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple g1(real t) {return (t,1/(1+t^2),0);} + path3 p1=graph(g1,0,1,operator ..); + draw(p1--(1,0,0),bluepen+.4mm); + draw(surface(p1 -- (1,0,0)--(0,0,0)--cycle), simplesurfacepen); + + // \filldraw [bottom color=colormaptwobottom,top color=colormaptwotop] + // (axis cs:.6,0,0) -- (axis cs:.7,0,0) -- (axis cs:.7,0,.7) -- (axis cs:.6,0,.7) -- cycle; + + path3 p2=(.6,0,0)--(.7,0,0)--(.7,.7,0)--(.6,.7,0)--cycle; + draw(surface(p2),simplesurfacepen2); + draw(p2,redpen+.4mm); + + label("$\displaystyle y=\frac{1}{1+x^2}$",(.75,1,0)); + + + + +
    + +
    + + + + + +

    + A three dimensional diagram of the area enclosed by the curve y=\frac{1}{1+x^2} , and the lines x=0 , x=1 and y=0 rotated about the y-axis. + The base of this shape is a unit circle lying on the xz plane. + This unit circle comes up from y=0 to y=0.5 creating a solid cylinder. + The region above y=0.5 forms a shape which looks like a perfectly symmetrical mountain whose peak occurs at the point (0,1,0). + The space below this region above the y-axis is contained in the shape which comes from rotating the curve. +

    +     + A 3d graph of the function in the first quadrant rotated fully about the y axis. + + + + + //ASY file for figshell_intro_a_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0,0)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-1.2,1.2); + pair ybounds=(-.2,1.2); + pair zbounds=(-1.2,1.2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple g1(real t) {return (t,1/(1+t^2),0);} + path3 p1=graph(g1,0,1,operator ..); + draw(p1--(1,0,0),bluepen+.4mm); + //draw(surface(p1 -- (1,0,0)--(0,0,0)--cycle), simplesurfacepen); + + // \filldraw [bottom color=colormaptwobottom,top color=colormaptwotop] + // (axis cs:.6,0,0) -- (axis cs:.7,0,0) -- (axis cs:.7,0,.7) -- (axis cs:.6,0,.7) -- cycle; + + pen p=apexmeshpen; + + triple f(pair t) { + return (cos(t.x),t.y,sin(t.x));} + surface s=surface(f,(0,0),(2*pi,.5),16,2,Spline); + draw(s,simplesurfacepen,meshpen=p); + + triple f(pair t) { + return (cos(t.x)*t.y,1/(1+t.y^2),t.y*sin(t.x));} + surface s=surface(f,(0,0),(2*pi,1),16,16,Spline); + draw(s,simplesurfacepen,meshpen=p); + + + + +
    + +
    + + + + + +

    + The red rectangle from figure (a) is rotated 360 degrees about the y-axis, creating a cylindrical shell. + Two circles, one of radius 0.5 and one of radius 0.6 are drawn on the xz-plane. + The area between the circles is shaded, creating a circular disc in the xz-plane. + The disc then goes up from y=0 until it meets the curve y=\frac{1}{1+x^2} , which creates the cylindrical shell. + Adding up many of these cylindrical shells will then allow us to approximate the volume of the solid described in the previous images. +

    +     + A three dimensional graph of the cylindrical shell which comes from rotating the rectangle in the first image about the y axis. + + + + + //ASY file for figshell_intro_d_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0,0)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-1.2,1.2); + pair ybounds=(-.2,1.2); + pair zbounds=(-1.2,1.2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple g1(real t) {return (t,1/(1+t^2),0);} + path3 p1=graph(g1,0,1,operator ..); + draw(p1--(1,0,0),bluepen+.4mm); + //draw(surface(p1 -- (1,0,0)--(0,0,0)--cycle), simplesurfacepen); + + // \filldraw [bottom color=colormaptwobottom,top color=colormaptwotop] + // (axis cs:.6,0,0) -- (axis cs:.7,0,0) -- (axis cs:.7,0,.7) -- (axis cs:.6,0,.7) -- cycle; + + pen p=apexmeshpen; + pen q=redmeshpen+.1mm; + + triple f(pair t) { + return (cos(t.x),t.y,sin(t.x));} + surface s=surface(f,(0,0),(2*pi,.5),8,2,Spline); + //draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (cos(t.x)*t.y,1/(1+t.y^2),t.y*sin(t.x));} + surface s=surface(f,(0,0),(2*pi,1),8,16,Spline); + //draw(s,surfacepen,meshpen=p); + + path3 p2=(.6,0,0)--(.7,0,0)--(.7,.7,0)--(.6,.7,0)--cycle; + //draw(surface(p2),simplesurfacepen2); + draw(p2,redpen+.4mm); + + triple f(pair t) { + return (.6*cos(t.x),t.y,.6*sin(t.x));} + surface s=surface(f,(0,0),(2*pi,.7),8,2,Spline); + draw(s,simplesurfacepen2,meshpen=q); + + triple f(pair t) { + return (.7*cos(t.x),t.y,.7*sin(t.x));} + surface s=surface(f,(0,0),(2*pi,.7),8,2,Spline); + draw(s,simplesurfacepen2,meshpen=q); + + triple g1(real t) {return (.6*cos(t),.7,.6*sin(t));} + path3 p1=graph(g1,0,2*pi,operator ..); + draw(p1,redpen+.3mm); + + triple g1(real t) {return (.7*cos(t),.7,.7*sin(t));} + path3 p1=graph(g1,0,2*pi,operator ..); + draw(p1,redpen+.3mm); + + triple g1(real t) {return (.6*cos(t),0,.6*sin(t));} + path3 p1=graph(g1,0,2*pi,operator ..); + draw(p1,redpen+.3mm); + + triple g1(real t) {return (.7*cos(t),0,.7*sin(t));} + path3 p1=graph(g1,0,2*pi,operator ..); + draw(p1,redpen+.3mm); + + triple g3(real t) {return (.7*cos(t),1/(1+.7^2),.7*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,purple+.3mm); + + triple g4(real t) {return (.655*cos(t),.7,.655*sin(t));} + path3 p4=graph(g4,0,2*pi,operator ..); + //draw(p4,purple+.3mm); + + triple f(pair t) { + return (t.y*cos(t.x),.7,t.y*sin(t.x));} + surface s=surface(f,(0,.6),(2*pi,.7),8,1,Spline); + draw(s,simplesurfacepen2,meshpen=q); + + + + +
    +     + +
    + +

    + To compute the volume of one shell, + first consider the paper label on a soup can with radius r and height h. + What is the area of this label? + A simple way of determining this is to cut the label and lay it out flat, + forming a rectangle with height h and length 2\pi r. + Thus the area is A = 2\pi rh; + see . +

    + +

    + Do a similar process with a cylindrical shell, with height h, + thickness \Delta x, and approximate radius r. + Cutting the shell and laying it flat forms a rectangular solid with length 2\pi r, + height h and depth \dx. + Thus the volume is V \approx 2\pi rh\dx; + see . + (We say approximately since our radius was an approximation.) +

    + +

    + By breaking the solid into n cylindrical shells, + we can approximate the volume of the solid as + + V \approx \sum_{i=1}^n 2\pi r_ih_i\dx_i + , + where r_i, h_i and \dx_i are the radius, + height and thickness of the ith shell, + respectively. +

    + +

    + This is a Riemann Sum. + Taking a limit as the thickness of the shells approaches 0 leads to a definite integral. +

    + +
    + Determining the volume of a thin cylindrical shell + + +
    + + + +

    + The left side of the image contains a cylinder, and the right side of the image shows a rectangle which comes from unraveling the side of the cylinder. + The cylinder has a height h , radius r and is pictured lying on its circular base. + Both the circular base, and the top of the cylinder are shaded, while the curved surface is unshaded, but contains a cut here instruction. + Once the curved surface of the cylinder is cut, it is unraveled into a rectangular region as seen on the right side of the image. + This rectangular region has a height h and a base 2\pi r . + The area of the rectangle is A=2\pi r h . +

    +     + A visualization of how we can unravel the outside shell of a cylinder into a rectangle. + + + \begin{tikzpicture}[scale=.75] + + \begin{scope}[xscale=1.5] + + \draw [left color=black!60,right color=black!20,thick] (0,0) circle (1cm); + \draw [left color=black!60,right color=black!20,thick] (-1,0) -- (-1,-3.5) arc (180:360:1) -- (1,0) arc (360:180:1); + \draw [fill=white](-1,-.2) -- node [pos=.5,left] { $h$} (-1,-3.3) arc (180:360:1) -- (1,-.2) arc (360:180:1); + \draw [dashed] (.71,-.91) -- node [pos=.5, above,rotate=90] { cut here} (.71,-4.01); + \draw (0,0) -- node [pos=.5,above] { $r$} (1,0); + + \end{scope} + + \draw [->,>=latex,ultra thick] (2,-2)--(3,-2); + + \begin{scope}[shift={(3.5,-.3)}] + + \draw (0,-.2) -- node [pos=.5,above] { $2\pi r$} (5,-.2) -- node [pos=.5,right] { $h$} (5,-3.3)--(0,-3.3) -- cycle; + \draw (2.5,-1.75) node { $A = 2\pi r h$}; + + \end{scope} + + \end{tikzpicture} + + + +
    + + + +
    + + + +

    + The left side of the image contains a cylindrical shell, which is just a cylinder with the middle part cut out. + The right side of the image shows a rectangular box which comes from unraveling the cylindrical shell. + The cylindrical shell has a height h , radius r which spans from the cut out center to the edge of the shell, and the shell has thickness \dx and is pictured lying on its circular base. + The entire cylindrical shell is shaded, and contains a cut here instruction along the curved surface as in the previous image. + Once the curved surface of the cylinderical shell is cut, the entire shell is unraveled into a rectangular box as seen on the right side of the image. + This rectangular box has a height h and a base length of 2\pi r and a thickness of \dx . + The volume of the unraveled cylindrical shell is approximated the volume of the box given by V=2\pi r h \dx . +

    +     + Unraveling the shell of a cylinder, but this time into a rectangular box with an arbitrary thickness. + + + \begin{tikzpicture}[scale=.75] + + \begin{scope}[xscale=1.5] + + \draw [left color=black!60,right color=black!20,thick] (0,0) circle (1cm); + \draw [left color=black!60,right color=black!20,thick] (-1,0) -- (-1,-3.5) arc (180:360:1) -- (1,0) arc (360:180:1); + \draw [left color=black!20,right color=black!60,thick] (0,0) circle (.8cm); + \draw (-1,-1.75) node [left] { $h$}; + \draw [dashed] (.565,-.565) -- (.71,-.71) -- node [pos=.5, above,rotate=90] { cut here} (.71,-4.2); + \draw (0,0) -- node [pos=.5,above] { $r$} (.9,0); + \draw [->,>=stealth](-1.2,.3) node [above] { $\dx$}-- (-.87,0); + + \end{scope} + + \draw [->,>=latex,ultra thick] (2,-2)--(3,-2); + + \begin{scope}[shift={(3.5,-.3)}] + + \draw [left color=black!60,right color=black!20,thick] (0,0) -- (.2,.2) -- node [pos=.5,above] { $2\pi r$} (5.2,.2) -- (5,0); + \draw [left color=black!60,right color=black!20,thick] (5,0) -- (5.2,.2) -- node [pos=.5,right] { $h$} (5.2,-3.3) -- (5,-3.5); + \draw [left color=black!60,right color=black!20,thick] (0,0) -- (5,0) -- (5,-3.5)--(0,-3.5) -- cycle; + \draw [>=stealth,->] (.5,0.3) node [above] { $\dx$} --(0.3,.1); + \draw (2.5,-1.75) node { $V\approx 2\pi r h\dx$}; + + \end{scope} + + \end{tikzpicture} + + + + +
    +     +
    + + + The Shell Method +

    + Let a solid be formed by revolving a region R, + bounded by x=a and x=b, around a vertical axis. + Let r(x) represent the distance from the axis of rotation to x (, the radius of a sample shell) and let h(x) represent the height of the solid at x (, the height of the shell). + The volume of the solid is + integrationvolume!Shell Method + Shell Method + + V = 2\pi\int_a^b r(x)h(x)\, dx + . +

    +     + +

    + Special Cases: +

    + +

    +

      +
    1. +

      + When the region R is bounded above by y=f(x) and below by y=g(x), + then h(x) = f(x)-g(x). +

      +
      2. + +
    2. +

      + When the axis of rotation is the y-axis (, x=0) then r(x) = x. +

      +
      3. +
    +

    + +

    + Let's practice using the Shell Method. +

    + + + Finding volume using the Shell Method + +

    + Find the volume of the solid formed by rotating the region bounded by y=0, + y=1/(1+x^2), x=0 and x=1 about the y-axis. +

    +     + +

    + This is the region used to introduce the Shell Method in , + but is sketched again in for closer reference. + A line is drawn in the region parallel to the axis of rotation representing a shell that will be carved out as the region is rotated about the y-axis. + (This is the differential element.) +

    + +
    + Graphing a region in + + + +

    + Graph of the region bounded by y=0, y=1/(1+x^2), x=0 and x=1. + The curve y=1/(1+x^2) begins at the y-axis at point (0,1) and slopes downwards before ending at the point (1,\frac12) . + The region contains the entire area below the curve, lying above the horizontal line y=0 . + On the x-axis, the graph contains an arbitrarily chosen point x . + The distance between the origin and the point x is given as the function r(x) , which will give us the radius of the cylindrical shell once the region is rotated about the y-axis. + Additionally, coming up from the point x is a red vertical line, which ends after meeting the curve y=1/(1+x^2). + This vertical line is labeled h(x), and will give the us the height of the cylindrical shell. +

    +     + Two dimensional graph of the region from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1}, + extra x ticks={.4}, + extra x tick labels={$x$}, + ytick={1}, + ymin=-.2,ymax=1.25, + xmin=-.15,xmax=1.1, + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1] {(1/(1+x^2))} \closedcycle; + \addplot [firstcurvestyle,domain=0:1] {(1/(1+x^2))}; + + \draw [thick,secondcolor] (axis cs:.4,0) -- (axis cs:.4,.86); + + \draw (axis cs:.4,.45) node [left] {$ h(x)\left\{\rule{0pt}{33pt}\right.$} + (axis cs:.2,-.12) node { $\underbrace{\rule{35pt}{0pt}}_{r(x)}$} + (axis cs: .7,1) node { $\ds y=\frac{1}{1+x^2}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + The distance this line is from the axis of rotation determines r(x); + as the distance from x to the y-axis is x, + we have r(x)=x. + The height of this line determines h(x); + the top of the line is at y=1/(1+x^2), + whereas the bottom of the line is at y=0. + Thus h(x) = 1/(1+x^2)-0 = 1/(1+x^2). + The region is bounded from x=0 to x=1, so the volume is + + V \amp = 2\pi\int_0^1 \frac{x}{1+x^2}\, dx. + This requires substitution. Let u=1+x^2, so du = 2x\, dx. We also change the bounds: u(0) = 1 and u(1) = 2. Thus we have: + \amp = \pi\int_1^2 \frac{1}{u}\, du + \amp = \pi\ln(u)\Big|_1^2 + \amp = \pi\ln(2) \approx 2.178 \,\text{units}^3 + . +

    + +

    + Note: in order to find this volume using the Disk Method, + two integrals would be needed to account for the regions above and below y=1/2. +

    +     + +     + +

    + With the Shell Method, + nothing special needs to be accounted for to compute the volume of a solid that has a hole in the middle, + as demonstrated next. +

    + + + Finding volume using the Shell Method + +

    + Find the volume of the solid formed by rotating the triangular region determined by the points (0,1), + (1,1) and (1,3) about the line x=3. +

    +     + +

    + The region is sketched in along with the differential element, + a line within the region parallel to the axis of rotation. + In , + we see the shell traced out by the differential element, + and in + the whole solid is shown. +

    + +
    + Graphing a region in + +
    + + + + +

    + Graph of the triangular region with edges on the points (0,1), (1,1) and (1,3). + The leftmost point of the triangular region begins at the y-axis at point (0,1) and linearly increases until the point (1,3) . + The equation of the line between the points (0,1) and (1,3) is given to be y=2x+1 . + The region contains the entire area below the curve, lying above the horizontal line y=1 . + Above the x-axis, the graph showcases the distance between an arbitrarily chosen value on the x-axis, labeled x , which is between 0 and 1, and the axis of rotation at x=3 . + The distance between these two x values is given as the function r(x) , which will give us the radius of the cylindrical shell once the region is rotated about the vertical line x=3. + Additionally, coming up from the point (x,1) is a red vertical line, which meets the upper bound of the triangular region, which is given by the line y=2x+1 . + This vertical line is labeled h(x), and will give the us the height of the cylindrical shell. +

    +     + Graph of the triangular region described in the example. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3}, + extra x ticks={.4}, + extra x tick labels={$x$}, + ytick={1,2,3}, + ymin=-.2,ymax=3.25, + xmin=-.15,xmax=3.2, + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(0,1)(1,1)(1,3)}; + \addplot [firstcurvestyle,domain=0:1] {2*x+1} node [rotate=57,pos=.5,above,black] { $y=2x+1$}; + \addplot [firstcurvestyle] coordinates {(0,1)(1,1)(1,3)}; + + \draw [thick,secondcolor] (axis cs: .4,1) -- (axis cs:.4,1.8); + + \draw [dashed] (axis cs:3,0) -- (axis cs:3,3.2); + + \draw (axis cs:1,1.4) node [right] {$ \left.\rule{0pt}{14pt}\right\} h(x).$} + (axis cs:1.7,.7) node { $\underbrace{\rule{100pt}{0pt}}_{r(x)}$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + +

    + A three dimensional graph of the same triangular region with edges on the points (0,1), (1,1) and (1,3) and an arbitrary cylindrical shell which lies inside the triangular region rotated about x=3. + The cylindrical shell is centered at the line x=3 and radius r(x) , which spans from arbitrarily chosen value on the x-axis between x=0 and x=1, and the axis of rotation at x=3 . + The height of the cylindrical shell is h(x) , which is the distance between the point (x,1) and upper bound of the triangular region at the particular value of x. +

    +     + Three dimensional depiction of an arbitrary cylindrical shell from rotating the region described in the example. + + + + + //ASY file for figshell2b_3D.asy in Chapter 7 + + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + //currentprojection=orthographic((6.1,13,42.3),(0,1,0),(0,0,0),.95,(.005,-.024)); + currentprojection=orthographic((2.6,4.6,47),(0,1,0),(0,0,0),.95,(0.004,0.004)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3,4,5,6}; + real[] myychoice={1,2,3}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.2,6.6); + pair ybounds=(-.2,3.4); + pair zbounds=(-3.4,3.4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); + + pen p=apexmeshpen; + + triple f2(pair t) {return ((3-t.x)*cos(t.y)+3,2*t.x+1,(3-t.x)*sin(t.y));} + surface s2=surface(f2,(0,0),(1,2*pi),2,16,Spline); + //draw(s2,simplesurfacepen); + + triple f3(pair t) {return (2*cos(t.y)+3,t.x,2*sin(t.y));} + surface s3=surface(f3,(1,0),(3,2*pi),2,16,Spline); + //draw(s3,simplesurfacepen); + + triple f4(pair t) {return (2.6*cos(t.y)+3,t.x,2.6*sin(t.y));} + surface s4=surface(f4,(1,0),(1.8,2*pi),2,16,Spline); + draw(s4,simplesurfacepen2); + + triple g3(real t) {return (2.6*cos(t)+3,1.8,2.6*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,redpen+.4mm); + + triple g3(real t) {return (2.6*cos(t)+3,1,2.6*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,redpen+.4mm); + + draw((0,1,0)--(1,3,0)--(1,1,0)--cycle,bluepen+.6mm); + + triple g3(real t) {return (2*cos(t)+3,3,2*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + draw((3,0,0)--(3,3.1,0),dashed+.2mm); + + draw((.4,1,0)--(.4,1.8,0),redpen+.4mm); + + triple g3(real t) {return (2*cos(t)+3,1,2*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + triple g3(real t) {return (3*cos(t)+3,1,3*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + //triple g3(real t) {return (2,3*cos(t),3*sin(t));} + //path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); + + + + +
    + +
    + + + +

    + A three dimensional plot showing the same triangular region rotated about x=3. + The resulting shape lies entirely above x=1 . + The leftmost point of the shape is the point (0,1) , which is the leftmost point of the original triangle. + The shape is then rotated in a full circle around the vertical line x=3 . + The resulting shape has a base of radius of 3 , whose center is an empty central cylinder of radius 2 centered at the axis of rotation. + As we move up the shape, the thickness tapers off as the line y=2x+1 reaches the point (3,1) in the original triangular region. + The point (3,1) rotated about the line x=3 then becomes the top of the resulting shape, while the base of the triangle, which is between the points (0,1) and (1,1) becomes the base. +

    +     + Three dimensional plot of the entire space which comes from rotating the triangular region. + + + + //ASY file for figshell2c_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + //currentprojection=orthographic((6.1,13,42.3),(0,1,0),(0,0,0),.95,(.005,-.024)); + currentprojection=orthographic((2.6,4.6,47),(0,1,0),(0,0,0),.95,(0.004,0.004)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3,4,5,6}; + real[] myychoice={1,2,3}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.2,6.6); + pair ybounds=(-.2,3.4); + pair zbounds=(-3.4,3.4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); + + pen p=apexmeshpen; + + triple f2(pair t) {return ((3-t.x)*cos(t.y)+3,2*t.x+1,(3-t.x)*sin(t.y));} + surface s2=surface(f2,(0,0),(1,2*pi),2,16,Spline); + draw(s2,simplesurfacepen); + + triple f3(pair t) {return (2*cos(t.y)+3,t.x,2*sin(t.y));} + surface s3=surface(f3,(1,0),(3,2*pi),2,16,Spline); + draw(s3,simplesurfacepen); + + triple f4(pair t) {return (2.6*cos(t.y)+3,t.x,2.6*sin(t.y));} + surface s4=surface(f4,(1,0),(1.8,2*pi),2,16,Spline); + //draw(s4,surfacepen2); + + triple g3(real t) {return (2.6*cos(t)+3,1.8,2.6*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,redpen+.4mm); + + draw((0,1,0)--(1,3,0)--(1,1,0)--cycle,bluepen+.6mm); + + triple g3(real t) {return (2*cos(t)+3,3,2*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.4mm); + + draw((3,0,0)--(3,3.1,0),dashed+.2mm); + + draw((.4,1,0)--(.4,1.8,0),redpen+.4mm); + + triple g3(real t) {return (2*cos(t)+3,1,2*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.4mm); + + triple g3(real t) {return (3*cos(t)+3,1,3*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.4mm); + + //triple g3(real t) {return (2,3*cos(t),3*sin(t));} + //path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); + + + +
    +     + +
    + +

    + The height of the differential element is the distance from y=1 to y=2x+1, + the line that connects the points (0,1) and (1,3). + Thus h(x) = 2x+1-1 = 2x. + The radius of the shell formed by the differential element is the distance from x to x=3; + that is, it is r(x)=3-x. + The x-bounds of the region are x=0 to x=1, giving + + V \amp = 2\pi\int_0^1 (3-x)(2x)\, dx + \amp = 2\pi\int_0^1 \big(6x-2x^2\big)\, dx + \amp = 2\pi\left(3x^2-\frac23x^3\right)\Big|_0^1 + \amp = \frac{14}{3}\pi\approx 14.66 \,\text{units}^3 + . +

    +     + +     + +

    + When revolving a region around a horizontal axis, + we must consider the radius and height functions in terms of y, + not x. +

    + + + Finding volume using the Shell Method + +

    + Find the volume of the solid formed by rotating the region given in about the x-axis. +

    +     + +

    + The region is sketched in with a sample differential element. + In the shell formed by the differential element is drawn, + and the solid is sketched in . (Note that the triangular region looks + short and wide here, + whereas in the previous example the same region looked + tall and narrow. + This is because the bounds on the graphs are different.) +

    + +

    + The height of the differential element is an x-distance, + between x=\frac12y-\frac12 and x=1. + Thus h(y) = 1-(\frac12y-\frac12) = -\frac12y+\frac32. + The radius is the distance from y to the x-axis, + so r(y) =y. + The y bounds of the region are y=1 and y=3, + leading to the integral +

    + +
    + Graphing a region in + +
    + + + + +

    + Graph of the triangular region with edges on the points (0,1), (1,1) and (1,3). + The leftmost point of the triangular region begins at the y-axis at point (0,1) and linearly increases until the point (1,3) . + The equation of the line between the points (0,1) and (1,3) is given as x=\frac12y-\frac12 . + The region contains the entire area to the right of the line x=\frac12y-\frac12 and to the left of the vertical line x=1 . + Coming up from the x-axis, the graph contains an arbitrarily chosen value on the y-axis, labeled y , which is between 1 and 3. + The graph showcases the distance between the point labeled y and the axis of rotation which is y=0 . + The distance between these two y values is given as the function r(y) , which will give us the radius of the cylindrical shell once the solid is formed by rotating about the x-axis. + Additionally, coming up from the line x=\frac12y-\frac12 is a red horizontal line, which meets the right side of the triangular region, which is given by the vertical line x=1 . + This vertical line is labeled h(y), and will give the us the height of the cylindrical shells. +

    +     + Graph of the triangular region described in the example. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3}, + ytick={1,2,3}, + extra y ticks={1.8}, + extra y tick labels={$y$}, + ymin=-.2,ymax=3.25, + xmin=-.15,xmax=1.5, + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(0,1)(1,1)(1,3)}; + \addplot [firstcurvestyle] coordinates {(0,1)(1,1)(1,3)(0,1)} node [rotate=41,shift={(75pt,0pt)},above,black] { $x=\frac12y-\frac12$}; + + \draw [thick,secondcolor] (axis cs: .4,1.8) -- node [black,pos=.5,below] {$\underbrace{\rule{42pt}{0pt}}_{h(y)}$} (axis cs:1,1.8); + + \draw (axis cs:1,.9) node [right] {$ \left.\rule{0pt}{29pt}\right\} r(y)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + +

    + A three dimensional graph of the same triangular region as the previous image and an arbitrary cylindrical shell which lies inside the triangular region rotated about the x-axis. + The cylindrical shell lies horizontally resting on its curved surface, and is centered at the line y=0 . + The cylindrical shell has radius r(y) , which spans from the y-axis to the arbitrarily chosen value on the y-axis between y=1 and y=3. + The height of the cylindrical shell is h(y) , which is the distance between the line x=\frac12y-\frac12 and the rightmost bound of the triangular region, which is the vertical line x=1. +

    +     + Three dimensional depiction of an arbitrary cylindrical shell from rotating the region described in the example. + + + + + //ASY file for figshell3b_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((3.13,8.3,49),(0,1,0),(0,0,0),.95,(0.0148,0.00673)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1,2,3}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.2,1.25); + pair ybounds=(-3.4,3.4); + pair zbounds=(-3.4,3.4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); + + pen p=apexmeshpen; + + triple f2(pair t) {return (t.x,(2*t.x+1)*cos(t.y),(2*t.x+1)*sin(t.y));} + surface s2=surface(f2,(0,0),(1,2*pi),2,16,Spline); + //draw(s2,simplesurfacepen); + + triple f3(pair t) {return (t.x,cos(t.y),sin(t.y));} + surface s3=surface(f3,(0,0),(1,2*pi),2,16,Spline); + //draw(s3,surfacepen); + + triple f4(pair t) {return (t.x,1.8*cos(t.y),1.8*sin(t.y));} + surface s4=surface(f4,(.4,0),(1,2*pi),2,16,Spline); + draw(s4,simplesurfacepen2); + + triple g3(real t) {return (.4,1.8*cos(t),1.8*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,redpen+.4mm); + + triple g3(real t) {return (1,1.8*cos(t),1.8*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,redpen+.4mm); + + draw((0,1,0)--(1,3,0)--(1,1,0)--cycle,bluepen+.6mm); + + draw((.4,1.8,0)--(1,1.8,0),redpen+.4mm); + + triple g3(real t) {return (0,cos(t),sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + triple g3(real t) {return (1,cos(t),sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + triple g3(real t) {return (1,3*cos(t),3*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + //triple g3(real t) {return (2,3*cos(t),3*sin(t));} + //path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); + + + + +
    + +
    + + + + + +

    + A three dimensional plot showing the same triangular region rotated about the x-axis. + The resulting shape lies entirely between x=0 and x=1 . + The highest point of the shape is the point (1,3) , which is the highest point of the original triangle. + The shape is then rotated in a full circle about the x-axis. + The resulting shape has a height given by the value h(y)= -\frac12y+\frac32 . + The radius of the shape is given by r(y)=y, with the y bounds between y=1 and y=3 as the center of the shape is an empty central cylinder of radius 1 centered at the axis of rotation. +

    +     + Three dimensional plot of the entire space which comes from rotating the triangular region. + + + + + //ASY file for figshell3c_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((3.13,8.3,49),(0,1,0),(0,0,0),.95,(0.0148,0.00673)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1,2,3}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.2,1.25); + pair ybounds=(-3.4,3.4); + pair zbounds=(-3.4,3.4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); + + pen p=apexmeshpen; + + triple f2(pair t) {return (t.x,(2*t.x+1)*cos(t.y),(2*t.x+1)*sin(t.y));} + surface s2=surface(f2,(0,0),(1,2*pi),2,16,Spline); + draw(s2,simplesurfacepen); + + triple f3(pair t) {return (t.x,cos(t.y),sin(t.y));} + surface s3=surface(f3,(0,0),(1,2*pi),2,16,Spline); + draw(s3,simplesurfacepen); + + triple f3(pair t) {return (1,t.x*cos(t.y),t.x*sin(t.y));} + surface s3=surface(f3,(1,0),(3,2*pi),2,16,Spline); + draw(s3,simplesurfacepen); + + triple f4(pair t) {return (t.x,1.8*cos(t.y),1.8*sin(t.y));} + surface s4=surface(f4,(.4,0),(1,2*pi),2,16,Spline); + //draw(s4,surfacepen2); + + triple g3(real t) {return (.4,1.8*cos(t),1.8*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,redpen+.4mm); + + draw((0,1,0)--(1,3,0)--(1,1,0)--cycle,bluepen+.6mm); + + draw((.4,1.8,0)--(1,1.8,0),redpen+.4mm); + + triple g3(real t) {return (0,cos(t),sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.4mm); + + triple g3(real t) {return (1,cos(t),sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.4mm); + + triple g3(real t) {return (1,3*cos(t),3*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.4mm); + + //triple g3(real t) {return (2,3*cos(t),3*sin(t));} + //path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); + + + + +
    +     + +
    + +

    + + V \amp = 2\pi\int_1^3\left[y\left(-\frac12y+\frac32\right)\right]\, dy + \amp = 2\pi\int_1^3\left[-\frac12y^2+\frac32y\right]\, dy + \amp = 2\pi\left[-\frac16y^3+\frac34y^2\right]\Big|_1^3 + \amp = 2\pi\left[\frac94-\frac7{12}\right] + \amp = \frac{10}{3}\pi \approx 10.472\,\text{units}^3 + . +

    +     + +     + +

    + At the beginning of this section it was stated that + it is good to have options. + The next example finds the volume of a solid rather easily with the Shell Method, + but using the Washer Method would be quite a chore. +

    + + + Finding volume using the Shell Method + +

    + Find the volume of the solid formed by revolving the region bounded by + y= \sin(x) and the x-axis from x=0 to x=\pi about the y-axis. +

    +     + +

    + The region and a differential element, + the shell formed by this differential element, + and the resulting solid are given in . +

    + +
    + Graphing a region in + +
    + + + + +

    + Graph of the region bounded by y= \sin(x) and the x-axis from x=0 to x=\pi. + The leftmost point of the triangular region begins at the origin after which we see a singular sine wave ending at the point (\pi,0) . + The region contains the entire area to below the curve y= \sin(x), lying above the x-axis. + The graph contains an arbitrarily chosen value on the x-axis, labeled x , which is between 0 and \pi. + The graph showcases the distance between the axis of rotation which is x=0 and the point labeled x as the function r(x) , which will be the radius of the cylindrical shells. + Additionally, coming up from the x-axis is a red vertical line, which meets the curve y= \sin(x) at the point (x,\sin(x)). + This vertical line is labeled h(x), and will give the us the height of the cylindrical shells. +

    +     + Graph of the region described in the example. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={1,1.57,3.14}, + extra x tick labels={$x$,$\frac{\pi}2$,$\pi$}, + ytick={1}, + ymin=-.2,ymax=1.25,% + xmin=-.1,xmax=3.5,% + ] + + \addplot [firstcurvestyle,areastyle,domain=0:3.14] {sin(deg(x))}; + \addplot [firstcurvestyle,domain=0:3.14,samples=50] {sin(deg(x))}; + + \draw [thick,secondcolor] (axis cs: 1,.84) -- node [black,pos=.49,right] {$ \left.\rule{0pt}{32pt}\right\} h(x)$} (axis cs:1,0); + + \draw (axis cs:.52,0) node [above] {$\overbrace{\rule{30pt}{0pt}}^{r(x)}$} ; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + +

    + A three dimensional graph of the same region as the previous image and an arbitrary cylindrical shell which will lie inside the region rotated about the x-axis. + The cylindrical shell lies vertically resting on its flat base and is centered on the y-axis. + The cylindrical shell has radius r(x)=x , which spans from the y-axis to the arbitrarily chosen value on the x-axis between x=0 and x=\pi. + The height of the cylindrical shell is h(x)=\sin(x) , which is the distance between the x-axis and the upper bound of the region, which curve y=\sin(x). +

    +     + Three dimensional depiction of an arbitrary cylindrical shell lying inside the rotated region described in the example. + + + + + //ASY file for figshell4b_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((23.8,4.1,64),(0,1,0),(0,0,0),1,(-.00343,0.004889)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-3.4,3.4); + pair ybounds=(-.2,1.2); + pair zbounds=(-3.4,3.4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); + + pen p=apexmeshpen; + + triple pt1=(pi,-.05,0); + draw((pi,0,0)--pt1,black+.2mm); + label("$\pi$",pt1,S); + + triple f2(pair t) {return (t.x*cos(t.y),sin(t.x),t.x*sin(t.y));} + surface s2=surface(f2,(0,0),(pi,2*pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + //draw(s2,surfacepen,meshpen=p); + + triple f4(pair t) {return (cos(t.y),t.x,sin(t.y));} + surface s4=surface(f4,(0,0),(.84,2*pi),2,16,Spline); + draw(s4,simplesurfacepen2); + + draw((1,0,0)--(1,.84,0),redpen+.4mm); + + triple g3(real t) {return (t,sin(t),0);} + path3 p3=graph(g3,0,pi,operator ..); + draw(p3,bluepen+.6mm); + + triple g3(real t) {return (cos(t),.84,sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,redpen+.4mm); + + triple g3(real t) {return (cos(t),0,sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,redpen+.4mm); + + triple g3(real t) {return (pi*cos(t),0,pi*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + //triple g3(real t) {return (2,3*cos(t),3*sin(t));} + //path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); + + + + +
    + +
    + + + + + +

    + A three dimensional plot showing the entire region rotated about the y-axis. + The base of the shape is a circle of radius \pi centered at the origin. + From the outside edge of the circle, the surface curves both upwards and inwards towards the positive y-axis until reaching a peak at y=1. + From here, the surface curves downwards and inwards until closing in at the origin. + The cylinderical shells lying inside the shape have a height given by h(x)= \sin(x) and radius r(x)=x. +

    +     + Three dimensional plot of the entire space which comes from rotating sine wave. + + + + + //ASY file for figshell4c_3D.asy in Chapter 7 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((23.8,4.1,64),(0,1,0),(0,0,0),1,(-.00343,0.004889)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-3.4,3.4); + pair ybounds=(-.2,1.2); + pair zbounds=(-3.4,3.4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] + // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); + + pen p=apexmeshpen; + + triple pt1=(pi,-.05,0); + draw((pi,0,0)--pt1,black+.2mm); + label("$\pi$",pt1,S); + + triple f2(pair t) {return (t.x*cos(t.y),sin(t.x),t.x*sin(t.y));} + surface s2=surface(f2,(0,0),(pi,2*pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s2,simplesurfacepen,meshpen=p); + + triple f4(pair t) {return (cos(t.y),t.x,sin(t.y));} + surface s4=surface(f4,(0,0),(.84,2*pi),2,16,Spline); + //draw(s4,emissive(rgb(1,.6,.6)+opacity(.2))); + + draw((1,0,0)--(1,.84,0),redpen+.4mm); + + triple g3(real t) {return (t,sin(t),0);} + path3 p3=graph(g3,0,pi,operator ..); + draw(p3,bluepen+.6mm); + + triple g3(real t) {return (cos(t),.84,sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,redpen+.4mm); + + triple g3(real t) {return (cos(t),0,sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,redpen+.4mm); + + triple g3(real t) {return (pi*cos(t),0,pi*sin(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.4mm); + //triple g3(real t) {return (2,3*cos(t),3*sin(t));} + //path3 p3=graph(g3,0,2*pi,operator ..); + //draw(p3,bluepen+.4mm); + + //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); + + + + +
    +     + +
    + +

    + The radius of a sample shell is r(x) = x; + the height of a sample shell is h(x) = \sin(x), + each from x=0 to x=\pi. + Thus the volume of the solid is + + V \amp = 2\pi\int_0^{\pi} x\sin(x) \, dx. + This requires Integration By Parts. Set u=x and dv=\sin(x) \, dx; we leave it to the reader to fill in the rest. We have: + \amp = 2\pi\Big[-x\cos(x) \Big|_0^{\pi} +\int_0^{\pi}\cos(x) \, dx \Big] + \amp = 2\pi\Big[\pi + \sin(x) \Big|_0^{\pi}\,\Big] + \amp = 2\pi\Big[\pi + 0 \Big] + \amp = 2\pi^2 \approx 19.74 \,\text{units}^3 + . +

    + +

    + Note that in order to use the Washer Method, + we would need to solve y=\sin(x) for x, + requiring the use of the arcsine function. + We leave it to the reader to verify that the outside radius function is + R(y) = \pi-\arcsin y and the inside radius function is r(y)=\arcsin y. + Thus the volume can be computed as + + \pi\int_0^1 \Big[ (\pi-\arcsin y)^2-(\arcsin y)^2\Big]\, dy + . +

    + +

    + This integral isn't terrible given that the \arcsin^2 y terms cancel, + but it is more onerous than the integral created by the Shell Method. +

    +     + +     + +

    + We end this section with a table summarizing the usage of the Washer and Shell Methods. +

    + + + Summary of the Washer and Shell Methods +

    + Let a region R be given with x-bounds x=a and x=b and y-bounds y=c and y=d. +

    + + + + + Washer Method + + Shell Method + + + Horizontal Axis + \ds \pi\int_a^b \big(R(x)^2-r(x)^2\big)\, dx + + \ds 2\pi\int_c^d r(y)h(y)\, dy + + + + + + Vertical Axis + \ds\pi \int_c^d\big(R(y)^2-r(y)^2\big)\, dy + + \ds 2\pi\int_a^b r(x)h(x)\, dx + + + +

    + integrationvolume!Washer Method + Washer Method + integrationvolume!Shell Method + Shell Method +

    +     + + + + + + + + + + + +

    + As in the previous section, + the real goal of this section is not to be able to compute volumes of certain solids. + Rather, it is to be able to solve a problem by first approximating, + then using limits to refine the approximation to give the exact value. + In this section, + we approximate the volume of a solid by cutting it into thin cylindrical shells. + By summing up the volumes of each shell, + we get an approximation of the volume. + By taking a limit as the number of equally spaced shells goes to infinity, + our summation can be evaluated as a definite integral, + giving the exact value. +

    + +

    + We use this same principle again in the next section, + where we find the length of curves in the plane. +

    + + + + Terms and Concepts + + + +

    + A solid of revolution is formed by revolving a shape around an axis. +

    +     + + +     + + + + +

    + The Shell Method can only be used when the Washer Method fails. +

    +     + +     + + + + +

    + The Shell Method works by integrating cross-sectional areas of a solid. +

    +     + + +     + + + + +

    + When finding the volume of a solid of revolution that was revolved around a vertical axis, + the Shell Method integrates with respect to x. +

    +     + + +     +     + + Problems + + + + +

    + Use the Shell Method to find the volume of the solid of revolution formed by + revolving the given region about the y-axis. +

    +     + + + + +

    + The region bounded by the curve y=3-x^2, + the x axis, and the y axis: +

    + + + +

    + Graph of the region bounded by the curve y=3-x^2, the x-axis, and the y-axis. + Note that this region can reference both the regions to the left or right of the y-axis, but we will consider the region to the right of the y-axis. + The curve y=3-x^2 begins at the point (0,3) from which it slopes down until reaching the x-axis. + The region then contains the entire area below this curve, and above the x-axis. +

    +     + Graph of the region lying in the first quadrant bounded by the curve and the two coordinate axes. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=3.5, + xmin=-2.1,xmax=2.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1.73] {3-x^2} \closedcycle; + \addplot [firstcurvestyle,domain=-1.73:1.73,samples=40] {3-x^2} node [shift={(-15pt,90pt)},black] { $y=3-x^2$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + 9\pi/2 units^3 +

    +     + +     + + + + +

    + The region between y=5x and the x axis, + for 1\leq x\leq 2: +

    + + + +

    + Graph of the region between y=5x and the x-axis, for 1\leq x\leq 2. + The line y=5x begins at the point (1,5) from which it linearly increases until reaching the rightmost bound which is at the point (2,10). + The region then contains the entire area below the line y=5x and above the x-axis for 1\leq x\leq 2. +

    +     + Graph of the region bounded by the the line y=5x and the x-axis for x between 1 and 2. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=11, + xmin=-.1,xmax=2.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=1:2] {5*x} \closedcycle; + \addplot [firstcurvestyle,domain=0:2.1] {5*x} node [shift={(-35pt,-10pt)},black] { $y=5x$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + 70\pi/3 units^3 +

    +     + +     + + + + +

    + The region between y=\cos(x) and the x axis, + for 0\leq x\leq \pi/2: +

    + + + +

    + Graph of the region between y=\cos(x) and the x-axis, for 0\leq x\leq \pi/2. + The curve y=\cos(x) begins at the point (0,1) from which it slopes downwards until ending after reaching the x-axis at the point (\pi/2,0). + The region then contains the entire area below the curve y=\cos(x) and above the x-axis for 0\leq x\leq \pi/2. + The bound 0\leq x\leq \pi/2 can also be interpreted as the leftmost bound of the region being the y-axis. +

    +     + Graph of the region bounded by the the cosine function and the two coordinate axes. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.7 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1.57] {cos(deg(x))} \closedcycle; + \addplot [firstcurvestyle,domain=0:1.57,samples=30] {cos(deg(x))} node [shift={(-20pt,65pt)},black] { $y=\cos(x) $}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \pi^2-2\pi units^3 +

    +     + +     + + + + +

    + The region between the curves y=x and y=\sqrt{x}: +

    + + + +

    + Graph of the region between the curves y=x and y=\sqrt{x}. + The curves y=x and y=\sqrt{x} both begin at the origin. + From this point the curve y=\sqrt{x} rises above the line y=x until reaching the point (1,1), where the curve once again intersects the line. + The region then contains the area below the curve y=\sqrt{x} and above the line y=x. +

    +     + Graph of the region bounded by the two curves. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1,samples=60] {sqrt(x)}; + + \addplot [firstcurvestyle,domain=0:.05] {sqrt(x)}; + \addplot [firstcurvestyle,domain=.05:1,samples=30] {sqrt(x)} node [shift={(-65pt,-15pt)} ,black] { $y=\sqrt{x}$}; + + \addplot [firstcurvestyle,domain=0:1] {x} node [shift={(-30pt,-45pt)},black] { $y=x$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + 2\pi/15 units^3 +

    +     + +     + +     + + + +

    + Use the Shell Method to find the volume of the solid of revolution formed by + revolving the given region about the x-axis. +

    +     + + + + +

    + The region between y=3-x^2 and the x axis: +

    + + + +

    + Graph of the region between y=3-x^2 and the x-axis. + The curve y=3-x^2 begins at x-axis at the point (-\sqrt{3},0). + From this point the curve rises until reaching the y-axis at the point (0,3). + From this point, the curve slopes downwards until once again reaching the x-axis at the point (\sqrt{3},0). + On both the left and right sides of the region, the area is bounded by the curve y=3-x^2. +

    +     + Graph of the region bounded by the curve and the x axis. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=3.5, + xmin=-2.1,xmax=2.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=-1.73:1.73] {3-x^2}; + \addplot [firstcurvestyle,domain=-1.73:1.73,samples=40] {3-x^2} node [shift={(-15pt,90pt)},black] { $y=3-x^2$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + 48\pi\sqrt{3}/5 units^3 +

    +     + +     + + + + +

    + The region between y=5x and the y axis, + for 5\leq y\leq 10: +

    + + + +

    + Graph of the region between y=5x and the y axis, for 5\leq y\leq 10. + The line y=5x begins at the point (1,5), from which it linearly increases until ending upper bound for y at the point (2,10). + The region then contains the entire area to the right of x=0 and to the left of the line y=5x for y between 5 and 10. +

    +     + Graph of the region bounded by the line and the y axis for y values between 5 and 10. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=11, + xmin=-.1,xmax=2.1 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(1,5) (2,10) (0,10) (0,5) (1,5)}; + \addplot [firstcurvestyle,domain=0:2.1] {5*x} node [shift={(-20pt,-40pt)},black] { $y=5x$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + 350\pi/3 units^3 +

    +     + +     + + + + +

    + The region between y=\cos(x) and the x axis, + for 0\leq x\leq \pi/2: +

    + + + +

    + Graph of the region between y=\cos(x) and the x axis, for 0\leq x\leq \pi/2. + The curve y=\cos(x) begins at the point (0,1) from which it slopes downwards until ending after reaching the x-axis at the point (\pi/2,0). + The region then contains the entire area to the right of x=0 and to the left of of the curve y=\cos(x) for y values between 0 and 1. +

    +     + Graph of the region bounded by the cosine curve and the coordinate axes. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.7 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1.57] {cos(deg(x))} \closedcycle; + \addplot [firstcurvestyle,domain=0:1.57,samples=30] {cos(deg(x))} node [shift={(-20pt,65pt)},black] { $y=\cos(x) $}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \pi^2/4 units^3 +

    +     + +     + + + + +

    + The region between the curves y=x and y=\sqrt{x}: +

    + + + +

    + Graph of the region between the curves y=x and y=\sqrt{x}. + The curves y=x and y=\sqrt{x} both begin at the origin. + From this point the curve y=\sqrt{x} rises above the line y=x until reaching the point (1,1), where the curve once again intersects the line. + The region consists of the area to the right of the curve y=\sqrt{x} and the to the left line y=x for y values between 0 and 1. +

    +     + Graph of the region bounded by the two curves. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:1,samples=60] {sqrt(x)}; + + \addplot [firstcurvestyle,domain=0:.05] {sqrt(x)}; + \addplot [firstcurvestyle,domain=.05:1,samples=30] {sqrt(x)} node [shift={(-65pt,-15pt)} ,black] { $y=\sqrt{x}$}; + + \addplot [firstcurvestyle,domain=0:1] {x} node [shift={(-30pt,-45pt)},black] { $y=x$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \pi/6 units^3 +

    +     + +     + +     + + + +

    + Use the Shell Method to find the volume of the solid of revolution formed + by revloving the given region about each of the given axes. +

    +     + + + + +

    + Region bounded by: y=\sqrt{x}, + y=0 and x=1. +

    +     + + + +

    + Rotate about the y axis. +

    +     + +

    + 4\pi/5 +

    +     +     + + + +

    + Rotate about x=1. +

    +     + +

    + 8\pi/15 +

    +     +     + + + +

    + Rotate about the x axis. +

    +     + +

    + \pi/2 +

    +     +     + + + +

    + Rotate about y=1. +

    +     + +

    + 5\pi/6 +

    +     +     + +     + + + + +

    + Region bounded by: y=4-x^2 and y=0. +

    +     + + + +

    + Rotate about x=2. +

    +     + +

    + 128\pi/3 +

    +     +     + + + +

    + Rotate about x=-2. +

    +     + +

    + 128\pi/3 +

    +     +     + + + +

    + Rotate about the x axis. +

    +     + +

    + 512\pi/15 +

    +     +     + + + +

    + Rotate about y=4. +

    +     + +

    + 256\pi/5 +

    +     +     + +     + + + + +

    + The triangle with vertices (1,1), + (1,2) and (2,1). +

    +     + + + +

    + Rotate about the y axis. +

    +     + +

    + 4\pi/3 +

    +     +     + + + +

    + Rotate about x=1. +

    +     + +

    + \pi/3 +

    +     +     + + + +

    + Rotate about the x axis. +

    +     + +

    + 4\pi/3 +

    +     +     + + + +

    + Rotate about y=2. +

    +     + +

    + 2\pi/3 +

    +     +     + +     + + + + +

    + Region bounded by y=x^2-2x+2 and y=2x-1. +

    +     + + + +

    + Rotate about the y axis. +

    +     + +

    + 16\pi/3 +

    +     +     + + + +

    + Rotate about x=1. +

    +     + +

    + 8\pi/3 +

    +     +     + + + +

    + Rotate about x=-1. +

    +     + +

    + 8\pi +

    +     +     + +     + + + + +

    + Region bounded by y=1/\sqrt{x^2+1}, + x=1 and the x and y axes. +

    +     + + + +

    + Rotate about the y axis. +

    +     + +

    + 2\pi(\sqrt{2}-1) +

    +     +     + + + +

    + Rotate about x=1. +

    +     + +

    + 2\pi(1-\sqrt{2}+\sinh^{-1}(1)) +

    +     +     + +     + + + + +

    + Region bounded by y=2x, + y=x and x=2. +

    +     + + + +

    + Rotate about the y axis. +

    +     + +

    + 16\pi/3 +

    +     +     + + + +

    + Rotate about x=2. +

    +     + +

    + 8\pi/3 +

    +     +     + + + +

    + Rotate about the x axis. +

    +     + +

    + 8\pi +

    +     +     + + + +

    + Rotate about y=4. +

    +     + +

    + 8\pi +

    +     +     + +     + +     +     +     +
    +
    + Arc Length and Surface Area + +

    + In previous sections we have used integration to answer the following questions: +

    + +

    +

      +
    1. +

      + Given a region, what is its area? +

      +
      2. + +
    2. +

      + Given a solid, what is its volume? +

      +
      3. +
    +

    + + + +

    + In this section, + we address two related questions: +

      +
    1. +

      + Given a curve, what is its length? + This is often referred to as arc length. +

      +
      2. + +
    2. +

      + Given a solid, what is its surface area? +

      +
      3. +
    +

    +     + + + Arc Length +

    + Consider the graph of y=\sin(x) on [0,\pi] given in . + How long is this curve? + That is, if we were to use a piece of string to exactly match the shape of this curve, + how long would the string be? +

    + +

    + As we have done in the past, we start by approximating; + later, we will refine our answer using limits to get an exact solution. +

    + +

    + The length of straight-line segments is easy to compute using the Distance Formula. + We can approximate the length of the given curve by approximating the curve with straight lines and measuring their lengths. +

    + +
    + Graphing y=\sin(x) on [0,\pi] and approximating the curve with line segments + + +
    + + + + + Graph of the function y=\sin(x) on [0,\pi]. + The curve y=\sin(x) begins at the point (0,0), from which it slopes upwards until reaching a peak at the point (\frac{\pi}{2},1). + From the point, the curve slopes downwards until reaching the x-axis at the point (\pi,0). + + Graph of the sine function for x between 0 and pi. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={.79,1.57,2.36,3.14}, + extra x tick labels={$\frac{\pi}4$,$\frac{\pi}2$,$\frac{3\pi}{4}$,$\pi$}, + ymin=-.2,ymax=1.25, + xmin=-.1,xmax=3.5, + ] + + \addplot [firstcurvestyle,closed,domain=0:3.14,samples=40] {sin(deg(x))}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + Graph of the function y=\sin(x) on [0,\pi], with four straight lines which will be used to approximate the length of this curve. + The four lines are evenly spaced out in intervals of \frac{\pi}{4} on the x-axis. + Each line begins at a point on the curve y=\sin(x), and ends at a point on the same curve after travelling a distance of \frac{\pi}{4} on the x-axis. + The first line begins at the same point (0,0) as the curve, from which it linearly increases until reaching the point (\frac{\pi}{4},\frac{sqrt2}{2}), which is also a point on the curve. + The second line begins at the same point the first line ends, given by (\frac{\pi}{4},\frac{sqrt2}{2}), from which it linearly increases until reaching the point (\frac{\pi}{2},1), which is the peak of the curve. + The third line begins at the same point the second line ends, given by (\frac{\pi}{2},1), from which it linearly decreases until reaching the point (\frac{3\pi}{4},\frac{sqrt2}{2}), which is also a point on the curve. + The fourth line begins at the same point the third line ends, given by (\frac{3\pi}{4},\frac{sqrt2}{2}), from which it linearly decreases until reaching the point (\pi,0), which is the end of the curve. + + Graph of the sine function for x between 0 and pi and four straight lines approximating this curve. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={.79,1.57,2.36,3.14}, + extra x tick labels={$\frac{\pi}4$,$\frac{\pi}2$,$\frac{3\pi}{4}$,$\pi$}, + extra y ticks={.71}, + extra y tick labels={$\frac{\sqrt{2}}2$}, + ymin=-.2,ymax=1.25, + xmin=-.1,xmax=3.5, + ] + + \addplot+ [domain=0:3.14,samples=40] {sin(deg(x))}; + + \draw [secondcolor,thick] (axis cs:0,0) -- (axis cs:.79,.71) -- (axis cs: 1.57,1) -- (axis cs:2.36,.71) -- (axis cs: 3.14,0); + + \filldraw (axis cs:0,0) circle (2pt) (axis cs:.79,.71) circle (2pt) (axis cs: 1.57,1) circle (2pt) (axis cs:2.36,.71) circle (2pt) (axis cs: 3.14,0) circle (2pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + In , + the curve y=\sin(x) has been approximated with 4 line segments + (the interval [0,\pi] has been divided into 4 subintervals of equal length). + It is clear that these four line segments approximate + y=\sin(x) very well on the first and last subinterval, + though not so well in the middle. + Regardless, the sum of the lengths of the line segments is 3.79, + so we approximate the arc length of + y=\sin(x) on [0,\pi] to be 3.79. +

    + +

    + In general, we can approximate the arc length of y=f(x) on [a,b] in the following manner. + Let a=x_0 \lt x_1 \lt \ldots \lt x_{n-1}\lt x_{n}=b be a partition of [a,b] into n subintervals. + Let \dx_i represent the length of the + ith subinterval [x_{i-1},x_{i}]. +

    + +
    + Zooming in on the ith subinterval [x_{i-1},x_{i}] of a partition of [a,b] + + + + Graph of the ith subinterval of the function y=f(x), which is graphed on the interval [x_{i-1},x_{i}]. + The graph contains a line between the start and endpoint of the curve, which will be used to approximate the length of the ith subinterval curve. + The ith subinterval of the curve y=f(x) begins at the point (x_{i-1},y_{i-1}) from which it heads upwards in a concave arc until reaching the point (x_{i},y_{i}). + The straight line then passes through the start and endpoints of the curve, (x_{i-1},y_{i-1}) and (x_{i},y_{i}) respectively. + The line lies below the curve for the entire interval that the curve is plotted on. + The graph also contains the measurements \dx_i and \dy_i, giving the respective length of the change in x and y between the start and endpoint of the subinterval of the curve. + + Graph of a portion of a curve with a line between the start and endpoint which is used to approximate the length of the curve. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={.3,1.37}, + extra x tick labels={$x_{i-1}$,$x_{i}$}, + ytick=\empty, + extra y ticks={.48,1}, + extra y tick labels={$y_{i-1}$,$y_{i}$}, + ymin=-.2,ymax=1.25,% + xmin=-.1,xmax=1.6,% + ] + + \addplot+ [domain=0.3:1.37,samples=30] {sin(deg(x+.2))}; + + \draw (axis cs: .25,1) -- (axis cs:.1,1) + (axis cs: .25,.48) -- (axis cs:.1,.48) + (axis cs: .175,.48) -- node [pos=.5,fill=white] { $\Delta y_i$} (axis cs:.175,1); + + \draw (axis cs: .3,.25) -- (axis cs: .3,.4) + (axis cs: 1.37,.25) -- (axis cs: 1.37,.4) + (axis cs: .3,.325) -- node [pos=.5,fill=white] { $\Delta x_i$} (axis cs: 1.37,.325); + + \draw [secondcolor,thick] (axis cs: .3,.48) -- (axis cs:1.37,1); + + \draw [dashed,thin] (axis cs: .3,.48) -| (axis cs:1.37,1); + + \filldraw (axis cs: .3,.48) circle (1pt) (axis cs:1.37,1) circle (1pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + + zooms in on the ith subinterval where y=f(x) is approximated by a straight line segment. + The dashed lines show that we can view this line segment as the hypotenuse of a right triangle whose sides have length \dx_i and \dy_i. + Using the Pythagorean Theorem, + the length of this line segment is + + \sqrt{\dx_i^2 + \Delta y_i^2} + . + Summing over all subintervals gives an arc length approximation + + L \approx \sum_{i=1}^n \sqrt{\dx_i^2 + \Delta y_i^2} + . +

    + +

    + As shown here, this is not a Riemann Sum. + While we could conclude that taking a limit as the subinterval length goes to zero gives the exact arc length, + we would not be able to compute the answer with a definite integral. + We need first to do a little algebra. +

    + +

    + In the above expression factor out a \dx_i^2 term: + + \sum_{i=1}^n \sqrt{\dx_i^2 + \Delta y_i^2} \amp = \sum_{i=1}^n \sqrt{\dx_i^2\left(1 + \frac{\Delta y_i^2}{\dx_i^2}\right)}. + \amp = \sum_{i=1}^n\sqrt{1 + \frac{\Delta y_i^2}{\dx_i^2}}\,\dx_i + + after pulling the \dx_i^2 term out of the square root. +

    + +

    + This is nearly a Riemann Sum. Consider the \Delta y_i^2/\dx_i^2 term. + The expression \Delta y_i/\dx_i measures the change in y/change in x, + that is, the rise over run of f on the ith subinterval. + The Mean Value Theorem of Differentiation () + states that there is a c_i in the ith subinterval where + \fp(c_i) = \Delta y_i/\dx_i. Thus we can rewrite our above expression as: + + L \approx \sum_{i=1}^n\sqrt{1+\fp(c_i)^2}\,\dx_i + . + This is a Riemann Sum. + As long as \fp is continuous, we can invoke + and conclude + + L = \int_a^b\sqrt{1+\fp(x)^2}\, dx + . +

    + + + Arc Length + +

    + Let f be differentiable on [a,b], + where \fp is also continuous on [a,b]. + Then the arc length of f from x=a to x=b is + integrationarc length + arc length + + L = \int_a^b \sqrt{1+\fp(x)^2}\, dx + . +

    +     +     + + + +

    + As the integrand contains a square root, + it is often difficult to use the formula in to find the length exactly. + When exact answers are difficult to come by, + we resort to using numerical methods of approximating definite integrals. + The following examples will demonstrate this. +

    + + + Finding arc length + +

    + Find the arc length of f(x) = x^{3/2} from x=0 to x=4. +

    +     + +

    + We find \fp(x)= \frac32x^{1/2}; + note that on [0,4], + f is differentiable and \fp is also continuous. + Using the formula, we find the arc length L as + + L \amp = \int_0^4 \sqrt{1+\left(\frac32x^{1/2}\right)^2}\, dx + \amp = \int_0^4 \sqrt{1+\frac94x} \, dx + \amp = \int_0^4 \left(1+\frac94x\right)^{1/2}\, dx + \amp = \frac23\cdot\frac49\cdot\left(1+\frac94x\right)^{3/2}\Big|_0^4 + \amp =\frac{8}{27}\left(10^{3/2}-1\right) \approx 9.07 \,\text{units} + . +

    + +
    + A graph of f(x) = x^{3/2} from + + + + Graph of the function f(x) = x^{3/2} on the interval between x=0 and x=4. + The curve f(x) = x^{3/2} begins at the point (0,0) from which it heads upwards in a convex arc until reaching the point (4,8). + A straight line plotted between the start and endpoints of the curve would lie entirely above the curve on the interval between x=0 and x=4 and would showcase the shortest distance between the two points. + + Graph of the function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.2,ymax=8.5, + xmin=-.1,xmax=4.5, + ] + + \addplot+ [closed,domain=0:4] {x^(3/2)}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + A graph of f is given in . +

    +     + +     + + + Finding arc length + +

    + Find the arc length of \ds f(x) =\frac18x^2-\ln(x) from x=1 to x=2. +

    +     + +

    + This function was chosen specifically because the resulting integral can be evaluated exactly. + We begin by finding \fp(x) = x/4-1/x. + The arc length is + + L \amp = \int_1^2 \sqrt{1+ \left(\frac x4-\frac1x\right)^2}\, dx + \amp = \int_1^2 \sqrt{1 + \frac{x^2}{16} -\frac12 + \frac1{x^2} } \, dx + \amp = \int_1^2 \sqrt{\frac{x^2}{16} +\frac12 + \frac1{x^2} } \, dx + \amp = \int_1^2 \sqrt{ \left(\frac x4 + \frac1x\right)^2}\, dx + + + \amp = \int_1^2 \left(\frac x4 + \frac1x\right) \, dx + \amp = \left.\left(\frac{x^2}8 + \ln(x)\right)\right|_1^2 + \amp = \frac38+\ln(2) \approx 1.07 \,\text{units} + . +

    + +
    + A graph of f(x) =\frac18x^2-\ln(x) from + + + + Graph of the function f(x) =\frac18x^2-\ln(x). + The curve is highlighted on the interval between x=1 and x=2. + The curve f(x) =\frac18x^2-\ln(x) begins near the point (0.4,1) from which it heads downwards in a convex arc until crossing the x-axis at approximately x=1.25. + The curve then continues in the convex arc, until it once again reaches the x-axis at approximately x=3. + + Graph of the function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.26,ymax=1.2, + xmin=-.1,xmax=3.1, + ] + + \addplot [firstcurvestyle,closed,semithick,domain=.2:3,samples=50] {(1/8)*x^2-ln(x)}; + \addplot [firstcurvestyle,-,very thick,domain=1:2] {(1/8)*x^2-ln(x)}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + A graph of f is given in ; + the portion of the curve measured in this problem is in bold. +

    +     + +     + + + +

    + The previous examples found the arc length exactly through careful choice of the functions. + In general, exact answers are much more difficult to come by and numerical approximations are necessary. +

    + + + Approximating arc length numerically + +

    + Find the length of the sine curve from x=0 to x=\pi. +

    +     + +

    + This is somewhat of a mathematical curiosity; + in + we found the area under one hump + of the sine curve is 2 square units; + now we are measuring its arc length. +

    + +

    + The setup is straightforward: + f(x) = \sin(x) and \fp(x) = \cos(x). + Thus + + L = \int_0^\pi \sqrt{1+\cos^2(x)}\, dx + . +

    + +

    + This integral cannot be evaluated in terms of elementary functions so we will approximate it with Simpson's Method with n=4. +

    + +
    + A table of values of y=\sqrt{1+\cos^2(x) } to evaluate a definite integral in + + + x\sqrt{1+\cos^2(x) } + + + 0\sqrt{2} + + + \pi/4\sqrt{3/2} + + + \pi/21 + + + 3 \pi/4\sqrt{3/2} + + + \pi\sqrt{2} + + +
    + +

    + + gives \sqrt{1+\cos^2(x) } evaluated at 5 evenly spaced points in [0,\pi]. + Simpson's Rule then states that + + \int_0^\pi \sqrt{1+\cos^2(x)}\, dx \amp \approx \frac{\pi-0}{4\cdot 3}\left(\sqrt{2}+4\sqrt{3/2}+2(1)+4\sqrt{3/2}+\sqrt{2}\right) + \amp =3.82918 + . +

    + +

    + Using a computer with n=100 the approximation is L\approx 3.8202; + our approximation with n=4 is quite good. +

    +     +     +     + + + Surface Area of Solids of Revolution +

    + We have already seen how a curve y=f(x) on [a,b] can be revolved around an axis to form a solid. + Instead of computing its volume, + we now consider its surface area. +

    + + + +
    + Establishing the formula for surface area + +
    + + + + + Graph of an arbitrary function y=f(x) on the interval [a,b]. + The curve is a concave arc starting at x=a at some arbitrary y value from which it slopes upwards until ending at x=b at some slightly higher y value. + The plot of the graph also contains a subinterval on the x-axis, given by [x_{i-1},x_{i}]. + A line is drawn through the points (x_{i-1},f(x_{i-1})) and (x_{i},f(x_{i})), which approximates the length of the curve y=f(x) on the interval [x_{i-1},x_{i}]. + + Graph of an arbitrary function on the interval from a to b, with a line approximating a small portion of the curve. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + axis y line=none, + axis lines=center, + y dir=reverse, + view={10}{30}, + xtick=\empty, + ztick=\empty, + extra x ticks={.5,.9,1.4,1.9}, + extra x tick labels={$a$,$x_{i-1}$,$x_{i}$,$b$}, + ymin=-1,ymax=1, + xmin=.25,xmax=2.1, + zmin=-.2,zmax=1.2 + ] + + \addplot3 [firstcurvestyle,closed,domain=.5:1.9,samples y=0] ({x},{0},{(sin(deg(x)))}); + + \draw [secondcolor,thick] (axis cs:.9,0,.78) -- (axis cs:1.4,0,.985); + \draw [fill=black] (axis cs:.9,.0,.78) circle (1pt) (axis cs:1.4,0,.985) circle (1pt); + + \end{axis} + + %\node [right] at (myplot.right of origin)[shift={(-10pt,-6pt)}] { $x$}; + %\node [above] at (myplot.above origin) [shift={(0,-17pt)}] { $y$}; + + \end{tikzpicture} + + + + +
    + +
    + + + + + + Graph of an arbitrary function y=f(x) on the interval [a,b]. + The line drawn through the points (x_{i-1},f(x_{i-1})) and (x_{i},f(x_{i})) is then rotated about the x-axis. + The resulting shape resembles a part of a cone which is lying horizontally, and can be used to approximate the surface area of the function y=f(x) being rotated about the x-axis on the interval [x_{i-1},x_{i}]. + The plot also contains two vertical measurements. + The first vertical measurement is r, which gives the radius of the cone at x=x_{i-1} and the second is R, which gives the radius of the cone at x=x_{i} + The plot also contains an additional measurement L which gives the length of the line connecting f(x_{i-1}) and f(x_{i}). + The measurement L is also the length of the part of the resulting part of a cone that comes from rotating the line about the x-axis. + + Graph of a function with the line approximating the length of a part of the curve being rotated about the x axis. + + + + + //ASY file for figarc4_3D.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((16.7,13.2,46),(0,1,0),(0,0,0),1,(-.0323,0.0012)); + //currentprojection=orthographic((16.7,13.2,46),(0,1,0),(0,0,0),1,(0.0148,0.00673)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.25,2.1); + pair ybounds=(-1.1,1.1); + pair zbounds=(-1.1,1.1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //\addplot3[domain=.9:1.4,y domain=-210:150,samples y=36,surf,shader=flat,colormap={mp2}{\colormapone}] + // ({x},{(.41*(x-.9)+.78)*sin(y)},{(.41*(x-.9)+.78)*cos(y)}); + + //\addplot3[domain=.5:1.9,,samples y=0,{\colorone},] ({x},{0},{(sin(deg(x)))}); + + //\addplot3[domain=0:360,,samples y=0,black,smooth] ({1.4},{.98*cos(x)},{.98*sin(x)}); + + //\addplot3[domain=130:330,,samples y=0,black,dashed,smooth] ({.9},{.78*cos(x)},{.78*sin(x)}); + + pen p=apexmeshpen; + + triple f2(pair t) {return (t.x,(.41*(t.x-.9)+.78)*sin(t.y),(.41*(t.x-.9)+.78)*cos(t.y));} + surface s2=surface(f2,(.9,0),(1.4,2*pi),2,16,Spline); + draw(s2,simplesurfacepen,meshpen=p); + + triple g3(real t) {return (t,sin(t),0);} + path3 p3=graph(g3,.5,1.9,operator ..); + draw(p3,bluepen+.4mm); + + triple g3(real t) {return (.9,.78*sin(t),.78*cos(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.2mm); + + triple g3(real t) {return (1.4,.985*sin(t),.985*cos(t));} + path3 p3=graph(g3,0,2*pi,operator ..); + draw(p3,bluepen+.2mm); + + draw((.9,.78,0)--(1.4,.985,0),redpen+.6mm); + + draw((.5,.05,0)--(.5,-.05,0),black+.2mm); + label("$a$",(.5,-.05,0),S); + + draw((1.9,.05,0)--(1.9,-.05,0),black+.2mm); + label("$b$",(1.9,-.05,0),S); + + draw((.9,.05,0)--(.9,-.05,0),black+.2mm); + label("$x_{i-1}$",(.9,-.05,0),S); + + draw((1.4,.05,0)--(1.4,-.05,0),black+.2mm); + label("$x_{i}$",(1.4,-.05,0),S); + + label(XY*"$\left.\rule{0pt}{40pt}\right\}\ R$",(1.4,.49,0),E,Embedded); + label(XY*"$r\left\{\rule{0pt}{30pt}\right.$",(.85,.37,0),W,Embedded); + label(rotate(20,(0,0,1))*"$\overbrace{\rule{41pt}{0pt}}^{\text{\normalsize \textit{L}}}$",(1.1,1.03,0),Embedded); + + + + +
    +     + +
    + +

    + We begin as we have in the previous sections: + we partition the interval [a,b] with n subintervals, + where the ith subinterval is [x_{i-1},x_{i}]. + On each subinterval, + we can approximate the curve y=f(x) with a straight line that connects f(x_{i-1}) and + f(x_{i}) as shown in . + Revolving this line segment about the x-axis creates part of a cone + (called a frustum of a cone) + as shown in . + The surface area of a frustum of a cone is + + 2\pi\cdot\,\text{length} \,\cdot\,\text{average of the two radii \(R\) and \(r\)} + . +

    + +

    + The length is given by L; + we use the material just covered by arc length to state that + + L\approx \sqrt{1+\fp(c_i)^2}\dx_i + + for some c_i in the ith subinterval. + The radii are just the function evaluated at the endpoints of the interval. + That is, + + R = f(x_{i}) \text{ and } r = f(x_{i-1}) + . +

    + +

    + Thus the surface area of this sample frustum of the cone is approximately + + 2\pi\frac{f(x_{i-1})+f(x_{i})}2\sqrt{1+\fp(c_i)^2}\dx_i + . +

    + +

    + Since f is a continuous function, + the Intermediate Value Theorem states there is some d_i in + [x_{i-1},x_{i}] such that \ds f(d_i) = \frac{f(x_{i-1})+f(x_{i})}2; + we can use this to rewrite the above equation as + + 2\pi f(d_i)\sqrt{1+\fp(c_i)^2}\dx_i + . +

    + +

    + Summing over all the subintervals we get the total surface area to be approximately + + \text{Surface Area}\, \approx \sum_{i=1}^n 2\pi f(d_i)\sqrt{1+\fp(c_i)^2}\dx_i + , + which is a Riemann Sum. + Taking the limit as the subinterval lengths go to zero gives us the exact surface area, + given in the following theorem. +

    + + + Surface Area of a Solid of Revolution + +

    + Let f be differentiable on [a,b], + where \fp is also continuous on [a,b]. + integrationsurface area + surface areasolid of revolution +

    + +

    +

      +
    1. +

      + The surface area of the solid formed by revolving the graph of y=f(x), + where f(x)\geq0, about the x-axis is + + \text{Surface Area}\, = 2\pi\int_a^b f(x)\sqrt{1+\fp(x)^2}\, dx + . +

      +
      2. + +
    2. +

      + The surface area of the solid formed by revolving the graph of y=f(x) about the y-axis, + where a,b\geq0, is + + \text{Surface Area}\, = 2\pi\int_a^b x\sqrt{1+\fp(x)^2}\, dx + . +

      +
      3. +
    +

    +     +     + +

    + (When revolving y=f(x) about the y-axis, + the radii of the resulting frustum are x_{i-1} and x_{i}; + their average value is simply the midpoint of the interval. + In the limit, this midpoint is just x. + This gives the second part of .) +

    + + + + + Finding surface area of a solid of revolution + +

    + Find the surface area of the solid formed by revolving + y=\sin(x) on [0,\pi] around the x-axis, + as shown in . +

    + +
    + Revolving y=\sin(x) on [0,\pi] about the x-axis + + + + + Three dimensional graph of the shape coming from revolving y=\sin(x) on [0,\pi] about the x-axis. + The curve y=\sin(x) is drawn on the interval [0,\pi]. + This concave curve begins at the point (0,0), from which it increases until reaching a maximum at the point (\frac{\pi}{2},1). + The curve then decreases until ending at the x-axis at the point (\pi,0). + The curve is then rotated about the x-axis, which creates the solid of revolution. + This solid has the largest diameter and is symmetric about x=\frac{pi}{2}, from which it shrinks down until closing in on itself at x=0 and x=\pi. + + Three dimensional graph of the sine curve being rotated about the x axis. + + + + + //ASY file for figarc4_3D.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((16.7,13.2,46),(0,1,0),(0,0,0),1,(-.0323,0.0012)); + //currentprojection=orthographic((16.7,13.2,46),(0,1,0),(0,0,0),1,(0.0148,0.00673)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={-1,1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.1,3.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //\addplot3[domain=.9:1.4,y domain=-210:150,samples y=36,surf,shader=flat,colormap={mp2}{\colormapone}] + // ({x},{(.41*(x-.9)+.78)*sin(y)},{(.41*(x-.9)+.78)*cos(y)}); + + //\addplot3[domain=.5:1.9,,samples y=0,{\colorone},] ({x},{0},{(sin(deg(x)))}); + + //\addplot3[domain=0:360,,samples y=0,black,smooth] ({1.4},{.98*cos(x)},{.98*sin(x)}); + + //\addplot3[domain=130:330,,samples y=0,black,dashed,smooth] ({.9},{.78*cos(x)},{.78*sin(x)}); + + pen p=apexmeshpen+.1mm; + + triple f2(pair t) {return (t.x,sin(t.x)*sin(t.y),sin(t.x)*cos(t.y));} + surface s2=surface(f2,(0,0),(pi,2*pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + draw(s2,simplesurfacepen,meshpen=p); + + triple g3(real t) {return (t,sin(t),0);} + path3 p3=graph(g3,0,pi,32,operator ..); + draw(p3,bluepen+.6mm); + + draw((pi,.05,0)--(pi,-.05,0),black+.2mm); + label("$\pi$",(pi,-.05,0),S); + + + + +
    +     + +

    + The setup is relatively straightforward. + Using , + we have the surface area SA is: + + SA \amp = 2\pi\int_0^\pi \sin(x) \sqrt{1+\cos^2(x) }\, dx + \amp = -2\pi\frac12\left.\left(\sinh^{-1}(\cos(x) )+\cos(x) \sqrt{1+\cos^2(x) }\right)\right|_0^\pi + \amp = 2\pi\left(\sqrt{2}+\sinh^{-1}(1) \right) + \amp \approx 14.42\,\text{units}^2 + . +

    + +

    + The integration step above is nontrivial, + utilizing the integration method of Trigonometric Substitution from . +

    + +

    + It is interesting to see that the surface area of a solid, + whose shape is defined by a trigonometric function, + involves both a square root and an inverse hyperbolic trigonometric function. +

    +     + +     + + + Finding surface area of a solid of revolution + +

    + Find the surface area of the solid formed by revolving the curve y=x^2 on [0,1] about: +

    + +

    +

      +
    1. +

      + the x-axis +

      +
      2. + +
    2. +

      + the y-axis. +

      +
      3. +
    +

    + +
    + The solids used in + + +
    + + + + + Three dimensional graph of the shape coming from revolving y=x^2 on [0,1] about the x-axis. + The quadratic function y=x^2 is drawn on the interval [0,1]. + This curve begins at the point (0,0), from which it quadratically increases until reaching a maximum at the point (1,1). + The curve is then rotated about the x-axis, creating a solid of revolution. + This shape has a circular vertical cross-section which has a radius of r(x)=x^2 and is hollow on the inside. + The shape also is not closed off on its rightmost boundary at x=1. + + Three dimensional graph of the quadratic function being rotated about the x axis. + + + + + //ASY file for figarc4_3D.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((7.4,20,47),(0,1,0),(0,0,0),1,(-.004432,-0.00537)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={-1,1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.1,1.5); + pair ybounds=(-1.1,1.1); + pair zbounds=(-1.1,1.1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + pen p=apexmeshpen+.1mm; + + triple f2(pair t) {return (t.x,t.x^2*sin(t.y),t.x^2*cos(t.y));} + surface s2=surface(f2,(0,0),(1,2*pi),8,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + draw(s2,simplesurfacepen,meshpen=p); + + triple g3(real t) {return (t,t^2,0);} + path3 p3=graph(g3,0,1,operator ..); + draw(p3,bluepen+.6mm); + + + + +
    + + +
    + + + + + Three dimensional graph of the shape coming from revolving y=x^2 on [0,1] about the y-axis. + The quadratic function y=x^2 is drawn on the interval [0,1]. + The curve is then rotated about the y-axis, creating a solid of revolution. + This shape has a circular horizontal cross-section, which has a radius of r(x)=x and is hollow on the inside. + The shape also is not closed off on its top boundary at y=1. + + Three dimensional graph of the quadratic function being rotated about the y axis. + + + + + //ASY file for figarc4_3D.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((10.4,3.7,64),(0,1,0),(0,0,0),.95,(0.00404,0.00136)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,-1}; + real[] myychoice={-1,1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(-.1,1.1); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + pen p=apexmeshpen+.1mm; + + triple f2(pair t) {return (t.x*sin(t.y),t.x^2,t.x*cos(t.y));} + surface s2=surface(f2,(0,0),(1,2*pi),8,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + draw(s2,simplesurfacepen,meshpen=p); + + triple g3(real t) {return (t,t^2,0);} + path3 p3=graph(g3,0,1,operator ..); + draw(p3,bluepen+.6mm); + + + + +
    +     +
    +     + +

    +

      +
    1. +

      + The integral is straightforward to setup: + + SA \amp = 2\pi\int_0^1 x^2\sqrt{1+(2x)^2}\, dx. + Like the integral in , this requires Trigonometric Substitution. + \amp = \left.\frac{\pi}{32}\left(2(8x^3+x)\sqrt{1+4x^2}-\sinh^{-1}(2x)\right)\right|_0^1 + \amp =\frac{\pi}{32}\left(18\sqrt{5}-\sinh^{-1}(2) \right) + \amp \approx 3.81\,\text{units}^2 + . + The solid formed by revolving y=x^2 around the x-axis is graphed in . +

      +
      2. + +
    2. +

      + Since we are revolving around the y-axis, + the radius of the solid is not f(x) but rather x. + Thus the integral to compute the surface area is: + + SA \amp = 2\pi\int_0^1x\sqrt{1+(2x)^2}\, dx. + This integral can be solved using substitution. Set u=1+4x^2; the new bounds are u=1 to u=5. We then have + \amp = \frac{\pi}4\int_1^5 \sqrt{u}\, du + \amp = \left.\frac{\pi}{4}\frac23 u^{3/2}\right|_1^5 + \amp = \frac{\pi}6\left(5\sqrt{5}-1\right) + \amp \approx 5.33\,\text{units}^2 + . + The solid formed by revolving y=x^2 about the y-axis is graphed in . +

      +
      3. +
    +

    +     + +     + +

    + Our final example is a famous mathematical paradox. +

    + + + The surface area and volume of Gabriel's Horn + +

    + Consider the solid formed by revolving y=1/x about the x-axis on [1,\infty). + Find the volume and surface area of this solid. + (This shape, + as graphed in , + is known as Gabriel's Horn + since it looks like a very long horn that only a supernatural person, + such as an angel, could play.) + Gabriel's Horn +

    + +
    + A graph of Gabriel's Horn + + + + + Three dimensional graph of the shape coming from revolving y=1/x on [1,\infty) about the x-axis. + The function y=1/x is drawn on the interval [1,\infty). + The curve is then rotated about the x-axis, creating the shape called Gabriel's Horn. + This shape has a circular vertical cross-section, which has a radius of r(x)=1/x and is hollow on the inside. + The shape also is not closed off on its leftmost boundary at x=1. + For calculating the volume, we consider how much space is enclosed by x=1 and Gabriel's Horn itself. + + Three dimensional graph of Gabriel's Horn. + + + + + //ASY file for figgabriel_3D.asy in Chapter 7 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((372,29,113),(0,1,0),(0,0,0),.95,(-0.024,-0.0032)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={5,10,15}; + real[] myychoice={-1,1}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-1,17); + pair ybounds=(-1.3,1.3); + pair zbounds=(-1.3,1.3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + pen p=apexmeshpen+.1mm; + + triple f2(pair t) {return (t.x,1/t.x*sin(t.y),1/t.x*cos(t.y));} + surface s2=surface(f2,(1,0),(15,2*pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + draw(s2,simplesurfacepen,meshpen=p); + + triple g3(real t) {return (t,1/t,0);} + path3 p3=graph(g3,1,15,16,operator ..); + draw(p3,bluepen+.6mm); + + draw((0,-2,0)--(0,2,0),invisible); + + + + +
    +     + +

    + To compute the volume it is natural to use the Disk Method. + We have: + + V \amp = \pi\int_1^\infty \frac{1}{x^2}\, dx + \amp = \lim_{b\to\infty}\pi\int_1^b\frac{1}{x^2}\, dx + \amp = \lim_{b\to\infty} \left.\pi\left(\frac{-1}{x}\right)\right|_1^b + \amp = \lim_{b\to\infty} \pi\left(1-\frac1b\right) + \amp = \pi \,\text{units}^3 + . +

    + +

    + Gabriel's Horn has a finite volume of \pi cubic units. + Since we have already seen that regions with infinite length can have a finite area, + this is not too difficult to accept. +

    + +

    + We now consider its surface area. + The integral is straightforward to setup: + + SA \amp = 2\pi\int_1^\infty \frac{1}{x}\sqrt{1+1/x^4}\, dx. + Integrating this expression is not trivial. We can, however, compare it to other improper integrals. Since 1\lt \sqrt{1+1/x^4} on [1,\infty), we can state that + 2\pi\int_1^\infty \frac{1}{x}\, dx \amp \lt 2\pi\int_1^\infty \frac{1}{x}\sqrt{1+1/x^4}\, dx + . +

    + +

    + By , + the improper integral on the left diverges. + Since the integral on the right is larger, + we conclude it also diverges, + meaning Gabriel's Horn has infinite surface area. +

    + +

    + Hence the paradox: + we can fill Gabriel's Horn with a finite amount of paint, + but since it has infinite surface area, we can never paint it. +

    + +

    + Somehow this paradox is striking when we think about it in terms of volume and area. + However, we have seen a similar paradox before, as referenced above. + We know that the area under the curve y=1/x^2 on [1,\infty) is finite, + yet the shape has an infinite perimeter. + Strange things can occur when we deal with the infinite. +

    +     + +     + +

    + A standard equation from physics is + Work = force distance, + when the force applied is constant. + In we learn how to compute work when the force applied is variable. +

    +     + + + + Terms and Concepts + + +

    + The integral formula for computing arc length was found by first approximating arc length with straight line segments. +

    +     + +     + + + +

    + The integral formula for computing arc length includes a square root, + meaning the integration is probably easy. +

    +     + +     +     + + Problems + + + +

    + Find the arc length of the function on the given interval. +

    +     + + + + + $f = Formula("x"); + package my::Function::numeric; + our @ISA = ('Parser::Function::numeric'); + sub sqrt { + my $self = shift; + my $value = $self->context->flag("setSqrt"); + return $value if $value; + return $self->SUPER::sqrt(@_); + } + package main; + Context("Numeric"); + Context()->functions->set(sqrt=>{class=>'my::Function::numeric'}); + Context()->flags->set(reduceConstantFunctions=>0); + + $ans = Compute("sqrt(2)"); + + +

    + \ds f(x) = on [0, 1]. +

    + +

    + +

    +     + +

    + \sqrt{2} +

    +     +     +     + + + + + + +

    + \ds f(x) = \sqrt{8}x on [-1, 1]. +

    +

    + +

    +     +     +     + + + + + + +

    + \ds f(x) = \frac13x^{3/2}-x^{1/2} on [0,1]. +

    + +

    + +

    +     +     +     + + + + + + +

    + \ds f(x) = \frac1{12}x^{3}+\frac1x on [1,4]. +

    + +

    + +

    +     +     +     + + + + + + +

    + \ds f(x) = 2x^{3/2}-\frac16\sqrt{x} on [1,4]. +

    +

    + +

    +     +     +     + + + + + + +

    + \ds f(x) = \cosh(x) on [-\ln(2) , \ln(2) ]. +

    + +

    + +

    +     +     +     + + + + + + +

    + \ds f(x) = \frac12\big(e^x+e^{-x}\big) on [0, \ln(5) ]. +

    + +

    + +

    +     +     +     + + + + + + +

    + \ds f(x) = \frac1{12}x^5+\frac1{5x^3} on [0.1, 1]. +

    + + +     +     +     + + + + +

    + \ds f(x) = \ln\big(\sin(x) \big) on [\pi/6, \pi/2]. +

    +     + +

    + -\ln(2-\sqrt{3}) \approx 1.31696 +

    +     + +     + + + + +

    + \ds f(x) = \ln\big(\cos(x) \big) on [0, \pi/4]. +

    +     + +

    + \sinh^{-1}(1) +

    +     + +     + +     + + + +

    + Set up the integral to compute the arc length of the function on the given interval. + Do not evaluate the integral. +

    +     + + + + +

    + \ds f(x) = x^2 on [0, 1]. +

    +     + +

    + \int_0^1 \sqrt{1+4x^2}\, dx +

    +     + +     + + + + +

    + \ds f(x) = x^{10} on [0, 1]. +

    +     + +

    + \int_0^1 \sqrt{1+100x^{18}}\, dx +

    +     + +     + + + + +

    + \ds f(x) = \ln(x) on [1, e]. +

    +     + +

    + \int_1^e \sqrt{1+\frac1{x^2}}\, dx +

    +     + +     + + + + +

    + \ds f(x) = \frac1x on [1,2]. +

    +     + +

    + \int_{1}^2 \sqrt{1+\frac1{x^4}}\, dx +

    +     + +     + + + + +

    + \ds f(x) = cos(x) on [0,\pi/2]. +

    +     + +

    + \int_0^{\pi/2}\sqrt{1+\sin^2(x)}\,dx +

    +     +     + + + + +

    + \ds f(x) = \sec(x) on [-\pi/4,\pi/4]. +

    +     + +

    + \int_{-\pi/4}^{\pi/4} \sqrt{1+\sec^2(x) \tan^2(x) }\, dx +

    +     + +     +     + + + +

    + Use Simpson's Rule, with n=4, + to approximate the arc length of the function on the given interval. + Note: these are the same problems as in Exercises. +

    +     + + + + +

    + \ds f(x) = x^2 on [0, 1]. +

    +     + +

    + 1.4790 +

    +     + +     + + + + +

    + \ds f(x) = x^{10} on [0, 1]. +

    +     + +

    + 1.8377 +

    +     + +     + + + + +

    + \ds f(x) = \ln(x) on [1, e]. +

    +     + +

    + 2.1300 +

    +     + +     + + + + +

    + \ds f(x) = \frac1x on [1,2]. +

    +     + +

    + 1.3254 +

    +     + +     + + + + +

    + \ds f(x) = \cos(x) on [0, \pi/2]. +

    +     + +

    + 1.00013 +

    +     + +     + + + + +

    + \ds f(x) = \sec(x) on [-\pi/4,\pi/4]. +

    +     + +

    + 1.7625 +

    +     + +     + +     + + + +

    + Find the surface area of the described solid of revolution. +

    +     + + + + +

    + The solid formed by revolving y=2x on [0,1] about the x-axis. +

    +     + +

    + 2\pi\int_0^1 2x\sqrt{5}\, dx = 2\pi\sqrt{5} +

    +     + +     + + + + +

    + The solid formed by revolving y=2x on [0,1] about the y-axis. +

    +     + +

    + 2\pi\int_0^1 x\sqrt{5}\, dx = \pi\sqrt{5} +

    +     + +     + + + + +

    + The solid formed by revolving y=x^2 on [0,1] about the y-axis. +

    +     + +

    + 2\pi\int_0^1 x\sqrt{1+4x^2}\, dx = \pi/6(5\sqrt{5}-1) +

    +     + +     + + + + +

    + The solid formed by revolving y=x^3 on [0,1] about the x-axis. +

    +     + +

    + 2\pi\int_0^1 x^3\sqrt{1+9x^4}\, dx = \pi/27(10\sqrt{10}-1) +

    +     + +     +     + + + +

    + The following arc length and surface area problems lead to improper integrals. + Although the hypotheses of and + are not satisfied, the improper integrals converge, + and formulas for arc length and surface area still give the correct result. +

    +     + + + + +

    + Find the length of the curve \ds f(x) = \sqrt{x} on [0, 1]. + (Note: this is the same as the length of f(x)=x^2 on [0,1]. Why?) +

    +     + +

    + \int_0^1 \sqrt{1+\frac{1}{4x}}\, dx +

    +     + +     + + + + +

    + Find the length of the curve + \ds f(x) = \sqrt{1-x^2} on [-1, 1]. + (Note: this describes the top half of a circle with radius 1.) +

    +     + +

    + \int_{-1}^1 \sqrt{1+\frac{x^2}{1-x^2}}\, dx +

    +     + +     + + + + +

    + Find the length of the curve \ds f(x) = \sqrt{1-x^2/9} on [-3, 3]. + (Note: this describes the top half of an ellipse with a major axis of length 6 and a minor axis of length 2.) +

    +     + +

    + \int_{-3}^3 \sqrt{1+\frac{x^2}{81-9x^2}}\, dx +

    +     + +     + + + + +

    + Find the surface area of the solid formed by revolving + y=\sqrt{x} on [0,1] about the x-axis. +

    +     + +

    + 2\pi\int_0^1 \sqrt{x}\sqrt{1+1/(4x)}\, dx = \pi/6(5\sqrt{5}-1) +

    +     + +     + + + + +

    + Find the surface area of the sphere formed by revolving + y=\sqrt{1-x^2} on [-1,1] about the x-axis. +

    +     + +

    + 2\pi\int_0^1 \sqrt{1-x^2}\sqrt{1+x/(1-x^2)}\, dx = 4\pi +

    +     + +     +     +     +     +
    +
    + Work + +

    + Work is the scientific term used to describe the action of a force which moves an object. + When a constant force F is applied to move an object a distance d, + the amount of work performed is W=F\cdot d. +

    + +

    + The SI unit of force is the newton; one newton is equal to one + , + and the SI unit of distance is a meter (m). + The fundamental unit of work is one newtonmeter, + or a joule (J). + That is, applying a force of one newton for one meter performs one joule of work. + In Imperial units + (as used in the United States), + force is measured in pounds (lb) and distance is measured in feet (ft), + hence work is measured in ftlb. +

    + +

    + When force is constant, the measurement of work is straightforward. + For instance, lifting a 200 + object 5 performs + 200\cdot 5 = 1000 ftlb of work. +

    + +

    + What if the force applied is variable? + For instance, + imagine a climber pulling a 200 rope up a vertical face. + The rope becomes lighter as more is pulled in, + requiring less force and hence the climber performs less work. +

    +     + + + Work Done by a Variable Force +

    + In general, let F(x) be a force function on an interval [a,b]. + We want to measure the amount of work done applying the force F from x=a to x=b. + We can approximate the amount of work being done by partitioning [a,b] into subintervals + a=x_0\lt x_1 \lt \cdots \lt x_{n}=b and assuming that F is constant on each subinterval. + Let c_i be a value in the + ith subinterval [x_{i-1},x_{i}]. + Then the work done on this interval is approximately W_i\approx F(c_i)\cdot(x_{i}-x_{i-1}) = F(c_i)\dx_i, + a constant force the distance over which it is applied. + The total work is + + W = \sum_{i=1}^n W_i \approx \sum_{i=1}^n F(c_i)\dx_i + . +

    + +

    + This, of course, is a Riemann sum. + Taking a limit as the subinterval lengths go to zero gives an exact value of work which can be evaluated through a definite integral. +

    + + + Work +

    + Let F(x) be a continuous function on [a,b] describing the amount of force being applied to an object in the direction of travel from distance x=a to distance x=b. + The total work W done on [a,b] is + integrationwork + work + + W = \int_a^b F(x)\, dx + . +

    +     + + + Computing work performed: applying variable force + +

    + A 60 climbing rope is hanging over the side of a tall cliff. + How much work is performed in pulling the rope up to the top, + where the rope has a linear mass density of + 66? +

    +     + +

    + We need to create a force function F(x) on the interval [0,60]. + To do so, we must first decide what x is measuring: + is it the length of the rope still hanging or is it the amount of rope pulled in? + As long as we are consistent, either approach is fine. + We adopt for this example the convention that x is the amount of rope pulled in. + This seems to match intuition better; + pulling up the first 10 meters of rope involves x=0 to x=10 instead of x=60 to x=50. +

    + +

    + As x is the amount of rope pulled in, + the amount of rope still hanging is 60-x. + This length of rope has a mass of 66 or + 0.066. + The mass of the rope still hanging is 0.066(60-x) ; + multiplying this mass by the acceleration of gravity, + 9.8, + gives our variable force function + + F(x) = (9.8)(0.066)(60-x) = 0.6468(60-x) + . +

    + +

    + Thus the total work performed in pulling up the rope is + + W = \int_0^{60} 0.6468(60-x)\, dx = 1,164.24 \,\text{J} + . +

    + +

    + By comparison, + consider the work done in lifting the entire rope 60 meters. + The rope weighs 60\times 0.066 \times 9.8 = 38.808 N, so the work applying this force for 60 meters is + 60\times 38.808 = 2,328.48 J. This is exactly twice the work calculated before (and we leave it to the reader to understand why.) +

    +     +     + + + Computing work performed: applying variable force + +

    + Consider again pulling a 60 rope up a cliff face, + where the rope has a mass of 66. + At what point is exactly half the work performed? +

    +     + +

    + From + we know the total work performed is 1,164.24 J. + We want to find a height h such that the work in pulling the rope from a height of x=0 + to a height of x=h is 582.12, or half the total work. + Thus we want to solve the equation + + \int_0^h 0.6468(60-x)\, dx = 582.12 + + for h. + + \int_0^h 0.6468(60-x)\, dx \amp = 582.12 + \left(38.808x-0.3234x^2\right)\Big|_0^h \amp =582.12 + 38.808h-0.3234h^2 \amp =582.12 + -0.3234h^2+38.808h-582.12 \amp =0 . + Apply the Quadratic Formula: + h\amp =17.57 \,\text{ and } \,102.43 + +

    + +

    + As the rope is only 60 long, + the only sensible answer is h=17.57. + Thus about half the work is done pulling up the first + 17.57; + the other half of the work is done pulling up the remaining + 42.43. +

    + + +     +     + + + Computing work performed: applying variable force + +

    + A box of 100 of sand is being pulled up at a uniform rate a distance of + 50 over 1 minute. + The sand is leaking from the box at a rate of 1. + The box itself weighs 5 and is pulled by a rope weighing + 0.2. +

    + +

    +

      +
    1. +

      + How much work is done lifting just the rope? +

      +
      2. + +
    2. +

      + How much work is done lifting just the box and sand? +

      +
      3. + +
    3. +

      + What is the total amount of work performed? +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + We start by forming the force function F_r(x) for the rope + (where the subscript denotes we are considering the rope). + As in the previous example, + let x denote the amount of rope, in feet, pulled in. + (This is the same as saying x denotes the height of the box.) + The weight of the rope with x feet pulled in is F_r(x) = 0.2(50-x) = 10-0.2x. + (Note that we do not have to include the acceleration of gravity here, + for the weight of the rope per foot is given, + not its mass per meter as before.) The work performed lifting the rope is + + W_r = \int_0^{50} (10-0.2x)\, dx = 250\,\text{ft--lb} + . +

      +
      2. + +
    2. +

      + The sand is leaving the box at a rate of + 1. + As the vertical trip is to take one minute, + we know that 60 + will have left when the box reaches its final height of + 50. + Again letting x represent the height of the box, + we have two points on the line that describes the weight of the sand: + when x=0, the sand weight is + 100, + producing the point (0,100); + when x=50, the sand in the box weighs + 40, + producing the point (50,40). + The slope of this line is \frac{100-40}{0-50} = -1.2, + giving the equation of the weight of the sand at height x as w(x) = -1.2x+100. + The box itself weighs a constant + 5, + so the total force function is F_b(x) = -1.2x+105. + Integrating from x=0 to x=50 gives the work performed in lifting box and sand: + + W_b = \int_0^{50} (-1.2x+105)\, dx = 3750\,\text{ft--lb} + . +

      +
      3. + +
    3. +

      + The total work is the sum of W_r and W_b: + 250+3750=4000 ftlb. + We can also arrive at this via integration: + + W \amp = \int_0^{50} (F_r(x)+F_b(x))\, dx + \amp = \int_0^{50} (10-0.2x-1.2x+105)\, dx + \amp = \int_0^{50} (-1.4x+115) \, dx + \amp = 4000 \,\text{ft--lb} + . +

      +
      4. +
    +

    +     +     +     + + + Hooke's Law and Springs + +

    + Hooke's Law states that the force required to compress or stretch a spring x + units from its natural length is proportional to x; + that is, this force is F(x) = kx for some constant k. + For example, if a force of 1 + stretches a given spring 2, + then a force of 5 + will stretch the spring 10. + Converting the distances to meters, + we have that stretching this spring 0.02 + requires a force of F(0.02) = k(0.02) = 1 + , + hence k = 1/0.02 = 50 + . + Hooke's Law +

    + + + Computing work performed: stretching a spring + +

    + A force of 20 + stretches a spring from a natural length of 7 inches to a length of 12 inches. + How much work was performed in stretching the spring to this length? +

    +     + +

    + In many ways, + we are not at all concerned with the actual length of the spring, + only with the amount of its change. + Hence, we do not care that 20 + of force stretches the spring to a length of 12 inches, + but rather that a force of + 20 + stretches the spring by 5 inches. + This is illustrated in ; + we only measure the change in the spring's length, + not the overall length of the spring. +

    + + +
    + Illustrating the important aspects of stretching a spring in computing work in + + + Image of a spring showing the spring both before and after streching. + In the first illustration, we see the unstreched spring with a block attached to it on its right side, with a rightward facing force vector labeled F, showing the direction of the force that is applied by the spring. + On the x-axis, the block that the spring is going to move is pictured to be centered at x=1. + In the second illustration, we see the streched spring, with the same block still attached on the right side. + This time, the block is centered at x=6, showing the spring moved the block a total of 5 inches to the right. + + Image showing the important aspects of strenching a spring which are used in computing work. + + + \begin{tikzpicture} + + \begin{scope}[xscale=.2,yscale=.5,shift={(.5,0)}] + \draw [thick,firstcolor] (-1,.5) -- (-.5,.5) -- (0,0); + + \foreach \x in {0,1,...,5} + { + \begin{scope}[shift={(\x*2,0)}] + \draw [thick,firstcolor] (0,0) -- (1,1) -- (2,0); + \end{scope} + } + + \begin{scope}[shift={(12,0)}] + \draw [thick,firstcolor] (0,0) -- (1,1) -- (1.5,.5) -- (2,.5); + \end{scope} + + \end{scope} + + \draw [shift={(.9,0)},thick,left color=firstcolor!70,right color=firstcolor!15,firstcolor] (2,0) rectangle (2.5,.5); + \draw [shift={(.9,0)},->,>=stealth] (2.75,.15)-= node [pos=.5,above] {$F$} (3.25,.15); + + \begin{scope}[shift={(0,-.25)}] + \draw [thick] (-.5,0) -- (5.5,0); + + \foreach \x in {0,...,6} + { + \draw (\x*.4+2.8,.1) -- (\x*.4+2.8,-.1) node [below] { $\x$}; + } + \end{scope} + + \begin{scope}[shift={(0,-1.75)}] + + \begin{scope}[xscale=.343,yscale=.5,shift={(.5,0)}] + \draw [thick,firstcolor] (-1,.5) -- (-.5,.5) -- (0,0); + \foreach \x in {0,1,...,5} + { + \begin{scope}[shift={(\x*2,0)}] + \draw [thick,firstcolor] (0,0) -- (1,1) -- (2,0); + \end{scope} + } + + \begin{scope}[shift={(12,0)}] + \draw [thick,firstcolor] (0,0) -- (1,1) -- (1.5,.5) -- (2,.5); + \end{scope} + \end{scope} + + \begin{scope}[shift={(0,-.25)}] + \draw [thick] (-.5,0) -- (5.5,0); + + \foreach \x in {0,...,6} + { + \draw (\x*.4+2.8,.1) -- (\x*.4+2.8,-.1) node [below] { $\x$}; + } + \end{scope} + + \draw [shift={(2.9,0)},thick,left color=firstcolor!70,right color=firstcolor!15,firstcolor] (2,0) rectangle (2.5,.5); + + \end{scope} + + \end{tikzpicture} + + + + +
    + +

    + Converting the units of length to feet, we have + + F(5/12) = 5/12k = 20\,\text{lb} + . +

    + +

    + Thus k = 48 and F(x) = 48x. +

    + +

    + We compute the total work performed by integrating F(x) from x=0 to x=5/12: + + W \amp = \int_0^{5/12} 48x \, dx + \amp = 24x^2\Big|_0^{5/12} + \amp = 25/6 \approx 4.1667\,\text{ft--lb} + . +

    +     +     +     + + + Pumping Fluids +

    + Another useful example of the application of integration to compute work comes in the pumping of fluids, + often illustrated in the context of emptying a storage tank by pumping the fluid out the top. + This situation is different than our previous examples for the forces involved are constant. + After all, the force required to move one cubic foot of water (about 62.4 ) + is the same regardless of its location in the tank. + What is variable is the distance that cubic foot of water has to travel; + water closer to the top travels less distance than water at the bottom, + producing less work. +

    + + + Weight and Mass densities + + + Fluid + lb/ft^3 + kg/m^3 + + + + + + + + Concrete + 150 + 2400 + + + Fuel Oil + 55.46 + 890.13 + + + Gasoline + 45.93 + 737.22 + + + Iodine + 307 + 4927 + + + Methanol + 49.3 + 791.3 + + + Mercury + 844 + 13546 + + + Milk + 63.665.4 + 10201050 + + + Water + 62.4 + 1000 + + + +
    + +

    + We demonstrate how to compute the total work done in pumping a fluid out of the top of a tank in the next two examples. +

    + + + Computing work performed: pumping fluids + +

    + A cylindrical storage tank with a radius of + 10 + and a height of 30 + is filled with water, which weighs approximately + 62.4. + Compute the amount of work performed by pumping the water up to a point 5 feet above the top of the tank. +

    +     + +

    + We will refer often to + which illustrates the salient aspects of this problem. +

    + +
    + Illustrating a water tank in order to compute the work required to empty it in + + + + Image of a cylindrical storage tank with a radius of 10 and a height of 30ft. + On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank. + The image also contains the ith subinterval of y, which corresponds to the shaded region between y_{i-1} and y_i on the y-axis. + The height of this subinterval of y given by y_{i}-y_{i-1} is also labeled \Delta y_i. + Additionally, the distance from the top of this subinterval, occuring at y_{i} and the point 5 feet above the tank is given on the y-axis as 35-y_i. + + Illustration of a cylindrical water tank with measurements used to compute the work required to empty it. + + + \begin{tikzpicture}[x=.1cm,y=.125cm,>=stealth] + + \draw [->] (0,-2) -- (0,37) node [above] { $y$}; + + \draw (-1,0) node [left] { 0} -- (1,0) + (-1,30) node [left] { 30} -- (1,30) + (-1,35) node [left] { 35} -- (1,35); + + \draw (2,12) -- (4,12) (2,35) -- (4,35) + (3,12) -- node [pos=2,rotate=90] { $35-y_i$} (3,18) + (3,29) -- (3,35); + + \begin{scope}[xscale=4,shift={(4.5,0)}] + + \draw (0,30) -- node [pos=.5,above,shift={(-4pt,-1pt)}] { $10$} (2.9,29.22); + \draw [thick] (3,30) --(3,0) arc (0:-180:3) -- (-3,30); + \draw [thick](0,30) circle (3); + \draw [thick,dashed] (3,0) arc (0:180:3); + + \foreach \y/\x in {12/{$y_{i-1}$},16/{$y_{i}$}} + { + \draw (3,\y) node [right] {\x} arc (0:-180:3); + \draw [dashed] (3,\y) arc (0:180:3); + } + + \draw (5.9,14) node { $\left.\rule{0pt}{.3cm}\right\}\Delta y_i$}; + \draw [left color=firstcolor,right color=firstcolor!15,thick] (0,16) circle (3); + \draw [left color=firstcolor,right color=firstcolor!15,thick] (-3,16) -- (-3,12) arc (180:360:3) -- (3,16) arc (360:180:3); + + \end{scope} + + \end{tikzpicture} + + + + +
    + +

    + We start as we often do: we partition an interval into subintervals. + We orient our tank vertically since this makes intuitive sense with the base of the tank at y=0. + Hence the top of the water is at y=30, + meaning we are interested in subdividing the y-interval [0,30] into n subintervals as + + 0 = y_0 \lt y_1 \lt \cdots \lt y_{n} = 30 + . +

    + +

    + Consider the work W_i of pumping only the water residing in the ith subinterval, + illustrated in . + The force required to move this water is equal to its weight which we calculate as volume density. + The volume of water in this subinterval is V_i = 10^2\pi \Delta y_i; + its density is 62.4. + Thus the required force is 6240\pi\Delta y_i . +

    + +

    + We approximate the distance the force is applied by using any y-value contained in the ith subinterval; + for simplicity, we arbitrarily use y_i for now + (it will not matter later on). + The water will be pumped to a point 5 feet above the top of the tank, that is, + to the height of y=35 . + Thus the distance the water at height y_i travels is 35-y_i . +

    + +

    + In all, the approximate work W_i performed in moving the water in the + ith subinterval to a point 5 feet above the tank is + + W_i \approx 6240\pi\Delta y_i(35-y_i) + . +

    + +

    + To approximate the total work performed in pumping out all the water from the tank, + we sum all the work W_i performed in pumping the water from each of the n subintervals of [0,30]: + + W \approx \sum_{i=1}^n W_i = \sum_{i=1}^n 6240\pi\Delta y_i(35-y_i) + . +

    + +

    + This is a Riemann sum. + Taking the limit as the subinterval length goes to 0 gives + + W \amp = \int_0^{30} 6240\pi(35-y)\, dy + \amp = 6240\pi\left(35y-1/2y^2\right)\Big|_0^{30} + \amp = 11,762,123 \,\text{ft--lb} + \amp \approx 1.176 \times 10^7 \,\text{ft--lb} + . +

    +     +     + +

    + We can streamline the above process a bit as we may now recognize what the important features of the problem are. + + shows the tank from + without the ith subinterval identified. +

    + +
    + A simplified illustration for computing work + + + + Image of the same cylindrical storage tank with a radius of 10 and a height of 30. + On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank. + Now instead of containing the ith subinterval of y, the image contains an arbitrarily chosen point y which is contained in the tank. + At this point y, the volume is given as V(y)=100\pi dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y. + Additionally, the distance from the arbitrary point y and the point 5 feet above the tank is given on the y-axis as 35-y. + + Illustration of a cylindrical water tank with simplified measurements used to compute the work required to empty it. + + + \begin{tikzpicture}[x=.1cm,y=.125cm,>=stealth] + + \draw [->] (0,-2) -- (0,37) node [above] { $y$}; + + \draw (-1,0) node [left] { 0} -- (1,0) + (-1,30) node [left] { 30} -- (1,30) + (-1,35) node [left] { 35} -- (1,35) + (-1,14) node [left] { $y$} -- (1,14); + + \draw (2,12) -- (4,12) (2,35) -- (4,35) + (3,12) -- node [pos=2,rotate=90] { $35-y_i$} (3,18) + (3,29) -- (3,35); + + \begin{scope}[xscale=4,shift={(4.5,0)}] + + \draw (0,30) -- node [pos=.5,above,shift={(-4pt,-1pt)}] { $10$} (2.9,29.22); + \draw [thick] (3,30) --(3,0) arc (0:-180:3) -- (-3,30); + \draw [thick] (0,30) circle (3); + \draw [thick,dashed] (3,0) arc (0:180:3); + \draw [left color=firstcolor,right color=firstcolor!15,thick,draw=firstcolor] (0,14) circle (3); + \draw [->] (4,16) node [above] { $V(y)=100\pi dy$} -- (2,14); + + \end{scope} + + \end{tikzpicture} + + + + +
    + +

    + Instead, we just draw one differential element. + This helps establish the height a small amount of water must travel along with the force required to move it + (where the force is volume density). +

    + +

    + We demonstrate the concepts again in the next examples. +

    + + + Computing work performed: pumping fluids + +

    + A conical water tank has its top at ground level and its base 10 feet below ground. + The radius of the cone at ground level is 2. + It is filled with water weighing 62.4 and is to be emptied by pumping the water to a spigot 3 feet above ground level. + Find the total amount of work performed in emptying the tank. +

    +     + +

    + The conical tank is sketched in . + We can orient the tank in a variety of ways; + we could let y=0 represent the base of the tank and y=10 represent the top of the tank, + but we choose to keep the convention of the wording given in the problem and let y=0 represent ground level and hence y=-10 represents the bottom of the tank. + The actual height of the water does not matter; + rather, we are concerned with the distance the water travels. +

    + +
    + A graph of the conical water tank in + + + + Image of a conical storage tank with its tip 10 below ground and the base at ground level with a radius of 2. + This means as we move further below ground level, the area of the circular cross-section of the conical storage tank decreases. + On the left side of the storage tank the y-axis is shown with measurements of y=3 which is 3 feet above the tank, y=0 which is the top of the tank, and y=-10 which is the tip of the cone, 10 feet below ground level. + The image contains an arbitrarily chosen point y between y=-10 and y=0 which is contained in the tank. + At this point y, the volume is given as V(y)=\pi (\frac{y}{5} + 2)^2 dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y. + Additionally, the distance from the arbitrary point y and the point 3 feet above the tank is given on the y-axis as 3-y. + + Illustration of a conical water tank with measurements used to compute the work required to empty it. + + + \begin{tikzpicture}[x=.1cm,y=.125cm,>=stealth] + + \draw [->] (0,-2) -- (0,37) node [above] { $y$}; + + \draw (-1,0) node [left] { $-10$} -- (1,0) + (-1,30) node [left] { 0} -- (1,30) + (-1,35) node [left] { 3} -- (1,35) + (-1,14) node [left] { $y$} -- (1,14); + + \draw (2,12) -- (4,12) (2,35) -- (4,35) + (3,12) -- node [pos=2,rotate=90] { $3-y_i$} (3,18) + (3,29) -- (3,35); + + \begin{scope}[xscale=4,shift={(5,0)}] + + \draw (0,30) -- node [pos=.5,above,shift={(-4pt,-1pt)}] { $2$} (2.9,29.22); + \draw [thick](3,30) -- (0,0) -- (-3,30); + \draw [thick] (0,30) circle (3); + \draw [left color=firstcolor,right color=firstcolor!15,thick,draw=firstcolor] (0,14) circle (1.4); + \draw [->] (5,6.5) node [below,shift={(10pt,0pt)}] { $V(y)=\pi(\frac y5+2)^2 dy$} -- (0,14); + + \end{scope} + + \end{tikzpicture} + + + + +
    + +

    + The figure also sketches a differential element, + a cross-sectional circle. + The radius of this circle is variable, depending on y. + When y=-10, the circle has radius 0; + when y=0, the circle has radius 2. + These two points, (-10,0) and (0,2), + allow us to find the equation of the line that gives the radius of the cross-sectional circle, + which is r(y) = 1/5y+2. + Hence the volume of water at this height is V(y)=\pi(1/5y+2)^2dy, + where dy represents a very small height of the differential element. + The force required to move the water at height y is F(y) = 62.4\times V(y). +

    + +

    + The distance the water at height y travels is given by h(y)=3-y. + Thus the total work done in pumping the water from the tank is + + W \amp = \int_{-10}^0 62.4\pi(1/5y+2)^2(3-y)\, dy + \amp = 62.4\pi\int_{-10}^0\left(-\frac1{25}y^3-\frac{17}{25}y^2-\frac85y+12\right)\, dy + \amp = 62.2\pi\cdot\frac{220}{3} \approx 14,376 \text{ft--lb} + . +

    +     +     + + + Computing work performed: pumping fluids + +

    + A rectangular swimming pool is 20 wide and has a 3 + shallow end and a 6 deep end. + It is to have its water pumped out to a point 2 above the current top of the water. + The cross-sectional dimensions of the water in the pool are given in ; + note that the dimensions are for the water, not the pool itself. + Compute the amount of work performed in draining the pool. +

    +
    + The cross-section of a swimming pool filled with water in + + + + Image of the swimming pool described in the example. + The pool is 25 long. + On the leftmost side of the image is the deep end of the pool, which is + 10 long having a depth of 6. + After the 10 mark, + the bottom of the pool slopes up until it rises to a depth of 3 + a distance of 5 away in the x dimension. + The shallow end of the pool has a depth of 3 + and length of 10. + The end of the shallow end marks the end of the pool. + + Illustration of the swimming pool from the example. + + + \begin{tikzpicture}[x=.18cm,y=.2cm,>=stealth] + + \draw [fill=firstcolor!15,draw=firstcolor!15] (0,0) -- node [below,pos=.5] { 10 ft.} (10,0) -- (15,3) -- node [below,pos=.5] { 10 ft.} (25,3)-- node [right,pos=.5] { 3 ft.} (25,6)--(0,6) -- node [left,pos=.5] { 6 ft.} (0,0); + + \draw [thick] (0,0) -- (10,0) -- (15,3) -- (25,3) -- (25,7) -- node [above,pos=.5] { 25 ft} (0,7) -- cycle; + + \end{tikzpicture} + + + + +
    +     + +

    + For the purposes of this problem we choose to set y=0 to represent the bottom of the pool, + meaning the top of the water is at y=6. +

    + +
    + Orienting the pool and showing differential elements for + + + + Image of the same swimming pool described in the example. + The image now contains the two coordinate axes x and y. + On the y axis are five markings. + The label y=0 marks the bottom of the deep end of the pool, the label y=3 marks the bottom of the shallow end of the pool. + The label y=6 marks the top of the water in the pool, and the label y=8 marks a point 2 above the top of the water. + The image also contains an arbitrary label y lying somewhere below the water level of the pool. + On the x-axis, the label x=0 marks the leftmost edge of the pool, x=10 marks the end of the deep end and after the pool slopes up to a depth of 3, the label x=15 marks the start of the shallow end. + The distance between the leftmost edge and rightmost edge of the pool when y is above y=3, which is the depth of shallow end, is constant, and is given by the length of the pool. + On the other hand, the distance between the same two edges of the pool when y is below y=3 is the length of the deep end, plus the additional distance in the x direction between the end of the deep end, x=10, and the point at which the arbitrary y level meets the sloped portion of the pool. + The sloped portion of the pool is given as the line which passes through the points (10,0) and (15,3), which are the end of the deep end, and the start of the shallow end, respectively. + + Illustration of the swimming pool from the example showing necessary differential elements. + + + \begin{tikzpicture}[x=.18cm,y=.2cm,>=stealth] + + \draw[>=stealth,->] (-2,-1) -- ( -2,10) node [above] { $y$}; + + \foreach \y/\x in {0/0,1.5/$y$,3/3,6/6,8/8} + { + \draw (-2.5,\y) node [left] { $\x$} -- (-1.5,\y); + } + + \draw [thick] (0,0) -- (10,0) -- (15,3) -- (25,3) -- (25,6) -- (0,6) -- (0,0); + + \draw [thick,firstcolor] (0,5) -- (25,5) + (0,1.5) -- (12.5,1.5); + + \draw [fill=black] (10,0) circle (2.4pt) node [shift={(15pt,-2pt)}] { $(10,0)$} + (15,3) circle (2.4pt) node [shift={(15pt,-6pt)}] { $(15,3)$}; + + \draw[>=stealth,->] (-2,-2) -- (20,-2) node [right] { $x$}; + + \foreach \y in {0,10,15} + { + \draw (\y,-2.5) node [below] { $\y$} -- (\y,-1.5); + } + + \end{tikzpicture} + + + + +
    + +

    + + shows the pool oriented with this y-axis, + along with 2 differential elements as the pool must be split into two different regions. +

    + +

    + The top region lies in the y-interval of [3,6], + where the length of the differential element is 25 as shown. + As the pool is 20 wide, + this differential element represents a thin slice of water with volume V(y) = 20\cdot25\cdot dy. + The water is to be pumped to a height of y=8, + so the height function is h(y) = 8-y. + The work done in pumping this top region of water is + + W_t = 62.4\int_3^6 500(8-y)\, dy = 327,600 \,\text{ft--lb} + . +

    + +

    + The bottom region lies in the y-interval of [0,3]; + we need to compute the length of the differential element in this interval. +

    + +

    + One end of the differential element is at x=0 and the other is along the line segment joining the points (10,0) and (15,3). + The equation of this line is y= 3/5(x-10); + as we will be integrating with respect to y, + we rewrite this equation as x=5/3y+10. + So the length of the differential element is a difference of x-values: + x=0 and x=5/3y+10, + giving a length of x=5/3y+10. +

    + +

    + Again, as the pool is 20 wide, + this differential element represents a thin slice of water with volume V(y) = 20\cdot(5/3y+10)\cdot dy; + the height function is the same as before at h(y)=8-y. + The work performed in emptying this part of the pool is + + W_b = 62.4\int_0^3 20(5/3y+10)(8-y)\, dy = 299,520\,\text{ft--lb} + . +

    + +

    + The total work in empyting the pool is + + W = W_b+W_t = 327,600+299,520 = 627,120\,\text{ft--lb} + . +

    + +

    + Notice how the emptying of the bottom of the pool performs almost as much work as emptying the top. + The top portion travels a shorter distance but has more water. + In the end, this extra water produces more work. +

    +     +     + +

    + The next section introduces one final application of the definite integral, + the calculation of fluid force on a plate. +

    +     + + + + Terms and Concepts + + +

    + What are the typical units of work? +

    +     + + + +

    + In SI units, it is one joule, + , one newtonmeter, + or + + + In Imperial Units, it is ftlb. +

    +     +     + + + +

    + If a man has a mass of + 80 + on Earth, + will his mass on the moon be bigger, smaller, or the same? +

    +     + + + +

    + Bigger +

    +     +     + + +

    + Smaller +

    +     +     + + +

    + The same +

    +     +     +     +     + + + +

    + If a woman weighs 130 on Earth, + will her weight on the moon be bigger, smaller, or the same? +

    +     + + + +

    + Bigger +

    +     +     + + +

    + Smaller +

    +     +     + + +

    + The same +

    +     +     +     +     + + + +

    + Fill in the blanks: +

    + +

    + Some integrals in this section are set up by multiplying a variable by a constant distance; + others are set up by multiplying a constant force by a variable . +

    +     + + + + + + + + + + + + +     +     + + Problems + + +

    + A 100 rope, weighing + 0.1, + hangs over the edge of a tall building. +

    +     + + + +

    + How much work is done pulling the entire rope to the top of the building? +

    +     + +

    + 500 ftlb +

    +     +     + + + +

    + How much rope is pulled in when half of the total work is done? +

    +     + +

    + 100-50\sqrt{2} \approx 29.29 ft +

    +     +     +     + + + +

    + A 50 rope, + with a mass density of 0.2, + hangs over the edge of a tall building. +

    +     + + + +

    + How much work is done pulling the entire rope to the top of the building? +

    +     + +

    + 2450 J +

    +     +     + + + +

    + How much work is done pulling in the first 20 m? +

    +     + +

    + 1568 J +

    +     +     +     + + + +

    + A rope of length \ell hangs over the edge of tall cliff. + (Assume the cliff is taller than the length of the rope.) + The rope has a weight density of d . +

    +     + + + +

    + How much work is done pulling the entire rope to the top of the cliff? +

    +     + +

    + \frac12\cdot d\cdot l^2 ftlb +

    +     +     + + + +

    + What percentage of the total work is done pulling in the first half of the rope? +

    +     + +

    + 75 % +

    +     +     + + + +

    + How much rope is pulled in when half of the total work is done? +

    +     + +

    + \ell(1-\sqrt{2}/2) \approx 0.2929\ell +

    +     +     +     + + + +

    + A 20 rope with mass density of 0.5 hangs over the edge of a 10 building. + How much work is done pulling the rope to the top? +

    +     + +

    + 735 J +

    +     +     + + + +

    + A crane lifts a 2000 + load vertically 30 + with a 1 cable weighing + 1.68. +

    +     + + + +

    + How much work is done lifting the cable alone? +

    +     + +

    + 756 ftlb +

    +     +     + + + +

    + How much work is done lifting the load alone? +

    +     + +

    + 60,000 ftlb +

    +     +     + + + +

    + Could one conclude that the work done lifting the cable is negligible compared to the work done lifting the load? +

    +     + +

    + Yes, for the cable accounts for about 1% of the total work. +

    +     +     +     + + + +

    + A100 bag of sand is lifted uniformly 120 in one minute. + Sand leaks from the bag at a rate of 1/4. + What is the total work done in lifting the bag? +

    +     + +

    + 11,100 ftlb +

    +     +     + + + +

    + A box weighing 2 lifts 10 of sand vertically 50. + A crack in the box allows the sand to leak out such that 9 of sand is in the box at the end of the trip. + Assume the sand leaked out at a uniform rate. + What is the total work done in lifting the box and sand? +

    +     + +

    + 575 ftlb +

    +     +     + + + +

    + A force of 1000 compresses a spring 3. + How much work is performed in compressing the spring? +

    +     + +

    + 125 ftlb +

    +     +     + + + +

    + A force of 2 stretches a spring 5. + How much work is performed in stretching the spring? +

    +     + +

    + 0.05 J +

    +     +     + + + +

    + A force of 50 compresses a spring from a natural length of 18 to 12. + How much work is performed in compressing the spring? +

    +     + +

    + 12.5 ftlb +

    +     +     + + + +

    + A force of 20 stretches a spring from a natural length of 6 to 8. + How much work is performed in stretching the spring? +

    +     + +

    + 5/3 ftlb +

    +     +     + + + +

    + A force of 7 stretches a spring from a natural length of 11 to 21. + How much work is performed in stretching the spring from a length of 16 to 21? +

    +     + +

    + 0.2625 = 21/80 J +

    +     +     + + + +

    + A force of f + stretches a spring d from its natural length. + How much work is performed in stretching the spring? +

    +     + +

    + f\cdot d/2 J +

    +     +     + + + +

    + A 20 weight is attached to a spring. + The weight rests on the spring, + compressing the spring from a natural length of 1 to 6. +

    + +

    + How much work is done in lifting the box 1.5 (i.e, the spring will be stretched 1 beyond its natural length)? +

    +     + +

    + 45 ftlb +

    +     +     + + + +

    + A 20 weight is attached to a spring. + The weight rests on the spring, + compressing the spring from a natural length of 1 to 6. +

    + +

    + How much work is done in lifting the box 6 (i.e, bringing the spring back to its natural length)? +

    +     + +

    + 5 ftlb +

    +     +     + + + +

    + A 5 tall cylindrical tank with radius of 2 is filled with 3 of gasoline, + with a mass density of 737.22. + Compute the total work performed in pumping all the gasoline to the top of the tank. +

    +     + +

    + 953,284 J +

    +     +     + + + +

    + A 6 cylindrical tank with a radius of 3 is filled with water, + which has a weight density of 62.4. + The water is to be pumped to a point 2 above the top of the tank. +

    +     + + + +

    + How much work is performed in pumping all the water from the tank? +

    +     + +

    + 52,929.6 ftlb +

    +     +     + + + +

    + How much work is performed in pumping 3 of water from the tank? +

    +     + +

    + 18,525.3 ftlb +

    +     +     + + + +

    + At what point is 1/2 of the total work done? +

    +     + +

    + When 3.83 of water have been pumped from the tank, + leaving about 2.17 in the tank. +

    +     +     +     + + + +

    + A gasoline tanker is filled with gasoline with a weight density of 45.93. + The dispensing valve at the base is jammed shut, + forcing the operator to empty the tank via pumping the gas to a point 1 above the top of the tank. + Assume the tank is a perfect cylinder, 20 long with a diameter of 7.5. + How much work is performed in pumping all the gasoline from the tank? +

    +     + +

    + 192,767 ftlb. + Note that the tank is oriented horizontally. + Let the origin be the center of one of the circular ends of the tank. + Since the radius is 3.75, the fluid is being pumped to y=4.75; + thus the distance the gas travels is h(y)=4.75-y. + A differential element of water is a rectangle, + with length 20 and width 2\sqrt{3.75^2-y^2}. + Thus the force required to move that slab of gas is F(y) = 40\cdot45.93\cdot\sqrt{3.75^2-y^2}dy. + Total work is \int_{-3.75}^{3.75} 40\cdot45.93\cdot(4.75-y)\sqrt{3.75^2-y^2}\, dy. + This can be evaluated without actual integration; + split the integral into \int_{-3.75}^{3.75} 40\cdot45.93\cdot(4.75)\sqrt{3.75^2-y^2}\, dy + \int_{-3.75}^{3.75} 40\cdot45.93\cdot(-y)\sqrt{3.75^2-y^2}\, dy. + The first integral can be evaluated as measuring half the area of a circle; + the latter integral can be shown to be 0 without much difficulty. + (Use substitution and realize the bounds are both 0.) +

    +     +     + + + +

    + A fuel oil storage tank is 10 deep with trapezoidal sides, 5 at the top and 2 at the bottom, + and is 15 wide + (see diagram below). + Given that fuel oil weighs 55.46, + find the work performed in pumping all the oil from the tank to a point 3 above the top of the tank. +

    + + + + Image of a fuel storage tank. + The storage tank has a base that has a 2 length and 15 width. + The forward, left and back walls of the tank are perpendicular to the base, while the right side of the tank slopes upward further away to the right side. + The top of the tank has a length of 5 due to the slope of the right wall of the fuel tank and the top same 15 width as the base. + The depth of the tank, which is the distance between the top and bottom of the tank is 10. + + Illustration of the fuel storage tank from the problem. + + + \begin{tikzpicture}[x={(1,0)},z={(0,1)},y={(.5,.87)},xscale=.38,yscale=.25] + + \draw [thick] (0,0,0) -- node [above,rotate=90,pos=.5] {10} (0,0,-10) -- node [below, pos=.5] {2} (2,0,-10) -- node [right, pos=.5] {15} (2,5,-10) -- (5,5,0) + (5,5,0) -- (5,0,0) + (5,0,0) -- (0,0,0) -- (0,5,0) -- node [above,pos=.5] { 5}(5,5,0) + (5,0,0) -- (2,0,-10); + + \draw [thick,dashed] (0,0,-10) -- (0,5,-10) -- (2,5,-10) + (0,5,-10) -- (0,5,0); + + \end{tikzpicture} + + + + +     + +

    + 212,135 ftlb +

    +     +     + + + +

    + A conical water tank is 5 deep with a top radius of 3. + (This is similar to .) + The tank is filled with pure water, + with a mass density of 1000. +

    +     + + + +

    + Find the work performed in pumping all the water to the top of the tank. +

    +     + +

    + approx. 577,000 J +

    +     +     + + + +

    + Find the work performed in pumping the top + 2.5 of water to the top of the tank. +

    +     + +

    + approx. 399,000 J +

    +     +     + + + +

    + Find the work performed in pumping the top half of the water, + by volume, to the top of the tank. +

    +     + +

    + approx 110,000 J (By volume, + half of the water is between the base of the cone and a height of 3.9685 m. + If one rounds this to 4, the work is approx 104,000 J.) +

    +     +     +     + + + +

    + A water tank has the shape of a truncated cone, + with dimensions given below, + and is filled with water with a weight density of 62.4. + Find the work performed in pumping all water to a point 1 above the top of the tank. +

    + + + + Image of a water tank in the shape of a circular truncated cone. + The water tank is laying on its base and is in the shape of a cone, from which a small portion including the top is sliced off paralell to the base. + The water tank has a circular base with a radius of 5. + From the base, the tank has a height of 10, and a circular top with a radius of 2. + + Illustration of the water tank in the shape of a truncated cone. + + + \begin{tikzpicture}[xscale=1.2,yscale=0.9] + + \draw [thick] (0,0) circle (1) + (-1,.1) -- (-1.5,-1.9) + (1,.1) -- (1.5,-1.9); + + \draw (0,0)-- node [above,pos=.5] {2 ft} (1,0) + (0,-2) -- node [above,pos=.5] {5 ft} (1.5,-2) + (1.7,0) -- (2.3,0) + (1.7,-2) -- (2.3,-2) + (2,0) -- node [pos=.5,right] {10 ft} (2,-2); + + \draw [thick] (-1.5,-2) arc (180:360:1.5); + \draw [thick,dashed] (-1.5,-2) arc (180:0:1.5); + + \end{tikzpicture} + + + + +     + +

    + 187,214 ftlb +

    +     +     + + + +

    + A water tank has the shape of an inverted pyramid, + with dimensions given below, + and is filled with water with a mass density of 1000. + Find the work performed in pumping all water to a point 5 above the top of the tank. +

    + + + + Image of the pyramidal water tank. + The water tank is in the shape of a pyramid with a square base which is laying on its tip, meaning that the square base is the top of the tank. + The top of the water tank has a square top with a side length of 2. + From the top, the tank has a depth of 7, which is the shortest distance between the square top of the tank and the peak of the pyramid, which is the bottom of the tank. + + Illustration of the water tank in the shape of an inverted pyramid. + + + \begin{tikzpicture}[scale=1.25] + + \draw [thick] (0,0) -- node [pos=.5,left]{2 m} (1,1) --node [above,pos=.5] {2 m} (3,1) -- (2,0) -- cycle + (0,0) -- (1.5,-2) + (3,1) -- (1.5,-2) + (2,0) -- (1.5,-2); + + \draw [thick,dashed] (1,1) -- (1.5,-2); + + \draw (3,.5) -- (3.6,.5) + (3,-2) -- (3.6,-2) + (3.3,.5) -- node [pos=.5,right] {7 m} (3.3,-2); + + \end{tikzpicture} + + + +     + +

    + 617,400 J +

    +     +     + + + +

    + A water tank has the shape of a truncated, + inverted pyramid, with dimensions given below, + and is filled with water with a mass density of 1000. + Find the work performed in pumping all water to a point 1 above the top of the tank. +

    + + + + Image of a water tank in the shape of a truncated pyramid. + The water tank has a square base with a side length of 2. + From the base, the pyramidal water tank expands on all sides until reaching the top of the tank. + The top of the water tank is a square with a side length of 5. + From the top, the tank has a depth of 9, which is the shortest distance between the bottom and top of the water tank. + + Illustration of the water tank in the shape of a truncated pyramid. + + + \begin{tikzpicture}[scale=1.25] + + \draw [thick] (0,0) node (A) {} -- node [pos=.5,left] {5 m} (1,1) node (B) {} -- node [above,pos=.5] {5 m} (3,1) node (C) {} -- (2,0) node (D) {} -- cycle; + + \begin{scope}[scale=.75,shift={(.5,-3)}] + \draw [thick] (0,0) node (AA) {} -- (1,1) node (BB) {} -- (3,1) node (CC) {} -- node [right,pos=.5] {2 m} (2,0) node (DD) {} -- node [pos=.5,below] {2 m} (0,0); + \end{scope} + + \draw [dashed,thick] (BB.center) -- (B.center); + + \draw [thick] (AA.center) -- (A.center) + (DD.center) -- (D.center) + (CC.center) -- (C.center); + + \draw (3.4,1) -- (4,1) + (3.4,-1.5) -- (4,-1.5) + (3.7,1) -- node [pos=.5,right] { 9 m} (3.7,-1.5); + + \end{tikzpicture} + + + + +     + +

    + 4,917,150 J +

    +     +     +     +     +
    +
    + Fluid Forces +

    + In the unfortunate situation of a car driving into a body of water, + the conventional wisdom is that the water pressure on the doors will quickly be so great that they will be effectively unopenable. + (Survival techniques suggest immediately opening the door, + rolling down or breaking the window, + or waiting until the water fills up the interior at which point the pressure is equalized and the door will open. + See Mythbusters episode #72 to watch Adam Savage test these options.) +

    + +

    + How can this be true? + How much force does it take to open the door of a submerged car? + In this section we will find the answer to this question by examining the forces exerted by fluids. +

    + +

    + We start with pressure, + which is related to force + by the following equations: + + \text{Pressure}\, = \,\frac{\text{Force}}{\text{Area}}\, \Leftrightarrow\, \text{Force}\, = \, \text{Pressure} \times\text{Area} + . +

    + +

    + In the context of fluids, we have the following definition. +

    + + + Fluid Pressure + +

    + Let w be the weight-density of a fluid. + The pressure p exerted on an object at depth d in the fluid is p = w\cdot d. + fluid pressure/force + integrationfluid force +

    +     +     + +

    + We use this definition to find the force + exerted on a horizontal sheet by considering the sheet's area. +

    + + + Computing fluid force + +

    +

      +
    1. +

      + A cylindrical storage tank has a radius of + 2 + and holds 10 + of a fluid with a weight-density of + 50. + (See .) + What is the force exerted on the base of the cylinder by the fluid? +

      + +
      + A cylindrical tank in + + + +

      + Illustration of a cylindrical storage tank with a radius of 2. + The cylindrical storage tank is then filled with water up to a marked level of 10 from the base of the tank. + The tank extends slightly above the 10 water level, but this region of the tank contains no water. +

      +       + Illustration of the cylindrical tank from the example. + + + \begin{tikzpicture}[>=stealth,scale=0.4] + + \begin{scope}[xscale=2] + + \draw [left color=firstcolor!15,right color=firstcolor!40,thick] (-2,10) -- (-2,0) arc (180:360:2) -- (2,10); + \draw [right color=firstcolor!15,left color=firstcolor!40,thick] (0,10) circle (2); + \draw [thick] (0,12) circle (2); + \draw [thick] (-2,12) -- (-2,10) + (2,12) -- (2,10); + + \draw [dashed,thick] (-2,0) arc (180:0:2); + \draw (0,0) -- node [above,pos=.5] { 2 ft} (2,0); + + \end{scope} + + \begin{scope}[shift={(5,0)}] + + \draw (-.5,0) -- (.5,0) + (-.5,10) -- (.5,10) + (0,0) -- node [pos=.5,fill=white,rotate=90] { 10 ft} (0,10); + + \end{scope} + + \end{tikzpicture} + + + + +
      +
      2. + +
    2. +

      + A rectangular tank whose base is a 5 + square has a circular hatch at the bottom with a radius of + 2. + The tank holds 10 + of a fluid with a weight-density of + 50. + (See .) + What is the force exerted on the hatch by the fluid? +

      + +
      + A rectangular tank in + + + +

      + Illustration of a rectangular storage tank with a square base. + The square base has a side length of 5. + The height of the storage tank extends slightly above the 10 mark, which is the amount of water contained in the tank. + On the square base of the tank is a centered circular hatch, having a radius of 2. +

      +       + Illustration of the rectangular storage tank with a circular hatch at the bottom. + + + \begin{tikzpicture}[x=.25cm,y=.25cm,>=stealth] + + \begin{scope}[xscale=2] + \begin{scope}[rotate=-30] + + \draw [thick] (-2.5,0) node (A) {} + (2.5,0) node (B) {} + (2.5,5) node (C) {}; + \draw [dashed,thick] (A.center) -- (-2.5,5) node (D) {} -- (C.center); + + \draw (0,2.5) node (O) {} + (-1,.77) node (OC) {}; + + \end{scope} + + \begin{scope}[shift={(0,10)}] + \begin{scope}[rotate=-30] + + \draw (-2.5,0) node (AAA) {} + (2.5,0) node (BBB) {} + (2.5,5) node (CCC) {} + (-2.5,5) node (DDD) {}; + + \end{scope} + \end{scope} + + \begin{scope}[shift={(0,12)}] + \begin{scope}[rotate=-30] + + \draw [thick] (-2.5,0) node (AA) {} + (2.5,0) node (BB) {} + (2.5,5) node (CC) {} + (-2.5,5) node (DD) {}; + + \end{scope} + \end{scope} + + \draw [left color=firstcolor!15,right color=firstcolor!40,thick] (AAA.center) -- (A.center)-- node [pos=.5,rotate=-15,below] { 5 ft} (B.center)--node [pos=.5,rotate=45,below] { 5 ft} (C.center) -- (CCC.center); + \draw [right color=firstcolor!15,left color=firstcolor!40,thick] (AAA.center) -- (BBB.center) -- (CCC.center)--(DDD.center) -- cycle; + + \draw [thick] (AA.center) -- (BB.center) -- (CC.center) -- (DD.center) -- cycle + (AA.center) -- (AAA.center) + (CC.center) -- (CCC.center) + (B.center) -- (BB.center) + (DD.center) --(DDD.center); + + \draw [thick,dashed] (DDD.center)--(D.center) -- (C.center) + (D.center) -- (A.center) + (O.center) circle (2) + (O.center) -- node [pos=.5,above,rotate=15] { 2 ft} (OC.center); + + \begin{scope}[shift={(5.5,3)}] + + \draw (-.5,0) -- (.5,0) + (-.5,10) -- (.5,10) + (0,0) -- node [pos=.5,fill=white,rotate=90] { 10 ft} (0,10); + + \end{scope} + \end{scope} + + \end{tikzpicture} + + + + +
      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + Using , + we calculate that the pressure exerted on the cylinder's base is + w\cdot d = 50 + 10=500. + The area of the base is \pi\cdot 2^2 = 4\pi . + So the force exerted by the fluid is + + F = 500\times 4\pi = 6283\,\text{lb} + . + Note that we effectively just computed the weight + of the fluid in the tank. +

      +
      2. + +
    2. +

      + The dimensions of the tank in this problem are irrelevant. + All we are concerned with are the dimensions of the hatch and the depth of the fluid. + Since the dimensions of the hatch are the same as the base of the tank in the previous part of this example, + as is the depth, we see that the fluid force is the same. + That is, F = 6283 . + A key concept to understand here is that we are effectively measuring the weight of a + 10 + column of water above the hatch. + The size of the tank holding the fluid does not matter. +

      +
      3. +
    +

    +     +     + +

    + The previous example demonstrates that computing the force exerted on a horizontally oriented plate is relatively easy to compute. + What about a vertically oriented plate? + For instance, + suppose we have a circular porthole located on the side of a submarine. + How do we compute the fluid force exerted on it? +

    + +

    + Pascal's Principle states that the pressure exerted by a fluid at a depth is equal in all directions. + Thus the pressure on any portion of a plate that is + 1 + below the surface of water is the same no matter how the plate is oriented. + (Thus a hollow cube submerged at a great depth will not simply be + crushed from above, + but the sides will also crumple in. + The fluid will exert force on all sides of the cube.) +

    + +

    + So consider a vertically oriented plate as shown in + submerged in a fluid with weight-density w. + What is the total fluid force exerted on this plate? + We find this force by first approximating the force on small horizontal strips. +

    + +
    + A thin, vertically oriented plate submerged in a fluid with weight-density w + + + +

    + Illustration of a thin, vertically oriented plate fully submerged in a fluid with weight-density w. + The plate is in the shape of an acute trapezoid. + On the plate, the image contains an arbitrarily placed small horizontal strip which stretches horizontally between the two edges of the submerged plate. + The small strip has a horizontal length measurement of \ell(c_i), and a vertical height measurement of \Delta y_i. + The vertical distance between the water level and the center of the small horizontal strip is given as d_i. +

    +     + Illustration of a thin, vertically oriented plate submerged in a fluid. + + + \begin{tikzpicture}[scale=1.32,>=stealth] + + \begin{scope} + + \draw [thick] (0,0) -- (1,2) -- (2,2) -- (4,0)--cycle; + \draw [secondcolor,fill=secondcolor!15,thick] (.5,.9) rectangle (3,1.1); + \draw (3,1) node [right] { $\left. \rule{0pt}{.2cm}\right\}\Delta y_i$}; + + \draw (.5,.5) -- (.5,.7) + (3,.5) -- (3,.7) + (.5,.6) -- node [pos=.5,below,] { $\ell(c_i)$} (3,.6); + + \foreach \x in {0.5,1.5,2.5,3.5} + { + \begin{scope}[shift={(\x*1,2.5)}] + \draw [firstcolor,thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw (.2,1) -- (.4,1) + (.2,2.5) -- (.4,2.5) + (.3,1) -- node [pos=.5,left] { $d_i$} (.3,2.5); + + \end{scope} + + \end{tikzpicture} + + + + +
    + +

    + Let the top of the plate be at depth b and let the bottom be at depth a. + (For now we assume that surface of the fluid is at depth 0, so if the bottom of the plate is + 3 under the surface, + we have a=-3. + We will come back to this later.) + We partition the interval [a,b] into n subintervals + + a = y_0 \lt y_1 \lt \cdots \lt y_{n} = b + , + with the ith subinterval having length \Delta y_i. + The force F_i exerted on the plate in the + ith subinterval is F_i = \text{Pressure} \times \text{Area}. +

    + +

    + The pressure is depth times the weight density w. + We approximate the depth of this thin strip by choosing any value d_i in [y_{i-1},y_{i}]; + the depth is approximately -d_i. + (Our convention has d_i being a negative number, + so -d_i is positive.) + For convenience, + we let d_i be an endpoint of the subinterval; + we let d_i = y_i. +

    + +

    + The area of the thin strip is approximately length width. + The width is \Delta y_i. + The length is a function of some y-value c_i in the ith subinterval. + We state the length is \ell(c_i). + Thus + + F_i \amp = \text{Pressure} \times \text{Area} + \amp = -y_i\cdot w \times \ell(c_i)\cdot\Delta y_i + . +

    + +

    + To approximate the total force, + we add up the approximate forces on each of the n thin strips: + + F = \sum_{i=1}^n F_i \approx \sum_{i=1}^n -w\cdot y_i\cdot\ell(c_i)\cdot\Delta y_i + . +

    + +

    + This is, of course, another Riemann Sum. + We can find the exact force by taking a limit as the subinterval lengths go to 0; + we evaluate this limit with a definite integral. +

    + + + Fluid Force on a Vertically Oriented Plate +

    + Let a vertically oriented plate be submerged in a fluid with weight-density w, + where the top of the plate is at y=b and the bottom is at y=a. + Let \ell(y) be the length of the plate at y. + fluid pressure/force + integrationfluid force + +

      +
    1. +

      + If y=0 corresponds to the surface of the fluid, + then the force exerted on the plate by the fluid is + + F=\int_a^b w\cdot(-y)\cdot\ell(y)\, dy + . +

      +
      2. + +
    2. +

      + In general, let d(y) represent the distance between the surface of the fluid and the plate at y. + Then the force exerted on the plate by the fluid is + + F=\int_a^b w\cdot d(y)\cdot\ell(y)\, dy + . +

      +
      3. +
    +

    +     + + + Finding fluid force + +

    + Consider a thin plate in the shape of an isosceles triangle as shown in , + submerged in water with a weight-density of + 62.4. + If the bottom of the plate is 10 + below the surface of the water, + what is the total fluid force exerted on this plate? +

    + +
    + A thin plate in the shape of an isosceles triangle in + + + +

    + Illustration of a thin plate in the shape of an isosceles triangle. + The triangle has one side having a length of 4. + The height of the triangle is also 4. + The length of the two remaining sides is equal but is not given in the image. + The triangle is oriented such that the side having a length of 4 is the top of the plate and lies parallel to the water level. + The distance between the top of the plate and the vertex lying across from the top plate is the height of the triangle, which is 4. +

    +     + Illustration of a thin plate in the shape of an isosceles triangle. + + + \begin{tikzpicture}[x=.8cm,y=.8cm,>=stealth] + + \draw[thick] (0,0) -- (2,4) -- node [pos=.5,above] { 4 ft} (-2,4) -- cycle; + + \draw (2.2,0) -- (2.6,0) + (2.2,4) -- (2.6,4) + (2.4,0) -- node [pos=.5,rotate=-90,above] { 4 ft} (2.4,4); + + \end{tikzpicture} + + + + +
    +     + +

    + We approach this problem in two different ways to illustrate the different ways can be implemented. + First we will let y=0 represent the surface of the water, + then we will consider an alternate convention. +

    + +

    +

      +
    1. +

      + We let y=0 represent the surface of the water; + therefore the bottom of the plate is at y=-10. + We center the triangle on the y-axis as shown in . + The depth of the plate at y is -y as indicated by the Key Idea. + We now consider the length of the plate at y. + + We need to find equations of the left and right edges of the plate. + The right hand side is a line that connects the points (0,-10) and (2,-6): + that line has equation x=1/2(y+10). + (Find the equation in the familiar y=mx+b format and solve for x.) + Likewise, the left hand side is described by the line x=-1/2(y+10). + The total length is the distance between these two lines: + \ell(y)=1/2(y+10) - (-1/2(y+10)) = y+10. +

      + +
      + Sketching the triangular plate in with the convention that the water level is at y=0 + + + +

      + Graph of the thin plate in the shape of an isosceles triangle with the convention that the water level is at y=0. + The triangle is oriented in the same way as described previously, with the top of the plate being the side having a length of 4 running parallel to the water level. + The graph contains the two coordinate axes. + The bottom of the plate, which is the vertex of the triangle lies 10 below the water level, which is marked as the point (0,-10). + The remaining two vertices of the triangle are the points (-2,-6) and (2,-6), with the line joining these two vertices making up the top of the plate. + The graph also contains an arbitrarily chosen thin slice of the plate, which spans the horizontal length of the triangular plate. + The red coloured thin slice occurs at the level y, which is an arbitrarily chosen value between y = -6 which marks the top of the plate and y = -10 which marks the bottom of the plate. + The distance between the water level and the thin horizontal slice is also given as d(y)=-y. +

      +       + Graph of a thin plate in the shape of an isosceles triangle. + + + \begin{tikzpicture}[scale=0.755,>=stealth] + + \draw[thick] (0,0) -- (2,4) -- (-2,4) -- cycle; + + \draw [fill=black] (2,4) circle (1pt) node [right] { $(2,-6)$} + (-2,4) circle (1pt) node [left] { $(-2,-6)$} + (0,0) circle (1pt) node [below right] { $(0,-10)$}; + + \draw [secondcolor,thick] (-1.5,3) -- (1.5,3) node [right,black] { $y$}; + \draw [->] (0,-1) -- (0,12) node [above] { $y$}; + \draw [->] (-3,10) -- (3,10) node [right] { $x$}; + + \foreach \x in {-2,-1,1,2} + { + \draw (\x,10.2) -- (\x,9.8) node [below] { $\x$}; + } + + \foreach \x in {-10,-8,-4,-2} + { + \draw (.2,{\x+10}) -- (-.2,{\x+10}) node [left] { $\x$}; + } + + \foreach \x in {-2,-1,0,1,2} + { + \begin{scope}[shift={(\x*1,10)}] + \draw [firstcolor,thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw (2,11) node { water line}; + + \draw (4.3,3) -- (4.7,3) + (4.3,10) -- (4.7,10) + (4.5,3) -- node [pos=.5,rotate=-90,above] { $d(y) = -y$} (4.5,10); + + \end{tikzpicture} + + + + +
      + +

      + The total fluid force is then: + + F \amp = \int_{-10}^{-6} 62.4(-y)(y+10)\, dy + \amp = 62.4\cdot \frac{176}{3} \approx 3660.8\,\text{lb} + . +

      +
      2. + +
    2. +

      + Sometimes it seems easier to orient the thin plate nearer the origin. + For instance, + consider the convention that the bottom of the triangular plate is at (0,0), + as shown in . + The equations of the left and right hand sides are easy to find. + They are y=2x and y=-2x, respectively, + which we rewrite as x= 1/2y and x=-1/2y. + Thus the length function is \ell(y) = 1/2y-(-1/2y) = y. +

      + +
      + Sketching the triangular plate in with the convention that the base of the triangle is at (0,0) + + + +

      + Graph of the thin plate in the shape of an isosceles triangle with the convention that the water level is at y=10. + The triangle is oriented in the same way as described previously, with the top of the plate being the side having a length of 4 running parallel to the water level. + The graph contains two the coordinate axes. + The bottom of the plate, which is the vertex of the triangle lies 10 below the water level, which is marked as the point (0,0). + The remaining two vertices of the triangle are the points (-2,4) and (2,4), with the line joining these two vertices making up the top of the plate. + The graph also contains an arbitrarily chosen thin slice of the plate, which spans the horizontal length of the triangular plate. + The red coloured thin slice occurs at the level y, which is an arbitrarily chosen value between y = 0 which marks the bottom of the plate and y = 4 which marks the top of the plate. + The distance between the water level and the thin horizontal slice is also given as d(y)=10-y. +

      +       + Graph of a thin plate in the shape of an isosceles triangle. + + + \begin{tikzpicture}[scale=0.755,>=stealth] + + \draw[thick] (0,0) -- (2,4) -- (-2,4) -- cycle; + + \draw [fill=black] (2,4) circle (1pt) node [right] { $(2,4)$} + (-2,4)circle (1pt) node [left] { $(-2,4)$} + (0,0)circle (1pt); + + \draw [secondcolor,thick] (-1.5,3) -- (1.5,3) node [right,black] { $y$}; + \draw [->] (0,-1) -- (0,12) node [above] { $y$}; + \draw [->] (-3,0) -- (3,0) node [right] { $x$}; + + \foreach \x in {-2,-1,1,2} + { + \draw (\x,0.2) -- (\x,-.2) node [below] { $\x$}; + } + + \foreach \x in {10,8,6,2} + { + \draw (.2,{\x}) -- (-.2,{\x}) node [left] { $\x$}; + } + + \foreach \x in {-2,-1,0,1,2} + { + \begin{scope}[shift={(\x*1,10)}] + \draw [firstcolor,thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw (2,11) node { water line}; + + \draw (4.1,3) -- (4.5,3) + (4.1,10) -- (4.5,10) + (4.3,3) -- node [pos=.5,rotate=-90,above] { $d(y) = 10-y$} (4.3,10); + + \end{tikzpicture} + + + + +
      + +

      + As the surface of the water is 10 above the base of the plate, + we have that the surface of the water is at y=10. + Thus the depth function is the distance between y=10 and y; + d(y) = 10-y. + We compute the total fluid force as: + + F \amp =\int_0^4 62.4(10-y)(y)\, dy + \amp \approx 3660.8\,\text{lb} + . +

      +
      3. +
    +

    + +

    + The correct answer is, of course, + independent of the placement of the plate in the coordinate plane as long as we are consistent. +

    +     +     + + + Finding fluid force + +

    + Find the total fluid force on a car door submerged up to the bottom of its window in water, + where the car door is a rectangle + 40 long and + 27 high + (based on the dimensions of a 2005 Fiat Grande Punto.) +

    +     + +

    + The car door, as a rectangle, + is drawn in . + Its length is 10/3 ft and its height is 2.25. + We adopt the convention that the top of the door is at the surface of the water, + both of which are at y=0. + Using the weight-density of water of 62.4, + we have the total force as + + F \amp = \int_{-2.25}^0 62.4(-y)10/3\, dy + \amp = \int_{-2.25}^0 -208y\, dy + \amp = -104y^2\Big|_{-2.25}^0 + \amp = 526.5 \,\text{lb} + . +

    + +

    + Most adults would find it very difficult to apply over 500 of force to a car door while seated inside, + making the door effectively impossible to open. + This is counter-intuitive as most assume that the door would be relatively easy to open. + The truth is that it is not, + hence the survival tips mentioned at the beginning of this section. +

    + +
    + Sketching a submerged car door in + + + +

    + Graph of a perfectly rectangular car door submerged in water with the convention that the water level is at y=0. + The coordinate axes are labeled in terms of feet. + The top of the rectangular door is 40 inches or 3.\overline{3} feet long and perfectly coincides with the water level at y=0. + The leftmost upper corner of the rectangular door is labeled to be the point (0,0), while the upper right corner is labeled by the point (3.\overline{3},0). + The leftmost bottom corner of the rectangular door is labeled to be the point (0,-2.25), while the upper right corner is labeled by the point (3.\overline{3},-2.25). + The graph also contains an arbitrarily chosen thin slice of the door, which spans the horizontal length of the rectangular door. + The red coloured thin slice occurs at the level y, which is an arbitrarily chosen value between y = -2.25 which marks the distance from the water level and the bottom of the door and y = 0 which marks the top of the door. + The distance between the water level and the thin horizontal slice will be given by d(y)=-y. +

    +     + Graph of a perfectly rectangular car door submerged in water. + + + \begin{tikzpicture}[scale=0.66,>=stealth] + + \draw [thick] (0,0) -- (3.33,0) -- (3.33,-2.25) -- (0,-2.25) -- cycle; + + \draw [fill=black] (3.33,0) circle (1pt) node [above right] { $(3.\overline{3},0)$} + (3.33,-2.25) circle (1pt) node [below right] { $(3.\overline{3},-2.25)$} + (0,-2.25) circle (1pt) node [below left] { $(0,-2.25)$}; + + \draw [fill=black] (0,0) circle (1pt) node [above left] { $(0,0)$}; + \draw [secondcolor,thick] (0,-1.5) -- (3.33,-1.5) node [right,black] { $y$}; + \draw [->] (0,-3) -- (0,1) node [above] { $y$}; + \draw [->] (-1,0) -- (6,0) node [right] { $x$}; + + \foreach \x in {-1,0,1,2,3,4} + { + \begin{scope}[shift={(\x*1,0)}] + \draw [firstcolor,thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \end{tikzpicture} + + + + +
    +     +     + + + Finding fluid force + +

    + An underwater observation tower is being built with circular viewing portholes enabling visitors to see underwater life. + Each vertically oriented porthole is to have a 3 + diameter whose center is to be located 50 underwater. + Find the total fluid force exerted on each porthole. + Also, compute the fluid force on a horizontally oriented porthole that is under + 50 of water. +

    + +
    + Measuring the fluid force on an underwater porthole in + + + +

    + Graph of an underwater circular porthole with the convention that the water level is at y=50. + The circular porthole having a diameter of 3 is centered at the origin. + The top of the circular porthole is at the point (0,1.5), while the bottom is at the point (0,-1.5). + The leftmost point of the circle occurs at the point (-1.5,0), whilet the rightmost point of the circle occurs at the point (-1.5,0). + The graph also contains an arbitrarily chosen thin slice of the circular porthole, which spans the horizontal length of the circle. + The red-coloured thin slice occurs at the level y, which is an arbitrarily chosen value between y = -1.5 which marks the bottom of the porthole and y = 1.5 which marks the top of the porthole. + The distance between the water level and the thin horizontal slice is given by d(y)=50-y. +

    +     + Graph of a circular underwater porthole. + + + \begin{tikzpicture}[scale=1.06,>=stealth] + + \draw [thick] (0,0) circle (1.5); + \draw [secondcolor,thick] (-1.3,.75) -- (1.3,.75) node [right,black] { $y$}; + \draw [->] (0,-2.2) -- (0,3)--(-.25,3.25) -- (0,3.5) -- (.25,3.75) -- (0,4) --(0,6) node [above] { $y$}; + \draw [->] (-2.5,0) -- (2.5,0) node [right] { $x$}; + + \foreach \x in {-2,-1,1,2} + { + \draw (\x,0.2) -- (\x,-.2) node [below] { $\x$}; + } + + \foreach \x in {-2,-1,1,2} + { + \draw (.2,{\x}) -- (-.2,{\x}) node [left] { $\x$}; + } + + \draw (.2,5) -- (-.2,5) node [left] { $50$}; + + \foreach \x in {-1.5,-0.5,0.5,1.5} + { + \begin{scope}[shift={(\x*1,5)}] + \draw [firstcolor,thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw (2,6) node { water line} + (2,-2.5) node { \textit{not to scale}}; + + \draw (2.1,.75) -- (2.5,.75) + (2.1,5) -- (2.5,5) + (2.3,.75) -- node [pos=.5,rotate=-90,above] { $d(y) = 50-y$} (2.3,5); + + \end{tikzpicture} + + + + +
    +     + +

    + We place the center of the porthole at the origin, + meaning the surface of the water is at y=50 and the depth function will be d(y)=50-y; + see +

    + +

    + The equation of a circle with a radius of 1.5 is x^2+y^2=2.25; + solving for x we have x=\pm \sqrt{2.25-y^2}, + where the positive square root corresponds to the right side of the circle and the negative square root corresponds to the left side of the circle. + Thus the length function at depth y is \ell(y) = 2\sqrt{2.25-y^2}. + Integrating on [-1.5,1.5] we have: + + F \amp = 62.4\int_{-1.5}^{1.5} 2(50-y)\sqrt{2.25-y^2}\, dy + \amp = 62.4\int_{-1.5}^{1.5} \big(100\sqrt{2.25-y^2} - 2y\sqrt{2.25-y^2}\big)\, dy + \amp = 6240\int_{-1.5}^{1.5} \big(\sqrt{2.25-y^2}\big)\, dy - 62.4\int_{-1.5}^{1.5} \big(2y\sqrt{2.25-y^2}\big)\, dy + . +

    + +

    + The second integral above can be evaluated using substitution. + Let u=2.25-y^2 with du = -2y\,dy. + The new bounds are: u(-1.5)=0 and u(1.5)=0; + the new integral will integrate from u=0 to u=0, + hence the integral is 0. +

    + +

    + The first integral above finds the area of half a circle of radius 1.5, thus the first integral evaluates to 6240\cdot\pi\cdot1.5^2/2 = 22,054. + Thus the total fluid force on a vertically oriented porthole is 22,054. +

    + +

    + Finding the force on a horizontally oriented porthole is more straightforward: + + F = \text{Pressure} \times\text{Area} = 62.4\cdot50\times \pi\cdot1.5^2 = 22,054\, \text{lb} + . +

    + +

    + That these two forces are equal is not coincidental; + it turns out that the fluid force applied to a vertically oriented circle whose center is at depth d is the same as force applied to a horizontally oriented circle at depth d. +

    +     +     + +

    + We end this chapter with a reminder of the true skills meant to be developed here. + We are not truly concerned with an ability to find fluid forces or the volumes of solids of revolution. + Work done by a variable force is important, + though measuring the work done in pulling a rope up a cliff is probably not. +

    + +

    + What we are actually concerned with is the ability to solve certain problems by first approximating the solution, + then refining the approximation, + then recognizing if/when this refining process results in a definite integral through a limit. + Knowing the formulas found inside the special boxes within this chapter is beneficial as it helps solve problems found in the exercises, + and other mathematical skills are strengthened by properly applying these formulas. + However, more importantly, + understand how each of these formulas was constructed. + Each is the result of a summation of approximations; + each summation was a Riemann sum, + allowing us to take a limit and find the exact answer through a definite integral. +

    + + + + + + Terms and Concepts + + + +

    + State in your own words Pascal's Principle. +

    +     + + +     + + + +

    + State in your own words how pressure is different from force. +

    +     + + +     +     + + Problems + + +

    + Find the fluid force exerted on the given plate, + submerged in water with a weight density of 62.4. +

    +     + + + + + +

    + Image of a submerged square plate having a side length of 2. + The top side of the square is parallel to the water level and lies 1 below the water level. +

    +     + Illustration of a submerged square plate. + + + \begin{tikzpicture}[scale=0.96] + + \draw [shift={(-1,0)},thick] (0,0) -- node [below,pos=.5] {2 ft} (2,0) -- node [right,pos=.5] {2 ft} (2,2) -- (0,2) -- cycle; + + \foreach \x in {-2,-1,0,1,2} + { + \begin{scope}[shift={(\x*1,3)}] + \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + \draw [dashed] (0,2.1) -- node [right,pos=.5] {1 ft} (0,2.9); + + \end{tikzpicture} + + + + +     + +

    + 499.2 +

    +     +     + + + + + +

    + Image of a submerged rectangular plate. + The rectangle has a horizontal length of 1 and a veritcal length of 2. + The shorter side of the rectangle is the top of the plate and is parallel to the water level and lies 1 below the water level. +

    +     + Illustration of a submerged rectangular plate. + + + \begin{tikzpicture}[scale=0.96] + + \draw [shift={(-.5,0)},thick] (0,0) -- node [below,pos=.5] {1 ft} (1,0)-- node [right,pos=.5] {2 ft} (1,2) -- (0,2) -- cycle; + + \foreach \x in {-2,-1,0,1,2} + { + \begin{scope}[shift={(\x*1,3)}] + \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw [dashed] (0,2.1) -- node [right,pos=.5] {1 ft} (0,2.9); + + \end{tikzpicture} + + + + +     + +

    + 249.6 +

    +     +     + + + + + +

    + Image of a submerged plate in the shape of an isosceles triangle. + The bottom of the triangle has a horizontal length of 4. + The distance from the bottom of the triangle to the vertex which is the nearest point on the plate to the water level is 6, which is the height of the triangle. + This vertex is the top of the plate and lies 5 below the water level. + The remaining two sides of the triangle have equal but unmarked lengths and connect the vertex to the base of the triangular plate. +

    +     + Illustration of a submerged triangular plate. + + + \begin{tikzpicture}[xscale=.8,yscale=.64] + + \draw [thick] (0,0) -- (-2,-6) -- node [below,pos=.5] {4 ft} (2,-6) -- cycle; + + \foreach \x in {-2,-1,0,1,2} + { + \begin{scope}[shift={(\x*1,3)}] + \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw [dashed] (0,0.1) -- node [right,pos=.5] {5 ft} (0,2.9); + + \draw (2.2,-6) -- (2.6,-6) + (2.2,0) -- (2.6,0) + (2.4,-6) -- node [pos=.5,right] {6 ft} (2.4,0); + + \end{tikzpicture} + + + + +     + +

    + 6739.2 +

    +     +     + + + + + +

    + Image of a submerged plate in the shape of an isosceles triangle. + The top of the triangle has a horizontal length of 4. + The distance from the top of the triangle to the vertex which is the lowest point below the water level on the plate is 6, which is the height of the triangle. + This top side of the plate having a horizontal length of 4 lies 5 below the water level. + The remaining two sides of the triangle have equal but unmarked lengths and connect the vertex to the top of the triangular plate. +

    +     + Illustration of a submerged triangular plate. + + + \begin{tikzpicture}[xscale=.8,yscale=.64] + + \draw [thick] (0,0) -- (-2,6) -- node [below,pos=.5] {4 ft} (2,6) -- cycle; + + \foreach \x in {-2,-1,0,1,2} + { + \begin{scope}[shift={(\x*1,9)}] + \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw [dashed] (0,6.1) -- node [right,pos=.5] {5 ft} (0,8.9); + + \draw (2.2,6) -- (2.6,6) + (2.2,0) -- (2.6,0) + (2.4,6) -- node [pos=.5,right] {6 ft} (2.4,0); + + \end{tikzpicture} + + + + +     + +

    + 5241.6 +

    +     +     + + + + + +

    + Image of a submerged circular plate with a radius of 2. + The center of the circular plate lies 5 below the water level. + The uppermost part of the circular plate lies 3 below the water level, and the lowest part lies 7 below the water level. +

    +     + Illustration of a submerged circular plate. + + + \begin{tikzpicture}[scale=.8] + + \draw [thick] (0,0) circle (2); + + \foreach \x in {-2,-1,0,1,2} + { + \begin{scope}[shift={(\x*1,5)}] + \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw [dashed] (0,0) -- node [above,pos=.5] {2 ft} (2,0); + + \draw (2.2,5) -- (2.6,5) + (2.2,0) -- (2.6,0) + (2.4,5) -- node [pos=.5,right] {5 ft} (2.4,0); + + \end{tikzpicture} + + + + +     + +

    + 3920.7 +

    +     +     + + + + + +

    + Image of a submerged circular plate with a radius of 4. + The center of the circular plate lies 5 below the water level. + The uppermost part of the circular plate lies 1 below the water level, and the lowest part lies 9 below the water level. +

    +     + Illustration of a submerged circular plate. + + + \begin{tikzpicture}[scale=0.5] + + \draw [thick] (0,0) circle (4); + + \foreach \x in {-3,-2,-1,0,1,2,3} + { + \begin{scope}[shift={(\x*1,5)}] + \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw [dashed] (0,0) -- node [above,pos=.5] {4 ft} (4,0); + + \draw (4.2,5) -- (4.6,5) + (4.2,0) -- (4.6,0) + (4.4,5) -- node [pos=.5,right] {5 ft} (4.4,0); + + \end{tikzpicture} + + + + +     + +

    + 15682.8 +

    +     +     + + + + + +

    + Image of a submerged rectangular plate with a horizontal length of 4 and a height of 2. + The horizontal line through the middle of the plate lies 5 below the water level. + The uppermost edge of the rectangular plate having a length of 4 lies 4 below the water level, and the lowest edge lies 6 below the water level. + Both the top and bottom edges lie parallel to the water level. +

    +     + Illustration of a submerged rectangular plate. + + + \begin{tikzpicture}[scale=0.688] + + \draw [thick] (-2,-1) -- node [below,pos=.5] {4 ft} (2,-1) -- (2,1) -- (-2,1) -- node [left,pos=.5] {2 ft} (-2,-1); + + \foreach \x in {-3,-2,-1,0,1,2,3} + { + \begin{scope}[shift={(\x*1,5)}] + \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw (2.2,5) -- (2.6,5) + (2.2,0) -- (2.6,0) + (2.4,5) -- node [pos=.5,right] {5 ft} (2.4,0); + + \end{tikzpicture} + + + + +     + +

    + 2496 +

    +     +     + + + + + +

    + Image of a submerged rectangular plate with a horizontal length of 2 and a height of 4. + The horizontal line through the middle of the plate lies 5 below the water level. + The uppermost edge of the rectangular plate having a length of 2 lies 3 below the water level, and the lowest edge lies 7 below the water level. + Both the top and bottom edges lie parallel to the water level. +

    +     + Illustration of a submerged rectangular plate. + + + \begin{tikzpicture}[scale=0.688] + + \draw [thick,rotate=90] (-2,-1) -- node [right,pos=.5] {4 ft} (2,-1) -- (2,1) -- (-2,1) -- node [below,pos=.5] {2 ft} (-2,-1); + + \foreach \x in {-3,-2,-1,0,1,2,3} + { + \begin{scope}[shift={(\x*1,5)}] + \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw (2.2,5) -- (2.6,5) + (2.2,0) -- (2.6,0) + (2.4,5) -- node [pos=.5,right] {5 ft} (2.4,0); + + \end{tikzpicture} + + + + +     + +

    + 2496 +

    +     +     + + + + + +

    + Image of a submerged rotated square plate with a side length of 2. + The square is rotated such that one of its vertices is the uppermost point of the plate, while the opposite vertex lying directly below the uppermost vertex is the lowest point of the plate. + The uppermost vertex of the square plate lies 1 below the water level. +

    +     + Illustration of a submerged rotated square plate. + + + \begin{tikzpicture}[scale=0.96] + + \draw [shift={(0,-.85)},thick,rotate=45] (0,0) -- node [shift={(8pt,-5pt)},pos=.5] {2 ft} (2,0)-- node [shift={(8pt,5pt)},pos=.5] {2 ft} (2,2) -- (0,2) -- cycle; + + \foreach \x in {-2,-1,0,1,2} + { + \begin{scope}[shift={(\x*1,3)}] + \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + + \draw [dashed] (0,2.1) -- node [right,pos=.5] {1 ft} (0,2.9); + + \end{tikzpicture} + + + + +     + +

    + 602.59 +

    +     +     + + + + + +

    + Image of a submerged plate in the shape of an equilateral right triangle. + The right angle occurs at the bottom left part of the triangle. + The two edges connected to the right angle lie parallel and perpendicular to the water level, and both have a side length of 2. + The edge that is perpendicular to the water level extends up from the right angle for 2, until reaching the uppermost vertex of the triangle. + This uppermost vertex lies 1 below the water level. + The edge that is connected to the right angle and parallel to the water level extends to the right of the right angle for a distance of 2, until reaching the rightmost vertex of the triangle. + The rightmost and uppermost vertices are then connected by an edge, which completes the equilateral right triangle. +

    +     + Illustration of a submerged plate in the shape of an equilateral right triangle. + + + \begin{tikzpicture}[scale=0.96] + + \draw [shift={(-.5,0)},thick] (0,0) -- node [below,pos=.5] {2 ft} (2,0) -- (0,2) -- node [left,pos=.5] {2 ft} (0,0); + + \foreach \x in {-2,-1,0,1,2} + { + \begin{scope}[shift={(\x*1,3)}] + \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); + \end{scope} + } + \draw [dashed] (0,2.1) -- node [right,pos=.5] {1 ft} (0,2.9); + + \end{tikzpicture} + + + + +     + +

    + 291.2 +

    +     +     + +     + + + +

    + The side of a container is pictured. + Find the fluid force exerted on this plate when the container is full of: +

    + +

    +

      +
    1. +

      + water, with a weight density of 62.4, and +

      +
      2. + +
    2. +

      + concrete, with a weight density of 150. +

      +
      3. +
    +

    + +     + + + + + +

    + Image of a rectangular container having horizontal length of 3 and a vertical height measurement of 5. +

    +     + Illustration of a rectangular container. + + + \begin{tikzpicture}[scale=1.2] + + \draw [thick] (0,0) -- node [below,pos=.5] {3 ft} (3,0) -- node [right,pos=.5] {5 ft} (3,5) -- (0,5) -- cycle; + + \end{tikzpicture} + + + + +     + +

    +

      +
    1. +

      + 2340 +

      +
      2. + +
    2. +

      + 5625 +

      +
      3. +
    +

    +     +     + + + + + +

    + Image of a container that is the combination of the parabola y=x^2 and a horizontal line having a length of 4. + The parabola is drawn such that the two sides of the parabola end when the horizontal distance between them measures 4, from which they are connected by the horizontal line. + The height, measured from the base of the parabola to the horizontal line measures 4. + A horizontal line which stretches from the left side of the container to the right side would lie entirely between the parabola given by y=x^2. +

    +     + Illustration of a container made with a parabola and a line. + + + \begin{tikzpicture}[scale=.8] + + \draw [thick] (-2,4) parabola bend (0,0) (2,4) -- node [above,pos=.5] {4 ft} (-2,4); + + \draw (0,0) node [below] {\(y=x^2\)}; + + \draw (2.2,0) -- (2.6,0) + (2.2,4) -- (2.6,4) + (2.4,0) -- node [pos=.5,right] {4 ft} (2.4,4); + + \end{tikzpicture} + + + + +     + +

    +

      +
    1. +

      + 1064.96 +

      +
      2. + +
    2. +

      + 2560 +

      +
      3. +
    +

    +     +     + + + + + +

    + Image of a container that is the combination of the parabola y=4-x^2 and a horizontal line having a length of 4. + The upsidedown parabola is drawn such that two sides of the parabola end when the horizontal distance between them measures 4, from which they are connected by the horizontal line. + The height, measured from the top of the parabola to the horizontal line which makes up the base of the container measures 4. + A horizontal line which stretches from the left side of the container to the right side would lie entirely between the parabola given by y=4-x^2. +

    +     + Illustration of a container made with a parabola and a line. + + + \begin{tikzpicture}[scale=.8] + + \draw [thick] (-2,0) parabola bend (0,4) (2,0) -- node [below,pos=.5] {4 ft} (-2,0); + + \draw (0,4) node [above] {\(y=4-x^2\)}; + + \draw (2.2,0) -- (2.6,0) + (2.2,4) -- (2.6,4) + (2.4,0) -- node [pos=.5,right] {4 ft} (2.4,4); + + \end{tikzpicture} + + + + +     + +

    +

      +
    1. +

      + 1597.44 +

      +
      2. + +
    2. +

      + 3840 +

      +
      3. +
    +

    +     +     + + + + + +

    + Image of a container in the shape of a semicircle having a radius of 1 that is the combination of the equation y=-\sqrt{1-x^2} and a horizontal line having a length of 2. + The circular arc coming from the equation y=-\sqrt{1-x^2} makes up the base of the container and is plotted for all x in the domain of y=-\sqrt{1-x^2}. + Connecting the two vertices of the circular arc is the horizontal line having a length of 2 which makes up the top of the container. + A horizontal line which stretches from the left side of the container to the right side would lie entirely between the circular arc given by y=-\sqrt{1-x^2}. +

    +     + Illustration of a container in the shape of a semicircle. + + + \begin{tikzpicture}[scale=2.16] + + \draw [thick] (-1,0) arc (180:360:1) -- node [above,pos=.5] {2 ft} (-1,0); + \draw (0,-1) node [below] {\(y=-\sqrt{1-x^2}\)}; + + \end{tikzpicture} + + + + +     + +

    +

      +
    1. +

      + 41.6 +

      +
      2. + +
    2. +

      + 100 +

      +
      3. +
    +

    +     +     + + + + + +

    + Image of a container in the shape of a semicircle having a radius of 1 that is the combination of the equation y=\sqrt{1-x^2} and a horizontal line having a length of 2. + The circular arc coming from the equation y=\sqrt{1-x^2} makes up the top of the container and is plotted for all x in the domain of y=\sqrt{1-x^2}. + Connecting the two vertices of the circular arc is the horizontal line having a length of 2 which makes up the bottom part of the container. + A horizontal line which stretches from the left side of the container to the right side would lie entirely between the circular arc given by y=\sqrt{1-x^2}. +

    +     + Illustration of a container in the shape of a semicircle. + + + \begin{tikzpicture}[scale=2] + + \draw [thick] (-1,0) arc (180:0:1) -- node [below,pos=.5] {2 ft} (-1,0); + \draw (0,1) node [above] {\(y=\sqrt{1-x^2}\)}; + + \end{tikzpicture} + + + + +     + +

    +

      +
    1. +

      + 56.42 +

      +
      2. + +
    2. +

      + 135.62 +

      +
      3. +
    +

    +     +     + + + + + +

    + Image of a container in the shape of a semicircle having a radius of 3 that is the combination of the equation y=-\sqrt{9-x^2} and a horizontal line having a length of 6. + The circular arc coming from the equation y=-\sqrt{9-x^2} makes up the top of the container and is plotted for all x in the domain of y=-\sqrt{9-x^2}. + Connecting the two vertices of the circular arc is the horizontal line having a length of 6 which makes up the bottom part of the container. + A horizontal line which stretches from the left side of the container to the right side would lie entirely between the circular arc given by y=-\sqrt{9-x^2}. +

    +     + Illustration of a container in the shape of a semicircle. + + + \begin{tikzpicture}[scale=.67] + + \draw [thick] (-3,0) arc (180:360:3) -- node [above,pos=.5] {6 ft} (-3,0); + \draw (0,-3) node [below] {\(y=-\sqrt{9-x^2}\)}; + + \end{tikzpicture} + + + + +     + +

    +

      +
    1. +

      + 1123.2 +

      +
      2. + +
    2. +

      + 2700 +

      +
      3. +
    +

    +     +     +     + + + +

    + How deep must the center of a vertically oriented circular plate with a radius of 1 be submerged in water, + with a weight density of 62.4, + for the fluid force on the plate to reach 1,000? +

    +     + +

    + 5.1 +

    +     +     + + + +

    + How deep must the center of a vertically oriented square plate with a side length of 2 be submerged in water, + with a weight density of 62.4, + for the fluid force on the plate to reach 1,000 lb? +

    +     + +

    + 4.1 +

    +     +     +     +     +
    +     + + + Differential Equations + +

    + One of the strengths of calculus is its ability to describe real-world phenomena. + We have seen hints of this in our discussion of the applications of derivatives and integrals in the previous chapters. + The process of formulating an equation or multiple equations to describe a physical phenomenon is called + mathematical modeling. As a simple example, + populations of bacteria are often described as + growing exponentially. + Looking in a biology text, we might see P(t) = P_0e^{kt}, + where P(t) is the bacteria population at time t, + P_0 is the initial population at time t=0, + and the constant k describes how quickly the population grows. + This equation for exponential growth arises from the assumption that the population of bacteria grows at a rate proportional to its size. + Recalling that the derivative gives the rate of change of a function, + we can describe the growth assumption precisely using the equation P' = kP. + This equation is called a differential equation, + and these equations are the subject of the current chapter. +

    +     +
    + Graphical and Numerical Solutions to Differential Equations + +

    + In , + we were introduced to the idea of a differential equation. + Given a function y = f(x), + we defined a differential equation + as an equation involving y, + x, and derivatives of y. + We explored the simple differential equation \yp = 2x, + and saw that a solution + to a differential equation is simply a function that satisfies the differential equation. +

    + +     + + Introduction and Terminology + + Differential Equation + +

    + Given a function y=f(x), + a differential equation is an equation relating + x, + y, and derivatives of y. +

      +
    • +

      + The variable x is called the + independent variable. +

      +
      • +
    • +

      + The variable y is called the + dependent variable. +

      +
      • +
    • +

      + The order of the differential equation is the order of the highest derivative of y that appears in the equation. +

      +
      • +
    + differential equationdefinition + orderof a differential equation + differential equationorder of +

    +     +     +

    + Let us return to the simple differential equation + + \yp = 2x + . +

    +

    + To find a solution, + we must find a function whose derivative is 2x. + In other words, we seek an antiderivative of 2x. + The function + + y = x^2 + + is an antiderivative of 2x, + and solves the differential equation. + So do the functions + + y = x^2 + 1 + + and + + y = x^2 - 2346 + . +

    +

    + We call the function + + y = x^2 + C + , + with C an arbitrary constant of integration, + the general solution to the differential equation. + general solutionof a differential equation + differential equationgeneral solution +

    +

    + In order to specify the value of the integration constant C, + we require additional information. + For example, if we know that y(1) = 3, + it follows that C=2. + This additional information is called an + initial condition. +

    + + Initial Value Problem + +

    + A differential equation paired with an initial condition + (or initial conditions) + is called an initial value problem. + initial value problemfor differential equations + initial condition +

    +

    + The solution to an initial value problem is called a + particular solution. + differential equationparticular solution + A particular solution does not include arbitrary constants. +

    +

    + The family of solutions to a differential equation that encompasses all possible solutions is called the + general solution to the differential equation. + differential equationgeneral solution +

    +     +     + + + + + + + + + A simple first-order differential equation + +

    + Solve the differential equation \yp = 2y. +

    +     + +

    + The solution is a function y such that differentiation yields twice the original function. + Unlike our starting example, + finding the solution here does not involve computing an antiderivative. + Notice that + integrating both sides + would yield the result y = \int 2y\,dx, which is not useful. + Without knowledge of the function y, + we can't compute the indefinite integral. + Later sections will explore systematic ways to find analytic solutions to simple differential equations. + For now, a bit of thought might let us guess the solution + + y = e^{2x} + . +

    +

    + Notice that application of the chain rule yields \yp = 2e^{2x} = 2y. + Another solution is given by + + y = -3e^{2x} + . +

    +

    + In fact, + + y = Ce^{2x} + , + where C is any constant, + is the general solution + to the differential equation because \yp = 2Ce^{2x} = 2y. +

    +

    + If we are provided with a single initial condition, + say y(0) = 3/2, we can identify C=3/2 so that + + y = \frac{3}{2}e^{2x} + + is the particular solution + to the initial value problem + + \yp = 2y, \text{ with } y(0) = \frac{3}{2} + . +

    +

    + + shows various members of the general solution to the differential equation \displaystyle \yp = 2y. + Each C value yields a different member of the family, + and a different function. + We emphasize the particular solution corresponding to the initial condition y(0)=3/2. +

    +
    + A representation of some of the members of general solution to the differential equation \yp = 2y, + including the particular solution to the initial value problem with y(0)=\displaystyle 3/2, + from + + + Graph of the members of the general solution to the equation \yp =2y, including a particular solution. + + +

    + The y axis is drawn from -10 to 10 and the x axis is drawn from -2 + to 2. A group of dashed lines emerge very close to the x axis on either side of it. + From left to right, these lines diverge with increasing values of x. At about x =1 the + lines diverge greatly, forming a trumpet shape. +

    +

    + The particular solution to the initial value problem with y(0) = 3/2 from the example is shown + as a curve in the second and the third quadrant that runs along one of the groups of dashed lines that + crosses the y axis at 3/2. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-10.5,ymax=10.75, + xmin=-2,xmax=2.15% + ] + + \foreach \n in {-6,...,6} + { + \addplot[secondcurvestyle,domain=-2:2,samples=50] {\n/2*exp(2*x)}; + } + \addplot[firstcurvestyle,domain=-2:2,samples=50] {3/2*exp(2*x)}; + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +
    +     + +     + + + A second-order differential equation + +

    + Solve the differential equation \yp' + 9y = 0. +

    +     + +

    + We seek a function whose second derivative is negative 9 multiplied by the original function. + Both \sin(3x) and \cos(3x) have this feature. + The general solution to the differential equation is given by + + y = C_1\sin(3x) + C_2\cos(3x) + , + where C_1 and C_2 are arbitrary constants. + To fully specify a particular solution, + we require two additional conditions. + For example, the initial conditions y(0)=1 and + \yp(0)=3 yield C_1 = C_2 = 1. +

    +     + +     + +

    + The differential equation in + is second order, because the equation involves a second derivative. + In general, the number of initial conditions required to specify a particular solution depends on the order of the differential equation. + For the remainder of the chapter, + we restrict our attention to first order differential equations and first order initial value problems. +

    + + + Verifying a solution to the differential equation + +

    + Which of the following is a solution to the differential equation + + \yp + \frac{y}{x} - \sqrt{y} = 0 + ? +

    +

    +

      +
    1. y = C \left( 1 + \ln(x) \right)^2
      2. +
    2. y = \left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right)^2
      3. +
    3. y = C e^{-3x} + \sqrt{\sin(x)}
      4. +
    +

    +     + +

    + Verifying a solution to a differential equation is simply an exercise in differentiation and simplification. + We substitute each potential solution into the differential equation to see if it satisfies the equation. +

    +

    +

      +
    1. +

      + Testing the potential solution y = C \left( 1 + \ln(x) \right)^2: +

      +

      + Differentiating, + we have \displaystyle \yp = \frac{2C(1 + \ln(x))}{x}. + Substituting into the differential equation, + + \amp \frac{2C(1+\ln(x))}{x} + \frac{C(1+\ln(x))^2}{x} -\sqrt{C}(1+\ln(x)) + \amp = (1+\ln(x))\left( \frac{2C}{x} + \frac{C(1+\ln(x))}{x} - \sqrt{C}\right) + \amp \neq 0 + . +

      +

      + Since it doesn't satisfy the differential equation, + y = C(1 + \ln(x))^2 is not a solution. +

      +
      2. + +
    2. +

      + Testing the potential solution \displaystyle y = \left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right)^2: +

      +

      + Differentiating, + we have \displaystyle \yp = 2\left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right)\left( \frac{1}{3} - \frac{C}{2x^{3/2}}\right). + Substituting into the differential equation, + + \amp 2\left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right)\left( \frac{1}{3} - \frac{C}{2x^{3/2}}\right) + \frac{1}{x}\left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right)^2 - \left(\frac{1}{3}x + \frac{C}{\sqrt{x}}\right) + \amp = \left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right) \left( \frac{2}{3} - \frac{C}{x^{3/2}} + \frac{1}{3} + \frac{C}{x^{3/2}} - 1 \right) + \amp = 0. \text{ (Note how the second parenthetical grouping above reduces to 0.) } + +

      +

      + Thus \displaystyle y = \left(\frac{1}{3}x + \frac{C}{\sqrt{x}} \right)^2 is + a solution to the differential equation. +

      +
      3. + +
    3. +

      + Testing the potential solution \displaystyle y = C e^{-3x} + \sqrt{\sin(x)}: +

      +

      + Differentiating, + \displaystyle \yp = -3Ce^{-3x} + \frac{\cos(x)}{2\sqrt{\sin(x)}}. + Substituting into the differential equation, + + -3Ce^{-3x} + \frac{\cos(x)}{2\sqrt{\sin(x)}} + \frac{C e^{-3x} + \sqrt{\sin(x)}}{x} - \sqrt{C e^{-3x} + \sqrt{\sin(x)}} \neq 0 + . +

      +

      + The function \displaystyle y = C e^{-3x} + \sqrt{\sin(x)} is not + a solution to the differential equation. +

      +
      4. +
    +

    +     + +     + + + + + Verifying a solution to a differential equation + +

    + Verify that x^2+y^2 = Cy is a solution to \displaystyle \yp = \frac{2xy}{x^2-y^2}. +

    +     + +

    + The solution in this example is called an + implicit solution. + That means the dependent variable y is a function of x, + but has not been explicitly solved for. + Verifying the solution still involves differentiation, + but we must take the derivatives implicitly. + Differentiating, we have + + 2x + 2y\yp = C\yp + . + differential equationimplicit soution +

    +

    + Solving for \yp, we have + + \yp = \frac{2x}{C-2y} + . +

    +

    + From the solution, + we know that \displaystyle C = \frac{x^2+y^2}{y}. + Then + + \yp \amp = \frac{2x}{\displaystyle \frac{x^2+y^2}{y} - 2y} + \amp =\frac{2xy}{x^2+y^2-2y^2} + \amp = \frac{2xy}{x^2-y^2} + . +

    +

    + We have verified that x^2+y^2 = Cy is a solution to \displaystyle \yp = \frac{2xy}{x^2-y^2}. +

    +     + +     +     + + + Graphical Solutions to Differential Equations +

    + In the examples we have explored so far, + we have found exact forms for the functions that solve the differential equations. + Solutions of this type are called + analytic solutions. + Many times a differential equation has a solution, + but it is difficult or impossible to find the solution analytically. + This is analogous to algebraic equations. + The algebraic equation x^2 + 3x - 1 = 0 has two real solutions that can be found analytically by using the quadratic formula. + The equation \cos(x) = x has one real solution, + but we can't find it analytically. + As shown in , + we can find an approximate solution graphically by plotting \cos(x) and x and observing the x-value of the intersection. + We can similarly use graphical tools to understand the qualitative behavior of solutions to a first order-differential equation. +

    + +
    + Graphically finding an approximate solution to \cos(x) = x + + + Graphically finding an approximate solution to cos(x)=x. + + +

    + The y and the x axes are drawn from 0 to 1. There is a curve and a + dashed line shown in the graph. The dashed straight line is drawn from the origin and has a + positive slope. The curve starts at point (0, 1) and curves down to about point (1, 0.5). +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=1, + %extra x ticks={.25,.75}, + ymin=-.1,ymax=1.1,% + xmin=-.1,xmax=1.1% + ] + + \addplot [firstcurvestyle,domain=0:1] {cos(deg(x))}; + \addplot [secondcurvestyle,-] coordinates {(0,0) (1,1)}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +
    +

    + Consider the first-order differential equation + + \yp = f(x,y) + . +

    +

    + The function f could be any function of the two variables x and y. + Written in this way, + we can think of the function f as providing a formula to find the slope of a solution at a given point in the xy-plane. + In other words, + suppose a solution to the differential equation passes through the point (x_0,y_0). + At the point (x_0,y_0), + the slope of the solution curve will be f(x_0,y_0). + Since this calculation of the slope is possible at any point (x,y) where the function f(x,y) is defined, + we can produce a plot called a slope field + (or direction field) + that shows the slope of a solution at any point in the xy-plane where the solution is defined. + Further, this process can be done purely by working with the differential equation itself. + In other words, + we can draw a slope field and use it to determine the qualitative behavior of solutions to a differential equation without having to solve the differential equation. + differential equationgraphical solution + direction fieldslope field +

    + + + Slope Field + +

    + A slope field + for a first-order differential equation + \yp = f(x,y) is a plot in the xy-plane made up of short line segments or arrows. + At each point (x_0,y_0) where f(x,y) is defined, + the slope of the line segment is given by f(x_0,y_0). + Plots of solutions to a differential equation are tangent to the line segments in the slope field. + slope field +

    +     +     + + + Sketching a slope field + +

    + Find a slope field for the differential equation \displaystyle \yp = x+y. +

    +     + +

    + Because the function f(x,y) = x+y is defined for all points (x,y), + every point in the xy-plane has an associated line segment. + It is not practical to draw an entire slope field by hand, + but many tools exist for drawing slope fields on a computer. + Here, we explicitly calculate a few of the line segments in the slope field. +

    + +

    +

      +
    • +

      + The slope of the line segment at (0,0) is f(0,0) = 0 + 0 = 0. +

      +
      • +
    • +

      + The slope of the line segment at (1,1) is f(1,1) = 1 + 1 = 2. +

      +
      • +
    • +

      + The slope of the line segment at (1,-1) is f(1,-1) = 1 - 1 = 0. +

      +
      • +
    • +

      + The slope of the line segment at (-2,-1) is f(-2,-1) = -2 - 1 = -3. +

      +
      • +
    +

    + +

    + Though it is possible to continue this process to sketch a slope field, + we usually use a computer to make the drawing. + Most popular computer algebra systems can draw slope fields. + There are also various online tools that can make the drawings. + The slope field for \yp = x+y is shown in . +

    +
    + Slope field for \yp = x+y from + + + Graph of slope field of \yp = y+x. + + +

    + The y and the x axis are shown, the graph shows the slope field as a group of + dashed lines. In the first quadrant, from left to right the slope fields are shown moving + from north-east to north. In the second quadrant the lines are south facing on the bottom + left and curving towards north-east to the top right. In the third quadrant all lines are + facing south-east. In the fourth quadrant the lines are facing south-east on the bottom left + and curving towards north-east to the top right, in the transition the lines are facing east. +

    +     + + \def\length{sqrt(1+(x+y)^2)} + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -2:2, + ytick={-2,-1,0,1,2}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(\length)}, v = {(x+y)/(\length)},scale arrows=.25},samples=9]{0}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + + +
    +     +     + + + A graphical solution to an initial value problem + +

    + Approximate, with a sketch, + the solution to the initial value problem + \displaystyle \yp = x+y, with y(1)=-1. +

    +     + +

    + The solution to the initial value problem should be a continuous smooth curve. + Using the slope field, + we can draw of a sketch of the solution using the following two criteria: +

      +
    1. +

      + The solution must pass through the point (1,-1). +

      +
      2. +
    2. +

      + When the solution passes through a point + (x_0,y_0) it must be tangent to the line segment at (x_0,y_0). +

      +
      3. +
    +

    +

    + Essentially, + we sketch a solution to the initial value problem by starting at the point (1,-1) and + following the lines + in either direction. + A sketch of the solution is shown in . +

    +
    + Solution to the initial value problem \yp = x+y, with y(1)=-1 from + + + Graph of slope field of \yp = y+x with initial value y(1) = -1. + + +

    + The y and the x axis are shown, the graph shows the slope field as a group + of dashed lines. In the first quadrant, from left to right the slope fields are shown + moving from north-east to north. In the second quadrant the lines are south facing on the + bottom left and curving towards north-east to the top right. In the third quadrant all + lines are facing south-east. In the fourth quadrant the lines are facing south-east on + the bottom left and curving towards north-east to the top right, in the transition the + lines are facing east. +

    +

    + The initial value at y(1) = -1 is a curve that moves down starting in the second + quadrant then crossing the third quadrant, in the fourth quadrant it curves up after passing + the point (1,-1) and moves up to touch the x axis. +

    +     + + \def\length{sqrt(1+(x+y)^2)} + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -2:2, + xtick={-2,-1,0,1,2}, + ytick={-2,-1,0,1,2}, + xmin=-2,xmax=2.3 + ] + + \addplot3 [secondcolor,quiver = {u = {1/(\length)}, v = {(x+y)/(\length)},scale arrows=.25},samples=9]{0}; + \addplot [firstcurvestyle,domain=-2.5:2.2,samples=51] {-(x+1)+exp(x-1)}; + + \fill[black,draw=black] (axis cs:1,-1) circle (2.4pt); + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +
    +     +     + + + + + Using a slope field to predict long term behavior + +

    + Use the slope field for the differential equation \yp = y(1-y), + shown in , + to predict long term behavior of solutions to the equation. +

    +
    + Slope field for the logistic differential equation \yp = y(1-y) from + + + Graph of slope field for the logistic differential equation y’=y(1-y) from the example. + + +

    + The y and the t axis are drawn. There are two positions where the slope field + runs parallel to the t axis, once along the t axis itself and again at some + positive value of y. The two horizontal lines appear to divide the slope files into + three distinct parts. Over the horizontal line about some y value, the field lines are + directed south-east. Between the line and the t axis the field lines are directed south-west. + Below the t axis the field lines are again directed south-east. +

    +     + + \def\length{sqrt(1+(y*(1-y))^2)} + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -1:3, + y domain=-.5:1.5, + xtick={-1,0,1,2,3}, + ytick={-.5,0,.5,1}, + ymin=-0.5,ymax=1.5, + xmin=-1,xmax=3, + xlabel={$t$} + ] + + \addplot3 [secondcolor,quiver = {u = {1/(\length)}, v = {(y*(1-y))/(\length)},scale arrows=.1},samples=20]{0}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $t$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +
    +     + +

    + This differential equation, + called the logistic differential equation, often appears in population biology to describe the size of a population. + For that reason, + we use t (time) as the independent variable instead of x. + We also often restrict attention to non-negative y-values because negative values correspond to a negative population. + differential equationlogistic +

    + +

    + Looking at the slope field in , + we can predict long term behavior for a given initial condition. +

      +
    • +

      + If the initial y-value is negative (y(0)\lt 0), + the solution curve must pass though the point + (0,y(0)) and follow the slope field. + We expect the solution y to become more and more negative as time increases. + Note that this result is not physically relevant when considering a population. +

      +
      • +
    • +

      + If the initial y-value is greater than 0 but less than 1, we expect the solution y to increase and level off at y=1. +

      +
      • +
    • +

      + If the initial y-value is greater than 1, we expect the solution y to decrease and level off at y=1. +

      +
      • +
    +

    + +

    + The slope field for the logistic differential equation, + along with representative solution curves, + is shown in . + Notice that any solution curve with positive initial value will tend towards the value y=1. + We call this the carrying capacity. + carrying capacity +

    +
    + Slope field for the logistic differential equation \yp = y(1-y) from + with a few representative solution curves + + + Graph of slope field for the logistic differential equation y’=y(1-y) with representative solution curves. + + +

    + The y and the t axis are drawn. There are two positions where the slope field runs + parallel to the t axis, once along the t axis itself and again at some positive + value of y. The two horizontal lines appear to divide the slope files into three distinct parts. +

    +

    + Over the horizontal line, about some y value, the field lines are directed south-east. + A curve that starts a little before the y axis in the second quadrant, enters the first + quadrant with a negative slope it bends and moves along the horizontal line. +

    +

    + Between the horizontal line and the t axis the field lines are directed south-west. + From left to right, a curve is drawn that moves with a positive slope from the second quadrant + to the first, then becomes parallel to the horizontal line. +

    +

    + Below the t axis the field lines are again directed south-east. From left to right, the + third curve starts in the third quadrant and moves parallel to the t axis then it moves away + from the t axis with a negative slope. +

    +     + + \def\length{sqrt(1+(y*(1-y))^2)} + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -1:3, + y domain=-.5:1.5, + xtick={-1,0,1,2,3}, + ytick={-.5,0,.5,1}, + xlabel={$t$}, + ymin=-0.5,ymax=1.5, + xmin=-1,xmax=3 + ] + + \addplot3 [secondcolor,quiver = {u = {1/(\length)}, v = {(y*(1-y))/(\length)},scale arrows=.1},samples=20]{0}; + \addplot [firstcurvestyle,domain = -1:2, samples=50] {1/(1-10*exp(-x))}; + \addplot [firstcurvestyle,domain = -1:3, samples=50] {1/(1+1*exp(-x))}; + \addplot [firstcurvestyle,domain = -1:3, samples=50] {1/(1-.3*exp(-x))}; + + \end{axis} + %\node [right] at (myplot.right of origin) { $t$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +
    +     +     + + +     + + + Numerical Solutions to Differential Equations: Euler's Method +

    + While the slope field is an effective way to understand the qualitative behavior of solutions to a differential equation, + it is difficult to use a slope field to make quantitative predictions. + For example, + if we have the slope field for the differential equation + \yp = x+y from + along with the initial condition y(0)=1, + we can understand the qualitative behavior of the solution to the initial value problem, + but will struggle to predict a specific value, + y(2) for example, with any degree of confidence. + The most straightforward way to predict y(2) is to find the analytic solution to the the initial value problem and evaluate it at x=2. + Unfortunately, + we have already mentioned that it is impossible to find analytic solutions to many differential equations. + In the absence of an analytic solution, + a numerical solution can serve as an effective tool to make quantitative predictions about the solution to an initial value problem. +

    +

    + There are many techniques for computing numerical solutions to initial value problems. + A course in numerical analysis will discuss various techniques along with their strengths and weaknesses. + The simplest technique is called + Euler's Method. + differential equationnumerical solution +

    + + + +

    + Consider the first-order initial value problem + + \yp = f(x,y), \text{ with } y(x_0) = y_0 + . +

    +

    + Using the definition of the derivative, + + \yp(x) = \lim_{h \to 0} \frac{y(x+h) - y(x)}{h} + . +

    +

    + This notation can be confusing at first, but + y(x) + simply means + the y-value of the solution when the x-value is x, and + y(x+h) + means + the y-value of the solution when the x-value is x+h. +

    +

    + If we remove the limit but restrict h to be + small, + we have + + \yp(x) \approx \frac{y(x+h) - y(x)}{h} + , + so that + + f(x,y) \approx \frac{y(x+h)-y(x)}{h} + , + because \yp = f(x,y) according to the differential equation. + Rearranging terms, + + y(x + h) \approx y(x) + h\,f(x,y) + . +

    +

    + This statement says that if we know the solution (y-value) to the initial value problem for some given x-value, + we can find an approximation for the solution at the value x+h by taking our y-value and adding h times the function f evaluated at the x and y values. + Euler's method uses the initial condition of an initial value problem as the starting point, + and then uses the above idea to find approximate values for the solution y at later x-values. + The algorithm is summarized in . +

    + + Euler's Method +

    + Consider the initial value problem + + \yp = f(x,y) \text{ with } y(x_0)=y_0 + . +

    +

    + Let h be a small positive number and N be an integer. +

      +
    1. +

      + For i = 0, 1, 2, \ldots, N, define + + x_i = x_0 + ih + . +

      +
      2. +
    2. +

      + The value y_0 is given by the initial condition. + For i = 0, 1, 2, \ldots, N-1, define + + y_{i+1} = y_i + hf(x_i,y_i) + . +

      +
      3. +
    +

    +

    + This process yields a sequence of N+1 points + (x_i,y_i) for i= 0,1,2,\ldots,N, + where (x_i, y_i) is an approximation for (x_i,y(x_i)). + Euler's Method +

    +     + + + +

    + Let's practice Euler's Method using a few concrete examples. +

    + + + Using Euler's Method 1 + +

    + Find an approximation at x=2 for the solution to + \yp = x + y with y(1)=-1 using Euler's Method with h=0.5. +

    +     + +

    + Our initial condition yields the starting values x_0 = 1 and y_0 = -1. + With h = 0.5, it takes N=2 steps to get to x=2. + Using steps 1 and 2 from the Euler's Method algorithm, + + x_0 \amp = 1 \amp y_0 \amp = -1 + x_1 \amp = x_0 + h \amp y_1 \amp = y_0 + hf(x_0,y_0) + \amp = 1 + 0.5 \amp \amp = -1 + 0.5(1 - 1) + \amp = 1.5 \amp \amp = -1 + x_2 \amp = x_0 + 2h \amp y_2 \amp = y_1 + hf(x_1,y_1) + \amp = 1 + 2(0.5) \amp \amp = -1 + 0.5(1.5 -1) + \amp = 2 \amp \amp = -0.75 + . +

    +

    + Using Euler's method, we find the approximation y(2) \approx -0.75. +

    + +
    + Euler's Method approximation to \yp = x + y with y(1) = -1 from , + along with the analytical solution to the initial value problem + + + Euler's Method approximation to \yp=x+y with y(1)=-1, with the analytical solution. + + +

    + The y axis is drawn from -1 to 0 and the x axis is drawn from 0 + to 2. The function \yp=x+y with y(1)=-1 is drawn in the fourth quadrant. From left to right from point (1, -1) + the function curves up till point (2, -0.25). There is a plot of three points which are joined + by straight lines; this plot starts at the same point (1,-1) as the curve. The second point is + a (1.5, -1) and the third point is at (2, -0.75). +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,1.5,2}, + ytick={-1,-.5,0}, + ymin=-1.25,ymax=0.1, + xmin=.9,xmax=2.1 + ] + + \addplot [secondcurvestyle,solid,domain=1:2,samples = 51] {-(x+1)+exp(x-1)}; + \addplot [firstcolor, mark = *] coordinates{ (1,-1) (1.5,-1) (2,-0.75)}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +
    + +

    + To help visualize the Euler's method approximation, these three points + (connected by line segments) + are plotted along with the analytical solution to the initial value problem in . +

    +     +     + +

    + This approximation doesn't appear terrific, + though it is better than merely guessing. + Let's repeat the previous example using a smaller h-value. +

    + + + Using Euler's Method 2 + +

    + Find an approximation on the interval [1,2] for the solution to + \yp = x + y with y(1)=-1 using Euler's Method with h=0.25. +

    +     + +

    + Our initial condition yields the starting values x_0 = 1 and y_0 = -1. + With h = 0.25, + we need N=4 steps on the interval [1,2] Using steps 1 and 2 from the Euler's Method algorithm + (and rounding to 4 decimal points), + we have + + x_0 \amp = 1 \amp y_0 \amp = -1 + x_1 \amp = 1.25 \amp y_1 \amp = -1 + 0.25(1 - 1) + \amp \amp \amp = -1 + x_2 \amp = 1.5 \amp y_2 \amp = -1 + 0.25(1.25-1) + \amp \amp \amp = -0.9375 + x_3 \amp = 1.75 \amp y_3 \amp = -0.9375 + 0.25(1.5-0.9375) + \amp \amp \amp = -0.7969 + x_4 \amp = 2 \amp y_4 \amp = -0.7969 + 0.25(1.75 - 0.7969) + \amp \amp \amp = -0.5586 + . +

    + +

    + Using Euler's method, we find y(2) \approx -0.5586. +

    + +

    + These five points, + along with the points from and the analytic solution, + are plotted in . +

    +
    + Euler's Method approximations to \yp = x + y with y(1) = -1 from Examples + and , along with the analytical solution + + + Euler's Method approximation to \yp=x+y with y(1)=-1 along with the analytical solution. + + +

    + The y axis is drawn from -1 to 0 and the x axis is drawn from 0 + to 2. The function is drawn in the fourth quadrant. From left to right from point (1, -1) + the function curves up till point (2, -0.25). +

    +

    + There is a plot of three points which are joined by straight lines; this plot starts at the same point + (1,-1) as the curve. The second point is a (1.5, -1) and the third point is at + (2, -0.75), this plot is marked as h=0.5. +

    +

    + A second plot is drawn with five points every 0.25 interval of the x value. It coincides + with the first plot from x=1 to x=1.25, after which the third, the fourth and the fifth + points of the second plot are all above the first plot at x=1.5, x= 1.75 and x=2. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,1.5,2}, + ytick={-1,-.5,0}, + ymin=-1.25,ymax=0.1, + xmin=.9,xmax=2.35 + ] + + \addplot [secondcolor,thick,domain=1:2,samples = 51] {-(x+1)+exp(x-1)}; + \addplot [firstcolor, mark = *] coordinates{ (1,-1) (1.5,-1) (2,-0.75)}; + \addplot [firstcolor, mark = *,dashed] coordinates{ (1,-1) (1.25,-1) (1.5,-.9375) (1.75,-.7969) (2,-.5586)}; + + \node [right] at (axis cs:2,-0.75) { $h = 0.5$}; + \node [right] at (axis cs:2,-0.5586) { $h = 0.25$}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +
    +     +     + + + +

    + Using the results from Examples + and , + we can make a few observations about Euler's method. + First, the Euler approximation generally gets worse as we get farther from the initial condition. + This is because Euler's method involves two sources of error. + The first comes from the fact that we're using a positive h-value in the derivative approximation instead of using a limit as h approaches zero. + Essentially, + we're using a linear approximation to the solution y + (similar to the process described in on Differentials.) + This error is often called the local truncation error. + The second source of error comes from the fact that every step in Euler's method uses the result of the previous step. + That means we're using an approximate y-value to approximate the next y-value. + Doing this repeatedly causes the errors to build on each other. + This second type of error is often called the propagated + or accumulated error. + accumulated errorusing Euler's method + Euler's methodaccumulated error +

    +

    + A second observation is that the Euler approximation is more accurate for smaller h-values. + This accuracy comes at a cost, though. + + is more accurate than , + but takes twice as many computations. + In general, numerical algorithms + (even when performed by a computer program) + require striking a balance between a desired level of accuracy and the amount of computational effort we are willing to undertake. +

    +

    + Let's do one final example of Euler's Method. +

    + + + Using Euler's Method 3 + +

    + Find an approximation for the solution to the logistic differential equation +

    +

    + \yp = y(1-y) with y(0) = 0.25, + for 0 \leq y \leq 4. + Use N = 10 steps. +

    +     + +

    + The logistic differential equation is what is called an + autonomous equation. + An autonomous differential equation has no explicit dependence on the independent variable + (t in this case). + This has no real effect on the application of Euler's method other than the fact that the function f(t,y) is really just a function of y. + To take steps in the y variable, we use + + y_{i+1} = y_i + hf(t_i,y_i) = y_i + hy_i(1-y_i) + . +

    +

    + Using N=10 steps requires \displaystyle h = \frac{4-0}{10} = 0.4. + Implementing Euler's Method, we have + + x_0 \amp = 0 \amp y_0 \amp = 0.25 + x_1 \amp = 0.4 \amp y_1 \amp = 0.25 + 0.4(0.25)(1-0.25) + \amp \amp \amp = 0.325 + x_2 \amp = 0.8 \amp y_2 \amp = 0.325 + 0.4(0.325)(1-0.325) + \amp \amp \amp = 0.41275 + x_3 \amp = 1.2 \amp y_3 \amp = 0.41275 + 0.4(0.41275)(1-0.41275) + \amp \amp \amp = 0.50970 + x_4 \amp = 1.6 \amp y_4 \amp = 0.50970 + 0.4(0.50970)(1 - 0.50970) + \amp \amp \amp = 0.60966 + x_5 \amp = 2.0 \amp y_5 \amp = 0.60966 + 0.4(0.60966)(1-0.60966) + \amp \amp \amp = 0.70485 + x_6 \amp = 2.4 \amp y_6 \amp = 0.70485 + 0.4(0.70485)(1 - 0.70485) + \amp \amp \amp = 0.78806 + x_7 \amp = 2.8 \amp y_7 \amp = 0.78806 + 0.4(0.78806)(1-0.78806) + \amp \amp \amp = 0.85487 + x_8 \amp = 3.2 \amp y_8 \amp = 0.85487 + 0.4(0.85487)(1-0.85487) + \amp \amp \amp = 0.90450 + x_9 \amp = 3.6 \amp y_9 \amp = 0.90450 + 0.4(0.90450)(1 - 0.90450) + \amp \amp \amp = 0.93905 + x_{10}\amp = 4.0 \amp y_{10}\amp = 0.93905 + 0.4(0.93905)(1 - 0.93905) + \amp \amp \amp = 0.96194 + . +

    +

    + These 11 points, along with the the analytic solution, + are plotted in . + Notice how well they seem to match the true solution. +

    +
    + Euler's Method approximation to \yp = y(1-y) with y(0) = 0.25 from , + along with the analytical solution + + + Euler's Method approximation to \yp = y(1-y) with y(0) = 0.25 along with its analytical solution. + + +

    + The y axis is drawn from 0 to 1 and the x axis is drawn from 0 + to 4. There is a curve and a plot of 11 points, the curve and the plot overlap indicating + the appropriateness of the approximation. + The curve starts at y intercept 0.25 and rises with a positive slope and ends in the graph at + point (4,1), the plot has the points distributed evenly along the curve, with few points slightly + away from the curve. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + xtick={0,1,2,3,4}, + ytick={0,0.5,1}, + xlabel={$t$}, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=4.1 + ] + + \addplot [firstcurvestyle,domain = 0:4, samples=50] {1/(1+3*exp(-x)}; + \addplot [firstcolor, mark = *] coordinates{(0,.25)(0.4,.325)(0.8,.41275)(1.2,.50970)(1.6,.60966)(2.0,.70485)(2.4,.78806)(2.8,.85487)(3.2,.90450)(3.6,.93905)(4.0,.96194)}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $t$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +
    +     +     + + + +

    + The study of differential equations is a natural extension of the study of derivatives and integrals. + The equations themselves involve derivatives, + and methods to find analytic solutions often involve finding antiderivatives. + In this section, + we focus on graphical and numerical techniques to understand solutions to differential equations. + We restrict our examples to relatively simple initial value problems that permit analytic solutions to the equations, + but we should remember that this is only for comparison purposes. + In reality, many differential equations, + even some that appear straightforward, + do not have solutions we can find analytically. + Even so, we can use the techniques presented in this section to understand the behavior of solutions. + In the next two sections, + we explore two techniques to find analytic solutions to two different classes of differential equations. +

    +     + + + + Terms and Concepts + + +

    + In your own words, what is an initial value problem, + and how is it different from a differential equation? +

    +     + + + +

    + An initial value problems is a differential equation that is paired with one or more initial conditions. + A differential equation is simply the equation without the initial conditions. +

    +     +     + + +

    + In your own words, + describe what it means for a function to be a solution to a differential equation. +

    +     + + +     + + +

    + How can we verify that a function is a solution to a differential equation? +

    +     + + + +

    + Substitute the proposed function into the differential equation, + and show the the statement is satisfied. +

    +     +     + + +

    + Describe the difference between a particular solution and a general solution. +

    +     + + + +

    + A particular solution is one specifica member of a family of solutions, + and has no arbitrary constants. + A general solution is a family of solutions, + includes all possible solutions to the differential equation, + and typically includes one or more arbitrary constants. +

    +     +     + + +

    + Why might we use a graphical or numerical technique to study solutions to a differential equation + instead of simply solving the differential equation to find an analytic solution? +

    +     + + + +

    + Many differential equations are impossible to solve analytically. +

    +     +     + + +

    + Describe the considerations that should be made when choosing an h value to use in a numerical method like Euler's Method. +

    +     + + + +

    + A smaller h value leads to a numerical solution that is closer to the true solution, + but decreasing the h value leads to more computational effort. +

    +     +     +     + + + Problems + + +

    + Verify that the given function is a solution to the differential equation or initial value problem. +

    +     + + + +

    + y = Ce^{-6x^2}; \yp = -12xy. +

    +     + + +

    + If y=Ce^{-6x^2}, then \yp = -12xCe^{-6x^2} = -12x(Ce^{-6x^2})=-12xy. +

    +     +     + + +

    + y = x\sin(x); + \yp - x\cos(x)= (x^2+1)\sin(x) - xy, with y(\pi) = 0. +

    +     + + +

    + Given y=x\sin(x), we find \yp = \sin(x)+x\cos(x), using the product rule. + Then, \yp-x\cos(x) = \sin(x); on the other hand, + + (x^2+1)\sin(x)-xy = x^2\sin(x)+\sin(x)-x^2\sin(x)=\sin(x) + . + Finally, we check that y(\pi) = \pi\sin(\pi)=0. +

    +     +     + + +

    + 2x^2-y^2=C; y\yp-2x = 0 +

    +     + + + +

    + Using implicit differentiation, 2x^2-y^2=C + implies that 4x - 2y\yp = 0. + Dividing by -2 gives us y\yp - 2x=0. +

    +     +     + + +

    + y=xe^x; \yp' - 2\yp + y = 0 +

    +     + + + +

    + If y=xe^x, we find \yp = e^x+xe^x, + and \yp' = 2e^x+xe^x. Therefore, + + \yp' - 2\yp + y = 2e^x+xe^x -2e^x-2xe^x + xe^x = 0 + . + +

    +     +     +     + + +

    + Verify that the given function is a solution to the differential equation and find the C + value required to make the function satisfy the initial condition. +

    +     + + +

    + y = 4e^{3x}\sin(x) + Ce^{3x}; + \yp - 3y = 4e^{3x}\cos(x), with y(0)=2 +

    +     + + + +

    + With y = 4e^{3x}\sin(x) + Ce^{3x}, we find + \yp = 12 e^{3x}\sin(x) + 4e^{3x}\cos(x) + 3Ce^{3x}. + This gives \yp - 3y = 4e^{3x}\cos(x), as required. +

    + +

    + We also find y(0) = 4(1)(0)+C(1) = C, so we must have C=2. +

    +     +     + + + +

    + y(x^2+y) = C; 2xy + (x^2+2y)\yp= 0, with y(1)=2 +

    +     + + + +

    + If y(x^2+y)=C, implicit differentiation gives us + + \yp(x^2+y)+y(2x+\yp) = 2xy + 2y\yp +x^2\yp = 0 + , + so the differential equation is satisfied. When x=1, y=2, + which gives 2(1+2)=C, or C=6. + +

    +     +     +     + + + +

    + Sketch a slope field for the given differential equation. + Let x and y range between -2 and 2. +

    +     + + + +

    + \yp = y-x +

    +     + + + + Graph showing slope field for the given differential equation. + + +

    + The x and y axes are uncalibrated.In the first quadrant in the top left, + the field lines are north-east facing and in the bottom right they are southeast facing. + In the second quadrant the field lines are all north-east facing. In the third quadrant + like in the first quadrant in the top left the field lines are northeast facing and in + the bottom right they are southeast facing. In the fourth quadrant all lines are + southeast facing. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -2:2, + ytick={-2,-1,0,1,2}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(y-x)^2))}, v = {(y-x)/(sqrt(1+(y-x)^2))},scale arrows=.25},samples=9]{0}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +     +     + + + +

    + \yp = \displaystyle \frac{x}{2y} +

    +     + + + + Graph showing slope field for the given differential equation. + + +

    + The x and y axes are uncalibrated. The field lines form concentric + ovals facing away from the origin on both positive and negative x and + y axes. The concentric shorter arcs are on either end of the x axis. + On the two ends of the y axis concentric wider arcs are drawn. + The field lines intermix to form an 'X' with centre at the origin. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -2:2, + ytick={-2,-1,0,1,2}, + ] + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(x/(2*y))^2))}, v = {(x/(2*y))/(sqrt(1+(x/(2*y))^2))},scale arrows=.2},samples=10]{0}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +     +     + + + +

    + \yp = \sin(\pi y) +

    +     + + + + Graph showing slope field for the given differential equation. + + +

    + The x and y axes are uncalibrated. There are five instances where + the field lines run parallel to the x axis. One of them is on the x + axis itself, other two pairs of such field lines are above and below the x + axis. In between the x axis and the first horizontal field line for some + positive y value, the field lines are all northeast facing. Above the + horizontal field line for some y value until another with a higher y + value, the field lines in between are southeast facing. +

    +

    + Similarly below the x axis till the first horizontal line with some negative + y value, the field lines in between are southeast facing. In between this + horizontal line and another horizontal line with a higher negative y value, + the field lines are northeast facing. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -2:2, + ytick={-2,-1,0,1,2}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(sin(deg pi*y))^2))}, v = {(sin(deg pi*y))/(sqrt(1+(sin(deg pi*y))^2))},scale arrows=.15},samples=20]{0}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +     +     + + + +

    + \yp = \frac{y}{4} +

    +     + + + + Graph showing slope field for the given differential equation. + + +

    + The x and y axes are uncalibrated. The field lines run + almost parallel to the x axis. Above the axis the field lines + are slightly facing north east. Below the x axis the lines are + directed facing southeast. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -2:2, + ytick={-2,-1,0,1,2}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(y/4)^2))}, v = {(y/4)/(sqrt(1+(y/4)^2))},scale arrows=.2},samples=10]{0}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +     +     +     + + + +

    + Match each slope field below with the appropriate differential equation. +

    + + + + + Graph of slope field used in the exercise. + + +

    + The y and the x axis are shown both uncalibrated. In the second and the third + quadrants the field lines towards larger negative values are south facing. The field lines + closer to the y axis are south-east facing. Around the y axis the field lines + are almost horizontal. In the first and the fourth quadrants the field lines closer to the + y axis are east facing but for greater values of x they are south-east facing. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -2:2, + y domain = -2:2, + ytick={-3,-2,-1,0,1,2,3}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(x*(1-x))^2))}, v = {(x*(1-x))/(sqrt(1+(x*(1-x))^2))},scale arrows=.1},samples=15]{0}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + + + + Graph of slope field used in the exercise. + + +

    + The field lines are concentric facing upwards in the second and the first quadrant along the + positive y axis. The lines furthest away from the origin are more circular and the lines + closest to the x axis are almost parallel to the x axis. Similarly in the third + and the fourth quadrant the field lines are concentric along the negative y axis and the + ones away from origin are more circular and the ones closest to the x axis are almost + parallel to the x axis. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -2:2, + y domain = -2:2, + ytick={-3,-2,-1,0,1,2,3}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(x*y)^2))}, v = {(x*y)/(sqrt(1+(x*y)^2))},scale arrows=.1},samples=15]{0}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +     + +

    + (a) +

    +

    + (b) +

    +     + + + + Graph of slope field used in the exercise. + + +

    + The field lines in the second and the first quadrant are south-east facing and appear to be east + facing when they come very close to the x axis. Similarly the field lines in the third + and the fourth quadrants are north-east facing and become east facing when they come very close + to the x axis. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -2:2, + y domain = -2:2, + ytick={-3,-2,-1,0,1,2,3}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(-y)^2))}, v = {(-y)/(sqrt(1+(-y)^2))},scale arrows=.1},samples=15]{0}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + + + + Graph of slope field used in the exercise. + + +

    + The field lines appear to be concentric dome shaped lines with peaks along the positive y + axis. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -2:2, + y domain = -2:2, + ytick={-3,-2,-1,0,1,2,3}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(-x)^2))}, v = {(-x)/(sqrt(1+(-x)^2))},scale arrows=.1},samples=15]{0}; + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +     + +

    + (c) +

    + +

    + (d) +

    +     +     +     + + + + \yp=xy + (b) + + + \yp = -y + (c) + + + \yp = -x + (d) + + + \yp = x(1-x) + (a) + + + + +     + + + +

    + Sketch the slope field for the differential equation, + and use it to draw a sketch of the solution to the initial value problem. +

    +     + + +

    + \yp = \displaystyle \frac{y}{x} - y, with y(0.5)=1. +

    +     + + + + Graph showing slope field for the given differential equation. + + +

    + The x and y axes are uncalibrated, the field lines in the first quadrant + are shown. The field lines very close to the y axis are almost north facing for + higher values of y and almost east facing for lower values of y. With + smaller values of x, the field lines, from left to right the lines first face + northeast then east and southeast after for greater values of x. +

    +

    + A curve is drawn that starts at a point for some small value of x and a high + value of y. The curve has a positive slope at first after reaching a peak it + declines almost close to the x axis. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + y domain = -.5:1.5, + domain = -.5:5, + minor x tick num=1, + extra x ticks={1,3,5}, + ymin=-.1,ymax=1.6, + xmin=-.1,xmax=5.2 + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(y/x-y)^2))}, v = {(y/x-y)/(sqrt(1+(y/x-y)^2))},scale arrows=.15},samples=15]{0}; + \addplot [firstcurvestyle,domain=0.5:5.1,samples = 51] {2*x*exp(1/2-x)}; + + \fill[black,draw=black] (axis cs:.5,1) circle (2.4pt); + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +     +     + + +

    + \yp = y\sin(x), with y(0)=1. +

    +     + + + + Graph showing slope field for the given differential equation. + + +

    + The x and y axes are uncalibrated, the field lines in the first + quadrant are shown. + Front left to right, a little away from the x axis the field lines are northeast + facing that transition to north facing. Moving further right then again become + northeast facing then transition to southeast facing, further right they become + south facing then east facing. The pattern then repeats. Very close to the x + axis the field lines are almost parallel to it. +

    +

    + A wave is drawn that starts at some y intercept above the origin. It has a high + positive slope, it reaches peak when the field lines change from northeast facing + to southeast facing, then it declines until the point the field lines are parallel + to the x axis. The curve continues to form a second wave. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + domain = -.5:10, + minor x tick num=1, + extra x ticks={2.5,7.5}, + minor y tick num=1, + extra y ticks={2.5,7.5}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(y*sin(deg x))^2))}, v = {(y*sin(deg x))/(sqrt(1+(y*sin(deg x))^2))},scale arrows=.4},samples=15]{0}; + \addplot [firstcurvestyle,domain=0:10,samples = 100] {exp(1-cos(deg x))}; + + \fill[black,draw=black] (axis cs:0,1) circle (2.4pt); + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +     +     + + +

    + \yp = y^2-3y+2, with y(0)=2. +

    +     + + + + Graph showing slope field for the given differential equation. + + +

    + The x and y axes are uncalibrated, the field lines in the first + quadrant are shown. There are two instances where the field lines are parallel + to the x axis. From under the x axis to the first such line the + field lines transition from almost north facing to northeast facing. Between + the horizontal field line for a small y value and a greater y value + the field lines are facing southeast. Above the line with a higher y value + the field lines transition from northeast facing to north facing. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + y domain = -.15:3, + domain = -.15:4, + minor x tick num=1, + extra x ticks={1,3}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(y^2-3*y+2)^2))}, v = {(y^2-3*y+2)/(sqrt(1+(y^2-3*y+2)^2))},scale arrows=.2},samples=15]{0}; + \addplot [firstcurvestyle,-] coordinates {(0,2) (4.1,2)}; + + \fill[black,draw=black] (axis cs:0,2) circle (2.4pt); + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +     +     + + +

    + \displaystyle \yp = -\frac{xy}{1+x^2}, with y(0)=1. +

    +     + + + + Graph showing slope field for the given differential equation. + + +

    + The x and y axes are uncalibrated, the field lines in the first quadrant are shown. + In the top right and the centre the field lines are southeast facing, very close to the x + and y axis the field lines are almost parallel to the x axis. + A curve is drawn that starts from a y intercept and decreases along the slope lines coming + close to the x axis. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + y domain = -.15:3, + domain = -.15:4, + minor x tick num=1, + extra x ticks={1,3}, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(-x*y/(1+x^2))^2))}, v = {(-x*y/(1+x^2))/(sqrt(1+(-x*y/(1+x^2))^2))},scale arrows=.2},samples=15]{0}; + \addplot [firstcurvestyle,domain=0:4,samples=50] {1/sqrt(1+x^2)}; + + \fill[black,draw=black] (axis cs:0,1) circle (2.4pt); + + \end{axis} + + %\node [right] at (myplot.right of origin) { $x$}; + %\node [above] at (myplot.above origin) { $y$}; + + \end{tikzpicture} + + +     +     +     + + + +

    + Use Euler's Method to make a table of values that approximates the solution to the initial value problem on the given interval. + Use the specified h or N value. +

    +     + + + +

    + \yp = x+2y +

    +

    + y(0)=1 +

    +

    + interval: [0,1] +

    +

    + h=0.25 +

    +     + +

    + + x_i \amp \quad \amp \quad \amp y_i + 0.00 \amp \quad \amp \quad \amp 1.0000 + 0.25 \amp \quad \amp \quad \amp 1.5000 + 0.50 \amp \quad \amp \quad \amp 2.3125 + 0.75 \amp \quad \amp \quad \amp 3.5938 + 1.00 \amp \quad \amp \quad \amp 5.5781 + +

    +     +     + + + +

    + \yp = xe^{-y} +

    +

    + y(0)=1 +

    +

    + interval: [0,0.5] +

    +

    + N=5 +

    +     + +

    + + x_i \amp \quad \amp \quad \amp y_i + 0.0 \amp \quad \amp \quad \amp 1.0000 + 0.1 \amp \quad \amp \quad \amp 1.0000 + 0.2 \amp \quad \amp \quad \amp 1.0037 + 0.3 \amp \quad \amp \quad \amp 1.0110 + 0.4 \amp \quad \amp \quad \amp 1.0219 + 0.5 \amp \quad \amp \quad \amp 1.0363 + +

    +     +     + + + +

    + \yp = y + \sin(x) +

    +

    + y(0)=2 +

    +

    + interval: [0,1] +

    +

    + h = 0.2 +

    +     + +

    + + x_i \amp \quad \amp \quad \amp y_i + 0.0 \amp \quad \amp \quad \amp 2.0000 + 0.2 \amp \quad \amp \quad \amp 2.4000 + 0.4 \amp \quad \amp \quad \amp 2.9197 + 0.6 \amp \quad \amp \quad \amp 3.5816 + 0.8 \amp \quad \amp \quad \amp 4.4108 + 1.0 \amp \quad \amp \quad \amp 5.4364 + +

    +     +     + + + +

    + \yp = e^{x-y} +

    +

    + y(0)=0 +

    +

    + interval: [0,2] +

    +

    + h = 0.5 +

    +     + +

    + + x_i \amp \quad \amp \quad \amp y_i + 0.0 \amp \quad \amp \quad \amp 0.0000 + 0.5 \amp \quad \amp \quad \amp 0.5000 + 1.0 \amp \quad \amp \quad \amp 1.8591 + 1.5 \amp \quad \amp \quad \amp 10.5824 + 2.0 \amp \quad \amp \quad \amp 88378.1190 + +

    +     +     +     + + + +

    + Use the provided solution y(x) and Euler's Method with the h=0.2 and h=0.1 to complete the following table. +

    + + + + + + + + + + x 0.0 0.20.4 0.6 0.8 1.0 + + + y(x) + + + h = 0.2 + + + h = 0.1 + + +     + + + +

    + \yp = xy^2 +

    +

    + y(0)=1 +

    +

    + Solution: \displaystyle y(x) = \frac{2}{1-x^2} +

    +     + + + + + + + + + + + x 0.0 0.20.4 0.6 0.8 1.0 + + + y(x)1.00001.02041.08701.21951.47062.0000 + + + h = 0.21.00001.00001.04001.12651.27881.5405 + + + h = 0.11.00001.01001.06231.16871.36011.7129 + + + +     + + +

    + \displaystyle \yp = xe^{x^2}+\frac{1}{2}xy +

    +

    + \displaystyle y(0)=\frac{1}{2} +

    +

    + Solution: \displaystyle y(x) = \frac{1}{2}(x^2+1)e^{x^2} +

    +     + + + + + + + + + + + x 0.0 0.20.4 0.6 0.8 1.0 + + + y(x)0.50000.54120.68060.97471.55512.7183 + + + h = 0.20.50000.50000.58160.76861.12501.7885 + + + h = 0.10.50000.52010.62820.86221.31322.1788 + + + +     +     +     +     +
    +
    + Separable Differential Equations + +

    + There are specific techniques that can be used to solve specific types of differential equations. + This is similar to solving algebraic equations. + In algebra, we can use the quadratic formula to solve a quadratic equation, + but not a linear or cubic equation. + In the same way, + techniques that can be used for a specific type of differential equation are often ineffective for a differential equation of a different type. + In this section, + we describe and practice a technique to solve a class of differential equations called + separable equations. +

    +     + + + Separation of Variables + + + Separable Differential Equation + +

    + A separable differential equation is one that can be written in the form + + \displaystyle n(y) \frac{dy}{dx} = m(x) + , + where n is a function that depends only on the dependent variable y, + and m is a function that depends only on the independent variable x. + differential equationseparable +

    +     +     +

    + Below, we show a few examples of separable differential equations, + along with similar looking equations that are not separable. +

    + + + Separable +
      +
    1. \displaystyle \frac{dy}{dx} = x^2y
      2. +
    2. \displaystyle y\sqrt{y^2-5} \frac{dy}{dx} - \sin(x) \cos(y) = 0
      3. +
    3. \displaystyle \frac{dy}{dx} = \frac{(x^2 + 1)e^{y}}{y}
      4. +
    +     + + Not Separable +
      +
    1. \displaystyle \frac{dy}{dx} = x^2 + y
      2. +
    2. \displaystyle y\sqrt{y^2-5} \frac{dy}{dx} - \sin(x) \cos(y) = 1
      3. +
    3. \displaystyle \frac{dy}{dx} = \frac{(xy + 1)e^{y}}{y}
      4. +
    +     +     +

    + Notice that a separable equation requires that the functions of the dependent and independent variables be multiplied, + not added + (like Item in List). + An alternate definition of a separable differential equation states that an equation is separable if it can be written in the form + + \frac{dy}{dx} = f(x)g(y) + , + for some functions f and g. +

    + + + +

    + separation of variables + Let's find a formal solution to the separable equation + + \displaystyle n(y) \frac{dy}{dx} = m(x) + . +

    +

    + Since the functions on the left and right hand sides of the equation are equal, + their antiderivatives should be equal up to an arbitrary constant of integration. + That is + + \displaystyle \int n(y) \frac{dy}{dx}\,dx = \int m(x)\, dx + C + . +

    +

    + Though the integral on the left may look a bit strange, + recall that y itself is a function of x. + Consider the substitution u = y(x). + The differential is du = \displaystyle \frac{dy}{dx}\,dx. + Using this substitution, the above equation becomes + + \int n(u)\,du = \int m(x)\,dx + C + . +

    +

    + Let N(u) and M(x) be antiderivatives of n(u) and m(x), + respectively. + Then + + N(u) = M(x) + C + . +

    +

    + Since u = y(x), this is + + N(y) = M(x) + C + . +

    +

    + This relationship between y and x is an implicit form of the solution to the differential equation. + Sometimes (but not always) it is possible to solve for y to find an explicit version of the solution. +

    +

    + Though the technique outlined above is formally correct, + what we did essentially amounts to integrating the function n with respect to its variable and integrating the function + m with respect to its variable. + The informal way to solve a separable equation is to treat the derivative + \displaystyle \frac{dy}{dx} as if it were a fraction. + The separated form of the equation is + + n(y)\,dy = m(x)\, dx + . +

    +

    + To solve, we integrate the left hand side with respect to y and the right hand side with respect to x and add a constant of integration. + As long as we are able to find the antiderivatives, + we can find an implicit form for the solution. + Sometimes we are able to solve for y in the implicit solution to find an explicit form of the solution to the differential equation. + We practice the technique by solving the three differential equations listed in the separable column above, + and conclude by revisiting and finding the general solution to the logistic differential equation from . +

    + + Solving a Separable Differential Equation + +

    + Find the general solution to the differential equation \yp = x^2y. +

    +     + +

    + Using the informal solution method outlined above, + we treat \displaystyle \frac{dy}{dx} as a fraction, + and write the separated form of the differential equation as + + \frac{dy}{y} = x^2 dx + . +

    + + + +

    + Integrating the left hand side of the equation with respect to y and the right hand side of the equation with respect to x yields + + \ln \abs{y} = \frac{1}{3}x^3 + C + . +

    +

    + This is an implicit form of the solution to the differential equation. + Solving for y yields an explicit form for the solution. + Exponentiating both sides, we have + + \abs{y} = e^{x^3/3 + C} = e^{x^3/3}e^C + . +

    + +

    + This solution is a bit problematic. + First, the absolute value makes the solution difficult to understand. + The second issue comes from our desire to find the + general solution. + Recall that a general solution includes all possible solutions to the differential equation. + In other words, for any given initial condition, + the general solution must include the solution to that specific initial value problem. + We can often satisfy any given initial condition by choosing an appropriate C value. + When solving separable equations, though, + it is possible to lose solutions that have the form y = \text{ constant}. + Notice that y=0 solves the differential equation, + but it is not possible to choose a finite C to make our solution look like y=0. + Our solution cannot solve the initial value problem + \displaystyle \frac{dy}{dx} = x^2y, with y(a) = 0 + (where a is any value). + Thus, we haven't actually found a general solution to the problem. + We can clean up the solution and recover the missing solution with a bit of clever thought. +

    + + + +

    + Recall the formal definition of the absolute value: + \abs{y} = y if y \geq 0 and \abs{y} = -y if y \lt 0. + Our solution is either y = e^C e^{\frac{x^3}{3}} or y = - e^C e^{{\frac{x^3}{3}}}. + Further, note that C is constant, + so e^C is also constant. + If we write our solution as y = Ae^{\frac{x^3}{3}}, + and allow the constant A to take on either positive or negative values, + we incorporate both cases of the absolute value. + Finally, if we allow A to be zero, + we recover the missing solution discussed above. + The best way to express the general solution to our differential equation is + + y = Ae^{\frac{x^3}{3}} + . +

    +     + +     + + + + + Solving a Separable Initial Value Problem + +

    + Solve the initial value problem + \displaystyle (y\sqrt{y^2-5}) \yp - \sin(x) \cos(x) = 0, + with y(0) = -3. +

    +     + +

    + We first put the differential equation in separated form + + y\sqrt{y^2-5}\,dy = \sin(x) \cos(x)\, dx + . +

    +

    + The indefinite integral \displaystyle \int y\sqrt{y^2-5}\,dy requires the substitution u = y^2-5. + Using this substitute yields the antiderivative \displaystyle \frac{1}{3} (y^2-5)^{3/2}. + The indefinite integral \displaystyle \int \sin(x) \cos(x)\,dx requires the substitution u = \sin(x). + Using this substitution yields the antiderivative \displaystyle \frac{1}{2} \sin^2 x. + Thus, we have an implicit form of the solution to the differential equation given by + + \frac{1}{3} (y^2-5)^{3/2} = \frac{1}{2} \sin^2 x + C + . +

    +

    + The initial condition says that y should be -3 when x is 0, or + + \frac{1}{3} ((-3)^2 - 5)^{3/2} = \frac{1}{2} \sin^2 0 + C + . +

    +

    + Evaluating the line above, we find C = 8/3, + yielding the particular solution to the initial value problem + + \frac{1}{3} (y^2-5)^{3/2} = \frac{1}{2} \sin^2 x + \frac{8}{3} + . +

    +     + +     + + + Solving a Separable Differential Equation + +

    + Find the general solution to the differential equation \displaystyle \frac{dy}{dx} = \frac{(x^2 + 1)e^{y}}{y}. +

    +     + +

    + We start by observing that there are no constant solutions to this differential equation because there are no constant y + values that make the right hand side of the equation identically zero. + Thus, we need not worry about losing solutions during the separation of variables process. + The separated form of the equation is given by + + ye^{-y}\,dy = (x^2+1)\,dx + . +

    +

    + The antiderivative of the left hand side requires Integration by Parts. + Evaluating both indefinite integrals yields the implicit solution + + -(y+1)e^{-y} = \frac{1}{3}x^3 + x + C + . +

    +

    + Since we cannot solve for y, + we cannot find an explicit form of the solution. +

    +     + +     + + + + + Solving the Logistic Differential Equation + +

    + Solve the logistic differential equation \displaystyle \frac{dy}{dt} = ky\left( 1 - \frac{y}{M}\right) +

    +     + +

    + We looked at a slope field for this equation in + in the specific case of k = M = 1. + Here, we use separation of variables to find an analytic solution to the more general equation. + Notice that the independent variable t does not explicitly appear in the differential equation. + We mentioned that an equation of this type is called autonomous. + All autonomous first order differential equations are separable. +

    +

    + We start by making the observation that both y=0 and y = M are constant solutions to the differential equation. + We must check that these solutions are not lost during the separation of variables process. + The separated form of the equation is + + \frac{1}{y \left(1-\displaystyle\frac{y}{M}\right)}\,dy = k\,dt + . +

    +

    + The antiderivative of the left hand side of the equation can be found by making use of partial fractions. + Using the techniques discussed in , we write + + \frac{1}{y\left(1-\frac{y}{M}\right)} = \frac{1}{y} + \frac{1}{M-y} + . +

    +

    + Then an implicit form of the solution is given by + + \ln \abs{y} - \ln \abs{M-y} = kt + C + . +

    +

    + Combining the logarithms, + + \ln \abs{\frac{y}{M-y}} = kt + C + . +

    +

    + Similarly to , + we can write + + \frac{y}{M-y} = Ae^{kt} + . +

    +

    + Letting A take on positive values or negative values incorporates both cases of the absolute value. + This is another implicit form of the solution. + Solving for y gives the explicit form + + y = \frac{M}{1 + be^{-kt}} + , + where b is an arbitrary constant. + Notice that b=0 recovers the constant solution y = M. + The constant solution y=0 cannot be produced with a finite b value, + and has been lost. + The general solution the logistic differential equation is the set containing \displaystyle y = \frac{M}{1 + be^{-kt}} and y=0. +

    +     + +     + + +     + + + + Problems + + +

    + Decide whether the differential equation is separable or not separable. + If the equation is separable, write it in separated form. +

    +     + + +

    + \displaystyle \yp = y^2 - y +

    +     + +

    + Separable. + \displaystyle \frac{1}{y^2-y}\,dy = dx +

    +     +     + + +

    + \displaystyle x\yp + x^2y = \frac{\sin(x)}{x-y} +

    +     + +

    + Not separable. +

    +     +     + + +

    + \displaystyle (y + 3)\yp + (\ln(x)) \yp - x\sin(y) = (y+3)\ln(x) +

    +     + +

    + Not separable. +

    +     +     + + +

    + \displaystyle \yp -x^2\cos(y) + y = \cos(y) - x^2 y +

    +     + +

    + Separable. + \displaystyle \frac{1}{\cos(y) - y}\,dy = (x^2+1)\,dx +

    +     +     +     + + +

    + Find the general solution to the separable differential equation. + Be sure to check for missing constant solutions. +

    +     + + +

    + \displaystyle \yp +1 - y^2 = 0 +

    +     + +

    + \left\{ \displaystyle y = \frac{1 + Ce^{2x}}{1 - Ce^{2x}}, y = -1\right\} +

    +     +     + + +

    + \displaystyle \yp = y-2 +

    +     + +

    + y = 2 + Ce^x +

    +     +     + + +

    + \displaystyle x \yp = 4y +

    +     + +

    + y = Cx^4 +

    +     +     + + +

    + \displaystyle y\yp = 4x +

    +     + +

    + y^2 - 4x^2 = C +

    +     +     + + +

    + \displaystyle e^xy \yp = e^{-y} + e^{-2x - y} +

    +     + +

    + \displaystyle (y-1)e^y = -e^{-x} - \frac{1}{3}e^{-3x} + C +

    +     +     + + +

    + \displaystyle (x^2 + 1) \yp = \frac{x}{y-1} +

    +     + +

    + \displaystyle (y-1)^2 = \ln(x^2+1) + C +

    +     +     + + +

    + \displaystyle \yp = \frac{x\sqrt{1-4y^2}}{x^4 + 2x^2 + 2} +

    +     + +

    + \left\{ \arcsin{2y} - \arctan(x^2+1) = C, y = \pm \displaystyle \frac{1}{2} \right\} +

    +     +     + + +

    + \displaystyle (e^x + e^{-x})\yp = y^2 +

    +     + +

    + \left\{ \displaystyle y = \frac{1}{C - \arctan x}, y = 0 \right\} +

    +     +     +     + + +

    + Find the particular solution to the separable initial value problem. +

    +     + + +

    + \displaystyle \yp = \frac{\sin(x)}{\cos(y)}, + with y(0) = \displaystyle \frac{\pi}{2} +

    +     + +

    + \sin(y) + \cos(x) = 2 +

    +     +     + + +

    + \displaystyle \yp = \frac{x^2}{1-y^2}, with y(0) = -2 +

    +     + +

    + -x^3 + 3y - y^3 = 2 +

    +     +     + + +

    + \displaystyle \yp = \frac{2x}{y+x^2y}, with y(0) = -4 +

    +     + +

    + \frac{1}{2}y^2 - \ln(1+x^2) = 8 +

    +     +     + + +

    + \displaystyle x + ye^{-x}\yp = 0, with y(0) = -2 +

    +     + +

    + y^2+2xe^x - 2e^x = 2 +

    +     +     + + +

    + \displaystyle \yp = \frac{x\ln(x^2+1)}{y-1}, with y(0) = 2 +

    +     + +

    + \displaystyle \frac{1}{2}y^2 - y = \frac{1}{2}\big ( (x^2+1)\ln(x^2+1) - (x^2 + 1)\big) + \frac{1}{2} +

    +     +     + + +

    + \displaystyle \sqrt{1-x^2}\,\yp - \frac{\arcsin x}{y\cos(y^2)}= 0, + with y(0) = \sqrt{\displaystyle\frac{7\pi}{6}} +

    +     + +

    + \sin(y^2)-(\arcsin x)^2 = -\frac{1}{2} +

    +     +     + + +

    + \displaystyle \yp = (\cos^2(x))(\cos^2 (2y)), with y(0) = 0 +

    +     + +

    + 2\tan(2y) = 2x + \sin(2x) +

    +     +     + + +

    + \displaystyle \yp = \frac{y^2\sqrt{1-y^2}}{x}, with y(0) = 1 +

    +     + +

    + x = exp \displaystyle \left( -\frac{\sqrt{1-y^2}}{y}\right) +

    +     +     +     +     +     +
    +
    + First Order Linear Differential Equations + +

    + In the previous section, + we explored a specific techique to solve a specific type of differential equation called a separable differential equation. + In this section, + we develop and practice a technique to solve a type of differential equation called a + first order linear differential equation. +

    + +

    + Recall than a linear algebraic equation in one variable is one that can be written ax + b = 0, + where a and b are real numbers. + Notice that the variable x appears to the first power. + The equations \sqrt{x}+1=0 and \sin(x)-3x = 0 are both nonlinear. + A linear differential equation is one in which the dependent variable and its derivatives appear only to the first power. + We focus on first order equations, which involve first + (but not higher order) + derivatives of the dependent variable. +

    + +     + + + Solving First Order Linear Equations + + + First Order Linear Differential Equation + +

    + A first order linear differential equation is a differential equation that can be written in the form + + \frac{dy}{dx} + p(x)y = q(x) + , + where p and q are arbitrary functions of the independent variable x. + differential equationfirst order linear +

    +     +     + + + Classifying Differential Equations + +

    + Classify each differential equation as first order linear, + separable, both, or neither. +

    +

    +

      +
    1. \displaystyle \yp = xy
      2. +
    2. \displaystyle \yp = e^y + 3x
      3. +
    3. \displaystyle \yp - (\cos(x))y = \cos(x)
      4. +
    4. \displaystyle y\yp -3xy = 4\ln(x)
      5. +
    +

    +     + +

    +

      +
    1. +

      Both. + We identify p(x) = -x and q(x) = 0. + The separated form of the equation is \displaystyle \frac{dy}{y} = x\,dx. +

      +
      2. +
    2. +

      + Neither. + The e^y term makes the equation nonlinear. + Because of the addition, + it is not possible to write the equation in separated form. +

      +
      3. +
    3. +

      + First order linear. + We identify p(x) = -\cos(x) and q(x) = \cos(x). + The equation cannot be written in separated form. +

      +
      4. +
    4. +

      + Neither. + Notice that dividing by y results in the nonlinear term \displaystyle \frac{4\ln(x)}{y}. + It is not possible to write the equation in separated form. +

      +
      5. +
    +

    +     +     + + + +

    + Notice that linearity depends on the dependent variable y, + not the independent variable x. + The functions p(x) and q(x) need not be linear, + as demonstrated in part (c) of . + Neither \cos(x) nor \sin(x) are linear functions of x, + but the differential equation is still linear. +

    + +

    + Before working out a general technique for solving first order linear differential equations, + we look at a specific example. + Consider the differential equation + + \frac{d}{dx}\bigl(xy\bigr) = \sin(x) \cos(x) + . +

    +

    + This is an easy differential equation to solve. + On the left, + the antiderivative of the derivative is simply the function xy. + Using the substitution u = \sin(x) on the right and integrating results in the implicit solution + + xy = \frac{1}{2}\sin^2 x + C + . +

    +

    + Solving for y yields the explicit solution + + y = \frac{\sin^2 x}{2x} + \frac{C}{x} + . +

    +

    + Though not obvious, + the differential equation above is actually a linear differential equation. + Using the product rule and implicit differentiation, + we can write \displaystyle \frac{d}{dx}\big(xy\big) = x\frac{dy}{dx} + y. + Our original differential equation can be written + + x\frac{dy}{dx} + y = \sin(x) \cos(x) + . +

    +

    + If we divide by x, we have + + \frac{dy}{dx} + \frac{1}{x} y = \frac{\sin(x) \cos(x)}{x} + , + which matches the form in . + Reversing our steps would lead us back to the original form our our differential equation. +

    + + +

    + As motivated by the problem we just explored, + the basic idea behind solving first order linear differential equations is to multiply both sides of the differential equation by a function, + called an integrating factor, + that makes the left hand side of the equation look like an expanded Product Rule. + We then condense the left hand side into the derivative of a product and integrate both sides. + An obvious question is, + How do you find this integrating factor? + differential equationintegrating factor +

    + + +

    + Consider the first order linear equation + + \frac{dy}{dx} + p(x)y = q(x) + . +

    +

    + Let's call the integrating factor \mu(x). + We multiply both sides of the differential equation by \mu(x) to get + + \mu(x) \left( \frac{dy}{dx} + p(x)y \right) = \mu(x)q(x) + . +

    + + + +

    + Our goal is to choose \mu(x) so that the left hand side of the differential equation looks like the result of a Product Rule. + The left hand side of the equation is + + \mu(x) \frac{dy}{dx} + \mu(x)p(x)y + . +

    +

    + Using the Product Rule and Implicit Differentiation, + + \frac{d}{dx} \bigl( \mu(x) y \bigr) = \frac{d\mu}{dx}y + \mu(x)\frac{dy}{dx} + . +

    + +

    + Equating + \frac{d}{dx} \big ( \mu(x) y \big ) and \mu(x) \left( \frac{dy}{dx} + p(x)y \right) gives + + \frac{d\mu}{dx}y + \mu(x)\frac{dy}{dx} = \mu(x) \frac{dy}{dx} + \mu(x)p(x)y + , + which is equivalent to + + \frac{d\mu}{dx} = \mu(x)p(x) + . +

    +

    + In order for the integrating factor \mu(x) to perform its job, + it must solve the differential equation above. + But that differential equation is separable, so we can solve it. + The separated form is + + \frac{d\mu}{\mu} = p(x)\,dx + . +

    +

    + Integrating, + + \ln(\mu) = \int p(x)\,dx + , + or + + \mu(x) = e^{\int p(x)\,dx} + . +

    + + + +

    + If \mu(x) is chosen this way, + after multiplying by \mu(x), + we can always write the differential equation in the form + + \frac{d}{dx} \bigl( \mu(x)y \bigr) = \mu(x)q(x) + . +

    +

    + Integrating and solving for y, the explicit solution is + + y = \frac{1}{\mu(x)}\int \bigl( \mu(x)q(x) \bigr)\,dx + . +

    +

    + Though this formula can be used to write down the solution to a first order linear equation, + we shy away from simply memorizing a formula. + The process is lost, and it's easy to forget the formula. + Rather, we always always follow the steps outlined in + when solving equations of this type. +

    + + + Solving First Order Linear Equations +

    +

      +
    1. +

      + Write the differential equation in the form + + \frac{dy}{dx} + p(x)y = q(x) + . +

      +
      2. +
    2. +

      + Compute the integrating factor + + \mu(x) = e^{\int p(x)\,dx} + . +

      +
      3. +
    3. +

      + Multiply both sides of the differential equation by \mu(x), + and condense the left hand side to get + + \frac{d}{dx}\big( \mu(x)y \big) = \mu(x)q(x) + . +

      +
      4. +
    4. +

      + Integrate both sides of the differential equation with respect to x, + taking care to remember the arbitrary constant. +

      +
      5. +
    5. +

      + Solve for y to find the explicit solution to the differential equation. +

      +
      6. +
    +

    +     + +

    + Let's practice the process by solving the two first order linear differential equations from . +

    + + + Solving a First Order Linear Equation + +

    + Find the general solution to \yp = xy. +

    +     + +

    + We solve by following the steps in . + Unlike the process for solving separable equations, + we need not worry about losing constant solutions. + The answer we find will + be the general solution to the differential equation. + We first write the equation in the form + + \frac{dy}{dx} - xy = 0 + . +

    +

    + By identifying p(x) = -x, + we can compute the integrating factor + + \mu(x) = e^{\int -x\,dx} = e^{-\frac{1}{2}x^2} + . +

    +

    + Multiplying both side of the differential equation by \mu(x), we have + + e^{-\frac{1}{2}x^2}\left( \frac{dy}{dx} - xy\right) = 0 + . +

    +

    + The left hand side of the differential equation condenses to yield + + \frac{d}{dx} \left( e^{-\frac{1}{2}x^2}y\right) = 0 + . +

    + + + +

    + We integrate both sides with respect to x to find the implicit solution + + e^{-\frac{1}{2}x^2}y = C + , + or the explicit solution + + y = Ce^{\frac{1}{2}x^2} + . +

    +     + +     + + + + + Solving a First Order Linear Equation + +

    + Find the general solution to \yp -(\cos(x))y = \cos(x). +

    +     + +

    + The differential equation is already in the correct form. + The integrating factor is given by + + \mu(x) = e^{-\int\cos(x)\,dx} = e^{-\sin(x)} + . +

    +

    + Multiplying both sides of the equation by the integrating factor and condensing, + + \frac{d}{dx}\left(e^{-\sin(x)}y \right) = (\cos(x)) e^{-\sin(x)} + +

    +

    + Using the substitution u = -\sin(x), + we can integrate to find the implicit solution + + e^{-\sin(x)} y = -e^{-\sin(x)} + C + . +

    +

    + The explicit form of the general solution is + + y = -1 + Ce^{\sin(x)} + . +

    +     + +     + +

    + We continue our practice by finding the particular solution to an initial value problem. +

    + + + Solving a First Order Linear Initial Value Problem + +

    + Solve the initial value problem \displaystyle x\yp - y = x^3\ln(x), + with y(1)=0. +

    +     + +

    + We first divide by x to get + + \frac{dy}{dx}-\frac{1}{x}y = x^2\ln(x) + . +

    +

    + The integrating factor is given by + + \mu(x) \amp = e^{\int-\frac{1}{x}\,dx} + \amp = e^{-\ln(x)} + \amp = e^{\ln(x)^{-1}} + \amp = x^{-1} + . +

    +

    + Multiplying both sides of the differential equation by the integrating factor and condensing the left hand side, we have + + \frac{d}{dx} \left( \frac{y}{x}\right) = x\ln(x) + . +

    +

    + Using Integrating by Parts to find the antiderivative of x\ln(x), + we find the implicit solution + + \frac{y}{x} = \frac{1}{2}x^2\ln(x) - \frac{1}{4}x^2 + C + . +

    +

    + Solving for y, the explicit solution is + + y = \frac{1}{2}x^3\ln(x) - \frac{1}{4}x^3 + Cx + . +

    +

    + The initial condition y(1) = 0 yields C = 1/4. + The solution to the initial value problem is + + y = \frac{1}{2}x^3\ln(x) - \frac{1}{4}x^3 + \frac{1}{4}x + . +

    +     + +     + +

    + Differential equations are a valuable tool for exploring various physical problems. + This process of using equations to describe real world situations is called mathematical modeling, + and is the topic of the next section. + The last two examples in this section begin our discussion of mathematical modeling. +

    + + + A Falling Object Without Air Resistance + +

    + Suppose an object with mass m is dropped from an airplane. + Find and solve a differential equation describing the vertical velocity of the object assuming no air resistance. +

    +     + +

    + The basic physical law at play is Newton's second law, + + \text{ mass } \times \text{ acceleration } = \text{ the sum of the forces } + . +

    + +

    + Using the fact that acceleration is the derivative of velocity, + mass acceleration can be writting mv'. + In the absence of air resistance, + the only force of interest is the force due to gravity. + This force is approximately constant, + and is given by mg, where g is the gravitational constant. + The word equation above can be written as the differential equation + + m\frac{dv}{dt} = mg + . +

    + +

    + Because g is constant, + this differential equation is simply an integration problem, + and we find + + v = gt + C + . +

    + +

    + Since v = C with t=0, + we see that the arbitrary constant here corresponds to the initial vertical velocity of the object. +

    +     +     + +

    + The process of mathematical modeling does not stop simply because we have found an answer. + We must examine the answer to see how well it can describe real world observations. + In the previous example, + the answer may be somewhat useful for short times, + but intuition tells us that something is missing. + Our answer says that a falling object's velocity will increase linearly as a function of time, + but we know that a falling object does not speed up indefinitely. + In order to more fully describe real world behavior, + our mathematical model must be revised. +

    + + + + + A Falling Object with Air Resistance + +

    + Suppose an object with mass m is dropped from an airplane. + Find and solve a differential equation describing the vertical velocity of the object, + taking air resistance into account. +

    +     + +

    + We still begin with Newon's second law, + but now we assume that the forces in the object come both from gravity and from air resistance. + The gravitational force is still given by mg. + For air resistance, + we assume the force is related to the velocity of the object. + A simple way to describe this assumption might be kv^{p}, + where k is a proportionality constant and p is a positive real number. + The value k depends on various factors such as the density of the object, + surface area of the object, and density of the air. + The value p affects how changes in the velocity affect the force. + Taken together, + a function of the form kv^{p} is often called a power law. + The differential equation for the velocity is given by + + m\frac{dv}{dt} = mg - kv^{p} + . +

    + +

    + (Notice that the force from air resistance opposes motion, + and points in the opposite direction as the force from gravity.) This differential equation is separable, + and can be written in the separated form + + \frac{m}{mg - kv^{p}}\,dv = dt + . +

    + +

    + For arbitrary positive p, + the integration is difficult, + making this problem hard to solve analytically. + In the case that p=1, the differential equation becomes linear, + and is easy to solve either using either separation of variables or integrating factor techniques. + We assume p=1, + and proceed with an integrating factor so we can continue practicing the process. + Writing + + \frac{dv}{dt} + \frac{k}{m}v = g + , + we identify the integrating factor + + \mu(t) = e^{\int \frac{k}{m}\,dt} = e^{\frac{k}{m}t} + . +

    + +

    + Then + + \frac{d}{dt}\left(e^{\frac{k}{m}t}v \right) = ge^{\frac{k}{m}t} + , + so + + e^{\frac{k}{m}t}v = \frac{mg}{k}e^{\frac{k}{m}t} + C + , + or + + v = \frac{mg}{k}+ Ce^{-\frac{k}{m}t} + . +

    +     +     + +
    + The velocity functions from Examples (dashed) and (solid) under the assumption that v(0)=0, with g=9.8, m=1, and k=1 + + + The velocity functions from the previous examples. + + +

    + The horizontal axis represents time and the vertical axis represents the velocity, + both axes are drawn from 0 to 10. The graph has assumptions that + v(0)=0 , g =9.8, m=1 and k=1. The velocity of a falling + object without air resistance is a straight line that is very close to the vertical + axis. The velocity function of a falling object with air resistance is along the + previous velocity till v=3 after which it diverges away, gets a bend at + y=10 and then runs parallel to the x axis after x=4. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + xtick={0,2,...,10}, + ymin=-.1,ymax=10.5, + xmin=-.1,xmax=10.5, + xlabel={$t$}, + ylabel={$v$} + ] + + \addplot [firstcurvestyle,domain = 0:10, samples=50] {9.8-9.8*exp(-x)}; + \addplot [secondcurvestyle,domain = 0:10, samples=50] {9.8*x}; + + \end{axis} + \end{tikzpicture} + + +
    + +

    + In the solution above, the exponential term decays as time increases, + causing the velocity to approach the constant value mg/k in the limit as t approaches infinity. + This value is called the terminal velocity. + If we assume a zero initial velocity + (the object is dropped, not thrown from the plane), + the velocities from Examples + and + are given by v = gt and v = \displaystyle \frac{mg}{k}\left(1 - e^{-\frac{k}{m}t}\right), + respectively. + These two functions are shown in , + with g = 9.8, m=1, and k=1. + Notice that the two curves agree well for short times, + but have dramatically different behaviors as t increases. + Part of the art in mathematical modeling is deciding on the level of detail required to answer the question of interest. + If we are only interested in the initial behavior of the falling object, + the simple model in may be sufficient. + If we are interested in the longer term behavior of the object, + the simple model is not sufficient, + and we should consider a more complicated model. +

    +     + + + + Problems + + +

    + Find the general solution to the first order linear differential equation. +

    +     + + +

    + \displaystyle \yp = 2y - 3 +

    +     + +

    + y = \displaystyle \frac{3}{2} + Ce^{2x} +

    +     +     + + +

    + \displaystyle x^2\yp + xy = 1 +

    +     + +

    + y = \displaystyle \frac{\ln \abs{x} + C}{x} +

    +     +     + + +

    + \displaystyle x^2\yp - xy = 1 +

    +     + +

    + y = \displaystyle -\frac{1}{2x} + Cx +

    +     +     + + +

    + \displaystyle x\yp +4y = x^3-x +

    +     + +

    + y = \displaystyle \frac{x^3}{7} - \frac{x}{5} + \frac{C}{x^4} +

    +     +     + + +

    + \displaystyle (\cos^2 x \sin(x))\yp + (\cos^3 x)y = 1 +

    +     + +

    + y = \sec x + C(\csc x) +

    +     +     + + +

    + \displaystyle \frac{\yp}{x} = 1-2y +

    +     + +

    + y = \displaystyle \frac{1}{2} + Ce^{-x^2} +

    +     +     + + +

    + \displaystyle x^3\yp-3x^3y=x^4e^{2x} +

    +     + +

    + y = \displaystyle Ce^{3x}-(x+1)e^{2x} +

    +     +     + + +

    + \displaystyle \yp + y = 5\sin(2x) +

    +     + +

    + y = sin(2x) - 2\cos(2x) + Ce^{-x} +

    +     +     +     + + +

    + Find the particular solution to the initial value problem. +

    +     + + +

    + \displaystyle \yp = y + 2xe^x, y(0) = 2 +

    +     + +

    + y = (x^2+2)e^x +

    +     +     + + +

    + \displaystyle x\yp + 2y = x^2 - x + 1, y(1) = 1 +

    +     + +

    + y = \displaystyle \frac{1}{4}x^2-\frac{1}{3}x+\frac{1}{2}+\frac{7}{12x^2} +

    +     +     + + +

    + \displaystyle x\yp + (x+2)y = x, y(1) = 0 +

    +     + +

    + y = \displaystyle 1 - \frac{2}{x} + \frac{2-e^{1-x}}{x^2} +

    +     +     + + +

    + \displaystyle \yp + 2y = 0, y(0) = 3 +

    +     + +

    + y = \displaystyle 3e^{-2x} +

    +     +     + + +

    + \displaystyle (x+1)\yp + (x+2)y = 2xe^{-x}, y(0) = 1 +

    +     + +

    + y = \displaystyle \frac{x^2+1}{x+1}e^{-x} +

    +     +     + + +

    + \displaystyle (\cos(x))\yp + (\sin(x))y = 1, y(0) = -3 +

    +     + +

    + y = \sin(x) - 3\cos(x) +

    +     +     + + +

    + \displaystyle (x^2-1)\yp + 2y = (x+1)^2, y(0) = 2 +

    +     + +

    + y = \displaystyle \frac{(x-2)(x+1)}{x-1} +

    +     +     + + +

    + \displaystyle x\yp-2y = \frac{x^3}{1+x^2}, y(1) = 0 +

    +     + +

    + y = \displaystyle x^2\left(\arctan x - \frac{\pi}{4}\right) +

    +     +     +     + + +

    + Classify the differential equation as separable, + first order linear, or both, + and solve the initial value problem using an appropriate method. +

    +     + + +

    + \displaystyle \yp = y + yx^2, y(0) = -5 +

    +     + +

    + Both; \displaystyle y = -5e^{x + \frac{1}{3}x^3} +

    +     +     + + +

    + \displaystyle xe^y \yp = x^2\sin(x), y(0) = 0 +

    +     + +

    + separable; \displaystyle e^y = \sin(x) - x\cos(x) + 1 +

    +     +     + + +

    + \displaystyle (x-1)\yp+y = x^2-1, y(0) = 2 +

    +     + +

    + linear; \displaystyle y = \frac{x^3-3x-6}{3(x-1)} +

    +     +     + + +

    + \displaystyle \yp = y^2+y-2, y(0) = 1 +

    +     + +

    + separable; \displaystyle y = 1 +

    +     +     +     + + +

    + Draw a slope field for the differential equation. + Use the slope field to predict the behavior of the solution to the initial value problem for large x values. + Solve the initial value problem, + and verify your prediction. +

    +     + + +

    + \displaystyle \yp = x-y, y(0) = 0 +

    +     + + + + Graph showing slope field for the given differential equation. + + +

    + The x and y axes are uncalibrated, the field lines + in the first quadrant are shown. + On the bottom right the field lines are facing northeast. On the + top left the field lines transition from southeast facing to east + facing moving downwards. + A curve is shown that almost represents a straight line with a + positive slope. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + y domain = -.15:4, + domain = -.15:4, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(x - y)^2))}, v = {(x-y)/(sqrt(1+(x - y)^2))},scale arrows=.2},samples=15]{0}; + \addplot [firstcurvestyle,domain=0:4.1,samples=50] {x-1+exp(-x)}; + + \end{axis} + + \end{tikzpicture} + + +

    + The solution will increase and begin to follow the line y=x-1. +

    +

    + y = x-1 + e^{-x} +

    +     +     + + +

    + \displaystyle (x+1)\yp + y = \frac{1}{x+1}, y(0) = 2 +

    +     + + + + Graph showing slope field for the given differential equation. + + +

    + The x and y axes are uncalibrated, the field lines in + the first quadrant are shown. The lines in the top are southeast facing, + for lower values of y from left to right the field lines are + northeast facing then they transition to east facing. + A downward sloping curve is shown on the field lines. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + view={0}{90}, + y domain = -.15:3, + domain = -.15:4, + ] + + \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(-y/(x+1)+1/(x+1)^2)^2))}, v = {(-y/(x+1)+1/(x+1)^2)/(sqrt(1+(-y/(x+1)+1/(x+1)^2)^2))},scale arrows=.2},samples=15]{0}; + \addplot [firstcurvestyle,domain=0:4.15,samples=50] {(2+ln(x+1))/(x+1)}; + + \end{axis} + + \end{tikzpicture} + + +

    + The solution will decrease and approach y=0. +

    +

    + \displaystyle y = \frac{2 + \ln(x+1)}{x+1} +

    +     +     +     +     +     +
    +
    + Modeling with Differential Equations + +

    + In the first three sections of this chapter, + we focused on the basic ideas behind differential equations and the mechanics of solving certain types of differential equations. + We have only hinted at their practical use. + In this section, + we use differential equations for mathematical modeling, + the process of using equations to describe real world processes. + We explore a few different mathematical models with the goal of gaining an introduction to this large field of applied mathematics. + differential equationmodeling +

    +     + + Models Involving Proportional Change +

    + Some of the simplest differential equation models involve one quantity that changes at a rate proportional to another quantity. + In the introduction to this chapter, + we considered a population that grows at a rate proportional to the current population. + The words in this assumption can be directly translated into a differential equation as shown below. +

    + +
    + Translating words into a differential equation + + + The formula showing the rate of population is proportional to the population. + + +

    + Image shows the equation of derivative of population with respect to time, and it being + proportional to the population. +

    +     + + \begin{tikzpicture} + + \draw (1,1) node {$\displaystyle \frac{dp}{dt} = kp$}; + + \draw [firstcolor] (-1.5,0) node [text width=60pt,align=center] (a) { \centering The rate of change of the population}; + \draw [firstcolor,->] (a) -- (.3,.7); + + \draw [firstcolor] (2,.25) node [text width=60pt,align=center] (b) { \centering the population.}; + \draw [firstcolor,->] (b) -- (1.6,.8); + + \draw [firstcolor] (.5,2) node [text width=32pt,align=center] (c) { \centering is}; + \draw [firstcolor,->] (c) -- (1,1.2); + + \draw [firstcolor] (2,2) node [text width=60pt,align=center] (d) { \centering proportional to}; + \draw [firstcolor,->] (d) -- (1.4,1.2); + + \end{tikzpicture} + + +
    + +

    + There are some key ideas that can be helpful when translating words into a differential equation. + Any time we see something about rates or changes, + we should think about derivatives. + The word + is + usually corresponds to an equal sign in the equation. + The words + proportional to + mean we have a constant multiplied by something. +

    +

    + The differential equation in + is easily solved using separation of variables. + We find + + p = Ce^{kt} + . +

    +

    + Notice that we need values for both C and k before we can use this formula to predict population size. + We require information about the population at two different times in order to fully determine the population model. +

    + + Bacterial Growth + +

    + Suppose a population of e-coli + bacteria grows at a rate proportional to the current population. + If an initial popluation of 200 bacteria has grown to 1600 three hours later, + find a function for the size of the population at time t, + and use it to predict when the population size will reach 10,000. + bacterial growth +

    +     + +

    + We already know that the population at time t is given by + p = Ce^{kt} for some C and k. + The information about the initial size of the population means that p(0)=200. + Thus C=200. + Our knowledge of the population size after three hours allows us to solve for k via the equation + + 1600 = 200e^{3k} + . +

    +

    + Solving this exponential equation yields k =\ln(8)/3 \approx 0.6931. + The popluation at time t is given by + + p = 200 e^{(\ln(8)/3)t} + . +

    +

    + Solving + + 10000 = 200e^{(\ln(8)/3)t} + + yields t =(3\ln(50))/\ln(8) \approx 5.644. + The population is predicted to reach 10,000 bacteria in slightly more than five and a half hours. +

    +     + +     + +

    + Another example of porportional change is + Newton's Law of Cooling. + The laws of thermodynamics state that heat flows from areas of higher temperature to areas of lower temperature. + A simple example is a hot object that cools down when placed in a cool room. + Newton's Law of Cooling is the simple assumption that the temperature of the object changes at a rate proportional to the difference between the temperature of the object and the ambient temperature of the room. + If T is the temperature of the object and A is the constant ambient temperature, + Newton's Law of Cooling can be expressed as the differential equation + + \frac{dT}{dt} = k(A - T) + . + Newton's Law of Cooling +

    +

    + This differential equation is both linear and separable. + The separated form is + + \frac{1}{A-T}\,dT = k\,dt + . +

    +

    + Then an implicit definition of the temperature is given by + + -\ln\abs{A-T} = kt + C + . +

    +

    + If we solve for T, we find the explicit temperature + + T = A-Ce^{-kt} + . +

    +

    + Though we didn't show the steps, + the explicit solution involves the typical process of renaming the constant \pm e^{-C} as C, + and allowing C to be positive, negative, + or zero to account for both cases of the absolution value and to catch the constant solution T=A. + Notice that the temperature of the object approaches the ambient temperature in the limit as t\to\infty. +

    + + + + + Hot Coffee + +

    + A freshly brewed cup of coffee is set on the counter and has a temperature of 200^\circ Fahrenheit. + After 3 minutes, it has cooled to 190^\circ, + but is still too hot to drink. + If the room is 72^\circ and the coffee cools according to Newton's Law of Cooling, + how long will the impatient coffee drinker have to wait until the coffee has cooled to 165^\circ? +

    +     + +

    + Since we have already solved the differential equation for Newton's Law of Cooling, + we can immediately use the function + + T = A - Ce^{-kt} + . +

    +

    + Since the room is 72^\circ, we know A = 72. + The initial temperature is 200^\circ, + which means C = -128. + At this point, we have + + T = 72 + 128e^{-kt} + +

    +

    + The information about the coffee cooling to 190^\circ in 3 minutes leads to the equation + + 190 = 72 + 128e^{-3k} + . +

    +

    + Solving the exponential equation for k, we have + + k = -\frac{1}{3}\ln \left(\frac{59}{64}\right) \approx 0.0271 + . +

    +

    + Finally, we finish the problem by solving the exponential equation + + 165 = 72 + 128e^{\frac{1}{3}\ln \left(\frac{59}{64}\right)t} + . +

    +

    + The coffee drinker must wait \displaystyle t = \frac{3 \ln \left(\frac{93}{128}\right)}{\ln \left(\frac{59}{64}\right)} \approx 11.78 minutes. +

    +     + +     +

    + We finish our discussion of models of proportional change by exploring three different models of disease spread through a population. + In all of the models, + we let y denote the proportion of the population that is sick + (0 \leq y \leq 1). + We assume a proportion of 0.05 is initially sick and that a proportion of 0.1 is sick 1 week later. +

    + + Disease Spread 1 + +

    + Suppose a disease spreads through a population at a rate proportional to the number of individuals who are sick. + If 5% of the population is sick initially and 10% of the population is sick one week later, + find a formula for the proportion of the popoulation that is sick at time t. +

    +     + +

    + The assumption here seems to have some merit because it matches our intuition that a disease should spread more rapidly when more individuals are sick. + The differential equation is simply + + \frac{dy}{dt} = ky + , + with solution + + y = Ce^{kt} + . +

    +

    + The conditions y(0)=0.05 and y(1) = 0.1 lead to + C = 0.05 a and k = \ln(2), so the function is + + y = 0.05e^{(\ln(2)t} + . +

    +

    + We should point out a glaring problem with this model. + The variable y is a proportion and should take on values between 0 and 1, but the function y = 0.05e^{2t} grows without bound. + After t \approx 4.32 weeks, + y exceeds 1, and the model ceases to make physical sense. +

    +     +     + + + + Disease Spread 2 + +

    + Suppose a disease spreads through a population at a rate proportional to the number of individuals who are not sick. + If 5% of the population is sick initially and 10% of the population is sick one week later, + find a formula for the proportion of the popoulation that is sick at time t. +

    +     + +

    + The intuition behind the assumption here is that a disease can only spread if there are individuals who are susceptible to the infection. + As fewer and fewer people are able to be infected, + the disease spread should slow down. + Since y is proportion of the population that is sick, + 1-y is the proportion who are not sick, + and the differential equation is + + \frac{dy}{dt} = k(1-y) + . +

    +

    + Though the context is quite different, + the differential equation is identical to the differential equation for Newton's Law of Cooling, + with A=1. + The solution is + + y = 1 - Ce^{-kt} + . +

    +

    + The conditions y(0)=0.05 and + y(1) = 0.1 yield C = 0.95 and + k = -\ln\left(\frac{18}{19}\right) \approx 0.0541, + so the final function is + + y = 1-.95e^{\ln\left(\frac{18}{19}\right)t} + . +

    +

    + Notice that this function approaches y=1 in the limit as t \to \infty, + and does not suffer from the non-physical behavior described in . +

    +     +     + +

    + In , + we assumed disease spread depends on the number of infected individuals. + In , + we assumed disease spread depends on the number of susceptible individuals who are able to become infected. + In reality, we would expect many diseases to require the interaction of both infected and susceptible individuals in order to spread. + One of the simplest ways to model this required interaction is to assume disease spread depends on the product of the proportions of infected and uninfected individuals. + This assumption + (regularly seen in the context of chemical reactions) + is often called the law of mass action. +

    + + + + Disease Spread 3 + +

    + Suppose a disease spreads through a population at a rate proportional to the product of the number of infected and uninfected individuals. + If 5% of the population is sick initially and 10% of the population is sick one week later, + find a formula for the proportion of the population that is sick at time t. + differential equationlogistic +

    +     + +

    + The differential equation is + + \frac{dy}{dt} = ky(1-y) + . +

    +

    + This is exactly the logistic equation with M = 1. + We solved this differential equation in , + and found + + y = \frac{1}{1 + be^{-kt}} + . +

    +

    + The conditions y(0)=0.05 and + y(1) = 0.1 yield b = 19 and k = -\ln\left(\frac{9}{19}\right) \approx 0.7472. + The final function is + + y = \frac{1}{1+19e^{\ln\left(\frac{9}{19}\right)t}} + . +

    +

    + Based on the three different assumptions about the rate of disease spread explored in the last three examples, + we now have three different functions giving the proportion of a population that is sick at time t. + Each of the three functions meets the conditions + y(0)=0.05 and y(1) = 0.1. + The three functions are shown in . +

    + +

    + Notice that the logistic function mimics specific parts of the functions from Examples + and . + We see in that the logistic and exponential functions are virtually indistinguishable for small t values. + When there are few infected individuals and lots of susceptible individuals, + the spread of a disease is largely determined by the number of sick people. + The logistic curve captures this feature, and is + almost exponential + early on. +

    + +

    + In , + we see that the logistic curve leaves the exponential curve from + and approaches the curve from . + This result implies that when most of the population is sick, + the spread of the disease is largely dependent on the number of susceptible individuals. + Though there are much more sophisticated mathematical models describing the spread of infections, + we could argue that the logistic model presented in this example is the + best + of the three. +

    + +
    + Plots of the functions from (dotted), + (dashed), + and (solid) + +
    + + + + Plots of functions from the last three examples for small values of t. + + +

    + The y axis is drawn from 0 to 0.1 and the t axis is drawn from + 0 to 1.2. The first function is a straight line that is positively inclined, + it starts from the point (0, 0.05) and moves away from the t axis with increasing + values of t. The exponential and logistic functions are indistinguishable and they + also begin from the point (0,0.05), they also have a positive slope but they have a dip + and separates from the line before merging again at point (1,1) after which they are + above the line. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ytick={0,0.05,.1}, + yticklabels = {0,0.05,0.1}, + ymin=-.01,ymax=.11, + xmin=-.1,xmax=1.25, + xlabel={$t$} + ] + + \addplot [firstcurvestyle,domain = 0:10, samples=50] {1/(1+19*exp(ln(9/19)*x))}; + \addplot [firstcurvestyle,dashed,domain = 0:10, samples=50] {1-.95*exp(ln(18/19)*x)}; + \addplot [firstcurvestyle,dotted,domain = 0:2, samples=50] {.05*exp(ln(2)*x)}; + + \fill[black,draw=black](axis cs:0,.05) circle (2.4pt); + \fill[black,draw=black](axis cs:1,.1) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + +
    +
    + + + + Plots of functions from the last three examples for large values of t. + + +

    + The y axis is drawn from 0 to 1 and the x axis is drawn from + 0 to 50. The three functions in the previous examples are shown. The three + curves are aligned till about x=2 after which they diverge. The exponential curve + moves vertically, very close to the y axis. The logistic curve diverges to the left + from the exponential curve at about point (3,0.3), it acquires a sharp bend at point + (10,1) after which it runs parallel to the x axis. The third curve is below the + other two and starts close to the origin and curves up to reach point (50, 0.9). +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.01,ymax=1.1, + xmin=-.1,xmax=50.1, + xlabel={$t$} + ] + + \addplot [firstcurvestyle,domain = 0:50, samples=101] {1/(1+19*exp(ln(9/19)*x))}; + \addplot [firstcurvestyle,dashed,domain = 0:50, samples=50] {1-.95*exp(ln(18/19)*x)}; + \addplot [firstcurvestyle,dotted,domain = 0:5, samples=50] {.05*exp(ln(2)*x)}; + + \fill[black,draw=black](axis cs:0,.05) circle (2.4pt); + \fill[black,draw=black](axis cs:1,.1) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + +
    +     +
    + +     + +     +     + + Rate-in Rate-out Problems +

    + One of the classic ways to build a mathematical model involves tracking the way the amount of something can change. + We sometimes say these models are based on + conservation laws. + Consider a box with some amount of a specific type of material inside. + (Some type of chemical, for example.) + The amount of material of the specific type in the box can only change in four ways; + we can add more to the box, we can remove some from the box, + some of the material can change into material of a different type, + or some other type of material can turn into the type we're tracking. + In the examples that follow, we assume material doesn't change type, + so we only need to keep track of material coming into the box and material leaving the box. + To derive a differential equation, we track rates: + + \text{ rate of change of some quantity } = \text{ rate in } - \text{ rate out } + . +

    + + +

    + Though we stick to relatively simple examples, + this basic idea can be used to derive some very important differential equations in mathematics and physics. +

    +

    + The examples to follow involve tracking the amount of a chemical in solution. + We assume liquid containing some chemical flows into a container at some rate. + That liquid mixes instantaneously with the liquid already in the container. + Then the liquid from the container flows out at some + (potentially different) + rate. +

    + + + + + Equal Flow Rates + +

    + Suppose a 10 liter tank has 5 liters of salt solution in it. + The initial concentration of the salt solution is 1 gram per liter. + A salt solution with concentration + 3 flows into the tank at a rate of + 2. + Suppose the salt solution mixes instantaneously with the solution already in the tank, + and that the mixed solution from the tank flows out at a rate of + 2. + Find a function that gives the amount of salt in the tank at time t. +

    +     + +

    + We use the rate in - rate out setup described above. + The quantity here is the amount + (in grams) + of salt in the tank at time t. + Let y denote the amount of salt. + In words, the differential equation is given by + + \frac{dy}{dt} = \text{ rate in } - \text{ rate out } + . +

    +

    + Thinking in terms of units can help fill in the details of the differential equation. + Since y has units of grams, + the left hand side of the equation has units g/min. + Both terms on the right hand side must have these same units. + Notice that the product of a concentration + (with units g/L) + and a flow rate + (with units L/min) + results in a quantity with units g/min. + Both terms on the right hand side of the equation will include a concentration multiplied by a flow rate. +

    +

    + For the rate in, + we multiply the inflow concentration by the rate that fluid is flowing into the bucket. + This is \displaystyle \left(3 \frac{\text{g} }{\text{L} }\right)\left(2 \frac{\text{L} }{\text{ min } }\right) = 6 g/min. +

    +

    + The rate out is more complicated. + The flow rate is still 2, + meaning that the overall volume of the fluid in the bucket is the constant + 5. + The salt concentration in the bucket is not constant though, + meaning that the outflow concentration is not constant. + In particular, the outflow concentration is not + the constant 1. + This is simply the initial concentration. + To find the concentration at any time, + we need the amount of salt in the bucket at that time and the volume of liquid in the bucket at that time. + The volume of liquid is the constant 5, + and the amount of salt is given by the dependent variable y. + Thus, the outflow concentration is + \displaystyle \frac{y}{5} g/L, yielding a rate out given by + + \left(\frac{y}{5}\frac{\text{ g } }{\text{ L } }\right)\left(2 \frac{\text{ L } }{\text{ min } }\right) = \frac{2y}{5}\text{ g/min } + . +

    +

    + The differential equation we wish to solve is given by + + \frac{dy}{dt} = 6 - \frac{2y}{5} + . +

    +

    + To furnish an initial condition, + we must convert the initial salt concentration into an initial amount of salt. + This is \left(1\displaystyle \frac{\text{ g } }{\text{ L } }\right)(5 \text{ L } ) = 5 g, so + y(0) = 5 is our initial condition. +

    +

    + Our differential equation is both separable and linear. + We solve using separation of variables. + The separated form of the differential equation is + + \frac{5}{30 - 2y}\,dy = dt + . +

    +

    + Integration yields the implicit solution + + -\frac{5}{2}\ln\abs{30 - 2y} = t+C + . +

    +

    + Solving for y + (and redefining the arbitrary constant C as necessary) + yields the explicit solution + + y = 15 + Ce^{-\frac{2}{5}t} + . +

    +

    + The initial condition y(0) = 5 means that C = -10 so that + + y = 15 - 10e^{-\frac{2}{5}t} + + is the particular solution to our initial value problem. +

    +

    + This function is plotted in . + Notice that in the limit as t\to\infty, + y approaches 15. + This corresponds to a bucket concentration of + 15/5 = 3 g/L. It should not be surprising that salt concentration inside the tank will move to match the inflow salt concentration. +

    +
    + Salt concentration at time t, from + + + Graph of salt concentration at time t used in this example. + + +

    + The y axis is drawn from 0 to 15 and the x axis is drawn from + 0 to 10. The function starts at point (0, 5) and curves up with a great + positive slope then the slope decreases. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=15.5, + xmin=-.1,xmax=10.3, + xlabel={$t$} + ] + + \addplot [firstcurvestyle,domain = 0:10, samples=50] {15-10*exp(-2/5*x)}; + + \end{axis} + + \end{tikzpicture} + + +
    +     + +     + + + Unequal Flow Rates + +

    + Suppose the setup is identical to the setup in + except that now liquid flows out of the bucket at a rate of 1 L/min. + Find a function that gives the amount of salt in the bucket at time t. + What is the salt concentration when the solution ceases to be valid? +

    +     + +

    + Because the inflow and outflow rates no longer match, + the volume of liquid in the bucket is not the constant 5 L. In general, + we can find the volume of liquid via the equation + + \text{ volume } = \text{ initial volume } + \text{ (inflow rate - outflow rate) } t + . +

    +

    + In this example, + the volume at time t is 5 + t liters. + Because the total volume of the bucket is only 10 L, it follows that our solution will only be valid for 0 \leq t \leq 5. + At that point it is no longer possible to have liquid flow into a the bucket at a rate of 2 L/min and out of the bucket at a rate of 1 L/min. +

    +

    + To update the differential equation, + we must modify the rate out. + Since the volume is 5 + t, + the concentration at time t is given by + \frac{y}{5+t} g/L. Thus for rate out, + we must use \left( \frac{y}{5+t}\right)(1) g/min. + The initial value problem is + + \frac{dy}{dt} = 6 - \frac{y}{5+t}, \text{ with } y(0)=5 + . +

    +

    + Unlike , + where we had equal flow rates, + this differential equation is no longer separable. + We must proceed with an integrating factor. + Writing the differential equation in the form + + \frac{dy}{dt} + \frac{1}{5+t}y = 6 + , + we identify the integrating factor + + \mu(t) = e^{\int \frac{1}{5+t}\,dt} = e^{\ln(5+t)} = 5+t + . +

    +

    + Then + + \frac{d}{dt}\big((5+t)y\big) = 6(5+t) + , + yielding the implicit solution + + (5+t)y = 30t + 3t^2 + C + . +

    +

    + The initial condition y(0) = 5 implies C = 25, + so the explicit solution to our initial value problem is given by + + y = \frac{3t^2 + 30t + 25}{5+t} + . +

    +

    + This solution ceases to be valid at t=5. + At that time, there are 25 g of salt in the tank. + The volume of liquid is 10 L, resulting in a salt concentration of 2.5 g/L. +

    +     + +     +

    + Differential equations are powerful tools that can be used to help describe the world around us. + Though relatively simple in concept, + the ideas of proportional change and matching rates can serve as building blocks in the development of more sophisticated mathematical models. + As we saw in this section, + some simple mathematical models can be solved analytically using the techniques developed in this chapter. + Most sophisticated mathematical models don't allow for analytic solutions. + Even so, there are an array of graphical and numerical techniques that can be used to analyze the model to make predictions and infer information about real world phenomena. +

    +     + + + + Problems + + +

    + Use the tools in the section to answer the questions presented. +

    +     + + +

    + Suppose the rate of change of y with respect to x is proportional to 10 - y. + Write down and solve a differential equation for y. +

    +     + +

    + y = 10 + Ce^{-kx} +

    +     +     + + +

    + A rumor is spreading through a middle school with 250 students. + Suppose the rumor spreads at a rate proportional to the number of students who haven't heard the rumor yet. + If 1 person starts the rumor, + and 75 students have heard the rumor 3 days later, + how many days will it take until 80% of the students in the school have heard the rumor? +

    +     + +

    + 13.66 days +

    +     +     + + +

    + A rumor is spreading through a middle school with 250 students. + Suppose the rumor spreads at a rate proportional to the product of number of students who have heard the rumor and the number who haven't heard the rumor. + If 1 person starts the rumor, + and 75 students have heard the rumor 3 days later, + how many days will it take until 80% of the students in the school have heard the rumor? +

    +     + +

    + 4.43 days +

    +     +     + + +

    + A feature of radioactive decay is that the amount of a radioactive substance decreases at a rate proportional to the current amount of the substance. + The half life + half life + of a substance is the amount of time it takes for half of a given amount of substance to decay. + The half life of carbon-14 is approximately 5730 years. + If an ancient object has a carbon-14 amount that is 20% of the original amount, + how old is the object? +

    +     + +

    + 13,304.65 years old +

    +     +     + + +

    + Consider a chemical reaction where molecules of type A combine with molecules of type B to form molecules of type C. Suppose one molecule of type A combines with one molecule of type B to form one molecule of type C, and that type C is produced at a rate proportional the product of the remaining number of molecules of types A and B. Let x denote moles of molecules of type C. + Find a function giving the number of moles of type C at time t if there are originally a moles of type A, b moles of type B, and zero moles of type C. +

    +     + +

    + x = \begin{cases}\displaystyle\frac{ab(1 - e^{(a-b)kt})}{b-ae^{(a-b)kt}} \amp \text{ if } a \neq b\\ \displaystyle \frac{a^2kt}{1+akt} \amp \text{ if } a = b \end{cases} +

    +     +     + + +

    + Suppose an object with a temperature of 100^\circ is introduced into a room with an ambient temperature of 70^\circ. + Suppose the temperature of the object changes at a rate proportional to the difference between the temperature of the object and the temperature of the room (Newton's Law of Cooling). + If the object has cooled to 92^\circ in 10 minutes, + how long until the object has cooled to 84^\circ? +

    +     + +

    + 24.57 minutes +

    +     +     + + +

    + Suppose an object with a temperature of 100^\circ is introduced into a room with an ambient temperature given by 60 + 20e^{-\frac{1}{4}t} degrees. + Suppose the temperature of the object changes at a rate proportional to the difference between the temperature of the object and the temperature of the room (Newton's Law of Cooling). + If the object is 80^\circ after 20 minutes, + find a formula giving the temperature of the object at time t. + (Note: This problem requires a numerical technique to solve for the unknown constants.) +

    +     + +

    + \displaystyle y = 60 - 3.69858e^{-\frac{1}{4}t} + 43.69858e^{-0.0390169 t} +

    +     +     + + +

    + A tank contains 5 gallons of salt solution with concentration 0.5 g/gal. + Pure water flows into the tank at a rate of 1 gallon per minute. + Salt solution flows out of the tank at a rate of 1 gallon per minute. + (Assume instantaneous mixing.) + Find the concentration of the salt solution at 10 minutes. +

    +     + +

    + 0.06767 g/gal +

    +     +     + + +

    + Dead leaves accumulate on the ground at a rate of 4 grams per square centimeter per year. + The dead leaves on the ground decompose at a rate of 50% per year. + Find a formula giving grams per square centimeter on the ground if there are no leaves on the ground at time t=0. +

    +     + +

    + y = 8(1-e^{-\frac{1}{2}t}) g/cm^2 +

    +     +     + + +

    + A pond initially contains 10 million gallons of fresh water. + Water containing an undesirable chemical flows into the pond at a rate of 5 million gallons per year, + and fluid from the pond flows out at the same rate. + (Assume instantaneous mixing.) + If the concentration + (in grams per million gallons) + of the incoming chemical varies periodically according to the expression 2 + \sin(2t), + find a formula giving the amount of chemical in the pond at time t. +

    +     + +

    + y = \displaystyle 20 - \frac{10}{17}\left(4\cos(2t)- \sin(2t)\right) - \frac{300}{17}e^{-\frac{1}{2}t} g +

    +     +     + + +

    + A large tank contains 1 gallon of a salt solution with concentration 2 g/gal. + A salt solution with concentration 1 g/gal flows into the tank at a rate of 4 gal/min. + Salt solution flows out of the tank at a rate of 3 gal/min. + (Assume instantaneous mixing.) + Find the amount of salt in the tank at 10 minutes. +

    +     + +

    + 11.00075 g +

    +     +     + + +

    + A stream flows into a pond containing 2 million gallons of fresh water at a rate of 1 million gallons per day. + The stream flows out of the first pond and into a second pond containing 3 million gallons of fresh water. + The stream then flows out of the second pond. + Suppose the inflow and outflow rates are the same so that both ponds maintain their volumes. + A factory upstream of the first pond starts polluting the stream. + Directly below the factory, + pollutant has a concentration of 55 grams per million gallons, + and this concentration starts to flow into the first pond. + Find the concentration of pollutant in the first and second ponds at 5 days. +

    +     + +

    + pond 1: 50.4853 grams per million gallons +

    +

    + pond 2: 32.8649 grams per million gallons +

    +     +     +     +     +     +
    +     + + + Sequences and Series + +

    + This chapter introduces sequences and series, + important mathematical constructions that are useful when solving a large variety of mathematical problems. + The content of this chapter is considerably different from the content of the chapters before it. + While the material we learn here definitely falls under the scope of calculus, + we will make very little use of derivatives or integrals. + Limits are extremely important, though, + especially limits that involve infinity. +

    + +

    + One of the problems addressed by this chapter is this: + suppose we know information about a function and its derivatives at a point, + such as f(1) = 3, \fp(1) = 1, + \fp'(1) = -2, \fp''(1) = 7, and so on. + What can I say about f(x) itself? + Is there any reasonable approximation of the value of f(2)? + The topic of Taylor Series addresses this problem, + and allows us to make excellent approximations of functions when limited knowledge of the function is available. +

    +     + +
    + Sequences +

    + We commonly refer to a set of events that occur one after the other as a + sequence of events. + In mathematics, we use the word sequence + to refer to an ordered set of numbers, + , a set of numbers that occur one after the other. +

    + + + +

    + For instance, + the numbers 2, 4, 6, 8, , form a sequence. + The order is important; the first number is 2, the second is 4, etc. + It seems natural to seek a formula that describes a given sequence, + and often this can be done. + For instance, + the sequence above could be described by the function a(n) = 2n, + for the values of n = 1, 2, \ldots To find the 10th term in the sequence, + we would compute a(10). + This leads us to the following, + formal definition of a sequence. +

    + + + Sequence + +

    + A sequence is a function a(n) whose domain is \mathbb{N}. + The range of a sequence is the set of all distinct values of a(n). + sequencesdefinition +

    + +

    + The terms of a sequence are the values a(1), + a(2), + , which are usually denoted with subscripts as a_1, a_2, + . +

    + +

    + A sequence a(n) is often denoted as \{a_n\}. +

    +     +     + + + + + +

    + A factorial refers to the product of a descending sequence of natural numbers. + For example, the expression 4! + (read as 4 factorial) + refers to the number 4\cdot 3\cdot2\cdot1 = 24. + + factorial + aa@"! +

    + +

    + In general, n! = n\cdot (n-1)\cdot(n-2)\cdots 2\cdot1, + where n is a natural number. +

    + +

    + We define 0! = 1. + While this does not immediately make sense, + it makes many mathematical formulas work properly. +

    +     +     + + + + + Listing terms of a sequence + +

    + List the first four terms of the following sequences. +

    + +

    +

      +
    1. \{a_n\} = \left\{\frac{3^n}{n!}\right\}
      2. + +
    2. \{a_n\} = \{4+(-1)^n\}
      3. + +
    3. + \{a_n\} = \left\{\frac{(-1)^{n(n+1)/2}}{n^2}\right\} +
      4. +
    +

    +     + +

    +

      +
    1. +

      + \ds a_1=\frac{3^1}{1!} = 3;\quad \ds a_2= \frac{3^2}{2!} = \frac92;\quad + \ds a_3 = \frac{3^3}{3!} = \frac92;\quad \ds a_4 = \frac{3^4}{4!} = \frac{27}8\quad +

      + +

      + We can plot the terms of a sequence with a scatter plot. + The horizontal axis is used for the values of n, + and the values of the terms are plotted on the vertical axis. + To visualize this sequence, see . +

      +
      + Plotting the sequence in of + + + Plot of the first four terms of the sequence in part 1 of this example. + +

      + A pair of coordinate axes are given, with the horizontal axis labeled n (rather than x), + and the origin at the bottom-left of the image. + The first four terms of the sequence 3^n/n! are plotted, + at coordinates (1,3),(2,4.5),(3,4.5), and (4,3.375). + The four points lie in an inverted U shape. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4,5}, + ymin=-.1,ymax=5.5, + xmin=-.1,xmax=4.5, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt}] coordinates {(1,3)(2,4.5)(3,4.5)(4,3.375)}; + + \draw (axis cs:2.5,1.5) node { $\ds a_n = \frac{3^n}{n!}$}; + + \end{axis} + \end{tikzpicture} + + + + +
      +
      2. + +
    2. +

      + a_1= 4+(-1)^1 = 3;\quad a_2 = 4+(-1)^2 = 5;\quad + + a_3=4+(-1)^3 = 3;\quad a_4 = 4+(-1)^4 = 5\quad. +

      + +

      + Note that the range of this sequence is finite, + consisting of only the values 3 and 5. + This sequence is plotted in . +

      +
      + Plotting the sequence in of + + + A plot of the first four terms of the sequence in part 2 of this example. + +

      + A pair of coordinate axes are shown, with the horizontal axis labeled n, + and the origin at the bottom-left of the image. + The first four terms of the sequence 4+(-1)^n are plotted, + illustrating the oscillatory nature of this sequence: + the four points make a sort of zig-zag configuration, with the y value + going back and forth between 3 and 5. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4,5}, + ymin=-.1,ymax=5.5, + xmin=-.1,xmax=4.5, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:4] {4+(-1)^x}; + + \draw (axis cs:2,1) node { $\ds a_n = 4+(-1)^n$}; + + \end{axis} + \end{tikzpicture} + + + + +
      +
      3. + +
    3. +

      + \ds a_1= \frac{(-1)^{1(2)/2}}{1^2} = -1;\quad \ds a_2 = \frac{(-1)^{2(3)/2}}{2^2} =-\frac14 ;\quad + + \ds a_3 = \frac{(-1)^{3(4)/2}}{3^2} = \frac19;\quad \ds a_4 = \frac{(-1)^{4(5)/2}}{4^2} = \frac1{16};\quad; + + \ds a_5 = \frac{(-1)^{5(6)/2}}{5^2}=-\frac1{25}\quad. +

      + +

      + We gave one extra term to begin to show the pattern of signs is -, + -, +, +, + -, -, \ldots, + due to the fact that the exponent of -1 is a special quadratic. + This sequence is plotted in . +

      +
      + Plotting the sequence in of + + + A plot of the first four terms of the sequence in part 3 of this example. + +

      + Another set of coordinate axes are shown; in this case, the horizontal axis (labeled n) + is about two thirds of the way from the bottom of the image. + The first five terms of the sequence (-1)^{n(n+1)/2}/n^2 are shown. + Since the triangular numbers n(n+1)/2 follow an odd, odd, even, even pattern, + the points plotted for the first two terms have negative y value, + the next two have positive y value, and the fifth term is again negative. + As n increases, the points get closer and closer to the n axis. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4,5}, + ytick={-1}, + extra y ticks={.5,.25}, + extra y tick labels={$1/2$,$1/4$}, + ymin=-1.1,ymax=0.6, + xmin=-.1,xmax=5.5, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt}] coordinates {(1,-1)(2,-.25)(3,.111)(4,0.0625)(5,-.04)}; + + \draw (axis cs:3,-.5) node { $\ds a_n = \frac{(-1)^{n(n+1)/2}}{n^2}$}; + + \end{axis} + \end{tikzpicture} + + + + +
      + +
      4. +
    +

    +     + +     + + + Determining a formula for a sequence + +

    + Find the nth term of the following sequences, + , find a function that describes each of the given sequences. +

    + +

    +

      +
    1. +

      + \{a_n\}=\{2, 5, 8, 11, 14, \ldots\} +

      +
      2. + +
    2. +

      + \{b_n\}=\{2,-5, 10, -17, 26, -37,\ldots\} +

      +
      3. + +
    3. +

      + \{c_n\}=\{1, 1, 2, 6, 24, 120, 720, \ldots\} +

      +
      4. + +
    4. +

      + \{d_n\}=\left\{\frac52, \frac52, \frac{15}8, \frac54,\frac{25}{32},\ldots\right\} +

      +
      5. +
    +

    +     + +

    + We should first note that there is never exactly one function that describes a finite set of numbers as a sequence. + There are many sequences that start with 2, then 5, as our first example does. + We are looking for a simple formula that describes the terms given, + knowing there is possibly more than one answer. +

    + +

    +

      +
    1. +

      + Note how each term is 3 more than the previous one. + This implies a linear function would be appropriate: + a(n) = a_n = 3n + b for some appropriate value of b. + If we were to think in terms of ordered pairs, + they would be of the form (n,a(n)). + So one such ordered pair would be (1,2). + As we want a_1=2, we set b=-1. + Thus a_n = 3n-1. +

      +
      2. + +
    2. +

      + First notice how the sign changes from term to term. + This is most commonly accomplished by multiplying the terms by either (-1)^n or (-1)^{n+1}. + Using (-1)^n multiplies the odd indexed terms by (-1). + Thus the first term would be negative and the second term would be positive. + Multiplying by (-1)^{n+1} multiplies the even indexed terms by (-1). + Thus the second term would be negative and the first term would be positive. + As this sequence has negative even indexed terms, + we will multiply by (-1)^{n+1}. +

      + +

      + After this, we might feel a bit stuck as to how to proceed. + At this point, we are just looking for a pattern of some sort: + what do the numbers 2, 5, 10, 17, etc., have in common? + There are many correct answers, + but the one that we'll use here is that each is one more than a perfect square. + That is, 2=1^2+1, 5=2^2+1, 10=3^2+1, etc. + Thus our formula is b_n= (-1)^{n+1}(n^2+1). +

      +
      3. + +
    3. +

      + One who is familiar with the factorial function will readily recognize these numbers. + They are 0!, 1!, + 2!, 3!, etc. + Since our sequences start with n=1, + we cannot write c_n = n!, + for this misses the 0! term. + Instead, we shift by 1, and write c_n = (n-1)!. +

      +
      4. + +
    4. +

      + This one may appear difficult, + especially as the first two terms are the same, + but a little sleuthing will help. + Notice how the terms in the numerator are always multiples of 5, and the terms in the denominator are always powers of 2. + Does something as simple as d_n = \frac{5n}{2^n} work? +

      + +

      + When n=1, we see that we indeed get 5/2 as desired. + When n=2, we get 10/4 = 5/2. + Further checking shows that this formula indeed matches the other terms of the sequence. +

      +
      5. +
    +

    +     + +     + +

    + A common mathematical endeavor is to create a new mathematical object + (for instance, a sequence) + and then apply previously known mathematics to the new object. + We do so here. + The fundamental concept of calculus is the limit, + so we will investigate what it means to find the limit of a sequence. +

    + + + Limit of a Sequence, Convergent, Divergent + +

    + Let \{a_n\} be a sequence and let L be a real number. + Given any \varepsilon \gt 0, + if an N can be found such that + \abs{a_n-L}\lt \varepsilon for all n\gt N, + then we say the limit of \{a_n\}, + as n approaches infinity, + is L, denoted + + \lim_{n\to\infty}a_n = L + . +

    + +

    + If \lim\limits_{n\to\infty} a_n exists, + we say the sequence converges; + otherwise, the sequence diverges. + limitof sequence + sequenceslimit + convergenceof sequence + divergenceof sequence + sequencesconvergent + sequencesdivergent +

    +     +     + +

    + This definition states, informally, + that if the limit of a sequence is L, + then if you go far enough out along the sequence, + all subsequent terms will be really close to L. + Of course, the terms far enough + and really close are subjective terms, + but hopefully the intent is clear. +

    + + + +

    + This definition is reminiscent of the + \varepsilon-\delta proofs of . + In that chapter we developed other tools to evaluate limits apart from the formal definition; + we do so here as well. +

    + + + Limit of Infinity, Divergent Sequence + +

    + Let \{a_n\} be a sequence. + We say \lim\limits_{n\to\infty} a_n=\infty if for all M \gt 0, + there exists a number N such that if n\ge N, + then a_n \gt M. + In this case, we say the sequence diverges to \infty. +

    +     +     + +

    + This definition states, informally, + that if the limit of a_n is \infty, + then you can guarantee that the terms of a_n will get arbitrarily large + (larger than any value of M that you think of), + by going out far enough in the sequence. +

    + + + Limit of a Sequence + +

    + Let \{a_n\} be a sequence, + let L be a real number, + and let f(x) be a function whose domain contains the positive real numbers where + f(n) = a_n for all n in \mathbb{N}. +

    + +

    +

      +
    1. +

      + If \lim\limits_{x\to\infty} f(x) = L, + then \lim\limits_{n\to\infty} a_n = L. +

      +
      2. + +
    2. +

      + If \lim\limits_{x\to\infty} f(x) = \infty, + then \lim\limits_{n\to\infty} a_n = \infty. +

      +
      3. +
    +

    +     +     + +

    + allows us, in certain cases, + to apply the tools developed in to limits of sequences. + Note two things not stated by the theorem: +

    + +

    +

      +
    1. +

      + If \lim\limits_{x\to\infty}f(x) does not exist, + we cannot conclude that \lim\limits_{n\to\infty} a_n does not exist. + It may, or may not, exist. + For instance, + we can define a sequence \{a_n\} = \{\cos(2\pi n)\}. + Let f(x) = \cos(2\pi x). + Since the cosine function oscillates over the real numbers, + the limit \lim\limits_{x\to\infty}f(x) does not exist. + + However, for every positive integer n, \cos(2\pi n) = 1, so \lim\limits_{n\to\infty} a_n = 1. +

      +
      2. + +
    2. +

      + If we cannot find a function f(x) whose domain contains the positive real numbers where + f(n) = a_n for all n in \mathbb{N}, + we cannot conclude \lim\limits_{n\to\infty} a_n does not exist. + It may, or may not, exist. +

      +
      3. +
    +

    + + + Determining convergence/divergence of a sequence + +

    + Determine the convergence or divergence of the following sequences. +

    + +

    +

      +
    1. + \ds\{a_n\} = \left\{\frac{3n^2-2n+1}{n^2-1000}\right\} +
      2. + +
    2. \{b_n\} = \{\cos(n) \}
      3. + +
    3. + \ds\{c_n\} = \left\{\frac{(-1)^n}{n}\right\} +
      4. +
    +

    +     + +

    +

      +
    1. +

      + Using , + we can state that \lim\limits_{x\to\infty} \frac{3x^2-2x+1}{x^2-1000} = 3. + (We could have also directly applied L'Hospital's Rule.) + Thus the sequence \{a_n\} converges, and its limit is 3. + A scatter plot of every 5 values of a_n is given in . + The values of a_n vary widely near n=30, + ranging from about -73 to 125, + but as n grows, the values approach 3. +

      +
      + Scatter plot for the sequence in of + + + A scatter plot showing a representative sample of points from the first sequence in this example. + +

      + A pair of coordinate axes are shown, with the horizontal axis (labeled n) in the center of the image. + The range for n is from 0 to 100, and every fourth point in the sequence has been plotted. + The scatter plot for the sequence a_n = \frac{3n^2-2n+1}{n^2-1000} follows the graph of + f(x)=\frac{3x^2-3x+1}{x^2-1000}. + The graph of this function has a vertical asymptote at x=\sqrt{1000}\approx 31.6. + Since this is not an integer value, a_n is defined for each n, + but we can see that the largest negative value of a_n occurs when n=31 + (to the left of the asymptote), and the largest positive value of a_n occurs when n=32 + (to the right of the asymptote). As n gets large, the y coordinate of each point gets closer to 3. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={20,40,60,80,100}, + ymin=-11,ymax=11, + xmin=-.1,xmax=110, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={1.75pt},domain=1:100,samples=21] {(3*x^2 - 2*x + 1)/(x^2 - 1000)}; + + \draw (axis cs:60,-8) node { $\ds a_n = \frac{3n^2 - 2n + 1}{n^2 - 1000}$}; + + \end{axis} + \end{tikzpicture} + + + + +
      +
      2. + +
    2. +

      + The limit \lim\limits_{x\to\infty}\cos(x) does not exist, + as \cos(x) oscillates + (and takes on every value in [-1,1] infinitely many times). + Thus we cannot apply . + + The fact that the cosine function oscillates strongly hints that \cos(n), + when n is restricted to + \mathbb{N}, will also oscillate. + , + where the sequence is plotted, shows that this is true. + Because only discrete values of cosine are plotted, + it does not bear strong resemblance to the familiar cosine wave. + The proof of the following statement is beyond the scope of this text, + but it is true: + there are infinitely many integers n that are arbitrarily (, + very) close to an even multiple of \pi, + so that \cos n \approx 1. + Similarly, there are infinitely many integers m that are arbitrarily close to an odd multiple of \pi, + so that \cos m \approx -1. + As the sequence takes on values near 1 and -1 infinitely many times, + we conclude that \lim\limits_{n\to\infty}a_n does not exist. +

      +
      + Scatter plot for the sequence in of + + + A scatter plot showing a representative sample of points from the second sequence in this example. + +

      + The scatter plot for this example shows the first 100 terms in the sequence a_n=\cos(n). + The plot shows a collection of points whose y values seem to be randomly scattered + between -1 and 1. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={20,40,60,80,100}, + ymin=-1.1,ymax=1.1, + xmin=-.1,xmax=110, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={1.5pt},domain=1:100,samples=101] {cos(deg(x))}; + + \end{axis} + + + \node at (myplot.north) { $\ds a_n = \cos(n) $}; + + \end{tikzpicture} + + + + +
      +
      3. + +
    3. +

      + We cannot actually apply here, + as the function f(x) = (-1)^x/x is not well defined. (What does (-1)^{\sqrt{2}} mean? + In actuality, there is an answer, + but it involves complex analysis, + beyond the scope of this text.) Instead, + we invoke the definition of the limit of a sequence. + By looking at the plot in , + we would like to conclude that the sequence converges to L=0. +

      + +
      + Scatter plot for the sequence in + + + A scatter plot showing a representative sample of points from the third sequence in this example. + +

      + The scatter plot for the sequence a_n = (-1)^n/n shows a pair of coordinate axes, + with the horizontal axis labeled n, and positioned in the center of the plot. +

      + +

      + When n is even, the points (n,a_n) follow the graph y=1/x, + and get closer and closer to the n axis as n increases. +

      + +

      + When n is odd, the points (n,a_n) follow the graph y=-1/x, + and approach the n axis from below. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-1.1,ymax=1.1, + xmin=-.1,xmax=22, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={1.75pt},domain=1:20,samples=20] {((-1)^x)/x}; + + \draw (axis cs:10,-.7) node { $\ds a_n = \frac{(-1)^n}{n}$}; + + \end{axis} + \end{tikzpicture} + + + + +
      + +

      + Let \epsilon\gt 0 be given. + We can find a natural number m such that 1/m \lt \varepsilon. + Let n\gt m, and consider \abs{a_n - L}: + + \abs{a_n - L} \amp = \left\lvert\frac{(-1)^n}{n} - 0\right\rvert + \amp = \frac1n + \amp \lt \frac1m \text{ (since \(n\gt m\)) } + \amp \lt \varepsilon + . + We have shown that by picking m large enough, + we can ensure that a_n is arbitrarily close to our limit, + L=0, + hence by the definition of the limit of a sequence, + we can say \lim\limits_{n\to\infty}a_n = 0. +

      +
      4. +
    +

    +     + +     + +

    + In the previous example we used the definition of the limit of a sequence to determine the convergence of a sequence as we could not apply . + In general, we like to avoid invoking the definition of a limit, + and the following theorem gives us tool that we could use in that example instead. +

    + + + Absolute Value Theorem + +

    + Let \{a_n\} be a sequence. + If \lim\limits_{n\to\infty} \abs{a_n} = 0, + then \lim\limits_{n\to\infty} a_n = 0 + + Absolute Value Theorem + limitAbsolute Value Theorem + sequenceAbsolute Value Theorem +

    +     + +

    + Let \lim\limits_{n\to\infty} \abs{a_n} = 0. + We start by noting that -\abs{a_n}\leq a_n \leq \abs{a_n}. + If we apply limits to this inequality: + + \lim\limits_{n \to \infty}\left(-\abs{a_n}\right) \leq \lim\limits_{n \to \infty} a_n \leq \lim\limits_{n \to \infty} \abs{a_n} + -\lim\limits_{n \to \infty}\abs{a_n} \leq \lim\limits_{n \to \infty} a_n \leq \lim\limits_{n \to \infty} \abs{a_n} + Using the fact that \lim\limits_{n\to\infty} \abs{a_n} = 0: + 0 \leq \lim\limits_{n \to \infty} a_n \leq 0 + + We conclude that the only possible answer for \lim\limits_{n \to \infty} a_n is 0. +

    +     +     + + + + + Determining the convergence/divergence of a sequence + +

    + Determine the convergence or divergence of the following sequences. +

    + +

    +

      +
    1. + \ds \{a_n\} = \left\{\frac{(-1)^n}{n}\right\} +
      2. + +
    2. + \ds \{a_n\} = \left\{\frac{(-1)^n(n+1)}{n}\right\} +
      3. +
    +

    +     + +

    +

      +
    1. +

      + This appeared in . + We want to apply , + so consider the limit of \{\abs{a_n}\}: + + \lim_{n\to\infty} \abs{a_n} \amp = \lim_{n\to\infty} \abs{\frac{(-1)^n}{n}} + \amp = \lim_{n\to\infty} \frac{1}{n} + \amp = 0 + . + Since this limit is 0, we can apply + and state that \lim\limits_{n\to\infty} a_n=0. +

      +
      2. + +
    2. +

      + Because of the alternating nature of this sequence (, every other term is multiplied by -1), + we cannot simply look at the limit \lim\limits_{x\to\infty} \frac{(-1)^x(x+1)}{x}. + We can try to apply the techniques of : + + \lim_{n\to\infty} \abs{a_n} \amp = \lim_{n\to\infty} \abs{\frac{(-1)^n(n+1)}{n}} + \amp = \lim_{n\to\infty} \frac{n+1}{n} + \amp = 1 + . + We have concluded that when we ignore the alternating sign, + the sequence approaches 1. + This means we cannot apply ; + it states the the limit must be 0 in order to conclude anything. +

      + +
      + A plot of a sequence in , part 2 + + + Scatter plot illustrating the first 20 points in the sequence from the second part of this example. + +

      + The scatter plot for a_n=(-1)^n(n+1)/n splits into two parts. + When n is even, the points (n,a_n) follow the graph y=1+1/x, + and approach the line y=1 from above as n gets large. + When n is odd, the points (n,a_n) follow the graph y=-1-1/x, + and approach the line y=-1 from below as n gets large. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ytick={-1,-2,1,2}, + ymin=-2.5,ymax=2.5, + xmin=-.1,xmax=22, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:20,samples=20] {((-1)^x*(x+1))/x}; + + \draw (axis cs:15,1.9) node { $\ds a_n = \frac{(-1)^n(n+1)}{n}$}; + + \end{axis} + \end{tikzpicture} + + + + +
      + +

      + Since we know that the signs of the terms alternate and + we know that the limit of \abs{a_n} is 1, we know that as n approaches infinity, + the terms will alternate between values close to 1 and -1, + meaning the sequence diverges. + A plot of this sequence is given in . +

      +
      3. +
    +

    +     +     + +

    + We continue our study of the limits of sequences by considering some of the properties of these limits. +

    + + + Properties of the Limits of Sequences + +

    + Let \{a_n\} and \{b_n\} be sequences such that \lim\limits_{n\to\infty} a_n = L, + \lim\limits_{n\to\infty} b_n = K, + and let c be a real number. +

    + +

    +

      +
    1. +

      + \lim\limits_{n\to\infty} (a_n\pm b_n) = L\pm K + + sequenceslimit properties + +

      +
      2. + +
    2. +

      + \lim\limits_{n\to\infty} (a_n\cdot b_n) = L\cdot K +

      +
      3. + +
    3. +

      + \lim\limits_{n\to\infty} \left(\frac{a_n}{b_n}\right) = \frac{L}{K}, K\neq 0 +

      +
      4. + +
    4. +

      + \lim\limits_{n\to\infty} c\cdot a_n = c\cdot L +

      +
      5. + +
    5. +

      + If g is continuous at L, then \lim\limits_{n\to\infty}g(a_n)=g(L). +

      +
      6. +
    +

    +     +     + + + + + Applying properties of limits of sequences + +

    + Let the following sequences, and their limits, be given: +

    + +

    +

      +
    • +

      + \ds \{a_n\} = \left\{\frac{n+1}{n^2}\right\}, + and \lim\limits_{n\to\infty} a_n = 0; +

      +
      • + +
    • +

      + \ds \{b_n\} = \left\{\left(1+\frac1n\right)^{n}\right\}, + and \lim\limits_{n\to\infty} b_n = e; and +

      +
      • + +
    • +

      + \ds \{c_n\} = \big\{n\cdot \sin(5/n)\big\}, + and \lim\limits_{n\to\infty} c_n = 5. +

      +
      • +
    +

    + +

    + Evaluate the following limits. +

    + +

    +

      +
    1. \lim\limits_{n\to\infty} (a_n+b_n)
      2. + +
    2. \lim\limits_{n\to\infty} (b_n\cdot c_n)
      3. + +
    3. \lim\limits_{n\to\infty} (1000\cdot a_n)
      4. +
    +

    +     + +

    + We will use to answer each of these. +

    + +

    +

      +
    1. +

      + Since \lim\limits_{n\to\infty} a_n = 0 and \lim\limits_{n\to\infty} b_n = e, + we conclude that \lim\limits_{n\to\infty} (a_n+b_n) = 0+e = e. + So even though we are adding something to each term of the sequence b_n, + we are adding something so small that the final limit is the same as before. +

      +
      2. + +
    2. +

      + Since \lim\limits_{n\to\infty} b_n = e and \lim\limits_{n\to\infty} c_n = 5, + we conclude that \lim\limits_{n\to\infty} (b_n\cdot c_n) = e\cdot 5 = 5e. +

      +
      3. + +
    3. +

      + Since \lim\limits_{n\to\infty} a_n = 0, + we have \lim\limits_{n\to\infty} 1000a_n =1000\cdot 0 = 0. + It does not matter that we multiply each term by 1000; + the sequence still approaches 0. (It just takes longer to get close to 0.) +

      +
      4. +
    +

    +     + +     + +

    + There is more to learn about sequences than just their limits. + We will also study their range and the relationships terms have with the terms that follow. + We start with some definitions describing properties of the range. +

    + + + Bounded and Unbounded Sequences + +

    + A sequence \{a_n\} is said to be bounded + if there exist real numbers m and M such that + m \leq a_n \leq M for all n in \mathbb{N}. + The number m is called a lower bound for the sequence, + and the number M is called an upper bound for the sequence. +

    + +

    + A sequence \{a_n\} is said to be + unbounded if it is not bounded. +

    + +

    + A sequence \{a_n\} is said to be + bounded above + if there exists an M such that + a_n \lt M for all n in \mathbb{N}; + it is bounded below if there exists an m such that + m\lt a_n for all n in \mathbb{N}. + sequencesboundedness + bounded sequence + unbounded sequence +

    +     +     + +

    + It follows from this definition that an unbounded sequence may be bounded above or bounded below; + a sequence that is both bounded above and below is simply a bounded sequence. +

    + + + Determining boundedness of sequences + +

    + Determine the boundedness of the following sequences. +

    + +

    +

      +
    1. \ds\{a_n\} = \left\{\frac1n\right\}
      2. + +
    2. \{a_n\} = \{2^n\}
      3. +
    +

    +     + +

    +

      +
    1. +

      + The terms of this sequence are always positive but are decreasing, + so we have 0\lt a_n\lt 2 for all n. + Thus this sequence is bounded. + illustrates this. +

      +
      2. + +
    2. +

      + The terms of this sequence obviously grow without bound. + However, it is also true that these terms are all positive, + meaning 0\lt a_n. + Thus we can say the sequence is unbounded, but also bounded below. + illustrates this. +

      +
      3. +
    +

    + +
    + A plot of \{a_n\} = \{1/n\} and \{a_n\} = \{2^n\} from + + +
    + + + + A scatter plot of the first 10 terms in the first sequence for this example. + +

    + The points (n,a_n) are plotted, for n ranging from 1 to 10. + These points follow the graph y=1/x, begining at (1,1), + and descending toward the n axis as n gets large. +

    + +

    + We can see that the value of a_n remains positive, but will continue to decrease as n increases. + This illustrates the fact that the sequence is bounded between 0 and 1. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,...,10}, + ytick={1}, + extra y ticks={.5,.25,.1}, + extra y tick labels={$1/2$,$1/4$,$1/10$}, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {1/x}; + + \draw (axis cs:6,.75) node { $\ds a_n = \frac{1}{n}$}; + + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + The scatter plot for the second sequence in this example, which shows exponential growth. + +

    + This time, the points (n,2^n) are plotted, for values of n between 1 and 8. + These points follow the exponential graph y=2^x; + we can see that the y value will continue to increase without bound, + illustrating the fact that this sequence is not bounded. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=275, + xmin=-.1,xmax=9, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:8,samples=8] {2^x}; + + \draw (axis cs:4,150) node { $\ds a_n = 2^{n}$}; + + \end{axis} + \end{tikzpicture} + + + + +
    +     +
    +     + +     + +

    + The previous example produces some interesting concepts. + First, we can recognize that the sequence \ds\left\{1/n\right\} converges to 0. + This says, informally, that most + of the terms of the sequence are + really close to 0. + This implies that the sequence is bounded, + using the following logic. + First, most terms are near 0, so we could find some sort of bound on these terms + (using , + the bound is \varepsilon). + That leaves a few terms that are not near 0 (, a + finite number of terms). + A finite list of numbers is always bounded. +

    + +

    + This logic implies that if a sequence converges, it must be bounded. + This is indeed true, as stated by the following theorem. +

    + + + Convergent Sequences are Bounded + +

    + Let \ds \left\{a_n\right\} be a convergent sequence. + Then \{a_n\} is bounded. + bounded sequenceconvergence + convergenceof sequence + sequencesconvergent +

    +     +     + + + + + +

    + In + we saw the sequence \ds \{b_n\} = \left\{\left(1+1/n\right)^{n}\right\}, + where it was stated that \lim\limits_{n\to\infty} b_n = e. + (Note that this is simply restating part of . + The limit can also be found using logarithms and L'Hospital's rule.) + Even though it may be difficult to intuitively grasp the behavior of this sequence, + we know immediately that it is bounded. +

    + + + + + +

    + Another interesting concept to come out of + again involves the sequence \{1/n\}. + We stated, without proof, that the terms of the sequence were decreasing. + That is, that a_{n+1} \lt a_n for all n. (This is easy to show. + Clearly n \lt n+1. + Taking reciprocals flips the inequality: + 1/n \gt 1/(n+1). + This is the same as a_n \gt a_{n+1}.) Sequences that either steadily increase or decrease are important, + so we give this property a name. +

    + + + Monotonic Sequences + +

    +

      +
    1. +

      + A sequence \{a_n\} is monotonically increasing + if a_n \leq a_{n+1} for all n, , + + a_1 \leq a_2 \leq a_3 \leq \cdots a_n \leq a_{n+1} \cdots + +

      +
      2. + +
    2. +

      + A sequence \{a_n\} is monotonically decreasing + if a_n \geq a_{n+1} for all n, , + + a_1 \geq a_2 \geq a_3 \geq \cdots a_n \geq a_{n+1} \cdots + +

      +
      3. + +
    3. +

      + A sequence is monotonic + if it is monotonically increasing or monotonically decreasing. + + sequencesmonotonic + monotonic sequence + +

      +
      4. +
    +

    +     +     + + + + + + + Determining monotonicity + +

    + Determine the monotonicity of the following sequences. +

    + +

    +

      +
    1. +

      + \ds \{a_n\} = \left\{\frac{n+1}n\right\} +

      +
      2. + +
    2. +

      + \ds \{a_n\} = \left\{\frac{n^2+1}{n+1}\right\} +

      +
      3. + +
    3. +

      + \ds \{a_n\} = \left\{\frac{n^2-9}{n^2-10n+26}\right\} +

      +
      4. + +
    4. +

      + \ds \{a_n\} = \left\{\frac{n^2}{n!}\right\} +

      +
      5. +
    +

    +     + +

    + In each of the following, we will examine a_{n+1}-a_n. + If a_{n+1}-a_n \geq 0, + we conclude that a_n\leq a_{n+1} and hence the sequence is increasing. + If a_{n+1}-a_n\leq 0, + we conclude that a_n\geq a_{n+1} and the sequence is decreasing. + Of course, a sequence need not be monotonic and perhaps neither of the above will apply. +

    + +

    + We also give a scatter plot of each sequence. + These are useful as they suggest a pattern of monotonicity, + but analytic work should be done to confirm a graphical trend. +

    + +

    +

      +
    1. +

      + + a_{n+1}-a_n \amp = \frac{n+2}{n+1} - \frac{n+1}{n} + \amp = \frac{(n+2)(n)-(n+1)^2}{(n+1)n} + \amp = \frac{-1}{n(n+1)} + \amp \lt 0 \text{ for all \(n\). } + + Since a_{n+1}-a_n\lt 0 for all n, + we conclude that the sequence is decreasing. +

      +
      2. + +
    2. +

      + + a_{n+1}-a_n \amp = \frac{(n+1)^2+1}{n+2} - \frac{n^2+1}{n+1} + \amp = \frac{\big((n+1)^2+1\big)(n+1)- (n^2+1)(n+2)}{(n+1)(n+2)} + \amp = \frac{n^2+3n}{(n+1)(n+2)} + \amp \gt 0 \text{ for all \(n\). } + + Since a_{n+1}-a_n\gt 0 for all n, + we conclude the sequence is increasing. +

      +
      3. + +
    3. +

      + We can clearly see in , + where the sequence is plotted, that it is not monotonic. + However, it does seem that after the first 4 terms it is decreasing. + To understand why, perform the same analysis as done before: + + + a_{n+1}-a_n \amp = \frac{(n+1)^2-9}{(n+1)^2-10(n+1)+26} - \frac{n^2-9}{n^2-10n+26} + \amp = \frac{n^2+2n-8}{n^2-8n+17}-\frac{n^2-9}{n^2-10n+26} + \amp = \frac{(n^2+2n-8)(n^2-10n+26)-(n^2-9)(n^2-8n+17)}{(n^2-8n+17)(n^2-10n+26)} + \amp = \frac{-10n^2+60n-55}{(n^2-8n+17)(n^2-10n+26)} + . +

      + +

      + We want to know when this is greater than, or less than, 0. + The denominator is always positive, + therefore we are only concerned with the numerator. + For small values of n, + the numerator is positive. + As n grows large, + the numerator is dominated by -10n^2, + meaning the entire fraction will be negative; + , for large enough n, a_{n+1}-a_n \lt 0. + Using the quadratic formula we can determine that the numerator is negative for n\geq 5. + In short, the sequence is simply not monotonic, + though it is useful to note that for n\geq 5, + the sequence is monotonically decreasing. +

      +
      4. + +
    4. + +

      + Again, the plot in + shows that the sequence is not monotonic, + but it suggests that it is monotonically decreasing after the first term. + We perform the usual analysis to confirm this. + + a_{n+1}-a_n \amp = \frac{(n+1)^2}{(n+1)!} - \frac{n^2}{n!} + \amp = \frac{(n+1)^2-n^2(n+1)}{(n+1)!} + \amp = \frac{-n^3+2n+1}{(n+1)!} + + When n=1, the above expression is \gt 0; + for n\geq 2, the above expression is \lt 0. + Thus this sequence is not monotonic, + but it is monotonically decreasing after the first term. +

      +
      5. +
    +

    + +
    + Plots of sequences in + + +
    + + + + Plot of the first sequence in this example. It is decreasing and bounded below. + +

    + The scatter plot for a_n = (n+1)/n shows a sequence that is decreasing, + and bounded below by y=1. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ytick={-1,-2,1,2}, + ymin=-.1,ymax=2.5, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {(x+1)/x}; + + \draw (axis cs:6,1.9) node { $\ds a_n = \frac{n+1}{n}$}; + + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + Scatter plot for the second sequence in this example. It is increasing but not bounded. + +

    + The points in the scatter plot for this example get larger as n increases, + illustrating the fact that this is an increasing sequence. + In fact, for large n we can see that a_n \approx n, + and the points in the plot do appear to follow close to the line y=x. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-1,ymax=11, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {(x^2+1)/(x+1)}; + + \draw (axis cs:7.5,1.9) node { $\ds a_n = \frac{n^2+1}{n+1}$}; + + \end{axis} + \end{tikzpicture} + + + + +
    +     + + +
    + + + + Scatter plot for the third sequence in this example. It is not monotonic. + +

    + The plot for a_n = \frac{n^2-9}{n^2-10n+26} shows a sequence that is initially increasing, + but begins to decrease after n=5. +

    + +

    + Although this sequence is not monotonic, it is eventually monotonic, + and bounded below, + which (as we will soon see) is sufficient for the determination of a limit. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-1,ymax=16, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {(x^2-9)/(x^2-10*x+26)}; + + \draw (axis cs:3.4,11) node { $\ds a_n = \frac{n^2-9}{n^2-10n+26}$}; + + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + + + Scatter plot for the last sequence in this example. It is not monotonic. + +

    + The plot for a_n =n^2/n! shows a sequence that is initially increasing, + but begins to decrease after n=2. +

    + +

    + Although this sequence is not monotonic, it is eventually monotonic, + and bounded below, + which (as we will soon see) is sufficient for the determination of a limit. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=2.1, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {(x^2)/(factorial(x))}; + + \draw (axis cs:7,1.25) node { $\ds a_n = \frac{n^2}{n!}$}; + + \end{axis} + + + + \end{tikzpicture} + + + + +
    +     +     +
    +     + +     + +

    + Knowing that a sequence is monotonic can be useful. + Consider, for example, + a sequence that is monotonically decreasing and is bounded below. + We know the sequence is always getting smaller, + but that there is a bound to how small it can become. + This is enough to prove that the sequence will converge, + as stated in the following theorem. +

    + + + Bounded Monotonic Sequences are Convergent + +

    +

      +
    1. +

      + Let \{a_n\} be a monotonically increasing sequence that is bounded above. + Then \{a_n\} converges. +

      +
      2. + +
    2. +

      + Let \{a_n\} be a monotonically decreasing sequence that is bounded below. + Then \{a_n\} converges. + sequencesconvergent + convergenceof monotonic sequences +

      +
      3. +
    +

    +     +     + +

    + Consider once again the sequence \{a_n\} = \{1/n\}. + It is easy to show it is monotonically decreasing and that it is always positive (, bounded below by 0). + Therefore we can conclude by that the sequence converges. + We already knew this by other means, + but in the following section this theorem will become very useful. +

    + +

    + We can replace + with the statement Let \{a_n\} be a bounded, + monotonic sequence. + Then \{a_n\} converges; + , \ds \lim_{n \to\infty}a_n exists. + We leave it to the reader in the exercises to show the theorem and the above statement are equivalent. +

    + + + +

    + Sequences are a great source of mathematical inquiry. + The On-Line Encyclopedia of Integer Sequences () contains thousands of sequences and their formulae. + (As of this writing, there are 328,977 sequences in the database.) + Perusing this database quickly demonstrates that a single sequence can represent several different + real life phenomena. +

    + +

    + Interesting as this is, our interest actually lies elsewhere. + We are more interested in the sum of a sequence. + That is, given a sequence \{a_n\}, + we are very interested in a_1+a_2+a_3+\cdots. + Of course, one might immediately counter with + Doesn't this just add up to infinity? Many times, + yes, but there are many important cases where the answer is no. + This is the topic of series, + which we begin to investigate in . +

    + + + + Terms and Concepts + + + +

    + Use your own words to define a sequence. +

    +     + + + +     + + + + +

    + The domain of a sequence is the numbers. +

    +     + + + + + + + + +     + + + + +

    + Use your own words to describe the + range of a sequence. +

    +     + + + +     + + + + +

    + Describe what it means for a sequence to be bounded. +

    +     + + + +     +     + + + Problems + + + +

    + Give the first five terms of the given sequence. +

    +     + + + + +

    + \ds \{a_n\} = \left\{\frac{4^n}{(n+1)!}\right\} +

    +     + +

    + 2,\frac{8}{3},\frac{8}{3},\frac{32}{15},\frac{64}{45} +

    +     + +     + + + + +

    + \ds \{b_n\} = \left\{\left(-\frac32\right)^n\right\} +

    +     + +

    + -\frac{3}{2},\frac{9}{4},-\frac{27}{8},\frac{81}{16}, + -\frac{243}{32} +

    +     + +     + + + + +

    + \ds \{c_n\} = \left\{-\frac{n^{n+1}}{n+2}\right\} +

    +     + +

    + -\frac{1}{3},-2,-\frac{81}{5},-\frac{512}{3},-\frac{15625}{7} +

    +     + +     + + + + +

    + \ds \{d_n\} = \left\{\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)\right\} +

    +     + +

    + 1, 1, 2, 3, 5 +

    +     + +     + +     + + + +

    + Determine the nth term of the given sequence. +

    +     + + + + +

    + 4, 7, 10, 13, 16, \ldots +

    +     + +

    + a_n = 3n+1 +

    +     + +     + + + + +

    + \ds 3,\, -\frac32,\, \frac34,\, -\frac38,\, \ldots +

    +     + +

    + a_n = (-1)^{n+1}\frac{3}{2^{n-1}} +

    +     + +     + + + + +

    + 10,\, 20,\, 40,\, 80,\, 160,\, \ldots +

    +     + +

    + a_n = 10\cdot 2^{n-1} +

    +     + +     + + + + +

    + \ds 1, 1,\, \frac12,\, \frac16,\, \frac1{24},\, \frac1{120},\, \ldots +

    +     + +

    + a_n = 1/(n-1)! +

    +     + +     + +     + + + +

    + Use the following information to determine the limit of the given sequences. +

    + +

    +

      +
    • +

      + \ds \{a_n\} = \left\{\frac{2^n-20}{2^n}\right\}; + \lim\limits_{n\to\infty} a_n = 1 +

      +
      • + +
    • +

      + \ds \{b_n\} = \left\{\left(1+\frac{2}{n}\right)^n\right\}; + \lim\limits_{n\to\infty} b_n = e^2 +

      +
      • + +
    • +

      + \ds \{c_n\} = \left\{\sin(3/n)\right\}; + \lim\limits_{n\to\infty} c_n = 0 +

      +
      • +
    +

    +     + + + + +

    + \ds\{a_n\} = \left\{ \frac{2^n-20}{7\cdot2^n} \right\} +

    +     + +

    + 1/7 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{ 3b_n-a_n \right\} +

    +     + +

    + 3e^2-1 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{ \sin(3/n)\left(1+\frac2n\right)^n \right\} +

    +     + +

    + 0 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{ \left(1+\frac2n\right)^{2n} \right\} +

    +     + +

    + e^4 +

    +     + +     + +     + + + +

    + Determine whether the sequence converges or diverges. + If convergent, give the limit of the sequence. +

    +     + + + + +

    + \ds\{a_n\} = \left\{(-1)^n\frac{n}{n+1}\right\} +

    +     + +

    + diverges +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\frac{4n^2-n+5}{3n^2+1}\right\} +

    +     + +

    + converges to 4/3 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\frac{4^n}{5^n}\right\} +

    +     + +

    + converges to 0 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\frac{n-1}{n}-\frac{n}{n-1}\right\}, + n\geq 2 +

    +     + +

    + converges to 0 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\ln(n)\right\} +

    +     + +

    + diverges +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\frac{3n}{\sqrt{n^2+1}}\right\} +

    +     + +

    + converges to 3 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\left(1+\frac1n\right)^n\right\} +

    +     + +

    + converges to e +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{5-\frac1n\right\} +

    +     + +

    + converges to 5 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\frac{(-1)^{n+1}}{n}\right\} +

    +     + +

    + converges to 0 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\frac{1.1^n}{n}\right\} +

    +     + +

    + diverges +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\frac{2n}{n+1}\right\} +

    +     + +

    + converges to 2 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{(-1)^n\frac{n^2}{2^n-1}\right\} +

    +     + +

    + converges to 0 +

    +     + +     + +     + + + +

    + Determine whether the sequence is bounded, + bounded above, bounded below, or none of the above. +

    +     + + + + +

    + \ds\{a_n\} = \left\{\sin(n) \right\} +

    +     + +

    + bounded +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\tan(n) \right\} +

    +     + +

    + neither bounded above or below +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{(-1)^n\frac{3n-1}{n}\right\} +

    +     + +

    + bounded +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\frac{3n^2-1}{n}\right\} +

    +     + +

    + bounded below +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{n\cos(n) \right\} +

    +     + +

    + neither bounded above or below +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{2^n-n!\right\} +

    +     + +

    + bounded above +

    +     + +     + +     + + + +

    + Determine whether the sequence is monotonically increasing or decreasing. + If it is not, + determine if there is an m such that it is monotonic for all n\geq m. +

    +     + + + + +

    + \ds\{a_n\} = \left\{\frac{n}{n+2}\right\} +

    +     + +

    + monotonically increasing +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\frac{n^2-6n+9}{n}\right\} +

    +     + +

    + monotonically increasing for n\geq 3 +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{(-1)^n\frac{1}{n^3}\right\} +

    +     + +

    + never monotonic +

    +     + +     + + + + +

    + \ds\{a_n\} = \left\{\frac{n^2}{2^n}\right\} +

    +     + +

    + monotonically decreasing for n\geq 3 +

    +     + +     +     + + + +

    + The following exercises explore further the theory of sequences. +

    +     + + + + +

    + Prove ; that is, + use the definition of the limit of a sequence to show that if \lim\limits_{n\to\infty} \abs{a_n} = 0, then \lim\limits_{n\to\infty} a_n = 0. +

    +     + +

    + Let \{a_n\} be given such that \lim\limits_{n\to\infty} \abs{a_n} = 0. + By the definition of the limit of a sequence, + given any \varepsilon \gt 0, + there is a m such that for all n \gt m, \abs{\abs{a_n} - 0} \lt \varepsilon. + Since \abs{\abs{a_n}-0} = \abs{a_n - 0}, + this directly implies that for all n \gt m, + \abs{a_n - 0} \lt \varepsilon, + meaning that \lim\limits_{n\to\infty} a_n = 0. +

    +     + +     + + + + +

    + Let \{a_n\} and \{b_n\} be sequences such that + \lim\limits_{n\to\infty} a_n = L and \lim\limits_{n\to\infty} b_n = K. +

    +     + + + +

    + Show that if a_n\lt b_n for all n, then L\leq K. +

    +     +     + + + +

    + Give an example where L = K. +

    +     + +

    + a_n = 1/3^n and b_n = 1/2^n +

    +     +     + +     + + + + +

    + Prove the Squeeze Theorem for sequences: Let \{a_n\} and \{b_n\} be such that + \lim\limits_{n\to\infty} a_n = L and \lim\limits_{n\to\infty} b_n = L, + and let \{c_n\} be such that a_n\leq c_n\leq b_n for all n. + Then \lim\limits_{n\to\infty} c_n = L +

    +     + +

    + A sketch of one proof method: +

    + +

    + Let any \epsilon>0 be given. + Since \{a_n\} and \{b_n\} converge, + there exists an N>0 such that for all n\geq N, + both a_n and b_n are within \epsilon/2 of L; + we can conclude that they are at most + \epsilon apart from each other. + Since a_n\leq c_n \leq b_n, + one can show that c_n is within \epsilon of L, + showing that \{c_n\} also converges to L. +

    +     + +     + + + + + +

    + Prove the statement Let \{a_n\} be a bounded, + monotonic sequence. + Then \{a_n\} converges; + , \ds \lim_{n \to\infty}a_n exists. + is equivalent to . + That is, + +

      +
    1. +

      + Show that if is true, + then above statement is true, and +

      +
      2. + +
    2. +

      + Show that if the above statement is true, + then is true. +

      +
      3. +
    +

    +     + +

    + A sketch of one proof method: + +

      +
    1. +

      + Assume that is true, + and let \{a_n\} be bounded and monotonic. + Since \{a_n\} is bounded, + it is bounded both above and below. + If it is increasing, + it is bounded above and we apply ; + if it is decreasing, it is bounded below and we apply the theorem. + Either way, \{a_n\} converges and the statement is true. +

      +
      2. + +
    2. +

      + Assume the statement is true, + and let \{a_n\} be a monotonically increasing sequence that is bounded above. + Since \{a_n\} is monotonically increasing, + a_1\leq a_2\leq \ldots; + that is, a_1 bounds \{a_n\} from below. + Therefore \{a_n\} is bounded and monotonic; + by the statement, \{a_n\} converges. + A similar statement can be made for when \{a_n\} is monotonically decreasing and bounded below. + Therefore is true. +

      +
      3. +
    +

    +     + +     +     +     +     +
    +
    + Infinite Series + + +

    + Given the sequence \{a_n\} = \{1/2^n\} = 1/2,\, 1/4,\, 1/8,\, \ldots, + consider the following sums: + + a_1 \amp= 1/2 \amp = \amp 1/2 + a_1+a_2\amp = 1/2+1/4 \amp = \amp 3/4 + a_1+a_2+a_3 \amp = 1/2+1/4+1/8 \amp =\amp 7/8 + a_1+a_2+a_3+a_4 \amp= 1/2+1/4+1/8+1/16 \amp = \amp 15/16 + +

    + +

    + In general, we can show that + + a_1+a_2+a_3+\cdots +a_n = \frac{2^n-1}{2^n} = 1-\frac{1}{2^n} + . +

    + +

    + Let S_n be the sum of the first n terms of the sequence \{1/2^n\}. + From the above, we see that S_1=1/2, S_2 = 3/4, etc. + Our formula at the end shows that S_n = 1-1/2^n. +

    + +

    + Now consider the following limit: + \lim\limits_{n\to\infty}S_n = \lim_{n\to\infty}\big(1-1/2^n\big) = 1. + This limit can be interpreted as saying something amazing: + the sum of all the terms of the sequence \{1/2^n\} is 1. +

    + +

    + This example illustrates some interesting concepts that we explore in this section. + We begin this exploration with some definitions. +

    +     + + + Convergence of series + + + Infinite Series, <m>n</m>th Partial Sums, Convergence, Divergence + +

    + Let \{a_n\} be a sequence, beginning at some index value n=k. +

    + +

    +

      +
    1. +

      + The sum \ds \sum_{n=k}^\infty a_n is called an + infinite series + (or, simply series). +

      +
      2. + +
    2. +

      + Let S_n denote the sum of the first n terms in the sequence \{a_n\}, + known as the nth partial sum of the sequence. + We can then define the sequence \{S_n\} of partial sums of \{a_n\}. +

      +
      3. + +
    3. +

      + If the sequence \{S_n\} converges to L, + we say the series \ds \sum_{n=k}^\infty a_n + converges to L, + and we write \ds \sum_{n=k}^\infty a_n = L. +

      +
      4. + +
    4. +

      + If the sequence \{S_n\} diverges, + the series \ds \sum_{n=k}^\infty a_n diverges. + seriesdefinition + seriespartial sums + seriesconvergent + seriesdivergent + convergenceof series + divergenceof series +

      +
      5. +
    +

    +     +     + +

    + Using our new terminology, + we can state that the series \ds \infser 1/2^n converges, + and \ds \infser 1/2^n = 1. +

    + +

    + Note that in the definition above, we do not necessarily assume that our sum begins with n=1. + In fact, it is quite common to have a series beginning at n=0, + and in some cases we may need to consider other values as well. + The nth partial sum S_n will always denote the sum of the first n terms: + For example, \infser 1/n has + + S_n = \overbrace{1+\frac12+\cdots + \frac1n}^{n \text{ terms}} + , + while \sum_{n=0}^\infty 3^{-n} has + + S_n = \overbrace{1+\frac13+\cdots + \frac{1}{3^{n-1}}}^{n \text{ terms}} + , + and \sum_{n=3}^\infty \frac{1}{n^2-2n} has + + S_n = \overbrace{\frac{1}{3}+\frac{1}{8}+\cdots + \frac{1}{(n+2)^2-2(n+2)}}^{n \text{ terms}} + . + In general, for the series \ds\sum_{n=k}^\infty a_n, + the nth partial sum will be \ds S_n = \sum_{i=k}^{k+n-1}a_i. +

    + +

    + We will explore a variety of series in this section. + We start with two series that diverge, + showing how we might discern divergence. +

    + + + Showing series diverge + +

    +

      +
    1. +

      + Let \{a_n\} = \{n^2\}. + Show \ds \infser a_n diverges. +

      +
      2. + +
    2. +

      + Let \{b_n\} = \{(-1)^{n+1}\}. + Show \ds \infser b_n diverges. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + Consider S_n, the nth partial sum. + + S_n \amp = a_1+a_2+a_3+\cdots+a_n + \amp = 1^2+2^2+3^2\cdots + n^2. + By , this is + \amp = \frac{n(n+1)(2n+1)}{6} + . + Since \lim\limits_{n\to\infty}S_n = \infty, + we conclude that the series \ds \infser n^2 diverges. + It is instructive to write \ds \infser n^2=\infty for this tells us + how the series diverges: + it grows without bound. + + A scatter plot of the sequences \{a_n\} and \{S_n\} is given in . + The terms of \{a_n\} are growing, + so the terms of the partial sums \{S_n\} are growing even faster, + illustrating that the series diverges. +

      +
      2. + +
    2. +

      + The sequence \{b_n\} starts with 1, -1, 1, -1, + \ldots. + Consider some of the partial sums S_n of \{b_n\}: + + S_1 \amp = 1 + S_2 \amp = 0 + S_3 \amp = 1 + S_4 \amp = 0 + + This pattern repeats; + we find that S_n = \begin{cases} 1 \amp n\, \text{ is odd } \\, 0 \amp n\, \text{ is even } \end{cases}. + As \{S_n\} oscillates, + repeating 1, 0, 1, 0, \ldots, + we conclude that \lim\limits_{n\to\infty}S_n does not exist, + hence \ds\infser (-1)^{n+1} diverges. + + A scatter plot of the sequence \{b_n\} and the partial sums \{S_n\} is given in . + When n is odd, + b_n = S_n so the marks for b_n are drawn oversized to show they coincide. +

      + +
      + Scatter plots relating to + +
      + + + + Scatter plots of the sequence, and corresponding partial sums, for the first part of this example. + +

      + The scatter plot shows the first 10 terms in the sequence a_n=n^2, which follow the graph y=x^2. + On the same plot, the first 10 terms in the sequence of partial sums S_n are shown. + As one might expect, the value of S_n grows much more rapidly than that of a_n. +

      +       + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + ymin=-10,%ymax=2.1, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,secondcolor,mark size={2.4pt},mark=square*,domain=1:10,samples=10] {x*(x+1)*(2*x+1)/6}; + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {x^2}; + + \end{axis} + + + \node[shift={(0,-25pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
      + +
      + + + + Scatter plots of the sequence, and corresponding partial sums, for the second part of this example. + +

      + This scatter plot shows the terms in the sequence b_n, along with the corresponding partial sums. + Since the value of b_n alternates between -1 and 1, + we see one sequence of dots along the line y=1, and another along the line y=-1. +

      + +

      + When n is even, S_n=0, so we also see a sequence of dots along the n axis. + When n is odd, S_n=a_n, so the dots for b_n and S_n overlap. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-1.1,ymax=1.1, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {(-1)^(x+1)}; + \addplot [only marks,secondcolor,mark=square*,mark size={2.5pt}] coordinates {(1,1)(2,0)(3,1)(4,0)(5,1)(6,0)(7,1)(8,0)(9,1)(10,0)}; + + \end{axis} + + + + \node[shift={(0,-15pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $b_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
      +       +
      +
      3. +
    +

    +     + +     + + + +

    + While it is important to recognize when a series diverges, + we are generally more interested in the series that converge. + In this section we will demonstrate a few general techniques for determining convergence; + later sections will delve deeper into this topic. +

    +     + + + Geometric Series +

    + One important type of series is a + geometric series. +

    + + + Geometric Series + +

    + A geometric series is a series of the form + + \infser[0] r^n = 1+r+r^2+r^3+\cdots+r^n+\cdots + +

    + +

    + Note that the index starts at n=0, not n=1. + seriesgeometric + geometric series +

    +     +     + + + +

    + We started this section with a geometric series, + although we dropped the first term of 1. + One reason geometric series are important is that they have nice convergence properties. +

    + + + Geometric Series Test + +

    + Consider the geometric series \ds \infser[0] r^n. +

    + +

    +

      +
    1. + +

      + For r\neq 1, the nth partial sum is: + + S_n = 1+r+r^2+\cdots + r^{n-1} = \frac{1-r^{n}}{1-r} + . + When r=1, S_n = n. +

      +
      2. + +
    2. +

      + The series converges if, and only if, \abs{r} \lt 1. + When \abs{r}\lt 1, + + seriesgeometric + geometric series + convergenceof geometric series + divergenceof geometric series + + \infser[0] r^n = \frac{1}{1-r} + . +

      +
      3. +
    +

    +     + +

    + We begin by proving the formula for the simplied form for the partial sums. + Consider the nth partial sum of the geometric series, + S_n=\sum_{i=0}^n r^i: + + S_n \amp = 1+r+r^2+\cdots+r^{n-2}+r^{n-1} + Multiply both sides by r: + r\cdot S_n \amp = r+r^2+r^3+\dots+r^{n-1}+r^{n} + Now subtract the second line from the first and solve for S_n: + S_n-r\cdot S_n \amp = 1-r^n + S_n(1-r) \amp = 1-r^{n} + S_n \amp = \frac{1-r^{n}}{1-r} + . + We have shown Part + of . +

    + +

    + Now, examining the partial sums, + we consider five cases to determine when S_n converges: + +

      +
    1. +

      + If \abs{r}\lt 1, then + r^n \to 0 as n \to \infty, + so we have \inflim S_n=\frac{1-0}{1-r}=\frac{1}{1-r}, + a convergent sequence of partial sums. +

      +
      2. + +
    2. +

      + If r \gt 1, then r^n \to \infty + as n \to \infty, so + + S_n = \frac{1-r^n}{1-r}=\frac{r^n}{r-1}-\frac{1}{r-1} + + diverges to infinity. (Note that r-1 is a positive constant.) +

      +
      3. + +
    3. +

      + If r \lt -1, then r^n will oscillate between large positive + and large negative values as n increases. + The same will be true of S_n, so \inflim S_n does not exist. +

      +
      4. + +
    4. + If r=1, + then S_n = \frac{1-1^{n+1}}{1-1} is undefined. + However, examining S_n = 1+r+r^2+\dots+r^n for r=1, + we can see that the partial sums simplify to S_n=n, + and this sequence diverges to \infty. +
      5. + +
    5. + If r=-1, then S_n = \frac{1-(-1)^{n}}{2}. + For even values of n, the partial sums are always 0. + For odd values of n, the partial sums are always 1. + So the sequence of partial sums diverges. +
      6. +
    + + Therefore, a geometric series converges if and only if \abs{r} \lt 1. +

    +     + +     + + + +

    + According to , + the series + + \ds\infser[0] \frac{1}{2^n} =\infser[0] \left(\frac 12\right)^2= 1+\frac12+\frac14+\cdots + + converges as r=1/2 \lt 1, + and \ds \infser[0] \frac{1}{2^n} = \frac{1}{1-1/2} = 2. + This concurs with our introductory example; + while there we got a sum of 1, we skipped the first term of 1. +

    + + + Exploring geometric series + +

    + Check the convergence of the following series. + If the series converges, find its sum. +

    + +

    +

      +
    1. +

      + \ds \sum_{n=2}^\infty \left(\frac34\right)^n +

      +
      2. + +
    2. +

      + \ds \infser[0] \left(\frac{-1}{2}\right)^n +

      +
      3. + +
    3. +

      + \ds \infser[0] 3^n +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + Since r=3/4\lt 1, this series converges. + By , we have that + + \infser[0] \left(\frac34\right)^n = \frac{1}{1-3/4} = 4 + . + However, note the subscript of the summation in the given series: + we are to start with n=2. + Therefore we subtract off the first two terms, giving: + + \sum_{n=2}^\infty \left(\frac34\right)^n = 4 - 1 - \frac34 = \frac94 + . + This is illustrated in . +

      +
      + Scatter plots for the series in + + + Scatter plots of the sequence, and corresponding partial sums, for the first part of this example. + +

      + Scatter plots for both a_n = (3/4)^n and S_n are shown together in the same image. + The point for a_2 is not visible, as it is covered up by S_2, which has the same value. +

      + +

      + The points in the plot for a_n illustrate a sequence that decreases toward 0, + while the points in the plot for S_n show that the partial sums are increasing toward a value slightly larger than 2. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={2,4,6,8,10}, + ymin=-.5,ymax=2.5, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt}] coordinates {(2,0.5625)(3,0.9844)(4,1.301)(5,1.538)(6,1.716)(7,1.85)(8,1.95)(9,2.025)(10,2.081)}; + \addplot [only marks,firstcolor,mark size={2.4pt},domain=2:10,samples=9] {(.75)^x}; + + \end{axis} + + + + \node[shift={(0,-10pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
      +
      2. + +
    2. +

      + Since \abs{r} = 1/2 \lt 1, + this series converges, and by , + + \infser[0] \left(\frac{-1}{2}\right)^n = \frac{1}{1-(-1/2)} = \frac23 + . + The partial sums of this series are plotted in . + Note how the partial sums are not purely increasing as some of the terms of the sequence \{(-1/2)^n\} are negative. +

      +
      + Scatter plots for the series in + + + Scatter plots of the sequence, and corresponding partial sums, for the second part of this example. + +

      + Scatter plots for a_n = (-1/2)^n and the corresponding partial sums are shown together in the same image. + The numbers a_n alternate in sign between positive and negative values, but get closer to 0 as n increases. +

      + +

      + The y coordinates for the points in the scatter plot for S_n also oscillate, + but the oscillations get steadily smaller, and appear to settle down toward a common y value. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={2,4,6,8,10}, + ymin=-1.1,ymax=1.1, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,secondcolor,mark=square*,mark size={2.4pt}] coordinates {(0,1.)(1,0.5)(2,0.75)(3,0.625)(4,0.6875)(5,0.6563)(6,0.6719)(7,0.6641)(8,0.668)(9,0.666)(10,0.667)}; + \addplot [only marks,firstcolor,mark size={2.4pt},domain=0:10,samples=11] {(-.5)^(x)}; + + \end{axis} + + + + \node[shift={(0,15pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
      +
      3. + +
    3. +

      + Since r \gt 1, the series diverges. + (This makes common sense; we expect the sum + + 1+3+9+27 + 81+243+\cdots + + to diverge.) + This is illustrated in . +

      +
      + Scatter plots for the series in + + + Scatter plots of the sequence, and corresponding partial sums, for the third part of this example. + +

      + Scatter plots for the sequence a_n=3^n and the corresponding partial sums are shown together in the same image. + Both plots follow curves that appear to be exponential in nature, with the points for S_n slightly above those for a_n. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-50,%ymax=1.1, + xmin=-1,xmax=7, + xlabel={$n$} + ] + + \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt}] coordinates {(0,1)(1,3)(2,12)(3,39)(4,120)(5,363)(6,1092)}; + \addplot [only marks,firstcolor,mark size={2.4pt},domain=0:6,samples=7] {3^(x)}; + + \end{axis} + + + + \node[shift={(0,-20pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
      +
      4. +
    +

    +     + +     +     + + + <m>p</m>-Series +

    + Another important type of series is the p-series. +

    + + + <m>p</m>-Series, General <m>p</m>-Series + +

    +

      +
    1. +

      + A p-series is a series of the form + + \infser \frac{1}{n^p}, \qquad \text{ where \(p \gt 0\). } + +

      +
      2. + +
    2. +

      + A general p-series is a series of the form + + seriesp@p-series + p@p-series + p@p-series + + + \infser \frac{1}{(an+b)^p} + , + where p \gt 0 and a, b are real numbers such that a\neq 0 and an+b\gt 0 for all n\geq 1. +

      +
      3. +
    +

    +     +     + +

    + Like geometric series, + one of the nice things about p-series is that they have easy to determine convergence properties. +

    + + + <m>p</m>-Series Test + +

    + A general p-series + \ds\infser \frac{1}{(an+b)^p} will converge if, + and only if, + p \gt 1. + seriesp@p-series + p@p-series + convergenceof p@of p-series + divergenceof p@of p-series +

    +     +     + + + + + Determining convergence of series + +

    + Determine the convergence of the following series. +

    + +

    +

      +
    1. +

      + \ds\infser \frac{1}{n} +

      +
      2. + +
    2. +

      + \ds\infser \frac{1}{n^2} +

      +
      3. + +
    3. +

      + \ds\infser \frac{1}{\sqrt{n}} +

      +
      4. + +
    4. +

      + \ds\infser \frac{(-1)^n}{n} +

      +
      5. + +
    5. +

      + \ds\sum_{n=11}^\infty \frac{1}{(\frac12n-5)^3} +

      +
      6. + +
    6. +

      + \ds\infser \frac{1}{2^n} +

      +
      7. +
    +

    +     + +

    +

      +
    1. +

      + This is a p-series with p=1. + By , this series diverges. + + This series is a famous series, + called the Harmonic Series, + so named because of its relationship to harmonics + in the study of music and sound. +

      +
      2. + +
    2. +

      + This is a p-series with p=2. + By , it converges. + Note that the theorem does not give a formula by which we can determine + what the series converges to; + we just know it converges. + A famous, unexpected result is that this series converges to \ds{\pi^2}/{6}. +

      +
      3. + +
    3. +

      + This is a p-series with p=1/2; + the theorem states that it diverges. +

      +
      4. + +
    4. +

      + This is not a p-series; + the definition does not allow for alternating signs. + Therefore we cannot apply . + (Another famous result states that this series, + the Alternating Harmonic Series, + converges to \ln(2).) +

      +
      5. + +
    5. +

      + This is a general p-series with p=3, + therefore it converges. +

      +
      6. + +
    6. +

      + This is not a p-series, + but a geometric series with r=1/2. + It converges. +

      +
      7. +
    +

    +     + +     + + + +

    + Later sections will provide tests by which we can determine whether or not a given series converges. + This, in general, is much easier than determining what + a given series converges to. + There are many cases, though, + where the sum can be determined. +

    + + + Telescoping series + +

    + Evaluate the sum \ds \infser \left(\frac1n-\frac1{n+1}\right). + + seriestelescoping + telescoping series +

    +     + +

    + It will help to write down some of the first few partial sums of this series. + + S_1 \amp = \frac11-\frac12 = 1-\frac12 + S_2 \amp = \left(\frac11-\frac12\right) + \left(\frac12-\frac13\right) = 1-\frac13 + S_3 \amp = \left(\frac11-\frac12\right) + \left(\frac12-\frac13\right)+\left(\frac13-\frac14\right) = 1-\frac14 + S_4 \amp = \left(\frac11-\frac12\right) + \left(\frac12-\frac13\right)+\left(\frac13-\frac14\right) +\left(\frac14-\frac15\right) = 1-\frac15 + +

    + +

    + Note how most of the terms in each partial sum are canceled out! + In general, we see that \ds S_n = 1-\frac{1}{n+1}. + The sequence \{S_n\} converges, + as \lim\limits_{n\to\infty}S_n = \lim_{n\to\infty}\left(1-\frac1{n+1}\right) = 1, + and so we conclude that \ds \infser \left(\frac1n-\frac1{n+1}\right) = 1. + Partial sums of the series are plotted in . +

    + +
    + Scatter plots relating to the series of + + + Scatter plots of the sequence, and corresponding partial sums, for this example. + +

    + A scatter plot for the sequence a_n = \frac1n-\frac{1}{n+1}=\frac{1}{n(n+1)} is shown. + These points begin at (1,1/2) and then descend toward the n axis. +

    + +

    + Also shown is the scatter plot for the sequence S_n of partial sums of this series. + These terms are shown to be given by S_n = 1-\frac{1}{n+1}, + and we see that the points on the scatter plot begin at (1,1/2) + and then rise toward the line y=1. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={2,4,6,8,10}, + ymin=-.1,ymax=1.1, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt},domain=1:10,samples=10] {1-1/(x+1)}; + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {1/x-1/(x+1)}; + + \end{axis} + + + + \node[shift={(0,-15pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
    +     + +     + +

    + The series in + is an example of a telescoping series. + Informally, a telescoping series is one in which most terms cancel with preceding or following terms, + reducing the number of terms in each partial sum. + The partial sum S_n did not contain n terms, + but rather just two: 1 and 1/(n+1). + seriestelescoping + telescoping series +

    + +

    + When possible, we seek a way to write an explicit formula for the + nth partial sum S_n. + This makes evaluating the limit + \lim\limits_{n\to\infty} S_n much more approachable. + We do so in the next example. +

    + + + Evaluating series + +

    + Evaluate each of the following infinite series. +

    + +

    +

      +
    1. +

      + \ds \infser \frac{2}{n^2+2n} +

      +
      2. + +
    2. +

      + \ds \infser \ln\left(\frac{n+1}{n}\right) +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + We can decompose the fraction 2/(n^2+2n) as + + \frac2{n^2+2n} = \frac1n-\frac1{n+2} + . + (See , Partial Fraction Decomposition, + to recall how this is done, if necessary.) + + + + Expressing the terms of \{S_n\} is now more instructive: + + S_1 \amp = 1-\frac13 + S_2 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right) + \amp = 1+\frac12-\frac13-\frac14 + S_3 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right) + \amp = 1+\frac12-\frac14-\frac15 + S_4 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\left(\frac14-\frac16\right) + \amp = 1+\frac12-\frac15-\frac16 + S_5 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\left(\frac14-\frac16\right)+\left(\frac15-\frac17\right) + \amp = 1+\frac12-\frac16-\frac17 + + We again have a telescoping series. + In each partial sum, + most of the terms cancel and we obtain the formula \ds S_n = 1+\frac12-\frac1{n+1}-\frac1{n+2}. + Taking limits allows us to determine the convergence of the series: + + \lim_{n\to\infty}S_n = \lim_{n\to\infty} \left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right) = \frac32 + , + so \infser \frac1{n^2+2n} = \frac32. + This is illustrated in . +

      +
      2. + +
    2. +

      + We begin by writing the first few partial sums of the series: + + S_1 \amp = \ln\left(2\right) + S_2 \amp = \ln\left(2\right)+\ln\left(\frac32\right) + S_3 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right) + S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) + + At first, this does not seem helpful, + but recall the logarithmic identity: + \ln(x) +\ln(y) = \ln(xy). + Applying this to S_4 gives: + + S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) + \amp = \ln\left(\frac21\cdot\frac32\cdot\frac43\cdot\frac54\right) = \ln\left(5\right) + . + We can conclude that \{S_n\} = \big\{\ln(n+1)\big\}. + This sequence does not converge, + as \lim\limits_{n\to\infty}S_n=\infty. + Therefore \ds\infser \ln\left(\frac{n+1}{n}\right)=\infty; + the series diverges. + Note in how the sequence of partial sums grows slowly; + after 100 terms, it is not yet over 5. + Graphically we may be fooled into thinking the series converges, + but our analysis above shows that it does not. +

      +
      3. +
    +

    + +
    + Scatter plots relating to the series in + + +
    + + + + Scatter plots of the sequence, and corresponding partial sums, for the first part of this example. + +

    + The scatter plots given appear very similar to the previous example. + Again, points plotted for the terms in the sequence a_n, + which begin at (1,2/3), descend toward the n axis, + while the points for the sequence of partial sums ascend from the same point, + in this case getting closer and closer to the line y=3/2. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={2,4,6,8,10}, + ymin=-.1,ymax=1.6, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt},domain=1:10,samples=10] {1.5-1/(x+1)-1/(x+2)}; + \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {2/(x^2+2*x)}; + + \end{axis} + + + + \node[shift={(0,-17pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
    + +
    + + + + Scatter plots of the sequence, and corresponding partial sums, for the second part of this example. + +

    + Scatter plots are shown for both a_n = \ln(1+1/n) and S_n = \ln(n+1), + for n=10,20,\ldots, 100. + The points for the sequence a_n are all very close to the n axis. + The points for the sequence S_n of partial sums follow the graph y=\ln(x+1); + although this may appear to be bounded, we know that the value of the logarithm + will eventually approach infinity, even if it does so very slowly. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=5, + xmin=-10,xmax=110, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=10:100,samples=10] {ln((x+1)/x)}; + \addplot [only marks,secondcolor,mark=square*,mark size={2.4pt},domain=10:100,samples=10] {ln(x+1)}; + + \end{axis} + + + + \node[shift={(0,-15pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
    +     +
    +     + +     + +

    + We are learning about a new mathematical object, the series. + As done before, we apply old + mathematics to this new topic. +

    + + + Properties of Infinite Series + +

    + Let \ds \infser a_n = L,\ds\infser b_n = K, + and let c be a constant. +

    + +

    +

      +
    1. +

      + Constant Multiple Rule: + \ds\infser c\cdot a_n = c\cdot\infser a_n = c\cdot L. + + Constant Multiple Ruleof series +

      +
      2. + +
    2. +

      + Sum/Difference Rule: + \ds\infser \big(a_n\pm b_n\big) = \infser a_n \pm \infser b_n = L \pm K. + + seriesproperties + Sum/Difference Ruleof series +

      +
      3. +
    +

    +     +     + +

    + Before using this theorem, we provide a few famous series. +

    + + + Important Series +

    +

      +
    1. +

      + \ds\infser[0] \frac1{n!} = e. + (Note that the index starts with n=0.) +

      +
      2. + +
    2. +

      + \ds\infser \frac1{n^2} = \frac{\pi^2}{6}. +

      +
      3. + +
    3. +

      + \ds\infser \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}. +

      +
      4. + +
    4. +

      + \ds\infser[0] \frac{(-1)^{n}}{2n+1} = \frac{\pi}{4}. +

      +
      5. + +
    5. +

      + \ds\infser \frac{1}{n} diverges. + (This is called the Harmonic Series.) + + Harmonic Series +

      +
      6. + +
    6. +

      + \ds\infser \frac{(-1)^{n+1}}{n} = \ln(2). + (This is called the Alternating Harmonic Series.) + Alternating Harmonic Series +

      +
      7. +
    +

    +     + + + Evaluating series + +

    + Evaluate the given series. +

    + +

    +

      +
    1. +

      + \ds\infser \frac{(-1)^{n+1}\big(n^2-n\big)}{n^3} +

      +
      2. + +
    2. +

      + \ds\infser \frac{1000}{n!} +

      +
      3. + +
    3. +

      + \ds \frac1{16}+\frac1{25}+\frac1{36}+\frac1{49}+\cdots +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + We start by using algebra to break the series apart: + + \infser \frac{(-1)^{n+1}\big(n^2-n\big)}{n^3} \amp = \infser\left(\frac{(-1)^{n+1}n^2}{n^3}-\frac{(-1)^{n+1}n}{n^3}\right) + \amp = \infser\frac{(-1)^{n+1}}{n}-\infser\frac{(-1)^{n+1}}{n^2} + \amp = \ln(2) - \frac{\pi^2}{12} \approx -0.1293 + . + This is illustrated in . +

      +
      2. + +
    2. +

      + This looks very similar to the series that involves e in . + Note, however, + that the series given in this example starts with n=1 and not n=0. + The first term of the series in the Key Idea is 1/0! = 1, + so we will subtract this from our result below: + + \infser \frac{1000}{n!} \amp = 1000\cdot\infser \frac{1}{n!} + \amp = 1000\cdot (e-1) \approx 1718.28 + . + This is illustrated in . + The graph shows how this particular series converges very rapidly. +

      + +
      + Scatter plots relating to the series in + + +
      + + + + Scatter plots of the sequence, and corresponding partial sums, for the first part of this example. + +

      + The points in the scatter plot for the sequence a_n + appear to alternate betwee positive and negative y values, + and they converge toward the n axis from both sides. +

      + +

      + The points in the scatter plot for the sequence of partial sums S_n + also appear to oscillate, although they all have negative y coordinates. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={10,20,30,40,50}, + ymin=-.3,ymax=.3, + xmin=-1,xmax=55, + xlabel={$n$} + ] + + \addplot [only marks,firstcolor,mark size={2.4pt},domain=5:50,samples=10] {((-1)^(x+1)*(x^2-x))/(x^3)}; + \addplot [only marks,secondcolor,mark=square*,mark size={2.4pt}] coordinates {(5,-0.05528)(10,-0.1723)(15,-0.09917)(20,-0.1525)(25,-0.1105)(30,-0.1452)(35,-0.1156)(40,-0.1414)(45,-0.1186)(50,-0.139)}; + + \end{axis} + + + + \node[shift={(0,0pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
      + +
      + + + + Scatter plots of the sequence, and corresponding partial sums, for the first part of this example. + +

      + Scatter plots are shown for the sequence a_n = 1000/n!, and the corresponding partial sums. + The value of a_n approaches zero quite rapidly, + and as a result, we also see that the points in the plot for S_n quickly converge toward the limiting value. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.3,ymax=2100, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt}] coordinates {(1,1000.)(2,1500.)(3,1667.)(4,1708.)(5,1717.)(6,1718.)(7,1718.)(8,1718.)(9,1718.)(10,1718.)}; + \addplot [only marks,firstcolor,mark size={2.4pt}] coordinates {(1,1000.)(2,500.)(3,166.7)(4,41.67)(5,8.333)(6,1.389)(7,0.1984)(8,0.0248)(9,0.002756)(10,0.0002756)}; + + \end{axis} + + + + \node[shift={(0,-25pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
      +       +
      +
      3. + +
    3. +

      + The denominators in each term are perfect squares; + we are adding \ds \sum_{n=4}^\infty \frac{1}{n^2} + (note we start with n=4, not n=1). + This series will converge. + Using the formula from , + we have the following: + + \infser \frac1{n^2} \amp = \sum_{n=1}^3 \frac1{n^2} +\sum_{n=4}^\infty \frac1{n^2} + \infser \frac1{n^2} - \sum_{n=1}^3 \frac1{n^2} \amp =\sum_{n=4}^\infty \frac1{n^2} + \frac{\pi^2}{6} - \left(\frac11+\frac14+\frac19\right) \amp = \sum_{n=4}^\infty \frac1{n^2} + \frac{\pi^2}{6} - \frac{49}{36} \amp = \sum_{n=4}^\infty \frac1{n^2} + 0.2838\amp \approx \sum_{n=4}^\infty \frac1{n^2} + +

      +
      4. +
    +

    +     + +     + +

    + It may take a while before one is comfortable with this statement, + whose truth lies at the heart of the study of infinite series: + it is possible that the sum of an infinite list of nonzero numbers is finite. + We have seen this repeatedly in this section, + yet it still may take some getting used to. +

    + +

    + As one contemplates the behavior of series, a few facts become clear. +

    + +

    +

      +
    1. +

      + In order to add an infinite list of nonzero numbers and get a finite result, + most of those numbers must be very near 0. +

      +
      2. + +
    2. +

      + If a series diverges, + it means that the sum of an infinite list of numbers is not finite + (it may approach \pm \infty or it may oscillate), + and: +

      + +

      +

        +
      1. +

        + The series will still diverge if the first term is removed. +

        +
        2. + +
      2. +

        + The series will still diverge if the first 10 terms are removed. +

        +
        3. + +
      3. +

        + The series will still diverge if the first 1,000,000 terms are removed. +

        +
        4. + +
      4. +

        + The series will still diverge if any finite number of terms from anywhere in the series are removed. +

        +
        5. +
      +

      +
      3. +
    +

    + +

    + These concepts are very important and lie at the heart of the next two theorems. +

    + + + <m>n</m>th-Term Test for Divergence + +

    + Consider the series \ds\infser a_n. + If \lim\limits_{n\to\infty}a_n \neq 0, + then \ds\infser a_n diverges. + + seriesnth@nth-term test + nth@nth-term test + convergencenth@nth-term test + divergencenth@nth-term test +

    +     +     + +

    + Important! This theorem does not state + that if \ds \lim_{n\to\infty} a_n = 0 then \ds \sum_{n=1}^\infty a_n converges. + The standard example of this is the Harmonic Series, + as given in . + The Harmonic Sequence, \{1/n\}, converges to 0; + the Harmonic Series, \ds \sum_{n=1}^\infty \frac1n, + diverges. +

    + +

    + Looking back, + we can apply this theorem to the series in . + In that example, + the nth terms of both sequences do not converge to 0, therefore we can quickly conclude that each series diverges. +

    + +

    + One can rewrite + to state If a series converges, + then the underlying sequence converges to 0. + While it is important to understand the truth of this statement, + in practice it is rarely used. + It is generally far easier to prove the convergence of a sequence than the convergence of a series. +

    + + + Infinite Nature of Series + +

    + The convergence or divergence of an infinite series remains unchanged by the addition or subtraction of any finite number of terms. + That is: +

    + +

    +

      +
    1. +

      + A divergent series will remain divergent with the addition or subtraction of any finite number of terms. +

      +
      2. + +
    2. +

      + A convergent series will remain convergent with the addition or subtraction of any finite number of terms. + (Of course, the sum will likely change.) +

      +
      3. +
    +

    +     +     + +

    + Consider once more the Harmonic Series \ds\infser \frac1n which diverges; + that is, the sequence of partial sums \{S_n\} grows (very, + very slowly) without bound. + One might think that by removing the large + terms of the sequence that perhaps the series will converge. + This is simply not the case. + For instance, + the sum of the first 10 million terms of the Harmonic Series is about 16.7. + Removing the first 10 million terms from the Harmonic Series changes the nth partial sums, + effectively subtracting 16.7 from the sum. + However, a sequence that is growing without bound will still grow without bound when 16.7 is subtracted from it. +

    + +

    + The equations below illustrate this. + The first line shows the infinite sum of the Harmonic Series split into the sum of the first 10 million terms plus the sum of + everything else. + The next equation shows us subtracting these first 10 million terms from both sides. + The final equation employs a bit of + psuedo-math: + subtracting 16.7 from infinity + still leaves one with infinity. + + \infser \frac1n \amp = \sum_{n=1}^{10,000,000}\frac1n + \ds\sum_{n=10,000,001}^\infty \frac1n + \infser \frac1n - \sum_{n=1}^{10,000,000}\frac1n\amp = \ds\sum_{n=10,000,001}^\infty \frac1n + \infty - 16.7 \amp = \infty + . +

    + +

    + Just for fun, + we can show that the Harmonic Series diverges algebraically + (without the use of ). +

    + + + Divergence of the harmonic series +

    + If you just consider the partial sums + + S_1, S_2, S_3, \dots, S_{1000}, S_{1001}, \dots + , + it is not apparent that the partial sums diverge. + Indeed they do diverge, but very, very slowly. + (If you graph them on a logarithmic scale however, + you can clearly see the divergence of the partial sums.) + Instead, we will consider the partial sums, + indexed by powers of 2. + That is, we will consider S_2,S_4, S_8, S_{16}, \dots. + + S_2=1+\frac12 + S_4=1+\frac12+\frac13+\frac14 + S_8=1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac18 + +

    + +

    + Next, we consider grouping together terms in each partial sum. + We will use these groupings to set up inequalities. + + S_2=1+\frac12 + S_4=1+\frac12+\left(\frac13+\frac14\right) + S_8=1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right) + +

    + +

    + In the partial sum S_4, + we note that since 1/3\gt 1/4, we can say + + S_4=1+\frac12+\left(\frac13+\frac14\right)\gt 1+\frac12+\underbrace{\left(\frac14+\frac14\right)}_{1/2}=1+\frac22 + . + Do the same in S_8 and also note that every term in the group + \left(\frac15+\frac16+\frac17+\frac18\right) is larger than 1/8. + So + + S_8 \amp = 1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right) + \amp \gt 1+\frac12+\underbrace{\left(\frac14+\frac14\right)}_{1/2}+\underbrace{\left(\frac18+\frac18+\frac18+\frac18\right)}_{1/2}=1+\frac32 + +

    + +

    + Generally, we can see that S_{2^n} \gt 1+\frac{n}2. + (In order to really show this, + we should employ proof by induction.) + Since the sequence of partial sums clearly diverges, + so does the series \infser 1/n. +

    +     + +

    + This section introduced us to series and defined a few special types of series whose convergence properties are well known: + we know when a p-series or a geometric series converges or diverges. + Most series that we encounter are not one of these types, + but we are still interested in knowing whether or not they converge. + The next three sections introduce tests that help us determine whether or not a given series converges. +

    +     + + + + Terms and Concepts + + + +

    + Use your own words to describe how sequences and series are related. +

    +     + +

    + Answers will vary. +

    +     + +     + + + + +

    + Use your own words to define a partial sum. +

    +     + + + +     + + + + +

    + Given a series \ds \infser a_n, + describe the two sequences related to the series that are important. +

    +     + + + +

    + One sequence is the sequence of terms \{a_\}. + The other is the sequence of nth partial sums, + \{S_n\} = \{\sum_{i=1}^n a_i\}. +

    +     + +     + + + + +

    + Use your own words to explain what a geometric series is. +

    +     + + + +     + + + + +

    + If \{a_n\} is convergent, + then \ds \infser a_n is also convergent. +

    +     + + +     + + + +

    + If \{a_n\} converges to 0, then \ds \sum_{n=0}^\infty a_n converges. +

    +     + +     +     + + + Problems + + + +

    + A series \ds\infser a_n is given. +

    + +

    +

      +
    1. +

      + Give the first 5 partial sums of the series. +

      +
      2. + +
    2. +

      + Give a graph of the first 5 terms of a_n and S_n on the same axes. +

      +
      3. +
    +

    +     + + + + +

    + \ds \infser \frac{(-1)^n}{n} +

    +     + +

    +

      +
    1. +

      + -1,-\frac{1}{2},-\frac{5}{6},-\frac{7}{12},-\frac{47}{60} +

      +
      2. + +
    2. +

      + Plot omitted +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{1}{n^2} +

    +     + +

    +

      +
    1. +

      + 1,\frac{5}{4},\frac{49}{36},\frac{205}{144},\frac{526 + 9}{3600} +

      +
      2. + +
    2. +

      + Plot omitted +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \cos(\pi n) +

    +     + +

    +

      +
    1. +

      + -1,0,-1,0,-1 +

      +
      2. + +
    2. +

      + Plot omitted +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser n +

    +     + +

    +

      +
    1. +

      + 1,3,6,10,15 +

      +
      2. + +
    2. +

      + Plot omitted +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{1}{n!} +

    +     + +

    +

      +
    1. +

      + 1,\frac{3}{2},\frac{5}{3},\frac{41}{24},\frac{103}{60} +

      +
      2. + +
    2. +

      + Plot omitted +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{1}{3^n} +

    +     + +

    +

      +
    1. +

      + \frac{1}{3},\frac{4}{9},\frac{13}{27},\frac{40}{81},\frac{121}{243} +

      +
      2. + +
    2. +

      + Plot omitted +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \left(-\frac{9}{10}\right)^n +

    +     + +

    +

      +
    1. +

      + -0.9,-0.09,-0.819,-0.1629,-0.75339 +

      +
      2. + +
    2. +

      + Plot omitted +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \left(\frac{1}{10}\right)^n +

    +     + +

    +

      +
    1. +

      + 0.1,0.11,0.111,0.1111,0.11111 +

      +
      2. + +
    2. +

      + Plot omitted +

      +
      3. +
    +

    +     + +     + +     + + + +

    + Use + to show the given series diverges. +

    +     + + + + +

    + \ds \infser \frac{3n^2}{n(n+2)} +

    +     + +

    + \lim\limits_{n\to\infty}a_n = 3; + by the series diverges. +

    +     + +     + + + + +

    + \ds \infser \frac{2^n}{n^2} +

    +     + +

    + \lim\limits_{n\to\infty}a_n = \infty; + by the series diverges. +

    +     + +     + + + + +

    + \ds \infser \frac{n!}{10^n} +

    +     + +

    + \lim\limits_{n\to\infty}a_n = \infty; + by the series diverges. +

    +     + +     + + + + +

    + \ds \infser \frac{5^n-n^5}{5^n+n^5} +

    +     + +

    + \lim\limits_{n\to\infty}a_n = 1; + by the series diverges. +

    +     + +     + + + + +

    + \ds \infser \frac{2^n+1}{2^{n+1}} +

    +     + +

    + \lim\limits_{n\to\infty}a_n = 1/2; + by the series diverges. +

    +     + +     + + + + +

    + \ds \infser \left(1+\frac1n\right)^n +

    +     + +

    + \lim\limits_{n\to\infty}a_n = e; + by the series diverges. +

    +     + +     + +     + + + +

    + State whether the given series converges or diverges. +

    +     + + + + +

    + \ds \infser \frac{1}{n^5} +

    +     + +

    + Converges; p-series with p=5. +

    +     + +     + + + + +

    + \ds \infser[0] \frac{1}{5^n} +

    +     + +

    + Converges; geometric series with r=1/5. +

    +     + +     + + + + +

    + \ds \infser[0] \frac{6^n}{5^n} +

    +     + +

    + Diverges; geometric series with r=6/5. +

    +     + +     + + + + +

    + \ds \infser n^{-4} +

    +     + +

    + Converges; p-series with p=4. +

    +     + +     + + + + +

    + \ds \infser \sqrt{n} +

    +     + +

    + Diverges; fails nth term test. +

    +     + +     + + + + +

    + \ds \infser \frac{10}{n!} +

    +     + +

    + Converges; by + and , + series converges to 10e. +

    +     + +     + + + + +

    + \ds \infser \left(\frac{1}{n!}+\frac1n\right) +

    +     + +

    + Diverges; although the series \ds\infser \frac{1}{n!} converges, \ds \infser \frac1n is the (divergent) harmonic series. We can only use the sum rule for series if both parts converge separately. +

    +     + +     + + + + +

    + \ds \infser \frac{2}{(2n+8)^2} +

    +     + +

    + Converges; general p-series with p=2. +

    +     + +     + + + + +

    + \ds \infser \frac{1}{2n} +

    +     + +

    + Diverges; by + this is half the Harmonic Series, + which diverges by growing without bound. + Half of growing without bound + is still growing without bound. +

    +     + +     + + + + +

    + \ds \infser \frac{1}{2n-1} +

    +     + +

    + Diverges; general p-series with p=1. +

    +     + +     + +     + + + +

    + A series is given. +

    + +

    +

      +
    1. +

      + Find a formula for S_n, + the nth partial sum of the series. +

      +
      2. + +
    2. +

      + Determine whether the series converges or diverges. + If it converges, state what it converges to. +

      +
      3. +
    +

    +     + + + + +

    + \ds \infser[0] \frac{1}{4^n} +

    +     + +

    +

      +
    1. +

      + S_n = \frac{1-(1/4)^{n+1}}{3/4} +

      +
      2. + +
    2. +

      + Converges to 4/3. +

      +
      3. +
    +

    +     + +     + + + +

    + \ds \sum_{n=1}^\infty 2 +

    +     + +

    +

      +
    1. +

      + S_n = 2n +

      +
      2. + +
    2. +

      + Diverges. +

      +
      3. +
    +

    +     +     + + + + + +

    + \ds 1^3+2^3+3^3+4^3+\cdots +

    +     + +

    +

      +
    1. +

      + S_n = \left(\frac{n(n+1)}{2}\right)^2 +

      +
      2. + +
    2. +

      + Diverges +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser (-1)^n n +

    +     + +

    +

      +
    1. +

      + + S_n = \left\{\begin{array}{cc} -\frac{n+1}{2} \amp n \text{ is odd } \\ + \frac{n}{2} \amp n \text{ is even } + \end{array} \right. + +

      +
      2. + +
    2. +

      + Diverges +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser[0] \frac{5}{2^n} +

    +     + +

    +

      +
    1. +

      + S_n = 5\frac{1-1/2^(n+1)}{1/2} +

      +
      2. + +
    2. +

      + Converges to 10. +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser[0] e^{-n} +

    +     + +

    +

      +
    1. +

      + S_n = \frac{1-(1/e)^{n+1}}{1-1/e}. +

      +
      2. + +
    2. +

      + Converges to 1/(1-1/e) = e/(e-1). +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds 1-\frac13+\frac19-\frac{1}{27}+\frac{1}{81}+\cdots +

    +     + +

    +

      +
    1. +

      + S_n = \frac{1-(-1/3)^(n+1)}{4/3} +

      +
      2. + +
    2. +

      + Converges to 3/4. +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{1}{n(n+1)} +

    +     + +

    +

      +
    1. +

      + With partial fractions, a_n = \frac1n-\frac1{n+1}. + Thus S_n = 1-\frac1{n+1}. +

      +
      2. + +
    2. +

      + Converges to 1. +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{3}{n(n+2)} +

    +     + +

    +

      +
    1. +

      + With partial fractions, + a_n = \frac32\left(\frac1n-\frac1{n+2}\right). + Thus S_n = \frac32\left(\frac32-\frac1{n+1}-\frac1{n+2}\right). +

      +
      2. + +
    2. +

      + Converges to 9/4 +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{1}{(2n-1)(2n+1)} +

    +     + +

    +

      +
    1. +

      + With partial fractions, + a_n = \frac12\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right). + Then S_n = \frac12\left(1-\frac1{2n+1}\right) = \frac{n}{2n+1}. +

      +
      2. + +
    2. +

      + Converges to 1/2. +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \ln\left(\frac{n}{n+1}\right) +

    +     + +

    +

      +
    1. +

      + S_n = \ln\big(1/(n+1)\big) +

      +
      2. + +
    2. +

      + Diverges + (to -\infty). +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{2n+1}{n^2(n+1)^2} +

    +     + +

    +

      +
    1. +

      + S_n = 1-\frac{1}{(n+1)^2} +

      +
      2. + +
    2. +

      + Converges to 1. +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \frac{1}{1\cdot 4}+\frac1{2\cdot5}+\frac1{3\cdot6}+\frac1{4\cdot7}+\cdots +

    +     + +

    +

      +
    1. +

      + a_n = \frac1{n(n+3)}; + using partial fractions, + the resulting telescoping sum reduces to S_n = \frac13\left(1+\frac12+\frac13-\frac1{n+1}-\frac1{n+2}-\frac1{n+3}\right) +

      +
      2. + +
    2. +

      + Converges to 11/18. +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds 2+\left(\frac12+\frac13\right) + \left(\frac14+\frac19\right) + \left(\frac18+\frac1{27}\right)+\cdots +

    +     + +

    +

      +
    1. +

      + a_n = 1/2^n+1/3^n for n\geq 0. + Thus S_n = \frac{1-1/2^2}{1/2}+\frac{1-1/3^n}{2/3}. +

      +
      2. + +
    2. +

      + Converges to 2+3/2 = 7/2. +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \sum_{n=2}^\infty \frac{1}{n^2-1} +

    +     + +

    +

      +
    1. +

      + With partial fractions, + a_n = \frac12\left(\frac1{n-1}-\frac1{n+1}\right). + Thus S_n = \frac12\left(3/2-\frac1n-\frac{1}{n+1}\right). +

      +
      2. + +
    2. +

      + Converges to 3/4. +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \infser[0] \big(\sin(1) \big)^n +

    +     + +

    +

      +
    1. +

      + S_n=\frac{1-(\sin(1) )^{n+1}}{1-\sin(1) } +

      +
      2. + +
    2. +

      + Converges to \frac{1}{1-\sin(1) }.. +

      +
      3. +
    +

    +     + +     +     + + + + +

    + Break the Harmonic Series into the sum of the odd and even terms: + + \infser \frac1n = \infser \frac{1}{2n-1}+\infser \frac{1}{2n} + . +

    + +

    + The goal is to show that each of the series on the right diverge. +

    +     + + + +

    + Show why \ds \infser \frac{1}{2n-1} \gt \infser \frac{1}{2n}. +

    + +

    + (Compare each nth partial sum.) +

    +     + +

    + The nth partial sum of the odd series is 1+\frac13+\frac15+\cdots+\frac{1}{2n-1}. + The nth partial sum of the even series is \frac12+\frac14 + \frac16 + \cdots +\frac1{2n}. + Each term of the even series is less than the corresponding term of the odd series, + giving us our result. +

    +     +     + + + +

    + Show why \ds\infser \frac{1}{2n-1}\lt 1+\infser \frac{1}{2n} +

    +     + +

    + The nth partial sum of the odd series is 1+\frac13+\frac15+\cdots+\frac1{2n-1}. + The nth partial sum of 1 plus the even series is 1+\frac12+\frac14+\cdots + \frac{1}{2(n-1)}. + Each term of the even series is now greater than or equal to the corresponding term of the odd series, + with equality only on the first term. + This gives us the result. +

    +     +     + + + +

    + Explain why (a) and (b) demonstrate that the series of odd terms is convergent, if, + and only if, the series of even terms is also convergent. + (That is, show both converge or both diverge.) +

    +     + +

    + If the odd series converges, + the work done in (a) shows the even series converges also. + (The sequence of the nth partial sum of the even series is bounded and monotonically increasing.) + Likewise, (b) shows that if the even series converges, + the odd series will, too. + Thus if either series converges, the other does. +

    + +

    + Similarly, (a) and (b) can be used to show that if either series diverges, + the other does, too. +

    +     +     + + + +

    + Explain why knowing the Harmonic Series is divergent determines that the even and odd series are also divergent. +

    +     + +

    + If both the even and odd series converge, + then their sum would be a convergent series. + This would imply that the Harmonic Series, their sum, is convergent. + It is not. + Hence each series diverges. +

    +     +     + +     + + + + +

    + Show the series \ds \infser \frac{n}{(2n-1)(2n+1)} diverges. +

    +     + +

    + Using partial fractions, + we can show that a_n = \frac14\left(\frac1{2n-1}+\frac{1}{2n+1}\right). + The series is effectively twice the sum of the odd terms of the Harmonic Series which was shown to diverge in . + Thus this series diverges. +

    +     + +     +     +     +
    +
    + Integral and Comparison Tests + +

    + Knowing whether or not a series converges is very important, + especially when we discuss Power Series in . + Theorems + and + give criteria for when Geometric and p-series converge, + and + gives a quick test to determine if a series diverges. + There are many important series whose convergence cannot be determined by these theorems, though, + so we introduce a set of tests that allow us to handle a broad range of series. + We start with the Integral Test. +

    +     + + + Integral Test +

    + We stated in + that a sequence \{a_n\} is a function a(n) whose domain is \mathN, + the set of natural numbers. + If we can extend a(n) to \mathbb{R}, + the real numbers, + and it is both positive and decreasing on [1,\infty), + then the convergence of \ds \infser a_n is the same as \ds\int_1^\infty a(x)\, dx. +

    + + + Integral Test + +

    + Let a sequence \{a_n\} be defined by a_n=a(n), + where a(n) is continuous, + positive and decreasing on [1,\infty). + Then \ds \infser a_n converges, if, + and only if, \ds\int_1^\infty a(x)\, dx converges. + seriesIntegral Test + Integral Test + convergenceIntegral Test + divergenceIntegral Test +

    +     +     + + + + +

    + We can demonstrate the truth of the Integral Test with two simple graphs. + In , + the height of each rectangle is + a(n)=a_n for n=1,2,\ldots, + and clearly the rectangles enclose more area than the area under y=a(x). + Therefore we can conclude that + + \ds \int_1^\infty a(x)\, dx \lt \infser a_n + . +

    + +
    + Illustrating the truth of the Integral Test + + +
    + + + + First of two graphs illustrating why the integral test works. + +

    + The graph of a positive, decreasing function y=a(x) is shown. + Also shown are the first four rectangles in a Riemann sum with partition points at each integer, + and using left endpoints. Since a(x) is decreasing, this is an overestimate, + which is illustrated by the fact that the rectangles all reach above the graph. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,2,3,4,5}, + ytick={1,2}, + ymin=-.1,ymax=2.5, + xmin=-.5,xmax=5.6 + ] + + \addplot+ [domain=-.5:5.6,samples=50] ({x},{2/(x+1)}); + + \draw [thick,secondcolor] (axis cs: 1,0) -- (axis cs:1,1) -- (axis cs: 2,1) -- (axis cs: 2,0) + (axis cs: 2,0) -- (axis cs:2,.667) -- (axis cs: 3,.667) -- (axis cs: 3,0) + (axis cs: 3,0) -- (axis cs:3,.5) -- (axis cs: 4,.5) -- (axis cs: 4,0) + (axis cs: 4,0) -- (axis cs:4,.4) -- (axis cs: 5,.4) -- (axis cs: 5,0); + + \draw (axis cs: .95,1.75) node { $y=a(x)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + Second of two graphs illustrating why the integral test works. + +

    + The graph of a positive, decreasing function y=a(x) is shown. + Also shown are the first four rectangles in a Riemann sum with partition points at each integer, + and using right endpoints. Since a(x) is decreasing, this is an underestimate, + which is illustrated by the fact that the rectangles all lie below the graph. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,2,3,4,5}, + ytick={1,2}, + ymin=-.1,ymax=2.5, + xmin=-.5,xmax=5.6 + ] + + \addplot+ [domain=-.5:5.6,samples=50] ({x},{2/(x+1)}); + + \draw [thick,secondcolor] (axis cs: 1,0) -- (axis cs:1,.667) -- (axis cs: 2,.667) -- (axis cs: 2,0) + (axis cs: 2,0) -- (axis cs:2,.5) -- (axis cs: 3,.5) -- (axis cs: 3,0) + (axis cs: 3,0) -- (axis cs:3,.4) -- (axis cs: 4,.4) -- (axis cs: 4,0) + (axis cs: 4,0) -- (axis cs:4,.333) -- (axis cs: 5,.333) -- (axis cs: 5,0); + + \draw (axis cs: .95,1.75) node { $y=a(x)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + In , + we draw rectangles under y=a(x) with the Right-Hand rule, + starting with n=2. + This time, the area of the rectangles is less than the area under y=a(x), + so \ds\sum_{n=2}^\infty a_n \lt \int_1^\infty a(x)\, dx. + Note how this summation starts with n=2; + adding a_1 to both sides lets us rewrite the summation starting with n=1: + + \infser a_n \lt a_1 +\int_1^\infty a(x)\, dx + . +

    + +

    + Combining Equations and , we have + + \infser a_n\lt a_1 +\int_1^\infty a(x)\, dx \lt a_1 + \infser a_n + . +

    + +

    + From Equation we can make the following two statements: +

    + +

    +

      +
    1. +

      + If \ds \infser a_n diverges, + so does \ds\int_1^\infty a(x)\, dx (because \ds \infser a_n \lt a_1 +\int_1^\infty a(x)\, dx) +

      +
      2. + +
    2. +

      + If \ds \infser a_n converges, + so does \ds\int_1^\infty a(x)\, dx (because \ds \ds \int_1^\infty a(x)\, dx \lt \infser a_n.) +

      +
      3. +
    +

    + +

    + Therefore the series and integral either both converge or both diverge. + + allows us to extend this theorem to series where a(n) is positive and decreasing on [b,\infty) for some b \gt 1. + A formal proof of the is shown below. +

    + + + Proof of the Integral Test +

    + Let a(x)=a_x be a postive, + continuous, decreasing function on [1,\infty). + We will consider how the partial sums of + \infser a_n compare to the integral \int_0^\infty a(x)\, dx . + We first consider the case where \int_1^{\infty}a(x)\, dx diverges. +

    + +

    +

      +
    1. +

      + Suppose that \int_1^{\infty}a(x)\, dx diverges. + Using , + we can say that S_n=\sum_{i=1}^{n}a_i\gt \int_1^{n+1}a(x)\, dx. + If we let n \to \infty in this inequality, + we know that \int_1^{n+1}a(x)\, dx will get arbitrarily large as n \to \infty (since + a(x) \gt 0 and \int_1^{\infty}a(x)\, dx diverges). + Therefore we conclude that + S_n=\sum_{i=1}^{n}a_i will also get arbitrarily large as n \to \infty, + and thus \infser a_n diverges. +

      +
      2. + +
    2. +

      + Now suppose that \int_1^{\infty}a(x)\, dx converges to M, + where M is some positive, finite number. + Using , + we can say that 0 \lt S_n=\sum_{i=1}^{n}a_i \lt \int_1^{\infty} a(x)\, dx=M. + Therefore our sequence of partial sums, S_n is bounded. + Furthermore, + S_n is a monotonically increasing sequence since all of the terms a_n are positive. + Since S_n is both bounded and monotonic, + S_n converges by and by , + the series \infser a_n converges as well. +

      +
      3. +
    +

    +     + + + Using the Integral Test + +

    + Determine the convergence of + \ds\infser \frac{\ln(n) }{n^2}. (The terms of the sequence + \{a_n\} = \{\ln(n) /n^2\} and the nth partial sums are given in .) +

    +     + +

    + + implies that a(n) = (\ln(n) )/n^2 is positive and decreasing on [2,\infty). + We can determine this analytically, too. + We know a(n) is positive as both \ln(n) and n^2 are positive on [2,\infty). + Treating a(n) as a continuous function of n defined on [1, \infty), + consider a'(n) = (1-2\ln(n) )/n^3, + which is negative for n\geq 2. + Since a'(n) is negative, + a(n) is decreasing for n\geq 2. + We can still use the integral test since a finite number of terms will not affect convergence of the series. +

    + +
    + Plotting the sequence and series in + + + Scatter plots of the sequence used in this example, and the corresponding sequence of partial sums. + +

    + A scatter plot of the sequence a_n=\ln(n)/n^2 is shown. + Aside from an initial point (1,0), this is a decreasing sequence, + descending from the point (2,\ln(2)/4) toward the n axis. +

    + +

    + The scatter plot for the sequence S_n of partial sums is also shown; + this is an increasing sequence, which appears to be bounded above by some value less than 1. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={2,4,...,20}, + ymin=-.05,ymax=.85, + xmin=-1,xmax=20.9, + xlabel={$n$} + ] + + \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt}] coordinates {(1.,0)(2.,0.1733)(3.,0.2954)(4.,0.382)(5.,0.4464)(6.,0.4961)(7.,0.5359)(8.,0.5684)(9.,0.5955)(10.,0.6185)(11.,0.6383)(12.,0.6556)(13.,0.6708)(14.,0.6842)(15.,0.6963)(16.,0.7071)(17.,0.7169)(18.,0.7258)(19.,0.734)(20.,0.7415)}; + \addplot [only marks,firstcolor,mark size={2.4pt}] coordinates {(1.,0)(2.,0.1733)(3.,0.1221)(4.,0.08664)(5.,0.06438)(6.,0.04977)(7.,0.03971)(8.,0.03249)(9.,0.02713)(10.,0.02303)(11.,0.01982)(12.,0.01726)(13.,0.01518)(14.,0.01346)(15.,0.01204)(16.,0.01083)(17.,0.009804)(18.,0.008921)(19.,0.008156)(20.,0.007489)}; + + \end{axis} + + \node[shift={(0,-20pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
    + +

    + Applying the Integral Test, + we test the convergence of \ds \int_1^\infty \frac{\ln(x) }{x^2}\, dx. + Integrating this improper integral requires the use of Integration by Parts, + with u = \ln(x) and dv = 1/x^2\, dx. + + \int_1^\infty \frac{\ln(x) }{x^2}\, dx \amp = \lim_{b\to\infty} \int_1^b \frac{\ln(x) }{x^2}\, dx + \amp = \lim_{b\to\infty} -\frac1x\ln(x) \Big|_1^b + \int_1^b\frac1{x^2}\, dx + \amp = \lim_{b\to\infty} -\frac1x\ln(x) -\frac 1x\Big|_1^b + \amp = \lim_{b\to\infty}1-\frac1b-\frac{\ln(b) }{b}. \text{ Apply L'Hospital's Rule: } + \amp = 1 + . +

    + +

    + Since \ds \int_1^\infty \frac{\ln(x) }{x^2}\, dx converges, + so does \ds \infser \frac{\ln(n) }{n^2}. +

    +     + +     + + + +

    + + was given without justification, + stating that the general p-series + \ds \infser \frac 1{(an+b)^p} converges if, and only if, + p \gt 1. + In the following example, + we prove this to be true by applying the Integral Test. +

    + + + Using the Integral Test to establish <xref ref="thm_pseries"/> + +

    + Let a, b be real numbers such that a\neq 0 and an+b\gt 0 for all n\geq 1. + Use the Integral Test to prove that + \ds \infser \frac1{(an+b)^p} converges if, and only if, + p \gt 1. +

    +     + +

    + Consider the integral \ds\int_1^\infty \frac1{(ax+b)^p}\, dx; + assuming p\neq 1 and a\neq 0, + + \int_1^\infty \frac1{(ax+b)^p}\, dx \amp = \lim_{c\to\infty} \int_1^c \frac1{(ax+b)^p}\, dx + \amp = \lim_{c\to\infty} \frac{1}{a(1-p)}(ax+b)^{1-p}\Big|_1^c + \amp = \lim_{c\to\infty} \frac{1}{a(1-p)}\big((ac+b)^{1-p}-(a+b)^{1-p}\big) + . +

    + +

    + This limit converges if, + and only if, p \gt 1 so that 1-p \lt 0. + It is easy to show that the integral also diverges in the case of p=1. + (This result is similar to the work preceding .) +

    + +

    + Therefore \ds \infser \frac 1{(an+b)^p} converges if, + and only if, + p \gt 1. +

    +     + +     + +

    + We consider two more convergence tests in this section, + both comparison tests. + That is, we determine the convergence of one series by comparing it to another series with known convergence. +

    +     + + + Direct Comparison Test + + Direct Comparison Test + +

    + Let \{a_n\} and \{b_n\} be positive sequences where + a_n\leq b_n for all n\geq N, for some N\geq 1. + seriesDirect Comparison Test + Direct Comparison Testfor series + convergenceDirect Comparison Test + divergenceDirect Comparison Test +

    + +

    +

      +
    1. +

      + If \ds \infser b_n converges, + then \ds \infser a_n converges. +

      +
      2. + +
    2. +

      + If \ds \infser a_n diverges, + then \ds \infser b_n diverges. +

      +
      3. +
    +

    +     + +

    + Let 0\lt a_n\leq b_n for all n\geq N \geq 1. + Note that both partial sums for both series are positive and increasing since the terms of both sequences are positive. +

    + +

    +

      +
    1. +

      + Suppose that \ds \infser b_n converges, + so \ds \infser b_n=S, where S is a finite, + positive number. + (S must be positive since b_n \gt 0.) +

      + +

      + Comparing the partial sums, + we must have \ds \sum_{i=N}^n a_i \leq \sum_{i=N}^n b_i since + a_n\leq b_n for all n\geq N. + Furthermore since \ds \infser b_n converges to S, + our partial sums for a_n are bounded + (note that the partial sums started at i=N, + but a finite number of terms will not affect the boundedness of the partial sums). + + 0\lt \sum_{i=N}^n a_i \leq \sum_{i=N}^n b_i \lt S + . + Since the sequence of partial sums, + s_n=\sum_{i=1}^n a_i is both monotonically increasing and bounded, + we can say that s_n converges + (by ), + and therefore so does \infser a_n . +

      +
      2. + +
    2. +

      + Suppose that \ds \infser a_n diverges, + so \ds \sum_{i=1}^n a_n=\infty. (We can say that the series diverges to \infty since the terms of the series are always positive). + Comparing the partial sums, we have + + \ds \sum_{i=N}^n a_i \leq \sum_{i=N}^n b_i + + Then applying limits, we get + + \ds \lim\limits_{n \to \infty}\sum_{i=N}^n a_i \leq \lim\limits_{n \to \infty}\sum_{i=N}^n b_i + . + Since the limit on the left side diverges to \infty, + we can say that \lim\limits_{n \to \infty}\sum_{i=N}^n b_i also diverges to \infty. +

      +
      3. +
    +

    +     + +     + + + + + + + Applying the Direct Comparison Test + +

    + Determine the convergence of \ds\infser \frac1{3^n+n^2}. +

    +     + +

    + This series is neither a geometric or p-series, + but seems related. + We predict it will converge, + so we look for a series with larger terms that converges. + (Note too that the Integral Test seems difficult to apply here.) +

    + +

    + Since 3^n \lt 3^n+n^2, + \ds \frac1{3^n} \gt \frac1{3^n+n^2} for all n\geq1. + The series \ds\infser \frac{1}{3^n} is a convergent geometric series; + by , + \ds \infser \frac1{3^n+n^2} converges. +

    +     + +     + + + Applying the Direct Comparison Test + +

    + Determine the convergence of \ds\infser \frac{1}{n-\ln(n) }. +

    +     + +

    + We know the Harmonic Series \ds\infser \frac1n diverges, + and it seems that the given series is closely related to it, + hence we predict it will diverge. +

    + +

    + Since n\geq n-\ln(n) for all n\geq 1, + \ds \frac1n \leq \frac1{n-\ln(n) } for all n\geq 1. +

    + +

    + The Harmonic Series diverges, + so we conclude that \ds\infser \frac{1}{n-\ln(n) } diverges as well. +

    +     + +     + + + +

    + The concept of direct comparison is powerful and often relatively easy to apply. + Practice helps one develop the necessary intuition to quickly pick a proper series with which to compare. + However, it is easy to construct a series for which it is difficult to apply the Direct Comparison Test. +

    + +

    + Consider \ds\infser \frac1{n+\ln(n) }. + It is very similar to the divergent series given in . + We suspect that it also diverges, + as \ds \frac 1n \approx \frac1{n+\ln(n) } for large n. + However, the inequality that we naturally want to use + goes the wrong way: + since n\leq n+\ln(n) for all n\geq 1, + \ds\frac1n \geq \frac{1}{n+\ln(n) } for all n\geq 1. + The given series has terms less than + the terms of a divergent series, + and we cannot conclude anything from this. +

    + +

    + Fortunately, + we can apply another test to the given series to determine its convergence. +

    + + +     + + + Limit Comparison Test + + Limit Comparison Test + +

    + Let \{a_n\} and \{b_n\} be positive sequences. + seriesLimit Comparison Test + Limit Comparison Testfor series + convergenceLimit Comparison Test + divergenceLimit Comparison Test +

    + +

    +

      +
    1. +

      + If \lim\limits_{n\to\infty} \dfrac{a_n}{b_n} = L, + where L is a positive real number, + then \ds \infser a_n and + \ds \infser b_n either both converge or both diverge. +

      +
      2. + +
    2. +

      + If \lim\limits_{n\to\infty} \dfrac{a_n}{b_n} = 0, + then if \ds \infser b_n converges, + then so does \ds \infser a_n. +

      +
      3. + +
    3. +

      + If \lim\limits_{n\to\infty} \dfrac{a_n}{b_n} = \infty, + then if \ds \infser b_n diverges, + then so does \ds \infser a_n. +

      +
      4. +
    +

    +     +     + +

    + + is most useful when the convergence of the series from \{b_n\} is known and we are trying to determine the convergence of the series from \{a_n\}. +

    + + + +

    + We use the Limit Comparison Test in the next example to examine the series + \ds\infser \frac1{n+\ln(n) } which motivated this new test. +

    + + + Applying the Limit Comparison Test + +

    + Determine the convergence of + \ds\infser \frac1{n+\ln(n) } using the Limit Comparison Test. +

    +     + +

    + We compare the terms of \ds\infser \frac1{n+\ln(n) } to the terms of the Harmonic Sequence \ds\infser \frac1{n}: + + \lim_{n\to\infty}\frac{1/(n+\ln(n) )}{1/n} \amp = \lim_{n\to\infty} \frac{n}{n+\ln(n) } + \amp = 1 \text{ (after applying L'Hospital's Rule) } + . +

    + +

    + Since the Harmonic Series diverges, + we conclude that \ds\infser \frac1{n+\ln(n) } diverges as well. +

    +     + +     + + + Applying the Limit Comparison Test + +

    + Determine the convergence of \ds\infser \frac1{3^n-n^2} +

    +     + +

    + This series is similar to the one in , + but now we are considering 3^n-n^2 + instead of 3^n+n^2. + This difference makes applying the Direct Comparison Test difficult. +

    + +

    + Instead, we use the Limit Comparison Test and compare with the series \ds\infser \frac1{3^n}: + + \lim_{n\to\infty}\frac{1/(3^n-n^2)}{1/3^n} \amp = \lim_{n\to\infty}\frac{3^n}{3^n-n^2} + \amp = 1 \text{ (after applying L'Hospital's Rule twice) } + . +

    + +

    + We know \ds\infser \frac1{3^n} is a convergent geometric series, + hence \ds\infser \frac1{3^n-n^2} converges as well. +

    +     + +     + +

    + As mentioned before, + practice helps one develop the intuition to quickly choose a series with which to compare. + A general rule of thumb is to pick a series based on the dominant term in the expression of \{a_n\}. + It is also helpful to note that factorials dominate exponentials, + which dominate algebraic functions (e.g., polynomials), + which dominate logarithms. + In the previous example, + the dominant term of \ds\frac{1}{3^n-n^2} was 3^n, + so we compared the series to \ds \infser \frac1{3^n}. + It is hard to apply the Limit Comparison Test to series containing factorials, though, + as we have not learned how to apply L'Hospital's Rule to n!. +

    + + + + + Applying the Limit Comparison Test + +

    + Determine the convergence of \ds\infser \frac{\sqrt{n}+3}{n^2-n+1}. +

    +     + +

    + We naïvely attempt to apply the rule of thumb given above and note that the dominant term in the expression of the series is 1/n^2. + Knowing that \ds \infser \frac1{n^2} converges, + we attempt to apply the Limit Comparison Test: + + \lim_{n\to\infty}\frac{(\sqrt{n}+3)/(n^2-n+1)}{1/n^2} \amp = \lim_{n\to\infty}\frac{n^2(\sqrt n+3)}{n^2-n+1} + \amp = \infty \text{ (Apply L'Hospital's Rule) } + . +

    + +

    + + part (3) only applies when \ds\infser b_n diverges; + in our case, it converges. + Ultimately, our test has not revealed anything about the convergence of our series. +

    + +

    + The problem is that we chose a poor series with which to compare. + Since the numerator and denominator of the terms of the series are both algebraic functions, + we should have compared our series to the dominant term of the numerator divided by the dominant term of the denominator. +

    + +

    + The dominant term of the numerator is n^{1/2} and the dominant term of the denominator is n^2. + Thus we should compare the terms of the given series to n^{1/2}/n^2 = 1/n^{3/2}: + + \lim_{n\to\infty}\frac{(\sqrt{n}+3)/(n^2-n+1)}{1/n^{3/2}} \amp = \lim_{n\to \infty} \frac{n^{3/2}(\sqrt n+3)}{n^2-n+1} + \amp = 1 \text{ (Apply L'Hospital's Rule) } + . +

    + +

    + Since the p-series \ds\infser \frac1{n^{3/2}} converges, + we conclude that \ds\infser \frac{\sqrt{n}+3}{n^2-n+1} converges as well. +

    +     + +     + +

    + We mentioned earlier that the Integral Test did not work well with series containing factorial terms. + The next section introduces the Ratio Test, + which does handle such series well. + We also introduce the Root Test, + which is good for series where each term is raised to a power. +

    +     + + + + Terms and Concepts + + + +

    + In order to apply the Integral Test to a sequence \{a_n\}, + the function a(n) = a_n must be + , , + and . +

    +     + + + + + + ["continuous","positive","decreasing"].includes(ans) + + + + + + + ["continuous","positive","decreasing"].includes(ans) && !ans_array.slice(0,1).includes(ans) + + + + + ans_array.slice(0,1).includes(ans) + + You already gave that answer. + + + + + + ["continuous","positive","decreasing"].includes(ans) && !ans_array.slice(0,2).includes(ans) + + + + + ans_array.slice(0,2).includes(ans) + + You already gave that answer. + + + + +     + + + + +

    + The Integral Test can be used to determine the sum of a convergent series. +

    +     + +     + + + + +

    + What test(s) in this section do not work well with factorials? +

    +     + + + +

    + The Integral Test + (we do not have a continuous definition of n! yet) + and the Limit Comparison Test + (same as above, hence we cannot take its derivative). +

    +     + +     + + + + +

    + Suppose \ds \infser[0] a_n is convergent, + and there are sequences \{b_n\} and \{c_n\} such that b_n \leq a_n \leq c_n for all n. + What can be said about the series + \ds \infser[0] b_n and \ds \infser[0] c_n? +

    +     + + + +

    + \ds \infser[0] b_n converges; + we cannot conclude anything about \ds \infser[0] c_n +

    +     + +     +     + + + Problems + + + +

    + Use the Integral Test to determine the convergence of the given series. +

    +     + + + + +

    + \ds \infser \frac{1}{2^n} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds \infser \frac{1}{n^4} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds \infser \frac{n}{n^2+1} +

    +     + +

    + Diverges +

    +     + +     + + + + +

    + \ds \sum_{n=2}^\infty \frac{1}{n\ln(n) } +

    +     + +

    + Diverges +

    +     + +     + + + + +

    + \ds \infser \frac{1}{n^2+1} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds \sum_{n=2}^\infty \frac{1}{n(\ln(n) )^2} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds \infser \frac{n}{2^n} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds \infser \frac{\ln(n) }{n^3} +

    +     + +

    + Converges +

    +     + +     + +     + + + +

    + Use the Direct Comparison Test to determine the convergence of the given series; + state what series is used for comparison. +

    +     + + + + +

    + \ds \infser \frac{1}{n^2+3n-5} +

    +     + +

    + Converges; compare to \ds \infser \frac{1}{n^2}, + as 1/(n^2+3n-5) \leq 1/n^2 for all n \gt 1. +

    +     + +     + + + + +

    + \ds \infser \frac{1}{4^n+n^2-n} +

    +     + +

    + Converges; compare to \ds \infser \frac{1}{4^n}, + as 1/(4^n+n^2-n) \leq 1/4^n for all n\geq 1. +

    +     + +     + + + + +

    + \ds \infser \frac{\ln(n) }{n} +

    +     + +

    + Diverges; compare to \ds \infser \frac{1}{n}, + as 1/n \leq \ln(n) /n for all n\geq 3. +

    +     + +     + + + + +

    + \ds \infser \frac{1}{n!+n} +

    +     + +

    + Converges; compare to \ds \infser \frac{1}{n!}, + as 1/(n!+n) \leq 1/n! for all n\geq 1. +

    +     + +     + + + + +

    + \ds \sum_{n=2}^\infty \frac{1}{\sqrt{n^2-1}} +

    +     + +

    + Diverges; compare to \ds \infser \frac{1}{n}. + Since n=\sqrt{n^2} \gt \sqrt{n^2-1}, + 1/n \leq 1/\sqrt{n^2-1} for all n\geq 2. +

    +     + +     + + + + +

    + \ds \sum_{n=5}^\infty \frac{1}{\sqrt{n}-2} +

    +     + +

    + Diverges; compare to \ds \infser \frac{1}{\sqrt n}, + as 1/\sqrt n \leq 1/(\sqrt{n}-2) for all n\geq 5. +

    +     + +     + + + + +

    + \ds \infser \frac{n^2+n+1}{n^3-5} +

    +     + +

    + Diverges; compare to \ds \infser \frac{1}{n}: + + \frac 1n = \frac{n^2}{n^3} \lt \frac{n^2+n+1}{n^3} \lt \frac{n^2+n+1}{n^3-5} + , + for all n\geq 1. +

    +     + +     + + + + +

    + \ds \infser \frac{2^n}{5^n+10} +

    +     + +

    + Converges; compare to \ds \infser \left(\frac{2}{5}\right)^n, + as 2^n/(5^n+10) \lt 2^n/5^n for all n\geq 1. +

    +     + +     + + + + +

    + \ds \sum_{n=2}^\infty \frac{n}{n^2-1} +

    +     + +

    + Diverges; compare to \ds \infser \frac 1n. + Note that + + \frac{n}{n^2-1} = \frac{n^2}{n^2-1}\cdot\frac1n \gt \frac 1n + , + as \frac{n^2}{n^2-1} \gt 1, for all n\geq 2. +

    +     + +     + + + + +

    + \ds \sum_{n=2}^\infty \frac{1}{n^2\ln(n) } +

    +     + +

    + Converges; compare to \ds \infser \frac 1{n^2}, + as 1/(n^2\ln(n) ) \leq 1/n^2 for all n\geq 3. +

    +     + +     + +     + + + +

    + Use the Limit Comparison Test to determine the convergence of the given series; + state what series is used for comparison. +

    +     + + + + +

    + \ds \infser \frac{1}{n^2 -3n+5} +

    +     + +

    + Converges; compare to \ds \infser \frac 1{n^2}. +

    +     + +     + + + + +

    + \ds \infser \frac{1}{4^n-n^2} +

    +     + +

    + Converges; compare to \ds \infser \frac 1{4^n}. +

    +     + +     + + + + +

    + \ds \sum_{n=4}^\infty \frac{\ln(n) }{n-3} +

    +     + +

    + Diverges; compare to \ds \infser \frac {\ln(n) }{n}. +

    +     + +     + + + + +

    + \ds \infser \frac{1}{\sqrt{n^2+n}} +

    +     + +

    + Diverges; compare to \ds \infser \frac {1}{n}. +

    +     + +     + + + + +

    + \ds \infser \frac{1}{n+\sqrt{n}} +

    +     + +

    + Diverges; compare to \ds \infser \frac {1}{n}. +

    +     + +     + + + + +

    + \ds \infser \frac{n-10}{n^2+10n+10} +

    +     + +

    + Diverges; compare to \ds \infser \frac {1}{n}. +

    +     + +     + + + + +

    + \ds \infser \sin\big(1/n\big) +

    +     + +

    + Diverges; compare to \ds \infser \frac {1}{n}. + Just as \lim\limits_{n\to0}\frac{\sin(n) }{n} = 1, + \lim\limits_{n\to\infty}\frac{\sin(1/n)}{1/n} = 1. +

    +     + +     + + + + +

    + \ds \infser \frac{n+5}{n^3-5} +

    +     + +

    + Converges; compare to \ds \infser \frac {1}{n^2}. +

    +     + +     + + + + +

    + \ds \infser \frac{\sqrt{n}+3}{n^2+17} +

    +     + +

    + Converges; compare to \ds \infser \frac {1}{n^{3/2}}. +

    +     + +     + + + + +

    + \ds \infser \frac{1}{\sqrt{n}+100} +

    +     + +

    + Diverges; compare to \ds \infser \frac {1}{n^{1/2}}. +

    +     + +     + +     + + + +

    + Determine the convergence of the given series. + State the test used; more than one test may be appropriate. +

    +     + + + + +

    + \ds \infser \frac{n^2}{2^n} +

    +     + +

    + Converges; Integral Test +

    +     + +     + + + + +

    + \ds \infser \frac{1}{(2n+5)^3} +

    +     + +

    + Converges; Integral Test, + p-Series Test, Direct & Limit Comparison Tests can all be used. +

    +     + +     + + + + +

    + \ds \infser \frac{n!}{10^n} +

    +     + +

    + Diverges; the nth Term Test and Direct Comparison Test can be used. +

    +     + +     + + + + +

    + \ds \infser \frac{\ln(n) }{n!} +

    +     + +

    + Converges; the Direct Comparison Test can be used with sequence 1/(n-1)!. +

    +     + +     + + + + +

    + \ds \infser \frac{1}{3^n+n} +

    +     + +

    + Converges; the Direct Comparison Test can be used with sequence 1/3^n. +

    +     + +     + + + + +

    + \ds \infser \frac{n-2}{10n+5} +

    +     + +

    + Diverges; the nth Term Test can be used, + along with the Limit Comparison Test + (compare with 1/10). +

    +     + +     + + + + +

    + \ds \infser \frac{3^n}{n^3} +

    +     + +

    + Diverges; the nth Term Test can be used, + along with the Integral Test. +

    +     + +     + + + + +

    + \ds \infser \frac{\cos(1/n)}{\sqrt{n}} +

    +     + +

    + Diverges; the Limit Comparison Test can be used with sequence 1/\sqrt{n}. +

    +     + +     +     + + + +

    + Given that \ds \infser a_n converges, + state which of the following series converges, + may converge, or does not converge. +

    +     + + + +

    + \ds \infser \frac{a_n}n +

    +     + +

    + Converges; use Direct Comparison Test as \frac{a_n}{n}\lt n. +

    +     +     + + + +

    + \ds \infser a_n a_{n+1} +

    +     + +

    + Converges; since original series converges, + we know \lim_{n\to\infty}a_n = 0. + Thus for large n, a_na_{n+1} \lt a_n. +

    +     +     + + + +

    + \ds \infser (a_n)^2 +

    +     + +

    + Converges; similar logic to so (a_n)^2\lt a_n. +

    +     +     + + + +

    + \ds \infser na_n +

    +     + +

    + May converge; + certainly na_n \gt a_n but that does not mean it does not converge. +

    +     +     + + + +

    + \ds \infser \frac{1}{a_n} +

    +     + +

    + Does not converge, using logic from part and n^{th} Term Test. +

    +     +     +     + + + +

    + In this exercise, we explore an approximation method for series to which the + applies. +

    +     + + +

    + Let a(x) be a function to which the + applies, + and for which the series \infser a_n converges. +

    +

    + Let R_n = \sum_{n+1}^\infty a_n denote the remainder; + that is, the difference between \infser a_n and the nth partial sum. + (Note that R_n is the size of the error that results if we approximate the series by the nth partial sum.) + Explain why we must have the following inequality: + + \int_n^\infty a(x)\,dx \leq R_n \leq \int_{n+1}^\infty a(x)\,dx + +

    +     + +

    + Sketch a graph to represent y=a(x). What property must a(x) have? + See if you can interpret R_n in terms of left and right handed Riemann sums. +

    +     + +

    + Since the Integral Test applies, we know that a(x) must be a decreasing function. + The sum R_n = a_{n+1}+a_{n+2}+\cdots can be viewed as a right-endpoint Riemann sum + for the integral \int_n^\infty a(x)\,dx, and this must be an underestimate, + since a(x) is a decreasing function. + But R_n can also be viewed as a left-endpoint Riemann sum for the integral + \int_{n+1}^\infty a(x)\,dx, and this must be an overestimate. +

    +     +     + + +

    + Estimate the error involved in using the first 12 terms to approximate the series \sum_{n=1}^\infty 1/n^4. + What is the approximate value of the series? +

    +     + +

    + For the first 12 terms, we take n=12. + We find: + + \int_{12}^\infty \frac{1}{x^4}\,dx \amp = \frac{1}{5184} \approx 0.000193 + \int_{13}^\infty \frac{1}{x^4}\,dx \amp = \frac{1}{6591} \approx 0.000152 + \sum_{1}^{12}\frac{1}{n^4} \amp \approx 1.082153 + . +

    +     +     + + +

    + How many terms must we take to ensure that the nth partial sum approximation for + \infser 1/n^4 is accurate to 5 decimal places? +

    +     + +

    + We have + + R_n\leq \int_n^\infty \frac{1}{x^4}\,dx = \frac{1}{3n^3} + . + For 5 decimal places of accuracy, we need R_n\leq 0.000005. + Setting 0.000005=1/(3n^3) and solving for n gives + n\approx 40.55, so we should take 41 terms. +

    +     +     +     +     +     +
    +
    + Ratio and Root Tests + +

    + The nth-Term Test of + states that in order for a series \ds \infser a_n to converge, + \lim\limits_{n\to\infty}a_n = 0. + That is, the terms of \{a_n\} must get very small. + Not only must the terms approach 0, they must approach 0 fast enough: + while \lim\limits_{n\to\infty}1/n=0, + the Harmonic Series \ds\infser \frac1n diverges as the terms of \{1/n\} do not approach 0 fast enough. +

    + +

    + The comparison tests of the previous section determine convergence by comparing terms of a series to terms of another series whose convergence is known. + This section introduces the Ratio and Root Tests, + which determine convergence by analyzing the terms of a series to see if they approach 0 fast enough. +

    +     + + + Ratio Test + + Ratio Test + +

    + Let \{a_n\} be a positive sequence and consider \lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}. + seriesRatio Comparison Test + Ratio Comparison Testfor series + convergenceRatio Comparison Test + divergenceRatio Comparison Test +

    + +

    +

      +
    1. +

      + If \lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}\lt 1, + then \ds\infser a_n converges. +

      +
      2. + +
    2. +

      + If \lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n} \gt 1 or \lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=\infty, + then \ds\infser a_n diverges. +

      +
      3. + +
    3. +

      + If \lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}=1, + the Ratio Test is inconclusive. +

      +
      4. +
    +

    +     +     + + + + +

    + The principle of the Ratio Test is this: + if \lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n} = L\lt 1, + then for large n, + each term of \{a_n\} is significantly smaller than its previous term which is enough to ensure convergence. +

    + + + Applying the Ratio Test + +

    + Use the Ratio Test to determine the convergence of the following series: +

    + +

    +

      +
    1. +

      + \ds \infser \frac{2^n}{n!} +

      +
      2. + +
    2. \ds \infser \frac{3^n}{n^3}
      3. + +
    3. \ds \infser \frac{1}{n^2+1}
      4. +
    +

    +     + +

    +

      +
    1. +

      + \ds \infser \frac{2^n}{n!}: + + \inflim{\frac{a_{n+1}}{a_n}} \amp = \lim_{n\to\infty}\frac{2^{n+1}/(n+1)!}{2^n/n!} + \amp = \lim_{n\to\infty} \frac{2^{n+1}n!}{2^n(n+1)!} + \amp = \lim_{n\to\infty} \frac{2}{n+1} + \amp =0 + . + Since the limit is 0\lt 1, + by the Ratio Test \ds\infser \frac{2^n}{n!} converges. + The fact that \inflim{\frac{a_{n+1}}{a_n}}=0 can be interpreted to mean that in the long run, + the term a_{n+1} is roughly 0 times as large as a_n. + In other words, not only is a_n decreasing to 0, + it is decreasing very quickly. + That is, the terms of a_n decrease to 0 sufficiently fast enough to guarantee the convergence of \infser a_n. +

      +
      2. + +
    2. +

      + \ds\infser \frac{3^n}{n^3}: + + \inflim{\frac{a_{n+1}}{a_n}} \amp = \lim_{n\to\infty} \frac{3^{n+1}/(n+1)^3}{3^n/n^3} + \amp = \lim_{n\to\infty}\frac{3^{n+1}n^3}{3^n(n+1)^3} + \amp = \lim_{n\to\infty} \frac{3n^3}{(n+1)^3} + \amp = 3 + . + Since the limit is 3 \gt 1, + by the Ratio Test \ds\infser \frac{3^n}{n^3} diverges. + The fact that \inflim{\frac{a_{n+1}}{a_n}}=3 can be interpreted to mean that in the long run, + the term a_{n+1} is roughly 3 times as large as a_n, + so a_n is increasing + by roughly a factor of 3 in the long run. + We could also use + to determine that this series diverges. + The exponential will dominate the polynomial in the long run, + so \inflim 3^n/n^3=\infty. +

      +
      3. + +
    3. +

      + \infser\frac{1}{n^2+1}: + + \inflim\frac{a_{n+1}}{a_n} \amp = \lim_{n\to\infty} \frac{1/\big((n+1)^2+1\big)}{1/(n^2+1)} + \amp = \lim_{n\to\infty} \frac{n^2+1}{(n+1)^2+1} + \amp = 1 + . + Since the limit is 1, the Ratio Test is inconclusive. + We can easily show this series converges using the . + We can also use or , + with each comparing to the series \ds \infser \frac{1}{n^2}. +

      +
      4. +
    +

    +     + +     + + + +

    + The Ratio Test is not effective when the terms of a series only + contain algebraic functions (e.g., polynomials). + It is most effective when the terms contain some factorials or exponentials. + The previous example also reinforces our developing intuition: + factorials dominate exponentials, + which dominate algebraic functions, + which dominate logarithmic functions. + In Part 1 of the example, + the factorial in the denominator dominated the exponential in the numerator, + causing the series to converge. + In Part 2, the exponential in the numerator dominated the algebraic function in the denominator, + causing the series to diverge. +

    + +

    + While we have used factorials in previous sections, + we have not explored them closely and one is likely to not yet have a strong intuitive sense for how they behave. + The following example gives more practice with factorials. +

    + + + Applying the Ratio Test + +

    + Determine the convergence of \ds\infser \frac{n!n!}{(2n)!}. +

    +     + +

    + Before we begin, + be sure to note the difference between (2n)! and 2n!. + When n=4, + the former is 8!=8\cdot7\cdot\ldots\cdot 2\cdot1=40,320, + whereas the latter is 2(4\cdot3\cdot2\cdot1) = 48. +

    + +

    + Applying the Ratio Test: + + \inflim\frac{a_{n+1}}{a_n} \amp = \lim_{n\to\infty} \frac{(n+1)!(n+1)!/\big(2(n+1)\big)!}{n!n!/(2n)!} + \amp = \lim_{n\to\infty}\frac{(n+1)!(n+1)!(2n)!}{n!n!(2n+2)!} + Noting that (n+1)!=(n+1)\cdot n! and (2n+2)! = (2n+2)\cdot(2n+1)\cdot(2n)!, we have + \amp = \lim_{n\to\infty}\frac{(n+1)(n+1)}{(2n+2)(2n+1)} + \amp = 1/4 + . +

    + +

    + Since the limit is 1/4\lt 1, + by the Ratio Test we conclude \ds \infser \frac{n!n!}{(2n)!} converges. +

    + +

    + To find the limit in the second to last line, + recall that we just need to examine the leading terms of the numerator and denominator, + which are n^2 and 4n^2 respectively. +

    +     + +     +     + + + Root Test +

    + The final test we introduce is the Root Test, + which works particularly well on series where each term is raised to a power, + and does not work well with terms containing factorials. +

    + + + + + Root Test + +

    + Let \{a_n\} be a positive sequence, + and consider \lim\limits_{n\to \infty} (a_n)^{1/n}. + seriesRoot Comparison Test + Root Comparison Testfor series + convergenceRoot Comparison Test + divergenceRoot Comparison Test +

    + +

    +

      +
    1. +

      + If \lim\limits_{n\to \infty} (a_n)^{1/n}\lt 1, + then \ds\infser a_n converges. +

      +
      2. + +
    2. +

      + If \lim\limits_{n\to \infty} (a_n)^{1/n} \gt1 or \lim\limits_{n\to \infty} (a_n)^{1/n}=\infty, + then \ds\infser a_n diverges. +

      +
      3. + +
    3. +

      + If \lim\limits_{n\to \infty} (a_n)^{1/n}=1, + the Root Test is inconclusive. +

      +
      4. +
    +

    +     +     + + + Applying the Root Test + +

    + Determine the convergence of the following series using the Root Test: +

    + +

    +

      +
    1. +

      + \ds \infser \left(\frac{3n+1}{5n-2}\right)^n +

      +
      2. + +
    2. +

      + \ds \infser\frac{n^4}{(\ln(n) )^n} +

      +
      3. + +
    3. +

      + \ds \infser \frac{2^n}{n^2} +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + + \inflim \left(a_n\right)^{1/n} \amp = \inflim \left(\left(\frac{3n+1}{5n-2}\right)^n\right)^{1/n} + \amp =\inflim \frac{3n+1}{5n-2} = \frac 35 + . + Since the limit is less than 1, we conclude the series converges. + Note: it is difficult to apply the Ratio Test to this series. +

      +
      2. + +
    2. + +

      + + \inflim \left(a_n \right)^{1/n} \amp =\inflim \left(\frac{n^4}{(\ln(n))^n}\right)^{1/n} + \amp = \inflim \frac{\big(n^{4/n}\big)}{\ln(n)} + The limit of the numerator must be found using L'Hospital's Rule for indeterminate powers + \inflim \left(n^{4/n}\right) \amp = \inflim e^{\ln\left(n^{4/n}\right)} + \amp = \inflim e^{{4\ln\left(n\right)}/n} + Now apply L'Hospital's to the expression in the exponent: + \amp \stackrel{\, \text{ by LHR } \, }{=} \inflim e^{4/n} + \amp = e^0=1 + . + Since the numerator approaches 1 + (by L'Hospital's Rule) + and the denominator + grows to infinity, we have + + \inflim \frac{\big(n^{4/n}\big)}{\ln(n)} =0 + . + Since the limit is less than 1, we conclude the series converges. +

      +
      3. + +
    3. +

      + \inflim \left(\frac{2^n}{n^2}\right)^{1/n} = \inflim \frac{2}{\big(n^{2/n}\big)} = 2. + + Since this is greater than 1, we conclude the series diverges. (Note: The is easy to apply to this series.) +

      + +

      + (Also note: The limit in the denominator is found in a similar fashion as was illustrated in Part. + In general \inflim (n)^{b/n}=1 for any real number b.) +

      +
      4. +
    +

    +     + +     + + + + + +

    + Each of the tests we have encountered so far has required that we analyze series from + positive sequences. + + relaxes this restriction by considering + alternating series, + where the underlying sequence has terms that alternate between being positive and negative. +

    + + +     + + + + Terms and Concepts + + + +

    + The Ratio Test is not effective when the terms of a sequence only contain functions. +

    +     + + + + polynomial|algebraic|rational + + +

    + polynomial, rational, or more generally, algebraic are all good answers here. +

    +     +     +     + +     + + + + +

    + The Ratio Test is most effective when the terms of a sequence contains + and/or functions. +

    +     + + + + + + ["factorial","exponential"].includes(ans) + + + + + + + ["factorial","exponential"].includes(ans) && !ans_array.slice(0,1).includes(ans) + + + + + ans_array.slice(0,1).includes(ans) + + You already gave that answer. + + + + +     + + + + +

    + What three convergence tests do not work well with terms containing factorials? +

    +     + + + +

    + Test for divergence +

    +     +     + + +

    + Comparison test +

    +     +     + + +

    + Limit comparison test +

    +     +     + + +

    + Integral test +

    +     +     + + +

    + Ratio test +

    +     +     + + +

    + Root test +

    +     +     +     + +

    + Integral Test, Limit Comparison Test, and Root Test +

    +     + +     + + + + +

    + The Root Test works particularly well on series where each term is . +

    +     + + + + a power|a power function|raised to a power + + + + +

    + Your answer includes the correct words but has extra or missing text. +

    +     +     +     +     + + +     +     + + Problems + + + +

    + Determine the convergence of the given series using the Ratio Test. + If the Ratio Test is inconclusive, + state so and determine convergence with another test. +

    +     + + + + +

    + \ds\infser[0] \frac{2n}{n!} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser[0] \frac{5^n-3n}{4^n} +

    +     + +

    + Diverges +

    +     + +     + + + + +

    + \ds\infser[0] \frac{n!10^n}{(2n)!} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser \frac{5^n+n^4}{7^n+n^2} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser \frac{1}{n} +

    +     + +

    + The Ratio Test is inconclusive; + the p-Series Test states it diverges. +

    +     + +     + + + + +

    + \ds\infser \frac{1}{3n^3+7} +

    +     + +

    + The Ratio Test is inconclusive; + the Direct Comparison Test with 1/n^3 shows it converges. +

    +     + +     + + + + +

    + \ds\infser \frac{10\cdot5^n}{7^n-3} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser n\cdot\left(\frac35\right)^n +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser \frac{2\cdot4\cdot6\cdot8\cdots 2n}{3\cdot6\cdot9\cdot12\cdots 3n} +

    +     + +

    + Converges; note the summation can be rewritten as \ds\infser \frac{2^nn!}{3^nn!}, + from which the Ratio Test or Geometric Series Test can be applied. +

    +     + +     + + + + +

    + \ds\infser \frac{n!}{5\cdot10\cdot15\cdots (5n)} +

    +     + +

    + Converges; rewrite the summation as + \ds\infser \frac{n!}{5^nn!} then apply the Ratio Test or Geometric Series Test. +

    +     + +     + +     + + + +

    + Determine the convergence of the given series using the Root Test. + If the Root Test is inconclusive, + state so and determine convergence with another test. +

    +     + + + + +

    + \ds\infser \left(\frac{2n+5}{3n+11}\right)^n +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser \left(\frac{0.9n^2-n-3}{n^2+n+3}\right)^n +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser \frac{2^nn^2}{3^n} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser \frac{1}{n^n} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser \frac{3^n}{n^22^{n+1}} +

    +     + +

    + Diverges +

    +     + +     + + + + +

    + \ds\infser \frac{4^{n+7}}{7^n} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser \left(\frac{n^2-n}{n^2+n}\right)^n +

    +     + +

    + Diverges. + The Root Test is inconclusive, + but the nth-Term Test shows divergence. + (The terms of the sequence approach e^{-2}, + not 0, as n\to\infty.) +

    +     + +     + + + + +

    + \ds\infser \left(\frac1n-\frac{1}{n^2}\right)^n +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser \frac1{\big(\ln(n) \big)^n} +

    +     + +

    + Converges +

    +     + +     + + + + +

    + \ds\infser \frac{n^2}{\big(\ln(n) \big)^n} +

    +     + +

    + Converges +

    +     + +     + +     + + + +

    + Determine the convergence of the given series. + State the test used; more than one test may be appropriate. +

    +     + + + + +

    + \ds\infser \frac{n^2+4n-2}{n^3+4n^2-3n+7} +

    +     + +

    + Diverges; Limit Comparison Test with the harmonic series 1/n. +

    +     + +     + + + + +

    + \ds\infser \frac{n^44^n}{n!} +

    +     + +

    + Converges; Ratio Test +

    +     + +     + + + + +

    + \ds\infser \frac{n^2}{3^n+n} +

    +     + +

    + Converges; Ratio Test or Limit Comparison Test with 1/3^n. +

    +     + +     + + + + +

    + \ds\infser \frac{3^n}{n^n} +

    +     + +

    + Converges; Root Test +

    +     + +     + + + + +

    + \ds\infser \frac{n}{\sqrt{n^2+4n+1}} +

    +     + +

    + Diverges; nth-Term Test or Limit Comparison Test with 1. +

    +     + +     + + + + +

    + \ds\infser \frac{n!n!n!}{(3n)!} +

    +     + +

    + Converges; Ratio Test +

    +     + +     + + + + +

    + \ds\infser[2] \frac{1}{\ln(n) } +

    +     + +

    + Diverges; Direct Comparison Test with 1/n +

    +     + +     + + + + +

    + \ds\infser \left(\frac{n+2}{n+1}\right)^n +

    +     + +

    + Diverges; nth-Term Test (nth term approaches e.) +

    +     + +     + + + + +

    + \ds\infser[2] \frac{n^3}{\big(\ln(n) \big)^n} +

    +     + +

    + Converges; Root Test +

    +     + +     + + + + +

    + \ds\infser \left(\frac1n-\frac1{n+2}\right) +

    +     + +

    + Converges; Limit Comparison Test with 1/n^2 + (get common denominator first). + It is also a Telescoping Series. +

    +     + +     + +     +     +     +
    +
    + Alternating Series and Absolute Convergence +

    + All of the series convergence tests we have used require that the underlying sequence \{a_n\} be a positive sequence. + (We can relax this with + and state that there must be an N \gt 0 such that a_n \gt 0 for all n \gt N; + that is, \{a_n\} is positive for all but a finite number of values of n.) +

    + +

    + In this section we explore series whose summation includes negative terms. + We start with a very specific form of series, + where the terms of the summation alternate between being positive and negative. +

    + + + Alternating Series + +

    + Let \{a_n\} be a positive sequence. + An alternating series + is a series of either the form + seriesalternating + + \infser (-1)^na_n\qquad \text{ or } \qquad \infser (-1)^{n+1}a_n + . +

    +     +     + +

    + Recall the terms of Harmonic Series come from the Harmonic Sequence \{a_n\} = \{1/n\}. + An important alternating series is the + Alternating Harmonic Series: + + \infser (-1)^{n+1}\frac1n = 1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots + +

    + +

    + Geometric Series can also be alternating series when r\lt 0. + For instance, if r=-1/2, the geometric series is + + \infser[0] \left(\frac{-1}{2}\right)^n = 1-\frac12+\frac14-\frac18+\frac1{16}-\frac1{32}+\cdots + +

    + +

    + + states that geometric series converge when \abs{r}\lt 1 and gives the sum: + \ds \infser[0] r^n = \frac1{1-r}. + When r=-1/2 as above, we find + + \infser[0] \left(\frac{-1}{2}\right)^n = \frac1{1-(-1/2)} = \frac 1{3/2} = \frac23 + . +

    + +

    + A powerful convergence theorem exists for other alternating series that meet a few conditions. +

    + + + Alternating Series Test + +

    + Let \{a_n\} be a positive, + decreasing sequence where \lim\limits_{n\to\infty}a_n=0. + Then + seriesAlternating Series Test + Alternating Series Test + convergenceAlternating Series Test + divergenceAlternating Series Test + + \infser (-1)^{n}a_n \qquad \text{ and } \qquad \infser (-1)^{n+1}a_n + + converge. +

    +     +     + + + + + + + + + +

    + The basic idea behind + is illustrated in . + A positive, decreasing sequence \{a_n\} is shown along with the partial sums + + S_n = \sum_{i=1}^n(-1)^{i+1}a_i =a_1-a_2+a_3-a_4+\cdots+(-1)^{n+1}a_n + . +

    + +

    + Because \{a_n\} is decreasing, + the amount by which S_n bounces up/down decreases. + Moreover, the odd terms of S_n form a decreasing, + bounded sequence, + while the even terms of S_n form an increasing, + bounded sequence. + Since bounded, monotonic sequences converge + (see ) + and the terms of \{a_n\} approach 0, one can show the odd and even terms of S_n converge to the same common limit L, + the sum of the series. +

    + +
    + Illustrating convergence with the Alternating Series Test + + + Scatter plots of a positive, decreasing sequence and its alternating sequence of partial sums. + +

    + On one set of coordinate axes, two scatter plots are shown. + The first is the plot of a positive, decreasing sequence a_n; + the second is the plot of the sequence of partial sums for the corresponding alternating sequence (-1)^na_n. +

    + +

    + The scatter plots illustrate why an alternating series converges: + as n increases, the partial sums oscillate back and forth across a horizontal line marked L + (the limiting value). + Since a_n is a decreasing sequence, the oscillations get smaller as n increases, + and the points in the scatter plot for S_n get closer and closer to the line y=L. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.25,ymax=1.25, + xmin=-1,xmax=11, + xlabel={$n$} + ] + + \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt}] coordinates{(1,1)}; + \addplot [only marks,firstcolor,mark size={2.4pt}] coordinates{(1,1)}; + \addplot [only marks,firstcolor,mark size={2.4pt},domain=2:10,samples=9] {2/(x+1)}; + \addplot [only marks,secondcolor,mark=square*,mark size={2.4pt},domain=2:10,samples=9] coordinates {(2., 0.333333)(3., 0.833333)(4., 0.433333)(5.,0.766667)(6., 0.480952)(7., 0.730952)(8., 0.50873)(9.,0.70873)(10., 0.526912)}; + \addplot [dashed,domain=-.1:10] {.614} node [pos=0,left] { $L$}; + + \end{axis} + + + \node[shift={(0,-10pt)},draw] at (myplot.south) + {\begin{tikzpicture} + + \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; + \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; + \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; + + \end{tikzpicture}}; + + \end{tikzpicture} + + + + +
    + +
    + A visual representation of adding terms of an alternating series. The arrows represent the length and direction of each term of the sequence. + + + An illustration of the partial sums in an alternating series, using line segments to demonstrate convergence. + +

    + On a set of coordinate axes, a sequence of horizontal line segments is shown. + At the top of the image is a line segment indicating in increase from 0 to a_1. + An arrow pointing to the right is at the end of the segment. +

    + +

    + Below this segment is a shorter segment with an arrow pointing to the left. + The right end of this segment aligns with the right end of the first segment, + indicating that we obtain a_1-a_2 by starting at a_1 and then moving to the left by a distance a_2. +

    + +

    + The third segment is below the second. Its left end aligns with the left end of the second segment, + and it points to the right, indicating the act of adding a_3 to the partial sum. + The length of this segment is shorter than that of the second, indicating the fact that the sequence a_n is decreasing. +

    + +

    + As we move down, additional segments are drawn, alternating between pointing left and right, + and getting shorter with each step. Below the segments is a horizontal axis, on which the values + S_1, S_2, \ldots of the partial sums are shown, as well as the limit L. + The illustration overall is intended to convey the idea that the line segments will shrink toward the limiting value L + as n increases. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + extra x ticks={1, 0.3333, 0.83333, 0.43333, 0.76667, 0.4809, 0.7309, 0.612}, + extra x tick labels={ $S_1$, $S_2$, $S_3$, $S_4$, $S_5$, $S_6$, $S_7$, $L$ }, + ytick=\empty, + xlabel=\empty, + ylabel=\empty, + ymin=-.2,ymax=2.25, + xmin=-.2,xmax=1.2, + x post scale=2, + ] + + \addplot[firstcurvestyle,infiniteright,domain=0:1] ({x},{2}) node [pos=1, right] {$a_1$}; + \addplot[firstcurvestyle,infiniteleft,domain=0.333:1] ({x},{1.75}) node [pos=0, left] {$-a_2$}; + \addplot[firstcurvestyle,infiniteright,domain=0.333:.833] ({x},{1.5}) node [pos=1, right] {$a_3$}; + \addplot[firstcurvestyle,infiniteleft,domain=0.433:.8333] ({x},{1.25}) node [pos=0, left] {$-a_4$}; + \addplot[firstcurvestyle,infiniteright,domain=0.4333:.76667] ({x},{1}) node [pos=1, right] {$a_5$}; + \addplot[firstcurvestyle,infiniteleft,domain=0.4809:.76667] ({x},{.75}) node [pos=0, left] {$-a_6$}; + \addplot[firstcurvestyle,infiniteright,domain=0.4809:.7309] ({x},{.5}) node [pos=1, right] {$a_7$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + + + Applying the Alternating Series Test + +

    + Determine if the Alternating Series Test applies to each of the following series. +

    + +

    +

      +
    1. +

      + \ds \infser (-1)^{n+1}\frac1n +

      +
      2. + +
    2. +

      + \ds \infser (-1)^n\frac{\ln(n) }{n} +

      +
      3. + +
    3. +

      + \ds \infser (-1)^{n+1}\frac{\abs{\sin(n) }}{n^2} +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + This is the Alternating Harmonic Series as seen previously. + The underlying sequence is \{a_n\} = \{1/n\}, + which is positive, decreasing, + and approaches 0 as n\to\infty. + Therefore we can apply the Alternating Series Test and conclude this series converges. + + While the test does not state what the series converges to, we will see later that \ds \infser (-1)^{n+1}\frac1n=\ln(2). +

      +
      2. + +
    2. +

      + The underlying sequence is \{a_n\} = \{\ln(n) /n\}. + This is positive and approaches 0 as n\to\infty + (use L'Hospital's Rule). + However, the sequence is not decreasing for all n. + It is straightforward to compute a_1=0, a_2\approx0.347, + a_3\approx 0.366, and a_4\approx 0.347: + the sequence is increasing for at least the first 3 terms. + + We do not immediately conclude that we cannot apply the Alternating Series Test. Rather, consider the long-term behavior of \{a_n\}. Treating a_n=a(n) as a continuous function of n defined on [1,\infty), we can take its derivative: + + a'(n) = \frac{1-\ln(n) }{n^2} + . + The derivative is negative for all n\geq 3 (actually, + for all n \gt e), + meaning a(n)=a_n is decreasing on [3,\infty). + We can apply the Alternating Series Test to the series when we start with n=3 and conclude that \ds \sum_{n=3}^\infty(-1)^n\frac{\ln(n) }{n} converges; + adding the terms with n=1 and n=2 do not change the convergence (, we apply ). + The important lesson here is that as before, + if a series fails to meet the criteria of the Alternating Series Test on only a finite number of terms, + we can still apply the test. +

      +
      3. + +
    3. +

      + The underlying sequence is \{a_n\} = \abs{\sin(n) }/n. + This sequence is positive and approaches 0 as n\to\infty. + However, it is not a decreasing sequence; + the value of \abs{\sin(n) } oscillates between 0 and 1 as n\to\infty. + We cannot remove a finite number of terms to make \{a_n\} decreasing, + therefore we cannot apply the Alternating Series Test. + + Keep in mind that this does not mean we conclude the series diverges; in fact, it does converge. We are just unable to conclude this based on . We will be able to show that this series converges shortly. +

      +
      4. +
    +

    +     + +     + +

    + + gives the sum of some important series. + Two of these are + + \infser \frac1{n^2} =\frac{\pi^2}6 \approx 1.64493 \text{ and } \infser \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}\approx 0.82247 + . +

    + +

    + These two series converge to their sums at different rates. + To be accurate to two places after the decimal, + we need 202 terms of the first series though only 13 of the second. + To get 3 places of accuracy, + we need 1069 terms of the first series though only 33 of the second. + Why is it that the second series converges so much faster than the first? +

    + +

    + While there are many factors involved when studying rates of convergence, + the alternating structure of an alternating series gives us a powerful tool when approximating the sum of a convergent series. +

    + + + The Alternating Series Approximation Theorem + +

    + Let \{a_n\} be a sequence that satisfies the hypotheses of the Alternating Series Test, + and let S_n and L be the + nth partial sums and sum, respectively, + of either \ds \infser (-1)^{n}a_n or \ds \infser (-1)^{n+1}a_n. + Then + + seriesalternating!Approximation Theorem + +

    + +

    +

      +
    1. +

      + E_n=\abs{S_n-L} \lt a_{n+1}, and +

      +
      2. + +
    2. +

      + L is between S_n and S_{n+1}. +

      +
      3. +
    +

    +     +     + + + +

    + Part 1 of + states that the nth partial sum of a convergent alternating series will be within a_{n+1} of its total sum. + You can see this visually in . + Look at the distance between S_6 and L. + Clearly this distance is less than the length of the arrow corresponding to a_7. +

    + +

    + Also consider the alternating series we looked at before the statement of the theorem, + \ds \infser \frac{(-1)^{n+1}}{n^2}. + Since a_{14} = 1/14^2 \approx 0.0051, + we know that S_{13} is within 0.0051 of the total sum. +

    + +

    + Moreover, Part 2 of the theorem states that since + S_{13} \approx 0.8252 and S_{14}\approx 0.8201, + we know the sum L lies between 0.8201 and 0.8252. + One use of this is the knowledge that S_{14} is accurate to two places after the decimal. +

    + +

    + Some alternating series converge slowly. + In + we determined the series \ds\infser (-1)^{n+1}\frac{\ln(n) }{n} converged. + With n=1001, we find \ln(n) /n \approx 0.0069, + meaning that S_{1000} \approx 0.1633 is accurate to one, + maybe two, places after the decimal. + Since S_{1001} \approx 0.1564, + we know the sum L is 0.1564\leq L\leq0.1633. +

    + + + Approximating the sum of convergent alternating series + +

    + Approximate the sum of the following series, + accurate to within 0.001. +

    + +

    +

      +
    1. +

      + \ds \infser (-1)^{n+1}\frac{1}{n^3} +

      +
      2. + +
    2. +

      + \ds \infser (-1)^{n+1}\frac{\ln(n) }{n} +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + Using , + we want to find n where 1/n^3 \leq 0.001. + That is, we want to find the the first time a term in the sequence a_n is smaller than the desired level of error: + + \frac1{n^3} \amp \leq 0.001=\frac{1}{1000} + n^3 \amp \geq 1000 + n \amp \geq \sqrt[3]{1000} + n \amp \geq 10 + . + Let L be the sum of this series. + By Part 1 of the theorem, + \abs{S_9-L}\lt a_{10} = 1/1000. (We found a_{10}=a_{n+1}\lt 0.0001, + so n=9). + We can compute S_9=0.902116, + which our theorem states is within 0.001 of the total sum. + + We can use Part 2 of the theorem to obtain an even more accurate result. As we know the 10th term of the series is (-1)^n/10^3=-1/1000, we can easily compute S_{10} = 0.901116. Part 2 of the theorem states that L is between S_9 and S_{10}, so 0.901116 \lt L\lt 0.902116. +

      +
      2. + +
    2. +

      + We want to find n where \ln(n)/n \lt 0.001. + We start by solving \ln(n)/n = 0.001 for n. + This cannot be solved algebraically, + so we will use Newton's Method to approximate a solution. + (Note: we can also use a Brute Force technique. + That is, we can guess and check numerically until we find a solution.) + + Let f(x) = \ln(x)/x-0.001; we want to know where f(x) = 0. We make a guess that x must be large, so our initial guess will be x_1=1000. Recall how Newton's Method works: given an approximate solution x_n, our next approximation x_{n+1} is given by + + x_{n+1} = x_n - \frac{f(x_n)}{\fp(x_n)} + . + We find \fp(x) = \big(1-\ln(x)\big)/x^2. + This gives + + x_2 \amp = 1000 - \frac{\ln(1000)/1000-0.001}{\big(1-\ln(1000)\big)/1000^2} + \amp = 2000 + . + Using a computer, + we find that Newton's Method seems to converge to a solution x=9118.01 after 8 iterations. + Taking the next integer higher, + we have n=9119, where \ln(9119)/9119 =0.000999903\lt 0.001. + + Again using a computer, we find S_{9118} = -0.160369. Part 1 of the theorem states that this is within 0.001 of the actual sum L. Already knowing the 9{,}119th term, we can compute S_{9119} = -0.159369, meaning -0.159369 \lt L \lt -0.160369. +

      +
      3. +
    +

    + +

    + Notice how the first series converged quite quickly, + where we needed only 10 terms to reach the desired accuracy, + whereas the second series took over 9,000 terms. +

    +     + +     + +

    + One of the famous results of mathematics is that the Harmonic Series, + \ds \infser \frac1n diverges, + yet the Alternating Harmonic Series, + \ds \infser (-1)^{n+1}\frac1n, converges. + The notion that alternating the signs of the terms in a series can make a series converge leads us to the following definitions. + Alternating Harmonic Series +

    + + + Absolute and Conditional Convergence + +

    +

      +
    1. +

      + A series \ds \infser a_n + converges absolutely + if \ds \infser \abs{a_n} converges. + + convergenceabsolute + convergenceconditional + seriesabsolute convergence + seriesconditional convergence +

      +
      2. + +
    2. +

      + A series \ds \infser a_n + converges conditionally + if \ds \infser a_n converges but \ds \infser \abs{a_n} diverges. +

      +
      3. +
    +

    +     +     + + +

    + Thus we say the Alternating Harmonic Series converges conditionally. +

    + + + + + Determining absolute and conditional convergence + +

    + Determine if the following series converge absolutely, + conditionally, or diverge. +

    + +

    +

      +
    1. \ds \infser (-1)^n\frac{n+3}{n^2+2n+5}
      2. + +
    2. + \ds \infser (-1)^n\frac{n^2+2n+5}{2^n} +
      3. + +
    3. + \ds \sum_{n=3}^\infty (-1)^n\frac{3n-3}{5n-10} +
      4. +
    +

    +     + +

    +

      +
    1. +

      + We can show the series + + \ds \infser \abs{(-1)^n\frac{n+3}{n^2+2n+5}}= \infser \frac{n+3}{n^2+2n+5} + + diverges using the Limit Comparison Test, + comparing with 1/n. + + The series \ds \infser (-1)^n\frac{n+3}{n^2+2n+5} converges using the Alternating Series Test; we conclude it converges conditionally. +

      +
      2. + +
    2. +

      + We can show the series + + \ds \infser \abs{(-1)^n\frac{n^2+2n+5}{2^n}}=\infser \frac{n^2+2n+5}{2^n} + + converges using the Ratio Test. + + Therefore we conclude \ds \infser (-1)^n\frac{n^2+2n+5}{2^n} converges absolutely. +

      +
      3. + +
    3. +

      + The series + + \ds \sum_{n=3}^\infty \abs{(-1)^n\frac{3n-3}{5n-10}} = \sum_{n=3}^\infty \frac{3n-3}{5n-10} + + diverges using the nth Term Test, + so it does not converge absolutely. + The series \ds \sum_{n=3}^\infty (-1)^n\frac{3n-3}{5n-10} fails the conditions of the Alternating Series Test as + (3n-3)/(5n-10) does not approach 0 as n\to\infty. + We can state further that this series diverges; + as n\to\infty, + the series effectively adds and subtracts 3/5 over and over. + This causes the sequence of partial sums to oscillate and not converge. + Therefore the series \ds \infser (-1)^n\frac{3n-3}{5n-10} diverges. +

      +
      4. +
    +

    +     + +     + +

    + Knowing that a series converges absolutely allows us to make two important statements, + given in below. + The first is that absolute convergence is stronger + than regular convergence. + That is, just because \infser a_n converges, + we cannot conclude that \infser \abs{a_n} will converge, + but knowing a series converges absolutely tells us that \infser a_n will converge. +

    + +

    + One reason this is important is that our convergence tests all require that the underlying sequence of terms be positive. + By taking the absolute value of the terms of a series where not all terms are positive, + we are often able to apply an appropriate test and determine absolute convergence. + This, in turn, + determines that the series we are given also converges. +

    + +

    + The second statement relates to + rearrangements + + rearrangements of series + seriesrearrangements of series. + When dealing with a finite set of numbers, + the sum of the numbers does not depend on the order which they are added. + (So 1+2+3 = 3+1+2.) + One may be surprised to find out that when dealing with an infinite set of numbers, + the same statement does not always hold true: + some infinite lists of numbers may be rearranged in different orders to achieve different sums. + The theorem states that the terms of an absolutely convergent series can be rearranged in any way without affecting the sum. +

    + + + + Absolute Convergence Theorem + +

    + Let \ds \infser a_n be a series that converges absolutely. + + convergenceabsolute + Absolute Convergence Theorem + seriesAbsolute Convergence Theorem + rearrangements of series + seriesrearrangements + +

    + +

    +

      +
    1. + +

      + \ds \infser a_n converges. +

      +
      2. + +
    2. +

      + Let \{b_n\} be any rearrangement of the sequence \{a_n\}. + Then + + \infser b_n = \infser a_n + . +

      +
      3. +
    +

    +     + +

    + We will provide a proof for Part + of . + Suppose that \infser \abs{a_n} converges. + We start by noting that for any sequence a_n, we have + + -\abs{a_n} \leq a_n \leq \abs{a_n} + If we add \abs{a_n} to all three sides: + 0\leq a_n +\abs{a_n} \leq 2\abs{a_n} + . + We are now in a position to apply the to the series \infser \left(a_n+\abs{a_n}\right). + Since \infser \abs{a_n} converges by our supposition, + so does \infser 2\abs{a_n} + (the scalar multiple of a convergent series also converges by ). + Therefore \infser \left(a_n+\abs{a_n}\right) converges by the . +

    + +

    + Now we turn our attention to \infser a_n. + We can say + + \infser a_n \amp =\infser \left(a_n+\abs{a_n}-\abs{a_n}\right) + \amp =\infser \left(a_n+\abs{a_n}\right)-\infser \abs{a_n} + . + The last line is the difference between two convergent series, + which is also convergent by . + Therefore \infser a_n converges. +

    +     + +     + +

    + In , + we determined the series in Part converges absolutely. + tells us the series converges + (which we could also determine using the Alternating Series Test). +

    + +

    + The theorem states that rearranging the terms of an absolutely convergent series does not affect its sum. + This implies that perhaps the sum of a conditionally convergent series can change based on the arrangement of terms. + Indeed, it can. + The Riemann Rearrangement Theorem + (named after Bernhard Riemann) + states that any conditionally convergent series can have its terms rearranged so that the sum is any desired value, + including \infty! +

    + +

    + As an example, + consider the Alternating Harmonic Series once more. + We have stated that + + \infser (-1)^{n+1}\frac1n = 1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17\cdots = \ln(2) + , +

    + +

    + (see + or ). +

    + +

    + Consider the rearrangement where every positive term is followed by two negative terms: + + 1-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac1{10}-\frac1{12}\cdots + +

    + +

    + (Convince yourself that these are exactly the same numbers as appear in the Alternating Harmonic Series, + just in a different order.) Now group some terms and simplify: + + \left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\frac1{12}+\cdots \amp = + \frac12-\frac14+\frac16-\frac18+\frac1{10}-\frac{1}{12}+\cdots \amp = + \frac12\left(1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots\right) \amp = \frac12\ln(2) + . +

    + +

    + By rearranging the terms of the series, + we have arrived at a different sum! (One could try + to argue that the Alternating Harmonic Series does not actually converge to \ln(2), + because rearranging the terms of the series + shouldn't change the sum. + However, the Alternating Series Test proves this series converges to L, + for some number L, + and if the rearrangement does not change the sum, + then L = L/2, implying L=0. + But the Alternating Series Approximation Theorem quickly shows that L \gt 0. + The only conclusion is that the rearrangement did + change the sum.) This is an incredible result. +

    + +

    + We end here our study of tests to determine convergence. + A table summarizing the tests can be found at the end of the text, + in . +

    + +

    + While series are worthy of study in and of themselves, + our ultimate goal within calculus is the study of Power Series, + which we will consider in the next section. + We will use power series to create functions where the output is the result of an infinite summation. +

    + + + + Terms and Concepts + + + +

    + Why is \ds\infser \sin(n) not an alternating series? +

    +     + + + +

    + The signs of the terms do not alternate; + in the given series, some terms are negative and the others positive, + but they do not necessarily alternate. +

    +     + +     + + + + +

    + A series \ds\infser (-1)^na_n converges when \{a_n\} is , + and \lim\limits_{n\to\infty}a_n =. +

    +     + + + + + ["positive", "decreasing"].includes(ans) + + + + + + + ["positive", "decreasing"].includes(ans) && !ans_array.slice(0,1).includes(ans) + + + + + ans_array.slice(0,1).includes(ans) + + You already gave that answer. + + + + + 0|zero + + + + +     + + + + +

    + Give an example of a series where + \ds \infser[0] a_n converges but \ds \infser[0] \abs{a_n} does not. +

    +     + + + +

    + Many examples exist; one common example is a_n = (-1)^n/n. +

    +     + +     + + + + +

    + The sum of a convergent + series can be changed by rearranging the order of its terms. +

    +     + + + + + + + + +

    + Your answer includes the correct word but has extra text. +

    +     +     +     +     + +     +     + + Problems + + + +

    + An alternating series \ds \sum_{n=i}^\infty a_n is given. +

    + +

    +

      +
    1. +

      + Determine if the series converges or diverges. +

      +
      2. + +
    2. +

      + Determine if \ds \infser[0] \abs{a_n} converges or diverges. +

      +
      3. + +
    3. +

      + If \ds \infser[0] a_n converges, + determine if the convergence is conditional or absolute. +

      +
      4. +
    +

    +     + + + + +

    + \ds \infser \frac{(-1)^{n+1}}{n^2} +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + converges (p-Series) +

      +
      3. + +
    3. +

      + absolute +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{(-1)^{n+1}}{\sqrt{n!}} +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + converges (use Ratio Test) +

      +
      3. + +
    3. +

      + absolute +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser[0] (-1)^{n}\frac{n+5}{3n-5} +

    +     + +

    +

      +
    1. +

      + diverges (limit of terms is not 0) +

      +
      2. + +
    2. +

      + diverges +

      +
      3. + +
    3. +

      + n/a; diverges +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser (-1)^{n}\frac{2^n}{n^2} +

    +     + +

    +

      +
    1. +

      + diverges (limit of terms is not 0) +

      +
      2. + +
    2. +

      + diverges +

      +
      3. + +
    3. +

      + n/a; diverges +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser[0] (-1)^{n+1}\frac{3n+5}{n^2-3n+1} +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + diverges (Limit Comparison Test with 1/n) +

      +
      3. + +
    3. +

      + conditional +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{(-1)^{n}}{\ln(n) +1} +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + diverges (Direct Comparison Test with 1/n) +

      +
      3. + +
    3. +

      + conditional +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \sum_{n=2}^\infty (-1)^n\frac{n}{\ln(n) } +

    +     + +

    +

      +
    1. +

      + diverges (limit of terms is not 0) +

      +
      2. + +
    2. +

      + diverges +

      +
      3. + +
    3. +

      + n/a; diverges +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{(-1)^{n+1}}{1+3+5+\cdots+(2n-1) } +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + converges (the sum in the denominator is n^2) +

      +
      3. + +
    3. +

      + absolute +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser \cos\big(\pi n\big) +

    +     + +

    +

      +
    1. +

      + diverges (terms oscillate between \pm 1) +

      +
      2. + +
    2. +

      + diverges +

      +
      3. + +
    3. +

      + n/a; diverges +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser[2] \frac{\sin\big((n+1/2)\pi\big)}{n\ln(n) } +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + diverges (Integral Test) +

      +
      3. + +
    3. +

      + conditional +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser[0] \left(-\frac23\right)^n +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + converges (Geometric Series with r=2/3) +

      +
      3. + +
    3. +

      + absolute +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser[0] (-e)^{-n} +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + converges (Geometric Series with r=1/e) +

      +
      3. + +
    3. +

      + absolute +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser[0] \frac{(-1)^nn^2}{n!} +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + converges (Ratio Test) +

      +
      3. + +
    3. +

      + absolute +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser[0] (-1)^n2^{-n^2} +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + converges (Ratio Test) +

      +
      3. + +
    3. +

      + absolute +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{(-1)^n}{\sqrt{n}} +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + diverges (p-Series Test with p=1/2) +

      +
      3. + +
    3. +

      + conditional +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \infser \frac{(-1000)^n}{n!} +

    +     + +

    +

      +
    1. +

      + converges +

      +
      2. + +
    2. +

      + converges (Ratio Test) +

      +
      3. + +
    3. +

      + absolute +

      +
      4. +
    +

    +     + +     + +     + + + +

    + Let S_n be the n^{ th } partial sum of a series. + A convergent alternating series is given and a value of n. + Compute S_n and S_{n+1} and use these values to find bounds on the sum of the series. +

    +     + + + + +

    + \ds \infser \frac{(-1)^n}{\ln(n+1)}, n=5 +

    +     + +

    + S_5 = -1.1906; S_{6} = -0.6767; +

    + +

    + \ds -1.1906 \leq \infser \frac{(-1)^n}{\ln(n+1)} \leq -0.6767 +

    +     + +     + + + + +

    + \ds \infser \frac{(-1)^{n+1}}{n^4}, n=4 +

    +     + +

    + S_4 = 0.9459; S_5 = 0.9475; +

    + +

    + \ds 0.9459 \leq \infser \frac{(-1)^n}{n^4} \leq 0.9475 +

    +     + +     + + + + +

    + \ds \infser[0] \frac{(-1)^{n}}{n!}, n=6 +

    +     + +

    + S_6 = 0.3681; S_7 = 0.3679; +

    + +

    + \ds 0.3679 \leq \infser[0] \frac{(-1)^{n}}{n!} \leq 0.3681 +

    +     + +     + + + + +

    + \ds \infser[0] \left(-\frac12\right)^n, n=9 +

    +     + +

    + S_9 = 0.666016; S_{10} = 0.666992; +

    + +

    + \ds 0.666016 \leq \infser[0] \left(-\frac12\right)^n \leq 0.666992 +

    +     + +     + +     + + + +

    + A convergent alternating series is given along with its sum and a value of \varepsilon. + Use + to find n such that the + nth partial sum of the series is within + \varepsilon of the sum of the series. +

    +     + + + + +

    + \ds \infser \frac{(-1)^{n+1}}{n^4} = \frac{7\pi^4}{720}, + \varepsilon = 0.001 +

    +     + +

    + n=5 +

    +     + +     + + + + +

    + \ds \infser[0] \frac{(-1)^{n}}{n!} = \frac1e, + \varepsilon = 0.0001 +

    +     + +

    + n=7 +

    +     + +     + + + + +

    + \ds \infser[0] \frac{(-1)^{n}}{2n+1}=\frac{\pi}4, + \varepsilon = 0.001 +

    +     + +

    + Using the theorem, + we find n=499 guarantees the sum is within 0.001 of \pi/4. + (Convergence is actually faster, + as the sum is within \varepsilon of \pi/24 when n\geq 249.) +

    +     + +     + + + + +

    + \ds \infser[0] \frac{(-1)^{n}}{(2n)!}=\cos(1), + \varepsilon = 10^{-8} +

    +     + +

    + n=5 ((2n)! \gt 10^8 when n\geq 6) +

    +     + +     + +     +     +     +
    +
    + Power Series +

    + So far, our study of series has examined the question of + Is the sum of these infinite terms finite?, + , Does the series converge? + We now approach series from a different perspective: as a function. + Given a value of x, + we evaluate f(x) by finding the sum of a particular series that depends on x + (assuming the series converges). + We start this new approach to series with a definition. +

    + + + + + Power Series + +

    + Let \{a_n\} be a sequence, + let x be a variable, and let c be a real number. + power series + seriespower +

    + +

    +

      +
    1. +

      + The power series in x is the series + + \infser[0] a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+\ldots + +

      +
      2. + +
    2. +

      + The power series in x centered at c is the series + + \infser[0] a_n(x-c)^n = a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\ldots + +

      +
      3. +
    +

    +     +     + + + Examples of power series + +

    + Write out the first five terms of the following power series: +

    + +

    +

      +
    1. \infser[0] x^n
      2. +
    2. \infser (-1)^{n+1}\frac{(x+1)^n}n
      3. +
    3. \infser[0] (-1)^{n+1} \frac{(x-\pi)^{2n}}{(2n)!}
      4. +
    +

    +     + +

    +

      +
    1. +

      + One of the conventions we adopt is that x^0=1 regardless of the value of x. + Therefore + + \infser[0] x^n = 1+x+x^2+x^3+x^4+\ldots + + This is a geometric series in x with r=x. +

      +
      2. + +
    2. +

      + This series is centered at c=-1. + Note how this series starts with n=1. + We could rewrite this series starting at n=0 with the understanding that a_0=0, + and hence the first term is 0. + + \amp \infser (-1)^{n+1}\frac{(x+1)^n}n + \amp =(x+1) - \frac{(x+1)^2}{2} + \frac{(x+1)^3}{3} - \frac{(x+1)^4}{4}+\frac{(x+1)^5}{5}\ldots + +

      +
      3. + +
    3. +

      + This series is centered at c=\pi. + Recall that 0!=1. + + \amp \infser[0] (-1)^{n+1} \frac{(x-\pi)^{2n}}{(2n)!} + \amp = -1+\frac{(x-\pi)^2}{2} - \frac{(x-\pi)^4}{24}+ \frac{(x-\pi)^6}{6!}-\frac{(x-\pi)^8}{8!}\ldots + +

      +
      4. +
    +

    +     + +     + +

    + We introduced power series as a type of function, + where a value of x is given and the sum of a series is returned. + Of course, not every series converges. + For instance, in part 1 of , + we recognized the series \ds \infser[0] x^n as a geometric series in x. + + states that this series converges only when \abs{x}\lt 1. +

    + +

    + This raises the question: + For what values of x will a given power series converge?, + which leads us to a theorem and definition. +

    + + + Convergence of Power Series + +

    + Let a power series \ds \infser[0] a_n(x-c)^n be given. + Then one of the following is true: + convergenceof power series + power seriesconvergence + seriespower + +

    + +

    +

      +
    1. +

      + The series converges only at x=c. +

      +
      2. + +
    2. +

      + There is an R \gt 0 such that the series converges for all x in + + (c-R,c+R) and diverges for all x\lt c-R and x \gt c+R. +

      +
      3. + +
    3. +

      + The series converges for all x. +

      +
      4. +
    +

    +     +     + + +

    + The value of R is important when understanding a power series, + hence it is given a name in the following definition. + Also, note that part 2 of + makes a statement about the interval (c-R,c+R), + but the not the endpoints of that interval. + A series may/may not converge at these endpoints. +

    + + + Radius and Interval of Convergence + +

    +

      +
    1. +

      + The number R given in + is the radius of convergence of a given series. + When a series converges for only x=c, + we say the radius of convergence is 0, , R=0. + When a series converges for all x, + we say the series has an infinite radius of convergence, + , R=\infty. +

      +
      2. + +
    2. +

      + The interval of convergence + is the set of all values of x for which the series converges. + + convergenceradius of + convergenceinterval of + radius of convergence + interval of convergence + seriesradius of convergence + seriesinterval of convergence +

      +
      3. +
    +

    +     +     + +

    + To find the interval of convergence, we start by using the ratio test to find the radius of convergence R. + If 0\lt R\lt\infty, we know the series converges on (c-R,c+R), + and it remains to check for convergence at the endpoints. +

    + +

    + Given \infser[0] a_n(x-c)^n we apply the ratio test to \infser[0]\abs{a_n(x-c)^n} + since the ratio test requires positive terms. + We find + + \lim_{n\to\infty}\frac{\abs{a_{n+1}(x-c)^{n+1}}}{\abs{a_n(x-c)^n}} = L\abs{x-c} + , + where L=\lim_{n\to\infty}\frac{\abs{a_{n+1}}}{\abs{a_n}}. + It follows that the series converges absolutely (and therefore converges) + for any x such that L\abs{x-c}\lt 1; + that is, for x in \left(c-\frac1L,c+\frac1L\right). +

    + +

    + On the other hand, suppose for some x that L\abs{x-c}\gt 1. + Then, for sufficiently large n, \abs{a_{n+1}}\gt \abs{a_n}. + This means that the terms of \infser[0] a_n(x-c)^n are growing in absolute value, + and therefore cannot converge to zero. This means that the series diverges, by . +

    + +

    + From the above observations, it follows that R=\frac{1}{L} must be the radius of convergence. +

    + + + Determining the Radius and Interval of Convergence + +

    + Given the power series \ds \infser[0] a_n(x-c)^n, + apply the ratio test to the series \ds \infser[0]\abs{a_n (x-c)^n}. + The result will be L\abs{x-c}, where \ds L=\lim_{n\to\infty}\frac{\abs{a_{n+1}}}{\abs{a_n}}. +

    + +

    +

      +
    1. +

      + If L=0, then the power series converges for every x + by the ratio test, since L\abs{x-c}=0\lt 1. +

      +
      2. + +
    2. +

      + If L=\infty, then power series converges only when x=c. +

      +
      3. + +
    3. +

      + If 0\lt L\lt \infty, then R=1/L is the radius of convergence: + by the ratio test, the series converges when \abs{x-c}\lt R. +

      + +

      + To determine the interval of convergence, plug the endpoints (x=c-R and x=c+R) + into the power series, and test the resulting series for convergence. + If the series converges, we include the endpoint. If it diverges, we exclude the endpoint. +

      +
      4. +
    +

    +     +     + + + +

    + + allows us to find the radius of convergence R of a series by applying the Ratio Test + (or any applicable test) + to the absolute value of the terms of the series. + We practice this in the following example. +

    + + + Determining the radius and interval of convergence + +

    + Find the radius and interval of convergence for each of the following series: +

    + +

    +

      +
    1. +

      + \ds \infser[0] \frac{x^n}{n!} +

      +
      2. + +
    2. +

      + \ds \infser (-1)^{n+1}\frac{x^n}{n} +

      +
      3. + +
    3. +

      + \ds \infser[0] 2^n(x-3)^n +

      +
      4. + +
    4. +

      + \ds \infser[0] n!x^n +

      +
      5. +
    +

    +     + +

    +

      +
    1. +

      + We apply the Ratio Test to the series \ds \infser \abs{\frac{x^n}{n!}}: + + \lim_{n\to\infty} \frac{\abs{x^{n+1}/(n+1)!}}{\abs{x^n/n!}} \amp = \lim_{n\to\infty} \abs{\frac{x^{n+1}}{x^n}\cdot\frac{n!}{(n+1)!}} + \amp = \lim_{n\to\infty} \abs{\frac x{n+1}} + \amp = 0 \text{ for all } x + . + The Ratio Test shows us that regardless of the choice of x, + the series converges. + Therefore the radius of convergence is R=\infty, + and the interval of convergence is (-\infty,\infty). +

      +
      2. + +
    2. +

      + We apply the Ratio Test to the series \ds \infser \abs{(-1)^{n+1}\frac{x^n}{n}} = \infser \abs{\frac{x^n}{n}}: + + \lim_{n\to\infty} \frac{\abs{x^{n+1}/(n+1)}}{\abs{x^n/n}} \amp = \lim_{n\to\infty} \abs{\frac{x^{n+1}}{x^n}\cdot \frac{n}{n+1}} + \amp = \lim_{n\to\infty} (\frac{n}{n+1})\abs{x} + \amp = \abs{x} + . + The Ratio Test states a series converges if the limit of \abs{a_{n+1}/a_n} = L\lt 1. + We found the limit above to be \abs{x}; + therefore, the power series converges when \abs{x} \lt 1, + or when x is in (-1,1). + Thus the radius of convergence is R=1. + + + To determine the interval of convergence, we need to check the endpoints of (-1,1). When x=-1, + we have the opposite of the Harmonic Series: + + \infser (-1)^{n+1}\frac{(-1)^n}{n} \amp = \infser \frac{-1}{n} + \amp = -\infty + . + The series diverges when x=-1. + + When x=1, we have the series \ds \infser (-1)^{n+1}\frac{(1)^n}{n}, + which is the Alternating Harmonic Series, which converges. Therefore the interval of convergence is (-1,1]. +

      +
      3. + +
    3. +

      + We apply the Ratio Test to the series \ds\infser[0] \abs{2^n(x-3)^n}: + + \lim_{n\to\infty} \frac{\abs{ 2^{n+1}(x-3)^{n+1}}}{\abs{2^n(x-3)^n}} \amp = \lim_{n\to\infty} \abs{\frac{2^{n+1}}{2^n}\cdot\frac{(x-3)^{n+1}}{(x-3)^n}} + \amp =\lim_{n\to\infty} \abs{2(x-3)} + . + According to the Ratio Test, + the series converges when \abs{2(x-3)}\lt 1 \implies \abs{x-3} \lt 1/2. + The series is centered at 3, and x must be within 1/2 of 3 in order for the series to converge. + Therefore the radius of convergence is R=1/2, + and we know that the series converges absolutely for all x in (3-1/2,3+1/2) = (2.5, 3.5). + We check for convergence at the endpoints to find the interval of convergence. + When x=2.5, we have: + + \infser[0] 2^n(2.5-3)^n \amp = \infser[0] 2^n(-1/2)^n + \amp =\infser[0] (-1)^n + , + which diverges. + A similar process shows that the series also diverges at x=3.5. + Therefore the interval of convergence is (2.5, 3.5). +

      +
      4. + +
    4. +

      + We apply the Ratio Test to \ds \infser[0] \abs{n!x^n}: + + \lim_{n\to\infty} \frac{\abs{ (n+1)!x^{n+1}}}{\abs{n!x^n}} \amp = \lim_{n\to\infty} \abs{(n+1)x} + \amp = \infty\, \text{ for all \(x\), except \(x=0\). } + + The Ratio Test shows that the series diverges for all x except x=0. + Therefore the radius of convergence is R=0. +

      +
      5. +
    +

    +     + +     + + + +

    + We can use a power series to define a function: + + f(x) = \infser[0] a_nx^n + + where the domain of f is a subset of the interval of convergence of the power series. + One can apply calculus techniques to such functions; + in particular, we can find derivatives and antiderivatives. +

    + + + Derivatives and Indefinite Integrals of Power Series Functions + +

    + Let \ds f(x) = \infser[0] a_n(x-c)^n be a function defined by a power series, + with radius of convergence R. +

    + +

    +

      +
    1. +

      + f(x) is continuous and differentiable on (c-R,c+R). +

      +
      2. + +
    2. +

      + \ds \fp(x) = \infser a_n\cdot n\cdot (x-c)^{n-1}, + with radius of convergence R. +

      +
      3. + +
    3. +

      + \ds \int f(x)\, dx = C+\infser[0] a_n\frac{(x-c)^{n+1}}{n+1}, + with radius of convergence R. +

      +
      4. +
    + seriespower!derivatives and integrals + integrationof power series + derivativepower series + power seriesderivatives and integrals + +

    +     +     + + +

    + A few notes about : +

    + +

    +

      +
    1. +

      + The theorem states that differentiation and integration do not change the radius of convergence. + It does not state anything about the + interval of convergence. + They are not always the same. +

      +
      2. + +
    2. +

      + Notice how the summation for \fp(x) starts with n=1. + This is because the constant term a_0 of f(x) becomes 0 through differentiation. +

      +
      3. + +
    3. +

      + Differentiation and integration are simply calculated term-by-term using the Power Rules. +

      +
      4. +
    +

    + + + Derivatives and indefinite integrals of power series + +

    + Let \ds f(x) = \infser[0] x^n. + Find \fp(x) and \ds F(x) =\int f(x)\, dx, + along with their respective intervals of convergence. +

    +     + +

    + We find the derivative and indefinite integral of f(x), + following . +

    + +

    +

      +
    1. +

      + + \fp(x) \amp = \infser nx^{n-1} = 1+2x+3x^2+4x^3+\cdots + \amp = \infser[0](n+1)x^n + . + In , + we recognized that \ds \infser[0] x^n is a geometric series in x. + We know that such a geometric series converges when \abs{x}\lt 1; + that is, the interval of convergence is (-1,1). + To determine the interval of convergence of \fp(x), + we consider the endpoints of (-1,1): + + \fp(-1) = 1-2+3-4+\cdots, \text{ which diverges. } + + + \fp(1) = 1+2+3+4+\cdots, \text{ which diverges. } + + Therefore, the interval of convergence of \fp(x) is (-1,1). +

      +
      2. + +
    2. +

      + \ds F(x) = \int f(x)\, dx = C+\infser[0] \frac{x^{n+1}}{n+1} = C+ x+\frac{x^2}{2}+\frac{x^3}3+\cdots + + To find the interval of convergence of F(x), + we again consider the endpoints of (-1,1): + + F(-1) = C-1+1/2-1/3+1/4+\cdots + + The value of C is irrelevant; + notice that the rest of the series is an Alternating Series that whose terms converge to 0. + By the Alternating Series Test, + this series converges. (In fact, + we can recognize that the terms of the series after C are the opposite of the Alternating Harmonic Series. + We can thus say that F(-1) = C-\ln(2).) + + F(1) = C+1+1/2+1/3+1/4+\cdots + + Notice that this summation is C\, + the Harmonic Series, + which diverges. + Since F converges for x=-1 and diverges for x=1, + the interval of convergence of F(x) is [-1,1). +

      +
      3. +
    +

    +     + +     + +

    + The previous example showed how to take the derivative and indefinite integral of a power series without motivation for why we care about such operations. + We may care for the sheer mathematical enjoyment that we can, + which is motivation enough for many. + However, we would be remiss to not recognize that we can learn a great deal from taking derivatives and indefinite integrals. +

    + +

    + Recall that \ds f(x) = \infser[0] x^n in is a geometric series. + According to , + this series converges to 1/(1-x) when \abs{x}\lt 1. + Thus we can say + + f(x) = \infser[0] x^n = \frac 1{1-x}, \text{ on } (-1,1) + . +

    + +

    + Integrating the power series, + (as done in ,) + we find + + F(x) = C_1+\infser[0] \frac{x^{n+1}}{n+1} + , + while integrating the function f(x) = 1/(1-x) gives + + F(x) = -\ln\abs{1-x} + C_2 + . +

    + +

    + Equating Equations and , we have + + F(x) = C_1+\infser[0] \frac{x^{n+1}}{n+1} = -\ln\abs{1-x} + C_2 + . +

    + +

    + Letting x=0, we have F(0) = C_1 = C_2. + This implies that we can drop the constants and conclude + + \infser[0] \frac{x^{n+1}}{n+1} = -\ln\abs{1-x} + . +

    + +

    + We established in + that the series on the left converges at x=-1; + substituting x=-1 on both sides of the above equality gives + + -1+\frac12-\frac13+\frac14-\frac15+\cdots = -\ln(2) + . +

    + +

    + On the left we have the opposite of the Alternating Harmonic Series; + on the right, we have -\ln(2). + We conclude that + + 1-\frac12+\frac13-\frac14+\cdots = \ln(2) + . +

    + +

    + Important: We stated in + (in ) + that the Alternating Harmonic Series converges to \ln(2), + and referred to this fact again in + of . + However, we never gave an argument for why this was the case. + The work above finally shows how we conclude that the Alternating Harmonic Series converges to \ln(2). + + Alternating Harmonic Series +

    + +

    + We use this type of analysis in the next example. +

    + + + Analyzing power series functions + +

    + Let \ds f(x) = \infser[0] \frac{x^n}{n!}. + Find \ds \fp(x) and \ds \int f(x)\, dx, + and use these to analyze the behavior of f(x). +

    +     + +

    + We start by making two notes: + first, in , + we found the interval of convergence of this power series is (-\infty,\infty). + Second, we will find it useful later to have a few terms of the series written out: + + \infser[0] \frac{x^n}{n!} = 1 + x + \frac{x^2}2+\frac{x^3}{6} + \frac{x^4}{24} +\cdots + +

    + +

    + We now find the derivative: + + \fp(x) \amp = \infser n\frac{x^{n-1}}{n!} + \amp =\infser \frac{x^{n-1}}{(n-1)!} = 1+x+\frac{x^2}{2!}+\cdots. + Since the series starts at n=1 and each term refers to (n-1), we can re-index the series starting with n=0: + \amp = \infser[0] \frac{x^{n}}{n!} + \amp = f(x) + . +

    + +

    + We found the derivative of f(x) is f(x). + The only functions for which this is true are of the form y=ce^x for some constant c. + As f(0) = 1 + (see Equation), + c must be 1. + Therefore we conclude that + + f(x) = \infser[0] \frac{x^n}{n!} = e^x + + for all x. +

    + +

    + We can also find \ds \int f(x)\, dx: + + \int f(x)\, dx \amp = C+\infser[0] \frac{x^{n+1}}{n!(n+1)} + \amp = C+ \infser[0] \frac{x^{n+1}}{(n+1)!} + +

    + +

    + We write out a few terms of this last series: + + C+ \infser[0] \frac{x^{n+1}}{(n+1)!} = C+ x+ \frac{x^2}2+\frac{x^3}{6}+\frac{x^4}{24}+\cdots + +

    + +

    + The integral of f(x) differs from f(x) only by a constant, + again indicating that f(x) = e^x. +

    +     + +     + +

    + + and the work following + established relationships between a power series function and regular + functions that we have dealt with in the past. + In general, given a power series function, it is difficult + (if not impossible) + to express the function in terms of elementary functions. + We chose examples where things worked out nicely. +

    + +

    + In this section's last example, + we show how to solve a simple differential equation with a power series. +

    + + + Solving a differential equation with a power series + +

    + Give the first 4 terms of the power series solution to \yp = 2y, + where y(0) = 1. +

    +     + +

    + The differential equation \yp = 2y describes a function y=f(x) where the derivative of y is twice y and y(0)=1. + This is a rather simple differential equation; + with a bit of thought one should realize that if y=Ce^{2x}, + then \yp = 2Ce^{2x}, and hence \yp = 2y. + By letting C=1 we satisfy the initial condition of y(0)=1. +

    + +

    + Let's ignore the fact that we already know the solution and find a power series function that satisfies the equation. + The solution we seek will have the form + + f(x) = \infser[0] a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+\cdots + + for unknown coefficients a_n. + We can find \fp(x) using : + + \fp(x) = \infser a_n\cdot n\cdot x^{n-1} = a_1+2a_2x+3a_3x^2+4a_4x^3\cdots + . +

    + +

    + Since \fp(x) = 2f(x), we have + + a_1+2a_2x+3a_3x^2+4a_4x^3\cdots \amp = 2\big(a_0+a_1x+a_2x^2+a_3x^3+\cdots\big) + \amp =2a_0+2a_1x+2a_2x^2+2a_3x^3+\cdots + +

    + +

    + The coefficients of like powers of x must be equal, + so we find that + + a_1 = 2a_0, 2a_2 = 2a_1, 3a_3 = 2a_2, 4a_4 = 2a_3, \text{ etc. } + +

    + +

    + The initial condition y(0) = f(0) = 1 indicates that a_0 = 1; + with this, we can find the values of the other coefficients: + + a_0 = 1 \text{ and } a_1=2a_0 \amp \Rightarrow a_1 = 2; + a_1 = 2 \text{ and } 2a_2 = 2a_1 \amp \Rightarrow a_2=4/2 =2; + a_2=2 \text{ and } 3a_3 = 2a_2 \amp \Rightarrow a_3=8/(2\cdot3)=4/3; + a_3=4/3 \text{ and } 4a_4 = 2a_3 \amp \Rightarrow a_4 =16/(2\cdot3\cdot4)= 2/3 + . +

    + +

    + Thus the first 5 terms of the power series solution to the differential equation \yp=2y is + + f(x) = 1+ 2x+2x^2 + \frac43x^3+\frac23x^4+\cdots + +

    + +

    + In , + as we study Taylor Series, + we will learn how to recognize this series as describing y=e^{2x}. +

    +     +     + +

    + Our last example illustrates that it can be difficult to recognize an elementary function by its power series expansion. + It is far easier to start with a known function, + expressed in terms of elementary functions, + and represent it as a power series function. + One may wonder why we would bother doing so, + as the latter function probably seems more complicated. + In the next two sections, + we show both how to do this and why + such a process can be beneficial. +

    + + + + Terms and Concepts + + + +

    + We adopt the convention that x^0=, + regardless of the value of x. +

    +     + + + + + + + + +

    + We define x^0=1 (even when x=0). +

    +     + +     + + + + +

    + What is the difference between the radius of convergence and the interval of convergence? +

    +     + + + +

    + The radius of convergence is a value + R such that a power series, centered at x=c, + converges for all values of x in (c-R,c+R). + The interval of convergence is an interval + on which the power series converges; + it may differ from (c-R,c+R) only at the endpoints. +

    +     + +     + + + + +

    + If the radius of convergence of + \ds\infser[0] a_nx^n is 5, what is the radius of convergence of \ds\infser n\cdot a_nx^{n-1}? +

    +     + +

    + It is still 5: the derivative of a power series has the same radius of convergence as the original power series. +

    +     + +     + + + + +

    + If the radius of convergence of + \ds\infser[0] a_nx^n is 5, + then the radius of convergence of \ds\infser[0] (-1)^na_nx^n is + + . +

    +     + + + + + + + + +

    + The radius of convergence will not change, since it is measured using absolute convergence. + However, it's possible that the endpoints of the interval of convergence might change. +

    +     +     +     +     + + +     +     + + Problems + + + +

    + Write out the sum of the first 5 terms of the given power series. +

    +     + + + + +

    + \ds\infser[0] 2^nx^n +

    +     + +

    + 1+2x+4x^2+8x^3+16x^4 +

    +     + +     + + + + +

    + \ds\infser \frac{1}{n^2}x^n +

    +     + +

    + x+\frac{x^2}{4}+\frac{x^3}{9}+\frac{x^4}{16}+\frac{x^5}{25} +

    +     + +     + + + + +

    + \ds\infser[0] \frac{1}{n!}x^n +

    +     + +

    + 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24} +

    +     + +     + + + + +

    + \ds\infser[0] \frac{(-1)^n}{(2n)!}x^{2n} +

    +     + +

    + 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320} +

    +     + +     + +     + + + +

    + A power series is given. +

    + +

    +

      +
    1. +

      + Find the radius of convergence. +

      +
      2. + +
    2. +

      + Find the interval of convergence. +

      +
      3. +
    +

    +     + + + + +

    + \ds\infser[0] \frac{(-1)^{n+1}}{n!}x^{n} +

    +     + +

    +

      +
    1. +

      + R=\infty +

      +
      2. + +
    2. +

      + (-\infty,\infty) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] nx^{n} +

    +     + +

    +

      +
    1. +

      + R=1 +

      +
      2. + +
    2. +

      + (-1,1) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser \frac{(-1)^n(x-3)^{n}}{n} +

    +     + +

    +

      +
    1. +

      + R=1 +

      +
      2. + +
    2. +

      + (2,4] +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] \frac{(x+4)^{n}}{n!} +

    +     + +

    +

      +
    1. +

      + R=\infty +

      +
      2. + +
    2. +

      + (-\infty,\infty) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] \frac{x^{n}}{2^n} +

    +     + +

    +

      +
    1. +

      + R=2 +

      +
      2. + +
    2. +

      + (-2,2) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] \frac{(-1)^n(x-5)^{n}}{10^n} +

    +     + +

    +

      +
    1. +

      + R=10 +

      +
      2. + +
    2. +

      + (-5,15) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] 5^n(x-1)^{n} +

    +     + +

    +

      +
    1. +

      + R=1/5 +

      +
      2. + +
    2. +

      + (4/5,6/5) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] (-2)^nx^{n} +

    +     + +

    +

      +
    1. +

      + R=1/2 +

      +
      2. + +
    2. +

      + (-1/2,1/2) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] \sqrt{n}x^{n} +

    +     + +

    +

      +
    1. +

      + R=1 +

      +
      2. + +
    2. +

      + (-1,1) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] \frac{n}{3^n}x^{n} +

    +     + +

    +

      +
    1. +

      + R=3 +

      +
      2. + +
    2. +

      + (-3,3) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] \frac{3^n}{n!}(x-5)^{n} +

    +     + +

    +

      +
    1. +

      + R=\infty +

      +
      2. + +
    2. +

      + (-\infty,\infty) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] (-1)^nn!(x-10)^{n} +

    +     + +

    +

      +
    1. +

      + R=0 +

      +
      2. + +
    2. +

      + x=10 +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser \frac{x^n}{n^2} +

    +     + +

    +

      +
    1. +

      + R=1 +

      +
      2. + +
    2. +

      + [-1,1] +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser \frac{(x+2)^n}{n^3} +

    +     + +

    +

      +
    1. +

      + R=1 +

      +
      2. + +
    2. +

      + [-3,-1] +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] n!\left(\frac x{10}\right)^n +

    +     + +

    +

      +
    1. +

      + R=0 +

      +
      2. + +
    2. +

      + x=0 +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] n^2\left(\frac{x+4}{4}\right)^n +

    +     + +

    +

      +
    1. +

      + R=4 +

      +
      2. + +
    2. +

      + x=(-8,0) +

      +
      3. +
    +

    +     + +     + +     + + + +

    + A function \ds f(x) = \infser[0] a_nx^n is given. +

    + +

    +

      +
    1. +

      + Give a power series for \fp(x) and its interval of convergence. +

      +
      2. + +
    2. +

      + Give a power series for \int f(x)\, dx and its interval of convergence. +

      +
      3. +
    +

    +     + + + + +

    + \ds\infser[0] nx^n +

    +     + +

    +

      +
    1. +

      + \ds \fp(x) = \infser n^2x^{n-1}; (-1,1) +

      +
      2. + +
    2. +

      + \ds \int f(x)\, dx = C+\infser[0] \frac{n}{n+1}x^{n+1}; + (-1,1) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser \frac{x^n}{n} +

    +     + +

    +

      +
    1. +

      + \ds \fp(x) = \infser x^{n-1}; (-1,1) +

      +
      2. + +
    2. +

      + \ds \int f(x)\, dx = C+\infser \frac{1}{n(n+1)}x^{n+1}; + [-1,1] +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] \left(\frac{x}{2}\right)^n +

    +     + +

    +

      +
    1. +

      + \ds \fp(x) = \infser \frac{n}{2^n}x^{n-1}; (-2,2) +

      +
      2. + +
    2. +

      + \ds \int f(x)\, dx = C+\infser[0] \frac{1}{(n+1)2^n}x^{n+1}; + [-2,2) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] (-3x)^n +

    +     + +

    +

      +
    1. +

      + \ds \fp(x) = \infser n(-3)^nx^{n-1}; (-1/3,1/3) +

      +
      2. + +
    2. +

      + \ds \int f(x)\, dx = C+\infser[0] \frac{(-3)^n}{n+1}x^{n+1}; + (-1/3,1/3] +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] \frac{(-1)^nx^{2n}}{(2n)!} +

    +     + +

    +

      +
    1. +

      + \ds \fp(x) = \infser \frac{(-1)^nx^{2n-1}}{(2n-1)!} =\infser[0] \frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}; + (-\infty,\infty) +

      +
      2. + +
    2. +

      + \ds \int f(x)\, dx = C+\infser[0] \frac{(-1)^nx^{2n+1}}{(2n+1)!}; + (-\infty,\infty) +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds\infser[0] \frac{(-1)^nx^{n}}{n!} +

    +     + +

    +

      +
    1. +

      + \ds \fp(x) = \infser \frac{(-1)^nx^{n-1}}{(n-1)!} =\infser[0] \frac{(-1)^{n+1}x^{n}}{n!}; + (-\infty,\infty) +

      +
      2. + +
    2. +

      + \ds \int f(x)\, dx = C+\infser[0] \frac{(-1)^{n}x^{n+1}}{(n+1)!}; + (-\infty,\infty) +

      +
      3. +
    +

    +     + +     + +     + + + +

    + Give the first 5 terms of the series that is a solution to the given differential equation. +

    +     + + + + +

    + y\,'=3y,y(0)=1 +

    +     + +

    + 1+3x+\frac92x^2+\frac92x^3+\frac{27}{8}x^4 +

    +     + +     + + + + +

    + y\,'=5y,y(0)=5 +

    +     + +

    + 5+25x+\frac{125}2x^2+\frac{625}{6}x^3+\frac{3125}{24}x^4 +

    +     + +     + + + + +

    + y\,'=y^2,y(0)=1 +

    +     + +

    + 1+x+x^2+x^3+x^4 +

    +     + +     + + + + +

    + y\,'=y+1,y(0)=1 +

    +     + +

    + 1+2x+x^2+\frac13{x^3}+\frac1{12}{x^4} +

    +     + +     + + + + +

    + y\,''=-y,y(0)=0, y\,'(0) = 1 +

    +     + +

    + 0+x+0x^2-\frac16x^3+0x^4 +

    +     + +     + + + + +

    + y\,''=2y,y(0)=1, y\,'(0) = 1 +

    +     + +

    + 1+x+x^2+\frac13x^3+\frac16x^4 +

    +     + +     + +     +     +     +
    +
    + Taylor Polynomials + + + +

    + Consider a function y=f(x) and a point \big(c,f(c)\big). + The derivative, \fp(c), + gives the instantaneous rate of change of f at x=c. + Of all lines that pass through the point \big(c,f(c)\big), + the line that best approximates f at this point is the tangent + line; that is, the line whose slope + (rate of change) + is \fp(c). +

    + +

    + In , + we see a function y=f(x) graphed. + The table in + shows that f(0)=2 and \fp(0) = 1; + therefore, the tangent line to f at x=0 is + p_1(x) = 1(x-0)+2 = x+2. + The tangent line is also given in the figure. + Note that near x=0, + p_1(x) \approx f(x); + that is, the tangent line approximates f well. +

    + + + +
    + + A graph of f(x) and its tangent line at 0 + + The graph of a generic function is shown, along with the tangent line to the graph at x=0. + +

    + The image shows the graph of a function f(x) and its tangent line at (0,f(0)). + The precise features of the graph of f(x) are unimportant for this illustration. + What is relevant is that points on the tangent line, which is the graph of a linear function labeled as p_1(x), + are close to the corresponding points on the graph of f(x), near the point (0,f(0)). + In other words, the image illustrates the fact that when x is close to zero, + the value of p_1(x) is close to the value of f(x). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-5.5,ymax=8.5, + xmin=-4.2,xmax=4.4 + ] + + \addplot [firstcurvestyle,infiniteleft,domain=-4:2,samples=60] {exp(x)*sin(deg(x))*cos(deg(x))+2}; + \addplot [firstcurvestyle,domain=2:3] {exp(x)*sin(deg(x))*cos(deg(x))+2}; + \addplot [firstcurvestyle,infiniteright,domain=3:3.4,samples=10] {exp(x)*sin(deg(x))*cos(deg(x))+2} node [pos=.8,left] { $y=f(x)$}; + + \addplot [secondcurvestyle,domain=-4:4] {x+2} node [pos=0,below right] { $y=p_1(x)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + Derivatives of f evaluated at 0 + +

    + + f(0) \amp = 2 \amp \fp''(0) \amp =-1 + \fp(0) \amp = 1 \amp f^{(4)}(0) \amp = -12 + \fpp(0) \amp = 2 \amp f^{(5)}(0) \amp = -19 + +

    +     + +
    +     + +

    + One shortcoming of this approximation is that the tangent line only + matches the slope of f; + it does not, for instance, match the concavity of f. + We can find a polynomial, p_2(x), + that does match the concavity near 0 without much difficulty, though. + The table in + gives the following information: + + f(0) = 2 \qquad \fp(0) = 1\qquad \fp'(0) = 2 + . +

    + +

    + Therefore, we want our polynomial p_2(x) to have these same + properties. That is, we need + + p_2(0) = 2 \qquad p_2'(0) = 1 \qquad p_2''(0) = 2 + . +

    + +

    + This is simply an initial-value problem. + We can solve this using the techniques first described in . + To keep p_2(x) as simple as possible, + we'll assume that not only p_2''(0)=2, + but that p_2''(x)=2. + That is, the second derivative of p_2 is constant, + meaning p_2 is a quadratic function. +

    + +

    + If p_2''(x) = 2, then + p_2'(x) = 2x+C for some constant C. + Since we have determined that p_2'(0) = 1, + we find that C=1 and so p_2'(x) = 2x+1. + Finally, we can compute p_2(x) = x^2+x+C. + Using our initial values, we know p_2(0) = 2 so C=2. + We conclude that p_2(x) = x^2+x+2. + This function is plotted with f in . +

    + + + +

    + We can repeat this approximation process by creating polynomials of + higher degree that match more of the derivatives of f at x=0. + In general, a polynomial of degree n can be created to match the + first n derivatives of f. + + shows p_4(x)= -x^4/2-x^3/6+x^2+x+2, + whose first four derivatives at 0 match those of f. (Using the table in , + start with p_4^{(4)}(x)=-12 and solve the related initial-value problem.) +

    + + + + + +
    + Plotting f, p_2 and p_4 + + + The graph of a function and two polynomials that approximate the function near x=0. + +

    + The graph of a function f(x) is shown. It is the same function as the first image in this section, + but again, the precise details of the graph are unimportant. +

    + +

    + Also shown are the graphs of two functions p_2(x) and p_4(x). + The function p_2(x) is quadratic, and its graph is a parabola that opens upward. + The function p_4(x) is a polynomial of degree 4. +

    + +

    + All three graphs intersect at the point (0,f(0)), + and the values of both p_2(x) and p_4(x) are close to the value of f(x) when x is close to 0. + Two observations are important: first, both of these polynomial graphs appear to lie more closely to the graph of f(x) + than the tangent line in the first image. Second, the graph of p_4(x) is a good approximation to f(x) + over a larger interval than the graph of p_2(x). +

    + +

    + In particular, the point (0,f(0)) appears to be a local minimum, + and there is a corresponding minimum in the graphs of both p_2(x) and p_4(x). + But the graph of f(x) also appears to have a local maximum near x=1. + Near x=1, the graph of p_2(x) separates from that of f(x): + the first continues to increase, while the second begins to decrease. + Near x=1, p_2(x) is no longer a good approximation to f(x). +

    + +

    + However, the graph of p_4(x) also has a maximum near x=1, + and p_4(x) appears to be a good approximation to f(x) at least until x=2. +

    +     + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-5.5,ymax=8.5, + xmin=-4.5,xmax=4.4 + ] + + \addplot+ [infinite,domain=-4:3.2,samples=120] {exp(x)*sin(deg(x))*cos(deg(x))+2} node [pos=0.6,left] { $y=f(x)$}; + \addplot+ [domain=-3:2] {x^2+x+2} node [pos=0.2, below left] { $y=p_2(x)$}; + \addplot+ [domain=-2:2] {-x^4/2-x^3/6+x^2+x+2} node [pos=0.1, below left] { $y=p_4(x)$}; + + \end{axis} + + \end{tikzpicture} + + + +
    + + + + + + + + + + + + + +

    + As we use more and more derivatives, + our polynomial approximation to f gets better and better. + In this example, the interval on which the approximation is + good gets bigger and bigger. + shows p_{13}(x); + we can visually affirm that this polynomial approximates f very + well on [-2,3]. + The polynomial p_{13}(x) is not + particularly nice. It is + + p_{13}(x)\amp=\frac{16901x^{13}}{6227020800}+\frac{13x^{12}}{1209600}-\frac{1321x^{11}}{39916800}-\frac{779x^{10}}{1814400} -\frac{359x^9}{362880} + \amp +\frac{x^8}{240}+\frac{139x^7}{5040}+\frac{11 x^6}{360}-\frac{19x^5}{120}-\frac{x^4}{2}-\frac{x^3}{6}+x^2+x+2 + . +

    + +
    + Plotting f and p_{13} + + + The graph of a function is shown, along with the graph of a degree 13 polynomial that closely approximates the function. + +

    + The graph of a function f(x) is shown. It is the same function as in the previous two images. + Also shown is the graph of a degree 13 polynomial p_{13}(x). + We can see that the values of f(x) and p_{13}(x) are very close over an interval + from -2 to 3. For x \lt -3, the graph of f(x) appears to approach a horizontal asymptote, + while the graph of p_{13}(x) approaches -\infty, so it ceases to be a good approximation to f(x) at this point. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-5.5,ymax=8.5, + xmin=-4.5,xmax=4.4 + ] + + \addplot+ [infinite,domain=-3.5:3.3,samples=120] {exp(x)*sin(deg(x))*cos(deg(x))+2} node [pos=0, above] { $y=f(x)$}; + \addplot+ [infinite,smooth] coordinates {(-3.52,-6.364)(-3.36,-2.334)(-3.2,-0.1706)(-3.04,0.9569)(-2.88,1.528)(-2.72,1.809)(-2.56,1.944)(-2.4,2.009)(-2.24,2.038)(-2.08,2.048)(-1.92,2.046)(-1.76,2.031)(-1.6,2.006)(-1.44,1.969)(-1.28,1.924)(-1.12,1.872)(-0.96,1.82)(-0.8,1.775)(-0.64,1.747)(-0.48,1.747)(-0.32,1.783)(-0.16,1.866)(0,2.)(0.16,2.185)(0.32,2.411)(0.48,2.662)(0.64,2.908)(0.8,3.112)(0.96,3.227)(1.12,3.202)(1.28,2.988)(1.44,2.546)(1.6,1.855)(1.76,0.9264)(1.92,-0.1926)(2.08,-1.406)(2.24,-2.565)(2.4,-3.474)(2.56,-3.891)(2.72,-3.542)(2.88,-2.133)(3.04,0.6461)(3.2,5.139)(3.36,11.79)} node [pos=0.2, right] {$p_{13}(x)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + The polynomials we have created are examples of + Taylor polynomials, + named after the British mathematician Brook Taylor who made important discoveries about such functions. + While we created the above Taylor polynomials by solving initial-value problems, + it can be shown that Taylor polynomials follow a general pattern that make their formation much more direct. + This is described in the following definition. +

    + + + + + Taylor Polynomial, Maclaurin Polynomial + +

    + Let f be a function whose first n derivatives exist at + x=c. + Taylor Polynomialdefinition + Maclaurin Polynomialdefinition + Maclaurin Polynomial|see{Taylor Polynomial} + +

      +
    1. +

      + The Taylor polynomial of degree n of f at + x=c is + + p_n(x) \amp = f(c) + \fp(c)(x-c) + \frac{\fpp(c)}{2!}(x-c)^2 + \amp \quad+\frac{\fp''(c)}{3!}(x-c)^3+\cdots+\frac{f\,^{(n)}(c)}{n!}(x-c)^n + . +

      +
      2. + +
    2. +

      + A special case of the Taylor polynomial is the Maclaurin + polynomial, where c=0. + That is, the Maclaurin polynomial of degree n of + f is + + p_n(x) = f(0) + \fp(0)x + \frac{\fpp(0)}{2!}x^2+\frac{\fp''(0)}{3!}x^3+\cdots+\frac{f\,^{(n)}(0)}{n!}x^n + . +

      +
      3. +
    +

    +     +     + + + + + + + + + +

    + We will practice creating Taylor and Maclaurin polynomials in the + following examples. +

    + + + Finding and using Maclaurin polynomials + +

    +

      +
    1. +

      + Find the nth Maclaurin polynomial for f(x) = e^x. +

      +
      2. + +
    2. +

      + Use p_5(x) to approximate the value of e. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + We start with creating a table of the derivatives of e^x + evaluated at x=0. + In this particular case, this is relatively simple, + as shown in . +

      + +
      + The derivatives of f(x)=e^x evaluated at x=0 + +

      + + f(x) \amp = e^x \amp f(0) \amp = 1 + \fp(x) \amp = e^x \amp \fp(0) \amp =1 + \fp'(x) \amp = e^x \amp \fp'(0) \amp =1 + \amp \vdots \amp \amp \vdots + f^{(n)}(x) \amp = e^x \amp f^{(n)}(0) \amp =1 + +

      +       + +
      + +

      + By the definition of the Maclaurin polynomial, we have + + p_n(x) \amp = f(0) + \fp(0)x + \frac{\fpp(0)}{2!}x^2+\frac{\fp''(0)}{3!}x^3+\cdots+\frac{f\,^{(n)}(0)}{n!}x^n + \amp = 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3 + \frac{1}{24}x^4 + \cdots + \frac{1}{n!}x^n + . +

      +
      2. + +
    2. +

      + Using our answer from part 1, we have + + e^x\approx p_5(x) = 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5 + . + To approximate the value of e, + note that e = e^1 = f(1) \approx p_5(1). + It is very straightforward to evaluate p_5(1): + + p_5(1) = 1+1+\frac12+\frac16+\frac1{24}+\frac1{120} = \frac{163}{60} \approx 2.71667 + . + A plot of f(x)=e^x and p_5(x) is given in + . + To 5 decimal places, the actual value of e is + 2.71828. So this approximation agrees to two decimal places. + +

      +
      3. +
    +

    +
    + A plot of f(x)=e^x and its 5^{th} degree + Maclaurin polynomial p_5(x) + + + The graph of the exponential function and its degree 5 Maclaurin polynomial. + +

    + The graph of f(x)=e^x is shown on the interval [-3,2]. + Also shown is the graph of p_5(x), the degree 5 Maclaurin polynomial of f(x). + For -2\lt x\lt 2, there is very little visible difference between the two graphs. + Near x=3 we can see that the two graphs begin to separate. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-3,ymax=11, + xmin=-3.75,xmax=2.9 + ] + + \addplot+ [infinite,domain=-3.2:2.2,samples=40] {e^x} node [pos=0,above right] { $y=f(x)$}; + \addplot+ [infinite,domain=-3.5:2.4,samples=40] {1+x+(1/2)*x^2+(1/6)*x^3+(1/24)*x^4+(1/120)*x^5} node [pos=0.4,below right] { $y=p_{5}(x)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +     + + + + + Finding and using Taylor polynomials + +

    +

      +
    1. +

      + Find the nth Taylor polynomial of y=\ln(x) at x=1. +

      +
      2. + +
    2. +

      + Use p_6(x) to approximate the value of \ln(1.5). +

      +
      3. + +
    3. +

      + Use p_6(x) to approximate the value of \ln(2). +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + We begin by creating a table of derivatives of \ln(x) + evaluated at x=1. + While this is not as straightforward as it was in the previous + example, a pattern does emerge (for n\ge 1), + as shown in . + + Notice in the table below that each time we take a derivative + (starting at the second derivative), + we apply the power rule and bring down + the exponent to multiply by the previous coefficent. + So the 6 in the 4^{th} derivative is actually + 1\cdot 2\cdot 3=3!. +

      + +
      + Derivatives of \ln(x) evaluated at x=1 + +

      + + f(x) \amp = \ln(x) \amp f(1) \amp = 0 + \fp(x) \amp = \frac1x \amp \fp(1) \amp = 1 + \fp'(x) \amp = -\frac{1}{x^2} \amp \fp'(1) \amp = -1 + \fp''(x) \amp = \frac{2}{x^3} \amp \fp''(1) \amp = 2 + f^{(4)}(x) \amp = -\frac{6}{x^4} \amp f^{(4)}(1) \amp = -6 + \amp \vdots \amp \amp \vdots + f^{(n)}(x) \amp = \amp f^{(n)}(1) \amp = + (-1)^{n+1}\amp\frac{(n-1)!}{x^n} \amp (-1)^{n+1}\amp(n-1)! + +

      +       + +
      + +

      + Notice that the coefficients alternate in sign starting at n=1. + Using , we have + + p_n(x) \amp = f(c) + \fp(c)(x-c) + \frac{\fpp(c)}{2!}(x-c)^2+\dots + \amp \dots\frac{\fp''(c)}{3!}(x-c)^3+\cdots+\frac{f\,^{(n)}(c)}{n!}(x-c)^n + \amp = 0 + \frac{0!}{1!}(x-1) - \frac{1!}{2!}(x-1)^2+\dots + \amp \dots\frac{2!}{3!}(x-1)^3+\cdots+\frac{(-1)^{n+1}\cdot (n-1)!}{n!}(x-1)^n + \amp = (x-1)-\frac12(x-1)^2+\frac13(x-1)^3-\dots + \amp \dots \frac14(x-1)^4+\cdots+\frac{(-1)^{n+1}}{n}(x-1)^n + . + Note how the coefficients of the (x-1) terms turn out to + be nice. +

      +
      2. + +
    2. +

      + We can compute p_6(x) using our work above: + + p_6(x) \amp = (x-1)-\frac12(x-1)^2+\frac13(x-1)^3 + \amp \quad\quad -\frac14(x-1)^4+\frac15(x-1)^5-\frac16(x-1)^6 + . + Since p_6(x) approximates \ln(x) well near x=1, + we approximate \ln(1.5) \approx p_6(1.5): + + p_6(1.5) \amp = (1.5-1)-\frac12(1.5-1)^2+\frac13(1.5-1)^3+\dots + \amp \dots -\frac14(1.5-1)^4 +\frac15(1.5-1)^5-\frac16(1.5-1)^6 + \amp =\frac{259}{640} + \amp \approx 0.404688 + . + This is a good approximation as a calculator shows that + \ln(1.5) \approx 0.4055. + below + plots y=\ln(x) with y=p_6(x). + We can see that \ln(1.5) \approx p_6(1.5). + + +

      +
      3. + +
    3. +

      + We approximate \ln 2 with p_6(2): + + p_6(2) \amp = (2-1)-\frac12(2-1)^2+\frac13(2-1)^3-\frac14(2-1)^4+\cdots + \amp \cdots +\frac15(2-1)^5-\frac16(2-1)^6 + \amp = 1-\frac12+\frac13-\frac14+\frac15-\frac16 + \amp = \frac{37}{60} + \amp \approx 0.616667 + . + This approximation is not terribly impressive: + a hand held calculator shows that \ln(2) \approx 0.693147. + The graph in + shows that p_6(x) provides less accurate approximations + of \ln(x) as x gets close to 0 or 2. + + Surprisingly enough, + even the 20th degree Taylor polynomial fails to + approximate \ln(x) for x\gt 2 very well, + as shown in . + We'll soon discuss why this is. + + + +

      +
      4. +
    +

    + +
    + A plot of y=\ln(x) and its 6th degree Taylor + polynomial at x=1 + + + A graph of the natural logarithm function and its degree 6 Taylor polynomial centered at x=1. + +

    + The graph of y=\ln(x) is shown on the interval (0,3). + Also shown is the graph of p_6(x), the degree 6 Taylor polynomial of \ln(x), + centered at x=1. + The graph of p_6(x) is very close to the graph of \ln(x) on the interval (0.5,1.5), + but does not approximate the logarithm very well outside of this interval. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-4.5,ymax=2.4, + xmin=-.5,xmax=3.2 + ] + + \addplot+ [infinite,domain=0.01:2.8,samples=101] {ln(x)} node [pos=1,above left] { $y=\ln(x)$}; + \addplot+ [infinite,domain=-0.22:2.8,samples=40] {(x-1)-.5*(x-1)^2+(1/3)*(x-1)^3-(1/4)*(x-1)^4+(1/5)*(x-1)^5-(1/6)*(x-1)^6} node [pos=1,left] { $y=p_{6}(x)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + A plot of y=\ln(x) and its 20^{th} degree + Taylor polynomial at x=1 + + + A graph of the natural logarithm function and its degree 20 Taylor polynomial centered at x=1. + +

    + The graph y=\ln(x) is shown again, this time with the degree 20 Taylor polynomial of \ln(x), centered at x=1. + Unlike with the example involving y=e^x, increasing the degree of the Taylor polynomial did not do much to improve + how well it approximates the logarithm. The approximation appears to be accurate over a slightly larger interval + than the degree 6 approximation, but there is not a significant difference between the two images. +

    +     + + \begin{tikzpicture} + \begin{axis}[ymax=3] + \addplot+[infinite,domain=-4.6:1.1,variable=t] ({e^t},{t}) node [pos=1,above left] {$y=\ln(x)$}; + \addplot+[infinite,domain=-0.1:2.3,smooth] {(x-1) - (x-1)^2/2 + (x-1)^3/3 - (x-1)^4/4 + (x-1)^5/5 - (x-1)^6/6 + (x-1)^7/7 - (x-1)^8/8 + (x-1)^9/9 - (x-1)^10/10 + (x-1)^11/11 - (x-1)^12/12 + (x-1)^13/13 - (x-1)^14/14 + (x-1)^15/15 - (x-1)^16/16 + (x-1)^17/17 - (x-1)^18/18 + (x-1)^19/19 - (x-1)^20/20} node [pos=0.8,below right] {$y=p_{20}(x)$}; + \end{axis} + \end{tikzpicture} + + + +
    +     +     + +     + + + +

    + Taylor polynomials are used to approximate functions f(x) in mainly + two situations: +

      +
    1. +

      + When f(x) is known, but perhaps + hard to compute directly. + For instance, we can define + the cosine of an angle as either the ratio of sides of a right + triangle (adjacent over hypotenuse) or using the definition + in terms of the unit circle. + However, neither of these provides a convenient way of computing + \cos(2). + A Taylor polynomial of sufficiently high degree can provide a + reasonable method of computing such values using only operations + usually hard-wired into a computer + (+, -, and \div). +

      +
      2. + +
    2. +

      + When f(x) is not known, + but information about its derivatives is known. + This occurs more often than one might think, + especially in the study of differential equations. +

      +
      3. +
    +

    + +

    + In both situations, a critical piece of information to have is + How good is my approximation? + If we use a Taylor polynomial to compute \cos(2), + how do we know how accurate the approximation is? +

    + + + +

    + We had the same problem when studying Numerical Integration. + + provided bounds on the error when using, + say, Simpson's Rule to approximate a definite integral. + These bounds allowed us to determine that, for instance, + using 10 subintervals provided an approximation within \pm 0.01 of the exact value. + The following theorem gives similar bounds for Taylor + (and hence Maclaurin) + polynomials. +

    + + + + + Taylor's Theorem + +

    +

      +
    1. +

      + Let f be a function whose + (n+1)\text{th } derivative exists on an interval I + and let c be in I. + Then, for each x in I, + there exists z_x between x and c such that + + f(x) = f(c) + \fp(c)(x-c) + \cdots + +\frac{f^{(n)}(c)}{n!}(x-c)^n+R_n(x) + , + where \ds R_n(x) = \frac{f^{(n+1)}(z_x)}{(n+1)!}(x-c)^{(n+1)}. + Taylor PolynomialTaylor's Theorem + Taylor's Theorem +

      +
      2. + +
    2. +

      + \ds \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{(x-c)^{(n+1)}}, + where z is in I. +

      +
      3. +
    +

    +     +     + + + + + + + +

    + The first part of Taylor's Theorem states that f(x) = p_n(x) + R_n(x), + where p_n(x) is the + nth order Taylor polynomial and R_n(x) is the remainder, + or error, in the Taylor approximation. + The second part gives bounds on how big that error can be. + If the (n+1)th derivative is large on I, + the error may be large; + if x is far from c, + the error may also be large. + However, the (n+1)! term in the denominator tends to ensure that + the error gets smaller as n increases. +

    + +

    + The following example computes error estimates for the approximations of + \ln(1.5) and \ln(2) made in . +

    + + + Finding error bounds of a Taylor polynomial + +

    + Use + to find error bounds when approximating + \ln(1.5) and \ln(2) with p_6(x), + the Taylor polynomial of degree 6 of f(x)=\ln(x) at x=1, + as calculated in . +

    +     + +

    +

      +
    1. +

      + We start with the approximation of + \ln(1.5) with p_6(1.5). + The theorem references an open interval I that contains + both x and c. + The smaller the interval we use the better; + it will give us a more accurate + (and smaller!) + approximation of the error. + We let I = (0.9,1.6), + as this interval contains both c=1 and x=1.5. + The theorem references \max\abs{f^{(n+1)}(z)}. + In our situation, this is asking + How big can the 7th derivative of + y=\ln(x) be on the interval (0.9,1.6)? + The seventh derivative is y = -6!/x^7. + The largest absolute value it attains on I is about 1506. + (There are no critical numbers of f^{(7)} in the interval + so we evaluate the endpoints: + f^{(7)}(0.9)\approx 1506 and f^{(7)}(1.6)\approx 27.) + In particular, + we are evaluating at x=1.5, so we let x=1.5. + Thus we can bound the error as: + + \abs{R_6(1.5)} \amp \leq \frac{\max\abs{f^{(7)}(z)}}{7!}\abs{(1.5-1)^7} + \amp \leq \frac{1506}{5040}\cdot\frac1{2^7} + \amp \approx 0.0023 + . + We computed p_6(1.5) = 0.404688; + using a calculator, we find \ln(1.5) \approx 0.405465, + so the actual error is about 0.000778, + which is less than our bound of 0.0023. + This affirms Taylor's Theorem; + the theorem states that our approximation would be within about + 2 thousandths of the actual value, + whereas the approximation was actually closer. + only gives an upper + bound on the error. +

      +
      2. + +
    2. +

      + We again find an interval I that contains both c=1 + and x=2; we choose I = (0.9,2.1). + The maximum value of the seventh derivative of f on this + interval is again about 1506 + (as the largest values come near x=0.9). + Thus + + \abs{ R_6(2)} \amp \leq \frac{\max\abs{f^{(7)}(z)}}{7!}\abs{(2-1)^7} + \amp \leq \frac{1506}{5040}\cdot1^7 + \amp \approx 0.30 + . + This bound is not as nearly as good as before. + Using the degree 6 Taylor polynomial at x =1 will bring + us within 0.3 of the correct answer. + As p_6(2)\approx 0.61667, + our error estimate guarantees that the actual value of + \ln(2) is somewhere between 0.31667 and 0.91667. + These bounds are not particularly useful. + In reality, our approximation was only off by about 0.07. + However, we are approximating ostensibly because we do not know + the real answer. + In order to be assured that we have a good approximation, + we would have to resort to using a polynomial of higher degree. +

      +
      3. +
    +

    +     + +     + +

    + We practice again. + This time, we use Taylor's theorem to find n that guarantees our + approximation is within a certain amount. +

    + + + Finding sufficiently accurate Taylor polynomials + +

    + Find n such that the nth Taylor polynomial of + f(x)=\cos(x) at x=0 approximates \cos(2) to + within 0.001 of the actual answer. + What is p_n(2)? +

    +     + +

    + Following Taylor's theorem, + we need bounds on the size of the derivatives of f(x)=\cos(x). + In the case of this trigonometric function, this is easy. + All derivatives of cosine are + \pm \sin(x) or \pm \cos(x). + In all cases, + these functions are never greater than 1 in absolute value. + We want the error to be less than 0.001. + To find the appropriate n, + consider the following inequalities: + + \frac{\max\abs{f^{(n+1)}(z)}}{(n+1)!}\abs{(2-0)^{(n+1)}} \amp \leq 0.001 + \frac1{(n+1)!}\cdot2^{(n+1)} \amp \leq 0.001 + . +

    + +

    + We find an n that satisfies this last inequality with + trial-and-error. + When n=8, + we have \ds \frac{2^{8+1}}{(8+1)!} \approx 0.0014; + when n=9, + we have \ds \frac{2^{9+1}}{(9+1)!} \approx 0.000282 \lt 0.001. + Thus we want to approximate \cos(2) with p_9(2). +

    + +

    + We now set out to compute p_9(x). + We again need a table of the derivatives of + f(x)=\cos(x) evaluated at x=0. + A table of these values is given in . +

    + + + +
    + A table of the derivatives of f(x)=\cos(x) evaluated at x=0 + +

    + + f(x) \amp = \cos(x) \amp f(0) \amp = 1 + f'(x) \amp = -\sin(x) \amp \fp(0) \amp = 0 + \fp'(x) \amp = -\cos(x) \amp \fp'(0) \amp = -1 + \fp''(x) \amp = \sin(x) \amp \fp''(0) \amp = 0 + f^{(4)}(x) \amp = \cos(x) \amp f^{(4)}(0) \amp =1 + f^{(5)}(x) \amp = -\sin(x) \amp f^{(5)}(0) \amp =0 + f^{(6)}(x) \amp = -\cos(x) \amp f^{(6)}(0) \amp =-1 + f^{(7)}(x) \amp = \sin(x) \amp f^{(7)}(0) \amp =0 + f^{(8)}(x) \amp = \cos(x) \amp f^{(8)}(0) \amp =1 + f^{(9)}(x) \amp = -\sin(x) \amp f^{(9)}(0) \amp =0 + +

    +     + +
    + + + +

    + Notice how the derivatives, evaluated at x=0, follow a certain + pattern. All the odd powers of x in the Taylor polynomial will + disappear as their coefficient is 0. + While our error bounds state that we need p_9(x), + our work shows that this will be the same as p_8(x). +

    + +

    + Since we are forming our polynomial at x=0, + we are creating a Maclaurin polynomial, and: + + p_8(x) \amp = f(0) + \fp(0)x + \frac{\fpp(0)}{2!}x^2 + \frac{\fp''(0)}{3!}x^3 + \cdots +\frac{f^{(8)}(0)}{8!}x^8 + \amp = 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+\frac{1}{8!}x^8 + . +

    + +

    + We finally approximate \cos(2): + + \cos(2) \approx p_8(2) = -\frac{131}{315} \approx -0.41587 + . +

    + +

    + Our error bound guarantee that this approximation is within + 0.001 of the correct answer. + Technology shows us that our approximation is actually within about + 0.0003 of the correct answer. +

    + +

    + + shows a graph of y=p_8(x) and y=\cos(x). + Note how well the two functions agree on about (-\pi,\pi). +

    + +
    + A graph of f(x)= \cos(x) and its degree 8 Maclaurin + polynomial + + + A graph of the cosine function is shown, along with its degree 8 Maclaurin polynomial approximation. + +

    + The graph y=\cos(x) is given on the interval [-5,5], + along with the graph of p_8(x), the degree 8 Maclaurin polynomial approximation of \cos(x). +

    + +

    + The image illustrates how well the Maclaurin polynomials approximate the cosine function. + With a degree 8 polynomial, there is little to no visible difference between the graphs over the interval (-\pi,\pi). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-5,-4,-3,-2,-1,1,2,3,4,5}, + ymin=-1.5,ymax=1.5, + xmin=-5.5,xmax=5.5 + ] + + \addplot+[infinite,domain=-5:5,samples=101] {cos(deg(x))} node[pos=0.6, right] { $y=cos(x)$}; + \addplot+[infinite,smooth] coordinates {(-5.,2.528)(-4.8,1.6)(-4.6,0.8894)(-4.4,0.343)(-4.2,-0.0768)(-4.,-0. + 3968)(-3.8,-0.6355)(-3.6,-0.8052)(-3.4,-0.9146)(-3.2,-0.9695)(-3.,-0. + 9748)(-2.8,-0.9345)(-2.6,-0.8532)(-2.4,-0.7357)(-2.2,-0.5878)(-2.,-0. + 4159)(-1.8,-0.2271)(-1.6,-0.02917)(-1.4,0.17)(-1.2,0.3624)(-1.,0.5403) + (-0.8,0.6967)(-0.6,0.8253)(-0.4,0.9211)(-0.2,0.9801)(0,1.)(0.2,0.9801) + (0.4,0.9211)(0.6,0.8253)(0.8,0.6967)(1.,0.5403)(1.2,0.3624)(1.4,0.17)( + 1.6,-0.02917)(1.8,-0.2271)(2.,-0.4159)(2.2,-0.5878)(2.4,-0.7357)(2.6,- + 0.8532)(2.8,-0.9345)(3.,-0.9748)(3.2,-0.9695)(3.4,-0.9146)(3.6,-0. + 8052)(3.8,-0.6355)(4.,-0.3968)(4.2,-0.0768)(4.4,0.343)(4.6,0.8894)(4. + 8,1.6)(5.,2.528)} node[pos=0.1,right] { $y=p_8(x)$} ; + + \end{axis} + + \end{tikzpicture} + + + + +
    + + +     + + +     + + + + + Finding and using Taylor polynomials + +

    +

      +
    1. +

      + Find the degree 4 Taylor polynomial, + p_4(x), for f(x)=\sqrt{x} at x=4. +

      +
      2. + +
    2. +

      + Use p_4(x) to approximate \sqrt{3}. +

      +
      3. + +
    3. +

      + Find bounds on the error when approximating \sqrt{3} + with p_4(3). +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + We begin by evaluating the derivatives of f at x=4. + This is done in . +

      + +
      + A table of the derivatives of f(x)=\sqrt{x} evaluated at x=4 + +

      + + f(x) \amp = \sqrt{x} \amp f(4) \amp = 2 + \fp(x) \amp = \frac{1}{2\sqrt{x}} \amp \fp(4) \amp = \frac14 + \fp'(x) \amp = \frac{-1}{4x^{3/2}} \amp \fp'(4) \amp = \frac{-1}{32} + \fp''(x) \amp = \frac{3}{8x^{5/2}} \amp \fp''(4) \amp = \frac{3}{256} + f^{(4)}(x) \amp = \frac{-15}{16x^{7/2}} \amp f^{(4)}(4) \amp = \frac{-15}{2048} + +

      +       + +
      + + + +

      + These values allow us to form the Taylor polynomial p_4(x): + + p_4(x) = 2 \amp + \frac14(x-4) +\frac{-1/32}{2!}(x-4)^2+\dots + \amp \dots \frac{3/256}{3!}(x-4)^3+\frac{-15/2048}{4!}(x-4)^4 + . +

      +
      2. + +
    2. +

      + As p_4(x) \approx \sqrt{x} near x=4, + we approximate \sqrt{3} with p_4(3) = 1.73212. +

      +
      3. + +
    3. +

      + To find a bound on the error, + we need an open interval that contains x=3 and x=4. + We set I = (2.9,4.1). + The largest value the fifth derivative of + f(x)=\sqrt{x} takes on this interval is near x=2.9, + at about 0.0273. (We often graph the + (n+1)^{th} derivative to find its extrema. + In this case is f^{(5)}(x)=105/(32x^{9/2}) is always + decreasing, so the maximum occurs at 2.9.) Thus + + \abs{R_4(3)} \leq \frac{0.0273}{5!}\abs{(3-4)^5} \approx 0.00023 + . + This shows our approximation is accurate to at least the first 2 + places after the decimal. + (It turns out that our approximation is actually accurate to 4 + places after the decimal.) + A graph of f(x)=\sqrt x and p_4(x) is given in + . + Note how the two functions are nearly indistinguishable on + (2,7). + + + + +

      +
      4. +
    +

    + +
    + A graph of f(x)=\sqrt{x} and its degree 4 Taylor + polynomial at x=4 + + + The graph of the square root function and its degree 4 Taylor polynomial, centered at x=4. + +

    + The graph y=\sqrt{x} is shown, for x in the interval [0,10]. + Also shown is the degree 4 Taylor polynomial p_4(x), centered at x=4. + The Taylor polynomial appears to approximate \sqrt{x} very well over the interval (2,7). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=3.6, + xmin=-1,xmax=11.2 + ] + + \addplot+ [rightarrow,domain=0:3.16,samples=40] ({x^2},{x}) node [pos=1,above left] { $y=\sqrt{x}$}; + \addplot+ [rightarrow] coordinates {(0,0.5469)(0.2,0.6536)(0.4,0.7551)(0.6,0.8518)(0.8,0.944)(1.,1.032)(1. + 2,1.116)(1.4,1.196)(1.6,1.273)(1.8,1.346)(2.,1.417)(2.2,1.485)(2.4,1. + 55)(2.6,1.613)(2.8,1.673)(3.,1.732)(3.2,1.789)(3.4,1.844)(3.6,1.897)( + 3.8,1.949)(4.,2.)(4.2,2.049)(4.4,2.098)(4.6,2.145)(4.8,2.191)(5.,2. + 236)(5.2,2.28)(5.4,2.324)(5.6,2.366)(5.8,2.408)(6.,2.448)(6.2,2.488)( + 6.4,2.527)(6.6,2.565)(6.8,2.602)(7.,2.637)(7.2,2.672)(7.4,2.705)(7.6, + 2.737)(7.8,2.768)(8.,2.797)(8.2,2.824)(8.4,2.849)(8.6,2.873)(8.8,2. + 894)(9.,2.913)(9.2,2.929)(9.4,2.942)(9.6,2.953)(9.8,2.96)(10.,2.964)} node [pos=0.8,below right] { $y=p_{4}(x)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +     + +

    + Our final example gives a brief introduction to using Taylor polynomials + to solve differential equations. +

    + + + Approximating an unknown function + +

    + A function y=f(x) is unknown save for the following two facts. + +

      +
    1. +

      + y(0) = f(0) = 1, and +

      +
      2. + +
    2. +

      + \yp= y^2 +

      +
      3. +
    +

    + +

    + (This second fact says that amazingly, + the derivative of the function is actually the function squared!) +

    + +

    + Find the degree 3 Maclaurin polynomial p_3(x) of y=f(x). +

    +     + +

    + One might initially think that not enough information is given to + find p_3(x). + However, note how the second fact above actually lets us know what + \yp(0) is: + + \yp = y^2 \Rightarrow \yp(0) = y^2(0) + . +

    + +

    + Since y(0) = 1, we conclude that \yp(0) = 1. +

    + +

    + Now we find information about \yp'. + Starting with \yp=y^2, + take derivatives of both sides, + with respect to x. + That means we must use implicit differentiation. + + \yp \amp = y^2 + \frac{d}{dx}\big(\yp\big) \amp = \frac{d}{dx}\big(y^2\big) + \yp' \amp = 2y\cdot \yp. + Now evaluate both sides at x=0: + \yp'(0) \amp = 2y(0)\cdot \yp(0) + \yp'(0) \amp = 2 + . +

    + +

    + We repeat this once more to find \yp''(0). + We again use implicit differentiation; + this time the Product Rule is also required. + + \frac{d}{dx}\big(\yp'\big) \amp = \frac{d}{dx} \big(2y\yp\big) + \yp'' \amp = 2\yp\cdot \yp + 2y\cdot \yp'. + Now evaluate both sides at x=0: + \yp''(0) \amp = 2\yp(0)^2 + 2y(0)\yp'(0) + \yp''(0) \amp = 2+4=6 + . +

    + +

    + In summary, we have: + + y(0) = 1 \qquad \yp(0) = 1 \qquad \yp'(0) = 2 \qquad \yp''(0) = 6 + . +

    + +

    + We can now form p_3(x): + + p_3(x) \amp = 1 + x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \amp = 1+x+x^2+x^3 + . + + + +

    + +

    + It turns out that the differential equation we started with, + \yp=y^2, + where y(0)=1, can be solved without too much difficulty: + y = \frac{1}{1-x}. + + shows this function plotted with p_3(x). + Note how similar they are near x=0. +

    + +
    + A graph of y=-1/(x-1) and y=p_3(x) from + + + + The graph of the exact solution to the differential equation in this example, and a polynomial approximation. + +

    + The graph y=1/(1-x) is shown, for x from -1 to about 0.6. + Also shown is the graph of p_3(x), the Maclaurin polynomial obtained as an approximate solution to the differential equation in this example. + As expected, the polynomial is a good approximation to the exact solution when x is close to 0. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=3.5, + xmin=-1.1,xmax=1.4 + ] + + \addplot+ [infinite,domain=-1:.65,samples=40] {-1/(x-1)} node[pos=1,left]{ $\displaystyle y= \frac{1}{1-x}$}; + \addplot+ [infinite,domain=-1:0.81,samples=40] {1+x+x^2+x^3} node [pos=1, below right]{ $y=p_3(x)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +     + +

    + It is beyond the scope of this text to pursue error analysis when using + Taylor polynomials to approximate solutions to differential equations. + This topic is often broached in introductory Differential Equations + courses and usually covered in depth in Numerical Analysis courses. + Such an analysis is very important; + one needs to know how good their approximation is. + We explored this example simply to demonstrate the usefulness of Taylor + polynomials. +

    + +

    + Most of this chapter has been devoted to the study of infinite series. + This section has taken a step back from this study, + focusing instead on finite summation of terms. + In the next section, we explore Taylor Series, + where we represent a function with an infinite series. +

    + + + + + + Terms and Concepts + + + + +

    + What is the difference between a Taylor polynomial and a + Maclaurin polynomial? +

    + + +     + + + +

    + The Maclaurin polynomial is a special case of Taylor polynomials. + Taylor polynomials are centered at a specific x-value; + when that x-value is 0, it is a Maclauring polynomial. +

    +     + +     + + + + +

    + In general, + p_n(x) approximates f(x) better and better as + n gets larger. + +

    +     + +

    + True: this is essentially the content of Taylor's Theorem. + For most well-behaved functions, the size of the remainder, + which quantifies the error in our approximation, gets smaller. +

    +     + +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $p = Formula("6+3*x-4*x^2"); + + +

    + For some function f(x), + the Maclaurin polynomial of degree 4 is + p_4(x) = 6+3x-4x^2+5x^3-7x^4. + What is p_2(x)? +

    + +

    + +

    +     + +

    + A higher-degree Maclaurin polynomial begins with the same terms + as any lower-degree + Maclaurin polynomial of the same function. Therefore, + p_2(x) = 6+3x-4x^2. +

    +     +     +     + + + + + + +

    + For some function f(x), + the Maclaurin polynomial of degree 4 is + p_4(x) = 6+3x-4x^2+5x^3-7x^4. + What is \fpp'(0)? +

    + +

    + +

    +     + +

    + Since the coefficient of x^3 in a Maclaurin polynomial + is equal to \frac{\fpp'(0)}{3!}, + we have 5=\frac{\fpp'(0)}{6}, so + \fpp'(0)=30. +

    +     +     +     +     + + + Problems + + + +

    + Find the Maclaurin polynomial of degree n for the given function. +

    +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $macpoly = Formula("1-x+(1/2)*x^2-(1/6)*x^3"); + + +

    + Degree n=3, + for f(x) = e^{-x}. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $macpoly = Formula("x-(1/6)*x^3+(1/120)*x^5-(1/5040)*x^7"); + + +

    + Degree n=8, + for f(x) = \sin(x). +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $macpoly = Formula("x+x^2+(1/2)*x^3+(1/6)*x^4+(1/24)*x^5"); + + +

    + Degree n=5, + for f(x) = x\cdot e^x. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $macpoly = Formula("x+(1/3)*x^3+(2/15)*x^5"); + + +

    + Degree n=6, + for f(x) = \tan(x). +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $macpoly = Formula("1+2*x+2*x^2+(4/3)*x^3+(2/3)*x^4"); + + +

    + Degree n=4, + for f(x) = e^{2x}. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $macpoly = Formula("1+x+x^2+x^3+x^4"); + + +

    + Degree n=4, + for \ds f(x) = \frac1{1-x}. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $macpoly = Formula("1-x+x^2-x^3+x^4"); + + +

    + Degree n=4, + for \ds f(x) = \frac1{1+x}. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $macpoly = Formula("1-x+x^2-x^3+x^4-x^5+x^6-x^7"); + + +

    + Degree n=7, + for \ds f(x) = \frac1{1+x}. +

    + +

    + +

    +     +     +     +     + + + +

    + Find the Taylor polynomial of degree n, + at x=c, for the given function. +

    +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $taypoly = Formula("1+(1/2)*(x-1)-(1/8)*(x-1)^2+(1/16)*(x-1)^3-(5/128)*(x-1)^4")->reduce(); + + +

    + f(x) = \sqrt x, + degree n=4, at c=1. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $c = random(2,15,1); + $c1 = $c +1; + $a0 = Compute("ln($c1)"); + + $f = Formula("ln(x+1)")->reduce(); + + $deg = 4; + + $taypoly = Formula("$a0 + (x - $c)/$c1 - (x-$c)^2/(2*$c1^2) + (x - $c)^3/(3*$c1^3) - (x-$c)^4/(4*$c1^4)")->reduce(); + + +

    + f(x) = \ln(x+1), degree , at c=. +

    + +

    + +

    + + + Note: if your answer is not accepted, + try replacing \ln() with its approximation to 5 decimal places. + +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $taypoly = Formula("(1/sqrt(2))-(1/sqrt(2))*(x-pi/4)-(1/(2*sqrt(2)))*(x-pi/4)^2+(1/(6*sqrt(2)))*(x-pi/4)^3+(1/(24*sqrt(2)))*(x-pi/4)^4-(1/(120*sqrt(2)))*(x-pi/4)^5-(1/(720*sqrt(2)))*(x-pi/4)^6")->reduce(); + + +

    + f(x) = \cos(x), degree 6, at c=\pi/4. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $taypoly = Formula("(1/2)+(sqrt(3)/2)*(x-pi/6)-(1/4)*(x-pi/6)^2-(sqrt(3)/12)*(x-pi/6)^3+(1/48)*(x-pi/6)^4+(sqrt(3)/240)*(x-pi/6)^5")->reduce(); + + +

    + f(x) = \sin(x), degree 5, at c=\pi/6. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $c = random(2,5,1); + $taypoly = Formula("(1/$c)-(1/$c^2)*(x-$c)+(1/$c^3)*(x-$c)^2-(1/$c^4)*(x-$c)^3+(1/$c^5)*(x-$c)^4-(1/$c^6)*(x-$c)^5")->reduce(); + + +

    + f(x) = \frac1x, degree 5, at c=. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $taypoly = Formula("1-2*(x-1)+3*(x-1)^2-4*(x-1)^3+5*(x-1)^4-6*(x-1)^5+7*(x-1)^6-8*(x-1)^7+9*(x-1)^8")->reduce(); + + +

    + \ds f(x) = \frac{1}{x^2}, degree 8, at c=1. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $taypoly = Formula("1/2+(1/2)*(x+1)+(1/4)*(x+1)^2")->reduce(); + + +

    + \ds f(x) = \frac{1}{x^2+1}, degree 3, at c=-1. +

    + +

    + +

    +     +     +     + + + + + Context("Fraction")->noreduce('(-x)+y') ; + Context()->flags->set( + reduceConstants => 0, + reduceConstantFunctions => 0, + ); + $taypoly = Formula("-pi^2-2*pi*(x-pi)+((pi^2-2)/2)*(x-pi)^2")->reduce(); + + +

    + f(x) = x^2\cos(x), degree 2, at c=\pi. +

    + +

    + +

    +     +     +     +     + + + +

    + Approximate the function value with the indicated Taylor polynomial + and give approximate bounds on the error. +

    +     + + + + +

    + Approximate \sin(0.1) with the Maclaurin polynomial of + degree 3. +

    +     + +

    + p_3(x) =x-\frac{x^3}{6}; + p_3(0.1) = 0.09983. + Error is bounded by + \pm \frac{1}{4!}\cdot0.1^4 \approx \pm 0.000004167. +

    +     + +     + + + + +

    + Approximate \cos(1) with the Maclaurin polynomial of + degree 4. +

    +     + +

    + p_4(x) =1-\frac{x^2}{2}+\frac{x^4}{24}; + p_4(1) = 13/24\approx 0.54167. + Error is bounded by + \pm \frac{1}{5!}\cdot1^5 \approx \pm 0.00833 +

    +     + +     + + + + +

    + Approximate \sqrt{10} with the Taylor polynomial of + degree 2 centered at x=9. +

    +     + +

    + p_2(x) =3+\frac{1}{6} (-9+x)-\frac{1}{216} (-9+x)^2; + p_2(10) = 3.16204. + The third derivative of f(x) =\sqrt x is bounded on + (8,11) by 0.003. + Error is bounded by \pm \frac{0.003}{3!}\cdot1^3 = \pm 0.0005. +

    +     + +     + + + + +

    + Approximate \ln(1.5) with the Taylor polynomial of + degree 3 centered at x=1. +

    +     + +

    + p_3(x) =-1+x-\frac{1}{2} (-1+x)^2+\frac{1}{3} (-1+x)^3; + p_3(1.5) = 0.41667. + The fourth derivative of f(x) =\ln(x) is bounded on + (0.9,2) by 10. + Error is bounded by \pm \frac{10}{4!}\cdot 0.5^4 = \pm 0.026. +

    +     + +     +     + + + +

    + The following exercises ask for an n to be found such that + p_n(x) approximates f(x) within a certain bound of accuracy. +

    +     + + + + +

    + Find n such that the Maclaurin polynomial of degree n of + f(x)= e^x approximates e within 0.0001 of the actual value. +

    +     + +

    + The nth derivative of + f(x)=e^x is bounded by 3 on intervals containing 0 and 1. + Thus \abs{R_n(1)}\leq \frac{3}{(n+1)!}1^{(n+1)}. + When n=7, this is less than 0.0001. +

    +     + +     + + + + +

    + Find n such that the Taylor polynomial of degree n of + f(x)= \sqrt x, centered at x=4, + approximates \sqrt 3 within 0.0001 of the actual value. +

    +     + +

    + The nth derivative of + f(x)=\sqrt x is bounded by 0.1 on intervals containing 3 and 4. + Thus \abs{R_n(\pi)}\leq \frac{0.1}{(n+1)!}(1)^{(n+1)}. + When n=4, this is less than 0.0001. +

    +     + +     + + + + +

    + Find n such that the Maclaurin polynomial of degree + n of f(x)= \cos(x) approximates + \cos(\pi/3) within 0.0001 of the actual value. +

    +     + +

    + The nth derivative of + f(x)=\cos(x) is bounded by 1 on intervals + containing 0 and \pi/3. + Thus \abs{R_n(\pi/3)}\leq \frac{1}{(n+1)!}(\pi/3)^{(n+1)}. + When n=7, this is less than 0.0001. + Since the Maclaurin polynomial of \cos(x) only uses even + powers, we can actually just use n=6. +

    +     + +     + + + + +

    + Find n such that the Maclaurin polynomial of degree + n of f(x)= \sin(x) approximates + \cos(\pi) within 0.0001 of the actual value. +

    +     + +

    + The nth derivative of + f(x)=\sin(x) is bounded by 1 on intervals + containing 0 and \pi. + Thus \abs{R_n(\pi)}\leq \frac{1}{(n+1)!}(\pi)^{(n+1)}. + When n=12, this is less than 0.0001. + Since the Maclaurin polynomial of \sin(x) only uses odd powers, + we can actually just use n=11. +

    +     + +     +     + + + +

    + Find the nth term of the indicated Taylor polynomial. +

    +     + + + + +

    + Find a formula for the nth term of the Maclaurin + polynomial for f(x)=e^x. +

    +     + +

    + The nth term is \frac{1}{n!}x^n. +

    +     + +     + + + + +

    + Find a formula for the nth term of the Maclaurin + polynomial for f(x)=\cos(x). +

    +     + +

    + The nth term is: + when n is even, \frac{(-1)^{n/2}}{n!}x^n; + when n is odd, 0. +

    +     + +     + + + +

    + Find a formula for the nth term of the Maclaurin polynomial + for f(x)=\sin(x). +

    +     + +

    + The nth term is: + when n even, 0; when n is odd, + \frac{(-1)^{(n-1)/2}}{n!}x^n. +

    +     +     + + + + +

    + Find a formula for the nth term of the Maclaurin + polynomial for \ds f(x)=\frac{1}{1-x}. +

    +     + +

    + The nth term is x^n. +

    +     + +     + + + + +

    + Find a formula for the nth term of the Maclaurin + polynomial for \ds f(x)=\frac{1}{1+x}. +

    +     + +

    + The nth term is (-1)^nx^n. +

    +     + +     + + + + +

    + Find a formula for the nth term of the Taylor + polynomial for \ds f(x)=\ln(x) + centered at x=1. +

    +     + +

    + The nth term is (-1)^n\frac{(x-1)^n}{n}. +

    +     + +     +     + + + +

    + Approximate the solution to the given differential equation with a + degree 4 Maclaurin polynomial. +

    +     + + + + +

    + \yp=y, y(0) = 1 +

    +     + +     + + + + +

    + \yp=5y, y(0) = 3 +

    +     + +     + + + + +

    + \ds \yp=\frac2y, y(0) = 1 +

    +     + +     + +     +     +     +
    +
    + Taylor Series +

    + In , + we showed how certain functions can be represented by a power series function. + In , + we showed how we can approximate functions with polynomials, + given that enough derivative information is available. + In this section we combine these concepts: + if a function f(x) is infinitely differentiable, + we show how to represent it with a power series function. +

    + + + + + Taylor and Maclaurin Series + +

    + Let f(x) have derivatives of all orders at x=c. + Taylor Seriesdefinition + Maclaurin Seriesdefinition + Maclaurin Series|see{Taylor Series} + seriesTaylor + seriesMaclaurin +

    + +

    +

      +
    1. +

      + The Taylor Series of f(x), + centered at c is + + \infser[0] \frac{f^{(n)}(c)}{n!}(x-c)^n + . +

      +
      2. + +
    2. +

      + Setting c=0 gives the Maclaurin Series of f(x): + + \infser[0] \frac{f^{(n)}(0)}{n!}x^n + . +

      +
      3. +
    +

    +     +     + +

    + If p_n(x) is the nth degree Taylor polynomial for f(x) centered at x=c, + we saw how f(x) is approximately equal + to p_n(x) near x=c. + We also saw how increasing the degree of the polynomial generally reduced the error. +

    + +

    + We are now considering series, + where we sum an infinite set of terms. + Our ultimate hope is to see the error vanish and claim a function is + equal to its Taylor series. +

    + +

    + When creating the Taylor polynomial of degree n for a function f(x) at x=c, + we needed to evaluate f, + and the first n derivatives of f, at x=c. + When creating the Taylor series of f, + it helps to find a pattern that describes the + nth derivative of f at x=c. + We demonstrate this in the next two examples. +

    + + + The Maclaurin series of <m>f(x) = \cos(x)</m> + +

    + Find the Maclaurin series of f(x)=\cos(x). +

    +     + +

    + In + we found the 8th degree Maclaurin polynomial of \cos(x). + In doing so, + we created the table shown in . +

    + +
    + Derivatives of f(x)=\cos (x) evaluated at x=0 + +

    + + f(x) \amp = \cos(x) \amp f(0) \amp = 1 + f'(x) \amp = -\sin(x) \amp \fp(0) \amp = 0 + \fp'(x) \amp = -\cos(x) \amp \fp'(0) \amp = -1 + \fp''(x) \amp = \sin(x) \amp \fp''(0) \amp = 0 + f^{(4)}(x) \amp = \cos(x) \amp f^{(4)}(0) \amp =1 + f^{(5)}(x) \amp = -\sin(x) \amp f^{(5)}(0) \amp =0 + f^{(6)}(x) \amp = -\cos(x) \amp f^{(6)}(0) \amp =-1 + f^{(7)}(x) \amp = \sin(x) \amp f^{(7)}(0) \amp =0 + f^{(8)}(x) \amp = \cos(x) \amp f^{(8)}(0) \amp =1 + f^{(9)}(x) \amp = -\sin(x) \amp f^{(9)}(0) \amp =0 + +

    +     + +
    + +

    + Notice how f^{(n)}(0)=0 when n is odd, + f^{(n)}(0)=1 when n is divisible by 4, + and f^{(n)}(0)=-1 when n is even but not divisible by 4. + Thus the Maclaurin series of \cos(x) is + + 1-\frac{x^2}2+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!} - \cdots + +

    + +

    + We can go further and write this as a summation. + The coefficients alternate between positive and negative. + Since we only need the terms where the power of x is even, + we write the power series in terms of x^{2n}: + + \infser[0] (-1)^{n}\frac{x^{2n}}{(2n)!} + . +

    + +

    + This Maclaurin series is a special type of power series. + As such, we should determine its interval of convergence. + Applying the Ratio Test, we have + + \lim_{n\to\infty}\frac{\abs{(-1)^{n+1}\dfrac{x^{2(n+1)}}{\big(2(n+1)\big)!}}}{\abs{(-1)^n\frac{x^{2n}}{(2n)!}}} \amp = \lim_{n\to\infty}\abs{\frac{x^{2n+2}}{x^{2n}}}\frac{(2n)!}{(2n+2)!} + \amp = \lim_{n\to\infty} \frac{\abs{x}^2}{(2n+2)(2n+1)} + . +

    + +

    + For any fixed x, this limit is 0. + Therefore this power series has an infinite radius of convergence, + converging for all x. + It is important to note what we have, and have not, determined: + we have determined the Maclaurin series for \cos(x) along with its interval of convergence. + We have not shown that \cos(x) is + equal to this power series. +

    +     + +     + +

    + + found the Taylor Series representation of \cos(x). + We can easily find the Taylor Series representation of \sin(x) by recognizing that + \int \cos(x)\, dx=\sin(x) and apply . +

    + + + The Taylor series of <m>f(x)=\ln(x)</m> at <m>x=1</m> + +

    + Find the Taylor series of f(x) = \ln(x) centered at x=1. +

    +     + +

    + + shows the nth derivative of \ln(x) evaluated at x=1 for n=0,\ldots,5, + along with an expression for the nth term: + + f^{(n)}(1) = (-1)^{n+1}(n-1)! \text{ for \(n\geq 1\). } + +

    + +

    + Remember that this is what distinguishes Taylor series from Taylor polynomials; + we are very interested in finding a pattern for the nth term, + not just finding a finite set of coefficients for a polynomial. +

    + +
    + Derivatives of \ln(x) evaluated at x=1 + +

    + + f(x) \amp = \ln(x) \amp f(1) \amp = 0 + \fp(x) \amp = \frac1x \amp \fp(1) \amp = 1 + \fp'(x) \amp = -\frac{1}{x^2} \amp \fp'(1) \amp = -1 + \fp''(x) \amp = \frac{2}{x^3} \amp \fp''(1) \amp = 2 + f^{(4)}(x) \amp = -\frac{6}{x^4} \amp f^{(4)}(1) \amp = -6 + f^{(5)}(x) \amp = \frac{24}{x^5} \amp f^{(5)}(1) \amp = 24 + \amp \vdots \amp \amp \vdots + f^{(n)}(x) \amp = \amp f^{(n)}(1) \amp = + (-1)^{n+1}\amp\frac{(n-1)!}{x^n} \amp (-1)^{n+1}\amp (n-1)! + +

    +     + +
    + +

    + Since f(1) = \ln(1) = 0, + we skip the first term and start the summation with n=1, + giving the Taylor series for \ln(x), + centered at x=1, as + + \infser (-1)^{n+1}(n-1)!\frac{1}{n!}(x-1)^n = \infser (-1)^{n+1}\frac{(x-1)^n}{n} + . +

    + +

    + We now determine the interval of convergence, using the Ratio Test. + + \lim_{n\to\infty} \frac{\abs{(-1)^{n+2}\dfrac{(x-1)^{n+1}}{n+1}}}{\abs{(-1)^{n+1}\dfrac{(x-1)^n}{n}}} \amp = \lim_{n\to\infty} \abs{\frac{(x-1)^{n+1}}{(x-1)^n}}\frac{n}{n+1} + \amp = \abs{x-1} + . +

    + +

    + By the Ratio Test, + we have convergence when \abs{x-1} \lt 1: + the radius of convergence is 1, and we have convergence on (0,2). + We now check the endpoints. +

    + +

    + At x=0, the series is + + \sum_{n=1}^\infty (-1)^{n+1}\frac{(-1)^n}{n} = -\sum_{n=1}^\infty \frac1n + , + which diverges (it is the Harmonic Series times (-1).) +

    + +

    + At x=2, the series is + + \sum_{n=1}^\infty (-1)^{n+1}\frac{(1)^n}{n} = \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n} + , + the Alternating Harmonic Series, which converges. +

    + +

    + We have found the Taylor series of \ln x centered at x=1, + and have determined the series converges on (0,2]. + We cannot (yet) say that \ln x is equal to this Taylor series on (0,2]. +

    +     + +     + +

    + It is important to note that + defines a Taylor series given a function f(x), but makes no claim about their equality. + We will find that most of the time they are equal, + but we need to consider the conditions that allow us to conclude this. +

    + +

    + + states that the error between a function f(x) and its + nth-degree Taylor polynomial p_n(x) is R_n(x), where + + \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{(x-c)^{(n+1)}} + . +

    + +

    + If R_n(x) goes to 0 for each x in an interval I as n approaches infinity, + we conclude that the function is equal to its Taylor series expansion. +

    + + + Function and Taylor Series Equality + +

    + Let f(x) have derivatives of all orders at x=c, + let R_n(x) be as stated in , + and let I be an interval on which the Taylor series of f(x) converges. + If \lim\limits_{n\to\infty} R_n(x) = 0 for all x in I, then + Taylor Seriesequality with generating function + + f(x) = \infser[0] \frac{f^{(n)}(c)}{n!}(x-c)^n\, \text{ on \(I\). } + +

    +     +     + + + +

    + We demonstrate the use of this theorem in an example. +

    + + + Establishing equality of a function and its Taylor series + +

    + Show that f(x) = \cos(x) is equal to its Maclaurin series, + as found in , for all x. +

    +     + +

    + Given a value x, + the magnitude of the error term R_n(x) is bounded by + + \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{x^{n+1}} + . +

    + +

    + Since all derivatives of \cos(x) are + \pm \sin(x) or \pm\cos(x), + whose magnitudes are bounded by 1, we can state + + \abs{R_n(x)} \leq \frac{1}{(n+1)!}\abs{x^{n+1}} + + which implies + + -\frac{\abs{x^{n+1}}}{(n+1)!} \leq R_n(x) \leq\frac{\abs{x^{n+1}}}{(n+1)!} + . +

    + +

    + For any x, + \lim\limits_{n\to\infty} \frac{x^{n+1}}{(n+1)!} = 0. + Applying the Squeeze Theorem to Equation, + we conclude that \lim\limits_{n\to\infty} R_n(x) = 0 for all x, + and hence + + \cos(x) = \infser[0] (-1)^{n}\frac{x^{2n}}{(2n)!} \text{ for all \(x\) } + . +

    +     +     + +

    + It is natural to assume that a function is equal to its Taylor series on the series' interval of convergence, + but this is not always the case. + In order to properly establish equality, + one must use . + This is a bit disappointing, + as we developed beautiful techniques for determining the interval of convergence of a power series, + and proving that R_n(x)\to 0 can be difficult. + For instance, + it is not a simple task to show that \ln x equals its Taylor series on (0,2] as found in ; + in the Exercises, + the reader is only asked to show equality on (1,2), + which is simpler. +

    + +

    + There is good news. + A function f(x) that is equal to its Taylor series, + centered at any point the domain of f(x), + is said to be an analytic function, + analytic function + and most, if not all, + functions that we encounter within this course are analytic functions. + Generally speaking, + any function that one creates with elementary functions (polynomials, + exponentials, trigonometric functions, + etc.) that is not piecewise defined is probably analytic. + For most functions, + we assume the function is equal to its Taylor series on the series' interval of convergence and only use + when we suspect something may not work as expected. +

    + +

    + We develop the Taylor series for one more important function, + then give a table of the Taylor series for a number of common functions. + Binomial Series + seriesBinomial +

    + + + The Binomial Series + +

    + Find the Maclaurin series of f(x) = (1+x)^k, k\neq 0. +

    +     + +

    + When k is a positive integer, + the Maclaurin series is finite. + For instance, when k=4, we have + + f(x) = (1+x)^4 = 1+4x+6x^2+4x^3+x^4 + . +

    + +

    + The coefficients of x when k is a positive integer are known as the + binomial coefficients, + giving the series we are developing its name. +

    + +

    + When k=1/2, we have f(x) = \sqrt{1+x}. + Knowing a series representation of this function would give a useful way of approximating \sqrt{1.3}, + for instance. +

    + +

    + To develop the Maclaurin series for + f(x) = (1+x)^k for any value of k\neq0, + we consider the derivatives of f evaluated at x=0: +

    + +

    + + f(x) \amp = (1+x)^k \amp f(0) \amp = 1 + \fp(x) \amp = k(1+x)^{k-1} \amp \fp(0) \amp =k + \fp'(x) \amp = k(k-1)(1+x)^{k-2} \amp \fp'(0) \amp =k(k-1) + \fp''(x) \amp = k(k-1)(k-2)(1+x)^{k-3} \amp \fp''(0) \amp =k(k-1)(k-2) + \amp \vdots \amp \amp \vdots + + For a general n, + + f^{(n)}(x) = k(k-1)\cdots\bigl(k-(n-1)\bigr)(1+x)^{k-n} + , + giving f^{(n)}(0) =k(k-1)\cdots\bigl(k-(n-1)\bigr). +

    +

    + Thus the Maclaurin series for f(x) = (1+x)^k is + + (1+x)^k\amp =1+ kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \ldots + \amp \dots + \frac{k(k-1)\cdots\big(k-(n-1)\big)}{n!}(x-c)^n+\ldots + +

    + +

    + It is important to determine the interval of convergence of this series. + With + + a_n = \frac{k(k-1)\cdots\big(k-(n-1)\big)}{n!}x^n + , + we apply the Ratio Test: + + \lim_{n\to\infty}\frac{\abs{a_{n+1}}}{\abs{a_n}}\amp =\lim_{n\to\infty} \frac{\abs{\dfrac{k(k-1)\cdots(k-(n-1))(k-n)}{(n+1)!}x^{n+1}}}{\abs{\dfrac{k(k-1)\cdots\big(k-(n-1)\big)}{n!}x^n}} + \amp =\lim_{n\to\infty} \abs{\frac{k-n}{n+1}x} + \amp = \abs{x} + . +

    + +

    + The series converges absolutely when the limit of the Ratio Test is less than 1; + therefore, we have absolute convergence when \abs{x}\lt 1. +

    + +

    + While outside the scope of this text, + the interval of convergence depends on the value of k. + When k \gt 0, the interval of convergence is [-1,1]. + When -1\lt k\lt 0, the interval of convergence is [-1,1). + If k\leq -1, the interval of convergence is (-1,1). +

    +     + +     + +

    + We learned that Taylor polynomials offer a way of approximating a + difficult to compute + function with a polynomial. + Taylor series offer a way of exactly representing a function with a series. + One probably can see the use of a good approximation; + is there any use of representing a function exactly as a series? +

    + +

    + While we should not overlook the mathematical beauty of Taylor series + (which is reason enough to study them), + there are practical uses as well. + They provide a valuable tool for solving a variety of problems, + including problems relating to integration and differential equations. +

    + +

    + In + we give a table of the Taylor series of a number of common functions. + We then give a theorem about the + algebra of power series, that is, + how we can combine power series to create power series of new functions. + This allows us to find the Taylor series of functions like + f(x) = e^x\cos(x) by knowing the Taylor series of e^x and \cos(x). +

    + +

    + Before we investigate combining functions, + consider the Taylor series for the arctangent function + (see ). + Knowing that \tan^{-1}(1) = \pi/4, + we can use this series to approximate the value of \pi: + + \frac{\pi}4 \amp = \tan^{-1}(1) = 1-\frac13+\frac15-\frac17+\frac19-\cdots + \pi \amp = 4\left(1-\frac13+\frac15-\frac17+\frac19-\cdots\right) + +

    + +

    + Unfortunately, + this particular expansion of \pi converges very slowly. + The first 100 terms approximate \pi as 3.13159, + which is not particularly good. +

    + +

    + + + Important Taylor Series Expansions + + + Function and Series + First Few Terms + Interval of + + + Convergence + + + \ds e^x = \infser[0] \frac{x^n}{n!} + \ds 1+ x+\frac{x^2}{2!} + \frac{x^3}{3!}+\cdots + (-\infty,\infty) + + + \ds \sin(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!} + \ds x-\frac{x^3}{3!}+\frac{x^5}{5!} - \frac{x^7}{7!}+\cdots + (-\infty,\infty) + + + \ds \cos(x) = \infser[0] (-1)^n\frac{x^{2n}}{(2n)!} + \ds 1-\frac{x^2}{2!}+\frac{x^4}{4!} - \frac{x^6}{6!} +\cdots + (-\infty,\infty) + + + \ds \ln(x) = \infser(-1)^{n+1}\frac{(x-1)^n}{n} + \ds (x-1)- \frac{(x-1)^2}{2} +\frac{(x-1)^3}{3}-\cdots + (0,2] + + + \ds \frac{1}{1-x} = \infser[0] x^n + \ds 1+x+x^2+x^3+\cdots + (-1,1) + + + \ds \tan^{-1}(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{2n+1} + \ds x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots + [-1,1] + + + \ds (1+x)^k=\infser[0] \binom{k}{n}x^n + \ds 1+kx+\frac{k(k-1)}{2!}x^2 + \cdots + (-1,1) + + + +

    + Note that for (1+x)^k, the interval of convergence may contain one or both endpoints, + depending on the value of k, and we are using the generalized binomial coefficients + + \binom{k}{n} = \frac{k(k-1)\cdots (k-(n-1))}{n!} + . + Taylor Seriescommon series +

    + + + + + + Algebra of Power Series + +

    + Let \ds f(x) = \infser[0] a_nx^n and + \ds g(x) = \infser[0] b_nx^n converge absolutely for \abs{x}\lt R, + and let h(x) be a polynomial function. + power seriesalgebra of +

    + +

    +

      +
    1. +

      + \ds f(x)\pm g(x) = \infser[0] (a_n\pm b_n)x^n for \abs{x}\lt R. +

      +
      2. + +
    2. +

      + \ds f(x)g(x) = \left(\infser[0] a_nx^n\right)\left(\infser[0] b_nx^n\right) = \infser[0]\big(a_0b_n+a_1b_{n-1}+\ldots a_nb_0\big)x^n for \abs{x}\lt R. +

      +
      3. + +
    3. +

      + \ds f\big(h(x)\big) = \infser[0] a_n\big(h(x)\big)^n for \abs{h(x)}\lt R. +

      +
      4. +
    +

    +     +     + + + + + Combining Taylor series + +

    + Write out the first 3 terms of the Taylor Series for + f(x) = e^x\cos(x) using + and . +

    +     + +

    + informs us that + + e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \text{ and } \cos(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots + . +

    + +

    + Applying , we find that + + e^x\cos(x) = \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right) + . + Distribute the right hand expression across the left: + + e^x\cos(x)\amp = 1\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)+x\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right) + \amp + \frac{x^2}{2!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)+\frac{x^3}{3!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right) + \amp + \frac{x^4}{4!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)+\cdots + + If we distribute again and collect like terms, we find + + e^x\cos(x) = 1 + x -\frac{x^3}{3}-\frac{x^4}{6} - \frac{x^5}{30}+\frac{x^7}{630}+\cdots + . +

    + +

    + While this process is a bit tedious, + it is much faster than evaluating all the necessary derivatives of + e^x\cos(x) and computing the Taylor series directly. +

    + +

    + Because the series for e^x and \cos(x) both converge on (-\infty,\infty), + so does the series expansion for e^x\cos(x). +

    +     + +     + + + + + + Creating new Taylor series + +

    + Use + to create series for y=\sin(x^2) and y=x^3/(3+x^4). +

    +     + +

    + Given that + + \sin(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!} = x-\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}+\cdots + , + we simply substitute x^2 for x in the series, giving + + \sin(x^2) \amp = \infser[0] (-1)^n\frac{(x^2)^{2n+1}}{(2n+1)!} + \amp \infser[0] (-1)^n\frac{(x^{4n+2}}{(2n+1)!} + \amp =x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!} -\frac{x^{14}}{7!}\cdots + . +

    + +

    + Since the Taylor series for \sin(x) has an infinite radius of convergence, + so does the Taylor series for \sin(x^2). +

    + +

    + For y=x^3/(3+x^4), we begin with the geometric series expansion + + \frac{1}{1-x} = \sum_{n=0}^\infty x^n + . + Note that we can write + + \frac{1}{3+x^4} = \frac13\cdot\frac{1}{1+x^4/3} = \frac13 \cdot \frac{1}{1-(-x^4/3)} + . + Substituting -x^4/3 into the geometric series expansion, we get + + \frac{1}{3+x^4} = \sum_{n=0}^\infty (-x^4/3)^n = \sum_{n=0}^\infty \frac{(-1^n)x^{4n}}{3^n} + . + Finally, we can multiply both sides of the above equation by x^3 to obtain + + \frac{x^3}{3+x^4} = x^3\sum_{n=0}^\infty \frac{(-1^n)x^{4n}}{3^n} = \sum_{n=0}^\infty \frac{(-1)^nx^{4n+3}}{3^n} + . +

    +     + +     + + + + + A (somewhat foolish) combination of Taylor series + +

    + Discuss possible methods for obtaining a Taylor series expansion for f(x)=\ln(\sqrt{x}). +

    +     + +

    + Since f(x) is a composition, our first instict might be to apply + to the problem. However, \sqrt{x} is not a polynomial function, + and neither \ln(x) nor \sqrt{x} have Maclaurin series expansions. +

    + +

    + You might already see a simple way to proceed, but let us first consider the following: + \sqrt{x}=(1+(x-1))^{1/2} can be expanded as a binomial series centered at x=1. + We also know the Taylor series for \ln(x) at x=1, and note that \sqrt{1}=1, + so when x is near 1, so is \sqrt{x}. +

    + +

    + What happens if we take the Taylor series + + \ln(x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{(x-1)^n}{n} = (x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\cdots + + and substitute in + + \sqrt{x} = \sum_{n=0}^\infty \binom{1/2}{n}(x-1)^n = 1+\frac12(x-1)-\frac14(x-1)^2+\cdots + , + where \binom{1/2}{n}=\frac{1/2(1/2-1)\cdots (1/2-(n-1))}{n!} denotes the binomial coefficient? +

    + +

    + Short answer: a mess. We have to replace each occurence of x-1 + in the power series for \ln(x) with \sqrt{x}-1 = \frac12(x-1)-\frac14(x-1)^2+\frac{1}{16}(x-1)^3+\cdots, + and then expand, and collect terms. + If we do this, keeping only terms up to (x-1)^3, we find: + + \ln(\sqrt{x}) \amp = \left(\frac12(x-1)-\frac14(x-1)^2+\frac{1}{16}(x-1)^3+\cdots\right) + \amp \quad -\frac12\left(\frac12(x-1)-\frac14(x-1)^2+\frac{1}{16}(x-1)^3+\cdots\right)^2 + \amp \quad + \frac13\left(\frac12(x-1)-\frac14(x-1)^2+\frac{1}{16}(x-1)^3+\cdots\right)+\cdots + \amp = \frac12(x-1)-\frac14(x-2)^2+\frac16(x-1)^3-\cdots + . +

    + +

    + But of course, there was a better way all along: + + \ln(\sqrt{x}) = \ln(x^{1/2}) = \frac12\ln(x) + + using properties of the logarithm, and indeed, + the result above is the same as the one we would have obtained by simply multiplying the Taylor series for \ln(x) by \frac12. + Power series manipulation is a powerful technique, but one should not apply it blindly. +

    +     +     + + + Using Taylor series to approximate a composition + +

    + Use Taylor series to determine a degree 5 Taylor polynomial approximation to f(x)=e^{\sin(x)}. +

    +     + +

    + Here we want to apply , but h(x)=\sin(x) is not a polynomial. + However, we are interested in approximation, so we replace \sin(x) + by the Maclaurin polynomial + + q(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + . + The Maclaurin series for f(x)=e^x is given by + + e^x = \sum_{n=0}^\infty \frac{x^n}{n!} + . + Next, we substitute q(x) into the series for e^x. + The algebra gets very messy, but we can simplify things: + since we want the degree 5 approximation, there is no need to write down terms involving x^6 or higher powers. + + e^{\sin(x)} \amp \approx 1 + q(x) + \frac{1}{2!}q(x)^2+\frac{1}{3!}q(x)^3+\frac{1}{4!}q(x)^4+\frac{1}{5!}q(x)^5 + \amp = 1 + \left(x-\frac{x^3}{6}+\frac{x^5}{120}\right) + \frac12\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right) + \amp \quad + \frac{1}{6}\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right)^3 + \frac{1}{24}\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right)^4 + \amp \quad\quad + \frac{1}{120}\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right)^5 + \amp = 1+x-\frac{x^3}{6}+\frac{x^5}{120}+\frac12\left(x^2-\frac13x^4+\cdots\right)+\frac16\left(x^3-\frac12x^5+\cdots\right) + \amp \quad + \frac{1}{24}\left(x^4+\cdots\right) + \frac{1}{120}\left(x^5+\cdots\right) + \amp = 1+x+\frac12x^2-\frac18x^4-\frac{1}{15}x^5+\cdots + . +

    +

    + While the algebra is a bit of a mess, it is often less work than computing the Taylor polynomial directly, + as the derivatives of a composite function quickly get complicated. + The function f(x)=e^{\sin(x)} and its approximation are plotted in below. + Note that our polynomial approximation is very good on [-1,1]. +

    +
    + A graph of f(x) and its degree 5 Maclaurin polynomial + + A graph of the function from this example, and its degree 5 Maclaurin polynomial approximation. + +

    + The image shows the graph of f(x)=e^{\sin(x)} on the interval [-2,2], + along with the graph of p_5(x), the degree 5 Maclaurin polynomial approximation of f(x). + As we have come to expect, there is very little noticeable difference between the two graphs near x=0, + and it appears that the polynomial approximation remains good over the interval (-1,1). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-0.5,ymax=3, + xmin=-2.1,xmax=2.2 + ] + + \addplot [firstcurvestyle,domain=-2:2,samples=60] {exp(sin(deg(x)))} node [pos=.8,left] { $y=f(x)$}; + + \addplot [secondcurvestyle,domain=-2:2] {1+x+x^2/2-x^4/8-x^5/15} node [pos=0,below right] { $y=p_5(x)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +     + +

    + In the previous example, the reader might be left wondering why we would bother with all that algebra, + when the computer could have given us the result in seconds. + One reason is simply that it lets us see how these different pieces fit together. + Computing a Taylor polynomial by combining existing results will give the same polynomial as computing derivatives. + Also we see that we can compute an approximation by replacing both parts of a composition with approximations. + In the last couple of examples in this chapter, we see another reason: + often we have to define functions in terms of power series derived through integration, + or the solution of a differential equation, where there is no known function we can simply plug into the computer. +

    + + + Using Taylor series to evaluate definite integrals + +

    + Use the Taylor series of e^{-x^2} to evaluate \ds \int_0^1e^{-x^2}\, dx. +

    +     + +

    + We learned, when studying Numerical Integration, + that e^{-x^2} does not have an antiderivative expressible in terms of elementary functions. + This means any definite integral of this function must have its value approximated, + and not computed exactly. +

    + +

    + We can quickly write out the Taylor series for + e^{-x^2} using the Taylor series of e^x: + + e^x \amp = \infser[0] \frac{x^n}{n!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots + and so + e^{-x^2} \amp = \infser[0] \frac{(-x^2)^n}{n!} + \amp = \infser[0] (-1)^n\frac{x^{2n}}{n!} + \amp = 1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots + . +

    + +

    + We use to integrate: + + \int e^{-x^2}\, dx = C + x - \frac{x^3}{3}+\frac{x^5}{5\cdot2!}-\frac{x^7}{7\cdot3!}+\cdots +(-1)^n\frac{x^{2n+1}}{(2n+1)n!}+\cdots + +

    + +

    + This is the antiderivative of e^{-x^2}; + while we can write it out as a series, + we cannot write it out in terms of elementary functions. + We can evaluate the definite integral + \ds \int_0^1e^{-x^2}\, dx using this antiderivative; + substituting 1 and 0 for x and subtracting gives + + \int_0^1e^{-x^2}\, dx = 1-\frac{1}{3}+\frac{1}{5\cdot 2!}-\frac{1}{7\cdot3!} + \frac{1}{9\cdot4!}\cdots + . +

    + +

    + Summing the 5 terms shown above give the approximation of 0.74749. + Since this is an alternating series, + we can use the Alternating Series Approximation Theorem, + (), + to determine how accurate this approximation is. + The next term of the series is 1/(11\cdot5!) \approx 0.00075758. + Thus we know our approximation is within + 0.00075758 of the actual value of the integral. + This is arguably much less work than using Simpson's Rule to approximate the value of the integral. +

    +     + +     + + + Using Taylor series to solve differential equations + +

    + Solve the differential equation \yp=2y in terms of a power series, + and use the theory of Taylor series to recognize the solution in terms of an elementary function. +

    +     + +

    + We found the first 5 terms of the power series solution to this differential equation in + in . + These are: + + a_0=1, a_1 = 2, a_2 = \frac42=2, a_3=\frac{8}{2\cdot3}=\frac43, a_4=\frac{16}{2\cdot3\cdot4} = \frac23 + . +

    + +

    + We include the unsimplified + expressions for the coefficients found in + as we are looking for a pattern. + It can be shown that a_n = 2^n/n!. + Thus the solution, written as a power series, is + + y = \infser[0] \frac{2^n}{n!}x^n = \infser[0] \frac{(2x)^n}{n!} + . +

    + +

    + Using + and , + we recognize f(x) = e^{2x}: + + e^x = \infser[0] \frac{x^n}{n!} \qquad \Rightarrow \qquad e^{2x} = \infser[0] \frac{(2x)^n}{n!} + . +

    +     + +     + +

    + Finding a pattern in the coefficients that match the series expansion of a known function, + such as those shown in , + can be difficult. + What if the coefficients in the previous example were given in their reduced form; + how could we still recover the function y=e^{2x}? +

    + +

    + Suppose that all we know is that + + a_0=1, a_1=2, a_2=2, a_3=\frac43, a_4=\frac23 + . +

    + +

    + + states that each term of the Taylor expansion of a function includes an n!. + This allows us to say that + + a_2=2=\frac{b_2}{2!}, a_3 = \frac43=\frac{b_3}{3!}, \text{ and } a_4 = \frac23=\frac{b_4}{4!} + + for some values b_2, b_3 and b_4. + Solving for these values, + we see that b_2=4, b_3 = 8 and b_4=16. + That is, we are recovering the pattern we had previously seen, + allowing us to write + + f(x) = \infser[0] a_nx^n \amp = \infser[0] \frac{b_n}{n!}x^n + \amp = 1+2x+ \frac{4}{2!}x^2 + \frac{8}{3!}x^3+\frac{16}{4!}x^4 + \cdots + +

    + +

    + From here it is easier to recognize that the series is describing an exponential function. +

    + +

    + There are simpler, + more direct ways of solving the differential equation \yp = 2y, + as discussed in . + We applied power series techniques to this equation to demonstrate its utility, + and went on to show how sometimes + we are able to recover the solution in terms of elementary functions using the theory of Taylor series. + Most differential equations faced in real scientific and engineering situations are much more complicated than this one, + but power series can offer a valuable tool in finding, + or at least approximating, the solution. +

    + +

    + This chapter introduced sequences, + which are ordered lists of numbers, + followed by series, wherein we add up the terms of a sequence. + We quickly saw that such sums do not always add up to + infinity, but rather converge. + We studied tests for convergence, + then ended the chapter with a formal way of defining functions based on series. + Such series-defined functions + are a valuable tool in solving a number of different problems throughout science and engineering. +

    + +

    + Coming in the next chapters are new ways of defining curves in the plane apart from using functions of the form y=f(x). + Curves created by these new methods can be beautiful, useful, + and important. +

    + + + + Terms and Concepts + + + +

    + What is the difference between a Taylor polynomial and a Taylor series? +

    +     + + + +

    + A Taylor polynomial is a polynomial, + containing a finite number of terms. + A Taylor series is a series, + the summation of an infinite number of terms. +

    +     + +     + + + + +

    + What theorem must we use to show that a function is equal to its Taylor series? +

    +     + + + +

    + , + entitled Function and Taylor Series Equality +

    +     + +     +     + + Problems + + +

    + + gives the nth term of the Taylor series of common functions. + Verify the formula given in the Key Idea by finding + the first few terms of the Taylor series of the given function and identifying a pattern. +

    +     + + + + +

    + f(x) = e^x; c=0 +

    +     + +

    + All derivatives of e^x are e^x which evaluate to 1 at x=0. +

    + +

    + The Taylor series starts 1+x+\frac12x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+\cdots; +

    + +

    + the Taylor series is \ds \infser[0] \frac{x^n}{n!} +

    +     + +     + + + + +

    + f(x) = \sin(x); c=0 +

    +     + +

    + All derivatives of \sin(x) are either + \pm\cos(x) or \pm \sin(x), + which evaluate to \pm 1 or 0 at x=0. + The Taylor series starts 0+x+0x^2-\frac16x^3+0x^4+\frac1{120}x^5; +

    + +

    + the Taylor series is \ds \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!} +

    +     + +     + + + + +

    + f(x) = 1/(1-x); c=0 +

    +     + +

    + The nth derivative of 1/(1-x) is f^{(n)}(x) = (n)!/(1-x)^{n+1}, + which evaluates to n! at x=0. +

    + +

    + The Taylor series starts 1+x+x^2+x^3+\cdots; +

    + +

    + the Taylor series is \ds \infser[0] x^n +

    +     + +     + + + + +

    + f(x) = \tan^{-1}(x); c=0 +

    +     + +

    + The derivative of \tan^{-1}(x) is 1/(1+x^2). + Taking successive derivatives using the Quotient Rule, + the derivatives of \tan^{-1}(x) fall into two categories in terms of their evaluation at x=0. +

    + +

    + When n is even, + \ds f^{(n)}(x) = (-1)^{(n-1)/2}\frac{p(x)}{(1+x^2)^n}, + where p(x) is a polynomial such that p(0) = 0. + Hence f^{(n)}(0) = 0 when n is even. +

    + +

    + When n is odd, + \ds f^{(n)}(x) = (-1)^{(n-1)/2}\frac{p(x)}{(1+x^2)^n}, + where p(x) is a polynomial such that p(0) = (n-1)!. + Hence f^{(n)}(0) = (-1)^{(n-1)/2}(n-1)! when n is odd. (The unusual power of (-1) is such that every other odd term is negative.) +

    + +

    + The Taylor series starts x-\frac13x^3+\frac15x^5+\cdots; + by reindexing to only obtain odd powers of x, we get +

    + +

    + the Taylor series is \ds \infser[0] (-1)^n\frac{x^{2n+1}}{2n+1}. +

    +     + +     + +     + + + +

    + Find a formula for the nth term of the Taylor series of f(x), + centered at c, + by finding the coefficients of the first few powers of x and looking for a pattern. + (The formulas for several of these are found in ; + show work verifying these formula.) +

    +     + + + + +

    + f(x) = \cos(x); c=\pi/2 +

    +     + +

    + The Taylor series starts 0-(x-\pi/2)+0x^2+\frac16(x-\pi/2)^3+0x^4-\frac1{120}(x-\pi/2)^5; +

    + +

    + the Taylor series is \ds \infser[0] (-1)^{n+1}\frac{(x-\pi/2)^{2n+1}}{(2n+1)!} +

    +     + +     + + + + +

    + f(x) = 1/x; c=1 +

    +     + +

    + The Taylor series starts 1-(x-1)+(x-1)^2-(x-1)^3+(x-1)^4-(x-1)^5; +

    + +

    + the Taylor series is \ds \infser[0] (-1)^{n}(x-1)^n +

    +     + +     + + + + +

    + f(x) = e^{-x}; c=0 +

    +     + +

    + f^{(n)}(x) = (-1)^ne^{-x}; + at x=0, f^{(n)}(0)=-1 when n is odd and + f^{(n)}(0)=1 when n is even. +

    + +

    + The Taylor series starts 1-x+\frac12x^2-\frac1{3!}x^3+\cdots; +

    + +

    + the Taylor series is \ds \infser[0] (-1)^n\frac{x^n}{n!}. +

    +     + +     + + + + +

    + f(x) = \ln(1+x); c=0 +

    +     + +

    + f^{(n)}(x) = (-1)^{n+1}\frac{(n-1)!}{(1+x)^n}; + at x=0, f^{(n)}(0)=(-1)^{n+1}(n-1)! +

    + +

    + The Taylor series starts x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots; +

    + +

    + the Taylor series is \ds \infser (-1)^{n+1}\frac{x^n}{n}. +

    +     + +     + + + + +

    + f(x) = x/(x+1); c=1 +

    +     + +

    + f^{(n)}(x) = (-1)^{n+1}\frac{n!}{(x+1)^{n+1}}; + at x=1, f^{(n)}(1)=(-1)^{n+1}\frac{n!}{2^{n+1}} +

    + +

    + The Taylor series starts \frac12+\frac14(x-1)-\frac18(x-1)^2+\frac1{16}(x-1)^3\cdots; +

    + +

    + the Taylor series is \ds \frac12+\infser (-1)^{n+1}\frac{(x-1)^n}{2^{n+1}}. +

    +     + +     + + + + +

    + f(x) = \sin(x); c=\pi/4 +

    +     + +

    + The derivatives of \sin(x) are + \pm \cos(x) and \pm \sin(x); + at x=\pi/4, + these derivatives evaluate to \pm \sqrt{2}/2. +

    + +

    + The Taylor series starts \frac{\sqrt{2}}2+\frac{\sqrt{2}}2(x-\pi/4) - \frac{\sqrt{2}}2\frac{(x-\pi/4)^2}{2}-\frac{\sqrt{2}}2\frac{(x-\pi/4)^3}{3!}+\frac{\sqrt{2}}2\frac{(x-\pi/4)^4}{4!}+\frac{\sqrt{2}}2\frac{(x-\pi/4)^5}{5!}\cdots. + Note how the signs are even, + even, odd, odd, even, even, odd, + odd,\ldots We saw signs like these in + of ; + one way of producing such signs is to raise (-1) to a special quadratic power. + While many possibilities exist, + one such quadratic is (n+3)(n+4)/2. +

    + +

    + Thus the Taylor series is \ds \infser[0] (-1)^{\frac{(n+3)(n+4)}{2}}\frac{\sqrt2}{2}\frac{(x-\pi/4)^n}{n!}. +

    +     + +     + +     + + + +

    + Show that the Taylor series for f(x), + as given in , + is equal to f(x) by applying ; + that is, show \lim\limits_{n\to\infty}R_n(x) =0. +

    +     + + + + +

    + f(x) = e^x +

    +     + +

    + Given a value x, + the magnitude of the error term R_n(x) is bounded by + + \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{x^{(n+1)}} + , + where z is between 0 and x. +

    + +

    + If x \gt 0, + then z\lt x and f^{(n+1)}(z) =e^z\lt e^x. + If x\lt 0, + then x\lt z\lt 0 and f^{(n+1)}(z) =e^z\lt 1. + So given a fixed x value, + let M = \max\{e^x,1\}; + f^{(n)}(z)\lt M. + This allows us to state + + \abs{R_n(x)} \leq \frac{M}{(n+1)!}\abs{x^{(n+1)}} + . +

    + +

    + For any x, + \lim\limits_{n\to\infty} \frac{M}{(n+1)!}\abs{x^{(n+1)}}= 0. + Thus by the Squeeze Theorem, + we conclude that \lim\limits_{n\to\infty} R_n(x) = 0 for all x, + and hence + + e^x = \infser[0] \frac{x^{n}}{n!} \text{ for all \(x\) } + . +

    +     + +     + + + + +

    + f(x) = \sin(x) +

    +     + +

    + The following argument is essentially the same as that given for + f(x) = \cos(x) in . +

    + +

    + Given a value x, + the magnitude of the error term R_n(x) is bounded by + + \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{x^{(n+1)}} + . +

    + +

    + Since all derivatives of \sin(x) are + \pm \cos(x) or \pm\sin(x), + whose magnitudes are bounded by 1, we can state + + \abs{R_n(x)} \leq \frac{1}{(n+1)!}\abs{x^{(n+1)}} + . +

    + +

    + For any x, + \lim\limits_{n\to\infty} \frac{x^{n+1}}{(n+1)!} = 0. + Thus by the Squeeze Theorem, + we conclude that \lim\limits_{n\to\infty} R_n(x) = 0 for all x, + and hence + + \sin(x) = \infser[0] (-1)^{n}\frac{x^{2n+1}}{(2n+1)!} \text{ for all \(x\) } + . +

    +     + +     + + + + +

    + f(x) = \ln(x) (show equality only on (1,2)) +

    +     + +

    + Per the statement of the problem, + we only consider the case 1\lt x\lt 2. +

    + +

    + If 1\lt x\lt 2, then + 1\lt z\lt x and f^{(n+1)}(z) =\frac{n!}{z^{n+1}}\lt n!. + Thus + + \abs{R_n(x)} \leq \frac{n!}{(n+1)!}\abs{(x-1)^{(n+1)}}= \frac{(x-1)^{n+1}}{n+1}\lt \frac1{n+1} + . +

    + +

    + Thus + + \lim_{n\to\infty} \abs{R_n(x)} \lt \lim_{n\to\infty} \frac1{n+1}=0 + , + hence + + \ln(x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{(x-1)^n}n\,\text{ on } \,(1,2) + . +

    +     + +     + + + + +

    + f(x) = 1/(1-x) (show equality only on (-1,0)) +

    +     + +

    + Given a value x, + the magnitude of the error term R_n(x) is bounded by + + \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{x^{(n+1)}} + , + where z is between 0 and x. +

    + +

    + Note that \abs{f^{(n+1)}(x)} = \frac{(n+1)!}{(1-x)^{n+2}}. +

    + +

    + If -1\lt x\lt 0, then + x\lt z\lt 0 and f^{(n+1)}(z) =\frac{(n+1)!}{(1-z)^{n+2}}\lt \frac{(n+1)!}{(1-x)^{n+2}}. + Thus + + \abs{R_n(x)} \leq \frac{(n+1)!}{(1-x)^{n+2}}\frac{1}{(n+1)!}\abs{x^{n+1}}= \frac{(x-1)^{n+1}}{n+1} + . +

    + +

    + For a fixed x, + + \lim_{n\to\infty} \frac{(x-1)^{n+1}}{n+1}=0 + + since \abs{x}\lt 1, hence + + \frac{1}{1-x} = \infser[0] x^n \text{ on } (-1,0) + . +

    +     + +     + +     + + + +

    + Use the Taylor series given in to verify the given identity. +

    +     + + + + +

    + \cos(-x) = \cos(x) +

    +     + +

    + Given \ds \cos(x) = \infser[0] (-1)^n\frac{x^{2n}}{(2n)!}, +

    + +

    + \ds\cos(-x) = \infser[0] (-1)^n\frac{(-x)^{2n}}{(2n)!}=\infser[0] (-1)^n\frac{x^{2n}}{(2n)!}=\cos(x), + as all powers in the series are even. +

    +     + +     + + + + +

    + \sin(-x) = -\sin(x) +

    +     + +

    + Given \ds \sin(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!}, +

    + +

    + \ds\sin(-x) = \infser[0] (-1)^n\frac{(-x)^{2n+1}}{(2n+1)!}=\infser[0] (-1)^n\frac{-x^{2n+1}}{(2n+1)!}=-\sin(x), + as all powers in the series are odd. +

    +     + +     + + + + +

    + \frac{d}{dx}\big(\sin(x) \big) = \cos(x) +

    +     + +

    + Given \ds \sin(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!}, +

    + +

    + \ds\frac{d}{dx}\big(\sin(x) \big) = \frac{d}{dx}\left(\infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!}\right)=\infser[0] (-1)^n\frac{(2n+1)x^{2n}}{(2n+1)!}=\infser[0] (-1)^n\frac{x^{2n}}{(2n)!}=\cos(x). (The summation still starts at n=0 as there was no constant term in the expansion of \sin(x)). +

    +     + +     + + + + +

    + \frac{d}{dx}\big(\cos(x) \big) = -\sin(x) +

    +     + +

    + Given \ds \cos(x) = \infser[0] (-1)^n\frac{x^{2n}}{(2n)!}, +

    + +

    + \ds\frac{d}{dx}\big(\cos(x) \big) = \frac{d}{dx}\left(\infser[0] (-1)^n\frac{x^{2n}}{(2n)!}\right)=\infser (-1)^n\frac{(2n)x^{2n-1}}{(2n)!}=\infser (-1)^n\frac{x^{2n-1}}{(2n-1)!}. + We can re-index this summation to start at n=0 by replacing n with n+1 in the summation: + + \infser (-1)^n\frac{x^{2n-1}}{(2n-1)!} =\infser[0] (-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!} + . +

    + +

    + Note that this series has the opposite sign of the Taylor series for \sin(x); + thus \frac{d}{dx}(\cos(x) ) = -\sin(x). +

    +     + +     + +     + + + +

    + Write out the first 5 terms of the Binomial series with the given k-value. +

    +     + + + + +

    + k=1/2 +

    +     + +

    + \ds 1+\frac x2-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128} +

    +     + +     + + + + +

    + k=-1/2 +

    +     + +

    + \ds 1-\frac x2+\frac{3x^2}{8}-\frac{5x^3}{16}+\frac{35x^4}{128} +

    +     + +     + + + + +

    + k=1/3 +

    +     + +

    + \ds 1+\frac x3-\frac{x^2}{9}+\frac{5x^3}{81}-\frac{10x^4}{243} +

    +     + +     + + + + +

    + k=4 +

    +     + +

    + \ds 1+4x+6x^2+4x^3+x^4 (note the series is finite, + and the formula still applies) +

    +     + +     + +     + + + +

    + Use the Taylor series given in + to create the Taylor series of the given functions. +

    +     + + + + +

    + f(x) = \cos\big(x^2\big) +

    +     + +

    + \ds \infser[0] (-1)^n\frac{(x^2)^{2n}}{(2n)!} = \infser[0] (-1)^n\frac{x^{4n}}{(2n)!}. +

    +     + +     + + + + +

    + f(x) = e^{-x} +

    +     + +

    + \ds \infser[0] \frac{(-x)^n}{n!}. +

    +     + +     + + + + +

    + f(x) = \sin\big(2x+3\big) +

    +     + +

    + \ds \infser[0] (-1)^n\frac{(2x+3)^{2n+1}}{(2n+1)!}. +

    +     + +     + + + + +

    + f(x) = \tan^{-1}\big(x/2\big) +

    +     + +

    + \ds \infser[0] (-1)^n\frac{(x/2)^{2n+1}}{(2n+1)}. +

    +     + +     + + + + +

    + f(x) = e^x\sin(x)(only find the first 4 terms) +

    +     + +

    + \ds x+x^2+\frac{x^3}{3}-\frac{x^5}{30} +

    +     + +     + + + + +

    + f(x) = (1+x)^{1/2}\cos(x)(only find the first 4 terms) +

    +     + +

    + \ds 1+\frac x2-\frac{5x^2}{8}-\frac{3x^3}{16} +

    +     + +     + +     + + + +

    + Approximate the value of the given definite integral by using the first 4 nonzero terms of the integrand's Taylor series. +

    +     + + + + +

    + \ds \int_0^{\sqrt{\pi}} \sin\big(x^2\big)\, dx +

    +     + +

    + \ds \int_0^{\sqrt{\pi}} \sin\big(x^2\big)\, dx \approx \int_0^{\sqrt{\pi}} \left(x^2-\frac{x^6}6+\frac{x^{10}}{120}-\frac{x^{14}}{5040}\right) dx = 0.8877 +

    +     + +     + + + + +

    + \ds \int_0^{\sqrt[3]{\pi}} \cos\left(x^3\right)\, dx +

    +     + +

    + \ds \int_0^{\sqrt[3]{\pi}} \cos\big(x^3\big)\, dx \approx \int_0^{\pi^2/4} \left(1-\frac{x^6}{2!}+\frac{x^{12}}{4!}-\frac{x^{18}}{6!}\right)\, dx = 0.7864. +

    +     + +     + +     +     +     +
    +     + + + Curves in the Plane + +

    + We have explored functions of the form y=f(x) closely throughout this text. + We have explored their limits, + their derivatives and their antiderivatives; + we have learned to identify key features of their graphs, + such as relative maxima and minima, + inflection points and asymptotes; + we have found equations of their tangent lines, + the areas between portions of their graphs and the x-axis, + and the volumes of solids generated by revolving portions of their graphs about a horizontal or vertical axis. +

    + +

    + Despite all this, + the graphs created by functions of the form y=f(x) are limited. + Since each x-value can correspond to only 1 y-value, + common shapes like circles cannot be fully described by a function in this form. + Fittingly, the vertical line test + excludes vertical lines from being functions of x, + even though these lines are important in mathematics. +

    + +

    + In this chapter we'll explore new ways of drawing curves in the plane. + We'll still work within the framework of functions, + as an input will still only correspond to one output. + However, our new techniques of drawing curves will render the vertical line test pointless, + and allow us to create important and beautiful new curves. + Once these curves are defined, + we'll apply the concepts of calculus to them, + continuing to find equations of tangent lines and the areas of enclosed regions. +

    +     + +
    + Conic Sections + +

    + The ancient Greeks recognized that interesting shapes can be formed by intersecting a plane with a + double napped cone (, two identical cones placed tip-to-tip as shown in the following figures). + As these shapes are formed as sections of conics, + they have earned the official name conic sections. +

    + + + +

    + The three most interesting + conic sections are given in the top row of . + They are the parabola, the ellipse + (which includes circles) + and the hyperbola. + In each of these cases, the plane does not intersect the tips of the cones + (usually taken to be the origin). +

    + +
    + Conic Sections + + +
    + Parabola + + A diagonal plane intersecting a double napped cone, forming a parabola + +

    + Two cones stacked tip-to-tip, intersected by a plane. + A diagonal plane intersects the top cone. + The plane does not touch the tip of either cones. + The top edge of the plane intersects the top edge of the cone. + The parts of the plane intersecting the cone are highlighted, which forms a parabola in the plane. +

    +     + +
    + +
    + Ellipse + + A diagonal plane intersecting a double napped cone, forming an ellipse. + +

    + Two cones stacked tip-to-tip, intersected by a plane. + A diagonal plane intersects the top cone. + The plane does not touch the tip of either cones. + The edge of the plane does not touch the top edge of the cone. + The parts of the plane intersecting the cone are highlighted, which forms an ellipse in the plane. +

    +     + +
    + +
    + Circle + + A horizontal plane intersecting a double napped cone, forming a circle. + +

    + Two cones stacked tip-to-tip, intersected by a plane. + A horizontal plane intersects the top cone. + The parts of the plane intersecting the cone are highlighted, which forms a circle in the plane. +

    +     + +
    + +
    + Hyperbola + + A vertical plane intersecting a double napped cone, forming a hyperbola. + +

    + Two cones stacked tip-to-tip, intersected by a vertical plane. + The plane does not pass through the tips of the cones. + The parts of the plane intersecting the cones are highlighted, which forms a hyperbola in the plane. + The hyperbola contains two seperate segments, with one placed vertically above the other with a space between them. + The top segment is in the shape of a rounded v, and the bottom segment is a vertical reflection of the top segment. +

    +     + +
    +     + + +
    + Point + + A horizontal plane intersecting the tips in a double napped cone. + +

    + Two cones stacked tip-to-tip, intersected by a horizontal plane. + The plane passes through the point where the tips of the two cones are touching. + The plane touches the cone at only that single point, forming a point in the plane. +

    +     + +
    + +
    + Line + + A diagonal plane intersecting a double napped cone, forming a line in the plane + +

    + Two cones stacked tip-to-tip, intersected by a diagonal plane. + The plane extends diagonally through the point where the tips of the cones are touching. + The plane only touches the outer edges of the cones. + The parts of the plane touching the cones are highlighted, forming a straight line in the plane. +

    +     + +
    + +
    + Crossed Lines + + A vertical plane intersecting a double napped cone, forming crossed lines in the plane. + +

    + Two cones stacked tip-to-tip, intersected by a vertical plane. + The plane extends vertically, passing through the points at which the tips of the cones touch. + The parts of the plane intersecting the cones are highlighted, forming crossed straight lines in the plane. + The image on the plane is straight lines in an X shape. +

    +     + +
    +     +     +
    + + +

    + When the plane does contain the origin, + three degenerate cones can be formed as shown the bottom row of : + a point, a line, and crossed lines. + We focus here on the nondegenerate cases. + conic sections + conic sectionsdegenerate +

    + +

    + While the above geometric constructs define the conics in an intuitive, + visual way, + these constructs are not very helpful when trying to analyze the shapes algebraically or consider them as the graph of a function. + It can be shown that all conics can be defined by the general second-degree equation + + Ax^2+Bxy+Cy^2+Dx+Ey+F=0 + . +

    + +

    + While this algebraic definition has its uses, + most find another geometric perspective of the conics more beneficial. +

    + +

    + Each nondegenerate conic can be defined as the locus, or set, + of points that satisfy a certain distance property. + These distance properties can be used to generate an algebraic formula, + allowing us to study each conic as the graph of a function. +

    +     + + + Parabolas + + + + + Parabola + +

    + A parabola is the locus of all points equidistant from a point + (called a focus) + and a line + (called the directrix) + that does not contain the focus. + conic sectionsparabola + paraboladefinition + directrix + focus +

    +     +     + +
    + Illustrating the definition of the parabola and establishing an algebraic formula + + + A sketch of a parabola with key components labeled. + +

    + An upward opening parabola with labels for key components. + At the bottom of the parabola a point is labeled as the vertex. + A dashed vertical line, labeled the axis of symmetry, separates the parabola into two halves. + The axis of symmetry crosses the parabola through the vertex. + Below the parabola, a horizontal line is labeled the Directrix. + The distance between the directrix and the vertex is labeled p. + Above the vertex, a point is drawn on the axis of symmetry and is labeled the "focus". + The distance between the vertex and the focus is p, the same distance between the vertex and the directrix. + On the right side of the parabola, a point is labeled (x,y). + The distance between the point and the focus, and the distance between the point and the directrix are equal, labeled d. +

    +     + + + \begin{tikzpicture}[scale=.88] + + \draw [thick,firstcolor] (-3,-1) node [above,black,shift={(12pt,0pt)}] { Directrix} -- (3,-1); + \filldraw [black] (0,1) circle (2.4pt) node [above left] { Focus}; + + \draw [thick,secondcolor] (-3,2.25) parabola bend (0,0) (3,2.25); + \filldraw (0,0) circle (2.4pt) node [below left] { Vertex}; + + \draw (.3,.5) node[] { $\left.\rule{0pt}{12pt}\right\}p$}; + \draw (.3,-.5) node[] { $\left.\rule{0pt}{12pt}\right\}p$}; + + \coordinate (A) at (2.5,1.5625); + + \filldraw [black] (A) circle (2.4pt) node [right] { $(x,y)$}; + \draw [thick,dashed] (0,1) -- (A) node [pos=.5,above] { $d$} -- (2.5,-1) node [pos=.5,right] { $d$}; + \draw [thick,dashed] (0,-1.5) -- node [pos=.85,above,rotate=90] { Axis of} node [pos=.85,below,rotate=90] { Symmetry} (0,3.2); + + \end{tikzpicture} + + + + +
    + +

    + illustrates this definition. + The point halfway between the focus and the directrix is the vertex. + The line through the focus, + perpendicular to the directrix, + is the axis of symmetry, + as the portion of the parabola on one side of this line is the mirror-image of the portion on the opposite side. +

    + +

    + The definition leads us to an algebraic formula for the parabola. + Let P=(x,y) be a point on a parabola whose focus is at F=(0,p) and whose directrix is at y=-p. (We'll assume for now that the focus lies on the y-axis; + by placing the focus p units above the x-axis and the directrix p units below this axis, + the vertex will be at (0,0).) +

    + +

    + We use the Distance Formula to find the distance d_1 between F and P: + + d_1=\sqrt{(x-0)^2+(y-p)^2} + . +

    + +

    + The distance d_2 from P to the directrix is more straightforward: + + d_2=y-(-p) = y+p + . +

    + +

    + These two distances are equal. + Setting d_1=d_2, we can solve for y in terms of x: + + d_1\amp = d_2 + \sqrt{x^2+(y-p)^2} \amp = y+p + Now square both sides. + x^2+(y-p)^2 \amp = (y+p)^2 + x^2+y^2-2yp+p^2 \amp = y^2+2yp+p^2 + x^2 \amp =4yp + y\amp = \frac{1}{4p}x^2 + . +

    + +

    + The geometric definition of the parabola has led us to the familiar quadratic function whose graph is a parabola with vertex at the origin. + When we allow the vertex to not be at (0,0), + we get the following standard form of the parabola. +

    + + + General Equation of a Parabola +

    +

      +
    1. +

      + Vertical Axis of Symmetry: + The equation of the parabola with vertex at (h,k) and directrix y=k-p in standard form is + + y=\frac{1}{4p}(x-h)^2+k + . + The focus is at (h,k+p). +

      +
      2. + +
    2. +

      + Horizontal Axis of Symmetry: + The equation of the parabola with vertex at (h,k) and directrix x=h-p in standard form is + + x=\frac{1}{4p}(y-k)^2+h + . + The focus is at (h+p,k). +

      +
      3. +
    +

    + +

    + Note: p is not necessarily a positive number. + parabolageneral equation +

    +     + + + Finding the equation of a parabola + +

    + Give the equation of the parabola with focus at (1,2) and directrix at y=3. +

    +     + +

    + The vertex is located halfway between the focus and directrix, + so (h,k) = (1,2.5). + This gives p=-0.5. + Using + we have the equation of the parabola as + + y=\frac{1}{4(-0.5)}(x-1)^2+2.5 = -\frac12(x-1)^2+2.5 + . +

    + +
    + The parabola described in + + + A downward opening parabola with a vertex in the first quadrant. + +

    + A downward opening parabola with a vertex in the first quadrant. + The vertex of the parabola is at (1,2.5). + The parabola crosses the x-axis at around x = -1.2 and x = 3.2. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-6.5,ymax=3.0, + xmin=-3.5,xmax=5.5 + ] + + \addplot+ [domain=-3:5,samples=40] {-.5*(x-1)^2+2.5}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + The parabola is sketched in . +

    +     + +     + + + Finding the focus and directrix of a parabola + +

    + Find the focus and directrix of the parabola x=\frac18y^2-y+1. + The point (7,12) lies on the graph of this parabola; + verify that it is equidistant from the focus and directrix. +

    +     + +

    + We need to put the equation of the parabola in its general form. + This requires us to complete the square: + + x \amp = \frac18y^2-y+1 + \amp = \frac18\big(y^2-8y+8\big) + \amp = \frac18\big(y^2-8y+16 -16+8\big) + \amp = \frac18\big((y-4)^2 - 8\big) + \amp = \frac18(y-4)^2 -1 + . +

    + +

    + Hence the vertex is located at (-1,4). + We have \frac18=\frac1{4p}, so p=2. + We conclude that the focus is located at (1,4) and the directrix is x=-3. + The parabola is graphed in , + along with its focus and directrix. +

    + +
    + The parabola described in . The distances from a point on the parabola to the focus and directrix are given. + + + A parabola opening to the right with its directrix and focus drawn. + +

    + A rightwards opening parabola with a vertex at (-1,4). + The focus of the parabola is drawn at the point (1,4). + The directrix is drawn as the vertical line at x=-3. + The point (7,12) is drawn lying on the parabola. + The distance between the point and the directrix is drawn. + The distance between the point and the focus is drawn. + Both distances are labeled as 10 units long. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-5.5,ymax=14.5, + xmin=-11,xmax=14 + ] + + \addplot+ [domain=-5:14,samples=40] ({x^2/8-x+1},x); + \addplot+ [solid,domain=-5:14] (-3,x); + + \draw [dashed,firstcolor!70,thick] (axis cs:-3,12) -- node [above,pos=.5,black] { 10} (axis cs:7,12) -- node [pos=.6,right,black] { 10} (axis cs: 1,4); + + \filldraw [secondcolor] (axis cs: 1,4) circle (2.4pt) (axis cs: 7,12) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + The point (7,12) lies on the graph and is + 7-(-3)=10 units from the directrix. + The distance from (7,12) to the focus is: + + \sqrt{(7-1)^2 + (12-4)^2} = \sqrt{100}=10 + . +

    + +

    + Indeed, the point on the parabola is equidistant from the focus and directrix. +

    +     + +     + + + Reflective Property +

    + One of the fascinating things about the nondegenerate conic sections is their reflective properties. + Parabolas have the following reflective property: +

    + +
    +

    + Any ray emanating from the focus that intersects the parabola reflects off along a line perpendicular to the directrix. +

    +
    + +

    + This is illustrated in . The following theorem states this more rigorously. +

    + +
    + Illustrating the parabola's reflective property + + + A rightward opening parabola demonstrating the reflective property. + +

    + A parabola opening towards the right. + The directrix is drawn as a vertical line to the left of the parabola. + The focus is drawn as a point to the right of the vertex. + Four different rays are emanating from the focus. + The first line extends towards the top right, where it then reflects off the parabola and extends to the right. + The second line extends upwards, where it then reflects off of the parabola and extends towards the right. + The third line extends towards the lower left, where it then reflects off the parabola and extends towards the right. + The fourth line exends twards the lower right, where it once again reflects off the parbola and extends towards the right. + All of the rays extending towards the right are perpendicular to the directrix. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis y line=none, + axis x line=none, + ymin=-6.5,ymax=14.5, + xmin=-4,xmax=12 + ] + + \coordinate (focus) at (axis cs: 1,4); + + \addplot+ [domain=-6:14,samples=40] ({x^2/8-x+1},x); + \addplot+ [solid,domain=-6:14] (-3,x); + + \draw [thick] (axis cs: 12,12) -- (axis cs: 7,12) -- (focus) + (axis cs: 12,8) -- (axis cs: 1,8) -- (focus) + (axis cs: 12,3) -- (axis cs: -.875,3) + (axis cs: -.875,3) -- (focus) + (axis cs: 12,-3) -- (axis cs: 5.125,-3) -- (focus); + + \filldraw [secondcolor] (focus) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    + + + Reflective Property of the Parabola + +

    + Let P be a point on a parabola. + The tangent line to the parabola at P makes equal angles with the following two lines: +

    + +

    +

      +
    1. +

      + The line containing P and the focus F, and +

      +
      2. + +
    2. +

      + The line perpendicular to the directrix through P. + parabolareflective property +

      +
      3. +
    +

    +     +     + +

    + Because of this reflective property, + paraboloids (the 3D analogue of parabolas) make for useful flashlight reflectors as the light from the bulb, + ideally located at the focus, + is reflected along parallel rays. + Satellite dishes also have paraboloid shapes. + Signals coming from satellites effectively approach the dish along parallel rays. + The dish then focuses these rays at the focus, + where the sensor is located. +

    +     +     + + + Ellipses + + + + Ellipse + +

    + An ellipse is the locus of all points whose sum of distances from two fixed points, + each a focus of the ellipse, is constant. + conic sectionsellipse + ellipsedefinition + focus +

    +     +     + +

    + An easy way to visualize this construction of an ellipse is to pin both ends of a string to a board. + The pins become the foci. + Holding a pencil tight against the string places the pencil on the ellipse; + the sum of distances from the pencil to the pins is constant: + the length of the string. + See . +

    + +
    + Illustrating the construction of an ellipse with pins, pencil and string + + + A demonstration of the creation of an ellipse from two foci. + +

    + Two points are drawn, around which an ellipse is sketched. + A pencil is drawn, representing the sketching of the ellipse. + The point of the pencil is touching the ellipse in the top right of the ellipse. + The distances between the point of the pencil and the two foci are drawn. + The distance between the left focus is labeled d_2, and the distance between the right focus is labeled d_1. + It is clear that d_2 is larger than d_1. + Beneath the foci, it is written that d_1 + d_2 = \mathrm{constant}. +

    +     + + + \begin{tikzpicture}[scale=1.32] + + \draw [thick,dashed] (0,0) circle [x radius=2,y radius=1.5]; + + \filldraw (1.3,0) circle (2.4pt) + (-1.3,0) circle (2.4pt); + + \draw (1.3,0) -- node [right,pos=.5] { $d_1$} (1.39,1.07) -- node [above,pos=.5] { $d_2$} (-1.3,0); + \draw (0,-.7) node { $d_1+d_2=$ constant}; + + \begin{scope}[shift={(1.9cm,1.225cm)}] + \begin{scope}[rotate=120] + \begin{scope}[xscale=.25,yscale=.5] + + \draw (0,0) -- (0,-2) -- (1,-2)--(1,0) -- (.5,1) -- cycle; + \draw [fill=black] (.3,.6) -- ( .5,1)--(.7,.6)--cycle; + \draw (0,0) cos (.5,-.1) sin (1,0); + + \end{scope} + \end{scope} + \end{scope} + + \end{tikzpicture} + + + + +
    + +

    + We can again find an algebraic equation for an ellipse using this geometric definition. + Let the foci be located along the x-axis, + c units from the origin. + Let these foci be labeled as + F_1 = (-c,0) and F_2=(c,0). + Let P=(x,y) be a point on the ellipse. + The sum of distances from F_1 to P (d_1) and from F_2 to P (d_2) is a constant d. + That is, d_1+d_2=d. + Using the Distance Formula, we have + + \sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = d + . +

    + +

    + Using a fair amount of algebra can produce the following equation of an ellipse (note that the equation is an implicitly defined function; + it has to be, as an ellipse fails the Vertical Line Test): + + \frac{x^2}{\left(\frac d2\right)^2} + \frac{y^2}{\left(\frac d2\right)^2-c^2} = 1 + . +

    + +

    + This is not particularly illuminating, + but by making the substitution a=d/2 and b=\sqrt{a^2-c^2}, + we can rewrite the above equation as + + \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 + . +

    + +

    + This choice of a and b is not without reason; + as shown in , + the values of a and b have geometric meaning in the graph of the ellipse. +

    + +
    + Labeling the significant features of an ellipse + + + An ellipse with labels for key components. + +

    + An ellipse is drawn with labels for key features. + The ellipse is separated into 4 equal sections by a vertical and a horizontal line, which both pass through the center of the ellipse. + The vertical line dividing the ellipse in half vertically is labeled as the "minor axis." + The horizontal dividing the ellipse in half horizontally is labeled as the "major axis." + The points on the major axis touching the ellipse are labeled as "vertices". + Two points on the major axis are labeled as "foci". The foci both lie a short distance away from the vertices. + The distance between the foci and their respective vertices is equal. + The distance between the foci and the center is also equal. + The distance between the left vertex and the minor axis is labeled a. The length of the major axis is 2a. + The length between the center and the top of the minor axis is labeled b. The length of the minor axis is 2b + The distance between the center and the right focus is labeled c. +

    +     + + + \begin{tikzpicture}[>=latex] + + \draw [thick,firstcolor] (0,0) circle [x radius=2,y radius=1.5]; + \draw [thick,dashed] (-2.4,0) -- (2.4,0) (0,1.8) -- (0,-1.9); + + \filldraw [secondcolor] (1.3,0) circle (2.4pt) (-1.3,0) circle (2.4pt); + + \draw [->] (-2,-2) node [fill=white] { Major axis} -- (-2.2,-.1); + \draw [->] (2,-2) node [fill=white] { Minor axis} -- (.1,-1.8); + \draw [->] (-2,2) -- (-2,.3); + \draw [->] (-2,2) node [fill=white] { Vertices} -- (1.9,.1); + \draw [->] (2,2) -- (-1.2,.1); + \draw [->] (2,2) node [fill=white] { Foci} -- (1.3,.1); + + \filldraw [firstcolor] (2,0) circle (2.4pt) + (-2,0) circle (2.4pt); + + \draw (-1,-.25) node { $\underbrace{\rule{1.8cm}{0pt}}_a$}; + \draw (.65,-.25) node { $\underbrace{\rule{1.1cm}{0pt}}_c$}; + \draw (-.25,.75) node [] { $b\left\{\rule[-.65cm]{0pt}{1.3cm}\right.$}; + + \end{tikzpicture} + + + + +
    + +

    + In general, the two foci of an ellipse lie on the + major axis of the ellipse, + and the midpoint of the segment joining the two foci is the center. + The major axis intersects the ellipse at two points, + each of which is a vertex. + The line segment through the center and perpendicular to the major axis is the minor axis. + The constant sum of distances + that defines the ellipse is the length of the major axis, + , 2a. +

    + +

    + Allowing for the shifting of the ellipse gives the following standard equations. +

    + + + Standard Equation of the Ellipse +

    + The equation of an ellipse centered at (h,k) with major axis of length 2a and minor axis of length 2b in standard form is: +

    + +

    +

      +
    1. +

      + Horizontal major axis: + \ds \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1. +

      +
      2. + +
    2. +

      + Vertical major axis: \ds \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1. +

      +
      3. +
    +

    + +

    + The foci lie along the major axis, + c units from the center, where c^2=a^2-b^2. + ellipsestandard equation +

    +     + + + Finding the equation of an ellipse + +

    + Find the general equation of the ellipse graphed in . +

    + +
    + The ellipse used in + + + An ellipse centered in the second quadrant, lying entirely in the second and third quadrants. + +

    + An ellipse with a center at (-3,1). + The ellipse has vertices at (-3,6) and (-3,4). + The ellipse appears to have a minor axis of length 4. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + ytick={-4,-2,2,4,6}, + ymin=-4.9,ymax=6.9, + xmin=-6.9,xmax=6.9 + ] + + \addplot+ [domain=0:360,samples=60] ({2*cos(x)-3},{5*sin(x)+1}); + \addplot [soliddot,color=black,fill=black] coordinates {(-3,1)}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + The center is located at (-3,1). + The distance from the center to a vertex is 5 units, hence a=5. + The minor axis seems to have length 4, so b=2. + Thus the equation of the ellipse is + + \frac{(x+3)^2}{4}+\frac{(y-1)^2}{25} = 1 + . +

    +     + +     + + + Graphing an ellipse + +

    + Graph the ellipse defined by 4x^2+9y^2-8x-36y=-4. +

    +     + +

    + It is simple to graph an ellipse once it is in standard form. + In order to put the given equation in standard form, + we must complete the square with both the x and y terms. + We first rewrite the equation by regrouping: + + 4x^2+9y^2-8x-36y=-4 \Rightarrow (4x^2-8x) + (9y^2-36y) = -4 + . +

    + +

    + Now we complete the squares. + + (4x^2-8x) + (9y^2-36y) \amp = -4 + 4(x^2-2x) + 9(y^2-4y) \amp = -4 + 4(x^2-2x +1 - 1) + 9(y^2-4y+4-4) \amp = - 4 + 4\big((x-1)^2-1\big) + 9\big((y-2)^2-4\big) \amp = -4 + 4(x-1)^2 -4 + 9(y-2)^2-36 \amp = -4 + 4(x-1)^2 + 9(y-2)^2 \amp = 36 + \frac{(x-1)^2}{9} + \frac{(y-2)^2}{4} \amp = 1 + . +

    + +

    + We see the center of the ellipse is at (1,2). + We have a=3 and b=2; + the major axis is horizontal, + so the vertices are located at (-2,2) and (4,2). + We find c=\sqrt{9-4} = \sqrt{5}\approx 2.24. + The foci are located along the major axis, + approximately 2.24 units from the center, at (1\pm 2.24,2). + This is all graphed in +

    + +
    + Graphing the ellipse in + + + An ellipse centered at (1,2), with vertices at (-2,2) and (4,2). + +

    + An ellipse centered at (1,2). + The vertices of the ellipse are drawn at (-2,2) and (4,2). + Two foci are drawn at (1 - 2.24, 2) and (1 + 2.24, 2). + The minor axis of the ellipse has a length of 4. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-2,-1,1,2,3,4}, + ytick={-1,1,2,3,4}, + ymin=-2,ymax=5, + xmin=-3,xmax=5 + ] + + \addplot+ [domain=0:360,samples=60] ({3*cos(x)+1},{2*sin(x)+2}); + + \filldraw (axis cs:1,2) circle (2.4pt) + (axis cs:3.24,2) circle (1pt) + (axis cs:-1.24,2) circle (1pt); + + \filldraw [firstcolor] (axis cs: 4,2) circle (2.4pt) + (axis cs: -2,2) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +     + + + Eccentricity +

    + When a=b, we have a circle. + The general equation becomes + + \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{a^2} = 1 \Rightarrow (x-h)^2 + (y-k)^2 = a^2 + , + the familiar equation of the circle centered at (h,k) with radius a. + Since a=b, c = \sqrt{a^2-b^2}=0. + The circle has two foci, + but they lie on the same point, the center of the circle. +

    + +

    + Consider , + where several ellipses are graphed with a=1. + In , we have c=0 and the ellipse is a circle. + As c grows, + the resulting ellipses look less and less circular. + A measure of this noncircularness + is eccentricity. +

    + +
    + Understanding the eccentricity of an ellipse + + +
    + + + + An ellipse of eccentricity 0. + +

    + A circle of radius 1 centered at the orgin. + The major and minor axis both have a length of 2. + Below the ellipse it is written that e = 0. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-1,1}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.35,xmax=1.35 + ] + + \addplot+ [domain=0:360,samples=60] ({cos(x)},{sin(x)}); + + \filldraw (axis cs:0,0) circle (2.4pt); + + \draw (axis cs:.8,-.9) node { $e=0$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + An ellipse of eccentricity 0.3. + +

    + An ellipse with foci at around (-0.2, 0) and (0.2,2). + The major axis has a length of 2 and the minor axis has a length of about 1.8. + It is written that e=0.3 +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-1,1}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.35,xmax=1.35 + ] + + \addplot+ [domain=0:360,samples=60] ({cos(x)},{.95*sin(x)}); + + \filldraw (axis cs:.3,0) circle (2.4pt) + (axis cs:-.3,0) circle (2.4pt); + + \draw (axis cs:.8,-.9) node { $e=0.3$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + + +
    + + + + An ellipse of eccentricity 0.8. + +

    + An ellipse with foci drawn at (-0.8,0) and (0.8,0). + The major axis has a length of 2, and the minor axis has a length of around 1. + Below the ellipse it is written that e = 0.8. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-1,1}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.35,xmax=1.35 + ] + + \addplot+ [domain=0:360,samples=60] ({cos(x)},{.6*sin(x)}); + + \filldraw (axis cs:.8,0) circle (2.4pt) + (axis cs:-.8,0) circle (2.4pt); + + \draw (axis cs:.8,-.9) node { $e=0.8$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + An ellipse of eccentricity 0.99. + +

    + An ellipse with eccentricity 0. + The vertices and the foci are at the same points, one at (-1,0) and (1,0). + The major axis has a length of 2, and the minor axis has a length of about 0.2. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-1,1}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.35,xmax=1.35 + ] + + \addplot+ [domain=0:360,samples=60] ({cos(x)},{.141*sin(x)}) ; + + \draw (axis cs:.8,-.9) node { $e=0.99$}; + + \filldraw (axis cs:.99,0) circle (2.4pt) + (axis cs:-.99,0) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +     + +
    + + + Eccentricity of an Ellipse + +

    + The eccentricity e of an ellipse is \ds e=\frac{c}{a}. + ellipseeccentricity + eccentricity +

    +     +     + +

    + The eccentricity of a circle is 0; + that is, a circle has no noncircularness. + As c approaches a, + e approaches 1, giving rise to a very noncircular ellipse, + as seen in . +

    + +

    + It was long assumed that planets had circular orbits. + This is known to be incorrect; + the orbits are elliptical. + Earth has an eccentricity of 0.0167 it has a nearly circular orbit. + Mercury's orbit is the most eccentric, with e=0.2056. + (Pluto's eccentricity is greater, at e=0.248, + the greatest of all the currently known dwarf planets.) + The planet with the most circular orbit is Venus, with e=0.0068. + The Earth's moon has an eccentricity of e=0.0549, + also very circular. +

    +     + + + Reflective Property +

    + The ellipse also possesses an interesting reflective property. + Any ray emanating from one focus of an ellipse reflects off the ellipse along a line through the other focus, + as illustrated in . + This property is given formally in the following theorem. +

    + +
    + Illustrating the reflective property of an ellipse + + + An ellipse with rays emanating from the foci, demonstrating the reflective property of an ellipse. + +

    + An ellipse with the foci drawn. + The left focus is labeled F_1, and the right focus is labeled F_2. + A ray is emanating from F_1 towards the bottom right. + It then reflects off the ellipse to the upper right, where it then passes through F_2. + A ray also emanates from F_2 towards the upper right. + It then reflects off the ellipse towards the bottom left, where it then passes through F_1. +

    +     + + + \begin{tikzpicture}[scale=1.2,>=latex,scale=1.32] + + \draw [thick] (0,0) circle [x radius=2cm,y radius = 1.2cm]; + + \filldraw (1.6,0) circle (2.4pt) node [above left] { $F_2$} + (-1.6,0) circle (2.4pt) node [above] { $F_1$}; + + \draw [->,firstcolor,thick] (1.6,0) -- ({2*cos(30)},{1.2*sin(30)}); + \draw [->,firstcolor,thick] ({2*cos(30)},{1.2*sin(30)}) -- (-1.55,0); + \draw [->,secondcolor,thick] (-1.6,0) -- ({2*cos(250)},{1.2*sin(250)}); + \draw [->,secondcolor,thick] ({2*cos(250)},{1.2*sin(250)}) -- (1.55,0); + + \end{tikzpicture} + + + + +
    + + + Reflective Property of an Ellipse + +

    + Let P be a point on a ellipse with foci F_1 and F_2. + The tangent line to the ellipse at P makes equal angles with the following two lines:ellipsereflective property +

    + +

    +

      +
    1. +

      + The line through F_1 and P, and +

      +
      2. + +
    2. +

      + The line through F_2 and P. +

      +
      3. +
    +

    +     +     + +

    + This reflective property is useful in optics and is the basis of the phenomena experienced in whispering halls. +

    +     +     + + + Hyperbolas + + + +

    + The definition of a hyperbola is very similar to the definition of an ellipse; + we essentially just change the word + sum to difference. +

    + + + Hyperbola + +

    + A hyperbola is the locus of all points where the absolute value of difference of distances from two fixed points, + each a focus of the hyperbola, is constant. + conic sectionshyperbola + hyperboladefinition + focus +

    +     +     + +

    + We do not have a convenient way of visualizing the construction of a hyperbola as we did for the ellipse. + The geometric definition does allow us to find an algebraic expression that describes it. + It will be useful to define some terms first. +

    + +

    + The two foci lie on the transverse axis of the hyperbola; + the midpoint of the line segment joining the foci is the + center of the hyperbola. + The transverse axis intersects the hyperbola at two points, + each a vertex of the hyperbola. + The line through the center and perpendicular to the transverse axis is the + conjugate axis. + This is illustrated in . + It is easy to show that the constant difference of distances used in the definition of the hyperbola is the distance between the vertices, + , 2a. +

    + +
    + Labeling the significant features of a hyperbola + + + A hyperbola with labels for key components + +

    + A hyperbola with labels for key components. + The hyperbola is composed of two seperate segments with a space between them. + The first shape is on the left. The shape begins in the upper left, moving almost linearly towards the lower right. + The curve then begins to bend gently backwards before moving linearly towards the bottom left. + The right curve is a reflection of the left curve. It opens towards the right. + The entire hyperbola is separated into 4 equal segments by two axes. + The axis dividing the hyperola in half vertically is labeled as the "conjugate axis." + The axis dividing th ehyperbola in half horizontally is labeled as the "transverse axis." + The point in which the axes touch is the center of the hyperbola. + The points in which the curves intersect the transverse axis are labeled as foci. + Two foci lie on the transverse axis. + The first focus is to the left of the left curve. + The right focus is to the right of the right curve. + The distance between the left vertex and the center is labeled as a. + The distance between the center and the right focus is labeled as c. +

    +     + + + \begin{tikzpicture}[>=latex] + + \begin{axis}[ + axis y line=none, + axis x line=none, + ymin=-3.2,ymax=3.2, + xmin=-3.2,xmax=3.2 + ] + + \addplot [firstcurvestyle,domain=-70:70,samples=40] ({sec(x)},{tan(x)}); + \addplot [firstcurvestyle,domain=-70:70,samples=40] ({-sec(x)},{tan(x)}); + + \filldraw (axis cs:0,0) circle (2.4pt); + + \draw [thick,dashed] (axis cs:-3.2,0) -- node [above,pos=.88] { Transverse} node [below,pos=.88] { axis} (axis cs:3.2,0) + (axis cs:0,-3.2) -- node [below,rotate=90,pos=.8] { axis }node [above,rotate=90,pos=.8] { Conjugate} (axis cs:0,3.2); + + \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt) + (axis cs:-1,0) circle (2.4pt); + + \filldraw [secondcolor] (axis cs:1.4,0) circle (2.4pt) + (axis cs:-1.4,0) circle (2.4pt); + + \draw [->] (axis cs: 1.25,-2.8) -- (axis cs: -1.35,-.1); + \draw [->] (axis cs: 1.25,-2.8) node [fill=white] { Foci } -- (axis cs: 1.4,-.1); + + \draw [->] (axis cs: -1.25,-2.8) -- (axis cs: -1,-.1); + \draw [->] (axis cs: -1.25,-2.8) node [fill=white] { Vertices} -- (axis cs: .95,-.1); + + \draw (axis cs:-.5,.4) node { $\overbrace{\rule{18pt}{0pt}}^a$}; + \draw (axis cs:.7,.4) node { $\overbrace{\rule{25pt}{0pt}}^c$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + + + Standard Equation of a Hyperbola +

    + The equation of a hyperbola centered at (h,k) in standard form is: + hyperbolastandard equation +

    + +

    +

      +
    1. +

      + Horizontal Transverse Axis: + \ds \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1. +

      +
      2. + +
    2. +

      + Vertical Transverse Axis: + \ds \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2} = 1. +

      +
      3. +
    +

    + +

    + The vertices are located a units from the center and the foci are located c units from the center, + where c^2 = a^2+b^2. +

    +     + + + Graphing Hyperbolas +

    + Consider the hyperbola \frac{x^2}9-\frac{y^2}1 = 1. + Solving for y, we find y=\pm\sqrt{x^2/9-1}. + As x grows large, + the -1 part of the equation for y becomes less significant and y\approx \pm\sqrt{x^2/9} = \pm x/3. + That is, as x gets large, + the graph of the hyperbola looks very much like the lines y=\pm x/3. + These lines are asymptotes of the hyperbola, + as shown in . +

    + +
    + Graphing the hyperbola \frac{x^2}9-\frac{y^2}1 = 1 along with its asymptotes, y=\pm x/3 + + + A graph of a hyperbola and its asymptotes. + +

    + A graph of the hyperbola \frac{x^2}{9} - \frac{y^2}{1} = 1, along with its two asymptotes. + The hyperbola is centered at the origin, with vertices at(-3,0) and (3,0). + Two asymptotes are drawn from the equations y = \frac{x}{3} and y = -\frac{x}{3}. + As the hyperbola extends outwards from the origin, it becomes close to the asymptotes. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-3.2,ymax=3.2, + xmin=-8.2,xmax=8.2 + ] + + \addplot [firstcurvestyle,domain=-70:70,samples=40] ({3*sec(x)},{tan(x)}); + \addplot [firstcurvestyle,domain=-70:70,samples=40] ({-3*sec(x)},{tan(x)}); + + \addplot [secondcurvestyle,domain=-8:8] {x/3}; + \addplot [secondcurvestyle,domain=-8:8] {-x/3}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + This is a valuable tool in sketching. + Given the equation of a hyperbola in general form, + draw a rectangle centered at (h,k) with sides of length 2a parallel to the transverse axis and sides of length 2b parallel to the conjugate axis. + (See + for an example with a horizontal transverse axis.) + The diagonals of the rectangle lie on the asymptotes. +

    + +
    + Using the asymptotes of a hyperbola as a graphing aid + + + An illustration of sketching a hyperbola with a rectangle. + +

    + A hyperbola sketched using a rectangle. + A center of the rectangle is drawn at the point (h,k). + The rectangle has a width of 2a and a height of 2b. + The vertices are drawn at the points (h-a,k) and (h+a,k). + The asymptotes are drawn as the lines passing through the center and the corners of the rectangle. + The hyperbola is then drawn, following the asymptotes, touching the vertices, and then moving away from the center to follow the asymptotes. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick=\empty, + extra x ticks={2,8,5}, + extra x tick labels={$h-a$, $h+a$,$h$}, + ytick=\empty, + extra y ticks={2,4,6}, + extra y tick labels={$k-b$, $k$, $k+b$}, + ymin=-2.2,ymax=10.2, + xmin=-3.2,xmax=13.2 + ] + + \addplot [firstcurvestyle,domain=-70:70,samples=40] ({3*sec(x)+5},{2*tan(x)+4}); + \addplot [firstcurvestyle,domain=-70:70,samples=40] ({-3*sec(x)+5},{2*tan(x)+4}); + + \draw [thick,dashed] (axis cs: 2,6) -- (axis cs:8,6) -- (axis cs:8,2) -- (axis cs: 2,2) -- cycle; + + \addplot [secondcurvestyle,domain=-3:13] {2*(x-5)/3+4}; + \addplot [secondcurvestyle,domain=-3:13] {-2*(x-5)/3+4}; + + \filldraw (axis cs:5,4) circle (2.4pt); + + \filldraw [firstcolor] (axis cs:2,4) circle (2.4pt); + \filldraw [firstcolor] (axis cs:8,4) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + These lines pass through (h,k). + When the transverse axis is horizontal, + the slopes are \pm b/a; + when the transverse axis is vertical, + their slopes are \pm a/b. + This gives equations: +

    + + + + + + +

    Horizontal Transverse Axis

    Vertical Transverse Axis +     + + \ds y=\pm\frac ba(x-h)+k + \ds y=\pm\frac ab(x-h)+k. + +     + + + Graphing a hyperbola + +

    + Sketch the hyperbola given by \ds \frac{(y-2)^2}{25}-\frac{(x-1)^2}{4}=1. +

    +     + +

    + The hyperbola is centered at (1,2); + a=5 and b=2. + In + we draw the prescribed rectangle centered at (1,2) along with the asymptotes defined by its diagonals. + The hyperbola has a vertical transverse axis, + so the vertices are located at (1,7) and (1,-3). + This is enough to make a good sketch. +

    + +
    + Graphing the hyperbola in + + + A graph of the hyperbola in this example + +

    + A graph of the hyperbola given in . + The hyperbola has the equation \frac{(y-2)^2}{25} - \frac{(x-1)^2}{4} = 1. + The center of the hyperbola is drawn at (1,2). + A rectangle is drawn around the center, with a height of 10 and a width of 4. + At the top and bottom of the rectangle, the vertices of the hyperbola are drawn at (1,7) and (1,-3). + The asymptotes of the hyperbola are drawn passing through the corners of the rectangle and the center. + The hyperbola is then drawn above and below the rectangle, opening upwards and downwards respectively. + The foci are also drawn a short distance above and below the vertices. + The top focus is near the point (1, 7.4) and the bottom focus is near the point (1,-3.4). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=4, + minor y tick num=4, + ymin=-7.2,ymax=11.2, + xmin=-5.2,xmax=7.2 + ] + + \addplot [firstcurvestyle,domain=-70:70,samples=40] ({2*tan(x)+1},{5*sec(x)+2}); + \addplot [firstcurvestyle,domain=-70:70,samples=40] ({2*tan(x)+1},{-5*sec(x)+2}); + + \draw [thick,dashed] (axis cs: -1,7) -- (axis cs:3,7) -- (axis cs:3,-3) -- (axis cs: -1,-3) -- cycle; + + \addplot [secondcurvestyle,domain=-3:13] {5*(x-1)/2+2}; + \addplot [secondcurvestyle,domain=-3:13] {-5*(x-1)/2+2}; + + \filldraw (axis cs:1,2) circle (2.4pt); + + \filldraw [firstcolor] (axis cs:1,7) circle (2.4pt); + \filldraw [firstcolor] (axis cs:1,-3) circle (2.4pt); + \filldraw [secondcolor] (axis cs:1,7.4) circle (2.4pt); + \filldraw [secondcolor] (axis cs:1,-3.4) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + We also find the location of the foci: + as c^2= a^2+b^2, we have c=\sqrt{29}\approx 5.4. + Thus the foci are located at + (1,2\pm 5.4) as shown in the figure. +

    +     + +     + + + Graphing a hyperbola + +

    + Sketch the hyperbola given by 9x^2-y^2+2y=10. +

    +     + +

    + We must complete the square to put the equation in general form. + (We recognize this as a hyperbola since it is a general quadratic equation and the x^2 and y^2 terms have opposite signs.) + + 9x^2-y^2+2y \amp =10 + 9x^2- (y^2-2y) \amp = 10 + 9x^2 - (y^2-2y+1-1) \amp = 10 + 9x^2 -\big((y-1)^2-1\big) \amp = 10 + 9x^2 - (y-1)^2 \amp = 9 + x^2 - \frac{(y-1)^2}{9} \amp =1 + +

    + +
    + Graphing the hyperbola in + + + The hyperbola described in this example + +

    + The hyperbola described in . + The hyperbola comes from the equation 9x^2 - y^2 + 2y = 10. + The hyperbola is centered at (0,1), around which a rectangle is drawn. + The rectangle has a height of 6 and a width of 2. + The vertices are drawn on the left and right sides of the rectangle, at the points (-1,1) and (1,1). + The hyperbola is the drawn, opening to the left and right and following the diagonal asymptotes, which are also drawn. + The foci are also drawn at the points (-3.2,1) and (3.2,1). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + minor y tick num=4, + ymin=-10.9,ymax=11.9, + xmin=-4.2,xmax=4.2 + ] + + \addplot [firstcurvestyle,domain=-75:75,samples=40] ({sec(x)},{3*tan(x)+1}); + \addplot [firstcurvestyle,domain=-75:75,samples=40] ({-sec(x)},{3*tan(x)+1}); + + \draw [thick,dashed] (axis cs: -1,4) -- (axis cs:1,4) -- (axis cs:1,-2) -- (axis cs: -1,-2) -- cycle; + + \addplot [secondcurvestyle,domain=-4:4] {3*(x-0)/1+1}; + \addplot [secondcurvestyle,domain=-4:4] {-3*(x-0)/1+1}; + + \filldraw (axis cs:0,1) circle (2.4pt); + + \filldraw [firstcolor] (axis cs:-1,1) circle (2.4pt); + \filldraw [firstcolor] (axis cs:1,1) circle (2.4pt); + \filldraw [secondcolor] (axis cs:3.2,1) circle (2.4pt); + \filldraw [secondcolor] (axis cs:-3.2,1) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + We see the hyperbola is centered at (0,1), + with a horizontal transverse axis, + where a=1 and b=3. + The appropriate rectangle is sketched in + along with the asymptotes of the hyperbola. + The vertices are located at (\pm 1,1). + We have c=\sqrt{10}\approx 3.2, + so the foci are located at + (\pm 3.2,1) as shown in the figure. +

    +     + +     +     + + + Eccentricity +

    + The eccentricity of a hyperbola is defined in the same way as an ellipse. +

    + + Eccentricity of a Hyperbola + +

    + The eccentricity of a hyperbola is \ds e=\frac ca. + hyperbolaeccentricity + eccentricity +

    +     +     + +

    + Note that this is the definition of eccentricity as used for the ellipse. + When c is close in value to a (, e\approx 1), + the hyperbola is very narrow + (looking almost like crossed lines). + + shows hyperbolas centered at the origin with a=1. + The graph in has c=1.05, + giving an eccentricity of e=1.05, which is close to 1. + As c grows larger, + the hyperbola widens and begins to look like parallel lines, + as shown in . +

    + +
    + Understanding the eccentricity of a hyperbola + + +
    + + + + A hyperbola with eccentricity 1.05 + +

    + A hyperbola with eccentricity 1.05. The hyperbola is centered at the origin. + The vertices are drawn to the left and right of the center. + The foci are also drawn, and are shown to be very close to the vertices. + The hyperbola is quite narrow, only slowing spreading out vertically as the curve moves away from the center. + Below the hyperbola it is written that e=1.05. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-10,-5,5,10}, + minor x tick num=4, + ytick={-10,-5,5,10}, + minor y tick num=4,%extra y ticks={-5,-3,...,7}, + ymin=-11.9,ymax=11.9, + xmin=-11.9,xmax=11.9 + ] + + \addplot [firstcurvestyle,domain=-85:85,samples=40] ({sec(x)},{.32*tan(x)}); + \addplot [firstcurvestyle,domain=-85:85,samples=40] ({-sec(x)},{.32*tan(x)}); + + \filldraw (axis cs:0,0) circle (2.4pt); + + \filldraw [firstcolor] (axis cs:-1,0) circle (2.4pt); + \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt); + \filldraw [secondcolor] (axis cs:1.05,0) circle (2.4pt); + \filldraw [secondcolor] (axis cs:-1.05,0) circle (2.4pt); + + \draw (axis cs: 9,-6) node { $e = 1.05$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + A hyperbola with eccentricity 1.5 + +

    + A hyperbola with eccentricity 1.5. + The hyperbola is centered at the origin. + The vertices are drawn to the left and right of the center. + The foci are drawn next to the vertices. + The foci are a short distance away from the vertices. + The hyperbola is of moderate width, increasing horizontally and vertically at a similar rate. + Below the hyperbola it is written that e=1.5 +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-10,-5,5,10}, + minor x tick num=4, + ytick={-10,-5,5,10}, + minor y tick num=4,%extra y ticks={-5,-3,...,7}, + ymin=-11.9,ymax=11.9, + xmin=-11.9,xmax=11.9 + ] + + \addplot [firstcurvestyle,domain=-85:85,samples=40] ({sec(x)},{1.12*tan(x)}); + \addplot [firstcurvestyle,domain=-85:85,samples=40] ({-sec(x)},{1.12*tan(x)}); + + \filldraw (axis cs:0,0) circle (2.4pt); + + \filldraw [firstcolor] (axis cs:-1,0) circle (2.4pt); + \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt); + \filldraw [secondcolor] (axis cs:1.5,0) circle (2.4pt); + \filldraw [secondcolor] (axis cs:-1.5,0) circle (2.4pt); + + \draw (axis cs: 9,-6) node { $e = 1.5$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + + +
    + + + + A hyperbola with eccentricity 3 + +

    + A hyperbola with eccentricity 3, centered at the origin. + The vertices are drawn to the left and right of the center. + The foci are drawn, and are a moderate distance away from the vertices. + The hyperbola is quite wide, quickly increasing vertically as the curve moves further from the origin. + Below the hyperbola it is written that e = 3. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-10,-5,5,10}, + minor x tick num=4, + ytick={-10,-5,5,10}, + minor y tick num=4,%extra y ticks={-5,-3,...,7}, + ymin=-11.9,ymax=11.9, + xmin=-11.9,xmax=11.9 + ] + + \addplot [firstcurvestyle,domain=-85:85,samples=40] ({sec(x)},{2.83*tan(x)}); + \addplot [firstcurvestyle,domain=-85:85,samples=40] ({-sec(x)},{2.83*tan(x)}); + + \filldraw (axis cs:0,0) circle (2.4pt); + + \filldraw [firstcolor] (axis cs:-1,0) circle (2.4pt); + \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt); + \filldraw [secondcolor] (axis cs:3,0) circle (2.4pt); + \filldraw [secondcolor] (axis cs:-3,0) circle (2.4pt); + + \draw (axis cs: 9,-6) node { $e = 3$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + A hyperbola with eccentricity 10 + +

    + A hyperbola of eccentricity 10, centered at the origin. + The vertices are drawn to the left and right of the center. + The foci are drawn a large distance form the vertices. + The hyperbola is very wide, increasing vertically significantly more quickly than horizontally. + Below the hyperbola it is written that e=10. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-10,-5,5,10}, + minor x tick num=4, + ytick={-10,-5,5,10}, + minor y tick num=4,%extra y ticks={-5,-3,...,7}, + ymin=-11.9,ymax=11.9, + xmin=-11.9,xmax=11.9 + ] + + \addplot [firstcurvestyle,domain=-85:85,samples=40] ({sec(x)},{9.95*tan(x)}); + \addplot [firstcurvestyle,domain=-85:85,samples=40] ({-sec(x)},{9.95*tan(x)}); + + \filldraw (axis cs:0,0) circle (2.4pt); + + \filldraw [firstcolor] (axis cs:-1,0) circle (2.4pt); + \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt); + \filldraw [secondcolor] (axis cs:10,0) circle (2.4pt); + \filldraw [secondcolor] (axis cs:-10,0) circle (2.4pt); + + \draw (axis cs: 9,-6) node { $e = 10$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +     +
    +     + + + Reflective Property +

    + Hyperbolas share a similar reflective property with ellipses. + However, in the case of a hyperbola, + a ray emanating from a focus that intersects the hyperbola reflects along a line containing the other focus, + but moving away from that focus. + This is illustrated in + (on the next page). + Hyperbolic mirrors are commonly used in telescopes because of this reflective property. + It is stated formally in the following theorem. +

    + + + Reflective Property of Hyperbolas + +

    + Let P be a point on a hyperbola with foci F_1 and F_2. + The tangent line to the hyperbola at P makes equal angles with the following two lines:hyperbolareflective property +

    + +

    +

      +
    1. +

      + The line through F_1 and P, and +

      +
      2. + +
    2. +

      + The line through F_2 and P. +

      +
      3. +
    +

    +     +     +     + + + Location Determination +

    + Determining the location of a known event has many practical uses + (locating the epicenter of an earthquake, an airplane crash site, + the position of the person speaking in a large room, etc.). +

    + +

    + To determine the location of an earthquake's epicenter, + seismologists use trilateration + (not to be confused with triangulation). + A seismograph allows one to determine how far away the epicenter was; + using three separate readings, + the location of the epicenter can be approximated. +

    + +

    + A key to this method is knowing distances. + What if this information is not available? + Consider three microphones at positions A, + B and C which all record a noise (a person's voice, + an explosion, etc.) created at unknown location D. + The microphone does not know + when the sound was created, + only when the sound was detected. + How can the location be determined in such a situation? +

    + +
    + Illustrating the reflective property of a hyperbola + + + A hyperbola demonstrating the reflective property. + +

    + A hyperbola with rays emanating from the left focus. + The ray moves towards the right, reflecting off a point on the right side of the hyperbola. + The reflected ray moves towards the upper left. + A dashed line is drawn between the right focus and the point where the ray touches the hyperbola. + The reflected ray is moving in the same direction as the dashed line. +

    +     + + + \begin{tikzpicture}[>=latex] + + \begin{axis}[ + axis y line=none, + axis x line=none, + ymin=-11.9,ymax=11.9, + xmin=-5,xmax=5 + ] + + \addplot [firstcurvestyle,domain=-85:85,samples=40] ({sec(x)},{2.83*tan(x)}); + \addplot [firstcurvestyle,domain=-85:85,samples=40] ({-sec(x)},{2.83*tan(x)}); + + \filldraw (axis cs:0,0) circle (2.4pt); + + \filldraw [firstcolor] (axis cs:-1,0) circle (2.4pt); + \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt); + + \draw [secondcolor,thick,->] (axis cs:-3,0) -- (axis cs: 1.4,2.83); + \draw [secondcolor,thick,->] (axis cs:1.4,2.83) -- (axis cs: -1.8,8.49); + + \draw [dashed,secondcolor,thick] (axis cs:1.4,2.83)--(axis cs: 3,0); + + \filldraw (axis cs:1.4,2.83) circle (2.4pt); + + \filldraw (axis cs:3,0) circle (2.4pt) node [below] { $F_2$}; + \filldraw (axis cs:-3,0) circle (2.4pt) node [below] { $F_1$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + If each location has a clock set to the same time, + hyperbolas can be used to determine the location. + Suppose the microphone at position A records the sound at exactly 12:00, + location B records the time exactly 1 second later, + and location C records the noise exactly 2 seconds after that. + We are interested in the difference of times. + Since the speed of sound is approximately 340 m/s, we can conclude quickly that the sound was created 340 meters closer to position A than position B. + If A and B are a known distance apart (as shown in ), + then we can determine a hyperbola on which D must lie. +

    + +

    + The difference of distances is 340; + this is also the distance between vertices of the hyperbola. + So we know 2a= 340. + Positions A and B lie on the foci, so 2c=1000. + From this we can find b\approx 470 and can sketch the hyperbola, + given in . + We only care about the side closest to A. (Why?) +

    + +

    + We can also find the hyperbola defined by positions B and C. + In this case, + 2a = 680 as the sound traveled an extra 2 seconds to get to C. + We still have 2c=1000, + centering this hyperbola at (-500,500). + We find b\approx 367. + This hyperbola is sketched in . + The intersection point of the two graphs is the location of the sound, + at approximately (188,-222.5). +

    + +
    + + + +
    + + + Three points drawn and labeled on a plane + +

    + Three points drawn and labeled on a plane. + A is at the point (500,0). + B is at the point (-500,0). + C is at the point (-500,1000). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1100,ymax=1100, + xmin=-1100,xmax=1100 + ] + + \filldraw (axis cs:500,0) circle (2.4pt) node [above] { $A$}; + \filldraw (axis cs:-500,0) circle (2.4pt) node [above] { $B$}; + \filldraw (axis cs:-500,1000) circle (2.4pt) node [left] { $C$}; + + \end{axis} + + \end{tikzpicture} + + + +
    + + +
    + + + A hyperbola drawn from points A and B to illustrate the location property + +

    + A hyperbola drawn from points A and B in . + A and B are the foci of the hyperbola. + The vertices of the hyperbola are 170 units away from the origin. + The left half of the hyperbola is drawn with a dashed line, while the right half of the parabola is drawn with a solid line. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1100,ymax=1100, + xmin=-1100,xmax=1100 + ] + + \addplot+ [domain=-85:85] ({170*sec(x)},{470*tan(x)}); + \addplot+ [domain=-85:85] ({-170*sec(x)},{470*tan(x)}); + + \filldraw (axis cs:500,0) circle (2.4pt) node [above] { $A$}; + \filldraw (axis cs:-500,0) circle (2.4pt) node [above] { $B$}; + \filldraw (axis cs:-500,1000) circle (2.4pt) node [left] { $C$}; + + \end{axis} + + \end{tikzpicture} + + + +
    + + +
    + + + A fourth point found from hyperbolas given by points in the previous figures + +

    + A hyperbola drawn from points B and C in , intersecting with the hyperbola drawn in . + The hyperbola uses B and C as the foci. The vertices of the hyperbola are 640 units apart. + The bottom hyperbola, with focus B, extends towards the right. + The point at which it intersects with the hyperbola with focus A is labeled as D. +

    +     + + + \begin{tikzpicture} + \begin{axis}[ + axis on top, + ymin=-1100,ymax=1100, + xmin=-1100,xmax=1100 + ] + + \addplot+ [domain=-85:85] ({170*sec(x)},{470*tan(x)}); + \addplot+ [domain=-85:85] ({367*tan(x)-500},{340*sec(x)+500}); + \addplot [secondcurvestyle,domain=-85:85] ({367*tan(x)-500},{-340*sec(x)+500}); + + \filldraw (axis cs:500,0) circle (2.4pt) node [above] { $A$}; + \filldraw (axis cs:-500,0) circle (2.4pt) node [above] { $B$}; + \filldraw (axis cs:-500,1000) circle (2.4pt) node [left] { $C$}; + \filldraw (axis cs:188,-222) circle (2.4pt) node [below left] { $D$}; + + \end{axis} + + \end{tikzpicture} + + + +
    + +     +
    +     + +

    + This chapter explores curves in the plane, + in particular curves that cannot be described by functions of the form y=f(x). + In this section, + we learned of ellipses and hyperbolas that are defined implicitly, + not explicitly. + In the following sections, + we will learn completely new ways of describing curves in the plane, + using parametric equations + and polar coordinates, + then study these curves using calculus techniques. +

    +     + + + + Terms and Concepts + + + + +

    + What is the difference between degenerate and nondegenerate conics? +

    +     + + + +

    + When defining the conics as the intersections of a plane and a double napped cone, + degenerate conics are created when the plane intersects the tips of the cones + (usually taken as the origin). + Nondegenerate conics are formed when this plane does not contain the origin. +

    +     + +     + + + + +

    + Use your own words to explain what the eccentricity of an ellipse measures. +

    +     + + + +     + + + + +

    + What has the largest eccentricity: + an ellipse or a hyperbola? +

    +     + + + +

    + Ellipse +

    +     +     + + +

    + Hyperbola +

    +     +     +     + +     + + + + +

    + Explain why the following is true: + If the coefficient of the x^2 term in the equation of an ellipse in standard form is smaller than the coefficient of the y^2 term, + then the ellipse has a horizontal major axis. +

    +     + + + +

    + With the equation \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1, + the ellipse has a horizontal major axis if a \gt b. + But the coefficient of the x^2 term is 1/a^2 + (not a^2), + so if 1/a^2\lt 1/b^2, then a \gt b and the major axis is horizontal. +

    +     + +     + + + + +

    + Explain how one can quickly look at the equation of a hyperbola in standard form + and determine whether the transverse axis is horizontal or vertical. +

    +     + + + +

    + With a horizontal transverse axis, + the x^2 term has a positive coefficient; + with a vertical transverse axis, + the y^2 term has a positive coefficient. +

    +     + +     + + + +

    + Fill in the blank: It can be said that ellipses and hyperbolas share the + same reflective property: + A ray emanating from one focus will reflect off the conic along a + that contains the other focus. +

    +     + + + + + + + + +

    + Your answer includes the correct word but has extra text. +

    +     +     +     +     +     +     + + + Problems + + + +

    + Find the equation of the parabola defined by the given information. + Sketch the parabola. +

    +     + + + + +

    + Focus: (3,2); directrix: y=1 +

    +     + +

    + y=\frac12(x-3)^2+\frac32 +

    +     + +     + + + + +

    + Focus: (-1,-4); directrix: y=2 +

    +     + +

    + y=\frac{-1}{12}(x+1)^2-1 +

    +     + +     + + + + +

    + Focus: (1,5); directrix: x=3 +

    +     + +

    + x=-\frac14(y-5)^2+2 +

    +     + +     + + + + +

    + Focus: (1/4,0); directrix: x=-1/4 +

    +     + +

    + x=y^2 +

    +     + +     + + + + +

    + Focus: (1,1); vertex: (1,2) +

    +     + +

    + y=-\frac14(x-1)^2+2 +

    +     + +     + + + + +

    + Focus: (-3,0); vertex: (0,0) +

    +     + +

    + x=-\frac1{12}y^2 +

    +     + +     + + + + +

    + Vertex: (0,0); directrix: y=-1/16 +

    +     + +

    + y=4x^2 +

    +     + +     + + + + +

    + Vertex: (2,3); directrix: x=4 +

    +     + +

    + x=-\frac18(y-3)^2+2 +

    +     + +     + +     + + + +

    + The equation of a parabola and a point on its graph are given. + Find the focus and directrix of the parabola, + and verify that the given point is equidistant from the focus and directrix. +

    +     + + + + +

    + y=\frac14x^2, P=(2,1) +

    +     + +

    + focus: (0,1); directrix: y=-1. + The point P is 2 units from each. +

    +     + +     + + + + +

    + x=\frac18(y-2)^2+3, P=(11,10) +

    +     + +

    + focus: (5,2); directrix: x=1. + The point P is 10 units from each. +

    +     + +     + +     + + + +

    + Sketch the ellipse defined by the given equation. + Label the center, foci and vertices. +

    +     + + + + +

    + \ds \frac{(x-1)^2}{3}+\frac{(y-2)^2}{5}=1 +

    +     + + +     + + + + +

    + \ds \frac{1}{25}x^2+\frac{1}{9}(y+3)^2=1 +

    +     + + +     + +     + + + +

    + Find the equation of the ellipse shown in the graph. + Give the location of the foci and the eccentricity of the ellipse. +

    +     + + + + + + + + An ellipse centered in the second quadrant. + +

    + An ellipse centered at the point (1,2). + The major axis has a length of 6. + The minor axis has a length of 4. + The vertices of the graph are at the points (-4,2) and (2,2). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-1.2,ymax=5.2, + xmin=-5.2,xmax=3.2 + ] + + \addplot+ [domain=0:360,samples=60] ({3*cos(x)-1},{2*sin(x)+2}); + + \end{axis} + + \end{tikzpicture} + + + + + +     + +

    + \frac{(x+1)^2}{9}+\frac{(y-2)^2}{4}=1; + foci at (-1\pm\sqrt{5},2); e=\sqrt{5}/3 +

    +     + +     + + + + + + + + A vertical ellipse centered on the x-axis. + +

    + An ellipse centered on the point (1,0). + The vertices of the ellipse are at the points (1,3) and (1,-3). + The major axis appears to have a length of 6. + The minor axis has a length of 1. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-3.5,ymax=3.5, + xmin=-1.2,xmax=2.2 + ] + + \addplot+ [domain=0:360,samples=60] ({.5*cos(x)+1},{3*sin(x))}); + + \end{axis} + + \end{tikzpicture} + + + + + +     + +

    + \frac{(x-1)^2}{1/4}+\frac{y^2}{9}=1; + foci at (1,\pm \sqrt{8.75}); + e=\sqrt{8.75}/3\approx 0.99 +

    +     + +     + +     + + + +

    + Find the equation of the ellipse defined by the given information. + Sketch the elllipse. +

    +     + + + + +

    + Foci: (\pm 2,0); vertices: (\pm 3,0) +

    +     + +

    + \frac{x^2}{9}+\frac{y^2}{5}=1 +

    +     + +     + + + + +

    + Foci: (-1,3) and (5,3); vertices: + (-3,3) and (7,3) +

    +     + +

    + \frac{(x-2)^2}{25}+\frac{(y-3)^2}{16}=1 +

    +     + +     + + + + +

    + Foci: (2,\pm 2); vertices: (2,\pm 7) +

    +     + +

    + \frac{(x-2)^2}{45}+\frac{y^2}{49}=1 +

    +     + +     + + + + +

    + Focus: (-1,5); vertex: + (-1,-4); center: (-1,1) +

    +     + +

    + \frac{(x+1)^2}{9}+\frac{(y-1)^2}{25}=1 +

    +     + +     + +     + + + +

    + Write the equation of the given ellipse in standard form. +

    +     + + + + +

    + x^2-2x+2y^2-8y=-7 +

    +     + +

    + \frac{(x-1)^2}{2}+(y-2)^2=1 +

    +     + +     + + + + +

    + 5x^2+3y^2=15 +

    +     + +

    + \frac{x^2}{3}+\frac{y^2}{5}=1 +

    +     + +     + + + + +

    + 3x^2+2y^2-12y+6=0 +

    +     + +

    + \frac{x^2}{4}+\frac{(y-3)^2}{6}=1 +

    +     + +     + + + + +

    + x^2+y^2-4x-4y+4=0 +

    +     + +

    + \frac{(x-2)^2}{4}+\frac{(y-2)^2}{4}=1 +

    +     + +     + +     + + + +

    + Find the equation of the hyperbola shown in the graph. +

    +     + + + + + + + + A hyperbola centered at the origin with a horizontal transverse axis. + +

    + A hyperbola centered at the origin with a horizontal transverse axis. + The foci of the hyperbola are at the points (-2,0) and (2,0). + The vertices of the hyperbola are at the points (-1,0) and (1,0) +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,1,2,-2}, + ymin=-3.5,ymax=3.5, + xmin=-3.5,xmax=3.5 + ] + + \addplot [firstcurvestyle,domain=-70:70] ({sec(x)},{1.73*tan(x))}); + \addplot [firstcurvestyle,domain=-70:70] ({-sec(x)},{1.73*tan(x))}); + + \filldraw (axis cs: 2,0) circle (2.4pt) + (axis cs: -2,0) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + + +     + +

    + x^2-\frac{y^2}{3}=1 +

    +     + +     + + + + + + + + A hyperbola centered at the origin with a vertical transverse axis. + +

    + A hyperbola centered at the origin with a vertical transverse axis. + The hyperbola has foci at the points (0,5) and (0,-5). + The vertices of the ellipse are at the points (0,1) and (0,-1). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + minor x tick num=4, + minor y tick num=4, + ymin=-6.5,ymax=6.5, + xmin=-8.5,xmax=8.5 + ] + + \addplot [firstcurvestyle,domain=-70:70] ({4.9*tan(x)},{sec(x))}); + \addplot [firstcurvestyle,domain=-70:70] ({4.9*tan(x)},{-sec(x))}); + + \filldraw (axis cs: 0,5) circle (2.4pt) + (axis cs: 0,-5) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + + +     + +

    + y^2-\frac{x^2}{24}=1 +

    +     + +     + + + + + + + + A hyperbola with center (1,3) and a vertical transverse axis. + +

    + A hyperbola with center (1,3) and a vertical transverse axis. + The foci of the hyperbola are not drawn. + The vertices of the hyperbola are at the points (1,5) and (1,1). + A box is drawn around the origin with a width of 6 and a height of 4. + The diagonal asymptotes are drawn as the lines passing through the corners of the box, passing through the center. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + minor x tick num=4, + ymin=-1.5,ymax=7.5, + xmin=-4.5,xmax=6.5 + ] + + \addplot [firstcurvestyle,domain=-70:70] ({3*tan(x)+1},{2*sec(x)+3}); + \addplot [firstcurvestyle,domain=-70:70] ({3*tan(x)+1},{-2*sec(x)+3}); + + \addplot [secondcurvestyle,domain=-8:8] (x,{2/3*(x-1)+3}); + \addplot [secondcurvestyle,domain=-8:8] (x,{-2/3*(x-1)+3}); + + \draw [thick,dashed,gray] (axis cs: -2,5) -- (axis cs:4,5) -- (axis cs:4,1) -- (axis cs: -2,1) --cycle; + + \filldraw (axis cs: 1,3) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + + +     + +

    + \frac{(y-3)^2}{4}-\frac{(x-1)^2}{9}=1 +

    +     + +     + + + + + + + + A hyperbola centered at the point (1,3) with a horizontal transverse axis + +

    + A hyperbola centered at the point (1,3) with a horizontal transverse axis. + The foci of the hyperbola are not drawn. + The vertices of the hyperbola are at the points (-2,3) and (4,3). + A box is drawn around the center with a width of 6 and a height of 4. + The diagonal asymptotes are drawn as the lines passing through the corners of the box, passing through the center. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=4, + ymin=-1.5,ymax=7.5, + xmin=-4.5,xmax=6.5 + ] + + \addplot [firstcurvestyle,domain=-70:70] ({3*sec(x)+1},{2*tan(x)+3}); + \addplot [firstcurvestyle,domain=-70:70] ({-3*sec(x)+1},{2*tan(x)+3}); + + \addplot [secondcurvestyle,domain=-8:8] (x,{2/3*(x-1)+3}); + \addplot [secondcurvestyle,domain=-8:8] (x,{-2/3*(x-1)+3}); + + \draw [thick,dashed,gray] (axis cs: -2,5) -- (axis cs:4,5) -- (axis cs:4,1) -- (axis cs: -2,1) --cycle; + + \filldraw (axis cs: 1,3) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + + +     + +

    + \frac{(x-1)^2}{9}-\frac{(y-3)^2}{4}=1 +

    +     + +     + +     + + + +

    + Sketch the hyperbola defined by the given equation. + Label the center and foci. +

    +     + + + + +

    + \ds \frac{(x-1)^2}{16}-\frac{(y+2)^2}{9}=1 +

    +     + + +     + + + + +

    + \ds (y-4)^2-\frac{(x+1)^2}{25}=1 +

    +     + + +     + +     + + + +

    + Find the equation of the hyperbola defined by the given information. + Sketch the hyperbola. +

    +     + + + + +

    + Foci: (\pm 3,0); vertices: (\pm 2, 0) +

    +     + +

    + \frac{x^2}{4}-\frac{y^2}{5}=1 +

    +     + +     + + + + + + +

    + Foci: (0,\pm 3); vertices: (0,\pm 2) +

    +     + +

    + \frac{y^2}{4}-\frac{x^2}{5}=1 +

    +     +     +     + + + + +

    + Foci: (-2,3) and (8,3); vertices: + (-1,3) and (7,3) +

    +     + +

    + \frac{(x-3)^2}{16}-\frac{(y-3)^2}{9}=1 +

    +     + +     + + + + +

    + Foci: (3,-2) and (3,8); vertices: + (3,0) and (3,6) +

    +     + +

    + \frac{(y-3)^2}{9}-\frac{(x-3)^2}{16}=1 +

    +     + +     + +     + + + +

    + Write the equation of the hyperbola in standard form. +

    +     + + + + +

    + 3x^2-4y^2=12 +

    +     + +

    + \frac{x^2}{4}-\frac{y^2}{3}=1 +

    +     + +     + + + + +

    + 3x^2-y^2+2y=10 +

    +     + +

    + \frac{x^2}{3}-\frac{(y-1)^2}{9}=1 +

    +     + +     + + + + +

    + x^2-10y^2+40y=30 +

    +     + +

    + (y-2)^2-\frac{x^2}{10}=1 +

    +     + +     + + + + +

    + (4y-x)(4y+x)=4 +

    +     + +

    + 4y^2-\frac{x^2}{4}=1 +

    +     + +     +     + + + + +

    + Consider the ellipse given by \ds \frac{(x-1)^2}{4}+\frac{(y-3)^2}{12}=1. +

    +     + + + +

    + Verify that the foci are located at (1,3\pm 2\sqrt{2}). +

    +     + +

    + c=\sqrt{12-4} = 2\sqrt{2}. +

    +     +     + + + +

    + The points P_1 = (2,6) and P_2 = (1+\sqrt{2},3+\sqrt{6}) \approx (2.414,5.449) lie on the ellipse. + Verify that the sum of distances from each point to the foci is the same. +

    +     + +

    + The sum of distances for each point is 2\sqrt{12}\approx 6.9282. +

    +     +     + +     + + + + +

    + Johannes Kepler discovered that the planets of our solar system have elliptical orbits with the Sun at one focus. + The Earth's elliptical orbit is used as a standard unit of distance; + the distance from the center of Earth's elliptical orbit to one vertex is 1 Astronomical Unit, or A.U. +

    + +

    + The following table gives information about the orbits of three planets. +

    + + + + Planet + Distance fromcenter to vertex + Orbiteccentricity + + + Mercury + 0.387 A.U. + 0.2056 + + + Earth + 1 A.U. + 0.0167 + + + Mars + 1.524 A.U. + 0.0934 + + +     + + + +

    + In an ellipse, + knowing c^2=a^2-b^2 and e=c/a allows us to find b in terms of a and e. + Show b=a\sqrt{1-e^2}. +

    +     + +

    + Solve for c in e=c/a: c=ae. + Thus a^2e^2=a^2-b^2, and b^2=a^2-a^2e^2. + The result follows. +

    +     +     + + + +

    + For each planet, + find equations of their elliptical orbit of the form + \ds\frac{x^2}{a^2}+\frac{y^2}{b^2}=1. (This places the center at (0,0), + but the Sun is in a different location for each planet.) +

    +     + +

    + Mercury: x^2/(0.387)^2 + y^2/(0.3787)^2=1 + + Earth: x^2+y^2/(0.99986)^2 = 1 + + Mars: x^2/(1.524)^2+y^2/(1.517)^2=1 +

    +     +     + + + +

    + Shift the equations so that the Sun lies at the origin. + Plot the three elliptical orbits. +

    +     + +

    + Mercury: (x-0.08)^2/(0.387)^2 + y^2/(0.3787)^2=1 + + Earth: (x-0.0167)^2+y^2/(0.99986)^2 = 1 + + Mars: (x-0.1423)^2/(1.524)^2+y^2/(1.517)^2=1 +

    +     +     + +     + + + + +

    + A loud sound is recorded at three stations that lie on a line as shown in the figure below. + Station A recorded the sound 1 second after Station B, + and Station C recorded the sound 3 seconds after B. + Using the speed of sound as 340m/s, determine the location of the sound's origination. +

    + + + A number line with 3 points drawn. + +

    + A number line with 3 seperate points drawn. + A is the left most point. + B is the point in the middle. The distance between A and B is 1000\mathrm{m}. + C is the left most point. The distance between B and C is 2000 \mathrm{m}. +

    +     + + + \begin{tikzpicture}[scale=1.3] + + \draw [fill=black] (0,0) circle (2.4pt) node [below] {\(A\)} -- node [pos=.5,above] {1000m} (1.5,0) circle (2.4pt) node [below] {\(B\)} -- node [pos=.5,above] {2000m} (4.5,0) circle (2.4pt) node [below] {\(C\)}; + + \end{tikzpicture} + + + +     + +

    + The sound originated from a point approximately 31m to the right of B and 1390m above or below it. (Since the three points are collinear, we cannot distinguish whether the sound originated above/below the line containing the points.) +

    +     + +     +     +     +
    +
    + Parametric Equations + + +

    + We are familiar with sketching shapes, + such as parabolas, by following this basic procedure: +

    + +
    + Plotting a graph y=f(x) + + Diagram outlining the steps involved in plotting a function. + +

    + A diagram containing mostly text, that outlines the process for plotting a function. + The steps illustrated are as follows: +

      +
    1. +

      + Choose x +

      +
      2. +
    2. +

      + Use function f to find y (y=f(x)) +

      +
      3. +
    3. +

      + Plot point (x,y) +

      +
      4. +
    +

    + +

    + This is not really indicative of how a person would sketch a graph by hand + (better techniques for doing so were covered in ), + but it is not far off from how a computer might generate a plot, + or how a student who is first learning about functions might use a table of values to create a plot. +

    +     + + + \begin{tikzpicture} + + \draw (0,0) node (A) {Choose \(x\)} + (5.25,0) node (B) {Use a function \(f\) to find \(y\) \(\big(y=f(x)\big)\)} + (11.25,0) node (C) {Plot point \((x,y)\)}; + + \draw [->] (A) -- (B); + \draw [->] (B) -- (C); + + \end{tikzpicture} + + + +
    + +

    + The rectangular equation + y=f(x) works well for some shapes like a parabola with a vertical axis of symmetry, + but in the previous section we encountered several shapes that could not be sketched in this manner. (To plot an ellipse using the above procedure, + we need to plot the top and bottom + separately.)curverectangular equation +

    + +

    + In this section we introduce a new sketching procedure: +

    + +
    + Plotting a curve (x(t),y(t)) + + Diagram illustrating the process for plotting a parametric curve. + +

    + This is another text-based diagram, which illustrates the method for plotting curves that will be covered in this section. + In this method, the coordinates x and y are both defined as functions of a third variable, t. + Four phrases are arranged in a diamond shape, with text left, right, top, and bottom. +

    + +

    + To the left of the diagram is the text, Choose t. + An arrow points up and to the left from this text to the phrase Use a function f to find x (x=f(t)), + which is located at the top of the diagram. + Another arrow points down and to the left from Choose t to the phrase + Use a function g to find y (y=g(t)), which is located at the bottom of the diagram. +

    + +

    + From the phrases at the top and bottom of the diagram there are two more arrows pointing to the right, + to a final phrase, which states, Plot point (x,y). +

    +     + + + \begin{tikzpicture}[every node/.append style={font=\large}] + + \draw (0,0) node (A) {Choose \(t\)} + (3,1) node (B1) {Use a function \(f\) to find \(x\) \(\big(x=f(t)\big)\)} + (3,-1) node (B2) {Use a function \(g\) to find \(y\) \(\big(y=g(t)\big)\)} + (6.25,0) node (C) {Plot point \((x,y)\)}; + + \draw [->] (A) -- (B1); + \draw [->] (A) -- (B2); + \draw [->] (B1) -- (C); + \draw [->] (B2) -- (C); + + \end{tikzpicture} + + + +
    + +

    + Here, x and y are found separately but then plotted together: + for each value of the input t, we plot the output - the point (x(t),y(t)). +

    +     + + + Plotting parametric curves +

    + The procedure outlined in leads us to a definition. +

    + + + Parametric Equations and Curves + +

    + Let f and g be continuous functions on an interval I. + The set of all points \big(x,y\big) = \big(f(t),g(t)\big) in the Cartesian plane, + as t varies over I, + is the graph of the + parametric equations + x=f(t) and y=g(t), + where t is the parameter. + A curve is a graph along with the parametric equations that define it. + curveparametrically defined + parametric equationsdefinition +

    +     +     + +

    + This is a formal definition of the word curve. + When a curve lies in a plane + (such as the Cartesian plane), + it is often referred to as a plane curve. + Examples will help us understand the concepts introduced in the definition. +

    + + + Plotting parametric functions + +

    + Plot the graph of the parametric equations x=t^2, + y=t+1 for t in [-2,2]. +

    +     + +

    + We plot the graphs of parametric equations in much the same manner as we plotted graphs of functions like y=f(x): + we make a table of values, plot points, + then connect these points with a + reasonable looking curve. + shows such a table of values; + note how we have 3 columns. +

    + +

    + The points (x,y) from the table are plotted in . + The points have been connected with a smooth curve. + Each point has been labeled with its corresponding t-value. + These values, along with the two arrows along the curve, + are used to indicate the orientation of the graph. + This information helps us determine the direction in which the graph is moving. +

    + +
    + A table of values of the parametric equations in along with a sketch of their graph + +
    + + + t + x + y + + + + + + + + -2 + 4 + -1 + + + -1 + 1 + 0 + + + 0 + 0 + 1 + + + 1 + 1 + 2 + + + 2 + 4 + 3 + + +
    +
    + + + + Plot of a parabola with its vertex at the point (0,1), opening to the right, with several marked points. + +

    + The parametric curve obtained from the table of values in is shown. + On a set of coordinate axes, the points corresponding to the values t=-2,-1,0,1,2 are plotted. + These are, respectively, (4,-1), (1,0), (0,1), (1,2), (4,3). +

    + +

    + The curve joining these points is plotted; it appears to be a parabola opening to the right, + with its vertex at the point (0,1). + Arrow heads on the curve indicate a direction of travel corresponding to an increasing value of t. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-2.2,ymax=4.2, + xmin=-.9,xmax=5.2 + ] + + \addplot+ [domain=-2:2,samples=60] ({x^2},{x+1}); + + \draw [->,>=latex] (axis cs:3.9204,2.98) -- (axis cs:3.96,2.99); + \draw [->,>=latex] (axis cs:2.25,-.5) -- (axis cs:2.22,-.49); + + \filldraw (axis cs:4,-1) circle (2.4pt) node [below] { $t=-2$}; + \filldraw (axis cs:1,0) circle (2.4pt) node [above] { $t=-1$}; + \filldraw (axis cs:0,1) circle (2.4pt) node [left] { $t=0$}; + \filldraw (axis cs:1,2) circle (2.4pt) node [above] { $t=1$}; + \filldraw (axis cs:4,3) circle (2.4pt) node [above ] { $t=2$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    +     + +     + +

    + We often use the letter t as the parameter as we often regard t as representing time. + Certainly there are many contexts in which the parameter is not time, + but it can be helpful to think in terms of time as one makes sense of parametric plots and their orientation (for instance, + At time t=0 the position is (1,2) and at time t=3 the position is (5,1).). +

    + + + Plotting parametric functions + +

    + Sketch the graph of the parametric equations x=\cos^2(t), + y=\cos(t) +1 for t in [0,\pi]. +

    +     + +

    + We again start by making a table of values in , + then plot the points (x,y) on the Cartesian plane in . +

    + +
    + A table of values of the parametric equations in along with a sketch of their graph + +
    + + + t + x + y + + + + + + + + 0 + 1 + 2 + + + \pi/4 + 1/2 + 1+\sqrt{2}/2 + + + \pi/2 + 0 + 1 + + + 3\pi/4 + 1/2 + 1-\sqrt{2}/2 + + + \pi + 1 + 0 + + +
    +
    + + + + Plot of a parabola with its vertex at the point (0,1), opening to the right, with several marked points. + +

    + The parametric curve obtained from the table of values in is shown. + The curve is very similar in appearance to the one in ; + again it appears to be a parabola opening to the right, with its vertex at the point (0,1). +

    + +

    + The primary difference is that the marked points now correspond to parameter values + t = 0, \pi/4, \pi/2, 3\pi/4, \pi. The direction of travel is opposite to that in . +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=2.1, + xmin=-.1,xmax=1.6 + ] + + \addplot+ [domain=0:180,samples=70] ({(cos(x))^2},{cos(x)+1}); + + \draw [->,>=latex] (axis cs:0.998271, 0.00086485) -- (axis cs:0.999534, 0.000233112); + \draw [->,>=latex] (axis cs:0.770151, 1.87758) -- (axis cs:0.75311, 1.86782); + + \filldraw (axis cs:1,2) circle (2.4pt) node [right] { $t=0$}; + \filldraw (axis cs:.5,1.71) circle (2.4pt) node [below right] { $t=\pi/4$}; + \filldraw (axis cs:0,1) circle (2.4pt) node [right] { $t=\pi/2$}; + \filldraw (axis cs:.5,.292) circle (2.4pt) node [above right] { $t=3\pi/4$}; + \filldraw (axis cs:1,0) circle (2.4pt) node [above right] { $t=\pi$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    + +

    + It is not difficult to show that the curves in Examples + and + are portions of the same parabola. + While the parabola is the same, + the curves are different. + In , + if we let t vary over all real numbers, + we'd obtain the entire parabola. + In this example, + letting t vary over all real numbers would still produce the same graph; + this portion of the parabola would be traced, + and re-traced, infinitely many times. + The orientation shown in + shows the orientation on [0,\pi], + but this orientation is reversed on [\pi,2\pi]. +

    + +

    + These examples begin to illustrate the powerful nature of parametric equations. + Their graphs are far more diverse than the graphs of functions produced by + y=f(x) functions. +

    +     +     + + + + + +

    + Technology Note: Most graphing utilities can graph functions given in parametric form. + Often the word parametric + is abbreviated as PAR + or PARAM in the options. + The user usually needs to determine the graphing window (i.e, the minimum and maximum x- and y-values), + along with the values of t that are to be plotted. + The user is often prompted to give a t minimum, + a t maximum, + and a t-step or \Delta t. + Graphing utilities effectively plot parametric functions just as we've shown here: + they plots lots of points. + A smaller t-step plots more points, + making for a smoother graph + (but may take longer). + In , the t-step is 1; + in , + the t-step is \pi/4. +

    + +

    + One nice feature of parametric equations is that their graphs are easy to shift. + While this is not too difficult in the + y=f(x) context, + the resulting function can look rather messy. + (Plus, to shift to the right by two, + we replace x with x-2, + which is counter-intuitive.) + The following example demonstrates this. +

    + + + Shifting the graph of parametric functions + +

    + Sketch the graph of the parametric equations x=t^2+t, + y=t^2-t. + Find new parametric equations that shift this graph to the right 3 places and down 2. +

    +     + +

    + The graph of the parametric equations is given in . + It is a parabola with a axis of symmetry along the line y=x; + the vertex is at (0,0). +

    + +

    + In order to shift the graph to the right 3 units, + we need to increase the x-value by 3 for every point. + The straightforward way to accomplish this is simply to add 3 to the function defining x: + x = t^2+t+3. + To shift the graph down by 2 units, + we wish to decrease each y-value by 2, so we subtract 2 from the function defining y: + y = t^2-t-2. + Thus our parametric equations for the shifted graph are x=t^2+t+3, + y=t^2-t-2. + This is graphed in . + Notice how the vertex is now at (3,-2). +

    + +
    + Illustrating how to shift graphs in + +
    + + + + Plot of the "unshifted" parametric curve in this example. + +

    + A plot of the curve x=t^2+1, y=t^2-t is shown. + The shape of the curve is similar to that of a parabola, but slightly distorted. +

    + +

    + The curve begins near the point (2,6), and then descends to a y intercept at (0,2). + It briefly enters the second quadrant before passing through the origin into the fourth quadrant. + A small portion of the curve lies in the fourth quarant before it crosses the x axis at (2,0); + after this point, it continues into the first quadrant, exiting the image near the point (10,5). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={2,4,6,8,10}, + ymin=-2.6,ymax=6.5, + xmin=-.6,xmax=10.5 + ] + + \addplot+ [domain=-3:3,samples=60] ({x^2+x},{x^2-x}); + + \draw [->,>=latex] (axis cs:0.96, 4.16) -- (axis cs:0.9381, 4.1181); + \draw [->,>=latex] (axis cs:6,2) -- (axis cs:6.05,2.03); + + \draw (axis cs: 4,5) node [align=left] { $x=t^2+t$\\ $y=t^2-t$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + The shifted version of the curve in this example. + +

    + The curve in this image is identical to the one beside it in ; + the only difference is that it has been shifted 3 units to the right, and 2 units down. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={2,4,6,8,10}, + ymin=-2.6,ymax=6.5, + xmin=-.6,xmax=10.5 + ] + + \addplot+ [domain=-3:3,samples=60] ({x^2+x+3},{x^2-x-2}); + + \draw [->,>=latex] (axis cs:3.96, 2.16) -- (axis cs:3.9381, 2.1181); + \draw [->,>=latex] (axis cs:9,0) -- (axis cs:9.05,0.03); + + \draw (axis cs: 2.5,5) node [align=left] { $x=t^2+t+3$\\ $y=t^2-t-2$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    +     +     + +

    + Because the x- and y-values of a graph are determined independently, + the graphs of parametric functions often possess features not seen on + y=f(x) type graphs. + The next example demonstrates how such graphs can arrive at the same point more than once. +

    + + + Graphs that cross themselves + +

    + Plot the parametric functions x=t^3-5t^2+3t+11 and + y=t^2-2t+3 and determine the t-values where the graph crosses itself. +

    +     + +

    + Using the methods developed in this section, + we again plot points and graph the parametric equations as shown in . + It appears that the graph crosses itself at the point (2,6), + but we'll need to analytically determine this. +

    + +
    + A graph of the parametric equations from + + + Plot of a more complicated parametric curve that has a self-intersection. + +

    + The curve in this image could describe the path of a fly that turns sharply in a loop, + before heading off in another direction. + It begins in the second quadrant, traveling down and to the right, + and crossing the y axis at approximately (0,6.5). +

    + +

    + The curve continues in this direction until it turns sharply near the point (12,2). + The curve then travels to the left and bends upwards, crossing itself at the point (2,6). + After this point of self-intersection, the curve continues in the first quadrant, + traveling up and to the right. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=4, + minor y tick num=4, + ymin=-.9,ymax=16, + xmin=-6,xmax=16 + ] + + \addplot+ [domain=-2:5,samples=70] ({x^3-5*x^2+3*x+11},{x^2-2*x+3}); + + \draw [->,>=latex] (axis cs:-4.62287, 7.5225) -- (axis cs:-4.4041, 7.4756); + \draw [->,>=latex] (axis cs:7,11) -- (axis cs:7.1107, 11.0601); + + \draw (axis cs: 7,14) node [align=left] { $x=t^3-5t^2+3t+11$\\ $y=t^2-2t+3$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + We are looking for two different values, say, + s and t, where + x(s) = x(t) and y(s) = y(t). + That is, the x-values are the same precisely when the y-values are the same. + This gives us a system of 2 equations with 2 unknowns: + + s^3-5s^2+3s+11 \amp = t^3-5t^2+3t+11 + s^2-2s+3 \amp = t^2-2t+3 + +

    + +

    + Solving this system is not trivial but involves only algebra. + Using the quadratic formula, + one can solve for t in the second equation and find that \ds t = 1\pm \sqrt{s^2-2s+1}. + This can be substituted into the first equation, + revealing that the graph crosses itself at t=-1 and t=3. + We confirm our result by computing + x(-1) = x(3)=2 and y(-1) = y(3) = 6. +

    +     +     + + +     + + + Converting between rectangular and parametric equations +

    + It is sometimes useful to rewrite equations in rectangular form (, y=f(x)) into parametric form, + and vice-versa. + Converting from rectangular to parametric can be very simple: + given y=f(x), the parametric equations x=t, + y=f(t) produce the same graph. + As an example, given y=x^2, + the parametric equations x=t, + y=t^2 produce the familiar parabola. + However, other parametrizations can be used. + The following example demonstrates one possible alternative. +

    + + + Converting from rectangular to parametric + +

    + Consider y=x^2. + Find parametric equations x=f(t), + y=g(t) for the parabola where t=\frac{dy}{dx}. + That is, t=a corresponds to the point on the graph whose tangent line has slope a. +

    +     + +

    + We start by computing \frac{dy}{dx}: \yp = 2x. + Thus we set t=2x. + We can solve for x and find x= t/2. + Knowing that y=x^2, we have y= t^2/4. + Thus parametric equations for the parabola y=x^2 are + + x=t/2 y=t^2/4 + . +

    + +

    + To find the point where the tangent line has a slope of -2, + we set t=-2. + This gives the point (-1, 1). + We can verify that the slope of the line tangent to the curve at this point indeed has a slope of -2. +

    +     + +     + +

    + We sometimes choose the parameter to accurately model physical behavior. +

    + + + Converting from rectangular to parametric + +

    + An object is fired from a height of 0 feet and lands 6 seconds later, 192 feet away. + Assuming ideal projectile motion, the height, in feet, + of the object can be described by h(x) = -x^2/64+3x, + where x is the distance in feet from the initial location. (Thus + h(0) = h(192) = 0 feet.) Find parametric equations x=f(t), + y=g(t) for the path of the projectile where x is the horizontal distance the object has traveled at time t + (in seconds) + and y is the height at time t. +

    +     + +

    + Physics tells us that the horizontal motion of the projectile is linear; + that is, the horizontal speed of the projectile is constant. + Since the object travels 192 in 6, + we deduce that the object is moving horizontally at a rate of 32, + giving the equation x=32t. + As y=-x^2/64+3x, we find y= -16t^2+96t. + We can quickly verify that \yp'=-32 , + the acceleration due to gravity, + and that the projectile reaches its maximum at t=3, + halfway along its path. +

    + +

    + These parametric equations make certain determinations about the object's location easy: + 2 seconds into the flight the object is at the point \big(x(2),y(2)\big) = \big(64,128\big). + That is, it has traveled horizontally 64 + and is at a height of 128, + as shown in . +

    + +
    + Graphing projectile motion in + + + A parabolic arc describing the motion of a projectile fired up and to the right from the origin. + +

    + The curve is a parabola that opens downward from a vertex at (96,144). + It illustrates the path of a projectile that is fired from the origin and travels up and to the right + before reaching its maximum height at the vertex of the parabola. + It then travels back down, returning to the x axis at the point (192,0). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=0,ymax=160, + xmin=0,xmax=210 + ] + + \addplot+ [domain=0:6,samples=40] ({32*x},{-16*x^2+96*x}); + + \draw [->,>=latex] (axis cs:32,80) -- (axis cs:35.2, 86.24); + + \filldraw (axis cs:64, 128) circle (2.4pt) node [above left] { $t=2$}; + + \draw (axis cs: 100,50) node [align=left] { $x=32t$\\ $y=-16t^2+96t$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +     + +

    + It is sometimes necessary to convert given parametric equations into rectangular form. + This can be decidedly more difficult, + as some simple looking parametric equations can have very + complicated rectangular equations. + This conversion is often referred to as + eliminating the parameter, + as we are looking for a relationship between x and y that does not involve the parameter t. +

    + + + Eliminating the parameter + +

    + Find a rectangular equation for the curve described by + + x= \frac{1}{t^2+1} \text{ and } y=\frac{t^2}{t^2+1} + . +

    +     + +

    + There is not a set way to eliminate a parameter. + One method is to solve for t in one equation and then substitute that value in the second. + We use that technique here, then show a second, simpler method. +

    + +

    + Starting with x= 1/(t^2+1), solve for t: + t = \pm\sqrt{1/x-1}. + Substitute this value for t in the equation for y: + + y \amp = \frac{t^2}{t^2 +1} + \amp = \frac{1/x-1}{1/x-1+1} + \amp = \frac{1/x - 1}{1/x} + \amp = \left(\frac1x-1\right)\cdot x + \amp = 1-x + . +

    + +
    + Graphing parametric and rectangular equations for a graph in + + + A graph of two overlapping lines illustrating rectangular and parametric descriptions of a curve. + +

    + Two lines are plotted. The first is the line y=1-x, + which is drawn from the point (-1,2) to the point (2,-1). +

    + +

    + A second, thicker line is drawn over the first line, from (0,1) to (1,0). + At (0,1) there is a hollow (white-filled) dot, + indicating that the second line approaches, but does not reach, this point. + At (1,0) there is a solid dot, indicating the fact that the parametric version of the line + reaches this point, but does not go past it. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.1,ymax=2.1, + xmin=-2.1,xmax=2.1 + ] + + \addplot+ [semithick,domain=-2:2] ({x},{1-x}); + \addplot [firstcurvestyle,very thick,domain=0:1] ({x},{1-x}); + + \draw [fill=white] (axis cs:0,1) circle (2.4pt); + \filldraw [fill=black] (axis cs:1,0) circle (2.4pt) node [align=left,shift={(5pt,30pt)}] { $\displaystyle x=\frac{1}{t^2+1}$\\[2pt] $\displaystyle y=\frac{t^2}{t^2+1}$}; + + \draw (axis cs: -1,1.4) node { $y=1-x$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Thus y=1-x. + One may have recognized this earlier by manipulating the equation for y: + + y = \frac{t^2}{t^2+1} = 1-\frac{1}{t^2+1} = 1-x + . +

    + +

    + This is a shortcut that is very specific to this problem; + sometimes shortcuts exist and are worth looking for. +

    + +

    + We should be careful to limit the domain of the function y=1-x. + The parametric equations limit x to values in (0,1], + thus to produce the same graph we should limit the domain of y=1-x to the same. +

    + +

    + The graphs of these functions is given in . + The portion of the graph defined by the parametric equations is given in a thick line; + the graph defined by y=1-x with unrestricted domain is given in a thin line. +

    +     + +     + + + Eliminating the parameter + +

    + Eliminate the parameter in + x=4\cos(t) +3, y= 2\sin(t) +1 +

    +     + +

    + We should not try to solve for t in this situation as the resulting algebra/trig would be messy. + Rather, we solve for \cos(t) and \sin(t) in each equation, + respectively. + This gives + + \cos(t) = \frac{x-3}{4} \text{ and } \sin(t) =\frac{y-1}{2} + . +

    + +

    + The Pythagorean Theorem gives \cos^2(t) +\sin^2(t) =1, so: + + \cos^2(t) +\sin^2(t) \amp =1 + \left(\frac{x-3}{4}\right)^2 +\left(\frac{y-1}{2}\right)^2 \amp =1 + \frac{(x-3)^2}{16}+\frac{(y-1)^2}{4} \amp =1 + +

    + +
    + Graphing the parametric equations x=4\cos(t) +3, y=2\sin(t) +1 in + + + Graph of the ellipse obtained from the parametric equations in this example. + +

    + The plot shows an ellipse. The major axis runs from (-1,1) to (7,1), + and the minor axis runs from (3,-1) to (3,5). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-3.1,ymax=5.1, + xmin=-1.1,xmax=8.5 + ] + + \addplot+ [domain=0:360,samples=70] ({4*cos(x)+3},{2*sin(x)+1}); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + This final equation should look familiar it is the equation of an ellipse! + + plots the parametric equations, + demonstrating that the graph is indeed of an ellipse with a horizontal major axis and center at (3,1). +

    +     + +     + + + + + +

    + The Pythagorean Theorem can also be used to identify parametric equations for hyperbolas. + We give the parametric equations for ellipses and hyperbolas in the following Key Idea. +

    + + + Parametric Equations of Ellipses and Hyperbolas +

    +

      +
    • +

      + The parametric equations + + x=a\cos(t) +h, y=b\sin(t) +k + + define an ellipse with horizontal axis of length 2a and vertical axis of length 2b, + centered at (h,k). + ellipseparametric equations +

      +
      • + +
    • +

      + The parametric equations + + x= a\tan(t) +h, y=\pm b\sec(t) +k + + define a hyperbola with vertical transverse axis centered at (h,k), and + + x=\pm a\sec(t) +h, y=b\tan(t) + k + + defines a hyperbola with horizontal transverse axis. + Each has asymptotes at y=\pm b/a(x-h)+k. + + + hyperbolaparametric equations + + +

      +
      • +
    +

    +     +     + + + Special Curves +

    + + gives a small gallery of interesting and famous + curves along with parametric equations that produce them. + Interested readers can begin learning more about these curves through internet searches. +

    + +

    + One might note a feature shared by two of these graphs: + sharp corners, or cusps. + We have seen graphs with cusps before and determined that such functions are not differentiable at these points. + This leads us to a definition. +

    + +
    + A gallery of interesting planar curves + + +
    + + Astroid where x=\cos^3(t) and y=\sin^3(t) + + Plot of an astroid: a curve like a 4-pointed star. + +

    + A plot of the parametric curve x=\cos^3(t), y=\sin^3(x), + known as an astroid. + The curve has four cusps, at the points (1,0), (0,1), (-1,0), and (0,-1). + The appearance is not unlike what one would get if one held these four points fixed on the unit circle, + and then bent the rest of the circle so that it curves inward (toward the origin) rather than outward. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.1 + ] + + \addplot+ [domain=0:360,samples=80] ({(cos(x))^3},{(sin(x))^3}); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + Rose Curve where x=\cos(t)\sin(4t) and y=\sin(t)\sin(4t) + + Plot of a rose curve with 8 loops. + +

    + The plot of the curve x=\cos(t)\sin(4t), y=\sin(t)\sin(4t) resembles a flower. + It consists of eight loops, equal in size and shape, each passing through the origin. + The shape of each loop resembles a narrow tear-drop. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-1,1}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.1 + ] + + \addplot+ [domain=0:360,samples=300] ({sin(4*x)*cos(x)},{(sin(4*x)*sin(x)}); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + + +
    + + Hypotrochoid where x=2\cos(t)+5\cos(2t/3) and y=2\sin(t)-5\sin(2t/3) + + Plot of a hypotrochoid, which resembles a smooth, five-pointed star. + +

    + The curve x=2\cos(t)+5\cos(2t/3), y=2\sin(t)-5\sin(2t/3), known as a hypotrochoid, is shown. + The curve is symmetric about the x axis, and resembles a five-pointed star, + with one point on the x axis at (7,0). +

    + +

    + The points of the star are rounded, rather than sharp. + The curve passes through itself several times, + in a manner similar to how a pentagram is drawn. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-7.1,ymax=7.1, + xmin=-7.1,xmax=7.5 + ] + + \addplot+[domain=0:1440,samples=120] ({2*cos(x)+5*cos(2*x/3)},{2*sin(x)-5*sin(2*x/3)}); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + Epicycloid where x=4\cos(t)-\cos(4t) and y=4\sin(t)-\sin(4t) + + Plot of an epicycloid, which looks somewhat like a puffy clover leaf. + +

    + The epicycloid x=4\cos(t)-\cos(4t), y=4\sin(t)-\sin(4t) is shown. + The curve travels once around the origin, and consists of three arcs, + with each pair of arcs meeting at a cusp. +

    + +

    + The effect is not unlike the appearance of a three-leaf clover, + if the leaves were much more puffy, and further from the center. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-7.1,ymax=7.1, + xmin=-7.1,xmax=7.5 + ] + + \addplot+ [domain=0:720,samples=240] ({4*cos(x)-cos(4*x)},{4*sin(x)-sin(4*x)}); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +     +
    + + + Smooth + +

    + A curve C defined by x=f(t), + y=g(t) is smooth + on an interval I if \fp and \gp are continuous on I and not simultaneously 0 + (except possibly at the endpoints of I). + A curve is piecewise smooth + on I if I can be partitioned into subintervals where C is smooth on each subinterval. + curvesmooth + smoothcurve + cusp +

    +     +     + +

    + Consider the astroid, given by x=\cos^3(t), y=\sin^3(t). + Taking derivatives, we have: + + x' = -3\cos^2(t) \sin(t) \text{ and } \yp = 3\sin^2(t) \cos(t) + . +

    + +

    + It is clear that each is 0 when t=0,\, \pi/2,\, \pi,\ldots. + Thus the astroid is not smooth at these points, + corresponding to the cusps seen in the figure. +

    + +

    + We demonstrate this once more. +

    + + + Determine where a curve is not smooth + +

    + Let a curve C be defined by the parametric equations + x=t^3-12t+17 and y=t^2-4t+8. + Determine the points, if any, where it is not smooth. +

    +     + +

    + We begin by taking derivatives. + + x' = 3t^2-12, \yp = 2t-4 + . +

    + +

    + We set each equal to 0: + + x' \amp = 0 \Rightarrow 3t^2-12=0 \Rightarrow t=\pm 2 + \yp \amp = 0 \Rightarrow 2t-4 = 0 \Rightarrow t=2 + +

    + +

    + We see at t=2 both x' and \yp are 0; + thus C is not smooth at t=2, + corresponding to the point (1,4). + The curve is graphed in , + illustrating the cusp at (1,4). +

    + +
    + Graphing the curve in ; note it is not smooth at (1,4) + + + Plot of a parametric curve with a sharp cusp. + +

    + A plot of the curve x=t^3-12t+17, y=t^2-4t+8 is shown. + Despite the fact that x and y are both differentiable functions of t, + there is a sharp cusp at the point corresponding to t=2, + which happens to be the point where the derivatives of x and y with respect to t are simultaneously zero. +

    + +

    + The appearance of the curve is not unlike that of a set of tweezers. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=4, + ymin=-1,ymax=9, + xmin=-1,xmax=11 + ] + + \addplot+ [domain=-2:3.5] ({x^3-12*x+17},{x^2-4*x+8}); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +     + +

    + If a curve is not smooth at t=t_0, + it means that x'(t_0)=\yp(t_0)=0 as defined. + This, in turn, means that rate of change of x + (and y) + is 0; + that is, at that instant, neither x nor y is changing. + If the parametric equations describe the path of some object, + this means the object is at rest at t_0. + An object at rest can make a sharp change in direction, + whereas moving objects tend to change direction in a smooth fashion. +

    + +

    + One should be careful to note that a sharp corner + does not have to occur when a curve is not smooth. + For instance, one can verify that x=t^3, + y=t^6 produce the familiar y=x^2 parabola. + However, in this parametrization, the curve is not smooth. + A particle traveling along the parabola according to the given parametric equations comes to rest at t=0, + though no sharp point is created. +

    + +

    + Our previous experience with cusps taught us that a function was not differentiable at a cusp. + This can lead us to wonder about derivatives in the context of parametric equations and the application of other calculus concepts. + Given a curve defined parametrically, + how do we find the slopes of tangent lines? + Can we determine concavity? + We explore these concepts and more in the next section. +

    +     + + + + Terms and Concepts + + + +

    + When sketching the graph of parametric equations, + the x and y values are found separately, + then plotted together. + +

    +     + +     + + + + +

    + The direction in which a graph is moving + is called the of the graph. +

    +     + + + + + + + + +     + + + + +

    + An equation written as y=f(x) is said to be written in + form. +

    +     + + + + + + + + +     + + + + +

    + Create parametric equations x=f(t), + y=g(t) and sketch their graph. + Explain any interesting features of your graph based on the functions f and g. +

    +     + + + +     +     + + + Problems + + + +

    + Sketch the graph of the given parametric equations by hand, + making a table of points to plot. + Be sure to indicate the orientation of the graph. +

    +     + + + + +

    + x=t^2+t,y=1-t^2,-3\leq t\leq 3 +

    +     + + + + Sketch of the parametric curve in this exercise. + +

    + The sketch for this exercise is a curve that lies mostly in the fourth quadrant. + It resembles part of a slingshot orbit for a comet passing around the sun: + the curve passes through the origin from below, turns quickly in the second quadrant, + crossing the y axis at (0,1), and then the x axis at (2,0), + where it returns to the fourth quadrant. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=4, + minor y tick num=4, + ymin=-9,ymax=2, + xmin=-1,xmax=13 + ] + + \addplot+ [domain=-3:3,samples=40] ({x^2+x},{1-x^2}); + + \draw [>=latex,->,thick] (axis cs: 6,-3) -- (axis cs:6.05,-3.04); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=1,y=5\sin(t),-\pi/2\leq t\leq \pi/2 +

    +     + + + + The vertical line x=1. + +

    + The curve for this exercise is the vertical line x=1. + An arrow on the line indicates that the direction of travel is up. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + minor y tick num=4, + ymin=-5.5,ymax=5.5, + xmin=-.5,xmax=1.6 + ] + + \addplot+ [domain=-5:5] ({1},{x}); + + \draw [>=latex,->,thick] (axis cs: 1,1) -- (axis cs:1,1.1); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=t^2,y=2,-2\leq t\leq 2 +

    +     + + + + The horizontal line y=2, marked with two arrows. + +

    + The horizontal line y=2. + On the line there are two arrows pointing in opposite directions. + These indicate that the direction of travel is to the left when t\lt 0, + and to the right when t\gt 0. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=4, + minor y tick num=4, + ymin=-.5,ymax=2.5, + xmin=-.5,xmax=2.5 + ] + + \addplot+ [domain=-2:2] ({x^2},{2}); + + \draw [>=latex,<->,thick] (axis cs: .9,2) -- (axis cs:1.5,2); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=t^3-t+3,y=t^2+1,-2\leq t\leq 2 +

    +     + + + + The solution curve for this exercise. + +

    + The curve begins to the left of the y axis, and crosses near (0,4). + It passes through the first quadrant to the point (3,2); + it then bends downward and makes a teardrop-shaped loop before passing through (3,2) a second time, + and then continuing up and to the right, through the first quadrant. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=4.5, + xmin=-.5,xmax=5.5 + ] + + \addplot+ [domain=-2:2,samples=50] ({x^3-x+3},{x^2+1}); + + \draw [>=latex,->,thick] (axis cs: 3.897,2.69) -- (axis cs:3.938,2.7161); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + +     + + + +

    + Sketch the graph of the given parametric equations; + using a graphing utility is advisable. + Be sure to indicate the orientation of the graph. +

    +     + + + + +

    + x=t^3-2t^2,y=t^2,-2\leq t \leq 3 +

    +     + + + + Computer-generated sketch of the parametric curve in this exercise. + +

    + A curve resembling a check mark, with a cusp at the origin. + Direction of travel is from the second quadrant toward the cusp, + and then up from the cusp to a y intercept at (0,4), + and then into the first quadrant. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=0.1,ymax=9, + xmin=-11,xmax=11 + ] + + \addplot+ [domain=-2:3,samples=40] ({x^3-2*x^2},{x^2}); + + \draw [>=latex,->,thick] (axis cs:-1.183,1.69)-- (axis cs:-1.18411,1.7161); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=1/t,y=\sin(t),0\lt t \leq 10 +

    +     + + + + Computer-generated sketch of the parametric curve in this exercise. + +

    + A curve resembling a sine wave with a period that gets longer for large values of x. + The direction of travel is that of decreasing x value, with the y value oscillating. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.1,ymax=1.1, + xmin=-0.5,xmax=1.5 + ] + + \addplot+ [domain=0.1:10,samples=101] ({1/x},{sin(deg(x))}); + + \draw [>=latex,->,thick] (axis cs:0.25,-.756802)-- (axis cs:0.249377,-.763301); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=3\cos(t),y=5\sin(t),0\leq t \leq 2\pi +

    +     + + + + Computer-generated sketch of the parametric curve in this exercise. + +

    + The curve is an ellipse, centered at the origin, with counter-clockwise direction of travel. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-5.5,ymax=5.5, + xmin=-5.5,xmax=5.5 + ] + + \addplot+ [domain=0:360,samples=60] ({3*cos(x)},{5*sin(x)}); + + \draw [>=latex,->,thick] (axis cs:2.12,3.5355)-- (axis cs:2.1,3.5707); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=3\cos(t) +2,y=5\sin(t) +3,0\leq t \leq 2\pi +

    +     + + + + Computer-generated sketch of the parametric curve in this exercise. + +

    + An ellipse with center (2,3) and counter-clockwise direction of travel. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=4, + minor y tick num=4, + ymin=-5.5,ymax=8.5, + xmin=-5.5,xmax=7.5 + ] + + \addplot+ [domain=0:360,samples=60] ({3*cos(x)+2},{5*sin(x)+3}); + + \draw [>=latex,->,thick] (axis cs:4.12,6.5355) -- (axis cs:4.1,6.5707); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=\cos(t),y=\cos(2t),0\leq t \leq \pi +

    +     + + + + Computer-generated sketch of the parametric curve in this exercise. + +

    + The curve resembles a parabola, with vertex at (0,-1). + The direction of travel is from right to left. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.1 + ] + + \addplot+ [domain=0:180,samples=60] ({cos(x)},{cos(x*2)}); + + \draw [>=latex,->,thick] (axis cs:0.8845,0.5646) -- (axis cs:0.8798,0.548024); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=\cos(t),y=\sin(2t),0\leq t \leq 2\pi +

    +     + + + + Computer-generated sketch of the parametric curve in this exercise. + +

    + A figure-eight curve, centered at the origin. + The orientation is counter-clockwise in the fourth and first quadrants; + once the curve passes through the origin (a point of self-intersection) + this direction becomes clockwise it the second and third quadrants. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.1] + + \addplot+ [domain=0:360,samples=120] ({cos(x)},{sin(x*2)}); + + \draw [>=latex,->,thick] (axis cs:0.8845,0.82534) -- (axis cs:0.8798,0.8365); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=2\sec(t),y=3\tan(t),-\pi/2\lt t \lt \pi/2 +

    +     + + + + Computer-generated sketch of the parametric curve in this exercise. + +

    + The curve resembles one branch of a hyperbola, opening to the right, with a vertex at (2,0). + The direction of travel is that of increasing y value. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=4, + minor y tick num=4, + ymin=-16.,ymax=16, + xmin=-1,xmax=11 + ] + + \addplot+ [domain=-80:80] ({2*sec(x)},{3*tan(x)}); + + \draw [>=latex,->,thick] (axis cs:5.32,7.395) -- (axis cs:5.4549,7.61254); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=\cosh(t),y=\sinh(t),-2\leq t \leq 2 +

    +     + + + Computer-generated sketch of the parametric curve in this exercise. + +

    + The curve resembles one branch of a hyperbola, opening to the right, with a vertex at (1,0). + The direction of travel is that of increasing y value. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + minor x tick num=4, + minor y tick num=4, + ymin=-4.5, + ymax=4.5, + xmin=-.1, + xmax=4.5 + ] + + \addplot+ [domain=-2:2,samples=60] ({cosh(x)},{sinh(x)}); + \draw[>=latex,->,thick] (axis cs:2.35241, -2.12928)-- (axis cs:2.33123, -2.10586); + \end{axis} + + \end{tikzpicture} + + +     +     + + + + + +

    + x=\cos(t) +\frac14\cos(8t),y=\sin(t) +\frac14\sin(8t),0\leq t \leq 2\pi +

    +     + + + + Computer-generated sketch of the parametric curve in this exercise. + +

    + A flower-shaped curve, with 7 petals. + Each petal is an arc that loops around and intersects itself before continuing to the next arc. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.5,ymax=1.5, + xmin=-1.5,xmax=1.5 + ] + + \addplot+ [domain=0:360,samples=280] ({cos(x)+cos(8*x)/4},{sin(x)+sin(8*x)/4}); + + \draw [>=latex,->,thick] (axis cs:.9571,.7071) -- (axis cs:.9534,.7341); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + x=\cos(t) +\frac14\sin(8t),y=\sin(t) +\frac14\cos(8t),0\leq t \leq 2\pi +

    +     + + + + Computer-generated sketch of the parametric curve in this exercise. + +

    + A curve that spirals around the origin, with several self-intersecting loops. + This curve has 9 arcs and 9 loops. It is similar to the curve in the previous exercise, + except that this time, the arcs bend inward toward the origin, rather than outward. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.5,ymax=1.5, + xmin=-1.5,xmax=1.5 + ] + + \addplot+ [domain=0:360,samples=260] ({cos(x)+sin(8*x)/4},{sin(x)+cos(8*x)/4}); + + \draw [>=latex,->,thick] (axis cs:0.3613,.6771) -- (axis cs:.33215,.68275); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + +     + + + +

    + Four sets of parametric equations are given. + Describe how their graphs are similar and different. + Be sure to discuss orientation and ranges. +

    +     + + + + + +

    + x=t y=t^2, -\infty\lt t\lt \infty +

    +     + +

    + Traces the parabola y=x^2, moves from left to right. +

    +     +     + + +

    + x=\sin(t) y=\sin^2(t), + -\infty\lt t\lt \infty +

    +     + +

    + Traces the parabola y=x^2, + but only from -1\leq x\leq 1; + traces this portion back and forth infinitely. +

    +     +     + + +

    + x=e^t y=e^{2t}, + -\infty\lt t\lt \infty +

    +     + +

    + Traces the parabola y=x^2, + but only for 0\lt x. + Moves left to right. +

    +     +     + + +

    + x=-t y=t^2, -\infty\lt t\lt \infty +

    +     + +

    + Traces the parabola y=x^2, moves from right to left. +

    +     +     + +     + + + + + +

    + x=\cos(t) y=\sin(t), 0\leq t\leq 2\pi +

    +     + +

    + Traces a circle of radius 1 counterclockwise once. +

    +     +     + + +

    + x=\cos(t^2) y=\sin(t^2), 0\leq t\leq 2\pi +

    +     + +

    + Traces a circle of radius 1 counterclockwise over 6 times. +

    +     +     + + +

    + x=\cos(1/t) y=\sin(1/t), 0\lt t\lt 1 +

    +     + +

    + Traces a circle of radius 1 clockwise infinite times. +

    +     +     + + +

    + x=\cos(\cos(t) ) y=\sin(\cos(t) ), 0\leq t\leq 2\pi +

    +     + +

    + Traces an arc of a circle of radius 1, from an angle of -1 radians to 1 radian, twice. +

    +     +     + +     + +     + + + +

    + Eliminate the parameter in the given parametric equations. +

    +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("3x+2y=17",limits=>[[-3,3],[6,11]]); + + +

    + x=2t+5, y=-3t+1 +

    + + + Eliminate the parameter t. + +

    + +

    +     +     +     + + + + +

    + x=\sec(t), y=\tan(t) +

    +     + +

    + x^2-y^2=1 +

    +     + +     + + + + +

    + x=4\sin(t) +1, y=3\cos(t) -2 +

    +     + +

    + \frac{(x-1)^2}{16}+\frac{(y+2)^2}{9}=1 +

    +     + +     + + + + +

    + x=t^2, y=t^3 +

    +     + +

    + y=x^{3/2} +

    +     + +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("y-2x=3",limits=>[[-3,3],[0,6]]); + + +

    + x=\frac{1}{t+1}, y=\frac{3t+5}{t+1} +

    + + + Eliminate the parameter t. + +

    + +

    +     +     +     + + + + +

    + \ds x=e^t, \ds y=e^{3t}-3 +

    +     + +

    + y=x^3-3 +

    +     + +     + + + + +

    + \ds x=\ln(t), \ds y=t^2-1 +

    +     + +

    + y=e^{2x}-1 +

    +     + +     + + + + +

    + \ds x=\cot(t), \ds y=\csc(t) +

    +     + +

    + y^2-x^2=1 +

    +     + +     + + + + +

    + \ds x=\cosh(t), \ds y=\sinh(t) +

    +     + +

    + x^2-y^2=1 +

    +     + +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("x=1-2y^2",limits=>[[-1,2],[-2,2]]); + + +

    + x=\cos(2t), y=\sin(t) +

    + + + Eliminate the parameter t. + +

    + +

    +     +     +     + +     + + + +

    + Eliminate the parameter in the given parametric equations. + Describe the curve defined by the parametric equations based on its rectangular form. +

    +     + + + + +

    + \ds x=at+x_0, \ds y=bt+y_0 +

    +     + +

    + y=\frac{b}{a}(x-x_0)+y_0; + line through (x_0,y_0) with slope b/a. +

    +     + +     + + + + +

    + \ds x=r\cos(t), \ds y=r\sin(t) +

    +     + +

    + x^2+y^2=r^2; + circle centered at (0,0) with radius r. +

    +     + +     + + + + +

    + \ds x=a\cos(t) +h, \ds y=b\sin(t) +k +

    +     + +

    + \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1; + ellipse centered at (h,k) with horizontal axis of length 2a and vertical axis of length 2b. +

    +     + +     + + + + +

    + \ds x=a\sec(t) +h, \ds y=b\tan(t) +k +

    +     + +

    + \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1; + hyperbola centered at (h,k) with horizontal transverse axis and asymptotes with slope b/a. + The parametric equations only give half of the hyperbola. + When a>0, the right half; + when a\lt 0, the left half. +

    +     + +     + +     + + + +

    + Find parametric equations for the given rectangular equation using the parameter \ds t=\frac{dy}{dx}. + Verify that at t=1, + the point on the graph has a tangent line with slope of 1. +

    +     + + + + + Context()->variables->are(t=>'Real'); + $x=Formula("(t+11)/6"); + $y=Formula("(t^2-97)/12"); + Context("Point"); + $point=Point("(2,-8)"); + Context("Numeric"); + $yp=Formula("6x-11"); + $yp1=Real("1"); + + +

    + y=3x^2-11x+2 +

    + + + Give the value for x(t): + + +

    + +

    + + + Give the value for y(t): + + +

    + +

    + + + Give the point on the curve when t=1. + + +

    + +

    + + + Determine the formula for \lz{y}{x}: + + +

    + +

    + + + Find the value of \lz{y}{x} when t=1: + + +

    + +

    +     +     +     + + + + + Context()->variables->are(t=>'Real'); + Context()->variables->set(t=>{limits=>[0.1,10]}); + $x=Formula("ln(t)"); + $y=Formula("t"); + Context("Point"); + $point=Point("(0,1)"); + Context("Numeric"); + $yp=Formula("e^x"); + $yp1=Real("1"); + + +

    + y=e^x +

    + + + Give the value for x(t): + + +

    + +

    + + + Give the value for y(t): + + +

    + +

    + + + Give the point on the curve when t=1. + + +

    + +

    + + + Determine the formula for \lz{y}{x}: + + +

    + +

    + + + Find the value of \lz{y}{x} when t=1: + + +

    + +

    +     +     +     + + + + + Context()->variables->are(t=>'Real'); + Context()->variables->set(t=>{limits=>[-0.9,0.9]}); + $x=Formula("cos^-1(t)"); + $y=Formula("sqrt(1-t^2)"); + Context("Point"); + $point=Point("(0,0)"); + Context("Numeric"); + $yp=Formula("cos(x)"); + $yp1=Real("1"); + + +

    + y=\sin(x) +

    + + + Give the value for x(t): + + +

    + +

    + + + Give the value for y(t): + + +

    + +

    + + + Give the point on the curve when t=1. + + +

    + +

    + + + Determine the formula for \lz{y}{x}: + + +

    + +

    + + + Find the value of \lz{y}{x} when t=1: + + +

    + +

    +     +     +     + + + + +

    + y=\sqrt{x} on [0,\infty) +

    +     + +

    + x=1/(4t^2), y=1/(2t). + At t=1, x=1/4, y=1/2. +

    + +

    + \yp=1/(2\sqrt{x}); when x=1/4, \yp=1. +

    +     + +     + +     + + + +

    + Find the values of t where the graph of the parametric equations crosses itself. +

    +     + + + + + + $crossings=List("-1,1"); + Context("Point"); + $points=List("(3,-2)"); + + +

    + x=t^3-t+3, y=t^2-3 +

    + + + Give the value(s) of t where the graph crosses itself. + Enter your answer as a comma-separated list of numbers. + + +

    + +

    + + + Give the point(s) in the plane where the crossings happen. + If there is more than one, use commas to separate them. + + +

    + +

    +     +     +     + + + + +

    + x=t^3-4t^2+t+7,y=t^2-t +

    +     + +

    + t=-1,\, 2 +

    +     + +     + + + + +

    + x=\cos(t),y=\sin(2t) on [0,2\pi] +

    +     + +

    + t=\pi/2, 3\pi/2 +

    +     + +     + + + + +

    + x=\cos(t) \cos(3t),y=\sin(t) \cos(3t) on [0,\pi] +

    +     + +

    + t=\pi/6, \pi/2, 5\pi/6 +

    +     + +     + +     + + + +

    + Find the value(s) of t where the curve defined by the parametric equations is not smooth. +

    +     + + + + +

    + x=t^3+t^2-t,y=t^2+2t+3 +

    +     + +

    + t=-1 +

    +     + +     + + + + + + $cusps=List("2"); + Context("Point"); + $points=List("(-4,-8)"); + + +

    + x=t^2-4t, y=t^3-2t^2-4t +

    + + + Give the values of t where the curve is not smooth. + (If there is more than one, use commas to separate them.) + + +

    + +

    + + + Give the corresponding points in the plane where the curve is not smooth. + (If there is more than one, use commas to separate them.) + + +

    + +

    +     +     +     + + + + +

    + x=\cos(t),y=2\cos(t) +

    +     + +

    + t=n\pi, where n is an integer +

    +     + +     + + + + + $cusps=List("0"); + Context("Point"); + $points=List("(1,0)"); + + +

    + x=2\cos(t)-\cos(2t), + y=2\sin(t)-\sin(2t) +

    + + + Give the values of t where the curve is not smooth. + (If there is more than one, use commas to separate them.) + + +

    + +

    + + + Give the corresponding points in the plane where the curve is not smooth. + (If there is more than one, use commas to separate them.) + + +

    + +

    +     + +

    + t=n\pi, where n is an even integer +

    +     +     +     + +     + + + +

    + Find parametric equations that describe the given situation. +

    +     + + + + +

    + A projectile is fired from a height of 0, + landing 16 away in 4. +

    +     + +

    + x=4t, y=-16t^2 + 64t +

    +     + +     + + + + +

    + A projectile is fired from a height of 0, + landing 200 away in 4. +

    +     + +

    + x=50t, y=-16t^2 + 64t +

    +     + +     + + + + +

    + A projectile is fired from a height of 0, + landing 200 away in 20. +

    +     + +

    + x=10t, y=-16t^2 + 320t +

    +     + +     + + + + + Context()->variables->are(t=>'Real'); + $x=Formula("2cos(t)"); + $y=Formula("-2sin(t)"); + Context()->flags->set(tolType=>'absolute',tolerance=>0.001); + $multians=MultiAnswer($x,$y)->with( + singleResult=>1, + checker=>sub{ + my ($correct,$student,$self)=@_; + my ($xcor,$ycor)=@{$correct}; + my ($xstu,$ystu)=@{$student}; + my $eliminated=Formula("($xstu/2)^2 + ($ystu/2)^2"); + return 0 unless ($eliminated == Formula("1")); + + #normalize answers + $xstu=$xstu/2; + $ystu=-$ystu/2; + $xcor=$xcor/2; + $ycor=-$ycor/2; + + #shift student answers + my $xstu0=$xstu->eval(t=>0); + my $ystu0=$ystu->eval(t=>0); + my $phi=($xstu0 >= 0)?Compute("arcsin($ystu0)"):Compute("pi-arcsin($ystu0)"); + $xstu=$xstu->substitute(t=>Formula("t-$phi")); + $ystu=$ystu->substitute(t=>Formula("t-$phi")); + + return ($xstu == $xcor and $ystu == $ycor); + } + ); + + +

    + Find parametric equations that describe a circle of radius 2, + centered at the origin, + that is traced clockwise once at constant speed on [0,2\pi]. +

    + + Enter the function for x. + +

    + +

    + + Enter the function for y. + +

    + +

    +     +     +     + + + + + Context()->variables->are(t=>'Real'); + $x=Formula("3cos(2 pi t)+1"); + $y=Formula("3sin(2 pi t)+1"); + Context()->flags->set(tolType=>'absolute',tolerance=>0.001); + $multians=MultiAnswer($x,$y)->with( + singleResult=>1, + checker=>sub{ + my ($correct,$student,$self)=@_; + my ($xcor,$ycor)=@{$correct}; + my ($xstu,$ystu)=@{$student}; + my $eliminated=Formula("(($xstu-1)/3)^2 + (($ystu-1)/3)^2"); + return 0 unless ($eliminated == Formula("1")); + + #normalize answers + $xstu=($xstu-1)/3; + $ystu=($ystu-1)/3; + $xcor=($xcor-1)/3; + $ycor=($ycor-1)/3; + + #shift student answers + my $xstu0=$xstu->eval(t=>0); + my $ystu0=$ystu->eval(t=>0); + my $phi=($xstu0 >= 0)?Compute("arcsin($ystu0)"):Compute("pi-arcsin($ystu0)"); + $xstu=$xstu->substitute(t=>Formula("t-$phi/(2pi) + ")); + $ystu=$ystu->substitute(t=>Formula("t-$phi/(2pi)")); + + return ($xstu == $xcor and $ystu == $ycor); + } + ); + + +

    + Find parametric equations that describe a circle of radius 3, + centered at (1,1), + that is traced once counter-clockwise at constant speed on [0,1]. +

    + + Enter the function for x. + +

    + +

    + + Enter the function for y. + +

    + +

    +     +     +     + + + + + Context()->variables->are(t=>'Real'); + $x=Formula("3cos(2 pi t)+1"); + $y=Formula("3sin(2 pi t)+1"); + $multians = MultiAnswer($x, $y)->with( + singleResult => 1, + checker => sub { + my ( $correct, $student, $self ) = @_; + my ( $xstu, $ystu ) = @{$student}; + my $dx = $xstu->D('t'); + my $dy = $ystu->D('t'); + return 0 if ($dx == Formula("0") or $dy == Formula("0")); + my $eliminated = Formula("($xstu-1)^2 + ($ystu/3 - 1)^2"); + return ($eliminated == Formula("1")); + } + ); + + +

    + Find parametric equations that describe an ellipse centered at (1,3), + with vertical major axis of length 6 and minor axis of length 2. +

    + + Enter the function for x. + +

    + +

    + + Enter the function for y. + +

    + +

    +     +     +     + + + + +

    + An ellipse with foci at (\pm 1,0) and vertices at (\pm 5,0). +

    +     + +

    + x=5\cos(t), y=\sqrt{24}\sin(t); + other answers possible +

    +     + +     + + + + +

    + A hyperbola with foci at (5,-3) and (-1,-3), + and with vertices at (1,-3) and (3,-3). +

    +     + +

    + x=\pm\sec(t) +2, y=\sqrt{8}\tan(t) -3; + other answers possible +

    +     + +     + + + + +

    + A hyperbola with vertices at + (0,\pm 6) and asymptotes y=\pm 3x. +

    +     + +

    + x=2\tan(t), y=\pm 6\sec(t); + other answers possible +

    +     + +     + +     +     +     +
    +
    + Calculus and Parametric Equations + +

    + The previous section defined curves based on parametric equations. + In this section we'll employ the techniques of calculus to study these curves. +

    + +

    + We are still interested in lines tangent to points on a curve. + They describe how the y-values are changing with respect to the x-values, + they are useful in making approximations, + and they indicate instantaneous direction of travel. +

    + + + +     + + + Tangent lines to parametric curves + +

    + The slope of the tangent line is still \frac{dy}{dx}, + and the Chain Rule allows us to calculate this in the context of parametric equations. + If x=f(t) and y=g(t), + the Chain Rule states that + + \frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt} + . +

    + +

    + Solving for \frac{dy}{dx}, we get + + \frac{dy}{dx} = \frac{dy}{dt}\Bigg/\frac{dx}{dt} = \frac{\gp(t)}{\fp(t)} + , + provided that \fp(t)\neq 0. + This is important so we label it a Key Idea. +

    + + + Finding <m>\frac{dy}{dx}</m> with Parametric Equations +

    + Let x=f(t) and y=g(t), + where f and g are differentiable on some open interval I and \fp(t)\neq 0 on I. + Then + parametric equationsfinding \frac{dy}{dx} + derivativeparametric equations + + \frac{dy}{dx} = \frac{\gp(t)}{\fp(t)} + . +

    +     + +

    + We use this to define the tangent line. +

    + + + Tangent and Normal Lines + +

    + Let a curve C be parametrized by x=f(t) and y=g(t), + where f and g are differentiable functions on some interval I containing t=t_0. + The tangent line to C at t=t_0 is the line through + \big(f(t_0),g(t_0)\big) with slope m=\gp(t_0)/\fp(t_0), + provided \fp(t_0)\neq 0. +

    + +

    + The normal line to C at t=t_0 is the line through + \big(f(t_0),g(t_0)\big) with slope m=-\fp(t_0)/\gp(t_0), + provided \gp(t_0)\neq 0. + tangent line + normal line + parametric equationstangent line + parametric equationsnormal line +

    +     +     + + + +

    + The definition leaves two special cases to consider. + When the tangent line is horizontal, + the normal line is undefined by the above definition as \gp(t_0)=0. + Likewise, when the normal line is horizontal, + the tangent line is undefined. + It seems reasonable that these lines be defined + (one can draw a line tangent to the right side + of a circle, for instance), + so we add the following to the above definition. +

    + +

    +

      +
    1. +

      + If the tangent line at t=t_0 has a slope of 0, the normal line to C at t=t_0 is the line x=f(t_0). +

      +
      2. + +
    2. +

      + If the normal line at t=t_0 has a slope of 0, the tangent line to C at t=t_0 is the line x=f(t_0). +

      +
      3. +
    +

    + + + Tangent and Normal Lines to Curves + +

    + Let x=5t^2-6t+4 and y=t^2+6t-1, + and let C be the curve defined by these equations. +

    + +

    +

      +
    1. +

      + Find the equations of the tangent and normal lines to C at t=3. +

      +
      2. + +
    2. +

      + Find where C has vertical and horizontal tangent lines. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + We start by computing \fp(t) = 10t-6 and \gp(t) =2t+6. + Thus + + \frac{dy}{dx} = \frac{2t+6}{10t-6} + . + Make note of something that might seem unusual: + \frac{dy}{dx} is a function of t, not x. + Just as points on the curve are found in terms of t, + so are the slopes of the tangent lines. + The point on C at t=3 is (31,26). + The slope of the tangent line is m=1/2 and the slope of the normal line is m=-2. + Thus, + +

        +
      • +

        + the equation of the tangent line is \ds y=\frac12(x-31)+26, and +

        +
        • + +
      • +

        + the equation of the normal line is \ds y=-2(x-31)+26. +

        +
        • +
      +

      + +

      + This is illustrated in . +

      + +
      + Graphing tangent and normal lines in + + + Plot of a parametric curve and its tangent and normal lines at one point. + +

      + The curve x=5t^2-6t+4, y=t^2+6t-1 is shown. + The shape of the curve is that of a distored parabola, opening to the right from a vertex near (2,3). + (The precise location is \left(\frac{11}{5},\frac{74}{25}\right).) + The curve begins below the x axis, traveling left until it reaches the vertex, + after which it continues up and to the right. +

      + +

      + At the point (31,26), the tangent and normal lines to the curve are shown. + The tangent line has a positive slope, while the normal line has a negative slope, + and as expected, the two lines are perpendicular. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-20,ymax=55, + xmin=-5,xmax=90 + ] + + \addplot+ [domain=-3.5:5,samples=40] ({5*x^2-6*x+4},{x^2+6*x-1}); + + \addplot [secondcurvestyle,solid,domain=-5:90] ({x},{.5*(x-31)+26}); + \addplot [secondcurvestyle,solid,domain=-20:55] ({x},{-2*(x-31)+26}); + + \draw [>=latex,->] (axis cs: 12.14,-5.204) -- (axis cs: 12,-5.16); + + \filldraw (axis cs: 67,-10) circle (2.4pt) + (axis cs: 2.2,2.96) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
      + +
      2. + +
    2. +

      + To find where C has a horizontal tangent line, + we set \frac{dy}{dx}=0 and solve for t. + In this case, + this amounts to setting \gp(t)=0 and solving for t (and making sure that \fp(t)\neq 0). + + \gp(t)=0 \Rightarrow 2t+6=0 \Rightarrow t=-3 + . + The point on C corresponding to t=-3 is (67,-10); + the tangent line at that point is horizontal + (hence with equation y=-10). + To find where C has a vertical tangent line, + we find where it has a horizontal normal line, + and set -\frac{\fp(t)}{\gp(t)}=0. + This amounts to setting \fp(t)=0 and solving for t (and making sure that \gp(t)\neq 0). + + \fp(t)=0 \Rightarrow 10t-6=0 \Rightarrow t=0.6 + . + The point on C corresponding to t=0.6 is (2.2,2.96). + The tangent line at that point is x=2.2. + The points where the tangent lines are vertical and horizontal are indicated on the graph in . +

      +
      3. +
    +

    +     + +     + + + Tangent and Normal Lines to a Circle + +

    +

      +
    1. +

      + Find where the unit circle, + defined by x=\cos(t) and y=\sin(t) on [0,2\pi], + has vertical and horizontal tangent lines. +

      +
      2. + +
    2. +

      + Find the equation of the normal line at t=t_0. +

      +
      3. +
    +

    +

    + + +

    +

      +
    1. +

      + We compute the derivative following : + + \frac{dy}{dx} = \frac{\gp(t)}{\fp(t)} = -\frac{\cos(t) }{\sin(t) } + . + The derivative is 0 when \cos(t) = 0; + that is, when t=\pi/2,\, 3\pi/2. + These are the points (0,1) and (0,-1) on the circle. + + The normal line is horizontal (and hence, the tangent line is vertical) when \sin(t) =0; that is, when t= 0,\,\pi,\,2\pi, corresponding to the points (-1,0) and (0,1) on the circle. These results should make intuitive sense. +

      +
      2. + +
    2. +

      + The slope of the normal line at t=t_0 is \ds m=\frac{\sin(t_0) }{\cos(t_0) } = \tan(t_0). + This normal line goes through the point (\cos(t_0) ,\sin(t_0) ), + giving the line + + y \amp =\frac{\sin(t_0) }{\cos(t_0) }(x-\cos(t_0) ) + \sin(t_0) + \amp = (\tan(t_0) )x + , + as long as \cos(t_0) \neq 0. + It is an important fact to recognize that the normal lines to a circle pass through its center, + as illustrated in . + Stated in another way, + any line that passes through the center of a circle intersects the circle at right angles. +

      + +
      + Illustrating how a circle's normal lines pass through its center + + + Sketch of the unit circle and one normal line, which passes through the circle's center. + +

      + A sketch of the unit circle is shown. + At a point on the circle in the first quadrant + (corresponding to an angle that appears to be slightly more than \pi/3), + a normal line is drawn. +

      + +

      + The normal line passes through the center of the circle, illustrating the conclusion of this example. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.3,xmax=1.3 + ] + + \addplot+ [domain=0:360,samples=70] ({cos(x)},{sin(x)}); + \addplot+ [solid,domain=-1:1] ({x},{2*x}); + + \filldraw (axis cs: .447,.894) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      3. +
    +

    + + +     + + + Tangent lines when <m>\frac{dy}{dx}</m> is not defined + +

    + Find the equation of the tangent line to the astroid x=\cos^3(t), + y=\sin^3(t) at t=0, + shown in . +

    +
    + A graph of an astroid + + + Graph of an astroid. + +

    + Graph of the astroid curve x^{2/3}+y^{2/3}=1. + It has cusps at the vertices (1,0),(0,1),(-1,0) and (0,-1). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-1,1}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.1 + ] + + \addplot+ [domain=0:360,samples=70] ({(cos(x))^3},{(sin(x))^3}); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + We start by finding x'(t) and \yp(t): + + x'(t) = -3\sin(t) \cos^2(t) , \qquad \yp(t) = 3\cos(t) \sin^2(t) + . +

    + +

    + Note that both of these are 0 at t=0; + the curve is not smooth at t=0 forming a cusp on the graph. + Evaluating \frac{dy}{dx} at this point returns the indeterminate form of 0/0. +

    + +

    + We can, however, + examine the slopes of tangent lines near t=0, + and take the limit as t\to 0. + + \lim_{t\to0} \frac{\yp(t)}{x'(t)} \amp =\lim_{t\to0} \frac{3\cos(t) \sin^2(t) }{-3\sin(t) \cos^2(t) } \text{ (We can cancel as \(t\neq 0\).) } + \amp = \lim_{t\to0} -\frac{\sin(t) }{\cos(t) } + \amp = 0 + . +

    + +

    + We have accomplished something significant. + When the derivative \frac{dy}{dx} returns an indeterminate form at t=t_0, + we can define its value by setting it to be + \lim\limits_{t\to t_0}\frac{dy}{dx}, if that limit exists. + This allows us to find slopes of tangent lines at cusps, + which can be very beneficial. +

    + +

    + We found the slope of the tangent line at t=0 to be 0; + therefore the tangent line is y=0, the x-axis. +

    +     + +     + + + + +     + + + Concavity +

    + We continue to analyze curves in the plane by considering their concavity; + that is, we are interested in \frac{d^2y}{dx^2}, + the second derivative of y with respect to x. To find this, + we need to find the derivative of + \frac{dy}{dx} with respect to x; that is, + + \frac{d^2y}{dx^2}=\frac{d}{dx}\left[\frac{dy}{dx}\right] + , + but recall that \frac{dy}{dx} is a function of t, + not x, making this computation not straightforward. + concavity + parametric equationsconcavity +

    + +

    + To make the upcoming notation a bit simpler, + let h(t) = \frac{dy}{dx}. + We want \frac{d}{dx}[h(t)]; + that is, we want \frac{dh}{dx}. + We again appeal to the Chain Rule. + Note: + + \frac{dh}{dt} = \frac{dh}{dx}\cdot\frac{dx}{dt} \Rightarrow \frac{dh}{dx} = \frac{dh}{dt}\Bigg/\frac{dx}{dt} + . +

    + +

    + In words, to find \frac{d^2y}{dx^2}, + we first take the derivative of \frac{dy}{dx} + with respect to t, + then divide by x'(t). + We restate this as a Key Idea. +

    + + + Finding <m>\frac{d^2y}{dx^2}</m> with Parametric Equations +

    + Let x=f(t) and y=g(t) be twice differentiable functions on an open interval I, + where \fp(t)\neq 0 on I. + Then + parametric equationsfinding \frac{d^2y}{dx^2} + + \frac{d^2y}{dx^2} = \frac{d}{dt}\left[\frac{dy}{dx}\right]\Bigg/\frac{dx}{dt} = \frac{d}{dt}\left[\frac{dy}{dx}\right]\Bigg/\fp(t) + . +

    +     + +

    + Examples will help us understand this Key Idea. +

    + + + Concavity of Plane Curves + +

    + Let x=5t^2-6t+4 and + y=t^2+6t-1 as in . + Determine the t-intervals on which the graph is concave up/down. +

    +     + +

    + Concavity is determined by the second derivative of y with respect to x, + \frac{d^2y}{dx^2}, + so we compute that here following . +

    + +

    + In , + we found \ds\frac{dy}{dx} = \frac{2t+6}{10t-6} and \fp(t) = 10t-6. + So: + + \frac{d^2y}{dx^2} \amp = \frac{d}{dt}\left[\frac{2t+6}{10t-6}\right]\Bigg/(10t-6) + \amp = -\frac{72}{(10t-6)^2}\Bigg/(10t-6) + \amp = -\frac{72}{(10t-6)^3} + \amp = -\frac{9}{(5t-3)^3} + +

    + +
    + Graphing the parametric equations in to demonstrate concavity + + + Sketch of a parametric curve illustrating concavity. + +

    + The curve shown is the same distorted parabola from , + but without the tangent and normal lines. +

    + +

    + Below the vertex, which corresponds to t=3/5, + the curve is concave up, and there is a label on the curve indicating that for t\lt 3/5, the curve is concave up. + Above the vertex, the curve is concave down, and there is a label on the curve indicating that for t\gt 3/5, the curve is concave down. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-20,ymax=55, + xmin=-5,xmax=90 + ] + + \addplot+ [domain=-3.5:5,samples=50] ({5*x^2-6*x+4},{x^2+6*x-1}); + + \draw [>=latex,->] (axis cs: 12.14,-5.204) -- (axis cs: 12,-5.16); + + \filldraw (axis cs: 2.2,2.96) circle (2.4pt); + + \draw (axis cs: 31,26) node [above,rotate=26.56] { $t>3/5$; concave down}; + \draw (axis cs: 50.25,-9.75) node [below] { $t<3/5$; concave up}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + The graph of the parametric functions is concave up when + \frac{d^2y}{dx^2} \gt 0 and concave down when \frac{d^2y}{dx^2} \lt 0. + We determine the intervals when the second derivative is greater/less than 0 by first finding when it is 0 or undefined. +

    + +

    + As the numerator of \ds -\frac{9}{(5t-3)^3} is never 0, \frac{d^2y}{dx^2} \neq 0 for all t. + It is undefined when 5t-3=0; + that is, when t= 3/5. + Following the work established in , + we look at values of t greater/less than 3/5 on a number line: +

    + + + A number line for the sign of the second derivative in this example. + +

    + A number line is shown, on which the value 3/5 is marked. + To the left of this point there is text indicating that \frac{d^2y}{dx^2}\gt 0, and that the curve is concave up. + To the right of this point there is text indicating that \frac{d^2y}{dx^2}\lt 0, and that the curve is concave down. +

    +     + + + \begin{tikzpicture} + \begin{axis}[numberline, + xmin=-0.9,xmax=0.9, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize $3/5$}; + },] + \addplot[guideline] coordinates {(0,0) (0,2)}; + \addplot[mark=none] coordinates {(-0.5,1)} node {\parbox{8em}{\centering \small $\frac{d^2y}{dx^2}> 0$\\concave up}}; + \addplot[mark=none] coordinates {(0.5,1)} node {\parbox{8em}{\centering \small $\frac{d^2y}{dx^2}<0$\\concave down}}; + \end{axis} + + \end{tikzpicture} + + + + +

    + Reviewing , + we see that when t=3/5=0.6, + the graph of the parametric equations has a vertical tangent line. + This point is also a point of inflection for the graph, + illustrated in . +

    + + + +     + +     + + + Concavity of Plane Curves + +

    + Find the points of inflection of the graph of the parametric equations x=\sqrt{t}, + y=\sin(t), for 0\leq t\leq 16. +

    +     + +

    + We need to compute \frac{dy}{dx} and \frac{d^2y}{dx^2}. + + \frac{dy}{dx} = \frac{\yp(t)}{x'(t)} = \frac{\cos(t) }{1/(2\sqrt{t})} = 2\sqrt{t}\cos(t) + . + + \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{x'(t)} = \frac{\cos(t) /\sqrt{t}-2\sqrt{t}\sin(t) }{1/(2\sqrt{t})}=2\cos(t) -4t\sin(t) + . +

    + +

    + The points of inflection are found by setting \frac{d^2y}{dx^2}=0. + This is not trivial, + as equations that mix polynomials and trigonometric functions generally do not have nice solutions. +

    + +

    + In we see a plot of the second derivative. + It shows that it has zeros at approximately t=0.5,\,3.5,\,6.5,\,9.5,\,12.5 and 16. + These approximations are not very good, + made only by looking at the graph. + Newton's Method provides more accurate approximations. + Accurate to 2 decimal places, we have: + + t=0.65,\,3.29,\,6.36,\,9.48,\,12.61\,\text{ and } \,15.74 + . +

    + +

    + The corresponding points have been plotted on the graph of the parametric equations in . + Note how most occur near the x-axis, + but not exactly on the axis. +

    + +
    + In (a), a graph of \frac{d^2y}{dx^2}, showing where it is approximately 0. In (b), graph of the parametric equations in along with the points of inflection + +
    + + + + Graph of the second derivative is a sinusoid with increasing amplitude. + +

    + The image shows the graph y = 2\cos(t)-4t\sin(t), which is a graph of \frac{d^2y}{dx^2}. + The curve is sinusoidal, but with an amplitude that increases as t increases. + The graph allows us to estimate the values of t where the second derivative is zero. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=4, + minor y tick num=4, + ymin=-59,ymax=59, + xmin=-.5,xmax=16.9, + xlabel={$t$} + ] + + \addplot+ [domain=0:16,samples=120] ({x},{2*cos(deg(x))-4*x*sin(deg(x))}); + + \draw (axis cs: 5,-40) node { $y=2\cos(t) -4t\sin(t) $}; + + \end{axis} + \end{tikzpicture} + + + + +
    + +
    + + + + Graph of the parametric curve in this example, with points of inflection marked. + +

    + The graph shows a curve that appears to be sinusoidal, but with a frequency that increases with x. + On the graph are several marked points, indicating where the inflection points occur. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={3,1,2,4}, + ymin=-1.1,ymax=1.1, + xmin=-.5,xmax=4.5 + ] + + \addplot+ [domain=0:.5,samples=40] ({sqrt(x)},{sin(deg(x))}); + \addplot [firstcurvestyle,domain=.5:3,samples=50] ({sqrt(x)},{sin(deg(x))}); + \addplot [firstcurvestyle,domain=3:18,samples=120] ({sqrt(x)},{sin(deg(x))}); + + \filldraw (axis cs: 0.808252, 0.607787) circle (2.4pt) + (axis cs: 1.81447, -0.150147) circle (2.4pt) + (axis cs: 2.52223, 0.0783547) circle (2.4pt) + (axis cs: 3.07855, -0.0526833) circle (2.4pt) + (axis cs: 3.55049, 0.0396324) circle (2.4pt) + (axis cs:3.96733, -0.0317508) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    +     +     + + +     + + + Arc Length +

    + We continue our study of the features of the graphs of parametric equations by computing their arc length. +

    + + +

    + Recall in + we found the arc length of the graph of a function, + from x=a to x=b, to be + + L = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx + . +

    + +

    + We can use this equation and convert it to the parametric equation context. + Letting x=f(t) and y=g(t), + we know that \frac{dy}{dx} = \gp(t)/\fp(t). + It will also be useful to calculate the differential of x: + + dx = \fp(t)dt \qquad \Rightarrow \qquad dt = \frac{1}{\fp(t)}\cdot dx + . +

    + +

    + Starting with the arc length formula above, consider: + + L \amp = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx + \amp = \int_a^b \sqrt{1+\frac{\gp(t)^2}{\fp(t)^2}}\, dx. + Factor out the \fp(t)^2: + \amp = \int_a^b \sqrt{\fp(t)^2+\gp(t)^2}\cdot\underbrace{\frac1{\fp(t)}\, dx}_{=dt} + \amp = \int_{t_1}^{t_2} \sqrt{\fp(t)^2+\gp(t)^2}\, dt + . +

    + +

    + Note the new bounds + (no longer x bounds, + but t bounds). + They are found by finding t_1 and t_2 such that a= f(t_1) and b=f(t_2). + This formula is important, so we restate it as a theorem. +

    + + + Arc Length of Parametric Curves + +

    + Let x=f(t) and y=g(t) be parametric equations with \fp and \gp continuous on [t_1,t_2], + on which the graph traces itself only once. + The arc length of the graph, from t=t_1 to t=t_2, + is + parametric equationsarc length + arc length + + L = \int_{t_1}^{t_2} \sqrt{\fp(t)^2+\gp(t)^2}\, dt + . +

    +     +     + + +

    + As before, these integrals are often not easy to compute. + We start with a simple example, + then give another where we approximate the solution. +

    + + + Arc Length of a Circle + +

    + Find the arc length of the circle parametrized by x=3\cos(t), + y=3\sin(t) on [0,3\pi/2]. +

    +     + +

    + By direct application of , we have + + L \amp = \int_0^{3\pi/2} \sqrt{(-3\sin(t) )^2 +(3\cos(t) )^2} \, dt. + Apply the Pythagorean identity. + \amp = \int_0^{3\pi/2} 3 \, dt + \amp = 3t\Big|_0^{3\pi/2} = 9\pi/2 + . +

    + +

    + This should make sense; + we know from geometry that the circumference of a circle with radius 3 is 6\pi; + since we are finding the arc length of 3/4 of a circle, + the arc length is 3/4\cdot 6\pi = 9\pi/2. +

    +     + +     + + + Arc Length of a Parametric Curve + +

    + The graph of the parametric equations x=t(t^2-1), + y=t^2-1 crosses itself as shown in , + forming a teardrop. Find the arc length of the teardrop. +

    +
    + A graph of the parametric equations in , where the arc length of the teardrop is calculated + + + Graph of a "teardrop" curve that intersects itself at the origin. + +

    + The curve given by x=t(t^2-1), y=t^2-1 forms a teardrop shape. + The curve crosses itself at the origin, and the teardrop portion of the graph lies below the x axis. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,-1}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.1 + ] + + \addplot+ [domain=-1.5:1.5,samples=60] ({x^3-x},{x^2-1}); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + We can see by the parametrizations of x and y that when t=\pm 1, + x=0 and y=0. + This means we'll integrate from t=-1 to t=1. + Applying , we have + + L \amp = \int_{-1}^1\sqrt{(3t^2-1)^2+(2t)^2}\, dt + \amp = \int_{-1}^1 \sqrt{9t^4-2t^2+1} \, dt + . +

    + +

    + Unfortunately, + the integrand does not have an antiderivative expressible by elementary functions. + We turn to numerical integration to approximate its value. + Using 4 subintervals, Simpson's Rule approximates the value of the integral as 2.65051. + Using a computer, more subintervals are easy to employ, + and n=20 gives a value of 2.71559. + Increasing n shows that this value is stable and a good approximation of the actual value. +

    +     + +     + + + + +     + + + Surface Area of a Solid of Revolution +

    + Related to the formula for finding arc length is the formula for finding surface area. + We can adapt the formula found in + from + in a similar way as done to produce the formula for arc length done before. +

    + + + Surface Area of a Solid of Revolution + +

    + Consider the graph of the parametric equations x=f(t) and y=g(t), + where \fp and \gp are continuous on an open interval I containing t_1 and t_2 on which the graph does not cross itself. + surface areasolid of revolution + integrationsurface area + parametric equationssurface area +

    + +

    +

      +
    1. +

      + The surface area of the solid formed by revolving the graph about the x-axis is (where + g(t)\geq 0 on [t_1,t_2]): + + \text{ Surface Area } = 2\pi\int_{t_1}^{t_2} g(t)\sqrt{\fp(t)^2+\gp(t)^2}\, dt + . +

      +
      2. + +
    2. +

      + The surface area of the solid formed by revolving the graph about the y-axis is (where + f(t)\geq 0 on [t_1,t_2]): + + \text{ Surface Area } = 2\pi\int_{t_1}^{t_2} f(t)\sqrt{\fp(t)^2+\gp(t)^2}\, dt + . +

      +
      3. +
    +

    +     +     + + + Surface Area of a Solid of Revolution + +

    + Consider the teardrop shape formed by the parametric equations x=t(t^2-1), + y=t^2-1 as seen in . + Find the surface area if this shape is rotated about the x-axis, + as shown in . +

    +
    + Rotating a teardrop shape about the x-axis in + + + + The surface obtained by revolving the teardrop shape from the previous example about the x axis. + +

    + The image, which is interactive, shows the surface in three dimensions that is obtained when the teardrop curve + from is revolved about the x axis. + The result is a donut-like shape, but with a pinch-point in the center rather than a hole. +

    +     + + + + + //ASY file for figparcalc8_3D.asy in Chapter 9 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,3,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.,1.); + pair zbounds=(-1.25,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$x$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$y$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //surface t(t^2-1), t^2-1 revolved around x axis //({(x^2-1)*cos(y)},{x^3-x},{(x^2-1)*sin(y)}); + triple f(pair t) { + return ((t.x^2-1)*cos(t.y),(t.x^3-t.x),(t.x^2-1)*sin(t.y)); + } + surface s=surface(f,(-1,0),(1,2*pi),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //curve on the surfaces + triple g(real t) {return (0,t^3-t,-1+t^2);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + + + +
    +     + +

    + The teardrop shape is formed between t=-1 and t=1. + Using , + we see we need for g(t)\geq 0 on [-1,1], + and this is not the case. + To fix this, we simplify replace g(t) with -g(t), + which flips the whole graph about the x-axis + (and does not change the surface area of the resulting solid). + The surface area is: + + \text{ Area } \,S \amp = 2\pi\int_{-1}^1 (1-t^2)\sqrt{(3t^2-1)^2+(2t)^2}\, dt + \amp = 2\pi\int_{-1}^1 (1-t^2)\sqrt{9t^4-2t^2+1} \, dt + . +

    + +

    + Once again we arrive at an integral that we cannot compute in terms of elementary functions. + Using Simpson's Rule with n=20, + we find the area to be S=9.44. + Using larger values of n shows this is accurate to 2 places after the decimal. +

    +     +     + +

    + After defining a new way of creating curves in the plane, + in this section we have applied calculus techniques to the parametric equation defining these curves to study their properties. + In the next section, + we define another way of forming curves in the plane. + To do so, we create a new coordinate system, + called polar coordinates, + that identifies points in the plane in a manner different than from measuring distances from the y- and x- axes. +

    +     + + + + Terms and Concepts + + + +

    + Given parametric equations x=f(t) and y=g(t), + \lz{y}{x} = \fp(t)/\gp(t), + as long as \gp(t) \neq 0. + +

    +     + +     + + + + +

    + Given parametric equations x=f(t) and y=g(t), + the derivative \frac{dy}{dx} as given in is a function of what variable? +

    +     + + + +

    + x +

    +     +     + + +

    + y +

    +     +     + + +

    + t +

    +     +     +     + +     + + + + +

    + Given parametric equations x=f(t) and y=g(t), + to find \lzn{2}{y}{x}, + one simply computes \lzoo{t}{\lz{y}{x}}. + +

    +     + +     + + + + +

    + If \lz{y}{x}=0 at t=t_0, + then the normal line to the curve at t=t_0 is a vertical line. + +

    +     + +     +     + + + Problems + + + +

    + Parametric equations for a curve are given. +

    + +

    +

      +
    1. +

      + Find \ds\frac{dy}{dx}. +

      +
      2. + +
    2. +

      + Find the equations of the tangent and normal line(s) at the point(s) given. +

      +
      3. + +
    3. +

      + Sketch the graph of the parametric functions along with the found tangent and normal lines. +

      +
      4. +
    +

    +     + + + + +

    + x=t, y=t^2;t=1 +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = 2t +

      +
      2. + +
    2. +

      + Tangent line: y= 2(x-1)+1; normal line: + y = -1/2(x-1)+1 +

      +
      3. +
    +

    +     + +     + + + + +

    + x=\sqrt{t}, y=5t+2;t=4 +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = 10\sqrt{t} +

      +
      2. + +
    2. +

      + Tangent line: y= 20(x-2)+22; normal line: + y = -1/20(x-2)+22 +

      +
      3. +
    +

    +     + +     + + + + +

    + x=t^2-t, y=t^2+t;t=1 +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = \frac{2 t+1}{2 t-1} +

      +
      2. + +
    2. +

      + Tangent line: y= 3x+2; + normal line: y = -1/3x+2 +

      +
      3. +
    +

    +     + +     + + + + +

    + x=t^2-1, y=t^3-t;t=0 and t=1 +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = \frac{3t^2}{2 t} +

      +
      2. + +
    2. +

      + t=0: Tangent line: x=-1; + normal line: y = 0 + + t=1: Tangent line: y=x; normal line: y=-x +

      +
      3. +
    +

    +     + +     + + + + +

    + x=\sec(t), + y=\tan(t) on (-\pi/2,\pi/2);t=\pi/4 +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = \csc(t) +

      +
      2. + +
    2. +

      + t=\pi/4: Tangent line: + y=\sqrt{2}(x-\sqrt{2})+1; normal line: + y = -1/\sqrt{2}(x-\sqrt{2})+1 +

      +
      3. +
    +

    +     + +     + + + + +

    + x=\cos(t), + y=\sin(2t) on [0,2\pi];t=\pi/4 +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = -2\cos(2t)\csc(t) +

      +
      2. + +
    2. +

      + t=\pi/4: Tangent line: + y=1; normal line: x=\sqrt{2}/2 +

      +
      3. +
    +

    +     + +     + + + + +

    + x=\cos(t) \sin(2t), + y=\sin(t) \sin(2t) on [0,2\pi]; t=3\pi/4 +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = \frac{\cos(t) \sin(2t)+2\sin(t) \cos(2t)}{-\sin(t) \sin(2t)+2\cos(t) \cos(2t)} +

      +
      2. + +
    2. +

      + Tangent line: y=x-\sqrt{2}; + normal line: y=-x +

      +
      3. +
    +

    +     + +     + + + + +

    + x=e^{t/10}\cos(t), y=e^{t/10}\sin(t); t=\pi/2 +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = \frac{\sin(t)+10 \cos(t)}{\cos(t)-10 \sin(t)} +

      +
      2. + +
    2. +

      + Tangent line: y=-x/10+e^{\pi/20}; normal line: + y=10x+e^{\pi/20} +

      +
      3. +
    +

    +     + +     + +     + + + +

    + Find the t-values where the curve defined by the given parametric equations has a horizontal tangent line. + Note: these are the same equations as in Exercises. +

    +     + + + + +

    + x=t, y=t^2 +

    +     + +

    + t=0 +

    +     + +     + + + + +

    + x=\sqrt{t}, y=5t+2 +

    +     + +

    + t=0 (though this uses a one-sided limit, + as x(t) is not defined for t\lt 0. +

    +     + +     + + + + + + $hort=List("-1/2"); + Context("Point"); + $points=List("(3/4,-1/4)"); + + +

    + x=t^2-t, + y=t^2+t +

    + + + Give the t values where the curve has a horizontal tangent line. + (If there is more than one, use commas to separate them.) + + +

    + +

    + + + Give the corresponding point(s) in the plane where the tangent line is horizontal. + (If there is more than one, use commas to separate them.) + + +

    + +

    +     +     +     + + + + +

    + x=t^2-1, y=t^3-t +

    +     + +

    + t=\pm 1/\sqrt{3} +

    +     + +     + + + + +

    + x=\sec(t), y=\tan(t) on (-\pi/2,\pi/2) +

    +     + +

    + The graph does not have a horizontal tangent line. +

    +     + +     + + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $hort=List(Formula("pi/4"),Formula("3pi/4"),Formula("5pi/4"),Formula("7pi/4")); + Context("Point"); + $points=Compute("(sqrt(2)/2,1), (-sqrt(2)/2,-1), (-sqrt(2)/2,1), (sqrt(2)/2,-1)"); + + +

    + x=\cos(t), y=\sin(2t), on [0,2\pi) +

    + + + Give the t values where the curve has a horizontal tangent line. + (If there is more than one, use commas to separate them.) + + +

    + +

    + + + Give the corresponding point(s) in the plane where the tangent line is horizontal. + (If there is more than one, use commas to separate them.) + + +

    + +

    +     +     +     + + + + +

    + x=\cos(t) \sin(2t), y=\sin(t) \sin(2t) on [0,2\pi] +

    +     + +

    + The solution is non-trivial; + use identities \sin(2t)=2\sin(t) \cos(t) and + \cos(2t)=\cos^2(t) -\sin^2(t) to rewrite \gp(t) = 2\sin(t) (2\cos^2(t) -\sin^2(t) ). + On [0,2\pi], \sin(t) = 0 when t=0,\pi,2\pi, + and 2\cos^2(t) -\sin^2(t) =0 when t=\tan^{-1}(\sqrt{2}),\,\pi \pm \tan^{-1}(\sqrt{2}),\,2\pi-\tan^{-1}(\sqrt{2}). +

    +     + +     + + + + +

    + x=e^{t/10}\cos(t), y=e^{t/10}\sin(t) +

    +     + +

    + t=\tan^{-1}(-10),\,\tan^{-1}(-10)+\pi +

    +     + +     + +     + + + +

    + Find the point t=t_0 where the graph of the given parametric equations is not smooth, + then find \lim\limits_{t\to t_0}\frac{dy}{dx}. +

    +     + + + + + + $t0 = Real("0"); + $L = Real("0"); + + +

    + x=\frac{1}{t^2+1}, + y=t^3 +

    + + + Find the point t_0 where the curve is not smooth. + + +

    + +

    + + + Find \lim\limits_{t\to t_0}\lz{y}{x}. + +

    + +

    +     +     +     + + + + + + $t0 = Real("2"); + $L = Real("1"); + + +

    + x=-t^3+7t^2-16t+13, + y=t^3-5t^2+8t-2 +

    + + + Find the point t_0 where the curve is not smooth. + + +

    + +

    + + + Find \lim\limits_{t\to t_0}\lz{y}{x}. + +

    + +

    +     +     +     + + + + +

    + x=t^3-3t^2+3t-1,y=t^2-2t+1 +

    +     + +

    + t_0=1; \lim_{t\to 1} \frac{dy}{dx} = \infty. +

    +     + +     + + + + +

    + \ds x=\cos^2(t),y=1-\sin^2(t) +

    +     + +

    + t_0=\ldots,-\pi/2,0,\pi/2,\pi,\ldots; + \lim_{t\to 0} \frac{dy}{dx} = 1. +

    +     + +     + +     + + + +

    + For the given parametric equations for a curve, + find \frac{d^2y}{dx^2}, + then determine the intervals on which the graph of the curve is concave up/down. + Note: these are the same equations as in Exercises. +

    +     + + + + +

    + x=t,y=t^2 +

    +     + +

    + \frac{d^2y}{dx^2}=2; always concave up +

    +     + +     + + + + +

    + x=\sqrt{t},y=5t+2 +

    +     + +

    + \frac{d^2y}{dx^2}=10; always concave up +

    +     + +     + + + + + + + Context()->variables->are(t=>'Real'); + $ypp=Formula("-(4/(2t-1)^3)"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up=List("(-inf,1/2]"); + $down=List("[1/2,inf)"); + + +

    + x=t^2-t, + y=t^2+t +

    + + + Find \lzn{2}{y}{x} as a function of t. + + +

    + +

    + + + Give the interval(s) where the curve is concave up. + (If there is more than one interval, + use commas to separate them.) + +

    + +

    + + + Give the interval(s) where the curve is concave down. + (If there is more than one interval, + use commas to separate them.) + + +

    + +

    +     +     +     + + + + +

    + x=t^2-1,y=t^3-t +

    +     + +

    + \frac{d^2y}{dx^2}=\frac{3t^2+1}{4t^3}; + concave down on (-\infty,0); + concave up on (0,\infty). +

    +     + +     + + + + +

    + x=\sec(t),y=\tan(t) on (-\pi/2,\pi/2) +

    +     + +

    + \frac{d^2y}{dx^2}=-\cot^3(t); + concave up on (-\pi/2,0); + concave down on (0,\pi/2). +

    +     + +     + + + + + + + Context()->variables->are(t=>'Real'); + $ypp=Formula("2 (sin(t)*-2sin(2t) - cos(2t)*cos(t)) / sin^3(t)"); + Context()->flags->set(ignoreEndpointTypes => 1); + $up=Compute("[pi/2,pi], [3pi/2,2pi]"); + $down=Compute("[0,pi/2], [pi,3pi/2]"); + + +

    + x=\cos(t), + y=\sin(2t), on + [0,2\pi) +

    + + + Find \lzn{2}{y}{x} as a function of t. + + +

    + +

    + + + Give the interval(s) where the curve is concave up. + (If there is more than one interval, + use commas to separate them.) + +

    + +

    + + + Give the interval(s) where the curve is concave down. + (If there is more than one interval, + use commas to separate them.) + + +

    + +

    +     +     +     + + + + +

    + \ds x=\cos(t) \sin(2t),y=\sin(t) \sin(2t) on [-\pi/2,\pi/2] +

    +     + +

    + \frac{d^2y}{dx^2} = \frac{4(13+3\cos(4t))}{(\cos(t) +3\cos(3t))^3}, + obtained with a computer algebra system; + concave up on \big(-\tan^{-1}(\sqrt{2}/2),\tan^{-1}(\sqrt{2}/2)\big), + concave down on \big(-\pi/2,-\tan^{-1}(\sqrt{2}/2)\big)\cup\big(\tan^{-1}(\sqrt{2}/2),\pi/2\big) +

    +     + +     + + + + +

    + x=e^{t/10}\cos(t),y=e^{t/10}\sin(t) +

    +     + +

    + \frac{d^2y}{dx^2}=\frac{1010}{e^{t/10}\big(\cos(t) -10\sin(t) \big)^3}; + concavity switches at t=\tan^{-1}(1/10)+n\pi, + where n is an integer. +

    +     + +     + +     + + + +

    + Find the arc length of the graph of the parametric equations on the given interval(s). +

    +     + + + + + Context()->flags->set(reduceConstants=>0); + $length=Formula("6pi"); + + +

    + x=-3\sin(2t), + y=3\cos(2t) on [0,\pi] +

    + +

    + +

    +     +     +     + + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $length1=Formula("sqrt(101)(e^(pi/5)-1)"); + $length2=Formula("sqrt(101)(e^(2pi/5)-e^(pi/5))"); + + +

    + x=e^{t/10}\cos(t), + y=e^{t/10}\sin(t) on + [0,2\pi] and [2\pi,4\pi]. +

    + + + Give the arc length on [0,2\pi]: + + +

    + +

    + + + Give the arc length on [2\pi,4\pi]: + + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $length=Formula("2sqrt(34)"); + + +

    + x=5t+2, + y=1-3t on [-1,1] +

    + +

    + +

    +     +     +     + + + + +

    + x=2t^{3/2},y=3t on [0,1] +

    +     + +

    + L=4\sqrt{2}-2 +

    +     + +     + +     + + + +

    + Numerically approximate the given arc length. +

    +     + + + + +

    + Approximate the arc length of one petal of the rose curve + x=\cos(t) \cos(2t),y=\sin(t) \cos(2t) using Simpson's Rule and n=4. +

    +     + +

    + L\approx 2.4416 (actual value: L=2.42211) +

    +     + +     + + + + +

    + Approximate the arc length of the bow tie curve + x=\cos(t),y=\sin(2t) using Simpson's Rule and n=6. +

    +     + +

    + L\approx 9.73004 (actual value: L=9.42943) +

    +     + +     + + + + +

    + Approximate the arc length of the parabola x=t^2-t,y=t^2+t on [-1,1] using Simpson's Rule and n=4. +

    +     + +

    + L\approx 4.19216 (actual value: L=4.18308) +

    +     + +     + + + + +

    + A common approximate of the circumference of an ellipse given by + x=a\cos(t),y=b\sin(t) is \ds C\approx 2\pi\sqrt{\frac{a^2+b^2}2}. + Use this formula to approximate the circumference of x=5\cos(t), + y=3\sin(t) and compare this to the approximation given by Simpson's Rule and n=6. +

    +     + +

    + Formula: C\approx 25.9062; Simpson's Rule: + C\approx 25.4786 (actual value: C=25.527) +

    +     + +     + +     + + + +

    + A solid of revolution is described. + Find or approximate its surface area as specified. +

    +     + + + + +

    + Find the surface area of the sphere formed by rotating the circle + x=2\cos(t),y=2\sin(t) about: +

    +     + + + +

    + The x-axis. +

    +     +     + + + +

    + The y-axis. +

    +     +     + +

    + The answer is 16\pi for both + (of course), + but the integrals are different. +

    +     + +     + + + + +

    + Find the surface area of the torus + (or donut) + formed by rotating the circle + x=\cos(t) +2,y=\sin(t) about the y-axis. +

    +     + +

    + 8\pi^2. +

    +     + +     + + + + +

    + Approximate the surface area of the solid formed by rotating the + upper right half + of the bow tie curve x=\cos(t),y=\sin(2t) on + [0,\pi/2] about the x-axis, + using Simpson's Rule and n=4. +

    +     + +

    + SA\approx 8.50101 (actual value SA = 8.02851 +

    +     + +     + + + + +

    + Approximate the surface area of the solid formed by rotating the one petal of the rose curve + x=\cos(t) \cos(2t),y=\sin(t) \cos(2t) on + [0,\pi/4] about the x-axis, + using Simpson's Rule and n=4. +

    +     + +

    + SA\approx 1.36751 (actual value SA = 1.36707) +

    +     + +     + +     +     +     +
    +
    + Introduction to Polar Coordinates + + + +

    + We are generally introduced to the idea of graphing curves by relating x-values to y-values through a function f. + That is, we set y=f(x), + and plot lots of point pairs (x,y) to get a good notion of how the curve looks. + This method is useful but has limitations, + not least of which is that curves that + fail the vertical line test + cannot be graphed without using multiple functions. +

    + +

    + The previous two sections introduced and studied a new way of plotting points in the x,y-plane. + Using parametric equations, + x and y values are computed independently and then plotted together. + This method allows us to graph an extraordinary range of curves. + This section introduces yet another way to plot points in the plane: + using polar coordinates. +

    +     + + + Polar Coordinates +

    + Start with a point O in the plane called the pole + (we will always identify this point with the origin). + From the pole, draw a ray, called the initial ray + (we will always draw this ray horizontally, + identifying it with the positive x-axis). + A point P in the plane is determined by the distance r that P is from O, + and the angle \theta formed between the initial ray and the segment \overline{OP} + (measured counter-clockwise). + We record the distance and angle as an ordered pair (r,\theta). + To avoid confusion with rectangular coordinates, + we will denote polar coordinates with the letter P, + as in P(r,\theta). + This is illustrated in + polar coordinates + polarcoordinates + coordinatespolar +

    + +
    + Illustrating polar coordinates + + Illustration of polar coordinates relative to a pole and initial ray. + +

    + A description of polar coordinates. + An initial point O is shown, and from O, + a ray pointing to the right, called the initial ray. +

    + +

    + An additional line segment is drawn from O in a direction up and to the right. + An angle is measured from the initial ray to the line segment, and labeled as \theta. + The length of the line segment is labeled r. +

    + +

    + At the other end of the line segment, a point P is labeled, and assigned the coordinates (r,\theta). +

    +     + + + \begin{tikzpicture}[scale=1.24] + + \draw[thick,->] (0,0) node [below] {\(O\)} -- (3,0) node [below] {initial ray}; + + \filldraw (0,0) circle (2.4pt); + \filldraw [rotate=55] (2,0) circle (2.4pt); + + \draw [thick,rotate=55] (0,0)-- node [rotate=55,pos=.5,above] {\(r\)} (2,0) node [above] {\(P=P(r,\theta)\)}; + \draw [->] (.75,0) arc(0:55:.75); + \draw [rotate=27.5] (1,0) node {\(\theta\)}; + + \end{tikzpicture} + + + + +
    + +

    + Practice will make this process more clear. +

    + + + Plotting Polar Coordinates + +

    + Plot the following polar coordinates: + polar coordinatesplotting points + + A = P(1,\pi/4),\, B=P(1.5,\pi),\, C = P(2,-\pi/3),\, D = P(-1,\pi/4) + +

    +     + +

    + To aid in the drawing, + a polar grid is provided below. + To place the point A, go out 1 unit along the initial ray + (putting you on the inner circle shown on the grid), + then rotate counter-clockwise \pi/4 radians + (or 45^\circ). + Alternately, one can consider the rotation first: + think about the ray from O that forms an angle of \pi/4 with the initial ray, + then move out 1 unit along this ray + (again placing you on the inner circle of the grid). +

    + + + Drawing of a polar grid: a set of concentric circles, and rays from their common center. + +

    + An illustration of the polar grid system. + It is intended as a polar adaptation of the Cartesian grid system. + Instead of x and y coordinate axes, + we see an origin O, and an initial ray, pointing horizontally to the right. + On the initial ray, points 1, 2, and 3 are marked. +

    + +

    + Through each of these points passes a circle centered at O, + representing the polar coordinate values r=1, r=2, and r=3. + Also drawn are several dashed lines through the origin. + Viewed as rays from the origin, each of these lines represents a fixed value for the polar coordinate \theta. +

    +     + + + \begin{tikzpicture}[scale=0.75] + + \draw [dashed,gray] (-3.1,0) -- (0,0); + \draw [thick,->,>=stealth] (0,0) node [below] {\(O\)} -- (3.5,0); + + \filldraw (0,0) circle (2.4pt); + + \foreach \x in {1,2,3} + { + \draw (0,0) circle (\x cm); + \draw (\x,0) node [below right] {\x}; + } + + \foreach \x in {30,45,60,90,120,135,150} + { + \draw [rotate=\x,dashed,gray] (-3.1,0) -- (3.1,0); + } + + \end{tikzpicture} + + + + +

    + To plot B, + go out 1.5 units along the initial ray and rotate \pi radians + (180^\circ). +

    + +

    + To plot C, go out 2 units along the initial ray then rotate + clockwise \pi/3 radians, + as the angle given is negative. +

    + +
    + Plotting polar points in + + + A plot of several points on a polar grid. + +

    + The polar grid from the previous image is shown, with the dashed lines removed. + On the circle of radius 1, points A and D are marked. + The segment OA makes an angle of \pi/4 with the initial ray, + while the point D is on the opposite end of the diameter through A. +

    + +

    + The point B lies between the circles of radius 1 and 2, + to the left of the point O. + The point C is on the circle of radius 2 + and lies below the initial ray, so that the segment OC + makes an angle of -\pi/3 with the initial ray. +

    +     + + + \begin{tikzpicture}[scale=.75,>=latex] + \draw[thick,->] (0,0) node [below] {$O$} -- (3.5,0) ; + \filldraw (0,0) circle (2.4pt); + \foreach \x in {1,2,3} + {\draw (0,0) circle (\x cm); + \draw (\x,0) node [below right] {\x}; + } + \filldraw [rotate=45] (1,0) circle (2.4pt) node [above right] {$A$}; + \filldraw [rotate=180] (1.5,0) circle (2.4pt) node [above] {$B$}; + \filldraw [rotate=-60] (2,0) circle (2.4pt) node [below right] {$C$}; + \filldraw [rotate=45] (-1,0) circle (2.4pt) node [below] {$D$}; + + \end{tikzpicture} + + + + +
    + +

    + To plot D, move along the initial ray -1 + units in other words, + back up 1 unit, then rotate counter-clockwise by \pi/4. + The results are given in . +

    +     +     + + + +

    + Consider the following two points: + A = P(1,\pi) and B = P(-1,0). + To locate A, + go out 1 unit on the initial ray then rotate \pi radians; + to locate B, + go out -1 units on the initial ray and don't rotate. + One should see that A and B are located at the same point in the plane. + We can also consider C=P(1,3\pi), or D = P(1,-\pi); + all four of these points share the same location. +

    + +

    + This ability to identify a point in the plane with multiple polar coordinates is both a + blessing and a curse. + We will see that it is beneficial as we can plot beautiful functions that intersect themselves + (much like we saw with parametric functions). + The unfortunate part of this is that it can be difficult to determine when this happens. + We'll explore this more later in this section. +

    +     + + + Polar to Rectangular Conversion +

    + It is useful to recognize both the rectangular (or, Cartesian) coordinates of a point in the plane and its polar coordinates. + + shows a point P in the plane with rectangular coordinates (x,y) and polar coordinates P(r,\theta). + Using trigonometry, + we can make the identities given in the following Key Idea. +

    + +
    + Converting between rectangular and polar coordinates + + + A triangle illustrating the conversion between rectangular and polar coordinates. + +

    + An right-angled triangle is shown, with the right angle in the bottom-right corner. + The vertex on the left is labeled O, and the angle at that vertex is labeled \theta. + The label on the hypotenuse indicates a length of r, + and the opposite end of the hypotenuse is labeled P. +

    + +

    + The bottom of the triangle (the side adjacent to the angle \theta) is labeled with the length x, + and the altitude of the triangle (the side oppsite the angle \theta) is labeled with the length y. +

    +     + + + \begin{tikzpicture}[scale=0.93] + + \draw [thick] (0,0) -- node [below,pos=.5] {$x$} (4,0) -- node [right,pos=.5] {$y$} (4,2) -- node [above,rotate=26.57,pos=.5] {$r$} (0,0); + + \draw [->,thick] (1.5,0) arc (0:26:1.5); + \draw [rotate=13] (1.8,0) node {$\theta$}; + + \filldraw (0,0) circle (2.4pt) node [below] {$O$} + (4,2) circle (2.4pt) node [above] {$P$}; + + \end{tikzpicture} + + + + +
    + + + Converting Between Rectangular and Polar Coordinates +

    + Given the polar point P(r,\theta), + the rectangular coordinates are determined by + + x=r\cos(\theta) \qquad y=r\sin(\theta) + . +

    + +

    + Given the rectangular coordinates (x,y), + the polar coordinates are determined by + + r^2=x^2+y^2\qquad \tan(\theta) = \frac yx + . +

    +     + + + Converting Between Polar and Rectangular Coordinates + +

    +

      +
    1. +

      + Convert the polar coordinates P(2,2\pi/3) and + P(-1,5\pi/4) to rectangular coordinates. +

      +
      2. + +
    2. +

      + Convert the rectangular coordinates (1,2) and (-1,1) to polar coordinates. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      +

        +
      1. +

        + We start with P(2,2\pi/3). + Using , we have + + x= 2\cos(2\pi/3) = -1\qquad y = 2\sin(2\pi/3) = \sqrt{3} + . + So the rectangular coordinates are (-1,\sqrt{3}) \approx (-1,1.732). +

        +
        2. + +
      2. +

        + The polar point P(-1,5\pi/4) is converted to rectangular with: + + x=-1\cos(5\pi/4) = \sqrt{2}/2\qquad y= -1\sin(5\pi/4) = \sqrt{2}/2 + . + So the rectangular coordinates are (\sqrt{2}/2,\sqrt{2}/2) \approx (0.707,0.707). +

        +
        3. +
      +

      + +

      + These points are plotted in . + The rectangular coordinate system is drawn lightly under the polar coordinate system so that the relationship between the two can be seen. +

      + +
      + Plotting rectangular and polar points in + +
      + + + + Overlaid rectangular and polar grid systems, with marked points. + +

      + A polar grid is shown, using a relatively thick line width. + Also shown is a rectangular grid, with a much thinner line width, so that the polar grid is emphasized. +

      + +

      + The point P(-1,5\pi/4) is marked on the circle of radius 1; + it lies in the first quadrant of the rectangular grid, + since 5\pi/4 is in the third quadrant, but r=-1 is negative. +

      + +

      + The point P(2,2\pi/3) is marked on the circle of radius 2 in the second quadrant. +

      +       + + + \begin{tikzpicture} + + \draw [thin,gray] (-2.5,-2.5) grid (2.5,2.5); + \draw [thick,->] (0,0) node [below] {$O$} -- (2.5,0); + + \foreach \x in {1,2} + { + \draw [very thick] (0,0) circle (\x cm); + } + + \filldraw (0,0) circle (2.4pt); + \filldraw [rotate=120] (2,0) circle (2.4pt) node [ left] { $P(2,\frac{2\pi}{3})$}; + \filldraw [rotate=225] (-1,0) circle (2.4pt) node [shift={(0pt,12pt)}] {$P(-1,\frac{5\pi}{4})$}; + + \end{tikzpicture} + + + + +
      + +
      + + + + Overlaid rectangular and polar grid systems, with marked points. + +

      + Rectangular and polar grids are shown; + this time the rectangular grid is emphasized with a thicker line width. +

      + +

      + The points (0,0), (1,2), and (-1,1) are marked on the rectangular grid. + Also indicated are the angles each point makes with the initial ray. +

      +       + + + \begin{tikzpicture} + + \draw [very thick] (-2.5,-2.5) grid (2.5,2.5); + \draw [thin,gray,->] (0,0) node [below left,black] {$(0,0)$} -- (2.5,0); + + \foreach \x in {1,2} + { + \draw [thin,gray] (0,0) circle (\x cm); + } + + \filldraw (0,0) circle (2.4pt); + + \filldraw [] (1,2) circle (2.4pt) node [above left] { $(1,2)$}; + \filldraw [] (-1,1) circle (2.4pt) node [above right] {$(-1,1)$}; + + \draw [gray,dashed] (0,0) -- ($(0,0)!2.2!(-1,1)$); + \draw [gray,->] (.5,0) arc (0:135:.5); + \draw [rotate=35,gray ] (.65,0) node [] {$\frac{3\pi}{4}$}; + + \draw [gray,dashed] (0,0) -- ($(0,0)!1.2!(1,2)$); + \draw [gray,->] (.5,0) arc (0:135:.5); + \draw [gray,->] (-.5,0) arc (180:135:.5); + \draw [rotate=35,gray ] (.65,0) node [] {$\frac{3\pi}{4}$}; + \draw [rotate=-30,gray ] (-.65,0) node [shift={(-3pt,0pt)}] {$\frac{-\pi}{4}$}; + + \draw [gray,->] (1.5,0) arc (0:63.4:1.5); + \draw [rotate=28,gray ] (1.75,0) node [] {$1.11$}; + + \end{tikzpicture} + + + + +
      +       +
      + +
      2. + +
    2. +

      +

        +
      1. +

        + To convert the rectangular point (1,2) to polar coordinates, + we use the Key Idea to form the following two equations: + + 1^2+2^2 = r^2 \qquad \tan(\theta) = \frac{2}{1} + . + The first equation tells us that r=\sqrt{5}. + Using the inverse tangent function, we find + + \tan(\theta) = 2 \Rightarrow \theta = \tan^{-1}(2) \approx 1.11\approx 63.43^\circ + . + Thus polar coordinates of (1,2) are P(\sqrt{5},1.11). +

        +
        2. + +
      2. +

        + To convert (-1,1) to polar coordinates, we form the equations + + (-1)^2+1^2=r^2 \qquad \tan(\theta) = \frac{1}{-1} + . + Thus r=\sqrt{2}. + We need to be careful in computing \theta: + using the inverse tangent function, we have + + \tan(\theta) = -1 \Rightarrow \theta = \tan^{-1}(-1) = -\pi/4 = -45^\circ + . + This is not the angle we desire. + The range of \tan^{-1}(x) is (-\pi/2,\pi/2); + that is, it returns angles that lie in the + 1st and 4th quadrants. + To find locations in the 2nd and 3rd quadrants, + add \pi to the result of \tan^{-1}(x). + So \pi+(-\pi/4) puts the angle at 3\pi/4. + Thus the polar point is P(\sqrt{2},3\pi/4). + + An alternate method is to use the angle \theta given by arctangent, but change the sign of r. Thus we could also refer to (-1,1) as + P(-\sqrt{2},-\pi/4). +

        +
        3. +
      +

      + +

      + These points are plotted in . + The polar system is drawn lightly under the rectangular grid with rays to demonstrate the angles used. +

      +
      3. +
    +

    +     +     + + + + +     + + + Polar Functions and Polar Graphs +

    + Defining a new coordinate system allows us to create a new kind of function, + a polar function. Rectangular coordinates lent themselves well to creating functions that related x and y, + such as y=x^2. + Polar coordinates allow us to create functions that relate r and \theta. + Normally these functions look like r=f(\theta), + although we can create functions of the form \theta = f(r). + The following examples introduce us to this concept. + polarfunctions + polarfunctions!graphing +

    + + + Introduction to Graphing Polar Functions + +

    + Describe the graphs of the following polar functions. +

    + +

    +

      +
    1. +

      + r = 1.5 +

      +
      2. + +
    2. +

      + \theta = \pi/4 +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + The equation r=1.5 describes all points that are 1.5 units from the pole; + as the angle is not specified, + any \theta is allowable. + All points 1.5 units from the pole describes a circle of radius 1.5. + We can consider the rectangular equivalent of this equation; + using r^2=x^2+y^2, we see that 1.5^2=x^2+y^2, + which we recognize as the equation of a circle centered at (0,0) with radius 1.5. + This is sketched in . +

      +
      2. + +
    2. +

      + The equation \theta = \pi/4 describes all points such that the line through them and the pole make an angle of \pi/4 with the initial ray. + As the radius r is not specified, it can be any value + (even negative). + Thus \theta = \pi/4 describes the line through the pole that makes an angle of + \pi/4 = 45^\circ with the initial ray. + We can again consider the rectangular equivalent of this equation. + Combine \tan(\theta) =y/x and \theta =\pi/4: + + \tan(\pi) /4 = y/x \Rightarrow x\tan(\pi) /4 = y \Rightarrow y = x + . + This graph is also plotted in . +

      + +
      + Plotting standard polar plots + + + Polar plot showing curves of constant radius and constant angle + +

      + A polar grid is shown. On the grid, two curves are drawn. + The curve r=1.5 is a circle centered at the origin of radius 1.5. + The curve \theta = \pi/4 is a line through the origin that makes an angle of \pi/4 with the initial ray. +

      +       + + + \begin{tikzpicture} + + \draw [thick,->] (0,0) node [below] {$O$} -- (2.5,0); + + \foreach \x in {1,2} + { + \draw [thin] (0,0) circle (\x cm); + \draw (\x,0) node [shift={(3pt,-4pt)}] {\x}; + } + + \filldraw (0,0) circle (2.4pt); + + \draw [very thick] (0,0) circle (1.5); + \draw[rotate=90] (1.7,0) node { $r=1.5$}; + + \draw [thick] (-2,-2) -- node [pos=1,left] { $\theta = \frac{\pi}4$} (2,2); + + \end{tikzpicture} + + + + +
      +
      3. +
    +

    +     + +     + +

    + The basic rectangular equations of the form x=h and y=k create vertical and horizontal lines, + respectively; + the basic polar equations r= h and + \theta =\alpha create circles and lines through the pole, + respectively. + With this as a foundation, + we can create more complicated polar functions of the form r=f(\theta). + The input is an angle; the output is a length, + how far in the direction of the angle to go out. +

    + +

    + We sketch these functions much like we sketch rectangular and parametric functions: + we plot lots of points and connect the dots with curves. + We demonstrate this in the following example. +

    + + + Sketching Polar Functions + +

    + Sketch the polar function r=1+\cos(\theta) on [0,2\pi] by plotting points. +

    +     + +

    + A common question when sketching curves by plotting points is + Which points should I plot? + With rectangular equations, + we often choose easy values integers, + then add more if needed. + When plotting polar equations, + start with the common angles multiples of \pi/6 and \pi/4. + + gives a table of just a few values of \theta in [0,\pi]. +

    + +

    + Consider the point P(2,0) determined by the first line of the table. + The angle is 0 radians we do not rotate from the initial ray then we go out 2 units from the pole. + When \theta=\pi/6, + r = 1.866 (actually, it is 1+\sqrt{3}/2); + so rotate by \pi/6 radians and go out 1.866 units. +

    + +

    + The graph shown uses more points, + connected with straight lines. + (The points on the graph that correspond to points in the table are signified with larger dots.) + Such a sketch is likely good enough to give one an idea of what the graph looks like. +

    + +
    + Graphing a polar function in by plotting points + +
    + + + \thetar=1+\cos(\theta) + + + 02 + + + \pi/61.86603 + + + \pi/21 + + + 4\pi/30.5 + + + 7 \pi/41.70711 + + +
    + +
    + + + + A rough sketch of a polar function. + +

    + On a polar grid, the points from the table in are plotted. + Each point is joined to an adjacent point by a line segement to form a rough sketch of the polar curve r=1+\cos(\theta). + The curve appears to be roughly heart-shaped, with a cusp at the origin. +

    +     + + + \begin{tikzpicture}[scale=1.06] + + \draw [dashed,gray] (-2.1,0) -- (0,0); + \draw[thick,->,>=stealth] (0,0) node [below] {$O$} -- (2.5,0); + + \filldraw (0,0) circle (2.4pt); + + \foreach \x in {1,2} + { + \draw (0,0) circle (\x cm); + \draw (\x,0) node [shift={(3pt,-4pt)}] {\x}; + } + + \foreach \x in {30,45,60,90,120,135,150} + { + \draw [rotate=\x,dashed,gray] (-2.3,0) -- (2.3,0); + } + + \draw [thick,firstcolor] (2,0)--(1.616,0.933)--(1.207,1.207) -- (0.75,1.299) -- (0,1.) -- (-0.25,0.433) -- (-0.2071,0.2071) -- (-0.116,0.06699) -- (0,0) -- (-0.116,-0.06699) -- (-0.2071,-0.2071) -- (-0.25,-0.433) -- (0,-1) -- (0.75,-1.299) -- (1.207,-1.207) -- (1.616,-0.933) -- (2,0); + + \filldraw (2.,0) circle (2pt) + (1.616,0.933) circle (2pt) + (1.207,1.207) circle (1pt) + (0.75,1.299) circle (1pt) + (0,1.) circle (2pt) + (-0.25,0.433) circle (1pt) + (-0.2071,0.2071) circle (1pt) + (-0.116,0.06699) circle (1pt) + (0,0) circle (1pt) + (-0.116,-0.06699) circle (1pt) + (-0.2071,-0.2071) circle (1pt) + (-0.25,-0.433) circle (2pt) + (0,-1.) circle (1pt) + (0.75,-1.299) circle (1pt) + (1.207,-1.207) circle (2pt) + (1.616,-0.933) circle (1pt); + + \end{tikzpicture} + + + + +
    +     +
    + +     + +     + + +

    + Technology Note: Plotting functions in this way can be tedious, + just as it was with rectangular functions. + To obtain very accurate graphs, + technology is a great aid. + Most graphing calculators can plot polar functions; + in the menu, + set the plotting mode to something like polar or POL, + depending on one's calculator. + As with plotting parametric functions, + the viewing window no longer determines the x-values that are plotted, + so additional information needs to be provided. + Often with the window settings are the settings for the beginning and ending \theta values + (often called \theta_{\text{ min } } and \theta_{\text{ max } }) + as well as the \theta_{\text{ step } } that is, + how far apart the \theta values are spaced. + The smaller the \theta_{\text{ step } } value, + the more accurate the graph + (which also increases plotting time). + Using technology, we graphed the polar function + r=1+\cos(\theta) from + in . +

    + +
    + Using technology to graph a polar function + + + A computer-generated sketch of a polar function. + +

    + On a polar grid, the curve from is plotted, using technology. + The result is a smooth version of the plot in . + The curve appears to be roughly heart-shaped, with a cusp at the origin. +

    +     + + + \begin{tikzpicture}[scale=1.1] + + \draw [dashed,gray] (-2.1,0) -- (0,0); + \draw [thick,->,>=stealth] (0,0) node [below] {$O$} -- (2.5,0); + + \filldraw (0,0) circle (2.4pt); + + \foreach \x in {1,2} + { + \draw (0,0) circle (\x cm); + \draw (\x,0) node [shift={(3pt,-4pt)}] {\x}; + } + + \foreach \x in {30,45,60,90,120,135,150} + { + \draw [rotate=\x,dashed,gray] (-2.3,0) -- (2.3,0); + } + + \draw [firstcolor,thick,domain=0:360,samples=60] plot ({cos(\x)*(1+cos(\x))},{sin(\x)*(1+cos(\x))}); + + \end{tikzpicture} + + + + +
    + + + Sketching Polar Functions + +

    + Sketch the polar function r=\cos(2\theta) on [0,2\pi] by plotting points. +

    +     + +

    + We start by making a table of + \cos(2\theta) evaluated at common angles \theta, + as shown in . + These points are then plotted in . + This particular graph moves + around quite a bit and one can easily forget which points should be connected to each other. + To help us with this, + we numbered each point in the table and on the graph. +

    + +
    + Table of points for plotting a polar curve in + + Pt.\theta\cos(2\theta) + 101 + 2\pi/60.5 + 3\pi/40 + 4\pi/3-0.5 + 5\pi/2-1 + 62\pi/3-0.5 + 73\pi/40 + 85\pi/60.5 + 9\pi1 + 107\pi/60.5 + 115\pi/40 + 124\pi/3-0.5 + 133\pi/2-1 + 145\pi/3-0.5 + 157\pi/40 + 1611\pi/60.5 + 172\pi1 + +
    + +

    + Using more points + (and the aid of technology) + a smoother plot can be made as shown in . + This plot is an example of a rose curve. +

    + +
    + Polar plots from + +
    + + + + An initial rough sketch of a polar curve obtained by connecting points on the curve. + +

    + On a polar grid, the points from the table in are plotted. + The points are labeled consecutively from 1 to 17, using the values from the table. + These points are then joined, in order, using straight lines, in the matter of a + connect-the-dots drawing. +

    + +

    + The result is four diamond-shaped leaves. + Each diamond has one point at the origin. + The diamonds then point left, right, up, and down. +

    +     + + + \begin{tikzpicture}[scale=1.1] + + \draw [dashed,gray] (-2.1,0) -- (0,0); + \draw [thick,->,>=latex] (0,0) node [below] {} -- (2.5,0); + + \filldraw (0,0) circle (2.4pt); + + \foreach \x/\y in {1/{},2/{}} + { + \draw (0,0) circle (\x cm); + \draw (\x,0) node [shift={(3pt,-4pt)}] { $\y$}; + } + + \foreach \x in {30,45,60,90,120,135,150} + { + \draw [rotate=\x,dashed,gray] (-2.3,0) -- (2.3,0); + } + + \draw [thick,firstcolor] plot coordinates {(2.,0)(0.866,0.5)(0,0)(-0.5,-0.866)(0,-2.)(0.5,-0.866)(0,0)(-0.866,0.5)(-2.,0)(-0.866,-0.5)(0,0)(0.5,0.866)(0,2.)(-0.5,0.866)(0,0)(0.866,-0.5)(2,0)}; + + \foreach \x/\y/\z/\w in { + 2./0/1/{above right},0.866/0.5/2/above,0/0/3/above,-0.5/-0.866/4/above, + 0/-2./5/above, 0.5/-0.866/6/above,0/0/7/left,-0.866/0.5/8/above,-2./0/9/above left, + -0.866/-0.5/10/above,0/0/11/right,0.5/0.866/12/above,0/2./13/above,-0.5/0.866/14/above, + 0/0/15/below,0.866/-0.5/16/above,2/0/17/{below right}} + { + \filldraw (\x,\y) circle (2.4pt) node [\w] { \z}; + } + + \end{tikzpicture} + + + + +
    + +
    + + + + A smoothed version of the plot for this example. + +

    + The plot in is replaced by a smoothed version. + The diamonds are replaced by rounded curves, each resembling a leaf or flower petal. + Each loop (or leaf) has a cusp at the origin; however, the curve as a whole is traced smoothly. +

    + +

    + Beginning at the rectangular point (1,0) (on the positive x axis, when \theta=0), + the curve bends up and to the left, before turning downward, and eventually reaching the origin. + The curve then continues down and to the left into the third quadrant, + until it bends to the right, reaching the y axis at the rectangular point (0,-1). + It then loops back up through the fourth quadrant to the origin; + it then continues into the second quadrant, intersecting the x axis at (-1,0), + and passing through the third quadrant to form the left-hand loop as it returns to the origin. +

    + +

    + From there, we get the upper loop in a similar fashion, in the first quadrant and then the second, + and finally, the curve passes through the origin one more time into the fourth quadrant, + where it completes bottom of the right-hand loop. +

    +     + + + \begin{tikzpicture}[scale=1.1] + + \draw [dashed,gray] (-2.1,0) -- (0,0); + \draw [thick,->,>=latex] (0,0) node [below] {$O$} -- (2.5,0); + + \filldraw (0,0) circle (2.4pt); + + \foreach \x/\y in {1/{},2/1} + { + \draw (0,0) circle (\x cm); + \draw (\x,0) node [shift={(3pt,-4pt)}] { $\y$}; + } + + \foreach \x in {30,45,60,90,120,135,150} + { + \draw [rotate=\x,dashed,gray] (-2.3,0) -- (2.3,0); + } + + \draw [thick,firstcolor,domain=0:360,samples=130] plot ({2*cos(\x)*cos(2*\x)},{2*sin(\x)*cos(2*\x)}); + + \foreach \x/\y in {2./0,0.866/0.5,0/0,-0.5/-0.866,0/-2.,0.5/-0.866,0/0, + -0.866/0.5,-2./0,-0.866/-0.5,0/0,0.5/0.866,0/2.,-0.5/0.866,0/0,0.866/-0.5} + { + \filldraw (\x,\y) circle (2.4pt); + } + + \end{tikzpicture} + + + + +
    +     + +
    +     + +     + +

    + It is sometimes desirable to refer to a graph via a polar equation, + and other times by a rectangular equation. + Therefore it is necessary to be able to convert between polar and rectangular functions, + which we practice in the following example. + We will make frequent use of the identities found in . +

    + + + Converting between rectangular and polar equations + + +

    + Convert from rectangular to polar. +

      +
    1. +

      + y=x^2 +

      +
      2. + +
    2. +

      + xy = 1 +

      +
      3. +
    +

    + +

    + Convert from polar to rectangular. +

      +
    1. +

      + \ds r=\frac{2}{\sin(\theta) -\cos(\theta) } +

      +
      2. + +
    2. +

      + r=2\cos(\theta) +

      +
      3. +
    +

    +     +     + +

    +

      +
    1. +

      + Replace y with r\sin(\theta) and replace x with r\cos(\theta), giving: + + y \amp =x^2 + r\sin(\theta) \amp = r^2\cos^2(\theta) + \frac{\sin(\theta) }{\cos^2(\theta) } \amp = r + + We have found that r=\sin(\theta) /\cos^2(\theta) = \tan(\theta) \sec(\theta). + The domain of this polar function is (-\pi/2,\pi/2); + plot a few points to see how the familiar parabola is traced out by the polar equation. +

      +
      2. + +
    2. +

      + We again replace x and y using the standard identities and work to solve for r: + + xy \amp = 1 + r\cos(\theta) \cdot r\sin(\theta) \amp = 1 + r^2 \amp = \frac{1}{\cos(\theta) \sin(\theta) } + r \amp = \frac{1}{\sqrt{\cos(\theta) \sin(\theta) }} + + This function is valid only when the product of \cos(\theta) \sin(\theta) is positive. + This occurs in the first and third quadrants, + meaning the domain of this polar function is (0,\pi/2) \cup (\pi,3\pi/2). + + We can rewrite the original rectangular equation xy=1 as y=1/x. This is graphed in ; note how it only exists in the first and third quadrants. +

      + +
      + Graphing xy=1 from + + + Graph of the hyperbola y=1/x. + +

      + The graph is that of the standard hyperbola y=1/x. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + minor x tick num=4, + minor y tick num=4, + ymin=-5.3,ymax=5.3, + xmin=-5.3,xmax=5.3 + ] + + \addplot [firstcurvestyle,domain=-5:-.1,samples=60] ({x},{1/x}); + \addplot [firstcurvestyle,domain=.1:5,samples=60] ({x},{1/x}); + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      3. + +
    3. +

      + There is no set way to convert from polar to rectangular; + in general, we look to form the products + r\cos(\theta) and r\sin(\theta), + and then replace these with x and y, respectively. + We start in this problem by multiplying both sides by \sin(\theta) -\cos(\theta): + + r \amp = \frac{2}{\sin(\theta) -\cos(\theta) } + r(\sin(\theta) -\cos(\theta) ) \amp = 2 + r\sin(\theta) -r\cos(\theta) \amp = 2. \qquad \text{ Now replace with \(y\) and \(x\): } + y-x \amp = 2 + y \amp = x+2 + . + The original polar equation, + r=2/(\sin(\theta) -\cos(\theta) ) does not easily reveal that its graph is simply a line. + However, our conversion shows that it is. + The upcoming gallery of polar curves gives the general equations of lines in polar form. +

      +
      4. + +
    4. +

      + By multiplying both sides by r, + we obtain both an r^2 term and an r\cos(\theta) term, + which we replace with x^2+y^2 and x, respectively. + + r \amp =2\cos(\theta) + r^2 \amp = 2r\cos(\theta) + x^2+y^2 \amp = 2x. + We recognize this as a circle; by completing the square we can find its radius and center. + x^2-2x+y^2 \amp = 0 + (x-1)^2 + y^2 \amp =1 + . + The circle is centered at (1,0) and has radius 1. + The upcoming gallery of polar curves gives the equations of + some circles in polar form; + circles with arbitrary centers have a complicated polar equation that we do not consider here. +

      +
      5. +
    +

    +     + +     + +

    + Some curves have very simple polar equations but rather complicated rectangular ones. + For instance, the equation + r=1+\cos(\theta) describes a cardioid + (an important shape for modeling the sensitivity of microphones, among other things; + one is graphed in the gallery in the Limaçon section). + It's rectangular form is not nearly as simple; + it is the implicit equation + x^4+y^4+2x^2y^2-2xy^2-2x^3-y^2=0. + The conversion is not hard, but takes several steps, + and is left as a problem in the Exercise section. +

    + + + Gallery of Polar Curves + +

    + There are a number of basic and + classic polar curves, + famous for their beauty and/or applicability to the sciences. + polarfunction!gallery of graphs + This section ends with a small gallery of some of these graphs. + We encourage the reader to understand how these graphs are formed, + and to investigate with technology other types of polar functions. +

    + +
    + Lines in polar coordinates + +
    + Through the origin: \theta = \alpha + + + A line through the origin with positive slope. + +

    + A line through the origin is shown on a set of rectangular axes. + Also shown is an angle \alpha, measured from the positive x axis to the line. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,rotate=50] (-2.1,0) -- (2.1,0); + + \draw [->] (.5,0) arc (0:50:.5); + \draw [rotate=25] (.7,0) node { $\alpha$}; + + \end{tikzpicture} + + + + +
    +
    + Horizontal line: r=a\csc(\theta) + + + A horizontal line. + +

    + The graph is that of a horizontal line, plotted on a set of rectangular axes. + The graph is labeled with the value a, which is the distance from the x axis to the line. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor] (-2,.6) -- (2,.6); + + \draw (-.2,.3) node { $a\left\{\rule[-.23cm]{0pt}{.23cm}\right.$}; + + \end{tikzpicture} + + + + +
    + +
    + Vertical line: r=a\sec(\theta) + + + A vertical line. + +

    + A vertical line is plotted against a set of rectangular axes. + The distance from the line to the y axis is labeled a. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor] (.6,-2) -- (.6,2); + + \draw (.3,.2) node { $\overbrace{\rule{.55cm}{0pt}}$}; + \draw (.3,.4) node { $a$}; + + \end{tikzpicture} + + + + +
    +
    + Not through origin: \ds r=\frac{b}{\sin(\theta) -m\cos(\theta)} + + + A line with positive slope m, and y intercept b. + +

    + A line is plotted against a set of rectangular axes. + The line has a positive slope m, which is indicated next to the line. + Also shown is a distance b from the origin to the y intercept of the line, + indicating that this particular line does not pass through the origin. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,shift={(0,.6)},rotate=50] (-2.1,0) -- (1.7,0) node [pos=.82,rotate=50,black,shift={(0,-5pt)}] { slope $=m$}; + + \draw (.2,.3) node { $\left.\rule[-.23cm]{0pt}{.23cm}\right\}b$}; + + \end{tikzpicture} + + + + +
    +     +
    + +
    + Circles and Spirals + +
    + Centered on x-axis: r=a\cos(\theta) + + + A circle centered on the positive x axis and passing through the origin. + +

    + A circle is shown, with its center on the positive x axis. + The circle passes through the origin, and the diameter of the circle is labeled as a. + This suggests that the circle has center (a/2,0) and radius a/2. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor] (.9,0) circle (.9); + + \draw (.9,.1) node { $\overbrace{\rule{1.7cm}{0cm}}$}; + \draw (.9,.3) node { $a$}; + + \end{tikzpicture} + + + + +
    +
    + Centered on y-axis: r=a\sin(\theta) + + + A circle centered on the positive y axis and passing through the origin. + +

    + A circle is shown, with its center on the positive y axis. + The circle passes through the origin, and the diameter of the circle is labeled as a. + This suggests that the circle has center (0,a/2) and radius a/2. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor] (0,.9) circle (.9); + + \draw (-.2,.9) node { $a\left\{\rule[-.8cm]{0cm}{0.8cm}\right.$}; + + \end{tikzpicture} + + + + +
    +
    + Centered on origin: r=a + + + A circle centered at the origin. + +

    + On a set of rectangular axes, a circle centered at the origin is plotted. + The radius a of the circle is labeled. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor] (0,0) circle (.9); + + \draw (.45,.1) node { $\overbrace{\rule{.8cm}{0cm}}$}; + \draw (.45,.3) node { $a$}; + + \end{tikzpicture} + + + + +
    +
    + Archimedean spiral: r=\theta + + + A counter-clockwise spiral that begins at the origin. + +

    + The curve begins at the origin and spirals outward, moving in a counter-clockwise direction. + The distance from the curve to the origin increases with the angle, + and three full revolutions of the spiral are shown. +

    + +

    + The appearance of the curve is not unlike that of a snail's shell. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:18.85,samples=200] plot ({cos(deg(\x))*(\x/9.5)},{sin(deg(\x))*(\x/9.5)}); + + \end{tikzpicture} + + + + +
    +     +
    + +
    + Limaçons + +
    + With inner loop: \ds \frac ab \lt 1 + + + A limaçon curve with a loop at the origin, and symmetric about the x axis. + +

    + The first of several curves in the limaçcon family. + The curve intersects the x axis three times: at the origin, which it passes through twice, + and at two other points on the positive x axis. +

    + +

    + The curve consists of two loops that are joined at the origin. + The larger, outer loop passes through the x intercept with the larger value of x. + It is somewhat heart-shaped, with a cusp at the origin. +

    + +

    + The smaller curve lies inside the larger curve, and is shaped like a teardrop. + The two curves join together in such a way that, although each individually has a cusp, + the curve as a whole is traced out smoothly, with the two cusps occurring at the origin, + where the curve intersects itself. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:360,samples=90] plot ({cos(\x)*(.6+1.2*cos(\x))},{sin(\x)*(.6+1.2*cos(\x))}); + + \end{tikzpicture} + + + + +
    +
    + Cardioid: \ds \frac ab=1 + + + A heart-shaped curve in the limaçon family, known as a cardioid. + +

    + This is also a limaçon curve, and it is again symmetric about the x axis. + Its appearance is similar to the previous curve, but with the inner loop removed. +

    + +

    + The resulting cusp at the origin gives this curve a heart-like shape; + as a result, the curve is also known as a cardioid. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:360,samples=90] plot ({cos(\x)*(.9+.9*cos(\x))},{sin(\x)*(.9+.9*cos(\x))}); + + \end{tikzpicture} + + + + +
    +
    + Dimpled: \ds 1\lt \frac ab \lt 2 + + + A dimpled limaçon curve. + +

    + A third curve in the limaçon family, also symmetric about the x axis. + This curve does not pass through the origin. + It has two x intercepts, one positive, and one negative. + The positive x intercept has a larger absolute value (it is further from the origin). +

    + +

    + To the right of the y axis, the curve appears to be almost oval-like in shape. + But as it crosses to the left of the y axis, it begins to bend more sharply toward the second x intercept. + On this side, the curve bends inward, forming a dimple, or dent. + The dent is smooth, however, and is not a cusp. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:360,samples=90] plot ({cos(\x)*(1+.8*cos(\x))},{sin(\x)*(1+.8*cos(\x))}); + + \end{tikzpicture} + + + + +
    +
    + Convex: \ds \frac ab \gt 2 + + + A convex limaçon curve. + +

    + A fourth limaçon curve. It is also symmetric about the x axis. + Like the dimpled limaçon, it has two x intercepts and does not pass through the origin. +

    + +

    + Unlike the dimpled limaçon, to the left of the y axis, the curve flattens out, + but does not bend inward. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:360,samples=80] plot ({cos(\x)*(1.3+.6*cos(\x))},{sin(\x)*(1.3+.6*cos(\x))}); + + \end{tikzpicture} + + + + +
    +     +
    + +

    + Symmetric about x-axis: r=a\pm b\cos(\theta) +

    +

    + Symmetric about y-axis: r=a\pm b\sin(\theta); a,b \gt 0 +

    + +
    + Rose curves + +
    + r=a\cos(2\theta) + + + A rose curve with four leaves along the coordinate axes. + +

    + The curve r=\cos(2\theta) is four-leaf rose curve; it appears to be the same as the one in . + Two of the leaves lie along the x axis, and two leaves lie along the y axis. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:360,samples=180] plot ({cos(\x)*(1.9*cos(2*\x))},{sin(\x)*(1.9*cos(2*\x))}); + + \end{tikzpicture} + + + + +
    +
    + r=a\sin(2\theta) + + + A rose curve with four leaves, each in one of the quadrants. + +

    + The curve r=\sin(2\theta) is a rose curve with four leaves. + It is has the same appearance as the rose curve in , + but it is rotated by 45 degrees, so that each leaf lies in one of the four quadrants. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:360,samples=180] plot ({cos(\x)*(1.9*sin(2*\x))},{sin(\x)*(1.9*sin(2*\x))}); + + \end{tikzpicture} + + + + +
    +
    + r=a\cos(3\theta) + + + A rose curve with three leaves, symmetric about the x axis. + +

    + The curve r=\cos(3\theta) is a rose curve with three leaves. + The leaves are narrower in appearance than those of the four-leaf rose curves. +

    + +

    + The curve overall is symmetric about the x axis. + One leaf lies along the positive x axis, + while the other two leaves are in the second and third quadrants. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:360,samples=160] plot ({cos(\x)*(1.9*cos(3*\x))},{sin(\x)*(1.9*cos(3*\x))}); + + \end{tikzpicture} + + + + +
    +
    + r=a\sin(3\theta) + + + A rose curve with three leaves, symmetric about the y axis. + +

    + Another rose curve with three leaves, this time given by r=\sin(3\theta). + The curve is symmetric about the y. + One leaf lies along the negative y axis, + while the other two leaves are in the first and second quadrants. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:360,samples=160] plot ({cos(\x)*(1.9*sin(3*\x))},{sin(\x)*(1.9*sin(3*\x))}); + + \end{tikzpicture} + + + + +
    +     +
    + +

    + Symmetric about x-axis: r=a \cos(n\theta) +

    + +

    + Symmetric about y-axis: r=a\sin(n\theta) +

    + +

    + Curve contains 2n petals when n is even and n petals when n is odd. +

    + +
    + Special Curves + +
    + Rose curve: r=a\sin(\theta/5) + + + A polar curve consisting of three loops. + +

    + The curve r=\sin(\theta/5) is also known as a rose curve, + but it is more similar to a limaçon in appearance, + except that it has more loops. +

    + +

    + The curve is symmetric about the y axis. + There are three loops in total, each of which intersects the y axis twice. + The smallest loop has a teardrop shape. It lies inside the middle loop, + which is heart-shaped. + The middle loop, in turn, lies inside of the largest loop, + which appears nearly circular, except that it has a slight cusp at the bottom, + where it joins with the middle loop. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:900,samples=200] plot ({cos(\x)*(1.9*sin(\x/5))},{sin(\x)*(1.9*sin(\x/5))}); + + \end{tikzpicture} + + + + +
    +
    + Rose curve: r=a\sin(2\theta/5) + + + An elaborate polar curve consisting of many intersecting loops. + +

    + The curve r=\sin(2\theta/5) is also known as a rose curve, + but it is more elaborate than any of the earlier examples. +

    + +

    + The curve is symmetric about both axes, and consists of many loops of varying size. + These loops intesect each other several times. + The overall appearance is similar to a sort of braided knot. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=0:1800,samples=500] plot ({cos(\x)*(1.9*cos(2*\x/5))},{sin(\x)*(1.9*cos(2*\x/5))}); + + \end{tikzpicture} + + + + +
    +
    + Lemniscate: r^2=a^2\cos(2\theta) + + + A lemniscate curve, which has the shape of a figure-eight. + +

    + The lemniscate r^2 = \cos^2(2\theta) is a figure-eight curve. + Since r^2 cannot be positive, + the points on the curve all correspond to angles where \cos(2\theta) is positive. + The curve is symmetric about the x axis, and looks much like the symbol for infinity. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=-45:45,samples=70] plot ({cos(\x)*(1.9*sqrt(cos(2*\x)))},{sin(\x)*(1.9*sqrt(cos(2*\x)))}); + \draw [thick,firstcolor,domain=-45:45,samples=70] plot ({-cos(\x)*(1.9*sqrt(cos(2*\x)))},{-sin(\x)*(1.9*sqrt(cos(2*\x)))}); + + \end{tikzpicture} + + + + +
    +
    + Eight Curve: r^2=a^2\sec^4(\theta) \cos(2\theta) + + + A polar curve in the shape of a figure-eight. + +

    + This final curve in the gallery is also a figure-eight curve. + Like the lemniscate, it is symmetric about the x axis, + but it is flatter at the ends than the lemniscate. +

    +     + + + \begin{tikzpicture} + + \draw [<->] (-2.1,0) -- (2.1,0); + \draw [<->] (0,-2.1) -- (0,2.1); + + \draw [thick,firstcolor,domain=-45:45,samples=80] plot ({cos(\x)*(1.9*sqrt(sec(\x)^4*cos(2*\x)))},{sin(\x)*(1.9*sqrt(sec(\x)^4*cos(2*\x)))}); + \draw [thick,firstcolor,domain=-45:45,samples=80] plot ({-cos(\x)*(1.9*sqrt(sec(\x)^4*cos(2*\x)))},{-sin(\x)*(1.9*sqrt(sec(\x)^4*cos(2*\x)))}); + + \end{tikzpicture} + + + + +
    +     +
    +     + +

    + Earlier we discussed how each point in the plane does not have a unique representation in polar form. + This can be a good thing, + as it allows for the beautiful and interesting curves seen in the preceding gallery. + However, it can also be a bad thing, + as it can be difficult to determine where two curves intersect. +

    + + + Finding points of intersection with polar curves + +

    + Determine where the graphs of the polar equations + r=1+3\cos(\theta) and r=\cos(\theta) intersect. +

    +     + +

    + As technology is generally readily available, + it is usually a good idea to start with a graph. + We have graphed the two functions in ; + to better discern the intersection points, + zooms in around the origin. +

    + +
    + Graphs to help determine the points of intersection of the polar functions given in + +
    + + + + Overlapping plots of a circle and a limaçon, showing their points of intersection. + +

    + The two curves for this example are plotted. + The curve r=1+3\cos(\theta) is a limaçon with an inner loop, + while r=\cos(\theta) is a circle with radius 1/2 and center (1/2,0). +

    + +

    + The circle is significantly smaller than the limaçon: + it intersects the x axis at (0,0) and (1,0), + while the inner loop of the limaçon intersects the x axis at (0,0) and (2,0). +

    + +

    + Near the origin, the inner loop of the limaçon narrows faster than the circle, + and in addition to the origin, two other points of intersection can be seen. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-2.6,ymax=2.6, + xmin=-1.5,xmax=4.5 + ] + + \addplot+ [domain=0:360,samples=120] ({cos(x)*(1+3*cos(x))},{sin(x)*(1+3*cos(x))}); + \addplot+ [solid,domain=0:180,samples=40] ({cos(x)*cos(x)},{sin(x)*cos(x)}); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
    + +
    + + + + Zoomed in view of the intersections between the two polar curves. + +

    + The second image is a zoomed in view of , + in which the three points of intersection can better be seen. + One intersection is at the origin, and the other two points of intersection lie on opposite sides of the x axis. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.6,ymax=.6, + xmin=-.72,xmax=.72 + ] + + \addplot+ [domain=0:360,samples=180] ({cos(x)*(1+3*cos(x))},{sin(x)*(1+3*cos(x))}); + \addplot+ [solid,domain=0:180,samples=80] ({cos(x)*cos(x)},{sin(x)*cos(x)}); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + To start we set the functions equal to each other and solve for \theta: + + 1+3\cos(\theta) \amp = \cos(\theta) + 2\cos(\theta) \amp = -1 + \cos(\theta) \amp = -\frac12 + \theta \amp = \frac{2\pi}{3}, \frac{4\pi}{3} + . +

    + +

    + (There are, of course, + infinitely many solutions to the equation \cos(\theta) =-1/2; + as the limaçon is traced out once on [0,2\pi], + we restrict our solutions to this interval.) +

    + +

    + We need to analyze this solution. + When \theta = 2\pi/3 we obtain the point of intersection that lies in the 4th quadrant. + When \theta = 4\pi/3, + we get the point of intersection that lies in the second quadrant. + There is more to say about this second intersection point, however. + The circle defined by r=\cos(\theta) is traced out once on [0,\pi], + meaning that this point of intersection occurs while tracing out the circle a second time. + It seems strange to pass by the point once and then recognize it as a point of intersection only when arriving there a second time. + The first time the circle arrives at this point is when \theta = \pi/3. + It is key to understand that these two points are the same: + (\cos(\pi/3),\pi/3) and (\cos(4\pi/3),4\pi/3). +

    + +

    + To summarize what we have done so far, + we have found two points of intersection: + when \theta=2\pi/3 and when \theta=4\pi/3. + When referencing the circle r=\cos(\theta), + the latter point is better referenced as when \theta=\pi/3. +

    + +

    + There is yet another point of intersection: the pole + (or, the origin). + We did not recognize this intersection point using our work above as each graph arrives at the pole at a different \theta value. +

    + +

    + A graph intersects the pole when r=0. + Considering the circle r=\cos(\theta), + r=0 when \theta = \pi/2 + (and odd multiples thereof, as the circle is repeatedly traced). + The limaçon intersects the pole when 1+3\cos(\theta) =0; + this occurs when \cos(\theta) = -1/3, + or for \theta = \cos^{-1}(-1/3). + This is a nonstandard angle, + approximately \theta = 1.9106 = 109.47^\circ. + The limaçon intersects the pole twice in [0,2\pi]; + the other angle at which the limaçon is at the pole is the reflection of the first angle across the x-axis. + That is, \theta = 4.3726 = 250.53^\circ. +

    +     + +     + +

    + If all one is concerned with is the (x,y) coordinates at which the graphs intersect, + much of the above work is extraneous. + We know they intersect at (0,0); + we might not care at what \theta value. + Likewise, using \theta =2\pi/3 and + \theta=4\pi/3 can give us the needed rectangular coordinates. + However, in the next section we apply calculus concepts to polar functions. + When computing the area of a region bounded by polar curves, + understanding the nuances of the points of intersection becomes important. +

    +     + + + + Terms and Concepts + + + +

    + In your own words, + describe how to plot the polar point P(r,\theta). +

    +     + + +     + + + + +

    + When plotting a point with polar coordinate P(r,\theta), + r must be positive. + +

    +     + +     + + + + +

    + Every point in the Cartesian plane can be represented by a polar coordinate. + +

    +     + +     + + + + +

    + Every point in the Cartesian plane can be represented uniquely by a polar coordinate. + +

    +     + +     +     + + + Problems + + + +

    + Plot the points with the given polar coordinates. +

    + +

    +

      +
    1. +

      + A=P(2,0) +

      +
      2. + +
    2. +

      + B=P(1,\pi) +

      +
      3. + +
    3. +

      + C=P(-2,\pi/2) +

      +
      4. + +
    4. +

      + D=P(1,\pi/4) +

      +
      5. +
    +

    +     + + + The four points plotted in this exercise. + +

    + On a polar grid, four points are plotted. + The point A is at the intersection of the initial ray and the circle of radius 2. + Points B and D are both on the circle of radius 1. + The point B is on the same line as the initial ray, but in the opposite direction. + The point D lies above the initial ray, making an angle of \pi/4. + Finally, the point C is at the bottom of the circle of radius 2. +

    +     + + + \begin{tikzpicture} + + \foreach \x in {1,2} + { + \draw (0,0) circle (\x); + \draw (\x,0) node [below right] {\(\x\)}; + } + \draw [thick,->,>=stealth] (0,0) node [below] {\(O\)} -- (2.5,0); + + \filldraw (xyz polar cs: angle=0,radius=2) circle (2pt) node [above right] {\(A\)} + (xyz polar cs: angle=180,radius=1)circle (2pt) node [right] {\(B\)} + (xyz polar cs: angle=90,radius=-2)circle (2pt) node [right] {\(C\)} + (xyz polar cs: angle=45,radius=1)circle (2pt) node [right] {\(D\)}; + + \end{tikzpicture} + + + +     + +     + + + + +

    + Plot the points with the given polar coordinates. +

    + +

    +

      +
    1. +

      + A=P(2,3\pi) +

      +
      2. + +
    2. +

      + B=P(1,-\pi) +

      +
      3. + +
    3. +

      + C=P(1,2) +

      +
      4. + +
    4. +

      + D=P(1/2,5\pi/6) +

      +
      5. +
    +

    +     + + + The four points plotted in this exercise. + +

    + On a polar grid, four points are plotted. + Points A and B are both on the same line as the initial ray, + but to the left of the origin O, with A on the circle of radius 2, + and B on the circle of radius 1. + The point C is on the circle of radius 1, + and makes an angle slightly greater than a right angle with the initial ray, + placing it above and just to the left of the origin O. + The point D lies almost immediately below the point C; + the circle of radius 1/2 is not marked as part of the grid. +

    +     + + + \begin{tikzpicture} + + \foreach \x in {1,2} + { + \draw (0,0) circle (\x); + \draw (\x,0) node [below right] {\(\x\)}; + } + \draw [thick,->,>=stealth] (0,0) node [below] {\(O\)} -- (2.5,0); + + \filldraw (xyz polar cs: angle=180,radius=2) circle (2pt) node [above] {\(A\)} + (xyz polar cs: angle=180,radius=1)circle (2pt) node [above] {\(B\)} + (xyz polar cs: angle=114.6,radius=1)circle (2pt) node [right] {\(C\)} + (xyz polar cs: angle=150,radius=.5)circle (2pt) node [right] {\(D\)}; + + \end{tikzpicture} + + + +     + +     + + + + +

    + For each of the given points give two sets of polar coordinates that identify it, + where 0\leq \theta\leq 2\pi. +

    + + + Four points plotted on a polar grid. + +

    + A standard polar grid is given, with circles of radius 1, 2, and 3. + Rays marking the angles \pi/6, \pi/4, \pi/3, and \pi/2 are also shown, + along with rays for the corresponding angles in the other quadrants. +

    + +

    + The point A lies on the ray \theta=\pi/4, + at a distance half way between the circles of radius 2 and 3. +

    + +

    + The point B lies on the ray \theta=-\pi/6, + and on the circle r=1. +

    + +

    + The point C lies on the ray \theta = 4\pi/3, + and on the circle r=3. +

    + +

    + The point D lies on the ray \theta = 2\pi/3, + at a distance half way between the circles of radius 1 and 2. +

    +     + + + \begin{tikzpicture} + + \draw [dashed,gray] (-3.1,0) -- (0,0); + \draw [thick,->,>=stealth] (0,0) node [below] {\(O\)} -- (3.5,0); + + \filldraw (0,0) circle (2.4pt); + + \foreach \x in {1,2,3} + { + \draw (0,0) circle (\x cm); + \draw (\x,0) node [below right] {\x}; + } + + \foreach \x in {30,45,60,90,120,135,150} + { + \draw [rotate=\x,dashed,gray] (-3.1,0) -- (3.1,0); + } + + \filldraw (xyz polar cs: angle=45,radius=2.5) circle (2pt) node [above] {\(A\)} + (xyz polar cs: angle=-30,radius=1)circle (2pt) node [below] {\(B\)} + (xyz polar cs: angle=240,radius=3)circle (2pt) node [right] {\(C\)} + (xyz polar cs: angle=120,radius=1.5)circle (2pt) node [right] {\(D\)}; + + \end{tikzpicture} + + + + +     + +

    + A=P(2.5,\pi/4) and P(-2.5,5\pi/4); +

    + +

    + B=P(-1,5\pi/6) and P(1,11\pi/6); +

    + +

    + C=P(3,4\pi/3) and P(-3,\pi/3); +

    + +

    + D=P(1.5,2\pi/3) and P(-1.5,5\pi/3); +

    +     + +     + + + + + Context()->variables->are(t=>"Real"); + @A=(Compute("(2,pi/6)"),Compute("(-2,-(5*pi)/6)")); + $Alist=List($A[0],$A[1]); + @B=(Compute("(1,-pi/3)"),Compute("(-1,(2*pi)/3)")); + $Blist=List($B[0],$B[1]); + @C=(Compute("(2,(3*pi)/4)"),Compute("(-2,-pi/4)")); + $Clist=List($C[0],$C[1]); + @D=(Compute("(5/2,pi)"),Compute("(5/2,-pi)")); + $Dlist=List($D[0],$D[1]); + + +

    + For each of the given points give two sets of polar coordinates that identify it, + where -\pi\lt \theta\leq \pi. +

    + + + Four points plotted on a polar grid. + +

    + A standard polar grid is given, with circles of radius 1, 2, and 3. + Rays marking the angles \pi/6, \pi/4, \pi/3, and \pi/2 are also shown, + along with rays for the corresponding angles in the other quadrants. +

    + +

    + The point A lies on the ray \theta=\pi/6, + and on the circle r=1. +

    + +

    + The point B lies on the ray \theta=-\pi/3, + and on the circle r=1. +

    + +

    + The point C lies on the ray \theta = 3\pi/4, + and on the circle r=2. +

    + +

    + The point D lies on the ray \theta = \pi, + at a distance half way between the circles of radius 2 and 3. +

    +     + + \begin{tikzpicture} + \draw [dashed,gray] (-3.1,0) -- (0,0); + \draw[thick,->,>=stealth] (0,0) node [below] {\(O\)} -- (3.5,0) ; + \filldraw (0,0) circle (2.4pt); + \foreach \x in {1,2,3} + {\draw (0,0) circle (\x cm); + \draw (\x,0) node [below right] {\x}; + } + \foreach \x in {30,45,60,90,120,135,150} + {\draw [rotate=\x,dashed,gray] (-3.1,0) -- (3.1,0); + } + \filldraw (xyz polar cs: angle=30,radius=2) circle (2pt) node [above] {\(A\)} + (xyz polar cs: angle=-60,radius=1)circle (2pt) node [below] {\(B\)} + (xyz polar cs: angle=135,radius=2)circle (2pt) node [right] {\(C\)} + (xyz polar cs: angle=180,radius=2.5)circle (2pt) node [below] {\(D\)}; + \end{tikzpicture} + + + + + Enter two sets of coordinates for the point A. Separate your answers by a comma. + +

    + +

    + + + Enter two sets of coordinates for the point B. Separate your answers by a comma. + +

    + +

    + + + Enter two sets of coordinates for the point C. Separate your answers by a comma. + +

    + +

    + + + Enter two sets of coordinates for the point D. Separate your answers by a comma. + +

    + +

    +     + +     +     + + + + + Context("Point"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + @xy=(Compute("(sqrt(2),sqrt(2))"), Compute("(sqrt(2),-sqrt(2))")); + @rt=(Compute("(sqrt(5),arctan(-1/2))"), Compute("(sqrt(5),pi+arctan(-1/2))")); + @rtev=map{$_->cmp(showCoordinateHints => 0, checker => sub { + my ($correct,$student,$ansHash) = @_; + my ($sr,$st) = $student->value; + my ($cr,$ct) = $correct->value; + return (($sr*cos($st) == $cr*cos($ct)) and ($sr*sin($st) == $cr*sin($ct))); + })}(@rt); + + +

    + Convert each of the following polar coordinates to rectangular, + and each of the following rectangular coordinates to polar. +

    + +

    +

      +
    1. +

      + A=P(2,\pi/4) +

      + +

      + (x,y)= +

      +
      2. + +
    2. +

      + B=P(2,-\pi/4) +

      + +

      + (x,y)= +

      +
      3. + +
    3. +

      + C=(2,-1) +

      + +

      + P(r,\theta)=P +

      +
      4. + +
    4. +

      + D=(-2,1) +

      + +

      + P(r,\theta)=P +

      +
      5. +
    +

    +     + +

    + A=(\sqrt{2},\sqrt{2}), + B=(\sqrt{2},-\sqrt{2}), + C=P(\sqrt{5},-0.46), + D=P(\sqrt{5},2.68). +

    +     +     +     + + + + + Context("Point"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + @xy=(Compute("(-3,0)"), Compute("(-1/2,sqrt(3)/2)")); + @rt=(Compute("(4,pi/2)"), Compute("(2,-pi/3)")); + @rtev=map{$_->cmp(showCoordinateHints => 0, checker => sub { + my ($correct,$student,$ansHash) = @_; + my ($sr,$st) = $student->value; + my ($cr,$ct) = $correct->value; + return (($sr*cos($st) == $cr*cos($ct)) and ($sr*sin($st) == $cr*sin($ct))); + })}(@rt); + + +

    + Convert each of the following polar coordinates to rectangular, + and each of the followingrectangular coordinates to polar. +

    + +

    +

      +
    1. +

      + A=P(3,\pi) +

      + +

      + (x,y)= +

      +
      2. + +
    2. +

      + B=P(1,2\pi/3) +

      + +

      + (x,y)= +

      +
      3. + +
    3. +

      + C=(0,4) +

      + +

      + P(r,\theta)=P +

      +
      4. + +
    4. +

      + D=(1,-\sqrt{3}) +

      + +

      + P(r,\theta)=P +

      +
      5. +
    +

    +     + +

    + A=(-3,0), + B=(-1/2,\sqrt{3}/2), + C=P(4,\pi/2), + D=P(2,-\pi/3). +

    +     +     +     + + + + +

    + Graph the polar function on the given interval. +

    +     + + + + +

    + r=2,0\leq \theta\leq \pi/2 +

    +     + + + + Portion of the circle of radius 2, centered at the origin, in the first quadrant. + +

    + An arc of the circle r=2 is shown, for 0\leq \theta\leq \pi/2. + This is the quarter of a circle of radius 2, centered at the origin, that lies in the first quadrant. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.5,ymax=2.1, + xmin=-.5,xmax=2.6 + ] + + \addplot+ [domain=0:90] ({2*cos(x)},{2*sin(x)}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + \theta=\pi/6,-1\leq r\leq 2 +

    +     + + + + A line segment through the origin with positive slope. + +

    + A line segment through the origin is shown. + The segment makes an angle of \pi/6 with the positive x axis, + and it extends a distance of 2 units into the first quadrant, + and one unit into the third quadrant. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,2,-1,-2}, + ymin=-2.1,ymax=2.1, + xmin=-2.5,xmax=2.5 + ] + + \addplot+ [domain=-.866:1.73] ({x},{.577*x}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=1-\cos(\theta),[0,2\pi] +

    +     + + + + A cardioid, symmetric about the x axis, with x intercepts at -2 and 0. + +

    + The curve is a cardioid that is symmetric about the x axis. + The cusp is at the origin, and the other x intercept is at (-2,0). + (It is in the opposite direction of the example in the gallery of polar curves.) +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-2.1,ymax=2.1, + xmin=-2.5,xmax=2.5 + ] + + \addplot+ [domain=0:360,samples=80] ({cos(x)*(1-cos(x))},{sin(x)*(1-cos(x))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=2+\sin(\theta),[0,2\pi] +

    +     + + + + A convex limaçon, symmetric about the y axis. + +

    + The curve is a convex limaçon. This is the fourth type of limaçon in the gallery of polar curves. + In this case, the limaçon is symmetric about the y axis, + with the flattened part of the curve at the bottom. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-2.6,ymax=3.6, + xmin=-3.7,xmax=3.7 + ] + + \addplot+ [domain=0:360,samples=80] ({cos(x)*(2+sin(x))},{sin(x)*(2+sin(x))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=2-\sin(\theta),[0,2\pi] +

    +     + + + + A convex limaçon, symmetric about the y axis. + +

    + The curve is a convex limaçon. This is the fourth type of limaçon in the gallery of polar curves. + In this case, the limaçon is symmetric about the y axis, + with the flattened part of the curve at the top. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-3.6,ymax=2.6, + xmin=-3.7,xmax=3.7 + ] + + \addplot+ [domain=0:360,samples=70] ({cos(x)*(2-sin(x))},{sin(x)*(2-sin(x))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=1-2\sin(\theta),[0,2\pi] +

    +     + + + + A limaçon with an inner loop, symmetric about the y axis. + +

    + The curve is a limaçon with an inner loop. + It is symmetric about the y axis. + The inner loop lies below the x axis, with y intercepts at y=0 and y=-1. + The outer loop has its other y intercept at y=-3. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-3.6,ymax=2.6, + xmin=-3.7,xmax=3.7 + ] + + \addplot+ [domain=0:360,samples=90] ({cos(x)*(1-2*sin(x))},{sin(x)*(1-2*sin(x))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=1+2\sin(\theta),[0,2\pi] +

    +     + + + + A limaçon with an inner loop, symmetric about the y axis. + +

    + The curve is a limaçon with an inner loop. + It is symmetric about the y axis. + The inner loop lies above the x axis, with y intercepts at y=0 and y=1. + The outer loop has its other y intercept at y=3. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-2.6,ymax=3.6, + xmin=-3.7,xmax=3.7 + ] + + \addplot+ [domain=0:360,samples=90] ({cos(x)*(1+2*sin(x))},{sin(x)*(1+2*sin(x))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=\cos(2\theta),[0,2\pi] +

    +     + + + + A rose curve with four petals. + +

    + The curve is a rose curve with four loops that all pass through the origin. + The loops are symmetric about the two coordinate axes, + with their second intercepts at (\pm 1,0) and (0,\pm 1). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.3,xmax=1.3 + ] + + \addplot+ [domain=0:360,samples=150] ({cos(x)*(cos(2*x))},{sin(x)*(cos(2*x))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=\sin(3\theta),[0,\pi] +

    +     + + + + A rose curve with three loops, symmetric about the y axis. + +

    + A rose curve with three loops that all pass through the origin. + One loop is along the negative y axis, with a y intercept at (-1,0). + The other two loops lie in the first and second quadrants. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.3,xmax=1.3 + ] + + \addplot+ [domain=0:180,samples=101] ({cos(x)*(sin(3*x))},{sin(x)*(sin(3*x))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=\cos(\theta/3),[0,3\pi] +

    +     + + + + A limaçon with an inner loop. + +

    + The curve is a limaçon with an inner loop, + symmetric about the x axis, + but it is shifted horizontally, to the left, + relative to the example in the gallery of polar curves. +

    + +

    + The point of self-intersection is at (-1/2, 0). + The other end of the inner loop is at the origin, + and the far end of the outer loop is at the point (1,0). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.3,xmax=1.3 + ] + + \addplot+ [domain=0:540,samples=120] ({cos(x)*(cos(x/3))},{sin(x)*(cos(x/3))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=\cos(2\theta/3),[0,6\pi] +

    +     + + + + An elaborate rose curve with many self-intersections and four primary loops. + +

    + This is a more complicated curve. + It passes several times through the origin, and has eight other points of self-intersection. + The largest loops in the curve are similar to cardioids; + there are four of these passing through the origin, with a second intercept at one of the four points (\pm 1, 0), (0,\pm 1). + As these loops intersect each other, they create four other loops of intermediate size, + and four smaller loops in the center. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.3,xmax=1.3 + ] + + \addplot+ [domain=0:1080,samples=300] ({cos(x)*(cos(2*x/3))},{sin(x)*(cos(2*x/3))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=\theta/2,[0,4\pi] +

    +     + + + + A counter-clockwise spiral + +

    + The curve is a counter-clockwise spiral that begins at the origin. + It makes two full revolutions, and ends at the point (2\pi, 0). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-5.9,ymax=4.9, + xmin=-6.5,xmax=6.5 + ] + + \addplot+ [domain=0:720,samples=150] ({cos(x)*(x*3.14/360)},{sin(x)*(x*3.14/360)}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=3\sin(\theta),[0,\pi] +

    +     + + + + A circle passing through the origin with its center on the positive y axis. + +

    + A circle of radius 3/2 with its center at (0,3/2). + It passes through the origin and the point (0,3). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ytick={2,3,1}, + ymin=-.9,ymax=3.9, + xmin=-2.9,xmax=2.9 + ] + + \addplot+ [domain=0:180,samples=60] ({cos(x)*(3*sin(x))},{sin(x)*(3*sin(x))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + +

    + r=2\cos(\theta),[0,\pi/2] +

    +     + + + A semi-circle in the first quadrant. + +

    + The curve is a semi-circle in the first quadrant. + Its endpoints are (0,0) and (2,0), + making the center (1,0) and radius 1. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-2.3,ymax=2.3,% + xmin=-2.5,xmax=2.5% + ] + + \addplot+ [domain=0:90,samples=60] ({cos(x)*(2*cos(x))},{sin(x)*(2*cos(x))}); + + \end{axis} + + \end{tikzpicture} + + +     +     + + + + + +

    + r=\cos(\theta) \sin(\theta),[0,2\pi] +

    +     + + + + A four-leaf rose with one petal in each quadrant. + +

    + The curve is a four-leafed rose that lies within the circle r=1/2. + One leaf lies in each of the four quadrants. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.1,ymax=1.1, + xmin=-1.3,xmax=1.3 + ] + + \addplot+ [domain=0:360,samples=120] ({cos(x)*(cos(x)*sin(x))},{sin(x)*(cos(x)*sin(x))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + r=\theta^2-(\pi/2)^2,[-\pi,\pi] +

    +     + + + + A curve with two loops: a circle inside a larger leaf shape. + +

    + The curve for this exercise is a fairly strange shape. + It is symmetric about the x axis. + There is a large, outer loop that looks like a leaf or a raindrop. + It has a cusp at (-7,0), and also passes through the origin. + There is a smaller, inner loop that looks almost like a circle. + It passes through the origin and the point (-3,0). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-3.75,ymax=3.75, + xmin=-8.1,xmax=.9 + ] + + \addplot+ [domain=-3.14:3.14,samples=120] ({cos(deg(x))*(x^2-2.4674)},{sin(deg(x))*(x^2-2.4674)}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + \ds r=\frac{3}{5\sin(\theta) -\cos(\theta) },[0,2\pi] +

    +     + + + + A straight line with positive slope. + +

    + The curve is a straight line with x intercept at (-3,0) + and y intercept (0,3/5). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-4.9,ymax=4.96, + xmin=-5.9,xmax=5.9 + ] + + \addplot+ [domain=.1:3] ({cos(deg(x))*(3/(5*sin(deg(x))-cos(deg(x)))},{sin(deg(x))*(3/(5*sin(deg(x))-cos(deg(x)))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + \ds r=\frac{-2}{3\cos(\theta) -2\sin(\theta) },[0,2\pi] +

    +     + + + + A straight line with positive slope. + +

    + The curve is a straight line with x intercept at (-2/3,0) + and y intercept (0,1). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-4.9,ymax=4.96, + xmin=-5.9,xmax=5.9 + ] + + \addplot+ [domain=.1:3] ({cos(deg(x))*(-2/(3*cos(deg(x))-2*sin(deg(x))))},{sin(deg(x))*(-2/(3*cos(deg(x))-2*sin(deg(x))))}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + \ds r=3\sec(\theta),(-\pi/2,\pi/2) +

    +     + + + + A vertical line with x=3. + +

    + The curve is the vertical line x=3. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-4.9,ymax=4.96, + xmin=-5.9,xmax=5.9 + ] + + \addplot+ [domain=-5:5] ({3},{x}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + +

    + \ds r=3\csc(\theta),(0,\pi) +

    +     + + + + The horizontal line y=4. + +

    + The curve is the horizontal line y=4. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-4.9,ymax=4.96, + xmin=-5.9,xmax=5.9 + ] + + \addplot+ [domain=-5:5] ({x},{4}); + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + +     + + + +

    + Convert the polar equation to a rectangular equation. +

    +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("(x-3)^2+y^2=9", limits=>[[-1,7],[-4,4]]); + + +

    + Convert the polar equation to a rectangular equation. +

    + +

    + r=6\cos(\theta) +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("x^2+(y+2)^2=4", limits=>[[-3,3],[-5,1]]); + + +

    + Convert the polar equation to a rectangular equation. +

    + +

    + r=-4\sin(\theta) +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("(x-1/2)^2+(y-1/2)^2=1/2", limits=>[[-1,2],[-1,2]]); + + +

    + Convert the polar equation to a rectangular equation. +

    + +

    + r=\cos(\theta) +\sin(\theta) +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("y=2/5x+7/5", limits=>[[-3,3],[-2,4]]); + + +

    + Convert the polar equation to a rectangular equation. +

    + +

    + r=\dfrac{7}{5\sin(\theta)-2\cos(\theta)} +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("x=3", limits=>[[0,6],[-3,3]]); + + +

    + Convert the polar equation to a rectangular equation. +

    + +

    + r=\dfrac{3}{\cos(\theta)} +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("y=4", limits=>[[-3,3],[1,7]]); + + +

    + Convert the polar equation to a rectangular equation. +

    + +

    + r=\dfrac{4}{\sin(\theta)} +

    + +

    + +

    +     +     +     + + + + +

    + \ds r=\tan(\theta) +

    +     + +

    + x^4+x^2y^2x^2-y^2=0 +

    +     + +     + + + +

    + \ds r=\cot\theta +

    +     + +

    + y^4+x^2y^2-x^2=0 +

    +     +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("x^2+y^2=4", limits=>[[-3,3],[-3,3]]); + + +

    + Convert the polar equation to a rectangular equation. +

    + +

    + r=2 +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + $eq=ImplicitEquation("y=x/sqrt(3)", limits=>[[-3,3],[-3,3]]); + + +

    + Convert the polar equation to a rectangular equation. +

    + +

    + \theta=\pi/6 +

    + +

    + +

    +     +     +     + +     + + + +

    + Convert the rectangular equation to a polar equation. +

    +     + + + + + Context("ImplicitEquation"); + Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); + $eq=ImplicitEquation("theta=pi/4", limits=>[[0,3],[0,3]]); + + +

    + Convert the rectangular equation to a polar equation. + Type theta for \theta. +

    + +

    + y=x +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); + $eq=ImplicitEquation("r=7/(sin(theta) -4cos(theta) )", limits=>[[1.5,3],[pi()/2,3*pi()/2]]); + + +

    + Convert the rectangular equation to a polar equation. + Type theta for \theta. +

    + +

    + y=4x+7 +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); + $eq=ImplicitEquation("r=5sec(theta)", limits=>[[4,10],[-pi()/4,pi()/4]]); + + +

    + Convert the rectangular equation to a polar equation. + Type theta for \theta. +

    + +

    + x=5 +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); + $eq=ImplicitEquation("r=5csc(theta)", limits=>[[4,10],[pi()/4,3*pi()/4]]); + + +

    + Convert the rectangular equation to a polar equation. + Type theta for \theta. +

    + +

    + y=5 +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); + $eq=ImplicitEquation("r=cos(theta)/sin^2(theta)", limits=>[[0,4],[-pi()/2,pi()/2]]); + + +

    + Convert the rectangular equation to a polar equation. + Type theta for \theta. +

    + +

    + x=y^2 +

    + +

    + +

    +     +     +     + + + + +

    + \ds x^2y=1 +

    +     + +

    + r=1/\sqrt[3]{\cos^2(\theta) \sin(\theta) } +

    +     + +     + + + + + Context("ImplicitEquation"); + Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $eq=ImplicitEquation("r=sqrt(7)", limits=>[[0,4],[0,2*pi()]]); + + +

    + Convert the rectangular equation to a polar equation. + Type theta for \theta. +

    + +

    + x^2+y^2=7 +

    + +

    + +

    +     +     +     + + + + +

    + \ds (x+1)^2+y^2=1 +

    +     + +

    + r=-2\cos(\theta) +

    +     + +     + +     + + + +

    + Find the points of intersection of the polar graphs. +

    +     + + + + + Context("Point"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->operators->add('P'=>{precedence=>6,associativity=>'left',type=>'unary',string=>"P",class=>'Parser::UOP::plus'}); + $rts=Compute("P(sqrt(3)/2,pi/6), P(0,pi/2), P(-sqrt(3)/2,5pi/6)"); + $rtsev=$rts->cmp(showCoordinateHints => 0, checker => sub { + my ($correct,$student,$ansHash) = @_; + my ($sr,$st) = $student->value; + my ($cr,$ct) = $correct->value; + return (($sr*cos($st) == $cr*cos($ct)) and ($sr*sin($st) == $cr*sin($ct))); + }); + + +

    + Find the points where r=\sin(2\theta) intersects r=\cos(\theta) on [0,\pi], + expressed in polar coordinates with notation P(r,\theta). +

    + +

    + +

    +     +     +     + + + + +

    + r=\cos(2\theta) and r=\cos(\theta) on [0,\pi] +

    +     + +

    + P(1,0), P(0,\pi/2)=P(0,\pi/4), P(-1/2,2\pi/3) +

    +     + +     + + + + + Context("Point"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->operators->add('P'=>{precedence=>6,associativity=>'left',type=>'unary',string=>"P",class=>'Parser::UOP::plus'}); + $rts=Compute("P(0,0), P(sqrt(2),pi/4)"); + $rtsev=$rts->cmp(showCoordinateHints => 0, checker => sub { + my ($correct,$student,$ansHash) = @_; + my ($sr,$st) = $student->value; + my ($cr,$ct) = $correct->value; + return (($sr*cos($st) == $cr*cos($ct)) and ($sr*sin($st) == $cr*sin($ct))); + }); + + +

    + Find the points where r=2\cos(\theta) intersects r=2\sin(\theta) on [0,\pi], + expressed in polar coordinates with notation P(r,\theta). +

    + +

    + +

    +     +     +     + + + + +

    + r=\sin(\theta) and r=\sqrt{3}+3\sin(\theta) on [0,2\pi] +

    +     + +

    + P(\sqrt{3}/2,\pi/3)=P(-\sqrt{3}/2,4\pi/3), + P(\sqrt{3}/2,2\pi/3)=P(-\sqrt{3}/2,5\pi/3), P(0,\pi/2) +

    +     + +     + + + + +

    + r=\sin(3\theta) and r=\cos(3\theta) on [0,\pi] +

    +     + +

    + P(0,0)=P(0,\pi/6), + P(\sqrt{2}/2,\pi/12), + P(-\sqrt{2}/2,5\pi/12), + P(\sqrt{2}/2,3\pi/4) +

    +     + +     + + + + + Context("Point"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->operators->add('P'=>{precedence=>6,associativity=>'left',type=>'unary',string=>"P",class=>'Parser::UOP::plus'}); + $rts=Compute("P(3/2,pi/3), P(3/2,-pi/3), P(0,pi)"); + $rtsev=$rts->cmp(showCoordinateHints => 0, checker => sub { + my ($correct,$student,$ansHash) = @_; + my ($sr,$st) = $student->value; + my ($cr,$ct) = $correct->value; + return (($sr*cos($st) == $cr*cos($ct)) and ($sr*sin($st) == $cr*sin($ct))); + }); + + +

    + Find the points where r=3\cos(\theta) intersects + r=1+\cos(\theta) on [-\pi,\pi], + expressed in polar coordinates with notation P(r,\theta). +

    + +

    + +

    +     +     +     + + + + +

    + r=1 and r=2\sin(2\theta) on [0,2\pi] +

    +     + +

    + For all points, r=1; + \theta = \pi/12,\,5\pi/12,\,7\pi/12,\,11\pi/12,\,13\pi/12,\,17\pi/12,\,19\pi/12,\,23\pi/12. +

    +     + +     + + + + +

    + r=1-\cos(\theta) and r=1+\sin(\theta) on [0,2\pi] +

    +     + +

    + P(0,0)=P(0,3\pi/2), + P(1+\sqrt{2}/2,3\pi/4), + P(1-\sqrt{2}/2,7\pi/4) +

    +     + +     + +     + + + + +

    + Pick a integer value for n, where n\neq 2,3, + and use technology to plot + \ds r=\sin\left(\frac mn\theta\right) for three different integer values of m. + Sketch these and determine a minimal interval on which the entire graph is shown. +

    +     + +

    + Answers will vary. + If m and n do not have any common factors, + then an interval of 2n\pi is needed to sketch the entire graph. +

    +     + +     + + + + +

    + Create your own polar function, + r=f(\theta) and sketch it. + Describe why the graph looks as it does. +

    +     + +

    + Answers will vary. +

    +     + +     +     +     +
    +
    + Calculus and Polar Functions + + +

    + The previous section defined polar coordinates, + leading to polar functions. + We investigated plotting these functions and solving a fundamental question about their graphs, namely, + where do two polar graphs intersect? +

    + +

    + We now turn our attention to answering other questions, + whose solutions require the use of calculus. + A basis for much of what is done in this section is the ability to turn a polar function + r=f(\theta) into a set of parametric equations. + Using the identities x=r\cos(\theta) and y=r\sin(\theta), + we can create the parametric equations x=f(\theta)\cos(\theta), + y=f(\theta)\sin(\theta) and apply the concepts of . +

    +     + + + Polar Functions and <m>dy/dx</m> +

    + We are interested in the lines tangent to a given graph, + regardless of whether that graph is produced by rectangular, + parametric, or polar equations. + In each of these contexts, + the slope of the tangent line is \frac{dy}{dx}. + Given r=f(\theta), + we are generally not concerned with r\,'=\fp(\theta); + that describes how fast r changes with respect to \theta. + Instead, we will use x=f(\theta)\cos(\theta), + y=f(\theta)\sin(\theta) to compute \frac{dy}{dx}. +

    + +

    + Using we have + + \frac{dy}{dx} = \frac{dy}{d\theta}\Big/\frac{dx}{d\theta} + . +

    + +

    + Each of the two derivatives on the right hand side of the equality requires the use of the Product Rule. + We state the important result as a Key Idea. +

    + + + Finding <m>\frac{dy}{dx}</m> with Polar Functions +

    + Let r=f(\theta) be a polar function. + With x=f(\theta)\cos(\theta) and y=f(\theta)\sin(\theta), + polarfunctions!finding \frac{dy}{dx} + tangent line + + \frac{dy}{dx} = \frac{\fp(\theta)\sin(\theta) +f(\theta)\cos(\theta) }{\fp(\theta)\cos(\theta) -f(\theta)\sin(\theta) } + . +

    +     + + + + + Finding <m>\frac{dy}{dx}</m> with polar functions + +

    + Consider the limaçon r=1+2\sin(\theta) on [0,2\pi]. +

    + +

    +

      +
    1. +

      + Find the equations of the tangent and normal lines to the graph at \theta=\pi/4. +

      +
      2. + +
    2. +

      + Find where the graph has vertical and horizontal tangent lines. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + We start by computing \frac{dy}{dx}. + With \fp(\theta) = 2\cos(\theta), we have + + \frac{dy}{dx} \amp = \frac{2\cos(\theta) \sin(\theta) + \cos(\theta) (1+2\sin(\theta) )}{2\cos^2(\theta) -\sin(\theta) (1+2\sin(\theta) )} + \amp = \frac{\cos(\theta) (4\sin(\theta) +1)}{2(\cos^2(\theta) -\sin^2(\theta) )-\sin(\theta) } + . + When \theta=\pi/4, \frac{dy}{dx}=-2\sqrt{2}-1 + (this requires a bit of simplification). + In rectangular coordinates, + the point on the graph at \theta=\pi/4 is (1+\sqrt{2}/2,1+\sqrt{2}/2). + Thus the rectangular equation of the line tangent to the limaçon at \theta=\pi/4 is + + y=(-2\sqrt{2}-1)\big(x-(1+\sqrt{2}/2)\big)+1+\sqrt{2}/2 \approx -3.83 x+8.24 + . + The limaçon and the tangent line are graphed in . + + The normal line has the opposite-reciprocal slope as the tangent line, so its equation is + + y \approx \frac{1}{3.83}x+1.26 + . +

      + +
      + The limaçon in with its tangent line at \theta=\pi/4 and points of vertical and horizontal tangency + + + A limaçon with an inner loops, symmetric about the y axis, and a tangent line. + +

      + The curve is a limaçon with an inner loop. + It is symmetric about the y axis, with intercepts on the positive y axis. + There are several marked points on the curve, indicating the location of horizontal and vertical tangent lines. +

      + +

      + Points marking horizontal tangent lines are at the top of both loops, + at the points (0,3) (outer loop) and (0,1) (inner loop). + There are also two points marking horizontal tangent lines at the bottom of the outer loop; + these lie below the x axis, on either side of the y axis. +

      + +

      + There are also four points marking vertical tangent lines. + These are on the left and right sides of each loop. +

      + +

      + Just above the point where the vertical tangent on the right side of the outer loop is marked, + a tangent line to the curve is drawn. The tangent line has a relatively large negative slope. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.35,ymax=3.4, + xmin=-2.1,xmax=2.1 + ] + + \addplot+ [domain=0:360,samples=101] ({cos(x)*(1+2*sin(x))},{sin(x)*(1+2*sin(x))}); + \addplot+ [solid,domain=0.1:2] ({x},{-3.82843*x+8.24264}); + + \filldraw (axis cs: 0,3) circle (2.4pt) + (axis cs: 0,1) circle (2.4pt) + (axis cs: -.493,-0.125) circle (2.4pt) + (axis cs: .483,-0.125) circle (2.4pt); + + \filldraw [fill=white,draw=black,thick] + (axis cs: 1.76011, 1.31048) circle (2.4pt) + (axis cs: -1.76011, 1.31048) circle (2.4pt) + (axis cs: 0.368977, 0.572639) circle (2.4pt) + (axis cs: -0.369009, 0.578743) circle (2.4pt); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
      +
      2. + +
    2. +

      + To find the horizontal lines of tangency, + we find where \frac{dy}{dx}=0; + thus we find where the numerator of our equation for \frac{dy}{dx} is 0. + + \cos(\theta) (4\sin(\theta) +1)=0 \Rightarrow \cos(\theta) =0 \text{ or } 4\sin(\theta) +1=0 + . + On [0,2\pi], + \cos(\theta) =0 when \theta=\pi/2,\,3\pi/2. + + Setting 4\sin(\theta) +1=0 gives \theta=\sin^{-1}(-1/4)\approx -0.2527 = -14.48^\circ. + We want the results in [0,2\pi]; + we also recognize there are two solutions, + one in the third quadrant and one in the fourth. + Using reference angles, the two solutions are + \theta =3.39 and 6.03 radians. + The four points we obtained where the limaçon has a horizontal tangent line are given in + with black-filled dots. + + To find the vertical lines of tangency, + we set the denominator of \frac{dy}{dx}=0. + + 2(\cos^2(\theta) -\sin^2(\theta) )-\sin(\theta) \amp = 0 . + Convert the \cos^2(\theta) term to 1-\sin^2(\theta): + 2(1-\sin^2(\theta) -\sin^2(\theta) )-\sin(\theta) \amp = 0 + 4\sin^2(\theta) + \sin(\theta) -2 \amp = 0. + Recognize this as a quadratic in the variable \sin(\theta). Using the quadratic formula, we have + \sin(\theta) \amp = \frac{-1\pm\sqrt{33}}{8} + . + We solve \sin(\theta) = \frac{-1+\sqrt{33}}8 and \sin(\theta) = \frac{-1-\sqrt{33}}8: + + \sin(\theta) \amp =\frac{-1+\sqrt{33}}8 \amp \sin(\theta) \amp = \frac{-1-\sqrt{33}}{8} + \theta \amp = \sin^{-1}\left(\frac{-1+\sqrt{33}}8\right) \amp \theta \amp = \sin^{-1}\left(\frac{-1-\sqrt{33}}8\right) + \theta \amp = 0.6349 \amp \theta \amp = -1.0030 + + In each of the solutions above, + we only get one of the possible two solutions as + \sin^{-1}(x) only returns solutions in [-\pi/2,\pi/2], + the 4th and 1st quadrants. + Again using reference angles, we have: + + \sin(\theta) = \frac{-1+\sqrt{33}}8 \Rightarrow \theta = 0.6349,\,2.5067 \text{ radians } + + and + + \sin(\theta) = \frac{-1-\sqrt{33}}8 \Rightarrow \theta = 4.1446,\,5.2802 \text{ radians. } + + These points are also shown in + with white-filled dots. +

      +
      3. +
    +

    +     + +     + + + +

    + When the graph of the polar function r=f(\theta) intersects the pole, + it means that f(\alpha)=0 for some angle \alpha. + Thus the formula for \frac{dy}{dx} in such instances is very simple, + reducing simply to + + \frac{dy}{dx} = \tan\alpha + . +

    + +

    + This equation makes an interesting point. + It tells us the slope of the tangent line at the pole is \tan\alpha; + some of our previous work (see, for instance, + ) shows us that the line through the pole with slope + \tan\alpha has polar equation \theta=\alpha. + Thus when a polar graph touches the pole at \theta=\alpha, + the equation of the tangent line at the pole is \theta=\alpha. +

    + + + Finding tangent lines at the pole + +

    + Let r=1+2\sin(\theta), a limaçon. + Find the equations of the lines tangent to the graph at the pole. +

    +     + +

    + We need to know when r=0. + + 1+2\sin(\theta) \amp = 0 + \sin(\theta) \amp = -1/2 + \theta \amp = \frac{7\pi}{6},\,\frac{11\pi}6 + . +

    + +

    + Thus the equations of the tangent lines, in polar, + are \theta = 7\pi/6 and \theta = 11\pi/6. + In rectangular form, the tangent lines are + y=\tan(7\pi/6)x and y=\tan(11\pi/6)x. + The full limaçon can be seen in ; + we zoom in on the tangent lines in . +

    + +
    + Graphing the tangent lines at the pole in + + + A zoomed in view of a limaçon near the origin, and two tangent lines at that point. + +

    + The curve is the same limaçon as in , but zoomed in near the origin. + The origin is a point of self-intersection for the curve, and there are two tangent lines: + one for the first time the curve passes through the origin, and one for the second. +

    + +

    + The two lines together make an X shape at the origin. + Due to the symmetry of the curve, one tangent line has positive slope, + and the other has a negative slope of the same absolute value. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=1.37, + xmin=-1.1,xmax=1.1 + ] + + \addplot+ [domain=0:360,samples=120] ({cos(x)*(1+2*sin(x))},{sin(x)*(1+2*sin(x))}); + \addplot [secondcurvestyle,solid,domain=-1:1] ({x},{.577*x}); + \addplot [secondcurvestyle,solid,domain=-1:1] ({x},{-.577*x}); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
    +     + +     +     + + + Area +

    + When using rectangular coordinates, + the equations x=h and y=k defined vertical and horizontal lines, + respectively, and combinations of these lines create rectangles + (hence the name rectangular coordinates). + It is then somewhat natural to use rectangles to approximate area as we did when learning about the definite integral. + polarfunctions!area +

    + +

    + When using polar coordinates, + the equations \theta=\alpha and r=c form lines through the origin and circles centered at the origin, + respectively, + and combinations of these curves form sectors of circles. + It is then somewhat natural to calculate the area of regions defined by polar functions by first approximating with sectors of circles. +

    + +

    + Consider + where a region defined by + r=f(\theta) on [\alpha,\beta] is given. + (Note how the sides of the region are the lines + \theta=\alpha and \theta=\beta, + whereas in rectangular coordinates the sides + of regions were often the vertical lines x=a and x=b.) +

    + + + +

    + Partition the interval [\alpha,\beta] into n equally spaced subintervals as \alpha = \theta_0 \lt \theta_1 \lt \cdots \lt \theta_{n}=\beta. + The length of each subinterval is \Delta\theta = (\beta-\alpha)/n, + representing a small change in angle. + The area of the region defined by the ith subinterval + [\theta_{i-1},\theta_{i}] can be approximated with a sector of a circle with radius f(c_i), + for some c_i in [\theta_{i-1},\theta_{i}]. + The area of this sector is \frac12f(c_i)^2\Delta\theta. + This is shown in , + where [\alpha,\beta] has been divided into 4 subintervals. + We approximate the area of the whole region by summing the areas of all sectors: + + \text{ Area } \approx \sum_{i=1}^n \frac12f(c_i)^2\Delta\theta + . +

    + +

    + This is a Riemann sum. + By taking the limit of the sum as n\to\infty, + we find the exact area of the region in the form of a definite integral. +

    + +
    + Computing the area of a polar region + +
    + + + + Plot of a generic polar function and the area it encloses between two angles. + +

    + The plot of some unknown polar curve r=f(\theta) is shown. + The curve lies entirely in the first quadrant, and has a somewhat wavy shape, + although the particular shape of the curve is unimportant. +

    + +

    + Two rays labeled \theta=\alpha and \theta=\beta are drawn, + and the area bounded by the two rays and the polar curve is shaded. + Several rays corresponding to angles between \alpha and \beta + are also shown as dashed lines. +

    +     + + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.1 + ] + + \addplot+ [areastyle,domain=18:72] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}) -- (axis cs:0,0) -- cycle; + \addplot [firstcurvestyle,domain=0:90,samples=50] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}); + + \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs: 0.905831, 0.294322) node [pos=.7,below,rotate=16,black] { $\theta=\alpha$}; + \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs:0.313792, 0.965751) node [pos=.7,above,rotate=72,black] { $\theta=\beta$}; + + \draw [thick,firstcolor,dashed] (axis cs:0,0) -- (axis cs: 0.862592, 0.528597) + (axis cs:0,0) -- (axis cs: 0.732107, 0.732107) + (axis cs:0,0) -- (axis cs: 0.497095, 0.811186); + + \draw (axis cs:.8,.85) node { $r=f(\theta)$}; + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
    + +
    + + + + Plot of a region bounded by a polar curve, and its approximation by cicular wedges. + +

    + A curve r=f(\theta) is shown; it is the same curve as . + The rays \theta=\alpha and \theta=\beta are also shown, + as well as the shaded region bounded by these rays and the polar curve. +

    + +

    + Overlaid on the shaded region are four circular wedges. + The angle between \theta=\alpha and \theta=\beta is divided into four smaller angles; + these correspond to the rays shown as dashed lines in . + Each wedge is a sector of a circle that spans one of these angles, + with radius corresponding some the value of f(\theta). +

    + +

    + Overall, the image illustrates the fact that the area bounded by the polar curve and the two rays + is approximated by the sum of the areas of the four circular wedges. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=18:72] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}) -- (axis cs:0,0) -- cycle; + + \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs:0.313792, 0.965751) node [pos=.7,above,rotate=72,black] { $\theta=\beta$}; + + \addplot [fill=white,domain=18:31.5] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; + \addplot [fill=white,domain=31.5:45] ({1.05*cos(x)},{1.05*sin(x)}) -- (axis cs:0,0) -- cycle; + \addplot [fill=white,domain=45:58.5] ({1.0*cos(x)},{1*sin(x)}) -- (axis cs:0,0) -- cycle; + \addplot [fill=white,domain=58.5:72] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; + + \addplot [secondcurvestyle,areastyle,domain=18:31.5] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; + \addplot [secondcurvestyle,areastyle,domain=31.5:45] ({1.05*cos(x)},{1.05*sin(x)}) -- (axis cs:0,0) -- cycle; + \addplot [secondcurvestyle,areastyle,domain=45:58.5] ({1.0*cos(x)},{1*sin(x)}) -- (axis cs:0,0) -- cycle; + \addplot [secondcurvestyle,areastyle,domain=58.5:72] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; + + \addplot [secondcurvestyle,solid,domain=18:31.5] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; + \addplot [secondcurvestyle,solid,domain=31.5:45] ({1.05*cos(x)},{1.05*sin(x)}) -- (axis cs:0,0) -- cycle; + \addplot [secondcurvestyle,solid,domain=45:58.5] ({1.0*cos(x)},{1*sin(x)}) -- (axis cs:0,0) -- cycle; + \addplot [secondcurvestyle,solid,domain=58.5:72] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; + + \draw (axis cs:.8,.85) node { $r=f(\theta)$}; + + \draw [thick,secondcolor] (axis cs:0,0) -- (axis cs: 0.905831, 0.294322) node [pos=.7,below,rotate=16,black] { $\theta=\alpha$}; + + \addplot [firstcurvestyle,domain=0:90,samples=50] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
    +     + +
    + + + + + Area of a Polar Region + +

    + Let f be continuous and non-negative on [\alpha,\beta], + where 0\leq \beta-\alpha\leq 2\pi. + The area A of the region bounded by the curve r=f(\theta) and the lines + \theta=\alpha and \theta=\beta is + + A \,=\, \frac12\int_\alpha^\beta f(\theta)^2 \, d\theta\, =\, \frac12\int_\alpha^\beta r^{\,2} \, d\theta + +

    +     +     + +

    + The theorem states that 0\leq \beta-\alpha\leq 2\pi. + This ensures that region does not overlap itself, + which would give a result that does not correspond directly to the area. +

    + + + Area of a polar region + +

    + Find the area of the circle defined by r=\cos(\theta). + (Recall this circle has radius 1/2.) +

    +     + +

    + This is a direct application of . + The circle is traced out on [0,\pi], leading to the integral + + \text{ Area } \amp = \frac12\int_0^\pi \cos^2(\theta)\, d\theta + \amp = \frac12\int_0^\pi \frac{1+\cos(2\theta)}{2}\, d\theta + \amp = \frac14\big(\theta +\frac12\sin(2\theta)\big)\Bigg|_0^\pi + \amp = \frac14\pi + . +

    + +

    + Of course, we already knew the area of a circle with radius 1/2. + We did this example to demonstrate that the area formula is correct. +

    +     + +     + + + + + Area of a polar region + +

    + Find the area of the cardioid r=1+\cos(\theta) bound between + \theta=\pi/6 and \theta=\pi/3, + as shown in . +

    + +
    + Finding the area of the shaded region of a cardioid in + + + A cardioid curve, within which is a shaded region bounded by the curve and two rays. + +

    + The polar curve r=1+\cos(\theta) is a cardioid that is symmetric about the x axis, + with a cusp at the origin, and second x intercept at (2,0). +

    + +

    + The rays \theta=\pi/6 and \theta=\pi/3 are drawn from the origin to where they meet the cardioid. + The region bounded by the cardioid and the two rays is shaded. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ytick={1}, + ymin=-.7,ymax=1.5, + xmin=-.5,xmax=2.1 + ] + + \addplot [firstcurvestyle,areastyle,domain=30:60] ({cos(x)*(1+cos(x))},{sin(x)*(1+cos(x))}) -- (axis cs:0,0) -- cycle; + \addplot [firstcurvestyle,domain=0:360,samples=101] ({cos(x)*(1+cos(x))},{sin(x)*(1+cos(x))}); + + \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs: 1.61603, 0.933013) node [pos=.6,below,rotate=30,black] { $\theta=\pi/6$}; + \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs:0.75, 1.29904) node [pos=.6,above,rotate=60,black] { $\theta=\pi/3$}; + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
    +     + +

    + This is again a direct application of . + + \text{ Area } \amp = \frac12\int_{\pi/6}^{\pi/3} (1+\cos(\theta) )^2\, d\theta + \amp = \frac12\int_{\pi/6}^{\pi/3} (1+2\cos(\theta) +\cos^2(\theta) )\, d\theta + \amp = \frac12\left(\theta+2\sin(\theta) +\frac12\theta+\frac14\sin(2\theta)\right)\Bigg|_{\pi/6}^{\pi/3} + \amp = \frac18\big(\pi+4\sqrt{3}-4\big) \approx 0.7587 + . +

    +     + +     + + + Area Between Curves + +

    + Our study of area in the context of rectangular functions led naturally to finding area bounded between curves. + We consider the same in the context of polar functions. + polarfunctions!area between curves +

    + +

    + Consider the shaded region shown in . + We can find the area of this region by computing the area bounded by + r_2=f_2(\theta) and subtracting the area bounded by + r_1=f_1(\theta) on [\alpha,\beta]. + Thus + + \text{ Area } \,= \,\frac12\int_\alpha^\beta r_2^{\,2}\, d\theta - \frac12\int_\alpha^\beta r_1^{\,2}\, d\theta = \frac12\int_\alpha^\beta \big(r_2^{\,2}-r_1^{\,2}\big)\, d\theta + . +

    + +
    + Illustrating area bound between two polar curves + + + Illustration of a region bounded by two polar curves and two rays. + +

    + Polar curves r=f_1(\theta) and r=f_2(\theta) are drawn in the first quadrant. + The functions are not specified, but the curves are intended to appear as generic polar curves: + both begin on the positive x axis and end on the positive y axis, + and both are somewhat wavy, with a value of r that oscillates on the interval [0,\pi/2]. +

    + +

    + The curve r=f_1(\theta) lies closer to the origin than r=f_2(\theta) at all points. + Two rays \theta=\alpha and \theta=\beta are shown. + The region bounded by the two rays and the two polar curves is shaded. + (The region corresponds to \alpha\leq\theta\leq \beta and f_1(\theta)\leq r\leq f_2(\theta).) +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=1.2, + xmin=-.1,xmax=1.2 + ] + + \addplot [firstcurvestyle,areastyle] coordinates { + (0.866,0.5)(0.8626,0.5286)(0.8577,0.557) + (0.8509,0.5848)(0.8417,0.6116)(0.83,0.6369) + (0.8155,0.6604)(0.7983,0.6818)(0.7785,0.7009) + (0.7563,0.7177)(0.7321,0.7321)(0.7063,0.7443) + (0.6795,0.7546)(0.652,0.7634)(0.6244,0.7711) + (0.5971,0.7782)(0.5705,0.7852)(0.5449,0.7928) + (0.5204,0.8013)(0.4971,0.8112)(0.475,0.8227) + (0.35,0.6062)(0.3617,0.5902)(0.3728,0.5741) + (0.3836,0.5582)(0.3942,0.5425)(0.4046,0.5273) + (0.4151,0.5126)(0.4257,0.4984)(0.4366,0.4849) + (0.4479,0.4719)(0.4596,0.4596)(0.4719,0.4479) + (0.4849,0.4366)(0.4984,0.4257)(0.5126,0.4151) + (0.5273,0.4046)(0.5425,0.3942)(0.5582,0.3836) + (0.5741,0.3728)(0.5902,0.3617)(0.6062,0.35)}; + + \addplot [firstcurvestyle,domain=0:90,samples=60] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}); + \addplot [secondcurvestyle,solid,domain=0:90,samples=40] ({cos(x)*(.7+.05*sin(6*x))},{sin(x)*(.7+.05*sin(6*x))}); + + \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs: 0.866025, 0.5) node [pos=.4,below,rotate=30,black] { $\theta=\alpha$}; + \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs:0.475, 0.822724) node [pos=.4,above,rotate=60,black] { $\theta=\beta$}; + + \draw (axis cs:.8,.85) node { $r_2=f_2(\theta)$} + (axis cs:.2,.8) node { $r_1=f_1(\theta)$}; + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
    + + + Area Between Polar Curves +

    + The area A of the region bounded by + r_1=f_1(\theta) and r_2=f_2(\theta), + \theta=\alpha and \theta=\beta, + where f_1(\theta)\leq f_2(\theta) on [\alpha,\beta], is + + A = \frac12\int_\alpha^\beta \big(r_2^{\,2}-r_1^{\,2}\big)\, d\theta + . +

    +     + + + + + Area between polar curves + +

    + Find the area bounded between the curves + r=1+\cos(\theta) and r=3\cos(\theta), + as shown in . +

    + +
    + Finding the area between polar curves in + + + A circle and a cardioid enclose a region that is inside the circle but outside the cardioid. + +

    + Two polar curves are plotted. The first is a circle with center at (3/2,0) and radius 3/2. + (The circle intercepts the x axis at (0,0) and (3,0) and is symmetric about the x axis.) + The second curve is a cardioid; it is symmetric about the x axis, + with its cusp at the origin, and a second x intercept at (2,0). +

    + +

    + The two curves intersect at two points on opposite sides of the x axis. + The cardioid covers up a significant portion of the circle, for x between 0 and 2. + The portion of the circle that is not covered by the cardioid is shaded. + This is the region that lies outside of the cardioid, but inside the circle. + It has a shape similar to that of a crescent moon. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-1.6,ymax=1.6, + xmin=-.6,xmax=3.3 + ] + + \addplot [firstcurvestyle,areastyle] coordinates { + (0.75,-1.299)(0.8719,-1.293)(0.9947,-1.273)(1.117,-1.24) + (1.237,-1.194)(1.353,-1.135)(1.464,-1.063)(1.567,-0.9793) + (1.663,-0.884)(1.748,-0.7783)(1.823,-0.6634)(1.885,-0.5406) + (1.935,-0.4113)(1.971,-0.277)(1.993,-0.1393)(2.,0)(1.993,0.1393) + (1.971,0.277)(1.935,0.4113)(1.885,0.5406)(1.823,0.6634) + (1.748,0.7783)(1.663,0.884)(1.567,0.9793)(1.464,1.063) + (1.353,1.135)(1.237,1.194)(1.117,1.24)(0.9947,1.273) + (0.8719,1.293)(0.75,1.299)(0.75,1.299)(0.9381,1.391) + (1.137,1.455)(1.343,1.492)(1.552,1.499)(1.76,1.477) + (1.964,1.427)(2.158,1.348)(2.339,1.244)(2.504,1.115) + (2.649,0.9642)(2.772,0.7949)(2.87,0.6101)(2.942,0.4135) + (2.985,0.2088)(3.,0)(2.985,-0.2088)(2.942,-0.4135) + (2.87,-0.6101)(2.772,-0.7949)(2.649,-0.9642)(2.504,-1.115) + (2.339,-1.244)(2.158,-1.348)(1.964,-1.427)(1.76,-1.477) + (1.552,-1.499)(1.343,-1.492)(1.137,-1.455)(0.9381,-1.391)(0.75,-1.299)}; + + \addplot [firstcurvestyle,domain=0:180,samples=70] ({cos(x)*(3*cos(x))},{sin(x)*(3*cos(x))}); + \addplot [secondcurvestyle,solid,domain=0:360,samples=80] ({cos(x)*(1+cos(x))},{sin(x)*(1+cos(x))}); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
    +     + +

    + We need to find the points of intersection between these two functions. + Setting them equal to each other, we find: + + 1+\cos(\theta) \amp = 3\cos(\theta) + \cos(\theta) \amp =1/2 + \theta \amp = \pm \pi/3 + +

    + +

    + Thus we integrate \frac12\big((3\cos(\theta) )^2-(1+\cos(\theta) )^2\big) on [-\pi/3,\pi/3]. + + \text{ Area } \amp = \frac12\int_{-\pi/3}^{\pi/3} \big((3\cos(\theta) )^2-(1+\cos(\theta) )^2\big)\, d\theta + \amp = \frac12\int_{-\pi/3}^{\pi/3} \big( 8\cos^2(\theta) -2\cos(\theta) -1\big)\, d\theta + \amp = \frac12\big(2\sin(2\theta) - 2\sin(\theta) +3\theta\big)\Bigg|_{-\pi/3}^{\pi/3} + \amp = \pi + . +

    + +

    + Amazingly enough, + the area between these curves has a nice value. +

    +     + +     + + + Area defined by polar curves + +

    + Find the area bounded between the polar curves r=1 and r=2\cos(2\theta), + as shown in . +

    +
    + The region bounded by the functions in + + A zoomed in view of a region bounded by a circle, a rose curve, and the x axis. + +

    + Two polar curves are plotted: a four-leaf rose curve, and a the unit circle. + The image is zoomed in to give a better view of the region whose area is being computed in this example. + One leaf of the rose curve is along the positive x axis. + It intersects the unit circle at a point in the first quadrant. +

    + +

    + A shaded region is also shown. The region is bounded below by the x axis. + To the left of the point where the rose curve intersects the circle, + the region is bounded above by the rose curve. + To the right of this point, until the circle meets the x axis at (1,0), + the region is bounded between the x axis and the circle. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-1.15,ymax=1.15, + xmin=-.5,xmax=2.2 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(1.,0)(0.9986,0.05234)(0.9945,0.1045)(0.9877,0.1564)(0.9781,0.2079)(0.9659,0.2588)(0.9511,0.309)(0.9336,0.3584)(0.9135,0.4067)(0.891,0.454)(0.866,0.5)(0.7742,0.4744)(0.6822,0.443)(0.5907,0.406)(0.5,0.3633)(0.4107,0.3151)(0.3232,0.2617)(0.2379,0.2032)(0.1554,0.1399)(0.07593,0.07205)(0,0)}; + + \addplot [firstcurvestyle,domain=0:360,samples=180] ({cos(x)*(2*cos(2*x))},{sin(x)*(2*cos(2*x))}); + \addplot [secondcurvestyle,solid,domain=0:360,samples=70] ({cos(x)*(1)},{sin(x)*(1)}); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + +
    +     + +

    + We need to find the point of intersection between the two curves. + Setting the two functions equal to each other, we have + + 2\cos(2\theta) = 1 \Rightarrow \cos(2\theta) = \frac12 \Rightarrow 2\theta = \pi/3 \Rightarrow \theta=\pi/6 + . +

    + +
    + Breaking the region bounded by the functions in into its component parts + + A zoomed in view of a polar region, showing it divided into two parts. + +

    + The region in is shown again, but zoomed in even further. + The point at which the rose curve intersects the circle is found to have polar coordinates (1,\pi/6). + The ray \theta=\pi/6 is shown as a dashed line; it divides the region into two parts. +

    + +

    + The first part corresponds to 0\leq \theta\leq \pi/6; + it is the part of the region that lies below the ray \theta=\pi/6. + It is bounded by the x axis, the unit circle, and the ray \theta=\pi/6. + The second part of the region lies above the ray \theta=\pi/6, + corresponding to angles \pi/6\leq \theta\leq \pi/4. + This part of the region lies between the rose curve and the ray \theta=\pi/6. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1, + xmin=-.1,xmax=1.2 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(0.866,0.5)(0.7742,0.4744)(0.6822,0.443)(0.5907,0.406)(0.5,0.3633)(0.4107,0.3151)(0.3232,0.2617)(0.2379,0.2032)(0.1554,0.1399)(0.07593,0.07205)(0,0)}; + \addplot [secondcurvestyle,areastyle] coordinates {(1.,0)(0.9986,0.05234)(0.9945,0.1045)(0.9877,0.1564)(0.9781,0.2079)(0.9659,0.2588)(0.9511,0.309)(0.9336,0.3584)(0.9135,0.4067)(0.891,0.454)(0.866,0.5)(0,0)}; + + \addplot [firstcurvestyle,domain=0:360,samples=230] ({cos(x)*(2*cos(2*x))},{sin(x)*(2*cos(2*x))}); + \addplot [secondcurvestyle,solid,domain=0:360,samples=80] ({cos(x)*(1)},{sin(x)*(1)}); + + \draw [thick,dashed] (axis cs: 0,0) -- (axis cs:.866,.5); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + +
    + +

    + In , + we zoom in on the region and note that it is not really bounded + between two polar curves, + but rather by two polar curves, + along with \theta=0. + The dashed line breaks the region into its component parts. + Below the dashed line, the region is defined by r=1, + \theta=0 and \theta = \pi/6. + (Note: the dashed line lies on the line \theta=\pi/6.) + Above the dashed line the region is bounded by + r=2\cos(2\theta) and \theta =\pi/6. + Since we have two separate regions, + we find the area using two separate integrals. +

    + +

    + Call the area below the dashed line A_1 and the area above the dashed line A_2. + They are determined by the following integrals: + + A_1 = \frac12\int_0^{\pi/6} (1)^2\, d\theta\qquad A_2 = \frac12\int_{\pi/6}^{\pi/4} \big(2\cos(2\theta)\big)^2\, d\theta + . +

    + +

    + (The upper bound of the integral computing A_2 is \pi/4 as + r=2\cos(2\theta) is at the pole when \theta=\pi/4.) +

    + +

    + We omit the integration details and let the reader verify that + A_1 = \pi/12 and A_2 = \pi/12-\sqrt{3}/8; + the total area is A = \pi/6-\sqrt{3}/8. +

    +     + +     +     +     + + + Arc Length +

    + As we have already considered the arc length of curves defined by rectangular and parametric equations, + we now consider it in the context of polar equations. + Recall that the arc length L of the graph defined by the parametric equations x=f(t), + y=g(t) on [a,b] is + arc length + polarfunction!arc length + + L = \int_a^b \sqrt{\fp(t)^2 + \gp(t)^2}\, dt = \int_a^b \sqrt{x'(t)^2+\yp(t)^2}\, dt + . +

    + +

    + Now consider the polar function r=f(\theta). + We again use the identities x=f(\theta)\cos(\theta) and + y=f(\theta)\sin(\theta) to create parametric equations based on the polar function. + We compute x'(\theta) and + \yp(\theta) as done before when computing \frac{dy}{dx}, + then apply Equation. +

    + +

    + The expression x'(\theta)^2+\yp(\theta)^2 can be simplified a great deal; + we leave this as an exercise and state that + + x'(\theta)^2+\yp(\theta)^2 = \fp(\theta)^2+f(\theta)^2 + . +

    + +

    + This leads us to the arc length formula. +

    + + + Arc Length of Polar Curves + +

    + Let r=f(\theta) be a polar function with \fp continuous on [\alpha,\beta], + on which the graph traces itself only once. + The arc length L of the graph on [\alpha,\beta] is + + L = \int_\alpha^\beta \sqrt{\fp(\theta)^2+f(\theta)^2}\, d\theta = \int_\alpha^\beta\sqrt{(r\,')^2+ r^2}\, d\theta + . +

    +     +     + + + Arc length of a limaçon + +

    + Find the arc length of the limaçon r=1+2\sin(\theta). +

    +     + +

    + With r=1+2\sin(\theta), we have r\,' = 2\cos(\theta). + The limaçon is traced out once on [0,2\pi], + giving us our bounds of integration. + Applying , we have + + L \amp = \int_0^{2\pi} \sqrt{(2\cos(\theta))^2+(1+2\sin(\theta))^2}\, d\theta + \amp = \int_0^{2\pi} \sqrt{4\cos^2\theta+4\sin^2\theta +4\sin(\theta)+1}\, d\theta + \amp = \int_0^{2\pi} \sqrt{4\sin(\theta)+5}\, d\theta + \amp \approx 13.3649 + . +

    + +
    + The limaçon in whose arc length is measured + + + A limaçon with an inner loop that is symmetric about the y axis. + +

    + The limaçon r=1+2\sin(\theta) has an inner loop, + and is symmetric about the y axis. + The point of self-intersection is at the origin; + the inner loop meets the y axis again at (0,1), + and the outer loop meets the y axis again at (0,3). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=3.3, + xmin=-2.22,xmax=2.22 + ] + + \addplot+ [domain=0:360,samples=101] ({cos(x)*(1+2*sin(x))},{sin(x)*(1+2*sin(x))}); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
    + +

    + The final integral cannot be solved in terms of elementary functions, + so we resorted to a numerical approximation. + (Simpson's Rule, with n=4, + approximates the value with 13.0608. + Using n=22 gives the value above, + which is accurate to 4 places after the decimal.) +

    +     + +     +     + + + Surface Area +

    + The formula for arc length leads us to a formula for surface area. + The following Theorem is based on . +

    + + + Surface Area of a Solid of Revolution + +

    + Consider the graph of the polar equation r=f(\theta), + where \fp is continuous on [\alpha,\beta], + on which the graph does not cross itself. + surface areasolid of revolution + polarfunction!surface area + integrationsurface area +

    + +

    +

      +
    1. +

      + The surface area of the solid formed by revolving the graph about the initial ray (\theta=0) is: + + \text{ Surface Area } = 2\pi\int_\alpha^\beta f(\theta)\sin(\theta)\sqrt{\fp(\theta)^2+f(\theta)^2}\, d\theta + . +

      +
      2. + +
    2. +

      + The surface area of the solid formed by revolving the graph about the line \theta=\pi/2 is: + + \text{ Surface Area } = 2\pi\int_\alpha^\beta f(\theta)\cos(\theta)\sqrt{\fp(\theta)^2+f(\theta)^2}\, d\theta + . +

      +
      3. +
    +

    +     +     + + + Surface area determined by a polar curve + +

    + Find the surface area formed by revolving one petal of the rose curve + r=\cos(2\theta) about its central axis, + as shown in . +

    +
    + Finding the surface area of a rose-curve petal that is revolved around its central axis + +
    + + + + A four leaf rose curve, with leaves along the coordinate axes. + +

    + The four leaf rose curve r=\cos(2\theta) is plotted. + Although the entire curve is plotted, the portion of the right-hand leaf + that lies in the first quadrant is highlighted. + This is the part of the curve corresponding to 0\leq \theta\leq \pi/4; + it begins at (1,0) and ends at the origin. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,1}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.3,xmax=1.3 + ] + + \addplot+ [domain=0:45,samples=30] ({cos(x)*(cos(2*x))},{sin(x)*(cos(2*x))}); + \addplot+ [solid,domain=45:360,samples=120] ({cos(x)*(cos(2*x))},{sin(x)*(cos(2*x))}); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + +
    + + + + +
    + + + The surface of revolution obtained by revolving one leaf of a rose curve about the x axis. + +

    + A three-dimensional plot shows the surface of revolution obtained when the portion of the rose curve + from the previous image is revolved about the x axis. + The surface is similar in appearance to a teardrop, + or perhaps a blimp. +

    +     + + + + + //ASY file for figparcalc8_3D.asy in Chapter 9 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,-8,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,1.25); + pair ybounds=(-.25,.25); + pair zbounds=(-.25,.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice));//,Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + //label("$x$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$y$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + label("\ ",(0,0,1.5)); + label("\ ",(0,-1.5,0)); + + //surface r=cos(2 theta) revolved around x axis // ({cos(y)*(cos(2*y))},{(sin(y)*(cos(2*y)))*cos(x)},{(sin(y)*(cos(2*y)))*sin(x)}); + triple f(pair t) { + return (cos(t.y)*cos(2*t.y),sin(t.y)*cos(2*t.y)*cos(t.x),sin(t.y)*cos(2*t.y)*sin(t.x)); + } + surface s=surface(f,(0,0),(2pi,pi/4),8,16,Spline); + //Should use apexmeshpen for p but that was causing a core dump somehow + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //curve on the surfaces //({cos(x)*cos(2*x)},{0},{sin(x)*cos(2*x)}); + triple g(real t) {return (cos(t)*cos(2*t),0,sin(t)*cos(2*t));} + path3 mypath=graph(g,0,pi/4,operator ..);draw(mypath,bluepen+linewidth(2)); + + + +
    + +     +
    +     + +

    + We choose, as implied by the figure, + to revolve the portion of the curve that lies on + [0,\pi/4] about the initial ray. + Using + and the fact that \fp(\theta) = -2\sin(2\theta), we have + + \text{ Surface Area } \amp = 2\pi\int_0^{\pi/4} \cos(2\theta)\sin(\theta)\sqrt{\big(-2\sin(2\theta)\big)^2+\big(\cos(2\theta)\big)^2}\, d\theta + \amp \approx 1.36707 + . +

    + +

    + The integral is another that cannot be evaluated in terms of elementary functions. + Simpson's Rule, with n=4, + approximates the value at 1.36751. +

    +     +     + +

    + This chapter has been about curves in the plane. + While there is great mathematics to be discovered in the two dimensions of a plane, + we live in a three dimensional world and hence we should also look to do mathematics in 3D that is, + in space. + The next chapter begins our exploration into space by introducing the topic of vectors, + which are incredibly useful and powerful mathematical objects. +

    +     + + + + Terms and Concepts + + + +

    + Given polar equation r=f(\theta), + how can one create parametric equations of the same curve? +

    +     + + + +

    + Using x=r\cos(\theta) and y=r\sin(\theta), + we can write x=f(\theta)\cos(\theta), + y=f(\theta)\sin(\theta). +

    +     + +     + + + + +

    + With rectangular coordinates, + it is natural to approximate area with ; + with polar coordinates, it is natural to approximate area with . +

    +     + + + + + + + + + wedges|circular wedges|sectors of circles|sections of circles|circular sectors|circular sections + + + + +     +     + + Problems + + + +

    + Find \lz{y}{x} + (in terms of \theta). + Then find the equations of the tangent and normal lines to the curve at the indicated \theta-value. +

    +     + + + + + + + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $dydx=Formula("-cot(theta)"); + Context("Numeric")->variables->add(y=>'Real'); + parser::Assignment->Allow; + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $tan=Formula("y=-(x-sqrt(2)/2)+sqrt(2)/2"); + $nor=Formula("y=x"); + + +

    + r=1, \theta=\pi/4 +

    + + + Find \lz{y}{x} in terms of \theta. + Type theta for \theta. + +

    + +

    + + + Give the equation of the tangent line. + + +

    + +

    + + + Give the equation of the normal line. + + +

    + +

    +     +     +     + + + + + + + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $dydx=Formula("1/2(tan(theta)-cot(theta))"); + Context("Numeric")->variables->add(y=>'Real'); + parser::Assignment->Allow; + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $tan=Formula("y=1/2"); + $nor=Formula("x=1/2"); + + +

    + r=\cos(\theta), \theta=\pi/4 +

    + + + Find \lz{y}{x} in terms of \theta. + Type theta for \theta. + +

    + +

    + + + Give the equation of the tangent line. + + +

    + +

    + + + Give the equation of the normal line. + + +

    + +

    +     +     +     + + + + +

    + r=1+\sin(\theta), \theta = \pi/6 +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = \frac{\cos(\theta) (1+2\sin(\theta) )}{\cos^2(\theta) -\sin(\theta) (1+\sin(\theta) )} +

      +
      2. + +
    2. +

      + tangent line: x=3\sqrt{3}/4; + normal line: y=3/4 +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds r=1-3\cos(\theta), \theta = 3\pi/4 +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = \frac{3 \sin^2(t)+(1-3 \cos(t)) \cos(t)}{3 \sin(t) \cos(t)-\sin(t) (1-3 \cos(t))} +

      +
      2. + +
    2. +

      + tangent line: + y=\frac{1}{1+3\sqrt{2}}(x+(1/\sqrt{2}+3/2))+1/\sqrt{2}+3/2 \approx y=0.19(x+2.21)+2.21; + normal line: + y=-(1+3\sqrt{2})(x+(1/\sqrt{2}+3/2))+1/\sqrt{2}+3/2 +

      +
      3. +
    +

    +     + +     + + + + + + + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $dydx=Formula("(theta cos(theta) + sin(theta))/(cos(theta)-theta sin(theta))"); + Context("Numeric")->variables->add(y=>'Real'); + parser::Assignment->Allow; + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $tan=Formula("y=-2/pi x + pi/2"); + $nor=Formula("y=pi/2 x + pi/2"); + + +

    + r=\theta, \theta=\pi/2 +

    + + + Find \lz{y}{x} in terms of \theta. + Type theta for \theta. + +

    + +

    + + + Give the equation of the tangent line. + + +

    + +

    + + + Give the equation of the normal line. + + +

    + +

    +     +     +     + + + + + + + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $dydx=Formula("(cos(theta)cos(3theta)-3sin(theta)sin(3theta))/(-cos(3theta)sin(theta)-3cos(theta)sin(3theta))"); + Context("Numeric")->variables->add(y=>'Real'); + parser::Assignment->Allow; + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $tan=Formula("y=x/sqrt(3)"); + $nor=Formula("y=-sqrt(3)x"); + + +

    + r=\cos(3\theta), \theta=\pi/6 +

    + + + Find \lz{y}{x} in terms of \theta. + Type theta for \theta. + +

    + +

    + + + Give the equation of the tangent line. + + +

    + +

    + + + Give the equation of the normal line. + + +

    + +

    +     +     +     + + + + + + + Context()->variables->are(theta=>['Real',TeX=>'\theta']); + $dydx=Formula("(4sin(theta)cos(4theta)+sin(4theta)cos(theta))/(4cos(theta)cos(4theta)-sin(theta)sin(4theta))"); + Context("Numeric")->variables->add(y=>'Real'); + parser::Assignment->Allow; + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $tan=Formula("y=5sqrt(3)(x+sqrt(3)/4)-3/4"); + $nor=Formula("y=-1/(5sqrt(3))(x+sqrt(3)/4)-3/4"); + + +

    + r=\sin(4\theta), \theta=\pi/3 +

    + + + Find \lz{y}{x} in terms of \theta. + Type theta for \theta. + +

    + +

    + + + Give the equation of the tangent line. + + +

    + +

    + + + Give the equation of the normal line. + + +

    + +

    +     +     +     + + + + +

    + \ds r=\frac1{\sin(\theta) -\cos(\theta) };\theta = \pi +

    +     + +

    +

      +
    1. +

      + \frac{dy}{dx} = 1 +

      +
      2. + +
    2. +

      + tangent line: y=x+1; normal line: y=-x-1 +

      +
      3. +
    +

    +     + +     + +     + + + +

    + Find the values of \theta in the given interval where the graph of the polar function has horizontal and vertical tangent lines. +

    +     + + + + +

    + \ds r=3; [0,2\pi] +

    +     + +

    + horizontal: \theta=\pi/2,3\pi/2; +

    + +

    + vertical: \theta = 0,\pi,2\pi +

    +     + +     + + + + +

    + \ds r=2\sin(\theta); [0,\pi] +

    +     + +

    + horizontal: \theta=0,\pi/2,\pi; +

    + +

    + vertical: \theta = \pi/4,3\pi/4 +

    +     + +     + + + + +

    + \ds r=\cos(2\theta); [0,2\pi] +

    +     + +

    + horizontal: \theta=\tan^{-1}(1/\sqrt{5}),\,\pi/2,\,\pi-\tan^{-1}(1/\sqrt{5}),\,\pi+\tan^{-1}(1/\sqrt{5}),\,3\pi/2,\,2\pi-\tan^{-1}(1/\sqrt{5}); +

    + +

    + vertical: \theta = 0,\,\tan^{-1}(\sqrt{5}),\,\pi-\tan^{-1}(\sqrt{5}),\,\pi,\,\pi+\tan^{-1}(\sqrt{5}),\,2\pi-\tan^{-1}(\sqrt{5}) +

    +     + +     + + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); + $hor=List(Formula("pi/3"), Formula("pi"), Formula("5pi/3")); + $ver=List(Formula("0"), Formula("2pi/3"), Formula("4pi/3")); + + +

    + r=1+\cos(\theta); [0, 2\pi) +

    + + + Give the values of \theta where the tangent line is horizontal. + Enter your answer as a comma-separated list. + +

    + +

    + + + Give the values of \theta where the tangent line is vertical. + Enter your answer as a comma-separated list. + +

    + +

    +     + +

    + horizontal: \theta=\pi/3,\,5\pi/3; +

    + +

    + vertical: \theta = 0,\,2\pi/3,\,4\pi/3,\,2\pi +

    + +

    + At \theta=\pi, \frac{dy}{dx} = 0/0; + apply L'Hopital's Rule to find that + \frac{dy}{dx}\rightarrow 0 as \theta\rightarrow \pi. +

    +     +     +     + +     + + + +

    + Find the equation of the lines tangent to the graph at the pole. +

    +     + + + + +

    + \ds r=\sin(\theta);[0,\pi] +

    +     + +

    + In polar: \theta = 0 \,\cong \,\theta = \pi +

    + +

    + In rectangular: y=0 +

    +     + +     + + + + +

    + \ds r=\sin(3\theta);[0,\pi] +

    +     + +

    + In polar: \theta = 0,\,\theta = \pi/3,\,\theta = 2\pi/3. +

    + +

    + In rectangular: y=0, + y=\sqrt{3}x, and y= -\sqrt{3}x. +

    +     + +     + +     + + + +

    + Find the area of the described region. +

    +     + + + + +

    + Enclosed by the circle: \ds r=4\sin(\theta) +

    +     + +

    + area = 4\pi +

    +     + +     + + + + +

    + Enclosed by the circle \ds r=5 +

    +     + +

    + area = 25\pi +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); + $area=Formula("pi/12"); + + +

    + Find the area enclosed by one petal of r=\sin(3\theta). +

    + +

    + +

    +     +     +     + + + +

    + Enclosed by one petal of the rose curve r=\cos (n\,\theta), + where n is a positive integer. +

    +     + +

    + area = \pi/(4n) +

    +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); + $area=Formula("3pi/2"); + + +

    + Find the area enclosed by the cardioid r=1-\sin(\theta). +

    + +

    + +

    +     +     +     + + + + + +

    + Enclosed by the inner loop of the limaçon \ds r=1+2\cos(\theta) +

    +     + +

    + area = \pi-3\sqrt{3}/2 +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); + $area=Formula("2pi+3sqrt(3)/2"); + + + +

    + Find the area enclosed by the outer loop of the limaçon r=1+2\cos(\theta) + (including area enclosed by the inner loop). +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); + $area=Formula("pi+3sqrt(3)"); + + + +

    + Find the area enclosed between the inner and outer loop of the limaçon r=1+2\cos(\theta). +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); + $area=Formula("1"); + + +

    + Find the area enclosed by r=2\cos(\theta), + r=2\sin(\theta), and the x-axis, as shown: +

    + + + Two circles, one along each axis, bound a region in the first quadrant. + +

    + There are two circles of radius 1. + The first circle lies along the x axis, with intercepts at (0,0) and (2,0). + The second circle lies along the y axis, with intercepts at (0,0) and (0,2). + The two circles intersect in the first quadrant, at the point (1,1). +

    + +

    + The shaded region consists of everything that is between the x axis and the first circle, + except for the portion of the first circle that overlaps with the second circle. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.1,ymax=2.1,% + xmin=-1.4,xmax=2.4% + ] + + \addplot [firstcurvestyle,areastyle,domain=0:45,samples=20] ({cos(x)*2*cos(x)},{sin(x)*2*cos(x)}) -- (axis cs:0,0); + \addplot [fill=white, smooth,domain=0:50,samples=20] ({cos(x)*2*sin(x)},{sin(x)*2*sin(x)}); + \addplot [firstcurvestyle,domain=0:180,samples=60] ({cos(x)*2*cos(x)},{sin(x)*2*cos(x)}); + \addplot [secondcurvestyle,domain=0:180,samples=40] ({cos(x)*2*sin(x)},{sin(x)*2*sin(x)}); + + \end{axis} + \end{tikzpicture} + + + +

    + The area is . +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); + $area=Formula("1/32(4pi-3sqrt(3))"); + + +

    + Find the area enclosed by r=\cos(\theta) and r=\sin(2\theta), + as shown: +

    + + + A semi-circle and one leaf of a rose curve, both in the first quadrant, overlapping in a shaded area. + +

    + Two curves are shown. One is a semi-circular arc, from (0,0) to (1,0), in the first quadrant. + The other curve the leaf of the four-leaf rose r=\sin(2\theta) that lies in the first quadrant. +

    + +

    + The two curves overlap in a region that lies above the rose curve, but below the circle. + This region is shaded. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xtick={1}, + ytick={1}, + ymin=-.1,ymax=1.1,% + xmin=-.1,xmax=1.34% + ] + + \addplot [firstcurvestyle,areastyle,domain=90:30,samples=20] ({cos(x)*cos(x)},{sin(x)*cos(x)}) -- (axis cs:0,0); + \addplot [firstcurvestyle,areastyle,domain=0:30,samples=20] ({cos(x)*sin(2*x)},{sin(x)*sin(2*x)}) -- (axis cs:0,0); + \addplot [firstcurvestyle,domain=0:90,samples=60] ({cos(x)*cos(x)},{sin(x)*cos(x)}); + \addplot [secondcurvestyle,domain=0:90,samples=40] ({cos(x)*sin(2*x)},{sin(x)*sin(2*x)}); + + \end{axis} + \end{tikzpicture} + + + +

    + The area is . +

    +     +     +     + + + + +

    + Enclosed by r=\cos(3 \theta) and r=\sin(3\theta), as shown: +

    + + + Overlapping leaves of two different three-leaf rose curves. + +

    + Two curves are shown. The first is the leaf of the three-leaf rose curve r=\sin(3\theta) that lies in the first quadrant. + The second is the leaf of the three-leaf rose curve r=\cos(3\theta) that lies along the positive x axis. + The shaded region is the interior of the leaf of r=\sin(3\theta), + except for the part that overlaps with the leaf of r=\cos(3\theta). +

    +     + + \begin{tikzpicture} + \begin{axis}[ + xtick={1}, + ytick={.5}, + ymin=-.25,ymax=.7,% + xmin=-.08,xmax=1.1% + ] + + \addplot [firstcurvestyle,areastyle,domain=15:60,samples=20] ({cos(x)*sin(3*x)},{sin(x)*sin(3*x)}) -- (axis cs:0,0); + \addplot [white,fill=white, smooth,domain=0:30,samples=20] ({cos(x)*cos(3*x)},{sin(x)*cos(3*x)}); + \addplot [firstcurvestyle, domain=0:60,samples=60] ({cos(x)*sin(3*x)},{sin(x)*sin(3*x)}); + \addplot [secondcurvestyle,domain=-30:30,samples=60] ({cos(x)*cos(3*x)},{sin(x)*cos(3*x)}); + + \end{axis} + \end{tikzpicture} + + +     + +

    + area = \ds \int_{\pi/12}^{\pi/3} \frac12 \sin^2(3\theta)\,d\theta - \int_{\pi/12}^{\pi/6}\frac12\cos^2(3\theta)\,d\theta = \frac1{12}+\frac{\pi}{24} +

    +     + +     + + + + +

    + Enclosed by r=\cos(\theta) and r=1-\cos(\theta), as shown: +

    + + + A cardioid and a circle, and a region of overlap in the first quadrant. + +

    + Two polar curves are plotted. The first is a circle with center (1/2,0) and radius 1/2; + it is symmetric about the x axis and passes through (0,0) and (1,0). +

    + +

    + The second curve is a cardioid. It is larger than the circle, and points in the opposite direction. + That is, it is symmetric about the x axis, with its cusp at the origin, + but its other x intercept lies on the negative x axis, at (-2,0). +

    + +

    + The two curves overlap in a small region in the first quadrant, which is shaded. + The region extends from the origin to the point in the first quadrant where the two curves intersect. + It is bounded above (and to the left) by the circle, and below (and to the right) by the cardioid. +

    +     + + \begin{tikzpicture} + \begin{axis}[ + ymin=-1.4,ymax=1.4,% + xmin=-2.2,xmax=1.2% + ] + + \addplot [firstcurvestyle,areastyle,domain=60:90,samples=60] ({cos(x)*cos(x)},{sin(x)*cos(x)}); + \addplot [firstcurvestyle,areastyle,domain=0:60,samples=40] ({cos(x)*(1-cos(x))},{sin(x)*(1-cos(x))}); + \addplot [firstcurvestyle,domain=0:180,samples=60] ({cos(x)*cos(x)},{sin(x)*cos(x)}); + \addplot [secondcurvestyle,domain=0:360,samples=60] ({cos(x)*(1-cos(x))},{sin(x)*(1-cos(x))}); + + \end{axis} + \end{tikzpicture} + + +     + +

    + area = \ds \int_{0}^{\pi/3} \frac12 (1-\cos(\theta) )^2\,d\theta +\int_{\pi/3}^{\pi/2} \frac12 (\cos(\theta) )^2\,d\theta =\frac{7\pi}{24}-\frac{\sqrt{3}}2\approx 0.0503 +

    +     + +     + +     + + + +

    + Answer the questions involving arc length. +

    +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); + $length=Formula("4pi"); + + +

    + Use the arc length formula to compute the arc length of the circle r=2. +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); + $length=Formula("4pi"); + + +

    + Use the arc length formula to compute the arc length of the circle r=4\sin(\theta). +

    + +

    + +

    +     +     +     + + + +

    + Use the arc length formula to compute the arc length of r=\cos(\theta)+\sin(\theta). +

    +     + +

    + \sqrt{2}\pi +

    +     +     + + + +

    + Use the arc length formula to compute the arc length of the cardioid r=1+\cos(\theta). + (Hint: apply the formula, simplify, + then use a Power-Reducing Formula to convert 1+\cos(\theta) into a square.) +

    +     + +

    + 8 +

    +     +     + + + + + $length=OneOf("2.2592,2.22748"); + + +

    + Approximate the arc length of one petal of the rose curve + r=\sin(3\theta) with Simpson's Rule and n=4. +

    + +

    + +

    +     +     +     + + + + + + + +

    + Let x(\theta) = f(\theta)\cos(\theta) and y(\theta)=f(\theta)\sin(\theta). + Show, as suggested by the text, that + + x\,'(\theta)^2+y\,'(\theta)^2 = \fp(\theta)^2+f(\theta)^2 + . +

    +     + +

    + x\,'(\theta) = \fp(\theta)\cos(\theta) -f(\theta)\sin(\theta), + y\,'(\theta) = \fp(\theta)\sin(\theta) + f(\theta)\cos(\theta). + Square each and add; + applying the Pythagoran Theorem twice achieves the result. +

    +     + +     + + +     + + + +

    + Answer the questions involving surface area. +

    +     + + + + +

    + Use + to find the surface area of the sphere formed by revolving the circle r=2 about the initial ray. +

    +     + +

    + SA = 16\pi +

    +     + +     + + + + +

    + Use + to find the surface area of the sphere formed by revolving the circle + r=2\cos(\theta) about the initial ray. +

    +     + +

    + SA = 4\pi +

    +     + +     + + + + +

    + Find the surface area of the solid formed by revolving the cardioid + r=1+\cos(\theta) about the initial ray. +

    +     + +

    + SA = 32\pi/5 +

    +     + +     + + + + +

    + Find the surface area of the solid formed by revolving the circle + r=2\cos(\theta) about the line \theta=\pi/2. +

    +     + +

    + SA = 4\pi^2 +

    +     + +     + + + + +

    + Find the surface area of the solid formed by revolving the line r=3\sec(\theta), + -\pi/4\leq\theta\leq\pi/4, + about the line \theta=\pi/2. +

    +     + +

    + SA = 36\pi +

    +     + +     + + + +

    + Find the surface area of the solid formed by revolving the line r=3\sec\theta, + 0\leq\theta\leq\pi/4, about the initial ray. +

    +     + +

    + SA = 9\pi +

    +     +     + +     +     +     +
    +     + + + Vectors + + + +

    + This chapter begins with moving our mathematics out of the plane and into space. That is, + we begin to think mathematically not only in two dimensions, + but in three. + With this foundation, + we can explore vectors both in the plane and in space. +

    +     + +
    + Introduction to Cartesian Coordinates in Space + +

    + Up to this point in this text we have considered mathematics in a 2-dimensional world. + We have plotted graphs on the xy-plane using rectangular and polar coordinates and found the area of regions in the plane. + We have considered properties of solid objects, + such as volume and surface area, + but only by first defining a curve in the plane and then rotating it out of the plane. +

    + +

    + While there is wonderful mathematics to explore in + 2D, we live in a 3D + world and eventually we will want to apply mathematics involving this third dimension. + In this section we introduce Cartesian coordinates in space and explore basic surfaces. + This will lay a foundation for much of what we do in the remainder of the text. +

    + + + +

    + Each point P in space can be represented with an ordered triple, + P=(a,b,c), where a, + b and c represent the relative position of P along the x-, y- and z-axes, + respectively. + Each axis is perpendicular to the other two. +

    + +

    + Visualizing points in space on paper can be problematic, + as we are trying to represent a 3-dimensional concept on a 2-dimensional medium. + We cannot draw three lines representing the three axes in which each line is perpendicular to the other two. + Despite this issue, + standard conventions exist for plotting shapes in space that we will discuss that are more than adequate. +

    + +

    + One convention is that the axes must conform to the + right hand rule. + This rule states that when the index finger of the right hand is extended in the direction of the positive x-axis, + and the middle finger + (bent inward so it is perpendicular to the palm) + points along the positive y-axis, + then the extended thumb will point in the direction of the positive z-axis. (It may take some thought to verify this, + but this system is inherently different from the one created by using the + left hand rule.)right hand ruleof Cartesian coordinates +

    + +

    + As long as the coordinate axes are positioned so that they follow this rule, + it does not matter how the axes are drawn on paper. + There are two popular methods that we briefly discuss. +

    + +
    + Plotting the point P=(2,1,3) in space + + + + Plot of point P = (2, 1, 3) in space. + + +

    + The x and y axes are drawn from -2 to 2 and the z + axis is drawn from -1 to 3. The point P= (2,1,3) is drawn in space. + A dashed cuboid is drawn with one vertex at the origin, three of its edges along the coordinate axes, + and the point P on the corner opposite the origin. +

    +     + + + + + //ASY file for figcartcoord1.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11,5,2.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1,2,3}; + real[] myychoice={-2,-1,1,2}; + real[] myzchoice={-2,-1,1,2}; + defaultpen(0.5mm); + pair xbounds=(-3,3.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$z$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$x$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$y$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // Draw the lines + draw((0,0,1)--(0,2,1)--(3,2,1)--(3,0,1)--(0,0,1), redpen+dashed+linewidth(.5));//top + draw((0,0,0)--(0,2,0)--(3,2,0)--(3,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//bottom + draw((3,0,0)--(3,0,1), redpen+dashed+linewidth(.5));//up1 + draw((0,2,0)--(0,2,1), redpen+dashed+linewidth(.5));//up2 + draw((3,2,0)--(3,2,1), redpen+dashed+linewidth(.5));//up3 + label("$P$",(3,2,1),E); + dotfactor=3; dot((3,2,1),bluepen); + + + + +
    + +

    + In + we see the point P=(2,1,3) plotted on a set of axes. + The basic convention here is that the xy-plane is drawn in its standard way, + with the z-axis down to the left. + The perspective is that the paper represents the xy-plane and the positive z axis is coming up, + off the page. + This method is preferred by many engineers. + Because it can be hard to tell where a single point lies in relation to all the axes, + dashed lines have been added to let one see how far along each axis the point lies. +

    + +

    + One can also consider the xy-plane as being a horizontal plane in, say, + a room, where the positive z-axis is pointing up. + When one steps back and looks at this room, + one might draw the axes as shown in . + The same point P is drawn, again with dashed lines. + This point of view is preferred by most mathematicians, + and is the convention adopted by this text. +

    + +

    + Just as the x- and y-axes divide the plane into four quadrants, + the x-, y-, and z-coordinate planes divide space into eight octants. + The octant in which x, y, + and z are positive is called the first octant. + We do not name the other seven octants in this text. + octantfirst + first octant +

    + +
    + Plotting the point P=(2,1,3) in space with a perspective used in this text + + + + Plot of point P = (2,1,3) in space with perspective used in the text. + + +

    + The x and y axes are drawn from -2 to 2 and the z + axis is drawn from -1 to 3. The point P= (2,1,3) is drawn in space. + A dashed cuboid is drawn with one vertex at the origin, three of its edges along the coordinate axes, + and the point P on the corner opposite the origin. + The z axis is shown as the vertical line. +

    +     + + + + + //ASY file for figcartcoord2.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={-2,3}; + defaultpen(0.5mm); + pair xbounds=(-2,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-1.5,4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // Draw the lines + //draw((2,0,0)--(2,1,0), redpen+dashed+linewidth(.5)); + //draw((2,1,0)--(2,1,3)--(0,0,3), redpen+dashed+linewidth(.5)); + //draw((0,1,0)--(2,1,0), redpen+dashed+linewidth(.5)); + draw((0,0,3)--(0,1,3)--(2,1,3)--(2,0,3)--(0,0,3), redpen+dashed+linewidth(.5));//top + draw((0,0,0)--(0,1,0)--(2,1,0)--(2,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//top + draw((2,0,0)--(2,0,3), redpen+dashed+linewidth(.5));//up1 + draw((0,1,0)--(0,1,3), redpen+dashed+linewidth(.5));//up2 + draw((2,1,0)--(2,1,3), redpen+dashed+linewidth(.5));//up + label("$P$",(2,1,3),E); + dotfactor=3; dot((2,1,3),bluepen); + + + + +
    +     + + + Measuring Distances +

    + It is of critical importance to know how to measure distances between points in space. + The formula for doing so is based on measuring distance in the plane, + and is known + (in both contexts) + as the Euclidean measure of distance. +

    + + + Distance In Space + +

    + Let P=(x_1,y_1,z_1) and Q = (x_2,y_2,z_2) be points in space. + The distance D between P and Q is distancebetween points in space + + D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} + . +

    +     +     + + + +

    + We refer to the line segment that connects points P and Q in space as \overline{PQ}, + and refer to the length of this segment as \norm{\overline{PQ}}. + The above distance formula allows us to compute the length of this segment. +

    + + + + + Length of a line segment + +

    + Let P = (1,4,-1) and let Q = (2,1,1). + Draw the line segment \overline{PQ} and find its length. +

    +     + +

    + The points P and Q are plotted in ; + no special consideration need be made to draw the line segment connecting these two points; + simply connect them with a straight line. + One cannot actually measure this line on the page and deduce anything meaningful; + its true length must be measured analytically. + Applying , we have + + \norm{\overline{PQ}} = \sqrt{(2-1)^2+(1-4)^2+(1-(-1))^2} = \sqrt{14}\approx 3.74 + . +

    + +
    + Plotting points P and Q in + + + + Plotting point P = (1, 4, -1) and point Q = (2, 1, 1). + + +

    + The x axis is drawn from 0 to 2, the z axis is + drawn from -2 to 2 and the y axis is drawn from 0 + to 4. Two points P = (1, 4, -1) and Q = (2, 1, 1) are + drawn in space and are connected by a straight line. +

    +     + + + + + //ASY file for figspace1.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,1); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2,4}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + pair xbounds=(-1,2.5); + pair ybounds=(-1,4.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // Draw the lines for Q=(2,1,1) + draw((2,0,0)--(2,1,0)--(0,1,0), redpen+dashed+linewidth(.5));//top Q + //draw((0,0,1)--(0,1,1)--(2,1,1)--(2,0,1)--(0,0,1), redpen+dashed+linewidth(.5));//top Q + //draw((0,0,0)--(0,1,0)--(2,1,0)--(2,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//bottom Q + //draw((2,0,0)--(2,0,1), redpen+dashed+linewidth(.5));//up1 Q + //draw((0,1,0)--(0,1,1), redpen+dashed+linewidth(.5));//up2 Q + draw((2,1,0)--(2,1,1), redpen+dashed+linewidth(.5));//up3 Q + label("$Q$",(2,1,1),E); + dotfactor=3; dot((2,1,1),bluepen); + + // Draw the lines for P=(1,4,-1) + //draw((0,0,-1)--(0,4,-1)--(1,4,-1)--(1,0,-1)--(0,0,-1), redpen+dashed+linewidth(.5));//top P + //draw((0,0,0)--(0,4,0)--(1,4,0)--(1,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//bottom P + //draw((1,0,0)--(1,0,-1), redpen+dashed+linewidth(.5));//up1 P + //draw((0,4,0)--(0,4,-1), redpen+dashed+linewidth(.5));//up2 P + draw((0,4,0)--(1,4,0)--(1,0,0), redpen+dashed+linewidth(.5));//bottom P + draw((1,4,0)--(1,4,-1), redpen+dashed+linewidth(.5));//up3 P + label("$P$",(1,4,-1),E); + dotfactor=3; dot((1,4,-1),bluepen); + + //line from P to Q + draw((1,4,-1)--(2,1,1), bluepen); + + + + +
    +     + +     +     + + + Spheres +

    + Just as a circle is the set of all points in the plane equidistant from a given point (its center), a sphere is the set of all points in space that are equidistant from a given point. + + allows us to write an equation of the sphere. + sphere +

    + +

    + We start with a point C = (a,b,c) which is to be the center of a sphere with radius r. + If a point P=(x,y,z) lies on the sphere, + then P is r units from C; that is, + + \norm{\overline{PC}} = \sqrt{(x-a)^2+(y-b)^2+(z-c)^2} = r + . +

    + +

    + Squaring both sides, + we get the standard equation of a sphere in space with center at + C=(a,b,c) with radius r, + as given in the following Key Idea. +

    + + + Standard Equation of a Sphere in Space +

    + The standard equation of the sphere with radius r, + centered at C=(a,b,c), is + + (x-a)^2+(y-b)^2+(z-c)^2=r^2 + . +

    +     + + + + + Equation of a sphere + +

    + Find the center and radius of the sphere defined by x^2+2x+y^2-4y+z^2-6z=2. +

    +     + +

    + To determine the center and radius, + we must put the equation in standard form. + This requires us to complete the square + (three times). + + x^2+2x+y^2-4y+z^2-6z\amp =2 + (x^2+2x+1) + (y^2-4y+4)+ (z^2-6z+9) - 14 \amp = 2 + (x+1)^2 + (y-2)^2 + (z-3)^2 \amp = 16 + +

    + +

    + The sphere is centered at (-1,2,3) and has a radius of 4. +

    +     + +     + +

    + The equation of a sphere is an example of an implicit function defining a surface in space. + In the case of a sphere, + the variables x, y and z are all used. + We now consider situations where surfaces are defined where one or two of these variables are absent. +

    +     + + + Introduction to Planes in Space +

    + The coordinate axes naturally define three planes + (shown in ), + the coordinate planes: + the xy-plane, + the yz-plane and the xz-plane. + The xy-plane is characterized as the set of all points in space where the z-value is 0. + planescoordinate plane + planesintroduction + This, in fact, gives us an equation that describes this plane: z=0. + Likewise, the xz-plane is all points where the y-value is 0, characterized by y=0. +

    + + + +
    + The xy-plane in (a), the yz-plane in (b) and the xz-plane in (c) + +
    + + + + + Graph of xy plane in space. + + +

    + The axes are uncalibrated. The plane is drawn along the x and y axes + at z=0, the normal is along the z axis. +

    +     + + + + + //ASY file for figspacexy.pdf in Chapter 10 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the xy-plane + triple f(pair t) { + return (t.x,t.y,0); + } + surface s=surface(f,(-2,-2),(2,2),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + + + +
    + +
    + + + + + Graph of yz plane in space. + + +

    + The axes are uncalibrated. The plane is drawn along the y and z + axes at x=0, the normal is along the x axis. +

    +     + + + + + //ASY file for figspaceyz.pdf in Chapter 10 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the yz-plane + triple f(pair t) { + return (0,t.x,t.y); + } + surface s=surface(f,(-2,-2),(2,2),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + + + +
    + +
    + + + + + Graph of xz plane in space. + + +

    + The axes are uncalibrated. The plane is drawn along the x and z axes at + y =0, the normal is along the y axis. +

    +     + + + + + //ASY file for figspacexz.pdf in Chapter 10 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the xz-plane + triple f(pair t) { + return (t.x,0,t.y); + } + surface s=surface(f,(-2,-2),(2,2),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + + + +
    +     +
    + +

    + The equation x=2 describes all points in space where the x-value is 2. + This is a plane, + parallel to the yz-coordinate plane, + shown in . +

    + +
    + The plane x=2 + + + + Graph of plane x=2 in space. + + +

    + The y and the z axes are uncalibrated, the x axis is + drawn from -2 to 2. The yz plane is drawn at x=-2. +

    +     + + + + + //ASY file for figspace2.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2.5,3.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the x=2 plane + triple f(pair t) { + return (2,t.x,t.y); + } + surface s=surface(f,(-2,-2),(2,2),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + + + +
    + + + Regions defined by planes + +

    + Sketch the region defined by the inequalities -1\leq y\leq 2. +

    +     + +

    + The region is all points between the planes y=-1 and y=2. + These planes are sketched in , + which are parallel to the xz-plane. + Thus the region extends infinitely in the x and z directions, + and is bounded by planes in the y direction. +

    + +
    + Sketching the boundaries of a region in + + + + Graph of two planes that are parallel to the xz plane. + + +

    + The y and z axes are uncalibrated, the x axis is drawn + from -2 and 2. Two planes are drawn parallel to the xz + plane at y=-1 and y =2, with normal along the y axis. +

    +     + + + + + //ASY file for figspace3.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,2.5,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the y=-1 plane + triple f(pair t) { + return (t.x,-1,t.y); + } + surface s=surface(f,(-2,-2),(2,2),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + //Draw the y=2 plane + triple f(pair t) { + return (t.x,2,t.y); + } + surface s=surface(f,(-2,-2),(2,2),8,8,Spline); + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + + + +
    +     + +     +     + + + Cylinders +

    + The equation x=1 obviously lacks the y and z variables, + meaning it defines points where the y and z coordinates can take on any value. + Now consider the equation x^2+y^2=1 + in space. In the plane, + this equation describes a circle of radius 1, centered at the origin. + In space, the z coordinate is not specified, + meaning it can take on any value. + In , + we show part of the graph of the equation + x^2+y^2=1 by sketching 3 circles: + the bottom one has a constant z-value of -1.5, + the middle one has a z-value of 0 and the top circle has a z-value of 1. + By plotting all possible z-values, + we get the surface shown in . + This surface looks like a tube, or a cylinder; + mathematicians call this surface a cylinder + for an entirely different reason. +

    + + + +
    + Sketching x^2+y^2=1 + +
    + + + + + Graph of three circles with radius 1 with centres on the z axis along different levels. + + +

    + The y and z axes are uncalibrated, the x axis is drawn from -2 + to 2. There are three circles with radius of 1 and centres all along the z + axis and are laid parallel to the xy plane. The circle in the middle has its centre on the origin. +

    +     + + + + + //ASY file for figspacecylinder1.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the circle at height z=1 + triple g(real t) {return (cos(t),sin(t),1);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + //Draw the circle at height z=0 + triple g(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + //Draw the circle at height z=-1.5 + triple g(real t) {return (cos(t),sin(t),-1.5);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + + + +
    + +
    + + + + + Graph of a hollow cylinder made with the three circles shown in the previous graph. + + +

    + The y and z axes are uncalibrated, the x axis is drawn from -2 to 2. + There are three circles with radius of 1 and centres all along the z axis and are laid + parallel to the xy plane. These circles form the area of cross-section and a cylinder of equation + x^2+y^2 =1 is drawn that includes all three circles. +

    +     + + + + + //ASY file for figspacecylinder1b.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the circle at height z=1 + triple g(real t) {return (cos(t),sin(t),1);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + //Draw the circle at height z=0 + triple g(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + //Draw the circle at height z=-1.5 + triple g(real t) {return (cos(t),sin(t),-1.5);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + //Draw the cylinder on top + triple f(pair t) { + return (cos(t.x),sin(t.x),t.y); + } + surface s=surface(f,(0,-1.5),(2*pi,1.2),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +
    +     + +
    + + + Cylinder + +

    + Let C be a curve in a plane and let L be a line not parallel to C. + A cylinder is the set of all lines parallel to L that pass through C. + The curve C is the directrix of the cylinder, + and the lines are the rulings. + cylinder + directrix +

    +     +     + +

    + In this text, + we consider curves C that lie in planes parallel to one of the coordinate planes, + and lines L that are perpendicular to these planes, + forming right cylinders. + Thus the directrix can be defined using equations involving 2 variables, + and the rulings will be parallel to the axis of the third variable. +

    + +

    + In the example preceding the definition, + the curve x^2+y^2=1 in the xy-plane is the directrix and the rulings are lines parallel to the z-axis. + (Any circle shown in + can be considered a directrix; + we simply choose the one where z=0.) + Sample rulings can also be viewed in . + More examples will help us understand this definition. +

    + + + Graphing cylinders + +

    + Graph the following cylinders. +

    + +

    +

      +
    1. z=y^2
      2. +
    2. x=\sin(z)
      3. +
    +

    +     + +

    +

      +
    1. +

      + We can view the equation z=y^2 as a parabola in the yz-plane, + as illustrated in . + As x does not appear in the equation, + the rulings are lines through this parabola parallel to the x-axis, + shown in . + These rulings give a general idea as to what the surface looks like, + drawn in . +

      + +
      + Sketching the cylinder defined by z=y^2 + +
      + + + + + Graph of cylinder in space with formula z = y^2. + + +

      + The x, y and z are uncalibrated. The graph shows a parabola z = y^2 + drawn on the zy plane. The parabola has its vertex on the origin. +

      +       + + + + + //ASY file for figspace4a.pdf in Chapter 10 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2.25,2.25); + pair ybounds=(-2.25,2.25); + pair zbounds=(-0.25,4.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the parabola z=y^2 + triple g(real t) {return (0,t,t^2);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,bluepen); + + //Draw the line for y=-1.9 on the parabola z=y^2 + triple g(real t) {return (t,-1.9,(1.9^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); + + //Draw the line for y=1.7 on the parabola z=y^2 + triple g(real t) {return (t,1.7,(1.7^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); + + + + +
      + +
      + + + + + Graph of parabola in space along with a group of lines on the parabola. + + +

      + The x, y and z axis are uncalibrated. The parabola is drawn + on the y and z axis, z being a function of y. There + is a group of parallel lines called rulings equidistant from each other that are + drawn on the parabola parallel to the xz plane, these lines give an idea + of the surface. +

      +       + + + + + //ASY file for figspace4b.pdf in Chapter 10.1 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2.25,2.25); + pair ybounds=(-2.25,2.25); + pair zbounds=(-0.25,4.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the parabola z=y^2 for t from -2 to 2 + triple g(real t) {return (0,t,t^2);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,bluepen); + + //Draw the line for y=-1.9 on the parabola z=y^2 + triple g(real t) {return (t,-1.9,(1.9^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=-1.6 on the parabola z=y^2 + triple g(real t) {return (t,-1.6,(1.6^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=-1.3 on the parabola z=y^2 + triple g(real t) {return (t,-1.3,(1.3^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=-1 on the parabola z=y^2 + triple g(real t) {return (t,-1,(1^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=-0.7 on the parabola z=y^2 + triple g(real t) {return (t,-0.7,(0.7^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=-0.4 on the parabola z=y^2 + triple g(real t) {return (t,-0.4,(0.4^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=-0.1 on the parabola z=y^2 + triple g(real t) {return (t,-0.1,(0.1^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=0.2 on the parabola z=y^2 + triple g(real t) {return (t,0.2,(0.2^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=0.5 on the parabola z=y^2 + triple g(real t) {return (t,0.5,(0.5^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=0.8 on the parabola z=y^2 + triple g(real t) {return (t,0.8,(0.8^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=1.1 on the parabola z=y^2 + triple g(real t) {return (t,1.1,(1.1^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=1.4 on the parabola z=y^2 + triple g(real t) {return (t,1.4,(1.4^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Draw the line for y=1.7 on the parabola z=y^2 + triple g(real t) {return (t,1.7,(1.7^2));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + + + +
      + +
      + + + + + Graph of surface with parabolic area of cross-section. + + +

      + The x, y and z axis are uncalibrated. The parabola is drawn on the y and z axis, + z being a function of y. A surface with a parabolic area of cross-section parallel to the yz + plane is shown. +

      +       + + + + + //ASY file for figspace4c.pdf in Chapter 10.1 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2.25,2.25); + pair ybounds=(-2.25,2.25); + pair zbounds=(-0.25,4.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the parabola z=y^2 for t from -2 to 2 + triple g(real t) {return (0,t,t^2);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,bluepen); + + //Draw the cylinder z=y^2 + triple f(pair t) { + return (t.x,t.y,(t.y)^2); + } + surface s=surface(f,(-2,-2),(2,2),8,32,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +
      +       +
      +
      2. + +
    2. + +

      + We can view the equation x=\sin(z) as a sine curve that exists in the xz-plane, + as shown in . + The rules are parallel to the y axis as the variable y does not appear in the equation x=\sin(z); + some of these are shown in . + The surface is shown in . +

      + +
      + Sketching the cylinder defined by x=\sin(z) + +
      + + + + + Graph of function x= sin (z) drawn in space. + + +

      + The x, y and z axes are uncalibrated. A function x= sin(z) + is drawn on the xz plane where x is a function of z. The z + axis is positioned vertically, and two sine waves are drawn on it one along the positive + z axis and one along the negative z axis. The two waves connect at the origin. +

      +       + + + + + //ASY file for figspace4d.pdf in Chapter 10.1 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,5,13); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2*pi-.5,2*pi+.75); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve x=sin z + triple g(real t) {return (sin(t),0,t);} + path3 mypath=graph(g,-2*pi,2*pi,operator ..); draw(mypath,bluepen); + + //Draw red lines on x=sin(z) for z in{-6.28,-4.71,-1.57,1.57,4.71,6.28} + triple g(real t) {return (sin(2*pi),t,2*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); + triple g(real t) {return (sin(1.5*pi),t,1.5*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); + triple g(real t) {return (sin(0.5*pi),t,0.5*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); + triple g(real t) {return (sin(-0.5*pi),t,-0.5*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); + triple g(real t) {return (sin(-1.5*pi),t,-1.5*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); + triple g(real t) {return (sin(-2*pi),t,-2*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); + + + + +
      + +
      + + + + + Graph of function x= sin (z) drawn in space along with the rules. + + +

      + The x, y and z axes are uncalibrated. A function x= sin(z) + is drawn on the xz plane where x is a function of z. The z + axis is positioned vertically, and two sine waves are drawn on it one along the positive + z axis and one along the negative z axis. The two waves connect at the origin. + The rules are drawn on the curve and are parallel to the y axis. +

      +       + + + + + //ASY file for figspace4e.pdf in Chapter 10.1 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,5,13); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2*pi-.5,2*pi+.75); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve x=sin z + triple g(real t) {return (sin(t),0,t);} + path3 mypath=graph(g,-2*pi,2*pi,operator ..); draw(mypath,bluepen); + + //Draw red lines on x=sin(z) for z in{-6.28,-4.71,-1.57,1.57,4.71,6.28} + triple g(real t) {return (sin(2*pi),t,2*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (sin(1.5*pi),t,1.5*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (sin(0.5*pi),t,0.5*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (sin(-0.5*pi),t,-0.5*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (sin(-1.5*pi),t,-1.5*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (sin(-2*pi),t,-2*pi);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + + + +
      + +
      + + + + + Graph of function x= sin (z) drawn in space along with the rules. + + +

      + The x, y and z axes are uncalibrated. The sine function + x=\sin(z) described previously is used as the area of cross-section to + form the surface. +

      +       + + + + + //ASY file for figspace4f.pdf in Chapter 10.1 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,5,13); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2*pi-.5,2*pi+.75); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve x=sin z + triple g(real t) {return (sin(t),0,t);} + path3 mypath=graph(g,-2*pi,2*pi,operator ..); draw(mypath,bluepen); + + //Draw the cylinder z=y^2 + triple f(pair t) { + return (sin(t.y),t.x,t.y); + } + surface s=surface(f,(-2,-2*pi),(2,2*pi),16,32,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +
      +       +
      +
      3. +
    +

    +     + +     +     + + + Surfaces of Revolution +

    + One of the applications of integration we learned previously was to find the volume of solids of revolution solids formed by revolving a curve about a horizontal or vertical axis. + We now consider how to find the equation of the surface of such a solid. +

    + + + +

    + Consider the surface formed by revolving + y=\sqrt{x} about the x-axis. + Cross-sections of this surface parallel to the yz-plane are circles, + as shown in . + Each circle has equation of the form + y^2+z^2=r^2 for some radius r. + The radius is a function of x; + in fact, it is r(x) = \sqrt{x}. + Thus the equation of the surface shown in is y^2+z^2=(\sqrt{x})^2. +

    + +
    + Introducing surfaces of revolution + +
    + + + + + Graph showing cross- section for surface of revolution. + + +

    + The y and x axes are drawn from -2 to 2 and the x axis is drawn from 0 to 4. + There are two planes drawn parallel to the yz plane and both of them have circles + outlined inside the plane. The first plane at x=1 has a smaller circle, while the + one at x=4 is bigger, both circles have formula y^2 + z^2 = r^2 for some radius r. + There is also half of a parabola drawn on the xy plane with x being a function of y, + this half parabola passes through both the circles intersecting them. +

    +     + + + + + //ASY file for figsurf_rev_intro.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,5,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={-2,2}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + pair xbounds=(-0.25,4.5); + pair ybounds=(-2.3,2.3); + pair zbounds=(-2.3,2.3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve y=sqrt(x) + triple g(real t) {return (t,sqrt(t),0);} + path3 mypath=graph(g,0,5,operator ..); draw(mypath,bluepen); + + //Draw red circles for x=1,4 + triple g(real t) {return (1,sin(t),cos(t));} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + triple g(real t) {return (4,2*sin(t),2*cos(t));} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Draw the x=1 plane with no spline + triple f(pair t) { + return (1,t.x,t.y); + } + surface s=surface(f,(-1.25,-1.25),(1.25,1.25)); + //pen p=apexmeshpen; + draw(s,surfacepen,nolight); + + //Draw the x=4 plane with no spline + triple f(pair t) { + return (4,t.x,t.y); + } + surface s=surface(f,(-2.25,-2.25),(2.25,2.25)); + //pen p=apexmeshpen; + draw(s,surfacepen,nolight); + + + + +
    + +
    + + + + + Graph showing a surface of revolution. + + +

    + The y and z axes are drawn from -2 to 2 and the + x axis is drawn from 0 to 4. There are two planes drawn + parallel to the yz plane and both of them have circles outlined inside + the plane. The first plane at x =1 has a smaller circle, while the one at + x=4 is bigger. There is also half of a parabola drawn on the xy + plane with x being a function of y, this half parabola passes + through both the circles intersecting the planes. This half parabola when rotated + over the x axis gives a hollow dome that opens along the positive x axis. +

    +     + + + + + //ASY file for figsurf_rev_introb.pdf in Chapter 10 + + size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,5,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={-2,2}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,4.5); + pair ybounds=(-2.3,2.3); + pair zbounds=(-2.3,2.3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve y=sqrt(x) + triple g(real t) {return (t,sqrt(t),0);} + path3 mypath=graph(g,0,5,operator ..); draw(mypath,bluepen); + + //Draw the revolved surface y^2 + z^2 = x + triple f(pair t) { + return (t.x,sqrt(t.x)*cos(t.y),sqrt(t.x)*sin(t.y)); + } + surface s=surface(f,(0,0),(5,2*pi),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +
    +     +
    + +

    + We generalize the above principles to give the equations of surfaces formed by revolving curves about the coordinate axes. +

    + + + Surfaces of Revolution, Part 1 +

    + Let r be a radius function. + surface of revolution +

      +
    1. +

      + The equation of the surface formed by revolving y=r(x) or z=r(x) about the x-axis is y^2+z^2=r(x)^2. +

      +
      2. + +
    2. +

      + The equation of the surface formed by revolving x=r(y) or z=r(y) about the y-axis is x^2+z^2=r(y)^2. +

      +
      3. + +
    3. +

      + The equation of the surface formed by revolving x=r(z) or y=r(z) about the z-axis is x^2+y^2=r(z)^2. +

      +
      4. +
    +

    +     + + + Finding equation of a surface of revolution + +

    + Let y=\sin(z) on [0,\pi]. + Find the equation of the surface of revolution formed by revolving + y=\sin(z) about the z-axis. +

    +     + +

    + Using , + we find the surface has equation x^2+y^2=\sin^2(z). + The curve is sketched in and the surface is drawn in . +

    + +

    + Note how the surface + (and hence the resulting equation) + is the same if we began with the curve x=\sin(z), + which is also drawn in . +

    +
    + Revolving y=\sin(z) about the z-axis in + +
    + + + + + Graph of two functions y = sin(z) and x = sin(z). + + +

    + The x and y axes are drawn from -1 to 1 and the + z axis is drawn from 0 to 3. Two functions x= \sin(z) + and y= \sin(z) are shown. The two curves are perpendicular to each other as + x=\sin(z) is drawn on the xz plane and y= \sin(z) is drawn on + the yz plane. Both curves start at the origin and end at the same point + (0, \pi/2, 0). +

    +     + + + + + //ASY file for figsurfrev1a.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,1.25); + pair zbounds=(-0.25,pi+.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve x=sin(z) in red + triple g(real t) {return (sin(t),0,t);} + path3 mypath=graph(g,0,pi,operator ..); draw(mypath,redpen); + label("$x=\sin(z)$",(.5,0,1.75)); + + //Draw the curve y=sin(z) in blue + triple g(real t) {return (0,sin(t),t);} + path3 mypath=graph(g,0,pi,operator ..); draw(mypath,bluepen); + label("$y=\sin(z)$",(0,.5,1.75)); + + + + +
    + +
    + + + + + Graph showing surface formed by revolving y = sin (z) about the z axis. + + +

    + The x and y axes are drawn from -1 to 1 and the z + axis is drawn from 0 to 3. The function y= \sin(z) is rotated + around the z axis and it forms a sphere with tapering top and bottom. +

    +     + + + + + //ASY file for figsurfrev1b.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,1.25); + pair zbounds=(-0.25,pi+.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve y=sin(z) in red + triple g(real t) {return (0,sin(t),t);} + path3 mypath=graph(g,0,pi,operator ..); draw(mypath,bluepen); + + //Draw the revolved surface x^2 + y^2 = sin^2(z) + triple f(pair t) { + return (sin(t.x)*cos(t.y),sin(t.x)*sin(t.y),t.x); + } + surface s=surface(f,(0,0),(pi,2*pi),16,32,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +
    +     +
    +     + +     + +

    + This particular method of creating surfaces of revolution is limited. + For instance, in + of + we found the volume of the solid formed by revolving + y=\sin(x) about the y-axis. + Our current method of forming surfaces can only rotate + y=\sin(x) about the x-axis. + Trying to rewrite y=\sin(x) as a function of y is not trivial, + as simply writing x=\sin^{-1}(y) only gives part of the region we desire. +

    + +

    + What we desire is a way of writing the surface of revolution formed by rotating y=f(x) about the y-axis. + We start by first recognizing this surface is the same as revolving z=f(x) about the z-axis. + This will give us a more natural way of viewing the surface. +

    + +

    + A value of x is a measurement of distance from the z-axis. + At the distance r, we plot a z-height of f(r). + When rotating f(x) about the z-axis, + we want all points a distance of r from the z-axis in the xy-plane to have a z-height of f(r). + All such points satisfy the equation r^2=x^2+y^2; + hence r=\sqrt{x^2+y^2}. + Replacing r with \sqrt{x^2+y^2} in f(r) gives z=f(\sqrt{x^2+y^2}). + This is the equation of the surface. +

    + + + Surfaces of Revolution, Part 2 +

    + Let z=f(x), x\geq 0, + be a curve in the xz-plane. + The surface formed by revolving this curve about the z-axis has equation z=f\big(\sqrt{x^2+y^2}\big). + surface of revolution +

    +     + + + Finding equation of surface of revolution + +

    + Find the equation of the surface found by revolving + z=\sin(x) about the z-axis. +

    +     + +

    + Using , + the surface has equation z=\sin\big(\sqrt{x^2+y^2}\big). + The curve and surface are graphed in . +

    + +
    + Revolving z=\sin(x) about the z-axis in + +
    + + + + + Graph of sine wave of amplitude 1 in the xz plane. + + +

    + The x and y axes are drawn from -5 to 5 and the z + axis is drawn from -1 to 1. The curve z= \sin(x) is drawn in the + xz plane, it is a wave with amplitude of z=1. The curve starts at the origin, + curves up and reaches a peak close to x=2, then it decreases and crosses the x + axis close to x=4, it decreases till it reaches a depth of z=-1, after which + it increases again to meet the x axis close to x=6. +

    +     + + + + + //ASY file for figsurfrev2a.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-5,5}; + real[] myychoice={-5,5}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-7,7); + pair ybounds=(-7,7); + pair zbounds=(-1.25,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve z=sin(x) in blue + triple g(real t) {return (t,0,sin(t));} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + + + +
    + +
    + + + + + Graph showing surface of revolution of function z= sin(x) around x axis. + + +

    + The surface of equation z= \sin(\sqrt {x^2 +y^2}) is made by revolving + the wave of formula z= \sin(x) about the z axis. +

    +     + + + + + //ASY file for figsurfrev2b.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-5,5}; + real[] myychoice={-5,5}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-7,7); + pair ybounds=(-7,7); + pair zbounds=(-1.25,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve z=sin(x) in blue + triple g(real t) {return (t,0,sin(t));} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen+.7mm); + + //Draw the revolved surface z = sin(sqrt(x^2 + y^2)) + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),sin(sqrt(((t.y*cos(t.x))^2+(t.y*sin(t.x))^2)))); + } + surface s=surface(f,(0,0),(2*pi,2*pi),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +
    +     +
    +     +     +     + + + Quadric Surfaces +

    + Spheres, planes and cylinders are important surfaces to understand. + We now consider one last type of surface, + a quadric surface. + The definition may look intimidating, + but we will show how to analyze these surfaces in an illuminating way. +

    + + + + + Quadric Surface + +

    + A quadric surface is the graph of the general second-degree equation in three variables: + quadric surfacedefinition + + Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0 + . +

    +     +     + +

    + When the coefficients D, + E or F are not zero, + the basic shapes of the quadric surfaces are rotated in space. + We will focus on quadric surfaces where these coefficients are 0; + we will not consider rotations. + There are six basic quadric surfaces: + the elliptic paraboloid, + elliptic cone, ellipsoid, hyperboloid of one sheet, + hyperboloid of two sheets, and the hyperbolic paraboloid. +

    + +
    + The elliptic paraboloid z=x^2/4+y^2 + + + + Graph of elliptic paraboloid z = x^2/4 +y^2. + + +

    + The axes are uncalibrated. There are two parabolas shown one in the plane x=0 + and the other in y=0. There is a circle drawn in the plane z=d. The + elliptical paraboloid z= x^2/4 +y^2 has both the parabolas and the circle included + in the surface. +

    +     + + + + + //ASY file for figquadric_parb.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-0.25,5.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z = x^2/4+y^2 + triple f(pair t) { + return (2*sqrt(t.y)*cos(t.x),sqrt(t.y)*sin(t.x),t.y);// + } + surface s=surface(f,(0,0),(2*pi,4),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the trace for y=0 in red + triple g(real t) {return (t,0,t^2/4);} + path3 mypath=graph(g,-4,4,operator ..); draw(mypath,redpen); + //Draw the trace for x=0 in red + triple g(real t) {return (0,t,t^2);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + //Draw the trace for z=2 in red + triple g(real t) {return (2*sqrt(2)*cos(t),sqrt(2)*sin(t),2);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent \centering In plane\\ $y=0$",(4,0,6),N);//label("$y=0$",(4,0,5.5),N); + draw((4.25,0,5.5)--(4,0,4.25),linewidth(.75),Arrow3); + label("\noindent \centering In plane\\ $x=0$",(0,2,6),N);//label("$x=0$",(0,2,5.5),N); + draw((0,2.25,5.5)--(0,2,4.25),linewidth(.75),Arrow3); + label("\noindent \centering In plane\\ $z=d$",(-1,3,1),N);//label("$z=d$",(-1,2,0.5),E); + draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +
    + +

    + We study each shape by considering traces, + trace + quadric surfacetrace + that is, intersections of each surface with a plane parallel to a coordinate plane. + For instance, + consider the elliptic paraboloid z= x^2/4+y^2, + shown in . + If we intersect this shape with the plane z=d (, replace z with d), + we have the equation: + + d \amp = \frac{x^2}4+y^2. + Divide both sides by d: + 1 \amp = \frac{x^2}{4d} + \frac{y^2}{d} + . +

    + +

    + This describes an ellipse so cross sections parallel to the xy-coordinate plane are ellipses. + This ellipse is drawn in the figure. +

    + +

    + Now consider cross sections parallel to the xz-plane. + For instance, + letting y=0 gives the equation z=x^2/4, clearly a parabola. + Intersecting with the plane x=0 gives a cross section defined by z=y^2, + another parabola. + These parabolas are also sketched in the figure. +

    + +

    + Thus we see where the elliptic paraboloid gets its name: + some cross sections are ellipses, and others are parabolas. +

    + +

    + Such an analysis can be made with each of the quadric surfaces. + We give a sample equation of each, + provide a sketch with representative traces, + and describe these traces. +

    + +

    +

    +
  • + Elliptic Paraboloid +

    + z=\frac{x^2}{a^2}+\frac{y^2}{b^2} +

    +
    • +
    +

    + + + + + + Graph of elliptic paraboloid. + + +

    + The axes are uncalibrated. The elliptical parabola opens along the positive z axis. +

    +     + + + + + //ASY file for figquadric_par.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-0.25,5.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z = x^2/4+y^2 + triple f(pair t) { + return (2*sqrt(t.y)*cos(t.x),sqrt(t.y)*sin(t.x),t.y);// + } + surface s=surface(f,(0,0),(2*pi,4),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + label("\noindent \centering In plane\\ $y=0$",(4,0,6),N,invisible); + label("\noindent \centering In plane\\ $x=0$",(0,2,6),N,invisible); + label("\noindent \centering In plane\\ $z=0$",(-1,3,1),N,invisible); + + + + + + + Plane + + Trace + + + + + + + + x=d + + Parabola + + + y=d + + Parabola + + + z=d + + Ellipse + + + + + + Graph of elliptic paraboloid. + + +

    + The axes are uncalibrated. There are two parabolas shown one in the plane x=0 + and the other in y=0. There is a circle drawn in the plane z=d. The + elliptical paraboloid has both the parabolas and the circle included in the surface. + It opens along the positive z axis. +

    +     + + + + + //ASY file for figquadric_parb.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-0.25,5.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z = x^2/4+y^2 + triple f(pair t) { + return (2*sqrt(t.y)*cos(t.x),sqrt(t.y)*sin(t.x),t.y);// + } + surface s=surface(f,(0,0),(2*pi,4),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the trace for y=0 in red + triple g(real t) {return (t,0,t^2/4);} + path3 mypath=graph(g,-4,4,operator ..); draw(mypath,redpen); + //Draw the trace for x=0 in red + triple g(real t) {return (0,t,t^2);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + //Draw the trace for z=2 in red + triple g(real t) {return (2*sqrt(2)*cos(t),sqrt(2)*sin(t),2);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent \centering In plane\\ $y=0$",(4,0,6),N);//label("$y=0$",(4,0,5.5),N); + draw((4.25,0,5.5)--(4,0,4.25),linewidth(.75),Arrow3); + label("\noindent \centering In plane\\ $x=0$",(0,2,6),N);//label("$x=0$",(0,2,5.5),N); + draw((0,2.25,5.5)--(0,2,4.25),linewidth(.75),Arrow3); + label("\noindent \centering In plane\\ $z=d$",(-1,3,1),N);//label("$z=d$",(-1,2,0.5),E); + draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +     + +

    + One variable in the equation of the elliptic paraboloid will be raised to the first power; + above, this is the z variable. The paraboloid will open in the direction of this variable's axis. + Thus x= y^2/a^2+z^2/b^2 is an elliptic paraboloid that opens along the x-axis. + Multiplying the right hand side by (-1) defines an elliptic paraboloid that opens in the opposite direction. + quadric surfacegallery + quadric surfaceelliptic paraboloid +

    + +

    +

    +
  • + Elliptic Cone +

    + z^2=\frac{x^2}{a^2}+\frac{y^2}{b^2} +

    +
    • +
    +

    + + + + + + + Graph showing elliptic cone opening along the z axis. + + +

    + The axes are uncalibrated. Two hollow elliptic cones are drawn with vertices at the origin, + one opening along the positive z axis and the other along the negative z axis. +

    +     + + + + + //ASY file for figquadric_cone.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,1); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-3,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the top half of the surface z^2 = x^2/(1.5)^2+y^2 + triple f(pair t) { + return (1.5*t.y*cos(t.x),t.y*sin(t.x),t.y);// + } + surface s=surface(f,(0,-2),(2*pi,2),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + label("\noindent\centering In plane\\ $y=d$",(4,0,-1),invisible); + label("\noindent\centering In plane\\ $y=0$",(0,4,1.5),invisible); + label("\noindent\centering In plane\\ $z=d$",(4,0,-1),invisible); + + + + + + + Plane + + Trace + + + + + + + + x=0 + + Crossed Lines + + + y=0 + + Crossed Lines + + + + + + x=d + + Hyperbola + + + y=d + + Hyperbola + + + z=d + + Ellipse + + +     + + + + + + Graph showing elliptic cone opening along the z axis. + + +

    + The axes are uncalibrated. Two hollow elliptic cones are drawn with vertices at the origin, + one opening along the positive z axis and the other along the negative z axis. + The graph shows three traces, on the plane y=0, the trace is a straight line passing through + the vertices of the cones, when the plane is z=d the trace is an ellipse. +

    +     + + + + //ASY file for figquadric_coneb.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,1); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-3,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the top half of the surface z^2 = x^2/(1.5)^2+y^2 + triple f(pair t) { + return (1.5*t.y*cos(t.x),t.y*sin(t.x),t.y);// + } + surface s=surface(f,(0,-2),(2*pi,2),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for y=0 in red + triple g(real t) {return (t,0,t/1.5);} + path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); + triple g(real t) {return (t,0,-t/1.5);} + path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); + + //Draw the trace for z=-2 in red + triple g(real t) {return (1.5*1.5*cos(t),1.5*sin(t),-1.5);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent\centering In plane\\ $y=0$",(0,4,1.5)); + draw((.1,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + draw((-.1,4,1.5)--(-3,0,2),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $z=d$",(4,0,-1)); + draw((4,0,-1)--(1.6,1.1,-1.5),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $y=d$",(4,0,-1),invisible); + + + + + + + + Graph showing elliptic cone opening along the z axis. + + +

    + The axes are uncalibrated. Two hollow elliptic cones are drawn with vertices at the origin, + one opening along the positive z axis and the other along the negative z axis. + A hyperbola is shown in the y=d plane, the two parts of the hyperbola are traced on the elliptic cone. +

    +     + + + + + //ASY file for figquadric_conec.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,1); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-3,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 = x^2/(1.5)^2+y^2 + triple f(pair t) { + return (1.5*t.y*cos(t.x),t.y*sin(t.x),t.y);// + } + surface s=surface(f,(0,-2),(2*pi,2),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for y=d in red + triple g(real t) {return (1.5*.5*tan(t),0.5,0.5/cos(t));} + path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); + triple g(real t) {return (1.5*.5*tan(t),0.5,-0.5/cos(t));} + path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $y=d$",(4,0,-1)); + draw((4,0,-1)--(1.3,.5,-1),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $y=0$",(0,4,1.5),invisible); + label("\noindent\centering In plane\\ $z=d$",(4,0,-1),invisible); + + + + +     +     +

    + One can rewrite the equation as z^2-x^2/a^2-y^2/{b^2} = 0. + The one variable with a positive coefficient corresponds to the axis that the cones open along. + quadric surfaceelliptic cone +

    + +

    +

    +
  • + Ellipsoid +

    + \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 +

    +
    • +
    +

    + + + + + + Graph of the three dimensional ellipsoid. + + +

    + The x, y and z axes are uncalibrated. Graph of an ellipsoid with + centre at origin. +

    +     + + + + + //ASY file for figquadric_ellipsoid.pdf in Chapter 10.1 + //STILL NEED TO FIX LABELS + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + xaxis3("",-3,3,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",-3,3,invisible,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",-3,3,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 = x^2/(1.5)^2+y^2 + triple f(pair t) { + return (1.5*cos(t.y)*cos(t.x),cos(t.y)*sin(t.x),sin(t.y));//({cos(x)*1.5*cos(y)},{sin(x)*cos(y)},{sin(y)}) + } + surface s=surface(f,(-pi,-pi/2),(1*pi,pi/2),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for y=d in red + //triple g(real t) {return (1.5*.5*tan(t),0.5,0.5/cos(t));} + //path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); + //triple g(real t) {return (1.5*.5*tan(t),0.5,-0.5/cos(t));} + //path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent\centering In plane\\ $y=0$",(2,0,-1),invisible); + label("\noindent\centering In plane\\ $x=0$",(0,2,-.5),invisible); + label("\noindent\centering In plane\\ $z=0$",(0,2,1.5),invisible); + + + + + + + Plane + + Trace + + + + + + + + x=d + + Ellipse + + + y=d + + Ellipse + + + z=d + + Ellipse + + + + + + Graph of the three dimensional ellipsoid. + + +

    + The x, y and z axes are uncalibrated. There are three ellipses drawn. + The first one is on the xz plane, y=0. The second is on the yz plane, + with x=0. The third on the xy plane, with z=0. + Filling in the traces gives the ellipsoid. +

    +     + + + + + //ASY file for figquadric_ellipsoidb.pdf in Chapter 10 + //STILL NEED TO FIX LABELS + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + xaxis3("",-3,3,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",-3,3,invisible,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",-3,3,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 = x^2/(1.5)^2+y^2 + triple f(pair t) { + return (1.5*cos(t.y)*cos(t.x),cos(t.y)*sin(t.x),sin(t.y));// + } + surface s=surface(f,(-pi,-pi/2),(1*pi,pi/2),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the trace for x=0 in red + triple g(real t) {return (0,sin(t),cos(t));} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + //Draw the trace for y=0 in red + triple g(real t) {return (1.5*sin(t),0,cos(t));} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + //Draw the trace for z=0 in red + triple g(real t) {return (1.5*sin(t),cos(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent\centering In plane\\ $y=0$",(2,0,-1),S); + draw((2,0,-1)--(1.3,0,-.5),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $x=0$",(0,2,-.5),S); + draw((0,2,-.5)--(0,.86,-.5),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $z=0$",(0,2,1.5),N); + draw((0,2,1.5)--(1.05,.75,0),linewidth(.75),Arrow3); + + + + +     + +

    + If a=b=c\neq0, the ellipsoid is a sphere with radius a; compare to . + quadric surfaceellipsoid + quadric surfacesphere +

    + +

    +

    +
  • + Hyperboloid of One Sheet +

    + \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1 +

    +
    • +
    +

    + + + + + + Graph showing hyperboloid of one sheet. + + +

    + The three axes are uncalibrated. Graph shows a hyperboloid of one sheet. +

    +     + + + + + //ASY file for figquadric_hyp_one_sheet.pdf in Chapter 10.1 + //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? + //STILL NEED TO FIX LABELS + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + xaxis3("",-3,3,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",-3,3,invisible,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",-3,3,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 = x^2/(1.5)^2+y^2 + triple f(pair t) { + return (cos(t.y)/cos(t.x),sin(t.y)/cos(t.x),tan(t.x));//({cos(y)*sec(x)},{sec(x)*sin(y)},{tan(x)}); + } + + bool cond(pair t) {return tan(t.x)<1;} + + surface s=surface(f,(-1,0),(1,2*pi),32,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for y=d in red + //triple g(real t) {return (1.5*.5*tan(t),0.5,0.5/cos(t));} + //path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); + //triple g(real t) {return (1.5*.5*tan(t),0.5,-0.5/cos(t));} + //path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent\centering In plane\\ $y=0$",(2,0,-1),S,invisible); + label("\noindent\centering In plane\\ $x=0$",(0,2,-.5),S,invisible); + label("\noindent\centering In plane\\ $z=0$",(-1.6,1.6,.4),N,invisible); + + + + + + + Plane + + Trace + + + + + + + + x=d + + Hyperbola + + + y=d + + Hyperbola + + + z=d + + Ellipse + + + + + + Graph showing hyperboloid of one sheet. + + +

    + The three axes are uncalibrated. Graph shows a hyperboloid of one sheet. The hyperboloid of one sheet + is drawn about the z axis. It appears to be a cylinder with a narrow middle. In the middle of + the sheet there is a circle drawn on the plane z=o. There are two hyperbolas drawn on the plane + x=0 and y=0. +

    +     + + + + + //ASY file for figquadric_hyp_one_sheetb.pdf in Chapter 10.1 + //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? + //STILL NEED TO FIX LABELS + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + xaxis3("",-3,3,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",-3,3,invisible,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",-3,3,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 = x^2/(1.5)^2+y^2 + triple f(pair t) { + return (cos(t.y)/cos(t.x),sin(t.y)/cos(t.x),tan(t.x));//({cos(y)*sec(x)},{sec(x)*sin(y)},{tan(x)}); + } + surface s=surface(f,(-1,0),(1,2*pi),32,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + triple g(real t) {return (0,1/cos(t),tan(t));} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (0,-1/cos(t),tan(t));} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for y=0 in red + triple g(real t) {return (1/cos(t),0,tan(t));} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (-1/cos(t),0,tan(t));} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for z=0 in red + triple g(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent\centering In plane\\ $y=0$",(2,0,-1),S); + draw((2,0,-1)--(1.2,0,-.5),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $x=0$",(0,2,-.5),S); + draw((0,2,-.5)--(0,1.2,-.5),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $z=0$",(-1.6,1.6,.4),N); + draw((-1.6,1.6,.4)--(-.75,.75,0),linewidth(.75),Arrow3); + + + + +     + +

    + The one variable with a negative coefficient corresponds to the axis that the hyperboloid opens along. + quadric surfacehyperboloid of one sheet +

    + +

    +

    +
  • + Hyperboloid of Two Sheets +

    + \frac{z^2}{c^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 +

    +
    • +
    +

    + + + + + + Graph showing hyperboloid of two sheets. + + +

    + The three axes are uncalibrated. Graph shows a hyperboloid of two sheets. +

    +     + + + + + //ASY file for figquadric_hyp_two_sheetb.pdf in Chapter 10.1 + //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? + //STILL NEED TO FIX LABELS + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 - x^2 - y^2=1 + triple f(pair t) { + return (cos(t.y)*tan(t.x),sin(t.y)*tan(t.x),1/cos(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) + } + surface s=surface(f,(-1,0),(1,pi),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + triple f(pair t) { + return (cos(t.y)*tan(t.x),sin(t.y)*tan(t.x),-1/cos(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) + } + + bool cond(pair t) {return 1/cos(t.x) <= 1.6;} + + surface s=surface(f,(-1,0),(1,pi),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + //triple g(real t) {return (0,1/cos(t),tan(t));} + //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + //triple g(real t) {return (0,-1/cos(t),tan(t));} + //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for y=0 in red + //triple g(real t) {return (1/cos(t),0,tan(t));} + //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + //triple g(real t) {return (-1/cos(t),0,tan(t));} + //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for z=0 in red + //triple g(real t) {return (cos(t),sin(t),0);} + //path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent\centering In plane\\ $y=0$",(1.5,0,-1),N,invisible); + label("\noindent\centering In plane\\ $x=0$",(0,1.5,-1),N,invisible); + label("\noindent\centering In plane\\ $z=d$",(1.1,1.1,-2),S,invisible); + + + + + + + Plane + + Trace + + + + + + + + x=d + + Hyperbola + + + y=d + + Hyperbola + + + z=d + + Ellipse + + + + + + Graph showing hyperboloid of two sheets. + + +

    + The three axes are uncalibrated. Graph shows a hyperboloid of two sheets. The hyperboloids of + two sheets are drawn about the z axis. The first sheet opens along the positive z + axis. The second sheet to the bottom opens along the negative z axis. Both plates have two + hyperbolas drawn one in the zy plane and one in the xz plane. In the bottom sheet + there is a circle drawn on the plane z=d. +

    +     + + + + + //ASY file for figquadric_hyp_two_sheetb.pdf in Chapter 10.1 + //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? + //STILL NEED TO FIX LABELS + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 - x^2 - y^2=1 + triple f(pair t) { + return (cos(t.y)*tan(t.x),sin(t.y)*tan(t.x),1/cos(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) + } + surface s=surface(f,(-1,0),(1,pi),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + triple f(pair t) { + return (cos(t.y)*tan(t.x),sin(t.y)*tan(t.x),-1/cos(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) + } + surface s=surface(f,(-1,0),(1,pi),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + triple g(real t) {return (0,tan(t),1/cos(t));} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (0,tan(t),-1/cos(t));} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for y=0 in red + triple g(real t) {return (tan(t),0,1/cos(t));} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (tan(t),0,-1/cos(t));} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for z=-1.7 in red + triple g(real t) {return (1.37*cos(t),1.37*sin(t),-1.7);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent\centering In plane\\ $y=0$",(1.5,0,-1),N); + draw((1.5,0,-1)--(1,0,-1.35),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $x=0$",(0,1.5,-1),N); + draw((0,1.5,-1)--(0,1,-1.35),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $z=d$",(1.1,1.1,-2),S); + draw((1.1,1.1,-2)--(.95,.95,-1.7),linewidth(.75),Arrow3); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +     + +

    + The one variable with a positive coefficient corresponds to the axis that the hyperboloid opens along. + In the case illustrated, when \abs{d}\lt \abs{c}, there is no trace. + quadric surfacehyperboloid of two sheets +

    + +

    +

    +
  • + Hyperbolic Paraboloid +

    + z=\frac{x^2}{a^2}-\frac{y^2}{b^2} +

    +
    • +
    +

    + + + + + + + Graph of the hyperbolic paraboloid. + + +

    + The three axes are uncalibrated. Graph shows a hyperbolic paraboloid. +

    +     + + + + + //ASY file for figquadric_hyp_par.pdf in Chapter 10.1 + //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? + //STILL NEED TO FIX LABELS + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(2,5,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 - x^2 - y^2=1 + triple f(pair t) { + return (t.x,t.y,(t.x)^2-(t.y)^2);//({x},{y},{x^2-y^2}); + } + surface s=surface(f,(-1,-1),(1,1),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + //triple g(real t) {return (0,1/cos(t),tan(t));} + //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + //triple g(real t) {return (0,-1/cos(t),tan(t));} + //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for y=0 in red + //triple g(real t) {return (1/cos(t),0,tan(t));} + //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + //triple g(real t) {return (-1/cos(t),0,tan(t));} + //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for z=0 in red + //triple g(real t) {return (cos(t),sin(t),0);} + //path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + + + + Plane + + Trace + + + + + + + + x=d + + Parabola + + + y=d + + Parabola + + + z=d + + Hyperbola + + +     + + + + + + Graph of the hyperbolic paraboloid. + + +

    + The three axes are uncalibrated. There are two parabolas drawn, one in plane + y=0 opening up along the positive z axis in the yz plane + and the other in x=0 opening down along the negative z axis in the + xz plane. Both parabolas have vertices at the origin. + Filling the traces gives the hyperbolic paraboloid. +

    +     + + + + + //ASY file for figquadric_hyp_parb.pdf in Chapter 10.1 + //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? + //STILL NEED TO FIX LABELS + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(2,5,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 - x^2 - y^2=1 + triple f(pair t) { + return (t.x,t.y,(t.x)^2-(t.y)^2);//({x},{y},{x^2-y^2}); + } + surface s=surface(f,(-1,-1),(1,1),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + triple g(real t) {return (0,t,-t^2);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for y=0 in red + triple g(real t) {return (t,0,t^2);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent\centering In plane\\ $y=0$",(.5,0,.6),N); + draw((.5,0,.6)--(.15,0,.05),linewidth(.75),Arrow3); + label("\noindent\centering In plane\\ $x=0$",(0,1.5,0.3),E); + draw((0,1.5,.3)--(0,.7,-.4),linewidth(.75),Arrow3); + //draw((-.2,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + + + + + Graph of the hyperbolic paraboloid. + + +

    + The three axes are uncalibrated. Graph shows the hyperbolic paraboloid along with two hyperbolas. There are two hyperbolas drawn, + in plane z=d. For d>0 the two hyperbolas opening up along the positive and negative x axis. For d<0 + the other hyperbola opens along the positive and negative y axis. +

    +     + + + + + //ASY file for figquadric_hyp_parb.pdf in Chapter 10.1 + //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? + //STILL NEED TO FIX LABELS + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(2,5,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //xaxis3("",-2,2,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); + //yaxis3("",-2,2,invisible,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",-2,2,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 - x^2 - y^2=1 + triple f(pair t) { + return (t.x,t.y,(t.x)^2-(t.y)^2);//({x},{y},{x^2-y^2}); + } + surface s=surface(f,(-1,-1),(1,1),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for z=1/4 in red + triple g(real t) {return (0.5/cos(t),0.5*tan(t),0.25);} + path3 mypath=graph(g,-1.05,1.05,operator ..); draw(mypath,redpen); + triple g(real t) {return (-0.5/cos(t),0.5*tan(t),0.25);} + path3 mypath=graph(g,-1.05,1.05,operator ..); draw(mypath,redpen); + + //Draw the traces for z=-1/4 in red + triple g(real t) {return (0.5*tan(t),0.5/cos(t),-0.25);} + path3 mypath=graph(g,-1.05,1.05,operator ..); draw(mypath,redpen); + triple g(real t) {return (0.5*tan(t),-0.5/cos(t),-0.25);} + path3 mypath=graph(g,-1.05,1.05,operator ..); draw(mypath,redpen); + + //Add labels + label("\noindent\centering Hyperbolas in \\plane $z=d$",(.7,0,1),N); + label("\noindent\centering $(d>0)$",(.5,0,.6),N); + draw((.52,0,.6)--(.5,0,.25),linewidth(.75),Arrow3); + draw((.48,0,.6)--(-.5,0,.25),linewidth(.75),Arrow3); + label("\noindent\centering $(d<0)$",(1,0,-0.5),S); + draw((1,.1,-0.5)--(.8,.9,-.25),linewidth(.75),Arrow3); + draw((1,-.1,-0.5)--(.8,-.9,-.25),linewidth(.75),Arrow3); + + + + +     +     +

    + The parabolic traces will open along the axis of the one variable that is raised to the first power. + quadric surfacegallery + quadric surfacehyperbolic paraboloid +

    + + + Sketching quadric surfaces + +

    + Sketch the quadric surface defined by the given equation. +

    + +

    +

      +
    1. y=\frac{x^2}{4}+\frac{z^2}{16}
      2. +
    2. x^2+\frac{y^2}{9}+\frac{z^2}{4}=1
      3. +
    3. \ds z=y^2-x^2
      4. +
    +

    +     + +

    +

      +
    1. +

      + \ds y=\frac{x^2}{4}+\frac{z^2}{16}: We first identify the quadric by pattern-matching with the equations given previously. + Only two surfaces have equations where one variable is raised to the first power, + the elliptic paraboloid and the hyperbolic paraboloid. + In the latter case, the other variables have different signs, + so we conclude that this describes a hyperbolic paraboloid. + As the variable with the first power is y, + we note the paraboloid opens along the y-axis. + To make a decent sketch by hand, + we need only draw a few traces. + In this case, + the traces x=0 and z=0 form parabolas that outline the shape. +

      +

      + x=0: The trace is the parabola + y=z^2/16 +

      +

      + z=0: The trace is the parabola y=x^2/4. +

      + +

      + Graphing each trace in the respective plane creates a sketch as shown in . + This is enough to give an idea of what the paraboloid looks like. + The surface is filled in in . +

      + +
      + Sketching an elliptic paraboloid + +
      + + + + + Graph of two parabolas that form the elliptic paraboloid. + + +

      + The x and z axes are drawn from -4 to 4 and the y axis + is drawn from 0 to 2. Two parabolas are drawn that are perpendicular to each other, + one in the xy plane and the other in the yz plane. Both have vertices at origin. +

      +       + + + + + //ASY file for figspace5ab.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(15,5,4.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-4,4}; + real[] myychoice={1,2}; + real[] myzchoice={-4,4}; + defaultpen(0.5mm); + pair xbounds=(-4.5,4.5); + pair ybounds=(-0.5,2.5); + pair zbounds=(-4.5,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface y=x^2/4+z^2/16 + //triple f(pair t) { + // return (1.5*cos(t.y)*cos(t.x),cos(t.y)*sin(t.x),sin(t.y));//({cos(x)*1.5*cos(y)},{sin(x)*cos(y)},{sin(y)}) + //} + //surface s=surface(f,(0,-2),(pi,2),32,16); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + triple g(real t) {return (0,t^2/16,t);} + path3 mypath=graph(g,-4,4,operator ..); draw(mypath,redpen); + + //Draw the traces for z=0 in red + triple g(real t) {return (t,t^2/4,0);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +
      + +
      + + + + + Graph of elliptic paraboloid. + + +

      + The three axes are uncalibrated. Two parabolas are drawn that are perpendicular to + each other, one in the xy plane and the other in the yz plane. Both have vertices at origin. + Filling in the trace gives the elliptic paraboloid. +

      +       + + + + + //ASY file for figspace5a.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(15,5,4.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-4.5,4.5); + pair ybounds=(-0.5,2.5); + pair zbounds=(-4.5,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface y=x^2/4+z^2/16 + triple f(pair t) { + return (2*t.y*cos(t.x),(t.y)^2,4*t.y*sin(t.x));//({2*cos(x)*y},{y^2},{4*y*sin(x)}); + } + surface s=surface(f,(0,-1),(pi,1),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + triple g(real t) {return (0,t^2/16,t);} + path3 mypath=graph(g,-4,4,operator ..); draw(mypath,redpen); + + //Draw the traces for z=0 in red + triple g(real t) {return (t,t^2/4,0);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +
      +       + +
      +
      2. + +
    2. +

      + \ds x^2+\frac{y^2}{9}+\frac{z^2}{4}=1: This is an ellipsoid. + We can get a good idea of its shape by drawing the traces in the coordinate planes. +

      + +

      + x=0: The trace is the ellipse \ds\frac{y^2}{9}+\frac{z^2}{4}=1. + The major axis is along the y-axis with length 6 (as b=3, + the length of the axis is 6); + the minor axis is along the z-axis with length 4. +

      + +

      + y=0: The trace is the ellipse \ds x^2+\frac{z^2}{4}=1. + The major axis is along the z-axis, + and the minor axis has length 2 along the x-axis. + z=0: The trace is the ellipse \ds x^2+\frac{y^2}{9}=1, + with major axis along the y-axis. +

      + +

      + Graphing each trace in the respective plane creates a sketch as shown in . Filling in the surface gives . +

      + +
      + Sketching an ellipsoid + +
      + + + + + Graph of trace for the ellipsoid. + + +

      + The x, y and z axes are drawn from -3 to 3. + There are three ellipses drawn. The first one is on the xz plane, y=0 + with equation x^2 + z^2/4 =1. The second is on the yz plane, with x=0 + with the equation y^2/9 + z^2/4 =1. The third on the xy plane, with z=0 + and has an equation x^2 +y^2 /9 =1. +

      +       + + + + + //ASY file for figspace5b.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-3,3}; + real[] myychoice={-3,3}; + real[] myzchoice={-3,3}; + defaultpen(0.5mm); + pair xbounds=(-4.5,4.5); + pair ybounds=(-4.5,4.5); + pair zbounds=(-4.5,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2+y^2/9+z^2/4=1 + //triple f(pair t) { + // return (cos(t.y)*cos(t.x),3*sin(t.y)*cos(t.x),2sin(t.x));// + //} + //surface s=surface(f,(0,0),(2*pi,2*pi),32,16,Spline); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + triple g(real t) {return (0,3*cos(t),2*sin(t));} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Draw the traces for y=0 in red + triple g(real t) {return (cos(t),0,2*sin(t));} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Draw the traces for z=0 in red + triple g(real t) {return (cos(t),3*sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +
      + +
      + + + + + Graph of the three dimensional ellipsoid. + + +

      + The x, y and z axes are drawn from -3 to 3. There are three ellipses drawn. The first one is on the + xz plane, y=0 with equation x^2 + z^2/4 =1. The second is on the yz plane, with x=0 with the equation + y^2/9 + z^2/4 =1. The third on the xy plane, with z=0 and has an equation x^2 +y^2 /9 =1. + Filling in the surface gives the ellipsoid. +

      +       + + + + + //ASY file for figspace5bb.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-4.5,4.5); + pair ybounds=(-4.5,4.5); + pair zbounds=(-4.5,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2+y^2/9+z^2/4=1 + triple f(pair t) { + return (cos(t.y)*cos(t.x),3*sin(t.y)*cos(t.x),2sin(t.x));// + } + surface s=surface(f,(-pi,-pi/2),(pi,pi/2),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + triple g(real t) {return (0,3*cos(t),2*sin(t));} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Draw the traces for y=0 in red + triple g(real t) {return (cos(t),0,2*sin(t));} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Draw the traces for z=0 in red + triple g(real t) {return (cos(t),3*sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +
      +       +
      +
      3. + +
    3. +

      + \ds z=y^2-x^2: This defines a hyperbolic paraboloid, + very similar to the one shown in the gallery of quadric sections. + Consider the traces in the y-z and x-z planes: +

      + +

      + x=0: The trace is z=y^2, + a parabola opening up in the y-z plane. +

      + +

      + y=0: The trace is z=-x^2, + a parabola opening down in the x-z plane. +

      + +

      + Sketching these two parabolas gives a sketch like that in , and filling in the surface gives a sketch like . +

      + +
      + Sketching a hyperbolic paraboloid + +
      + + + + + Graph showing traces for a hyperbolic paraboloid. + + +

      + The x, y and z axes are drawn from -1 to 1. + There are two parabolas drawn, one in plane x=0 with equation z= y^2 + opening up in the yz plane and the other in y=0 with equation z=-x^2 + opening down in the xz plane. Both parabolas have vertices at the origin. +

      +       + + + + + //ASY file for figspace5c.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4.5,3,2.7); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 = y^2 - x^2 + //triple f(pair t) { + // return (t.x,t.y,-(t.x)^2+(t.y)^2);// + //} + //surface s=surface(f,(-1,-1),(1,1),32,16,Spline); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + triple g(real t) {return (0,t,t^2);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for y=0 in red + triple g(real t) {return (t,0,-t^2);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +
      + +
      + + + + + Graph of the hyperbolic paraboloid. + + +

      + The three axes are uncalibrated. There are two parabolas drawn, one in + plane x=0 with equation z= y^2 opening up in the yz + plane and the other in y=0 with equation z=-x^2 opening down + in the xz plane. Both parabolas have vertices at the origin. + Filling the traces gives the hyperbolic paraboloid. +

      +       + + + + + //ASY file for figspace5cb.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4.5,3,2.7); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2 = y^2 - x^2 + triple f(pair t) { + return (t.x,t.y,-(t.x)^2+(t.y)^2);// + } + surface s=surface(f,(-1,-1),(1,1),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces for x=0 in red + triple g(real t) {return (0,t,t^2);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the traces for y=0 in red + triple g(real t) {return (t,0,-t^2);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +
      +       +
      +
      4. +
    +

    +     + +     + + + Identifying quadric surfaces + +

    + Consider the quadric surface shown in . + Which of the following equations best fits this surface? +

    + +

    +

      +
    1. \ds x^2-y^2-\frac{z^2}{9}=0
      2. +
    2. \ds x^2-y^2-z^2=1
      3. +
    3. \ds z^2-x^2-y^2=1
      4. +
    4. 4x^2-y^2-\frac{z^2}9=1
      5. +
    +

    + +
    + A possible equation of this quadric surface is found in + + + + Graph showing hyperboloid of two sheets. + + +

    + The z and y axes are drawn from -3 to 3 and the x axis is + drawn from -1 to 1. The y and z axes are drawn from -3 to + 3. The hyperboloids of two sheets are drawn about the x axis. The first sheet + has a centre at x =0.5 and opens along the positive y axis. The second sheet has + a centre at x=-0.5 and opens along the negative y axis. +

    +     + + + + + //ASY file for figspace6.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,9.5,5.7); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-3,3}; + real[] myzchoice={-3,3}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-4.5,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface 4x^2 - y^2 - z^2/9 = 1 + triple f(pair t) { + return (0.5/cos(t.x),cos(t.y)*tan(t.x),3*sin(t.y)*tan(t.x));// + } + surface s=surface(f,(-pi/3,0),(pi/3,pi),16,24); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + triple f(pair t) { + return (-0.5/cos(t.x),cos(t.y)*tan(t.x),3*sin(t.y)*tan(t.x));// + } + surface s=surface(f,(-pi/3,0),(pi/3,pi),16,24); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +
    +     + +

    + The image clearly displays a hyperboloid of two sheets. + The gallery informs us that the equation will have a form similar to \frac{z^2}{c^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. +

    + +

    + We can immediately eliminate option (a), + as the constant in that equation is not 1. +

    + +

    + The hyperboloid opens along the x-axis, + meaning x must be the only variable with a positive coefficient, + eliminating (c). +

    + +

    + The hyperboloid is wider in the z-direction than in the y-direction, + so we need an equation where c \gt b. + This eliminates (b), leaving us with (d). + We should verify that the equation given in (d), + 4x^2-y^2-\frac{z^2}9=1, fits. +

    + +

    + We already established that this equation describes a hyperboloid of two sheets that opens in the x-direction and is wider in the z-direction than in the y. + Now note the coefficient of the x-term. + Rewriting 4x^2 in standard form, we have: + \ds 4x^2 = \frac{x^2}{(1/2)^2}. + Thus when y=0 and z=0, + x must be 1/2; + , each hyperboloid starts at x=1/2. + This matches our figure. +

    + +

    + We conclude that \ds 4x^2-y^2-\frac{z^2}9=1 best fits the graph. +

    +     +     + + + +

    + This section has introduced points in space and shown how equations can describe surfaces. + The next sections explore vectors, + an important mathematical object that we'll use to explore curves in space. +

    +     + + + + Terms and Concepts + + + + +

    + Axes drawn in space must conform to the rule. +

    +     + + + + + + + + +     + + + + + + + + +

    + In the plane, + the equation x=2 defines a ; + in space, x=2 defines a . +

    +     + + + + + + + + + + + + + +     + + + + +

    + In the plane, the equation y=x^2 defines a ; + in space, y=x^2 defines a . +

    +     + + + + curve|parabola + + + + + surface|cylinder|parabolic cylinder + + + + +     + + + + + +

    + Which quadric surface looks like a Pringleschip? +

    + +     + + + +

    + Elliptic paraboloid +

    +     +     + + +

    + Elliptic cone +

    +     +     + + +

    + Ellipsoid +

    +     +     + + +

    + Hyperboloid of one sheet +

    +     +     + + +

    + Hyperboloid of two sheets +

    +     +     + + +

    + Hyperbolic paraboloid +

    +     +     +     + +     + + + + +

    + Consider the hyperbola x^2-y^2=1 in the plane. + If this hyperbola is rotated about the x-axis, + what quadric surface is formed? +

    +     + + + +

    + A hyperboloid of two sheets +

    +     + +     + + + + +

    + Consider the hyperbola x^2-y^2=1 in the plane. + If this hyperbola is rotated about the y-axis, + what quadric surface is formed? +

    +     + + + +

    + A hyperboloid of one sheet +

    +     + +     +     + + + Problems + + + + parserPopUp.pl + + + Context("Point"); + $A=Point("(1,4,2)"); + $B=Point("(2,6,3)"); + $C=Point("(4,3,1)"); + $AB=$B-$A; + $BC=$C-$B; + $CA=$A-$C; + @ABcoords=$AB->value; + @BCcoords=$BC->value; + @CAcoords=$CA->value; + Context()->flags->set(reduceConstantFunctions=>0); + $magAB=Formula("sqrt(($ABcoords[0])^2+($ABcoords[1])^2+($ABcoords[2])^2)"); + $magBC=Formula("sqrt(($BCcoords[0])^2+($BCcoords[1])^2+($BCcoords[2])^2)"); + $magCA=Formula("sqrt(($CAcoords[0])^2+($CAcoords[1])^2+($CAcoords[2])^2)"); + ($a,$b,$c)=num_sort($magAB,$magBC,$magCA); + $dodonot=(($a)**2+($b)**2==($c)**2)?'do':'do not'; + $yn=DropDown(['do','do not'],$dodonot,showInStatic=>0); + + + +

    + The points A=(1,4,2), + B=(2,6,3) and C=(4,3,1) form a triangle in space. + Find the distances between each pair of points and determine if the triangle is a right triangle. +

    + + Find the distance \left\lVert\overline{AB}\right\rVert + +

    + +

    + + Find the distance \left\lVert\overline{BC}\right\rVert + +

    + +

    + + Find the distance \left\lVert\overline{CA}\right\rVert + +

    + +

    + + Use the drop-down to indicate whether or not the three points form a right triangle. + +

    + +

    +     + +     +     + + + + +

    + The points A=(1,1,3), B=(3,2,7), + C=(2,0,8) and D = (0,-1,4) form a quadrilateral ABCD in space. + Is this a parallelogram? +

    +     + +

    + Yes, as opposite sides have equal length. + \norm{\overline{AB}} = \sqrt{21}=\norm{\overline{CD}}; + \norm{\overline{BC}} = \sqrt{6}=\norm{\overline{AD}}. +

    +     + +     + + + + + Context("Point"); + $center=Point("(4,-1,0)"); + $radius=Real("3"); + + + +

    + Find the center and radius of the sphere defined by + + x^2-8x+y^2+2y+z^2+8=0: + +

    + + Enter the center of the sphere. +

    + +

    + + Enter the radius of the sphere. + +

    + +

    +     +     +     + + + + + Context("Point"); + $center=Point("(-2,1,2)"); + Context()->flags->set(reduceConstantFunctions=>0); + $radius=Formula("sqrt(5)"); + + + +

    + Find the center and radius of the sphere defined by + + x^2+y^2+z^2+4x-2y-4z+4=0: + +

    + + Enter the center of the sphere. +

    + +

    + + Enter the radius of the sphere. + +

    + +

    +     +     +     + + + + +

    + Describe the region in space defined by the inequalities. +

    +     + + + + +

    + x^2+y^2+z^2\lt 1 +

    +     + +

    + Interior of a sphere with radius 1 centered at the origin. +

    +     + +     + + + + + + + +

    + 0\leq x\leq 3 +

    +     + +

    + Region bounded between the planes x=0 + (the y-z coordinate plane) + and x=3. +

    +     +     +     + + + + +

    + x\geq 0,\ y\geq0, \ z\geq0 +

    +     + +

    + The first octant of space; + all points (x,y,z) where each of x, + y and z are positive. + (Analogous to the first quadrant in the plane.) +

    +     + +     + + + + +

    + y\geq 3 +

    +     + +

    + All points in space where the y value is greater than 3; + viewing space as often depicted in this text, + this is the region to the right + of the plane y=3 (which is parallel to the x-z coordinate plane.) +

    +     + +     + +     + + + + +

    + Sketch the cylinder in space. +

    +     + + + + +

    + z=x^3 +

    +     + + + + A surface obtained by translating the cubic curve along the y axis + +

    + Viewed with the y axis pointing out the page, + the surface looks like the graph z=x^3 in the xz plane. + The rest of the surface is generated by sliding this curve back and forth along the y axis. + The result is a sheet that looks something like a lounge chair. +

    +     + + + + + //ASY file for fig10_01_ex_153D.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,1.25); + pair zbounds=(-1.25,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the parabola z=x^3 for t from -2 to 2 + triple g(real t) {return (t,0,t^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,bluepen); + + //Draw the cylinder z=x^3 + triple f(pair t) { + return (t.x,t.y,(t.x)^3); + } + surface s=surface(f,(-1,-1),(1,1),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces (in this case lines) + triple g(real t) {return (-0.2,t,(-0.2)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (-0.4,t,(-0.4)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (-0.6,t,(-0.6)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (-0.8,t,(-0.8)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (-1,t,(-1)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (0.2,t,(0.2)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (0.4,t,(0.4)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (0.6,t,(0.6)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (0.8,t,(0.8)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (1,t,(1)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + triple g(real t) {return (t,0,(t)^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,bluepen); + + + + +     + +     + + + + +

    + y=\cos(z) +

    +     + + + + A surface generated by translating the curve y=cos(z) along the x axis + +

    + In the yz plane, the surface appears to be a cosine curve. + This curve is translated back and forth along the x axis, + so that when viewed from other angles, it appears to be a wavy sheet. +

    +     + + + + + //ASY file for fig10_01_ex_173D.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={-3,3}; + real[] myzchoice={-6,6}; + defaultpen(0.5mm); + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2*pi,2*pi); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the cylinder on top + triple f(pair t) { + return (t.x,cos(t.y),t.y); + } + surface s=surface(f,(-2,-2*pi),(2,2*pi),8,32,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces (in this case vertical lines) + triple g(real t) {return (t,cos(-3*pi/2),-3*pi/2);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (t,cos(-pi/2),-pi/2);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (t,cos(0),0);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (t,cos(3*pi/2),3*pi/2);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (t,cos(pi/2),pi/2);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (0,cos(t),t);} + path3 mypath=graph(g,-2*pi,pi*2,operator ..); draw(mypath,bluepen); + + + + +     + +     + + + + +

    + \ds \frac{x^2}{4}+\frac{y^2}{9}=1 +

    +     + + + + A cylindrical tube with cross sections in the shape of an ellipse + +

    + A classic cylinder, in the shape of a tube. + Cross-sections parallel to the xy plane are ellipses. +

    +     + + + + + //ASY file for fig10_01_ex_173D.pdf in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-3,3}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-4,4); + pair ybounds=(-4,4); + pair zbounds=(-4,4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the cylinder on top + triple f(pair t) { + return (2*cos(t.x),3*sin(t.x),t.y); + } + surface s=surface(f,(0,-2),(2*pi,2),32,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces (in this case vertical lines) + triple g(real t) {return (2*cos(0),3*sin(0),t);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (2*cos(2*pi/10),3*sin(2*pi/10),t);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (2*cos(2*2*pi/10),3*sin(2*2*pi/10),t);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (2*cos(3*2*pi/10),3*sin(3*2*pi/10),t);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (2*cos(4*2*pi/10),3*sin(4*2*pi/10),t);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (2*cos(5*2*pi/10),3*sin(5*2*pi/10),t);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (2*cos(6*2*pi/10),3*sin(6*2*pi/10),t);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (2*cos(7*2*pi/10),3*sin(7*2*pi/10),t);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (2*cos(8*2*pi/10),3*sin(8*2*pi/10),t);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + triple g(real t) {return (2*cos(9*2*pi/10),3*sin(9*2*pi/10),t);} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); + + triple g(real t) {return (2*cos(t),3*sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + + + +     + +     + + + + +

    + \ds y=\frac1x +

    +     + + + + A pair of sheets that look like a hyperbola when viewed from above + +

    + Viewed along the z axis, the surface appears to be the hyperbola y=1/x. + From other angles, it is a pair of curved sheets, bent in the shape of a hyperbola. +

    +     + + + + + //ASY file for fig10_01_ex_20.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(-8,41,2.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-10,10}; + real[] myychoice={-10,10}; + real[] myzchoice={-5,5}; + defaultpen(0.5mm); + + pair xbounds=(-12,12); + pair ybounds=(-12,12); + pair zbounds=(-6,6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface y=1/x + triple f(pair t) { + return (t.x,1/t.x,t.y);//({y^2},{cos(x)*y},{3*sin(x)*y}); + } + surface s=surface(f,(0.1,-3),(10,3),16,8); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (t.x,1/t.x,t.y);//({y^2},{cos(x)*y},{3*sin(x)*y}); + } + surface s=surface(f,(-10,-3),(-.1,3),16,8); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the traces (in this case vertical lines) + triple g(real t) {return (1,1,t);} + path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); + triple g(real t) {return (-1,-1,t);} + path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); + triple g(real t) {return (5,.2,t);} + path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); + triple g(real t) {return (-5,-.2,t);} + path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); + triple g(real t) {return (10,.1,t);} + path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); + triple g(real t) {return (-10,-.1,t);} + path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); + triple g(real t) {return (t,1/t,0);} + path3 mypath=graph(g,.1,10,operator ..); draw(mypath,bluepen); + triple g(real t) {return (t,1/t,0);} + path3 mypath=graph(g,-10,-.1,operator ..); draw(mypath,bluepen); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +     + +     + +     + + + +

    + Give the equation of the surface of revolution described. +

    +     + + + + + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + $eq=ImplicitEquation("x^2+z^2=(1/(1+y^2))^2"); + + + +

    + Give the equation of the surface formed by revolving + z=\frac{1}{1+y^2} in the yz-plane about the y-axis. +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + $eq=ImplicitEquation("y^2+z^2=x^4"); + + + +

    + Give the equation of the surface formed by revolving y=x^2 in the xy-plane about the x-axis. +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + $eq=ImplicitEquation("x^2+y^2=z"); + + + +

    + Give the equation of the surface formed by revolving z=x^2 in the xz-plane about the z-axis. +

    + +

    + +

    +     +     +     + + + + + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + $eq=ImplicitEquation("x^2+y^2=1/z^2"); + + + +

    + Give the equation of the surface formed by revolving z=1/x in the xz-plane about the z-axis. +

    + +

    + +

    +     +     +     + +     + + + +

    + A quadric surface is sketched. + Determine which of the given equations best fits the graph. +

    +     + + + + + + + + Graph showing a paraboloid opening along the x axis. + + +

    + The z and y axes are drawn from -3 to 3 and the x axis is + drawn from -1 to 1. The y and z axes are drawn from -3 to 3. The elliptic paraboloid is shown with centre at the origin and it opens along the positive x axis. +

    +     + + + + + //ASY file for fig10_01_ex_19.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(3,12,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={-3,3}; + real[] myzchoice={-3,3}; + defaultpen(0.5mm); + + pair xbounds=(-.5,1.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-3.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2 + y^2/9 + z^2/4 = 1 + triple f(pair t) { + return (t.y^2,cos(t.x)*t.y,3*sin(t.x)*t.y);//({y^2},{cos(x)*y},{3*sin(x)*y}); + } + surface s=surface(f,(0,0),(2*pi,1),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +

    + (a) \ds x=y^2+\frac{z^2}{9} (b) \ds x=y^2+\frac{z^2}{3} +

    +     + +

    + (a)\ds x=y^2+\frac{z^2}{9} +

    +     + +     + + + + + + + + Graph showing elliptic cone opening along the y axis. + + +

    + The axes are drawn from -1 to 1. Two hollow elliptic cones are drawn with + vertices at the origin, one opening along the positive y axis and the other along + the negative y axis. +

    +     + + + + + //ASY file for fig10_01_ex_20.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,3,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2 + y^2/9 + z^2/4 = 1 + triple f(pair t) { + return (cos(t.x)*(t.y)^2,(t.y)^2,sin(t.x)*(t.y)^2);//({cos(x)*(y)^2},{y^2},{sin(x)*(y)^2}); + } + surface s=surface(f,(0,0),(2*pi,1),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (cos(t.x)*(t.y)^2,-(t.y)^2,sin(t.x)*(t.y)^2);//({cos(x)*(y)^2},{y^2},{sin(x)*(y)^2}); + } + surface s=surface(f,(0,0),(2*pi,1),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +

    + (a) \ds x^2-y^2-z^2=0 (b) x^2-y^2+z^2=0 +

    +     + +

    + (b) x^2-y^2+z^2=0 +

    +     + +     + + + + + + + + Graph of an ellipsoid. + + +

    + The three axes are uncalibrated. An ellipsoid is shown. +

    +     + + + + + //ASY file for fig10_01_ex_21.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-3,3}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-3.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2 + y^2/9 + z^2/4 = 1 + triple f(pair t) { + return (cos(t.x)*cos(t.y),3*sin(t.x)*cos(t.y),2*sin(t.y));//({cos(x)*cos(y)},{sin(x)*3*cos(y)},{2*sin(y)}); + } + surface s=surface(f,(-pi,-pi/2),(pi,pi/2),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +

    + (a) \ds x^2+\frac{y^2}3+\frac{z^2}2=1 (b) \ds x^2+\frac{y^2}9+\frac{z^2}4=1 +

    +     + +

    + (b) \ds x^2+\frac{y^2}9+\frac{z^2}4=1 +

    +     + +     + + + + + + + + Graph showing hyperboloid with two sheets. + + +

    + The x, y and z axes are drawn from -2 to 2. + The hyperboloid of two sheets is drawn about the y axis. The first sheet + has a centre at y =1 and opens along the positive y axis. The second + sheet has a centre at y=-1 and opens along the negative y axis. +

    +     + + + + + //ASY file for fig10_01_ex_22.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface 4x^2 - y^2 - z^2/9 = 1 + triple f(pair t) { + return (cos(t.y)*tan(t.x),1/cos(t.x),sin(t.y)*tan(t.x));//({cos(y)*tan(x)},{sec(x)},{tan(x)*sin(y)}); + } + surface s=surface(f,(0,0),(1,2*pi),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (cos(t.y)*tan(t.x),-1/cos(t.x),sin(t.y)*tan(t.x));//({cos(y)*tan(x)},{sec(x)},{tan(x)*sin(y)}); + } + surface s=surface(f,(0,0),(1,2*pi),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Add labels + //label("In plane $y=0$",(0,4,1.5),E); + //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); + //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); + //label("In plane $z=d$",(4,0,-1),W); + //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); + + + + +

    + (a) y^2-x^2-z^2=1 (b) y^2+x^2-z^2=1 +

    +     + +

    + (a) y^2-x^2-z^2=1 +

    +     + +     + +     + + + +

    + Sketch the quadric surface. +

    +     + + + + +

    + \ds z-y^2+x^2=0 +

    +     + + + + A hyperbolic paraboloid (the classic "Pringles chip") + +

    + The surface is a standard hyperbolic paraboloid, opening upward along the y axis, + and downward along the x axis. +

    +     + + + + + //ASY file for fig10_01_ex_283D.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,3,1); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.25,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the hyp par on top + //Draw the surface z^2 - y^2 - x^2=0 + triple f(pair t) { + return (t.x,t.y,-(t.x)^2+(t.y)^2); + } + surface s=surface(f,(-1,-1),(1,1),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +     + +     + + + + +

    + \ds z^2=x^2+\frac{y^2}4 +

    +     + + + + An elliptical cone + +

    + The surface is a cone, opening along the z axis. + Cross-sections parallel to the xy plane are ellipses. +

    +     + + + + + //ASY file for fig10_01_ex_263D.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,5,1); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-4,4}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + pair xbounds=(-5,5); + pair ybounds=(-5,5); + pair zbounds=(-3,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2/9 - y^2 + z^2/25 = 1 + triple f(pair t) { + return (t.y*cos(t.x),2*t.y*sin(t.x),t.y);// + } + surface s=surface(f,(0,-2),(2*pi,2),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +     + +     + + + + +

    + x=-y^2-z^2 +

    +     + + + + A circular paraboloid, opening along the negative x axis + +

    + The surface is a circular paraboloid, opening along the negative x axis. +

    +     + + + + + //ASY file for fig10_01_ex_233D.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.25,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x=-y^2-z^2 + triple f(pair t) { + return (-t.x,sqrt(t.x)*cos(t.y),sqrt(t.x)*sin(t.y)); + } + surface s=surface(f,(0,0),(1,2*pi),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +     + +     + + + + +

    + \ds 16x^2-16y^2-16z^2=1 +

    +     + + + + A hyperboloid of two sheets, opening along the x axis. + +

    + A hyperboloid of two sheets with circular cross sections, opening along the x axis. +

    +     + + + + + //ASY file for fig10_01_ex_263D.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,5,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-.25,.25}; + real[] myychoice={-.5,.5}; + real[] myzchoice={-.5,.5}; + defaultpen(0.5mm); + pair xbounds=(-1,1); + pair ybounds=(-1,1); + pair zbounds=(-1,1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2/9 - y^2 + z^2/25 = 1 + triple f(pair t) { + return (.25/cos(t.x),.25*cos(t.y)*tan(t.x),.25*sin(t.y)*tan(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) + } + surface s=surface(f,(-pi/3,0),(pi/3,pi),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (-.25/cos(t.x),.25*cos(t.y)*tan(t.x),.25*sin(t.y)*tan(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) + } + surface s=surface(f,(-pi/3,0),(pi/3,pi),32,16); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +     + +     + + + + +

    + \ds \frac{x^2}9-y^2+\frac{z^2}{25}=1 +

    +     + + + + A hyperboloid of one sheet, opening along the y axis + +

    + A hyperboloid of one sheet. It opens along the y axis, + and cross sections parallel to the xz plane are ellipses. +

    +     + + + + + //ASY file for fig10_01_ex_263D.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(13,5,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-3,3}; + real[] myychoice={-1,1}; + real[] myzchoice={-5,5}; + defaultpen(0.5mm); + pair xbounds=(-6,6); + pair ybounds=(-1.5,1.5); + pair zbounds=(-6,6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2/9 - y^2 + z^2/25 = 1 + triple f(pair t) { + return (3*cos(t.y)/cos(t.x),tan(t.x),5*sin(t.y)/cos(t.x)); + } + surface s=surface(f,(-1,0),(1,2*pi),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +     + +     + + + + +

    + \ds 4x^2+2y^2+z^2=4 +

    +     + + + + An ellpsoid, somewhat in the shape of a squashed olive. + +

    + An ellpsoid, somewhat in the shape of a squashed olive. +

    +     + + + + + //ASY file for fig10_01_ex_263D.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,5,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-sqrt(2),sqrt(2)}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2/9 - y^2 + z^2/25 = 1 + triple f(pair t) { + return (cos(t.y)*cos(t.x),sqrt(2)*cos(t.y)*sin(t.x),2*sin(t.y));//({cos(x)*1.5*cos(y)},{sin(x)*cos(y)},{sin(y)}) + } + surface s=surface(f,(-pi,-pi/2),(1*pi,pi/2),32,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +     + +     + +     +     +     +
    +
    + An Introduction to Vectors +

    + Many quantities we think about daily can be described by a single number: + temperature, speed, cost, weight and height. + There are also many other concepts we encounter daily that cannot be described with just one number. + For instance, + a weather forecaster often describes wind with its speed and its direction + (\ldots with winds from the southeast gusting up to 30 mph \ldots). + When applying a force, + we are concerned with both the magnitude and direction of that force. + In both of these examples, direction is important. + Because of this, we study vectors, + mathematical objects that convey both magnitude and direction information. +

    + + + +

    + One bare-bones definition of a vector is based on what we wrote above: + a vector is a mathematical object with magnitude and direction parameters. + This definition leaves much to be desired, + as it gives no indication as to how such an object is to be used. + Several other definitions exist; + we choose here a definition rooted in a geometric visualization of vectors. + It is very simplistic but readily permits further investigation. +

    + + + Vector + +

    + A vector is a directed line segment. +

    + +

    + Given points P and Q + (either in the plane or in space), + we denote with \overrightarrow{PQ} the vector + from P to Q. + The point P is said to be the + initial point of the vector, + and the point Q is the terminal point. +

    + +

    + The magnitude, + length or norm + of \overrightarrow{PQ} is the length of the line segment \overline{PQ}: + \norm{\overrightarrow{PQ}} = \norm{\overline{PQ}}. +

    + +

    + Two vectors are equal + if they have the same magnitude and direction. + vectors + vectorsdefinition + vectorsnorm + vectorsmagnitude + norm + magnitude of vector + terminal point + initial point +

    +     +     + +

    + + shows multiple instances of the same vector. + Each directed line segment has the same direction and length (magnitude), + hence each is the same vector. +

    + +
    + Drawing the same vector with different initial points + + + + Graph shows the same vectors drawn at different initial points. + + +

    + The x and y axes are drawn from -4 to 4. There are four vectors drawn. + In the first quadrant is the first vector drawn from point (0,0) to (3,1). The second + vector is drawn in the second quadrant from point (-4,1) to point (-1,2). In the fourth + quadrant the vector is drawn from (2, -4) to (5, -3). The last vector is drawn from + (-2, -3) to (1, -2) and it crosses the third . +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + ymin=-5.5,ymax=5.5, + xmin=-5.5,xmax=5.5 + ] + + \draw [very thick,->] (axis cs:0,0) -- (axis cs:3,1); + \draw [very thick,->] (axis cs:-2,-3) -- (axis cs:1,-2); + \draw [very thick,->] (axis cs:-4,1) -- (axis cs:-1,2); + \draw [very thick,->] (axis cs:2,-4) -- (axis cs:5,-3); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + We use \mathbb{R}^2 + (pronounced r two) + to represent all the vectors in the plane, + and use \mathbb{R}^3 + (pronounced r three) + to represent all the vectors in space. + r@\mathbb{R} +

    + +
    + Illustrating how equal vectors have the same displacement + + + + Graphs showing equal vectors have the same displacement. + + +

    + The x and y axes are drawn from -4 to 4. There are two vectors + PQ and RS. The vector PQ is drawn in the first quadrant from point + P=(1,0) and Q=(3,1). The second vector is in the second quadrant from point + R=(-3, 1) to S=(-1,2). +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + ymin=-4.5,ymax=4.5, + xmin=-4.5,xmax=4.5 + ] + + \draw [very thick,->] (axis cs:1,0) node [above] { $P$} -- (axis cs:3,1) node [above] { $Q$}; + \draw [very thick,->] (axis cs:-3,1) node [above] { $R$} -- (axis cs:-1,2) node [above] { $S$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Consider the vectors \overrightarrow{PQ} and \overrightarrow{RS} as shown in . + The vectors look to be equal; + that is, they seem to have the same length and direction. + Indeed, they are. + Both vectors move 2 units to the right and 1 unit up from the initial point to reach the terminal point. + One can analyze this movement to measure the magnitude of the vector, + and the movement itself gives direction information + (one could also measure the slope of the line passing through P and Q or R and S). + Since they have the same length and direction, + these two vectors are equal. +

    + +

    + This demonstrates that inherently all we care about is displacement; + that is, how far in the x, + y and possibly z directions the terminal point is from the initial point. + Both the vectors \overrightarrow{PQ} and \overrightarrow{RS} in + have an x-displacement of 2 and a y-displacement of 1. + This suggests a standard way of describing vectors in the plane. + A vector whose x-displacement is a and whose y-displacement is b will have terminal point (a,b) when the initial point is the origin, + (0,0). + This leads us to a definition of a standard and concise way of referring to vectors. +

    + + + Component Form of a Vector + +

    +

      +
    1. +

      + The component form of a vector \vec{v} in \mathbb{R}^2, + whose terminal point is (a,b) when its initial point is (0,0), + is \la a,b\ra. +

      +
      2. + +
    2. +

      + The component form of a vector \vec{v} in \mathbb{R}^3, + whose terminal point is (a,b,c) when its initial point is (0,0,0), + is \la a,b,c\ra. +

      +
      3. +
    +

    + +

    + The numbers a, b + (and c, respectively) + are the components of \vec v. + vectorscomponent form +

    +     +     + +

    + It follows from the definition that the component form of the vector \overrightarrow{PQ}, + where P=(x_1,y_1) and Q=(x_2,y_2) is + + \overrightarrow{PQ} = \la x_2-x_1, y_2-y_1\ra; + + in space, where P=(x_1,y_1,z_1) and Q=(x_2,y_2,z_2), + the component form of \overrightarrow{PQ} is + + \overrightarrow{PQ} = \la x_2-x_1, y_2-y_1,z_2-z_1\ra + . +

    + +

    + We practice using this notation in the following example. +

    + + + Using component form notation for vectors + +

    +

      +
    1. +

      + Sketch the vector \vec v=\la 2,-1\ra starting at P=(3,2) and find its magnitude. +

      +
      2. + +
    2. +

      + Find the component form of the vector \vec w whose initial point is + R=(-3,-2) and whose terminal point is S=(-1,2). +

      +
      3. + +
    3. +

      + Sketch the vector \vec u = \la 2,-1,3\ra starting at the point Q = (1,1,1) and find its + magnitude. +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + Using P as the initial point, + we move 2 units in the positive x-direction and -1 units in the positive y-direction to arrive at the terminal point P\,'=(5,1), + as drawn in . + + The magnitude of \vec v is determined directly from the component form: + + \norm{\vec v} =\sqrt{2^2+(-1)^2} = \sqrt{5} + . +

      +
      + Graphing vectors in + +
      + + + + + Graph showing two vectors RS and PP';. + + +

      + The x and y axes are drawn from -4 to 4. Two vectors are drawn. + The vector PP’ is in the first quadrant it is downward facing, it starts from + P=(3, 2) and ends at P'=(5, 1). The second vector RS from (-3, -2) + to (-1, 2). This vector lies midway between the second and the third quadrant. +

      +       + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + minor x tick num=4, + minor y tick num=4, + ymin=-5.5,ymax=5.5, + xmin=-5.5,xmax=5.9 + ] + + \draw [very thick,->] (axis cs:3,2) node [left] { $P$} -- (axis cs:5,1) node [shift={(5pt,0pt)}] { $P\,'$}; + \draw [very thick,->] (axis cs:-3,-2) node [below left] { $R$} -- (axis cs:-1,2) node [above] { $S$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      + +
      + + + + + Graph showing vector QQ'. + + +

      + The x and y axes are drawn from 0 to 2. The z axis is drawn + from 0 to 4. The points Q = (1, 1, 1) and Q'= (3, 0, 4) along with + the vector are drawn in space. + A dashed cube is drawn with Q at the vertex furthest away from the origin. +

      +       + + + + + //ASY file for figvectintro3b.pdf in Chapter 10.1 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,9,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={-2,4}; + defaultpen(0.5mm); + + pair xbounds=(-1,3.5); + pair ybounds=(-1,2.5); + pair zbounds=(-1,5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // Draw the lines for Q=(1,1,1) + draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1), redpen+dashed+linewidth(.5));//top Q + draw((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//bottom Q + draw((1,0,0)--(1,0,1), redpen+dashed+linewidth(.5));//up1 Q + draw((0,1,0)--(0,1,1), redpen+dashed+linewidth(.5));//up2 Q + draw((1,1,0)--(1,1,1), redpen+dashed+linewidth(.5));//up3 Q + label("$Q$",(1,1,1),N); + //dotfactor=3; dot((2,1,1),bluepen); + + // Draw the lines for Q'=(3,0,4) + draw((0,0,4)--(0,0,4)--(3,0,4)--(3,0,4)--(3,0,4), redpen+dashed+linewidth(.5));//top P + draw((0,0,0)--(3,0,0)--(3,0,0)--(3,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//bottom P + draw((3,0,0)--(3,0,4), redpen+dashed+linewidth(.5));//up1 P + draw((0,0,0)--(0,0,3), redpen+dashed+linewidth(.5));//up2 P + draw((3,0,0)--(3,0,3), redpen+dashed+linewidth(.5));//up3 P + label("$Q'$",(3,0,4),N); + //dotfactor=3; dot((1,4,-1),bluepen); + + //line from P to Q + draw((1,1,1)--(3,0,4), black,Arrow3(size=2mm)); + + + + +
      +       +
      +
      2. + +
    2. +

      + Using the note following , we have + + \overrightarrow{RS} = \la -1-(-3), 2-(-2)\ra = \la 2,4\ra + . + One can readily see from that the x- and y-displacement of \overrightarrow{RS} is 2 and 4, respectively, + as the component form suggests. +

      +
      3. + +
    3. +

      + Using Q as the initial point, + we move 2 units in the positive x-direction, + -1 unit in the positive y-direction, + and 3 units in the positive z-direction to arrive at the terminal point Q' = (3,0,4), + illustrated in . + + The magnitude of \vec u is: + + \norm{\vec u} = \sqrt{2^2+(-1)^2+3^2} = \sqrt{14} + . +

      +
      4. +
    +

    +     + +     + + + +

    + Now that we have defined vectors, + and have created a nice notation by which to describe them, + we start considering how vectors interact with each other. + That is, we define an algebra on vectors. +

    + + + Vector Algebra + +

    +

      +
    1. +

      + Let \vec u = \la u_1,u_2\ra and + \vec v = \la v_1,v_2\ra be vectors in \mathbb{R}^2, + and let c be a scalar. + vectorsalgebra of +

      + +

      +

        +
      1. +

        + The addition, or sum, + of the vectors \vec u and \vec v is the vector + + \vec u+\vec v = \la u_1+v_1, u_2+v_2\ra + . +

        +
        2. + +
      2. +

        + The multiplication of a scalar c and a vector \vec v is the vector + + c\vec v = c\la v_1,v_2\ra = \la cv_1,cv_2\ra + . +

        +
        3. +
      +

      +
      2. + +
    2. +

      + Let \vec u = \la u_1,u_2,u_3\ra and + \vec v = \la v_1,v_2,v_3\ra be vectors in \mathbb{R}^3, + and let c be a scalar. +

      + +

      +

        +
      1. +

        + The addition, or sum, + of the vectors \vec u and \vec v is the vector + + \vec u+\vec v = \la u_1+v_1, u_2+v_2, u_3+v_3\ra + . +

        +
        2. + +
      2. +

        + The multiplication of a scalar c and a vector \vec v is the vector + + c\vec v = c\la v_1,v_2,v_3\ra = \la cv_1,cv_2,cv_3\ra + . +

        +
        3. +
      +

      +
      3. +
    +

    +     +     + +

    + In short, we say addition and scalar multiplication are computed + component-wise. +

    + + + + + Adding vectors + +

    + Sketch the vectors \vec u = \la1,3\ra, + \vec v = \la 2,1\ra and + \vec u+\vec v all with initial point at the origin. +

    +     + +

    + We first compute \vec u +\vec v. + + \vec u+\vec v \amp = \la 1,3\ra + \la 2,1\ra + \amp = \la 3,4\ra + . +

    + +
    + Graphing the sum of vectors in + + + + Graph showing sum of two vectors u and v. + + +

    + The x and y axes are drawn from 0 to 4. Two vectors \vec u and \vec v + are shown along with the vector addition of the two. The \vec u vector is drawn from point + (0,0) to (1,3). The \vec v vector is drawn from origin to (2,1). The vector + \vec u is longer than the \vec v vector. The \vec u + \vec v vector is drawn from + the origin to (3, 4). + The \vec u + \vec v vector is the longest and is in between the two vectors. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=4.1, + xmin=-.5,xmax=4.1 + ] + + \draw [very thick,->] (axis cs:0,0) -- (axis cs:1,3) node [pos=.5,above,rotate=70] { $\vec u$}; + \draw [very thick,->] (axis cs:0,0) -- (axis cs:2,1) node [pos=.5,above,rotate=40] { $\vec v$}; + \draw [very thick,->] (axis cs:0,0) -- (axis cs:3,4) node [pos=.5,above,rotate=50] { $\vec u+\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + These are all sketched in . +

    +     + +     + +

    + As vectors convey magnitude and direction information, + the sum of vectors also convey length and magnitude information. + Adding \vec u+\vec v suggests the following idea: +

    + +
    +

    + Starting at an initial point, go out \vec u, then go out \vec v. +

    +
    + +

    + This idea is sketched in , + where the initial point of \vec v is the terminal point of \vec u. + This is known as the Head to Tail Rule of adding vectors. + Vector addition is very important. + For instance, + if the vectors \vec u and \vec v represent forces acting on a body, + the sum \vec u+\vec v gives the resulting force. + Because of various physical applications of vector addition, + the sum \vec u+\vec v is often referred to as the + resultant vector, + or just the resultant. + vectorsresultant + Parallelogram Law + vectorsParallelogram Law + Head To Tail Rule + vectorsHead To Tail Rule + vectorszero vector +

    + +
    + Illustrating how to add vectors using the Head to Tail Rule and Parallelogram Law + + + + Graph showing sum of two vectors u and v. + + +

    + The x and y axes are drawn from 0 to 4. Two vectors \vec u and + \vec v are shown along with the vector addition of the two. The \vec u vector is drawn + from point (0,0) to (1,3). The \vec v vector is drawn from origin to (2,1). +

    +

    + The \vec v vector is translated to start from the point (1, 3) to (3, 4) and it + forms a triangle with \vec u and u+v. The \vec u vector is translated to start + from point (2, 1) to point (3, 4). The vector \vec u is longer than the \vec v + vector and it forms a triangle with \vec u and u+v. The \vec u + \vec v vector is drawn + from the origin to (3, 4). + The \vec u + \vec v vector is the longest and is in between the two vectors. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=4.1, + xmin=-.5,xmax=4.1 + ] + + \draw [very thick,->] (axis cs:0,0) -- (axis cs:1,3) node [pos=.5,above,rotate=70] { $\vec u$}; + \draw [very thick,->] (axis cs:0,0) -- (axis cs:2,1) node [pos=.5,above,rotate=30] { $\vec v$}; + + \draw [very thick,->,gray] (axis cs:1,3) -- (axis cs:3,4) node [pos=.5,above,rotate=30] { $\vec v$}; + \draw [very thick,->,gray] (axis cs:2,1) -- (axis cs:3,4) node [pos=.4,above,rotate=70] { $\vec u$}; + + \draw [very thick,->] (axis cs:0,0) -- (axis cs:3,4) node [pos=.5,above,rotate=50] { $\vec u+\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Analytically, + it is easy to see that \vec u+\vec v = \vec v+\vec u. + + also gives a graphical representation of this, using gray vectors. + Note that the vectors \vec u and \vec v, + when arranged as in the figure, form a parallelogram. + Because of this, + the Head to Tail Rule is also known as the Parallelogram Law: + the vector \vec u+\vec v is defined by forming the parallelogram defined by the vectors \vec u and \vec v; + the initial point of \vec u+\vec v is the common initial point of parallelogram, + and the terminal point of the sum is the common terminal point of the parallelogram. +

    + +

    + While not illustrated here, + the Head to Tail Rule and Parallelogram Law hold for vectors in \mathbb{R}^3 as well. +

    + +

    + It follows from the properties of the real numbers and that + + \vec u-\vec v = \vec u + (-1)\vec v + . +

    + +

    + The Parallelogram Law gives us a good way to visualize this subtraction. + We demonstrate this in the following example. +

    + + + Vector Subtraction + +

    + Let \vec u = \la 3,1\ra and \vec v=\la 1,2\ra. + Compute and sketch \vec u-\vec v. +

    +     + +

    + The computation of \vec u-\vec v is straightforward, + and we show all steps below. + Usually the formal step of multiplying by (-1) is omitted and we just subtract. + + \vec u-\vec v \amp = \vec u + (-1)\vec v + \amp = \la 3,1\ra + \la -1,-2\ra + \amp = \la 2,-1\ra + . +

    + +
    + Illustrating how to subtract vectors graphically + + + + Graph showing subtraction of two vectors u from v. + + +

    + The x axis is drawn from 0 to 4 and the y axis is drawn from -1 to + 3. The \vec u vector is drawn from origin to point (1, 2). The \vec v vector + is drawn from the origin to (2,1). The vector \vec u - \vec v is drawn from origin to + (2, -1) and it lies in the fourth quadrant. +

    +

    + The \vec u - \vec v vector is translated to start from point (1, 2) and ends at point + (3, 1). The \vec v vector is also translated but in the opposite direction and it starts + from (3, 1) and ends at point (2, -1). +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + ymin=-1.5,ymax=3.1, + xmin=-.5,xmax=4.1 + ] + + \draw [very thick,->] (axis cs:0,0) -- (axis cs:3,1) node [pos=.5,above,rotate=20] { $\vec u$}; + \draw [very thick,->] (axis cs:0,0) -- (axis cs:1,2) node [pos=.5,above,rotate=60] { $\vec v$}; + \draw [very thick,->] (axis cs:0,0) -- (axis cs:2,-1) node [pos=.5,below,rotate=-20] { $\vec u-\vec v$}; + + \draw [very thick,->,gray] (axis cs:3,1) -- (axis cs:2,-1) node [black,pos=.7,below,rotate=60] { $-\vec v$}; + \draw [very thick,->,gray] (axis cs:1,2) -- (axis cs:3,1) node [black,pos=.5,above,rotate=-20] { $\vec u-\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + illustrates, + using the Head to Tail Rule, + how the subtraction can be viewed as the sum \vec u + (-\vec v). + The figure also illustrates how + \vec u-\vec v can be obtained by looking only at the terminal points of \vec u and \vec v + (when their initial points are the same). +

    +     + +     + + + Scaling vectors + +

    +

      +
    1. +

      + Sketch the vectors \vec v = \la 2,1\ra and 2\vec v with initial point at the origin. +

      +
      2. + +
    2. +

      + Compute the magnitudes of \vec v and 2\vec v. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + We compute 2\vec v: + + 2\vec v \amp = 2\la 2,1\ra + \amp = \la 4,2\ra + . +

      + +
      + Graphing vectors \vec v and 2\vec v in + + + + Graph showing two similar vectors of different magnitudes. + + +

      + The x axis is drawn from 0 to 4 and the y axis is drawn from -1 + to 3. The \vec v vector is drawn from origin to point (2,1), another vector + 2\vec v is also drawn from origin to point (4, 2). The two vectors have the same + direction. +

      +       + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=3.1, + xmin=-.5,xmax=4.1 + ] + + \draw [gray,very thick,->] (axis cs:0,0) -- (axis cs:4,2) node [pos=.7,above,rotate=30,black] { $2\vec v$}; + + \draw [very thick,->] (axis cs:0,0) -- (axis cs:2,1) node [pos=.5,above,rotate=30] { $\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      + +

      + Both \vec v and 2\vec v are sketched in . + Make note that 2\vec v does not start at the terminal point of \vec v; + rather, its initial point is also the origin. +

      +
      2. + +
    2. +

      + The figure suggests that 2\vec v is twice as long as \vec v. + We compute their magnitudes to confirm this. + + \norm{\vec v} \amp = \sqrt{2^2+1^2} + \amp = \sqrt{5}. + \norm{2\vec v}\amp =\sqrt{4^2+2^2} + \amp = \sqrt{20} + \amp = \sqrt{4\cdot 5} = 2\sqrt{5} + . + As we suspected, 2\vec v is twice as long as \vec v. +

      +
      3. +
    +

    +     + +     + +

    + The zero vector is the vector whose initial point is also its terminal point. + It is denoted by \vec 0. + Its component form, in \mathbb{R}^2, is \la 0,0\ra; + in \mathbb{R}^3, it is \la 0,0,0\ra. + Usually the context makes is clear whether \vec 0 is referring to a vector in the plane or in space. +

    + + + +

    + Our examples have illustrated key principles in vector algebra: + how to add and subtract vectors and how to multiply vectors by a scalar. + The following theorem states formally the properties of these operations. +

    + + + Properties of Vector Operations + +

    + The following are true for all scalars c and d, + and for all vectors \vec u, + \vec v and \vec w, where \vec u, + \vec v and \vec w are all in + \mathbb{R}^2 or where \vec u, + \vec v and \vec w are all in + \mathbb{R}^3: + vectorsalgebraic properties +

    + +

    +

      +
    1. +

      + \vec u+\vec v = \vec v+\vec u Commutative Property +

      +
      2. + +
    2. +

      + (\vec u+\vec v)+\vec w = \vec u+(\vec v+\vec w) Associative Property +

      +
      3. + +
    3. +

      + \vec v+\vec 0 = \vec v Additive Identity +

      +
      4. + +
    4. +

      + (cd)\vec v= c(d\vec v) +

      +
      5. + +
    5. +

      + c(\vec u+\vec v) = c\vec u+c\vec v Distributive Property +

      +
      6. + +
    6. +

      + (c+d)\vec v = c\vec v+d\vec v Distributive Property +

      +
      7. + +
    7. +

      + 0\vec v = \vec 0 +

      +
      8. + +
    8. + +

      + \norm{c\vec v} = \abs{c}\cdot\norm{\vec v} +

      +
      9. + +
    9. + +

      + \vnorm u = 0 if, and only if, \vecu = \vec 0. +

      +
      10. +
    +

    +     +     + + + +

    + As stated before, + each nonzero vector \vec v conveys magnitude and direction information. + We have a method of extracting the magnitude, + which we write as \norm{\vec v}. + Unit vectors are a way of extracting just the direction information from a vector. +

    + + + Unit Vector + +

    + A unit vector is a vector \vec v with a magnitude of 1; + that is, + vectorsunit vector + unit vector + + \norm{\vec v}=1 + . +

    +     +     + + + +

    + Consider this scenario: + you are given a vector \vec v and are told to create a vector of length 10 in the direction of \vec v. + How does one do that? + If we knew that \vec u was the unit vector in the direction of \vec v, + the answer would be easy: 10\vec u. + So how do we find \vec u? +

    + +

    + Property + of holds the key. + If we divide \vec v by its magnitude, + it becomes a vector of length 1. + Consider: + + \snorm{\frac{1}{\norm{\vec v}}\vec v} \amp = \frac{1}{\norm{\vec v}}\norm{\vec v} \amp \text{ (we can pull out \(\ds \frac{1}{\norm{\vec v}}\) as it is a positive scalar)} + \amp = 1 + . +

    + +

    + So the vector of length 10 in the direction of \vec v is \ds 10\frac{1}{\norm{\vec v}}\vec v. + An example will make this more clear. +

    + + + Using Unit Vectors + +

    + Let \vec v= \la 3,1\ra and let \vec w = \la 1,2,2\ra. +

    + +

    +

      +
    1. +

      + Find the unit vector in the direction of \vec v. +

      +
      2. + +
    2. +

      + Find the unit vector in the direction of \vec w. +

      +
      3. + +
    3. +

      + Find the vector in the direction of \vec v with magnitude 5. +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + We find \norm{\vec v} = \sqrt{10}. + So the unit vector \vec u in the direction of \vec v is + + \vec u = \frac{1}{\sqrt{10}}\vec v = \la \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\ra + . +

      +
      2. + +
    2. +

      + We find \norm{\vec w} = 3, + so the unit vector \vec z in the direction of \vec w is + + \vec u = \frac13\vec w = \la \frac13,\frac23,\frac23\ra + . +

      +
      3. + +
    3. +

      + To create a vector with magnitude 5 in the direction of \vec v, + we multiply the unit vector \vec u by 5. + Thus 5\vec u = \la 15/\sqrt{10},5/\sqrt{10}\ra is the vector we seek. + This is sketched in . +

      + +
      + Graphing vectors in . All vectors shown have their initial point at the origin + + + + Graph showing three similar vectors of different magnitudes. + + +

      + The x axis is drawn from 0 to 5 and the y axis is drawn from 0 + to 3. The \vec u vector is drawn from origin to point (1, 0.4), another vector + \vec v is also drawn from origin to point (3, 1). The third vector 5\vec u is + also drawn from the origin to point (5, 1.5). The three vectors have the same direction. +

      +       + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=3.1, + xmin=-.5,xmax=5.1 + ] + + \draw [gray,very thick,->] (axis cs:0,0) -- (axis cs:4.74,1.58) node [pos=.8,above,rotate=30,black] { $5\vec u$}; + + \draw [very thick,->] (axis cs:0,0) -- (axis cs:3,1) node [pos=.6,above,rotate=30] { $\vec v$}; + + \draw [very thick,->,gray] (axis cs:0,0) -- (axis cs:.949,.316) node [pos=.5,above,rotate=30,black] { $\vec u$}; + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      4. +
    +

    +     + +     + +

    + The basic formation of the unit vector \vec u in the direction of a vector \vec v leads to a interesting equation. + It is: + + \vec v = \norm{\vec v}\frac{1}{\norm{\vec v}}\vec v + . +

    + +

    + We rewrite the equation with parentheses to make a point: + + \vec v = \underbrace{\norm{\vec v}}_{\text{magnitude } }\cdot\underbrace{\left(\frac{1}{\norm{\vec v}}\vec v\right)}_{\text{direction } } + . +

    + +

    + This equation illustrates the fact that a nonzero vector has both magnitude and direction, + where we view a unit vector as supplying + only direction information. + Identifying unit vectors with direction allows us to define + parallel vectors. +

    + + + Parallel Vectors + +

    +

      +
    1. +

      + Unit vectors \vec u_1 and + \vec u_2 are parallel + if \vec u_1 = \pm \vec u_2. + vectorsparallel + parallel vectors +

      +
      2. + +
    2. +

      + Nonzero vectors \vec v_1 and + \vec v_2 are parallel + if their respective unit vectors are parallel. +

      +
      3. +
    +

    +     +     + + + +

    + It is equivalent to say that vectors \vec v_1 and + \vec v_2 are parallel if there is a scalar c\neq 0 such that \vec v_1 = c\vec v_2 + (see marginal note). +

    + +

    + If one graphed all unit vectors in + \mathbb{R}^2 with the initial point at the origin, + then the terminal points would all lie on the unit circle. + Based on what we know from trigonometry, + we can then say that the component form of all unit vectors in \mathbb{R}^2 is + \la \cos(\theta) ,\sin(\theta) \ra for some angle \theta. +

    + +

    + A similar construction in \mathbb{R}^3 shows that the terminal points all lie on the unit sphere. + These vectors also have a particular component form, + but its derivation is not as straightforward as the one for unit vectors in \mathbb{R}^2. + Important concepts about unit vectors are given in the following Key Idea. +

    + + + Unit Vectors +

    +

      +
    1. +

      + The unit vector in the direction of a nonzero vector \vec v is + + \vec u = \frac1{\norm{\vec v}} \vec v + . +

      +
      2. + +
    2. +

      + A vector \vec u in \mathbb{R}^2 is a unit vector if, + and only if, + its component form is \la \cos\theta,\sin\theta\ra for some angle \theta. + unit vectorproperties + vectorsunit vector +

      +
      3. + +
    3. +

      + A vector \vec u in \mathbb{R}^3 is a unit vector if, + and only if, + its component form is \la \sin(\theta) \cos(\varphi) ,\sin(\theta) \sin(\varphi) ,\cos(\theta) \ra for some angles \theta and \varphi. +

      +
      4. +
    +

    +     + + + +

    + These formulas can come in handy in a variety of situations, + especially the formula for unit vectors in the plane. +

    + + + Finding Component Forces + +

    + Consider a weight of 50lb hanging from two chains, + as shown in . + One chain makes an angle of 30^\circ with the vertical, + and the other an angle of 45^\circ. + Find the force applied to each chain. +

    + +
    + A diagram of a weight hanging from 2 chains in + + + + Image shows weight suspended with two chains. + + +

    + Image is of a weight of 50 pounds suspended by two chains. The chain on the left forms an angle + of 30 degrees with the vertical and the chain on the left forms a degree of 45 with the + vertical. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.29] + + \filldraw[thick,black,fill=gray!30] (-.5,0) -- (.5,0) -- (1,-1) -- (-1,-1) -- cycle; + \draw (0,-.5) node {50lb}; + + \draw [thick] (-1.5,1.5) -- (2,1.5); + \clip (-1.5,1.5) rectangle (2,-1.25); + + \draw [thick,rotate=120] (0,0) -- (3,0); + \draw [thick,rotate=45] (0,0) -- (3,0); + \draw [dashed] (0,0) -- (0,2); + \draw [rotate=45,->] (.8,0) arc (0:45:.8); + \draw [rotate=67] (1,0) node { $45^\circ$}; + \draw [rotate=90,->] (1,0) arc (0:30:1); + \draw [rotate=105] (1.2,0) node { $30^\circ$}; + + \end{tikzpicture} + + + + +
    +     + +

    + Knowing that gravity is pulling the 50lb weight straight down, + we can create a vector \vec F to represent this force. + + \vec F = 50\la 0,-1\ra = \la 0,-50\ra + . +

    + +

    + We can view each chain as pulling the weight up, + preventing it from falling. + We can represent the force from each chain with a vector. + Let \vec F_1 represent the force from the chain making an angle of 30^\circ with the vertical, + and let \vec F_2 represent the force form the other chain. + Convert all angles to be measured from the horizontal + (as shown in ), + and apply . + As we do not yet know the magnitudes of these vectors, + (that is the problem at hand), + we use m_1 and m_2 to represent them. + + \vec F_1 = m_1\la \cos(120^\circ) ,\sin(120^\circ) \ra + + + \vec F_2 = m_2\la \cos(45^\circ) ,\sin(45^\circ) \ra + +

    + +

    + As the weight is not moving, we know the sum of the forces is \vec 0. + This gives: + + \vec F + \vec F_1 + \vec F_2 \amp = \vec 0 + \la 0,-50\ra + m_1\la \cos(120^\circ) ,\sin(120^\circ) \ra + m_2\la \cos(45^\circ) ,\sin(45^\circ) \ra \amp =\vec 0 + +

    + +
    + A diagram of the force vectors from + + + + Image showing force vectors for this example. + + +

    + Image shows the force vectors from the exercise. The vector \vec F_2 is at an angle 45 + from the horizontal and the vector \vec F_1 forms an angle of 120 from the horizontal. + A third vector representing the downward pull by gravity marked as \vec F. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.75] + + \draw [thick,rotate=120,->] (0,0) -- (1.5,0) node [left] { $\vec F_1$}; + \draw [thick,rotate=45,->] (0,0) -- (1.75,0) node [right] { $\vec F_2$}; + \draw [thick,rotate=-90,->] (0,0) -- (1,0) node [right] { $\vec F$}; + + \draw [dashed] (0,0) -- (0,1.5); + \draw [dashed] (-1.25,0) -- (1.25,0); + + \filldraw [black] (0,0) circle (2.4pt); + + \draw [dashed,->] (.6,0) arc (0:120:.6); + \draw [rotate=20] (.9,0) node { $120^\circ$}; + \draw [dashed,->] (1.5,0) arc (0:45:1.5); + \draw [rotate=15] (1.8,0) node { $45^\circ$}; + + \end{tikzpicture} + + + + +
    + +

    + The sum of the entries in the first component is 0, and the sum of the entries in the second component is also 0. + This leads us to the following two equations: + + m_1\cos(120^\circ) + m_2\cos(45^\circ) \amp =0 + m_1\sin(120^\circ) + m_2\sin(45^\circ) \amp =50 + +

    + +

    + This is a simple 2-equation, 2-unknown system of linear equations. + We leave it to the reader to verify that the solution is + + m_1=50(\sqrt{3}-1) \approx 36.6;\qquad m_2=\frac{50\sqrt{2}}{1+\sqrt{3}} \approx 25.88 + . +

    + +

    + It might seem odd that the sum of the forces applied to the chains is more than 50lb. + We leave it to a physics class to discuss the full details, + but offer this short explanation. + Our equations were established so that the vertical + components of each force sums to 50lb, + thus supporting the weight. + Since the chains are at an angle, + they also pull against each other, + creating an additional horizontal force while holding the weight in place. +

    +     + +     + +

    + Unit vectors were very important in the previous calculation; + they allowed us to define a vector in the proper direction but with an unknown magnitude. + Our computations were then computed component-wise. + Because such calculations are often necessary, + the standard unit vectors can be useful. +

    + + + Standard Unit Vectors + +

    +

      +
    1. +

      + In \mathbb{R}^2, + the standard unit vectors are + vectorsstandard unit vectorunit vectorstandard unit vector + + \veci = \la 1,0\ra \text{ and } \vecj = \la 0,1\ra + . +

      +
      2. + +
    2. +

      + In \mathbb{R}^3, the standard unit vectors are + + \veci = \la 1,0,0\ra \text{ and } \vecj = \la 0,1,0\ra \text{ and } \veck = \la 0,0,1\ra + . +

      +
      3. +
    +

    +     +     + + + Using standard unit vectors + +

    +

      +
    1. +

      + Rewrite \vec v = \la 2,-3\ra using the standard unit vectors. +

      +
      2. + +
    2. +

      + Rewrite \vec w = 4\veci - 5\vecj +2\veck in component form. +

      +
      3. +
    +

    +

    + + +

    +

      +
    1. +

      + \displaystyle \begin{aligned}\vec v \amp = \la 2,-3\ra \\ + \amp = \la 2,0\ra + \la 0,-3\ra \\ + \amp = 2\la 1,0\ra -3\la 0,1\ra\\ + \amp = 2\veci - 3\vecj + \end{aligned} +

      +
      2. + +
    2. +

      + \displaystyle \begin{aligned}\vec w \amp = 4\veci - 5\vecj +2\veck\\ + \amp = \la 4,0,0\ra +\la 0,-5,0\ra + \la 0,0,2\ra \\ + \amp = \la 4,-5,2\ra + \end{aligned} +

      +
      3. +
    +

    + +

    + These two examples demonstrate that converting between component form and the standard unit vectors is rather straightforward. + Many mathematicians prefer component form, + and it is the preferred notation in this text. + Many engineers prefer using the standard unit vectors, + and many engineering text use that notation. +

    + +     + + + Finding Component Force + +

    + A weight of 25lb is suspended from a chain of length 2ft while a wind pushes the weight to the right with constant force of 5lb as shown in . + What angle will the chain make with the vertical as a result of the wind's pushing? + How much higher will the weight be? +

    + +
    + A figure of a weight being pushed by the wind in + + + + Image showing weight suspended by a chain and wind action on the weight. + + +

    + Image shows a weight of 25 pounds being suspended by a chain. The chain forms an angle of + \theta with the vertical, and an angle \varphi with the horizontal. The wind is pushing + the weight to the right with force \vec F_w. The chain is of length 2 feet. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.29] + + \draw [dashed] (-1.5,-.2) -- (2,-.2); + + \draw (1.75,.65) node { 2ft $\left\}\rule{0pt}{1.2cm}\right.$}; + + \filldraw[thick,black,fill=gray!30] (-.5,0) -- (.5,0) -- (1,-1) -- (-1,-1) -- cycle; + \draw (0,-.5) node {25lb}; + + \draw [thick] (-1.5,1.5) -- (2,1.5); + \clip (-1.5,1.5) rectangle (2,-1.25); + + \draw [thick,rotate=110] (0,0) -- (3,0); + \draw [dashed] (0,0) -- (0,2); + \draw [rotate=90,->] (1,0) arc (0:20:1); + \draw [->] (.6,0) arc (0:109:.6); + \draw [rotate=50] (.75,0) node { $\varphi$}; + \draw [rotate=99] (1.2,0) node { $\theta$}; + \draw [thick,->] (-1.45,-.5) -- (-.8,-.5) node [pos=.18,below] { $\vec F_w$}; + + \end{tikzpicture} + + + + +
    +     + +

    + The force of the wind is represented by the vector \vec F_w = 5\veci. + The force of gravity on the weight is represented by \vec F_g = -25\vecj. + The direction and magnitude of the vector representing the force on the chain are both unknown. + We represent this force with + + \vec F_c = m\la \cos(\varphi) ,\sin(\varphi) \ra = m\cos(\varphi) \, \veci + m\sin(\varphi) \,\vecj + + for some magnitude m and some angle with the horizontal \varphi. + (Note: \theta is the angle the chain makes with the vertical; + \varphi is the angle with the horizontal.) +

    + +

    + As the weight is at equilibrium, + the sum of the forces is \vec0: + + \vec F_c + \vec F_w + \vec F_g \amp = \vec 0 + m\cos(\varphi) \, \veci + m\sin(\varphi) \,\vecj + 5\veci - 25\vecj \amp =\vec 0 + +

    + +

    + Thus the sum of the \veci and \vecj components are 0, leading us to the following system of equations: + + \begin{split} + 5+m\cos\varphi \amp = 0\\ + -25+m\sin\varphi \amp = 0 + \end{split} + +

    + +

    + This is enough to determine \vec F_c already, + as we know m\cos(\varphi) = -5 and m\sin(\varphi) =25. + Thus F_c = \la -5,25\ra. + We can use this to find the magnitude m: + + m = \sqrt{(-5)^2+25^2} = 5\sqrt{26}\approx 25.5\text{ lb } + . +

    + +

    + We can then use either equality from Equation to solve for \varphi. + We choose the first equality as using arccosine will return an angle in the 2nd quadrant: + + 5 + 5\sqrt{26}\cos(\varphi) = 0 \Rightarrow \varphi = \cos^{-1}\left(\frac{-5}{5\sqrt{26}}\right) \approx 1.7682\approx 101.31^\circ + . +

    + +

    + Subtracting 90^\circ from this angle gives us an angle of 11.31^\circ with the vertical. +

    + +

    + We can now use trigonometry to find out how high the weight is lifted. + The diagram shows that a right triangle is formed with the 2ft chain as the hypotenuse with an interior angle of 11.31^\circ. + The length of the adjacent side + (in the diagram, the dashed vertical line) + is 2\cos(11.31^\circ) \approx 1.96ft. + Thus the weight is lifted by about 0.04ft, almost 1/2in. +

    +     +     + +

    + The algebra we have applied to vectors is already demonstrating itself to be very useful. + There are two more fundamental operations we can perform with vectors, + the dot product and the cross product. + The next two sections explore each in turn. +

    + + + + Terms and Concepts + + + + +

    + Name two different things that cannot be described with just one number, + but rather need 2 or more numbers to fully describe them. +

    +     + + + +     + + + + +

    + What is the difference between (1,2) and \la 1,2\ra? +

    +     + + + +

    + (1,2) is a point; + \la 1,2\ra is a vector that describes a displacement of 1 unit in the x-direction and 2 units in the y-direction. +

    +     + +     + + + + +

    + What is a unit vector? +

    +     + + + +

    + A vector with magnitude 1. +

    +     + +     + + + +

    + Unit vectors can be thought of as conveying what type of information? +

    +     + + + +

    + Direction +

    +     +     + + + + + +

    + What does it mean for two vectors to be parallel? +

    +     + + + +

    + Their respective unit vectors are parallel; + unit vectors \vec u_1 and + \vec u_2 are parallel if \vec u_1 = \pm \vec u_2. +

    + +

    + Alternatively, two vectors are parallel if one is a scalar multiple of the other. +

    +     + +     + + + + +

    + What effect does multiplying a vector by -2 have? +

    +     + + + +

    + It stretches the vector by a factor of 2, and points it in the opposite direction. +

    +     + +     + +     + + Problems + + + +

    + Points P and Q are given. + Write the vector \overrightarrow{PQ} in component form and using the standard unit vectors. +

    +     + + + + + + Context("Vector2D"); + Context()->constants->undefine(i,j); + $comp=Vector("<1,6>"); + Context("Vector2D"); + Context()->flags->set(ijk=>1); + Context()->parens->undefine('<','(','{'); + $stan=Vector("i+6j"); + + + +

    + If P=(2,-1) and Q = (3,5), + write the vector \overrightarrow{PQ}: +

    +     + + + +

    + in component form. +

    + +

    + +

    +     +     + + + +

    + using the standard unit vectors. +

    + +

    + +

    +     +     +     +     + + + + + + Context("Vector2D"); + Context()->constants->undefine(i,j); + $comp=Vector("<4,-4>"); + Context("Vector2D"); + Context()->flags->set(ijk=>1); + Context()->parens->undefine('<','(','{'); + $stan=Vector("4i-4j"); + + + +

    + If P=(3,2) and Q = (7,-2), + write the vector \overrightarrow{PQ}: +

    +     + + + +

    + in component form. +

    + +

    + +

    +     +     + + + +

    + using the standard unit vectors. +

    + +

    + +

    +     +     +     +     + + + + + + Context("Vector"); + Context()->constants->undefine(i,j,k); + $comp=Vector("<6,-1,6>"); + Context("Vector"); + Context()->flags->set(ijk=>1); + Context()->parens->undefine('<','(','{'); + $stan=Vector("6i-j+6k"); + + + +

    + If P=(0,3,-1) and Q = (6,2,5), + write the vector \overrightarrow{PQ}: +

    +     + + + +

    + in component form. +

    + +

    + +

    +     +     + + + +

    + using the standard unit vectors. +

    + +

    + +

    +     +     +     +     + + + + + + Context("Vector"); + Context()->constants->undefine(i,j,k); + $comp=Vector("<2,2,0>"); + Context("Vector"); + Context()->flags->set(ijk=>1); + Context()->parens->undefine('<','(','{'); + $stan=Vector("2i+2j"); + + + +

    + If P=(2,1,2) and Q = (4,3,2), + write the vector \overrightarrow{PQ}: +

    +     + + + +

    + in component form. +

    + +

    + +

    +     +     + + + +

    + using the standard unit vectors. +

    + +

    + +

    +     +     +     +     + +     + + + + +

    + Let \vec u = \la 1,-2\ra and \vec v= \la 1,1\ra. +

    +     + + + +

    + Find \vec u+\vec v, + \vec u-\vec v, 2\vec u-3\vec v. +

    +     + +

    + \vec u+\vec v = \la 2,-1\ra; + \vec u -\vec v = \la0,-3\ra; + 2\vec u-3\vec v = \la -1,-7\ra. +

    +     +     + + + +

    + Sketch the above vectors on the same axes, + along with \vec u and \vec v. +

    +     +     + + + +

    + Find \vec x where \vec u+\vec x = 2\vec v-\vec x. +

    +     + +

    + \vec x = \la 1/2,2\ra. +

    +     +     + +     + + + + +

    + Let \vec u = \la 1,1,-1\ra and \vec v= \la 2,1,2\ra. +

    +     + + + +

    + Find \vec u+\vec v, + \vec u-\vec v, \pi\vec u-\sqrt{2}\vec v. +

    +     + +

    + \vec u+\vec v = \la 3,2,1\ra; + \vec u -\vec v = \la-1,0,-3\ra; + \pi\vec u-\sqrt{2}\vec v = \la \pi-2\sqrt{2},\pi-\sqrt{2},-\pi-2\sqrt{2}\ra. +

    +     +     + + + +

    + Sketch the above vectors on the same axes, + along with \vec u and \vec v. +

    +     +     + + + +

    + Find \vec x where \vec u+\vec x = \vec v+2\vec x. +

    +     + +

    + \vec x = \la-1,0,-3\ra. +

    +     +     + +     + + + + +

    + Sketch \vec u, \vec v, + \vec u+\vec v and \vec u-\vec v on the same axes. +

    +     + + + + + + + + Graph shows two vectors in the plane. + + +

    + The x and y axes are uncalibrated. Two vectors \vec u and \vec v + are shown, both starting at the origin and facing away. The vector \vec u is in the + first quadrant, and is bent close to the positive x axis, the \vec v vector is + in the third quadrant and is bent close to the negative y axis. The vector \vec v + appears to be slightly longer than \vec u. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + xtick=\empty, + ytick=\empty, + ymin=-4.5,ymax=4.5, + xmin=-4.5,xmax=4.5 + ] + + \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: 3,1) node [above,black] { $\vec u$}; + \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: -1,-4) node [black, left] { $\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + + + + Solution image for this exercise + +

    + The solution image shows the same two vectors as in the statement of the problem. + The vector \vec{u}-\vec{v} points from the tip of \vec{v} to the tip of \vec{u}. + The vector \vec{u}+\vec{v} points between the two original vectors; + the tail is at the origin, and in this case the tip is in the 4th quadrant. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + xtick=\empty, + ytick=\empty, + ymin=-4.5,ymax=4.5, + xmin=-4.5,xmax=4.5 + ] + + \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: 3,1) node [above,black] { $\vec u$}; + \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: -1,-4)node [black, left] { $\vec v$}; + + \draw [->,thick,gray] (axis cs:0,0) -- (axis cs: 2,-3) node [black,below right] { $\vec u+\vec v$}; + \draw [->,thick,gray] (axis cs:-1,-4) -- (axis cs: 3,1) node [black,below,pos=.3,sloped] { $\vec u-\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + + + + + Graph shows two vectors in the plane. + + +

    + The x and y axes are uncalibrated. Two vectors \vec u and \vec v are shown, + both start at the origin and are facing away from each other. The \vec u vector is in the first + quadrant while the \vec v is in the third quadrant, the \vec v vector appears to be 1 / 4 + th \vec u. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + xtick=\empty, + ytick=\empty, + ymin=-4.5,ymax=4.5, + xmin=-4.5,xmax=4.5 + ] + + \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: 3,3) node [above,black] { $\vec u$}; + \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: -1,-1) node [black, below] { $\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + + + + Solution image for this exercise + +

    + The solution image shows the same two vectors as in the statement of the problem. + The vector \vec{u}+\vec{v} points in the same direction as \vec{u}, but is not as long. + The vector \vec{u}-\vec{v} is drawn below the two original vectors. + It points in the same direction as \vec{u}, but its length is the sum of the lengths of \vec{u} and \vec{v}. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + xtick=\empty, + ytick=\empty, + ymin=-4.5,ymax=4.5, + xmin=-4.5,xmax=4.5 + ] + + \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: 3,3) node [above,black] { $\vec u$}; + \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: -1,-1) node [black, left] { $\vec v$}; + + \draw [->,thick,gray] (axis cs:0,0) -- (axis cs: 2,2) node [black,above left] { $\vec u+\vec v$}; + \draw [->,thick,gray] (axis cs:-1,-2) -- (axis cs: 3,2) node [black,below,pos=.3,sloped] { $\vec u-\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +

    + Sketch of \vec u-\vec v shifted for clarity. +

    +     + +     + + + + + + + + Graph shows two vectors in space. + + +

    + The x, y and z axes are uncalibrated. Two vectors \vec u and + \vec v are shown, both start at the origin and face away from each other. The vector + \vec u is longer and appears to be in the zy plane, and the vector \vec v + is shorter and appears to be in the xz plane. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + axis lines=center, + view={135}{35}, + xtick=\empty, + ytick=\empty, + ztick=\empty, + ymin=-1.5,ymax=1.5, + xmin=-1.5,xmax=1.5, + zmin=-0.5, zmax=1.5, + every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, + xlabel={ $x$}, + every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, + ylabel={ $y$}, + every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, + zlabel={ $z$} + ] + + \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:-1,1,0) node [black,right] { $\vec u$}; + \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:1,0,1) node [black,left] { $\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + + + + Solution image for this exercise + +

    + The solution image shows the same two vectors as in the statement of the problem. + The vector \vec{u}-\vec{v} points from the tip of \vec{v} to the tip of \vec{u}. + The vector \vec{u}+\vec{v} points between the two original vectors, in the plane containing them. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + axis lines=center, + view={135}{35}, + xtick=\empty, + ytick=\empty, + ztick=\empty, + ymin=-1.5,ymax=1.5, + xmin=-1.5,xmax=1.5, + zmin=-0.5, zmax=1.5, + every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, + xlabel={ $x$}, + every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, + ylabel={ $y$}, + every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, + zlabel={ $z$} + ] + + \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:-1,1,0) node [black,right] { $\vec u$}; + \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:1,0,1) node [black,left] { $\vec v$}; + + \draw [thick,->,gray] (axis cs:0,0,0) -- (axis cs:0,1,1) node [black,above] { $\vec u+\vec v$}; + \draw [thick,->,gray] (axis cs:1,0,1) -- (axis cs:-1,1,0) node [black,above,pos=.9] { $\vec u-\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + + + + + + + + + Graph shows two vectors in space. + + +

    + The x, y and z axes are uncalibrated. Two vectors \vec u and + \vec v are shown, both start at the origin. The \vec u vector is along the + positive z axis and the \vec v vector is along the positive y axis. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis lines=center, + view={135}{35}, + xtick=\empty, + ytick=\empty, + ztick=\empty, + ymin=-1.5,ymax=1.5, + xmin=-1.5,xmax=1.5, + zmin=-0.5, zmax=1.5, + every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, + xlabel={ $x$}, + every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, + ylabel={ $y$}, + every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, + zlabel={ $z$} + ] + + \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:0,0,1) node [black,left] { $\vec u$}; + \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:0,1,0) node [black,below] { $\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + + + + Solution image for this exercise + +

    + The solution image shows the same two vectors as in the statement of the problem. + The vector \vec{u}-\vec{v} points from the tip of \vec{v} to the tip of \vec{u}. + The vector \vec{u}+\vec{v} points between the two original vectors, in the plane containing them. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + width=175pt, + axis lines=center, + view={135}{35}, + xtick=\empty, + ytick=\empty, + ztick=\empty, + ymin=-1.5,ymax=1.5, + xmin=-1.5,xmax=1.5, + zmin=-0.5, zmax=1.5, + every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, + xlabel={ $x$}, + every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, + ylabel={ $y$}, + every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, + zlabel={ $z$} + ] + + \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:0,0,1) node [black,left] { $\vec u$}; + \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:0,1,0) node [black,below] { $\vec v$}; + + \draw [thick,->,gray] (axis cs:0,0,0) -- (axis cs:0,1,1) node [black,above] { $\vec u+\vec v$}; + \draw [thick,->,gray] (axis cs:0,1,0) -- (axis cs:0,0,1) node [black,right,pos=.2] { $\vec u-\vec v$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +     + +     + + + +

    + Find + \norm{\vec u}, \norm{\vec v}, + \norm{\vec u+\vec v} and \norm{\vec u-\vec v}. +

    +     + + + + + Context()->flags->set(reduceConstantFunctions=>0); + $normu=Formula("sqrt(5)"); + $normv=Formula("sqrt(13)"); + $normupv=Formula("sqrt(26)"); + $normumv=Formula("sqrt(10)"); + + + +

    + \vec u=\la 2,1\ra, \vec v = \la 3,-2\ra. +

    + + + Enter the value of \norm{\vec u}. + + +

    + +

    + + + Enter the value of \norm{\vec v}. + + +

    + +

    + + + Enter the value of \norm{\vec{u} +\vec{v}}. + + +

    + +

    + + + Enter the value of \norm{\vec{u}-\vec{v}}. + + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstantFunctions=>0); + $normu=Formula("sqrt(17)"); + $normv=Formula("sqrt(3)"); + $normupv=Formula("sqrt(14)"); + $normumv=Formula("sqrt(26)"); + + + +

    + \vec u=\la -3,2,2\ra, \vec v = \la 1,-1,1\ra. +

    + + + Enter the value of \norm{\vec u}. + + +

    + +

    + + + Enter the value of \norm{\vec v}. + + +

    + +

    + + + Enter the value of \norm{\vec{u} +\vec{v}}. + + +

    + +

    + + + Enter the value of \norm{\vec{u}-\vec{v}}. + + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstantFunctions=>0); + $normu=Formula("sqrt(5)"); + $normv=Formula("3sqrt(5)"); + $normupv=Formula("2sqrt(5)"); + $normumv=Formula("4sqrt(5)"); + + + +

    + \vec u=\la 1,2\ra, \vec v = \la -3,-6\ra. +

    + + + Enter the value of \norm{\vec u}. + + +

    + +

    + + + Enter the value of \norm{\vec v}. + + +

    + +

    + + + Enter the value of \norm{\vec{u} +\vec{v}}. + + +

    + +

    + + + Enter the value of \norm{\vec{u}-\vec{v}}. + + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstantFunctions=>0); + $normu=Formula("7"); + $normv=Formula("35"); + $normupv=Formula("42"); + $normumv=Formula("28"); + + + +

    + \vec u=\la 2,-3,6\ra, \vec v = \la 10,-15,30\ra. +

    + + + Enter the value of \norm{\vec u}. + + +

    + +

    + + + Enter the value of \norm{\vec v}. + + +

    + +

    + + + Enter the value of \norm{\vec{u} +\vec{v}}. + + +

    + +

    + + + Enter the value of \norm{\vec{u}-\vec{v}}. + + +

    + +

    +     +     +     + +     + + + + +

    + Under what conditions is \norm{\vec u}+\norm{\vec v} = \norm{\vec u+\vec v}? +

    +     + +

    + When \vec u and \vec v have the same direction. + (Note: parallel is not enough.) +

    +     + +     + + + + +

    + Find the unit vector \vec u in the direction of \vec v. +

    +     + + + + + Context("Vector2D"); + Context()->flags->set(reduceConstantFunctions=>0); + $u=Compute("<3/sqrt(58), 7/sqrt(58)>"); + + + +

    + Find the unit vector \vec u in the direction of \vec v = \la 3,7\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector2D"); + Context()->flags->set(reduceConstantFunctions=>0); + $u=Compute("<0.6, 0.8>"); + + + +

    + Find the unit vector \vec u in the direction of \vec v = \la 6,8\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstantFunctions=>0,reduceConstants=>0); + $u=Compute("<1/3,-2/3,2/3>"); + + + +

    + Find the unit vector \vec u in the direction of \vec v = \la 1,-2,2\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstantFunctions=>0,reduceConstants=>0); + $u=Compute("<1/sqrt(3),-1/sqrt(3),1/sqrt(3)>"); + + + +

    + Find the unit vector \vec u in the direction of \vec v = \la 2,-2,2\ra. +

    + +

    + +

    +     +     +     + +     + + + + + Context("Vector2D"); + Context()->flags->set(reduceConstantFunctions=>0,reduceConstants=>0); + $u=Compute("<cos(50*pi/180),sin(50*pi/180)>"); + + + +

    + Find the unit vector in the first quadrant of \mathbb{R}^2 that makes a + 50^{\circ} angle with the x-axis. +

    + +

    + +

    +     +     +     + + + + + Context("Vector2D"); + Context()->flags->set(reduceConstantFunctions=>0,reduceConstants=>0); + $u=Compute("<-1/2,sqrt(3)/2>"); + + + +

    + Find the unit vector in the second quadrant of \mathbb{R}^2 that makes a + 30^{\circ} angle with the y-axis. +

    + +

    + +

    +     +     +     + + + + +

    + Verify, from , + that + + \vec u=\la \sin(\theta) \cos(\varphi) ,\sin(\theta) \sin(\varphi) ,\cos(\theta) \ra + + is a unit vector for all angles \theta and \varphi. +

    +     + +

    + + \norm{\vec u} \amp = \sqrt{\sin^2(\theta) \cos^2(\varphi) +\sin^2(\theta) \sin^2(\varphi) +\cos^2(\theta) } + \amp = \sqrt{\sin^2(\theta) (\cos^2(\varphi) +\sin^2(\varphi) )+\cos^2(\theta) } + \amp = \sqrt{\sin^2(\theta) +\cos^2(\theta) } + \amp =1 + . +

    +     + +     + + + + +

    + A weight of 100lb is suspended from two chains, + making angles with the vertical of \theta and \varphi as shown in the figure below. +

    + + + + Image shows weight suspended with two chains. + + +

    + Image is of a weight of 100 pounds suspended by two chains. The chain on the left forms an + angle of \varphi degrees with the vertical and the chain on the right forms a degree of + \theta with the vertical. +

    +     + + \begin{tikzpicture} + \filldraw[thick,black,fill=gray!30] (-.5,0) -- (.5,0) -- (1,-1) -- (-1,-1)--cycle; + \draw (0,-.5) node {100lb}; + + \draw [thick] (-1.75,1.5) -- (1.75,1.5); + \clip (-1.5,1.5) rectangle (2,-1.25); + \draw [thick,rotate=135] (0,0) -- (3,0);% node[below,sloped,pos=.5] { $\vec F_2$}; + \draw [thick,rotate=45] (0,0) -- (3,0); %node[below,sloped,pos=.5] { $\vec F_1$};; + \draw [dashed] (0,0) -- (0,2); + \draw [rotate=45,->,>=stealth] (.8,0) arc (0:45:.8); + \draw [rotate=67] (1,0) node { $\theta$}; + \draw [rotate=90,->,>=stealth] (1,0) arc (0:45:1); + \draw [rotate=105] (1.2,0) node { $\varphi$}; + \end{tikzpicture} + + + +

    + Angles \theta and \varphi are given. + Find the magnitude of the force applied to each chain. +

    +     + + + + +

    + \theta = 30^\circ,\varphi=30^\circ +

    +     + +

    + The force on each chain is 100/\sqrt{3}\approx 57.735lb. +

    +     + +     + + + + +

    + \theta = 60^\circ,\varphi=60^\circ +

    +     + +

    + The force on each chain is 100lb. +

    +     + +     + + + + +

    + \theta = 20^\circ,\varphi=15^\circ +

    +     + +

    + The force on the chain with angle \theta is approx. + 45.124lb; + the force on the chain with angle \varphi is approx. + 59.629lb. +

    +     + +     + + + + +

    + \theta = 0^\circ,\varphi=0^\circ +

    +     + +

    + The force on each chain is 50lb. +

    +     + +     + +     + + + +

    + A weight of plb is suspended from a chain of length \ell while a constant force of + \vec F_w pushes the weight to the right, + making an angle of \theta with the vertical, + as shown in the figure below. +

    + + + + Image showing weight suspended by a chain and wind action on the weight. + + +

    + Image shows a weight of p pounds being suspended by a chain. The chain forms an angle of + \theta with the vertical. The wind is pushing the weight to the right with force \vec F_w. + The chain is of length \ell feet. +

    +     + + \begin{tikzpicture} + \draw [dashed] (-1.5,-.2) -- (2,-.2); + \draw (1.75,.65) node { $\ell$\ ft $\left.\rule{0pt}{.8cm}\right\}$}; + \filldraw[thick,black,fill=gray!30] (-.5,0) -- (.5,0) -- (1,-1) -- (-1,-1)--cycle; + \draw (0,-.5) node {$p$ lb}; + \draw [thick] (-1.5,1.5) -- (2,1.5); + \clip (-1.5,1.5) rectangle (2,-1.25); + \draw [thick,rotate=110] (0,0) -- (3,0); + \draw [dashed] (0,0) -- (0,2); + \draw [rotate=90,->,>=stealth] (1,0) arc (0:20:1); + \draw [rotate=99] (1.2,0) node { $\theta$}; + \draw [thick,->,>=stealth] (-1.45,-.5) -- (-.8,-.5) node [pos=.15,below] { $\vec F_w$}; + \end{tikzpicture} + + + +

    + A force \vec F_w and length \ell are given. + Find the angle \theta and the height the weight is lifted as it moves to the right. +

    +     + + + + +

    + \vec F_w=1lb, \ell = 1ft, p = 1lb +

    +     + +

    + \theta = 45^\circ; the weight is lifted 0.29 ft + (about 3.5in). +

    +     + +     + + + + +

    + \vec F_w=1lb, \ell = 1ft, p = 10lb +

    +     + +

    + \theta = 5.71^\circ; the weight is lifted 0.005 ft + (about 1/16th of an inch). +

    +     + +     + + + + +

    + \vec F_w=1lb, \ell = 10ft, p = 1lb +

    +     + +

    + \theta = 45^\circ; the weight is lifted 2.93 ft. +

    +     + +     + + + + +

    + \vec F_w=10lb, \ell = 10ft, p = 1lb +

    +     + +

    + \theta = 84.29^\circ; + the weight is lifted 9 ft. +

    +     + +     +     +     +     +
    +
    + The Dot Product +

    + The previous section introduced vectors and described how to add them together and how to multiply them by scalars. + This section introduces a + multiplication on vectors called the dot product. +

    + + + + + Dot Product + +

    +

      +
    1. +

      + Let \vec u = \la u_1,u_2\ra and + \vec v = \la v_1,v_2\ra in \mathbb{R}^2. + The dot product of \vec u and \vec v, + denoted \dotp uv, is + + \dotp uv = u_1v_1+u_2v_2 + . +

      +
      2. + +
    2. +

      + Let \vec u = \la u_1,u_2,u_3\ra and + \vec v = \la v_1,v_2,v_3\ra in \mathbb{R}^3. + The dot product of \vec u and \vec v, + denoted \dotp uv, is + + \dotp uv = u_1v_1+u_2v_2+u_3v_3 + . +

      +
      3. +
    + dot productdefinition + vectorsdot product +

    +     +     + + + +

    + Note how this product of vectors returns a + scalar, not another vector. + We practice evaluating a dot product in the following example, + then we will discuss why this product is useful. +

    + + + Evaluating dot products + +

    +

      +
    1. +

      + Let \vec u=\la 1,2\ra, + \vec v=\la 3,-1\ra in \mathbb{R}^2. + Find \dotp uv. +

      +
      2. + +
    2. +

      + Let \vec x = \la 2,-2,5\ra and + \vec y = \la -1, 0, 3\ra in \mathbb{R}^3. + Find \dotp xy. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + Using , we have + + \dotp uv = 1(3)+2(-1) = 1 + . +

      +
      2. + +
    2. +

      + Using the definition, we have + + \dotp xy = 2(-1) -2(0) + 5(3) = 13 + . +

      +
      3. +
    +

    +     + +     + +

    + The dot product, as shown by the preceding example, + is very simple to evaluate. + It is only the sum of products. + While the definition gives no hint as to why we would care about this operation, + there is an amazing connection between the dot product and angles formed by the vectors. + Before stating this connection, + we give a theorem stating some of the properties of the dot product. +

    + + + Properties of the Dot Product + +

    + Let \vec u, + \vec v and \vec w be vectors in \mathbb{R}^2 or + \mathbb{R}^3 and let c be a scalar. + dot productproperties + vectorsdot product +

    + +

    +

      +
    1. +

      + \dotp uv = \dotp vu (Commutative Property) +

      +
      2. + +
    2. +

      + \vec u\cdot(\vec v+\vec w) = \dotp uv + \dotp uw (Distributive Property) +

      +
      3. + +
    3. +

      + c(\dotp uv) = (c\vec u)\cdot \vec v = \vec u \cdot (c\vec v) +

      +
      4. + +
    4. +

      + \dotp 0v = 0 +

      +
      5. + +
    5. +

      + \dotp vv=\norm{\vec v}^2 +

      +
      6. +
    +

    +     +     + +

    + The last statement of the theorem makes a handy connection between the magnitude of a vector and the dot product with itself. + Our definition and theorem give properties of the dot product, + but we are still likely wondering + What does the dot product mean? + It is helpful to understand that the dot product of a vector with itself is connected to its magnitude. +

    + + + +

    + The next theorem extends this understanding by connecting the dot product to magnitudes and angles. + Given vectors \vec u and \vec v in the plane, + an angle \theta is clearly formed when \vec u and \vec v are drawn with the same initial point as illustrated in . (We always take \theta to be the angle in [0,\pi] as two angles are actually created.) +

    + +
    + Illustrating the angle formed by two vectors with the same initial point + +
    + + + + + The angle formed by two vectors. + + +

    + Image shows the angle \theta formed by two vectors, \vec u + and \vec v. The angle is formed from \vec u to \vec v. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.48] + + \draw [rotate=-15,->] (0,0) -- (2,0) node (A) [right] { $\vec u$}; + \draw [rotate=35,->] (0,0) -- (2.5,0) node (B) [right] { $\vec v$}; + \draw [rotate=-15,->] (.75,0) arc (0:50:.75); + \draw [rotate=10] (.9,0) node { $\theta$}; + + \end{tikzpicture} + + + + +
    + +
    + + + + + + Graph showing angle formed by two vectors on a plane containing both vectors. + + +

    + The three axes are shown and all are uncalibrated. Two vectors \vec u and + \vec v are shown, along with the place that contains both vectors. The two + vectors form an angle and it is labeled as \theta. The pane lies at an angle + in the positive values of all axes. +

    +     + + + + + //ASY file for figdotpangle3D.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-0.5,2.5); + pair ybounds=(-0.5,2.5); + pair zbounds=(-0.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw theplane z=.1*x+y+1; + triple f(pair t) { + return (t.x,t.y,-.1*t.x+.3*t.y+.5); + } + surface s=surface(f,(0,0),(2,2),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + // Draw the vectors + draw((0.5,0.5,.6)--(1.5,0.25,.425), black+linewidth(2),Arrow3(size=3mm));//u + draw((0.5,0.5,.6)--(0.25,1.8,1.015), black+linewidth(2),Arrow3(size=3mm));//v + label("$\vec{u}$",(1.5,0.25,.425),N); + label("$\vec{v}$",(0.25,1.8,1.015),N); + dotfactor=3; dot((0.5,0.5,.6),black); + + //Draw the arc with theta + draw((.99,.38,.51)..(.81,.78,.65)..(.41,.97,.75)); + label("$\theta$",(0.5,0.5,.7),W); + + + + +
    +     +
    + +

    + The same is also true of 2 vectors in space: + given \vec u and \vec v in + \mathbb{R}^3 with the same initial point, + there is a plane that contains both \vec u and \vec v. + (When \vec u and \vec v are co-linear, + there are infinitely many planes that contain both vectors.) + In that plane, + we can again find an angle \theta between them + (and again, 0\leq \theta\leq \pi). + This is illustrated in . +

    + +

    + The following theorem connects this angle \theta to the dot product of \vec u and \vec v. +

    + + + The Dot Product and Angles + +

    + Let \vec u and \vec v be nonzero vectors in + \mathbb{R}^2 or \mathbb{R}^3. + Then + + \dotp uv = \norm{\vec u}\,\norm{\vec v} \cos(\theta) + , + where \theta, 0\leq\theta\leq \pi, + is the angle between \vec u and \vec v. + dot productproperties + vectorsdot product +

    +     +     + + + +

    + Using , + we can rewrite this theorem as + + \frac{\vec u}{\norm{\vec u}}\cdot \frac{\vec v}{\norm{\vec v}} = \cos(\theta) + . +

    + + + +

    + Note how on the left hand side of the equation, + we are computing the dot product of two unit vectors. + Recalling that unit vectors essentially only provide direction information, + we can informally restate + as saying The dot product of two directions gives the cosine of the angle between them. +

    + +

    + When \theta is an acute angle (, 0\leq \theta \lt \pi/2), + \cos(\theta) is positive; + when \theta = \pi/2, \cos(\theta) = 0; + when \theta is an obtuse angle (\pi/2\lt \theta \leq \pi), + \cos(\theta) is negative. + Thus the sign of the dot product gives a general indication of the angle between the vectors, + illustrated in . +

    + +
    + Illustrating the relationship between the angle between vectors and the sign of their dot product + + + + + Image shows the relation between the angle between two vectors and the sign of their dot product. + + +

    + Each image has the two vectors \vec u and \vec v along with \theta shown. + The dot product along with three possible cases of <, > and equal to 0 are shown. + In the first image, the dot product has a positive value, the angle between the two vectors is acute. + In the second image, the dot product is equal to 0, the angle is 90 degrees. + In the third image, the dot product has a negative value and the angle is obtuse. +

    +     + + + \begin{tikzpicture}[scale=1.5,>=stealth] + + \begin{scope} + \draw[->] (0,0) -- (1,0) node [pos=.5,below] { $\vec u\cdot \vec v >0$} node [right] { $\vec u$}; + \draw[->] (0,0) -- (.5,.866) node [above right] { $\vec v$}; + \draw[->] (.3,0) arc (0:60:.3); + \draw[rotate=30] (.45,0) node { $\theta$}; + \end{scope} + + \begin{scope}[shift={(3cm,0)}] + \draw[->] (0,0) -- (1,0) node [pos=.5,below] { $\vec u\cdot \vec v =0$}node [right] { $\vec u$}; + \draw[->] (0,0) -- (0,1) node [above ] { $\vec v$}; + \draw[->] (.3,0) arc (0:90:.3); + \draw[rotate=22.5] (.7,.2) node { $\theta=\pi/2$}; + \end{scope} + + \begin{scope}[shift={(6.5cm,0)}] + \draw[->] (0,0) -- (1,0) node [pos=.5,below] { $\vec u\cdot \vec v <0$}node [right] { $\vec u$}; + \draw[->] (0,0) -- (-.707,.707) node [above left] { $\vec v$}; + \draw[->] (.3,0) arc (0:135:.3); + \draw[rotate=67.5] (.45,0) node { $\theta$}; + \end{scope} + + \end{tikzpicture} + + + + +     +
    + +

    + We can use to compute the dot product, + but generally this theorem is used to find the angle between known vectors + (since the dot product is generally easy to compute). + To this end, we rewrite the theorem's equation as + + \cos(\theta) = \frac{\dotp uv}{\norm{\vec u}\norm{\vec v}} \Leftrightarrow \theta = \cos^{-1}\left(\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right) + . +

    + +

    + We practice using this theorem in the following example. +

    + + + Using the dot product to find angles + +

    + Let \vec u = \la 3,1\ra, + \vec v = \la -2,6\ra and \vec w = \la -4,3\ra, + as shown in . + Find the angles \alpha, + \beta and \theta. +

    +
    + Vectors used in + + + + Graph shows three vectors and the angles between them. + + +

    + The x axis is drawn from -4 to 4 and the y + axis is drawn from 0 to 6. The \vec u, \vec v + and \vec w are shown. +

    +

    + The \vec u starts at the origin and ends at point (3, 1). + The \vec v also starts at the origin and ends at point (-2, 6) + and \vec w is also drawn from the origin and ends at point (-4, 3). +

    +

    + The angle \alpha is drawn between \vec u and \vec v, + angle \beta is drawn between \vec v and \vec w and angle + \theta is drawn between \vec u and \vec w. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + minor x tick num=4, + ymin=-1.8,ymax=6.99, + xmin=-5.5,xmax=5.5, + ] + + \draw [thick,->] (axis cs:0,0) -- (axis cs: 3,1) node [above] {$\vec u$}; + \draw [thick,->] (axis cs:0,0) -- (axis cs: -2,6) node [above] {$\vec v$}; + \draw [thick,->] (axis cs:0,0) -- (axis cs: -4,3) node [above] {$\vec w$}; + + \draw [->] (axis cs: .949,.316) arc (18.4:108.4:12pt); + \draw (axis cs:0.680986, 1.33651) node { $\alpha$}; + + \draw [->] (axis cs: -0.474342, 1.42302) arc (108.4:155.1:12pt); + \draw (axis cs:-1.13929, 1.58257) node { $\beta$}; + + \draw [->] (axis cs: 2.3725, 0.79) arc (18.4:143.1:30pt); + \draw (axis cs:1.5, 2.59808) node { $\theta$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + We start by computing the magnitude of each vector. + + \norm{\vec u} = \sqrt{10}; \norm{\vec v} = 2\sqrt{10}; \norm{\vec w} = 5 + . +

    + +

    + We now apply to find the angles. + + \alpha \amp = \cos^{-1}\left(\frac{\dotp uv}{(\sqrt{10})(2\sqrt{10})}\right) + \amp = \cos^{-1}(0) = \frac{\pi}2 = 90^\circ + . + + \beta \amp = \cos^{-1}\left(\frac{\dotp vw}{(2\sqrt{10})(5)}\right) + \amp = \cos^{-1}\left(\frac{26}{10\sqrt{10}}\right) + \amp \approx 0.6055 \approx 34.7^\circ. + \theta \amp = \cos^{-1}\left(\frac{\dotp uw}{(\sqrt{10})(5)}\right) + \amp = \cos^{-1}\left(\frac{-9}{5\sqrt{10}}\right) + \amp \approx 2.1763 \approx 124.7^\circ + +

    +     + +     + +

    + We see from our computation that \alpha + \beta = \theta, + as indicated by . + While we knew this should be the case, + it is nice to see that this non-intuitive formula indeed returns the results we expected. +

    + +

    + We do a similar example next in the context of vectors in space. +

    + + + Using the dot product to find angles + +

    + Let \vec u = \la 1,1,1\ra, + \vec v = \la -1,3,-2\ra and \vec w = \la -5,1,4\ra, + as illustrated in . + Find the angle between each pair of vectors. +

    + +
    + Vectors used in + + + + Graph showing three vectors drawn in space. + + +

    + The x axis is drawn from -5 to 5, + the y axis is drawn from 0 to 4 and + the z axis is drawn from -2 to 4. + Three vectors \vec u, \vec v and \vec w are + shown all starting at the origin, along with separate hollow dashed + cuboids that represent their three axes values and its three sides. + The \vec u has value, \vec v has value and \vec w has value. +

    +     + + + + + //ASY file for figdotp3.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(23,7.4,3.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-5,5}; + real[] myychoice={2,4}; + real[] myzchoice={-2,2,4}; + defaultpen(0.5mm); + pair xbounds=(-5.5,5.5); + pair ybounds=(-1,4.5); + pair zbounds=(-2.5,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // Draw the lines for vec{u}=<1,1,1> + draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1), dashed+linewidth(.5)+bluepen);//top + draw((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0), dashed+linewidth(.5)+bluepen);//bottom + draw((1,0,0)--(1,0,1), dashed+linewidth(.5)+bluepen);//up1 + draw((0,1,0)--(0,1,1), dashed+linewidth(.5)+bluepen);//up2 + draw((1,1,0)--(1,1,1), dashed+linewidth(.5)+bluepen);//up3 + draw((0,0,0)--(1,1,1), bluepen,Arrow3(size=2mm)); + label("$\vec{u}$",(1,1,1),E); + //dotfactor=3; dot((2,1,1),bluepen); + + // Draw the lines for \vec{v}=<-1,3,-2> + draw((0,0,-2)--(0,3,-2)--(-1,3,-2)--(-1,0,-2)--(0,0,-2), dashed+linewidth(.5)+redpen);//top + draw((0,0,0)--(0,3,0)--(-1,3,0)--(-1,0,0)--(0,0,0), dashed+linewidth(.5)+redpen);//bottom + draw((-1,0,0)--(-1,0,-2), dashed+linewidth(.5)+redpen);//up1 + draw((0,3,0)--(0,3,-2), dashed+linewidth(.5)+redpen);//up2 + draw((-1,3,0)--(-1,3,-2), dashed+linewidth(.5)+redpen);//up3 + draw((0,0,0)--(-1,3,-2),redpen,Arrow3(size=2mm)); + label("$\vec{v}$",(-1,3,-2),E); + + // Draw the lines for \vec{w}=<-5,1,4> + draw((0,0,4)--(0,1,4)--(-5,1,4)--(-5,0,4)--(0,0,4), dashed+linewidth(.5)+orange);//top + draw((0,0,0)--(0,1,0)--(-5,1,0)--(-5,0,0)--(0,0,0), dashed+linewidth(.5)+orange);//bottom + draw((-5,0,0)--(-5,0,4), dashed+linewidth(.5)+orange);//up1 + draw((0,1,0)--(0,1,4), dashed+linewidth(.5)+orange);//up2 + draw((-5,1,0)--(-5,1,4), dashed+linewidth(.5)+orange);//up3 + draw((0,0,0)--(-5,1,4), orange,Arrow3(size=2mm)); + label("$\vec{w}$",(-5,1,4),E); + + + + +
    +     + +

    +

      +
    1. +

      + Between \vec u and \vec v: + + \theta \amp = \cos^{-1}\left(\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right) + \amp = \cos^{-1}\left(\frac{0}{\sqrt{3}\sqrt{14}}\right) + \amp = \frac{\pi}2 + . +

      +
      2. + +
    2. +

      + Between \vec u and \vec w: + + \theta \amp = \cos^{-1}\left(\frac{\dotp uw}{\norm{\vec u}\norm{\vec w}}\right) + \amp = \cos^{-1}\left(\frac{0}{\sqrt{3}\sqrt{42}}\right) + \amp = \frac{\pi}2 + . +

      +
      3. + +
    3. +

      + Between \vec v and \vec w: + + \theta \amp = \cos^{-1}\left(\frac{\dotp vw}{\norm{\vec v}\norm{\vec w}}\right) + \amp = \cos^{-1}\left(\frac{0}{\sqrt{14}\sqrt{42}}\right) + \amp = \frac{\pi}2 + . +

      +
      4. +
    +

    + +

    + While our work shows that each angle is \pi/2, , 90^\circ, + none of these angles looks to be a right angle in . + Such is the case when drawing three-dimensional objects on the page. +

    +     +     + +

    + All three angles between these vectors was \pi/2, + or 90^\circ. + We know from geometry and everyday life that + 90^\circ angles are nice + for a variety of reasons, + so it should seem significant that these angles are all \pi/2. + Notice the common feature in each calculation (and also the calculation of \alpha in ): + the dot products of each pair of angles was 0. + We use this as a basis for a definition of the term orthogonal, + which is essentially synonymous to perpendicular. +

    + + + Orthogonal + +

    + Nonzero vectors \vec u and \vec v are orthogonal + if their dot product is 0. + orthogonal + perpendicular|see{orthogonal} + vectorsorthogonal +

    +     +     + + + + + Finding orthogonal vectors + +

    + Let \vec u = \la 3,5\ra and \vec v = \la 1,2,3\ra. +

    + +

    +

      +
    1. +

      + Find two vectors in \mathbb{R}^2 that are orthogonal to \vec u. +

      +
      2. + +
    2. +

      + Find two non-parallel vectors in + \mathbb{R}^3 that are orthogonal to \vec v. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + Recall that a line perpendicular to a line with slope m has slope -1/m, + the opposite reciprocal slope. + We can think of the slope of \vec u as 5/3, + its rise over run. A vector orthogonal to \vec u will have slope -3/5. + There are many such choices, though all parallel: + + \la -5,3\ra \text{ or } \la 5,-3\ra \text{ or } \la -10,6\ra \text{ or } \la 15,-9\ra,\text{ etc. } + +

      +
      2. + +
    2. +

      + There are infinitely many directions in space orthogonal to any given direction, + so there are an infinite number of non-parallel vectors orthogonal to \vec v. + Since there are so many, we have great leeway in finding some. + One way is to arbitrarily pick values for the first two components, + leaving the third unknown. + For instance, let \vec v_1 = \la 2,7,z\ra. + If \vec v_1 is to be orthogonal to \vec v, + then \vec v_1\cdot\vec v = 0, so + + 2+14+3z=0 \Rightarrow z = \frac{-16}{3} + . + So \vec v_1 = \la 2, 7, -16/3\ra is orthogonal to \vec v. + We can apply a similar technique by leaving the first or second component unknown. + Another method of finding a vector orthogonal to \vec v mirrors what we did in part 1. + Let \vec v_2 = \la-2,1,0\ra. + Here we switched the first two components of \vec v, + changing the sign of one of them + (similar to the opposite reciprocal concept before). + Letting the third component be 0 effectively ignores the third component of \vec v, + and it is easy to see that + + \vec v_2\cdot\vec v = \la -2,1,0\ra\cdot\la 1,2,3\ra = 0 + . + Clearly \vec v_1 and \vec v_2 are not parallel. +

      +
      3. +
    +

    +     + +     + +

    + An important construction is illustrated in , + where vectors \vec u and \vec v are sketched. + In , + a dotted line is drawn from the tip of \vec u to the line containing \vec v, + where the dotted line is orthogonal to \vec v. + In , + the dotted line is replaced with the vector \vec z and \vec w is formed, + parallel to \vec v. + It is clear by the diagram that \vec u = \vec w+\vec z. + What is important about this construction is this: + \vec u is decomposed + as the sum of two vectors, + one of which is parallel to \vec v and one that is perpendicular to \vec v. + It is hard to overstate the importance of this construction + (as we'll see in upcoming examples). +

    + +

    + The vectors \vec w, + \vec z and \vec u as shown in form a right triangle, + where the angle between \vec v and \vec u is labeled \theta. + We can find \vec w in terms of \vec v and \vec u. +

    + +

    + Using trigonometry, we can state that + + \norm{\vec w} = \norm{\vec u}\cos(\theta) + . +

    + +
    + Developing the construction of the orthogonal projection + +
    + + + + + Diagram shows two vectors and the angle between them. + + +

    + Two vectors \vec u and \vec v are shown that start from the + same position, the angle between them is marked \theta, the vector + \vec u is shorter. From the end of \vec u a dashed perpendicular + is drawn on \vec v. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=0.93] + + \draw [thick,->] (0,0) -- (4,2) node [right] { $\vec v$}; + \draw [thick,->] (0,0) -- (1,3) node [above ] { $\vec u$}; + \draw [dotted,thick] (1,3) -- (2,1); + \draw (1.82111, 0.910557) -- ({1.73167, 1.08944}) -- (1.91056, 1.17889); + \draw (.4,.2) arc (26.5:71.6:.45); + \draw [rotate=49] (.6,0) node { $\theta$}; + + \end{tikzpicture} + + + + +
    + +
    + + + + + Diagram shows two vectors and the angle between them. + + +

    + Two vectors \vec u and \vec v are shown that + start from the same position, the angle between them is marked + \theta, the vector \vec u is shorter. From the end + of \vec u a dashed perpendicular is drawn on \vec v. + The vector \vec w and \vec z are also added. From the + initial point the vector \vec w is drawn in the same direction + as \vec v but it ends at the start of the perpendicular. The + perpendicular starts at the tip of \vec w and ends at the tip + of \vec u and is labeled \vec z. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=0.93] + + \draw [thick,->] (2,1) -- (4,2) node [right] { $\vec v$}; + \draw [thick,->] (0,0) -- (1,3) node [above ] { $\vec u$}; + \draw [thick,->,gray] (0,0) -- (2,1) node [below,black] { $\vec w$}; + \draw [<-,thick] (1,3) -- (2,1) node [right,pos=.5] { $\vec z$}; + \draw (1.82111, 0.910557) -- ({1.73167, 1.08944})--(1.91056, 1.17889); + \draw (.4,.2) arc (26.5:71.6:.45); + \draw [rotate=49] (.6,0) node { $\theta$}; + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + We also know that \vec w is parallel to to \vec v; that is, + the direction of \vec w is the direction of \vec v, + described by the unit vector \vec v/\norm{\vec v}. + The vector \vec w is the vector in the direction + \vec v/\norm{\vec v} with magnitude \norm{\vec u}\cos(\theta): + + \vec w \amp = \Big(\norm{\vec u}\cos(\theta) \Big)\frac{1}{\norm{\vec v}}\vec v \amp \amp + \amp = \left(\norm{\vec u}\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right)\frac{1}{\norm{\vec v}} \vec v \amp \quad \amp \text{ (Replacing } \cos(\theta) \text{ using } \text{)} + \amp = \frac{\dotp uv}{\norm{\vec v}^2}\vec v \amp \amp + \amp = \frac{\dotp uv}{\dotp vv}\vec v \amp \amp \text{ (Applying }\text{)} + . +

    + +

    + Since this construction is so important, + it is given a special name. +

    + + + Orthogonal Projection + +

    + Let nonzero vectors \vec u and \vec v be given. + The orthogonal projection of \vec u onto \vec v, + denoted \proj uv, is + orthogonal projection + vectorsorthogonal projection + + \proj uv = \frac{\dotp uv}{\dotp vv}\vec v + . +

    +     +     + + + + + Computing the orthogonal projection + +

    +

      +
    1. +

      + Let \vec u= \la -2,1\ra and \vec v=\la 3,1\ra. + Find \proj uv, + and sketch all three vectors with initial points at the origin. +

      +
      2. + +
    2. +

      + Let \vec w = \la 2,1,3\ra and \vec x = \la 1,1,1\ra. + Find \proj wx, + and sketch all three vectors with initial points at the origin. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + Applying , we have + + \proj uv \amp = \frac{\dotp uv}{\dotp vv}\vec v + \amp = \frac{-5}{10}\la 3,1\ra + \amp = \la -\frac32,-\frac12\ra + . + Vectors \vec u, \vec v and + \proj uv are sketched in . + Note how the projection is parallel to \vec v; + that is, it lies on the same line through the origin as \vec v, + although it points in the opposite direction. + That is because the angle between \vec u and \vec v is obtuse (, greater than 90^\circ). +

      + +
      + Sketching the three vectors in Part of + + + Graph shows two vectors u and v, and the projection of u on v. + + +

      + The x axis is drawn from -2 to 3 and the + y axis is drawn from -2 to 2. Two vectors + \vec u and \vec v are shown, both start at the origin. + The vector u ends in point (-2, 1) and lies in the + second quadrant and \vec v ends in point ( 3, 1) and + lies in the first quadrant. The projection of \vec u on + \vec vis shown and it lies in the third quadrant. +

      +       + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + xtick={-3,-2,-1,1,2,3}, + ytick={1,2,-1,-2}, + ymin=-2.8,ymax=2.8, + xmin=-2.9,xmax=3.9, + ] + + \draw [thick,->] (axis cs:0,0) -- (axis cs: -2,1) node [above] {$\vec u$}; + \draw [thick,->] (axis cs:0,0) -- (axis cs: 3,1) node [above] {$\vec v$}; + \draw [thick,->,gray] (axis cs:0,0) -- (axis cs: -1.5,-.5) node [below,black] {$\text{proj}_{\, \vec v}\, \vec u$}; + + \draw [dotted,thick] (axis cs: -2,1) -- (axis cs:-1.5,-.5); + + \end{axis} + + \end{tikzpicture} + + + +
      +
      2. + +
    2. +

      + Apply the definition: + + \proj wx \amp = \frac{\dotp wx}{\dotp xx}\vec x + \amp = \frac{6}{3}\la 1,1,1\ra + \amp = \la 2,2,2\ra + . + These vectors are sketched in , + and again in from a different perspective. + Because of the nature of graphing these vectors, + the sketch in makes it difficult to recognize that the drawn projection has the geometric properties it should. + The graph shown in illustrates these properties better. +

      + +
      + Sketching the three vectors in Part of + +
      + + + + + Graph shows two vectors w and x in space, and the projection of w on x. + + +

      + The x, y and z axes are drawn from 0 to + 2. Two vectors \vec w and \vec x are drawn, + both start at the origin, \vec w ends at (2, 1, 3) + and \vec x ends at (1, 1, 1). The projection of + \vec w on \vec x is shown, it also starts at the + origin and ends at (2, 2, 2). +

      +       + + + + + //ASY file for figdotp4b.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + + // The text calls for two viewpoints of this picture to show orthogonality + // The two projections below give those two views + // view "b", saved as figdotp4b_3D + currentprojection=orthographic(8,2,5); + + // view "c", saved as figdotp4c_3D + //currentprojection=orthographic(2,8,6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-0.5,2.5); + pair ybounds=(-1,2.5); + pair zbounds=(-0.5,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // Draw the lines for vec{x}=<1,1,1> + //draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1), dashed+linewidth(.5));//top + draw((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0), dashed+linewidth(.5));//bottom + //draw((1,0,0)--(1,0,1), dashed+linewidth(.5));//up1 + //draw((0,1,0)--(0,1,1), dashed+linewidth(.5));//up2 + draw((1,1,0)--(1,1,1), dashed+linewidth(.5));//up3 + draw((0,0,0)--(1,1,1), Arrow3(size=3mm)); + label("$\vec{x}$",(1,1,1),N); + //dotfactor=3; dot((2,1,1),bluepen); + + // Draw the lines for projection of w onto x as <2,2,2> + //draw((0,0,2)--(0,2,2)--(2,2,2)--(2,0,2)--(0,0,2), dashed+linewidth(.5));//top + draw((0,0,0)--(0,2,0)--(2,2,0)--(2,0,0)--(0,0,0), dashed+linewidth(.5));//bottom + //draw((2,0,0)--(2,0,2), dashed+linewidth(.5));//up1 + //draw((0,2,0)--(0,2,2), dashed+linewidth(.5));//up2 + draw((2,2,0)--(2,2,2), dashed+linewidth(.5));//up3 + draw((0,0,0)--(2,2,2), gray,Arrow3(size=3mm)); + label("$\textrm{proj}_{\vec{x}} \vec{w}$",(2,2,2),N); + + // Draw the lines for \vec{w}=<2,1,3> + //draw((0,0,3)--(0,1,3)--(2,1,3)--(2,0,3)--(0,0,3), dashed+linewidth(.5));//top + draw((0,0,0)--(0,1,0)--(2,1,0)--(2,0,0)--(0,0,0), dashed+linewidth(.5));//bottom + //draw((2,0,0)--(2,0,3), dashed+linewidth(.5));//up1 + //draw((0,1,0)--(0,1,3), dashed+linewidth(.5));//up2 + draw((2,1,0)--(2,1,3), dashed+linewidth(.5));//up3 + draw((0,0,0)--(2,1,3), Arrow3(size=3mm)); + label("$\vec{w}$",(2,1,3),N); + + + + +
      + +
      + + + + + + Graph shows two vectors w and x in space, and the projection of w on x. + + +

      + The x, y and z axes are drawn from 0 to + 2. Two vectors \vec w and \vec x are drawn, + both start at the origin, \vec w ends at (2, 1, 3) + and \vec x ends at (1, 1, 1). The projection of + \vec w on \vec x is shown, it also starts at the + origin and ends at (2, 2, 2). +

      +       + + + + + //ASY file for figdotp4b.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + + // The text calls for two viewpoints of this picture to show orthogonality + // The two projections below give those two views + // view "b", saved as figdotp4b_3D + //currentprojection=orthographic(8,2,5); + + // view "c", saved as figdotp4c_3D + currentprojection=orthographic(2,8,6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-0.5,2.5); + pair ybounds=(-1,2.5); + pair zbounds=(-0.5,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // Draw the lines for vec{x}=<1,1,1> + //draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1), dashed+linewidth(.5));//top + draw((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0), dashed+linewidth(.5));//bottom + //draw((1,0,0)--(1,0,1), dashed+linewidth(.5));//up1 + //draw((0,1,0)--(0,1,1), dashed+linewidth(.5));//up2 + draw((1,1,0)--(1,1,1), dashed+linewidth(.5));//up3 + draw((0,0,0)--(1,1,1), Arrow3(size=3mm)); + label("$\vec{x}$",(1,1,1),N); + //dotfactor=3; dot((2,1,1),bluepen); + + // Draw the lines for projection of w onto x as <2,2,2> + //draw((0,0,2)--(0,2,2)--(2,2,2)--(2,0,2)--(0,0,2), dashed+linewidth(.5));//top + draw((0,0,0)--(0,2,0)--(2,2,0)--(2,0,0)--(0,0,0), dashed+linewidth(.5));//bottom + //draw((2,0,0)--(2,0,2), dashed+linewidth(.5));//up1 + //draw((0,2,0)--(0,2,2), dashed+linewidth(.5));//up2 + draw((2,2,0)--(2,2,2), dashed+linewidth(.5));//up3 + draw((0,0,0)--(2,2,2), gray,Arrow3(size=3mm)); + label("$\textrm{proj}_{\vec{x}} \vec{w}$",(2,2,2),N); + + // Draw the lines for \vec{w}=<2,1,3> + //draw((0,0,3)--(0,1,3)--(2,1,3)--(2,0,3)--(0,0,3), dashed+linewidth(.5));//top + draw((0,0,0)--(0,1,0)--(2,1,0)--(2,0,0)--(0,0,0), dashed+linewidth(.5));//bottom + //draw((2,0,0)--(2,0,3), dashed+linewidth(.5));//up1 + //draw((0,1,0)--(0,1,3), dashed+linewidth(.5));//up2 + draw((2,1,0)--(2,1,3), dashed+linewidth(.5));//up3 + draw((0,0,0)--(2,1,3), Arrow3(size=3mm)); + label("$\vec{w}$",(2,1,3),N); + + + + +
      +       +
      +
      3. +
    +

    +     + +     + +

    + We can use the properties of the dot product found in + to rearrange the formula found in : + + \proj uv \amp = \frac{\dotp uv}{\dotp vv}\vec v + \amp = \frac{\dotp uv}{\norm{\vec v}^2}\vec v + \amp = \left(\vec u \cdot \frac{\vec v}{\norm{\vec v}}\right) \frac{\vec v}{\norm{\vec v}} + . +

    + +

    + The above formula shows that the orthogonal projection of \vec u onto \vec v is only concerned with the + direction of \vec v, + as both instances of \vec v in the formula come in the form \vec v/\norm{\vec v}, + the unit vector in the direction of \vec v. +

    + +

    + A special case of orthogonal projection occurs when \vec v is a unit vector. + In this situation, + the formula for the orthogonal projection of a vector \vec u onto \vec v reduces to just \proj uv = (\vec u\cdot\vec v)\vec v, + as \vec v\cdot\vec v = 1. +

    + +

    + This gives us a new understanding of the dot product. + When \vec v is a unit vector, + essentially providing only direction information, + the dot product of \vec u and \vec v gives + how much of \vec u is in the direction of \vec v. + This use of the dot product will be very useful in future sections. +

    + +

    + Now consider + where the concept of the orthogonal projection is again illustrated. + It is clear that + + \vec u = \proj uv + \vec z + . +

    + +
    + Illustrating the orthogonal projection + + + + Diagram shows two vectors and the angle between them. + + +

    + Two vectors \vec u and \vec v are shown that start from the + same position, the angle between them is marked \theta, the vector + u is shorter. From the end of \vec u a dashed perpendicular + is drawn on \vec v. The projection of \vec u and \vec z + are also added. The projection of \vec u is in the same direction as + \vec v but it ends at the start of the perpendicular. The perpendicular + starts at the tip of the projection of \vec u and ends at the tip of + \vec u and is labeled \vec z. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.32] + + \draw [thick,->] (2,1) -- (4,2) node [below] { $\vec v$}; + \draw [thick,->] (0,0) -- (1,3) node [above] { $\vec u$}; + \draw [thick,->,gray] (0,0) -- (2,1) node [pos=.5,below right,black] { $\text{proj}_{\, \vec v}\, \vec u$}; + \draw [<-,thick] (1,3) -- (2,1) node [right,pos=.5] { $\vec z$}; + \draw (1.82111, 0.910557) -- ({1.73167, 1.08944}) -- (1.91056, 1.17889); + + \end{tikzpicture} + + + + +
    + +

    + As we know what \vec u and \proj uv are, + we can solve for \vec z and state that + + \vec z = \vec u - \proj uv + . +

    + +

    + This leads us to rewrite Equation in a seemingly silly way: + + \vec u = \proj uv + (\vec u - \proj uv) + . +

    + +

    + This is not nonsense, as pointed out in the following Key Idea. + (Notation note: the expression \parallel \vec y + means is parallel to \vec y. + We can use this notation to state + \vec x\parallel\vec y which means + \vec x is parallel to \vec y. + The expression \perp \vec y + means is orthogonal to \vec y, and is used similarly.) +

    + + + Orthogonal Decomposition of Vectors +

    + Let nonzero vectors \vec u and \vec v be given. + Then \vec u can be written as the sum of two vectors, + one of which is parallel to \vec v, + and one of which is orthogonal to \vec v: + orthogonal decomposition of vectors + orthogonaldecomposition + vectorsorthogonal decomposition + + \vec u = \underbrace{\proj uv}_{\parallel\ \vec v}\ +\ (\underbrace{\vec u-\proj uv}_{\perp\ \vec v}) + . +

    +     + + + +

    + We illustrate the use of this equality in the following example. +

    + + + Orthogonal decomposition of vectors + +

    +

      +
    1. +

      + Let \vec u = \la -2,1\ra and + \vec v = \la 3,1\ra as in . + Decompose \vec u as the sum of a vector parallel to \vec v and a vector orthogonal to \vec v. +

      +
      2. + +
    2. +

      + Let \vec w =\la 2,1,3\ra and + \vec x =\la 1,1,1\ra as in . + Decompose \vec w as the sum of a vector parallel to \vec x and a vector orthogonal to \vec x. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + In , + we found that \proj uv = \la -1.5,-0.5\ra. + Let + + \vec z = \vec u - \proj uv = \la -2,1\ra - \la -1.5,-0.5\ra = \la-0.5, 1.5\ra + . + Is \vec z orthogonal to \vec v? (, is + \vec z \perp\vec v?) We check for orthogonality with the dot product: + + \dotp zv = \la -0.5,1.5\ra \cdot \la 3,1\ra =0 + . + Since the dot product is 0, we know \vec z \perp \vec v. + Thus: + + \vec u \amp = \proj uv\ +\ (\vec u - \proj uv) + \la -2,1\ra \amp = \underbrace{\la -1.5,-0.5\ra}_{\parallel\ \vec v}\ +\ \underbrace{\la -0.5,1.5\ra}_{\perp \ \vec v} + . +

      +
      2. + +
    2. +

      + We found in + that \proj wx = \la 2,2,2\ra. + Applying the Key Idea, we have: + + \vec z = \vec w - \proj wx = \la 2,1,3\ra - \la 2,2,2\ra = \la 0,-1,1\ra + . + We check to see if \vec z \perp \vec x: + + \dotp zx = \la 0,-1,1\ra \cdot \la 1,1,1\ra = 0 + . + Since the dot product is 0, we know the two vectors are orthogonal. + We now write \vec w as the sum of two vectors, + one parallel and one orthogonal to \vec x: + + \vec w \amp = \proj wx\ +\ (\vec w - \proj wx) + \la 2,1,3\ra \amp = \underbrace{\la 2,2,2\ra}_{\parallel\ \vec x}\ +\ \underbrace{\la 0,-1,1\ra}_{\perp \ \vec x} + +

      +
      3. +
    +

    +     + +     + +

    + We give an example of where this decomposition is useful. +

    + + + Orthogonally decomposing a force vector + +

    + Consider , + showing a box weighing 50lb on a ramp that rises 5ft over a span of 20ft. + Find the components of force, + and their magnitudes, + acting on the box (as sketched in ): +

    + +
    + Sketching the ramp and box in . Note: The vectors are not drawn to scale. + +
    + + + + + Diagram showing a box on a ramp. + + +

    + Image of a box being placed on an inclined plane. The downward + force for gravity is marked as \vec g. The surface of the + ramp is labeled as \vec r base of the ramp is labeled 20 + and the height is marked as 5. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{scope}[scale=.25] + \draw [thick,->] (20,5) -- node [right,pos=.5] { $5$} (20,0) -- node [below,pos=.6] { 20} (0,0) -- (20,5) node [above] { $\vec r$}; + \draw [thick] (10,2.5) -- (9.25,5.5) -- (12.25,6.25) -- (13,3.25); + \draw [thick,->] (11.125,4.375) -- (11.125,-3.625) node [below] { $\vec g$}; + \end{scope} + + \end{tikzpicture} + + + + +
    + +
    + + + + + Diagram showing a box on a ramp. + + +

    + Image of a box being placed on an inclined plane. The downward force for + gravity is marked as \vec g. The surface of the ramp is labeled as + \vec r base of the ramp is labeled 20 and the height is marked + as 5. The projection of \vec g on \vec r is shown. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{scope}[scale=.25] + \draw [thick,->] (20,5) -- node [right,pos=.5] { $5$} (20,0) -- node [below,pos=.6] { 20}(0,0) -- (20,5) node [above] { $\vec r$}; + \draw [thick] (10,2.5) -- (9.25,5.5) -- (12.25,6.25) -- (13,3.25); + \draw [thick,->] (11.125,4.375) -- (11.125,-3.625) node [below] { $\vec g$}; + \draw [gray,thick,->] (11.125,4.375) -- (13,-3.15) node [right,pos=.4,black] { $\vec z$}; + \draw [gray,thick,->] (13,-3.15) -- (11.125,-3.625) node [shift={(12pt,-8pt)} ,black,pos=0] { $\proj gr$}; + \end{scope} + + \end{tikzpicture} + + + + +
    +     +
    + +

    +

      +
    1. +

      + in the direction of the ramp, and +

      +
      2. + +
    2. +

      + orthogonal to the ramp. +

      +
      3. +
    +

    +     + +

    + As the ramp rises 5ft over a horizontal distance of 20ft, + we can represent the direction of the ramp with the vector \vec r= \la 20,5\ra. + Gravity pulls down with a force of 50lb, + which we represent with \vec g = \la 0,-50\ra. +

    + +

    +

      +
    1. +

      + To find the force of gravity in the direction of the ramp, + we compute \proj gr: + + \proj gr \amp = \frac{\dotp gr}{\dotp rr}\vec r + \amp = \frac{-250}{425}\la 20,5\ra + \amp = \la -\frac{200}{17},-\frac{50}{17}\ra \approx \la -11.76,-2.94\ra + . + The magnitude of \proj gr is \norm{\proj gr} = 50/\sqrt{17} \approx 12.13\text{ lb }. + Though the box weighs 50lb, + a force of about 12lb is enough to keep the box from sliding down the ramp. +

      +
      2. + +
    2. +

      + To find the component \vec z of gravity orthogonal to the ramp, + we use . + + \vec z \amp = \vec g - \proj gr + \amp = \la \frac{200}{17},-\frac{800}{17}\ra \approx \la 11.76,-47.06\ra + . + The magnitude of this force is \norm{\vec z} \approx 48.51lb. + In physics and engineering, + knowing this force is important when computing things like static frictional force. + (For instance, + we could easily compute if the static frictional force alone was enough to keep the box from sliding down the ramp.) +

      +
      3. +
    +

    +     + +     + + + Application to Work +

    + In physics, the application of a force F to move an object in a straight line a distance d produces work; + the amount of work W is W=Fd, + (where F is in the direction of travel). + The orthogonal projection allows us to compute work when the force is not in the direction of travel. +

    + +
    + Finding work when the force and direction of travel are given as vectors + + + + Diagram showing the two vectors for force and displacement used to find work. + + +

    + A box is shown that is being pushed to the right by a force. The displacement is + shown as a horizontal vector and is labeled \vec d. Force is drawn from the + middle of the box at an angle and is labeled as \vec F. The projection of + the \vec F is also shown, it is parallel to the displacement vector. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \draw [thick] (0,0) rectangle (2,2); + \draw [thick,->] (0,-.25) -- (5,-.25) node [right] { $\vec d$}; + \draw [thick,->] (1,1) -- (3,2) node [shift={(5pt,2.5pt)}] { $\vec F$}; + \draw [thick,gray,->] (1,1) -- (3,1) node [right,black] { $\proj Fd$}; + + \end{tikzpicture} + + + + +
    + +

    + Consider , + where a force \vec F is being applied to an object moving in the direction of \vec d. + (The distance the object travels is the magnitude of \vec d.) + The work done is the amount of force in the direction of \vec d, + \norm{\proj Fd}, times \vnorm d: + + \norm{\proj Fd}\cdot\vnorm d \amp = \norm{\frac{\dotp Fd}{\dotp dd}\vec d}\cdot \vnorm d + \amp = \abs{\frac{\dotp Fd}{\norm{\vec{d}}^2}}\cdot \vnorm d\cdot\vnorm d + \amp = \frac{\abs{\dotp Fd}}{\norm{\vec{d}}^2}\norm{\vec{d}}^2 + \amp = \abs{\dotp Fd} + . +

    + +

    + The expression \dotp Fd will be positive if the angle between \vec F and \vec d is acute; + when the angle is obtuse + (hence \dotp Fd is negative), + the force is causing motion in the opposite direction of \vec d, + resulting in negative work. + We want to capture this sign, + so we drop the absolute value and find that W = \dotp Fd. +

    + + + Work + +

    + Let \vec F be a constant force that moves an object in a straight line from point P to point Q. + Let \vec d = \overrightarrow{PQ}. + The work W done by \vec F along \vec d is W = \dotp Fd. + work +

    +     +     + + + Computing work + +

    + A man slides a box along a ramp that rises 3ft over a distance of 15ft by applying 50lb of force as shown in . + Compute the work done. +

    +
    + Computing work when sliding a box up a ramp in + + + + Diagram showing a box on a ramp. + + +

    + Image of a box being placed on an inclined plane or ramp. The base of the + ramp is labeled 15 and the height is marked as 3. Force is + drawn from the middle of the box at an angle of 30 from the horizontal + and is labeled as \vec F. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=.35] + + \draw [thick] (0,0) -- node [below,pos=.5] { 15} (15,0) -- node [right, pos=.5] { 3} (15,3) -- cycle; + \draw [thick] (5,1) -- (4.55,3.25) -- (6.8,3.7) -- (7.25,1.45); + + \begin{scope}[shift={(5.9,2.35)}] + \draw [thick,rotate=30,->] (0,0) -- (5,0) node [right] { $\vec F$}; + \draw [thick,dashed] (-2,0) -- (4,0); + \draw (2,0) arc (0:30:2); + \draw [rotate=15] (3,0) node { $30^\circ$}; + \end{scope} + + \end{tikzpicture} + + + + +
    +     + +

    + The figure indicates that the force applied makes a + 30^\circ angle with the horizontal, + so \vec F = 50\la \cos(30^\circ) ,\sin(30^\circ) \ra \approx \la 43.3,25\ra. + The ramp is represented by \vec d = \la 15,3\ra. + The work done is simply + + \dotp Fd = 50\la \cos(30^\circ) ,\sin(30^\circ) \ra \cdot \la 15,3\ra \approx 724.5 \text{ ft--lb } + . +

    + +

    + Note how we did not actually compute the distance the object traveled, + nor the magnitude of the force in the direction of travel; + this is all inherently computed by the dot product! +

    +     +     + +

    + The dot product is a powerful way of evaluating computations that depend on angles without actually using angles. + The next section explores another + product on vectors, + the cross product. Once again, + angles play an important role, + though in a much different way. +

    +     + + + + Terms and Concepts + + + +

    + The dot product of two vectors is a , not a vector. +

    +     + + + + + + + + +     + + + + +

    + How are the concepts of the dot product and vector magnitude related? +

    +     + + + +

    + The magnitude of a vectors is the square root of the dot product of a vector with itself; that is, + \norm{\vec v} = \sqrt{\vec v\cdot \vec v}. +

    +     + +     + + + + +

    + How can one quickly tell if the angle between two vectors is acute or obtuse? +

    +     + + + +

    + By considering the sign of the dot product of the two vectors. + If the dot product is positive, the angle is acute; + if the dot product is negative, the angle is obtuse. +

    +     + +     + + + + +

    + Give a synonym for orthogonal. +

    +     + + + +

    + Perpendicular is one answer, + although we tend to use perpendicular to refer to lines, + and orthogonal to refer to vectors. +

    +

    + Note that the zero vector is orthogonal to any vector, + but it doesn't really make sense to say it is perpendicular. +

    +     + +     +     + + + Problems + + + +

    + Find the dot product of the given vectors. +

    +     + + + + + Context("LimitedNumeric"); + $dot=Real("-22"); + + + +

    + \vec u = \la 2,-4\ra, \vec v = \la 3,7\ra +

    + +

    + +

    +     +     +     + + + + + Context("LimitedNumeric"); + $dot=Real("33"); + + + +

    + \vec u = \la 5,3\ra, \vec v = \la 6,1\ra +

    + +

    + +

    +     +     +     + + + + + Context("LimitedNumeric"); + $dot=Real("3"); + + + +

    + \vec u = \la 1,-1,2\ra, + \vec v = \la 2,5,3\ra +

    + +

    + +

    +     +     +     + + + + + Context("LimitedNumeric"); + $dot=Real("0"); + + + +

    + \vec u = \la 3,5,-1\ra, + \vec v = \la 4,-1,7\ra +

    + +

    + +

    +     +     +     + + + + +

    + \vec u = \la 1,1\ra, \vec v = \la 1,2,3\ra +

    +     + +

    + not defined +

    +     + +     + + + + + Context("LimitedNumeric"); + $dot=Real("0"); + + + +

    + \vec u = \la 1,2,3\ra, + \vec v = \la 0,0,0\ra +

    + +

    + +

    +     +     +     + +     + + + + +

    + Create your own vectors \vec u, + \vec v and \vec w in + \mathbb{R}^2 and show that \vec u\cdot (\vec v+\vec w) = \vec u\cdot \vec v + \vec u\cdot \vec w. +

    +     + +

    + Answers will vary. +

    +     + +     + + + + +

    + Create your own vectors \vec u and \vec v in + \mathbb{R}^3 and scalar c and show that c(\vec u\cdot \vec v) = \vec u\cdot (c\vec v). +

    +     + +

    + Answers will vary. +

    +     + +     + + + + +

    + Find the measure of the angle between the two vectors in radians. +

    +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $theta=Formula("arccos(3/sqrt(10))"); + + + +

    + \vec u = \la 1,1\ra and + \vec v = \la 1,2\ra. +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $theta=Formula("arccos(-1/sqrt(170))"); + + + +

    + \vec u = \la -2,1\ra and + \vec v = \la 3,5\ra. +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $theta=Formula("pi/4"); + + + +

    + \vec u = \la 8,1,-4\ra and + \vec v = \la 2,2,0\ra. +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $theta=Formula("pi/2"); + + + +

    + \vec u = \la 1,7,2\ra and + \vec v = \la 4,-2,5\ra. +

    + +

    + +

    +     +     +     + +     + + + +

    + A vector \vec v is given. + Give two vectors that are orthogonal to \vec v. +

    +     + + + + + Context("Vector"); + $v=Vector("<4,7>"); + $u=Vector("<-7,4>"); + $w=Vector("<14,-8>"); + $multians = MultiAnswer($u,$v)->with( + singleResult => 0, + checker => sub { + my ($correct,$student,$self) = @_; + my ($s1, $s2) = @{$student}; + $ret[0] = (($s1 . $v == 0) and ($s1 != Vector("<0,0>"))); + $ret[1] = (($s2 . $v == 0) and ($s2 != Vector("<0,0>")) and ($s1 != $s2)); + return ~~@ret; + } + ); + + + +

    + Find two nonzero vectors orthogonal to \vec v = \la 4,7\ra. +

    + +

    + +

    + +

    + +

    +     + +

    + Answers will vary; two possible answers are + \la -7,4\ra and \la 14,-8\ra. +

    +     +     +     + + + + + Context("Vector"); + $v=Vector("<-3,5>"); + $u=Vector("<5,3>"); + $w=Vector("<-15,-9>"); + $multians = MultiAnswer($u,$v)->with( + singleResult => 0, + checker => sub { + my ($correct,$student,$self) = @_; + my ($s1, $s2) = @{$student}; + $ret[0] = (($s1 . $v == 0) and ($s1 != Vector("<0,0>"))); + $ret[1] = (($s2 . $v == 0) and ($s2 != Vector("<0,0>")) and ($s1 != $s2)); + return ~~@ret; + } + ); + + + +

    + Find two nonzero vectors orthogonal to \vec v = \la -3,5\ra. +

    + +

    + +

    + +

    + +

    +     + +

    + Answers will vary; two possible answers are + \la 5,3\ra and \la -15,-9\ra. +

    +     +     +     + + + + + Context("Vector"); + $v=Vector("<1,1,1>"); + $u=Vector("<1,0,-1>"); + $w=Vector("<4,5,-9>"); + $multians = MultiAnswer($u,$v)->with( + singleResult => 0, + checker => sub { + my ($correct,$student,$self) = @_; + my ($s1, $s2) = @{$student}; + $ret[0] = (($s1 . $v == 0) and ($s1 != Vector("<0,0>"))); + $ret[1] = (($s2 . $v == 0) and ($s2 != Vector("<0,0>")) and ($s1 != $s2)); + return ~~@ret; + } + ); + + + +

    + Find two nonzero vectors orthogonal to \vec v = \la 1,1,1\ra. +

    + +

    + +

    + +

    + +

    +     + +

    + Answers will vary; two possible answers are + \la 1,0,-1\ra and \la 4,5,-9\ra. +

    +     +     +     + + + + + Context("Vector"); + $v=Vector("<1,-2,3>"); + $u=Vector("<2,1,0>"); + $w=Vector("<1,1,1/3>"); + $multians = MultiAnswer($u,$v)->with( + singleResult => 0, + checker => sub { + my ($correct,$student,$self) = @_; + my ($s1, $s2) = @{$student}; + $ret[0] = (($s1 . $v == 0) and ($s1 != Vector("<0,0>"))); + $ret[1] = (($s2 . $v == 0) and ($s2 != Vector("<0,0>")) and ($s1 != $s2)); + return ~~@ret; + } + ); + + + +

    + Find two nonzero vectors orthogonal to \vec v = \la 1,-2,3\ra. +

    + +

    + +

    + +

    + +

    +     + +

    + Answers will vary; two possible answers are + \la 2,1,0\ra and \la 1,1,1/3\ra. +

    +     +     +     + +     + + + +

    + Vectors \vec u and \vec v are given. + Find \proj uv, + the orthogonal projection of \vec u onto \vec v, + and sketch all three vectors with the same initial point. +

    +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<1,2>"); + $v=Vector("<-1,3>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); + + + +

    + \vec u = \la 1,2\ra and \vec v = \la -1,3\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<5,5>"); + $v=Vector("<1,3>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); + + + +

    + \vec u = \la 5,5\ra and \vec v = \la 1,3\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<-3,2>"); + $v=Vector("<1,1>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); + + + +

    + \vec u = \la -3,2\ra and \vec v = \la 1,1\ra +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<-3,2>"); + $v=Vector("<2,3>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); + + + +

    + \vec u = \la -3,2\ra and \vec v = \la 2,3\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<1,5,1>"); + $v=Vector("<1,2,3>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den,$nums[2]/$den>"); + + + +

    + \vec u = \la 1,5,1\ra and \vec v = \la 1,2,3\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<3,-1,2>"); + $v=Vector("<2,2,1>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den,$nums[2]/$den>"); + + + +

    + \vec u = \la 3,-1,2\ra and \vec v = \la 2,2,1\ra. +

    + +

    + +

    +     +     +     +     + + + +

    + Vectors \vec u and \vec v are given. + Write \vec u as the sum of two vectors, + one of which is parallel to \vec v + (or is zero) + and one of which is orthogonal to \vec v. + Note: these are the same pairs of vectors as found in Exercises. +

    +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<1,2>"); + $v=Vector("<-1,3>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); + @uc=$u->value; + @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1); + $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den>"); + $multians = MultiAnswer($projuv,$comp)->with( + singleResult => 0, + checker => sub { + my ($correct,$student,$self) = @_; + my ($s1, $s2) = @{$student}; + my ($c1, $c2) = @{$correct}; + if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) + {return [1,1];} + elsif ($s1+$s2 != $u) + {$self->setMessage(1,"These vectors do not sum to u."); + $self->setMessage(2,"These vectors do not sum to u."); + return [0,0]; + } + elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is parallel to v."); + $self->setMessage(2,"But this vector is not orthogonal to v."); + return [1,0]; + } + elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is not orthogonal to v."); + $self->setMessage(2,"This vector is parallel to v."); + return [0,1]; + } + elsif ($s1 . $v == 0) + {$self->setMessage(1,"This vector is orthogonal to v."); + $self->setMessage(2,"But this vector is not parallel to v."); + return [1,0]; + } + elsif ($s2 . $v == 0) + {$self->setMessage(1,"This vector is not parallel to v."); + $self->setMessage(2,"This vector is orthogonal to v."); + return [0,1]; + } + else { + return [0,0]; + }; + } + ); + + + +

    + \vec u = \la 1,2\raand \vec v = \la -1,3\ra. +

    + +

    + \vec u={}+{} +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<5,5>"); + $v=Vector("<1,3>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); + @uc=$u->value; + @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1); + $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den>"); + $multians = MultiAnswer($projuv,$comp)->with( + singleResult => 0, + checker => sub { + my ($correct,$student,$self) = @_; + my ($s1, $s2) = @{$student}; + my ($c1, $c2) = @{$correct}; + if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) + {return [1,1];} + elsif ($s1+$s2 != $u) + {$self->setMessage(1,"These vectors do not sum to u."); + $self->setMessage(2,"These vectors do not sum to u."); + return [0,0]; + } + elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is parallel to v."); + $self->setMessage(2,"But this vector is not orthogonal to v."); + return [1,0]; + } + elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is not orthogonal to v."); + $self->setMessage(2,"This vector is parallel to v."); + return [0,1]; + } + elsif ($s1 . $v == 0) + {$self->setMessage(1,"This vector is orthogonal to v."); + $self->setMessage(2,"But this vector is not parallel to v."); + return [1,0]; + } + elsif ($s2 . $v == 0) + {$self->setMessage(1,"This vector is not parallel to v."); + $self->setMessage(2,"This vector is orthogonal to v."); + return [0,1]; + } + else { + return [0,0]; + }; + } + ); + + + +

    + \vec u = \la 5,5\ra and \vec v = \la 1,3\ra. +

    + +

    + \vec u={}+{} +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<-3,2>"); + $v=Vector("<1,1>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); + @uc=$u->value; + @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1); + $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den>"); + $multians = MultiAnswer($projuv,$comp)->with( + singleResult => 0, + checker => sub { + my ($correct,$student,$self) = @_; + my ($s1, $s2) = @{$student}; + my ($c1, $c2) = @{$correct}; + if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) + {return [1,1];} + elsif ($s1+$s2 != $u) + {$self->setMessage(1,"These vectors do not sum to u."); + $self->setMessage(2,"These vectors do not sum to u."); + return [0,0]; + } + elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is parallel to v."); + $self->setMessage(2,"But this vector is not orthogonal to v."); + return [1,0]; + } + elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is not orthogonal to v."); + $self->setMessage(2,"This vector is parallel to v."); + return [0,1]; + } + elsif ($s1 . $v == 0) + {$self->setMessage(1,"This vector is orthogonal to v."); + $self->setMessage(2,"But this vector is not parallel to v."); + return [1,0]; + } + elsif ($s2 . $v == 0) + {$self->setMessage(1,"This vector is not parallel to v."); + $self->setMessage(2,"This vector is orthogonal to v."); + return [0,1]; + } + else { + return [0,0]; + }; + } + ); + + + +

    + \vec u = \la -3,2\ra and \vec v = \la 1,1\ra. +

    + +

    + \vec u={}+{} +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<-3,2>"); + $v=Vector("<2,3>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); + @uc=$u->value; + @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1); + $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den>"); + $multians = MultiAnswer($projuv,$comp)->with( + singleResult => 0, + checker => sub { + my ($correct,$student,$self) = @_; + my ($s1, $s2) = @{$student}; + my ($c1, $c2) = @{$correct}; + if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) + {return [1,1];} + elsif ($s1+$s2 != $u) + {$self->setMessage(1,"These vectors do not sum to u."); + $self->setMessage(2,"These vectors do not sum to u."); + return [0,0]; + } + elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is parallel to v."); + $self->setMessage(2,"But this vector is not orthogonal to v."); + return [1,0]; + } + elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is not orthogonal to v."); + $self->setMessage(2,"This vector is parallel to v."); + return [0,1]; + } + elsif ($s1 . $v == 0) + {$self->setMessage(1,"This vector is orthogonal to v."); + $self->setMessage(2,"But this vector is not parallel to v."); + return [1,0]; + } + elsif ($s2 . $v == 0) + {$self->setMessage(1,"This vector is not parallel to v."); + $self->setMessage(2,"This vector is orthogonal to v."); + return [0,1]; + } + else { + return [0,0]; + }; + } + ); + + + +

    + \vec u = \la -3,2\ra and \vec v = \la 2,3\ra. +

    + +

    + \vec u={}+{} +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<1,5,1>"); + $v=Vector("<1,2,3>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den,$nums[2]/$den>"); + @uc=$u->value; + @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1,2); + $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den,$compnums[2]/$den>"); + $multians = MultiAnswer($projuv,$comp)->with( + singleResult => 0, + checker => sub { + my ($correct,$student,$self) = @_; + my ($s1, $s2) = @{$student}; + my ($c1, $c2) = @{$correct}; + if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) + {return [1,1];} + elsif ($s1+$s2 != $u) + {$self->setMessage(1,"These vectors do not sum to u."); + $self->setMessage(2,"These vectors do not sum to u."); + return [0,0]; + } + elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is parallel to v."); + $self->setMessage(2,"But this vector is not orthogonal to v."); + return [1,0]; + } + elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is not orthogonal to v."); + $self->setMessage(2,"This vector is parallel to v."); + return [0,1]; + } + elsif ($s1 . $v == 0) + {$self->setMessage(1,"This vector is orthogonal to v."); + $self->setMessage(2,"But this vector is not parallel to v."); + return [1,0]; + } + elsif ($s2 . $v == 0) + {$self->setMessage(1,"This vector is not parallel to v."); + $self->setMessage(2,"This vector is orthogonal to v."); + return [0,1]; + } + else { + return [0,0]; + }; + } + ); + + + +

    + \vec u = \la 1,5,1\ra and \vec v = \la 1,2,3\ra. +

    + +

    + \vec u={}+{} +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $u=Vector("<3,-1,2>"); + $v=Vector("<2,2,1>"); + $num=$u . $v; + $den=$v . $v; + @vc=$v->value; + @nums=map{$_*$num}(@vc); + $projuv=Compute("<$nums[0]/$den,$nums[1]/$den,$nums[2]/$den>"); + @uc=$u->value; + @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1,2); + $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den,$compnums[2]/$den>"); + $multians = MultiAnswer($projuv,$comp)->with( + singleResult => 0, + checker => sub { + my ($correct,$student,$self) = @_; + my ($s1, $s2) = @{$student}; + my ($c1, $c2) = @{$correct}; + if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) + {return [1,1];} + elsif ($s1+$s2 != $u) + {$self->setMessage(1,"These vectors do not sum to u."); + $self->setMessage(2,"These vectors do not sum to u."); + return [0,0]; + } + elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is parallel to v."); + $self->setMessage(2,"But this vector is not orthogonal to v."); + return [1,0]; + } + elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) + {$self->setMessage(1,"This vector is not orthogonal to v."); + $self->setMessage(2,"This vector is parallel to v."); + return [0,1]; + } + elsif ($s1 . $v == 0) + {$self->setMessage(1,"This vector is orthogonal to v."); + $self->setMessage(2,"But this vector is not parallel to v."); + return [1,0]; + } + elsif ($s2 . $v == 0) + {$self->setMessage(1,"This vector is not parallel to v."); + $self->setMessage(2,"This vector is orthogonal to v."); + return [0,1]; + } + else { + return [0,0]; + }; + } + ); + + + +

    + \vec u = \la 3,-1,2\ra and \vec v = \la 2,2,1\ra. +

    + +

    + \vec u={}+{} +

    +     +     +     + +     + + + + +

    + A 10lb box sits on a ramp that rises 4ft over a distance of 20ft. + How much force is required to keep the box from sliding down the ramp? +

    +     + +

    + 1.96lb +

    +     + +     + + + + +

    + A 10lb box sits on a 15ft ramp that makes a + 30^\circ angle with the horizontal. + How much force is required to keep the box from sliding down the ramp? +

    +     + +

    + 5lb +

    +     + +     + + + + +

    + How much work is performed in moving a box horizontally 10ft with a force of 20lb applied at an angle of 45^\circ to the horizontal? +

    +     + +

    + 141.42ftlb +

    +     + +     + + + + +

    + How much work is performed in moving a box horizontally 10ft with a force of 20lb applied at an angle of 10^\circ to the horizontal? +

    +     + +

    + 196.96ftlb +

    +     + +     + + + + +

    + How much work is performed in moving a box up the length of a ramp that rises 2ft over a distance of 10ft, + with a force of 50lb applied horizontally? +

    +     + +

    + 500ftlb +

    +     + +     + + + + +

    + How much work is performed in moving a box up the length of a ramp that rises 2ft over a distance of 10ft, + with a force of 50lb applied at an angle of 45^\circ to the horizontal? +

    +     + +

    + 424.26ftlb +

    +     + +     + + + + +

    + How much work is performed in moving a box up the length of a 10ft ramp that makes a 5^\circ angle with the horizontal, + with 50lb of force applied in the direction of the ramp? +

    +     + +

    + 500ftlb +

    +     + +     +     +     +
    +
    + The Cross Product + +

    + Orthogonality is immensely important. + A quick scan of your current environment will undoubtedly reveal numerous surfaces and edges that are perpendicular to each other + (including the edges of this page). + The dot product provides a quick test for orthogonality: + vectors \vec u and \vec v are perpendicular if, + and only if, \dotp uv=0. +

    + + + +

    + Given two non-parallel, + nonzero vectors \vec u and \vec v in space, + it is very useful to find a vector \vec w that is perpendicular to both \vec u and \vec v. + There is an operation, called the cross product, + that creates such a vector. + This section defines the cross product, + then explores its properties and applications. +

    + + + Cross Product + +

    + Let \vec u =\la u_1,u_2,u_3\ra and + \vec v = \la v_1,v_2,v_3\ra be vectors in \mathbb{R}^3. + The cross product of \vec u and \vec v, + denoted \crossp uv, + is the vector vectorscross productcross productdefinition + + \crossp uv = \la u_2v_3-u_3v_2,-(u_1v_3-u_3v_1),u_1v_2-u_2v_1\ra + . +

    +     +     + +

    + This definition can be a bit cumbersome to remember. + After an example we will give a convenient method for computing the cross product. + For now, careful examination of the products and differences given in the definition should reveal a pattern that is not too difficult to remember. + (For instance, + in the first component only 2 and 3 appear as subscripts; + in the second component, only 1 and 3 appear as subscripts. + Further study reveals the order in which they appear.) +

    + +

    + Let's practice using this definition by computing a cross product. +

    + + + Computing a cross product + +

    + Let \vec u = \la 2,-1,4\ra and \vec v = \la 3,2,5\ra. + Find \crossp uv, + and verify that it is orthogonal to both \vec u and \vec v. +

    +     + +

    + Using , we have + + \crossp uv = \la (-1)5-(4)2,-\big((2)5-(4)3\big), (2)2-(-1)3\ra = \la -13,2,7\ra + . +

    + +

    + (We encourage the reader to compute this product on their own, + then verify their result.) +

    + +

    + We test whether or not \crossp uv is orthogonal to \vec u and \vec v using the dot product: + + \big(\crossp uv\big) \cdot \vec u = \la -13,2,7\ra \cdot \la 2,-1,4\ra = 0 + , + + \big(\crossp uv\big) \cdot \vec v = \la -13,2,7\ra \cdot \la 3,2,5 \ra = 0 + . +

    + +

    + Since both dot products are zero, + \crossp uv is indeed orthogonal to both \vec u and \vec v. +

    +     + +     + +

    + A convenient method of computing the cross product starts with forming a particular 3\times 3 + matrix, or rectangular array. + The first row comprises the standard unit vectors \veci, + \vecj, and \veck. + The second and third rows are the vectors \vec u and \vec v, + respectively. + Using \vec u and \vec v from , + we begin with: + + \begin{matrix} \veci\amp \vecj\amp \veck \\ 2\amp -1\amp 4\\3\amp 2\amp 5 + \end{matrix} + + +

    + +

    + Now repeat the first two columns after the original three: + + \begin{matrix} \veci\amp \vecj\amp \veck\amp \veci\amp \vecj \\ 2\amp -1\amp 4\amp 2\amp -1\\3\amp 2\amp 5\amp 3\amp 2 + \end{matrix} + + This gives three full upper left to lower right diagonals, + and three full upper right to lower left diagonals, + as shown. + Compute the products along each diagonal, + then add the products on the right and subtract the products on the left: +

    + + + Schematic diagram for computing the cross product by multiplying along diagonal arrows. + +

    + The image illustrates how to use the array given above, with the two repeated columns, + to compute the cross product. + From the top row, diagonal arrows are drawn down and to the right, + starting in the first three columns from \veci, \vecj, and vec k. + Diagonal arrows are also drawn down and to the left, + starting in the last three columns from \veck, \veci, and \vecj. + (Recall that the last two columns are copies of the first two.) +

    + +

    + The cross product is computed by multiplying along each arrow. + The quantities obtained from the right-pointing arrows are added, + while the quantities obtained from the left-pointing arrows are subtracted. + The result of the product along each arrow is indicated at the tip of that arrow. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.32] + + \node at (0,0) {\(\begin{array}{ccccc} \ \veci\ \amp \ \vecj\ \amp \ \veck\ \amp \ \veci\ \amp \ \vecj\ \\ 2\amp -1\amp 4\amp 2\amp -1\\3\amp 2\amp 5\amp 3\amp 2 + \end{array} \)}; + + \draw[->, thin] (-1.3,.4) -- (.5,-.8) node[below] {\(-5\veci\)}; + \draw[->, thin] (-.7,.4) -- (1.3,-.8) node[below] {\(12\vecj\)}; + \draw[->, thin] (0,.4) -- (2,-.8) node[below] {\(4\veck\)}; + \draw[->, thin] (0,.4) -- (-2,-.8) node[below] {\(-3\veck\)}; + \draw[->, thin] (.7,.4) -- (-1.3,-.8) node[below] {\(8\veci\)}; + \draw[->, thin] (1.3,.4) -- (-.5,-.8) node[below] {\(10\vecj\)}; + + \end{tikzpicture} + + + + +

    + + \crossp uv = \big(-5\veci+12\vecj+4\veck\,\big) - \big(-3\veck+8\veci+10\vecj\,\big) = -13\veci+2\vecj+7\veck = \la -13,2,7\ra + . +

    + + + +

    + We practice using this method. +

    + + + Computing a cross product + +

    + Let \vecu=\la 1,3,6\ra and \vec v = \la -1,2,1\ra. + Compute both \crossp uv and \crossp vu. +

    +     + +

    + To compute \crossp uv, + we form the matrix as prescribed above, + complete with repeated first columns: + + \begin{matrix} \ \veci\ \amp \ \vecj\ \amp \ \veck\ \amp \ \veci\ \amp \ \vecj\ \\ 1\amp 3\amp 6\amp 1\amp 3\\-1\amp 2\amp 1\amp -1\amp 2 \end{matrix} + +

    + +

    + We let the reader compute the products of the diagonals; + we give the result: + + \crossp uv = \big(3\veci-6\vecj+2\veck\,\big) - \big(-3\veck + 12\veci+\vecj\,\big) = \la -9,-7,5\ra + . +

    + +

    + To compute \crossp vu, + we switch the second and third rows of the above matrix, + then multiply along diagonals and subtract: + + \begin{matrix} \ \veci\ \amp \ \vecj\ \amp \ \veck\ \amp \ \veci\ \amp \ \vecj\ \\-1\amp 2\amp 1\amp -1\amp 2\\ 1\amp 3\amp 6\amp 1\amp 3 \end{matrix} + +

    + +

    + Note how with the rows being switched, + the products that once appeared on the right now appear on the left, + and vice-versa. + Thus the result is: + + \crossp vu = \big(12\veci+\vecj-3\veck\,\big) - \big(2\veck + 3\veci-6\vecj\,\big) = \la 9,7,-5\ra + , + which is the opposite of \crossp uv. + We leave it to the reader to verify that each of these vectors is orthogonal to \vec u and \vec v. +

    +     + +     +     + + + Properties of the Cross Product +

    + It is not coincidence that + \crossp vu = -(\crossp uv) in the preceding example; + one can show using + that this will always be the case. + The following theorem states several useful properties of the cross product, + each of which can be verified by referring to the definition. +

    + + + Properties of the Cross Product + +

    + Let \vecu, \vecv and \vecw be vectors in + \mathbb{R}^3 and let c be a scalar. + The following identities hold: + vectorscross product + cross productproperties +

    + +

    +

      +
    1. +

      + \crossp uv = -(\crossp vu) (Anticommutative Property) +

      +
      2. + +
    2. +

      +

        +
      1. +

        + (\vec u+\vec v)\times \vecw = \crossp uw+\crossp vw (Distributive Properties) +

        +
        2. + +
      2. +

        + \vec u \times (\vec v+\vec w) = \crossp uv+\crossp uw +

        +
        3. +
      +

      +
      3. + +
    3. +

      + c(\crossp uv) = (c\vecu) \times \vec v = \vecu \times (c\vecv) +

      +
      4. + +
    4. +

      +

        +
      1. +

        + (\crossp uv)\cdot \vecu = 0 (Orthogonality Properties) +

        +
        2. + +
      2. +

        + (\crossp uv)\cdot \vecv = 0 +

        +
        3. +
      +

      +
      5. + +
    5. +

      + \crossp uu = \vec 0 +

      +
      6. + +
    6. +

      + \crossp u0 = \vec 0 +

      +
      7. + +
    7. +

      + \vecu \cdot (\vecv\times\vecw) = (\crossp uv)\cdot \vecw (Triple Scalar Product) +

      +
      8. +
    +

    +     +     + + + +

    + We introduced the cross product as a way to find a vector orthogonal to two given vectors, + but we did not give a proof that the construction given in satisfies this property. + asserts this property holds; + we leave it as a problem in the Exercise section to verify this. +

    + + + +

    + Property 5 from the theorem is also left to the reader to prove in the Exercise section, + but it reveals something more interesting than + the cross product of a vector with itself is \vec 0. + Let \vec u and \vec v be parallel vectors; + that is, let there be a scalar c such that \vecv = c\vecu. + Consider their cross product: + + \crossp uv \amp = \vecu \times (c\vec u)\amp \amp + \amp = c(\crossp uu) \amp \amp\text{ (by Property 3 of } \text{)} + \amp = \vec 0 \amp \amp\text{ (by Property 5 of } \text{)} + . +

    + +

    + We have just shown that the cross product of parallel vectors is \vec 0. + This hints at something deeper. + + related the angle between two vectors and their dot product; + there is a similar relationship relating the cross product of two vectors and the angle between them, + given by the following theorem. +

    + + + The Cross Product and Angles + +

    + Let \vec u and \vec v be nonzero vectors in \mathbb{R}^3. + Then + + \norm{\crossp uv} = \vnorm u\, \vnorm v \sin(\theta) + , + where \theta, 0\leq \theta \leq \pi, + is the angle between \vecu and \vecv. + vectorscross product + cross productproperties +

    +     +     + + + +

    + Note that this theorem makes a statement about the + magnitude of the cross product. + When the angle between \vecu and \vecv is 0 or \pi (, the vectors are parallel), + the magnitude of the cross product is 0. + The only vector with a magnitude of 0 is \vec 0 + (see Property + of ), + hence the cross product of parallel vectors is \vec 0. +

    + + + + + +

    + We demonstrate the truth of this theorem in the following example. +

    + + + The cross product and angles + +

    + Let \vec u = \la 1,3,6\ra and + \vec v = \la -1,2,1\ra as in . + Verify by finding \theta, + the angle between \vecu and \vecv, + and the magnitude of \crossp uv. +

    +     + +

    + We use + to find the angle between \vecu and \vecv. + + \theta \amp = \cos^{-1}\left(\frac{\dotp uv}{\vnorm u\, \vnorm v}\right) + \amp = \cos^{-1}\left(\frac{11}{\sqrt{46}\sqrt{6}}\right) + \amp \approx 0.8471 = 48.54^\circ + . +

    + +

    + Our work in + showed that \crossp uv = \la -9,-7,5\ra, + hence \norm{\crossp uv} = \sqrt{155}. + Is \norm{\crossp uv} = \vnorm u\, \vnorm v\sin(\theta)? + Using numerical approximations, we find: + + \norm{\crossp uv} \amp =\sqrt{155} \amp \vnorm u\,\vnorm v \sin(\theta) \amp = \sqrt{46}\sqrt{6}\sin(0.8471) + \amp \approx 12.45. \amp \amp \approx 12.45 + . +

    + +

    + Numerically, they seem equal. + Using a right triangle, one can show that + + \sin\left(\cos^{-1}\left(\frac{11}{\sqrt{46}\sqrt{6}}\right)\right) = \frac{\sqrt{155}}{\sqrt{46}\sqrt{6}} + , + which allows us to verify the theorem exactly. +

    +     +     + + + Right Hand Rule +

    + The anticommutative property of the cross product demonstrates that \crossp uv and + \crossp vu differ only by a sign these vectors have the same magnitude but point in the opposite direction. + When seeking a vector perpendicular to \vec u and \vec v, + we essentially have two directions to choose from, + one in the direction of \crossp uv and one in the direction of \crossp vu. + Does it matter which we choose? + How can we tell which one we will get without graphing, etc.? +

    + +

    + Another wonderful property of the cross product, as defined, + is that it follows the right hand rule. + Given \vec u and \vec v in + \mathbb{R}^3 with the same initial point, + point the index finger of your right hand in the direction of \vecu and let your middle finger point in the direction of \vecv + (much as we did when establishing the right hand rule for the 3-dimensional coordinate system). + Your thumb will naturally extend in the direction of \crossp uv. + One can practice this using . + If you switch, + and point the index finder in the direction of \vecv and the middle finger in the direction of \vecu, + your thumb will now point in the opposite direction, + allowing you to visualize + the anticommutative property of the cross product. + right hand ruleof the cross product +

    + +
    + Illustrating the Right Hand Rule of the cross product + + + + Three-dimensional image illustrating the right-hand rule for the direction of the cross product. + +

    + On a three-dimensional coordinate system, with x, y, and z axes, + vectors \vec u and \vec v are plotted with their tails at the origin. + The plane passing through the origin that contains these two vectors is also plotted, + along with a normal vector to the plane. +

    + +

    + The normal vector is upward-pointing, and is equal to the cross product \vec{u}\times\vec{v}. + When the plane is viewed from the side on which the normal vector is placed, + the position of the vector \vec v corresponds to a counter-clockwise rotation from the position of \vec u. +

    +     + + + + + //ASY file for figcrossp_rhr.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-0.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw theplane z=(-2*x+5*y)/27; + triple f(pair t) { + return (t.x,t.y,(-2*t.x+5*t.y)/27); + } + surface s=surface(f,(-1.5,-1.5),(1.5,1.5),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + // Draw vec{u}=<-1/5,1,1/5> + draw((0,0,0)--(-1/5,1,1/5), bluepen,Arrow3(size=3mm)); + label("$\vec{u}$",(-1/5,1,1/5),N); + + // Draw \vec{v}=<-6/5,3/5,1/5> + draw((0,0,0)--(-6/5,3/5,1/5), bluepen,Arrow3(size=3mm)); + label("$\vec{v}$",(-6/5,3/5,1/5),N); + + // Draw the cross product of u and v <2,-5,27> but scale it as <2,-5,27>/25 + draw((0,0,0)--(2/25,-5/25,27/25), redpen,Arrow3(size=3mm)); + label("$\vec{u} \times \vec{v}$",(2/25,-5/25,27/25),NW); + + + + +
    +     +     + + + Applications of the Cross Product +

    + There are a number of ways in which the cross product is useful in mathematics, + physics and other areas of science beyond just + finding a vector perpendicular to two others. + We highlight a few here. + cross productapplications +

    + + + Area of a Parallelogram +

    + It is a standard geometry fact that the area of a parallelogram is A = bh, + where b is the length of the base and h is the height of the parallelogram, + as illustrated in . + As shown when defining the Parallelogram Law of vector addition, + two vectors \vecu and \vecv define a parallelogram when drawn from the same initial point, + as illustrated in . + Trigonometry tells us that h = \vnorm u \sin(\theta), + hence the area of the parallelogram is + + A = \vnorm u\,\vnorm v\sin(\theta) = \norm{\crossp uv} + , + where the second equality comes from . +

    + +
    + Using the cross product to find the area of a parallelogram + +
    + + + + A parallelogram with base b and height h labeled. + +

    + A parallelogram with a horizontal base is shown. + The base is labeled with the quantity b. + A dashed line is drawn from the top-left vertex of the parallelogram to the base; + this line is perpendicular to the base, + and its length is equal to the height of the parallelogram, + which is labeled h. +

    +     + + + \begin{tikzpicture}[scale=1.24] + + \draw (0,0) -- node [below,pos=.5] { $b$} (2,0) -- (3,1.5) -- (1,1.5) -- (0,0); + \draw [dashed] (1,1.5) -- (1,0) node [pos=.5,right] {$h$}; + \draw (.8,0) -- (.8,.2) -- (1,.2); + + \end{tikzpicture} + + + + +
    + +
    + + + + Another parallelogram, with two adjacent sides labeled as vectors. + +

    + The parallelogram is the same one as , + but with different labeling. + This time, the two sides adjacent to the bottom-right vertex are labeled with vectors. + The bottom of the parallelogram is labeled with a vector \vec v, + and the left side is labeled with a vector \vec u. + The angle between these vectors is labeled \theta. + The same altitude of the parellelogram is drawn as a dashed line with height h. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.24] + + \draw [->](0,0) -- node [below,pos=1] { $\vec v$} (2,0); + \draw (2,0) -- (3,1.5) -- (1,1.5); + \draw (.3,.175) node { $\theta$}; + \draw [->](0,0) -- (1,1.5) node [above] { $\vec u$}; + \draw [dashed] (1,1.5) -- (1,0) node [pos=.5,right] {$h$}; + \draw (.8,0) -- (.8,.2) -- (1,.2); + + \end{tikzpicture} + + + + +
    +     +
    + +

    + We illustrate using Equation in the following example. + cross productapplications!area of parallelogram +

    + + + Finding the area of a parallelogram + +

    +

      +
    1. +

      + Find the area of the parallelogram defined by the vectors + \vecu = \la 2,1\ra and \vecv = \la 1,3\ra. +

      +
      2. + +
    2. +

      + Verify that the points A = (1,1,1), B = (2,3,2), + C = (4,5,3) and D = (3,3,2) are the vertices of a parallelogram. + Find the area of the parallelogram. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + sketches the parallelogram defined by the vectors \vec u and \vec v. + We have a slight problem in that our vectors exist in + \mathbb{R}^2, not \mathbb{R}^3, + and the cross product is only defined on vectors in \mathbb{R}^3. + We skirt this issue by viewing \vec u and \vecv as vectors in the x-y plane of \mathbb{R}^3, + and rewrite them as \vec u = \la 2,1,0\ra and \vecv =\la 1,3,0\ra. + We can now compute the cross product. + It is easy to show that \crossp uv = \la 0,0,5\ra; therefore the area of the parallelogram is A = \norm{\crossp uv} = 5. +

      + +
      + Sketching the parallelograms in + +
      + + + + A parallelogram in the plane, spanned by vectors u and v. + +

      + The first quadrant in \R^2 is shown. + Two vectors a drawn with their tails at the origin. + The vector \vec u ends at the point (2,1), + while the vector \vec v ends at the point (1,3). +

      + +

      + These vectors make up two of the four sides of the parallelogram. + The side from (1,3) to (3,4) is parallel to \vec u, + and the side from (2,1) to (3,4) is parallel to \vec v. +

      +       + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4,5}, + ymin=-.5,ymax=5.5, + xmin=-.5,xmax=4.5 + ] + + \draw [thick,->] (axis cs:0,0) -- (axis cs:2,1) node [right,] { $\vec u$}; + \draw [thick,->] (axis cs:0,0) -- (axis cs:1,3) node [above] { $\vec v$}; + + \draw [thick] (axis cs:2,1) -- (axis cs:3,4) -- (axis cs: 1,3); + + \end{axis} + + \end{tikzpicture} + + + + +
      + +
      + + + + + A three-dimensional image of a parallelogram in space, with vertices A, B, C, and D. + +

      + A three-dimensional coordinate system is shown, with x, y, and z axes. + A parallelogram is shown in space relative to these axes, + with vertices labeled A, B, C, and D. +

      +       + + + + + //ASY file for figcrossp4a.asy in Chapter 10 + // CAN'T FILL IT IN WITH ''fill'' (only in 2d) + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + //currentprojection=orthographic(4,4,2); + currentprojection=orthographic(7.5,14,4); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-0.5,4.5); + pair ybounds=(-0.5,4.5); + pair zbounds=(-0.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the edges of the parallelogram with corners at (1,1,1),(2,3,2),(4,5,3),(3,3,2) + draw((1,1,1)--(2,3,2)--(4,5,3)--(3,3,2)--cycle); + label("A",(1,1,1),S); + label("B",(2,3,2),N); + label("C",(4,5,3),N); + label("D",(3,3,2),W); + + // Fill in the parallelogram + //I changed the viewpoint to (1,4,1), then added://Hartman comment + triple f(pair t) {return (1+t.x+2t.y,1+2t.x+2t.y,1+t.x+t.y);} + surface s=surface(f,(0,0),(1,1),1,1); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +
      +       +
      +
      2. + +
    2. +

      + To show that the quadrilateral ABCD is a parallelogram (shown in ), + we need to show that the opposite sides are parallel. + We can quickly show that \overrightarrow{AB} =\overrightarrow{DC} = \la 1,2,1\ra and \overrightarrow{BC} = \overrightarrow{AD} = \la 2,2,1\ra. + We find the area by computing the magnitude of the cross product of \overrightarrow{AB} and \overrightarrow{BC}: + + \overrightarrow{AB} \times \overrightarrow{BC} = \la 0,1,-2\ra \Rightarrow \norm{\overrightarrow{AB}\times\overrightarrow{BC}} = \sqrt{5} \approx 2.236 + . +

      +
      3. +
    +

    +     + +     + +

    + This application is perhaps more useful in finding the area of a triangle + (in short, triangles are used more often than parallelograms). + We illustrate this in the following example. +

    + + + Area of a triangle + +

    + Find the area of the triangle with vertices A=(1,2), + B=(2,3) and C=(3,1), + as pictured in . +

    +
    + Finding the area of a triangle in + + + A triangle in the plane, with vertices A(1,2), B(2,3), and C(3,1). + +

    + The triangle for this problem is sketched in the first quadrant of the plane. + The vertices are labeled, with A at (1,2), + B at (2,3), and C at (3,1). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3}, + ymin=-.1,ymax=3.5, + xmin=-.1,xmax=3.9 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(1,2) (2,3) (3,1) (1,2)}; + \addplot [firstcurvestyle,-] coordinates {(1,2) (2,3) (3,1) (1,2)}; + + \draw (axis cs:1,2) node [left] { $A$} (axis cs:2,3) node [above] { $B$} (axis cs:3,1) node [below] { $C$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + We found the area of this triangle in + to be 1.5 using integration. + There we discussed the fact that finding the area of a triangle can be inconvenient using the + \frac12bh + formula as one has to compute the height, + which generally involves finding angles, etc. + Using a cross product is much more direct. +

    + +

    + We can choose any two sides of the triangle to use to form vectors; + we choose \overrightarrow{AB} = \la 1,1\ra and \overrightarrow{AC}=\la 2,-1\ra. + As in the previous example, + we will rewrite these vectors with a third component of 0 so that we can apply the cross product. + The area of the triangle is + + \frac12\norm{\overrightarrow{AB}\times\overrightarrow{AC}} = \frac12\norm{\la 1,1,0\ra \times \la 2,-1,0\ra} = \frac12\norm{\la 0,0,-3\ra} = \frac32 + . +

    + +

    + We arrive at the same answer as before with less work. +

    +     + +     + + +     + + + Volume of a Parallelepiped +

    + The three dimensional analogue to the parallelogram is the + parallelepiped. + Each face is parallel to the opposite face, + as illustrated in . + By crossing \vec v and \vec w, + one gets a vector whose magnitude is the area of the base. + Dotting this vector with \vecu computes the volume + of the parallelepiped! (Up to a sign; + take the absolute value.) +

    + +
    + A parallelepiped is the three dimensional analogue to the parallelogram + + + + A parellelepiped in three dimensions. It is the box-like object spanned by three vectors. + +

    + Three vectors \vec u, \vec v, and \vec w are plotted in three dimensions. + These vectors make up three edges of a three-dimensional solid known as a parallelepiped. + Each side of the parallelepiped is a parallelogram, and opposite sides are equal parallelograms lying in parallel planes. + One pair of opposite sides comes from the parallelogram spanned by \vec u and \vec v, + another from the parallelogram spanned by \vec v and \vec w, + and the last pair from the parallelogram spanned by \vec u and \vec w. +

    +     + + + + + //ASY file for figcrosspparallelpiped.asy in Chapter 10 + // NOT SHADED!! + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + //real[] myxchoice={}; + //real[] myychoice={}; + //real[] myzchoice={}; + defaultpen(0.5mm); + //xaxis3("$x$",-0.5,1,black,OutTicks(myxchoice),Arrow3(size=3mm)); + //yaxis3("$y$",-1,1,black,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("$z$",-0.5,1.5,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //Draw the parallelepiped with u=<1,1,0>, v=<-1,1,0>, w=<0,1,1> + draw((1,1,0)--(0,0,0), Arrow3(size=3mm));// u + label("$\vec{u}$",(0,0,0),W); + draw((1,1,0)--(0,2,0), Arrow3(size=3mm));// v + label("$\vec{v}$",(0,2,0),S); + draw((1,1,0)--(1,2,1), Arrow3(size=3mm));// w + label("$\vec{w}$",(1,2,1),W); + //shifted u to get the other edges of the box + draw((-1,1,0)--(0,2,0),bluepen);// u shifted to v + draw((0,1,1)--(1,2,1),bluepen);// u shifted to w + draw((-1,2,1)--(0,3,1),bluepen);// u shifted to v+w + //shifted v to get the other edges of the box + draw((0,0,0)--(-1,1,0),bluepen);// v shifted to u + draw((0,1,1)--(-1,2,1),bluepen);// v shifted to w + draw((1,2,1)--(0,3,1),bluepen);// v shifted to u+w + //shifted w to get the other edges of the box + draw((0,0,0)--(0,1,1),bluepen);// w shifted to u + draw((-1,1,0)--(-1,2,1),bluepen);// w shifted to v + draw((0,2,0)--(0,3,1),bluepen);// w shifted to u+v + + // MIGHT BE ABLE TO USE THIS TO FILL IN THE 6 SIDES + // Fill in the parallelogram + //I changed the viewpoint to (1,4,1), then added://Hartman comment + //triple f(pair t) {return (1+t.x+2t.y,1+2t.x+2t.y,1+t.x+t.y);} + //surface s=surface(f,(0,0),(1,1),1,1); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + //OR THIS FROM THE WEB???? + //http://tex.stackexchange.com/questions/137153/fill-a-region-between-two-coplanar-paths-in-asymptote + //settings.outformat="pdf"; + //settings.render=0; + //import three; + //size(5cm); + //path3 p = (0,-2,-2)-- (0,2,-2) -- (0,2,2) -- (0,-2,2) -- (0,-2,-2) -- (0,-2,-2); + //path3 q = (0,-.25,-1.1) -- (0,.25,-1.1) -- (0,.25,1.1) -- (0,-.25,1.1) -- (0,-.25,-1.1); + //draw(surface(p -- reverse(q) -- cycle), emissive(yellow)); + //draw(p ^^ q, black); + + //my try + import three; + path3 p = (0,0,0)-- (1,1,0) -- (0,2,0) -- (-1,1,0); //Left + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0,0,0)-- (1,1,0) -- (1,2,1) -- (0,1,1); //bottom + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (-1,1,0)-- (0,2,0) -- (0,3,1) -- (-1,2,1); //right + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0,0,0)-- (-1,1,0) -- (-1,2,1) -- (0,1,1); //back + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (1,1,0)-- (0,2,0) -- (0,3,1) -- (1,2,1); //front + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0,1,1)-- (1,2,1) -- (0,3,1) -- (-1,2,1); //top + draw(surface(p -- cycle), simplesurfacepen); + + + + +
    + +

    + cross productapplications!volume of parallelepiped +

    + +

    + Thus the volume of a parallelepiped defined by vectors \vecu, + \vecv and \vec w is + + V = \abs{\vecu\cdot (\crossp vw)} + . +

    + +

    + Note how this is the Triple Scalar Product, + first seen in . + Applying the identities given in the theorem shows that we can apply the Triple Scalar Product in any order + we choose to find the volume. + That is, + + V = \abs{\vecu\cdot(\crossp vw)} = \abs{\vec u\cdot (\crossp wv)} = \abs{(\crossp uv)\cdot \vecw}, \text{ etc. } + +

    + + + Finding the volume of parallelepiped + +

    + Find the volume of the parallelepiped defined by the vectors \vecu = \la 1,1,0\ra, + \vecv = \la -1,1,0\ra and \vecw = \la 0,1,1\ra. +

    +     + +

    + We apply Equation. + We first find \crossp vw =\la 1,1,-1\ra. + Then + + \abs{\vec u\cdot(\crossp vw)} = \abs{\la 1,1,0\ra \cdot \la1,1,-1\ra} = 2 + . +

    + +

    + So the volume of the parallelepiped is 2 cubic units. +

    + +
    + A parallelepiped in + + + + The parallelepiped whose volume is computed in this example. + +

    + A parallelpiped is plotted against a three-dimensional coordinate system, + with x, y, and z axes. + Vectors \vec{u}, \vec{v}, \vec{w} are plotted with their tails at the origin, + although the vector \vec{v} is not visible in the default view of the parallelepiped. + (It can be seen once the image is rotated.) +

    + +

    + These vectors make up the three edges of the parallelepiped that are adjacent to the origin. +

    +     + + + + + //ASY file for figcrossp6.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={2}; + real[] myzchoice={0.5,1}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(0,3); + pair zbounds=(-0.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the parallelepiped with u=<1,1,0>, v=<-1,1,0>, w=<0,1,1> + draw((0,0,0)--(1,1,0), Arrow3(size=3mm));// u + label("$\vec{u}$",(1,1,0),W); + draw((0,0,0)--(-1,1,0), Arrow3(size=3mm));// v + label("$\vec{v}$",(-1,1,0),N); + draw((0,0,0)--(0,1,1), Arrow3(size=3mm));// w + label("$\vec{w}$",(0,1,1),W); + //shifted u to get the other edges of the box + draw((-1,1,0)--(0,2,0),bluepen);// u shifted to v + draw((0,1,1)--(1,2,1),bluepen);// u shifted to w + draw((-1,2,1)--(0,3,1),bluepen);// u shifted to v+w + //shifted v to get the other edges of the box + draw((1,1,0)--(0,2,0),bluepen);// v shifted to u + draw((0,1,1)--(-1,2,1),bluepen);// v shifted to w + draw((1,2,1)--(0,3,1),bluepen);// v shifted to u+w + //shifted w to get the other edges of the box + draw((1,1,0)--(1,2,1),bluepen);// w shifted to u + draw((-1,1,0)--(-1,2,1),bluepen);// w shifted to v + draw((0,2,0)--(0,3,1),bluepen);// w shifted to u+v + + //my try at shading. + import three; + path3 p = (0,0,0)-- (1,1,0) -- (0,2,0) -- (-1,1,0); //Left + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0,0,0)-- (1,1,0) -- (1,2,1) -- (0,1,1); //bottom + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (-1,1,0)-- (0,2,0) -- (0,3,1) -- (-1,2,1); //right + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0,0,0)-- (-1,1,0) -- (-1,2,1) -- (0,1,1); //back + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (1,1,0)-- (0,2,0) -- (0,3,1) -- (1,2,1); //front + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0,1,1)-- (1,2,1) -- (0,3,1) -- (-1,2,1); //top + draw(surface(p -- cycle), simplesurfacepen); + + + + +
    +     + +     +     + +

    + While this application of the Triple Scalar Product is interesting, + it is not used all that often: + parallelepipeds are not a common shape in physics and engineering. + The last application of the cross product is very applicable in engineering. +

    + + + Torque +

    + Torque is a measure of the turning force applied to an object. + A classic scenario involving torque is the application of a wrench to a bolt. + When a force is applied to the wrench, the bolt turns. + When we represent the force and wrench with vectors \vec F and \vec \ell, + we see that the bolt moves + (because of the threads) + in a direction orthogonal to \vec F and \vec \ell. + Torque is usually represented by the Greek letter \tau, or tau, + and has units of N\cdotm, + a Newtonmeter, or ft\cdotlb, a footpound. + cross productapplications!torque + torque +

    + +

    + While a full understanding of torque is beyond the purposes of this book, + when a force \vec F is applied to a lever arm + \vec \ell, the resulting torque is + + \vec \tau = \crossp \ell F + . +

    + + + Computing torque + +

    + A lever of length 2ft makes an angle with the horizontal of 45^\circ. + Find the resulting torque when a force of 10lb is applied to the end of the level where: +

    + +
    + Showing a force being applied to a lever in + + + Two adjacent images, each showing a pair of vectors, one of which represents a lever, and the other, a force. + +

    + Two images are shown side-by-side. + Both images show a pair of vectors, one of which is labeled \vec l (the lever), + and the other, which is labeled \vec F (the force). +

    + +

    + In each image, the vector \vec l has its tail at a pivot point. + A circular arrow is drawn around this point to show the direction of rotation of the lever about the pivot. + The direction is clockwise in each image. +

    + +

    + The force vector \vec F is drawn with its tip at the tip of the vector \vec l. + In both images, the vector \vec l points up and to the right, + and the vector \vec F points down and to the right. + The difference is that in the first image, the vectors meet at a 90^\circ angle, + while in the second image, the angle is 60^\circ. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.1] + + \draw [thick,->,rotate=45] (-2,0) -- (0,0) node [below,pos=.5] { $\vec\ell$}; + \draw [rotate=45,->] (-1.8,0) arc (0:-270:.2); + \draw [rotate=45] (-.4,0) arc (180:90:.4); + \draw [rotate=0] (-.6,0) node { $90^\circ$}; + + \filldraw [black,rotate=45] (-2,0) circle (2.4pt); + + \begin{scope}[shift={(-.05,.05)}] + \draw [thick,->,rotate=-45] (-1.5,0)node [left] { $\vec F$} -- (0,0); + \end{scope} + + \begin{scope}[shift={(3,0)}] + + \draw [thick,->,rotate=45] (-2,0) -- (0,0) node [below,pos=.5] { $\vec\ell$}; + \draw [rotate=45,->] (-1.8,0) arc (0:-270:.2); + \draw [rotate=45] (-.4,0) arc (180:120:.4); + \draw [rotate=15] (-.6,0) node { $60^\circ$}; + + \filldraw [black,rotate=45] (-2,0) circle (2.4pt); + + \begin{scope}[shift={(-.05,.05)}] + \draw [thick,->,rotate=-15] (-1.5,0) node [above] { $\vec F$} -- (0,0); + \end{scope} + + \end{scope} + + \end{tikzpicture} + + + + +
    + +

    +

      +
    1. +

      + the force is perpendicular to the lever, and +

      +
      2. + +
    2. +

      + the force makes an angle of 60^\circ with the lever, + as shown in . +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + We start by determining vectors for the force and lever arm. + Since the lever arm makes a + 45^\circ angle with the horizontal and is 2ft long, + we can state that \vec \ell = 2\la \cos(45^\circ) ,\sin(45^\circ) \ra = \la \sqrt2,\sqrt2\ra. + + Since the force vector is perpendicular to the lever arm (as seen in the left hand side of ), + we can conclude it is making an angle of -45^\circ with the horizontal. + As it has a magnitude of 10lb, we can state \vec F = 10\la \cos(-45^\circ), \sin(-45^\circ)\ra = \la 5\sqrt2,-5\sqrt2\ra. + + Using Equation to find the torque requires a cross product. We again let the third component of each vector be 0 and compute the cross product: + + \vec\tau \amp = \crossp \ell F + \amp = \la \sqrt2,\sqrt2,0\ra \times \la 5\sqrt2,-5\sqrt2,0\ra + \amp = \la 0,0,-20\ra + + This clearly has a magnitude of 20 ft-lb. + We can view the force and lever arm vectors as lying on the page; + our computation of \vec\tau shows that the torque goes into the page. + This follows the Right Hand Rule of the cross product, + and it also matches well with the example of the wrench turning the bolt. + Turning a bolt clockwise moves it in. +

      +
      2. + +
    2. +

      + Our lever arm can still be represented by \vec \ell = \la \sqrt2,\sqrt2\ra. + As our force vector makes a + 60^\circ angle with \vec \ell, we can see + (referencing the right hand side of the figure) + that \vec F makes a + -15^\circ angle with the horizontal. + Thus + + \vec F = 10\la \cos-15^\circ,\sin-15^\circ\ra \amp = \la \frac{5(1+\sqrt3)}{\sqrt2},\frac{5(-1+\sqrt3)}{\sqrt2}\ra + \amp \approx \la 9.659,-2.588\ra + . + We again make the third component 0 and take the cross product to find the torque: + + \vec\tau \amp = \crossp \ell F + \amp = \la \sqrt2,\sqrt2,0\ra \times \la \frac{5(1+\sqrt3)}{\sqrt2},\frac{5(-1+\sqrt3)}{\sqrt2},0\ra + \amp = \la 0,0,-10\sqrt3\ra + \amp \approx \la 0,0,-17.321\ra + . + As one might expect, when the force and lever arm vectors + are orthogonal, + the magnitude of force is greater than when the vectors + are not orthogonal. +

      +
      3. +
    +

    +     +     +     + +

    + While the cross product has a variety of applications + (as noted in this chapter), + its fundamental use is finding a vector perpendicular to two others. + Knowing a vector is orthogonal to two others is of incredible importance, + as it allows us to find the equations of lines and planes in a variety of contexts. + The importance of the cross product, + in some sense, relies on the importance of lines and planes, + which see widespread use throughout engineering, physics and mathematics. + We study lines and planes in the next two sections. +

    +     + + + + Terms and Concepts + + + + +

    + The cross product of two vectors is a , not a scalar. +

    +     + + + + + + + + +     + + + + +

    + One can visualize the direction of \vec u\times\vec v using the rule. +

    +     + + + + right hand|right-hand + + + + +     + + + + +

    + Give a synonym for orthogonal. +

    +     + + + +

    + Perpendicular is one answer, + although we tend to use perpendicular to refer to lines, + and orthogonal to refer to vectors. +

    +

    + Note that the zero vector is orthogonal to any vector, + but it doesn't really make sense to say it is perpendicular. +

    +     + +     + + + + + +

    + A fundamental principle of the cross product is that + \vec u\times\vec v is orthogonal to \vec u and \vec v. + +

    +     + +     + + + + +

    + is a measure of the turning force applied to an object. +

    +     + + + + + + + + + +     + + + +

    + If \vec u and \vec v are parallel, + then \vec u\times\vec v=\vec 0. +

    +     +     + +     + + Problems + + + +

    + Vectors \vec u and \vec v are given. + Compute \vec u\times\vec v and check that this vector is orthogonal to both \vec u and \vec v. +

    +     + + + + + Context("Vector"); + $cp=Vector("<12, -15, 3>"); + + + +

    + Let \vec u = \la 3,2,-2\ra, + \vec v = \la 0,1,5\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + $cp=Vector("<11, 1, -17>"); + + + +

    + Let \vec u = \la 5, -4, 3\ra, + \vec v = \la 2, -5, 1\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + $cp=Vector("<-5, -31, 27>"); + + + +

    + Let \vec u = \la 4, -5, -5\ra, + \vec v = \la 3, 3, 4\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + $cp=Vector("<47, -36, -44>"); + + + +

    + Let \vec u = \la -4, 7, -10\ra, + \vec v = \la 4, 4, 1\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + $cp=Vector("<0, -2, 0>"); + + + +

    + Let \vec u = \la 1, 0, 1\ra, + \vec v = \la 5, 0, 7\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + $cp=Vector("<0, 0, 0>"); + + + +

    + Let \vec u = \la 1, 5, -4\ra, + \vec v = \la -2, -10, 8\ra. +

    + +

    + +

    +     +     +     + + + +

    + \vec u = \langle a,b,0\rangle, + \vec v = \langle c,d,0\rangle +

    +     + +

    + \vec u\times \vec v = \langle 0,0,ad-bc\rangle +

    +     +     + + + + + + Context("Vector"); + Context()->flags->set(ijk=>1); + $cp=Vector("k"); + + + +

    + \vec u = \hat\imath, + \vec v = \hat\jmath. +

    + +

    + +

    + +

    + Check this is orthogonal to both \vec u and \vec v. +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(ijk=>1); + $cp=Vector("-j"); + + + +

    + \vec u = \hat\imath, \vec v = \hat{k}. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(ijk=>1); + $cp=Vector("i"); + + + +

    + \vec u = \hat\jmath, \vec v = \hat{k}. +

    + +

    + \vec u\times\vec v= +

    +     +     +     + +     + + + + +

    + Pick any vectors \vec u, + \vec v and \vec w in + \mathbb{R}^3 and show that \vec u \times (\vec v+\vec w) = \vec u\times \vec v+\vec u\times \vec w. +

    +     + +

    + Answers will vary. +

    +     + +     + + + + +

    + Pick any vectors \vec u, + \vec v and \vec w in + \mathbb{R}^3 and show that \vec u \cdot (\vec v\times\vec w) = (\vec u\times \vec v)\cdot \vec w. +

    +     + +

    + Answers will vary. +

    +     + +     + + + + +

    + The magnitudes of vectors \vec u and \vec v in \mathbb{R}^3 are given, + along with the angle \theta between them. + Use this information to find the magnitude of \vec u\times\vec v. +

    +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $normucv=Formula("5"); + + + +

    + If \vnorm{u} = 2, \vnorm{v} = 5, + and \theta = 30^{\circ} is the angle between \vec u and \vec v, + then \norm{\vec u\times\vec v}= +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $normucv=Formula("21"); + + + +

    + If \vnorm{u} = 3, \vnorm{v} = 7, + and \theta = \pi/2 is the angle between \vec u and \vec v, + then \norm{\vec u\times\vec v}= +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $normucv=Formula("0"); + + + +

    + If \vnorm{u} = 3, \vnorm{v} = 4, + and \theta = \pi is the angle between \vec u and \vec v, + then \norm{\vec u\times\vec v}= +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $normucv=Formula("5"); + + + +

    + If \vnorm{u} = 2, \vnorm{v} = 5, + and \theta = 5\pi/6 is the angle between \vec u and \vec v, + then \norm{\vec u\times\vec v}= +

    +     +     +     + +     + + + +

    + Find the area of the parallelogram defined by the given vectors. +

    +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $area=Formula("sqrt(14)"); + + + +

    + Find the area of the parallelogram defined by \vec u = \la 1,1,2\ra, + and \vec v = \la 2,0,3\ra. +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $area=Formula("sqrt(230)"); + + + +

    + Find the area of the parallelogram defined by \vec u = \la -2,1,5\ra, + and \vec v = \la -1,3,1\ra. +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $area=Formula("3"); + + + +

    + Find the area of the parallelogram defined by \vec u = \la 1,2\ra, + and \vec v = \la 2,1\ra. +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $area=Formula("6"); + + + +

    + Find the area of the parallelogram defined by \vec u = \la 2,0\ra, + and \vec v = \la 0,3\ra. +

    + +

    + +

    +     +     +     + +     + + + +

    + Find the area of the triangle with the given vertices. +

    +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $area=Formula("5sqrt(2)/2"); + + + +

    + Find the area of the triangle with vertices (0,0,0), + (1,3,-1) and (2,1,1). +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $area=Formula("3sqrt(30)"); + + + +

    + Find the area of the triangle with vertices (5,2,-1), + (3,6,2) and (1,0,4). +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $area=Formula("1"); + + + +

    + Find the area of the triangle with vertices (1,1), + (1,3) and (2,2). +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $area=Formula("5/2"); + + + +

    + Find the area of the triangle with vertices (3,1), + (1,2) and (4,3). +

    + +

    + +

    +     +     +     + +     + + + +

    + Find the area of the quadrilateral with the given vertices. + (Hint: break the quadrilateral into two triangles.) +

    +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $area=Formula("7"); + + + +

    + Find the area of the quadrilateral with vertices (0,0), + (1,2), + (3,0), and (4,3). +

    + +

    + +

    +     + +

    + Break the quadrilateral into two triangles. +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $area=Formula("4sqrt(14)"); + + + +

    + Find the area of the quadrilateral with vertices (0,0,0), + (2,1,1), (-1,2,-8), and (1,-1,5). +

    + +

    + +

    +     + +

    + Break the quadrilateral into two triangles. +

    +     +     +     + +     + + + +

    + Find the volume of the parallelepiped defined by the given vectors. +

    +     + + + + + $volume=Compute("2"); + + + +

    + Find the volume of the parallelepiped defined by \vec u = \la 1,1,1\ra, + \vec v=\la 1,2,3\ra, and \vec w = \la 1,0,1\ra. +

    + +

    + +

    +     +     +     + + + + + $volume=Compute("15"); + + + +

    + Find the volume of the parallelepiped defined by \vec u = \la -1,2,1\ra, + \vec v=\la 2,2,1\ra, and \vec w = \la 3,1,3\ra. +

    + +

    + +

    +     +     +     + +     + + + +

    + Find a unit vector orthogonal to both \vec u and \vec v. +

    +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $w=Compute("<1/sqrt(6),1/sqrt(6),-2/sqrt(6)>"); + $w=OneOf("$w,-$w"); + + + +

    + Find a unit vector orthogonal to both \vec u = \la 1,1,1\ra, + and \vec v=\la 2,0,1\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $w=Compute("<-2/sqrt(21),1/sqrt(21),4/sqrt(21)>"); + $w=OneOf("$w,-$w"); + + + +

    + Find a unit vector orthogonal to both \vec u = \la 1,-2,1\ra, + and \vec v=\la 3,2,1\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $w=Compute("<0,1,0>"); + $w=OneOf("$w,-$w"); + + + +

    + Find a unit vector orthogonal to both \vec u = \la 5,0,2\ra, + and \vec v=\la -3,0,7\ra. +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); + $u=Compute("<1,-2,1>"); + $w=Compute("<2/sqrt(5),1/sqrt(5),0>"); + $ev=$w->cmp(checker => sub { + my ($correct,$student,$self) = @_; + return 1 if (($student . $student == 1) and ($student . $u == 0)); + }); + + + +

    + Find a unit vector orthogonal to both \vec u = \la 1,-2,1\ra, + and \vec v=\la -2,4,-2\ra. +

    + +

    + +

    +     + +

    + Since \vec u and \vec v are parallel, + any unit vector orthogonal to \vec u works + (such as \frac{1}{\sqrt{2}}\la 1,0,-1\ra). +

    +     +     +     + +     + + + + +

    + A bicycle rider applies 150lb of force, straight down, + onto a pedal that extends 7in horizontally from the crankshaft. + Find the magnitude of the torque applied to the crankshaft. +

    +     + +

    + 87.5ftlb +

    +     + +     + + + + +

    + A bicycle rider applies 150lb of force, + straight down, onto a pedal that extends 7in from the crankshaft, + making a 30^\circ angle with the horizontal. + Find the magnitude of the torque applied to the crankshaft. +

    +     + +

    + 43.75\sqrt{3}\approx 75.78ftlb +

    +     + +     + + + + +

    + To turn a stubborn bolt, 80lb of force is applied to a 10in wrench. + What is the maximum amount of torque that can be applied to the bolt? +

    +     + +

    + 200/3\approx 66.67ftlb +

    +     + +     + + + + +

    + To turn a stubborn bolt, 80lb of force is applied to a 10in wrench in a confined space, + where the direction of applied force makes a + 10^\circ angle with the wrench. + How much torque is subsequently applied to the wrench? +

    +     + +

    + 11.58ftlb +

    +     + +     + + + + +

    + Show, using the definition of the Cross Product, + that \vec u\cdot(\vec u\times\vec v)=0; + that is, that \vec u is orthogonal to the cross product of \vec u and \vec v. +

    +     + +

    + With \vec u = \la u_1,u_2,u_3\ra and \vec v = \la v_1,v_2,v_3\ra, we have + + \vec u\cdot(\vec u\times\vec v) \amp = \la u_1,u_2,u_3\ra\cdot (\la u_2v_3-u_3v_2,-(u_1v_3-u_3v_1),u_1v_2-u_2v_1\ra) + \amp = u_1(u_2v_3-u_3v_2) - u2(u_1v_3-u_3v_1)+u3(u_1v_2-u_2v_1) + \amp =0 + . +

    +     + +     + + + + +

    + Show, using the definition of the Cross Product, + that \vec u\times\vec u=\vec 0. +

    +     + +

    + With \vec u = \la u_1,u_2,u_3\ra, we have + + \vec u\times\vec u \amp = \la u_2u_3-u_3u_2,-(u_1u_3-u_3u_1),u_1u_2-u_2u_1\ra) + \amp = \la 0,0,0\ra + \amp =\vec 0 + . +

    +     + +     +     +     +
    +
    + Lines + +

    + lines + To find the equation of a line in the xy-plane, we need two pieces of information: a point and the slope. + The slope conveys direction information. + As vertical lines have an undefined slope, + the following statement is more accurate: +

    + +

    + To define a line, one needs a point on the line and the direction of the line. +

    + +

    + This holds true for lines in space. +

    + + +     + + + Lines in space +

    + Let P be a point in space, + let \vec p be the vector with initial point at the origin and terminal point at P (, \vec p + points to P), + and let \vec d be a vector. + Consider the points on the line through P in the direction of \vec d. +

    + +

    + Clearly one point on the line is P; + we can say that the vector + \vec p lies at this point on the line. + To find another point on the line, + we can start at \vec p and move in a direction parallel to \vec d. + For instance, + starting at \vec p and traveling one length of \vec d places one at another point on the line. + Consider + where certain points along the line are indicated. +

    + +
    + Defining a line in space + + + + In three dimensions, vectors p (position), d (direction), and p+d are plotted. A line passes through the tips of p and p+d. + +

    + In a three-dimensional coordinate system, vectors \vec d and \vec p are plotted with their tails at the origin, + along with their sum, \vec{p}+\vec{d}. + The vector \vec{p} is the position vector for a line, which is also shown. + The line passes through the tips of \vec{p} and \vec{p}+\vec{d}, + since the difference (\vec{p}+\vec{d})-\vec{p}=\vec d is the direction vector of the line. + Also plotted is the vector \vec{p}-2\vec{d}, which points to another point on the line. +

    +     + + + + + //ASY file for figlines_intro.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-1,1); + pair ybounds=(-1,3.5); + pair zbounds=(-1,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //draw vector d=<-1,0,1> + draw((0,0,0)--(-1,0,1),redpen,Arrow3(size=2mm)); + label("$\vec{d}$",(-1,0,1),N); + //draw vector p=<0,2,1> + draw((0,0,0)--(0,2,1),black,Arrow3(size=2mm)); + label("$\vec{p}$",(0,2,1),N); + //draw vector p+d + draw((0,0,0)--(-1,2,2),black,Arrow3(size=2mm)); + label("$\vec{p}+\vec{d}$",(-1,2,2),W); + //draw vector p-2d + draw((0,0,0)--(2,2,-1),black,Arrow3(size=2mm)); + label("$\vec{p}-2\vec{d}$",(2,2,-1),W); + //draw the line ({-x},{2},{1+x}); + draw((2.2,2,-1.2)--(-1.5,2,2.5),bluepen); + + + + +
    + +

    + The figure illustrates how every point on the line can be obtained by starting with \vec p and moving a certain distance in the direction of \vec d. + That is, we can define the line as a function of t: + + \vec\ell(t) = \vec p + t\ \vec d + . +

    + +

    + In many ways, this is not a new concept. + Compare Equation to the familiar + y=mx+b equation of a line: +

    + +
    + Understanding the vector equation of a line + + A text description, comparing the elements the equations of a line in the plane, and in space. + +

    + On the left is the equation y=mx+b for a line in the plane with slope m and y intercept b. + On the right is the equation \vec{\ell}(t) = \vec p + t\vec d for a line in space + through the point \vec p with direction vector \vec d. +

    + +

    + Above the two equations is the text Starting Point. + From this text are two arrows, pointing to the value b in the plane equation, and the value \vec p in the space equation. + Also above the equations is the text Direction. + From this text, two arrows point to the value m in the plane equation, and the value \vec d in the space equation. +

    + +

    + Below the two equations is the text How Far To Go In That Direction. + Arrows point from this text to the value x in the plane equation, and the value t in the space equation. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=0.75] + + \draw (0,0) node (L) {\(y\ =\ b\ +\ m x\)}; + \draw (5,0) node (R) {\(\vec \ell(t)\ =\ \vec p\ +\ t \vec d\)}; + + \node (A) at (1,1.5) {Starting Point}; + \node (B) at (4,1.5) {Direction}; + \node (C) at (2.5,-1.5) {How Far To Go In That Direction}; + + \draw [->,thick] (A) -- (5,.2); + \draw [->,thick] (A) -- (0,.2); + \draw [->,thick] (B) -- (.9,.2); + \draw [->,thick] (B) -- (6.45,.25); + \draw [->,thick] (C) -- (1.3,-.15); + \draw [->,thick] (C) -- (6.1,-.3); + + \end{tikzpicture} + + + +
    + +

    + The equations exhibit the same structure: + they give a starting point, + define a direction, and state how far in that direction to travel. +

    + + + +

    + Equation is an example of a + vector-valued function; + the input of the function is a real number and the output is a vector. + We will cover vector-valued functions extensively in the next chapter. +

    + +

    + There are other ways to represent a line. + Let P = (x_0,y_0,z_0), \vec p = \la x_0,y_0,z_0\ra, and let \vec d = \la a,b,c\ra. + Then the equation of the line through P in the direction of \vec d is: + + \vec\ell(t) \amp = \vec p + t\vec d + \amp = \la x_0,y_0,z_0\ra + t\la a,b,c\ra + \amp = \la x_0 + at, y_0+bt, z_0+ct\ra + . +

    + +

    + The last line states that the x values of the line are given by x=x_0+at, + the y values are given by y = y_0+bt, + and the z values are given by z = z_0 + ct. + These three equations, taken together, + are the parametric equations of the line + through \vec p in the direction of \vec d. +

    + +

    + Finally, each of the equations for x, + y and z above contain the variable t. + We can solve for t in each equation: + + x = x_0+at \amp \Rightarrow t=\frac{x-x_0}{a}, + y=y_0+bt \amp \Rightarrow t = \frac{y-y_0}{b}, + z = z_0+ct \amp \Rightarrow t = \frac{z-z_0}{c} + , + assuming a,b,c\neq 0. + Since t is equal to each expression on the right, + we can set these equal to each other, + forming the symmetric equations of the line + through \vec p in the direction of \vec d: + + \frac{x-x_0}{a} = \frac{y-y_0}{b}=\frac{z-z_0}{c} + . +

    + +

    + Each representation has its own advantages, depending on the context. + We summarize these three forms in the following definition, + then give examples of their use. +

    + + + Equations of Lines in Space + +

    + Let P = (x_0,y_0,z_0) and let \vec p = \la x_0,y_0,z_0\ra. Consider the line in space that passes through + P in the direction of \vec d = \la a,b,c\ra. + linesequations for +

    + +

    +

      +
    1. +

      + The vector equation of the line is + + \vec \ell(t) = \vec p+t\vec d + . +

      +
      2. + +
    2. +

      + The parametric equations of the line are + + x = x_0+at, y=y_0+bt, z = z_0+ct + . +

      +
      3. + +
    3. +

      + The symmetric equations of the line are + + \frac{x-x_0}{a} = \frac{y-y_0}{b}=\frac{z-z_0}{c} + . +

      +
      4. +
    +

    +     +     + + + Finding the equation of a line + +

    + Give all three equations, + as given in , + of the line through P = (2,3,1) in the direction of \vec d = \la -1,1,2\ra. + Does the point Q=(-1,6,6) lie on this line? +

    +     + +

    + We identify the point P=(2,3,1) with the vector \vec p =\la 2,3,1\ra. + Following the definition, we have +

    + +

    +

      +
    • +

      + the vector equation of the line is \vec\ell(t) = \la 2,3,1\ra + t\la -1,1,2\ra; +

      +
      • + +
    • +

      + the parametric equations of the line are + + x = 2-t, y = 3+t, z = 1+2t; \text{ and } + +

      +
      • + +
    • +

      + the symmetric equations of the line are + + \frac{x-2}{-1}=\frac{y-3}{1} = \frac{z-1}{2} + . +

      +
      • +
    +

    + +
    + Graphing a line in + + + + A line in space with direction vector d passes through a point P. + +

    + A set three-dimensional coordinate axes is shown. + A vector \vec d is plotted with its tail at the origin of this coordinate system; + it is the direction vector for a line. The line is also plotted, + and is shown passing through a point P. + An additional point Q is in the image, but it is not on the line. +

    +     + + + + + //ASY file forfiglines1.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={5}; + real[] myzchoice={5}; + defaultpen(0.5mm); + pair xbounds=(-1,2.5); + pair ybounds=(-1,5.5); + pair zbounds=(-1,5.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //draw vector d=<-1,1,2> + draw((0,0,0)--(-1,1,2),black,Arrow3(size=2mm)); + label("$\vec{d}$",(-1,0,1),N); + // dot at P=(2,3,1) + dotfactor=3;dot((2,3,1)); + label("P",(2,3,1),N); + // dot at Q=(-1,6,6) + dotfactor=3;dot((-1,6,6)); + label("Q",(-1,6,6),S); + //draw the line ({2-t},{3+t},{1+2t}); + draw((3,2,-1)--(-1,6,7),bluepen); + + + + +
    + +

    + The first two equations of the line are useful when a t value is given: + one can immediately find the corresponding point on the line. + These forms are good when calculating with a computer; + most software programs easily handle equations in these formats. (For instance, + the graphics program that made + can be given the input (2-t,3+t,1+2*t) + for -1\leq t\leq 3.). +

    + +

    + Does the point Q = (-1,6,6) lie on the line? + The graph in + makes it clear that it does not. + We can answer this question without the graph using any of the three equation forms. + Of the three, + the symmetric equations are probably best suited for this task. + Simply plug in the values of x, + y and z and see if equality is maintained: + + \frac{-1-2}{-1} \stackrel{?}{=} \frac{6-3}{1} \stackrel{?}{=} \frac{6-1}{2} \Rightarrow 3=3\neq2.5 + . +

    + +

    + We see that Q does not lie on the line as it did not satisfy the symmetric equations. +

    +     + +     + + + Finding the equation of a line through two points + +

    + Find the parametric equations of the line through the points + P=(2,-1,2) and Q = (1,3,-1). +

    +     + +

    + Recall the statement made at the beginning of this section: + to find the equation of a line, + we need a point and a direction. + We have two points; + either one will suffice. + The direction of the line can be found by the vector with initial point P and terminal point Q: + \overrightarrow{PQ} = \la -1,4,-3\ra. +

    + +

    + The parametric equations of the line \ell through P in the direction of \overrightarrow{PQ} are: + + \ell: x= 2-t y=-1+4t z=2-3t + . +

    + +
    + A graph of the line in + + + + A line in space passes through points P and Q. The vector PQ between these points is also shown. + +

    + Two points P and Q are plotted in space, against a set of three-dimensional coordinate axes. + The vector \overrightarrow{PQ} is plotted as an arrow pointing from P to Q. + The line passing through P and Q is also plotted; + between P and Q, it overlaps with the vector \overrightarrow{PQ}. +

    +     + + + + + //ASY file for figlines6.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,17,9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={-4,4}; + real[] myzchoice={-4,4}; + defaultpen(0.5mm); + pair xbounds=(-1,3); + pair ybounds=(-4.5,4.5); + pair zbounds=(-4.5,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //draw vector PQ=<-1,4,-3> + draw((2,-1,2)--(1,3,-1),redpen,Arrow3(size=2mm)); + label("$\overrightarrow{PQ}$",(1.5,1,0.5),S); + // dot at P=(2,-1,2) + dotfactor=3;dot((2,-1,2));label("P",(2,-1,2),N); + // dot at Q=(1,3,-1) + dotfactor=3;dot((1,3,-1));label("Q",(1,3,-1),N); + //draw the line ({2-t},{-1+4t},{2-3t}) in two parts, before P and after Q + draw((2.5,-3,3.5)--(2,-1,2),bluepen); + draw((1,3,-1)--(0.5,5,-2.5),bluepen); + + + + +
    + +

    + A graph of the points and line are given in . + Note how in the given parametrization of the line, + t=0 corresponds to the point P, + and t=1 corresponds to the point Q. + This relates to the understanding of the vector equation of a line described in . + The parametric equations start at the point P, + and t determines how far in the direction of \overrightarrow{PQ} to travel. + When t=0, we travel 0 lengths of \overrightarrow{PQ}; + when t=1, we travel one length of \overrightarrow{PQ}, + resulting in the point Q. +

    +     + +     +     + + + Parallel, Intersecting and Skew Lines + +

    + In the plane, two distinct + lines can either be parallel or they will intersect at exactly one point. + In space, given equations of two lines, + it can sometimes be difficult to tell whether the lines are distinct or not (, the same line can be represented in different ways). + Given lines \vec\ell_1(t) = \vec p_1 + t\vec d_1 and \vec \ell_2(t) = \vec p_2+t\vec d_2, + we have four possibilities: + \vec \ell_1 and \vec \ell_2 are + linesskew + linesparallel + linesintersecting +

    + +

    +

    +
  • + the same line +

    they share all points

    +
    • + +
  • + intersecting lines +

    they share only 1 point;

    +
    • + +
  • + parallel lines +

    \vec d_1\parallel \vec d_2, no points in common;

    +
    • + +
  • + skew lines +

    \vec d_1\nparallel \vec d_2, no points in common.

    +
    • +
    +

    + + + +

    + The next two examples investigate these possibilities. +

    + + + Comparing lines + +

    + Consider lines \ell_1 and \ell_2, + given in parametric equation form: + + \ell_1:\, \begin{matrix} + x\amp =\amp 1+3t \\ + y\amp =\amp 2-t\\ + z\amp =\amp t + \end{matrix}\qquad\qquad + \ell_2:\, \begin{matrix} + x\amp =\amp -2+4s\\ + y\amp =\amp 3+s\\ + z\amp =\amp 5+2s + \end{matrix} + . +

    + +

    + Determine whether \ell_1 and \ell_2 are the same line, + intersect, are parallel, or skew. +

    +     + +

    + We start by looking at the directions of each line. + Line \ell_1 has the direction given by + \vec d_1=\la 3,-1,1\ra and line \ell_2 has the direction given by \vec d_2 = \la 4,1,2\ra. + It should be clear that \vec d_1 and \vec d_2 are not parallel, + hence \ell_1 and \ell_2 are not the same line, + nor are they parallel. + verifies this fact + (where the points and directions indicated by the equations of each line are identified). +

    + +

    + We next check to see if they intersect + (if they do not, they are skew lines). + To find if they intersect, + we look for t and s values such that the respective x, + y and z values are the same. + That is, we want s and t such that: + + \begin{matrix} 1+3t \amp =\amp -2+4s\\ 2-t\amp =\amp 3+s\\ t\amp =\amp 5+2s \end{matrix} + . +

    + +
    + Sketching the lines from + + + + A three dimensional plot of two lines in space. The lines are not parallel, and they do not intersect. + +

    + A set of three-dimensional coordinate axes are shown, + with labels x, y, and z, as usual. + Two points P_1 and P_2 are plotted. + The point P_1 is in the xy plane, + while the point P_2 is higher up, near the top of the figure. +

    + +

    + Through each of the two points, a line is drawn. + The line through P_1 has a direction vector \vec{d}_1 + that points in a different direction from that of the direction vector \vec{d}_2 of the line through P_2, + so the lines are not parallel. It also appears from the figure that the lines do not intersect. +

    +     + + + + + //ASY file for figlines2.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-5,5}; + real[] myychoice={-5,5}; + real[] myzchoice={-5,5}; + defaultpen(0.5mm); + pair xbounds=(-5.5,5.5); + pair ybounds=(-5.5,5.5); + pair zbounds=(-1,6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //draw P1 and vector d1 at P1 + dotfactor=3;dot((1,2,0));label("$P_1$",(1,2,0),S); + draw((1,2,0)--(4,1,1),black,Arrow3(size=2mm)); + label("$\vec{d}_1$",(4,1,1),N); + + //draw P2 and vector d2 at P2 + dotfactor=3;dot((-2,3,5));label("$P_2$",(-2,3,5),N); + draw((-2,3,5)--(2,4,7),black,Arrow3(size=2mm)); + label("$\vec{d}_2$",(2,4,7),S); + + //draw the line L1 ({1+3t},{2-t},{t}) in two parts, before P1 and after P1+d1 + draw((-2,3,-1)--(1,2,0),bluepen); + draw((4,1,1)--(7,0,2),bluepen); + label("$\ell_1$",(7,0,2),S); + + //draw the line L2 ({-2+4t},{3+t},{5+2t}) in two parts, before P2 and after P2+d2 + draw((-6,2,3)--(-2,3,5),bluepen); + draw((2,4,7)--(6,5,9),bluepen); + label("$\ell_2$",(-6,2,3),S); + + + + +
    + +

    + This is a relatively simple system of linear equations. + Since the last equation is already solved for t, + substitute that value of t into the equation above it: + + 2-(5+2s) = 3+s \Rightarrow s=-2,\ t=1 + . +

    + +

    + A key to remember is that we have + three equations; + we need to check if s=-2,\ t=1 satisfies the first equation as well: + + 1+3(1) \neq -2+4(-2) + . +

    + +

    + It does not. + Therefore, we conclude that the lines \ell_1 and \ell_2 are skew. +

    +     + +     + + + Comparing lines + +

    + Consider lines \ell_1 and \ell_2, + given in parametric equation form: + + \ell_1:\, \begin{matrix} x\amp =\amp -0.7+1.6t \\ y\amp =\amp 4.2+2.72t\\z\amp =\amp 2.3-3.36t \end{matrix} \qquad\qquad \ell_2:\,\begin{matrix} x\amp =\amp 2.8-2.9s\\y\amp =\amp 10.15-4.93s\\z\amp =\amp -5.05+6.09s. \end{matrix} + +

    + +

    + Determine whether \ell_1 and \ell_2 are the same line, + intersect, are parallel, or skew. +

    +     + +

    + It is obviously very difficult to simply look at these equations and discern anything. + This is done intentionally. + In the real world, most equations that are used do not have nice, + integer coefficients. + Rather, there are lots of digits after the decimal and the equations can look messy. +

    + +

    + We again start by deciding whether or not each line has the same direction. + The direction of \ell_1 is given by + \vec d_1 = \la 1.6,2.72,-3.36\ra and the direction of \ell_2 is given by \vec d_2 = \la -2.9,-4.93,6.09\ra. + When it is not clear through observation whether two vectors are parallel or not, + the standard way of determining this is by comparing their respective unit vectors. + Using a calculator, we find: + + \vec u_1 \amp = \frac{\vec d_1}{\norm{\vec d_1}} = \la 0.3471,0.5901,-0.7289\ra + \vec u_2 \amp = \frac{\vec d_2}{\norm{\vec d_2}} = \la -0.3471,-0.5901,0.7289\ra + . +

    + +

    + The two vectors seem to be parallel + (at least, their components are equal to 4 decimal places). + In most situations, + it would suffice to conclude that the lines are at least parallel, + if not the same. + One way to be sure is to rewrite \vec d_1 and + \vec d_2 in terms of fractions, not decimals. + We have + + \vec d_1 =\la \frac{16}{10},\frac{272}{100},-\frac{336}{100}\ra \qquad \vec d_2 = \la -\frac{29}{10},-\frac{493}{100},\frac{609}{100}\ra + . +

    + +

    + One can then find the magnitudes of each vector in terms of fractions, + then compute the unit vectors likewise. + After a lot of manual arithmetic + (or after briefly using a computer algebra system), + one finds that + + \vec u_1 = \la \sqrt{\frac{10}{83}},\frac{17}{\sqrt{830}},-\frac{21}{\sqrt{830}}\ra \qquad \vec u_2 = \la -\sqrt{\frac{10}{83}},-\frac{17}{\sqrt{830}},\frac{21}{\sqrt{830}}\ra + . +

    + +

    + We can now say without equivocation that these lines are parallel. +

    + +

    + Are they the same line? + The parametric equations for a line describe one point that lies on the line, + so we know that the point P_1 = (-0.7,4.2,2.3) lies on \ell_1. + To determine if this point also lies on \ell_2, + plug in the x, + y and z values of P_1 into the symmetric equations for \ell_2: + + \frac{(-0.7)-2.8}{-2.9} \amp \stackrel{?}{=} \frac{(4.2)-10.15}{-4.93} \stackrel{?}{=} \frac{(2.3)-(-5.05)}{6.09} + 1.2069\amp=1.2069=1.2069 + . +

    + +
    + Graphing the lines in + + + + A plot in space of a single line through two points, with two parallel direction vectors shown next to it. + +

    + Two points P_1 and P_2 are plotted in a three-dimensional coordinate system. + A line is shown passing through these two points. + Next to the line, two vectors are plotted. + The vector \vec{d}_1 points downward, in the direction of the line. + The vector \vec{d}_2 points in the opposite direction, and is twice as long. +

    + +

    + The diagram shows that the line through P_1 in the direction of \vec{d}_1 + is in fact the same as the line through P_2 in the direction of \vec{d}_2. +

    +     + + + + + //ASY file for figlines3.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(10,26,7); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={10}; + real[] myzchoice={-5,5}; + defaultpen(0.5mm); + pair xbounds=(-1,4.5); + pair ybounds=(-1,11); + pair zbounds=(-6,6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //draw P1 and vector d1 at P1 + dotfactor=3;dot((-0.7,4.2,2.3));label("$P_1$",(-0.7,4.2,2.3),E); + //draw((-0.7,4.2,2.3)--(0.9,6.92,-1.06),black,Arrow3(size=2mm)); + label("$\vec{d}_1$",(0.9+.5,6.92,-1.06),W); + + //draw P2 and vector d2 at P2 + //dotfactor=3;dot((5.7,15.08,-11.14));label("$P_2$",(5.7,15.08,-11.14),E); + //draw((5.7,15.08,-11.14)--(2.8,10.15,-5.05),black,Arrow3(size=2mm)); + //label("$\vec{d}_2$",(2.8,10.15,-5.05),E); + + dotfactor=3;dot((2.8,10.15,-5.05));label("$P_2$",(2.8,10.15,-5.05),W); + //draw((2.8,10.15,-5.05)--(-.1,5.22,1.04),black,Arrow3(size=2mm)); + label("$\vec{d}_2$",(-.1-.5,5.22,1.04),E); + + //draw the line L ({-0.7+1.6t},{4.2+2.72t},{2.3-3.36t}) + draw((3.3,11,-6.1)--(-2.3,1.48,5.66),bluepen); + + draw((2.8-.5,10.15,-5.05)--(-.1-.5,5.22,1.04),redpen,Arrow3(size=2mm)); + draw((-0.7+.5,4.2,2.3)--(0.9+.5,6.92,-1.06),redpen,Arrow3(size=2mm)); + + + + +
    + +

    + The point P_1 lies on both lines, + so we conclude they are the same line, + just parametrized differently. + + graphs this line along with the points and vectors described by the parametric equations. + Note how \vec d_1 and \vec d_2 are parallel, + though point in opposite directions + (as indicated by their unit vectors above). +

    +     +     +     + + + Distances +

    + Given a point Q and a line \vec\ell(t) = \vec p+t\vec d in space, + it is often useful to know the distance from the point to the line. + (Here we use the standard definition of distance, + , the length of the shortest line segment from the point to the line.) + Identifying \vec p with the point P, + + will help establish a general method of computing this distance h. +

    + +
    + Establishing the distance from a point to a line + + + A line through a point P, with direction d. Near the line is a point Q, whose distance to the line is shown as the height h of a triangle. + +

    + A generic line is shown, without coordinates. + On the line is a point P and a direction vector \vec{d}. + Near the line is a point Q. + The vector \overrightarrow{PQ} points from the point P on the line to the point Q off the line. + This vector forms the hypotenuse of a right-angled triangle whose base is part of the line, + and whose height, h, is shown as a dashed line segment from Q to the line. + The angle \theta between \vec d and \overrightarrow{PQ} is also shown. + The height h is the perpendicular distance from Q to the line. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.32] + + \draw [thin,firstcolor] (-2,-1) -- (2,1); + \draw [thick,->,secondcolor] (0,0) -- (1,.5) node [below,black] { $\vec d$}; + + \filldraw [black] (0,3) circle (2.4pt) node [above] { $Q$} + (0,0) circle (2.4pt) node [below] { $P$}; + + \draw [thick,dashed] (0,3) -- (1.2,.6) node [right,pos=.5] { $h$}; + \draw [thick,->] (0,0) -- (0,3) node [pos=.5,left] { $\overrightarrow{PQ}$}; + \draw [rotate=58.3] (.4,0) node { $\theta$}; + + \end{tikzpicture} + + + + +
    + +

    + From trigonometry, we know h = \norm{\overrightarrow{PQ}}\sin(\theta). + We have a similar identity involving the cross product: + \norm{\overrightarrow{PQ}\times \vec d} = \norm{\overrightarrow{PQ}}\, \vnorm{d}\sin(\theta). + Divide both sides of this latter equation by \vnorm{d} to obtain h: + + h = \frac{\norm{\overrightarrow{PQ}\times \vec d}}{\vnorm{d}} + . +

    + + + +

    + It is also useful to determine the distance between lines, + which we define as the length of the shortest line segment that connects the two lines + (an argument from geometry shows that this line segments is perpendicular to both lines). + Let lines \vec\ell_1(t) = \vec p_1 + t\vec d_1 and \vec\ell_2(t) = \vec p_2 + t\vec d_2 be given, + as shown in . + To find the direction orthogonal to both \vec d_1 and \vec d_2, + we take the cross product: + \vec c = \vec d_1\times \vec d_2. + The magnitude of the orthogonal projection of + \overrightarrow{P_1P_2} onto \vec c is the distance h we seek: + + h\amp = \norm{\operatorname{proj}_{\vec c}\overrightarrow{P_1P_2}} + \amp = \norm{\frac{\overrightarrow{P_1P_2}\cdot\vec c}{\dotp cc}\vec c} + \amp =\frac{\abs{\overrightarrow{P_1P_2}\cdot \vec c}}{\vnorm c^2}\vnorm c + \amp =\frac{\abs{\overrightarrow{P_1P_2}\cdot \vec c}}{\vnorm c} + . +

    + +
    + Establishing the distance between lines + + + + Two skew lines in space are plotted with respective points and direction vectors. The distance h between them is indicated. + +

    + Two lines in space are shown, without a coordinate system. + The lines pass, respectively, through points P_1 and P_2, + and have respective direction vectors \vec{d}_1 and \vec{d}_2. +

    + +

    + The vector \overrightarrow{P_1P_2} is shown pointing from the first line to the second. + Also shown is the cross product \vec{c}=\vec{d}_1\times\vec{d}_2; + this vector is perpendicular to both lines. + The distance h between the lines is shown as a segment parallel to \vec c. +

    +     + + + + + //ASY file for figlines_dist23D.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((-.14,-9.67,3.98),(0.003,0.01,0.026),(0,0,0),1,(-0.09,-0.2)); + defaultrender.merge=true; + + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-3,3); + pair ybounds=(-3,3); + pair zbounds=(-3,3); + + xaxis3("",xbounds.x,xbounds.y,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,invisible,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + triple p1=(0,0,0); + triple p2=(0,0,2); + triple p12=p2-p1; + triple d1=(1,0,0); + triple d2=(1,1,-1); + + // line 1 + triple f(real t) { + return (p1+t*d1);} + + // line 2 + triple g(real t) { + return (p2+t*d2);} + + //draw P1 (t=0) and vector d1 at P1 (t=1) + dotfactor=3; + dot(p1);label("$P_1$",p1,S); + draw(f(0)--f(.6),redpen+.7mm,Arrow3(size=2mm)); + label("$\vec{d}_1$",f(.6),S); + + //draw P2 (t=-1.5) and vector d2 at P2 (t=-0.5) + dotfactor=3; + dot(p2);label("$P_2$",p2,N); + draw(g(0)--g(.6),redpen+.7mm,Arrow3(size=2mm)); + label("$\vec{d}_2$",g(.6),N); + + //draw vector P1 to P2 + draw(p1--p2,black,Arrow3(size=2mm)); + //label("$\overrightarrow{P_1 P_2}$",(p1+p2)/2,W); + label("$\overrightarrow{P_1 P_2}$",(p1+p2)/2,W); + + //draw the lines + draw(f(-1)--f(0),bluepen);//L1 + draw(f(.6)--f(2),bluepen);//L1 + draw(g(-1)--g(0),bluepen);//L2 + draw(g(.6)--g(2),bluepen);//L2 + + triple c=cross(d1,d2); + c=c/sqrt(dot(c,c)); + + //draw perp line + draw(f(1)--f(1)+c*abs(dot(p12,c)),black); + + // draw perp vector + draw(f(.9)--f(.9)+c*abs(dot(p12,c))*.4,redpen,Arrow3(size=2mm)); + label("$\vec c$",f(.9)+c*abs(dot(p12,c))*.4,W); + + label("$h$",(f(1)+f(1)+c*abs(dot(p12,c)))/2,E); + + // draw squares on perp line + triple sq1a = f(1)+c*abs(dot(p12,c))*.15; + triple sq1b = sq1a+.15*d1; + triple sq1c = sq1b+(f(1)-sq1a); + draw(sq1a -- sq1b -- sq1c); + + triple sq2a = f(1)+c*abs(dot(p12,c))*.85; + triple sq2b = sq2a+.15*d2/length(d2); + triple sq2c = sq2b+(f(1)+c*abs(dot(p12,c))-sq2a); + draw(sq2a -- sq2b -- sq2c); + + + + +
    + +

    + A problem in the Exercise section is to show that this distance is 0 when the lines intersect. + Note the use of the Triple Scalar Product: + \overrightarrow{P_1P_2}\cdot \vec c = \overrightarrow{P_1P_2}\cdot (\vec d_1\times \vec d_2). +

    + + + +

    + The following Key Idea restates these two distance formulas. +

    + + + Distances to Lines +

    +

      +
    1. +

      + Let P be a point on a line \ell that is parallel to \vec d. + The distance h from a point Q to the line \ell is: + distancebetween point and line + distancebetween lines + linesdistances between + + h =\frac{\norm{\overrightarrow{PQ}\times \vec d}}{\vnorm{d}} + . +

      +
      2. + +
    2. +

      + Let P_1 be a point on line \ell_1 that is parallel to \vec d_1, + and let P_2 be a point on line \ell_2 parallel to \vec d_2, + and let \vec c = \vec d_1\times \vec d_2, + where lines \ell_1 and \ell_2 are not parallel. + The distance h between the two lines is: + + h=\frac{\abs{\overrightarrow{P_1P_2}\cdot \vec c}}{\vnorm c} + . +

      +
      3. +
    +

    +     + + + + + Finding the distance from a point to a line + +

    + Find the distance from the point + Q=(1,1,3) to the line \vec\ell(t) = \la 1,-1,1\ra+t\la 2,3,1\ra. +

    +     + +

    + The equation of the line gives us the point + P=(1,-1,1) that lies on the line, + hence \overrightarrow{PQ} = \la 0,2,2\ra. + The equation also gives \vec d= \la 2,3,1\ra. + Following , + we have the distance as + + h \amp = \frac{\norm{\overrightarrow{PQ}\times \vec d}}{\vnorm{d}} + \amp = \frac{\norm{\la -4,4,-4\ra}}{\sqrt{14}} + \amp =\frac{4\sqrt{3}}{\sqrt{14}} \approx 1.852 + . +

    + +

    + The point Q is approximately 1.852 units from the line \vec\ell(t). +

    +     + +     + + + Finding the distance between lines + +

    + Find the distance between the lines + + \ell_1: \begin{matrix} x\amp =\amp 1+3t \\ y\amp =\amp 2-t\\z\amp =\amp t \end{matrix} \qquad\qquad + \ell_2:\begin{matrix} x\amp =\amp -2+4s\\y\amp =\amp 3+s\\z\amp =\amp 5+2s. \end{matrix} + +

    +     + +

    + These are the sames lines as given in , + where we showed them to be skew. + The equations allow us to identify the following points and vectors: + + P_1 = (1,2,0) P_2 = (-2,3,5) \Rightarrow \overrightarrow{P_1P_2} = \la -3,1,5\ra + . + + \vec d_1 = \la 3,-1,1\ra \vec d_2 = \la 4,1,2\ra \Rightarrow \vec c = \vec d_1\times \vec d_2 = \la -3,-2,7\ra + . +

    + +

    + From + we have the distance h between the two lines is + + h \amp = \frac{\abs{\overrightarrow{P_1P_2}\cdot \vec c}}{\vnorm c} + \amp =\frac{42}{\sqrt{62}} \approx 5.334 + . +

    + +

    + The lines are approximately 5.334 units apart. +

    +     + +     + +

    + One of the key points to understand from this section is this: + to describe a line, we need a point and a direction. + Whenever a problem is posed concerning a line, + one needs to take whatever information is offered and glean point and direction information. + Many questions can be asked + (and are asked in the Exercise section) + whose answer immediately follows from this understanding. +

    + +

    + Lines are one of two fundamental objects of study in space. + The other fundamental object is the plane, + which we study in detail in the next section. + Many complex three dimensional objects are studied by approximating their surfaces with lines and planes. +

    +     + + + + Terms and Concepts + + + +

    + To find an equation of a line, + what two pieces of information are needed? +

    +     + + + +

    + A point on the line and the direction of the line. +

    +     + +     + + + + +

    + Two distinct lines in the plane can intersect or be . +

    +     + + + + + + + + +     + + + + +

    + Two distinct lines in space can intersect, + be or be . +

    +     + + + + + + ["parallel","skew"].includes(ans) + + + + + + + ["parallel","skew"].includes(ans) && !ans_array.slice(0,1).includes(ans) + + + + + ans_array.slice(0,1).includes(ans) + + You already gave that answer. + + + + +     + + + + +

    + Use your own words to describe what it means for two lines in space to be skew. +

    +     + + + +     +     + + + Problems + + + +

    + Write the vector, + parametric and symmetric equations of the lines described. +

    +     + + + + +

    + Passes through P=(2,-4,1), + parallel to \vec d=\la 9,2,5\ra. +

    +     + +

    + vector: \ell(t) = \la 2,-4,1\ra + t\la 9,2,5\ra +

    + +

    + parametric: x= 2+9t, y=-4+2t, z = 1+5t +

    + +

    + symmetric: (x-2)/9 =(y+4)/2 = (z-1)/5 +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(t=>"Real"); + $v=Compute("(6,1,7)+t<-3,2,5>"); + $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; + $ds=Formula($st)->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; + $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return ($par and $tch); + }); + Context("Numeric"); + Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); + parser::Assignment->Allow; + $p=List(Formula("x=6-3t"),Formula("y=1+2t"),Formula("z=7+5t")); + $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + }} + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + }} + if(!$aH->{isPreview}){ + push(@err,"You need to provide more parametrizations")if $i<3; + push(@err,"You have given too many parametrizations")if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + Context("Numeric"); + Context()->variables->add(y=>'Real',z=>'Real'); + Context()->operators->redefine('=',using=>',',from=>'Numeric'); + Context()->operators->set('='=>{string=>' = ',TeX=>'='}); + Context()->lists->set(List=>{separator=>" = "}); + $s=List("(x-6)/-3","(y-1)/2","(z-7)/5"); + $sev=$s->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$dv;my$incpt;my%uses; + if($entry->class ne "Formula"){push(@err,"Your $ith quantity $entry is not a formula");next;} + else{$uses{'x'}=$entry->usesOneOf("x");$uses{'y'}=$entry->usesOneOf("y");$uses{'z'}=$entry->usesOneOf("z"); + if($uses{'x'}+$uses{'y'}+$uses{'z'}!=1){push(@err,"Your $ith quantity is not a formula in only x, y, or z");next;}; + $dv=$entry->D('x')->eval(x=>0,y=>0,z=>0)+$entry->D('y')->eval(x=>0,y=>0,z=>0)+$entry->D('z')->eval(x=>0,y=>0,z=>0); + $incpt=$entry->eval(x=>0,y=>0,z=>0)/$dv; + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->class eq "Formula"){ + $uj{'x'}=$st->[$j]->usesOneOf("x"); + $uj{'y'}=$st->[$j]->usesOneOf("y"); + $uj{'z'}=$st->[$j]->usesOneOf("z"); + if(($uses{'x'} and $uj{'x'}) or ($uses{'y'} and $uj{'y'}) or ($uses{'z'} and $uj{'z'})){ + push(@err,"Your $ith quantity uses the same variable as a previous one")unless $aH->{isPreview}; + $used=1;last; + }}} + if(!$used){ + if($uses{'x'}){$sV=$sV+Formula("<t/$dv-$incpt,0,0>")}; + if($uses{'y'}){$sV=$sV+Formula("<0,t/$dv-$incpt,0>")}; + if($uses{'z'}){$sV=$sV+Formula("<0,0,t/$dv-$incpt>")}; + }} + if(!$aH->{isPreview}){ + push(@err,"You need to provide more symmetric quantities") if $i<3; + push(@err,"You have given too many symmetric quantities") if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant) {$ds = Vector("$ds");} else {return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + + + + +

    + \ell is a line that passes through P=(6,1,7), + parallel to \vec d=\la -3,2,5\ra. +

    + + + Give the right-hand side (without \ell(t)=) of a vector equation for \ell. + +

    + +

    + + + Give the parametric equations for \ell + (separated using commas). + +

    + +

    + + + Give the symmetric equations for \ell. + (Enter DNE if they are not defined.) + +

    + +

    +     +     +     + + + + +

    + Passes through P=(2,1,5) and Q = (7,-2,4). +

    +     + +

    + Answers can vary: + vector: \ell(t) = \la 2,1,5\ra + t\la 5,-3,-1\ra +

    + +

    + parametric: x= 2+5t, y=1-3t, z = 5-t +

    + +

    + symmetric: (x-2)/5 =-(y-1)/3 = -(z-5) +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(t=>"Real"); + $v=Compute("(1,-2,3)+t<4,7,2>"); + $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; + $ds=Formula($st)->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; + $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return ($par and $tch); + }); + Context("Numeric"); + Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); + parser::Assignment->Allow; + $p=List(Formula("x=1+4t"),Formula("y=-2+7t"),Formula("z=3+2t")); + $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{ + ($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + } + } + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + } + } + if(!$aH->{isPreview}){ + push(@err,"You need to provide more parametrizations")if $i<3; + push(@err,"You have given too many parametrizations")if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + Context("Numeric"); + Context()->variables->add(y=>'Real',z=>'Real'); + Context()->operators->redefine('=',using=>',',from=>'Numeric'); + Context()->operators->set('='=>{string=>' = ',TeX=>'='}); + Context()->lists->set(List=>{separator=>" = "}); + $s=List("(x-1)/4","(y+2)/7","(z-3)/2"); + $sev=$s->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$dv;my$incpt;my%uses; + if($entry->class ne "Formula"){push(@err,"Your $ith quantity $entry is not a formula");next;} + else { + $uses{'x'}=$entry->usesOneOf("x"); + $uses{'y'}=$entry->usesOneOf("y"); + $uses{'z'}=$entry->usesOneOf("z"); + if($uses{'x'}+$uses{'y'}+$uses{'z'}!=1){push(@err,"Your $ith quantity is not a formula in only x, y, or z");next;}; + $dv=$entry->D('x')->eval(x=>0,y=>0,z=>0)+$entry->D('y')->eval(x=>0,y=>0,z=>0)+$entry->D('z')->eval(x=>0,y=>0,z=>0); + $incpt=$entry->eval(x=>0,y=>0,z=>0)/$dv; + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->class eq "Formula"){ + $uj{'x'}=$st->[$j]->usesOneOf("x"); + $uj{'y'}=$st->[$j]->usesOneOf("y"); + $uj{'z'}=$st->[$j]->usesOneOf("z"); + if(($uses{'x'} and $uj{'x'}) or ($uses{'y'} and $uj{'y'}) or ($uses{'z'} and $uj{'z'})){ + push(@err,"Your $ith quantity uses the same variable as a previous one")unless $aH->{isPreview}; + $used=1;last; + } + } + } + if(!$used){ + if($uses{'x'}){$sV=$sV+Formula("<t/$dv-$incpt,0,0>")}; + if($uses{'y'}){$sV=$sV+Formula("<0,t/$dv-$incpt,0>")}; + if($uses{'z'}){$sV=$sV+Formula("<0,0,t/$dv-$incpt>")}; + } + } + if(!$aH->{isPreview}){ + push(@err,"You need to provide more symmetric quantities") if $i<3; + push(@err,"You have given too many symmetric quantities") if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant) {$ds = Vector("$ds");} else {return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + + + + +

    + \ell is a line that passes through + P=(1,-2,3) and Q = (5,5,5). +

    + + + Give the right-hand side (without \ell(t)=) of a vector equation for \ell. + +

    + +

    + + + Give the parametric equations for \ell + (separated using commas). + +

    + +

    + + + Give the ymmetric equations for \ell. + (Enter DNE if they are not defined.) + +

    + +

    +     +     +     + + + + +

    + Passes through P=(0,1,2) and orthogonal to both +

    + +

    + \vec d_1=\la 2,-1,7\ra and \vec d_2=\la 7,1,3\ra. +

    +     + +

    + Answers can vary; + here the direction is given by \vec d_1\times \vec d_2: + vector: \ell(t) = \la 0,1,2\ra + t\la -10,43,9\ra +

    + +

    + parametric: x= -10t, y=1+43t, z = 2+9t +

    + +

    + symmetric: -x/10 =(y-1)/43 = (z-2)/9 +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(t=>"Real"); + $v=Compute("(5,1,9)+t<0,-1,0>"); + $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; + $ds=Formula($st)->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; + $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return ($par and $tch); + }); + Context("Numeric"); + Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); + parser::Assignment->Allow; + $p=List(Formula("x=5"),Formula("y=1-t"),Formula("z=9")); + $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{ + ($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + } + } + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + } + } + if(!$aH->{isPreview}){ + push(@err,"You need to provide more parametrizations")if $i<3; + push(@err,"You have given too many parametrizations")if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + + + +

    + \ell is a line that passes through + P=(5,1,9) and is orthogonal to both + \vec d_1=\la 1,0,1\ra and \vec d_2=\la 2,0,3\ra. +

    + + + Give the right-hand side (without \ell(t)=) of a vector equation for \ell. + +

    + +

    + + + Give parametric equations for \ell + (separated using commas). + +

    + +

    + + + Give symmetric equations for \ell. + (Enter DNE if they are not defined.) + +

    + +

    +     +     +     + + + + + Context("Vector");Context()->variables->are(t=>"Real"); + $v=Compute("(7,2,-1)+t<1,-1,2>"); + $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; + $ds=Formula($st)->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; + $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return($par and $tch); + }); + Context("Numeric");Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); + parser::Assignment->Allow; + $p=List(Formula("x=7+t"),Formula("y=2-t"),Formula("z=-1+2t")); + $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + }} + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + }} + if(!$aH->{isPreview}){push(@err,"You need to provide more parametrizations")if $i<3;push(@err,"You have given too many parametrizations")if $i>3;} + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return(3*($par and $tch),@err); + }); + Context("Numeric");Context()->variables->add(y=>'Real',z=>'Real'); + Context()->operators->redefine('=',using=>',',from=>'Numeric'); + Context()->operators->set('='=>{string=>' = ',TeX=>'='}); + Context()->lists->set(List=>{separator=>" = "}); + $s=List("x-7","2-y","(z+1)/2"); + $sev=$s->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector");Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$dv;my$incpt;my%uses; + if($entry->class ne "Formula"){push(@err,"Your $ith quantity $entry is not a formula");next;} + else{$uses{'x'}=$entry->usesOneOf("x");$uses{'y'}=$entry->usesOneOf("y");$uses{'z'}=$entry->usesOneOf("z"); + if($uses{'x'}+$uses{'y'}+$uses{'z'}!=1){push(@err,"Your $ith quantity is not a formula in only x, y, or z");next;}; + $dv=$entry->D('x')->eval(x=>0,y=>0,z=>0)+$entry->D('y')->eval(x=>0,y=>0,z=>0)+$entry->D('z')->eval(x=>0,y=>0,z=>0); + $incpt=$entry->eval(x=>0,y=>0,z=>0)/$dv;}; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->class eq "Formula"){ + $uj{'x'}=$st->[$j]->usesOneOf("x");$uj{'y'}=$st->[$j]->usesOneOf("y");$uj{'z'}=$st->[$j]->usesOneOf("z"); + if(($uses{'x'} and $uj{'x'}) or ($uses{'y'} and $uj{'y'}) or ($uses{'z'} and $uj{'z'})){push(@err,"Your $ith quantity uses the same variable as a previous one")unless $aH->{isPreview};$used=1;last;} + }} + if(!$used){ + if($uses{'x'}){$sV=$sV+Formula("<t/$dv-$incpt,0,0>")}; + if($uses{'y'}){$sV=$sV+Formula("<0,t/$dv-$incpt,0>")}; + if($uses{'z'}){$sV=$sV+Formula("<0,0,t/$dv-$incpt>")}; + }} + if(!$aH->{isPreview}){push(@err,"You need to provide more symmetric quantities") if $i<3;push(@err,"You have given too many symmetric quantities") if $i>3;} + $sV=Formula($sV);$ds=$sV->D('t')->reduce; + if($ds->isConstant) {$ds = Vector("$ds");} else {return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0);$tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return(3*($par and $tch),@err); + }); + + + +

    + \ell is a line that passes through the intersection of + \vec\ell_1(t)=\la 2,1,1\ra+t\la 5,1,-2\ra and \vec\ell_2(t)=\la -2,-1,2\ra+t\la 3,1,-1\ra, + and is orthogonal to both lines. +

    + + + Give the right-hand side (without \ell(t)=) of a vector equation for \ell. + +

    + +

    + + + Give parametric equations for \ell + (separated using commas). + +

    + +

    + + + Give symmetric equations for \ell. + (Enter DNE if they are not defined.) + +

    + +

    +     +     +     + + + + + Context("Vector");Context()->variables->are(t=>"Real"); + $v=Compute("(2,2,3)+t<5,-1,-3>"); + $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; + $ds=Formula($st)->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; + $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return($par and $tch); + }); + Context("Numeric");Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); + parser::Assignment->Allow; + $p=List(Formula("x=2+5t"),Formula("y=2-t"),Formula("z=3-3t")); + $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + }} + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + }} + if(!$aH->{isPreview}){push(@err,"You need to provide more parametrizations")if $i<3;push(@err,"You have given too many parametrizations")if $i>3;} + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return(3*($par and $tch),@err); + }); + Context("Numeric");Context()->variables->add(y=>'Real',z=>'Real'); + Context()->operators->redefine('=',using=>',',from=>'Numeric'); + Context()->operators->set('='=>{string=>' = ',TeX=>'='}); + Context()->lists->set(List=>{separator=>" = "}); + $s=List("(x-2)/5","-(y-2)","-(z-3)/3"); + $sev=$s->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector");Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$dv;my$incpt;my%uses; + if($entry->class ne "Formula"){push(@err,"Your $ith quantity $entry is not a formula");next;} + else{$uses{'x'}=$entry->usesOneOf("x");$uses{'y'}=$entry->usesOneOf("y");$uses{'z'}=$entry->usesOneOf("z"); + if($uses{'x'}+$uses{'y'}+$uses{'z'}!=1){push(@err,"Your $ith quantity is not a formula in only x, y, or z");next;}; + $dv=$entry->D('x')->eval(x=>0,y=>0,z=>0)+$entry->D('y')->eval(x=>0,y=>0,z=>0)+$entry->D('z')->eval(x=>0,y=>0,z=>0); + $incpt=$entry->eval(x=>0,y=>0,z=>0)/$dv;}; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->class eq "Formula"){ + $uj{'x'}=$st->[$j]->usesOneOf("x");$uj{'y'}=$st->[$j]->usesOneOf("y");$uj{'z'}=$st->[$j]->usesOneOf("z"); + if(($uses{'x'} and $uj{'x'}) or ($uses{'y'} and $uj{'y'}) or ($uses{'z'} and $uj{'z'})){push(@err,"Your $ith quantity uses the same variable as a previous one")unless $aH->{isPreview};$used=1;last;} + }} + if(!$used){ + if($uses{'x'}){$sV=$sV+Formula("<t/$dv-$incpt,0,0>")}; + if($uses{'y'}){$sV=$sV+Formula("<0,t/$dv-$incpt,0>")}; + if($uses{'z'}){$sV=$sV+Formula("<0,0,t/$dv-$incpt>")}; + }} + if(!$aH->{isPreview}){push(@err,"You need to provide more symmetric quantities") if $i<3;push(@err,"You have given too many symmetric quantities") if $i>3;} + $sV=Formula($sV);$ds=$sV->D('t')->reduce; + if($ds->isConstant) {$ds = Vector("$ds");} else {return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0);$tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return(3*($par and $tch),@err); + }); + + + +

    + \ell is a line that passes through the intersection of + \vec\ell_1(t)=\begin{cases}x\amp=t\\y\amp=-2+2t\\z\amp=1+t\end{cases} and \vec\ell_2(t)=\begin{cases}x\amp=2+t\\y\amp=2-t\\z\amp=3+2t\end{cases}, + and is orthogonal to both lines. +

    + + + Give the right-hand side (without \ell(t)=) of a vector equation for \ell. + +

    + +

    + + + Give parametric equations for \ell + (separated using commas). + +

    + +

    + + + Give symmetric equations for \ell. + (Enter DNE if they are not defined.) + +

    + +

    +     +     +     + + + + +

    + Passes through P=(1,1), + parallel to \vec d = \la 2,3\ra. +

    +     + +

    + vector: \ell(t) = \la 1,1\ra + t\la 2,3\ra +

    + +

    + parametric: x= 1+2t, y=1+3t +

    + +

    + symmetric: (x-1)/2=(y-1)/3 +

    +     + +     + + + + + Context("Vector");Context()->variables->are(t=>"Real"); + $v=Compute("(-2,5)+t<0,1>"); + $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; + $ds=Formula($st)->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; + $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return($par and $tch); + }); + Context("Numeric");Context()->variables->add(y=>"Real",t=>"Real"); + parser::Assignment->Allow; + $p=List(Formula("x=-2"),Formula("y=5+t")); + $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0)+t*<0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y")) or $for->usesOneOf("x","y")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + }} + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0>")} + else{$sV=$sV+Formula("<0,$for>")}; + }} + if(!$aH->{isPreview}){push(@err,"You need to provide more parametrizations")if $i<2;push(@err,"You have given too many parametrizations")if $i>2;} + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return(2*($par and $tch),@err); + }); + + + +

    + \ell is a line that passes through P=(-2,5), + parallel to \vec d = \la 0,1\ra. +

    + + + Give the right-hand side (without \ell(t)=) of a vector equation for \ell. + +

    + +

    + + + Give parametric equations for \ell + (separated using commas) + +

    + +

    + + + Give symmetric equations for \ell. + (Enter DNE if they are not defined.) + +

    + +

    +     +     +     + +     + + + +

    + Determine if the described lines are the same line, + parallel lines, intersecting or skew lines. + If intersecting, give the point of intersection. +

    +     + + + + + Context("Point"); + Context()->strings->add(same=>{},parallel=>{},skew=>{}); + $answer=Compute("parallel"); + + + +

    + \vec\ell_1(t) = \la 1,2,1\ra + t\la 2,-1,1\ra and + \vec\ell_2(t) = \la 3,3,3\ra + t\la -4,2,-2\ra. +

    + + You can answer with same, + parallel, + skew, + or give the point of intersection. + + +

    + +

    +     +     +     + + + + + Context("Point"); + Context()->strings->add(same=>{},parallel=>{},skew=>{}); + $answer=Compute("(12,3,7)"); + + + +

    + \vec\ell_1(t) = \la 2,1,1\ra + t\la 5,1,3\ra and + \vec\ell_2(t) = \la 14,5,9\ra + t\la 1,1,1\ra. +

    + + You can answer with same, + parallel, + skew, + or give the point of intersection. + + +

    + +

    +     +     +     + + + + +

    + \vec\ell_1(t) = \la 3,4,1\ra + t\la 2,-3,4\ra, +

    + +

    + \vec\ell_2(t) = \la -3,3,-3\ra + t\la 3,-2,4\ra. +

    +     + +

    + intersecting; + \vec\ell_1(3) = \vec\ell_2(4) = \la 9,-5,13\ra +

    +     + +     + + + + + Context("Point"); + Context()->strings->add(same=>{},parallel=>{},skew=>{}); + $answer=Compute("same"); + + + +

    + \vec\ell_1(t) = \la 1,1,1\ra + t\la 3,1,3\ra and + \vec\ell_2(t) = \la 7,3,7\ra + t\la 6,2,6\ra. +

    + + You can answer with same, + parallel, + skew, + or give the point of intersection. + + +

    + +

    +     +     +     + + + + + Context("Point"); + Context()->strings->add(same=>{},parallel=>{},skew=>{}); + $answer=Compute("skew"); + + + +

    + \vec\ell_1(t) = \begin{cases}x\amp = 1+2t\\ y\amp = 3-2t\\ z\amp = t\end{cases} and + \vec\ell_2(t) = \begin{cases}x\amp = 3-t\\ y\amp = 3+5t\\ z\amp = 2+7t\end{cases}. +

    + + You can answer with same, + parallel, + skew, + or give the point of intersection. + + +

    + +

    +     +     +     + + + + + Context("Point"); + Context()->strings->add(same=>{},parallel=>{},skew=>{}); + $answer=Compute("parallel"); + + + +

    + \vec\ell_1(t) = \begin{cases}x\amp = 1.1+0.6t\\ y\amp = 3.77+0.9t\\ z\amp = -2.3+1.5t\end{cases} and + \vec\ell_2(t) = \begin{cases}x\amp = 3.11+3.4t\\ y\amp = 2+5.1t\\ z\amp = 2.5+8.5t\end{cases}. +

    + + You can answer with same, + parallel, + skew, + or give the point of intersection. + + +

    + +

    +     +     +     + + + + +

    + \ell_1 = \left\{\begin{aligned}x\amp = 0.2+0.6t\\ y\amp = 1.33-0.45t\\ z\amp = -4.2+1.05t + \end{aligned} \right. and + \ell_2 = \left\{\begin{aligned}x\amp = 0.86+9.2t\\ y\amp = 0.835-6.9t\\ z\amp = -3.045+16.1t + \end{aligned} \right. +

    +     + +

    + same +

    +     + +     + + + + + Context("Point"); + Context()->strings->add(same=>{},parallel=>{},skew=>{}); + $answer=Compute("skew"); + + + +

    + \vec\ell_1(t) = \begin{cases}x\amp = 0.1+1.1t\\ y\amp = 2.9-1.5t\\ z\amp = 3.2+1.6t\end{cases} and + \vec\ell_2(t) = \begin{cases}x\amp = 4-2.1t\\ y\amp = 1.8+7.2t\\ z\amp = 3.1+1.1t\end{cases}. +

    + + You can answer with same, + parallel, + skew, + or give the point of intersection. + + +

    + +

    +     +     +     + +     + + + +

    + Find the distance from the point to the line. +

    +     + + + + +

    + Q=(1,1,1), \vec\ell(t) = \la 2,1,3\ra + t\la 2,1,-2\ra +

    +     + +

    + \sqrt{41}/3 +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $distance=Formula("3sqrt(2)"); + + + +

    + Find the distance from the point + Q=(2,5,6) to the line \vec\ell(t) = \la -1,1,1\ra + t\la 1,0,1\ra. +

    + +

    + +

    +     +     +     + + + + +

    + Q=(0,3), \vec\ell(t) = \la 2,0\ra + t\la 1,1\ra +

    +     + +

    + 5\sqrt{2}/2 +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $distance=Formula("5"); + + + +

    + Find the distance from the point Q=(1,1) to the line \vec\ell(t) = \la 4,5\ra + t\la -4,3\ra. +

    + +

    + +

    +     +     +     + +     + + + +

    + Find the distance between the two lines. +

    +     + + + + +

    + \vec\ell_1(t) = \la 1,2,1\ra + t\la 2,-1,1\ra, +

    + +

    + \vec\ell_2(t) = \la 3,3,3\ra + t\la 4,2,-2\ra. +

    +     + +

    + 3/\sqrt{2} +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $distance=Formula("2"); + + + +

    + Find the distance between the line + \vec\ell_1(t) = \la 0,0,1\ra + t\la 1,0,0\ra and the line \vec\ell_2(t) = \la 0,0,3\ra + t\la 0,1,0\ra. +

    + +

    + +

    +     +     +     + +     + + + +

    + The following exercises explore special cases of the distance formulas found in . +

    +     + + + + +

    + Let Q be a point on the line \vec\ell(t). + Show why the distance formula correctly gives the distance from the point to the line as 0. +

    +     + +

    + Since both P and Q are on the line, + \overrightarrow{PQ} is parallel to \vec d. + Thus \overrightarrow{PQ}\times \vec d = \vec 0, + giving a distance of 0. +

    +     + +     + + + + +

    + Let lines \vec\ell_1(t) and + \vec\ell_2(t) be intersecting lines. + Show why the distance formula correctly gives the distance between these lines as 0. +

    +     + +

    + (Note: this solution is easier once one has studied .) Since the two lines intersect, + we can state P_2= P_1 + a\vec d_1+b\vec d_2 for some scalars a and b. + (Here we abuse notation slightly and add points to vectors.) + Thus \overrightarrow{P_1P_2} = a\vec d_1+b\vec d_2. + Vector \vec c is the cross product of \vec d_1 and \vec d_2, + hence is orthogonal to both, and hence is orthogonal to \overrightarrow{P_1P_2}. + Thus \overrightarrow{P_1P_2}\cdot\vec c = 0, + and the distance between lines is 0. +

    +     + +     + + + + +

    + Let lines \vec\ell_1(t) and \vec\ell_2(t) be parallel. +

    +     + + + +

    + Show why the distance formula for distance between lines cannot be used as stated to find the distance between the lines. +

    +     + +

    + The distance formula cannot be used because since + \vec d_1 and \vec d_2 are parallel, + \vec c is \vec 0 and we cannot divide by \vnorm{0}. +

    +     +     + + + +

    + Show why letting \vec c=(\overrightarrow{P_1P_2}\times\vec d_2)\times\vec d_2 allows one to the use the formula. +

    +     + +

    + Since \vec d_1 and \vec d_2 are parallel, + \overrightarrow{P_1P_2} lies in the plane formed by the two lines. + Thus \overrightarrow{P_1P_2}\times\vec d_2 is orthogonal to this plane, + and \vec c=(\overrightarrow{P_1P_2}\times\vec d_2)\times \vec d_2 is parallel to the plane, + but still orthogonal to both \vec d_1 and \vec d_2. + We desire the length of the projection of \overrightarrow{P_1P_2} onto \vec c, + which is what the formula provides. +

    +     +     + + + +

    + Show how one can use the formula for the distance between a point and a line to find the distance between parallel lines. +

    +     + +

    + Since the lines are parallel, + one can measure the distance between the lines at any location on either line + (just as to find the distance between straight railroad tracks, + one can use a measuring tape anywhere along the track, + not just at one specific place.) + Let P=P_1 and Q=P_2 as given by the equations of the lines, + and apply the formula for distance between a point and a line. +

    +     +     + +     +     +     +     +
    +
    + Planes + +

    + Any flat surface, such as a wall, + table top or stiff piece of cardboard can be thought of as representing part of a plane. + Consider a piece of cardboard with a point P marked on it. + One can take a nail and stick it into the cardboard at P such that the nail is perpendicular to the cardboard; + see . +

    + +
    + Illustrating defining a plane with a sheet of cardboard and a nail + + + + A nail sticks into a flat, rectangular sheet of cardboard. + +

    + A flat, rectangular surface represents a sheet of cardboard. + The image of a nail is drawn sticking into the sheet, at a point P, which is marked. + The nail is standing upright, in a position perpendicular to the sheet of cardboard. +

    +     + + + + + //ASY file for figlines3.asy in Chapter 10 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((-4,27,7),(.012,-0.002,0.015),(0,0,0),.9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2,2); + + xaxis3("",xbounds.x,xbounds.y,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,invisible,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); + + defaultpen(0.5mm); + + real f(pair z) {return 0;} + surface s=surface(f,(-2,-2),(2,2),1,1); + pen p=bluepen+.3mm; + draw(s,surfacepen,meshpen=p,render(merge=true)); + + draw(scale(.1,.1,.2)*rotate(180,Y)*shift(-1Z)*unitcone,surfacepen=white); + draw(scale(.1,.1,.5)*shift(.4Z)*unitcylinder,surfacepen=white); + draw(scale(.15,.15,.05)*shift(14Z)*unitcylinder,surfacepen=white); + + //real f(pair z) {return 0;} + //surface s=surface(f,(-2,-2),(2,2)); + //draw(s,white,meshpen=p,render(merge=true)); + draw(scale3(.15)*shift(4.7Z)*unitdisk,surfacepen=white); + draw(scale3(.15)*shift(5Z)*unitdisk,surfacepen=white); + + draw(scale3(.05)*unitdisk); + label("$P$",(0,0,0),NW); + + //draw((0,0,.9)--(0,0,0),gray+1mm,Arrow3(size=2mm)); + //draw((0,0,.9)--(0,0,1),gray+1.2mm); + + + + +
    + +

    + This nail provides a handle for the cardboard. + Moving the cardboard around moves P to different locations in space. + Tilting the nail + (but keeping P fixed) + tilts the cardboard. + Both moving and tilting the cardboard defines a different plane in space. + In fact, we can define a plane by: 1) the location of P in space, + and 2) the direction of the nail. +

    + + + +

    + The previous section showed that one can define a line given a point on the line and the direction of the line + (usually given by a vector). + One can make a similar statement about planes: + we can define a plane in space given a point on the plane and the direction the plane faces + (using the description above, + the direction of the nail). + Once again, the direction information will be supplied by a vector, + called a normal vector, + that is orthogonal to the plane. + vectorsnormal vector + normal vector + planesnormal vector +

    + +

    + What exactly does orthogonal to the plane mean? + Choose any two points P and Q in the plane, + and consider the vector \overrightarrow{PQ}. + We say a vector \vec n is orthogonal to the plane if \vec n is perpendicular to \overrightarrow{PQ} for all choices of P and Q; + that is, if \vec n\cdot \overrightarrow{PQ}=0 for all P and Q. +

    + +

    + This gives us way of writing an equation describing the plane. + Let P=(x_0,y_0,z_0) be a point in the plane and let + \vec n = \la a,b,c\ra be a normal vector to the plane. + A point Q = (x,y,z) lies in the plane defined by P and \vec n if, + and only if, \overrightarrow{PQ} is orthogonal to \vec n. + Knowing \overrightarrow{PQ} = \la x-x_0,y-y_0,z-z_0\ra, consider: + + \overrightarrow{PQ}\cdot\vec n \amp = 0 + \la x-x_0,y-y_0,z-z_0\ra\cdot \la a,b,c\ra \amp =0 + a(x-x_0)+b(y-y_0)+c(z-z_0) \amp =0 + . + Equation defines an implicit function describing the plane. More algebra produces: + + ax+by+cz = ax_0+by_0+cz_0 + . + The right hand side is just a number, so we replace it with d: + + ax+by+cz = d + . + As long as c\neq 0, we can solve for z: + + z = \frac1c(d-ax-by) + . +

    + +

    + Equation is especially useful as many computer programs can graph functions in this form. + Equations and have specific names, + given next. +

    + + + Equations of a Plane in Standard and General Forms + +

    + The plane passing through the point P=(x_0,y_0,z_0) with normal vector + \vec n=\la a,b,c\ra can be described by an equation with + standard form + + a(x-x_0)+b(y-y_0)+c(z-z_0) =0; + + the equation's general form + is planesequations of + + ax+by+cz = d + . +

    +     +     + + + +

    + A key to remember throughout this section is this: + to find the equation of a plane, + we need a point and a normal vector. + We will give several examples of finding the equation of a plane, + and in each one different types of information are given. + In each case, + we need to use the given information to find a point on the plane and a normal vector. +

    + + + Finding the equation of a plane + +

    + Write the equation of the plane that passes through the points P=(1,1,0), + Q = (1,2,-1) and R = (0,1,2) in standard form. +

    +     + +

    + We need a vector \vec n that is orthogonal to the plane. + Since P, Q and R are in the plane, + so are the vectors \overrightarrow{PQ} and \overrightarrow{PR}; + \overrightarrow{PQ}\times\overrightarrow{PR} is orthogonal to \overrightarrow{PQ} and \overrightarrow{PR} and hence the plane itself. +

    + +

    + It is straightforward to compute \vec n = \overrightarrow{PQ}\times\overrightarrow{PR} = \la 2,1,1\ra. + We can use any point we wish in the plane + (any of P, Q or R will do) + and we arbitrarily choose P. + Following , + the equation of the plane in standard form is + + 2(x-1) + (y-1)+z = 0 + . +

    + +

    + The plane is sketched in . +

    + +
    + Sketching the plane in + + + + Three points P, Q, and R are plotted in space, along with the plane containing them. + +

    + In a three-dimensional coordinate system, three points P, Q, and R are plotted. + Three vectors are drawn with their tails at P: + \overrightarrow{PQ}, \overrightarrow{PR}, and the cross product \overrightarrow{PQ}\times\overrightarrow{PR}. + The plane containing P, Q, and R is also shown. + The vectors \overrightarrow{PQ} and \overrightarrow{PR} are parallel to the plane, + while their cross product is perpendicular to the plane. +

    +     + + + + + //ASY file for figlines_dist2.asy in Chapter 10 + //ASY file for figplanes1.asy in Chapter 10 + + //size(200,200,IgnoreAspect); + size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((1.92,19.3,4),(0,-0.015,0.06),(0,0,0),1); + defaultrender.merge=true; + + + // setup and draw the axes + real[] myxchoice={-4,4}; + real[] myychoice={-4,4}; + real[] myzchoice={-4,4}; + defaultpen(0.5mm); + pair xbounds=(-3.5,3.5); + pair ybounds=(-4.5,4.5); + pair zbounds=(-4.5,6.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0),S); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane + triple f(pair t) { + return (t.x,t.y,-2*t.x-t.y+3); + } + surface s=surface(f,(-1,-2),(2,3),8,8,Spline); + pen p=apexmeshpen+.2mm; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + //Draw points P=(1,1,0), Q=(1,2,-1), R=(0,1,2) + dotfactor=3; + dot((1,1,0));label("$P$",(1,1,0),N); + dot((1,2,-1));label("$Q$",(1,2,-1),W); + dot((0,1,2));label("$R$",(0,1,2),W); + //Draw Vectors + draw((1,1,0)--(1,2,-1),bluepen,Arrow3(size=2mm));//PQ + draw((1,1,0)--(0,1,2),bluepen,Arrow3(size=2mm));//PR + draw((1,1,0)--(3,2,1),bluepen,Arrow3(size=2mm));//P to PQxPR + label("$\overrightarrow{PQ}\times \overrightarrow{PR}$",(3,2,1),N); + + //test i,j,k moved to P + //draw((1,1,0)--(2,1,0),bluepen,Arrow3(size=2mm));//P by i + //draw((1,1,0)--(1,2,0),bluepen,Arrow3(size=2mm));//P by j + //draw((1,1,0)--(1,1,1),bluepen,Arrow3(size=2mm));//P by k + + // Use lines L1 1+3t,2-t,t and L2 -2+4t,3+t,5+2t + //draw P1 (t=0) and vector d1 at P1 (t=1) + //dotfactor=3;dot((1,2,0));label("$P_1$",(1,2,0),N); + //draw((1,2,0)--(4,1,1),redpen,Arrow3(size=2mm)); + //label("$\vec{d}_1$",(4,1,1),N); + //draw P2 (t=-1.5) and vector d2 at P2 (t=-0.5) + //dotfactor=3;dot((-8,1.5,2));label("$P_2$",(-8,1.5,2),N); + //draw((-8,1.5,2)--(-4,2.5,4),redpen,Arrow3(size=2mm)); + //label("$\vec{d}_2$",(-4,2.5,4),N); + + //draw vector P1 to P2 + //draw((1,2,0)--(-8,1.5,2),black,Arrow3(size=2mm)); + //label("$\overrightarrow{P_1 P_2}$",(-3.5,1.75,1),W); + + //draw the lines 1+3t,2-t,t and -2+4t,3+t,5+2t + //draw((-5,4,-2)--(7,0,2),bluepen);//L1 + //draw((-14,0,-1)--(2,4,7),bluepen);//L2 + + + + +
    +     + +     + +

    + We have just demonstrated the fact that any three non-collinear points define a plane. + (This is why a three-legged stool does not rock; + it's three feet always lie in a plane. + A four-legged stool will rock unless all four feet lie in the same plane.) +

    + + + Finding the equation of a plane + +

    + Verify that lines \ell_1 and \ell_2, + whose parametric equations are given below, intersect, + then give the equation of the plane that contains these two lines in general form. + + \ell_1: \begin{matrix} x\amp =\amp -5+2s \\ y\amp =\amp 1+s \\ z\amp =\amp -4+2s \end{matrix} \qquad\qquad \ell_2: \begin{matrix} x \amp =\amp 2+3t\\ y\amp =\amp 1-2t \\ z\amp =\amp 1+t \end{matrix} + +

    +     + +

    + The lines clearly are not parallel. + If they do not intersect, they are skew, + meaning there is not a plane that contains them both. + If they do intersect, there is such a plane. +

    + +

    + To find their point of intersection, we set the x, + y and z equations equal to each other and solve for s and t: + + \begin{matrix} -5+2s \amp =\amp 2+3t \\ 1+s \amp =\amp 1-2t \\ -4+2s \amp =\amp 1+t \end{matrix} \Rightarrow s=2, t=-1 + . +

    + +

    + When s=2 and t=-1, + the lines intersect at the point P= (-1,3,0). +

    + +

    + Let \vec d_1 = \la 2,1,2\ra and + \vec d_2=\la 3,-2,1\ra be the directions of lines \ell_1 and \ell_2, + respectively. + A normal vector to the plane containing these the two lines will also be orthogonal to \vec d_1 and \vec d_2. + Thus we find a normal vector \vec n by computing \vec n = \vec d_1 \times \vec d_2= \la 5,4-7\ra. +

    + +

    + We can pick any point in the plane with which to write our equation; + each line gives us infinite choices of points. + We choose P, the point of intersection. + We follow + to write the plane's equation in general form: + + 5(x+1) +4(y-3) -7z \amp = 0 + 5x + 5 + 4y-12 -7z \amp = 0 + 5x+4y-7z \amp = 7 + . +

    + +

    + The plane's equation in general form is 5x+4y-7z=7; + it is sketched in . +

    + +
    + Sketching the plane in + + + + In three dimensions, two lines are plotted, intersecting at a point P, along with the plane containing both lines. + +

    + Three-dimensional coordinate axes are shown, with the points (5,0,0), (0,5,0), and (0,0,-5) indicated for scale. + A point P is plotted in space. Two lines \ell_1 and \ell_2 are shown intersecting at this point. + The plane containing both lines is also plotted, along with a normal vector, located with its tail at P. +

    +     + + + + + //ASY file for figplanes2.asy in Chapter 10 + + size(282,282,Aspect); + //size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((-16,18.6,6.76),(0.014,-0.015,0077),(0,0,0),.99); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={5}; + real[] myychoice={5}; + real[] myzchoice={-5}; + defaultpen(0.5mm); + pair xbounds=(-4,6); + pair ybounds=(-1,6); + pair zbounds=(-6,5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane + triple f(pair t) { + return (t.x,t.y,(5/7)*t.x+(4/7)*t.y-1); + } + surface s=surface(f,(-5,-6),(5,6),8,8,Spline); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + //Draw points P=(-1,3,0) + dotfactor=3;dot((-1,3,0));label("$P$",(-1,3,0),E); + + //Draw normal vector at P, n=(5,4,-7) + draw((-1,3,0)--(4,7,-7),black,Arrow3(size=2mm));//n at P + label("$\vec{n}$",(4,7,-7),W); + + //draw the lines -5+2t,1+t,-4+2t and 2+3t,1-2t,1+t + draw((5,6,6)--(-5,1,-4),bluepen);label("$\ell_1$",(5,6,6),N);//L1 + draw((5,-1,2)--(-4.9,5.6,-1.3),bluepen);label("$\ell_2$",(5,-1,2),N);//L2 + + + + +
    +     +     + + + Finding the equation of a plane + +

    + Give the equation, in standard form, + of the plane that passes through the point + P=(-1,0,1) and is orthogonal to the line with vector equation \vec \ell(t) = \la -1,0,1\ra + t\la 1,2,2\ra. +

    +     + +

    + As the plane is to be orthogonal to the line, + the plane must be orthogonal to the direction of the line given by \vec d = \la 1,2,2\ra. + We use this as our normal vector. + Thus the plane's equation, in standard form, is + + (x+1) +2y+2(z-1)=0 + . +

    + +

    + The line and plane are sketched in . +

    + +
    + The line and plane in + + + + A plane is plotted in a three-dimensional coordinate system, along with a line passing through it at a point P. + +

    + A plane is plotted against a set of three-dimensional coordinate axes. + A line passes through the plane, intersecting it at a point P. + The line is perpendicular to the plane. +

    +     + + + + + //ASY file for figplanes3.asy in Chapter 10 + + size(282,282,Aspect); + //size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-3,3}; + real[] myychoice={-3,3}; + real[] myzchoice={-3,3}; + defaultpen(0.5mm); + pair xbounds=(-4,4); + pair ybounds=(-4,4); + pair zbounds=(-4,4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane z=-(x+1)/2-y+1 + triple f(pair t) { + return (t.x,t.y,(-1/2)*(t.x+1)-t.y+1); + } + surface s=surface(f,(-3,-3),(3,3),8,8,Spline); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + //Draw points P=(-1,0,1) + dotfactor=3;dot((-1,0,1));label("$P$",(-1,0,1),N); + + //draw the line -1+t,2t,1+2t + draw((1,4,5)--(-3,-4,-3),bluepen);//L t=2 to t=-2 + + //Draw normal vector at P, n=(1,2,2) + //draw((-1,3,0)--(4,7,-7),black,Arrow3(size=2mm));//n at P + //label("$\vec{n}$",(4,7,-7),W); + + + + +
    +     + +     + + + Finding the intersection of two planes + +

    + Give the parametric equations of the line that is the intersection of the planes p_1 and p_2, where: + + p_1: x-(y-2)+(z-1) =0 + p_2: -2(x-2)+(y+1)+(z-3)=0 + +

    +     + +

    + To find an equation of a line, + we need a point on the line and the direction of the line. +

    + +

    + We can find a point on the line by solving each equation of the planes for z: + + p_1: z = -x+y-1 + p_2: z = 2x-y-2 + +

    + +

    + We can now set these two equations equal to each other (, we are finding values of x and y where the planes have the same z value): + + -x+y-1 \amp = 2x-y-2 + 2y \amp = 3x-1 + y \amp = \frac12(3x-1) + +

    + +

    + We can choose any value for x; + we choose x=1. + This determines that y=1. + We can now use the equations of either plane to find z: + when x=1 and y=1, + z=-1 on both planes. + We have found a point P on the line: P= (1,1,-1). +

    + +

    + We now need the direction of the line. + Since the line lies in each plane, + its direction is orthogonal to a normal vector for each plane. + Considering the equations for p_1 and p_2, + we can quickly determine their normal vectors. + For p_1, + \vec n_1 = \la 1,-1,1\ra and for p_2, + \vec n_2 = \la -2,1,1\ra. + A direction orthogonal to both of these directions is their cross product: + \vec d = \vec n_1\times \vec n_2 = \la -2,-3,-1\ra. +

    + +

    + The parametric equations of the line through + P=(1,1,-1) in the direction of d=\la -2,-3,-1\ra is: + + \ell: x= -2t+1 y = -3t+1 z=-t-1 + . +

    + +

    + The planes and line are graphed in . +

    + +
    + Graphing the planes and their line of intersection in + + + + Two planes are shown intersecting along a common line. + +

    + Two planes are plotted in a three-dimensional coordinate system. + The planes intersect along a line, which is also plotted, + along with a point P that lies on the line, as well as on both planes. +

    +     + + + + + //ASY file for figplanes43D.asy in Chapter 10 + + //size(282,282,Aspect); + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8,14.7,14); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={-5}; + defaultpen(0.5mm); + pair xbounds=(-3,3); + pair ybounds=(-3,3); + pair zbounds=(-6,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the planes z=-x+y-1 and z=2x-y-2 + triple f(pair t) { + return (t.x,t.y,-t.x+t.y-1); + } + surface s=surface(f,(-3,-3),(3,3),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p+.2mm,nolight,render(merge=true)); + triple f(pair t) { + return (t.x,t.y,2*t.x-t.y-2); + } + surface s=surface(f,(-3,-3),(3,3),8,8,Spline); + pen pp=apexmeshpen+.2mm; + //draw(s,rgb(1,.4,.7)+opacity(.7),meshpen=pp,nolight,render(merge=true)); // HOT PINK + + pen q=redcurvepen+.1mm; + draw(s,surfacepen2,meshpen=q,nolight,render(merge=true)); // better red + + //Draw point P=(1,1,-1) + dotfactor=3;dot((1,1,-1));label("$P$",(1,1,-1),N); + + //draw the line 1-2t,1-3t,-1-t + draw((4,5.5,0.5)--(-2,-3.5,-2.5),bluepen);//L t=-1.5 to t=1.5 + + + + +
    + +     + +     + + + Finding the intersection of a plane and a line + +

    + Find the point of intersection, if any, + of the line \ell(t) = \la 3,-3,-1\ra +t\la-1,2,1\ra and the plane with equation in general form 2x+y+z=4. +

    +     + +

    + The equation of the plane shows that the vector + \vec n = \la 2,1,1\ra is a normal vector to the plane, + and the equation of the line shows that the line moves parallel to \vec d = \la -1,2,1\ra. + Since these are not orthogonal, + we know there is a point of intersection. + (If there were orthogonal, + it would mean that the plane and line were parallel to each other, + either never intersecting or the line was in the plane itself.) +

    + +

    + To find the point of intersection, + we need to find a t value such that \ell(t) satisfies the equation of the plane. + Rewriting the equation of the line with parametric equations will help: + + \ell(t) = \left\{\begin{aligned}x\amp = 3-t\\ y\amp =-3+2t\\ z\amp = -1+t \end{aligned} \right. + . +

    + +

    + Replacing x, + y and z in the equation of the plane with the expressions containing t found in the equation of the line allows us to determine a t value that indicates the point of intersection: + + 2x+y+z \amp =4 + 2(3-t) + (-3+2t) + (-1+t) \amp = 4 + t\amp =2 + . +

    + +

    + When t=2, + the point on the line satisfies the equation of the plane; + that point is \ell(2) = \la 1,1,1\ra. + Thus the point (1,1,1) is the point of intersection between the plane and the line, + illustrated in . +

    + +
    + Illustrating the intersection of a line and a plane in + + + + A line in three dimensions passes through a plane, intersecting it at a point. + +

    + A plane is plotted against a three-dimensional coordinate system. + A line, labeled \ell(t), is also plotted. + The line appears to be almost, but not quite, parallel to the plane. + It passes through the plane, intersecting it at a point that is plotted, but not labeled. +

    +     + + + + + //ASY file for figplanes53D.asy in Chapter 10 + + //size(282,282,Aspect); + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-3,4); + pair ybounds=(-3,4); + pair zbounds=(-2,7); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane z=4-2x-y + triple f(pair t) { + return (t.x,t.y,4-2*t.x-t.y); + } + surface s=surface(f,(-2,-2),(3,3),8,8,Spline); + pen p=apexmeshpen+.2mm; + draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + //Draw point P=(1,1,1) + dotfactor=3;dot((1,1,1),redpen);//label("$P$",(1,1,1),N); + + //draw the line 3-t,-3+2t,-1+t + draw((3,-3,-1)--(0,3,2),redpen);//L t=0 to t=3 + label("$\ell(t)$",(3,-3,-1),N); + + + + +
    +     + +     +     + + + Distances +

    + Just as it was useful to find distances between points and lines in the previous section, + it is also often necessary to find the distance from a point to a plane. +

    + +

    + Consider , + where a plane with normal vector \vec n is sketched containing a point P and a point Q, + not on the plane, is given. + We measure the distance from Q to the plane by measuring the length of the projection of \overrightarrow{PQ} onto \vec n. + That is, we want: + + \snorm{\text{ proj } _{\,\vec n}\,{\overrightarrow{PQ}}} = \snorm{\frac{\vec n\cdot \overrightarrow{PQ}}{\vnorm n^2}\vec n} = \frac{\abs{\vec n\cdot \overrightarrow{PQ}}}{\vnorm n} + +

    + + + +

    + Equation is important as it does more than just give the distance between a point and a plane. + We will see how it allows us to find several other distances as well: + the distance between parallel planes and the distance from a line and a plane. + Because Equation is important, + we restate it as a Key Idea. +

    + +
    + Illustrating finding the distance from a point to a plane + + + + A plane, on which a point P and normal vector n are marked, along with a point Q not on the plane, and its distance from the plane. + +

    + A generic plane is shown, along with a point P on the plane. + The normal vector \vec n is plotted with its tail at P. + A point Q is shown above the plane, and the vector \overrightarrow{PQ} from P to Q is drawn. + A perpendicular is drawn from the point Q to the plane, and labeled with the distance h. +

    +     + + + + + //ASY file for figplanes_dist3D.asy in Chapter 10 + + size(282,282,Aspect); + //size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4.5,4.5,1.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-4,4}; + real[] myychoice={-4,4}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-5,5); + pair ybounds=(-5,5); + pair zbounds=(-5,5); + + //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane xy-plane + triple f(pair t) { + return (t.x,t.y,0); + } + surface s=surface(f,(-2,-1),(2,1.5),1,1,Spline); + pen p=blackmeshpen; + //pen p=apexmeshpen; + draw(s,rgb(1,1,1)+opacity(0),meshpen=p+thick(),nolight,render(merge=true)); + //draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); + + //Draw origin point and P=(1,1,1) + dotfactor=3;dot((0,0,0),black);dot((0,1,0),black);label("$P$",(0,1,0),E); + + //Draw the normal at P + draw((0,1,0)--(0,1,1),linewidth(.75),Arrow3(size=2mm));label("$\vec{n}$",(0,1,1),E); + + //Draw the dashed line from origin to Q=(0,0,2) with label h + draw((0,0,0)--(0,0,2),dashed+linewidth(.75));dot((0,0,2),black);label("$Q$",(0,0,2),W); + label("$h$",(0,0,1),W); + + //Draw the vector PQ stopping just short of Q + draw((0,1,0)--(0,.05,1.95),linewidth(.75),Arrow3(size=2mm));//label("$\vec{n}$",(0,1,1),E); + + //Draw the perpendicular symbol as two small lines + draw((0,0,0.2)--(0,0.2,0.2),linewidth(.75)); + draw((0,0.2,0)--(0,0.2,0.2),linewidth(.75)); + + + + +
    + + + Distance from a Point to a Plane +

    + Let a plane with normal vector \vec n be given, + and let Q be a point. + The distance h from Q to the plane is + + h = \frac{\abs{\vec n\cdot \overrightarrow{PQ}}}{\vnorm n} + , + where P is any point in the plane. + planesdistance between point and plane + distancebetween point and plane +

    +     + + + + + Distance between a point and a plane + +

    + Find the distance between the point + Q = (2,1,4) and the plane with equation 2x-5y+6z=9. +

    +     + +

    + Using the equation of the plane, + we find the normal vector \vec n = \la 2,-5,6\ra. + To find a point on the plane, + we can let x and y be anything we choose, + then let z be whatever satisfies the equation. + Letting x and y be 0 seems simple; + this makes z = 1.5. + Thus we let P = \la 0,0,1.5\ra, + and \overrightarrow{PQ} = \la 2,1,2.5\ra. +

    + +

    + The distance h from Q to the plane is given by : + + h \amp = \frac{\abs{\vec n\cdot \overrightarrow{PQ}}}{\vnorm n} + \amp = \frac{\abs{\la 2,-5,6\ra \cdot \la 2,1,2.5\ra}}{\norm{\la 2,-5,6\ra}} + \amp = \frac{ \abs{14}}{\sqrt{65}} + \amp \approx 1.74 + . +

    +     + +     + +

    + We can use to find other distances. + Given two parallel planes, + we can find the distance between these planes by letting P be a point on one plane and Q a point on the other. + If \ell is a line parallel to a plane, + we can use the Key Idea to find the distance between them as well: + again, let P be a point in the plane and let Q be any point on the line. + (One can also use .) + The Exercise section contains problems of these types. +

    + +

    + These past two sections have not explored lines and planes in space as an exercise of mathematical curiosity. + However, there are many, many applications of these fundamental concepts. + Complex shapes can be modeled (or, + approximated) using planes. + For instance, + part of the exterior of an aircraft may have a complex, + yet smooth, shape, + and engineers will want to know how air flows across this piece as well as how heat might build up due to air friction. + Many equations that help determine air flow and heat dissipation are difficult to apply to arbitrary surfaces, + but simple to apply to planes. + By approximating a surface with millions of small planes one can more readily model the needed behavior. +

    +     + + + + Terms and Concepts + + + +

    + In order to find the equation of a plane, + what two pieces of information must one have? +

    +     + + + +

    + A point in the plane and a normal vector (, a direction orthogonal to the plane). +

    +     + +     + + + + +

    + What is the relationship between a plane and one of its normal vectors? +

    +     + + + +

    + A normal vector is orthogonal to the plane. +

    +     + +     +     + + Problems + + + +

    + Give any two points in the given plane. +

    +     + + + + +

    + 2x-4y+7z=2 +

    +     + +

    + Answers will vary. +

    +     + +     + + + + + Context("Point"); + $points=List("(-2,9,0),(2,9,3)"); + $pev=$points->cmp(list_checker => sub { + my ($correct,$student,$ansHash,$value) = @_; + my $n = scalar(@$student); + my $score = 0; + my @errors = (); + my $i, $j; + for ($i = 0; $i < $n; $i++) { + my $ith = Value::List->NameForNumber($i+1); + my $p = $student->[$i]; # i-th student answer + if ($p->type ne "Point") { + push(@errors,"Your $ith entry is not a point"); + next; + } + for ($j = 0, $used = 0; $j < $i; $j++) { + if ($student->[$j]->type eq "Point" and $student->[$j] == $p) { + push(@errors,"Your $ith point is the same as a previous one") unless $ansHash->{isPreview}; + $used = 1; last; + } + } + if (!$used) { + my ($a,$b,$c) = $p->value; + if (3*($a+2)+5*($b-9)-4*$c==0) {$score++} else { + push(@errors,"Your $ith point is not correct") unless $ansHash->{isPreview} + } + } + } + if (!$ansHash->{isPreview}) { + push(@errors,"You need to provide more points") if $i < 2; + push(@errors,"You have given too many points") if $score > 2 and $i != $score; + } + return ($score,@errors); + }); + + + +

    + List any two points in the plane with equation 3(x+2)+5(y-9)-4z=0. +

    + +

    + +

    +     +     +     + + + + +

    + x=2 +

    +     + +

    + Answers will vary. +

    +     + +     + + + + + Context("Point"); + $points=List("(0,-2,6),(1,-2,6)"); + $pev=$points->cmp(list_checker => sub { + my ($correct,$student,$ansHash,$value) = @_; + my $n = scalar(@$student); + my $score = 0; + my @errors = (); + my $i, $j; + for ($i = 0; $i < $n; $i++) { + my $ith = Value::List->NameForNumber($i+1); + my $p = $student->[$i]; # i-th student answer + if ($p->type ne "Point") { + push(@errors,"Your $ith entry is not a point"); + next; + } + for ($j = 0, $used = 0; $j < $i; $j++) { + if ($student->[$j]->type eq "Point" and $student->[$j] == $p) { + push(@errors,"Your $ith point is the same as a previous one") unless $ansHash->{isPreview}; + $used = 1; last; + } + } + if (!$used) { + my ($a,$b,$c) = $p->value; + if (4*($b+2)-($c-6)==0) {$score++} else { + push(@errors,"Your $ith point is not correct") unless $ansHash->{isPreview} + } + } + } + if (!$ansHash->{isPreview}) { + push(@errors,"You need to provide more points") if $i < 2; + push(@errors,"You have given too many points") if $score > 2 and $i != $score; + } + return ($score,@errors); + }); + + + +

    + List any two points in the plane with equation 4(y+2)-(z-6)=0. +

    + +

    + +

    +     +     +     + +     + + + +

    + Give the equation of the described plane in standard and general forms. +

    +     + + + + +

    + Passes through (2,3,4) and has normal vector +

    + +

    + \vec n= \la 3,-1,7\ra. +

    +     + +

    + Standard form: 3(x-2)-(y-3)+7(z-4)=0 +

    + +

    + general form: 3x-y+7z=31 +

    +     + +     + + + + + sub grouped { + my $op = shift; + return 1 if ($op->{name} or $op->{isConstant}); + if ($op->{uop}) {return grouped($op->{op});}; + my $context=Context(); + my $lft=$context->Package("Formula")->new($context,$op->{lop}); + my $rgt=$context->Package("Formula")->new($context,$op->{rop}); + return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); + if ($op->{bop} eq '+' or $op->{bop} eq '-') { + return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); + }; + return (grouped($op->{lop}) and grouped($op->{rop})); + }; + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); + $st=ImplicitEquation("2(y-3)+4(z-5)=0"); + $stev=$st->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + #return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); + return $correct == $student; + }); + $ge=ImplicitEquation("2y+4z=26"); + $geev=$ge->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); + Value::Error("Your answer is not in general form") unless $right->isConstant; + return $correct == $student; + }); + + + +

    + A plane passes through (1,3,5) and has normal vector \vec n= \la 0,2,4\ra. +

    + + + An equation for this plane in standard form is: + +

    + +

    + + + An equation for this plane in general form is: + +

    + +

    +     +     +     + + + + +

    + Passes through the points (1,2,3), + (3,-1,4) and (1,0,1). +

    +     + +

    + Answers may vary; +

    + +

    + Standard form: 8(x-1)+4(y-2)-4(z-3)=0 +

    + +

    + general form: 8x+4y-4z=4 +

    +     + +     + + + + + + sub grouped { + my $op = shift; + return 1 if ($op->{name} or $op->{isConstant}); + if ($op->{uop}) {return grouped($op->{op});}; + my $context=Context(); + my $lft=$context->Package("Formula")->new($context,$op->{lop}); + my $rgt=$context->Package("Formula")->new($context,$op->{rop}); + return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); + if ($op->{bop} eq '+' or $op->{bop} eq '-') { + return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); + }; + return (grouped($op->{lop}) and grouped($op->{rop})); + }; + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); + $st=ImplicitEquation("-5(x-5)+3(y-3)+2(z-8)=0"); + $stev=$st->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + #return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); + return $correct == $student; + }); + $ge=ImplicitEquation("-5x+3y+2z=0"); + $geev=$ge->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); + Value::Error("Your answer is not in general form") unless $right->isConstant; + return $correct == $student; + }); + + + +

    + A plane passes through the points (5,3,8), + (6,4,9) and (3,3,3). +

    + + + An equation for this plane in standard form is: + +

    + +

    + + + An equation for this plane in general form is: + +

    + +

    +     +     +     + + + + +

    + Contains the intersecting lines +

    + +

    + \vec\ell_1(t) = \la 2,1,2\ra + t\la 1,2,3\ra and +

    + +

    + \vec\ell_2(t) = \la 2,1,2\ra + t\la 2,5,4\ra. +

    +     + +

    + Answers may vary; +

    + +

    + Standard form: -7(x-2)+2(y-1)+(z-2)=0 +

    + +

    + general form: -7x+2y+z=-10 +

    +     + +     + + + + + + sub grouped { + my $op = shift; + return 1 if ($op->{name} or $op->{isConstant}); + if ($op->{uop}) {return grouped($op->{op});}; + my $context=Context(); + my $lft=$context->Package("Formula")->new($context,$op->{lop}); + my $rgt=$context->Package("Formula")->new($context,$op->{rop}); + return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); + if ($op->{bop} eq '+' or $op->{bop} eq '-') { + return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); + }; + return (grouped($op->{lop}) and grouped($op->{rop})); + }; + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); + $st=ImplicitEquation("3(x-5)+3(z-3)=0"); + $stev=$st->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + #return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); + return $correct == $student; + }); + $ge=ImplicitEquation("3x+3z=24"); + $geev=$ge->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); + Value::Error("Your answer is not in general form") unless $right->isConstant; + return $correct == $student; + }); + + + +

    + A plane contains the intersecting lines + \vec\ell_1(t) = \la 5,0,3\ra + t\la -1,1,1\ra and \vec\ell_2(t) = \la 1,4,7\ra + t\la 3,0,-3\ra. +

    + + + An equation for this plane in standard form is: + +

    + +

    + + + An equation for this plane in general form is: + +

    + +

    +     +     +     + + + + +

    + Contains the parallel lines +

    + +

    + \vec\ell_1(t) = \la 1,1,1\ra + t\la 1,2,3\ra and +

    + +

    + \vec\ell_2(t) = \la 1,1,2\ra + t\la 1,2,3\ra. +

    +     + +

    + Answers may vary; +

    + +

    + Standard form: 2(x-1)-(y-1)=0 +

    + +

    + general form: 2x-y=1 +

    +     + +     + + + + + + sub grouped { + my $op = shift; + return 1 if ($op->{name} or $op->{isConstant}); + if ($op->{uop}) {return grouped($op->{op});}; + my $context=Context(); + my $lft=$context->Package("Formula")->new($context,$op->{lop}); + my $rgt=$context->Package("Formula")->new($context,$op->{rop}); + return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); + if ($op->{bop} eq '+' or $op->{bop} eq '-') { + return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); + }; + return (grouped($op->{lop}) and grouped($op->{rop})); + }; + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); + $st=ImplicitEquation("2(x-1)+(y-1)-3(z-1)=0"); + $stev=$st->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + #return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); + return $correct == $student; + }); + $ge=ImplicitEquation("2x+y-3z=0"); + $geev=$ge->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); + Value::Error("Your answer is not in general form") unless $right->isConstant; + return $correct == $student; + }); + + + +

    + A plane contains the parallel lines + \vec\ell_1(t) = \la 1,1,1\ra + t\la 4,1,3\ra and \vec\ell_2(t) = \la 2,2,2\ra + t\la 4,1,3\ra. +

    + + + An equation for this plane in standard form is: + +

    + +

    + + + An equation for this plane in general form is: + +

    + +

    +     +     +     + + + + +

    + Contains the point (2,-6,1) and the line +

    + +

    + \vec\ell(t) = \left\{\begin{aligned}x\amp =2+5t \\ + y\amp =2+2t \\ + z\amp =-1+2t + \end{aligned} \right. +

    +     + +

    + Answers may vary; +

    + +

    + Standard form: 2(x-2)-(y+6)-4(z-1)=0 +

    + +

    + general form: 2x-y-4z=6 +

    +     + +     + + + + + + sub grouped { + my $op = shift; + return 1 if ($op->{name} or $op->{isConstant}); + if ($op->{uop}) {return grouped($op->{op});}; + my $context=Context(); + my $lft=$context->Package("Formula")->new($context,$op->{lop}); + my $rgt=$context->Package("Formula")->new($context,$op->{rop}); + return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); + if ($op->{bop} eq '+' or $op->{bop} eq '-') { + return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); + }; + return (grouped($op->{lop}) and grouped($op->{rop})); + }; + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); + $st=ImplicitEquation("4(x-5)-2(y-7)-2(z-3)=0"); + $stev=$st->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + #return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); + return $correct == $student; + }); + $ge=ImplicitEquation("4x-2y-2z=0"); + $geev=$ge->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); + Value::Error("Your answer is not in general form") unless $right->isConstant; + return $correct == $student; + }); + + + +

    + A plane contains the point (5,7,3) and the line \vec\ell(t) = \begin{cases}x\amp=t\\y\amp=t\\z\amp=t\end{cases}. +

    + + + An equation for this plane in standard form is: + +

    + +

    + + + An equation for this plane in general form is: + +

    + +

    +     +     +     + + + + + + sub grouped { + my $op = shift; + return 1 if ($op->{name} or $op->{isConstant}); + if ($op->{uop}) {return grouped($op->{op});}; + my $context=Context(); + my $lft=$context->Package("Formula")->new($context,$op->{lop}); + my $rgt=$context->Package("Formula")->new($context,$op->{rop}); + return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); + if ($op->{bop} eq '+' or $op->{bop} eq '-') { + return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); + }; + return (grouped($op->{lop}) and grouped($op->{rop})); + }; + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); + $st=ImplicitEquation("(x-5)+(y-7)+(z-3)=0"); + $stev=$st->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + #return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); + return $correct == $student; + }); + $ge=ImplicitEquation("x+y+z=15"); + $geev=$ge->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); + Value::Error("Your answer is not in general form") unless $right->isConstant; + return $correct == $student; + }); + + + +

    + A plane contains the point (5,7,3) and is orthogonal to the line \vec\ell(t) = \la 4,5,6\ra+ t\la 1,1,1\ra. +

    + + + An equation for this plane in standard form is: + +

    + +

    + + + An equation for this plane in general form is: + +

    + +

    +     +     +     + + + + + + sub grouped { + my $op = shift; + return 1 if ($op->{name} or $op->{isConstant}); + if ($op->{uop}) {return grouped($op->{op});}; + my $context=Context(); + my $lft=$context->Package("Formula")->new($context,$op->{lop}); + my $rgt=$context->Package("Formula")->new($context,$op->{rop}); + return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); + if ($op->{bop} eq '+' or $op->{bop} eq '-') { + return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); + }; + return (grouped($op->{lop}) and grouped($op->{rop})); + }; + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); + $st=ImplicitEquation("4(x-4)+(y-1)+(z-1)=0"); + $stev=$st->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + #return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); + return $correct == $student; + }); + $ge=ImplicitEquation("4x+y+z=18"); + $geev=$ge->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); + Value::Error("Your answer is not in general form") unless $right->isConstant; + return $correct == $student; + }); + + + +

    + A plane contains the point (4,1,1) and is orthogonal to the line \begin{cases}x\amp=4+4t\\y\amp=1+t\\z\amp=1+t\end{cases}. +

    + + + An equation for this plane in standard form is: + +

    + +

    + + + An equation for this plane in general form is: + +

    + +

    +     +     +     + + + + + + sub grouped { + my $op = shift; + return 1 if ($op->{name} or $op->{isConstant}); + if ($op->{uop}) {return grouped($op->{op});}; + my $context=Context(); + my $lft=$context->Package("Formula")->new($context,$op->{lop}); + my $rgt=$context->Package("Formula")->new($context,$op->{rop}); + return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); + if ($op->{bop} eq '+' or $op->{bop} eq '-') { + return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); + }; + return (grouped($op->{lop}) and grouped($op->{rop})); + }; + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); + $st=ImplicitEquation("3(x+4)+8(y-7)-10(z-2)=0"); + $stev=$st->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + #return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); + return $correct == $student; + }); + $ge=ImplicitEquation("3x+8y-10z=24"); + $geev=$ge->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); + Value::Error("Your answer is not in general form") unless $right->isConstant; + return $correct == $student; + }); + + + +

    + A plane contains the point + (-4,7,2) and is parallel to the plane 3(x-2)+8(y+1) -10z=0. +

    + + + An equation for this plane in standard form is: + +

    + +

    + + + An equation for this plane in general form is: + +

    + +

    +     +     +     + + + + + + sub grouped { + my $op = shift; + return 1 if ($op->{name} or $op->{isConstant}); + if ($op->{uop}) {return grouped($op->{op});}; + my $context=Context(); + my $lft=$context->Package("Formula")->new($context,$op->{lop}); + my $rgt=$context->Package("Formula")->new($context,$op->{rop}); + return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); + if ($op->{bop} eq '+' or $op->{bop} eq '-') { + return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); + }; + return (grouped($op->{lop}) and grouped($op->{rop})); + }; + Context("ImplicitEquation"); + Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); + Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); + $st=ImplicitEquation("x-1=0"); + $stev=$st->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + #return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); + return $correct == $student; + }); + $ge=ImplicitEquation("x=1"); + $geev=$ge->cmp(checker=>sub{ + my ( $correct, $student, $ansHash ) = @_; + return 0 if $ansHash->{isPreview} || $correct != $student; + my $context = Context(); + my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); + my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); + Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); + Value::Error("Your answer is not in general form") unless $right->isConstant; + return $correct == $student; + }); + + + +

    + A plane contains the point (1,2,3) and is parallel to the plane x=5. +

    + + + An equation for this plane in standard form is: + +

    + +

    + + + An equation for this plane in general form is: + +

    + +

    +     +     +     + +     + + + +

    + Give the equation of the line that is the intersection of the given planes. +

    +     + + + + +

    + p1:\ 3 (x - 2) + (y - 1) + 4 z=0, and +

    + +

    + p2:\ 2 (x - 1) - 2 (y + 3) + 6 (z - 1)=0. +

    +     + +

    + Answers may vary: +

    + +

    + \ell = \left\{\begin{aligned}x \amp = 14t\\ + y \amp = -1-10t\\ + z\amp = 2-8t + \end{aligned} \right. +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(t=>"Real"); + $v=Compute("(1,3,3.5)+t<20,2,-26>"); + $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; + $ds=Formula($st)->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; + $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return ($par and $tch); + }); + + + +

    + Give the equation of the line + (in vector form) + that is the intersection of the planes 5 (x - 5) + 2 (y + 2) + 4 (z - 1)=0, + and 3 x - 4 (y - 1) + 2 (z - 1)=0. +

    + +

    + +

    +     +     +     + +     + + + +

    + Find the point of intersection between the line and the plane. +

    +     + + + + +

    +

      +
    • +

      + line: \la 5,1,-1\ra + t\la 2,2,1\ra +

      +
      • +
    • +

      + plane: 5x-y-z=-3 +

      +
      • +
    +

    +     + +

    + (-3,-7,-5) +

    +     + +     + + + + + Context("Point"); + Context()->strings->add('the entire line'=>{}); + $point=Point("(3,1,1)"); + + + +

    +

      +
    • +

      + line: \la 4,1,0\ra + t\la 1,0,-1\ra +

      +
      • +
    • +

      + plane: 3x+y-2z=8 +

      +
      • +
    +

    + + If the plane contains the line, + enter the phrase the entire line. + If the line does not intersect the plane at all, enter none. + + +

    + +

    +     +     +     + + + + +

    +

      +
    • +

      + line: \la 1,2,3\ra + t\la 3,5,-1\ra +

      +
      • +
    • +

      + plane: 3x-2y-z=4 +

      +
      • +
    +

    +     + +

    + No point of intersection; the plane and line are parallel. +

    +     + +     + + + + + Context("Point"); + Context()->strings->add('the entire line'=>{}); + $point=Compute('the entire line'); + + + +

    +

      +
    • +

      + line: \la 1,2,3\ra + t\la 3,5,-1\ra +

      +
      • +
    • +

      + plane: 3x-2y-z=-4 +

      +
      • +
    +

    + + If the plane contains the line, + enter the phrase the entire line. + If the line does not intersect the plane at all, enter none. + + +

    + +

    +     +     +     + +     + + + +

    + Find the indicated distance. +

    +     + + + + +

    + The distance from the point (1,2,3) to the plane +

    + +

    + 3(x-1)+(y-2)+5(z-2)=0. +

    +     + +

    + \sqrt{5/7} +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $distance=Formula("8/sqrt(21)"); + + + +

    + Find the distance from the point (2,6,2) to the plane 2(x-1)-y+4(z+1)=0. +

    + +

    + +

    +     +     +     + + + + +

    + The distance between the parallel planes +

    + +

    + x+y+z=0 and +

    + +

    + (x-2)+(y-3)+(z+4)=0 +

    +     + +

    + 1/\sqrt{3} +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $distance=Formula("3"); + + + +

    + Find the distance between the parallel planes + 2(x-1)+2(y+1)+(z-2)=0 and 2(x-3)+2(y-1)+(z-3)=0. +

    + +

    + +

    +     +     +     + +     + + + + +

    + Show why if the point Q lies in a plane, + then the distance formula correctly gives the distance from the point to the plane as 0. +

    +     + +

    + If P is any point in the plane, + and Q is also in the plane, + then \overrightarrow{PQ} lies parallel to the plane and is orthogonal to \vec n, + the normal vector. + Thus \vec n\cdot \overrightarrow{PQ}=0, giving the distance as 0. +

    +     + +     + + + + +

    + How is + in + easier to answer once we have an understanding of planes? +

    +     + +

    + The intersecting lines define a plane with normal vector \vec n = \vec c = \vec d_1\times \vec d_2. + Since points P_1 and P_2 lie in the plane, + \vec c is orthogonal to \overrightarrow{P_1P_2}, + hence \overrightarrow{P_1P_2}\cdot\vec c = 0, giving a distance of 0. + Knowing the principles of planes, + especially their normal vectors, makes this simpler. +

    +     + +     +     +     +
    +     + + + Vector Valued Functions + +

    + In the previous chapter, + we learned about vectors and were introduced to the power of vectors within mathematics. + In this chapter, + we'll build on this foundation to define functions whose input is a real number and whose output is a vector. + We'll see how to graph these functions and apply calculus techniques to analyze their behavior. + Most importantly, we'll see why + we are interested in doing this: + we'll see beautiful applications to the study of moving objects. +

    +     + +
    + Vector-Valued Functions + +

    + We are very familiar with real valued functions, that is, + functions whose output is a real number. + This section introduces vector-valued functions + functions whose output is a vector. +

    + + + + + Vector-Valued Functions + +

    + A vector-valued function + is a function of the form + + \vec r(t) = \la\, f(t),g(t)\,\ra \text{ or } \vec r(t) = \la \,f(t),g(t),h(t)\,\ra + , + where f, g and h are real valued functions. +

    + +

    + The domain of \vec r is the set of all values of t for which \vec r(t) is defined. + The range of \vec r is the set of all possible output vectors \vec r(t). + vector-valued functiondefinition + functionvector-valued +

    +     +     +     + + + Evaluating and Graphing Vector-Valued Functions +

    + Evaluating a vector-valued function at a specific value of t is straightforward; + simply evaluate each component function at that value of t. + For instance, if \vec r(t) = \la t^2,t^2+t-1\ra, + then \vec r(-2) = \la 4,1\ra. + We can sketch this vector, + as is done in . + Plotting lots of vectors is cumbersome, though, + so generally we do not sketch the whole vector but just the terminal point. + The graph of a vector-valued function is the set of all terminal points of \vec r(t), + where the initial point of each vector is always the origin. + In we sketch the graph of \vec r; we can indicate individual points on the graph with their respective vector, + as shown. + vector-valued functiongraphing +

    + +
    + Sketching the graph of a vector-valued function + +
    + + + + + Graph of the vector \vec r(-2) =\la 4,1\ra. + This vector starts at the origin and ends at the point (4,1). + + Graph of a vector whose terminal point corresponds to a point on the curve. + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,2,3,4,5}, + ytick={-3,-2,-1,1,2,3}, + ymin=-3.5,ymax=3.5, + xmin=-.5,xmax=5.5 + ] + + \draw [->,thick,secondcolor] (axis cs: 0,0) -- (axis cs:4,1) node [black,pos=.5,above] { $\vec r(-2)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + The image contains the plot of the vector valued function \vec r(t) = \la t^2,t^2+t-1\ra. + The image also contains the graph of the vector \vec r(-2) =\la 4,1\ra, which ends at a point on the function \vec r(t). + The function \vec r(t) resembles a parabola which has been rotated about 45 degrees clockwise. + The function is plotted for t approximately between -2.5 and 1.5. + The function begins near the point (5,2) and then slopes down crossing the x-axis at approximately x=2.5. + The slanted parabola then curves upwards around the point (0,-1), from which point it begins increasing in both x and y as t increases. + + Graph of the vector and vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,2,3,4,5}, + ytick={-3,-2,-1,1,2,3}, + ymin=-3.5,ymax=3.5, + xmin=-.5,xmax=5.5 + ] + + \addplot+ [domain=-2.5:2.5,samples=50] ({x^2},{x^2+x-1}); + + \draw [->,thick] (axis cs:1,-1) -- (axis cs: 0.9025,-1.0475); + \draw [->,thick,secondcolor] (axis cs: 0,0) -- (axis cs:4,1) node [black,pos=.5,above] { $\vec r(-2)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + Vector-valued functions are closely related to parametric equations of graphs. + While in both methods we plot points + \big(x(t), + y(t)\big) or \big(x(t),y(t),z(t)\big) to produce a graph, + in the context of vector-valued functions each such point represents a vector. + The implications of this will be more fully realized in the next section as we apply calculus ideas to these functions. +

    + + + Graphing vector-valued functions + +

    + Graph \ds \vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra, + for -2\leq t\leq 2. + Sketch \vec r(-1) and \vec r(2). +

    +     + +

    + We start by making a table of t, + x and y values as shown in . + Plotting these points gives an indication of what the graph looks like. + In , + we indicate these points and sketch the full graph. + We also highlight \vec r(-1) and \vec r(2) on the graph. +

    + +
    + Sketching the vector-valued function of + +
    + + + + t + t^3-t + \ds \frac{1}{t^2+1} + + + + + + + + -2 + -6 + 1/5 + + + -1 + 0 + 1/2 + + + 0 + 0 + 1 + + + 1 + 0 + 1/2 + + + 2 + 6 + 1/5 + + +
    + +
    + + + + + The image contains the plot of the vector valued function \vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra for -2\leq t\leq 2. + The image also contains the vectors \vec r(-1) =\la 0,\frac12 \ra and \vec r(2) =\la 6,\frac15 \ra, which both end at a point on the function \vec r(t). + The function \vec r(t) begins at the point (-6,\frac15) corresponding to the lower bound of t=-2, from which it slowly slopes upwards until crossing the y-axis at y=\frac12. + From here, the function slopes upwards and slightly outwards away from the y-axis until curving back and crossing the y-axis once again at y=1. + The function then slopes downwards and slightly away from the y-axis until curving back and crossing the y-axis again at y=\frac12, completing a loop. + After crossing the y-axis, the function continues slowly sloping downwards until reaching the point (6,\frac15) which corresponds to the upper bound of t=2. + The function is also symmetric about the y-axis. + + Graph of two vectors and vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[% + xtick={-6,-4,-2,2,4,6}, + ymin=-.1,ymax=1.1,% + xmin=-7,xmax=7% + ] + + \addplot+[domain=-2:2] ({x^3-x},{1/(x^2+1)}); + + \filldraw [black] (axis cs: -6,.2) circle (2.4pt); + \filldraw [black] (axis cs: 6,.2) circle (2.4pt); + \filldraw [black] (axis cs: 0,.5) circle (2.4pt); + \filldraw [black] (axis cs: 0.,1) circle (2.4pt); + + \draw [->,thick,secondcolor] (axis cs:0,0)--(axis cs: 0,.5) node [,black,pos=.4,rotate=90,below] { $\vec r(-1)$}; + \draw [->,thick,secondcolor] (axis cs:0,0)--(axis cs: 6,.2) node [black,pos=.4,above] { $\vec r(2)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    +     + +     + + + Graphing vector-valued functions + +

    + Graph \vec r(t) = \la \cos(t) ,\sin(t) ,t\ra for 0\leq t\leq 4\pi. +

    +     + +

    + We can again plot points, + but careful consideration of this function is very revealing. + Momentarily ignoring the third component, + we see the x and y components trace out a circle of radius 1 centered at the origin. + Noticing that the z component is t, + we see that as the graph winds around the z-axis, + it is also increasing at a constant rate in the positive z direction, + forming a spiral. + This is graphed in . + In the graph \vec r(7\pi/4)\approx (0.707,-0.707,5.498) is highlighted to help us understand the graph. +

    + +
    + The graph of \vec r(t) in + + + + + The image contains the plot of the vector valued function \vec r(t) = \la \cos(t) ,\sin(t) ,t\ra for 0\leq t\leq 4\pi. + The image also contains the vector \vec r(7\pi/4)\approx (0.707,-0.707,5.498), which corresponds to a point on the function \vec r(t). + The function \vec r(t) begins at the point (1,0,0) corresponding to the lower bound of t=0. + Ignoring the z coordinate of the vector valued function, the curve is simply a circle centered at the origin in the xy plane. + The addition of the z coordinate given by t then makes this curve linearly increase at t increases. + Accounting for the z coordinate, the function \vec r(t) resembles a linearly increasing circular spiral, which completes two full revolutions. + The spiral ends at the same x and y coordinates as it started, but is 4\pi units above in the z direction, ending at the point (1,0,4\pi). + + Graph of a vector and the three-dimensional vector valued function from the example. + + + + + //ASY file for figvvf23D.asy in Chapter 11 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4.5,4.5,20); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={5,10}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-0.5,15); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the helix <cos t,sin t,t> for t from 0 to 4pi + triple g(real t) {return (cos(t),sin(t),t);} + path3 mypath=graph(g,0,4*pi,operator ..); draw(mypath,bluepen); + + //Draw the vector for r(7pi/4) + draw((0,0,0)--(0.707,-0.707,5.498),redpen+linewidth(2),Arrow3(size=3mm)); + label("$\vec{r}(7 \pi/4)$",(0.707,-0.707,5.498),W); + + + + +
    +     + +     +     + + + Algebra of Vector-Valued Functions + + Operations on Vector-Valued Functions + +

    + Let \vec r_1(t)=\la f_1(t),g_1(t)\ra and + \vec r_2(t)=\la f_2(t),g_2(t)\ra be vector-valued functions in + \mathbb{R}^2 and let c be a scalar. + Then: +

    + +

    +

      +
    1. +

      + \vec r_1(t) \pm \vec r_2(t) = \la\, f_1(t)\pm f_2(t),g_1(t)\pm g_2(t)\,\ra. +

      +
      2. + +
    2. +

      + c\vec r_1(t) = \la\, cf_1(t),cg_1(t)\,\ra. +

      +
      3. +
    +

    + +

    + A similar definition holds for vector-valued functions in \mathbb{R}^3. + vector-valued functionalgebra of +

    +     +     + +

    + This definition states that we add, + subtract and scale vector-valued functions component-wise. + Combining vector-valued functions in this way can be very useful + (as well as create interesting graphs). +

    + + + + + Adding and scaling vector-valued functions + +

    + Let \vec r_1(t) = \la\,0.2t,0.3t\,\ra, + \vec r_2(t) = \la\,\cos(t) ,\sin(t) \,\ra and \vec r(t) = \vec r_1(t)+\vec r_2(t). + Graph \vec r_1(t), \vec r_2(t), + \vec r(t) and 5\vec r(t) on -10\leq t\leq10. +

    +     + +

    + We can graph \vec r_1 and + \vec r_2 easily by plotting points + (or just using technology). + Let's think about each for a moment to better understand how vector-valued functions work. +

    + +

    + We can rewrite \vec r_1(t) = \la\, 0.2t,0.3t\,\ra as \vec r_1(t) = t\la 0.2,0.3\ra. + That is, the function \vec r_1 scales the vector \la 0.2,0.3\ra by t. + This scaling of a vector produces a line in the direction of \la 0.2,0.3\ra. +

    + +

    + We are familiar with \vec r_2(t) = \la\, \cos(t) ,\sin(t) \,\ra; + it traces out a circle, centered at the origin, of radius 1. + graphs + \vec r_1(t) and \vec r_2(t). +

    + +

    + Adding \vec r_1(t) to + \vec r_2(t) produces \vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra, + graphed in . + The linear movement of the line combines with the circle to create loops that move in the direction of + \la 0.2,0.3\ra. (We encourage the reader to experiment by changing + \vec r_1(t) to \la 2t,3t\ra, + etc., and observe the effects on the loops.) +

    + +
    + Graphing the functions in + +
    + + + + + The image contains the plot of the vector valued functions \vec r_1(t) = \la 0.2t,0.3t \ra, \vec r_2(t) = \la \cos(t) ,\sin(t) \ra on -10\leq t\leq10. + The function \vec r_1(t) = \la 0.2t,0.3t \ra is a line which begins at the point (-2,-3) corresponding to t=-10 and ends when t=10 at the point (2,3). + The second function \vec r_2(t) = \la \cos(t) ,\sin(t) \ra looks like a circle of radius 1, but as -10\leq t\leq10 the function completes \frac{20}{2\pi} full circular rotations, which cannot be seen in the graph. + + Graph of the two vector valued functions prior to adding them together. + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-4.1,ymax=4.1, + xmin=-4.9,xmax=4.9 + ] + + \addplot+ [domain=-10:10] ({.2*x},{.3*x}); + \addplot+ [solid,domain=0:360,samples=60] ({cos(x)},{sin(x)}); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + The image contains the plot of the vector valued function \vec r(t) = \vec r_1(t)+\vec r_2(t). + The function is given by \vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra on -10\leq t\leq10. + The function begins in the third quadrant, near the point (-3,-2). + Near this point, the function is concave up, starting with a downward slope and later beginning to slope upwards. + The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate. + The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant. + The function nearly completes a third loop, before ending near the point (3,1.5). + + Graph of the vector valued function coming from adding the circle and line vector valued functions. + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-4.1,ymax=4.1, + xmin=-4.9,xmax=4.9 + ] + + \addplot+ [domain=-10:10,samples=120] ({.2*x+cos(deg(x))},{.3*x+sin(deg(x))}); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + The image contains the plot of the vector valued function 5\vec r(t) = 5\vec r_1(t)+5\vec r_2(t). + The function is now given by \vec r(t) = \la\,5\cos(t) + t,5\sin(t) +1.5t\,\ra on -10\leq t\leq10. + Compared to the unscaled function \vec r(t), the function 5\vec r(t) is exactly 5 times larger than the original function. + The function begins in the third quadrant, near the point (-15,-10). + Like the unscaled function, near this point 5\vec r(t) is concave up, starting with a downwards slope and later beginning to slope upwards. + The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate. + The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant. + The function nearly completes a third loop, before ending near the point (15,7.5). + The entirety of the original function \vec r(t) can be seen to fit between the two loops of the scaled function 5\vec r(t). + + Graph of the vector valued function coming from scaling the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-20,-10,10,20}, + ymin=-20.5,ymax=20.5, + xmin=-24.5,xmax=24.5 + ] + + \addplot+ [domain=-10:10,samples=60] ({.2*x+cos(deg(x))},{.3*x+sin(deg(x))}); + \addplot+ [solid,domain=-10:10,samples=120] ({x+5*cos(deg(x))},{1.5*x+5*sin(deg(x))}); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + Multiplying \vec r(t) by 5 scales the function by 5, producing 5\vec r(t) = \langle 5\cos(t) +t,5\sin(t) +1.5t\rangle, + which is graphed in along with \vec r(t). + The new function is 5 times bigger than \vec r(t). + Note how the graph of 5\vec r(t) in looks identical to the graph of \vec r(t) in . + This is due to the fact that the x and y bounds of the plot in + are exactly 5 times larger than the bounds in . +

    +     + +     + + + Adding and scaling vector-valued functions + +

    + A cycloid is a graph traced by a point p on a rolling circle, + as shown in . + Find an equation describing the cycloid, + where the circle has radius 1. + cycloid +

    + + +
    + Tracing a cycloid + + + The image shows the cycloid which comes from tracking a point p on a rolling circle of radius 1 on a flat surface. + The point p is initially at the top of the circle of radius 1. + Once the circle rolls, the point p is tracked to create a graph of a cycloid. + The graph coming from tracking the point p first begins to decrease, until the point p is at the bottom of the circle, at which point the point p touches the surface the ball is rolling on. + After this point, the graph begins to increase. + The circle continues to roll, with the point p once again becoming the top of the circle, after which the graph begins to decrease. + The circle continues rolling, with the point p becoming the lowest point on the circle two more times, after which the graph stops. + Between the starting point and the point at which p is at the bottom of the circle, the graph resembles a slightly stretched quarter of a circle. + The part of the graph that comes from the remaining two full revolutions of the circle resembles two horizontally stretched semi-circles. + + Image showing the cycloid which comes from tracking a point on a rolling circle. + + + \begin{tikzpicture} + + \draw [thick] (0,1) circle (1); + + \filldraw (1,1) circle (2pt) node [right] {\large $p$}; + + \draw [firstcolor,thick,smooth,domain=-1:14,samples=60] plot ({cos(\x r)+\x},{-sin(\x r)+1}); + + \draw [thick] (-1,0) -- (15,0); + \draw [thick,->,>=stealth] (-1.3,1) arc (180:90:1.3); + \draw [thick,white] (0,2.5) -- (1,2.5); + + \end{tikzpicture} + + + +
    + +     + +

    + This problem is not very difficult if we approach it in a clever way. + We start by letting \vec p(t) describe the position of the point p on the circle, + where the circle is centered at the origin and only rotates clockwise (, it does not roll). + This is relatively simple given our previous experiences with parametric equations; + \vec p(t) = \la \cos(t) , -\sin(t) \ra. +

    + +

    + We now want the circle to roll. + We represent this by letting + \vec c(t) represent the location of the center of the circle. + It should be clear that the y component of \vec c(t) should be 1; + the center of the circle is always going to be 1 if it rolls on a horizontal surface. +

    + +

    + The x component of \vec c(t) is a linear function of t: + f(t) = mt for some scalar m. + When t=0, f(t) = 0 + (the circle starts centered on the y-axis). + When t=2\pi, the circle has made one complete revolution, + traveling a distance equal to its circumference, + which is also 2\pi. + This gives us a point on our line f(t) = mt, + the point (2\pi, 2\pi). + It should be clear that m=1 and f(t) = t. + So \vec c(t) = \la t, 1\ra. +

    + +

    + We now combine \vec p and \vec c together to form the equation of the cycloid: + \vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra, + which is graphed in . +

    +
    + The cycloid in + + + + The graph shows the function \vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra, which is the graph of a cycloid which comes from tracking a point p on a rolling a circle of radius 1 on a flat surface. + The curve begins near the point (0,2), after which it begins to decrease until it reaches its lowest point near the point (2,0). + After this point, the curve begins increasing, until it reaches its highest point when p is at the top of the circle, after which it decreases until reaching another minimum near the point (8,0). + The curve continues in the same fashion, reaching another minimum near the point (14,0), after which it continues for a slight duration in the same fashion as before. + + Graph showing the cycloid which comes from tracking a point on a rolling circle. + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=14, + xmin=-.5,xmax=16 + ] + + \addplot+ [domain=-1:16,samples=80] ({cos(deg(x))+x},{-sin(deg(x))+1}); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +     +     + + + Displacement + +

    + A vector-valued function + \vec r(t) is often used to describe the position of a moving object at time t. + At t=t_0, the object is at \vec r(t_0); + at t=t_1, the object is at \vec r(t_1). + Knowing the locations \vec r(t_0) and + \vec r(t_1) give no indication of the path taken between them, + but often we only care about the difference of the locations, + \vec r(t_1)-\vec r(t_0), + the displacement. + displacement + vector-valued functiondisplacement +

    + + + Displacement + +

    + Let \vec r(t) be a vector-valued function and let + t_0\lt t_1 be values in the domain. + The displacement \vec d of \vec r, + from t=t_0 to t=t_1, is + + \vec d=\vec r(t_1)-\vec r(t_0) + . +

    +     +     + + + +

    + When the displacement vector is drawn with initial point at \vec r(t_0), + its terminal point is \vec r(t_1). + We think of it as the vector which points from a starting position to an ending position. +

    + + + Finding and graphing displacement vectors + +

    + Let \vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra. + Graph \vec r(t) on -1\leq t\leq 1, + and find the displacement of \vec r(t) on this interval. +

    +     + +

    + The function \vec r(t) traces out the unit circle, + though at a different rate than the usual + \la \cos(t) ,\sin(t) \ra parametrization. + At t_0=-1, we have \vec r(t_0) = \la 0,-1\ra; + at t_1=1, we have \vec r(t_1) = \la 0,1\ra. + The displacement of \vec r(t) on [-1,1] is thus \vec d = \la 0,1\ra - \la 0,-1\ra = \la 0,2\ra. +

    + +
    + Graphing the displacement of a position function in + + + + Graph of the function \vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra on -1\leq t\leq 1. + The function \vec r(t) is the right half of the unit circle, beginning at the point (0,-1), crossing the x-axis at the point (1,0), and ending at the point (0,-1). + The graph also contains the displacement vector, which begins at the start of the curve at the point (0,-1) and heads directly upwards until reaching the endpoint of the curve at the point (0,1). + + Graph of the semicircle coming from plotting the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.32,xmax=1.32 + ] + + \addplot+ [domain=-90:90,samples=40] ({cos(x)},{sin(x)}); + + \draw [thick,->,secondcolor,>=stealth] (axis cs: 0,-1) -- (axis cs:0,1) node [left,pos=.7,black]{ $\vec d$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + A graph of \vec r(t) on [-1,1] is given in , + along with the displacement vector \vec d on this interval. +

    +     + +     + +

    + Measuring displacement makes us contemplate related, + yet very different, concepts. + Considering the semi-circular path the object in took, + we can quickly verify that the object ended up a distance of 2 units from its initial location. + That is, we can compute \vnorm{d} = 2. + However, measuring distance from the starting point + is different from measuring distance traveled. + Being a semi-circle, + we can measure the distance traveled by this object as \pi\approx 3.14 units. + Knowing distance from the starting point + allows us to compute average rate of change. +

    + + + Average Rate of Change + +

    + Let \vec r(t) be a vector-valued function, + where each of its component functions is continuous on its domain, + and let t_0\lt t_1. + The average rate of change + of \vec r(t) on [t_0,t_1] is + average rate of change + vector-valued functionaverage rate of change + + \text{ average rate of change } = \frac{\vec r(t_1) - \vec r(t_0)}{t_1-t_0} + . +

    +     +     + + + Average rate of change + +

    + Let \vec r(t) = \la \cos(\frac{\pi}2t),\sin(\frac{\pi}2t)\ra as in . + Find the average rate of change of + \vec r(t) on [-1,1] and on [-1,5]. +

    +     + +

    + We computed in + that the displacement of \vec r(t) on [-1,1] was \vec d = \la 0,2\ra. + Thus the average rate of change of \vec r(t) on [-1,1] is: + + \frac{\vec r(1) -\vec r(-1)}{1-(-1)} = \frac{\la 0,2\ra}{2} = \la 0,1\ra + . +

    + +

    + We interpret this as follows: + the object followed a semi-circular path, + meaning it moved towards the right then moved back to the left, + while climbing slowly, then quickly, then slowly again. + On average, however, + it progressed straight up at a constant rate of \la 0,1\ra per unit of time. +

    + +

    + We can quickly see that the displacement on [-1,5] is the same as on [-1,1], + so \vec d = \la 0,2\ra. + The average rate of change is different, though: + + \frac{\vec r(5)-\vec r(-1)}{5-(-1)} = \frac{\la 0,2\ra}{6} = \la 0,1/3\ra + . +

    + +

    + As it took 3 times as long + to arrive at the same place, + this average rate of change on [-1,5] is 1/3 the average rate of change on [-1,1]. +

    +     + +     + +

    + We considered average rates of change in Sections + and + as we studied limits and derivatives. + The same is true here; + in the following section we apply calculus concepts to vector-valued functions as we find limits, + derivatives, and integrals. + Understanding the average rate of change will give us an understanding of the derivative; + displacement gives us one application of integration. +

    +     + + + + Terms and Concepts + + + +

    + Vector-valued functions are closely related to + of curves. +

    +     + + + + + + + + + + + + + +     + + + + +

    + When sketching vector-valued functions, + technically one isn't graphing points, but rather . +

    +     + + + + + + + + +     + + + + +

    + It can be useful to think of as a vector that points from a starting position to an ending position. +

    +     + + + + + + + + +     + + + +

    + In the context of vector-valued functions, + average rate of change is divided by time. +

    +     + + + + + + + +     +     + + + Problems + + + +

    + Sketch the vector-valued function on the given interval. +

    +     + + + + +

    + \vec r(t) = \la t^2,t^2-1\ra, + for -2\leq t\leq 2. +

    +     + + + + + Graph of the function \vec r(t) = \la t^2,t^2-1\ra for -2\leq t\leq 2. + The graph of the function \vec r(t) is a line which begins at the point (0,-1) and ends at the point (4,3). + The function begins at the point (4,3) when t=-2, linearly decreases until the point (0,-1) and once again follows the same linear path until ending at the point (4,3). + + Graph of the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,2,3,4}, + ytick={-1,1,2,3,4}, + ymin=-1.5,ymax=4.5, + xmin=-.5,xmax=4.5 + ] + + \addplot+ [domain=-2:2] ({x^2},{x^2-1}); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + + + + +

    + \vec r(t) = \la t^2,t^3\ra, + for -2\leq t\leq 2. +

    +     + + + + + Graph of the function \vec r(t) = \la t^2,t^3\ra, for -2\leq t\leq 2. + The graph of the function \vec r(t) begins at the point (4,-8), and upwards and to the left in a concave down fashion until reaching the origin. + After reaching the origin, the function is concave and ends when it reaches the point (4,8). + The function is also symmetric about the x-axis, meaning that the upper and lower half of the curves are mirror images of each other. + + Graph of the vector valued function from the example. + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + xtick={1,2,3,4}, + ymin=-8.9,ymax=8.9, + xmin=-.5,xmax=4.5 + ] + + \addplot+ [domain=-2:2] ({x^2},{x^3}); + + \draw [->,thick,firstcolor] (axis cs: 1,-1) -- (axis cs:0.98,-.97); + \draw [->,thick,firstcolor] (axis cs: 0.98,.97) -- (axis cs: 1,1); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + + + + +

    + \vec r(t) = \la 1/t,1/t^2\ra, + for -2\leq t\leq 2. +

    +     + + + + + Graph of the function \vec r(t) = \la 1/t,1/t^2\ra, for -2\leq t\leq 2. + The graph of the function \vec r(t) begins at the point (-\frac12,\frac14) corresponding to t=2, heads upwards and to the left as t increases, following the path of the left side of the parabola y=x^2. + At t=2 the function starts at the point (\frac12,\frac14) and as t decreases follows the path of the right side of the parabola given by y=x^2. + The function is also mirrored about y-axis, and is equivalent to the parabola y=x^2 outside of the region -\frac12 \leq x \frac12. + + Graph of the vector valued function from the example. + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-.5,ymax=10.5, + xmin=-5.5,xmax=5.5 + ] + + \addplot+ [domain=-2:-.1,samples=40] ({1/x},{1/x^2}); + \addplot+ [solid,domain=.1:2,samples=40] ({1/x},{1/x^2}); + + \draw [->,thick,firstcolor] (axis cs: -1,1) -- (axis cs:-1.01,1.02); + \draw [->,thick,secondcolor] (axis cs: 1.01,1.02) -- (axis cs: 1,1); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + + + + +

    + \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, + for -2\pi\leq t\leq 2\pi. +

    +     + + + + + Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi. + The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi. + As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease. + The function continues to decrease, crossing the xaxis at x=1. + After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin. + The function now increases upwards and to the right until reaching a maximum near point (0.25,1). + After this point, the curve decreases, crossing the x-axis once again at x=1. + The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0). + The function is also symmetric about the x-axis. + + Graph of the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.1,ymax=1.1, + xmin=-.5,xmax=4.5 + ] + + \addplot+ [domain=-6.28:6.28,samples=120] ({.1*x^2},{sin(deg(x))}); + + \draw [->,thick,firstcolor,>=stealth] (axis cs: 0.1, -0.841471) -- (axis cs:0.09801, -0.836026); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + + + + +

    + \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, + for -2\pi\leq t\leq 2\pi. +

    +     + + + + + Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi. + The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi. + As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease. + The function continues to decrease, crossing the xaxis at x=1. + After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin. + The function now increases upwards and to the right until reaching a maximum near point (0.25,1). + After this point, the curve decreases, crossing the x-axis once again at x=1. + The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0). + The function is also symmetric about the x-axis. + + Graph of the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,2,3,4}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-0.1,xmax=4.1 + ] + + \addplot+ [domain=-6.28:6.28,samples=70] ({x*x/10},{sin(deg(x))}); + + \draw [->,thick,firstcolor,>=stealth] (axis cs:3.61201, 0.2698) -- (axis cs: 3.6, 0.279415); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + + + + +

    + \vec r(t) = \la 3\sin(\pi t),2\cos(\pi t)\ra, on [0,2]. +

    +     + + + + + Graph of the function \vec r(t) = \la 3\sin(\pi t),2\cos(\pi t)\ra, on [0,2]. + The graph of the function \vec r(t) is an oval having a horizontal width of 6 and a height of 4 centered at the origin. + The rightmost point of the oval is the point (3,0). + The leftmost point of the oval is the point (-3,0). + The highest point of the oval is the point (0,2), while the lowest is the point (0,-2). + + Graph of the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-3,-2,-1,1,2,3}, + ytick={-1,-2,1,2}, + ymin=-2.5,ymax=2.5, + xmin=-3.5,xmax=3.5 + ] + + \addplot+ [domain=0:2,samples=70] ({3*sin(deg(3.14159*x))},{2*cos(deg(3.14159*x))}); + + \draw [->,thick,firstcolor,>=stealth] (axis cs:1.64607, -1.67205)--(axis cs: 1.62091, -1.68294); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + + + + +

    + \vec r(t) = \la 3\cos(t) ,2\sin(2 t)\ra, on [0,2\pi]. +

    +     + + + + + Graph of the function \vec r(t) = \la 3\cos(t) ,2\sin(2 t)\ra, on [0,2\pi]. + The graph of the function \vec r(t) resembles the symbol \infty. + The curve begins at the point (3,0) which corresponds to t=0. + The curve then begins going upwards and to the left, until it reaches a maximum near the point (2,2). + From here the curve begins going downwards and to the left, passing through the origin, until reaching a minimum near (-2,-2). + The curve then begins going upwards to the left, passing through the point (-3,0), after which it begins going to the right. + The curve reaches a maximum near the point (-2,2), after which it begins decreasing and going to the right, once again crossing the origin. + Finally, the curve reaches another minimum near the point (2,-2), after which it begins going upwards to the right, until reaching its starting point given by (3,0). + The curve is symmetric about both the x and y axes. + + Graph of the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-3,-2,-1,1,2,3}, + ytick={-1,-2,1,2}, + ymin=-2.5,ymax=2.5, + xmin=-3.5,xmax=3.5 + ] + + \addplot+ [domain=0:6.28,samples=120] ({3*cos(deg(x))},{2*sin(deg(2*x))}); + + \draw [->,thick,firstcolor] (axis cs:1.62091, -1.81859) -- (axis cs: 1.64607, -1.83488); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + + + + +

    + \vec r(t) = \la 2\sec(t) ,\tan(t) \ra, on [-\pi,\pi]. +

    +     + + + + + Graph of the function \vec r(t) = \la 2\sec(t) ,\tan(t) \ra, on [-\pi,\pi]. + The graph of the function is symmetric about both the x and y axes. + The curve crosses the x-axis at the point x=2 corresponding to t=\pi. + After this point the upper part of the curve goes upwards and to the right in a mostly linear fashion. + Since the curve is symmetric about the y-axis, the lower part of the curve is the mirrored version of the upper part, meaning that it goes downwards and to the left. + As the curve is symmetric about the x-axis, the left side of the curve is the mirrored version of the right part, meaning that it crosses the x-axis at x=-2. + From this point, the upper part of the left side of the curve goes upwards and to the left, while the lower part goes downwards and to the left. + + Graph of the vector valued function from the example. + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-5.5,ymax=5.5, + xmin=-11,xmax=11 + ] + + \addplot+ [domain=-1.4708:1.4708,samples=40] ({2*sec(deg(x))},{tan(deg(x))}); + \addplot+ [solid,domain=1.6708:4.61239,samples=40] ({2*sec(deg(x))},{tan(deg(x))}); + + \draw [->,thick,firstcolor] (axis cs:2.27899, 0.546302) -- (axis cs: 2.29162, 0.559359); + \draw [->,thick,secondcolor] (axis cs: -2.02022, -0.1425477) -- (axis cs: -2.01744, -0.132358); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + +     + + + +

    + Sketch the vector-valued function on the given interval in \mathbb{R}^3. + Technology may be useful in creating the sketch. +

    +     + + + + +

    + \vec r(t) = \la 2\cos(t) , t, 2\sin(t) \ra, on [0,2\pi]. +

    +     + + + + + Graph of the function \vec r(t) = \la 2\cos(t) , t, 2\sin(t) \ra, on [0,2\pi]. + The graph of the function is a circular spiral centered about the y-axis. + Ignoring the y-axis, the curve is simply a singular circle of radius 2 in the xz-plane. + Incorporating the y coordinate then creates the linearly increasing spiral, which begins at the point (2,0,0), completes precisely one full revolution and ends at the point (2,2\pi,0). + + Graph of the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + axis lines=center, + view={145}{25}, + xtick={1}, + ymin=-.1,ymax=6.5, + xmin=-2.1,xmax=2.1, + zmin=-2.1,zmax=2.1, + every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, + xlabel={ $x$}, + every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, + ylabel={ $y$}, + every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, + zlabel={ $z$} + ] + + \addplot3 [firstcurvestyle,domain=0:6.28,samples y=0] ({2*cos(deg(x))},{x},{2*sin(deg(x))}); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + + + + +

    + \vec r(t) = \la 3\cos(t) , \sin(t) , t/\pi\ra on [0,2\pi]. +

    +     + + + + + Graph of the function \vec r(t) = \la 3\cos(t) , \sin(t) , t/\pi\ra on [0,2\pi]. + The graph of the function is an oval-shaped spiral centered about the z-axis. + Ignoring the z-axis, the curve is simply an oval having a horizontal width of 6 and a height of 2 in the xy-plane. + Incorporating the z coordinate then creates the linearly increasing oval spiral, which begins at the point (3,0,0), completes precisely one full revolution and ends at the point (3,0,2). + + Graph of the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + axis lines=center, + view={145}{25}, + ymin=-1.1,ymax=1.1, + xmin=-3.5,xmax=3.5, + zmin=-0.1, zmax=2.1, + every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, + xlabel={ $x$}, + every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, + ylabel={ $y$}, + every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, + zlabel={ $z$} + ] + + \addplot3 [firstcurvestyle,domain=0:6.28,samples y=0,samples=60] ({3*cos(deg(x))},{sin(deg(x))},{x/3.14159}); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + + + + +

    + \vec r(t) = \la \cos(t) , \sin(t) ,\sin(t) \ra on [0,2\pi]. +

    +     + + + + + Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(t) \ra on [0,2\pi]. + The graph of the function is an oval lying in the plane coming from rotating the xy plane 45 degrees towards the z-axis. + The oval lying in this plane has a horizontal width of \sqrt{2} and a height of 1. + Ignoring the z coordinate, the curve is a unit circle in the xy plane. + Similarly ignoring the y coordinate, the curve is a unit circle in the xz plane. + If we now ignore the x coordinate, the resulting curve is a diagonal line given by z=y in the yz plane. + This line turns back on itself, which can be seen in the image of the oval when considering all three coordinate axes. + + Graph of the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + axis lines=center, + view={145}{25}, + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.1, + zmin=-1.1, zmax=1.1, + every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, + xlabel={ $x$}, + every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, + ylabel={ $y$}, + every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, + zlabel={ $z$} + ] + + \addplot3 [firstcurvestyle,domain=0:6.28,samples y=0,samples=60] ({cos(deg(x))},{sin(deg(x))},{sin(deg(x))}); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + + + + +

    + \vec r(t) = \la \cos(t) , \sin(t) ,\sin(2t)\ra on [0,2\pi]. +

    +     + + + + + Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(2t)\ra on [0,2\pi]. + The graph of the function resembles a saddle centered at the origin whose height is defined by the z-axis. + The two sides of the saddle that taper off fall into negative z and lie in the second and third quadrants in the xy plane. + Ignoring the z coordinate, the curve is a unit circle in the xy plane. + Ignoring the x or y coordinates individually, the curve looks like the \infty symbol in the yz and the xz planes, respectively. + We now describe the z coordinate with respect to travelling along the unit circle in the xy plane. + Starting at t=, the function begins at the point (1,0,0). + As t increases and we travel along the unit circle in the x and y coordinates, z increases until we get to t=\frac{\pi}{2} at which z=1. + Then, continuing along the unit circle, z decreases until it reaches a minimum of z=-1 when t=\frac{3\pi}{4}. + Continuing along the circle, z begins to increase once again, reaching one more maximum of z=1 when t=\frac{5\pi}{4}. + Finally, z begins to decrease, reaching its last minimum of z=1 when t=\frac{7\pi}{4}, after which z increases, and the curve ends where it began, at the point (1,0,0). + + Graph of the vector valued function from the example. + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + axis lines=center, + view={155}{25}, + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.1, + zmin=-1.1, zmax=1.1, + every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, + xlabel={ $x$}, + every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, + ylabel={ $y$}, + every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, + zlabel={ $z$} + ] + + \addplot3 [firstcurvestyle,domain=0:6.28,samples y=0,samples=120] ({cos(deg(x))},{sin(deg(x))},{sin(deg(2*x))}); + + \end{axis} + + \end{tikzpicture} + + + + + + +     + +     + + + +

    + Find \norm{\vec r(t)}. +

    +     + + + + + Context()->variables->are(t=>'Real'); + $norm=Formula("|t| sqrt(1+t^2)"); + + + +

    + If \vec r(t) = \la t,t^2\ra, + then \norm{\vec r(t)}=. +

    +     +     +     + + + + +

    + \vec r(t) = \la 5\cos(t) ,3\sin(t) \ra. +

    +     + +

    + \norm {\vec r(t)} = \sqrt{25\cos^2(t) +9\sin^2(t) }. +

    +     + +     + + + + + Context()->variables->are(t=>'Real'); + $norm=Formula("sqrt(4+t^2)"); + + + +

    + If \vec r(t) = \la 2\cos(t) ,2\sin(t) ,t\ra, + then \norm{\vec r(t)}=. +

    +     +     +     + + + + +

    + \vec r(t) = \la \cos(t) ,t,t^2\ra. +

    +     + +

    + \norm {\vec r(t)} = \sqrt{\cos^2(t) +t^2+t^4}. +

    +     + +     + +     + + + +

    + Create a vector-valued function whose graph matches the given description. +

    +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + $vvf=Formula("<2 cos(t)+1, 2 sin(t) + 2>"); + Context()->flags->set(tolType=>'absolute',tolerance=>0.001); + $vvfev=$vvf->cmp(checker=>sub{ + my ($correct,$student,$self)=@_; + my ($sx,$sy)=$student->value; + my ($cx,$cy)=$correct->value; + my $eliminated=Formula("(($sx-1)/2)^2 + (($sy-2)/2)^2"); + return 0 unless ($eliminated == Formula("1")); + #normalize answers + $sx=($sx-1)/2; + $sy=($sy-2)/2; + $cx=($cx-1)/2; + $cy=($cy-2)/2; + #shift student answers + my $sx0=$sx->eval(t=>0); + my $sy0=$sy->eval(t=>0); + my $phi=($sx0 >= 0)?Compute("arcsin($sy0)"):Compute("pi-arcsin($sy0)"); + $sx=$sx->substitute(t=>Formula("t-$phi")); + $sy=$sy->substitute(t=>Formula("t-$phi")); + return ($sx == $cx and $sy == $cy); + } + ); + + + +

    + A circle of radius 2, + centered at (1,2), + traced counter-clockwise once at constant speed on [0,2\pi). +

    + + + \vec r(t)= + +

    + +

    +     +     +     + + + + +

    + A circle of radius 3, centered at (5,5), + traced clockwise once on [0,2\pi]. +

    +     + +

    + Answers may vary; three solutions are +

    + +

    + \vec r(t) = \la 3\sin(t) +5,3\cos(t) +5\ra, +

    + +

    + \vec r(t) = \la -3\cos(t) +5,3\sin(t) +5\ra and +

    + +

    + \vec r(t) = \la 3\cos(t) +5,-3\sin(t) +5\ra. +

    +     + +     + + + + +

    + An ellipse, centered at (0,0) with vertical major axis of length 10 and minor axis of length 3, traced once counter-clockwise on [0,2\pi]. +

    +     + +

    + Answers may vary, though most direct solution is +

    + +

    + \vec r(t) = \la 1.5\cos(t) ,5\sin(t) \ra. +

    +     + +     + + + + +

    + An ellipse, centered at (3,-2) with horizontal major axis of length 6 and minor axis of length 4, traced once clockwise on [0,2\pi]. +

    +     + +

    + Answers may vary, though most direct solutions are +

    + +

    + \vec r(t) = \la -3\cos(t) +3,2\sin(t) -2\ra, +

    + +

    + \vec r(t) = \la 3\cos(t) +3,-2\sin(t) -2\ra and +

    + +

    + \vec r(t) = \la 3\sin(t) +3,2\cos(t) -2\ra. +

    +     + +     + + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + $vvf=ParametricLine("<t+2, 5t+3>"); + + + +

    + A line through (2,3) with a slope of 5. +

    + + + \vec r(t)= + +

    + +

    +     +     +     + + + + +

    + A line through (1,5) with a slope of -1/2. +

    +     + +

    + Answers may vary, though most direct solutions are +

    + +

    + \vec r(t) = \la t,-1/2(t-1)+5\ra, +

    + +

    + \vec r(t) = \la t+1,-1/2t+5\ra, +

    + +

    + \vec r(t) = \la -2t+1,t+5\ra and +

    + +

    + \vec r(t) = \la 2t+1,-t+5\ra. +

    +     + +     + + + +

    + The line through points (1,2,3) and (4,5,6), where +

    + +

    + \vec r(0) = \la 1,2,3\ra and \vec r(1) = \la 4,5,6\ra. +

    +     + +

    + Specific forms may vary, though most direct solutions are +

    + +

    + \vec r(t) = \la 1,2,3\ra +t\la 3,3,3\ra and +

    + +

    + \vec r(t) = \la 3t+1, 3t+2, 3t+3\ra. +

    +     +     + + + +

    + The line through points (1,2) and (4,4), where +

    + +

    + \vec r(0) = \la 1,2\ra and \vec r(1) = \la 4,4\ra. +

    +     + +

    + Specific forms may vary, though most direct solutions are +

    + +

    + \vec r(t) = \la 1,2\ra +t\la 3,2\ra and +

    + +

    + \vec r(t) = \la 3t+1, 2t+2\ra. +

    +     +     + + + + + Context("Vector"); + Context()->variables->are(t=>'Real'); + $vvf=Formula("<2 cos(t), 2 sin(t), 2t>"); + Context()->flags->set(tolType=>'absolute',tolerance=>0.001); + $vvfev=$vvf->cmp(checker=>sub{ + my ($correct,$student,$self)=@_; + my ($sx,$sy,$sz)=$student->value; + my ($cx,$cy,$cz)=$correct->value; + return 0 unless ($sx == $cx and $sy == $cy); + $szd=$sz->reduce->D('t'); + return 0 unless ($szd == Formula("2")); + my $r = ($sz->eval(t=>0))/(4*pi); + return 0 unless ($r == int($r)); + } + ); + + + +

    + A vertically oriented helix with radius of 2 that starts at (2,0,0) and ends at + (2,0,4\pi) after one revolution on [0,2\pi]. +

    + + + \vec r(t)= + +

    + +

    +     +     +     + + + + +

    + A vertically oriented helix with radius of 3 that starts at (3,0,0) and ends at (3,0,3) after 2 revolutions on [0,1]. +

    +     + +

    + Answers may vary, though most direct solution is +

    + +

    + \vec r(t) = \la 3\cos(4\pi t),3\sin(4\pi t),3t\ra. +

    +     + +     + +     + + + +

    + Find the average rate of change of + \vec r(t) on the given interval. +

    +     + + + + + Context("Vector2D"); + $arc=Vector("<1,0>"); + + + +

    + \vec r(t) = \la t,t^2\ra on [-2,2]. +

    + +

    + +

    +     +     +     + + + + +

    + \vec r(t) = \la t,t+\sin(t) \ra on [0,2\pi]. +

    +     + +

    + \la 1,1\ra +

    +     + +     + + + + + Context("Vector"); + $arc=Vector("<0,0,1>"); + + + +

    + \vec r(t) = \la 3\cos(t) ,2\sin(t) ,t\ra on [0,2\pi]. +

    + +

    + +

    +     +     +     + + + + +

    + \vec r(t) = \la t,t^2,t^3\ra on [-1,3]. +

    +     + +

    + \la 1,2,7\ra +

    +     + +     + +     +     +     +
    +
    + Calculus and Vector-Valued Functions + +

    + The previous section introduced us to a new mathematical object, + the vector-valued function. + We now apply calculus concepts to these functions. + We start with the limit, + then work our way through derivatives to integrals. +

    +     + + + Limits of Vector-Valued Functions +

    + The initial definition of the limit of a vector-valued function is a bit intimidating, + as was the definition of the limit in . + The theorem following the definition shows that in practice, + taking limits of vector-valued functions is no more difficult than taking limits of real-valued functions. +

    + + + + Limits of Vector-Valued Functions + +

    + Let I be an open interval containing c, + and let \vec r(t) be a vector-valued function defined on I, + except possibly at c. + The limit of \vec r(t), + as t approaches c, + is \vec L, expressed as + + \lim_{t\to c} \vec r(t) = \vec L + , + means that given any \varepsilon \gt 0, + there exists a \delta \gt 0 such that for all t\neq c, + if \abs{t-c} \lt \delta, + we have \norm{\vec r(t) - \vec L} \lt \varepsilon. + + vector-valued functionlimits + limitof vector-valued functions +

    +     +     + +

    + Note how the measurement of distance between real numbers is the absolute value of their difference; + the measure of distance between vectors is the vector norm, + or magnitude, of their difference. +

    + +

    + + states that we can compute limits of vector-valued functions component-wise. +

    + + + Limits of Vector-Valued Functions + +

    +

      +
    1. +

      + Let \vec r(t) = \la \,f(t),g(t)\,\ra be a vector-valued function in + \mathbb{R}^2 defined on an open interval I containing c, + except possibly at c. + Then + vector-valued functionlimits + limitof vector-valued functions + + + \lim_{t\to c} \vec r(t) = \la \lim_{t\to c}f(t)\, , \,\lim_{t\to c} g(t)\ra + . +

      +
      2. + +
    2. +

      + Let \vec r(t) = \la \,f(t),g(t),h(t)\,\ra be a vector-valued function in + \mathbb{R}^3 defined on an open interval I containing c, + except possibly at c. + Then + + \lim_{t\to c} \vec r(t) = \la \lim_{t\to c}f(t)\, , \,\lim_{t\to c} g(t)\,, \,\lim_{t\to c} h(t)\ra + +

      +
      3. +
    +

    +     +     + + + + + Finding limits of vector-valued functions + +

    + Let \ds\vec r(t) = \la \frac{\sin(t) }{t},\, t^2-3t+3,\,\cos(t) \ra. + Find \lim\limits_{t\to 0}\vec r(t). +

    +     + +

    + We apply the theorem and compute limits component-wise. + + \lim_{t\to0} \vec r(t) \amp = \la \lim_{t\to 0}\frac{\sin(t) }{t}\, , \, \lim_{t\to 0} t^2-3t+3\, , \, \lim_{t\to 0} \cos(t) \ra + \amp = \la 1,3,1\ra + . +

    +     + +     +     + + + Continuity + + Continuity of Vector-Valued Functions + +

    + Let \vec r(t) be a vector-valued function defined on an open interval I containing c. + vector-valued functioncontinuity + continuous functionvector-valued + +

      +
    1. +

      + \vec r(t) is continuous at c + if \lim\limits_{t\to c} \vec r(t) = r(c). +

      +
      2. + +
    2. +

      + If \vec r(t) is continuous at all c in I, + then \vec r(t) is continuous on I. +

      +
      3. +
    +

    +     +     + + +

    + We again have a theorem that lets us evaluate continuity component-wise. +

    + + + Continuity of Vector-Valued Functions + +

    + Let \vec r(t) be a vector-valued function defined on an open interval I containing c. + Then \vec r(t) is continuous at c if, and only if, + each of its component functions is continuous at c. + vector-valued functioncontinuity + continuous functionvector-valued +

    +     +     + + + + + Evaluating continuity of vector-valued functions + +

    + Let \ds\vec r(t) = \la \frac{\sin(t) }{t},\, t^2-3t+3,\,\cos(t) \ra. + Determine whether \vec r is continuous at t=0 and t=1. +

    +     + +

    + While the second and third components of + \vec r(t) are defined at t=0, + the first component, (\sin(t) )/t, is not. + Since the first component is not even defined at t=0, + \vec r(t) is not defined at t=0, + and hence it is not continuous at t=0. +

    + +

    + At t=1 each of the component functions is continuous. + Therefore \vec r(t) is continuous at t=1. +

    +     +     +     + + + Derivatives +

    + Consider a vector-valued function \vec r defined on an open interval I containing t_0 and t_1. + We can compute the displacement of \vec r on [t_0,t_1], + as shown in . + Recall that dividing the displacement vector by t_1-t_0 gives the average rate of change on [t_0,t_1], + as shown in . +

    + +
    + Illustrating displacement, leading to an understanding of the derivative of vector-valued functions + +
    + + + + + Graph of an arbitrary vector-valued \vec r on the interval which includes [t_0,t_1]. + The function \vec r is a small part of a concave down circular arc. + The graph includes the vectors \vec r (t_0) and \vec r (t_1), which begin from the origin and end at the corresponding point of the function \vec r . + The graph also includes the vector \vec r (t_1) - \vec r (t_0), which begins where \vec r (t_0) ends, and then terminates at the same termination point as \vec r (t_1). + The three vectors \vec r (t_0), \vec r (t_1) and \vec r (t_1) - \vec r (t_0) for a triangle, where following the path of \vec r (t_0) and \vec r (t_1) - \vec r (t_0) takes you to the same point as \vec r (t_1). + The vector \vec r (t_0) terminates on the left side of the circular arc, while \vec r (t_1) terminates further on the right side of the circular arc of given by the function \vec r. + + Illustration of a vector-valued function on a given interval. + + + \begin{tikzpicture}[>=stealth] + + \draw [thick,firstcolor] (2,1) arc [x radius=4,y radius=3,start angle=145,end angle=135] node (A) {}; + \draw [thick,firstcolor] (A.center) arc [x radius=4,y radius=3,start angle=135,end angle=70] node (B) {}; + \draw [thick,firstcolor] (B.center) arc [x radius=4,y radius=3,start angle=70,end angle=60]; + + \draw [thick,->] (0,0) -- (A.center) node [left,pos=.6] { $\vec r(t_0)$}; + \draw [thick,->] (0,0) -- (B.center) node [below,pos=.5] { $\vec r(t_1)$}; + \draw [->,thick,secondcolor] (A.center) -- (B.center) node [above,black,pos=.5,sloped] { $\vec r(t_1)-\vec r(t_0)$}; + + \end{tikzpicture} + + + + +
    + +
    + + + + + Graph of the same arbitrary vector-valued \vec r as well as the vectors \vec r (t_0) and \vec r (t_1) as described in the previous image. + This time the graph includes two additional vectors \vrp (t_0) and \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} . + The vector \vrp (t_0) begins at the termination point of \vec r (t_0), and is tangent to the function \vec r at this point. + The vector \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} also begins at the termination point of \vec r (t_0), and follows the path of the vector \vec r (t_1) - \vec r (t_0) from the previous image. + However, this vector does not end at the termination point of \vec r (t_1) but instead terminates at some point further away in the same direction as the vector \vec r (t_1) - \vec r (t_0). + + Illustration of a vector-valued function on a given interval showcasing a derivative vector. + + + \begin{tikzpicture}[>=stealth] + + \draw [thick,firstcolor] (2,1) arc [x radius=4,y radius=3,start angle=145,end angle=135] node (A) {}; + \draw [thick,firstcolor] (A.center) arc [x radius=4,y radius=3,start angle=135,end angle=70] node (B) {}; + \draw [thick,firstcolor] (B.center) arc [x radius=4,y radius=3,start angle=70,end angle=60]; + + \draw [thick,->] (0,0) -- (A.center) node [left,pos=.6] { $\vec r(t_0)$}; + \draw [thick,->] (0,0) -- (B.center) node [below,pos=.5] { $\vec r(t_1)$}; + \draw [->,thick,secondcolor] (A.center) -- ($(A)!1.2!(B)$) node [above,black,pos=.99] { $\displaystyle\frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0}$}; + + \draw [->,thick,secondcolor] (A.center)--($(A)!.4!29:(B)$) node [above,black] { $\vrp(t_0)$}; + + \end{tikzpicture} + + + + +
    +     +
    + +

    + The derivative of a vector-valued function is a measure of the + instantaneous rate of change, + measured by taking the limit as the length of [t_0,t_1] goes to 0. + Instead of thinking of an interval as [t_0,t_1], + we think of it as [c,c+h] for some value of h + (hence the interval has length h). + The average rate of change is + + \frac{\vec r(c+h)-\vec r(c)}{h} + + for any value of h\neq0. + We take the limit as h\to0 to measure the instantaneous rate of change; + this is the derivative of \vec r. +

    + + + Derivative of a Vector-Valued Function + +

    + Let \vec r(t) be continuous on an open interval I containing c. + vector-valued functionderivatives + derivativevector-valued functions + +

      +
    1. +

      + The derivative of \vec r at t=c is + + \vrp (c) = \lim_{h\to 0} \frac{\vec r(c+h) - \vec r(c)}{h} + . +

      +
      2. + +
    2. +

      + The derivative of \vec r is + + \vrp (t) = \lim_{h\to 0} \frac{\vec r(t+h) - \vec r(t)}{h} + . +

      +
      3. +
    +

    +     +     + + +

    + If a vector-valued function has a derivative for all c in an open interval I, + we say that \vec r(t) is + differentiable on I. +

    + +

    + Once again we might view this definition as intimidating, + but recall that we can evaluate limits component-wise. + The following theorem verifies that this means we can compute derivatives component-wise as well, + making the task not too difficult. +

    + + + + Derivatives of Vector-Valued Functions + +

    +

      +
    1. +

      + Let \vec r(t) = \la \, f(t), g(t)\,\ra. + Then + + \vrp(t) = \la\, \fp(t), \gp(t)\, \ra + . +

      +
      2. + +
    2. +

      + Let \vec r(t) = \la \, f(t), g(t), h(t)\,\ra. + Then + vector-valued functionderivatives + derivativevector-valued functions + + \vrp(t) = \la\, \fp(t), \gp(t), h'(t)\, \ra + . +

      +
      3. +
    +

    +     +     + + + + + Derivatives of vector-valued functions + +

    + Let \vec r(t) = \la t^2,t\ra. +

    + +

    +

      +
    1. +

      + Sketch \vec r(t) and \vrp(t) on the same axes. +

      +
      2. + +
    2. +

      + Compute \vrp(1) and sketch this vector with its initial point at the origin and at \vec r(1). +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + + allows us to compute derivatives component-wise, so + + \vrp(t) = \la 2t, 1\ra + . + \vec r(t) and \vrp(t) are graphed together in . + Note how plotting the two of these together, + in this way, is not very illuminating. + When dealing with real-valued functions, + plotting f(x) with \fp(x) gave us useful information as we were able to compare f and \fp at the same x-values. + When dealing with vector-valued functions, + it is hard to tell which points on the graph of \vrp correspond to which points on the graph of \vec r. +

      +
      2. + +
    2. +

      + We easily compute \vrp(1) = \la 2,1\ra, + which is drawn in + with its initial point at the origin, + as well as at \vec r(1) = \la 1,1\ra. + These are sketched in . +

      + +
      + Graphing the derivative of a vector-valued function in + +
      + + + + + Graph of the vector-valued function \vec r(t) = \la t^2,t\ra. + The graph of the function \vec r(t) is simply the graph of the parabola y=x^2, but instead of opening towards the positive y-axis, the function \vec r(t) opens towards the positive x-axis. + The function \vec r(t) is plotted on the interval [-2,2], where \vec r(-2)= \la 4,-2\ra and \vec r(2)= \la 4,2\ra. + The graph also includes the derivative function \vrp(t) = \la 2t, 1\ra which takes the path of the horizontal line y=1 going from left to right in the standard coordinate axes. + + Graph of the vector-valued function from the example and its derivative. + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-2.2,ymax=2.2, + xmin=-4.54,xmax=4.54 + ] + + \addplot+ [domain=-2:2,samples=40] ({x^2},{x}); + + \draw (axis cs:2,1.7) node { $\vec r(t)$}; + + \draw [thick,secondcolor] (axis cs: -4,1) -- (axis cs:4,1) node [below,pos=.9,black]{ $\vrp(t)$}; + \draw [thick,->,secondcolor] (axis cs: -2,1) -- (axis cs:-1.99,1); + + \draw [thick,->,firstcolor] (axis cs:1,-1) -- (axis cs:0.9801,-.99); + + \end{axis} + + \end{tikzpicture} + + + + +
      + +
      + + + + + Graph of the vector-valued function \vec r(t) = \la t^2,t\ra described in the previous image. + The graph also includes two copies of the vector \vrp(1) = \la 2,1\ra. + The first copy of the vector \vrp(1) = \la 2,1\ra begins at the origin, and ends at the point (2,1), which is also a point on the derivative function \vrp(t) = \la 2t, 1\ra from the previous image. + The second copy of the vector \vrp(1) = \la 2,1\ra begins at the point (1,1), which corresponds to the termination point of \vec r(1) = \la 1,1 \ra. + The second copy of the vector \vrp(1) = \la 2,1\ra is tangent to the function \vec r(t) = \la t^2,t\ra at the point (1,1) corresponding to when t=1 in the function \vec r(t). + + Graph of the vector-valued function from the example and its derivative. + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-2.2,ymax=2.2, + xmin=-4.54,xmax=4.54 + ] + + \addplot+ [domain=-2:2,samples=40] ({x^2},{x}); + + \draw [thick,->,secondcolor] (axis cs: 0,0) -- (axis cs:2,1) node [shift={(15pt,0)},pos=.6,black] { $\vrp(1)$}; + \draw [thick,->,secondcolor] (axis cs: 1,1) -- (axis cs:3,2) node [left,pos=.8,black] { $\vrp(1)$}; + + \draw [thick,->,firstcolor] (axis cs:1,-1) -- (axis cs:0.9801,-.99); + + \end{axis} + + \end{tikzpicture} + + + + +
      +       +
      +
      3. +
    +

    +     + +     + + + Derivatives of vector-valued functions + +

    + Let \vec r(t) = \la \cos(t) , \sin(t) , t\ra. + Compute \vrp(t) and \vrp(\pi/2). + Sketch \vrp(\pi/2) with its initial point at the origin and at \vec r(\pi/2). +

    +     + +

    + We compute \vrp as \vrp(t) = \la -\sin(t) , \cos(t) , 1\ra. + At t= \pi/2, we have \vrp(\pi/2) = \la -1,0,1\ra. + + shows a graph of \vec r(t), + with \vrp(\pi/2) plotted with its initial point at the origin and at \vec r(\pi/2). +

    + +
    + Viewing a vector-valued function and its derivative at one point + + + Graph of the vector-valued function \vec r(t) = \la \cos(t) , \sin(t) , t\ra. + The function is a circular spiral which climbs the z-axis. + Looking at the graph from above the z-axis, the function resembles a unit circle in the xy-plane. + Adding the z coordinate then creates the linearly increasing spiral. + The graph also includes two copies of the vector \vrp(\pi/2) = \la -1,0,1\ra + The first copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the origin, and ends at the point (-1,0,1). + The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the point (0,1,\pi/2), which corresponds to the termination point of \vec r(\pi/2). + The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra is also tangent to the function \vec r(t) at the point (0,1,\pi/2) corresponding to when t=\pi/2 in the function \vec r(t). + + Graph of the vector-valued function from the example and its derivative at a point. + + + + + //ASY file for figvvf23D.asy in Chapter 11 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,5,10); + //currentprojection=orthographic(-5.6,3.8,9.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myzchoice={0,2,4,6}; + real[] myychoice={-1,1}; + + defaultpen(0.5mm); + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,1.25); + pair zbounds=(-0.5,7); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple g(real t) {return (cos(t),sin(t),t);} + path3 mypath=graph(g,0,2pi,operator ..); + + defaultpen(0.75mm); + draw(O--(-1,0,1),redpen,Arrow3(size=4mm)); + + draw(g(pi/2)--g(pi/2)+(-1,0,1),redpen,Arrow3(size=4mm)); + + draw(mypath,bluepen,Arrow3(size=4mm)); + + + + +
    +     +     + +

    + In Examples + and , + sketching a particular derivative with its initial point at the origin did not seem to reveal anything significant. + However, when we sketched the vector with its initial point on the corresponding point on the graph, + we did see something significant: + the vector appeared to be tangent to the graph. + We have not yet defined what tangent + means in terms of curves in space; + in fact, we use the derivative to define this term. +

    + + + Tangent Vector, Tangent Line + +

    + Let \vec r(t) be a differentiable vector-valued function on an open interval I containing c, + where \vrp(c)\neq \vec 0. + tangent line + vector-valued functiontangent line +

    + +

    +

      +
    1. +

      + A vector \vec v is tangent to the graph of \vec r(t) at t=c + if \vec v is parallel to \vrp(c). +

      +
      2. + +
    2. +

      + The tangent line to the graph of + \vec r(t) at t=c is the line through + \vec r(c) with direction parallel to \vrp(c). + An equation of the tangent line is + + \vec \ell(t) = \vec r(c) + t\,\vrp(c) + . +

      +
      3. +
    +

    +     +     + + + + + Finding tangent lines to curves in space + +

    + Let \vec r(t) = \la t,t^2,t^3\ra on [-1.5,1.5]. + Find the vector equation of the line tangent to the graph of \vec r at t=-1. +

    +     + + +

    + To find the equation of a line, + we need a point on the line and the line's direction. + The point is given by \vec r(-1) = \la -1,1,-1\ra. + (To be clear, \la -1,1,-1\ra is a + vector, not a point, + but we use the point pointed to by this vector.) +

    + +

    + The direction comes from \vrp(-1). + We compute, component-wise, + \vrp(t) = \la 1,2t, 3t^2\ra. + Thus \vrp(-1) = \la 1,-2,3\ra. +

    + +
    + Graphing a curve in space with its tangent line + + + + + Graph of the vector-valued function \vec r(t) = \la t,t^2,t^3\ra on [-1.5,1.5]. + The function begins at the point (-1.5,2.25,-3.375), from which it begins to increase linearly in the x and z coordinates, and decrease in the y coordinate. + The function then curves towards the origin. + After passing through the origin, the function begins to increase in all x,y and z coordinates until it reaches the point (1.5,2.25,3.375), at which it ends. + The graph also contains the line \ell(t) = \la -1,1,-1\ra + t\la 1,-2,3\ra, which is the tangent line to the function at \vec r(-1), which is the point (-1,1,-1). + The line can be described as the line which passes through the point (-1,1,-1), moving in the direction of the vector \la 1,-2,3\ra. + Additionally, the line is defined for all t, so it also moves in the opposite direction of the vector \la 1,-2,3\ra, or in other words in the direction of \la -1,2,-3\ra from the point (-1,1,-1). + + Graph of the vector-valued function from the example and the tangent line to a point on the curve. + + + + + //ASY file for figvvfderiv1.asy in Chapter 11 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + pair xbounds=(-2.75,2.75); + pair ybounds=(-2.75,2.75); + pair zbounds=(-2.75,2.75); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve <t,t^2,t^3> for t from -1.5 to 1.5 + triple g(real t) {return (t,t^2,t^3);} + path3 mypath=graph(g,-1.5,1.5,operator ..); draw(mypath,bluepen); + + //Draw the line <-1,1,-1>+t<1,-2,3> for t=-0.75 to t=1 + draw((-1.75,2.5,-3.25)--(0,-1,2),redpen+linewidth(1)); + //label("$\vec{r}\,'(\pi/2)$",(-1,0,1),N); + + + + +
    + +

    + The vector equation of the line is \ell(t) = \la -1,1,-1\ra + t\la 1,-2,3\ra. + This line and \vec r(t) are sketched in . +

    +     +     + + + Finding tangent lines to curves + +

    + Find the equations of the lines tangent to + \vec r(t) = \la t^3,t^2\ra at t=-1 and t=0. +

    +     + +

    + We find that \vrp(t) = \la 3t^2,2t\ra. + At t=-1, we have + + \vec r(-1) = \la -1,1\ra \text{ and } \vrp(-1) = \la 3,-2\ra + , + so the equation of the line tangent to the graph of \vec r(t) at t=-1 is + + \ell(t) = \la -1,1\ra + t\la 3,-2\ra + . +

    + +

    + This line is graphed with \vec r(t) in . +

    + +
    + Graphing \vec r(t) and its tangent line in + + + + Graph of the vector-valued function \vec r(t) = \la t^3,t^2\ra. + The function begins near the point (-3,2), from which it is concave down and sloping downwards towards the origin. + After passing through the origin, the curve is concave down and begins increasing in both x and y coordinates. + The curve is also symmetric about the y-axis. + The graph also contains the line \ell(t) = \la -1,1\ra + t\la 3,-2\ra , which is tangent to the function \vec r(t) = \la t^3,t^2\ra at the point (-1,1) corresponding to when t=1. + + Graph of the vector-valued function from the example and the tangent line to a point on the curve. + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-2.96,ymax=2.96, + xmin=-3.5,xmax=3.5 + ] + + \addplot+ [domain=-1.5:1.5,samples=40] ({x^3},{x^2}); + \addplot+ [solid,domain=-1:1.5] ({3*x-1},{-2*x+1}); + + \draw (axis cs:2,2) node { $\vec r(t)$}; + \draw (axis cs:2,-1.7) node { $\vec \ell(t)$}; + + \draw [thick,firstcolor,->] (axis cs: 1,1) -- (axis cs:1.03,1.02); + \draw [thick,secondcolor,->] (axis cs:2,-1)--(axis cs:2.03,-1.02); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + At t=0, we have \vrp(0) = \la 0,0\ra=\vec 0! + This implies that the tangent line + has no direction. + We cannot apply , + hence cannot find the equation of the tangent line. +

    +     + +     + +

    + We were unable to compute the equation of the tangent line to + \vec r(t)= \la t^3,t^2\ra at t=0 because \vrp(0) = \vec 0. + The graph in + shows that there is a cusp at this point. + This leads us to another definition of smooth, + previously defined by + in . +

    + + + Smooth Vector-Valued Functions + +

    + Let \vec r(t) be a differentiable vector-valued function on an open interval I where \vrp(t) is continuous on I. + \vec r(t) is smooth + on I if \vrp(t)\neq \vec 0 on I. + smooth + vector-valued functionsmooth +

    +     +     + + + +

    + Having established derivatives of vector-valued functions, + we now explore the relationships between the derivative and other vector operations. + The following theorem states how the derivative interacts with vector addition and the various vector products. +

    + + + Properties of Derivatives of Vector-Valued Functions + +

    + Let \vec r and \vec s be differentiable vector-valued functions, + let f be a differentiable real-valued function, + and let c be a real number. + vector-valued functionderivatives + derivativevector-valued functions + dot productand derivatives + cross productand derivatives +

    + +

    +

      +
    1. +

      + \ds \frac{d}{dt}\Big(\vec r(t) \pm \vec s(t)\Big) = \vrp(t) \pm \vec s\,'(t) +

      +
      2. + +
    2. +

      + \ds \frac{d}{dt}\Big(c\vec r(t)\Big) = c\vrp(t) +

      +
      3. + +
    3. +

      + \ds \frac{d}{dt}\Big(f(t)\vec r(t)\Big) = \fp(t)\vec r(t) + f(t)\vrp(t) Product Rule +

      +
      4. + +
    4. +

      + \ds \frac{d}{dt}\Big(\vec r(t)\cdot \vec s(t) \Big) = \vrp(t)\cdot \vec s(t) + \vec r(t)\cdot \vec s\,'(t) Product Rule +

      +
      5. + +
    5. +

      + \ds \frac{d}{dt}\Big(\vec r(t)\times \vec s(t) \Big) = \vrp(t)\times \vec s(t) + \vec r(t)\times \vec s\,'(t) Product Rule +

      +
      6. + +
    6. +

      + \ds \frac{d}{dt}\Big(\vec r\big(f(t)\big)\Big) = \vrp\big(f(t)\big)\fp(t) Chain Rule +

      +
      7. +
    +

    +     +     + + + + + Using derivative properties of vector-valued functions + +

    + Let \vec r(t) = \la t, t^2-1\ra and let + \vec u(t) be the unit vector that points in the direction of \vec r(t). +

    + +

    +

      +
    1. +

      + Graph \vec r(t) and + \vec u(t) on the same axes, on [-2,2]. +

      +
      2. + +
    2. +

      + Find \vec u\,'(t) and sketch \vec u\,'(-2), + \vec u\,'(-1) and \vec u\,'(0). + Sketch each with initial point the corresponding point on the graph of \vec u. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + To form the unit vector that points in the direction of \vec r, + we need to divide \vec r(t) by its magnitude. + + \norm{\vec r(t)} = \sqrt{t^2+(t^2-1)^2} \Rightarrow \vec u(t) = \frac{1}{\sqrt{t^2+(t^2-1)^2}}\la t,t^2-1\ra + . + \vec r(t) and \vec u(t) are graphed in . + Note how the graph of \vec u(t) forms part of a circle; + this must be the case, as the length of + \vec u(t) is 1 for all t. +

      + +
      + Graphing \vec r(t) and \vec u(t) in + + + + Graph of the vector-valued function \vec r(t) = \la t,t^2 -1\ra on [-2,2]. + The function looks like the parabola y=x^2 -1, which takes the path going towards positive x. + The function begins at the point (-2,3), decreases until reaching the point (0,-1), and then increases until ending at the point (2,3). + The graph also contains the function \vec u(t) which is the unit vector that points in the direction of \vec r(t). + The function \vec u(t) looks like the unit circle which is missing a piece of the top. + The circular arc from the graph of \vec u(t) begins at the point where the vector \vec r(-2)=\la -2,3\ra crosses the unit circle. + The circular arc then goes counterclockwise following the path of the unit circle, until ending at the point where the vector\vec r(2)=\la 2,3\ra crosses the unit circle. + + Graph of the vector-valued function with the unit vector function which points in the direction of the function. + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-1.1,ymax=3.1, + xmin=-2.5,xmax=2.5 + ] + + \addplot+ [domain=-2:2,samples=40] ({x},{x^2-1}); + \addplot+ [solid,domain=-2:2,samples=60] ({x/sqrt(x^2+(x^2-1)^2)},{(x^2-1)/sqrt(x^2+(x^2-1)^2)}); + + \draw (axis cs:-2,1.5) node { $\vec r(t)$}; + \draw (axis cs:.7,1.1) node { $\vec u(t)$}; + + \draw [thick,->,secondcolor] (axis cs: -0.768,.64) -- (axis cs:-.7737,.633); + \draw [thick,->,firstcolor] (axis cs:-1.5,1.25) -- (axis cs:-1.49,1.22); + + \end{axis} + + \end{tikzpicture} + + + + +
      +
      2. + +
    2. +

      + To compute \vec u\,'(t), + we use , writing + + \vec u(t) = f(t)\vec r(t), \text{ where } f(t) = \frac{1}{\sqrt{t^2+(t^2-1)^2}}=\big(t^2+(t^2-1)^2\big)^{-1/2} + . + (We could write + + \vec u(t) = \la \frac{t}{\sqrt{t^2+(t^2-1)^2}}, \frac{t^2-1}{\sqrt{t^2+(t^2-1)^2}}\ra + + and then take the derivative. + It is a matter of preference; + this latter method requires two applications of the Quotient Rule where our method uses the Product and Chain Rules.) We find \fp(t) using the Chain Rule: + + \fp(t) \amp = -\frac12\big(t^2+(t^2-1)^2\big)^{-3/2}\big(2t+2(t^2-1)(2t)\big) + \amp = -\frac{2t(2t^2-1)}{2\big(\sqrt{t^2+(t^2-1)^2}\,\big)^3} + + We now find \vec u\,'(t) using part 3 of : + + \vec u\,'(t) \amp = \fp(t)\vec u(t) + f(t)\vec u\,'(t) + \amp = -\frac{2t(2t^2-1)}{2\big(\sqrt{t^2+(t^2-1)^2}\,\big)^3}\la t,t^2-1\ra + \frac{1}{\sqrt{t^2+(t^2-1)^2}}\la 1,2t\ra + . + This is admittedly very messy; + such is usually the case when we deal with unit vectors. + We can use this formula to compute \vec u\,'(-2), + \vec u\,'(-1) and \vec u\,'(0): + + \vec u\,'(-2) \amp = \la-\frac{15}{13 \sqrt{13}},-\frac{10}{13 \sqrt{13}}\ra \approx \la -0.320,-0.213\ra + \vec u\,'(-1) \amp = \la 0,-2\ra + \vec u\,'(0) \amp = \la 1,0\ra + +

      + +
      + Graphing some of the derivatives of \vec u(t) in + + + + Graph of the function \vec u(t) which is the unit vector that points in the direction of \vec r(t). + The graph also contains three unit vectors, \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0). + The vector \vec u\,'(-2)\approx \la -0.320,-0.213\ra begins at the starting point of the function \vec u(t) and points in the direction tangent to the circular arc at t=-2. + The vector \vec u\,'(-1)= \la 0,-2\ra begins at the point (-1,0) corresponding to the termination point of \vec u(-1) + From here, the vector points straight down, as it is tangent to the leftmost point of the circular arc. + The vector \vec u\,'(0)= \la 1,0\ra begins at the point (0,-1) corresponding to the termination point of \vec u(0). + From here the vector points to the right, as it is tangent to the bottom point of the circular arc. + If the vectors \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0) were extended to lines, all three lines would be tangent to a point on the nearly complete unit circle given by \vec u(t). + + Graph of the circular unit vector function along with three derivative vectors of the function. + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-2.1,ymax=1.1, + xmin=-1.92,xmax=1.92 + ] + + \addplot+ [domain=-2:2,samples=80] ({x/sqrt(x^2+(x^2-1)^2)},{(x^2-1)/sqrt(x^2+(x^2-1)^2)}); + + \draw (axis cs:.5,.6) node { $\vec u(t)$}; + \draw [thick,firstcolor,->] (axis cs: -0.98226, 0.187522) -- (axis cs:-0.985435, 0.170055); + \draw [thick,firstcolor,->] (axis cs:-1.5,1.25) -- (axis cs:-1.49,1.22); + \draw [thick,secondcolor,->] (axis cs:-0.5547, 0.83205) -- (axis cs:-0.87472, 0.618704); + \draw [thick,secondcolor,->] (axis cs:-1., 0) -- (axis cs:-1., -2.); + \draw [thick,secondcolor,->] (axis cs:0., -1.) -- (axis cs:1., -1.); + + \end{axis} + + \end{tikzpicture} + + + + +
      + +

      + Each of these is sketched in . + Note how the length of the vector gives an indication of how quickly the circle is being traced at that point. + When t=-2, the circle is being drawn relatively slow; + when t=-1, the circle is being traced much more quickly. +

      +
      3. +
    +

    +     + +     + +

    + It is a basic geometric fact that a line tangent to a circle at a point P is perpendicular to the line passing through the center of the circle and P. + This is illustrated in ; + each tangent vector is perpendicular to the line that passes through its initial point and the center of the circle. + Since the center of the circle is the origin, + we can state this another way: + \vec u\,'(t) is orthogonal to \vec u(t). +

    + +

    + Recall that the dot product serves as a test for orthogonality: + if \vec u\cdot \vec v = 0, + then \vec u is orthogonal to \vec v. + Thus in the above example, \vec u(t)\cdot \vec u\,'(t)=0. +

    + +

    + This is true of any vector-valued function that has a constant length, that is, + that traces out part of a circle. + It has important implications later on, + so we state it as a theorem (and leave its formal proof as an Exercise.) +

    + + + Vector-Valued Functions of Constant Length + +

    + Let \vec r(t) be a vector-valued function of constant length that is differentiable on an open interval I. + That is, \norm{\vec r(t)} = c for all t in I; equivalently, + \vec r(t)\cdot \vec r(t) = c^2 for all t in I. + Then \vec r(t)\cdot\vrp(t) = 0 for all t in I. + vector-valued functionof constant length +

    +     +     +     + + + Integration +

    + Before formally defining integrals of vector-valued functions, + consider the following equation that our calculus experience tells us + should be true: + + \int_a^b \vrp(t)\, dt = \vec r(b) - \vec r(a) + . +

    + +

    + That is, the integral of a rate of change function should give total change. + In the context of vector-valued functions, + this total change is displacement. + The above equation is true; + we now develop the theory to show why. +

    + +

    + We can define antiderivatives and the indefinite integral of vector-valued functions in the same manner we defined indefinite integrals in . + However, we cannot define the definite integral of a vector-valued function as we did in . + That definition was based on the signed area between a function y=f(x) and the x-axis. + An area-based definition will not be useful in the context of vector-valued functions. + Instead, we define the definite integral of a vector-valued function in a manner similar to that of , + utilizing Riemann sums. +

    + + + Antiderivatives, Indefinite and Definite Integrals of Vector-Valued Functions + +

    + Let \vec r(t) be a continuous vector-valued function on [a,b]. + An antiderivative of \vec r(t) is a function + \vec R(t) such that \vec R'(t) = \vec r(t). +

    + +

    + The set of all antiderivatives of \vec r(t) is the + indefinite integral of \vec r(t), denoted by + + \int \vec r(t)\, dt + . +

    + +

    + The definite integral of \vec r(t) on [a,b] is + + \int_a^b \vec r(t)\, dt =\lim_{\norm{\Delta t}\to 0} \sum_{i=1}^n\vec r(c_i)\Delta t_i + , + where \Delta t_i is the length of the + ith subinterval of a partition of [a,b], + \norm{\Delta t} is the length of the largest subinterval in the partition, + and c_i is any value in the + ith subinterval of the partition. + antiderivativeof vector-valued function + definite integralof vector-valued function + indefinite integralof vector-valued function + integrationof vector-valued function +

    +     +     + +

    + It is probably difficult to infer meaning from the definition of the definite integral. + The important thing to realize from the definition is that it is built upon limits, + which we can evaluate component-wise. +

    + +

    + The following theorem simplifies the computation of definite integrals; + the rest of this section and the following section will give meaning and application to these integrals. +

    + + + Indefinite and Definite Integrals of Vector-Valued Functions + +

    + Let \vec r(t) = \la f(t),g(t)\ra be a vector-valued function in + \mathbb{R}^2 that is continuous on [a,b]. +

    + +

    +

      +
    1. +

      + \ds \int \vec r(t)\, dt = \la \int f(t)\, dt, \int g(t)\, dt\ra +

      +
      2. + +
    2. +

      + \ds \int_a^b \vec r(t)\, dt = \la \int_a^b f(t)\, dt, \int_a^b g(t)\, dt\ra +

      +
      3. +
    +

    + +

    + A similar statement holds for vector-valued functions in \mathbb{R}^3. + vector-valued functionintegration + integrationof vector-valued functions +

    +     +     + + + + + Evaluating a definite integral of a vector-valued function + +

    + Let \vec r(t) = \la e^{2t},\sin(t) \ra. + Evaluate \ds \int_0^1 \vec r(t) \,dt. +

    +     + +

    + We follow . + + \int_0^1 \vec r(t) \,dt \amp = \int_0^1 \la e^{2t},\sin(t) \ra \,dt + \amp = \la \int_0^1 e^{2t}\,dt\,, \int_0^1 \sin(t) \,dt \ra + \amp = \la \frac12e^{2t}\Big|_0^1\,, -\cos(t) \Big|_0^1\ra + \amp = \la \frac12(e^2-1)\,, -\cos(1)+1\ra + \amp \approx \la 3.19,0.460\ra + . +

    +     + +     + + + Solving an initial value problem + +

    + Let \vrp'(t) = \la 2, \cos(t) , 12t\ra. + Find \vec r(t) where: +

    + +

    +

      +
    • +

      + \vec r(0) = \la-7,-1,2\ra and +

      +
      • + +
    • +

      + \vrp(0) = \la 5,3,0\ra. +

      +
      • +
    +

    +     + +

    + Knowing \vrp'(t) = \la 2,\cos(t) , 12t\ra, + we find \vrp(t) by evaluating the indefinite integral. + + \int \vrp'(t)\,dt \amp = \la \int 2\,dt\,, \int \cos(t) \,dt\,, \int 12t\,dt\ra + \amp = \la 2t+C_1, \sin(t) + C_2, 6t^2 + C_3\ra + \amp = \la 2t,\sin(t) ,6t^2 \ra + \la C_1,C_2,C_3\ra + \amp = \la 2t,\sin(t) ,6t^2 \ra + \vec C + . +

    + +

    + Note how each indefinite integral creates its own constant which we collect as one constant vector \vec C. + Knowing \vrp(0) = \la 5,3,0\ra allows us to solve for \vec C: + + \vrp(t) \amp = \la 2t,\sin(t) ,6t^2 \ra + \vec C + \vrp(0) \amp = \la 0,0,0 \ra + \vec C + \la 5,3,0\ra \amp = \vec C + . +

    + +

    + So \vrp(t) = \la 2t,\sin(t) ,6t^2\ra + \la 5,3,0\ra = \la 2t+5, \sin(t) + 3, 6t^2\ra. + To find \vec r(t), we integrate once more. + + \int \vrp(t)\,dt \amp = \la \int 2t+5\,dt, \int \sin(t) + 3\,dt, \int 6t^2\,dt \ra + \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \vec C + . +

    + +

    + With \vec r(0) = \la -7,-1,2\ra, + we solve for \vec C: + + \vec r(t) \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \vec C + \vec r(0) \amp = \la 0,-1,0\ra + \vec C + \la -7,-1,2\ra \amp = \la 0,-1,0\ra + \vec C + \la -7,0,2\ra \amp = \vec C + . +

    + +

    + So + + \vec r(t) \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \la -7,0,2\ra + \amp = \la t^2+5t-7,-\cos(t) +3t,2t^3+2\ra + . +

    +     + +     + +

    + What does the integration of a vector-valued function mean? + There are many applications, + but none as direct as the area under the curve + that we used in understanding the integral of a real-valued function. +

    + +

    + A key understanding for us comes from considering the integral of a derivative: + + \int_a^b \vrp(t)\, dt = \vec r(t)\Big|_a^b = \vec r(b)-\vec r(a) + . +

    + +

    + Integrating a rate of change + function gives displacement. + displacement +

    + +

    + Noting that vector-valued functions are closely related to parametric equations, + we can describe the arc length of the graph of a vector-valued function as an integral. + Given parametric equations x=f(t), + y=g(t), the arc length on [a,b] of the graph is + + \text{ Arc Length } = \int_a^b\sqrt{\fp(t)^2+\gp(t)^2}\, dt + , + as stated in + in . + If \vrt = \la f(t), g(t)\ra, + note that \sqrt{\fp(t)^2+\gp(t)^2} = \norm{\vrp(t)}. + Therefore we can express the arc length of the graph of a vector-valued function as an integral of the magnitude of its derivative. +

    + + + Arc Length of a Vector-Valued Function + +

    + Let \vrt be a vector-valued function where \vrp(t) is continuous on [a,b]. + The arc length L of the graph of \vrt is + vector-valued functionarc length + arc length + + L = \int_a^b \norm{\vrp(t)}\, dt + . +

    +     +     + +

    + Note that we are actually integrating a scalar-function here, + not a vector-valued function. +

    + +

    + The next section takes what we have established thus far and applies it to objects in motion. + We will let \vrt describe the path of an object in the plane or in space and will discover the information provided by \vrp(t) and \vrp'(t). +

    +     + + + + Terms and Concepts + + + +

    + Limits, derivatives and integrals of vector-valued functions are all evaluated -wise. +

    +     + + + + + + + + +     + + + + +

    + The definite integral of a velocity function gives . +

    +     + + + + + + + + +     + + + + +

    + Why is it generally not useful to graph both + \vec r(t) and \vrp(t) on the same axes? +

    +     + + + +

    + It is difficult to identify the points on the graphs of + \vec r(t) and \vrp(t) that correspond to each other. +

    +     + +     + + + +

    + contains three product rules. + What are the three different types of products used in these rules? +

    +     + + + +

    + A scalar-vector product, a dot product and a cross product. +

    +     +     +     + + + Problems + + + + +

    + Evaluate the given limit. +

    +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $L=Compute("<11,74,sin(5)>"); + + + +

    + \lim\limits_{t\to 5} \la 2t+1,3t^2-1,\sin(t) \ra +

    +

    + +

    +     +     +     + + + + +

    + \lim\limits_{t\to 3} \la e^t,\frac{t^2-9}{t+3}\ra +

    +     + +

    + \la e^3,0\ra +

    +     + +     + + + + + Context("Vector2D"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $L=Compute("<1,e>"); + + + +

    + \lim\limits_{t\to 0} \la \frac{t}{\sin(t) }, (1+t)^{\frac1t}\ra +

    +

    + +

    +     +     +     + + + + +

    + \lim\limits_{h\to 0} \frac{\vec r(t+h)-\vec r(t)}{h}, where + \vec r(t) = \la t^2,t,1\ra. +

    +     + +

    + \la2t,1,0\ra +

    +     + +     + +     + + + +

    + Identify the interval or union of intervals on which \vec r(t) is continuous. +

    +     + + + + + Context("Interval"); + $dom=Compute("(-inf,0)U(0,inf)"); + + + +

    + \vec r(t) = \la t^2,1/t\ra +

    + +

    + +

    +     +     +     + + + + +

    + \vec r(t) = \la \cos(t) , e^t, \ln(t) \ra +

    +     + +

    + (0,\infty) +

    +     + +     + +     + + + +

    + Find the derivative of the given function. +

    +     + + + + + Context("Vector"); + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $d = Compute("< -sin(t), e^t, 1/t>"); + + + +

    + \vec r(t) = \la \cos(t) , e^t, \ln(t) \ra +

    + +

    + +

    +     +     +     + + + + +

    + \ds \vec r(t) = \la \frac 1t, \frac {2t-1}{3t+1}, \tan(t) \ra +

    +     + +

    + \vrp(t) = \la -1/t^2, 5/(3t+1)^2, \sec^2(t) \ra +

    +     + +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $d = Compute("< 2t sin(t)+t^2cos(t), 6t^2+10t>"); + + + +

    + \vec r(t) = (t^2)\la \sin(t) , 2t+5\ra +

    + +

    + +

    +     +     +     + + + + + Context("Numeric"); + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $d = Compute("(t^2+1)cos(t) +2tsin(t) + 4t+3"); + + + +

    + \vec r(t) = \la t^2+1, t-1\ra\cdot \la \sin(t) , 2t+5\ra +

    + +

    + +

    +     +     +     + + + + + Context("Vector"); + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $d = Compute("<-1,cos(t)-2t,6t^2+10t+2+cos(t)-sin(t)-tcos(t)>"); + + + +

    + \vec r(t) = \la t^2+1, t-1,1\ra\times \la \sin(t) , 2t+5,1\ra +

    + +

    + +

    +     +     +     + + + +

    + \ds \vec r(t) = \la \cosh t, \sinh t\ra +

    +     + +

    + \vrp(t) = \la \sinh t,\cosh t\ra +

    +     +     + +     + + + +

    + First, find \vrp (t). + Then sketch \vec r(t) and \vrp(1), + with the initial point of \vrp(1) at \vec r(1). +

    +     + + + + +

    + \ds \vec r(t) = \la t^2+t, t^2-t\ra +

    +     + + + + + Graph of the function \vec r(t) = \la t^2+t, t^2-t\ra. + The function \vec r(t) resembles a standard parabola which has been rotated 45 degrees clockwise. + The curve passes through the point (2,0), the origin, and the point (0,2) in the given order. + The graph also contains \vrp(1) beggining at the point (2,0) corresponding to the termination point of \vec r(1). + The vector \vrp(1)= \la 3,1 \ra and is tangent to \vec r(t) = \la t^2+t, t^2-t\ra corresponding to the point where t=1. + + Graph of the vector-valued function and its derivative at a point. + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.5,ymax=6.5, + xmin=-1.5,xmax=6.5 + ] + + \addplot+ [domain=-2:2,samples=40] ({x^2+x},{x^2-x}); + + \draw [thick,secondcolor,->,>=stealth] (axis cs: 2,0) -- (axis cs:5,1) node [black,right] { $\vrp(1)$}; + + \end{axis} + + \end{tikzpicture} + + + + + +

    + \vrp(t) = \la 2t+1,2t-1\ra +

    +     + +     + + + + +

    + \ds \vec r(t) = \la t^2-2t+2,t^3-3t^2+2t\ra +

    +     + + + + + Graph of the function \vec r(t) = \la t^2-2t+2,t^3-3t^2+2t\ra. + The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). + After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). + From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. + The graph of the function is also symmetric about the y-axis. + The graph also contains \vrp(1) beginning at the point (1,0) corresponding to the termination point of \vec r(1). + The vector \vrp(1)= \la 0,-1 \ra and is tangent to \vec r(t) corresponding to the point where t=1. + + Graph of the vector-valued function and its derivative at a point. + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.5,ymax=1.5, + xmin=-.5,xmax=3.5 + ] + + \addplot+ [domain=-.5:2.5,samples=60] ({x^2-2*x+2},{x^3-3*x^2+2*x}); + + \draw [thick,secondcolor,->,>=stealth] (axis cs: 1,0) -- (axis cs:1,-1) node [black,right] { $\vrp(1)$}; + + \end{axis} + + \end{tikzpicture} + + + + + +

    + \vrp(t) = \la 2t-2,3t^2-6t+2\ra +

    +     + +     + + + + +

    + \ds \vec r(t) = \la t^2+1,t^3-t\ra +

    +     + + + + + Graph of the function \vec r(t) = \la t^2+1,t^3-t\ra. + The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). + After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). + From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. + The graph of the function is also symmetric about the y-axis. + The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1). + The vector \vrp(1)= \la 2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1. + + Graph of the vector-valued function and its derivative at a point. + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-2.5,ymax=2.5, + xmin=-.5,xmax=4.5 + ] + + \addplot+[domain=-.5:2.5,samples=50] ({x^2-2*x+2},{x^3-3*x^2+2*x}); + + \draw [thick,secondcolor,->,>=stealth] (axis cs: 2,0) -- (axis cs:4,2) node [black,above] { $\vrp(1)$}; + + \end{axis} + + \end{tikzpicture} + + + + + +

    + \vrp(t) = \la 2t,3t^2-1\ra +

    +     + +     + + + + +

    + \ds \vec r(t) = \la t^2-4t+5,t^3-6t^2+11t-6\ra +

    +     + + + + + Graph of the function \vec r(t) = \la t^2-4t+5,t^3-6t^2+11t-6\ra. + The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). + After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). + From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. + The graph of the function is also symmetric about the y-axis. + The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1). + The vector \vrp(1)= \la -2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1. + + Graph of the vector-valued function and its derivative at a point. + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-2.5,ymax=2.5, + xmin=-.5,xmax=4.5 + ] + + \addplot+ [domain=-.5:2.5,samples=40] ({x^2-2*x+2},{x^3-3*x^2+2*x}); + + \draw [thick,secondcolor,->,>=stealth] (axis cs: 2,0) -- (axis cs:0,2) node [black,right] { $\vrp(1)$}; + + \end{axis} + + \end{tikzpicture} + + + + + +

    + \vrp(t) = \la 2t-4,3t^2-12t+11\ra +

    +     + +     + +     + + + +

    + Give the equation of the line tangent to the graph of + \vec r(t) at the given t value. +

    +     + + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + $L=ParametricLine("<2+3t, t>"); + + + +

    + \vec r(t) = \la t^2+t, t^2-t\ra, at t=1 +

    + + + \vec\ell(t)= + +

    + +

    +     +     +     + + + + + + Context("Vector2D"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->are(t=>'Real'); + $L=ParametricLine("(3sqrt(2)/2,sqrt(2)/2) + t<-3sqrt(2)/2,sqrt(2)/2 >"); + + + +

    + \vec r(t) = \la 3\cos(t) ,\sin(t) \ra, at t=\pi/4 +

    + + + \vec\ell(t)= + +

    + +

    +     +     +     + + + + +

    + \ds \vec r(t) = \la 3\cos(t) ,3\sin(t) ,t\ra at t=\pi. +

    +     + +

    + \ell(t) = \la -3,0,\pi\ra + t\la0,-3,1\ra +

    +     + +     + + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->are(t=>'Real'); + $L=ParametricLine("(1,0,0) + t<1,1,1 >"); + + + +

    + \vec r(t) = \la e^t,\tan(t) ,t\ra, at t=0. +

    + + + \vec\ell(t)= + +

    + +

    +     +     +     + +     + + + +

    + Find the value(s) of t for which \vec r(t) is not smooth. +

    +     + + + + +

    + \ds \vec r(t) = \la \cos(t) ,\sin(t) - t\ra +

    +     + +

    + t=2n\pi, where n is an integer; + so t = \ldots-4\pi,-2\pi,0,2\pi,4\pi,\ldots +

    +     + +     + + + + + $notsmooth=List("1"); + + + +

    + \vec r(t) = \la t^2-2t+1,t^3+t^2-5t+3\ra +

    + + + Enter the value(s) of t for which \vec r(t) is not smooth. + +

    + +

    +     +     +     + + + + +

    + \ds \vec r(t) = \la \cos(t) -\sin(t) , \sin(t) - \cos(t) ,\cos(4t)\ra +

    +     + +

    + \vec r(t) is not smooth at t=3\pi/4+n\pi, + where n is an integer +

    +     + +     + + + + + $notsmooth=List("1,-1"); + + + +

    + \vec r(t) = \la t^3-3t+2, -\cos(\pi t),\sin^2(\pi t) \ra +

    + + + Enter the value(s) of t for which \vec r(t) is not smooth. + +

    + +

    +     +     +     + +     + + + +

    + The following exercises ask you to verify parts of . + In each let f(t) = t^3, + \vec r(t) =\la t^2,t-1,1\ra and \vec s(t) = \la \sin(t) , e^t,t\ra. + Compute the various derivatives as indicated. +

    +     + + + + +

    + Simplify f(t)\vec r(t), then find its derivative; + show this is the same as \fp(t)\vec r(t) + f(t)\vrp(t). +

    +     + +

    + Both derivatives return \la 5t^4,4t^3-3t^2,3t^2\ra. +

    +     + +     + + + + +

    + Simplify \vec r(t)\cdot\vec s(t), then find its derivative; + show this is the same as \vrp(t)\cdot\vec s(t) + \vec r(t)\cdot\vec s\,'(t). +

    +     + +

    + Both derivatives return 2\sin(t) +t^2\cos(t) + te^t+1. +

    +     + +     + + + + +

    + Simplify \vec r(t)\times\vec s(t), then find its derivative; + show this is the same as \vrp(t)\times\vec s(t) + \vec r(t)\times\vec s\,'(t). +

    +     + +

    + Both derivatives return \la 2t-e^t-1,\cos(t) -3t^2,(t^2+2t)e^t-(t-1)\cos(t) -\sin(t) \ra. +

    +     + +     + + + +

    + Simplify \ds \vec r\big(f(t)\big), then find its derivative; + show this is the same as \ds \vrp\big(f(t)\big)\fp(t). +

    +     + +

    + Both derivatives return \la 6t^5,3t^2,0\ra +

    +     +     + +     + + + +

    + Evaluate the given definite or indefinite integral. +

    +     + + + + +

    + \ds \int \la t^3,\cos(t) , te^t\ra\,dt +

    +     + +

    + \la \frac14t^4,\sin(t) ,te^t-e^t\ra + \vec C +

    +     + +     + + + + +

    + \ds \int \la \frac{1}{1+t^2},\sec^2(t) \ra\,dt +

    +     + +

    + \la \tan^{-1}(t) ,\tan(t) \ra + \vec C +

    +     + +     + + + + + Context("Vector2D"); + $int=Compute("<-2,0>"); + + + +

    + \displaystyle\int_0^{\pi} \la -\sin(t) ,\cos(t) \ra\,dt=. +

    +     +     +     + + + + +

    + \ds \int_{-2}^{2} \la 2t+1,2t-1\ra\,dt +

    +     + +

    + \la4,-4 \ra +

    +     + +     + +     + + + +

    + Solve the given initial value problems. +

    +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + $r=Compute("<t^2/2+2,-cos(t)+3>"); + + + +

    + Find \vec r(t), given that + \vrp(t) = \la t,\sin(t) \ra and \vec r(0) = \la 2,2\ra. +

    + +

    + \vec r(t)= +

    +     +     +     + + + + +

    + Find \vec r(t), + given that \vrp(t) = \la 1/(t+1),\tan(t) \ra and +

    + +

    + \vec r(0) = \la 1,2\ra. +

    +     + +

    + \vec r(t) = \la \ln\abs{t+1} + 1, -\ln\abs{\cos(t) } + 2\ra +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(t=>'Real'); + $r=Compute("<t^4/12+t+4,t^3/6+2t+5,t^2/2+3t+6>"); + + + +

    + Find \vec r(t), given that \vrp'(t) = \la t^2,t,1\ra, + \vrp(0) = \la 1,2,3\ra and \vec r(0) = \la 4,5,6\ra. +

    + +

    + \vec r(t)= +

    +     +     +     + + + + +

    + Find \vec r(t), + given that \vrp'(t) = \la \cos(t) ,\sin(t) ,e^t\ra, +

    + +

    + \vrp(0) = \la 0,0,0\ra and \vec r(0) = \la 0,0,0\ra. +

    +     + +

    + \vec r(t) = \la-\cos(t) +1,t-\sin(t) ,e^t-t-1 \ra +

    +     + +     + +     + + + +

    + Find the arc length of \vec r(t) on the indicated interval. +

    +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstatnFunctions=>0); + $l=Formula("2sqrt(13)pi"); + + + +

    + \vec r(t) = \la 2\cos(t) , 2\sin(t) , 3t \ra on [0,2\pi]. +

    + +

    + +

    +     +     +     + + + + +

    + \vec r(t) = \la 5\cos(t) , 3\sin(t) , 4\sin(t) \ra on [0,2\pi]. +

    +     + +

    + 10\pi +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstatnFunctions=>0); + $l=Formula("1/54(22^(3/2)-8)"); + + + +

    + \vec r(t) = \la t^3,t^2,t^3 \ra on [0,1]. +

    + +

    + +

    +     +     +     + + + + +

    + \vec r(t) = \la e^{-t}\cos(t) ,e^{-t}\sin(t) \ra on [0,1]. +

    +     + +

    + \sqrt{2}(1-e^{-1}) +

    +     + +     + +     + + + + +

    + Prove ; + that is, show if \vec r(t) has constant length and is differentiable, + then \vec r(t)\cdot \vrp(t)=0. (Hint: + use the Product Rule to compute \frac{d}{dt}\big(\vec r(t)\cdot\vec r(t)\big).) +

    +     + +

    + As \vec r(t) has constant length, + \vec r(t)\cdot\vec r(t)=c^2 for some constant c. + Thus + + \vec r(t)\cdot\vec r(t) \amp = c^2 + \frac{d}{dt}\big(\vec r(t)\cdot\vec r(t)\big) \amp = \frac{d}{dt}\big(c^2\big) + \vrp(t)\cdot \vec r(t)+\vec r(t)\cdot \vrp(t) \amp = 0 + 2\vec r(t)\cdot \vrp(t) \amp =0 + \vec r(t)\cdot \vrp(t) \amp =0 + . +

    +     + +     +     +     +
    +
    + The Calculus of Motion + +

    + A common use of vector-valued functions is to describe the motion of an object in the plane or in space. + A position function + \vec r(t) gives the position of an object at time t. More formally, let O = \vec 0 (either in the plane or in space) and suppose an object is at point P_c at time t=t_c. Then \vec r\big(t_c\big) = \overrightarrow{OP_c}; that is, the vector \vec r\big(t_c\big) points to the location of the object at a given time. + This section explores how derivatives and integrals are used to study the motion described by such a function. +

    + + + + + Velocity, Speed and Acceleration + +

    + Let \vec r(t) be a position function in + \mathbb{R}^2 or \mathbb{R}^3. +

    + +

    +

    +
  • + Velocity +

    + The instantaneous rate of position change, denoted \vec v(t); + that is, \vec v(t) = \vrp(t). +

    +
    • + +
  • + Speed +

    + The magnitude of velocity: \norm{\vec v(t)}. +

    +
    • + +
  • + Acceleration +

    + The instantaneous rate of velocity change, denoted \vec a(t); + that is, \vec a(t) = \vec v\,'(t) = \vrp'(t). + velocity + speed + acceleration + vector-valued functiondescribing motion +

    +
    • +
    +

    +     +     + + + Finding velocity and acceleration + +

    + An object is moving with position function + \vec r(t) = \la t^2-t,t^2+t\ra, -3\leq t\leq 3, + where distances are measured in feet and time is measured in seconds. +

    + +

    +

      +
    1. +

      + Find \vvt and \vat. +

      +
      2. + +
    2. +

      + Sketch \vrt; plot \vec v(-1), + \vec a(-1), \vec v(1) and \vec a(1), + each with their initial point at their corresponding point on the graph of \vrt. +

      +
      3. + +
    3. +

      + When is the object's speed minimized? +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + Taking derivatives, we find + + \vvt = \vrp(t) =\la 2t-1,2t+1\ra \text{ and } \vat = \vrp'(t) = \la 2,2\ra + . + Note that acceleration is constant. +

      +
      2. + +
    2. +

      + \vec v(-1) = \la -3,-1\ra, + \vec a(-1) = \la 2,2\ra; + \vec v(1) = \la 1,3\ra, + \vec a(1) = \la 2,2\ra. + These are plotted with \vrt in . + + We can think of acceleration as pulling + the velocity vector in a certain direction. + At t=-1, the velocity vector points down and to the left; + at t=1, + the velocity vector has been pulled in the \la 2,2\ra + direction and is now pointing up and to the right. + In we plot more velocity/acceleration vectors, + making more clear the effect acceleration has on velocity. +

      + +
      + Graphing the position, velocity and acceleration of an object in + +
      + + + + The curve corresponding the position function for this example, with velocity and acceleration vectors shown at two points. + +

      + The curve given by \vec r(t) = \la t^2-t,t^2+t\ra is plotted, + for -3\leq t\leq 3. + The shape of the curve is parabolic, but rotated, so that it is symmetric about the line y=x. + There are intercepts at (2,0), (0,0), and (0,2) (given in order of increasing t). +

      + +

      + At each of the points (2,0) and (0,2), a pair of vectors is plotted. + One vector in each pair corresponds to velocity, and is tangent to the curve. + The other vector represents acceleration; this vector is always parallel to the line of symmetry for the curve. +

      +       + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-1.9,ymax=13, + xmin=-1.9,xmax=13 + ] + + \addplot+ [domain=-3:3,samples=60] ({x^2-x},{x^2+x}); + + \filldraw (axis cs:2,0) circle (2.4pt); + \filldraw (axis cs:0,2) circle (2.4pt); + + \draw [thick,->,secondcolor] (axis cs:2,0) -- (axis cs:-1,-1); + \draw [thick,->,secondcolor] (axis cs:0,2) -- (axis cs:1,5); + \draw [thick,->] (axis cs:2,0) -- (axis cs:4,2); + \draw [thick,->] (axis cs:0,2) -- (axis cs:2,4); + \draw [thick,->,firstcolor] (axis cs:8.75,3.75) -- (axis cs: 8.69,3.71); + + \end{axis} + + \end{tikzpicture} + + + + +
      + +
      + + + + The same curve as the previous image, with more vectors plotted. + +

      + The curve in is the same as the curve from . + The difference in this image is that additional pairs of velocity and acceleration vectors are plotted. + Since the acceleration is constant, this vector is the same at every point. + The velocity vector is always tangent to the curve, + and is smaller in magnitude the closer the curve is to the origin. +

      +       + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-1.9,ymax=13, + xmin=-1.9,xmax=13 + ] + + \addplot+ [domain=-3:3,samples=60] ({x^2-x},{x^2+x}); + + \filldraw (axis cs:6,2) circle (2.4pt); + \filldraw (axis cs:2,0) circle (2.4pt); + \filldraw (axis cs:0,0) circle (2.4pt); + \filldraw (axis cs:0,2) circle (2.4pt); + \filldraw (axis cs:2,6) circle (2.4pt); + \filldraw (axis cs:12,6) circle (2.4pt); + \filldraw (axis cs:6,12) circle (2.4pt); + + \draw [thick,->,secondcolor] (axis cs:6,2) -- (axis cs:1,-1); + \draw [thick,->,secondcolor] (axis cs:2,0) -- (axis cs:-1,-1); + \draw [thick,->,secondcolor] (axis cs:0,0) -- (axis cs:-1,1); + \draw [thick,->,secondcolor] (axis cs:0,2) -- (axis cs:1,5); + \draw [thick,->,secondcolor] (axis cs:2,6) -- (axis cs:5,11); + + \draw [thick,->] (axis cs:6,2) -- (axis cs:8,4); + \draw [thick,->] (axis cs:2,0) -- (axis cs:4,2); + \draw [thick,->] (axis cs:0,0) -- (axis cs:2,2); + \draw [thick,->] (axis cs:0,2) -- (axis cs:2,4); + \draw [thick,->] (axis cs:2,6) -- (axis cs:4,8); + + \draw [thick,->,firstcolor] (axis cs:8.75,3.75) -- (axis cs: 8.69,3.71); + + \end{axis} + + \end{tikzpicture} + + + + +
      +       +
      + +

      + Since \vat is constant in this example, + as t grows large \vvt becomes almost parallel to \vat. + For instance, when t=10, + \vec v(10) = \la 19,21\ra, + which is nearly parallel to \la 2,2\ra. +

      +
      3. + +
    3. +

      + The object's speed is given by + + \norm{\vvt} = \sqrt{(2t-1)^2+(2t+1)^2} =\sqrt{8t^2+2} + . + To find the minimal speed, we could apply calculus techniques + (such as set the derivative equal to 0 and solve for t, etc.) + but we can find it by inspection. + Inside the square root we have a quadratic which is minimized when t=0. + Thus the speed is minimized at t=0, + with a speed of \sqrt{2} . + + The graph in also implies speed is minimized here. + The filled dots on the graph are located at integer values of t between -3 and 3. + Dots that are far apart imply the object traveled a far distance in 1 second, indicating high speed; + dots that are close together imply the object did not travel far in 1 second, + indicating a low speed. The dots are closest together near t=0, + implying the speed is minimized near that value. +

      +
      4. +
    +

    +     + +     + + + Analyzing Motion + +

    + Two objects follow an identical path at different rates on [-1,1]. + The position function for Object 1 is \vec r_1(t) = \la t, t^2\ra; + the position function for Object 2 is \vec r_2(t) = \la t^3, t^6\ra, + where distances are measured in feet and time is measured in seconds. + Compare the velocity, + speed and acceleration of the two objects on the path. +

    +     + +

    + We begin by computing the velocity and acceleration function for each object: + + \vec v_1(t) \amp = \la 1,2t\ra \amp \vec v_2(t) \amp = \la 3t^2,6t^5\ra + \vec a_1(t) \amp = \la 0,2\ra \amp \vec a_2(t) \amp =\la 6t,30t^4\ra + +

    + +

    + We immediately see that Object 1 has constant acceleration, + whereas Object 2 does not. +

    + +

    + At t=-1, + we have \vec v_1(-1) = \la 1,-2\ra and \vec v_2(-1) = \la 3,-6\ra; + the velocity of Object 2 is three times that of Object 1 and so it follows that the speed of Object 2 is three times that of Object 1 (3\sqrt{5} ft/s compared to \sqrt{5} ft/s.) +

    + +
    + Plotting velocity and acceleration vectors for Object 1 in + + + A parabola with vertex at the origin, opening upward. Several pairs of velocity and acceleration vectors are plotted. + +

    + The plot shows the parabola y=x^2, for -1\leq x\leq 1. + At five points along the curve, velocity and acceleration vectors are plotted. + The acceleration vector is the same at every point, always pointing straight up. + The velocity vector is tangent to the curve at each point, + but always points somewhat to the right, indicating motion along the curve from x=-1 to x=1. + The magnitude of the velocity vector is largest at the two ends of the parabola, and smallest at the origin. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-1.1,ymax=3.1, + xmin=-2.1,xmax=2.1 + ] + + \addplot+ [domain=-1:1,samples=40] ({x},{x^2}); + + \filldraw (axis cs:-1.,1.) circle (2.4pt); + \filldraw (axis cs:-0.5,0.25) circle (2.4pt); + \filldraw (axis cs:0.,0.) circle (2.4pt); + \filldraw (axis cs:0.5,0.25) circle (2.4pt); + \filldraw (axis cs:1.,1.) circle (2.4pt); + + \draw [thick,->,secondcolor] (axis cs:-1.,1.) -- (axis cs:0.,-1.); + \draw [thick,->,secondcolor] (axis cs:-0.5,0.25) -- (axis cs:0.5,-0.75); + \draw [thick,->,secondcolor] (axis cs:0.,0.) -- (axis cs:1.,0.); + \draw [thick,->,secondcolor] (axis cs:0.5,0.25) -- (axis cs:1.5,1.25); + \draw [thick,->,secondcolor] (axis cs:1.,1.) -- (axis cs:2.,3.); + + \draw [thick,->] (axis cs:-1.,1.) -- (axis cs:-1.,3.); + \draw [thick,->] (axis cs:-0.5,0.25) -- (axis cs:-0.5,2.25); + \draw [thick,->] (axis cs:0.,0.) -- (axis cs:0.,2.); + \draw [thick,->] (axis cs:0.5,0.25) -- (axis cs:0.5,2.25); + \draw [thick,->] (axis cs:1.,1.) -- (axis cs:1.,3.); + + \draw [thick,->,firstcolor] (axis cs:-.9,.81) -- (axis cs: -.89,.7921); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + At t=0, the velocity of Object 1 is + \vec v(1) = \la 1,0\ra and the velocity of Object 2 is \vec 0! + This tells us that Object 2 comes to a complete stop at t=0. +

    + +

    + In , + we see the velocity and acceleration vectors for Object 1 plotted for t=-1, -1/2, 0, 1/2 and t=1. + Note again how the constant acceleration vector seems to pull + the velocity vector from pointing down, right to up, right. + We could plot the analogous picture for Object 2, but the velocity and acceleration vectors are rather large (\vec a_2(-1) = \la -6,30\ra!) +

    + +

    + Instead, we simply plot the locations of Object 1 and 2 on intervals of 1/5th of a second, + shown in and . + Note how the x-values of Object 1 increase at a steady rate. + This is because the x-component of \vec a(t) is 0; + there is no acceleration in the x-component. + The dots are not evenly spaced; + the object is moving faster near t=-1 and t=1 than near t=0. +

    + +
    + Comparing the positions of Objects 1 and 2 in + +
    + + + + The parabola y equals x squared, with several points plotted, corresponding to equally-spaced values of t. + +

    + Another plot of the curve y=x^2 is shown, for -1\leq x\leq 1, + this time with points plotted on the curve, + corresponding to equally-spaced values of t, in increments of 0.2. + The points are closer together near the origin, and further apart near the ends of the curve. + This illustrates the fact that the object described by the vector-valued function \vec{r}(t) = \la t,t^2\ra + is moving slowest near the origin: for equal intervals of time, + points will be closer together when the object is moving slowly, + and further apart when the object is moving quickly. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=1.1, + xmin=-1.1,xmax=1.1 + ] + + \addplot+ [domain=-1:1,samples=40] ({x},{x^2}); + + \filldraw (axis cs:-1.,1.) circle (2.4pt); + \filldraw (axis cs:-0.8,0.64) circle (2.4pt); + \filldraw (axis cs:-0.6,0.36) circle (2.4pt); + \filldraw (axis cs:-0.4,0.16) circle (2.4pt); + \filldraw (axis cs:-0.2,0.04) circle (2.4pt); + \filldraw (axis cs:0.,0.) circle (2.4pt); + \filldraw (axis cs:0.2,0.04) circle (2.4pt); + \filldraw (axis cs:0.4,0.16) circle (2.4pt); + \filldraw (axis cs:0.6,0.36) circle (2.4pt); + \filldraw (axis cs:0.8,0.64) circle (2.4pt); + \filldraw (axis cs:1.,1.) circle (2.4pt); + + \draw [thick,->,firstcolor,>=stealth] (axis cs:-.9,.81) -- (axis cs: -.89,.7921); + \draw (axis cs: .5,.6) node { $\vec r_1(t)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + The parabola y equals x squared, with several points plotted, corresponding to equally-spaced values of t. + +

    + The curve is the same as the one plotted in . + However, the distribution of points corresponding to equally-spaced values of t is very different. + Since the motion of this object is given by \vec{r}(t)=\la t^3,t^6\ra, + the points closest to the origin are much closer together. + For example, when t changes from 0 to 0.1, + x changes from 0 to 0.001, which is a very small change. + However, when t changes from 0.9 to 1, + x changes from 0.729 to 1, a difference that is 271 times as large! +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=1.1, + xmin=-1.1,xmax=1.1 + ] + + \addplot+ [domain=-1:1,samples=40] ({x},{x^2}); + + \filldraw (axis cs:-1.,1.) circle (2.4pt); + \filldraw (axis cs:-0.512,0.262144) circle (2.4pt); + \filldraw (axis cs:-0.216,0.046656) circle (2.4pt); + \filldraw (axis cs:-0.064,0.004096) circle (2.4pt); + \filldraw (axis cs:-0.008,0.000064) circle (2.4pt); + \filldraw (axis cs:0.,0.) circle (2.4pt); + \filldraw (axis cs:0.008,0.000064) circle (2.4pt); + \filldraw (axis cs:0.064,0.004096) circle (2.4pt); + \filldraw (axis cs:0.216,0.046656) circle (2.4pt); + \filldraw (axis cs:0.512,0.262144) circle (2.4pt); + \filldraw (axis cs:1.,1.) circle (2.4pt); + + \draw [thick,->,firstcolor,>=stealth] (axis cs:-.9,.81) -- (axis cs: -.89,.7921); + \draw (axis cs: .5,.6) node { $\vec r_2(t)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + In , we see the points plotted for Object 2. + Note the large change in position from t=-1 to t=-0.8; + the object starts moving very quickly. + However, it slows considerably at it approaches the origin, + and comes to a complete stop at t=0. + While it looks like there are 3 points near the origin, + there are in reality 5 points there. +

    + +

    + Since the objects begin and end at the same location, + they have the same displacement. + Since they begin and end at the same time, + with the same displacement, + they have the same average rate of change (, they have the same average velocity). + Since they follow the same path, + they have the same distance traveled. + Even though these three measurements are the same, + the objects obviously travel the path in very different ways. +

    +     +     + + + Analyzing the motion of a whirling ball on a string + +

    + A young boy whirls a ball, + attached to a string, above his head in a counter-clockwise circle. + The ball follows a circular path and makes 2 revolutions per second. + The string has length 2. +

    + +

    +

      +
    1. +

      + Find the position function + \vec r(t) that describes this situation. +

      +
      2. + +
    2. +

      + Find the acceleration of the ball and give a physical interpretation of it. +

      +
      3. + +
    3. +

      + A tree stands 10 in front of the boy. + At what t-values should the boy release the string so that the ball hits the tree? +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + The ball whirls in a circle. + Since the string is 2ft long, + the radius of the circle is 2. + The position function \vrt = \la 2\cos(t) , 2\sin(t) \ra describes a circle with radius 2, + centered at the origin, + but makes a full revolution every 2\pi seconds, + not two revolutions per second. + We modify the period of the trigonometric functions to be 1/2 by multiplying t by 4\pi. + The final position function is thus + + \vrt = \la 2\cos(4\pi t), 2\sin(4\pi t)\ra + . + (Plot this for 0\leq t\leq 1/2 to verify that one revolution is made in 1/2 a second.) +

      +
      2. + +
    2. +

      + To find \vat, we take the derivative of \vrt twice. + + \vvt = \vrp(t) \amp = \la -8\pi \sin(4\pi t), 8\pi \cos(4\pi t)\ra + \vat =\vrp'(t) \amp = \la -32\pi^2 \cos(4\pi t), -32\pi^2 \sin(4\pi t) \ra + \amp = -32\pi^2\la \cos(4\pi t), \sin(4\pi t)\ra + . + Note how \vat is parallel to \vrt, + but has a different magnitude and points in the opposite direction. + Why is this? +

      + +

      + Recall the classic physics equation, + Force = mass acceleration. + A force acting on a mass induces acceleration (, the mass moves); + a mass that is accelerating is being acted upon by a force. + Thus force and acceleration are closely related. + A moving ball wants to travel in a straight line. + Why does the ball in our example move in a circle? + It is attached to the boy's hand by a string. + The string applies a force to the ball, affecting its motion: + the string accelerates the ball. + This is not acceleration in the sense of + it travels faster; rather, + this acceleration is changing the velocity of the ball. + In what direction is this force/acceleration being applied? + In the direction of the string, towards the boy's hand. +

      + +

      + The magnitude of the acceleration is related to the speed at which the ball is traveling. + A ball whirling quickly is rapidly changing direction/velocity. + When velocity is changing rapidly, + the acceleration must be large. +

      +
      3. + +
    3. +

      + When the boy releases the string, + the string no longer applies a force to the ball, + meaning acceleration is \vec 0 and the ball can now move in a straight line in the direction of \vec v(t). +

      + +

      + Let t=t_0 be the time when the boy lets go of the string. + The ball will be at \vec r(t_0), + traveling in the direction of \vec v(t_0). + We want to find t_0 so that this line contains the point (0,10) + (since the tree is 10 directly in front of the boy). +

      + +
      + Modeling the flight of a ball in + + + A diagram showing a coniferous tree, a circle, and a tangent line from the circle to the tree. + +

      + A large tree is at the top of the image; it appears to be pine or spruce. + Below the tree, a circle is shown, with a diameter of 2 feet labeled. + A line is drawn, tangent to the circle, from the circle to a point on the trunk of the tree. + The line is labeled with the vector \la 0, 10\ra - \vrp(t_0). +

      +       + + + \begin{tikzpicture} + + \begin{scope}[scale=.5,shift={(0,10)}] + + \draw[fill=treestump,ultra thick] + (.75,-1) + .. controls (.5,.5) and (.5,3) .. (0.5,4) -- (-0.5,4) + .. controls (-.5,3) and (-.5,.5) .. (-.75,-1) + ; + + \draw[ultra thick,fill=treetop] + (0,10) + .. controls (0,8) and (1,7) .. (1.5,7) + .. controls (1,7) and (1,7) .. (0.5,7.25) + .. controls (1.5,5) and (2.5,4) .. (3,4) + .. controls (2,4) and (1.25,4) .. (1,4.5) + .. controls (2,2) and (3.5,2) .. (4,2) + .. controls (1,1) and (-1,1) .. (-4,2) + .. controls (-3.5,2) and (-2,2) .. (-1,4.5) + .. controls (-1.25,4) and (-2,4) .. (-3,4) + .. controls (-2.5,4) and (-1.5,5) .. (-0.5,7.25) + .. controls (-1,7) and (-1,7) .. (-1.5,7) + .. controls (-1,7) and (0,8) .. (0,10) + ; + + \end{scope} + + \begin{scope}[scale=.5] + + \coordinate (ball) at (215:2cm); + \coordinate (release) at (11.5:2cm); + + \draw [rotate=11.5] (0,0) -- (1.9,0) node [above,sloped,pos=.5] { 2 ft}; + \draw [thick,dashed] (0,0) circle (2cm); + \draw [thick,->] (ball) arc (215:225:2cm); + + \draw (0,10) circle (2pt); + \draw [dotted,thick] (release) -- (0,10) node [sloped, pos=.5, above] { $\langle 0,10\rangle - \vec r(t_0)$}; + + \filldraw [draw=black,fill=white,thick] (release) circle (.1); + + \end{scope} + + \end{tikzpicture} + + + + +
      + +

      + There are many ways to find this time value. + We choose one that is relatively simple computationally. + As shown in , + the vector from the release point to the tree is \la 0,10\ra - \vec r(t_0). + This line segment is tangent to the circle, + which means it is also perpendicular to \vec r(t_0) itself, + so their dot product is 0. + + \amp \vec r(t_0) \cdot \big(\la 0,10\ra - \vec r(t_0)\big) =0 + \amp \la 2\cos(4\pi t_0), 2\sin(4\pi t_0)\ra \cdot \la -2\cos(4\pi t_0),10-2\sin(4\pi t_0)\ra =0 + \amp -4\cos^2(4\pi t_0) + 20\sin(4\pi t_0)-4\sin^2(4\pi t_0) = 0 + \amp 20\sin(4\pi t_0) - 4 =0 + \amp \sin(4\pi t_0) =1/5 + \amp 4\pi t_0 = \sin^{-1}(1/5) + \amp 4\pi t_0 \approx 0.2 + 2\pi n, + where n is an integer. Solving for t_0 we have: + \amp t_0 \approx 0.016 + n/2 + + This is a wonderful formula. + Every 1/2 second after t=0.016\,\text{s} the boy can release the string + (since the ball makes 2 revolutions per second, + he has two chances each second to release the ball). +

      +
      4. +
    +

    +     + +     + + + Analyzing motion in space + +

    + An object moves in a helix with position function \vrt = \la \cos(t) , \sin(t) , t\ra, + where distances are measured in meters and time is in minutes. + Describe the object's speed and acceleration at time t. +

    +     + +

    + With \vrt = \la \cos(t) ,\sin(t) , t\ra, we have: + + \vvt \amp = \la -\sin(t) , \cos(t) , 1\ra \text{ and } + \vat \amp = \la -\cos(t) , -\sin(t) , 0\ra + . +

    + +

    + The speed of the object is \norm{\vvt} = \sqrt{(-\sin(t) )^2+\cos^2(t) +1} = \sqrt{2} + ; + it moves at a constant speed. + Note that the object does not accelerate in the z-direction, + but rather moves up at a constant rate of 1. +

    +     +     + +

    + The objects in Examples + and traveled at a constant speed. + That is, \norm{\vvt} = c for some constant c. + Recall , + which states that if a vector-valued function \vrt has constant length, + then \vrt is perpendicular to its derivative: + \vrt\cdot\vrp(t) = 0. + In these examples, the velocity function has constant length, + therefore we can conclude that the velocity is perpendicular to the acceleration: + \vvt\cdot\vat = 0. + A quick check verifies this. + vector-valued functionof constant length +

    + +

    + There is an intuitive understanding of this. + If acceleration is parallel to velocity, + then it is only affecting the object's speed; + it does not change the direction of travel. + (For example, consider a dropped stone. + Acceleration and velocity are parallel straight down and the direction of velocity never changes, + though speed does increase.) + If acceleration is not perpendicular to velocity, + then there is some acceleration in the direction of travel, + influencing the speed. + If speed is constant, then acceleration must be orthogonal to velocity, + as it then only affects direction, and not speed. +

    + + + Objects With Constant Speed +

    + If an object moves with constant speed, + then its velocity and acceleration vectors are orthogonal. + That is, \vvt\cdot\vat=0. + vector-valued functionof constant length +

    +     +     + + + Projectile Motion +

    + An important application of vector-valued position functions is + projectile motion: + the motion of objects under only the influence of gravity. + We will measure time in seconds, + and distances will either be in meters or feet. + We will show that we can completely describe the path of such an object knowing its initial position and initial velocity (, where it is + and where it is going.) + vector-valued functionprojectile motion + projectile motion +

    + + + +

    + Suppose an object has initial position + \vec r(0) = \la x_0,y_0\ra and initial velocity \vec v(0) = \la v_x,v_y\ra. + It is customary to rewrite + \vec v(0) in terms of its speed v_0 and direction \vec u, + where \vec u is a unit vector. + Recall all unit vectors in + \mathbb{R}^2 can be written as \la \cos(\theta) ,\sin(\theta) \ra, + where \theta is an angle measure counter-clockwise from the x-axis. + (We refer to \theta as the angle of elevation.) + angle of elevation + Thus \vec v(0) = v_0\la \cos(\theta),\sin(\theta)\ra. +

    + +

    + Since the acceleration of the object is known, + namely \vat = \la 0,-g\ra, + where g is the gravitational constant, + we can find \vrt knowing our two initial conditions. + We first find \vvt: +

    + +

    + + \vec v(t) \amp = \int \vat \, dt + \vvt \amp = \int \la 0,-g\ra \, dt + \vvt \amp = \la 0,-gt\ra + \vec C + . +

    + +

    + Knowing \vec v(0) = v_0\la \cos(\theta) ,\sin(\theta) \ra, + we have \vec C = v_0\la \cos(t) ,\sin(t) \ra and so + + \vec v(t) = \la v_0\cos(\theta) , -gt+v_0\sin(\theta) \ra + . +

    + +

    + We integrate once more to find \vrt: + + \vrt \amp = \int \vvt\,dt + \vrt \amp = \int \la v_0\cos(\theta) , -gt+v_0\sin(\theta) \ra\,dt + \vrt \amp = \la \big(v_0\cos(\theta) \big)t, -\frac12gt^2+\big(v_0\sin(\theta) \big)t\ra + \vec C. + Knowing \vec r(0) = \la x_0,y_0\ra, we conclude \vec C = \la x_0,y_0\ra and + \vrt \amp = \la \big(v_0\cos(\theta) \big)t+x_0\,, -\frac12gt^2+\big(v_0\sin(\theta) \big)t+y_0\,\ra + . +

    + + + Projectile Motion +

    + The position function of a projectile propelled from an initial position of \vec r_0=\la x_0,y_0\ra, + with initial speed v_0, + with angle of elevation \theta and neglecting all accelerations but gravity is + vector-valued functionprojectile motion + projectile motion + + \vrt = \la \big(v_0\cos(\theta) \big)t+x_0\,, -\frac12gt^2+\big(v_0\sin(\theta) \big)t+y_0\,\ra + . +

    + +

    + Letting \vec v_0 = v_0\la \cos(\theta) ,\sin(\theta) \ra, + \vrt can be written as + + \vrt = \la 0,-\frac12gt^2\ra + \vec v_0t+\vec r_0 + . +

    +     + +

    + We demonstrate how to use this position function in the next two examples. +

    + + + Projectile Motion + +

    + Sydney shoots her Red Ryderbb gun across level ground from an elevation of 4, + where the barrel of the gun makes a 5^\circ angle with the horizontal. + Find how far the bb travels before landing, + assuming the bb is fired at the advertised rate of 350 and ignoring air resistance. +

    +     + +

    + A direct application of gives + + \vrt \amp = \la (350\cos(5^\circ) )t, -16t^2 + (350\sin(5^\circ) )t + 4\ra + \amp \approx \la 346.67t, -16t^2+30.50t+4\ra + , + where we set her initial position to be \la 0,4\ra. + We need to find when the bb lands, + then we can find where. + We accomplish this by setting the y-component equal to 0 and solving for t: + + -16t^2+30.50t+4 \amp = 0 + t \amp = \frac{-30.50 \pm \sqrt{30.50^2-4(-16)(4)}}{-32} + t \amp \approx 2.03\,\text{s} + . +

    + +

    + (We discarded a negative solution that resulted from our quadratic equation.) +

    + +

    + We have found that the bb lands 2.03 after firing; + with t=2.03, + we find the x-component of our position function is 346.67(2.03) = 703.74\,\text{ft}. + The bb lands about 704 feet away. +

    +     + +     + + + Projectile Motion + +

    + Alex holds his sister's bb gun at a height of 3 + and wants to shoot a target that is 6 above the ground, + 25 away. + At what angle should he hold the gun to hit his target? + (We still assume the muzzle velocity is 350.) +

    +     + +

    + The position function for the path of Alex's bb is + + \vrt = \la (350\cos(\theta) )t, -16t^2+(350\sin(\theta) )t+3\ra + . +

    + +

    + We need to find \theta so that + \vrt =\la 25,6\ra for some value of t. + That is, we want to find \theta and t such that + + (350\cos(\theta) )t = 25 \text{ and } -16t^2+(350\sin(\theta) )t+3 = 6 + . +

    + +

    + This is not trivial + (though not hard). + We start by solving each equation for + \cos(\theta) and \sin(\theta), respectively. + + \cos(\theta) = \frac{25}{350t} \text{ and } \sin(\theta) = \frac{3+16t^2}{350t} + . +

    + +

    + Using the Pythagorean Identity \cos^2(\theta) +\sin^2(\theta) =1, we have + + \left(\frac{25}{350t}\right)^2 + \left(\frac{3+16t^2}{350t}\right)^2 \amp =1 + Multiply both sides by (350t)^2: + 25^2 + (3+16t^2)^2 \amp =350^2t^2 + 256t^4-122,404t^2+634 \amp =0. + This is a quadratic in t^2. That is, we can apply the quadratic formula to find t^2, then solve for t itself. + t^2 \amp = \frac{122,404\pm\sqrt{122,404^2-4(256)(634)}}{512} + t^2 \amp = 0.0052,\,478.135 + t \amp = \pm 0.072,\,\pm 21.866 + +

    + +

    + Clearly the negative t values do not fit our context, + so we have t=0.072 and t=21.866. + Using \cos(\theta) = 25/(350 t), + we can solve for \theta: + + \theta \amp = \cos^{-1}\left(\frac{25}{350\cdot 0.072}\right) \text{ and } \cos^{-1}\left(\frac{25}{350\cdot 21.866}\right) + \theta \amp = 7.03^\circ \text{ and } 89.8^\circ + . +

    + +

    + Alex has two choices of angle. + He can hold the rifle at an angle of about 7^\circ with the horizontal and hit his target + 0.07 after firing, + or he can hold his rifle almost straight up, + with an angle of 89.8^\circ, + where he'll hit his target about 22 later. + The first option is clearly the option he should choose. +

    +     +     +     + + + Distance Traveled +

    + Consider a driver who sets her cruise-control to 60, + and travels at this speed for an hour. + We can ask: +

    + +

    +

      +
    1. +

      + How far did the driver travel? +

      +
      2. + +
    2. +

      + How far from her starting position is the driver? +

      +
      3. +
    +

    + +

    + The first is easy to answer: + she traveled 60 miles. + The second is impossible to answer with the given information. + We do not know if she traveled in a straight line, + on an oval racetrack, or along a slowly-winding highway. +

    + +

    + This highlights an important fact: + to compute distance traveled, + we need only to know the speed, + given by \norm{\vvt}. +

    + + + Distance Traveled + +

    + Let \vvt be a velocity function for a moving object. + The distance traveled by the object on [a,b] is: + distancetraveled + vector-valued functiondistance traveled + integrationdistance traveled + + \text{ distance traveled } = \int_a^b \norm{\vvt}\, dt + . +

    +     +     + +

    + Note that this is just a restatement of : + arc length is the same as distance traveled, + just viewed in a different context. +

    + + + + + Distance Traveled, Displacement, and Average Speed + +

    + A particle moves in space with position function \vrt = \la t,t^2,\sin(\pi t)\ra on [-2,2], + where t is measured in seconds and distances are in meters. + Find: +

    + +

    +

      +
    1. +

      + The distance traveled by the particle on [-2,2]. +

      +
      2. + +
    2. +

      + The displacement of the particle on [-2,2]. +

      +
      3. + +
    3. +

      + The particle's average speed. +

      +
      4. +
    +

    +     + +

    +

      +
    1. +

      + We use to establish the integral: + + \text{ distance traveled } \amp = \int_{-2}^2 \norm{\vvt}\, dt + \amp = \int_{-2}^2 \sqrt{1+(2t)^2+ \pi^2\cos^2(\pi t)}\, dt + . + This cannot be solved in terms of elementary functions so we turn to numerical integration, + finding the distance to be 12.88. +

      +
      2. + +
    2. +

      + The displacement is the vector + + \vec r(2)-\vec r(-2) = \la 2,4,0\ra - \la -2,4,0\ra = \la 4,0,0\ra + . + That is, the particle ends with an x-value increased by 4 and with y- and z-values the same + (see ). +

      +
      3. + +
    3. +

      + We found above that the particle traveled 12.88 over 4 seconds. + We can compute average speed by dividing: 12.88/4 = 3.22\,\text{m/s}. +

      + +
      + The path of the particle in + + + + An illustration of the path followed by the particle in this example. + +

      + This three-dimensional image illustrates the path of the particle in . + In the default view, the path appears to be quite complicated. + But if the image is rotated so that the view is along the y axis, + the curve appears sinusoidal, while it appears parabolic when viewed from above. +

      +       + + + + + //ASY file for figmotion63D.asy in Chapter 11 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5.6,10.7,4); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={2,4}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-3,3); + pair ybounds=(-1,5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve <t,t^2,sin(pi*t)> for t from -2 to 2 + triple g(real t) {return (t,t^2,sin(pi*t));} + path3 mypath=graph(g,-2,2,operator ..); draw(mypath,bluepen); + path3 mypath=graph(g,-2,-1.2,operator ..); draw(mypath,bluepen,Arrow3(size=5mm)); + //path3 mypath=graph(g,-2,-0.4,operator ..); draw(mypath,bluepen,Arrow3(size=5mm)); + //path3 mypath=graph(g,-2,0.8,operator ..); draw(mypath,bluepen,Arrow3(size=5mm)); + //path3 mypath=graph(g,-2,1.6,operator ..); draw(mypath,bluepen,Arrow3(size=5mm)); + + + + +
      + +

      + We should also consider + of , + which says that the average value of a function f on [a,b] is \frac{1}{b-a}\int_a^b f(x)\, dx. + In our context, the average value of the speed is + + \text{average speed}\, = \frac{1}{2-(-2)}\int_{-2}^2 \norm{\vvt}\, dt \approx \frac14 12.88 = 3.22\,\text{m/s} + . + Note how the physical context of a particle traveling gives meaning to a more abstract concept learned earlier. +

      +
      4. +
    +

    +     + +     + +

    + In + of + we defined the average value of a function f(x) on [a,b] to be + + \frac{1}{b-a}\int_a^bf(x)\, dx + . +

    + +

    + Note how in + we computed the average speed as + + \frac{\text{ distance traveled } }{\text{ travel time } } = \frac1{2-(-2)}\int_{-2}^2\norm{\vvt}\, dt + ; + that is, we just found the average value of \norm{\vvt} on [-2,2]. +

    + +

    + Likewise, given position function \vrt, + the average velocity on [a,b] is + + \frac{\text{displacement}}{\text{travel time}}\, = \frac1{b-a}\int_a^b \vec{r}\,'(t)\, dt = \frac{\vec r(b)-\vec r(a)}{b-a} + ; + that is, it is the average value of \vrp(t), + or \vvt, on [a,b]. +

    + + + Average Speed, Average Velocity +

    + Let \vec r(t) be a differentiable position function on [a,b]. +

    + +

    + The average speed is: + + \frac{\text{distance traveled}}{\text{travel time}}\, = \frac{\int_a^b \norm{\vec{r}\,'(t)}\, dt}{b-a} = \frac1{b-a}\int_a^b\norm{\vvt}\, dt + . +

    + +

    + The average velocity is: + + \frac{\text{ displacement}}{\text{travel time}}\, = \frac{\int_a^b \vec{r}\,'(t)\, dt}{b-a} = \frac1{b-a}\int_a^b\vec{r}\,'(t)\, dt + . +

    +     + +

    + The next two sections investigate more properties of the graphs of vector-valued functions and we'll apply these new ideas to what we just learned about motion. +

    +     + + + + Terms and Concepts + + + +

    + How is velocity different from speed? +

    +     + + + +

    + Velocity is a vector, + indicating an objects direction of travel and its rate of distance change (, its speed). + Speed is a scalar. +

    +     + +     + + + + +

    + What is the difference between displacement + and distance traveled? +

    +     + + + +

    + Displacement is a vector, + indicating the difference between the starting and ending positions of an object. + Distance traveled is a scalar, + indicating the arc length of the path followed. +

    +     + +     + + + + +

    + What is the difference between average velocity + and average speed? +

    +     + + + +

    + The average velocity is found by dividing the displacement by the time traveled it is a vector. + The average speed is found by dividing the distance traveled by the time traveled it is a scalar. +

    +     + +     + + + + +

    + Distance traveled is the same as , just viewed in a different context. +

    +     + + + + arc length|arclength + + + + +     + + + + +

    + Describe a scenario where an object's average speed is a large number, + but the magnitude of the average velocity is not a large number. +

    +     + + + +

    + One example is traveling at a constant speed s in a circle, + ending at the starting position. + Since the displacement is \vec 0, + the average velocity is \vec 0, hence \vnorm 0=0. + But traveling at constant speed s means the average speed is also s \gt 0. +

    +     + +     + + + + +

    + Explain why it is not possible to have an average velocity with a large magnitude but a small average speed. +

    +     + + + +

    + Distance traveled is always greater than or equal to the magnitude of displacement, + therefore average speed will always be at least as large as the magnitude of the average velocity. +

    +     + +     +     + + Problems + + + +

    + A position function \vrt is given. + Find \vvt and \vat. +

    +     + + + + +

    + \vrt = \la 2t+1, 5t-2, 7\ra +

    +     + +

    + \vvt = \la 2,5,0\ra, \vat = \la 0,0,0\ra +

    +     + +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + $velocity=Vector("<6t-2,-2t+1>"); + $acceleration=Vector("<6,-2>"); + + + +

    + \vrt = \la 3t^2-2t+1, -t^2+t+14\ra +

    + + + The velocity function \vvt is: + +

    + +

    + + + The acceleration function \vat is: + +

    + +

    +     +     +     + + + + +

    + \vrt = \la \cos(t) ,\sin(t) \ra +

    +     + +

    + \vvt = \la -\sin(t) , \cos(t) \ra, + \vat = \la -\cos(t) , -\sin(t) \ra +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $velocity=Vector("<1/10,sin(t),cos(t)>"); + $acceleration=Vector("<0,cos(t),-sin(t)>"); + + + +

    + \vrt = \la t/10,-\cos(t) ,\sin(t) \ra +

    + + + The velocity function \vvt is: + +

    + +

    + + + The acceleration function \vat is: + +

    + +

    +     +     +     + +     + + + +

    + A position function \vrt is given. + Sketch \vrt on the indicated interval. + Find \vvt and \vat, + then add \vec v(t_0) and \vec a(t_0) to your sketch, + with their initial points at \vec r(t_0), + for the given value of t_0. +

    +     + + + + +

    + \ds \vrt = \la t,\sin(t) \ra on [0,\pi/2]; + t_0= \pi/4 +

    +     + +

    + \vvt = \la 1,\cos(t) \ra, + \vat = \la 0,-\sin(t) \ra +

    + +     + +     + + + + +

    + \ds \vrt = \la t^2,\sin(t^2) \ra on [0,\pi/2]; + t_0=\sqrt{\pi/4} +

    +     + +

    + \vvt = \la 2 t,2 t \cos\left(t^2\right)\ra, + \vat = \la 2,2 \left(\cos\left(t^2\right)-2 t^2 \sin\left(t^2\right)\right)\ra +

    + +     + +     + + + + +

    + \ds \vrt = \la t^2+t,-t^2+2t \ra on [-2,2]; + t_0=1 +

    +     + +

    + \vvt = \la 2t+1,-2t+2\ra, \vat = \la 2,-2\ra +

    + +     + +     + + + + +

    + \ds \vrt = \la \frac{2t+3}{t^2+1},t^2\ra on [-1,1]; + t_0= 0 +

    +     + +

    + \vvt = \la -\frac{2 \left(t^2+3 \ + t-1\right)}{\left(t^2+1\right)^2},2 t\ra, + \vat = \la \frac{2 \left(2 t^3+9 t^2-6 t-3\right)}{\left(t^2+1\right)^3},2\ra +

    + +     + +     + +     + + + +

    + A position function \vrt of an object is given. + Find the speed of the object in terms of t, + and find where the speed is minimized/maximized on the indicated interval. +

    +     + + + + +

    + \ds \vrt = \la t^2,t \ra on [-1,1] +

    +     + +

    + \norm{\vvt} = \sqrt{4t^2+1}. +

    + +

    + Min at t=0; Max at t=\pm 1. +

    +     + +     + + + + + Context()->variables->are(t=>'Real'); + $speed=Formula("|t| sqrt(9t^2-12t+8)"); + $mins=List("0"); + $maxs=List("-1"); + + + +

    + \vrt = \la t^2,t^2-t^3 \ra on [-1,1] +

    + + Find the speed of the object in terms of t. + +

    + +

    + + + Find where the speed is minimized on the given interval. + +

    + +

    + + + Find where the speed is maximized on the given interval. + +

    + +

    +     +     +     + + + + +

    + \ds \vrt = \la 5\cos(t) ,5\sin(t) \ra on [0,2\pi] +

    +     + +

    + \norm{\vvt} = 5. +

    + +

    + Speed is constant, so there is no difference between min/max +

    +     + +     + + + + +

    + \ds \vrt = \la 2\cos(t) ,5\sin(t) \ra on [0,2\pi] +

    +     + +

    + \norm{\vvt} = \sqrt{4\sin^2(t) +25\cos^2(t) }. +

    + +

    + min: t=\pi/2,\,3\pi/2; + max: t=0,\,2\pi +

    +     + +     + + + + + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $speed=Formula("|sec(t)| sqrt(tan^2(t)+sec^2(t))"); + $mins=List("0"); + $maxs=Compute("pi/4"); + + + +

    + \vrt=\la \sec(t) ,\tan(t) \ra on [0,\pi/4]. +

    + + Find the speed of the object in terms of t. + +

    + +

    + + + Find where the speed is minimized on the given interval. + +

    + +

    + + + Find where the speed is maximized on the given interval. + +

    + +

    +     +     +     + + + + +

    + \ds \vrt = \la t+\cos(t) ,1-\sin(t) \ra on [0,2\pi] +

    +     + +

    + \norm{\vvt} = \sqrt{2-2\sin(t) }. +

    + +

    + min: t=\pi/2; max: t=3\pi/2 +

    +     + +     + + + + +

    + \ds \vrt = \la 12t,5\cos(t) ,5\sin(t) \ra on [0,4\pi] +

    +     + +

    + \norm{\vvt} = 13. +

    + +

    + speed is constant, so there is no difference between min/max +

    +     + +     + + + + + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $speed=Formula("sqrt(8t^2+3)"); + $mins=List("0"); + $maxs=Compute("1"); + + + +

    + \vrt=\la t^2-t,t^2+t,t \ra on [0,1]. +

    + + Find the speed of the object in terms of t. + +

    + +

    + + + Find where the speed is minimized on the given interval. + +

    + +

    + + + Find where the speed is maximized on the given interval. + +

    + +

    +     +     +     + + + + +

    + \ds \vrt = \la t,t^2,\sqrt{1-t^2}\ra on [-1,1] +

    +     + +

    + \norm{\vvt} = \sqrt{4t^2+1+t^2/(1-t^2)}. +

    + +

    + min: t=0; max: there is no max; + speed approaches \infty as t\to\pm 1 +

    +     + +     + + + + +

    + Projectile Motion: + \ds \vrt = \la (v_0\cos(\theta) )t,-\frac12gt^2+(v_0\sin(\theta) )t \ra on \ds \left[0,\frac{2v_0\sin(\theta) }g\right] +

    +     + +

    + \norm{\vvt} = \sqrt{g^2t^2-(2gv_0\sin(\theta) )t+v_0^2}. +

    + +

    + min: t=(v_0\sin(\theta) )/g; max: + t=0, t=(2v_0\sin(\theta) )/g +

    +     + +     + +     + + + +

    + Position functions \vec r_1(t) and + \vec r_2(s) for two objects are given that follow the same path on the respective intervals. +

    + +

    +

      +
    1. +

      + Show that the positions are the same at the indicated t_0 and s_0 values; + , show \vec r_1(t_0) = \vec r_2(s_0). +

      +
      2. + +
    2. +

      + Find the velocity, + speed and acceleration of the two objects at t_0 and s_0, + respectively. +

      +
      3. +
    +

    +     + + + + +

    + \vec r_1(t) = \la t,t^2\ra on [0,1]; t_0 = 1 +

    + +

    + \vec r_2(s) = \la s^2,s^4\ra on [0,1]; s_0 = 1 +

    +     + +

    +

      +
    1. +

      + \vec r_1(1) = \la 1,1\ra; + \vec r_2(1) = \la 1,1\ra +

      +
      2. + +
    2. +

      + \vec v_1(1) = \la 1,2\ra; + \norm{\vec v_1(1)} = \sqrt{5}; + \vec a_1(1) = \la 0,2\ra + + \vec v_2(1) = \la 2,4\ra; \norm{\vec v_2(1)} = 2\sqrt{5}; \vec a_2(1) = \la 2,12\ra +

      +
      3. +
    +

    +     + +     + + + + +

    + \vec r_1(t) = \la 3\cos(t) ,3\sin(t) \ra on [0,2\pi]; + t_0 = \pi/2 +

    + +

    + \vec r_2(s) = \la 3\cos(4s),3\sin(4s)\ra on [0,\pi/2]; + s_0 = \pi/8 +

    +     + +

    +

      +
    1. +

      + \vec r_1(\pi/2) = \la0,3\ra; + \vec r_2(\pi/8) = \la 0,3\ra +

      +
      2. + +
    2. +

      + \vec v_1(\pi/2) = \la -3,0\ra; + \norm{\vec v_1(\pi/2)} = 3; + \vec a_1(\pi/2) = \la 0,-3\ra + + \vec v_2(\pi/8) = \la -12,0\ra; \norm{\vec v_2(\pi/8)} = 12; \vec a_2(\pi/8) = \la 0,-48\ra +

      +
      3. +
    +

    +     + +     + + + + +

    + \vec r_1(t) = \la 3t,2t\ra on [0,2]; t_0 = 2 +

    + +

    + \vec r_2(s) = \la 6s-6,4s-4\ra on [1,2]; + s_0 = 2 +

    +     + +

    +

      +
    1. +

      + \vec r_1(2) = \la6,4\ra; + \vec r_2(2) = \la 6,4\ra +

      +
      2. + +
    2. +

      + \vec v_1(2) = \la 3,2\ra; + \norm{\vec v_1(2)} = \sqrt{13}; + \vec a_1(2) = \la 0,0\ra + + \vec v_2(2) = \la 6,4\ra; \norm{\vec v_2(2)} = 2\sqrt{13}; \vec a_2(2) = \la 0,0\ra +

      +
      3. +
    +

    +     + +     + + + + +

    + \vec r_1(t) = \la t,\sqrt{t}\ra on [0,1]; + t_0 = 1 +

    + +

    + \vec r_2(s) = \la \sin(s) ,\sqrt{\sin(s) }\ra on [0,\pi/2]; + s_0 = \pi/2 +

    +     + +

    +

      +
    1. +

      + \vec r_1(1) = \la 1,1\ra; + \vec r_2(\pi/2) = \la 1,1\ra +

      +
      2. + +
    2. +

      + \vec v_1(1) = \la 1,1/2\ra; + \norm{\vec v_1(1)} = \sqrt{5}/2; + \vec a_1(1) = \la 0,-1/4\ra + + \vec v_2(\pi/2) = \la 0,0\ra; \norm{\vec v_2(\pi/2)} = 0; \vec a_2(\pi/2) = \la -1,-1/2\ra +

      +
      3. +
    +

    +     + +     + +     + + + +

    + Find the position function of an object given its acceleration and initial velocity and position. +

    +     + + + + +

    + \vat = \la 2,3\ra;\vec v(0) = \la 1,2\ra,\vec r(0) = \la 5,-2\ra +

    +     + +

    + \vvt = \la 2t+1,3t+2\ra, + \vrt = \la t^2+t+5,3t^2/2+2t-2\ra +

    +     + +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + $pos=Compute("<t^2-t+5,3t^2/2-t-5/2>"); + + + +

    + Given \vat = \la 2,3\ra, + \vec v(1) = \la 1,2\ra, + and \vec r(1) = \la 5,-2\ra, + find the position function \vrt. +

    + +

    + +

    +     +     +     + + + + +

    + \vat = \la \cos(t) ,-\sin(t) \ra;\vec v(0) = \la 0,1\ra,\vec r(0) = \la 0,0\ra +

    +     + +

    + \vvt = \la \sin(t) ,\cos(t) \ra, + \vrt = \la 1-\cos(t) ,\sin(t) \ra +

    +     + +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + $pos=Compute("<10t,-16t^2+50t>"); + + + +

    + Given \vat =\la 0,-32\ra, + \vec v(0) = \la 10,50\ra, + and \vec r(0) = \la 0,0\ra, + find the position function \vrt. +

    + +

    + +

    +     +     +     + +     + + + +

    + Find the displacement, distance traveled, + average velocity and average speed of the described object on the given interval. +

    +     + + + + +

    + An object with position function \vrt = \la 2\cos(t) ,2\sin(t) , 3t\ra, + where distances are measured in feet and time is in seconds, + on [0,2\pi]. +

    +     + +

    + Displacement: \la 0,0,6\pi\ra; distance traveled: + 2\sqrt{13}\pi \approx 22.65ft; average velocity: + \la 0,0,3\ra; average speed: + \sqrt{13} \approx 3.61ft/s +

    +     + +     + + + + + Context("Vector2D"); + $disp=Compute("<-10,0>"); + $dist=Compute("5pi"); + $avvel=Compute("<-10/pi,0>"); + $avspe=Compute("5"); + + + +

    + An object has position function \vrt = \la 5\cos(t) ,-5\sin(t) \ra, + where distances are measured in feet and time is in seconds. + Over [0,\pi]. +

    + + What is the displacement? + +

    + +

    + + What is the distance traveled? + +

    + +

    + + What is the average velocity? + +

    + +

    + + What is the average speed? + +

    + +

    +     +     +     + + + + +

    + An object with velocity function \vvt = \la \cos(t) ,\sin(t) \ra, + where distances are measured in feet and time is in seconds, + on [0,2\pi]. +

    +     + +

    + Displacement: \la 0,0\ra; distance traveled: + 2\pi \approx 6.28ft; average velocity: + \la 0,0\ra; average speed: 1ft/s +

    +     + +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $disp=Compute("<10,20,-10>"); + $dist=Compute("10sqrt(6)"); + $avvel=Compute("<1,2,-1>"); + $avspe=Compute("sqrt(6)"); + + + +

    + An object has velocity function \vvt = \la 1,2,-1 \ra, + where distances are measured in feet and time is in seconds. + Over [0,10]. +

    + + What is the displacement? + +

    + +

    + + What is the distance traveled? + +

    + +

    + + What is the average velocity? + +

    + +

    + + What is the average speed? + +

    + +

    +     +     +     + +     + + + +

    + The following exercises ask you to solve a variety of problems based on the principles of projectile motion. +

    +     + + + + +

    + A boy whirls a ball, attached to a 3 ft string, + above his head in a counter-clockwise circle. + The ball makes 2 revolutions per second. +

    + +

    + At what t-values should the boy release the string so that the ball heads directly for a tree standing 10 ft in front of him? +

    +     + +

    + At t-values of \sin^{-1}(9/30)/(4\pi) + n/2 \approx 0.024 + n/2 seconds, + where n is an integer. +

    +     + +     + + + + +

    + David faces Goliath with only a stone in a 3 ft sling, + which he whirls above his head at 4 revolutions per second. + They stand 20 ft apart. +

    +     + + + +

    + At what t-values must David release the stone in his sling in order to hit Goliath? +

    +     + +

    + t=\sin^{-1}(3/20)/(8\pi) + n/4 \approx 0.006 + n/4, + where n is an integer +

    +     + +

    + The stone, while whirling, + can be modeled by \vrt = \la 3\cos(8\pi t),3\sin(8\pi t)\ra. +

    + +

    + He can releast the stone for t-values t=\sin^{-1}(3/20)/(8\pi) + n/4 \approx 0.006 + n/4, + where n is an integer. +

    +     +     + + + +

    + What is the speed at which the stone is traveling when released? +

    +     + +

    + \norm{\vrp(t)} = 24\pi\approx 51.4 ft/s +

    +     +     + + + +

    + Assume David releases the stone from a height of 6ft and Goliath's forehead is 9 ft above the ground. + What angle of elevation must David apply to the stone to hit Goliath's head? +

    +     + +

    + 0.27 radians, or 15.69^\circ +

    +     + +

    + At t=0.006, + the stone is approximately 19.77 ft from Goliath. + Using the formula for projectile motion, + we want the angle of elevation that lets a projectile starting at + \la 0,6\ra with a initial velocity of 51.4 ft/s arrive at \la 19.77,9\ra. + The desired angle is 0.27 radians, or 15.69^\circ. +

    +     +     + +     + + + + + Context()->flags->set(tolerance=>0.005); + $angle=NumberWithUnits("0.013 radians"); + $lead=NumberWithUnits("11.7 ft"); + + + +

    + A hunter aims at a deer which is 40 yards away. + Her crossbow is at a height of 5 ft, + and she aims for a spot on the deer 4 ft above the ground. + The crossbow fires her arrows at 300 ft/s. +

    +     + + + +

    + At what angle of elevation should she hold the crossbow to hit her target? +

    + +

    + +

    +     +     + + + +

    + If the deer is moving perpendicularly to her line of sight at a rate of 20 mph, + by approximately how much should she lead the deer in order to hit it in the desired location? (How far ahead of the deer should she aim?) +

    + +

    + +

    +     +     +     +     + + + + +

    + A baseball player hits a ball at 100 mph, + with an initial height of 3 ft and an angle of elevation of + 20^\circ, at Boston's Fenway Park. + The ball flies towards the famed Green Monster, + a wall 37 ft high located 310 ft from home plate. +

    +     + + + +

    + Show that as hit, the ball hits the wall. +

    +     + +

    + The position function of the ball is \vrt = \la (146.67\cos(\theta) )t,-16t^2+(146.67\sin(\theta) )t+3\ra, + where \theta is the angle of elevation. +

    + +

    + With \theta=20^\circ, + the ball reaches 310 ft from home plate in 2.25 seconds; + at this time, the height of the ball is 34.9 ft, + not enough to clear the Green Monster. +

    +     +     + + + +

    + Show that if the angle of elevation is 21^\circ, + the ball clears the Green Monster. +

    +     + +

    + With \theta=21^\circ, + the ball reaches 310 ft from home plate in 2.26 s, + with a height of 40 ft, clearing the wall. +

    +     +     + +     + + + + +

    + A Cessna flies at 1000 ft at 150 mph and drops a box of supplies to the professor + (and his wife) + on an island. + Ignoring wind resistance, + how far horizontally will the supplies travel before they land? +

    +     + +

    + The position function is \vrt = \la 220t,-16t^2+1000\ra. + The y-component is 0 when t=7.9; + \vec r(7.9) = \la 1739.25,0\ra, + meaning the box will travel about 1740 ft horizontally before it lands. +

    +     + +     + + + + +

    + A football quarterback throws a pass from a height of 6 ft, + intending to hit his receiver 20 yds away at a height of 5 ft. +

    +     + + + +

    + If the ball is thrown at a rate of 50mph, + what angle of elevation is needed to hit his intended target? +

    +     + +

    + The position function of the ball is \vrt = \la (v_0\cos(\theta) )t,-16t^2+(v_0\sin(\theta) )t+6\ra, + where \theta is the angle of elevation and v_0 is the initial ball speed. +

    + +

    + With v_0 = 73.33 ft/s, there are two angles of elevation possible. + An angle of \theta = 9.47^\circ delivers the ball in 0.83 s, + while an angle of 79.57^\circ delivers the ball in 4.5 s. +

    +     +     + + + +

    + If the ball is thrown at with an angle of elevation of 8^\circ, + what initial ball speed is needed to hit his target? +

    +     + +

    + With \theta=8^\circ, + the initial speed must be 53.8 mph\approx 78.9 ft/s. +

    +     +     + +     +     +     +     +
    +
    + Unit Tangent and Normal Vectors + + Unit Tangent Vector +

    + Given a smooth vector-valued function \vrt, + we defined in + that any vector parallel to \vrp(t_0) is tangent + to the graph of \vrt at t=t_0. + It is often useful to consider just the direction + of \vrp(t) and not its magnitude. + Therefore we are interested in the unit vector in the direction of \vrp(t). + This leads to a definition. +

    + + + Unit Tangent Vector + +

    + Let \vrt be a smooth function on an open interval I. + The unit tangent vector \unittangent(t) is + unit tangent vectordefinition + unit vectorunit tangent vector + + \unittangent(t) = \frac{1}{\norm{\vrp(t)}}\vrp(t) + . +

    +     +     + + + + + Computing the unit tangent vector + +

    + Let \vrt = \la 3\cos(t) , 3\sin(t) , 4t\ra. + Find \unittangent(t) and compute + \unittangent(0) and \unittangent(1). +

    +     + +

    + We apply to find \unittangent(t). + + \unittangent(t) \amp = \frac{1}{\norm{\vrp(t)}}\vrp(t) + \amp =\frac{1}{\sqrt{\big(-3\sin(t) \big)^2+\big(3\cos(t) \big)^2+ 4^2}}\la -3\sin(t) ,3\cos(t) , 4\ra + \amp = \la -\frac35\sin(t) ,\frac35\cos(t) ,\frac45\ra + . +

    + +

    + We can now easily compute \unittangent(0) and \unittangent(1): + + \unittangent(0) = \la 0,\frac35,\frac45\ra\,; \unittangent(1) = \la -\frac35\sin(1) ,\frac35\cos(1) ,\frac45\ra \approx \la -0.505,0.324,0.8\ra + . +

    + +

    + These are plotted in + with their initial points at + \vec r(0) and \vec r(1), respectively. + (They look rather short + since they are only length 1.) +

    + +
    + Plotting unit tangent vectors in + + + + A 3D plot of a portion of a helix, with two unit tangent vectors shown. + +

    + The curve \vec{r}(t) = \la 3\cos(t), 3\sin(t), 4t\ra is a circular helix of radius 3. + One revolution of the helix is shown, representing the interval -\pi\leq t\leq \pi. + Two unit tangent vectors are shown, plotted at the points corresponding to \vec{r}(0) + and \vec{r}(1). +

    + +

    + The length of the unit tangent vectors in the plot are short relative to the length of the curve, + so one has to look closely to tell that the curve does indeed bend away from the tip of the unit tangent vector. +

    +     + + //ASY file for figtannorm13D.asy in Chapter 11 + //ASPECT RATIO IS MAKING THE UNIT NORMAL LOOK LONGER THAN THE UNIT TANGENT + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(10,10,43); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={-10,10}; + defaultpen(0.5mm); + pair xbounds=(-4,4); + pair ybounds=(-4,4); + pair zbounds=(-15,15); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve <3cos t,3sin t,4t> for t from -pi to pi + triple g(real t) {return (3*cos(t),3*sin(t),4*t);} + path3 mypath=graph(g,-pi,pi,operator ..); draw(mypath,bluepen); + //Draw the unit tangent and norma on the curve at t=pi/2 + draw((3,0,0)--(3,.6,.8),redpen+linewidth(2),Arrow3(size=3mm));//T + draw((1.62,2.52,4)--(1.12,2.85,4.8),redpen+linewidth(2),Arrow3(size=3mm));//N + + + + +
    + +

    + The unit tangent vector \unittangent(t) always has a magnitude of 1, though it is sometimes easy to doubt that is true. + We can help solidify this thought in our minds by computing \norm{\unittangent(1)}: + + \norm{\unittangent(1)} \approx \sqrt{(-0.505)^2+0.324^2+0.8^2} = 1.000001 + . +

    + +

    + We have rounded in our computation of \unittangent(1), + so we don't get 1 exactly. + We leave it to the reader to use the exact representation of + \unittangent(1) to verify it has length 1. +

    +     + +     + +

    + In many ways, the previous example was too nice. + It turned out that \vrp(t) was always of length 5. + In the next example the length of \vrp(t) is variable, + leaving us with a formula that is not as clean. +

    + + + Computing the unit tangent vector + +

    + Let \vrt=\la t^2-t,t^2+t\ra. + Find \unittangent(t) and compute + \unittangent(0) and \unittangent(1). +

    +     + +

    + We find \vrp(t) = \la 2t-1,2t+1\ra, and + + \norm{\vrp(t)} = \sqrt{(2t-1)^2+(2t+1)^2} = \sqrt{8t^2+2} + . +

    + +

    + Therefore + + \unittangent(t) = \frac{1}{\sqrt{8t^2+2}}\la 2t-1,2t+1\ra = \la \frac{2t-1}{\sqrt{8t^2+2}},\frac{2t+1}{\sqrt{8t^2+2}}\ra + . +

    + +

    + When t=0, + we have \unittangent(0) = \la -1/\sqrt{2},1/\sqrt{2}\ra; + when t=1, + we have \unittangent(1) = \la 1/\sqrt{10}, 3/\sqrt{10}\ra. + We leave it to the reader to verify each of these is a unit vector. + They are plotted in +

    + +
    + Plotting unit tangent vectors in + + + A rotated parabola, along with two of its unit tangent vectors. + +

    + The curve in this example is the same one that we saw in . + It is a parabola that has been rotated so that it is symmetric about the line y=x, + and it has intercepts at the points (2,0), (0,0), and (0,2). +

    + +

    + There are unit tangent vectors plotted at (0,0) (corresponding to t=0), + and (0,2) (corresponding to t=1). + The tangent vector at the origin points up and to the left, + and the tangent vector at (0,2) points up and to the right. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-1.1,ymax=6.5, + xmin=-2.1,xmax=7 + ] + + \addplot+ [domain=-2:2,samples=40] ({x^2-x},{x^2+x}); + + \draw [thick,->,secondcolor] (axis cs:0,0) -- (axis cs:-.707,.707); + \draw [thick,->,secondcolor] (axis cs:0,2) -- (axis cs:0.32,2.95); + \draw [thick,->,firstcolor] (axis cs:3.79,.77) -- (axis cs:3.75,.75); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +     +     + + + Unit Normal Vector +

    + Just as knowing the direction tangent to a path is important, + knowing a direction orthogonal to a path is important. + When dealing with real-valued functions, + we defined the normal line at a point to the be the line through the point that was perpendicular to the tangent line at that point. + We can do a similar thing with vector-valued functions. + Given \vrt in \mathbb{R}^2, + we have 2 directions perpendicular to the tangent vector, + as shown in . + It is good to wonder Is one of these two directions preferable over the other? +

    + +
    + Given a direction in the plane, there are always two directions orthogonal to it + + + A generic plane curve with a tangent vector, along with two possible normal vectors. + +

    + A portion of a curve is shown in the first quadrant. + The curve is smooth, and forms a large arc that is parabolic in shape. + A tangent vector is shown at one point on the curve. + At that point, two possible normal vectors are plotted. + One points to the inside of the curve, in the direction in which it is turning. + The other points to the outside of the curve, + away from the direction in which the curve is turning. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ytick=\empty, + ymin=-.1,ymax=1.9, + xmin=-.1,xmax=2.3 + ] + + \addplot+ [domain=-1:1,samples=40] ({x^2-x+.6},{x^2+x+.3}); + + \draw[thick,->,black] (axis cs:.51,.41) -- (axis cs:.233,.83); + \draw[thick,->,secondcolor] (axis cs:.51,.41) -- (axis cs:.1,0.13); + \draw[thick,->,secondcolor] (axis cs:.51,.41) -- (axis cs:.93,.69); + \draw[thick,->,firstcolor] (axis cs:0.71, 0.21) -- (axis cs:.698,.218); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Given \vrt in \mathbb{R}^3, + there are infinitely many vectors orthogonal to the tangent vector at a given point. + Again, we might wonder Is one out of this infinite number of choices preferable over the others? + Is one of these the right choice? +

    + +

    + The answer in both \mathbb{R}^2 and \mathbb{R}^3 is Yes, + there is one vector that is not only preferable, + it is the right one to choose. + Recall , + which states that if \vrt has constant length, + then \vrt is orthogonal to \vrp(t) for all t. + We know \unittangent(t), the unit tangent vector, has constant length. + Therefore \unittangent(t) is orthogonal to + \unittangentprime(t). + vector-valued functionof constant length +

    + +

    + We'll see that \unittangentprime(t) is more than just a convenient choice of vector that is orthogonal to \vrp(t); + rather, it is the right choice. + Since all we care about is the direction, + we define this newly found vector to be a unit vector. +

    + + + + Unit Normal Vector + +

    + Let \vrt be a vector-valued function where the unit tangent vector, + \unittangent(t), is smooth on an open interval I. + The unit normal vector + \unitnormal(t) is + unit normal vectordefinition + unit vectorunit normal vector + + \unitnormal(t) = \frac1{\norm{\unittangentprime(t)}}\unittangentprime(t) + . +

    +     +     + + + + + Computing the unit normal vector + +

    + Let \vrt = \la 3\cos(t) , 3\sin(t) , 4t\ra as in . + Sketch both \unittangent(\pi/2) and + \unitnormal(\pi/2) with initial points at \vec r(\pi/2). +

    +     + +

    + In , + we found + + \unittangent(t) = \la (-3/5)\sin(t) ,(3/5)\cos(t) ,4/5\ra + . + Therefore + + \unittangentprime(t) = \la -\frac35\cos(t) ,-\frac35\sin(t) ,0\ra \text{ and } \norm{\unittangentprime(t)} = \frac35 + . +

    + +

    + Thus + + \unitnormal(t) = \frac{\unittangentprime(t)}{3/5} = \la -\cos(t) ,-\sin(t) ,0\ra + . +

    + +

    + We compute \unittangent(\pi/2) = \la -3/5,0,4/5\ra and \unitnormal(\pi/2) = \la 0,-1,0\ra. + These are sketched in . +

    + +
    + Plotting unit tangent and normal vectors in + + + + A three-dimensional helix. At one point, the unit tangent and normal vectors are shown. + +

    + The plot is three-dimensional, and shows a helix revolving about the z axis. + At a point above the y axis, the unit tangent and normal vectors are plotted. + The unit tangent vector points in the direction of travel, + while the unit normal vector points inward, toward the z axis. +

    +     + + //ASY file for figtannorm13D.asy in Chapter 11 + //ASPECT RATIO IS MAKING THE UNIT NORMAL LOOK LONGER THAN THE UNIT TANGENT + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(10,10,34); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-3,3}; + real[] myychoice={-3,3}; + real[] myzchoice={-10,10}; + defaultpen(0.5mm); + //pair xbounds=(-5,5); + //pair ybounds=(-5,5); + //pair zbounds=(-1,9); + + pair xbounds=(-4,4); + pair ybounds=(-4,4); + pair zbounds=(-15,15); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve <3cos t,3sin t,4t> for t from -pi to pi + triple g(real t) {return (3*cos(t),3*sin(t),4*t);} + //path3 mypath=graph(g,0,pi/2+pi/4,operator ..); draw(mypath,bluepen); + path3 mypath=graph(g,-pi,pi,operator ..); draw(mypath,bluepen); + + //Draw the unit tangent and norma on the curve at t=pi/2 + //draw((0,3,2*pi)--(-3/5,3,2*pi+4/5),redpen+linewidth(2),Arrow3(size=3mm));//T + draw((0,3,2*pi)--(-.942,3,2*pi+1.256),redpen+linewidth(2),Arrow3(size=3mm));//T + draw((0,3,2*pi)--(0,2,2*pi),redpen+linewidth(2),Arrow3(size=3mm));//N + + + + +
    +     + +     + + + +

    + The previous example was once again + too nice. In general, + the expression for \unittangent(t) contains fractions of square roots, + hence the expression of \unittangentprime(t) is very messy. + We demonstrate this in the next example. +

    + + + Computing the unit normal vector + +

    + Let \vrt=\la t^2-t,t^2+t\ra as in . + Find \unitnormal(t) and sketch \vrt with the unit tangent and normal vectors at t=-1,0 and 1. +

    +     + +

    + In , we found + + \unittangent(t) = \la \frac{2t-1}{\sqrt{8t^2+2}},\frac{2t+1}{\sqrt{8t^2+2}}\ra + . +

    + +

    + Finding \unittangentprime(t) requires two applications of the Quotient Rule: + + T\,'(t) \amp = \la \frac{\sqrt{8t^2+2}(2)-(2t-1)\left(\frac12(8t^2+2)^{-1/2}(16t)\right)}{8t^2+2},\right. + \amp \left.\frac{\sqrt{8t^2+2}(2)-(2t+1)\left(\frac12(8t^2+2)^{-1/2}(16t)\right)}{8t^2+2} \ra + \amp = \la \frac{4 (2 t+1)}{\left(8 t^2+2\right)^{3/2}},\frac{4 + (1-2 t)}{\left(8 t^2+2\right)^{3/2}}\ra + +

    + +

    + This is not a unit vector; to find \unitnormal(t), + we need to divide \unittangentprime(t) by its magnitude. + + \norm{\unittangentprime(t)} \amp = \sqrt{\frac{16(2t+1)^2}{(8t^2+2)^3}+\frac{16(1-2t)^2}{(8t^2+2)^3}} + \amp = \sqrt{\frac{16(8t^2+2)}{(8t^2+2)^3}} + \amp = \frac{4}{8t^2+2} + . +

    + +

    + Finally, + + \unitnormal(t) \amp = \frac1{4/(8t^2+2)}\la \frac{4 (2 t+1)}{\left(8 t^2+2\right)^{3/2}},\frac{4 + (1-2 t)}{\left(8 t^2+2\right)^{3/2}}\ra + \amp = \la \frac{2t+1}{\sqrt{8t^2+2}},-\frac{2t-1}{\sqrt{8t^2+2}}\ra + . +

    + +

    + Using this formula for \unitnormal(t), + we compute the unit tangent and normal vectors for t=-1,0 and 1 and sketch them in . +

    + +
    + Plotting unit tangent and normal vectors in + + + A rotated parabola, with unit tangent and normal vectors plotted at three points. + +

    + The curve is the same rotated parabola seen previously, + with symmetry about the line y=x, + and intercepts at the points (2,0), (0,0), and (0,2). + At each of the intercepts, the unit tangent and normal vectors are plotted. + The tangent vectors point along the direction of travel, + while the normal vectors point inward, toward the line of symmetry. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-1.1,ymax=6.5, + xmin=-2.1,xmax=7 + ] + + \addplot+ [domain=-2:2,samples=60] ({x^2-x},{x^2+x}); + + \draw [thick,->,secondcolor] (axis cs:0,0) -- (axis cs:-.707,.707); + \draw [thick,->,secondcolor] (axis cs:0,2) -- (axis cs:0.32,2.95); + \draw [thick,->,secondcolor] (axis cs:2,0) -- (axis cs:1.05,-.32); + + \draw [thick,->,secondcolor] (axis cs:0,0) -- (axis cs:.707,.707); + \draw [thick,->,secondcolor] (axis cs:0,2) -- (axis cs:.95,1.68); + \draw [thick,->,secondcolor] (axis cs:2,0) -- (axis cs:1.68,0.95); + + \draw [thick,->,firstcolor] (axis cs:3.79,.77) -- (axis cs:3.75,.75); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +     + +

    + The final result for \unitnormal(t) in + is suspiciously similar to \unittangent(t). + There is a clear reason for this. + If \vec u = \la u_1,u_2\ra is a unit vector in \mathbb{R}^2, + then the only unit vectors orthogonal to \vec u are + \la -u_2,u_1\ra and \la u_2,-u_1\ra. + Given \unittangent(t), we can quickly determine + \unitnormal(t) if we know which term to multiply by (-1). +

    + +

    + Consider again , + where we have plotted some unit tangent and normal vectors. + Note how \unitnormal(t) always points + inside the curve, + or to the concave side of the curve. + This is not a coincidence; this is true in general. + Knowing the direction that \vec r(t) turns + allows us to quickly find \unitnormal(t). +

    + + + Unit Normal Vectors in <m>\mathbb{R}^2</m> + +

    + Let \vec r(t) be a vector-valued function in \mathbb{R}^2 where + \unittangentprime(t) is smooth on an open interval I. + Let t_0 be in I and + \unittangent(t_0) = \la t_1,t_2\ra Then \unitnormal(t_0) is either + + \unitnormal(t_0) = \la -t_2,t_1\ra \text{ or } \unitnormal(t_0) = \la t_2,-t_1\ra + , + whichever is the vector that points to the concave side of the graph of \vec r. + unit tangent vectorin \mathbb{R}^2 + unit normal vectorin \mathbb{R}^2 +

    +     +     +     + + + Application to Acceleration +

    + Let \vrt be a position function. + It is a fact (stated later in ) + that acceleration, \vat, + lies in the plane defined by \unittangent and \unitnormal. + That is, there are scalar functions + a_\text{T}(t) and a_\text{N}(t) such that + + \vat = a_\text{T}(t)\unittangent(t) + a_\text{N}(t)\unitnormal(t) + . +

    + +

    + We generally drop the of t + part of the notation and just write + a_\text{T} and a_\text{N}. +

    + +

    + The scalar a_\text{T} measures how much + acceleration is in the direction of travel, that is, + it measures the component of acceleration that affects the speed. + The scalar a_\text{N} measures how much + acceleration is perpendicular to the direction of travel, that is, + it measures the component of acceleration that affects the direction of travel. + unit tangent vectorand acceleration + unit normal vectorand acceleration +

    + +

    + We can find a_\text{T} using the orthogonal projection of + \vec a(t) onto \unittangent(t) + (review + in if needed). + Recalling that since \unittangent(t) is a unit vector, + \unittangent(t)\cdot\unittangent(t)=1, so we have + + \proj{a(t)}{T(t)} = \frac{\vec a(t)\cdot\unittangent(t)}{\unittangent(t)\cdot\unittangent(t)}\unittangent(t) = \underbrace{\big(\vec a(t)\cdot\unittangent(t)\big)}_{a_\text{T}}\unittangent(t) + . +

    + +

    + Thus the amount of \vat in the direction of + \unittangent(t) is a_\text{T}=\vat\cdot\unittangent(t). + The same logic gives a_\text{N} = \vat\cdot\unitnormal(t). +

    + +

    + While this is a fine way of computing a_\text{T}, + there are simpler ways of finding a_\text{N} + (as finding \unitnormal itself can be complicated). + The following theorem gives alternate formulas for + a_\text{T} and a_\text{N}. +

    + + + + Acceleration in the Plane Defined by <m>\unittangent</m> and <m>\unitnormal</m> + +

    + Let \vrt be a position function with acceleration \vat and unit tangent and normal vectors + \unittangent(t) and \unitnormal(t). + Then \vat lies in the plane defined by + \unittangent(t) and \unitnormal(t); + that is, there exists scalars + a_\text{T} and a_\text{N} such that + + \vat = a_\text{T} \unittangent(t) + a_\text{N} \unitnormal(t) + . +

    + +

    + Moreover, + + a_\text{T} \amp = \vat\cdot\unittangent(t) = \frac{d}{dt}\Big(\norm{\vec v(t)}\Big) + a_\text{N} \amp = \vat\cdot \unitnormal(t) = \sqrt{\norm{\vec a(t)}^2-a_\text{T} ^2} = \frac{\norm{\vat\times\vvt}}{\norm{\vvt}} = \norm{\vvt}\,\norm{\unittangentprime(t)} + +

    +     +     + +

    + unit tangent vectorand acceleration + unit normal vectorand acceleration + an@a_\text{N} + at@a_\text{T} + unit normal vectoran@a_\text{N} + unit tangent vectorat@a_\text{T} +

    + +

    + Note the second formula for a_\text{T}: + \ds \frac{d}{dt}\Big(\norm{\vvt}\Big). + This measures the rate of change of speed, + which again is the amount of acceleration in the direction of travel. +

    + + + + + Computing <m>a_T</m> and <m>a_N</m> + +

    + Let \vrt = \la 3\cos(t) , 3\sin(t) , 4t\ra as in Examples + and . + Find a_\text{T} and a_\text{N}. +

    +     + +

    + The previous examples give \vat = \la -3\cos(t) ,-3\sin(t) ,0\ra and + + \unittangent(t) = \la -\frac35\sin(t) ,\frac35\cos(t) ,\frac45\ra \text{ and } \unitnormal(t) = \la -\cos(t) ,-\sin(t) ,0\ra + . +

    + +

    + We can find a_\text{T} and + a_\text{N} directly with dot products: + + a_\text{T} \amp = \vat\cdot \unittangent(t) = \frac95\cos(t) \sin(t) -\frac95\cos(t) \sin(t) +0 = 0. + a_\text{N} \amp = \vat\cdot \unitnormal(t) = 3\cos^2(t) +3\sin^2(t) + 0 = 3 + . +

    + +

    + Thus \vat = 0\unittangent(t) + 3\unitnormal(t) = 3\unitnormal(t), + which is clearly the case. +

    + +

    + What is the practical interpretation of these numbers? + a_\text{T} =0 means the object is moving at a constant speed, + and hence all acceleration comes in the form of direction change. +

    +     + +     + + + Computing <m>a_T</m> and <m>a_N</m> + +

    + Let \vrt=\la t^2-t,t^2+t\ra as in Examples + and . + Find a_\text{T} and a_\text{N}. +

    +     + +

    + The previous examples give \vat = \la 2,2\ra and + + \unittangent(t) = \la \frac{2t-1}{\sqrt{8t^2+2}},\frac{2t+1}{\sqrt{8t^2+2}}\ra \text{ and } \unitnormal(t) = \la \frac{2t+1}{\sqrt{8t^2+2}},-\frac{2t-1}{\sqrt{8t^2+2}}\ra + . +

    + +

    + While we can compute a_\text{N} using \unitnormal(t), + we instead demonstrate using another formula from . + + a_\text{T} \amp = \vat\cdot\unittangent(t) = \frac{4t-2}{\sqrt{8t^2+2}}+\frac{4t+2}{\sqrt{8t^2+2}} = \frac{8t}{\sqrt{8t^2+2}}. + a_\text{N} \amp = \sqrt{\norm{\vat}^2-a_\text{T} ^2} = \sqrt{8-\left(\frac{8t}{\sqrt{8t^2+2}}\right)^2}=\frac{4}{\sqrt{8t^2+2}} + . +

    + +

    + When t=2, + \ds a_\text{T} = \frac{16}{\sqrt{34}}\approx 2.74 and \ds a_\text{N} = \frac{4}{\sqrt{34}} \approx 0.69. + We interpret this to mean that at t=2, + the particle is accelerating mostly by increasing speed, + not by changing direction. + As the path near t=2 is relatively straight, + this should make intuitive sense. + + gives a graph of the path for reference. +

    + +
    + Graphing \vec r(t) in + + + A rotated parabola, with two points marked, corresponding to parameter values t=0 and t=2. + +

    + The curve is again a rotated parabola that is symmetric about y=x. + There are two marked points: the origin, which is labeled with the parameter value t=0, + and the point (2,6), which is labeled with the parameter value t=2. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.1,ymax=6.5, + xmin=-2.1,xmax=7 + ] + + \addplot+ [domain=-2.1:2.1,samples=60] ({x^2-x},{x^2+x}); + + \filldraw [black] (axis cs: 2,6) circle (2.4pt) node [below right] { $t=2$}; + \filldraw [black] (axis cs: 0,0) circle (2.4pt) node [above right] { $t=0$}; + + \draw (axis cs: 4.2,1.8) node { $\vec r(t)$}; + \draw [thick,->,firstcolor,>=stealth] (axis cs:3.79,.77) -- (axis cs:3.75,.75); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Contrast this with t=0, + where a_\text{T} = 0 and a_\text{N} = 4/\sqrt{2}\approx 2.82. + Here the particle's speed is not changing and all acceleration is in the form of direction change. +

    +     + +     + + + Analyzing projectile motion + +

    + A ball is thrown from a height of 240 + with an initial speed of 64 + and an angle of elevation of 30^\circ. + Find the position function \vrt of the ball and analyze + a_\text{T} and a_\text{N}. + projectile motion +

    +     + +

    + Using + of + we form the position function of the ball: + + \vrt = \la \big(64\cos(30^\circ) \big)t, -16t^2+\big(64\sin(30^\circ) \big)t+240\ra + , + which we plot in . +

    + +
    + Plotting the position of a thrown ball, with 1s increments shown + + + An inverted parabola with vertex in the first quadrant. Points corresponding to several parameter values are marked. + +

    + The path of a ball is shown; the ball travels in a parabolic arc, + beginning, when t=0 at the point (0,240). + It reaches its vertex when t=1, at the point (32\sqrt{3},256), + and then travels down and to the right, + meeting the x axis at the point (160\sqrt{3},0), when t=5. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-10,ymax=260, + xmin=-10,xmax=310 + ] + + \addplot+ [domain=0:5,samples=40] ({64*cos(30)*x},{-16*x^2+64*sin(30)*x+240}); + + \filldraw [black] (axis cs: 0,240) circle (2.4pt) node [below right] { $t=0$}; + \filldraw [black] (axis cs: 55.4,256) circle (2.4pt) node (A)[ ] {}; + \filldraw [black] (axis cs: 111,240) circle (2.4pt) node [above right] { $t=2$}; + \filldraw [black] (axis cs: 166,192) circle (2.4pt) node [above right] { $t=3$}; + \filldraw [black] (axis cs: 222,112) circle (2.4pt) node [above right] { $t=4$}; + \filldraw [black] (axis cs: 277,0) circle (2.4pt) node (B) [] {}; + + \draw [thick,->,firstcolor] (axis cs:149.649, 209.76) -- (axis cs:152,207); + + \end{axis} + + \draw (A) node[ above] {{ $t=1$}}; + \draw (B) node [shift={(-16pt,8pt)}] { $t=5$}; + + \end{tikzpicture} + + + + +
    + +

    + From this we find \vvt = \la 64\cos(30^\circ) , -32t+64\sin(30^\circ) \ra and \vat = \la 0,-32\ra. + Computing \unittangent(t) is not difficult, + and with some simplification we find + + \unittangent(t) = \la \frac{\sqrt{3}}{\sqrt{t^2-2t+4}}, \frac{1-t}{\sqrt{t^2-2t+4}}\ra + . +

    + +

    + With \vat as simple as it is, + finding a_\text{T} is also simple: + + a_\text{T} = \vat\cdot \unittangent(t) = \frac{32t-32}{\sqrt{t^2-2t+4}} + . +

    + +

    + We choose to not find \unitnormal(t) and find + a_\text{N} through the formula a_\text{N} = \sqrt{\norm{\vat}^2-a_\text{T}^2\,}: + + a_\text{N} = \sqrt{32^2-\left(\frac{32t-32}{\sqrt{t^2-2t+4}}\right)^2} = \frac{32\sqrt{3}}{\sqrt{t^2-2t+4}} + . +

    + +

    + + gives a table of values of + a_\text{T} and a_\text{N}. + When t=0, we see the ball's speed is decreasing; + when t=1 the speed of the ball is unchanged. + This corresponds to the fact that at t=1 the ball reaches its highest point. +

    + +

    + After t=1 we see that + a_\text{N} is decreasing in value. + This is because as the ball falls, + its path becomes straighter and most of the acceleration is in the form of speeding up the ball, + and not in changing its direction. +

    + +
    + A table of values of a_T and a_N in + + + t + a_\text{T} + a_\text{N} + + + + + + + + 0 + -16 + 27.7 + + + 1 + 0 + 32 + + + 2 + 16 + 27.7 + + + 3 + 24.2 + 20.9 + + + 4 + 27.7 + 16 + + + 5 + 29.4 + 12.7 + + +
    +     +     + +

    + Our understanding of the unit tangent and normal vectors is aiding our understanding of motion. + The work in + gave quantitative analysis of what we intuitively knew. +

    + +

    + The next section provides two more important steps towards this analysis. + We currently describe position only in terms of time. + In everyday life, + though, we often describe position in terms of distance + (The gas station is about 2 miles ahead, on the left.). + The arc length parameter + allows us to reference position in terms of distance traveled. +

    + +

    + We also intuitively know that some paths are straighter than others and some are curvier than others, + but we lack a measurement of curviness. + The arc length parameter provides a way for us to compute curvature, + a quantitative measurement of how curvy a curve is. +

    +     + + + + Terms and Concepts + + + +

    + If \unittangent(t) is a unit tangent vector, + what is \norm{\unittangent(t)}? +

    + +

    + +

    +     + + + + + + + +

    + What is the length of any unit vector? +

    +     +     +     +     + +     + + + + +

    + If \unitnormal(t) is a unit normal vector, + what is \unitnormal(t)\cdot \vrp(t)? +

    + +

    + +

    +     + + + + + +

    + Any vector of constant magnitude is orthogonal to its derivative. +

    +     +     +     +     + +     + + + + +

    + The acceleration vector \vat lies in the plane defined by what two vectors? +

    +     + + + +

    + \unittangent(t) and \unitnormal(t). +

    +     + +     + + + + +

    + a_\text{T} measures how much the acceleration is affecting the of an object. +

    +     + + + + + + + + +     +     + + + Problems + + + +

    + Given \vrt, + find \unittangent(t) and evaluate it at the indicated value of t. +

    +     + + + + +

    + \vrt = \la 2t^2,t^2-t\ra,t=1 +

    +     + +

    + \unittangent(t) = \la\frac{4 t}{\sqrt{20 t^2-4t+1}},\frac{2 t-1}{\sqrt{20 t^2-4t+1}}\ra; + \unittangent(1) = \la 4/\sqrt{17},1/\sqrt{17}\ra +

    +     + +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $T=Compute("<1/sqrt(1+sin^2(t)),-sin(t)/sqrt(1+sin^2(t))>"); + $Ta=Compute("<sqrt(2/3),-1/sqrt(3)>"); + + + +

    + \vrt = \la t,\cos(t) \ra, t=\pi/4. +

    + + Find \unittangent(t). + +

    + +

    + + Find \unittangent(\pi/4). + +

    + +

    +     +     +     + + + + +

    + \vrt = \la \cos^3(t) ,\sin^3(t) \ra,t=\pi/4 +

    +     + +

    + \unittangent(t) = \frac{\cos(t) \sin(t) }{\sqrt{\cos^2(t) \sin^2(t) }}\la -\cos(t) ,\sin(t) \ra. (Be careful; + this cannot be simplified as just + \la -\cos(t) ,\sin(t) \ra as \sqrt{\cos^2(t) \sin^2(t) }\neq \cos(t) \sin(t), + but rather \abs{\cos(t) \sin(t) }.) + \unittangent(\pi/4) = \la -\sqrt{2}/2,\sqrt{2}/2\ra +

    +     + +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $T=Compute("<-sin(t), cos(t)>"); + $Ta=Compute("<0,-1>"); + + + +

    + \vrt = \la \cos(t) , \sin(t) \ra, t=\pi. +

    + + Find \unittangent(t). + +

    + +

    + + Find \unittangent(\pi). + +

    + +

    +     +     +     + +     + + + +

    + Find the equation of the line tangent to the curve at the indicated t-value using the unit tangent vector. + Note: these are the same problems as in Exercises. +

    +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + $L=Compute("(2,0) + t*<4/sqrt(17),1/sqrt(17)>"); + + + +

    + Find the vector equation of the line tangent to + \vrt = \la 2t^2,t^2-t \ra at t=1 using the unit tangent vector. +

    + + + \ell(t)= + +

    + +

    +     +     +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + $L=Compute("(pi/4,sqrt(2)/2) + t*<sqrt(2/3),-1/sqrt(3)>"); + + + +

    + Find the vector equation of the line tangent to + \vrt = \la t,\cos(t) \ra at t=\pi/4 using the unit tangent vector. +

    + + + \ell(t)= + +

    + +

    +     +     +     + + + + +

    + \vrt = \la \cos^3(t) ,\sin^3(t) \ra,t=\pi/4 +

    +     + +

    + \ell(t) = \la \sqrt{2}/4,\sqrt{2}/4\ra + t\la -\sqrt{2}/2,\sqrt{2}/2\ra; + in parametric form, +

    + +

    + \ell(t) = \left\{\begin{array}{ccc} x\amp =\amp \sqrt{2}/4-\sqrt{2}t/2 \\ y \amp =\amp \sqrt{2}/4+\sqrt{2}t/2 + \end{array} \right. +

    +     + +     + + + + + Context("Vector2D"); + Context()->variables->are(t=>'Real'); + $L=Compute("(-1,0) + t*<0,-1>"); + + + +

    + Find the vector equation of the line tangent to + \vrt = \la \cos(t) ,\sin(t) \ra at t=\pi using the unit tangent vector. +

    + + + \ell(t)= + +

    + +

    +     +     +     + +     + + + +

    + Find \unitnormal(t) using . + Confirm the result using . +

    +     + + + + +

    + \vrt = \la 3\cos(t) , 3\sin(t) \ra +

    +     + +

    + \unittangent(t) = \la -\sin(t) ,\cos(t) \ra; + \unitnormal(t) = \la -\cos(t) ,-\sin(t) \ra +

    +     + +     + + + + +

    + \vrt = \la t, t^2 \ra +

    +     + +

    + \unittangent(t) = \la \frac{1}{\sqrt{1+4t^2}},\frac{2t}{\sqrt{1+4t^2}}\ra; + \unitnormal(t) = \la -\frac{2t}{\sqrt{1+4t^2}},\frac{1}{\sqrt{1+4t^2}}\ra +

    +     + +     + + + + +

    + \vrt = \la \cos(t) ,2\sin(t) \ra +

    +     + +

    + \unittangent(t) = \la -\frac{\sin(t) }{\sqrt{4\cos^2(t) +\sin^2(t) }},\frac{2\cos(t) }{\sqrt{4\cos^2(t) +\sin^2(t) }}\ra; + \unitnormal(t) = \la -\frac{2\cos(t) }{\sqrt{4\cos^2(t) +\sin^2(t) }},-\frac{\sin(t) }{\sqrt{4\cos^2(t) +\sin^2(t) }}\ra +

    +     + +     + + + + +

    + \vrt = \la e^t,e^{-t} \ra +

    +     + +

    + \unittangent(t) = \la \frac{e^t}{\sqrt{e^{2t}+e^{-2t}}},-\frac{e^{-t}}{\sqrt{e^{2t}+e^{-2t}}}\ra; + \unitnormal(t) = \la \frac{e^{-t}}{\sqrt{e^{2t}+e^{-2t}}},\frac{e^{t}}{\sqrt{e^{2t}+e^{-2t}}}\ra +

    +     + +     + +     + + + +

    + A position function \vrt is given along with its unit tangent vector + \unittangent(t) evaluated at t=a, + for some value of a. +

    + +

    +

      +
    1. +

      + Confirm that \unittangent(a) is as stated. +

      +
      2. + +
    2. +

      + Using a graph of \vrt and , + find \unitnormal(a). +

      +
      3. +
    +

    +     + + + + +

    + \vrt = \la 3\cos(t) ,5\sin(t) \ra;\ds \unittangent(\pi/4) = \la -\frac3{\sqrt{34}}, \frac5{\sqrt{34}}\ra. +

    +     + +

    +

      +
    1. +

      + Be sure to show work +

      +
      2. + +
    2. +

      + \unitnormal(\pi/4) = \la -5/\sqrt{34}, -3/\sqrt{34}\ra +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \vrt = \la t,\frac{1}{t^2+1}\ra;\ds \unittangent(1) = \la \frac2{\sqrt{5}}, -\frac1{\sqrt{5}}\ra. +

    +     + +

    +

      +
    1. +

      + Be sure to show work +

      +
      2. + +
    2. +

      + \unitnormal(1) = \la \frac1{\sqrt{5}}, \frac2{\sqrt{5}}\ra +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \vrt = (1+2\sin(t) )\la \cos(t) ,\sin(t) \ra;\ds \unittangent(0) = \la \frac2{\sqrt{5}}, \frac1{\sqrt{5}}\ra. +

    +     + +

    +

      +
    1. +

      + Be sure to show work +

      +
      2. + +
    2. +

      + \unitnormal(0) = \la -\frac1{\sqrt{5}}, \frac2{\sqrt{5}}\ra +

      +
      3. +
    +

    +     + +     + + + + +

    + \ds \vrt = \la \cos^3(t) ,\sin^3(t) \ra;\ds \unittangent(\pi/4) = \la -\frac1{\sqrt{2}}, \frac1{\sqrt{2}}\ra. +

    +     + +

    +

      +
    1. +

      + Be sure to show work +

      +
      2. + +
    2. +

      + \unitnormal(\pi/4) = \la \frac1{\sqrt{2}}, \frac1{\sqrt{2}}\ra +

      +
      3. +
    +

    +     + +     + +     + + + +

    + Find \unitnormal(t). +

    +     + + + + +

    + \vrt = \la 4t,2\sin(t) ,2\cos(t) \ra +

    +     + +

    + \unittangent(t) = \frac{1}{\sqrt{5}}\la 2,\cos(t) ,-\sin(t) \ra; + \unitnormal(t) = \la 0,-\sin(t) ,-\cos(t) \ra +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $N=Compute("<-cos(t), -3/5 sin(t), -4/5sin(t)>"); + + + +

    + If \vrt = \la 5\cos(t) ,3\sin(t) ,4\sin(t) \ra, + find \unitnormal(t). +

    + +

    + +

    +     +     +     + + + + +

    + \vrt = \la a\cos(t) ,a\sin(t) ,b t \ra; a \gt 0 +

    +     + +

    + \unittangent(t) = \frac{1}{\sqrt{a^2+b^2}}\la -a\sin(t) ,a\cos(t) ,b \ra; + \unitnormal(t) = \la -\cos(t) ,-\sin(t) ,0\ra +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(t=>'Real',a=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $N=Compute("<-cos(at), -sin(at), 0>"); + + + +

    + If \vrt = \la \cos(at),\sin(at),t \ra, find \unitnormal(t). +

    + +

    + +

    +     +     +     + +     + + + +

    + Find a_\text{T} and a_\text{N} given \vrt. + Be sure you can sketch \vrt on the indicated interval, + and comment on the relative sizes of a_\text{T} and + a_\text{N} at the indicated t values. +

    +     + + + + +

    + \vrt = \la t,t^2 \ra on [-1,1]; + consider t=0 and t=1. +

    +     + +

    + a_\text{T} = \frac{4t}{\sqrt{1+4t^2}} and a_\text{N} = \sqrt{4-\frac{16t^2}{1+4t^2}} +

    + +

    + At t=0, + a_\text{T} = 0 and a_\text{N} = 2; +

    + +

    + At t=1, + a_\text{T} = 4/\sqrt{5} and a_\text{N} = 2/\sqrt{5}. +

    + +

    + At t=0, + all acceleration comes in the form of changing the direction of velocity and not the speed; + at t=1, + more acceleration comes in changing the speed than in changing direction. +

    +     + +     + + + + + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->set(t=>{limits=>[0.1,4]}); + $aT=Compute("-2/t^5/sqrt(1+1/t^4)"); + $aN=Compute("2/t^3/sqrt(1+1/t^4)"); + $aT1=Formula("-sqrt(2)"); + $aN1=Formula("sqrt(2)"); + $aT2=Formula("-1/(4sqrt(17))"); + $aN2=Formula("1/sqrt(17)"); + + + +

    + \vrt = \la t,1/t \ra on (0,4]; + consider t=1 and t=2. +

    + + Find a_{T}: + +

    + +

    + + Find a_{N}: + +

    + +

    + + Find a_{T}(1): + +

    + +

    + + Find a_{N}(1): + +

    + +

    + + Find a_{T}(2): + +

    + +

    + + Find a_{N}(2): + +

    + +

    +     + +

    + At t=1, + acceleration comes from changing speed and changing direction in + equal measure; at t=2, + acceleration is nearly \vec 0 as it is; + the low value of a_\text{T} shows that the speed is nearly constant and the low value of + a_\text{N} shows the direction is not changing quickly. +

    +     +     +     + + + + +

    + \vrt = \la 2\cos(t) ,2\sin(t) \ra on [0,2\pi]; + consider t=0 and t=\pi/2. +

    +     + +

    + a_\text{T} = 0 and a_\text{N} = 2 +

    + +

    + At t=0, + a_\text{T} = 0 and a_\text{N} = 2; +

    + +

    + At t=\pi/2, + a_\text{T} = 0 and a_\text{N} = 2. +

    + +

    + The object moves at constant speed, + so all acceleration comes from changing direction, + hence a_\text{T}=0. + \vat is always parallel to \unitnormal(t), + but twice as long, hence a_\text{N}=2. +

    +     + +     + + + + + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $aT=Compute("2"); + $aN=Compute("4t^2"); + $aT1=Formula("2"); + $aN1=Formula("2pi"); + $aT2=Formula("2"); + $aN2=Formula("4pi"); + + + +

    + \vrt = \la \cos(t^2),\sin(t^2) \ra on (0,2\pi]; + consider t=\pi/2 and t=\pi. +

    + + Find a_{T}: + +

    + +

    + + Find a_{N}: + +

    + +

    + + Find a_{T}\mathopen{}\left(\sqrt{\pi/2}\right)\mathclose{}: + +

    + +

    + + Find a_{N}\mathopen{}\left(\sqrt{\pi/2}\right)\mathclose{}: + +

    + +

    + + Find a_{T}\mathopen{}\left(\sqrt{\pi}\right)\mathclose{}: + +

    + +

    + + a_{N}\mathopen{}\left(\sqrt{\pi}\right)\mathclose{}: + +

    + +

    +     + +

    + The object moves at increasing speed + (increasing at a constant rate of acceleration), + hence a_\text{T}=2. + Since the object is increasing speed yet always traveling in a circle of radius 1, the direction must change more quickly; + the amount of acceleration that changes direction increases over time. +

    +     +     +     + + + + +

    + \vrt = \la a\cos(t) ,a\sin(t) , bt \ra on [0,2\pi], + where a,b \gt 0; + consider t=0 and t=\pi/2. +

    +     + +

    + a_\text{T} = 0 and a_\text{N} = a +

    + +

    + At t=0, + a_\text{T} = 0 and a_\text{N} = a; +

    + +

    + At t=\pi/2, + a_\text{T} = 0 and a_\text{N} = a. +

    + +

    + The object moves at constant speed, + meaning that a_\text{T} is always 0. + The object rises along the z-axis at a constant rate, + so all acceleration comes in the form of changing direction circling the z-axis. + The greater the radius of this circle the greater the acceleration, + hence a_\text{N}=a. +

    +     + +     + + + + + Context()->variables->are(t=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $aT=Compute("0"); + $aN=Compute("5"); + $aT1=Formula("0"); + $aN1=Formula("5"); + $aT2=Formula("0"); + $aN2=Formula("5"); + + + +

    + \vrt = \la 5\cos(t) ,4\sin(t) , 3\sin(t) \ra on [0,2\pi]; + consider t=0 and t=\pi/2. +

    + + Find a_{T}: + +

    + +

    + + Find a_{N}: + +

    + +

    + + Find a_{T}(0): + +

    + +

    + + Find a_{N}(0): + +

    + +

    + + Find a_{T}(\pi/2): + +

    + +

    + + Find a_{N}(\pi/2): + +

    + +

    +     + +

    + The object moves at constant speed, + meaning that a_\text{T} is always 0. + Acceleration is thus always perpendicular to the direction of travel; + in this particular case, + it is always 5 times the unit vector pointing orthogonal to the direction of travel. +

    +     +     +     + +     +     +     +
    +
    + The Arc Length Parameter and Curvature + + The Arc Length Parametrization +

    + In normal conversation we describe position in terms of both + time and distance. + For instance, imagine driving to visit a friend. + If she calls and asks where you are, + you might answer I am 20 minutes from your house, + or you might say I am 10 miles from your house. + Both answers provide your friend with a general idea of where you are. +

    + + + +

    + Currently, our vector-valued functions have defined points with a parameter t, + which we often take to represent time. + Consider , + where \vrt = \la t^2-t,t^2+t\ra is graphed and the points corresponding to t=0,\ 1 and 2 are shown. + Note how the arc length between t=0 and t=1 is smaller than the arc length between t=1 and t=2; + if the parameter t is time and \vec r is position, + we can say that the particle traveled faster on [1,2] than on [0,1]. +

    + +
    + Introducing the arc length parameter + +
    + + + + A parabola, rotated to be symmetric about the line y=x, with three marked points. + +

    + It is our old friend, the parabolic curve, rotated so that it lies primarily in the first quadrant, + with y=x as its line of symmetry. + There are three marked points: (0,0), (0,2), and (2,6), + corresponding to parameter values t=0, t=1, and t=2, respectively. +

    + +

    + The distance between the first and second points is much less than the distance between the second and third. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.1,ymax=6.5, + xmin=-2.1,xmax=7 + ] + + \addplot+ [domain=-2:2.1,samples=40] ({x^2-x},{x^2+x}); + + \filldraw [secondcolor] (axis cs: 0,0) circle (2.4pt) node [black,above right] { $t=0$}; + \filldraw [secondcolor] (axis cs: 0,2) circle (2.4pt) node [black, right] { $t=1$}; + \filldraw [secondcolor] (axis cs: 2,6) circle (2.4pt) node [black, right] { $t=2$}; + + \draw [thick,->,firstcolor,>=stealth] (axis cs:3.79,.77) -- (axis cs:3.75,.75) node [above ,black] { $\vrt$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + The same parabola as the previous image, this time with six marked points, all equally spaced along the curve. + +

    + The curve generated by \vec{r}(s) is the same as the curve generated by \vec{r}(t) in . + However, as a description of particle motion, the vector-valued function is quite different. + In this image we see six marked points, corresponding to parameter values s=0 through s=6. + The distance between successive points is equal, illustrating the fact that the particle speed is constant in this case. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-1.1,ymax=6.5, + xmin=-2.1,xmax=7 + ] + + \addplot+ [domain=-2:2.1,samples=40] ({x^2-x},{x^2+x}); + + \filldraw [secondcolor] (axis cs: 0,0) circle (2.4pt) node [black,above right] { $s=0$}; + \filldraw [secondcolor] (axis cs: -.24,.94) circle (2.4pt) node [black, right] { $s=1$}; + \filldraw [secondcolor] (axis cs: -.03,1.91) circle (2.4pt) node [black, right] { $s=2$}; + \filldraw [secondcolor] (axis cs: .33,2.84) circle (2.4pt) node [black, right] { $s=3$}; + \filldraw [secondcolor] (axis cs: .75,3.75) circle (2.4pt) node [black, right] { $s=4$}; + \filldraw [secondcolor] (axis cs: 1.21,4.63) circle (2.4pt) node [black, right] { $s=5$}; + \filldraw [secondcolor] (axis cs: 1.71, 5.51) circle (2.4pt) node [black, right] { $s=6$}; + + \draw [thick,->,firstcolor,>=stealth] (axis cs:3.79,.77)--(axis cs:3.75,.75)node [above,black] { $\vec r(s)$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + Now consider , + where the same graph is parametrized by a different variable s. + Points corresponding to s=0 through s=6 are plotted. + The arc length of the graph between each adjacent pair of points is 1. + We can view this parameter s as distance; + that is, the arc length of the graph from s=0 to s=3 is 3, the arc length from s=2 to s=6 is 4, etc. + If one wants to find the point 2.5 units from an initial location (, s=0), + one would compute \vec r(2.5). + This parameter s is very useful, + and is called the arc length parameter. + arc length parameter + arc length +

    + +

    + How do we find the arc length parameter? +

    + +

    + Start with any parametrization of \vec r. + We can compute the arc length of the graph of \vec r on the interval [0,t] with + + \text{arc length}\, = \int_0^t\norm{\vrp(u)}\, du + . +

    + +

    + We can turn this into a function: + as t varies, + we find the arc length s from 0 to t. + This function is + + s(t) = \int_0^t \norm{\vrp(u)}\, du + . +

    + +

    + This establishes a relationship between s and t. + Knowing this relationship explicitly, + we can rewrite \vrt as a function of s: + \vec r(s). + We demonstrate this in an example. +

    + + + Finding the arc length parameter + +

    + Let \vrt = \la 3t-1,4t+2\ra. + Parametrize \vec r with the arc length parameter s. +

    +     + +

    + Using Equation, we write + + s(t) = \int_0^t \norm{\vrp(u)}\, du + . +

    + +

    + We can integrate this, + explicitly finding a relationship between s and t: + + s(t) \amp = \int_0^t \norm{\vrp(u)}\, du + \amp = \int_0^t \sqrt{3^2+4^2}\, du + \amp = \int_0^t 5\, du + \amp = 5t + . +

    + +

    + Since s=5t, + we can write t=s/5 and replace t in \vrt with s/5: + + \vec r(s) = \la 3(s/5)-1, 4(s/5)+2\ra = \la \frac35s-1,\frac45s+2\ra + . +

    + +

    + Clearly, as shown in , + the graph of \vec r is a line, + where t=0 corresponds to the point (-1,2). + What point on the line is 2 units away from this initial point? + We find it with \vec r(2) = \la 1/5, 18/5\ra. +

    + +
    + Graphing \vec r in with parameters t and s + + + A straight line, with positive slope, and several equally-spaced marked points. + +

    + The curve generated by \vec{r}(t) is shown; + it is a straight line with slope 3/4 and y intercept 10/3. + There are six equally-spaced marked points on the line. + The first, at (-1,2), corresponds to both t=0 and s=0. + The last, at (2,6), corresponds to both t=1 and s=5. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,2,-1,-2}, + ymin=-1.1,ymax=6.5, + xmin=-2.1,xmax=2.9 + ] + + \addplot+ [domain=-2.5:2.5] ({3*x-1},{4*x+2}); + + \filldraw (axis cs:-1,2) circle (2.4pt) node [left] { $t=0$} + (axis cs:2,6) circle (2.4pt) node [left] { $t=1$} + (axis cs: -.4,2.8) circle (2.4pt) node [right,secondcolor] { $s=1$} + (axis cs: .2,3.6) circle (2.4pt) node [right,secondcolor] { $s=2$} + (axis cs: .8,4.4) circle (2.4pt) node [right,secondcolor] { $s=3$} + (axis cs: 1.4,5.2) circle (2.4pt) node [right,secondcolor] { $s=4$} + (axis cs: 2,6) circle (2.4pt) node [right,secondcolor] { $s=5$}; + + \draw (axis cs:-1,2) node [secondcolor,shift={(10pt,-5pt)}] { $s=0$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Is the point (1/5,18/5) really 2 units away from (-1,2)? + We use the Distance Formula to check: + + d = \sqrt{\left(\frac15-(-1)\right)^2+ \left(\frac{18}5-2\right)^2} = \sqrt{\frac{36}{25}+\frac{64}{25}} = \sqrt{4}=2 + . +

    + +

    + Yes, \vec r(2) is indeed 2 units away, + in the direction of travel, from the initial point. +

    +     + +     + +

    + Things worked out very nicely in ; + we were able to establish directly that s=5t. + Usually, the arc length parameter is much more difficult to describe in terms of t, + a result of integrating a square root. + There are a number of things that we can learn about the arc length parameter from Equation, though, + that are incredibly useful. +

    + +

    + First, take the derivative of s with respect to t. + The Fundamental Theorem of Calculus + (see ) + states that + + \frac{ds}{dt}=s\primeskip'(t) = \norm{\vrp(t)} + . +

    + +

    + Letting t represent time and \vrt represent position, + we see that the rate of change of s with respect to t is speed; + that is, the rate of change of distance traveled is speed, + which should match our intuition. +

    + +

    + The Chain Rule states that + + \frac{d\vec r}{dt} \amp = \frac{d\vec r}{ds}\cdot\frac{ds}{dt} + \vrp(t) \amp = \vrp(s)\cdot \norm{\vrp(t)} + . +

    + +

    + Solving for \vrp(s), we have + + \vrp(s) = \frac{\vrp(t)}{\norm{\vrp(t)}} = \unittangent(t) + , + where \unittangent(t) is the unit tangent vector. + Equation is often misinterpreted, + as one is tempted to think it states \vrp(t) = \unittangent(t), + but there is a big difference between \vrp(s) and \vrp(t). + The key to take from it is that \vrp(s) is a unit vector. + In fact, the following theorem states that this characterizes the arc length parameter. +

    + + + Arc Length Parameter + +

    + Let \vec r(s) be a vector-valued function. + The parameter s is the arc length parameter if, + and only if, \norm{\vrp(s)} = 1. + arc length parameter +

    +     +     + + +     + + + Curvature +

    + Consider points A and B on the curve graphed in . + One can readily argue that the curve curves more sharply at A than at B. + It is useful to use a number to describe how sharply the curve bends; + that number is the curvature of the curve. +

    + +
    + Establishing the concept of curvature + +
    + + + + Plot of a parametric curve that bends rapidly at one point, but is otherwise relatively flat. + +

    + An L-shaped curve is shown. There are two marked points, labeled A and B. + The point A is at the corner of the curve. + The corner is smooth, not sharp, but represents a significant bend in the curve. + Near the point B, the curve bends much more gently. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=2.9, + xmin=-.1,xmax=3.5 + ] + + \addplot+ [domain=-1:1.5,samples=40] ({x^2-x+1},{x^3+1}); + + \draw [->,firstcolor,thick,>=stealth] (axis cs:2.44, 0.488) -- (axis cs: 2.4141, 0.506961); + + \filldraw (axis cs:.77,1.04) circle (2.4pt) node [below] { $A$} + (axis cs: 1.75, 0.875) circle (2.4pt) node [below] { $B$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + Plot of a parametric curve that bends rapidly at one point, but is otherwise relatively flat, along with some tangent vectors. + +

    + The same L-shaped curve is shown as in , along with the marked points A and B. + Additional points are marked on either side of A and B, + and tangent vectors are plotted at each of these points. + The image illustrates how the directions of the tangent vectors are not that different near B, + but the directions near A are significantly different. +

    +     + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + ymin=-.1,ymax=2.9, + xmin=-.1,xmax=3.5 + ] + + \addplot+ [domain=-1:1.5,samples=40] ({x^2-x+1},{x^3+1}); + + \draw [->,secondcolor,thick] (axis cs:0.8461, 1.53144) -- (axis cs:1.14654, 2.48524); + \draw [->,secondcolor,thick] (axis cs:1.2684, 0.989352) -- (axis cs:0.273445, 1.08968); + \draw [->,secondcolor,thick] (axis cs:2.20203, 0.649597) -- (axis cs:1.35163, 1.17574); + \draw [->,firstcolor,thick] (axis cs:2.44, 0.488) -- (axis cs: 2.4141, 0.506961); + + \filldraw (axis cs:.77,1.04) circle (2.4pt) node [below] { $A$} + (axis cs:0.8461, 1.53144) circle (2.4pt) + (axis cs: 1.2684, 0.989352) circle (2.4pt) + (axis cs: 1.75, 0.875) circle (2.4pt) node [below] { $B$} + (axis cs: 2.20203, 0.649597) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + We derive this number in the following way. + Consider , + where unit tangent vectors are graphed around points A and B. + Notice how the direction of the unit tangent vector changes quite a bit near A, + whereas it does not change as much around B. + This leads to an important concept: + measuring the rate of change of the unit tangent vector with respect to arc length gives us a measurement of curvature. +

    + + + Curvature + +

    + Let \vec r(s) be a vector-valued function where s is the arc length parameter. + The curvature \kappa of the graph of \vec r(s) is + curvature + unit tangent vectorand curvature + + \kappa = \snorm{\frac{d\unittangent}{ds}} = \snorm{\unittangentprime(s)} + . +

    +     +     + + + +

    + If \vec r(s) is parametrized by the arc length parameter, then + + \unittangent(s) = \frac{\vrp(s)}{\norm{\vrp(s)}} \text{ and } \unitnormal(s) = \frac{\unittangentprime(s)}{\norm{\unittangentprime(s)}} + . +

    + +

    + Having defined \norm{\unittangentprime(s)} =\kappa, + we can rewrite the second equation as + + \unittangentprime(s) = \kappa\unitnormal(s) + . +

    + +

    + We already knew that \unittangentprime(s) is in the same direction as \unitnormal(s); + that is, we can think of \unittangent(s) as being pulled + in the direction of \unitnormal(s). + How hard is it being pulled? + By a factor of \kappa. + When the curvature is large, + \unittangent(s) is being pulled hard + and the direction of \unittangent(s) changes rapidly. + When \kappa is small, + T(s) is not being pulled hard and hence its direction is not changing rapidly. +

    + +

    + We use + to find the curvature of the line in . +

    + + + Finding the curvature of a line + +

    + Use + to find the curvature of \vrt = \la 3t-1,4t+2\ra. +

    +     + +

    + In , + we found that the arc length parameter was defined by s=5t, + so \vec r(s) =\la 3s/5-1, 4s/5+2\ra parametrized \vec r with the arc length parameter. + To find \kappa, we need to find \unittangentprime(s). + + \unittangent(s) \amp = \vrp(s) \text{ (recall this is a unit vector) } + \amp = \la 3/5, 4/5\ra. + Therefore + \unittangentprime(s) \amp = \la 0,0\ra + and + \kappa\amp =\snorm{\unittangentprime(s)} = 0 + . +

    + +

    + It probably comes as no surprise that the curvature of a line is 0. (How + curvy is a line? + It is not curvy at all.) +

    +     + +     + +

    + While the definition of curvature is a beautiful mathematical concept, + it is nearly impossible to use most of the time; + writing \vec r in terms of the arc length parameter is generally very hard. + Fortunately, + there are other methods of calculating this value that are much easier. + There is a tradeoff: the definition is easy + to understand though hard to compute, + whereas these other formulas are easy to compute though it may be hard to understand why they work. +

    + + + Formulas for Curvature + +

    + Let C be a smooth curve in the plane or in space. + curvatureequations for +

      +
    1. +

      + If C is defined by y=f(x), then + + \kappa = \frac{\abs{\fp'(x)}}{\Big(1+\big(\fp(x)\big)^2\Big)^{3/2}} + . +

      +
      2. + +
    2. +

      + If C is defined as a vector-valued function in the plane, + \vrt = \la x(t), y(t)\ra, then + + \kappa = \frac{\abs{x'\yp'-x''\yp}}{\big((x')^2+(\yp)^2\big)^{3/2}} + . +

      +
      3. + +
    3. +

      + If C is defined in space by a vector-valued function \vrt, then + + \kappa = \frac{\norm{\unittangentprime(t)}}{\norm{\vrp(t)}} = \frac{\norm{\vrp(t)\times\vrpp(t)}}{\norm{\vrp(t)}^3} = \frac{\vec a(t)\cdot \unitnormal(t)}{\norm{\vec v(t)}^2} + . +

      +
      4. +
    +

    +     +     + +

    + We practice using these formulas. +

    + + + Finding the curvature of a circle + +

    + Find the curvature of a circle with radius r, + defined by \vec c(t) = \la r\cos(t), r\sin(t)\ra. + curvatureof circle +

    +     + +

    + Before we start, + we should expect the curvature of a circle to be constant, + and not dependent on t. (Why?) +

    + +

    + We compute \kappa using the second part of . + + \kappa \amp = \frac{\abs{(-r\sin(t) )(-r\sin(t) ) - (-r\cos(t) )(r\cos(t) )}}{\big( (-r\sin(t) )^2+(r\cos(t) )^2\big)^{3/2}} + \amp = \frac{r^2(\sin^2(t) +\cos^2(t) )}{\big(r^2(\sin^2(t) +\cos^2(t) )\big)^{3/2}} + \amp = \frac{r^2}{r^3} = \frac1r + . +

    + +

    + We have found that a circle with radius r has curvature \kappa = 1/r. +

    +     + +     + +

    + gives a great result. + Before this example, if we were told + The curve has a curvature of 5 at point A, + we would have no idea what this really meant. + Is 5 big does is correspond to a really sharp turn, + or a not-so-sharp turn? + Now we can think of 5 in terms of a circle with radius 1/5. + Knowing the units + (inches vs.miles, for instance) + allows us to determine how sharply the curve is curving. +

    + +

    + Let a point P on a smooth curve C be given, + and let \kappa be the curvature of the curve at P. + A circle that: +

    + +

    +

      +
    • +

      + passes through P, +

      +
      • + +
    • +

      + lies on the concave side of C, +

      +
      • + +
    • +

      + has a common tangent line as C at P and +

      +
      • + +
    • +

      + has radius r=1/\kappa (hence has curvature \kappa) +

      +
      • +
    +

    + +

    + is the osculating circle, + or circle of curvature, + to C at P, and r is the + radius of curvature. + curvatureof circle + osculating circle + circle of curvature + radius of curvature + curvatureradius of + shows the graph of the curve seen earlier in + and its osculating circles at A and B. + A sharp turn corresponds to a circle with a small radius; + a gradual turn corresponds to a circle with a large radius. + Being able to think of curvature in terms of the radius of a circle is very useful. +

    + +
    + Illustrating the osculating circles for the curve seen in + + + A curve with two marked points, corresponding to points of large and small curvature, along with osculating circles at those points. + +

    + The curve shown is the same L-shaped curve as in , + with the same two marked points, labeled A and B. + The osculating circle at A is relatively small, + indicating a large curvature at that point. + At the point B, where the curvature is small, + the osculating circle is large. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=2.9, + xmin=-.1,xmax=3.5 + ] + + \addplot+ [domain=-1:1.5,samples=40] ({x^2-x+1},{x^3+1}); + \addplot [secondcurvestyle,solid,domain=0:360] ({.08*cos(x)+.83},{.08*sin(x)+1.09}); + \addplot [secondcurvestyle,solid,domain=0:360,samples=101] ({2.17*cos(x)+.99},{2.17*sin(x)+-1.15}); + + \draw [->,firstcolor,thick,>=stealth] (axis cs:2.44, 0.488) -- (axis cs: 2.4141, 0.506961); + + \filldraw (axis cs:.77,1.04) circle (2.4pt) node [below] { $A$} + (axis cs: 1.75, 0.875) circle (2.4pt) node [below] { $B$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + + + +

    + (The word osculating comes from a Latin word related to kissing; + an osculating circle kisses + the graph at a particular point. + Many beautiful ideas in mathematics have come from studying the osculating circles to a curve.) +

    + + + Finding curvature + +

    + Find the curvature of the parabola defined by y=x^2 at the vertex and at x=1. +

    +     + +

    + We use the first formula found in . + + \kappa(x) \amp = \frac{\abs{2}}{\big(1+(2x)^2\big)^{3/2}} + \amp = \frac2{\big(1+4x^2\big)^{3/2}} + . +

    + +
    + Examining the curvature of y=x^2 + + + A parabola along with osculating circles at two points. + +

    + The parabola y=x^2 is shown, along with two osculating circles. + At the origin, where the curvature is largest, + the osculating circle is small, and sits completely within the parabola. + At the point (1,1) the curvature is much less, + leading to a large osculating circle that intercepts the parabola twice. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-2.5,ymax=10.1, + xmin=-10.2,xmax=4.9 + ] + + \addplot+ [domain=-3.2:3.2,samples=40] ({x},{x^2}); + \addplot [secondcurvestyle,solid,domain=0:360] ({.5*cos(x)},{.5*sin(x)+.5}); + \addplot [secondcurvestyle,solid,domain=0:360,samples=70] ({5.6*cos(x)-4},{5.6*sin(x)+3.5}); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + At the vertex (x=0), the curvature is \kappa = 2. + At x=1, + the curvature is \kappa = 2/(5)^{3/2} \approx 0.179. + So at x=0, + the curvature of y=x^2 is that of a circle of radius 1/2; + at x=1, + the curvature is that of a circle with radius \approx 1/0.179 \approx 5.59. + This is illustrated in . + At x=3, the curvature is 0.009; + the graph is nearly straight as the curvature is very close to 0. +

    +     +     + + + Finding curvature + +

    + Find where the curvature of \vrt = \la t, t^2, 2t^3\ra is maximized. +

    +     + +

    + We use the third formula in + as \vrt is defined in space. + We leave it to the reader to verify that + + \vrp(t) =\la 1,2t,6t^2\ra, \vrp'(t) = \la 0,2,12t\ra, \text{ and } \vrp(t)\times \vrpp(t) = \la 12t^2,-12t,2\ra + . +

    + +

    + Thus + + \kappa(t) \amp = \frac{\norm{\vrp(t)\times\vrp'(t)}}{\norm{\vrp(t)}^3} + \amp = \frac{\norm{\la 12t^2,-12t,2\ra}}{\norm{\la 1,2t,6t^2\ra}^3} + \amp = \frac{\sqrt{144t^4+144t^2+4}}{\left(\sqrt{1+4t^2+36t^4\ }\right)^3} + +

    + +

    + While this is not a particularly nice formula, + it does explicitly tell us what the curvature is at a given t value. + To maximize \kappa(t), + we should solve \kappa'(t)=0 for t. + This is doable, but very time consuming. + Instead, consider the graph of + \kappa(t) as given in . + We see that \kappa is maximized at two t values; + using a numerical solver, we find these values are t\approx\pm 0.189. + In we graph \vrt and indicate the points where curvature is maximized. +

    + +
    + Understanding the curvature of a curve in space + +
    + The curvature of \vec{r}(t) + + + A plot of the curvature as a function of the parameter t. + +

    + The curve illustrates the value of the curvature \kappa as a function of t. + The graph is symmetric about the y axis, and has the shape of a tall cowboy hat, + with local maxima on either side of the y axis, and a local minimum in between. + The t axis is a horizontal asymptote, + showing that the curve becomes almost straight when t is large in absolute value. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-.1,ymax=2.5, + xmin=-2.1,xmax=2.1, + xlabel = {$t$} + ] + + \addplot+ [smooth] coordinates {(-1.5,0.0122)(-1.45,0.01404)(-1.4,0.01623)(-1.35,0.01887)(-1.3,0.02207)(-1.25,0.02596)(-1.2,0.03074)(-1.15,0.03666)(-1.1,0.04403)(-1.05,0.05331)(-1.,0.06509)(-0.95,0.08019)(-0.9,0.09974)(-0.85,0.1253)(-0.8,0.159)(-0.75,0.204)(-0.7,0.2643)(-0.65,0.3457)(-0.6,0.4557)(-0.55,0.6035)(-0.5,0.7989)(-0.45,1.049)(-0.4,1.352)(-0.35,1.686)(-0.3,2.006)(-0.25,2.246)(-0.2,2.353)(-0.15,2.318)(-0.1,2.191)(-0.05,2.057)(0,2.)(0.05,2.057)(0.1,2.191)(0.15,2.318)(0.2,2.353)(0.25,2.246)(0.3,2.006)(0.35,1.686)(0.4,1.352)(0.45,1.049)(0.5,0.7989)(0.55,0.6035)(0.6,0.4557)(0.65,0.3457)(0.7,0.2643)(0.75,0.204)(0.8,0.159)(0.85,0.1253)(0.9,0.09974)(0.95,0.08019)(1.,0.06509)(1.05,0.05331)(1.1,0.04403)(1.15,0.03666)(1.2,0.03074)(1.25,0.02596)(1.3,0.02207)(1.35,0.01887)(1.4,0.01623)(1.45,0.01404)(1.5,0.0122)}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + A plot of the curve \vec{r}(t)=\la t, t^2, 2t^3\ra + + + + A plot of the vector-valued function in this example, with points of maximum curvature marked. + +

    + A three-dimensional plot of the curve generated by the vector-valued function \vec{r}(t) = \la t, t^2, 3t^3\ra. + In the default viewing angle, the curve sweeps in from the foreground, bending toward the origin. + At the origin, the curve bends sharply and appears to tend upward toward the z axis. +

    + +

    + Rotating the viewpoint reveals that the curve begins below the xy plane, + bends in toward the origin, and then heads upward, above the xy plane. + It actually moves outward, away from the z axis. +

    +     + + + + + //ASY file for figcurvature43D.asy in Chapter 11 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-2,2); + pair ybounds=(-1,2); + pair zbounds=(-2,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the curve <t, t^2, 2*t^3> for t from -1 to 1 + triple g(real t) {return (t, t^2, 2*t^3);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,bluepen); + //path3 mypath=graph(g,-1,-0.9,operator ..); draw(mypath,bluepen,Arrow3(size=4mm)); + path3 mypath=graph(g,-1,-0.7,operator ..); draw(mypath,bluepen,Arrow3(size=4mm)); + //path3 mypath=graph(g,-1,0.7,operator ..); draw(mypath,bluepen,Arrow3(size=4mm)); + //path3 mypath=graph(g,-1,0.9,operator ..); draw(mypath,bluepen,Arrow3(size=4mm)); + + //Draw points of maximum curvature (0.189,(0.189)^2,2*(0.189)^3) + // cooresponding to values t=+/-0.189 + dotfactor=4; + dot(((0.189,(0.189)^2,2*(0.189)^3)),redpen); + dot(((-0.189,(-0.189)^2,2*(-0.189)^3)),redpen); + + + + +
    +     + +
    + +     +     +     + + + Curvature and Motion +

    + Let \vrt be a position function of an object, + with velocity \vvt = \vrp(t) and acceleration \vec a(t)=\vrpp(t). + In + we established that acceleration is in the plane formed by + \unittangent(t) and \unitnormal(t), + and that we can find scalars a_\text{T} and + a_\text{N} such that + curvatureand motion + unit tangent vectorand curvature + unit normal vectorand curvature + at@a_\text{T} + an@a_\text{N} + + \vat = a_\text{T} \unittangent(t) + a_\text{N} \unitnormal(t) + . +

    + +

    + + gives formulas for a_\text{T} and a_\text{N}: + + a_\text{T} = \frac{d}{dt}\Big(\norm{\vvt}\Big) \text{ and } a_\text{N} = \frac{\norm{\vvt\times \vat}}{\norm{\vvt}} + . +

    + +

    + We understood that the amount of acceleration in the direction of \unittangent relates only to how the speed of the object is changing, + and that the amount of acceleration in the direction of \unitnormal relates to how the direction of travel of the object is changing. + (That is, if the object travels at constant speed, a_\text{T} =0; + if the object travels in a constant direction, a_\text{N} =0.) +

    + +

    + In Equation at the beginning of this section, we found + s\primeskip'(t) = \norm{\vvt}. + We can combine this fact with the above formula for a_\text{T} to write + + a_\text{T} = \frac{d}{dt}\Big(\norm{\vvt}\Big) = \frac{d}{dt}\big( s\primeskip'(t)\big) = s\primeskip''(t) + . +

    + +

    + Since s\primeskip'(t) is speed, + s\primeskip''(t) is the rate at which speed is changing with respect to time. + We see once more that the component of acceleration in the direction of travel relates only to speed, + not to a change in direction. +

    + +

    + Now compare the formula for + a_\text{N} above to the formula for curvature in : + + a_\text{N} = \frac{\norm{\vvt\times \vat}}{\norm{\vvt}} \text{ and } \kappa = \frac{\norm{\vrp(t)\times\vrpp(t)}}{\norm{\vrp(t)}^3}=\frac{\norm{\vvt\times \vat}}{\norm{\vvt}^3} + . +

    + +

    + Thus + + a_\text{N} \amp = \kappa \norm{\vvt}^2 + \amp = \kappa\Big(s\primeskip'(t)\Big)^2 + +

    + +

    + This last equation shows that the component of acceleration that changes the object's direction is dependent on two things: + the curvature of the path and the speed of the object. +

    + +

    + Imagine driving a car in a clockwise circle. + You will naturally feel a force pushing you towards the door + (more accurately, + the door is pushing you as the car is turning and you want to travel in a straight line). + If you keep the radius of the circle constant but speed up (, increasing s\primeskip'(t)), + the door pushes harder against you + (a_\text{N} has increased). + If you keep your speed constant but tighten the turn (, increase \kappa), + once again the door will push harder against you. +

    + +

    + Putting our new formulas for + a_\text{T} and a_\text{N} together, we have + + \vat = s\primeskip''(t)\unittangent(t) + \kappa\norm{\vvt}^2\unitnormal(t) + . +

    + +

    + This is not a particularly practical way of finding + a_\text{T} and a_\text{N}, + but it reveals some great concepts about how acceleration interacts with speed and the shape of a curve. +

    + + + Curvature and road design + +

    + The minimum radius of the curve in a highway cloverleaf is determined by the operating speed, + as given in the table in . + For each curve and speed, compute a_\text{N}. +

    +     + + + Operating speed and minimum radius in highway cloverleaf design + + + + OperatingSpeed (mph) + MinimumRadius (ft) + + + 35 + 310 + + + 40 + 430 + + + 45 + 540 + + + +
    + +

    + Using Equation, + we can compute the acceleration normal to the curve in each case. + We start by converting each speed from + miles per hour to feet per second + by multiplying by 5280/3600. + + \text{35 mph, 310 ft} \amp \Rightarrow 51.33\,\text{ft/s},\, \kappa = 1/310 + a_\text{N} \amp = \kappa\, \norm{\vvt}^2 + \amp = \frac1{310}\big(51.33\big)^2 + \amp = 8.50\,\text{ft/s}^2 + . + + \text{40 mph, 430 ft} \amp \Rightarrow 58.67\,\text{ft/s},\, \kappa = 1/430 + a_\text{N} \amp = \frac1{430}\big(58.67\big)^2 + \amp = 8.00\,\text{ft/s}^2 + . + + \text{45 mph, 540 ft} \amp \Rightarrow 66\,\text{ft/s},\, \kappa = 1/540 + a_\text{N} \amp = \frac1{540}\big(66\big)^2 + \amp = 8.07\,\text{ft/s}^2 + . +

    + +

    + Note that each acceleration is similar; this is by design. + Considering the classic Force = mass acceleration formula, + this acceleration must be kept small in order for the tires of a vehicle to keep a + grip on the road. + If one travels on a turn of radius 310 + at a rate of 50, + the acceleration is double, at 17.35. + If the acceleration is too high, + the frictional force created by the tires may not be enough to keep the car from sliding. + Civil engineers routinely compute a + safe design speed, + then subtract 5-10 mph to create the posted speed limit for additional safety. +

    +     +     + +

    + We end this chapter with a reflection on what we've covered. + We started with vector-valued functions, + which may have seemed at the time to be just another way of writing parametric equations. + However, we have seen that the vector perspective has given us great insight into the behavior of functions and the study of motion. + Vector-valued position functions convey displacement, + distance traveled, speed, velocity, + acceleration and curvature information, + each of which has great importance in science and engineering. +

    +     + + + + Terms and Concepts + + + +

    + It is common to describe position in terms of both + and/or . +

    +     + + + + + ["time","distance"].includes(ans) + + + + + + + ["time","distance"].includes(ans) && !ans_array.slice(0,1).includes(ans) + + + + + ans_array.slice(0,1).includes(ans) + + You already gave that answer. + + + + +     + + + + +

    + A measure of the curviness + of a curve is . +

    +     + + + + + + + + +     + + + + +

    + Give two shapes with constant curvature. +

    +     + + + +

    + Answers may include lines, circles, helixes +

    +     + +     + + + + +

    + Describe in your own words what an + osculating circle is. +

    +     + + + +

    + Answer should mention the circle is tangent to the curve and has the same curvature as the curve at that point. +

    +     + +     + + + +

    + Rearrange the blocks to form a valid identity. +

    +     + + \unittangentprime(s) + = + \kappa + \unitnormal(s) + \unittangentprime(t) + \gamma + \tau + \pi + \unitnormal(t) + +     + + + + + + +

    + Given a position function \vrt, + how are a_\text{T} and + a_\text{N} affected by the curvature? +

    +     + + + +

    + a_\text{T} is not affected by curvature; + the greater the curvature, the larger a_\text{N} becomes. +

    +     + +     +     + + Problems + + + +

    + A position function \vrt is given, + where t=0 corresponds to the initial position. + Find the arc length parameter s, + and rewrite \vrt in terms of s; + that is, find \vec r(s). +

    +     + + + + +

    + \vrt = \la 2t, t, -2t\ra +

    +     + +

    + s = 3t, so \vec r(s) = \la 2s/3, s/3, -2s/3\ra +

    +     + +     + + + + + Context("Vector2D"); + Context()->variables->are(s=>'Real',t=>'Real'); + $s=Formula("7t"); + $rs=Compute("<7cos(s/7),7sin(s/7)>"); + + + +

    + \vrt = \la 7\cos(t) ,7\sin(t) \ra. +

    + + Find the arc length paramter s. + +

    + +

    + + + Rewrite \vrt in terms of s. + +

    + +

    +     +     +     + + + + +

    + \vrt = \la 3\cos(t) ,3\sin(t) , 2t\ra +

    +     + +

    + s = \sqrt{13}t, + so \vec r(s) = \la 3\cos(s/\sqrt{13}), 3\sin(s/\sqrt{13}), 2s/\sqrt{13}\ra +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(s=>'Real',t=>'Real'); + $s=Formula("13t"); + $rs=Compute("<5cos(s/13),13sin(s/13),12cos(s/13)>"); + + + +

    + \vrt = \vrt = \la 5\cos(t) ,13\sin(t) , 12\cos(t) \ra. +

    + + Find the arc length paramter s. + +

    + +

    + + + Rewrite \vrt in terms of s. + + +

    + +

    +     +     +     + +     + + + +

    + A curve C is described along with 2 points on C. +

    + +

    +

      +
    1. +

      + Using a sketch, + determine at which of these points the curvature is greater. +

      +
      2. + +
    2. +

      + Find the curvature \kappa of C, + and evaluate \kappa at each of the 2 given points. +

      +
      3. +
    +

    +     + + + + +

    + C is defined by y = x^3-x; + points given at x=0 and x=1/2. +

    +     + +

    + \kappa = \frac{\abs{6x}}{\left(1+(3x^2-1)^2\right)^{3/2}}; +

    + +

    + \kappa(0) = 0, + \kappa(1/2) = \frac{192}{17\sqrt{17}} \approx 2.74. +

    +     + +     + + + + + parserPopUp.pl + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $comp=DropDown(['greater than', 'equal to', 'less than'],0,showInStatic=>0); + $kappa=Compute("|(6x^2-2)/(x^2+1)^3|/(1+4x^2/(x^2+1)^4)^(3/2)"); + $k[0]=Compute("2"); + $k[1]=Formula("2750/641^(3/2)"); + + + +

    + C is defined by y=\frac{1}{x^2+1}; points given at + x=0 and x=2. +

    + +

    + The curvature at x=0 is + the curvature at x=2. +

    + + + Give the curvature as a function of x. + + +

    + +

    + + + Give the value of \kappa(0). + +

    + \kappa(0)= +

    + + + Give the value of \kappa(2). + +

    + +

    +     +     +     + + + + +

    + C is defined by \ds y = \cos(x); + points given at x=0 and x=\pi/2. +

    +     + +

    + \kappa = \frac{\abs{\cos(x) }}{\left(1+\sin^2(x) \right)^{3/2}}; +

    + +

    + \kappa(0) = 1, \kappa(\pi/2) = 0 +

    +     + +     + + + + +

    + C is defined by \ds y = \sqrt{1-x^2} on (-1,1); + points given at x=0 and x=1/2. +

    +     + +

    + \kappa = 1; +

    + +

    + \kappa(0) = 1, \kappa(1/2) = 1 +

    +     + +     + + + + + parserPopUp.pl + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->are(t=>'Real'); + $comp=DropDown(['greater than', 'equal to', 'less than'],2,showInStatic=>0); + $kappa=Compute("|2cos(t)cos(2t)+4sin(t)sin(2t)|/(4cos^2(2t)+sin^2(t))^(3/2)"); + $k[0]=Formula("1/4"); + $k[1]=Formula("8"); + + + +

    + C is defined by \vrt = \la \cos(t) , \sin(2t)\ra; + points given at t=0 and t=\pi/4. +

    + +

    + The curvature at t=0 is + the curvature at t=\pi/4. +

    + + + Give the curvature as a function of t. + + +

    + +

    + + + Give the value of \kappa(0). + +

    + \kappa(0)= +

    + + + Give the value of \kappa(pi/4). + +

    + +

    +     +     +     + + + + +

    + C is defined by \ds \vrt = \la \cos^2(t) , \sin(t) \cos(t) \ra; + points given at t=0 and t=\pi/3. +

    +     + +

    + \kappa = 2; +

    + +

    + \kappa(0) = 2, \kappa(\pi/3) = 2 +

    +     + +     + + + + +

    + C is defined by \ds \vrt = \la t^2-1,t^3-t\ra; + points given at t=0 and t=5. +

    +     + +

    + \kappa = \frac{\abs{6t^2+2}}{\left(4t^2+(3t^2-1)^2\right)^{3/2}}; +

    + +

    + \kappa(0) = 2, + \kappa(5) = \frac{19}{1394\sqrt{1394}}\approx 0.0004 +

    +     + +     + + + + + parserPopUp.pl + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->are(t=>'Real'); + $comp=DropDown(['greater than', 'equal to', 'less than'],0,showInStatic=>0); + $kappa=Compute("|sec^3(t)|/(sec^4(t)+sec^2(t)tan^2(t))^(3/2)"); + $k[0]=Formula("1"); + $k[1]=Formula("3sqrt(3)/(5sqrt(5))"); + + + +

    + C is defined by \vrt = \la \tan(t) ,\sec(t) \ra; + points given at t=0 and t=\pi/6. +

    + +

    + The curvature at t=0 is + the curvature at t=\pi/6. +

    + + + Give the curvature as a function of t. + + +

    + +

    + + + Give the value of \kappa(0). + +

    + \kappa(0)= +

    + + + Give the value of \kappa(\pi/6). + +

    + +

    +     +     +     + + + + +

    + C is defined by \ds \vrt = \la 4t+2,3t-1,2t+5\ra; + points given at t=0 and t=1. +

    +     + +

    + \kappa = 0; +

    + +

    + \kappa(0) = 0, \kappa(1) = 0 +

    +     + +     + + + + + parserPopUp.pl + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->are(t=>'Real'); + $comp=DropDown(['greater than', 'equal to', 'less than'],0,showInStatic=>0); + $kappa=Compute("2sqrt(18t^4+15t^2+1)/(18t^4-2t^2+1)^(3/2)"); + $k[0]=Formula("2"); + $k[1]=Formula("2sqrt(2)/17"); + + + +

    + C is defined by \vrt = \la t^3-t,t^3-4,t^2-1\ra; + points given at t=0 and t=1. +

    + +

    + The curvature at t=0 is + the curvature at t=1. +

    + + + Give the curvature as a function of t. + + +

    + +

    + + + Give the value of \kappa(0). + +

    + \kappa(0)= +

    + + + Give the value of \kappa(1). + +

    + +

    +     +     +     + + + + +

    + C is defined by \ds \vrt = \la 3\cos(t) ,3\sin(t) , 2t\ra; + points given at t=0 and t=\pi/2. +

    +     + +

    + \kappa = \frac{3}{13}; +

    + +

    + \kappa(0) = 3/13, \kappa(\pi/2) = 3/13 +

    +     + +     + + + + + parserPopUp.pl + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + Context()->variables->are(t=>'Real'); + $comp=DropDown(['greater than', 'equal to', 'less than'],1,showInStatic=>0); + $kappa=Formula("1/13"); + $k[0]=Formula("1/13"); + $k[1]=Formula("1/13"); + + + +

    + C is defined by \vrt = \la 5\cos(t) ,13\sin(t) , 12\cos(t) \ra. + Points given at t=0 and t=\pi/2. +

    + +

    + The curvature at t=0 is + the curvature at t=\pi/2. +

    + + + Give the curvature as a function of t. + + +

    + +

    + + + Give the value of \kappa(0). + +

    + \kappa(0)= +

    + + + Give the value of \kappa(\pi/2). + +

    + +

    +     +     + +     + + + +

    + Find the value of x or t where curvature is maximized. +

    +     + + + + + + parser::Root->Enable; + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $max=List(Formula("sqrt(2)/root(4,5)"), Formula("-sqrt(2)/root(4,5)")); + + + +

    + y=\frac{1}{6}x^3 +

    + +

    + +

    +     +     +     + + + + +

    + \ds y=\sin(x) +

    +     + +

    + maximized at x=\ldots -3\pi/2, -\pi/2, \pi/2, \ldots +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $max=Formula("1/4"); + + + +

    + \vrt = \la t^2+2t,3t-t^2\ra +

    + +

    + +

    +     +     +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $max=List(Formula("sqrt(5)"),Formula("-sqrt(5)")); + + + +

    + \vrt = \la t, 4/t, 3/t\ra +

    + +

    + +

    +     +     +     + +     + + + +

    + Find the radius of curvature at the indicated value. +

    +     + + + + +

    + y=\tan(x), at x=\pi/4 +

    +     + +

    + radius of curvature is 5\sqrt{5}/4. +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $radius=Formula("5sqrt(10)"); + + + +

    + y=x^2+x-3 at x=1 +

    + +

    + +

    +     +     +     + + + + +

    + \vrt = \la \cos(t) , \sin(3t)\ra, at t=0 +

    +     + +

    + radius of curvature is 9. +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $radius=Formula("1/45"); + + + +

    + \vrt = \la 5\cos(3 t), t\ra at t=0 +

    + +

    + +

    +     +     +     + +     + + + +

    + Find the equation of the osculating circle to the curve at the indicated t-value. +

    +     + + + + +

    + \vrt = \la t,t^2\ra, at t=0 +

    +     + +

    + x^2+(y-1/2)^2 = 1/4, + or \vec c(t) = \la 1/2\cos(t) , 1/2\sin(t) + 1/2\ra +

    +     + +     + + + + + Context("ImplicitEquation"); + Context()->variables->set(x=>{limits=>[2,4]}); + Context()->variables->set(y=>{limits=>[-1,1]}); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $osc=ImplicitEquation("(x-8/3)^2+y^2 = 1/9"); + + + +

    + \vrt = \la 3\cos(t) , \sin(t) \ra at t=0 +

    + +

    + +

    +     +     +     + + + + +

    + \vrt = \la 3\cos(t) , \sin(t) \ra, at t=\pi/2 +

    +     + +

    + x^2+(y+8)^2 = 81, + or \vec c(t) = \la 9\cos(t) , 9\sin(t) -8\ra +

    +     + +     + + + + + Context("ImplicitEquation"); + Context()->variables->set(x=>{limits=>[-1,2]}); + Context()->variables->set(y=>{limits=>[-1,2]}); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $osc=ImplicitEquation("(x-1/2)^2+(y-1/2)^2 = 1/2"); + + + +

    + \vrt = \la t^2-t,t^2+t\ra at t=0 +

    + +

    + +

    +     +     +     + +     +     +     +
    +     + + + Functions of Several Variables + +

    + A function of the form y=f(x) is a function of a single variable; + given a value of x, we can find a value y. + Even the vector-valued functions of + are single-variable functions; + the input is a single variable though the output is a vector. +

    + +

    + There are many situations where a desired quantity is a function of two or more variables. + For instance, + wind chill is measured by knowing the temperature and wind speed; + the volume of a gas can be computed knowing the pressure and temperature of the gas; + to compute a baseball player's batting average, + one needs to know the number of hits and the number of at-bats. +

    + +

    + This chapter studies multivariable functions, that is, + functions with more than one input. +

    +     + +
    + Introduction to Multivariable Functions + + + Function of Two Variables + +

    + Let D be a subset of \mathbb{R}^2. + A function f of two variables + is a rule that assigns each pair (x,y) in D a value + z=f(x,y) in \mathbb{R}. + D is the domain of f; + the set of all outputs of f is the range. + multivariable function + multivariable functiondomain + multivariable functionrange + functionof two variables +

    +     +     + + + + + + + Understanding a function of two variables + +

    + Let z=f(x,y) = x^2-y. + Evaluate f(1,2), f(2,1), and f(-2,4); + find the domain and range of f. +

    +     + +

    + Using the definition f(x,y) = x^2-y, we have: + + f(1,2) \amp = 1^2-2 = -1 + f(2,1) \amp = 2^2-1 = 3 + f(-2,4) \amp = (-2)^2-4 = 0 + +

    + +

    + The domain is not specified, + so we take it to be all possible pairs in + \mathbb{R}^2 for which f is defined. + In this example, f is defined for + all pairs (x,y), + so the domain D of f is \mathbb{R}^2. +

    + +

    + The output of f can be made as large or small as possible; + any real number r can be the output. (In fact, + given any real number r, + f(0,-r)=r.) So the range R of f is \mathbb{R}. +

    +     +     + + + Understanding a function of two variables + +

    + Let f(x,y) = \sqrt{1-\frac{x^2}9-\frac{y^2}4}. + Find the domain and range of f. +

    +     + +

    + The domain is all pairs (x,y) allowable as input in f. + Because of the square root, + we need (x,y) such that 0\leq1-\frac{x^2}9-\frac{y^2}4: + + 0\amp \leq 1-\frac{x^2}9-\frac{y^2}4 + \frac{x^2}9+\frac{y^2}4 \amp \leq 1 + +

    + +

    + The above equation describes an ellipse and its interior as shown in . + We can represent the domain D graphically with the figure; + in set notation, + we can write D = \{(x,y)|\,\frac{x^2}9+\frac{y^2}4 \leq 1\}. +

    + +
    + Illustrating the domain of f(x,y) in + + + An ellipse, centered at the origin, with shaded interior, illustrating the domain of a function. + +

    + The ellipse \frac{x^2}{9}+\frac{y^2}{4}=1 is plotted in the xy plane. + It is centered at the origin, with intercepts at (\pm 3,0) and (0,\pm 2). + The interior of the ellipse is shaded, to illustrate the domain of + f(x,y) = \sqrt{1-\frac{x^2}{9}-\frac{y^2}{4}}. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-5.9,ymax=5.9, + xmin=-5.9,xmax=5.9 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:360] ({3*cos(x)},{2*sin(x)}); + \addplot [firstcurvestyle,domain=0:360,samples=60] ({3*cos(x)},{2*sin(x)}); + + \draw (axis cs: 3,3) node { $\displaystyle \frac{x^2}9+\frac{y^2}4=1$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + The range is the set of all possible output values. + The square root ensures that all output is \geq 0. + Since the x and y terms are squared, then subtracted, + inside the square root, + the largest output value comes at x=0, + y=0: f(0,0) = 1. + Thus the range R is the interval [0,1]. +

    +     +     +     + + + Graphing Functions of Two Variables +

    + The graph of a function f of two variables is the set of all points + \big(x,y,f(x,y)\big) where (x,y) is in the domain of f. + This creates a surface in space. +

    + +
    + Graphing a function of two variables + +
    + + + + + A collection of points plotted in space, against a set of three-dimensional coordinate axes. + +

    + About two dozen points are plotted in space, + along with a set of three-dimensional coordinate axes. + The location of the points can be better observed by rotating the figure, + but the precise arrangement of the points is unimportant. + The main observation is that these points all lie on the surface plotted in the next image. +

    +     + + + + + //ASY file for figmultigraph_intro3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8,8,1.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={0.5,1}; + defaultpen(0.5mm); + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.25,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw 25 points on the surface z=1/(1+x^2+y^2) + dotfactor=3; + real[] A={-2,-1,0,1,2}; + for (int i=0; i<5; ++i) + { + for (int j=0; j<5; ++j) + { + dot((A[i],A[j],1/(1+(A[i])^2+(A[j])^2)),dotblue); + } + } + + + + +
    + +
    + + + + + A bell-shaped surface in three dimensions. It is symmetric about the z axis and has a peak at the point (0,0,1) + +

    + A three-dimensional plot the surface given by a graph z=f(x,y). + The surface is a bell-shaped hill, with its peak at (0,0,1) on the z axis. + There is rotational symmetry about the z axis, + and as the suface descends, it flattens out and tapers off toward the xy plane. +

    +     + + + + + //ASY file for figmultigraph_introb3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8,8,1.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={0.5,1}; + defaultpen(0.5mm); + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.25,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z=1/(1+x^2+y^2) + triple f(pair t) { + return (t.x,t.y,1/(1+(t.x)^2+(t.y)^2)); + } + surface s=surface(f,(-2,-2),(2,2),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +
    +     + +
    + +

    + One can begin sketching a graph by plotting points, but this has limitations. + Consider where 25 points have been plotted of f(x,y) = \frac1{x^2+y^2+1}. + More points have been plotted than one would reasonably want to do by hand, + yet it is not clear at all what the graph of the function looks like. + Technology allows us to plot lots of points, + connect adjacent points with lines and add shading to create a graph like which does a far better job of illustrating the behavior of f. +

    + +

    + While technology is readily available to help us graph functions of two variables, + there is still a paper-and-pencil approach that is useful to understand and master as it, + combined with high-quality graphics, + gives one great insight into the behavior of a function. + This technique is known as sketching level curves. +

    +     + + + Level Curves +

    + It may be surprising to find that the problem of representing a three dimensional surface on paper is familiar to most people + (they just don't realize it). + multivariable functionlevel curves + level curves + contour lines + Topographical maps, + like the one shown in , + represent the surface of Earth by indicating points with the same elevation with contour lines. + The elevations marked are equally spaced; + in this example, + each thin line indicates an elevation change in 50ft increments and each thick line indicates a change of 200ft. + When lines are drawn close together, elevation changes rapidly + (as one does not have to travel far to rise 50ft). + When lines are far apart, + such as near Aspen Campground, + elevation changes more gradually as one has to walk farther to rise 50ft. +

    + + + +
    + A topographical map displays elevation by drawing contour lines, along with the elevation is constant. + USGS 1:24000-scale Quadrangle for Chrome Mountain, MT 1987. + + + A topographical map of Chrome Mountain in Montana. + +

    + An excerpt from a topographical map of Chrome Mountain in Montana. + The map illustrates the concept of contour lines through its use of lines of elevation. + There is a river that runs through the middle of the map, from top to bottom. + To the right of the river, the lines of elevation are more spread out, + indicating that the land is more gently-sloped in this region, + which includes an area marked as Aspen Campground. +

    + +

    + Near the left and right edges of the map, the lines of elevation are much closer together, + which indicates that these regions consist of steeply-sloped mountainsides. +

    +     + + +
    + +

    + Given a function f(x,y), + we can draw a topographical map + of the graph z=f(x,y) by drawing level curves + (or, contour lines). + A level curve at z=c is a curve in the xy-plane such that for all points (x,y) on the curve, + f(x,y) = c. +

    + +

    + When drawing level curves, + it is important that the c values are spaced equally apart as that gives the best insight to how quickly the + elevation is changing. + Examples will help one understand this concept. +

    + + + Drawing Level Curves + +

    + Let f(x,y) = \sqrt{1-\frac{x^2}9-\frac{y^2}4}. + Find the level curves of f for c=0, + 0.2, 0.4, 0.6, 0.8 and 1. +

    +     + +

    + Consider first c=0. + The level curve for c=0 is the set of all points (x,y) such that 0=\sqrt{1-\frac{x^2}9-\frac{y^2}4}. + Squaring both sides gives us + + \frac{x^2}9+\frac{y^2}4=1 + , + an ellipse centered at (0,0) with horizontal major axis of length 6 and minor axis of length 4. + Thus for any point (x,y) on this curve, f(x,y) = 0. +

    + +

    + Now consider the level curve for c=0.2 + + 0.2 \amp = \sqrt{1-\frac{x^2}9-\frac{y^2}4} + 0.04 \amp = 1-\frac{x^2}9-\frac{y^2}4 + \frac{x^2}9+\frac{y^2}4 \amp =0.96 + \frac{x^2}{8.64}+\frac{y^2}{3.84} \amp =1 + . +

    + +

    + This is also an ellipse, + where a = \sqrt{8.64}\approx 2.94 and b=\sqrt{3.84}\approx 1.96. +

    + +

    + In general, for z=c, the level curve is: + + c \amp = \sqrt{1-\frac{x^2}9-\frac{y^2}4} + c^2 \amp = 1-\frac{x^2}9-\frac{y^2}4 + \frac{x^2}9+\frac{y^2}4 \amp =1-c^2 + \frac{x^2}{9(1-c^2)}+\frac{y^2}{4(1-c^2)} \amp =1 + , + ellipses that are decreasing in size as c increases. + A special case is when c=1; + there the ellipse is just the point (0,0). +

    + +

    + The level curves are shown in . + Note how the level curves for c=0 and c=0.2 are very, + very close together: + this indicates that f is growing rapidly along those curves. +

    + +
    + Graphing the level curves in + +
    + + + + A family of concentric ellipses, centered at the origin. + +

    + This plot, in the xy plane, illustrates several level curves for the function f(x,y) = \sqrt{1-\frac{x^2}{9}-\frac{y^2}{4}}. + The level curve c=1 is a single point: the origin (0,0). + The other level curves form a family of concentric ellipses centered about the origin. + The largest ellipse is the boundary of the domain of f, given by \frac{x^2}{9}+\frac{y^2}{4}=1. + Level curves near the boundary are close together, while those nearer to the origin are further apart. + This illustrates the fact that the surface is relative flat near the top, + while it becomes steeply-sloped along its sides. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-1,1,2,-2,3,-3}, + ytick={-1,1,-2,2,3,-3}, + ymin=-2.5,ymax=2.5, + xmin=-3.2,xmax=3.2 + ] + + \addplot [firstcurvestyle,domain=0:360,samples=80] ({3*(cos(x))},{2*(sin(x))}); + \addplot [firstcurvestyle,domain=0:360,samples=80] ({2.93*(cos(x))},{1.96*(sin(x))}); + \addplot [firstcurvestyle,domain=0:360,samples=80] ({2.75*(cos(x))},{1.83*(sin(x))}); + \addplot [firstcurvestyle,domain=0:360,samples=70] ({2.4*(cos(x))},{1.6*(sin(x))}); + \addplot [firstcurvestyle,domain=0:360,samples=60] ({1.8*(cos(x))},{1.2*(sin(x))}); + + \filldraw [firstcolor] (axis cs:0,0) circle (1pt) node [black, above right] { $c=1$}; + + \draw [->,>=stealth] (axis cs:1.7,1.13) -- (axis cs:2.6,2) node [above] { $c=0.6$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + A surface in the shape of an elliptical dome. + +

    + The surface given by the graph z=\sqrt{1--\frac{x^2}{9}-\frac{y^2}{4}} is plotted. + Its shape is that of an elliptical dome, with its peak on the z axis at the point (0,0,1). + The bottom of the dome lies in the xy plane, and its intersection with this plane is an ellipse. + Near the top, the surface is relatively flat, + but the sides become close to vertical near the xy plane. +

    +     + + + + + //ASY file for figlevelcurve13D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(9,9,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={1,2}; + defaultpen(0.5mm); + pair xbounds=(-3.5,3.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.25,2.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw top half of surface x^2/9+y^2/4+z^2=1 + triple f(pair t) { + //return (cos(t.x)*3*cos(t.y),sin(t.x)*2*cos(t.y),sin(t.y)); + return (3*sin(t.y),sin(t.x)*2*cos(t.y),cos(t.x)*cos(t.y)); + } + surface s=surface(f,(-pi/2,-pi/2),(pi/2,pi/2),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw level curves for surface x^2/9+y^2/4+z^2=1 at c=0,0.2,0.4,0.6,0.8,1 + triple g(real t) {return (3*cos(t)*sqrt(1-(0.8)^2),2*sin(t)*sqrt(1-(0.8)^2),0.8);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + triple g(real t) {return (3*cos(t)*sqrt(1-(0.6)^2),2*sin(t)*sqrt(1-(0.6)^2),0.6);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + triple g(real t) {return (3*cos(t)*sqrt(1-(0.4)^2),2*sin(t)*sqrt(1-(0.4)^2),0.4);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + triple g(real t) {return (3*cos(t)*sqrt(1-(0.2)^2),2*sin(t)*sqrt(1-(0.2)^2),0.2);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + triple g(real t) {return (3*cos(t)*sqrt(1-(0)^2),2*sin(t)*sqrt(1-(0)^2),0);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + dotfactor=3; + dot((0,0,1),bluepen); + + + + +
    +     + +
    + +

    + In , + the curves are drawn on a graph of f in space. + Note how the elevations are evenly spaced. + Near the level curves of c=0 and c=0.2 we can see that f indeed is growing quickly. +

    +     +     + + + Analyzing Level Curves + +

    + Let f(x,y) = \frac{x+y}{x^2+y^2+1}. + Find the level curves for z=c. +

    +     + +

    + We begin by setting f(x,y)=c for an arbitrary c and seeing if algebraic manipulation of the equation reveals anything significant. + + \frac{x+y}{x^2+y^2+1} \amp = c + x+y \amp = c(x^2+y^2+1). + We recognize this as a circle, though the center and radius are not yet clear. By completing the square, we can obtain: + \left(x-\frac{1}{2c}\right)^2+\left(y-\frac1{2c}\right)^2\amp =\frac{1}{2c^2}-1 + , + a circle centered at \big(1/(2c),1/(2c)\big) with radius \sqrt{1/(2c^2)-1}, + where \abs{c}\lt 1/\sqrt{2}. + The level curves for c=\pm 0.2,\,\pm 0.4 and \pm0.6 are sketched in . + To help illustrate elevation, + we use thicker lines for c values near 0, and dashed lines indicate where c\lt 0. +

    + +

    + There is one special level curve, when c=0. + The level curve in this situation is x+y=0, the line y=-x. +

    + +

    + In we see a graph of the surface. + Note how the y-axis is pointing away from the viewer to more closely resemble the orientation of the level curves in . +

    + +
    + Graphing the level curves in + +
    + + + + A line through the origin in the plane, and two sets of nested circles, one on either side of the line. + +

    + This plot in the xy plane shows several level curves for the function f(x,y) = \frac{x+y}{x^2+y^2+1}. + The level curve c=0 is the line y=-x passing through the origin. + Above this line there is a family of nested, but not concentric, circles. + The smallest circles are closest to the origin. These represent level curves for positive values of c. + Across the line y=-x is another set of circles that is the mirror image of the first. + These are the level curves for negative values of c. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-5.1,ymax=5.1, + xmin=-6.1,xmax=6.1 + ] + + \addplot [thin,firstcolor,smooth] coordinates {(1.457,0.8333)(1.443,0.963)(1.403,1.087)(1.338,1.2)(1.251,1.297)(1.145,1.373)(1.026,1.426)(0.8985,1.454)(0.7681,1.454)(0.6406,1.426)(0.5215,1.373)(0.4161,1.297)(0.3288,1.2)(0.2636,1.087)(0.2234,0.963)(0.2097,0.8333)(0.2234,0.7037)(0.2636,0.5797)(0.3288,0.4668)(0.4161,0.3699)(0.5215,0.2933)(0.6406,0.2402)(0.7681,0.2131)(0.8985,0.2131)(1.026,0.2402)(1.145,0.2933)(1.251,0.3699)(1.338,0.4668)(1.403,0.5797)(1.443,0.7037)(1.457,0.8333)}; + \addplot [thick,firstcolor,smooth] coordinates {(2.708,1.25)(2.676,1.553)(2.582,1.843)(2.429,2.107)(2.225,2.333)(1.979,2.512)(1.7,2.636)(1.402,2.7)(1.098,2.7)(0.7995,2.636)(0.5211,2.512)(0.2746,2.333)(0.07067,2.107)(-0.08171,1.843)(-0.1759,1.553)(-0.2077,1.25)(-0.1759,0.9469)(-0.08171,0.6571)(0.07067,0.3932)(0.2746,0.1667)(0.5211,-0.01244)(0.7995,-0.1364)(1.098,-0.1998)(1.402,-0.1998)(1.7,-0.1364)(1.979,-0.01244)(2.225,0.1667)(2.429,0.3932)(2.582,0.6571)(2.676,0.9469)(2.708,1.25)}; + \addplot [very thick,firstcolor,smooth] coordinates {(5.891,2.5)(5.817,3.205)(5.598,3.879)(5.244,4.493)(4.769,5.02)(4.196,5.437)(3.548,5.725)(2.854,5.873)(2.146,5.873)(1.452,5.725)(0.8044,5.437)(0.2309,5.02)(-0.2435,4.493)(-0.598,3.879)(-0.8171,3.205)(-0.8912,2.5)(-0.8171,1.795)(-0.598,1.121)(-0.2435,0.5067)(0.2309,-0.02013)(0.8044,-0.4368)(1.452,-0.7252)(2.146,-0.8726)(2.854,-0.8726)(3.548,-0.7252)(4.196,-0.4368)(4.769,-0.02013)(5.244,0.5067)(5.598,1.121)(5.817,1.795)(5.891,2.5)}; + \addplot [thin,firstcolor,dashed,smooth] coordinates {(-0.2097,-0.8333)(-0.2234,-0.7037)(-0.2636,-0.5797)(-0.3288,-0.4668)(-0.4161,-0.3699)(-0.5215,-0.2933)(-0.6406,-0.2402)(-0.7681,-0.2131)(-0.8985,-0.2131)(-1.026,-0.2402)(-1.145,-0.2933)(-1.251,-0.3699)(-1.338,-0.4668)(-1.403,-0.5797)(-1.443,-0.7037)(-1.457,-0.8333)(-1.443,-0.963)(-1.403,-1.087)(-1.338,-1.2)(-1.251,-1.297)(-1.145,-1.373)(-1.026,-1.426)(-0.8985,-1.454)(-0.7681,-1.454)(-0.6406,-1.426)(-0.5215,-1.373)(-0.4161,-1.297)(-0.3288,-1.2)(-0.2636,-1.087)(-0.2234,-0.963)(-0.2097,-0.8333)}; + \addplot [thick,firstcolor,dashed,smooth] coordinates {(0.2077,-1.25)(0.1759,-0.9469)(0.08171,-0.6571)(-0.07067,-0.3932)(-0.2746,-0.1667)(-0.5211,0.01244)(-0.7995,0.1364)(-1.098,0.1998)(-1.402,0.1998)(-1.7,0.1364)(-1.979,0.01244)(-2.225,-0.1667)(-2.429,-0.3932)(-2.582,-0.6571)(-2.676,-0.9469)(-2.708,-1.25)(-2.676,-1.553)(-2.582,-1.843)(-2.429,-2.107)(-2.225,-2.333)(-1.979,-2.512)(-1.7,-2.636)(-1.402,-2.7)(-1.098,-2.7)(-0.7995,-2.636)(-0.5211,-2.512)(-0.2746,-2.333)(-0.07067,-2.107)(0.08171,-1.843)(0.1759,-1.553)(0.2077,-1.25)}; + \addplot [very thick,firstcolor,dashed,smooth] coordinates {(0.8912,-2.5)(0.8171,-1.795)(0.598,-1.121)(0.2435,-0.5067)(-0.2309,0.02013)(-0.8044,0.4368)(-1.452,0.7252)(-2.146,0.8726)(-2.854,0.8726)(-3.548,0.7252)(-4.196,0.4368)(-4.769,0.02013)(-5.244,-0.5067)(-5.598,-1.121)(-5.817,-1.795)(-5.891,-2.5)(-5.817,-3.205)(-5.598,-3.879)(-5.244,-4.493)(-4.769,-5.02)(-4.196,-5.437)(-3.548,-5.725)(-2.854,-5.873)(-2.146,-5.873)(-1.452,-5.725)(-0.8044,-5.437)(-0.2309,-5.02)(0.2435,-4.493)(0.598,-3.879)(0.8171,-3.205)(0.8912,-2.5)}; + \addplot [very thick,firstcolor] {-x}; + + \draw (axis cs:5,-3.5) node { $c=0$}; + \draw (axis cs:3.8,4.6) node { $c=0.2$}; + \draw (axis cs:1.5,3) node { $c=0.4$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + The surface given by the graph whose level curves were plotted in the previous image. + +

    + The surface given by the graph z=\frac{x+y}{x^2+y^2+1} is plotted in three dimensions. + For points that are far from the z axis, + the surface lies close to the xy plane and is relatively flat. + Near the origin, the surface rises sharply to a peak that lies above the line y=x in the xy plane. + On the opposite side of the origin, the surface drops into a well that is the mirror image of the peak. +

    +     + + + + + //ASY file for figlevelcurve23D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(16,-12,1.4); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-5,5}; + real[] myychoice={-5,5}; + real[] myzchoice={-0.7,0.7}; + defaultpen(0.5mm); + pair xbounds=(-6,6); + pair ybounds=(-6,6); + pair zbounds=(-0.8,.8); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=(x+y)/(1+x^2+y^2) + triple f(pair t) { + return (t.x,t.y,(t.x+t.y)/(1+(t.x)^2+(t.y)^2)); + } + surface s=surface(f,(-6,-6),(6,6),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw level curves for surface x^2/9+y^2/4+z^2=1 at c=0.2,0.4,0.6 + real[] A={0.2,0.4,0.6}; + for (int i=0; i<3; ++i) + { + triple g(real t) { + return (1/(2*A[i])+(cos(t))*sqrt(-1+(1/(2*A[i]^2))),1/(2*A[i])+(sin(t))*sqrt(-1+(1/(2*A[i]^2))),A[i]);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + //Negative values too + triple g(real t) { + return (1/(-2*A[i])+(cos(t))*sqrt(-1+(1/(2*A[i]^2))),1/(-2*A[i])+(sin(t))*sqrt(-1+(1/(2*A[i]^2))),-A[i]);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + } + + draw((-6,6,0)--(6,-6,0),bluepen); + + //triple g(real t) { + //return (1/(2*0.6)+(cos(t))*sqrt(-1+(1/(2*0.6^2))),1/(2*0.6)+(sin(t))*sqrt(-1+(1/(2*0.6^2))),0.6);} + //path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + + + +
    +     + +
    + +

    + Seeing the level curves helps us understand the graph. + For instance, + the graph does not make it clear that one can walk + along the line y=-x without elevation change, + though the level curve does. +

    +     +     +     + + + Functions of Three Variables +

    + We extend our study of multivariable functions to functions of three variables. + (One can make a function of as many variables as one likes; + we limit our study to three variables.) +

    + + + Function of Three Variables + +

    + Let D be a subset of \mathbb{R}^3. + A function f of three variables + is a rule that assigns each triple (x,y,z) in D a value + w=f(x,y,z) in \mathbb{R}. + D is the domain of f; + the set of all outputs of f is the range. + multivariable function + functionof three variables + multivariable functiondomain + multivariable functionrange +

    +     +     + +

    + Note how this definition closely resembles that of . +

    + + + Understanding a function of three variables + +

    + Let f(x,y,z) = \frac{x^2+z+3\sin(y) }{x+2y-z}. + Evaluate f at the point (3,0,2) and find the domain and range of f. +

    +     + +

    + To evaluate the function we simply set x=3, y=0, and z=3 in the definition of f: + + f(3,0,2) = \frac{3^2+2+3\sin(0) }{3+2(0)-2} = 11 + . +

    + +

    + As the domain of f is not specified, + we take it to be the set of all triples (x,y,z) for which f(x,y,z) is defined. + As we cannot divide by 0, + we find the domain D is + + D = \{(x,y,z)\,|\,x+2y-z\neq 0\} + . +

    + +

    + We recognize that the set of all points in + \mathbb{R}^3 that are not + in D form a plane in space that passes through the origin + (with normal vector \la 1,2,-1\ra). +

    + +

    + We determine the range R is \mathbb{R}; + that is, all real numbers are possible outputs of f. + There is no set way of establishing this. + Rather, to get numbers near 0 we can let y=0 and choose z \approx -x^2. + To get numbers of arbitrarily large magnitude, + we can let z\approx x+2y. +

    +     +     +     + + + Level Surfaces +

    + It is very difficult to produce a meaningful graph of a function of three variables. + A function of one variable is a curve + drawn in 2 dimensions; + a function of two variables is a surface + drawn in 3 dimensions; + a function of three variables is a hypersurface + drawn in 4 dimensions. + multivariable functionlevel surface + level surface +

    + +

    + There are a few techniques one can employ to try to picture + a graph of three variables. + One is an analogue of level curves: + level surfaces. + Given w=f(x,y,z), + the level surface at w=c is the surface in space formed by all points (x,y,z) where f(x,y,z)=c. +

    + + + Finding level surfaces + +

    + If a point source S is radiating energy, + the intensity I at a given point P in space is inversely proportional to the square of the distance between S and P. + That is, when S=(0,0,0), + I(x,y,z) = \frac{k}{x^2+y^2+z^2} for some constant k. +

    + +

    + Let k=1; find the level surfaces of I. +

    +     + +

    + We can (mostly) answer this question using + common sense. If energy (say, + in the form of light) is emanating from the origin, + its intensity will be the same at all points equidistant from the origin. + That is, at any point on the surface of a sphere centered at the origin, + the intensity should be the same. + Therefore, the level surfaces are spheres. +

    + +

    + We now find this mathematically. + The level surface at I=c is defined by + + c \amp = \frac{1}{x^2+y^2+z^2}. + A small amount of algebra reveals + x^2+y^2+z^2 \amp = \frac1c + . +

    + +

    + Given an intensity c, + the level surface I=c is a sphere of radius + 1/\sqrt{c}, centered at the origin. +

    + +
    + A table of c values and the corresponding radius r of the spheres of constant value in + + + c + r + + + + + + + 16 + 0.25 + + + 8 + 0.35 + + + 4 + 0.5 + + + 2 + 0.71 + + + 1 + 1 + + + 0.5 + 1.41 + + + 0.25 + 2 + + + 0.125 + 2.83 + + + 0.0625 + 4 + + +
    + +

    + + gives a table of the radii of the spheres for given c values. + Normally one would use equally spaced c values, + but these values have been chosen purposefully. + At a distance of 0.25 from the point source, the intensity is 16; + to move to a point of half that intensity, + one just moves out 0.1 to 0.35 not much at all. + To again halve the intensity, + one moves 0.15, a little more than before. +

    + +

    + Note how each time the intensity if halved, + the distance required to move away grows. + We conclude that the closer one is to the source, + the more rapidly the intensity changes. +

    + +     +     + +

    + In the next section we apply the concepts of limits to functions of two or more variables. +

    +     + + + + Terms and Concepts + + + + +

    + Give two examples + (other than those given in the text) + of real world functions that require more than one input. +

    +     + + +     + + + + +

    + The graph of a function of two variables is a . +

    +     + + + + + + + + +     + + + + +

    + Most people are familiar with the concept of level curves in the context of maps. +

    +     + + + + + + + +

    + Double-check your spelling before you go hunting for the answer. +

    +     +     +     +     + +     + + + + +

    + Along a level curve, the output of a function does not change. +

    +     + +     + + + + +

    + The analogue of a level curve for functions of three variables is a level . +

    +     + + + + + + + + +     + + + + +

    + What does it mean when level curves are close together? + Far apart? +

    +     + + + +

    + When level curves are close together, + it means the function is changing z-values rapidly. + When far apart, it changes z-values slowly. +

    +     + +     +     + + + Problems + + + +

    + Give the domain and range of the multivariable function. +

    +     + + + + +

    + f(x,y) = x^2+y^2+2 +

    +     + +

    + domain: \mathbb{R}^2 +

    + +

    + range: z\geq 2 +

    +     + +     + + + + +

    + f(x,y) = x+2y +

    +     + +

    + domain: \mathbb{R}^2 +

    + +

    + range: \mathbb{R} +

    +     + +     + + + + +

    + f(x,y) = x-2y +

    +     + +

    + domain: \mathbb{R}^2 +

    + +

    + range: \mathbb{R} +

    +     + +     + + + + +

    + \ds f(x,y) = \frac{1}{x+2y} +

    +     + +

    + domain: x\neq 2y; in set notation, + \{(x,y)\,|\, x\neq 2y\} +

    + +

    + range: z\neq 0 +

    +     + +     + + + + +

    + \ds f(x,y) = \frac{1}{x^2+y^2+1} +

    +     + +

    + domain: \mathbb{R}^2 +

    + +

    + range: 0\lt z\leq 1 +

    +     + +     + + + + +

    + \ds f(x,y) = \sin(x) \cos(y) +

    +     + +

    + domain: \mathbb{R}^2 +

    + +

    + range: -1\leq z\leq 1 +

    +     + +     + + + + +

    + \ds f(x,y) = \sqrt{9-x^2-y^2} +

    +     + +

    + domain: \{(x,y)\,|\, x^2+y^2\leq 9\}, + , the domain is the circle and interior of a circle centered at the origin with radius 3. +

    + +

    + range: 0\leq z\leq 3 +

    +     + +     + + + + +

    + \ds f(x,y) = \frac1{\sqrt{x^2+y^2-9}} +

    +     + +

    + domain: \{(x,y)\,|\, x^2+y^2\geq 9\}, + , the domain is the exterior of the circle + (not including the circle itself) + centered at the origin with radius 3. +

    + +

    + range: 0\lt z\lt \infty, or (0,\infty) +

    +     + +     + +     + + + +

    + Describe in words and sketch the level curves for the function and given c values. +

    +     + + + + +

    + \ds f(x,y) = 3x-2y; c = -2,0,2 +

    +     + +

    + Level curves are lines y = (3/2)x-c/2. +

    + +     + +     + + + + +

    + \ds f(x,y) = x^2-y^2; c = -1,0,1 +

    +     + +

    + Level curves are hyperbolas + \frac{x^2}{c}-\frac{y^2}{c}=1, except for c=0, + where the level curve is the pair of lines y=x, + y=-x. +

    + +     + +     + + + + +

    + \ds f(x,y) = x-y^2; c = -2,0,2 +

    +     + +

    + Level curves are parabolas x=y^2+c. +

    + +     + +     + + + + +

    + \ds f(x,y) = \frac{1-x^2-y^2}{2y-2x}; c = -2,0,2 +

    +     + +

    + Level curves are hyperbolas (x-c)^2-(y-c)^2=1, + drawn in graph in different styles to differentiate the curves. +

    + +     + +     + + + + +

    + \ds f(x,y) = \frac{2x-2y}{x^2+y^2+1}; c = -1,0,1 +

    +     + +

    + When c\neq 0, the level curves are circles, + centered at (1/c,-1/c) with radius \sqrt{2/c^2-1}. + When c=0, the level curve is the line y=x. +

    + +     + +     + + + + +

    + \ds f(x,y) = \frac{y-x^3-1}{x}; + c = -3,-1,0,1,3 +

    +     + +

    + Level curves are cubics of the form y=x^3+cx+1. + Note how each curve passes through (0,1) and that the function is not defined at x=0. +

    + +     + +     + + + + +

    + \ds f(x,y) = \sqrt{x^2+4y^2}; c = 1,2,3,4 +

    +     + +

    + Level curves are ellipses of the form \frac{x^2}{c^2}+\frac{y^2}{c^2/4}=1, + , a=c and b=c/2. +

    + +     + +     + + + + +

    + \ds f(x,y) = x^2+4y^2; c = 1,2,3,4 +

    +     + +

    + Level curves are ellipses of the form \frac{x^2}{c}+\frac{y^2}{c/4}=1, + , a=\sqrt{c} and b=\sqrt{c}/2. +

    + +     + +     + +     + + + +

    + Give the domain and range of the functions of three variables. +

    +     + + + + +

    + \ds f(x,y,z) = \frac{x}{x+2y-4z} +

    +     + +

    + domain: x+2y-4z\neq 0; + the set of points in \mathbb{R}^3 NOT in the domain form a plane through the origin. +

    + +

    + range: \mathbb{R} +

    +     + +     + + + + +

    + \ds f(x,y,z) = \frac{1}{1-x^2-y^2-z^2} +

    +     + +

    + domain: x^2+y^2+z^2\neq 1; + the set of points in \mathbb{R}^3 NOT in the domain form a sphere of radius 1. +

    + +

    + range: (-\infty,0)\cup[1,\infty) +

    +     + +     + + + + +

    + \ds f(x,y,z) = \sqrt{z-x^2+y^2} +

    +     + +

    + domain: z\geq x^2-y^2; + the set of points in \mathbb{R}^3 above + (and including) + the hyperbolic paraboloid z=x^2-y^2. +

    + +

    + range: [0,\infty) +

    +     + +     + + + + +

    + \ds f(x,y,z) = z^2\sin(x) \cos(y) +

    +     + +

    + domain: \mathbb{R}^3 +

    + +

    + range: \mathbb{R} +

    +     + +     + +     + + + +

    + Describe the level surfaces of the given functions of three variables. +

    +     + + + + +

    + \ds f(x,y,z) = x^2+y^2+z^2 +

    +     + +

    + The level surfaces are spheres, + centered at the origin, with radius \sqrt{c}. +

    +     + +     + + + + +

    + \ds f(x,y,z) = z-x^2+y^2 +

    +     + +

    + The level surfaces are hyperbolic paraboloids of the form z=x^2-y^2+c; + each is shifted up/down by c. +

    +     + +     + + + + +

    + \ds f(x,y,z) = \frac{x^2+y^2}{z} +

    +     + +

    + The level surfaces are paraboloids of the form z=\frac{x^2}{c}+\frac{y^2}{c}; + the larger c, the wider the paraboloid. +

    +     + +     + + + + +

    + \ds f(x,y,z) = \frac{z}{x-y} +

    +     + +

    + The level surfaces are planes through the origin of the form cx-cy-z=0, that is, + planes through the origin with normal vector \la c,-c,-1\ra. +

    +     + +     + + + + +

    + Compare the level curves of Exercises + and . + How are they similar, and how are they different? + Each surface is a quadric surface; + describe how the level curves are consistent with what we know about each surface. +

    +     + +

    + The level curves for each surface are similar; + for z=\sqrt{x^2+4y^2} the level curves are ellipses of the form \frac{x^2}{c^2}+\frac{y^2}{c^2/4}=1, + , a=c and b=c/2; + whereas for z=x^2+4y^2 the level curves are ellipses of the form \frac{x^2}{c}+\frac{y^2}{c/4}=1, + , a=\sqrt{c} and b=\sqrt{c}/2. + The first set of ellipses are spaced evenly apart, + meaning the function grows at a constant rate; + the second set of ellipses are more closely spaced together as c grows, + meaning the function grows faster and faster as c increases. +

    + +

    + The graph z=\sqrt{x^2+4y^2} can be rewritten as z^2=x^2+4y^2, + an elliptic cone; + the graph z=x^2+4y^2 is a paraboloid, + each matching the description above. +

    +     + +     + +     +     +     +
    +
    + Limits and Continuity of Multivariable Functions + +

    + We continue with the pattern we have established in this text: + after defining a new kind of function, + we apply calculus ideas to it. + The previous section defined functions of two and three variables; + this section investigates what it means for these functions to be continuous. +

    + +

    + We begin with a series of definitions. + We are used to open intervals such as (1,3), + which represents the set of all x such that 1\lt x\lt 3, + and closed intervals such as [1,3], + which represents the set of all x such that 1\leq x\leq 3. + We need analogous definitions for open and closed sets in the xy-plane. +

    + + +     + + + Open and Closed Subsets in Higher Dimensions + + + + Open Disk, Boundary and Interior Points, Open and Closed Sets, Bounded Sets + +

    + An open disk B in \mathbb{R}^2 centered at + (x_0,y_0) with radius r is the set of all points (x,y) such that \ds\sqrt{(x-x_0)^2+(y-y_0)^2} \lt r. +

    + +

    + Let S be a set of points in \mathbb{R}^2. + A point P in \mathbb{R}^2 is a + boundary point + of S if all open disks centered at P contain both points in S and points not in S. +

    + +

    + A point P in S is an + interior point + of S if there is an open disk centered at P that contains only points in S. +

    + +

    + A set S is open + if every point in S is an interior point. +

    + +

    + A set S is closed + if it contains all of its boundary points. +

    + +

    + A set S is bounded + if there is an M \gt 0 such that the open disk, + centered at the origin with radius M, contains S. + A set that is not bounded is unbounded. + open + closed + open disk + closed disk + boundary point + interior point + bounded set + unbounded set +

    +     +     + +

    + + shows several sets in the xy-plane. + In each set, + point P_1 lies on the boundary of the set as all open disks centered there contain both points in, + and not in, the set. + In contrast, + point P_2 is an interior point for there is an open disk centered there that lies entirely within the set. +

    + +
    + Illustrating open and closed sets in the xy-plane + +
    + + + + A region in the first quadrant is shaded, and its boundary is shown as a solid curve. + +

    + A region in the first quadrant of the xy plane is shown. + It is somewhat in the shape of a pond, although the shape is not important. + The interior of the region is shaded, indicating that all the interior points are part of the set. + The boundary of the region is drawn as a solid curve, indicating that all points on the boundary are also part of the set. + Two points are plotted: a point P_1 on the boundary, and a point P_2 in the interior. + Both points are part of the set. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + ytick=\empty, + ymin=-1,ymax=5, + xmin=-1,xmax=7.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; + \addplot [firstcurvestyle,smooth] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; + + \filldraw (axis cs: 3.7,4.062) circle (2.4pt) node [shift={(0,9pt)}] { $P_1$}; + \draw [dashed] (axis cs: 3.7,4.062) circle (5pt); + + \filldraw (axis cs: 2,2) circle (2.4pt) node [shift={(0,9pt)}] { $P_2$}; + \draw [dashed] (axis cs: 2,2) circle (5pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + A region in the first quadrant is shaded, and its boundary is shown as a dashed curve. + +

    + A region in the first quadrant of the xy plane is shown. + It is the same region as the one depicted in . + The difference is that the boundary of the region is drawn as a dashed curve, + indicating that none of the points on the boundary are part of the set. + The same two points are plotted: a point P_1 on the boundary, and a point P_2 in the interior. + This time, P_1 is not part of the set, but P_2 is. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + ytick=\empty, + ymin=-1,ymax=5, + xmin=-1,xmax=7.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; + \addplot [firstcurvestyle,dashed,smooth] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; + + \filldraw (axis cs: 3.7,4.062) circle (2.4pt) node [shift={(0,9pt)}] { $P_1$}; + \draw [dashed] (axis cs: 3.7,4.062) circle (5pt); + + \filldraw (axis cs: 2,2) circle (2.4pt) node [shift={(0,9pt)}] { $P_2$}; + \draw [dashed] (axis cs: 2,2) circle (5pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + A region in the first quadrant is shaded, and its boundary is shown as a partially solid, partially dashed curve. + +

    + A region in the first quadrant of the xy plane is shown. + It is the same region as the one depicted in . + This time, part of the boundary is drawn as a solid curve, and part of the boundary is drawn as a dashed curve. + This indicates that some of the points on the boundary are part of the set, while other points are not. + The same two points are plotted: a point P_1 on the boundary, and a point P_2 in the interior. + This time, P_1 is not part of the set, but P_2 is. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + ytick=\empty, + ymin=-1,ymax=5, + xmin=-1,xmax=7.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; + \addplot [firstcurvestyle,dashed,smooth] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)}; + + \addplot [firstcurvestyle,-,smooth] coordinates {(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; + + \filldraw (axis cs: 3.7,4.062) circle (2.4pt) node [shift={(0,9pt)}] { $P_1$}; + \draw [dashed] (axis cs: 3.7,4.062) circle (5pt); + + \filldraw (axis cs: 2,2) circle (2.4pt) node [shift={(0,9pt)}] { $P_2$}; + \draw [dashed] (axis cs: 2,2) circle (5pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     +
    + +

    + The set depicted in is a closed set as it contains all of its boundary points. + The set in is open, for all of its points are interior points + (or, equivalently, it does not contain any of its boundary points). + The set in is neither open nor closed as it contains some of its boundary points. +

    + + + Determining open/closed, bounded/unbounded + +

    + Determine if the domain of the function f(x,y)=\sqrt{1-x^2/9-y^2/4} is open, closed, + or neither, and if it is bounded. +

    +     + +

    + This domain of this function was found in + to be D = \{(x,y)\,|\,\frac{x^2}9+\frac{y^2}4\leq 1\}, + the region bounded by the ellipse \frac{x^2}9+\frac{y^2}4=1. + Since the region includes the boundary + (indicated by the use of \leq), + the set contains all of its boundary points and hence is closed. + The region is bounded as a disk of radius 4, centered at the origin, + contains D. +

    +     +     + + + Determining open/closed, bounded/unbounded + +

    + Determine if the domain of + f(x,y) = \frac1{x-y} is open, closed, or neither. +

    +     + +

    + As we cannot divide by 0, we find the domain to be D = \{(x,y)\,|\,x-y\neq 0\}. + In other words, the domain is the set of all points (x,y) + not on the line y=x. +

    + +
    + Sketching the domain of the function in + + + An image of the xy plane. It is almost entirely shaded, except for the line y=x. + +

    + The image shows the xy plane with its coordinate axes. + The entire plane is shaded, except for a very thin strip along the line y=x. + This is intended as an illustration of the set of all points (x,y) for which y\neq x. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + ytick=\empty, + ymin=-1,ymax=1, + xmin=-1,xmax=1 + ] + + \filldraw [firstcolor!15,fill=firstcolor!15] (axis cs:-1,-1) rectangle (axis cs: 1,1); + + \addplot [ultra thick,white] coordinates {(-1.,-1.)(1,1)}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + The domain is sketched in . + Note how we can draw an open disk around any point in the domain that lies entirely inside the domain, + and also note how the only boundary points of the domain are the points on the line y=x. + We conclude the domain is an open set. + The set is unbounded. +

    +     +     +     + + + Limits +

    + Recall a pseudo-definition of the limit of a function of one variable: +

    +

    + \lim\limits_{x\to c}f(x) = L +

    +

    + means that if x is really close to c, + then f(x) is really close to L. + A similar pseudo-definition holds for functions of two variables. + We'll say that +

    + +

    + \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y) = L +

    + +

    + means if the point (x,y) is really close to the point (x_0,y_0), + then f(x,y) is really close to L. + The formal definition is given below. +

    + + + + Limit of a Function of Two Variables + +

    + Let S be a set containing + P=(x_0,y_0) where every open disk centered at P contains points in S other than P, + let f be a function of two variables defined on S, + except possibly at P, and let L be a real number. + The limit of f(x,y) as (x,y) approaches + (x_0,y_0) is L, denoted + + \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y) = L + , + means that given any \varepsilon \gt 0, + there exists \delta \gt 0 such that for all (x,y) in S, + where (x,y)\neq (x_0,y_0), + if (x,y) is in the open disk centered at + (x_0,y_0) with radius \delta, + then \abs{f(x,y) - L}\lt \varepsilon. + limitof multivariable function + multivariable functionlimit +

    +     +     + +

    + The concept behind + is sketched in . + Given \varepsilon \gt 0, + find \delta \gt 0 such that if (x,y) is any point in the open disk centered at + (x_0,y_0) in the xy-plane with radius \delta, + then f(x,y) should be within \varepsilon of L. +

    + +
    + Illustrating the definition of a limit. The open disk in the xy-plane has radius \delta. Let (x,y) be any point in this disk; f(x,y) is within \varepsilon of L. + + + + An image that illustrates the concept of the limit of a function of two variables. + +

    + A three-dimensional image, showing the three coordinate axes, + along with a surface, and some additional details. + The surface is plotted in the first octant, and lies above the xy plane. + The shape of the surface is similar to a portion of an umbrella. +

    + +

    + Below the surface, in the xy plane, a small circle with dashed boundary is drawn. + The center of the circle is at a point marked as (x_0,y_0,0). + Above the circle, on the surface, is a circular curve consisting of all the points of the form (x,y,f(x,y)) + on the surface, where the point (x,y,0) lies on the circle in the xy plane. + At the center of this curve is a point marked (x_0,y_0,L). +

    + +

    + Along the z axis there are three points marked, + with labels L-\varepsilon, L, and L+\varepsilon. + This illustrates that for each point (x,y,0) interior to the circle in the xy plane, + the value z=f(x,y) lies between L-\varpsilon and L+\varepsilon. + In other words, the z coordinate of each point on the surface that lies on the + interior of the circular curve must be within \varepsilon of L. +

    +     + + + + + //ASY file for figmultilimit_def3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic((10.3,-3.3,2.4),(-0.004,0.001,0.016),(0,0,0),1,(0,0)); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-0.5,3); + pair ybounds=(-0.5,3); + pair zbounds=(-0.1,2.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=2-x^2/4-y^2/4 + triple f(pair t) { + return (sqrt(4*(2-t.y))*cos(t.x),sqrt(4*(2-t.y))*sin(t.x),t.y); + } + surface s=surface(f,(0,0.75),(pi/2,2),10,10,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //draw the dot on the surface and in the xy-plane + dotfactor=3;dot((1,1,1.5));dot((1,1,0)); + + //draw the dashed circle in the xy-plane of radius 0.5 with label + triple g(real t) {return (1+.24*cos(t),1+.24*sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,linetype(new real[] {4,4})+bluepen+linewidth(1.5)); + label("$(x_0,y_0,0)$",(2,2,0)); + draw((1.6,1.6,0)--(1.1,1.1,0),linewidth(.75),Arrow3(size=3mm)); + + //draw the dashed circle on the surface of radius 0.5 with label + triple g(real t) { + return (1+.24*cos(t),1+.24*sin(t),2-((1+.24*cos(t))^2)/4-((1+.24*sin(t))^2)/4); + } + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,linetype(new real[] {4,4})+bluepen+linewidth(1.5)); + label("$(x_0,y_0,L)$",(2.,2,2.5)); + draw((1.8,1.8,2.2)--(1.1,1.1,1.6),linewidth(0.75),Arrow3(size=3mm)); + + draw((0,-.1,1.5)--(0,0,1.5)); + label("$L$",(0,-.1,1.5),W); + + draw((0,-.1,1.7)--(0,0,1.7)); + label("$L+\varepsilon$",(0,-.1,1.7),W); + draw((0,-.1,1.3)--(0,0,1.3)); + label("$L-\varepsilon$",(0,-.1,1.3),W); + + //draw((1,1,0)--(1+.5*cos(3*pi/2),1+.5*sin(3*pi/2),0),dashed+bluepen+linewidth(1)); + //label("$\delta$",((1,1,0)+(1+.5*cos(3*pi/2),1+.5*sin(3*pi/2),0))/2,N); + + + + +
    + +

    + Computing limits using this definition is rather cumbersome. + The following theorem allows us to evaluate limits much more easily. +

    + + + Basic Limit Properties of Functions of Two Variables + +

    + Let b, x_0, y_0, + L and K be real numbers, + let n be a positive integer, + and let f and g be functions with the following limits: + + \lim_{(x,y)\to (x_0,y_0)}f(x,y) = L \textit{ and } \lim_{(x,y)\to (x_0,y_0)} g(x,y) = K + . +

    + +

    + The following limits hold. + limitof multivariable function + limitproperties + multivariable functionlimit + +

      +
    1. +

      + Constants: \lim\limits_{(x,y)\to (x_0,y_0)} b = b +

      +
      2. + +
    2. +

      + Identity \lim\limits_{(x,y)\to (x_0,y_0)} x = x_0; + \lim\limits_{(x,y)\to (x_0,y_0)} y = y_0 +

      +
      3. + +
    3. +

      + Sums/Differences: + \lim\limits_{(x,y)\to (x_0,y_0)}\big(f(x,y)\pm g(x,y)\big) = L\pm K +

      +
      4. + +
    4. +

      + Scalar Multiples: \lim\limits_{(x,y)\to (x_0,y_0)} b\cdot f(x,y) = bL +

      +
      5. + +
    5. +

      + Products: \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y)\cdot g(x,y) = LK +

      +
      6. + +
    6. +

      + Quotients: \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y)/g(x,y) = L/K, (K\neq 0) +

      +
      7. + +
    7. +

      + Powers: \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y)^n = L^n +

      +
      8. +
    +

    +     +     + +

    + This theorem, + combined with Theorems + and + of , + allows us to evaluate many limits. +

    + + + Evaluating a limit + +

    + Evaluate the following limits: +

      +
    1. \lim_{(x,y)\to (1,\pi)} \left(\frac yx + \cos(xy)\right)
      2. +
    2. \lim_{(x,y)\to (0,0)} \frac{3xy}{x^2+y^2}
      3. +
    +

    +     + +

    +

      +
    1. +

      + The aforementioned theorems allow us to simply evaluate + y/x+\cos(xy) when x=1 and y=\pi. + If an indeterminate form is returned, + we must do more work to evaluate the limit; + otherwise, the result is the limit. + Therefore + + \lim_{(x,y)\to (1,\pi)} \frac yx + \cos(xy) \amp = \frac\pi{1}+\cos(\pi) + \amp = \pi -1 + . +

      +
      2. + +
    2. +

      + We attempt to evaluate the limit by substituting 0 in for x and y, + but the result is the indeterminate form + 0/0. To evaluate this limit, + we must do more work, but we have not yet learned what + kind of work to do. + Therefore we cannot yet evaluate this limit. +

      +
      3. +
    +

    +     +     + +

    + When dealing with functions of a single variable we also considered one-sided limits and stated + + \lim_{x\to c}f(x) = L \text{ if, and only if, } \lim_{x\to c^+}f(x) =L \textbf{ and} \lim_{x\to c^-}f(x) =L + . +

    + +

    + That is, the limit is L if and only if f(x) approaches L when x approaches c from either + direction, the left or the right. +

    + +

    + In the plane, + there are infinitely many directions from which (x,y) might approach (x_0,y_0). + In fact, we do not have to restrict ourselves to approaching + (x_0,y_0) from a particular direction, + but rather we can approach that point along a path that is not a straight line. + It is possible to arrive at different limiting values by approaching + (x_0,y_0) along different paths. + If this happens, + we say that \lim\limits_{(x,y)\to(x_0,y_0) } f(x,y) does not exist + (this is analogous to the left and right hand limits of single variable functions not being equal). +

    + +

    + Our theorems tell us that we can evaluate most limits quite simply, + without worrying about paths. + When indeterminate forms arise, + the limit may or may not exist. + If it does exist, + it can be difficult to prove this as we need to show the same limiting value is obtained regardless of the path chosen. + The case where the limit does not exist is often easier to deal with, + for we can often pick two paths along which the limit is different. +

    + + + Showing limits do not exist + +

    +

      +
    1. +

      + Show \lim\limits_{(x,y)\to (0,0)} \frac{3xy}{x^2+y^2} does not exist by finding the limits along the lines y=mx. +

      +
      2. + +
    2. +

      + Show \lim\limits_{(x,y)\to (0,0)} \frac{\sin(xy)}{x+y} does not exist by finding the limit along the path y=-\sin(x). +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + Evaluating \lim\limits_{(x,y)\to (0,0)} \frac{3xy}{x^2+y^2} along the lines y=mx means replace all y's with mx and evaluating the resulting limit: + + \lim_{(x,mx)\to (0,0)} \frac{3x(mx)}{x^2+(mx)^2} \amp =\lim_{x\to 0} \frac{3mx^2}{x^2(m^2+1)} + \amp = \lim_{x\to 0} \frac{3m}{m^2+1} + \amp = \frac{3m}{m^2+1} + . + While the limit exists for each choice of m, + we get a different limit for each choice of m. + That is, along different lines we get differing limiting values, + meaning the limit does not exist. +

      +
      2. + +
    2. +

      + Let f(x,y) = \frac{\sin(xy)}{x+y}. + We are to show that \lim\limits_{(x,y)\to (0,0)} f(x,y) does not exist by finding the limit along the path y=-\sin(x). + First, however, + consider the limits found along the lines y=mx as done above. + + \lim_{(x,mx)\to (0,0)} \frac{\sin\big(x(mx)\big)}{x+mx} \amp = \lim_{x\to 0} \frac{\sin(mx^2)}{x(m+1)} + \amp = \lim_{x\to 0} \frac{\sin(mx^2)}{x}\cdot\frac1{m+1} + . + By applying L'Hospital's Rule, + we can show this limit is 0 except when m=-1, + that is, along the line y=-x. + This line is not in the domain of f, + so we have found the following fact: + along every line y=mx in the domain of f, + \lim\limits_{(x,y)\to(0,0)} f(x,y)=0. + Now consider the limit along the path y=-\sin(x): + + \lim_{(x,-\sin(x) )\to (0,0)} \frac{\sin\big(-x\sin(x) \big)}{x-\sin(x) } \amp = \lim_{x\to0} \frac{\sin\big(-x\sin(x) \big)}{x-\sin(x) } + + Now apply L'Hospital's Rule twice: + + \amp = \lim_{x\to 0}\frac{\cos\big(-x\sin(x) \big)(-\sin(x) -x\cos(x) )}{1-\cos(x) } \quad \left(0/0\right) + \amp = \lim_{x\to 0}\frac{\begin{matrix}-\sin\big(-x\sin(x) \big)(-\sin(x) -x\cos(x) )^2\phantom{\cos(x)}\\ + \phantom{\cos(x)}+\cos\big(-x\sin(x) \big)(-2\cos(x) +x\sin(x) )\end{matrix}}{\sin(x) } + + . + This last limit is of the form 2/0, + which suggests that the limit does not exist. + Step back and consider what we have just discovered. + Along any line y=mx in the domain of the f(x,y), + the limit is 0. + However, along the path y=-\sin(x), + which lies in the domain of f(x,y) for all x\neq 0, + the limit does not exist. + Since the limit is not the same along every path to (0,0), + we say \lim\limits_{(x,y)\to (0,0)}\frac{\sin(xy)}{x+y} does not exist. +

      +
      3. +
    +

    +     +     + + + Finding a limit + +

    + Let f(x,y) = \frac{5x^2y^2}{x^2+y^2}. + Find \lim\limits_{(x,y)\to (0,0)} f(x,y). +

    +     + +

    + It is relatively easy to show that along any line y=mx, + the limit is 0. + This is not enough to prove that the limit exists, + as demonstrated in the previous example, + but it tells us that if the limit does exist then it must be 0. +

    + +

    + To prove the limit is 0, we apply . + Let \varepsilon \gt 0 be given. + We want to find \delta \gt 0 such that if \sqrt{(x-0)^2+(y-0)^2} \lt \delta, + then \abs{f(x,y)-0} \lt \varepsilon. +

    + +

    + Set \delta \lt \sqrt{\varepsilon/5}. + Note that \abs{\frac{5y^2}{x^2+y^2}} \lt 5 for all (x,y)\neq (0,0), + and that if \sqrt{x^2+y^2} \lt \delta, + then x^2\lt \delta^2. +

    + +

    + Let \sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2}\lt \delta. + Consider \abs{f(x,y)-0}: + + \abs{f(x,y)-0} \amp = \abs{\frac{5x^2y^2}{x^2+y^2}-0} + \amp = \abs{x^2\cdot\frac{5y^2}{x^2+y^2}} + \amp \lt \delta^2\cdot 5 + \amp \lt \frac{\varepsilon}{5}\cdot 5 + \amp = \varepsilon + . +

    + +

    + Thus if \sqrt{(x-0)^2+(y-0)^2}\lt \delta then \abs{f(x,y)-0}\lt \varepsilon, + which is what we wanted to show. + Thus \lim\limits_{(x,y)\to(0,0)} \frac{5x^2y^2}{x^2+y^2} = 0. +

    +     +     +     + + + Continuity +

    + + defines what it means for a function of one variable to be continuous. + In brief, it meant that the graph of the function did not have breaks, holes, + jumps, etc. + We define continuity for functions of two variables in a similar way as we did for functions of one variable. +

    + + + Continuous + +

    + Let a function f(x,y) be defined on a set S containing the point (x_0,y_0). +

    + +

    +

      +
    1. +

      + f is continuous + at (x_0,y_0) if \lim\limits_{(x,y)\to(x_0,y_0)} f(x,y) = f(x_0,y_0). + continuous function + multivariable functioncontinuity +

      +
      2. + +
    2. +

      + f is continuous on S + if f is continuous at all points in S. + If f is continuous at all points in \mathbb{R}^2, + we say that f is continuous everywhere. +

      +
      3. +
    +

    +     +     + + + Continuity of a function of two variables + +

    + Let f(x,y) = \left\{ \begin{array}{rl} \frac{\cos(y) \sin(x) }{x} \amp x\neq 0 \\ + \cos(y) \amp x=0 + \end{array} \right.. + Is f continuous at (0,0)? + Is f continuous everywhere? +

    +     + +

    + To determine if f is continuous at (0,0), + we need to compare \lim\limits_{(x,y)\to (0,0)} f(x,y) to f(0,0). +

    + +

    + Applying the definition of f, + we see that f(0,0) = \cos(0) = 1. +

    + +

    + We now consider the limit \lim\limits_{(x,y)\to (0,0)} f(x,y). + Substituting 0 for x and y in + (\cos(y) \sin(x) )/x returns the indeterminate form 0/0, + so we need to do more work to evaluate this limit. +

    + +

    + Consider two related limits: + \lim\limits_{(x,y)\to (0,0)} \cos(y) and \lim\limits_{(x,y)\to(0,0)} \frac{\sin(x) }x. + The first limit does not contain x, + and since \cos(y) is continuous, + + \lim\limits_{(x,y)\to (0,0)} \cos(y) =\lim_{y\to 0} \cos(y) = \cos(0) = 1 + . +

    + +

    + The second limit does not contain y. + By we can say + + \lim_{(x,y)\to (0,0)} \frac{\sin(x) }{x} = \lim_{x\to 0} \frac{\sin(x) }{x} = 1 + . +

    + +

    + Finally, + of this section states that we can combine these two limits as follows: + + \lim_{(x,y)\to (0,0)} \frac{\cos(y) \sin(x) }{x} \amp = \lim_{(x,y)\to (0,0)} (\cos(y) )\left(\frac{\sin(x) }{x}\right) + \amp =\left(\lim_{(x,y)\to (0,0)} \cos(y) \right)\left(\lim_{(x,y)\to (0,0)} \frac{\sin(x) }{x}\right) + \amp = (1)(1) + \amp =1 + . +

    + +

    + We have found that \lim\limits_{(x,y)\to (0,0)} \frac{\cos(y) \sin(x) }{x} = f(0,0), + so f is continuous at (0,0). +

    + +

    + A similar analysis shows that f is continuous at all points in \mathbb{R}^2. + As long as x\neq0, we can evaluate the limit directly; + when x=0, + a similar analysis shows that the limit is \cos(y). + Thus we can say that f is continuous everywhere. + A graph of f is given in . + Notice how it has no breaks, jumps, etc. +

    + +
    + A graph of f(x,y) in + + + + A graph of the piecewise-defined function in this example, illustrating its continuity. + +

    + A plot of the surface given by the graph z=f(x,y) for the function f in this example. + The surface consists of many peaks and valleys, with the largest peaks lying along the y axis. + Viewing the graph shows us that although the surface is quite bumpy, + there are no jumps or breaks in the surface, illustrating the fact that the function f is continuous. +

    +     + + + + + //ASY file for figmulticont13D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(31,31,2.3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-10,10}; + real[] myychoice={-10,10}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-12,12); + pair ybounds=(-12,12); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=cos(y)sin(x)/x + triple f(pair t) { + return (t.x,t.y,cos(t.y)*sin(t.x)/t.x); + } + surface s=surface(f,(-10,-10),(10,10),15,15,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //triple f(pair t) { + // return (t.x,t.y,cos(t.y)*sin(t.x)/t.x); + //} + //surface s=surface(f,(-11,-11),(0.1,11),16,16,Spline); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + + + +
    +     +     + +

    + The following theorem is very similar to , + giving us ways to combine continuous functions to create other continuous functions. +

    + + + Properties of Continuous Functions + +

    + Let f and g be continuous on a set S, + let c be a real number, + and let n be a positive integer. + The following functions are continuous on S. + continuous functionproperties + multivariable functioncontinuity + +

      +
    1. +

      + Sums/Differences: f\pm g +

      +
      2. + +
    2. +

      + Constant Multiples: c\cdot f +

      +
      3. + +
    3. +

      + Products: f\cdot g +

      +
      4. + +
    4. +

      + Quotients: f/g { (as longs as g\neq 0 on S)} +

      +
      5. + +
    5. +

      + Powers: f^n +

      +
      6. + +
    6. +

      + Roots: \sqrt[n]{f} (if n is even then f\geq 0 on S; + if n is odd, then true for all values of f on S.) +

      +
      7. + +
    7. +

      + Compositions:Adjust the definitions of f and g to: Let f be continuous on S, + where the range of f on S is J, + and let g be a single variable function that is continuous on J. + Then g\circ f, , g(f(x,y)), + is continuous on S. +

      +
      8. +
    +

    +     +     + + + Establishing continuity of a function + +

    + Let f(x,y) = \sin(x^2\cos(y) ). + Show f is continuous everywhere. +

    +     + +

    + We will apply both Theorems + and . + Let f_1(x,y) = x^2. + Since y is not actually used in the function, + and polynomials are continuous + (by ), + we conclude f_1 is continuous everywhere. + A similar statement can be made about f_2(x,y) = \cos(y). + Part 3 of + states that f_3=f_1\cdot f_2 is continuous everywhere, + and Part 7 of the theorem states the composition of sine with f_3 is continuous: + that is, \sin(f_3) = \sin(x^2\cos(y) ) is continuous everywhere. +

    +     +     +     + + + Functions of Three Variables +

    + The definitions and theorems given in this section can be extended in a natural way to definitions and theorems about functions of three + (or more) + variables. + We cover the key concepts here; + some terms from Definitions + and + are not redefined but their analogous meanings should be clear to the reader. +

    + + + Open Balls, Limit, Continuous + +

    +

      +
    1. +

      + An open ball in \mathbb{R}^3 centered at + (x_0,y_0,z_0) with radius r is the set of all points (x,y,z) such that + + \sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2} = r + . + multivariable functionlimit + limitof multivariable function + multivariable functioncontinuity + open ball +

      +
      2. + +
    2. +

      + Let D be a set in \mathbb{R}^3 containing + (x_0,y_0,z_0) where every open ball centered at + (x_0,y_0,z_0) contains points of D other than (x_0,y_0,z_0), + and let f(x,y,z) be a function of three variables defined on D, + except possibly at (x_0,y_0,z_0). + The limit of f(x,y,z) as (x,y,z) approaches (x_0,y_0,z_0) is L, denoted + + \lim_{(x,y,z)\to (x_0,y_0,z_0)} f(x,y,z) = L + , + means that given any \varepsilon \gt 0, + there is a \delta \gt 0 such that for all (x,y,z) in D, + (x,y,z)\neq(x_0,y_0,z_0), + if (x,y,z) is in the open ball centered at + (x_0,y_0,z_0) with radius \delta, + then \abs{f(x,y,z) - L}\lt \varepsilon. +

      +
      3. + +
    3. +

      + Let f(x,y,z) be defined on a set D containing (x_0,y_0,z_0). + We say f is continuous + at (x_0,y_0,z_0) if + + \lim\limits_{(x,y,z)\to (x_0,y_0,z_0)} f(x,y,z) = f(x_0,y_0,z_0) + . + If f is continuous at all points in D, + we say f is continuous on D. +

      +
      4. +
    +

    +     +     + +

    + These definitions can also be extended naturally to apply to functions of four or more variables. + + also applies to function of three or more variables, + allowing us to say that the function + + f(x,y,z) = \frac{e^{x^2+y}\sqrt{y^2+z^2+3}}{\sin(xyz)+5} + + is continuous everywhere. +

    + +

    + When considering single variable functions, + we studied limits, then continuity, then the derivative. + In our current study of multivariable functions, + we have studied limits and continuity. + In the next section we study derivation, + which takes on a slight twist as we are in a multivariable context. +

    +     + + + + Terms and Concepts + + + + +

    + Describe in your own words the difference between the boundary and interior points of a set. +

    +     + + + +     + + + + +

    + Use your own words to describe (informally) what \lim\limits_{(x,y)\to (1,2)} f(x,y) = 17 means. +

    +     + + + +

    + Answers will vary. + One answer is As (x,y) gets close to (1,2), + f(x,y) gets close to 17. +

    +     + +     + + + + +

    + Give an example of a closed, bounded set. +

    +     + + + +

    + Answers will vary. +

    + +

    + One possible answer: \{(x,y) | x^2+y^2\leq 1\} +

    +     + +     + + + + +

    + Give an example of a closed, unbounded set. +

    +     + + + +

    + Answers will vary. +

    + +

    + One possible answer: \{(x,y) | y\geq x^2 \} +

    +     + +     + + + + +

    + Give an example of a open, bounded set. +

    +     + + + +

    + Answers will vary. +

    + +

    + One possible answer: \{(x,y) | x^2+y^2\lt 1 \} +

    +     + +     + + + + +

    + Give an example of a open, unbounded set. +

    +     + + + +

    + Answers will vary. +

    + +

    + One possible answer: \{(x,y) | y \gt x^2 \} +

    +     + +     +     + + Problems + + + +

    + A set S is given. +

    + +

    +

      +
    1. +

      + Give one boundary point and one interior point, + when possible, of S. +

      +
      2. + +
    2. +

      + State whether S is open, closed, or neither. +

      +
      3. + +
    3. +

      + State whether S is bounded or unbounded. +

      +
      4. +
    +

    +     + + + + +

    + \ds S = \left\{(x,y)\,\left| \, \frac{(x-1)^2}{4}+\frac{(y-3)^2}{9}\leq 1\right.\right\} +

    +     + +

    +

      +
    1. +

      + Answers will vary. + + interior point: (1,3) + + boundary point: (3,3) +

      +
      2. + +
    2. +

      + S is a closed set +

      +
      3. + +
    3. +

      + S is bounded +

      +
      4. +
    +

    +     + +     + + + + + +

    + S = \left\{(x,y)\mid y\neq x^2\right\} +

    + + +     + +

    +

      +
    1. +

      + Answers will vary. Interior point: (1,0) (any point with y\neq x^2 will do). + Boundary point: (1,1) (any point with y=x^2 will do). +

      +
      2. +
    2. +

      + S is an open set. +

      +
      3. +
    3. +

      + S is unbounded. +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds S = \left\{(x,y)\,| \, x^2+y^2=1\right\} +

    +     + +

    +

      +
    1. +

      + Answers will vary. + + interior point: none + + boundary point: (0,-1) +

      +
      2. + +
    2. +

      + S is a closed set, consisting only of boundary points +

      +
      3. + +
    3. +

      + S is bounded +

      +
      4. +
    +

    +     + +     + + + + + +

    + S = \left\{(x,y)\mid y \gt \sin(x)\right\}. +

    + + +     + +     + +     + + + +

    + For the given function: +

    + +

    +

      +
    1. +

      + Find the domain D of the function. +

      +
      2. + +
    2. +

      + State whether D is an open or closed set. +

      +
      3. + +
    3. +

      + State whether D is bounded or unbounded. +

      +
      4. +
    +

    +     + + + + +

    + \ds f(x,y) = \sqrt{9-x^2-y^2} +

    +     + +

    +

      +
    1. +

      + D = \left\{(x,y)\, |\, 9-x^2-y^2\geq 0\right\}. +

      +
      2. + +
    2. +

      + D is a closed set. +

      +
      3. + +
    3. +

      + D is bounded. +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds f(x,y) = \sqrt{y-x^2} +

    +     + +

    +

      +
    1. +

      + D = \left\{(x,y)\, |\, y\geq x^2\right\}. +

      +
      2. + +
    2. +

      + D is a closed set. +

      +
      3. + +
    3. +

      + D is unbounded. +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds f(x,y) = \frac{1}{\sqrt{y-x^2}} +

    +     + +

    +

      +
    1. +

      + D = \left\{(x,y)\, |\, y \gt x^2\right\}. +

      +
      2. + +
    2. +

      + D is an open set. +

      +
      3. + +
    3. +

      + D is unbounded. +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds f(x,y) = \frac{x^2-y^2}{x^2+y^2} +

    +     + +

    +

      +
    1. +

      + D = \left\{(x,y)\, |\, (x,y)\neq (0,0) \right\}. +

      +
      2. + +
    2. +

      + D is an open set. +

      +
      3. + +
    3. +

      + D is unbounded. +

      +
      4. +
    +

    +     + +     + +     + + + +

    + A limit is given. + Evaluate the limit along the paths given, + then state why these results show the given limit does not exist. +

    +     + + + + +

    + \lim\limits_{(x,y)\to(0,0)} \frac{x^2-y^2}{x^2+y^2} +

    +     + + + +

    + Along the path y=0. +

    +     +     + + + +

    + Along the path x=0. +

    +     +     + +

    +

      +
    1. +

      + Along y=0, the limit is 1. +

      +
      2. + +
    2. +

      + Along x=0, the limit is -1. +

      +
      3. +
    +

    + +

    + Since the above limits are not equal, the limit does not exist. +

    +     + +     + + + + +

    + \lim\limits_{(x,y)\to(0,0)} \frac{x+y}{x-y} +

    + +

    + Along the path y=mx. +

    +     + +

    +

      +
    1. +

      + Along y=mx, the limit is \frac{m+1}{1-m}. +

      +
      2. +
    +

    + +

    + Since the above limit varies according to what m is used, + each limit is different, meaning the overall limit does not exist. +

    +     + +     + + + + +

    + \lim\limits_{(x,y)\to(0,0)} \frac{xy-y^2}{y^2+x} +

    +     + + + +

    + Along the path y=mx. +

    +     +     + + + +

    + Along the path x=0. +

    +     +     + +

    +

      +
    1. +

      + Along y=mx, the limit is \ds\lim_{x\to 0}\frac{mx(1-m)}{m^2x+1}=0 for all m +

      +
      2. + +
    2. +

      + Along x=0, the limit is -1. +

      +
      3. +
    +

    + +

    + Since the above limits are not equal, the limit does not exist. +

    +     + +     + + + + +

    + \lim\limits_{(x,y)\to(0,0)} \frac{\sin(x^2)}{y} +

    +     + + + +

    + Along the path y=mx. +

    +     +     + + + +

    + Along the path y=x^2. +

    +     +     + +

    +

      +
    1. +

      + Along y=mx, the limit is: + + \lim_{(x,y)\to(0,0)} \frac{\sin(x^2)}{y} \amp = \lim_{x\to 0} \frac{\sin(x^2)}{mx} + apply L'Hospital's Rule + \amp = \lim_{x\to 0} \frac{2x\cos(x^2)}{m} + \amp = 0 + . +

      +
      2. +
    2. +

      + Along y=x^2, the limit is: + + \lim_{(x,y)\to(0,0)} \frac{\sin(x^2)}{y} = \lim_{x\to 0} \frac{\sin(x^2)}{x^2} + . + This can be evaluated with L'Hospital's Rule or from known limits; it is 1. +

      +
      3. +
    +

    + +

    + Since the limits along the lines y=mx are not the same as the limit along y=x^2, + the overall limit does not exist. +

    +     + +     + + + + +

    + \lim\limits_{(x,y)\to(1,2)} \frac{x+y-3}{x^2-1} +

    +     + + + +

    + Along the path y=2. +

    +     +     + + + +

    + Along the path y=x+1. +

    +     +     + +

    +

      +
    1. +

      + Along y=2, the limit is: + + \lim_{(x,y)\to(1,2)} \frac{x+y-3}{x^2-1} \amp = \lim_{x\to 1} \frac{x-1}{x^2-1} + \amp = \lim_{x\to 1} \frac{1}{x+1} + \amp = 1/2 + . +

      +
      2. + +
    2. +

      + Along y=x+1, the limit is: + + \lim_{(x,y)\to(1,2)} \frac{x+y-3}{x^2-1} \amp = \lim_{x\to 1} \frac{2(x-1)}{x^2-1} + \amp = \lim_{x\to 1} \frac{2}{x+1} + \amp = 1 + . +

      +
      3. +
    +

    + +

    + Since the limits along the lines y=2 and y=x+1 differ, + the overall limit does not exist. +

    +     + +     + + + + +

    + \lim\limits_{(x,y)\to(\pi,\pi/2)} \frac{\sin(x) }{\cos(y) } +

    +     + + + +

    + Along the path x=\pi. +

    +     +     + + + +

    + Along the path y=x-\pi/2. +

    +     +     + +

    +

      +
    1. +

      + Along x=\pi, the limit is: + + \lim_{(x,y)\to(\pi,\pi/2)} \frac{\sin(x) }{\cos(y) } \amp = \lim_{y\to \pi/2} \frac{0}{\cos(y) } + \amp = 0 + . +

      +
      2. + +
    2. +

      + Along y=x-\pi/2, the limit is: + + \lim_{(x,y)\to(\pi,\pi/2)} \frac{\sin(x) }{\cos(y) } \amp = \lim_{x\to \pi} \frac{\sin(x) }{\cos(x-\pi/2)} + Apply L'Hospital's Rule: + \amp = \lim_{x\to \pi} \frac{\cos(x) }{\sin(x-\pi/2)} + \amp = 1 + . +

      +
      3. +
    +

    + +

    + Since the limits along the lines x=\pi and y=x-\pi differ, + the overall limit does not exist. +

    +     + +     + +     +     +     +
    +
    + Partial Derivatives + +

    + Let y be a function of x. + We have studied in great detail the derivative of y with respect to x, + that is, \frac{dy}{dx}, + which measures the rate at which y changes with respect to x. + Consider now z=f(x,y). + It makes sense to want to know how z changes with respect to x and/or y. + This section begins our investigation into these rates of change. +

    + + +     + + + First-order partial derivatives +

    + Consider the function f(x,y) = x^2+2y^2, + as graphed in . + By fixing y=2, + we focus our attention to all points on the surface where the y-value is 2, shown in both and . + These points form a curve in the plane y=2: + z = f(x,2) = x^2+8 which defines z as a function of just one variable. + We can take the derivative of z with respect to x along this curve and find equations of tangent lines, etc. +

    + +
    + By fixing y=2, the surface z=f(x,y) = x^2+2y^2 is a curve in space + +
    + + + + + An elliptic paraboloid plotted over a rectangular domain. The trace y=2 is highlighted on the surface. + +

    + The graph z=x^2+2y^2 is an elliptic paraboloid, opening upward, with its vertex at the origin. + The plot uses a rectangular domain, so we see peaks at the corners of the domain. + Mesh curves corresponding to the traces where either x or y are constant are shown, + and one of these curves, the trace y=2, is highlighted. +

    +     + + + + + //ASY file for figpartialintro3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(16,16,36); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-4,-2,2,4}; + real[] myychoice={-4,-2,2,4}; + real[] myzchoice={10,20}; + defaultpen(0.5mm); + pair xbounds=(-5,5); + pair ybounds=(-5,5); + pair zbounds=(-1,22); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=x^2+2*y^2 + triple f(pair t) { + return (t.x,t.y,t.x^2+2*t.y^2); + } + surface s=surface(f,(-3,-3),(3,3),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}, + vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the trace for y=2 + triple g(real t) {return (t,2,t^2+8);} + path3 mypath=graph(g,-3,3,operator ..); draw(mypath,bluepen); + + + + +
    + +
    + + + + + The trace y=2 on the surface of the ellipic paraboloid from the previous image, with the surface removed. + +

    + The curve given by the trace y=2 in the surface z=x^2+2y^2 is plotted in three dimensions. + This is the same curve shown in , + but this time only the curve is plotted. The surface has been removed, to help visualize the curve. +

    + +

    + The curve itself has the shape of a parabola, opening upwards. + It is hanging in space, with its vertex above the mark on the y axis for y=2. + The curve is viewed in perspective but is clearly a parabola. +

    +     + + + + + //ASY file for figpartialintrob3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(16,16,36); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-4,-2,2,4}; + real[] myychoice={-4,-2,2,4}; + real[] myzchoice={10,20}; + defaultpen(0.5mm); + pair xbounds=(-5,5); + pair ybounds=(-5,5); + pair zbounds=(-1,22); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the trace for y=2 + triple g(real t) {return (t,2,t^2+8);} + path3 mypath=graph(g,-3,3,operator ..); draw(mypath,bluepen); + + + + +
    +     +
    + +

    + The key notion to extract from this example is: + by treating y as constant + (it does not vary) + we can consider how z changes with respect to x. + In a similar fashion, + we can hold x constant and consider how z changes with respect to y. + This is the underlying principle of + partial derivatives. + We state the formal, limit-based definition first, + then show how to compute these partial derivatives without directly taking limits. +

    + + + + + + Partial Derivative + +

    + Let z=f(x,y) be a continuous function on a set S in \mathbb{R}^2. +

    + +

    +

      +
    1. +

      + The partial derivative of f with respect to x is: + + f_x(x,y) = \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}h + . +

      +
      2. + +
    2. +

      + The partial derivative of f with respect to y is: + + f_y(x,y) = \lim_{h\to 0} \frac{f(x,y+h) - f(x,y)}h + . +

      +
      3. +
    +

    + +

    + partial derivative + derivativepartial +

    +     +     + + + Computing partial derivatives with the limit definition + +

    + Let f(x,y) = x^2y + 2x+y^3. + Find f_x(x,y) using the limit definition. +

    +     + +

    + Using , we have: + + f_x(x,y) \amp = \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}{h} + \amp = \lim_{h\to 0} \frac{(x+h)^2y+2(x+h)+y^3 - (x^2y+2x+y^3)}{h} + \amp = \lim_{h\to 0} \frac{x^2y+2xhy+h^2y+2x+2h+y^3-(x^2y+2x+y^3)}{h} + \amp = \lim_{h\to 0} \frac{2xhy+h^2y+2h}{h} + \amp =\lim_{h\to 0} 2xy+hy+2 + \amp = 2xy+2 + . +

    + +

    + We have found f_x(x,y) = 2xy+2. +

    +     +     + +

    + + found a partial derivative using the formal, + limit-based definition. + Using limits is not necessary, though, + as we can rely on our previous knowledge of derivatives to compute partial derivatives easily. + When computing f_x(x,y), + we hold y fixed it does not vary. + Therefore we can compute the derivative with respect to x by treating y as a constant or coefficient. +

    + +

    + Just as \frac{d}{dx}\big(5x^2\big) = 10x, + we compute \frac{\partial}{\px}\big(x^2y\big) = 2xy. + Here we are treating y as a coefficient. +

    + +

    + Just as \frac{d}{dx}\big(5^3\big) = 0, + we compute \frac{\partial}{\px}\big(y^3\big) = 0. + Here we are treating y as a constant. + More examples will help make this clear. +

    + + + Finding partial derivatives + +

    + Find f_x(x,y) and f_y(x,y) in each of the following. +

    + +

    +

      +
    1. f(x,y) = x^3y^2+ 5y^2-x+7
      2. +
    2. f(x,y) = \cos(xy^2)+\sin(x)
      3. +
    3. f(x,y) = e^{x^2y^3}\sqrt{x^2+1}
      4. +
    +

    +     + +

    +

      +
    1. +

      + We have f(x,y) = x^3y^2+ 5y^2-x+7. + Begin with f_x(x,y). + Keep y fixed, + treating it as a constant or coefficient, as appropriate: + + f_x(x,y) = 3x^2y^2-1 + . + Note how the 5y^2 and 7 terms go to zero. + To compute f_y(x,y), we hold x fixed: + + f_y(x,y) = 2x^3y+10y + . + Note how the -x and 7 terms go to zero. +

      +
      2. + +
    2. +

      + We have f(x,y) = \cos(xy^2)+\sin(x). + + Begin with f_x(x,y). We need to apply the Chain Rule with the cosine term; y^2 is the coefficient of the x-term inside the cosine function. + + f_x(x,y) = -\sin(xy^2)(y^2)+\cos(x) = -y^2\sin(xy^2)+\cos(x) + . + To find f_y(x,y), + note that x is the coefficient of the y^2 term inside of the cosine term; + also note that since x is fixed, + \sin(x) is also fixed, + and we treat it as a constant. + + f_y(x,y) = -\sin(xy^2)(2xy) = -2xy\sin(xy^2) + . +

      +
      3. + +
    3. +

      + We have f(x,y) = e^{x^2y^3}\sqrt{x^2+1}. + Beginning with f_x(x,y), + note how we need to apply the Product Rule. + + f_x(x,y) \amp = e^{x^2y^3}(2xy^3)\sqrt{x^2+1} + e^{x^2y^3}\frac12\big(x^2+1\big)^{-1/2}(2x) + \amp = 2xy^3e^{x^2y^3}\sqrt{x^2+1}+\frac{xe^{x^2y^3}}{\sqrt{x^2+1}} + . + Note that when finding f_y(x,y) we do not have to apply the Product Rule; + since \sqrt{x^2+1} does not contain y, + we treat it as fixed and hence becomes a coefficient of the e^{x^2y^3} term. + + f_y(x,y) = e^{x^2y^3}(3x^2y^2)\sqrt{x^2+1} = 3x^2y^2e^{x^2y^3}\sqrt{x^2+1} + . +

      +
      4. +
    +

    +     +     + + + +

    + We have shown how to compute a partial derivative, + but it may still not be clear what a partial derivative means. + Given z=f(x,y), + f_x(x,y) measures the rate at which z changes as only x varies: + y is held constant. + partial derivativemeaning +

    + +

    + Imagine standing in a rolling meadow, + then beginning to walk due east. + Depending on your location, you might walk up, + sharply down, or perhaps not change elevation at all. + This is similar to measuring z_x: + you are moving only east + (in the x-direction) + and not north/south at all. + Going back to your original location, + imagine now walking due north + (in the y-direction). + Perhaps walking due north does not change your elevation at all. + This is analogous to z_y=0: + z does not change with respect to y. + We can see that z_x and z_y do not have to be the same, + or even similar, + as it is easy to imagine circumstances where walking east means you walk downhill, + though walking north makes you walk uphill. +

    + + + +

    + The following example helps us visualize this more. +

    + + + Evaluating partial derivatives + +

    + Let z=f(x,y)=-x^2-\frac12y^2+xy+10. + Find f_x(2,1) and f_y(2,1) and interpret their meaning. +

    +     + +

    + We begin by computing f_x(x,y) = -2x+y and f_y(x,y) = -y+x. + Thus + + f_x(2,1) = -3 \text{ and } f_y(2,1) = 1 + . +

    + +

    + It is also useful to note that f(2,1) = 7.5. + What does each of these numbers mean? +

    + +

    + Consider f_x(2,1)=-3, + along with . + If one stands on the surface at the point + (2,1,7.5) and moves parallel to the x-axis (, only the x-value changes, + not the y-value), + then the instantaneous rate of change is -3. + Increasing the x-value will decrease the z-value; + decreasing the x-value will increase the z-value. +

    + +
    + Illustrating the meaning of partial derivatives + +
    + + + + + On a surface in space, a trace curve is highlighted. At one pointon this curve, a tangent line is drawn. + +

    + The surface given by the graph z=f(x,y) is shown, + for f(x,y) = -x^2-\frac12 y^2+xy+10. + It is a portion of an elliptic paraboloid, opening downward. +

    + +

    + Along the surface a curve is drawn, corresponding to the trace y=1. + This is a curve that moves along the surface as x varies, while y is held constant. + At the point (2,1,f(2,1)) on the surface, a line is drawn, tangent to the curve. + This is a line in space, but its slope, relative to x, + is given by the partial derivative f_x(2,1). +

    +     + + + + + //ASY file for figpartial3a3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8,8,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={5,10}; + defaultpen(0.5mm); + pair xbounds=(-1,3); + pair ybounds=(-1,3); + pair zbounds=(-1,12); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=-x^2-(1/2)*y^2+xy+10 + triple f(pair t) { + return (t.x,t.y,-t.x^2-(1/2)*t.y^2+t.x*t.y+10); + } + surface s=surface(f,(0,0),(3,3),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the point (2,1,7.5) trace for y=1 + dotfactor=3;dot((2,1,7.5)); + triple g(real t) {return (t,1,-t^2-(1/2)+t+10);} + path3 mypath=graph(g,0,3,operator ..); draw(mypath,bluepen); + + //Draw the tan line (y=1) L=(2,1,7.5)+t(1,0,-3) for t=-1.5,1 + draw((0.5,1,12)--(3,1,4.5),redpen); + + + + +
    + +
    + + + + + On a surface in space, a trace curve is highlighted. At one pointon this curve, a tangent line is drawn. + +

    + The surface given by the graph z=f(x,y) is shown, + for f(x,y) = -x^2-\frac12 y^2+xy+10. + It is a portion of an elliptic paraboloid, opening downward. +

    + +

    + Along the surface a curve is drawn, corresponding to the trace x=. + This is a curve that moves along the surface as y varies, while x is held constant. + At the point (2,1,f(2,1)) on the surface, a line is drawn, tangent to the trace x=2. + This tangent line is perpendicular to the one drawn in . + It is a line in space, but its slope, relative to y, + is given by the partial derivative f_y(2,1). +

    +     + + + + + //ASY file for figpartial3a3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8,8,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={5,10}; + defaultpen(0.5mm); + pair xbounds=(-1,3); + pair ybounds=(-1,3); + pair zbounds=(-1,12); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=-x^2-(1/2)*y^2+xy+10 + triple f(pair t) { + return (t.x,t.y,-t.x^2-(1/2)*t.y^2+t.x*t.y+10); + } + surface s=surface(f,(0,0),(3,3),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the point (2,1,7.5) trace for x=2 + dotfactor=3;dot((2,1,7.5)); + triple g(real t) {return (2,t,-4-(1/2)*t^2+2*t+10);} + path3 mypath=graph(g,0,3,operator ..); draw(mypath,bluepen); + + //Draw the tan line (x=2) L=(2,1,7.5)+t(0,1,1) for t=-1,2 + draw((2,0,6.5)--(2,3,9.5),redpen); + + + + +
    +     +
    + +

    + Now consider f_y(2,1)=1, + illustrated in . + Moving along the curve drawn on the surface, + , parallel to the y-axis and not changing the x-values, + increases the z-value instantaneously at a rate of 1. + Increasing the y-value by 1 would increase the z-value by approximately 1. +

    + +

    + Since the magnitude of f_x is greater than the magnitude of f_y at (2,1), + it is steeper in the x-direction than in the y-direction. +

    +     +     +     + + + Tangent Planes +

    + Another way to interpret partial derivatives is in terms of the tangent plane. + Consider the graph of a function f(x,y), such as the one in . + Setting x=a, y=b defines a point (a,b,f(a,b)) on the graph. + Through the point (a,b), we have the lines x=a+s, y=b, and x=a, y=b+t, + parallel to the x and y axes, respectively (where s,t are parameters). +

    + +

    + Using the function f(x,y) we define two vector-valued functions: + + \vec{r}_1(s) \amp = \la a+s, b, f(a+s,b)\ra + \vec{r}_2(t) \amp = \la a, b+t, f(a,b+t)\ra + . + Both vector-valued functions define space curves that lie on the surface z=f(x,y), + and these curves intersect at the point (a,b,f(a,b)), when s=t=0. +

    + +

    + Now consider computing \vec{r}_1'(s). + The first two components of this derivative are found in a straightforward manner: + they are 1 and 0, respectively. + To find the third component of the derivative, + notice that in \vec{r}_1(s) we vary the x-component of f while holding the y-component constant. + Using the Chain Rule and , + we find that the third component is f_x(a+s,b). Altogether, we have + + \vec{r}_1'(s) = \la 1,0,f_x(a+s,b)\ra + . + Evaluating this at s=0 gives + + \vec{v}=\vec{r}_1'(0) = \la 1,0,f_x(a,b)\ra + . + We can perform a similar process with \vec{r}_2(t), ultimately leading to + + \vec{w}=\vec{r}_2'(0) = \la 0,1,f_y(a,b)\ra + . + From , + we know that \vec{r}_1'(0) defines a tangent vector to the curve \vec{r}_1(s) when s=0, + and similarly, \vec{r}_2'(0) defines a tangent vector to the curve \vec{r}_2(t) when t=0. +

    + +

    + It seems reasonable that any vector that is tangent to these curves, + which lie on our surface, should also be considered tangent to that surface. + The vectors \vec{v} and \vec{w} are therefore tangent to z=f(x,y) at (a,b,f(a,b)), + and they are definitely not parallel. + From we know that any two non-parallel vectors at a point define a plane through that point. + We also know that taking the cross product of these two vectors gives us a normal vector: + the cross product gives us + + \vec{n}=\vec{v}\times\vec{w}=\la -f_x(a,b), -f_y(a,b), 1\ra + . +

    + +

    + The equation of the plane through (a,b,f(a,b)) with normal vector \vec{n}=\la -f_x(a,b),-f_y(a,b),1\ra is + + -f_x(a,b)(x-a)-f_y(a,b)(y-b)+(z-f(a,b))=0 + . + It is customary to solve for z in this equation and make the following definition. +

    + + + +

    + Let f(x,y) be a function whose first-order partial derivatives exist at (a,b). + The tangent plane to the surface z=f(x,y) at the point (a,b,f(a,b)) is the plane defined by the equation + + z = f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) + . + tangent plane + tangent planeto a graph +

    +     +     + + + Finding a tangent plane equation + +

    + Find the equation of tangent plane to the surface z=x^2+3y^2 at (x,y)=(1,-1). +

    +     + +

    + Our function is f(x,y)=x^2+3y^2, and we have f(1,-1)=4, + so the point on the surface is (1,-1,4). + The partial derivatives are f_x(x,y)=2x and f_y(x,y)=6y, + so f_x(1,-1)=2, f_y(1,-1)=-6. + Using , our plane is given by + + z = 4+2(x-1)-6(y+1) + . +

    +     +     + +

    + Notice the similarity between the tangent plane equation in + and the single variable tangent line equation y = f(c)+f'(c)(x-c). + As with functions of one variable, this suggests a connection between derivatives and linear approximation. + We explore this connection in , + where we'll see that should be strengthed to require that the partial derivatives of f be continuous. +

    +     + + + Second-order partial derivatives +

    + Let z=f(x,y). + We have learned to find the partial derivatives f_x(x,y) and f_y(x,y), + which are each functions of x and y. + Therefore we can take partial derivatives of them, + each with respect to x and y. + We define these second partials along with the notation, + give examples, then discuss their meaning. +

    + + + Second Partial Derivative, Mixed Partial Derivative + +

    + Let z=f(x,y) be continuous on a set S. +

    + +

    +

      +
    1. +

      + The second partial derivative of f with respect to x then x is + + \frac{\partial}{\partial x}\left(\frac{\partial f}{\px}\right) = \frac{\partial^2 f}{\px^2} = \big(\,f_x\,\big)_x = f_{xx} + +

      +
      2. + +
    2. +

      + The second partial derivative of f with respect to x then y is + + \frac{\partial}{\partial y}\left(\frac{\partial f}{\px}\right) = \frac{\partial^2f}{\py\px} = \big(\,f_x\,\big)_y = f_{xy} + +

      +
      3. +
    +

    + +

    + Similar definitions hold for + \frac{\partial^2f}{\py^2} = f_{yy} and \frac{\partial^2f}{\px\py} = f_{yx}. +

    + +

    + The second partial derivatives f_{xy} and f_{yx} are + mixed partial derivatives. + partial derivativemixed + partial derivativesecond derivative + derivativemixed partial +

    +     +     + +

    + The notation of second partial derivatives gives some insight into the notation of the second derivative of a function of a single variable. + If y=f(x), then \fp'(x) = \frac{d^2 y}{dx^2}. + The d^2y portion means + take the derivative of y twice, while dx^2 + means with respect to x both times. + When we only know of functions of a single variable, + this latter phrase seems silly: + there is only one variable to take the derivative with respect to. + Now that we understand functions of multiple variables, + we see the importance of specifying which variables we are referring to. +

    + + + + Second partial derivatives + +

    + For each of the following, find all six first and second partial derivatives. + That is, find + + f_x, f_y, f_{xx}, f_{yy}, f_{xy} \text{ and } f_{yx}\, + . +

    + +

    +

      +
    1. f(x,y) = x^3y^2 + 2xy^3+\cos(x)
      2. + +
    2. f(x,y) = \frac{x^3}{y^2}
      3. + +
    3. f(x,y)=e^{x}\sin(x^2y)
      4. +
    +

    +     + +

    + In each, we give f_x and f_y immediately and then spend time deriving the second partial derivatives. +

    + +

    +

      +
    1. +

      + + f(x,y) \amp = x^3y^2+2xy^3+\cos(x) + f_x(x,y) \amp = 3x^2y^2+2y^3-\sin(x) + f_y(x,y) \amp = 2x^3y+6xy^2 + f_{xx}(x,y) \amp = \frac{\partial}{\px}\big(f_x\big) = \frac{\partial}{\px}\big(3x^2y^2+2y^3-\sin(x) \big) + \amp = 6xy^2-\cos(x) + f_{yy}(x,y) \amp = \frac{\partial}{\py}\big(f_y\big) = \frac{\partial}{\py}\big(2x^3y+6xy^2\big) + \amp = 2x^3+12xy + f_{xy}(x,y) \amp = \frac{\partial}{\py}\big(f_x\big) = \frac{\partial}{\py}\big(3x^2y^2+2y^3-\sin(x) \big) + \amp = 6x^2y+6y^2 + f_{yx}(x,y) \amp = \frac{\partial}{\px}\big(f_x\big) = \frac{\partial}{\px}\big(2x^3y+6xy^2\big) + \amp = 6x^2y+6y^2 + +

      +
      2. + +
    2. +

      + + f(x,y) \amp = \frac{x^3}{y^2} = x^3y^{-2} + f_x(x,y) \amp = \frac{3x^2}{y^2} + f_y(x,y) \amp = -\frac{2x^3}{y^3} + f_{xx}(x,y) \amp = \frac{\partial}{\px}\big(f_x\big) = \frac{\partial}{\px}\big(\frac{3x^2}{y^2}\big) + \amp = \frac{6x}{y^2} + f_{yy}(x,y) \amp = \frac{\partial}{\py}\big(f_y\big) = \frac{\partial}{\py}\big(-\frac{2x^3}{y^3}\big) + \amp = \frac{6x^3}{y^4} + f_{xy}(x,y) \amp = \frac{\partial}{\py}\big(f_x\big) = \frac{\partial}{\py}\big(\frac{3x^2}{y^2}\big) + \amp = -\frac{6x^2}{y^3} + f_{yx}(x,y) \amp = \frac{\partial}{\px}\big(f_x\big) = \frac{\partial}{\px}\big(-\frac{2x^3}{y^3}\big) + \amp = -\frac{6x^2}{y^3} + +

      +
      3. + +
    3. +

      + f(x,y) = e^x\sin(x^2y) Because the following partial derivatives get rather long, + we omit the extra notation and just give the results. + In several cases, + multiple applications of the Product and Chain Rules will be necessary, + followed by some basic combination of like terms. + + f_x(x,y) \amp = e^x\sin(x^2y) + 2xye^x\cos(x^2y) + f_y(x,y) \amp = x^2e^x\cos(x^2y) + f_{xx}(x,y) \amp = e^x\sin(x^2y)+4xye^x\cos(x^2y)+2ye^x\cos(x^2y)-4x^2y^2e^x\sin(x^2y) + f_{yy}(x,y) \amp = -x^4e^x\sin(x^2y) + f_{xy}(x,y) \amp = x^2e^x\cos(x^2y)+2xe^x\cos(x^2y)-2x^3ye^x\sin(x^2y) + f_{yx}(x,y) \amp = x^2e^x\cos(x^2y)+2xe^x\cos(x^2y)-2x^3ye^x\sin(x^2y) + +

      +
      4. +
    +

    +     +     + + + +

    + Notice how in each of the three functions in , + f_{xy} = f_{yx}. + Due to the complexity of the examples, + this likely is not a coincidence. + The following theorem states that it is not. +

    + + + Mixed Partial Derivatives + +

    + Let f be defined such that f_{xy} and f_{yx} are continuous on a set S. + Then for each point (x,y) in S, + f_{xy}(x,y) = f_{yx}(x,y). +

    +     +     + +

    + Finding f_{xy} and f_{yx} independently and comparing the results provides a convenient way of checking our work. +

    +     + + + Understanding Second Partial Derivatives +

    + Now that we know how to find second partials, + we investigate what they tell us. +

    + +

    + Again we refer back to a function y=f(x) of a single variable. + The second derivative of f is + the derivative of the derivative, + or the rate of change of the rate of change. + The second derivative measures how much the derivative is changing. + If \fp'(x)\lt 0, + then the derivative is getting smaller (so the graph of f is concave down); + if \fp'(x) \gt 0, then the derivative is growing, + making the graph of f concave up. +

    + +

    + Now consider z=f(x,y). + Similar statements can be made about f_{xx} and f_{yy} as could be made about \fp'(x) above. + When taking derivatives with respect to x twice, + we measure how much f_x changes with respect to x. + If f_{xx}(x,y)\lt 0, + it means that as x increases, f_x decreases, + and the graph of f will be concave down + in the x-direction. + Using the analogy of standing in the rolling meadow used earlier in this section, + f_{xx} measures whether one's path is concave up/down when walking due east. +

    + +

    + Similarly, f_{yy} measures the concavity in the y-direction. + If f_{yy}(x,y) \gt 0, + then f_y is increasing with respect to y and the graph of f will be concave up in the y-direction. + Appealing to the rolling meadow analogy again, + f_{yy} measures whether one's path is concave up/down when walking due north. +

    + +

    + We now consider the mixed partials f_{xy} and f_{yx}. + The mixed partial f_{xy} measures how much f_x changes with respect to y. + Once again using the rolling meadow analogy, + f_{x} measures the slope if one walks due east. + Looking east, begin walking north + (side-stepping). + Is the path towards the east getting steeper? + If so, f_{xy} \gt 0. + Is the path towards the east not changing in steepness? + If so, then f_{xy}=0. + A similar thing can be said about f_{yx}: + consider the steepness of paths heading north while side-stepping to the east. +

    + +

    + The following example examines these ideas with concrete numbers and graphs. +

    + + + Understanding second partial derivatives + +

    + Let z=x^2-y^2+xy. + Evaluate the 6 first and second partial derivatives at + (-1/2,1/2) and interpret what each of these numbers mean. +

    +     + +

    + We find that: +

    + +

    + f_x(x,y) = 2x+y,f_y(x,y) = -2y+x,f_{xx}(x,y) = 2, + f_{yy}(x,y) = -2 and f_{xy}(x,y) = f_{yx}(x,y) = 1. + Thus at (-1/2,1/2) we have + + f_x(-1/2,1/2) = -1/2,\qquad f_y(-1/2,1/2) = -3/2 + . +

    + +

    + The slope of the tangent line at + (-1/2, 1/2, -1/4) in the direction of x is -1/2: + if one moves from that point parallel to the x-axis, + the instantaneous rate of change will be -1/2. + The slope of the tangent line + at this point in the direction of y is -3/2: + if one moves from this point parallel to the y-axis, + the instantaneous rate of change will be -3/2. + These tangents lines are graphed in and , + respectively, where the tangent lines are drawn in a solid line. +

    + +
    + Understanding the second partial derivatives in + +
    + + + + + A hyperbolic paraboloid, or saddle surface. A trace of constant x value is shown, which has the shape of a downward parabola. + +

    + The surface z=x^2-y^2+xy is a hyperbolic paraboloid, or saddle surface. + It is plotted along with the trace x=-\frac12. + This is a curve lying on the surface that has the shape of a downward-opening parabola. + The fact that this curve is concave down, when viewed along the x axis, + corresponds to the fact that f_{yy} is negative. +

    + +

    + At three points along this trace, tangent lines are drawn. + These lines are tangent to curves on the surface given by traces of constant y, + for three different values of y. Each one has negative slope, relative to x. + For larger values of y, the slope of these lines, relative to x, becomes less negative. + This suggests that the slope, given by f_x is increasing with y, + and this corresponds to the fact that f_{xy} is positive. +

    +     + + + + + //ASY file for figpartial63D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(2.5,8.4,3.4); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-2,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=x^2-y^2+xy + triple f(pair t) { + return (t.x,t.y,t.x^2-t.y^2+t.x*t.y); + } + surface s=surface(f,(-1,-1),(1,1),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the point (-0.5,0.5,-0.25) trace for x=-0.5 + dotfactor=3;dot((-0.5,0.5,-0.25)); + triple g(real t) {return (-0.5,t,0.25-t^2-0.5*t);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the tangent lines in x direction, slope -0.5 + // L=(-0.5,0.5,-0.25)+t(1,0,-0.5) for t=-1,1 + draw((-1.5,0.5,0.25)--(0.5,0.5,-0.75),bluepen); + + //Draw the tangent lines in x direction, slope -0.75 and dot + // L=(-0.5,0.25,0.065)+t(1,0,-0.75) for t=-1,1 + dot((-0.5,0.25,0.065)); + draw((-1.5,0.25,0.815)--(0.5,0.25,-0.685),dashed+bluepen); + + //Draw the tangent lines in x direction, slope -0.25 and dot + // L=(-0.5,0.75,-0.6875)+t(1,0,-0.25) for t=-1,1 + dot((-0.5,0.75,-0.6875)); + draw((-1.5,0.75,-0.4375)--(0.5,0.75,-0.9375),dashed+bluepen); + + + + +
    + +
    + + + + + A hyperbolic paraboloid, or saddle surface. A trace of constant y value is shown, which has the shape of an upward parabola. + +

    + The surface z=x^2-y^2+xy is a hyperbolic paraboloid, or saddle surface. + It is plotted along with the trace y=\frac12. + This is a curve lying on the surface that has the shape of an upward-opening parabola. + The fact that this curve is concave up, when viewed along the y axis, + corresponds to the fact that f_{xx} is positive. +

    + +

    + At three points along this trace, tangent lines are drawn. + These lines are tangent to curves on the surface given by traces of constant x, + for three different values of x. Each one has negative slope, relative to y. + For larger values of x, the slope of these lines, relative to y, becomes less negative. + This suggests that the slope, given by f_y is increasing with x, + and this corresponds to the fact that f_{yx} is positive. +

    +     + + + + + //ASY file for figpartial6b3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(2.5,8.4,3.); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-2,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=x^2-y^2+xy + triple f(pair t) { + return (t.x,t.y,t.x^2-t.y^2+t.x*t.y); + } + surface s=surface(f,(-1,-1),(1,1),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the point (-0.5,0.5,-0.25) and trace for y=0.5 + dotfactor=3;dot((-0.5,0.5,-0.25)); + triple g(real t) {return (t,0.5,t^2-0.25+0.5*t);} + path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); + + //Draw the tangent lines in y direction, slope -1.5 + // L=(-0.5,0.5,-0.25)+t(0,1,-1.5) for t=-1,1 + draw((-0.5,-0.3,.95)--(-0.5,1.3,-1.45),bluepen); + + //Draw the tangent lines in y direction, slope -1.75 and dot + // L=(-0.75,0.5,-0.0625)+t(0,1,-1.75) for t=-1,1 + dot((-0.75,0.5,-0.0625)); + draw((-0.75, -0.3, 1.3375)--(-0.75, 1.3, -1.4625),dashed+bluepen); + + //Draw the tangent lines in y direction, slope -1.25 and dot + // L=(-0.25,0.5,-0.3125)+t(0,1,-1.25) for t=-1,1 + dot((-0.25,0.5,-0.3125)); + draw((-0.25, -0.3, 0.6875)--(-0.25, 1.3, -1.3125),dashed+bluepen); + + + + +
    +     +
    + +

    + Now consider only . + Three directed tangent lines are drawn + (two are dashed), + each in the direction of x; + that is, each has a slope determined by f_x. + Note how as y increases, + the slope of these lines get closer to 0. + Since the slopes are all negative, + getting closer to 0 means the slopes are increasing. + The slopes given by f_x are increasing as y increases, + meaning f_{xy} must be positive. +

    + +

    + Since f_{xy}=f_{yx}, + we also expect f_y to increase as x increases. + Consider where again three directed tangent lines are drawn, + this time each in the direction of y with slopes determined by f_y. + As x increases, the slopes become less steep + (closer to 0). + Since these are negative slopes, + this means the slopes are increasing. +

    + +

    + Thus far we have a visual understanding of f_x, + f_y, and f_{xy}=f_{yx}. + We now interpret f_{xx} and f_{yy}. + In , + we see a curve drawn where x is held constant at x=-1/2: + only y varies. + This curve is clearly concave down, + corresponding to the fact that f_{yy}\lt 0. + In part of the figure, + we see a similar curve where y is constant and only x varies. + This curve is concave up, corresponding to the fact that f_{xx} \gt 0. +

    +     +     +     + + + Partial Derivatives and Functions of Three Variables +

    + The concepts underlying partial derivatives can be easily extend to more than two variables. + We give some definitions and examples in the case of three variables and trust the reader can extend these definitions to more variables if needed. +

    + + + Partial Derivatives with Three Variables + +

    + Let w=f(x,y,z) be a continuous function on a set D in \mathbb{R}^3. +

    + +

    + The partial derivative of f with respect to x is: + + f_x(x,y,z) = \lim_{h\to 0} \frac{f(x+h,y,z)-f(x,y,z)}{h} + . +

    + +

    + Similar definitions hold for + f_y(x,y,z) and f_z(x,y,z). + derivativepartial + partial derivative +

    +     +     + +

    + By taking partial derivatives of partial derivatives, + we can find second partial derivatives of f with respect to z then y, + for instance, just as before. +

    + + + Partial derivatives of functions of three variables + +

    + For each of the following, + find f_x, f_y, f_z, f_{xz}, f_{yz}, + and f_{zz}. +

    + +

    +

      +
    1. f(x,y,z) = x^2y^3z^4+x^2y^2+x^3z^3+y^4z^4
      2. + +
    2. f(x,y,z) = x\sin(yz)
      3. +
    +

    +     + +

    +

      +
    1. +

      + + f_x(x,y,z) \amp = 2xy^3z^4+2xy^2+3x^2z^3 + f_y(x,y,z) \amp = 3x^2y^2z^4+2x^2y+4y^3z^4 + f_z(x,y,z) \amp = 4x^2y^3z^3+3x^3z^2+4y^4z^3 + f_{xz}(x,y,z) \amp = 8xy^3z^3+9x^2z^2 + f_{yz}(x,y,z) \amp = 12x^2y^2z^3+16y^3z^3 + f_{zz}(x,y,z) \amp = 12x^2y^3z^2+6x^3z+12y^4z^2 + +

      +
      2. + +
    2. +

      + f_x = \sin(yz); f_y = xz\cos(yz); f_z = xy\cos(yz), and + + f_{xz}(x,y,z) \amp = y\cos(yz) + f_{yz}(x,y,z) \amp = x\cos(yz) - xyz\sin(yz) + f_{zz}(x,y,z) \amp = -xy^2\sin(yz) + +

      +
      3. +
    +

    +     +     +     + + + Higher Order Partial Derivatives +

    + We can continue taking partial derivatives of partial derivatives of partial derivatives of ; + we do not have to stop with second partial derivatives. + These higher order partial derivatives do not have a tidy graphical interpretation; + nevertheless they are not hard to compute and worthy of some practice. + partial derivativehigh order +

    + +

    + We do not formally define each higher order derivative, + but rather give just a few examples of the notation. + + f_{xyx}(x,y) \amp = \frac{\partial}{\px}\left(\frac{\partial}{\py}\left(\frac{\pf}{\px}\right)\right) \text{ and } + f_{xyz}(x,y,z) \amp =\frac{\partial}{\partial z}\left(\frac{\partial}{\py}\left(\frac{\pf}{\px}\right)\right) + . +

    + + + Higher order partial derivatives + +

    +

      +
    1. +

      + Let f(x,y) = x^2y^2+\sin(xy). + Find f_{xxy} and f_{yxx}. +

      +
      2. + +
    2. +

      + Let f(x,y,z) = x^3e^{xy}+\cos(z). + Find f_{xyz}. +

      +
      3. +
    +

    +

    + + +

    +

      +
    1. +

      + To find f_{xxy}, we first find f_x, + then f_{xx}, then f_{xxy}: + + f_x(x,y) \amp = 2xy^2+y\cos(xy) + f_{xx}(x,y) \amp = 2y^2-y^2\sin(xy) + f_{xxy}(x,y) \amp = 4y-2y\sin(xy) - xy^2\cos(xy) + . + To find f_{yxx}, we first find f_y, + then f_{yx}, then f_{yxx}: + + f_y(x,y) \amp = 2x^2y+x\cos(xy) + f_{yx}(x,y) \amp = 4xy + \cos(xy) - xy\sin(xy) + f_{yxx}(x,y) \amp = 4y-y\sin(xy) - \big(y\sin(xy) + xy^2\cos(xy)\big) + \amp = 4y-2y\sin(xy)-xy^2\cos(xy) + . + Note how f_{xxy} = f_{yxx}. +

      +
      2. + +
    2. +

      + To find f_{xyz}, we find f_x, + then f_{xy}, then f_{xyz}: + + f_x(x,y,z) \amp = 3x^2e^{xy}+ x^3ye^{xy} + f_{xy}(x,y,z) \amp= 3x^3e^{xy}+x^3e^{xy}+x^4ye^{xy} + \amp = 4x^3e^{xy}+x^4ye^{xy} + f_{xyz}(x,y,z) \amp = 0 + . +

      +
      3. +
    +

    + +     + +

    + In the previous example we saw that f_{xxy} = f_{yxx}; + this is not a coincidence. + While we do not state this as a formal theorem, + as long as each partial derivative is continuous, + it does not matter the order in which the partial derivatives are taken. + For instance, f_{xxy} = f_{xyx} = f_{yxx}. +

    + +

    + This can be useful at times. + Had we known this, + the second part of + would have been much simpler to compute. + Instead of computing f_{xyz} in the x, + y then z orders, + we could have applied the z, + then x then y order + (as f_{xyz} = f_{zxy}). + It is easy to see that f_z = -\sin(z); + then f_{zx} and f_{zxy} are clearly 0 as f_z does not contain an x or y. +

    + +

    + A brief review of this section: + partial derivatives measure the instantaneous rate of change of a multivariable function with respect to one variable. + With z=f(x,y), + the partial derivatives f_x and f_y measure the instantaneous rate of change of z when moving parallel to the x- and y-axes, + respectively. + How do we measure the rate of change at a point when we do not move parallel to one of these axes? + What if we move in the direction given by the vector \la 2,1\ra? + Can we measure that rate of change? + The answer is, of course, yes, we can. + This is the topic of . + First, we need to define what it means for a function of two variables to be + differentiable. +

    +     + + + + Terms and Concepts + + + +

    + What is the difference between a constant and a coefficient? +

    +     + + + +

    + A constant is a number that is added or subtracted in an expression; + a coefficient is a number that is being multiplied by a nonconstant function. +

    + +

    + Coefficients are often constant, but they don't have to be. +

    +     + +     + + + + +

    + Given a function f(x,y), + explain in your own words how to compute f_x. +

    +     + + + +

    + Answers will vary; + each should include something about treating y as a constant or a coefficient. +

    +     + +     + + + + + +

    + In the mixed partial fraction f_{xy}, + which is computed first, f_x or f_y? +

    + + +     + + + +

    + f_x +

    +     +     + + +

    + f_y +

    +     +     +     + +     + + + + + + +

    + In the mixed partial fraction \frac{\partial^2f}{\partial x\partial y}, + which is computed first, f_x or f_y? +

    + + +     + + + +

    + f_x +

    +     +     + + +

    + f_y +

    +     +     +     + +     +     + + + Problems + + + +

    + Evaluate f_x(x,y) and + f_y(x,y) at the indicated point. +

    +     + + + + +

    + f(x,y) = x^2y-x+2y+3 at (1,2) +

    +     + +

    + f_x=2xy-1, f_y=x^2+2 +

    + +

    + f_x(1,2) = 3, f_y(1,2) = 3 +

    +     + +     + + + + + $fx=0; + $fy=0; + + + +

    + f(x,y) = x^3-3x+y^2-6y at (-1,3). +

    + + + Find f_x(-1,3). + +

    + +

    + + + Find f_y(-1,3). + +

    + +

    +     +     +     + + + + +

    + f(x,y) = \sin(y) \cos(x) at (\pi/3,\pi/3) +

    +     + +

    + f_x=-\sin(x) \sin(y), f_y=\cos(x) \cos(y) +

    + +

    + f_x(\pi/3,\pi/3) = -3/4, + f_y(\pi/3,\pi/3) = 1/4 +

    +     + +     + + + + + Context("Fraction"); + $fx=Fraction(-1,2); + $fy=Fraction(-1,3); + + + +

    + f(x,y) = \ln(xy) at (-2,-3) + Find: +

    + + + Find f_x(-2,-3). + +

    + +

    + + + Find f_y(-2,-3). + +

    + +

    +     +     +     + +     + + + +

    + Find f_x, f_y, + f_{xx}, f_{yy}, + f_{xy} and f_{yx}. +

    +     + + + + +

    + f(x,y) = x^2y+3x^2+4y-5 +

    +     + +

    + f_x=2xy+6x, f_y=x^2+4 +

    + +

    + f_{xx}=2y+6, f_{yy}=0 +

    + +

    + f_{xy}=2x, f_{yx}=2x +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $fx=Compute("3x^2+6xy+3y^2"); + $fy=Compute("3x^2+6xy+3y^2"); + $fxx=Compute("6x+6y"); + $fxy=Compute("6x+6y"); + $fyx=Compute("6x+6y"); + $fyy=Compute("6x+6y"); + + + +

    + f(x,y) = y^3+3xy^2+3x^2y+x^3 +

    + + + Find f_{x}(x,y) + +

    + +

    + + + Find f_{y}(x,y). + +

    + +

    + + + Find f_{xx}(x,y). + +

    + +

    + + + Find f_{xy}(x,y). + +

    + +

    + + + Find f_{yx}(x,y) + +

    + +

    + + + Find f_{yy}(x,y). + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) = \frac xy +

    +     + +

    + f_x=1/y, f_y=-x/y^2 +

    + +

    + f_{xx}=0, f_{yy}=2x/y^3 +

    + +

    + f_{xy}=-1/y^2, f_{yx}=-1/y^2 +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $fx=Compute("-4/(x^2y)"); + $fy=Compute("-4/(xy^2)"); + $fxx=Compute("8/(x^3y)"); + $fxy=Compute("4/(x^2y^2)"); + $fyx=Compute("4/(x^2y^2)"); + $fyy=Compute("8/(xy^3)"); + + + +

    + f(x,y) = \frac{4}{xy} +

    + + + Find f_{x}(x,y) + +

    + +

    + + + Find f_{y}(x,y). + +

    + +

    + + + Find f_{xx}(x,y). + +

    + +

    + + + Find f_{xy}(x,y). + +

    + +

    + + + Find f_{yx}(x,y) + +

    + +

    + + + Find f_{yy}(x,y). + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) = e^{x^2+y^2} +

    +     + +

    + f_x=2xe^{x^2+y^2}, f_y=2ye^{x^2+y^2} +

    + +

    + f_{xx}=2e^{x^2+y^2}+4x^2e^{x^2+y^2}, + f_{yy}=2e^{x^2+y^2}+4y^2e^{x^2+y^2} +

    + +

    + f_{xy}=4xye^{x^2+y^2}, + f_{yx}=4xye^{x^2+y^2} +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $fx=Compute("e^(x+2y)"); + $fy=Compute("2e^(x+2y)"); + $fxx=Compute("e^(x+2y)"); + $fxy=Compute("2e^(x+2y)"); + $fyx=Compute("2e^(x+2y)"); + $fyy=Compute("4e^(x+2y)"); + + + +

    + f(x,y) = e^{x+2y} +

    + + + Find f_{x}(x,y) + +

    + +

    + + + Find f_{y}(x,y). + +

    + +

    + + + Find f_{xx}(x,y). + +

    + +

    + + + Find f_{xy}(x,y). + +

    + +

    + + + Find f_{yx}(x,y) + +

    + +

    + + + Find f_{yy}(x,y). + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) = \sin(x) \cos(y) +

    +     + +

    + f_x=\cos(x) \cos(y), f_y=-\sin(x) \sin(y) +

    + +

    + f_{xx}=-\sin(x) \cos(y), + f_{yy}=-\sin(x) \cos(y) +

    + +

    + f_{xy}=-\sin(y) \cos(x), + f_{yx}=-\sin(y) \cos(x) +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $fx=Compute("3(x+y)^2"); + $fy=Compute("3(x+y)^2"); + $fxx=Compute("6(x+y)"); + $fxy=Compute("6(x+y)"); + $fyx=Compute("6(x+y)"); + $fyy=Compute("6(x+y)"); + + + +

    + f(x,y) = (x+y)^3 +

    + + + Find f_{x}(x,y) + +

    + +

    + + + Find f_{y}(x,y). + +

    + +

    + + + Find f_{xx}(x,y). + +

    + +

    + + + Find f_{xy}(x,y). + +

    + +

    + + + Find f_{yx}(x,y) + +

    + +

    + + + Find f_{yy}(x,y). + +

    + +

    +     +     +     + + + + +

    + f(x,y) = \cos(5xy^3) +

    +     + +

    + f_x=-5y^3\sin(5xy^3), f_y=-15xy^2\sin(5xy^3) +

    + +

    + f_{xx}=-25y^6\cos(5xy^3), + f_{yy}=-225x^2y^4\cos(5xy^3)-30xy\sin(5xy^3) +

    + +

    + f_{xy}=-75xy^5\cos(5xy^3)-15y^2\sin(5xy^3), + f_{yx}=-75xy^5\cos(5xy^3)-15y^2\sin(5xy^3) +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $fx=Compute("10x cos(5x^2+2y^3)"); + $fy=Compute("6y^2 cos(5x^2+2y^3)"); + $fxx=Compute("10 cos(5x^2+2y^3)-100x^2 sin(5x^2+2y^3)"); + $fxy=Compute("-60xy^2 sin(5x^2+2y^3)"); + $fyx=Compute("-60xy^2 sin(5x^2+2y^3)"); + $fyy=Compute("12y cos(5x^2+2y^3)-36y^4 sin(5x^2+2y^3)"); + + + +

    + f(x,y) = \sin\mathopen{}\left(5x^2+2y^3\right)\mathclose{} +

    + + + Find f_{x}(x,y) + +

    + +

    + + + Find f_{y}(x,y). + +

    + +

    + + + Find f_{xx}(x,y). + +

    + +

    + + + Find f_{xy}(x,y). + +

    + +

    + + + Find f_{yx}(x,y) + +

    + +

    + + + Find f_{yy}(x,y). + +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real'); + $fx=Compute("2y^2/sqrt(4xy^2+1)"); + $fy=Compute("4xy/sqrt(4xy^2+1)"); + $fxx=Compute("-4y^4/sqrt(4xy^2+1)^3"); + $fxy=Compute("-8xy^3/sqrt(4xy^2+1)^3+4y/sqrt(4xy^2+1)"); + $fyx=Compute("-8xy^3/sqrt(4xy^2+1)^3+4y/sqrt(4xy^2+1)"); + $fyy=Compute("-16x^2y^2/sqrt(4xy^2+1)^3+4x/sqrt(4xy^2+1)"); + + + +

    + f(x,y) = \sqrt{4xy^2+1} +

    + + + Find f_{x}(x,y) + +

    + +

    + + + Find f_{y}(x,y). + +

    + +

    + + + Find f_{xx}(x,y). + +

    + +

    + + + Find f_{xy}(x,y). + +

    + +

    + + + Find f_{yx}(x,y) + +

    + +

    + + + Find f_{yy}(x,y). + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) =(2x+5y)\sqrt{y} +

    +     + +

    + f_x=2\sqrt{y}, f_y=5\sqrt{y}+\frac{2x+5y}{2\sqrt{y}} +

    + +

    + f_{xx}=0, + f_{yy}=\frac{5}{\sqrt{y}}-\frac{2x+5y}{4y^{3/2}} +

    + +

    + f_{xy}=\frac{1}{\sqrt{y}}, + f_{yx}=\frac{1}{\sqrt{y}} +

    +     + +     + + + + +

    + \ds f(x,y) =\frac{1}{x^2+y^2+1} +

    +     + +

    + f_x=-\frac{2x}{(x^2+y^2+1)^2}, + f_y=-\frac{2y}{(x^2+y^2+1)^2} +

    + +

    + f_{xx}=\frac{8x^2}{(x^2+y^2+1)^3}-\frac{2}{(x^2+y^2+1)^2}, + f_{yy}=\frac{8y^2}{(x^2+y^2+1)^3}-\frac{2}{(x^2+y^2+1)^2} +

    + +

    + f_{xy}=\frac{8xy}{(x^2+y^2+1)^3}, + f_{yx}=\frac{8xy}{(x^2+y^2+1)^3} +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $fx=Compute("5"); + $fy=Compute("-17"); + $fxx=Compute("0"); + $fxy=Compute("0"); + $fyx=Compute("0"); + $fyy=Compute("0"); + + + +

    + f(x,y) = 5x-17y +

    + + + Find f_{x}(x,y) + +

    + +

    + + + Find f_{y}(x,y). + +

    + +

    + + + Find f_{xx}(x,y). + +

    + +

    + + + Find f_{xy}(x,y). + +

    + +

    + + + Find f_{yx}(x,y) + +

    + +

    + + + Find f_{yy}(x,y). + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) =3x^2+1 +

    +     + +

    + f_x=6x, f_y=0 +

    + +

    + f_{xx}=6, f_{yy}=0 +

    + +

    + f_{xy}=0, f_{yx}=0 +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $fx=Compute("2x/(x^2+y)"); + $fy=Compute("1/(x^2+y)"); + $fxx=Compute("-4x^2/(x^2+y)^2+2/(x^2+y)"); + $fxy=Compute("-2x/(x^2+y)^2"); + $fyx=Compute("-2x/(x^2+y)^2"); + $fyy=Compute("-1/(x^2+y)^2"); + + + +

    + f(x,y) = \ln(x^2+y) +

    + + + Find f_{x}(x,y) + +

    + +

    + + + Find f_{y}(x,y). + +

    + +

    + + + Find f_{xx}(x,y). + +

    + +

    + + + Find f_{xy}(x,y). + +

    + +

    + + + Find f_{yx}(x,y) + +

    + +

    + + + Find f_{yy}(x,y). + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) =\frac{\ln(x) }{4y} +

    +     + +

    + f_x=\frac{1}{4xy}, f_y=-\frac{\ln(x) }{4y^2} +

    + +

    + f_{xx}=-\frac{1}{4x^2y}, + f_{yy}=\frac{\ln(x) }{2y^3} +

    + +

    + f_{xy}=-\frac{1}{4xy^2}, + f_{yx}=-\frac{1}{4xy^2} +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $fx=Compute("5e^x sin(y)"); + $fy=Compute("5e^x cos(y)"); + $fxx=Compute("5e^x sin(y)"); + $fxy=Compute("5e^x cos(y)"); + $fyx=Compute("5e^x cos(y)"); + $fyy=Compute("-5e^x sin(y)"); + + + +

    + f(x,y) = 5e^x\sin(y)+9 +

    + + + Find f_{x}(x,y) + +

    + +

    + + + Find f_{y}(x,y). + +

    + +

    + + + Find f_{xx}(x,y). + +

    + +

    + + + Find f_{xy}(x,y). + +

    + +

    + + + Find f_{yx}(x,y) + +

    + +

    + + + Find f_{yy}(x,y). + +

    + +

    +     +     +     + +     + + + +

    + Form a function f(x,y) such that f_x and f_y match those given. +

    +     + + + + +

    + f_x = \sin(y) +1,f_y = x\cos(y) +

    +     + +

    + f(x,y) = x\sin(y) + x+ C, + where C is any constant. +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $f=Compute("1/2x^2+xy+1/2y^2"); + $fev=$f->cmp(upToConstant=>1); + + + +

    + f_x = x+y and f_y = x+y +

    + +

    + +

    +     +     +     + + + + +

    + f_x = 6xy-4y^2,f_y = 3x^2-8xy+2 +

    +     + +

    + f(x,y) = 3x^2y-4xy^2+2y +C, + where C is any constant. +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $f=Compute("ln(x^2+y^2)"); + $fev=$f->cmp(upToConstant=>1); + + + +

    + f_x = \frac{2x}{x^2+y^2} and f_y = \frac{2y}{x^2+y^2} +

    + +

    + +

    +     +     +     + +     + + + +

    + Find f_x, + f_y, f_z, f_{yz} and f_{zy}. +

    +     + + + + +

    + \ds f(x,y,z) = x^2e^{2y-3z} +

    +     + +

    + f_x = 2xe^{2y-3z}, + f_y = 2x^2e^{2y-3z}, f_z = -3x^2e^{2y-3z} +

    + +

    + f_{yz} = -6x^2e^{2y-3z}, + f_{zy} = -6x^2e^{2y-3z} +

    +     + +     + + + + + Context()->variables->add(y=>'Real',z=>'Real'); + $fx=Compute("3x^2y^2+3x^2z"); + $fy=Compute("2x^3y+2yz"); + $fz=Compute("x^3+y^2"); + $fyz=Compute("2y"); + $fzy=Compute("2y"); + + + +

    + f(x,y,z) = x^3y^2+x^3z+y^2z +

    + + + Find f_{x}(x,y,z). + +

    + +

    + + + Find f_{y}(x,y,z). + +

    + +

    + + + Find f_{z}(x,y,z). + +

    + +

    + + + Find f_{yz}(x,y,z). + +

    + +

    + + + Find f_{zy}(x,y,z). + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y,z) = \frac{3x}{7y^2z} +

    +     + +

    + f_x = \frac{3}{7y^2z}, + f_y = -\frac{6x}{7y^3z}, + f_z = -\frac{3x}{7y^2z^2} +

    + +

    + f_{yz} = \frac{6x}{7y^3z^2}, + f_{zy} = \frac{6x}{7y^3z^2} +

    +     + +     + + + + + Context()->variables->add(y=>'Real',z=>'Real'); + $fx=Compute("1/x"); + $fy=Compute("1/y"); + $fz=Compute("1/z"); + $fyz=Compute("0"); + $fzy=Compute("0"); + + + +

    + f(x,y,z) = \ln(xyz) +

    + + + Find f_{x}(x,y,z). + +

    + +

    + + + Find f_{y}(x,y,z). + +

    + +

    + + + Find f_{z}(x,y,z). + +

    + +

    + + + Find f_{yz}(x,y,z). + +

    + +

    + + + Find f_{zy}(x,y,z). + +

    + +

    +     +     +     + +     +     +     +
    +
    + Differentiability and the Total Differential + +

    + We studied differentials + in , + where + states that if y=f(x) and f is differentiable, + then dy=\fp(x)dx. + One important use of this differential is in Integration by Substitution. + Another important application is approximation. + Let \dx = dx represent a change in x. + When dx is small, dy\approx \dy, + the change in y resulting from the change in x. + Fundamental in this understanding is this: + as dx gets small, + the difference between \dy and dy goes to 0. + Another way of stating this: + as dx goes to 0, the error + in approximating \dy with dy goes to 0. +

    + +

    + We extend this idea to functions of two variables. + Let z=f(x,y), and let + \dx = dx and \dy=dy represent changes in x and y, + respectively. + Let \ddz = f(x+dx,y+dy) - f(x,y) be the change in z over the change in x and y. + Recalling that f_x and f_y give the instantaneous rates of z-change in the x- and y-directions, + respectively, + we can approximate \ddz with dz = f_xdx+f_ydy; + in words, the total change in z is approximately the change caused by changing x plus the change caused by changing y. + In a moment we give an indication of whether or not this approximation is any good. + First we give a name to dz. +

    +     + + + The Total Differential + + + Total Differential + +

    + Let z=f(x,y) be continuous on a set S. + Let dx and dy represent changes in x and y, + respectively. + Where the partial derivatives f_x and f_y exist, + the total differential of z is + total differential + partial derivativetotal differential + + dz = f_x(x,y)\,dx + f_y(x,y)\,dy + . +

    +     +     + + + Finding the total differential + +

    + Let z = x^4e^{3y}. + Find dz. +

    +     + +

    + We compute the partial derivatives: + f_x = 4x^3e^{3y} and f_y = 3x^4e^{3y}. + Following , we have + + dz = 4x^3e^{3y}dx+3x^4e^{3y}dy + . +

    +     +     + + + +

    + We can approximate \ddz with dz, + but as with all approximations, there is error involved. + A good approximation is one in which the error is small. + At a given point (x_0,y_0), + let E_x and E_y be functions of dx and dy such that + E_xdx+E_ydy describes this error. + Then + + \ddz \amp = dz + E_xdx+ E_ydy + \amp = f_x(x_0,y_0)dx+f_y(x_0,y_0)dy + E_xdx+E_ydy + . +

    + +

    + If the approximation of \ddz by dz is good, + then as dx and dy get small, + so does E_xdx+E_ydy. + The approximation of \ddz by dz is even better if, + as dx and dy go to 0, so do E_x and E_y. + This leads us to our definition of differentiability. +

    + + + Multivariable Differentiability + +

    + Let z=f(x,y) be defined on a set S containing (x_0,y_0) where + f_x(x_0,y_0) and f_y(x_0,y_0) exist. + Let dz be the total differential of z at (x_0,y_0), + let \ddz = f(x_0+dx,y_0+dy) - f(x_0,y_0), + and let E_x and E_y be functions of dx and dy such that + + \ddz = dz + E_xdx + E_ydy + . +

    + +

    +

      +
    1. +

      + We say f is differentiable at (x_0,y_0) if, + given \varepsilon \gt 0, + there is a \delta \gt 0 such that if \norm{\la dx,dy\ra} \lt \delta, + then \norm{\la E_x,E_y\ra} \lt \varepsilon. + That is, as dx and dy go to 0, so do E_x and E_y. +

      +
      2. + +
    2. +

      + We say f is differentiable on S + if f is differentiable at every point in S. + If f is differentiable on \mathbb{R}^2, + we say that f is differentiable everywhere. + differentiable + derivativemultivariable differentiability + multivariable functiondifferentiability +

      +
      3. +
    +

    +     +     + + + + + Showing a function is differentiable + +

    + Show f(x,y) = xy+3y^2 is differentiable using . +

    +     + +

    + We begin by finding f(x+dx,y+dy), + \ddz, f_x and f_y. + + f(x+dx,y+dy) \amp = (x+dx)(y+dy) + 3(y+dy)^2 + \amp = xy + xdy+ydx+dxdy + 3y^2+6ydy+3dy^2 + . +

    + +

    + \ddz = f(x+dx,y+dy) - f(x,y), so + + \ddz = xdy + ydx + dxdy + 6ydy+3dy^2 + . +

    + +

    + It is straightforward to compute f_x = y and f_y = x+6y. + Consider once more \ddz: + + \ddz \amp = xdy + ydx + dxdy + 6ydy+3dy^2 \qquad \text{ (now reorder) } + \amp = ydx + xdy+6ydy+ dxdy + 3dy^2 + \amp = \underbrace{(y)}_{f_x}dx + \underbrace{(x+6y)}_{f_y}dy + \underbrace{(dy)}_{E_x}dx+\underbrace{(3dy)}_{E_y}dy + \amp = f_xdx + f_ydy + E_xdx+E_ydy + . +

    + +

    + With E_x = dy and E_y = 3dy, + it is clear that as dx and dy go to 0, E_x and E_y also go to 0. + Since this did not depend on a specific point (x_0,y_0), + we can say that f(x,y) is differentiable for all pairs (x,y) in \mathbb{R}^2, + or, equivalently, that f is differentiable everywhere. +

    +     +     + + + +

    + Our intuitive understanding of differentiability of functions y=f(x) of one variable was that the graph of f was smooth. + A similar intuitive understanding of functions + z=f(x,y) of two variables is that the surface defined by f is also smooth, + not containing cusps, edges, breaks, etc. + The following theorem states that differentiable functions are continuous, + followed by another theorem that provides a + more tangible way of determining whether a great number of functions are differentiable or not. +

    + + + Continuity and Differentiability of Multivariable Functions + +

    + Let z=f(x,y) be defined on a set S containing (x_0,y_0). + If f is differentiable at (x_0,y_0), + then f is continuous at (x_0,y_0). + multivariable functiondifferentiability + multivariable functioncontinuity +

    +     +     + + + Differentiability of Multivariable Functions + +

    + Let z=f(x,y) be defined on a set S. + If f_x and f_y are both continuous on S, + then f is differentiable on S. + multivariable functiondifferentiability +

    +     +     + +

    + The above theorems assure us that essentially all functions that we see in the course of our studies here are differentiable + (and hence continuous) + on their natural domains. + There is a difference between + and , though: + it is possible for a function f to be differentiable yet f_x and/or f_y is + not continuous. + Such strange behavior of functions is a source of delight for many mathematicians, + but in practical situations we want to avoid it, leading to the following definition. +

    + + + Continouously Differentiable Function + + +

    + Let U be an open subset of \R^2. + We say that a function f is continuously differentiable on U + if f_x and f_y are defined and continuous at each point in U. + continuously differentiable +

    + +

    + A similar statement applies for functions of three variables in \R^3. +

    +     +     + +

    + When f_x and f_y exist at a point but are not continuous at that point, + we need to use other methods to determine whether or not f is differentiable at that point. +

    + +

    + For instance, consider the function + + f(x,y) = \begin{cases} + \frac{xy}{x^2+y^2} \amp (x,y)\neq (0,0) \\ + 0 \amp (x,y) = (0,0) + \end{cases} + . +

    + +

    + We can find f_x(0,0) and + f_y(0,0) using : + + f_x(0,0) \amp = \lim_{h\to 0} \frac{f(0+h,0) - f(0,0)}{h} + \amp = \lim_{h\to 0} \frac{0}{h^2} = 0; + f_y(0,0) \amp = \lim_{h\to 0} \frac{f(0,0+h) - f(0,0)}{h} + \amp = \lim_{h\to 0} \frac{0}{h^2} = 0 + . +

    + +

    + Both f_x and f_y exist at (0,0), + but they are not continuous at (0,0), as + + f_x(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)^2} \qquad \text{ and } \qquad f_y(x,y) = \frac{x(x^2-y^2)}{(x^2+y^2)^2} + + are not continuous at (0,0). (Take the limit of f_x as + (x,y)\to(0,0) along the x- and y-axes; + they give different results.) So even though f_x and f_y exist + at every point in the xy-plane, they are not continuous. + Therefore it is possible, + by , + for f to not be differentiable. +

    + +

    + Indeed, it is not. + One can show that f is not continuous at (0,0) + (see ), + and by , + this means f is not differentiable at (0,0). +

    +     + + + Approximating with the Total Differential +

    + By the definition, + when f is differentiable dz is a good approximation for \ddz when dx and dy are small. + We give some simple examples of how this is used here. +

    + + + Approximating with the total differential + +

    + Let z = \sqrt{x}\sin(y). + Approximate f(4.1,0.8). +

    +     + +

    + Recognizing that \pi/4 \approx 0.785\approx 0.8, + we can approximate f(4.1,0.8) using f(4,\pi/4). + We can easily compute f(4,\pi/4) = \sqrt{4}\sin(\pi/4) = 2\left(\frac{\sqrt{2}}2\right) = \sqrt{2}\approx 1.414. + Without calculus, + this is the best approximation we could reasonably come up with. + The total differential gives us a way of adjusting this initial approximation to hopefully get a more accurate answer. +

    + +

    + We let \ddz = f(4.1,0.8) - f(4,\pi/4). + The total differential dz is approximately equal to \ddz, so + + f(4.1,0.8) - f(4,\pi/4) \approx dz \Rightarrow f(4.1,0.8) \approx dz + f(4,\pi/4) + . +

    + +

    + To find dz, we need f_x and f_y. + + f_x(x,y) \amp = \frac{\sin(y) }{2\sqrt{x}} \Rightarrow\amp + f_x(4,\pi/4) \amp = \frac{\sin(\pi) /4}{2\sqrt{4}} + \amp \amp \amp = \frac{\sqrt{2}/2}{4} = \sqrt{2}/8. + f_y(x,y) \amp = \sqrt{x}\cos(y) \Rightarrow\amp + f_y(4,\pi/4) \amp = \sqrt{4}\frac{\sqrt{2}}2 + \amp \amp \amp = \sqrt{2} + . +

    + +

    + Approximating 4.1 with 4 gives dx = 0.1; + approximating 0.8 with \pi/4 gives dy \approx 0.015. + Thus + + dz \amp = f_x(4,\pi/4)(0.1) + f_y(4,\pi/4)(0.015) + \amp = \frac{\sqrt{2}}8(0.1) + \sqrt{2}(0.015) + \amp \approx 0.039 + . +

    + +

    + Returning to Equation, we have + + f(4.1,0.8) \approx 0.039 + 1.414 = 1.4531 + . +

    + +

    + We, of course, + can compute the actual value of f(4.1,0.8) with a calculator; + the actual value, accurate to 5 places after the decimal, + is 1.45254. + Obviously our approximation is quite good. +

    +     +     + +

    + The point of the previous example was not + to develop an approximation method for known functions. + After all, we can very easily compute + f(4.1,0.8) using readily available technology. + Rather, it serves to illustrate how well this method of approximation works, + and to reinforce the following concept: +

    + +

    + New position = old position + amount of change, so +

    + +

    + New position \approx old position + approximate amount of change. +

    + +

    + In the previous example, + we could easily compute f(4,\pi/4) and could approximate the amount of z-change when computing f(4.1,0.8), + letting us approximate the new z-value. +

    + +

    + It may be surprising to learn that it is not uncommon to know the values of f, + f_x and f_y at a particular point without actually knowing the function f. + The total differential gives a good method of approximating f at nearby points. +

    + + + Approximating an unknown function + +

    + Given that f(2,-3) = 6, + f_x(2,-3) = 1.3 and f_y(2,-3) = -0.6, + approximate f(2.1,-3.03). +

    +     + +

    + The total differential approximates how much f changes from the point (2,-3) to the point (2.1,-3.03). + With dx = 0.1 and dy = -0.03, we have + + dz \amp = f_x(2,-3)dx + f_y(2,-3)dy + \amp = 1.3(0.1) + (-0.6)(-0.03) + \amp = 0.148 + . +

    + +

    + The change in z is approximately 0.148, + so we approximate f(2.1,-3.03)\approx 6.148. +

    +     +     +     + + + Tangent Plane Approximation +

    + Recall from that in one variable, + the essence of differentiability is the tangent line approximation. + This idea is emphasized in , + where we first introduced the differential. +

    + +

    + In we saw that the partial derivatives of a function f(x,y) + can be used to define the tangent plane to a graph z=f(x,y). + We will now see that this plane plays the same role for functions of two variables as the tangent line to a graph y=f(x) for a function of one variable. +

    + +

    + Recall from that for a function f(x), + when x is near c we have the linear approximation f(x)\approx \ell(x), + where + + \ell(x) = f(c)+f'(c)(x-c) + + is the linearization of f at c. If we set dx=\dx = x-c, + and evaluate the differential dy = f'(x)\,dx at c, + then we have + + \dy \amp = f(x)-f(c) + dy \amp = \ell(x)-f(c) + . +

    + +

    + Given the graph y=f(x), we know that y=\ell(x) gives the tangent line to the graph at c. + For the graph z=f(x,y) of a function of two variables, we similarly have the tangent plane + + z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) + + defined in , + suggesting that we define the two variable linearization + + \ell(x,y) = f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) + . +

    + +

    + Consider the total differential dz at (a,b): + + dz = f_x(a,b)\,dx + f_y(a,b)\,dy + . + If we assume that (x,y) is close to (a,b), + and set dx = x-a, dy = y-b, + then we have + + dz = f_x(a,b)\,dx + f_y(a,b)\,dy = f_x(a,b)(x-a)+f_y(a,b)(y-b) + . + Since \ell(a,b)=f(a,b), we have + \ell(x,y)-\ell(a,b) = dz, + which agrees with the one-variable situation, + and reinforces the concept of the differential as the linear change in a function. +

    + +

    + If we recast in the language of tangent planes, + we can more easily see the analogy with functions of a single variable. + We can now say that f(x,y) is differentiable at (a,b) if it has a valid tangent plane approximation at (a,b). + Note that f(x,y)-\ell(x,y) is equal to the error term E_x\,dx+E_y\,dy. +

    + +

    + By , we know that the tangent plane at (a,b,f(a,b)) + exists, and gives a good approximation to the graph z=f(x,y), + as long as the partial derivatives of f exist and are continuous at (a,b). +

    +     + + + Error/Sensitivity Analysis +

    + The total differential gives an approximation of the change in z given small changes in x and y. + We can use this to approximate error propagation; + that is, if the input is a little off from what it should be, + how far from correct will the output be? + We demonstrate this in an example. +

    + +

    + sensitivity analysis + total differentialsensitivity analysis +

    + + + Sensitivity analysis + +

    + A cylindrical steel storage tank is to be built that is 10ft tall and 4ft across in diameter. + It is known that the steel will expand/contract with temperature changes; + is the overall volume of the tank more sensitive to changes in the diameter or in the height of the tank? +

    +     + +

    + A cylindrical solid with height h and radius r has volume V = \pi r^2h. + We can view V as a function of two variables, r and h. + We can compute partial derivatives of V: + + \frac{\partial V}{\partial r} = V_r(r,h) = 2\pi rh \qquad \text{ and } \qquad \frac{\partial V}{\partial h} = V_h(r,h) = \pi r^2 + . +

    + +

    + The total differential is dV = (2\pi rh)dr + (\pi r^2)dh. + When h = 10 and r = 2, + we have dV = 40\pi dr + 4\pi dh. + Note that the coefficient of dr is 40\pi\approx 125.7; + the coefficient of dh is a tenth of that, + approximately 12.57. + A small change in radius will be multiplied by 125.7, whereas a small change in height will be multiplied by 12.57. + Thus the volume of the tank is more sensitive to changes in radius than in height. +

    +     +     + +

    + The previous example showed that the volume of a particular tank was more sensitive to changes in radius than in height. + Keep in mind that this analysis only applies to a tank of those dimensions. + A tank with a height of 1ft and radius of 5ft would be more sensitive to changes in height than in radius. +

    + +

    + One could make a chart of small changes in radius and height and find exact changes in volume given specific changes. + While this provides exact numbers, + it does not give as much insight as the error analysis using the total differential. +

    +     + + + Differentiability of Functions of Three Variables +

    + The definition of differentiability for functions of three variables is very similar to that of functions of two variables. + We again start with the total differential. +

    + + + Total Differential + +

    + Let w=f(x,y,z) be continuous on a set D. + Let dx, + dy and dz represent changes in x, + y and z, respectively. + Where the partial derivatives f_x, + f_y and f_z exist, + the total differential of w is + total differential + partial derivativetotal differential + + dw = f_x(x,y,z)\,dx + f_y(x,y,z)\,dy+f_z(x,y,z)\,dz + . +

    +     +     + +

    + This differential can be a good approximation of the change in w when w = f(x,y,z) is + differentiable. +

    + + + Multivariable Differentiability + +

    + Let w=f(x,y,z) be defined on a set D containing + (x_0,y_0,z_0) where f_x(x_0,y_0,z_0), + f_y(x_0,y_0,z_0) and f_z(x_0,y_0,z_0) exist. + Let dw be the total differential of w at (x_0,y_0,z_0), + let \Delta w = f(x_0+dx,y_0+dy,z_0+dz) - f(x_0,y_0,z_0), + and let E_x, + E_y and E_z be functions of dx, + dy and dz such that + differentiable + derivativemultivariable differentiability + multivariable functiondifferentiability + + \Delta w = dw + E_xdx + E_ydy + E_zdz + . +

    + +

    +

      +
    1. +

      + We say f is differentiable at (x_0,y_0,z_0) if, + given \varepsilon \gt 0, + there is a \delta \gt 0 such that if \norm{\la dx,dy,dz\ra} \lt \delta, + then \norm{\la E_x,E_y,E_z\ra} \lt \varepsilon. +

      +
      2. + +
    2. +

      + We say f is differentiable on B + if f is differentiable at every point in B. + If f is differentiable on \mathbb{R}^3, + we say that f is differentiable everywhere. +

      +
      3. +
    +

    +     +     + +

    + Just as before, + this definition gives a rigorous statement about what it means to be differentiable that is not very intuitive. + We follow it with a theorem similar to . +

    + + + Continuity and Differentiability of Functions of Three Variables + +

    + Let w=f(x,y,z) be defined on a set D containing (x_0,y_0,z_0). +

    + +

    +

      +
    1. +

      + If f is differentiable at (x_0,y_0,z_0), + then f is continuous at (x_0,y_0,z_0). +

      +
      2. + +
    2. +

      + If f_x, + f_y and f_z are continuous on D, + then f is differentiable on D. + multivariable functiondifferentiability + multivariable functioncontinuity +

      +
      3. +
    +

    +     +     + + + +

    + This set of definition and theorem extends to functions of any number of variables. + The theorem again gives us a simple way of verifying that most functions that we encounter are differentiable on their natural domains. +

    + +

    + This section has given us a formal definition of what it means for a functions to be differentiable, + along with a theorem that gives a more accessible understanding. + The following sections return to notions prompted by our study of partial derivatives that make use of the fact that most functions we encounter are differentiable. +

    +     + + + + Terms and Concepts + + + + +

    + If f(x,y) is differentiable on S, + the f is continuous on S. +

    +     + +     + + + + +

    + If f_x and f_y are continuous on S, + then f is differentiable on S. +

    +     + +     + + + + +

    + If z=f(x,y) is differentiable, + then the change in z over small changes dx and dy in x and y is approximately dz. +

    +     + +     + + + + +

    + Finish the sentence: + The new z-value is approximately the old z-value plus the approximate . +

    +     + + + + change|amount of change|change in z + + + + +     +     + + + Problems + + + +

    + Find the total differential dz. +

    +     + + + + +

    + z = x\sin(y) + x^2 +

    +     + +

    + dz = (\sin(y) + 2x)dx + (x\cos(y) )dy +

    +     + +     + + + + +

    + z = (2x^2+3y)^2 +

    +     + +

    + dz = 8x(2x^2+3y)dx + 6(2x^2+3y)dy +

    +     + +     + + + + +

    + z = 5x-7y +

    +     + +

    + dz = 5dx -7dy +

    +     + +     + + + + +

    + z = xe^{x+y} +

    +     + +

    + dz = (e^{x+y}+xe^{x+y})dx +xe^{x+y}dy +

    +     + +     + +     + + + +

    + A function f(x,y) is given. + Give the indicated approximation using the total differential. +

    +     + + + + +

    + f(x,y) = \sqrt{x^2+y}. + Approximate f(2.95,7.1) knowing f(3,7) = 4. +

    +     + +

    + dz = \frac{x}{\sqrt{x^2+y}}dx + \frac{1}{2\sqrt{x^2+y}}dy, + with dx = -0.05 and dy = .1. + At (3,7), dz = 3/4(-0.05) + 1/8(.1) = -0.025, + so f(2.95,7.1) \approx -0.025+4 = 3.975. +

    +     + +     + + + + +

    + f(x,y) = \sin(x) \cos(y). + Approximate f(0.1,-0.1) knowing f(0,0) = 0. +

    +     + +

    + dz = (\cos(x) \cos(y) )dx - (\sin(x) \sin(y) )dy, + with dx = 0.1 and dy = -0.1. + At (0,0), dz = 1(.1) - (0)(-0.1) = 0.1, + so f(0.1,-0.1) \approx 0.1+0 = 0.1. +

    +     + +     + + + + +

    + f(x,y) = x^2y-xy^2. + Approximate f(2.04,3.06) knowing f(2,3) = -6. +

    +     + +

    + dz = (2xy-y^2)dx + (x^2-2xy)dy, + with dx = 0.04 and dy = 0.06. + At (2,3), dz = 3(0.04) + (-8)(0.06) = -0.36, + so f(2.04,3.06) \approx -0.36-6 = -6.36. +

    +     + +     + + + + +

    + f(x,y) = \ln(x-y). + Approximate f(5.1,3.98) knowing f(5,4) = 0. +

    +     + +

    + dz = \frac1{x-y}dx -\frac{1}{x-y}dy, + with dx = 0.1 and dy = -0.02. + At (5,4), dz = 1(0.1) + (-1)(-0.02) = 0.12, + so f(5.1,3.98) \approx 0.12+0 = 0.12. +

    +     + +     + +     + + + + +

    + Find the total differential dw. +

    +     + + + + +

    + w= x^2yz^3 +

    +     + +

    + dw = 2xyz^3\,dx + x^2z^3\,dy + 3x^2yz^2\,dz +

    +     + +     + + + + +

    + w= e^x\sin(y) \ln(z) +

    +     + +

    + dw = e^x\sin(y) \ln(z) \,dx + e^x\cos(y) \ln(z) \,dy + e^x\sin(y) \frac1z\,dz +

    +     + +     + +     + + + +

    + Use the information provided and the total differential to make the given approximation. +

    +     + + + + +

    + f(3,1) = 7, + f_x(3,1) = 9, f_y(3,1) = -2. + Approximate f(3.05, 0.9). +

    +     + +

    + dx = 0.05, dy = -0.1. + dz = 9(.05)+(-2)(-0.1) = 0.65. + So f(3.05,0.9) \approx 7+0.65=7.65. +

    +     + +     + + + + +

    + f(-4,2) = 13, + f_x(-4,2) = 2.6, f_y(-4,2) = 5.1. + Approximate f(-4.12, 2.07). +

    +     + +

    + dx = -0.12, dy = 0.07. + dz = 2.6(-.12)+(5.1)(0.07) = 0.045. + So f(-4.12, 2.07) \approx 13+0.045=13.045. +

    +     + +     + + + + +

    + f(2,4,5) = -1, f_x(2,4,5) = 2, + f_y(2,4,5) = -3, f_z(2,4,5) = 3.7. + Approximate f(2.5,4.1,4.8). +

    +     + +

    + dx = 0.5, dy = 0.1, dz = -0.2. +

    + +

    + dw = 2(0.5) + (-3)(0.1) + 3.7(-0.2) = -0.04, + so f(2.5, 4.1, 4.8) \approx -1-0.04 = -1.04. +

    +     + +     + + + + +

    + f(3,3,3) = 5, f_x(3,3,3) = 2, + f_y(3,3,3) = 0, f_z(3,3,3) = -2. + Approximate f(3.1, 3.1,3.1). +

    +     + +

    + dx = 0.1, dy = 0.1, dz = 0.1. +

    + +

    + dw = 2(0.1) + (0)(0.1) + (-2)(.1) = 0, + so f(3.1,3.1,3.1) \approx 5+0=5. +

    +     + +     + +     + + + + +

    + The following exercises ask a variety of questions dealing with approximating error and sensitivity analysis. +

    +     + + + + +

    + A cylindrical storage tank is to be 2ft tall with a radius of 1ft. + Is the volume of the tank more sensitive to changes in the radius or the height? +

    +     + +

    + The total differential of volume is dV = 4\pi dr + \pi dh. + The coefficient of dr is greater than the coefficient of dh, + so the volume is more sensitive to changes in the radius. +

    +     + +     + + + + +

    + Projectile Motion: The x-value of an object moving under the principles of projectile motion is x(\theta,v_0,t)= (v_0\cos(\theta) )t. + A particular projectile is fired with an initial velocity of v_0=250ft/s and an angle of elevation of \theta = 60^\circ. + It travels a distance of 375ft in 3 seconds. +

    + +

    + Is the projectile more sensitive to errors in initial speed or angle of elevation? +

    +     + +

    + Distance of the projectile is a function of two variables (leaving t=3): + D(v_0,\theta) = 3v_0\cos(\theta). + The total differential of D is dD = 3\cos(\theta) dv_0-3v_0\sin(\theta) d\theta. + The coefficient of d\theta has a much greater magnitude than the coefficient of dv_0, + so a small change in the angle of elevation has a much greater effect on distance traveled than a small change in initial velocity. +

    +     + +     + + + + +

    + The length \ell of a long wall is to be approximated. + The angle \theta, as shown in the diagram + (not to scale), + is measured to be 85^\circ, + and the distance x is measured to be 30'. Assume that the triangle formed is a right triangle. +

    + +

    + Is the measurement of the length of \ell more sensitive to errors in the measurement of x or in \theta? +

    + + + A right-angled triangle illustrating a wall of unknown height, a horizontal distance, and an angle. + +

    + The diagram illustrates a simple right-angled triangle. + The base of the triangle is labeled x. The height of the triangle is labeled \ell. + The angle opposite the wall is labeled \theta. +

    +     + + + \begin{tikzpicture} + + \draw [ultra thick] (1,-1) -- node [pos=.5,right] { \(\ell=\)?}(1,1); + \draw [dashed] (1,1) -- (-1,-1) node [xshift=10pt,yshift=5pt] { \(\theta\)} -- node [pos=.5,below] { \(x\)} (1,-1); + + \end{tikzpicture} + + + +     + +

    + Using trigonometry, \ell = x\tan(\theta), + so d\ell = \tan(\theta) dx + x\sec^2(\theta) d\theta. + With \theta = 85^\circ and x=30, + we have d\ell = 11.43dx+3949.38d\theta. + The measured length of the wall is much more sensitive to errors in \theta than in x. + While it can be difficult to compare sensitivities between measuring feet and measuring degrees + (it is somewhat like comparing apples to oranges), + here the coefficients are so different that the result is clear: + a small error in degree has a much greater impact than a small error in distance. +

    +     + +     + + + + +

    + It is common sense that it is far better to measure a long distance with a long measuring tape rather than a short one. + A measured distance D can be viewed as the product of the length \ell of a measuring tape times the number n of times it was used. + For instance, + using a 3' tape 10 times gives a length of 30'. To measure the same distance with a 12' tape, + we would use the tape 2.5 times. (, + 30=12\times 2.5.) Thus D = n\ell. +

    + +

    + Suppose each time a measurement is taken with the tape, + the recorded distance is within 1/16'' of the actual distance. (, d\ell = 1/16'' \approx 0.005ft). + Using differentials, + show why common sense proves correct in that it is better to use a long tape to measure long distances. +

    +     + +

    + With D = n\ell, + the total differential is dD = \ell\, dn+ n\,d\ell. + If one measures with a short tape, + n must be large and hence + n\,d\ell is going to be greater than when a large tape is used + (wherein n will be small). +

    +     + +     +     +     +     +
    +
    + The Multivariable Chain Rule + +

    + Consider driving an off-road vehicle along a dirt road. + As you drive, your elevation likely changes. + What factors determine how quickly your elevation rises and falls? + After some thought, + generally one recognizes that one's velocity + (speed and direction) + and the terrain influence your rise and fall. +

    + +

    + One can represent the terrain as the surface defined by a multivariable function f(x,y); + one can represent the path of the off-road vehicle, as seen from above, + with a vector-valued function \vec r(t) = \langle x(t), + y(t)\rangle; + the velocity of the vehicle is thus \vrp(t) = \langle x'(t),\yp(t)\rangle. +

    + +

    + Consider + in which a surface z=f(x,y) is drawn, + along with a dashed curve in the xy-plane. + Restricting f to just the points on this circle gives the curve shown on the surface (, + the path of the off-road vehicle.) The derivative + \frac{df}{dt} gives the instantaneous rate of change of f with respect to t. + If we consider an object traveling along this path, + \frac{df}{dt}=\frac{dz}{dt} gives the rate at which the object rises/falls (, + the rate of elevation change + of the vehicle.) Conceptually, + the Multivariable Chain Rule combines terrain and velocity information properly to compute this rate of elevation change. +

    + +
    + Understanding the application of the Multivariable Chain Rule + + + + A plot of a surface in space, and a parametric curve on the surface. + +

    + A surface is plotted in space against a set of three-dimensional coordinate axes. + In the xy plane, a dashed circle is shown. + Each point (x,y,0) on this circle determines a point (x,y,f(x,y)) on the surface. + Together, these points form a curve on the surface; this is also illustrated in the image. +

    +     + + + + + //ASY file for figmchain_intro3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-1,5); + pair ybounds=(-1,5); + pair zbounds=(0,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=-0.2*(x-1)^2-0.05*y^2+2 //{-.2*(x-1)^2-.05*y^2+2}; + triple f(pair t) { + return (t.x,t.y,-0.2*(t.x-1)^2-0.05*t.y^2+2); + } + surface s=surface(f,(0,0),(4,4),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the circle in xy-plane + triple g(real t) {return (cos(t)+2,sin(t)+2,0);} + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,dashed+bluepen+linewidth(.75)); + + //Draw the circle on surface + triple g(real t) { + return (cos(t)+2,sin(t)+2,-0.2*(cos(t)+2-1)^2-0.05*(sin(t)+2)^2+2); + } + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + + + +
    + +

    + Abstractly, let z be a function of x and y; + that is, z=f(x,y) for some function f, + and let x and y each be functions of t. + By choosing a t-value, + x- and y-values are determined, + which in turn determine z: + this defines z as a function of t. + The Multivariable Chain Rule gives a method of computing \frac{dz}{dt}. +

    + + +     + + + Multivariable Chain Rule, Part I + + + + Multivariable Chain Rule, Part I + +

    + Let z=f(x,y), x=g(t) and y=h(t), where f, + g and h are differentiable functions. + Then z = f(x,y) = f\big(g(t),h(t)\big) is a function of t, + and + derivativeChain Rule + Chain Rulemultivariable + + \frac{dz}{dt} = \frac{df}{dt} \amp = f_x(x,y)\frac{dx}{dt}+f_y(x,y)\frac{dy}{dt} + \amp = \frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt} + \amp = \langle\, f_x,f_y\rangle \cdot \langle x',\yp\rangle + . +

    +     +     + +

    + The Chain Rule of + states that + + \frac{d}{dx}\Big(f\big(g(x)\big)\Big) = \fp\big(g(x)\big)\gp(x) + . + If t=g(x), we can express the Chain Rule as + + \frac{df}{dx} = \frac{df}{dt}\frac{dt}{dx}; + + recall that the derivative notation is deliberately chosen to reflect their fraction-like properties. + A similar effect is seen in . + In the second line of equations, + one can think of the dx and \partial x as + sort of canceling out, + and likewise with dy and \partial y. +

    + +

    + Notice, too, + the third line of equations in . + The vector \langle\,f_x,f_y\rangle contains information about the surface (terrain); + the vector \langle x',\yp\rangle can represent velocity. + In the context measuring the rate of elevation change of the off-road vehicle, + the Multivariable Chain Rule states it can be found through a product of terrain and velocity information. +

    + +

    + We now practice applying the Multivariable Chain Rule. +

    + + + + + Using the Multivariable Chain Rule + +

    + Let z=x^2y+x, where x=\sin(t) and y=e^{5t}. + Find \frac{dz}{dt} using the Chain Rule. +

    +     + +

    + Following , we find + + f_x(x,y) = 2xy+1,\qquad f_y(x,y) = x^2,\qquad \frac{dx}{dt} = \cos(t) ,\qquad \frac{dy}{dt}= 5e^{5t} + . +

    + +

    + Applying the theorem, we have + + \frac{dz}{dt} = (2xy+1)\cos(t) + 5x^2e^{5t} + . +

    + +

    + This may look odd, as it seems that + \frac{dz}{dt} is a function of x, y and t. + Since x and y are functions of t, + \frac{dz}{dt} is really just a function of t, + and we can replace x with \sin(t) and y with e^{5t}: + + \frac{dz}{dt} = (2xy+1)\cos(t) + 5x^2e^{5t} = (2\sin(t)e^{5t}+1)\cos(t) +5e^{5t}\sin^2(t) + . +

    +     +     + +

    + The previous example can make us wonder: + if we substituted for x and y at the end to show that + \frac{dz}{dt} is really just a function of t, + why not substitute before differentiating, + showing clearly that z is a function of t? +

    + +

    + That is, z = x^2y+x = (\sin(t) )^2e^{5t}+\sin(t) . + Applying the Chain and Product Rules, we have + + \frac{dz}{dt} = 2\sin(t) \cos(t) \, e^{5t}+ 5\sin^2(t) \,e^{5t}+\cos(t) + , + which matches the result from the example. +

    + +

    + This may now make one wonder What's the point? + If we could already find the derivative, + why learn another way of finding it? In some cases, + applying this rule makes deriving simpler, + but this is hardly the power of the Chain Rule. + Rather, in the case where z=f(x,y), + x=g(t) and y=h(t), + the Chain Rule is extremely powerful when + we do not know what f, + g and/or h are. + It may be hard to believe, + but often in the real world + we know rate-of-change information (, information about derivatives) without explicitly knowing the underlying functions. + The Chain Rule allows us to combine several rates of change to find another rate of change. + The Chain Rule also has theoretic use, + giving us insight into the behavior of certain constructions + (as we'll see in the next section). +

    + +

    + We demonstrate this in the next example. +

    + + + Applying the Multivariable Chain Rule + +

    + An object travels along a path on a surface. + The exact path and surface are not known, + but at time t=t_0 it is known that : + + \frac{\partial z}{\partial x} = 5,\qquad \frac{\partial z}{\partial y}=-2,\qquad \frac{dx}{dt}=3\qquad \text{ and } \qquad \frac{dy}{dt}=7 + . +

    + +

    + Find \frac{dz}{dt} at time t_0. +

    +     + +

    + The Multivariable Chain Rule states that + + \frac{dz}{dt} \amp = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} + \amp = 5(3)+(-2)(7) + \amp =1 + . +

    + +

    + By knowing certain rates-of-change information about the surface and about the path of the particle in the xy-plane, + we can determine how quickly the object is rising/falling. +

    +     +     + +

    + We next apply the Chain Rule to solve a max/min problem. +

    + + + Applying the Multivariable Chain Rule + +

    + Consider the surface z=x^2+y^2-xy, a paraboloid, + on which a particle moves with x and y coordinates given by + x=\cos(t) and y=\sin(t). + Find \frac{dz}{dt} when t=0, + and find where the particle reaches its maximum/minimum z-values. +

    +     + +

    + It is straightforward to compute + + f_x(x,y) \amp = 2x-y \amp f_y(x,y) \amp = 2y-x + \frac{dx}{dt} \amp = -\sin(t) \amp \frac{dy}{dt} \amp = \cos(t) + . +

    + +

    + Combining these according to the Chain Rule gives: + + \frac{dz}{dt} = -(2x-y)\sin(t) + (2y-x)\cos(t) + . +

    + +
    + Plotting the path of a particle on a surface in + + + + A circular paraboloid, opening upward, plotted over a rectangular domain. A curve lying on the surface is also plotted. + +

    + The graph z=x^2+y^2-xy is shown; this is a circular paraboloid. + The surface is plotted using a rectangular domain, resulting in peaks at the corners of the domain. + A parametric curve is also plotted; this curve lies on the surface. + When viewed from above, the curve is circular, + but the height of the curve varies as it follows the contours of the surface. +

    +     + + + + + //ASY file for figmchain23D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(6.5,3.4,1.9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={1,2,3}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(0,4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=x^2+y^2-xy + triple f(pair t) { + return (t.x,t.y,t.x^2+t.y^2-t.x*t.y); + } + surface s=surface(f,(-1.1,-1.1),(1.1,1.1),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the circle on surface + triple g(real t) { + return (cos(t),sin(t),(cos(t))^2+(sin(t))^2-cos(t)*sin(t)); + } + path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); + + //plot points for min and max along the curve + dotfactor=3; + dot((cos(pi/4),sin(pi/4),1-cos(pi/4)*sin(pi/4))); + dot((cos(5*pi/4),sin(5*pi/4),1-cos(5*pi/4)*sin(5*pi/4))); + dot((cos(3*pi/4),sin(3*pi/4),1-cos(3*pi/4)*sin(3*pi/4))); + dot((cos(7*pi/4),sin(7*pi/4),1-cos(7*pi/4)*sin(7*pi/4))); + + + + +
    + +

    + When t=0, x=1 and y=0. + Thus \ds\frac{dz}{dt} = -(2)(0)+ (-1)(1) = -1. + When t=0, the particle is moving down, + as shown in . +

    + +

    + To find where z-value is maximized/minimized on the particle's path, + we set \frac{dz}{dt}=0 and solve for t: + + \frac{dz}{dt} =0 \amp = -(2x-y)\sin(t) + (2y-x)\cos(t) + 0\amp = -(2\cos(t) -\sin(t) )\sin(t) +(2\sin(t) -\cos(t) )\cos(t) + 0\amp = \sin^2(t) -\cos^2(t) + \cos^2(t) \amp =\sin^2(t) + t\amp = n\frac{\pi}4 \text{ (for odd \(n\)) } + +

    + +

    + We can use the First Derivative Test to find that on [0,2\pi], + z has reaches its absolute minimum at t=\pi/4 and 5\pi/4; + it reaches its absolute maximum at t=3\pi/4 and 7\pi/4, + as shown in . +

    +     +     + +

    + We can extend the Chain Rule to include the situation where z is a function of more than one variable, + and each of these variables is also a function of more than one variable. + The basic case of this is where z=f(x,y), + and x and y are functions of two variables, + say s and t. +

    + + + Multivariable Chain Rule, Part II + +

    +

      +
    1. +

      + Let z=f(x,y), + x=g(s,t) and y=h(s,t), where f, + g and h are differentiable functions. + Then z is a function of s and t, and +

      + +

      +

        +
      • \frac{\partial z}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial s}
        • + +
      • \frac{\partial z}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}
        • +
      + derivativeChain Rule + Chain Rulemultivariable +

      +
      2. + +
    2. +

      + Let z = f(x_1,x_2,\ldots,x_m) be a differentiable function of m variables, + where each of the x_i is a differentiable function of the variables t_1,t_2,\ldots,t_n. + Then z is a function of the t_i, and + + \frac{\partial z}{\partial t_i} = \frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial t_i} + \frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial t_i} + \cdots + \frac{\partial f}{\partial x_m}\frac{\partial x_m}{\partial t_i} + . +

      +
      3. +
    +

    +     +     + + + Using the Multivariable Chain Rule, Part II + +

    + Let z=x^2y+x, x=s^2+3t and y=2s-t. + Find \frac{\partial z}{\partial s} and \frac{\partial z}{\partial t}, + and evaluate each when s=1 and t=2. +

    +     + +

    + Following , + we compute the following partial derivatives: + + \frac{\partial f}{\partial x} = 2xy+1\qquad\qquad \frac{\partial f}{\partial y} = x^2 + , + + \frac{\partial x}{\partial s} = 2s \qquad\qquad \frac{\partial x}{\partial t} = 3\qquad\qquad \frac{\partial y}{\partial s} = 2 \qquad\qquad \frac{\partial y}{\partial t} = -1 + . +

    + +

    + Thus + + \ds \frac{\partial z}{\partial s} = (2xy+1)(2s) + (x^2)(2) = 4xys+2s + 2x^2, \text{ and } + + + \ds \frac{\partial z}{\partial t} = (2xy+1)(3) + (x^2)(-1) = 6xy-x^2+3 + . +

    + +

    + When s=1 and t=2, + x= 7 and y= 0, so + + \frac{\partial z}{\partial s} = 100\qquad \text{ and } \qquad \frac{\partial z}{\partial t} = -46 + . +

    +     +     + + + + + Using the Multivariable Chain Rule, Part II + +

    + Let w = xy+z^2, where x= t^2e^s, + y= t\cos(s), and z=s\sin(t). + Find \frac{\partial w}{\partial t} when s=0 and t=\pi. +

    +     + +

    + Following , + we compute the following partial derivatives: + + \frac{\partial f}{\partial x} \amp = y \amp \frac{\partial f}{\partial y} \amp = x \amp \frac{\partial f}{\partial z} \amp = 2z + \frac{\partial x}{\partial t} \amp = 2te^s \amp \frac{\partial y}{\partial t} \amp = \cos(s) \amp \frac{\partial z}{\partial t} \amp = s\cos(t) + . +

    + +

    + Thus + + \frac{\partial w}{\partial t} = y(2te^s) + x(\cos(s) ) + 2z(s\cos(t) ) + . +

    + +

    + When s=0 and t=\pi, + we have x=\pi^2, y=\pi and z=0. + Thus + + \frac{\partial w}{\partial t} = \pi(2\pi) + \pi^2 = 3\pi^2 + . +

    +     +     +     + + + Implicit Differentiation +

    + We studied finding \frac{dy}{dx} when y is given as an implicit function of x in detail in . + We find here that the Multivariable Chain Rule gives a simpler method of finding \frac{dy}{dx}. +

    + +

    + For instance, consider the implicit function x^2y-xy^3=3. + We learned to use the following steps to find \frac{dy}{dx}: + + \frac{d}{dx}\Big(x^2y-xy^3\Big) \amp = \frac{d}{dx}\Big(3\Big) + 2xy + x^2\frac{dy}{dx}-y^3-3xy^2\frac{dy}{dx} \amp = 0 + \frac{dy}{dx} = -\frac{2xy-y^3}{x^2-3xy^2} + . +

    + +

    + Instead of using this method, + consider z=x^2y-xy^3. + The implicit function above describes the level curve z=3. + Considering x and y as functions of x, + the Multivariable Chain Rule states that + + \frac{dz}{dx} = \frac{\partial z}{\partial x}\frac{dx}{dx}+\frac{\partial z}{\partial y}\frac{dy}{dx} + . +

    + +

    + Since z is constant + (in our example, z=3), + \frac{dz}{dx} = 0. + We also know \frac{dx}{dx} = 1. + Equation becomes + + 0 \amp = \frac{\partial z}{\partial x}(1) + \frac{\partial z}{\partial y}\frac{dy}{dx} \Rightarrow + \frac{dy}{dx} \amp = -\frac{\partial z}{\partial x}\Big/\frac{\partial z}{\partial y} + \amp = -\frac{\,f_x\,}{f_y} + . +

    + +

    + Note how our solution for \frac{dy}{dx} in Equation is just the partial derivative of z with respect to x, + divided by the partial derivative of z with respect to y, + all multiplied by (-1). +

    + +

    + We state the above as a theorem. +

    + + + Implicit Differentiation + +

    + Let f be a differentiable function of x and y, + where f(x,y)=c defines y as an implicit function of x, + for some constant c. + Then + derivativeimplicit + implicit differentiation + + \frac{dy}{dx} = - \frac{f_x(x,y)}{f_y(x,y)} + . +

    +     +     + +

    + We practice using + by applying it to a problem from . +

    + + + Implicit Differentiation + +

    + Given the implicitly defined function \sin(x^2y^2)+y^3=x+y, + find \yp. + Note: this is the same problem as given in + of , + where the solution took about a full page to find. +

    +     + +

    + Let f(x,y) = \sin(x^2y^2)+y^3-x-y; + the implicitly defined function above is equivalent to f(x,y)=0. + We find \frac{dy}{dx} by applying . + We find + + f_x(x,y) \amp = 2xy^2\cos(x^2y^2)-1 + f_y(x,y) \amp = 2x^2y\cos(x^2y^2)+3y^2-1 + , + so + + \frac{dy}{dx} = -\frac{2xy^2\cos(x^2y^2)-1}{2x^2y\cos(x^2y^2)+3y^2-1} + , + which matches our solution from . +

    +     +     + + + +

    + We can also do implicit differentiation for functions of three variables. + In the same way that a level curve f(x,y)=c is used to implicitly define y as a function of x, + a level surface f(x,y,z)=c can be viewed as implicitly defining z as a function of x and y. +

    + +

    + Suppose the equation f(x,y,z)=c, where c is a constant, + defines the function z=g(x,y). + Then we can use the chain rule to compute the derivatives of f(x,y,z) + with respect to x and y, where we set x=x, y=y, and z=g(x,y). + Since f(x,y,z) is constant, we have + + 0 \amp = \frac{\partial}{\partial x}f(x,y,z) + \amp = f_x(x,y,z)\plz{x}{x}+f_y(x,y,z)\plz{y}{x}+f_z(x,y,z)\plz{z}{x} + \amp = f_x(x,y,z)(1)+f_y(x,y,z)(0)+f_z(x,y,z)\plz{z}{x} + . + Solving for \plz{z}{x} gives us + + \plz{z}{x} = -\frac{f_x(x,y,z)}{f_z(x,y,z)} + , + and similarly, + + \plz{z}{y} = -\frac{f_y(x,y,z)}{f_z(x,y,z)} + . +

    + + + +

    + In we saw that we can use partial + derivatives to determine the equation of the tangent plane to a graph z=f(x,y). + Using implicit differentiation, we can do the same for a level surface f(x,y,z)=c. +

    + + Implicit Differentiation with three variables + +

    + Given that the equation + + x^2yz^3-\sin(x-3z)+4xy^2-3yz=0 + + defines z implicitly as a function of x and y, + compute \plz{z}{x} and \plz{z}{y} using implicit differentiation. + Then, determine the equation of the tangent plane to the surface at the point (3,0,1). +

    +     + +

    + There are two ways to proceed. One is to use implicit differentiation as before, + but using partial derivatives. Whenever we differentiate a function of z, + we multiply by the appropriate partial derivative of z. + The other option is to use the formula derived above. + We will use the first method for the x derivative, and the second for y. +

    +

    + We first take the partial derivative of both sides of Equation with respect to x: + + \frac{\partial}{\partial x}(x^2yz^3-\sin(x-3z)+4xy^2-3yz) \amp = 0 + 2xyz^3 + x^2y(3z^2)\plz{z}{x}-\cos(x-3z)\left(1-3\plz{z}{x}\right)+4y^2-3y\plz{z}{x} \amp = 0 + . + Note that we treated y as a constant, since the derivative is with respect to x. + Next, we collect terms: + + \plz{z}{x}\left(3x^2yz^2+3\cos(x-3z)-3y\right) = -2xyz^3+\cos(x-3z)-4y^2 + . + Lastly, we solve for \plz{z}{x}: + + \plz{z}{x} = \frac{-2xyz^3+\cos(x-3z)-4y^2}{3x^2yz^2+3\cos(x-3z)-3y} + . +

    +

    + For the y derivative, we will use the result given above. + Setting f(x,y,z) = x^2yz^3-\sin(x-3z)+4xy^2-3yz, we have \plz{z}{y} = -\frac{f_y(x,y,z)}{f_z(x,y,z)}. + Therefore, + + \plz{z}{y} = -\frac{x^2z^3+8xy-3z}{3x^2yz^2+3\cos(x-3z)-3y} + . + The second method certainly seems simpler! + The reader is invited to try each part with the other method, and compare answers. +

    +

    + Finally, we consider the problem of the tangent plane. + First, we check that the point (3,0,1) is indeed on the surface: f(3,0,1)=0, + as required. Next we note that z=1 is given to us from this point. + So if f(x,y,z)=c implicitly defines the graph z=g(x,y), + then we must have g(3,0)=1. Next, we have + + g_x(3,0) \amp = \plzoa{z}{x}{(3,0)} = \frac{0+1-0}{0+3-0}=\frac13 + g_y(3,0) \amp = \plzoa{z}{y}{(3,0)} = -\frac{9+0-3}{0+3-0} = -2 + . + The equation of the tangent plane is therefore + + z = g(3,0)+g_x(3,0)(x-3)+g_y(3,0)(y-0) = 1+\frac13(x-3)-2y + . +

    +     +     + +

    + In + we learned how partial derivatives give certain instantaneous rate of change information about a function f(x,y). + In that section, + we measured the rate of change of f by holding one variable constant and letting the other vary + (such as, holding y constant and letting x vary gives f_x). + We can visualize this change by considering the surface defined by f at a point and moving parallel to the x-axis. +

    + +

    + What if we want to move in a direction that is not parallel to a coordinate axis? + Can we still measure instantaneous rates of change? + Yes; we find out how in . + In doing so, + we'll see how the Multivariable Chain Rule informs our understanding of these + directional derivatives. +

    +     + + + + Terms and Concepts + + + +

    + Let a level curve of z=f(x,y) be described by x=g(t), + y = h(t). + Explain why \frac{dz}{dt}=0. +

    +     + + + +

    + Because the parametric equations describe a level curve, + z is constant for all t. + Therefore \frac{dz}{dt}=0. +

    +     + +     + + + + +

    + Fill in the blank: The single variable Chain Rule states + \ds\frac{d}{dx}\Big(f\big(g(x)\big)\Big) = \fp\big(g(x)\big)\cdot. +

    +     + + + + + + + +

    + Type in the correct mathematical expression using plain text. +

    +     +     +     +     + +     + + + + +

    + Put the blocks in order to form a correct chain rule statement. +

    +     + + \frac{df}{dt} + = + \frac{\partial f}{\partial x} + \frac{dx}{dt} + \frac{dt}{dx} + + + \frac{\partial f}{\partial y} + \frac{dy}{dt} + \frac{\partial y}{\partial t} + + +     + + + + + + +

    + If z=f(x,y), where x=g(t) and y=h(t), + we can substitute and write z as an explicit function of t. +

    + +

    + Using the Multivariable Chain Rule to find + \frac{dz}{dt} is sometimes easier than first substituting and then taking the derivative. +

    +     + +     + + + + +

    + The Multivariable Chain Rule is only useful when all the related functions are known explicitly. +

    +     + + +     + + + + + +

    + The Multivariable Chain Rule allows us to compute implicit derivatives easily by just computing two derivatives. +

    +     + + + + + + + + +     +     + + + Problems + + + +

    + Given the functions z=f(x,y), x=g(t) and y=h(t): +

    + +

    +

      +
    1. +

      + Use the Multivariable Chain Rule to compute \lz{z}{t}. +

      +
      2. + +
    2. +

      + Evaluate \lz{z}{t} at the indicated t-value. +

      +
      3. +
    +

    +     + + + + +

    + z=3x+4y, x=t^2, y=2t; t=1 +

    +     + +

    +

      +
    1. +

      + \frac{dz}{dt} = 3(2t)+4(2) = 6t+8. +

      +
      2. + +
    2. +

      + At t=1, \frac{dz}{dt} = 14. +

      +
      3. +
    +

    +     + +     + + + + + Context()->variables->add(y=>'Real',t=>'Real'); + $x=Compute("t"); + $y=Compute("t^2-1"); + $dzdt=Compute("2x-4yt"); + $reduced=$dzdt->substitute(x=>$x,y=>$y); + $dzdt1=$reduced->eval(t=>1,x=>0,y=>0); + $ev=$dzdt->cmp(checker => sub { + my ($correct,$student,$ansHash) = @_; + return 1 if ($correct == $student); + my $sreduced=$student->substitute(x=>$x,y=>$y); + Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($reduced == $sreduced); + return 0; + }); + + + +

    + z=x^2-y^2, x=t, + and y=t^2-1; t=1 +

    + + + Use the Multivariable Chain Rule to compute \lz{z}{t}. + +

    + +

    + + + Evaluate \lz{z}{t} at the indicated t-value. + +

    + +

    +     +     +     + + + + +

    + \ds z=5x+2y, x=2\cos(t) +1, + y=\sin(t) -3; t=\pi/4 +

    +     + +

    +

      +
    1. +

      + \frac{dz}{dt} = 5(-2\sin(t) )+2(\cos(t) ) = -10\sin(t) +2\cos(t) +

      +
      2. + +
    2. +

      + At t=\pi/4, \frac{dz}{dt} = -4\sqrt{2}. +

      +
      3. +
    +

    +     + +     + + + + + Context()->variables->add(y=>'Real',t=>'Real'); + $x=Compute("cos(t)"); + $y=Compute("sin(t)"); + $dzdt=Compute("-sin(t)/(1+y^2)-2xycos(t)/(y^2+1)^2"); + $reduced=$dzdt->substitute(x=>$x,y=>$y); + $ev=$dzdt->cmp(checker => sub { + my ($correct,$student,$ansHash) = @_; + return 1 if ($correct == $student); + my $sreduced=$student->substitute(x=>$x,y=>$y); + Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($reduced == $sreduced); + return 0; + }); + Context("Fraction"); + $dzdt1=Fraction($reduced->eval(t=>pi/2,x=>0,y=>0)); + + + +

    + z=\frac{x}{y^2+1}, x=\cos(t), + and y=\sin(t); t=\pi/2 +

    + + Use the Multivariable Chain Rule to compute \lz{z}{t}. + +

    + +

    + + + Evaluate \lz{z}{t} at the indicated t-value. + +

    + +

    +     +     +     + + + + +

    + \ds z=x^2+2y^2, x=\sin(t), + y=3\sin(t); t=\pi/4 +

    +     + +

    +

      +
    1. +

      + \ds\frac{dz}{dt} = 2x(\cos(t) ) + 4y(3\cos(t) ). +

      +
      2. + +
    2. +

      + At t=\pi/4, x=\sqrt{2}/2, + y=3\sqrt{2}/2, and \frac{dz}{dt} = 19. +

      +
      3. +
    +

    +     + +     + + + + + Context()->variables->add(y=>'Real',t=>'Real'); + $x=Compute("pi t"); + $y=Compute("2pi t + pi/2"); + $dzdt=Compute("-sin(x) sin(y) (pi) + cos(x) cos(y) (2pi)"); + $reduced=$dzdt->substitute(x=>$x,y=>$y); + $ev=$dzdt->cmp(checker => sub { + my ($correct,$student,$ansHash) = @_; + return 1 if ($correct == $student); + my $sreduced=$student->substitute(x=>$x,y=>$y); + Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($reduced == $sreduced); + return 0; + }); + Context("Fraction"); + $dzdt1=Fraction($reduced->eval(t=>3,x=>0,y=>0)); + + + +

    + z=\cos(x) \sin(y), x=\pi t, + and y=2\pi t+\pi/2; t=3 +

    + + + Use the Multivariable Chain Rule to compute \lz{z}{t}. + +

    + +

    + + + Evaluate \lz{z}{t} at the indicated t-value. + +

    + +

    +     +     +     + +     + + + +

    + Functions z=f(x,y), x=g(t) and y=h(t) are given. + Find the values of t where \frac{dz}{dt}=0. + Note: these are the same surfaces/curves as found in Exercises. +

    +     + + + + +

    + \ds z=3x+4y, x=t^2, y=2t +

    +     + +

    + t=-4/3; this corresponds to a minimum +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $zeros=Compute("-sqrt(3/2), 0, sqrt(3/2)"); + + + +

    + Given z=x^2-y^2, x=t, and y=t^2-1, + at what values of t does \lz{z}{t}=0? +

    + +

    + +

    +     +     +     + + + + +

    + \ds z=5x+2y, x=2\cos(t) +1, y=\sin(t) -3 +

    +     + +

    + t=\tan^{-1}(1/5) +n\pi, + where n is an integer +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $zeros=Compute("0, pi"); + + + +

    + Given z=\frac{x}{y^2+1}, + x=\cos(t), and y=\sin(t), + at what values of t in + [0,2\pi) does \lz{z}{t}=0? +

    + +

    + +

    +     + +

    + We find that + + \frac{dz}{dt} = -\frac{2\cos^2(t) \sin(t) }{(1+\sin^2(t) )^2}-\frac{\sin(t) }{1+\sin^2(t) } + . +

    + +

    + Setting this equal to 0, + finding a common denominator and factoring out \sin(t), we get + + \sin(t) \left(\frac{2\cos^2(t) +\sin^2(t) +1}{(1+\sin^2(t) )^2}\right)=0 + . +

    + +

    + We have \sin(t) = 0 when t = \pi n, + where n is an integer. + The expression in the parenthesis above is always positive, + and hence never equal 0. + So all solutions are t=\pi n, n is an integer. + In [0,2\pi), the only solutions are 0 and \pi. +

    +     +     +     + + + + +

    + \ds z=x^2+2y^2, x=\sin(t), y=3\sin(t) +

    +     + +

    + We find that + + \frac{dz}{dt} = 38\cos(t) \sin(t) + . +

    + +

    + Thus \frac{dz}{dt} = 0 when t=\pi n or \pi n+\pi/2, + where n is any integer. +

    +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $zeros=Compute("0, 1/pi*arctan(sqrt(5)), 1- 1/pi*arctan(sqrt(5)), 1, 1+1/pi*arctan(sqrt(5)), 2-1/pi*arctan(sqrt(5))"); + + + +

    + Given z=\cos(x) \sin(y), + x=\pi t, and y=2\pi t+\pi/2, + at what values of t in [0,2) does \lz{z}{t}=0? +

    + +

    + +

    +     + +

    + We find that + + \frac{dz}{dt} = -\pi\sin(\pi t)\sin(2\pi t+\pi/2)+2\pi\cos(\pi t)\cos(2\pi t+\pi/2) + . +

    + +

    + One can easily see that when t is an integer, + \sin(\pi t) =0 and \cos(2\pi t+\pi/2)=0, + hence \frac{dz}{dt}=0 when t is an integer. + There are other places where z has a relative max/min that require more work. + First, verify that \sin(2\pi t+\pi/2) = \cos(2\pi t), + and \cos(2\pi t+\pi/2) = -\sin(2\pi t). + This lets us rewrite \frac{dz}{dt} = 0 as + + -\sin(\pi t)\cos(2\pi t)-2\cos(\pi t)\sin(2\pi t)=0 + . +

    + +

    + By bringing one term to the other side of the equality then dividing, + we can get + + 2\tan(2\pi t) = -\tan(\pi t) + . +

    + +

    + Using the angle sum/difference formula for tangent, we know + + \tan(2\pi t) = \tan(\pi t + \pi t) = \frac{\tan(\pi t)+\tan(\pi t)}{1-\tan^2(\pi t)} + . +

    + +

    + Thus we write + + 2\frac{\tan(\pi t)+\tan(\pi t)}{1-\tan^2(\pi t)} = -\tan(\pi t) + . +

    + +

    + Solving for \tan^2(\pi t), we find + + \tan^2(\pi t) = 5 \implies \tan(\pi t) = \pm\sqrt{5} + , + and so + + \pi t = \tan^{-1}(\pm\sqrt{5}) = \pm\tan^{-1}(\sqrt{5}) + . +

    + +

    + Since the period of tangent is \pi, + we can adjust our answer to be + + \pi t = \pm\tan^{-1}(\sqrt{5})+ n\pi,\text{ where \(n\) is an integer. } + +

    + +

    + Dividing by \pi, we find + + t = \pm\frac1\pi\tan^{-1}(\sqrt{5})+ n,\text{ where \(n\) is an integer. } + +

    + +

    + So the solutions in [0,2) are 0, + \frac{1}{pi}\arctan(sqrt{5}), + 1-\frac{1}{pi}\arctan(sqrt{5}), 1, + 1+\frac{1}{pi}\arctan(sqrt{5}), + and 2-\frac{1}{pi}\arctan(sqrt{5}). +

    +     +     +     + +     + + + +

    + Given the + functions z=f(x,y), + x=g(s,t) and y=h(s,t): +

    + +

    +

      +
    1. +

      + Use the Multivariable Chain Rule to compute + \frac{\partial z}{\partial s} and \frac{\partial z}{\partial t}. +

      +
      2. + +
    2. +

      + Evaluate \frac{\partial z}{\partial s} and + \frac{\partial z}{\partial t} at the indicated s and t values. +

      +
      3. +
    +

    +     + + + + +

    + \ds z=x^2y, x=s-t, y=2s+4t; + s=1, t=0 +

    +     + +

    +

      +
    1. +

      + \frac{\partial z}{\partial s} = 2xy(1) + x^2(2) = 2xy+2x^2; + + \frac{\partial z}{\partial t} = 2xy(-1) + x^2(4) = -2xy+4x^2 +

      +
      2. + +
    2. +

      + With s=1, t=0, x=1 and y=2. + Thus \frac{\partial z}{\partial s} = 6 and \frac{\partial z}{\partial t} = 0 +

      +
      3. +
    +

    +     + +     + + + + + Context()->variables->add(y=>'Real',s=>'Real', t=>'Real'); + $x=Compute("st^2"); + $y=Compute("s^2t"); + $dzds=Compute("-pi sin(pi x + pi y/2)(t^2)-1/2pi sin(pi x + pi y/2)(2st)"); + $dzdt=Compute("-pi sin(pi x + pi y/2)(2st)-1/2pi sin(pi x + pi y/2)(s^2)"); + $dzdsreduced=$dzds->substitute(x=>$x,y=>$y); + $dzdtreduced=$dzdt->substitute(x=>$x,y=>$y); + $dzds1=Compute("2pi"); + $dzdt1=Compute("5pi/2"); + $dzdsev=$dzds->cmp(checker => sub { + my ($correct,$student,$ansHash) = @_; + return 1 if ($correct == $student); + my $sreduced=$student->substitute(x=>$x,y=>$y); + Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdsreduced == $sreduced); + return 0; + }); + $dzdtev=$dzdt->cmp(checker => sub { + my ($correct,$student,$ansHash) = @_; + return 1 if ($correct == $student); + my $sreduced=$student->substitute(x=>$x,y=>$y); + Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdtreduced == $sreduced); + return 0; + }); + + + +

    + z=\cos\mathopen{}\left(\pi x+\frac{\pi}2y\right)\mathclose{}, + x=st^2, and y=s^2t; s=1, t=0 +

    + + + Use the Multivariable Chain Rule to compute \plz{z}{s}. + +

    + +

    + + + Use the Multivariable Chain Rule to compute \plz{z}{t}. + +

    + +

    + + + Evaluate \plz{z}{s} at s=1, t=0. + +

    + +

    + + + Evaluate \plz{z}{t} at s=1, t=0. + +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real',s=>'Real', t=>'Real'); + $x=Compute("st^2"); + $y=Compute("s^2t"); + $dzds=Compute("2x cos(t) + 2y sin(t)"); + $dzdt=Compute("-2xs sin(t) + 2ys cos(t)"); + $dzdsreduced=$dzds->substitute(x=>$x,y=>$y); + $dzdtreduced=$dzdt->substitute(x=>$x,y=>$y); + $dzds1=Compute("4"); + $dzdt1=Compute("0"); + $dzdsev=$dzds->cmp(checker => sub { + my ($correct,$student,$ansHash) = @_; + return 1 if ($correct == $student); + my $sreduced=$student->substitute(x=>$x,y=>$y); + Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdsreduced == $sreduced); + return 0; + }); + $dzdtev=$dzdt->cmp(checker => sub { + my ($correct,$student,$ansHash) = @_; + return 1 if ($correct == $student); + my $sreduced=$student->substitute(x=>$x,y=>$y); + Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdtreduced == $sreduced); + return 0; + }); + + + +

    + z=x^2+y^2, x=s\cos(t), + and y=s\sin(t); s=2, t=\pi/4 +

    + + + Use the Multivariable Chain Rule to compute \plz{z}{s}. + +

    + +

    + + + Use the Multivariable Chain Rule to compute \plz{z}{t}. + +

    + +

    + + + Evaluate \plz{z}{s} at s=2, t=\pi/4. + +

    + +

    + + + Evaluate \plz{z}{t} at s=2, t=\pi/4. + +

    + +

    +     +     +     + + + + + Context()->variables->add(y=>'Real',s=>'Real', t=>'Real'); + $x=Compute("st^2"); + $y=Compute("s^2t"); + $dzds=Compute("-2 y t^2 e^(-(x^2+y^2))"); + $dzdt=Compute("-2 x e^(-(x^2+y^2)) - 4 s t y e^(-(x^2+y^2))"); + $dzdsreduced=$dzds->substitute(x=>$x,y=>$y); + $dzdtreduced=$dzdt->substitute(x=>$x,y=>$y); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $dzds1=Compute("-2/e^2"); + $dzdt1=Compute("-6/e^2"); + $dzdsev=$dzds->cmp(checker => sub { + my ($correct,$student,$ansHash) = @_; + return 1 if ($correct == $student); + my $sreduced=$student->substitute(x=>$x,y=>$y); + Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdsreduced == $sreduced); + return 0; + }); + $dzdtev=$dzdt->cmp(checker => sub { + my ($correct,$student,$ansHash) = @_; + return 1 if ($correct == $student); + my $sreduced=$student->substitute(x=>$x,y=>$y); + Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdtreduced == $sreduced); + return 0; + }); + + + +

    + z=e^{-(x^2+y^2)}, x=t, + and y=st^2, s=1, t=1 +

    + + + Use the Multivariable Chain Rule to compute \plz{z}{s}. + +

    + +

    + + + Use the Multivariable Chain Rule to compute \plz{z}{t}. + +

    + +

    + + + Evaluate \plz{z}{s} at s=1, t=1. + +

    + +

    + + + Evaluate \plz{z}{t} at s=1, t=1. + +

    + +

    +     +     +     + +     + + + +

    + The given equation defines y implicitly as a function of x. + Find \lz{y}{x} using Implicit Differentiation and . +

    +     + + + + +

    + x^2\tan(y) = 50 +

    +     + +

    + f_x = 2x\tan(y), f_y = x^2\sec^2(y); +

    + +

    + \ds\frac{dy}{dx} = -\frac{2\tan(y) }{x\sec^2(y) } +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $dydx=Compute("-x/y^2"); + + + +

    + \left(3x^2+2y^3\right)^4=2 +

    + + Find \lz{y}{x}: + +

    + +

    +     +     +     + + + + +

    + \ds \frac{x^2+y}{x+y^2}=17 +

    +     + +

    + \ds f_x = \frac{(x+y^2)(2x)-(x^2+y)(1)}{(x+y^2)^2}, + \ds f_y = \frac{(x+y^2)(1)-(x^2+y)(2y)}{(x+y^2)^2}; +

    + +

    + \ds\frac{dy}{dx} = -\frac{2x(x+y^2)-(x^2+y)}{x+y^2-2y(x^2+y)} +

    +     + +     + + + + + Context()->variables->add(y=>'Real'); + $dydx=Compute("-(2x+y)/(2y+x)"); + + + +

    + \ln\mathopen{}\left(x^2+xy+y^2\right)\mathclose{}=1 +

    + + Find \lz{y}{x}: + + +

    + +

    +     +     +     + +     + + + +

    + Find \lz{z}{t}, + or \plz{z}{s} and \plz{z}{t}, + using the supplied information. +

    +     + + + + +

    + \ds\frac{\partial z}{\partial x} = 2,\ds\frac{\partial z}{\partial y} = 1,\ds\frac{dx}{dt} = 4,\ds\frac{dy}{dt} = -5 +

    +     + +

    + \frac{dz}{dt} = 2(4)+1(-5) = 3. +

    +     + +     + + + + + $dzdt=Compute("0"); + + + +

    + \plz{z}{x} = 1, + \plz{z}{y} = -3, \lz{x}{t} = 6, and \lz{y}{t} = 2. +

    + + Find \lz{z}{t}: + +

    + +

    +     +     +     + + + + +

    + \ds\frac{\partial z}{\partial x} = -4,\ds\frac{\partial z}{\partial y} = 9, +

    + +

    + \ds\frac{\partial x}{\partial s} = 5,\ds\frac{\partial x}{\partial t} = 7,\ds \frac{\partial y}{\partial s} = -2,\ds \frac{\partial y}{\partial t} = 6 +

    +     + +

    + \frac{\partial z}{\partial s} = -4(5)+9(-2) = -38, +

    + +

    + \frac{\partial z}{\partial t} = -4(7)+9(6) = 26. +

    +     + +     + + + + + + $dzds=Compute("-2"); + $dzdt=Compute("5"); + + + +

    + \plz{z}{x} = 2, + \plz{z}{y} = 1, \plz{x}{s} = -2, + \plz{x}{t} = 3, \plz{y}{s} = 2 and \plz{y}{t} = -1 +

    + + + Find \plz{z}{s}: + +

    + +

    + + + Find \plz{z}{t}: + +

    + +

    +     +     +     + +     +     +     +
    +
    + Directional Derivatives + +

    + Partial derivatives give us an understanding of how a surface changes when we move in the x and y directions. + We made the comparison to standing in a rolling meadow and heading due east: + the amount of rise/fall in doing so is comparable to f_x. + Likewise, the rise/fall in moving due north is comparable to f_y. + The steeper the slope, the greater in magnitude f_y. +

    + +

    + But what if we didn't move due north or east? + What if we needed to move northeast and wanted to measure the amount of rise/fall? + Partial derivatives alone cannot measure this. + This section investigates directional derivatives, + which do measure this rate of change. +

    + + +     + + + Functions of Two Variables +

    + We begin with a definition. +

    + + + Directional Derivatives + +

    + Let z=f(x,y) be continuous on a set S and let \vec u = \la u_1,u_2\ra be a unit vector. + For all points (x,y), + the directional derivative of f at (x,y) in the direction of \vec u + is derivativedirectional + directional derivative + + D_{\vec u\,}f(x,y) = \lim_{h\to 0} \frac{f(x+hu_1,y+hu_2) - f(x,y)}h + . +

    +     +     + +

    + The partial derivatives f_x and f_y are defined with similar limits, + but only x or y varies with h, not both. + Here both x and y vary with a weighted h, + determined by a particular unit vector \vec u. + This may look a bit intimidating but in reality it is not too difficult to deal with; + it often just requires extra algebra. + However, the following theorem reduces this algebraic load. +

    + + + Directional Derivatives + +

    + Let z=f(x,y) be differentiable on a set S containing (x_0,y_0), + and let \vec u = \la u_1,u_2\ra be a unit vector. + The directional derivative of f at + (x_0,y_0) in the direction of \vec u is derivativedirectional + directional derivative + + D_{\vec u\,}f(x_0,y_0)=f_x(x_0,y_0)u_1 + f_y(x_0,y_0)u_2 + . +

    +     +     + + + Computing directional derivatives + +

    + Let z= 14-x^2-y^2 and let P=(1,2). + Find the directional derivative of f, + at P, in the following directions: +

    + +

    +

      +
    1. +

      + toward the point Q=(3,4), +

      +
      2. + +
    2. +

      + in the direction of \la 2,-1\ra, and +

      +
      3. + +
    3. +

      + toward the origin. +

      +
      4. +
    +

    +     + +

    + The surface is plotted in , + where the point P=(1,2) is indicated in the x,y-plane as well as the point (1,2,9) which lies on the surface of f. + We find that f_x(x,y) = -2x and f_x(1,2) = -2; + f_y(x,y) = -2y and f_y(1,2) = -4. +

    + +
    + Understanding the directional derivative in + + + + A surface is plotted in 3D. Three vectors in the plane are shown corresponding to three tangent vectors on the surface. + +

    + A surface in space is shown, lying above the xy plane in the first octant. + In the xy plane a grid is drawn, along with points P and Q. + At the point P there are also three vectors drawn, parallel to the xy plane. + The vector \vec{u}_1 points toward the point Q. +

    + +

    + A point on the surface is marked, directly above the point P. + At this point, we see three vectors drawn. Each vector is tangent to the surface. + The x and y components of these vectors match the corresponding vectors \vec{u}_1,\vec{u}_2,\vec{u}_3 in the plane, + but they also have a z component given by the directional derivative. + We see that, depending on what direction one moves in the plane, + our path on the surface might climb, or descend, or remain level. +

    +     + + + + + //ASY file for figdirect13D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(16,5,25); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={4}; + real[] myychoice={4}; + real[] myzchoice={10}; + defaultpen(0.5mm); + pair xbounds=(-0.5,4.75); + pair ybounds=(-0.5,4.75); + pair zbounds=(0,15); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=14-x^2-y^2 + triple f(pair t) { + return (t.x,t.y,14-t.x^2-t.y^2); + } + surface s=surface(f,(0,0),(2,2.5),6,5,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point P and on curve and point Q + dotfactor=3;dot((1,2,9)); + dot((1,2,0)); + label("$P$",(1,2,0),N); + dot((3,4,0)); + label("$Q$",(3,4,0),N); + + //draw grid lines + draw((0,1,0)--(4.5,1,0),gray+linewidth(.5)); + draw((0,2,0)--(4.5,2,0),gray+linewidth(.5)); + draw((0,3,0)--(4.5,3,0),gray+linewidth(.5)); + draw((0,4,0)--(4.5,4,0),gray+linewidth(.5)); + draw((1,0,0)--(1,4.5,0),gray+linewidth(.5)); + draw((2,0,0)--(2,4.5,0),gray+linewidth(.5)); + draw((3,0,0)--(3,4.5,0),gray+linewidth(.5)); + draw((4,0,0)--(4,4.5,0),gray+linewidth(.5)); + + //Draw the vectors in the plane from P + draw((1,2,0)--(1.707,2.707,0),Arrow3(size=2mm));//P to u1 + label("$\vec{u}_1$",(1.707,2.707,0),E);//u1 + draw((1,2,0)--(1.894,1.553,0),Arrow3(size=2mm));//P to u2 + label("$\vec{u}_2$",(1.894,1.553,0),S);//u2 + draw((1,2,0)--(0.552,1.106,0),Arrow3(size=2mm));//P to u3 + label("$\vec{u}_3$",(0.552,1.106,0),W);//u3 + + //Draw the vectors on the surface above P + draw((1,2,9)--(1.707,2.707,4.76),Arrow3(size=2mm));//P to u1 + draw((1,2,9)--(1.894,1.553,9),Arrow3(size=2mm));//P to u2 + draw((1,2,9)--(0.552,1.106,13.47),Arrow3(size=2mm));//P to u3 + + + + +
    + +

    +

      +
    1. +

      + Let \vec u_1 be the unit vector that points from the point (1,2) to the point Q=(3,4), + as shown in the figure. + The vector \overrightarrow{PQ} = \la 2,2\ra; + the unit vector in this direction is \vec u_1=\la 1/\sqrt{2}, 1/\sqrt{2}\ra. + Thus the directional derivative of f at (1,2) in the direction of \vec u_1 is + + D_{\vec u_1}f(1,2) = -2(1/\sqrt{2}) +(-4)(1/\sqrt{2}) = -6/\sqrt{2}\approx -4.24 + . + Thus the instantaneous rate of change in moving from the point (1,2,9) on the surface in the direction of \vec u_1 + (which points toward the point Q) + is about -4.24. + Moving in this direction moves one steeply downward. +

      +
      2. + +
    2. +

      + We seek the directional derivative in the direction of \la 2,-1\ra. + The unit vector in this direction is \vec u_2 = \la 2/\sqrt{5},-1/\sqrt{5}\ra. + Thus the directional derivative of f at (1,2) in the direction of \vec u_2 is + + D_{\vec u_2}f(1,2) = -2(2/\sqrt{5})+(-4)(-1/\sqrt{5}) = 0 + . + Starting on the surface of f at (1,2) and moving in the direction of \la 2,-1\ra + (or \vec u_2) + results in no instantaneous change in z-value. + This is analogous to standing on the side of a hill and choosing a direction to walk that does not change the elevation. + One neither walks up nor down, + rather just along the side of the hill. + Finding these directions of no elevation change is important. +

      +
      3. + +
    3. +

      + At P=(1,2), + the direction towards the origin is given by the vector \la -1,-2\ra; + the unit vector in this direction is \vec u_3=\la -1/\sqrt{5},-2/\sqrt{5}\ra. + The directional derivative of f at P in the direction of the origin is + + D_{\vec u_3}f(1,2) = -2(-1/\sqrt{5}) + (-4)(-2/\sqrt{5}) = 10/\sqrt{5} \approx 4.47 + . + Moving towards the origin means + walking uphill quite steeply, + with an initial slope of about 4.47. +

      +
      4. +
    +

    +     +     + + + +

    + As we study directional derivatives, + it will help to make an important connection between the unit vector + \vec u = \la u_1,u_2\ra that describes the direction and the partial derivatives f_x and f_y. + We start with a definition and follow this with a Key Idea. +

    + + + Gradient + +

    + Let z=f(x,y) be differentiable on a set S that contains the point (x_0,y_0). + gradient +

      +
    1. +

      + The gradient of f + is \nabla f(x,y) = \la f_x(x,y),f_y(x,y)\ra. +

      +
      2. + +
    2. +

      + The gradient of f at + (x_0,y_0) is \nabla f(x_0,y_0) = \la f_x(x_0,y_0),f_y(x_0,y_0)\ra. +

      +
      3. +
    +

    +     +     + + +

    + To simplify notation, + we often express the gradient as \nabla f = \la f_x, f_y\ra. + The gradient allows us to compute directional derivatives in terms of a dot product. +

    + + + The Gradient and Directional Derivatives +

    + The directional derivative of + z=f(x,y) in the direction of \vec u is + gradient + directional derivative + derivativedirectional + + D_{\vec u\,}f = \nabla f\cdot \vec u + . +

    +     + +

    + The properties of the dot product previously studied allow us to investigate the properties of the directional derivative. + Given that the directional derivative gives the instantaneous rate of change of z when moving in the direction of \vec u, + three questions naturally arise: +

    + +

    +

      +
    1. +

      + In what direction(s) is the change in z the greatest (, the + steepest uphill)? +

      +
      2. + +
    2. +

      + In what direction(s) is the change in z the least (, the + steepest downhill)? +

      +
      3. + +
    3. +

      + In what direction(s) is there no change in z? +

      +
      4. +
    +

    + +

    + Using the key property of the dot product, we have + + \nabla f\cdot \vec u = \norm{\nabla f}\,\vnorm u \cos(\theta) = \norm{\nabla f}\cos(\theta) + , + where \theta is the angle between the gradient and \vec u. + (Since \vec u is a unit vector, \vnorm{u} = 1.) + This equation allows us to answer the three questions stated previously. +

    + +

    +

      +
    1. +

      + Equation + is maximized when \cos(\theta) =1, + , when the gradient and \vec u have the same direction. + We conclude the gradient points in the direction of greatest z change. +

      +
      2. + +
    2. +

      + Equation + is minimized when \cos(\theta) = -1, + , when the gradient and \vec u have opposite directions. + We conclude the gradient points in the opposite direction of the least z change. +

      +
      3. + +
    3. +

      + Equation + is 0 when \cos(\theta) = 0, + , when the gradient and \vec u are orthogonal to each other. + We conclude the gradient is orthogonal to directions of no z change. +

      +
      4. +
    +

    + +

    + This result is rather amazing. + Once again imagine standing in a rolling meadow and face the direction that leads you steepest uphill. + Then the direction that leads steepest downhill is directly behind you, + and side-stepping either left or right (, moving perpendicularly to the direction you face) does not change your elevation at all. +

    + +

    + Recall that a level curve is defined as a curve in the xy-plane along which the z-values of a function do not change. + Let a surface z=f(x,y) be given, + and let's represent one such level curve as a vector-valued function, + \vrt = \la x(t), y(t)\ra. + As the output of f does not change along this curve, + f\big(x(t),y(t)\big) = c for all t, + for some constant c. +

    + +

    + Since f is constant for all t, \frac{df}{dt} = 0. + By the Multivariable Chain Rule, we also know + + \frac{df}{dt} \amp = f_x(x,y)x'(t) + f_y(x,y)\yp(t) + \amp = \la f_x(x,y),f_y(x,y)\ra \cdot \la x'(t),\yp(t)\ra + \amp = \nabla f \cdot \vrp(t) + \amp =0 + . +

    + +

    + This last equality states \nabla f \cdot \vrp(t) = 0: + the gradient is orthogonal to the derivative of \vec r, + meaning the gradient is orthogonal to the graph of \vec r. + Our conclusion: at any point on a surface, + the gradient at that point is orthogonal to the level curve that passes through that point. +

    + +

    + We restate these ideas in a theorem, + then use them in an example. +

    + + + The Gradient and Directional Derivatives + +

    + Let z=f(x,y) be differentiable on a set S with gradient \nabla f, + let P=(x_0,y_0) be a point in S and let \vec u be a unit vector. +

    + +

    +

      +
    1. +

      + The maximum value of D_{\vec u\,}f(x_0,y_0) is \norm{\nabla f(x_0,y_0)}; + the direction of maximal z increase is \nabla f(x_0,y_0). +

      +
      2. + +
    2. +

      + The minimum value of D_{\vec u\,}f(x_0,y_0) is -\norm{\nabla f(x_0,y_0)}; + the direction of minimal z increase is -\nabla f(x_0,y_0). +

      +
      3. + +
    3. +

      + At P, + \nabla f(x_0,y_0) is orthogonal to the level curve passing through \big(x_0,y_0\big). + gradient + directional derivative + derivativedirectional + level curves + gradientand level curves +

      +
      4. +
    +

    +     +     + + + Finding directions of maximal and minimal increase + +

    + Let f(x,y) = \sin(x) \cos(y) and let P=(\pi/3,\pi/3). + Find the directions of maximal/minimal increase, + and find a direction where the instantaneous rate of z change is 0. +

    +     + +

    + We begin by finding the gradient. + f_x = \cos(x) \cos(y) and f_y = -\sin(x) \sin(y), thus + + \nabla f = \la \cos(x) \cos(y) ,-\sin(x) \sin(y) \ra \text{ and, at \(P\), } \nabla f\left(\frac{\pi}3,\frac{\pi}3\right) = \la\frac14,-\frac34\ra + . +

    + +

    + Thus the direction of maximal increase is \la 1/4, -3/4\ra. + In this direction, + the instantaneous rate of z change is \norm{\la 1/4,-3/4\ra} = \sqrt{10}/4 \approx 0.79. +

    + +

    + + shows the surface plotted from two different perspectives. + In each, the gradient is drawn at P with a dashed line + (because of the nature of this surface, + the gradient points into the surface). + Let \vec u = \la u_1, u_2\ra be the unit vector in the direction of \nabla f at P. + Each graph of the figure also contains the vector \la u_1, u_2, \norm{\nabla f\,}\ra. + This vector has a run of 1 + (because in the xy-plane it moves 1 unit) + and a rise of \norm{\nabla f}, + hence we can think of it as a vector with slope of + \norm{\nabla f} in the direction of \nabla f, + helping us visualize how steep + the surface is in its steepest direction. +

    + +
    + Graphing the surface and important directions in + +
    + + + + + A portion of the graph f(x,y) = sin(x)cos(y), along with a trace curve, and tangent and gradient vectors. + +

    + The surface given by the graph z=\sin(x)\cos(y) resembles rolling hills, + with sinusoidal oscillations in both the x and y directions. + One of the peaks is shown; near the top of the peak a trace of constant z value is drawn. + At one point on this curve, a point is marked, and at this point there are three vectors plotted: +

      +
    • +

      + A vector tangent to the curve given by the trace. +

      +
      • +
    • +

      + A copy of the gradient vector. This vector is dashed and is covered by the surface, + making it somewhat hard to see in the default perspective. + (Note that the actual gradient vector is a vector in the plane; + a copy of it is drawn at ths point on the surface, parallel to the xy plane.) +

      +
      • +
    • +

      + A vector whose x and y components correspond to the gradient, + but whose z component is given by the magnitude of the gradient. + This vector lies tangent to the surface, but is orthogonal to the tangent vector. + It points in a direction moving up along the surface. +

      +
      • +
    +

    +     + + + + + //ASY file for figdirect23D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + + // for the default version, figdirect2 + currentprojection=orthographic(12,13,3); + + // for the second, zoomed in view, figdirect2b + //currentprojection=orthographic((5.6,16.3,2.6),(0,0,1),(0,0,0),1.6,(-0.03,-0.34)); + + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-1,6); + pair ybounds=(-1,5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=sin(x)*cos(y) + triple f(pair t) { + return (t.x,t.y,sin(t.x)*cos(t.y)); + } + surface s=surface(f,(-1.5,-1.5),(4,4),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3;dot((pi/3,pi/3,sqrt(3)/4)); + + //Draw the vectors on the surface + draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+3/sqrt(10),pi/3+1/sqrt(10),sqrt(3)/4),Arrow3(size=2mm));//T at point + draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+1/sqrt(10),pi/3-3/sqrt(10),sqrt(3)/4),linetype(new real[] {2,2}),Arrow3(size=2mm));//N at point + //draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+1/sqrt(10),pi/3-3/sqrt(10),sqrt(3)/4),linetype(new real[] {2,2}),Arrow3(size=2mm));//N at point + draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+1/sqrt(10),pi/3-3/sqrt(10),0.79+sqrt(3)/4),Arrow3(size=2mm));//tangent + + //plot level curve + draw((0.4485,0,0.4336)..(0.4693,0.2857,0.434)..(0.5417,0.5714,0.4337)..(0.7143,0.8481,0.4333)..(0.9331,1.,0.4341)..(1.143,1.075,0.4331)..(1.429,1.118,0.4333)..(1.714,1.118,0.4333)..(2.,1.074,0.4331)..(2.209,1.,0.4341)..(2.429,0.8467,0.4333)..(2.6,0.5714,0.4333)..(2.673,0.2857,0.4333)..(2.694,0,0.4332)..(2.673,-0.2857,0.4333)..(2.615,-0.5276,0.4341)..(2.531,-0.7143,0.4333)..(2.347,-0.9187,0.433)..(2.098,-1.044,0.4341)..(1.821,-1.107,0.4333)..(1.551,-1.122,0.4335)..(1.237,-1.094,0.4338)..(0.9783,-1.022,0.4329)..(0.7185,-0.8529,0.433)..(0.5714,-0.639,0.4341)..(0.4838,-0.3734,0.4331)..(0.4517,-0.1197,0.4334)..(0.4485,-0.01989,0.4335),bluepen+linewidth(2)); + + + + +
    + +
    + + + + + A zoomed-in view of the same surface, curve, and vectors as the previous image. + +

    + This is the same image as . + The only difference is that this image is zoomed in closer to the point at which the three vectors are drawn, + and the perspective is changed slightly. +

    +     + + + + + //ASY file for figdirect23D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + + // for the default version, figdirect2 + //currentprojection=orthographic(12,13,3); + + // for the second, zoomed in view, figdirect2b + currentprojection=orthographic((5.6,16.3,2.6),(0,0,1),(0,0,0),1.6,(-0.03,-0.25)); + + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-1,6); + pair ybounds=(-1,5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=sin(x)*cos(y) + triple f(pair t) { + return (t.x,t.y,sin(t.x)*cos(t.y)); + } + surface s=surface(f,(-1.5,-1.5),(4,4),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3;dot((pi/3,pi/3,sqrt(3)/4)); + + //Draw the vectors on the surface + draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+3/sqrt(10),pi/3+1/sqrt(10),sqrt(3)/4),Arrow3(size=2mm));//T at point + draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+1/sqrt(10),pi/3-3/sqrt(10),sqrt(3)/4),linetype(new real[] {2,2}),Arrow3(size=2mm));//N at point + draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+1/sqrt(10),pi/3-3/sqrt(10),0.79+sqrt(3)/4),Arrow3(size=2mm));//tangent + + //plot level curve + draw((0.4485,0,0.4336)..(0.4693,0.2857,0.434)..(0.5417,0.5714,0.4337)..(0.7143,0.8481,0.4333)..(0.9331,1.,0.4341)..(1.143,1.075,0.4331)..(1.429,1.118,0.4333)..(1.714,1.118,0.4333)..(2.,1.074,0.4331)..(2.209,1.,0.4341)..(2.429,0.8467,0.4333)..(2.6,0.5714,0.4333)..(2.673,0.2857,0.4333)..(2.694,0,0.4332)..(2.673,-0.2857,0.4333)..(2.615,-0.5276,0.4341)..(2.531,-0.7143,0.4333)..(2.347,-0.9187,0.433)..(2.098,-1.044,0.4341)..(1.821,-1.107,0.4333)..(1.551,-1.122,0.4335)..(1.237,-1.094,0.4338)..(0.9783,-1.022,0.4329)..(0.7185,-0.8529,0.433)..(0.5714,-0.639,0.4341)..(0.4838,-0.3734,0.4331)..(0.4517,-0.1197,0.4334)..(0.4485,-0.01989,0.4335),bluepen+linewidth(2)); + + + + +
    +     + +
    + +

    + The direction of minimal increase is \la -1/4,3/4\ra; + in this direction the instantaneous rate of z change is -\sqrt{10}/4 \approx -0.79. +

    + +

    + Any direction orthogonal to + \nabla f is a direction of no z change. + We have two choices: the direction of + \la 3,1\ra and the direction of \la -3,-1\ra. + The unit vector in the direction of + \la 3,1\ra is shown in each graph of the figure as well. + The level curve at z=\sqrt{3}/4 is drawn: + recall that along this curve the z-values do not change. + Since \la 3,1\ra is a direction of no z-change, + this vector is tangent to the level curve at P. +

    +     +     + + + Understanding when <m>\nabla f = \vec 0</m> + +

    + Let f(x,y) = -x^2+2x-y^2+2y+1. + Find the directional derivative of f in any direction at P=(1,1). +

    +     + +

    + We find \nabla f = \la -2x+2, -2y+2\ra. + At P, we have \nabla f(1,1) = \la 0,0\ra. + According to , + this is the direction of maximal increase. + However, \la 0,0\ra is directionless; + it has no displacement. + And regardless of the unit vector \vec u chosen, + D_{\vec u\,}f = 0. +

    + +

    + + helps us understand what this means. + We can see that P lies at the top of a paraboloid. + In all directions, the instantaneous rate of change is 0. +

    + +

    + So what is the direction of maximal increase? + It is fine to give an answer of \vec 0 = \la 0,0\ra, + as this indicates that all directional derivatives are 0. +

    + +
    + At the top of a paraboloid, all directional derivatives are 0 + + + + A downward-opening circular paraboloid, with a point P marked at the vertex. + +

    + The image shows the graph of a circular paraboloid, opening downward. + A point P is marked at the vertex of the paraboloid. + Since this is the highest point, there is no direction in which f can increase. +

    +     + + + + + //ASY file for figdirect93D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(9,3.3,3.2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={1,2}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-0.5,2.5); + pair ybounds=(-0.5,2.5); + pair zbounds=(-0.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=-x^2+2x-y^2+2y+1 -x^2-y^2+2*x+2*y+1 + triple f(pair t) { + return (t.x,t.y,-t.x^2+2*t.x-t.y^2+2*t.y+1); + } + surface s=surface(f,(0,0),(2,2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=4;dot((1,1,3)); + + label("$P$",(1,1,3),N); + + + + +
    +     +     + +

    + The fact that the gradient of a surface always points in the direction of steepest increase/decrease is very useful, + as illustrated in the following example. +

    + + + The flow of water downhill + +

    + Consider the surface given by the graph of f(x,y)= 20-x^2-2y^2. + Water is poured on the surface at (1,1/4). + What path does it take as it flows downhill? +

    +     + +

    + Let \vrt = \la x(t), + y(t)\ra be the vector-valued function describing the path of the water in the xy-plane; + we seek x(t) and y(t). + We know that water will always flow downhill in the steepest direction; + therefore, at any point on its path, + it will be moving in the direction of -\nabla f. + (We ignore the physical effects of momentum on the water.) + Thus \vrp(t) will be parallel to \nabla f, + and there is some constant c such that c\nabla f = \vrp(t) = \la x'(t), + \yp(t)\ra. +

    + +

    + We find \nabla f = \la -2x, -4y\ra and write x'(t) as + \frac{dx}{dt} and \yp(t) as \frac{dy}{dt}. + Then + + c\nabla f \amp = \la x'(t), \yp(t)\ra + \la -2cx, -4cy \ra \amp = \la \frac{dx}{dt}, \frac{dy}{dt}\ra + . +

    + +

    + This implies + + -2cx = \frac{dx}{dt} \text{ and } -4cy =\frac{dy}{dt}, \text{i.e.,} + + + c = -\frac{1}{2x}\frac{dx}{dt} \text{ and } c =-\frac{1}{4y}\frac{dy}{dt} + . +

    + +

    + As c equals both expressions, we have + + \frac{1}{2x}\frac{dx}{dt} =\frac{1}{4y}\frac{dy}{dt} + . +

    + +

    + To find an explicit relationship between x and y, + we can integrate both sides with respect to t. + Recall from our study of differentials that \frac{dx}{dt}dt = dx. + Thus: + + \int \frac{1}{2x}\frac{dx}{dt}\,dt \amp =\int \frac{1}{4y}\frac{dy}{dt}\,dt + \int \frac{1}{2x}\,dx \amp =\int\frac{1}{4y}\,dy + \frac 12\ln\abs{x} \amp = \frac14\ln\abs{y} +C_1 + 2\ln\abs{x} \amp = \ln\abs{y} +C_1 + \ln\abs{x^2} \amp = \ln\abs{y}+C_1 + Now raise both sides as a power of e: + x^2 \amp = e^{\ln\abs{y}+C_1} + x^2 \amp = e^{\ln\abs{y}}e^{C_1}\qquad \text{(Note that \(e^{C_1}\) is just a constant.)} + x^2 \amp = yC_2 + \frac1{C_2}x^2 \amp =y \qquad \text{ (Note that \(1/C_2\) is just a constant.) } + Cx^2 \amp = y + . +

    + +

    + As the water started at the point (1,1/4), + we can solve for C: + + C(1)^2 = \frac14 \Rightarrow C = \frac14 + . +

    + +
    + A sketch of the surface described in along with the path in the xy-plane with the level curves + +
    + + + + + A sector of a downward-opening elliptic paraboloid, along with a path that begins near the top, and travels downward. + +

    + The surface is an elliptic paraboloid, opening downward, with a vertex at (0,0,20). + A point is marked on the surface, a little bit below the vertex. + A curve originating at this point shows the trajectory, down along the surface, + followed by water flowing downhill due to gravity. +

    +     + + + + + //ASY file for figdirect33D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(3.6,18.5,24); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={10,20}; + defaultpen(0.5mm); + pair xbounds=(-1,5); + pair ybounds=(-1,5); + pair zbounds=(-0.5,22); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=20-x^2-2y^2 + triple f(pair t) { + return (t.x,t.y,20-t.x^2-2*t.y^2); + } + + surface s=surface(f,(0,0),(4,4),16,16,Spline); + + triple g(pair t) { + return (sqrt(20)*cos(t.x)*t.y,sqrt(10)*sin(t.x)*t.y,20-(sqrt(20)*cos(t.x)*t.y)^2-2*(sqrt(10)*sin(t.x)*t.y)^2); + } + + surface ss=surface(g,(0,0),(pi/2,1),16,16,Spline); + + pen p=apexmeshpen; + draw(ss,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3;dot((1,1/4,18.875)); + + //Draw a trace on the surface + triple g(real t) {return (t,t^2/4,20-t^2-t^4/8);} + path3 mypath=graph(g,1,3.044,operator ..); draw(mypath,bluepen); + + + + +
    + +
    + + + + A contour plot of the paraboloid in this example, and the corresponding trajectory in the plane. + +

    + The contour plot for the function f(x,y)=20-x^2-2y^2 + consists of a family of concentric ellipses, centered at the origin. + The path followed along the surface by the water projects to a path in the xy plane; + we see that this path appears to be parabolic, and intersects each level curve orthogonally. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + ymin=-3.5,ymax=3.5, + xmin=-4.2,xmax=4.2 + ] + + \addplot [secondcurvestyle,solid,domain=0:360,samples=60] ({2*cos(x)},{sqrt(2)*sin(x)}); + \addplot [secondcurvestyle,solid,domain=0:360,samples=60] ({2*sqrt(2)*cos(x)},{2*sin(x)}); + \addplot [secondcurvestyle,solid,domain=0:360,samples=80] ({2*sqrt(3)*cos(x)},{sqrt(6)*sin(x)}); + \addplot [secondcurvestyle,solid,domain=0:360,samples=80] ({4*cos(x)},{2*sqrt(2)*sin(x)}); + + \addplot [firstcurvestyle,-,domain=1:3.1] ({x},{x^2/4}); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + Thus the water follows the curve y=x^2/4 in the xy-plane. + The surface and the path of the water is graphed in . + In , + the level curves of the surface are plotted in the xy-plane, + along with the curve y=x^2/4. + Notice how the path intersects the level curves at right angles. + As the path follows the gradient downhill, + this reinforces the fact that the gradient is orthogonal to level curves. +

    +     +     + + +     + + + Functions of Three Variables +

    + The concepts of directional derivatives and the gradient are easily extended to three + (and more) + variables. + We combine the concepts behind Definitions + and + and into one set of definitions. +

    + + + Directional Derivatives and Gradient with Three Variables + +

    + Let w=F(x,y,z) be differentiable on a set D and let \vec u be a unit vector in \mathbb{R}^3. + gradient + directional derivative + derivativedirectional + +

      +
    1. +

      + The gradient of F is \nabla F = \la F_x,F_y,F_z\ra. +

      +
      2. + +
    2. +

      + The directional derivative of F in the direction of \vec u is + + D_{\vec u\,}F=\nabla F\cdot \vec u + . +

      +
      3. +
    +

    +     +     + +

    + The same properties of the gradient given in , + when f is a function of two variables, + hold for F, a function of three variables. +

    + + + The Gradient and Directional Derivatives with Three Variables + +

    + Let w=F(x,y,z) be differentiable on a set D, + let \nabla F be the gradient of F, + and let \vec u be a unit vector. + gradient + directional derivative + derivativedirectional + +

      +
    1. +

      + The maximum value of D_{\vec u\,}F is \norm{\nabla F}, + obtained when the angle between + \nabla F and \vec u is 0, , the direction of maximal increase is \nabla F. +

      +
      2. + +
    2. +

      + The minimum value of D_{\vec u\,}F is -\norm{\nabla F}, + obtained when the angle between + \nabla F and \vec u is \pi, + , the direction of minimal increase is -\nabla F. +

      +
      3. + +
    3. +

      + D_{\vec u\,}F = 0 when + \nabla F and \vec u are orthogonal. +

      +
      4. +
    +

    +     +     + +

    + We interpret the third statement of the theorem as + the gradient is orthogonal to level surfaces, + the three-variable analogue to level curves. +

    + + + Finding directional derivatives with functions of three variables + +

    + If a point source S is radiating energy, + the intensity I at a given point P in space is inversely proportional to the square of the distance between S and P. + That is, when S=(0,0,0), + I(x,y,z) = \frac{k}{x^2+y^2+z^2} for some constant k. +

    + +

    + Let k=1, + let \vec u = \la 2/3, 2/3, 1/3\ra be a unit vector, + and let P = (2,5,3). + Measure distances in inches. + Find the directional derivative of I at P in the direction of \vec u, + and find the direction of greatest intensity increase at P. +

    +     + +

    + We need the gradient \nabla I, + meaning we need I_x, I_y and I_z. + Each partial derivative requires a simple application of the Quotient Rule, giving + + \nabla I \amp = \la \frac{-2x}{(x^2+y^2+z^2)^2},\frac{-2y}{(x^2+y^2+z^2)^2},\frac{-2z}{(x^2+y^2+z^2)^2}\ra + \nabla I(2,5,3) \amp = \la \frac{-4}{1444},\frac{-10}{1444},\frac{-6}{1444}\ra \approx \la -0.003,-0.007,-0.004\ra + D_{\vec u\,}I \amp = \nabla I(2,5,3)\cdot \vec u + \amp = -\frac{17}{2166} \approx -0.0078 + . +

    + +

    + The directional derivative tells us that moving in the direction of \vec u from P results in a decrease in intensity of about -0.008 units per inch. + (The intensity is decreasing as \vec u moves one farther from the origin than P.) +

    + +

    + The gradient gives the direction of greatest intensity increase. + Notice that + + \nabla I(2,5,3) \amp = \la \frac{-4}{1444},\frac{-10}{1444},\frac{-6}{1444}\ra + \amp = \frac{2}{1444}\la -2,-5,-3\ra + . +

    + +

    + That is, the gradient at (2,5,3) is pointing in the direction of \la -2,-5,-3\ra, + that is, towards the origin. + That should make intuitive sense: + the greatest increase in intensity is found by moving towards to source of the energy. +

    +     +     + +

    + The directional derivative allows us to find the instantaneous rate of z change in any direction at a point. + We can use these instantaneous rates of change to define lines and planes that are + tangent to a surface at a point, + which is the topic of the next section. +

    +     + + + + Terms and Concepts + + + +

    + What is the difference between a directional derivative and a partial derivative? +

    +     + + + +

    + A partial derivative is essentially a special case of a directional derivative; + it is the directional derivative in the direction of x or y, + , \la 1,0\ra or \la 0,1\ra. +

    +     + +     + + + + + +

    + For f(x,y), + for what \vec u is D_{\vec u}\, f = f_x? +

    + + +     + + + +

    + \veci +

    +     +     + + +

    + \vecj +

    +     +     + + +

    + \veck +

    +     +     +     + +     + + + + + +

    + For f(x,y), + for what \vec u is D_{\vec u}\, f = f_y? +

    + + +     + + + +

    + \veci +

    +     +     + + +

    + \vecj +

    +     +     + + +

    + \veck +

    +     +     +     + +     + + + + + +

    + The gradient is to level curves. +

    +     + + + + + + + perpendicular + +

    + There is a synonym for this that is more appropriate when referring to vectors. +

    +     +     +     +     + +     + + + + +

    + The gradient points in the direction of increase. +

    +     + + + + maximal|maximum|greatest|largest + + + + +     + + + + + +

    + It is generally more informative to view the directional derivative not as the result of a limit, + but rather as the result of a product. +

    +     + + + + + + + + +     +     + + + Problems + + + +

    + A function f(x,y) is given. + Find \nabla f. +

    +     + + + + +

    + f(x,y) = -x^2y+xy^2+xy +

    +     + +

    + \nabla f = \la -2xy+y^2+y, -x^2+2xy+x\ra +

    +     + +     + + + + + Context("Vector2D"); + $gradient=Compute("<cos(x)cos(y),-sin(x)sin(y)>"); + + + +

    + Find \nabla f, where f(x,y) = \sin(x)\cos(y). +

    + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) = \frac{1}{x^2+y^2+1} +

    +     + +

    + \nabla f = \la \frac{-2x}{(x^2+y^2+1)^2}, \frac{-2y}{(x^2+y^2+1)^2}\ra +

    +     + +     + + + + + Context("Vector2D"); + $gradient=Compute("<-4,3>"); + + + +

    + Find \nabla f, where f(x,y) = -4x+3y. +

    + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) = x^2+2y^2-xy-7x +

    +     + +

    + \nabla f = \la 2x-y-7,4y-x\ra +

    +     + +     + + + + + Context("Vector2D"); + $gradient=Compute("<2xy^3-2,3x^2y^2>"); + + + +

    + Find \nabla f, where f(x,y) = x^2y^3-2x. +

    + +

    + +

    +     +     +     + +     + + + +

    + A function f(x,y) and a point P are given. + Find the directional derivative of f in the indicated directions. + Note: these are the same functions as in Exercises. +

    +     + + + + +

    + f(x,y) = -x^2y+xy^2+xy, P= (2,1) +

    +     + + +

    + In the direction of \vec v = \la 3,4\ra +

    +     + +

    + 2/5 +

    +     + +

    + \nabla f = \la -2xy+y^2+y, -x^2+2xy+x\ra; + \nabla f(2,1) = \la -2,2\ra. + The unit vector in the direction of \vec v is \vec u = \la 3/5,4/5\ra. + Therefore, D_{\vec u}f(2,1) = -2(3/5)+2(4/5)=2/5. +

    +     +     + + +

    + In the direction toward the point Q = (1,-1). +

    +     + +

    + -2/\sqrt{5} +

    +     + +

    + \nabla f = \la -2xy+y^2+y, -x^2+2xy+x\ra; + \nabla f(2,1) = \la -2,2\ra. + The vector from P to Q is \vec v = \la 1-2, -1-1\ra = \la -1,-2\ra. + The unit vector in the direction of \vec v is \vec u = \la -1/\sqrt{5},-2/\sqrt{5}\ra. + Therefore, D_{\vec{u}}f(2,1)=-2(-1/\sqrt{5})+2(-2/\sqrt{5})=-2/\sqrt{5}. +

    +     +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $a=Compute("1/4 (1-sqrt(3))"); + $b=Compute("(4sqrt(3)-3)/(10 sqrt(2))"); + + + +

    + Consider f(x,y) = \sin(x)\cos(y), + at P = \left(\frac{\pi}{4},\frac{\pi}{3}\right). +

    +     + + +

    + In the direction of \vec v=\la 1,1\ra. +

    + +

    + +

    +     +     + + + +

    + In the direction toward the point Q = (0,0). +

    + +

    + +

    +     +     +     +     + + + + +

    + \ds f(x,y) = \frac{1}{x^2+y^2+1}, P = (1,1). +

    +     + + +

    + In the direction of \vec v = \la 1,-1\ra. +

    +     + +

    + 0 +

    +     + +

    + \nabla f = \la \frac{-2x}{(x^2+y^2+1)^2}, \frac{-2y}{(x^2+y^2+1)^2}\ra; + \nabla f(1,1) = \la -2/9,-2/9\ra. + The unit vector in the direction of \vec v is \vec u = \la 1/\sqrt{2}, -1/\sqrt{2}\ra. + The directional derivative is D_{\vec u}f(1,1) = -2/9(1\/sqrt{2})-2/9(-1/\sqrt{2})=0. +

    +     +     + + +

    + In the direction toward the point Q = (-2,-2). +

    +     + +

    + 2\sqrt{2}/9 +

    +     + +

    + \nabla f = \la \frac{-2x}{(x^2+y^2+1)^2}, \frac{-2y}{(x^2+y^2+1)^2}\ra; + \nabla f(1,1) = \la -2/9,-2/9\ra. + The vector from P to Q is \vec v = \la -2-1,-2-1\ra = \la -3, -3\ra. + The unit vector in the direction of \vec v is \vec u = \la -1/\sqrt{2}, -1/\sqrt{2}\ra. + The directional derivative is D_{\vec u}f(1,1) = -2/9(-1\/sqrt{2})-2/9(-1/\sqrt{2})=2\sqrt{2}/9. +

    +     +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $a=Compute("-9/sqrt(10)"); + $b=Compute("27/sqrt(34)"); + + + +

    + Consider f(x,y) = -4x+3y, at P = (5,2). +

    +     + + +

    + In the direction of \vec v=\la 3,1\ra. +

    + +

    + +

    +     +     + + +

    + In the direction toward the point Q = (2,7). +

    + +

    + +

    +     +     +     +     + + + + +

    + \ds f(x,y) = x^2+2y^2-xy-7x, P = (4,1) +

    +     + + +

    + In the direction of \vec v = \la -2,5\ra +

    +     + +

    + 0 +

    +     + +

    + \nabla f = \la 2x-y-7,4y-x\ra; + \nabla f(4,1) = \la 0,0\ra. + Since the gradient is zero, so is the directional derivative. +

    +     +     + + +

    + In the direction toward the point Q = (4,0). +

    +     + +

    + 0 +

    +     + +

    + \nabla f = \la 2x-y-7,4y-x\ra; + \nabla f(4,1) = \la 0,0\ra. + Since the gradient is zero, so is the directional derivative. +

    +     +     + +     + + + + + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $a=Compute("3/sqrt(2)"); + $b=Compute("3"); + + + +

    + Consider f(x,y) = x^2y^3-2x, at P = (1,1). +

    +     + + + +

    + Find the directional derivative in the direction of \vec v=\la 3,3\ra. +

    + +

    + +

    +     +     + + + +

    + Find the directional derivative in the direction toward the point Q = (1,2). +

    + +

    + +

    +     +     +     +     + +     + + + +

    + A function f(x,y) and a point P are given. + Investigate the directions of maximal increase and decrease, as indicated. +

    + +

    + Note: these are the same functions and points as in Exercises. +

    +     + + + + +

    + f(x,y) = -x^2y+xy^2+xy, P= (2,1) +

    +     + + +

    + Find the direction of maximal increase of f at P. +

    +     + +

    + \nabla f(2,1) = \la -2,2\ra +

    +     + +

    + The gradient of f is \nabla f(x,y) = \la -2xy+y^2+y, -x^2+2xy+x\ra. + The direction of maximal increase is \nabla f(2,1) = \la -2,2\ra. +

    +     +     + + +

    + What is the maximal value of D_{\vec u}\,f at P? +

    +     + +

    + \sqrt{8} +

    +     + +

    + The maximal value is the norm of the gradient at P, + which is \norm{\nabla f(2,1)} = \norm{\la -2,2\ra} = \sqrt{8}. +

    +     +     + + +

    + Find the direction of maximal decrease in f at P. +

    +     + +

    + \la 2, -2\ra +

    +     + +

    + This is the direction opposite the direction of maximal increase; therefore, \la 2,-2\ra. +

    +     +     + + +

    + Give a direction \vec u such that D_{\vec u}\,f=0 at P. +

    +     + +

    + \vec u = \la 1/\sqrt{2},1/\sqrt{2}\ra +

    +     + +

    + We want \vec u = \la a, b\ra such that \nabla f(2,1)\cdot \vec u = -2a+2b=0. + One such unit vector is \vec u = \la 1/\sqrt{2},1/\sqrt{2}\ra. +

    +     +     + +     + + + + + Context("Vector2D"); + Context()->flags->set(reduceConstants=>0,redcuceConstantFunctions=>0); + $maxdir=Compute("<1/(2sqrt(2)), -1/2 sqrt(3/2)>"); + $maxdirev=$maxdir->cmp(parallel=>1,sameDirection=>1); + $maxval=Compute("1/sqrt(2)"); + $mindir=Compute("<-1/(2sqrt(2)), 1/2 sqrt(3/2)>"); + $mindirev=$mindir->cmp(parallel=>1,sameDirection=>1); + $nochange=Compute("<1/2 sqrt(3/2), 1/(2 sqrt(2))>"); + $nochangeev=$nochange->cmp(parallel=>1); + + + +

    + f(x,y) = \sin(x)\cos(y), + P = \left(\frac{\pi}{4},\frac{\pi}{3}\right): +

    +     + + + +

    + Find the direction of maximal increase of f at P. +

    + +

    + +

    +     +     + + +

    + What is the maximal value of D_{\vec u}\,f at P? +

    + +

    + +

    +     +     + + +

    + Find the direction of maximal decrease in f at P. +

    + +

    + +

    +     +     + + +

    + Give a direction \vec u such that D_{\vec u}\,f=0 at P. +

    + +

    + +

    +     +     +     +     + + + + +

    + \ds f(x,y) = \frac{1}{x^2+y^2+1}, P = (1,1). +

    +     + + + +

    + Find the direction of maximal increase of f at P. +

    +     + +

    + \nabla f(1,1) = \la -2/9,-2/9\ra +

    +     + +

    + The gradient of f is \nabla f = \la \frac{-2x}{(x^2+y^2+1)^2}, \frac{-2y}{(x^2+y^2+1)^2}\ra. + The direction of maximal increase is \nabla f(1,1) = \la -2/9,-2/9\ra. +

    +     +     + + +

    + What is the maximal value of D_{\vec u}\,f at P? +

    +     + +

    + 2\sqrt{2}/9 +

    +     + +

    + The maximal value is the norm of the gradient at P, + which is \norm{\nabla f(1,1)} = \norm{\la -2/9,-2/9\ra}=2\sqrt{2}/9. +

    +     +     + + +

    + Find the direction of maximal decrease in f at P. +

    +     + +

    + \la 2/9,2/9\ra +

    +     + +

    + This is the direction opposite the direction of maximal increase; therefore, \la 2/9,2/9\ra. +

    +     +     + + +

    + Give a direction \vec u such that D_{\vec u}\,f=0 at P. +

    +     + +

    + \vec u = \la 1/\sqrt{2},-1/\sqrt{2}\ra +

    +     + +

    + We want \vec u = \la a, b\ra such that \nabla f(1,1)\cdot \vec u = -\frac29 a-\frac29 b=0. + One such unit vector is \vec u = \la 1/\sqrt{2},-1/\sqrt{2}\ra. +

    +     +     + +     + + + + + Context("Vector2D"); + Context()->flags->set(reduceConstants=>0,redcuceConstantFunctions=>0); + $maxdir=Compute("<-4, 3>"); + $maxdirev=$maxdir->cmp(parallel=>1,sameDirection=>1); + $maxval=Compute("5"); + $mindir=Compute("<4,-3>"); + $mindirev=$mindir->cmp(parallel=>1,sameDirection=>1); + $nochange=Compute("<3,4>"); + $nochangeev=$nochange->cmp(parallel=>1); + + + +

    + f(x,y) = -4x+3y, P = (5,4) +

    +     + + + +

    + Find the direction of maximal increase of f at P. +

    + +

    + +

    +     +     + + +

    + What is the maximal value of D_{\vec u}\,f at P? +

    + +

    + +

    +     +     + + +

    + Find the direction of maximal decrease in f at P. +

    + +

    + +

    +     +     + + +

    + Give a direction \vec u such that D_{\vec u}\,f=0 at P. +

    + +

    + +

    +     +     +     +     + + + + +

    + \ds f(x,y) = x^2+2y^2-xy-7x, P = (4,1) +

    +     + + + +

    + Find the direction of maximal increase of f at P. +

    +     + +

    + No such direction +

    +     + +

    + The gradient of f is \nabla f(x,y) = \la 2x-y-7,4y-x\ra. + At the point P we get \nabla f(4,1) = \la 0,0\ra, + so P is a critical point; there is no such direction. +

    +     +     + + +

    + What is the maximal value of D_{\vec u}\,f at P? +

    +     + +

    + 0 +

    +     + +

    + The maximal value is the norm of the gradient at P, + which is \norm{\nabla f(4,1)} = 0. +

    +     +     + + +

    + Find the direction of maximal decrease in f at P. +

    +     + +

    + No such direction +

    +     + +

    + There is no direction of maximal decrease, since P is a critical point. +

    +     +     + + +

    + Give a direction \vec u such that D_{\vec u}\,f=0 at P. +

    +     + +

    + All directions +

    +     + +

    + Since \nabla f(4,1)=\la 0,0\ra, this is true in all directions. +

    +     +     + +     + + + + + Context("Vector2D"); + Context()->flags->set(reduceConstants=>0,redcuceConstantFunctions=>0); + $maxdir=Compute("<0, 3>"); + $maxdirev=$maxdir->cmp(parallel=>1,sameDirection=>1); + $maxval=Compute("3"); + $mindir=Compute("<0,-3>"); + $mindirev=$mindir->cmp(parallel=>1,sameDirection=>1); + $nochange=Compute("<1,0>"); + $nochangeev=$nochange->cmp(parallel=>1); + + + +

    + Given f(x,y) = x^2y^3-2x, P = (1,1): +

    +     + + + +

    + Find the direction of maximal increase of f at P. +

    + +

    + +

    +     +     + + +

    + What is the maximal value of D_{\vec u}\,f at P? +

    + +

    + +

    +     +     + + +

    + Find the direction of maximal decrease in f at P. +

    + +

    + +

    +     +     + + +

    + Give a direction \vec u such that D_{\vec u}\,f=0 at P. +

    + +

    + +

    +     +     +     +     + +     + + + +

    + A function F(x,y,z), + a vector \vec v and a point P are given. +

    + +

    + Compute the gradient of F, and the derivative of F in the direction of \vec v at P. +

    +     + + + + +

    + \ds F(x,y,z) = 3x^2z^3+4xy-3z^2, + \vec v = \la 1,1,1\ra, P = (3,2,1) +

    +     + + + +

    + Compute the gradient of F. +

    +     + +

    + \nabla F(x,y,z) = \la 6xz^3+4y, 4x, 9x^2z^2-6z\ra +

    +     +     + + +

    + Find the derivative of F at P in the direction of \vec v. +

    +     + +

    + 113/\sqrt{3} +

    +     + +

    + The gradient of F at P is \nabla F(3,2,1) = \la 26, 12, 75\ra. + A unit vector in the direction of \vec v is \vec u = \la 1/\sqrt{3},1/\sqrt{3},1/\sqrt{3}\ra. + The directional derivative is D_{\vec u}f(3,2,1) = \nabla F(3,2,1)\cdot \vec u = 113/\sqrt{3}. +

    +     +     + +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,redcuceConstantFunctions=>0); + $gradient=Compute("<cos(x) cos(y) e^z, -sin(x) sin(y) e^z, sin(x) cos(y) e^z>"); + $directional=Compute("2/3"); + + + +

    + F(x,y,z) = \sin(x)\cos(y)e^z, + \vec v = \la 2,2,1\ra, P = (0,0,0). +

    +     + + + +

    + Find \nabla F(x,y,z). +

    + +

    + +

    +     +     + + + +

    + Find D_{\vec u}\,F at P. +

    + +

    + +

    +     +     +     +     + + + + +

    + \ds F(x,y,z) = x^2y^2-y^2z^2, + \vec v = \la -1,7,3\ra, P = (1,0,-1) +

    +     + + + +

    + Compute the gradient of F. +

    +     + +

    + \nabla F(x,y,z) = \la 2xy^2, 2y(x^2-z^2), -2y^2z\ra +

    +     +     + + +

    + Find the derivative of F at P in the direction of \vec v. +

    +     + +

    + 0 +

    +     + +

    + The gradient of F at P is \nabla F(1,0,-1) = \la 0,0,0\ra. + Therefore, the derivative of F at P in any direction is 0. +

    +     +     + +     + + + + + Context("Vector"); + Context()->flags->set(reduceConstants=>0,redcuceConstantFunctions=>0); + $gradient=Compute("<-4x/(x^2+y^2+z^2)^2, -4y/(x^2+y^2+z^2)^2, -4z/(x^2+y^2+z^2)^2>"); + $directional=Compute("0"); + + + +

    + Given F(x,y,z) = \frac{2}{x^2+y^2+z^2}, + \vec v = \la 1,1,-2\ra, P = (1,1,1): +

    +     + + + +

    + Find \nabla F(x,y,z). +

    + +

    + +

    +     +     + + + +

    + Find D_{\vec u}\,F at P. +

    + +

    + +

    +     +     +     +     + +     +     +     +
    +
    + Tangent Lines, Normal Lines, and Tangent Planes + + Tangent Lines + +

    + Derivatives and tangent lines go hand-in-hand. + Given y=f(x), + the line tangent to the graph of f at x=x_0 is the line through + \big(x_0,f(x_0)\big) with slope \fp(x_0); + that is, the slope of the tangent line is the instantaneous rate of change of f at x_0. +

    + +

    + When dealing with functions of two variables, + the graph is no longer a curve but a surface. + At a given point on the surface, + it seems there are many lines that fit our intuition of being + tangent to the surface. +

    + +

    + In we introduced the concept of the tangent plane, + which could be thought of as consisting of all possible lines tangent to the surface at a given point. + In this section, we explore this idea in more detail. +

    + +
    + Showing various lines tangent to a surface + + + + A surface in three dimensions, including curves on the surface, and tangent lines to those curves. + +

    + A curved surface in space is plotted against a three-dimensional coordinate system. + It appears to be a portion of a downward-opening elliptic paraboloid. + A point is plotted on the surface, and there are several curves on the surface passing through that point. + At this point where the curves meet, a tangent line is drawn to each of the curves. + Since the curves lie in the surface, we can say that these lines are tangent to the surface itself, + and it appears that they all lie in a common plane. +

    +     + + + + + //ASY file for figspace_tangent_intro3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8,8,13); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={10}; + defaultpen(0.5mm); + pair xbounds=(-0.5,3); + pair ybounds=(-0.5,3); + pair zbounds=(-0.5,11); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=-x^2-.5*(y^2)+x*y+10 //(x,y,{-x^2-.5*(y^2)+x*y+10}); + triple f(pair t) { + return (t.x,t.y,-t.x^2-.5*(t.y^2)+t.x*t.y+10); + } + surface s=surface(f,(0,0),(3,3),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3;dot((2,1,7.5)); + + //Draw traces on the surface + triple g(real t) {return (2,t,-4-0.5*t^2+2*t+10);}//-x^2-.5*(y^2)+x*y+10 + path3 mypath=graph(g,0,3,operator ..); draw(mypath,bluepen); + triple g(real t) {return (t,1,-t^2-0.5+t+10);}//-x^2-.5*(y^2)+x*y+10 + path3 mypath=graph(g,0,3,operator ..); draw(mypath,bluepen); + triple g(real t) {return (t,3-t,-t^2-0.5*(3-t)^2+(3-t)*t+10);} + path3 mypath=graph(g,0,3,operator ..); draw(mypath,bluepen); + + //Draw tangent lines on the surface with gradient (-3,1) + //L1 = (2,1,7.5)+t(1,0,-3) for t=-1,1 + draw((1,1,10.5)--(3,1,4.5),redpen); + //L2 = (2,1,7.5)+t(0,1,1) for t=-1,1 + draw((2,0,6.5)--(2,2,8.5),redpen); + //L3 = (2,1,7.5)+t(0.707,-0.707,-2.828) for t=-1,1 + draw((1.293,1.707,10.328)--(2.707,0.293,4.672),redpen); + + + + +
    + +

    + In + we see lines that are tangent to curves in space. + Since each curve lies on a surface, + it makes sense to say that the lines are also tangent to the surface. + The next definition formally defines what it means to be + tangent to a surface. +

    + + + Directional Tangent Line + +

    + Let z=f(x,y) be differentiable on a set S containing + (x_0,y_0) and let \vec u = \la u_1, u_2\ra be a unit vector. + tangent linedirectional +

    + +

    +

      +
    1. +

      + The line \ell_x through + \big(x_0,y_0,f(x_0,y_0)\big) parallel to \la 1,0,f_x(x_0,y_0)\ra is the + tangent line to f in the direction of x at (x_0,y_0). +

      +
      2. + +
    2. +

      + The line \ell_y through + \big(x_0,y_0,f(x_0,y_0)\big) parallel to \la 0,1,f_y(x_0,y_0)\ra is the + tangent line to f in the direction of y at (x_0,y_0). +

      +
      3. + +
    3. +

      + The line \ell_{\vec u} through + \big(x_0,y_0,f(x_0,y_0)\big) parallel to \la u_1,u_2,D_{\vec u\,}f(x_0,y_0)\ra is the + tangent line to f in the direction of \vec u at (x_0,y_0). +

      +
      4. +
    +

    +     +     + +

    + It is instructive to consider each of three directions given in the definition in terms of slope. + The direction of \ell_x is \la 1,0,f_x(x_0,y_0)\ra; + that is, the run is one unit in the x-direction and the rise + is f_x(x_0,y_0) units in the z-direction. + Note how the slope is just the partial derivative with respect to x. + A similar statement can be made for \ell_y. + The direction of \ell_{\vec u} is \la u_1,u_2,D_{\vec u\,}f(x_0,y_0)\ra; + the run is one unit in the \vec u direction + (where \vec u is a unit vector) + and the rise is the directional derivative of z in that direction. +

    + +

    + + leads to the following parametric equations of directional tangent lines: +

    + +

    + + \ell_x(t) \amp = \left\{\begin{array}{l} x=x_0+t \\ y=y_0\\z=z_0+f_x(x_0,y_0)t + \end{array} \right. + \ell_y(t) \amp = \left\{\begin{array}{l} x=x_0 \\ y=y_0+t\\z=z_0+f_y(x_0,y_0)t + \end{array} \right. + \ell_{\vec u}(t) \amp = \left\{\begin{array}{l} x=x_0+u_1t \\ y=y_0+u_2t\\z=z_0+D_{\vec u\,}f(x_0,y_0)t + \end{array} \right. + . +

    + + + Finding directional tangent lines + +

    + Find the lines tangent to the surface z=\sin(x) \cos(y) at + (\pi/2,\pi/2) in the x and y directions and also in the direction of \vec v = \la -1,1\ra. +

    +     + +

    + The partial derivatives with respect to x and y are: + + f_x(x,y) = \cos(x) \cos(y) \amp \Rightarrow f_x(\pi/2,\pi/2) = 0 + f_y(x,y) = -\sin(x) \sin(y) \amp \Rightarrow f_y(\pi/2,\pi/2)=-1 + . +

    + +

    + At (\pi/2,\pi/2), the z-value is 0. +

    + +

    + Thus the parametric equations of the line tangent to f at + (\pi/2,\pi/2) in the directions of x and y are: + + \ell_x(t) = \left\{\begin{array}{l} x=\pi/2 + t\\ y=\pi/2 \\z=0 + \end{array} \right. \text{ and } + \ell_y(t) = \left\{\begin{array}{l} x=\pi/2 \\ y=\pi/2+t \\z=-t + \end{array} \right. + . +

    + +

    + The two lines are shown with the surface in . +

    + +
    + A surface and directional tangent lines in + +
    + + + + + A bumpy surface. At a point on the surface, the tangent lines given by the partial derivatives are shown. + +

    + One of the hills in the bumpy surface given by f(x,y) = \sin(x)\cos(y) is shown, + as well parts of adjacent hills and valleys. + A point is plotted on one face of this hill. + Two lines are drawn through this point, tangent to the surface. + One line appears to be parallel to the x axis; + its slope relative to z is given by the partial derivative with respect to x. + The other line lies in a plane parallel to the yz plane, + and has slope given by the partial derivative with respect to y. +

    +     + + + + + //ASY file for figpartial4a3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-0.5,5); + pair ybounds=(-0.5,4.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=sin(x)cos(y) + triple f(pair t) { + return (t.x,t.y,sin(t.x)*cos(t.y)); + } + surface s=surface(f,(-1,-1),(4,4),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3;dot((pi/2,pi/2,0)); + + //Draw lines on the surface + //L1 = (pi/2,pi/2,0)+t(1,0,0) for t=-1,1 + draw((0.57,1.57,0)--(2.57,1.57,0),redpen); + //L2 = (pi/2,pi/2,0)+t(0,1,-1) for t=-1,1 + draw((1.57,.57,1)--(1.57,2.57,-1),redpen); + + + + +
    + +
    + + + + + A curve lies along a bumpy surface. At one point on the curve, a tangent line to the curve is drawn. + +

    + The same bumpy surface as is shown. + This time, we see a curve that lies in the surface; + the curve lies above the line x+y=\pi in the xy plane, + and passes through the same point as in the previous image. + The tangent line to the curve at this point is plotted; + it is the line tangent to the surface at (\pi/2, \pi/2) in the direction of \langle -1, 1\rangle. +

    +     + + + + + //ASY file for figpartial4b3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-0.5,5); + pair ybounds=(-0.5,4.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=sin(x)cos(y) + triple f(pair t) { + return (t.x,t.y,sin(t.x)*cos(t.y)); + } + surface s=surface(f,(-1,-1),(4,4),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3;dot((pi/2,pi/2,0)); + + //Draw traces on the surface + triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} + path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); + + //Draw lines on the surface + //L1 = (pi/2,pi/2,0)+t(-0.707,0.707,-0.707) for t=-1,1 + draw((2.28,.864,0.707)--(.864,2.28,-0.707),redpen); + + + + +
    +     +
    + +

    + To find the equation of the tangent line in the direction of \vec v, + we first find the unit vector in the direction of \vec v: + \vec u = \la -1/\sqrt{2},1/\sqrt{2}\ra. + The directional derivative at + (\pi/2,\pi,2) in the direction of \vec u is + + D_{\vec u\,}f(\pi/2,\pi,2) = \la 0,-1\ra \cdot \la -1/\sqrt{2},1/\sqrt 2\ra = -1/\sqrt 2 + . +

    + +

    + Thus the directional tangent line is + + \ell_{\vec u}(t) = \left\{\begin{array}{l} x= \pi/2 -t/\sqrt{2}\\ y = \pi/2 + t/\sqrt{2} \\ z= -t/\sqrt{2} + \end{array} \right. + . +

    + +

    + The curve through (\pi/2,\pi/2,0) in the direction of \vec v is shown in along with \ell_{\vec u}(t). +

    +     +     + + + Finding directional tangent lines + +

    + Let f(x,y) = 4xy-x^4-y^4. + Find the equations of all + directional tangent lines to f at (1,1). +

    +     + +

    + First note that f(1,1) = 2. + We need to compute directional derivatives, + so we need \nabla f. + We begin by computing partial derivatives. + + f_x = 4y-4x^3 \Rightarrow f_x(1,1) = 0; f_y = 4x-4y^3\Rightarrow f_y(1,1) = 0 + . +

    + +

    + Thus \nabla f(1,1) = \la 0,0\ra. + Let \vec u = \la u_1,u_2\ra be any unit vector. + The directional derivative of f at (1,1) will be D_{\vec u\,}f(1,1) = \la 0,0\ra\cdot \la u_1,u_2\ra = 0. + It does not matter what direction we choose; + the directional derivative is always 0. + Therefore + + \ell_{\vec u}(t) = \left\{\begin{array}{l} x= 1 +u_1t\\ y = 1+ u_2 t\\ z= 2 + \end{array} \right. + . +

    + +

    + + shows a graph of f and the point (1,1,2). + Note that this point comes at the top of a hill, + and therefore every tangent line through this point will have a slope of 0. +

    + +
    + Graphing f in + + + + A graph of the function in this example; it has the shape of a steep hill. + +

    + The graph z = 4xy-x^4-y^4 is plotted against a set of three-dimensional coordinate axes. + The shape of the surface given by the graph is that of a steep hill: + the degree-four terms in the function result in a surface that descends more rapidly than a paraboloid. +

    + +

    + The peak of the hill appears to be at the point (1,1,2). +

    +     + + + + + //ASY file for figpartial4b3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8,2.6,4.7); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={1,2}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-0.5,2.5); + pair ybounds=(-0.5,2.5); + pair zbounds=(-3.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=\addplot3[domain=-.1:2,y domain=-.1:2,surf,colormap={mp2}{\colormapplaneone}//,shader=faceted,faceted color=black!40,samples y=20,very thin,z buffer=sort,%opacity=.6, + //samples=20,] (x,y,{4*x*y-x^4-y^4}); + + bool cond(pair t) {return 4*t.x*t.y-(t.x)^4-(t.y)^4 >-3;} + + //triple f(pair t) { + // return (t.x,t.y,4*t.x*t.y-(t.x)^4-(t.y)^4); + //} + //surface s=surface(f,(-.1,-.1),(1.5,1.5),32,32,Spline); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (cos(t.x)*t.y+1,sin(t.x)*t.y+1,4*(cos(t.x)*t.y+1)*(sin(t.x)*t.y+1)-(cos(t.x)*t.y+1)^4-(sin(t.x)*t.y+1)^4); + } + surface s=surface(f,(0,0),(2*pi,.75),32,32,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3; + dot((1,1,2)); + + //Draw traces on the surface + //triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} + //path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); + + + + +
    + +

    + That is, consider any curve on the surface that goes through this point. + Each curve will have a relative maximum at this point, + hence its tangent line will have a slope of 0. + The following section investigates the points on surfaces where all tangent lines have a slope of 0. +

    +     +     +     + + + Normal Lines +

    + When dealing with a function y=f(x) of one variable, + we stated that a line through (c,f(c)) was tangent + to f if the line had a slope of \fp(c) and was + normal (or, perpendicular, + orthogonal) to f if it had a slope of -1/\fp(c). + We extend the concept of normal, + or orthogonal, to functions of two variables. +

    + +

    + Let z=f(x,y) be a differentiable function of two variables. + By , + at (x_0,y_0), + \ell_x(t) is a line parallel to the vector \vec d_x=\la 1,0,f_x(x_0,y_0)\ra and + \ell_y(t) is a line parallel to \vec d_y=\la 0,1,f_y(x_0,y_0)\ra. + Since lines in these directions through \big(x_0,y_0,f(x_0,y_0)\big) are + tangent to the surface, + a line through this point and orthogonal to these directions would be orthogonal, + or normal, to the surface. + We can use this direction to create a normal line. +

    + +

    + The direction of the normal line is orthogonal to \vec d_x and \vec d_y, + hence the direction is parallel to \vec d_n = \vec d_x\times \vec d_y. + It turns out this cross product has a very simple form: + + \vec d_x\times \vec d_y = \la 1,0,f_x\ra \times \la 0,1,f_y\ra = \la -f_x,-f_y,1\ra + . +

    + +

    + It is often more convenient to refer to the opposite of this direction, + namely \la f_x,f_y,-1\ra. + This leads to a definition. +

    + + + Normal Line + +

    + Let z=f(x,y) be differentiable on a set S containing (x_0,y_0) where + + a = f_x(x_0,y_0) \text{ and } b=f_y(x_0,y_0) + + are defined. + normal line + orthogonal +

    + +

    +

      +
    1. +

      + A nonzero vector parallel to \vec n=\la a,b,-1\ra is + orthogonal to f at P=\big(x_0,y_0,f(x_0,y_0)\big). +

      +
      2. + +
    2. +

      + The line \ell_n through P with direction parallel to \vec n is the + normal line to f at P. +

      +
      3. +
    +

    +     +     + +

    + Thus the parametric equations of the normal line to a surface z=f(x,y) at \big(x_0,y_0,f(x_0,y_0)\big) is: + + \ell_{n}(t) = \left\{\begin{array}{l} x= x_0+at\\ y = y_0 + bt \\ z = f(x_0,y_0) - t + \end{array} \right. + . +

    + + + Finding a normal line + +

    + Find the equation of the normal line to z=-x^2-y^2+2 at (0,1). +

    +     + +

    + We find z_x(x,y) = -2x and z_y(x,y) = -2y; + at (0,1), we have z_x = 0 and z_y = -2. + We take the direction of the normal line, + following , + to be \vec n=\la 0,-2,-1\ra. + The line with this direction going through the point (0,1,1) is + + \ell_n(t) = \left\{\begin{array}{l} x=0\\y=-2t+1\\z=-t+1 + \end{array} \right. \text{ or } \ell_n(t)=\la 0,-2,-1\ra t+\la 0,1,1\ra + . +

    + +
    + Graphing a surface with a normal line from + + + + A circular paraboloid, opening downward. At one point on the surface, a normal line is drawn. + +

    + The graph z=-x^2-y^2+2 is a circular paraboloid with vertex at (0,0,2), + and opening downward. + At the point (0,1,1) on the surface, a point is plotted, + and through this point, a normal line is drawn. + We can see that the normal line is perpendicular to the surface at the point where they meet. +

    +     + + + + + //ASY file for figpartial4b3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8,4.8,2.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=\addplot3[domain=-.1:2,y domain=-.1:2,surf,colormap={mp2}{\colormapplaneone}//,shader=faceted,faceted color=black!40,samples y=20,very thin,z buffer=sort,%opacity=.6, + //samples=20,] (x,y,{4*x*y-x^4-y^4}); + + bool cond(pair t) {return 2-(t.x)^2-(t.y)^2 >-2;} + + //triple f(pair t) { + // return (t.x,t.y,2-t.x^2-t.y^2); + //} + + triple f(pair t) { + return (cos(t.x)*t.y,sin(t.x)*t.y,2-t.y^2); + } + surface s=surface(f,(0,0),(2*pi,2),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + draw ((0,2.5,1.75)--(0,-1,0),redpen+2bp); + + //triple f(pair t) { + // return (cos(t.x)*t.y,sin(t.x)*t.y,4*cos(t.x)*t.y*sin(t.x)*t.y-(cos(t.x)*t.y)^4-(sin(t.x)//*t.y)^4); + //} + //surface s=surface(f,(-.1,0),(pi/2+.1,1.6),32,32,cond); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3; + dot((0,1,1)); + + //Draw traces on the surface + //triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} + //path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); + + + + +
    + +

    + The surface z=-x^2-y^2+2, + along with the found normal line, + is graphed in . +

    +     +     + +

    + The direction of the normal line has many uses, + one of which is the definition of the + tangent plane which we define shortly. + Another use is in measuring distances from the surface to a point. + Given a point Q in space, + it is a general geometric concept to define the distance from Q to the surface as being the length of the shortest line segment + \overline{PQ} over all points P on the surface. + This, in turn, + implies that \overrightarrow{PQ} will be orthogonal to the surface at P. + Therefore we can measure the distance from Q to the surface z=f(x,y) by finding a point P on the surface such that \overrightarrow{PQ} is parallel to the normal line to f at P. +

    + + + Finding the distance from a point to a surface + +

    + Let f(x,y) = 2-x^2-y^2 and let Q = (2,2,2). + Find the distance from Q to the surface defined by f. +

    +     + +

    + This surface is used in , + so we know that at (x,y), + the direction of the normal line will be \vec d_n = \la -2x,-2y,-1\ra. + A point P on the surface will have coordinates (x,y,2-x^2-y^2), + so \overrightarrow{PQ} = \la 2-x,2-y,x^2+y^2\ra. + To find where \overrightarrow{PQ} is parallel to + \vec d_n, we need to find x, + y and c such that c\overrightarrow{PQ} = \vec d_n. + + c\overrightarrow{PQ} \amp = \vec d_n + c\la 2-x,2-y,x^2+y^2\ra \amp = \la -2x,-2y,-1\ra. + This implies + c(2-x) \amp = -2x + c(2-y) \amp = -2y + c(x^2+y^2) \amp = -1 + +

    + +

    + In each equation, we can solve for c: + + c = \frac{-2x}{2-x} = \frac{-2y}{2-y} = \frac{-1}{x^2+y^2} + . +

    + +

    + The first two fractions imply x=y, + and so the last fraction can be rewritten as c=-1/(2x^2). + Then + + \frac{-2x}{2-x} \amp = \frac{-1}{2x^2} + -2x(2x^2) \amp = -1(2-x) + 4x^3 \amp = 2-x + 4x^3+x-2 \amp =0 + . +

    + +

    + This last equation is a cubic, + which is not difficult to solve with a numeric solver. + We find that x= 0.689, + hence P = (0.689,0.689, 1.051). + We find the distance from Q to the graph of f is + + \norm{\overrightarrow{PQ}} = \sqrt{(2-0.689)^2 +(2-0.689)^2+(2-1.051)^2} = 2.083 + . +

    +     +     + +

    + We can take the concept of measuring the distance from a point to a surface to find a point Q a particular distance from a surface at a given point P on the surface. +

    + + + Finding a point a set distance from a surface + +

    + Let f(x,y) = x-y^2+3. + Let P = \big(2,1,f(2,1)\big) = (2,1,4). + Find points Q in space that are 4 units from the graph of f at P. + That is, find Q such that + \norm{\overrightarrow{PQ}}=4 and \overrightarrow{PQ} is orthogonal to f at P. +

    +     + +

    + We begin by finding partial derivatives: + + f_x(x,y) =1 \qquad \amp \Rightarrow \qquad f_x(2,1) = 1 + f_y(x,y) = -2y \qquad \amp \Rightarrow \qquad f_y(2,1) = -2 + +

    + +

    + The vector \vec n=\la 1,-2,-1\ra is orthogonal to f at P. + For reasons that will become more clear in a moment, + we find the unit vector in the direction of \vec n: + + \vec u = \frac{\vec n}{\vnorm n} = \la 1/\sqrt{6},-2/\sqrt{6},-1/\sqrt{6}\ra \approx \la 0.408,-0.816,-0.408\ra + . +

    + +

    + Thus a the normal line to f at P can be written as + + \ell_n(t) = \la 2,1,4\ra + t\la 0.408,-0.816,-0.408 \ra + . +

    + +

    + An advantage of this parametrization of the line is that letting t=t_0 gives a point on the line that is \abs{t_0} units from P. + (This is because the direction of the line is given in terms of a unit vector.) + There are thus two points in space 4 units from P: + + Q_1 \amp = \ell_n(4) \amp Q_2 \amp = \ell_n(-4) + \amp \approx \la 3.63, -2.27, 2.37\ra \amp \amp \approx\la 0.37, 4.27, 5.63\ra + +

    + +
    + Graphing the surface in along with points 4 units from the surface + + + + A parabolic cylinder, and a normal line to the surface at one point. + +

    + The surface is a parabolic cylinder: + the shape one would obtain by bending a sheet of paper so that its cross-sections are parabolas. + At one point on the surface, a normal line has been drawn. + The normal line extends a distance of four units on either side of the surface; + additional points are plotted at the two ends. +

    +     + + + + + //ASY file for figpartial4b3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(15,10,9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={2,4}; + defaultpen(0.5mm); + pair xbounds=(-1,5.5); + pair ybounds=(-2,4.5); + pair zbounds=(-1,5.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // -y^2+x+3 + + bool cond(pair t) {return 4*t.x*t.y-(t.x)^4-(t.y)^4 >-3;} + + triple f(pair t) { + return (t.x,t.y,3+t.x-t.y^2); + } + surface s=surface(f,(-.5,-2.1),(4,2),8,16,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic},usplinetype=Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3; + dot((2,1,4)); + dot((3.63,-2.27,2.37)); + dot((.37,4.27,5.63)); + + draw ((.37,4.27,5.63)--(3.63,-2.27,2.37),redpen+2bp); + + //Draw traces on the surface + //triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} + //path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); + + + + +
    + +

    + The surface is graphed along with points P, Q_1, + Q_2 and a portion of the normal line to f at P. +

    +     +     +     + + + Tangent Planes +

    + We can use the direction of the normal line to define a plane. + With a=f_x(x_0,y_0), + b=f_y(x_0,y_0) and P = \big(x_0,y_0,f(x_0,y_0)\big), + the vector \vec n=\la a,b,-1\ra is orthogonal to f at P. + (See .) + The plane through P with normal vector \vec n is therefore + tangent to f at P. +

    + + + + + Tangent Plane + +

    + Let z=f(x,y) be differentiable on a set S containing (x_0,y_0), + where a = f_x(x_0,y_0), b=f_y(x_0,y_0), + \vec n= \la a,b,-1\ra and P=\big(x_0,y_0,f(x_0,y_0)\big). +

    + +

    + The plane through P with normal vector \vec n is the + tangent plane to f at P. + The standard form of this plane is + tangent plane + planestangent + + a(x-x_0) + b(y-y_0) - \big(z-f(x_0,y_0)\big) = 0 + . +

    +     +     + + + Finding tangent planes + +

    + Find the equation of the tangent plane to z=-x^2-y^2+2 at (0,1). +

    +     + +

    + Note that this is the same surface and point used in . + There we found \vec n = \la 0,-2,-1\ra and P = (0,1,1). + Therefore the equation of the tangent plane is + + -2(y-1)-(z-1)=0 + . +

    + +

    + The surface z=-x^2-y^2+2 and tangent plane are graphed in . +

    + +
    + Graphing a surface with tangent plane from + + + + A downward-opening circular paraboloid, and a tangent plane to this surface at one point. + +

    + The surface is a circular paraboloid, opening downward. + One point is marked on the surface, and at this point, a tangent plane is drawn. + The tangent plane is illustrated as a rectangle pinned to the surface. +

    + +

    + By rotating the image, we are able to observe how the tangent plane lies close to the surface, + near the point of tangency. +

    +     + + + + + //ASY file for figpartial4b3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(13,10,9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + pair xbounds=(-3,3); + pair ybounds=(-3,3); + pair zbounds=(-3,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // -x^2-y^2+2 + + triple f(pair t) { + return (cos(t.x)*t.y,sin(t.x)*t.y,2-t.y^2); + } + surface s=surface(f,(0,0),(2*pi,2),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3; + dot((0,1,1)); + + // -2*y+3 + triple g(pair t) { + return (t.x,t.y,-2*t.y+3);} + surface ss=surface(g,(-1,.5),(1,1.5),1,1); + pen q=redcurvepen+.1mm; + draw(ss,surfacepen2,meshpen=q,nolight,render(merge=true)); + + //Draw traces on the surface + //triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} + //path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); + + + + +
    +     +     + + + Using the tangent plane to approximate function values + +

    + The point (3,-1,4) lies on the graph of an unknown differentiable function f where + f_x(3,-1) = 2 and f_y(3,-1) = -1/2. + Find the equation of the tangent plane to f at P, + and use this to approximate the value of f(2.9,-0.8). +

    +     + +

    + Knowing the partial derivatives at (3,-1) allows us to form the normal vector to the tangent plane, + \vec n = \la 2,-1/2,-1\ra. + Thus the equation of the tangent line to f at P is: + + 2(x-3)-1/2(y+1) - (z-4) = 0 \Rightarrow z = 2(x-3)-1/2(y+1)+4 + . +

    + +

    + Just as tangent lines provide excellent approximations of curves near their point of intersection, + tangent planes provide excellent approximations of surfaces near their point of intersection. + So f(2.9,-0.8) \approx z(2.9,-0.8) = 3.7. +

    + +

    + This is not a new method of approximation. + Compare the right hand expression for z in Equation to the total differential: + + dz = f_xdx + f_ydy \text{ and } z = \underbrace{\underbrace{2}_{f_x}\underbrace{(x-3)}_{dx}+\underbrace{-1/2}_{f_y}\underbrace{(y+1)}_{dy}}_{dz}+4 + . +

    + +

    + Thus the new z-value + is the sum of the change in z (, dz) and the old z-value (4). + As mentioned when studying the total differential, + it is not uncommon to know partial derivative information about a unknown function, + and tangent planes are used to give accurate approximations of the function. +

    +     +     +     + + + The Gradient and Normal Lines, Tangent Planes +

    + The methods developed in this section so far give a straightforward method of finding equations of normal lines and tangent planes for surfaces with explicit equations of the form z=f(x,y). + However, they do not handle implicit equations well, + such as x^2+y^2+z^2=1. + There is a technique that allows us to find vectors orthogonal to these surfaces based on the gradient. +

    + + + Gradient + +

    + Let w=F(x,y,z) be differentiable on a set D that contains the point (x_0,y_0,z_0). + gradient +

    + +

    +

      +
    1. +

      + The gradient of F + is \nabla F(x,y,z) = \la f_x(x,y,z),f_y(x,y,z),f_z(x,y,z)\ra. +

      +
      2. + +
    2. +

      + The gradient of F at (x_0,y_0,z_0) is + + \nabla F(x_0,y_0,z_0) = \la f_x(x_0,y_0,z_0),f_y(x_0,y_0,z_0),f_z(x_0,y_0,z_0)\ra + . +

      +
      3. +
    +

    +     +     + +

    + Recall that when z=f(x,y), + the gradient \nabla f = \la f_x,f_y\ra is orthogonal to level curves of f. + An analogous statement can be made about the gradient + \nabla F, where w= F(x,y,z). + Given a point (x_0,y_0,z_0), + let c = F(x_0,y_0,z_0). + Then F(x,y,z) = c is a level surface + that contains the point (x_0,y_0,z_0). + The following theorem states that + \nabla F(x_0,y_0,z_0) is orthogonal to this level surface. +

    + + + The Gradient and Level Surfaces + +

    + Let w=F(x,y,z) be differentiable on a set D containing + (x_0,y_0,z_0) with gradient \nabla F, + where F(x_0,y_0,z_0) = c. +

    + +

    + The vector \nabla F(x_0,y_0,z_0) is orthogonal to the level surface + F(x,y,z)=c at (x_0,y_0,z_0). + gradientand level surfaces + level surface +

    +     +     + +

    + The gradient at a point gives a vector orthogonal to the surface at that point. + This direction can be used to find tangent planes and normal lines. +

    + + + Using the gradient to find a tangent plane + +

    + Find the equation of the plane tangent to the ellipsoid + \frac{x^2}{12} +\frac{y^2}{6}+\frac{z^2}{4}=1 at P = (1,2,1). +

    +     + +

    + We consider the equation of the ellipsoid as a level surface of a function F of three variables, + where F(x,y,z) = \frac{x^2}{12} +\frac{y^2}{6}+\frac{z^2}{4}. + The gradient is: + + \nabla F(x,y,z) \amp = \la F_x, F_y,F_z\ra + \amp = \la \frac x6, \frac y3, \frac z2\ra + . +

    + +

    + At P, + the gradient is \nabla F(1,2,1) = \la 1/6, 2/3, 1/2\ra. + Thus the equation of the plane tangent to the ellipsoid at P is + + \frac 16(x-1) + \frac23(y-2) + \frac 12(z-1) = 0 + . +

    + +

    + The ellipsoid and tangent plane are graphed in . +

    + +
    + An ellipsoid and its tangent plane at a point + + + + An ellipsoid centered at the origin in space, and a tangent plane at one point. + +

    + An ellipsoid is plotted against a set of three-dimensional coordinate axes, + with its center at the origin. The ellipsoid is egg-like in shape. +

    + +

    + At one point on the ellipsoid, a point is marked, and a tangent plane is drawn at that point. + The tangent plane is illustrated as a flat rectangle stuck to the surface at the point of tangency. + By rotating the image, we are able to observe how the tangent plane lies close to the surface, + as long as we are not far from the point of tangency. +

    +     + + + + + //ASY file for figpartial4b3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14,11,4.7); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-3.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + // ({cos(y)*(3.46*cos(x))},{cos(y)*(2.45*sin(x))},{2*sin(y)}) + + triple f(pair t) { + return (cos(t.y)*(3.46*cos(t.x)),cos(t.y)*(2.45*sin(t.x)),2*sin(t.y)); + } + surface s=surface(f,(-pi,-pi/2),(pi,pi/2),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=3; + dot((1,2,1)); + + // 1-(x-1)/3-4*(y-2)/3 1. - 0.980581 x, 2. + 0.156893 x + 0.6 y, 1. + 0.11767 x - 0.8 y + //triple g(pair t) { + // return (t.x,t.y,1-(t.x-1)/3-4*(t.y-2)/3);} + //surface ss=surface(g,(.5,1.5),(1.5,2.5),1,1); + triple g(pair t) { + return (1.-0.980581*t.x, 2.+0.156893*t.x+0.6*t.y,1.+0.11767*t.x-0.8*t.y);} + surface ss=surface(g,(-.75,-.75),(.75,.75),1,1); + pen q=redcurvepen+.1mm; + draw(ss,surfacepen2,meshpen=q,nolight,render(merge=true)); + + //Draw traces on the surface + //triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} + //path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); + + + + +
    +     +     + +

    + To understand why is true, + recall the method of implicit differentiation given in . + A level surface f(x,y,z)=0 can be viewed as defining z=g(x,y) implicitly. + We found that the partial derivatives of z with respect to x and y are then given by + + \plz{z}{x} = g_x(x,y) = -\frac{f_x(x,y,z)}{f_z(x,y,z)}\quad \plz{z}{y} = g_y(x,y) = -\frac{f_y(x,y,z)}{f_z(x,y,z)} + . + If we plug these values into the tangent plane equation + + z = g(a,b)+g_x(a,b)(x-a)+g_y(a,b)(y-b) + + we get, with c=g(a,b), + + z = c - \frac{f_x(a,b,c)}{f_z(a,b,c)}(x-a)-\frac{f_y(a,b,c)}{f_z(a,b,c)}(y-b) + . + If we move everything to the left-hand side of the equation and multiply through by f_z(a,b,c), + we get + + f_x(a,b,c)(x-a)+f_y(a,b,c)(y-b)+f_z(a,b,c)(z-c)=0 + , + which is the equation of a plane with normal vector \nabla f(a,b,c). +

    + + + Finding the tangent plane of a level surface + +

    + Determine the equation of the tangent plane to the level surface + x^2yz^3-\sin(x-3z)+4xy^2-3yz=0 at the point (3,0,1). + (Note that this is the same problem as .) +

    +     + +

    + With f(x,y,z)=x^2yz^3-\sin(x-3z)+4xy^2-3yz we have + + f_x(x,y,z) \amp = 2xyz^3-\cos(x-3z)+4y^2 \amp f_x(3,0,1)\amp = -1 + f_y(x,y,z) \amp = x^2z^3+8xy-3z \amp f_y(3,0,1)\amp = 6 + f_z(x,y,z) \amp = 3x^2yz^2+3\cos(x-3z)-3y \amp f_z(3,0,1) \amp = 3 + . +

    +

    + The equation of the tangent plane is therefore + + -1(x-3)+6y+3(z-1)=0 + . + Note that solving for z gives z=1+\frac13(x-3)-2y, + which is the same result as . +

    +     +     + +

    + Tangent lines and planes to surfaces have many uses, + including the study of instantaneous rates of changes and making approximations. + Normal lines also have many uses. + In this section we focused on using them to measure distances from a surface. + Another interesting application is in computer graphics, + where the effects of light on a surface are determined using normal vectors. +

    + +

    + The next section investigates another use of partial derivatives: + determining relative extrema. + When dealing with functions of the form y=f(x), + we found relative extrema by finding x where \fp(x) = 0. + We can start finding relative extrema of + z=f(x,y) by setting f_x and f_y to 0, but it turns out that there is more to consider. +

    +     + + + + Terms and Concepts + + + + +

    + Explain how the vector \vec v=\la 1,0,3\ra can be thought of as having a slope of 3. +

    +     + + + +

    + Answers will vary. + The displacement of the vector is one unit in the x-direction and 3 units in the z-direction, + with no change in y. + Thus along a line parallel to \vec v, + the change in z is 3 times the change in x , a slope of 3. + Specifically, + the line in the xz-plane parallel to z has a slope of 3. +

    +     + +     + + + + +

    + Explain how the vector \vec v=\la 0.6,0.8, -2\ra can be thought of as having a + slope of -2. +

    +     + + + +

    + Answers will vary. + Let \vec u = \la 0.6,0.8\ra; + this is a unit vector. + The displacement of the vector is one unit in the \vec u-direction and -2 units in the z-direction. + In the plane containing the z-axis and the vector \vec u, + the line parallel to \vec v has slope -2. +

    +     + +     + + + + + +

    + Let z=f(x,y) be differentiable at P. + If \vec n is a normal vector to the tangent plane of f at P, + then \vec n is orthogonal to \ell_x and \ell_y at P. + +

    +     + +     + + + + +

    + Explain in your own words why we do not refer to the + tangent line to a surface at a point, + but rather to directional + tangent lines to a surface at a point. +

    +     + + + +

    + On a surface through a point, + there are many different smooth curves, + each with a tangent line at the point. + Each of these tangent lines is also + tangent to the surface. + There is not just one tangent line, + but many, each in a different direction. + Therefore we refer to directional tangent lines, + not just the tangent line. +

    +     + +     +     + + + Problems + + + +

    + A function f(x,y), + a vector \vec v and a point P are given. + Give the parametric equations of the following directional tangent lines to z=f(x,y) at P: +

    + +

    +

      +
    1. +

      + \ell_x(t) +

      +
      2. + +
    2. +

      + \ell_y(t) +

      +
      3. + +
    3. +

      + \ell_{\vec u\,}(t), + where \vec u is the unit vector in the direction of \vec v. +

      +
      4. +
    +

    +     + + + + +

    + f(x,y) = 2x^2y-4xy^2, \vec v = \la 1,3\ra, + P=(2,3). +

    +     + +

    +

      +
    1. +

      + \ell_x(t) = \left\{\begin{array}{l} x=2+t\\ y=3 \\ z = -48-12t + \end{array} \right. +

      +
      2. + +
    2. +

      + \ell_y(t) = \left\{\begin{array}{l} x=2\\ y=3+t \\ z = -48-40t + \end{array} \right. +

      +
      3. + +
    3. +

      + \ell_{\vec u\,}(t) = \left\{\begin{array}{l} x=2+t/\sqrt{10}\\ y=3+3t/\sqrt{10} \\ z = -48-66\sqrt{2/5}t + \end{array} \right. +

      +
      4. +
    +

    +     + +     + + + + + + Context("Vector"); + Context()->variables->add(t=>"Real"); + $v[0]=Compute("(pi/3,pi/6,3/4)+t<1,0,-3sqrt(3)/4>"); + $v[1]=Compute("(pi/3,pi/6,3/4)+t<0,1,3sqrt(3)/4>"); + $v[2]=Compute("(pi/3,pi/6,3/4)+t<1/sqrt(5),2/sqrt(5),3sqrt(3/5)/4>"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + parser::Assignment->Allow; + $p[0]=List(Formula("x=pi/3+t"),Formula("y=pi/6"),Formula("z=3/4-3sqrt(3)/4t")); + $p[1]=List(Formula("x=pi/3"),Formula("y=pi/6+t"),Formula("z=3/4+3sqrt(3)/4t")); + $p[2]=List(Formula("x=pi/3+t/sqrt(5)"),Formula("y=pi/6+2t/sqrt(5)"),Formula("z=3/4+3sqrt(3/5)/4t")); + for my $k (0..2) { + $pev[$k]=$p[$k]->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + }} + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + }} + if(!$aH->{isPreview}){ + push(@err,"You need to provide more parametrizations")if $i<3; + push(@err,"You have given too many parametrizations")if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v[$k]->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v[$k]->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + }; + + + +

    + f(x,y) = 3\cos(x) \sin(y), \vec v = \la 1,2\ra, P=(\pi/3, \pi/6) +

    + + + Find parametric equations for the directional tangent line \ell_{x} at P. + +

    + +

    + + + Find parametric equations for the directional tangent line \ell_{y} at P. + +

    + +

    + + Find parametric equations for the directional tangent line \ell_{\vec{v}} at P. + +

    + +

    +     +     +     + + + + +

    + f(x,y) = 3x-5y, \vec v = \la 1,1\ra, P=(4,2). +

    +     + +

    +

      +
    1. +

      + \ell_x(t) = \left\{\begin{array}{l} x = 4+t\\ y = 2 \\ z = 2 + 3t + \end{array} \right. +

      +
      2. + +
    2. +

      + \ell_y(t) = \left\{\begin{array}{l} x = 4\\ y = 2+t\\ z = 2-5t + \end{array} \right. +

      +
      3. + +
    3. +

      + \ell_{\vec u\,}(t) = \left\{\begin{array}{l} x = 4+t/\sqrt{2}\\ y = 2+t/\sqrt{2} \\ z = 2 -\sqrt{2}t + \end{array} \right. +

      +
      4. +
    +

    +     + +     + + + + + + Context("Vector"); + Context()->variables->add(t=>"Real"); + $v[0]=Compute("(1,2,3)+t<1,0,0>"); + $v[1]=Compute("(1,2,3)+t<0,1,0>"); + $v[2]=Compute("(1,2,3)+t<1/sqrt(2),1/sqrt(2),0>"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + parser::Assignment->Allow; + $p[0]=List(Formula("x=1+t"),Formula("y=2"),Formula("z=3")); + $p[1]=List(Formula("x=1"),Formula("y=2+t"),Formula("z=3")); + $p[2]=List(Formula("x=1+t/sqrt(2)"),Formula("y=2+t/sqrt(2)"),Formula("z=3")); + for my $k (0..2) { + $pev[$k]=$p[$k]->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + }} + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + }} + if(!$aH->{isPreview}){ + push(@err,"You need to provide more parametrizations")if $i<3; + push(@err,"You have given too many parametrizations")if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v[$k]->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v[$k]->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + }; + + + +

    + f(x,y) = x^2-2x-y^2+4y, \vec v = \la 1,1\ra, P=(1, 2) +

    + + + Find parametric equations for the directional tangent line \ell_{x} at P. + +

    + +

    + + Find parametric equations for the directional tangent line \ell_{y} at P. + +

    + +

    + + Find parametric equations for the directional tangent line \ell_{\vec{v}} at P. + +

    + +

    +     +     +     + +     + + + +

    + A function f(x,y) and a point P are given. + Find the equation of the normal line to z=f(x,y) at P. + Note: these are the same functions as in Exercises. +

    +     + + + + +

    + f(x,y) = 2x^2y-4xy^2, P=(2,3). +

    +     + +

    + \ell_{\vec n}(t) = \left\{\begin{array}{l} x=2-12t\\ y=3-40t \\ z = -48-t + \end{array} \right. +

    +     + +     + + + + + + Context("Vector"); + Context()->variables->are(t=>"Real"); + $v=Compute("(pi/3,pi/6,3/4)+t<-3sqrt(3)/4,3sqrt(3)/4,-1>"); + Context("Numeric"); + Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + parser::Assignment->Allow; + $n=List(Formula("x=pi/3-3sqrt(3)/4t"),Formula("y=pi/6+3sqrt(3)/4t"),Formula("z=3/4-t")); + $nev=$n->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + }} + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + }} + if(!$aH->{isPreview}){ + push(@err,"You need to provide more parametrizations")if $i<3; + push(@err,"You have given too many parametrizations")if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + + + +

    + f(x,y) = 3\cos(x) \sin(y) and P=(\pi/3, \pi/6) +

    + + Find parametric equations for the normal line at P. + + +

    + +

    +     +     +     + + + + +

    + f(x,y) = 3x-5y, P=(4,2). +

    +     + +

    + \ell_{\vec n}(t) = \left\{\begin{array}{l} x = 4+3t\\ y = 2-5t \\ z = 2 -t + \end{array} \right. +

    +     + +     + + + + + + Context("Vector"); + Context()->variables->are(t=>"Real"); + $v=Compute("(1,2,3)+t<0,0,-1>"); + Context("Numeric"); + Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + parser::Assignment->Allow; + $n=List(Formula("x=1"),Formula("y=2"),Formula("z=3-t")); + $nev=$n->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + }} + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + }} + if(!$aH->{isPreview}){ + push(@err,"You need to provide more parametrizations")if $i<3; + push(@err,"You have given too many parametrizations")if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + + + +

    + f(x,y) = x^2-2x-y^2+4y and P=(1,2) +

    + + Find parametric equations for the normal line at P. + +

    + +

    +     +     +     + +     + + + +

    + A function f(x,y) and a point P are given. + Find the two points that are 2 units from the surface z=f(x,y) at P. + Note: these are the same functions as in Exercises. +

    +     + + + + +

    + f(x,y) = 2x^2y-4xy^2, P=(2,3). +

    +     + +

    + (1.425, 1.085, -48.078), + (2.575, 4.915, -47.952) +

    +     + +     + + + + +

    + f(x,y) = 3\cos(x) \sin(y), P=(\pi/3, \pi/6). +

    +     + +

    + (-0.195,1.766,-0.206) and (2.289,-0.719, 1.706) +

    +     + +     + + + + +

    + f(x,y) = 3x-5y, P=(4,2). +

    +     + +

    + (5.014, 0.31, 1.662) and (2.986, 3.690, 2.338) +

    +     + +     + + + + +

    + f(x,y) = x^2-2x-y^2+4y, P=(1,2). +

    +     + +

    + (1,2,1) and (1,2,5) +

    +     + +     + +     + + + +

    + A function f(x,y) and a point P are given. + Find an equation of the tangent plane to z=f(x,y) at P. + Note: these are the same functions as in Exercises. +

    +     + + + + +

    + f(x,y) = 2x^2y-4xy^2, P=(2,3). +

    +     + +

    + -12(x-2)-40(y-3) -(z+48) = 0 +

    +     + +     + + + + + Context("ImplicitPlane"); + $t=Compute("-3sqrt(3)/4(x-pi/3)+3sqrt(3)/4(y-pi/6)-(z-3/4)=0"); + + + +

    + f(x,y) = 3\cos(x) \sin(y), P=(\pi/3, \pi/6). +

    + + + Find an equation of the tangent plane to z=f(x,y) at P. + +

    + +

    +     +     +     + + + + +

    + f(x,y) = 3x-5y, P=(4,2). +

    +     + +

    + 3(x-4)-5(y-2) - (z-2) = 0 (Note that this tangent plane is the same as the original function, + a plane.) +

    +     + +     + + + + + Context("ImplicitPlane"); + $t=Compute("z=3"); + + + +

    + f(x,y) = x^2-2x-y^2+4y, P=(1,2) +

    + + Find an equation of the tangent plane to z=f(x,y) at P. + +

    + +

    +     +     +     + +     + + + +

    + An implicitly defined function of x, + y and z is given, along with a point P that lies on the surface. + Use the gradient \nabla F to: + +

      +
    1. +

      + find the equation of the normal line to the surface at P, and +

      +
      2. + +
    2. +

      + find the equation of the plane tangent to the surface at P. +

      +
      3. +
    +

    +     + + + + +

    + \ds \frac{x^2}{8}+\frac{y^2}4+\frac{z^2}{16}=1, + at P = (1,\sqrt{2},\sqrt{6}) +

    +     + +

    + \nabla F = \la x/4, y/2, z/8\ra; + at P, + \nabla F = \la 1/4, \sqrt{2}/2, \sqrt{6}/8\ra +

    + +

    +

      +
    1. +

      + \ell_{\vec n}(t) = \left\{\begin{array}{l} x= 1+ t/4 \\ y = \sqrt{2}+ \sqrt{2}t/2\\ z = \sqrt{6} + \sqrt{6}t/8 + \end{array} \right. +

      +
      2. + +
    2. +

      + \frac14(x-1) + \frac{\sqrt{2}}{2}(y-\sqrt{2}) + \frac{\sqrt{6}}8(z-\sqrt{6}) = 0. +

      +
      3. +
    +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(t=>"Real"); + $v=Compute("(4,-3,sqrt(5))+t<-2,2/3,2sqrt(5)>"); + Context("Numeric"); + Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + parser::Assignment->Allow; + $n=List(Formula("x=4-2t"),Formula("y=-3+2/3t"),Formula("z=sqrt(5)+2sqrt(5)t")); + $nev=$n->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + }} + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + }} + if(!$aH->{isPreview}){ + push(@err,"You need to provide more parametrizations")if $i<3; + push(@err,"You have given too many parametrizations")if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + Context("ImplicitPlane"); + $t=Compute("-2(x-4) + 2/3(y+3) + 2sqrt(5)(z-sqrt(5)) = 0"); + + + +

    + z^2-\frac{x^2}{4} - \frac{y^2}9=0, + at P = (4,-3,\sqrt{5}) +

    + + Find parametric equations for the normal line. + +

    + +

    + + Find an equation of the tangent plane. + +

    + +

    +     +     +     + + + + +

    + \ds xy^2-xz^2=0, at P = (2,1,-1) +

    +     + +

    + \nabla F = \la y^2-z^2, 2xy, -2xz\ra; + at P, \nabla F = \la 0, 4, 4\ra +

    + +

    +

      +
    1. +

      + \ell_{\vec n}(t) = \left\{\begin{array}{l} x= 2 \\ y = 1+4t\\ z = -1+4t + \end{array} \right. +

      +
      2. + +
    2. +

      + 4(y-1) + 4(z+1) = 0. +

      +
      3. +
    +

    +     + +     + + + + + Context("Vector"); + Context()->variables->are(t=>"Real"); + $v=Compute("(2,pi/12,4)+t<pi/(8sqrt(3)),-sqrt(3),-pi/(8sqrt(3))>"); + Context("Numeric"); + Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + parser::Assignment->Allow; + $n=List(Formula("x=2+pi/(8sqrt(3))t"),Formula("y=pi/12-sqrt(3)t"),Formula("z=4-pi/(8sqrt(3))t")); + $nev=$n->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; + my$n=scalar(@$st); + my@err=();my$i,$j; + Context("Vector"); + Context()->variables->add(t=>"Real"); + my$sV=Formula("(0,0,0)+t*<0,0,0>"); + for($i=0;$i<$n;$i++){ + my$ith=Value::List->NameForNumber($i+1); + my$entry=$st->[$i]; + my$var;my$for; + if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} + else{($var,$for)=split('=',"$entry"); + $var=Formula("$var");$for=Formula("$for"); + if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) + {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} + }; + for($j=0,$used=0;$j<$i;$j++){ + if($st->[$j]->type eq "Assignment"){ + my($vj,$fj)=split('=',$st->[$j]->string); + $vj=Formula("$vj"); + if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} + }} + if(!$used){ + if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} + elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} + else{$sV=$sV+Formula("<0,0,$for>")}; + }} + if(!$aH->{isPreview}){ + push(@err,"You need to provide more parametrizations")if $i<3; + push(@err,"You have given too many parametrizations")if $i>3; + } + $sV=Formula($sV); + $ds=$sV->D('t')->reduce; + if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; + $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); + $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); + $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); + return (3*($par and $tch),@err); + }); + Context("ImplicitPlane"); + $t=Compute("pi/(8sqrt(3))(x-2) -sqrt(3)(y-pi/12) -pi/(8sqrt(3))(z-4) = 0"); + + + +

    + \sin(xy)+\cos(yz)=1, + at P = (2, \pi/12, 4) +

    + + + Find parametric equations for the normal line. + +

    + +

    + + Find an equation of the tangent plane. + +

    + +

    +     +     +     + +     +     +     +
    +
    + Extreme Values + + Critical Points of Functions of Two Variables + +

    + Given a function f(x,y), + we are often interested in points where z=f(x,y) takes on the largest or smallest values. + For instance, if f represents a cost function, + we would likely want to know what (x,y) values minimize the cost. + If f represents the ratio of a volume to surface area, + we would likely want to know where f is greatest. + This leads to the following definition. +

    + + + Relative and Absolute Extrema + +

    + Let z=f(x,y) be defined on a set S containing the point P=(x_0,y_0). + maximumrelative/local + minimumrelative/local + extremarelative + extremaabsolute + maximumabsolute + minimumabsolute + +

      +
    1. +

      + If f(x_0,y_0)\geq f(x,y) for all (x,y) in S, + then f has an absolute maximum + at P If f(x_0,y_0)\leq f(x,y) for all (x,y) in S, + then f has an absolute minimum at P. +

      +
      2. + +
    2. +

      + If there is an open disk D containing P such that + f(x_0,y_0) \geq f(x,y) for all points (x,y) that are in both D and S, + then f has a relative maximum at P. + If there is an open disk D containing P such that + f(x_0,y_0) \leq f(x,y) for all points (x,y) that are in both D and S, + then f has a relative minimum at P. +

      +
      3. + +
    3. +

      + If f has an absolute maximum or minimum at P, + then f has an absolute extremum at P. + If f has a relative maximum or minimum at P, + then f has a relative extremum at P. +

      +
      4. +
    +

    +     +     + +

    + If f has a relative or absolute maximum at (x_0,y_0), + it means every curve on the graph of f through (x_0,y_0,f(x_0,y_0)) will also have a relative or absolute maximum at P. + Recalling what we learned in , + the slopes of the tangent lines to these curves at P must be 0 or undefined. + Since directional derivatives are computed using f_x and f_y, + we are led to the following definition and theorem. +

    + + + Critical Point + +

    + Let z = f(x,y) be continuous on a set S. + A critical point + P=(x_0,y_0) of f is a point in S such that, + at P, + critical point +

      +
    • +

      + f_x(x_0,y_0) = 0 and f_y(x_0,y_0) = 0, or +

      +
      • + +
    • +

      + f_x(x_0,y_0) and/or f_y(x_0,y_0) is undefined. +

      +
      • +
    +

    +     +     + + + Critical Points and Relative Extrema + +

    + Let z=f(x,y) be defined on an open set S containing P=(x_0,y_0). + If f has a relative extrema at P, + then P is a critical point of f. + extremarelative + critical point +

    +     +     + +

    + Therefore, to find relative extrema, + we find the critical points of f and determine which correspond to relative maxima, + relative minima, or neither. + The following examples demonstrate this process. +

    + + + + + Finding critical points and relative extrema + +

    + Let f(x,y) = x^2+y^2-xy-x-2. + Find the relative extrema of f. +

    +     + +

    + We start by computing the partial derivatives of f: + + f_x(x,y) = 2x-y-1 \qquad \text{ and } \qquad f_y(x,y) = 2y-x + . +

    + +

    + Each is never undefined. + A critical point occurs when f_x and f_y are simultaneously 0, leading us to solve the following system of linear equations: + + 2x-y-1 = 0\qquad \text{ and } \qquad -x+2y = 0 + . +

    + +

    + This solution to this system is x=2/3, + y=1/3. (Check that at (2/3,1/3), + both f_x and f_y are 0.) +

    + +
    + The surface in with its absolute minimum indicated + + + + A circular paraboloid plotted over a rectangular domain. It is bowl-shaped, with peaks at the corners of the domain. + +

    + The image shows a three-dimensional plot of the surface z = x^2+y^2-xy-x-2. + This is a circular paraboloid, opening upward. + It is bowl-shaped, and because it is plotted over a rectangular domain, + we see peaks at points on the surface over the four corners of the domain. + The bottom of the bowl shape is the location of the absolute minimum. +

    +     + + + + + //ASY file for figmulti_extreme13D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={5}; + defaultpen(0.5mm); + pair xbounds=(-3,3); + pair ybounds=(-3,3); + pair zbounds=(-1,6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=x^2+y^2-xy-x-2 + triple f(pair t) { + return (t.x,t.y,t.x^2+t.y^2-t.x*t.y-t.x-2); + } + surface s=surface(f,(-1,-1),(2,2),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=4;dot((2/3,1/3,-7/3)); + + + + +
    + +

    + The graph in + shows f along with this critical point. + It is clear from the graph that this is a relative minimum; + further consideration of the function shows that this is actually the absolute minimum. +

    +     +     + + + Finding critical points and relative extrema + +

    + Let f(x,y) = -\sqrt{x^2+y^2}+2. + Find the relative extrema of f. +

    +     + +

    + We start by computing the partial derivatives of f: + + f_x(x,y) = \frac{-x}{\sqrt{x^2+y^2}}\qquad \text{ and } \qquad f_y(x,y) = \frac{-y}{\sqrt{x^2+y^2}} + . +

    + +

    + It is clear that f_x=0 when x=0 & y\neq0, + and that f_y=0 when y=0 & x\neq0. + At (0,0), both f_x and f_y are not + 0, but rather undefined. + The point (0,0) is still a critical point, + though, because the partial derivatives are undefined. + This is the only critical point of f. +

    + +
    + The surface in with its absolute maximum indicated + + + + An inverted cone, with a maximum corresponding to a sharp peak on the z axis. + +

    + The surface z = -\sqrt{x^2+y^2}+2 is an inverted circular cone. + The image shows a plot of this graph over a rectangular domain. + At the point (0,0,2) there is a sharp peak; + this is an absolute maximum corresponding to a point where the partial derivatives are undefined. +

    +     + + + + + //ASY file for figmulti_extreme23D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-3,3); + pair ybounds=(-3,3); + pair zbounds=(-1,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=2-sqrt(x^2+y^2) + triple f(pair t) { + return (t.x,t.y,2-sqrt(t.x^2+t.y^2)); + } + surface s=surface(f,(-2,-2),(2,2),32,32); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=4;dot((0,0,2)); + + + + +
    + +

    + The graph of f is plotted in + along with the point (0,0,2). + The graph shows that this point is the absolute maximum of f. +

    +     +     + +

    + In each of the previous two examples, + we found a critical point of f and then determined whether or not it was a relative + (or absolute) + maximum or minimum by graphing. + It would be nice to be able to determine whether a critical point corresponded to a max or a min without a graph. + Before we develop such a test, + we do one more example that sheds more light on the issues our test needs to consider. +

    + + + Finding critical points and relative extrema + +

    + Let f(x,y) = x^3-3x-y^2+4y. + Find the relative extrema of f. +

    +     + +

    + Once again we start by finding the partial derivatives of f: + + f_x(x,y) = 3x^2-3\qquad \text{ and } \qquad f_y(x,y) = -2y+4 + . +

    + +

    + Each is always defined. + Setting each equal to 0 and solving for x and y, we find + + f_x(x,y) = 0 \amp \Rightarrow x=\pm 1 + f_y(x,y) = 0 \amp \Rightarrow y = 2 + . +

    + +

    + We have two critical points: + (-1,2) and (1,2). + To determine if they correspond to a relative maximum or minimum, + we consider the graph of f in . +

    + +
    + The surface in with both critical points marked + + + + The cubic surface studied in this example. + +

    + The surface plotted is the graph of the cubic function f(x,y) = x^3-3x-y^2+4y. + It shape is somewhat wave-like, with some interesting features worth noting. + To the right (in the perspective used for the image) there is a peak; + near this peak, the surface looks similar to a downward-opening elliptic paraboloid. +

    + +

    + To the left of the image, the surface dips down to what initially appears to be a minimum, + if we are thinking of the graph of a cubic function in two dimensions. + However, this is only the case if we view the surface along the y axis. + Rotating the image shows that while the surface curves upward in the x direction, + it curves downward in the y direction. + Near this point, the shape of the surface is more like the saddle surface defined by a hyperbolic paraboloid. +

    +     + + + + + //ASY file for figmulti_extreme33D.asy in Chapter 12 + + size(300,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(3,10,9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={2}; + real[] myzchoice={5}; + defaultpen(0.5mm); + pair xbounds=(-1.5,2); + pair ybounds=(-.5,3.5); + pair zbounds=(-2,6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=x^3-3x-y^2+4y + triple f(pair t) { + return (t.x,t.y,t.x^3-3*t.x-t.y^2+4*t.y); + } + surface s=surface(f,(-2,1),(2,3),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=4;dot((-1,2,6)); + dot((1,2,2)); + + + + +
    + +

    + The critical point (-1,2) clearly corresponds to a relative maximum. + However, the critical point at (1,2) is neither a maximum nor a minimum, + displaying a different, interesting characteristic. +

    + +

    + If one walks parallel to the y-axis towards this critical point, + then this point becomes a relative maximum along this path. + But if one walks towards this point parallel to the x-axis, + this point becomes a relative minimum along this path. + A point that seems to act as both a max and a min is a saddle point. + A formal definition follows. +

    +     +     + + + Saddle Point + +

    + Let P=(x_0,y_0) be in the domain of f where f_x=0 and f_y=0 at P. + We say P is a saddle point of f if, + for every open disk D containing P, + there are points (x_1,y_1) and + (x_2,y_2) in D such that + f(x_0,y_0) \gt f(x_1,y_1) and f(x_0,y_0)\lt f(x_2,y_2). + saddle point + critical point +

    +     +     + +

    + At a saddle point, + the instantaneous rate of change in all directions is 0 and there are points nearby with z-values both less than and greater than the z-value of the saddle point. +

    + +

    + Before + we mentioned the need for a test to differentiate between relative maxima and minima. + We now recognize that our test also needs to account for saddle points. + To do so, we consider the second partial derivatives of f. +

    + +

    + Recall that with single variable functions, + such as y=f(x), if \fp'(c) \gt 0, + then f is concave up at c, and if \fp(c) =0, + then f has a relative minimum at x=c. + (We called this the Second Derivative Test.) + Note that at a saddle point, + it seems the graph is both + concave up and concave down, + depending on which direction you are considering. +

    + +

    + It would be nice if the following were true: +

    + + + + f_{xx} and f_{yy} \gt 0 + \Rightarrow + relative minimum + + + f_{xx} and f_{yy} \lt 0 + \Rightarrow + relative maximum + + + f_{xx} and f_{yy} have opposite signs + \Rightarrow + saddle point. + + + +

    + However, this is not the case. + Functions f exist where f_{xx} and f_{yy} are both positive but a saddle point still exists. + In such a case, + while the concavity in the x-direction is up (, + f_{xx} \gt 0) and the concavity in the y-direction is also up (, f_{yy} \gt 0), + the concavity switches somewhere in between the x- and y-directions. +

    + +

    + To account for this, consider D = f_{xx}f_{yy}-f_{xy}f_{yx}. + Since f_{xy} and f_{yx} are equal when continuous + (refer back to ), + we can rewrite this as D = f_{xx}f_{yy}-f_{xy}^{\,2}. + D can be used to test whether the concavity at a point changes depending on direction. + If D \gt 0, + the concavity does not switch (, at that point, + the graph is concave up or down in all directions). + If D\lt 0, the concavity does switch. + If D=0, + our test fails to determine whether concavity switches or not. + We state the use of D in the following theorem. +

    + + + Second Derivative Test + +

    + Let R be an open set on which a function + z=f(x,y) and all its first and second partial derivatives are defined, + let P = (x_0,y_0) be a critical point of f in R, and let + Second Derivative Test + maximumrelative/local + minimumrelative/local + saddle point + + D = f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^{\,2}(x_0,y_0) + . +

    + +

    +

      +
    1. +

      + If D \gt 0 and f_{xx}(x_0,y_0) \gt 0, + then f has a relative minimum at P. +

      +
      2. + +
    2. +

      + If D \gt 0 and f_{xx}(x_0,y_0)\lt 0, + then f has a relative maximum at P. +

      +
      3. + +
    3. +

      + If D\lt 0, then f has a saddle point at P. +

      +
      4. + +
    4. +

      + If D=0, the test is inconclusive. +

      +
      5. +
    +

    +     +     + + + +

    + We first practice using this test with the function in the previous example, + where we visually determined we had a relative maximum and a saddle point. +

    + + + Using the Second Derivative Test + +

    + Let f(x,y) = x^3-3x-y^2+4y as in . + Determine whether the function has a relative minimum, maximum, + or saddle point at each critical point. +

    +     + +

    + We determined previously that the critical points of f are (-1,2) and (1,2). + To use the Second Derivative Test, + we must find the second partial derivatives of f: + + f_{xx} = 6x;\qquad f_{yy} = -2;\qquad f_{xy} = 0 + . +

    + +

    + Thus D(x,y) = -12x. +

    + +

    + At (-1,2): D(-1,2) = 12 \gt 0, + and f_{xx}(-1,2) = -6. + By the Second Derivative Test, + f has a relative maximum at (-1,2). +

    + +

    + At (1,2): D(1,2) = -12 \lt 0. + The Second Derivative Test states that f has a saddle point at (1,2). +

    + +

    + The Second Derivative Test confirmed what we determined visually. +

    +     +     + + + Using the Second Derivative Test + +

    + Find the relative extrema of f(x,y) = x^2y+y^2+xy. +

    +     + +

    + We start by finding the first and second partial derivatives of f: + + f_x \amp = 2xy+y \amp f_y \amp = x^2+2y+x + f_{xx} \amp = 2y \amp f_{yy} \amp = 2 + f_{xy} \amp = 2x+1 \amp f_{yx} \amp = 2x+1 + . +

    + +

    + We find the critical points by finding where f_x and f_y are simultaneously 0 + (they are both never undefined). + Setting f_x=0, we have: + + f_x=0 \Rightarrow 2xy+y=0 \Rightarrow y(2x+1)=0 + . +

    + +

    + This implies that for f_x=0, + either y=0 or 2x+1=0. +

    + +

    + Assume y=0 then consider f_y=0: + + f_y \amp = 0 + x^2+2y+x \amp = 0, \qquad \text{ and since \(y=0\), we have } + x^2+x \amp = 0 + x(x+1) \amp = 0 + . +

    + +

    + Thus if y=0, we have either x=0 or x=-1, + giving two critical points: + (-1,0) and (0,0). +

    + +

    + Going back to f_x, now assume 2x+1=0, + , that x=-1/2, then consider f_y=0: + + f_y \amp = 0 + x^2+2y+x \amp = 0, \qquad \text{ and since \(x=-1/2\), we have } + 1/4+2y-1/2 \amp = 0 + y\amp = 1/8 + . +

    + +

    + Thus if x=-1/2, + y=1/8 giving the critical point (-1/2,1/8). +

    + +

    + With D = 4y-(2x+1)^2, + we apply the Second Derivative Test to each critical point. +

    + +

    + At (-1,0), D \lt 0, + so (-1,0) is a saddle point. +

    + +

    + At (0,0), D\lt 0, + so (0,0) is also a saddle point. +

    + +

    + At (-1/2,1/8), D \gt 0 and f_{xx} \gt 0, + so (-1/2,1/8) is a relative minimum. +

    + +
    + Graphing f from and its relative extrema + + + + A plot of the graph of the function used in this example. It is a more complicated-looking surface. + +

    + The surface given by the graph of f(x,y)=x^2y+y^2+xy in is rather hard to describe. + It resembles what one would obtain if one started with a rectangular sheet of metal (or some sufficiently flexible material), + bent the two corners on one side upwards, and bent the two corners on the opposite side downwards. + In the middle the surface appears to be relatively flat, + although there are three points plotted there, where the critical points occur. + At the point where the relative minimum occurs, it is very difficult to see that it is indeed a relative minimum. +

    +     + + + + + //ASY file for figmulti_extreme53D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(24,.7,11.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2}; + real[] myychoice={-0.2,0.2,0.4}; + real[] myzchoice={}; + defaultpen(0.5mm); + pair xbounds=(-3,3); + pair ybounds=(-0.3,0.5); + pair zbounds=(-1,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=x^2y+y^2+xy + triple f(pair t) { + return (t.x,t.y,t.x^2*t.y+t.y^2+t.x*t.y); + } + surface s=surface(f,(-2,-0.3),(1.5,0.5),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //plot point on surface + dotfactor=4;dot((0,0,0));dot((-1,0,0));dot((-1/2,1/8,-.016)); + + + + +
    + +

    + + shows a graph of f and the three critical points. + Note how this function does not vary much near the critical points that is, + visually it is difficult to determine whether a point is a saddle point or relative minimum + (or even a critical point at all!). + This is one reason why the Second Derivative Test is so important to have. +

    +     +     + + +     + + + Constrained Optimization +

    + When optimizing functions of one variable such as y=f(x), + we made use of , + the Extreme Value Theorem, that said that over a closed interval I=[a,b], + a continuous function has both a maximum and minimum value. + To find these maximum and minimum values, + we evaluated f at all critical points in the interval, + as well as at the endpoints + (the boundary) + of the interval. + constrained optimization + optimizationconstrained +

    + +

    + A similar theorem and procedure applies to functions of two variables. + A continuous function over a closed set also attains a maximum and minimum value + (see the following theorem). + We can find these values by evaluating the function at the critical values in the set and over the boundary of the set. + After formally stating this extreme value theorem, we give examples. +

    + + + Extreme Value Theorem + +

    + Let z=f(x,y) be a continuous function on a closed, + bounded set S. + Then f has a maximum and minimum value on S. + Extreme Value Theorem +

    +     +     + + + Finding extrema on a closed set + +

    + Let f(x,y) = x^2-y^2+5 and let S be the triangle with vertices (-1,-2), + (0,1) and (2,-2). + Find the maximum and minimum values of f on S. +

    +     + +

    + It can help to see a graph of f along with the set S. + In the triangle defining S is shown in the xy-plane in a dashed line. + Above it is the graph of f; + we are only concerned with the portion of the surface z=f(x,y) enclosed by the + triangle. +

    + +
    + Plotting the graph of f along with the restricted domain S in + +
    + + + + + A saddle surface on which a triangular curve is plotted. + +

    + There are several things going on in this image. + We have the usual three-dimensional coordinate axes; + against these axes, we see the plot of a hyperbolic paraboloid (saddle surface). + Below the surface, in the xy plane, there is a triangle plotted with dashed lines. + Each point on this triangle corresponds to a point on the surface, + and these points together form a triangular curve on the surface, which is also plotted. +

    +     + + + + + //ASY file for figconopt1.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5.6,-7.8,20.3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,2}; + real[] myychoice={-2,1,2}; + real[] myzchoice={5}; + defaultpen(0.5mm); + pair xbounds=(-1.5,3); + pair ybounds=(-2.5,2.5); + pair zbounds=(-1,10); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=x^2 - y^2 + 5 + triple f(pair t) { + return (t.x,t.y,t.x^2-t.y^2+5); + } + surface s=surface(f,(-1.5,-2.25),(2.25,1.75),16,16,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic},usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw triangle in plane + draw((-1,-2,0)--(0,1,0)--(2,-2,0)--cycle,redpen+dashed+linewidth(1)); + + //Draw triangle on surface + triple g(real t) {return (t,-2,t^2+1);} + path3 mypath=graph(g,-1,2,operator ..); draw(mypath,redpen); //side 1 + triple g(real t) {return (t,3*t+1,t^2-(3*t+1)^2+5);} + path3 mypath=graph(g,-1,0,operator ..); draw(mypath,redpen); //side 2 + triple g(real t) {return (t,-3*t/2+1,t^2-(-3*t/2+1)^2+5);} + path3 mypath=graph(g,0,2,operator ..); draw(mypath,redpen); //side 3 + + + + +
    + +
    + + + + A triangle in the plane representing the domain of the function in this example. + +

    + This plot in the xy plane shows the domain used in . + The domain is a triangle, consisting of three lines: +

      +
    1. +

      + The line y=-2, from the point (-1,-2) to the point (2,-2) +

      +
      2. +
    2. +

      + The line y=-\frac32 x+1, from the point (2,-2) to the point (0,1) +

      +
      3. +
    3. +

      + The line y = 3x+1, from the point (-1,-2) to the point (0,1) +

      +
      4. +
    +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-2,2}, + extra x ticks={-1,1}, + ymin=-2.5,ymax=1.5, + xmin=-2.1,xmax=2.1 + ] + + \addplot [firstcurvestyle,domain=-1:0] {3*x+1} node [black,pos=.2,above,sloped] { $y=3x+1$}; + \addplot [firstcurvestyle,domain=0:2] {-(3/2)*x+1} node [black,pos=.6,below,sloped] { $y=-3/2x+1$}; + \addplot [firstcurvestyle,domain=-1:2] {-2} node [black,pos=.5,below] { $y=-2$}; + + \fill [black] (axis cs:-1,-2) circle (2.4pt); + \fill [black] (axis cs:2,-2) circle (2.4pt); + \fill [black] (axis cs:0,1) circle (2.4pt); + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + We begin by finding the critical points of f. + With f_x = 2x and f_y = -2y, + we find only one critical point, at (0,0). +

    + +

    + We now find the maximum and minimum values that f attains along the boundary of S, that is, + along the edges of the triangle. + In we see the triangle sketched in the plane with the equations of the lines forming its edges labeled. +

    + +

    + Start with the bottom edge, along the line y=-2. + If y is -2, + then on the surface, we are considering points f(x,-2); + that is, our function reduces to f(x,-2) = x^2-(-2)^2+5 = x^2+1=f_1(x). + We want to maximize/minimize + f_1(x)=x^2+1 on the interval [-1,2]. + To do so, we evaluate f_1(x) at its critical points and at the endpoints. +

    + +

    + The critical points of f_1 are found by setting its derivative equal to 0: + + \fp_1(x)=0\qquad \Rightarrow x=0 + . +

    + +

    + Evaluating f_1 at this critical point, + and at the endpoints of [-1,2] gives: + + f_1(-1) = 2 \qquad\amp \Rightarrow\qquad f(-1,-2) = 2 + f_1(0) = 1 \qquad\amp \Rightarrow \qquad f(0,-2) = 1 + f_1(2) = 5 \qquad\amp \Rightarrow \qquad f(2,-2) = 5 + . +

    + +

    + Notice how evaluating f_1 at a point is the same as evaluating f at its corresponding point. +

    + +

    + We need to do this process twice more, + for the other two edges of the triangle. +

    + +

    + Along the left edge, along the line y=3x+1, + we substitute 3x+1 in for y in f(x,y): + + f(x,y) = f(x,3x+1) = x^2-(3x+1)^2+5 = -8x^2-6x+4 = f_2(x) + . +

    + +

    + We want the maximum and minimum values of f_2 on the interval [-1,0], + so we evaluate f_2 at its critical points and the endpoints of the interval. + We find the critical points: + + \fp_2(x) = -16x-6=0 \qquad \Rightarrow \qquad x=-3/8 + . +

    + +

    + Evaluate f_2 at its critical point and the endpoints of [-1,0]: + + f_2(-1) = 2 \qquad\amp \Rightarrow\qquad f(-1,-2) = 2 + f_2(-3/8) = 41/8=5.125 \qquad\amp \Rightarrow \qquad f(-3/8,-0.125) = 5.125 + f_2(0) = 4 \qquad\amp \Rightarrow \qquad f(0,1) = 4 + . +

    + +

    + Finally, we evaluate f along the right edge of the triangle, + where y = -3/2x+1. + + f(x,y) = f(x,-3/2x+1) = x^2-(-3/2x+1)^2+5 = -\frac54x^2+3x+4=f_3(x) + . +

    + +

    + The critical points of f_3(x) are: + + \fp_3(x) = 0 \qquad \Rightarrow \qquad x=6/5=1.2 + . +

    + +

    + We evaluate f_3 at this critical point and at the endpoints of the interval [0,2]: + + f_3(0) = 4 \qquad\amp \Rightarrow\qquad f(0,1) = 4 + f_3(1.2) = 5.8 \qquad\amp \Rightarrow \qquad f(1.2,-0.8) = 5.8 + f_3(2) = 5 \qquad\amp \Rightarrow \qquad f(2,-2) = 5 + . +

    + +

    + One last point to test: the critical point of f, + (0,0). + We find f(0,0) = 5. +

    + +
    + The graph of f along with important points along the boundary of S and the interior in + + + Another view of the hyperbolic paraboloid in this example; this time with points of interest indicated. + +

    + This image is another view of the hyperbolic paraboloid in . + We again see the triangular domain in the xy plane, + and the corresponding curve along the surface, outlining the portion of the surface given by the domain. +

    + +

    + There are several points added to the plot this time: one point where the surface meets the z axis; + this corresponds to the critical point in the interior of the domain. + There are also three points corresponding to the three corners of the triangles, + as well as three points on the sides of the triangular curve, + where potential boundary maxima or minima are located. +

    +     + + + + + //ASY file for figconopt1c3D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(2.6,-7.,24); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,2}; + real[] myychoice={-2,1,2}; + real[] myzchoice={5}; + defaultpen(0.5mm); + pair xbounds=(-1.5,3); + pair ybounds=(-2.5,2.5); + pair zbounds=(-1,10); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw surface z=x^2 - y^2 + 5 + triple f(pair t) { + return (t.x,t.y,t.x^2-t.y^2+5); + } + surface s=surface(f,(-1.5,-2.25),(2.25,1.75),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw triangle in plane + draw((-1,-2,0)--(0,1,0)--(2,-2,0)--cycle,redpen+dashed+linewidth(1)); + + //Draw triangle on surface + triple g(real t) {return (t,-2,t^2+1);} + path3 mypath=graph(g,-1,2,operator ..); draw(mypath,redpen); //side 1 + triple g(real t) {return (t,3*t+1,t^2-(3*t+1)^2+5);} + path3 mypath=graph(g,-1,0,operator ..); draw(mypath,redpen); //side 2 + triple g(real t) {return (t,-3*t/2+1,t^2-(-3*t/2+1)^2+5);} + path3 mypath=graph(g,0,2,operator ..); draw(mypath,redpen); //side 3 + + //Dots at 7 points + dotfactor=4; + dot((-1,-2,2));dot((2,-2,5)); dot((0,1,4)); //at corner of triangles + dot((0,0,5)); //maximum + dot((0,-2,1)); //minimum + dot((1.2,-0.8,5.8));dot((-3/8,-1/8,5.125)); //along edges of triangles + + + + +
    + +

    + We have evaluated f at a total of 7 different places, + all shown in . + We checked each vertex of the triangle twice, + as each showed up as the endpoint of an interval twice. + Of all the z-values found, + the maximum is 5.8, found at (1.2,-0.8); + the minimum is 1, found at (0,-2). +

    +     +     + +

    + This portion of the text is entitled + Constrained Optimization + because we want to optimize a function (, find its maximum and/or minimum values) subject to a constraint + some limit to what values the function can attain. + In the previous example, + we constrained ourselves by considering a function only within the boundary of a triangle. + This was largely arbitrary; + the function and the boundary were chosen just as an example, + with no real meaning behind the function or the chosen constraint. +

    + +

    + However, solving constrained optimization problems is a very important topic in applied mathematics. + The techniques developed here are the basis for solving larger problems, + where more than two variables are involved. +

    + + + +

    + We illustrate the technique once more with a classic problem. +

    + + + Constrained Optimization + +

    + The U.S. Postal Service states that the girth+length of Standard Post Package must not exceed 130''. Given a rectangular box, + the length is the longest side, + and the girth is twice the width+height. +

    + +

    + Given a rectangular box where the width and height are equal, + what are the dimensions of the box that give the maximum volume subject to the constraint of the size of a Standard Post Package? +

    +     + +

    + Let w, h and \ell denote the width, + height and length of a rectangular box; + we assume here that w=h. + The girth is then 2(w+h) = 4w. + The volume of the box is V(w,\ell) = wh\ell = w^2\ell. + We wish to maximize this volume subject to the constraint 4w+\ell\leq 130, + or \ell\leq 130-4w. + (Common sense also indicates that \ell \gt 0, w \gt 0.) +

    + +

    + We begin by finding the critical values of V. + We find that V_w = 2w\ell and V_\ell = w^2; + these are simultaneously 0 at points of the form (0,\ell). + These give a volume of 0, so we can ignore these critical points. +

    + +

    + We now consider the volume along the constraint \ell=130-4w. + Along this line, we have: + + V(w,\ell) = V(w,130-4w) = w^2(130-4w) = 130w^2-4w^3 = V_1(w) + . +

    + +

    + The constraint is applicable on the w-interval + [0,32.5] as indicated in the figure. + Thus we want to maximize V_1 on [0,32.5]. +

    + +

    + Finding the critical values of V_1, + we take the derivative and set it equal to 0: + + V\,'_1(w) = 260w-12w^2 = 0 \Rightarrow w(260-12w)= 0 \Rightarrow w=0,\frac{260}{12}\approx 21.67 + . +

    + +

    + We found two critical values: + when w=0 and when w=21.67. + We again ignore the w=0 solution; + the maximum volume, subject to the constraint, + comes at w=h=21.67, \ell = 130-4(21.6) =43.33. + This gives a volume of V(21.67,43.33) \approx 20,343in^3. +

    + +
    + Graphing the volume of a box with girth 4w and length \ell, subject to a size constraint + + + + A three-dimensional plot of the volume in this example, as a function of length and width. + +

    + A set of three-dimensional coordinate axes are given. + The vertical axis is labeled V (in thousands). + The two horizontal axes are labeled w and \ell. +

    + +

    + The surface has the appearance of what one would obtain by taking a rectangular sheet of rubber, + located in the first quadrant of the \ell w plane, + and stretching the corner furthers from the origin upwards. +

    + +

    + In the \ell w plane, a dashed line corresponding to the contraint curve 4w+\ell=130 is shown. + A corresponding curve lies along the surface, + and we see that the height along this curve is zero where it meets the two axes, + but it rises to a maximum value in the middle of the surface. +

    +     + + + + + //ASY file for figconopt23D.asy in Chapter 12 + + size(200,200,IgnoreAspect); + //size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(176,-2075,580); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={20}; + real[] myychoice={50,100}; + real[] myzchoice={20,40,60,80}; + defaultpen(0.5mm); + pair xbounds=(-5,40); + pair ybounds=(-2,150); + //pair zbounds=(0,100000); + pair zbounds=(0,100); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$w$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$\ell$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("\centering $V$\\ (in thousands)",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x)),N); + + //Draw surface z=x^2*y + triple f(pair t) { + return (t.x,t.y,t.x^2*t.y/1000); + } + surface s=surface(f,(0,0),(32.5,130),16,16,Spline); + //s=crop(s,(-1,-1,-1),(33,135,100)); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw path in plane + draw((32.5,0,0)--(0,130,0),redpen+dashed+linewidth(1)); + + //Draw path on surface + triple g(real t) {return (t,130-4*t,t^2*(130-4*t)/1000);} + path3 mypath=graph(g,0,32.5,operator ..); draw(mypath,redpen); //side 1 + + //Dots at max point + dotfactor=4; dot((21.67,43.33,19.4)); + + + + +
    + +

    + The volume function V(w,\ell) is shown in + along with the constraint \ell = 130-4w. + As done previously, + the constraint is drawn dashed in the xy-plane and also along the graph of the function. + The point where the volume is maximized is indicated. +

    +     +     + +

    + It is hard to overemphasize the importance of optimization. + In the real world, we routinely seek to make + something better. + By expressing the something + as a mathematical function, + making something better + means optimize some function. +

    + +

    + The techniques shown here are only the beginning of an incredibly important field. + Many functions that we seek to optimize are incredibly complex, + making the step of find the gradient and set it equal to \vec 0 highly nontrivial. + Mastery of the principles here are key to being able to tackle these more complicated problems. +

    +     + + + + Terms and Concepts + + + + + +

    + + states that if f has a critical point at P, + then f has a relative extrema at P. + +

    +     + +     + + + + + +

    + A point P is a critical point of f if f_x and f_y are both 0 at P. + +

    +     + +     + + + + + +

    + A point P is a critical point of f if f_x or f_y are undefined at P. + +

    +     + +     + + + + +

    + Explain what it means to solve a constrained optimization problem. +

    +     + + + +

    + Answers will vary. + A good answer will state that we are optimizing a function subject to a constraint, + or limit, on the domain of the function. + We are looking to maximize/minimize the function while looking + at only a certain part of the domain. +

    +     + +     +     + + Problems + + + + +

    + Find the critical points of the given function. + Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, + or saddle point. +

    +     + + + + +

    + f(x,y) = \frac12x^2+2y^2-8y+4x +

    +     + +

    + One critical point at (-4,2); + f_{xx} = 1 and D = 4, + so this point corresponds to a relative minimum. +

    +     + +     + + + + + Context("Point"); + $max=List("None"); + $min=List("None"); + $sad=List("(7,-6)"); + $inc=List("None"); + + + +

    + f(x,y) = x^2+4x+y^2-9y+3xy +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test indicates the critical point is a relative maximum. + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test + indicates the critical point is a relative minimum. + + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test + indicates the critical point is a saddle point. + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test is inconclusive. + + +

    + +

    +     +     +     + + + + +

    + f(x,y) = x^2+3y^2-6y+4xy +

    +     + +

    + One critical point at (6,-3); + D = -4, so this point corresponds to a saddle point. +

    +     + +     + + + + + Context("Point"); + $max=List("(0,0)"); + $min=List("None"); + $sad=List("None"); + $inc=List("None"); + + + +

    + f(x,y) = \frac{1}{x^2+y^2+1} +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test indicates the critical point is a relative maximum. + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test + indicates the critical point is a relative minimum. + + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test + indicates the critical point is a saddle point. + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test is inconclusive. + + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) = x^2+y^3-3y+1 +

    +     + +

    + Two critical points: at (0,-1); + f_{xx} = 2 and D = -12, + so this point corresponds to a saddle point; +

    + +

    + at (0,1), f_{xx} = 2 and D = 12, + so this corresponds to a relative minimum. +

    +     + +     + + + + + Context("Point"); + $max=List("(-1,-2)"); + $min=List("(1,2)"); + $sad=List("(1,-2), (-1,2)"); + $inc=List("None"); + + + +

    + f(x,y) = \frac13x^3-x+\frac13y^3-4y +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test indicates the critical point is a relative maximum. + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test + indicates the critical point is a relative minimum. + + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test + indicates the critical point is a saddle point. + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test is inconclusive. + + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) = x^2y^2 +

    +     + +

    + There are infinitly many critical points, + whenever x=0 or y=0. + With D = -12x^2y^2, + at each critical point D = 0 and the test is inconclusive. + (Some elementary thought shows that each is the absolute minimum, + since f(x,y)\geq 0 for all (x,y).) +

    +     + +     + + + + + Context("Point"); + $max=List("(0,-3)"); + $min=List("(-1,3), (1,3)"); + $sad=List("(-1,-3), (0,3), (1,-3)"); + $inc=List("None"); + + + +

    + f(x,y) = x^4-2x^2+y^3-27y-15 +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test indicates the critical point is a relative maximum. + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test + indicates the critical point is a relative minimum. + + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test + indicates the critical point is a saddle point. + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test is inconclusive. + + +

    + +

    +     +     +     + + + + +

    + \ds f(x,y) = \sqrt{16-(x-3)^2-y^2} +

    +     + +

    + One critical point: f_x = 0 when x=3; + f_y = 0 when y = 0, + so one critical point at (3,0), + which is a relative maximum, where + f_{xx} = \frac{y^2-16}{(16-(x-3)^2-y^2)^{3/2}} and D = \frac{16}{(16-(x-3)^2-y^2)^{2}}. +

    + +

    + Both f_x and f_y are undefined along the circle (x-3)^2+y^2=16; + at any point along this curve, + f(x,y)=0, the absolute minimum of the function. +

    +     + +     + + + + + Context("Point"); + $max=List("None"); + $min=List("None"); + $sad=List("None"); + $inc=List("(0,0)"); + + + +

    + f(x,y) = \sqrt{x^2+y^2} +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test indicates the critical point is a relative maximum. + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test + indicates the critical point is a relative minimum. + + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test + indicates the critical point is a saddle point. + +

    + +

    + + List all critical points + (or report that there are none) + of f where the Second Derivative Test is inconclusive. + + +

    + +

    +     +     +     + +     + + + +

    + Find the absolute maximum and minimum of the function subject to the given constraint. +

    +     + + + + + Context("Point"); + $maxout = Compute("3"); + $maxin = Compute("(0,1)"); + $minout = Compute("3/4"); + $minin = Compute("(0,-1/2)"); + + + +

    + Let f(x,y) = x^2+y^2+y+1, + constrained to the triangle with vertices (0,1), + (-1,-1) and (1,-1). +

    + + + The absolute maximum value is: + +

    + +

    + + The absolute maximum happens for input: + +

    + +

    + + + The absolute minimum value is: + +

    + +

    + + + The absolute minimum happens for input: + +

    + +

    +     + +

    + The triangle is bound by the lines y=-1, + y=2x+1 and y=-2x+1. +

    + +

    + Along y=-1, there is a critical point at (0,-1). +

    + +

    + Along y=2x+1, there is a critical point at (-3/5,-1/5). +

    + +

    + Along y=-2x+1, there is a critical point at (3/5,-1/5). +

    + +

    + The function f has one critical point, + irrespective of the constraint, at (0,-1/2). +

    + +

    + Checking the value of f at these four points, + along with the three vertices of the triangle, + we find the absolute maximum is at (0,1,3) and the absolute minimum is at (0,-1/2,3/4). +

    +     +     +     + + + + + Context("Point"); + $maxout = Compute("25/28"); + $maxin = Compute("(5/14,25/196)"); + $minout = Compute("-12"); + $minin = Compute("(-1,1)"); + + + +

    + Let f(x,y) = 5x-7y, + constrained to the region bounded by y=x^2 and y=1. +

    + + + The absolute maximum value is: + +

    + +

    + + The absolute maximum happens for input: + +

    + +

    + + + The absolute minimum value is: + +

    + +

    + + + The absolute minimum happens for input: + +

    + +

    +     + +

    + The region has two corners + at (1,1) and (-1,1). +

    + +

    + Along y=1, there is no critical point. +

    + +

    + Along y=x^2, + there is a critical point at (5/14,25/196)\approx (0.357,0.128). +

    + +

    + The function f itself has no critical points. + Checking the value of f at the corners (1,1), + (-1,1) and the critical point (5/14,25/196), + we find the absolute maximum is at + (5/14,25/196,25/28) \approx (0.357,0.128,0.893) and the absolute minimum is at (-1,1,-12). +

    +     +     +     + + + + +

    + \ds f(x,y) = x^2+2x+y^2+2y, + constrained to the region bounded by the circle x^2+y^2=4. +

    +     + +

    + The region has no corners + or vertices, just a smooth edge. +

    + +

    + To find critical points along the circle x^2+y^2=4, + we solve for y^2: y^2=4-x^2. + We can go further and state y=\pm\sqrt{4-x^2}. +

    + +

    + We can rewrite f as f(x)=x^2+2x + (4-x^2) + 2\sqrt{4-x^2} = 2x+4+2\sqrt{4-x^2}. + (We will return and use -\sqrt{4-x^2} later.) + Solving f\,'(x)=0, we get x=\sqrt{2} \Rightarrow y=\sqrt{2}. + f\,'(x) is also undefined at x=\pm 2, where y=0. +

    + +

    + Using y=-\sqrt{4-x^2}, + we rewrite f(x,y) as f(x) = 2x+4-2\sqrt{4-x^2}. + Solving \fp(x) =0, we get x=-\sqrt{2},\, y=-\sqrt{2}. + Again, \fp(x) is undefined at x=\pm 2. +

    + +

    + The function f(x,y) itself has a critical point at (-1,-1). +

    + +

    + Checking the value of f at (-1,-1), + (\sqrt{2},\sqrt{2}), + (-\sqrt{2},-\sqrt{2}), + (2,0) and (-2,0), + we find the absolute maximum is at (\sqrt2,\sqrt2,4+4\sqrt2) and the absolute minimum is at (-1,-1,-2). +

    +     + +     + + + + +

    + \ds f(x,y) = 3y-2x^2, + constrained to the region bounded by the parabola + y=x^2+x-1 and the line y=x. +

    +     + +

    + The region has two corners + at (-1,-1) and (1,1). +

    + +

    + Along the line y=x, f(x,y) becomes f(x) = 3x-2x^2. + Along this line, we have a critical point at (3/4,3/4). +

    + +

    + Along the curve y=x^2+x-1, + f(x,y) becomes f(x) =x^2+3x-3. + There is a critical point along this curve at (-3/2, -1/4). + Since x=-3/2 lies outside our bounded region, + we ignore this critical point. +

    + +

    + The function f itself has no critical points. +

    + +

    + Checking the value of f at (-1,-1), + (1,1), (3/4,3/4), + we find the absolute maximum is at + (3/4,3/4,9/8) and the absolute minimum is at (-1,-1,-5). +

    +     + +     + +     +     +     +
    + + + +     + + + Multiple Integration + +

    + introduced multivariable functions and we applied concepts of differential calculus to these functions. + We learned how we can view a function of two variables as a surface in space, + and learned how partial derivatives convey information about how the surface is changing in any direction. +

    + +

    + In this chapter we apply techniques of integral calculus to multivariable functions. + In + we learned how the definite integral of a single variable function gave us + area under the curve. + In this chapter we will see that integration applied to a multivariable function gives us + volume under a surface. + And just as we learned applications of integration beyond finding areas, + we will find applications of integration in this chapter beyond finding volume. +

    +     + +
    + Iterated Integrals and Area + +

    + In + we found that it was useful to differentiate functions of several variables with respect to one variable, + while treating all the other variables as constants or coefficients. + We can integrate functions of several variables in a similar way. + For instance, if we are told that f_x(x,y) = 2xy, + we can treat y as staying constant and integrate to obtain f(x,y): + + f(x,y) \amp = \int f_x(x,y)\, dx + \amp = \int 2xy\, dx + \amp = x^2y + C + . +

    + +

    + Make a careful note about the constant of integration, + C. + This constant is something with a derivative of 0 with respect to x, + so it could be any expression that contains only constants and functions of y. + For instance, if f(x,y) = x^2y+ \sin(y) + y^3 + 17, + then f_x(x,y) = 2xy. + To signify that C is actually a function of y, + we write: + integrationof multivariable functions + + f(x,y) = \int f_x(x,y)\, dx = x^2y+C(y) + . +

    + +

    + Using this process we can even evaluate definite integrals. +

    +     + + + Iterated integrals + + + Integrating functions of more than one variable + +

    + Evaluate the integral \ds \int_1^{2y} 2xy\, dx. +

    +     + +

    + We find the indefinite integral as before, + then apply the Fundamental Theorem of Calculus to evaluate the definite integral: + + \int_1^{2y} 2xy\, dx \amp = x^2y\Big|_1^{2y} + \amp = (2y)^2y - (1)^2y + \amp = 4y^3-y + . +

    +     +     + +

    + We can also integrate with respect to y. + In general, + + \int_{h_1(y)}^{h_2(y)} f_x(x,y)\, dx = f(x,y)\Big|_{h_1(y)}^{h_2(y)} = f\big(h_2(y),y\big)-f\big(h_1(y),y\big) + , + and + + \int_{g_1(x)}^{g_2(x)} f_y(x,y)\, dy = f(x,y)\Big|_{g_1(x)}^{g_2(x)} = f\big(x,g_2(x)\big)-f\big(x,g_1(x)\big) + . +

    + +

    + Note that when integrating with respect to x, + the bounds are functions of y (of the form x=h_1(y) and + x=h_2(y)) and the final result is also a function of y. + When integrating with respect to y, + the bounds are functions of x (of the form y=g_1(x) and + y=g_2(x)) and the final result is a function of x. + Another example will help us understand this. +

    + + + Integrating functions of more than one variable + +

    + Evaluate \ds \int_1^x\big(5x^3y^{-3}+6y^2\big)\, dy. +

    +     + +

    + We consider x as staying constant and integrate with respect to y: + + \int_1^x\big(5x^3y^{-3}+6y^2\big)\, dy \amp = \left(\frac{5x^3y^{-2}}{-2}+\frac{6y^3}{3}\right)\Bigg|_1^x + \amp = \left(-\frac52x^3x^{-2}+2x^3\right) - \left(-\frac52x^3+2\right) + \amp = \frac92x^3-\frac52x-2 + . +

    + +

    + Note how the bounds of the integral are from y=1 to y=x and that the final answer is a function of x. +

    +     +     + +

    + In the previous example, + we integrated a function with respect to y and ended up with a function of x. + We can integrate this as well. + This process is known as iterated integration, + or multiple integration. +

    + + + Evaluating an integral + +

    + Evaluate \ds \int_1^2\left(\int_1^x\big(5x^3y^{-3}+6y^2\big)\, dy\right)\, dx. +

    +     + +

    + We follow a standard order of operations + and perform the operations inside parentheses first (which is the integral evaluated in .) + + \int_1^2\left(\int_1^x\big(5x^3y^{-3}+6y^2\big)\, dy\right)\, dx \amp = \int_1^2 \left(\left[\frac{5x^3y^{-2}}{-2}+\frac{6y^3}{3}\right]\Bigg|_1^x\right)\, dx + \amp = \int_1^2 \left(\frac92x^3-\frac52x-2\right)\, dx + \amp = \left(\frac98x^4-\frac54x^2-2x\right)\Bigg|_1^2 + \amp = \frac{89}8 + . +

    + +

    + Note how the bounds of x were x=1 to x=2 and the final result was a number. +

    +     +     + +

    + The previous example showed how we could perform something called an iterated integral; + we do not yet know why + we would be interested in doing so nor what the result, + such as the number 89/8, means. + Before we investigate these questions, + we offer some definitions. +

    + + + Iterated Integration + +

    + Iterated integration + is the process of repeatedly integrating the results of previous integrations. + Evaulating one integral is denoted as follows. +

    + +

    + Let a, b, + c and d be numbers and let g_1(x), + g_2(x), + h_1(y) and h_2(y) be functions of x and y, + respectively. + Then: + integrationiterated + integrationmultiple + iterated integration + multiple integration|see{iterated integration} + integrationnotation +

    + +

    +

      +
    1. +

      + \ds \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy = \int_c^d\left(\int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\right) dy. +

      +
      2. + +
    2. +

      + \ds \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\right) dx. +

      +
      3. +
    +

    +     +     + +

    + Again make note of the bounds of these iterated integrals. +

    + +

    + With \ds \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy, + x varies from h_1(y) to h_2(y), + whereas y varies from c to d. + That is, the bounds of x are curves, + the curves x=h_1(y) and x=h_2(y), + whereas the bounds of y are constants, + y=c and y=d. + It is useful to remember that when setting up and evaluating such iterated integrals, + we integrate from curve to curve, + then from point to point. +

    + +

    + We now begin to investigate why + we are interested in iterated integrals and + what they mean. +

    +     + + + Area of a plane region +

    + Consider the plane region R bounded by + a\leq x\leq b and g_1(x)\leq y\leq g_2(x), + shown in . + We learned in + that the area of R is given by + integrationarea + + \int_a^b \big(g_2(x)-g_1(x)\big)\, dx + . +

    + +
    + Calculating the area of a plane region R with an iterated integral + + + A region in the first quadrant is bounded above and below by graphs, and left and right by vertical lines. + +

    + Two graphs are plotted in the first quadrant of the plane. The graph y=g_1(x) lies entirely below the graph y=g_2(x). + There are also two vertical lines: x=a on the left, and x=b on the right. + Together, the two graphs and two lines bound a region in the plane, labeled R. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + ytick=\empty, + extra x ticks={.1,.9}, + extra x tick labels={$a$,$b$}, + ymin=-.1,ymax=1, + xmin=-.1,xmax=1 + ] + + \addplot [firstcurvestyle,domain=0:1] ({x},{.2*cos(deg(3*x))+.7}) node [pos=.4,above right,black] { $y=g_2(x)$}; + \addplot [firstcurvestyle,domain=0:1] ({x},{.3*x^2+.1}) node [pos=.4,below right,black] { $y=g_1(x)$}; + + \draw [thick,secondcolor] (axis cs: .1,0) -- (axis cs: .1,1); + \draw [thick,secondcolor] (axis cs: .9,0) -- (axis cs: .9,1); + + \draw (axis cs:.5,.5) node { $R$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + We can view the expression \big(g_2(x)-g_1(x)\big) as + + \big(g_2(x)-g_1(x)\big) = \int_{g_1(x)}^{g_2(x)} 1\, dy =\int_{g_1(x)}^{g_2(x)} \, dy + , + meaning we can express the area of R as an iterated integral: + + \text{ area of } R = \int_a^b \big(g_2(x)-g_1(x)\big)\, dx = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} \, dy\right) dx =\int_a^b\int_{g_1(x)}^{g_2(x)} \, dy\, dx + . +

    + +

    + In short: a certain iterated integral can be viewed as giving the area of a plane region. +

    + +

    + A region R could also be defined by + c\leq y\leq d and h_1(y)\leq x\leq h_2(y), + as shown in . + Using a process similar to that above, we have + + \text{ the area of } R = \int_c^d\int_{h_1(y)}^{h_2(y)} \, dx\, dy + . +

    + +
    + Calculating the area of a plane region R with an iterated integral + + + A region in the first quadrant is bounded left and right by graphs, and above and below by horizontal lines. + +

    + Two curves are plotted in the first quadrant of the plane. + Both curves are graphs, but with x as a function of y. + The graph x=h_1(y) lies entirely to the left of the graph x=h_2(y). + There are also two horizontal lines: y=c at the bottom, and y=d at the top. + Together, the two graphs and two lines bound a region in the plane, labeled R. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick=\empty, + ytick=\empty, + extra y ticks={.1,.9}, + extra y tick labels={$c$,$d$}, + ymin=-.1,ymax=1, + xmin=-.1,xmax=1 + ] + + \addplot [firstcurvestyle,domain=0:1] ({.3*x^2+.1},x) node [pos=.7,right,black] { $x=h_1(y)$}; + \addplot [firstcurvestyle,domain=0:1] ({.2*sin(deg(3*x))+.7},{x}) node [pos=.7,left,black] { $x=h_2(y)$}; + + \draw [thick,secondcolor] (axis cs: 0,.1) -- (axis cs: 1,.1); + \draw [thick,secondcolor] (axis cs: 0,.9) -- (axis cs: 1,.9); + + \draw (axis cs:.5,.5) node { $R$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + We state this formally in a theorem. +

    + + + Area of a plane region + +

    +

      +
    1. +

      + Let R be a plane region bounded by + a\leq x\leq b and g_1(x)\leq y\leq g_2(x), + where g_1 and g_2 are continuous functions on [a,b]. + The area A of R is + + A = \int_a^b\int_{g_1(x)}^{g_2(x)} \, dy\, dx + . +

      +
      2. + +
    2. +

      + Let R be a plane region bounded by + c\leq y\leq d and h_1(y)\leq x\leq h_2(y), + where h_1 and h_2 are continuous functions on [c,d]. + The area A of R is + + integrationarea + + + A = \int_c^d\int_{h_1(y)}^{h_2(y)} \, dx\, dy + . +

      +
      3. +
    +

    +     +     + +

    + The following examples should help us understand this theorem. +

    + + + Area of a rectangle + +

    + Find the area A of the rectangle with corners (-1,1) and (3,3), + as shown in . +

    +
    + Calculating the area of a rectangle with an iterated integral in + + + A rectangle in the plane, spanning x values from -1 to 3, and y values from 1 to 3. + +

    + A rectangle, labeled R, is plotted in the xy plane. + It is defined by the inequalities -1\leq x\leq 3 and 1\leq y\leq 3. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,1,2,3}, + ymin=-.1,ymax=3.5, + xmin=-1.5,xmax=3.5 + ] + + \draw [thick,firstcolor] (axis cs: -1,1) -- (axis cs:3,1)--(axis cs:3,3) -- (axis cs:-1,3) -- cycle; + \draw (axis cs:1,2) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + Multiple integration is obviously overkill in this situation, + but we proceed to establish its use. +

    + +

    + The region R is bounded by x=-1, + x=3, y=1 and y=3. + Choosing to integrate with respect to y first, we have + + A = \int_{-1}^3\int_1^3 1\, dy\, dx = \int_{-1}^3 \left(y\, \Big|_1^3\right)\, dx = \int_{-1}^3 2\, dx = 2x\Big|_{-1}^3=8 + . +

    + +

    + We could also integrate with respect to x first, giving: + + A = \int_1^3\int_{-1}^3 1\, dx \, dy =\int_1^3 \left(x\, \Big|_{-1}^3\right)\, dy = \int_1^3 4\, dy = 4y\Big|_1^3 = 8 + . +

    + +

    + Clearly there are simpler ways to find this area, + but it is interesting to note that this method works. +

    +     +     + + + Area of a triangle + +

    + Find the area A of the triangle with vertices at (1,1), + (3,1) and (5,5), + as shown in . +

    + +
    + Calculating the area of a triangle with iterated integrals in + + + An obtuse triangle in the plane. The base is horizontal, and the other sides are lines with positive slope. + +

    + The image shows a triangle in the first quadrant. + It is an obtuse triangle, with a horizontal base. + The base lies along the line y=1, for 1\leq x\leq 3. + The other two sides of triangle are lines with positive slope: + the line y=x, from the point (1,1) to the point (5,5), + and the line y=2x-5, from the point (1,1) to the point (5,5). +

    + +

    + Note that the upper vertex lies to the right of the right-most vertex at the bottom. + This means that an attempt to integrate first with respect to y will require dividing the triangle into two regions. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={4,5,1,2,3}, + ytick={1,2,3,4,5}, + ymin=-.5,ymax=5.9, + xmin=-.5,xmax=5.9 + ] + + \addplot [firstcurvestyle,domain=1:3] {1} node [pos=.5,black,below] { $y=1$}; + \addplot [firstcurvestyle,domain=1:5] {x} node [pos=.5,black,above,sloped] { $y=x$}; + \addplot [firstcurvestyle,domain=3:5] {2*x-5} node [pos=.5,black,below,sloped] { $y=2x-5$}; + + \draw (axis cs:2.5,1.75) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + The triangle is bounded by the lines as shown in the figure. + Choosing to integrate with respect to x first gives that x is bounded by x=y to x = \frac{y+5}2, + while y is bounded by y=1 to y=5. (Recall that since x-values increase from left to right, + the leftmost curve, x=y, + is the lower bound and the rightmost curve, x=(y+5)/2, + is the upper bound.) The area is + + A \amp = \int_1^5\int_{y}^{\frac{y+5}2}\, dx\, dy + \amp = \int_1^5\left(x\, \Big|_y^{\frac{y+5}2}\right)\, dy + \amp = \int_1^5 \left(-\frac12y+\frac52\right)\, dy + \amp = \left(-\frac14y^2+\frac52y\right)\Big|_1^5 + \amp =4 + . +

    + +

    + We can also find the area by integrating with respect to y first. + In this situation, though, + we have two functions that act as the lower bound for the region R, + y=1 and y=2x-5. + This requires us to use two iterated integrals. + Note how the x-bounds are different for each integral: + + A \amp = \int_1^3\int_1^x 1\, dy \, dx \amp +\amp \amp \amp \int_3^5\int_{2x-5}^x1\, dy\, dx + \amp = \int_1^3\big(y\big)\Big|_1^x\, dx \amp + \amp \amp \amp \int_3^5\big(y\big)\Big|_{2x-5}^x\, dx + \amp = \int_1^3\big(x-1\big)\, dx \amp + \amp \amp \amp \int_3^5\big(-x+5\big)\, dx + \amp = 2 \amp + \amp \amp \amp 2 + \amp =4 + . +

    + +

    + As expected, we get the same answer both ways. +

    +     +     + + + Area of a plane region + +

    + Find the area of the region enclosed by y=2x and y=x^2, + as shown in . +

    + +
    + Calculating the area of a plane region with iterated integrals in + + + A region in the first quadrant of the plane, bounded by a line and a parabola. + +

    + A region R in the first quadrant of the plane is shown. +

      +
    • +

      + It is bounded by two curves. +

      +
      • +
    • +

      + The curve that lies above (or alternatively, to the left of) the region is the line y=2x. +

      +
      • +
    • +

      + The curve that lies below (or alternatively, to the right of) the region is the parabola y=x^2. +

      +
      • +
    +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={4,5,1,2,3}, + ytick={1,2,3,4,5}, + ymin=-.5,ymax=4.5, + xmin=-.5,xmax=2.5 + ] + + \addplot [firstcurvestyle,domain=0:2.5] ({x},{2*x}) node [pos=.4,sloped,above,black] { $y=2x$}; + \addplot [firstcurvestyle,domain=0:2.5] ({x},{x^2}) node [pos=.3,sloped,below,black] { $y=x^2$}; + + \draw (axis cs:1,1.5) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + Once again we'll find the area of the region using both orders of integration. +

    + +

    + Using dy\, dx: + + \int_0^2\int_{x^2}^{2x}1\, dy \, dx = \int_0^2(2x-x^2)\, dx = \big(x^2-\frac13x^3\big)\Big|_0^2 = \frac43 + . +

    + +

    + Using dx\, dy: + + \int_0^4\int_{y/2}^{\sqrt{y}} 1\, dx\, dy = \int_0^4 (\sqrt{y}-y/2)\, dy = \left(\frac23y^{3/2} - \frac14y^2\right)\Big|_0^4 = \frac43 + . +

    +     +     +     + + + Changing Order of Integration + +

    + In each of the previous examples, + we have been given a region R and found the bounds needed to find the area of R using both orders of integration. + We integrated using both orders of integration to demonstrate their equality. + iterated integrationchanging order +

    + +

    + We now approach the skill of describing a region using both orders of integration from a different perspective. + Instead of starting with a region and creating iterated integrals, + we will start with an iterated integral and rewrite it in the other integration order. + To do so, we'll need to understand the region over which we are integrating. +

    + +

    + The simplest of all cases is when both integrals are bound by constants. + The region described by these bounds is a rectangle + (see ), + and so: + + \int_a^b\int_c^d 1\, dy\, dx = \int_c^d\int_a^b1\, dx\, dy + . +

    + +

    + When the inner integral's bounds are not constants, + it is generally very useful to sketch the bounds to determine what the region we are integrating over looks like. + From the sketch we can then rewrite the integral with the other order of integration. +

    + +

    + Examples will help us develop this skill. +

    + + + Changing the order of integration + +

    + Rewrite the iterated integral + \ds \int_0^6\int_0^{x/3} 1\, dy\, dx with the order of integration dx\, dy. +

    +     + +

    + We need to use the bounds of integration to determine the region we are integrating over. +

    + +

    + The bounds tell us that y is bounded by 0 and x/3; + x is bounded by 0 and 6. + We plot these four curves: y=0, y=x/3, + x=0 and x=6 to find the region described by the bounds. + shows these curves, + indicating that R is a triangle. +

    + +
    + Sketching the region R described by the iterated integral in + + + A triangular region in the plane, bounded by lines y=0, x=6, and y=x/3. + +

    + The region R in the plane is a right-angled triangle. +

      +
    • +

      + The base of the triangle is the x axis, from the origin to (6,0) +

      +
      • +
    • +

      + The vertical side of the triangle is the line x=6, from (6,0) to (6.2) +

      +
      • +
    • +

      + The hypotenuse of the triangle is the line y=x/3, from (0,0) to (6,2) +

      +
      • +
    +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=2.5, + xmin=-.5,xmax=6.9 + ] + + \draw [very thick,secondcolor] (axis cs:-.5,0) -- (axis cs:6.9,0) + (axis cs:-.5,-.1666) -- node [above,black,pos=.5,sloped] { $y=x/3$} + (axis cs: 6.9, 2.3); + + \draw [very thick,firstcolor] (axis cs:0,-.5) -- (axis cs:0,2.5) + (axis cs:6,-.5) -- (axis cs:6,2.5); + + \draw (axis cs:4,.5) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + To change the order of integration, + we need to consider the curves that bound the x-values. + We see that the lower bound is x=3y and the upper bound is x=6. + The bounds on y are 0 to 2. + Thus we can rewrite the integral as + \ds \int_0^2\int_{3y}^6 1\, dx \, dy. +

    +     +     + + + Changing the order of integration + +

    + Change the order of integration of \ds\int_0^4\int_{y^2/4}^{(y+4)/2}1\, dx\, dy. +

    +     + +

    + We sketch the region described by the bounds to help us change the integration order. + x is bounded below and above (, to the left and right) by x=y^2/4 and x=(y+4)/2 respectively, + and y is bounded between 0 and 4. + Graphing the previous curves, + we find the region R to be that shown in . +

    + +
    + Drawing the region determined by the bounds of integration in + + + A sketch of the region of integration for this example. + +

    + A region R is plotted in the first quadrant of the plane. + It is bounded by several curves. +

      +
    • +

      + The left boundary of the region is the parabola x=y^2/4. + This is a parabola with its vertex at the origin that opens to the right. +

      +
      • +
    • +

      + The right boundary of the region is the line x=(y+4)/2. + This line intersects the parabola at the point (4,4), + and intersects the x axis at the point (2,0). +

      +
      • +
    • +

      + There is also a bottom boundary, given by the portion of the x axis from (0,0) to (2,0). +

      +
      • +
    +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=4.5, + xmin=-.5,xmax=4.5 + ] + + \addplot [firstcurvestyle,domain=-.5:4.5,samples=30] ({x^2/4},{x}) node [pos=.4,sloped,above,black] {$x=y^2/4$}; + \addplot [firstcurvestyle,domain=-.5:4.5] ({((x+4))/2},{x}) node [pos=.4,sloped,below,black] { $x= (y+4)/2$}; + + \draw [very thick,secondcolor] (axis cs:-.5,0) -- (axis cs:4.5,0) + (axis cs:-.5,4) -- (axis cs:4.5,4); + + \draw (axis cs:2,1.75) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + To change the order of integration, + we need to establish curves that bound y. + The figure makes it clear that there are two lower bounds for y: + y=0 on 0\leq x\leq 2, + and y=2x-4 on 2\leq x\leq 4. + Thus we need two double integrals. + The upper bound for each is y=2\sqrt{x}. + Thus we have + + \int_0^4\int_{y^2/4}^{(y+4)/2}1\, dx\, dy = \int_0^2\int_0^{2\sqrt{x}} 1\, dy\, dx + \int_2^4\int_{2x-4}^{2\sqrt{x}}1\, dy\, dx + . +

    +     +     + +

    + This section has introduced a new concept, the iterated integral. + We developed one application for iterated integration: + area between curves. + However, this is not new, + for we already know how to find areas bounded by curves. +

    + +

    + In the next section we apply iterated integration to solve problems we currently do not know how to handle. + The real goal of this section was not to learn a new way of computing area. + Rather, our goal was to learn how to define a region in the plane using the bounds of an iterated integral. + That skill is very important in the following sections. +

    +     + + + + Terms and Concepts + + + + +

    + When integrating f_x(x,y) with respect to x, + the constant of integration C is really which: + C(x) or C(y)? + What does this mean? +

    +     + + + +

    + C(y), meaning that instead of being just a constant, + like the number 5, it is a function of y, + which acts like a constant when taking derivatives with respect to x. +

    +     + +     + + + + +

    + Evaluating a double integral in steps is called + + . +

    +     + + + + + + + + + + + + + +     + + + + +

    + When evaluating an iterated integral, + we integrate from to , + then from to . +

    +     + + + + + + + + + + + + + + + + + + + + + + + +     + + + + +

    + One understanding of an iterated integral is that + \ds \int_a^b\int_{g_1(x)}^{g_2(x)} \, dy\, dx gives the of a plane region. +

    +     + + + + + + + + +     +     + + + Problems + + + +

    + Evaluate the integral and subsequent iterated integral. +

    +     + + + + + +

    + \ds \int_2^{5} \big(6x^2+4xy-3y^2\big)\, dy +

    +     + +

    + 18x^2+42x-117 +

    +     +     + + + +

    + \ds \int_{-3}^{-2} \int_2^{5} \big(6x^2+4xy-3y^2\big)\, dy\, dx +

    +     + +

    + -108 +

    +     +     + +     + + + + + + Context()->variables->are(x=>'Real',y=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $a=Compute("2+pi^2cos(y)"); + $b=Formula("pi^2+pi"); + + + + +

    + \ds \int_0^{\pi} (2x\cos(y) + \sin(x) )\,dx +

    + +

    + +

    +     +     + + + +

    + \ds \int_{0}^{\pi/2} \int_0^{\pi} (2x\cos(y) + \sin(x) )\,dx\,dy +

    +

    + +

    +     +     +     +     + + + + + +

    + \ds \int_1^x \big(x^2y - y+2\big)\, dy +

    +     + +

    + x^4/2-x^2+2x-3/2 +

    +     +     + + + +

    + \ds \int_0^2\int_1^x \big(x^2y - y+2\big)\, dy\, dx +

    +     + +

    + 23/15 +

    +     +     + +     + + + + + + Context()->variables->are(x=>'Real',y=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $a=Compute("y^4/2-y^3+y^2/2"); + $b=Formula("8/15"); + + + + +

    + \ds \int_y^{y^2} (x-y)\,dx +

    +

    + +

    +     +     + + + +

    + \ds \int_{-1}^1\int_y^{y^2} (x-y)\,dx\,dy +

    +

    + +

    +     +     +     +     + + + + + + +

    + \ds \int_0^{y} \big(\cos(x) \sin(y) \big)\, dx +

    +     + +

    + \sin^2(y) +

    +     +     + + + +

    + \ds \int_0^\pi \int_0^{y} \big(\cos(x) \sin(y) \big)\, dx\, dy +

    +     + +

    + \pi/2 +

    +     +     + +     + + + + + + Context()->variables->are(x=>'Real',y=>'Real'); + Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); + $a=Compute("x/(1+x^2)"); + $b=Formula("1/2 ln(5/2)"); + + + + +

    + \ds \int_0^{x} \left(\frac{1}{1+x^2}\right)\,dy +

    +

    + +

    +     +     + + + +

    + \ds \int_1^2 \int_0^{x} \left(\frac{1}{1+x^2}\right)\,dy\,dx +

    +

    + +

    +     +     +     +     + +     + + + +

    + A graph of a planar region R is given. + Give the iterated integrals, + with both orders of integration dy\, dx and dx\, dy, + that give the area of R. + Evaluate one of the iterated integrals to find the area. +

    +     + + + + + + + The region R is a rectangle, with x from 1 to 4, and y from -2 to 1. + +

    + A sketch of the region R for this exercise. + R is a rectangle, with bounds 1\leq x\leq 4 and -2\leq y\le1 1. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={-2,-1,1}, + ymin=-2.5,ymax=1.5, + xmin=-.5,xmax=4.5 + ] + + \draw [very thick,firstcolor] (axis cs:1,-2) -- (axis cs: 4,-2) -- (axis cs: 4,1) -- (axis cs: 1,1) -- cycle; + + \draw (axis cs: 2.5,-1) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \ds \int_1^4\int_{-2}^1\, dy\, dx and \ds\int_{-2}^1\int_1^4\, dx\, dy. +

    + +

    + area of R = 9u^2 +

    +     + +     + + + + + + + The region R is a right triangle with vertices (1,1), (4,1), and (4,3) + +

    + A sketch of the region R for this exercise. + The boundary of R is a right-angle triangle, + with vertices at (1,1), (4,1), and (4,3). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3}, + ymin=-.5,ymax=3.5, + xmin=-.5,xmax=4.5 + ] + + \draw [very thick,firstcolor] (axis cs:1,1) -- (axis cs: 4,1) -- (axis cs: 4,3) -- cycle; + + \draw (axis cs: 3,1.75) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \ds\int_1^4\int_{1}^{\frac23x+\frac13}\, dy\, dx and \ds\int_{1}^3\int_{\frac32y-\frac12}^4\, dx\, dy. +

    + +

    + area of R = 3u^2 +

    +     + +     + + + + + + + A triangular region R. The vertices are (2,5), (2,1), and (4,3). + +

    + The region R for this exercise is a triangle. + The left side of the triangle is vertical, from (2,1) to (2,5). + The other vertex is at (4,3) +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3,4}, + ytick={1,2,3,4,5}, + ymin=-.5,ymax=5.5, + xmin=-.5,xmax=4.5% + ] + + \draw [very thick,firstcolor] (axis cs:2,1) -- (axis cs: 4,3) -- (axis cs: 2,5) -- cycle; + + \draw (axis cs: 2.75,3) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \ds\int_2^4\int_{x-1}^{7-x}\, dy\, dx. + The order dx\, dy needs two iterated integrals as x is bounded above by two different functions. + This gives: + + \int_{1}^3\int_{2}^{y+1}\, dx\, dy+\int_{3}^5\int_{2}^{7-y}\, dx\, dy + . + area of R = 4u^2 +

    +     + +     + + + + + + + The region R is bounded to the left by a parabola opening along the x axis, and the line x=12. + +

    + The region R for this exercise is bounded by a parabola and a line. +

      +
    • +

      + The parabola is x=y^2/3, which has vertex (0,0) and opens to the right. +

      +
      • +
    • +

      + The line is the vertical line x=12. +

      +
      • +
    +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={2,4,...,12}, + ytick={-2,-4,-6,6,4,2}, + ymin=-6.9,ymax=6.9, + xmin=-.5,xmax=12.9 + ] + + \draw (axis cs: 6,2) node {$R$}; + + \addplot+ [domain=-6:6,samples=60] ({x^2/3},{x}) node [pos=.7,sloped,above,black] { $x=y^2/3$} -- cycle; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \ds\int_0^{12}\int_{-\sqrt{3x}}^{\sqrt{3x}}\, dy\, dx and \ds \int_{-6}^6\int_{y^2/3}^{12}\, dx\, dy +

    + +

    + area of R = 96u^2 +

    +     + +     + + + + + + + A region in the first quadrant bounded by two curves that intersect at the points (0,0) and (1,1) + +

    + The region R is bounded by two intersecting curves. +

      +
    • +

      + The curve that lies below (and to the right) of the region is y=x^4. + This curve is increasing and concave up. +

      +
      • +
    • +

      + The curve that lies above (and to the left) of the region is y=\sqrt{x}. + This curve is increasing and concave down. +

      +
      • +
    • +

      + The two curves intersect at the origin, and again at the point (1,1). +

      +
      • +
    +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=1.1, + xmin=-.5,xmax=1.1 + ] + + \addplot [firstcurvestyle,domain=0:1.05] ({x^2},{x}) node [pos=.5,sloped,above,black] { $y=\sqrt{x}$}; + \addplot [firstcurvestyle,domain=-.1:1.05,samples=40] ({x},{x^4}) node [pos=.5,sloped,below,black] { $y=x^4$}; + + \draw (axis cs: .5,.4) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \ds\int_0^{1}\int_{x^4}^{\sqrt{x}}\, dy\, dx and \ds\int_{0}^1\int_{y^2}^{\sqrt[4]{y}}\, dx\, dy +

    + +

    + area of R = 7/15u^2 +

    +     + +     + + + + + + + A region R bounded above by a line, and below by a cubic curve. + +

    + The region R lies in the first quadrant. It is bounded by two curves that intersect twice. +

      +
    • +

      + Above (and to the left of) the region is the line y=4x. +

      +
      • +
    • +

      + Below (and to the right of) the region is the curve y=x^3. +

      +
      • +
    • +

      + The two curves meet at the origin. The curve y=x^3 begins below the line y=4x, + but bends upward to meet the line again at the point (2,8). +

      +
      • +
    +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=8.9, + xmin=-.5,xmax=2.5 + ] + + \addplot [firstcurvestyle,domain=-.1:2.1] ({x},{x^3}) node [pos=.2,sloped,below right,black] { $y=x^3$}; + \addplot [firstcurvestyle,domain=-.1:2.1] {4*x} node [pos=.4,sloped,above,black] { $y=4x$}; + + \draw (axis cs: 1,2.25) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +     + +

    + \ds\int_0^{2}\int_{x^3}^{4{x}}\, dy\, dx and \ds\int_{0}^8\int_{y/4}^{\sqrt[3]{y}}\, dx\, dy +

    + +

    + area of R = 4u^2 +

    +     + +     + +     + + + +

    + Iterated integrals are given that compute the area of a region R in the xy-plane. + Sketch the region R, + and give the iterated integral(s) that give the area of R with the opposite order of integration. +

    +     + + + + +

    + \ds \int_{-2}^2\int_0^{4-x^2} \, dy \, dx +

    +     + + + + A region R bounded by a downward-opening parabola and the x axis. + +

    + The curve y=4-x^2 is a parabola centered on the y axis. + Its vertex is at (0,4), and the parabola opens downward, meeting the x axis at (-2,0) and (2,0). + The parabola forms the upper boundary of a region R; + the lower boundary is the portion of the x axis between the two intercepts at (-2,0) and (2,0). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=4.5, + xmin=-2.5,xmax=2.5 + ] + + \addplot+ [domain=-2:2,samples=60] ({x},{4-x^2}) node [pos=.1,right,black] { $y=4-x^2$} -- cycle; + + \draw (axis cs: 1,1) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +

    + area of R = \ds\int_{0}^{4}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\, dx\, dy +

    +     + +     + + + + +

    + \ds \int_{0}^{1}\int_{5-5x}^{5-5x^2} \, dy \, dx +

    +     + + + + The region R is in the first quadrant, between a parabola and a line. + +

    + The upper boundary of the region R is the parabola y=5-5x^2. + The vertex of this parabola is at (0,5), and the parabola opens downward. +

    + +

    + The lower boundary of R is the line y=5-5x. + This line passes through the vertex at (0,5), + and meets the parabola again at (1,0). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ytick={1,2,3,4,5}, + ymin=-.5,ymax=5.9, + xmin=-.1,xmax=1.1 + ] + + \addplot [firstcurvestyle,domain=0:1] ({x},{5-5*x^2}) node [pos=.05,above right,black] { $y=5-5x^2$}; + \addplot [firstcurvestyle,domain=0:1] {-5*x+5} node [pos=.5,below,sloped,black] { $y=-5x+5$}; + + \draw (axis cs: .5,3.3) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +

    + area of R = \ds \int_0^5\int_{1-y/5}^{\sqrt{1-y/5}} \, dx \, dy +

    +     + +     + + + + +

    + \ds \int_{-2}^2\int_{0}^{2\sqrt{4-y^2}} \, dx \, dy +

    +     + + + + A region bounded by the y axis and the right half of an ellipse centered at the origin. + +

    + The curve x=2\sqrt{4-y^2} is the right half of the ellipse \frac{x^2}{16}+\frac{y^2}{4}=1. + The ellipse intersects the y axis at (0,2) and (0,-2), + and the x axis at (4,0). +

    + +

    + The region R is bounded to the right by the ellipse, and to the left by the portion of the y axis + between the points (0,-2) and (0,2). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-2.5,ymax=2.5, + xmin=-.5,xmax=4.5 + ] + + \draw (axis cs: 1,1) node {$R$}; + + \addplot+ [domain=-90:90,samples=40] ({4*cos(x)},{2*sin(x)}) node [pos=.25,above left,black] { $x^2/16+y^2/4=1$} -- cycle; + + \end{axis} + + \end{tikzpicture} + + + + +

    + area of R = \ds \int_0^4\int_{-\sqrt{4-x^2/4}}^{\sqrt{4-x^2/4}} \, dy \, dx +

    +     + +     + + + + +

    + \ds \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \, dy \, dx +

    +     + + + + A circle of radius 3, centered at the origin. + +

    + The boundary of the region described by the bounds in this integral + is the circle x^2+y^2=9. + Its center is the origin, and its radius is 3. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-3.1,ymax=3.1, + xmin=-3.7,xmax=3.7 + ] + + \draw (axis cs: 1,1) node {$R$}; + + \addplot+ [domain=0:360,samples=80] ({3*cos(x)},{3*sin(x)}) node [pos=.4,below right,black] { $x^2+y^2=9$}; + + \end{axis} + + \end{tikzpicture} + + + + +

    + area of R = \ds \int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \, dx \, dy +

    +     + +     + + + + +

    + \ds \int_{0}^{1}\int_{-\sqrt{y}}^{\sqrt{y}} \, dx \, dy+\int_{1}^{4}\int_{y-2}^{\sqrt{y}} \, dx \, dy +

    +     + + + + The region between a right-opening parabola and a vertical line. + +

    + The region of integration for this exercise is bounded to the left by the curve x=y^2, + which is a parabola opening to the right, with its vertex at the origin. + The region is bounded to the right by the vertical line x=1. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,1,2}, + ytick={1,2,3,4}, + ymin=-.1,ymax=4.5, + xmin=-1.1,xmax=2.1 + ] + + \draw (axis cs: .5,1.5) node {$R$}; + + \addplot [firstcurvestyle,domain=-1:2,] ({x},{x^2}) node [pos=.5,below right,black] { $y=x^2$}; + \addplot [firstcurvestyle,domain=-1:2] {x+2} node [pos=.5,sloped,above,black] { $y=x+2$}; + + \end{axis} + + \end{tikzpicture} + + + + +

    + area of R = \ds \int_{-1}^2\int_{x^2}^{x+2} \, dy \, dx +

    +     + +     + + + + +

    + \ds \int_{-1}^{1}\int_{(x-1)/2}^{(1-x)/2} \, dy \, dx +

    +     + + + + A region bounded by a triangle, with vertices at (-1,1), (-1,-1), and (1,0). + +

    + The boundary of the region of integration for this problem is a triangle. + It is an isosceles triangle, with a vertical side from (-1,-1) to (-1,1), + and remaining vertex on the x axis at (1,0). +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,1}, + ytick={-1,1}, + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.1 + ] + + \draw (axis cs: -.5,.25) node {$R$}; + + \draw [firstcolor,very thick] (axis cs: -1,1) -- (axis cs:1,0) -- (axis cs: -1,-1) -- (axis cs: -1,1); + + \end{axis} + + \end{tikzpicture} + + + + +

    + area of R = \ds \int_{-1}^0\int_{0}^{2y+1} \, dx \, dy + \int_{0}^1\int_{0}^{1-2y} \, dx \, dy +

    +     + +     + +     +     +     +
    +
    + Double Integration and Volume +

    + The definite integral of f over [a,b], \int_a^b f(x)\, dx, + was introduced as the signed area under the curve. + We approximated the value of this area by first subdividing [a,b] into n subintervals, + where the ith subinterval has length \dx_i, + and letting c_i be any value in the ith subinterval. + We formed rectangles that approximated part of the region under the curve with width \dx_i, + height f(c_i), and hence with area f(c_i)\dx_i. + Summing all the rectangle's areas gave an approximation of the definite integral, + and stated that + + \int_a^bf(x)\, dx = \lim_{\norm{\Delta x}\to 0}\sum f(c_i)\dx_i + , + connecting the area under the curve with sums of the areas of rectangles. +

    + +

    + We use a similar approach in this section to find volume under a surface. +

    + + + +

    + Let R be a closed, + bounded region in the xy-plane and let + z=f(x,y) be a continuous function defined on R. + We wish to find the signed volume under the graph of f over R. + (We use the term signed volume + to denote that space above the xy-plane, + under f, will have a positive volume; + space above f and under the xy-plane will have a + negative volume, + similar to the notion of signed area used before.) +

    + +

    + We start by partitioning R into n rectangular subregions as shown in . + For simplicity's sake, + we let all widths be \dx and all heights be \dy. + Note that the sum of the areas of the rectangles is not equal to the area of R, + but rather is a close approximation. + Arbitrarily number the rectangles 1 through n, + and pick a point (x_i,y_i) in the ith subregion. +

    + +
    + Developing a method for finding signed volume under a surface + +
    + + + + An illustration of the partition of a plane region into small rectangles. + +

    + A region in the plane is plotted against x and y coordinate axes. +

      +
    • +

      + The region is symmetric about the x axis, + and extends from the origin to the point (2,0). +

      +
      • +
    • +

      + It is close to a circle in shape, but it is not a true circle: + it is more egg-like, being more sharply curved near the origin. +

      +
      • +
    • +

      + The precise shape is not that important in this context. +

      +
      • +
    +

    + +

    + A grid of small rectangles is drawn over the region, + to illustrate how it can be approximated by subdividing it into rectangles. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.85,ymax=.85, + xmin=-.3,xmax=2.2 + ] + + \addplot+ [domain=-90:90,samples=80] ({cos(x)*(1+cos(2*x))},{sin(x)*(1+cos(2*x))}); + + \draw (axis cs: 0,-.3) -- (axis cs:0,.3) + (axis cs: .2,-.4) -- (axis cs:.2,.4) + (axis cs: 0.4,-.6) -- (axis cs:.4,.6) + (axis cs: 0.6,-.7) -- (axis cs:.6,.7) + (axis cs: .8,-.8) -- (axis cs:.8,.8) + (axis cs: 1,-.8) -- (axis cs:1,.8) + (axis cs: 1.2,-.8) -- (axis cs:1.2,.8) + (axis cs: 1.4,-.8) -- (axis cs:1.4,.8) + (axis cs: 1.6,-.7) -- (axis cs:1.6,.7) + (axis cs: 1.8,-.6) -- (axis cs:1.8,.6) + (axis cs: 2,-.4) -- (axis cs:2,.4); + + \draw (axis cs: .8,.8) -- (axis cs:1.4,.8) + (axis cs: .6,.7) -- (axis cs:1.6,.7) + (axis cs: .4,.6) -- (axis cs:1.8,.6) + (axis cs: .4,.5) -- (axis cs:1.8,.5) + (axis cs: .2,.4) -- (axis cs:2,.4) + (axis cs: 0,.3) -- (axis cs:2,.3) + (axis cs: 0,.2) -- (axis cs:2,.2) + (axis cs: 0,.1) -- (axis cs:2,.1) + (axis cs: .8,-.8) -- (axis cs:1.4,-.8) + (axis cs: .6,-.7) -- (axis cs:1.6,-.7) + (axis cs: .4,-.6) -- (axis cs:1.8,-.6) + (axis cs: .4,-.5) -- (axis cs:1.8,-.5) + (axis cs: .2,-.4) -- (axis cs:2,-.4) + (axis cs: 0,-.3) -- (axis cs:2,-.3) + (axis cs: 0,-.2) -- (axis cs:2,-.2) + (axis cs: 0,-.1) -- (axis cs:2,-.1); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + A surface in three dimensions lies over a partitioned region in the xy plane. A rectangular prism extends from one rectangle to the surface. + +

    + A surface is plotted in space, against three-dimensional coordinate axes. + In the xy plane, there is a copy of the region from , + including a rendering of the rectangular grid representing the partition of the region into small rectangles. +

    + +

    + The surface lies above the xy plane. + On the surface, we can see a curve that corresponds to the boundary of the region in the xy plane. + The grid lines on the surface also appear to correspond to the partition of the region below the surface. +

    + +

    + Over one of the small rectangles in the xy plane, + there is a rectangular box. The base of the box is the rectangle in the plane. + The box extends upward to the surface, + with its height equal to the z coordinate on the surface at one point in the rectangle. +

    +     + + + + + //ASY file for figdouble_intro23D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.5,3.1,3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={-1,1}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-1,2.5); + pair ybounds=(-1.25,1.25); + pair zbounds=(0,2.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface//{z=-.5*(x-1)^2-.5*(y)^2+2}; + triple f(pair t) { + return (t.x,t.y,-.5*(t.x-1)^2-.5*(t.y)^2+2); + } + surface s=surface(f,(-0.221,-1),(2.2,1),12,20,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic},usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + pen pp=linewidth(.25mm); + //draw the grid in the xy-plane + //along fixed x + draw((0,-.3,0) -- (0,.3,0),pp); + draw((.2,-.4,0) -- (.2,.4,0),pp); + draw((0.4,-.6,0) -- (.4,.6,0),pp); + draw((0.6,-.7,0) -- (.6,.7,0),pp); + draw((.8,-.8,0) -- (.8,.8,0),pp); + draw((1,-.8,0) -- (1,.8,0),pp); + draw((1.2,-.8,0) -- (1.2,.8,0),pp); + draw((1.4,-.8,0) -- (1.4,.8,0),pp); + draw((1.6,-.7,0) -- (1.6,.7,0),pp); + draw((1.8,-.6,0) -- (1.8,.6,0),pp); + draw((2,-.4,0) -- (2,.4,0),pp); + //along fixed y + draw((.8,.8,0) -- (1.4,.8,0),pp); + draw((.6,.7,0) -- (1.6,.7,0),pp); + draw((.4,.6,0) -- (1.8,.6,0),pp); + draw((.4,.5,0) -- (1.8,.5,0),pp); + draw((.2,.4,0) -- (2,.4,0),pp); + draw((0,.3,0) -- (2,.3,0),pp); + draw((0,.2,0) -- (2,.2,0),pp); + draw((0,.1,0) -- (2,.1,0),pp); + draw((.8,-.8,0) -- (1.4,-.8,0),pp); + draw((.6,-.7,0) -- (1.6,-.7,0),pp); + draw((.4,-.6,0) -- (1.8,-.6,0),pp); + draw((.4,-.5,0) -- (1.8,-.5,0),pp); + draw((.2,-.4,0) -- (2,-.4,0),pp); + draw((0,-.3,0) -- (2,-.3,0),pp); + draw((0,-.2,0) -- (2,-.2,0),pp); + draw((0,-.1,0) -- (2,-.1,0),pp); + + //Draw curve on top of the grid in xy plane ({cos(x)*(1+cos(2*x))},{sin(x)*(1+cos(2*x))},0); + triple g(real t) {return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),0);} + path3 mypath=graph(g,-pi/2,pi/2,operator ..); + draw(mypath,bluepen); + + // draw curve on surface + triple g(real t) { + return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),-.5*(cos(t)*(1+cos(2*t))-1)^2-.5*(sin(t)*(1+cos(2*t)))^2+2); + } + path3 mypath=graph(g,-pi/2,pi/2,operator ..); + draw(mypath,bluepen); + + //surface s=surface(f,(-pi/2,0),(pi/2,2*pi),8,8,Spline); + //pen p=apexmeshpen; + //draw(s,surfacepen,bluepen+linewidth(2)); + //path3 mypath=graph(g,0,32.5,operator ..); draw(mypath,bluepen); + + //Draw the top of the column on the surface + //draw((1.6,0.3,1.865)--(1.8,.3,1.725),bluepen+linewidth(2)); + //draw((1.8,0.3,1.7)--(1.8,.4,1.7),bluepen+linewidth(2)); + //draw((1.8,0.4,1.7)--(1.6,.4,1.7),bluepen+linewidth(2)); + //draw((1.6,0.4,1.7)--(1.6,.3,1.7),bluepen+linewidth(2)); + draw((1.6,.3,1.775) -- (1.6,.4,1.74) -- (1.8,.4,1.6) -- (1.8,.3,1.635) --cycle,bluepen+.4mm); + + //Draw the rectangular column + //base + draw((1.6,0.3,0)--(1.8,.3,0)--(1.8,.4,0)--(1.6,.4,0)--cycle,redpen+.4mm); + + // //verticals plotted to average of 4 function heights + //draw((1.6,0.3,0)--(1.6,.3,1.7),redpen+.4mm);//base + //draw((1.8,0.3,0)--(1.8,.3,1.7),redpen+.4mm);//base + //draw((1.8,0.4,0)--(1.8,.4,1.7),redpen+.4mm);//base + //draw((1.6,0.4,0)--(1.6,.4,1.7),redpen+.4mm);//base + //top + draw((1.6,0.3,1.7)--(1.8,.3,1.7)--(1.8,.4,1.7)--(1.6,.4,1.7)--cycle,redpen); + + pen q=redcurvepen+.4mm; + // draw(ss,surfacepen2,meshpen=q,nolight,render(merge=true)); + // emissive(redpen+opacity(0.7)) + + //Shade the column now + //import three; + + path3 p = (1.6,0.3,0)--(1.8,.3,0) -- (1.8,0.3,1.7)--(1.6,.3,1.7); //Left + draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); + path3 p = (1.6,0.3,1.7)--(1.8,.3,1.7)--(1.8,.4,1.7)--(1.6,.4,1.7); //top + draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); + //draw(surface(p -- cycle), emissive(redpen+opacity(0.7))); + path3 p = (1.6,0.4,0)--(1.8,.4,0) -- (1.8,0.4,1.7)--(1.6,.4,1.7); //right + draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); + //draw(surface(p -- cycle), emissive(redpen+opacity(0.7))); + path3 p = (1.6,0.3,0)--(1.6,.4,0) -- (1.6,0.4,1.7)--(1.6,.3,1.7); //back + draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); + //draw(surface(p -- cycle), emissive(redpen+opacity(0.7))); + path3 p = (1.8,0.3,0)--(1.8,.4,0) -- (1.8,0.4,1.7)--(1.8,.3,1.7); //front + draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); + //draw(surface(p -- cycle), emissive(redpen+opacity(0.7))); + path3 p = (1.6,0.3,0)--(1.8,.3,0)--(1.8,.4,0)--(1.6,.4,0); //bottom + draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); + //draw(surface(p -- cycle), emissive(redpen+opacity(0.7))); + + + + +
    +     + +
    + +

    + The volume of the rectangular solid whose base is the + ith subregion and whose height is + f(x_i,y_i) is V_i=f(x_i,y_i)\dx\dy. + Such a solid is shown in . + Note how this rectangular solid only approximates the true volume under the surface; + part of the solid is above the surface and part is below. +

    + +

    + For each subregion R_i used to approximate R, + create the rectangular solid with base area \dx\dy and height f(x_i,y_i). + The sum of all rectangular solids is + + \ds \sum_{i=1}^n f(x_i,y_i)\dx\dy + . +

    + +

    + This approximates the signed volume under f over R. + As we have done before, + to get a better approximation we can use more rectangles to approximate the region R. +

    + +

    + In general, each rectangle could have a different width \dx_j and height \dy_k, + giving the ith rectangle an area \Delta A_i = \dx_j\dy_k and the + ith rectangular solid a volume of f(x_i,y_i)\Delta A_i. + Let \norm{\Delta A} denote the length of the longest diagonal of all rectangles in the subdivision of R; + \norm{\Delta A}\to 0 means each rectangle's width and height are both approaching 0. + If f is a continuous function, + as \norm{\Delta A} shrinks + (and hence n\to\infty) + the summation \ds \sum_{i=1}^n f(x_i,y_i)\Delta A_i approximates the signed volume better and better. + This leads to a definition. +

    + + + + + Double Integral, Signed Volume + +

    + Let z=f(x,y) be a continuous function defined over a closed, + bounded region R in the xy-plane. + The signed volume V under f over R is denoted by the + double integral + + V = \iint_R f(x,y)\, dA + . +

    + +

    + Alternate notations for the double integral are + integrationdouble + double integral + iterated integration + signed volume + volume + + \iint_R f(x,y)\, dA=\iint_R f(x,y)\, dx\, dy=\iint_R f(x,y)\, dy\, dx + . +

    +     +     + + + +

    + does not state how to find the signed volume, + though the notation offers a hint. + We need the next two theorems to evaluate double integrals to find volume. +

    + + + Double Integrals and Signed Volume + +

    + Let z=f(x,y) be a continuous function defined over a closed , bounded region R in the xy-plane. + Then the signed volume V under f over R is + + V = \iint_R f(x,y)\, dA = \lim_{\norm{\Delta A}\to 0}\sum_{i=1}^n f(x_i,y_i)\Delta A_i + . +

    +     +     + +

    + This theorem states that we can find the exact signed volume using a limit of sums. + The partition of the region R is not specified, + so any partitioning where the diagonal of each rectangle shrinks to 0 results in the same answer. +

    + +

    + This does not offer a very satisfying way of computing volume, though. + Our experience has shown that evaluating the limits of sums can be tedious. + We seek a more direct method. +

    + +

    + Recall + in . + This stated that if A(x) gives the cross-sectional area of a solid at x, + then \int_a^b A(x)\, dx gave the volume of that solid over [a,b]. +

    + +

    + Consider , + where a surface z=f(x,y) is drawn over a region R. + Fixing a particular x value, + we can consider the area under f over R where x has that fixed value. + That area can be found with a definite integral, namely + + A(x)=\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy + . +

    + +
    + Finding volume under a surface by sweeping out a cross-sectional area + + + + A parabolic surface in space is intersected by a plane of constant x value. + +

    + The plot begins with a set of three-dimensional coordinate axes. + In the xy plane, there is a curve that appears to be a circle of radius 1, centered at (1,0,0). +

    + +

    + The circle is the boundary of a region R. + Above this region, there is a surface; the surface has the shape of a downward-opening circular paraboloid. +

    + +

    + A portion of a vertical plane is also drawn. + The plane corresponds to a fixed value of x, and extends in the y direction across the circle. + The plane is shown extending upward to the surface, which it intersects along a curve. + We can think of this portion of the plane as a slice of the solid that lies below the paraboloid and above the circle. +

    +     + + + + + //ASY file for figdouble_intro23D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5,5.5,1.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={-1,1}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-1,2.5); + pair ybounds=(-1.25,1.25); + pair zbounds=(0,2.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Create a shaded slice + + pen q=redcurvepen+.4mm; + path3 p = (1.6,-0.65,0)--(1.6,0.65,0) -- (1.6,0.65,1.65)..(1.6,0,1.8)..(1.6,-0.65,1.65)--cycle; + draw(surface(p), surfacepen2,meshpen=invisible,nolight,render(merge=true)); + draw(p,q); + + //Draw the surface//{z=-.5*(x-1)^2-.5*(y)^2+2}; + triple f(pair t) { + return (t.x,t.y,-.5*(t.x-1)^2-.5*(t.y)^2+2); + } + surface s=surface(f,(-0.221,-1),(2.2,1),12,20,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic},usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw curve in xy plane ({cos(x)*(1+cos(2*x))},{sin(x)*(1+cos(2*x))},0); + triple g(real t) {return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),0);} + path3 mypath=graph(g,-pi/2,pi/2,operator ..); + draw(mypath,.4mm+bluepen); + label("$R$",(0.5,0.6,0),E); + + //Draw curve on the surface //{z=-.5*(cos(t)*(1+cos(2*t))-1)^2-.5*(sin(t)*(1+cos(2*t)))^2+2}; + triple gg(real t) { + return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),-.5*(cos(t)*(1+cos(2*t))-1)^2-.5*(sin(t)*(1+cos(2*t)))^2+2); + } + path3 mypath=graph(gg,-pi/2,pi/2,operator ..); + draw(mypath,.4mm+bluepen); + + //draw the edge in thick red pen q=redcurvepen+.4mm; + //draw((1.6,-0.65,0)--(1.6,0.65,0)--(1.6,0.65,1.625)..(1.6,0,1.825)..(1.6,-0.65,1.625)--(1.6,-0.65,0),redcurvepen+.4mm); + + + + +
    + +

    + Remember that though the integrand contains x, + we are viewing x as fixed. + Also note that the bounds of integration are functions of x: + the bounds depend on the value of x. +

    + +

    + As A(x) is a cross-sectional area function, + we can find the signed volume V under f by integrating it: + + V = \int_a^b A(x)\, dx = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\right)dx = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx + . +

    + +

    + This gives a concrete method for finding signed volume under a surface. + We could do a similar procedure where we started with y fixed, + resulting in an iterated integral with the order of integration dx\, dy. + The following theorem states that both methods give the same result, + which is the value of the double integral. + It is such an important theorem it has a name associated with it. +

    + + + Fubini's Theorem + +

    + Let R be a closed, + bounded region in the xy-plane and let + z=f(x,y) be a continuous function on R. + double integral + iterated integration + signed volume + volume + Fubini's Theorem +

    + +

    +

      +
    1. +

      + If R is bounded by a\leq x\leq b and g_1(x)\leq y\leq g_2(x), + where g_1 and g_2 are continuous functions on [a,b], then + + \iint_R f(x,y)\, dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx + . +

      +
      2. + +
    2. +

      + If R is bounded by c\leq y\leq d and h_1(y)\leq x\leq h_2(y), + where h_1 and h_2 are continuous functions on [c,d], then + + \iint_R f(x,y)\, dA = \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy + . +

      +
      3. +
    +

    +     +     + +

    + Note that once again the bounds of integration follow the curve to curve, + point to point pattern discussed in the previous section. + In fact, one of the main points of the previous section is developing the skill of describing a region R with the bounds of an iterated integral. + Once this skill is developed, + we can use double integrals to compute many quantities, + not just signed volume under a surface. +

    + + + + + Evaluating a double integral + +

    + Let f(x,y) = xy+e^y. + Find the signed volume under f on the region R, + which is the rectangle with corners (3,1) and (4,2) pictured in , + using Fubini's Theorem and both orders of integration. +

    +
    + Finding the signed volume under a surface in + + + + A ramp-like surface in space, with the portion over a rectangular domain highlighted. + +

    + The surface given by the graph of f(x,y)=xy+e^y has the appearance of a curved, twisted, and very steep ramp. + Along the x axis we have the constant value z=1. + For small, fixed values of x, the traces are exponential curves that increase with y. + For larger values of x, the xy term contributes, and the rate of growth is larger. +

    + +

    + In the xy plane, a rectangular region R is shown. + Dashed lines move upward from the boundary of this region to the surface. + Along the surface, a corresponding rectangle-shaped curve is shown. + The two rectangular curves, and the dashed lines between corresponding corners, + illustrate the region whose volume is computed in this example. +

    +     + + + + + //ASY file for figdouble13D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(10,-8,25); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3,4}; + real[] myychoice={1,2}; + real[] myzchoice={10}; + defaultpen(0.5mm); + + pair xbounds=(0,5); + pair ybounds=(0,2.5); + pair zbounds=(0,15); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface//{z=xy+e^y}; + triple f(pair t) { + return (t.x,t.y,t.x*t.y+exp(t.y)); + } + surface s=surface(f,(0,0),(4.25,2.25),12,20,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw rectangle in plane + draw((3,1,0)--(4,1,0)--(4,2,0)--(3,2,0)--(3,1,0),bluepen+linewidth(2)); + label("$R$",(3.5,1.5,0)); + + //Draw rectangle on surface + draw((3,1,3+exp(1))--(4,1,4+exp(1))..(4,1.5,6+exp(1.5))..(4,2,8+exp(2))--(3,2,6+exp(2))..(3,1.5,4.5+exp(1.5))..(3,1,3+exp(1)),bluepen+linewidth(2)); + + //Draw vertical lines + draw((3,1,0)--(3,1,3+exp(1)),bluepen+dashed+linewidth(0.75)); + draw((3,2,0)--(3,2,6+exp(2)),bluepen+dashed+linewidth(0.75)); + draw((4,1,0)--(4,1,4+exp(1)),bluepen+dashed+linewidth(0.75)); + draw((4,2,0)--(4,2,8+exp(2)),bluepen+dashed+linewidth(0.75)); + + + + +
    +     + +

    + We wish to evaluate \iint_R \big(xy+e^y\big)\, dA. + As R is a rectangle, + the bounds are easily described as + 3\leq x\leq 4 and 1\leq y\leq 2. +

    + +

    + Using the order dy\, dx: + + \iint_R\big(xy+e^y\big) \, dA \amp = \int_3^4\int_1^2\big(xy+e^y\big)\, dy \, dx + \amp = \int_3^4 \left(\left.\left[\frac12xy^2+e^y\right]\right|_1^2\, \right) dx + \amp = \int_3^4\left(\frac 32x + e^2-e\right)dx + \amp = \left.\left(\frac 34x^2 + \big(e^2-e\big)x\right)\right|_3^4 + \amp = \frac {21}4+ e^2-e\approx 9.92 + . +

    + +

    + Now we check the validity of Fubini's Theorem by using the order dx\, dy: + + \iint_R\big(xy+e^y\big) \, dA \amp = \int_1^2\int_3^4\big(xy+e^y\big)\, dx \, dy + \amp = \int_1^2\left(\left.\left[\frac12x^2y+xe^y\right]\right|_3^4\right)dy + \amp = \int_1^2\left(\frac72y+e^y\right)\, dy + \amp = \left.\left(\frac74y^2+e^y\right)\right|_1^2 + \amp =\frac{21}4+e^2-e\approx 9.92 + . +

    + +

    + Both orders of integration return the same result, as expected. +

    +     +     + + + Evaluating a double integral + +

    + Evaluate \iint_R \big(3xy-x^2-y^2+6\big)\, dA, + where R is the triangle bounded by x=0, + y=0 and x/2+y=1, + as shown in . +

    +
    + Finding the signed volume under the surface in + + + + A surface in space, plotted over a triangular domain. + +

    + The graph z = 3xy-x^2-y^2+6 is plotted in the first octant of a three-dimensional coordinate system. + It appears to be a hyperbolic paraboloid, + with slight upward curvature when viewed along the plane y=1-x, + and a larger downward curvature when viewed along the plane y=x. +

    + +

    + In the xy plane, a triangular region is plotted. + This region is bounded by the x and y axes, + and a line segment running from (2,0,0) to (0,1,0). + There are also dashed, vertical lines running from these two points up to the surface. +

    + +

    + On the surface, there are curves corresponding to the three edges of the triangle in the xy plane. + These curves illustrate the portion of the surface under which the solid whose volume we are finding lies. + Each curve has the shape of a downward-opening parabola. +

    +     + + + + + //ASY file for figdouble23D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(1.6,-5.5,12.4); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3,4}; + real[] myychoice={1}; + real[] myzchoice={2,4,6,8}; + defaultpen(0.5mm); + + pair xbounds=(0,2.5); + pair ybounds=(0,1.5); + pair zbounds=(0,9); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface//{z=3xy-x^2-y^2+6}; + triple f(pair t) { + return (t.x,t.y,3*t.x*t.y-t.x^2-t.y^2+6); + } + surface s=surface(f,(-0.1,-0.1),(2.1,1.1),12,20,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw triangle in plane + draw((0,0,0)--(2,0,0)--(0,1,0)--(0,0,0),bluepen+linewidth(2)); + label("$R$",(0,0.5,0),E); + + //Draw triangle on surface + triple g(real t) {return (t,0,6-t^2);} + path3 mypath=graph(g,0,2,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (0,t,6-t^2);} + path3 mypath=graph(g,0,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (2-2*t,t,3*(2-2*t)*(t)-(2-2*t)^2-(t)^2+6);} + path3 mypath=graph(g,0,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //Draw vertical lines + draw((2,0,0)--(2,0,2),bluepen+dashed+linewidth(0.75)); + draw((0,1,0)--(0,1,5),bluepen+dashed+linewidth(0.75)); + + + + +
    +     + +

    + While it is not specified which order we are to use, + we will evaluate the double integral using both orders to help drive home the point that it does not matter which order we use. +

    + +

    + Using the order dy\, dx: + The bounds on y go from curve to curve, + , 0\leq y\leq 1-x/2, + and the bounds on x go from + point to point, , 0\leq x\leq 2. + + \iint_R \big(3xy-x^2-y^2+6\big)\, dA \amp = \int_0^2\int_0^{-\frac x2+1} \big(3xy-x^2-y^2+6\big)\, dy\, dx + \amp = \int_0^2\left.\left(\frac32xy^2-x^2y-\frac13y^3+6y\right)\right|_0^{-\frac x2+1}dx + \amp = \int_0^2 \left(\frac{11}{12}x^3-\frac{11}{4}x^2-x+\frac{17}3\right)dx + \amp = \left.\left(\frac{11}{48}x^4-\frac{11}{12}x^3-\frac12x^2+\frac{17}3x\right)\right|_0^2 + \amp = \frac{17}3=5.\overline{6} + . +

    + +

    + Now lets consider the order dx \, dy. + Here x goes from curve to curve, 0\leq x\leq 2-2y, + and y goes from point to point, 0\leq y\leq 1: + + \iint_R \big(3xy-x^2-y^2+6\big)\, dA \amp = \int_0^1\int_0^{2-2y} \big(3xy-x^2-y^2+6\big)\, dx\, dy + \amp = \int_0^1\left.\left(\frac32x^2y-\frac13x^3-xy^2+6x\right)\right|_0^{2-2y} dy + \amp = \int_0^1\left(\frac{32}3y^3-22y^2+2y+\frac{28}3\right)dy + \amp =\left.\left(\frac83y^4-\frac{22}3y^3+y^2+\frac{28}3y\right)\right|_0^1 + \amp =\frac{17}3=5.\overline{6} + . +

    + +

    + We obtained the same result using both orders of integration. +

    +     +     + + +

    + Note how in these examples that the bounds of integration depend only on R; + the bounds of integration have nothing to do with f(x,y). + This is an important concept, + so we include it as a Key Idea. +

    + + + Double Integration Bounds +

    + When evaluating \iint_R f(x,y)\, dA using an iterated integral, + the bounds of integration depend only on R. + The function f does not determine the bounds of integration. +

    +     + +

    + Before doing another example, + we give some properties of double integrals. + Each should make sense if we view them in the context of finding signed volume under a surface, + over a region. +

    + + + Properties of Double Integrals + +

    + Let f and g be continuous functions over a closed, + bounded plane region R, + and let c be a constant. + double integralproperties + iterated integrationproperties +

    + +

    +

      +
    1. +

      + \ds \iint_Rc\,f(x,y)\, dA = c\iint_Rf(x,y)\, dA. +

      +
      2. + +
    2. +

      + \ds \iint_R \big(f(x,y)\pm g(x,y)\big)\, dA = \iint_R f(x,y)\, dA \pm \iint_R g(x,y)\, dA +

      +
      3. + +
    3. +

      + If f(x,y)\geq 0 on R, + then \ds \iint_R f(x,y)\, dA\geq 0. +

      +
      4. + +
    4. +

      + If f(x,y)\geq g(x,y) on R, + then \ds \iint_R f(x,y)\, dA\geq \iint_R g(x,y)\, dA. +

      +
      5. + +
    5. + +

      + Let R be the union of two nonoverlapping regions, + R = R_1\bigcup R_2 + (see ). + Then + + \iint_R f(x,y)\, dA = \iint_{R_1}f(x,y)\, dA+ \iint_{R_2}f(x,y)\, dA + . +

      + +
      + R is the union of two nonoverlapping regions, R_1 and R_2 + + + Illustration of a pond-like region in the plane that has been divided into two sub-regions. + +

      + The sketch shows a generic region R in the plane. + The precise shape of the region is unimportant, + but its shape is like that of a pond with a rounded but uneven boundary. +

      + +

      + A straight line passes through the middle of the region, + dividing it into two smaller, nonoverlapping regions R_1 and R_2. +

      +       + + + \begin{tikzpicture}[scale=0.88] + + \draw [firstcolor,thick,fill=firstcolor!15,smooth] plot coordinates {(0,2.)(0.06639,2.527)(0.2407,2.839)(0.4857,3.009)(0.7641,3.112)(1.039, + 3.221)(1.294,3.367)(1.532,3.53)(1.759,3.689)(1.98,3.823)(2.2,3.915)(2. + 42,3.967)(2.64,3.992)(2.86,4.)(3.08,4.)(3.305,3.987)(3.543,3.93)(3. + 808,3.797)(4.107,3.558)(4.45,3.185)(4.815,2.703)(5.17,2.152)(5.483,1. + 577)(5.723,1.018)(5.872,0.5022)(5.938,0.02083)(5.93,-0.4363)(5.861,-0. + 88)(5.741,-1.319)(5.58,-1.742)(5.389,-2.128)(5.179,-2.456)(4.96,-2. + 705)(4.737,-2.865)(4.502,-2.953)(4.243,-2.991)(3.95,-3.)(3.614,-2.999) + (3.244,-2.984)(2.86,-2.935)(2.484,-2.829)(2.137,-2.646)(1.832,-2.378)( + 1.558,-2.06)(1.299,-1.734)(1.039,-1.443)(0.7641,-1.222)(0.4857,-0. + 9779)(0.2407,-0.5015)(0.06639,0.4201)(0,2.)}; + + \draw [thick,firstcolor] (3,4) -- (4,-3); + + \draw (2,1) node {$R_1$}; + \draw (5,.5) node {$R_2$}; + \draw (1,-2) node {$R$}; + + \end{tikzpicture} + + + + +
      +
      6. +
    +

    +     +     + + + + + Evaluating a double integral + +

    + Let f(x,y) = \sin(x) \cos(y) and R be the triangle with vertices (-1,0), + (1,0) and (0,1) + (see ). + Evaluate the double integral \iint_Rf(x,y)\, dA. +

    + +
    + Finding the signed volume under a surface in + + + + A wave-like surface on which a triangular curve is drawn, corresponding to a domain in the plane z=0 below. + +

    + The graph z=\sin(x)\cos(y) is plotted against a set of three-dimensional coordinate axes. + It is a wave-like surface, but no crests or troughs are visible, + as we only see the portion of the surface with -1\leq x\leq 1 and 0\leq y\leq 1. +

    + +

    + In the xy plane, the domain of f(x,y) for this integral is shown as a triangle, + with vertices at (-1,0,0), (1,0,0), and (0,1,0). + In the default view of the image, this triangle is only partially visible below the surface. +

    + +

    + On the surface, we can see a triangular curve given by the points on the graph that lie above the boundary of the domain. +

    +     + + + + + //ASY file for figdouble33D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(.92,4,3.7); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={1}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(0,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface//{z=sin(x)cos(y)}; + triple f(pair t) { + return (t.x,t.y,sin(t.x)*cos(t.y)); + } + surface s=surface(f,(-1.25,-0.25),(1.25,1.5),12,20,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw triangle in plane + draw((-1,0,0)--(1,0,0)--(0,1,0)--(-1,0,0),bluepen+dashed+linewidth(0.75)); + label("$R$",(0.8,0.25,0),E); + + //Draw triangle on surface + triple g(real t) {return (t,0,sin(t));} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t-1,t,sin(t-1)*cos(t));} + path3 mypath=graph(g,0,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,1-t,sin(t)*cos(1-t));} + path3 mypath=graph(g,0,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //Draw vertical lines + //draw((-1,0,0)--(-1,0,sin(-1)),bluepen+dashed+linewidth(0.75)); + //draw((1,0,0)--(1,0,sin(1)),bluepen+dashed+linewidth(0.75)); + + + + +
    +     + +

    + If we attempt to integrate using an iterated integral with the order dy\, dx, + note how there are two upper bounds on R meaning we'll need to use two iterated integrals. + We would need to split the triangle into two regions along the y-axis, + then use , + Part. +

    + +

    + Instead, let's use the order dx\, dy. + The curves bounding x are y-1\leq x\leq 1-y; + the bounds on y are 0\leq y\leq 1. + This gives us: + + \iint_R f(x,y)\, dA \amp = \int_0^1\int_{y-1}^{1-y}\sin(x) \cos(y) \, dx\, dy + \amp = \int_0^1\left.\Big( -\cos(x) \cos(y) \Big)\right|_{y-1}^{1-y}\,dy + \amp = \int_0^1 \cos(y) \Big(-\cos(1-y) + \cos(y-1)\Big)dy + . +

    + +

    + Recall that the cosine function is an even function; + that is, \cos(x) = \cos(-x). + Therefore, from the last integral above, + we have \cos(y-1) = \cos(1-y). + Thus the integrand simplifies to 0, and we have + + \iint_R f(x,y)\, dA \amp = \int_0^1 0\, dy + \amp = 0 + . +

    + +

    + It turns out that over R, + there is just as much volume above the xy-plane as below + (look again at ), + giving a final signed volume of 0. +

    +     +     + + + + + Evaluating a double integral + +

    + Evaluate \iint_R (4-y)\, dA, + where R is the region bounded by the parabolas y^2=4x and x^2=4y, + graphed in . +

    + +
    + Finding the volume under the surface in + + + + A rectangular portion of a plane in space, and a petal-like curve on the plane. + +

    + The graph z=4-y is a plane. + It is plotted as a rectangular region in the first octant, + intersecting the xy plane along the line y=4, + and the xz plane along the line z=4. +

    + +

    + The parabolas y^2=4x and x^2=4y are sketched in the xy plane, + from the origin to their intersection at (4,4,0). + On the plane, we see the corresponding curves, + which are also parabolas. The region between them has the shape of a narrow leaf or petal. +

    +     + + + + + //ASY file for figdouble43D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(16,6.2,5.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={2,4}; + defaultpen(0.5mm); + + pair xbounds=(0,5); + pair ybounds=(0,5); + pair zbounds=(0,5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface//{z=4-y}; + triple f(pair t) { + return (t.x,t.y,4-t.y); + } + surface s=surface(f,(-0.25,-0.25),(4.5,4.25),12,20,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw parabolas in plane + triple g(real t) {return (t^2/4,t,0);} + path3 mypath=graph(g,0,4.25,operator ..); + draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,t^2/4,0);} + path3 mypath=graph(g,0,4.25,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //Draw parabolas on surface + triple g(real t) {return (t^2/4,t,4-t);} + path3 mypath=graph(g,0,4.,operator ..); + draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,t^2/4,4-t^2/4);} + path3 mypath=graph(g,0,4.,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //Draw vertical lines + //draw((-1,0,0)--(-1,0,sin(-1)),bluepen+dashed+linewidth(0.75)); + //draw((1,0,0)--(1,0,sin(1)),bluepen+dashed+linewidth(0.75)); + + + + +
    +     + +

    + Graphing each curve can help us find their points of intersection. + Solving analytically, the second equation tells us that y=x^2/4. + Substituting this value in for y in the first equation gives us x^4/16 = 4x. + Solving for x: + + \frac{x^4}{16} \amp = 4x + x^4-64x \amp =0 + x(x^3-64) \amp =0 + x\amp = 0,\, 4 + . +

    + +

    + Thus we've found analytically what was easy to approximate graphically: + the regions intersect at (0,0) and (4,4), + as shown in . +

    + +

    + We now choose an order of integration: + dy\, dx or dx\, dy? + Either order works; since the integrand does not contain x, + choosing dx\, dy might be simpler at least, + the first integral is very simple. +

    + +

    + Thus we have the following curve to curve, + point to point bounds: + + y^2/4\leq x\leq 2\sqrt y, \text{ and } 0\leq y\leq 4 + . + Therefore, + + \iint_R (4-y)\, dA \amp = \int_0^4\int_{y^2/4}^{2\sqrt{y}}(4-y)\, dx\, dy + \amp = \int_0^4 \big(x(4-y)\big)\Big|_{y^2/4}^{2\sqrt{y}} dy + \amp = \int_0^4 \Big(\big(2\sqrt{y}-\frac{y^2}{4}\big)\big(4-y\big)\Big)\, dy + \amp = \int_0^4 \Big( \frac{y^3}{4}-y^2-2y^{3/2}+8y^{1/2}\Big)\, dy + \amp = \left.\left(\frac{y^4}{16}-\frac{y^3}{3}-\frac{4y^{5/2}}5+\frac{16y^{3/2}}3\right)\right|_0^4 + \amp = \frac{176}{15} = 11.7\overline{3} + . +

    + +

    + The signed volume under the surface z=f(x,y) is about 11.7 cubic units. +

    +     +     + + +

    + In the previous section we practiced changing the order of integration of a given iterated integral, + where the region R was not explicitly given. + Changing the bounds of an integral is more than just an test of understanding. + Rather, there are cases where integrating in one order is really hard, + if not impossible, + whereas integrating with the other order is feasible. +

    + + + Changing the order of integration + +

    + Rewrite the iterated integral + \ds \int_0^3\int_y^3 e^{-x^2}\, dx\, dy with the order dy\, dx. + Comment on the feasibility to evaluate each integral. +

    +     + +

    + Once again we make a sketch of the region over which we are integrating to facilitate changing the order. + The bounds on x are from x=y to x=3; + the bounds on y are from y=0 to y=3. + These curves are sketched in , + enclosing the region R. +

    + +
    + Determining the region R determined by the bounds of integration in + + + Several lines are plotted in the plane. Three of these lines form a triangle. + +

    + The x and y coordinate axes are plotted in the plane, + with the origin to the bottom-left of the image, + so that the first quadrant is shown. +

    + +

    + Four lines are plotted: the horizontal lines y=0 (along the x axis) and y=3, + the vertical line x=3, and the diagonal line y=x. + The line y=x divides the square 0\leq x,y\leq 3 into two triangles, + one below the line, and one above. +

    + +

    + The line y=3 illustrates that the upper bound for y is 3, + but this line is otherwise not really needed. + The remaining lines y=0, x=3, and y=x form a triangle with vertices at + (0,0), (3,0), and (3,3). +

    + +

    + A label R placed inside the triangle hepls to illustrate that the region lies below y=x, + and not above. +

    +     + + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.5,ymax=3.5, + xmin=-.5,xmax=3.5 + ] + + \addplot+ [very thick,domain=-.5:3.5] {x} node [pos=.5,sloped,above,black] { $y=x$}; + + \draw [very thick,firstcolor] (axis cs:3,-.5) -- (axis cs:3,3.5); + + \draw [very thick,secondcolor] (axis cs:-.5,0) -- (axis cs:3.5,0) + (axis cs:-.5,3) -- (axis cs:3.5,3); + + \draw (axis cs:2,1) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + To change the bounds, + note that the curves bounding y are y=0 up to y=x; + the triangle is enclosed between x=0 and x=3. + Thus the new bounds of integration are + 0\leq y\leq x and 0\leq x\leq 3, + giving the iterated integral \ds \int_0^3\int_0^x e^{-x^2}\, dy\, dx. +

    + +

    + How easy is it to evaluate each iterated integral? + Consider the order of integrating dx\, dy, + as given in the original problem. + The first indefinite integral we need to evaluate is \int e^{-x^2}\, dx; + we have stated before + (see ) + that this integral cannot be evaluated in terms of elementary functions. + We are stuck. +

    + +

    + Changing the order of integration makes a big difference here. + In the second iterated integral, + we are faced with \int e^{-x^2}\, dy; + integrating with respect to y gives us ye^{-x^2}+C, + and the first definite integral evaluates to + + \int_0^x e^{-x^2}\, dy = xe^{-x^2} + . +

    + +

    + Thus + + \int_0^3\int_0^x e^{-x^2}\, dy\, dx = \int_0^3\Big(xe^{-x^2}\Big)dx + . +

    + +

    + This last integral is easy to evaluate with substitution, + giving a final answer of \frac12(1-e^{-9})\approx 0.5. + + shows the surface over R. +

    + +
    + Showing the surface z=f(x,y) defined in over its region R + + + + A wave-like surface in three dimensions, and a curve on the surface illustrating a triangular domain. + +

    + In two dimensions, the graph y=e^{-x^2} is a bell curve, with a peak at (0,1). + The surface z=e^{-x^2} is a cylinder through this curve. + We are shown a portion of this surface with x\geq 0, + which has the appearance of a wave parallel to the y axis that reaches its crest when x=0. +

    + +

    + The triangular domain is sketched in the xy plane, + with vertices at (0,0,0), (3,0,0), and (3,3,0), + but it is mostly obscured by the surface. +

    + +

    + On the surface itself we see the corresponding triangular curve, + illustrating the region under which we are computing a volume with this integral. +

    +     + + + + + //ASY file for figdouble6b3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(6.3,9.6,2.4); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(0,4); + pair ybounds=(0,3); + pair zbounds=(0,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface//{z=exp(-x^2)}; + triple f(pair t) { + return (t.x,t.y,exp(-t.x^2)); + } + surface s=surface(f,(-0.25,0),(3.5,3.5),12,20,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw lines in plane + draw((0,0,0)--(3,0,0)--(3,3,0)--cycle,bluepen+dashed+linewidth(0.75)); + + //Draw lines on surface + draw((3,0,0)--(3,3,0),bluepen+linewidth(2)); + triple g(real t) {return (t,0,exp(-t^2));} + path3 mypath=graph(g,0,3,operator ..); + draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,t,exp(-t^2));} + path3 mypath=graph(g,0,3,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //Draw vertical lines + //draw((-1,0,0)--(-1,0,sin(-1)),bluepen+dashed+linewidth(0.75)); + //draw((1,0,0)--(1,0,sin(1)),bluepen+dashed+linewidth(0.75)); + + + + +
    + +

    + In short, evaluating one iterated integral is impossible; + the other iterated integral is relatively simple. +

    +     +     + + + +

    + + defines the average value of a single-variable function f(x) on the interval [a,b] as + + \text{ average value of \(f(x)\) on \([a,b]\) } = \frac1{b-a}\int_a^b f(x)\, dx; + + that is, it is the area under f over an interval divided by the length of the interval. + We make an analogous statement here: + the average value of z=f(x,y) over a region R is the volume under f over R divided by the area of R. +

    + + + The Average Value of <m>f</m> on <m>R</m> + +

    + Let z=f(x,y) be a continuous function defined over a closed, + bounded region R in the xy-plane. + The average value of f on R is + average value of a function + + \text{ average value of \(f\) on \(R\) } = \frac{\ds \iint_R f(x,y)\, dA}{\ds\iint_R \, dA} + . +

    +     +     + + + Finding average value of a function over a region <m>R</m> + +

    + Find the average value of f(x,y) = 4-y over the region R, + which is bounded by the parabolas y^2=4x and x^2=4y. + Note: this is the same function and region as used in . +

    +     + +

    + In we found + + \iint_R f(x,y)\, dA = \int_0^4\int_{y^2/4}^{2\sqrt{y}}(4-y)\, dx\, dy = \frac{176}{15} + . +

    + +

    + We find the area of R by computing \iint_R \, dA: + + \iint_R \, dA = \int_0^4\int_{y^2/4}^{2\sqrt{y}} \, dx\, dy = \frac{16}{3} + . +

    + +

    + Dividing the volume under the surface by the area gives the average value: + + \text{ average value of \(f\) on \(R\) } = \frac{176/15}{16/3} = \frac{11}5 = 2.2 + . +

    + +

    + While the surface, as shown in , + covers z-values from z=0 to z=4, + the average z-value on R is 2.2. +

    + +
    + Finding the average value of f in + + + + A rectangular portion of a plane in space, and a petal-like curve on the plane. + +

    + This is the same image as the one used in . + We repeat the description here for convenience. +

    + +

    + The graph z=4-y is a plane. + It is plotted as a rectangular region in the first octant, + intersecting the xy plane along the line y=4, + and the xz plane along the line z=4. +

    + +

    + The parabolas y^2=4x and x^2=4y are sketched in the xy plane, + from the origin to their intersection at (4,4,0). + On the plane, we see the corresponding curves, + which are also parabolas. The region between them has the shape of a narrow leaf or petal. +

    +     + + + + + //ASY file for figdouble43D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(16,6.2,5.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={2,4}; + defaultpen(0.5mm); + + pair xbounds=(0,5); + pair ybounds=(0,5); + pair zbounds=(0,5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface//{z=4-y}; + triple f(pair t) { + return (t.x,t.y,4-t.y); + } + surface s=surface(f,(-0.25,-0.25),(4.5,4.25),12,20,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw parabolas in plane + triple g(real t) {return (t^2/4,t,0);} + path3 mypath=graph(g,0,4.25,operator ..); + draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,t^2/4,0);} + path3 mypath=graph(g,0,4.25,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //Draw parabolas on surface + triple g(real t) {return (t^2/4,t,4-t);} + path3 mypath=graph(g,0,4.,operator ..); + draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,t^2/4,4-t^2/4);} + path3 mypath=graph(g,0,4.,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //Draw vertical lines + //draw((-1,0,0)--(-1,0,sin(-1)),bluepen+dashed+linewidth(0.75)); + //draw((1,0,0)--(1,0,sin(1)),bluepen+dashed+linewidth(0.75)); + + + + +
    +     +     + +

    + The previous section introduced the iterated integral in the context of finding the area of plane regions. + This section has extended our understanding of iterated integrals; + now we see they can be used to find the signed volume under a surface. +

    + +

    + This new understanding allows us to revisit what we did in the previous section. + Given a region R in the plane, + we computed \iint_R 1\, dA; + again, our understanding at the time was that we were finding the area of R. + However, we can now view the graph z=1 as a surface, + a flat surface with constant z-value of 1. + The double integral \iint_R 1\, dA finds the volume, + under z=1, over R, + as shown in . + Basic geometry tells us that if the base of a general right cylinder has area A, + its volume is A\cdot h, + where h is the height. + In our case, the height is 1. + We were actually computing the volume of a solid, + though we interpreted the number as an area. +

    + +
    + Showing how an iterated integral used to find area also finds a certain volume + + + + A circular cylinder, bounded below by the xy plane, and above by the plane z=1. + +

    + A circular cylinder is plotted in three dimensions. + The cylinder appears to lie over a circle with center (1,0,0) and radius 1 in the xy plane. +

    + +

    + The plane z=1 sits at the top of the cylinder. +

    + +

    + The image is used to illustrate that for any region in the plane, + there is a cylinder that lies over the region, between the planes z=0 and z=1. + A double integral of the form \iint_R 1 \, dA can be thought of as either computing the area of the region R, + or the volume of the corresponding cylinder of height 1 over R. +

    +     + + + + + //ASY file for figdouble_intro23D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8.2,4.6,1.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={-1,1}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-1,2.5); + pair ybounds=(-1.25,1.25); + pair zbounds=(-.05,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface//{z=-.5*(x-1)^2-.5*(y)^2+2}; + //triple f(pair t) { + // return (t.x,t.y,-.5*(t.x-1)^2-.5*(t.y)^2+2); + //} + //surface s=surface(f,(-0.221,-1),(2.2,1),12,20,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic},usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + //Draw curve on top of the grid in xy plane ({cos(x)*(1+cos(2*x))},{sin(x)*(1+cos(2*x))},0); + triple g(real t) {return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),0);} + path3 mypath=graph(g,-pi/2,pi/2,operator ..); + draw(mypath,bluepen); + + // draw curve on surface + triple g(real t) { + return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),1); + } + path3 mypath=graph(g,-pi/2,pi/2,operator ..); + draw(mypath,bluepen); + + pen q=redcurvepen; + triple gg(pair t) {return (cos(t.x)*(1+cos(2*t.x)),sin(t.x)*(1+cos(2*t.x)),t.y);} + surface s=surface(gg,(-pi/2,0),(pi/2,1),16,4,Spline); + draw(s,surfacepen2,meshpen=q,nolight,render(merge=true)); + + triple f(pair t) { + return (t.x,t.y,1); + } + surface s=surface(f,(-.1,ybounds.x),(xbounds.y,ybounds.y),8,8); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + + +
    + +

    + The next section extends our abilities to find + volumes under surfaces. Currently, + some integrals are hard to compute because either the region R we are integrating over is hard to define with rectangular curves, + or the integrand itself is hard to deal with. + Some of these problems can be solved by converting everything into polar coordinates. +

    + + + + Terms and Concepts + + + + +

    + An integral can be interpreted as giving the signed area over an interval; + a double integral can be interpreted as giving the signed over a region. +

    +     + + + + + + + + +     + + + + +

    + Explain why the following statement is false: + Fubini's Theorem states that + + \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx = \int_a^b\int_{g_1(y)}^{g_2(y)} f(x,y)\, dx\, dy + . +

    +     + + + +

    + When switching the order of integration, + the bounds integrals must change to reflect the bounds of the region of integration. + You cannot merely change the letters x and y in a few places. +

    +     + +     + + + + +

    + Explain why if f(x,y) \gt 0 over a region R, then +

    + +

    + \iint_Rf(x,y)\, dA \gt 0. +

    +     + + + +

    + The double integral gives the signed volume under the surface. + Since the surface is always positive, + it is always above the xy-plane and hence produces only + positive volume. +

    +     + +     + + + + +

    + If \iint_R f(x,y)\, dA = \iint_R g(x,y)\, dA, + does this imply f(x,y) = g(x,y)? +

    +     + + + +

    + No. + It means that there is the same amount of signed volume under f and g over R, + but the functions could be very different. +

    +     + +     +     + + + Problems + + + +

    + For the given integral, + +

      +
    1. +

      + Evaluate the given iterated integral, and +

      +
      2. + +
    2. +

      + rewrite the integral using the other order of integration. +

      +
      3. +
    +

    +     + + + + +

    + \ds \int_1^2\int_{-1}^1\left(\frac xy+3\right)\, dx\, dy +

    +     + +

    + 6; \ds \int_{-1}^1\int_{1}^2\left(\frac xy+3\right)\, dy\, dx +

    +     + +     + + + + + +

    + \ds \int_{-\pi/2}^{\pi/2}\int_{0}^\pi\left(\sin(x) \cos(y) \right)\,dx\,dy +

    + + + + +     + +     + + + + +

    + \ds \int_{0}^{4}\int_{0}^{-x/2+2}\left(3x^2-y+2\right)\, dy\, dx +

    +     + +

    + 112/3; \ds \int_{0}^{2}\int_{0}^{4-2y}\left(3x^2-y+2\right)\, dx\, dy +

    +     + +     + + + + +

    + \ds \int_{1}^{3}\int_{y}^{3}\left(x^2y-xy^2\right)\,dx\,dy +

    + + + + +     + +     + + + + +

    + \ds \int_{0}^{1}\int_{-\sqrt{1-y}}^{\sqrt{1-y}}\left(x+y+2\right)\, dx\, dy +

    +     + +

    + 16/5; \ds \int_{-1}^{1}\int_{0}^{1-x^2}\left(x+y+2\right)\, dy\, dx +

    +     + +     + + + + +

    + \ds \int_{0}^{9}\int_{y/3}^{\sqrt{y}}\left(xy^2\right)\,dx\,dy +

    + + + + + +     + +     + +     + + + +

    + For the given integral: + +

      +
    1. +

      + Sketch the region R given by the problem. +

      +
      2. + +
    2. +

      + Set up the iterated integrals, in both orders, + that evaluate the given double integral for the described region R. +

      +
      3. + +
    3. +

      + Evaluate one of the iterated integrals to find the signed volume under the surface + z=f(x,y) over the region R. +

      +
      4. +
    +

    +     + + + + +

    + \ds \iint_R x^2y\, dA, + where R is bounded by y=\sqrt{x} and y=x^2. +

    +     + +

    +

      +
    1. + + + + A region in the first quadrant of the plane, bounded by two parabolas. + +

      + The correct region is in the first quadrant of the xy plane. + It is a leaf-shaped region, bounded by the parabolas y=x^2 and x=y^2. + The region is symmetric about the line y=x. + The two parabolas intersect at (0,0) and (1,1). + The parabola y=x^2 lies below the line y=x, + and the parabola y=\sqrt{x} lies above. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1}, + ytick={1}, + ymin=-.1,ymax=1.5, + xmin=-.1,xmax=1.5 + ] + + \addplot [firstcurvestyle,domain=-.1:1.1] ({x^2},{x}) node [pos=.5,sloped,above,black] {$y=\sqrt{x}$}; + \addplot [firstcurvestyle,domain=-.1:1.1] ({x},{x^2}) node [pos=.5,sloped,below,black] {$y=x^2$}; + + \draw (axis cs: .5,.5) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + + +
      2. + +
    2. +

      + \ds \int_0^1\int_{x^2}^{\sqrt{x}}x^2y\, dy\, dx = \int_0^1\int_{y^2}^{\sqrt{y}}x^2y\, dx\, dy. +

      +
      3. + +
    3. +

      + \frac3{56} +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \iint_R x^2y\, dA, + where R is bounded by y=\sqrt[3]{x} and y=x^3. +

    +     + +

    +

      +
    1. + + + + A region in the first quadrant of the plane, bounded by a cubic function and its inverse. + +

      + The region for this integral lies in the first quadrant. +

      + +

      + The functions given by y=x^3 and y=\sqrt[3]{x} are inverses of each other, + so the region is symmetric about the line y=x. +

      + +

      + The curve y=x^3 lies below the line y=x, + and the curve y=\sqrt[3]{x} lies above this line. +

      +       + + + \begin{tikzpicture}[>=stealth] + + \begin{axis}[ + axis on top, + xtick={1}, + ytick={1}, + ymin=-1.5,ymax=1.5, + xmin=-1.5,xmax=1.5 + ] + + \addplot [firstcurvestyle,domain=-1.1:1.1,samples=60] ({x^3},{x}) node [pos=.85,above left,black] {$y=\sqrt[3]{x}$}; + \addplot [firstcurvestyle,domain=-1.1:1.1,samples=60] ({x},{x^3}) node [pos=.75,below right,black] {$y=x^3$};; + + \draw (axis cs: -.5,.5) node (A) {$R$}; + + \draw [->] (A) -- (axis cs: .5,.5); + \draw [->] (A) -- (axis cs: -.5,-.5); + + \end{axis} + + \end{tikzpicture} + + + + + +
      2. + +
    2. +

      + \ds \int_0^1\int_{x^3}^{\sqrt[3]{x}}x^2y\, dy\, dx + \int_{-1}^0\int_{\sqrt[3]{x}}^{x^3}x^2y\, dy\, dx \ds = \int_0^1\int_{y^3}^{\sqrt[3]{y}}x^2y\, dx\, dy +\int_{-1}^0\int_{\sqrt[3]{y}}^{y^3}x^2y\, dx\, dy. +

      +
      3. + +
    3. +

      + 0 +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \iint_R x^2-y^2\, dA, + where R is the rectangle with corners (-1,-1), + (1,-1), + (1,1) and (-1,1). +

    +     + +

    +

      +
    1. + + + + A square of side length 2, with its center at the origin. + +

      + A rectangle in the plane. In fact, it is a square, with side length 2. +

      + +

      + The vertices are at (\pm 1,0) and (0,\pm 1), + so that the square is centered at the origin. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-1,1}, + ytick={1,-1}, + ymin=-1.1,ymax=1.1, + xmin=-1.1,xmax=1.1 + ] + + \draw [very thick,firstcolor] (axis cs:-1,-1) -- (axis cs:1,-1) -- (axis cs:1,1) -- (axis cs:-1,1) -- cycle; + + \draw (axis cs: .5,.5) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + + +
      2. + +
    2. +

      + \ds \int_{-1}^1\int_{-1}^{1}x^2-y^2\, dy\, dx = \int_{-1}^1\int_{-1}^{1}x^2-y^2\, dx\, dy. +

      +
      3. + +
    3. +

      + 0 +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \iint_R ye^x\, dA, + where R is bounded by x=0, + x=y^2 and y=1. +

    +     + +

    +

      +
    1. + + + + A region in the plane that lies above a rightward-opening parabola, and below a horizontal line. + +

      + The parabola x=y^2 has its vertex at the origin, and opens toward the right. +

      + +

      + The region for this integral is in the first quadrant. + It lies above the parabola, and below the line y=1. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1}, + ytick={1}, + ymin=-.5,ymax=1.2, + xmin=-.5,xmax=1.2 + ] + + \addplot [firstcurvestyle,domain=-.25:1.1] ({x},{1}); + \addplot [firstcurvestyle,domain=-.25:1.1] ({x^2},{x}) node [pos=.5,below right,black] { $x=y^2$}; + \addplot [firstcurvestyle,domain=-.25:1.1] ({0},{x}); + + \draw (axis cs: .25,.75) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + + +
      2. + +
    2. +

      + \ds \int_{0}^1\int_{0}^{y^2}ye^x\, dx\, dy = \int_{0}^1\int_{\sqrt{x}}^{1}ye^x\, dy\, dx. +

      +
      3. + +
    3. +

      + e/2-1 +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \iint_R \big(6-3x-2y\big)\, dA, + where R is bounded by x=0, + y=0 and 3x+2y=6. +

    +     + +

    +

      +
    1. + + + + A triangular region in the plane, with vertices at (0,0), (2,0), and (0,3). + +

      + The region for this integral is a triangle in the first quadrant, + bounded by the coordinate axes and the line 3x+3y=6. +

      + +

      + The vertices of the triangle are at (0,0), (2,0), and (0,3). +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2}, + ytick={1,2,3}, + ymin=-.5,ymax=3.5, + xmin=-.5,xmax=2.5 + ] + + \addplot [firstcurvestyle,domain=-.25:2.25] ({x},{0}); + \addplot [firstcurvestyle,domain=-.25:2.25] ({x},{3-3/2*x}) node [pos=.4,sloped,above,black] { $3x+2y=6$}; + \addplot [firstcurvestyle,domain=-.25:3.25] ({0},{x}); + + \draw (axis cs: .75,1) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + + +
      2. + +
    2. +

      + \ds \int_{0}^2\int_{0}^{3-3/2x}\big(6-3x-2y\big)\, dy\, dx = \int_{0}^3\int_{0}^{2-2/3y}\big(6-3x-2y\big)\, dx\, dy. +

      +
      3. + +
    3. +

      + 6 +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \iint_R e^y\, dA, where R is bounded by y=\ln(x) and +

    + +

    + \ds y=\frac{1}{e-1}(x-1). +

    +     + +

    +

      +
    1. + + + + A region in the plane bounded by the graph of the natural logarithm and a straight line. + +

      + The graph y=\ln(x) meets the x axis at (1,0). + It tends toward -\infty as it approaches the y axis, + and grows slowly upward as x increases past x=1. +

      + +

      + The line y=\frac{1}{e-1}(x-1) also meets the x axis at (1,0), + and it intercepts the graph y=\ln(x) a second time when x=e. +

      + +

      + The region for this integral therefore lies below y=\ln(x) + and above y=\frac{1}{e-1}(x-1), for 1\leq x\leq e. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3}, + ytick={1,2,3}, + ymin=-.5,ymax=1.25, + xmin=-.5,xmax=3.25 + ] + + \addplot [firstcurvestyle,domain=.5:3.25] ({x},{ln(x)}) node [pos=.7,sloped,above,black] { $y=\ln(x)$}; + \addplot [firstcurvestyle,domain=-.25:3.25] ({x},{1/e*(x-1)}) node [pos=.6,below right,black] { $y=\frac{1}{1-e}(x-1)$}; + + \draw (axis cs: 1,.5) node (A) {$R$}; + + \draw[->,>=stealth] (A) -- (axis cs:1.8,.5); + + \end{axis} + + \end{tikzpicture} + + + + + +
      2. + +
    2. +

      + \ds \int_{1}^e\int_{\frac{x-1}{e-1}}^{\ln(x) }e^y\, dy\, dx = \int_{0}^1\int_{e^y}^{y(e-1)+1}e^y\, dx\, dy. +

      +
      3. + +
    3. +

      + -\frac12e^2+2e-\frac32 +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \iint_R \big(x^3y-x\big)\, dA, + where R is the half of the circle + x^2+y^2=9 in the first and second quadrants. +

    +     + +

    +

      +
    1. + + + + A region bounded by the x axis and the upper half of a semicircle of radius 3. + +

      + The region for this integral is bounded above by the semicircle y=\sqrt{9-x^2}, + which has center (0,0) and radius 3. + The region is bounded below by the x axis, for -3\leq x\leq 3. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={-3,3}, + ytick={-3,3}, + ymin=-3.1,ymax=3.1, + xmin=-3.7,xmax=3.7 + ] + + \addplot+ [domain=0:180,samples=40] ({3*cos(x)},{3*sin(x)}) --cycle; + + \draw (axis cs: 1,1) node (A) {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + + +
      2. + +
    2. +

      + \ds \int_{-3}^3\int_{0}^{\sqrt{9-x^2}}\big(x^3y-x\big)\, dy\, dx = \int_{0}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\big(x^3y-x\big)\, dx\, dy. +

      +
      3. + +
    3. +

      + 0 +

      +
      4. +
    +

    +     + +     + + + + +

    + \ds \iint_R \big(4-3y\big)\, dA, + where R is bounded by y=0, + y=x/e and y=\ln(x). +

    +     + +

    +

      +
    1. + + + + The region above the graph of the natural logarithm, but below its tangent line at x=e. + +

      + For 0\leq x\leq 1, the region is bounded below by the x axis. + For 1\leq x\leq e, the lower bound of the region is y=\ln(x). +

      + +

      + The upper bound of the region is the line y=x/e. + This line is tangent to y=ln(x) at the point (e,1), + and lies above y=\ln(x). + The region is bounded by the portion of this line from the point of tangency to the origin. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2}, + extra x ticks={2.718}, + extra x tick labels={$e$}, + ytick={1}, + %%extra y ticks={-5,-3,...,7},% + ymin=-.5,ymax=1.2,% + xmin=-.5,xmax=3.5% + ] + + \addplot [firstcurvestyle,domain=.5:3.25] ({x},{ln(x)}); + \addplot [firstcurvestyle,domain=-.5:3.25] ({x},{0}); + \addplot [firstcurvestyle,domain=-.5:3.25] ({x},{x/e}); + + \draw (axis cs: .8,.1) node (A) {$R$}; + + \filldraw (axis cs:2.718,1) circle (2.4pt); + + \draw (axis cs:2.718,1) node [below right] { $(e,1)$}; + + \end{axis} + + \end{tikzpicture} + + + + + +
      2. + +
    2. +

      + \ds \int_{0}^1\int_{ey}^{e^y}\big(4-3y\big)\, dx\, dy = \int_{0}^1\int_{0}^{x/e}\big(4-3y\big)\, dy\, dx+\int_{1}^e\int_{\ln(x) }^{x/e}\big(4-3y\big)\, dy\, dx. +

      +
      3. + +
    3. +

      + 3e-7 +

      +
      4. +
    +

    +     + +     + +     + + + +

    + State why it is difficult/impossible to integrate the iterated integral in the given order of integration. + Change the order of integration and evaluate the new iterated integral. +

    +     + + + + +

    + \ds \int_0^4\int_{y/2}^2 e^{x^2}\, dx\, dy +

    +     + +

    + Integrating e^{x^2} with respect to x is not possible in terms of elementary functions. + \ds \int_0^2\int_0^{2x}e^{x^2}\, dy\, dx = e^4-1. +

    +     + +     + + + + +

    + \ds \int_0^{\sqrt{\pi/2}}\int_{x}^{\sqrt{\pi/2}} \cos\big(y^2\big)\, dy\, dx +

    +     + +

    + Integrating \cos(y^2) with respect to y is not possible in terms of elementary functions. + \ds \int_0^{\sqrt{\pi/2}}\int_0^{y}\cos(y^2)\, dx\, dy = \frac12. +

    +     + +     + + + + +

    + \ds \int_0^{1}\int_y^{1} \frac{2y}{x^2+y^2}\, dx\, dy +

    +     + +

    + Integrating \ds\int_y^{1}\frac{2y}{x^2+y^2}\, dx gives \tan^{-1}(1/y)-\pi/4; + integrating \tan^{-1}(1/y) is hard. +

    + +

    + \ds\int_0^1\int_0^x\frac{2y}{x^2+y^2}\, dy \, dx = \ln(2). +

    +     + +     + + + + +

    + \ds \int_{-1}^{1}\int_1^{2} \frac{x\tan^2(y) }{1+\ln(y) }\, dy\, dx +

    +     + +

    + Integrating in the order shown is hard/impossible. + By changing the order of integration, + we have \ds \int_{1}^{2}\int_{-1}^{1} \frac{x\tan^2(y) }{1+\ln(y) }\, dx\, dy = 0, + since the integrand is an odd function with respect to x. + Thus the iterated integral evaluates to 0. +

    +     + +     + +     + + + +

    + Find the average value of f over the region R. + Notice how these functions and regions are related to the iterated integrals given in + Exercises. +

    +     + + + + +

    + \ds f(x,y) =\frac xy+3;R is the rectangle with opposite corners (-1,1) and (1,2). +

    +     + +

    + average value of f = 6/2 = 3 +

    +     + +     + + + + +

    + \ds f(x,y) = \sin(x) \cos(y); + R is bounded by x=0, x=\pi, + y=-\pi/2 and y=\pi/2. +

    +     + +

    + average value of f= 4/\pi^2 +

    +     + +     + + + + +

    + \ds f(x,y) = 3x^2-y+2;R is bounded by the lines y=0, + y=2-x/2 and x=0. +

    +     + +

    + average value of f= \frac{112/3}{4} =28/3 +

    +     + +     + + + + +

    + \ds f(x,y) =x^2y-xy^2; + R is bounded by y=x, + y=1 and x=3. +

    +     + +

    + average value of f= \frac{76/15}{2} = \frac{38}{15} \approx 2.53 +

    +     + +     + +     +     +     +
    +
    + Double Integration with Polar Coordinates +

    + We have used iterated integrals to evaluate double integrals, + which give the signed volume under a surface, z=f(x,y), + over a region R of the xy-plane. + The integrand is simply f(x,y), + and the bounds of the integrals are determined by the region R. +

    + +

    + Some regions R are easy to describe using rectangular coordinates that is, + with equations of the form y=f(x), x=a, etc. + However, some regions are easier to handle if we represent their boundaries with polar equations of the form r=f(\theta), + \theta = \alpha, etc. +

    + + + +

    + The basic form of the double integral is \iint_R f(x,y)\, dA. + We interpret this integral as follows: + over the region R, + sum up lots of products of heights (given by f(x_i,y_i)) and areas + (given by \Delta A_i). + That is, dA represents a little bit of area. + In rectangular coordinates, + we can describe a small rectangle as having area dx\, dy or dy\, dx the area of a rectangle is simply lengthwidth a small change in x times a small change in y. + Thus we replace dA in the double integral with dx\, dy or dy\, dx. +

    + +
    + Approximating a region R with portions of sectors of circles + +
    + + + + A polar grid in the first quadrant, with one grid sector highlighted. + +

    + A polar grid is shown in the first quadrant of the plane. + Concentric circles are plotted with radii 0.2, 0.4, 0.6, and 0.8. + Rays are drawn at angles \pi/12, \pi/6, \pi/4, \pi/3, and 5\pi/12. +

    + +

    + A polar curve is shown. It is somewhat wavy, + but lies close to the circle r=1, from the x axis to the y axis. + Instead of using the circle r=1 as part of the polar grid, + circular sectors are drawn with radius slightly more or less than 1, + so that they align with this curve. +

    + +

    + One of the sectors in the polar grid is highlighted. + It corresponds to the values 0.6\leq r\leq 0.8, and \pi/6\leq \theta\leq \pi/4. + The sector is approximately rectangular in shape, + but it is not a true rectangle, as two of its sides are circular arcs, + and the other two sides are rays from the origin. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[axis equal image, + ymin=-.1,ymax=1.25, + xmin=-.2,xmax=1.24 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(0.6928,0.4)(0.6857,0.412)(0.6784,0.4239)(0.6709,0.4357)(0.6632,0.4474)(0.6553,0.4589)(0.6472,0.4702)(0.6389,0.4815)(0.6304,0.4925)(0.6217,0.5035)(0.6128,0.5142)(0.6038,0.5248)(0.5945,0.5353)(0.5851,0.5456)(0.5755,0.5557)(0.5657,0.5657)(0.4243,0.4243)(0.4316,0.4168)(0.4388,0.4092)(0.4459,0.4015)(0.4528,0.3936)(0.4596,0.3857)(0.4663,0.3776)(0.4728,0.3694)(0.4792,0.3611)(0.4854,0.3527)(0.4915,0.3441)(0.4974,0.3355)(0.5032,0.3268)(0.5088,0.318)(0.5143,0.309)(0.5196,0.3)(0.6928,0.4)}; + + \addplot [secondcurvestyle,solid,domain=0:15] ({cos(x)*(1.05)},{sin(x)*(1.05)}); + \addplot [secondcurvestyle,solid,domain=15:30] ({cos(x)*(.95)},{sin(x)*(.95)}); + \addplot [secondcurvestyle,solid,domain=30:45] ({cos(x)*(1.05)},{sin(x)*(1.05)}); + \addplot [secondcurvestyle,solid,domain=45:60] ({cos(x)*(.97)},{sin(x)*(.97)}); + \addplot [secondcurvestyle,solid,domain=60:75] ({cos(x)*(1)},{sin(x)*(1)}); + \addplot [secondcurvestyle,solid,domain=75:90] ({cos(x)*(1.05)},{sin(x)*(1.05)}); + \addplot [secondcurvestyle,solid,domain=0:90] ({cos(x)*(.8)},{sin(x)*(.8)}); + \addplot [secondcurvestyle,solid,domain=0:90] ({cos(x)*(.6)},{sin(x)*(.6)}); + \addplot [secondcurvestyle,solid,domain=0:90] ({cos(x)*(.4)},{sin(x)*(.4)}); + \addplot [secondcurvestyle,solid,domain=0:90] ({cos(x)*(.2)},{sin(x)*(.2)}); + \addplot [secondcurvestyle,solid,domain=0:1.05] ({cos(15)*(x)},{sin(15)*(x)}); + \addplot [secondcurvestyle,solid,domain=0:1.05] ({cos(30)*(x)},{sin(30)*(x)}); + \addplot [secondcurvestyle,solid,domain=0:1.05] ({cos(45)*(x)},{sin(45)*(x)}); + \addplot [secondcurvestyle,solid,domain=0:1] ({cos(60)*(x)},{sin(60)*(x)}); + \addplot [secondcurvestyle,solid,domain=0:1.05] ({cos(75)*(x)},{sin(75)*(x)}); + + \addplot [firstcurvestyle,-,domain=0:90,samples=80] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}); + + \end{axis} + + \node [right] at (myplot.right of origin) { $0$}; + \node [above] at (myplot.above origin) { $\pi/2$}; + + \end{tikzpicture} + + + + +
    + +
    + + + + A close-up view of one region in a polar grid + +

    + A single sector within a polar grid. + Two rays from the origin are shown; they are separated by an angle marked as \Delta\theta. + The rays form two of the four sides of the sector; + the other two sides are circular arcs. +

    + +

    + The radii of the two circular arcs are also labeled. + The smaller radius is labeled r_1, and the larger r_2. + The region bounded by the rays and the two arcs is shaded. +

    + +

    + One way to think of this region is to imagine the crust at the end of a narrow slice of pizza. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis y line=none, + axis x line=none, + ymin=-.05,ymax=0.65, + xmin=-.1,xmax=.74 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(0.6928,0.4)(0.6857,0.412)(0.6784,0.4239)(0.6709,0.4357)(0.6632,0.4474)(0.6553,0.4589)(0.6472,0.4702)(0.6389,0.4815)(0.6304,0.4925)(0.6217,0.5035)(0.6128,0.5142)(0.6038,0.5248)(0.5945,0.5353)(0.5851,0.5456)(0.5755,0.5557)(0.5657,0.5657)(0.4243,0.4243)(0.4316,0.4168)(0.4388,0.4092)(0.4459,0.4015)(0.4528,0.3936)(0.4596,0.3857)(0.4663,0.3776)(0.4728,0.3694)(0.4792,0.3611)(0.4854,0.3527)(0.4915,0.3441)(0.4974,0.3355)(0.5032,0.3268)(0.5088,0.318)(0.5143,0.309)(0.5196,0.3)(0.6928,0.4)}; + + \addplot [secondcurvestyle,solid,domain=30:45] ({cos(x)*(.8)},{sin(x)*(.8)}); + \addplot [secondcurvestyle,solid,domain=30:45] ({cos(x)*(.8)},{sin(x)*(.8)}); + \addplot [secondcurvestyle,solid,domain=30:45] ({cos(x)*(.6)},{sin(x)*(.6)}); + \addplot [secondcurvestyle,solid,domain=0:.8] ({cos(30)*(x)},{sin(30)*(x)}); + \addplot [secondcurvestyle,solid,domain=0:.8] ({cos(45)*(x)},{sin(45)*(x)}); + + \draw [rotate=30] (axis cs:.31,-.12) node [rotate=30] {$\underbrace{\rule{140pt}{0pt}}_{r_1}$}; + \draw [rotate=45] (axis cs:.41,-.03) node [rotate=45] {$\overbrace{\rule{180pt}{0pt}}^{r_2}$}; + + \draw [thick,->,>=stealth,rotate=24] (axis cs:.45,0) arc (0:13:120pt); + \draw (axis cs: .4,.3) node { $\Delta \theta$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + Now consider representing a region R with polar coordinates. + Consider . + Let R be the region in the first quadrant bounded by the curve. + We can approximate this region using the natural shape of polar coordinates: + portions of sectors of circles. + In the figure, one such region is shaded, + shown again in . +

    + +

    + As the area of a sector of a circle with radius r, + subtended by an angle \theta, + is A = \frac12r^2\theta, + we can find the area of the shaded region. + The whole sector has area \frac12r_2^2\Delta \theta, + whereas the smaller, + unshaded sector has area \frac12r_1^2\Delta \theta. + The area of the shaded region is the difference of these areas: + + \Delta A_i = \frac12r_2^2\Delta\theta-\frac12r_1^2\Delta\theta = \frac12\big(r_2^2-r_1^2\big)\big(\Delta\theta\big) = \frac{r_2+r_1}{2}\big(r_2-r_1\big)\Delta\theta + . +

    + +

    + Note that (r_2+r_1)/2 is just the average of the two radii. +

    + +

    + To approximate the region R, + we use many such subregions; + doing so shrinks the difference r_2-r_1 between radii to 0 and shrinks the change in angle \Delta \theta also to 0. + We represent these infinitesimal changes in radius and angle as dr and d\theta, + respectively. + Finally, as dr is small, + r_2\approx r_1, and so (r_2+r_1)/2\approx r_1. + Thus, when dr and d\theta are small, + + \Delta A_i \approx r_i\, dr\, d\theta + . +

    + +

    + Taking a limit, + where the number of subregions goes to infinity and both r_2-r_1 and \Delta\theta go to 0, we get + + dA = r\, dr\, d\theta + . +

    + +

    + So to evaluate \iint_Rf(x,y)\, dA, + replace dA with r\, dr\, d\theta. + Convert the function f(x,y) to a function with polar coordinates with the substitutions + x=r\cos(\theta), y=r\sin(\theta). + Finally, find bounds g_1(\theta)\leq r\leq g_2(\theta) and + \alpha\leq\theta\leq\beta that describe R. + This is the key principle of this section, + so we restate it here as a Key Idea. +

    + + + Evaluating Double Integrals with Polar Coordinates +

    + Let z=f(x,y) be a continuous function defined over a closed, + bounded region R in the xy-plane, + where R is bounded by the polar equations + \alpha\leq\theta\leq\beta and g_1(\theta)\leq r\leq g_2(\theta). + Then + double integralin polar + + \iint_Rf(x,y)\, dA = \int_\alpha^\beta\int_{g_1(\theta)}^{g_2(\theta)} f\big(r\cos(\theta) ,r\sin(\theta) \big)\, r\, dr\, d\theta + . +

    +     + +

    + Examples will help us understand this Key Idea. +

    + + + Evaluating a double integral with polar coordinates + +

    + Find the signed volume under the plane + z= 4-x-2y over the disk bounded by the circle with equation x^2+y^2=1. +

    +     + +

    + The bounds of the integral are determined solely by the region R over which we are integrating. + In this case, it is a disk with boundary x^2+y^2=1. + We need to find polar bounds for this region. + It may help to review ; + bounds for this disk are 0\leq r\leq 1 and 0\leq \theta\leq 2\pi. +

    + +

    + We replace f(x,y) with f(r\cos(\theta) ,r\sin(\theta) ). + That means we make the following substitutions: + + 4-x-2y \Rightarrow 4-r\cos(\theta) -2r\sin(\theta) + . +

    + +

    + Finally, we replace dA in the double integral with r\, dr\, d\theta. + This gives the final iterated integral, which we evaluate: + + \iint_Rf(x,y)\, dA \amp = \int_0^{2\pi}\int_0^1\big(4-r\cos(\theta) -2r\sin(\theta) \big)r\, dr\, d\theta + \amp = \int_0^{2\pi}\int_0^1\big(4r-r^2(\cos(\theta) -2\sin(\theta) )\big)\, dr\, d\theta + \amp = \int_0^{2\pi}\left.\left(2r^2-\frac13r^3(\cos(\theta) -2\sin(\theta) )\right)\right|_0^1d\theta + \amp = \int_0^{2\pi} \left(2-\frac13\big(\cos(\theta) -2\sin(\theta) \big)\right)\, d\theta + \amp = \left.\left(2\theta -\frac13\big(\sin(\theta) +2\cos(\theta) \big)\right)\right|_0^{2\pi} + \amp = 4\pi \approx 12.566 + . +

    + +
    + Evaluating a double integral with polar coordinates in + + + + An ellipse on a plane in space corresponds to a circle drawn below it in the xy plane. + +

    + Three-dimensional coordinate axes are drawn in space. + In the xy plane, a unit circle is drawn. + The plane z=4-x-2y lies above the xy plane. + On the plane, we see the curve corresponding to the unit circle; + since the plane lies at an angle relative to the xy plane, + the image of the unit circle on the plane is an ellipse. +

    +     + + + + + //ASY file for figdoublepol13D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.2,-1.9,13); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={5}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(0,10); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane //({z=4-x-2y); + triple f(pair t) { + return (t.x,t.y,4-t.x-2*t.y); + } + surface s=surface(f,(-1.5,-1.5),(1.5,1.5),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //draw curve on xy plane + triple g(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //draw curve on plane + triple g(real t) {return (cos(t),sin(t),4-cos(t)-2*sin(t));} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + + + +
    + +

    + The surface and region R are shown in . +

    +     +     + + + + + Evaluating a double integral with polar coordinates + +

    + Find the volume under the paraboloid + z=4-(x-2)^2-y^2 over the region bounded by the circles + (x-1)^2+y^2=1 and (x-2)^2+y^2=4. +

    +     + +

    + At first glance, + this seems like a very hard volume to compute as the region R (shown in ) has a hole in it, + cutting out a strange portion of the surface, + as shown in . + However, by describing R in terms of polar equations, + the volume is not very difficult to compute. +

    + +
    + Showing the region R and surface used in + +
    + + + + A region in the plane between that lies between two circles. + +

    + Two circles are plotted in the plane. The smaller of the two circles lies inside the larger. + The circles meet, and are tangent to each other, at the origin. +

    + +

    + The larger circle is centered at (2,0) and has radius 2. + The smaller circle is centered at (1,0) and has radius 1. + The region that lies between the two circles is shaded. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={2,1,3,4}, + ymin=-2.1,ymax=2.1, + xmin=-.54,xmax=4.5 + ] + + \addplot [firstcurvestyle,areastyle,domain=0:360,samples=40] ({2+2*cos(x)},{2*sin(x)}); + \addplot [firstcurvestyle,domain=0:360,samples=80] ({2+2*cos(x)},{2*sin(x)}); + + \addplot [firstcolor,thick,fill=white,area style,domain=0:360,samples=60] ({1+cos(x)},{sin(x)}); + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + + A region in space that lies outside of a cylinder, and below a circular paraboloid. + +

    + The image shows a region in space between two surfaces. + The first surface is a circular paraboloid; + it has its vertext at the point (2,0,4), + and it meets the xy plane along the larger of the two circles in +

    + +

    + The second surface is a circular cylinder that lies above the smaller of the two circles in . + Overall, the shape is like a circular hill that has had a hole drilled from above on one side of its peak. +

    +     + + + + + //ASY file for figdoublepol2b3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(-2.7,-11,16); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={-2,2}; + real[] myzchoice={5}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,4.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(0,7); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface //({z=4-(x-2)^2-y^2); + //triple f(pair t) { + // return (t.x,t.y,4-(t.x-2)^2-t.y^2); + //} + + //surface s=surface(f,(-0.25,-2.25),(4.5,2.25),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + //,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + //); + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + //draw(s,surfacepen); + + real r(real t) { return sqrt((2*cos(t)+1)^2+(2*sin(t))^2);} + real ang(real tt) {return atan2(2*sin(tt),(2*cos(tt)+1));} + real ra(pair s) {return 1+s.y*(r(s.x)-1);} + + triple g(pair t) { return (cos(ang(t.x))*ra(t)+1,sin(ang(t.x))*ra(t),4-(cos(ang(t.x))*ra(t)+1-2)^2-(sin(ang(t.x))*ra(t))^2);} + + surface s=surface(g,(0,0),(2*pi,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + draw(s,surfacepen,meshpen=p); + + triple g(pair t) { return (cos(ang(t.x))*ra(t)+1,sin(ang(t.x))*ra(t),0);} + + surface s=surface(g,(0,0),(2*pi,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + draw(s,surfacepen,meshpen=p); + + triple ff(pair t) { + return (cos(t.x)+1,sin(t.x),t.y*(4-(cos(t.x)+1-2)^2-(sin(t.x))^2));} + surface ss= surface(ff,(0,0),(2*pi,1),16,4,Spline); + pen q=redcurvepen; + //draw(ss,surfacepen2,meshpen=q,nolight,render(merge=true)); + draw(ss,surfacepen,meshpen=p); + + //draw curves on xy plane + triple g(real t) {return (2*cos(t)+2,2*sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (cos(t)+1,sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+linewidth(1)); + + //draw curve on plane + triple g(real t) {return (cos(t)+1,sin(t),4-(cos(t)-1)^2-sin(t)^2);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+linewidth(1)); + + + + +
    +     + +
    + +

    + It is straightforward to show that the circle + (x-1)^2+y^2=1 has polar equation r=2\cos(\theta), + and that the circle (x-2)^2+y^2=4 has polar equation r=4\cos(\theta). + Each of these circles is traced out on the interval 0\leq\theta\leq\pi. + The bounds on r are 2\cos(\theta) \leq r\leq 4\cos(\theta). +

    + +

    + Replacing x with r\cos(\theta) in the integrand, + along with replacing y with r\sin(\theta), + prepares us to evaluate the double integral \iint_Rf(x,y)\, dA: + + \iint_Rf(x,y)\, dA \amp = \int_0^{\pi}\int_{2\cos(\theta) }^{4\cos(\theta) } \Big(4-\big(r\cos(\theta) -2\big)^2-\big(r\sin(\theta) \big)^2\Big)r\, dr\, d\theta + \amp = \int_0^{\pi}\int_{2\cos(\theta) }^{4\cos(\theta) } \big(-r^3+4r^2\cos(\theta) \big)\, dr\, d\theta + \amp = \int_0^\pi \left.\left(-\frac14r^4+\frac43r^3\cos(\theta) \right)\right|_{2\cos(\theta) }^{4\cos(\theta) }d\theta + \amp =\int_0^\pi \left(\left[-\frac14(256\cos^4(\theta) )+\frac43(64\cos^4(\theta) )\right]-\right. + \amp \quad \left.\left[-\frac14(16\cos^4(\theta) )+\frac43(8\cos^4(\theta) )\right]\right)\, d\theta + \amp =\int_0^\pi\frac{44}3\cos^4(\theta) \, d\theta + + To integrate \cos^4(\theta), + rewrite it as \cos^2(\theta) \cos^2(\theta) and employ the power-reducing formula twice: + + \cos^4(\theta) \amp =\cos^2(\theta) \cos^2(\theta) + \amp = \frac12\big(1+\cos(2\theta)\big)\frac12\big(1+\cos(2\theta)\big) + \amp = \frac14\big(1+2\cos(2\theta)+\cos^2(2\theta)\big) + \amp =\frac14\Big(1+2\cos(2\theta)+\frac12\big(1+\cos(4\theta)\big)\Big) + \amp = \frac38+\frac12\cos(2\theta)+\frac18\cos(4\theta) + . + Picking up from where we left off above, we have + + \iint_R f(x,y)\,dA \amp =\int_0^\pi\frac{44}3\cos^4(\theta) \, d\theta + \amp =\int_0^\pi \frac{44}3\left(\frac38+\frac12\cos(2\theta)+\frac18\cos(4\theta)\right)d\theta + \amp = \left.\frac{44}3\left(\frac{3}8\theta+\frac14\sin(2\theta)+\frac{1}{32}\sin(4\theta)\right)\right|_0^\pi + \amp =\frac{11}2\pi\approx 17.279 + . +

    + +

    + While this example was not trivial, + the double integral would have been much + harder to evaluate had we used rectangular coordinates. +

    +     +     + + + Evaluating a double integral with polar coordinates + +

    + Find the volume under the surface given by the graph of + \ds f(x,y) =\frac1{x^2+y^2+1} over the sector of the circle with radius a centered at the origin in the first quadrant, + as shown in . +

    +
    + The surface and region R used in + + + + A portion of a bell-shaped curve, and a triangular curve on the surface that corresponds to a quarter circle in the xy plane. + +

    + The surface given by the graph z=\frac{1}{1+x^2+y^2} is shaped like a steep mountain, + or a witch's hat, + and has circular symmetry about the z axis. + However, only the portion of the surface in the first octant is shown. +

    + +

    + In the xy plane, a quarter-circle is sketched in the first quadrant. + On the surface, we see a curve illustrating the portion of the surface used in the integral. + The curve appears to be triangular; its sides correspond to the x and y coordinate axes, + and the quarter circle. +

    +     + + + + + //ASY file for figdoublepol53D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8,-13,2.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={0.5,1}; + defaultpen(0.5mm); + + pair xbounds=(0,5.5); + pair ybounds=(0,5.5); + pair zbounds=(0,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface //(z=1/(1+x^2+y^2); + triple f(pair t) { + return (t.x,t.y,1/(1+t.x^2+t.y^2)); + } + surface s=surface(f,(-0.25,-0.25),(5,5),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //draw curves on xy plane + triple g(real t) {return (2.5*cos(t),2.5*sin(t),0);} + path3 mypath=graph(g,0,pi/2,operator ..); + draw(mypath,bluepen+dashed+linewidth(0.75)); + label("$a$",(2.7,-0.2,0),S); + + //draw curves on surface + triple g(real t) {return (2.5*cos(t),2.5*sin(t),1/(1+2.5^2));} + path3 mypath=graph(g,0,pi/2,operator ..); + draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,0,1/(1+t^2));} + path3 mypath=graph(g,0,2.5,operator ..); + draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (0,t,1/(1+t^2));} + path3 mypath=graph(g,0,2.5,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //draw connecting lines + draw((2.5,0,0) -- (2.5,0,1/(1+2.5^2)),bluepen+dashed+linewidth(0.75)); + draw((0,2.5,0) -- (0,2.5,1/(1+2.5^2)),bluepen+dashed+linewidth(0.75)); + + + + +
    +     + +

    + The region R we are integrating over is a circle with radius a, + restricted to the first quadrant. + Thus, in polar, the bounds on R are 0\leq r\leq a, + 0\leq\theta\leq\pi/2. + The integrand is rewritten in polar as + + \frac{1}{x^2+y^2+1} \Rightarrow \frac{1}{r^2\cos^2(\theta) +r^2\sin^2(\theta) +1} = \frac1{r^2+1} + . +

    + +

    + We find the volume as follows: + + \iint_Rf(x,y)\, dA \amp = \int_0^{\pi/2}\int_0^a\frac{r}{r^2+1}\, dr\, d\theta + \amp = \int_0^{\pi/2} \frac12\big(\ln\abs{r^2+1}\big)\Big|_0^a\, d\theta + \amp =\int_0^{\pi/2} \frac12\ln(a^2+1)\, d\theta + \amp = \left.\left(\frac12\ln(a^2+1)\theta\right)\right|_0^{\pi/2} + \amp = \frac{\pi}{4}\ln(a^2+1) + . +

    + + +

    + + shows that f shrinks to near 0 very quickly. + Regardless, as a grows, + so does the volume, without bound. +

    +     +     + + + Finding the volume of a sphere + +

    + Find the volume of a sphere with radius a. +

    +     + +

    + The sphere of radius a, + centered at the origin, has equation x^2+y^2+z^2=a^2; + solving for z, we have z=\sqrt{a^2-x^2-y^2}. + This gives the upper half of a sphere. + We wish to find the volume under this top half, + then double it to find the total volume. +

    + +

    + The region we need to integrate over is the disk of radius a, + centered at the origin. + Polar bounds for this equation are 0\leq r\leq a, + 0\leq\theta\leq2\pi. +

    + +

    + All together, the volume of a sphere with radius a is: + + 2\iint_R\sqrt{a^2-x^2-y^2}\, dA \amp = 2\int_0^{2\pi}\int_0^a\sqrt{a^2-(r\cos(\theta) )^2-(r\sin(\theta) )^2}r\, dr\, d\theta + \amp =2\int_0^{2\pi}\int_0^ar\sqrt{a^2-r^2}\, dr\, d\theta. + We can evaluate this inner integral with substitution. + With u=a^2-r^2, du = -2r\, dr. + The new bounds of integration are u(0) = a^2 to u(a)=0. Thus we have: + + \amp = \int_0^{2\pi}\int_{a^2}^0\big(-u^{1/2}\big)\, du\, d\theta + \amp = \int_0^{2\pi}\left.\left(-\frac23u^{3/2}\right)\right|_{a^2}^0 d\theta + \amp = \int_0^{2\pi}\left(\frac23a^3\right)\, d\theta + \amp = \left.\left(\frac23a^3\theta\right)\right|_0^{2\pi} + \amp = \frac43\pi a^3 + . +

    + +

    + Generally, the formula for the volume of a sphere with radius r is given as 4/3\pi r^3; + we have justified this formula with our calculation. +

    +     +     + + + + + Finding the volume of a solid + +

    + A sculptor wants to make a solid bronze cast of the solid shown in , + where the base of the solid has boundary, + in polar coordinates, r=\cos(3\theta), + and the top is defined by the plane z=1-x+0.1y. + Find the volume of the solid. +

    + +
    + Visualizing the solid used in + + + + A solid whose cross-sections are like a three-leaf clover; it has been sliced at an angle. + +

    + A three-dimensional solid is shown, plotted against a set of three-dimensional coordinate axes. + The base of the solid is a three-leaf rose curve. +

    + +

    + The solid is a cylinder, except that it is cut at an angle, + where it meets the plane z=1-x+0.1y. +

    +     + + + + + //ASY file for figdoublepol43D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2,2); + + //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //plane is z=1-x+0.1y + //Draw the surface + //({cos(3*x)*cos(x)},{cos(3*x)*sin(x)},{(1-cos(3*x)*cos(x)+.1*cos(3*x)*sin(x))*y}); + triple f(pair t) { + return (cos(3*t.x)*cos(t.x),cos(3*t.x)*sin(t.x),(1-cos(3*t.x)*cos(t.x)+.1*cos(3*t.x)*sin(t.x))*t.y); + } + surface s=surface(f,(0,0),(2*pi,1),32,32,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //draw curve on xy plane //({cos(3*x)*cos(x)},{cos(3*x)*sin(x)},{0}) + triple g(real t) {return (cos(3*t)*cos(t),cos(3*t)*sin(t),0);} + path3 mypath=graph(g,0,pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //draw curves on surface + triple g(real t) {return (cos(3*t)*cos(t),cos(3*t)*sin(t),1-(cos(3*t)*cos(t))+0.1*(cos(3*t)*sin(t)));} + path3 mypath=graph(g,0,pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //Shade the bottom + import three; + path3 p = (0,0,0)..(.175,0.064,0)..(.264,.119,0)..(0.683,0.183,0)..(.916,0.121,0)..(1,0,0)..(.916,-0.121,0)..(0.683,-0.183,0)..(.264,-.119,0)..(.175,-0.086,0); + draw(surface(p -- cycle), simplesurfacepen); + + path3 p = (0,0,0)..(-0.1429, 0.1196,0)..(-0.2351, 0.1691,0)..(-0.5000, 0.5000,0)..(-0.5628, 0.7328,0)..(-0.5000, 0.8660,0)..(-0.3532, 0.8538,0)..(-0.1830, 0.6830,0)..(-0.0289, 0.2881,0)..(-0.0130, 0.1946,0); + draw(surface(p -- cycle), simplesurfacepen); + + path3 p = (0,0,0)..(-0.0321,-0.1836,0)..(-0.0289,-0.2881,0)..(-0.1830,-0.6830,0)..(-0.3532,-0.8538,0)..(-0.5000,-0.8660,0)..(-0.5628,-0.7328,0)..(-0.5000,-0.5000,0)..(-0.2351,-0.1691,0)..(-0.1620,-0.1086,0); + draw(surface(p -- cycle), simplesurfacepen); + + //Shade the top + path3 p = (0,0,1.0000)..(-0.1429,0.1196,1.1549)..(-0.2351,0.1691,1.2520)..(-0.5000,0.5000,1.5500)..(-0.5628,0.7328,1.6361)..(-0.5000,0.8660,1.5866)..(-0.3532,0.8538,1.4386)..(-0.1830 ,0.6830 ,1.2513)..(-0.0289,0.2881,1.0578)..(-0.0130,0.1946,1.0325); + + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0 , 0 , 1.0000)..(-0.0321 , -0.1836 , 1.0137)..(-0.0289, -0.2881, 1.0001)..(-0.1830, -0.6830, 1.1147)..(-0.3532 , -0.8538 , 1.2678)..(-0.5000, -0.8660 , 1.4134)..(-0.5628 , -0.7328, 1.4895)..(-0.5000 , -0.5000 , 1.4500)..(-0.2351 , -0.1691 , 1.2181)..(-0.1620 , -0.1086, 1.1511); + draw(surface(p -- cycle), simplesurfacepen); + + path3 p = (0 , 0, 1.0000)..( 0.1750, 0.0640 , 0.8314)..( 0.2640, 0.1190 , 0.7479)..( 0.6830, 0.1830 , 0.3353)..( 0.9160, 0.1210 , 0.0961)..( 1.0000, 0 , 0)..( 0.9160, -0.1210 , 0.0719)..( 0.6830, -0.1830 , 0.2987)..( 0.2640, -0.1190 , 0.7241)..( 0.1750, -0.0860 , 0.8164); + draw(surface(p -- cycle), simplesurfacepen); + + //path3 p = (0,0,0)-- (1,1,0) -- (1,2,1) -- (0,1,1); //bottom + //draw(surface(p -- cycle), simplesurfacepen); + //path3 p = (-1,1,0)-- (0,2,0) -- (0,3,1) -- (-1,2,1); //right + //draw(surface(p -- cycle), simplesurfacepen); + //path3 p = (0,0,0)-- (-1,1,0) -- (-1,2,1) -- (0,1,1); //back + //draw(surface(p -- cycle), simplesurfacepen); + //path3 p = (1,1,0)-- (0,2,0) -- (0,3,1) -- (1,2,1); //front + //draw(surface(p -- cycle), simplesurfacepen); + //path3 p = (0,1,1)-- (1,2,1) -- (0,3,1) -- (-1,2,1); //top + //draw(surface(p -- cycle), simplesurfacepen); + + + + +
    +     + +

    + From the outset, + we should recognize that knowing how to set up + this problem is probably more important than knowing + how to compute the integrals. + The iterated integral to come is not + hard to evaluate, + though it is long, requiring lots of algebra. + Once the proper iterated integral is determined, + one can use readily available technology to help compute the final answer. +

    + +

    + The region R that we are integrating over is bound by 0\leq r\leq \cos(3\theta), + for 0\leq \theta\leq\pi + (note that this rose curve is traced out on the interval [0,\pi], + not [0,2\pi]). + This gives us our bounds of integration. + The integrand is z=1-x+0.1y; + converting to polar, we have that the volume V is: + + V = \iint_R f(x,y)\, dA = \int_0^\pi\int_0^{\cos(3\theta)}\big(1-r\cos(\theta) +0.1r\sin(\theta) \big)r\, dr\, d\theta + . +

    + +

    + Distributing the r, the inner integral is easy to evaluate, + leading to + + \int_0^\pi \left(\frac12\cos^2(3\theta)-\frac13\cos^3(3\theta)\cos(\theta) +\frac{0.1}3\cos^3(3\theta)\sin(\theta) \right)\, d\theta + . +

    + +

    + This integral takes time to compute by hand; + it is rather long and cumbersome. + The powers of cosine need to be reduced, + and products like \cos(3\theta)\cos(\theta) need to be turned to sums using the Product To Sum formulas in the back cover of this text. +

    + +

    + We rewrite \frac12\cos^2(3\theta) as \frac14(1+\cos(6\theta)). + We can also rewrite \frac13\cos^3(3\theta)\cos(\theta) as: + + \frac13\cos^3(3\theta)\cos(\theta) \amp = \frac13\cos^2(3\theta)\cos(3\theta)\cos(\theta) + \amp = \frac13\frac{1+\cos(6\theta)}2\big(\cos(4\theta)+\cos(2\theta)\big) + . +

    + +

    + This last expression still needs simplification, + but eventually all terms can be reduced to the form a\cos(m\theta) or + a\sin(m\theta) for various values of a and m. +

    + +

    + We forgo the algebra and recommend the reader employ technology, + such as WolframAlpha, + to compute the numeric answer. + Such technology gives: + + \int_0^\pi\int_0^{\cos(3\theta)}\big(1-r\cos(\theta) +0.1r\sin(\theta) \big)r\, dr\, d\theta = \frac{\pi}{4} \approx 0.785u^3 + . +

    + +

    + Since the units were not specified, + we leave the result as almost 0.8 cubic units (meters, + feet, etc.) Should the artist want to scale the piece uniformly, + so that each rose petal had a length other than 1, she should keep in mind that scaling by a factor of k scales the volume by a factor of k^3. +

    +     +     + + + +

    + We have used iterated integrals to find areas of plane regions and volumes under surfaces. + Just as a single integral can be used to compute much more than + area under the curve, + iterated integrals can be used to compute much more than we have thus far seen. + The next two sections show two, among many, + applications of iterated integrals. +

    + + + + + Terms and Concepts + + + +

    + Match the correct elements on the left to the corresponding elements on the right, + so that \iint_R f(x,y)\, dA is correctly converted to polar coordinates. +

    +     + + + r\cos(\theta) + x + + + r\sin(\theta) + y + + + r\,dr\,d\theta + dA + + + dr\,d\theta + + + \cos(\theta) + + + \sin(\theta) + + + + +     + + + + +

    + Why would one be interested in evaluating a double integral with polar coordinates? +

    +     + + + +

    + Some regions in the xy-plane are easier to describe using polar coordinates than using rectangular coordinates. + Also, some integrals are easier to evaluate one the polar substitutions have been made. +

    +     + +     +     + + + Problems + + + +

    + A function f(x,y) is given and a region R of the xy-plane is described. + Set up and evaluate \iint_Rf(x,y)\, dA using polar coordinates. +

    +     + + + + + $int=Compute("4pi"); + + + +

    + f(x,y) = 3x-y+4 and R is the region enclosed by the circle x^2+y^2=1. +

    + +

    + +

    +     +     +     + + + + +

    + f(x,y) = 4x+4y; + R is the region enclosed by the circle x^2+y^2=4. +

    +     + +

    + \ds \int_0^{2\pi}\int_0^2 \big(4r\cos(\theta) +4r\sin(\theta) \big)r\, dr\, d\theta = 0 +

    +     + +     + + + + + $int=Compute("16pi"); + + + +

    + f(x,y) = 8-y and R is the region enclosed by the circles with polar equations + r=\cos(\theta) and r=3\cos(\theta). +

    + +

    + +

    +     +     +     + + + + +

    + f(x,y) = 4; + R is the region enclosed by the petal of the rose curve + r=\sin(2\theta) in the first quadrant. +

    +     + +

    + \ds \int_0^{\pi/2}\int_{0}^{\sin(2\theta)} \big(4\big)r\, dr\, d\theta = \pi/2 +

    +     + +     + + + + +

    + f(x,y) = \ln(x^2+y^2); + R is the annulus enclosed by the circles + x^2+y^2=1 and x^2+y^2=4. +

    +     + +

    + \ds \int_0^{2\pi}\int_{1}^{2} \big(\ln(r^2)\big)r\, dr\, d\theta = 2\pi\big(\ln(16) -3/2\big) +

    +     + +     + + + + + $int=Compute("pi/2"); + + + +

    + f(x,y) = 1-x^2-y^2 and R is the region enclosed by the circle x^2+y^2=1. +

    + +

    + +

    +     +     +     + + + + +

    + f(x,y) = x^2-y^2; + R is the region enclosed by the circle + x^2+y^2=36 in the first and fourth quadrants. +

    +     + +

    + \ds \int_{-\pi/2}^{\pi/2}\int_{0}^{6} \big(r^2\cos^2(\theta) -r^2\sin^2(\theta) \big)r\, dr\, d\theta = \int_{-\pi/2}^{\pi/2}\int_{0}^{6} \big(r^2\cos(2\theta)\big)r\, dr\, d\theta= 0 +

    +     + +     + + + + +

    + f(x,y) = (x-y)/(x+y); + R is the region enclosed by the lines y=x, + y=0 and the circle x^2+y^2=1 in the first quadrant. +

    +     + +

    + \ds \int_{0}^{\pi/4}\int_{0}^{1} \left(\frac{\cos(\theta) -\sin(\theta) }{\cos(\theta) +\sin(\theta) }\right)r\, dr\, d\theta = \ln(2) +

    +     + +     + +     + + + +

    + An iterated integral in rectangular coordinates is given. + Rewrite the integral using polar coordinates and evaluate the new double integral. +

    +     + + + + +

    + \ds \int_{0}^{5}\int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}} \sqrt{x^2+y^2}\, dy\, dx +

    +     + +

    + \ds \int_{-\pi/2}^{\pi/2}\int_{0}^{5} \big(r^2\big)\, dr\, d\theta=125\pi/3 +

    +     + +     + + + + + $int=Compute("128/3"); + + + +

    + \ds \int_{-4}^{4}\int_{-\sqrt{16-y^2}}^{0} \big(2y-x\big)\, dx\, dy +

    + +

    + +

    +     +     +     + + + + +

    + \ds \int_{0}^{2}\int_{y}^{\sqrt{8-y^2}} \big(x+y\big)\, dx\, dy +

    +     + +

    + \ds \int_{0}^{\pi/4}\int_{0}^{\sqrt8} \big(r\cos(\theta) +r\sin(\theta) \big)r\, dr\, d\theta=16\sqrt2/3 +

    +     + +     + + + + +

    + \ds \int_{-2}^{-1}\int_{0}^{\sqrt{4-x^2}} \big(x+5\big)\, dy\, dx+\int_{-1}^{1}\int_{\sqrt{1-x^2}}^{\sqrt{4-x^2}} \big(x+5\big)\, dy\, dx +\int_{1}^{2}\int_{0}^{\sqrt{4-x^2}} \big(x+5\big)\, dy\, dx +

    + +

    + Hint: draw the region of each integral carefully and see how they all connect. +

    +     + +

    + \ds \int_{0}^{\pi}\int_{1}^{2} \big(r\cos(\theta) +5\big)r\, dr\, d\theta=15\pi/2 +

    +     + +     + +     + + + + +

    + The next two exercises present special double integrals that are especially well suited for evaluation in polar coordinates. +

    +     + + + + +

    + Consider \ds \iint_R e^{-(x^2+y^2)}\, dA. +

    +     + + + +

    + Why is this integral difficult to evaluate in rectangular coordinates, + regardless of the region R? +

    +     + +

    + This is impossible to integrate with rectangular coordinates as + e^{-(x^2+y^2)} does not have an antiderivative in terms of elementary functions. +

    +     +     + + + +

    + Let R be the region bounded by the circle of radius a centered at the origin. + Evaluate the double integral using polar coordinates. +

    +     + +

    + \ds \int_0^{2\pi}\int_0^a re^{r^2}\, dr\, d\theta = \pi(1-e^{-a^2}). +

    +     +     + + + +

    + Take the limit of your answer from (b), as a\to\infty. + What does this imply about the volume under the surface + z= e^{-(x^2+y^2)} over the entire xy-plane? +

    +     + +

    + \lim\limits_{a\to\infty} \pi(1-e^{-a^2})=\pi. + This implies that there is a finite volume under the surface + z=e^{-(x^2+y^2)} over the entire xy-plane. +

    +     +     + +     + + + + +

    + The surface of a right circular cone with height h and base radius a can be described by the equation \ds f(x,y) = h-h\sqrt{\frac{x^2}{a^2}+\frac{y^2}{a^2}}, + where the tip of the cone lies at (0,0,h) and the circular base lies in the xy-plane, + centered at the origin. +

    + +

    + Confirm that the volume of a right circular cone with height h and base radius a is \ds V = \frac13\pi a^2h by evaluating + \ds \iint_R f(x,y)\, dA in polar coordinates. +

    +     + +

    + + \iint_R f(x,y)\, dA \amp = \int_0^{2\pi}\int_0^a \left(h-h\sqrt{\frac{r^2\cos^2(\theta) }{a^2}+\frac{r^2\sin^2(\theta) }{a^2}}\right)r\, dr\, d\theta + \amp = \int_0^{2\pi}\int_0^a \left(hr-h\frac{r^2}{a}\right)\, dr\, d\theta + \amp = \int_0^{2\pi}\left.\left(\frac12hr^2-\frac{h}{3a}r^3\right)\right|_0^a d\theta + \amp = \int_0^{2\pi} \left(\frac16a^2h\right)\, d\theta + \amp = \frac13\pi a^2h + . +

    +     + +     + +     +     +     +
    +
    + Center of Mass + +

    + We have used iterated integrals to find areas of plane regions and signed volumes under surfaces. + A brief recap of these uses will be useful in this section as we apply iterated integrals to compute the mass + and center of mass of planar regions. +

    + +

    + To find the area of a planar region, + we evaluated the double integral \iint_R\, dA. + That is, summing up the areas of lots of little subregions of R gave us the total area. + Informally, we think of \iint_R\, dA as meaning + sum up lots of little areas over R. +

    + +

    + To find the signed volume under a surface, + we evaluated the double integral \iint_R f(x,y)\, dA. + Recall that the dA + is not just a bookend + at the end of an integral; + rather, it is multiplied by f(x,y). + We regard f(x,y) as giving a height, + and dA still giving an area: + f(x,y)\, dA gives a volume. + Thus, informally, \iint_Rf(x,y)\, dA means + sum up lots of little volumes over R. +

    + +

    + We now extend these ideas to other contexts. +

    +     + + + Mass and Weight +

    + Consider a thin sheet of material with constant thickness and finite area. + Mathematicians + (and physicists and engineers) + call such a sheet a lamina. + So consider a lamina, + as shown in , + with the shape of some planar region R, as shown in . + lamina + mass +

    + +
    + Illustrating the concept of a lamina + +
    + + + + A region in the plane with two vertical sides, a curved top, and a slanted bottom. + +

    + An illustration of a lamina as a region in the plane. + This particular region has a curved top, like a downward-opening parabola, + and two sides given by vertical lines. + The right side is shorter than the left side, + and the bottom is a line segment with positive slope. +

    + +

    + There is no coordinate system or other means of reference in the image. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis y line=none, + axis x line=none, + axis on top, + ymin=-.1,ymax=3.5, + xmin=-.1,xmax=3.5 + ] + + \addplot [name path=A,firstcurvestyle,-,domain=1:3] {-(x-2)^2+3}; + \addplot [name path=B,firstcurvestyle,-,domain=1:3] {x/2}; + \addplot [firstcurvestyle,domain=.5:2] (1,x); + \addplot [firstcurvestyle,domain=1.5:2] (3,x); + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B]; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +
    + + + + A region in the plane, with the same shape as the previous image, this time plotted against coordinate axes. + +

    + The first quadrant in the plane is shown, + with the x axis at the bottom of the image, and the y axis on the left. + The same region as is shown, + but this time with the coordinate axes and labels for reference. +

    + +

    + Now it is specified that the sides of the region are the vertical lines x=1 and x=3. + The line along the bottom is labeled with the equation y=f_1(x), + the curve along the top is labeled y=f_2(x), and the region itself is labeled R. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=3.5, + xmin=-.1,xmax=3.5 + ] + + \addplot [name path=A,firstcurvestyle,domain=-.1:3.5,samples=60] {-(x-2)^2+3} node [pos=.65,above,black] { $y=f_2(x)$}; + \addplot [name path=B,firstcurvestyle,domain=-.1:3.5] {x/2} node [pos=.6,sloped,above,black] { $y=f_1(x)$}; + + \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=1:3}]; + + \addplot [secondcurvestyle,solid,domain=0:3] (1,x); + \addplot [secondcurvestyle,solid,domain=0:3] (3,x); + + \draw (axis cs:2,2) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +
    + +

    + We can write a simple double integral that represents the mass of the lamina: + \iint_R\, dm, where dm + means a little mass. That is, + the double integral states the total mass of the lamina can be found by + summing up lots of little masses over R. +

    + +

    + To evaluate this double integral, + partition R into n subregions as we have done in the past. + The ith subregion has area \Delta A_i. + A fundamental property of mass is that + mass=densityarea. + If the lamina has a constant density \delta, + then the mass of this ith subregion is \Delta m_i=\delta\Delta A_i. + That is, we can compute a small amount of mass by multiplying a small amount of area by the density. +

    + +

    + If density is variable, with density function \delta= \delta(x,y), + then we can approximate the mass of the + ith subregion of R by multiplying + \Delta A_i by \delta(x_i,y_i), + where (x_i,y_i) is a point in that subregion. + That is, for a small enough subregion of R, + the density across that region is almost constant. +

    + +

    + The total mass M of the lamina is approximately the sum of approximate masses of subregions: + + M \approx \sum_{i=1}^n \Delta m_i = \sum_{i=1}^n \delta(x_i,y_i)\Delta A_i + . +

    + +

    + Taking the limit as the size of the subregions shrinks to 0 gives us the actual mass; + that is, integrating \delta(x,y) over R gives the mass of the lamina. +

    + + + Mass of a Lamina with Variable Density + +

    + Let \delta(x,y) be a continuous density function of a lamina corresponding to a closed, + bounded plane region R. + The mass M of the lamina is + mass + + \text{ mass } M = \iint_R\, dm = \iint_R \delta(x,y)\, dA + . +

    +     +     + + + + + Finding the mass of a lamina with constant density + +

    + Find the mass of a square lamina, + with side length 1, with a density of \delta = 3\,\text{g/cm}^2. +

    +     + +

    + We represent the lamina with a square region in the plane as shown in . + As the density is constant, it does not matter where we place the square. +

    + +
    + A region R representing a lamina in + + + The unit square in the first quadrant, plotted against x and y coordinate axes. + +

    + The x axis is at the bottom of the image, and the y axis along the left, + so that the first quadrant in the plane is shown. +

    + +

    + The unit square, given by 0\leq x\leq 1 and 0\leq y\leq 1 + is drawn and shaded. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.1 + ] + + \filldraw [fill=firstcolor!15,draw=firstcolor,thick] (axis cs:0,0) -- (axis cs:1,0) -- (axis cs: 1,1) -- (axis cs:0,1) -- cycle; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Following , + the mass M of the lamina is + + M = \iint_R 3\, dA = \int_0^1\int_0^1 3\, dx\, dy = 3\int_0^1\int_0^1 \, dx\, dy=3\,\text{g} + . +

    + +

    + This is all very straightforward; + note that all we really did was find the area of the lamina and multiply it by the constant density of + 3. +

    +     +     + + + Finding the mass of a lamina with variable density + +

    + Find the mass of a square lamina, + represented by the unit square with lower lefthand corner at the origin + (see ), + with variable density \delta(x,y) = (x+y+2)\,\text{g/cm}^2. +

    +
    + Graphing the density functions in and + + + + Two planes intersect along a line in three dimensions. + +

    + Two planes are plotted against a set of three-dimensional coordinate axes. + Both planes are drawn sitting about the xy plane, + and they intersect along a line. +

    + +

    + The line of intersection is not important: + each plane is illustrating the graph of a different density function. + One is the constant function \delta(x,y)=3, + while the other is the linear function \delta(x,y)=x+y+2. +

    +     + + + + import labelpath3; + //ASY file for figmass2.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(2.9,-4.6,5.2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={2,4}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,1.25); + pair ybounds=(-0.25,1.25); + pair zbounds=(0,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane z=3 with thick line + triple f(pair t) { + return (t.x,t.y,3); + } + surface s=surface(f,(0,0),(1,1),4,4,Spline); + pen q=redcurvepen; + draw(s,surfacepen2,meshpen=q); + draw((0,0,3)--(1,0,3)--(1,1,3)--(0,1,3)--cycle,redpen+linewidth(1)); + label("$z=3$",(1,0.5,3),S); + path3 gg=(1,0,2.8)--(1,1,2.8); + //draw(labelpath("$z=3$",subpath(gg,.35,.65),angle=-90)); + + //draw plane z=x+y+2 + triple f(pair t) { + return (t.x,t.y,t.x+t.y+2); + } + surface s=surface(f,(0,0),(1,1),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + draw((0,0,2)--(1,0,3)--(1,1,4)--(0,1,3)--cycle,bluepen+linewidth(1)); + label("$z=x+y+2$",(.5,0,2.5),S); + + draw((1,0,3)--(0,1,3),black+dashed+linewidth(1)); + + + + +
    +     + +

    + The variable density \delta, + in this example, is very uniform, + giving a density of 3 in the center of the square and changing linearly. + A graph of \delta(x,y) can be seen in ; + notice how same amount of density is above z=3 as below. + We'll comment on the significance of this momentarily. +

    + +

    + The mass M is found by integrating \delta(x,y) over R. + The order of integration is not important; + we choose dx\, dy arbitrarily. + Thus: + + M = \iint_R(x+y+2)\, dA \amp = \int_0^1\int_0^1 (x+y+2)\, dx\, dy + \amp = \int_0^1\left.\left(\frac 12x^2+x(y+2)\right)\right|_0^1dy + \amp = \int_0^1 \left(\frac52+y\right)\, dy + \amp = \left.\left(\frac52y+\frac12y^2\right)\right|_0^1 + \amp = 3\,\text{g} + . +

    + +

    + It turns out that since the density of the lamina is so uniformly distributed above and below + z=3 that the mass of the lamina is the same as if it had a constant density of 3. + The density functions in + and + are graphed in , + which illustrates this concept. +

    +     +     + + + Finding the weight of a lamina with variable density + +

    + Find the weight of the lamina represented by the disk with radius + 2, + centered at the origin, + with density function \delta(x,y) = (x^2+y^2+1)\,\text{lb/ft}^2. + Compare this to the weight of the lamina with the same shape and density + \delta(x,y) = (2\sqrt{x^2+y^2}+1)\,\text{lb/ft}^2. +

    +     + +

    + A direct application of + states that the weight of the lamina is \iint_R\delta(x,y)\, dA. + Since our lamina is in the shape of a circle, + it makes sense to approach the double integral using polar coordinates. +

    + +

    + The density function \delta(x,y) = x^2+y^2+1 becomes + + \delta(r,\theta) = (r\cos(\theta) )^2+(r\sin(\theta) )^2+1 = r^2+1 + . + The circle is bounded by 0\leq r\leq 2 and 0\leq\theta\leq2\pi. + Thus the weight W is: + + W \amp = \int_0^{2\pi}\int_0^2 (r^2+1)r\, dr\, d\theta + \amp = \int_0^{2\pi} \left.\left(\frac14r^4+\frac12r^2\right)\right|_0^2d\theta + \amp = \int_0^{2\pi} \left(6\right) d\theta + \amp = 12\pi \approx 37.70\text{ lb } + . +

    + +

    + Now compare this with the density function \delta(x,y) = 2\sqrt{x^2+y^2}+1. + Converting this to polar coordinates gives + + \delta(r,\theta) = 2\sqrt{(r\cos(\theta) )^2+(r\sin(\theta) )^2}+1 = 2r+1 + . + Thus the weight W is: + + W \amp = \int_0^{2\pi}\int_0^2 (2r+1)r\, dr\, d\theta + \amp = \int_0^{2\pi} (\frac23r^3+\frac12r^2)\Big|_0^2d\theta + \amp = \int_0^{2\pi} \left(\frac{22}3\right)\, d\theta + \amp = \frac{44}3\pi \approx 46.08\text{ lb } + . +

    + +

    + One would expect different density functions to return different weights, + as we have here. + The density functions were chosen, though, to be similar: + each gives a density of 1 at the origin and a density of 5 at the outside edge of the circle, + as seen in . +

    + +
    + Graphing the density functions in . In (a) is the density function \delta(x,y) = x^2+y^2+1; in (b) is \delta(x,y) = 2\sqrt{x^2+y^2}+1 + +
    + + + + + A circular paraboloid, opening upward, plotted over a circular domain. + +

    + The graph of \delta(x,y)=x^2+y^2+1 is a circular paraboloid, + opening upward, with vertex at (0,0,1). + A circle of radius 2, centered at the origin, is also plotted in the xy plane; + this is the boundary of the domain for the graph. +

    +     + + + + + //ASY file for figmass3a3D.asy in Chapter 13 + + size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={5}; + defaultpen(0.5mm); + + pair xbounds=(-2.75,2.75); + pair ybounds=(-2.75,2.75); + pair zbounds=(0,6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane z=3 with thick line + triple f(pair t) { + return (cos(t.x)*t.y,sin(t.x)*t.y,t.y^2+1); + } + surface s=surface(f,(0,0),(2*pi,2),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //draw circle in plane + triple g(real t) {return (2*cos(t),2*sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + + + +
    + +
    + + + + + A circular cone, opening upward, plotted over a circular domain. + +

    + The graph of \delta(x,y)=2\sqrt{x^2+y^2}+1 is a circular cone, + opening upward, with vertex at (0,0,1). + A circle of radius 2, centered at the origin, is also plotted in the xy plane; + this is the boundary of the domain for the graph. +

    +     + + + + + //ASY file for figmass3b3D.asy in Chapter 13 + + size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={5}; + defaultpen(0.5mm); + + pair xbounds=(-2.75,2.75); + pair ybounds=(-2.75,2.75); + pair zbounds=(0,6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane z=3 with thick line + triple f(pair t) { + return (cos(t.x)*t.y,sin(t.x)*t.y,2*t.y+1); + } + surface s=surface(f,(0,0),(2*pi,2),16,16,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //draw circle in plane + triple g(real t) {return (2*cos(t),2*sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + + + +
    +     +
    + +

    + Notice how x^2+y^2+1 \leq 2\sqrt{x^2+y^2}+1 over the circle; + this results in less weight. +

    +     +     + +

    + Plotting the density functions can be useful as our understanding of mass can be related to our understanding of + volume under a surface. + We interpreted \iint_R f(x,y)\, dA as giving the volume under f over R; + we can understand \iint_R\delta(x,y)\, dA in the same way. + The volume under \delta over R is actually mass; + by compressing the volume + under \delta onto the xy-plane, + we get more mass in some areas than others , areas of greater density. +

    + +

    + Knowing the mass of a lamina is one of several important measures. + Another is the center of mass, which we discuss next. +

    +     + + + Center of Mass +

    + Consider a disk of radius 1 with uniform density. + It is common knowledge that the disk will balance on a point if the point is placed at the center of the disk. + What if the disk does not have a uniform density? + Through trial-and-error, + we should still be able to find a spot on the disk at which the disk will balance on a point. + This balance point is referred to as the + center of mass, + or center of gravity. + It is though all the mass is centered there. + In fact, if the disk has a mass of 3, + the disk will behave physically as though it were a point-mass of + 3 located at its center of mass. + For instance, + the disk will naturally spin with an axis through its center of mass + (which is why it is important to + balance the tires of your car: + if they are out of balance, + their center of mass will be outside of the axle and it will shake terribly). + center of mass + masscenter of +

    + +

    + We find the center of mass based on the principle of a + weighted average. + Consider a college class in which your homework average is 90%, + your test average is 73%, and your final exam grade is an 85%. + Experience tells us that our final grade + is not the average + of these three grades: that is, it is not: + + \frac{0.9+0.73+0.85}{3} \approx 0.837 = 83.7% + . +

    + +

    + That is, you are probably not pulling a B in the course. + Rather, your grades are weighted. + Let's say the homework is worth 10% of the grade, + tests are 60% and the exam is 30%. + Then your final grade is: + + (0.1)(0.9) + (0.6)(0.73)+(0.3)(0.85) = 0.783 = 78.3% + . +

    + +

    + Each grade is multiplied by a weight. +

    + +

    + In general, given values x_1,x_2,\ldots,x_n and weights w_1,w_2,\ldots,w_n, + the weighted average of the n values is + + \sum_{i=1}^n w_ix_i\Bigg/\sum_{i=1}^n w_i + . +

    + +

    + In the grading example above, + the sum of the weights 0.1, 0.6 and 0.3 is 1, so we don't see the division by the sum of weights in that instance. +

    + +

    + How this relates to center of mass is given in the following theorem. +

    + + + Center of Mass of Discrete Linear System + +

    + Let point masses m_1,m_2,\ldots,m_n be distributed along the x-axis at locations x_1,x_2,\ldots,x_n, + respectively. + The center of mass \overline{x} of the system is located at + center of mass + + \overline{x} = \sum_{i=1}^nm_ix_i\Bigg/\sum_{i=1}^n m_i + . +

    +     +     + + + Finding the center of mass of a discrete linear system + +

    +

      +
    1. +

      + Point masses of 2 are located at x=-1, + x=2 and x=3 are connected by a thin rod of negligible weight. + Find the center of mass of the system. +

      +
      2. + +
    2. +

      + Point masses of 10, + 2 and 1 are located at x=-1, + x=2 and x=3, + respectively, are connected by a thin rod of negligible weight. + Find the center of mass of the system. +

      +
      3. +
    +

    +     + +

    +

      +
    1. +

      + Following , + we compute the center of mass as: + + \overline{x}=\frac{2(-1) + 2(2)+2(3)}{2+2+2} = \frac43 = 1.\overline{3} + . + So the system would balance on a point placed at x=4/3, + as illustrated in . +

      + +
      + Illustrating point masses along a thin rod and the center of mass + +
      + + + + A number line with three points marked at -1, 2, and 3, and a marking for the center of mass. + +

      + A number line is marked as the x axis. + There are markings for x values from x=-1 to x=3. + Solid dots are placed at x=-1, x=2, and x=3 to illustrate the presence of point masses. +

      + +

      + A small triangle marks the point 4/3, the location of the center of mass, + which is also labeled as \overline{x}. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + axis y line=none, + xtick={-1,0,1,2,3}, + ymin=-.5,ymax=.5, + xmin=-1.5,xmax=3.5 + ] + + \filldraw (axis cs:-1,0) circle (2pt) + (axis cs: 2,0) circle (2pt) + (axis cs: 3,0) circle (2pt) + (axis cs: 1.3,0) node [above] {$\overline{x}$} -- (axis cs: 1.2,-.05) -- (axis cs: 1.4,-.05) -- cycle; + + \draw [very thick] (axis cs: -1,0) -- (axis cs:3,0); + + \end{axis} + + \end{tikzpicture} + + + + +
      + +
      + + + + A number line with three points marked at -1, 2, and 3, and a marking for the center of mass. + +

      + A number line is marked as the x axis. + There are markings for x values from x=-1 to x=3. + Solid dots are placed at x=-1, x=2, and x=3 to illustrate the presence of point masses. + The dots have different sizes to represent the fact that the correspond to different masses. +

      + +

      + A small triangle just to the left of 0 marks the location of the center of mass, + which is also labeled as \overline{x}. +

      +       + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + axis y line=none, + xtick={-1,0,1,2,3}, + ymin=-.5,ymax=.5, + xmin=-1.5,xmax=3.5 + ] + + \filldraw (axis cs:-1,0) circle (4.5pt) + (axis cs: 2,0) circle (2pt) + (axis cs: 3,0) circle (1.4pt) + (axis cs: -.23,0) node [above] {$\overline{x}$} -- (axis cs: -.33,-.05) -- (axis cs: -.13,-.05) -- cycle; + + \draw [very thick] (axis cs: -1,0) -- (axis cs:3,0); + + \end{axis} + + \end{tikzpicture} + + + + +
      +       + +
      +
      2. + +
    2. +

      + Again following , we find: + + \overline{x} = \frac{10(-1)+2(2)+1(3)}{10+2+1} = \frac{-3}{13} \approx -0.23 + . + Placing a large weight at the left hand side of the system moves the center of mass left, + as shown in . +

      +
      3. +
    +

    +     +     + +

    + In a discrete system (, mass is located at individual points, + not along a continuum) we find the center of mass by dividing the mass into a + moment of the system. + In general, a moment is a weighted measure of distance from a particular point or line. + In the case described by , + we are finding a weighted measure of distances from the y-axis, + so we refer to this as the moment about the y-axis, + represented by M_y. + Letting M be the total mass of the system, + we have \overline{x} = M_y/M. +

    + +

    + We can extend the concept of the center of mass of discrete points along a line to the center of mass of discrete points in the plane rather easily. + To do so, we define some terms then give a theorem. +

    + + + Moments about the <m>x</m> and <m>y</m> Axes + +

    + Let point masses m_1, + m_2,\ldots,m_n be located at points + + (x_1,y_1), (x_2,y_2), \ldots, (x_n,y_n) + , + respectively, in the xy-plane. + moment +

    + +

    +

      +
    1. +

      + The moment about the y-axis, M_y, is + \ds M_y = \sum_{i=1}^n m_ix_i. +

      +
      2. + +
    2. +

      + The moment about the x-axis, M_x, is + \ds M_x = \sum_{i=1}^n m_iy_i. +

      +
      3. +
    +

    +     +     + +

    + One can think that these definitions are backwards + as M_y sums up x distances. + But remember, + x distances are measurements of distance from the y-axis, + hence defining the moment about the y-axis. +

    + +

    + We now define the center of mass of discrete points in the plane. +

    + + + Center of Mass of Discrete Planar System + +

    + Let point masses m_1, + m_2,\ldots,m_n be located at points + + (x_1,y_1), (x_2,y_2),\ldots,(x_n,y_n) + , + respectively, in the xy-plane, + and let \ds M = \sum_{i=1}^n m_i. + center of mass +

    + +

    + The center of mass of the system is at (\overline{x},\overline{y}), where + + \overline{x}= \frac{M_y}{M} \text{ and } \overline{y} = \frac{M_x}{M} + . +

    +     +     + + + Finding the center of mass of a discrete planar system + +

    + Let point masses of 1, + 2 + and 5 be located at points (2,0), + (1,1) and (3,1), respectively, + and are connected by thin rods of negligible weight. + Find the center of mass of the system. +

    +     + +

    + We follow + and to find M, + M_x and M_y: +

    + +

    + M = 1+2+5 = 8\,\text{kg}. +

    + + +

    + + M_x \amp = \sum_{i=1}^n m_iy_i + \amp = 1(0) + 2(1) + 5(1) + \amp = 7 + . +

    +

    + + M_y \amp = \sum_{i=1}^n m_ix_i + \amp = 1(2) + 2(1) + 5(3) + \amp = 19 + . +

    +     + +
    + Illustrating the center of mass of a discrete planar system in + + + A triangle in the plane. The vertices are marked with dots of different sizes, and an interior point marks the center of mass. + +

    + The first quadrant in the plane is shown, with x and y coordinate axes. + A triangle is plotted, with vertices at (1,1), (3,1), and (2,0). + Each vertex is marked with a dot, and the size of each dot corresponds to the mass it represents. +

    + +

    + The center of mass is also marked, as a point in the interior of the triangle, + closest to the vertex (3,1), where the largest of the three masses is located. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + xtick={1,2,3}, + ytick={1,2}, + ymin=-.5,ymax=1.5, + xmin=-.5,xmax=3.5 + ] + + \filldraw (axis cs:2,0) circle (1.4pt) + (axis cs: 1,1) circle (2pt) + (axis cs: 3,1) circle (3.16pt) + (axis cs: 2.375,0.875) circle (1pt) node [below] { $(\overline{x},\overline{y})$}; + + + \draw [very thick] (axis cs: 2,0) -- (axis cs:3,1) -- (axis cs:1,1) -- cycle; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Thus the center of mass is \ds (\overline{x},\overline{y}) = \left(\frac{M_y}{M},\frac{M_x}M\right) = \left(\frac{{19}}8,\frac78\right) =(2.375,0.875), + illustrated in . +

    +     +     + +

    + We finally arrive at our true goal of this section: + finding the center of mass of a lamina with variable density. + While the above measurement of center of mass is interesting, + it does not directly answer more realistic situations where we need to find the center of mass of a contiguous region. + However, understanding the discrete case allows us to approximate the center of mass of a planar lamina; + using calculus, we can refine the approximation to an exact value. +

    + +

    + We begin by representing a planar lamina with a region R in the xy-plane with density function \delta(x,y). + Partition R into n subdivisions, + each with area \Delta A_i. + As done before, we can approximate the mass of the + ith subregion with \delta(x_i,y_i)\Delta A_i, + where (x_i,y_i) is a point inside the ith subregion. + We can approximate the moment of this subregion about the y-axis with x_i\delta(x_i,y_i)\Delta A_i that is, + by multiplying the approximate mass of the region by its approximate distance from the y-axis. + Similarly, we can approximate the moment about the x-axis with y_i\delta(x_i,y_i)\Delta A_i. + By summing over all subregions, we have: + + \text{ mass: } M \amp \approx \sum_{i=1}^n \delta(x_i,y_i)\Delta A_i \text{ (as seen before) } + \text{ moment about the \(x\)-axis: } M_x \amp \approx \sum_{i=1}^n y_i\delta(x_i,y_i)\Delta A_i + \text{ moment about the \(y\)-axis: } M_y \amp \approx \sum_{i=1}^n x_i\delta(x_i,y_i)\Delta A_i + +

    + +

    + By taking limits, + where size of each subregion shrinks to 0 in both the x and y directions, + we arrive at the double integrals given in the following theorem. +

    + + + Center of Mass of a Planar Lamina, Moments + +

    + Let a planar lamina be represented by a closed, + bounded region R in the xy-plane with density function \delta(x,y). + center of mass + moment +

    + +

    +

      +
    1. +

      + \ds \text{ mass: } M = \iint_R\delta(x,y)\, dA +

      +
      2. + +
    2. +

      + \ds \text{ moment about the \(x\)-axis: } M_x = \iint_Ry\delta(x,y)\, dA +

      +
      3. + +
    3. +

      + \ds \text{ moment about the \(y\)-axis: } M_y = \iint_Rx\delta(x,y)\, dA +

      +
      4. + +
    4. +

      + The center of mass of the lamina is + + (\overline{x},\overline{y}) = \left(\frac{M_y}{M},\frac{M_x}M\right) + . +

      +
      5. +
    +

    +     +     + +

    + We start our practice of finding centers of mass by revisiting some of the lamina used previously in this section when finding mass. + We will just set up the integrals needed to compute M, + M_x and M_y and leave the details of the integration to the reader. +

    + + + Finding the center of mass of a lamina + +

    + Find the center mass of a square lamina, + with side length 1, with a density of \delta = 3\,\text{g/cm}^2. + (Note: this is the lamina from .) +

    +     + +

    + We represent the lamina with a square region in the plane as shown in as done previously. +

    + +
    + A region R representing a lamina in + + + The unit square in the first quadrant of the plane. + +

    + The unit square, given by 0\leq x\leq 1 and 0\leq y\leq 1, + plotted in the first quadrant, relative to x and y coordinate axes. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-.1,ymax=1.1, + xmin=-.1,xmax=1.1 + ] + + \filldraw [fill=firstcolor!15,draw=firstcolor,thick] (axis cs:0,0) -- (axis cs:1,0) -- (axis cs: 1,1) -- (axis cs:0,1) -- cycle; + + \end{axis} + + \end{tikzpicture} + + + + +
    + +

    + Following , + we find M, M_x and M_y: + + M \amp = \iint_R 3\, dA = \int_0^1\int_0^1 3\, dx\, dy =3\,\text{g} + M_x \amp = \iint_R 3y\, dA = \int_0^1\int_0^1 3y\, dx\, dy =3/2 = 1.5 + M_y \amp = \iint_R 3x\, dA = \int_0^1\int_0^1 3x\, dx\, dy =3/2 = 1.5 + . +

    + +

    + Thus the center of mass is \ds (\overline{x},\overline{y}) = \left(\frac{M_y}M,\frac{M_x}M\right) = (1.5/3,1.5/3) = (0.5,0.5). + This is what we should have expected: + the center of mass of a square with constant density is the center of the square. +

    +     +     + + + Finding the center of mass of a lamina + +

    + Find the center of mass of a square lamina, + represented by the unit square with lower lefthand corner at the origin + (see ), + with variable density \delta(x,y) = (x+y+2)\,\text{g/cm}^2. + (Note: this is the lamina from .) +

    +     + +

    + We follow , + to find M, M_x and M_y: + + M \amp = \iint_R (x+y+2)\, dA = \int_0^1\int_0^1 (x+y+2)\, dx\, dy =3\,\text{g} + M_x \amp = \iint_R y(x+y+2)\, dA = \int_0^1\int_0^1 y(x+y+2)\, dx\, dy =\frac{19}{12} + M_y \amp = \iint_R x(x+y+2)\, dA = \int_0^1\int_0^1 x(x+y+2)\, dx\, dy =\frac{19}{12} + . +

    + +

    + Thus the center of mass is + + (\overline{x},\overline{y}) = \left(\frac{M_y}M,\frac{M_x}M\right) = \left(\frac{19}{36},\frac{19}{36}\right) \approx (0.528,0.528) + . + While the mass of this lamina is the same as the lamina in the previous example, + the greater density found with greater x and y values pulls the center of mass from the center slightly towards the upper righthand corner. +

    +     +     + + + Finding the center of mass of a lamina + +

    + Find the center of mass of the lamina represented by the circle with radius 2, + centered at the origin, + with density function \delta(x,y) = (x^2+y^2+1)\,\text{lb/ft}^2. + (Note: this is one of the lamina used in .) +

    +     + +

    + As done in , + it is best to describe R using polar coordinates. + Thus when we compute M_y, + we will integrate not x\delta(x,y) = x(x^2+y^2+1), + but rather \big(r\cos(\theta) \big)\delta(r\cos(\theta) ,r\sin(\theta) ) = \big(r\cos(\theta) \big)\big(r^2+1\big). + We compute M, M_x and M_y: + + M \amp = \int_0^{2\pi}\int_0^2 (r^2+1)r\, dr\, d\theta = 12\pi\approx 37.7\,\text{lb} + M_x \amp = \int_0^{2\pi}\int_0^2 (r\sin(\theta) )(r^2+1)r \, dr\, d\theta = 0 + M_y \amp = \int_0^{2\pi}\int_0^2 (r\cos(\theta) )(r^2+1)r \, dr\, d\theta = 0 + . +

    + +

    + Since R and the density of R are both symmetric about the x and y axes, + it should come as no big surprise that the moments about each axis is 0. + Thus the center of mass is (\overline{x},\overline{y})=(0,0). +

    +     +     + + + Finding the center of mass of a lamina + +

    + Find the center of mass of the lamina represented by the region R shown in , + half an annulus with outer radius 6 + and inner radius 5, + with constant density 2. +

    +
    + Illustrating the region R in + + + An annular region, between semi-circles of radii 5 and 6 in the upper half-plane. + +

    + A sketch of the region occupied by the lamina for this example. + It is bounded between two semi-circles and the x axis. + The semi-circles have radii 5 and 6, and lie above the x axis. +

    + +

    + The overall shape is that of a relatively thin arch. + The center of mass is also marked in the diagram. + What is interesting in this example is that the center of mass does not lie within the region of the lamina. + Instead, it is located on the y axis, below the region. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis on top, + ymin=-2,ymax=8.8, + xmin=-6.5,xmax=6.5 + ] + + \addplot [firstcurvestyle,areastyle] coordinates {(6.,0)(5.967,0.6272)(5.869,1.247)(5.706,1.854)(5.481,2.44)(5.196,3.)(4.854,3.527)(4.459,4.015)(4.015,4.459)(3.527,4.854)(3.,5.196)(2.44,5.481)(1.854,5.706)(1.247,5.869)(0.6272,5.967)(0,6.)(-0.6272,5.967)(-1.247,5.869)(-1.854,5.706)(-2.44,5.481)(-3.,5.196)(-3.527,4.854)(-4.015,4.459)(-4.459,4.015)(-4.854,3.527)(-5.196,3.)(-5.481,2.44)(-5.706,1.854)(-5.869,1.247)(-5.967,0.6272)(-6.,0)(-5.,0)(-4.973,0.5226)(-4.891,1.04)(-4.755,1.545)(-4.568,2.034)(-4.33,2.5)(-4.045,2.939)(-3.716,3.346)(-3.346,3.716)(-2.939,4.045)(-2.5,4.33)(-2.034,4.568)(-1.545,4.755)(-1.04,4.891)(-0.5226,4.973)(0,5.)(0.5226,4.973)(1.04,4.891)(1.545,4.755)(2.034,4.568)(2.5,4.33)(2.939,4.045)(3.346,3.716)(3.716,3.346)(4.045,2.939)(4.33,2.5)(4.568,2.034)(4.755,1.545)(4.891,1.04)(4.973,0.5226)(5.,0)(6,0)}; + \addplot [firstcurvestyle] coordinates {(6.,0)(5.967,0.6272)(5.869,1.247)(5.706,1.854)(5.481,2.44)(5.196,3.)(4.854,3.527)(4.459,4.015)(4.015,4.459)(3.527,4.854)(3.,5.196)(2.44,5.481)(1.854,5.706)(1.247,5.869)(0.6272,5.967)(0,6.)(-0.6272,5.967)(-1.247,5.869)(-1.854,5.706)(-2.44,5.481)(-3.,5.196)(-3.527,4.854)(-4.015,4.459)(-4.459,4.015)(-4.854,3.527)(-5.196,3.)(-5.481,2.44)(-5.706,1.854)(-5.869,1.247)(-5.967,0.6272)(-6.,0)(-5.,0)(-4.973,0.5226)(-4.891,1.04)(-4.755,1.545)(-4.568,2.034)(-4.33,2.5)(-4.045,2.939)(-3.716,3.346)(-3.346,3.716)(-2.939,4.045)(-2.5,4.33)(-2.034,4.568)(-1.545,4.755)(-1.04,4.891)(-0.5226,4.973)(0,5.)(0.5226,4.973)(1.04,4.891)(1.545,4.755)(2.034,4.568)(2.5,4.33)(2.939,4.045)(3.346,3.716)(3.716,3.346)(4.045,2.939)(4.33,2.5)(4.568,2.034)(4.755,1.545)(4.891,1.04)(4.973,0.5226)(5.,0)(6,0)}; + + \filldraw (axis cs:0,3.51) circle (2.4pt) node [below right] { $(\overline{x},\overline{y})$}; + + \end{axis} + + \end{tikzpicture} + + + + +
    +     + +

    + Once again it will be useful to represent R in polar coordinates. + Using the description of R and/or the illustration, + we see that R is bounded by + 5\leq r\leq 6 and 0\leq\theta\leq\pi. + As the lamina is symmetric about the y-axis, + we should expect M_y=0. + We compute M, M_x and M_y: + + M \amp = \int_0^{\pi}\int_5^6 (2)r\, dr\, d\theta = 11\pi\,\text{lb} + M_x \amp = \int_0^{\pi}\int_5^6 (r\sin(\theta) )(2)r\, dr\, d\theta = \frac{364}3\approx 121.33 + M_y \amp = \int_0^{\pi}\int_5^6 (r\cos(\theta) )(2)r\, dr\, d\theta = 0 + . +

    + +

    + Thus the center of mass is (\overline{x},\overline{y}) = \left(0,\frac{364}{33\pi}\right) \approx (0,3.51). + The center of mass is indicated in ; + note how it lies outside of R! +

    +     +     + +

    + This section has shown us another use for iterated integrals beyond finding area or signed volume under the curve. + While there are many uses for iterated integrals, + we give one more application in the following section: + computing surface area. +

    +     + + + + Terms and Concepts + + + + +

    + Why is it easy to use mass + and weight interchangeably, + even though they are different measures? +

    +     + + + +

    + Because they are scalar multiples of each other. +

    +     + +     + + + + +

    + Given a point (x,y), + the value of x is a measure of distance from the -axis. +

    +     + + + + + + + + +     + + + + +

    + We can think of \iint_R\, dm as meaning + sum up lots of +

    +     + + + + little masses|small masses + + + + +     + + + + +

    + What is a discrete planar system? +

    +     + + + +

    + A collection of individual masses in the plane. + Each mass is a point mass, + , mass located at a point, not across a region. +

    +     + +     + + + + +

    + Why does M_x use \iint_R y\delta(x,y)\, dA instead of \iint_R x\delta(x,y)\, dA; + that is, why do we use y and not x? +

    +     + + + +

    + M_x measures the moment about the x-axis, + meaning we need to measure distance from the x-axis. + Such measurements are measures in the y-direction. +

    +     + +     + + + + +

    + Describe a situation where the center of mass of a lamina does not lie within the region of the lamina itself. +

    +     + + + +

    + If the lamina is an annulus, + the center of mass will likely be in the middle, outside of the region. + (See .) +

    +     + +     +     + + + Problems + + + +

    + Point masses are given along a line or in the plane. + Find the center of mass \overline{x} or (\overline{x},\overline{y}), + as appropriate. + (All masses are in grams and distances are in cm.) +

    +     + + + + +

    + m_1=4 at x=1; m_2=3 at x=3; + m_3=5 at x=10 +

    +     + +

    + \overline{x}=5.25 +

    +     + +     + + + + +

    + m_1=2 at x=-3; + m_2=2 at x=-1; +

    + +

    + m_3=3 at x=0;m_4=3 at x=7 +

    +     + +

    + \overline{x}=1.3 +

    +     + +     + + + + +

    + m_1=2 at (-2,-2); + m_2=2 at (2,-2); +

    + +

    + m_3=20 at (0,4) +

    +     + +

    + (\overline{x},\overline{y}) = (0,3) +

    +     + +     + + + + +

    + m_1=1 at (-1,-1); + m_2=2 at (-1,1); +

    + +

    + m_3=2 at (1,1);m_4=1 at (1,-1) +

    +     + +

    + (\overline{x},\overline{y}) = (0,1/3) +

    +     + +     + +     + + + +

    + Find the mass/weight of the lamina described by the region R in the plane and its density function \delta(x,y). +

    +     + + + + +

    + R is the rectangle with corners (1,-3), (1,2), + (7,2) and (7,-3); + \delta(x,y) = 5\,\text{g/cm}^2 +

    +     + +

    + M = 150\,text{g}; +

    +     + +     + + + + +

    + R is the rectangle with corners (1,-3), (1,2), + (7,2) and (7,-3); + \delta(x,y) = (x+y^2)\,\text{g/cm}^2 +

    +     + +

    + M = 190\,\text{g} +

    +     + +     + + + + +

    + R is the triangle with corners (-1,0), + (1,0), and (0,1); + \delta(x,y) = 2\,\text{lb/in}^2 +

    +     + +

    + M = 2\,text{lb} +

    +     + +     + + + + +

    + R is the triangle with corners (0,0), + (1,0), and (0,1); + \delta(x,y) = (x^2+y^2+1)\,\text{lb/in}^2 +

    +     + +

    + M = 2/3\,\text{lb} +

    +     + +     + + + + +

    + R is the disk centered at the origin with radius 2; + \delta(x,y) = (x+y+4)\,\text{kg/m}^2 +

    +     + +

    + M = 16\pi\approx 50.27\,\text{kg} +

    +     + +     + + + + +

    + R is the circle sector bounded by + x^2+y^2=25 in the first quadrant; + \delta(x,y) = (\sqrt{x^2+y^2}+1)\,\text{kg/m}^2 +

    +     + +

    + M = 325\pi/12\approx 85\,\text{kg} +

    +     + +     + + + + +

    + R is the annulus in the first and second quadrants bounded by + x^2+y^2=9 and x^2+y^2=36; + \delta(x,y) = 4\,\text{lb/ft}^2 +

    +     + +

    + M = 54\pi\approx 169.65\,\text{lb} +

    +     + +     + + + + +

    + R is the annulus in the first and second quadrants bounded by + x^2+y^2=9 and x^2+y^2=36; + \delta(x,y) = \sqrt{x^2+y^2}\,\text{lb/ft}^2 +

    +     + +

    + M = 63\pi\approx 197.92\,\text{lb} +

    +     + +     + +     + + + +

    + Find the center of mass of the lamina described by the region R in the plane and its density function \delta(x,y). +

    + +

    + Note: these are the same lamina as in + . +

    +     + + + + +

    + R is the rectangle with corners (1,-3), (1,2), + (7,2) and (7,-3); + \delta(x,y) = 5\,\text{g/cm}^2 +

    +     + +

    + M = 150\,\text{g}; M_y=600; M_x=-75; + (\overline{x},\overline{y}) = (4,-0.5) +

    +     + +     + + + + +

    + R is the rectangle with corners (1,-3), (1,2), + (7,2) and (7,-3); + \delta(x,y) = (x+y^2)\,\text{g/cm}^2 +

    +     + +

    + M = 190\,\text{g}; M_y= 850; M_x = -315/2; + (\overline{x},\overline{y}) = (4.47,-0.83) +

    +     + +     + + + + +

    + R is the triangle with corners (-1,0), + (1,0), and (0,1); + \delta(x,y) = 2\,\text{lb/in}^2 +

    +     + +

    + M = 2\,\text{lb}; M_y= 0; M_x = 2/3; + (\overline{x},\overline{y}) = (0,1/3) +

    +     + +     + + + + +

    + R is the triangle with corners (0,0), + (1,0), and (0,1); + \delta(x,y) = (x^2+y^2+1)\,\text{lb/in}^2 +

    +     + +

    + M = 2/3\,\text{lb}; M_y= 7/30; M_x = 7/30; + (\overline{x},\overline{y}) = (0.35,0.35) +

    +     + +     + + + + +

    + R is the disk centered at the origin with radius 2; + \delta(x,y) = (x+y+4)\,\text{kg/m}^2 +

    +     + +

    + M = 16\pi\approx 50.27\,\text{kg}; + M_y= 4\pi; M_x = 4\pi; + (\overline{x},\overline{y}) = (1/4,1/4) +

    +     + +     + + + + +

    + R is the circle sector bounded by + x^2+y^2=25 in the first quadrant; + \delta(x,y) = (\sqrt{x^2+y^2}+1)\,\text{kg/m}^2 +

    +     + +

    + M = 325\pi/12\approx 85\,\text{kg}; M_y= 2375/12; + M_x = 2375/12; + (\overline{x},\overline{y}) = (2.33,2.33) +

    +     + +     + + + + +

    + R is the annulus in the first and second quadrants bounded by + x^2+y^2=9 and x^2+y^2=36; + \delta(x,y) = 4\,\text{lb/ft}^2 +

    +     + +

    + M = 54\pi\approx 169.65\,\text{lb}; + M_y= 0; M_x = 504; + (\overline{x},\overline{y}) = (0,2.97) +

    +     + +     + + + + +

    + R is the annulus in the first and second quadrants bounded by + x^2+y^2=9 and x^2+y^2=36; + \delta(x,y) = \sqrt{x^2+y^2}\,\text{lb/ft}^2 +

    +     + +

    + M = 63\pi\approx 197.92\,\text{lb}; + M_y= 0; M_x = 1215/2; + (\overline{x},\overline{y}) = (0,3.07) +

    +     + +     + +     + + + +

    + The moment of inertia + I is a measure of the tendency of a lamina to resist rotating about an axis or continue to rotate about an axis. + I_x is the moment of inertia about the x-axis, + I_y is the moment of inertia about the y-axis, + and I_O is the moment of inertia about the origin. + These are computed as follows: +

    + +

    +

      +
    • I_x = \iint_R y^2\, dm
      • + +
    • I_y = \iint_R x^2\, dm
      • + +
    • I_O = \iint_R \big(x^2+y^2\big)\, dm
      • +
    +

    + +

    + A lamina corresponding to a planar region R is given with a mass of 16 units. + For each, compute I_x, + I_y and I_O. +

    +     + + + + +

    + R is the 4\times 4 square with corners at (-2,-2) and (2,2) with density \delta(x,y) = 1. +

    +     + +

    + I_x = 64/3; I_y = 64/3; I_O = 128/3 +

    +     + +     + + + + +

    + R is the 8\times 2 rectangle with corners at (-4,-1) and (4,1) with density \delta(x,y) = 1. +

    +     + +

    + I_x = 16/3; I_y = 256/3; I_O = 272/3 +

    +     + +     + + + + +

    + R is the 4\times 2 rectangle with corners at (-2,-1) and (2,1) with density \delta(x,y) = 2. +

    +     + +

    + I_x = 16/3; I_y = 64/3; I_O = 80/3 +

    +     + +     + + + + +

    + R is the disk with radius 2 centered at the origin with density \delta(x,y) = 4/\pi. +

    +     + +

    + I_x = 16; I_y = 16; I_O = 32 +

    +     + +     + +     +     +     +
    +
    + Surface Area +

    + In + we used definite integrals to compute the arc length of plane curves of the form y=f(x). + We later extended these ideas to compute the arc length of plane curves defined by parametric or polar equations. +

    + +

    + The natural extension of the concept of + arc length over an interval + to surfaces is surface area over a region. +

    + +

    + Consider the surface z=f(x,y) over a region R in the xy-plane, + shown in . + Because of the domed shape of the surface, + the surface area will be greater than that of the area of the region R. + We can find this area using the same basic technique we have used over and over: + we'll make an approximation, + then using limits, we'll refine the approximation to the exact value. +

    + +
    + Developing a method of computing surface area + +
    + + + + + A surface is plotted above a circular domain. A partition of the domain corresponds to a grid system on the surface. + +

    + In space, a surface is plotted against a set of three-dimensional coordinate axes. + The surface appears to be a downward-opening elliptic paraboloid. +

    + +

    + A domain is shown in the xy plane; it appears to be bounded by a circle of radius 1, + centered at (1,0,0). + The domain is divided into many small rectangles, illustrating a partition like the ones used to define the double integral. +

    + +

    + The grid of rectangles in the domain corresponds to a grid of curves on the surface. + One of the rectangles in the domain is highlighted, along with the corresponding region on the surface. +

    +     + + + + + //ASY file for figsurfacearea_intro13D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.5,3.1,3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={-1,1}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-1,2.5); + pair ybounds=(-1.25,1.25); + pair zbounds=(0,2.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface//{z=-.5*(x-1)^2-.5*(y)^2+2}; + triple f(pair t) { + return (t.x,t.y,-.5*(t.x-1)^2-.5*(t.y)^2+2); + } + surface s=surface(f,(-0.221,-1),(2.2,1),12,20,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + pen pp=linewidth(.25mm); + //draw the grid in the xy-plane + //along fixed x + draw((0,-.3,0) -- (0,.3,0),pp); + draw((.2,-.4,0) -- (.2,.4,0),pp); + draw((0.4,-.6,0) -- (.4,.6,0),pp); + draw((0.6,-.7,0) -- (.6,.7,0),pp); + draw((.8,-.8,0) -- (.8,.8,0),pp); + draw((1,-.8,0) -- (1,.8,0),pp); + draw((1.2,-.8,0) -- (1.2,.8,0),pp); + draw((1.4,-.8,0) -- (1.4,.8,0),pp); + draw((1.6,-.7,0) -- (1.6,.7,0),pp); + draw((1.8,-.6,0) -- (1.8,.6,0),pp); + draw((2,-.4,0) -- (2,.4,0),pp); + //along fixed y + draw((.8,.8,0) -- (1.4,.8,0),pp); + draw((.6,.7,0) -- (1.6,.7,0),pp); + draw((.4,.6,0) -- (1.8,.6,0),pp); + draw((.4,.5,0) -- (1.8,.5,0),pp); + draw((.2,.4,0) -- (2,.4,0),pp); + draw((0,.3,0) -- (2,.3,0),pp); + draw((0,.2,0) -- (2,.2,0),pp); + draw((0,.1,0) -- (2,.1,0),pp); + draw((.8,-.8,0) -- (1.4,-.8,0),pp); + draw((.6,-.7,0) -- (1.6,-.7,0),pp); + draw((.4,-.6,0) -- (1.8,-.6,0),pp); + draw((.4,-.5,0) -- (1.8,-.5,0),pp); + draw((.2,-.4,0) -- (2,-.4,0),pp); + draw((0,-.3,0) -- (2,-.3,0),pp); + draw((0,-.2,0) -- (2,-.2,0),pp); + draw((0,-.1,0) -- (2,-.1,0),pp); + + //Draw curve on top of the grid in xy plane ({cos(x)*(1+cos(2*x))},{sin(x)*(1+cos(2*x))},0); + triple g(real t) {return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),0);} + path3 mypath=graph(g,-pi/2,pi/2,operator ..); + draw(mypath,bluepen); + + // draw curve on surface + triple g(real t) { + return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),-.5*(cos(t)*(1+cos(2*t))-1)^2-.5*(sin(t)*(1+cos(2*t)))^2+2); + } + path3 mypath=graph(g,-pi/2,pi/2,operator ..); + draw(mypath,bluepen); + + //Draw the box in the plane and on the surface + draw((1.6,.3,1.775) -- (1.6,.4,1.74) -- (1.8,.4,1.6) -- (1.8,.3,1.635) --cycle,bluepen); + draw((1.6,.3,0) -- (1.6,.4,0) -- (1.8,.4,0) -- (1.8,.3,0) --cycle,bluepen); + + + + +
    + +
    + + + + + A zoomed-in view of a portion of the surface from the previous image, and one of the rectangular grid pieces. + +

    + This image is a zoomed-in view of the rectangular region that was highlighted on the surface in . + Below the surface, we see a single rectangle from the partition of the domain. + The sides of this rectangle are marked with lengths \Delta x_i and \Delta y_j. +

    + +

    + The rectangle in the xy plane corresponds to a small, rectangular patch on the surface. + This patch is approximated by a piece of the tangent plane at one point, + spanned by vectors \vec{u} and \vec{v}. + These vectors are labeled, but the point is not. +

    +     + + + + + //ASY file for figsurfacearea_intro23D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(3,3.5,4.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(0,1.25); + pair ybounds=(0,1.2); + pair zbounds=(0,2.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface//{-x^2+2}; + triple f(pair t) { + return (t.x,t.y,-t.x^2+2); + } + surface s=surface(f,(0,0),(1,1.2),12,12,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the box in the plane and on the surface + path3 pp=(.6,.4,0)--(.6,.6,0)--(.8,.6,0)--(.8,.4,0)--cycle; + draw(pp,bluepen); + label("$\Delta y_j$",(.9,.5,0));label("$\Delta x_i$",(.7,.7,0)); + path3 ppp=(.6,.4,2-.6^2)--(.6,.6,2-.6^2)--(.8,.6,2-.8^2)--(.8,.4,2-.8^2)--cycle; + draw(ppp,bluepen); + draw(surface(ppp),darksurfacepen,meshpen=p); + label("$\vec{v}$",(.9,.5,2-.85^2));label("$\vec{u}$",(.7,.7,2-.7^2)); + + + + +
    +     + +
    + +

    + As done to find the volume under a surface or the mass of a lamina, + we subdivide R into n subregions. + Here we subdivide R into rectangles, as shown in the figure. + One such subregion is outlined in the figure, + where the rectangle has dimensions \dx_i and \dy_j, + along with its corresponding region on the surface. +

    + +

    + In part of the figure, we zoom in on this portion of the surface. + When \dx_i and \dy_j are small, + the function is approximated well by the tangent plane at any point (x_i,y_j) in this subregion, + which is graphed in part . + In fact, the tangent plane approximates the function so well that in this figure, + it is virtually indistinguishable from the surface itself! + Therefore we can approximate the surface area S_{ij} of this region of the surface with the area + T_{ij} of the corresponding portion of the tangent plane. +

    + +

    + This portion of the tangent plane is a parallelogram, + defined by sides \vec u and \vec v, as shown. + One of the applications of the cross product from + is that the area of this parallelogram is \norm{\vec u\times \vec v}. + Once we can determine \vec u and \vec v, + we can determine the area. +

    + +

    + The vector \vec u is tangent to the surface in the direction of x, + therefore, from , + \vec u is parallel to \la 1,0,f_x(x_i,y_j)\ra. + The x-displacement of \vec u is \dx_i, + so we know that \vec u = \dx_i\la 1,0,f_x(x_i,y_j)\ra. + Similar logic shows that \vec v = \dy_j\la 0,1,f_y(x_i,y_j)\ra. + Thus: + + \text{surface area \(S_{ij}\)} \amp \approx \,\text{area of \(T_{ij}\)} + \amp = \norm{\vec u\times \vec v} + \amp = \norm{\dx_i\la 1,0,f_x(x_i,y_j)\ra\times\dy_j\la 0,1,f_y(x_i,y_j)\ra} + \amp =\sqrt{1+f_x(x_i,y_j)^2+f_y(x_i,y_j)^2}\dx_i\dy_j + . +

    + +

    + Note that \dx_i\dy_j = \Delta A_{ij}, + the area of the i,jth subregion. +

    + +

    + Summing up all n of the approximations to the surface area gives + + \text{surface area over \(R\)} \approx \sum_{i=1}^m\sum_{j=1}^n \sqrt{1+f_x(x_i,y_j)^2+f_y(x_i,y_j)^2}\Delta A_{ij} + . +

    + +

    + Once again take a limit as all of the \dx_i and \dy_j shrink to 0; + this leads to a double integral. +

    + + + Surface Area + +

    + Let z=f(x,y) where f_x and f_y are continuous over a closed, + bounded region R. + The surface area S over R is + surface area + + S \amp = \iint_R \, dS + \amp =\iint_R \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\, dA + . +

    +     +     + + +

    + We test this definition by using it to compute surface areas of known surfaces. + We start with a triangle. +

    + + + Finding the surface area of a plane over a triangle + +

    + Let f(x,y) = 4-x-2y, + and let R be the region in the plane bounded by x=0, + y=0 and y=2-x/2, + as shown in . + Find the surface area of f over R. +

    + +
    + Finding the area of a triangle in space in + + + + A triangle drawn on a plane in space. + +

    + Three-dimensional coordinate axes are drawn in space. + In the xy plane there is a triangle that illustrates the domain for this problem. + The triangle is bounded by the x and y axes, and the line from (4,0,0) to (0,2,0). +

    + +

    + This line also happens to be where the plane z=4-x-2y intersects the xy plane, + so it makes up one of the three sides of the triangle on the surface for this problem. + The other two sides are the intersections of z=4-x-2y with the xz and yz planes. +

    + +

    + Also shown is a line illustrating the altitude of the triangle, + as measured within the plane z=4-x-2y. +

    +     + + + + + //ASY file for figsurfacearea13D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(10.5,6.1,12); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={2,4}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,4.5); + pair ybounds=(-0.25,4.25); + pair zbounds=(0,5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane//{z=4-x-2y}; + triple f(pair t) { + return (t.x,t.y,4-t.x-2*t.y); + } + surface s=surface(f,(-0.25,-0.25),(4,2),4,4); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //draw the triangle on plane + draw((4,0,0)--(0,2,0)--(0,0,4)--(4,0,0),bluepen); + + //draw the triangle on xy plane + draw((4,0,0)--(0,2,0)--(0,0,0)--(4,0,0),bluepen+dashed+linewidth(2)); + + //draw red line on plane + //draw((0.8,1.6,0)--(0,0,4),redpen+linewidth(1)); + + + + +
    +     + +

    + We follow . + We start by noting that f_x(x,y) = -1 and f_y(x,y) = -2. + To define R, we use bounds + 0\leq y\leq 2-x/2 and 0\leq x\leq 4. + Therefore + + S \amp = \iint_R\, dS + \amp = \int_0^4\int_0^{2-x/2} \sqrt{1+(-1)^2+(-2)^2}\, dy\, dx + \amp = \int_0^4 \sqrt{6}\left(2-\frac x2\right)\, dx + \amp = 4\sqrt{6} + . +

    + +

    + Because the surface is a triangle, + we can figure out the area using geometry. + Considering the base of the triangle to be the side in the xy-plane, + we find the length of the base to be \sqrt{20}. + We can find the height using our knowledge of vectors: + let \vec u be the side in the xz-plane and let \vec v be the side in the xy-plane. + The height is then \norm {\vec u - \proj{u}{v}} = 4\sqrt{6/5}; + this is indicated by the additional line drawn in . + Geometry states that the area is thus + + \frac 12\cdot4\sqrt{6/5}\cdot\sqrt{20} = 4\sqrt{6} + . +

    + +

    + We affirm the validity of our formula. +

    +     +     + +

    + It is common knowledge that the surface area of a sphere of radius r is 4\pi r^2. + We confirm this in the following example, + which involves using our formula with polar coordinates. +

    + + + The surface area of a sphere + +

    + Find the surface area of the sphere with radius a centered at the origin, + whose top hemisphere has equation f(x,y)=\sqrt{a^2-x^2-y^2}. +

    +     + +

    + We start by computing partial derivatives and find + + f_x(x,y) = \frac{-x}{\sqrt{a^2-x^2-y^2}} \text{ and } f_y(x,y) = \frac{-y}{\sqrt{a^2-x^2-y^2}} + . +

    + +

    + As our function f only defines the top upper hemisphere of the sphere, + we double our surface area result to get the total area: + + S \amp = 2\iint_R \sqrt{1+ f_x(x,y)^2+f_y(x,y)^2}\, dA + \amp = 2\iint_R \sqrt{1+ \frac{x^2+y^2}{a^2-x^2-y^2}}\, dA + . +

    + +

    + The region R that we are integrating over is bounded by the circle, + centered at the origin, with radius a: x^2+y^2=a^2. + Because of this region, + we are likely to have greater success with our integration by converting to polar coordinates. + Using the substitutions x=r\cos(\theta), y=r\sin(\theta), + dA = r\, dr\, d\theta and bounds + 0\leq\theta\leq2\pi and 0\leq r\leq a, we have: + + S \amp = 2\int_0^{2\pi}\int_0^a \sqrt{1+\frac{r^2\cos^2(\theta) +r^2\sin^2(\theta) }{a^2-r^2\cos^2(\theta) -r^2\sin^2(\theta) }}\, r\, dr\, d\theta + \amp =2\int_0^{2\pi}\int_0^ar\sqrt{1+\frac{r^2}{a^2-r^2}}\, dr\, d\theta + \amp =2\int_0^{2\pi}\int_0^ar\sqrt{\frac{a^2}{a^2-r^2}}\, dr\, d\theta. + Apply substitution u=a^2-r^2 and integrate the inner integral, giving + \amp = 2\int_0^{2\pi} a^2\, d\theta + \amp = 4\pi a^2 + . +

    + +

    + Our work confirms our previous formula. +

    +     +     + + + + + Finding the surface area of a cone + +

    + The general formula for a right cone with height h and base radius a is + + \ds f(x,y) = h-\frac{h}a\sqrt{x^2+y^2} + , + shown in . + Find the surface area of this cone. +

    + +
    + Finding the surface area of a cone in + + + + A downward-opening circular cone, with vertex on the z axis. + +

    + A circular cone, with vertex on the z axis at a point (0,0,h). + The cone opens downward toward the xy plane, + and ends at the xy plane with a circular base of radius a. +

    +     + + + + + //ASY file for figsurfacearea33D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.25,3.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the cone {x*cos(y)}, {x*sin(y)},{1-x} + triple f(pair t) { + return (t.x*cos(t.y),t.x*sin(t.y),1.5*(2-t.x)); + } + surface s=surface(f,(0,0),(2,2*pi),12,12,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //labels a and h + label("$h$",(.2,-.2,3)); + label("$a$",(2.2,.2,0)); + label("$a$",(.2,2.2,0)); + + + + +
    +     + +

    + We begin by computing partial derivatives. + + f_x(x,y) = -\frac{xh}{a\sqrt{x^2+y^2}} \text{ and } f_y(x,y)=-\frac{yh}{a\sqrt{x^2+y^2}} + . +

    + +

    + Since we are integrating over the disk bounded by x^2+y^2=a^2, + we again use polar coordinates. + Using the standard substitutions, our integrand becomes + + \sqrt{1+ \left(\frac{hr\cos(\theta) }{a\sqrt{r^2}}\right)^2 + \left(\frac{hr\sin(\theta) }{a\sqrt{r^2}}\right)^2} + . +

    + +

    + This may look intimidating at first, + but there are lots of simple simplifications to be done. + It amazingly reduces to just + + \sqrt{1+\frac{h^2}{a^2}} = \frac1a\sqrt{a^2+h^2} + . +

    + +

    + Our polar bounds are 0\leq\theta\leq2\pi and 0\leq r\leq a. + Thus + + S \amp = \int_0^{2\pi}\int_0^ar\frac1a\sqrt{a^2+h^2}\, dr\, d\theta + \amp = \int_0^{2\pi} \left.\left(\frac12r^2\frac1a\sqrt{a^2+h^2}\right)\right|_0^ad\theta + \amp = \int_0^{2\pi} \frac12a\sqrt{a^2+h^2} \, d\theta + \amp = \pi a\sqrt{a^2+h^2} + . +

    + +

    + This matches the formula found in the back of this text. +

    + + +     +     + + + Finding surface area over a region + +

    + Find the area of the surface + f(x,y) = x^2-3y+3 over the region R bounded by + -x\leq y\leq x, 0\leq x\leq 4, + as pictured in . +

    + +
    + Graphing the surface in + + + + A triangular region is shown on a trough-like surface + +

    + The surface z=f(x,y) is a parabolic cylinder; + it curves upward, and has a downward slope relative to the yz plane. +

    + +

    + The surface is plotted against the usual three-dimensional coordinate axes. + In the xy plane we see the triangular outline of the domain for this problem. + The vertices of this triangle are at (4,4,0), (4,-4,0), and (0,0,0). +

    + +

    + On the surface, the curves corresponding to the three sides of this triangle are shown, + illustrating the triangular region whose surface area is computed in this example. +

    +     + + + + + //ASY file for figsurfacearea43D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5.3,17,74); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={-4,-2,2,4}; + real[] myzchoice={20}; + defaultpen(0.5mm); + + pair xbounds=(-.25,5); + pair ybounds=(-5,5); + pair zbounds=(-1,25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2-3y+3 + triple f(pair t) { + return (t.x,t.y,t.x^2-3*t.y+3); + } + surface s=surface(f,(-0.25,-4.5),(4.5,4.5),12,12,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //triangle in xy plane + draw((0,0,0)--(4,4,0)--(4,-4,0)--cycle,bluepen+dashed+linewidth(1)); + + //triangle on surface + triple g(real t) {return (t,t,t^2-3*t+3);} + path3 mypath=graph(g,0,4,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,-t,t^2+3*t+3);} + path3 mypath=graph(g,0,4,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (4,t,16-3*t+3);} + path3 mypath=graph(g,-4,4,operator ..);draw(mypath,bluepen+linewidth(2)); + + + + +
    +     + +

    + It is straightforward to compute + f_x(x,y) = 2x and f_y(x,y) = -3. + Thus the surface area is described by the double integral + + \iint_R \sqrt{1+(2x)^2+(-3)^2}\, dA = \iint_R \sqrt{10+4x^2}\, dA + . +

    + +

    + As with integrals describing arc length, + double integrals describing surface area are in general hard to evaluate directly because of the square-root. + This particular integral can be easily evaluated, + though, with judicious choice of our order of integration. +

    + +

    + Integrating with order dx\, dy requires us to evaluate \int \sqrt{10+4x^2}\, dx. + This can be done, + though it involves Integration By Parts and \sinh^{-1}(x). + Integrating with order dy\, dx has as its first integral \int \sqrt{10+4x^2}\, dy, + which is easy to evaluate: + it is simply y\sqrt{10+4x^2}+C. + So we proceed with the order dy\, dx; + the bounds are already given in the statement of the problem. + + \iint_R\sqrt{10+4x^2}\, dA \amp = \int_0^4\int_{-x}^x\sqrt{10+4x^2}\, dy \, dx + \amp = \int_0^4\left.\big(y\sqrt{10+4x^2}\big)\right|_{-x}^x dx + \amp =\int_0^4\big(2x\sqrt{10+4x^2}\big)\, dx. + Apply substitution with u = 10+4x^2: + \amp = \left.\left(\frac16\big(10+4x^2\big)^{3/2}\right)\right|_0^4 + \amp = \frac13\big(37\sqrt{74}-5\sqrt{10}\big) \approx 100.825\,\text{units}^2 + . +

    + +

    + So while the region R over which we integrate has an area of 16 square units, + the surface has a much greater area as its z-values change dramatically over R. +

    +     +     + +

    + In practice, + technology helps greatly in the evaluation of such integrals. + High powered computer algebra systems can compute integrals that are difficult, + or at least time consuming, by hand, + and can at the least produce very accurate approximations with numerical methods. + In general, just knowing how + to set up the proper integrals brings one very close to being able to compute the needed value. + Most of the work is actually done in just describing the region R in terms of polar or rectangular coordinates. + Once this is done, technology can usually provide a good answer. +

    + +

    + We have learned how to integrate integrals; + that is, we have learned to evaluate double integrals. + In the next section, + we learn how to integrate double integrals that is, + we learn to evaluate triple integrals, + along with learning some uses for this operation. +

    + + + + Terms and Concepts + + + + +

    + Surface area is related to what previously studied concept? +

    +     + + + +     + + + + +

    + To approximate the area of a small portion of a surface, + we computed the area of its plane. +

    +     + + + + + + + + +     + + + + +

    + We interpret \ds \iint_R\, dS as + sum up lots of little . +

    +     + + + + + + + + + surface area + + +

    + Your answer should be plural. +

    +     +     +     + +     + + + + +

    + Why is it important to know how to set up a double integral to compute surface area, + even if the resulting integral is hard to evaluate? +

    +     + + + +

    + Technology makes good approximations accessible, if not exact answers. +

    +     + +     + + + + +

    + Why do z=f(x,y) and z=g(x,y)=f(x,y)+h, + for some real number h, + have the same surface area over a region R? +

    +     + + + +

    + Intuitively, + adding h to f only shifts f up (, parallel to the z-axis) and does not change its shape. + Therefore it will not change the surface area over R. +

    + +

    + Analytically, f_x = g_x and f_y=g_y; + therefore, the surface area of each is computed with identical double integrals. +

    +     + +     + + + + +

    + Let f(x,y) be a function defined over a region R and let g(x,y)=2f(x,y). + Why is the surface area of z=g(x,y) over R not twice the surface area of z=f(x,y) over R? +

    +     + + + +

    + Analytically, g_x = 2f_x and g_y=2f_y. + The double integral to compute the surface area of f over R is \ds \iint_R \sqrt{1+f_x^2+f_y^2}\, dA; + the double integral to compute the surface area of g over R is \ds \iint_R\sqrt{1+4f_x^2+4f_y^2}\, dA, + which is not twice the double integral used to calculate the surface area of f. +

    +     + +     +     + + + Problems + + + +

    + Set up the iterated integral that computes the surface area of the graph of the given function over the region R. +

    +     + + + + +

    + f(x,y) = \sin(x) \cos(y);R is the rectangle with bounds + 0\leq x\leq 2\pi, 0\leq y\leq2\pi. +

    + + + + + A bumpy surface with several peaks and valleys. + +

    + The surface z=\sin(x)\cos(y) is plotted over a rectangular domain. + The domain is large enough that several of the peaks and valleys typical of this surface can be seen. +

    + +

    + Note that the image is somewhat decorative in this case: + the formula for surface area can be applied to the given function without knowing the appearance of the surface. +

    +     + + + + + //ASY file for fig13_05_ex_053D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={5}; + real[] myychoice={5}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,7); + pair ybounds=(-0.5,7); + pair zbounds=(-1.25,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface sin(x)cos(y) + triple f(pair t) { + return (t.x,t.y,sin(t.x)*cos(t.y)); + } + surface s=surface(f,(0,0),(2*pi,2*pi),12,12,Spline); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p); + + //lines in xy plane + draw((0,0,0)--(0,2*pi,0),bluepen+dashed+linewidth(2)); + draw((2*pi,0,0)--(2*pi,2*pi,0),bluepen+dashed+linewidth(2)); + + //lines on surface + triple g(real t) {return (t,0,sin(t));} + path3 mypath=graph(g,0,2*pi,operator ..);draw(mypath,bluepen+dashed+linewidth(2)); + triple g(real t) {return (t,2*pi,sin(t));} + path3 mypath=graph(g,0,2*pi,operator ..);draw(mypath,bluepen+dashed+linewidth(2)); + + + + +     + +

    + \ds SA = \int_0^{2\pi}\int_0^{2\pi} \sqrt{1+ \cos^2(x) \cos^2(y) +\sin^2(x) \sin^2(y) }\, dx\, dy +

    +     + +     + + + + +

    + \ds f(x,y) = \frac{1}{x^2+y^2+1};R is bounded by the circle x^2+y^2=9. +

    + + + + + A steep surface with a single peak, resembling a witch's hat. + +

    + The surface z=\frac{1}{x^2+y^2+1} has rotational symmetry about the z axis. + It resembles a steep mountain with a smooth peak, or perhaps the hat of a witch. +

    + +

    + Again, the image is primarily decorative; it is sufficient to know the function and the region to determine the surface area. +

    +     + + + + + //ASY file for fig13_05_ex_063D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(13,13,1.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-4,-2,2,4}; + real[] myychoice={-4,-2,2,4}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-4.25,4.25); + pair ybounds=(-4.25,4.25); + pair zbounds=(-0.25,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface 1/(1+x^2+y^2) + triple f(pair t) { + return (t.x,t.y,1/(1+t.x^2+t.y^2)); + } + surface s=surface(f,(-4,-4),(4,4),12,12,Spline); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p); + + //circle + triple g(real t) {return (3*cos(t),3*sin(t),1/10);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+dashed+linewidth(1)); + + + + +     + +

    + \ds SA = \int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \sqrt{1+ \frac{4x^2+4y^2}{(1+x^2+y^2)^4}}\, dx\, dy +

    + +

    + Polar offers simpler bounds: +

    + +

    + \ds SA = \int_0^{2\pi}\int_0^3 r\sqrt{1+\frac{4r^2}{(1+r^2)^4}}\, dr\, d\theta +

    +     + +     + + + + +

    + \ds f(x,y) = x^2-y^2;R is the rectangle with opposite corners (-1,-1) and (1,1). +

    + + + + + A saddle surface, with its saddle point at the origin in three dimensions. + +

    + A standard hyperbolic paraboloid, or saddle surface. + The saddle point is at the origin. + The surface curves upward along the x axis, and downward along the y axis. +

    +     + + + + + //ASY file for fig13_05_ex_073D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(2.2,4.1,3.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface x^2-y^2 + triple f(pair t) { + return (t.x,t.y,t.x^2-t.y^2); + } + surface s=surface(f,(-1.2,-1.2),(1.2,1.2),12,12,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p); + + //lines on surface + triple g(real t) {return (t,1,t^2-1);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); + triple g(real t) {return (t,-1,t^2-1);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); + triple g(real t) {return (1,t,1-t^2);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); + triple g(real t) {return (-1,t,1-t^2);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); + + + + +     + +

    + \ds SA = \int_{-1}^{1}\int_{-1}^{1} \sqrt{1+ 4x^2+4y^2}\, dx\, dy +

    +     + +     + + + + +

    + \ds f(x,y) = \frac{1}{e^{x^2}+1};R is the rectangle bounded by +

    + +

    + -5\leq x\leq 5 and 0\leq y\leq 1. +

    + + + + + A mostly flat surface with a ridge along the y axis + +

    + A graph of the function f(x,y) = \frac{1}{e^{x^2}+1} in three dimensions. + The surface is mostly flat, with a ridge along the y axis of height 1/2. +

    +     + + + + + //ASY file for fig13_05_ex_083D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.6,3.9,3.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-5,5}; + real[] myychoice={-1,1}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-6,6); + pair ybounds=(-1,1.5); + pair zbounds=(-0.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface 1/(exp(x^2)+1) + triple f(pair t) { + return (t.x,t.y,1/(exp(t.x^2)+1)); + } + surface s=surface(f,(-6,-0.5),(6,1.5),16,12,Spline); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p); + + //lines on surface + draw((5,0,0)--(5,1,0),bluepen+dashed+linewidth(1.5)); + draw((-5,0,0)--(-5,1,0),bluepen+dashed+linewidth(1.5)); + triple g(real t) {return (t,0,1/(exp(t^2)+1));} + path3 mypath=graph(g,-5,5,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); + triple g(real t) {return (t,1,1/(exp(t^2)+1));} + path3 mypath=graph(g,-5,5,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); + + + + +     + +

    + \ds SA = \int_{-5}^{5}\int_{0}^{1} \sqrt{1+ \frac{4x^2e^{2x^2}}{\big(1+e^{x^2}\big)^4}}\, dy\, dx +

    +     + +     + +     + + + +

    + Find the area of the given surface over the region R. +

    +     + + + + +

    + z = 3x-7y+2; + R is the rectangle with opposite corners (-1,0) and (1,3). +

    +     + +

    + \ds SA = \int_{0}^{3}\int_{-1}^{1} \sqrt{1+ 9+49}\, dx\, dy = 6\sqrt{59} \approx 46.09 +

    +     + +     + + + + +

    + z = 2x+2y+2; + R is the triangle with corners (0,0), + (1,0) and (0,1). +

    +     + +

    + \ds SA = \int_{0}^{1}\int_{0}^{1-x} \sqrt{1+ 4+4}\, dy\, dx = 18 +

    +     + +     + + + + +

    + z = x^2+y^2+10; R is bounded by the circle x^2+y^2=16. +

    +     + +

    + This is easier in polar: + + SA \amp = \int_{0}^{2\pi}\int_{0}^{4} r\sqrt{1+ 4r^2\cos^2(t) +4r^2\sin^2(t) }\, dr\, d\theta + \amp = \int_0^{2\pi}\int_0^4r\sqrt{1+4r^2}\, dr\, d\theta + \amp = \frac{\pi}{6}\big(65\sqrt{65}-1\big) \approx 273.87 + +

    +     + +     + + + + +

    + z = -2x+4y^2+7 over R, + the triangle bounded by y=-x, + y=x, 0\leq y\leq 1. +

    +     + +

    + + SA \amp = \int_{0}^{1}\int_{-y}^{y} \sqrt{1+ 4+64y^2}\, dx\, dy + \amp = \int_0^{1}\big(2y\sqrt{5+64y^2}\big)\, dy + \amp = \frac1{96}\big(69\sqrt{69}-5\sqrt{5}\big)\approx 5.85 + +

    +     + +     + + + + +

    + z = x^2+y over R, + the triangle bounded by y=2x, + y=0 and x=2. +

    +     + +

    + + SA \amp = \int_{0}^{2}\int_{0}^{2x} \sqrt{1+ 1+4x^2}\, dy\, dx + \amp = \int_0^{2}\big(2x\sqrt{2+4x^2}\big)\, dx + \amp = \frac{26}{3}\sqrt{2}\approx 12.26 + +

    +     + +     + + + + +

    + z = \frac23x^{3/2}+2y^{3/2} over R, + the rectangle with opposite corners (0,0) and (1,1). +

    +     + +

    + + SA \amp = \int_{0}^{1}\int_{0}^{1} \sqrt{1+ x+9y}\, dx\, dy + \amp = \int_0^{1}\frac23\Big((9y+2)^{3/2}-(9y+1)^{3/2}\Big)\, dy + \amp = \frac{4}{135}\big(121\sqrt{11}-100\sqrt{10}-4\sqrt{2}+1\big)\approx 2.383 + +

    +     + +     + + + + +

    + z = 10-2\sqrt{x^2+y^2} over the region R + bounded by the circle x^2+y^2=25. + (This is the cone with height 10 and base radius 5; + be sure to compare your result with the known formula.) +

    +     + +

    + This is easier in polar: + + SA \amp = \int_{0}^{2\pi}\int_{0}^{5} r\sqrt{1+ \frac{4r^2\cos^2(t) +4r^2\sin^2(t) }{r^2\sin^2(t) +r^2\cos^2(t) }}\, dr\, d\theta + \amp = \int_0^{2\pi}\int_0^5r\sqrt{5}\, dr\, d\theta + \amp = 25\pi\sqrt{5}\approx 175.62 + +

    +     + +     + + + + +

    + Find the surface area of the sphere with radius 5 by doubling the surface area of f(x,y) = \sqrt{25-x^2-y^2} over the region R + bounded by the circle x^2+y^2=25. + (Be sure to compare your result with the known formula.) +

    +     + +

    + This is easier in polar: + + SA \amp = 2\int_{0}^{2\pi}\int_{0}^{5} r\sqrt{1+ \frac{r^2\cos^2(t) +r^2\sin^2(t) }{25-r^2\sin^2(t) -r^2\cos^2(t) }}\, dr\, d\theta + \amp = 2\int_0^{2\pi}\int_0^5r\sqrt{\frac{1}{25-r^2}}\, dr\, d\theta + \amp = 100\pi\approx 314.16 + +

    +     + +     + + + + +

    + Find the surface area of the ellipse formed by restricting the plane + f(x,y) = cx+dy+h to the region R bounded by + the circle x^2+y^2=1, + where c, d and h are some constants. + Your answer should be given in terms of c and d; + why does the value of h not matter? +

    +     + +

    + Integrating in polar is easiest considering R: + + SA \amp = \int_{0}^{2\pi}\int_{0}^{1} r\sqrt{1+ c^2+d^2}\, dr\, d\theta + \amp = \int_0^{2\pi}\frac12\Big(\sqrt{1+c^2+d^2}\Big)\, dy + \amp = \pi\sqrt{1+c^2+d^2} + . +

    + +

    + The value of h does not matter as it only shifts the plane vertically (, parallel to the z-axis). + Different values of h do not create different ellipses in the plane. +

    +     + +     + +     +     +     +
    +
    + Volume Between Surfaces and Triple Integration + + Volume between surfaces +

    + We learned in + how to compute the signed volume V under a surface + z=f(x,y) over a region R: + V = \iint_R f(x,y)\, dA. + It follows naturally that if f(x,y)\geq g(x,y) on R, + then the volume between f(x,y) and g(x,y) on R is + + V = \iint_R f(x,y)\, dA - \iint_R g(x,y)\, dA = \iint_R \big(f(x,y)-g(x,y)\big)\, dA + . +

    + + + Volume Between Surfaces + +

    + Let f and g be continuous functions on a closed, + bounded region R, + where f(x,y)\geq g(x,y) for all (x,y) in R. + The volume V between f and g over R is + volume + + V =\iint_R \big(f(x,y)-g(x,y)\big)\, dA + . +

    +     +     + + + Finding volume between surfaces + +

    + Find the volume of the space region bounded by the planes z=3x+y-4, + z=8-3x-2y, x=0 and y=0. + + In the planes are drawn; + in , only the defined region is given. +

    + +
    + Finding the volume between the planes given in + +
    + + + + + Two planes in space intersect along a line. + +

    + Two planes are shown in space relative to a set of three-dimensional coordinate axes. + Each plane is plotted over the same rectangular domain in the xy plane, + and drawn in the shape of a parallelogram. +

    + +

    + The planes intersect along a line. + With respect to the parallelogram shapes used in the image, + one pair of opposite corners for each plane lies along the line of intersection. +

    + +

    + When viewed from above, the line of intersection divides the rectangular domain into two triangles. +

    +     + + + + + //ASY file for figtrip13D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5.4,-12.7,20.2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2,4}; + real[] myzchoice={-5,5}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,3); + pair ybounds=(-0.5,5); + pair zbounds=(-5.25,10); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane z=3x+y-4 + triple f(pair t) { + return (t.x,t.y,3*t.x+t.y-4); + } + surface s=surface(f,(0,0),(2,4),1,1,Spline); + pen p=apexmeshpen; + draw(s,simplesurfacepen,meshpen=bluepen+linewidth(1.5)); + + //Draw the plane z=8-3x-2y + triple f(pair t) { + return (t.x,t.y,8-3*t.x-2*t.y); + } + surface s=surface(f,(0,0),(2,4),1,1,Spline); + pen p=apexmeshpen; + draw(s,simplesurfacepen2,meshpen=redpen+linewidth(1.5)); + + //line + draw((0,4,0)--(2,0,2),dashed+linewidth(1.5)); + + //lines on surface + //draw((5,0,0)--(5,1,0),bluepen+dashed+linewidth(1.5)); + //draw((-5,0,0)--(-5,1,0),bluepen+dashed+linewidth(1.5)); + //triple g(real t) {return (t,0,1/(exp(t^2)+1));} + //path3 mypath=graph(g,-5,5,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); + //triple g(real t) {return (t,1,1/(exp(t^2)+1));} + //path3 mypath=graph(g,-5,5,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); + + + + +
    + +
    + + + + + Two planes in space plotted over a triangular domain. The planes intersect along one edge of the triangle. + +

    + This is another drawing of the same two planes as , + but this time, the planes are plotted over a triangular domain. + The sides of the triangle correspond to the x and y axes, + and the line of intersection of the two planes. +

    + +

    + Each plane is therefore drawn as a triangle in space. + The two triangles meet along an edge, + and the opposite vertex of each triangle is along the z axis. +

    + +

    + One triangle lies below the other, if we measure height relative to the z axis. + These two triangles make up two of the four faces of a tetrahedron. + The other two faces lie in the xz and yz coordinate planes, respectively. +

    +     + + + + + //ASY file for figtrip1b3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(5.4,-12.7,20.2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2,4}; + real[] myzchoice={-5,5}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,3); + pair ybounds=(-0.5,5); + pair zbounds=(-5.25,10); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //lines of the triangles + draw((0,4,0)--(2,0,2)--(0,0,8)--cycle,redpen+linewidth(1.5)); + draw((0,0,-4)--(0,0,8)--(2,0,2)--cycle,black+linewidth(1.5)); + draw((0,0,-4)--(0,4,0),bluepen+linewidth(1.5)); + + //shade + import three; + path3 p = (0,4,0)--(2,0,2)--(0,0,8); //Left + //draw(surface(p -- cycle), emissive(rgb(1,.4,.7)+opacity(0.7))); + draw(surface(p -- cycle), simplesurfacepen2); + + path3 p = (0,0,-4)--(0,4,0)--(2,0,2); //Left + draw(surface(p -- cycle), simplesurfacepen); + + path3 p = (0,0,-4)--(0,0,8)--(2,0,2); //Left + draw(surface(p -- cycle), rgb(.7,.7,.7)+opacity(.2)); + + path3 p = (0,0,-4)--(0,0,8)--(0,4,0); //Left + draw(surface(p -- cycle), rgb(.7,.7,.7)+opacity(.2)); + + //surfacepen2 + + // to get perspectives right with figtrip1a, draw with invisible pen + + draw((2,0,2)--(2,4,6),invisible); + draw((2,0,2)--(2,4,-6),invisible); + + + + +
    +     + +
    +     + +

    + We need to determine the region R over which we will integrate. + To do so, we need to determine where the planes intersect. + They have common z-values when 3x+y-4=8-3x-2y. + Applying a little algebra, we have: + + 3x+y-4 \amp = 8-3x-2y + 6x+3y \amp =12 + 2x+y \amp =4 + +

    + +

    + The planes intersect along the line 2x+y=4. + Therefore the region R is bounded by x=0, + y=0, and y=4-2x; + we can convert these bounds to integration bounds of + 0\leq x\leq 2, 0\leq y\leq 4-2x. + Thus + + V \amp = \iint_R \big(8-3x-2y-(3x+y-4)\big)\, dA + \amp = \int_0^2\int_0^{4-2x} \big(12-6x-3y\big)\, dy\, dx + \amp = 16\,\text{units}^3 + . +

    + +

    + The volume between the surfaces is 16 cubic units. +

    +     +     + +

    + In the preceding example, we found the volume by evaluating the integral + + \ds \int_0^2\int_0^{4-2x} \big(8-3x-2y-(3x+y-4)\big)\, dy\, dx + . +

    + +

    + Note how we can rewrite the integrand as an integral, + much as we did in : + + 8-3x-2y-(3x+y-4) = \int_{3x+y-4}^{8-3x-2y}\, dz + . +

    + +

    + Thus we can rewrite the double integral that finds volume as + + \int_0^2\int_0^{4-2x} \big(8-3x-2y-(3x+y-4)\big)\, dy\, dx = \int_0^2\int_0^{4-2x}\left(\int_{3x+y-4}^{8-3x-2y}\, dz\right)\, dy\, dx + . +

    + +

    + This no longer looks like a double integral, + but more like a triple integral. + Just as our first introduction to double integrals was in the context of finding the area of a plane region, + our introduction into triple integrals will be in the context of finding the volume of a space region. +

    + +
    + Approximating the volume of a region D in space + +
    + + + + + An ellipsoid in space, centered at the origin. + +

    + An ellipsoid in space, plotted against three-dimensional coordinate axes. + It is egg-shaped, and is longest along the x axis. +

    + +

    + The grid lines on the ellipsoid are like lines of latitude and longitude. +

    +     + + + + + //ASY file for figtripintro3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + + pair xbounds=(-2.25,2.25); + pair ybounds=(-2.25,2.25); + pair zbounds=(-2.25,2.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Ellipsoid + triple f(pair t) { + return (cos(t.x)*2*cos(t.y),sin(t.x)*cos(t.y),sin(t.y)); + } + surface s=surface(f,(-pi,-pi/2),(1*pi,pi/2),16,16,Spline); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p); + + + + +
    + +
    + + + + + A zoomed-in view of a surface, showing grid lines. A small rectangular prism illustrates a small piece of volume. + +

    + A zoomed in view of a portion of the ellipsoid in where it meets the positive y axis. + The grid lines on the surface are horizontal (traces with constant z value) and vertical (traces by planes through the z axis), + like lines of latitude and longitude. +

    + +

    + On one of the rectangles formed by the grid, + a small rectangular box is drawn. + The image illustrates how we can approximate the volume of an object like an ellipsoid by dividing it up + into small cube-like pieces. +

    +     + + + + + //ASY file for figtripintroa3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(0,2.25); + pair ybounds=(0.25,1.75); + pair zbounds=(-0.5,0.5); + + //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Ellipsoid + triple f(pair t) { + return (cos(t.x)*2*cos(t.y),sin(t.x)*cos(t.y),sin(t.y)); + } + surface s=surface(f,(pi/6,-pi/3),(2*pi/3,pi/3),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p); + + //draw red cube + //draw((0., 1.05, -0.105) -- (0,1.05,.105) -- (-.345,1.05,0.105) -- (-0.345, 1.05, -0.105)--cycle,redpen);//front + + draw((0.13, 1.05, 0.26) -- (0.13, 1.05, 0) -- (-0.26, 1.05, 0) -- (-0.25, 1.05, 0.26)--cycle,redpen);//front + + draw((0.13, 0.9, 0.26) -- (0.13, 0.9, 0) -- (-0.26, 0.9, 0) -- (-0.25, 0.9, 0.26)--cycle,redpen);//back + + //draw((-.345,0.979,0.105)--(0,0.979,0.105)--(0,0.979,-0.105)--(-.345,0.979,-0.105)--cycle,redpen);//back + //sides + draw((0.13, 0.9, 0.26)--(0.13, 1.05, 0.26),redpen); + draw((0.13, 0.9, 0)--(0.13, 1.05, 0),redpen); + draw((-0.26, 0.9, 0)--(-0.26, 1.05, 0),redpen); + draw((-0.25, 0.9, 0.26)--(-0.25, 1.05, 0.26),redpen); + + + + +
    +     + +
    + +

    + To formally find the volume of a closed, + bounded region D in space, + such as the one shown in , + we start with an approximation. + Break D into n rectangular solids; + the solids near the boundary of D may possibly not include portions of D and/or include extra space. + In , + we zoom in on a portion of the boundary of D to show a rectangular solid that contains space not in D; + as this is an approximation of the volume, + this is acceptable and this error will be reduced as we shrink the size of our solids. +

    + +

    + The volume \Delta V_i of the + ith solid D_i is \Delta V_i = \dx_i\dy_i\ddz_i, + where \dx_i, + \dy_i and \ddz_i give the dimensions of the rectangular solid in the x, + y and z directions, respectively. + By summing up the volumes of all n solids, + we get an approximation of the volume V of D: + + V \approx \sum_{i=1}^n \Delta V_i = \sum_{i=1}^n \dx_i\dy_i\ddz_i + . +

    + +

    + Let \norm{\Delta D} represent the length of the longest diagonal of rectangular solids in the subdivision of D. + As \norm{\Delta D}\to 0, + the volume of each solid goes to 0, as do each of \dx_i, + \dy_i and \ddz_i, for all i. + Our calculus experience tells us that taking a limit as + \norm{\Delta D}\to 0 turns our approximation of V into an exact calculation of V. + Before we state this result in a theorem, + we use a definition to define some terms. +

    + + + Triple Integrals, Iterated Integration (Part I) + +

    + Let D be a closed, bounded region in space. + Let a and b be real numbers, + let g_1(x) and g_2(x) be continuous functions of x, + and let f_1(x,y) and + f_2(x,y) be continuous functions of x and y. + integrationtriple + triple integral + iterated integration +

    + +

    +

      +
    1. +

      + The volume V of D is denoted by a + triple integral, + + V = \iiint_D dV + . +

      +
      2. + +
    2. +

      + The iterated integral \ds \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \, dz\, dy\, dx is evaluated as + + \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \, dz\, dy\, dx=\int_a^b\int_{g_1(x)}^{g_2(x)}\left(\int_{f_1(x,y)}^{f_2(x,y)} \, dz\right)\, dy\, dx + . + Evaluating the above iterated integral is + triple integration. +

      +
      3. +
    +

    +     +     + +

    + Our informal understanding of the notation \iiint_D\, dV is + sum up lots of little volumes over D, + analogous to our understanding of + \iint_R\, dA and \iint_R\, dm. +

    + + + +

    + We now state the major theorem of this section. +

    + + + Triple Integration (Part I) + +

    + Let D be a closed, + bounded region in space and let + \Delta D be any subdivision of D into n rectangular solids, + where the ith subregion D_i has dimensions + \dx_i\times\dy_i\times\ddz_i and volume \Delta V_i. + integrationtriple + triple integral + iterated integration +

    + +

    +

      +
    1. +

      + The volume V of D is + + V = \iiint_D\, dV = \lim_{\norm{\Delta D}\to0} \sum_{i=1}^n \Delta V_i = \lim_{\norm{\Delta D}\to0} \sum_{i=1}^n \dx_i\dy_i\ddz_i + . +

      +
      2. + +
    2. +

      + If D is defined as the region bounded by the planes x=a and x=b, + the cylinders y=g_1(x) and y=g_2(x), + and the surfaces z=f_1(x,y) and + z=f_2(x,y), where a\lt b, + g_1(x)\leq g_2(x) and f_1(x,y)\leq f_2(x,y) on D, then + + \iiint_D \, dV = \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \, dz\, dy\, dx + . +

      +
      3. + +
    3. +

      + V can be determined using iterated integration with other orders of integration + (there are 6 total), + as long as D is defined by the region enclosed by a pair of planes, + a pair of cylinders, and a pair of surfaces. +

      +
      4. +
    +

    +     +     + +

    + We evaluated the area of a plane region R by iterated integration, + where the bounds were from curve to curve, + then from point to point. + allows us to find the volume of a space region with an iterated integral with bounds + from surface to surface, then from curve to curve, + then from point to point. In the iterated integral + + \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \, dz\, dy\, dx + , + the bounds a\leq x\leq b and + g_1(x)\leq y\leq g_2(x) define a region R in the xy-plane over which the region D exists in space. + However, these bounds are also defining surfaces in space; + x=a is a plane and y=g_1(x) is a cylinder. + The combination of these 6 surfaces enclose, and define, + D. +

    + +

    + Examples will help us understand triple integration, + including integrating with various orders of integration. +

    + + + + + Finding the volume of a space region with triple integration + +

    + Find the volume of the space region in the first octant bounded by the plane z=2-y/3-2x/3, + shown in , + using the order of integration dz\, dy\, dx. + Set up the triple integrals that give the volume in the other 5 orders of integration. +

    +
    + The region D used in + + + + A triangular portion of a plane in space. It lies in the first octant and has vertices on the coordinate axes. + +

    + A plane in space is illustrated, by showing the portion of the plane the lies in the first octant. + This portion of the plane is a triangle, whose vertices lie on the coordinate axes. + Using the image (or the equation z=2-y/3-2x/3), + we can determine that the plane meets the coordinate axes at the points (3,0,0), (0,6,0), and (0,0,2). +

    + +

    + The region in space whose volume is computed in this problem is a tetrahedron, + whose faces are the plane described above, + as well as the triangles formed in the three coordinate planes by the coordinate axes and the edges of the first triangle. +

    +     + + + + + //ASY file for figtrip23D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.2,10.1,6.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={5}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,5); + pair ybounds=(-0.5,7); + pair zbounds=(-0.5,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //triangle in space + draw((3,0,0)--(0,6,0)--(0,0,2)--cycle,bluepen+linewidth(1.5)); + + //shade + import three; + path3 p = (3,0,0)--(0,6,0)--(0,0,2); //Left + draw(surface(p -- cycle), simplesurfacepen); + + + + +
    +     + +

    + Starting with the order of integration dz\, dy\, dx, + we need to first find bounds on z. + The region D is bounded below by the plane z=0 + (because we are restricted to the first octant) + and above by z=2-y/3-2x/3; + 0\leq z\leq 2-y/3-2x/3. +

    + +

    + To find the bounds on y and x, + we collapse the region onto the xy-plane, + giving the triangle shown in . (We know the equation of the line y=6-2x in two ways. + First, by setting z=0, + we have 0 = 2-y/3-2x/3 \Rightarrow y=6-2x. + Secondly, we know this is going to be a straight line between the points (3,0) and (0,6) in the xy-plane.) +

    + +
    + The region found by collapsing D onto the xy-plane + + + + A triangle in the xy plane, plotted in space relative to 3D coordinate axes. + +

    + A set of three-dimensional coordinate axes is shown, with the z axis pointing upward. + In the xy plane a triangle is drawn. + The sides of the triangle are formed by the x axis, the y axis, + and the line from (3,0,0) to (0,6,0), + which is one edge of the triangle in space shown in . + This line is labeled with the equation y-6-2x. +

    + +

    + This triangle represents the shadow of the plane in space on the xy plane, + when viewed directly from above. + It will be used to determine the bounds in the triple integral when we integrate first with respect to z. +

    +     + + + + + //ASY file for figtrip2b3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.2,10.1,6.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={5}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,5); + pair ybounds=(-0.5,7); + pair zbounds=(-0.5,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //triangle in space + draw((3,0,0)--(0,6,0)--(0,0,0)--cycle,bluepen+linewidth(1.5)); + + //shade + import three; + path3 p = (3,0,0)--(0,6,0)--(0,0,0); //Left + draw(surface(p -- cycle), simplesurfacepen); + label("$y=6-2x$",(2.5,3.5,0)); + + + + +
    + +

    + We define that region R, + in the integration order of dy\, dx, + with bounds 0\leq y\leq 6-2x and 0\leq x\leq 3. + Thus the volume V of the region D is: + + V \amp = \iiint_D \, dV + \amp = \int_0^3\int_0^{6-2x}\int_0^{2-\frac 13y-\frac 23x}\, dz\, dy\, dx + \amp = \int_0^3\int_0^{6-2x}\left(\int_0^{2-\frac 13y-\frac 23x}\, dz\right)\, dy\, dx + \amp =\int_0^3\int_0^{6-2x}z\Big|_0^{2-\frac 13y-\frac 23x}\, dy\, dx + \amp = \int_0^3\int_0^{6-2x}\left(2-\frac 13y-\frac 23x\right)\, dy\, dx + . + From this step on, we are evaluating a double integral as done many times before. We skip these steps and give the final volume, V=6. +

    + +

    + The order dz\, dx\, dy: +

    + +

    + Now consider the volume using the order of integration dz\, dx\, dy. + The bounds on z are the same as before, + 0\leq z\leq 2-y/3-2x/3. + Collapsing the space region on the xy-plane as shown in , + we now describe this triangle with the order of integration dx\, dy. + This gives bounds 0\leq x\leq 3-y/2 and 0\leq y\leq 6. + Thus the volume is given by the triple integral + + V = \int_0^6\int_0^{3-\frac12y}\int_0^{2-\frac13y-\frac23x}\, dz\, dx\, dy + . +

    + +

    + The order dx\, dy\, dz: +

    + +

    + Following our surface to surface\ldots strategy, + we need to determine the x-surfaces + that bound our space region. + To do so, approach the region from behind, + in the direction of increasing x. + The first surface we hit as we enter the region is the yz-plane, + defined by x=0. + We come out of the region at the plane z=2-y/3-2x/3; + solving for x, we have x= 3-y/2-3z/2. + Thus the bounds on x are: + 0\leq x\leq 3-y/2-3z/2. +

    + +

    + Now collapse the space region onto the yz-plane, + as shown in . (Again, + we find the equation of the line z=2-y/3 by setting x=0 in the equation + x=3-y/2-3z/2.) We need to find bounds on this region with the order dy\, dz. + The curves that bound y are y=0 and y=6-3z; + the points that bound z are 0 and 2. + Thus the triple integral giving volume is: +

    + + +

    + + 0 \amp \leq x\leq 3-y/2-3z/2 + 0 \amp \leq y\leq 6-3z + 0 \amp \leq z\leq 2 + +

    + +

    + \Rightarrow +

    + +

    + + \int_0^2\int_0^{6-3z}\int_0^{3-y/2-3z/2}\, dx\, dy\, dz + . +

    +     + +

    + The order dx\, dz\, dy: +

    + +
    + The region D in is collapsed onto the yz-plane in (a); in (b), the region is collapsed onto the xz-plane + +
    + + + + + A triangle plotted in the yz plane, relative to a set of 3D coordinate axes. + +

    + A three-dimensional coordinate system is shown, with the z axis pointing upward. + A triangle is plotted in the yz plane. + The edges of the triangle are the y and z coordinate axes, + and the line from (0,6,0) to (0,0,2), + where the plane in space from meets the yz plane. + This line is labeled with the equation z=2-y/3. +

    + +

    + This is the shadow of the region of integration for this problem on the yz plane, + when viewed along the x axis. + It is used to set up the bounds of integration when we integrate first with respect to x. +

    +     + + + + + //ASY file for figtrip2c3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.2,10.1,6.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={5}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,5); + pair ybounds=(-0.5,7); + pair zbounds=(-0.5,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //triangle in space + draw((0,0,2)--(0,6,0)--(0,0,0)--cycle,bluepen+linewidth(1.5)); + + //shade + import three; + path3 p = (0,0,2)--(0,6,0)--(0,0,0); //Left + draw(surface(p -- cycle), simplesurfacepen); + label("$z=2-y/3$",(0,2,1.5),E); + + + + +
    + +
    + + + + + A triangle plotted in the xz plane, relative to a set of 3D coordinate axes. + +

    + A three-dimensional coordinate system is shown, with the z axis pointing upward. + A triangle is plotted in the xz plane. + The edges of the triangle are the x and z coordinate axes, + and the line from (3,0,0) to (0,0,2), + where the plane in space from meets the xz plane. + This line is labeled with the equation z=2-2x/3. +

    + +

    + This is the shadow of the region of integration for this problem on the xz plane, + when viewed along the y axis. + It is used to set up the bounds of integration when we integrate first with respect to y. +

    +     + + + + + //ASY file for figtrip2d3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.2,10.1,6.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={5}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,5); + pair ybounds=(-0.5,7); + pair zbounds=(-0.5,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //triangle in space + draw((0,0,2)--(3,0,0)--(0,0,0)--cycle,bluepen+linewidth(1.5)); + + //shade + import three; + path3 p = (0,0,2)--(3,0,0)--(0,0,0); //Left + draw(surface(p -- cycle), simplesurfacepen); + label("$z=2-2x/3$",(1,0,1.5),W); + + + + +
    +     + +
    + +

    + The x-bounds are the same as the order above. + We now consider the triangle in and describe it with the order dz\, dy: + 0\leq z\leq 2-y/3 and 0\leq y\leq 6. + Thus the volume is given by: +

    + + +

    + + 0 \amp \leq x\leq 3-y/2-3z/2 + 0 \amp \leq z\leq 2-y/3 + 0 \amp \leq y\leq 6 + +

    + +

    + \Rightarrow +

    + +

    + + \int_0^6\int_0^{2-y/3}\int_0^{3-y/2-3z/2}\, dx\, dz\, dy + . +

    +     + +

    + The order dy\, dz\, dx: +

    + +

    + We now need to determine the y-surfaces that determine our region. + Approaching the space region from behind + and moving in the direction of increasing y, + we first enter the region at y=0, + and exit along the plane z= 2-y/3-2x/3. + Solving for y, this plane has equation y = 6-2x-3z. + Thus y has bounds 0\leq y\leq 6-2x-3z. +

    + +

    + Now collapse the region onto the xz-plane, + as shown in . + The curves bounding this triangle are z=0 and z=2-2x/3; + x is bounded by the points x=0 to x=3. + Thus the triple integral giving volume is: +

    + + +

    + + 0 \amp \leq y\leq 6-2x-3z + 0 \amp \leq z\leq 2-2x/3 + 0 \amp \leq x\leq 3 + +

    + +

    + \Rightarrow +

    + +

    + + \int_0^3\int_0^{2-2x/3}\int_0^{6-2x-3z}\, dy\, dz\, dx + . +

    +     + +

    + The order dy\, dx\, dz: +

    + +

    + The y-bounds are the same as in the order above. + We now determine the bounds of the triangle in using the order dy\, dx\, dz. + x is bounded by x=0 and x=3-3z/2; + z is bounded between z=0 and z=2. + This leads to the triple integral: +

    + + +

    + + 0 \amp \leq y\leq 6-2x-3z + 0 \amp \leq x\leq 3-3z/2 + 0 \amp \leq z\leq 2 + +

    + +

    + \Rightarrow +

    + +

    + + \int_0^2\int_0^{3-3z/2}\int_0^{6-2x-3z}\, dy\, dx\, dz + . +

    +     + +

    + This problem was long, but hopefully useful, + demonstrating how to determine bounds with every order of integration to describe the region D. + In practice, + we only need 1, but being able to do them all gives us flexibility to choose the order that suits us best. +

    +     +     + +

    + In the previous example, + we collapsed the surface into the x-y, x-z, + and yz-planes as we determined the curve to curve, + point to point bounds of integration. + Since the surface was a triangular portion of a plane, + this collapsing, or projecting, was simple: + the projection of a straight line in space onto a coordinate plane is a line. +

    + +

    + The following example shows us how to do this when dealing with more complicated surfaces and curves. +

    + + + Finding the projection of a curve in space onto the coordinate planes + +

    + Consider the surfaces z=3-x^2-y^2 and z=2y, + as shown in . + The curve of their intersection is shown, + along with the projection of this curve into the coordinate planes, + shown dashed. + Find the equations of the projections into the coordinate planes. +

    +
    + Finding the projections of the curve of intersection in + +
    + + + + + A surface in space intersects with a plane through the origin, forming a parabolic curve in space. + +

    + Two surfaces are plotted relative to a set of three-dimensional coordinate axes. + One surface is a portion of a downward-opening circular paraboloid; + the other is a plane through the origin. +

    + +

    + The plane cuts the paraboloid along a curve, which is a parabola in space. +

    +     + + + + + //ASY file for fig3d_proj3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(6.7,1.9,11.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={1,2}; + real[] myzchoice={2,4}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,2.5); + pair ybounds=(-0.5,2.25); + pair zbounds=(-0.5,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //plane z=2y + triple f(pair t) { + return (t.x,t.y,2*t.y); + } + surface s=surface(f,(0,0),(2,1.5),1,1,Spline); + pen p=redpen+.1mm; + draw(s,surfacepen2,meshpen=p); + + //z=3-x^2-y^2 + triple f(pair t) { + return (t.x,t.y,3-t.x^2-t.y^2); + } + surface s=surface(f,(-.1,-.1),(2,1.5),8,8,Spline); + //triple f(pair t) { + // return (cos(t.x)*t.y,sin(t.x)*t.y,3-t.y^2-t.y^2); + //} + // + //surface s=surface(f,(-.1,0),(pi/2+.1,2),8,8,Spline); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p); + + //lines and curves on the surfaces + draw((0,0,0)--(0,1,2),bluepen+dashed+linewidth(1)); + triple g(real t) {return (t,0,2*(1-(t/2)^2));} + path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+dashed+linewidth(1)); + triple g(real t) {return (t,1-(t/2)^2,2*(1-(t/2)^2));} + path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,1-(t/2)^2,0);} + path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+dashed+linewidth(1)); + + + + +
    + +
    + + + + + A curve in space, and its projections onto the three coordinate planes. + +

    + The curve of intersection of the two surfaces in + is plotted relative to a set of three dimensional coordinate axes, + with the surfaces removed to highlight the curve. +

    + +

    + The curve is a parabolic arc in space, from the point (2,0,0) on the x axis, + to the point (0,1,2) on the yz plane. +

    + +

    + Three other curves are shown, these are the projections of the original curve onto the coordinate planes. +

      +
    • +

      + In the xy plane, there is a parabolic arc from (2,0,0) on the x axis, to (0,1,0) on the y axis. +

      +
      • +
    • +

      + In the xz plane, there is a parabolic arc from (2,0,0) on the x axis, to (0,0,2) on the z axis. +

      +
      • +
    • +

      + The projection onto the yz plane is not a parabola; + rather, it is a straight line segment, from the origin to the point (0,1,2). +

      +
      • +
    +

    +     + + + + + //ASY file for fig3d_projb3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(6.7,1.9,11.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={1,2}; + real[] myzchoice={2,4}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,2.5); + pair ybounds=(-0.5,2.25); + pair zbounds=(-0.5,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //plane z=2y + triple f(pair t) { + return (t.x,t.y,2*t.y); + } + surface s=surface(f,(0,0),(2,1.5),1,1,Spline); + pen p=apexmeshpen; + //draw(s,rgb(1,.4,.7)+opacity(.7),meshpen=p); + + //z=3-x^2-y^2 + triple f(pair t) { + return (t.x,t.y,3-t.x^2-t.y^2); + } + surface s=surface(f,(0,0),(2,1.5),8,8,Spline); + pen p=apexmeshpen; + draw(s,invisible,meshpen=invisible); + + //lines and curves on the surfaces + draw((0,0,0)--(0,1,2),bluepen+dashed+linewidth(1)); + triple g(real t) {return (t,0,2*(1-(t/2)^2));} + path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+dashed+linewidth(1)); + triple g(real t) {return (t,1-(t/2)^2,2*(1-(t/2)^2));} + path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,1-(t/2)^2,0);} + path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+dashed+linewidth(1)); + + + + +
    +     + +
    +     + +

    + The two surfaces are z=3-x^2-y^2 and z=2y. + To find where they intersect, + it is natural to set them equal to each other: 3-x^2-y^2=2y. + This is an implicit function of x and y that gives all points (x,y) in the xy-plane where the z values of the two surfaces are equal. +

    + +

    + We can rewrite this implicit function by completing the square: + + 3-x^2-y^2=2y \Rightarrow y^2+2y+x^2=3 \Rightarrow (y+1)^2+x^2=4 + . +

    + +

    + Thus in the xy-plane the projection of the intersection is a circle with radius 2, centered at (0,-1). +

    + +

    + To project onto the xz-plane, + we do a similar procedure: + find the x and z values where the y values on the surface are the same. + We start by solving the equation of each surface for y. + In this particular case, it works well to actually solve for y^2: +

    + +

    + z=3-x^2-y^2 \Rightarrow y^2=3-x^2-z +

    + +

    + z=2y \Rightarrow y^2=z^2/4. +

    + +

    + Thus we have (after again completing the square): + + 3-x^2-z = z^2/4 \Rightarrow \frac{(z+2)^2}{16}+\frac{x^2}4=1 + , + and ellipse centered at (0,-2) in the xz-plane with a major axis of length 8 and a minor axis of length 4. +

    + +

    + Finally, to project the curve of intersection into the yz-plane, + we solve equation for x. + Since z=2y is a cylinder that lacks the variable x, + it becomes our equation of the projection in the yz-plane. +

    + +

    + All three projections are shown in . +

    +     +     + + + + + Finding the volume of a space region with triple integration + +

    + Set up the triple integrals that find the volume of the space region D bounded by the surfaces x^2+y^2=1, + z=0 and z=-y, + as shown in , + with the orders of integration dz\, dy\, dx, + dy\, dx\, dz and dx\, dz\, dy. +

    + +
    + The region D in is shown in (a); in (b), it is collapsed onto the xy-plane + +
    + + + + + A cylindrical wedge in space, resembling a wedge cut from a tree. + +

    + An illustration of the region of integration in . + The region is a three-dimensional solid cut from a cylinder. +

    + +

    + The plane z=-y meets the xy plane along the x axis. + The region that lies between these planes, and within the cylinder x^2+y^2=1, + is a semi-circular wedge. + It resembles the shape of a piece cut from a tree that is being cut down. +

    +     + + + + + //ASY file for figtrip33D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.1,1.9,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,-0.5}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,0.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple f(pair t) {return (cos(t.x),sin(t.x),-sin(t.x)*t.y);} + surface s=surface(f,(-pi,0),(0,1),16,2,Spline); + pen p=apexmeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + triple f(pair t) {return (cos(t.x)*t.y,sin(t.x)*t.y,-sin(t.x)*t.y);} + surface s=surface(f,(-pi,0),(0,1),16,2,Spline); + pen p=redpen+.1mm; + draw(s,simplesurfacepen2,meshpen=p); + + //disk x^2+y^2=1 in 3rd and 4th quadrants + triple g(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(g,pi,2*pi,operator ..);draw(mypath,bluepen+linewidth(1.5)); + + //surface above disk x^2+y^2=1 in 3rd and 4th quadrants + triple g(real t) {return (cos(t),sin(t),-sin(t));} + path3 mypath=graph(g,pi,2*pi,operator ..); + draw(mypath,bluepen+linewidth(1.5)); + + //label the surfaces + label("$x^2+y^2=1$",(1.4,-.5,0)); + draw((1.1,-.6,0)--(.7,-.7,.35),Arrow3(size=2mm)); + label("$z=-y$",(0,-.5,1.25)); + draw((0,-.5,1.15)--(0,-.5,.5),Arrow3(size=2mm)); + + + + +
    + +
    + + + + + The projection of the cylindrical wedge in the previous image onto the xy plane. It is a semi-circular region. + +

    + In three dimensions, the projection of the solid in onto the xy plane is shown. + The projection forms a semi-circular region in the plane, + between the x axis and the half of the circle x^2+y^2=1 with y\leq 0. +

    +     + + + + + //ASY file for figtrip3b3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.1,1.9,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,-0.5}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,0.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //disk x^2+y^2=1 in 3rd and 4th quadrants + triple g(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(g,pi,2*pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + //line + draw((-1,0,0)--(1,0,0),bluepen+linewidth(2)); + + int k=24;//number of panels + + //shade bottom + import three; + for (int i=k; i<2*k; ++i) + { + path3 p = (0,0,0)--(cos(i*pi/k),sin(i*pi/k),0)--(cos((i+1)*pi/k),sin((i+1)*pi/k),0); //Left + draw(surface(p -- cycle), simplesurfacepen); + } + + + + +
    +     + +
    +     + +

    + The order dz\, dy\, dx: +

    + +

    + The region D is bounded below by the plane z=0 and above by the plane z=-y. + The cylinder x^2+y^2=1 does not offer any bounds in the z-direction, + as that surface is parallel to the z-axis. + Thus 0\leq z\leq -y. +

    + +

    + Collapsing the region into the xy-plane, + we get part of the circle with equation + x^2+y^2=1 as shown in . + As a function of x, + this half circle has equation y=-\sqrt{1-x^2}. + Thus y is bounded below by + -\sqrt{1-x^2} and above by y=0: + -\sqrt{1-x^2}\leq y\leq 0. + The x bounds of the half circle are -1\leq x\leq 1. + All together, + the bounds of integration and triple integral are as follows: +

    + + +

    + + 0 \amp \leq z\leq -y + -\sqrt{1-x^2}\leq \amp y\leq 0 + -1 \amp \leq x\leq 1 + +

    + +

    + \Rightarrow +

    + +

    + + \int_{-1}^1\int_{-\sqrt{1-x^2}}^{0}\int_0^{-y}\, dz\, dy\, dx + . +

    +     + +

    + We evaluate this triple integral: + + \int_{-1}^1\int_{-\sqrt{1-x^2}}^{0}\int_0^{-y}\, dz\, dy\, dx \amp = \int_{-1}^1\int_{-\sqrt{1-x^2}}^{0}\big(-y\big)\, dy\, dx + \amp =\int_{-1}^1\big(-\frac12y^2\big)\Big|_{-\sqrt{1-x^2}}^{0}\, dx + \amp = \int_{-1}^1 \frac12\big(1-x^2\big)\, dx + \amp = \left.\left(\frac12\left(x-\frac13x^3\right)\right)\right|_{-1}^1 + \amp = \frac23\text{ units } ^3 + . +

    + +

    + With the order dy\, dx\, dz: +

    + +

    + The region is bounded below + in the y-direction by the surface x^2+y^2=1 \Rightarrow y=-\sqrt{1-x^2} and above + by the surface y=-z. + Thus the y bounds are -\sqrt{1-x^2}\leq y\leq -z. +

    + +

    + Collapsing the region onto the xz-plane gives the region shown in ; + this half disk is bounded by z=0 and x^2+z^2=1. + (We find this curve by solving each surface for y^2, + then setting them equal to each other. + We have y^2=1-x^2 and y=-z\Rightarrow y^2=z^2. + Thus x^2+z^2=1.) + It is bounded below by x=-\sqrt{1-z^2} and above by x=\sqrt{1-z^2}, + where z is bounded by 0\leq z\leq 1. + All together, we have: +

    + + +

    + + -\sqrt{1-x^2} \amp \leq y\leq -z + -\sqrt{1-z^2} \amp \leq x\leq \sqrt{1-z^2} + 0 \amp \leq z\leq 1 + +

    + +

    + \Rightarrow +

    + +

    + + \int_{0}^1\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\int_{-\sqrt{1-x^2}}^{-z}\, dy\, dx\, dz + . +

    +     + +
    + The region D in is shown collapsed onto the xz-plane in (a); in (b), it is collapsed onto the yz-plane + +
    + + + + + The projection of the solid in this example onto the xy plane; it is a semi-circular region. + +

    + On a set of three-dimensional coordinate axes, + the projection of the solid in onto the xz plane is shown. + It is a semi-circular region, bounded below by the x axis, + and bounded above by the semi-circle x^2+z^2=1, z\geq 0. +

    +     + + + + + //ASY file for figtrip3c3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.1,1.9,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,-0.5}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,0.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //disk x^2+y^2=1 in 3rd and 4th quadrants + triple g(real t) {return (cos(t),0,-sin(t));} + path3 mypath=graph(g,pi,2*pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + //line + draw((-1,0,0)--(1,0,0),bluepen+linewidth(2)); + + int k=24;//number of panels + + //shade bottom + import three; + for (int i=k; i<2*k; ++i) + { + path3 p = (0,0,0)--(cos(i*pi/k),0,-sin(i*pi/k))--(cos((i+1)*pi/k),0,-sin((i+1)*pi/k)); //Left + draw(surface(p -- cycle), simplesurfacepen); + } + + + + +
    + +
    + + + + + The projection of the solid in this example onto the yz plane; it is a triangle. + +

    + Using the same three-dimensional coordinate system, + the projection of the solid in onto the yz plane is shown. + This projection is a triangle. The base of the triangle lies along the negative y axis, + from the origin to (0,-1,0). + Another side is the vertical line y=-1 in the yz plane, from (0,-1,0) to (0,-1,1). + The remaining side is the line z=-y, from (-1,0,1) to the origin. +

    +     + + + + + //ASY file for figtrip3d3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.1,1.9,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,-0.5}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,0.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //triangle in space + draw((0,0,0)--(0,-1,0)--(0,-1,1)--cycle,bluepen+linewidth(2)); + + //shade triangle + import three; + + path3 p = (0,0,0)--(0,-1,0)--(0,-1,1); //Left + draw(surface(p -- cycle), simplesurfacepen); + + + + +
    +     + +
    + +

    + With the order dx\, dz\, dy: +

    + +

    + D is bounded below by the surface + x=-\sqrt{1-y^2} and above by \sqrt{1-y^2}. + We then collapse the region onto the yz-plane and get the triangle shown in . (The hypotenuse is the line z=-y, + just as the plane.) Thus z is bounded by + 0\leq z\leq -y and y is bounded by -1\leq y\leq 0. + This gives: +

    + + +

    + + -\sqrt{1-y^2} \amp \leq x\leq \sqrt{1-y^2} + 0 \amp \leq z\leq -y + -1 \amp \leq y\leq 0 + +

    + +

    + \Rightarrow +

    + +

    + + \int_{-1}^0\int_{0}^{-y}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\, dx\, dz\, dy + . +

    +     +     +     + +

    + The following theorem states two things that should make + common sense to us. + First, using the triple integral to find volume of a region D should always return a positive number; + we are computing volume here, + not signed volume. + Secondly, to compute the volume of a + complicated region, + we could break it up into subregions and compute the volumes of each subregion separately, + summing them later to find the total volume. +

    + + + Properties of Triple Integrals + +

    + Let D be a closed, bounded region in space, + and let D_1 and D_2 be non-overlapping regions such that D=D_1\bigcup D_2. + triple integralproperties + iterated integrationproperties +

    + +

    +

      +
    1. +

      + \ds \iiint_D \, dV \geq 0 +

      +
      2. + +
    2. +

      + \ds \iiint_D\, dV = \iiint_{D_1}\, dV + \iiint_{D_2}\, dV. +

      +
      3. +
    +

    +     +     + +

    + We use this latter property in the next example. +

    + + + Finding the volume of a space region with triple integration + +

    + Find the volume of the space region D bounded by the coordinate planes, + z=1-x/2 and z=1-y/4, + as shown in . + Set up the triple integrals that find the volume of D in all 6 orders of integration. +

    + +
    + The region D in is shown in (a); in (b), it is collapsed onto the xy-plane + +
    + + + + + A pyramid with a rectangular base in the plane z=0 and its peak on the z axis. + +

    + The region of integration for is a pyramid with a rectangular base. + The base lies in the xy plane, and the peak is on the z axis at (0,0,1). + The other four sides are triangles that lie above the xy plane: +

      +
    • +

      + One side is in the xz plane, with vertices (0,0,0), (2,0,0) and (0,0,1). +

      +
      • +
    • +

      + One side is in the yz plane, with vertices (0,0,0), (0,4,0), and (0,0,1) +

      +
      • +
    • +

      + Another side lies in a plane labeled with the equation z=1-\frac12 x. + The vertices of this triangle are (2,0,0), (2,4,0), and (0,0,1). +

      +
      • +
    • +

      + The last side lies in a plane labeled with the equation z=1-\frac14 y. + The vertices of this triangle are (0,4,0), (2,4,0), and (0,0,1). +

      +
      • +
    +

    +     + + + + + //ASY file for figtrip43D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11.2,8.3,2.7); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.1,4.25); + pair ybounds=(-0.1,4.25); + pair zbounds=(-0.1,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //lines for tetrahedron + draw((0,0,0)--(2,0,0)--(2,4,0)--(0,4,0)--cycle,bluepen+linewidth(2)); + draw((0,0,0)--(0,0,1),bluepen+linewidth(2)); + draw((2,0,0)--(0,0,1),bluepen+linewidth(2)); + draw((0,4,0)--(0,0,1),bluepen+linewidth(2)); + draw((2,4,0)--(0,0,1),bluepen+linewidth(2)); + + //shade rectangle and sides + import three; + path3 p = (0,0,0)--(2,0,0)--(2,4,0)--(0,4,0); + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0,0,0)--(2,0,0)--(0,0,1); + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0,0,0)--(0,4,0)--(0,0,1); + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0,4,0)--(2,4,0)--(0,0,1); + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (2,0,0)--(2,4,0)--(0,0,1); + draw(surface(p -- cycle), simplesurfacepen); + + //labels and arrows + label("$z=1-\frac{1}{2}x$",(3.5,1.65,0)); + draw((3,1.5,0)--(1.8,2,.1),Arrow3(size=2mm)); + label("$z=1-\frac{1}{4}y$",(0,3,1)); + draw((0,3,0.8)--(1,3,0.25),Arrow3(size=2mm)); + + + + +
    + +
    + + + + + The base of the pyramid shown in the previous image. It is a rectangle, divided into two triangles along one diagonal. + +

    + The base of the pyramid in is shown in the xy plane, + in a three-dimensional coordinate system. +

    + +

    + A dashed line in the plane runs from the origin to the opposite corner of the rectangle, at the point (2,4,0). + This line divides the rectangle into two triangles. + One triangle lies between the x axis and the dashed line, and is labeled R_1. + The other triangle lies between the y axis and the dashed line, and is labeled R_2. +

    +     + + + + + //ASY file for figtrip4d3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11.2,8.3,2.7); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={4}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.1,4.25); + pair ybounds=(-0.1,4.25); + pair zbounds=(-0.1,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //rectangle in xy plane w diagonal and labels + draw((0,0,0)--(2,0,0)--(2,4,0)--(0,4,0)--cycle,bluepen+linewidth(2)); + draw((0,0,0)--(2,4,0),bluepen+dashed+linewidth(1)); + label("$R_1$",(1.5,1,.1)); + label("$R_2$",(1,3,.1)); + + //shade rectangle + import three; + path3 p = (0,0,0)--(2,0,0)--(2,4,0)--(0,4,0); //Left + draw(surface(p -- cycle), simplesurfacepen); + + + + +
    +     + +
    +     + +

    + Following the bounds-determining strategy of + surface to surface, curve to curve, + and point to point, we can see that the most difficult orders of integration are the two in which we integrate with respect to z first, + for there are two upper + surfaces that bound D in the z-direction. + So we start by noting that we have + + 0\leq z\leq 1-\frac12x \text{ and } 0\leq z\leq 1-\frac14y + . +

    + +

    + We now collapse the region D onto the xy-axis, + as shown in . + The boundary of D, + the line from (0,0,1) to (2,4,0), + is shown in as a dashed line; + it has equation y=2x. (We can recognize this in two ways: + one, in collapsing the line from (0,0,1) to (2,4,0) onto the xy-plane, + we simply ignore the z-values, + meaning the line now goes from (0,0) to (2,4). + Secondly, the two surfaces meet where z=1-x/2 is equal to z=1-y/4: + thus 1-x/2=1-y/4 \Rightarrow y=2x.) +

    + +

    + We use the second property of to state that + + \iiint_D \, dV = \iiint_{D_1}\, dV + \iiint_{D_2}\, dV + , + where D_1 and D_2 are the space regions above the plane regions R_1 and R_2, + respectively. + Thus we can say + + \iiint_D\, dV = \iint_{R_1}\left(\int_0^{1-x/2}\, dz\right)dA + \iint_{R_2}\left(\int_0^{1-y/4}\, dz\right)dA + . +

    + +

    + All that is left is to determine bounds of R_1 and R_2, + depending on whether we are integrating with order dx\, dy or dy\, dx. + We give the final integrals here, + leaving it to the reader to confirm these results. +

    + +

    + dz\, dy\, dx: +

    + +

    + + \amp\quad 0 \leq z\leq 1-x/2 \amp \amp\quad 0 \leq z\leq 1-y/4 + \amp\quad 0 \leq y\leq 2x \amp \amp\quad 2x \leq y\leq 4 + \amp\quad 0 \leq x\leq 2 \amp \amp\quad 0 \leq x\leq 2 + \amp\amp\amp + \iiint_D\, dV = \amp \int_0^2\int_0^{2x}\int_0^{1-x/2}\, dz\, dy\, dx \amp +\quad \amp \int_0^2\int_{2x}^4\int_0^{1-y/4}\, dz\, dy\, dx + +

    + +

    + dz\, dx\, dy: +

    + +

    + + \amp\quad 0 \leq z\leq 1-x/2\amp \amp\quad 0 \leq z\leq 1-y/4 + \amp\quad y/2 \leq x\leq 2\amp \amp\quad 0 \leq x\leq y/2 + \amp\quad 0 \leq y\leq 4\amp \amp\quad 0 \leq y\leq 4 + \amp\amp\amp + \iiint_D\, dV = \amp \int_0^4\int_{y/2}^{2}\int_0^{1-x/2}\, dz\, dx\, dy \amp +\quad \amp \int_0^4\int_{0}^{y/2}\int_0^{1-y/4}\, dz\, dx\, dy + +

    + +

    + The remaining four orders of integration do not require a sum of triple integrals. + In + we show D collapsed onto the other two coordinate planes. + Using these graphs, we give the final orders of integration here, + again leaving it to the reader to confirm these results. +

    + +
    + The region D in is shown collapsed onto the xz-plane in (a); in (b), it is collapsed onto the yz-plane + +
    + + + + + The projection of the pyramid onto the plane y=0 is a triangle. + +

    + In a three-dimensional coordinate system, + the side of the pyramid in in the xz plane is shown. + This represents the projection of the solid onto the xz plane, + and it is a triangle, with vertices (0,0,0), (2,0,0), and (0,0,1). +

    +     + + + + + //ASY file for figtrip4b3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11.2,8.3,2.7); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.1,4.25); + pair ybounds=(-0.1,4.25); + pair zbounds=(-0.1,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //lines for tetrahedron + draw((0,0,0)--(2,0,0)--(0,0,1)--cycle,bluepen+linewidth(2)); + + //shade rectangle and sides + import three; + path3 p = (0,0,0)--(2,0,0)--(0,0,1); + draw(surface(p -- cycle), simplesurfacepen); + + + + +
    + +
    + + + + + The projection of the pyramid onto the plane x=0 is a triangle. + +

    + In a three-dimensional coordinate system, + the side of the pyramid in in the yz plane is shown. + This represents the projection of the solid onto the yz plane, + and it is a triangle, with vertices (0,0,0), (0,4,0), and (0,0,1). +

    +     + + + + + //ASY file for figtrip4c3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={2,4}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.1,4.25); + pair ybounds=(-0.1,4.25); + pair zbounds=(-0.1,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //lines for tetrahedron + draw((0,0,0)--(0,4,0)--(0,0,1)--cycle,bluepen+linewidth(2)); + + //shade rectangle and sides + import three; + path3 p = (0,0,0)--(0,4,0)--(0,0,1); + draw(surface(p -- cycle), simplesurfacepen); + + + + +
    +     + +
    + +

    + dy\, dx\, dz: +

    + + +

    + + 0 \amp \leq y\leq 4-4z + 0 \amp \leq x\leq 2-2z + 0 \amp \leq z\leq 1 + +

    + +

    + \Rightarrow +

    + +

    + + \int_0^1\int_{0}^{2-2z}\int_0^{4-4z}\, dy\, dx\, dz + . +

    +     + +

    + dy\, dz\, dx: +

    + + +

    + + 0 \amp \leq y\leq 4-4z + 0 \amp \leq z\leq 1-x/2 + 0 \amp \leq x\leq 2 + +

    + +

    + \Rightarrow +

    + +

    + + \int_0^2\int_{0}^{1-x/2}\int_0^{4-4z}\, dy\, dx\, dz + . +

    +     + +

    + dx\, dy\, dz: +

    + + +

    + + 0 \amp \leq x\leq 2-2z + 0 \amp \leq y\leq 4-4z + 0 \amp \leq z\leq 1 + +

    + +

    + \Rightarrow +

    + +

    + + \int_0^1\int_{0}^{4-4z}\int_0^{2-2z}\, dx\, dy\, dz + . +

    +     + +

    + dx\, dz\, dy: +

    + + +

    + + 0 \amp \leq x\leq 2-2z + 0 \amp \leq z\leq 1-y/4 + 0 \amp \leq y\leq 4 + +

    + +

    + \Rightarrow +

    + +

    + + \int_0^4\int_{0}^{1-y/4}\int_0^{2-2z}\, dx\, dz\, dy + . +

    +     +     +     + +

    + We give one more example of finding the volume of a space region. +

    + + + Finding the volume of a space region + +

    + Set up a triple integral that gives the volume of the space region D bounded by + z= 2x^2+2 and z=6-2x^2-y^2. + These surfaces are plotted in and , + respectively; + the region D is shown in . +

    + +
    + The region D is bounded by the surfaces shown in (a) and (b); D is shown in (c) + +
    + + + + + An elliptic paraboloid, opening downward, is plotted over a rectangular domain. + +

    + A three-dimensional sketch of a surface in space. + The surface is the elliptic paraboloid z=6-2x^2-y^2, + which opens downward and has its vertex at (0,0,6). +

    +     + + + + + //ASY file for figtrip5b3D.asy in Chapter 13 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8.4,6.8,12.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={5}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.5,7); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple f(pair t) { + return (t.x,t.y,6-2*t.x^2-t.y^2); + } + surface s=surface(f,(-1,-2),(1,2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p); + + //lines for tetrahedron + //draw((0,0,0)--(0,4,0)--(0,0,1)--cycle,bluepen+linewidth(2)); + + //shade rectangle and sides + //import three; + //path3 p = (0,0,0)--(0,4,0)--(0,0,1); + //draw(surface(p -- cycle), simplesurfacepen); + + + + +
    + +
    + + + + + A parabolic cylinder in space. It opens upward, and is symmetric about the y axis. + +

    + The surface z=2x^2+2 is a parabolic cylinder: + it has the shape of a long trough with parabolic cross-sections. + It opens upward, and is symmetric about the y axis. +

    +     + + + + + //ASY file for figtrip5c3D.asy in Chapter 13 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8.4,6.8,12.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={5}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.5,7); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple f(pair t) { + return (t.x,t.y,2*t.x^2+2); + } + surface s=surface(f,(-1,-2),(1,2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=redpen+.1mm; + draw(s,surfacepen2,meshpen=p); + + //lines for tetrahedron + //draw((0,0,0)--(0,4,0)--(0,0,1)--cycle,bluepen+linewidth(2)); + + //shade rectangle and sides + //import three; + //path3 p = (0,0,0)--(0,4,0)--(0,0,1); + //draw(surface(p -- cycle), simplesurfacepen); + + + + +
    + +
    + + + + + The region in space bounded by the paraboloid and cylinder in the previous two images. + +

    + A solid in space. + It is bounded above by the elliptic paraboloid in , + and below by the parabolic cylinder in . +

    + +

    + The overall shape of the solid is interesting. + The portion of the parabolic cylinder on the bottom looks like a taco shell. + The paraboloid on top is like an elongated dome. + Perhaps the shape is what you get if you completely fill a taco with a scoop of ice cream, + or make a really thick madeline cookie. +

    +     + + + + + //ASY file for figtrip53D.asy in Chapter 13 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8.4,6.8,12.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={5}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.5,7); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //paraboloid 6-2*t.x^2-t.y^2 + triple f(pair t) { + return (cos(t.x)*t.y,2*sin(t.x)*t.y,6-(2*(cos(t.x)*t.y)^2+(2*sin(t.x)*t.y)^2)); + } + surface s=surface(f,(0,0),(2*pi,1),16,8,Spline); + pen p=apexmeshpen+.1mm; + draw(s,surfacepen,meshpen=p); + + //parametric trough + triple f(pair t) { + return (cos(t.x)*t.y,2*sin(t.x)*t.y,(2*(cos(t.x)*t.y)^2+2)); + } + surface s=surface(f,(0,0),(2*pi,1),16,16,Spline); + pen p=redpen+.1mm; + draw(s,surfacepen2,meshpen=p); + + //ellipse in xy plane + triple g(real t) {return (cos(t),2*sin(t),0);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + //ellipse up on the surface intersection + triple g(real t) {return (cos(t),2*sin(t),2*cos(t)^2+2);} + path3 mypath=graph(g,0,2*pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + + + +
    +     + +
    +     + +

    + The main point of this example is this: + integrating with respect to z first is rather straightforward; + integrating with respect to x first is not. +

    + +

    + The order dz\, dy\, dx: +

    + +

    + The bounds on z are clearly 2x^2+2\leq z\leq 6-2x^2-y^2. + Collapsing D onto the xy-plane gives the ellipse shown in . + The equation of this ellipse is found by setting the two surfaces equal to each other: + + 2x^2+2 = 6-2x^2-y^2 \Rightarrow 4x^2+y^2=4 \Rightarrow x^2+\frac{y^2}4=1 + . +

    + +

    + We can describe this ellipse with the bounds + + -\sqrt{4-4x^2} \leq y\leq \sqrt{4-4x^2} \text{ and } -1\leq x\leq 1 + . +

    + +

    + Thus we find volume as +

    + + +

    + + 2x^2+2 \amp \leq z\leq 6-2x^2-y^2 + -\sqrt{4-4x^2} \amp \leq y\leq \sqrt{4-4x^2} + -1 \amp \leq x\leq 1 + +

    + +

    + \Rightarrow +

    + +

    + + \int_{-1}^1\int_{-\sqrt{4-4x^2}}^{\sqrt{4-4x^2}}\int_{2x^2+2}^{6-2x^2-y^2}\, dz\, dy\, dx + . +

    +     + +

    + The order dy\, dz\, dx: +

    + +

    + Integrating with respect to y is not too difficult. + Since the surface z=2x^2+2 is a cylinder whose directrix is the y-axis, + it does not create a border for y. + The paraboloid z=6-2x^2-y^2 does; + solving for y, we get the bounds + + -\sqrt{6-2x^2-z}\leq y\leq \sqrt{6-2x^2-z} + . +

    + +

    + Collapsing D onto the xz-plane gives the region shown in ; + the lower curve is from the cylinder, + with equation z=2x^2+2. + The upper curve is from the paraboloid; + with y=0, the curve is z=6-2x^2. + Thus bounds on z are 2x^2+2\leq z\leq 6-2x^2; + the bounds on x are -1\leq x\leq 1. + Thus we have: +

    + + +

    + + -\sqrt{6-2x^2-z} \amp \leq y\leq \sqrt{6-2x^2-z} + 2x^2+2 \amp \leq z\leq 6-2x^2 + -1 \amp \leq x\leq 1 + +

    + +

    + \Rightarrow +

    + +

    + + \int_{-1}^1\int_{2x^2+2}^{6-2x^2}\int_{-\sqrt{6-2x^2-z}}^{\sqrt{6-2x^2-z}}\, dy\, dz\, dx + . +

    +     + +
    + The region D in is collapsed onto the xz-plane in (a); in (b), it is collapsed onto the yz-plane + +
    + + + + + The shadow of the solid for this example on the plane y=0. + +

    + In a three-dimensional coordinate system, + a sketch is given of the shadow obtained by projecting the solid in + onto the xz plane. +

    + +

    + It is a plane region, drawn in perspective. + The region is bounded above by a parabola opening downward, with its vertex at (0,0,6), + and is bounded below by a parabola opening upward, with its vertex at (0,0,2). +

    +     + + + + + //ASY file for figtrip5e3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8.4,6.8,12.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={2,4,6}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.5,7); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //curves in xy plane + triple g(real t) {return (t,0,2*t^2+2);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,0,6-2*t^2);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + //shade area inbetween + int k=12; + import three; + for (int i=0; i<k; ++i) + { + path3 p = (i/k,0,2*(i/k)^2+2)-- ((i+1)/k,0,2*((i+1)/k)^2+2)--((i+1)/k,0,6-2*((i+1)/k)^2) -- ((i)/k,0,6-2*(i/k)^2); //Left + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (-i/k,0,2*(i/k)^2+2)-- (-(i+1)/k,0,2*((i+1)/k)^2+2)--(-(i+1)/k,0,6-2*((i+1)/k)^2) -- (-(i)/k,0,6-2*(i/k)^2); //Left + draw(surface(p -- cycle), simplesurfacepen); + } + + + + +
    + +
    + + + + + The shadow of the solid for this example on the plane x=0. It is divided into two regions. + +

    + In a three-dimensional coordinate system, + a sketch is given of the shadow obtained by projecting the solid in + onto the yz plane. +

    + +

    + It is a plane region, drawn in perspective. + The region is divided into two parts. + The bottom part is labeled R_1. + It is bounded below by a horizontal line parallel to the y axis, with z=2. + This corresponds to the bottom of the parabolic cylinder. + The upper bound of R_1 is a parabola that opens downward, with its vertex at (0,0,4). + This parabola comes from the curve that is formed by the intersection of the two surfaces; + it is labeled with the equation z=4-y^2/2. +

    + +

    + The upper part is labeled R_2. This region is bounded below by the parabola that forms the upper bound of R_1. + It is bounded above by a second parabola that also opens downward, with its vertex at (0,0,6). + This parabola corresponds to the x=0 trace through the elliptic paraboloid. + It is labeled with the equation z=6-y^2. +

    +     + + + + + //ASY file for figtrip5d3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8.4,6.8,12.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={2,4,6}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.5,7); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //curves in xy plane + triple g(real t) {return (0,t,6-t^2);} + path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (0,t,4-t^2/2);} + path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + draw((0,-2,2)--(0,2,2),bluepen+linewidth(2)); + + //shade area inbetween + int k=12; + import three; + for (int i=-2*k; i<2*k; ++i) + { + path3 p = (0,i/k,6-(i/k)^2)-- (0,(i+1)/k,6-((i+1)/k)^2)--(0,(i+1)/k,4-(0.5)*((i+1)/k)^2) -- (0,(i)/k,4-(0.5)*((i)/k)^2); //top in blue + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0,i/k,2)-- (0,(i+1)/k,2)--(0,(i+1)/k,4-(0.5)*((i+1)/k)^2) -- (0,(i)/k,4-(0.5)*((i)/k)^2); //bottom in pink + draw(surface(p -- cycle), simplesurfacepen2); + } + + //labels + label("$R_1$",(0.1,.5,3)); + label("$R_2$",(0.1,.5,5)); + label("$z=4-y^2/2$",(0,-1.5,5.5)); + label("$z=6-y^2$",(0,2,6)); + draw((0.2,-1.5,5.3)--(0,-1,3.55),Arrow3(size=2mm)); + draw((0.2,2,5.7)--(0,1,5.05),Arrow3(size=2mm)); + + + + +
    +     + +
    + +

    + The order dx\, dz\, dy: +

    + +

    + This order takes more effort as D must be split into two subregions. + The two surfaces create two sets of upper/lower bounds in terms of x; + the cylinder creates bounds + + -\sqrt{z/2-1}\leq x\leq \sqrt{z/2-1} + + for region D_1 and the paraboloid creates bounds + + -\sqrt{3-y^2/2-z^2/2}\leq x\leq \sqrt{3-y^2/2-z^2/2} + + for region D_2. +

    + +

    + Collapsing D onto the yz-axes gives the regions shown in . + We find the equation of the curve + z=4-y^2/2 by noting that the equation of the ellipse seen in has equation + + x^2+y^2/4=1 \Rightarrow x = \sqrt{1-y^2/4} + . +

    + +

    + Substitute this expression for x in either surface equation, + z=6-2x^2-y^2 or z=2x^2+2. + In both cases, we find + + z=4-\frac12y^2 + . +

    + +

    + Region R_1, corresponding to D_1, has bounds + + 2\leq z\leq 4-y^2/2, -2\leq y\leq 2 + + and region R_2, corresponding to D_2, has bounds + + 4-y^2/2\leq z\leq 6-y^2, -2\leq y\leq 2 + . +

    + +

    + Thus the volume of D is given by: + + \int_{-2}^2\int_2^{4-y^2/2}\int_{-\sqrt{z/2-1}}^{\sqrt{z/2-1}}\, dx\, dz\, dy \, +\, \int_{-2}^2\int_{4-y^2/2}^{6-y^2}\int_{-\sqrt{3-y^2/2-z^2/2}}^{\sqrt{3-y^2/2-z^2/2}}\, dx\, dz\, dy + . +

    +     +     + +

    + If all one wanted to do in + was find the volume of the region D, + one would have likely stopped at the first integration setup + (with order dz\, dy\, dx) + and computed the volume from there. + However, we included the other two methods 1) to show that it could be done, + messy or not, + and 2) because sometimes we have + to use a less desirable order of integration in order to actually integrate. +

    +     + + + Triple Integration and Functions of Three Variables +

    + There are uses for triple integration beyond merely finding volume, + just as there are uses for integration beyond + area under the curve. + These uses start with understanding how to integrate functions of three variables, + which is effectively no different than integrating functions of two variables. + This leads us to a definition, followed by an example. +

    + +

    + + + Iterated Integration, (Part II) + +

    + Let D be a closed, + bounded region in space, + over which g_1(x), g_2(x), f_1(x,y), + f_2(x,y) and h(x,y,z) are all continuous, + and let a and b be real numbers. +

    + +

    + The iterated integral + \ds \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\, dz\, dy\, dx is evaluated as + integrationtriple + triple integral + iterated integration + + \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\, dz\, dy\, dx = \int_a^b\int_{g_1(x)}^{g_2(x)}\left(\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\, dz\right) dy\, dx + . +

    + + + + + Evaluating a triple integral of a function of three variables + +

    + Evaluate \ds \int_0^1\int_{x^2}^x\int_{x^2-y}^{2x+3y} \big(xy+2xz\big)\, dz\, dy\, dx. +

    +     + +

    + We evaluate this integral according to . +

    + +

    + + \amp \int_0^1\int_{x^2}^x\int_{x^2-y}^{2x+3y} \big(xy+2xz\big)\, dz\, dy\, dx + \amp = \int_0^1\int_{x^2}^x\left(\int_{x^2-y}^{2x+3y} \big(xy+2xz\big)\, dz\right)\, dy\, dx + \amp = \int_0^1\int_{x^2}^x\left(\big(xyz+ xz^2\big)\Big|_{x^2-y}^{2x+3y}\right)\, dy\, dx + \amp = \int_0^1\int_{x^2}^x\Bigg(xy(2x+3y)+x(2x+3y)^2-\Big(xy(x^2-y)+x(x^2-y)^2\Big)\Bigg)\, dy\, dx + \amp =\int_0^1\int_{x^2}^x\Big(-x^5+x^3y+4x^3+14x^2y+12xy^2\Big)\, dy\, dx + . +

    + +

    + We continue as we have in the past, showing fewer steps. + + \amp = \int_0^1\Bigg(-\frac72x^7-8x^6-\frac72x^5+15x^4\Bigg)\, dx + \amp = \frac{281}{336}\approx 0.836 + . +

    +     +     + +

    + We now know how to evaluate a triple integral of a function of three variables; + we do not yet understand what it means. + We build up this understanding in a way very similar to how we have understood integration and double integration. +

    + +

    + Let h(x,y,z) be a continuous function of three variables, + defined over some space region D. + We can partition D into n rectangular-solid subregions, + each with dimensions \dx_i\times\dy_i\times\ddz_i. + Let (x_i,y_i,z_i) be some point in the ith subregion, + and consider the product h(x_i,y_i,z_i)\dx_i\dy_i\ddz_i. + It is the product of a function value (that's the + h(x_i,y_i,z_i) part) and a small volume \Delta V_i + (that's the \dx_i\dy_i\ddz_i part). + One of the simplest understanding of this type of product is when h describes the density of an object, + for then h\times\text{ volume } =\text{ mass }. +

    + +

    + We can sum up all n products over D. + Again letting \norm{\Delta D} represent the length of the longest diagonal of the n rectangular solids in the partition, + we can take the limit of the sums of products as \norm{\Delta D}\to 0. + That is, we can find + + S = \lim_{\norm{\Delta D}\to 0} \sum_{i=1}^n h(x_i,y_i,z_i)\Delta V_i=\lim_{\norm{\Delta D}\to 0} \sum_{i=1}^n h(x_i,y_i,z_i)\dx_i\dy_i\ddz_i + . +

    + +

    + While this limit has lots of interpretations depending on the function h, + in the case where h describes density, + S is the total mass of the object described by the region D. +

    + +

    + We now use the above limit to define the + triple integral, + give a theorem that relates triple integrals to iterated iteration, + followed by the application of triple integrals to find the centers of mass of solid objects. +

    + + + Triple Integral + +

    + Let w=h(x,y,z) be a continuous function over a closed, + bounded region D in space, + and let \Delta D be any partition of D into n rectangular solids with volume \Delta V_i. + The triple integral of h over D + is + integrationtriple + triple integral + iterated integration + + \iiint_Dh(x,y,z)\, dV = \lim_{\norm{\Delta D}\to 0}\sum_{i=1}^n h(x_i,y_i,z_i)\Delta V_i + . +

    +     +     + +

    + The following theorem assures us that the above limit exists for continuous functions h and gives us a method of evaluating the limit. +

    + + + Triple Integration (Part II) + +

    + Let w=h(x,y,z) be a continuous function over a closed, + bounded region D in space, + and let \Delta D be any partition of D into n rectangular solids with volume V_i. + integrationtriple + triple integral + iterated integration +

      +
    1. +

      + The limit \lim\limits_{\norm{\Delta D}\to 0}\sum_{i=1}^n h(x_i,y_i,z_i)\Delta V_i exists. +

      +
      2. + +
    2. +

      + If D is defined as the region bounded by the planes x=a and x=b, + the cylinders y=g_1(x) and y=g_2(x), + and the surfaces z=f_1(x,y) and + z=f_2(x,y), where a\lt b, + g_1(x)\leq g_2(x) and f_1(x,y)\leq f_2(x,y) on D, then + + \iiint_D h(x,y,z)\, dV = \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\, dz\, dy\, dx + . +

      +
      3. +
    +

    +     +     + + +

    + Note: In an aside in , + we showed how the summation of rectangles over a region R in the plane could be viewed as a double sum, + leading to the double integral. + Likewise, we can view the sum + + \sum_{i=1}^nh(x_i,y_i,z_i)\dx_i\dy_i\ddz_i + + as a triple sum, + + \sum_{k=1}^p\sum_{j=1}^n\sum_{i=1}^mh(x_i,y_j,z_k)\dx_i\dy_j\ddz_k + , + which we evaluate as + + \sum_{k=1}^p\left(\sum_{j=1}^n\left(\sum_{i=1}^mh(x_i,y_j,z_k)\dx_i\right)\dy_j\right)\ddz_k + . +

    + +

    + Here we fix a k value, + which establishes the z-height of the rectangular solids on one level + of all the rectangular solids in the space region D. + The inner double summation adds up all the volumes of the rectangular solids on this level, + while the outer summation adds up the volumes of each level. +

    + +

    + This triple summation understanding leads to the + \iiint_D notation of the triple integral, + as well as the method of evaluation shown in . +

    + +

    + We now apply triple integration to find the centers of mass of solid objects. +

    +     + + + Mass and Center of Mass +

    + One may wish to review + for a reminder of the relevant terms and concepts. +

    + + + Mass, Center of Mass of Solids + +

    + Let a solid be represented by a closed, + bounded region D in space with variable density function \delta(x,y,z). + moment + center of mass + mass +

    + +

    +

      +
    1. +

      + The mass of the object is \ds M= \iiint_D \, dm=\iiint_D \delta(x,y,z)\, dV. +

      +
      2. + +
    2. +

      + The moment about the yz-plane + is \ds M_{yz}=\iiint_D x\delta(x,y,z)\, dV. +

      +
      3. + +
    3. +

      + The moment about the xz-plane + is \ds M_{xz}=\iiint_D y\delta(x,y,z)\, dV. +

      +
      4. + +
    4. +

      + The moment about the xy-plane + is \ds M_{xy}=\iiint_D z\delta(x,y,z)\, dV. +

      +
      5. + +
    5. +

      + The center of mass of the object is + + \big(\overline{x},\overline{y},\overline{z}\big) = \left(\frac{M_{yz}}M,\frac{M_{xz}}M,\frac{M_{xy}}M\right) + . +

      +
      6. +
    +

    +     +     + + + Finding the center of mass of a solid + +

    + Find the mass and center of mass of the solid represented by the space region bounded by the coordinate planes and z=2-y/3-2x/3, + shown in , + with constant density \delta(x,y,z)=3\,\text{g/cm}^3. + (Note: this space region was used in .) +

    + +
    + Finding the center of mass of the solid in + + + + A triangle in space, formed by the portion of a plane that lies in the first octant. + +

    + This is the same surface shown in . + We repeat the description given there. +

    + +

    + A plane in space is illustrated, by showing the portion of the plane the lies in the first octant. + This portion of the plane is a triangle, whose vertices lie on the coordinate axes. + Using the image (or the equation z=2-y/3-2x/3), + we can determine that the plane meets the coordinate axes at the points (3,0,0), (0,6,0), and (0,0,2). +

    + +

    + The region in space whose volume is computed in this problem is a tetrahedron, + whose faces are the plane described above, + as well as the triangles formed in the three coordinate planes by the coordinate axes and the edges of the first triangle. +

    +     + + + + + //ASY file for figtrip23D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.2,10.1,6.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2,4}; + real[] myychoice={5}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,5); + pair ybounds=(-0.5,7); + pair zbounds=(-0.5,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //triangle in space + draw((3,0,0)--(0,6,0)--(0,0,2)--cycle,bluepen+linewidth(1.5)); + + //shade + import three; + path3 p = (3,0,0)--(0,6,0)--(0,0,2); //Left + draw(surface(p -- cycle), simplesurfacepen); + + + + +
    +     + +

    + We apply . + In , + we found bounds for the order of integration + dz\, dy\, dx to be 0\leq z\leq 2-y/3-2x/3, + 0\leq y\leq 6-2x and 0\leq x\leq 3. + We find the mass of the object: + + M \amp = \iiint_D \delta(x,y,z)\, dV + \amp = \int_0^3\int_0^{6-2x}\int_0^{2-y/3-2x/3} \big(3\big)\, dz\, dy\, dx + \amp = 3\int_0^3\int_0^{6-2x}\int_0^{2-y/3-2x/3} \, dz\, dy\, dx + \amp = 3(6) = 18\,\text{g} + . +

    + +

    + The evaluation of the triple integral is done in , + so we skipped those steps above. + Note how the mass of an object with constant density is simply + densityvolume. +

    + +

    + We now find the moments about the planes. + + M_{xy} \amp = \iiint_D 3z\, dV + \amp = \int_0^3\int_0^{6-2x}\int_0^{2-y/3-2x/3} \big(3z\big)\, dz\, dy\, dx + \amp = \int_0^3\int_0^{6-2x} \frac32\big(2-y/3-2x/3\big)^2\, dy\, dx + \amp = \int_0^3 -\frac49\big(x-3\big)^3\, dx + \amp = 9 + . +

    + +

    + We omit the steps of integrating to find the other moments. + + M_{yz} \amp = \iiint_D 3x\, dV + \amp = \frac{27}2. + M_{xz} \amp = \iiint_D 3y\, dV + \amp = 27 + . +

    + +

    + The center of mass is + + \big(\overline{x},\overline{y},\overline{z}\big) = \left(\frac{27/2}{18},\frac{27}{18},\frac{9}{18}\right) = \big(0.75,1.5,0.5\big) + . +

    +     +     + + + Finding the center of mass of a solid + +

    + Find the center of mass of the solid represented by the region bounded by the planes z=0 and z=-y and the cylinder x^2+y^2=1, + shown in , + with density function \delta(x,y,z) = 10+x^2+5y-5z. + (Note: this space region was used in .) +

    + +
    + Finding the center of mass of the solid in + + + + A cylindrical wedge in space, resembling a wedge cut from a tree. + +

    + This is the same solid in space that was depicted in . + We repeat the description that was given there. +

    +

    + An illustration of the region of integration in . + The region is a three-dimensional solid cut from a cylinder. +

    + +

    + The plane z=-y meets the xy plane along the x axis. + The region that lies between these planes, and within the cylinder x^2+y^2=1, + is a semi-circular wedge. + It resembles the shape of a piece cut from a tree that is being cut down. +

    +     + + + + + //ASY file for figtrip33D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.1,1.9,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,-0.5}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,0.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple f(pair t) {return (cos(t.x),sin(t.x),-sin(t.x)*t.y);} + surface s=surface(f,(-pi,0),(0,1),16,2,Spline); + pen p=apexmeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + triple f(pair t) {return (cos(t.x)*t.y,sin(t.x)*t.y,-sin(t.x)*t.y);} + surface s=surface(f,(-pi,0),(0,1),16,2,Spline); + pen p=redpen+.1mm; + draw(s,simplesurfacepen2,meshpen=p); + + //disk x^2+y^2=1 in 3rd and 4th quadrants + triple g(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(g,pi,2*pi,operator ..);draw(mypath,bluepen+linewidth(1.5)); + + //surface above disk x^2+y^2=1 in 3rd and 4th quadrants + triple g(real t) {return (cos(t),sin(t),-sin(t));} + path3 mypath=graph(g,pi,2*pi,operator ..); + draw(mypath,bluepen+linewidth(1.5)); + + //label the surfaces + label("$x^2+y^2=1$",(1.4,-.5,0)); + draw((1.1,-.6,0)--(.7,-.7,.35),Arrow3(size=2mm)); + label("$z=-y$",(0,-.5,1.25)); + draw((0,-.5,1.15)--(0,-.5,.5),Arrow3(size=2mm)); + + + + +
    +     + +

    + As we start, consider the density function. + It is symmetric about the yz-plane, + and the farther one moves from this plane, the denser the object is. + The symmetry indicates that \overline x should be 0. +

    + +

    + As one moves away from the origin in the y or z directions, + the object becomes less dense, + though there is more volume in these regions. +

    + +

    + Though none of the integrals needed to compute the center of mass are particularly hard, + they do require a number of steps. + We emphasize here the importance of knowing how to set up the proper integrals; + in complex situations we can appeal to technology for a good approximation, + if not the exact answer. + We use the order of integration dz\, dy\, dx, + using the bounds found in . + (As these are the same for all four triple integrals, + we explicitly show the bounds only for M.) + + M \amp = \iiint_D \big(10+x^2+5y-5z\big)\, dV + \amp = \int_{-1}^1\int_{-\sqrt{1-x^2}}^0\int_0^{-y} \big(10+x^2+5y-5z\big)\, dV + \amp = \frac{64}5-\frac{15\pi}{16} \approx 3.855. + M_{yz} \amp = \iiint_D x\big(10+x^2+5y-5z\big)\, dV + \amp =0. + M_{xz} \amp = \iiint_D y\big(10+x^2+5y-5z\big)\, dV + \amp = 2-\frac{61\pi}{48}\approx -1.99. + M_{xy} \amp = \iiint_D z\big(10+x^2+5y-5z\big)\, dV + \amp = \frac{61\pi}{96}-\frac{10}9\approx 0.885 + . +

    + +

    + Note how M_{yz}=0, as expected. + The center of mass is + + \big(\overline{x},\overline{y},\overline{z}\big) = \left(0,\frac{-1.99}{3.855},\frac{0.885}{3.855}\right) \approx \big(0,-0.516, 0.230\big) + . +

    +     +     + +

    + As stated before, + there are many uses for triple integration beyond finding volume. + When h(x,y,z) describes a rate of change function over some space region D, + then \ds \iiint_D h(x,y,z)\, dV gives the total change over D. + Our one specific example of this was computing mass; + a density function is simply a rate of mass change per volume function. + Integrating density gives total mass. +

    + +

    + While knowing how to integrate is important, + it is arguably much more important to know + how to set up integrals. + It takes skill to create a formula that describes a desired quantity; + modern technology is very useful in evaluating these formulas quickly and accurately. +

    + +

    + In , we learn about two new coordinate systems + (each related to polar coordinates) + that allow us to integrate over closed regions in space more easily than when using rectangular coordinates. +

    +     + + + + Terms and Concepts + + + +

    + The strategy for establishing bounds for triple integrals is + to , + to and + to . +

    +     + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

    + surface to surface, curve to curve and point to point +

    +     + +     + + + + +

    + Give an informal interpretation of what + \ds \iiint_D\, dV means. +

    +     + + + +

    + One possible answer is sum up lots of little volumes over D. +

    +     + +     + + + + +

    + Give two uses of triple integration. +

    +     + + + +

    + Answers can vary. + From this section we used triple integration to find the volume of a solid region, + the mass of a solid, and the center of mass of a solid. +

    +     + +     + + + + +

    + If an object has a constant density \delta and a volume V, + what is its mass? +

    +     + + + +

    + \delta V. +

    +     + +     +     + + + Problems + + + +

    + Two functions f_1(x,y) and + f_2(x,y) and a region R in the x, + y plane are given. + Set up and evaluate the double integral that finds the volume + between the surfaces given by the graphs of these two functions over R. +

    +     + + + + +

    + f_1(x,y) = 8-x^2-y^2, f_2(x,y) = 2x+y; +

    + +

    + R is the square with corners (-1,-1) and (1,1). +

    +     + +

    + V = \int_{-1}^1\int_{-1}^1 \big(8-x^2-y^2-(2x+y)\big)\, dx\, dy = 88/3 +

    +     + +     + + + + + $int=Compute("52"); + + + +

    + z=f_1(x,y) = x^2+y^2 and z=f_2(x,y) = -x^2-y^2; +

    +

    + R is the square with corners (0,0) and (2,3). +

    + +

    + +

    +     +     +     + + + + +

    + f_1(x,y) = \sin(x) \cos(y), + f_2(x,y) = \cos(x) \sin(y) +2; +

    + +

    + R is the triangle with corners (0,0), + (\pi,0) and (\pi,\pi). +

    +     + +

    + V = \int_{0}^{\pi}\int_{0}^x \big(\cos(x) \sin(y) +2-\sin(x) \cos(y) \big)\, dy\, dx = \pi^2-\pi\approx 6.728 +

    +     + +     + + + + + $int=Compute("3pi/2"); + + + +

    + f_1(x,y) = 2x^2+2y^2+3 and f_2(x,y) = 6-x^2-y^2; +

    + +

    + R is the disc x^2+y^2\leq1. +

    + +

    + +

    +     +     +     + +     + + + +

    + A domain D is described by its bounding surfaces, + along with a graph. + Set up the triple integrals that give the volume of D in all 6 orders of integration, + and find the volume of D by evaluating the indicated triple integral. +

    +     + + + + +

    + D is bounded by the coordinate planes and +

    + +

    + z=2-2x/3-2y. +

    + +

    + Evaluate the triple integral with order dz\, dy\, dx. +

    + + + + + A tetrahedron in the first octant, with one vertex at the origin, and the others on the coordinate axes. + +

    + The solid is a tetrahedron, plotted in the first octant relative to the usual three-dimensional coordinate axes. + The vertices of the tetrahedron are at (0,0,0), (3,0,0), (0,1,0), and (0,0,2). + Three of the four faces lie in the coordinate planes; + the remaining face is in the first octant, and is labeled with the equation z=2-\frac23 x-2y. +

    +     + + + + + //ASY file for fig13_06_ex_07.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3}; + real[] myychoice={1,2,3}; + real[] myzchoice={1,2}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,3.5); + pair ybounds=(-0.5,3.5); + pair zbounds=(-0.5,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //edges of tetrahedron + draw((0,0,0)--(3,0,0)--(0,1,0)--cycle,bluepen+linewidth(2)); + draw((3,0,0)--(0,0,2),bluepen+linewidth(2)); + draw((0,0,0)--(0,0,2),bluepen+linewidth(2)); + draw((0,1,0)--(0,0,2),bluepen+linewidth(2)); + + //shade faces + import three; + path3 p=(0,0,0)--(3,0,0)--(0,1,0); + draw(surface(p -- cycle), simplesurfacepen); + path3 p=(0,0,0)--(3,0,0)--(0,0,2); + draw(surface(p -- cycle), simplesurfacepen); + path3 p=(0,0,0)--(0,1,0)--(0,0,2); + draw(surface(p -- cycle), simplesurfacepen); + path3 p=(0,0,0)--(3,0,0)--(0,0,2); + draw(surface(p -- cycle), simplesurfacepen); + + //labels and arrow + label("$z=2-\frac{2}{3}x-2y$",(3,3,0)); + draw((2.5,2.5,.25)--(1.5,0.5,0.7),Arrow3(size=2mm)); + + + + +     + +

    + dz\, dy\, dx: + \ds\int_0^3\int_0^{1-x/3}\int_0^{2-2x/3-2y}\, dz\, dy\, dx +

    + +

    + dz\, dx\, dy: + \ds\int_0^1\int_0^{3-3y}\int_0^{2-2x/3-2y}\, dz\, dx\, dy +

    + +

    + dy\, dz\, dx: + \ds\int_0^3\int_0^{2-2x/3}\int_0^{1-x/3-z/2}\, dy\, dz\, dx +

    + +

    + dy\, dx\, dz: + \ds\int_0^2\int_0^{3-3z/2}\int_0^{1-x/3-z/2}\, dy\, dx\, dz +

    + +

    + dx\, dz\, dy: + \ds\int_0^1\int_0^{2-2y}\int_0^{3-3y-3z/2}\, dx\, dz\, dy +

    + +

    + dx\, dy\, dz: + \ds\int_0^2\int_0^{1-z/2}\int_0^{3-3y-3z/2}\, dx\, dy\, dz +

    + +

    + \ds V = \int_0^3\int_0^{1-x/3}\int_0^{2-2x/3-2y}\, dz\, dy\, dx =1. +

    +     + +     + + + + +

    + D is bounded by the planes y=0, + y=2, x=1, z=0 and +

    + +

    + z=(3-x)/2. +

    + +

    + Evaluate the triple integral with order dx\, dy\, dz. +

    + + + + + A triangular prism. The two triangular sides are parallel to the xz plane. + +

    + The solid is a triangular prism. The base is a square, and it lies in the xy plane. + The corners of the square are at (1,0,0), (3,0,0), (3,2,0), and (1,2,0). +

    + +

    + The two triangular faces are parallel to the xz plane. + One face is in the plane y=0, with vertices (1,0,0), (3,0,0), and (3,0,1). + The other face is in the plane y=2, with vertices (1,2,0), (3,2,0), and (3,2,1). +

    + +

    + The front face (according to how the solid is oriented in the image) + is a rectangle in the plane z=\frac12(3-x). + Two edges are parallel to the y axis, and two edges are parallel to the hypotenuse of the triangular faces. +

    + +

    + The last face is at the back, in the plane x=1. + This is a vertical rectangle, with 0\leq y\leq 2 and 0\leq z\leq 1. +

    +     + + + + + //ASY file for fig13_06_ex_083D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8.8,7.8,3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3}; + real[] myychoice={1,2,3}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,3.5); + pair ybounds=(-0.5,3.5); + pair zbounds=(-0.5,1.75); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //edges of object + draw((1,0,0)--(1,2,0)--(3,2,0)--(3,0,0)--cycle,bluepen+linewidth(2)); + draw((3,0,0)--(1,0,1)--(1,2,1)--(3,2,0),bluepen+linewidth(2)); + draw((3,0,0)--(1,0,1),bluepen+linewidth(2)); + draw((1,0,0)--(1,0,1),bluepen+linewidth(2)); + draw((1,2,0)--(1,2,1),bluepen+linewidth(2)); + + //shade faces + import three; + path3 p=(1,0,0)--(1,2,0)--(3,2,0)--(3,0,0); + draw(surface(p -- cycle), simplesurfacepen); + path3 p=(1,0,1)--(1,2,1)--(3,2,0)--(3,0,0); + draw(surface(p -- cycle), simplesurfacepen); + path3 p=(1,0,0)--(1,2,0)--(1,2,1)--(1,0,1); + draw(surface(p -- cycle), simplesurfacepen); + path3 p=(1,0,0)--(3,0,0)--(1,0,1); + draw(surface(p -- cycle), simplesurfacepen); + path3 p=(1,2,0)--(3,2,0)--(1,2,1); + draw(surface(p -- cycle), simplesurfacepen); + + //labels and arrow + label("$z=\frac{1}{2}(3-x)$",(3,0,1.1)); + draw((3,0.25,1)--(2,1,0.45),Arrow3(size=2mm)); + + + + +     + +

    + dz\, dy\, dx: + \ds\int_1^3\int_0^{2}\int_0^{(3-x)/2}\, dz\, dy\, dx +

    + +

    + dz\, dx\, dy: + \ds\int_0^2\int_1^{3}\int_0^{(3-x)/2}\, dz\, dx\, dy +

    + +

    + dy\, dz\, dx: + \ds\int_1^3\int_0^{(3-x)/2}\int_0^{2}\, dy\, dz\, dx +

    + +

    + dy\, dx\, dz: + \ds\int_0^1\int_1^{3-2z}\int_0^{2}\, dy\, dx\, dz +

    + +

    + dx\, dz\, dy: + \ds\int_0^2\int_0^{1}\int_1^{3-2z}\, dx\, dz\, dy +

    + +

    + dx\, dy\, dz: + \ds\int_0^1\int_0^{2}\int_1^{3-2z}\, dx\, dy\, dz +

    + +

    + \ds V = \int_0^1\int_0^{2}\int_1^{3-2z}\, dx\, dy\, dz =2. +

    +     + +     + + + + +

    + D is bounded by the planes x=0, + x=2, z=-y and by +

    + +

    + z=y^2/2. +

    + +

    + Evaluate the triple integral with the order dy\, dz\, dx. +

    + + + + + A solid bounded above by a plane, and below by a parabolic cylinder. + +

    + On a set of three-dimensional coordinate axes, a solid is given between two surfaces. +

    + +

    + One surface is a plane. It is illustrated as a rectangle. +

      +
    • +

      + The bottom of the rectangle lies along the x axis, for x between 0 and 2. +

      +
      • +
    • +

      + The sides are negatively sloped, relative to the yz: + the equation of the plane is z=-y, + and setting x=0 or x=2 in this plane produces the sides. +

      +
      • +
    • +

      + The top of the rectangle is parallel to the x axis; + it is a line segment with y=-2, z=2, and 0\leq x\leq 2. +

      +
      • +
    +

    + +

    + The other surface is a parabolic cylinder. + Cross-sections are the half of the cylinder z=\frac12 y^2 with y\leq 0. + This surface meets the rectangle in the plane at the top and bottom edges of the rectangle. + It has the appearance of a sheet that has been hung below the rectangle and attached at either end. +

    + +

    + Projecting the solid onto either the xy or xz planes produces a rectangle, + while projection onto the yz plane produces a region that is bounded above by the line z=-y, + and below by the parabola z=\frac12 y^2. +

    +     + + + + + //ASY file for fig13_06_ex_093D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={}; + real[] myzchoice={1,2}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,2.5); + pair ybounds=(-2.5,0.5); + pair zbounds=(-0.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //edges of plane + draw((0,0,0)--(2,0,0)--(2,-2,2)--(0,-2,2)--cycle,bluepen+linewidth(2)); + + //shade plane + import three; + path3 p=(0,0,0)--(2,0,0)--(2,-2,2)--(0,-2,2); + draw(surface(p -- cycle), simplesurfacepen); + + //edges of parabolic cylinder + triple g(real t) {return (2,t,0.5*t^2);} + path3 mypath=graph(g,-2,0,operator ..); + draw(mypath,redpen+linewidth(2)); + triple g(real t) {return (0,t,0.5*t^2);} + path3 mypath=graph(g,-2,0,operator ..); + draw(mypath,redpen+linewidth(2)); + + //Shade the cylinder + int k=10; + for (int i=0; i<2*k; ++i) + { + path3 p=(2,-i/k,0.5*(i/k)^2)--(2,-(i+1)/k,0.5*((i+1)/k)^2)--(0,-(i+1)/k,0.5*((i+1)/k)^2)--(0,-i/k,0.5*(i/k)^2); + draw(surface(p -- cycle), surfacepen2); + } + + //labels and arrow + label("$z=-y$",(.5,0.5,2),E);draw((.5,.75,1.8)--(.75,-1,1),Arrow3(size=2mm)); + label("$z=\frac{1}{2}y^2$",(2.25,-1.5,.75),W); + draw((2.25,-1.5,.7)--(2,-0.85,.32),Arrow3(size=2mm)); + + + + +     + +

    + dz\, dy\, dx: + \ds\int_0^2\int_{-2}^{0}\int_{y^2/2}^{-y}\, dz\, dy\, dx +

    + +

    + dz\, dx\, dy: + \ds\int_{-2}^0\int_0^{2}\int_{y^2/2}^{-y}\, dz\, dx\, dy +

    + +

    + dy\, dz\, dx: + \ds\int_0^2\int_0^{2}\int_{-\sqrt{2z}}^{-z}\, dy\, dz\, dx +

    + +

    + dy\, dx\, dz: + \ds\int_0^2\int_0^{2}\int_{-\sqrt{2z}}^{-z}\, dy\, dx\, dz +

    + +

    + dx\, dz\, dy: + \ds\int_{-2}^0\int_{y^2/2}^{-y}\int_0^{2}\, dx\, dz\, dy +

    + +

    + dx\, dy\, dz: + \ds\int_0^2\int_{-\sqrt{2z}}^{-z}\int_0^{2}\, dx\, dy\, dz \ds V = \int_0^2\int_0^{2}\int_{-\sqrt{2z}}^{-z}\, dy\, dz\, dx =4/3. +

    +     + +     + + + + +

    + D is bounded by the planes z=0, + y=9, x=0 and by +

    + +

    + z=\sqrt{y^2-9x^2}. +

    + +

    + Do not evaluate any triple integral. +

    + + + + + One quarter of an elliptic cone that opens along the y axis. + +

    + The solid for this exercise is bounded by the xy and yz coordinate planes, + and the elliptic cone z=\sqrt{y^2-9x^2}. +

    + +

    + The cone opens along the y axis. + Only a quarter of the cone is shown in the image; + namely, that portion that lies in the first octant. +

    + +

    + The projection of the solid onto the xy plane is a triangle, + with vertices at (0,0,0), (0,9,0), and (3,9,0). +

    + +

    + The solid also meets the yz plane in a triangle, + with vertices at (0,0,0), (0,9,0), and (0,9,9). +

    + +

    + The projection of the solid onto the xz plane is a quarter ellipse. + It is the region bounded by the ellipse 9x^2+z^2=81, + and the positive x and z axes. +

    +     + + + + + //ASY file for fig13_06_ex_103D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14,10,21); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={3}; + real[] myychoice={3,6,9}; + real[] myzchoice={3,6,9}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,5); + pair ybounds=(-0.25,10); + pair zbounds=(-0.25,10); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //edges of object + draw((0,0,0)--(0,9,0)--(3,9,0)--cycle,bluepen+linewidth(2)); + draw((0,9,0)--(0,9,9)--(0,0,0)--cycle,bluepen+linewidth(2)); + triple g(real t) {return (t,9,sqrt(81-9*t^2));} + path3 mypath=graph(g,0,3,operator ..);draw(mypath,bluepen+linewidth(2)); + + triple f(pair t) { + return (t.x/3*t.y,t.x,sqrt(t.x^2-9(t.x/3*t.y)^2)); + } + surface s=surface(f,(0,0),(9,1),8,12,Spline); + pen p=apexmeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + //label and arrow + label("$z=\sqrt{y^2-9x^2}$",(4,4.5,0)); + draw((3.3,4,.2)--(1.25,4,1.4),Arrow3(size=2mm)); + + + + +     + +

    + dz\, dy\, dx: + \ds\int_0^3\int_{3x}^{9}\int_{0}^{\sqrt{y^2-9x^2}}\, dz\, dy\, dx +

    + +

    + dz\, dx\, dy: + \ds\int_{0}^9\int_0^{y/3}\int_{0}^{\sqrt{y^2-9x^2}}\, dz\, dx\, dy +

    + +

    + dy\, dz\, dx: + \ds\int_0^3\int_0^{\sqrt{81-9x^2}}\int_{\sqrt{z^2+9x^2}}^{9}\, dy\, dz\, dx +

    + +

    + dy\, dx\, dz: + \ds\int_0^9\int_0^{\sqrt{9-z^2/9}}\int_{\sqrt{z^2+9x^2}}^{9}\, dy\, dx\, dz +

    + +

    + dx\, dz\, dy: + \ds\int_{0}^9\int_{0}^{y}\int_0^{\frac13\sqrt{y^2-z^2}}\, dx\, dz\, dy +

    + +

    + dx\, dy\, dz: + \ds\int_0^9\int_{z}^{9}\int_0^{\frac13\sqrt{y^2-z^2}}\, dx\, dy\, dz +

    +     + +     + + + + +

    + D is bounded by the planes x=2, + y=1, z=0 and +

    + +

    + z=2x+4y-4. +

    + +

    + Evaluate the triple integral with the order dx\, dy\, dz. +

    + + + + + A tetrahedron in space, plotted with respect to a three-dimensional coordinate system. + +

    + The image contains the usual three-dimensional coordinate axes, and a tetrahedron. + Three of the faces of the tetrahedron are parallel to the coordinate planes. +

      +
    • +

      + The bottom is a triangle in the xy plane with vertices (2,0,0), (2,1,0), and (0,1,0). +

      +
      • +
    • +

      + Another face lies in the plane x=2, forming a triangle with vertices (2,0,0), (2,1,0), and (2,1,4). +

      +
      • +
    • +

      + A third face lies in the plane y=1; it is a triangle with vertices (0,1,0), (2,1,0), and (2,1,4). +

      +
      • +
    • +

      + The last face lies in the plane z=2x+4y-4; it is a triangle with vertices (2,0,0), (0,1,0), and (2,1,4). +

      +
      • +
    +

    +     + + + + + //ASY file for fig13_06_ex_113D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.2,-5.1,8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={1,2}; + real[] myzchoice={2,4}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,2.5); + pair ybounds=(-0.25,2.5); + pair zbounds=(-0.25,5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //edges of object + draw((0,1,0)--(2,1,0)--(2,0,0)--cycle,bluepen+linewidth(2)); + draw((0,1,0)--(2,1,4)--(2,1,0)--cycle,bluepen+linewidth(2)); + draw((0,1,0)--(2,1,4)--(2,0,0)--cycle,bluepen+linewidth(2)); + + //shade plane + import three; + path3 p=(0,1,0)--(2,1,4)--(2,0,0); + draw(surface(p -- cycle), simplesurfacepen); + + //label and arrow + label("$z=2x+4y-4$",(1,0.5,4)); + draw((1,0.5,3.6)--(1.2,.65,1.6),Arrow3(size=2mm)); + + + + +     + +

    + dz\, dy\, dx: + \ds\int_0^2\int_{1-x/2}^{1}\int_{0}^{2x+4y-4}\, dz\, dy\, dx +

    + +

    + dz\, dx\, dy: + \ds\int_{0}^1\int_{2-2y}^{2}\int_{0}^{2x+4y-4}\, dz\, dx\, dy +

    + +

    + dy\, dz\, dx: + \ds\int_0^2\int_0^{2x}\int_{z/4-x/2+1}^{1}\, dy\, dz\, dx +

    + +

    + dy\, dx\, dz: + \ds\int_0^4\int_{z/2}^{2}\int_{z/4-x/2+1}^{1}\, dy\, dx\, dz +

    + +

    + dx\, dz\, dy: + \ds\int_{0}^1\int_{0}^{4y}\int_{z/2-2y+2}^2\, dx\, dz\, dy +

    + +

    + dx\, dy\, dz: + \ds\int_0^4\int_{z/4}^{1}\int_{z/2-2y+2}^2\, dx\, dy\, dz + \ds V = \int_0^4\int_{z/4}^{1}\int_{z/2-2y-2}^2\, dx\, dy\, dz = 4/3. +

    +     + +     + + + + +

    + D is bounded by the planes z=0 and z=2y, and by y=4-x^2. +

    + +

    + Evaluate the triple integral with the order dz\, dy\, dx. +

    + + + + + A cylindrical wedge with its edge along the x axis. + +

    + A cylindrical wedge, between the planes z=0 and z=2y. + These planes intersect along the x axis, forming the sharp edge of the wedge. + The round face of the wedge is the parabolic cylinder y=4-x^2, for y\geq 0. +

    + +

    + The projection of the surface onto the xy plane is the region bounded below by the x axis, + and above by the parabola y=4-x^2. +

    + +

    + The projection of the surface onto the yz plane is a triangle, + with vertices at (0,0,0), (0,4,0), and (0,4,8). +

    + +

    + The projection of the surface onto the xz plane is the region bounded below by the x axis, + and above by the parabola z = 8-2x^2. +

    +     + + + + + //ASY file for fig13_06_ex_123D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12.1,-7.1,16); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,-1,1,2}; + real[] myychoice={1,2,3,4}; + real[] myzchoice={2,4,6,8}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-0.25,5); + pair zbounds=(-0.25,10); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + triple g(real t) {return (t,4-t^2,0);} + path3 mypath=graph(g,-2,2,operator ..);draw(mypath,redpen+linewidth(2)); + triple g(real t) {return (t,4-t^2,2*(4-t^2));} + path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + //shade object + import three; + int k=12; + for (int i=-2*k; i<2*k; ++i) + { + path3 p=(i/k,4-(i/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i)/k,4-((i)/k)^2,2*(4-((i)/k)^2)); + draw(surface(p -- cycle), simplesurfacepen2); + path3 p=(i/k,0,0)--(i/k,4-(i/k)^2,2*(4-(i/k)^2))--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i+1)/k,0,0); + draw(surface(p -- cycle), simplesurfacepen); + } + + //label and arrow + label("$z=2y$",(-2,2,7)); + draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + label("$y=4-x^2$",(2.5,2,0)); + draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + + +     + +

    + dz\, dy\, dx: + \ds\int_{-2}^2\int_{0}^{4-x^2}\int_{0}^{2y}\, dz\, dy\, dx +

    + +

    + dz\, dx\, dy: + \ds\int_{0}^4\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\int_{0}^{2x+4y-4}\, dz\, dx\, dy +

    + +

    + dy\, dz\, dx: + \ds\int_{-2}^2\int_0^{8-2x^2}\int_{z/2}^{4-x^2}\, dy\, dz\, dx +

    + +

    + dy\, dx\, dz: + \ds\int_0^8\int_{-\sqrt{4-z/2}}^{\sqrt{4-z/2}}\int_{z/2}^{4-x^2}\, dy\, dx\, dz +

    + +

    + dx\, dz\, dy: + \ds\int_{0}^4\int_{0}^{2y}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\, dx\, dz\, dy +

    + +

    + dx\, dy\, dz: + \ds\int_0^8\int_{z/2}^{4}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\, dx\, dy\, dz + \ds V = \int_{-2}^2\int_{0}^{4-x^2}\int_{0}^{2y}\, dz\, dy\, dx = 512/15. +

    +     + +     + + + + +

    + D is bounded by the coordinate planes and by +

    + +

    + y=1-x^2 and y=1-z^2. +

    + +

    + Do not evaluate any triple integral. + Which order is easier to evaluate: + dz\, dy\, dx or dy\, dz\, dx? + Explain why. +

    + + + + + A region in the first octant bounded by the planes x=0 and z=0, and two parabolic cylinders. + +

    + This solid is bounded by two planes, and two parabolic cylinders. +

      +
    • +

      + The bottom of the solid is in the xy plane, + in a region bounded by the x and y coordinate axes and the parabola y=1-x^2. +

      +
      • +
    • +

      + Another side of the solid lies in the yz plane, + forming a region bounded by the y and z coordinate axes and the parabola y=1-z^2. +

      +
      • +
    • +

      + The solid also has a face in the xz plane. This face is a square, with x and z between 0 and 1. +

      +
      • +
    • +

      + The solid is bounded above by the parabolic cylinder y=1-z^2, + and the remaining face is the parabolic cylinder y=1-x^2. +

      +
      • +
    +

    + +

    + When setting up integrals where the integration is first with respect to y, + it is important to note that the square in the xz plane needs to be divided into two triangles along the line x=z. + For z\geq x, y is bounded above by the cylinder y=1-z^2. + For z\leq x, y is bounded above by the cylinder y=1-x^2. +

    +     + + + + + //ASY file for fig13_06_ex_133D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,1.5); + pair ybounds=(-0.25,1.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + triple g(real t) {return (t,1-t^2,0);} + path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (0,1-t^2,t);} + path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,1-t^2,t);} + path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + //draw square and sideline + draw((0,0,0)--(1,0,0)--(1,0,1)--(0,0,1)--(0,0,0),bluepen+linewidth(2)); + draw((0,0,0)--(0,1,0),bluepen+linewidth(2)); + + //shade object + import three; + int k=12; + for (int i=0; i<k; ++i) + { + path3 p=(0,1-(i/k)^2,i/k)--(0,1-((i+1)/k)^2,(i+1)/k)--((i+1)/k,1-((i+1)/k)^2,(i+1)/k)--((i)/k,1-((i)/k)^2,(i)/k); + draw(surface(p -- cycle), simplesurfacepen2);//pink + + path3 p=(i/k,1-(i/k)^2,0)--((i+1)/k,1-((i+1)/k)^2,0)--((i+1)/k,1-((i+1)/k)^2,(i+1)/k)--((i)/k,1-((i)/k)^2,(i)/k); + draw(surface(p -- cycle), simplesurfacepen);//blue + } + + //label and arrow + label("$y=1-x^2$",(1,1,0)); + draw((0.9,0.9,.1)--(.65,.6,.4),Arrow3(size=2mm)); + label("$y=1-z^2$",(0,1,.85)); + draw((.1,.9,.75)--(.3,.8,.45),Arrow3(size=2mm)); + + + + +     + +

    + dz\, dy\, dx: + \ds\int_{0}^1\int_{0}^{1-x^2}\int_{0}^{\sqrt{1-y}}\, dz\, dy\, dx +

    + +

    + dz\, dx\, dy: + \ds\int_{0}^1\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\, dz\, dx\, dy +

    + +

    + dy\, dz\, dx: + \ds\int_{0}^1\int_0^{x}\int_{0}^{1-x^2}\, dy\, dz\, dx + \int_{0}^1\int_x^{1}\int_{0}^{1-z^2}\, dy\, dz\, dx +

    + +

    + dy\, dx\, dz: + \ds\int_0^1\int_{0}^{z}\int_{0}^{1-z^2}\, dy\, dx\, dz+\int_0^1\int_{z}^{1}\int_{0}^{1-x^2}\, dy\, dx\, dz +

    + +

    + dx\, dz\, dy: + \ds\int_{0}^1\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\, dx\, dz\, dy +

    + +

    + dx\, dy\, dz: + \ds\int_0^1\int_{0}^{1-z^2}\int_{0}^{\sqrt{1-y}}\, dx\, dy\, dz + Answers will vary. Neither order is particularly hard. The order dz\, dy\, dx requires integrating a square root, so powers can be messy; the order dy\, dz\, dx requires two triple integrals, but each uses only polynomials. +

    +     + +     + + + + +

    + D is bounded by the coordinate planes and by +

    + +

    + z=1-y/3 and z=1-x. +

    + +

    + Evaluate the triple integral with order dx\, dy\, dz. +

    + + + + + A pyramid with a rectangular base and vertex on the z axis. + +

    + The solid for this exercise is a pyramid. The base in the xy plane is a rectangle, + with x between 0 and 1, and y between 0 and 3. +

    + +

    + One side of the pyramid lies in the plane z=1-x; this is the upper bound when integrating first with respect to y. +

    + +

    + Another side of the pyramid lies in the plane z=1-y/3; this is the upper bound when integrating first with respect to x. +

    + +

    + The remaining sides are in the other two coordinate planes: +

      +
    • +

      + In the xz plane, we have the triangle with vertices (0,0,0), (1,0,0), and (0,0,1) +

      +
      • +
    • +

      + In the yz plane, we have the triangle with vertices (0,0,0), (0,2,0), and (0,0,1) +

      +
      • +
    +

    + +

    + When integrating first with respect to z, + note that the rectangle in the xy plane needs to be divided into two triangles along the line y=3x. + For y\geq 3x, the upper bound is the plane z=1-y/3. + For y\leq 3x, the upper bound is the plane z=1-x. +

    +     + + + + + //ASY file for fig13_06_ex_143D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={1,2,3}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,2.5); + pair ybounds=(-0.25,3.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //edges of object + draw((0,0,0)--(1,0,0)--(1,3,0)--(0,3,0)--cycle,bluepen+linewidth(2)); + draw((1,0,0)--(0,0,1),bluepen+linewidth(2)); + draw((0,3,0)--(0,0,1),bluepen+linewidth(2)); + draw((0,0,0)--(0,0,1),bluepen+linewidth(2)); + draw((1,3,0)--(0,0,1),bluepen+linewidth(2)); + + //shade object + import three; + path3 p=(1,0,0)--(1,3,0)--(0,0,1); + draw(surface(p -- cycle), simplesurfacepen); + path3 p=(1,3,0)--(0,3,0)--(0,0,1); + draw(surface(p -- cycle), simplesurfacepen2); + + //label and arrow + label("$z=1-x$",(2,1.5,0)); + draw((1.75,1.5,0.1)--(0.75,1,.35),Arrow3(size=2mm)); + label("$z=1-y/3$",(0,2.5,1)); + draw((0.15,2.5,0.85)--(0.35,2.25,.35),Arrow3(size=2mm)); + + + + +     + +

    + dz\, dy\, dx: + \ds\int_{0}^1\int_{0}^{3x}\int_{0}^{1-x}\, dz\, dy\, dx+\int_{0}^1\int_{3x}^{3}\int_{0}^{1-y/3}\, dz\, dy\, dx +

    + +

    + dz\, dx\, dy: + \ds\int_{0}^3\int_{0}^{y/3}\int_{0}^{1-y/3}\, dz\, dy\, dx+\int_{0}^3\int_{y/3}^{1}\int_{0}^{1-x}\, dz\, dx\, dy +

    + +

    + dy\, dz\, dx: + \ds\int_{0}^1\int_0^{1-x}\int_{0}^{3-3z}\, dy\, dz\, dx +

    + +

    + dy\, dx\, dz: + \ds\int_0^1\int_{0}^{1-z}\int_{0}^{3-3z}\, dy\, dx\, dz +

    + +

    + dx\, dz\, dy: + \ds\int_{0}^3\int_{0}^{1-y/3}\int_{0}^{1-z}\, dx\, dz\, dy +

    + +

    + dx\, dy\, dz: + \ds\int_0^1\int_{0}^{3-3z}\int_{0}^{1-z}\, dx\, dy\, dz + \ds V = \int_0^1\int_{0}^{3-3z}\int_{0}^{1-z}\, dx\, dy\, dz = 1. +

    +     + +     + +     + + + +

    + Evaluate the triple integral. +

    +     + + + + +

    + \ds \int_{-\pi/2}^{\pi/2}\int_0^\pi\int_0^\pi\big(\cos(x) \sin(y) \sin(z) \big)\, dz\, dy\, dx +

    +     + +     + + + + + $int=Compute("7/8"); + + + +

    + Evaluate \ds \int_{0}^{1}\int_0^x\int_0^{x+y}(x+y+z)\,dz\,dy\,dx. +

    + +

    + +

    +     +     +     + + + + +

    + \ds \int_{0}^{\pi}\int_{0}^{1}\int_{0}^{z}\big(\sin(yz)\big)\, dx\, dy\, dz +

    +     + +

    + \pi +

    +     + +     + + + + + $int=Compute("0"); + + + +

    + Evaluate \ds \int_{\pi}^{\pi^2}\int_{x}^{x^3}\int_{-y^2}^{y^2}\left(z\frac{x^2y+y^2x}{e^{x^2+y^2}}\right)\,dz\,dy\,dx. +

    + +

    + +

    +     +     +     + +     + + + +

    + Find the center of mass of the solid represented by the indicated space region D with density function \delta(x,y,z). +

    +     + + + + +

    + D is bounded by the coordinate planes and +

    + +

    + z=2-2x/3-2y; \delta(x,y,z) = 10\,\text{g/cm}^3. +

    + +

    + (Note: this is the same region as used in .) +

    +     + +

    + M = 10, M_{yz} = 15/2, + M_{xz}=5/2, M_{xy}=5; +

    + +

    + (\overline{x},\overline{y},\overline{z}) = (3/4, 1/4, 1/2) +

    +     + +     + + + + +

    + D is bounded by the planes y=0, + y=2, x=1, z=0 and +

    + +

    + z=(3-x)/2; \delta(x,y,z) = 2\,\text{g/cm}^3. +

    + +

    + (Note: this is the same region as used in .) +

    +     + +

    + M = 4, M_{yz} = 20/3, + M_{xz}=4, M_{xy}=4/3; +

    + +

    + (\overline{x},\overline{y},\overline{z}) = (5/3, 1, 1/3) +

    +     + +     + + + + +

    + D is bounded by the planes x=2, + y=1, z=0 and +

    + +

    + z=2x+4y-4;\delta(x,y,z) = x^2lb/in^3. +

    + +

    + (Note: this is the same region as used in .) +

    +     + +

    + M = 16/5, M_{yz} = 16/3, + M_{xz}=104/45, M_{xy}=32/9; +

    + +

    + (\overline{x},\overline{y},\overline{z}) = (5/3,13/18,10/9) \approx (1.67,0.72,1.11) +

    +     + +     + + + + +

    + D is bounded by the plane z=2y and by y=4-x^2. +

    + +

    + \delta(x,y,z) = y^2lb/in^3. +

    + +

    + (Note: this is the same region as used in .) +

    +     + +

    + M = \frac{65,536}{15}\approx 208.05, + M_{yz} = 0, + M_{xz}=\frac{2,097,152}{3465}\approx 605.24, + M_{xy}=\frac{2,097,152}{3465}\approx 605.24; +

    + +

    + (\overline{x},\overline{y},\overline{z}) = (0,32/11,32/11) \approx (0,2.91,2.91) +

    +     + +     + +     +     +     +
    +
    + Triple Integration with Cylindrical and Spherical Coordinates + +

    + Just as polar coordinates gave us a new way of describing curves in the plane, + in this section we will see how + cylindrical and spherical + coordinates give us new ways of desribing surfaces and regions in space. +

    + + + + +     + + + Cylindrical Coordinates +

    + In short, cylindrical coordinates can be thought of as a combination of the polar and rectangular coordinate systems. + One can identify a point (x_0,y_0,z_0), + given in rectangular coordinates, + with the point (r_0,\theta_0,z_0), + given in cylindrical coordinates, + where the z-value in both systems is the same, + and the point (x_0,y_0) in the xy-plane is identified with the polar point P(r_0,\theta_0); + see . + So that each point in space that does not lie on the z-axis is defined uniquely, + we will restrict r\geq 0 and 0\leq \theta\leq 2\pi. + cylindrical coordinates + coordinatescylindrical +

    + +
    + Illustrating the principles behind cylindrical coordinates + + A schematic diagram of the cylindrical coordinate system in three dimensions. + +

    + A diagram illustrating how a point is located in the cylindrical coordinate system: +

      +
    • +

      + The usual three-dimensional x,y,z coordinate axes are drawn in space, + with the z axis pointing up. +

      +
      • +
    • +

      + In the xy plane, a line segment is drawn from the origin to a point in the first quadrant. + The line segment is labeled r; it is the same r used for polar coordinates in two dimensions. +

      +
      • +
    • +

      + An angle is drawn in the xy plane, from the positive x axis to the ray labeled r. + The angle is labeled \theta; it is the same \theta used for polar coordinates in two dimensions. +

      +
      • +
    • +

      + From the end of the ray labeled r, a vertical line segment is drawn, + from the xy plane to a point in the first octant. + The line segment is labeled z, and the point is labeled (r,\theta,z). +

      +
      • +
    +

    +     + + + //import apexstyle; + + + //ASY file for figstokes2_3D.asy in Chapter 13 + + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(24,17,11.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.1,1.1); + pair ybounds=(-.1,1.1); + pair zbounds=(-.1,1.1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + draw((.5,sqrt(3)/2,1)--(.5,sqrt(3)/2,0)--(0,0,0),dashed); + + draw(arc((0,0,0),(.5,0,0),(.25,sqrt(3)/4,0)),Arrow3(size=2mm)); + + dot((.5,sqrt(3)/2,1),dotblue); + + label("$\theta$",(.55*cos(pi/6),.55*sin(pi/6),.1)); + + label("$(r,\theta,z)$",(.5,sqrt(3)/2,1.1)); + + label("$z$",(.5,sqrt(3)/2+.1,.5)); + + label("$r$",(.20,sqrt(3)/4,0.05)); + + //draw(s,emissive(gray+opacity(.3)),meshpen=invisible); + + + //draw(s,simplesurfacepen,meshpen=apexmeshpen); + + + + //label("$\mathcal{S}$",(1,1,5)); + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + + +
    + +

    + We use the identity z=z along with the identities found in + to convert between the rectangular coordinate (x,y,z) and the cylindrical coordinate (r,\theta,z), namely: + + \begin{array}{l} \text{ From rectangular to cylindrical: } r=\sqrt{x^2+y^2}, \tan(\theta) = y/x \text{ and } z=z;\\ + \text{ From cylindrical to rectangular: } x=r\cos(\theta), y=r\sin(\theta) \text{ and } z=z. \end{array} + +

    + + + +

    + These identities, + along with conversions related to spherical coordinates, + are given later in . +

    + + + + + Converting between rectangular and cylindrical coordinates + +

    + Convert the rectangular point + (2,-2,1) to cylindrical coordinates, + and convert the cylindrical point (4,3\pi/4,5) to rectangular. +

    +     + +

    + Following the identities given above (and, + later in ), + we have r = \sqrt{2^2+(-2)^2} = 2\sqrt{2}. + Using \tan(\theta) = y/x, + we find \theta = \tan^{-1}(-2/2) =-\pi/4. + As we restrict \theta to being between 0 and 2\pi, + we set \theta = 7\pi/4. + Finally, z = 1, + giving the cylindrical point (2\sqrt2,7\pi/4,1). +

    + +

    + In converting the cylindrical point (4,3\pi/4,5) to rectangular, + we have x = 4\cos\big(3\pi/4\big) = -2\sqrt{2}, + y = 4\sin\big(3\pi/4\big) = 2\sqrt{2} and z=5, + giving the rectangular point (-2\sqrt{2},2\sqrt{2},5). +

    +     +     + +

    + Setting each of r, + \theta and z equal to a constant defines a surface in space, + as illustrated in the following example. +

    + + + Canonical surfaces in cylindrical coordinates + +

    + Describe the surfaces r=1, + \theta = \pi/3 and z=2, + given in cylindrical coordinates. +

    +     + +

    + The equation r=1 describes all points in space that are 1 unit away from the z-axis. + This surface is a tube or cylinder + of radius 1, centered on the z-axis, + as graphed in + (which describes the cylinder x^2+y^2=1 in space). +

    + +

    + The equation \theta=\pi/3 describes the plane formed by extending the line \theta=\pi/3, + as given by polar coordinates in the xy-plane, + parallel to the z-axis. +

    + +

    + The equation z=2 describes the plane of all points in space that are 2 units above the xy-plane. + This plane is the same as the plane described by z=2 in rectangular coordinates. +

    + +
    + Graphing the canonical surfaces in cylindrical coordinates from + + Three surfaces in space, corresponding to fixed values of each of the three cylindrical coordinates. + +

    + Three surfaces are drawn in space relative to a set of three-dimensional coordinate axes. + Each surface is obtained by setting one of the three cylindrical coordinates equal to a constant. +

      +
    • +

      + The surface r=1 is a circular cylinder centered about the z axis. + Its intersection with the xy plane is the unit circle. +

      +
      • +
    • +

      + The surface \theta=\pi/3 is a half-plane that terminates on the z axis. + It cuts the cylinder r=1 in a vertical line that passes through the point given by \theta=\pi/3 + on the unit circle. +

      +
      • +
    • +

      + The surface z=2 is a horizontal plane. It intersects the cylinder r=1 + in a circle. +

      +
      • +
    +

    +     + + + //import apexstyle; + + + //ASY file for figcylindrical1_3D.asy in Chapter 13 + + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(27,12,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={1,2}; + defaultpen(0.5mm); + + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-.5,2.75); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (t.x,t.y,2);// + } + surface s=surface(f,(-1.5,-1.5),(1.5,1.5),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=bluemeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + draw((-1.5,-1.5,2) -- (-1.5,1.5,2) -- (1.5,1.5,2) -- (1.5,-1.5,2) -- (-1.5,-1.5,2),apexmeshpen+.25mm); + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (t.x,t.x*sqrt(3),t.y);// + } + surface s=surface(f,(0,-.5),(1.25,2.25),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen q=redmeshpen2+.1mm; + draw(s,simplesurfacepen2,meshpen=q); + + draw((0,0,-.5) -- (1.25,1.25*sqrt(3),-.5) -- (1.25,1.25*sqrt(3),2.25) -- (0,0,2.25) -- (0,0,-.5),redcurvepen+.25mm); + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (cos(t.x),sin(t.x),t.y);// + } + surface s=surface(f,(0,-.5),(2*pi,2.25),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen q=greenmeshpen+.1mm; + + draw(s,simplesurfacepen3,meshpen=q); + + //dot((.5,sqrt(3)*.5,2),rgb(.1,.1,.1)); + + draw((.5,sqrt(3)*.5,-.5)--(.5,sqrt(3)*.5,2.25)); + + triple g(real t) {return (cos(t),sin(t),2);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,blackmeshpen); + + triple g(real t) {return (cos(t),sin(t),2.25);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,greencurvepen+.25mm); + + triple g(real t) {return (cos(t),sin(t),-.5);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,greencurvepen+.25mm); + + triple g(real t) {return (t,sqrt(3)*t,2);} + path3 mypath=graph(g,0,.866,operator ..); + + draw(mypath,blackmeshpen); + + //draw(s,emissive(gray+opacity(.3)),meshpen=invisible); + + + //draw(s,simplesurfacepen,meshpen=apexmeshpen); + + + + label("$z=2$",(-1.25,1.25,2.3)); + label("$r=1$",(1.1*cos(pi/6),1.1*sin(pi/6),-.75)); + + label("$\theta=\frac{\pi}{3}$",(1.25,1.25*sqrt(3),1)); + + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + +
    + +

    + All three surfaces are graphed in . + Note how their intersection uniquely defines the point P=(1,\pi/3,2). +

    +     +     + +

    + Cylindrical coordinates are useful when describing certain domains in space, + allowing us to evaluate triple integrals over these domains more easily than if we used rectangular coordinates. +

    + +

    + + shows how to evaluate \iiint_Dh(x,y,z)\, dV using rectangular coordinates. + In that evaluation, we use dV = dz\,dy\,dx + (or one of the other five orders of integration). + Recall how, in this order of integration, + the bounds on y are curve to curve + and the bounds on x are point to point: + these bounds describe a region R in the xy-plane. + We could describe R using polar coordinates as done in . + In that section, we saw how we used + dA = r\,dr\,d\theta instead of dA = dy\,dx. +

    + +

    + Considering the above thoughts, + we have dV = dz\big(r\,dr\,d\theta\big) = r\,dz\,dr\,d\theta. + We set bounds on z as surface to surface + as done in the previous section, + and then use curve to curve + and point to point + bounds on r and \theta, respectively. + Finally, using the identities given above, + we change the integrand h(x,y,z) to h(r,\theta,z). +

    + +

    + This process should sound plausible; + the following theorem states it is truly a way of evaluating a triple integral. +

    + + + Triple Integration in Cylindrical Coordinates + +

    + Let w=h(r,\theta,z) be a continuous function on a closed, + bounded region D in space, + bounded in cylindrical coordinates by \alpha \leq \theta \leq \beta, + g_1(\theta)\leq r \leq g_2(\theta) and f_1(r,\theta) \leq z \leq f_2(r,\theta). + Then integrationwith cylindrical coordinates + + \iiint_D h(r,\theta,z)\, dV = \int_\alpha^\beta\int_{g_1(\theta)}^{g_2(\theta)}\int_{f_1(r,\theta)}^{f_2(r,\theta)}h(r,\theta,z) r\,dz\,dr\,d\theta + . +

    +     +     + + + + + Evaluating a triple integral with cylindrical coordinates + +

    + Find the mass of the solid represented by the region in space bounded by z=0, + z=\sqrt{4-x^2-y^2}+3 and the cylinder x^2+y^2=4 + (as shown in ), + with density function \delta(x,y,z) = x^2+y^2+z+1, + using a triple integral in cylindrical coordinates. + Distances are measured in centimeters and density is measured in grams per cubic centimeter. +

    +
    + Visualizing the solid used in + + A circular cylinder capped by a spherical dome. + +

    + The surface depicted in this image looks like the top of a grain silo, + or like the end of a pill capsule. + It consists of a circular cylinder, centered on the z axis, + extending from the xy plane to the plane z=3. +

    + +

    + When z=3, the cylinder meets a hemisphere of the same radius as the cylinder. + The peak of the hemisphere lies on the z axis, at the point (0,0,5). +

    +     + + + //import apexstyle; + + + //ASY file for figcylindrical2_3D.asy in Chapter 13 + + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(22,22,11); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={1,3,5}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-.5,6); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (2*t.y*cos(t.x),2*t.y*sin(t.x),sqrt(4-4*t.y^2)+3);// + } + surface s=surface(f,(0,0),(2pi,1),16,16,Spline); + pen p=bluemeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + //draw((-1.5,-1.5,2) -- (-1.5,1.5,2) -- (1.5,1.5,2) -- (1.5,-1.5,2) -- (-1.5,-1.5,2),apexmeshpen+.25mm); + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (2*cos(t.x),2*sin(t.x),t.y);// + } + surface s=surface(f,(0,0),(2*pi,3),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=bluemeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + + + //draw(s,emissive(gray+opacity(.3)),meshpen=invisible); + + + //draw(s,simplesurfacepen,meshpen=apexmeshpen); + + + + + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + +
    +     + +

    + We begin by describing this region of space with cylindrical coordinates. + The plane z=0 is left unchanged; + with the identity r=\sqrt{x^2+y^2}, + we convert the hemisphere of radius 2 to the equation z=\sqrt{4-r^2}; + the cylinder x^2+y^2=4 is converted to r^2=4, + or, more simply, r=2. + We also convert the density function: + \delta(r,\theta,z) = r^2+z+1. +

    + +

    + To describe this solid with the bounds of a triple integral, + we bound z with 0\leq z\leq \sqrt{4-r^2}+3; + we bound r with 0 \leq r \leq 2; + we bound \theta with 0 \leq \theta \leq 2\pi. +

    + +

    + Using + and , + we have the mass of the solid is + + M=\iiint_D\delta(x,y,z)\, dV \amp = \int_0^{2\pi}\int_0^2\int_0^{\sqrt{4-r^2}+3}\big(r^2+z+1\big)r\,dz\,dr\,d\theta + \amp = \int_0^{2\pi}\int_0^2\big((r^3+4r)\sqrt{4-r^2}+\frac52r^3+\frac{19}2r\big)\,dr\,d\theta + \amp = \frac{1318\pi}{15} \approx 276.04\,\text{g} + , + where we leave the details of the remaining double integral to the reader. +

    +     +     + + + Finding the center of mass using cylindrical coordinates + +

    + Find the center of mass of the solid with constant density whose base can be described by the polar curve + r=\cos(3\theta) and whose top is defined by the plane z=1-x+0.1y, + where distances are measured in feet, + as seen in . + (The volume of this solid was found in .) +

    +
    + Visualizing the solid used in + + A solid whose cross-sections are like a three-leaf clover; it has been sliced at an angle. + +

    + A three-dimensional solid is shown, plotted against a set of three-dimensional coordinate axes. + The base of the solid is a three-leaf rose curve. +

    + +

    + The solid is a cylinder, except that it is cut at an angle, + where it meets the plane z=1-x+0.1y. + We first saw this solid in . +

    +     + + + + //ASY file for figdoublepol43D.asy in Chapter 13 + + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-2,2); + + //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //plane is z=1-x+0.1y + //Draw the surface + //({cos(3*x)*cos(x)},{cos(3*x)*sin(x)},{(1-cos(3*x)*cos(x)+.1*cos(3*x)*sin(x))*y}); + triple f(pair t) { + return (cos(3*t.x)*cos(t.x),cos(3*t.x)*sin(t.x),(1-cos(3*t.x)*cos(t.x)+.1*cos(3*t.x)*sin(t.x))*t.y); + } + surface s=surface(f,(0,0),(2*pi,1),32,32,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //draw curve on xy plane //({cos(3*x)*cos(x)},{cos(3*x)*sin(x)},{0}) + triple g(real t) {return (cos(3*t)*cos(t),cos(3*t)*sin(t),0);} + path3 mypath=graph(g,0,pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + + //draw curves on surface + triple g(real t) {return (cos(3*t)*cos(t),cos(3*t)*sin(t),1-(cos(3*t)*cos(t))+0.1*(cos(3*t)*sin(t)));} + path3 mypath=graph(g,0,pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + + //Shade the bottom + import three; + path3 p = (0,0,0)..(.175,0.064,0)..(.264,.119,0)..(0.683,0.183,0)..(.916,0.121,0)..(1,0,0)..(.916,-0.121,0)..(0.683,-0.183,0)..(.264,-.119,0)..(.175,-0.086,0); + draw(surface(p -- cycle), simplesurfacepen); + + path3 p = (0,0,0)..(-0.1429, 0.1196,0)..(-0.2351, 0.1691,0)..(-0.5000, 0.5000,0)..(-0.5628, 0.7328,0)..(-0.5000, 0.8660,0)..(-0.3532, 0.8538,0)..(-0.1830, 0.6830,0)..(-0.0289, 0.2881,0)..(-0.0130, 0.1946,0); + draw(surface(p -- cycle), simplesurfacepen); + + path3 p = (0,0,0)..(-0.0321,-0.1836,0)..(-0.0289,-0.2881,0)..(-0.1830,-0.6830,0)..(-0.3532,-0.8538,0)..(-0.5000,-0.8660,0)..(-0.5628,-0.7328,0)..(-0.5000,-0.5000,0)..(-0.2351,-0.1691,0)..(-0.1620,-0.1086,0); + draw(surface(p -- cycle), simplesurfacepen); + + //Shade the top + path3 p = (0,0,1.0000)..(-0.1429,0.1196,1.1549)..(-0.2351,0.1691,1.2520)..(-0.5000,0.5000,1.5500)..(-0.5628,0.7328,1.6361)..(-0.5000,0.8660,1.5866)..(-0.3532,0.8538,1.4386)..(-0.1830 ,0.6830 ,1.2513)..(-0.0289,0.2881,1.0578)..(-0.0130,0.1946,1.0325); + + draw(surface(p -- cycle), simplesurfacepen); + path3 p = (0 , 0 , 1.0000)..(-0.0321 , -0.1836 , 1.0137)..(-0.0289, -0.2881, 1.0001)..(-0.1830, -0.6830, 1.1147)..(-0.3532 , -0.8538 , 1.2678)..(-0.5000, -0.8660 , 1.4134)..(-0.5628 , -0.7328, 1.4895)..(-0.5000 , -0.5000 , 1.4500)..(-0.2351 , -0.1691 , 1.2181)..(-0.1620 , -0.1086, 1.1511); + draw(surface(p -- cycle), simplesurfacepen); + + path3 p = (0 , 0, 1.0000)..( 0.1750, 0.0640 , 0.8314)..( 0.2640, 0.1190 , 0.7479)..( 0.6830, 0.1830 , 0.3353)..( 0.9160, 0.1210 , 0.0961)..( 1.0000, 0 , 0)..( 0.9160, -0.1210 , 0.0719)..( 0.6830, -0.1830 , 0.2987)..( 0.2640, -0.1190 , 0.7241)..( 0.1750, -0.0860 , 0.8164); + draw(surface(p -- cycle), simplesurfacepen); + + +
    +     + +

    + We convert the equation of the plane to use cylindrical coordinates: + z= 1-r\cos(\theta)+0.1r\sin(\theta). + Thus the region is space is bounded by 0 \leq z \leq 1-r\cos(\theta) + 0.1r\sin(\theta), + 0 \leq r \leq \cos(3\theta), + 0 \leq \theta \leq \pi (recall that the rose curve + r=\cos(3\theta) is traced out once on [0,\pi]. +

    + +

    + Since density is constant, + we set \delta = 1 and finding the mass is equivalent to finding the volume of the solid. + We set up the triple integral to compute this but do not evaluate it; + we leave it to the reader to confirm it evaluates to the same result found in . + + M = \iiint_D\delta \, dV = \int_0^{\pi}\int_0^{\cos(3\theta)}\int_0^{1-r\cos(\theta)+0.1r\sin(\theta)} r\,dz\,dr\,d\theta = \frac{\pi}{4} + . +

    + +

    + From + we set up the triple integrals to compute the moments about the three coordinate planes. + The computation of each is left to the reader (using technology is recommended): + + M_{yz} = \iiint_D x\,dV \amp = \int_0^{\pi}\int_0^{\cos(3\theta)}\int_0^{1-r\cos(\theta)+0.1r\sin(\theta)} (r\cos(\theta)) r\,dz\,dr\,d\theta + \amp = \frac{-3\pi}{64} \approx -0.147. + + + M_{xz} = \iiint_D y\,dV \amp = \int_0^{\pi}\int_0^{\cos(3\theta)}\int_0^{1-r\cos(\theta)+0.1r\sin(\theta)} (r\sin(\theta)) r\,dz\,dr\,d\theta + \amp = \frac{3\pi}{640} \approx 0.015. + M_{xy} = \iiint_D z\,dV \amp = \int_0^{\pi}\int_0^{\cos(3\theta)}\int_0^{1-r\cos(\theta)+0.1r\sin(\theta)} (z) r\,dz\,dr\,d\theta + \amp = \frac{1903\pi}{12800} \approx 0.467 + . +

    + +

    + The center of mass in rectangular coordinates, + found by dividing the respective moments by the mass, + is approximately located at (-0.188,0.019,0.595), + which lies outside the bounds of the solid. +

    +     +     +     + + + Spherical Coordinates +

    + spherical coordinates + coordinatesspherical +

    + +

    + In short, spherical coordinates can be thought of as a + double application + of the polar coordinate system. + In spherical coordinates, + a point P is identified with (\rho,\theta,\varphi), + where \rho is the distance from the origin to P, + \theta is the same angle as would be used to describe P in the cylindrical coordinate system, + and \varphi is the angle between the xy-plane and the ray from the origin to P; + see . + So that each point in space that does not lie on the z-axis is defined uniquely, + we will restrict \rho \geq 0, + 0 \leq \theta \leq 2\pi and -\pi/2 \leq \varphi \leq \pi/2. +

    + + + + +

    + Note that most mathematics textbooks define \varphi to be measured + from the positive z-axis, with values in [0,\pi], + rather than from the xy-plane. +

    + +

    + We have chosen our convention with a number of considerations in mind: +

      +
    • +

      + The coordinates (\rho,\theta,\varphi) form a right-handed + coordinate system: one in which the orientation matches that of our usual (x,y,z) + coordinates, where the right-hand rule applies. + If \varphi is measured from the z-axis, + the order (\rho, \varphi, \theta) is needed to get a right-handed system. +

      +
      • +
    • +

      + Points of the form (a,\alpha,0) are the same in both cylindrical and spherical coordinates. +

      +
      • +
    • +

      + Some integration problems become slightly easier: + we will see soon that the volume element in spherical coordinates involves \cos(\varphi), + which integrates to \sin(\varphi). + In the usual convention, the volume element involves \sin(\varphi), + which integrates to -\cos(\varphi) a source of many common sign errors. +

      +
      • +
    +

    + +

    + Students of Physics will encounter yet another convention. + In Physics, the variable r is preferred as the radial coordinate, + and spherical coordinates are given as (r,\theta,\varphi); + however, in Physics, \varphi becomes the angle in the xy-plane, + while \theta is the angle measured from the positive z-axis. +

    + +

    + Note that the angle in the xy-plane (\theta, in our case) + is known as the azimuthal angle. + Our angle \varphi is known as the elevation angle. + The angle used in other conventions that is measured from the positive z-axis + (often identified with the north pole) is known as the polar angle. + For further discussion, the Wikipedia article + is quite useful. +

    +     + +
    + Illustrating the principles behind spherical coordinates + + A schematic diagram illustrating the spherical coordinate system relative to the rectangular coordinate axes. + +

    + A diagram illustrating how a point is located in the spherical coordinate system: +

      +
    • +

      + The usual three-dimensional x,y,z coordinate axes are drawn, with the z axis pointing up. +

      +
      • +
    • +

      + A point in space is plotted somewhere in the first octant; it is labelled with the coordinates (\rho,\varphi,\theta). + A line segment is drawn from the origin to this point; its length is labeled with the spherical coordinate \rho. +

      +
      • +
    • +

      + Another line segement is drawn from the point in space, vertically down to the xy plane: + if the rectangular coordinates of the point in space are (x,y,z), + then the point in the plane has rectangular coordinates (x,y,0). + (Note that this is the same vertical segment used to define the cylindrical coordinate z.) +

      +
      • +
    • +

      + Finally, a third line segment is drawn from the origin to the point in the plane that lies directly below the original point in space. + This line segment lies in the xy plane. The three line segments drawn in this diagram form a right angle triangle, + with the segment of length \rho as hypotenuse. +

      +
      • + +
    • +

      + An angle is drawn from the segment in the xy plane to the segment of length \rho. + This angle is labeled with the spherical coordinate \varphi; + it is the angle at the vertex in the right-angled triangle that is at the origin. +

      +
      • + +
    • +

      + Another angle is drawn from the positive x axis to the side of the triangle in the xy plane, + and labeled with the spherical coordinate \theta. +

      +
      • + +
    • +

      + Note that the angle \theta is the same in polar, cylindrical, and spherical coordinates. +

      +
      • +
    +

    +     + + + //import apexstyle; + + + //ASY file for figsphericalintro.asy in Chapter 13 + + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(24,17,11.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.1,1.1); + pair ybounds=(-.1,1.1); + pair zbounds=(-.1,1.1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + draw((.5,sqrt(3)/2,1)--(.5,sqrt(3)/2,0)--(0,0,0)--cycle,dashed); + + draw(arc((0,0,0),(.5,0,0),(.25,sqrt(3)/4,0)),Arrow3(size=2mm)); + + draw(arc((0,0,0),(0.25,sqrt(3)/4,0),(.5,sqrt(3)/2,1)*.5/1.414),Arrow3(size=2mm)); + + guide3 myarc=arc((0,0,0),(0.25,sqrt(3)/4,0),(0.5,sqrt(3)/2,1)*.5/1.414); + + label("$\varphi$",myarc,NE); + + + dot((.5,sqrt(3)/2,1),dotblue); + + label("$\theta$",(.55*cos(pi/6),.55*sin(pi/6),.1)); + + label("$(\rho,\theta,\varphi)$",(.5,sqrt(3)/2,1.1)); + + //label("$z$",(.5,sqrt(3)/2+.1,.5)); + + label("$\rho$",(.5,sqrt(3)/2,1)*.5+(0.0,0.0,.07)); + + //draw(s,emissive(gray+opacity(.3)),meshpen=invisible); + + + //draw(s,simplesurfacepen,meshpen=apexmeshpen); + + +
    + + + + + +

    + The following Key Idea gives conversions to/from our three spatial coordinate systems. +

    + + + Converting Between Rectangular, Cylindrical and Spherical Coordinates +

    +

      +
    • + Rectangular and Cylindrical +

      + + \amp r^2 = x^2+y^2, \amp \amp \tan(\theta) = y/x,\amp z\amp=z + \amp x=r\cos(\theta), \amp \amp y =r\sin(\theta),\amp z\amp =z + +

      +
      • +
    • + Rectangular and Spherical +

      + + \amp \rho = \sqrt{x^2+y^2+z^2}, \amp \amp \tan(\theta) = y/x, \amp \amp \sin(\varphi) = z/\sqrt{x^2+y^2+z^2} + \amp x=\rho\cos(\varphi)\cos(\theta),\amp \amp y=\rho\cos(\varphi)\sin(\theta),\amp \amp z=\rho\sin(\varphi) + +

      +
      • +
    • + Cylindrical and Spherical +

      + + \amp \rho =\sqrt{r^2+z^2},\amp \amp \theta = \theta,\amp \amp \tan(\varphi) = z/r + \amp r=\rho \cos(\varphi),\amp \amp \theta = \theta,\amp \amp z=\rho\sin(\varphi) + +

      +
      • +
    +

    +     + + + + + Converting between rectangular and spherical coordinates + +

    + Convert the rectangular point + (2,-2,1) to spherical coordinates, + and convert the spherical point + (6,\pi/3,0) to rectangular and cylindrical coordinates. +

    +     + +

    + This rectangular point is the same as used in . + Using , + we find \rho = \sqrt{2^2+(-1)^2+1^2} = 3. + Using the same logic as in , + we find \theta = 7\pi/4. + Finally, \sin(\varphi) = 1/3, + giving \varphi = \sin^{-1}(1/3) \approx 0.34, + or about 19.47^\circ. + Thus the spherical coordinates are approximately (3,7\pi/4,0.34). +

    + +

    + Converting the spherical point (6,\pi/3,0) to rectangular, + we have x = 6\cos(0)\cos(\pi/3) = 3, + y = 6\cos(0)\sin(\pi/3) = 3\sqrt{3} and z = 6\sin(0) = 0. + Thus the rectangular coordinates are (3,3\sqrt{3},0). +

    + +

    + To convert this spherical point to cylindrical, + we have r = 6\cos(0) = 6, + \theta = \pi/3 and z = 6\sin(0) =0, + giving the cylindrical point (6,\pi/3,0). +

    +     +     + + + Canonical surfaces in spherical coordinates + +

    + Describe the surfaces \rho=1, + \theta = \pi/3 and \varphi = \pi/3, + given in spherical coordinates. +

    +     + +

    + The equation \rho = 1 describes all points in space that are 1 unit away from the origin: + this is the sphere of radius 1, centered at the origin. +

    + +

    + The equation \theta = \pi/3 describes the same surface in spherical coordinates as it does in cylindrical coordinates: + beginning with the line \theta = \pi/3 in the xy-plane as given by polar coordinates, + extend the line parallel to the z-axis, forming a plane. +

    + +

    + The equation \varphi=\pi/3 describes all points P + in space where the ray from the origin to P makes an angle of \pi/3 with the xy-plane. + This describes a cone, + with the positive z-axis its axis of symmetry, with point at the origin. +

    + +
    + Graphing the canonical surfaces in spherical coordinates from + + A three-dimensional plot showing three surfaces, each of which is obtained by fixing a value of one of the three spherical coordinates. + +

    + Three surfaces are plotted relative to the usual three-dimensional coordinate axes. +

      +
    • +

      + The surface \rho=1 is the unit sphere: a sphere of radius 1, centered at the origin. +

      +
      • +
    • +

      + The surface \varphi = \pi/3 is the top half of a circular cone that opens along the z axis. + The angle between any line through the origin that lies on the cone and the xy plane is \pi/3. + (Equivalently, the angle between any line through the origin that lies on the cone and the z axis is \pi/6.) +

      +
      • +
    • +

      + The surface \theta=\pi/3 is a half-plane that terminates on the z axis, + the same as it was in cylindrical coordinates. +

      +
      • +
    +

    +     + + + //import apexstyle; + + + //ASY file for figspherical1_3D.asy in Chapter 13 + + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(27,13,9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-2,2); + pair ybounds=(-2,2); + pair zbounds=(-1.5,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),sqrt(3)*t.y);// + } + surface s=surface(f,(0,0),(2*pi,1.5/sqrt(3)),16,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=bluemeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + //draw((-1.5,-1.5,2) -- (-1.5,1.5,2) -- (1.5,1.5,2) -- (1.5,-1.5,2) -- (-1.5,-1.5,2),apexmeshpen+.25mm); + + triple g(real t) {return (1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5);} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,apexmeshpen); + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (t.x,t.x*sqrt(3),t.y);// + } + surface s=surface(f,(0,-1.25),(1.25,1.5),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen q=redmeshpen2+.1mm; + draw(s,simplesurfacepen2,meshpen=q); + + draw((0,0,-1.25) -- (1.25,1.25*sqrt(3),-1.25) -- (1.25,1.25*sqrt(3),1.5) -- (0,0,1.5) -- (0,0,-1.5),redcurvepen+.25mm); + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (sin(t.y)*cos(t.x),sin(t.y)*sin(t.x),cos(t.y));// + } + surface s=surface(f,(0,0),(2*pi,pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen q=greenmeshpen+.1mm; + draw(s,simplesurfacepen3,meshpen=q); + + //dot((.5,sqrt(3)*.5,2),rgb(.1,.1,.1)); + + //draw((.5,sqrt(3)*.5,-.5)--(.5,sqrt(3)*.5,2.25)); + + triple g(real t) {return (1/2*cos(t),1/2*sin(t),1/2*sqrt(3));} + path3 mypath=graph(g,0,2*pi,operator ..); + draw(mypath,blackmeshpen); + + //(1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5) + draw((0,0,0) -- (1.5/sqrt(3)*cos(pi/3),1.5/sqrt(3)*sin(pi/3),1.5)); + + triple g(real t) {return (sin(t)*cos(pi/3),sin(t)*sin(pi/3),cos(t));} + path3 mypath=graph(g,0,pi,operator ..); + draw(mypath,blackmeshpen); + + triple g(real t) {return (cos(t),sin(t),-.5);} + path3 mypath=graph(g,0,2*pi,operator ..); + //draw(mypath,greencurvepen+.25mm); + + triple g(real t) {return (t,sqrt(3)*t,2);} + path3 mypath=graph(g,-.866,.866,operator ..); + //draw(mypath,blackmeshpen); + + //draw(s,emissive(gray+opacity(.3)),meshpen=invisible); + + + //draw(s,simplesurfacepen,meshpen=apexmeshpen); + + + + label("$\varphi=\frac{\pi}{3}$",(1.5/sqrt(3)*cos(2pi/3),1.5/sqrt(3)*sin(2pi/3),1.5),E); + label("$\rho=1$",(1.1*cos(pi/200),1.1*sin(pi/200),-.9)); + label("$\theta=\frac{\pi}{3}$",(1.25,1.25*sqrt(3),1)); + + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + +
    + +

    + All three surfaces are graphed in . + Note how their intersection uniquely defines the point P=(1,\pi/3,\pi/6). +

    +     +     + +

    + Spherical coordinates are useful when describing certain domains in space, + allowing us to evaluate triple integrals over these domains more easily than if we used rectangular coordinates or cylindrical coordinates. + The crux of setting up a triple integral in spherical coordinates is appropriately describing the + small amount of volume, + dV, used in the integral. +

    + +

    + Considering , + we can make a small spherical wedge by varying \rho, + \theta and \varphi each a small amount, \Delta\rho, + \Delta\theta and \Delta\varphi, respectively. + This wedge is approximately a rectangular solid when the change in each coordinate is small, + giving a volume of about + + \Delta V \approx \Delta\rho\ \times\ \rho\Delta\varphi\ \times\ \rho\cos(\varphi)\Delta\theta + . +

    + +
    + Approximating the volume of a standard region in space using spherical coordinates + + A schematic diagram illustrating how the spherical volume element is computed. + +

    + An illustration of the volume obtained from a small spherical wedge. + The diagram is quite complicated: +

      +
    • +

      + Along the z axis, a segment of length \rho is indicated. +

      +
      • +
    • +

      + This segement is rotated in a vertical plane (corresponding to a fixed value of \theta). + It sweeps out a circular arc spanned by an angle \Delta \varphi; + the length of the arc is \rho\Delta\varphi. +

      +
      • +
    • +

      + Two rays from the origin are drawn in the xy plane; + the angle between them is labeled \Delta\theta. +

      +
      • +
    • +

      + Above one of these two rays, a right triangle is drawn. + The hypotenuse of the triangle is a ray from the origin to one end of the arc of length \rho\Delta\varphi. + The base of the triangle is the side adjacent to the angle \varphi, so the length of the base is \rho\cos\varphi. +

      +
      • +
    • +

      + The circular arc between the two rays in the xy plane therefore has length \rho\cos\varphi\Delta\theta. +

      +
      • +
    • +

      + The spherical wedge that is depicted looks like a distorted rectangular box. + The dimensions of the box, based on the observations above, are + + \Delta\rho\times\rho\Delta\varphi\times\rho\cos\varphi\Delta\theta + . +

      +
      • +
    +

    +     + + + //import apexstyle; + + + //ASY file for figspherical1_3D.asy in Chapter 13 + + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14.4,2.6,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.1,.6); + pair ybounds=(-.1,.6); + pair zbounds=(-.1,1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + + real t1=pi/3-.2; + real t2=pi/3+.2; + real p1=pi/6-.2; + real p2=pi/6+.2; + real r1=.75; + real r2=.9; + + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (r1*cos(t.x)*sin(t.y),r1*sin(t.x)*sin(t.y),r1*cos(t.y));// + } + surface s=surface(f,(t1,p1),(t2,p2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + + triple f(pair t) { + return (r2*cos(t.x)*sin(t.y),r2*sin(t.x)*sin(t.y),r2*cos(t.y));// + } + surface s=surface(f,(t1,p1),(t2,p2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + triple f(pair t) { + return (t.y*cos(t.x)*sin(p1),t.y*sin(t.x)*sin(p1),t.y*cos(p1));// + } + surface s=surface(f,(t1,r1),(t2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + triple f(pair t) { + return (t.y*cos(t.x)*sin(p2),t.y*sin(t.x)*sin(p2),t.y*cos(p2));// + } + surface s=surface(f,(t1,r1),(t2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + triple f(pair t) { + return (t.y*cos(t1)*sin(t.x),t.y*sin(t1)*sin(t.x),t.y*cos(t.x));// + } + surface s=surface(f,(p1,r1),(p2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + triple f(pair t) { + return (t.y*cos(t2)*sin(t.x),t.y*sin(t2)*sin(t.x),t.y*cos(t.x));// + } + surface s=surface(f,(p1,r1),(p2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + // + // lines for phi + // + draw((r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1)) -- (0,0,0) -- (r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)),black+.25mm+dashed); + + draw(arc((0,0,0),(r1*cos(t1)*sin(p2)/2.5,r1*sin(t1)*sin(p2)/2.5,r1*cos(p2)/2.5),(r1*cos(t1)*sin(p1)/2.5,r1*sin(t1)*sin(p1)/2.5,r1*cos(p1)/2.5)),black+.25mm,Arrow3(size=1.5mm)); + + label("$\Delta\varphi$",.95*(r1*cos(t1)*sin((p1+p2)/2)/2,r1*sin(t1)*sin((p1+p2)/2)/2,r1*cos((p1+p2)/2)/2)); + + + + // + // lines for theta + // + + draw((0,0,0) -- (r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0) -- (r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)),black+.25mm+dashed); + + draw((0,0,0) -- (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),0) -- (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)),black+.25mm+dashed); + + draw(arc((0,0,0),.5*(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0),.5*(r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),0)),black+.25mm,Arrow3(size=1.5mm)); + + label("$\Delta\theta$",1.2*.5*(r1*cos((t1+t2)/2)*sin(p2),r1*sin((t1+t2)/2)*sin(p2),0)); + + // + // lines for rho + // + + draw(arc((0,0,0),(r1*cos(t1)*sin(0),r1*sin(t1)*sin(0),r1*cos(0)),(r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1))),black+.25mm+dashed); + + draw(arc((0,0,0),(r2*cos(t1)*sin(0),r2*sin(t1)*sin(0),r2*cos(0)),(r2*cos(t1)*sin(p1),r2*sin(t1)*sin(p1),r2*cos(p1))),black+.25mm+dashed); + + + label("$\Delta\rho$",((r1+r2)/2*cos(t1)*sin(.65*(p1)),(r1+r2)/2*sin(t1)*sin(.65*(p1)),(r1+r2)/2*cos(.65*(p1)))); + + real poff = -0.05; + + draw((r1*cos(t1)*sin(p1+poff),r1*sin(t1)*sin(p1+poff),r1*cos(p1+poff)) -- (r2*cos(t1)*sin(p1+poff),r2*sin(t1)*sin(p1+poff),r2*cos(p1+poff)),black+.25mm,Arrows3(size=1.5mm)); + + label("$\rho$",(0,-.04,.5*r1)); + + draw((0,-.02,.01) -- (0,-.02,.99*r1),black+.25mm,Arrows3(size=1.5mm)); + + + // + // lines for theta length + // + + draw((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) -- (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05),black+.25mm,Arrows3(size=1.5mm)); + + label("$\rho\cos(\varphi)\Delta\theta$",1.5*((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05)+(0,0,-.05))/2+(0,0,-.25),S); + + draw(1.5*((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05)+(0,0,-.05))/2 + (0,0,-.25) -- 1.05((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05))/2+(0,0,-.01),black+.25mm+dashed,Arrow3(size=1.5mm)); + + // + // lines for phi length + // + + label("$\rho\Delta\varphi$",.85*(r1*cos(t1)*sin((p1+p2)/2),r1*sin(t1)*sin((p1+p2)/2),r1*cos((p1+p2)/2))); + + draw(arc((0,0,0),.95*(r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1)),.95*(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2))),black+.25mm,Arrows3(size=1.5mm)); + + +
    + +

    + Given a region D in space, + we can approximate the volume of D with many such wedges. + As the size of each of \Delta\rho, + \Delta\theta and \Delta\varphi goes to zero, + the number of wedges increases to infinity and the volume of D is more accurately approximated, giving + + dV = d\rho\ \times\ \rho\, d\varphi\ \times\ \rho\cos(\varphi)d\theta = \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi + . +

    + +

    + Again, this development of dV should sound reasonable, + and the following theorem states it is the appropriate manner by which triple integrals are to be evaluated in spherical coordinates. +

    + + + + Triple Integration in Spherical Coordinates + +

    + Let w=h(\rho,\theta,\varphi) be a continuous function on a closed, + bounded region D in space, + bounded in spherical coordinates by \alpha_1 \leq \varphi \leq \alpha_2, + \beta_1 \leq \theta \leq \beta_2 and f_1(\theta,\varphi) \leq \rho \leq f_2(\theta,\varphi). + Then integrationwith spherical coordinates + + \iiint_D h(\rho,\theta,\varphi)\, dV = \int_{\alpha_1}^{\alpha_2}\int_{\beta_1}^{\beta_2}\int_{f_1(\theta,\varphi)}^{f_2(\theta,\varphi)} h(\rho,\theta,\varphi) \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi + . +

    +     +     + + + Establishing the volume of a sphere + +

    + Let D be the region in space bounded by the sphere, + centered at the origin, of radius r. + Use a triple integral in spherical coordinates to find the volume V of D. +

    +     + +

    + The sphere of radius r, + centered at the origin, has equation \rho = r. + To obtain the full sphere, + the bounds on \theta and \varphi are + 0\leq \theta \leq 2\pi and -\pi/2 \leq \varphi \leq \pi/2. + This leads us to: + + V \amp = \iiint_D\, dV + \amp = \int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\int_0^r\big(\rho^2\cos(\varphi)\big)\, d\rho\, d\theta\, d\varphi + \amp = \int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\left(\frac13\rho^3\cos(\varphi)\Big|_0^r\right)\, d\theta\, d\varphi + \amp = \int_{-\pi/2}^{\pi/2}\int_0^{2\pi} \left(\frac13r^3\cos(\varphi)\right)\, d\theta\, d\varphi + \amp = \int_{-\pi/2}^{\pi/2} \left(\frac{2\pi}3r^3\cos(\varphi)\right)\, d\varphi + \amp = \left.\left(\frac{2\pi}3r^3\sin(\varphi)\right)\right|_{-\pi/2}^{\pi/2} + \amp = \frac{4\pi}3r^3 + , + the familiar formula for the volume of a sphere. + Note how the integration steps were easy, + not using square roots nor integration steps such as Substitution. +

    +     +     + + + + + Finding the center of mass using spherical coordinates + +

    + Find the center of mass of the solid with constant density enclosed above by \rho=4 and below by \varphi = \pi/3, + as illustrated in . +

    +
    + Graphing the solid, and its center of mass, from + + A gem-like solid, bounded below by the top half of a circular cone, and above by a spherical cap. + +

    + The solid depicted in this image is reminiscent of a gemstone. + It is bounded below by a circular cone with its vertex at the origin, that opens upwards along the z axis. + The solid is bounded above by a spherical cap. + (Perhaps the solid also resembles a snow cone.) +

    + +

    + Also depicted (but only barely visible) is a point marked on the z axis to indicate the location of the center of mass. +

    +     + + + //import apexstyle; + + + //ASY file for figspherical1_3D.asy in Chapter 13 + + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12,13,3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={4}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-.1,5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (t.y*cos(t.x)*sin(pi/6),t.y*sin(t.x)*sin(pi/6),t.y*cos(pi/6));// + } + surface s=surface(f,(0,0),(2*pi,4),16,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (4*cos(t.x)*sin(t.y),4*sin(t.x)*sin(t.y),4*cos(t.y));// + } + surface s=surface(f,(0,0),(2*pi,pi/6),16,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + dot((0,0,2.799),blackmeshpen+1.2mm); + + + //draw((-1.5,-1.5,2) -- (-1.5,1.5,2) -- (1.5,1.5,2) -- (1.5,-1.5,2) -- (-1.5,-1.5,2),apexmeshpen+.25mm); + + triple g(real t) {return (1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5);} + path3 mypath=graph(g,0,2*pi,operator ..); + //draw(mypath,apexmeshpen); + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (t.x,t.x*sqrt(3),t.y);// + } + surface s=surface(f,(-1.25,-1.5),(1.25,1.5),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen q=redmeshpen; + //draw(s,simplesurfacepen2,meshpen=q); + + //draw((-1.25,-1.25*sqrt(3),-1.5) -- (1.25,1.25*sqrt(3),-1.5) -- (1.25,1.25*sqrt(3),1.5) -- (-1.25,-1.25*sqrt(3),1.5) -- (-1.25,-1.25*sqrt(3),-1.5),redcurvepen+.25mm); + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (sin(t.y)*cos(t.x),sin(t.y)*sin(t.x),cos(t.y));// + } + surface s=surface(f,(0,0),(2*pi,pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen q=greencurvepen; + //draw(s,simplesurfacepen3,meshpen=q); + + //dot((.5,sqrt(3)*.5,2),rgb(.1,.1,.1)); + + //draw((.5,sqrt(3)*.5,-.5)--(.5,sqrt(3)*.5,2.25)); + + triple g(real t) {return (1/2*cos(t),1/2*sin(t),1/2*sqrt(3));} + path3 mypath=graph(g,0,2*pi,operator ..); + //draw(mypath,blackmeshpen); + + //(1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5) + //draw((1.5/sqrt(3)*cos(4*pi/3),1.5/sqrt(3)*sin(4*pi/3),1.5) -- (0,0,0) -- (1.5/sqrt(3)*cos(pi/3),1.5/sqrt(3)*sin(pi/3),1.5)); + + triple g(real t) {return (sin(t)*cos(pi/3),sin(t)*sin(pi/3),cos(t));} + path3 mypath=graph(g,0,2*pi,operator ..); + //draw(mypath,blackmeshpen); + + +
    +     + +

    + We will set up the four triple integrals needed to find the center of mass (, to compute M, + M_{yz}, + M_{xz} and M_{xy}) and leave it to the reader to evaluate each integral. + Because of symmetry, + we expect the x- and y- coordinates of the center of mass to be 0. +

    + +

    + While the surfaces describing the solid are given in the statement of the problem, + to describe the full solid D, + we use the following bounds: + 0 \leq \rho \leq 4, + 0 \leq \theta \leq 2\pi and 0 \leq \varphi \leq \pi/3. + Since density \delta is constant, + we assume \delta =1. +

    + +

    + The mass of the solid: + + M \amp = \iiint_D\, dm = \iiint_D\, dV + \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4\big(\rho^2\cos(\varphi)\big)\, d\rho\, d\theta\, d\varphi + \amp = \frac{64}3\big(2-\sqrt{3}\big)\pi \approx 17.958 + . +

    + +

    + To compute M_{yz}, the integrand is x; + using , + we have x = \rho\cos(\varphi)\cos(\theta). + This gives: + + M_{yz} \amp = \iiint_D x\, dm + \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big((\rho\cos(\varphi)\cos(\theta))\rho^2\cos(\varphi)\big) \, d\rho\, d\theta\, d\varphi + \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big(\rho^3\cos^2(\varphi)\cos(\theta)\big) \, d\rho\, d\theta\, d\varphi + \amp =0 + , + which we expected as we expect \overline{x} = 0. +

    + +

    + To compute M_{xz}, the integrand is y; + using , + we have y = \rho\cos(\varphi)\sin(\theta). + This gives: + + M_{xz} \amp = \iiint_D y\, dm + \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big((\rho\cos(\varphi)\sin(\theta))\rho^2\cos(\varphi)\big) \, d\rho\, d\theta\, d\varphi + \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big(\rho^3\cos^2(\varphi)\sin(\theta)\big) \, d\rho\, d\theta\, d\varphi + \amp = 0 + , + which we also expected as we expect \overline{y} = 0. +

    + +

    + To compute M_{xy}, the integrand is z; + using , + we have z = \rho\sin(\varphi). + This gives: + + M_{xy} \amp = \iiint_D z\, dm + \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big((\rho\sin(\varphi))\rho^2\cos(\varphi)\big) \, d\rho\, d\theta\, d\varphi + \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big(\rho^3\sin(\varphi)\cos(\varphi)\big) \, d\rho\, d\theta\, d\varphi + \amp =16\pi \approx 50.266 + . +

    + +

    + Thus the center of mass is (0,0,M_{xy}/M) \approx (0,0,2.799), + as indicated in . +

    +     +     + + + + + +

    + This section has provided a brief introduction into two new coordinate systems useful for identifying points in space. + Each can be used to define a variety of surfaces in space beyond the canonical surfaces graphed as each system was introduced. +

    + +

    + However, the usefulness of these coordinate systems does not lie in the variety of surfaces that they can describe nor the regions in space these surfaces may enclose. + Rather, cylindrical coordinates are mostly used to describe cylinders and spherical coordinates are mostly used to describe spheres. + These shapes are of special interest in the sciences, especially in physics, + and computations on/inside these shapes is difficult using rectangular coordinates. + For instance, in the study of electricity and magnetism, + one often studies the effects of an electrical current passing through a wire; + that wire is essentially a cylinder, + described well by cylindrical coordinates. +

    + +

    + This chapter investigated the natural follow-on to partial derivatives: + iterated integration. + We learned how to use the bounds of a double integral to describe a region in the plane using both rectangular and polar coordinates, + then later expanded to use the bounds of a triple integral to describe a region in space. + We used double integrals to find volumes under surfaces, + surface area, + and the center of mass of lamina; + we used triple integrals as an alternate method of finding volumes of space regions and also to find the center of mass of a region in space. +

    + +

    + Integration does not stop here. + We could continue to iterate our integrals, + next investigating quadruple integrals + whose bounds describe a region in 4-dimensional space + (which are very hard to visualize). + We can also look back to regular + integration where we found the area under a curve in the plane. + A natural analogue to this is finding the + area under a curve, + where the curve is in space, not in a plane. + These are just two of many avenues to explore under the heading of integration. +

    +     + + + + Terms and Concepts + + +

    + Explain the difference between the roles r, + in cylindrical coordinates, and \rho, + in spherical coordinates, play in determining the location of a point. +

    +     + + + +

    + In cylindrical, + r determines how far from the origin one goes in the xy-plane before considering the z-component. + Equivalently, + if on projects a point in cylindrical coordinates onto the xy-plane, + r will be the distance of this projection from the origin. +

    + +

    + In spherical, + \rho is the distance from the origin to the point. +

    +     +     + + + +

    + Why are points on the z-axis not determined uniquely when using cylindrical and spherical coordinates? +

    +     + + + +

    + If r=0 or \rho=0, + then the point in each coordinate system lies on the z-axis regardless of the value of \theta. +

    +     +     + + + +

    + What surfaces are naturally defined using cylindrical coordinates? +

    +     + + + +

    + Cylinders (tubes) centered at the origin, + parallel to the z-axis; + planes parallel to the z-axis that intersect the z-axis; + planes parallel to the xy-plane. +

    +     +     + + + +

    + What surfaces are naturally defined using spherical coordinates? +

    +     + + + +

    + Spheres centered at the origin; + planes parallel to the z-axis that intersect the z-axis; + cones centered on the z-axis with point at the origin. +

    +     +     +     + + + Problems + + + +

    + Points are given in either the rectangular, + cylindrical or spherical coordinate systems. + Find the coordinates of the points in the other systems. +

    +     + + + + +

    + Points in rectangular coordinates: + (2,2,1) and (-\sqrt{3},1,0) +

    +     + +

    + Cylindrical: + (2\sqrt 2,\pi/4,1) and (2,5\pi/6,0) +

    + +

    + Spherical: + (3,\pi/4,\cos^{-1}(3)) and (2,5\pi/6,0) +

    +     +     + + +

    + Points in cylindrical coordinates: + (2,\pi/4,2) and (3,3\pi/2,-4) +

    +     + +

    + Rectangular: + (\sqrt 2,\sqrt 2,2) and (0,-3,-4) +

    + +

    + Spherical: + (2\sqrt 2,\pi/4,\pi/4) and (5,3\pi/2,-\tan^{-1}(4/3)) +

    +     +     + + +

    + Points in spherical coordinates: + (2,\pi/4,\pi/4) and (1,0,0) +

    +     + +

    + Rectangular: + (1,1,\sqrt{2}) and (1,0,0) +

    + +

    + Cylindrical: + (\sqrt{2},\pi/4,\sqrt{2}) and (1,0,0) +

    +     +     +     + + + + +

    + Points in rectangular coordinates: + (0,1,1) and (-1,0,1) +

    +     + +

    + Cylindrical: (1,\pi/2,1) and (1,\pi,1) +

    +

    + Spherical: + (\sqrt 2,\pi/2,\pi/4) and (\sqrt{2}, \pi, \pi/4) +

    +     +     + + + +

    + Points in cylindrical coordinates: + (0,\pi,1) and (2,4\pi/3,0) +

    +     + +

    + Rectangular: + (0,0,1) and (-1,-\sqrt 3,0) +

    +

    + Spherical: + (1,\pi,\pi/2) and (2,4\pi/3,0) +

    +     +     + + +

    + Points in spherical coordinates: + (2,\pi/6,0) and (3,\pi,-\pi/2) +

    +     + +

    + Rectangular: + (\sqrt 3,1,0) and (0,0,-3) +

    +

    + Cylindrical: + (2,\pi/6,0) and (0,\pi,-3) +

    +     +     +     +     + + + +

    + Describe the curve, + surface or region in space determined by the given bounds in cylindrical coordinates. +

    +     + + + + +

    + r=1, 0\leq \theta\leq 2\pi, 0\leq z\leq 1 +

    +     + +

    + A cylindrical surface or tube, + centered along the z-axis of radius 1, extending from the xy-plane up to the plane z=1 (, the tube has a length of 1). +

    +     +     + + +

    + 1\leq r\leq 2, 0\leq \theta\leq \pi, 0\leq z\leq 1 +

    +     + +

    + This is a region of space, + being half of a tube with thick + walls of inner radius 1 and outer radius 2, centered along the z-axis with a length of 1, where the half below + the xz-plane is removed. +

    +     +     +     + + + + +

    + 1\leq r\leq 2, \theta= \pi/2, 0\leq z\leq 1 +

    +     + +

    + A square portion of the yz-plane with corners at (0,1,0), + (0,1,1), + (0,2,1) and (0,2,0). +

    +     +     + + +

    + r= 2, 0\leq \theta\leq 2\pi, z=5 +

    +     + +

    + This is a curve, a circle of radius 2, centered at (0,0,5), + lying parallel to the xy-plane (, in the plane z=5). +

    +     +     +     +     + + + +

    + Describe the curve, + surface or region in space determined by the given bounds in spherical coordinates. +

    +     + + + + +

    + \rho=3, 0\leq \theta\leq2\pi, + 0\leq\varphi\leq \pi/2 +

    +     + +

    + This is upper half of the sphere of radius 3 centered at the origin (, the upper hemisphere). +

    +     +     + + +

    + 2\leq\rho\leq3, 0\leq \theta\leq2\pi, + -\pi/2\leq\varphi\leq \pi/2 +

    +     + +

    + This is a region of space, + where the ball of radius 2, centered at the origin, + is removed from the ball of radius 3, centered at the origin. +

    +     +     +     + + + + +

    + 0\leq\rho\leq2, 0\leq \theta\leq\pi, \varphi = \pi/4 +

    +     + +

    + This is a region of space, a half of a solid cone with rounded top, + where the rounded top is a portion of the ball of radius 2 centered at the origin and the sides of the cone make an angle of \pi/4 with the positive z-axis. + The bounds on \theta mean only the portion above + the xz-plane are retained. +

    +     +     + + +

    + \rho=2, + 0\leq \theta\leq2\pi, \varphi = \pi/3 +

    +     +     + + +

    + This is a curve, + a circle of radius 1 centered at (0,0,\sqrt 3), + lying parallel to the xy-plane. +

    +     +     +     +     + + + +

    + Standard regions in space, + as defined by cylindrical and spherical coordinates, are shown. + Set up the triple integral that integrates the given function over the graphed region. +

    +     + + + + +

    + Cylindrical coordinates, integrating h(r,\theta,z): +

    + + + A region in space given by constant bounds in cylindrical coordinates. + +

    + The image depicts a solid region in space given by the following inequalities: + + r_1 \amp \leq r \leq r_2 + \theta_1 \amp \leq \theta\leq \theta_2 + z_1 \amp \leq z\leq z_2 + . +

    + +

    + The shape of the solid resembles that of a circular wall. +

    +     + + //import apexstyle; + //ASY file for figspherical1_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(16.5,1,4.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + pair zbounds=(-.1,1.1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + real r1=.4; + real r2=.6; + real t1=pi/6; + real t2=4pi/3; + real z1=.4; + real z2=.8; + + // (t.x*cos(t.y),t.x*sin(t.y),z) + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (r1*cos(t.y),r1*sin(t.y),t.x);// + } + surface s=surface(f,(z1,t1),(z2,t2),4,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + triple f(pair t) { + return (r2*cos(t.y),r2*sin(t.y),t.x);// + } + surface s=surface(f,(z1,t1),(z2,t2),4,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + triple f(pair t) { + return (t.x*cos(t1),t.x*sin(t1),t.y);// + } + surface s=surface(f,(r1,z1),(r2,z2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + triple f(pair t) { + return (t.x*cos(t2),t.x*sin(t2),t.y);// + } + surface s=surface(f,(r1,z1),(r2,z2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + triple f(pair t) { + return (t.x*cos(t.y),t.x*sin(t.y),z1);// + } + surface s=surface(f,(r1,t1),(r2,t2),4,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + triple f(pair t) { + return (t.x*cos(t.y),t.x*sin(t.y),z2);// + } + surface s=surface(f,(r1,t1),(r2,t2),4,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + draw((0,0,0) -- .8*(cos(t1),sin(t1),0),dashed+.25mm); + draw((0,0,0) -- .8*(cos(t2),sin(t2),0),dashed+.25mm); + + draw(arc((0,0,0),.3*(1,0,0),.3*(cos(t2),sin(t2),0),(0,0,1)),black+.25mm,Arrow3(size=2mm)); + draw(arc((0,0,0),.5*(1,0,0),.3*(cos(t1),sin(t1),0),(0,0,1)),black+.25mm,Arrow3(size=2mm)); + + label("$\theta_1$",.6(cos(t1/2),sin(t1/2),-.05)); + label("$\theta_2$",.5(cos(t2/1.5),sin(t2/1.5),0)); + + draw ((r2*cos(t2),r2*sin(t2),z1) -- (r2*cos(t2),r2*sin(t2),0),dashed+.25mm); + draw ((r2*cos(t1),r2*sin(t1),z1) -- (r2*cos(t1),r2*sin(t1),0),dashed+.25mm); + + draw ((0,0,z2) -- (r2*cos(t2/1.5),r2*sin(t2/1.5),z2),dashed+.25mm); + draw ((0,0,z2) -- (r1*cos(t2/3),r1*sin(t2/3),z2),dashed+.25mm); + + label("$r_2$",(r2*.95*cos(t2/1.5),r2*.95*sin(t2/1.5),z2+.05)); + label("$r_1$",(r1*.8*cos(t2/3),r1*.8*sin(t2/3),z2+.05)); + + draw((0,0,z1) -- (r1*cos(t1),r1*sin(t1),z1),dashed+.25mm); + draw((0,0,z2) -- (r1*cos(t1),r1*sin(t1),z2),dashed+.25mm); + + label("$z_1$",(0,0,z1),W); + label("$z_2$",(0,0,z2),W); + + //dot((0,0,2.799),blackmeshpen+1.2mm); + //draw((-1.5,-1.5,2) -- (-1.5,1.5,2) -- (1.5,1.5,2) -- (1.5,-1.5,2) -- (-1.5,-1.5,2),apexmeshpen+.25mm); + + triple g(real t) {return (1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5);} + path3 mypath=graph(g,0,2*pi,operator ..); + //draw(mypath,apexmeshpen); + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (t.x,t.x*sqrt(3),t.y);// + } + surface s=surface(f,(-1.25,-1.5),(1.25,1.5),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen q=redmeshpen; + //draw(s,simplesurfacepen2,meshpen=q); + + //draw((-1.25,-1.25*sqrt(3),-1.5) -- (1.25,1.25*sqrt(3),-1.5) -- (1.25,1.25*sqrt(3),1.5) -- (-1.25,-1.25*sqrt(3),1.5) -- (-1.25,-1.25*sqrt(3),-1.5),redcurvepen+.25mm); + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (sin(t.y)*cos(t.x),sin(t.y)*sin(t.x),cos(t.y));// + } + surface s=surface(f,(0,0),(2*pi,pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen q=greencurvepen; + //draw(s,simplesurfacepen3,meshpen=q); + + //dot((.5,sqrt(3)*.5,2),rgb(.1,.1,.1)); + + //draw((.5,sqrt(3)*.5,-.5)--(.5,sqrt(3)*.5,2.25)); + + triple g(real t) {return (1/2*cos(t),1/2*sin(t),1/2*sqrt(3));} + path3 mypath=graph(g,0,2*pi,operator ..); + //draw(mypath,blackmeshpen); + + //(1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5) + //draw((1.5/sqrt(3)*cos(4*pi/3),1.5/sqrt(3)*sin(4*pi/3),1.5) -- (0,0,0) -- (1.5/sqrt(3)*cos(pi/3),1.5/sqrt(3)*sin(pi/3),1.5)); + + triple g(real t) {return (sin(t)*cos(pi/3),sin(t)*sin(pi/3),cos(t));} + path3 mypath=graph(g,0,2*pi,operator ..); + //draw(mypath,blackmeshpen); + + +     + +

    + \ds\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}\int_{z_1}^{z_2}h(r,\theta,z)r\, dz\, dr\, d\theta +

    +     +     + + + +

    + Spherical coordinates, integrating h(\rho,\theta,\varphi): +

    + + A region in space given by constant bounds in spherical coordinates. + +

    + A solid region in space is depicted; it illustrates the following inequalities: + + \rho_1 \amp \leq \rho \leq \rho_2 + \varphi_1 \amp \leq \varphi\leq \varphi_2 + \theta_1 \amp \leq \theta \leq \theta_2 + . +

    + +

    + The shape of the solid looks similar to a thick piece of rind peeled from the top of an orange. +

    +     + + + //import apexstyle; + + + //ASY file for figspherical1_3D.asy in Chapter 13 + + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(16,.9,8.1); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.6,.6); + pair ybounds=(-.6,.6); + pair zbounds=(-.1,1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + + real t1=pi/3; + real t2=7pi/6; + real p1=pi/6-.2; + real p2=pi/6+.2; + real r1=.75; + real r2=.9; + + + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (r1*cos(t.x)*sin(t.y),r1*sin(t.x)*sin(t.y),r1*cos(t.y));// + } + surface s=surface(f,(t1,p1),(t2,p2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + + triple f(pair t) { + return (r2*cos(t.x)*sin(t.y),r2*sin(t.x)*sin(t.y),r2*cos(t.y));// + } + surface s=surface(f,(t1,p1),(t2,p2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + triple f(pair t) { + return (t.y*cos(t.x)*sin(p1),t.y*sin(t.x)*sin(p1),t.y*cos(p1));// + } + surface s=surface(f,(t1,r1),(t2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + triple f(pair t) { + return (t.y*cos(t.x)*sin(p2),t.y*sin(t.x)*sin(p2),t.y*cos(p2));// + } + surface s=surface(f,(t1,r1),(t2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + triple f(pair t) { + return (t.y*cos(t1)*sin(t.x),t.y*sin(t1)*sin(t.x),t.y*cos(t.x));// + } + surface s=surface(f,(p1,r1),(p2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + triple f(pair t) { + return (t.y*cos(t2)*sin(t.x),t.y*sin(t2)*sin(t.x),t.y*cos(t.x));// + } + surface s=surface(f,(p1,r1),(p2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,simplesurfacepen,meshpen=p); + + + // + // lines for phi + // + draw((r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1)) -- (0,0,0) -- (r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)),black+.25mm+dashed); + + //draw(arc((0,0,0),(r1*cos(t1)*sin(p1)/2.5,r1*sin(t1)*sin(p1)/2.5,r1*cos(p1)/2.5),(r1*cos(t1)*sin(p2)/2.5,r1*sin(t1)*sin(p2)/2.5,r1*cos(p2)/2.5)),black+.25mm,Arrow3(size=1.5mm)); + + //label("$\Delta\varphi$",.95*(r1*cos(t1)*sin((p1+p2)/2)/2,r1*sin(t1)*sin((p1+p2)/2)/2,r1*cos((p1+p2)/2)/2)); + + + + // + // lines for theta + // + + draw((0,0,0) -- (r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0),black+.25mm+dashed); + + draw((0,0,0) -- (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),0),black+.25mm+dashed); + + draw(arc((0,0,0),.25*(r1,0,0),.5*(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0)),black+.25mm,Arrow3(size=1.5mm)); + + draw(arc((0,0,0),.4*(r1,0,0),.4*(r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),0),(0,0,1)),black+.25mm,Arrow3(size=1.5mm)); + + label("$\theta_1$",.25*(r1*cos(t1/2)*sin(p2),r1*sin(t1/2)*sin(p2),0)); + label("$\theta_2$",.5*(r1*cos(t2/2)*sin(p2),r1*sin(t2/2)*sin(p2),0)); + + // + // lines for rho + // + + draw(arc((0,0,0),(r1*cos(t1)*sin(0),r1*sin(t1)*sin(0),r1*cos(0)),(r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1))),black+.25mm+dashed); + + draw(arc((0,0,0),(r2*cos(t1)*sin(0),r2*sin(t1)*sin(0),r2*cos(0)),(r2*cos(t1)*sin(p1),r2*sin(t1)*sin(p1),r2*cos(p1))),black+.25mm+dashed); + + + label("$\rho_1$",(0,0,r1),W); + label("$\rho_2$",(0,0,r2),W); + + real poff = -0.05; + + //draw((r1*cos(t1)*sin(p1+poff),r1*sin(t1)*sin(p1+poff),r1*cos(p1+poff)) -- (r2*cos(t1)*sin(p1+poff),r2*sin(t1)*sin(p1+poff),r2*cos(p1+poff)),black+.25mm,Arrows3(size=1.5mm)); + + //label("$\rho$",(0,-.04,.5*r1)); + + //draw((0,-.02,.01) -- (0,-.02,.99*r1),black+.25mm,Arrows3(size=1.5mm)); + + + // + // lines for theta length + // + + //draw((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) -- (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05),black+.25mm,Arrows3(size=1.5mm)); + + //label("$\rho\sin(\varphi)\Delta\theta$",1.5*((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05)+(0,0,-.05))/2+(0,0,-.25),S); + + //draw(1.5*((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05)+(0,0,-.05))/2 + (0,0,-.25) -- 1.05((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05))/2+(0,0,-.01),black+.25mm+dashed,Arrow3(size=1.5mm)); + + // + // lines for phi length + // + + draw(arc((0,0,0),(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0),.95*(r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1))),black+.25mm,Arrows3(size=1.5mm)); +label("$\varphi_2$",.8*(r1*cos(t1)*sin(pi/2-p1/2-.5),r1*sin(t1)*sin(pi/2-p1/2-.5),r1*cos(pi/2-p1/2-.5)),S); + +label("$\varphi_1$",.63*(r1*cos(t1)*sin(pi/2-p1-(p2-p1)/2),r1*sin(t1)*sin(pi/2-p1-(p2-p1)/2),r1*cos(pi/2-p1-(p2-p1)/2)),S); +draw(arc((0,0,0),.8*(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0),.7*(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2))),black+.25mm,Arrows3(size=1.5mm)); + + +     + +

    + \ds\int_{\varphi_1}^{\varphi_2}\int_{\theta_1}^{\theta_2}\int_{\rho_1}^{\rho_2}h(\rho,\theta,\varphi)\rho^2\sin(\varphi)\, d\rho\, d\theta\, d\varphi +

    +     +     +     + + + +

    + A triple integral in cylindrical coordinates is given. + Describe the region in space defined by the bounds of the integral. +

    +     + + + + +

    + \ds \int_0^{\pi/2}\int_0^2\int_0^2 r\, dz\, dr\, d\theta +

    +     + +

    + The region in space is bounded between the planes z=0 and z=2, + inside of the cylinder x^2+y^2=4, + and the planes \theta = 0 and \theta = \pi/2: + describes a wedge of a cylinder of height 2 and radius 2; + the angle of the wedge is \pi/2, or 90^\circ. +

    +     +     + + + +

    + \ds \int_0^{2\pi}\int_3^4\int_0^5 r\, dz\, dr\, d\theta +

    +     + +

    + Bounded between the planes z=0 and z=5, + between the cylinders x^2+y^2=9 and x^2+y^2=16: + describes a pipe or tube + of length 5, an inner radius of 3 and outer radius of 4. +

    +     +     + + + +

    + \ds \int_0^{2\pi}\int_0^1\int_0^{1-r} r\, dz\, dr\, d\theta +

    +     + +

    + Bounded between the plane z=0 and the cone z=1-\sqrt{x^2+y^2}: + describes an inverted cone, + with height of 1, point at (0,0,1) and base radius of 1. +

    +     +     + + + +

    + \ds \int_0^{\pi}\int_0^1\int_0^{2-r} r\, dz\, dr\, d\theta +

    +     + +

    + Bounded between y\geq 0, + inside the cylinder x^2+y^2=1, + above the plane z=0 and below the cone z = 2-\sqrt{x^2+y^2}: + describes cylindrical solid of height 1 and radius 2, topped with an inverted cone of height 1 and base radius 1 with point at (0,0,2). +

    +     +     + + + +

    + \ds \int_0^{\pi}\int_0^3\int_0^{\sqrt{9-r^2}} r\, dz\, dr\, d\theta +

    +     + +

    + Describes a quarter of a ball of radius 3, centered at the origin; + the quarter resides above the xy-plane and above the xz-plane. +

    +     +     + + + +

    + \ds \int_0^{2\pi}\int_0^a\int_0^{\sqrt{a^2-r^2}+b} r\, dz\, dr\, d\theta +

    +     + +

    + Bounded between the plane z=0, + inside the cylinder x^2+y^2 = a^2, + and below the upper hemisphere z= \sqrt{a^2-x^2-y^2}+b, + with radius a and centered at (0,0,b): + describes a cylindrical solid of radius a and height b, + topped with the upper hemisphere of radius a. +

    +     +     +     + + + +

    + A triple integral in spherical coordinates is given. + Describe the region in space defined by the bounds of the integral. +

    +     + + + + +

    + \ds \int_0^{\pi/2}\int_0^{\pi/2}\int_0^{1} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi +

    +     + +

    + Describes the portion of the unit ball that resides in the first octant. +

    +     +     + + + +

    + \ds \int_{-\pi/2}^{\pi/2}\int_0^{\pi}\int_1^{1.1} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi +

    +     + +

    + Describes half of a spherical shell (, y\geq 0) with inner radius of 1 and outer radius of 1.1 centered at the origin. +

    +     +     + + + +

    + \ds \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_0^{2} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi +

    +     + +

    + Bounded below by the cone z=\sqrt{x^2+y^2} and above by the sphere x^2+y^2+z^2=4: + describes a shape that is somewhat diamond-like; + some think of it as looking like an ice cream cone + (see ). + It describes a cone, + where the side makes an angle of \pi/4 with the xy-plane, + topped by the portion of the ball of radius 2, centered at the origin. +

    +     +     + + + +

    + \ds \int_{\pi/4}^{\pi/3}\int_0^{2\pi}\int_0^{2} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi +

    +     + +

    + It is the region is space bounded below by + z=\sqrt{x^2+y^2} and above by the sphere x^2+y^2+z^2=4, + with the portion above the cone z=\sqrt3\sqrt{x^2+y^2} removed: + it describes a cone, + where the side makes an angle of \pi/4 with the xy-plane, + topped by the portion of the ball of radius 2, centered at the origin, + with the inner cone with angle \pi/3 removed, + along with corresponding portion of the ball of radius 2. +

    +     +     + + + +

    + \ds \int_{\pi/3}^{\pi/2}\int_0^{2\pi}\int_0^{\csc(\varphi)} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi +

    +     + +

    + The region in space is bounded below by the cone + z=\sqrt{3}\sqrt{x^2+y^2} and above by the plane z=1: + it describes a cone, with point at the origin, + centered along the positive z-axis, + with height of 1 and base radius of \cot(\pi/3) = \tan(\pi/6) = 1/\sqrt{3}. +

    +     +     + + + +

    + \ds \int_{\pi/3}^{\pi/2}\int_0^{2\pi}\int_0^{a\csc(\varphi)} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi +

    +     + +

    + The region in space is bounded below by the cone + z=\sqrt{3}\sqrt{x^2+y^2} and above by the plane z=a: + it describes a cone, with point at the origin, + centered along the positive z-axis, + with height of a and base radius of a\cot(\pi/3). +

    +     +     +     + + + +

    + A solid is described along with its density function. + Find the mass of the solid using cylindrical coordinates. +

    +     + + + +

    + Bounded by the cylinder x^2+y^2=4 and the planes z=0 and z=4 with density function \delta(x,y,z) =\sqrt{x^2+y^2}+1. +

    +     + +

    + In cylindrical coordinates, the density is \delta(r,\theta,z) = r+1. + Thus mass is + + \int_0^{2\pi}\int_0^2\int_0^4 (r+1)r\, dz\, dr\, d\theta = 112\pi/3 + . +

    +     +     + + + +

    + Bounded by the cylinders x^2+y^2=4 and x^2+y^2=9, + between the planes z=0 and z=10 with density function \delta(x,y,z) =z. +

    +     + +

    + In cylindrical coordinates, the density is \delta(r,\theta,z) = z. + Thus mass is + + \int_0^{2\pi}\int_2^3\int_0^{10} zr\, dz\, dr\, d\theta = 250\pi + . +

    +     +     + + + +

    + Bounded by y\geq 0, the cylinder x^2+y^2=1, + and between the planes z=0 and z=4-y with density function \delta(x,y,z) =1. +

    +     + +

    + In cylindrical coordinates, the density is \delta(r,\theta,z) = 1. + Thus mass is + + \int_0^{\pi}\int_0^1\int_0^{4-r\sin(\theta)} r\, dz\, dr\, d\theta = 2\pi-2/3\approx 5.617 + . +

    +     +     + + + +

    + The upper half of the unit ball, + bounded between z= 0 and z=\sqrt{1-x^2-y^2}, + with density function \delta(x,y,z) =1. +

    +     + +

    + In cylindrical coordinates, the density is \delta(r,\theta,z) = 1. + Thus mass is + + \int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} r\, dz\, dr\, d\theta = 4\pi/3 + . +

    +     +     +     + + + +

    + A solid is described along with its density function. + Find the center of mass of the solid using cylindrical coordinates. + (Note: these are the same solids and density functions as found in Exercises.) +

    +     + + + +

    + Bounded by the cylinder x^2+y^2=4 and the planes z=0 and z=4 with density function \delta(x,y,z) =\sqrt{x^2+y^2}+1. +

    +     + +

    + In cylindrical coordinates, the density is \delta(r,\theta,z) = r+1. + Thus mass is + + M=\int_0^{2\pi}\int_0^2\int_0^4 (r+1)r\, dz\, dr\, d\theta = 112\pi/3 + . +

    + +

    + We find M_{yz} = 0, M_{xz} = 0, + and M_{xy} = 224\pi/3, + placing the center of mass at (0,0,2). +

    +     +     + + + +

    + Bounded by the cylinders x^2+y^2=4 and x^2+y^2=9, + between the planes z=0 and z=10 with density function \delta(x,y,z) =z. +

    +     + +

    + In cylindrical coordinates, the density is \delta(r,\theta,z) = z. + Thus mass is + + M=\int_0^{2\pi}\int_2^3\int_0^{10} zr\, dz\, dr\, d\theta = 250\pi + . +

    + +

    + We find M_{yz} = 0, M_{xz} = 0, + and M_{xy} = 5000\pi/3, + placing the center of mass at (0,0,20/3). +

    +     +     + + + +

    + Bounded by y\geq 0, the cylinder x^2+y^2=1, + and between the planes z=0 and z=4-y with density function \delta(x,y,z) =1. +

    +     + +

    + In cylindrical coordinates, the density is \delta(r,\theta,z) = 1. + Thus mass is + + \int_0^{\pi}\int_0^1\int_0^{4-r\sin(\theta)} r\, dz\, dr\, d\theta = 2\pi-2/3\approx 5.617 + . +

    + +

    + We find M_{yz} = 0, + M_{xz} = 8/3-\pi/8, and M_{xy} = 65\pi/16-8/3, + placing the center of mass at \approx (0,0.405,1.80). +

    +     +     + + + +

    + The upper half of the unit ball, + bounded between z= 0 and z=\sqrt{1-x^2-y^2}, + with density function \delta(x,y,z) =1. +

    +     + +

    + In cylindrical coordinates, the density is \delta(r,\theta,z) = 1. + Thus mass is + + \int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} r\, dz\, dr\, d\theta = 2\pi/3 + . +

    + +

    + We find M_{yz} = 0, M_{xz} = 0, + and M_{xy} = \pi/4, placing the center of mass at (0,0,3/8). +

    +     +     +     + + + +

    + A solid is described along with its density function. + Find the mass of the solid using spherical coordinates. +

    +     + + + +

    + The upper half of the unit ball, + bounded between z= 0 and z=\sqrt{1-x^2-y^2}, + with density function \delta(x,y,z) =1. +

    +     + +

    + In spherical coordinates, + the density is \delta(\rho,\theta,\varphi) = 1. + Thus mass is + + \int_0^{\pi/2}\int_0^{2\pi}\int_{0}^{1} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = 2\pi/3 + . +

    +     +     + + + +

    + The spherical shell bounded between x^2+y^2+z^2=16 and + x^2+y^2+z^2=25 with density function \delta(x,y,z) = \sqrt{x^2+y^2+z^2}. +

    +     + +

    + In spherical coordinates, + the density is \delta(\rho,\theta,\varphi) = \rho. + Thus mass is + + \int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\int_{4}^{5} (\rho)\rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = 369\pi + . +

    +     +     + + + +

    + The conical region bounded below by z=\sqrt{x^2+y^2} and above by the sphere + x^2+y^2+z^2=1 with density function \delta(x,y,z) = z. +

    +     + +

    + In spherical coordinates, + the density is \delta(\rho,\theta,\varphi) = \rho\sin(\varphi). + Thus mass is + + \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_{0}^{1} \big(\rho\sin(\varphi)\big)\rho^2\sin(\varphi)\, d\rho\, d\theta\, d\varphi = \pi/8 + . +

    +     +     + + + +

    + The cone that lies above the cone z=\sqrt{x^2+y^2} and below the plane z=1 with density function \delta(x,y,z) = z. +

    +     + +

    + In spherical coordinates, + the density is \delta(\rho,\theta,\varphi) = \rho\sin(\varphi). + Thus mass is + + \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_{0}^{\csc(\varphi)} \big(\rho\sin(\varphi)\big)\rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = \pi/4 + . +

    +     +     +     + + + +

    + A solid is described along with its density function. + Find the center of mass of the solid using spherical coordinates. + (Note: these are the same solids and density functions as found in Exercises.) +

    +     + + + +

    + The upper half of the unit ball, + bounded between z= 0 and z=\sqrt{1-x^2-y^2}, + with density function \delta(x,y,z) =1. +

    +     + +

    + In spherical coordinates, + the density is \delta(\rho,\theta,\varphi) = 1. + Thus mass is + + \int_0^{\pi/2}\int_0^{2\pi}\int_{0}^{1} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = 2\pi/3 + . +

    + +

    + We find M_{yz} = 0, M_{xz} = 0, + and M_{xy} = \pi/4, placing the center of mass at (0,0,3/8). +

    +     +     + + + +

    + The spherical shell bounded between x^2+y^2+z^2=16 and + x^2+y^2+z^2=25 with density function \delta(x,y,z) = \sqrt{x^2+y^2+z^2}. +

    +     + +

    + In spherical coordinates, + the density is \delta(\rho,\theta,\varphi) = \rho. + Thus mass is + + \int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\int_{4}^{5} (\rho)\rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = 369\pi + . +

    + +

    + We find M_{yz} = 0, M_{xz} = 0, + and M_{xy} = 0, placing the center of mass at (0,0,0). +

    +     +     + + + +

    + The conical region bounded above z=\sqrt{x^2+y^2} and below the sphere + x^2+y^2+z^2=1 with density function \delta(x,y,z) = z. +

    +     + +

    + In spherical coordinates, + the density is \delta(\rho,\theta,\varphi) = \rho\sin(\varphi). + Thus mass is + + \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_{0}^{1} \big(\rho\sin(\varphi)\big)\rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = \pi/8 + . +

    + +

    + We find M_{yz} = 0, + M_{xz} = 0, and M_{xy} = (4-\sqrt{2})\pi/30, + placing the center of mass at (0,0,4(4-\sqrt2)/15). +

    +     +     + + + +

    + The cone bounded above z=\sqrt{x^2+y^2} and below the plane z=1 with density function \delta(x,y,z) = z. +

    +     + +

    + In spherical coordinates, + the density is \delta(\rho,\theta,\varphi) = \rho\sin(\varphi). + Thus mass is + + \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_{0}^{\csc(\varphi)} \big(\rho\sin(\varphi)\big)\rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = \pi/4 + . +

    + +

    + We find M_{yz} = 0, M_{xz} = 0, + and M_{xy} = \pi/5, placing the center of mass at (0,0,4/5). +

    +     +     +     + + + +

    + A region is space is described. + Set up the triple integrals that find the volume of this region using rectangular, + cylindrical and spherical coordinates, + then comment on which of the three appears easiest to evaluate. +

    +     + + + +

    + The region enclosed by the unit sphere, x^2+y^2+z^2=1. +

    +     + +

    + Rectangular: + \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}}\, dz\, dy\, dx +

    + +

    + Cylindrical: + \int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}r\, dz\, dr\, d\theta +

    + +

    + Spherical: \int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\int_0^1 \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi +

    + +

    + Spherical appears simplest, + avoiding the integration of square-roots and using techniques such as Substitution; + all bounds are constants. +

    +     +     + + + +

    + The region enclosed by the cylinder + x^2+y^2=1 and planes z=0 and z=1. +

    +     + +

    + Rectangular: + \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{0}^{1}\, dz\, dy\, dx +

    + +

    + Cylindrical: + \int_0^{2\pi}\int_0^1\int_{0}^{1}r\, dz\, dr\, d\theta +

    + +

    + Spherical: \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_0^{\csc(\varphi)} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi + \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_0^{\csc(\varphi)} \rho^2\sin(\varphi)\, d\rho\, d\theta\, d\varphi +

    + +

    + Cylindrical appears simplest, + avoiding the integration of square-roots and two triple integrals; + all bounds are constants. +

    +     +     + + + +

    + The region enclosed by the cone + z=\sqrt{x^2+y^2} and plane z=1. +

    +     + +

    + Rectangular: + \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{\sqrt{x^2+y^2}}^{1}\, dz\, dy\, dx +

    + +

    + Cylindrical: + \int_0^{2\pi}\int_0^1\int_{r}^{1}r\, dz\, dr\, d\theta +

    + +

    + Spherical: \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_0^{\csc(\varphi)} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi +

    + +

    + Cylindrical appears simplest, + avoiding the integration of square-roots that rectangular uses. + Spherical is not difficult, though it requires Substitution, + an extra step. +

    +     +     + + + +

    + The cube enclosed by the planes x=0, + x=1, y=0, y=1, + z=0 and z=1. + (Hint: in spherical, + use order of integration d\rho\, d\varphi\, d\theta.) +

    +     + +

    + Rectangular: + \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\, dz\, dy\, dx +

    + +

    + Cylindrical: + \int_0^{\pi/4}\int_0^{\sec(\theta)}\int_{0}^{1}r\, dz\, dr\, d\theta + \int_{\pi/4}^{\pi/2}\int_0^{\csc(\theta)}\int_{0}^{1}r\, dz\, dr\, d\theta +

    + +

    + Spherical: + + \amp \int_0^{\pi/4}\int_0^{\tan^{-1}(\sec(\theta))}\int_0^{\csc(\varphi)}\rho^2 \cos(\varphi)\, d\rho\, d\varphi\, d\theta + \amp\quad+\int_0^{\pi/4}\int_{\tan^{-1}(\sec(\theta))}^{\pi/2}\int_0^{\sec(\theta)\sec(\varphi)}\rho^2 \cos(\varphi)\, d\rho\, d\varphi\, d\theta + \amp\quad+\int_{\pi/4}^{\pi/2}\int_0^{\tan^{-1}(\csc(\theta))}\int_0^{\csc(\varphi)}\rho^2 \cos(\varphi)\, d\rho\, d\varphi\, d\theta + \amp\quad+\int_{\pi/4}^{\pi/2}\int_{\tan^{-1}(\csc(\theta))}^{\pi/2}\int_0^{\csc(\theta)\sec(\varphi)}\rho^2 \cos(\varphi)\, d\rho\, d\varphi\, d\theta + . +

    + +

    + Rectangular is clearly the simplest. +

    +     +     +     +     +     +
    + +     + + + Vector Analysis + +

    + This chapter explores completely different relationships between vectors and integration. + These relationships will enable us to compute the work done by a magnetic field in moving an object along a path and find how much air moves through an oddly-shaped screen in space, + among other things. +

    + +

    + Our upcoming work with integration will benefit from a review. + We are not concerned here with techniques of integration, + but rather what an integral does + and how that relates to the notation we use to describe it. +

    + + + Integration review +

    + Recall from + that when R is a region in the xy-plane, + \iint_R dA gives the area of the region R. + The integral symbols are elongated esses meaning sum + and dA represents a small amount of area. Taken together, + \iint_R dA means sum up, over R, + small amounts of area. This sum then gives the total area of R. + We use two integral symbols since R is a two-dimensional region. +

    + +

    + Now let z=f(x,y) represent a surface. + The integral \iint_R f(x,y)\, dA means + sum up, over R, + function values (heights) given by f times small amounts of area. + Since height area = volume, + we are summing small amounts of volume over R, + giving the total signed volume under the surface + z=f(x,y) and above the xy-plane. +

    + +

    + This notation does not directly inform us how + to evaluate the double integrals to find an area or a volume. + With additional work, + we recognize that a small amount of area dA can be measured as the area of a small rectangle, + with one side length a small change in x and the other side length a small change in y. + That is, dA = dx\,dy or dA = dy\,dx. + We could also compute a small amount of area by thinking in terms of polar coordinates, + where dA = r\,dr\,d\theta. + These understandings lead us to the iterated integrals we used in . +

    + +

    + Let us back our review up farther. + Note that \int_1^3\, dx = x\big|_1^3 = 3-1 = 2. + We have simply measured the length of the interval [1,3]. + We could rewrite the above integral using syntax similar to the double integral syntax above: + + \int_1^3\, dx = \int_Idx, \text{ where \(I\) = \([1,3]\) } + . +

    + +

    + We interpret \int_I dx as meaning sum up, + over the interval I, + small changes in x. + A change in x is a length along the x-axis, + so we are adding up along I small lengths, + giving the total length of I. +

    + +

    + We could also write \int_1^3f(x)\, dx as \int_I f(x)\, dx, + interpreted as sum up, over I, + heights given by y = f(x) times small changes in x. + Since heightlength = area, + we are summing up areas and finding the total signed area between + y = f(x) and the x-axis. +

    + +

    + This method of referring to the process of integration can be very powerful. + It is the core of our notion of the Riemann Sum. + When faced with a quantity to compute, + if one can think of a way to approximate its value through a sum, + the one is well on their way to constructing an integral (or, + double or triple integral) that computes the desired quantity. + We will demonstrate this process throughout this chapter, + starting with the next section. +

    +     +     + + +
    + Introduction to Line Integrals + +

    + We first used integration to find + area under a curve. In this section, + we learn to do this (again), but in a different context. +

    +     + + + Line Integrals of Functions + +

    + Consider the surface and curve shown in . + The surface is given by f(x,y)=1-\cos(x)\sin(y). + The dashed curve lies in the xy-plane and is the familiar y=x^2 parabola from -1\leq x\leq 1; + we'll call this curve C. + The curve drawn with a solid line in the graph is the curve in space that lies on our surface with x and y values that lie on C. +

    + +

    + The question we want to answer is this: + what is the area that lies below the curve drawn with the solid line? + In other words, + what is the area of the region above C and under the the surface z=f(x,y)? + This region is shown in . +

    + +

    + We suspect the answer can be found using an integral, + but before trying to figure out what that integral is, + let us first try to approximate its value. +

    + +
    + Finding area under a curve in space + +
    + + + A curve in the x,y plane, and the corresponding curve on a surface lying above it. + +

    + A set of three-dimensional coordinate axes are drawn in space. + A surface lies above the xy plane; it is the graph z=1-\cos(x)\sin(y). + The surface looks a little bit like a camping chair, but the precise shape is unimportant. +

    + +

    + In the xy plane, the parabola y=x^2 is plotted.. + On the surface is the corresponding curve, given by parametric equations + x=t, y=t^2, z = 1-\cos(t)\sin(t^2). + These are the points on the surface that lie directly above the parabola. +

    +     + + + + + //ASY file for fig10_01_ex_233D.asy in Chapter 10 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(1.17,4.69,2.35); + //currentprojection=orthographic(2.5,4.6,1.9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={1}; + real[] myzchoice={1,2}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-.2,1.5); + pair zbounds=(-.1,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z=x^2-y^2+1 + triple f(pair t) { + return (t.x,t.y,-sin(t.y)*cos(t.x)+1); + } + surface s=surface(f,(-1,-.1),(1,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return (t,t^2,-cos(t)*sin(t^2)+1);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,t^2,0);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)+dashed); + + + +
    + +
    + + + A cylindrical surface in space that lies above a parabola in the plane, and below a space curve. + +

    + This image shows the surface created by the portion of the cylinder y=x^2 in space + that lies above the plane z=0 and below the surface z=1-\cos(x)\sin(y). +

    + +

    + The surface is like a wall that is bent in the shape of a parabola. + The top of the wall forms a curve that is sinusoidal in shape. + The wall is highest where it meets the z axis, + and then it slopes gently downward toward the ends of the wall, which are of a lower height. +

    +     + + + + + //ASY file for fig10_01_ex_233D.asy in Chapter 10 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(1.17,4.69,2.35); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={1}; + real[] myzchoice={1,2}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-.2,1.5); + pair zbounds=(-.1,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z=x^2-y^2+1 + triple f(pair t) { + return (t.x,t.x^2,t.y*(-sin(t.x^2)*cos(t.x)+1)); + } + surface s=surface(f,(-1,0),(1,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=apexmeshpen; + draw(s,simplesurfacepen,meshpen=p); + + triple g(real t) {return (t,t^2,-cos(t)*sin(t^2)+1);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,t^2,0);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)+dashed); + + + +
    + +
    + + + Approximating the area under a curve in space using planar rectangles. + +

    + The same curve as in is plotted in space. + This time, the cylindrical surface has been removed, + and replaced by four rectangles used to approximate the area under the curve. +

    + +

    + Each rectangle lies in a vertical plane. + The base of each rectangle lies along a secant line between two points of the parabola y=x^2 in the xy plane. + The height of each rectangle is determined by the height of the curve in space at some point above the base. +

    +     + + + + + //ASY file for fig10_01_ex_233D.asy in Chapter 10 + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(1.17,4.69,2.35); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={1}; + real[] myzchoice={1,2}; + defaultpen(0.5mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-.2,1.5); + pair zbounds=(-.1,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z=x^2-y^2+1 + triple f1(pair t) { + return (t.x,-.75/.5*(t.x+1)+1,t.y*(.664)); + } + surface s1=surface(f1,(-1,0),(-.5,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=redpen+linewidth(1); + draw(s1,simplesurfacepen2,meshpen=p); + + //Draw the surface z=x^2-y^2+1 + triple f2(pair t) { + return (t.x,-.5*(t.x+.5)+.25,t.y*(.89)); + } + surface s2=surface(f2,(-.5,0),(0,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=redpen+linewidth(1); + draw(s2,simplesurfacepen2,meshpen=p); + + //Draw the surface z=x^2-y^2+1 + triple f3(pair t) { + return (t.x,.5*(t.x-.5)+.25,t.y*(.89)); + } + surface s3=surface(f3,(0,0),(.5,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=redpen+linewidth(1); + draw(s3,simplesurfacepen2,meshpen=p); + + //Draw the surface z=x^2-y^2+1 + triple f4(pair t) { + return (t.x,.75/.5*(t.x-1)+1,t.y*(.664)); + } + surface s4=surface(f4,(.5,0),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=redpen+linewidth(1); + draw(s4,simplesurfacepen2,meshpen=p); + + triple g(real t) {return (t,t^2,-cos(t)*sin(t^2)+1);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,t^2,0);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)+dashed); + + + +
    +     + +
    + +

    + In , + four rectangles have been drawn over the curve C. + The bottom corners of each rectangle lie on C, + and each rectangle has a height given by the function f(x,y) for some (x,y) pair along C between the rectangle's bottom corners. +

    + +

    + As we know how to find the area of each rectangle, + we are able to approximate the area above C and under f. + Clearly, our approximation will be + an approximation. + The heights of the rectangles do not match exactly with the surface f, + nor does the base of each rectangle follow perfectly the path of C. +

    + +

    + In typical calculus fashion, + our approximation can be improved by using more rectangles. + The sum of the areas of these rectangles gives an approximate value of the true area above C and under f. + As the area of each rectangle is + height width, we assert that the + + \text{ area above \(C\) } \approx \sum (\text{ heights } \times\text{ widths } ) + . +

    + +

    + When first learning of the integral, + and approximating areas with heights widths, + the width was a small change in x: dx. + That will not suffice in this context. + Rather, each width of a rectangle is actually approximating the arc length of a small portion of C. + In , + we used s to represent the arc-length parameter of a curve. + A small amount of arc length will thus be represented by ds. +

    + +

    + The height of each rectangle will be determined in some way by the surface f. + If we parametrize C by s, + an s-value corresponds to an (x,y) pair that lies on the parabola C. + Since f is a function of x and y, + and x and y are functions of s, + we can say that f is a function of s. + Given a value s, we can compute f(s) and find a height. + Thus + + \text{ area under \(f\) and above \(C\) } \amp \approx \sum (\text{ heights } \times\text{ widths } ); + \text{ area under \(f\) and above \(C\) } \amp =\lim_{\norm{\Delta s}\to0}\sum f(c_i)\Delta s_i + \amp =\int_Cf(s)\, ds + . +

    + +

    + Here we have introduce a new notation, + the integral symbol with a subscript of C. + It is reminiscent of our usage of \iint_R. + Using the train of thought found in the Integration Review preceding this section, + we interpret \int_C f(s)\, ds + as meaning sum up, along a curve C, + function values f(s)\timessmall arc lengths. + It is understood here that s represents the arc-length parameter. +

    + +

    + All this leads us to a definition. + The integral found in Equation + is called a line integral. + We formally define it below, but note that the definition is very abstract. + On one hand, + one is apt to say the definition makes sense, while on the other, + one is equally apt to say but I don't know what I'm supposed to do with this definition. + We'll address that after the definition, + and actually find an answer to the area problem we posed at the beginning of this section. +

    + + + Line Integral Over A Scalar Field + +

    + Let C be a smooth curve parametrized by s, + the arc-length parameter, + and let f be a continuous function of s. + A line integral is an integral of the form + + \int_C f(s)\, ds = \lim_{\norm{\Delta s}\to 0}\sum_{i=1}^n f(c_i)\Delta s_i + , + where s_0\lt s_1\lt \ldots\lt s_n is any partition of the s-interval over which C is defined, + c_i is any value in the ith subinterval, + \Delta s_i is the width of the ith subinterval, + and \norm{\Delta s} is the length of the longest subinterval in the partition. + line integralover scalar field +

    +     +     + + +

    + When C is a closed curve, + , a curve that ends at the same point at which it starts, we use + + \oint_C f(s)\, ds \text{ instead of } \int_C f(s)\, ds + . +

    + +

    + The definition of the line integral does not specify whether C is a curve in the plane or space + (or hyperspace), + as the definition holds regardless. + For now, we'll assume C lies in the xy-plane. +

    + +

    + This definition of the line integral doesn't really say anything new. + If C is a curve and s is the arc-length parameter of C on a\leq s\leq b, then + + \int_Cf(s)\, ds = \int_a^bf(s)\, ds + . +

    + +

    + The real difference with this integral from the standard + \int_a^bf(x)\, dx + we used in the past is that of context. + Our previous integrals naturally summed up values over an interval on the x-axis, + whereas now we are summing up values over a curve. + If we can parametrize the curve with the arc-length parameter, + we can evaluate the line integral just as before. + Unfortunately, + parametrizing a curve in terms of the arc-length parameter is usually very difficult, + so we must develop a method of evaluating line integrals using a different parametrization. +

    + +

    + Given a curve C, find any parametrization of C: + x = g(t) and y=h(t), + for continuous functions g and h, + where a\leq t\leq b. + We can represent this parametrization with a vector-valued function, + \vrt = \langle g(t),h(t)\rangle. +

    + +

    + In , + we defined the arc-length parameter in Equation as + + s(t) = \int_0^t \norm{\vec r\,'(u)}\, du + . +

    + +

    + By the Fundamental Theorem of Calculus, + ds = \norm{\vec r\,'(t)}\, dt. + We can substitute the right hand side of this equation for ds in the line integral definition. +

    + +

    + We can view f as being a function of x and y since it is a function of s. + Thus f(s) =f(x,y) =f\big(g(t),h(t)\big). + This gives us a concrete way to evaluate a line integral: + + \int_C f(s)\, ds = \int_a^bf\big(g(t),h(t)\big)\norm{\vec r\,'(t)}\, dt + . +

    + +

    + We restate this as a theorem, + along with its three-dimensional analogue, + followed by an example where we finally evaluate an integral and find an area. +

    + + + Evaluating a Line Integral Over A Scalar Field + +

    +

      +
    • +

      + Let C be a curve parametrized by \vrt =\langle g(t), + h(t)\rangle, + a\leq t\leq b, + where g and h are continuously differentiable, + and let z=f(x,y), where f is continuous over C. + Thenline integralover scalar field + + \int_C f(s)\, ds = \int_a^b f\big(g(t),h(t)\big)\norm{\vec r\,'(t)}\, dt + . +

      +
      • + +
    • +

      + Let C be a curve parametrized by + \vrt =\langle g(t), + h(t),k(t)\rangle, a\leq t\leq b, + where g, + h and k are continuously differentiable, + and let w=f(x,y,z), where f is continuous over C. + Then + + \int_Cf(s)\, ds = \int_a^bf\big(g(t),h(t),k(t)\big)\norm{\vec r\,'(t)}\, dt + . +

      +
      • +
    +

    +     +     + +

    + To be clear, + the first point of + can be used to find the area under a surface + z=f(x,y) and above a curve C. + We will later give an understanding of the line integral when C is a curve in space. +

    + +

    + Let's do an example where we actually compute an area. +

    + + + Evaluating a line integral: area under a surface over a curve + +

    + Find the area under the surface + f(x,y) =\cos(x)+\sin(y)+2 over the curve C, + which is the segment of the line y=2x+1 on -1\leq x\leq 1, + as shown in . +

    +
    + Finding area under a curve in + +
    + + + A curve lies along a surface in space. The curve corresponds to a line drawn in the x,y plane. + +

    + The surface z=\cos(x)+\sin(y)+2 is plotted in space against three-dimensional coordinate axes. + It has the appearance of a saddle surface in the region that is plotted. +

    + +

    + A curve lies along this surface; it is the set of points on the surface + that lie above the line y=2x+1 in the xy plane. + The line in the plane is also plotted for reference. +

    +     + + + + + // ASY file for figlinescalarfield2_3D.asy in Chapter 10 + // The surface is z=cos(x)+sin(y)+2; the path is the line + // y=2x+1. + // This draws the path, the curve in space, and the surface. + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(10,7.25,3.75); + //(7.63,8.65,5); + //currentprojection=orthographic(2.5,4.6,1.9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,3}; + real[] myychoice={-1,3}; + real[] myzchoice={3}; + defaultpen(0.5mm); + pair xbounds=(-1.2,3.4); + pair ybounds=(-1.2,3.4); + pair zbounds=(-.1,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z=cos(x)+sin(y)=2 + triple f(pair t) { + return (t.x,t.y,-sin(t.y)+cos(t.x)+2); + } + surface s=surface(f,(-1,-1),(pi,pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return (t,2*t+1,-sin(2t+1)+cos(t)+2);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,2*t+1,0);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)+dashed); + + + +
    + +
    + + + A curve in space lies above the line y=2x+1 in the x,y plane. The surface that lies between the line and curve is shown. + +

    + The curve in space from is plotted again with respect to three-dimensional coordinate axes. + This time, the surface on which the curve lies is not shown. + Instead, we again see the line y=2x+1 in the xy plane. +

    + +

    + Between the curve in space and the line in the plane there is a surface, which is plotted. + This surface lies in the plane defined by the equation y=2x+1. + It is bounded below by the xy plane, and above by the curve in space. +

    + +

    + From the perspective of a direction normal to the plane y=2x+1, + this is just the area under a curve. +

    +     + + + + + // ASY file for figlinescalarfield2_3D.asy in Chapter 10 + // The surface is z=cos(x)+sin(y)=2; the path is the line + // y=2x+1. + // This draws the surface which we are finding the area of. + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(10,7.25,3.75); + //currentprojection=orthographic(2.5,4.6,1.9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,3}; + real[] myychoice={-1,3}; + real[] myzchoice={3}; + defaultpen(0.5mm); + pair xbounds=(-1.2,3.4); + pair ybounds=(-1.2,3.4); + pair zbounds=(-.1,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z=cos(x)+sin(y)=2 + triple f(pair t) { + return (t.x,2*t.x+1,t.y*(-sin(t.x*2+1)+cos(t.x)+2)); + } + surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=bluecurvepen; + draw(s,simplesurfacepen,meshpen=p); + + triple g(real t) {return (t,2*t+1,-sin(2t+1)+cos(t)+2);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,2*t+1,0);} + path3 mypath=graph(g,-1,1,operator ..); + draw(mypath,bluepen+linewidth(2)+dashed); + + + +
    +     + +
    +     + +

    + Our first step is to represent C with a vector-valued function. + Since C is a simple line, + and we have a explicit relationship between y and x (namely, + that y is 2x+1), + we can let x = t, y = 2t+1, + and write \vrt = \langle t, 2t+1\rangle for -1\leq t\leq 1. +

    + +

    + We find the values of f over C as f(x,y) = f(t,2t+1) = \cos(t)+\sin(2t+1) + 2. + We also need \norm{\vec r\,'(t)}; + with \vrp(t) = \langle 1,2\rangle, + we have \norm{\vrp(t)} = \sqrt{5}. + Thus ds = \sqrt{5}\, dt. +

    + +

    + The area we seek is + + \int_Cf(s)\, ds \amp = \int_{-1}^1 \big(\cos(t)+\sin(2t+1) + 2\big)\sqrt{5}\, dt + \amp = \left.\sqrt{5}\big(\sin(t) - \frac12\cos(2t+1)+2t\big)\right|_{-1}^1 + \amp \approx 14.418\ \text{ units } ^2 + . +

    +     +     + +

    + We will practice setting up and evaluating a line integral in another example, + then find the area described at the beginning of this section. +

    + + + Evaluating a line integral: area under a surface over a curve + +

    + Find the area over the unit circle in the xy-plane and under the graph of f(x,y) = x^2-y^2+3, + shown in . +

    +
    + Finding area under a curve in + +
    + + + A curve lies on a hyperbolic paraboloid above a circle in the plane. + +

    + The hyperbolic paraboloid z=x^2-y^2+3 is plotted in space, + against a set of three-dimensional coordinate axes. + Below the surface, in the xy plane, the unit circle is plotted. +

    + +

    + The points on the surface that lie above the circle form a curve on the surface. + The curve cuts out a portion of the surface that looks exactly like a Pringles chip. +

    +     + + + + + // ASY file for figlinescalarfield2_3D.asy in Chapter 10 + // The surface is z=x^2-y^2+3; the path is the unit circle. + // + // This draws the path, the curve in space, and the surface. + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.5,9,8); + //(7.63,8.65,5); + //currentprojection=orthographic(2.5,4.6,1.9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={4}; + defaultpen(0.5mm); + pair xbounds=(-1.2,1.2); + pair ybounds=(-1.2,1.2); + pair zbounds=(-.1,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z=x^2-y^2+3 + triple f(pair t) { + return (t.x,t.y,t.x^2-t.y^2+3); + } + surface s=surface(f,(-1.1,-1.1),(1.1,1.1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return (cos(t),sin(t),cos(t)^2-sin(t)^2+3);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(2)+dashed); + + + +
    + +
    + + + A cylindrical surface between the x,y plane and a hyperbolic paraboloid. + +

    + A plot of the portion of the cylinder x^2+y^2=1 + that lies between the xy plane and the hyperbolic paraboloid z=x^2-y^2+3. + The surface looks something like a roller coaster track, along with the scaffolding beneath it. +

    +     + + + + + // ASY file for figlinescalarfield2_3D.asy in Chapter 10 + // The surface is z=x^2-y^2+3; the path is the unit circle. + // + // This draws the surface of the area we seek. + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.5,9,8); + //(7.63,8.65,5); + //currentprojection=orthographic(2.5,4.6,1.9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={4}; + defaultpen(0.5mm); + pair xbounds=(-1.2,1.2); + pair ybounds=(-1.2,1.2); + pair zbounds=(-.1,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z=x^2-y^2+3 + triple f(pair t) { + return (cos(t.x),sin(t.x),t.y*(cos(t.x)^2-sin(t.x)^2+3)); + } + surface s=surface(f,(0,0),(2pi,1),16,8,Spline); + pen p=apexmeshpen; + draw(s,simplesurfacepen,meshpen=p); + + triple g(real t) {return (cos(t),sin(t),cos(t)^2-sin(t)^2+3);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(2)+dashed); + + + +
    +     + +
    +     + +

    + The curve C is the unit circle, + which we will describe with the parametrization + \vrt = \langle \cos(t), \sin(t)\rangle for 0\leq t\leq 2\pi. + We find \norm{\vrp(t)} = 1, so ds = 1 dt. +

    + +

    + We find the values of f over C as f(x,y) = f(\cos(t), \sin(t)) = \cos^2t-\sin^2t+3. + Thus the area we seek is (note the use of the \oint f(s) ds notation): + + \oint_C f(s)\, ds \amp = \int_0^{2\pi}\big(\cos^2t-\sin^2t+3\big)\, dt + \amp = 6\pi + . +

    + +

    + (Note: we may have approximated this answer from the start. + The unit circle has a circumference of 2\pi, + and we may have guessed that due to the apparent symmetry of our surface, + the average height of the surface is 3.) +

    +     +     + +

    + We now consider the example that introduced this section. +

    + + + Evaluating a line integral: area under a surface over a curve + +

    + Find the area under f(x,y) = 1-\cos(x)\sin(y) and over the parabola y = x^2, + from -1\leq x\leq 1. +

    +     + +

    + We parametrize our curve C as + \vrt = \langle t,t^2\rangle for -1\leq t\leq 1; + we find \norm{\vrp(t)} = \sqrt{1+4t^2}, + so ds = \sqrt{1+4t^2}\, dt. +

    + +

    + Replacing x and y with their respective functions of t, + we have f(x,y) = f(t,t^2) = 1-\cos(t)\sin(t^2). + Thus the area under f and over C is found to be + + \int_C f(s)\, ds \amp = \int_{-1}^1 \Big(1-\cos(t)\sin\big(t^2\big)\Big)\sqrt{1+t^2}\, dt. + This integral is impossible to evaluate using the techniques developed in this text. We resort to a numerical approximation; accurate to two places after the decimal, we find the area is + \amp = 2.17 + . +

    +     +     + +

    + We give one more example of finding area. +

    + + + Evaluating a line integral: area under a curve in space + +

    + Find the area above the xy-plane and below the helix parametrized by \vrt = \langle \cos(t),2\sin(t),t/\pi\rangle, + for 0\leq t\leq 2\pi, + as shown in . +

    +
    + Finding area under a curve in + + A helical ramp in space. + +

    + A cylindrical surface that lies between the xy plane and a circular helix. + The surface looks somewhat like a spiral staircase that makes one revolution. +

    +     + + + + + // ASY file for figlinescalarfield2_3D.asy in Chapter 10 + // The path is an elliptical helix ⟨ cos t, 2sin t, t/pi ⟩; no specific surface. + // + // This draws the surface of the area we seek. + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(9.8,10.4,5.5); + //(7.63,8.65,5); + //currentprojection=orthographic(2.5,4.6,1.9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-2.2,2.2); + pair ybounds=(-2.2,2.2); + pair zbounds=(-.1,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z=x^2-y^2+3 + triple f(pair t) { + return (cos(t.x),2sin(t.x),t.y*(t.x/pi)); + } + surface s=surface(f,(0,0),(2pi,1),16,4,Spline); + pen p=apexmeshpen; + draw(s,simplesurfacepen,meshpen=p); + + triple g(real t) {return (cos(t),2sin(t),t/pi);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (cos(t),2sin(t),0);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(2)+dashed); + + + +
    +     + +

    + Note how this is problem is different than the previous examples: + here, the height is not given by a surface, but by the curve itself. +

    + +

    + We use the given vector-valued function \vec r(t) to determine the curve C in the xy-plane by simply using the first two components of \vec r(t): + \vec c(t) = \langle \cos(t),2\sin(t)\rangle. + Thus ds = \norm{\vec c\,'(t)}\,dt = \sqrt{\sin^2t + 4\cos^2t}\,dt. +

    + +

    + The height is not found by evaluating a surface over C, + but rather it is given directly by the third component of \vec r(t): + t/\pi. + Thus + + \oint_C f(s)\, ds = \int_0^{2\pi} \frac{t}{\pi}\sqrt{\sin^2t + 4\cos^2t}\,dt \approx 9.69 + , + where the approximation was obtained using numerical methods. +

    +     +     + + + +

    + Note how in each of the previous examples we are effectively finding + area under a curve, + just as we did when first learning of integration. + We have used the phrase area over + a curve C and under a surface, + but that is because of the important role C plays in the integral. + The figures show how the curve C defines another curve on the surface z=f(x,y), + and we are finding the area under that curve. +

    +     + + + Properties of Line Integrals +

    + Many properties of line integrals can be inferred from general integration properties. + For instance, if k is a scalar, + then \int_C k\,f(s)ds = k\int_Cf(s)ds. +

    + +

    + One property in particular of line integrals is worth noting. + If C is a curve composed of subcurves C_1 and C_2, + where they share only one point in common (see , + then the line integral over C is the sum of the line integrals over C_1 and C_2: + + \int_Cf(s)\, ds = \int_{C_1}f(s)\, ds+\int_{C_2}f(s)\, ds + . +

    + +
    + Illustrating properties of line integrals + +
    + + + Two oriented curves in the plane that are joined at a point. + +

    + Two oriented curves in the plane are drawn without the use of coordinate axes. + The first curve C_1 goes from a point A to a point D. + The second curve C_2 goes from the point D to a point B. +

    + +

    + The two curves have the same tangent line at D, + so that the curve obtained by combining them is smooth. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.35] + + \draw [thick,firstcolor] (0,1) arc [start angle=90,end angle=270,radius=1]; + \draw [thick,firstcolor] (0,-1) parabola (1,1); + + \filldraw (0,1) circle [radius=2pt] node [below] {$A$}; + \filldraw (1,1) circle [radius=2pt] node [right] {$B$}; + \filldraw (0,-1) circle [radius=2pt] node [above] {$D$}; + + \draw (-1,0) node [right] {$C_1$}; + \draw (.5,-.5) node [right] {$C_2$}; + \draw [thick,->,firstcolor] (-1,0) -- (-1,-.01); + \draw [thick,->,firstcolor] (.5,-.5) -- (.51,-.48); + \draw (-1,0) node {\phantom{M}}; + + \end{tikzpicture} + + + +
    + +
    + + + Two oriented curves in the plane that are joined at a point. + +

    + Two oriented curves in the plane are drawn without the use of coordinate axes. + The first curve C_1 goes from a point A to a point D. + The second curve C_2 goes from the point D to a point B. +

    + +

    + The curve C_1 is circular, while C_2 is a line segment. + The two curves have the different tangent lines at D, + so that the curve obtained by combining them is not smooth, but is piecewise smooth. +

    +     + + + \begin{tikzpicture}[>=stealth,scale=1.35] + + \draw [thick,firstcolor] (0,1) arc [start angle=90,end angle=270,radius=1]; + \draw [thick,firstcolor] (0,-1) -- (1,1); + + \filldraw (0,1) circle [radius=2pt] node [below] {$A$}; + \filldraw (1,1) circle [radius=2pt] node [right] {$B$}; + \filldraw (0,-1) circle [radius=2pt] node [above left] {$D$}; + + \draw (-1,0) node [right] {$C_1$}; + \draw (.5,0) node [right] {$C_2$}; + \draw [thick,->,firstcolor] (-1,0) -- (-1,-.01); + \draw [thick,->,firstcolor] (.5,0) -- (.51,.02); + \draw (-1,0) node {\phantom{M}}; + + \end{tikzpicture} + + + +
    +     + +
    + +

    + This property allows us to evaluate line integrals over some curves C that are not smooth. + Note how in the curve is not smooth at D, + so by our definition of the line integral we cannot evaluate \int_C f(s)ds. + However, one can evaluate line integrals over C_1 and C_2 and their sum will be the desired quantity. +

    + +

    + A curve C that is composed of two or more smooth curves is said to be + piecewise smooth. + In this chapter, + any statement that is made about smooth curves also holds for piecewise smooth curves. + smooth curvepiecewise + piecewise smooth curve +

    + +

    + We state these properties as a theorem. +

    + + + Properties of Line Integrals Over Scalar Fields + +

    +

      +
    1. +

      + Let C be a smooth curve parametrized by the arc-length parameter s, + let f and g be continuous functions of s, + and let k_1 and k_2 be scalars. + Then line integralproperties over a scalar field + + \ds \int_C\big(k_1f(s)+k_2g(s)\big)\, ds = k_1\int_Cf(s)\, ds +k_2\int_Cg(s)\, ds + . +

      +
      2. + +
    2. +

      + Let C be piecewise smooth, + composed of smooth components C_1 and C_2. + Then + + \int_Cf(s)\, ds = \int_{C_1}f(s)\, ds + \int_{C_2}f(s)\, ds + . +

      +
      3. +
    +

    +     +     +     + + + Mass and Center of Mass +

    + We first learned integration as a method to find area under a curve, + then later used integration to compute a variety of other quantities, + such as arc length, volume, force, etc. + In this section, + we also introduced line integrals as a method to find area under a curve, + and now we explore one more application. +

    + +

    + Let a curve C + (either in the plane or in space) + represent a thin wire with variable density \delta(s). + We can approximate the mass of the wire by dividing the wire (, the curve) into small segments of length + \Delta s_i and assume the density is constant across these small segments. + The mass of each segment is density of the segment its length; + by summing up the approximate mass of each segment we can approximate the total mass: + + \text{ Total Mass of Wire } = \sum \delta(s_i)\Delta s_i + . +

    + +

    + By taking the limit as the length of the segments approaches 0, we have the definition of the line integral as seen in . + When learning of the line integral, + we let f(s) represent a height; + now we let f(s) = \delta(s) represent a density. +

    + +

    + We can extend this understanding of computing mass to also compute the center of mass of a thin wire. + (As a reminder, + the center of mass can be a useful piece of information as objects rotate about that center.) + We give the relevant formulas in the next definition, followed by an example. + Note the similarities between this definition and , + which gives similar properties of solids in space. +

    + + + Mass, Center of Mass of Thin Wire + +

    + Let a thin wire lie along a smooth curve C with continuous density function \delta(s), + where s is the arc length parameter. + masscenter of + mass + +

      +
    1. +

      + The mass of the thin wire is \ds M = \int_C \delta(s)\, ds. +

      +
      2. + +
    2. +

      + The moment about the yz-plane + is \ds M_{yz} = \int_C x\delta(s)\, ds. +

      +
      3. + +
    3. +

      + The moment about the xz-plane + is \ds M_{xz} = \int_C y\delta(s)\, ds. +

      +
      4. + +
    4. +

      + The moment about the xy-plane + is \ds M_{xy} = \int_C z\delta(s)\, ds. +

      +
      5. + +
    5. +

      + The center of mass of the wire is + + (\overline{x},\overline{y},\overline{z}) = \left(\frac{M_{yz}}M, \frac{M_{xz}}M,\frac{M_{xy}}M\right) + . +

      +
      6. +
    +

    +     +     + + + Evaluating a line integral: calculating mass + +

    + A thin wire follows the path + \vrt = \langle 1+\cos(t),1+\sin(t), 1+ \sin(2t)\rangle, + 0\leq t\leq 2\pi. + The density of the wire is determined by its position in space: + \delta(x,y,z) = y+z gm/cm. + The wire is shown in , + where a light color indicates low density and a dark color represents high density. + Find the mass and center of mass of the wire. +

    + +
    + Finding the mass of a thin wire in + + A curve in space representing a wire, and a point representing its center of mass. + +

    + A curve in space is drawn in the first octant, + relative to a set of three-dimensional coordinate axes. +

    + +

    + The curve looks like it could be the intersection of a hyperbolic paraboloid with a cylinder. + From above, the curve appears to be circular, but from the side, + we can see that the curve oscillates up and down. +

    + +

    + The center of mass is shown as a point in space. + This point is not on the curve, but it is surrounded by the curve. +

    +     + + + + import palette; + + // ASY file for figlinescalarfield2_3D.asy in Chapter 10 + // The path is an elliptical helix ⟨ cos t, 2sin t, t/pi ⟩; no specific surface. + // + // This draws the surface of the area we seek. + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14.4,-6.86,4); + //(7.63,8.65,5); + //currentprojection=orthographic(2.5,4.6,1.9); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={2}; + real[] myychoice={2}; + real[] myzchoice={2}; + defaultpen(0.5mm); + pair xbounds=(-.2,2.5); + pair ybounds=(-.2,2.5); + pair zbounds=(-.1,2.1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + real r(real t) {return 3exp(-0.1*t);} + real x(real t) {return 1+cos(t);} + real y(real t) {return 1+sin(t);} + real z(real t) {return 1+sin(2*t);} + + path3 p=graph(x,y,z,0,2*pi,50,operator ..); + + tube T=tube(p,.05); + surface s=T.s; + s.colors(palette(s.map(ypart)+s.map(zpart),Gradient(palered,heavyblue))); + draw(s,render(merge=true)); + //dot((1,1.25,1.2),black); + draw(shift(1,1.25,1.2)*scale3(.05)*unitsphere); + //draw(T.center,thin()); + + + +
    +     + +

    + We compute the density of the wire as + + \delta(x,y,z) = \delta\big(1+\cos(t),1+\sin(t), 1+\sin(2t)\big) = 2+\sin(t)+\sin(2t) + . +

    + +

    + We compute ds as + + ds = \norm{\vrp(t)}\, dt = \sqrt{\sin^2t+\cos^2t+4\cos^2(2t)}\, dt = \sqrt{1+4\cos^2(2t)}\, dt + . +

    + +

    + Thus the mass is + + M = \oint_C \delta(s)\, ds = \int_0^{2\pi} \big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt \approx 21.08\text{ gm } + . +

    + +

    + We compute the moments about the coordinate planes: + + M_{yz} \amp = \oint_C x\delta(s)\, ds + \amp = \int_0^{2\pi}(1+\cos(t))\big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt + \amp \approx 21.08. + M_{xz} \amp = \oint_C y\delta(s)\, ds + \amp = \int_0^{2\pi}(1+\sin(t))\big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt + \amp \approx 26.35 + M_{xy} \amp = \oint_C z\delta(s)\, ds + \amp = \int_0^{2\pi}\big(1+\sin(2 t)\big)\big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt + \amp \approx 25.40 + +

    + +

    + Thus the center of mass of the wire is located at + + (\overline{x},\overline{y},\overline{z}) = \left(\frac{M_{yz}}M, \frac{M_{xz}}M,\frac{M_{xy}}M\right) \approx (1,1.25,1.20) + , + as indicated by the dot in . + Note how in this example, + the curve C is centered + about the point (1,1,1), + though the variable density of the wire pulls the center of mass out along the y and z axes. +

    +     +     + +

    + We end this section with a callback to the Integration Review that preceded this section. + A line integral looks like: \int_C f(s)\, ds. + As stated before the definition of the line integral, + this means sum up, along a curve C, + function values f(s) small arc lengths. + When f(s) represents a height, + we have height length = area. + When f(s) is a density (and we use + \delta(s) by convention), we have density + (mass per unit length) + length = mass. +

    + +

    + In the next section, we investigate a new mathematical object, + the vector field. + The remaining sections of this chapter are devoted to understanding integration in the context of vector fields. +

    +     + + + + Terms and Concepts + + +

    + Explain how a line integral can be used to find the area under a curve. +

    +     + + + +

    + When C is a curve in the plane and f is a function defined over C, + then \int_C f(s)\, ds describes the area under the spatial curve that lies on f, + over C. +

    +     +     + + + +

    + How does the evaluation of a line integral given as + \int_C f(s)\, ds differ from a line integral given as \oint_C f(s)\, ds? +

    +     + + + +

    + The evaluation is the same. + The \oint notation signifies that the curve C is a closed curve, + though the evaluation is the same. +

    +     +     + + + +

    + Why are most line integrals evaluated using instead of directly + as \int_C f(s)\, ds? +

    +     + + + +

    + The variable s denotes the arc-length parameter, + which is generally difficult to use. + allows one to parametrize a curve using another, + ideally easier-to-use, parameter. +

    +     +     + + + +

    + Sketch a closed, + piecewise smooth curve composed of three subcurves. +

    +     + + +     +     + + + Problems + + +

    + A planar curve C is given along with a function f that is defined over C. + Evaluate the line integral \ds \int_Cf(s)\, ds. +

    +     + + + +

    + C is the line segment joining the points (-2,-1) and (1,2); + the function is f(x,y)=x^2+y^2+2. +

    +     + +

    + 12\sqrt{2} +

    +     +     + + + +

    + C is the segment of y=3x+2 on [1,2]; + the function is f(x,y)=5x+2y. +

    +     + +

    + 41\sqrt{10}/2 +

    +     +     + + + +

    + C is the circle with radius 2 centered at the point (4,2); + the function is f(x,y)=3x-y. +

    +     + +

    + 40\pi +

    +     +     + + + +

    + C is the curve given by \vec r(t) = \langle \cos(t)+t\sin(t), \sin(t)-t\cos(t)\rangle on [0,2\pi]; + the function is f(x,y)=5. +

    +     + +

    + 10\pi^2 +

    +     +     + + + +

    + C is the piecewise curve composed of the line segments that connect (0,1) to (1,1), + then connect (1,1) to (1,0); + the function is f(x,y)=x+y^2. +

    +     + +

    + Over the first subcurve of C, + the line integral has a value of 3/2; + over the second subcurve, the line integral has a value of 4/3. + The total value of the line integral is thus 17/6. +

    +     +     + + + +

    + C is the piecewise curve composed of the line segment joining the points (0,0) and (1,1), + along with the quarter-circle parametrized by \langle \cos(t),-\sin(t)+1\rangle on + [0,\pi/2](which starts at the point (1,1) and ends at (0,0); + the function is f(x,y)=x^2+y^2. +

    +     + +

    + Over the first subcurve of C, + the line integral has a value of 2\sqrt{2}/3; + over the second subcurve, the line integral has a value of \pi-2. + The total value of the line integral is thus \pi+2\sqrt{2}/3-2. +

    +     +     +     + + + +

    + A planar curve C is given along with a function f that is defined over C. + Set up the line integral \ds \int_Cf(s)\, ds, + then approximate its value using technology. +

    +     + + + +

    + C is the portion of the parabola y=2x^2+x+1 on [0,1]; + the function is f(x,y)=x^2+2y. +

    +     + +

    + \int_0^1(5t^2+_2t+2)\sqrt{(4t+1)^2+1}\, dt \approx 17.071 +

    +     +     + + + +

    + C is the portion of the curve y=\sin(x) on [0,\pi]; + the function is f(x,y)=x. +

    +     + +

    + \int_0^\pi t\sqrt{1+\cos^2t}\, dt \approx 6.001 +

    +     +     + + + +

    + C is the ellipse given by \vec r(t) = \langle 2\cos(t),\sin(t)\rangle on [0,2\pi]; + the function is f(x,y)=10-x^2-y^2. +

    +     + +

    + \oint_0^{2\pi} \big(10-4\cos^2t-\sin^2t\big)\sqrt{\cos^2t+4\sin^2t}\, dt \approx 74.986 +

    +     +     + + + +

    + C is the portion of y=x^3 on [-1,1]; + the function is f(x,y)=2x+3y+5. +

    +     + +

    + \int_{-1}^{1} \big(3t^3+2t+5\big)\sqrt{9t^4+1}\, dt \approx 15.479 +

    +     +     +     + + + +

    + A parametrized curve C in space is given. + Find the area above the xy-plane that is under C. +

    +     + + + +

    + C: \vec r(t) = \langle 5t,t,t^2\rangle for 1\leq t\leq 2. +

    +     + +

    + 7\sqrt{26}/3 +

    +     +     + + + +

    + C: \vec r(t) = \langle \cos(t),\sin(t), \sin(2t)+1\rangle for 0\leq t\leq 2\pi. +

    +     + +

    + 2\pi +

    +     +     + + + +

    + C: \vec r(t) = \langle 3\cos(t),3\sin(t), t^2\rangle for 0\leq t\leq 2\pi. +

    +     + +

    + 8\pi^3 +

    +     +     + + + +

    + C: \vec r(t) = \langle 3t,4t, + t\rangle for 0\leq t\leq 1. +

    +     + +

    + 5/2 +

    +     +     +     + + + +

    + A parametrized curve C is given that represents a thin wire with density \delta. + Find the mass and center of mass of the thin wire. +

    +     + + + +

    + C: \vec r(t) = \langle \cos(t),\sin(t), t\rangle for 0\leq t\leq 4\pi; + \delta(x,y,z) = z. +

    +     + +

    + M=8\sqrt{2}\pi^2; center of mass is (0,-1/(2\pi), 8\pi/3). +

    +     +     + + + +

    + C: \vec r(t) = \langle t-t^2,t^2-t^3,t^3-t^4\rangle for 0\leq t\leq 1; + \delta(x,y,z) = x+2y+2z. + Use technology to approximate the value of each integral. +

    +     + +

    + M\approx 0.237; + center of mass is approximately (0.173, 0.099,0.065). +

    +     +     +     +     +     +
    +
    + Vector Fields + +

    + We have studied functions of two and three variables, + where the input of such functions is a point + (either a point in the plane or in space) + and the output is a number. +

    + +

    + We could also create functions where the input is a point (again, + either in the plane or in space), + but the output is a vector. + For instance, we could create the following function: + \vec F(x,y) = \langle x+y, x-y\rangle, + where \vec F(2,3) = \langle 5,-1\rangle. + We are to think of \vec F assigning the vector + \langle 5,-1\rangle to the point (2,3); + in some sense, + the vector \langle 5,-1\rangle lies at the point (2,3). +

    + + + +

    + Such functions are extremely useful in any context where magnitude and direction are important. + For instance, + we could create a function \vec F that represents the electromagnetic force exerted at a point by a electromagnetic field, + or the velocity of air as it moves across an airfoil. +

    + +

    + Because these functions are so important, + we need to formally define them. +

    + + + Vector Field + +

    +

      +
    1. +

      + A vector field in the plane + is a function \vec F(x,y) whose domain is a subset of + \mathbb{R}^2 and whose output is a two-dimensional vector: + vector field + + \vec F(x,y) = \langle M(x,y), N(x,y)\rangle + . +

      +
      2. + +
    2. +

      + A vector field in space + is a function \vec F(x,y,z) whose domain is a subset of + \mathbb{R}^3 and whose output is a three-dimensional vector: + + \vec F(x,y,z) = \langle M(x,y,z), N(x,y,z), P(x,y,z)\rangle + . +

      +
      3. +
    +

    +     +     + +

    + This definition may seem odd at first, + as a special type of function is called a field. However, + as the function determines a field of vectors, + we can say the field is defined by the function, + and thus the field is a function. +

    + +
    + Demonstrating methods of graphing vector fields + +
    + + + Several arrows are drawn in the plane to illustrate the concept of a vector field. + +

    + We tend to think of a vector field as a collection of vectors, + with each vector drawn with its tail at the point where it is defined. +

    + +

    + This image illustrates the fact that this conception does not work well in practice. + Eight different vectors are plotted, each with its tail at the point used to compute the vector. + The vectors end up in a couple of configurations resembling tridents, + and the overall effect is not very informative. +

    +     + + + import graph; + size(282,282); + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + + if(incolor) { + colorone = bluepen+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + } else + { + colorone = black+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + }; + + real f(real x) {return exp(x);} + pair F(real x) {return (x,f(x));} + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + real[] myxchoice={-3,-2,-1,1,2,3}; + real[] myychoice={-3,-2,-1,1,2,3}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-2,-2); + pair b=(2,2); + + int[] array={-1,0,1}; + for(int j : array){ + for(int k : array){ + draw((j,k) -- (j+j+k,k+j-k),arrow=Arrow(DefaultHead,size=4),colorone); + } + } + + //pair vectfunction(pair z) {return (z.y+z.x,z.x-z.y);} + //path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + //add(vectorfield(vector,a,b,5,5,bluepen)); + + + +
    + +
    + + + Several arrows are plotted in the plane to illustrate the concept of a vector field. + +

    + This plot shows the same vectors that were plotted in . + The difference is that this time, the vectors have been shifted so that their midpoint, + not their tail, is at the point used to compute each vector. +

    + +

    + This image gives a better idea of the directions represented by the vector field; + however, it is still somewhat confusing due to the length of the vectors. + To help with visualization, the vectors in a vector field are often drawn much shorter than their true length. +

    +     + + + import graph; + size(282,282); + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + + if(incolor) { + colorone = bluepen+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + } else + { + colorone = black+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + }; + + real f(real x) {return exp(x);} + pair F(real x) {return (x,f(x));} + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + real[] myxchoice={-3,-2,-1,1,2,3}; + real[] myychoice={-3,-2,-1,1,2,3}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-2,-2); + pair b=(2,2); + + int[] array={-1,0,1}; + for(int j : array){ + for(int k : array){ + draw((j-(j+k)/2,k-(j-k)/2) -- (j+(j+k)/2,k+(j-k)/2),arrow=Arrow(DefaultHead,size=4),bluepen); + } + } + + //pair vectfunction(pair z) {return (z.y+z.x,z.x-z.y);} + //path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + //add(vectorfield(vector,a,b,5,5,bluepen)); + + + +
    +     + +
    + +

    + Visualizing vector fields helps cement this connection. + When graphing a vector field in the plane, + the general idea is to draw the vector + \vec F(x,y) at the point (x,y). + For instance, + using \vec F(x,y) = \langle x+y,x-y\rangle as before, + at (1,1) we would draw \langle 2,0\rangle. +

    + +

    + In , + one can see that the vector \langle 2,0\rangle is drawn + starting from the point (1,1). + A total of 8 vectors are drawn, + with the x- and y-values of -1,0,1. + In many ways, the resulting graph is a mess; + it is hard to tell what this field looks like. +

    + +

    + In , + the same field is redrawn with each vector \vec F(x,y) drawn + centered on the point (x,y). + This makes for a better looking image, + though the long vectors can cause confusion: + when one vector intersects another, + the image looks cluttered. +

    + +

    + A common way to address this problem is limit the length of each arrow, + and represent long vectors with thick arrows, + as done in . + Usually we do not use a graph of a vector field to determine exactly the magnitude of a particular vector. + Rather, we are more concerned with the relative magnitudes of vectors: + which are bigger than others? + Thus limiting the length of the vectors is not problematic. +

    + +
    + Demonstrating methods of graphing vector fields + +
    + + + Eight vectors are plotted in the plane to represent a vector field, using relative length to decrease clutter. + +

    + This is another representation of the vector field in . + In this version, the vectors plotted are shorter than their true length. + Instead, relative length is used to indicate varying magnitude. +

    + +

    + This change reduces clutter and makes it easier to understand the vector field. + Eight vectors are plotted at the points (1,1), (1,0), (1,-1), + (0,-1), (-1,-1), (-1,0), (-1,1), and (0,1). +

    + +

    + As we move counter-clockwise about the square, + the vectors rotate clockwise. + The vector at (1,1) points to the right, + and the vector at each adjacent point is rotated by 45 degrees as we move around the square. +

    +     + + + import graph; + size(282,282); + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + + if(incolor) { + colorone = bluepen+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + } else + { + colorone = black+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + }; + + real f(real x) {return exp(x);} + pair F(real x) {return (x,f(x));} + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + real[] myxchoice={-3,-2,-1,1,2,3}; + real[] myychoice={-3,-2,-1,1,2,3}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-1,-1); + pair b=(1,1); + + pair vectfunction(pair z) {return (z.y+z.x,z.x-z.y);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,3,3,bluepen)); + + + +
    + +
    + + + Another plot of the same vector field used in every image so far, but this time with many more vectors. + +

    + We get one more plot of the same vector field use in + and the previous images in this section. + This time, instead of eight vectors being plotted, there are about eighty. +

    + +

    + Using more vectors, we can see that magnitude increases with distance from the origin, + and the directions of the vectors appear to follow hyperbolic curves in the plane. +

    +     + + + import graph; + size(282,282); + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen+ linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + + real f(real x) {return exp(x);} + pair F(real x) {return (x,f(x));} + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + real[] myxchoice={-3,-2,-1,1,2,3}; + real[] myychoice={-3,-2,-1,1,2,3}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-3,-3); + pair b=(3,3); + + pair vectfunction(pair z) {return (z.y+z.x,z.x-z.y);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,8,8,mainvect)); + + + +
    +     + +
    + +

    + Drawing arrows with variable thickness is best done with technology; + search the documentation of your favorite graphing program for terms like vector fields + or slope fields to learn how. + Technology obviously allows us to plot many vectors in a vector field nicely; + in , + we see the same vector field drawn with many vectors, + and finally get a clear picture of how this vector field behaves. + (If this vector field represented the velocity of air moving across a flat surface, + we could see that the air tends to move either to the upper-right or lower-left, + and moves very slowly near the origin.) +

    + +

    + We can similarly plot vector fields in space, + as shown in , + though it is not often done. + The plots get very busy very quickly, + as there are lots of arrows drawn in a small amount of space. + In + the field \vec F = \langle -y,x,z\rangle is graphed. + If one could view the graph from above, + one could see the arrows point in a circle about the z-axis. + One should also note how the arrows far from the origin are larger than those close to the origin. +

    + +

    + It is good practice to try to visualize certain vector fields in one's head. + For instance, + consider a point mass at the origin and the vector field that represents the gravitational force exerted by the mass at any point in the room. + The field would consist of arrows pointing toward the origin, + increasing in size as they near the origin + (as the gravitational pull is strongest near the point mass). +

    + +
    + Graphing a vector field in space + + A very chaotic plot of a three-dimensional vector field. + +

    + This plot shows a vector field in three dimensions, relative to the usual coordinate axes. + It is a very cluttered plot. + With the default perspective, it appears to be a large jumble of arrows, + pointing every which way in space. +

    + +

    + Rotating the image reveals a bit more structure: + from above, the arrows appear to follow a circular trajectory. + In fact, the vectors in this vector field are all tangent to helical curves, + with those above the xy plane pointing upward, and those below pointing downward. +

    +     + + + + + size(200,200,IgnoreAspect); + currentprojection=orthographic(13.5,-13,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={1}; + defaultpen(0.25mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + path3 gradient1(triple z){ + return O--(-z.y,z.x,z.z); + } + + triple A=(-1,-1,-1); + triple B=(1,1,1); + + picture VectorPlot3D(path3 vector(triple t), triple a, triple b, + int nx=nmesh, int ny=nx, int nz=nx,bool truesize=false, + real maxlength=truesize ? 0 : min(abs(b.x-a.x)/nx,abs(b.y-a.y)/ny,abs(b.z-a.z)/nz), + // bool cond(pair z)=null, + pen p=currentpen, + arrowbar3 arrow=Arrow3(6), margin3 margin=PenMargin3, + string name="", render render=defaultrender) + { + picture pic; + real dx=1/nx; + real dy=1/ny; + real dz=1/nz; + real scale; + if(maxlength > 0) { + real size(triple t) { + path3 g=vector(t); + return abs(point(g,size(g)-1)-point(g,0)); + } + real max=size((0,0,0)); + + for(int i=0; i <= nx; ++i) { + real x=interp(a.x,b.x,i*dx); + for(int j=0; j <= ny; ++j) + { + real y=interp(a.y,b.y,j*dy); + for(int k=0; k <= nz; ++k) + max=max(max,size((x,y,interp(a.z,b.z,k*dz)))); + }} + scale=max > 0 ? maxlength/max : 1; + } else scale=1; + bool group=name != "" || render.defaultnames; + if(group) + begingroup3(pic,name == "" ? "vectorfield" : name,render); + for(int i=0; i <= nx; ++i) { + real x=interp(a.x,b.x,i*dx); + for(int j=0; j <= ny; ++j) { + real y=interp(a.y,b.y,j*dy); + for(int k=0; k <= nz; ++k) + { triple z=(x,y,interp(a.z,b.z,k*dz)); + { + path3 g=scale3(scale)*vector(z); + string name="vector"; + if(truesize) { + picture opic; + draw(opic,g,p,arrow,margin,name,render); + add(pic,opic,z); + } else + draw(pic,shift(z)*g,p,arrow,margin,name,render); + } + } + }} + if(group) + endgroup3(pic); + return pic; + + } + add(VectorPlot3D(gradient1,A,B,3,3,3,bluepen)); + + + +
    +     + + + Vector Field Notation and Del Operator +

    + + defines a vector field \vec F using the notation + + \vec F(x,y) = \langle M(x,y), N(x,y)\rangle \text{ and } \vec F(x,y,z) = \langle M(x,y,z), N(x,y,z),P(x,y,z)\rangle + . +

    + +

    + That is, the components of \vec F are each functions of x and y + (and also z in space). + As done in other contexts, we will drop the of x, + y and z portions of the notation and refer to vector fields in the plane and in space as + + \vec F = \langle M, N\rangle \text{ and } \vec F = \langle M,N,P\rangle + , + respectively, as this shorthand is quite convenient. +

    + +

    + Another item of notation will become useful: + the del operator. + del operator + Recall in + how we used the symbol \nabla + (pronounced del) + to represent the gradient of a function of two variables. + That is, if z = f(x,y), + then del f = \nabla f = \langle f_x, f_y\rangle. +

    + + + +

    + We now define \nabla to be the del operator. + It is a vector whose components are partial derivative operations. +

    + +

    + In the plane, + \ds\nabla = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right\rangle; + in space, \ds\nabla = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle. +

    + +

    + With this definition of \nabla, + we can better understand the gradient \nabla f. + As f returns a scalar, + the properties of scalar and vector multiplication gives + + \nabla f = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right\rangle f = \left\langle \frac{\partial}{\partial x}\,f, \frac{\partial}{\partial y}\,f\right\rangle = \langle f_x, f_y\rangle + . +

    + + + +

    + Now apply the del operator \nabla to vector fields. + Let \vec F = \langle x+\sin(y),y^2+z,x^2\rangle. + We can use vector operations and find the dot product of \nabla and \vec F: + + \nabla \cdot \vec F \amp = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle\cdot \langle x+\sin(y),y^2+z,x^2\rangle + \amp = \frac{\partial}{\partial x}(x+\sin(y))+ \frac{\partial}{\partial y}(y^2+z) + \frac{\partial}{\partial z}(x^2) + \amp =1+2y + . +

    + +

    + We can also compute their cross products: + + \nabla\times \vec F \amp = \left\langle \frac{\partial}{\partial y}\big(x^2\big)-\frac{\partial}{\partial z}\big(y^2+z\big),\frac{\partial}{\partial z}\big(x+\sin(y)\big)-\frac{\partial}{\partial x}\big(x^2\big),\right. + \amp \phantom{=\frac{\partial}{\partial y}\big(x^2\big)-\frac{\partial}{\partial z}\big(y^2+z\big)}\left.\frac{\partial}{\partial x}\big(y^2+z\big)-\frac{\partial}{\partial y}\big(x+\sin(y)\big)\right\rangle + \amp =\langle -1,-2x,-\cos(y)\rangle + . +

    + +

    + We do not yet know why we would want to compute the above. + However, as we next learn about properties of vector fields, + we will see how these dot and cross products with the del operator are quite useful. +

    +     + + + Divergence and Curl +

    + Two properties of vector fields will prove themselves to be very important: + divergence and curl. + Each is a special derivative of a vector field; + that is, each measures an instantaneous rate of change of a vector field. +

    + + + +

    + If the vector field represents the velocity of a fluid or gas, + then the divergence + divergence + vector fielddivergence of + of the field is a measure of the + compressibility of the fluid. + If the divergence is negative at a point, + it means that the fluid is compressing: + more fluid is going into the point than is going out. + If the divergence is positive, + it means the fluid is expanding: + more fluid is going out at that point than going in. + A divergence of zero means the same amount of fluid is going in as is going out. + If the divergence is zero at all points, + we say the field is incompressible. + incompressible vector field +

    + +

    + It turns out that the proper measure of divergence is simply \nabla \cdot \vec F, + as stated in the following definition. +

    + + + Divergence of a Vector Field + +

    + The divergence of a vector field \vec F is + divergence + vector fielddivergence of + + \divv \vec F = \nabla \cdot \vec F + . +

      +
    • +

      + In the plane, with \vec F = \langle M,N\rangle, + \divv \vec F = M_x+N_y. +

      +
      • + +
    • +

      + In space, with \vec F = \langle M,N,P\rangle, + \divv \vec F = M_x+N_y+P_z. +

      +
      • +
    +

    +     +     + + + +

    + Curlvector fieldcurl of + curl + is a measure of the spinning action of the field. + Let \vec F represent the flow of water over a flat surface. + If a small round cork were held in place at a point in the water, + would the water cause the cork to spin? + No spin corresponds to zero curl; + counterclockwise spin corresponds to positive curl and clockwise spin corresponds to negative curl. +

    + +

    + In space, things are a bit more complicated. + Again let \vec F represent the flow of water, + and imagine suspending a tennis ball in one location in this flow. + The water may cause the ball to spin along an axis. + If so, the curl of the vector field is a vector + (not a scalar, as before), + parallel to the axis of rotation, + following a right hand rule: + when the thumb of one's right hand points in the direction of the curl, + the ball will spin in the direction of the curling fingers of the hand. +

    + +

    + In space, it turns out the proper measure of curl is \nabla \times \vec F, + as stated in the following definition. + To find the curl of a planar vector field \vec F = \langle M,N\rangle, + embed it into space as \vec F = \langle M, N, 0\rangle and apply the cross product definition. + Since M and N are functions of just x and y + (and not z), + all partial derivatives with respect to z become 0 and the result is simply \langle 0,0,N_x-M_y\rangle. + The third component is the measure of curl of a planar vector field. +

    + + + + + Curl of a Vector Field + +

    +

      +
    • +

      + Let \vec F = \langle M,N\rangle be a vector field in the plane. + The curl of \vec F is \curl \vec F = N_x - M_y. + curl +

      +
      • + +
    • +

      + Let \vec F = \langle M,N,P\rangle be a vector field in space. + The curl of \vec F is \curl \vec F = \nabla \times \vec F = \langle P_y-N_z,M_z-P_x,N_x - M_y\rangle. +

      +
      • +
    +

    +     +     + +

    + We adopt the convention of referring to curl as \nabla \times \vec F, + regardless of whether \vec F is a vector field in two or three dimensions. + (Some people prefer to write (\nabla\times \vec F)\cdot \vec k in two dimensions.) +

    + + + +

    + We now practice computing these quantities. +

    + + + Computing divergence and curl of planar vector fields + +

    + For each of the planar vector fields given below, + view its graph and try to visually determine if its divergence and curl are 0. + Then compute the divergence and curl. + +

      +
    1. +

      + \vec F = \langle y,0\rangle (see ) +

      +
      2. + +
    2. +

      + \vec F = \langle -y,x\rangle (see ) +

      +
      3. + +
    3. +

      + \vec F = \langle x,y\rangle (see ) +

      +
      4. + +
    4. +

      + \vec F = \langle \cos(y), \sin(x)\rangle (see ) +

      +
      5. +
    +

    +

    + + +

    +

      +
    1. +

      + The arrow sizes are constant along any horizontal line, + so if one were to draw a small box anywhere on the graph, + it would seem that the same amount of fluid would enter the box as exit. + Therefore it seems the divergence is zero; it is, as + + \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(0) = 0 + . +

      + +
      + The vector fields in parts 1 and 2 in + +
      + + + A vector field of horizontal arrows: to the right above the x axis, and to the left below. + +

      + A two-dimensional vector field is plotted against x and y coordinate axes. + Several things are notable in this image: +

        +
      • +

        + All of the vectors in the vector field are horizontal +

        +
        • +
      • +

        + Vectors above the x axis point to the right, + and vectors below the x axis point to the left. +

        +
        • +
      • +

        + Near the x axis, the magnitude of the vectors is very small, + and the vector field vanishes completely along the x axis. + The vectors get larger as they get further from the x axis. +

        +
        • +
      +

      +       + + + import graph; + size(282,282); + + // vector field F = ⟨ y, 0 ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen+ linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + + real f(real x) {return exp(x);} + pair F(real x) {return (x,f(x));} + + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-1,-1); + pair b=(1,1); + + pair vectfunction(pair z) {return (z.y,0);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,8,12,mainvect)); + + + +
      + +
      + + + A rotational vector field. The vectors appear to describe counter-clockwise circular motion. + +

      + A two-dimensional vector field is plotted against x and y coordinate axes. + The vectors in this vector field appear to describe circular motion. + Each vector could be tangent to a circle centered at the origin, although no circles are depicted in the image. + The directions of the vectors correspond to counter-clockwise motion. +

      + +

      + The magnitudes of the vectors depend on their distance from the origin. + Vectors near the centre of the image are small, while those near the edges are larger. +

      +       + + + import graph; + size(282,282); + + // vector field F = ⟨ -y, x ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen+ linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + + real f(real x) {return exp(x);} + pair F(real x) {return (x,f(x));} + + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-1,-1); + pair b=(1,1); + + pair vectfunction(pair z) {return (-z.y,z.x);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,mainvect)); + + + +
      +       +
      + +

      + At any point on the x-axis, + arrows above it move to the right and arrows below it move to the left, + indicating that a cork placed on the axis would spin clockwise. + A cork placed anywhere above the x-axis would have water above it moving to the right faster than the water below it, + also creating a clockwise spin. + A clockwise spin also appears to be created at points below the x-axis. + Thus it seems the curl should be negative + (and not zero). + Indeed, it is: + + \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(0) - \frac{\partial}{\partial y}(y) = -1 + . +

      +
      2. + +
    2. +

      + It appears that all vectors that lie on a circle of radius r, + centered at the origin, have the same length + (and indeed this is true). + That implies that the divergence should be zero: + draw any box on the graph, + and any fluid coming in will lie along a circle that takes the same amount of fluid out. + Indeed, the divergence is zero, as + + \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(-y) + \frac{\partial}{\partial y}(x) = 0 + . + Clearly this field moves objects in a circle, + but would it induce a cork to spin? + It appears that yes, it would: + place a cork anywhere in the flow, + and the point of the cork closest to the origin would feel less flow than the point on the cork farthest from the origin, + which would induce a counterclockwise flow. + Indeed, the curl is positive: + + \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(x) - \frac{\partial}{\partial y}(-y) = 1-(-1) = 2 + . + Since the curl is constant, + we conclude the induced spin is the same no matter where one is in this field. +

      +
      3. + +
    3. +

      + At the origin, + there are many arrows pointing out but no arrows pointing in. + We conclude that at the origin, + the divergence must be positive + (and not zero). + If one were to draw a box anywhere in the field, + the edges farther from the origin would have larger arrows passing through them than the edges close to the origin, + indicating that more is going from a point than going in. + This indicates a positive + (and not zero) + divergence. + This is correct: + + \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) = 1+1=2 + . + One may find this curl to be harder to determine visually than previous examples. + One might note that any arrow that induces a clockwise spin on a cork will have an equally sized arrow inducing a counterclockwise spin on the other side, + indicating no spin and no curl. + This is correct, as + + \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(y) - \frac{\partial}{\partial y}(x) = 0 + . +

      + +
      + The vector fields in parts 3 and 4 in + +
      + + + A radial vector field. Each vector points away from the origin, and vectors further from the origin are larger. + +

      + A two-dimensional vector field is plotted against x and y coordinate axes. + The vectors in this vector field could be tangent to lines through the origin: + each one points directly away from the origin, + in the same direction as the line from the origin to the point where the vector is plotted. +

      + +

      + The magnitude of each vector depends on its distance from the origin. + Those close to the origin are small, while those further out are larger. +

      +       + + + import graph; + size(282,282); + + // vector field F = ⟨ x,y ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen+ linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + + real f(real x) {return exp(x);} + pair F(real x) {return (x,f(x));} + + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-1,-1); + pair b=(1,1); + + pair vectfunction(pair z) {return (z.x,z.y);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,mainvect)); + + + +
      + +
      + + + A vector field that appears to describe swirling motion, with several vortices. + +

      + A two-dimensional vector field is plotted against x and y coordinate axes. + This vector field is quite chaotic, as one might expect given the oscillatory nature of the sine and cosine functions. +

      + +

      + There appear to be several vortices, where the vectors circulate around certain points. + Between each vortex is an area where the vectors could be tangent to a hyperbola, + like contour lines from a hyperbolic paraboloid. +

      +       + + + import graph; + size(282,282); + + // vector field F = ⟨ cos y, sin x ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen+ linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + real f(real x) {return exp(x);} + pair F(real x) {return (x,f(x));} + + pair xbounds=(-7,7); + pair ybounds=(-7,7); + real[] myxchoice={-6,-3,3,6}; + real[] myychoice={-6,-3,3,6}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-2*pi,-2*pi); + pair b=(2*pi,2*pi); + + pair vectfunction(pair z) {return (cos(z.y),sin(z.x));} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,mainvect)); + + + +
      +       +
      +
      4. + +
    4. +

      + One might find this divergence hard to determine visually as large arrows appear in close proximity to small arrows, + each pointing in different directions. + Instead of trying to rationalize a guess, + we compute the divergence: + + \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(\cos(y)) + \frac{\partial}{\partial y}(\sin(x)) = 0 + . + Perhaps surprisingly, the divergence is 0. + With all the loops of different directions in the field, + one is apt to reason the curl is variable. + Indeed, it is: + + \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(\sin(x)) - \frac{\partial}{\partial y}(\cos(y)) = \cos(x) + \sin(y) + . + Depending on the values of x and y, + the curl may be positive, negative, or zero. +

      +
      5. +
    +

    + +     + + + Computing divergence and curl of vector fields in space + +

    + Compute the divergence and curl of each of the following vector fields. + +

      +
    1. +

      + \vec F = \langle x^2+y+z, -x-z, x+y\rangle +

      +
      2. + +
    2. +

      + \vec F = \langle e^{xy}, \sin(x+z),x^2+y\rangle +

      +
      3. +
    +

    +     + +

    + We compute the divergence and curl of each field following the definitions. + +

      +
    1. +

      + + \divv \vec F \amp = \nabla \cdot \vec F = M_x+N_y+P_z = 2x+0+0= 2x + \curl\vec F \amp = \nabla \times \vec F = \langle P_y-N_z,M_z-P_x,N_x - M_y\rangle + \amp = \langle 1 - (-1), 1-1,-1-(1)\rangle = \langle 2,0,-2\rangle + . + For this particular field, + no matter the location in space, + a spin is induced with axis parallel to \langle 2,0,-2\rangle. +

      +
      2. + +
    2. +

      + + \divv \vec F \amp = \nabla \cdot \vec F = M_x+N_y+P_z = ye^{xy}+0+0= ye^{xy} + \curl\vec F \amp = \nabla \times \vec F = \langle P_y-N_z,M_z-P_x,N_x - M_y\rangle + \amp = \langle 1-\cos(x+z), -2x, \cos(x+z) - xe^{xy}\rangle + +

      +
      3. +
    +

    +     +     + + + + + Creating a field representing gravitational force + +

    + The force of gravity between two objects is inversely proportional to the square of the distance between the objects. + Locate a point mass at the origin. + Create a vector field \vec F that represents the gravitational pull of the point mass at any point (x,y,z). + Find the divergence and curl of this field. +

    +     + +

    + The point mass pulls toward the origin, so at (x,y,z), + the force will pull in the direction of \langle -x, -y, -z\rangle. + To get the proper magnitude, + it will be useful to find the unit vector in this direction. + Dividing by its magnitude, we have + + \vec u = \left\langle \frac{-x}{\sqrt{x^2+y^2+z^2}}, \frac{-y}{\sqrt{x^2+y^2+z^2}},\frac{-z}{\sqrt{x^2+y^2+z^2}}\right\rangle + . +

    + +

    + The magnitude of the force is inversely proportional to the square of the distance between the two points. + Letting k be the constant of proportionality, + we have the magnitude as \ds\frac{k}{x^2+y^2+z^2}. + Multiplying this magnitude by the unit vector above, + we have the desired vector field: + + \vec F = \left\langle \frac{-kx}{(x^2+y^2+z^2)^{3/2}}, \frac{-ky}{(x^2+y^2+z^2)^{3/2}},\frac{-kz}{(x^2+y^2+z^2)^{3/2}}\right\rangle + . +

    + +

    + We leave it to the reader to confirm that + \divv \vec F = 0 and \curl \vec F = \vec 0. +

    + +
    + A vector field representing a planar gravitational force + + A radial vector field. Each vector points toward the origin, and vectors near the origin are larger. + +

    + A two-dimensional vector field is plotted relative to the x and y axes, + with the origin at the center of the image. + This is a radial vector field, like the one in , + but there are some important differences. +

    + +

    +

      +
    • +

      + There are fewer vectors plotted than the previous example of a radial field +

      +
      • +
    • +

      + The vectors point inward, toward the origin +

      +
      • +
    • +

      + The magnitudes of the vectors close to the origin are larger than those further out. +

      +
      • +
    +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ cos y, sin x ⟩ + + defaultpen(fontsize(8)); + + real f(real x) {return exp(x);} + pair F(real x) {return (x,f(x));} + + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(.3,.3); + pair b=(1,1); + + pair vectfunction(pair z) {if ((z.x==0) && (z.y==0)) {return (0,0);} + return (-z.x/(z.x^2+z.y^2)^(3/2),-z.y/(z.x^2+z.y^2)^(3/2));} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,3,3,bluepen)); + + pair aa=(-1,.3); + pair bb=(-.3,1); + + pair vectfunction(pair z) {if ((z.x==0) && (z.y==0)) {return (0,0);} + return (-z.x/(z.x^2+z.y^2)^(3/2),-z.y/(z.x^2+z.y^2)^(3/2));} + path vectorb(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vectorb,aa,bb,3,3,bluepen)); + + pair aa=(-1,-1); + pair bb=(-.3,-.3); + + pair vectfunction(pair z) {if ((z.x==0) && (z.y==0)) {return (0,0);} + return (-z.x/(z.x^2+z.y^2)^(3/2),-z.y/(z.x^2+z.y^2)^(3/2));} + path vectorb(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vectorb,aa,bb,3,3,bluepen)); + + pair aa=(.3,-1); + pair bb=(1,-.3); + + pair vectfunction(pair z) {if ((z.x==0) && (z.y==0)) {return (0,0);} + return (-z.x/(z.x^2+z.y^2)^(3/2),-z.y/(z.x^2+z.y^2)^(3/2));} + path vectorb(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vectorb,aa,bb,3,3,bluepen)); + + + +
    + +

    + The analogous planar vector field is given in . + Note how all arrows point to the origin, + and the magnitude gets very small when + far from the origin. +

    +     +     + +

    + A function f(x,y) naturally induces a vector field, + \vec F = \nabla f = \langle f_x,f_y\rangle. + Given what we learned of the gradient in , + we know that the vectors of \vec F point in the direction of greatest increase of f. + Because of this, f is said to be the + potential function + potential function + vector fieldpotential function of + of \vec F. + Vector fields that are the gradient of potential functions will play an important role in the next section. +

    + + + A vector field that is the gradient of a potential function + +

    + Let f(x,y) = 3-x^2-2y^2 and let \vec F = \nabla f. + Graph \vec F, and find the divergence and curl of \vec F. +

    +     + +

    + Given f, + we find \vec F = \nabla f = \langle -2x,-4y\rangle. + A graph of \vec F is given in . + In , + the vector field is given along with a graph of the surface itself; + one can see how each vector is pointing in the direction of + steepest uphill, + which, in this case, is not simply just + toward the origin. +

    + +
    + A graph of a function f(x,y) and the vector field \vec F = \nabla f in + +
    + + + The gradient vector field of a potential fuction. Vectors point toward the origin, but appear to be tangent to parabolic paths. + +

    + A vector field is plotted in two dimensions relative to the x and y coordinate axes. + The origin is at the center of the image. +

    + +

    + The vectors in this vector field point toward the origin, but it is not a radial vector field. +

    + +

    + The vectors appear to lie tangent to curved paths. + In particular, vectors above the x axis lie tangent to paths that could be a family of parabolas of the form y=kx^2, + where k\gt 0. + Vectors near the x axis are nearly horizontal, and could be tangent to a very wide parabola (with k very small). + Vectors near the y axis are nearly vertical, and could be tangent to a very steep parabola (with k very large). +

    + +

    + Vectors below the x axis describe similar trajectories, + except that they follow parths of downward-opening parabolas (ones where k\lt 0). +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ cos y, sin x ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen+ linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + real f(real x) {return exp(x);} + pair F(real x) {return (x,f(x));} + + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-.9,-.9); + pair b=(.9,.9); + + pair vectfunction(pair z) {return (-2*z.x,-4*z.y);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,mainvect)); + + + +
    + +
    + + + A three-dimensional image showing the graph of a function of two variables, with its gradient vector field plotted in the x,y plane below. + +

    + The image is three-dimensional, with x, y, and z coordinate axes, and the origin at the center of the image. + The vector field from is plotted in the xy plane. + Above the plane is the graph of the function f(x,y)=3-x^2-2y^2; + the graph is a downward-opening elliptic paraboloid. +

    +     + + + + import palette; + + // ASY file for figlinescalarfield2_3D.asy in Chapter 10 + // The path is an elliptical helix ⟨ cos t, 2sin t, t/pi ⟩; no specific surface. + // + // This draws the surface of the area we seek. + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(10.5,-9,18.3); + //(7.63,8.65,5); + //currentprojection=orthographic(10.5,-9,18.3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={3}; + defaultpen(0.5mm); + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + pair zbounds=(-1,4); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + + //currentlight=(1,-1,0.5); + + real f(pair z) {return -z.x^2-2*z.y^2+3;} + + path3 gradient(pair z) { + static real dx=sqrtEpsilon, dy=dx; + return (0,0,0)--((f(z+dx)-f(z-dx))/2dx,(f(z+I*dy)-f(z-I*dy))/2dy,0); + } + + pair a=(-.9,-.9); + pair b=(.9,.9); + + triple F(pair z) {return (z.x,z.y,0);} + + add(vectorfield(gradient,F,a,b,12,12,bluepen)); + + pen p=apexmeshpen; + draw(surface(f,a,b,Spline),surfacepen,meshpen=p); + //draw(surface(f,a,b,Spline),gray+opacity(0.5)); + + //pen p=apexmeshpen; + //draw(s,surfacepen,meshpen=p); + + + +
    +     + +
    + +

    + We leave it to the reader to confirm that + \divv \vec F = -6 and \curl \vec F = 0. +

    +     +     + + + +

    + There are some important concepts visited in this section that will be revisited in subsequent sections and again at the very end of this chapter. + One is: given a vector field \vec F, + both \divv\vec F and + \curl\vec F are measures of rates of change of \vec F. + The divergence measures how much the field spreads (diverges) at a point, + and the curl measures how much the field twists (curls) at a point. + Another important concept is this: + given z=f(x,y), the gradient + \nabla f is also a measure of a rate of change of f. + We will see how the integrals of these rates of change produce meaningful results. +

    + +

    + This section introduces the concept of a vector field. + The next section applies calculus to vector fields. + A common application is this: + let \vec F be a vector field representing a force + (hence it is called a force field, + though this name has a decidedly comic-book feel) + and let a particle move along a curve C under the influence of this force. + What work is performed by the field on this particle? + The solution lies in correctly applying the concepts of line integrals in the context of vector fields. +

    +     + + + + Terms and Concepts + + +

    + Give two quantities that can be represented by a vector field in the plane or in space. +

    +     + + + +

    + Answers will vary. + Appropriate answers include velocities of moving particles (air, + water, ); + gravitational or electromagnetic forces. +

    +     +     + + + +

    + In your own words, + describe what it means for a vector field to have a negative divergence at a point. +

    +     + + + +

    + Specific answers will vary, + though should relate to the idea that + more of the vector field is moving into that point than out of that point. +

    +     +     + + + +

    + In your own words, + describe what it means for a vector field to have a negative curl at a point. +

    +     + + + +

    + Specific answers will vary, + though should relate to the idea that the vector field is spinning clockwise at that point. +

    +     +     + + + +

    + The divergence of a vector field \vec F at a particular point is 0. + Does this mean that \vec F is incompressible? + Why/why not? +

    +     + + + +

    + No; to be incompressible, + the divergence needs to be 0 everywhere, not just at one point. +

    +     +     +     + + + Problems + + +

    + Sketch the given vector field over the rectangle with opposite corners (-2,-2) and (2,2), + sketching one vector for every point with integer coordinates (, at (0,0), + (1,2), etc.). +

    +     + + + +

    + \vec F = \langle x,0\rangle +

    +     + +

    + Correct answers should look similar to +

    + + + A vector field of horizontal vectors, pointing away from the y axis. + +

    + A two-dimmensional vector field is plotted relative to x and y coordinate axes, + with the origin at the center. All of the vectors are horizontal. +

    + +

    + The vectors all point away from the y axis: + those with x\gt 0 point to the right, while those with x\lt 0 point to the left. + Vectors close to the y axis are small, + and those near the left and right edges of the image are largest. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ x,0 ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + pen fgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen+ linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + fgvect = bluepen+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + fgvect = rgb(.6,.6,.6)+linejoin(0); + }; + pair xbounds=(-2.8,2.8); + pair ybounds=(-2.8,2.8); + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-2,-2); + pair b=(2,2); + + pair vectfunction(pair z) {return (z.x,0);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,5,5,fgvect)); + + +     +     + + + +

    + \vec F = \langle 0,x\rangle +

    +     + +

    + Correct answers should look similar to +

    + + + A vertical vector field. Vectors right of the y axis point up, and those to the left point down. + +

    + A two-dimmensioanl vector field is plotted relative to x and y coordinate axes, + with the origin at the center. +

    + +

    + All of the vectors are vertical. + Those with x\gt 0 point up, while those with x\lt 0 point down. + Vectors close to the y axis are small, + and those near the left and right edges of the image are largest. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ 0,x ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + pen fgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen+ linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + fgvect = bluepen+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + fgvect = rgb(.6,.6,.6)+linejoin(0); + }; + pair xbounds=(-2.8,2.8); + pair ybounds=(-2.8,2.8); + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-2,-2); + pair b=(2,2); + + pair vectfunction(pair z) {return (0,z.x);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,5,5,fgvect)); + + +     +     + + + +

    + \vec F = \langle 1,-1\rangle +

    +     + +

    + Correct answers should look similar to +

    + + + A constant vector field. All vectors point down and to the right. + +

    + A two-dimensional vector field is plotted relative to x and y coordinate axes, + with the origin at the center of the image. +

    + +

    + The vector field in this image is constant: + many vectors are plotted, but they all have the same magnitude and direction. + In particular, each vector points down and to the right. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ 1,-1 ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + pen fgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen+ linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + fgvect = bluepen+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + fgvect = rgb(.6,.6,.6)+linejoin(0); + }; + pair xbounds=(-2.8,2.8); + pair ybounds=(-2.8,2.8); + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-2,-2); + pair b=(2,2); + + pair vectfunction(pair z) {return (1,-1);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + add(vectorfield(vector,a,b,5,5,fgvect)); + + +     +     + + + +

    + \vec F = \langle y^2,1\rangle +

    +     + +

    + Correct answers should look similar to +

    + + + A two dimensional vector field. The vectors point up and to the right. Vectors near the x axis are shorter and steeper. + +

    + A two-dimensional vector field is plotted relative to x and y coordinate axes, + with the origin at the center. +

    + +

    + Along the lines y=2 and y=-2, + vectors point up and to the right, with a relatively small slope. + These vectors have the largest magnitude. +

    + +

    + Along the lines y=1 and y=-1 + vectors point up and to the right, but with a steeper slope. + The magnitudes of these vectors are less than the ones at the top and bottom of the image. +

    + +

    + Along the x axis, the vectors are vertical, and have the smallest magnitude. +

    + +

    + It looks like the vectors could lie tangent to a family of cubic curves of the form y = a\sqrt[3]{x}+b. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ y^2,1 ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + pen fgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen+ linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + fgvect = bluepen+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + fgvect = rgb(.6,.6,.6)+linejoin(0); + }; + pair xbounds=(-2.8,2.8); + pair ybounds=(-2.8,2.8); + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-2,-2); + pair b=(2,2); + + pair vectfunction(pair z) {return (z.y^2,1);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,5,5,fgvect)); + + +     +     +     + + + +

    + Find the divergence and curl of the given vector field. +

    +     + + + +

    + \vec F = \langle x,y^2\rangle +

    +     + +

    + \divv \vec F = 1+2y +

    + +

    + \curl \vec F = 0 +

    +     +     + + + +

    + \vec F = \langle -y^2,x\rangle +

    +     + +

    + \divv \vec F = 0 +

    + +

    + \curl \vec F = 1+2y +

    +     +     + + + +

    + \vec F = \langle \cos(xy), \sin(xy)\rangle +

    +     + +

    + \divv \vec F = x\cos(xy)-y\sin(xy) +

    + +

    + \curl \vec F = y\cos(xy)+x\sin(xy) +

    +     +     + + + +

    + \ds\vec F = \la \frac{-2x}{(x^2+y^2)^2},\frac{-2y}{(x^2+y^2)^2}\ra +

    +     + +

    + \divv \vec F = \frac{4}{(x^2+y^2)^2} +

    + +

    + \curl \vec F = 0 +

    +     +     + + + +

    + \ds\vec F = \la x+y,y+z,x+z\ra +

    +     + +

    + \divv \vec F = 3 +

    + +

    + \curl \vec F = \la -1,-1,-1\ra +

    +     +     + + + +

    + \ds\vec F = \la x^2+z^2,x^2+y^2,y^2+z^2\ra +

    +     + +

    + \divv \vec F = 2x+2y+2z +

    + +

    + \curl \vec F = \la 2y,2z,2x\ra +

    +     +     + + + +

    + \vec F = \nabla f, where f(x,y) = \frac12x^2+\frac13y^3. +

    +     + +

    + \divv \vec F = 1+2y +

    + +

    + \curl\vec F = 0 +

    +     +     + + + +

    + \vec F = \nabla f, where f(x,y) = x^2y. +

    +     + +

    + \divv \vec F = 2y +

    + +

    + \curl\vec F = 0 +

    +     +     + + + +

    + \vec F = \nabla f, where f(x,y,z) = x^2y+\sin(z). +

    +     + +

    + \divv \vec F = 2y-\sin(z) +

    + +

    + \curl\vec F = \vec 0 +

    +     +     + + + +

    + \vec F = \nabla f, where \ds f(x,y,z) = \frac1{x^2+y^2+z^2}. +

    +     + +

    + \divv \vec F = \frac{2}{(x^2+y^2+z^2)^2} +

    + +

    + \curl\vec F = \vec 0 +

    +     +     +     +     +     +
    +
    + Line Integrals over Vector Fields + +

    + Suppose a particle moves along a curve C under the influence of an electromagnetic force described by a vector field \vec F. + Since a force is inducing motion, work is performed. + How can we calculate how much work is performed? +

    + +

    + Recall that when moving in a straight line, + if \vec F represents a constant force and \vec d represents the direction and length of travel, + then work is simply W = \vec F\cdot \vec d. + However, we generally want to be able to calculate work even if \vec F is not constant and C is not a straight line. +

    + +

    + As we have practiced many times before, + we can calculate work by first approximating, + then refining our approximation through a limit that leads to integration. +

    + + + +

    + Assume as we did in + that C can be parametrized by the arc length parameter s. + Over a short piece of the curve with length ds, + the curve is approximately straight and our force is approximately constant. + The straight-line direction of this short length of curve is given by \vec T, + the unit tangent vector; + let \vec d = \vec T\, ds, + which gives the direction and magnitude of a small section of C. + Thus work over this small section of C is \vec F \cdot \vec d = \vec F\cdot \vec T\, ds. +

    + +

    + Summing up all the work over these small segments gives an approximation of the work performed. + By taking the limit as ds goes to zero, + and hence the number of segments approaches infinity, + we can obtain the exact amount of work. + Following the logic presented at the beginning of this chapter in the Integration Review, + we see that + + W = \int_C \vec F\cdot \vec T\, ds + , + a line integral. +

    + +

    + This line integral is beautiful in its simplicity, + yet is not so useful in making actual computations + (largely because the arc length parameter is so difficult to work with). + To compute actual work, + we need to parametrize C with another parameter t via a vector-valued function \vec r(t). + As stated in , + ds = \norm{\vrp(t)}\, dt, + and recall that \vec T = \vrp(t)/\norm{\vrp(t)}. + Thus + + W \amp = \int_C \vec F\cdot\vec T\, ds = \int_C \vec F\cdot \frac{\vrp(t)}{\norm{\vrp(t)}}\norm{\vrp(t)}\, dt + \amp = \int_C\vec F\cdot \vrp(t)\, dt = \int_C \vec F\cdot d\vec r + , + where the final integral uses the differential d\vec r for \vrp(t)\,dt. +

    +     + + + Evaluating Line Integrals over Vector Fields +

    + These integrals are known as line integrals over vector fields. + By contrast, + the line integrals we dealt with in + are sometimes referred to as line integrals over scalar fields. + line integralover scalar field + Just as a vector field is defined by a function that returns a vector, + a scalar field is a function that returns a scalar, + such as z = f(x,y). + We waited until now to introduce this terminology so we could contrast the concept with vector fields. +

    + +

    + We formally define this line integral, + then give examples and applications. +

    + + + Line Integral Over A Vector Field + +

    + Let \vec F be a vector field with continuous components defined on a smooth curve C, + parametrized by \vrt, + and let \vec T be the unit tangent vector of \vrt. + The line integral over \vec F along C + is + line integralover vector field + + \int_C \vec F\cdot d\vec r = \int_C \vec F\cdot\vec T\, ds + . +

    +     +     + +

    + In , + note how the dot product \vec F \cdot \vec T is just a scalar. + Therefore, this new line integral is really just a special kind of line integral found in ; + letting f(s) = \vec F(s)\cdot \vec T(s), + the right-hand side simply becomes \int_C f(s)\, ds, + and we can use the techniques of that section to evaluate the integral. + We combine those techniques, + along with parts of Equation, + to clearly state how to evaluate a line integral over a vector field in the following Key Idea. +

    + + + + + Evaluating a Line Integral Over A Vector Field +

    + Let \vec F be a vector field with continuous components defined on a smooth curve C, + parametrized by \vrt, a\leq t\leq b, + where \vec r is continuously differentiable. + Thenvector fieldover vector field + + \int_C\vec F\cdot\vec T\, ds = \int_C \vec F\cdot d\vec r =\int_a^b \vec F\big(\vec r(t)\big) \cdot \vrp(t)\, dt + . +

    +     + +

    + An important concept implicit in this Key Idea: + we can use any continuously differentiable parametrization \vrt of C that preserves the orientation of C: + there isn't a right one. + In practice, choose one that seems easy to work with. +

    + +

    + Notation note: the above Definition and Key Idea implicitly evaluate \vec F along the curve C, + which is parametrized by \vrt. + For instance, + if \vec F = \langle x+y, x-y\rangle and \vrt = \langle t^2,\cos(t)\rangle, + then evaluating \vec F along C means substituting the x- and y-components of \vrt in for x and y, + respectively, in \vec F. + Therefore, along C, + \vec F = \langle x+y,x-y\rangle = \la t^2+\cos(t), t^2-\cos(t)\ra. + Since we are substituting the output + of \vrt for the input of \vec F, + we write this as \vec F\big(\vrt\big). + This is a slight abuse of notation as technically the input of \vec F is to be a point, + not a vector, but this shorthand is useful. +

    + +

    + We use an example to practice evaluating line integrals over vector fields. +

    + + + Evaluating a line integral over a vector field: computing work + +

    + Two particles move from (0,0) to (1,1) under the influence of the force field \vec F = \langle x, x+y\rangle. + One particle follows C_1, the line y=x; + the other follows C_2, + the curve y=x^4, as shown in . + Force is measured in newtons and distance is measured in meters. + Find the work performed by each particle. +

    +
    + Paths through a vector field in + + A vector field is plotted in two dimensions, along with two different paths that begin and end at the same points. + +

    + A two dimensional vector field is plotted relative to x and y coordinate axes. + The origin is at the bottom-left of the image, with the first quadrant visible. +

    + +

    + The vectors in the vector field all point up and to the right. + Those close to the y axis are nearly vertical; + the vectors close to the x axis are at an angle of approximately 45 degrees, + and the directions of the rest lie somewhere in between. + The magnitude of the vectors shrinks to zero as they get close to the origin. +

    + +

    + Two curves are also plotted in the plane. + Both curves meet at the points (0,0) and (1,1). + The curve y=x^4 lies below the line y=x. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ x, x+y ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + + + pen colorone; + pen colortwo; + + if(incolor) { + colorone = bluepen+1.5pt; + colortwo = redpen+1.5pt; + } else + { + colorone = black+1.5pt; + colortwo = gray+1.5pt; + }; + + real f(real x) {return x;} + pair F(real x) {return (x,f(x));} + + real g(real x) {return x^4;} + pair G(real x) {return (x,g(x));} + + + pair xbounds=(-.1,1.1); + pair ybounds=(-.1,1.1); + real[] myxchoice={1}; + real[] myychoice={1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(0,0); + pair b=(1,1); + + pair vectfunction(pair z) {return (z.x,z.x+z.y);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,bgvect)); + + draw(graph(f,0,1,operator ..),colorone); + draw(graph(g,0,1,operator ..),colortwo); + + label("$y=x$",(.4,.55),filltype=Fill(white)); + label("$y=x^4$",(.55,.25),filltype=Fill(white)); + + + +
    +     + +

    + To compute work, we need to parametrize each path. + We use \vec r_1(t) = \langle t,t\rangle to parametrize y=x, + and let \vec r_2(t) =\langle t,t^4\rangle parametrize y=x^4; + for each, 0\leq t\leq 1. +

    + +

    + Along the straight-line path, + \vec F\big(\vec r_1(t)\big) = \langle x, x+y\rangle = \langle t, t+t\rangle = \langle t,2t\rangle. + We find \vrp_1(t) =\langle 1,1\rangle. + The integral that computes work is: + + \int_{C_1} \vec F\cdot d\vec r \amp = \int_0^1 \langle t,2t\rangle\cdot\la 1,1\ra\, dt + \amp = \int_0^1 3t\, dt + \amp = \frac32t^2\Big|_0^1 = 1.5 \text{ joules } + . +

    + +

    + Along the curve y = x^4, + \vec F\big(\vec r_2(t)\big) = \la x, x+y\ra = \la t, t+t^4\ra. + We find \vrp_2(t) = \la 1, 4t^3\ra. + The work performed along this path is + + \int_{C_2} \vec F\cdot d\vec r \amp = \int_0^1 \la t,t+t^4\ra\cdot\la 1,4t^3\ra\, dt + \amp = \int_0^1 \big(t + 4t^4+ 4t^7\big)\, dt + \amp = \big(\frac12t^2 + \frac45t^5 + \frac12t^8\big)\Big|_0^1 = 1.8 \text{ joules } + . +

    + +

    + Note how differing amounts of work are performed along the different paths. + This should not be too surprising: + the force is variable, one path is longer than the other, etc. +

    +     +     + + + Evaluating a line integral over a vector field: computing work + +

    + Two particles move from (-1,1) to (1,1) under the influence of a force field \vec F = \la y, x\ra. + One moves along the curve C_1, + the parabola defined by y = 2x^2-1. + The other particle moves along the curve C_2, + the bottom half of the circle defined by x^2+(y-1)^2=1, + as shown in . + Force is measured in pounds and distances are measured in feet. + Find the work performed by moving each particle along its path. +

    +
    + Paths through a vector field in + + A two-dimensional vector field, and two curves in the plane. The vectors appear to lie tangent to a family of hyperbolas. + +

    + A two-dimensional vector field is plotted relative to x and y coordinate axes, + with the origin at the center of the image. +

    + +

    + The vectors in the vector field appear to follow hyperbolic trajectories: + if we drew a contour plot for the function f(x,y)=x^2-y^2, + the vector field would lie tangent to the curves in that plot. +

    + +

    + Two curves are plotted against this vector field. +

      +
    • +

      + The curves meet at the points (-1,1) and (1,1). +

      +
      • +
    • +

      + One curve is a parabola with its vertex at (0,-1). + It opens upward, and ends at the points (-1,1) and (1,1). +

      +
      • +
    • +

      + The other curve is the bottom half of a circle of radius 1, + with its center at (1,0). +

      +
      • +
    • +

      + The two curves also meet at the points \left(\pm\frac{\sqrt{3}}{2},\frac12\right), + but these intersections are not relevant to the solution of the problem. +

      +
      • +
    +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ y, x ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + + real f(real x) {return 2*x^2-1;} + pair F(real x) {return (x,f(x));} + + //real g(real x) {return x^4;} + //pair G(real x) {return (x,g(x));} + + pair G(real t) {return (cos(t),sin(t)+1);} + + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-1,-1); + pair b=(1,1); + + pair vectfunction(pair z) {return (z.y,z.x);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,bgvect)); + + draw(graph(F,-1,1,operator ..),colorone); + draw(graph(G,pi,2*pi,operator ..),colortwo); + + label("$y=2x^2-1$",(.7,-.9),filltype=Fill(white+opacity(.7))); + label("$x^2+(y-1)^2=1$",(0,.25),filltype=Fill(white+opacity(.7))); + + + +
    +     + +

    + We start by parametrizing C_1: + the parametrization \vec r_1(t) = \la t, 2t^2-1\ra is straightforward, + giving \vrp_1 = \la 1,4t\ra. + On C_1, + \vec F\big(\vec r_1(t)\big) = \la y,x\ra = \la 2t^2-1,t\ra. +

    + +

    + Computing the work along C_1, we have: + + \int_{C_1} \vec F\cdot d\vec r_1 \amp = \int_{-1}^1 \la 2t^2-1,t\ra\cdot\la 1,4t\ra\, dt + \amp = \int_{-1}^1 \big(2t^2-1+4t^2\big)\, dt = 2 \text{ ft-lbs } + . +

    + +

    + For C_2, + it is probably simplest to parametrize the half circle using sine and cosine. + Recall that \vec r(t) = \la \cos(t), \sin(t)\ra is a parametrization of the unit circle on 0\leq t\leq 2\pi; + we add 1 to the second component to shift the circle up one unit, + then restrict the domain to + \pi\leq t\leq 2\pi to obtain only the lower half, + giving \vec r_2(t) = \la \cos(t), \sin(t)+1\ra, + \pi\leq t\leq 2\pi, + and hence \vrp_2(t) = \la -\sin(t), \cos(t)\ra and \vec F\big(\vec r_2(t)\big) = \la y,x\ra = \la \sin(t)+1,\cos(t)\ra. +

    + +

    + Computing the work along C_2, we have: + + \int_{C_2} \vec F\cdot d\vec r_2 \amp = \int_{\pi}^{2\pi} \la \sin(t)+1,\cos(t)\ra\cdot\la -\sin(t),\cos(t)\ra\, dt + \amp = \int_{\pi}^{2\pi} \big(-\sin^2t-\sin(t)+\cos^2t\big)\, dt = 2 \text{ ft-lbs } + . +

    + +

    + Note how the work along C_1 and C_2 in this example is the same. + We'll address why later in this section when + conservative fields + and path independence are discussed. +

    +     +     +     + + + Properties of Line Integrals Over Vector Fields +

    + Line integrals over vector fields share the same properties as line integrals over scalar fields, + with one important distinction. + The orientation of the curve C matters with line integrals over vector fields, + whereas it did not matter with line integrals over scalar fields. +

    + + +

    + It is relatively easy to see why. + Let C be the unit circle. + The area under a surface over C is the same whether we traverse the circle in a clockwise or counterclockwise fashion, + hence the line integral over a scalar field on C is the same irrespective of orientation. + On the other hand, if we are computing work done by a force field, + direction of travel definitely matters. + Opposite directions create opposite signs when computing dot products, + so traversing the circle in opposite directions will create line integrals that differ by a factor of -1. +

    + + + Properties of Line Integrals Over Vector Fields + +

    +

      +
    1. +

      + Let \vec F and \vec G be vector fields with continuous components defined on a smooth curve C, + parametrized by \vrt, and let k_1 and k_2 be scalars. + Thenline integralproperties over a vector field + + \ds \int_C\big(k_1\vec F+k_2\vec G\big)\cdot d\vec r = k_1\int_C\vec F\cdot d\vec r +k_2\int_C\vec G\cdot d\vec r + . +

      +
      2. + +
    2. +

      + Let C be piecewise smooth, + composed of smooth components C_1 and C_2. + Then + + \int_C\vec F\cdot d\vec r = \int_{C_1}\vec F\cdot d\vec r + \int_{C_2}\vec F\cdot d\vec r + . +

      +
      3. + +
    3. +

      + Let C^* be the curve C with opposite orientation, + parametrized by \vec r\,^*. + Then + + \int_C\vec F\cdot d\vec r = -\int_{C^*}\vec F\cdot d\vec r\,^* + . +

      +
      4. +
    +

    +     +     + + + +

    + We demonstrate using these properties in the following example. +

    + + + Using properties of line integrals over vector fields + +

    + Let \vec F = \la 3(y-1/2),1\ra and let C be the path that starts at (0,0), + goes to (1,1) along the curve y=x^3, + then returns to (0,0) along the line y=x, + as shown in . + Evaluate \oint_C \vec F\cdot d\vec r. +

    + +
    + The vector field and curve in + + A two-dimensional vector field of vectors tangent to a family of parabolas, and a pair of curves in the plane. + +

    + A two dimensional vector field is plotted relative to x and y coordinate axes. + The origin is at the bottom-left of the image, with the first quadrant visible. +

    + +

    + The vectors in the vector field appear to lie tangent to a family of parabolas. + These parabolas are not shown, but if they were, + they would all open to the right, with their vertices along the line y=\frac12. +

    + +

    + Two curves are also plotted in the plane; they are the same curves as the ones in . + Both curves meet at the points (0,0) and (1,1). + The curve y=x^4 lies below the line y=x. +

    + +

    + There are arrows on the curves indicating direction of motion. + Along the y=x^4 curve is an arrow pointing away from the origin in the direction of travel toward (1,1). + The arrow on the line y=x points away from (1,1) and back toward the origin. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ 3(y-.5), 1 ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + + pair xbounds=(-.2,1.2); + pair ybounds=(-.2,1.2); + real[] myxchoice={1}; + real[] myychoice={1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-.1,-.1); + pair b=(1.1,1.1); + + pair vectfunction(pair z) {return (3(z.y-.5),1);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,bgvect)); + + real f(real x) {return 2*x^2-1;} + pair F(real x) {return (x,x^3);} + + //real g(real x) {return x^4;} + //pair G(real x) {return (x,g(x));} + + pair G(real t) {return (t,t);} + + draw(graph(F,0,1,operator ..),colorone,arrow=MidArrow(5)); + draw(graph(G,1,0,operator ..),colortwo,arrow=MidArrow(5)); + + //label("$y=2x^2-1$",(.7,-.9),filltype=Fill(white+opacity(.7))); + //label("$x^2+(y-1)^2=1$",(0,.25),filltype=Fill(white+opacity(.7))); + + + +
    +     + +

    + As C is piecewise smooth, + we break it into two components C_1 and C_2, + where C_1 follows the curve y=x^3 and C_2 follows the curve y=x. +

    + +

    + We parametrize C_1 with + \vec r_1(t) = \la t, t^3\ra on 0\leq t\leq 1, + with \vrp_1(t) = \la 1,3t^2\ra. + We will use \vec F\big(\vec r_1(t)\big) = \la 3(t^3-1/2),1\ra. +

    + +

    + While we always have unlimited ways in which to parametrize a curve, + there are 2 direct methods to choose from when parametrizing C_2. + The parametrization \vec r_2(t)=\la t,t\ra, + 0\leq t\leq 1 traces the correct line segment but with the wrong orientation. + Using Property 3 of , + we can use this parametrization and negate the result. +

    + +

    + Another choice is to use the techniques of + to create the line with the orientation we desire. + We wish to start at ( 1,1) and travel in the + \vec d = \la -1,-1\ra direction for one length of \vec d, + giving equation \vec \ell(t) = \la 1,1\ra + t\la -1,-1\ra = \la 1-t,1-t\ra on 0\leq t\leq 1. +

    + +

    + Either choice is fine; we choose + \vec r_2(t) to practice using line integral properties. + We find \vrp_2(t) = \la 1,1\ra and \vec F\big(\vec r_2(t)\big) = \la 3(t-1/2),1\ra. +

    + +

    + Evaluating the line integral (note how we subtract the integral over C_2 as the orientation of \vec r_2(t) is opposite): + + \oint_C \vec F\cdot d\vec r \amp = \int_{C_1}\vec F\cdot d\vec r_1 - \int_{C_2}\vec F\cdot d\vec r_2 + \amp = \int_0^1 \la 3(t^3-1/2),1\ra\cdot \la 1,3t^2\ra dt- \int_0^1 \la 3(t-1/2),1\ra\cdot\la 1,1\ra dt + \amp = \int_0^1\Big(3t^3+3t^2-3/2\Big)\, dt - \int_0^1 \Big(3t-1/2\Big)\, dt + \amp = \big(1/4 \big) - \big(1\big) + \amp = -3/4 + . +

    + +

    + If we interpret this integral as computing work, + the negative work implies that the motion is mostly against + the direction of the force, + which seems plausible when we look at . +

    +     +     + + + + + Evaluating a line integral over a vector field in space + +

    + Let \vec F = \la -y, x, 1\ra, + and let C be the portion of the helix given by \vrt = \langle \cos(t),\sin(t), t/(2\pi)\rangle on [0,2\pi], + as shown in . + Evaluate \int_C\vec F\cdot d\vec r. +

    + +
    + The graph of \vec r(t) in + + A three-dimensional vector field and a portion of a helix curve. + +

    + A set of three-dimensional coordinate axes is plotted, with the origin at the center. + Relative to these axes there is a vector field and a curve. +

    + +

    + From the default perspective, the vector field appears to be rather chaotic. + It is difficult to tell what is going on; + we can tell that all vectors have an upward slope, but the left and right orientation of the vectors seems random. + When the image is rotated to show the perspective from the positive z axis, + the vectors in the vector field appear to follow circular trajectories. +

    + +

    + Also plotted is the helix given by the vector-valued function \vec{r}(t) = \la \cos(t),\sin(t),t/(2\pi)\ra. + The image shows one revolution of the helix, beginning on the x axis at (1,0,0) + and ending at (1,0,1). +

    + +

    + Collectively, the vectors in the vector field appear as though they could be tangent vectors + to a family of such helix curves. +

    +     + + + + + size(200,200,IgnoreAspect); + currentprojection=orthographic(16.8,9.4,8.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={1}; + defaultpen(0.25mm); + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-1.5,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + path3 gradient1(triple z){ + return O--(-z.y,z.x,1); + } + + triple A=(-1,-1,-1); + triple B=(1,1,1); + + picture VectorPlot3D(path3 vector(triple t), triple a, triple b, + int nx=nmesh, int ny=nx, int nz=nx,bool truesize=false, + real maxlength=truesize ? 0 : min(abs(b.x-a.x)/nx,abs(b.y-a.y)/ny,abs(b.z-a.z)/nz), + // bool cond(pair z)=null, + pen p=currentpen, + arrowbar3 arrow=Arrow3(6), margin3 margin=PenMargin3, + string name="", render render=defaultrender) + { + picture pic; + real dx=1/nx; + real dy=1/ny; + real dz=1/nz; + real scale; + if(maxlength > 0) { + real size(triple t) { + path3 g=vector(t); + return abs(point(g,size(g)-1)-point(g,0)); + } + real max=size((0,0,0)); + + for(int i=0; i <= nx; ++i) { + real x=interp(a.x,b.x,i*dx); + for(int j=0; j <= ny; ++j) + { + real y=interp(a.y,b.y,j*dy); + for(int k=0; k <= nz; ++k) + max=max(max,size((x,y,interp(a.z,b.z,k*dz)))); + }} + scale=max > 0 ? maxlength/max : 1; + } else scale=1; + bool group=name != "" || render.defaultnames; + if(group) + begingroup3(pic,name == "" ? "vectorfield" : name,render); + for(int i=0; i <= nx; ++i) { + real x=interp(a.x,b.x,i*dx); + for(int j=0; j <= ny; ++j) { + real y=interp(a.y,b.y,j*dy); + for(int k=0; k <= nz; ++k) + { triple z=(x,y,interp(a.z,b.z,k*dz)); + { + path3 g=scale3(scale)*vector(z); + string name="vector"; + if(truesize) { + picture opic; + draw(opic,g,p,arrow,margin,name,render); + add(pic,opic,z); + } else + draw(pic,shift(z)*g,p,arrow,margin,name,render); + } + } + }} + if(group) + endgroup3(pic); + return pic; + + } + add(VectorPlot3D(gradient1,A,B,3,3,3,bluepen)); + + triple g(real t) {return(cos(t),sin(t),t/(2*pi));} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,curvepen2,MidArrow3(6)); + + //triple g(real t) {return (t,t,-2*t^4+4*t^2);} + //path3 mypath=graph(g,-1,1,operator ..); + + + +
    +     + +

    + A parametrization is already given for C, + so we just need to find \vec F\big(\vec r(t)\big) and \vec r '(t). +

    + +

    + We have \vec F\big(\vec r(t)\big) = \la -\sin(t), \cos(t), 1\ra and \vrp(t) = \la -\sin(t), \cos(t), 1/(2\pi)\ra. + Thus + + \int_C \vec F\cdot d\vec r \amp = \int_0^{2\pi} \la -\sin(t), \cos(t), 1\ra\cdot \la -\sin(t), \cos(t), 1/(2\pi)\ra dt + \amp = \int_0^{2\pi} \Big(\sin^2t+\cos^2t + \frac1{2\pi}\Big)dt + \amp = 2\pi + 1 \approx 7.28 + +

    +     +     + + +     + + + The Fundamental Theorem of Line Integrals +

    + We are preparing to make important statements about the value of certain line integrals over special vector fields. + Before we can do that, + we need to define some terms that describe the domains over which a vector field is defined. + line integralFundamental Theorem + Fundamental Theorem of Line Integrals +

    + + +

    + A region in the plane is connected + connected + if any two points in the region can be joined by a piecewise smooth curve that lies entirely in the region. + In , + sets R_1 and R_2 are connected; + set R_3 is not connected, + though it is composed of two connected subregions. +

    + +

    + A region is simply connected + simply connected + connectedsimply + if every simple closed curve that lies entirely in the region can be continuously deformed (shrunk) to a single point without leaving the region. (A curve is simple + simple curve + if it does not cross itself.) In , + only set R_1 is simply connected. + Region R_2 is not simply connected as any closed curve that goes around the hole + in R_2 cannot be continuously shrunk to a single point. + As R_3 is not even connected, + it cannot be simply connected, + though again it consists of two simply connected subregions. +

    + +

    + We have applied these terms to regions of the plane, + but they can be extended intuitively to domains in space + (and hyperspace). + In , + the domain bounded by the sphere + (at left) + and the domain with a subsphere removed + (at right) + are both simply connected. + Any simple closed path that lies entirely within these domains can be continuously deformed into a single point. + In , + neither domain is simply connected. + A left, the ball has a hole that extends its length and the pictured closed path cannot be deformed to a point. + At right, two paths are illustrated on the torus that cannot be shrunk to a point. +

    + +

    + We will use the terms connected and simply connected in subsequent definitions and theorems. +

    + +
    + R_1 is simply connected; R_2 is connected, but not simply connected; R_3 is not connected + + Several regions in the plane are used to illustrate the concepts of connected, and simply connected. + +

    + Three different regions in the plane are plotted, without the use of coordinate axes. + The regions are pond-like, with boundaries resembling bumpy, poorly-drawn circles. +

    + +

    + The region R_1 is both connected and simply connected. + It is a single pond-like shape, with its interior completely shaded. +

    + +

    + The region R_2 is connected but not simply connected. + It is like a pond that contains an island: there is a region in its interior that is not shaded, + indicating that this part of the interior is not part of R_2. +

    + +

    + The region R_3 is not connected. + It consists of two pond-like shapes, each separately resembling R_1. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + axis y line=none, + axis x line=none, + ymin=-.75,ymax=3, + xmin=-2,xmax=2, + ] + + \draw (axis cs: -1.6,1.4) node {$R_1$} + (axis cs: 1.6,1.1) node {$R_2$} + (axis cs: 0,.6) node {$R_3$}; + + \addplot [firstcurvestyle,areastyle] coordinates {(-0.45,2.)(-0.4429,2.07)(-0.4788,2.134)(-0.5466,2.18)(-0.6142,2.212)(-0.6552,2.251)(-0.6695,2.31)(-0.6802,2.387)(-0.7106,2.456)(-0.766,2.497)(-0.8346,2.509)(-0.9034,2.506)(-0.9684,2.502)(-1.031,2.496)(-1.091,2.476)(-1.144,2.442)(-1.192,2.407)(-1.246,2.388)(-1.318,2.384)(-1.399,2.374)(-1.464,2.337)(-1.49,2.27)(-1.476,2.189)(-1.447,2.115)(-1.435,2.055)(-1.45,2.)(-1.477,1.94)(-1.491,1.874)(-1.482,1.809)(-1.457,1.749)(-1.426,1.691)(-1.389,1.634)(-1.339,1.59)(-1.273,1.57)(-1.2,1.576)(-1.135,1.586)(-1.081,1.573)(-1.03,1.528)(-0.9669,1.474)(-0.8941,1.445)(-0.8255,1.463)(-0.7738,1.519)(-0.737,1.586)(-0.702,1.64)(-0.6605,1.681)(-0.6167,1.722)(-0.5804,1.769)(-0.5524,1.823)(-0.5227,1.877)(-0.4848,1.935)(-0.45,2.)}; 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+ \addplot [firstcurvestyle,smooth] coordinates {(-0.05,0)(-0.0447,0.07015)(-0.07696,0.1343)(-0.1383,0.1828)(-0.2054,0.217)(-0.2552,0.2505)(-0.2804,0.3001)(-0.2931,0.3709)(-0.3146,0.4497)(-0.3593,0.5116)(-0.4255,0.5371)(-0.4999,0.5248)(-0.5691,0.4919)(-0.629,0.461)(-0.6849,0.4453)(-0.7436,0.442)(-0.8063,0.4384)(-0.8688,0.4236)(-0.9266,0.3947)(-0.9786,0.3555)(-1.026,0.3093)(-1.066,0.2559)(-1.09,0.1942)(-1.093,0.1266)(-1.075,0.06001)(-1.05,0)(-1.037,-0.05518)(-1.046,-0.1144)(-1.068,-0.1853)(-1.082,-0.2648)(-1.064,-0.3373)(-1.009,-0.3844)(-0.9306,-0.3996)(-0.8505,-0.3947)(-0.785,-0.3932)(-0.7345,-0.414)(-0.6873,-0.4575)(-0.6318,-0.5062)(-0.5662,-0.537)(-0.4976,-0.537)(-0.4346,-0.5091)(-0.3805,-0.4664)(-0.333,-0.4208)(-0.2891,-0.3758)(-0.2496,-0.329)(-0.2167,-0.2785)(-0.1892,-0.2258)(-0.1607,-0.1739)(-0.1245,-0.1221)(-0.08292,-0.06532)(-0.05,0)}; + + \addplot [firstcurvestyle,areastyle] coordinates {(1.15,0)(1.157,0.07033)(1.143,0.1393)(1.107,0.2007)(1.054,0.2495)(0.992,0.2848)(0.9305,0.3104)(0.8756,0.3332)(0.8286,0.3602)(0.786,0.3953)(0.742,0.4371)(0.6914,0.4793)(0.6323,0.5134)(0.5665,0.5321)(0.4985,0.5319)(0.433,0.514)(0.3727,0.4829)(0.3182,0.444)(0.2685,0.4007)(0.223,0.354)(0.183,0.303)(0.1517,0.2464)(0.1328,0.185)(0.128,0.1212)(0.1355,0.05868)(0.15,0)(0.1646,-0.055)(0.174,-0.1094)(0.177,-0.1675)(0.1775,-0.2323)(0.183,-0.303)(0.2016,-0.3742)(0.2382,-0.4373)(0.2928,-0.4841)(0.3602,-0.5095)(0.433,-0.514)(0.504,-0.503)(0.5695,-0.4846)(0.6293,-0.4659)(0.6859,-0.4504)(0.742,-0.4371)(0.7985,-0.4219)(0.8541,-0.4004)(0.9059,-0.3698)(0.952,-0.3305)(0.992,-0.2848)(1.028,-0.2353)(1.063,-0.1831)(1.097,-0.1275)(1.128,-0.06665)(1.15,0)}; + \addplot [firstcurvestyle,smooth] coordinates {(1.15,0)(1.157,0.07033)(1.143,0.1393)(1.107,0.2007)(1.054,0.2495)(0.992,0.2848)(0.9305,0.3104)(0.8756,0.3332)(0.8286,0.3602)(0.786,0.3953)(0.742,0.4371)(0.6914,0.4793)(0.6323,0.5134)(0.5665,0.5321)(0.4985,0.5319)(0.433,0.514)(0.3727,0.4829)(0.3182,0.444)(0.2685,0.4007)(0.223,0.354)(0.183,0.303)(0.1517,0.2464)(0.1328,0.185)(0.128,0.1212)(0.1355,0.05868)(0.15,0)(0.1646,-0.055)(0.174,-0.1094)(0.177,-0.1675)(0.1775,-0.2323)(0.183,-0.303)(0.2016,-0.3742)(0.2382,-0.4373)(0.2928,-0.4841)(0.3602,-0.5095)(0.433,-0.514)(0.504,-0.503)(0.5695,-0.4846)(0.6293,-0.4659)(0.6859,-0.4504)(0.742,-0.4371)(0.7985,-0.4219)(0.8541,-0.4004)(0.9059,-0.3698)(0.952,-0.3305)(0.992,-0.2848)(1.028,-0.2353)(1.063,-0.1831)(1.097,-0.1275)(1.128,-0.06665)(1.15,0)}; + + \end{axis} + + \end{tikzpicture} + + + +
    + +
    + The domains in (a) are simply connected, while the domains in (b) are not + +
    + + + Two spherical surfaces in space, plotted without coordinate axes. The second sphere has a smaller sphere in its interior. + +

    + Two spheres are plotted side by side in three dimensions. + There are no coordinate axes used in the image. +

    + +

    + On the left is a single sphere; it represents the domain consisting of all points on and inside the sphere. +

    + +

    + The sphere on the right has another smaller sphere inside it. + It represents the domain consisting of all points outside the smaller sphere, but inside the larger sphere. + Despite this hole in the middle of the domain, it is still simply connected. + The sphere on the inside does not obstruct us from transforming one curve inside the large sphere to another, + as we can stretch the curve around the small sphere. +

    +     + + + + + //ASY file for fig10_01_ex_233D.asy in Chapter 10 + + size(200,200); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(0,-5,3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={1}; + real[] myzchoice={1,2}; + defaultpen(0.5mm); + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + pair zbounds=(-1.1,1.1); + + //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2+x^2+y^2=1, a sphere shifted + triple f(pair t) { + return (cos(t.x)*sin(t.y)-1.1,sin(t.x)*sin(t.y),cos(t.y)); + } + surface s=surface(f,(0,0),(2pi,pi),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=apexmeshpen+.09mm; + draw(s,simplesurfacepen,meshpen=p); + + //Draw the surface z^2+x^2+y^2=1, a sphere shifted + triple f(pair t) { + return (cos(t.x)*sin(t.y)+1.1,sin(t.x)*sin(t.y),cos(t.y)); + } + surface s=surface(f,(0,0),(2pi,pi),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + //pen p=rgb(0,0,.2); + draw(s,simplesurfacepen,meshpen=p); + + //Draw the surface z^2+x^2+y^2=1, a smaller sphere shifted + triple f(pair t) { + return (.3*cos(t.x)*sin(t.y)+1.1,.3*sin(t.x)*sin(t.y),.3*cos(t.y)); + } + surface s=surface(f,(0,0),(2pi,pi),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + //pen p=rgb(0,0,.2); + draw(s,emissive(surfacepen),meshpen=p); + + triple g(real t) {return (t,t^2,-cos(t)*sin(t^2)+1);} + path3 mypath=graph(g,-1,1,operator ..); + //draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,t^2,0);} + path3 mypath=graph(g,-1,1,operator ..); + //draw(mypath,bluepen+linewidth(2)+dashed); + + + +
    + +
    + + + Two surfaces in three dimensions. One is a sphere with a cylinder through its center; the other is a torus. + +

    + Two surfaces are shown side by side in three dimensions, without the use of coordinate axes. + Each represents a solid domain in space that is not simply connected. +

    + +

    + The domain on the left consists of the region on and inside a sphere, + except that a cylindrical hole has been cut along one diameter of the sphere. + This surface is not simply connected: + any curve that wraps around the cylinder on the inside cannot be shrunk to a point. +

    + +

    + The domain on the right is a torus. + A torus is a ring-shaped surface, like a donut or a bagel. + The torus is not simply connected because any curve that travels all the way around the hole in the middle cannot be shrunk to a point. +

    + +

    + An additional fun fact: these surfaces may be geometrically different, but topologically, they are the same. + If we squish the sphere vertically, until the cylinder through the middle is shrunk to a circle, + we end up with the torus. +

    +     + + + + + //ASY file for fig10_01_ex_233D.asy in Chapter 10 + + size(200,200); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(0,-5,3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={1}; + real[] myzchoice={1,2}; + defaultpen(0.5mm); + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + pair zbounds=(-1.1,1.1); + + //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the surface z^2+x^2+y^2=1, a sphere + triple f(pair t) { + return (cos(t.x)*sin(t.y)-1.1,sin(t.x)*sin(t.y),cos(t.y)); + } + surface s=surface(f,(0,1/4),(2pi,pi-1/4),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + pen p=apexmeshpen+.09mm; + draw(s,simplesurfacepen,meshpen=p); + + //Draw the surface z^2+x^2+y^2=1, a sphere + triple f(pair t) { + return (cos(t.x)*sin(.25)-1.1,sin(t.x)*sin(.25),cos(.25)+t.y*(cos(pi-.25)-cos(.25))); + } + surface s=surface(f,(0,0),(2pi,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + //pen p=rgb(0,0,.2); + draw(s,simplesurfacepen,meshpen=p); + + triple g(real t) {return (cos(t)-1.1, sin(t),0);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(1)); + + //Draw the torus + real cc=.75; + real aa=.25; + triple f(pair t) { + return ((cc+aa*cos(t.y))*cos(t.x)+1.1,(cc+aa*cos(t.y))*sin(t.x),aa*sin(t.y)); + } + surface s=surface(f,(0,0),(2pi,2pi),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} + ); + //pen p=apexmeshpen; + draw(s,simplesurfacepen,meshpen=p); + + triple g(real t) {return (cos(t)+1.1,sin(t),0);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(1)); + + real yy = 5pi/4; + + triple g(real t) {return ((cc+aa*cos(t))*cos(yy)+1.1,(cc+aa*cos(t))*sin(yy),aa*sin(t));} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(1)); + + + +
    +     + +
    + +

    + Recall how in + particles moved from A = (-1,1) to + B = (1,1) along two different paths, + wherein the same amount of work was performed along each path. + It turns out that regardless of the choice of path from A to B, + the amount of work performed under the field \vec F = \la y, x\ra is the same. + Since our expectation is that differing amounts of work are performed along different paths, + we give such special fields a name. +

    + + + Conservative Field, Path Independent + +

    + Let \vec F be a vector field defined on an open, + connected domain D in the plane or in space containing points A and B. + If the line integral \int_C \vec F\cdot d\vec r has the same value for all choices of paths C starting at A and ending at B, then + conservative field + vector fieldconservative + path independent + line integralpath independent + +

      +
    • +

      + \vec F is a conservative field and +

      +
      • + +
    • +

      + The line integral \int_C \vec F\cdot d\vec r is + path independent and can be written as + + \int_C \vec F\cdot d\vec r = \int_A^B \vec F\cdot \, d\vec r + . +

      +
      • +
    +

    +     +     + +

    + When \vec F is a conservative field, + the line integral from points A to B is sometimes written as + \int_A^B\vec F\cdot d\vec r to emphasize the independence of its value from the choice of path; + all that matters are the beginning and ending points of the path. +

    + +

    + How can we tell if a field is conservative? + To show a field \vec F is conservative using the definition, + we need to show that all + line integrals from points A to B have the same value. + It is equivalent to show that all + line integrals over closed paths C are 0. + Each of these tasks are generally nontrivial. +

    + +

    + There is a simpler method. + Consider the surface defined by z = f(x,y) = xy. + We can compute the gradient of this function: + \nabla f = \la f_x, f_y\ra = \la y, x\ra. + Note that this is the field from , + which we have claimed is conservative. + We will soon give a theorem that states that a field \vec F is conservative if, + and only if, + it is the gradient of some scalar function f. + To show \vec F is conservative, + we need to determine whether or not + \vec F = \nabla f for some function f. (We'll later see that there is a yet simpler method). + To recognize the special relationship between \vec F and f in this situation, + f is given a name. +

    + + + Potential Function + +

    + Let f be a differentiable function defined on a domain D in the plane or in space (, + z = f(x,y) or w = f(x,y,z)) and let + \vec F = \nabla f, the gradient of f. + Then f is a potential function of \vec F. + potential function + vector fieldpotential function of +

    +     +     + +

    + We now state the Fundamental Theorem of Line Integrals, + which connects conservative fields and path independence to fields with potential functions. +

    + + + Fundamental Theorem of Line Integrals + +

    + Let \vec F be a vector field whose components are continuous on a connected domain D in the plane or in space, + let A and B be any points in D, + and let C be any path in D starting at A and ending at B. + line integralFundamental Theorem + Fundamental Theorem of Line Integrals + conservative field + vector fieldconservative + path independent + line integralpath independent + +

      +
    1. +

      + \vec F is conservative if and only if there exists a differentiable function f such that \vec F = \nabla f. +

      +
      2. + +
    2. +

      + If \vec F is conservative, then + + \int_C\vec F\cdot d\vec r = \int_A^B \vec F\cdot d\vec r = f(B) - f(A) + . +

      +
      3. +
    +

    +     +     + +

    + Once again considering , + we have A = (-1,1), + B = (1,1) and \vec F = \la y,x\ra. + In that example, + we evaluated two line integrals from A to B and found the value of each was 2. + Note that f(x,y) = xy is a potential function for \vec F. + Following the Fundamental Theorem of Line Integrals, + consider f(B) - f(A): + + f(B) - f(A) = f(1,1) - f(-1,1) = 1 - (-1) = 2 + , + the same value given by the line integrals. +

    + +

    + We practice using this theorem again in the next example. +

    + + + Using the Fundamental Theorem of Line Integrals + +

    + Let \vec F = \la 3x^2y+2x, + x^3+1\ra, A = (0,1) and B = (1,4). + Use the first part of the Fundamental Theorem of Line Integrals to show that \vec F is conservative, + then choose any path from A to B and confirm the second part of the theorem. +

    +     + +

    + To show \vec F is conservative, + we need to find z = f(x,y) such that \vec F = \nabla f = \la f_x, f_y\ra. + That is, we need to find f such that + f_x = 3x^2y+2x and f_y = x^3+1. + As all we know about f are its partial derivatives, + we recover f by integration: + + \int \frac{\partial f}{\partial x}\, dx = f(x,y) + C(y) + . +

    + +

    + Note how the constant of integration is more than + just a constant: + it is anything that acts as a constant when taking a derivative with respect to x. + Any function that is a function of y + (containing no x's) + acts as a constant when deriving with respect to x. +

    + +

    + Integrating f_x in this example gives: + + \int \frac{\partial f}{\partial x}\, dx = \int (3x^2y+2x)\, dx = x^3y+x^2 + C_1(y) + . +

    + +

    + Likewise, integrating f_y with respect to y gives: + + \int \frac{\partial f}{\partial y}\, dy = \int( x^3+1)\, dy = x^3y+ y + C_2(x) + . +

    + +

    + These two results should be equal with appropriate choices of C_1(y) and C_2(x): + + x^3y+x^2 + C_1(y) = x^3y+ y + C_2(x) \Rightarrow C_2(x) = x^2 \text{ and } C_1(y) = y + . +

    + +

    + We find f(x,y) = x^3y+x^2+y, + a potential function of \vec F. (If \vec F were not conservative, + no choice of C_2(x) and C_1(y) would give equality.) +

    + +

    + By the Fundamental Theorem of Line Integrals, + regardless of the path from A to B, + + \int_A^B\vec F\cdot d\vec r \amp = f(B) - f(A) + \amp = f(1,4) - f(0,1) + \amp = 9 - 1 = 8 + . +

    + +

    + To illustrate the validity of the Fundamental Theorem, + we pick a path from A to B. + The line between these two points would be simple to construct; + we choose a slightly more complicated path by choosing the parabola y = x^2+2x+1. + This leads to the parametrization \vrt = \la t, t^2+2t+1\ra, + 0\leq t\leq 1, with \vrp(t) = \la t, 2t+2\ra. + Thus + + \int_C \vec F\cdot d\vec r \amp = \int_C\vec F\big(\vrt\big)\cdot\vrp(t)\, dt + \amp = \int_0^1\la 3(t)(t^2+2t+1)+2t, t^3+1\ra\cdot\la t,2t+2\ra\, dt + \amp = \int_0^1 \big(5t^4+8t^3+3t^2+4t+2\big)\, dt + \amp = \big(t^5+2t^4+t^3+2t^2+2t\big)\Big|_0^1 + \amp = 8 + , + which matches our previous result. +

    +     +     + +

    + The Fundamental Theorem of Line Integrals states that we can determine whether or not \vec F is conservative by determining whether or not \vec F has a potential function. + This can be difficult. + A simpler method exists if the domain of \vec F is simply connected + (not just connected as needed in the Fundamental Theorem of Line Integrals), + which is a reasonable requirement. + We state this simpler method as a theorem. +

    + + + + + Curl of Conservative Fields + +

    + Let \vec F be a vector field whose components have continuous partial + derivatives on a simply connected domain D in the plane or in space. + Then \vec F is conservative if and only if + \curl \vec F = 0 or \vec 0, in 2D or 3D, respectively. + curlof conservative fields + conservative field +

    +     +     + +

    + In , + we showed that \vec F =\langle 3x^2y+2x,x^3+1\rangle is conservative by finding a potential function for \vec F. + Using the above theorem, + we can show that \vec F is conservative much more easily by computing its curl: + + \curl \vec F = N_x - M_y = 3x^2 - 3x^2 = 0 + . +

    + +     + + + + Terms and Concepts + + + +

    + In practice, + the evaluation of line integrals over vector fields involves computing the magnitude of a vector-valued function. +

    +     + +

    + Although magnitude might appear in the definition, it cancels out once we parametrize. + It is true for line integrals over scalar fields, though. +

    +     +     + + + +

    + Let \vec F(x,y) be a vector field in the plane and let + \vec r(t) be a two-dimensional vector-valued function. + Why is \vec F\big(\vec r(t)\big) + an abuse of notation? +

    +     + + + +

    + The input of \vec F should be a point in the plane, + not a two dimensional vector. +

    +     +     + + + +

    + The orientation of a curve C matters when computing a line integral over a vector field. +

    +     + +

    + Much like swapping the bounds in a definite integral, + changing the orientation of a curve introduces a minus sign. +

    +     +     + + + +

    + The orientation of a curve C matters when computing a line integral over a scalar field. +

    +     + +

    + For scalar fields, we multiply by the magnitude of the tangent vector, + and the minus sign introduced by a change in orientation does not change the magnitude. +

    +     +     + + + +

    + Under reasonable conditions, + if \curl \vec F = \vec 0, + what can we conclude about the vector field \vec F? +

    +     + + + +

    + We can conclude that \vec F is conservative. +

    +     +     + + + +

    + Let \vec F be a conservative field and let C be a closed curve. + Why are we able to conclude that \oint _C \vec F\cdot d\vec r = 0? +

    +     + + + +

    + By the Fundamental Theorem of Line Integrals, + since \vec F is conservative, + \oint_C \vec F\cdot d\vec r = f(B) - f(A), + where f is a potential function for \vec F and A and B are the initial and terminal points of C, + respectively. + Since C is a closed curve, A = B, + and hence f(B) - f(A) = 0. +

    +     +     +     + + Problems + + +

    + A vector field \vec F and a curve C are given. + Evaluate \ds\int_C\vec F\cdot d\vec r. +

    +     + + + +

    + \vec F = \langle y, y^2\rangle; + C is the line segment from (0,0) to (3,1). +

    +     + +

    + 11/6. (One parametrization for C is + \vec r(t) = \langle 3t,t\rangle on 0\leq t\leq 1.) +

    +     +     + + + +

    + \vec F = \langle x,x+y\rangle; + C is the portion of the parabola y=x^2 from (0,0) to (1,1). +

    +     + +

    + 5/3. (One parametrization for C is + \vec r(t) = \langle t,t^2\rangle on 0\leq t\leq 1.) +

    +     +     + + + +

    + \vec F = \langle y,x\rangle; + C is the top half of the unit circle, + beginning at (1,0) and ending at (-1,0). +

    +     + +

    + 0. (One parametrization for C is + \vec r(t) = \langle \cos(t),\sin(t)\rangle on 0\leq t\leq \pi.) +

    +     +     + + + +

    + \vec F = \langle xy,x\rangle; + C is the portion of the curve y=x^3 on -1\leq x\leq 1. +

    +     + +

    + 2/5. (One parametrization for C is + \vec r(t) = \langle t,t^3\rangle on -1\leq t\leq 1.) +

    +     +     + + + +

    + \vec F = \langle z,x^2,y\rangle; + C is the line segment from (1,2,3) to (4,3,2). +

    +     + +

    + 12. (One parametrization for C is + \vec r(t) = \langle 1,2,3\rangle+t\langle 3,1,-1\rangle on 0\leq t\leq 1.) +

    +     +     + + + +

    + \vec F = \langle y+z,x+z,x+y\rangle; + C is the helix \vec r(t) = \langle \cos(t),\sin(t),t/(2\pi)\rangle on 0\leq t\leq 2\pi. +

    +     + +

    + 1. +

    +     +     +     + + + +

    + Find the work performed by the force field \vec F moving a particle along the path C. +

    +     + + + +

    + \vec F = \langle y,x^2\rangle N; C is the segment of the line y=x from (0,0) to (1,1), + where distances are measured in meters. +

    +     + +

    + 5/6 joules. (One parametrization for C is + \vec r(t) = \langle t,t\rangle on 0\leq t\leq 1.) +

    +     +     + + + +

    + \vec F = \langle y,x^2\rangle N; C is the portion of + y=\sqrt x from (0,0) to (1,1), + where distances are measured in meters. +

    +     + +

    + 13/15 joules. (One parametrization for C is + \vec r(t) = \langle t,\sqrt t\rangle on 0\leq t\leq 1.) +

    +     +     + + + +

    + \vec F = \langle 2xy,x^2,1\rangle lbs; + C is the path from (0,0,0) to (2,4,8) via + \vec r(t) = \langle t, t^2, t^3\rangle on 0\leq t\leq 2, + where distance are measured in feet. +

    +     + +

    + 24 ft-lbs. +

    +     +     + + + +

    + \vec F = \langle 2xy,x^2,1\rangle lbs; + C is the path from (0,0,0) to (2,4,8) via + \vec r(t) = \langle t,2t, 4t\rangle on 0\leq t\leq 2, + where distance are measured in feet. +

    +     + +

    + 24 ft-lbs. +

    +     +     +     + + + +

    + A conservative vector field \vec F and a curve C are given. + +

      +
    1. +

      + Find a potential function f for \vec F. +

      +
      2. + +
    2. +

      + Compute \curl \vec F. +

      +
      3. + +
    3. +

      + Evaluate \ds\int_C \vec F\cdot d\vec r directly, + , using . +

      +
      4. + +
    4. +

      + Evaluate \ds\int_C \vec F\cdot d\vec r using the Fundamental Theorem of Line Integrals. +

      +
      5. +
    +

    +     + + + +

    + \vec F = \langle y+1,x\rangle, + C is the line segment from (0,1) to (1,0). +

    +     + +

    +

      +
    1. +

      + f(x,y) = xy+x +

      +
      2. + +
    2. +

      + \curl \vec F = 0. +

      +
      3. + +
    3. +

      + 1. (One parametrization for C is + \vec r(t) = \langle t,t-1\rangle on 0\leq t\leq 1.) +

      +
      4. + +
    4. +

      + 1 (with A = (0,1) and + B = (1,0), f(B) - f(A) = 1.) +

      +
      5. +
    +

    +     +     + + + +

    + \vec F = \langle 2x+y, 2y+x\rangle, + C is curve parametrized by + \vec r(t) = \langle t^2-t, t^3-t\rangle on 0\leq t\leq 1. +

    +     + +

    +

      +
    1. +

      + f(x,y) = x^2+xy+y^2 +

      +
      2. + +
    2. +

      + \curl \vec F = 0. +

      +
      3. + +
    3. +

      + 0. +

      +
      4. + +
    4. +

      + 0 (with A = (0,0) and + B = (0,0), f(B) - f(A) = 0.) +

      +
      5. +
    +

    +     +     + + + +

    + \vec F = \langle 2xyz,x^2z,x^2y\rangle, + C is curve parametrized by + \vec r(t) = \langle 2t+1,3t-1,t\rangle on 0\leq t\leq 2. +

    +     + +

    +

      +
    1. +

      + f(x,y) = x^2yz +

      +
      2. + +
    2. +

      + \curl \vec F = \vec 0. +

      +
      3. + +
    3. +

      + 250. +

      +
      4. + +
    4. +

      + 250 (with A = (1,-1,0) and B = (5,5,2), + f(B) - f(A) = 250.) +

      +
      5. +
    +

    +     +     + + + +

    + \vec F = \langle 2x, 2y, 2z\rangle, + C is curve parametrized by + \vec r(t) = \langle \cos(t),\sin(t), \sin (2t)\rangle on 0\leq t\leq 2\pi. +

    +     + +

    +

      +
    1. +

      + f(x,y) = x^2+y^2+z^2 +

      +
      2. + +
    2. +

      + \curl \vec F = \vec 0. +

      +
      3. + +
    3. +

      + 0. +

      +
      4. + +
    4. +

      + 0 (with A = (1,0,0) and + B = (1,0,0), f(B) - f(A) = 0.) +

      +
      5. +
    +

    +     +     +     + + + +

    + Prove part of : + let \vec F =\langle M,N,P\rangle be a conservative vector field. + Show that \curl \vec F = 0. +

    +     + +

    + Since \vec F is conservative, + it is the gradient of some potential function. + That is, \nabla f = \langle f_x,f_y,f_z\rangle = \vec F = \langle M, N, P\rangle. + In particular, M = f_x, + N = f_y and P=f_z. +

    + +

    + Note that \curl \vec F = \langle P_y - N_z, M_z-P_x, N_x-M_y \rangle = \langle f_{zy} - f_{yz}, f_{xz} - f_{zx}, f_{yx} - f_{xy}\rangle, which, + by , + is \langle 0,0,0\rangle. +

    +     +     +     +     +
    +
    + Flow, Flux, Green's Theorem and the Divergence Theorem + + Flow and Flux +

    + Line integrals over vector fields have the natural interpretation of computing work when \vec F represents a force field. + It is also common to use vector fields to represent velocities. + In these cases, the line integral + \int_C \vec F\cdot d\vec r is said to represent flow. + flow + flux + circulation +

    + +
    + Illustrating the principles of flow and flux + + A constant vector field in the plane, pointing to the right, and a triangular path in the first quadrant. + +

    + A two-dimensional vector field is plotted relative to x and y coordinate axes. + The origin is at the bottom-left of the image, so that the first quadrant is displayed. +

    + +

    + The vector field is constant, with all arrows of equal magnitude, and pointing to the right. +

    + +

    + A triangular path is drawn in the first quadrant. +

      +
    • +

      + The bottom of the triangle is a line labeled C_1, traveling from left to right. +

      +
      • +
    • +

      + The left side of the triangle is vertical; this line is labeled C_2, + and oriented to travel from bottom to top. +

      +
      • +
    • +

      + The hypotenuse of the triangle is labeled C_3, + and it is oriented to travel from top-left to bottom-right. +

      +
      • +
    +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ 1,0 ⟩ + + defaultpen(fontsize(8)); + + pen colorone; + pen colortwo; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + }; + + pair xbounds=(-.2,1.2); + pair ybounds=(-.2,1.2); + real[] myxchoice={1}; + real[] myychoice={1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-.1,-.1); + pair b=(1.1,1.1); + + pair vectfunction(pair z) {return (1,0);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,6,6,rgb(.6,.6,.6))); + + pair f(real x) {return (.2+x,1-x);} + pair F(real x) {return (x,.2);} + + //real g(real x) {return x^4;} + //pair G(real x) {return (x,g(x));} + + pair G(real t) {return (.2,t);} + + draw(graph(F,0.2,1,operator ..),colorone,arrow=MidArrow(5)); + draw(graph(G,0.2,1,operator ..),colorone,arrow=MidArrow(5)); + draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); + + label("$C_1$",(.8,.15),filltype=Fill(white+opacity(.7))); + label("$C_2$",(.13,.8),filltype=Fill(white+opacity(.7))); + label("$C_3$",(.6,.5),filltype=Fill(white+opacity(.7))); + + + +
    + +

    + Let the vector field \vec F = \la 1,0\ra represent the velocity of water as it moves across a smooth surface, + depicted in . + A line integral over C will compute + how much water is moving + along the path C. +

    + +

    + In the figure, + all of the water above C_1 is moving along that curve, + whereas none of the water above C_2 is moving along that curve + (the curve and the flow of water are at right angles to each other). + Because C_3 has nonzero horizontal and vertical components, + some of the water above that curve is moving along the curve. +

    + +

    + When C is a closed curve, + we call flow circulation, + represented by \oint_C \vec F\cdot d\vec r. +

    + +

    + The opposite of flow is flux, + a measure of how much water is moving + across the path C. + If a curve represents a filter in flowing water, + flux measures how much water will pass through the filter. + Considering again , + we see that a screen along C_1 will not filter any water as no water passes across that curve. + Because of the nature of this field, + C_2 and C_3 each filter the same amount of water per second. +

    + +

    + The terms flow and flux + are used apart from velocity fields, too. + Flow is measured by \int_C \vec F\cdot d\vec r, + which is the same as \int_C \vec F\cdot\vec T\, ds by . + That is, flow is a summation of the amount of \vec F that is + tangent to the curve C. +

    + +

    + By contrast, + flux is a summation of the amount of \vec F that is orthogonal + to the direction of travel. + To capture this orthogonal amount of \vec F, + we use \int_C \vec F \cdot \vec n\, ds to measure flux, + where \vec n is a unit vector orthogonal to the curve C. + (Later, we'll measure flux across surfaces, too. + For example, + in physics it is useful to measure the amount of a magnetic field that passes through a surface.) +

    + + + +

    + How is \vec n determined? + We'll later see that if C is a closed curve, + we'll want \vec n to point to the outside of the curve + (measuring how much is going out). + We'll also adopt the convention that closed curves should be traversed counterclockwise. +

    + +

    + (If C is a complicated closed curve, + it can be difficult to determine what + counterclockwise means. + Consider . + Seeing the curve as a whole, we know which way + counterclockwise is. + If we zoom in on point A, + one might incorrectly choose to traverse the path in the wrong direction. + So we offer this definition: + a closed curve is being traversed counterclockwise if the outside is to the right of the path and the inside is to the left.) +

    + +
    + Determining counterclockwise is not always simple without a good definition + + A closed curve in the plane, in the shape of a cashew or boomerang. + +

    + A set of two-dimensional coordinate axes are plotted, with the origin at the center of the image. + A closed curve is plotted in the plane. + The curve is symmetric about the y axis, + and its shape is like that of a cashew nut (or perhaps a crescent moon, croissant, or boomerang). +

    + +

    + There are intercepts at (1,0), (0,1), (-1,0), + and a fourth point, labeled A, that lies below the point (0,1) on the positive y axis. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ 1,0 ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + + if(incolor) { + colorone = bluepen+.8pt+linejoin(0); + colortwo = redpen+.8pt+linejoin(0); + } else + { + colorone = black+.8pt+linejoin(0); + colortwo = gray+.8pt+linejoin(0); + }; + + pair xbounds=(-1.2,1.2); + pair ybounds=(-1.2,1.2); + real[] myxchoice={-1,1}; + real[] myychoice={1,-1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-.1,-.1); + pair b=(1.1,1.1); + + pair vectfunction(pair z) {return (1,0);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + //add(vectorfield(vector,a,b,5,5,rgb(.6,.6,.6))); + + pair f(real x) {return (cos(x),sin(x))*(1+.5*cos(10x));} + pair F(real x) { + if(x<pi) {return (cos(x),sin(x));} + return (cos(x),-sin(3(x-pi))/3); + } + //real g(real x) {return x^4;} + //pair G(real x) {return (x,g(x));} + + pair G(real t) {return (.2,t);} + + //draw(graph(F,0.2,1,operator ..),colorone,arrow=MidArrow(5)); + draw(graph(F,0,2pi,operator ..),colorone); + //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); + + filldraw(shift(0,.33)*scale(.03)*unitcircle,black); + label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); + //label("$C_2$",(.13,.8),filltype=Fill(white+opacity(.7))); + //label("$C_3$",(.6,.5),filltype=Fill(white+opacity(.7))); + + + +
    + +

    + When a curve C is traversed counterclockwise by \vrt = \la f(t), + g(t)\ra, + we rotate \vec T clockwise 90^\circ to obtain \vec n: + + \vec T = \frac{\la \fp(t),\gp(t)\ra}{\norm{\vrp(t)}} \Rightarrow \vec n = \frac{\la \gp(t),-\fp(t)\ra}{\norm{\vrp(t)}} + . +

    + +

    + Letting \vec F = \la M, N\ra, we calculate flux as: + + \int_C \vec F\cdot \vec n\, ds \amp = \int_C \vec F\cdot \frac{\la \gp(t),-\fp(t)\ra}{\norm{\vrp(t)}} \norm{\vrp(t)}\, dt + \amp = \int_C \la M, N\ra \cdot \la \gp(t),-\fp(t)\ra\, dt + \amp = \int_C \Big(M\,\gp(t) - N\,\fp(t)\Big)dt + \amp = \int_C M\,\gp(t)\, dt - \int_C N\,\fp(t)\, dt. + As the x and y components of \vrt are f(t) and g(t) respectively, the differentials of x and y are dx = \fp(t)dt and dy=\gp(t)dt. We can then write the above integrals as: + \amp = \int_C M\, dy - \int_C N\, dx. + This is often written as one integral (not incorrectly, though somewhat confusingly, as this one integral has two d 's): + \amp =\int_CM\, dy -N\, dx + . +

    + +

    + We summarize the above in the following definition. +

    + + + Flow, Flux + +

    + Let \vec F=\la M,N\ra be a vector field with continuous components defined on a smooth curve C, + parametrized by \vrt =\la f(t),g(t)\ra, + let \vec T be the unit tangent vector of \vrt, + and let \vec n be the clockwise 90^\circdegree rotation of \vec T. + flow + flux + +

      +
    • +

      + The flow of \vec F along C is + + \int_C \vec F\cdot\vec T\, ds=\int_C \vec F\cdot d\vec r + . +

      +
      • + +
    • +

      + The flux of \vec F across C is + + \int_C \vec F\cdot \vec n\, ds = \int_C M\, dy -N\, dx = \int_C\Big(M\,\gp(t) - N\,\fp(t)\Big)dt + . +

      +
      • +
    +

    +     +     + +

    + This definition of flow also holds for curves in space, + though it does not make sense to measure + flux across a curve in space. +

    + +

    + Measuring flow is essentially the same as finding work performed by a force as done in the previous examples. + Therefore we practice finding only flux in the following example. +

    + + + Finding flux across curves in the plane + +

    + Curves C_1 and C_2 each start at (1,0) and end at (0,1), + where C_1 follows the line y=1-x and C_2 follows the unit circle, + as shown in . + Find the flux across both curves for the vector fields + \vec F_1 = \la y, -x+1\ra and \vec F_2 = \la -x, 2y-x\ra. +

    +
    + Illustrating the curves and vector fields in . In (a) the vector field is \vec F_1, and in (b) the vector field is \vec F_2. + +
    + + + Two paths in the plane are plotted relative to a vector field. + +

    + The first quadrant in a two-dimensional coordinate system is shown, + with x and y axes, and the origin at the bottom-left of the image. +

    + +

    + There are two curves plotted; both are paths from the point (1,0) to the point (0,1). + The curve C_1 is a straight line, while C_2 is a quarter of the unit circle. +

    + +

    + A vector field is also shown in the image. + The vectors in the vector field appear to lie tangent to a family of circles centered at the point (1,0). + The vectors indicate a clockwise direction of travel, and their magnitude is smallest near (1,0). + Near the two curves, the vectors all cross the curves in a direction that is up and to the right. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ y,-x+1 ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + pair xbounds=(-.2,1.2); + pair ybounds=(-.2,1.2); + real[] myxchoice={1}; + real[] myychoice={1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-.1,-.1); + pair b=(1.1,1.1); + + pair vectfunction(pair z) {return (z.y,-z.x+1);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,bgvect)); + + pair f(real x) {return (cos(x),sin(x));} + pair F(real x) {return (x, 1-x);} + //real g(real x) {return x^4;} + //pair G(real x) {return (x,g(x));} + + pair G(real t) {return (.2,t);} + + draw(graph(F,1,0,operator ..),colorone,arrow=MidArrow(5)); + draw(graph(f,0,pi/2,operator ..),colortwo,arrow=MidArrow(5)); + //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); + + //filldraw(shift(0,.33)*scale(.03)*unitcircle,black); + //label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); + label("$C_2$",(.75,.77),filltype=Fill(white+opacity(.7))); + label("$C_1$",(.43,.45),filltype=Fill(white+opacity(.7))); + + + +
    + +
    + + + Two paths in the plane are plotted relative to a vector field. + +

    + The first quadrant in a two-dimensional coordinate system is shown, + with x and y axes, and the origin at the bottom-left of the image. +

    + +

    + There are two curves plotted; both are paths from the point (1,0) to the point (0,1). + The curve C_1 is a straight line, while C_2 is a quarter of the unit circle. + (These are same curves as in .) +

    + +

    + A vector field is also shown in the image. + The vectors in the vector field appear to lie tangent to a family of hyperbolas centered at the origin. + Unlike in , the orientations of the vectors relative to the curves change along the curve. + Near (1,0), the vectors are pointing down and to the left, + while near the point (0,1), the vectors are pointing up and to the left. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ -x, 2y-x ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + + pair xbounds=(-.2,1.2); + pair ybounds=(-.2,1.2); + real[] myxchoice={1}; + real[] myychoice={1}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-.1,-.1); + pair b=(1.1,1.1); + + pair vectfunction(pair z) {return (-z.x,2z.y-z.x);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,bgvect)); + + pair f(real x) {return (cos(x),sin(x));} + pair F(real x) {return (x, 1-x);} + //real g(real x) {return x^4;} + //pair G(real x) {return (x,g(x));} + + pair G(real t) {return (.2,t);} + + draw(graph(F,1,0,operator ..),colorone,arrow=MidArrow(5)); + draw(graph(f,0,pi/2,operator ..),colortwo,arrow=MidArrow(5)); + //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); + + //filldraw(shift(0,.33)*scale(.03)*unitcircle,black); + //label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); + label("$C_2$",(.75,.77),filltype=Fill(white+opacity(.7))); + label("$C_1$",(.43,.45),filltype=Fill(white+opacity(.7))); + + + +
    +     + +
    +     + +

    + We begin by finding parametrizations of C_1 and C_2. + As done in , + parametrize C_1 by creating the line that starts at (1,0) and moves in the \la -1,1\ra direction: + \vec r_1(t) = \la 1,0\ra + t\la -1,1\ra = \la 1-t, t\ra, + for 0\leq t\leq 1. + We parametrize C_2 with the familiar + \vec r_2(t) = \la \cos(t),\sin(t)\ra on 0\leq t\leq \pi/2. + For reference later, we give each function and its derivative below: + + \vec r_1(t) = \la 1-t, t\ra, \vrp_1(t) = \la -1,1\ra + . + + \vec r_2(t) = \la \cos(t), \sin(t)\ra, \vrp_2(t) = \la -\sin(t) ,\cos(t)\ra + . +

    + +

    + When \vec F = \vec F_1 = \la y, -x+1\ra (as shown in ), + over C_1 we have M = y =t and N = -x+1 = -(1-t)+1 = t. + Using , we compute the flux: + + \int_{C_1} \vec F\cdot \vec n\, ds \amp = \int_{C_1} \Big(M\,\gp(t) - N\,\fp(t)\Big)\, dt + \amp = \int_0^1 \Big( t(1) - t(-1)\Big)\, dt + \amp = \int_0^1 2t\, dt + \amp = 1 + . +

    + +

    + Over C_2, + we have M = y = \sin(t) and N = -x+1 = 1-\cos(t). + Thus the flux across C_2 is: + + \int_{C_1} \vec F\cdot \vec n\, ds \amp = \int_{C_1} \Big(M\,\gp(t) - N\,\fp(t)\Big)\, dt + \amp = \int_0^{\pi/2}\Big((\sin(t))(\cos(t)) - (1-\cos(t))(-\sin(t))\Big)\, dt + \amp = \int_0^{\pi/2} \sin(t)\, dt + \amp =1 + . +

    + +

    + Notice how the flux was the same across both curves. + This won't hold true when we change the vector field. +

    + +

    + When \vec F = \vec F_2 = \la -x,2y-x\ra (as shown in ), + over C_1 we have M = -x = t-1 and N = 2y-x = 2t-(1-t) = 3t-1. + Computing the flux across C_1: + + \int_{C_1} \vec F\cdot \vec n\, ds \amp = \int_{C_1} \Big(M\,\gp(t) - N\,\fp(t)\Big)\, dt + \amp = \int_0^1 \Big( (t-1)(1) - (3t-1)(-1)\Big)\, dt + \amp = \int_0^1 (4t-2)\, dt + \amp = 0 + . +

    + +

    + Over C_2, + we have M = -x = -\cos(t) and N = 2y-x = 2\sin(t)-\cos(t). + Thus the flux across C_2 is: + + \int_{C_1} \vec F\cdot \vec n\, ds \amp = \int_{C_1} \Big(M\,\gp(t) - N\,\fp(t)\Big)\, dt + \amp = \int_0^{\pi/2}\Big((-\cos(t))(\cos(t)) - (2\sin(t)-\cos(t))(-\sin(t))\Big)\, dt + \amp = \int_0^{\pi/2} \big(2\sin^2 t-\sin(t)\cos(t)-\cos^2t\big)\, dt + \amp =\pi/4 - 1/2\approx 0.285 + . +

    + +

    + We analyze the results of this example below. +

    +     +     + +

    + In , + we saw that the flux across the two curves was the same when the vector field was \vec F_1 = \la y, -x+1\ra. + This is not a coincidence. + We show why they are equal in . + In short, the reason is this: + the divergence of \vec F_1 is 0, and when \divv \vec F = 0, + the flux across any two paths with common beginning and ending points will be the same. +

    + +

    + We also saw in the example that the flux across C_1 was 0 when the field was \vec F_2 = \la -x, 2y-x\ra. + Flux measures how much of the field crosses the path from left to right + (following the conventions established before). + Positive flux means most of the field is crossing from left to right; + negative flux means most of the field is crossing from right to left; + zero flux means the same amount crosses from each side. + When we consider , + it seems plausible that the same amount of + \vec F_2 was crossing C_1 from left to right as from right to left. +

    +     + + + Green's Theorem +

    + There is an important connection between the circulation around a closed region R and the curl of the vector field inside of R, + as well as a connection between the flux across the boundary of R and the divergence of the field inside R. + These connections are described by Green's Theorem and the Divergence Theorem, + respectively. + We'll explore each in turn. + Green's Theorem +

    + + +

    + Green's Theorem states the counterclockwise circulation around a closed region R is equal to the sum of the curls over R. +

    + + + Green's Theorem + +

    + Let R be a closed, + bounded region of the plane whose boundary C is composed of finitely many smooth curves, + let \vec r(t) be a counterclockwise parametrization of C, + and let \vec F =\la M,N\ra where N_x and M_y are continuous over R. + Then + Green's Theorem + + \oint_C \vec F\cdot d\vec r = \iint_R\curl \vec F\, dA + . +

    +     +     + +

    + We'll explore Green's Theorem through an example. +

    + + + Confirming Green's Theorem + +

    + Let \vec F =\la -y,x^2+1\ra and let R be the region of the plane bounded by the triangle with vertices (-1,0), + (1,0) and (0,2), + shown in . + Verify Green's Theorem; that is, + find the circulation of \vec F around the boundary of R and show that is equal to the double integral of \curl \vec F over R. +

    + +
    + The vector field and planar region used in + + A triangular path in the plane is plotted against a two-dimensional vector field. + +

    + A set of coordinate axes in the plane is shown, with the x axis at the bottom of the image, + and the y axis in the center. +

    + +

    + A two-dimensional vector field is plotted. + Near the bottom of the image, the vectors in the field are nearly vertical. + As we move up the image, the vectors rotate; + while they maintain an upward trajectory, the vectors also point more to the left the higher up they are. +

    + +

    + Over the vector field, a triangular path is plotted. + The path is oriented counter-clockwise, + from the point (-1,0), along the x axis to (1,0), + then up to (0,2), and then back down to (-1,0). +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ -y,x^2+1 ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + pair xbounds=(-1.2,1.2); + pair ybounds=(-.2,2.2); + real[] myxchoice={-1,1}; + real[] myychoice={1,2}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-1.1,-.1); + pair b=(1.1,2.1); + + pair vectfunction(pair z) {return (-z.y,z.x^2+1);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,bgvect)); + + //pair f(real x) {return (cos(x),sin(x));} + //real g(real x) {return x^4;} + //pair G(real x) {return (x,g(x));} + + //pair G(real t) {return (.2,t);} + + pair F(real x) {return (x, 0);} + draw(graph(F,-1,1,operator ..),colorone,arrow=MidArrow(5)); + + pair F(real x) {return (x, 2-2x);} + draw(graph(F,1,0,operator ..),colorone,arrow=MidArrow(5)); + + pair F(real x) {return (x, 2+2x);} + draw(graph(F,0,-1,operator ..),colorone,arrow=MidArrow(5)); + + //draw(graph(f,0,pi/2,operator ..),colortwo,arrow=MidArrow(5)); + //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); + + //filldraw(shift(0,.33)*scale(.03)*unitcircle,black); + //label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); + //label("$C_2$",(.75,.77),filltype=Fill(white+opacity(.7))); + label("$R$",(0,.5),filltype=Fill(white+opacity(.7))); + + + +
    +     + +

    + The curve C that bounds R is composed of 3 lines. + While we need to traverse the boundary of R in a counterclockwise fashion, + we may start anywhere we choose. + We arbitrarily choose to start at (-1,0), + move to (1,0), + etc., with each line parametrized by \vec r_1(t), + \vec r_2(t) and \vec r_3(t), respectively. +

    + +

    + We leave it to the reader to confirm that the following parametrizations of the three lines are accurate: +

    + + + + \vec r_1(t) = \la 2t-1,0\ra, + for 0\leq t\leq 1, + with \vrp_1(t) = \la 2,0\ra, + + + \vec r_2(t) = \la 1-t,2t\ra, + for 0\leq t\leq 1, + with \vrp_2(t) = \la -1,2\ra, and + + + \vec r_3(t) = \la -t,2-2t\ra, + for 0\leq t\leq 1, + with \vrp_3(t) = \la -1,-2\ra. + + + +

    + The circulation around C is found by summing the flow along each of the sides of the triangle. + We again leave it to the reader to confirm the following computations: + + \int_{C_1}\vec F\cdot d\vec r_1 \amp = \int_0^1 \la 0,(2t-1)^2+1\ra\cdot \la 2,0\ra dt = 0, + \int_{C_2}\vec F\cdot d\vec r_2 \amp = \int_0^1 \la -2t,(1-t)^2+1\ra\cdot \la -1,2\ra dt = 11/3, \text{ and } + \int_{C_3}\vec F\cdot d\vec r_3 \amp = \int_0^1 \la 2t-2,t^2+1\ra\cdot \la -1,-2\ra dt = -5/3 + . +

    + +

    + The circulation is the sum of the flows: 2. +

    + +

    + We confirm Green's Theorem by computing \iint_R \curl \vec F\, dA. + We find \curl \vec F = 2x+1. + The region R is bounded by the lines y = 2x+2, + y=-2x+2 and y=0. + Integrating with the order dx\, dy is most straightforward, + leading to + + \int_0^2\int_{y/2-1}^{1-y/2} (2x+1)\, dx\, dy = \int_0^2 (2-y)\, dy = 2 + , + which matches our previous measurement of circulation. +

    +     +     + + + + + Using Green's Theorem + +

    + Let \vec F = \la \sin(x),\cos(y)\ra and let R be the region enclosed by the curve C parametrized by + \vec r(t) = \la 2\cos(t)+ \frac1{10}\cos(10t),2\sin(t)+\frac1{10}\sin(10t)\ra on 0\leq t\leq 2\pi, + as shown in . + Find the circulation around C. +

    + +
    + The vector field and planar region used in + + A curve in the plane, like a bumpy circle or flower, and a vector field. + +

    + A coordinate plane with x and y axes, and the origin at the center of the image. +

    + +

    + The vector field is fairly complicated. +

      +
    • +

      + It appears to have two sources at (0,\pi/2) and (0,-\pi/2), where the vector field vanishes. +

      +
      • +
    • +

      + Along the lines y=\pi/2 and y=-\pi/2, the vector field is horizontal, and points away from the y axis. +

      +
      • +
    • +

      + Below y=-\pi/2, the vectors point down and away from the y axis. +

      +
      • +
    • +

      + Above y=\pi/2, the vecors point down and away from the y axis. +

      +
      • +
    • +

      + Between the lines y=\pm \pi/2, the vectors point up and away from the y axis +

      +
      • +
    +

    + +

    + The curve is a bumpy circle, resembling the outline of a flower. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ sin x, cos y ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + + pair xbounds=(-2.8,2.8); + pair ybounds=(-2.8,2.8); + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-2.5,-2.5); + pair b=(2.5,2.5); + + pair vectfunction(pair z) {return (sin(z.x),cos(z.y));} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,bgvect)); + + //pair f(real x) {return (cos(x),sin(x));} + //real g(real x) {return x^4;} + //pair G(real x) {return (x,g(x));} + + //pair G(real t) {return (.2,t);} + + pair F(real x) {return (2*cos(x)+cos(10x)/10,2*sin(x)+sin(10x)/10);} + draw(graph(F,0+.1,2pi+.1,operator ..),colorone,arrow=MidArrow(5)); + + //draw(graph(f,0,pi/2,operator ..),colortwo,arrow=MidArrow(5)); + //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); + + //filldraw(shift(0,.33)*scale(.03)*unitcircle,black); + //label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); + //label("$C_2$",(.75,.77),filltype=Fill(white+opacity(.7))); + label("$R$",(0,.5),filltype=Fill(white+opacity(.7))); + + + +
    +     + +

    + Computing the circulation directly using the line integral looks difficult, + as the integrand will include terms like + \sin\big(2\cos(t) + \frac1{10}\cos(10t)\big). +

    + +

    + Green's Theorem states that \oint_C\vec F\cdot d\vec r = \iint_R \curl\vec F\, dA; + since \curl \vec F = 0 in this example, + the double integral is simply 0 and hence the circulation is 0. +

    + +

    + Since \curl \vec F = 0, + we can conclude that the circulation is 0 in two ways. + One method is to employ Green's Theorem as done above. + The second way is to recognize that \vec F is a conservative field, + hence there is a function f(x,y) wherein \vec F = \nabla f. + Let A be any point on the curve C; + since C is closed, + we can say that C begins + and ends at A. + By the Fundamental Theorem of Line Integrals, + \oint_C \vec F\, d\vec r = f(A)-f(A) = 0. +

    +     +     + + + + + +

    + One can use Green's Theorem to find the area of an enclosed region by integrating along its boundary. + Let C be a closed curve, + enclosing the region R, + parametrized by \vec r(t) = \la f(t),g(t)\ra. + We know the area of R is computed by the double integral \iint_R \, dA, + where the integrand is 1. + By creating a field \vec F where \curl \vec F =1, + we can employ Green's Theorem to compute the area of R as \oint_C \vec F\cdot d\vec r. +

    + +

    + One is free to choose any field \vec F to use as long as \curl\vec F = 1. + Common choices are \vec F = \la 0,x\ra, + \vec F = \la -y,0\ra and \vec F = \la -y/2,x/2\ra. + We demonstrate this below. +

    + + + Using Green's Theorem to find area + +

    + Let C be the closed curve parametrized by + \vrt = \la t-t^3,t^2\ra on -1\leq t\leq 1, + enclosing the region R, + as shown in . + Find the area of R. +

    + +
    + The region R, whose area is found in + + A teardrop-shaped region in the plane. + +

    + The region R is plotted in two dimensions, + with the x axis at the bottom of the image, + and the y axis in the center. +

    + +

    + The region R is a teardrop shape. + It is symmetric about the y axis, + with a cusp at (0,1), and a fairly flat bottom at the origin. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={-1,1}, + ytick={1}, + ymin=-.5,ymax=1.5, + xmin=-1.1,xmax=1.1, + ] + + \addplot+ [domain=-1:1,samples=40] ({x*(x^2-1)},{x^2}); + + \draw (axis cs:.2,.4) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + +
    +     + +

    + We can choose any field \vec F, + as long as \curl \vec F = 1. + We choose \vec F = \la -y,0\ra. + We also confirm + (left to the reader) + that \vrt traverses the region R in a counterclockwise fashion. + Thus + + \text{ Area of \(R\) } \amp = \iint_R\, dA + \amp = \oint_C \vec F\cdot d\vec r + \amp = \int_{-1}^1 \la -t^2,0\ra\cdot \la 1-3t^2,2t\ra\, dt + \amp = \int_{-1}^1 (-t^2)(1-3t^2)\, dt + \amp = \frac8{15} + . +

    +     +     + + + + + + + + +     + + + The Divergence Theorem +

    + Green's Theorem makes a connection between the circulation around a closed region R and the sum of the curls over R. + The Divergence Theorem makes a somewhat + opposite connection: + the total flux across the boundary of R is equal to the sum of the divergences over R. +

    + + + The Divergence Theorem (in the plane) + +

    + Let R be a closed, + bounded region of the plane whose boundary C is composed of finitely many smooth curves, + let \vec r(t) be a counterclockwise parametrization of C, + and let \vec F =\la M,N\ra where M_x and N_y are continuous over R. + ThenDivergence Theoremin the plane + + \oint_C \vec F\cdot \vec n\, ds = \iint_R\divv \vec F\, dA + . +

    +     +     + + + Confirming the Divergence Theorem + +

    + Let \vec F = \la x-y,x+y\ra, + let C be the circle of radius 2 centered at the origin and define R to be the interior of that circle, + as shown in . + Verify the Divergence Theorem; + that is, find the flux across C and show it is equal to the double integral of \divv \vec F over R. +

    + +
    + The region R used in + + A circle in the plane, plotted against a spiral vector field. + +

    + Two-dimensional coordinate axes are plotted, with the origin at the center. + A circle is shown, with its center at the origin, and radius 2. +

    + +

    + The vector field appears to follow paths that spriral outward from the origin. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ x-y,x+y ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + + pair xbounds=(-2.8,2.8); + pair ybounds=(-2.8,2.8); + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + + xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-2.5,-2.5); + pair b=(2.5,2.5); + + pair vectfunction(pair z) {return (z.x-z.y,z.x+z.y);} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,bgvect)); + + +
    +     + +

    + We parametrize the circle in the usual way, with \vrt = +

    + +

    + \la 2\cos(t),2\sin(t)\ra, 0\leq t\leq 2\pi. + The flux across C is + + \oint_C \vec F\cdot \vec n\, ds \amp = \oint_C\big(M\gp(t)-N\fp(t)\big)\, dt + \amp = \int_0^{2\pi} \big((2\cos(t)-2\sin(t))(2\cos(t)) - (2\cos(t)+2\sin(t))(-2\sin(t))\big)\, dt + \amp = \int_0^{2\pi} 4\, dt = 8\pi + . +

    + +

    + We compute the divergence of \vec F as \divv \vec F = M_x+N_y = 2. + Since the divergence is constant, + we can compute the following double integral easily: + + \iint_R \divv \vec F\, dA = \iint_R 2\, dA = 2\iint_R\, dA = 2(\text{ area of \(R\) } ) = 8\pi + , + which matches our previous result. +

    +     +     + + + Flux when <m>\divv \vec F = 0</m> + +

    + Let \vec F be any field where \divv \vec F = 0, + and let C_1 and C_2 be any two nonintersecting paths, + except that each begin at point A and end at point B + (see ). + Show why the flux across C_1 and C_2 is the same. +

    +     + +

    + By referencing , + we see we can make a closed path C that combines C_1 with C_2, + where C_2 is traversed with its opposite orientation. + We label the enclosed region R. + Since \divv \vec F = 0, + the Divergence Theorem states that + + \oint_C \vec F\cdot \vec n\, ds = \iint_R \divv \vec F\, dA = \iint_R 0\, dA = 0 + . +

    + +

    + Using the properties and notation given in , + consider: +

    + +
    + As used in , the vector field has a divergence of 0 and the two paths only intersect at their initial and terminal points. + + Two curves in the plane between points A and B, and a vector field with many vortices. + +

    + Two curves in the plane are plotted without reference to coordinate axes. + Both curves travel from a point A at the bottom right of the image to a point B at the top left. +

    + +

    + The vector field is quite complicated, and appears to circle about many different vortices. +

    +     + + + import graph; + size(282,282); + + // vector field F = ⟨ y+2,-2x+1 ⟩ + + defaultpen(fontsize(8)); + + + pen colorone; + pen colortwo; + pen mainvect; + pen bgvect; + + if(incolor) { + colorone = bluepen+1.5pt+linejoin(0); + colortwo = redpen+1.5pt+linejoin(0); + mainvect = bluepen + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + } else + { + colorone = black+1.5pt+linejoin(0); + colortwo = gray+1.5pt+linejoin(0); + mainvect = black + linejoin(0); + bgvect = rgb(.6,.6,.6)+linejoin(0); + }; + pair xbounds=(-.2,1.2); + pair ybounds=(-.2,1.2); + real[] myxchoice={1}; + real[] myychoice={1}; + + //xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); + //yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); + ////label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); + //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); + ////label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); + + pair a=(-.1,-.1); + pair b=(1.1,1.1); + + pair vectfunction(pair z) {return (cos(pi*3*z.y),sin(pi*3*z.x));} + path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} + + add(vectorfield(vector,a,b,12,12,bgvect)); + + pair f(real x) {return (cos(x)+.05*sin(10x),sin(x)+.05*sin(10x));} + pair F(real x) {return (x, (1-x)^2);} + //real g(real x) {return x^4;} + //pair G(real x) {return (x,g(x));} + + pair G(real t) {return (.2,t);} + + draw(graph(F,1,0,operator ..),colorone,arrow=MidArrow(5)); + draw(graph(f,0,pi/2,operator ..),colortwo,arrow=MidArrow(5)); + //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); + + dot((1,0),black); + dot((0,1),black); + label("$A$",(1,-.05),filltype=Fill(white+opacity(.7))); + label("$B$",(-.05,1),filltype=Fill(white+opacity(.7))); + + //filldraw(shift(0,.33)*scale(.03)*unitcircle,black); + //label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); + label("$C_1$",(.75,.85),filltype=Fill(white+opacity(.7))); + label("$C_2$",(.43,.45),filltype=Fill(white+opacity(.7))); + + + +
    +

    + + 0 \amp = \oint_C \vec F\cdot \vec n\, ds + \amp = \int_{C_1} \vec F\cdot \vec n\, ds +\int_{C_2^*} \vec F\cdot \vec n\, ds + (where C_2^* is the path C_2 traversed with opposite orientation) + \amp = \int_{C_1} \vec F\cdot \vec n\, ds -\int_{C_2} \vec F\cdot \vec n\, ds. + \int_{C_2} \vec F\cdot \vec n\, ds\amp = \int_{C_1} \vec F\cdot \vec n\, ds + . + Thus the flux across each path is equal. +

    +     +     + +

    + In this section, we have investigated flow and flux, + quantities that measure interactions between a vector field and a planar curve. + We can also measure flow along spatial curves, + though as mentioned before, + it does not make sense to measure flux across spatial curves. +

    + +

    + It does, however, + make sense to measure the amount of a vector field that passes across a surface in space i.e, the flux across a surface. + We will study this, + though in the next section we first learn about a more powerful way to describe surfaces than using functions of the form z=f(x,y). +

    +     + + + + Terms and Concepts + + + +

    + Let \vec F be a vector field and let C be a curve. + Flow is a measure of the amount of \vec F going C; + flux is a measure of the amount of \vec F going C. +

    +     + + + + + + + + + + + + +     + + + +

    + What is circulation? +

    +     + + + +

    + It is the measure of flow around the entirety of a closed curve C. +

    +     +     + + + +

    + Green's Theorem states, informally, + that the circulation around a closed curve that bounds a region R is equal to the sum of the + of \vec{F} across R. +

    +     + + + + + + + +     + + + +

    + The Divergence Theorem states, informally, + that the outward flux across a closed curve that bounds a region R + is equal to the sum of the of \vec{F} across R. +

    +     + + + + + + + + +

    + the divergence of \vec F, or \divv \vec F +

    +     +     + + + +

    + Let \vec F be a vector field and let C_1 and C_2 + be any nonintersecting paths except that each starts at point A and ends at point B. + If the of \vec{F} is 0, + then \int_{C_1} \vec F\cdot \vec T\, ds = \int_{C_2} \vec F\cdot \vec T\, ds. +

    +     + + + + + + + +     + + + +

    + Let \vec F be a vector field and let C_1 and C_2 + be any nonintersecting paths except that each starts at point A and ends at point B. + If the of \vec{F} is 0, + then \int_{C_1} \vec F\cdot \vec n\, ds = \int_{C_2} \vec F\cdot \vec n\, ds. +

    +     + + + + + + + +     +     + + + Problems + + +

    + A vector field \vec F and a curve C are given. + Evaluate \int_C \vec F\cdot\vec n\, ds, + the flux of \vec F over C. +

    +     + + + +

    + \vec F = \langle x+y,x-y\rangle; + C is the curve with initial and terminal points (3,-2) and (3,2), + respectively, + parametrized by \vec r(t) = \langle 3t^2,2t\rangle on -1\leq t\leq 1. +

    +     + +

    + 12 +

    +     +     + + + +

    + \vec F = \langle x+y,x-y\rangle; + C is the curve with initial and terminal points (3,-2) and (3,2), + respectively, + parametrized by \vec r(t) = \langle 3,t\rangle on -2\leq t\leq 2. +

    +     + +

    + 12 +

    +     +     + + + +

    + \vec F = \langle x^2,y+1\rangle; + C is line segment from (0,0) to (2,4). +

    +     + +

    + -2/3 +

    +     +     + + + +

    + \vec F = \langle x^2,y+1\rangle; + C is the portion of the parabola y=x^2 from (0,0) to (2,4). +

    +     + +

    + 10/3 +

    +     +     + + + +

    + \vec F = \langle y,0\rangle; + C is the line segment from (0,0) to (0,1). +

    +     + +

    + 1/2 +

    +     +     + + + +

    + \vec F = \langle y,0\rangle; + C is the line segment from (0,0) to (1,1). +

    +     + +

    + 1/2 +

    +     +     +     + + + +

    + A vector field \vec F and a closed curve C, + enclosing a region R, are given. + Verify Green's Theorem by evaluating \oint_C\vec F\cdot d\vec r and + \iint_R \curl \vec F\, dA, showing they are equal. +

    +     + + + +

    + \vec F = \langle x-y,x+y\rangle; + C is the closed curve composed of the parabola y=x^2 on + 0\leq x\leq 2 followed by the line segment from (2,4) to (0,0). +

    +     + +

    + The line integral \oint_C\vec F\cdot d\vec r, + over the parabola, is 38/3; + over the line, it is -10. + The total line integral is thus 38/3-10 = 8/3. + The double integral of \curl \vec F = 2 over R also has value 8/3. +

    +     +     + + + +

    + \vec F = \langle -y,x\rangle; + C is the unit circle. +

    +     + +

    + Both the line integral and double integral have value of 2\pi. +

    +     +     + + + +

    + \vec F = \langle 0,x^2\rangle; + C the triangle with corners at (0,0), + (2,0) and (1,1). +

    +     + +

    + Three line integrals need to be computed to compute \oint_C \vec F\cdot d\vec r. + It does not matter which corner one starts from first, + but be sure to proceed around the triangle in a counterclockwise fashion. +

    + +

    + From (0,0) to (2,0), + the line integral has a value of 0. + From (2,0) to (1,1) the integral has a value of 7/3. + From (1,1) to (0,0) the line integral has a value of -1/3. + Total value is 2. +

    + +

    + The double integral of \curl\vec F over R also has value 2. +

    +     +     + + + +

    + \vec F = \langle x+y,2x\rangle; + C the curve that starts at (0,1), + follows the parabola y=(x-1)^2 to (3,4), + then follows a line back to (0,1). +

    +     + +

    + Two line integrals need to be computed to compute \oint_C \vec F\cdot d\vec r. + Along the parabola, the line integral has value 25.5. + Along the line, the line integral has value -21. + Together, the total value is 4.5 +

    + +

    + The double integral of \curl\vec F over R also has value 4.5. +

    +     +     +     + + + +

    + A closed curve C enclosing a region R is given. + Find the area of R by computing + \oint_C \vec F\cdot d\vec r for an appropriate choice of vector field \vec F. +

    +     + + + +

    + C is the ellipse parametrized by + \vec r(t) = \langle 4\cos(t),3\sin(t)\rangle on 0\leq t\leq 2\pi. +

    +     + +

    + Any choice of \vec F is appropriate as long as \curl \vec F = 1. + When \vec F = \langle -y/2,x/2\rangle, + the integrand of the line integral is simply 6. + The area of R is 12\pi. +

    +     +     + + + +

    + C is the curve parametrized by + \vec r(t) = \langle \cos(t),\sin (2t)\rangle on -\pi/2\leq t\leq \pi/2. +

    +     + +

    + Any choice of \vec F is appropriate as long as \curl \vec F = 1. + The choices of \vec F = \langle -y,0\rangle and + \langle 0,x\rangle each lead to reasonable integrands. + The area of R is 4/3. +

    +     +     + + + +

    + C is the curve parametrized by + \vec r(t) = \langle -t^3+3t^2-2t,2(t-1)^2\rangle on 0\leq t\leq 2. +

    +     + +

    + Any choice of \vec F is appropriate as long as \curl \vec F = 1. + The choices of \vec F = \langle -y,0\rangle, + \langle 0,x\rangle and + \langle -y/2,x/2\rangle each lead to reasonable integrands. + The area of R is 16/15. +

    +     +     + + + +

    + C is the curve parametrized by + \vec r(t) = \langle 2\cos(t)+\frac1{10}\cos(10t),2\sin(t)+\frac1{10}\sin (10t)\rangle on 0\leq t\leq 2\pi. +

    +     + +

    + Any choice of \vec F is appropriate as long as \curl \vec F = 1. + The choice of \vec F = \langle -y/2,x/2\rangle leads to a reasonable integrand after simplification. + The area of R is 41\pi/10. +

    +     +     +     + + + +

    + A vector field \vec F and a closed curve C, + enclosing a region R, are given. + Verify the Divergence Theorem by evaluating \oint_C\vec F\cdot \vec n\, ds and + \iint_R \divv \vec F\, dA, showing they are equal. +

    +     + + + +

    + \vec F = \langle x-y,x+y\rangle; + C is the closed curve composed of the parabola y=x^2 on + 0\leq x\leq 2 followed by the line segment from (2,4) to (0,0). +

    +     + +

    + The line integral \oint_C\vec F\cdot \vec n\, ds, + over the parabola, is -22/3; + over the line, it is 10. + The total line integral is thus -22/3+10 = 8/3. + The double integral of \divv \vec F = 2 over R also has value 8/3. +

    +     +     + + + +

    + \vec F = \langle -y,x\rangle; + C is the unit circle. +

    +     + +

    + Both the line integral and double integral have value of 0. +

    +     +     + + + +

    + \vec F = \langle 0,y^2\rangle; + C the triangle with corners at (0,0), + (2,0) and (1,1). +

    +     + +

    + Three line integrals need to be computed to compute \oint_C \vec F\cdot \vec n\, ds. + It does not matter which corner one starts from first, + but be sure to proceed around the triangle in a counterclockwise fashion. +

    + +

    + From (0,0) to (2,0), + the line integral has a value of 0. + From (2,0) to (1,1) the integral has a value of 1/3. + From (1,1) to (0,0) the line integral has a value of 1/3. + Total value is 2/3. +

    + +

    + The double integral of \divv\vec F over R also has value 2/3. +

    +     +     + + + +

    + \vec F = \langle x^2/2,y^2/2\rangle; + C the curve that starts at (0,1), + follows the parabola y=(x-1)^2 to (3,4), + then follows a line back to (0,1). +

    +     + +

    + Two line integrals need to be computed to compute \oint_C \vec F\cdot \vec n\, ds. + Along the parabola, the line integral has value 159/20. + Along the line, the line integral has value 6. + Together, the total value is 279/20. +

    + +

    + The double integral of \divv\vec F over R also has value 279/20. +

    +     +     +     +     +     +
    +
    + Parametrized Surfaces and Surface Area + +

    + Thus far we have focused mostly on 2-dimensional vector fields, + measuring flow and flux along/across curves in the plane. + Both Green's Theorem and the Divergence Theorem make connections between planar regions and their boundaries. + We now move our attention to 3-dimensional vector fields, + considering both curves and surfaces in space. +

    +     + + + Parametrizing surfaces + + +

    + We are accustomed to describing surfaces as functions of two variables, + usually written as z=f(x,y). + For our coming needs, + this method of describing surfaces will prove to be insufficient. + Instead, we will parametrize our surfaces, + describing them as the set of terminal points of some vector-valued function \vec r(u,v) =\langle f(u,v),g(u,v),h(u,v)\rangle. + The bulk of this section is spent practicing the skill of describing a surface \surfaceSusing a vector-valued function. + Once this skill is developed, + we'll show how to find the surface area S of a parametrically-defined surface \surfaceS, a skill needed in the remaining sections of this chapter. +

    + + + + Parametrized Surface + +

    + Let \vec r(u,v) = \langle\, f(u,v),g(u,v),h(u,v)\rangle + be a vector-valued function that is continuous and one to one on the interior of its domain R in the u-v plane. + The set of all terminal points of \vec r (, the range + of \vec r ) is the surface + \surfaceS, and \vec r along with its domain R form a + parametrization of \surfaceS. +

    + +

    + This parametrization is smooth + on R if \vec r_u and \vec r_v are continuous and + \vec r_u\times \vec r_v is never \vec 0 on the interior of R. + surface + parametric equationsof a surface + parametrized surface + smoothsurface + surfacesmooth +

    +     +     + +

    + Given a point (u_0,v_0) in the domain of a vector-valued function \vec r, + the vectors \vec r_u(u_0,v_0) and + \vec r_v(u_0,v_0) are tangent to the surface \surfaceS at + \vec r(u_0,v_0) (a proof of this is developed later in this section). + The definition of smoothness dictates that \vec r_u\times \vec r_v \neq \vec 0; + this ensures that neither \vec r_u nor \vec r_v are \vec 0, + nor are they ever parallel. + Therefore smoothness guarantees that \vec r_u and + \vec r_v determine a plane that is tangent to \surfaceS. +

    + + + +

    + A surface \surfaceS is said to be orientable + orientable + if a field of normal vectors can be defined on \surfaceS that vary continuously along \surfaceS. This definition may be hard to understand; + it may help to know that orientable surfaces are often called two sided. + A sphere is an orientable surface, + and one can easily envision an inside + and outside of the sphere. + A paraboloid is orientable, + where again one can generally envision + inside and outside sides + (or top and bottom sides) + to this surface. + Just about every surface that one can imagine is orientable, + and we'll assume all surfaces we deal with in this text are orientable. +

    + +

    + Möbius band + + It is enlightening to examine a classic non-orientable surface: + the Möbius band, shown in . + Vectors normal to the surface are given, + starting at the point indicated in the figure. + These normal vectors vary continuously + as they move along the surface. + Letting each vector indicate the + top side of the band, + we can easily see near any vector which side is the top. +

    + +

    + However, if as we progress along the band, + we recognize that we are labeling + both sides of the band as the top; + in fact, there are not two sides to this band, but one. + The Möbius band is a non-orientable surface. +

    + +
    + A Möbius band, a non-orientable surface + + A depction of a Möbius band, showing how the normal vector changes direction while traveling around the surface. + +

    + A Möbius band is the surface one obtains by taking a strip of paper, + giving it a half-twist, and gluing the ends of the strip together. + It is a famous example of a surface with only one side. +

    + +

    + This is illustrated in the image by plotting a collection of normal vectors along the central curve of the surface. + (This is the curve obtained if you draw a line down the middle of your strip of paper before connecting the two ends.) + The normal vector begins its trip pointing up, + but as it travels around the curve it rotates, and ends up pointing down when we return to the starting point of the curve. +

    + +

    + This illustrates that it is impossible to assign a consistent choice of normal vector across the surface, + so that the Möbius band is non-orientable. +

    +     + + + + + //ASY file for fig3d_proj3D.asy in Chapter 13 + + size(282,282,Aspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(6.7,-2,1.9); + defaultrender.merge=true; + + // setup and draw the axes + //real[] myxchoice={2}; + //real[] myychoice={1,2}; + //real[] myzchoice={2,4}; + //defaultpen(0.5mm); + + //pair xbounds=(-0.5,2.5); + //pair ybounds=(-0.5,2.25); + //pair zbounds=(-0.5,4.5); + + //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //plane z=2y + triple f(pair t) { + return ((1+t.y/2*cos(t.x/2))*cos(t.x),(1+t.y/2*cos(t.x/2))*sin(t.x),t.y/2*sin(t.x/2)); + } + surface s=surface(f,(0,-.3),(2*pi,.3),50,5); + pen p=bluepen+.1mm; + draw(s,simplesurfacepen,meshpen=p); + + triple g(real t) { + return f((t,.3)); + } + + path3 mypath=graph(g,-2pi,2pi,operator ..); + draw(mypath,bluepen+linewidth(1)); + + // + // Attempted various ways of getting a normal vector. Turns out all + // were probably ok, it was just that the aspect ratio was off. + // Used ``Aspect'' in the orthographic line above to fix it. + // + + triple n(real t) { + return (.5*cos(t)*sin(t/2),(1/2)*(sin(t/2)*sin(t)),-.5*cos(t/2)); + } + //(.5*cos(t)*sin(t/2),(1/4)*(cos(t/2)-cos(3*t/2)),-.5*cos(t/2)) + + //for(real i=0; i<=2*pi;i=i+2pi/10) { + //draw(f((i,0.)) -- (f((i,0.))+n(i)),rgb(1-i/(2pi),0,i/(2pi))+.2mm,Arrow3); + //} + + //draw((f((0,0)).x,0,0) -- (1,0,-.5),redpen+.2mm,Arrow3); + //draw((0.309017, 0.951057, 0.) -- (0.399835, 1.23057, -0.404508),rgb(.9,0,.1)+.2mm,Arrow3); + //draw((-0.809017, 0.587785, 0.) -- (-1.19373, 0.867294, -0.154508),rgb(.5,0,.5)+.2mm,Arrow3); + //draw((0.309017, -0.951057, 0.) -- (0.399835, -1.23057, 0.404508),rgb(.1,0,.9)+.2mm,Arrow3); + + // + // Final method of drawing normal vectors. + // + + if(incolor) { + for(real i=0; i<=2pi; i=i+2pi/20) { + triple a = f((i+.1,0))-f((i-.1,0)); + triple b = f((i,.1))-f((i,-.1)); + triple c = .3*(a.y*b.z-a.z*b.y,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x)/sqrt(((a.y*b.z-a.z*b.y)^2+(a.z*b.x-a.x*b.z)^2+(a.x*b.y-a.y*b.x)^2)); + draw(f((i,0)) -- (f((i,0))+c),rgb(1-i/(2pi),0,i/(2pi))+.2mm,Arrow3); + } + } else { + for(real i=0; i<=2pi; i=i+2pi/20) { + triple a = f((i+.1,0))-f((i-.1,0)); + triple b = f((i,.1))-f((i,-.1)); + triple c = .3*(a.y*b.z-a.z*b.y,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x)/sqrt(((a.y*b.z-a.z*b.y)^2+(a.z*b.x-a.x*b.z)^2+(a.x*b.y-a.y*b.x)^2)); + draw(f((i,0)) -- (f((i,0))+c),rgb(.75*i/(2pi),.75*i/(2pi),.75*i/(2pi))+.2mm,Arrow3); + } + } + + triple a = f((0+.1,0))-f((0-.1,0)); + triple b = f((0,.1))-f((0,-.1)); + triple c = .3*(a.y*b.z-a.z*b.y,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x)/sqrt(((a.y*b.z-a.z*b.y)^2+(a.z*b.x-a.x*b.z)^2+(a.x*b.y-a.y*b.x)^2)); + label("end",(f((0,0))+c-(0,0,.1))); + + triple a = f((2pi+.1,0))-f((2pi-.1,0)); + triple b = f((2pi,.1))-f((2pi,-.1)); + triple c = .3*(a.y*b.z-a.z*b.y,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x)/sqrt(((a.y*b.z-a.z*b.y)^2+(a.z*b.x-a.x*b.z)^2+(a.x*b.y-a.y*b.x)^2)); + label("start",(f((2pi,0))+c+(0,0,.1))); + + + +
    + +

    + We now practice parametrizing surfaces. +

    + + + Parametrizing a surface over a rectangle + +

    + Parametrize the surface z=x^2+2y^2 over the rectangular region R defined by + -3\leq x\leq 3, -1\leq y\leq 1. +

    +     + +

    + There is a straightforward way to parametrize a surface of the form + z=f(x,y) over a rectangular domain. + We let x=u and y=v, + and let \vec r(u,v) = \langle u,v, f(u,v)\rangle. + In this instance, + we have \vec r(u,v) = \langle u,v,u^2+2v^2\rangle, + for -3\leq u\leq 3, -1\leq v\leq 1. + This surface is graphed in . +

    + +
    + The surface parametrized in + + A portion of an elliptic paraboloid with a rectangular domain. + +

    + A set of three-dimensional coordinate axes are shown, with the z axis pointing up. + In the xy plane a shaded rectangle is plotted; this represents the parameter domain. + Sitting above the rectangle is a portion of an elliptic paraboloid, opening upward. + The appearance of the surface is similar to that of a hammock or sling. +

    +     + + + + + //ASY file for figparsurf1_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11.9,9.6,17.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-3,3}; + real[] myychoice={-3,3}; + real[] myzchoice={5,10}; + defaultpen(0.5mm); + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-1,12); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the top half of the surface z^2 = x^2+2y^2 + triple f(pair t) { + return (t.x,t.y,t.x^2+2*t.y^2);// + } + surface s=surface(f,(-3,-1),(3,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //triple g(real t) {return(cos(t),sin(t),t/(2*pi));} + //path3 mypath=graph(g,0,2pi,operator ..); + draw(surface((3,1,0)--(3,-1,0)--(-3,-1,0)--(-3,1,0)--cycle),curvepen+opacity(.5)); + draw((3,1,0)--(3,-1,0)--(-3,-1,0)--(-3,1,0)--cycle,curvepen); + + + +
    +     +     + + + Parametrizing a surface over a circular disk + +

    + Parametrize the surface z=x^2+2y^2 over the circular region R enclosed by the circle of radius 2 that is centered at the origin. +

    +     + +

    + We can parametrize the circular boundary of R with the vector-valued function \la 2\cos(u),2\sin(u)\ra, + where 0\leq u\leq 2\pi. + We can obtain the interior of R by scaling this function by a variable amount, + , by multiplying by v: + \la 2v\cos(u),2v\sin(u)\ra, + where 0\leq v\leq 1. +

    + +

    + It is important to understand the role of v in the above function. + When v=1, + we get the boundary of R, a circle of radius 2. + When v=0, we simply get the point (0,0), + the center of R + (which can be thought of as a circle with radius of 0). + When v=1/2, + we get the circle of radius 1 that is centered at the origin, + which is the circle halfway + between the boundary and the center. + As v varies from 0 to 1, we create a series of concentric circles that fill out all of R. +

    + +
    + The surface parametrized in + + An elliptic paraboloid plotted over a circular domain. + +

    + A set of three-dimensional coordinate axes are shown, with the z axis pointing up. + In the xy plane a shaded circle is plotted, with its center at the origin; this represents the parameter domain. + Sitting above the circle is a portion of an elliptic paraboloid, opening upward. + Because the domain is circular, the surface appears more bowl-shaped than in the previous example. +

    +     + + + + + //ASY file for figparsurf1_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11.9,9.6,17.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-3,3}; + real[] myychoice={-3,3}; + real[] myzchoice={5,10}; + defaultpen(0.5mm); + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-1,12); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + + //Draw the top half of the surface z^2 = x^2+2y^2 + triple f(pair t) { + return (2*t.y*cos(t.x),2*t.y*sin(t.x),(2*t.y*cos(t.x))^2+2*(2*t.y*sin(t.x))^2);// + } + surface s=surface(f,(0,0),(2pi,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + triple g(real t) { + return (2*cos(t),2*sin(t),0); + } + + //triple g(real t) {return(cos(t),sin(t),t/(2*pi));} + path3 mypath=graph(g,0,2pi,operator ..); + draw(surface(mypath--cycle),curvepen+opacity(.5)); + draw(mypath,curvepen); + + + + +
    + +

    + Thus far, we have determined the x and y components of our parametrization of the surface: + x=2v\cos(u) and y=2v\sin(u). + We find the z component simply by using z = f(x,y) = x^2+2y^2: + + z = (2v\cos(u))^2+2(2v\sin(u))^2 = 4v^2\cos^2u+8v^2\sin^2u + . +

    + +

    + Thus \vec r(u,v) = \langle 2v\cos(u),2v\sin(u),4v^2\cos^2u+8v^2\sin^2u\rangle, + 0\leq u\leq 2\pi, 0\leq v\leq 1, + which is graphed in . + The way that this graphic was generated highlights how the surface was parametrized. + When viewing from above, one can see lines emanating from the origin; + they represent different values of u as u sweeps from an angle of 0 up to 2\pi. + One can also see concentric circles, + each corresponding to a different value of v. +

    +     +     + +

    + Examples + and + demonstrate an important principle when parametrizing surfaces given in the form + z=f(x,y) over a region R: + if one can determine x and y in terms of u and v, + then z follows directly as z=f(x,y). +

    + +

    + In the following two examples, + we parametrize the same surface over triangular regions. + Each will use v as a scaling factor + as done in . +

    + + + Parametrizing a surface over a triangle + +

    + Parametrize the surface z=x^2+2y^2 over the triangular region R enclosed by the coordinate axes and the line y=2-2x/3, + as shown in . +

    +
    + Part (a) shows a graph of the region R, and part (b) shows the surface over R, as defined in + +
    + + + A triangular domain is plotted in the first quadrant of the plane. + +

    + The first quadrant in the plane is shown, with the x axis at the bottom of the image, and the y axis to the left. + A triangular region is shaded and labeled as R. + It is a right triangle, with its base along the x axis, from (0,0) to (3,0), + and another side along the y axis, from (0,0) to (0,2). +

    + +

    + The hypotenuse, from (0,2) to (3,0), is labeled with the equation y=2-2x/3. + A dashed line is also drawn through the region, from (0,1) to (3,0). + This illustrates the parameter value v=\frac12. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,2,3}, + ytick={1,2,3}, + ymin=-.5,ymax=3.5, + xmin=-.5,xmax=3.5, + ] + + \filldraw [thick,draw=firstcolor,fill=firstcolor!15] (axis cs: 0,0) -- (axis cs: 0,2) -- (axis cs: 3,0) node [pos=.5,sloped,above] { $y = 2-2x/3$} -- cycle; + + \draw [thick,dashed,draw=secondcolor] (axis cs: 3,0) -- (axis cs: 0,1); + \draw (axis cs: .5,.5) node {$R$}; + + \end{axis} + + \end{tikzpicture} + + + +
    + +
    + + + An elliptic paraboloid is plotted in space over a triangular domain. + +

    + A set of three-dimensional coordinate axes are drawn, with the z axis pointing up. + The surface z=x^2+2y^2 is plotted over the triangular domain illustrated in . + This is the same surface as the previous two examples, plotted again with a differently-shaped domain. + The domain, as usual, is rendered in the xy plane. +

    + +

    + The shape of the surface is hard to tell from the default perspective of the image. + Rotating the image, so that the viewpoint is from above, + shows that the surface has a cusp at the origin, which is one corner of the domain, + and the surface rises up from there into the first octant. +

    + +

    + The sides of the domain along the x and y axes lead to parabolic curves in the xz and yz planes. + These curves rise to cusps at the other corners of the domain, + and they are joined by a parabolic curve in space that corresponds to the hypotenuse of the triangle. +

    +     + + + + + //ASY file for figparsurf1_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11.9,9.6,17.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-3,3}; + real[] myychoice={-3,3}; + real[] myzchoice={5,10}; + defaultpen(0.5mm); + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-1,12); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the top half of the surface z^2 = x^2+2y^2 + triple f(pair t) { + return (t.x,t.y*(2-.666*t.x),(t.x)^2+2*(t.y*(2-.666*t.x))^2);// + } + surface s=surface(f,(0,0),(3,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //triple g(real t) {return(cos(t),sin(t),t/(2*pi));} + //path3 mypath=graph(g,0,2pi,operator ..); + draw(surface((0,0,0)--(0,2,0)--(3,0,0)--cycle),curvepen+opacity(.5)); + draw((0,0,0)--(0,2,0)--(3,0,0)--cycle,curvepen); + + + +
    +     +
    +     + +

    + We may begin by letting x=u, + 0\leq u\leq 3, and y = 2-2u/3. + This gives only the line on the + upper side of the triangle. + To get all of the region R, + we can once again scale y by a variable factor, + v. +

    + +

    + Still letting x = u, + 0\leq u\leq 3, we let + y = v(2-2u/3), 0\leq v\leq 1. + When v=0, + all y-values are 0, and we get the portion of the x-axis between x=0 and x=3. + When v=1, we get the upper side of the triangle. + When v=1/2, we get the line y=1/2(2-2u/3) = 1-u/3, + which is the line halfway up the triangle, + shown in the figure with a dashed line. +

    + +

    + Letting z = f(x,y) = x^2+2y^2, + we have \vec r(u,v) = \langle u, v(2-2u/3), + u^2+2\big(v(2-2u/3)\big)^2\rangle, + 0\leq u\leq 3, 0\leq v\leq 1. + This surface is graphed in . + Again, when one looks from above, + we can see the scaling effects of v: + the series of lines that run to the point (3,0) each represent a different value of v. +

    + +

    + Another common way to parametrize the surface is to begin with y=u, + 0\leq u\leq 2. + Solving the equation of the line y=2-2x/3 for x, + we have x = 3-3y/2, + leading to using x=v(3-3u/2), 0\leq v\leq 1. + With z=x^2+2y^2, + we have \vec r(u,v) = \langle v(3-3u/2),u, \big(v(3-3u/2)\big)^2+2v^2\rangle, + 0\leq u\leq 2, 0\leq v\leq 1. +

    +     +     + + + Parametrizing a surface over a triangle + +

    + Parametrize the surface z=x^2+2y^2 over the triangular region R enclosed by the lines y=3-2x/3, + y=1 and x=0 as shown in . +

    +
    + Part (a) shows a graph of the region R, and part (b) shows the surface over R, as defined in + +
    + + + A triangular domain in the plane, with vertices at (0,1), (0,3), and (3,1). + +

    + A set of two-dimensional coordinate axes is shown, with the origin in the bottom-left of the image. + A region R is plotted as a shaded rectangle. + The rectangle has vertices (0,1), (3,1), and (0,3). + The hypotenuse of the triangle is labeled with the equation y=3-2x/3. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick={1,2,3}, + ytick={1,2,3}, + ymin=-.5,ymax=3.5, + xmin=-.5,xmax=3.5, + ] + + \filldraw [thick,draw=firstcolor,fill=firstcolor!15] (axis cs: 0,1) -- (axis cs: 0,3) -- (axis cs: 3,1) node [pos=.5,sloped,above] { $y = 3-2x/3$} -- cycle; + + \draw (axis cs: 1,1.75) node {$R$}; + + \end{axis} + \end{tikzpicture} + + + +
    + +
    + + + An elliptic paraboloid is plotted in space over a triangular domain. + +

    + A set of three-dimensional coordinate axes are drawn, with the z axis pointing up. + The surface z=x^2+2y^2 is plotted over the triangular domain illustrated in . + This is the same surface as the previous three examples, plotted again with a different domain. + The domain, as usual, is rendered in the xy plane. + The image is very similar to the one in , + with the bottom cusp on the surface at the point (0,1,2) instead of the origin. +

    + +

    + In the default view, the surface looks something like a triangular tortilla chip. +

    + +

    + The sides of the domain along the x and y axes lead to parabolic curves in the planes x=0 and y=1. + These curves rise to cusps at the other corners of the domain, + and they are joined by a parabolic curve in space that corresponds to the hypotenuse of the triangle. +

    +     + + + + + //ASY file for figparsurf4_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11.9,9.6,17.8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-3,3}; + real[] myychoice={-3,3}; + real[] myzchoice={5,10}; + defaultpen(0.5mm); + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-1,12); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the top half of the surface z^2 = x^2+2y^2 + triple f(pair t) { + return (t.x,1+t.y*(2-.666*t.x),(t.x)^2+2*(1+t.y*(2-.666*t.x))^2);// + } + surface s=surface(f,(0,0),(3,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //triple g(real t) {return(cos(t),sin(t),t/(2*pi));} + //path3 mypath=graph(g,0,2pi,operator ..); + draw(surface((0,1,0)--(0,3,0)--(3,1,0)--cycle),curvepen+opacity(.5)); + draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + + +
    +     + +
    +     + +

    + While the region R in this example is very similar to the region R in the previous example, + and our method of parametrizing the surface is fundamentally the same, + it will feel as though our answer is much different than before. +

    + +

    + We begin with letting x=u, 0\leq u\leq 3. + We may be tempted to let y = v(3-2u/3), + 0\leq v\leq 1, but this is incorrect. + When v = 1, + we obtain the upper line of the triangle as desired. + However, when v=0, + the y-value is 0, which does not lie in the region R. +

    + +

    + We will describe the general method of proceeding following this example. + For now, consider y = 1+v(2-2u/3), 0\leq v\leq 1. + Note that when v=1, + we have y=3-2u/3, the upper line of the boundary of R. + Also, when v=0, we have y=1, + which is the lower boundary of R. + With z=x^2+2y^2, + we determine \vec r(u,v) = \langle u, 1+v(2-2u/3), + u^2+2\big(1+v(2-2u/3)\big)^2\rangle, + 0\leq u\leq 3, 0\leq v\leq 1. +

    + +

    + The surface is graphed in . +

    +     +     + +

    + Given a surface of the form z=f(x,y), + one can often determine a parametrization of the surface over a region R in a manner similar to determining bounds of integration over a region R. + Using the techniques of , + suppose a region R can be described by a\leq x\leq b, + g_1(x) \leq y\leq g_2(x), + , the area of R can be found using the iterated integral + + \int_a^b\int_{g_1(x)}^{g_2(x)}\, dy\, dx + . +

    + +

    + When parametrizing the surface, + we can let x=u, a\leq u\leq b, + and we can let y = g_1(u)+v\big(g_2(u)-g_1(u)\big), + 0\leq v\leq 1. + The parametrization of x is straightforward, + but look closely at how y is determined. + When v=0, y=g_1(u) = g_1(x). + When v=1, y= g_2(u)=g_2(x). +

    + +

    + As a specific example, + consider the triangular region R from , + shown in . + Using the techniques of , + we can find the area of R as + + \int_0^3\int_1^{3-2x/3} dy\, dx + . +

    + +

    + Following the above discussion, + we can set x=u, where 0\leq u\leq 3, + and set y = 1+ v\big(3-2u/3-1\big) = 1+v(2-2u/3), + 0\leq v\leq 1, as used in that example. +

    + +

    + One can do a similar thing if R is bounded by c\leq y\leq d, + h_1(y)\leq x\leq h_2(y), + but for the sake of simplicity we leave it to the reader to flesh out those details. + The principles outlined above are given in the following Key Idea for reference. +

    + + + Parametrizing Surfaces +

    + Let a surface \surfaceS be the graph of a function f(x,y), + where the domain of f is a closed, + bounded region R in the xy-plane. + Let R be bounded by a\leq x\leq b, + g_1(x)\leq y\leq g_2(x), + , the area of R can be found using the iterated integral \int_a^b\int_{g_1(x)}^{g_2(x)}\, dy\, dx, + and let h(u,v) = g_1(u)+v\big(g_2(u)-g_1(u)\big). +

    + +

    + \surfaceS can be parametrized as + + \vec r(u,v) = \la u, h(u,v), f\big(u,h(u,v)\big)\ra, a\leq u\leq b,\ 0\leq v\leq 1 + . +

    +     + + + Parametrizing a cylindrical surface + +

    + Find a parametrization of the cylinder x^2 + z^2/4=1, + where -1\leq y\leq 2, + as shown in . +

    +
    + The cylinder parametrized in + + A cylindrical surface centered along the y axis. + +

    + The surface is an elliptical cylinder. It is plotted relative to a set of three-dimensional coordinate axes, + with the y axis running down the center of the cylinder. +

    +     + + + + + //ASY file for figparsurf5_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(20.4,10,4.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-2.5,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the cylinder x^2+z^2/4=1 + triple f(pair t) { + return (cos(t.x),t.y,2sin(t.x));// + } + surface s=surface(f,(0,-1),(2pi,2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return(cos(t),-1,2sin(t));} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,curvepen); + + triple g(real t) {return(cos(t),2,2sin(t));} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,curvepen); + + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + + +
    +     + +

    + The equation x^2+z^2/4=1 can be envisioned to describe an ellipse in the xz-plane; + as the equation lacks a y-term, + the equation describes a cylinder + (recall ) + that extends without bound parallel to the y-axis. + This ellipse has a vertical major axis of length 4, a horizontal minor axis of length 2, and is centered at the origin. + We can parametrize this ellipse using sines and cosines; + our parametrization can begin with + + \vec r(u,v) = \la \cos(u), \text{ ??? } , 2\sin(u)\ra, 0\leq u\leq 2\pi + , + where we still need to determine the y component. +

    + +

    + While the cylinder x^2+z^2/4=1 is satisfied by any y value, + the problem states that all y values are to be between y=-1 and y=2. + Since the value of y does not depend at all on the values of x or z, + we can use another variable, v, to describe y. + Our final answer is + + \vec r(u,v) = \la \cos(u), v, 2\sin(u)\ra, 0\leq u\leq 2\pi, -1\leq v\leq 2 + . +

    +     +     + + + Parametrizing an elliptic cone + +

    + Find a parametrization of the elliptic cone z^2 = \frac{x^2}{4}+\frac{y^2}{9}, + where -2\leq z\leq 3, + as shown in . +

    +
    + The elliptic cone as described in + + An elliptic cone with its cusp at the origin, including portions above and below the x,y plane. + +

    + An elliptic cone is plotted relative to three-dimensional coordinate axes. + The cone is centered along the z axis, and has its cusp at the origin. + Portions of both halves of the cone (both above and below the xy plane) are included. +

    +     + + + + + //ASY file for figparsurf6a_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(20.4,10,4.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-10,-5,5,10}; + real[] myychoice={-10,-5,5,10}; + real[] myzchoice={-3,3}; + defaultpen(0.5mm); + + pair xbounds=(-10.5,10.5); + pair ybounds=(-10.5,10.5); + pair zbounds=(-3.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the cone z^2=x^2/4+y^2/9 + triple f(pair t) { + return (2*t.y*cos(t.x),3*t.y*sin(t.x),t.y);// + } + surface s=surface(f,(0,-2),(2pi,3),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return(-4cos(t),-6sin(t),-2);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,curvepen); + + triple g(real t) {return(6cos(t),9sin(t),3);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,curvepen); + + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + + +
    +     + +

    + One way to parametrize this cone is to recognize that given a z value, + the cross section of the cone at that z value is an ellipse with equation \frac{x^2}{(2z)^2} + \frac{y^2}{(3z)^2}=1. + We can let z=v, for + -2\leq v\leq 3 and then parametrize the above ellipses using sines, + cosines and v. +

    + +

    + We can parametrize the x component of our surface with + x=2z\cos(u) and the y component with y=3z\sin(u), + where 0\leq u\leq 2\pi. + Putting all components together, we have + + \vec r(u,v) = \la 2v\cos(u), 3v\sin(u), v\ra, 0\leq u\leq 2\pi, -2\leq v\leq 3 + . +

    + +

    + When v takes on negative values, + the radii of the cross-sectional ellipses become negative, + which can lead to some surprising results. + Consider , + where the cone is graphed for 0\leq u\leq \pi. + Because v is negative below the xy-plane, + the radii of the cross-sectional ellipses are negative, + and the opposite side of the cone is sketched below the xy-plane. +

    + +
    + The elliptic cone as described in with restricted domain + + A cone plotted parametrically with restricted domain. + +

    + A plot of part of the cone illustrated in . + This image shows the effect of both the restricted domain, + and the fact that the cross-sectional ellipses have negative radius when z\lt 0. +

    + +

    + With 0\leq u\leq \pi, the result is that above the xy plane, + we get the half of the cone where y\geq 0, + while below the xy plane, we get the half of the cone where y\leq 0. +

    +     + + + + + //ASY file for figparsurf6a_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(20.4,10,4.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-10,-5,5,10}; + real[] myychoice={-10,-5,5,10}; + real[] myzchoice={-3,3}; + defaultpen(0.5mm); + + pair xbounds=(-10.5,10.5); + pair ybounds=(-10.5,10.5); + pair zbounds=(-3.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the cone z^2=x^2/4+y^2/9 + triple f(pair t) { + return (2*t.y*cos(t.x),3*t.y*sin(t.x),t.y);// + } + surface s=surface(f,(0,-2),(pi,3),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return(-4cos(t),-6sin(t),-2);} + path3 mypath=graph(g,0,pi,operator ..); + draw(mypath,curvepen); + + triple g(real t) {return(6cos(t),9sin(t),3);} + path3 mypath=graph(g,0,pi,operator ..); + draw(mypath,curvepen); + + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + + +
    +     +     + + + Parametrizing an ellipsoid + +

    + Find a parametrization of the ellipsoid + \frac{x^2}{25}+y^2+\frac{z^2}{4}=1 as shown in . +

    +
    + An ellipsoid in (a), drawn again in (b) with its domain restricted, as described in + +
    + + + The ellipsoid to be parametrized in this example. + +

    + A set of three-dimensional coordinate axes are shown, with the origin in the center of the image. + The ellipsoid has the shape of a smooth, rounded pebble. + It is narrowest in the y direction, and longest in the x direction. +

    +     + + + + + //ASY file for figparsurf7a_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(20.4,10,4.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-5,5}; + real[] myychoice={-1,1}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + + pair xbounds=(-6,6); + pair ybounds=(-3,3); + pair zbounds=(-3,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the ellipoid x^2/25+y^2+z^2/4=1 + triple f(pair t) { + return (5*cos(t.y)*sin(t.x),sin(t.y)*sin(t.x),2cos(t.x));// + } + surface s=surface(f,(0,0),(pi,2pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + + +
    + +
    + + + A portion of an ellipsoid in space corresponding to a restricted parameter domain. + +

    + A cutout of part of the ellipsoid in . + It is a U-shaped surface obtained by cutting off the top and bottom of the ellipsoid, + as well as a good part of the end that meets the positive x axis. +

    +     + + + + + //ASY file for figparsurf7b_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(20.4,10,4.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-5,5}; + real[] myychoice={-1,1}; + real[] myzchoice={-2,2}; + defaultpen(0.5mm); + + pair xbounds=(-6,6); + pair ybounds=(-3,3); + pair zbounds=(-3,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the ellipoid x^2/25+y^2+z^2/4=1, restricted domain + triple f(pair t) { + return (5*cos(t.y)*sin(t.x),sin(t.y)*sin(t.x),2cos(t.x));// + } + surface s=surface(f,(pi/4,pi/4),(2pi/3,3pi/2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + + +
    +     + +
    +     + +

    + Recall + from , + which states that all unit vectors in space have the form + \langle \sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta\rangle for some angles \theta and \varphi. + If we choose our angles appropriately, + this allows us to draw the unit sphere. + To get an ellipsoid, + we need only scale each component of the sphere appropriately. +

    + +

    + The x-radius of the given ellipsoid is 5, the y-radius is 1 and the z-radius is 2. + Substituting u for \theta and v for \varphi, + we have + + \vec r(u,v) = \langle 5\sin(u)\cos(v), \sin(u)\sin(v),2\cos(u)\rangle + , + where we still need to determine the ranges of u and v. +

    + +

    + Note how the x and y components of \vec r have \cos(v) and \sin(v) terms, + respectively. + This hints at the fact that ellipses are drawn parallel to the xy-plane as v varies, + which implies we should have v range from 0 to 2\pi. +

    + +

    + One may be tempted to let 0\leq u\leq 2\pi as well, + but note how the z component is 2\cos(u). + We only need \cos(u) to take on values between -1 and 1 once, + therefore we can restrict u to 0\leq u\leq \pi. +

    + +

    + The final parametrization is thus + + \vec r(u,v) = \langle 5\sin(u)\cos(v), \sin(u)\sin(v),2\cos(u)\rangle, 0\leq u\leq\pi, 0\leq v\leq 2\pi + . +

    + +

    + In , + the ellipsoid is graphed on \frac{\pi}{4}\leq u\leq \frac{2\pi}{3}, + \frac{\pi}4\leq v\leq \frac{3\pi}2 to demonstrate how each variable affects the surface. +

    +     +     + +

    + Parametrization is a powerful way to represent surfaces. + One of the advantages of the methods of parametrization described in this section is that the domain of + \vec r(u,v) is always a rectangle; + that is, the bounds on u and v are constants. + This will make some of our future computations easier to evaluate. +

    + +

    + Just as we could parametrize curves in more than one way, + there will always be multiple ways to parametrize a surface. + Some ways will be more natural than others, + but these other ways are not incorrect. + Because technology is often readily available, + it is often a good idea to check one's work by graphing a parametrization of a surface to check if it indeed represents what it was intended to. +

    +     + + + Surface Area + +

    + It will become important in the following sections to be able to compute the surface area of a surface \surfaceS given a smooth parametrization \vec r(u,v), + a\leq u\leq b, c\leq v\leq d. + Following the principles given in the integration review at the beginning of this chapter, + we can say thatsurface areaof parametrized surface + + \text{ Surface Area of }\surfaceS\, =S = \iint_{\surfaceS}\, dS + , + where dS represents a small amount of surface area. + That is, to compute total surface area S, + add up lots of small amounts of surface area dS across the entire surface \surfaceS. The key to finding surface area is knowing how to compute dS. + We begin by approximating. +

    + +

    + In + we used the area of a plane to approximate the surface area of a small portion of a surface. + We will do the same here. +

    + +

    + Let R be the region of the u-v plane bounded by a\leq u\leq b, + c\leq v\leq d as shown in . + Partition R into rectangles of width + \Delta u = \frac{b-a}n and height \Delta v = \frac{d-c}n, + for some n. + Let p=(u_0,v_0) be the lower left corner of some rectangle in the partition, + and let m and q be neighboring corners as shown. +

    + +

    + The point p maps to a point + P = \vec r(u_0,v_0) on the surface \surfaceS, and the rectangle with corners p, + m and q maps to some region + (probably not rectangular) + on the surface as shown in , + where M = \vec r(m) and Q = \vec r(q). + We wish to approximate the surface area of this mapped region. +

    + +

    + Let \vec u = M-P and \vec v = Q-P. + These two vectors form a parallelogram, + illustrated in , + whose area approximates + the surface area we seek. + In this particular illustration, + we can see that parallelogram does not particularly match well the region we wish to approximate, + but that is acceptable; + by increasing the number of partitions of R, + \Delta u and \Delta v shrink and our approximations will become better. +

    + +
    + Illustrating the process of finding surface area by approximating with planes + +
    + + + A rectangular region in the plane, with a smaller subrectangle highlighted in its interior. + +

    + A rectangle R is drawn and shaded in the first quadrant of the uv plane. + Within the rectangle, points p, q, and m are marked. + These points make up three of the four corners of a smaller subrectangle inside of R. +

    + +

    + Along the u axis (the horizontal axis), there are points marked a and b corresponding to the left and right edges of R. + Also marked is a point u_0, which is the u coordinate of both p and q, + and a point u_0+\Delta u, which is the u coordinate of m. +

    + +

    + Along the v axis (the vertical axis), points c and d are marked to indicate the top and bottom of R. + Another point is marked with the value v_0; this is the v coordinate of both p and m. + Also marked is a point with the value v_0+\Delta v, which is the v coordinate of q. +

    + +

    + A smaller subrectangle is drawn inside of R. + It has its bottom left vertex at p, with coordinates (u_0,v_0), + and sides of length \Delta u and \Delta v. +

    +     + + + \begin{tikzpicture} + + \begin{axis}[ + xtick=\empty, + ytick=\empty, + extra x ticks={.5,2.25,1,1.5}, + extra x tick labels={$a$,$b$,$u_0$,$u_0+\Delta u$}, + extra y ticks={.1,2,.75,1}, + extra y tick labels={$c$,$d$,$v_0$,$v_0+\Delta v$}, + ymin=-.1,ymax=2.5, + xmin=-.1,xmax=2.5, + xlabel={$u$}, + ylabel={$v$} + ] + + \filldraw [thick,draw=firstcolor,fill=firstcolor!15] (axis cs: .5,.1) -- (axis cs: .5,2) -- (axis cs: 2.25,2) -- (axis cs: 2.25,.1) -- cycle; + + \coordinate (P) at (axis cs: 1,.75); + \coordinate (Q) at (axis cs: 1.5,.75); + \coordinate (R) at (axis cs: 1,1); + \coordinate (d) at (axis cs: 1.5,1); + + \draw [thick,draw=secondcolor,fill=secondcolor!15] (P) -- (Q) -- (d) -- (R) -- cycle; + + \filldraw (P) circle (1pt) node [below left] {$p$}; + \filldraw (R) circle (1pt) node [above left] {$q$}; + \filldraw (Q) circle (1pt) node [below right] {$m$}; + + \draw (axis cs: .5,2) node [below right] {$R$}; + + \end{axis} + + + \end{tikzpicture} + + + +
    + +
    + + + A surface in space with parameter domain R, highlighting the portion corresponding to a particular subrectangle. + +

    + A surface is drawn in space, relative to a set of three-dimensional coordinate axes. + The precise shape of the surface is unimportant, + but it appears to be a portion of an elliptic paraboloid, + corresponding to the rectangular parameter domain R in . +

    + +

    + There are points on the surface marked P, Q, and M, + corresponding to the points p, q, and m in the parameter domain. + There is also a shaded patch on the surface, representing the portion that corresponds to the subrectangle + drawn in . +

    + +

    + The sides of this patch are curved, since the surface is, but the patch is approximately rectangular. + To emphasize this point, the image also shows the vectors \overrightarrow{PQ} (from P to Q) + and \overrightarrow{PM} (from P to M). + These vectors are straight line approximations of the corresponding sides of the patch on the surface. +

    +     + + + + + //ASY file for figparsurfareaA_3D.asy in Chapter 14, developing surface area + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(15.5,18.6,18.1,true,true); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-.1,1.5); + pair ybounds=(-.1,1.5); + pair zbounds=(-.1,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the top half of the surface z = x^2+2y^2 + triple ff(pair t) { + return (t.y*(cos(t.x)+.1*sin(3t.x)),t.y*sin(t.x),2-(t.y*(cos(t.x)+.1*sin(3t.x)))^2-(t.y*sin(t.x))^2);// + } + + triple f(pair t) { + return -ff(t)+(1,1,3); + } + + surface s=surface(f,(0,0),(2pi/4,1),4,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + surface s=surface(f,(2pi/16,9/16),(4pi/16,11/16),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen2,meshpen=invisible); + + real u0=2pi/16; + real v0=9/16; + real du=2pi/16; + real dv=2/16; + triple ru=f((u0+du,v0))-f((u0,v0)); + triple rv=f((u0+du,v0))-f((u0,v0+dv)); + + draw(f((u0+du,v0))--f((u0,v0)),curvepen2,Arrow3(5)); + draw(f((u0+du,v0))--f((u0+du,v0+dv)),curvepen2,Arrow3(5)); + + label("$P$",f((u0+du,v0))+(.05,-.05,.1)); + label("$M$",f((u0,v0))+(0,+.1,.1)); + label("$Q$",f((u0+du,v0+dv))+(.05,-.05,.1)); + + dot(f((u0+du,v0)),.7mm+black); + dot(f((u0,v0)),.7mm+black); + dot(f((u0+du,v0+dv)),.7mm+black); + + +
    + +
    + + + A zoomed in view of the parallelogram approximation of a small patch on a surface. + +

    + This is a zoomed-in view of the patch illustrated in . + We see how the vectors \vec{u} = \overrightarrow{PQ} and \vec{v} = \overrightarrow{PM} + form two sides of a parallelogram with one vertex at P. +

    + +

    + This parallelogram overlaps with most (but not all) of the patch on the surface given by + u_0\leq u\leq u_0+\Delta u and v_0\leq v\leq v_0+\Delta v, + illustrating how the area of the parallelogram (which we know how to compute) + is a good approximation of the area of the patch on the surface. +

    +     + + + + + //ASY file for figparsurfareaA_3D.asy in Chapter 14, developing surface area + + size(198,198,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(15.5,18.6,18.1,Z,(0,0,0),1,(0,-.15),true,true); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(.3,.8); + pair ybounds=(.3,.8); + pair zbounds=(.5,1.5); + + //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the top half of the surface z = x^2+2y^2 + triple ff(pair t) { + return (t.y*(cos(t.x)+.1*sin(3t.x)),t.y*sin(t.x),2-(t.y*(cos(t.x)+.1*sin(3t.x)))^2-(t.y*sin(t.x))^2);// + } + + triple f(pair t) { + return -ff(t)+(1,1,3); + } + + surface s=surface(f,(pi/16,8/16),(2pi/4*3/4*10/12,12/16),8,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + surface s=surface(f,(2pi/16,9/16),(4pi/16,11/16),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + //draw(s,rgb(1,.5,.5)+opacity(.7),meshpen=invisible); + + real u0=2pi/16; + real v0=9/16; + real du=2pi/16; + real dv=2/16; + triple ru=f((u0+du,v0))-f((u0,v0)); + triple rv=f((u0+du,v0))-f((u0+du,v0+dv)); + + triple aa=f((u0+du,v0)); + triple bb=aa-ru; + triple cc=bb-rv; + triple dd=cc+ru; + + draw(f((u0+du,v0))--f((u0,v0)),curvepen2,Arrow3(5)); + draw(f((u0+du,v0))--f((u0+du,v0+dv)),curvepen2,Arrow3(5)); + draw(f((u0+du,v0+dv))--f((u0+du,v0+dv))-ru--f((u0+du,v0+dv))-ru+rv,curvepen2); + + label("$P$",f((u0+du,v0))+(.005,-.005,.05)); + label("$M$",f((u0,v0))+(0,+.025,.005)); + label("$Q$",f((u0+du,v0+dv))+(.005,-.005,.05)); + + label("$\vec u$",(f((u0+du,v0))+f((u0,v0)))/2+(.05,.005,0)); + label("$\vec v$",(f((u0+du,v0))+f((u0+du,v0+dv)))/2+(.005,-.005,.05)); + + dot(f((u0+du,v0)),.7mm+black); + dot(f((u0,v0)),.7mm+black); + dot(f((u0+du,v0+dv)),.7mm+black); + + draw(surface(aa--bb--cc--dd--cycle),surfacepen2); + + //draw(surface(f((u0,v0))--f((u0+du,v0))--f((u0+du,v0+dv))--f((u0,v0+dv))--cycle),surfacepen2); + + dot(f((u0+du,v0)),.7mm+black); + + surface s=surface(f,(2pi/16,9/16),(4pi/16,9/16),8,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen2,meshpen=curvepen2); + + surface s=surface(f,(2pi/16,11/16),(4pi/16,11/16),8,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen2,meshpen=curvepen2); + + surface s=surface(f,(2pi/16,9/16),(2pi/16,11/16),1,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen2,meshpen=curvepen2); + + surface s=surface(f,(4pi/16,9/16),(4pi/16,11/16),1,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen2,meshpen=curvepen2); + + +
    +     + +
    + +

    + From + we know the area of this parallelogram is \snorm{\vec u\times \vec v}. + If we repeat this approximation process for each rectangle in the partition of R, + we can sum the areas of all the parallelograms to get an approximation of the surface area S: + + \text{ Surface area of } \surfaceS \, =S \approx \sum_{j=1}^n\sum_{i=1}^n \snorm{\vec u_{i,j}\times \vec v_{i,j}} + , + where \vec u_{i,j} = \vec r(u_i+\Delta u,v_j) - \vec r(u_i,v_j) and \vec v_{i,j} = \vec r(u_i,v_j+\Delta v)-\vec r(u_i,v_j). +

    + +

    + From our previous calculus experience, + we expect that taking a limit as + n\to \infty will result in the exact surface area. + However, the current form of the above double sum makes it difficult to realize what the result of that limit is. + The following rewriting of the double summation will be helpful: + + \amp \sum_{j=1}^n\sum_{i=1}^n \snorm{\vec u_{i,j}\times \vec v_{i,j}}= + \amp \sum_{j=1}^n\sum_{i=1}^n \snorm{\big(\vec r(u_i+\Delta u,v_j) - \vec r(u_i,v_j)\big) \times \big(\vec r(u_i,v_j+\Delta v)-\vec r(u_i,v_j)\big)}= + \amp \sum_{j=1}^n\sum_{i=1}^n \snorm{\frac{\vec r(u_i+\Delta u,v_j) - \vec r(u_i,v_j)}{\Delta u} \times \frac{\vec r(u_i,v_j+\Delta v)-\vec r(u_i,v_j)}{\Delta v}}\Delta u\Delta v + . +

    + +

    + We now take the limit as n\to\infty, + forcing \Delta u and \Delta v to 0. + As \Delta u\to 0, + + \frac{\vec r(u_i+\Delta u,v_j) - \vec r(u_i,v_j)}{\Delta u} \to \vec r_u(u_i,v_j) \text{ and } + + + \frac{\vec r(u_i,v_j+\Delta v)-\vec r(u_i,v_j)}{\Delta v} \to \vec r_v(u_i,v_j) + . +

    + +

    + (This limit process also demonstrates that \vec r_u(u,v) and + \vec r_v(u,v) are tangent to the surface \surfaceS at \vec r(u,v). + We don't need this fact now, but it will be important in the next section.) +

    + +

    + Thus, in the limit, the double sum leads to a double integral: + + \lim_{n\to\infty} \sum_{j=1}^n\sum_{i=1}^n \snorm{\vec u_{i,j}\times \vec v_{i,j}}= \int_c^d\int_a^b \snorm{\vec r_u\times\vec r_v}\, du\, dv + . +

    + + + Surface Area of Parametrically Defined Surfaces + +

    + Let \vec r(u,v) be a smooth parametrization of a surface \surfaceS over a closed, + bounded region R of the u-v plane. + surface areaof parametrized surface +

      +
    • +

      + The surface area differential dS is: + dS = \snorm{\vec r_u\times \vec r_v}\, dA. +

      +
      • + +
    • +

      + The surface area S of \surfaceS is + + S = \iint_\surfaceS\, dS = \iint_R \snorm{\vec r_u\times \vec r_v}\, dA + . +

      +
      • +
    +

    +     +     + + + + + Finding the surface area of a parametrized surface + +

    + Using the parametrization found in , + find the surface area of z=x^2+2y^2 over the circular disk of radius 2, centered at the origin. +

    +     + +

    + In , + we parametrized the surface as \vec r(u,v) = \la 2v\cos(u), 2v\sin(u), 4v^2\cos^2u+8v^2\sin^2u\ra, + for 0\leq u\leq 2\pi, 0\leq v\leq 1. + To find the surface area using , + we need \snorm{\vec r_u\times\vec r_v}. + We find: + + \vec r_u \amp = \la -2v\sin(u), 2v\cos(u), 8v^2\cos(u)\sin(u)\ra + \vec r_v \amp = \la 2\cos(u), 2\sin(v), 8v\cos^2 u+16v\sin^2u\ra + \vec r_u\times\vec r_v \amp = \la 16v^2\cos(u), 32v^2\sin(u), -4v\ra + \snorm{\vec r_u\times\vec r_v} \amp = \sqrt{256v^4\cos^2u+1024v^4\sin^2u+16v^2} + . +

    + +

    + Thus the surface area is + + S = \iint_\surfaceS \, dS \amp = \iint_R\snorm{\vec r_u\times \vec r_v}\, dA + \amp = \int_0^1\int_0^{2\pi} \sqrt{256v^4\cos^2u+1024v^4\sin^2u+16v^2}\, du\, dv + \amp \approx 53.59 + . +

    + +

    + There is a lot of tedious work in the above calculations and the final integral is nontrivial. + The use of a computer-algebra system is highly recommended. +

    +     +     + + + +

    + In , + we recalled the arc length differential ds=\snorm{\vrp(t)}dt. + In subsequent sections, we used that differential, + but in most applications the \snorm{\vrp(t)} + part of the differential canceled out of the integrand + (to our benefit, + as integrating the square roots of functions is generally difficult). + We will find a similar thing happens when we use the surface area differential dS in the following sections. + That is, our main goal is not to be able to compute surface area; + rather, surface area is a tool to obtain other quantities that are more important and useful. + In our applications, we will use dS, + but most of the time the \snorm{\vec r_u\times \vec r_v} + part will cancel out of the integrand, + making the subsequent integration easier to compute. +

    +     + + + + Terms and Concepts + + +

    + In your own words, describe what an orientable surface is. +

    +     + + + +

    + Answers will vary, + though generally should meaningfully include terms like two sided. +

    +     +     + + + +

    + Give an example of a non-orientable surface. +

    +     + + + +

    + Many possible answers exist; + the one given by the book is the Möbius band. +

    +     +     +     + + + Problems + + +

    + Parametrize the surface defined by the function + z=f(x,y) over each of the given regions R of the xy-plane. +

    +     + + + +

    + z = 3x^2y; + +

      +
    1. +

      + R is the rectangle bounded by + -1\leq x\leq 1 and 0\leq y\leq 2. +

      +
      2. + +
    2. +

      + R is the circle of radius 3, centered at (1,2). +

      +
      3. + +
    3. +

      + R is the triangle with vertices (0,0), + (1,0) and (0,2). +

      +
      4. + +
    4. +

      + R is the region bounded by the x-axis and the graph of y = 1-x^2. +

      +
      5. +
    +

    +     + +

    +

      +
    1. +

      + \vec r(u,v) = \langle u, v, 3u^2v\rangle on + -1\leq u\leq 1, 0\leq v\leq 2. +

      +
      2. + +
    2. +

      + \vec r(u,v) = \langle 3v\cos(u)+1, 3v\sin(u)+2, 3(3v\cos(u)+1)^2(3v\sin(u)+2)\rangle, + on 0\leq u\leq 2\pi, 0\leq v\leq 1. +

      +
      3. + +
    3. +

      + \vec r(u,v) = \langle u, v(2-2u), 3u^2v(2-2u)\rangle on 0\leq u, v\leq 1. +

      +
      4. + +
    4. +

      + \vec r(u,v) = \langle u, v(1-u^2), 3u^2v(1-u^2)\rangle on + -1\leq u\leq 1, 0\leq v\leq 1. +

      +
      5. +
    +

    +     +     + + + +

    + z = 4x+2y^2; + +

      +
    1. +

      + R is the rectangle bounded by + 1\leq x\leq 4 and 5\leq y\leq 7. +

      +
      2. + +
    2. +

      + R is the ellipse with major axis of length 8 parallel to the x-axis, + and minor axis of length 6 parallel to the y-axis, + centered at the origin. +

      +
      3. + +
    3. +

      + R is the triangle with vertices (0,0), + (2,2) and (0,4). +

      +
      4. + +
    4. +

      + R is the annulus bounded between the circles, + centered at the origin, with radius 2 and radius 5. +

      +
      5. +
    +

    +     + +

    +

      +
    1. +

      + \vec r(u,v) = \langle u, v, 4u+2u^2\rangle on + 1\leq u\leq 4, 5\leq v\leq 7. +

      +
      2. + +
    2. +

      + \vec r(u,v) = \langle 4v\cos(u), 3v\sin(u), 16v\cos(u)+2(3v\sin(u))^2\rangle, + on 0\leq u\leq 2\pi, 0\leq v\leq 1. +

      +
      3. + +
    3. +

      + \vec r(u,v) = \langle u, u+v(4-2u), 4u+2\big(u+v(4-2u)\big)^2\rangle on + 0\leq u\leq 2, 0\leq v\leq 1. +

      +
      4. + +
    4. +

      + \vec r(u,v) = \langle v\cos(u), v\sin(u), 4v\cos(u) + 2(v\sin(u))^2\rangle on + 0\leq u\leq 2\pi, 2\leq v\leq 5. +

      +
      5. +
    +

    +     +     +     + + + +

    + A surface \surfaceS in space is described that cannot be defined as the graph of a function f(x,y). + Give a parametrization of \surfaceS. +

    +     + + + +

    + \surfaceS is the rectangle in space with corners at (0,0,0), + (0,2,0), + (0,2,1) and (0,0,1). +

    +     + +

    + \vec r(u,v) = \langle 0, u, v\rangle with + 0\leq u\leq 2, 0\leq v\leq 1. +

    +     +     + + + +

    + \surfaceS is the triangle in space with corners at (1,0,0), + (1,0,1) and (0,0,1). +

    +     + +

    + \vec r(u,v) = \langle u, 0, 1-u+vu\rangle with + 0\leq u\leq 1, 0\leq v\leq 1. +

    +     +     + + + +

    + \surfaceS is the ellipsoid \ds\frac{x^2}{9} + \frac{y^2}{4}+\frac{z^2}{16} = 1. +

    +     + +

    + \vec r(u,v) = \langle 3\sin(u)\cos(v), 2\sin(u)\sin(v), 4\cos(u)\rangle with + 0\leq u\leq \pi, 0\leq v\leq 2\pi. +

    +     +     + + + +

    + \surfaceS is the elliptic cone \ds y^2= x^2+\frac{z^2}{16}, + for -1 \leq y \leq 5. +

    +     + +

    + Answers may vary; + one solution is \vec r(u,v) = \langle v\cos(u), v, 4v\sin(u)\rangle with + 0\leq u\leq 2\pi, -1\leq v\leq 5. +

    +     +     +     + + + +

    + A domain D in space is given. + Parametrize each of the bounding surfaces of D. +

    +     + + + +

    + D is the domain bounded by the planes z = \frac12(3-x), + x=1, + y=0, y=2 and z=0. +

    + + + A triangular prism plotted in three dimensions. + +

    + A set of three-dimensional coordinate axes are drawn, with the z axis pointing up. + The domain D for this problem is a triangular prism. + Cross-sections parallel to the xz plane are triangles, + while cross-sections parallel to the other coordinate planes are rectangles. +

    + +

    +

      +
    • +

      + The base of the prism is a square in the xy plane, with 1\leq x\leq 3 and 0\leq y\leq 2. +

      +
      • +
    • +

      + There is a vertical face in the plane x=1 that is also a square, with 0\leq y\leq 2 and 0\leq z\leq 2. +

      +
      • +
    • +

      + Two other vertical faces are triangles. The triangle in the plane y=0 has vertices (1,0,0), + (3,0,0), and (1,0,2). The triangle in the plane y=2 has vertices + (1,2,0), (3,2,0), and (3,2,2). +

      +
      • +
    • +

      + The remaining face corrsponds to the hypotenuse in each triangular cross-section; + it lies in the plane z=\frac12(3-x). +

      +
      • +
    +

    +     + + + + + //ASY file for fig13_06_ex_083D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8.8,7.8,3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3}; + real[] myychoice={1,2,3}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,3.5); + pair ybounds=(-0.5,3.5); + pair zbounds=(-0.5,1.75); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //edges of object + draw((1,0,0)--(1,2,0)--(3,2,0)--(3,0,0)--cycle,bluepen+linewidth(2)); + draw((3,0,0)--(1,0,1)--(1,2,1)--(3,2,0),bluepen+linewidth(2)); + draw((3,0,0)--(1,0,1),bluepen+linewidth(2)); + draw((1,0,0)--(1,0,1),bluepen+linewidth(2)); + draw((1,2,0)--(1,2,1),bluepen+linewidth(2)); + + //shade faces + import three; + path3 p=(1,0,0)--(1,2,0)--(3,2,0)--(3,0,0); + draw(surface(p -- cycle), surfacepen); + path3 p=(1,0,1)--(1,2,1)--(3,2,0)--(3,0,0); + draw(surface(p -- cycle), surfacepen); + path3 p=(1,0,0)--(1,2,0)--(1,2,1)--(1,0,1); + draw(surface(p -- cycle), surfacepen); + path3 p=(1,0,0)--(3,0,0)--(1,0,1); + draw(surface(p -- cycle), surfacepen); + path3 p=(1,2,0)--(3,2,0)--(1,2,1); + draw(surface(p -- cycle), surfacepen); + + //labels and arrow + label("$z=\frac{1}{2}(3-x)$",(3,0,1.1)); + draw((3,0.25,1)--(2,1,0.45),Arrow3(size=2mm)); + + + +     + +

    + Answers may vary. +

    + +

    + For z = \frac12(3-x): + \vec r(u,v) = \langle u, v , \frac12(3-u)\rangle, + with 1\leq u\leq 3 and 0\leq v\leq 2. +

    + +

    + For x=1: \vec r(u,v) = \langle 1,u,v\rangle, + with 0\leq u\leq 2, 0\leq v\leq 1 +

    + +

    + For y=0: \vec r(u,v) = \langle u,0,v/2(3-u)\rangle, + with 1\leq u\leq 3, 0\leq v\leq 1 +

    + +

    + For y=2: \vec r(u,v) = \langle u,2,v/2(3-u)\rangle, + with 1\leq u\leq 3, 0\leq v\leq 1 +

    + +

    + For z=0: \vec r(u,v) = \langle u,v,0\rangle, + with 1\leq u\leq 3, 0\leq v\leq 2 +

    +     +     + + + +

    + D is the domain bounded by the planes z=2x+4y-4, + x=2, y=1 and z=0. +

    + + + A tetrahedron with vertices (2,0,0), (0,1,0), (2,1,0), and (2,1,4). + +

    + A tetrahedron is plotted with respect to three-dimensional coordinate axes. +

      +
    • +

      + One face lies in the xy plane, with vertices (2,0,0), (0,1,0), and (2,1,0). +

      +
      • +
    • +

      + One face lies in the plane x=2, with vertices (2,0,0), (2,1,0), and (2,1,4). +

      +
      • +
    • +

      + Another face lies in the plane y=1, with vertices (0,1,0), (2,1,0), and (2,1,4) +

      +
      • +
    • +

      + The last face lies in the plane z=2x+4y-4. + It meets the xy plane along the line from (2,0,0) to (0,1,0), + and has its remaining vertex at (2,1,4). +

      +
      • +
    +

    +     + + + + + //ASY file for fig14_05_ex_10_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(7.2,-5.1,8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={1,2}; + real[] myzchoice={2,4}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,2.5); + pair ybounds=(-0.25,2.5); + pair zbounds=(-0.25,5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //edges of object + draw((0,1,0)--(2,1,0)--(2,0,0)--cycle,bluepen+linewidth(2)); + draw((0,1,0)--(2,1,4)--(2,1,0)--cycle,bluepen+linewidth(2)); + draw((0,1,0)--(2,1,4)--(2,0,0)--cycle,bluepen+linewidth(2)); + + //shade plane + import three; + path3 p=(0,1,0)--(2,1,4)--(2,0,0); + draw(surface(p -- cycle), surfacepen); + + path3 p=(2,0,0)--(2,1,4)--(2,1,0); + draw(surface(p -- cycle), surfacepen); + + path3 p=(0,1,0)--(2,1,4)--(2,1,0); + draw(surface(p -- cycle), surfacepen); + + path3 p=(0,1,0)--(2,1,0)--(2,0,0); + draw(surface(p -- cycle), surfacepen); + + //label and arrow + label("$z=2x+4y-4$",(1,0.5,4)); + draw((1,0.5,3.6)--(1.2,.65,1.6),Arrow3(size=2mm)); + + + +     + +

    + Answers may vary. +

    + +

    + For z=2x+4y-4: + \vec r(u,v) = \langle u, 1-u/2+uv/2, 2u+4(1-u/2+uv/2)-4\rangle, + with 0\leq u\leq 2, 0\leq v\leq 1. +

    + +

    + For x=2: \vec r(u,v) = \langle 2,u,4uv\rangle, + with 0\leq u\leq 1, 0\leq v\leq 1 +

    + +

    + For y=1: \vec r(u,v) = \langle u,1,2uv\rangle, + with 0\leq u\leq 2, 0\leq v\leq 1 +

    + +

    + For z=0: + \vec r(u,v) = \langle u, 1-u/2+uv/2,0\rangle, + with 0\leq u\leq 2, 0\leq v\leq 1 +

    +     +     + + + +

    + D is the domain bounded by z=2y, + y=4-x^2 and z=0. +

    + + + A cylindrical wedge with its edge along the x axis. + +

    + A cylindrical wedge, between the planes z=0 and z=2y. + These planes intersect along the x axis, forming the sharp edge of the wedge. + The round face of the wedge is the parabolic cylinder y=4-x^2, for y\geq 0. +

    + +

    + The face of the surface in the xy plane is the region bounded below by the x axis, + and above by the parabola y=4-x^2. +

    + +

    + Another face is the portion of the plane z=2y that lies above the face in the xy plane. +

    + +

    + The last face is the parabolic cylinder. It is the portion of the cylinder y=4-x^2 + that lies above the plane z=0 and below the plane z=2y. +

    +     + + + + + //ASY file for fig13_06_ex_123D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12.1,-7.1,16); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,-1,1,2}; + real[] myychoice={1,2,3,4}; + real[] myzchoice={2,4,6,8}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-0.25,5); + pair zbounds=(-0.25,10); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + triple g(real t) {return (t,4-t^2,0);} + path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,4-t^2,2*(4-t^2));} + path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + //shade object + //import three; + //int k=12; + //for (int i=-2*k; i<2*k; ++i) + //{ + //path3 p=(i/k,4-(i/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i)/k,4-((i)/k)^2,2*(4-((i)/k)^2)); + //draw(surface(p -- cycle), simplesurfacepen2); + //path3 p=(i/k,0,0)--(i/k,4-(i/k)^2,2*(4-(i/k)^2))--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i+1//)/k,0,0); + //draw(surface(p -- cycle), simplesurfacepen); + //} + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y*(4-t.x^2),2*t.y*(4-t.x^2));// + } + surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,(4-t.x^2),2*t.y*(4-t.x^2));// + } + surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y*(4-t.x^2),0);// + } + surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + label("$z=2y$",(-2,2,7)); + draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + label("$y=4-x^2$",(2.5,2,0)); + draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + Answers may vary. +

    + +

    + For z=2y: \vec r(u,v) = \langle u, v(4-u^2), 2v(4-u^2)\rangle with + -2\leq u\leq 2 and 0\leq v\leq 1. +

    + +

    + For y=4-x^2: \vec r(u,v) = \langle u, 4-u^2, 2v(4-u^2)\rangle with + -2\leq u\leq 2 and 0\leq v\leq 1. +

    + +

    + For z=0: + \vec r(u,v) = \langle u, v(4-u^2), 0\rangle with + -2\leq u\leq 2 and 0\leq v\leq 1. +

    +     +     + + + +

    + D is the domain bounded by y=1-z^2, + y=1-x^2, x=0, y=0 and z=0. +

    + + + A region in the first octant bounded by the planes x=0 and z=0, and two parabolic cylinders. + +

    + This surface has three faces that lie in planes, and two faces that are parts of parabolic cylinders. +

      +
    • +

      + The bottom face is in the xy plane, + in a region bounded by the x and y coordinate axes and the parabola y=1-x^2. +

      +
      • +
    • +

      + Another face lies in the yz plane, + forming a region bounded by the y and z coordinate axes and the parabola y=1-z^2. +

      +
      • +
    • +

      + There is another face in the xz plane. This face is a square, with x and z between 0 and 1. +

      +
      • +
    • +

      + Another part of the surface is the portion parabolic cylinder y=1-z^2, + and the remaining face is the parabolic cylinder y=1-x^2. + Both of these can be viewed as graphs over the xz plane. + The surface y=1-z^2 lies over the triangle bounded by x=z, x=0, and z=1. + The surface y=1-x^2 lies over the triangle bounded by x=z, x=1, and z=0. +

      +
      • +
    +

    +     + + + + + //ASY file for fig13_06_ex_133D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(4,4,2); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={1}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.25,1.5); + pair ybounds=(-0.25,1.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + triple g(real t) {return (t,1-t^2,0);} + path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (0,1-t^2,t);} + path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,1-t^2,t);} + path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + //draw square and sideline + draw((0,0,0)--(1,0,0)--(1,0,1)--(0,0,1)--(0,0,0),bluepen+linewidth(2)); + draw((0,0,0)--(0,1,0),bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,(1-t.x^2),t.y*t.x);// + } + surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.y*t.x,(1-t.x^2),t.x);// + } + surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,0,t.y);// + } + surface s=surface(f,(0,0),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (0,t.y*(1-t.x^2),t.x);// + } + surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y*(1-t.x^2),0);// + } + surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + label("$y=1-x^2$",(1,1,0)); + draw((0.9,0.9,.1)--(.65,.6,.4),Arrow3(size=2mm)); + label("$y=1-z^2$",(0,1,.85)); + draw((.1,.9,.75)--(.3,.8,.45),Arrow3(size=2mm)); + + //shade object + //import three; + //int k=12; + //for (int i=0; i<k; ++i) + //{ + //path3 p=(0,1-(i/k)^2,i/k)--(0,1-((i+1)/k)^2,(i+1)/k)--((i+1)/k,1-((i+1)/k)^2,(i+1)/k)--((i)/k,1-((i)/k)^2,(i)/k); + //draw(surface(p -- cycle), simplesurfacepen2);//pink + + //path3 p=(i/k,1-(i/k)^2,0)--((i+1)/k,1-((i+1)/k)^2,0)--((i+1)/k,1-((i+1)/k)^2,(i+1)/k)--((i)/k,1-((i)/k)^2,(i)/k); + //draw(surface(p -- cycle), simplesurfacepen);//blue + //} + + + +     + +

    + Answers may vary. +

    + +

    + For y=1-z^2: + \vec r(u,v) = \langle u, v(1-u^2), \sqrt{1-v(1-u^2)}\rangle with + 0\leq u\leq 1 and 0\leq v\leq 1. +

    + +

    + For y=1-x^2: + \vec r(u,v) = \langle u, 1-u^2, uv\rangle with + 0\leq u\leq 1 and 0\leq v\leq 1. +

    + +

    + For x=0: + \vec r(u,v) = \langle 0, v(1-u^2),u\rangle with + 0\leq u\leq 1 and 0\leq v\leq 1. +

    + +

    + For y=0: \vec r(u,v) = \langle u, 0,v\rangle with + 0\leq u\leq 1 and 0\leq v\leq 1. +

    + +

    + For z=0: + \vec r(u,v) = \langle u, v(1-u^2), 0rangle with + 0\leq u\leq 1 and 0\leq v\leq 1. +

    +     +     + + + +

    + D is the domain bounded by the cylinder + x^2+y^2/9=1 and the planes z=1 and z=3. +

    + + + An elliptical cylinder, centered on the z axis, capped by planes z=1 and z=3. + +

    + This surface has three components. + Two are faces in the planes z=1 and z=3. + Each of these is bounded by the ellipse x^2+y^2/9=1. +

    + +

    + The remaining component is the portion of the elliptical cylinder x^2+y^2/9=1 + that lies between the planes z=1 and z=3. +

    +     + + + + + //ASY file for fig14_05_ex_21_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14,14,4); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-3,3}; + real[] myzchoice={1,3}; + defaultpen(0.5mm); + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-0.25,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple g(real t) {return (cos(t),3sin(t),3);} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (cos(t),3sin(t),1);} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (cos(t.x),3*sin(t.x),t.y);// + } + surface s=surface(f,(0,1),(2pi,3),16,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (t.y*cos(t.x),t.y*3*sin(t.x),1);// + } + surface s=surface(f,(0,0),(2pi,1),16,2,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (t.y*cos(t.x),t.y*3*sin(t.x),3);// + } + surface s=surface(f,(0,0),(2pi,1),16,2,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + label("$x^2+y^2/9=1$",(2.5,0,1)); + draw((2.25,0,1.25)--(1,0,2),Arrow3(size=2mm)); + + //triple g(real t) {return (0,t,6-t^2);} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + + +     + +

    + Answers may vary. +

    + +

    + For x^2+y^2/9=1: + \vec r(u,v) = \langle \cos(u), 3\sin(u), v\rangle with + 0\leq u\leq 2\pi and 1\leq v\leq 3. +

    + +

    + For z=1: + \vec r(u,v) = \langle v\cos(u), 3v\sin(u), 1\rangle with + 0\leq u\leq 2\pi and 0\leq v\leq 1. +

    + +

    + For z=3: + \vec r(u,v) = \langle v\cos(u), 3v\sin(u), 3\rangle with + 0\leq u\leq 2\pi and 0\leq v\leq 1. +

    +     +     + + + +

    + D is the domain bounded by the cone + x^2+y^2=(z-1)^2 and the plane z=0. +

    + + + An inverted circular cone, together with a disk in the x,y plane + +

    + This surface has two components. + One is the cone x^2+y^2=(z-1)^2; + this is a circular cone; the vertex is at (0,0,1), + and our surface is the part that opens downward. +

    + +

    + The cone meets the xy plane along the unit circle. + The other part of the surface is the disk in the xy plane bounded by the unit circle. +

    +     + + + + + //ASY file for fig14_05_ex_21_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14,14,4); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + triple g(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),1-t.y);// + } + surface s=surface(f,(0,0),(2pi,1),16,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),0);// + } + surface s=surface(f,(0,0),(2pi,1),16,2,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + label("$x^2+y^2=(z-1)^2$",(1,0,1)); + draw((.9,0,.9)--(.55,0,.5),Arrow3(size=2mm)); + + //triple g(real t) {return (0,t,6-t^2);} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + + +     + +

    + Answers may vary. +

    + +

    + For x^2+y^2=(z-1)^2: + \vec r(u,v) = \langle v\cos(u), v\sin(u), 1-v\rangle with + 0\leq u\leq 2\pi and 0\leq v\leq 1. +

    + +

    + For z=0: + \vec r(u,v) = \langle v\cos(u), v\sin(u), 0\rangle with + 0\leq u\leq 2\pi and 0\leq v\leq 1. +

    +     +     + + + +

    + D is the domain bounded by the cylinder z=1-x^2 and the planes y=-1, + y=2 and z=0. +

    + + + A solid that looks like a barn with a parabolic roof. + +

    + The boundary of this solid consists of four surface components. + Overall, the shape looks something like a barn or greenhouse with a curved roof. +

    + +

    + The base is a rectangle in the xy plane, with -1\leq x\leq 1 and -1\leq y\leq 2. +

    + +

    + The two ends lie in the planes y=-1 and y=2. + They are bounded above by the parabola z=1-x^2, and below by z=0. +

    + +

    + The last part is the parabolic cylinder z=1-x^2, for -1\leq x\leq 1, and -1\leq y\leq 2. +

    +     + + + + + //ASY file for fig14_05_ex_23_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14,14,4); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-2,2}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y,1-t.x^2);// + } + surface s=surface(f,(-1,-1),(1,2),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,-1,t.y*(1-t.x^2));// + } + surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,2,t.y*(1-t.x^2));// + } + surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (t.x,t.y,0);// + } + surface s=surface(f,(-1,-1),(1,2),2,2,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return (t,-1,1-t^2);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,2,1-t^2);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + draw((1,-1,0)--(1,2,0)--(-1,2,0)--(-1,-1,0)--cycle,bluepen+linewidth(2)); + + //label and arrow + label("$z=1-x^2$",(0,1,1.55)); + draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); + + //triple g(real t) {return (0,t,6-t^2);} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + + +     + +

    + Answers may vary. +

    + +

    + For z=1-x^2: \vec r(u,v) = \langle u,v,1-u^2\rangle with + -1\leq u\leq 1 and -1\leq v\leq 2. +

    + +

    + For y=-1: + \vec r(u,v) = \langle u,-1,v(1-u^2)\rangle with + -1\leq u\leq 1 and 0\leq v\leq 1. +

    + +

    + For y=2: + \vec r(u,v) = \langle u,2,v(1-u^2)\rangle with + -1\leq u\leq 1 and 0\leq v\leq 1. +

    + +

    + For z=0: \vec r(u,v) = \langle u,v,0\rangle with + -1\leq u\leq 1 and -1\leq v\leq 2. +

    +     +     + + + +

    + D is the domain bounded by the paraboloid + z=4-x^2-4y^2 and the plane z=0. +

    + + + A region bounded above by an elliptic paraboloid, and below by the x,y plane. + +

    + The solid looks something like half of a seed. + It is bounded above by the elliptic paraboloid z=4-x^2-4y^2, + which opens downward from a vertex at (0,0,4), forming a sort of domed roof. +

    + +

    + This paraboloid forms one component of the bounding surface for the solid. + The other part is the elliptical disk formed when the surface meets the xy plane. + It is the region in the xy plane that lies on and inside the ellipse x^2+4y^2=4. +

    +     + + + + + //ASY file for fig14_05_ex_23_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14,14,8); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={4}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.25,4.75); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (2*t.y*cos(t.x),t.y*sin(t.x),4-(2*t.y*cos(t.x))^2-4(t.y*sin(t.x))^2);// + } + surface s=surface(f,(0,0),(2pi,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (2*t.y*cos(t.x),t.y*sin(t.x),0);// + } + surface s=surface(f,(0,0),(2pi,1),16,2,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return (2cos(t),sin(t),0);} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + //label and arrow + label("$z=4-x^2-4y^2$",(0,1.5,3)); + draw((0,1.45,2.9)--(0,1,1),Arrow3(size=2mm)); + + //triple g(real t) {return (0,t,6-t^2);} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + + +     + +

    + Answers may vary. +

    + +

    + For z=4-x^2-4y^2: + \vec r(u,v) = \langle 2v\cos(u),v\sin(u),4-(2v\cos(u))^2-4(v\sin(u))^2\rangle with + 0\leq u\leq 2\pi and 0\leq v\leq 1. +

    + +

    + For z=0: + \vec r(u,v) = \langle 2v\cos(u),v\sin(u),0\rangle with + 0\leq u\leq 2\pi and 0\leq v\leq 1. +

    +     +     +     + + + +

    + Find the surface area S of the given surface \surfaceS. (The associated integrals are computable without the assistance of technology.) +

    +     + + + + +

    + \surfaceS is the plane z=2x+3y over the rectangle + -1\leq x\leq 1, 2\leq y \leq 3. +

    +     + +

    + S = 2\sqrt{14}. +

    +     +     + + + +

    + \surfaceS is the plane z=x+2y over the triangle with vertices at (0,0), + (1,0) and (0,1). +

    +     + +

    + S = \sqrt{6}/2. +

    +     +     + + + +

    + \surfaceS is the plane z=x+y over the circular disk, + centered at the origin, with radius 2. +

    +     + +

    + S = 4\sqrt{3}\pi. +

    +     +     + + + +

    + \surfaceS is the plane z=x+y over the annulus bounded by the circles, + centered at the origin, with radius 1 and radius 2. +

    +     + +

    + S = 3\sqrt{3}\pi. +

    +     +     +     + + + +

    + Set up the double integral that finds the surface area S of the given surface \surfaceS, then use technology to approximate its value. +

    +     + + + +

    + \surfaceS is the paraboloid + z=x^2+y^2 over the circular disk of radius 3 centered at the origin. +

    +     + +

    + S =\int_0^3\int_0^{2\pi}\sqrt{v^2+4v^4}\, du\, dv= (37\sqrt{37}-1)\pi/6 \approx 117.319. +

    +     +     + + + +

    + \surfaceS is the paraboloid + z=x^2+y^2 over the triangle with vertices at (0,0), + (0,1) and (1,1). +

    +     + +

    + S = \int_0^1\int_0^1\sqrt{v^2+4u^2v^2+4v^4}\, du\, dv \approx 0.931. +

    +     +     + + + +

    + \surfaceS is the plane z=5x-y over the region enclosed by the parabola y=1-x^2 and the x-axis. +

    +     + +

    + S =\int_0^1\int_{-1}^{1}\sqrt{(5u^2-5)^2+2(1-u^2)^2}\, du\, dv = 4\sqrt{3}\approx 6.9282. +

    +     +     + + + +

    + \surfaceS is the hyperbolic paraboloid + z=x^2-y^2 over the circular disk of radius 1 centered at the origin. +

    +     + +

    + S =\int_0^1\int_{0}^{2\pi}\sqrt{v^2+4v^4}\, du\, dv = (5\sqrt{5}-1)\pi/6 \approx 5.330. +

    +     +     +     +     +     +
    +
    + Surface Integrals + +

    + Consider a smooth surface \surfaceS that represents a thin sheet of metal. + How could we find the mass of this metallic object? +

    + +

    + If the density of this object is constant, + then we can find mass via mass= density surface area, + and we could compute the surface area using the techniques of the previous section. +

    + +

    + What if the density were not constant, + but variable, described by a function \delta(x,y,z)? + We can describe the mass using our general integration techniques as + + \text{ mass } = \iint_\surfaceS \, dm + , + where dm represents a little bit of mass. That is, + to find the total mass of the object, + sum up lots of little masses over the surface. +

    + +

    + How do we find the little bit of mass dm? + On a small portion of the surface with surface area \Delta S, + the density is approximately constant, + hence dm \approx \delta(x,y,z)\Delta S. + As we use limits to shrink the size of + \Delta S to 0, we get dm = \delta(x,y,z)dS; + that is, a little bit of mass is equal to a density times a small amount of surface area. + Thus the total mass of the thin sheet is + + \text{ mass } =\iint_\surfaceS \delta(x,y,z)\, dS + . +

    + +

    + To evaluate the above integral, + we would seek \vec r(u,v), + a smooth parametrization of \surfaceS over a region R of the u-v plane. + The density would become a function of u and v, + and we would integrate \iint_R \delta(u,v)\snorm{\vec r_u\times \vec r_v}\, dA. +

    + +

    + The integral in Equation is a specific example of a more general construction defined below. +

    +     + + + Surface integrals of scalar fields + + + + + Surface Integral + +

    + Let G(x,y,z) be a continuous function defined on a surface \surfaceS. The + surface integral of G on \surfaceS is + surface integral + + \iint_\surfaceS G(x,y,z)\, dS + . +

    +     +     + +

    + Surface integrals can be used to measure a variety of quantities beyond mass. + If G(x,y,z) measures the static charge density at a point, + then the surface integral will compute the total static charge of the sheet. + If G measures the amount of fluid passing through a screen + (represented by \surfaceS) + at a point, then the surface integral gives the total amount of fluid going through the screen. +

    + + + Finding the mass of a thin sheet + +

    + Find the mass of a thin sheet modeled by the plane + 2x+y+z=3 over the triangular region of the xy-plane bounded by the coordinate axes and the line y=2-2x, + as shown in , + with density function \delta(x,y,z) = x^2+5y+z, + where all distances are measured in cm and the density is given as gm/cm^2. +

    +
    + The surface whose mass is computed in + + A triangular portion of a plane in space. It is the graph of a linear function over a triangular domain in the plane. + +

    + The positive x, y, and z axes are shown in space, with the z axis pointing up. + In the xy plane, a triangular domain is shaded. + It is the region in the plane z=0 bounded by the x and y axes, + and the line segment from (1,0,0) to (0,2,0). +

    + +

    + Above the xy plane is another triangle. + This triangle lies in the plane 2x+y+z=3, + and can be viewed as the graph z=3-2x-y over the triangular domain in the plane. +

    + +

    + The vertices of this triangular surface in space are + (1,0,1), (0,2,1), and (0,0,3). +

    +     + + + + + //ASY file for figsurfint1_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(16,12.5,14); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={2}; + real[] myzchoice={3}; + defaultpen(0.5mm); + + pair xbounds=(-.1,2.5); + pair ybounds=(-.1,2.5); + pair zbounds=(-.1,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane 2x+y+z=3 over the triangle bounded + // by coordinate planes and y=2-2x + triple f(pair t) { + return (t.x,t.y*(2-2*t.x),3-2*t.x-t.y*(2-2t.x));// + } + surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=invisible); + + draw(surface((1,0,0)--(0,0,0)--(0,2,0)--cycle),curvepen+opacity(.5)); + draw((1,0,0)--(0,0,0)--(0,2,0)--cycle,curvepen); + draw(f((0,0))--f((1,0))--f((0,1))--cycle,curvepen); + + +
    +     + +

    + We begin by parametrizing the planar surface \surfaceS. Using the techniques of the previous section, + we can let x=u and y=v(2-2u), + where 0\leq u\leq 1 and 0\leq v\leq 1. + Solving for z in the equation of the plane, + we have z=3-2x-y, hence z = 3-2u-v(2-2u), + giving the parametrization \vec r(u,v) = \langle u, v(2-2u), 3-2u-v(2-2u)\rangle. +

    + +

    + We need dS=\snorm{\vec r_u\times \vec r_v}dA, + so we need to compute \vec r_u, + \vec r_v and the norm of their cross product. + We leave it to the reader to confirm the following: + + \vec r_u = \langle 1,-2v,2v-2\rangle, \vec r_v = \langle 0,2-2u, 2u-2\rangle + , + + \vec r_u\times \vec r_v = \langle 4-4u,2-2u,2-2u\rangle \text{ and } \snorm{\vec r_u\times \vec r_v} = 2\sqrt{6}\sqrt{(u-1)^2} + . +

    + +

    + We need to be careful to not simplify + \snorm{\vec r_u\times \vec r_v} = 2\sqrt{6}\sqrt{(u-1)^2} as 2\sqrt{6}(u-1); + rather, it is 2\sqrt{6}|u-1|. + In this example, u is bounded by 0\leq u\leq 1, + and on this interval |u-1| = 1-u. + Thus dS = 2\sqrt{6}(1-u)dA. +

    + +

    + The density is given as a function of x, y and z, + for which we'll substitute the corresponding components of \vec r (with the slight abuse of notation that we used in previous sections): + + \delta(x,y,z) \amp = \delta\big(\vec r(u,v)\big) + \amp = u^2 + 5v(2-2u)+3-2u-v(2-2u) + \amp = u^2-8uv-2u+8v+3 + . +

    + +

    + Thus the mass of the sheet is: + + M \amp = \iint_\surfaceS\, dm + \amp = \iint_R \delta\big(\vec r(u,v)\big)\snorm{\vec r_u\times \vec r_v}\, dA + \amp = \int_0^1\int_0^1 \big(u^2-8uv-2u+8v+3\big)\big(2\sqrt{6}(1-u)\big)\, du\, dv + \amp = \frac{31}{\sqrt{6}} \approx 12.66 \text{ gm. } + +

    +     +     +     + + + Flux +

    + Let a surface \surfaceS lie within a vector field \vec F. + One is often interested in measuring the flux + of \vec F across \surfaceS; that is, + measuring how much of the vector field passes across \surfaceS. For instance, + if \vec F represents the velocity field of moving air and \surfaceS represents the shape of an air filter, + the flux will measure how much air is passing through the filter per unit time. + flux +

    + +

    + As flux measures the amount of \vec F passing across \surfaceS, we need to find the + amount of \vec F orthogonal to \surfaceS. + Similar to our measure of flux in the plane, + this is equal to \vec F\cdot \vec n, + where \vec n is a unit vector normal to \surfaceS at a point. + We now consider how to find \vec n. +

    + + + +

    + Given a smooth parametrization + \vec r(u,v) of \surfaceS, the work in the previous section showing the development of our method of computing surface area also shows that \vec r_u(u,v) and + \vec r_v(u,v) are tangent to \surfaceS at \vec r(u,v). + Thus \vec r_u\times \vec r_v is orthogonal to \surfaceS, and we let + + \vec n = \frac{\vec r_u\times \vec r_v}{\snorm{\vec r_u\times \vec r_v}} + , + which is a unit vector normal to \surfaceS at \vec r(u,v). +

    + +

    + The measurement of flux across a surface is a surface integral; + that is, to measure total flux we sum the product of + \vec F\cdot\vec n times a small amount of surface area: + \vec F\cdot \vec n\, dS. +

    + +

    + A nice thing happens with the actual computation of flux: + the \snorm{\vec r_u\times \vec r_v} terms go away. + Consider: + + \text{ Flux } \amp = \iint_\surfaceS \vec F\cdot \vec n\, dS + \amp = \iint_R \vec F\cdot \frac{\vec r_u\times \vec r_v}{\snorm{\vec r_u\times \vec r_v}}\snorm{\vec r_u\times \vec r_v}\, dA + \amp = \iint_R \vec F\cdot (\vec r_u\times \vec r_v)\, dA + . +

    + +

    + The above only makes sense if \surfaceS is orientable; + the normal vectors \vec n must vary continuously across \surfaceS. We assume that \vec n does vary continuously. + (If the parametrization \vec r of \surfaceS is smooth, + then our above definition of \vec n will vary continuously.) +

    + + + Flux over a surface + +

    + Let \vec F be a vector field with continuous components defined on an orientable surface \surfaceS with normal vector \vec n. + The flux of \vec F across \surfaceS is + flux + + \text{ Flux } = \iint_\surfaceS \vec F\cdot \vec n\, dS + . +

    + +

    + If \surfaceS is parametrized by \vec r(u,v), + which is smooth on its domain R, then + + \text{ Flux } = \iint_R \vec F\big(\vec r(u,v)\big)\cdot (\vec r_u\times \vec r_v)\, dA + . +

    +     +     + +

    + Since \surfaceS is orientable, + we adopt the convention of saying one passes from the back + side of \surfaceS to the front + side when moving across the surface parallel to the direction of \vec n. + Also, when \surfaceS is closed, + it is natural to speak of the regions of space + inside and outside + \surfaceS. We also adopt the convention that when \surfaceS is a closed surface, + \vec n should point to the outside of \surfaceS. If + \vec n = \vec r_u\times\vec r_v points inside \surfaceS, use \vec n = \vec r_v\times \vec r_u instead. +

    + +

    + When the computation of flux is positive, + it means that the field is moving from the back side of \surfaceS to the front side; + when flux is negative, + it means the field is moving opposite the direction of \vec n, + and is moving from the front of \surfaceS to the back. + When \surfaceS is not closed, + there is not a right and wrong + direction in which \vec n should point, + but one should be mindful of its direction to make full sense of the flux computation. +

    + +

    + We demonstrate the computation of flux, + and its interpretation, in the following examples. +

    + + + Finding flux across a surface + +

    + Let \surfaceS be the surface given in , + where \surfaceS is parametrized by + \vec r(u,v) = \langle u, v(2-2u),3-2u-v(2-2u)\rangle on 0\leq u\leq 1, + 0\leq v\leq 1, and let \vec F = \langle 1, x,-y\rangle, + as shown in . + Find the flux of \vec F across \surfaceS. +

    + +
    + The surface and vector field used in + + A triangular surface in space, and a vector field flowing across the surface. + +

    + The positive x, y, and z coordinate axes are drawn in space. + A triangular surface lies in the first octant above the xy plane; + it is the same surface that was plotted in +

    + +

    + A vector field in space is also plotted. + From the perspective used in the book, + the vectors appear to point mainly downward, + and are largest in magnitude to the right of the image, + where the value of y is largest. +

    +     + + + + + //ASY file for figsurfint1_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(16,12.5,14); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={2}; + real[] myzchoice={3}; + defaultpen(0.5mm); + + pair xbounds=(-.1,2.5); + pair ybounds=(-.1,2.5); + pair zbounds=(-.1,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane 2x+y+z=3 over the triangle bounded + // by coordinate planes and y=2-2x + triple f(pair t) { + return (t.x,t.y*(2-2*t.x),3-2*t.x-t.y*(2-2t.x));// + } + surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=invisible); + + path3 gradient1(triple z){ + return O--(1,z.x,-z.y); + } + + triple A=(0,0,1); + triple B=(1,2,3); + + picture VectorPlot3D(path3 vector(triple t), triple a, triple b, + int nx=nmesh, int ny=nx, int nz=nx,bool truesize=false, + real maxlength=truesize ? 0 : min(abs(b.x-a.x)/nx,abs(b.y-a.y)/ny,abs(b.z-a.z)/nz), + // bool cond(pair z)=null, + pen p=currentpen, + arrowbar3 arrow=Arrow3(6), margin3 margin=PenMargin3, + string name="", render render=defaultrender) + { + picture pic; + real dx=1/nx; + real dy=1/ny; + real dz=1/nz; + real scale; + if(maxlength > 0) { + real size(triple t) { + path3 g=vector(t); + return abs(point(g,size(g)-1)-point(g,0)); + } + real max=size((0,0,0)); + + for(int i=0; i <= nx; ++i) { + real x=interp(a.x,b.x,i*dx); + for(int j=0; j <= ny; ++j) + { + real y=interp(a.y,b.y,j*dy); + for(int k=0; k <= nz; ++k) + max=max(max,size((x,y,interp(a.z,b.z,k*dz)))); + }} + scale=max > 0 ? maxlength/max : 1; + } else scale=1; + bool group=name != "" || render.defaultnames; + if(group) + begingroup3(pic,name == "" ? "vectorfield" : name,render); + for(int i=0; i <= nx; ++i) { + real x=interp(a.x,b.x,i*dx); + for(int j=0; j <= ny; ++j) { + real y=interp(a.y,b.y,j*dy); + for(int k=0; k <= nz; ++k) + { triple z=(x,y,interp(a.z,b.z,k*dz)); + { + path3 g=scale3(scale)*vector(z); + string name="vector"; + if(truesize) { + picture opic; + draw(opic,g,p,arrow,margin,name,render); + add(pic,opic,z); + } else + draw(pic,shift(z)*g,p,arrow,margin,name,render); + } + } + }} + if(group) + endgroup3(pic); + return pic; + + } + add(VectorPlot3D(gradient1,A,B,3,3,3,bluepen)); + + + +
    +     + +

    + Using our work from the previous example, + we have \vec n = \vec r_u\times\vec r_v = \langle 4-4u,2-2u,2-2u\rangle. + We also need \vec F\big(\vec r(u,v)\big) = \langle 1, u, -v(2-2u)\rangle. +

    + +

    + Thus the flux of \vec F across \surfaceS is: + + \text{ Flux } \amp = \iint_\surfaceS \vec F\cdot \vec n\, dS + \amp = \iint_R \langle 1,u,-v(2-2u)\rangle\cdot\langle 4-4u,2-2u,2-2u\rangle\, dA + \amp = \int_0^1\int_0^1 \big(-4u^2v-2u^2+8uv-2u-4v+4\big)\, du\, dv + \amp = 5/3 + . +

    + +

    + To make full use of this numeric answer, + we need to know the direction in which the field is passing across \surfaceS. The graph in helps, + but we need a method that is not dependent on a graph. +

    + +

    + Pick a point (u,v) in the interior of R and consider \vec n(u,v). + For instance, + choose (1/2,1/2) and look at \vec n(1/2,1/2) = \langle 2,1,1\rangle/\sqrt{6}. + This vector has positive x, + y and z components. + Generally speaking, one has some + idea of what the surface \surfaceS looks like, + as that surface is for some reason important. + In our case, + we know \surfaceS is a plane with z-intercept of z=3. + Knowing \vec n and the flux measurement of positive 5/3, + we know that the field must be passing from behind + \surfaceS, , the side the origin is on, + to the front of \surfaceS. +

    +     +     + + + + + Flux across surfaces with shared boundaries + +

    + Let \surfaceS_1 be the unit disk in the xy-plane, + and let \surfaceS_2 be the paraboloid z=1-x^2-y^2, + for z\geq 0, + as graphed in . + Note how these two surfaces each have the unit circle as a boundary. +

    + +
    + The surfaces used in + + A closed surface consisting of a downward-opening circular paraboloid, and a disc in the x,y plane. + +

    + The surface \surfaceS_2 is a circular paraboloid, opening downward, + with its vertex on the z axis at (0,0,1). + It intersects the xy plane in the unit circle. +

    + +

    + The unit circle and its interior form the disk that is \surfaceS_1. + The two surfaces are therefore joined along the unit circle in the xy plane, + forming a closed surface like a dome or a hill. +

    +     + + + + + //ASY file for figsurfint1_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(22.8,20,5.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={2}; + real[] myzchoice={3}; + defaultpen(0.5mm); + + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + pair zbounds=(-.1,1.1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane unit disk + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),0);// + } + surface s=surface(f,(-pi/2,0),(3pi/2,1),8,8,usplinetype=Spline,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=redmeshpen; + draw(s,surfacepen2,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),1-t.y^2);// + } + surface s=surface(f,(0,0),(2pi,1),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(f,0,2pi,operator ..); + draw(mypath,curvepen); + + + +
    + +

    + Let \vec F_1 = \langle 0,0,1\rangle and \vec F_2 = \langle 0,0,z\rangle. + Using normal vectors for each surface that point upward, + , with a positive z-component, + find the flux of each field across each surface. +

    +     + +

    + We begin by parametrizing each surface. +

    + +

    + The boundary of the unit disk in the xy-plane is the unit circle, + which can be described with + \langle \cos(u),\sin(u),0\rangle, 0\leq u\leq 2\pi. + To obtain the interior of the circle as well, + we can scale by v, giving + + \vec r_1(u,v) = \langle v\cos(u),v\sin(u), 0\rangle, 0\leq u\leq 2\pi 0\leq v\leq 1 + . +

    + +

    + As the boundary of \surfaceS_2 is also the unit circle, + the x and y components of + \vec r_2 will be the same as those of \vec r_1; + we just need a different z component. + With z = 1-x^2-y^2, we have + + \vec r_2(u,v) = \langle v\cos(u),v\sin(u), 1-v^2\cos^2u-v^2\sin^2u\rangle = \langle v\cos(u),v\sin(u), 1-v^2\rangle + , + where 0\leq u\leq 2\pi and 0\leq v\leq 1. +

    + +

    + We now compute the normal vectors \vec n_1 and \vec n_2. +

    + +

    + For \vec n_1: + \vec r_{1u}= \langle -v\sin(u), v\cos(u),0\rangle, + \vec r_{1v} = \langle \cos(u),\sin(u),0\rangle, so + + \vec n_1 = \vec r_{1u}\times \vec r_{1v} = \langle 0,0,-v\rangle + . +

    + +

    + As this vector has a negative z-component, we instead use + + \vec n_1 = \vec r_{1v}\times \vec r_{1u} = \langle 0,0,v\rangle + . +

    + +

    + Similarly, \vec n_2: + \vec r_{2u}= \langle -v\sin(u), v\cos(u),0\rangle, + \vec r_{2v} = \langle \cos(u),\sin(u),-2v\rangle, so + + \vec n_2 = \vec r_{2u}\times \vec r_{2v} = \langle -2v^2\cos(u),-2v^2\sin(u),-v\rangle + . +

    + +

    + Again, this normal vector has a negative z-component so we use + + \vec n_2 = \vec r_{2v}\times \vec r_{2u} = \langle 2v^2\cos(u),2v^2\sin(u),v\rangle + . +

    + +

    + We are now set to compute flux. + Over field \vec F_1=\langle 0,0,1\rangle: + + \text{ Flux across } \surfaceS_1 \amp = \iint_{\surfaceS_1} \vec F_1\cdot \vec n_1\, dS + \amp = \iint_R\langle 0,0,1\rangle\cdot\langle 0,0,v\rangle\, dA + \amp = \int_0^1\int_0^{2\pi} (v)\, du\, dv + \amp = \pi + . + + \text{ Flux across } \surfaceS_2 \amp = \iint_{\surfaceS_2} \vec F_1\cdot \vec n_2\, dS + \amp = \iint_R\langle 0,0,1\rangle\cdot\langle 2v^2\cos(u),2v^2\sin(u),v\rangle\, dA + \amp = \int_0^1\int_0^{2\pi} (v)\, du\, dv + \amp = \pi + . +

    + +

    + These two results are equal and positive. + Each are positive because both normal vectors are pointing in the positive z-directions, + as does \vec F_1. + As the field passes through each surface in the direction of their normal vectors, + the flux is measured as positive. +

    + +

    + We can also intuitively understand why the results are equal. + Consider \vec F_1 to represent the flow of air, + and let each surface represent a filter. + Since \vec F_1 is constant, + and moving straight up, + it makes sense that all air passing through \surfaceS_1 also passes through + \surfaceS_2, and vice-versa. +

    + +

    + If we treated the surfaces as creating one piecewise-smooth surface \surfaceS, we would find the total flux across \surfaceS by finding the flux across each piece, + being sure that each normal vector pointed to the outside of the closed surface. + Above, \vec n_1 does not point outside the surface, + though \vec n_2 does. + We would instead want to use -\vec n_1 in our computation. + We would then find that the flux across \surfaceS_1 is -\pi, + and hence the total flux across \surfaceS is -\pi + \pi = 0. + (As 0 is a special number, + we should wonder if this answer has special significance. + It does, which is briefly discussed following this example and will be more fully developed in the next section.) +

    + +

    + We now compute the flux across each surface with \vec F_2=\langle 0,0,z\rangle: + + \text{ Flux across } \surfaceS_1 \amp = \iint_{\surfaceS_1} \vec F_2\cdot \vec n_1\, dS . + Over \surfaceS_1, \vec F_2 = \vec F_2\big(\vec r_2(u,v)\big) = \langle 0,0,0\rangle. Therefore, + \amp = \iint_R\langle 0,0,0\rangle\cdot\langle 0,0,v\rangle\, dA + \amp = \int_0^1\int_0^{2\pi} (0)\, du\, dv + \amp = 0 + . + + \text{ Flux across } \surfaceS_2 \amp = \iint_{\surfaceS_2} \vec F_2\cdot \vec n_2\, dS . + Over \surfaceS_2, \vec F_2 = \vec F_2\big(\vec r_2(u,v)\big) = \langle 0,0,1-v^2\rangle. Therefore, + \amp = \iint_R\langle 0,0,1-v^2\rangle\cdot\langle 2v^2\cos(u),2v^2\sin(u),v\rangle\, dA + \amp = \int_0^1\int_0^{2\pi} (v^3-v)\, du\, dv + \amp = \pi/2 + . +

    + +

    + This time the measurements of flux differ. + Over \surfaceS_1, + the field \vec F_2 is just \vec 0, hence there is no flux. + Over \surfaceS_2, + the flux is again positive as + \vec F_2 points in the positive z direction over + \surfaceS_2, as does \vec n_2. +

    +     +     + +

    + In the previous example, + the surfaces \surfaceS_1 and + \surfaceS_2 form a closed surface that is piecewise smooth. + That the measurement of flux across each surface was the same for some fields + (and not for others) + is reminiscent of a result from , + where we measured flux across curves. + The quick answer to why the flux was the same when considering + \vec F_1 is that \divv \vec F_1 = 0. + In the next section, + we'll see the second part of the Divergence Theorem, + which will more fully explain this occurrence. + We will also explore Stokes' Theorem, + the spatial analogue to Green's Theorem. +

    + + + + + + +     + + + + Terms and Concepts + + + +

    + In the plane, + flux is a measurement of how much of the vector field passes across a ; + in space, flux is a measurement of how much of the vector field passes across a . +

    +     + + + + + + + + + + + + +     + + + +

    + When computing flux, + what does it mean when the result is a negative number? +

    +     + + + +

    + Answers will vary; in general, + it means that more of the vector field passes through the surface opposite the + direction of the normal vector than in the same direction of the normal vector. +

    +     +     + + + +

    + When \surfaceS is a closed surface, + we choose the normal vector so that it points to the of the surface. +

    +     + + + + outside|exterior + + + +     + + + +

    + If \surfaceS is a plane, + and \vec F is always parallel to \surfaceS, + then the flux of \vec F across \surfaceS will be . +

    +     + + + + 0|zero + + + +     +     + + + Problems + + + +

    + A surface \surfaceS that represents a thin sheet of material with density \delta is given. + Find the mass of each thin sheet. +

    +     + + + + +

    + \surfaceS is the plane + z=x+y on -2\leq x\leq 2, + -3\leq y\leq 3, with \delta(x,y,z) = z+10. +

    +     + +

    + 240\sqrt{3} +

    +     +     + + + +

    + \surfaceS is the unit sphere, + with \delta(x,y,z) = x+y+z+10. +

    +     + +

    + 40\pi +

    +     +     +     + + + +

    + A surface \surfaceS and a vector field \vec F are given. + Compute the flux of \vec F across \surfaceS. + (If \surfaceS is not a closed surface, + choose \vec n so that it has a positive z-component, + unless otherwise indicated.) +

    +     + + + +

    + \surfaceS is the plane z = 3x+y on + 0\leq x\leq 1, 1\leq y\leq 4; + \vec F = \langle x^2,-z,2y\rangle. +

    +     + +

    + 24 +

    +     +     + + + +

    + \surfaceS is the plane + z = 8-x-y over the triangle with vertices at (0,0), + (1,0) and (1,5); + \vec F = \langle 3,1,2\rangle. +

    +     + +

    + 15 +

    +     +     + + + +

    + \surfaceS is the paraboloid z = x^2+y^2 over the unit disk; + \vec F = \langle 1,0,0\rangle. +

    +     + +

    + 0 +

    +     +     + + + +

    + \surfaceS is the unit sphere; + \vec F = \langle y-z,z-x,x-y\rangle. +

    +     + +

    + 0 +

    +     +     + + + +

    + \surfaceS is the square in space with corners at (0,0,0), + (1,0,0), + (1,0,1) and (0,0,1) (choose \vec n such that it has a positive y-component); + \vec F = \langle 0,-z,y\rangle. +

    +     + +

    + -1/2 +

    +     +     + + + +

    + \surfaceS is the disk in the yz-plane with radius 1, centered at (0,1,1) (choose \vec n such that it has a positive x-component); + \vec F = \langle y,z,x\rangle. +

    +     + +

    + \pi +

    +     +     + + + +

    + \surfaceS is the closed surface composed of \surfaceS_1, + whose boundary is the ellipse in the xy-plane described by + \frac{x^2}{25}+\frac{y^2}9 = 1 and \surfaceS_2, + part of the elliptical paraboloid f(x,y) = 1-\frac{x^2}{25}-\frac{y^2}9 (see graph); + \vec F = \langle 5,2,3\rangle. +

    + + + An elliptical paraboloid opens downward, intersecting the x,y plane in an ellipse. + +

    + An elliptical paraboloid is plotted relative to the usual three-dimensional coorindate axes. + The paraboloid has its vertex on the z axis at (0,0,1), and opens downward. +

    + +

    + The paraboloid intersects the xy plane along an ellipse; + the surface \surfaceS_2 consists of the portion of the paraboloid with 0\leq z\leq 1. +

    + +

    + The surface \surfaceS_1 is illustrated as the region on and inside the ellipse where \surfaceS_2 meets the xy plane. +

    +     + + + + + //ASY file for + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(19,17,1); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-5,5}; + real[] myychoice={-3,3}; + real[] myzchoice={1}; + defaultpen(0.5mm); + pair xbounds=(-5.75,5.75); + pair ybounds=(-5.75,5.75); + pair zbounds=(-0.25,1.1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the parabola z=y^2 for t from -2 to 2 + triple g(real t) {return (5cos(t),3sin(t),0);} + path3 mypath=graph(g,0,2pi,operator ..); draw(mypath,bluepen); + + //Draw the cylinder z=y^2 + triple f(pair t) { + return (5t.y*cos(t.x),3t.y*sin(t.x),0); + } + surface s=surface(f,(0,0),(2pi,1),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the cylinder z=y^2 + triple f(pair t) { + return (5t.y*cos(t.x),3t.y*sin(t.x),1-t.y^2); + } + surface s=surface(f,(0,0),(2pi,1),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + +     + +

    + 0; the flux over \surfaceS_1 is -45\pi and the flux over \surfaceS_2 is 45\pi. +

    +     +     + + + +

    + \surfaceS is the closed surface composed of \surfaceS_1, + part of the unit sphere and \surfaceS_2, + part of the plane z=1/2 (see graph); + \vec F = \langle x,-y,z\rangle. +

    + + + Approximately three quarters of the unit sphere. The top has been removed and replaced by a disk. + +

    + The unit sphere is plotted with respect to the usual three-dimensional coordnate axes. + However, it is not the entire sphere, but the portion with -1\leq z\leq \frac12. +

    + +

    + The top portion of the sphere has been removed, + an in its place is a disk in the plane z=\frac12. + The center of the disk is at \left(0,0,\frac12\right), + and the radius is \frac{\sqrt{3}}{2}. +

    + +

    + The surface looks something like what you'd get by slicing the top off of an orange. +

    +     + + + + + //ASY file for + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(17,17,11); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,1.25); + pair zbounds=(-1.25,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the parabola z=y^2 for t from -2 to 2 + triple g(real t) {return (sqrt(3)/2*cos(t),sqrt(3)/2*sin(t),.5);} + path3 mypath=graph(g,0,2pi,operator ..); draw(mypath,bluepen); + + //Draw the cylinder z=y^2 + triple f(pair t) { + return (sin(t.x)*cos(t.y),sin(t.x)*sin(t.y),cos(t.x)); + } + surface s=surface(f,(pi/3,0),(pi,2pi),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw the cylinder z=y^2 + triple f(pair t) { + return (t.y*sqrt(3)/2*cos(t.x),t.y*sqrt(3)/2*sin(t.x),.5); + } + surface s=surface(f,(0,0),(2pi,1),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + + +     + +

    + 9\pi/8; the flux over + \surfaceS_1 is 3\pi/4 (use + \vec r(u,v) = \langle \sin(u)\cos(v),\sin(u)\sin(v),\cos(u)\rangle on \pi/3\leq u\leq \pi, + 0\leq v\leq 2\pi) and the flux over + \surfaceS_2 is 3\pi/8 (use \vec r(u,v) = \langle v\sqrt{3}\cos (u)/2, v\sqrt{3}\sin(u)/2,1/2\rangle for + 0\leq u\leq 2\pi, 0\leq v\leq 1. +

    +     +     +     +     +     +
    +
    + The Divergence Theorem and Stokes' Theorem + + The Divergence Theorem +

    + + gives the Divergence Theorem in the plane, + which states that the flux of a vector field across a closed curve + equals the sum of the divergences over the region enclosed by the curve. + Recall that the flux was measured via a line integral, + and the sum of the divergences was measured through a double integral. +

    + +

    + We now consider the three-dimensional version of the Divergence Theorem. + It states, in words, that the flux across a closed surface + equals the sum of the divergences over the domain enclosed by the surface. + Since we are in space + (versus the plane), + we measure flux via a surface integral, + and the sums of divergences will be measured through a triple integral. +

    + + + + + The Divergence Theorem (in space) + +

    + Let D be a closed domain in space whose boundary is an orientable, + piecewise smooth surface \surfaceS with outer unit normal vector \vec n, + and let \vec F be a vector field whose components are differentiable on D. + ThenDivergence Theoremin space + + \iint_\surfaceS \vec F\cdot\vec n\, dS =\iiint_D \divv \vec F\, dV + . +

    +     +     + + + + + Using the Divergence Theorem in space + +

    + Let D be the domain in space bounded by the planes z=0 and z=2x, + along with the cylinder x=1-y^2, + as graphed in , + let \surfaceS be the boundary of D, + and let \vec F = \langle x+y,y^2, 2z\rangle. +

    + +
    + The surfaces used in + + A parabolic wedge, bounded by two planes and a parabolic cylinder. + +

    + The x, y, and z axes are plotted in three dimensions. + The z axis points up, the x axis points back and to the left, + and the y axis points toward us and to the left. +

    + +

    + The surface consists of three components: +

      +
    • +

      + In the xy plane is the region bounded by the y axis + and the parabola x=1-y^2. +

      + +

      + The parabola has its vertex at (1,0,0) and it meets the + y axis at (0,1,0) and (0,-1,0) +

      +
      • + +
    • +

      + Another parabolic region lies in the plane z=2x. + This plane intersects the y axis along the segment -1\leq y\leq 1, + which is the common bottom of the two parabolic regions; + it forms a sharp edge where the two planes meet. +

      + +

      + The parabolic boundary of this region is the curve given by + \vec{r}(t) = \la 1-t^2, t, 2-2t^2\ra. + This is a parabola in the plane z=2x that lies directly above the parabola x=1-y^2 in the xy plane. +

      +
      • + +
    • +

      + The remaining surface is the partion of the parabolic cylinder x=1-y^2 + that lies between the planes z=0 and z=2x. +

      +
      • +
    +

    + +

    + Overall, the surfaces form a wedge shape similar to a segment that might be cut from a tree being chopped down. +

    +     + + + + + //ASY file for divthm_space2_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(-17,32,12.6); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={-1,1}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-.1,1.2); + pair ybounds=(-1.3,1.3); + pair zbounds=(-.1,2.2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the planar parabola + triple f(pair t) { + return (t.y*(1-t.x^2),t.x,0);// + } + surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=redmeshpen; + draw(s,surfacepen2,meshpen=p); + + //Draw plane over parabola + triple f(pair t) { + return (t.y*(1-t.x^2),t.x,2*t.y*(1-t.x^2));// + } + surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw parabolic cylinder side + triple f(pair t) { + return ((1-t.x^2),t.x,2*t.y*(1-t.x^2));// + } + surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(real t) {return (1-t^2,t,0);} + path3 mypath=graph(f,-1,1,operator ..); + draw(mypath,curvepen); + + triple f(real t) {return (1-t^2,t,2*(1-t^2));} + path3 mypath=graph(f,-1,1,operator ..); + draw(mypath,curvepen); + + + +
    + +

    + Verify the Divergence Theorem by finding the total outward flux of \vec F across \surfaceS, and show this is equal to \iiint_D \divv\vec F\, dV. +

    +     + +

    + The surface \surfaceS is piecewise smooth, + comprising surfaces \surfaceS_1, + which is part of the plane z=2x, + surface \surfaceS_2, which is part of the cylinder x=1-y^2, + and surface \surfaceS_3, + which is part of the plane z=0. + To find the total outward flux across \surfaceS, we need to compute the outward flux across each of these three surfaces. +

    + +

    + We leave it to the reader to confirm that surfaces \surfaceS_1, + \surfaceS_2 and \surfaceS_3 can be parametrized by \vec r_1, + \vec r_2 and \vec r_3 respectively as + + \vec r_1(u,v) \amp = \la v(1-u^2), u, 2v(1-u^2)\ra, + \vec r_2(u,v) \amp = \la (1-u^2), u, 2v(1-u^2)\ra, + \vec r_3(u,v) \amp = \la v(1-u^2), u, 0\ra + , + where -1\leq u\leq 1 and + 0\leq v\leq 1 for all three functions. +

    + +

    + We compute a unit normal vector \vec n for each as \frac{\vec r_u\times\vec r_v}{\snorm{\vec r_u\times\vec r_v}}, + though recall that as we are integrating \vec F\cdot \vec n\, dS, + we actually only use \vec r_u\times\vec r_v. + Finally, in previous flux computations, + it did not matter which direction \vec n pointed as long as we made note of its direction. + When using the Divergence Theorem, + we need \vec n to point to the outside of the closed surface, + so in practice this means we'll either use + \vec r_u\times\vec r_v or \vec r_v\times\vec r_u, + depending on which points outside of the closed surface \surfaceS. +

    + +

    + We leave it to the reader to confirm the following cross products and integrations are correct. +

    + +

    + For \surfaceS_1, + we need to use \vec r_{1v}\times\vec r_{1u} = \langle 2(u^2-1),0,1-u^2\rangle. + (Note the z-component is nonnegative as u\leq 1, + therefore this vector always points up, + meaning to the outside, of \surfaceS.) + The flux across \surfaceS_1 is: + + \amp \text{ Flux across }\surfaceS_1: + \amp \quad = \iint_{\surfaceS_1} \vec F\cdot \vec n_1\, dS + \amp \quad = \int_0^1\int_{-1}^1 \vec F\big(\vec r_1(u,v)\big)\cdot \big(\vec r_{1v}\times\vec r_{1u}\big)\, du\, dv + \amp \quad = \int_0^1\int_{-1}^1 \la v(1-u^2)+u, u^2,4v(1-u^2)\ra \cdot \la 2(u^2-1),0,1-u^2\ra\, du\, dv + \amp \quad = \int_0^1\int_{-1}^1 \big(2u^4v+2u^3-4u^2v-2u+2v\big)\, du\, dv + \amp \quad = \frac{16}{15} + . +

    + +

    + For \surfaceS_2, + we use \vec r_{2u}\times\vec r_{2v} = \langle 2(1-u^2), 4u(1-u^2),0\rangle. + (Note the x-component is always nonnegative, + meaning this vector points outside \surfaceS.) + The flux across \surfaceS_2 is: + + \amp \text{ Flux across } \surfaceS_2: + \amp \quad = \iint_{\surfaceS_2} \vec F\cdot \vec n_2\, dS + \amp \quad = \int_0^1\int_{-1}^1 \vec F\big(\vec r_2(u,v)\big)\cdot \big(\vec r_{2u}\times\vec r_{2v}\big)\, du\, dv + \amp \quad = \int_0^1\int_{-1}^1 \la 1-u^2+u, u^2, 4v(1-u^2)\ra \cdot \la 2(1-u^2), 4u(1-u^2),0\ra\, du\, dv + \amp \quad = \int_0^1\int_{-1}^1 \big(4u^5-2u^4-2u^3+4u^2-2u-2\big)\, du\, dv + \amp \quad = \frac{32}{15} + . +

    + +

    + For \surfaceS_3, + we use \vec r_{3u}\times\vec r_{3v} = \langle 0,0,u^2-1\rangle. + (Note the z-component is never positive, + meaning this vector points down, outside of \surfaceS.) + The flux across \surfaceS_3 is: + + \amp\text{ Flux across } \surfaceS_3: + \amp \quad = \iint_{\surfaceS_3} \vec F\cdot \vec n_3\, dS + \amp \quad = \int_0^1\int_{-1}^1 \vec F\big(\vec r_3(u,v)\big)\cdot \big(\vec r_{3u}\times\vec r_{3v}\big)\, du\, dv + \amp \quad = \int_0^1\int_{-1}^1 \la v(1-u^2)+u,u^2,0\ra \cdot \la 0,0,u^2-1\ra\, du\, dv + \amp \quad = \int_0^1\int_{-1}^1 0\, du\, dv + \amp \quad = 0 + . +

    + +

    + Thus the total outward flux, + measured by surface integrals across all three component surfaces of \surfaceS, is 16/15+32/15+0 = 48/15 = 16/5 = 3.2. + We now find the total outward flux by integrating \divv \vec F over D. +

    + +

    + Following the steps outlined in , + we see the bounds of x, + y and z can be set as (thinking surface to surface, + curve to curve, point to point): + + 0\leq z\leq 2x; 0\leq x\leq 1-y^2; -1\leq y\leq 1 + . +

    + +

    + With \divv \vec F = 1+2y+2 = 2y+3, + we find the total outward flux of \vec F over \surfaceS as: + + \text{ Flux = } \iiint_D\divv \vec F\, dV = \int_{-1}^1\int_0^{1-y^2}\int_0^{2x}\big(2y+3\big)\, dz\, dx\, dy = 16/5 + , + the same result we obtained previously. +

    +     +     + +

    + In + we see that the total outward flux of a vector field across a closed surface can be found two different ways because of the Divergence Theorem. + One computation took far less work to obtain. + In that particular case, + since \surfaceS was comprised of three separate surfaces, + it was far simpler to compute one triple integral than three surface integrals + (each of which required partial derivatives and a cross product). + In practice, if outward flux needs to be measured, + one would choose only one method. + We will use both methods in this section simply to reinforce the truth of the Divergence Theorem. +

    + +

    + We practice again in the following example. +

    + + + Using the Divergence Theorem in space + +

    + Let \surfaceS be the surface formed by the paraboloid z=1-x^2-y^2, + z\geq 0, + and the unit disk centered at the origin in the xy-plane, + graphed in , + and let \vec F = \langle 0,0,z\rangle. + (This surface and vector field were used in .) +

    + +
    + The surfaces used in + + A circular paraboloid, opening downward, meets the x,y plane along the unit circle. The interior of the circle is a disk. + +

    + This is the same image used in . + A circular paraboloid has its vertex at (0,0,1) on the z axis and opens downward, + meeting the xy plane along the unit circle. +

    + +

    + The interior of the unit circle is a disk that serves as a cap at the bottom of the paraboloid. +

    +     + + + + + //ASY file for figsurfint1_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(22.8,20,5.5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1}; + real[] myychoice={2}; + real[] myzchoice={3}; + defaultpen(0.5mm); + + pair xbounds=(-1.1,1.1); + pair ybounds=(-1.1,1.1); + pair zbounds=(-.1,1.1); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane unit disk + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),0);// + } + surface s=surface(f,(-pi/2,0),(3pi/2,1),8,8,usplinetype=Spline,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=redmeshpen; + draw(s,surfacepen2,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),1-t.y^2);// + } + surface s=surface(f,(0,0),(2pi,1),8,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(real t) {return (cos(t),sin(t),0);} + path3 mypath=graph(f,0,2pi,operator ..); + draw(mypath,curvepen); + + + +
    + +

    + Verify the Divergence Theorem; + find the total outward flux across \surfaceS and evaluate the triple integral of \divv \vec F, + showing that these two quantities are equal. +

    +     + +

    + We find the flux across \surfaceS first. + As \surfaceS is piecewise-smooth, + we decompose it into smooth components \surfaceS_1, + the disk, + and \surfaceS_2, the paraboloid, + and find the flux across each. +

    + +

    + In , + we found the flux across \surfaceS_1 is 0. + We also found that the flux across \surfaceS_2 is \pi/2. + (In that example, + the normal vector had a positive z component hence was an outer normal.) + Thus the total outward flux is 0+\pi/2 = \pi/2. +

    + +

    + We now compute \iiint_D \divv \vec F\, dV. + We can describe D as the domain bounded by (think surface to surface, + curve to curve, point to point): + + 0\leq z\leq 1-x^2-y^2, -\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}, -1\leq x\leq 1 + . +

    + +

    + This description of D is not very easy to integrate. + With polar, we can do better. + Let R represent the unit disk, + which can be described in polar simply as r, + where 0\leq r\leq 1 and 0\leq \theta\leq 2\pi. + With x=r\cos \theta and y=r\sin\theta, + the surface \surfaceS_2 becomes + + z=1-x^2-y^2 \Rightarrow 1-(r\cos\theta)^2-(r\sin\theta)^2 \Rightarrow 1-r^2 + . +

    + +

    + Thus D can be described as the domain bounded by: + + 0\leq z\leq 1-r^2, 0\leq r\leq 1, 0\leq \theta\leq 2\pi + . +

    + +

    + With \divv \vec F = 1, + we can integrate, recalling that dV = r\, dz\, dr\, d\theta: + + \iiint_D\divv \vec F\, dV = \int_0^{2\pi}\int_0^1\int_0^{1-r^2} r\, dz\, dr\, d\theta = \frac{\pi}2 + , + which matches our flux computation above. +

    +     +     + + + A <q>paradox</q> of the Divergence Theorem and Gauss's Law + +

    + The magnitude of many physical quantities + (such as light intensity or electromagnetic and gravitational forces) + follow an inverse square law: + the magnitude of the quantity at a point is inversely proportional to the square of the distance to the source of the quantity. + Gauss's Law +

    + +

    + Let a point light source be placed at the origin and let \vec F be the vector field which describes the intensity and direction of the emanating light. + At a point (x,y,z), + the unit vector describing the direction of the light passing through that point is \langle x,y,z\rangle/\sqrt{x^2+y^2+z^2}. + As the intensity of light follows the inverse square law, + the magnitude of \vec F at (x,y,z) is + k/(x^2+y^2+z^2) for some constant k. + Taken together, + + \vec F(x,y,z) = \frac{k}{(x^2+y^2+z^2)^{3/2}}\langle x,y,z\rangle + . +

    + +

    + Consider the cube, centered at the origin, + with sides of length 2a for some a \gt 0 (hence corners of the cube lie at (a,a,a), + (-a,-a,-a), + etc., as shown in ). + Find the flux across the six faces of the cube and compare this to \iint_D \divv\vec F\, dV. +

    + +
    + The cube used in + + A cube centered at the origin in space. + +

    + A cube is drawn in space, relative to three-dimensional coordinate axes. + It is centered at the origin, and its faces are parallel to the coordinate planes. +

    + +

    + One corner of the cube (the nearest one in the perspective used for the image) + is labeled with the coordinates (a,a,a). + The opposite corner is labeled with coordinates (-a,-a,-a). +

    +     + + + + + //ASY file for divthm_space3_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(30,20,12); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={}; + real[] myychoice={}; + real[] myzchoice={}; + defaultpen(0.5mm); + + pair xbounds=(-1.2,1.2); + pair ybounds=(-1.2,1.2); + pair zbounds=(-1.2,1.2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the faces of the cube + pen p=apexmeshpen; + + triple f(pair t) { + return (t.x,t.y,1);// + } + surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (t.x,t.y,-1);// + } + surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (t.x,1,t.y);// + } + surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (t.x,-1,t.y);// + } + surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (1,t.x,t.y);// + } + surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (-1,t.x,t.y);// + } + surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + draw(s,surfacepen,meshpen=p); + + label("$(a,a,a)$",(1.05,1.05,1.1)); + label("$(-a,-a,-a)$",(-1.05,-1.05,-1.1)); + dot((1,1,1),black+1mm); + + dot((-1,-1,-1),black+1mm); + + + +
    +     + +

    + Let \surfaceS_1 be the top face of the cube, + which can be parametrized by \vec r(u,v) = \langle u,v,a\rangle for + -a\leq u\leq a, -a\leq v\leq a. + We leave it to the reader to confirm that \vec r_u\times \vec r_v = \langle 0,0,1\rangle, + which points outside of the cube. +

    + +

    + The flux across this face is: + + \text{ Flux } \amp = \iint_{\surfaceS_1} \vec F\cdot \vec n\, dS + \amp = \int_{-a}^a\int_{-a}^a \vec F\big(\vec r(u,v)\big)\cdot \big(\vec r_u\times \vec r_v\big)\, du\, dv + \amp = \int_{-a}^a\int_{-a}^a \frac{k\ a}{(u^2+v^2+a^2)^{3/2}}\, du\, dv. + This double integral is not trivial to compute, requiring multiple trigonometric substitutions. This example is not meant to stress integration techniques, so we leave it to the reader to confirm the result is + \amp = \frac{2k\pi}3 + . +

    + +

    + Note how the result is independent of a; + no matter the size of the cube, + the flux through the top surface is always 2k\pi/3. +

    + +

    + An argument of symmetry shows that the flux through each of the six faces is 2k\pi/3, + thus the total flux through the faces of the cube is 6\times2k\pi/3 = 4k\pi. +

    + +

    + It takes a bit of algebra, but we can show that \divv\vec F = 0. + Thus the Divergence Theorem would seem to imply that the total flux through the faces of the cube should be + + \text{ Flux } =\iiint_D \divv \vec F\, dV = \iiint_D 0\, dV = 0 + , + but clearly this does not match the result from above. + What went wrong? +

    + +

    + Revisit the statement of the Divergence Theorem. + One of the conditions is that the components of \vec F must be differentiable on the domain enclosed by the surface. + In our case, \vec F is not + differentiable at the origin it is not even defined! + As \vec F does not satisfy the conditions of the Divergence Theorem, + it does not apply, + and we cannot expect \iint_\surfaceS \vec F\cdot\vec n\, dA = \iiint_D \divv\vec F\, dV. +

    + +

    + Since \vec F is differentiable everywhere except the origin, + the Divergence Theorem does apply over any domain that does not include the origin. + Let \surfaceS_2 be any surface that encloses the cube used before, + and let \hat D be the domain between + the cube and \surfaceS_2; + note how \hat D does not include the origin and so the Divergence Theorem does apply over this domain. + The total outward flux over \hat D is thus \iint_{\hat D}\divv \vec F\, dV = 0, + which means the amount of flux coming out of + \surfaceS_2 is the same as the amount of flux coming out of the cube. + The conclusion: the flux across any + surface enclosing the origin will be 4k\pi. +

    + +

    + This has an important consequence in electrodynamics. + Let q be a point charge at the origin. + The electric field generated by this point charge is + + \vec E = \frac{q}{4\pi \epsilon_0}\frac{\la x,y,z\ra}{(x^2+y^2+z^2)^{3/2}} + , + , it is \vec F with k = q/(4\pi \epsilon_0), + where \epsilon_0 is a physical constant + (the permittivity of free space). + Gauss's Law states that the outward flux of \vec E across any surface enclosing the origin is q/\epsilon_0. +

    +     +     + + + +

    + Our interest in the Divergence Theorem is twofold. + First, its truth alone is interesting: + to study the behavior of a vector field across a closed surface, + one can examine properties of that field within the surface. + Secondly, it offers an alternative way of computing flux. + When there are multiple methods of computing a desired quantity, + one has power to select the easiest computation as illustrated next. +

    + + + + + + Using the Divergence Theorem to compute flux + +

    + Let \surfaceS be the cube bounded by the planes x=\pm 1, + y=\pm 1, z=\pm 1, + and let \vec F = \langle x^2y,2yz,x^2z^3\rangle. + Compute the outward flux of \vec F over \surfaceS. +

    +     + +

    + We compute \divv \vec F = 2xy+2z+3x^2z^2. + By the Divergence Theorem, + the outward flux is the triple integral over the domain D enclosed by \surfaceS: + + \text{ Outward flux: } \int_{-1}^1\int_{-1}^1\int_{-1}^1(2xy+2z+3x^2z^2)\, dz\, dy\, dx = \frac83 + . +

    + +

    + The direct flux computation requires six surface integrals, + one for each face of the cube. + The Divergence Theorem offers a much more simple computation. +

    +     +     +     + + + Stokes' Theorem +

    + Just as the spatial Divergence Theorem of this section is an extension of the planar Divergence Theorem, Stokes' Theorem is the spatial extension of Green's Theorem. + Recall that Green's Theorem states that the circulation of a vector field around a closed curve in the plane is equal to the sum of the curl of the field over the region enclosed by the curve. + Stokes' Theorem effectively makes the same statement: + given a closed curve that lies on a surface \surfaceS, the circulation of a vector field around that curve is the same as the sum of + the curl of the field + across the enclosed surface. + We use quotes around the curl of the field + to signify that this statement is not quite correct, + as we do not sum \curl \vec F, + but \curl \vec F\cdot\vec n, + where \vec n is a unit vector normal to \surfaceS. That is, + we sum the portion of \curl \vec F that is orthogonal to \surfaceS at a point. +

    + + + +

    + Green's Theorem dictated that the curve was to be traversed counterclockwise when measuring circulation. + Stokes' Theorem will follow a right hand rule: + when the thumb of one's right hand points in the direction of \vec n, + the path C will be traversed in the direction of the curling fingers of the hand + (this is equivalent to traversing counterclockwise in the plane). +

    + + + Stokes' Theorem + +

    + Let \surfaceS be a piecewise smooth, + orientable surface whose boundary is a piecewise smooth curve C, + let \vec n be a unit vector normal to \surfaceS, let C be traversed with respect to \vec n according to the right hand rule, + and let the components of \vec F have continuous first partial derivatives over \surfaceS. Then + Stokes' Theorem + + \oint_C \vec F\cdot \, d\vec r = \iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS + . +

    +     +     + +

    + In general, the best approach to evaluating the surface integral in Stokes' Theorem is to parametrize the surface \surfaceS with a function \vec r(u,v). + We can find a unit normal vector \vec n as + + \vec n = \frac{\vec r_u\times\vec r_v}{\snorm{\vec r_u\times\vec r_v}} + . +

    + +

    + Since dS = \snorm{\vec r_u\times\vec r_v}\, dA, + the surface integral in practice is evaluated as + + \iint_\surfaceS (\curl \vec F)\cdot (\vec r_u\times\vec r_v)\, dA + , + where \vec r_u\times\vec r_v may be replaced by + \vec r_v\times\vec r_u to properly match the direction of this vector with the orientation of the parametrization of C. +

    + + + Verifying Stokes' Theorem + +

    + Considering the planar surface f(x,y) = 7-2x-2y, + let C be the curve in space that lies on this surface above the circle of radius 1 and centered at (1,1) in the xy-plane, + let \surfaceS be the planar region enclosed by C, + as illustrated in , + and let \vec F = \langle x+y,2y, y^2\rangle. + Verify Stoke's Theorem by showing \oint_C \vec F\cdot \, d\vec r = \iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS. +

    + +
    + As given in , the surface \surfaceSis the portion of the plane bounded by the curve + + An ellipse drawn on a plane in space. It lies over a circle in the x,y plane. + +

    + The usual three-dimensional coordinate axes are shown, + with the x axis toward us and to the left, + the y axis toward us and to the right, and the z axis pointing up. +

    + +

    + A circle is drawn in the xy plane. It lies in the first quadrant; + its center is at (1,1,0), and its radius is 1. + It intercepts the x axis at (1,0,0), and the y axis at (0,1,0). +

    + +

    + A plane in space is drawn in the first octant. + It is shaped like a rectangle; its highest corner is on the z axis at (0,0,7); + the opposite corner is lowest, at (2,2,-1). +

    + +

    + On the plane, a curve C is drawn. + The curve is an ellipse, consisting of all points in the plane that lie above the circle in the xy plane. + The surface \surfaceS for this problem is the region on and inside this ellipse. +

    +     + + + + + //ASY file for figstokes1_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11,19,27); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={1,2}; + real[] myzchoice={2,4,6,8}; + defaultpen(0.5mm); + + pair xbounds=(-.1,2.5); + pair ybounds=(-.1,2.5); + pair zbounds=(-.1,9); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (t.x,t.y,-2t.x-2t.y+7);// + } + surface s=surface(f,(-.1,-.1),(2.2,2.2),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return (cos(t)+1,sin(t)+1,7-2(cos(t)+1)-2(sin(t)+1));} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (cos(t)+1,sin(t)+1,0);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+dashed+linewidth(1)); + + label("$\mathcal{S}$",(1,1,4)); + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + + +
    +     + +

    + We begin by parametrizing C and then find the circulation. + A unit circle centered at (1,1) can be parametrized with x=\cos(t)+1, + y=\sin(t)+1 on 0\leq t\leq 2\pi; + to put this curve on the surface f, + make the z component equal f(x,y): + z = 7-2(\cos(t)+1)-2(\sin(t)+1) = 3-2\cos(t) - 2\sin(t). + All together, + we parametrize C with \vec r(t) = \la \cos(t)+1, \sin(t)+1, 3-2\cos(t)-2\sin(t)\ra. +

    + +

    + The circulation of \vec F around C is + + \oint_C\vec F\cdot \, d\vec r \amp = \int_0^{2\pi}\vec F\big(\vec r(t)\big)\cdot \vrp(t)\, dt + \amp = \int_0^{2\pi}\big(2\sin^3t-2\cos(t)\sin^2t+3\sin^2t-3\cos(t)\sin(t)\big)\, dt + \amp = 3\pi + . +

    + +

    + We now parametrize \surfaceS. (We reuse the letter r + for our surface as this is our custom.) Based on the parametrization of C above, + we describe \surfaceS with \vec r(u,v) = \la v\cos(u)+1, v\sin(u)+1, 3-2v\cos(u)-2v\sin(u)\ra, + where 0\leq u\leq 2\pi and 0\leq v\leq 1. +

    + +

    + We leave it to the reader to confirm that \vec r_u\times \vec r_v = \langle 2v,2v,v\rangle. + As 0\leq v\leq 1, + this vector always has a non-negative z-component, + which the right-hand rule requires given the orientation of C used above. + We also leave it to the reader to confirm \curl\vec F = \langle 2y,0,-1\rangle. +

    + +

    + The surface integral of Stokes' Theorem is thus + + \iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS \amp = \iint_\surfaceS (\curl\vec F)\cdot (\vec r_u\times \vec r_v)\, dA + \amp = \int_0^1\int_0^{2\pi} \langle 2v\sin(u)+2,0,-1\rangle\cdot\langle 2v,2v,v\rangle\, du\, dv + \amp = 3\pi + , + which matches our previous result. +

    +     +     + + +

    + One of the interesting results of Stokes' Theorem is that if two surfaces \surfaceS_1 and + \surfaceS_2 share the same boundary, + then \iint_{\surfaceS_1} (\curl \vec F)\cdot \vec n\, dS = \iint_{\surfaceS_2} (\curl \vec F)\cdot \vec n\, dS. + That is, the value of these two surface integrals is somehow independent of the interior of the surface. + We demonstrate this principle in the next example. +

    + + + Stokes' Theorem and surfaces that share a boundary + +

    + Let C be the curve given in + and note that it lies on the surface z = 6-x^2-y^2. + Let \surfaceS be the region of this surface bounded by C, + and let \vec F = \langle x+y,2y,y^2\rangle as in the previous example. + Compute \iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS to show it equals the result found in the previous example. +

    + +
    + As given in , the surface \surfaceSis the portion of the plane bounded by the curve + +
    + + + The portion of a circular paraboloid in the first octant, and an elliptical curve that lies on the surface. + +

    + This image has elements in common with . + We have the same three-dimensional coordinate axes, + and the same circle in the xy plane, centered at (1,1,0), of radius 1. +

    + +

    + The portion of the circular paraboloid z=6-x^2-y^2 that lies in the first octant is plotted. + It has its vertex at (0,0,6), and opens downward. + On the surface is an ellipse. + Although the surface is curved, the ellipse is the same ellipse that was drawn in . + The surface \surfaceS is the portion of the paraboloid on and inside the ellipse. +

    +     + + + + + //ASY file for figstokes2_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11,19,27); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={1,2}; + real[] myzchoice={2,4,6,8}; + defaultpen(0.5mm); + + pair xbounds=(-.1,2.5); + pair ybounds=(-.1,2.5); + pair zbounds=(-.1,9); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane z=6-x^2-y^2 + triple f(pair t) { + return (t.x,t.y,6-t.x^2-t.y^2);// + } + surface s=surface(f,(-.1,-.1),(2.2,2.2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return (cos(t)+1,sin(t)+1,6-(cos(t)+1)^2-(sin(t)+1)^2);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (cos(t)+1,sin(t)+1,0);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+dashed+linewidth(1)); + + label("$\mathcal{S}$",(1,1,5)); + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + + +
    + +
    + + + The parabolic surface from the previous image is shown intersecting a plane along a common elliptical curve. + +

    + This image shows both the circular paraboloid from + and the plane from . +

    + +

    + The two surfaces are shown intersecting along the same ellipse. + This illustrates how the curve C is the boundary of both surfaces. +

    +     + + + + + //ASY file for figstokes2_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11,19,27); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2}; + real[] myychoice={1,2}; + real[] myzchoice={2,4,6,8}; + defaultpen(0.5mm); + + pair xbounds=(-.1,2.5); + pair ybounds=(-.1,2.5); + pair zbounds=(-.1,9); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw the plane z=7-2x-2y + triple f(pair t) { + return (t.x,t.y,-2t.x-2t.y+7);// + } + surface s=surface(f,(-.1,-.1),(2.2,2.2),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=redmeshpen; + draw(s,surfacepen2,meshpen=p); + + //Draw the plane z=6-x^2-y^2 + triple f(pair t) { + return (t.x,t.y,6-t.x^2-t.y^2);// + } + surface s=surface(f,(-.1,-.1),(2.2,2.2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return (cos(t)+1,sin(t)+1,6-(cos(t)+1)^2-(sin(t)+1)^2);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (cos(t)+1,sin(t)+1,0);} + path3 mypath=graph(g,0,2pi,operator ..); + draw(mypath,bluepen+dashed+linewidth(1)); + + //label("$\mathcal{S}$",(1,1,5)); + //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); + + + +
    +     + +
    +     + +

    + We begin by demonstrating that C lies on the surface z=6-x^2-y^2. + We can parametrize the x and y components of C with x=\cos(t)+1, + y=\sin(t)+1 as before. + Lifting these components to the surface z=6-x^2-y^2 gives the z component as z = 6-x^2-y^2 = 6-(\cos(t)+1)^2-(\sin(t)+1)^2 = 3-2\cos(t)-2\sin(t), + which is the same z component as found in . + Thus the curve C lies on the surface z=6-x^2-y^2, + as illustrated in . +

    + +

    + Since C and \vec F are the same as in the previous example, + we already know that \oint_C\vec F\cdot\, d\vec r = 3\pi. + We confirm that this is also the value of \iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS. +

    + +

    + We parametrize \surfaceS with + + \vec r(u,v) = \langle v\cos(u)+1,v\sin(u)+1, 6-(v\cos(u)+1)^2-(v\sin(u)+1)^2\rangle + , + where 0\leq u\leq 2\pi and 0\leq v\leq 1, + and leave it to the reader to confirm that + + \vec r_u\times \vec r_v = \la 2v\big(v\cos(u)+1\big), 2v\big(v\sin(u)+1\big),v\ra + , + which also conforms to the right-hand rule with regard to the orientation of C. + With \curl \vec F = \langle 2y,0,-1\rangle as before, we have + + \amp\iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS + \quad\quad\amp = \int_0^1\int_0^{2\pi} \la 2v\sin(u)+2,0,-1\ra\cdot \la 2v\big(v\cos(u)+1\big), 2v\big(v\sin(u)+1\big),v\ra\, du\, dv + \quad\quad\amp =3\pi + . +

    + +

    + Even though the surfaces used in this example and in are very different, + because they share the same boundary, Stokes' Theorem guarantees they have equal sum of curls + across their respective surfaces. +

    +     +     + + +     + + + A Common Thread of Calculus +

    + We have threefold interest in each of the major theorems of this chapter: + the Fundamental Theorem of Line Integrals, Green's, Stokes' and the Divergence Theorems. + First, we find the beauty of their truth interesting. + Second, each provides two methods of computing a desired quantity, + sometimes offering a simpler method of computation. +

    + +

    + There is yet one more reason of interest in the major theorems of this chapter. + These important theorems also all share an important principle with the Fundamental Theorem of Calculus, + introduced in . +

    + +

    + Revisit this fundamental theorem, + adopting the notation used heavily in this chapter. + Let I be the interval [a,b] and let y=F(x) be differentiable on I, + with F\,'(x) = f(x). + The Fundamental Theorem of Calculus states that + + \int_I f(x)\, dx = F(b) - F(a) + . +

    + +

    + That is, the sum of the rates of change of a function F over an interval I can also be calculated with a certain sum of F itself on the boundary of I + (in this case, at the points x=a and x=b). +

    + +

    + Each of the named theorems above can be expressed in similar terms. + Consider the Fundamental Theorem of Line Integrals: + given a function f(x,y), + the gradient \nabla f is a type of rate of change of f. + Given a curve C with initial and terminal points A and B, + respectively, + this fundamental theorem states that + + \int_C \nabla f\, ds = f(B) - f(A) + , + where again the sum of a rate of change of f along a curve C can also be evaluated by a certain sum of f at the boundary of C (, the points A and B). +

    + +

    + Green's Theorem is essentially a special case of Stokes' Theorem, + so we consider just Stokes' Theorem here. + Recalling that the curl of a vector field \vec F is a measure of a rate of change of \vec F, Stokes' Theorem states that over a surface \surfaceS bounded by a closed curve C, + + \iint_\surfaceS \big(\curl \vec F\big)\cdot \vec n\, dS = \oint_C \vec F\cdot d\vec r + , + , the sum of a rate of change of \vec F can be calculated with a certain sum of \vec F itself over the boundary of \surfaceS. In this case, + the latter sum is also an infinite sum, requiring an integral. +

    + +

    + Finally, the Divergence Theorems state that the sum of divergences of a vector field + (another measure of a rate of change of \vec F) + over a region can also be computed with a certain sum of \vec F over the boundary of that region. + When the region is planar, the latter sum of \vec F is an integral; + when the region is spatial, + the latter sum of \vec F is a double integral. +

    + +

    + The common thread among these theorems: + the sum of a rate of change of a function over a region can be computed as another sum of the function itself on the boundary of the region. + While very general, this is a very powerful and important statement. +

    +     + + + + Terms and Concepts + + +

    + What are the differences between the Divergence Theorems of and this section? +

    +     + + + +

    + Answers will vary; in , + the Divergence Theorem connects outward flux over a closed curve in the plane to the divergence of the vector field, + whereas in this section the Divergence Theorem connects outward flux over a closed surface in space to the divergence of the vector field. +

    +     +     + + + +

    + What property of a vector field does the Divergence Theorem relate to flux? +

    +     + + + +

    + Divergence. +

    +     +     + + + +

    + What property of a vector field does Stokes' Theorem relate to circulation? +

    +     + + + +

    + Curl. +

    +     +     + + + +

    + Stokes' Theorem is the spatial version of what other theorem? +

    +     + + + +

    + Green's Theorem. +

    +     +     +     + + + Problems + + + +

    + A closed surface \surfaceS enclosing a domain D and a vector field \vec F are given. + Verify the Divergence Theorem on \surfaceS; that is, + show \iint_\surfaceS \vec F\cdot \vec n\, dS = \iiint_D\divv \vec F\, dV. +

    +     + + + +

    + \surfaceS is the surface bounding the domain D enclosed by the plane + z=2-x/2-2y/3 and the coordinate planes in the first octant; + \vec F = \langle x^2,y^2,x\rangle. +

    + + + A tetrahedron with vertices (0,0,0), (4,0,0), (0,3,0), and (0,0,2). + +

    + A surface consisting of the four faces of a tetrahedron is plotted relative to three-dimensional coordinate axes. + Each face is a triangle; there is one in each coordinate plane, + and the fourth face lies in the plane z=2-x/2-2y/3. +

    + +

    + The plane z=2-x/2-2y/3 is plotted in the first octant. + It meets the coordinate axes at the points (4,0,0), (0,3,0), and (0,0,2). + These are three of the four vertices of the tetrahedron; the remaining vertex is at the origin. +

    +     + + + + + //ASY file for fig14_07_ex_05_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12,12,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={4}; + real[] myychoice={3}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-.5,4.5); + pair ybounds=(-.5,4.5); + pair zbounds=(-0.25,3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + //triple g(real t) {return (2cos(t),2sin(t),0);} + //path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + //triple g(real t) {return (t,4-t^2,2*(4-t^2));} + //path3 mypath=graph(g,-2,2,operator ..); + draw((4,0,0)--(0,3,0)--(0,0,2)--cycle,bluepen+linewidth(2)); + draw((4,0,0)--(0,0,0),bluepen+linewidth(2)); + draw((0,3,0)--(0,0,0),bluepen+linewidth(2)); + draw((0,0,2)--(0,0,0),bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y*(3-3t.x/4),2-t.x/2-2t.y*(3-3t.x/4)/3);// + } + surface s=surface(f,(0,0),(4,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y*(3-3t.x/4),0);// + } + surface s=surface(f,(0,0),(4,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,0,t.y*(2-t.x/2));// + } + surface s=surface(f,(0,0),(4,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (0,t.x,t.y*(2-2t.x/3));// + } + surface s=surface(f,(0,0),(3,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + //label("$z=2y$",(-2,2,7)); + //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + //label("$y=4-x^2$",(2.5,2,0)); + //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + Outward flux across the plane z=2-x/2-2y/3 is 14; + across the plane z=0 the outward flux is -8; + across the planes x=0 and y=0 the outward flux is 0. +

    + +

    + Total outward flux: 14. +

    + +

    + \iint_D\divv\vec F\, dV = \int_0^{4}\int_0^{3-3x/4}\int_0^{2-x/2-2y/3}(2x+2y)\, dz\, dy\, dx = 14. +

    +     +     + + + +

    + \surfaceS is the surface bounding the domain D enclosed by the cylinder + x^2+y^2=1 and the planes z=-3 and z=3; + \vec F = \langle -x,y,z\rangle. +

    + + + A circular cylinder, centered on the z axis, capped by disks in the planes z=3 and z=-3. + +

    + The surface consists of three parts: +

      +
    1. +

      + The circular cylinder x^2+y^2=1, which is symmetric about the z axis, + from z=-3 to z=3. +

      +
      2. +
    2. +

      + The disk bounded by x^2+y^2=1 in the plane z=3; + this is the top cap of the cylinder. +

      +
      3. +
    3. +

      + The disk bounded by x^2+y^2=1 in the plane z=-3; + this is the bottom cap of the cylinder. +

      +
      4. +
    +

    +     + + + + + //ASY file for fig14_07_ex_06_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12,12,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={-3,3}; + defaultpen(0.5mm); + + pair xbounds=(-3.5,3.5); + pair ybounds=(-3.5,3.5); + pair zbounds=(-3.5,3.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + triple g(real t) {return (cos(t),sin(t),3);} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (cos(t),sin(t),-3);} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (cos(t.x),sin(t.x),t.y);// + } + surface s=surface(f,(0,-3),(2pi,3),16,2,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),3);// + } + surface s=surface(f,(0,0),(2pi,1),8,2,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),-3);// + } + surface s=surface(f,(0,0),(2pi,1),8,2,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + //label("$z=2y$",(-2,2,7)); + //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + //label("$y=4-x^2$",(2.5,2,0)); + //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + Outward flux across the cylinder x^2+y^2=1 is 0; + across the plane z=3 the outward flux is 3\pi; + across the plane z=-3 the outward flux is 3\pi. +

    + +

    + Total outward flux: 6\pi. +

    + +

    + \iint_D\divv\vec F\, dV = \int_0^{2\pi}\int_0^{1}\int_{-3}^{3}r\, dz\, dr\, d\theta = 6\pi. +

    +     +     + + + +

    + \surfaceS is the surface bounding the domain D enclosed by + z=xy(3-x)(3-y) and the plane z=0; + \vec F = \langle 3x,4y,5z+1\rangle. +

    + + + A steep hill with a square base in the x-y plane. + +

    + The graph z=xy(3-x)(3-y) is plotted relative to three-dimensional coordinate axes. + This surface meets the xy plane along a square, + with sides given by the x and y axes, + and the lines x=3 and y=3 in the xy plane. +

    + +

    + This square in the plane is part of the surface, + and the other part is the graph. + The graph is shaped like a steep hill, with a single peak in the center of the square. +

    +     + + + + + //ASY file for fig14_07_ex_07_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12,12,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={3}; + real[] myychoice={3}; + real[] myzchoice={4}; + defaultpen(0.5mm); + + pair xbounds=(-.5,3.5); + pair ybounds=(-.5,3.5); + pair zbounds=(-.5,4.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + //triple g(real t) {return (cos(t),sin(t),3);} + //path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + //triple g(real t) {return (cos(t),sin(t),-3);} + //path3 mypath=graph(g,0,2pi,operator ..); + draw((0,0,0) -- (3,0,0) -- (3,3,0) -- (0,3,0)--cycle,bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y,t.x*t.y*(3-t.x)*(3-t.y));// + } + surface s=surface(f,(0,0),(3,3),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y,0);// + } + surface s=surface(f,(0,0),(3,3),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + //label("$z=2y$",(-2,2,7)); + //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + //label("$y=4-x^2$",(2.5,2,0)); + //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + Outward flux across the surface z=xy(3-x)(3-y) is 252; + across the plane z=0 the outward flux is -9. +

    + +

    + Total outward flux: 243. +

    + +

    + \iint_D\divv\vec F\, dV = \int_0^{3}\int_0^{3}\int_{0}^{xy(3-x)(3-y)}12\, dz\, dy\, dx = 243. +

    +     +     + + + +

    + \surfaceS is the surface composed of \surfaceS_1, + the paraboloid z=4-x^2-y^2 for z\geq 0, + and \surfaceS_2, the disk of radius 2 centered at the origin; + \vec F = \langle x,y,z^2\rangle. +

    + + + A familiar dome-shaped surface between a circular paraboloid and the x,y plane. + +

    + The surface consists of the circular paraboloid z=4-x^2-y^2, + which opens downward from (0,0,4) and meets the xy plane along the circle x^2+y^2=4, + as well as the disk in the xy plane that lies on and inside the circle of intersection + between the plane and the paraboloid. +

    +     + + + + + //ASY file for fig14_07_ex_08_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12,12,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={4}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.25,5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + triple g(real t) {return (2cos(t),2sin(t),0);} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + //triple g(real t) {return (t,4-t^2,2*(4-t^2));} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (2*t.y*cos(t.x),t.y*2*sin(t.x),4-4*t.y^2);// + } + surface s=surface(f,(0,0),(2pi,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (2*t.y*cos(t.x),2*t.y*sin(t.x),0);// + } + surface s=surface(f,(0,0),(2pi,1),8,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + //label("$z=2y$",(-2,2,7)); + //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + //label("$y=4-x^2$",(2.5,2,0)); + //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + Outward flux across the paraboloid is 112\pi/3; + across the disk the outward flux is 0. +

    + +

    + Total outward flux: 112\pi/3. +

    + +

    + \iint_D\divv\vec F\, dV = \int_0^{2\pi}\int_0^2\int_0^{4-r^2}(2z+2)r\, dz\, dr\, d\theta = 112\pi/3. +

    +     +     +     + + + +

    + A closed curve C that is the boundary of a surface \surfaceS is given along with a vector field \vec F. + Verify Stokes' Theorem on C; + that is, show \oint_C \vec F\cdot d\vec r = \iint_\surfaceS\big(\curl \vec F\,\big)\cdot\vec n\, dS. +

    +     + + + +

    + C is the curve parametrized by + \vec r(t) = \langle \cos(t), \sin(t), 1\rangle and \surfaceS is the portion of + z=x^2+y^2 enclosed by C; + \vec F = \langle z,-x,y\rangle. +

    + + + A bowl-shaped surface given by a circular paraboloid plotted over the unit disk. + +

    + The surface z=x^2+y^2 is plotted relative to a set of three-dimensional coordinate axes. + The domain used for the graph is the unit disk, so that the surface is shaped like a bowl. + The boundary of the surface is a circle in the plane z=1. +

    +     + + + + + //ASY file for fig14_07_ex_09_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12,12,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-0.25,2); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + triple g(real t) {return (cos(t),sin(t),1);} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + //triple g(real t) {return (t,4-t^2,2*(4-t^2));} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),t.y^2);// + } + surface s=surface(f,(0,0),(2pi,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + //label("$z=2y$",(-2,2,7)); + //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + //label("$y=4-x^2$",(2.5,2,0)); + //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + Circulation on C: \oint_C \vec F\cdot d\vec r = -\pi +

    + +

    + \iint_\surfaceS\big(\curl \vec F\big)\cdot\vec n\, dS = -\pi. +

    +     +     + + + +

    + C is the curve parametrized by + \vec r(t) = \langle \cos(t), \sin(t), e^{-1}\rangle and \surfaceS is the portion of + z=e^{-x^2-y^2} enclosed by C; + \vec F = \langle -y,x,1\rangle. +

    + + + A bell-shaped surface in space, with a circular boundary. + +

    + The graph z=e^{-x^2-y^2} is plotted relative to a set of three-dimensional coordinate axes, + for 0\leq x^2+y^2\leq 1. + The surface is shaped like a bell or a dome, opening downward from a vertext at (0,0,1). +

    + +

    + The boundary of the surface is a circle; + the circle lies in the horizontal plane z=e^{-1}, + has radius 1, and is centered on the z axis. +

    +     + + + + + //ASY file for fig14_07_ex_10_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12,12,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + triple g(real t) {return (cos(t),sin(t),exp(-1));} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + //triple g(real t) {return (t,4-t^2,2*(4-t^2));} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),exp(-t.y^2));// + } + surface s=surface(f,(0,0),(2pi,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + //label("$z=2y$",(-2,2,7)); + //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + //label("$y=4-x^2$",(2.5,2,0)); + //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + Circulation on C: \oint_C \vec F\cdot d\vec r = \pi +

    + +

    + \iint_\surfaceS\big(\curl \vec F\big)\cdot\vec n\, dS = \pi. +

    +     +     + + + +

    + C is the curve that follows the triangle with vertices at (0,0,2), + (4,0,0) and (0,3,0), + traversing the the vertices in that order and returning to (0,0,2), + and \surfaceS is the portion of the plane + z=2-x/2-2y/3 enclosed by C; + \vec F = \langle y,-z,y\rangle. +

    + + + A triangular surface in the first octant, with intercepts at (4,0,0), (0,3,0), and (0,0,2). + +

    + The curve C is plotted in space relative to three-dimensional coordinate axes. + It is a triangle, given by the path from (0,0,2) on the z axis, + to (4,0,0) on the x axis, to (0,3,0) on the y axis, and then back to (0,0,2). +

    + +

    + The surface \surfaceS is the portion of the plane in the first octant that lies on and inside this triangle. +

    +     + + + + + //ASY file for fig14_07_ex_10_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12,12,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={4}; + real[] myychoice={3}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-.5,4.5); + pair ybounds=(-.5,4.5); + pair zbounds=(-0.25,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + //triple g(real t) {return (cos(t),sin(t),exp(-1));} + //path3 mypath=graph(g,0,2pi,operator ..); + draw((0,0,2)--(4,0,0)--(0,3,0)--cycle,bluepen+linewidth(2)); + //triple g(real t) {return (t,4-t^2,2*(4-t^2));} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y*(3-3t.x/4),2-t.x/2-2*(t.y*(3-3t.x/4))/3);// + } + surface s=surface(f,(0,0),(4,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + //label("$z=2y$",(-2,2,7)); + //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + //label("$y=4-x^2$",(2.5,2,0)); + //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + Circulation on C: The flow along the line from (0,0,2) to (4,0,0) is 0; + from (4,0,0) to (0,3,0) it is -6, + and from (0,3,0) to (0,0,2) it is 6. + The total circulation is 0+(-6)+6=0. +

    + +

    + \iint_\surfaceS\big(\curl \vec F\big)\cdot\vec n\, dS = \iint_\surfaceS 0 \, dS = 0. +

    +     +     + + + +

    + C is the curve whose x and y coordinates follow the parabola y=1-x^2 from x=1 to x=-1, + then follow the line from (-1,0) back to (1,0), + where the z coordinates of C are determined by f(x,y) = 2x^2+y^2, + and \surfaceS is the portion of + z=2x^2+y^2 enclosed by C; + \vec F = \langle y^2+z,x,x^2-y\rangle. +

    + + + Graph of an elliptic paraboloid over a parabolic domain. + +

    + The usual three-dimensional coordinate axes are drawn. + The surface is the graph z=2x^2+y^2, which is an elliptic paraboloid. +

    + +

    + The domain for the graph is not illustrated, + but it is the region between the x axis and the parabola y=1-x^2. +

    + +

    + The surface is shaped like a hammock or a sail. + Its boundary consists of two parabolic curves that meet at two cusps. + The cusps are the highest points on the surface; + they lie in the xz plane at (1,0,2) and (-1,0,2). +

    + +

    + One part of the boundary is a parabola in the xz plane, + given by z=2x^2 for -1\leq x\leq 1. + The other part is the portion of the elliptic paraboloid z=2x^2+y^2 + that lies above the curve y=1-x^2 in the xy plane. +

    +     + + + + + //ASY file for fig14_07_ex_10_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12,12,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={1}; + real[] myzchoice={2}; + defaultpen(0.5mm); + + pair xbounds=(-1.5,1.5); + pair ybounds=(-1.5,1.5); + pair zbounds=(-0.25,2.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + triple g(real t) {return (t,1-t^2,2t^2+(1-t^2)^2);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,0,2t^2);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + //triple g(real t) {return (t,4-t^2,2*(4-t^2));} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y*(1-t.x^2),2t.x^2+(t.y*(1-t.x^2))^2);// + } + surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + //label("$z=2y$",(-2,2,7)); + //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + //label("$y=4-x^2$",(2.5,2,0)); + //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + Circulation on C: The flow along the parabola is -32/15; + the flow along the line is 4/3. + The total circulation is 4/3-32/15 = -4/5. +

    + +

    + \iint_\surfaceS\big(\curl \vec F\big)\cdot\vec n\, dS = -4/5. +

    +     +     +     + + + +

    + A closed surface \surfaceS and a vector field \vec F are given. + Find the outward flux of \vec F over \surfaceS either through direct computation or through the Divergence Theorem. +

    +     + + + +

    + \surfaceS is the surface formed by the intersections of z=0 and z=(x^2-1)(y^2-1); + \vec F = \langle x^2+1,yz,xz^2\rangle. +

    + + + A domed hill with a square base in the x-y plane. + +

    + The graph z=(x^2-1)(y^2-1) is plotted relative to three-dimensional coordinate axes. + This surface meets the xy plane along a square, + with sides given by the lines x=\pm 1 and y=\pm 1 in the xy plane. +

    + +

    + The graph is shaped like a hill, with its peak on the z axis at (0,0,1). + The boundary of the graph is the square in the xy plane. +

    + +

    + The surface consists of this graph and the region in the xy + plane bounded by the square -1\leq x,y\leq 1. +

    +     + + + + + //ASY file for fig14_07_ex_13_3D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12,12,5); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={-1,1}; + defaultpen(0.5mm); + + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,1.25); + pair zbounds=(-0.25,1.25); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + //triple g(real t) {return (cos(t),sin(t),3);} + //path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + //triple g(real t) {return (cos(t),sin(t),-3);} + //path3 mypath=graph(g,0,2pi,operator ..); + draw((-1,-1,0) -- (-1,1,0) -- (1,1,0) -- (1,-1,0)--cycle,bluepen+linewidth(2)); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y,(t.x^2-1)*(t.y^2-1));// + } + surface s=surface(f,(-1,-1),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y,0);// + } + surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + //label("$z=2y$",(-2,2,7)); + //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + //label("$y=4-x^2$",(2.5,2,0)); + //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + 128/225 +

    +     +     + + + +

    + \surfaceS is the surface formed by the intersections of the planes z=\frac12(3-x), + x=1, + y=0, y=2 and z=0; + \vec F = \langle x,y^2,z\rangle. +

    + + + A triangular prism, with two triangular faces and three rectangular faces. + +

    + The surface \surfaceS consists of all five faces of a triangular prism. + The prism is plotted relative to a set of three-dimensional coordinate axes. +

    + +

    + Two of the faces are triangles. These lie in the planes y=0 and y=2. + The vertices of the triangles are at (1,0,0), (3,0,0), and (1,0,1) for the face in the xz plane, + and at (1,2,0), (3,2,0), and (1,2,1) for the face in the plane y=2. +

    + +

    + One face is in the xy plane; + it is a square given by 1\leq x\leq 3 and 0\leq y\leq 2. +

    + +

    + Another face is in the plane x=1; + it is a rectangle with 0\leq y\leq 2 and 0\leq z\leq 1. +

    + +

    + The final face is in the plane z=\frac12(3-x). + This face lies on the hypotenuse of each triangular face. + It is a rectangle, with vertices (3,0,0), (3,2,0), (1,2,1), and (1,0,1). +

    +     + + + + + //ASY file for fig13_06_ex_083D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(8.8,7.8,3); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={1,2,3}; + real[] myychoice={1,2,3}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-0.5,3.5); + pair ybounds=(-0.5,3.5); + pair zbounds=(-0.5,1.75); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //edges of object + draw((1,0,0)--(1,2,0)--(3,2,0)--(3,0,0)--cycle,bluepen+linewidth(2)); + draw((3,0,0)--(1,0,1)--(1,2,1)--(3,2,0),bluepen+linewidth(2)); + draw((3,0,0)--(1,0,1),bluepen+linewidth(2)); + draw((1,0,0)--(1,0,1),bluepen+linewidth(2)); + draw((1,2,0)--(1,2,1),bluepen+linewidth(2)); + + //shade faces + import three; + path3 p=(1,0,0)--(1,2,0)--(3,2,0)--(3,0,0); + draw(surface(p -- cycle), surfacepen); + path3 p=(1,0,1)--(1,2,1)--(3,2,0)--(3,0,0); + draw(surface(p -- cycle), surfacepen); + path3 p=(1,0,0)--(1,2,0)--(1,2,1)--(1,0,1); + draw(surface(p -- cycle), surfacepen); + path3 p=(1,0,0)--(3,0,0)--(1,0,1); + draw(surface(p -- cycle), surfacepen); + path3 p=(1,2,0)--(3,2,0)--(1,2,1); + draw(surface(p -- cycle), surfacepen); + + //labels and arrow + label("$z=\frac{1}{2}(3-x)$",(3,0,1.1)); + draw((3,0.25,1)--(2,1,0.45),Arrow3(size=2mm)); + + + +     + +

    + 8 +

    +     +     + + + +

    + \surfaceS is the surface formed by the intersections of the planes z=2y, + y=4-x^2 and z=0; + \vec F = \langle xz,0,xz\rangle. +

    + + + A wedge-shaped surface given by two intersecting planes and a parabolic cylinder. + +

    + This is another wedge-shaped surface, similar to one in . +

    + +

    + There are three components to the surface. +

      +
    • +

      + The first is the region in the xy plane bounded by the x axis and the parabola y=4-x^2. + It meets the x axis along the interval -2\leq x\leq 2. +

      +
      • +
    • +

      + The second part lies in the plane z=2y. + This is a planar surface, bounded below by the segment -2\leq x\leq 2 along the x axis, + and above by a parabola that lies in the plane z=2y. +

      + +

      + This part of the surface lies directly above the part in the xy plane, + and meets it along the x axis. +

      +
      • +
    • +

      + The last part is the portion of the parabolic cylinder y=4-x^2 + that lies between the planes z=0 and z=2y. +

      +
      • +
    +

    +     + + + + + //ASY file for fig13_06_ex_123D.asy in Chapter 13 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(12.1,-7.1,16); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,-1,1,2}; + real[] myychoice={1,2,3,4}; + real[] myzchoice={2,4,6,8}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-0.25,5); + pair zbounds=(-0.25,10); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //parabola in plane + triple g(real t) {return (t,4-t^2,0);} + path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + triple g(real t) {return (t,4-t^2,2*(4-t^2));} + path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + //shade object + //import three; + //int k=12; + //for (int i=-2*k; i<2*k; ++i) + //{ + //path3 p=(i/k,4-(i/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i)/k,4-((i)/k)^2,2*(4-((i)/k)^2)); + //draw(surface(p -- cycle), simplesurfacepen2); + //path3 p=(i/k,0,0)--(i/k,4-(i/k)^2,2*(4-(i/k)^2))--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i+1//)/k,0,0); + //draw(surface(p -- cycle), simplesurfacepen); + //} + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y*(4-t.x^2),2*t.y*(4-t.x^2));// + } + surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,(4-t.x^2),2*t.y*(4-t.x^2));// + } + surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y*(4-t.x^2),0);// + } + surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //label and arrow + label("$z=2y$",(-2,2,7)); + draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); + label("$y=4-x^2$",(2.5,2,0)); + draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); + + + +     + +

    + 8192/105\approx 78.019 +

    +     +     + + + +

    + \surfaceS is the surface formed by the intersections of the cylinder z=1-x^2 and the planes y=-2, + y=2 and z=0; + \vec F = \langle 0,y^3,0\rangle. +

    + + + A surface resembling a barn or greenhouse with an arched roof. + +

    + There are four components to this surface. + Overall, the surface looks like the exterior of a barn or greenhouse with an arched roof. +

    + +

    +

      +
    • +

      + The graph z=1-x^2 is a parabolic cylinder, opening downwards. + It is shaped like an arch, with its peak along the line with z=1 directly above the y axis. + The cylinder extends to the xy plane, and is plotted for -2\leq y\leq 2. +

      +
      • + +
    • +

      + The ends of the surface lie in the planes y=2 and y=-2. + They are bounded above by the parabola z=1-x^2, + and below by the xy plane, for -1\leq x\leq 1. +

      +
      • + +
    • +

      + The bottom of the surface is a rectangle in the xy plane, + given by -1\leq x\leq 1 and -2\leq y\leq 2. + This is where the parabolic cylinder, and the two ends, meet the xy plane. +

      +
      • +
    +

    +     + + + + + //ASY file for fig14_05_ex_23_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14,14,4); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-2,2}; + real[] myzchoice={1}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-0.25,1.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y,1-t.x^2);// + } + surface s=surface(f,(-1,-2),(1,2),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,-2,t.y*(1-t.x^2));// + } + surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,2,t.y*(1-t.x^2));// + } + surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple f(pair t) { + return (t.x,t.y,0);// + } + surface s=surface(f,(-1,-2),(1,2),2,2,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + triple g(real t) {return (t,-2,1-t^2);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,2,1-t^2);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + draw((1,-2,0)--(1,2,0)--(-1,2,0)--(-1,-2,0)--cycle,bluepen+linewidth(2)); + + //label and arrow + label("$z=1-x^2$",(0,1,1.55)); + draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); + + //triple g(real t) {return (0,t,6-t^2);} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + + +     + +

    + 64/3 +

    +     +     +     + + + +

    + A closed curve C that is the boundary of a surface \surfaceS is given along with a vector field \vec F. + Find the circulation of \vec F around C either through direct computation or through Stokes' Theorem. +

    +     + + + +

    + C is the curve whose x- and y-values are determined by the three sides of a triangle with vertices at (-1,0), + (1,0) and (0,1), traversed in that order, + and the z-values are determined by the function z=xy; + \vec F = \langle z-y^2,x,z\rangle. +

    + + + A twisted-looking surface given by a portion of a hyperbolic paraboloid. + +

    + The graph z=xy is a hyperbolic paraboloid; + it is plotted here relative to the usual three-dimensional coordinate axes, over a triangular domain. +

    + +

    + The surface has a twisted, wavy shape. + It has one edge that lies along the x axis, for -1\leq x\leq 1. + For 0\leq x\leq 1, the surface bends upward, with its highest points above the line y=x. + It then bends back down to meet the y axis. +

    + +

    + The portion for -1\leq x\leq 0 is like a mirror image of the other side, + but it lies below the xy plane. + The lowest points lie below the line y=-x, + and the surface dips down from the x and y axes to meet these points. +

    + +

    + The other part of the boundary begins on the x axis at (1,0,0). + It curves upward in a parabolic arch, meeting the y axis at (0,1,0). + The boundary then dips downward in another parabolic curve + that rises up to meet the x axis again at (-1,0,0). +

    +     + + + + + //ASY file for fig14_07_ex_23_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(-10,22,1); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={1}; + real[] myzchoice={.25}; + defaultpen(0.5mm); + + pair xbounds=(-1.25,1.25); + pair ybounds=(-.25,1.25); + pair zbounds=(-.25,.3); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x-1+t.y*(2-2*t.x),t.x,(t.x-1+t.y*(2-2*t.x))*t.x);// + } + surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //triple g(real t) {return (t,-2,1-t^2);} + //path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,0,0);} + path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,1-t,t*(1-t));} + path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (t,t+1,t*(t+1));} + path3 mypath=graph(g,-1,0,operator ..);draw(mypath,bluepen+linewidth(2)); + + //draw((-1,0,0)--(1,0,0)--(0,1,0)--cycle,bluepen+linewidth(2)); + + //label and arrow + //label("$z=1-x^2$",(0,1,1.55)); + //draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); + + //triple g(real t) {return (0,t,6-t^2);} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + + +     + +

    + 5/3 +

    +     +     + + + +

    + C is the curve whose x- and y-values are given by + \vec r(t) = \langle 2\cos(t),2\sin(t)\rangle and the z-values are determined by the function z=x^2+y^3-3y+1; + \vec F = \langle -y,x,z\rangle. +

    + + + A surface that looks something like a ladle, if the handle had melted and drooped downward. + +

    + This might be one of the more complicated surfaces in the book. + It is plotted relative to the usual set of coordinate axes in three dimensions. + There is a local minimum that lies along the y axis. + The surface is bowl-shaped near the minimum. +

    + +

    + The boundary of the surface is wavy. + There is a peak on the boundary at the point (0,2,6). + If we travel along the boundary in either direction from this point, + the surface dips down when y is close to 1, and then climbs back up again. +

    + +

    + Above the y axis, the surface climbs from the minimum to a saddle point when y=-1. + The surface climbs up if we move away from the saddle point in the x direction, + and it dips back down if we move in the y direction. +

    + +

    + If the surface were flipped over, it would look like a saddle with a back rest, + and a hump in the front. +

    +     + + + + + //ASY file for fig14_07_ex_18_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14,14,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={6}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-.25,6.5); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.y*cos(t.x),t.y*sin(t.x),(t.y*cos(t.x))^2+(t.y*sin(t.x))^3-3t.y*sin(t.x)+1);// + } + surface s=surface(f,(0,0),(2pi,2),16,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //triple g(real t) {return (t,-2,1-t^2);} + //path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (2cos(t),2sin(t),4cos(t)^2+8sin(t)^3-6sin(t)+1);} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + //triple g(real t) {return (t,1-t,t*(1-t));} + //path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + //triple g(real t) {return (t,t+1,t*(t+1));} + //path3 mypath=graph(g,-1,0,operator ..);draw(mypath,bluepen+linewidth(2)); + + //draw((-1,0,0)--(1,0,0)--(0,1,0)--cycle,bluepen+linewidth(2)); + + //label and arrow + //label("$z=1-x^2$",(0,1,1.55)); + //draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); + + //triple g(real t) {return (0,t,6-t^2);} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + + +     + +

    + 8\pi +

    +     +     + + + +

    + C is the curve whose x- and y-values are given by + \vec r(t) = \langle \cos(t),3\sin(t)\rangle and the z-values are determined by the function z=5-2x-y; + \vec F = \langle -\frac13y,3x,\frac23y-3x\rangle. +

    + + + A region in a plane in space, bounded by an ellipse in that plane. + +

    + The surface is planar; it is bounded by an ellipse that lies in the plane z=5-2x-y. + The ellipse is centered on the z axis. + The surface consists of all points in the plane that lie on or inside this ellipse. +

    +     + + + + + //ASY file for fig14_07_ex_18_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(14,14,10); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-2,2}; + real[] myychoice={-2,2}; + real[] myzchoice={10}; + defaultpen(0.5mm); + + pair xbounds=(-2.5,2.5); + pair ybounds=(-2.5,2.5); + pair zbounds=(-.25,11); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.y*cos(t.x),3*t.y*sin(t.x),-(t.y*cos(t.x))*2-3*(t.y*sin(t.x))+5);// + } + surface s=surface(f,(0,0),(2pi,1),16,8,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //triple g(real t) {return (t,-2,1-t^2);} + //path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + triple g(real t) {return (cos(t),3sin(t),-2cos(t)-3sin(t)+5);} + path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + //triple g(real t) {return (t,1-t,t*(1-t));} + //path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + //triple g(real t) {return (t,t+1,t*(t+1));} + //path3 mypath=graph(g,-1,0,operator ..);draw(mypath,bluepen+linewidth(2)); + + //draw((-1,0,0)--(1,0,0)--(0,1,0)--cycle,bluepen+linewidth(2)); + + //label and arrow + //label("$z=1-x^2$",(0,1,1.55)); + //draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); + + //triple g(real t) {return (0,t,6-t^2);} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + + +     + +

    + 23\pi +

    +     +     + + + +

    + C is the curve whose x- and y-values are sides of the square with vertices at (1,1), + (-1,1), + (-1,-1) and (1,-1), traversed in that order, + and the z-values are determined by the function z=10-5x-2y; + \vec F = \langle 5y^2,2y^2,y^2\rangle. +

    + + + A rectangle lying within a plane in space. + +

    + The image shows a rectangle in space, viewed in perspective, + relative to a set of three-dimensional coordinate axes. +

    + +

    + The sides of the rectangle lie above the square given by -1\leq x\leq 1 and -1\leq y\leq 1. + The z coordinates of the points on the rectangle are given by the plane equation z=10-5x-2y. +

    +     + + + + + //ASY file for fig14_07_ex_20_3D.asy in Chapter 14 + + size(200,200,IgnoreAspect); + //currentprojection=perspective(7,2,1); + currentprojection=orthographic(11,11,80); + defaultrender.merge=true; + + // setup and draw the axes + real[] myxchoice={-1,1}; + real[] myychoice={-1,1}; + real[] myzchoice={5,10,15}; + defaultpen(0.5mm); + + pair xbounds=(-1.25,1.25); + pair ybounds=(-1.25,1.25); + pair zbounds=(-.25,18); + + xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); + yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); + zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); + + label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); + label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); + label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); + + //Draw paraboloid over unit disk + triple f(pair t) { + return (t.x,t.y,10-5t.x-2t.y);// + } + surface s=surface(f,(-1,-1),(1,1),3,3,Spline); + pen p=apexmeshpen; + draw(s,surfacepen,meshpen=p); + + //triple g(real t) {return (t,-2,1-t^2);} + //path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + //triple g(real t) {return (cos(t),3sin(t),-2cos(t)-3sin(t)+5);} + //path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); + + //triple g(real t) {return (t,1-t,t*(1-t));} + //path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); + + //triple g(real t) {return (t,t+1,t*(t+1));} + //path3 mypath=graph(g,-1,0,operator ..);draw(mypath,bluepen+linewidth(2)); + + draw((1,1,3)--(-1,1,13)--(-1,-1,17)--(1,-1,7)--cycle,bluepen+linewidth(2)); + + //label and arrow + //label("$z=1-x^2$",(0,1,1.55)); + //draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); + + //triple g(real t) {return (0,t,6-t^2);} + //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); + + + +     + +

    + 0 +

    +     +     +     + + + +

    + The following exercisesare designed to challenge your understanding and require no computation. +

    +     + + + + +

    + Let \surfaceS be any closed surface enclosing a domain D. + Consider \vec F_1 = \langle x,0,0\rangle and \vec F_2=\langle y,y^2,z-2yz\rangle. +

    + +

    + These fields are clearly very different. + Why is it that the total outward flux of each field across \surfaceS is the same? +

    +     + +

    + Each field has a divergence of 1; + by the Divergence Theorem, + the total outward flux across \surfaceS is \iint_D 1\, dS for each field. +

    +     +     + + + + +

    + Green's Theorem can be used to find the area of a region enclosed by a curve by evaluating a line integral with the appropriate choice of vector field \vec F. + What condition on \vec F makes this possible? +

    +     + +

    + \curl\vec F = 1. +

    +     +     + + + +

    + Likewise, Stokes' Theorem can be used to find the surface area of a region enclosed by a curve in space by evaluating a line integral with the appropriate choice of vector field \vec F. + What condition on \vec F makes this possible? +

    +     + +

    + \curl\vec F\cdot \vec n = 1, + where \vec n is a unit vector normal to \surfaceS. +

    +     +     +     + + + +

    + The Divergence Theorem establishes equality between a particular double integral and a particular triple integral. + What types of circumstances would lead one to choose to evaluate the triple integral over the double integral? +

    +     + +

    + Answers will vary. + Often the closed surface \surfaceS is composed of several smooth surfaces. + To measure total outward flux, + this may require evaluating multiple double integrals. + Each double integral requires the parametrization of a surface and the computation of the cross product of partial derivatives. + One triple integral may require less work, + especially as the divergence of a vector field is generally easy to compute. +

    +     +     + + + +

    + Stokes' Theorem establishes equality between a particular line integral and a particular double integral. + What types of circumstances would lead one to choose to evaluate the double integral over the line integral? +

    +     + +

    + Answers will vary. + Often the closed curve C is composed of several smooth curves. + To measure the total circulation, + one may have to evaluate line integrals along each curve. + Each line integral requires the parametrization of its curve. + It may be less work to evaluate one single double (, surface) integral. +

    +     +     +     +     +     +
    +     + + + + + Answers to Selected Exercises + + + Quick Reference + +
    + Differentiation Formulas + + Derivative Rules +
      +
    1. + \lzo{x}(cx)=c +
      2. + +
    2. + \lzo{x}(u\pm v)=u'\pm v' +
      3. + +
    3. + \lzo{x}(u\cdot v)=uv'+ u'v +
      4. + +
    4. + \lzo{x}(\frac uv)=\frac{vu'-uv'}{v^2} +
      5. + +
    5. + \lzo{x}(u(v))=u'(v)v' +
      6. + +
    6. + \lzo{x}(c)=0 +
      7. + +
    7. + \lzo{x}(x)=1 +
      8. +
    +     + + + Derivatives of Elementary Functions +
      +
    1. + \lzo{x}(x^n)=nx^{n-1} +
      2. + +
    2. + \lzo{x}(e^x)=e^x +
      3. + +
    3. + \lzo{x}(a^x)=\ln a\cdot a^x +
      4. + +
    4. + \lzo{x}(\ln x)=\frac{1}{x} +
      5. + +
    5. + \lzo{x}(\log_a x)=\frac{1}{\ln a}\cdot \frac1x +
      6. + +
    6. + \lzo{x}(\sin(x))=\cos(x) +
      7. + +
    7. + \lzo{x}(\cos(x))=-\sin(x) +
      8. + +
    8. + \lzo{x}(\csc(x))=-\csc(x)\cot(x) +
      9. + +
    9. + \lzo{x}(\sec(x))=\sec(x)\tan(x) +
      10. + +
    10. + \lzo{x}(\tan(x))=\sec^2(x) +
      11. + +
    11. + \lzo{x}(\cot(x))=-\csc^2(x) +
      12. + +
    12. + \lzo{x}(\cosh=(x))=\sinh(x) +
      13. + +
    13. + \lzo{x}(\sinh(x))=\cosh(x) +
      14. + +
    14. + \lzo{x}(\sech(x))=-\sech(x)\tanh(x) +
      15. + +
    15. + \lzo{x}(\tanh(x))=\sech^2(x) +
      16. + +
    16. + \lzo{x}(\csch(x))=-\csch(x)\coth(x) +
      17. + +
    17. + \lzo{x}(\coth(x))=-\csch^2(x) +
      18. +
    +     + + + Derivatives of Inverse Functions +
      +
    1. + \lzo{x}(\sin^{-1}(x))=\frac{1}{\sqrt{1-x^2}} +
      2. + +
    2. + \lzo{x}(\cos^{-1}(x))=\frac{-1}{\sqrt{1-x^2}} +
      3. + +
    3. + \lzo{x}(\csc^{-1}(x))=\frac{-1}{\abs{x}\sqrt{x^2-1}} +
      4. + +
    4. + \lzo{x}(\sec^{-1}(x))=\frac{1}{\abs{x}\sqrt{x^2-1}} +
      5. + +
    5. + \lzo{x}(\tan^{-1}(x))=\frac{1}{1+x^2} +
      6. + +
    6. + \lzo{x}(\cot^{-1}(x))=\frac{-1}{1+x^2} +
      7. + +
    7. + \lzo{x}(\cosh^{-1}(x))=\frac1{\sqrt{x^2-1}} +
      8. + +
    8. + \lzo{x}(\sinh^{-1}(x))=\frac1{\sqrt{x^2+1}} +
      9. + +
    9. + \lzo{x}(\sech^{-1}(x))=\frac{-1}{x\sqrt{1-x^2}} +
      10. + +
    10. + \lzo{x}(\csch^{-1}(x))=\frac{-1}{\abs{x}\sqrt{1+x^2}} +
      11. + +
    11. + \lzo{x}(\tanh^{-1}(x))=\frac1{1-x^2} +
      12. + +
    12. + \lzo{x}(\coth^{-1}(x))=\frac1{1-x^2} +
      13. +
    +     +
    + +
    + Integration Formulas + + + Basic Rules +
      +
    1. + \int c\cdot f(x)\,dx=c\int f(x)\,dx +
      2. + +
    2. + \int \bigl(f(x)\pm g(x)\bigr)\,dx = \int f(x)\, dx \pm \int g(x)\, dx +
      3. + +
    3. + \int 0\,dx = C +
      4. + +
    4. + \int 1\,dx=x+C +
      5. +
    +     + + + Integrals of Elementary (non-Trig) Functions +
      +
    1. + \int e^x\,dx=e^x+C +
      2. + +
    2. + \int \ln(x)\,dx=x\ln(x) -x +C +
      3. + +
    3. + \int a^x\,dx=\frac{1}{\ln(a)}\cdot a^x+C +
      4. + +
    4. + \int \frac{1}{x}\,dx =\ln \abs{x} + C +
      5. + +
    5. + \int x^n\,dx=\frac{1}{n+1}x^{n+1}+C, n\neq -1 +
      6. +
    +     + + + Integrals Involving Trigonometric Functions +
      +
    1. + \int \cos(x)\,dx=\sin(x)+C +
      2. + +
    2. + \int \sin(x)\,dx=-\cos(x)+C +
      3. + +
    3. + \int \tan(x)\,dx=-\ln \abs{\cos(x)}+C +
      4. + +
    4. + \int \sec(x)\,dx=\ln \abs{\sec(x)+\tan(x)}+C +
      5. + +
    5. + \int \csc(x)\,dx=-\ln \abs{\csc(x)+\cot(x)}+C +
      6. + +
    6. + \int \cot(x)\,dx=\ln \abs{\sin(x)}+C +
      7. + +
    7. + \int \sec^2(x)\,dx=\tan(x)+C +
      8. + +
    8. + \int \csc^2(x)\,dx=-\cot(x)+C +
      9. + +
    9. + \int \sec(x)\tan(x)\,dx=\sec(x)+C +
      10. + +
    10. + \int \csc(x)\cot(x)\,dx=-\csc(x)+C +
      11. + +
    11. + \int \cos^2(x)\,dx=\frac12x+\frac14\sin\big(2x\big)+C +
      12. + +
    12. + \int \sin^2(x)\,dx=\frac12x-\frac14\sin\big(2x\big)+C +
      13. + +
    13. + \int \frac{1}{x^2+a^2}\,dx = \frac1a\tan^{-1}\left(\frac xa\right)+C +
      14. + +
    14. + \int \frac{1}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac xa\right)+C +
      15. + +
    15. + \int \frac{1}{x\sqrt{x^2-a^2}} = \frac1a\sec^{-1}\left(\frac{\abs{x}}{a}\right)+C +
      16. +
    +     + + + Integrals Involving Hyperbolic Functions +
      +
    1. + \int \cosh(x)\,dx=\sinh(x)+C +
      2. + +
    2. + \int \sinh(x)\,dx=\cosh(x)+C +
      3. + +
    3. + \int \tanh(x)\,dx=\ln(\cosh(x))+C +
      4. + +
    4. + \int \coth(x)\,dx=\ln \abs{\sinh(x)}+C +
      5. + +
    5. + \int \frac{1}{\sqrt{x^2-a^2}}\, dx =\ln\abs{x+\sqrt{x^2-a^2}}+C +
      6. + +
    6. + \int \frac{1}{\sqrt{x^2+a^2}}\, dx=\ln\abs{x+\sqrt{x^2+a^2}}+C +
      7. + +
    7. + \int \frac{1}{a^2-x^2}\, dx =\frac{1}{2a}\ln\abs{\frac{a+x}{a-x}}+C +
      8. + +
    8. + \int \frac{1}{x\sqrt{a^2-x^2}}\, dx = \frac{1}{a}\ln\left(\frac{x}{a+\sqrt{a^2-x^2}}\right)+C +
      9. + +
    9. + \int \frac{1}{x\sqrt{x^2+a^2}}\, = \frac{1}{a}\ln\abs{\frac{x}{a+\sqrt{x^2+a^2}}}+C +
      10. +
    +     +
    + +
    + Trigonometry Reference + + + + The Unit Circle + + A detailed plot of the unit circle, showing angles in both degrees and radians, and the coordinates of the corresponding points on the circle. + +

    + The unit circle x^2+y^2=1 is plotted, along with the x and y coordinate axes. + There are sixteen marked points on the circle, showing various angles and the coordinates of the corresponding points on the circle. + This can be used to evaluate the trigonometric functions at these special angles: + for a point (x,y) on the circle corresponding to an angle \theta, we have x=\cos(\theta) and y=\sin(\theta). +

    + +

    + The values given in the diagram are as follows: +

      +
    • +

      + 0^\circ, 0 radians, point (1,0) +

      +
      • +
    • +

      + 30^\circ, \pi/6 radians, point \left(\frac{\sqrt{3}}{2},\frac12\right) +

      +
      • +
    • +

      + 45^\circ, \pi/4 radians, point \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) +

      +
      • +
    • +

      + 60^\circ, \pi/3 radians, point \left(\frac12,\frac{\sqrt{3}}{2}\right) +

      +
      • +
    • +

      + 90^\circ, \pi/2 radians, point (0,1) +

      +
      • +
    • +

      + 120^\circ, 2\pi/3 radians, point \left(-\frac12,\frac{\sqrt{3}}{2}\right) +

      +
      • +
    • +

      + 135^\circ, 3\pi/4radians, point \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) +

      +
      • +
    • +

      + 150^\circ, 5\pi/6 radians, point \left(-\frac{\sqrt{3}}{2},\frac12\right) +

      +
      • +
    • +

      + 180^\circ, \pi radians, point (-1,0) +

      +
      • +
    • +

      + 210^\circ, 7\pi/6 radians, point \left(-\frac{\sqrt{3}}{2},-\frac12\right) +

      +
      • +
    • +

      + 225^\circ, 5\pi/4 radians, point \left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right) +

      +
      • +
    • +

      + 240^\circ, 4\pi/3 radians, point \left(-\frac12,-\frac{\sqrt{3}}{2}\right) +

      +
      • +
    • +

      + 270^\circ, 3\pi/2radians, point (0,-1) +

      +
      • +
    • +

      + 300^\circ, 5\pi/3 radians, point \left(\frac{\sqrt{3}}{2},-\frac12\right) +

      +
      • +
    • +

      + 315^\circ, 7\pi/4 radians, point \left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right) +

      +
      • +
    • +

      + 330^\circ, 11\pi/6 radians, point \left(\frac{\sqrt{3}}{2},-\frac12\right) +

      +
      • +
    +

    +     + + + \begin{tikzpicture}[scale=3, every node/.append style={font=\footnotesize}] + + \draw [<->,>=latex] (-1.5,0) -- (1.4,0) node [right] { $x$}; + \draw [<->,>=latex] (0,-1.3) -- (0,1.3) node [above] { $y$}; + + \foreach \x / \y / \z / \w / \v in { + 0/0/{1,0}/right/white, + 30/{\pi/6}/{\frac{\sqrt{3}}2,\frac 12}/above right/none,% + 45/{\pi/4}/{\frac{\sqrt{2}}2,\frac{\sqrt{2}}2}/above right/none, + 60/{\pi/3}/{\frac{1}2,\frac{\sqrt{3}}2}/{above right}/none, + 90/{\pi/2}/{0,1}/above/white,% + 120/{2\pi/3}/{-\frac{1}2,\frac{\sqrt{3}}2}/above left/none, + 135/{3\pi/4}/{-\frac{\sqrt{2}}2,\frac{\sqrt{2}}2}/above left/none, + 150/{5\pi/6}/{-\frac{\sqrt{3}}2,\frac{1}2}/above left/none,% + 180/{\pi}/{-1,0}/left/white, + 210/{7\pi/6}/{-\frac{\sqrt{3}}2,-\frac{1}2}/below left/none, + 225/{5\pi/4}/{-\frac{\sqrt{2}}2,-\frac{\sqrt{2}}2}/below left/none, + 240/{4\pi/3}/{-\frac{1}2,-\frac{\sqrt{3}}2}/below left/none, + 270/{3\pi/2}/{0,-1}/below/white, + 300/{5\pi/3}/{\frac{1}2,-\frac{\sqrt{3}}2}/below right/none, + 315/{7\pi/4}/{\frac{\sqrt{2}}2,-\frac{\sqrt{2}}2}/below right/none, + 330/{11\pi/6}/{\frac{\sqrt{3}}2,-\frac{1}2}/below right/none% + } + { + \draw (\x:.65cm) node [fill=\v] { \x$^\circ$}; + \draw (\x:.85cm) node [fill=\v] { $\y$}; + \draw (\x:1cm) node [\w,fill=\v] { $\left(\z\right)$}; + \draw [fill=black] (\x:1) circle (.5pt); + } + + \draw [thick] (0,0) circle (1); + + \end{tikzpicture} + + + +     +     + + + Definitions of the Trigonometric Functions + + Unit Circle Definition + + + An illustration of the correspondence between an angle and a point on the unit circle. + +

    + The unit circle is drawn over the x and y coordinate axes in the plane. + No scale markings are given. +

    + +

    + A point on the circle in the second quadrant is labeled with coordintes (x,y). + A line is drawn from the origin to this point. + There is an arc from the positive x axis to the line segment from the origin to the point; + the arc is labeled with the angle \theta. +

    + +

    + A dashed line is drawn vertically from the point (x,y) to the x axis, and labeled with the coordinate y. + The portion of the x axis between the origin and the point where the dashed line meets the axis is labeled x. +

    +     + + + \begin{tikzpicture}[>=latex,scale=1.73,thick] + + \draw [<->](-1.3,0)--(1.3,0) node [right] {\(x\)}; + \draw [<->] (0,-1.3) -- (0,1.3) node [above] {\(y\)}; + \draw (0,0) circle (1); + \draw [fill= black] (-.6,.8) circle (1pt); + \draw (0,0) -- (-.6,.8) node [above left] {\((x,y)\)}; + \draw [->] (.5,0) arc (0:127:.5); + \draw [dashed,thin] (-.6,.8) -- (-.6,0) node [pos=.5,left] {\(y\)}; + \draw (-.3,0) node [below] {\(x\)}; + \draw (.45,.45) node {\(\theta\)}; + + \end{tikzpicture} + + + + + + + \sin(\theta) = y + \cos(\theta) = x + + + + + + \ds\csc(\theta) = \frac1y + \ds\sec(\theta) = \frac1x + + + + + + \ds\tan(\theta) = \frac yx + \ds\cot(\theta) = \frac xy + + + +     +     + + + Right Triangle Definition + + + A right angle triangle, with sides labeled "opposite", "adjacent", and "hypotenuse". + +

    + A right angle triangle is drawn without reference to a coordinate system. + The angle at the bottom-left vertex is labeled \theta. + The bottom of the triangle is labeled Adjacent. + The right side of the triangle, which is vertical, is labeled Opposite. + The diagonal from bottom-left to top-right is labeled Hypotenuse. +

    +     + + + \begin{tikzpicture}[thick,scale=1.35] + + \draw (0,0) -- (2.5,0) node [below,pos=.5] {Adjacent} -- (2.5,2) node [pos=.5,rotate=-90,shift={(0pt,7pt)}] {Opposite} -- (0,0) node [pos=.5,above,rotate=38.7] {Hypotenuse} node [shift={(20pt,8pt)}] {\(\theta\)}; + \draw[->,>=latex] (1,0) arc (0:38.7:1); + \draw (2.2,0) -- (2.2,.3) -- (2.5,.3); + + \end{tikzpicture} + + + + + + + \ds\sin(\theta) = \frac{\text{O} }{\text{H} } + \ds\csc(\theta) = \frac{\text{H} }{\text{O} } + + + + + + \ds\cos(\theta) = \frac{\text{A} }{\text{H} } + \ds\sec(\theta) = \frac{\text{H} }{\text{A} } + + + + + + \ds\tan(\theta) = \frac{\text{O} }{\text{A} } + \ds\cot(\theta) = \frac{\text{A} }{\text{O} } + + +     +     +     + + + Common Trigonometric Identities + + + Pythagorean Identities +
      +
    1. \sin^2(x)+\cos^2(x)= 1
      2. +
    2. \tan^2(x)+ 1 = \sec^2(x)
      3. +
    3. 1 + \cot^2(x)=\csc^2(x)
      4. +
    +     + + + Double Angle Formulas +
      +
    1. \sin(2x) = 2\sin(x)\cos(x)
      2. + +
    2. +

      + + \cos(2x) \amp = \cos^2(x) - \sin^2(x) \amp \amp + \amp = 2\cos^2(x)-1 \amp \amp + \amp = 1-2\sin^2(x) \amp \amp + +

      +
      3. + +
    3. \tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}
      4. +
    +     +     + + + + Cofunction Identities +
      +
    1. \sin\left(\frac{\pi}{2}-x\right) = \cos(x)
      2. + +
    2. \cos\left(\frac{\pi}{2}-x\right) = \sin(x)
      3. + +
    3. \tan\left(\frac{\pi}{2}-x\right) = \cot(x)
      4. +
    4. \csc\left(\frac{\pi}{2}-x\right) = \sec(x)
      5. + +
    5. \sec\left(\frac{\pi}{2}-x\right) = \csc(x)
      6. + +
    6. \cot\left(\frac{\pi}{2}-x\right) = \tan(x)
      7. +
    +     + + + Even/Odd Identities +
      +
    1. \sin(-x) = -\sin(x)
      2. + +
    2. \cos (-x) = \cos(x)
      3. + +
    3. \tan (-x) = -\tan(x)
      4. + +
    4. \csc(-x) = -\csc(x)
      5. + +
    5. \sec (-x) = \sec(x)
      6. + +
    6. \cot (-x) = -\cot(x)
      7. +
    +     +     + + + + Power-Reducing Formulas +
      +
    1. \sin^2(x) = \frac{1-\cos(2x)}{2}
      2. + +
    2. \cos^2(x) = \frac{1+\cos(2x)}{2}
      3. + +
    3. \tan^2(x) = \frac{1-\cos(2x)}{1+\cos(2x)}
      4. +
    +     + + + Sum to Product Formulas +
      +
    1. \sin(x)+\sin(y) = 2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)
      2. + +
    2. \sin(x)-\sin(y) = 2\sin\left(\frac{x-y}2\right)\cos\left(\frac{x+y}2\right)
      3. + +
    3. \cos(x)+\cos(y) = 2\cos\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)
      4. + +
    4. \cos(x)-\cos(y) = -2\sin\left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)
      5. +
    +     +     + + + Product to Sum Formulas +
      +
    1. \sin(x)\sin(y) = \frac12 \big(\cos(x-y) - \cos (x+y)\big)
      2. + +
    2. \cos(x)\cos(y) = \frac12\big(\cos (x-y) +\cos (x+y)\big)
      3. + +
    3. \sin(x)\cos(y) = \frac12 \big(\sin(x+y) + \sin (x-y)\big)
      4. +
    +     + + + Angle Sum/Difference Formulas +
      +
    1. \sin (x\pm y) = \sin(x)\cos(y) \pm \cos(x)\sin(y)
      2. + +
    2. \cos (x\pm y) = \cos(x)\cos(y) \mp \sin(x)\sin(y)
      3. + +
    3. \tan (x\pm y) = \frac{\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}
      4. +
    +     +     +
    + + +
    + Areas and Volumes + + +

    + Triangles +

      +
    • +

      + h=a\sin(\theta) +

      +
      • +
    • +

      + Area = \frac12bh +

      +
      • +
    • +

      + Law of Cosines: + c^2=a^2+b^2-2ab\cos(\theta) +

      +
      • +
    +

    + + + A schematic diagram of a triangle, labeling three sides, an angle, and an altitude. + +

    + A triangle is drawn without reference to a coordinate system. + The bottom of the triangle is horizontal, and the other two sides are diagonal, + with the upper vertex positioned about two thirds of the way from left to right. +

    + +

    + The right diagonal is labeled a, + the bottom is labeled b, and the left diagonal is labeled c. + A dashed line is drawn from the upper vertex to the bottom, which it meets at a right angle. + This line is the altitude of the triangle, and it is labeled h, for height. + The angle at the bottom-right vertex of the triangle is labeled \theta. +

    +     + + + \begin{tikzpicture}[x=30pt,y=30pt,thick,scale=0.75] + + \draw (0,0) -- node [below,pos=.5] { \(b\)} (3,0) node [shift={(-15pt,8pt)}] {\(\theta\)} -- node [pos=.5,right] { \(a\)} (2,1.5) -- node [pos=.5,above left] { \(c\)} (0,0); + \draw (2.7,0) arc (180:125:.3); + \draw [dashed] (2,1.5) -- (2,0) node [pos=.5,left] {\(h\)}; + \draw (2,.2) -- (1.8,.2) -- (1.8,0); + + \end{tikzpicture} + + + + +

    + Right Circular Cone +

      +
    • +

      + Volume = \frac 13 \pi r^2h +

      +
      • +
    • +

      + Surface Area = + \pi r\sqrt{r^2+h^2} +\pi r^2 +

      +
      • +
    +

    + + + + A schematic diagram of a right circular cone, showing the height and radius. + +

    + A right circular cone is drawn without reference to a coordinate system. + The apex of the cone is at the top, and the circular base is at the bottom. + A dashed line from the center of the base to the edge is labeled r, for radius. + Another dashed line from the center of the base to the apex is labeled h, for height. +

    +     + + + \begin{tikzpicture}[x=13pt,y=15pt,thick,scale=1.13] + + \begin{scope}[xscale=2] + \draw (-1,0) arc (-180:0:1); + \draw [dashed] (1,0) arc (0:180:1); + \draw (-1,.1) -- (0,3) -- (1,.15); + \draw [dashed] (0,3) -- node [pos=.5,left] { \(h\)} (0,0); + \draw [dashed] (0,0) -- (1,0) node [pos=.5,above] { \(r\)}; + \end{scope} + \draw [fill=black] (0,0) circle (1pt); + + \end{tikzpicture} + + + +     + + +

    + Parallelograms +

      +
    • +

      + Area = bh +

      +
      • +
    +

    + + + A generic parallelogram, with base and height labeled. + +

    + A parallelogram is drawn without reference to a coordinate system. + The top and bottom sides are horizontal, while the left and right sides are diagonal, with positive slope. + The bottom side of the parallelogram is labeled b, for base. +

    + +

    + A dashed line is drawn from the top-left vertex to the base, which it meets at a right angle. + This line is labeled h, for the height of the parallelogram. +

    +     + + + \begin{tikzpicture}[x=30pt,y=25pt,thick,scale=0.75] + + \draw (0,0) -- node [below,pos=.5] { \(b\)} (2,0) -- (3,1.5) -- (1,1.5) -- (0,0); + \draw [dashed] (1,1.5) -- (1,0) node [pos=.5,right] {\(h\)}; + \draw (.8,0) -- (.8,.2) -- (1,.2); + + \end{tikzpicture} + + + + +

    + Right Circular Cylinder +

      +
    • +

      + Volume = \pi r^2h +

      +
      • +
    • +

      + Surface Area = + 2\pi rh +2\pi r^2 +

      +
      • +
    +

    + + + A diagram of a right circular cylinder, labeling the radius and height. + +

    + A generic right circular cylinder is drawn without reference to a coordinate system. + The cylinder is oriented vertically, with a circular base at the bottom. +

    + +

    + One side of the cylinder is labeled h, for the height, + and a line segment is drawn from the center of the circular top to the edge, + and labeled r, for radius. +

    +     + + + \begin{tikzpicture}[x=13pt,y=14pt,thick,scale=1.13] + + \begin{scope}[xscale=2] + \draw (-1,0) arc (-180:0:1); + \draw [dashed] (1,0) arc (0:180:1); + \draw (0,2.5) circle (1); + \draw (-1,0) -- (-1,2.5) (1,0)-- (1,2.5) node [right,pos=.5] {\(h\)}; + \draw (0,2.5) -- (1,2.5) node [above,pos=.5] {\(r\)}; + \end{scope} + + \draw [fill=black] (0,2.5) circle (1pt); + + \end{tikzpicture} + + + +     + + +

    + Trapezoids +

      +
    • +

      + Area = \frac12(a+b)h +

      +
      • +
    +

    + + + A schematic diagram of a generic trapezoid. + +

    + A trapezoid is drawn without reference to a coordinate system. + Its two parallel sides are drawn horizontally. +

    + +

    + The top side is shorter, and labeled with its length, a. + The longer bottom side is labeled with the length b. + The other two sides are slanted. +

    + +

    + A dashed line is drawn from the top-left vertex to the base, + perpendicular to the two parallel sides. + This line is labeled h, for the height of the trapezoid. +

    +     + + + \begin{tikzpicture}[x=30pt,y=25pt,thick,scale=0.75] + + \draw (0,0) -- node [below,pos=.7] { \(b\)} (3,0) -- (2.5,1.5) -- node [above,pos=.5] {\(a\)} (1.5,1.5) -- (0,0); + \draw [dashed] (1.5,1.5) -- (1.5,0) node [pos=.5,right] {\(h\)}; + \draw (1.3,0) -- (1.3,.2) -- (1.5,.2); + + \end{tikzpicture} + + + + +

    + Sphere +

      +
    • +

      + Volume = \frac43\pi r^3 +

      +
      • +
    • +

      + Surface Area =4\pi r^2 +

      +
      • +
    +

    + + + A image of a sphere, showing one circumference and its radius. + +

    + A sphere is drawn without reference to a coordinate system. + A line from the center of the sphere to its right edge is labeled r, for the radius. + A great circle is drawn around the middle of the sphere, + corresponding to what would be the equator on Earth. +

    +     + + + \begin{tikzpicture}[x=13pt,y=13pt,thick,scale=1.12] + + \begin{scope}[xscale=2] + \draw (-1,0) arc (-180:0:1); + \draw [dashed] (1,0) arc (0:180:1); + \end{scope} + + \draw (0,0) circle (2); + \draw [dashed] (0,0) -- (2,0) node [pos=.5,above] {\(r\)}; + \draw [fill=black] (0,0) circle (1pt); + + \end{tikzpicture} + + + +     + + +

    + Circles +

      +
    • +

      + Area = \pi r^2 +

      +
      • +
    • +

      + Circumference = 2\pi r +

      +
      • +
    +

    + + + A generic circle with its radius indicated. + +

    + A simple sketch of a circle, without reference to coordinates. + A line segment from the center of the circle to the circumference indicates the radius, r. +

    +     + + + \begin{tikzpicture}[x=30pt,y=30pt,thick] + + \draw (0,0) circle (1); + \draw [dashed] (0,0) -- (1,0) node [pos=.5,above] {\(r\)}; + \draw [fill=black] (0,0) circle (1pt); + + \end{tikzpicture} + + + + +

    + General Cone +

      +
    • +

      + Area of Base = A +

      +
      • +
    • +

      + Volume = \frac13Ah +

      +
      • +
    +

    + + + A drawing of a general cone, with an arbitrary plane region for its base. + +

    + A sketch of a general cone: the base is a closed curve in a plane, + drawn in perspective, but without reference to a coordinate system. +

    + +

    + The base is labeled with its area, A. + A dashed line is drawn from the apex of the cone to the base, and labeled h, for height. +

    +     + + + \begin{tikzpicture}[x=13pt,y=10pt,thick,scale=1.12] + + \begin{scope} + \clip (0,0) rectangle (4,-2.5); + \draw [smooth] plot coordinates {(0,0) (1,1.5) (2,1.5) (4,0) (3,-1) (2,-1.5) (1,-2) (0,0)}; + \end{scope} + + \begin{scope} + \clip (0,0) rectangle (4,2.5); + \draw [smooth,dashed] plot coordinates {(0,0) (1,1.5) (2,1.5) (4,0) (3,-1) (2,-1.5) (1,-2) (0,0)}; + \end{scope} + + \draw (0,0) -- (2,4)--(4,0); + \draw [dashed] (2,0)--(2,4) node [pos=.5,right] {\(h\)}; + \draw [fill=black](2,0) circle (1pt); + \draw (1.5,-.75) node {\(A\)}; + + \end{tikzpicture} + + + +     + + +

    + Sectors of Circles +

      +
    • +

      + \theta in radians +

      +
      • +
    • +

      + Area = \frac12\theta r^2 +

      +
      • +
    • +

      + s=r\theta +

      +
      • +
    +

    + + + A pie-shaped sector of a circle, labeled with angle, radius, and arc length. + +

    + A drawing of a sector of a circle, without reference to a coordinate system. + The sector is shaped like a slice of pie, corresponding to an acute angle. + The angle is labeled \theta, and the line segment on one side of the angle is labeled r, for the radius. + The circular arc opposite the angle is labeled s, for its arc length. +

    +     + + + \begin{tikzpicture}[x=30pt,y=30pt,thick] + + \draw (2,0) arc (0:50:2) -- (0,0); + \draw [] (0,0) -- (2,0) node [pos=.5,below] {\(r\)}; + \draw [fill=black] (0,0) circle (1pt); + \draw (1.95,1.0) node {\(s\)}; + \draw (0,0) node [shift={(15pt,8pt)}] {\(\theta\)}; + + \end{tikzpicture} + + + + +

    + General Right Cylinder +

      +
    • +

      + Area of Base = A +

      +
      • +
    • +

      + Volume = Ah +

      +
      • +
    +

    + + + A sketch of a right cylinder with an arbitrary base. + +

    + A drawing of a general right cylinder without reference to a coordinate system. + The base of the cylinder is a general plane region bounded by a simple, closed curve. + The base is labeled with its area, A. + The sides of the cylinder are vertical, and labeled with a height, h. + The top of the cylinder is another copy of the base. +

    +     + + + \begin{tikzpicture}[x=13pt,y=10pt,thick,scale=1.12] + + \begin{scope} + \clip (0,0) rectangle (4,-2.5); + \draw [smooth] plot coordinates {(0,0) (1,1.5) (2,1.5) (4,0) (3,-1) (2,-1.5) (1,-2) (0,0)}; + \end{scope} + + \begin{scope} + \clip (0,0) rectangle (4,2.5); + \draw [smooth,dashed] plot coordinates {(0,0) (1,1.5) (2,1.5) (4,0) (3,-1) (2,-1.5) (1,-2) (0,0)}; + \end{scope} + + \begin{scope}[shift={(0,4)}] + \draw [smooth] plot coordinates {(0,0) (1,1.5) (2,1.5) (4,0) (3,-1) (2,-1.5) (1,-2) (0,0)}; + \end{scope} + + \draw (0,0) -- (0,4) (4,0) -- (4,4) node [pos=.5,right] {\(h\)}; + \draw (2,0) node {\(A\)}; + + \end{tikzpicture} + + + +     +     +
    + +
    + Algebra + + + Factors and Zeros of Polynomials + +

    + Let p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 be a polynomial. + If p(a)=0, + then a is a zero of the polynomial and a solution of the equation p(x)=0. + Furthermore, (x-a) is a factor of the polynomial. +

    +     + + + Fundamental Theorem of Algebra + +

    + An nth degree polynomial has n + (not necessarily distinct) + zeros. + Although all of these zeros may be imaginary, + a real polynomial of odd degree must have at least one real zero. +

    +     + + + Quadratic Formula + +

    + If p(x) = ax^2 + bx + c, + and 0 \le b^2 - 4ac, + then the real zeros of p are x=(-b\pm \sqrt{b^2-4ac})/2a +

    +     + + + Special Factors + +

    + + x^2 - a^2 \amp = (x-a)(x+a) + x^3 - a^3 \amp= (x-a)(x^2+ax+a^2) + x^3 + a^3 \amp= (x+a)(x^2-ax+a^2) + x^4 - a^4 \amp= (x^2-a^2)(x^2+a^2) + (x+y)^n \amp=x^n + nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2+\cdots +nxy^{n-1}+y^n + (x-y)^n \amp=x^n - nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2-\cdots \pm nxy^{n-1}\mp y^n + +

    +     + + + Binomial Theorem + +

    + + (x+y)^2 \amp= x^2 + 2xy + y^2 + (x-y)^2 \amp= x^2 -2xy +y^2 + (x+y)^3 \amp= x^3 + 3x^2y + 3xy^2 + y^3 + (x-y)^3 \amp= x^3 -3x^2y + 3xy^2 -y^3 + (x+y)^4 \amp= x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 + (x-y)^4 \amp= x^4 - 4x^3y + 6x^2y^2 - 4xy^3 + y^4 + +

    +     + + + Rational Zero Theorem + +

    + If p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 has integer coefficients, + then every rational zero of p is of the form x=r/s, + where r is a factor of a_0 and s is a factor of a_n. +

    +     + + + Factoring by Grouping + +

    + ac x^3 + adx^2 + bcx + bd = ax^2(cx+d)+b(cx+d)=(ax^2+b)(cx+d) +

    +     + + + Arithmetic Operations + +

    + + ab+ac\amp=a(b+c) \amp \frac{a}{b}+\frac{c}{d} \amp= \frac{ad+bc}{bd} \amp \frac{a+b}{c} \amp = \frac{a}{c} + \frac{b}{c} + \frac{\left(\displaystyle\frac{a}{b}\right)}{\left(\displaystyle\frac{c}{d}\right)}\amp=\left(\frac{a}{b}\right)\left(\frac{d}{c}\right)=\frac{ad}{bc} \amp \frac{\left(\displaystyle\frac{a}{b}\right)}{c} \amp = \frac{a}{bc} \amp \frac{a}{\left(\displaystyle\frac{b}{c}\right)} \amp= \frac{ac}{b} + a\left(\frac{b}{c}\right)\amp= \frac{ab}{c}\amp \frac{a-b}{c-d}\amp=\frac{b-a}{d-c}\amp \frac{ab+ac}{a}\amp=b+c + +

    +     + + + Exponents and Radicals + +

    + + a^0\amp =1, \, a \ne 0 \amp (ab)^x\amp=a^xb^x \amp a^xa^y \amp = a^{x+y} \amp \sqrt{a}\amp=a^{1/2} + \frac{a^x}{a^y}\amp=a^{x-y} \amp \sqrt[n]{a}\amp =a^{1/n} \amp \left(\frac{a}{b}\right)^x\amp=\frac{a^x}{b^x} \amp \sqrt[n]{a^m}\amp=a^{m/n} + a^{-x}\amp=\displaystyle\frac{1}{a^x} \amp \sqrt[n]{ab}\amp=\sqrt[n]{a}\sqrt[n]{b} \amp (a^x)^y\amp=a^{xy} \amp \sqrt[n]{\frac{a}{b}}\amp=\frac{\sqrt[n]{a}}{\sqrt[n]{b}} + +

    +     +
    + +
    + Additional Formulas + + + Summation Formulas: + +

    + + + \sum^n_{i=1}{c} \amp = cn \amp \sum^n_{i=1}{i} \amp= \frac{n(n+1)}{2} + + + \sum^n_{i=1}{i^2} \amp = \frac{n(n+1)(2n+1)}{6} \amp \sum^n_{i=1}{i^3} \amp = \left(\frac{n(n+1)}{2}\right)^2 + + +

    +     + + + Trapezoidal Rule: +

    + \int_a^b{f(x)}\, dx \approx \frac{\Delta x}{2}\big[f(x_0)+2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_{n})\big] +

    + +

    + with Error \ds \leq \frac{(b-a)^3}{12n^2}\big[\max \abs{f\,''(x)}\big] +

    +     + + + Simpson's Rule: +

    + \int_a^b{f(x)}\, dx \approx \frac{\Delta x}{3}\big[f(x_0)+4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_{n})\big] +

    + +

    + with Error \ds \leq \frac{(b-a)^5}{180n^4}\big[ \max\abs{f\,^{(4)}(x)}\big] +

    +     + + + Arc Length: +

    + L = \int_a^b\sqrt{1+ f\,'(x)^2}\,dx +

    +     + + + Surface of Revolution: +

    + 2\pi \int_a^b{f(x) \sqrt{1+ f\,'(x)^2}}dx +

    + +

    + (where f(x)\geq 0) +

    + +

    + S = 2\pi \int_a^b{x \sqrt{1+ f\,'(x)^2}}dx +

    + +

    + (where a,b \geq 0) +

    +     + + + Work Done by a Variable Force: +

    + W = \int_a^b{F(x)}dx +

    +     + + + Force Exerted by a Fluid: +

    + F = \int_a^b{w\,d(y)\,\ell(y)}dy +

    +     + + + Taylor Series Expansion for <m>f(x)</m>: +

    + p_n(x) = f(c) + f\,'(c)(x-c) + \frac{f\,''(c)}{2!}(x-c)^2 + \cdots + \frac{f\,^{(n)}(c)}{n!}(x-c)^n + \cdots +

    +     + + + Maclaurin Series Expansion for <m>f(x)</m>, where <m>c=0</m>: +

    + p_n(x) = f(0) + f\,'(0)x + \frac{f\,''(0)}{2!}x^2 + \frac{f\,'''(0)}{3!}x^3 + \cdots + \frac{f\,^{(n)}(0)}{n!}x^n+\cdots +

    +     +
    + +
    + Summary of Tests for Series + + + <tabular> + <col/> + <col/> + <col width="19%"/> + <col width="19%"/> + <col width="22%"/> + <row bottom="minor"> + <cell>Test</cell> + <cell>Series</cell> + <cell><p>Condition(s) of Convergence</p></cell> + <cell><p>Condition(s) of Divergence</p></cell> + <cell>Comment</cell> + </row> + <row bottom="minor"> + <cell><m>n</m>th-Term</cell> + <cell><m>\displaystyle{\sum^\infty_{n=1}{a_n}}</m></cell> + <cell/> + <cell><p><m>\displaystyle{\lim_{n \to \infty} a_n \neq 0}</m></p></cell> + <cell><p>Cannot be used to show convergence.</p></cell> + </row> + <row bottom="minor"> + <cell>Geometric Series</cell> + <cell><m>\displaystyle{\sum^\infty_{n=0}{r^n}}</m></cell> + <cell><p><m>\abs{r} \lt 1</m></p></cell> + <cell><p><m>\abs{r} \geq 1</m></p></cell> + <cell><p><m>\displaystyle{\text{ Sum } = \frac{1}{1-r}}</m></p></cell> + </row> + <row bottom="minor"> + <cell>Telescoping Series</cell> + <cell><m>\displaystyle{\sum^\infty_{n=1}{(b_n-b_{n+a})}}</m></cell> + <cell><p><m>\displaystyle{\lim_{n \to \infty} b_n = L}</m></p></cell> + <cell/> + <cell><p><m>\displaystyle\text{ Sum } = \left(\sum^a_{n=1}b_n\right) -L</m></p></cell> + </row> + <row bottom="minor"> + <cell><m>p</m>-Series</cell> + <cell><m>\displaystyle{\sum^\infty_{n=1}{\frac{1}{(an+b)^p}}}</m></cell> + <cell><p><m>p \gt 1</m></p></cell> + <cell><p><m>p\leq 1</m></p></cell> + <cell/> + </row> + <row bottom="minor"> + <cell>Integral Test</cell> + <cell><m>\displaystyle{\sum^\infty_{n=0}{a_n}}</m></cell> + <cell><p><m>\displaystyle \int_1^\infty a(n)\, dn</m> converges</p></cell> + <cell><p><m>\displaystyle \int_1^\infty a(n)\, dn</m> diverges</p></cell> + <cell><p><m>a_n = a(n)</m> must be continuous</p></cell> + </row> + <row bottom="minor"> + <cell>Direct Comparison</cell> + <cell><m>\displaystyle{\sum^\infty_{n=0}{a_n}}</m></cell> + <cell><p><m>\displaystyle \sum_{n=0}^\infty b_n</m> converges and <m>0\leq a_n\leq b_n</m></p></cell> + <cell><p><m>\displaystyle \sum_{n=0}^\infty b_n</m> diverges and <m>0\leq b_n\leq a_n</m></p></cell> + <cell/> + </row> + <row bottom="minor"> + <cell>Limit Comparison</cell> + <cell><m>\displaystyle{\sum^\infty_{n=0}{a_n}}</m></cell> + <cell><p><m>\displaystyle \sum_{n=0}^\infty b_n</m> converges and <m>\lim\limits_{n\to\infty}\frac{a_n}{b_n} \geq 0</m></p></cell> + <cell><p><m>\displaystyle \sum_{n=0}^\infty b_n</m> diverges and <m>\lim\limits_{n\to\infty}\frac{a_n}{b_n} \gt 0</m></p></cell> + <cell><p>Also diverges if <m>\lim\limits_{n\to\infty}\frac{a_n}{b_n}=\infty</m></p></cell> + </row> + <row> + <cell>Ratio Test</cell> + <cell><m>\displaystyle{\sum^\infty_{n=0}{a_n}}</m></cell> + <cell><p><m>\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n} \lt 1</m></p></cell> + <cell><p><m>\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n} \gt 1</m></p></cell> + <cell><p><m>\{a_n\}</m> must be positive</p></cell> + </row> + <row bottom="minor"> + <cell/> + <cell/> + <cell/> + <cell><p>Also diverges if</p></cell> + <cell><p><m>\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}=\infty</m></p></cell> + </row> + <row> + <cell>Root Test</cell> + <cell><m>\displaystyle{\sum^\infty_{n=0}{a_n}}</m></cell> + <cell><p><m>\displaystyle \lim_{n\to\infty} \big(a_n\big)^{1/n} \lt 1</m></p></cell> + <cell><p><m>\displaystyle \lim_{n\to\infty} \big(a_n\big)^{1/n} \gt 1</m></p></cell> + <cell><p><m>\{a_n\}</m> must be positive</p></cell> + </row> + <row bottom="minor"> + <cell/> + <cell/> + <cell/> + <cell><p>Also diverges if</p></cell> + <cell><p><m>\lim\limits_{n\to\infty} (a_n)^{1/n}=\infty</m></p></cell> + </row> + </tabular> + </table> + </section> +</appendix> + <index label="terminology-index"> + <title>Index + + + + + \ No newline at end of file diff --git a/apex-validation.txt b/apex-validation.txt new file mode 100644 index 000000000..3c8d4bab0 --- /dev/null +++ b/apex-validation.txt @@ -0,0 +1,30735 @@ +Validation Report +================= + +Two tools have examined an assembled version of your source: + + (1) "jing" checked conformance with the RELAX-NG schema at + /home/sean/github/pretext/schema/pretext.rng + (a schema can only describe parent-child relationships, + plus the attributes of each element) + (2) the PreTeXt "validation-plus" stylesheet made checks that + no RELAX-NG schema could ever express, and offers advice + besides + +Locations refer to the ASSEMBLED version of your source: your +modular source files have been knitted together, and any version +support has been applied (as elected by a "version" element +within the "source" element of the publication file you +supplied). In particular, an element excluded from the version +being built cannot raise a message here. The assembled source +has been deposited at + /home/sean/github/APEXCalculusPTX/apex-assembled.xml + +Each message locates its problem four ways: + + file: the source file where the problem lies + path: the location within the assembled source + line: the line number within the assembled source + text: an excerpt of the offending content + +Only "file" points into your own source files. In particular, +"line" is a line number of the deposited assembled source named +above, never of one of your files. + +To read a "path", count elements of each name. For example, + + /pretext[1]/book[1]/chapter[7]/section[2]/p[13]/em[2] + +is the second "em" within the thirteenth "p" (paragraph) of the +second "section" of the seventh "chapter" of the book. Counts +are of elements with the same name: that paragraph is the +thirteenth "p" of its section, though other elements may precede +it or intervene. + + +====================================================================== +Messages: RELAX-NG schema validation, from "jing" +====================================================================== + +error: element "document-id" not allowed anywhere; expected the element end-tag or element "asymptote-preamble", "author-biographies", "brandlogo", "cross-references", "feedback", "images", "initialism", "latex-image-preamble", "macros", "math-package", "ns:prefigure-preamble", "numbering", "parsons", "programs" or "rename" (with xmlns:ns="https://prefigure.org") + file: docinfo.ptx + path: /pretext[1]/docinfo[1]/document-id[1] + line: 11 + text: APEX + +error: element "blurb" not allowed anywhere; expected the element end-tag or element "asymptote-preamble", "author-biographies", "brandlogo", "cross-references", "feedback", "images", "initialism", "latex-image-preamble", "macros", "math-package", "ns:prefigure-preamble", "numbering", "parsons", "programs" or "rename" (with xmlns:ns="https://prefigure.org") + file: docinfo.ptx + path: /pretext[1]/docinfo[1]/blurb[1] + line: 14 + text: + +error: found attribute "component", but no attributes allowed here + file: docinfo.ptx + path: /pretext[1]/docinfo[1]/latex-image-preamble[1] + line: 118 + text: + +error: found attribute "component", but no attributes allowed here + file: docinfo.ptx + path: /pretext[1]/docinfo[1]/asymptote-preamble[1] + line: 243 + text: + +error: found attribute "label", but no attributes allowed here + file: apex.ptx + path: /pretext[1]/book[1]/frontmatter[1]/titlepage[1] + line: 328 + text: + +error: attribute "vshift" not allowed here; expected attribute "component", "label", "landscape" or "xml:lang" + file: sec_limit_intro.ptx + path: /pretext[1]/book[1]/chapter[1]/section[1]/figure[1] + line: 767 + text:
    + +error: attribute "vshift" not allowed here; expected attribute "component", "label", "landscape" or "xml:lang" + file: sec_limit_intro.ptx + path: /pretext[1]/book[1]/chapter[1]/section[1]/figure[2] + line: 869 + text:
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    + +error: attribute "hskip" not allowed here; expected attribute "component", "label", "landscape" or "xml:lang" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[2]/p[2]/m[3] + line: 73734 + text:
    + +error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[1]/exercise[1]/statement[1]/p[1]/ie[1] + line: 74288 + text: + +error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[1]/exercise[2]/statement[1]/p[1]/xref[1] + line: 74314 + text: + +error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/p[1]/m[1] + line: 74349 + text: + +error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/statement[1]/p[1]/m[1] + line: 74377 + text: + +error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/p[1]/m[1] + line: 74406 + text: + +error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/statement[1]/p[1]/m[1] + line: 74435 + text: + +error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/statement[1]/p[1]/m[1] + line: 74464 + text: + +error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/statement[1]/p[1]/m[1] + line: 74492 + text: + +error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/statement[1]/p[1]/m[1] + line: 74520 + text: + +error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/statement[1]/p[1]/m[1] + line: 74555 + text: + +error: attribute "vshift" not allowed here; expected attribute "component", "label", "xml:id" or "xml:lang" + file: sec_lhopitals_rule.ptx + path: /pretext[1]/book[1]/chapter[6]/section[7]/subsection[1]/theorem[1]/statement[1]/p[1]/m[11] + line: 75476 + text: elements, if present, or else against the first row. Results may be unpredictable. + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/paragraphs[2]/example[1]/solution[1]/table[1]/tabular[1] + line: 12541 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/statement[1]/image[1] + line: 12915 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/statement[1]/image[1] + line: 12971 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/webwork[1]/statement[1]/image[1] + line: 13029 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/webwork[1]/statement[1]/image[1] + line: 13086 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/webwork[1]/statement[1]/image[1] + line: 13147 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/webwork[1]/statement[1]/image[1] + line: 13204 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/webwork[1]/statement[1]/image[1] + line: 13266 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/webwork[1]/statement[1]/image[1] + line: 13342 + text: + +PTX:WARNING: 'medium' or 'major' table rule attributes will be handled as 'minor' in the output of a WeBWorK PG table produced by WeBWorK's hardcopy production engine + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[1]/webwork[1]/statement[1]/tabular[1]/row[1] + line: 14080 + text: + +PTX:WARNING: 'medium' or 'major' table rule attributes will be handled as 'minor' in the output of a WeBWorK PG table produced by WeBWorK's hardcopy production engine + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[2]/webwork[1]/statement[1]/tabular[1]/row[1] + line: 14170 + text: + +PTX:WARNING: 'medium' or 'major' table rule attributes will be handled as 'minor' in the output of a WeBWorK PG table produced by WeBWorK's hardcopy production engine + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[3]/webwork[1]/statement[1]/tabular[1]/row[1] + line: 14260 + text: + +PTX:WARNING: 'medium' or 'major' table rule attributes will be handled as 'minor' in the output of a WeBWorK PG table produced by WeBWorK's hardcopy production engine + file: sec_limit_continuity.ptx + path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[4]/webwork[1]/statement[1]/tabular[1]/row[1] + line: 14350 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/introduction[1]/figure[1]/image[1] + line: 14445 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[1]/example[1]/statement[1]/figure[1]/image[1] + line: 14622 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[1]/example[2]/statement[1]/figure[1]/image[1] + line: 14699 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] + line: 14806 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[2]/figure[1]/image[1] + line: 14882 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 15082 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 15186 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 15217 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 15247 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 15487 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 15519 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 15558 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/introduction[1]/image[1] + line: 15866 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/introduction[1]/image[1] + line: 15952 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/webwork[1]/introduction[1]/image[1] + line: 16079 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/webwork[1]/introduction[1]/image[1] + line: 16179 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/webwork[1]/introduction[1]/image[1] + line: 16271 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_limit_infty.ptx + path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/webwork[1]/introduction[1]/image[1] + line: 16346 + text: + +PTX:WARNING: A run of text contains Unicode characters for double quotation marks (U+201C, decimal 8220; U+201D, decimal 8221). Likely this was introduced in a conversion of source material authored in a word-processor. A matching pair U+201C, U+201D should be replaced by the "" element enclosing content. In rare cases, U+201C might be replaced by the empty element "". In rare cases, U+201D might be replaced by the empty element "". + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/p[4]/md[1] + line: 17355 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] + line: 17499 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] + line: 17534 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] + line: 17571 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/sbsgroup[1]/sidebyside[2]/figure[2]/image[1] + line: 17608 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 17797 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 17900 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[4]/solution[1]/figure[1]/image[1] + line: 18062 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[8]/statement[1]/figure[1]/image[1] + line: 18355 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[8]/solution[1]/figure[1]/image[1] + line: 18463 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[9]/statement[1]/figure[1]/image[1] + line: 18519 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[9]/solution[1]/figure[1]/image[1] + line: 18622 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] + line: 18800 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercise[1]/webwork[1]/introduction[1]/image[1] + line: 19519 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercise[2]/webwork[1]/introduction[1]/image[1] + line: 19609 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/statement[1]/image[1] + line: 19698 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/statement[1]/image[1] + line: 19728 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/statement[1]/image[1] + line: 19757 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/statement[1]/image[1] + line: 19789 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[1]/webwork[1]/statement[1]/image[1] + line: 19878 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_intro.ptx + path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[2]/webwork[1]/statement[1]/image[1] + line: 19962 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_interpret.ptx + path: /pretext[1]/book[1]/chapter[2]/section[2]/subsection[4]/example[1]/statement[1]/figure[1]/image[1] + line: 20467 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_interpret.ptx + path: /pretext[1]/book[1]/chapter[2]/section[2]/subsection[4]/example[1]/solution[1]/figure[1]/image[1] + line: 20510 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_interpret.ptx + path: /pretext[1]/book[1]/chapter[2]/section[2]/subsection[4]/example[2]/statement[1]/figure[1]/image[1] + line: 20571 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_interpret.ptx + path: /pretext[1]/book[1]/chapter[2]/section[2]/subsection[4]/example[2]/solution[1]/figure[1]/image[1] + line: 20639 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_interpret.ptx + path: /pretext[1]/book[1]/chapter[2]/section[2]/subsection[4]/example[3]/solution[1]/figure[1]/image[1] + line: 20713 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_interpret.ptx + path: /pretext[1]/book[1]/chapter[2]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/statement[1]/image[1] + line: 21362 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_interpret.ptx + path: /pretext[1]/book[1]/chapter[2]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/statement[1]/image[1] + line: 21431 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_interpret.ptx + path: /pretext[1]/book[1]/chapter[2]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/webwork[1]/statement[1]/image[1] + line: 21486 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_interpret.ptx + path: /pretext[1]/book[1]/chapter[2]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/webwork[1]/statement[1]/image[1] + line: 21549 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_basic_rules.ptx + path: /pretext[1]/book[1]/chapter[2]/section[3]/example[1]/solution[1]/figure[1]/image[1] + line: 21813 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_basic_rules.ptx + path: /pretext[1]/book[1]/chapter[2]/section[3]/example[2]/solution[1]/sidebyside[1]/figure[1]/image[1] + line: 22015 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_basic_rules.ptx + path: /pretext[1]/book[1]/chapter[2]/section[3]/example[2]/solution[1]/sidebyside[1]/figure[2]/image[1] + line: 22052 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_prodquot.ptx + path: /pretext[1]/book[1]/chapter[2]/section[4]/example[1]/solution[1]/figure[1]/image[1] + line: 23696 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_prodquot.ptx + path: /pretext[1]/book[1]/chapter[2]/section[4]/example[6]/solution[1]/figure[1]/image[1] + line: 24050 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_chainrule.ptx + path: /pretext[1]/book[1]/chapter[2]/section[5]/figure[1]/image[1] + line: 25631 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_chainrule.ptx + path: /pretext[1]/book[1]/chapter[2]/section[5]/example[4]/solution[1]/figure[1]/image[1] + line: 26034 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_chainrule.ptx + path: /pretext[1]/book[1]/chapter[2]/section[5]/paragraphs[1]/figure[1]/image[1] + line: 26423 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/introduction[1]/figure[1]/image[1] + line: 27590 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 27784 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[3]/solution[1]/figure[1]/image[1] + line: 27910 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[4]/solution[1]/figure[1]/image[1] + line: 28019 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[4]/solution[1]/figure[2]/image[1] + line: 28070 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[5]/solution[1]/figure[1]/image[1] + line: 28152 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[6]/solution[1]/figure[1]/image[1] + line: 28290 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[6]/solution[1]/figure[2]/image[1] + line: 28342 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[3]/figure[1]/image[1] + line: 28454 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] + line: 28539 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/webwork[1]/introduction[1]/image[1] + line: 29398 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/webwork[1]/introduction[1]/image[1] + line: 29458 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/webwork[1]/introduction[1]/image[1] + line: 29517 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/webwork[1]/introduction[1]/image[1] + line: 29589 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[5]/webwork[1]/introduction[1]/image[1] + line: 29657 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_implicit.ptx + path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[6]/webwork[1]/introduction[1]/image[1] + line: 29721 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_inverse_function.ptx + path: /pretext[1]/book[1]/chapter[2]/section[7]/figure[1]/image[1] + line: 30170 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_inverse_function.ptx + path: /pretext[1]/book[1]/chapter[2]/section[7]/figure[2]/image[1] + line: 30232 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_inverse_function.ptx + path: /pretext[1]/book[1]/chapter[2]/section[7]/example[1]/solution[1]/figure[1]/image[1] + line: 30415 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_inverse_function.ptx + path: /pretext[1]/book[1]/chapter[2]/section[7]/figure[3]/sidebyside[1]/image[1] + line: 30484 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_deriv_inverse_function.ptx + path: /pretext[1]/book[1]/chapter[2]/section[7]/figure[3]/sidebyside[1]/image[2] + line: 30512 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 31769 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 31802 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 31841 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/example[1]/statement[1]/figure[1]/image[1] + line: 31921 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/example[2]/statement[1]/figure[1]/image[1] + line: 32051 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/example[3]/statement[1]/figure[1]/image[1] + line: 32108 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/figure[2]/image[1] + line: 32221 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/example[4]/statement[1]/figure[1]/image[1] + line: 32305 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/example[5]/solution[1]/sidebyside[1]/figure[2]/image[1] + line: 32481 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/example[6]/solution[1]/sidebyside[1]/figure[2]/image[1] + line: 32590 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/example[7]/solution[1]/sidebyside[1]/figure[1]/image[1] + line: 32641 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/statement[1]/image[1] + line: 32871 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/statement[1]/image[1] + line: 32958 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/webwork[1]/statement[1]/image[1] + line: 33044 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/webwork[1]/statement[1]/image[1] + line: 33089 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/webwork[1]/statement[1]/image[1] + line: 33146 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/webwork[1]/statement[1]/image[1] + line: 33199 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[5]/webwork[1]/statement[1]/image[1] + line: 33260 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[6]/webwork[1]/statement[1]/image[1] + line: 33316 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[7]/webwork[1]/statement[1]/image[1] + line: 33370 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[8]/webwork[1]/statement[1]/image[1] + line: 33420 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_mvt.ptx + path: /pretext[1]/book[1]/chapter[3]/section[2]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 33911 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_mvt.ptx + path: /pretext[1]/book[1]/chapter[3]/section[2]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 33945 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_mvt.ptx + path: /pretext[1]/book[1]/chapter[3]/section[2]/figure[1]/image[1] + line: 34047 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_mvt.ptx + path: /pretext[1]/book[1]/chapter[3]/section[2]/example[2]/solution[1]/figure[1]/image[1] + line: 34229 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_incr_decr.ptx + path: /pretext[1]/book[1]/chapter[3]/section[3]/figure[1]/image[1] + line: 34972 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_incr_decr.ptx + path: /pretext[1]/book[1]/chapter[3]/section[3]/figure[2]/sidebyside[1]/image[1] + line: 35094 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_incr_decr.ptx + path: /pretext[1]/book[1]/chapter[3]/section[3]/example[1]/solution[1]/figure[1]/image[1] + line: 35328 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_incr_decr.ptx + path: /pretext[1]/book[1]/chapter[3]/section[3]/example[1]/solution[1]/figure[2]/image[1] + line: 35408 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_incr_decr.ptx + path: /pretext[1]/book[1]/chapter[3]/section[3]/example[1]/solution[1]/figure[3]/image[1] + line: 35480 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_incr_decr.ptx + path: /pretext[1]/book[1]/chapter[3]/section[3]/remark[1]/figure[1]/image[1] + line: 35631 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_incr_decr.ptx + path: /pretext[1]/book[1]/chapter[3]/section[3]/example[2]/solution[1]/figure[1]/image[1] + line: 35792 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_incr_decr.ptx + path: /pretext[1]/book[1]/chapter[3]/section[3]/example[2]/solution[1]/figure[2]/image[1] + line: 35878 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_incr_decr.ptx + path: /pretext[1]/book[1]/chapter[3]/section[3]/example[3]/solution[1]/figure[1]/image[1] + line: 35966 + text: + +PTX:WARNING: You have an image without a description and do not declare the image to be decorative. Because of this, output may not be accessible. If the image does not add information that is not already present, use @decorative="yes". Otherwise, provide a . + file: sec_graph_incr_decr.ptx + path: /pretext[1]/book[1]/chapter[3]/section[3]/example[3]/solution[1]/figure[2]/image[1] + line: 36111 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 37374 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 37401 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[2]/image[1] + line: 37533 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[3]/image[1] + line: 37600 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[4]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] + line: 37674 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[4]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] + line: 37694 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[4]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] + line: 37716 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[4]/sbsgroup[1]/sidebyside[2]/figure[2]/image[1] + line: 37736 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[5]/image[1] + line: 37783 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 37886 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[1]/solution[1]/figure[2]/image[1] + line: 37921 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 38046 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[2]/solution[1]/figure[2]/image[1] + line: 38123 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[3]/statement[1]/figure[1]/image[1] + line: 38191 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[3]/solution[1]/figure[1]/image[1] + line: 38258 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[6]/image[1] + line: 38312 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[2]/figure[1]/image[1] + line: 38384 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_concavity.ptx + path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] + line: 38453 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[1]/solution[1]/figure[1]/image[1] + line: 41502 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[1]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 41588 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[1]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 41615 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[1]/solution[1]/figure[2]/sidebyside[1]/figure[3]/image[1] + line: 41643 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[2]/solution[1]/figure[1]/image[1] + line: 41782 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 41852 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 41884 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[3]/image[1] + line: 41913 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[3]/solution[1]/figure[1]/image[1] + line: 42053 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[3]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 42152 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[3]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 42183 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/example[3]/solution[1]/figure[2]/sidebyside[1]/figure[3]/image[1] + line: 42215 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 42307 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_graph_sketch.ptx + path: /pretext[1]/book[1]/chapter[3]/section[5]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 42331 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_newton.ptx + path: /pretext[1]/book[1]/chapter[4]/section[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 42998 + text: + +PTX:WARNING: A run of text contains Unicode characters for single quotation marks (U+2018, decimal 8216; U+2019, decimal 8217). Likely this was introduced in a conversion of source material authored in a word-processor. A U+2019 in isolation could be an apostrophe. Replace it with the keyboard version: U+0027. A matching pair U+2018, U+2019 should be replaced by the "" element enclosing content. In rare cases, U+2018 might be replaced by the empty element "". In rare cases, U+2019 might be replaced by the empty element "". + file: sec_newton.ptx + path: /pretext[1]/book[1]/chapter[4]/section[1]/figure[1]/sidebyside[1]/figure[1]/image[1]/description[1]/p[2] + line: 43009 + text:

    + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_newton.ptx + path: /pretext[1]/book[1]/chapter[4]/section[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 43035 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_newton.ptx + path: /pretext[1]/book[1]/chapter[4]/section[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 43064 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_newton.ptx + path: /pretext[1]/book[1]/chapter[4]/section[1]/example[1]/solution[1]/figure[1]/image[1] + line: 43275 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_newton.ptx + path: /pretext[1]/book[1]/chapter[4]/section[1]/example[2]/solution[1]/figure[1]/image[1] + line: 43362 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_newton.ptx + path: /pretext[1]/book[1]/chapter[4]/section[1]/paragraphs[1]/figure[1]/image[1] + line: 43485 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_newton.ptx + path: /pretext[1]/book[1]/chapter[4]/section[1]/paragraphs[1]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 43554 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_newton.ptx + path: /pretext[1]/book[1]/chapter[4]/section[1]/paragraphs[1]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 43594 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_newton.ptx + path: /pretext[1]/book[1]/chapter[4]/section[1]/paragraphs[1]/figure[2]/sidebyside[1]/figure[3]/image[1] + line: 43634 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_related_rates.ptx + path: /pretext[1]/book[1]/chapter[4]/section[2]/example[3]/statement[1]/figure[1]/image[1] + line: 44679 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_related_rates.ptx + path: /pretext[1]/book[1]/chapter[4]/section[2]/example[4]/statement[1]/figure[1]/image[1] + line: 44830 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_related_rates.ptx + path: /pretext[1]/book[1]/chapter[4]/section[2]/exercises[1]/subexercises[2]/exercise[5]/webwork[1]/introduction[1]/image[1] + line: 45267 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_related_rates.ptx + path: /pretext[1]/book[1]/chapter[4]/section[2]/exercises[1]/subexercises[2]/exercise[7]/webwork[1]/introduction[1]/image[1] + line: 45452 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_related_rates.ptx + path: /pretext[1]/book[1]/chapter[4]/section[2]/exercises[1]/subexercises[2]/exercise[8]/webwork[1]/introduction[1]/image[1] + line: 45559 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_related_rates.ptx + path: /pretext[1]/book[1]/chapter[4]/section[2]/exercises[1]/subexercises[2]/exercise[10]/webwork[1]/introduction[1]/image[1] + line: 45743 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_optimization.ptx + path: /pretext[1]/book[1]/chapter[4]/section[3]/example[1]/solution[1]/figure[1]/sidebyside[1]/image[1] + line: 46050 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_optimization.ptx + path: /pretext[1]/book[1]/chapter[4]/section[3]/example[1]/solution[1]/figure[1]/sidebyside[1]/image[2] + line: 46100 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_optimization.ptx + path: /pretext[1]/book[1]/chapter[4]/section[3]/example[2]/solution[1]/p[2]/ol[1]/li[1]/figure[1]/sidebyside[1]/image[1] + line: 46322 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_optimization.ptx + path: /pretext[1]/book[1]/chapter[4]/section[3]/example[2]/solution[1]/p[2]/ol[1]/li[1]/figure[1]/sidebyside[1]/image[2] + line: 46369 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_optimization.ptx + path: /pretext[1]/book[1]/chapter[4]/section[3]/example[3]/statement[1]/figure[1]/image[1] + line: 46512 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_optimization.ptx + path: /pretext[1]/book[1]/chapter[4]/section[3]/example[3]/solution[1]/figure[1]/image[1] + line: 46590 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_optimization.ptx + path: /pretext[1]/book[1]/chapter[4]/section[3]/exercises[1]/subexercises[2]/exercise[11]/webwork[1]/statement[1]/image[1] + line: 47137 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_differentials.ptx + path: /pretext[1]/book[1]/chapter[4]/section[4]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 47422 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_differentials.ptx + path: /pretext[1]/book[1]/chapter[4]/section[4]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 47458 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_differentials.ptx + path: /pretext[1]/book[1]/chapter[4]/section[4]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/webwork[1]/introduction[1]/image[1] + line: 48983 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_differentials.ptx + path: /pretext[1]/book[1]/chapter[4]/section[4]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[2]/webwork[1]/introduction[1]/image[1] + line: 49061 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_differentials.ptx + path: /pretext[1]/book[1]/chapter[4]/section[4]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[3]/webwork[1]/introduction[1]/image[1] + line: 49139 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_differentials.ptx + path: /pretext[1]/book[1]/chapter[4]/section[4]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/webwork[1]/statement[1]/image[1] + line: 49215 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_antider.ptx + path: /pretext[1]/book[1]/chapter[5]/section[1]/figure[1]/image[1] + line: 49494 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/figure[1]/image[1] + line: 51076 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/figure[2]/image[1] + line: 51158 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/example[1]/solution[1]/figure[1]/image[1] + line: 51315 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/example[2]/statement[1]/figure[1]/image[1] + line: 51498 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/example[2]/solution[1]/figure[1]/image[1] + line: 51605 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/example[3]/statement[1]/figure[1]/image[1] + line: 51762 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/example[4]/solution[1]/sidebyside[1]/figure[1]/image[1] + line: 51916 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/example[4]/solution[1]/sidebyside[1]/figure[2]/image[1] + line: 51958 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/example[5]/statement[1]/figure[1]/image[1] + line: 52004 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/figure[3]/image[1] + line: 52089 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/introduction[1]/image[1] + line: 52221 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/introduction[1]/image[1] + line: 52319 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/webwork[1]/introduction[1]/image[1] + line: 52422 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/webwork[1]/introduction[1]/image[1] + line: 52528 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/webwork[1]/introduction[1]/image[1] + line: 52629 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/webwork[1]/introduction[1]/image[1] + line: 52711 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/webwork[1]/introduction[1]/image[1] + line: 52801 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/webwork[1]/introduction[1]/image[1] + line: 52898 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/webwork[1]/introduction[1]/image[1] + line: 52988 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/webwork[1]/introduction[1]/image[1] + line: 53081 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/webwork[1]/introduction[1]/image[1] + line: 53185 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_def_int.ptx + path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[2]/webwork[1]/introduction[1]/image[1] + line: 53259 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/figure[1]/image[1] + line: 53618 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/figure[2]/image[1] + line: 53660 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/figure[3]/image[1] + line: 53726 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 53878 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 53932 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 53986 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/figure[1]/image[1] + line: 54230 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/image[1] + line: 54283 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] + line: 54448 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/figure[2]/image[1] + line: 54620 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] + line: 54760 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_riemann.ptx + path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/example[4]/solution[1]/figure[1]/image[1] + line: 54981 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/introduction[1]/figure[1]/image[1] + line: 56694 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/introduction[1]/example[1]/statement[1]/figure[1]/image[1] + line: 56747 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/introduction[1]/example[1]/solution[1]/figure[1]/image[1] + line: 56799 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/introduction[1]/example[1]/solution[1]/figure[2]/image[1] + line: 56877 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[2]/example[1]/solution[1]/sidebyside[1]/figure[1]/image[1] + line: 57299 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[2]/example[1]/solution[1]/sidebyside[1]/figure[2]/image[1] + line: 57349 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[4]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 57548 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[4]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 57598 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[4]/example[1]/solution[1]/figure[1]/image[1] + line: 57696 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[1]/image[1] + line: 57772 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 57839 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 57880 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[2]/sidebyside[1]/figure[3]/image[1] + line: 57921 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/example[1]/solution[1]/figure[1]/image[1] + line: 58034 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[3]/sidebyside[1]/image[1] + line: 58111 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[3]/sidebyside[1]/image[2] + line: 58161 + text: + +PTX:WARNING: A run of text contains Unicode characters for double quotation marks (U+201C, decimal 8220; U+201D, decimal 8221). Likely this was introduced in a conversion of source material authored in a word-processor. A matching pair U+201C, U+201D should be replaced by the "" element enclosing content. In rare cases, U+201C might be replaced by the empty element "". In rare cases, U+201D might be replaced by the empty element "". + file: sec_FTC.ptx + path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[3]/sidebyside[1]/image[2]/shortdescription[1] + line: 58162 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/introduction[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 59796 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/introduction[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 59832 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/introduction[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 59871 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 59985 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 60025 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[1]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 60227 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[1]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 60274 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[2]/figure[1]/image[1] + line: 60359 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[2]/figure[2]/image[1] + line: 60404 + text: + +PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of

    elements, if present, or else against the first row. Results may be unpredictable. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[3]/example[2]/solution[1]/figure[1]/tabular[1] + line: 60705 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[4]/figure[1]/image[1] + line: 60892 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[4]/figure[2]/image[1] + line: 60970 + text: + +PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of elements, if present, or else against the first row. Results may be unpredictable. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[4]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/tabular[1] + line: 61046 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[4]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 61085 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[4]/example[2]/solution[1]/figure[2]/image[1] + line: 61236 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[5]/example[1]/solution[1]/figure[1]/image[1] + line: 61551 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[5]/example[1]/solution[1]/figure[2]/image[1] + line: 61635 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/webwork[1]/statement[1]/image[1] + line: 63296 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_numerical_integration.ptx + path: /pretext[1]/book[1]/chapter[5]/section[5]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/webwork[1]/statement[1]/image[1] + line: 63372 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_substitution.ptx + path: /pretext[1]/book[1]/chapter[6]/section[1]/subsection[5]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 64607 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_substitution.ptx + path: /pretext[1]/book[1]/chapter[6]/section[1]/subsection[5]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 64647 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_substitution.ptx + path: /pretext[1]/book[1]/chapter[6]/section[1]/subsection[5]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 64742 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_substitution.ptx + path: /pretext[1]/book[1]/chapter[6]/section[1]/subsection[5]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 64785 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_trigint.ptx + path: /pretext[1]/book[1]/chapter[6]/section[3]/subsection[1]/figure[1]/image[1] + line: 68891 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_trig_sub.ptx + path: /pretext[1]/book[1]/chapter[6]/section[4]/insight[1]/p[1]/ol[1]/li[1]/sidebyside[1]/figure[1]/image[1] + line: 70191 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_trig_sub.ptx + path: /pretext[1]/book[1]/chapter[6]/section[4]/insight[1]/p[1]/ol[1]/li[2]/sidebyside[1]/figure[1]/image[1] + line: 70234 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_trig_sub.ptx + path: /pretext[1]/book[1]/chapter[6]/section[4]/insight[1]/p[1]/ol[1]/li[3]/sidebyside[1]/figure[1]/image[1] + line: 70281 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/introduction[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 73018 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/introduction[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 73070 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 73198 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 73252 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[1]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 73326 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[1]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 73384 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[2]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] + line: 73807 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[2]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] + line: 73858 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[2]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] + line: 73918 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_hyperbolic.ptx + path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[2]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[2]/image[1] + line: 73979 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_lhopitals_rule.ptx + path: /pretext[1]/book[1]/chapter[6]/section[7]/subsection[3]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] + line: 75900 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/introduction[1]/figure[1]/image[1] + line: 77256 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] + line: 77416 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] + line: 77477 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] + line: 77532 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[4]/figure[1]/image[1] + line: 77594 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 77667 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] + line: 77802 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] + line: 77860 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] + line: 77969 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[3]/example[2]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] + line: 78142 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[3]/example[2]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] + line: 78206 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_improper_integration.ptx + path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[3]/example[3]/solution[1]/figure[1]/image[1] + line: 78350 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 79767 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 79809 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 79850 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/example[1]/statement[1]/figure[1]/image[1] + line: 79944 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/example[2]/statement[1]/figure[1]/image[1] + line: 80006 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/example[3]/statement[1]/figure[1]/image[1] + line: 80096 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/example[3]/solution[1]/figure[1]/image[1] + line: 80173 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/example[4]/statement[1]/figure[1]/image[1] + line: 80269 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/example[5]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 80374 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/example[5]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 80397 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/statement[1]/image[1] + line: 80611 + text: Graph of the region enclosed by the functions y=\frac12 x +3 and y=\frac12\co... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/statement[1]/image[1] + line: 80676 + text: Graph of the region enclosed by the functions y=-3x^3+3x+2 and y=x^2+x-1 ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/webwork[1]/statement[1]/image[1] + line: 80804 + text: Graph of the region enclosed by the functions y=\sin(x)+1 and y=\sin(x) b... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/webwork[1]/statement[1]/image[1] + line: 80871 + text: Graph of the region enclosed by the functions y=\sin(4x) and y=\sec^2(x) ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/webwork[1]/statement[1]/image[1] + line: 80937 + text: Graph of the region enclosed by the functions y=\sin(x) and y=\cos(x) bet... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/webwork[1]/statement[1]/image[1] + line: 81003 + text: Graph of the region enclosed by the functions y=2^x and y=4^x between ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/statement[1]/image[1] + line: 81049 + text: Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1Graph of the region enclosed by the functions y=x^2+1, y=\frac14(x-3)^2+1Graph of the region enclosed by the functions y=\sqrt{x}, y=-2x+3 and Graph of the region enclosed by the functions y=x+2 and y=x^2. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/webwork[1]/statement[1]/image[1] + line: 81579 + text: Graph of the region enclosed by the functions x=-\frac12 y+1 and x=\frac12 y^... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[5]/webwork[1]/statement[1]/image[1] + line: 81651 + text: Graph of the region enclosed by the functions y=x^{1/3}, y=\sqrt{x-1/2} a... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[6]/statement[1]/image[1] + line: 81702 + text: Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1 + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_ABC.ptx + path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercise[4]/webwork[1]/statement[1]/image[1] + line: 81950 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/figure[1]/image[1] + line: 82009 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[1]/solution[1]/figure[1]/image[1] + line: 82158 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[1]/solution[1]/figure[2]/image[1] + line: 82266 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 82487 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 82564 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 82715 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 82794 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 82907 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 83005 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/figure[3]/image[1] + line: 83144 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 83303 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 83366 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 83442 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 83563 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 83648 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 83730 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/image[1] + line: 83958 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/statement[1]/image[1] + line: 84010 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/image[1] + line: 84059 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/statement[1]/image[1] + line: 84107 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/statement[1]/image[1] + line: 84171 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/statement[1]/image[1] + line: 84221 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/statement[1]/image[1] + line: 84275 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/statement[1]/image[1] + line: 84323 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/statement[1]/image[1] + line: 84759 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/statement[1]/image[1] + line: 84816 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/statement[1]/image[1] + line: 84871 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_disk.ptx + path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/statement[1]/image[1] + line: 84922 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 85012 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 85077 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 85146 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 85300 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 85345 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/example[1]/solution[1]/figure[1]/image[1] + line: 85454 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 85555 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 85607 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 85703 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 85855 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 85904 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 85999 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 86136 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 86183 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 86271 + text: + +PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of elements, if present, or else against the first row. Results may be unpredictable. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/insight[2]/tabular[1] + line: 86404 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/image[1] + line: 86543 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/statement[1]/image[1] + line: 86592 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/image[1] + line: 86640 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/statement[1]/image[1] + line: 86688 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/statement[1]/image[1] + line: 86750 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/statement[1]/image[1] + line: 86800 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/statement[1]/image[1] + line: 86848 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_shell_method.ptx + path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/statement[1]/image[1] + line: 86895 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 87372 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 87406 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[1]/figure[2]/image[1] + line: 87472 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 87633 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 87696 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 87828 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 87876 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] + line: 88086 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/example[2]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 88212 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/example[2]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 88272 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_arc_length.ptx + path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/example[3]/statement[1]/figure[1]/image[1] + line: 88390 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_work.ptx + path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] + line: 89462 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_work.ptx + path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] + line: 89657 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_work.ptx + path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[3]/figure[1]/image[1] + line: 89773 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_work.ptx + path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] + line: 89848 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_work.ptx + path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[3]/example[3]/statement[1]/figure[1]/image[1] + line: 89930 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_work.ptx + path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[3]/example[3]/solution[1]/figure[1]/image[1] + line: 89968 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_work.ptx + path: /pretext[1]/book[1]/chapter[7]/section[5]/exercises[1]/subexercises[2]/exercise[19]/statement[1]/image[1] + line: 90640 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_work.ptx + path: /pretext[1]/book[1]/chapter[7]/section[5]/exercises[1]/subexercises[2]/exercise[21]/statement[1]/image[1] + line: 90737 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_work.ptx + path: /pretext[1]/book[1]/chapter[7]/section[5]/exercises[1]/subexercises[2]/exercise[22]/statement[1]/image[1] + line: 90784 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_work.ptx + path: /pretext[1]/book[1]/chapter[7]/section[5]/exercises[1]/subexercises[2]/exercise[23]/statement[1]/image[1] + line: 90828 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/example[1]/statement[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] + line: 90938 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/example[1]/statement[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] + line: 90995 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/figure[1]/image[1] + line: 91143 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/example[2]/statement[1]/figure[1]/image[1] + line: 91288 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/example[2]/solution[1]/p[2]/ol[1]/li[1]/figure[1]/image[1] + line: 91346 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/example[2]/solution[1]/p[2]/ol[1]/li[2]/figure[1]/image[1] + line: 91428 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/example[3]/solution[1]/figure[1]/image[1] + line: 91547 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/example[4]/statement[1]/figure[1]/image[1] + line: 91607 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/image[1] + line: 91773 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/statement[1]/image[1] + line: 91810 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/image[1] + line: 91849 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/statement[1]/image[1] + line: 91894 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/statement[1]/image[1] + line: 91939 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/statement[1]/image[1] + line: 91982 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/statement[1]/image[1] + line: 92025 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/statement[1]/image[1] + line: 92067 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[9]/statement[1]/image[1] + line: 92109 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[10]/statement[1]/image[1] + line: 92148 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/statement[1]/image[1] + line: 92217 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/statement[1]/image[1] + line: 92257 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/statement[1]/image[1] + line: 92306 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/statement[1]/image[1] + line: 92355 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[5]/statement[1]/image[1] + line: 92399 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_fluid_force.ptx + path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[6]/statement[1]/image[1] + line: 92443 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 92734 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/figure[1]/image[1] + line: 92967 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] + line: 93090 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] + line: 93166 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[3]/statement[1]/figure[1]/image[1] + line: 93227 + text: + +PTX:WARNING: A run of text contains Unicode characters for single quotation marks (U+2018, decimal 8216; U+2019, decimal 8217). Likely this was introduced in a conversion of source material authored in a word-processor. A U+2019 in isolation could be an apostrophe. Replace it with the keyboard version: U+0027. A matching pair U+2018, U+2019 should be replaced by the "" element enclosing content. In rare cases, U+2018 might be replaced by the empty element "". In rare cases, U+2019 might be replaced by the empty element "". + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[3]/statement[1]/figure[1]/image[1]/shortdescription[1] + line: 93228 + text: Graph of slope field for the logistic differential equation y’=y(1-y) from the example. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[3]/solution[1]/figure[1]/image[1] + line: 93315 + text: + +PTX:WARNING: A run of text contains Unicode characters for single quotation marks (U+2018, decimal 8216; U+2019, decimal 8217). Likely this was introduced in a conversion of source material authored in a word-processor. A U+2019 in isolation could be an apostrophe. Replace it with the keyboard version: U+0027. A matching pair U+2018, U+2019 should be replaced by the "" element enclosing content. In rare cases, U+2018 might be replaced by the empty element "". In rare cases, U+2019 might be replaced by the empty element "". + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[3]/solution[1]/figure[1]/image[1]/shortdescription[1] + line: 93316 + text: Graph of slope field for the logistic differential equation y’=y(1-y) with representative solution c... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] + line: 93530 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] + line: 93620 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[3]/example[3]/solution[1]/figure[1]/image[1] + line: 93764 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/answer[1]/image[1] + line: 94060 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[2]/answer[1]/image[1] + line: 94103 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[3]/answer[1]/image[1] + line: 94144 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/answer[1]/image[1] + line: 94195 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercise[1]/statement[1]/sbsgroup[1]/sidebyside[1]/image[1] + line: 94237 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercise[1]/statement[1]/sbsgroup[1]/sidebyside[1]/image[2] + line: 94270 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercise[1]/statement[1]/sbsgroup[1]/sidebyside[3]/image[1] + line: 94314 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercise[1]/statement[1]/sbsgroup[1]/sidebyside[3]/image[2] + line: 94346 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/answer[1]/image[1] + line: 94425 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/answer[1]/image[1] + line: 94478 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/answer[1]/image[1] + line: 94533 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Graphical_Numerical.ptx + path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/answer[1]/image[1] + line: 94581 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Linear.ptx + path: /pretext[1]/book[1]/chapter[8]/section[3]/subsection[1]/figure[1]/image[1] + line: 96129 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Linear.ptx + path: /pretext[1]/book[1]/chapter[8]/section[3]/exercises[1]/subexercises[1]/exercisegroup[4]/exercise[1]/answer[1]/image[1] + line: 96469 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Linear.ptx + path: /pretext[1]/book[1]/chapter[8]/section[3]/exercises[1]/subexercises[1]/exercisegroup[4]/exercise[2]/answer[1]/image[1] + line: 96516 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Modeling.ptx + path: /pretext[1]/book[1]/chapter[8]/section[4]/subsection[1]/figure[1]/image[1] + line: 96583 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Modeling.ptx + path: /pretext[1]/book[1]/chapter[8]/section[4]/subsection[1]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 96957 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Modeling.ptx + path: /pretext[1]/book[1]/chapter[8]/section[4]/subsection[1]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 96999 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_Modeling.ptx + path: /pretext[1]/book[1]/chapter[8]/section[4]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] + line: 97206 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_sequences.ptx + path: /pretext[1]/book[1]/chapter[9]/section[1]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] + line: 97679 + text: Plot of the first four terms of the sequence in part 1 of this example.A plot of the first four terms of the sequence in part 2 of this example.A plot of the first four terms of the sequence in part 3 of this example.A scatter plot showing a representative sample of points from the first sequence i... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_sequences.ptx + path: /pretext[1]/book[1]/chapter[9]/section[1]/example[3]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] + line: 98198 + text: A scatter plot showing a representative sample of points from the second sequence ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_sequences.ptx + path: /pretext[1]/book[1]/chapter[9]/section[1]/example[3]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] + line: 98249 + text: A scatter plot showing a representative sample of points from the third sequence i... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_sequences.ptx + path: /pretext[1]/book[1]/chapter[9]/section[1]/example[4]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] + line: 98406 + text: Scatter plot illustrating the first 20 points in the sequence from the second part... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_sequences.ptx + path: /pretext[1]/book[1]/chapter[9]/section[1]/example[6]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 98678 + text: A scatter plot of the first 10 terms in the first sequence for this example.The scatter plot for the second sequence in this example, which shows exponential ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_sequences.ptx + path: /pretext[1]/book[1]/chapter[9]/section[1]/example[7]/solution[1]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] + line: 99024 + text: Plot of the first sequence in this example. It is decreasing and bounded below.Scatter plot for the second sequence in this example. It is increasing but not bou... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_sequences.ptx + path: /pretext[1]/book[1]/chapter[9]/section[1]/example[7]/solution[1]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] + line: 99097 + text: Scatter plot for the third sequence in this example. It is not monotonic.Scatter plot for the last sequence in this example. It is not monotonic.Scatter plots of the sequence, and corresponding partial sums, for the first part ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 100406 + text: Scatter plots of the sequence, and corresponding partial sums, for the second part... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] + line: 100680 + text: Scatter plots of the sequence, and corresponding partial sums, for the first part ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] + line: 100742 + text: Scatter plots of the sequence, and corresponding partial sums, for the second part... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] + line: 100804 + text: Scatter plots of the sequence, and corresponding partial sums, for the third part ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] + line: 101080 + text: Scatter plots of the sequence, and corresponding partial sums, for this example.Scatter plots of the sequence, and corresponding partial sums, for the first part ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[3]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 101301 + text: Scatter plots of the sequence, and corresponding partial sums, for the second part... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[3]/example[4]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 101510 + text: Scatter plots of the sequence, and corresponding partial sums, for the first part ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[3]/example[4]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 101562 + text: Scatter plots of the sequence, and corresponding partial sums, for the first part ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_int_comp_tests.ptx + path: /pretext[1]/book[1]/chapter[9]/section[3]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 103227 + text: First of two graphs illustrating why the integral test works. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_int_comp_tests.ptx + path: /pretext[1]/book[1]/chapter[9]/section[3]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 103269 + text: Second of two graphs illustrating why the integral test works. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_int_comp_tests.ptx + path: /pretext[1]/book[1]/chapter[9]/section[3]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 103430 + text: Scatter plots of the sequence used in this example, and the corresponding sequence... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_alt_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[5]/figure[1]/image[1] + line: 106080 + text: Scatter plots of a positive, decreasing sequence and its alternating sequence of p... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_alt_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[5]/figure[2]/image[1] + line: 106136 + text: An illustration of the partial sums in an alternating series, using line segments ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/sidebyside[1]/figure[1]/image[1] + line: 109495 + text: The graph of a generic function is shown, along with the tangent line to the graph... + +PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/sidebyside[1]/figure[2]/sidebyside[1] + line: 109534 + text:

    + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/figure[1]/image[1] + line: 109625 + text: The graph of a function and two polynomials that approximate the function near x=0... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/figure[2]/image[1] + line: 109713 + text: The graph of a function is shown, along with the graph of a degree 13 polynomial t... + +PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1] + line: 109840 + text:

    + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/example[1]/solution[1]/figure[1]/image[1] + line: 109909 + text: The graph of the exponential function and its degree 5 Maclaurin polynomial. normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/example[2]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1] + line: 109992 + text:

    + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/example[2]/solution[1]/sidebyside[1]/figure[1]/image[1] + line: 110116 + text: A graph of the natural logarithm function and its degree 6 Taylor polynomial cente... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/example[2]/solution[1]/sidebyside[1]/figure[2]/image[1] + line: 110153 + text: A graph of the natural logarithm function and its degree 20 Taylor polynomial cent... + +PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/example[4]/solution[1]/figure[1]/sidebyside[1] + line: 110455 + text:

    + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/example[4]/solution[1]/figure[2]/image[1] + line: 110558 + text: A graph of the cosine function is shown, along with its degree 8 Maclaurin polynom... + +PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/example[5]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1] + line: 110649 + text:

    + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/example[5]/solution[1]/figure[1]/image[1] + line: 110737 + text: The graph of the square root function and its degree 4 Taylor polynomial, centered... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_taylor_poly.ptx + path: /pretext[1]/book[1]/chapter[9]/section[7]/example[6]/solution[1]/figure[1]/image[1] + line: 110888 + text: The graph of the exact solution to the differential equation in this example, and ... + +PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the + file: sec_taylor_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[8]/example[1]/solution[1]/figure[1]/sidebyside[1] + line: 111905 + text:

    + +PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the + file: sec_taylor_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[8]/example[2]/solution[1]/figure[1]/sidebyside[1] + line: 112040 + text:

    + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_taylor_series.ptx + path: /pretext[1]/book[1]/chapter[9]/section[8]/example[8]/solution[1]/figure[1]/image[1] + line: 112721 + text: A graph of the function from this example, and its degree 5 Maclaurin polynomial a... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/introduction[1]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] + line: 113876 + text: A diagonal plane intersecting a double napped cone, forming a parabolaA diagonal plane intersecting a double napped cone, forming an ellipse.A horizontal plane intersecting a double napped cone, forming a circle.A vertical plane intersecting a double napped cone, forming a hyperbola.A horizontal plane intersecting the tips in a double napped cone.A diagonal plane intersecting a double napped cone, forming a line in the planeA vertical plane intersecting a double napped cone, forming crossed lines in the p... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[1]/figure[1]/image[1] + line: 114043 + text: A sketch of a parabola with key components labeled. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 114191 + text: A downward opening parabola with a vertex in the first quadrant.A parabola opening to the right with its directrix and focus drawn.A rightward opening parabola demonstrating the reflective property.A demonstration of the creation of an ellipse from two foci. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/figure[2]/image[1] + line: 114536 + text: An ellipse with labels for key components. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] + line: 114639 + text: An ellipse centered in the second quadrant, lying entirely in the second and third... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] + line: 114728 + text: An ellipse centered at (1,2), with vertices at (-2,2) and (4,2).An ellipse of eccentricity 0. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/paragraphs[1]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] + line: 114839 + text: An ellipse of eccentricity 0.3. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/paragraphs[1]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] + line: 114880 + text: An ellipse of eccentricity 0.8. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/paragraphs[1]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[2]/image[1] + line: 114919 + text: An ellipse of eccentricity 0.99. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/paragraphs[2]/figure[1]/image[1] + line: 115004 + text: An ellipse with rays emanating from the foci, demonstrating the reflective propert... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/figure[1]/image[1] + line: 115115 + text: A hyperbola with labels for key components + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[1]/figure[1]/image[1] + line: 115226 + text: A graph of a hyperbola and its asymptotes. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[1]/figure[2]/image[1] + line: 115272 + text: An illustration of sketching a hyperbola with a rectangle. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[1]/example[1]/solution[1]/figure[1]/image[1] + line: 115364 + text: A graph of the hyperbola in this example + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[1]/example[2]/solution[1]/figure[1]/image[1] + line: 115448 + text: The hyperbola described in this example + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[2]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] + line: 115548 + text: A hyperbola with eccentricity 1.05 + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[2]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] + line: 115596 + text: A hyperbola with eccentricity 1.5 + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[2]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] + line: 115648 + text: A hyperbola with eccentricity 3 + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[2]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[2]/image[1] + line: 115696 + text: A hyperbola with eccentricity 10 + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[4]/figure[1]/image[1] + line: 115817 + text: A hyperbola demonstrating the reflective property. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[4]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 115908 + text: Three points drawn and labeled on a plane + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[4]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 115942 + text: A hyperbola drawn from points A and B to illustrate the location property A fourth point found from hyperbolas given by points in the previous figuresAn ellipse centered in the second quadrant. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/statement[1]/image[1] + line: 116473 + text: A vertical ellipse centered on the x-axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/exercises[1]/subexercises[2]/exercisegroup[7]/exercise[1]/statement[1]/image[1] + line: 116687 + text: A hyperbola centered at the origin with a horizontal transverse axis.A hyperbola centered at the origin with a vertical transverse axis.A hyperbola with center (1,3) and a vertical transverse axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_conic_sections.ptx + path: /pretext[1]/book[1]/chapter[10]/section[1]/exercises[1]/subexercises[2]/exercisegroup[7]/exercise[4]/statement[1]/image[1] + line: 116844 + text: A hyperbola centered at the point (1,3) with a horizontal transverse axisA number line with 3 points drawn. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/introduction[1]/figure[1]/image[1] + line: 117277 + text: Diagram outlining the steps involved in plotting a function. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/introduction[1]/figure[2]/image[1] + line: 117340 + text: Diagram illustrating the process for plotting a parametric curve.Plot of a parabola with its vertex at the point (0,1), opening to the right, with ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[1]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 117608 + text: Plot of a parabola with its vertex at the point (0,1), opening to the right, with ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[1]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 117750 + text: Plot of the "unshifted" parametric curve in this example. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[1]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 117794 + text: The shifted version of the curve in this example. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[1]/example[4]/solution[1]/figure[1]/image[1] + line: 117860 + text: Plot of a more complicated parametric curve that has a self-intersection.A parabolic arc describing the motion of a projectile fired up and to the right fr... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[2]/example[3]/solution[1]/figure[1]/image[1] + line: 118102 + text: A graph of two overlapping lines illustrating rectangular and parametric descripti... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[2]/example[4]/solution[1]/figure[1]/image[1] + line: 118203 + text: Graph of the ellipse obtained from the parametric equations in this example.Plot of an astroid: a curve like a 4-pointed star. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[3]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] + line: 118348 + text: Plot of a rose curve with 8 loops. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[3]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] + line: 118384 + text: Plot of a hypotrochoid, which resembles a smooth, five-pointed star.Plot of an epicycloid, which looks somewhat like a puffy clover leaf.Plot of a parametric curve with a sharp cusp. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/answer[1]/image[1] + line: 118720 + text: Sketch of the parametric curve in this exercise. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/answer[1]/image[1] + line: 118768 + text: The vertical line x=1. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/answer[1]/image[1] + line: 118813 + text: The horizontal line y=2, marked with two arrows. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/answer[1]/image[1] + line: 118860 + text: The solution curve for this exercise. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_param_eqs.ptx + path: /pretext[1]/book[1]/chapter[10]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/answer[1]/image[1] + line: 118917 + text: Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Plot of a parametric curve and its tangent and normal lines at one point.Sketch of the unit circle and one normal line, which passes through the circle's c... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_par_calc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[3]/subsection[1]/example[3]/statement[1]/figure[1]/image[1] + line: 120805 + text: Graph of an astroid. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_par_calc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[3]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] + line: 120967 + text: Sketch of a parametric curve illustrating concavity. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_par_calc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[3]/subsection[2]/example[1]/solution[1]/image[1] + line: 121022 + text: A number line for the sign of the second derivative in this example.Graph of the second derivative is a sinusoid with increasing amplitude.Graph of the parametric curve in this example, with points of inflection marked.Graph of a "teardrop" curve that intersects itself at the origin.The surface obtained by revolving the teardrop shape from the previous example abo... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[1]/figure[1]/image[1] + line: 122747 + text: Illustration of polar coordinates relative to a pole and initial ray.Drawing of a polar grid: a set of concentric circles, and rays from their common c... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 122873 + text: A plot of several points on a polar grid. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[2]/figure[1]/image[1] + line: 122958 + text: A triangle illustrating the conversion between rectangular and polar coordinates.<... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 123071 + text: Overlaid rectangular and polar grid systems, with marked points.Overlaid rectangular and polar grid systems, with marked points.Polar plot showing curves of constant radius and constant angle + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 123425 + text: A rough sketch of a polar function. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/figure[1]/image[1] + line: 123514 + text: A computer-generated sketch of a polar function. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[3]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 123608 + text: An initial rough sketch of a polar curve obtained by connecting points on the curv... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[3]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 123665 + text: A smoothed version of the plot for this example. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[4]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] + line: 123814 + text: Graph of the hyperbola y=1/x. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 123923 + text: A line through the origin with positive slope. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 123952 + text: A horizontal line. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 123981 + text: A vertical line. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[1]/sidebyside[1]/figure[4]/image[1] + line: 124010 + text: A line with positive slope m, and y intercept b. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 124046 + text: A circle centered on the positive x axis and passing through the origin.A circle centered on the positive y axis and passing through the origin.A circle centered at the origin. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[2]/sidebyside[1]/figure[4]/image[1] + line: 124134 + text: A counter-clockwise spiral that begins at the origin. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[3]/sidebyside[1]/figure[1]/image[1] + line: 124171 + text: A limaçon curve with a loop at the origin, and symmetric about the x axis.A heart-shaped curve in the limaçon family, known as a cardioid.A dimpled limaçon curve. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[3]/sidebyside[1]/figure[4]/image[1] + line: 124277 + text: A convex limaçon curve. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[4]/sidebyside[1]/figure[1]/image[1] + line: 124321 + text: A rose curve with four leaves along the coordinate axes. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[4]/sidebyside[1]/figure[2]/image[1] + line: 124347 + text: A rose curve with four leaves, each in one of the quadrants. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[4]/sidebyside[1]/figure[3]/image[1] + line: 124374 + text: A rose curve with three leaves, symmetric about the x axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[4]/sidebyside[1]/figure[4]/image[1] + line: 124406 + text: A rose curve with three leaves, symmetric about the y axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[5]/sidebyside[1]/figure[1]/image[1] + line: 124452 + text: A polar curve consisting of three loops. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[5]/sidebyside[1]/figure[2]/image[1] + line: 124489 + text: An elaborate polar curve consisting of many intersecting loops. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[5]/sidebyside[1]/figure[3]/image[1] + line: 124521 + text: A lemniscate curve, which has the shape of a figure-eight. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[5]/sidebyside[1]/figure[4]/image[1] + line: 124550 + text: A polar curve in the shape of a figure-eight. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 124610 + text: Overlapping plots of a circle and a limaçon, showing their points of intersection.... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 124657 + text: Zoomed in view of the intersections between the two polar curves.The four points plotted in this exercise. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercise[2]/answer[1]/image[1] + line: 124951 + text: The four points plotted in this exercise. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercise[3]/statement[1]/image[1] + line: 125000 + text: Four points plotted on a polar grid. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercise[4]/webwork[1]/statement[1]/image[1] + line: 125099 + text: Four points plotted on a polar grid. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/answer[1]/image[1] + line: 125349 + text: Portion of the circle of radius 2, centered at the origin, in the first quadrant.<... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/answer[1]/image[1] + line: 125390 + text: A line segment through the origin with positive slope. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/answer[1]/image[1] + line: 125434 + text: A cardioid, symmetric about the x axis, with x intercepts at -2 and 0.A convex limaçon, symmetric about the y axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/answer[1]/image[1] + line: 125518 + text: A convex limaçon, symmetric about the y axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/answer[1]/image[1] + line: 125560 + text: A limaçon with an inner loop, symmetric about the y axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/answer[1]/image[1] + line: 125603 + text: A limaçon with an inner loop, symmetric about the y axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/answer[1]/image[1] + line: 125646 + text: A rose curve with four petals. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[9]/answer[1]/image[1] + line: 125690 + text: A rose curve with three loops, symmetric about the y axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[10]/answer[1]/image[1] + line: 125733 + text: A limaçon with an inner loop. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[11]/answer[1]/image[1] + line: 125784 + text: An elaborate rose curve with many self-intersections and four primary loops.A counter-clockwise spiral + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[13]/answer[1]/image[1] + line: 125871 + text: A circle passing through the origin with its center on the positive y axis.A semi-circle in the first quadrant. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[15]/answer[1]/image[1] + line: 125947 + text: A four-leaf rose with one petal in each quadrant. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[16]/answer[1]/image[1] + line: 125988 + text: A curve with two loops: a circle inside a larger leaf shape. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[17]/answer[1]/image[1] + line: 126033 + text: A straight line with positive slope. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[18]/answer[1]/image[1] + line: 126074 + text: A straight line with positive slope. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[19]/answer[1]/image[1] + line: 126115 + text: A vertical line with x=3. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polar.ptx + path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[20]/answer[1]/image[1] + line: 126155 + text: The horizontal line y=4. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polarcalc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] + line: 126932 + text: A limaçon with an inner loops, symmetric about the y axis, and a tangent line.A zoomed in view of a limaçon near the origin, and two tangent lines at that point... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polarcalc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[2]/aside[1]/image[1] + line: 127184 + text: Illustration of a pie-shaped sector of a circle. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polarcalc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[2]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 127237 + text: Plot of a generic polar function and the area it encloses between two angles.Plot of a region bounded by a polar curve, and its approximation by cicular wedges... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polarcalc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[2]/example[2]/statement[1]/figure[1]/image[1] + line: 127442 + text: A cardioid curve, within which is a shaded region bounded by the curve and two ray... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polarcalc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[2]/paragraphs[1]/figure[1]/image[1] + line: 127520 + text: Illustration of a region bounded by two polar curves and two rays.A circle and a cardioid enclose a region that is inside the circle but outside the... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polarcalc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[2]/paragraphs[1]/example[2]/statement[1]/figure[1]/image[1] + line: 127708 + text: A zoomed in view of a region bounded by a circle, a rose curve, and the x axis.A zoomed in view of a polar region, showing it divided into two parts.A limaçon with an inner loop that is symmetric about the y axis.A four leaf rose curve, with leaves along the coordinate axes. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polarcalc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[4]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 128070 + text: The surface of revolution obtained by revolving one leaf of a rose curve about the... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polarcalc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[5]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[9]/webwork[1]/statement[1]/image[1] + line: 128875 + text: Two circles, one along each axis, bound a region in the first quadrant.A semi-circle and one leaf of a rose curve, both in the first quadrant, overlappin... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polarcalc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[5]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[11]/statement[1]/image[1] + line: 128974 + text: Overlapping leaves of two different three-leaf rose curves. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_polarcalc.ptx + path: /pretext[1]/book[1]/chapter[10]/section[5]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[12]/statement[1]/image[1] + line: 129020 + text: A cardioid and a circle, and a region of overlap in the first quadrant. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/introduction[1]/figure[2]/image[1] + line: 129490 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 129606 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[3]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 129774 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[3]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 129828 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[3]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 129882 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[3]/figure[2]/image[1] + line: 129945 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] + line: 130015 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 130108 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 130167 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 130302 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 130360 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 130465 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 130540 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 130606 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 130673 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 130765 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 130841 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 130964 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 131024 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 131145 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 131199 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/figure[1]/image[1] + line: 131299 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[1]/image[1] + line: 131426 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[1]/image[2] + line: 131506 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[1]/sidebyside[1]/image[1] + line: 131600 + text: + +PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of

    elements, if present, or else against the first row. Results may be unpredictable. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[1]/sidebyside[1]/tabular[1] + line: 131653 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[1]/sidebyside[2]/image[1] + line: 131697 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[1]/sidebyside[2]/image[2] + line: 131766 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[2]/image[1] + line: 131852 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[2]/image[2] + line: 131946 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[3]/image[2] + line: 132138 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[4]/image[2] + line: 132347 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[2]/sidebyside[2]/image[1] + line: 132558 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[2]/sidebyside[2]/image[2] + line: 132630 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 132767 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 132837 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 132939 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 133015 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 133117 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 133188 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[2]/statement[1]/figure[1]/image[1] + line: 133285 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/answer[1]/image[1] + line: 133861 + text: A surface obtained by translating the cubic curve along the y axisA surface generated by translating the curve y=cos(z) along the x axisA cylindrical tube with cross sections in the shape of an ellipseA pair of sheets that look like a hyperbola when viewed from above + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/statement[1]/image[1] + line: 134382 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/statement[1]/image[1] + line: 134465 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/statement[1]/image[1] + line: 134539 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[1]/answer[1]/image[1] + line: 134637 + text: A hyperbolic paraboloid (the classic "Pringles chip") + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[2]/answer[1]/image[1] + line: 134700 + text: An elliptical cone + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[3]/answer[1]/image[1] + line: 134762 + text: A circular paraboloid, opening along the negative x axis + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[4]/answer[1]/image[1] + line: 134823 + text: A hyperboloid of two sheets, opening along the x axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[5]/answer[1]/image[1] + line: 134891 + text: A hyperboloid of one sheet, opening along the y axis + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_space_coord.ptx + path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[6]/answer[1]/image[1] + line: 134953 + text: An ellpsoid, somewhat in the shape of a squashed olive. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/figure[1]/image[1] + line: 135084 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/figure[2]/image[1] + line: 135134 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 135289 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 135328 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/example[2]/solution[1]/figure[1]/image[1] + line: 135538 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/figure[3]/image[1] + line: 135617 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/example[3]/solution[1]/figure[1]/image[1] + line: 135717 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/example[4]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] + line: 135806 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/example[5]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] + line: 136076 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/example[6]/statement[1]/figure[1]/image[1] + line: 136277 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/example[6]/solution[1]/figure[1]/image[1] + line: 136354 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/example[8]/statement[1]/figure[1]/image[1] + line: 136525 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/statement[1]/image[1] + line: 137077 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/solution[1]/image[1] + line: 137115 + text: Solution image for this exercise + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/statement[1]/image[1] + line: 137161 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/solution[1]/image[1] + line: 137198 + text: Solution image for this exercise + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/statement[1]/image[1] + line: 137247 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/solution[1]/image[1] + line: 137294 + text: Solution image for this exercise + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/statement[1]/image[1] + line: 137350 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/solution[1]/image[1] + line: 137395 + text: Solution image for this exercise + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[5]/introduction[1]/image[1] + line: 137843 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_intro.ptx + path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[6]/introduction[1]/image[1] + line: 137964 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 138258 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 138288 + text: + +PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("image") and remove the + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[2]/sidebyside[1] + line: 138422 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[2]/sidebyside[1]/image[1] + line: 138423 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/example[2]/statement[1]/figure[1]/image[1] + line: 138496 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/example[3]/statement[1]/figure[1]/image[1] + line: 138601 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[3]/sidebyside[1]/figure[1]/image[1] + line: 138878 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[3]/sidebyside[1]/figure[2]/image[1] + line: 138911 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/example[5]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] + line: 139031 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/example[5]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 139094 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/example[5]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 139180 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[4]/image[1] + line: 139315 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/example[7]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 139492 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/example[7]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 139524 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/paragraphs[1]/figure[1]/image[1] + line: 139631 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_dot_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[3]/paragraphs[1]/example[1]/statement[1]/figure[1]/image[1] + line: 139705 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cross_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[4]/introduction[1]/image[1] + line: 141118 + text: Schematic diagram for computing the cross product by multiplying along diagonal ar... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cross_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[1]/paragraphs[1]/figure[1]/image[1] + line: 141477 + text: Three-dimensional image illustrating the right-hand rule for the direction of the ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cross_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 141581 + text: A parallelogram with base b and height h labeled. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cross_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 141611 + text: Another parallelogram, with two adjacent sides labeled as vectors.A parallelogram in the plane, spanned by vectors u and v. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cross_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[1]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 141738 + text: A three-dimensional image of a parallelogram in space, with vertices A, B, C, and ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cross_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[1]/example[2]/statement[1]/figure[1]/image[1] + line: 141833 + text: A triangle in the plane, with vertices A(1,2), B(2,3), and C(3,1).A parellelepiped in three dimensions. It is the box-like object spanned by three v... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cross_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[2]/example[1]/solution[1]/figure[1]/image[1] + line: 142063 + text: The parallelepiped whose volume is computed in this example. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cross_product.ptx + path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[3]/example[1]/statement[1]/figure[1]/image[1] + line: 142195 + text: Two adjacent images, each showing a pair of vectors, one of which represents a lev... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_lines.ptx + path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[1]/figure[1]/image[1] + line: 143364 + text: In three dimensions, vectors p (position), d (direction), and p+d are plotted. A l... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_lines.ptx + path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[1]/figure[2]/image[1] + line: 143440 + text: A text description, comparing the elements the equations of a line in the plane, a... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_lines.ptx + path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 143635 + text: A line in space with direction vector d passes through a point P.A line in space passes through points P and Q. The vector PQ between these points ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_lines.ptx + path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] + line: 143922 + text: A three dimensional plot of two lines in space. The lines are not parallel, and th... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_lines.ptx + path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] + line: 144103 + text: A plot in space of a single line through two points, with two parallel direction v... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_lines.ptx + path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[3]/figure[1]/image[1] + line: 144201 + text: A line through a point P, with direction d. Near the line is a point Q, whose dist... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_lines.ptx + path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[3]/figure[2]/image[1] + line: 144271 + text: Two skew lines in space are plotted with respective points and direction vectors. ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_planes.ptx + path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/figure[1]/image[1] + line: 145926 + text: A nail sticks into a flat, rectangular sheet of cardboard. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_planes.ptx + path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/example[1]/solution[1]/figure[1]/image[1] + line: 146122 + text: Three points P, Q, and R are plotted in space, along with the plane containing the... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_planes.ptx + path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/example[2]/solution[1]/figure[1]/image[1] + line: 146285 + text: In three dimensions, two lines are plotted, intersecting at a point P, along with ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_planes.ptx + path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/example[3]/solution[1]/figure[1]/image[1] + line: 146377 + text: A plane is plotted in a three-dimensional coordinate system, along with a line pas... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_planes.ptx + path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/example[4]/solution[1]/figure[1]/image[1] + line: 146514 + text: Two planes are shown intersecting along a common line. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_planes.ptx + path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/example[5]/solution[1]/figure[1]/image[1] + line: 146638 + text: A line in three dimensions passes through a plane, intersecting it at a point.A plane, on which a point P and normal vector n are marked, along with a point Q n... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 148255 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 148286 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 148399 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 148469 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 148616 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 148647 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 148681 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[2]/example[2]/statement[1]/figure[1]/image[1] + line: 148745 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] + line: 148818 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] + line: 148912 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/solution[1]/image[1] + line: 149146 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/solution[1]/image[1] + line: 149189 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/solution[1]/image[1] + line: 149235 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/solution[1]/image[1] + line: 149281 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/solution[1]/image[1] + line: 149330 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/solution[1]/image[1] + line: 149380 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/solution[1]/image[1] + line: 149426 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/solution[1]/image[1] + line: 149476 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/solution[1]/image[1] + line: 149534 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/answer[1]/image[1] + line: 149586 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/answer[1]/image[1] + line: 149637 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf.ptx + path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/answer[1]/image[1] + line: 149691 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 150414 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 150446 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 150644 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 150682 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] + line: 150744 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[3]/solution[1]/figure[1]/image[1] + line: 150882 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[4]/solution[1]/figure[1]/image[1] + line: 150969 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[5]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] + line: 151150 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[5]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] + line: 151226 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/solution[1]/image[1] + line: 151937 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/solution[1]/image[1] + line: 151985 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/solution[1]/image[1] + line: 152035 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_calc.ptx + path: /pretext[1]/book[1]/chapter[12]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/solution[1]/image[1] + line: 152085 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_motion.ptx + path: /pretext[1]/book[1]/chapter[12]/section[3]/introduction[1]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 152799 + text: The curve corresponding the position function for this example, with velocity and ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_motion.ptx + path: /pretext[1]/book[1]/chapter[12]/section[3]/introduction[1]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 152847 + text: The same curve as the previous image, with more vectors plotted.A parabola with vertex at the origin, opening upward. Several pairs of velocity an... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_motion.ptx + path: /pretext[1]/book[1]/chapter[12]/section[3]/introduction[1]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 153055 + text: The parabola y equals x squared, with several points plotted, corresponding to equ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_motion.ptx + path: /pretext[1]/book[1]/chapter[12]/section[3]/introduction[1]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 153107 + text: The parabola y equals x squared, with several points plotted, corresponding to equ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_motion.ptx + path: /pretext[1]/book[1]/chapter[12]/section[3]/introduction[1]/example[3]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] + line: 153295 + text: A diagram showing a coniferous tree, a circle, and a tangent line from the circle ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vvf_motion.ptx + path: /pretext[1]/book[1]/chapter[12]/section[3]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] + line: 153802 + text: An illustration of the path followed by the particle in this example.A 3D plot of a portion of a helix, with two unit tangent vectors shown.A rotated parabola, along with two of its unit tangent vectors. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_tan_norm.ptx + path: /pretext[1]/book[1]/chapter[12]/section[4]/subsection[2]/figure[1]/image[1] + line: 155479 + text: A generic plane curve with a tangent vector, along with two possible normal vector... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_tan_norm.ptx + path: /pretext[1]/book[1]/chapter[12]/section[4]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] + line: 155609 + text: A three-dimensional helix. At one point, the unit tangent and normal vectors are s... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_tan_norm.ptx + path: /pretext[1]/book[1]/chapter[12]/section[4]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] + line: 155740 + text: A rotated parabola, with unit tangent and normal vectors plotted at three points.<... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_tan_norm.ptx + path: /pretext[1]/book[1]/chapter[12]/section[4]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] + line: 156016 + text: A rotated parabola, with two points marked, corresponding to parameter values t=0 ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_tan_norm.ptx + path: /pretext[1]/book[1]/chapter[12]/section[4]/subsection[3]/example[3]/solution[1]/figure[1]/image[1] + line: 156086 + text: An inverted parabola with vertex in the first quadrant. Points corresponding to se... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_curvature.ptx + path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 157230 + text: A parabola, rotated to be symmetric about the line y=x, with three marked points.<... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_curvature.ptx + path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 157273 + text: The same parabola as the previous image, this time with six marked points, all equ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_curvature.ptx + path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 157407 + text: A straight line, with positive slope, and several equally-spaced marked points.Plot of a parametric curve that bends rapidly at one point, but is otherwise relat... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_curvature.ptx + path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 157580 + text: Plot of a parametric curve that bends rapidly at one point, but is otherwise relat... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_curvature.ptx + path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[2]/figure[2]/image[1] + line: 157871 + text: A curve with two marked points, corresponding to points of large and small curvatu... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_curvature.ptx + path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[2]/example[3]/solution[1]/figure[1]/image[1] + line: 157938 + text: A parabola along with osculating circles at two points. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_curvature.ptx + path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 158031 + text: A plot of the curvature as a function of the parameter t. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_curvature.ptx + path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 158067 + text: A plot of the vector-valued function in this example, with points of maximum curva... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_intro.ptx + path: /pretext[1]/book[1]/chapter[13]/section[1]/introduction[1]/example[2]/solution[1]/figure[1]/image[1] + line: 159382 + text: An ellipse, centered at the origin, with shaded interior, illustrating the domain ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_intro.ptx + path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 159444 + text: A collection of points plotted in space, against a set of three-dimensional coordi... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_intro.ptx + path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 159503 + text: A bell-shaped surface in three dimensions. It is symmetric about the z axis and ha... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_intro.ptx + path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[2]/figure[1]/image[1] + line: 159602 + text: A topographical map of Chrome Mountain in Montana. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_intro.ptx + path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 159701 + text: A family of concentric ellipses, centered at the origin. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_intro.ptx + path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 159748 + text: A surface in the shape of an elliptical dome. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_intro.ptx + path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[2]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 159868 + text: A line through the origin in the plane, and two sets of nested circles, one on eit... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_intro.ptx + path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[2]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 159915 + text: The surface given by the graph whose level curves were plotted in the previous ima... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_limit.ptx + path: /pretext[1]/book[1]/chapter[13]/section[2]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 160990 + text: A region in the first quadrant is shaded, and its boundary is shown as a solid cur... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_limit.ptx + path: /pretext[1]/book[1]/chapter[13]/section[2]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 161035 + text: A region in the first quadrant is shaded, and its boundary is shown as a dashed cu... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_limit.ptx + path: /pretext[1]/book[1]/chapter[13]/section[2]/subsection[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 161080 + text: A region in the first quadrant is shaded, and its boundary is shown as a partially... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_limit.ptx + path: /pretext[1]/book[1]/chapter[13]/section[2]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 161173 + text: An image of the xy plane. It is almost entirely shaded, except for the line y=x.An image that illustrates the concept of the limit of a function of two variables.... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_limit.ptx + path: /pretext[1]/book[1]/chapter[13]/section[2]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] + line: 161805 + text: A graph of the piecewise-defined function in this example, illustrating its contin... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_partial_derivatives.ptx + path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 162937 + text: An elliptic paraboloid plotted over a rectangular domain. The trace y=2 is highlig... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_partial_derivatives.ptx + path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 162998 + text: The trace y=2 on the surface of the ellipic paraboloid from the previous image, wi... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_partial_derivatives.ptx + path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[1]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 163320 + text: On a surface in space, a trace curve is highlighted. At one pointon this curve, a ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_partial_derivatives.ptx + path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[1]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 163391 + text: On a surface in space, a trace curve is highlighted. At one pointon this curve, a ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_partial_derivatives.ptx + path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[4]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 163877 + text: A hyperbolic paraboloid, or saddle surface. A trace of constant x value is shown, ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_partial_derivatives.ptx + path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[4]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 163962 + text: A hyperbolic paraboloid, or saddle surface. A trace of constant y value is shown, ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_total_differential.ptx + path: /pretext[1]/book[1]/chapter[13]/section[4]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[3]/statement[1]/image[1] + line: 166781 + text: A right-angled triangle illustrating a wall of unknown height, a horizontal distan... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_chain.ptx + path: /pretext[1]/book[1]/chapter[13]/section[5]/introduction[1]/figure[1]/image[1] + line: 166895 + text: A plot of a surface in space, and a parametric curve on the surface.A circular paraboloid, opening upward, plotted over a rectangular domain. A curve ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_directional_derivative.ptx + path: /pretext[1]/book[1]/chapter[13]/section[6]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 168818 + text: A surface is plotted in 3D. Three vectors in the plane are shown corresponding to ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_directional_derivative.ptx + path: /pretext[1]/book[1]/chapter[13]/section[6]/subsection[1]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 169227 + text: A portion of the graph f(x,y) = sin(x)cos(y), along with a trace curve, and tangen... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_directional_derivative.ptx + path: /pretext[1]/book[1]/chapter[13]/section[6]/subsection[1]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 169324 + text: A zoomed-in view of the same surface, curve, and vectors as the previous image.A downward-opening circular paraboloid, with a point P marked at the vertex.A sector of a downward-opening elliptic paraboloid, along with a path that begins ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_directional_derivative.ptx + path: /pretext[1]/book[1]/chapter[13]/section[6]/subsection[1]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 169670 + text: A contour plot of the paraboloid in this example, and the corresponding trajectory... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_tangent.ptx + path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[1]/figure[1]/image[1] + line: 171151 + text: A surface in three dimensions, including curves on the surface, and tangent lines ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_tangent.ptx + path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 171344 + text: A bumpy surface. At a point on the surface, the tangent lines given by the partial... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_tangent.ptx + path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 171413 + text: A curve lies along a bumpy surface. At one point on the curve, a tangent line to t... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_tangent.ptx + path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 171549 + text: A graph of the function in this example; it has the shape of a steep hill.A circular paraboloid, opening downward. At one point on the surface, a normal lin... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_tangent.ptx + path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[2]/example[3]/solution[1]/figure[1]/image[1] + line: 171952 + text: A parabolic cylinder, and a normal line to the surface at one point.A downward-opening circular paraboloid, and a tangent plane to this surface at one... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_tangent.ptx + path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[4]/example[1]/solution[1]/figure[1]/image[1] + line: 172326 + text: An ellipsoid centered at the origin in space, and a tangent plane at one point.A circular paraboloid plotted over a rectangular domain. It is bowl-shaped, with p... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[13]/section[8]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 173706 + text: An inverted cone, with a maximum corresponding to a sharp peak on the z axis.The cubic surface studied in this example. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[13]/section[8]/subsection[1]/example[5]/solution[1]/figure[1]/image[1] + line: 174166 + text: A plot of the graph of the function used in this example. It is a more complicated... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[13]/section[8]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 174298 + text: A saddle surface on which a triangular curve is plotted. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[13]/section[8]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 174366 + text: A triangle in the plane representing the domain of the function in this example.Another view of the hyperbolic paraboloid in this example; this time with points o... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_multi_extreme_values.ptx + path: /pretext[1]/book[1]/chapter[13]/section[8]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] + line: 174710 + text: A three-dimensional plot of the volume in this example, as a function of length an... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/subsection[2]/figure[1]/image[1] + line: 175800 + text: A region in the first quadrant is bounded above and below by graphs, and left and ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/subsection[2]/figure[2]/image[1] + line: 175868 + text: A region in the first quadrant is bounded left and right by graphs, and above and ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] + line: 175964 + text: A rectangle in the plane, spanning x values from -1 to 3, and y values from 1 to 3... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/subsection[2]/example[2]/statement[1]/figure[1]/image[1] + line: 176036 + text: An obtuse triangle in the plane. The base is horizontal, and the other sides are l... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/subsection[2]/example[3]/statement[1]/figure[1]/image[1] + line: 176130 + text: A region in the first quadrant of the plane, bounded by a line and a parabola.A triangular region in the plane, bounded by lines y=0, x=6, and y=x/3.A sketch of the region of integration for this example. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/statement[1]/image[1] + line: 176773 + text: The region R is a rectangle, with x from 1 to 4, and y from -2 to 1.The region R is a right triangle with vertices (1,1), (4,1), and (4,3)A triangular region R. The vertices are (2,5), (2,1), and (4,3).The region R is bounded to the left by a parabola opening along the x axis, and th... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[5]/statement[1]/image[1] + line: 176988 + text: A region in the first quadrant bounded by two curves that intersect at the points ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[6]/statement[1]/image[1] + line: 177055 + text: A region R bounded above by a line, and below by a cubic curve. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/solution[1]/image[1] + line: 177137 + text: A region R bounded by a downward-opening parabola and the x axis.The region R is in the first quadrant, between a parabola and a line.A region bounded by the y axis and the right half of an ellipse centered at the or... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/solution[1]/image[1] + line: 177294 + text: A circle of radius 3, centered at the origin. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_iterated_integrals.ptx + path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[5]/solution[1]/image[1] + line: 177342 + text: The region between a right-opening parabola and a vertical line.A region bounded by a triangle, with vertices at (-1,1), (-1,-1), and (1,0).An illustration of the partition of a plane region into small rectangles.A surface in three dimensions lies over a partitioned region in the xy plane. A re... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_volume.ptx + path: /pretext[1]/book[1]/chapter[14]/section[2]/figure[2]/image[1] + line: 177891 + text: A parabolic surface in space is intersected by a plane of constant x value.A ramp-like surface in space, with the portion over a rectangular domain highlight... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_volume.ptx + path: /pretext[1]/book[1]/chapter[14]/section[2]/example[2]/statement[1]/figure[1]/image[1] + line: 178186 + text: A surface in space, plotted over a triangular domain. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_volume.ptx + path: /pretext[1]/book[1]/chapter[14]/section[2]/theorem[3]/statement[1]/p[2]/ol[1]/li[5]/figure[1]/image[1] + line: 178390 + text: Illustration of a pond-like region in the plane that has been divided into two sub... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_volume.ptx + path: /pretext[1]/book[1]/chapter[14]/section[2]/example[3]/statement[1]/figure[1]/image[1] + line: 178453 + text: A wave-like surface on which a triangular curve is drawn, corresponding to a domai... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_volume.ptx + path: /pretext[1]/book[1]/chapter[14]/section[2]/example[4]/statement[1]/figure[1]/image[1] + line: 178592 + text: A rectangular portion of a plane in space, and a petal-like curve on the plane.Several lines are plotted in the plane. Three of these lines form a triangle.A wave-like surface in three dimensions, and a curve on the surface illustrating a... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_volume.ptx + path: /pretext[1]/book[1]/chapter[14]/section[2]/example[6]/solution[1]/figure[1]/image[1] + line: 179014 + text: A rectangular portion of a plane in space, and a petal-like curve on the plane.A circular cylinder, bounded below by the xy plane, and above by the plane z=1.A region in the first quadrant of the plane, bounded by two parabolas.A region in the first quadrant of the plane, bounded by a cubic function and its i... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_volume.ptx + path: /pretext[1]/book[1]/chapter[14]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/solution[1]/p[1]/ol[1]/li[1]/image[1] + line: 179763 + text: A square of side length 2, with its center at the origin. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_volume.ptx + path: /pretext[1]/book[1]/chapter[14]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/solution[1]/p[1]/ol[1]/li[1]/image[1] + line: 179835 + text: A region in the plane that lies above a rightward-opening parabola, and below a ho... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_volume.ptx + path: /pretext[1]/book[1]/chapter[14]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[5]/solution[1]/p[1]/ol[1]/li[1]/image[1] + line: 179909 + text: A triangular region in the plane, with vertices at (0,0), (2,0), and (0,3).A region in the plane bounded by the graph of the natural logarithm and a straight... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_volume.ptx + path: /pretext[1]/book[1]/chapter[14]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[7]/solution[1]/p[1]/ol[1]/li[1]/image[1] + line: 180067 + text: A region bounded by the x axis and the upper half of a semicircle of radius 3.The region above the graph of the natural logarithm, but below its tangent line at... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_polar.ptx + path: /pretext[1]/book[1]/chapter[14]/section[3]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 180417 + text: A polar grid in the first quadrant, with one grid sector highlighted.A close-up view of one region in a polar grid + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_polar.ptx + path: /pretext[1]/book[1]/chapter[14]/section[3]/example[1]/solution[1]/figure[1]/image[1] + line: 180665 + text: An ellipse on a plane in space corresponds to a circle drawn below it in the xy pl... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_polar.ptx + path: /pretext[1]/book[1]/chapter[14]/section[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 180762 + text: A region in the plane between that lies between two circles. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_polar.ptx + path: /pretext[1]/book[1]/chapter[14]/section[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 180805 + text: A region in space that lies outside of a cylinder, and below a circular paraboloid... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_polar.ptx + path: /pretext[1]/book[1]/chapter[14]/section[3]/example[3]/statement[1]/figure[1]/image[1] + line: 180969 + text: A portion of a bell-shaped curve, and a triangular curve on the surface that corre... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_double_int_polar.ptx + path: /pretext[1]/book[1]/chapter[14]/section[3]/example[5]/statement[1]/figure[1]/image[1] + line: 181155 + text: A solid whose cross-sections are like a three-leaf clover; it has been sliced at a... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_center_of_mass.ptx + path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 181803 + text: A region in the plane with two vertical sides, a curved top, and a slanted bottom.... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_center_of_mass.ptx + path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 181849 + text: A region in the plane, with the same shape as the previous image, this time plotte... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_center_of_mass.ptx + path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 181981 + text: The unit square in the first quadrant, plotted against x and y coordinate axes.Two planes intersect along a line in three dimensions. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_center_of_mass.ptx + path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[1]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 182218 + text: A circular paraboloid, opening upward, plotted over a circular domain.A circular cone, opening upward, plotted over a circular domain.A number line with three points marked at -1, 2, and 3, and a marking for the cent... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_center_of_mass.ptx + path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 182537 + text: A number line with three points marked at -1, 2, and 3, and a marking for the cent... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_center_of_mass.ptx + path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] + line: 182728 + text: A triangle in the plane. The vertices are marked with dots of different sizes, and... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_center_of_mass.ptx + path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[2]/example[3]/solution[1]/figure[1]/image[1] + line: 182877 + text: The unit square in the first quadrant of the plane. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_center_of_mass.ptx + path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[2]/example[6]/statement[1]/figure[1]/image[1] + line: 183002 + text: An annular region, between semi-circles of radii 5 and 6 in the upper half-plane.<... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_area.ptx + path: /pretext[1]/book[1]/chapter[14]/section[5]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 183771 + text: A surface is plotted above a circular domain. A partition of the domain correspond... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_area.ptx + path: /pretext[1]/book[1]/chapter[14]/section[5]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 183884 + text: A zoomed-in view of a portion of the surface from the previous image, and one of t... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_area.ptx + path: /pretext[1]/book[1]/chapter[14]/section[5]/example[1]/statement[1]/figure[1]/image[1] + line: 184066 + text: A triangle drawn on a plane in space. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_area.ptx + path: /pretext[1]/book[1]/chapter[14]/section[5]/example[3]/statement[1]/figure[1]/image[1] + line: 184271 + text: A downward-opening circular cone, with vertex on the z axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_area.ptx + path: /pretext[1]/book[1]/chapter[14]/section[5]/example[4]/statement[1]/figure[1]/image[1] + line: 184391 + text: A triangular region is shown on a trough-like surface + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_area.ptx + path: /pretext[1]/book[1]/chapter[14]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/image[1] + line: 184689 + text: A bumpy surface with several peaks and valleys. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_area.ptx + path: /pretext[1]/book[1]/chapter[14]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/statement[1]/image[1] + line: 184772 + text: A steep surface with a single peak, resembling a witch's hat. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_area.ptx + path: /pretext[1]/book[1]/chapter[14]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/image[1] + line: 184857 + text: A saddle surface, with its saddle point at the origin in three dimensions.A mostly flat surface with a ridge along the y axis + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 185276 + text: Two planes in space intersect along a line. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 185360 + text: Two planes in space plotted over a triangular domain. The planes intersect along o... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 185515 + text: An ellipsoid in space, centered at the origin. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 185573 + text: A zoomed-in view of a surface, showing grid lines. A small rectangular prism illus... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[2]/statement[1]/figure[1]/image[1] + line: 185838 + text: A triangular portion of a plane in space. It lies in the first octant and has vert... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 185919 + text: A triangle in the xy plane, plotted in space relative to 3D coordinate axes.A triangle plotted in the yz plane, relative to a set of 3D coordinate axes.A triangle plotted in the xz plane, relative to a set of 3D coordinate axes.A surface in space intersects with a plane through the origin, forming a parabolic... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[3]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 186423 + text: A curve in space, and its projections onto the three coordinate planes.A cylindrical wedge in space, resembling a wedge cut from a tree.The projection of the cylindrical wedge in the previous image onto the xy plane. I... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 186845 + text: The projection of the solid in this example onto the xy plane; it is a semi-circul... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 186910 + text: The projection of the solid in this example onto the yz plane; it is a triangle.A pyramid with a rectangular base in the plane z=0 and its peak on the z axis.The base of the pyramid shown in the previous image. It is a rectangle, divided in... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 187314 + text: The projection of the pyramid onto the plane y=0 is a triangle. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 187370 + text: The projection of the pyramid onto the plane x=0 is a triangle. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[6]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 187545 + text: An elliptic paraboloid, opening downward, is plotted over a rectangular domain.A parabolic cylinder in space. It opens upward, and is symmetric about the y axis.... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[6]/statement[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 187670 + text: The region in space bounded by the paraboloid and cylinder in the previous two ima... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[6]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 187853 + text: The shadow of the solid for this example on the plane y=0. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[6]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 187923 + text: The shadow of the solid for this example on the plane x=0. It is divided into two ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[3]/example[1]/statement[1]/figure[1]/image[1] + line: 188363 + text: A triangle in space, formed by the portion of a plane that lies in the first octan... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[3]/example[2]/statement[1]/figure[1]/image[1] + line: 188493 + text: A cylindrical wedge in space, resembling a wedge cut from a tree.A tetrahedron in the first octant, with one vertex at the origin, and the others o... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/statement[1]/image[1] + line: 189014 + text: A triangular prism. The two triangular sides are parallel to the xz plane.A solid bounded above by a plane, and below by a parabolic cylinder.One quarter of an elliptic cone that opens along the y axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[5]/statement[1]/image[1] + line: 189448 + text: A tetrahedron in space, plotted with respect to a three-dimensional coordinate sys... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[6]/statement[1]/image[1] + line: 189575 + text: A cylindrical wedge with its edge along the x axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[7]/statement[1]/image[1] + line: 189712 + text: A region in the first octant bounded by the planes x=0 and z=0, and two parabolic ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_triple_int.ptx + path: /pretext[1]/book[1]/chapter[14]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[8]/statement[1]/image[1] + line: 189868 + text: A pyramid with a rectangular base and vertex on the z axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cylindrical_spherical.ptx + path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[1]/figure[1]/image[1] + line: 190255 + text: A schematic diagram of the cylindrical coordinate system in three dimensions.Three surfaces in space, corresponding to fixed values of each of the three cylind... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cylindrical_spherical.ptx + path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[1]/example[3]/statement[1]/figure[1]/image[1] + line: 190646 + text: A circular cylinder capped by a spherical dome. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cylindrical_spherical.ptx + path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[1]/example[4]/statement[1]/figure[1]/image[1] + line: 190772 + text: A solid whose cross-sections are like a three-leaf clover; it has been sliced at a... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cylindrical_spherical.ptx + path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[2]/figure[1]/image[1] + line: 190994 + text: A schematic diagram illustrating the spherical coordinate system relative to the r... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cylindrical_spherical.ptx + path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] + line: 191230 + text: A three-dimensional plot showing three surfaces, each of which is obtained by fixi... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cylindrical_spherical.ptx + path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[2]/figure[2]/image[1] + line: 191390 + text: A schematic diagram illustrating how the spherical volume element is computed.A gem-like solid, bounded below by the top half of a circular cone, and above by a... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_cylindrical_spherical.ptx + path: /pretext[1]/book[1]/chapter[14]/section[7]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/statement[1]/image[1] + line: 192277 + text: A region in space given by constant bounds in cylindrical coordinates.A region in space given by constant bounds in spherical coordinates.A curve in the x,y plane, and the corresponding curve on a surface lying above it.... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_intro.ptx + path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 193562 + text: A cylindrical surface in space that lies above a parabola in the plane, and below ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_intro.ptx + path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 193629 + text: Approximating the area under a curve in space using planar rectangles.A curve lies along a surface in space. The curve corresponds to a line drawn in th... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_intro.ptx + path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[1]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 194042 + text: A curve in space lies above the line y=2x+1 in the x,y plane. The surface that lie... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_intro.ptx + path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[1]/example[2]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 194165 + text: A curve lies on a hyperbolic paraboloid above a circle in the plane.A cylindrical surface between the x,y plane and a hyperbolic paraboloid.A helical ramp in space. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_intro.ptx + path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[2]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 194497 + text: Two oriented curves in the plane that are joined at a point. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_intro.ptx + path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 194536 + text: Two oriented curves in the plane that are joined at a point. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_intro.ptx + path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[3]/example[1]/statement[1]/figure[1]/image[1] + line: 194736 + text: A curve in space representing a wire, and a point representing its center of mass.... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/introduction[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 195304 + text: Several arrows are drawn in the plane to illustrate the concept of a vector field.... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/introduction[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 195376 + text: Several arrows are plotted in the plane to illustrate the concept of a vector fiel... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/introduction[1]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 195494 + text: Eight vectors are plotted in the plane to represent a vector field, using relative... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/introduction[1]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 195566 + text: Another plot of the same vector field used in every image so far, but this time wi... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/introduction[1]/figure[3]/image[1] + line: 195674 + text: A very chaotic plot of a three-dimensional vector field. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 196071 + text: A vector field of horizontal arrows: to the right above the x axis, and to the lef... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 196157 + text: A rotational vector field. The vectors appear to describe counter-clockwise circul... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 196304 + text: A radial vector field. Each vector points away from the origin, and vectors furthe... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 196377 + text: A vector field that appears to describe swirling motion, with several vortices.A radial vector field. Each vector points toward the origin, and vectors near the ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 196712 + text: The gradient vector field of a potential fuction. Vectors point toward the origin,... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 196794 + text: A three-dimensional image showing the graph of a function of two variables, with i... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/answer[1]/image[1] + line: 197001 + text: A vector field of horizontal vectors, pointing away from the y axis.A vertical vector field. Vectors right of the y axis point up, and those to the le... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_vector_fields.ptx + path: /pretext[1]/book[1]/chapter[15]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/answer[1]/image[1] + line: 197163 + text: A constant vector field. All vectors point down and to the right.A two dimensional vector field. The vectors point up and to the right. Vectors nea... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_vf.ptx + path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[1]/example[1]/statement[1]/figure[1]/image[1] + line: 197676 + text: A vector field is plotted in two dimensions, along with two different paths that b... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_vf.ptx + path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[1]/example[2]/statement[1]/figure[1]/image[1] + line: 197832 + text: A two-dimensional vector field, and two curves in the plane. The vectors appear to... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_vf.ptx + path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] + line: 198071 + text: A two-dimensional vector field of vectors tangent to a family of parabolas, and a ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_vf.ptx + path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[2]/example[2]/statement[1]/figure[1]/image[1] + line: 198235 + text: A three-dimensional vector field and a portion of a helix curve.Several regions in the plane are used to illustrate the concepts of connected, and... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_vf.ptx + path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[3]/figure[2]/sidebyside[1]/figure[1]/image[1] + line: 198510 + text: Two spherical surfaces in space, plotted without coordinate axes. The second spher... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_line_int_vf.ptx + path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[3]/figure[2]/sidebyside[1]/figure[2]/image[1] + line: 198602 + text: Two surfaces in three dimensions. One is a sphere with a cylinder through its cent... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_greensthm.ptx + path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[1]/figure[1]/image[1] + line: 199501 + text: A constant vector field in the plane, pointing to the right, and a triangular path... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_greensthm.ptx + path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[1]/figure[2]/image[1] + line: 199679 + text: A closed curve in the plane, in the shape of a cashew or boomerang.Two paths in the plane are plotted relative to a vector field. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_greensthm.ptx + path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[1]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 199937 + text: Two paths in the plane are plotted relative to a vector field. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_greensthm.ptx + path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] + line: 200181 + text: A triangular path in the plane is plotted against a two-dimensional vector field.<... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_greensthm.ptx + path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[2]/example[2]/statement[1]/figure[1]/image[1] + line: 200354 + text: A curve in the plane, like a bumpy circle or flower, and a vector field.A teardrop-shaped region in the plane. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_greensthm.ptx + path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[3]/example[1]/statement[1]/figure[1]/image[1] + line: 200635 + text: A circle in the plane, plotted against a spiral vector field. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_greensthm.ptx + path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] + line: 200757 + text: Two curves in the plane between points A and B, and a vector field with many vorti... + +PTX:WARNING: You have an image without a description and do not declare the image to be decorative. Because of this, output may not be accessible. If the image does not add information that is not already present, use @decorative="yes". Otherwise, provide a . + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/aside[1]/image[1] + line: 201395 + text: A rendering of a script S with serifs. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/figure[1]/image[1] + line: 201513 + text: A depction of a Möbius band, showing how the normal vector changes direction while... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] + line: 201658 + text: A portion of an elliptic paraboloid with a rectangular domain. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] + line: 201749 + text: An elliptic paraboloid plotted over a circular domain. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[3]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 201862 + text: A triangular domain is plotted in the first quadrant of the plane.An elliptic paraboloid is plotted in space over a triangular domain.A triangular domain in the plane, with vertices at (0,1), (0,3), and (3,1).An elliptic paraboloid is plotted in space over a triangular domain.A cylindrical surface centered along the y axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[6]/statement[1]/figure[1]/image[1] + line: 202351 + text: An elliptic cone with its cusp at the origin, including portions above and below t... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[6]/solution[1]/figure[1]/image[1] + line: 202443 + text: A cone plotted parametrically with restricted domain. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[7]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 202523 + text: The ellipsoid to be parametrized in this example. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[7]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 202577 + text: A portion of an ellipsoid in space corresponding to a restricted parameter domain.... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[2]/figure[1]/sidebyside[1]/figure[1]/image[1] + line: 202762 + text: A rectangular region in the plane, with a smaller subrectangle highlighted in its... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[2]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 202832 + text: A surface in space with parameter domain R, highlighting the portion corresponding... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[2]/figure[1]/sidebyside[1]/figure[3]/image[1] + line: 202925 + text: A zoomed in view of the parallelogram approximation of a small patch on a surface.... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/statement[1]/image[1] + line: 203454 + text: A triangular prism plotted in three dimensions. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[2]/statement[1]/image[1] + line: 203588 + text: A tetrahedron with vertices (2,0,0), (0,1,0), (2,1,0), and (2,1,4).A cylindrical wedge with its edge along the x axis. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/statement[1]/image[1] + line: 203842 + text: A region in the first octant bounded by the planes x=0 and z=0, and two parabolic ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[5]/statement[1]/image[1] + line: 204022 + text: An elliptical cylinder, centered on the z axis, capped by planes z=1 and z=3.An inverted circular cone, together with a disk in the x,y planeA solid that looks like a barn with a parabolic roof. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_parametric_surfaces.ptx + path: /pretext[1]/book[1]/chapter[15]/section[5]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[8]/statement[1]/image[1] + line: 204368 + text: A region bounded above by an elliptic paraboloid, and below by the x,y plane.A triangular portion of a plane in space. It is the graph of a linear function ove... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_integral.ptx + path: /pretext[1]/book[1]/chapter[15]/section[6]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] + line: 204931 + text: A triangular surface in space, and a vector field flowing across the surface.A closed surface consisting of a downward-opening circular paraboloid, and a disc ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_integral.ptx + path: /pretext[1]/book[1]/chapter[15]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[7]/statement[1]/image[1] + line: 205583 + text: An elliptical paraboloid opens downward, intersecting the x,y plane in an ellipse.... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_surface_integral.ptx + path: /pretext[1]/book[1]/chapter[15]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[8]/statement[1]/image[1] + line: 205667 + text: Approximately three quarters of the unit sphere. The top has been removed and repl... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/subsection[1]/example[1]/statement[1]/figure[1]/image[1] + line: 205814 + text: A parabolic wedge, bounded by two planes and a parabolic cylinder.A circular paraboloid, opening downward, meets the x,y plane along the unit circle... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/subsection[1]/example[3]/statement[1]/figure[1]/image[1] + line: 206248 + text: A cube centered at the origin in space. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] + line: 206539 + text: An ellipse drawn on a plane in space. It lies over a circle in the x,y plane.The portion of a circular paraboloid in the first octant, and an elliptical curve ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/subsection[2]/example[2]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] + line: 206762 + text: The parabolic surface from the previous image is shown intersecting a plane along ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/image[1] + line: 207044 + text: A tetrahedron with vertices (0,0,0), (4,0,0), (0,3,0), and (0,0,2).A circular cylinder, centered on the z axis, capped by disks in the planes z=3 and... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/image[1] + line: 207284 + text: A steep hill with a square base in the x-y plane. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/statement[1]/image[1] + line: 207386 + text: A familiar dome-shaped surface between a circular paraboloid and the x,y plane.A bowl-shaped surface given by a circular paraboloid plotted over the unit disk.A bell-shaped surface in space, with a circular boundary. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/statement[1]/image[1] + line: 207661 + text: A triangular surface in the first octant, with intercepts at (4,0,0), (0,3,0), and... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/statement[1]/image[1] + line: 207752 + text: Graph of an elliptic paraboloid over a parabolic domain. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/statement[1]/image[1] + line: 207863 + text: A domed hill with a square base in the x-y plane. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[2]/statement[1]/image[1] + line: 207959 + text: A triangular prism, with two triangular faces and three rectangular faces.A wedge-shaped surface given by two intersecting planes and a parabolic cylinder.<... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/statement[1]/image[1] + line: 208191 + text: A surface resembling a barn or greenhouse with an arched roof. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/statement[1]/image[1] + line: 208330 + text: A twisted-looking surface given by a portion of a hyperbolic paraboloid.A surface that looks something like a ladle, if the handle had melted and drooped ... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: sec_stokes_divergence.ptx + path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/statement[1]/image[1] + line: 208539 + text: A region in a plane in space, bounded by an ellipse in that plane.A rectangle lying within a plane in space. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[3]/introduction[1]/assemblage[1]/image[1] + line: 209144 + text: A detailed plot of the unit circle, showing angles in both degrees and radians, an... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[3]/subsection[1]/assemblage[1]/sidebyside[1]/image[1] + line: 209286 + text: An illustration of the correspondence between an angle and a point on the unit cir... + +PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of elements, if present, or else against the first row. Results may be unpredictable. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[3]/subsection[1]/assemblage[1]/sidebyside[1]/tabular[1] + line: 209325 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[3]/subsection[1]/assemblage[2]/sidebyside[1]/image[1] + line: 209352 + text: A right angle triangle, with sides labeled "opposite", "adjacent", and "hypotenuse... + +PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of elements, if present, or else against the first row. Results may be unpredictable. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[3]/subsection[1]/assemblage[2]/sidebyside[1]/tabular[1] + line: 209376 + text: + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[1]/image[1] + line: 209544 + text: A schematic diagram of a triangle, labeling three sides, an angle, and an altitude... + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[1]/image[2] + line: 209593 + text: A schematic diagram of a right circular cone, showing the height and radius.A generic parallelogram, with base and height labeled. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[2]/image[2] + line: 209678 + text: A diagram of a right circular cylinder, labeling the radius and height.A schematic diagram of a generic trapezoid. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[3]/image[2] + line: 209773 + text: A image of a sphere, showing one circumference and its radius. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[4]/image[1] + line: 209819 + text: A generic circle with its radius indicated. + +PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. + file: appendix_back_reference.ptx + path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[4]/image[2] + line: 209856 + text: A drawing of a general cone, with an arbitrary plane region for its base.A pie-shaped sector of a circle, labeled with angle, radius, and arc length.A sketch of a right cylinder with an arbitrary base. diff --git a/ptx/sec_conic_sections.ptx b/ptx/sec_conic_sections.ptx index e4670b6d2..bcbf4bbe4 100644 --- a/ptx/sec_conic_sections.ptx +++ b/ptx/sec_conic_sections.ptx @@ -1540,7 +1540,7 @@ - A graph of the hyperbola given in + A graph of the hyperbola in this example

    A graph of the hyperbola given in . @@ -1627,7 +1627,7 @@

    - The hyperbola described in . + The hyperbola described in this example

    The hyperbola described in . @@ -2124,7 +2124,7 @@

    @@ -839,7 +846,7 @@ -
    +
    diff --git a/ptx/sec_graph_sketch.ptx b/ptx/sec_graph_sketch.ptx index cbbebf77b..e6f00d72b 100644 --- a/ptx/sec_graph_sketch.ptx +++ b/ptx/sec_graph_sketch.ptx @@ -1113,7 +1113,7 @@
    - A graph of y = \sin(x) generated by Sage. + A graph of the sine function generated by Sage.

    The plot features a solid blue curve representing the sine wave. diff --git a/ptx/sec_hyperbolic.ptx b/ptx/sec_hyperbolic.ptx index 136165557..e6c1ef962 100644 --- a/ptx/sec_hyperbolic.ptx +++ b/ptx/sec_hyperbolic.ptx @@ -909,8 +909,8 @@ The y and the x axes are drawn from -10 to 10. The functions y=\sinh(x) and y=\sinh^{-1}(x). The axis y=x is shown. -

    +

    From left to right, the \sinh(x) function starts in the third quadrant and it rises steeply, very closely to the y axis. It crosses the origin along the y=x line, has a dip then increases diff --git a/ptx/sec_limit_intro.ptx b/ptx/sec_limit_intro.ptx index a367877bb..60a0e2580 100644 --- a/ptx/sec_limit_intro.ptx +++ b/ptx/sec_limit_intro.ptx @@ -47,8 +47,7 @@

    - Graph of sin(x)/x, shows values for the domain -7 <=x <=7, - undefined at x = 0. + Graph of sin(x)/x, shows values for the domain -7 <=x <=7, undefined at x = 0. \begin{tikzpicture}[ @@ -81,8 +80,7 @@

    - Graph of sin(x)/x zoomed in on values where x is near 1. Shows that for x = 1, - sin(x)/x is approx. 0.84. + Graph of sin(x)/x zoomed in on values where x is near 1. Shows that for x = 1, sin(x)/x is approx. 0.84. \begin{tikzpicture} diff --git a/ptx/sec_line_int_intro.ptx b/ptx/sec_line_int_intro.ptx index 8b7de69e1..d0bc30c0c 100644 --- a/ptx/sec_line_int_intro.ptx +++ b/ptx/sec_line_int_intro.ptx @@ -725,7 +725,7 @@

    The points on the surface that lie above the circle form a curve on the surface. - The curve cuts out a portion of the surface that looks exactly like a Pringles chip. + The curve cuts out a portion of the surface that looks exactly like a Pringles chip.

    diff --git a/ptx/sec_multi_limit.ptx b/ptx/sec_multi_limit.ptx index edd11bb8d..5d3a6a2fd 100644 --- a/ptx/sec_multi_limit.ptx +++ b/ptx/sec_multi_limit.ptx @@ -1434,7 +1434,7 @@ \ds S = \left\{(x,y)\,| \, x^2+y^2=1\right\}

    - answer> +

    1. @@ -1460,7 +1460,7 @@

    - answer> +     diff --git a/ptx/sec_optimization.ptx b/ptx/sec_optimization.ptx index 8ffd093cc..8948ed26d 100644 --- a/ptx/sec_optimization.ptx +++ b/ptx/sec_optimization.ptx @@ -61,7 +61,7 @@ - A rectangular enclosure with green grass and a fence with x and y dimensions. + A rectangular enclosure with green grass and a fence with x and y dimensions.

    @@ -111,7 +111,7 @@ - A rectangle shape is drawn with sides labeled x and y. + A rectangle shape is drawn with sides labeled x and y.

    @@ -737,7 +737,7 @@

    The horizontal line of the image is labelled 5000-x which is the distance of the power line laid along land,and x from the middle to the pont of the offshore facility which is the distance the power line is not laid. - The hypotenuse of the right angle triangle measures underwater distance with the equation x^2-1000^2. + The hypotenuse of the right angle triangle measures underwater distance with the equation \sqrt{x^2-1000^2}.

    diff --git a/ptx/sec_polar.ptx b/ptx/sec_polar.ptx index 71bb5da23..9a61b34a8 100644 --- a/ptx/sec_polar.ptx +++ b/ptx/sec_polar.ptx @@ -1826,7 +1826,7 @@     - A rose curve with three leaves, symmetric about the y axis. + A rose curve with three leaves, symmetric about the y axis.

    Another rose curve with three leaves, this time given by r=\sin(3\theta). diff --git a/ptx/sec_space_coord.ptx b/ptx/sec_space_coord.ptx index d572f19fe..6e361dcbe 100644 --- a/ptx/sec_space_coord.ptx +++ b/ptx/sec_space_coord.ptx @@ -4251,7 +4251,7 @@

    - Which quadric surface looks like a Pringles(TM)chip? + Which quadric surface looks like a Pringleschip?

    diff --git a/ptx/sec_taylor_poly.ptx b/ptx/sec_taylor_poly.ptx index ba993e962..f23e82763 100644 --- a/ptx/sec_taylor_poly.ptx +++ b/ptx/sec_taylor_poly.ptx @@ -143,7 +143,7 @@ Finally, we can compute p_2(x) = x^2+x+C. Using our initial values, we know p_2(0) = 2 so C=2. We conclude that p_2(x) = x^2+x+2. - This function is plotted with f in . + This function is plotted with f in .

    @@ -172,7 +172,7 @@ higher degree that match more of the derivatives of f at x=0. In general, a polynomial of degree n can be created to match the first n derivatives of f. - + shows p_4(x)= -x^4/2-x^3/6+x^2+x+2, whose first four derivatives at 0 match those of f. (Using the table in , start with p_4^{(4)}(x)=-12 and solve the related initial-value problem.) @@ -223,7 +223,7 @@ whose first four derivatives at 0 match those of f.

    -
    +
    @@ -1288,11 +1288,18 @@ Thus we want to approximate \cos(2) with p_9(2).

    -

    +

    + We now set out to compute p_9(x). + We again need a table of the derivatives of + f(x)=\cos(x) evaluated at x=0. + A table of these values is given in . +

    + +

    We now set out to compute p_9(x). We again need a table of the derivatives of f(x)=\cos(x) evaluated at x=0. - A table of these values is given in . + A table of these values is given in .

    diff --git a/ptx/sec_vector_fields.ptx b/ptx/sec_vector_fields.ptx index d85197e3b..0ee61ef8f 100644 --- a/ptx/sec_vector_fields.ptx +++ b/ptx/sec_vector_fields.ptx @@ -1800,7 +1800,7 @@

    - A vector field of horizontal vectors, pointing away from the y axis. + A vector field of horizontal vectors, pointing away from the y axis.

    A two-dimmensional vector field is plotted relative to x and y coordinate axes, From cb05851d06d68e1be9cc8cfb54fa5f5ade91455f Mon Sep 17 00:00:00 2001 From: sean-fitzpatrick Date: Mon, 6 Jul 2026 15:18:09 -0600 Subject: [PATCH 3/5] more validation fixes --- ptx/appendix_back_reference.ptx | 8 ++--- ptx/review-exercises-limits.ptx | 2 +- ptx/sec_graph_incr_decr.ptx | 18 +++++++++- ptx/sec_limit_continuity.ptx | 10 +++--- ptx/sec_limit_intro.ptx | 60 +++++++++++++++---------------- ptx/sec_numerical_integration.ptx | 4 +-- ptx/sec_shell_method.ptx | 2 +- ptx/sec_space_coord.ptx | 2 +- ptx/sec_taylor_poly.ptx | 8 +++-- 9 files changed, 67 insertions(+), 47 deletions(-) diff --git a/ptx/appendix_back_reference.ptx b/ptx/appendix_back_reference.ptx index a5ea0ef8f..371bf7b97 100644 --- a/ptx/appendix_back_reference.ptx +++ b/ptx/appendix_back_reference.ptx @@ -513,14 +513,14 @@ \cos(\theta) = x - + \ds\csc(\theta) = \frac1y \ds\sec(\theta) = \frac1x - + \ds\tan(\theta) = \frac yx @@ -564,14 +564,14 @@ \ds\csc(\theta) = \frac{\text{H} }{\text{O} } - + \ds\cos(\theta) = \frac{\text{A} }{\text{H} } \ds\sec(\theta) = \frac{\text{H} }{\text{A} } - + \ds\tan(\theta) = \frac{\text{O} }{\text{A} } diff --git a/ptx/review-exercises-limits.ptx b/ptx/review-exercises-limits.ptx index 9bb21c03a..8c14ddf4f 100644 --- a/ptx/review-exercises-limits.ptx +++ b/ptx/review-exercises-limits.ptx @@ -272,7 +272,7 @@ For a numerical approximation, make a table:

    - + x diff --git a/ptx/sec_graph_incr_decr.ptx b/ptx/sec_graph_incr_decr.ptx index 6ebb4d4ab..3fd812454 100644 --- a/ptx/sec_graph_incr_decr.ptx +++ b/ptx/sec_graph_incr_decr.ptx @@ -544,6 +544,7 @@ + A number line showing the two critical points for a function

    A number line is shown with two marked points. @@ -1256,7 +1257,22 @@

    - + A graph with a curved W shape. + +

    + The graph of the function f(x) in is plotted, + along with its derivative. + The graph of f(x) has the shape of a curved W. + There are two local minima with horizontal tangent lines at x=1 and x=-1. + There is a local maximum that is a cusp at the origin. +

    +

    + The graph of \fp(x) is shown to be decreasing when x is negative, + with an intercept at x=-1. There is a break in the graph at the y axis, + since the derivative is undefined there. + For positive x, the graph of \fp(x) is increasing, with an intercept at x=1. +

    +     \begin{tikzpicture} \begin{axis}[ diff --git a/ptx/sec_limit_continuity.ptx b/ptx/sec_limit_continuity.ptx index b7b850ae9..612bf9227 100644 --- a/ptx/sec_limit_continuity.ptx +++ b/ptx/sec_limit_continuity.ptx @@ -1496,7 +1496,7 @@ - + @@ -3274,7 +3274,7 @@ If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. - + Iteration Interval Midpoint Sign @@ -3364,7 +3364,7 @@ If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. - + Iteration Interval Midpoint Sign @@ -3454,7 +3454,7 @@ If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. - + Iteration Interval Midpoint Sign @@ -3544,7 +3544,7 @@ If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. - + Iteration Interval Midpoint Sign diff --git a/ptx/sec_limit_intro.ptx b/ptx/sec_limit_intro.ptx index 60a0e2580..2a49b0a3d 100644 --- a/ptx/sec_limit_intro.ptx +++ b/ptx/sec_limit_intro.ptx @@ -1994,7 +1994,7 @@ For a numerical approximation, make a table:

    - + x @@ -2087,7 +2087,7 @@ For a numerical approximation, make a table:

    - + x @@ -2183,7 +2183,7 @@ For a numerical approximation, make a table:

    - + x @@ -2294,7 +2294,7 @@ For a numerical approximation, make a table:

    - + x @@ -2405,7 +2405,7 @@ For a numerical approximation, make a table:

    - + x @@ -2512,7 +2512,7 @@ For a numerical approximation, make a table:

    - + x @@ -2619,7 +2619,7 @@ For a numerical approximation, make a table:

    - + x @@ -2724,7 +2724,7 @@ For a numerical approximation, make a table:

    - + x @@ -2829,7 +2829,7 @@ For a numerical approximation, make a table:

    - + x @@ -2937,7 +2937,7 @@ For a numerical approximation, make a table:

    - + x @@ -3028,7 +3028,7 @@ For a numerical approximation, make a table:

    - + x @@ -3117,7 +3117,7 @@ For a numerical approximation, make a table:

    - + x @@ -3218,7 +3218,7 @@ For a numerical approximation, make a table:

    - + x \big\lfloor\lvert x\rvert\big\rfloor ! @@ -3333,7 +3333,7 @@ For a numerical approximation, make a table:

    - + x \big\lfloor\lvert x\rvert\big\rfloor ! @@ -3482,7 +3482,7 @@     - + h \frac{f(a+h)-f(a)}{h} @@ -3518,7 +3518,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -3602,7 +3602,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -3638,7 +3638,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -3725,7 +3725,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -3761,7 +3761,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -3846,7 +3846,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -3882,7 +3882,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -3973,7 +3973,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -4009,7 +4009,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -4090,7 +4090,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -4126,7 +4126,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -4208,7 +4208,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -4244,7 +4244,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -4326,7 +4326,7 @@ - + h \frac{f(a+h)-f(a)}{h} @@ -4362,7 +4362,7 @@ - + h \frac{f(a+h)-f(a)}{h} diff --git a/ptx/sec_numerical_integration.ptx b/ptx/sec_numerical_integration.ptx index 2c9a824b3..f79155539 100644 --- a/ptx/sec_numerical_integration.ptx +++ b/ptx/sec_numerical_integration.ptx @@ -1045,7 +1045,7 @@ 0.0740 - + @@ -1342,7 +1342,7 @@ 0.368 - + diff --git a/ptx/sec_shell_method.ptx b/ptx/sec_shell_method.ptx index e3b76a511..2185229f5 100644 --- a/ptx/sec_shell_method.ptx +++ b/ptx/sec_shell_method.ptx @@ -1460,7 +1460,7 @@ \ds 2\pi\int_c^d r(y)h(y)\, dy - + Vertical Axis diff --git a/ptx/sec_space_coord.ptx b/ptx/sec_space_coord.ptx index 6e361dcbe..967e484ca 100644 --- a/ptx/sec_space_coord.ptx +++ b/ptx/sec_space_coord.ptx @@ -2382,7 +2382,7 @@ Crossed Lines - + x=d diff --git a/ptx/sec_taylor_poly.ptx b/ptx/sec_taylor_poly.ptx index f23e82763..9915759c6 100644 --- a/ptx/sec_taylor_poly.ptx +++ b/ptx/sec_taylor_poly.ptx @@ -1636,9 +1636,13 @@

    1. -

      +

      + We begin by evaluating the derivatives of f at x=4. + This is done in . +

      +

      We begin by evaluating the derivatives of f at x=4. - This is done in . + This is done in .

      From 69dec7577bf6dc4122642eb45240d5eb6b482499 Mon Sep 17 00:00:00 2001 From: sean-fitzpatrick Date: Fri, 10 Jul 2026 11:56:10 -0600 Subject: [PATCH 4/5] fix accidental file commit --- apex-assembled.xml | 210304 ----------------------------------------- apex-validation.txt | 30735 ------ 2 files changed, 241039 deletions(-) delete mode 100644 apex-assembled.xml delete mode 100644 apex-validation.txt diff --git a/apex-assembled.xml b/apex-assembled.xml deleted file mode 100644 index e040583ba..000000000 --- a/apex-assembled.xml +++ /dev/null @@ -1,210304 +0,0 @@ - - - - - APEX - Exercise - Part - Discussion Questions - - - APEX - - - - A traditional calculus textbook with many exercises and few proofs, - covering calculus from limits to vector calculus. - - - - - - \newcommand\blank[2]{\,\colorbox{gray}{$\phantom{\rule{#1pt}{#2pt}}$}\,} - \newcommand{\highlight}[1]{{\color{blue}{#1}}} - \newcommand{\ds}{\displaystyle} - \newcommand{\fp}{f\hskip.75pt '} - \newcommand{\fpp}{f\hskip.75pt ''} - - % Leibniz notation - % Usage: \lz{y}{x} - \newcommand{\lz}[2]{\frac{d#1}{d#2}} - % - % higher Leibniz notation - % Usage: \lzn{n}{y}{x} - \newcommand{\lzn}[3]{\frac{d^{#1}#2}{d#3^{#1}}} - % - % Leibniz operator - % Usage: \lzo{x} - \newcommand{\lzo}[1]{\frac{d}{d#1}} - % - % Leibniz operator on .... - % Usage: \lzoo{x}{y} - \newcommand{\lzoo}[2]{{\frac{d}{d#1}}{\left(#2\right)}} - % - % higher Leibniz operator - % Usage: \lzon{n}{x}{y} - \newcommand{\lzon}[2]{\frac{d^{#1}}{d#2^{#1}}} - % - % Leibniz operator at .... - % Usage: \lzoa{y}{x}{a} - \newcommand{\lzoa}[3]{\left.{\frac{d#1}{d#2}}\right|_{#3}} - % - % partial Leibniz notation - % Usage: \plz{y}{x} - \newcommand{\plz}[2]{\frac{\partial#1}{\partial#2}} - % - % partial Leibniz operator at .... - % Usage: \plzoa{y}{x}{a} - \newcommand{\plzoa}[3]{\left.{\frac{\partial#1}{\partial#2}}\right|_{#3}} - - %Limit at infinity for sequences - %Usage: \inflim e^{-n} or \inflim[m] e^{-m} - \newcommand{\inflim}[1][n]{\lim\limits_{#1 \to \infty}} - 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}% - \fi - }% - \let\pgfplots@path@@tikz@do@circle@orig=\tikz@do@circle - \let\pgfplots@path@@tikz@do@ellipse@orig=\tikz@do@ellipse - - \let\pgfplots@path@@tikz@do@circle@oldandbroken=\pgfplots@path@@tikz@do@circle - \def\pgfplots@path@@tikz@do@circle#1{\pgfplots@path@@tikz@do@circle@oldandbroken{#1}{#1}} - \def\pgfplots@path@@tikz@do@ellipse#1#2{\pgfplots@path@@tikz@do@circle@oldandbroken{#1}{#2}} - \makeatother - - - - - - - import graph3; - bool incolor; - incolor = true; - pen apexmeshpen=rgb(0,0,.7); - pen blackmeshpen=rgb(0,0,0); - pen surfacepen=rgb(.6,.6,1)+opacity(.7); - pen surfacepen2=rgb(1,.6,.6)+opacity(1); - material simplesurfacepen=emissive(rgb(.6,.6,1)+opacity(0.7)); - material simplesurfacepen2=emissive(rgb(1,.6,.6)+opacity(0.7)); - material simplesurfacepen3=emissive(rgb(.5,.9,.5)+opacity(0.7)); - pen bluepen=blue; - pen bluemeshpen=rgb(0,0,.5); - pen bluecurvepen=rgb(.1,.1,.7); - pen dotblue=rgb(.6,.6,1); - pen redpen=red; - pen redmeshpen=rgb(.7,0,0); - pen redmeshpen2=rgb(.5,0,0); - pen redcurvepen=rgb(.9,0,0); - pen greenmeshpen=rgb(0,.5,0); - pen greencurvepen=rgb(0,.7,0); - pen curvepen=.4mm+bluepen; - pen curvepen2=.4mm+redpen; - pen darksurfacepen=rgb(.2,.2,1)+opacity(.7); - if(settings.outformat == "html") currentlight.background=opacity(0.0); - - - - - Key Idea - - - - APEX Calculus - - - - - - Gregory Hartman, Ph.D. - Department of Applied Mathematics - Virginia Military Institute - - - - Sean Fitzpatrick, Ph.D. - Department of Mathematics and Computer Science - University of Lethbridge - - - - Alex Jordan, Ph.D. - Department of Mathematics - Portland Community College - - - - Carly Vollet, M.S. - Department of Mathematics - Portland Community College - - - - - - Contributors to the 4th Edition - Jennifer Bowen, Troy Siemers, Brian Heinold, Dimplekumar Chalishajar - - - 5 - - - apexcalculus.com - - - - 2021 - Gregory Hartman - Creative Commons BY NC - - Licensed to the public under Creative Commons Attribution-Noncommercial 4.0 International Public License - - - - - - - - - - - - - Thanks -

      - There are many people who deserve recognition for the important role they have played in the development of this text. - First, I thank Michelle for her support and encouragement, - even as this project from work occupied my time and attention at home. - Many thanks to Troy Siemers, whose most important contributions extend far beyond the sections he wrote or the 227 figures he coded in Asymptote for 3D interaction. - He provided incredible support, advice and encouragement for which I am very grateful. - My thanks to Brian Heinold and Dimplekumar Chalishajar for their contributions and to Jennifer Bowen for reading through so much material and providing great feedback early on. - Thanks to Troy, Lee Dewald, Dan Joseph, Meagan Herald, Bill Lowe, John David, - Vonda Walsh, Geoff Cox, Jessica Libertini and other faculty of VMI who have given me numerous suggestions and corrections based on their experience with teaching from the text. - (Special thanks to Troy, Lee and Dan for their patience in teaching Calc III while I was still writing the Calc III material.) - Thanks to Randy Cone for encouraging his tutors of VMI's Open Math Lab to read through the text and check the solutions, - and thanks to the tutors for spending their time doing so. - A very special thanks to Kristi Brown and Paul Janiczek who took this opportunity far above and beyond what I expected, - meticulously checking every solution and carefully reading every example. - Their comments have been extraordinarily helpful. I am also thankful for the support provided by Wane Schneiter, - who as my Dean provided me with extra time to work on this project. - I am blessed to have so many people give of their time to make this book better. -

      -       - - - - A Note on Using this Text -

      - Thank you for reading this short preface. - Allow us to share a few key points about the text so that you may better understand what you will find beyond this page. -

      - -

      - This text comprises a threevolume series on Calculus. - The first part covers material taught in many Calc 1 courses: - limits, derivatives, and the basics of integration, found in Chapters 1 through 6.1. - The second text covers material often taught in Calc 2: integration and its applications, - including an introduction to differential equations, - along with an introduction to sequences, series and Taylor Polynomials, - found in Chapters 5 through 8. - The third text covers topics common in Calc 3 or multivariable calc: - parametric equations, polar coordinates, vector-valued functions, and functions of more than one variable, - found in Chapters 10 through 15. - All three are available separately for free at apexcalculus.com, - and HTML versions of the book can be found at opentext.uleth.ca. -

      - -

      - These three texts are intended to work together and make one cohesive text, - APEX Calculus, which can also be downloaded from the website. -

      - -

      - Printing the entire text as one volume makes for a large, heavy, cumbersome book. - One can certainly only print the pages they currently need, but some prefer to have a nice, - bound copy of the text. Therefore this text has been split into these three manageable parts, - each of which can be purchased for about $15 at Amazon.com. -

      -       - - - For Students: How to Read this Text -

      - Mathematics textbooks have a reputation for being hard to read. - Highlevel mathematical writing often seeks to say much with few words, - and this style often seeps into texts of lowerlevel topics. - This book was written with the goal of being easier to read than many other calculus textbooks, - without becoming too verbose. -

      - -

      - Each chapter and section starts with an introduction of the coming material, - hopefully setting the stage for why you should care, - and ends with a look ahead to see how the justlearned material helps address future problems. -

      - -

      -

        -
      • Please read the text -

        - It is written to explain the concepts of Calculus. - There are numerous examples to demonstrate the meaning of definitions, - the truth of theorems, and the application of mathematical techniques. - When you encounter a sentence you don't understand, read it again. - If it still doesn't make sense, read on anyway, - as sometimes confusing sentences are explained by later sentences. -

        -
        • -
      • You don't have to read every equation -

        - The examples generally show all the steps needed to solve a problem. - Sometimes reading through each step is helpful; sometimes it is confusing. - When the steps are illustrating a new technique, - one probably should follow each step closely to learn the new technique. - When the steps are showing the mathematics needed to find a number to be used later, - one can usually skip ahead and see how that number is being used, - instead of getting bogged down in reading how the number was found. -

        -
        • -
      • Most proofs have been omitted -

        - In mathematics, proving something is always true is extremely important, - and entails much more than testing to see if it works twice. - However, students often are confused by the details of a proof, - or become concerned that they should have been able to construct this proof on their own. - To alleviate this potential problem, we do not include the proofs to most theorems in the text. - The interested reader is highly encouraged to find proofs online or from their instructor. - In most cases, one is very capable of understanding what a theorem means - and how to apply it without knowing fully why it is true. -

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        • -
      -

      -       - - - Interactive, 3D Graphics -

      - Versions 3.0 and 4.0 of the textbook include interactive, - 3D graphics in the PDF version. Nearly all graphs of objects in space can be rotated, - shifted, and zoomed in/out so the reader can better understand the object illustrated. - However, the only pdf viewers that support these 3D graphics are Adobe Reader Acrobat - (and only the versions for PC/Mac/Unix/Linux computers, not tablets or smartphones). -

      - -

      - The latest version of the book, which is authored in , - is available in HTML. - In HTML, the 3D graphics are rendered using WebGL, - and should work in any modern web browser. -

      - -

      - Interactive graphics are no longer supported within the PDF, - but clicking on any 3D graphic within the PDF - will take you directly to the interactive version on the web. -

      -       - - - APEX <ndash/> Affordable Print and Electronic teXts -

      - APEX is a consortium of authors who collaborate to produce high quality, low cost textbooks. - The current textbookwriting paradigm is facing a potential revolution as desktop publishing and electronic formats increase in popularity. - However, writing a good textbook is no easy task, as the time requirements alone are substantial. - It takes countless hours of work to produce text, write examples and exercises, edit and publish. - Through collaboration, however, the cost to any individual can be lessened, - allowing us to create texts that we freely distribute electronically and sell in printed form for an incredibly low cost. - Having said that, nothing is entirely free; someone always bears some cost. - This text cost the authors of this book their time, and that was not enough. - APEX Calculus would not exist had not the Virginia Military Institute, - through a generous JacksonHope grant, - given the lead author significant time away from teaching so he could focus on this text. -

      - -

      - Each text is available as a free .pdf, protected by a Creative Commons Attribution - Noncommercial 4.0 copyright. - That means you can give the .pdf to anyone you like, print it in any form you like, - and even edit the original content and redistribute it. - If you do the latter, you must clearly reference this work and you cannot sell your edited work for money. -

      - -

      - We encourage others to adapt this work to fit their own needs. - One might add sections that are missing or remove sections that your students won't need. - The source files can be found at github.com/APEXCalculus. -

      - -

      - You can learn more at www.vmi.edu/APEX. -

      -       - - - First <pretext/> Edition (Version 5.0) -

      - Key changes from Version 4.0 to 5.0: -

        -
      • -

        - The underlying source code has been completely rewritten, - to use the language, - instead of the original \LaTeX. -

        -
        • -
      • -

        - Using allows us to produce the books in multiple formats, - including HTML, which is both more accessible and more interactive than the original PDF. - HTML versions of the book can be found at opentext.uleth.ca. -

        -
        • -
      • -

        - The appendix on differential equations from the Calculus for Quarters - version of the book has been included as Chapter 8, just after applications of integration. - Chapters 8 14 are now numbered 9 15 as a result. -

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        • -
      • -

        - In the HTML version of the book, - many of the exercises are now interactive, and powered by WeBWorK. -

        -
        • -
      -

      - -

      - Key changes from Version 3.0 to 4.0: -

        -
      • -

        - Numerous typographical and small mathematical corrections (again, thanks to all my close readers!). -

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        • -
      • -

        - Large mathematical corrections and adjustments. - There were a number of places in Version 3.0 where a definition/theorem was not correct as stated. - See www.apexcalculus.com for more information. -

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        • -
      • -

        - More useful numbering of Examples, Theorems, . - Definition 11.4.2 refers to the second definition of Chapter 11, Section 4. -

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        - The addition of Section 13.7: Triple Integration with Cylindrical and Spherical Coordinates -

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        - The addition of Chapter 14: Vector Analysis. -

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        • -
      -

      -       -       - - A Brief History of Calculus -

      - Calculus means a method of calculation or reasoning. - When one computes the sales tax on a purchase, - one employs a simple calculus. - When one finds the area of a polygonal shape by breaking it up into a set of triangles, - one is using another calculus. - Proving a theorem in geometry employs yet another calculus. -

      - -

      - Despite the wonderful advances in mathematics that had taken place into the first half of the 17th century, - mathematicians and scientists were keenly aware of what they could not do. - (This is true even today.) In particular, - two important concepts eluded mastery by the great thinkers of that time: - area and rates of change. -

      - -

      - Area seems innocuous enough; - areas of circles, rectangles, parallelograms, - etc., are standard topics of study for students today just as they were then. - However, the areas of arbitrary - shapes could not be computed, - even if the boundary of the shape could be described exactly. -

      - -

      - Rates of change were also important. - When an object moves at a constant rate of change, - then \text{distance} = \text{rate}\times\text{time}. - But what if the rate is not constantcan distance still be computed? - Or, if distance is known, can we discover the rate of change? -

      - -

      - It turns out that these two concepts were related. - Two mathematicians, Sir Isaac Newton and Gottfried Leibniz, - are credited with independently formulating a system of computing that solved the above problems and showed how they were connected. - Their system of reasoning was a calculus. - However, as the power and importance of their discovery took hold, - it became known to many as the calculus. - Today, we generally shorten this to discuss calculus. -

      -       -       - - - Limits - -

      - The foundation of the calculus - is the limit. - It is a tool to describe a particular behavior of a function. - This chapter begins our study of the limit by approximating its value graphically and numerically. - After a formal definition of the limit, - properties are established that make - finding limits tractable. - Once the limit is understood, - then the problems of area and rates of change can be approached. -

      - - -       - -
      - An Introduction To Limits -

      - We begin our study of limits - by considering examples that demonstrate key concepts that will be explained as we progress. -

      - - - -

      - Consider the function y = \frac{\sin(x) }{x}. - When x is near the value 1, what value (if any) - is y near? -

      - -

      - While our question is not precisely formed - (what constitutes near the value 1?), - the answer does not seem difficult to find. - One might think first to look at a graph of this function to approximate the appropriate - y values. - Consider , - where y = \frac{\sin(x) }{x} is graphed. - For values of x near 1, it seems that y takes on values near 0.85. - In fact, when x=1, then y=\frac{\sin(1) }{1} \approx 0.84, - so it makes sense that when x is near - 1, y will be near 0.84. -

      - - -
      -
    - - -

    - Graph of \sin(x)/x, shown for x between - -7 and 7, and y between 0 and - 1. The x intercepts are at - x=-2\pi, -\pi, \pi, and 2\pi, and a y intercept is at y = 1. - The graph has a downward curve for -\pi \lt x \lt \pi and an - upward curve for -2\pi \lt x \lt -\pi, and - \pi \lt x \lt 2\pi. The graph is undefined for x = 0. -

    -     - - Graph of sin(x)/x, shows values for the domain -7 <=x <=7, undefined at x = 0. - - - \begin{tikzpicture}[ - declare function = {func(\x) = (\x != 0) * (sin(x*180/pi)/x) + (\x == 0) * (1);} - ] - \begin{axis}[ - ymin=-0.25, - ymax=1.2, - xmin=-7, - xmax=7, - ] - \addplot+[infinite,domain=-7:7,samples=50] {func(x)}; - \addplot[hollowdot] coordinates {(0,1)}; - \end{axis} - \end{tikzpicture} - - - - -
    -
    - - - -

    - Graph of \sin(x)/x zoomed in on values where x is near 1. - This view of the graph shows x from 0.5 to 1.5. - The graph has only a slight downward curve. It shows that for x = 1, - \sin(x)/x is approximately 0.84 -

    -     - - Graph of sin(x)/x zoomed in on values where x is near 1. Shows that for x = 1, sin(x)/x is approx. 0.84. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=.3, - xmax=1.6, - ymin=.4, - ymax=1.1, - xtick={0.5,1,1.5}, - ytick={0.6,0.8,1}, - xdiscontinuity, - ydiscontinuity, - grid=both, - ] - \addplot+[infinite,domain=0.5:1.5] {sin(x*180/pi)/x}; - \end{axis} - \end{tikzpicture} - - - - - - -

    - Consider this same function again at a different value for x. - When x is near 0, what value - (if any) - is y near? - By considering , - one can see that it seems that y takes on values near 1. - But what happens when x=0? - We have - - y \rightarrow \frac{\sin(0) }{0} \rightarrow {\genfrac{}{}{0pt}{0}{\text{“}}{}}\frac{0}{0}{\genfrac{}{}{0pt}{0}{\text{”}}{}} - . - The expression 0/0 has no value; it is - - limitindeterminate form - indeterminate form - - indeterminate. - Such an expression gives no information about what is going on with the function nearby. - We cannot find out how y behaves near x=0 - for this function simply by letting x=0. -

    - -
    -
    - - - -

    - Graph of \sin(x)/x zoomed in on values where x is near 0. - The image shows the portion of the graph where x is from -1 to 1. - The graph has a downward curve and is symmetric about x=0. - The height of the graph approaches y = 1 when x is near 0. - A hollow dot at the point (0,1) shows that the function is undefined when x = 0; - that is, f(0) = undefined. -

    -     - - Graph of sin(x)/x zoomed in on values where x is near 0. Shows that when x = 0 - sin(x)/x is undefined. - - - \begin{tikzpicture}[ - declare function = {func(\x) = (\x != 0) * (sin(x*180/pi)/x) + (\x == 0) * (1);} - ] - \begin{axis}[ - xmin=-1.1, - xmax=1.2, - ymin=.75, - ymax=1.05, - ytick={.8,.9,1}, - extra y ticks={.1,.3,...,.9}, - extra y tick labels={}, - ] - \addplot+[infinite,domain=-1:1] {func(x)}; - \addplot[hollowdot] coordinates {(0,1)}; - \end{axis} - \end{tikzpicture} - - - - - - - -

    - Finding a limit entails understanding how a function behaves near a particular value of - x. - Before continuing, it will be useful to establish some notation. - Let y=f(x); that is, - let y be a function of x for some function f. - The expression the limit of y as x approaches - 1 describes a number, often referred to as L, - that y nears as x nears 1. - We write all this as - - \lim_{x\to 1} y = \lim_{x\to 1} f(x) = L - . - This is not a complete definition (that will come in the next section); - this is a pseudo-definition that will allow us to explore the idea of a limit. - limitpseudo-definition -

    - -

    - Above, where f(x) = \sin(x)/x, we approximated - - \lim_{x\to 1} \frac{\sin(x) }{x} \approx 0.84 \quad \text{ and } \quad \lim_{x\to 0}\frac{\sin(x) }{x} \approx 1 - . - (We approximated these limits, - hence used the \approx symbol, - since we are working with the pseudo-definition of a limit, - not the actual definition.) -

    - - - -

    - Once we have the true definition of a limit, - we will find limits analytically; - that is, determining exact values using a variety of mathematical tools. - For now, we will approximate - limits both graphically and numerically. - Graphing a function can provide a good approximation, - though often not very precise. - Numerical methods can provide a more accurate approximation. - We have already approximated limits graphically, - so we now turn our attention to numerical approximations. -

    - -

    - Consider again \lim_{x\to 1}\frac{\sin(x)}{x}. - To approximate this limit numerically, - we can create a table of x and f(x) values where x is near 1. - This is done in . -

    - -

    - Notice that for values of x near 1, - we have \sin(x)/x near 0.841. - The x=1 row is included, - but we stress the fact that when considering limits, - we are not concerned with the value of the function at that particular x value; - we are only concerned with the values of the function when x is near 1. -

    - -
    -
    - - - x - \sin(x)/x - - - 0.9 - 0.870363 - - - 0.99 - 0.844471 - - - 0.999 - 0.841772 - - - 1 - 0.841471 - - - 1.001 - 0.841170 - - - 1.01 - 0.838447 - - - 1.1 - 0.810189 - - - - -

    - Now approximate \lim_{x\to 0} \frac{\sin(x)}{x} numerically. - We already approximated the value of this limit as 1 graphically in - . - - shows the value of \sin(x)/x for values of x near 0. - Ten places after the decimal point are shown to highlight how close to 1 the value of - \sin(x)/x gets as x takes on values very near 0. - We include the x=0 row but again stress that we are not concerned with the value of our function at x=0, - only on the behavior of the function near 0. -

    - -
    -
    - - - x - \sin(x)/x - - - -0.1 - 0.9983341665 - - - -0.01 - 0.9999833334 - - - -0.001 - 0.9999998333 - - - 0 - not defined - - - 0.001 - 0.9999998333 - - - 0.01 - 0.9999833334 - - - 0.1 - 0.9983341665 - - - - -

    - This numerical method gives confidence to say that 1 is a good approximation of - \lim_{x\to 0} \frac{\sin(x)}{x}; that is, - - \lim_{x\to 0} \frac{\sin(x)}{x} \approx 1 - . -

    - -

    - Later we will be able to prove that the limit is - exactly 1. -

    - - - -

    - We now consider several examples that allow us explore different aspects of the limit concept. -

    - - - Approximating the value of a limit - -

    - Use graphical and numerical methods to approximate - - \lim_{x\to 3} \frac{x^2-x-6}{6x^2-19x+3} - . -

    -     - -

    - To graphically approximate the limit, graph - - y = \frac{x^2-x-6}{6x^2-19x+3} - - on a small interval that contains 3. - To numerically approximate the limit, - create a table of values where the x values are near 3. - This is done in - and , respectively. -

    - - -
    -
    - - - -

    - Graph of f(x)=\frac{x^2 - x - 6}{6x^2 - 19x + 3}, - zoomed on values near x = 3, and showing the portion of the graph - for x from 2.5 to 3.5. -

    -

    - There is a slight upward curve to the graph. - The graph suggests that the limit of the function - as x approaches 3 is 0.294. The graph also - shows that the function is undefined for x = 3. -

    -     - - Graph of the function for this example, which shows that when x = 3, f(x) is undefined, - but near 0.294. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=2.3, - xmax=3.7, - ymin=0.2, - ymax=.35, - xdiscontinuity, - ydiscontinuity, - extra y tick labels={}, - grid=both, - ] - \addplot+[infinite,domain=2.5:3.45] {(x^2 - x - 6)/(6*x^2 - 19*x + 3)}; - \addplot [hollowdot] coordinates{(3,0.294)}; - \end{axis} - \end{tikzpicture} - - - - - -
    -
    - - - x - \frac{x^2-x-6}{6x^2-19x+3} - - - 2.9 - 0.29878 - - - 2.99 - 0.294569 - - - 2.999 - 0.294163 - - - 3 - not defined - - - 3.001 - 0.294073 - - - 3.01 - 0.293669 - - - 3.1 - 0.289773 - - - - - -

    - The graph shows that when x is near 3, - the value of y is very near 0.3. - By considering values of x near 3, - we see that y=0.294 is a better approximation. - The graph and the table imply that - - \lim_{x\to 3} \frac{x^2-x-6}{6x^2-19x+3} \approx 0.294 - . -

    - - - - -

    - This example may bring up a few questions about approximating limits - (and the nature of limits themselves). -

      -
    1. -

      - If a graph does not produce as good an approximation as a table, - why bother with it? -

      -
      2. -
    2. -

      - How many values of x in a table are - enough? In the previous example, - could we have just used x=3.001 and found a fine approximation? -

      -
      3. -
    -

    - -

    - Graphs are useful since they give a visual understanding concerning the behavior of a function. - Sometimes a function may act erratically - near certain x values which is hard to discern numerically but very plain graphically - (see ). - Since graphing utilities are very accessible, - it makes sense to make proper use of them. -

    - -

    - Since tables and graphs are used only to - approximate the value of a limit, - there is not a firm answer to how many data points are enough. - Include enough so that a trend is clear, and use values - (when possible) - both less than and greater than the value in question. - In , - we used both values less than and greater than 3. - Had we used just x=3.001, - we might have been tempted to conclude that the limit had a value of 0.3. - While this is not far off, we could do better. - Using values on both sides of 3 - helps us identify trends. -

    - - - Approximating the value of a limit - -

    - Graphically and numerically approximate the limit of - f(x) as x approaches 0, where - - f(x) = \begin{cases}x+1 \amp x\lt 0 \\ -x^2+1 \amp x \gt 0\end{cases} - . -

    -     - -

    - Again we graph f(x) and create a table of its values near - x=0 to approximate the limit. - Note that this is a piecewise defined function, - so it behaves differently on either side of 0. - shows a graph of f(x), - and on either side of 0 it seems the y values approach 1. - Note that f(0) is not actually defined, - as indicated in the graph with the open circle. -

    - - -
    -
    - - - -

    - Graph of the piecewise-defined function in . - For values of x \lt 0 the graph is straight - with a slope of 1 and for values of x \gt 0 the graph - curves downward. A hollow dot at the point (0,1) shows that at x = 0, - f(x) is undefined. - However, both parts of the graph, for x\lt 0 and for x\gt 0, - get close to the point (0,1) as x gets close to 0. -

    -     - - Graph of the piecewise function, shows that at x = 0, y is undefined, - but is near 1. - - - \begin{tikzpicture}[ - declare function = {func(\x) = (\x < 0) * (x+1) + (\x > 0) * (-x^2+1);} - ] - \begin{axis}[ - xmin=-1.1, - xmax=1.2, - ymin=-.1, - ymax=1.1, - ] - \addplot+[infinite,domain=-1:1] {func(x)}; - \addplot[hollowdot] coordinates {(0,1)}; - \end{axis} - \end{tikzpicture} - - - - - -
    -
    - - - x - f(x) - - - -0.1 - 0.9 - - - -0.01 - 0.99 - - - -0.001 - 0.999 - - - 0.001 - 0.999999 - - - 0.01 - 0.9999 - - - 0.1 - 0.99 - - - - - -

    - - shows values of f(x) for values of x near 0. - It is clear that as x takes on values very near 0, - f(x) takes on values very near 1. - It turns out that if we let x=0 for either piece - of f(x), 1 is returned; - this is significant and we'll return to this idea later. -

    - -

    - The graph and table allow us to say that \lim_{x\to 0}f(x) \approx 1; - in fact, we are probably very sure it equals 1. -

    - - - - - - Identifying When Limits Do Not Exist -

    - A function may not have a limit for all values of x. - That is, we cannot write that \lim_{x\to c}f(x)=L - (where L is some real number) - for all values of c, - for there may not be a number that f(x) is approaching. - There are three common ways in which a limit may fail to exist. - - limitdoes not exist -

      -
    1. -

      - The function f(x) may approach different values on either side of c. -

      -
      2. - -
    2. -

      - The function may grow without upper or lower bound as x approaches c. -

      -
      3. - -
    3. -

      - The function may oscillate as x approaches c - without approaching a specific value. -

      -
      4. -
    - We'll explore each of these in turn. -

    - - - - - Different Values Approached From Left and Right - -

    - Explore why \lim_{x\to 1} f(x) does not exist, where - - f(x) = \begin{cases}x^2-2x+3 \amp x\leq 1 \\ x \amp x \gt 1\end{cases} - . -

    -     - -

    - A graph of f(x) around x=1 and a table are given in - Figures and - , respectively. - It is clear that as x approaches 1, - f(x) does not seem to approach a single number. - Instead, it seems as though f(x) approaches two different numbers. - When considering values of x less than 1 - (approaching 1 from the left), - it seems that f(x) is approaching 2; - when considering values of x greater than 1 - (approaching 1 from the right), - it seems that f(x) is approaching 1. - Recognizing this behavior is important; - we'll study this in greater depth later. - Right now, it suffices to say that the limit does not exist since - f(x) is approaching two different - values as x approaches 1. -

    - - -
    -
    - - - -

    - Graph of piecewise function in . - For values of x \leq 1 the graph - has a upward curve, and the graph ends at the point (1,2), - illustrating the fact that f(1)=2. -

    -

    - For values of x \gt 1 the graph is a straight line with a positive slope. - Moving left to right, the line begins at the point (1,1), - at which there is a hollow dot, indicating that to the right of x=1, - the value of f(x) approaches 1. -

    -

    - The most important feature of the graph is that it shows how f(x) - approaches two different values as x approaches 1, - depending on whether x\lt 1 or x\gt 1. -

    -     - - Graph of the piecewise function, which shows that as x approaches 1 there - is no limit. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.1, - xmax=2.1, - ymin=-.1, - ymax=3.3, - ] - \addplot[firstcurvestyle,leftarrow, domain=0:1] {x^2 - 2*x + 3}; - \addplot[firstcurvestyle,rightarrow, domain=1:2] {x}; - \addplot[soliddot] coordinates {(1,2)}; - \addplot[hollowdot] coordinates {(1,1)}; - \end{axis} - \end{tikzpicture} - - - - - -
    -
    - - - x - f(x) - - - 0.9 - 2.01 - - - 0.99 - 2.0001 - - - 0.999 - 2.000001 - - - 1.001 - 1.001 - - - 1.01 - 1.01 - - - 1.1 - 1.1 - - - - - - - - - - - The Function Grows Without Bound - -

    - Explore why \lim_{x\to 1} \frac{1}{(x-1)^2} does not exist. -

    -     - -

    - A graph and table of f(x) = \frac{1}{(x-1)^2} are given in - - and , respectively. - Both show that as x approaches 1, - f(x) grows larger and larger. -

    - - -
    -
    - - - -

    - Graph of the function for . - The graph hows a horizontal - asymptote at y = 0 and a vertical asymptote at x = 1. - Because of the vertical asymptote at x = 1 the function - has no limit as x approaches 1. -

    -     - - Graph of the function f(x), showing that as x approaches 1, there is a vertical asymptote. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.1, - xmax=2.1, - ymin=-1, - ymax=110 - ] - \addplot[firstcurvestyle,infinite,domain=0:0.9] {1/(x - 1)^2}; - \addplot[firstcurvestyle,infinite,domain=1.1:2] {1/(x - 1)^2}; - \addplot[asymptote,rightarrow] coordinates {(1,1) (1,100)}; - \end{axis} - \end{tikzpicture} - - - - - -
    -
    - - - x - f(x) - - - 0.9 - 100. - - - 0.99 - 10000. - - - 0.999 - 1.\times 10^6 - - - 1.001 - 1.\times 10^6 - - - 1.01 - 10000. - - - 1.1 - 100. - - - - - -

    - We can deduce this on our own, - without the aid of the graph and table. - If x is near 1, then (x-1)^2 is very small, and: - - \frac{1}{\text{ very small number } } = \text{ very large number } - . -

    - -

    - Since f(x) is not approaching a single number, we conclude that - \lim_{x\to 1}\frac{1}{(x-1)^2} does not exist. -

    - - - - - The Function Oscillates - -

    - Explore why \lim_{x\to 0}\sin(1/x) does not exist. -

    -     - -

    - Two graphs of f(x) = \sin(1/x) are given in . - - shows f(x) on the interval [-1,1]; - notice how f(x) seems to oscillate near x=0. - One might think that despite the oscillation, - as x approaches 0, - f(x) approaches 0. - However, zooms in on \sin(1/x), - on the interval [-0.1,0.1]. - Here the oscillation is even more pronounced. - Finally, in , - we see \sin(1/x) evaluated for values of x near 0. - As x approaches 0, - f(x) does not appear to approach any value. -

    - -
    -
    - -
    -
    - - - x - \sin(1/x) - - - 0.1 - -0.544021 - - - 0.01 - -0.506366 - - - 0.001 - 0.82688 - - - 0.0001 - -0.305614 - - - 1.\times 10^{-5} - 0.0357488 - - - 1.\times 10^{-6} - -0.349994 - - - 1.\times 10^{-7} - 0.420548 - - - - -

    - It can be shown that in reality, - as x approaches 0, \sin(1/x) takes on all values between - -1 and 1 infinitely many times! - Because of this oscillation, - \lim_{x\to 0}\sin(1/x) does not exist. -

    - - - - - - - - - Limits of Difference Quotients -

    - We have approximated limits of functions as x approached a particular number. - We will consider another important kind of limit after explaining a few key ideas. - - limitdifference quotient - -

    - - - -

    - Let f(x) represent the position function, in feet, - of some particle that is moving in a straight line, - where x is measured in seconds. - Let's say that when x=1, - the particle is at position 10 ft., and when x=5, - the particle is at 20 ft. - Another way of expressing this is to say -

    - -
    -

    - f(1)=10 and f(5) = 20. -

    -
    - -

    - Since the particle traveled 10 feet in 4 seconds, - we can say the particle's - - average velocity - - velocityaverage velocity - - average velocity was 2.5 ft/s. - We write this calculation using a - quotient of differences, or, a - - difference quotient - - difference quotient: - - \frac{f(5) - f(1)}{5-1} \frac{\text{ ft}}{\text{s}} = \frac{10 \text{ ft}}{4\text{ s}} = 2.5 \text{ ft/s } - . -

    - -

    - This difference quotient can be thought of as the familiar - rise over run used to compute the slopes of lines. - In fact, that is essentially what we are doing: - given two points on the graph of f, - we are finding the slope of the secant line - through those two points. - See . -

    - -
    -
    - - - -

    - The image shows the graph of a function, along with a line that intersects the graph at two points. - The graph has the shape of a parabola that opens downward, - and is displayed over the region 0\leq x\leq 6, - with a y range from 0 to 25. - There are two points plotted on the graph at coordinates - (1, 10) and (5, 20), and the line through these points is an example of a secant line. -

    -     - - A downward curved graph, with marked points at (1,10) and (5,20), and a - line intercepting both points. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1, - xmax=6.5, - ymin=-1, - ymax=25, - ] - \addplot+[infinite,domain=0:6] {-1.5*x^2+11.5*x}; - \addplot+[infinite,domain=0:6] {2.5*(x-1)+10}; - \addplot[soliddot] coordinates {(1,10) (5,20)}; - \end{axis} - \end{tikzpicture} - - - - - -

    - Now consider finding the average speed on another time interval. - We again start at x=1, - but consider the position of the particle h seconds later. - That is, consider the positions of the particle when x=1 - and when x=1+h. The difference quotient (excluding units) - is now - - \frac{f(1+h)-f(1)}{(1+h)-1} = \frac{f(1+h)-f(1)}{h} - . -

    - -

    - Let f(x) = -1.5x^2+11.5x; - note that f(1)=10 and - f(5) = 20, as in our discussion. - We can compute this difference quotient for all values of h - (even negative values!) - except h=0, for then we get 0/0, - the indeterminate form introduced earlier. - For all values h\neq 0, - the difference quotient computes the average velocity of the particle over an interval of time of length h starting at x=1. -

    - -

    - For small values of h, - , values of h close to 0, - we get average velocities over very short time periods and compute secant lines over small intervals. - See . - This leads us to wonder what the limit of the difference quotient is as - h approaches 0. - That is, - - \lim_{h\to 0} \frac{f(1+h)-f(1)}{h} = \text{ ? } - -

    - - - -
    -
    - -
    -
    - - - -

    - Graph of the function from , with the points on the graph (1,10) - and (3,21) marked. A secant line is drawn through these points; it has a steeper slope than in - . Here the value of h is 2. -

    -     - - Graph of the same function as the previous figure, with the points (1,10) - and (3,21). The secant line has a steeper slope, equal to 5.5. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1, - xmax=6.5, - ymin=-1, - ymax=25, - ] - \addplot+[infinite,domain=0:6] {-1.5*x^2+11.5*x}; - \addplot+[infinite,domain=0:3.5454] {5.5*(x-1)+10}; - \addplot[soliddot] coordinates {(1,10) (3,21)}; - \end{axis} - \end{tikzpicture} - - - - - -
    -
    - - - -

    - Graph of the function from , but - with the points (1,10) and (2,17) on the graph marked. - These points correspond to a value of h=1, - and the secant line through these points has a steeper slope than in . -

    -     - - Graph of the same function as the previous figure, with the points (1,10) - and (2,17) marked. The secant line has a steeper slope equal to 7. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1, - xmax=6.5, - ymin=-1, - ymax=25, - ] - \addplot+[infinite,domain=0:6] {-1.5*x^2+11.5*x}; - \addplot+[infinite,domain=0:3] {7*(x-1)+10}; - \addplot[soliddot] coordinates {(1,10) (2,17)}; - \end{axis} - \end{tikzpicture} - - - - - -
    -
    - - - -

    - Graph of the function from , but - with the points (1,10) and (1.5,13.875) on the graph marked, corresponding to the value h=0.5. - The secant line through these points again has a steeper slope than in the previous figures. -

    -     - - Graph of the same function, with the points (1,10) and - (1.5,13.875). The secant line has a steeper slope equal to 7.75. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1, - xmax=6.5, - ymin=-1, - ymax=25, - ] - \addplot+[infinite,domain=0:6] {-1.5*x^2+11.5*x}; - \addplot+[infinite,domain=0:2.80645] {7.75*(x-1)+10}; - \addplot[soliddot] coordinates {(1,10) (1.5,13.875)}; - \end{axis} - \end{tikzpicture} - - - - - - - -

    - As we do not yet have a true definition of a limit nor an exact method for computing it, - we settle for approximating the value. - While we could graph the difference quotient - (where the x-axis would represent h values and the - y-axis would represent values of the difference quotient) - we settle for making a table. - See . - The table gives us reason to assume the value of the limit is about 8.5. -

    - -
    -
    - - - h - \frac{f(1+h)-f(1)}{h} - - - -0.5 - 9.25 - - - -0.1 - 8.65 - - - -0.01 - 8.515 - - - 0.01 - 8.485 - - - 0.1 - 8.35 - - - 0.5 - 7.75 - - - - - -

    - Proper understanding of limits is key to understanding calculus. - With limits, - we can accomplish seemingly impossible mathematical things, - like adding up an infinite number of numbers - (and not get infinity) - and finding the slope of a line between two points, - where the two points are actually the same point. - These are not just mathematical curiosities; - they allow us to link position, - velocity and acceleration together, - connect cross-sectional areas to volume, - find the work done by a variable force, and much more. -

    - - - -

    - In the next section we give the formal definition of the limit and begin our study of finding limits analytically. - In the following exercises, - we continue our introduction and approximate the value of limits. -

    - - - - - - - Terms and Concepts - - - - -

    - In your own words, what does it mean to - find the limit of f(x) as x approaches 3? -

    - - -     - - -     - - - - -

    - An expression of the form \frac{0}{0} is called . - -

    -     - - - - indeterminate - - - undefined - -

    - Yes, but there is a more specific answer. -

    -     -     -     -     - -     - - - - - -

    - - The limit of f(x) as x approaches 5 is f(5). -

    -     - -

    - The limit considers values of x near to 5, - but not equal to 5. -

    - -

    - The case where the limit is equal to f(5) has a special name; - we will consider this in detail in . -

    -     - -     - - - - -

    - Describe three situations where \lim\limits_{x\to c}f(x) does not exist. -

    - -     - - -

    - The function may approach different values from the left and right, - the function may grow without bound, - or the function might oscillate. -

    -     - -     - - - - -

    - In your own words, what is a difference quotient? -

    - -     - - -     - - - - -

    - When x is near 0, - \dfrac{\sin(x)}{x} is near the value . -

    - -     - - - - - - - - - -

    - That is a reasonable guess, but no. -

    -     -     -     -     - -

    - Try values of x close to 0, such as 0.0001. - A calculator reveals that \dfrac{\sin 0.0001}{0.0001}\approx0.999999998\ldots. - This is near 1. -

    -     - -     -     - - - Problems - - -

    - Approximate the limit numerically and graphically. -

    -     - - - - - do { - $a = list_random(-1,1); - $b = non_zero_random(-5,5,1); - $c = non_zero_random(-5,5,1); - if ($envir{problemSeed} == 1){ - ($a,$b,$c) = (1,3,-5); - }; - $f = Formula("x^2 + $b x + $c")->reduce; - $vx = $b/2; - $vy = $f->eval(x => $vx); - $l = $f->eval(x => $a); - } until($l != 0); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - - -

    - \lim\limits_{x\to }\left(\right) -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph shows a parabola opening upward, - with vertex at (,), and domain [-2,2]. - When x is close to , the value of f(x) appears to be close to . -

    -     - A parabola opening upward, with vertex at (,), - also passing through (,) - - \begin{tikzpicture} - \begin{axis}[grid=both] - \addplot+[infinite,domain=-2:2] {$f}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , - that is close to . - So \lim_{x\to }\left(\right)=. -

    -     -     -     - - - - - do { - $a = random(-1,1,1); - $b = non_zero_random(-5,5,1); - $c = non_zero_random(-5,5,1); - $d = non_zero_random(-5,5,1); - if ($envir{problemSeed} == 1){ - ($a,$b,$c,$d) = (0,-3,1,-5); - }; - $f = Formula("x^3 + $b x^2 + $c x + $d")->reduce; - $l = $f->eval(x => $a); - } until ($l != 0); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - - -

    - \lim\limits_{x\to }\left(\right) -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph of . - When x is close to , - the value of f(x) appears to be close to ; - however, this is difficult to tell at the scale used for the graph. -

    -     - The graph of the function for this exercise. When x is close to , y appears to be close to - - \begin{tikzpicture} - \begin{axis}[grid=both] - \addplot+[infinite,domain=-2:2] {$f}; - % Be sure to include some area above and below x-axis - \addplot[mark=none] coordinates {(0,5) (0,-5)}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , - that is close to . - So \lim_{x\to }\left(\right)=. -

    -     -     -     - - - - - $a = 0; - ($b,$c) = random_subset(2,-5..-1,1..5); - if ($envir{problemSeed} == 1){ - ($b,$c) = (-1,-3); - }; - ($left,$right) = num_sort(0,$c); - # for graphing - $xmin = $left - 2; - $xmax = $right + 2; - @d = ($left - 0.02, $left + 0.02, $right - 0.02, $right + 0.02); - $f = Formula("(x - $b)/(x^2 - $c x)")->reduce; - $l = Compute("DNE"); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - - -

    - \lim\limits_{x\to }\left(\right) -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The image shows the graph of a function with vertical asymptotes at x=0 - and at x=. The x axis is also a horizontal asymptote. -

    -

    - On one side of x=, values of f(x) appear large and positive. - On the other side, the values of f(x) are large and negative. -

    -     - Graph of the function for this problem, showing a vertical asymptote at the limit point - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - xmin=$xmin, - xmax=$xmax, - ymin=-25, - ymax=25, - ] - \addplot[firstcurvestyle,infinite, domain=$xmin:$d[0]] {$f}; - \addplot[firstcurvestyle,infinite, domain=$d[1]:$d[2]] {$f}; - \addplot[firstcurvestyle,infinite, domain=$d[3]:$xmax] {$f}; - \addplot[asymptote] coordinates {($left,-25) ($left,25)}; - \addplot[asymptote] coordinates {($right,-25) ($right,25)}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , - that grows without bound. - So \lim_{x\to }\left(\right) does not exist (DNE). -

    -     -     -     - - - - - ($a,$b,$c) = random_subset(3,-5..-1,1..5); - if ($envir{problemSeed} == 1){ - ($a,$b,$c) = (3,-1,1); - }; - $f = Formula("(x^2 - ($a + $b)x + $a*$b)/(x^2 - ($a + $c)x + $a*$c)")->reduce; - ($left,$right) = num_sort($a,$c); - # for graphing - @d = ($c - 0.5, $c + 0.5); - Context("Fraction"); - $l = Fraction(($a - $b)/($a - $c)); - $lreal= ($a - $b)/($a - $c); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - $xmin = -abs($a) - 2; - $xmax = abs($a) + 2; - $xmid1 = $c - 0.01; - $xmid2 = $c + 0.01; - $ymin = -ceil(abs(Real($l))) - 2; - $ymax = ceil(abs(Real($l))) + 2; - - -

    - \lim\limits_{x\to }\left(\right) -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph for this problem has a vertical asymptote at x=, - and a hole in the graph at x=. -

    - -

    - When x is close to , the value of f(x) is close to . -

    -     - A graph with a hole at and a vertical asymptote at - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - xmin=$xmin, - xmax=$xmax, - ymin=$ymin, - ymax=$ymax, - ] - \addplot[firstcurvestyle,infinite, domain=$xmin:$d[0]] {$f}; - \addplot[firstcurvestyle,infinite, domain=$d[1]:$xmax] {$f}; - \addplot[asymptote] coordinates {($c,$ymin) ($c,$ymax)}; - \addplot[hollowdot] coordinates {($a,$lreal)}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , - that is close to . - So \lim_{x\to }\left(\right)=. -

    -     -     -     - - - - - ($a,$c) = random_subset(2,-5..-1,1..5); - $b = list_random(-9..-6,6..9); - if ($envir{problemSeed} == 1){ - ($a,$b,$c) = (-1,-7,-5); - }; - $f = Formula("(x^2 - ($a + $b)x + $a*$b)/(x^2 - ($a + $c)x + $a*$c)")->reduce; - ($left,$right) = num_sort($a,$c); - # for graphing - @d = ($c - 0.1, $c + 0.1); - Context("Fraction"); - $l = Fraction(($a - $b)/($a - $c)); - $lreal = ($a - $b)/($a - $c); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - $xmin = $left - 2; - $xmax = $right + 2; - $xmid1 = $c - 0.01; - $xmid2 = $c + 0.01; - $ymin = -ceil(abs(Real($l))) - 2; - $ymax = ceil(abs(Real($l))) + 2; - - -

    - \lim\limits_{x\to }\left(\right) -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph for this problem has a vertical asymptote at x=, - and a hole in the graph at x=. -

    - -

    - When x is close to , the value of f(x) is close to . -

    -     - A graph with a hole at and a vertical asymptote at - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - xmin=$xmin, - xmax=$xmax, - ymin=$ymin, - ymax=$ymax, - ] - \addplot[firstcurvestyle,infinite, domain=$xmin:$d[0]] {$f}; - \addplot[firstcurvestyle,infinite, domain=$d[1]:$xmax] {$f}; - \addplot[asymptote] coordinates {($c,$ymin) ($c,$ymax)}; - \addplot[hollowdot] coordinates {($a,$lreal)}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , - that is close to . - So \lim_{x\to }\left(\right)=. -

    -     -     -     - - - - - ($a,$b,$c) = random_subset(3,-9..-1,1..9); - if ($envir{problemSeed} == 1){ - ($a,$b,$c) = (2,-5,-2); - }; - $f = Formula("(x^2-($c+$b)x+$a*$b)/(x^2-(2*$a)x+($a)^2)")->reduce; - $l = (($a-$c)*($a-$b) > 0) ? OneOf("DNE","inf") : OneOf("DNE","-inf"); - # for graphing - @d = ($a - 0.1, $a + 0.1); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - $xmin = -abs($a) - 2; - $xmax = abs($a) + 2; - $xmid1 = $a - 0.01; - $xmid2 = $a + 0.01; - $ymin = -100; - $ymax = 100; - - -

    - \lim\limits_{x\to }\left(\right) -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph for this problem has a vertical asymptote at x=. -

    - -

    - When x is close to , the value of f(x) increases without bound. -

    -     - A graph with a vertical asymptote at - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - xmin=$xmin, - xmax=$xmax, - ymin=$ymin, - ymax=$ymax, - ] - \addplot[firstcurvestyle,infinite, domain=$xmin:$d[0], samples=40] {$f}; - \addplot[firstcurvestyle,infinite, domain=$d[1]:$xmax, samples=40] {$f}; - \addplot[asymptote] coordinates {($a,$ymin) ($a,$ymax)}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , - that grows without bound. - So \lim_{x\to }\left(\right) does not exist (DNE). -

    -     -     -     - - - - - $a = non_zero_random(-5,5,1); - $b = non_zero_random(-5,5,1); - $c = list_random(-4..-1,2..4); - $diff = non_zero_random(-3,3,1); - $d = $a + $b - ($c*$a) - $diff; - if ($envir{problemSeed} == 1){ - ($a,$b,$c,$d) = (2,2,3,-5); - }; - $f1 = Formula("x + $b")->reduce; - $f2 = Formula("$c*x + $d")->reduce; - Context("PiecewiseFunction"); - Context()->flags->set(tolType=>"absolute",tolerance => 0.00001); - $f = PiecewiseFunction("x <= $a" => "$f1", "x > $a" => "$f2"); - $l = Compute("DNE"); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - $xmin = -abs($a) - 2; - $xmax = abs($a) + 2; - - -

    - \lim\limits_{x\to }f(x), where f(x)= -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph shows a piecewise-defined function. - At x=, the y value jumps, - with values of y on one side of - different from those on the other side. -

    -     - A piecewise-linear graph with a jump at x= - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - xmin=$xmin, - xmax=$xmax, - ] - \addplot[firstcurvestyle, domain=$xmin:$a, infiniteopen] {$f1}; - \addplot[firstcurvestyle, domain=$a:$xmax, closedinfinite] {$f2}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , - that f(x) approaches different values from the left and right. - So \lim_{x\to }f(x) does not exist. -

    -     -     -     - - - - - do { - $a = non_zero_random(-3,3,1); - $b = random(-4,-1,1); - $c = non_zero_random(-5,5,1); - $d = non_zero_random(-3,3,1); - if ($envir{problemSeed} == 1){ - ($a,$b,$c,$d) = (3,-1,1,2); - }; - $e = $a**2 + $b*$a + $c - $d*$a; - $f1 = Formula("x^2+$b*x+$c")->reduce; - $f2 = Formula("$d*x+$e")->reduce; - Context("PiecewiseFunction"); - Context()->flags->set(tolType=>"absolute",tolerance => 0.00001); - $f = PiecewiseFunction("x <= $a" => "$f1", "x > $a" => "$f2"); - $l = $f1->eval(x => $a); - } until ($l!=0); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - $xmin = -4; - $xmax = 4; - - -

    - \lim\limits_{x\to }f(x), where f(x)= -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph of a piecewise-defined function. - To the left of x=, - the graph is a parabola. To the right, the graph is a line. -

    - -

    - The two parts of the graph appear to have the same y value when x=. -

    -     - A piecewise-defined graph. One part is a parabola, and the other, a line. The two parts meet when x= - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - xmin=$xmin, - xmax=$xmax, - ] - \addplot[firstcurvestyle, domain=$xmin:$a, infiniteleft] {$f1}; - \addplot[firstcurvestyle, domain=$a:$xmax, infiniteright] {$f2}; - % Be sure to include some area above and below x-axis - \addplot[mark=none] coordinates {(0,5) (0,-5)}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , - that f(x) approaches . - So \lim_{x\to }f(x)=. -

    -     -     -     - - - - - $a=0; - $b = non_zero_random(-5,5,1); - if ($envir{problemSeed} == 1){ - $b = 3; - }; - $f1 = Formula("cos(x)")->reduce; - $f2 = Formula("x^2+$b*x+1")->reduce; - Context("PiecewiseFunction"); - Context()->flags->set(tolType=>"absolute",tolerance => 0.00001); - $f = PiecewiseFunction("x <= $a" => "$f1", "x > $a" => "$f2"); - $l = $f1->eval(x => $a); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - $xmin = -3; - $xmax = 3; - - -

    - \lim\limits_{x\to }f(x), where f(x)= -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph of a piecewise-defined function. - To the left of x=, - the graph is a cosine function. To the right, the graph is a parabola. -

    - -

    - The two parts of the graph appear to have the same y value when x=. -

    -     - A piecewise-defined graph. One part is a parabola, and the other, a cosine wave. The two parts meet when x= - - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - xmin=$xmin, - xmax=$xmax, - ] - \addplot[firstcurvestyle, domain=$xmin:$a, infiniteleft] {cos(x*180/pi)}; - \addplot[firstcurvestyle, domain=$a:$xmax, infiniteright] {$f2}; - % Be sure to include some area above and below x-axis - \addplot[mark=none] coordinates {(0,2) (0,-2)}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , - that f(x) approaches . - So \lim_{x\to }f(x)=. -

    -     -     -     - - - - - Context()->flags->set(reduceConstants => 0); - $a = list_random(Formula("pi/2"),Formula("pi/3"),Formula("pi/6")); - if ($envir{problemSeed} == 1){ - $a = Formula("pi/2") - }; - $f1=Formula("sin(x)")->reduce; - $f2=Formula("cos(x)")->reduce; - Context("PiecewiseFunction"); - Context()->flags->set(tolType=>"absolute",tolerance => 0.00001); - $areal = $a->eval(x => 0); - $f = PiecewiseFunction("x <= $areal" => "$f1", "x > $areal" => "$f2"); - $l = Compute("DNE"); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_->eval(x => 0))} (@tabin); - $xmin = -1; - $xmax = 3; - - -

    - \lim\limits_{x\to }f(x), - where f(x)=\begin{cases}\sin(x)\amp x\leq \\\cos(x)\amp x\gt \end{cases} -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph of a piecewise-defined function. - The two parts are portions of sine and cosine graphs, - and they do not approach the same y values near x=. -

    -     - A piecewise-defined graph consisting of two sine waves that approach different values near x= - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - xmin=$xmin, - xmax=$xmax, - ] - \addplot[firstcurvestyle, domain=$xmin:$a, infiniteopen] {sin(x*180/pi)}; - \addplot[firstcurvestyle, domain=$a:$xmax, closedinfinite] {cos(x*180/pi)}; - % Be sure to include some area above and below x-axis - \addplot[mark=none] coordinates {(0,2) (0,-2)}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , - that f(x) approaches different values from the left and right. - So \lim_{x\to }f(x) does not exist. -

    -     -     -     - - - - - $f = Formula("abs(x)^x"); - $a = 0; - $l = 1; - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - - -

    - \lim\limits_{x\to 0}\lvert x\rvert^x -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph shows a wave-like curve that approaches the point (0,1) - when x is close to zero, for both positive and negative values of x. - However, a hollow dot at the point (0,1) indicates that the function is undefined at this point. -

    -     - A wave-like graph with a hole at the point (0,1). - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - ymin=-1, - ymax=2, - ] - \addplot[firstcurvestyle,domain=-1.5:-0.005,infiniteleft] {(-x)^x}; - \addplot[only marks,mark=o,color=firstcolor] coordinates { (0,1) }; - \addplot[firstcurvestyle,domain=0.005:1.5,infiniteright] {x^x}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to 0, that \lvert x \rvert^x is close to 1. - So \lim_{x\to 0}\lvert x\rvert^x=1. -

    -     -     -     - - - - - $f = Formula("e^(-e^(1/x))"); - $a = 0; - $l = Compute("DNE"); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{$f->eval(x => $_)} (@tabin); - - -

    - \lim\limits_{x\to0}e^{-e^{1/x}} -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph of a piecewise-defined function. - For negative values of x, - the graph climbs slowly, from values near 0.5 to values near 1 as x approaches 0. - For positive values of x, the graph is close to 0 when x is close to 0, - and then climbs slowly as x increases. -

    -     - The graph of a piecewise-defined function that jumps from 1 to 0 at x=0. - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - ymin=-2, - ymax=2, - ] - \addplot[firstcurvestyle, domain=-2:-0.001,infiniteleft] {e^(-e^(1/x))}; - \addplot[only marks,mark=o,color=firstcolor] coordinates { (0,1) }; - \addplot[firstcurvestyle, domain=0.2:2, infiniteright] {e^(-e^(1/x))}; - \addplot[only marks,mark=o,color=firstcolor] coordinates { (0,0) }; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to 0, that e^{-e^{1/x}} approaches different values from the left and right. - So \lim_{x\to 0}e^{-e^{1/x}} does not exist. -

    -     -     -     - - - - - %fact = (0 => 1, 1 => 1, 2 => 2, 3 => 6, 4 => 24, 5 => 120, 6 => 720); - sub f { - my $x = shift; - return $fact{int(abs($x))}; - } - $a = list_random(-5..-2,2..5); - $l = (abs($a) <= 1) ? Real(1) : Compute("DNE"); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{f($_)} (@tabin); - $ymax = f(abs($a)+0.5)+10; - $ymin = max(-10, -1*f(abs($a)+0.5)); - - -

    - \lim\limits_{x\to}\big\lfloor\lvert x\rvert\big\rfloor !, - where \lvert x\rvert is the absolute value of x, - \lfloor x\rfloor is the floor of x (the greatest integer less than or equal to x), - and x! is x factorial. -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - \big\lfloor\lvert x\rvert\big\rfloor ! - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph is piecewise-defined, and consists of a collection of horizontal line segments. - For negative values of x, the segments include the right endpoint, but not the left endpoint. - For positive value of x, the segments include the left endpoint, but not the left endpoint. -

    - -

    - The y values for the segments are small when \abs{x} is small, - and then get larger for larger values of \abs{x}. -

    - -

    - Near x=, the y value is different just to the left and right of . -

    -     - A piecewise-defined graph consisting of many short horizontal line segments - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - xmin=$xmin, - xmax=$xmax, - ymin=$ymin, - ymax=$ymax, - ] - \addplot[firstcurvestyle, open] coordinates {(-2,1) (2,1)}; - \addplot[firstcurvestyle, closedopen] coordinates {(2,2) (3,2)}; - \addplot[firstcurvestyle, closedopen] coordinates {(3,6) (4,6)}; - \addplot[firstcurvestyle, closedopen] coordinates {(4,24) (5,24)}; - \addplot[firstcurvestyle, closedopen] coordinates {(5,120) (6,120)}; - \addplot[firstcurvestyle, closedopen] coordinates {(-2,2) (-3,2)}; - \addplot[firstcurvestyle, closedopen] coordinates {(-3,6) (-4,6)}; - \addplot[firstcurvestyle, closedopen] coordinates {(-4,24) (-5,24)}; - \addplot[firstcurvestyle, closedopen] coordinates {(-5,120) (-6,120)}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , that \big\lfloor\lvert x\rvert\big\rfloor ! approaches different values from the left and right. So - - \lim_{x\to}\big\lfloor\lvert x\rvert\big\rfloor !\text{ does not exist (DNE).} - -

    -     -     -     - - - - - %fact = (0 => 1, 1 => 1, 2 => 2, 3 => 6, 4 => 24, 5 => 120, 6 => 720); - sub f { - my $x = shift; - return $fact{int(abs($x))}; - } - $a = random(-1,1,1); - $l = (abs($a) <= 1) ? Real(1) : Compute("DNE"); - @tabin = ($a - 0.1, $a - 0.01, $a - 0.001, $a + 0.001, $a + 0.01, $a + 0.1); - @tabout = map{f($_)} (@tabin); - - -

    - \lim\limits_{x\to}\big\lfloor\lvert x\rvert\big\rfloor !, - where \lvert x\rvert is the absolute value of x, - \lfloor x\rfloor is the floor of x (the greatest integer less than or equal to x), - and x! is x factorial. -

    - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -     - -

    - For a numerical approximation, make a table: -

    - - - x - \big\lfloor\lvert x\rvert\big\rfloor ! - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - For a graphical approximation: -

    - - -

    - The graph is piecewise-defined, and consists of several horizontal line segments. - For negative values of x, the segments include the right endpoint, but not the left endpoint. - For positive value of x, the segments include the left endpoint, but not the left endpoint. -

    - -

    - The y values for the segments are small when \abs{x} is small, - and then get larger for larger values of \abs{x}. -

    - -

    - Near x=, the y appears to be equal to 1. -

    -     - A piecewise-defined graph consisting of several short horizontal line segments - - \begin{tikzpicture} - \begin{axis}[ - grid=both, - xmin=$xmin, - xmax=$xmax, - ymin=-1, - ymax=3, - ] - \addplot[firstcurvestyle, open] coordinates {(-2,1) (2,1)}; - \addplot[firstcurvestyle, closedopen] coordinates {(2,2) (3,2)}; - \addplot[firstcurvestyle, closedopen] coordinates {(3,6) (4,6)}; - \addplot[firstcurvestyle, closedopen] coordinates {(4,24) (5,24)}; - \addplot[firstcurvestyle, closedopen] coordinates {(5,120) (6,120)}; - \addplot[firstcurvestyle, closedopen] coordinates {(-2,2) (-3,2)}; - \addplot[firstcurvestyle, closedopen] coordinates {(-3,6) (-4,6)}; - \addplot[firstcurvestyle, closedopen] coordinates {(-4,24) (-5,24)}; - \addplot[firstcurvestyle, closedopen] coordinates {(-5,120) (-6,120)}; - \end{axis} - \end{tikzpicture} - - -

    - It appears that when x is close to , that \big\lfloor\lvert x\rvert\big\rfloor ! is always equal to 1. So - - \lim_{x\to}\big\lfloor\lvert x\rvert\big\rfloor !=1 - -

    -     -     -     -     - - - -

    - Approximate the limit of the difference quotient, - \lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h}, - using h = \pm 0.1, \pm 0.01. -

    -     - - - - #$m = non_zero_random(-9,9,1); - #do { - # $b = non_zero_random(-9,9,1); - #} until (abs($m) != abs($b)); - #$a = random(1,5,1); - #if ($envir{problemSeed} == 1){ - $a=3; - $m=-7; - $b=2; - #}; - $f = Formula("$m*x + $b")->reduce; - Context()->variables->add(h => 'Real'); - $dq = ($f->substitute(x => Formula("$a + h")) - ($f->eval(x => $a)))/Formula("h"); - @tabin = (Real(-0.1), Real(-0.01), Real(0.01), Real(0.1)); - @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); - @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); - $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ - my ($correct,$student,$self)=@_; - my @cor = @{$correct}; - my @stu = @{$student}; - my @return = (1,1,1,1,1,1,1,1); - for my $i (0,2,4,6) { - do { - $return[$i+1] = 0; - $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); - } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); - do { - $return[$i] = 0; - $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); - } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); - }; - do { - $return[2] = 0; - $self->setMessage(3,'You already used this h-value.'); - } unless ($stu[2] != $stu[0] or $stu[2] eq ""); - do { - $return[4] = 0; - $self->setMessage(5,'You already used this h-value.'); - } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); - do { - $return[6] = 0; - $self->setMessage(7,'You already used this h-value.'); - } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); - return [@return]; - }); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x)=, - a= -

    - -     - - - h - \frac{f(a+h)-f(a)}{h} - - - - - - - - - - - - - - - - - - - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -

    - -

    - - - - - - - h - \frac{f(a+h)-f(a)}{h} - - - -0.1 - - - - -0.01 - - - - 0.01 - - - - 0.1 - - - -

    - \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. -

    - - - - - - - - #$m = non_zero_random(-9,9,1); - #$b = random(0.01,0.09,0.01) + random(0.1,0.9,0.1) + random(-9,9,1); - #$a = random(-5,5,1); - #if ($envir{problemSeed} == 1){ - $a = -1; - $m = 9; - $b = 0.06; - #}; - $f = Formula("$m*x + $b")->reduce; - Context()->variables->add(h => 'Real'); - $dq = ($f->substitute(x => Formula("$a + h")) - ($f->eval(x => $a)))/Formula("h"); - @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); - @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); - @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); - $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ - my ($correct,$student,$self)=@_; - my @cor = @{$correct}; - my @stu = @{$student}; - my @return = (1,1,1,1,1,1,1,1); - for my $i (0,2,4,6) { - do { - $return[$i+1] = 0; - $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); - } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); - do { - $return[$i] = 0; - $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); - } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); - }; - do { - $return[2] = 0; - $self->setMessage(3,'You already used this h-value.'); - } unless ($stu[2] != $stu[0] or $stu[2] eq ""); - do { - $return[4] = 0; - $self->setMessage(5,'You already used this h-value.'); - } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); - do { - $return[6] = 0; - $self->setMessage(7,'You already used this h-value.'); - } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); - return [@return]; - }); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x)=, - a= -

    - -     - - - h - \frac{f(a+h)-f(a)}{h} - - - - - - - - - - - - - - - - - - - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -

    - -

    - - - - - - - h - \frac{f(a+h)-f(a)}{h} - - - -0.1 - - - - -0.01 - - - - 0.01 - - - - 0.1 - - - -

    - It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. -

    - - - - - - - - #do { - # $b = non_zero_random(-5,5,1); - # $c = non_zero_random(-9,9,1); - # $a = random(-5,5,1); - # if ($envir{problemSeed} == 1){ - $a = 1; - $b = 3; - $c = -7; - # }; - $f = Formula("x^2+$b*x+$c")->reduce; - $m = $f->D('x')->eval(x => $a); - #} until ($m!=0); - Context()->variables->add(h => 'Real'); - $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a)))/Formula("h"); - @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); - @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); - @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); - $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ - my ($correct,$student,$self)=@_; - my @cor = @{$correct}; - my @stu = @{$student}; - my @return = (1,1,1,1,1,1,1,1); - for my $i (0,2,4,6) { - do { - $return[$i+1] = 0; - $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); - } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); - do { - $return[$i] = 0; - $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); - } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); - }; - do { - $return[2] = 0; - $self->setMessage(3,'You already used this h-value.'); - } unless ($stu[2] != $stu[0] or $stu[2] eq ""); - do { - $return[4] = 0; - $self->setMessage(5,'You already used this h-value.'); - } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); - do { - $return[6] = 0; - $self->setMessage(7,'You already used this h-value.'); - } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); - return [@return]; - }); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x)=, - a= -

    - -     - - - h - \frac{f(a+h)-f(a)}{h} - - - - - - - - - - - - - - - - - - - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -

    - -

    - - - - - - - h - \frac{f(a+h)-f(a)}{h} - - - -0.1 - - - - -0.01 - - - - 0.01 - - - - 0.1 - - - -

    - It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. -

    - - - - - - - - #do { - # $b = non_zero_random(-5,5,1); - # $a = random(-5,5,1); - #} until ($a + $b!= 0); - #if ($envir{problemSeed} == 1){ - $a = 2; - $b = 1; - #}; - $f = Formula("1/(x+$b)")->reduce; - $m = $f->D('x')->eval(x => $a)->with(tolerance => .01); - Context()->variables->add(h => 'Real'); - $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a)))/Formula("h"); - @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); - @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); - @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); - $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ - my ($correct,$student,$self)=@_; - my @cor = @{$correct}; - my @stu = @{$student}; - my @return = (1,1,1,1,1,1,1,1); - for my $i (0,2,4,6) { - do { - $return[$i+1] = 0; - $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); - } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); - do { - $return[$i] = 0; - $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); - } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); - }; - do { - $return[2] = 0; - $self->setMessage(3,'You already used this h-value.'); - } unless ($stu[2] != $stu[0] or $stu[2] eq ""); - do { - $return[4] = 0; - $self->setMessage(5,'You already used this h-value.'); - } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); - do { - $return[6] = 0; - $self->setMessage(7,'You already used this h-value.'); - } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); - return [@return]; - }); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x)=, - a= -

    - -     - - - h - \frac{f(a+h)-f(a)}{h} - - - - - - - - - - - - - - - - - - - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -

    - -

    - - - - - - - h - \frac{f(a+h)-f(a)}{h} - - - -0.1 - - - - -0.01 - - - - 0.01 - - - - 0.1 - - - -

    - It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. -

    - - - - - - - - #do { - # do { - # $d = non_zero_random(-5,5,1); - # $b = non_zero_random(-5,5,1); - # $c = non_zero_random(-5,5,1); - # } until (gcd($d,gcd($b,$c))==1); - # $a = random(-5,5,1); - # if ($envir{problemSeed} == 1){ - $a = -3; - $d = -4; - $b = 5; - $c = -1; - # }; - $f = Formula("$d*x^2+$b*x+$c")->reduce; - $m = $f->D('x')->eval(x => $a); - #} until ($m!=0); - Context()->variables->add(h => 'Real'); - $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a)))/Formula("h"); - @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); - @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); - @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); - $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ - my ($correct,$student,$self)=@_; - my @cor = @{$correct}; - my @stu = @{$student}; - my @return = (1,1,1,1,1,1,1,1); - for my $i (0,2,4,6) { - do { - $return[$i+1] = 0; - $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); - } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); - do { - $return[$i] = 0; - $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); - } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); - }; - do { - $return[2] = 0; - $self->setMessage(3,'You already used this h-value.'); - } unless ($stu[2] != $stu[0] or $stu[2] eq ""); - do { - $return[4] = 0; - $self->setMessage(5,'You already used this h-value.'); - } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); - do { - $return[6] = 0; - $self->setMessage(7,'You already used this h-value.'); - } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); - return [@return]; - }); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x)=, - a= -

    - -     - - - h - \frac{f(a+h)-f(a)}{h} - - - - - - - - - - - - - - - - - - - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -

    - -

    - - - - - - - h - \frac{f(a+h)-f(a)}{h} - - - -0.1 - - - - -0.01 - - - - 0.01 - - - - 0.1 - - - -

    - It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. -

    - - - - - - - - #$a = random(2,9,1); - #if ($envir{problemSeed} == 1){ - $a = 5; - #}; - $f = Formula("ln(x)"); - $m = $f->D('x')->eval(x => $a)->with(tolerance => .01); - Context()->variables->add(h => 'Real'); - $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a)))/Formula("h"); - @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); - @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); - @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); - $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ - my ($correct,$student,$self)=@_; - my @cor = @{$correct}; - my @stu = @{$student}; - my @return = (1,1,1,1,1,1,1,1); - for my $i (0,2,4,6) { - do { - $return[$i+1] = 0; - $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); - } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); - do { - $return[$i] = 0; - $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); - } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); - }; - do { - $return[2] = 0; - $self->setMessage(3,'You already used this h-value.'); - } unless ($stu[2] != $stu[0] or $stu[2] eq ""); - do { - $return[4] = 0; - $self->setMessage(5,'You already used this h-value.'); - } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); - do { - $return[6] = 0; - $self->setMessage(7,'You already used this h-value.'); - } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); - return [@return]; - }); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x)=, - a= -

    - -     - - - h - \frac{f(a+h)-f(a)}{h} - - - - - - - - - - - - - - - - - - - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -

    - -

    - - - - - - - h - \frac{f(a+h)-f(a)}{h} - - - -0.1 - - - - -0.01 - - - - 0.01 - - - - 0.1 - - - -

    - It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. -

    - - - - - - - - Context()->flags->set(reduceConstants => 0); - #$a = list_random(Formula("pi/3"),Formula("2pi/3"),Formula("pi"),Formula("4pi/3"),Formula("5pi/3"),); - #if ($envir{problemSeed} == 1){ - $a = Formula("pi"); - #}; - $f = Formula("sin(x)"); - $m = $f->D('x')->eval(x => $a->eval(x => 0))->with(tolerance => .01); - Context()->variables->add(h => 'Real'); - $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a->eval(x => 0))))/Formula("h"); - @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); - @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); - @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); - $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ - my ($correct,$student,$self)=@_; - my @cor = @{$correct}; - my @stu = @{$student}; - my @return = (1,1,1,1,1,1,1,1); - for my $i (0,2,4,6) { - do { - $return[$i+1] = 0; - $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); - } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); - do { - $return[$i] = 0; - $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); - } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); - }; - do { - $return[2] = 0; - $self->setMessage(3,'You already used this h-value.'); - } unless ($stu[2] != $stu[0] or $stu[2] eq ""); - do { - $return[4] = 0; - $self->setMessage(5,'You already used this h-value.'); - } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); - do { - $return[6] = 0; - $self->setMessage(7,'You already used this h-value.'); - } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); - return [@return]; - }); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x)=, - a= -

    - -     - - - h - \frac{f(a+h)-f(a)}{h} - - - - - - - - - - - - - - - - - - - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -

    - -

    - - - - - - - h - \frac{f(a+h)-f(a)}{h} - - - -0.1 - - - - -0.01 - - - - 0.01 - - - - 0.1 - - - -

    - It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. -

    - - - - - - - - Context()->flags->set(reduceConstants => 0); - #$a = list_random(Formula("pi/3"),Formula("2pi/3"),Formula("pi"),Formula("4pi/3"),Formula("5pi/3"),); - #if ($envir{problemSeed} == 1){ - $a = Formula("pi"); - #}; - $f = Formula("cos(x)"); - $m = $f->D('x')->eval(x => $a->eval(x => 0))->with(tolerance => .01); - Context()->variables->add(h => 'Real'); - $dq = ($f->substitute(x => Formula("$a+h"))-($f->eval(x => $a->eval(x => 0))))/Formula("h"); - @tabin = (Real(-0.1),Real(-0.01),Real(0.01),Real(0.1)); - @tabout = map{$dq->eval(h => $_, x => 1)} (@tabin); - @table = map{($tabin[$_], $tabout[$_])} (0 .. $#tabin); - $tableanswers = MultiAnswer(@table)->with(allowBlankAnswers => 1,checker => sub{ - my ($correct,$student,$self)=@_; - my @cor = @{$correct}; - my @stu = @{$student}; - my @return = (1,1,1,1,1,1,1,1); - for my $i (0,2,4,6) { - do { - $return[$i+1] = 0; - $self->setMessage($i+2, 'This is not the difference quotient when h='."$stu[$i]".'.'); - } unless ($stu[$i] ne "" and $dq->eval(h => $stu[$i]) == $stu[$i+1]); - do { - $return[$i] = 0; - $self->setMessage($i+1, 'You should use h-values '."$cor[0]".', '."$cor[2]".', '."$cor[4]".', and '."$cor[6]".'.'); - } unless ($stu[$i] == $cor[0] or $stu[$i] == $cor[2] or $stu[$i] == $cor[4] or $stu[$i] == $cor[6]); - }; - do { - $return[2] = 0; - $self->setMessage(3,'You already used this h-value.'); - } unless ($stu[2] != $stu[0] or $stu[2] eq ""); - do { - $return[4] = 0; - $self->setMessage(5,'You already used this h-value.'); - } unless ($stu[4] != $stu[0] and $stu[4] != $stu[2] or $stu[2] eq ""); - do { - $return[6] = 0; - $self->setMessage(7,'You already used this h-value.'); - } unless ($stu[6] != $stu[0] and $stu[6] != $stu[2] and $stu[6] != $stu[4] or $stu[2] eq ""); - return [@return]; - }); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x)=, - a= -

    - -     - - - h - \frac{f(a+h)-f(a)}{h} - - - - - - - - - - - - - - - - - - - - If the limit does not exist, you may type DNE. - If you need to enter \infty, you may type inf. - -

    - -

    -

    - -

    - - - - - - - h - \frac{f(a+h)-f(a)}{h} - - - -0.1 - - - - -0.01 - - - - 0.01 - - - - 0.1 - - - -

    - It appears that \lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=. -

    - - - - - - - -
    - Epsilon-Delta Definition of a Limit -

    - This section introduces the formal definition of a limit. - Many refer to this as the epsilon-delta definition, - referring to the letters \varepsilon and \delta - of the Greek alphabet. -

    - - - -

    - Before we give the actual definition, - let's consider a few informal ways of describing a limit. - Given a function y=f(x) and an x-value, - c, we say that the limit of the function f, - as x approaches c, is a value L if: -

    -
  • - Tends -

    - y tends to L - as x tends to c. -

    -
    • - -
  • - Approaches -

    - y approaches L - as x approaches c. -

    -
    • - -
  • - Near -

    - y is near L - whenever x is near c. -

    -
    • -
    -

    - -

    - The problem with these definitions is that the words - tends, approach, - and especially near are not exact. - In what way does the variable x tend to, or approach, - c? - How near do x and y have to be to c and L, - respectively? -

    - -

    - The definition we describe in this section comes from formalizing - Near. - A quick restatement gets us closer to what we want: -

    -
  • - Tolerance Levels -

    - If x is within a certain - tolerance level of c, - then the corresponding value y=f(x) is within a certain - tolerance level of L. -

    -
    • -
    -

    - -

    - The traditional notation for the x-tolerance is the lowercase Greek letter delta, - or \delta, - and the y-tolerance is denoted by lowercase epsilon, - or \varepsilon. - One more rephrasing of Tolerance Levels - nearly gets us to the actual definition: -

    -
  • - Named Tolerance Levels -

    - If x is within \delta units of c, - then the corresponding value of y is within - \varepsilon units of L. -

    -
    • -
    -

    - -

    - We can write x is within \delta units of c - mathematically as - - \abs{x-c} \lt \delta - , - which is equivalent to - - c-\delta \lt x \lt c+\delta - . -

    - - - -

    - Letting the symbol \implies - represent the word implies, - we can rewrite Named Tolerance Levels as - - \abs{x - c} \lt \delta \implies \abs{y - L} \lt \varepsilon - - or - - c - \delta \lt x \lt c + \delta \implies L - \varepsilon \lt y \lt L + \varepsilon - . -

    - -

    - The point is that \delta and \varepsilon, - being tolerances, can be any positive - (but typically small) - values satisfying this implication. - Finally, we have the formal definition of the limit with the notation seen in the previous section. -

    - - - - - The Limit of a Function <m>f</m> at a point - -

    - Let I be an open interval containing c, - and let f be a function defined on I, - except possibly at c. - The statement that the limit of f(x), - as x approaches c, - is L is denoted by - - \lim_{x\to c} f(x) = L - , - and means that given any \varepsilon \gt 0, - there exists \delta \gt 0 such that for all x in I, - where x \neq c, - if \abs{x - c} \lt \delta, - then \abs{f(x) - L} \lt \varepsilon. - - limitdefinition - -

    -     -     - - - - - - -

    - Mathematicians often enjoy writing ideas without using any words. - Here is the wordless definition of the limit: - - \lim_{x\to c} f(x) = L - \iff - \forall \, \varepsilon \gt 0, \exists \, \delta \gt 0 \text{ s.t. }0\lt \abs{x - c} \lt \delta \implies \abs{f(x) - L} \lt \varepsilon - . -

    - -

    - Note the order in which \varepsilon and \delta are given. - In the definition, the y-tolerance - \varepsilon is given first - and then the limit will exist if - we can find an x-tolerance \delta that works. -

    - -

    - An example will help us understand this definition. - Note that the explanation is long, - but it will take one through all steps necessary to understand the ideas. -

    - - - Evaluating a limit using the definition - -

    - Show that \lim\limits_{x\to 4} \sqrt{x} = 2. -

    -     - -

    - Before we use the formal definition, - let's try some numerical tolerances. - What if the y tolerance is 0.5, - or in other words \varepsilon =0.5? - How close to 4 does x have to be so that y - is within 0.5 units of 2? - That is, 1.5 \lt y \lt 2.5? - In this case, we can proceed as follows: - - 1.5 \amp\lt y \lt 2.5 - 1.5 \amp\lt \sqrt{x} \lt 2.5 \amp\amp \text{(Let } y=\sqrt{x} \text{)} - 1.5^2 \amp\lt x \lt 2.5^2\amp\amp \text{(Square the inequality)} - 2.25 \amp\lt x \lt 6.25 - 2.25-4 \amp\lt x-4 \lt 6.25-4\amp\amp \text{(Subtract }4\text{ from both sides)} - -1.75 \amp\lt x-4 \lt 2.25 - -

    - -

    - So, what is the desired x tolerance? - Remember, we want to find a \delta so that - \abs{x-4} is smaller than \delta. - Since 1.75\lt2.25, - then if we require \abs{x-4}\lt 1.75, then we have - - \abs{x-4}\lt1.75 - \implies-1.75\lt x-4\lt1.75\lt2.25 - - Therefore we can have \delta \leq 1.75. - See . -

    - -
    -
    - - -
    -
    - - - -

    - Graph of x^2 zoomed in where x = 2. There are three points on - the graph: -

      -
    1. (1.73, 3), since when x = 1.73 we get f(1.73) = 3
      2. -
    2. (2, 4), because when x = 2, f(2) = 4
      3. -
    3. (2.24, 5) when x = 2.24 we get f(2.24) = 5
      4. -
    -

    -

    - There is a vertial line showing the point (1.73, 3) is \varepsilon - from the point (2, 4), because 3 + \varepsilon = 4. Another vertical - line is used to show the point (2.24, 5) is \varepsilon from the - point 2, 4, since 5 - \varepsilon = 4. -

    -

    - There are also vertical lines at x = 2.24, x = 1.73, - x = 2.2, and x = 1.8. This creates two sets of vertical lines, - the outer set is at x = 1.73 and x = 2.24. The inner set - is at x = 1.8 and x = 2.2. The out set of lines mark where - \delta = 1, while the inner set marks - \delta = \epsilon / 5. -

    -

    - The inner set of vertical lines shows that the points at (1.73, 3) - and 2.24, 5 are not within \delta of the point 2, 4. -

    -

    - There are horizontal lines that are between the inner vertical lines - marking 2 \pm (\varepsilon / 5). -

    -     - - Graph of x squared, has 3 points. Shows that two points are within epsilon - and delta of the third point. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=0, - xmax=4, - ymin=2, - ymax=6, - xtick={2}, - ytick={4}, - ydiscontinuity - ] - \addplot+[infinite,domain=0:2.5] {x^2}; - \addplot[soliddot] coordinates {(2,4)}; - \addplot[soliddot] coordinates {(1.73,3)}; - \addplot[soliddot] coordinates {(2.24,5)}; - \addplot[guideline] coordinates {(0,3) (1.73,3)}; - \addplot[guideline] coordinates {(0,5) (2.24,5)}; - \addplot[guideline,dashed] coordinates {(2.24,0) (2.24,5)}; - \addplot[guideline,dashed] coordinates {(1.73,0) (1.73,3)}; - \addplot[guideline,dotted] coordinates {(2.2,0) (2.2,4.84)}; - \addplot[guideline,dotted] coordinates {(1.8,0) (1.8,3.24)}; - \addplot[guideline,|->] coordinates {(0.2,4) (0.2,5)} node [below right]{$\varepsilon$}; - \addplot[guideline,|->] coordinates {(0.2,4) (0.2,3)} node [above right]{$\varepsilon$}; - \addplot[guideline,<-|] coordinates {(2.2,2.2) (2,2.2)}; - \addplot[guideline,<-|] coordinates {(1.8,2.2) (2,2.2)}; - \node[pin={[pin distance=2em, pin edge={inner sep=0pt}]60:{$\delta=\varepsilon/5$}}] at (axis cs:2.1,2.2) {}; - \end{axis} - \end{tikzpicture} - - - - - -

    - In summary, given \varepsilon \gt 0, - set \delta=\varepsilon/5. - Then \abs{x-2} \lt \delta implies - \abs{x^2 - 4}\lt\varepsilon (\abs{y-4}\lt\varepsilon) as desired. - This shows that \lim_{x\to 2} x^2 = 4. - - gives a visualization of this; - by restricting x to values within \delta = \varepsilon/5 of 2, - we see that f(x) is within \varepsilon of 4. -

    - - - - -

    - Make note of the general pattern exhibited in these last two examples. - In some sense, each starts out backwards. - That is, while we want to - -

      -
    1. -

      - start with \abs{x-c}\lt \delta and conclude that -

      -
      2. - -
    2. -

      - \abs{f(x)-L}\lt \varepsilon, -

      -
      3. -
    -

    -

    - we actually start by doing what is essentially some - scratch-work first: - -

      -
    1. -

      - assume \abs{f(x)-L}\lt \varepsilon, - then perform some algebraic manipulations to give an inequality of the form -

      -
      2. - -
    2. -

      - \abs{x-c}\lt something. -

      -
      3. -
    -

    -

    - When we have properly done this, - the something on the greater than - side of the inequality becomes our \delta. - We can refer to this as the scratch-work phase of our proof. - Once we have \delta, - we can formally start the actual proof with - \abs{x-c}\lt \delta and use algebraic manipulations to conclude that - \abs{f(x)-L}\lt \varepsilon, - usually by using the same steps of our - scratch-work in reverse order. -

    - -

    - We highlight this process in the following example. -

    - - - Evaluating a limit using the definition - -

    - Prove that \lim\limits_{x\to 1}(x^3-2x) = -1. -

    -     - -

    - We start our scratch-work by considering \abs{f(x) - (-1)} \lt \varepsilon: - - \abs{f(x)-(-1)} \amp \lt \varepsilon - \abs{x^3-2x + 1}\amp \lt \varepsilon \amp\amp \text{(Now factor)} - \abs{(x-1)(x^2+x-1)}\amp \lt \varepsilon - \abs{x-1}\amp\lt\frac{\varepsilon}{\abs{x^2+x-1}} - . -

    - -

    - We are at the phase of saying that \abs{x-1}\lt something, - where \textit{something}=\varepsilon/\abs{x^2+x-1}. - We want to turn that something into \delta. -

    - -

    - Since x is approaching 1, - we are safe to assume that x is between 0 and 2. - But we need to be careful! The quadratic x^2+x-1 has roots \frac{-1\pm\sqrt{5}}{2}, - and \frac{-1+\sqrt{5}}{2}\approx 0.62 is in the interval [0,2]! - We want to divide by x^2+x-1, so to avoid any division by zero issues, - we choose a smaller interval containing 1: 0.75\lt x\lt 1.25. - So - - \frac34 \amp \lt x\lt \frac54 - \frac{9}{16}\amp \lt x^2\lt \frac{25}{16}\amp\amp \text{(Squared each term.)} - Adding the above two inequalities, we get: - \frac{21}{16}\amp \lt x^2+x\lt \frac{45}{16} - \frac{5}{16}\amp \lt x^2+x-1\lt \frac{29}{16}\amp\amp \text{(Subtracted 1 from each part.)} - -

    - -

    - In Inequality, - we wanted \abs{x-1}\lt \varepsilon/\abs{x^2+x-1}. - The above shows that given any x in (0.75,1.25), - we know that - - \abs{x^2+x-1} \amp \lt \frac{29}{16} \amp\amp \text{which implies that} - \frac{16}{29} \amp \lt \frac{1}{\abs{x^2+x-1}} \amp\amp \text{which implies that} - \frac{16\varepsilon}{29} \amp \lt \frac{\varepsilon}{\abs{x^2+x-1}}\text{.}\amp\amp - -

    - -

    - So we set \delta \lt 16\epsilon/29, on the interval (0.75, 1.25). - This ends our scratch-work, and we begin the formal proof - (which also helps us understand why this was a good choice of \delta). -

    - -

    - Given \varepsilon\gt 0, let \delta \lt \min\{1/4, 16\varepsilon/29\}. - We want to show that when \abs{x-1}\lt \delta, - then \abs{(x^3-2x)-(-1)}\lt \varepsilon. - We start with \abs{x-1}\lt \delta: - - \abs{x-1} \amp \lt \delta - \abs{x-1} \amp \lt \frac{16\varepsilon}{29} - \abs{x-1} \amp \lt \frac{\varepsilon}{\abs{x^2+x-1}} \amp\amp \text{(}Inequality, x\text{ near 1)} - \abs{x-1}\cdot \abs{x^2+x-1} \amp \lt \varepsilon - \abs{x^3-2x+1} \amp \lt \varepsilon - \abs{(x^3-2x)-(-1)} \amp \lt \varepsilon - , - which is what we wanted to show. - (Note that the condition x near 1 is satisfied by the fact that \delta\lt 1/4, - so that 3/4\lt x\lt 5/4.) - Thus \lim_{x\to 1}(x^3-2x) = -1. -

    -     - -     - - - - - -

    - We illustrate evaluating limits once more. -

    - - - Evaluating a limit using the definition - -

    - Prove that \lim\limits_{x\to 0} e^x = 1. -

    -     - -

    - Symbolically, we want to take the inequality - \abs{e^x - 1} \lt \varepsilon and unravel it to the form \abs{x-0} \lt \delta. - Here is our scratch-work: - - \amp\abs{e^x - 1}\lt \varepsilon - -\varepsilon \amp\lt e^x - 1 \lt \varepsilon\amp\amp \text{(Definition of absolute value)} - 1-\varepsilon \amp\lt e^x \lt 1+\varepsilon \amp\amp \text{(Add 1)} - \ln(1-\varepsilon) \amp\lt x \lt \ln(1+\varepsilon) \amp\amp \text{(Take natural logs)} - -

    - -

    - Making the safe assumption that - \varepsilon\lt 1 ensures the last inequality is valid (, so that \ln(1-\varepsilon) is defined). - We can then set \delta to be the minimum of - \abs{\ln(1-\varepsilon)} and \ln(1+\varepsilon); , - - \delta = \min\{\abs{\ln(1-\varepsilon)}, \ln(1+\varepsilon)\} = \ln(1+\varepsilon)\text{.} \quad \text{(See marginal note.)} - -

    - -

    - Now, we work through the actual the proof: - - \amp\abs{x - 0}\lt \delta - -\delta \amp \lt x \lt \delta \amp\amp \text{(Definition of absolute value)} - -\ln(1+\varepsilon) \amp \lt x \lt \ln(1+\varepsilon) - \ln(1-\varepsilon) \amp \lt x \lt \ln(1+\varepsilon) \amp \amp \text{(since \(\ln(1-\varepsilon) \lt -\ln(1+\varepsilon)\)). } - The above line is true by our choice of \delta and by the fact that since \abs{\ln(1-\varepsilon)} \gt \ln(1+\varepsilon) and \ln(1-\varepsilon)\lt 0, we know \ln(1-\varepsilon) \lt -\ln(1+\varepsilon ). - 1-\varepsilon \amp \lt e^x \lt 1+\varepsilon \amp \amp \text{(Exponentiate)} - -\varepsilon \amp \lt e^x - 1 \lt \varepsilon \amp\amp \text{(Subtract 1)} - -

    - -

    - In summary, given \varepsilon \gt 0, - let \delta = \ln(1+\varepsilon). - Then \abs{x - 0} \lt \delta implies \abs{e^x - 1}\lt \varepsilon as desired. - We have shown that \lim_{x\to 0} e^x = 1. -

    -     -     - -

    - We note that we could actually show that - \lim_{x\to c} e^x = e^c for any constant c. - We do this by factoring out e^c from both sides, - leaving us to show \lim_{x\to c} e^{x-c} = 1 instead. - By using the substitution u=x-c, - this reduces to showing \lim_{u\to 0} e^u = 1 which we just did in the last example. - As an added benefit, this shows that in fact the function - f(x)=e^x is continuous - at all values of x, - an important concept we will define in . -

    - -

    - This formal definition of the limit is not an easy concept grasp. - Our examples are actually easy examples, - using simple functions like polynomials, - square roots and exponentials. - It is very difficult to prove, - using the techniques given above, - that \lim_{x\to 0}\frac{\sin(x)}{x} = 1, - as we approximated in . -

    -

    - There is hope. - - shows how one can evaluate complicated limits using certain basic limits as building blocks. - While limits are an incredibly important part of calculus - (and hence much of higher mathematics), - rarely are limits evaluated using the definition. - Rather, the techniques of are employed. -

    - - - - - - - - Terms and Concepts - - - - -

    - What is wrong with the following - definition of a limit? -

    - - -
    -

    - The limit of f(x), as x approaches a, - is K means that given any \delta \gt 0 there exists - \varepsilon \gt 0 such that whenever \abs{f(x)-K}\lt\varepsilon, - we have \abs{x-a}\lt \delta. -

    -
    - -     - - - -

    - \varepsilon should be given first, - and the restriction \abs{x-a}\lt \delta implies - \abs{f(x)-K}\lt\varepsilon, not the other way around. -

    -     - -     - - - - - -

    - Which is given first in establishing a limit? -

    - -     - - - -

    - x-tolerance -

    -     -     - - -

    - y-tolerance -

    -     -     -     - -     - - - - - -

    - - \varepsilon must always be positive. -

    -     - -     - - - - - -

    - - \delta must always be positive. -

    -     - -     -     - - - Problems - - - -

    - Prove the given limit using an \varepsilon-\delta proof. -

    -     - - - - -

    - \lim\limits_{x\to 4} (2x+5) = 13 -

    - -     - - - -

    - Let \varepsilon \gt 0 be given. - We wish to find \delta \gt 0 such that when |x-4|\lt \delta, - \abs{f(x)-13}\lt \epsilon. -

    -

    - Consider \abs{f(x)-13}\lt \varepsilon: - - \abs{f(x) -13} \lt \varepsilon - \abs{(2x+5)-13}\lt \varepsilon - \abs{2x-8} \lt \varepsilon - 2\abs{x-4} \lt \varepsilon - -\varepsilon/2 \lt x-4 \lt \varepsilon/2 - . -

    -

    - This implies we can let \delta =\varepsilon/2. - Then: - - \abs{x-4}\lt \delta - -\delta \lt x-4 \lt \delta - -\varepsilon/2 \lt x-4 \lt \varepsilon/2 - -\varepsilon \lt 2x-8 \lt \varepsilon - -\varepsilon \lt (2x+5)-13 \lt \varepsilon - \abs{(2x+5) - 13} \lt \epsilon - , - which is what we wanted to prove. -

    -     - -     - - - - -

    - \lim\limits_{x\to 5}(3-x)=-2 -

    - -     - - - -

    - Let \varepsilon \gt 0 be given. - We wish to find \delta \gt 0 such that when \abs{x-5}\lt\delta, - \abs{f(x)-(-2)}\lt\varepsilon. -

    -

    - First, some preliminary investigation to find a suitable \delta. - Consider \abs{f(x)-(-2)}\lt\varepsilon: - - \abs{f(x)+2}\amp\lt\varepsilon - \abs{(3-x)+2}\amp\lt\varepsilon - \abs{5-x}\amp\lt\varepsilon - \abs{x-5}\amp\lt\varepsilon - - Since we want to start with \abs{x-5}\lt\delta, - this suggests we let \delta=\varepsilon. -

    -

    - Now we can apply the definition. - - \abs{x-5}\amp\lt\delta - \abs{x-5}\amp\lt\varepsilon - -\varepsilon\lt x-5\amp\lt\varepsilon - -\varepsilon\lt (x-3)-2\amp\lt\varepsilon - \varepsilon\gt(-x+3)-(-2)\gt\amp-\varepsilon - \abs{(3-x)-(-2)}\amp\lt\varepsilon - . - In other words, - \abs{x-5}\lt\delta implies \abs{(3-x)-(-2)}\lt\varepsilon. - This is what we needed to prove. -

    -     - -     - - - - -

    - \lim\limits_{x\to 3}\left(x^2-3\right)=6 -

    - -     - - - -

    - Let \varepsilon \gt 0 be given. - We wish to find \delta \gt 0 such that when \abs{x-3}\lt\delta, - \abs{f(x)-6}\lt\varepsilon. -

    -

    - First, some preliminary investigation to find a suitable \delta. - Consider \abs{f(x)-6}\lt\varepsilon: - - \abs{\left(x^2-3\right)-6}\amp\lt\varepsilon - \abs{x^2-9}\amp\lt\varepsilon - \abs{x-3}\cdot\abs{x+3}\amp\lt\varepsilon - \abs{x-3}\amp\lt\frac{\varepsilon}{\abs{x+3}} - - Since x is near 3, - we can safely assume that, for instance, 2\lt x\lt 4. - Thus - - 2+3\lt x+3\lt4+3 - 5\lt x+3\lt 7 - \frac{1}{7}\lt\frac{1}{x+3}\lt\frac{1}{5} - \frac{\varepsilon}{7}\lt\frac{\varepsilon}{x+3}\lt\frac{\varepsilon}{5} - - Since we need to begin the actual proof with \abs{x-3}\lt\delta, - this suggests that we take \delta=\frac{\varepsilon}{7}. -

    -

    - Now we can apply the definition. - - \abs{x-3}\amp\lt\delta - \abs{x-3}\amp\lt\frac{\varepsilon}{7} - \abs{x-3}\amp\lt\frac{\varepsilon}{\abs{x+3}} - \abs{x-3}\cdot\abs{x+3}\amp\lt\varepsilon - \abs{x^2-9}\amp\lt\varepsilon - \abs{\left(x^2-3\right)-6}\amp\lt\varepsilon - - In other words, - \abs{x-3}\lt\delta implies \abs{\left(x^2-3\right)-6}\lt\varepsilon. - This is what we needed to prove. -

    -     - -     - - - - -

    - \lim\limits_{x\to 4}\left(x^2+x-5\right)=15 -

    - -     - - - -

    - Let \varepsilon \gt 0 be given. - We wish to find \delta \gt 0 such that when \abs{x-4}\lt\delta, - \abs{f(x)-15}\lt\varepsilon. -

    -

    - First, some preliminary investigation to find a suitable \delta. - Consider \abs{f(x)-15}\lt\varepsilon: - - \abs{\left(x^2+x-5\right)-15}\amp\lt\varepsilon - \abs{x^2+x-20}\amp\lt\varepsilon - \abs{x-4}\cdot\abs{x+5}\amp\lt\varepsilon - \abs{x-4}\amp\lt\frac{\varepsilon}{\abs{x+5}} - - Since x is near 4, - we can safely assume that, for instance, 3\lt x\lt 5. - Thus - - 3+5\amp\lt x+5\lt5+5 - 8\amp\lt x+5\lt 10 - \frac{1}{10}\amp\lt\frac{1}{x+5}\lt\frac{1}{8} - \frac{\varepsilon}{10}\amp\lt\frac{\varepsilon}{x+5}\lt\frac{\varepsilon}{8} - - Since we need to begin the actual proof with \abs{x-4}\lt\delta, - this suggests that we take \delta=\frac{\varepsilon}{10}. -

    -

    - Now we can apply the definition. - - \abs{x-4}\amp\lt\delta - \abs{x-4}\amp\lt\frac{\varepsilon}{10} - \abs{x-4}\amp\lt\frac{\varepsilon}{\abs{x+5}} - \abs{x-4}\cdot\abs{x+5}\amp\lt\varepsilon - \abs{x^2+x-20}\amp\lt\varepsilon - \abs{\left(x^2+x-5\right)-15}\amp\lt\varepsilon - - In other words, - \abs{x-4}\lt\delta implies \abs{\left(x^2+x-5\right)-15}\lt\varepsilon. - This is what we needed to prove. -

    -     - -     - - - - -

    - \lim\limits_{x\to 1} \left(2x^2+3x+1\right) = 6 -

    - -     - - - -

    - Let \varepsilon \gt 0 be given. - We wish to find \delta \gt 0 such that when \abs{x-1}\lt \delta, - \abs{f(x)-6}\lt \varepsilon. -

    -

    - First, some prelimary investigation to find a suitable \delta. - Consider \abs{f(x)-6}\lt \varepsilon, - keeping in mind we want to make a statement about \abs{x-1}: - - \abs{f(x) -6} \lt \varepsilon - \abs{(2x^2+3x+1)-6}\lt \varepsilon - \abs{2x^2+3x-5} \lt \varepsilon - \abs{2x+5}\cdot\abs{x-1} \lt \varepsilon - \abs{x-1} \lt \varepsilon/\abs{2x+5} - -

    -

    - Since x is near 1, we can safely assume that, - for instance, 0\lt x\lt 2. - Thus - - 0+5\lt 2x+5\lt 4+5 - 5 \lt 2x+5 \lt 9 - \frac{1}{9} \lt \frac{1}{2x+5} \lt \frac{1}{5} - \frac{\varepsilon}{9} \lt \frac{\varepsilon}{2x+5} \lt \frac{\varepsilon}{5} - -

    -

    - Let \delta =\frac{\varepsilon}{9}. - Then: - - \abs{x-1}\lt \delta - \abs{x-1} \lt \frac{\varepsilon}{9} - \abs{x-1} \lt \frac{\varepsilon}{2x+5} - \abs{x-1}\cdot\abs{2x+5} \lt \frac{\epsilon}{2x+5}\cdot\abs{2x+5} - -

    -

    - Assuming x is near 1, 2x+5 is positive and we can drop the absolute value signs on the right. - - \abs{x-1}\cdot\abs{2x+5} \lt \frac{\varepsilon}{2x+5}\cdot(2x+5) - \abs{2x^2+3x-5} \lt \varepsilon - \abs{(2x^2+3x+1) -6} \lt \varepsilon - , - which is what we wanted to prove. -

    -     - -     - - - - -

    - \lim\limits_{x\to 2}\left(x^3-1\right)=7 -

    - -     - - - -

    - Let \varepsilon \gt 0 be given. - We wish to find \delta \gt 0 such that when \abs{x-2}\lt\delta, - \abs{f(x)-7}\lt\varepsilon. -

    -

    - First, some preliminary investigation to find a suitable \delta. - Consider \abs{f(x)-7}\lt\varepsilon: - - \abs{\left(x^3-1\right)-7}\amp\lt\varepsilon - \abs{x^3-8}\amp\lt\varepsilon - \abs{x-2}\cdot\abs{x^2+2x+4}\amp\lt\varepsilon - \abs{x-2}\amp\lt\frac{\varepsilon}{\abs{x^2+2x+4}} - - Since x is near 2, - we can safely assume that, for instance, 1\lt x\lt 3. - Thus - - 1^2+2(1)+4\amp\lt x^2+2x+4\lt3^2+2(3)+4 - 7\amp\lt x^2+2x+4\lt 19 - \frac{1}{19}\amp\lt\frac{1}{x^2+2x+4}\lt\frac{1}{7} - \frac{\varepsilon}{19}\amp\lt\frac{\varepsilon}{x^2+2x+4}\lt\frac{\varepsilon}{7} - - Since we need to begin the actual proof with \abs{x-2}\lt\delta, - this suggests that we take \delta=\frac{\varepsilon}{19}. -

    -

    - Now we can apply the definition. - - \abs{x-2}\amp\lt\delta - \abs{x-2}\amp\lt\frac{\varepsilon}{19} - \abs{x-2}\amp\lt\frac{\varepsilon}{\abs{x^2+2x+4}} - \abs{x-2}\cdot\abs{x^2+2x+4}\amp\lt\varepsilon - \abs{x^3-8}\amp\lt\varepsilon - \abs{\left(x^3-1\right)-7}\amp\lt\varepsilon - - In other words, - \abs{x-2}\lt\delta implies \abs{\left(x^3-1\right)-7}\lt\varepsilon. - This is what we needed to prove. -

    -     - -     - - - - -

    - \lim\limits_{x\to 2}5=5 -

    - -     - - - -

    - Let \varepsilon \gt 0 be given. - We wish to find \delta \gt 0 such that when \abs{x-2}\lt\delta, - \abs{f(x)-5}\lt\varepsilon. -

    -

    - First, some preliminary investigation to find a suitable \delta. - Consider \abs{f(x)-5}\lt\varepsilon: - - \abs{5-5}\amp\lt\varepsilon - \abs{0}\amp\lt\varepsilon - - Well, this is just plain true no matter what \varepsilon is - (as long as its positive). - So it really doesn't matter what \delta is. -

    -

    - Now we can apply the definition. - - \abs{x-2}\amp\lt\delta - \abs{0}\amp\lt\varepsilon - \abs{5-5}\amp\lt\varepsilon - - In other words, - \abs{x-2}\lt\delta (vacuously) implies \abs{5-5}\lt\varepsilon. - This is what we needed to prove. -

    -     - -     - - - - -

    - \lim\limits_{x\to 0}\left(e^{2x}-1\right)=0 -

    - -     - - - -

    - Let \varepsilon \gt 0 be given. - We wish to find \delta \gt 0 such that when \abs{x-0}\lt\delta, - \abs{f(x)-0}\lt\varepsilon. -

    -

    - First, some preliminary investigation to find a suitable \delta. - Consider \abs{f(x)-0}\lt\varepsilon: - - \abs{\left(e^{2x}-1\right)-0}\amp\lt\varepsilon - \abs{e^{2x}-1}\amp\lt\varepsilon - -\varepsilon\amp\lt e^{2x}-1\lt\varepsilon - 1-\varepsilon\amp\lt e^{2x}\lt1+\varepsilon - \ln(1-\varepsilon)\amp\lt 2x\lt\ln(1+\varepsilon) - \frac{1}{2}\ln(1-\varepsilon)\amp\lt x\lt\frac{1}{2}\ln(1+\varepsilon) - \frac{1}{2}\ln(1-\varepsilon)\amp\lt x-0\lt\frac{1}{2}\ln(1+\varepsilon) - - Since we will need to start with \abs{x-0}\lt\delta, - this suggests we take \delta=\min\left\{\frac{1}{2}\abs{\ln(1-\varepsilon)},\frac{1}{2}\ln(1+\varepsilon)\right\}=\frac{1}{2}\ln(1+\varepsilon). -

    -

    - Now we can apply the definition. - - \abs{x-0}\amp\lt\delta - \abs{x}\amp\lt\frac{1}{2}\ln(1+\varepsilon)\lt\frac{1}{2}\abs{\ln(1-\varepsilon)} - \frac{1}{2}\ln(1-\varepsilon)\amp\lt x\lt\frac{1}{2}\ln(1+\varepsilon) - \ln(1-\varepsilon)\amp\lt 2x\lt\ln(1+\varepsilon) - 1-\varepsilon\amp\lt e^{2x}\lt1+\varepsilon - -\varepsilon\amp\lt e^{2x}-1\lt\varepsilon - \abs{e^{2x}-1}\amp\lt\varepsilon - \abs{\left(e^{2x}-1\right)-0}\amp\lt\varepsilon - - In other words, - \abs{x-0}\lt\delta implies \abs{\left(e^{2x}-1\right)-0}\lt\varepsilon. - This is what we needed to prove. -

    -     - -     - - - - -

    - \lim\limits_{x\to 1} \frac1x = 1 -

    - -     - - - -

    - Let \epsilon \gt 0 be given. - We wish to find \delta \gt 0 such that when |x-1|\lt \delta, - |f(x)-1|\lt \epsilon. -

    -

    - Consider |f(x)-1|\lt \epsilon, - keeping in mind we want to make a statement about |x-1|: - - |f(x) -1 | \lt \epsilon - |1/x-1 |\lt \epsilon - | (1-x)/x | \lt \epsilon - | x-1 |/|x| \lt \epsilon - | x-1 | \lt \epsilon\cdot|x| - -

    -

    - Since x is near 1, we can safely assume that, - for instance, 1/2\lt x\lt 3/2. - Thus - \epsilon/2 \lt \epsilon\cdot x. -

    -

    - Let \delta =\frac{\epsilon}{2}. - Then: - - \abs{x-1}\lt \delta - \abs{x-1} \lt \frac{\epsilon}{2} - \abs{x-1} \lt \epsilon\cdot x - \abs{x-1}/x \lt \epsilon - -

    -

    - Assuming x is near 1, x is positive and we can bring it into the absolute value signs on the left. - - \abs{(x-1)/x} \lt \epsilon - \abs{1-1/x} \lt \epsilon - \abs{(1/x) -1} \lt \epsilon - , - which is what we wanted to prove. -

    -     - -     - - - - -

    - \lim\limits_{x\to 0} \sin(x)= 0 -

    - -     - - - -

    - Use the fact that \abs{\sin(x)} \leq \abs{x}, - with equality only when x=0. -

    -     - -

    - Let \varepsilon \gt 0 be given. - We wish to find \delta \gt 0 such that when \abs{x-0}\lt\delta, - \abs{f(x)-0}\lt\varepsilon. -

    -

    - First, some preliminary investigation to find a suitable \delta. - Consider \abs{f(x)-0}\lt\varepsilon: - - \abs{\sin(x)-0}\amp\lt\varepsilon - \abs{\sin(x)}\amp\lt\varepsilon - - Using the hint that \abs{\sin(x)} \leq \abs{x}, - then if \abs{x}\lt0, it is automatic that \abs{\sin(x)}\lt0. - This suggests we take \delta=\varepsilon. -

    -

    - Now we can apply the definition. - - \abs{x-0}\amp\lt\delta - \abs{x}\amp\lt\varepsilon - \abs{\sin(x)}\amp\lt\varepsilon - \abs{\sin(x)-0}\amp\lt\varepsilon - - In other words, - \abs{x-0}\lt\delta implies \abs{\sin(x)-0}\lt\varepsilon. - This is what we needed to prove. -

    -     - -     -     -     -     - -
    - Finding Limits Analytically - -

    - In - we explored the concept of the limit without a strict definition, - meaning we could only make approximations. - In the previous section we gave the definition of the limit and demonstrated how to use it to verify our approximations were correct. - Thus far, our method of finding a limit is -

      -
    1. - make a really good approximation either graphically or numerically, and -
      2. -
    2. - verify our approximation is correct using a - \varepsilon-\delta proof. -
      3. -
    -

    - -

    - Recognizing that \varepsilon-\delta proofs are cumbersome, - this section gives a series of theorems which allow us to find limits much more quickly and intuitively. -

    - -

    - Suppose that \lim_{x\to 2} f(x)=2 and \lim_{x\to 2} g(x) = 3. - What is \lim_{x\to 2}(f(x)+g(x))? - Intuition tells us that the limit should be 5, - as we expect limits to behave in a nice way. - The following theorem states that already established limits do behave nicely. -

    - - - Basic Limit Properties - -

    - Let b, c, L and K be real numbers, - let n be a positive integer, - and let f and g be functions defined on an open interval I containing c with the following limits: - - limitproperties - - - \lim_{x\to c}f(x)\amp = L\amp \lim_{x\to c} g(x)\amp = K - . -

    - -

    - The following limits hold. -

    -
  • - Constants -

    - \lim\limits_{x\to c} b = b -

    -
    • -
  • - Identity -

    - \lim\limits_{x\to c} x = c -

    -
    • -
  • - Sums/Differences -

    - \lim\limits_{x\to c}(f(x)\pm g(x)) = L\pm K -

    -
    • -
  • - Scalar Multiples -

    - \lim\limits_{x\to c}(b\cdot f(x)) = bL -

    -
    • -
  • - Products -

    - \lim\limits_{x\to c} (f(x)\cdot g(x)) = LK -

    -
    • -
  • - Quotients -

    - \lim\limits_{x\to c} (f(x)/g(x)) = L/K, when K\neq 0 -

    -
    • -
  • - Powers -

    - \lim\limits_{x\to c} f(x)^n = L^n -

    -
    • -
  • - Roots -

    - \lim\limits_{x\to c} \sqrt[n]{f(x)} = \sqrt[n]{L} -

    - -

    - (If n is even then require f(x)\geq 0 on I.) -

    -
    • -
  • - Compositions -

    - Adjust the limit requirements to - - \lim_{x\to c}f(x)\amp=L\amp\lim_{x\to L}g(x)\amp=K\amp g(L)\amp=K - . - Then \lim\limits_{x\to c}g(f(x)) = K. -

    -
    • -
    -

    -     -     - - - - - - -

    - We apply the theorem to an example. -

    - - - Using basic limit properties - -

    - Let - - \lim_{x\to 2} f(x)\amp=2\amp\lim_{x\to 2} g(x)\amp= 3\amp p(x)\amp = 3x^2-5x+7 - . - Find the following limits: -

      -
    1. \lim\limits_{x\to 2}(f(x) + g(x))
      2. -
    2. \lim\limits_{x\to 2}(5f(x) + g(x)^2)
      3. -
    3. -

      - \lim\limits_{x\to 2}p(x) -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - Using the property, - we know that - - \lim_{x\to 2}(f(x) + g(x)) \amp = \lim_{x\to 2}f(x) + \lim_{x\to 2}g(x) - \amp = 2+3 =5 - . -

      -
      2. -
    2. -

      - Using the , - , - and properties, - we find that - - \lim_{x\to 2}(5f(x) + g(x)^2) \amp = \lim_{x\to 2}(5f(x))+\lim_{x\to 2}(g(x)^2) - \amp = 5\lim_{x\to 2}f(x) + \mathopen{}\left(\lim_{x\to 2}g(x)\right)^2\mathclose{} - \amp = 5\cdot 2 + 3^2 = 19 - . -

      -
      3. -
    3. -

      - Here we combine the , - , - and properties. - We show quite a few steps, but in general these can be omitted: - - \lim_{x\to 2} p(x) \amp = \lim_{x\to 2}\left(3x^2-5x+7\right) - \amp = \lim_{x\to 2}\mathopen{}\left(3x^2\right)\mathclose{}-\lim_{x\to 2}(5x)+\lim_{x\to 2}7 - \amp = 3\bigl(\lim_{x\to 2}x\bigr)^2-5\lim_{x\to 2}(x) +7 - \amp = 3\cdot 2^2 - 5\cdot 2+7 - \amp = 9 - . -

      -
      4. -
    -

    -     - -     - -

    - Part - of the previous example demonstrates how the limit of a quadratic polynomial - can be determined using the properties of - . - Not only that, recognize that - - \lim_{x\to 2} p(x) = 9 = p(2); - - , the limit at 2 could have been found just by plugging 2 - into the function. This holds true for all polynomials, - and also for rational functions - (which are quotients of polynomials), - as stated in the following theorem. -

    - - - - Limits of Polynomial and Rational Functions - -

    - Let p(x) and q(x) be polynomials and c a real number. - Then: - - limitof polynomial functions - limitof rational functions - continuityof polynomial functions - continuityof rational functions - polynomial functioncontinuity of - rational functioncontinuity of - -

      -
    1. -

      - \lim\limits_{x\to c} p(x) = p(c) -

      -
      2. -
    2. -

      - \lim\limits_{x\to c} \frac{p(x)}{q(x)} = \frac{p(c)}{q(c)}, - when q(c) \neq 0. -

      -
      3. -
    -

    -     -     - - - - - - - Finding a limit of a rational function - -

    - Using , find - - \lim_{x\to -1} \frac{3x^2-5x+1}{x^4-x^2+3} - . -

    -     - -

    - Using , we can quickly state that - - \lim_{x\to -1}\frac{3x^2-5x+1}{x^4-x^2+3} \amp = \frac{3(-1)^2-5(-1)+1}{(-1)^4-(-1)^2+3} - \amp = \frac{9}{3} =3 - . -

    -     -     - -

    - It was likely frustrating in - to do a lot of work with \varepsilon and \delta to prove that - - \lim_{x\to 2} x^2 = 4 - - as it seemed fairly obvious. - The previous theorems state that many functions behave in such an - obvious fashion, - as demonstrated by the rational function in - . -

    - -

    - Polynomial and rational functions are not the only functions to behave in such a predictable way. - The following theorem gives a list of functions whose behavior is particularly - nice in terms of limits. - In , - we will give a formal name to these functions that behave nicely. -

    - - - - - - Limits of Common Functions - -

    - Let c be a real number in the domain of the given function and let - n be a positive integer. - The following limits hold: - - limitof trigonometric functions - limitof exponential functions - limitof logarithmic functions - continuityof trigonometric functions - continuityof exponential functions - continuityof logarithmic functions - trigonometric functioncontinuity of - exponential functioncontinuity of - logarithmic functioncontinuity of - -

      -
    1. \lim\limits_{x\to c}\sin(x) = \sin(c)
      2. - -
    2. \lim_{x\to c}\cos(x) = \cos(c)
      3. - -
    3. \lim_{x\to c}\tan(x)= \tan(c)
      4. - -
    4. \lim_{x\to c}\csc(x) = \csc(c)
      5. - -
    5. \lim_{x\to c}\sec(x) = \sec(c)
      6. - -
    6. \lim_{x\to c}\cot(x) = \cot(c)
      7. - -
    7. \lim_{x\to c}a^x = a^c, if a \gt 0
      8. - -
    8. \lim_{x\to c}\ln(x) = \ln(c)
      9. - -
    9. \lim_{x\to c}\sqrt[n]{x} = \sqrt[n]{c}
      10. -
    -

    -     -     - - - Evaluating limits analytically - -

    - Evaluate the following limits. -

      -
    1. \lim\limits_{x\to \pi} \cos(x)
      2. -
    2. \lim\limits_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right)
      3. -
    3. \lim\limits_{x\to \pi/2}(\cos(x)\sin(x))
      4. -
    4. \lim\limits_{x\to 1} e^{\ln(x)}
      5. -
    5. \lim\limits_{x\to 0} \dfrac{\sin(x)}{x}
      6. -
    -

    -     - -

    -

      -
    1. -

      - This is a straightforward application of - : - \lim\limits_{x\to \pi} \cos(x) = \cos(\pi) = -1. -

      -
      2. -
    2. -

      - We can approach this in at least two ways. - First, by directly applying - , we have: - - \lim_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right) = \sec^2(3)-\tan^2(3) - . - Using the Pythagorean Theorem, - this last expression is 1; therefore - - \lim_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right) = 1 - . -

      -

      - We can also use the Pythagorean Theorem from the start. - - \lim_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right) = \lim_{x\to 3} 1 = 1 - , - using the rule. - Either way, we find the limit is 1. -

      -
      3. -
    3. -

      - Applying the rule and - gives - - \lim\limits_{x\to \pi/2} \cos(x)\sin(x) = \cos(\pi/2)\sin(\pi/2) = 0\cdot 1 = 0 - . -

      -
      4. -
    4. -

      - Again, we can approach this in two ways. - First, we can use the exponential/logarithmic identity that - e^{\ln(x)} = x and evaluate - \lim\limits_{x\to 1} e^{\ln(x)} = \lim\limits_{x\to 1} x = 1. -

      - -

      - We can also use the rule. - Using , - we have \lim\limits_{x\to 1}\ln(x) = \ln(1) = 0 and - \lim_{x\to 0} e^x= e^0=1, - satisfying the conditions of the - rule. - Applying this rule, - - \lim_{x\to 1} e^{\ln(x)} = e^{\lim_{x\to 1} \ln(x)}=e^{\ln(1)} = e^0 = 1 - . - Both approaches are valid, giving the same result. -

      -
      5. - -
    5. -

      - We encountered this limit in . - Applying our theorems, we attempt to find the limit as - - \lim_{x\to 0}\frac{\sin(x)}{x}\rightarrow \frac{\sin(0) }{0} - , - which is of the form \frac{0}{0}. - This, of course, - violates a condition of the - rule, - as the limit of the denominator is not allowed to be 0. - Therefore, we are still unable to evaluate this limit with tools we currently have at hand. -

      -
      6. -
    -

    -     -     - -

    - Based on what we've done so far, this section could have been titled - Using Known Limits to Find Unknown Limits. - By knowing certain limits of functions, - we can find limits involving sums, products, - powers, , of these functions. - We further the development of such comparative tools with the Squeeze Theorem, - a clever and intuitive way to find the value of some limits. -

    - -

    - Before stating this theorem formally, - suppose we have functions f, - g, and h where g always takes on values between f and h; - that is, for all x in an interval, - - f(x) \leq g(x) \leq h(x) - . -

    - -

    - If f and h have the same limit at c, - and g is always squeezed between them, - then g must have the same limit as well. - That is what the Squeeze Theorem states. - This is illustrated in . -

    - - - Squeeze Theorem - -

    - Let f, g, and h be functions on an open interval I - containing c such that for all x in I, - - f(x)\leq g(x) \leq h(x) - . - - limitSqueeze Theorem - Squeeze Theorem - - If - - \lim_{x\to c} f(x) = L = \lim_{x\to c} h(x) - , - then - - \lim_{x\to c} g(x) = L - . -

    -     -     - -
    -
    - - -

    - An illustration of the squeeze theorem. There are graphs of three functions shown, labeled h(x), - g(x), and f(x). On the yaxis, there is a marker at y = 4, - labeled L and on the xaxis there is a marker at x = 5, labeled - c. -

    -

    - For all values of x \leq c f(x) \leq L and h(x) \geq L. - For all values of x f(x) \leq g(x) \leq h(x), that is, the graph - of the function g(x) lies between the graphs of the functions g(x) - and f(x). -

    -

    - The image shows that where x = c, f(x) and h(x) converge - on y = L = 4. Because f(x) \leq g(x) \leq h(x), we can extrapolate - that \lim\limits_{x\to \c} g(x) = L too. -

    -     - - Graph that illustrates the squeeze theorem using functions h, g, and f. - - - \begin{tikzpicture} - \begin{axis}[ - xtick={5}, - ytick={4}, - xticklabel={$c$}, - yticklabel={$L$} - ] - \addplot+[smooth] coordinates {(1,3) (2,5) (3,5) (4,4.3) (5,4) (6,3.7) (7,4.5) (8,5) (9,6)} node[pos=0.3, above] {$g$}; - \addplot+[smooth] coordinates {(1,1) (2,1.5) (3,2.5) (4,3.9) (5,4) (6,3.5) (7,4) (8,4.1) (9,5.3)} node[pos=0.25, above left] {$f$}; - \addplot+[smooth] coordinates {(1,6.5) (2,6.9) (3,6.5) (4,5) (5,4) (6,4.2) (7,5) (8,5.3) (9,7)} node[pos=0.25, above right] {$h$}; - \end{axis} - \end{tikzpicture} - - - - - - -

    - It can take some work to figure out appropriate functions by which to - squeeze a given function. - However, that is generally the only place where work is necessary; - the theorem makes the evaluating the limit part very simple. -

    - - - - - -

    - We use the - in the following example to finally prove that - \lim\limits_{x\to 0} \frac{\sin(x)}{x} = 1. -

    - - - Using the Squeeze Theorem - -

    - Use the to show that - - \lim_{x\to 0} \frac{\sin(x)}{x} = 1 - . -

    -     - -

    - We begin by considering the unit circle. - Each point on the unit circle has coordinates - (\cos(\theta),\sin(\theta)) for some angle \theta as shown in - . - Using similar triangles, - we can extend the line from the origin through the point to the point - (1,\tan(\theta)), as shown. - (Here we are assuming that 0\leq \theta \leq \pi/2. - Later we will show that we can also consider \theta \leq 0.) -

    - -
    -
    - - - -

    - Illustration of a unit circle, with a right and acute triangle in - quadrant one. The right triangle purtudes out of the unit circle, - while the acute triangle stays within the bounds of the circle. - The acute triangle is also contained within the right triangle. - There are horizontal lines and vertical lines inside the circle, - these lines act only as guides. - The horizontal and vertical lines intersect at (0, 0). - The horizontal line starts at (-1, 0) and ends at (1, 0), - the vertical line starts at (0, 1) and ends at (0, -1). -

    -

    - The right triangle is made up of the points (0, 0), - (1, \tan(\theta)), and (1, 0). The acute Triangle is made - of the points (0, 0), (\cos(\theta), \sin(\theta)), and - (1, 0). The angle at point (0, 0) is labeled as \theta -

    -     - - Illustration of a unit circle, with a right and acute triangle in - the first quadrant. - - - \begin{tikzpicture} - \begin{axis}[ - compat=1.5.1, - clip=false, - ticks=none, - xmin=-1.1, xmax=1.1, - ymin=-1.1, ymax=1.1, - xlabel={}, - ylabel={}, - axis equal image - ] - \draw (axis cs:0,0) circle [radius=1]; - \draw (axis cs:0,0) node[shift={(22.5:25)}]{$\theta$} -- (axis cs:1,0.839) node[above]{$(1,\tan(\theta) )$} -- (axis cs:1,0) node[below right]{$(1,0)$} -- (axis cs:0.766,0.643) node [left]{$(\cos(\theta) ,\sin(\theta))$}; - \addplot[soliddot] coordinates {(1,0.839) (1,0) (0.766,0.643)}; - \end{axis} - \end{tikzpicture} - - - - - -

    - - shows three regions have been constructed in the first quadrant, - two triangles and a sector of a circle, - which are also drawn below. - The area of the large triangle is \frac{1}{2}\tan(\theta); - the area of the sector is \theta/2; - the area of the triangle contained inside the sector is - \frac{1}{2}\sin(\theta). - It is then clear from that - - \frac{\tan(\theta)}{2}\geq\frac{\theta}{2}\geq\frac{\sin(\theta)}{2} - . - (You may need to recall that the area of a sector of a circle is - \frac{1}{2}r^2 \theta with \theta measured in radians.) -

    - -
    -
    - - -
    -
    - - - -

    - Graph of the function (x^2-1)/(x-1), showing the region with - y from 0 to 3 and x from 0 - to 2. The graph is a straight line, with slope 1 and a y intercept at y = 1, - except for a hole at the point (1,2), since the function is undefined at x = 1. -

    -     - - Graph of the polynomial x squared minus 1 divided by the polynomial x minus 1. - It is the same as the line y=x+1, except that it is undefined at x = 1. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.1, - xmax=2.2, - ymin=-.1, - ymax=3.2, - ] - \addplot+[] coordinates {(0,1) (2,3)}; - \addplot[hollowdot] coordinates {(1,2)}; - \end{axis} - \end{tikzpicture} - - - - - -

    - Clearly \lim\limits_{x\to 1}(x+1) = 2. - Recall that when considering limits, - we are not concerned with the value of the function at 1, - only the value the function approaches as x approaches 1. - Since (x^2-1)/(x-1) and x+1 are the same at all points except at x=1, - they both approach the same value as x approaches 1. - Therefore we can conclude that - - \lim_{x \to 1} \frac{x^2-1}{x-1} \amp =\lim_{x \to 1} (x+1) - \amp =2 - -

    - - - -

    - The key to - is that the functions y=(x^2-1)/(x-1) and y=x+1 are identical except at x=1. - Since limits describe a value the function is approaching, - not the value the function actually attains, - the limits of the two functions are always equal. -

    - - - Limits of Functions Equal At All But One Point - -

    - Let g(x) = f(x) for all x in an open interval, - except possibly at c, - and let \lim\limits_{x\to c} g(x) = L for some real number L. - Then - - \lim_{x\to c}f(x) = L - . -

    -     -     - -

    - The Fundamental Theorem of Algebra tells us that when dealing with a rational function of the form - g(x)/f(x) and directly evaluating the limit - \lim\limits_{x\to c} \frac{g(x)}{f(x)} returns 0/0, - then (x-c) is a factor of both g(x) and f(x). - One can then use algebra to factor this binomial out, - cancel, then apply . - We demonstrate this once more. -

    - - - Evaluating a limit using <xref ref="thm_limit_allbut1"/> - -

    - Evaluate - - \lim\limits_{x\to 3} \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15} - . -

    -     - -

    - We attempt to apply - by substituting 3 for x. - This returns the familiar indeterminate form of 0/0. - Since the numerator and denominator are each polynomials, - we know that (x-3) is factor of each. - Using whatever method is most comfortable to you, - factor out (x-3) from each - (using polynomial division, - synthetic division, a computer algebra system, etc.). - We find that - - \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15} = \frac{(x-3)\left(x^2+x-2\right)}{(x-3)\left(2 x^2+9 x-5\right)} - . - We can cancel the (x-3) factors as long as x\neq 3. - Using we conclude: - - \lim_{x\to 3} \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15} \amp = \lim_{x\to 3}\frac{(x-3)\left(x^2+x-2\right)}{(x-3)\left(2 x^2+9 x-5\right)} - \amp =\lim_{x\to 3} \frac{x^2+x-2}{2 x^2+9 x-5} - \amp = \frac{10}{40} - \amp = \frac{1}{4} - . -

    -     - -     - - - Evaluating a Limit with a Hole - -

    - Evaluate - - \lim\limits_{x\to 9} \frac{\sqrt{x}-3}{x-9} - . -

    -     - -

    - We begin by trying to apply the limit rule, - but the denominator evaluates to zero. - In fact, this limit is of the indeterminate form 0/0. - We will do some algebra to resolve the indeterminate form. - In this case, - we multiply the numerator and denominator by the conjugate of the numerator. - - \frac{\sqrt{x}-3}{x-9} \amp = \frac{\sqrt{x}-3}{x-9} \cdot \frac{\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)} - \amp= \frac{x-9}{(x-9)(\sqrt{x}+3)} - - We can cancel the (x-9) factors as long as x\neq 9. - Using we conclude: - - \lim_{x\to 9} \frac{\sqrt{x}-3}{x-9} \amp =\lim_{x\to 9} \frac{x-9}{(x-9)\left(\sqrt{x}+3\right)} - \amp = \lim_{x\to 9 }\frac{1}{\sqrt{x}+3} - \amp = \frac{1}{\lim_{x\to 9}\sqrt{x}+\lim_{x \to 9}3} - \amp = \frac{1}{\sqrt{\lim_{x\to 9}x}+3} - \amp = \frac{1}{\sqrt{3+3}} - \amp = \frac{1}{6} - . -

    -     - -     - - - -

    - We end this section by revisiting a limit first seen in - , - a limit of a difference quotient. - Let f(x) = -1.5x^2+11.5x; - we approximated the limit - \lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h}\approx 8.5. - We formally evaluate this limit in the following example. -

    - - - Evaluating the limit of a difference quotient - -

    - Let f(x) = -1.5x^2+11.5x; - find \lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h}. -

    -     - -

    - Since f is a polynomial, - our first attempt should be to employ - and substitute 0 for h. - However, we see that this gives us 0/0. - Knowing that we have a rational function hints that some algebra will help. - Consider the following steps: - - \lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\amp = \lim_{h\to 0}\frac{-1.5(1+h)^2 + 11.5(1+h) - \left(-1.5(1)^2+11.5(1)\right)}{h} - \amp =\lim_{h\to 0}\frac{-1.5(1+2h+h^2) + 11.5+11.5h - 10}{h} - \amp =\lim_{h\to 0}\frac{-1.5h^2 +8.5h}{h} - \amp =\lim_{h\to 0}\frac{h(-1.5h+8.5)}h - \amp =\lim_{h\to 0}(-1.5h+8.5) \quad (\text{using } , \text{ as } h\neq 0 ) - \amp =8.5 \quad (\text{using } ) - - This matches our previous approximation. -

    -     -     - -

    - This section contains several valuable tools for evaluating limits. - One of the main results of this section is - ; - it states that many functions that we use regularly behave in a very nice, - predictable way. - In - we give a name to this nice behavior; - we label such functions as continuous. - Defining that term will require us to look again at what a limit is and what causes limits to not exist. -

    - - - - - Terms and Concepts - - - - -

    - Explain in your own words, - without using \varepsilon-\delta formality, - why \lim\limits_{x\to c}b=b. -

    - -     - - -     - - - - -

    - Explain in your own words, - without using \varepsilon-\delta formality, - why \lim\limits_{x\to c}x=c. -

    - -     - - -     - - - - -

    - What does the text mean when it says that certain functions' - behavior is nice in terms of limits? - What, in particular, is nice? -

    - -     - - -     - - - -

    - Sketch a graph that visually demonstrates the Squeeze Theorem. -

    -     - -     - - - - -

    - You are given the following information: - - \lim_{x\to 1}f(x)\amp=0\amp\lim_{x\to 1}g(x)\amp=0\amp\lim_{x\to 1}\frac{f(x)}{g(x)}\amp=2 - - What can be said about the relative sizes of f(x) and g(x) as x approaches 1? -

    - -     - - -

    - As x is near 1, - both f and g are near 0, - but f is approximately twice the size of g. (That is, - f(x)\approx2g(x).) -

    -     - -     - - - - - -

    - Consider the following logarithmic limit: -

    -     - - -

    - - \lim\limits_{x\to 1}\ln(x) = 0. -

    -     - -

    - Look for a list of limits of common functions in this section. - Then, review the known values for the logarithm. -

    -     -     - - -

    - Use a theorem to defend your answer to the first part. -

    -     - - - - - -

    - True. We know that limits involving \ln(x) - can be evaluated by direct substitution; therefore, - \lim_{x\to 1}\ln x = \ln 1 =0. -

    -     -     - -     -     - - - Problems - - - -

    - Use the following information to evaluate the given limit, when possible. - - \lim\limits_{x\to9}f(x)\amp=6\amp\lim\limits_{x\to6}f(x)\amp=9\amp f(9)\amp=6 - \lim\limits_{x\to9}g(x)\amp=3\amp\lim\limits_{x\to6}g(x)\amp=3\amp g(6)\amp=3 - -

    -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 9; - $q = 6; - $r = 3; - %Lf = ($p => $q, $q => $p); - %Lg = ($p => $r, $q => $r); - %f = ($p => $q); - %g = ($q => $r); - $a = $p; - if (exists $Lf{$a} and exists $Lg{$a}) - {$L=$Lf{$a}+$Lg{$a};} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }(f(x)+g(x)) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 9; - $q = 6; - $r = 3; - %Lf = ($p => $q, $q => $p); - %Lg = ($p => $r, $q => $r); - %f = ($p => $q); - %g = ($q => $r); - $a = $p; - if (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} != 0)) - {$L=3*$Lf{$a}/$Lg{$a};} - elsif (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} == 0) and ($Lf{$a} != 0)) - {$L=Compute('does not exist');} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }\left(\frac{3f(x)}{g(x)}\right) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     - -

    - Using the constant and quotient rules, - - \lim\limits_{x\to }\left(\frac{3f(x)}{g(x)}\right)\amp = - \frac{\lim\limits_{x\to }(3f(x))}{\lim\limits_{x\to }g(x)} - \amp = \frac{3\lim\limits_{x\to }f(x)}{\lim\limits_{x\to }g(x)} - \amp \frac{3(6)}{3} = 6 - . -

    -     -     -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 9; - $q = 6; - $r = 3; - %Lf = ($p => $q, $q => $p); - %Lg = ($p => $r, $q => $r); - %f = ($p => $q); - %g = ($q => $r); - $a = $p; - if (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} != 0)) - {$L=($Lf{$a}-2*$Lg{$a})/$Lg{$a};} - elsif (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} == 0) and ($Lf{$a}-2*$Lg{$a} != 0)) - {$L=Compute('does not exist');} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }\left(\frac{f(x)-2g(x)}{g(x)}\right) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 9; - $q = 6; - $r = 3; - %Lf = ($p => $q, $q => $p); - %Lg = ($p => $r, $q => $r); - %f = ($p => $q); - %g = ($q => $r); - $a = $q; - if (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} != 3)) - {$L=$Lf{$a}/(3-$Lg{$a});} - elsif (exists $Lf{$a} and exists $Lg{$a} and ($Lg{$a} == 3) and ($Lf{$a} != 0)) - {$L=Compute('does not exist');} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }\left(\frac{f(x)}{3-g(x)}\right) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 9; - $q = 6; - $r = 3; - %Lf = ($p => $q, $q => $p); - %Lg = ($p => $r, $q => $r); - %f = ($p => $q); - %g = ($q => $r); - $a = $p; - if (exists $Lf{$a} and exists $Lg{$Lf{$a}} and ($g{$Lf{$a}} == $Lg{$Lf{$a}})) - {$L=$Lg{$Lf{$a}};} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }g(f(x)) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 9; - $q = 6; - $r = 3; - %Lf = ($p => $q, $q => $p); - %Lg = ($p => $r, $q => $r); - %f = ($p => $q); - %g = ($q => $r); - $a = $q; - if (exists $Lg{$a} and exists $Lf{$Lg{$a}} and ($f{$Lg{$a}} == $Lf{$Lg{$a}})) - {$L=$Lf{$Lg{$a}};} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }f(g(x)) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 9; - $q = 6; - $r = 3; - %Lf = ($p => $q, $q => $p); - %Lg = ($p => $r, $q => $r); - %f = ($p => $q); - %g = ($q => $r); - $a = $q; - if (exists $Lf{$a} and exists $Lf{$Lf{$a}} and exists $Lg{$Lf{$Lf{$a}}} and ($f{$Lf{$a}} == $Lf{$Lf{$a}}) and ($g{$Lf{$Lf{$a}}} == $Lg{$Lf{$Lf{$a}}})) - {$L=$Lg{$Lf{$Lf{$a}}};} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }g(f(f(x))) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 9; - $q = 6; - $r = 3; - %Lf = ($p => $q, $q => $p); - %Lg = ($p => $r, $q => $r); - %f = ($p => $q); - %g = ($q => $r); - $a = $q; - if (exists $Lf{$a} and exists $Lg{$a}) - {$L=$Lf{$a}*$Lg{$a} - ($Lf{$a})**2 + ($Lg{$a})**2;} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }\left(f(x)g(x)-f(x)^2+g(x)^2\right) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     -     - - - -

    - Use the following information to evaluate the given limit, when possible. - If it is not possible to determine the limit, state why not. - - \lim_{x\to1}f(x)\amp=2\amp\lim_{x\to10}f(x)\amp=1\amp f(1)\amp=1/5 - \lim_{x\to1}g(x)\amp=0\amp\lim_{x\to10}g(x)\amp=\pi\amp g(10)\amp=\pi - -

    -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 1; - $q = 10; - $r = 2; - $s = 0; - $t = Compute("pi"); - Context("Fraction"); - $u = Fraction("1/5"); - %Lf = ($p => $r, $q => $p); - %Lg = ($p => $s, $q => $t); - %f = ($p => $u); - %g = ($q => $t); - $a = $p; - if (exists $Lf{$a} and exists $Lg{$a}) - {$L=$Lf{$a}*$Lg{$a}} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }\left(f(x)g(x)\right) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 1; - $q = 10; - $r = 2; - $s = 0; - $t = Compute("pi"); - Context("Fraction"); - $u = Fraction("1/5"); - %Lf = ($p => $r, $q => $p); - %Lg = ($p => $s, $q => $t); - %f = ($p => $u); - %g = ($q => $t); - $a = $q; - if (exists $Lg{$a}) - {$L=Compute("cos($Lg{$a})")} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }\cos(g(x)) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 1; - $q = 10; - $r = 2; - $s = 0; - $t = Compute("pi"); - Context("Fraction"); - $u = Fraction("1/5"); - %Lf = ($p => $r, $q => $p); - %Lg = ($p => $s, $q => $t); - %f = ($p => $u); - %g = ($q => $t); - $a = $p; - if (exists $Lf{$a} and exists $Lg{5*$Lf{$a}} and ($Lg{5*$Lf{$a}} == $g{5*$Lf{$a}})) - {$L=$Lg{5*$Lf{$a}}} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }g(5f(x)) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('not possible to know'=>{}); - Context()->strings->add('NPK'=>{alias=>'not possible to know'}); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $p = 1; - $q = 10; - $r = 2; - $s = 0; - $t = Compute("pi"); - Context("Fraction"); - $u = Fraction("1/5"); - %Lf = ($p => $r, $q => $p); - %Lg = ($p => $s, $q => $t); - %f = ($p => $u); - %g = ($q => $t); - $a = $p; - if (exists $Lf{$a} and exists $Lg{$a} and ($Lf{$a} > 0)) - {$L=($Lf{$a})**($Lg{$a});} - else - {$L=Compute('not possible to know');} - $showwork = '[@ explanation_box(message => "If it is not possible to determine the limit, state why not.") @]*'; - - -

    - \lim\limits_{x\to }5^{g(x)} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - If it is not possible to know whether or not it exists, you may type not possible to know, or just NPK. - -

    - -

    -

    - -

    -     -     -     - -     - - - -

    - Evaluate the given limit. -

    -     - - - - - $a=random(2,8,1); - $b=non_zero_random(-5,5,1); - $c=non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=3;$b=-3;$c=7}; - $f=Formula("x^2+$b x+$c")->reduce; - $L=$f->eval(x=>$a); - - -

    - \lim\limits_{x\to }\left(\right) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     - -

    - - \lim\limits_{x\to }\left(\right)\amp=\lim\limits_{x\to }x^2+\lim\limits_{x\to }x+\lim\limits_{x\to } - \amp=\left(\lim\limits_{x\to }x\right)^2+\lim\limits_{x\to }x+\lim\limits_{x\to } - \amp=\left(\right)^2+ \left(\right)+ - \amp= - -

    -     -     -     - - - - - ($b,$c) = random_subset(2,-9..-1,1..9); - $n=random(3,9,1); - if($envir{problemSeed}==1){$b=3;$c=5;$n=7}; - $f=Formula("((x-$b)/(x-$c))^($n)")->reduce; - $L=$f->substitute(x=>Formula("pi")); - - -

    - \lim\limits_{x\to\pi} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     - -

    - - \lim\limits_{x\to\pi}\amp=\left(\lim\limits_{x\to\pi}\frac{x- }{x- }\right)^{} - \amp=\left(\frac{\lim\limits_{x\to\pi}(x- )}{\lim\limits_{x\to\pi}(x- )}\right)^{} - \amp=\left(\frac{\lim\limits_{x\to\pi}x-\lim\limits_{x\to\pi}}{\lim\limits_{x\to\pi}x-\lim\limits_{x\to\pi}}\right)^{} - \amp= - -

    -     -     -     - - - - - $d=list_random(3,4,6); - if($envir{problemSeed}==1){$d=4;}; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $a=Formula("pi/$d"); - %f=(3=>Formula("sqrt(3)/4"),4=>Formula("1/2"),6=>Formula("sqrt(3)/4")); - $L=$f{$d}; - %p=(3=>Formula("1/2"),4=>Formula("sqrt(2)/2"),6=>Formula("sqrt(3)/2")); - $l=$p{$d}; - $r=$p{$d**2/2-5.5*$d+18}; - - -

    - \lim\limits_{x\to }\cos(x)\sin(x) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     - -

    - - \lim_{x\to }\cos(x)\sin(x)\amp=\lim_{x\to }\cos(x)\cdot\lim_{x\to }\sin(x) - \amp=\cos\mathopen{}\left(\right)\mathclose{}\cdot\sin\mathopen{}\left(\right)\mathclose{} - \amp=\cdot - \amp= - -

    -     -     -     - - - - - ($a,$d) = random_subset(2,-6..-1,1..6); - $b=non_zero_random(-5,5,1); - $c=non_zero_random(-5,5,1); - if($envir{problemSeed}==1){$a=1;$b=2;$c=2;$d=4}; - $f=Formula("$b*x-$c")->reduce; - $g=Formula("x-$d")->reduce; - $h=$f/$g; - Context("Fraction"); - $L=Fraction($b*$a-$c,$a-$d); - - -

    - \lim\limits_{x\to} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     - -

    - - \lim_{x\to} \amp = \frac{-}{-} - \amp = - -

    -     -     -     - - - - - $L=Compute("DNE"); - - -

    - \lim\limits_{x\to0}\ln(x) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     - -

    - This limit does not exist for at least two reasons. - For one reason, - \ln(x) can be arbitrarily large negative if you allow x to be positive and close enough to 0. - Also, since \ln(x) is undefined for negative x and - \lim\limits_{x\to0}\ln(x) is an expression that depends on outputs of \ln using inputs - both slightly smaller and - slightly larger than 0, - we cannot hope to give meaning to \lim\limits_{x\to0}\ln(x). -

    -     -     -     - - - - - ($a,$b) = random_subset(2,2,3,4); - $c=$a**2-random(1,2,1); - if($envir{problemSeed}==1){$a=3;$b=4;$c=8;}; - $f=Formula("($b)^(x^3-$c x)"); - $L=$f->eval(x=>$a); - - -

    - \lim\limits_{x\to } -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     - -

    - - \lim\limits_{x\to }\amp=^{\lim\limits_{x\to }\left(x^3- x\right)} - \amp=^{\lim\limits_{x\to }x^3-\lim\limits_{x\to }x} - \amp=^{\left(\lim\limits_{x\to }x\right)^3- \lim\limits_{x\to }x} - \amp=^{^3- \cdot } - \amp= - -

    -     -     -     - - - - - $d=list_random(3,4,6); - if($envir{problemSeed}==1){$d=6;}; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $a=Formula("pi/$d"); - %f=(3=>Formula("2sqrt(3)/3"),4=>Formula("sqrt(2)"),6=>Formula("2")); - $L=$f{$d}; - - -

    - \lim\limits_{x\to }\csc(x) -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     - -

    - - \lim_{x\to }\csc(x)\amp=\csc\mathopen{}\left(\right)\mathclose{} - \amp= - -

    -     -     -     - - - - - Context()->flags->set(reduceConstantFunctions=>0); - $a=random(1,9,1); - if($envir{problemSeed}==1){$a=1;}; - $f=Formula("ln($a+x)"); - $L=($a==1)?0:Formula("ln($a)"); - - -

    - \lim\limits_{x\to0} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - #ensure rational function doesn't reduce - do{$a=non_zero_random(-5,5,1); - $b=non_zero_random(-5,5,1); - $c=random(2,5,1); - do{$d=non_zero_random(-5,5,1);}until($d!=-$a*$c); - $e=non_zero_random(-5,5,1); - }until((($b*$c+$e)/(-$a*$c-$d))**2+$a*(($b*$c+$e)/(-$a*$c-$d))+$b); - if($envir{problemSeed}==1){$a=3;$b=5;$c=5;$d=2;$e=2}; - $f=Formula("x^2+$a x+$b")->reduce; - $g=Formula("$c x^2-$d x-$e")->reduce; - $r=$f/$g; - $L=$r->substitute(x=>Formula("pi")); - - -

    - \lim\limits_{x\to\pi} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     - -

    - - \lim\limits_{x\to\pi}\amp=\frac{\lim\limits_{x\to\pi}\left(\right)}{\lim\limits_{x\to\pi}\left(\right)} - \amp=\frac{\lim\limits_{x\to\pi}x^2+\lim\limits_{x\to\pi}(x)+\lim\limits_{x\to\pi}()}{\lim\limits_{x\to\pi}(x^2)+\lim\limits_{x\to\pi}(- x)+\lim\limits_{x\to\pi}- } - \amp=\frac{\lim\limits_{x\to\pi}x^2+ \lim\limits_{x\to\pi}x+\lim\limits_{x\to\pi}()}{\lim\limits_{x\to\pi}x^2- \lim\limits_{x\to\pi}x-\lim\limits_{x\to\pi}} - \amp= - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $a=non_zero_random(-5,5,1); - $b=non_zero_random(-5,5,1); - $c=non_zero_random(-5,5,1); - do{$d=non_zero_random(-5,5,1);}until($a*$c!=-$b*$d); - if($envir{problemSeed}==1){$a=3;$b=1;$c=1;$d=-1;}; - $f=Formula("$a x+$b")->reduce; - $g=Formula("$c-$d x")->reduce; - $r=$f/$g; - $L=$r->substitute(x=>Formula("pi")); - - -

    - \lim\limits_{x\to\pi} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - ($a,$b,$c) = random_subset(3,-9..-1,1..9); - if($envir{problemSeed}==1){$a=6;$b=-2;$c=7;}; - $f=Formula("x^2-($a+$b)x+$a*$b")->reduce; - $g=Formula("x^2-($a+$c)x+$a*$c")->reduce; - $r=$f/$g; - Context("Fraction"); - $L=Fraction($a-$b,$a-$c); - - -

    - \lim\limits_{x\to } -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - ($b,$c) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$b=-2;$c=2;}; - $f=Formula("x^2-($b)x")->reduce; - $g=Formula("x^2-($c)x")->reduce; - $r=$f/$g; - Context("Fraction"); - $L=Fraction($b,$c); - - -

    - \lim\limits_{x\to0} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - ($a,$b,$c) = random_subset(3,-9..-1,1..9); - if($envir{problemSeed}==1){$a=2;$b=-8;$c=1;}; - $f=Formula("x^2-($a+$b)x+$a*$b")->reduce; - $g=Formula("x^2-($a+$c)x+$a*$c")->reduce; - $r=$f/$g; - Context("Fraction"); - $L=Fraction($a-$b,$a-$c); - - -

    - \lim\limits_{x\to } -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - ($a,$b,$c) = random_subset(3,-9..-1,1..9); - if($envir{problemSeed}==1){$a=2;$b=8;$c=-1;}; - $f=Formula("x^2-($a+$b)x+$a*$b")->reduce; - $g=Formula("x^2-($a+$c)x+$a*$c")->reduce; - $r=$f/$g; - Context("Fraction"); - $L=Fraction($a-$b,$a-$c); - - -

    - \lim\limits_{x\to } -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - ($a,$b,$c) = random_subset(3,-9..-1,1..9); - if($envir{problemSeed}==1){$a=-2;$b=7;$c=-8;}; - $f=Formula("x^2-($a+$b)x+$a*$b")->reduce; - $g=Formula("x^2-($a+$c)x+$a*$c")->reduce; - $r=$f/$g; - Context("Fraction"); - $L=Fraction($a-$b,$a-$c); - - -

    - \lim\limits_{x\to } -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - ($a,$b,$c) = random_subset(3,-9..-1,1..9); - if($envir{problemSeed}==1){$a=-1;$b=-8;$c=7;}; - $f=Formula("x^2-($a+$b)x+$a*$b")->reduce; - $g=Formula("x^2-($a+$c)x+$a*$c")->reduce; - $r=$f/$g; - Context("Fraction"); - $L=Fraction($a-$b,$a-$c); - - -

    - \lim\limits_{x\to } -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     -     - - - -

    - Use the Squeeze Theorem to evaluate the limit. -

    -     - - - - $showwork = '[@ explanation_box(message=>"Show your work.") @]*'; - - -

    - \lim\limits_{x\to0}\left(x\sin\mathopen{}\left(\frac{1}{x}\right)\mathclose{}\right) -

    -

    - -

    -

    - -

    -     -     -     - - - - - $showwork = '[@ explanation_box(message=>"Show your work.") @]*'; - - -

    - \lim\limits_{x\to0}\left(\sin(x)\cos\mathopen{}\left(\frac{1}{x^2}\right)\mathclose{}\right) -

    -

    - -

    -

    - -

    -     -     -     - - - - - $showwork = '[@ explanation_box(message=>"Show your work.") @]*'; - - -

    - \lim\limits_{x\to1} f(x), - where 3x-2\leq f(x) \leq x^3 -

    -

    - -

    -

    - -

    -     -     -     - - - - - $showwork = '[@ explanation_box(message=>"Show your work.") @]*'; - - -

    - \lim\limits_{x\to3} f(x), - where 6x-9\leq f(x) \leq x^2 -

    -

    - -

    -

    - -

    -     -     -     -     - - - -

    - The following exercises challenge your understanding of limits but can be evaluated using the knowledge gained in . -

    -     - - - - - $L=random(2,9,1); - if($envir{problemSeed}==1){$L=3;}; - $f=Formula("sin($L x)/x"); - - -

    - \lim\limits_{x\to0} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - ($a,$b) = random_subset(2,2..9); - if($envir{problemSeed}==1){$a=5;$b=8}; - $f=Formula("sin($a x)/($b x)"); - Context("Fraction"); - $L=Fraction($a,$b); - - -

    - \lim\limits_{x\to0} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - $L=1; - - -

    - \lim\limits_{x\to0}\frac{\ln(1+x)}{x} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0); - $L=Formula("pi/180"); - - -

    - \lim\limits_{x\to0}\frac{\sin(x)}{x}, - where x is measured in degrees, not radians. -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     -     - - - - -

    - Let f(x)=0 and g(x)=\frac{x}{x}. -

    -     - - - -

    - Explain why \lim\limits_{x\to2}f(x)=0. -

    - -     - - - -

    - Apply Part 1 of . -

    -     -     - - - -

    - Explain why \lim\limits_{x\to0}g(x)=1. -

    - -     - - - -

    - Apply ; - g(x)=\frac{x}{x} is the same as g(x)=1 everywhere except at x=0. - Thus \lim\limits_{x\to0}g(x)=\lim_{x\to0}1=1. -

    -     -     - - - -

    - Explain why \lim\limits_{x\to2} g(f(x)) does not exist. -

    - -     - - - -

    - The function f always gives output 0, - so g(f(x)) is never defined as g is not defined for an input of 0. - Therefore the limit does not exist. -

    -     -     - - - -

    - Explain why the previous statement does not violate the Composition Rule of . -

    - -     - - - -

    - The Composition Rule requires that - \lim\limits_{x\to0}g(x) be equal to g(0). - They are not equal, - so the conditions of the Composition Rule are not satisfied, - and hence the rule is not violated. -

    -     -     - -     -     -     - -
    - One-Sided Limits - -

    - We introduced the concept of a limit gently, - approximating their values graphically and numerically. - Next came the rigorous definition of the limit, - along with an admittedly tedious method for evaluating them. - gave us tools - (which we call theorems) - that allow us to compute limits with greater ease. - Chief among the results were the facts that polynomials and - rational, trigonometric, exponential and logarithmic functions - (and their sums, products, etc.) - all behave nicely. - In this section we rigorously define what we mean by nicely. -

    -

    - In - we saw three ways in which limits of functions can fail to exist: -

      -
    1. -

      - The function approaches different values from the left - and right. -

      -
      2. -
    2. -

      - The function grows without bound. -

      -
      3. -
    3. -

      - The function oscillates. -

      -
      4. -
    -

    - -

    - In this section we explore in depth the concepts behind - by introducing the one-sided limit. - We begin with formal definitions that are very similar to the - definition of the limit given in , - but the notation is slightly different and x\neq c - is replaced with either x\lt c or x \gt c. -

    -

    - There is a slightly different definition for a left-hand limit, - than for a right-hand limit, - but both have a lot in common with . -

    - - - One Sided Limits: Left- and Right-Hand Limits - -

    -

    -
  • - Left-Hand Limit - limitone-sided - limitleft-handed - -

    - Let f be a function defined on (a,c) - for some a\lt c and let L be a real number. - The statement that the limit of f(x), - as x approaches c from the left, - is L, (alternatively, - that the left-hand limit of f at - c is L) is denoted by - - \lim_{x\to c^-} f(x) = L - , - and means that for any \varepsilon \gt 0, - there exists \delta \gt 0 such that for all - x\in (a,c), if \abs{x - c} \lt \delta, - then \abs{f(x) - L} \lt \varepsilon. -

    -
    • - -
  • - Right-Hand Limit - limitone-sided - limitright-handed - -

    - Let f be a function defined on (c,b) - for some b \gt c and let L be a real number. - The statement that the limit of f(x), - as x approaches c from the right, - is L, (alternatively, - that the right-hand limit of f at - c is L) is denoted by - - \lim_{x\to c^+} f(x) = L - , - and means that for any \varepsilon \gt 0, - there exists \delta \gt 0 such that for all - x\in (c,b), if \abs{x - c} \lt \delta, - then \abs{f(x) - L} \lt \varepsilon. -

    -
    • -
    -

    -     -     - - - - - -

    - Practically speaking, when evaluating a left-hand limit, - we consider only values of x - to the left of c, - , where x\lt c. - The admittedly imperfect notation x\to c^- is used to - imply that we look at values of x to the left of c. - The notation has nothing to do with positive or negative values - of either x or c. - It's more like you are adding very small negative values to - c to get values for x. - A similar statement holds for evaluating right-hand limits; - there we consider only values of x to the right of c, - , x \gt c. - We can use the theorems from previous sections to help us evaluate these limits; - we just restrict our view to one side of c. -

    - - - - - - - -

    - We practice evaluating left- and right-hand - limits through a series of examples. -

    - - - Evaluating one-sided limits - -

    - Let f(x) = \begin{cases} - x \amp 0\leq x\leq 1 \\ - 3-x \amp 1\lt x\lt 2 - \end{cases}, - as shown in . - Find each of the following: - -

      -
    1. - \lim_{x\to 1^-} f(x) -
      2. -
    2. - \lim_{x\to 1^+} f(x) -
      3. -
    3. - \lim_{x\to 1} f(x) -
      4. -
    4. - f(1) -
      5. -
    5. - \lim_{x\to 0^+} f(x) -
      6. -
    6. - f(0) -
      7. -
    7. - \lim_{x\to 2^-} f(x) -
      8. -
    8. - f(2) -
      9. -
    -

    - -
    -
    - - - -

    - Graph of the piecewise function - f(x) = \begin{cases} x \amp 0\leq x\leq 1 \\ 3-x \amp 1\lt x\lt 2 \end{cases}. - There are two line segments: for 0\leq x\leq 1 we have a line segment - with positive slope, and for 1\lt x\lt 2 we have a line segment with negative slope. -

    -

    - The line segment with a positive slope starts at the point (0, 0) - and ends at (1, 1). The line segment with a negative slope starts - at (1, 2) and ends at (2, 1). -

    -

    - The start and end points of the line segment with a positive slope - are solid dots, indicating that those points are part of the graph. - The start and end points of the line segment - with a negative slope are hollow dots. - This tells that although the second line segment gets arbitrarily close - to the points (1,2) and (2,1), these points are not part of the graph. -

    - -

    - Since f(x) is close to 1 when x is close to 1, but x\lt 1, - while f(x) is close to 2 when x is close to 1, but x\gt 1, - we can conclude that the left and right hand limits are different. -

    -     - - Graph of a piecewise function that has different left and right hand limits when x=1. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.4, - xmax=2.4, - ymin=-.4, - ymax=2.4, - ] - \addplot[firstcurvestyle] coordinates {(0,0) (1,1)}; - \addplot[firstcurvestyle] coordinates {(1,2) (2,1)}; - \addplot[soliddot] coordinates {(0,0) (1,1)}; - \addplot[hollowdot] coordinates {(1,2) (2,1)}; - \end{axis} - \end{tikzpicture} - - - - - - -

    - For these problems, - the visual aid of the graph is likely more effective in - evaluating the limits than using f itself. - Therefore we will refer often to the graph. -

    - -

    -

      -
    1. -

      - As x goes to 1 from the left, - we see that f(x) is approaching the value of 1. -

      -

      - Therefore \lim\limits_{x\to 1^-} f(x) =1. -

      - -
      2. -
    2. -

      - As x goes to 1 from the right, - we see that f(x) is approaching the value of 2. - Recall that it does not matter that there is an - open circle there; - we are evaluating a limit, not the value of the function. -

      -

      - Therefore \lim\limits_{x\to 1^+} f(x)=2. -

      -
      3. -
    3. -

      - The limit of f as x approaches - 1 does not exist, as discussed in - . - The function does not approach one particular value, - but two different values from the left and the right. -

      -
      4. -
    4. -

      - Using the definition, and by looking at the graph, - we see that f(1) = 1. -

      -
      5. -
    5. -

      - As x goes to 0 from the right, - we see that f(x) is approaching 0. - Therefore \lim_{x\to 0^+} f(x)=0. - Note we cannot consider a left-hand limit at 0 - as f is not defined for values of x\lt 0. -

      -
      6. -
    6. -

      - Using the definition and the graph, f(0) = 0. -

      -
      7. -
    7. -

      - As x goes to 2 from the left, - we see that f(x) is approaching the value of 1. -

      -

      - Therefore \lim\limits_{x\to 2^-} f(x)=1. -

      -
      8. -
    8. -

      - The graph and the definition of the function show - that f(2) is not defined. -

      -
      9. -
    -

    -     - - - -

    - Note how the left- and right-hand limits were different at x=1. - This, of course, causes the limit to not exist. - The following theorem states what is fairly intuitive: - the limit exists precisely when the left- and right-hand limits are equal. -

    - - - Limits and One-Sided Limits - -

    - Let f be a function defined on an open interval I - containing c, except possibly at c. - - limitdoes not exist - - Then - - \lim_{x\to c}f(x) = L - - if, and only if, - - \lim_{x\to c^-}f(x) = L \text{ and } \lim_{x\to c^+}f(x) = L - . -

    -     -     - -

    - The phrase if, and only if - means the two statements are equivalent: - they are either both true or both false. - If the limit equals L, - then the left and right hand limits both equal L. - If the limit is not equal to L, - then at least one of the left and right-hand limits is not equal to - L (it may not even exist). -

    - - - -

    - One thing to consider in Examples - is that the value of the function may/may not be equal to the value(s) - of its left/right-hand limits, even when these limits agree. -

    - - - Evaluating limits of a piecewise-defined function - -

    - Let f(x) =\begin{cases} - 2-x \amp 0\lt x\lt 1 \\ - (x-2)^2 \amp 1\lt x\lt 2 - \end{cases} - . - - Evaluate the following: -

      -
    1. - \lim_{x\to 1^-} f(x) -
      2. -
    2. - \lim_{x\to 1^+} f(x) -
      3. -
    3. - \lim_{x\to 1} f(x) -
      4. -
    4. - f(1) -
      5. -
    5. - \lim_{x\to 0^+} f(x) -
      6. -
    6. - f(0) -
      7. -
    7. - \lim_{x\to 2^-} f(x) -
      8. -
    8. - f(2) -
      9. -
    -

    -     - -

    - In this example, we evaluate each expression using just the definition of f, - without using a graph as we did in the previous example. -

    - -

    -

      -
    1. -

      - As x approaches 1 from the left, - we consider a limit where all x-values are less than 1. - This means we use the 2-x piece of the piecewise-defined function f. - As the x-values near 1, 2-x approaches 1; - that is, f(x) approaches 1. -

      -

      - Therefore \lim\limits_{x\to 1^-} f(x)=1. -

      -

      - A concise mathematical presentation of the above argument could be written as follows: - - \lim_{x\to 1^-}f(x) \amp = \lim_{x\to 1^-}(2-x) \quad (f(x)=x-2 \text{ for } 0\lt x\lt 1) - \amp = 2-1 = 1 \quad (\text{ properties of limits }) - - - -

      -
      2. -
    2. -

      - As x approaches 1 from the right, - we consider a limit where all x-values are greater than 1. - This means we use the (x-2)^2 piece of f. - As the x-values near 1, (x-2)^2 approaches 1; - that is, we see that again f(x) approaches 1. -

      -

      - Therefore \lim\limits_{x\to 1+} f(x)=1. -

      -

      - Once again, we can present our work computationally as follows: - - \lim_{x\to 1^+}f(x) \amp = \lim_{x\to 1^+}(x-2)^2 \quad (f(x)=(x-2)^2 \text{ for } 1\lt x\lt 2) - \amp = (1-2)^2=1 \quad (\text{ properties of limits }) - - -

      -
      3. -
    3. -

      - The limit of f as x approaches 1 exists and is 1, - as f approaches 1 from both the right and left. -

      -

      - Therefore \lim\limits_{x\to 1} f(x)=1. -

      -
      4. -
    4. -

      - Neither piece of f is defined for the x-value of 1; in other words, - 1 is not in the domain of f. Therefore f(1) is not defined. - -

      -
      5. -
    5. -

      - As x approaches 0 from the right, we consider a limit where all x-values are greater than 0. - This means we use the 2-x piece of f. As the x-values near 0, 2-x approaches 2; - that is, f(x) approaches 2. -

      -

      - So \lim\limits_{x\to 0^+} f(x)=2. - -

      -
      6. -
    6. -

      - f(0) is not defined as 0 is not in the domain of f. -

      -
      7. -
    7. -

      - As x approaches 2 from the left, we consider a limit where all x-values are less than 2. - This means we use the (x-2)^2 piece of f. As the x-values near 2, (x-2)^2 nears 0; - that is, f(x) approaches 0. -

      -

      - So \lim\limits_{x\to 2^-} f(x)=0. - -

      -
      8. -
    8. -

      - f(2) is not defined as 2 is not in the domain of f. -

      -
      9. -
    -

    - -

    - We can confirm our analytic result by consulting the graph of f shown in . - Note the open circles on the graph at x=0, 1 and 2, where f is not defined. -

    -
    -
    - - - -

    - Graph of f from . - The graph consists of two parts. - The first part is a line that starts at the point (0, 2) and ends at - (1, 1). The second part is a curve that starts at (1, 1) and ends - at (2, 0). - The points (0,2), (1,1), and (2,0) are all marked with hollow dots, - indicating that although the graph gets close to these points, they are not part of the graph. -

    -

    - The function is undefined for x = 1, - but the graph shows that f(x) approaches the same value (namely, 1) - from both the left and the right, allowing us to conclude that \lim_{x\to 1}f(x) exists, - and is equal to 1. -

    -     - - Graph of a piecewise-defined function. It is undefined when x=1, but has a limit at this point. - - - \begin{tikzpicture}[declare function = {func(\x) = (\x < 1) * (2 - x) + (\x > = 1) * ((x - 2)^2);}] - \begin{axis}[ - xmin=-.4, - xmax=2.4, - ymin=-.4, - ymax=2.4 - ] - \addplot+[domain=0:2] {func(x)}; - \addplot[hollowdot] coordinates {(0,2) (1,1) (2,0)}; - \end{axis} - \end{tikzpicture} - - - - - - - - - - Evaluating limits of a piecewise-defined function - -

    - Let f(x) =\begin{cases} (x-1)^2 \amp 0\leq x\leq 2, x\neq 1\\ 1 \amp x=1 - \end{cases} as shown in . - Evaluate the following: -

      -
    1. - \lim_{x\to 1^-} f(x) -
      2. -
    2. - \lim_{x\to 1^+} f(x) -
      3. -
    3. - \lim_{x\to 1} f(x) -
      4. -
    4. - f(1) -
      5. -
    -

    - -
    -
    - - - -

    - The graph of f(x) in . - The graph is a parabola, opening upward, plotted from - (0, 1) to (2, 1), with its vertex at (1,0). - However, there is a hole in the graph at the vertex, indicated by a hollow dot. - The function is still defined at x = 1, - because there is a solid dot at (1, 1). This shows that - f(1) = 1, but the limit of f as x approaches 1 - is 0. -

    -     - - Graph of f(x), showing an upward curved line and a solid dot at (1, 1). - Line of the equation is undefined at x = 1. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.4, - xmax=2.4, - ymin=-.4, - ymax=1.4 - ] - \addplot+[domain=0:2] {(x-1)^2}; - \addplot[soliddot] coordinates {(0,1) (1,1) (2,1)}; - \addplot[hollowdot] coordinates {(1,0)}; - \end{axis} - \end{tikzpicture} - - - - - - -

    - It is clear by looking at the graph that both the left- and right-hand limits of f, - as x approaches 1, are 0. - Thus it is also clear that the limit is 0; - , \lim_{x\to 1} f(x) = 0. - It is also clearly stated that f(1) = 1. -

    -     - - - - - Evaluating limits of a piecewise-defined function - -

    - Let f(x) = \begin{cases} x^2 \amp 0\leq x\leq 1 \\ 2-x \amp 1\lt x\leq 2 - \end{cases} as shown in . - Evaluate the following: -

      -
    1. - \lim_{x\to 1^-} f(x) -
      2. -
    2. - \lim_{x\to 1^+} f(x) -
      3. -
    3. - \lim_{x\to 1} f(x) -
      4. -
    4. - f(1) -
      5. -
    -

    - -
    -
    - - - -

    - Graph of a piecewise-defined function, on the interval [0,2]. - For x values between 0 - and 1 (inclusive) the graph is an upward curved parabola. - For x values 1 to 2 (inclusive) the - graph is a straight line with a negative slope. -

    - -

    - The parabola and the line meet at the point (1,1). -

    -

    - There are three solid dots plotted on the graph at the points, - (0, 0), (1, 1), and (2, 0), to indicate where each part of the graph begins and ends. -

    -     - - Graph of the piecewise function for the above example. - - - \begin{tikzpicture}[declare function = {func(\x) = (\x < 1) * (x^2) + (\x > 1) * (2-x);}] - \begin{axis}[ - xmin=-.4, - xmax=2.4, - ymin=-.4, - ymax=1.4 - ] - \addplot+[domain=0:2] {func(x)}; - \addplot[soliddot] coordinates {(0,0) (1,1) (2,0)}; - \end{axis} - \end{tikzpicture} - - - - - - -

    - It is clear from the definition of the function and its graph that all of the following are equal: - - \lim_{x\to 1^-} f(x) = \lim_{x\to 1^+} f(x) =\lim_{x\to 1} f(x) =f(1) = 1 - . -

    -     - - - -

    - In Examples - we were asked to find both \lim_{x\to 1}f(x) and f(1). - Consider the following table: -

    - - - - - \lim\limits_{x\to 1}f(x) - f(1) - - - - does not exist - 1 - - - - 1 - not defined - - - - 0 - 1 - - - - 1 - 1 - - - -

    - Only in - do both the function and the limit exist and agree. - This seems nice; in fact, - it seems normal. This is in fact an important situation which we explore in - entitled Continuity. In short, - a continuous function - is one in which when a function approaches a value as x\to c (, when \lim_{x\to c} f(x) = L), - it actually attains that value at c. - Such functions behave nicely as they are very predictable. -

    - - - - - - Terms and Concepts - - - - -

    - What are the three ways in which a limit may fail to exist? -

    - -     - - - -

    - The function approaches different values from the left and right; - the function grows without bound; - the function oscillates. -

    -     - -     - - - - - -

    - - If \lim\limits_{x\to 1^-}f(x)=5, - then \lim\limits_{x\to 1}f(x)=5. -

    -     - -

    - We aren't told the value of the right-hand limit, - and the right-hand limit might not equal 5. -

    -     - -     - - - - - -

    - - If \lim\limits_{x\to 1^-}f(x)=5, - then \lim\limits_{x\to 1^+}f(x)=5. -

    -     - -

    - Just because the left-hand limit equals some number, - that doesn't mean the right-hand limit exists or equals the same number. -

    -     - -     - - - - - -

    - - If \lim\limits_{x\to 1}f(x)=5, - then \lim\limits_{x\to 1^-}f(x)=5. -

    -     - -

    - When the limit exists, both the left-hand and right-hand limits also exist - and have the same value. -

    -     - -     - -     - - - Problems - - - -

    - Evaluate each expression using the given graph of f. -

    -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - @x=num_sort((0,random_subset(2,1..5))); - @y=random_subset(3,1..5); - $z=list_random(1..$y[1]-1,$y[1]+1..5); - $b=list_random($x[0],$x[2]); - if($envir{problemSeed}==1){@x=(0,1,2);@y=(1,2,0);$z=1;$b=0;}; - $a=$x[1]; - $f=Formula("($y[1]-$y[0])/($x[1]-$x[0])*(x-$x[0])+$y[0]"); - $g=Formula("($y[1]-$y[2])/($x[1]-$x[2])**2*(x-$x[2])^2+$y[2]"); - $xmin=min(@x)-1;$xmax=max(0,@x)+1; - $ymin=min(0,@y,$z)-1;$ymax=max(@y,$z)+1; - @L=($y[1],$y[1],$y[1],$z); - ($L[4],$L[5])=($b==0)?(Compute("DNE"),$y[0]):($y[2],Compute("DNE")); - - - - -

    - Graph that shows the domain to - . For x values less than - the graph is a straight line and for - x values where \lt x \leq - the graph is slightly curved. When x equals - the graph is defined by the point - (, ). -

    -

    - The straight line is defined by the points - (, ) and - (, ). - The curved line is defined by the points - (, ) and - (, ). -

    -     - - Graph for exercise problem 5. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = $ymin, - ymax = $ymax, - ] - \addplot[firstcurvestyle, domain=$x[0]:$x[1]] {$f}; - \addplot[firstcurvestyle, domain=$x[1]:$x[2]] {$g}; - \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[1],$z) ($x[2],$y[2])}; - \addplot[hollowdot] coordinates {($x[1],$y[1])}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - @x=num_sort((0,random_subset(2,1..5))); - @y=random_subset(2,1..5); - $y[2]=random(0,5,1); - $z=list_random(1..$y[1]-1,$y[1]+1..5); - $b=list_random($x[0],$x[2]); - $a=$x[1]; - $f=Formula("($y[1]-$y[0])/($x[1]-$x[0])*(x-$x[0])+$y[0]"); - $g=Formula("($z-$y[2])/($x[1]-$x[2])*(x-$x[2])+$y[2]"); - $xmin=min(0,@x)-1;$xmax=max(0,@x)+1;$ymin=min(0,@y,$z)-1;$ymax=max(0,@y,$z)+1; - @L=($y[1],$z,Compute("DNE"),$z); - ($L[4],$L[5])=($b==0)?(Compute("DNE"),$y[0]):($y[2],Compute("DNE")); - - - - -

    - Graph that shows the domain to . - For x values in the interval \leq x \lt - the graph is a line segment from (,) (a solid dot) - to (, ) (a hollow dot). - At x equals there is a jump in the y value. - For \leq x\leq the graph is the line segment - from (, ) to (,) - (both endpoints are solid dots). -

    -     - - Graph for exercise problem 6. See long description for details. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = $ymin, - ymax = $ymax, - ] - \addplot[firstcurvestyle, domain=$x[0]:$x[1]] {$f}; - \addplot[firstcurvestyle, domain=$x[1]:$x[2]] {$g}; - \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[1],$z) ($x[2],$y[2])}; - \addplot[hollowdot] coordinates {($x[1],$y[1])}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - @x=num_sort((0,random_subset(2,1..5))); - $y[0]=random(0,5,1); - $y[1]=Compute("inf"); - $y[2]=random(0,5,1); - $b=list_random($x[0],$x[2]); - if($envir{problemSeed}==1){@x=(0,1,2);@y=(0,Compute("inf"),0);$b=2;}; - $a=$x[1]; - $c=($b==$x[0])?$x[2]:$x[0]; - $f=Formula("sec((x-$x[0])/($x[1]-$x[0])*90)-1+$y[0]"); - $g=Formula("sec((x-$x[2])/($x[1]-$x[2])*90)-1+$y[2]"); - $xmin=min(0,@x)-1;$xmax=max(0,@x)+1;$ymin=-1;$ymax=max(0,$y[0],$y[2])+4; - $xmidlow = asec($ymax + 1 - $y[0])/90*180/pi * ($x[1]-$x[0]) + $x[0]; - $xmidhigh = asec($ymax + 1 - $y[2])/90*180/pi * ($x[1]-$x[2]) + $x[2]; - @L=(OneOf(Compute("DNE"),Compute("INF")),OneOf(Compute("DNE"),Compute("INF")),OneOf(Compute("DNE"),Compute("INF")),Compute("DNE")); - ($L[4],$L[5])=($b==0)?(Compute("DNE"),Compute("DNE")):($y[2],$y[0]); - - - - -

    - Graph that shows the domain to . There is a - vertical asymptote at . The graph on either side has - an upward curve. The line on the left of the asymptote starts at - (, ) and has a positive slope, - as x gets closer to the slope of the line - gets steeper. The line to the right of the asymptote starts at - (, ) and has a negative slope, - as x gets closer to the slope of the line - gets steeper. -

    -     - - Graph for exercise problem 7. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = $ymin, - ymax = $ymax, - ] - \addplot[firstcurvestyle,infiniteright, domain=$x[0]:$xmidlow] {$f}; - \addplot[firstcurvestyle, infiniteleft, domain=$xmidhigh:$x[2]] {$g}; - \addplot[asymptote, -] coordinates {($x[1],$ymin) ($x[1],$ymax)}; - \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[2],$y[2])}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - @x=num_sort((0,random_subset(2,1..5))); - # actual y values the function takes - @y=random_subset(3,1..5); - # y values for limit points (w from left, z from right) - ($w,$z)=random_subset(2,1..$y[1]-1,$y[1]+1..5); - $a=$x[1]; - $f=Formula("($w-$y[0])/($x[1]-$x[0])*(x-$x[0])+$y[0]"); - $g=Formula("($y[2]-$z)/($x[2]-$x[1])^2*(x-$x[1])^2+$z"); - $xmin=min(0,@x)-1;$xmax=max(0,@x)+1;$ymin=min(0,@y,$z,$w)-1;$ymax=max(0,@y,$z,$w)+1; - @L=($w,$z,Compute("DNE"),$y[1]); - - - - -

    - Graph that shows the domain to . - The graph is piecewise, with two parts. - The first part is a line from a solid dot at - (, ) to a hollow dot at - (, ). - The second part is a parabolic arc from a hollow dot at (, ) - to a solid dot at (, ). - There is also a point marked by a solid dot at (, ). -

    -     - - Graph for exercise problem 8. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = $ymin, - ymax = $ymax, - ] - \addplot[firstcurvestyle, domain=$x[0]:$x[1]] {$f}; - \addplot[firstcurvestyle, domain=$x[1]:$x[2]] {$g}; - \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[1],$y[1]) ($x[2],$y[2])}; - \addplot[hollowdot] coordinates {($x[1],$w) ($x[1],$z)}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - @x=num_sort((0,random_subset(2,1..5))); - @y=random_subset(2,1..5); - $y[2]=$y[0]; - if($envir{problemSeed}==1){@x=(0,1,2);@y=(0,2,0);}; - $a=$x[1]; - $f=Formula("($y[0]-$y[1])/($x[0]-$x[1])^2*(x-$x[1])^2+$y[1]"); - $g=Formula("($y[2]-$y[1])/($x[2]-$x[1])*(x-$x[1])+$y[1]"); - $xmin=min(0,@x)-1;$xmax=max(0,@x)+1;$ymin=min(0,@y,0)-1;$ymax=max(0,@y,0)+1; - @L=($y[1],$y[1],$y[1],$y[1]); - - - - -

    - Graph that shows the domain to . The graph - is defined on the interval \leq x \leq - . At (, ) - there is a change in the graphs curve, it goes from a curved line - to a straight line. -

    -     - - Graph for exercise problem 9. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = $ymin, - ymax = $ymax, - ] - \addplot[firstcurvestyle, domain=$x[0]:$x[1]] {$f}; - \addplot[firstcurvestyle, domain=$x[1]:$x[2]] {$g}; - \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[2],$y[2])}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $x[0]=random(-6,-2,1); - $x[1]=$x[0]+random(3,5,1); - $x[2]=$x[1]+random(3,5,1); - ($y[1],$y[2],$y[3])=random_subset(3,-6..6); - $y[0]=list_random(-6..$y[1]-1,$y[1]+1..6); - $y[4]=list_random(-6..$y[3]-1,$y[3]+1..6); - if($envir{problemSeed}==1){@x=(-4,0,4);@y=(-4,4,0,-4,4);}; - $a=$x[1]; - $f=Formula("cos((x-$x[1])/($x[0]-$x[1])*180)*($y[1]-$y[0])/2+($y[0]+$y[1])/2"); - $g=Formula("cos((x-$x[2])/($x[1]-$x[2])*180)*($y[4]-$y[3])/2+($y[3]+$y[4])/2"); - $xmin=min(0,@x)-1;$xmax=max(0,@x)+1;$ymin=min(0,@y)-1;$ymax=max(0,@y)+1; - @L=($y[1],$y[3],Compute("DNE"),$y[2]); - - - - -

    - Graph that shows the domain to . There are two - lines, one on the interval \leq x \lt - .The second line is on the interval - \lt x \leq . -

    -

    - The first line is defined by the solid point (, ) - and hollow point (, ). - The second line is defined by the hollow point (, ) - and solid point (, ). - For x = 0 the graph is defined by the point - (, ). -

    -     - - Graph for exercise problem 10. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = $ymin, - ymax = $ymax, - ] - \addplot[firstcurvestyle, domain=$x[0]:$x[1]] {cos((x-$x[1])/($x[0]-$x[1])*180)*($y[1]-$y[0])/2+($y[0]+$y[1])/2}; - \addplot[firstcurvestyle, domain=$x[1]:$x[2]] {$g}; - \addplot[soliddot] coordinates {($x[0],$y[0]) ($x[1],$y[2]) ($x[2],$y[4])}; - \addplot[hollowdot] coordinates {($x[1],$y[1]) ($x[1],$y[3])}; - \end{axis} - \end{tikzpicture} - - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - @L=(2,2,2,0,2,2,2,Compute("DNE")); - - - - -

    - Graph that shows the domain -4 to - 4. - There are four lines that make up this graph. The first - going from left to right is defined by the solid point - (-4, 0) and hollow point (-2. 2), and has - a positive slope. The second line is defined by the points - (-2, 2), 0, 0 with a negative slope. The - third line by (0, 0), 2, 2 with a positive slope. - And the last line is defined by the hollow point (2, 2) - and solid point (4, 0), with a negative slope. -

    -

    - For x = -2, F(x) = 0 -

    -     - - Graph for exercise problem 11. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = -5, - xmax = 5, - ymin = -4, - ymax = 4, - ] - \addplot[firstcurvestyle, domain=-4:-2] {x+4}; - \addplot[firstcurvestyle, domain=-2:0] {-x}; - \addplot[firstcurvestyle, domain=0:2] {x}; - \addplot[firstcurvestyle, domain=2:4] {4-x}; - \addplot[soliddot] coordinates {(-4,0) (-2,0) (4,0)}; - \addplot[hollowdot] coordinates {(-2,2) (2,2)}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to -2^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to -2^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to -2} f(x) -

    -

    - -

    -     -     - - - -

    - f(-2) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to 2^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to 2^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to 2} f(x) -

    -

    - -

    -     -     - - - -

    - f(2) -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - Context()->variables->are(a=>'Real'); - @L=(Formula("a-1"),Formula("a"),Compute("DNE"),Formula("a")); - - - - -

    - The graph is - made up of eight separate lines that do not connect to each - other. Each of the lines is parallel with the x axis, defined - by a solid dot on its left and a hollow dot on its right, and - has a length of 1 unit. The first line is inside quadrant 3 - of the graph and the last is in quadrant 1, there are no - lines in quadrants 2 or 4. Each of the other six lines is between - the first and last line moving along a straight diagonal path. -

    -

    - The following are the points that define each line: -

      -
    1. solid dot at (-4, -4) and hollow dot at (-3, -4)
      2. -
    2. solid dot at (-3, -3) and hollow dot at (-2, -3)
      3. -
    3. solid dot at (-2, -2) and hollow dot at (-1, -2)
      4. -
    4. solid dot at (-1, -1) and hollow dot at (0, -1)
      5. -
    5. solid dot at (0, 0) and hollow dot at (1, 0)
      6. -
    6. solid dot at (1, 1) and hollow dot at (2, 1)
      7. -
    7. solid dot at (2, 2) and hollow dot at (3, 2)
      8. -
    8. solid dot at (3, 3) and hollow dot at (4, 3)
      9. -
    -

    -     - - Graph for exercise problem 12. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = -5, - xmax = 5, - ymin = -5, - ymax = 5, - ] - \foreach \i in {-4,...,3} { - \addplot[firstcurvestyle, domain=\i:\i+1] {\i}; - \addplot[soliddot] coordinates {(\i,\i)}; - \addplot[hollowdot] coordinates {(\i+1,\i)}; - - } - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - - -

    - Let a be an integer with -3\leq a\leq3. -

    -     - - - -

    - \lim\limits_{x\to a^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to a^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to a} f(x) -

    -

    - -

    -     -     - - - -

    - f(a) -

    -

    - -

    -     -     -     -     -     - - - -

    - Evaluate the given limits of the piecewise defined function. -

    -     - - - - - $a=random(-2,4,1); - $b=non_zero_random(-5,5,1); - do{$c=non_zero_random(-5,5,1);}until($a+$b!=$a**2+$c); - if($envir{problemSeed}==1){$a=1;$b=1;$c=-5}; - @f=(Formula("x+$b")->reduce,Formula("x^2+$c")->reduce); - Context("PiecewiseFunction"); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $g=PiecewiseFunction("x <= $a" => "$f[0]", "x > $a" => "$f[1]"); - @L=($f[0]->eval(x=>$a),$f[1]->eval(x=>$a)); - $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); - $L[3]=$g->eval(x=>$a); - - -

    - f(x) = -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     -     -     - - - - - $a=random(-2,4,1); - $b=non_zero_random(-2,2,1); - ($c,$d)=random_subset(2,-5..-1,1..5); - if($envir{problemSeed}==1){$a=0;$b=2;$c=5;$d=-1;}; - @f=(Formula("$b x^2 + $c x+$d")->reduce,Formula("sin(x-$a)")->reduce); - Context("PiecewiseFunction"); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $g=PiecewiseFunction("x < $a" => "$f[0]", "x >= $a" => "$f[1]"); - @L=($f[0]->eval(x=>$a),$f[1]->eval(x=>$a)); - $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); - $L[3]=$g->eval(x=>$a); - - -

    - f(x) = -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     -     -     - - - - - ($a,$b)=num_sort(random_subset(2,-5..-1,1..5)); - if($envir{problemSeed}==1){$a=-1;$b=1;}; - @f=(Formula("x^2 + ($a+1) x+(($a)^3+1-($a)^2-($a+1)*$a)")->reduce,Formula("x^3+1")->reduce,Formula("x^2 + ($b-1) x+(($b)^3+1-($b)^2-($b-1)*$b)")->reduce); - Context("PiecewiseFunction"); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $g=PiecewiseFunction("x < $a" => "$f[0]", "$a <= x <= $b" => "$f[1]", "x > $b" => $f[2]); - @L=($f[0]->eval(x=>$a),$f[1]->eval(x=>$a)); - $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); - $L[3]=$g->eval(x=>$a); - ($L[4],$L[5])=($f[1]->eval(x=>$b),$f[2]->eval(x=>$b)); - $L[6]=($L[4]==$L[5])?$L[4]:Compute("DNE"); - $L[7]=$g->eval(x=>$b); - - -

    - f(x) = -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - @f=(Formula("cos(x)"),Formula("sin(x)")); - @L=(-1,0,Compute("DNE"),0); - - -

    - f(x) = \begin{cases}\amp x\lt\pi\\ \amp x\geq\pi\end{cases} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to\pi^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to\pi^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to\pi} f(x) -

    -

    - -

    -     -     - - - -

    - f(\pi) -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - Context()->variables->add(a=>'Real'); - @L=(Formula("1-cos^2(a)"),Formula("sin^2(a)"),OneOf(Formula("1-cos^2(a)"),Formula("sin^2(a)")),Formula("sin^2(a)")); - - -

    - f(x) = \begin{cases}1-\cos^2(x)\amp x<a\\\sin^2(x)\amp x\geq a\end{cases} - where a is a real number. -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to a^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to a^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to a} f(x) -

    -

    - -

    -     -     - - - -

    - f(a) -

    -

    - -

    -     -     -     -     - - - - - $a=random(-3,3,1); - @y= random_subset(3,-2..2); - if($envir{problemSeed}==1){$a=1;@y=(1,0,-1);}; - @f=(Formula("x+$y[0]")->reduce,Formula("x+$y[1]")->reduce,Formula("x+$y[2]")->reduce); - Context("PiecewiseFunction"); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $g=PiecewiseFunction("x < $a" => "$f[0]", "x = $a" => "$f[1]", "x > $a" => $f[2]); - @L=($f[0]->eval(x=>$a),$f[2]->eval(x=>$a)); - $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); - $L[3]=$g->eval(x=>$a); - - -

    - f(x) = -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     -     -     - - - - - $a=random(-3,3,1); - @y=random_subset(2,-5..5); - if($envir{problemSeed}==1){$a=2;@y=(4,3);}; - @f=(Formula("x^2+($y[0]-2*$a)x+($y[0]-($a)^2-($y[0]-2*$a)*$a)")->reduce->reduce,Formula("x+($y[1]-$a)")->reduce,Formula("-x^2+$a*x+$y[0]")->reduce); - Context("PiecewiseFunction"); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - $g=PiecewiseFunction("x < $a" => "$f[0]", "x = $a" => "$f[1]", "x > $a" => $f[2]); - @L=($f[0]->eval(x=>$a),$f[2]->eval(x=>$a)); - $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); - $L[3]=$g->eval(x=>$a); - - -

    - f(x) = -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - f() -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - Context()->variables->add(c=>'Real'); - $L=Formula("c"); - - -

    - f(x) = \begin{cases}a(x-b)^2+c\amp x\lt b\\a(x-b)+c\amp x\geq b\end{cases} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to b^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to b^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to b} f(x) -

    -

    - -

    -     -     - - - -

    - f(b) -

    -

    - -

    -     -     -     -     - - - - - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - @L=(-1,1,Compute("DNE"),0); - - -

    - f(x) = \begin{cases}\frac{\abs{x}}{x}\amp x\neq0\\0\amp x=0\end{cases} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to 0^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to 0^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to 0} f(x) -

    -

    - -

    -     -     - - - -

    - f(0) -

    -

    - -

    -     -     -     -     -     -     -     - -
    - Continuity - -

    - As we have studied limits, - we have gained the intuition that limits measure - where a function is heading. That is, - if \lim\limits_{x\to 1} f(x) = 3, - then as x is close to 1, - f(x) is close to 3. - We have seen, though, - that this is not necessarily a good indicator of what f(1) actually is. - This can be problematic; - functions can tend to one value but attain another. - This section focuses on functions that - do not exhibit such behavior. -

    - - - Continuous Function - -

    - Let f be a function whose domain contains an open interval I. - - continuous function - functioncontinuous - -

      -
    1. -

      - f is continuous at a point c in I - if \lim\limits_{x\to c}f(x) = f(c). - - continuousat a point - -

      -
      2. -
    2. -

      - f is continuous on the open interval I - if f is continuous at c for all values of - c in I. - - continuouson an interval - - If f is continuous on (-\infty,\infty), - we say f is continuous everywhere - (or everywhere continuous). - - continuouseverywhere - everywhere continuous - -

      -
      3. -
    -

    -     -     - -

    - Note that in , - a function f can only be continuous at a point c if c is in the domain of f. -

    - - - -

    - A useful way to establish whether or not a function f - is continuous at c is to verify the following three things: - -

      -
    1. -

      - \lim\limits_{x\to c} f(x) exists, -

      -
      2. -
    2. -

      - f(c) is defined, and -

      -
      3. -
    3. -

      - \lim\limits_{x\to c} f(x) = f(c). -

      -
      4. -
    -

    - - - - - Finding intervals of continuity - -

    - Let f be defined as shown in . - Give the interval(s) on which f is continuous. -

    - -
    -
    - - - -

    - The graph of a piecewise-defined function is shown, for x from 0 to 3. - For 0 \leq x \lt 1 - the graph looks like a parabola opening downward. - This part of the graph approaches, but does not reach, the point (1,1). - There is a hollow dot at (1,1), indicating that f(1) is undefined. - For 1 \lt x \leq 2 the graph - is a horizontal line segment, with y=1. - For 2 \leq x \leq 3 the graph again has the appearance of a downward-facing parabola - that begins at (2,1) and ends at (3,1). -

    -     - - Graph of a function with a discontinuity when x=1. Although the limit at 1 exists, f(1) is undefined. - - - \begin{tikzpicture}[declare function = {func(\x) = (\x >= 0)*(\x <= 1)*(-(\x-1/4)^2+1/16+1.5) + (\x > 1)*(\x <= 2) + (\x > 2)*(\x <= 3)*((2-\x)*(\x-3)+1);}] - \begin{axis}[ - xmin=-.4, - xmax=3.4, - ymin=-.4, - ymax=1.7 - ] - \addplot+[domain=0:3,samples=48] {func(x)}; - \addplot[soliddot] coordinates {(0,1.5) (2,1) (3,1)}; - \addplot[hollowdot] coordinates {(1,1)}; - \end{axis} - \end{tikzpicture} - - - - - - -

    - We proceed by examining the three criteria for continuity. -

      -
    1. -

      - The limits \lim\limits_{x\to c} f(x) exists for all - c between 0 and 3. -

      -
      2. -
    2. -

      - f(c) is defined for all c between 0 and 3, - except for c=1. - We know immediately that f cannot be continuous at x=1. -

      -
      3. -
    3. -

      - The limit \lim\limits_{x\to c} f(x) = f(c) for all - c between 0 and 3, - except, of course, for c=1. -

      -
      4. -
    -

    - -

    - We conclude that f is continuous at every point of the interval - (0,3) except at x=1. - Therefore f is continuous on (0,1) and (1,3). -

    - -     - - - - Finding intervals of continuity - -

    - The floor function, - - floor function - functionfloor - - f(x) = \lfloor x \rfloor, - returns the largest integer smaller than, or equal to, - the input x. (For example, - f(\pi) = \lfloor \pi \rfloor = 3.) The graph of f in - - demonstrates why this is often called a step function. -

    - -

    - Give the intervals on which f is continuous. -

    - -
    -
    - - - -

    - Shows the graph of the greatest integer function, for x from -2 to 3. There are five - horizontal line segments in a staircase configuration, ascending from left to right. Each segment is - one unit in length and includes its left endpoint, but the right endpoint of each segment is not included. - The first segment is from (-2, -2) to (-1, -2), - the second from (-1, -1) to (0, -1), the third from (0, 0) to (1, 0), - the fourth from (1, 1) to (2, 1), and the fifth from (2, 2) to (3, 2). -

    -     - - Graph of a step function whose value at x is the greatest integer less than or equal to x. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-2.4, - xmax=3.4, - ymin=-2.4, - ymax=2.4 - ] - \addplot[firstcurvestyle,-] coordinates {(-2,-2) (-1,-2)}; - \addplot[firstcurvestyle,-] coordinates {(-1,-1) (0,-1)}; - \addplot[firstcurvestyle,-] coordinates {(0,0) (1,0)}; - \addplot[firstcurvestyle,-] coordinates {(1,1) (2,1)}; - \addplot[firstcurvestyle,-] coordinates {(2,2) (3,2)}; - \addplot[soliddot] coordinates {(-2,-2) (-1,-1) (0,0) (1,1) (2,2)}; - \addplot[hollowdot] coordinates {(-1,-2) (0,-1) (1,0) (2,1) (3,2)}; - \end{axis} - \end{tikzpicture} - - - - - - -

    - We examine the three criteria for continuity. -

      -
    1. -

      - The limits \lim\limits_{x\to c} f(x) - do not exist at the jumps from one step to the next, - which occur at all integer values of c. - Therefore the limits exist for all c except when - c is an integer. -

      -
      2. -
    2. -

      - The function is defined for all values of c. -

      -
      3. -
    3. -

      - The limit \lim\limits_{x\to c} f(x) = f(c) for all values of - c where the limit exist, - since each step consists of just a line. -

      -
      4. -
    -

    - -

    - We conclude that f is continuous everywhere except at integer values of c. - So the intervals on which f is continuous are - - \ldots, (-2,-1), (-1,0), (0,1), (1,2), \ldots - . - We could also say that f is continuous on all intervals of the form - (n,n+1) where n is an integer. -

    -     - - - - -

    - Our definition of continuity on an interval specifies the interval is an open interval. - We can extend the definition of continuity to closed intervals of the form [a,b] - by considering the appropriate one-sided limits at the endpoints. -

    - - - - - Continuity on Closed Intervals - -

    - Let f be defined on the closed interval - [a,b] for some real numbers a\lt b. -

    - -

    - We say f is continuous on the closed interval - [a,b] if: -

      -
    1. -

      - f is continuous on (a,b), -

      -
      2. -
    2. -

      - \lim\limits_{x\to a^+} f(x) = f(a) and -

      -
      3. -
    3. -

      - \lim\limits_{x\to b^-} f(x) = f(b). -

      -
      4. -
    -

    -     -     - -

    - We can make the appropriate adjustments to talk about continuity on half-open intervals such as - [a,b) or (a,b] if necessary. -

    - -

    - If the domain of f includes values less than a, - we say that Item in - indicates that f is continuous from the right at a. - But if f is undefined for x\lt a, - we can say that f is continuous at a without ambiguity. -

    - -

    - Similarly, Item indcates that f - is continuous from the left at b, - and if f is not defined for x\gt b, - we can simply say that f is continuous at b. -

    - -

    - For example, it makes sense to say that the function f(x)=\sqrt{1-x^2} - is continuous at 1 and -1, - while the floor function in is continuous from the left at 1 and -1, - but is not continuous at these points. -

    - -

    - Using this new definition, - we can adjust our answer in - by stating that f is continuous on [0,1) and (1,3], - as mentioned in that example. - We can also revisit - and state that the floor function is continuous on the following half-open intervals - - \ldots, [-2,-1), [-1,0), [0,1), [1,2), \ldots - . -

    - - - -

    - This can tempt us to conclude that f is continuous everywhere; - after all, if f is continuous on [0,1) and [1,2), - isn't f also continuous on [0,2)? - Of course, the answer is no, - and the graph of the floor function immediately confirms this. -

    -

    - Continuous functions are important as they behave in a predictable fashion: - functions attain the value they approach. - Because continuity is so important, - most of the functions you have likely seen in the past are continuous on their domains. - This is demonstrated in the following example where we examine the intervals of continuity of a variety of common functions. -

    - - - Determining intervals on which a function is continuous - -

    - For each of the following functions, - give the domain of the function and the interval(s) on which it is continuous. -

      -
    1. -

      - f(x) = 1/x -

      -
      2. -
    2. -

      - f(x) = \sin(x) -

      -
      3. -
    3. -

      - f(x) = \sqrt{x} -

      -
      4. -
    4. -

      - f(x) = \sqrt{1-x^2} -

      -
      5. -
    5. -

      - f(x) = \abs{x} -

      -
      6. -
    -

    -     - -

    - We examine each in turn. -

      -
    1. -

      - The domain of f(x) = 1/x is (-\infty,0) \cup (0,\infty). - As it is a rational function, - we apply - to recognize that f is continuous on all of its domain. -

      -
      2. -
    2. -

      - The domain of f(x) = \sin(x) is all real numbers, - or (-\infty,\infty). - Applying - shows that \sin(x) is continuous everywhere. -

      -
      3. -
    3. -

      - The domain of f(x) = \sqrt{x} is [0,\infty). - Applying - shows that f(x) = \sqrt{x} is continuous on its domain of - [0,\infty). -

      -
      4. -
    4. -

      - The domain of f(x) = \sqrt{1-x^2} is [-1,1]. - Applying Theorems - and - shows that f is continuous on all of its domain, - [-1,1]. -

      -
      5. -
    5. -

      - The domain of f(x) = \abs{x} is (-\infty,\infty). - We can define the absolute value function as - - f(x) = \begin{cases}-x \amp x\lt 0 \\ x \amp x\geq 0\end{cases} - . - Each piece of this piecewise defined function is continuous on all of its domain, - giving that f is continuous on - (-\infty,0) and [0,\infty). - We cannot assume this implies that f is continuous on (-\infty,\infty); - we need to check that \lim\limits_{x\to 0}f(x) = f(0), - as x=0 is the point where f transitions from one piece - of its definition to the other. - It is easy to verify that this is indeed true, - hence we conclude that f(x) = \abs{x} is continuous everywhere. -

      -
      6. -
    -

    -     - -     - -

    - Continuity is inherently tied to the properties of limits. - Because of this, - the properties of limits found in Theorems - and apply to continuity as well. - Further, now knowing the definition of continuity we can re-read - - as giving a list of functions that are continuous on their domains. - The following theorem states how continuous functions can be combined to form other continuous functions, - followed by a theorem which formally lists functions that we know are continuous on their domains. -

    - - - Properties of Continuous Functions - -

    - Let f and g be continuous functions on an interval I, - let c be a real number and let n be a positive integer. - The following functions are continuous on I. -

    -
  • - Sums/Difference -

    - f\pm g -

    -
    • -
  • - Constant Multiple -

    - c\cdot f -

    -
    • -
  • - Product -

    - f\cdot g -

    -
    • -
  • - Quotient -

    - f/g (as long as g\neq 0 on I) -

    -
    • -
  • - Power -

    - f^n -

    -
    • -
  • - Root -

    - \sqrt[n]{f} (If n is even then require f(x)\geq 0 on I.) -

    -
    • -
  • - Compositions -

    - Adjust the definitions of f and g to: Let f be continuous on I, - where the range of f on I is J, - and let g be continuous on J. - Then g\circ f, , g(f(x)), - is continuous on I. -

    -
    • -
    -

    -     -     - - - - - - - - - Continuous Functions - -

    - Let n be a positive integer. - The following functions are continuous on their domains. - - continuous functionproperties - -

      -
    1. -

      - f(x) = \sin(x) -

      -
      2. -
    2. -

      - f(x) = \tan(x) -

      -
      3. -
    3. -

      - f(x) = \sec(x) -

      -
      4. -
    4. -

      - f(x) = \ln(x) -

      -
      5. -
    5. -

      - f(x) = a^x (a\gt 0) -

      -
      6. -
    6. -

      - f(x) = \cos(x) -

      -
      7. -
    7. -

      - f(x) = \cot(x) -

      -
      8. -
    8. -

      - f(x) = \csc(x) -

      -
      9. -
    9. -

      - f(x) = \sqrt[n]{x}, where n is a positive integer. -

      -
      10. -
    -

    -     -     - -

    - Note also that we can interpret as telling us that polynomial - and rational functions are also continuous on their domains. -

    - - - - - Checking continuity at a given point - -

    - Let f(x) = \begin{cases}x^2+2x+1,\amp \text{ if } x\lt 0\\2, \amp \text{ if } x=0\\5x+\cos(x^3), \amp \text{ if } x\gt 0\end{cases}. - For each point below, determine whether or not f is continuous at that point, and explain why. -

      -
    1. x=0
      2. -
    2. x=-2
      3. -
    3. x=2
      4. -
    -

    -     - -

    -

      -
    1. -

      - At x=0, we must rely on . - If x\lt 0, then f(x) = x^2+2x+1; therefore, - - \lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(x^2+2x+1)=1 - . - Similarly, for x\gt 0, f(x)=5x+\cos(x^3), so - - \lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}(5x+\cos(x^3)) = 0+\cos(0)=1 - . - Therefore, \lim_{x\to 1}f(x)=1 exists. -

      - -

      - However, f is not continuous at 0, since - - f(0)=2 \neq 1 = \lim_{x\to 0}f(x) - . -

      -
      2. - -
    2. -

      - If x is near -2, then we can assume that x\lt 0, - and therefore f(x)=x^2+2x+1. - This is a polynomial function, and by , - every polynomial function is continuous on its domain. - Therefore, f is continuous at -2. -

      -
      3. - -
    3. -

      - If x is near 2, then we can assume x\gt 0. - Since 2x and x^3 are polynomial, - we know that these functions are continuous, by . - We know that \cos(x) is continuous by , - and by , the composition of continuous functions is continuous, - so \cos(x^3) is continuous. - Also by , the sum of continuous functions is continuous. - Therefore, for any x\gt 0, f(x)=2x+\cos(x^3) is continuous, and in particular, - f is continuous at x=2. -

      -
      4. -
    -

    -     -     - - - - - - -

    - We apply these theorems in the following Example. -

    - - - Determining intervals on which a function is continuous - -

    - State the interval(s) on which each of the following functions is continuous. -

      -
    1. -

      - f(x) = \sqrt{x-1} + \sqrt{5-x} -

      -
      2. -
    2. -

      - f(x) = x\sin(x) -

      -
      3. -
    3. -

      - f(x) = \tan(x) -

      -
      4. -
    4. -

      - f(x) = \sqrt{\ln(x)} -

      -
      5. -
    -

    -     - -

    - We examine each in turn, - applying Theorems - and as appropriate. -

    - -
    -
    - - - -

    - Shows the graph of f(x)=\sqrt{x-1}+\sqrt{5-x} on its domain [1,5]. - The graph looks somewhat like the top of a slice of bread, - rising from the point (1,2) to it highest point near x=3, - and then falling to the point (5,2). -

    -     - - Graph of function that is the sum of two square root functions. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.4, - xmax=5.4, - ymin=-.4, - ymax=3.2, - ] - \addplot+[domain=0:180] ({3+2*cos(x)},{sqrt(2+2*cos(x))+sqrt(2-2*cos(x))}); - \addplot[soliddot] coordinates {(1,2) (5,2)}; - \end{axis} - \end{tikzpicture} - - - - - -

    -

      -
    1. -

      - The square root terms are continuous on the intervals - [1,\infty) and (-\infty,5], respectively. - As f is continuous only where each term is continuous, - f is continuous on [1,5], - the intersection of these two intervals. - A graph of f is given in - . -

      -
      2. -
    2. -

      - The functions y=x and - y=\sin(x) are each continuous everywhere, - hence their product is, too. -

      -
      3. -
    3. -

      - - states that f(x) = \tan(x) is continuous on its domain. - Its domain includes all real numbers except odd multiples of \pi/2. - Thus the intervals on which f(x) = \tan(x) is continuous are - - \ldots \left(-\frac{3\pi}{2},-\frac{\pi}2\right),\ \left(-\frac{\pi}2,\frac{\pi}2\right),\ \left(\frac{\pi}2,\frac{3\pi}2\right),\ldots, - . -

      -
      4. -
    4. -

      - Here, f(x) is the composition g(h(x)), where - g(x) = \sqrt{x} and h(x)=\ln(x). - The domain of g is [0,\infty), while - the range of h is (-\infty,\infty). - If we restrict the domain to - [1,\infty), then the output from h(x)=\ln(x) is restricted to - [0,\infty), on which g(x) = \sqrt{x} is defined. - Thus the domain of f(x) = \sqrt{\ln(x)} is [1,\infty). -

      -
      5. -
    -

    - - - - - - - Classification of discontinuities -

    - We now know what it means for a function to be continuous, - so of course we can easily say what it means for a function to be discontinuous; - namely, not continuous. - However, to better understand continuity, - it is worth our time to discuss the different ways in which a function can fail to be discontinuous. - By definition, a function f is continuous at a point - a in its domain if \lim\limits_{x\to a}f(x) = f(a). - If this equality fails to hold, then f is not continuous. - We note, however, that there are a number of different things that can go wrong with this equality. -

    - - - -

    -

      -
    1. -

      - \lim\limits_{x\to a}f(x)=L exists, - but L\neq f(a), or f(a) is undefined. - Such a discontinuity is called a - removable discontinuity - discontinuityremovable. -

      - -

      - A removable discontinuity can be pictured as a - hole in the graph of f. - The term removable refers to the fact that - by simply redefining f(a) to equal L - (that is, changing the value of f at a single - point), we can create a new function that is - continuous at x=a, and agrees with f - at all x\neq a. -

      -
      2. - -
    2. -

      - \lim\limits_{x\to a^+}f(x) = L and - \lim\limits_{x\to a^-}f(x)=M exist, - but L\neq M. - In this case the left and right hand limits both exist, - but since they are not equal, the limit of f as - x\to a does not exist. - Such a discontinuity is called a - jump discontinuity. - discontinuityjump -

      - -

      - The phrase jump discontinuity is meant to - represent the fact that visually, - the graph of f jumps from one value - to another as we cross the value x=a. -

      -
      3. - -
    3. -

      - The function f is unbounded - near x=a. This means that the value of - f becomes arbitrarily large (or large and - negative) as x approaches a. - Such a discontinuity is called an - infinite discontinuity. - discontinuityinfinite -

      - -

      - Infinite discontinuities are most easily understood - in terms of infinite limits, which are - discussed in . -

      -
      4. -
    -

    - -
    -
    - -
    -
    - - -

    - A portion of the graph of a function is shown, for x from 0 to 4. - The graph has the shape of a parabola opening downward, - but at x=2 there is a hole in the graph, - and instead the point (2,1) (which is not on the graph) is plotted. - The graph of this function illustrates a removable discontinuity because - \lim_{x\to 2}f(x) exists, but does not equal f(2). -

    -     - - Graph showing a removable discontinuity: a hole in the graph when x=2 shows that the limit and function values disagree. - - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.1, - ymax=3.5, - xmin=-0.1, - xmax=4.3, - axis equal - ] - \addplot+[leftarrow,domain=0:1.98]{-0.35*x^2+x+2}; - \addplot [firstcurvestyle,rightarrow,domain=2.02:4]{-0.35*x^2+x+2}; - \addplot[soliddot] coordinates{(2,1)}; - \addplot[hollowdot] coordinates{(2,2.6)}; - \end{axis} - \end{tikzpicture} - - - - -
    -
    - - -

    - The graph of a function is shown for x from 0 to 4. - As x approaches 2 from the left, - the graph of f approaches a point that is not part of the graph, - as indicated by a hollow dot. - As x approaches 2 from the right, - the graph of f approaches a point that is part of the graph, - as indicated by a solid dot. - The point marked by the solid dot lies below the point marked by the hollow dot, - illustrating that the left and right hand limits are different as x\to 2. -

    - -

    - On the interval 0 \lt x \leq 2 the graph is curved downward and on the - interval 2 \leq x \leq 4 the graph is a straight line with a - positive slope. -

    -     - - Graph of a function with a jump discontinuity at x = 2. - - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.1, - ymax=3.5, - xmin=-0.1, - xmax=4.3, - axis equal - ] - \addplot+[leftarrow,domain=0:1.98]{-0.35*x^2+x+2}; - \addplot [firstcurvestyle,rightarrow,domain=2.02:4]{x-1}; - \addplot[soliddot] coordinates{(2,1)}; - \addplot[hollowdot] coordinates{(2,2.6)}; - \end{axis} - \end{tikzpicture} - - - - -
    -
    - - -

    - The graph of a function is shown for x from 0 to 4. - There is a - vertical dotted line at x = 2 illustrating a vertical asymptote. - As x approaches 2 from either side, - the graph of f extends upward along the asymptote, - indicating that the value of f(x) is increasing without bound. -

    -     - - Graph of a function with an infinite discontinuity at x = 2, - illustrated by a vertical asymtote. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.1, - xmax=4.3, - ymin=-1, - ymax=110, - ] - \addplot[firstcurvestyle,domain=0:1.9] {1/(x-2)^2}; - \addplot[firstcurvestyle,domain=2.1:4] {1/(x-2)^2}; - \addplot[asymptote,rightarrow] coordinates {(2,1) (2,100)}; - \end{axis} - \end{tikzpicture} - - - - - - - - - - - Consequences of continuity -

    - A common way of thinking of a continuous function is that - its graph can be sketched without lifting your pencil. That is, - its graph forms a continuous curve, - without holes, breaks or jumps. - This pseudo-definition glosses over some of the finer points of continuity. - There are some very strange continuous functions that one would be hard pressed to actually sketch by hand. -

    - -

    - However, this intuitive notion of continuity does help us understand another important concept as follows. - Suppose f is defined on [1,2], and f(1) = -10 and f(2) = 5. - If f is continuous on [1,2] (, its graph can be sketched as a continuous curve - from (1,-10) to (2,5)) then we know intuitively that somewhere on the interval [1,2] - f must be equal to -9, and -8, and -7,-6,\ldots,0,1/2, etc. - In short, f takes on all intermediate - values between -10 and 5. - It may take on more values; - f may actually equal 6 at some time, for instance, - but we are guaranteed all values between -10 and 5. This concept is illustrated in . -

    - -
    -
    - - -

    - The image shows the graph of three functions - defined on the interval 1 \leq x \leq 2. - One function is plotted in blue, using a solid line style. - Two other functions are plotted in red, with a dash-dot line style. - All three functions are continuous, and satisfy f(1)=-10, and f(2)=5. -

    - -

    - There are other lines marking certain values on the graphs. - A horizontal line at y=-10 indicates that all graphs start at the point (1,-10). - Another horizontal line at y=5 indicates that all graphs end at the point (2,5). - A third horizontal line indicates the value y=3, which is a value between -10 and 5. - There are also three shorter vertical lines, marking an x value on each of the three graphs where f(x)=3. -

    -     - - Graph illustrating the Intermediate Value Theorem. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=0.5, - xmax=2.5, - ymin=-15, - ymax=10, - minor ytick={-15,-14,...,10} - ] - \addplot+[smooth] coordinates {(1,-10) (1.5,-7) (1.8,3) (2,5)}; - \addplot+[smooth] coordinates {(1,-10) (1.2,6) (1.5,3) (2,5)}; - \addplot+[smooth] coordinates {(1,-10) (1.1,-12) (1.2,-5) (1.3,-7) (1.4,0) (1.5,-3) (1.6,3) (1.7,0) (1.8,5) (1.9,3) (2,5)}; - \addplot[soliddot] coordinates {(1,-10) (2,5)}; - \addplot[guideline] coordinates {(1,-10) (0.5,-10)}; - \addplot[guideline] coordinates {(2,5) (0.5,5)}; - \addplot[guideline] coordinates {(1.8,0) (1.8,3) (0.5,3)}; - \addplot[guideline] coordinates {(1.5,0) (1.5,3) (0.5,3)}; - \addplot[guideline] coordinates {(1.6,0) (1.6,3) (0.5,3)}; - \end{axis} - \end{tikzpicture} - - - - -

    - While this notion seems intuitive, - it is not trivial to prove and its importance is profound. - Therefore the concept is stated in the form of a theorem. -

    - - - Intermediate Value Theorem - -

    - Let f be a continuous function on [a,b] and, - without loss of generality, let f(a) \lt f(b). - Then for every value y, - where f(a) \lt y \lt f(b), - there is at least one value c in (a,b) such that f(c) = y. - - Intermediate Value Theorem - theoremIntermediate Value - -

    -     -     - - - - -

    - One important application of the is root finding. - Given a function f, - we are often interested in finding values of x where f(x) = 0. - These roots may be very difficult to find exactly. - Good approximations can be found through successive applications of this theorem. - Suppose through direct computation we find that - f(a) \lt 0 and f(b)\gt 0, where a\lt b. - The states that there is at least one c in (a,b) such that f(c) = 0. - The theorem does not give us any clue as to where to find such a value in the interval (a,b), - just that at least one such value exists. -

    - -

    - There is a technique that produces a good approximation of c. - Let d be the midpoint of the interval [a,b], - with f(a) \lt 0 and - f(b) \gt 0 and consider f(d). - There are three possibilities: -

      -
    1. -

      - f(d) = 0: We got lucky and stumbled on the actual value. - We stop as we found a root. -

      -
      2. -
    2. -

      - f(d) \lt 0: Then we know there is a root of f on the interval [d,b] we have halved the size of our interval, - hence are closer to a good approximation of the root. -

      -
      3. -
    3. -

      - f(d) \gt 0: Then we know there is a root of f on the interval [a,d] again,we have halved the size of our interval, - hence are closer to a good approximation of the root. -

      -
      4. -
    -

    - -

    - Successively applying this technique is called the - Bisection Method - - Bisection Method - - of root finding. - We continue until the interval is sufficiently small. - We demonstrate this in the following example. -

    - - - Using the Bisection Method - -

    - Approximate the root of f(x) = x-\cos(x), - accurate to three places after the decimal. -

    -     - -

    - Consider the graph of f(x) = x-\cos(x), - shown in . - It is clear that the graph crosses the x-axis somewhere near x=0.8. - To start the Bisection Method, - pick an interval that contains 0.8. - We choose [0.7,0.9]. - Note that all we care about are signs of f(x), - not their actual value, so this is all we display. -

    - -

    -

    -
  • - Iteration 1: -

    - f(0.7) \lt 0, f(0.9) \gt 0, and f(0.8) \gt 0. - So replace 0.9 with 0.8 and repeat. -

    -
    • -
  • - Iteration 2: -

    - f(0.7)\lt 0, f(0.8) \gt 0, - and at the midpoint, 0.75, - we have f(0.75) \gt 0. - So replace 0.8 with 0.75 and repeat. - Note that we don't need to continue to check the endpoints, - just the midpoint. - Thus we put the rest of the iterations in . -

    -
    • -
    -

    - -
    -
    - - - -

    - Graph of the function f(x)=x-\cos(x) on [0,1]. - The graph has an upward curve and intersects the x axis around - x = 0.8. There are two vertical guide lines, one at - x = 0.7, the other at x = 0.9. The guide lines - mark the interval in which the intercept occurs. -

    -     - - Graph of f(x) = x - cos(x), showing the x intercept between - values x = 0.7 and x = 0.9. - - - \begin{tikzpicture} - \begin{axis}[ - minor xtick={0,0.1,...,1}, - xmin=-.05, - xmax=1.07, - ymin=-1.1, - ymax=.7 - ] - \addplot+[domain=0:1] {x - cos(deg(x))}; - \addplot[guideline] coordinates {(0.7,0) (0.7,-0.0648)}; - \addplot[guideline] coordinates {(0.9,0) (0.9,0.278)}; - \end{axis} - \end{tikzpicture} - - - - - -
    Graphing the hyperbola in Graphing the hyperbola in - A hyperbola drawn from points A and B in + A hyperbola drawn from points A and B to illustrate the location property

    A hyperbola drawn from points A and B in . @@ -2161,7 +2161,7 @@

    - A fourth point found from hyperbolas given by points in + A fourth point found from hyperbolas given by points in the previous figures

    A hyperbola drawn from points B and C in , intersecting with the hyperbola drawn in . diff --git a/ptx/sec_deriv_intro.ptx b/ptx/sec_deriv_intro.ptx index 4ebefc72b..b3826eac5 100644 --- a/ptx/sec_deriv_intro.ptx +++ b/ptx/sec_deriv_intro.ptx @@ -795,13 +795,20 @@ .

    -

    +

    Thus our approximation of the equation of the tangent line is y = 0.9983(x-0) +0 = 0.9983x; it is graphed in . The graph seems to imply the approximation is rather good.

    +

    + Thus our approximation of the equation of the tangent line is + y = 0.9983(x-0) +0 = 0.9983x; + it is graphed in . + The graph seems to imply the approximation is rather good. +

    +
    f(x) = \sin(x) graphed with an approximation to its tangent line at x=0 f(x) = \sin(x) graphed with an approximation to its tangent line at x=0 Sage output r=a\sin(3\theta) Plotting f, p_2 and p_4 A graph of f(x) in , showing where f is increasing and decreasing \sin(x)/x\sin(x)/x near x=1\sin(x)/x near x=0Values of \sin(x)/x with x near 1Values of \sin(x)/x with x near 0Graphically approximating a limit in Numerically approximating a limit in Graphically approximating a limit in Numerically approximating a limit in Observing no limit as x\to 1 in Values of f(x) near x=1 in Observing no limit as x\to 1 in Values of f(x) near x=1 in Observing that f(x)=\sin(1/x) has no limit as x\to0 in - - - -

    - The graph of f(x)=sin(1/x) is shown, for x values between -1 and 1. - Like any sinusoidal graph, the curve oscillates back and forth between y=1 and y=-1. - However, as x gets close to 0, the argument of the sine function increases rapidly, - causing the distance between successive peaks to get smaller and smaller as the graph nears the y axis. - As x gets close to zero, the oscillations get so close together that it is no longer possible to distinguish them, - and the curve appears to become a solid, vertical strip. -

    -     - - Graph of the function sin(1/x), showing oscillations that become so rapid near the origin that they blur together. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1.1, - xmax=1.1, - ymin=-1.1, - ymax=1.2, - ] - \addplot[firstcurvestyle,leftarrow, - domain=-1.018591636:-0.318309886, % 1/(15/48 pi) to 1/(48/48 pi) - samples=34, % sample every pi/48 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=-0.318309886:-0.106103295, % 1/pi to 1/(3 pi) - samples=25, % sample every pi/12 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=-0.106103295:-0.063661977, % 1/(3 pi) to 1/(5 pi) - samples=49, % sample every pi/24 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=-0.063661977:-0.045472841, % 1/(5 pi) to 1/(7 pi) - samples=97, % sample every pi/48 through phase - ] {sin(1/x * 180 / pi)}; - \draw [color=firstcolor,fill=firstcolor] (axis cs:-0.045472841,-1) rectangle (axis cs:0.045472841,1); - %\addplot[firstcurvestyle,-,smooth, - % domain=-0.045472841:-0.0106103295, % 1/(7 pi) to 1/(30 pi) - % samples=93, % sample every pi/2 through phase - %] {sin(1/x * 180 / pi)}; - %\addplot[firstcurvestyle,-,smooth, - % domain=0.0106103295:0.045472841, % 1/(30 pi) to 1/(7 pi) - % samples=93, % sample every pi/2 through phase - %] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=0.045472841:0.063661977, % 1/(7 pi) to 1/(5 pi) - samples=97, % sample every pi/48 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=0.063661977:0.106103295, % 1/(5 pi) to 1/(3 pi) - samples=49, % sample every pi/24 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=0.106103295:0.318309886, % 1/(3 pi) to 1/pi - samples=25, % sample every pi/12 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,rightarrow, - domain=0.318309886:1.018591636, % 1/(48/48 pi) to 1/(15/48 pi) - samples=34, % sample every pi/48 through phase - ] {sin(1/x * 180 / pi)}; - \end{axis} - \end{tikzpicture} - - - - - -
    -
    - - - -

    - Another graph of f(x)=\sin(1/x) is shown, - this time zoomed in to show only the x interval from -0.1 to 0.1. - The features of the graph are the similar to what is visible over the larger interval: - further from the origin, we see the graph oscillating (rapidly) between y=1 and y=-1. - Near the orgin, the oscillations become so rapid that we can no longer tell them apart. - What we conclude from the graph is that on any interval containing x=0, - f(x)=\sin(1/x) takes on every y value between -1 and 1. - (In fact, f(x) attains every value infinitely many times!) -

    -     - - Graph of the same function, sin(1/x), shown with a smaller x interval, -0.1 to 0.1. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.11, - xmax=.11, - ymin=-1.1, - ymax=1.1 - ] - \addplot[firstcurvestyle,leftarrow, - domain=-0.100518911:-0.063661977, % 1/(152/48 pi) to 1/(240/48 pi) - samples=89, % sample every pi/48 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=-0.063661977:-0.045472841, % 1/(5 pi) to 1/(7 pi) - samples=25, % sample every pi/12 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=-0.045472841:-0.035367765, % 1/(7 pi) to 1/(9 pi) - samples=49, % sample every pi/24 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=-0.035367765:-0.024485376, % 1/(9 pi) to 1/(13 pi) - samples=145, % sample every pi/36 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=-0.024485376:-0.012732395, % 1/(13 pi) to 1/(25 pi) - samples=73, % sample every pi/6 through phase - ] {sin(1/x * 180 / pi)}; - \draw [color=firstcolor,fill=firstcolor] (axis cs:-0.012732395,-1) rectangle (axis cs:0.012732395,1); - %\addplot[firstcurvestyle,-,smooth, - % domain=-0.0031830988618379:-0.0010610329539459, % 1/(100 pi) to 1/(300 pi) - % samples=401, % sample every pi/2 through phase - %] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,leftarrow, - domain=0.100518911:0.063661977, % 1/(152/48 pi) to 1/(240/48 pi) - samples=89, % sample every pi/48 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=0.063661977:0.045472841, % 1/(5 pi) to 1/(7 pi) - samples=25, % sample every pi/12 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=0.045472841:0.035367765, % 1/(7 pi) to 1/(9 pi) - samples=49, % sample every pi/24 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=0.035367765:0.024485376, % 1/(9 pi) to 1/(13 pi) - samples=145, % sample every pi/36 through phase - ] {sin(1/x * 180 / pi)}; - \addplot[firstcurvestyle,-,smooth, - domain=0.024485376:0.012732395, % 1/(13 pi) to 1/(25 pi) - samples=73, % sample every pi/6 through phase - ] {sin(1/x * 180 / pi)}; - %\addplot[firstcurvestyle,-,smooth, - % domain=0.0031830988618379:0.0010610329539459, % 1/(100 pi) to 1/(300 pi) - % samples=401, % sample every pi/2 through phase - %] {sin(1/x * 180 / pi)}; - \end{axis} - \end{tikzpicture} - - - - - - - -
    -
    Observing that f(x)=\sin(1/x) has no limit as x\to0 in Interpreting a difference quotient as the slope of a secant lineSecant lines of f(x) at x=1 and x=1+h, for shrinking values of h (, h\rightarrow 0)h=2h=1h=0.5The difference quotient evaluated at values of h near 0 Illustrating the \varepsilon-\delta process. - With \varepsilon=0.5, we pick any \delta \lt 1.75 - - -

    - Graph of the function \sqrt{x}. - There are three points on the graph, the first is at x = 2.25, - f(2.25) = 1.5, giving the point (2.25, 1.5). For the - second point we have x = 4, making f(4) = 2, which gives - the point (4, 2). The third point found at x = 6.25 - which gives f(6.25) = 2.5, giving the point (6.25, 2.5). -

    -

    - Taking f(2.25) and adding \varepsilon we get 2, which is - f(4) = 2 and taking f(6.25) and subtracting \varepsilon - we again get 2. This shows the range given by \varepsilon=0.5. -

    -

    - There are two vertical lines near the y axis, at y = 2. One which goes - down and the other goes up. The length of the lines represents - \varepsilon = 0.5 and are used to show that the points (2.25, 1.5) - and (6.25, 2.5) are within \varepsilon of the point - (4, 2). -

    -     - - Graph of square root x with 3 points, shows that two points are within epsilon - of the other point. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.1, - xmax=7.1, - ymin=-.1, - ymax=3.1, - xtick={2,4,6}, - ytick={1,2}, - ] - \addplot+[infiniteright,domain=0:2.64575] ({x^2},{x}); - \addplot[soliddot] coordinates {(4,2)}; - \addplot[guideline] coordinates {(0,1.5) (2.25,1.5)}; - \addplot[guideline] coordinates {(0,2.5) (6.25,2.5)}; - \addplot[guideline,|->] coordinates {(0.2,2) (0.2,2.5)} node [below right]{$\varepsilon=0.5$} ; - \addplot[guideline,|->] coordinates {(0.2,2) (0.2,1.5)} node [above right]{$\varepsilon=0.5$} ; - \addplot[soliddot] coordinates {(2.25,1.5)}; - \addplot[soliddot] coordinates {(6.25,2.5)}; - \addplot[mark=none] coordinates {(0,2.5)} node [above right]{Choose $\varepsilon>0$. Then \ldots} ; - \end{axis} - \end{tikzpicture} - - - - - -
    -
    - - -

    - Graph of the equation sqrt{x}. - There are three points on the graph, the first is at x = 2.25, - f(2.25) = 1.5, giving the point (2.25, 1.5). For the - second point we have x = 4, making f(4) = 2, which - gives the point (4, 2). The third point found at x = 6.25 - which gives f(6.25) = 2.5, giving the point (6.25, 2.5). -

    -

    - This graph has the same vertical lines as - There are also horizontal lines near the x axis, showing the distance from - 2.25 to 4 is 1.75 and the distance from 6.25 to - 4 is 2.25. -

    -     - - Graph of square root x with 3 points, shows that two points are within epsilon - of the other point. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.1, - xmax=7.1, - ymin=-.1, - ymax=3.1, - xtick={2,4,6}, - ytick={1,2}, - ] - \addplot+[infiniteright,domain=0:2.64575] ({x^2},{x}); - \addplot[soliddot] coordinates {(4,2)}; - \addplot[guideline] coordinates {(0,1.5) (2.25,1.5)}; - \addplot[guideline] coordinates {(0,2.5) (6.25,2.5)}; - \addplot[guideline] coordinates {(2.25,0) (2.25,1.5)}; - \addplot[guideline] coordinates {(6.25,0) (6.25,2.5)}; - \addplot[guideline,|->] coordinates {(0.2,2) (0.2,2.5)} node [below right]{$\varepsilon=0.5$} ; - \addplot[guideline,|->] coordinates {(0.2,2) (0.2,1.5)} node [above right]{$\varepsilon=0.5$} ; - \addplot[guideline,|->] coordinates {(4,0.2) (2.25,0.2) } node [above right]{\parbox{3em}{width\\ 1.75}} ; - \addplot[guideline,|->] coordinates {(4,0.2) (6.25,0.2)} node [above left]{\parbox{3em}{\raggedleft width\\ 2.25}} ; - \addplot[soliddot] coordinates {(2.25,1.5)}; - \addplot[soliddot] coordinates {(6.25,2.5)}; - \addplot[mark=none] coordinates {(2.25,1.5)} node [below right]{\parbox{9em}{\ldots choose $\delta$ smaller\\ than each of these:}}; - \end{axis} - \end{tikzpicture} - - - - - - - -

    - Given the y tolerance \varepsilon =0.5, - we have found an x tolerance, \delta \lt 1.75, - such that whenever x is within \delta units of 4, - then y is within \varepsilon units of 2. - That's what we were trying to find. -

    - -

    - Let's try another value of \varepsilon. -

    - -

    - What if the y tolerance is 0.01, - \varepsilon =0.01? - How close to 4 does x have to be in order for - y to be within 0.01 units of 2? - (In other words for 1.99 \lt y \lt 2.01?) - Again, we just square these values to get 1.99^2 \lt x \lt 2.01^2, or - - 3.9601 \amp\lt x \lt 4.0401 - -0.0399\amp\lt x-4 \lt 0.0401 - -

    - -

    - What is the desired x tolerance? - In this case we must have \delta \lt 0.0399, - which is the minimum distance from 4 of the two bounds given above. -

    - -

    - What we have so far: if \varepsilon =0.5, - then \delta \lt 1.75 leads to f(x) being less than - \varepsilon from f(4) and if \varepsilon = 0.01, - then \delta \lt 0.0399 being less than \varepsilon from f(4). - A pattern is not easy to see, - so we switch to general \varepsilon try to determine an adequate \delta symbolically. - We start by assuming y=\sqrt{x} is within \varepsilon units of 2: - - \amp\abs{y - 2} \lt \varepsilon - -\varepsilon \amp\lt y - 2 \lt \varepsilon - -\varepsilon \amp\lt \sqrt{x} - 2 \lt \varepsilon \amp\amp (y=\sqrt{x}) - 2 - \varepsilon \amp\lt \sqrt{x} \lt 2+ \varepsilon \amp\amp \text{(Add 2)} - (2 - \varepsilon)^2 \amp\lt x \lt (2+ \varepsilon) ^2 \amp\amp \text{(Square all)} - 4 - 4\varepsilon + \varepsilon^2 \amp\lt x \lt 4 + 4\varepsilon + \varepsilon^2 \amp\amp \text{(Expand)} - -4\varepsilon + \varepsilon^2 \amp\lt x-4 \lt 4\varepsilon + \varepsilon^2\amp\amp \text{(Subtract 4)} - -

    - -

    - The desired form in the last step is - 4-\textit{something} \lt x \lt 4 + \textit{something}. - Since we want this last interval to describe an x tolerance around 4, - we have that either \delta \lt 4\varepsilon - \varepsilon^2 or - \delta \lt 4\varepsilon + \varepsilon^2, whichever is smaller: - - \delta \lt \min\{4\varepsilon - \varepsilon^2, 4\varepsilon + \varepsilon^2\} - . -

    - -

    - Since \varepsilon \gt 0, - we have 4\varepsilon - \varepsilon^2 \lt 4\varepsilon + \varepsilon^2, - the minimum is \delta \leq 4\varepsilon - \varepsilon^2. - That's the formula: given an \varepsilon, - set \delta \leq 4\varepsilon-\varepsilon^2. -

    - -

    - We can check this for our previous values. - If \varepsilon=0.5, the formula gives - \delta \lt 4(0.5) - (0.5)^2 = 1.75 and when \varepsilon=0.01, - the formula gives \delta \lt 4(0.01) - (0.01)^2 = 0.0399. -

    - -

    - So given any \varepsilon \gt 0, - set \delta \lt 4\varepsilon - \varepsilon^2. - Then if \abs{x-4}\lt \delta - (and x\neq 4), - then \abs{f(x) - 2} \lt \varepsilon, - satisfying the definition of the limit. - We have shown formally - (and finally!) - that \lim_{x\to 4} \sqrt{x} = 2. -

    - - - - -

    - The previous example was a little long in that we sampled a few specific cases of - \varepsilon before handling the general case. - Normally this is not done. - The previous example is also a bit unsatisfying in that \sqrt{4}=2; - why work so hard to prove something so obvious? - Many \varepsilon-\delta proofs are long and difficult to do. - In this section, - we will focus on examples where the answer is, frankly, - obvious, because the non-obvious examples are even harder. - In the next section we will learn some theorems that allow us to evaluate limits - analytically, that is, - without using the \varepsilon-\delta definition. -

    - - - - - Evaluating a limit using the definition - -

    - Show that \lim\limits_{x\to 2} x^2 = 4. -

    -     - -

    - Let's do this example symbolically from the start. - Let \varepsilon \gt 0 be given; - we want \abs{y-4} \lt \varepsilon, - , \abs{x^2-4} \lt \varepsilon. - How do we find \delta such that when \abs{x-2} \lt \delta, - we are guaranteed that \abs{x^2-4}\lt \varepsilon? -

    - -

    - This is a bit trickier than the previous example, - but let's start by noticing that \abs{x^2-4} = \abs{x-2}\cdot\abs{x+2}. - Consider: - - \abs{x^2-4} \lt \varepsilon \implies \abs{x-2}\cdot\abs{x+2} \lt \varepsilon \implies \abs{x-2} \lt \frac{\varepsilon}{\abs{x+2}} - . -

    - -

    - Could we not set \delta = \frac{\varepsilon}{\abs{x+2}}? -

    - -

    - We are close to an answer, - but the catch is that \delta must be a constant value - (so it can't depend on x). - There is a way to work around this, - but we do have to make an assumption. - Remember that \varepsilon is supposed to be a small number, - which implies that \delta will also be a small value. - In particular, we can (probably) assume that \delta \lt 1. - If this is true, - then \abs{x-2} \lt \delta would imply that \abs{x-2} \lt 1, - giving 1 \lt x \lt 3. -

    - -

    - Now, back to the fraction \frac{\varepsilon}{\abs{x+2}}. - If 1\lt x\lt 3, then 3\lt x+2\lt 5 - (add 2 to all terms in the inequality). - Taking reciprocals, we have - - \frac15 \lt \frac1{\abs{x+2}} \lt \frac 13 - , - which implies - - \frac15 \lt \frac1{\abs{x+2}} - , - which implies - - \frac{\varepsilon}{5}\lt \frac{\varepsilon}{\abs{x+2}} - . -

    - -

    - This suggests that we set \delta \lt \frac{\varepsilon}{5}. - To see why, let consider what follows when we assume \abs{x-2}\lt \delta: - - \abs{x-2}\amp\lt\delta - \abs{x-2}\amp\lt\frac{\varepsilon}{5}\amp\amp\text{(Our choice of } \delta\text{)} - \abs{x-2}\cdot\abs{x+2}\amp\lt\abs{x+2}\cdot\frac{\varepsilon}{5}\amp\amp\text{(Multiply by }\abs{x+2}\text{)} - \abs{x^2-4}\amp\lt\abs{x+2}\cdot\frac{\varepsilon}{5}\amp\amp\text{(Simplify left side)} - \abs{x^2-4}\amp\lt\abs{x+2}\cdot\frac{\varepsilon}{\abs{x+2}}\amp\amp\text{(}Inequality, \delta \lt 1\text{)} - \abs{x^2-4}\amp\lt\varepsilon - -

    - -

    - We have arrived at \abs{x^2-4}\lt \varepsilon as desired. - Note again, in order to make this happen we needed \delta - to first be less than 1. That is a safe assumption; - we want \varepsilon to be arbitrarily small, - forcing \delta to also be small. -

    - -

    - We have also picked \delta to be smaller than necessary. - We could get by with a slightly larger \delta, - as shown in . - The outer lines show the boundaries defined by our choice of \varepsilon. - The inner lines show the boundaries defined by setting \delta = \varepsilon/5. - Note how these dotted lines are within the dashed lines. - That is perfectly fine; - by choosing x within the dotted lines we are guaranteed that - f(x) will be within \varepsilon of 4. -

    - -
    -
    Choosing \delta = \varepsilon/5 in An illustration of the Squeeze TheoremThe unit circle and related trianglesBounding the sector between two triangles - - -

    - A right triangle with two dotted lines inside it. One is curved and - represents a sector of a circle inside the triangle. - The sector of a circle has an area of \frac{1}{2}r^2 \theta. - The other dotted line represents an acute triangle housed within the - larger right triangle. -

    -

    - The right triangle has the side lengths 1 for its base and \tan(\theta) - for the vertical line. Where these lines connect is the right angle. There is - an angle \theta that is shared by both triangles and the circle sector. -

    -     - - Right triangle housing an acute triangle and sector of a circle. - - - \begin{tikzpicture}[x=100pt,y=100pt,scale=0.9] - \fill [draw=black,thick,fill=firstcolor!20] (0,0) node [shift={(22.5:0.2)}] {$\theta$} -- (1,.839) -- node [pos=.5,below,rotate=90] {$\tan(\theta)$} (1,0) -- cycle; - \draw (.5,0) node [below] {$1$}; - \draw [black,dashed,thick] (1,0) arc (0:40:1); - \draw [black,dashed,thick] (1,0) -- (.766,.643); - \end{tikzpicture} - - - - - -
    -
    - - -

    - Sector of a circle from with dotted lines - creating the right triangle from , there is - also a dotted line showing the same acute triangle from - . The acute triangle is contained inside - the circle sector, while the right triangle purtrudes out of it. -

    -

    - There is an angle \theta that is shared by both triangles and - the circle sector. They also share the length 1 for the horizontal line. -

    -     - - Sector of a circle from previous image with dotted lines - creating the right and acute triangles. - - - \begin{tikzpicture}[x=100pt,y=100pt] - \fill [draw=black,thick,fill=firstcolor!20] (0,0) node [shift={(22.5:0.2)}] {$\theta$} -- (1,0) arc(0:40:1) -- cycle; - \draw (.5,0) node [below] {$1$}; - \draw [black,dashed,thick] (1,0) arc (0:40:1); - \draw [black,dashed,thick] (1,0) -- (.766,.643) -- (1,.839) -- node [pos=.5,below,rotate=90,opacity=0] {$\tan(\theta)$} cycle; - \end{tikzpicture} - - - - - -
    -
    - - -

    - Acute triangle from with dotted lines - representing the circle sector from and - right triangle from . There is an additional - dotted line that creates a second right triangle within the acute triangle. - Both the sector of a circle and right triangle fall outside the bounds of the - acute triangle. -

    -

    - The second right triangle a height of \sin(\theta). As with - and the horizontal - line of the triangle is 1. -

    -     - - Acute triangle from previous image with dotted lines creating the - two right triangle an circle sector. - - - \begin{tikzpicture}[x=100pt,y=100pt] - \fill [draw=black,thick,fill=firstcolor!20] (0,0) node [shift={(22.5:0.2)}] {$\theta$} -- (1,0) --(.766,.643) -- cycle; - \draw [dashed,thick] (.766,0) -- node [pos=.4,above,rotate=90] {$\sin(\theta)$}(.766,.643); - \draw (.766,5pt) -- ++(5pt,0) -- ++(0,-5pt); - \draw (.5,0) node [below] {$1$}; - \draw [black,dashed,thick] (1,0) arc(0:40:1) -- (1,.839) -- node [pos=.5,below,rotate=90,opacity=0] {$\tan(\theta)$} cycle; - \end{tikzpicture} - - - - - - - -

    - Multiply all terms by \frac{2}{\sin(\theta)}, giving - - \frac{1}{\cos(\theta)} \geq \frac{\theta}{\sin(\theta)} \geq 1 - . -

    - -

    - Taking reciprocals reverses the inequalities, giving - - \cos(\theta) \leq \frac{\sin(\theta)}{\theta} \leq 1 - . - (These inequalities hold for all values of \theta near 0, - even negative values, - since \cos(-\theta) = \cos(\theta) and - \sin(-\theta) = -\sin(\theta).) -

    - -

    - Now take limits. - - \lim_{\theta\to 0} \cos(\theta)\amp \leq \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} \leq \lim_{\theta\to 0} 1 - \cos(0) \amp\leq \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} \leq 1 - 1\amp \leq \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} \leq 1 - -

    - -

    - Clearly this means that - \lim\limits_{\theta\to 0} \frac{\sin(\theta)}{\theta}=1. -

    - - - - - - - - -

    - Two notes about the are worth mentioning. - First, one might be discouraged by this application, - thinking I would never - have come up with that on my own. - This is too hard! Don't be discouraged; - within this text we will guide you in your use of the . - As one gains mathematical maturity, - clever proofs like this are easier and easier to create. -

    - -

    - Second, this limit tells us more than just that as x approaches 0, - \sin(x)/x approaches 1. - Both x and \sin(x) are approaching 0, - but the ratio of x and \sin(x) approaches 1, - meaning that they are approaching 0 in essentially the same way. - Another way of viewing this is: - for small x, the functions y=x and - y=\sin(x) are essentially indistinguishable. -

    - - - -

    - We include this special limit, - along with three others, in the following theorem. -

    - - - Special Limits - -

    -

      -
    1. \lim\limits_{x\to 0} \dfrac{\sin(x)}{x} = 1
      2. - -
    2. \lim\limits_{x\to 0} \dfrac{\cos(x)-1}{x} = 0
      3. - -
    3. \lim\limits_{x\to 0} (1+x)^{1/x} = e
      4. - -
    4. \lim\limits_{x\to 0} \dfrac{e^x-1}{x} = 1
      5. -
    -

    -     -     - -

    - A short word on how to interpret the latter three limits. - We know that as x goes to 0, - \cos(x) goes to 1. - So, in the second limit, - both the numerator and denominator are approaching 0. - However, since the limit is 0, - we can interpret this as saying that - \cos(x) is approaching 1 faster than x is approaching 0. -

    - -

    - In the third limit, - inside the parentheses we have an expression that is approaching 1 - (though never equaling 1), - and we know that 1 raised to any power is still 1. - At the same time, the power is growing toward infinity. - What happens to a number near 1 raised to a very large power? - In this particular case, - the result approaches Euler's number, - e, approximately 2.718. -

    - -

    - In the fourth limit, we see that as x\to 0, - e^x approaches 1 just as fast as x\to 0, - resulting in a limit of 1. -

    - - - -

    - The special limits stated in - are called indeterminate forms; - - limitindeterminate form - - in this case they are of the form 0/0, - except the third limit, which is of a different form. - You'll learn techniques to find these limits exactly using calculus in - . -

    - -

    - Our final theorem for this section will be motivated by the following example. -

    - - - Using algebra to evaluate a limit - -

    - Evaluate the following limit: - - \lim_{x\to 1}\frac{x^2-1}{x-1} - . -

    -     - -

    - We begin by attempting to apply - and substituting 1 for x in the quotient. - This gives: - - \lim_{x\to 1}\frac{x^2-1}{x-1} = \frac{1^2-1}{1-1} - - which is of the form \frac{0}{0}, an indeterminate form. - We cannot apply the theorem. -

    - -

    - By graphing the function, - as in , - we see that the function seems to be linear, - implying that the limit should be easy to evaluate. - Recognize that the numerator of our quotient can be factored: - - \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} - . - The function is not defined when x=1, - but for all other x, - - \frac{x^2-1}{x-1}\amp = \frac{(x-1)(x+1)}{x-1} - \amp = \frac{\cancel{(x-1)}(x+1)}{\cancel{(x-1)} } - \amp = x+1, \quad \text{if } x \neq 1 - -

    - -
    -
    Graphing f in to understand a limitA graph of f in A graph of f from Graphing f in Graphing f in A graph of f in A graph of the step function in A graph of f(x)=\sqrt{x-1}+\sqrt{5-x}Illustrating three common types of discontinuityThe graph of a function with a removable discontinuity at x=2The graph of a function with a jump discontinuity at x=2The graph of a function with an infinite discontinuity at x=2Illustration of the Intermediate Value Theorem: the output 3 is in between -10 and 5, - and therefore any continuous function on [1,2] with f(1) = -10 and f(2) = 5 - will achieve the output 3 somewhere in [1,2]Graphing a root of f(x) = x-\cos(x)
    - Iterations of the Bisection Method of Root Finding - - - Iteration # - Interval - Midpoint Sign - - - - - - - - 1 - [\highlight{0.7},0.9] - f(0.8) \gt 0 - - - 2 - [\highlight{0.7},0.8] - f(0.75) \gt 0 - - - 3 - [\highlight{0.7},\highlight{0.75}] - f(0.725)\lt 0 - - - 4 - [0.725,\highlight{0.75}] - f(0.7375)\lt 0 - - - 5 - [\highlight{0.7375},\highlight{0.75}] - f(0.7438)\gt - - - 6 - [\highlight{0.7375},0.7438] - f(0.7407)\gt 0 - - - 7 - [\highlight{0.7375},0.7407] - f(0.7391)\gt 0 - - - 8 - [\highlight{0.7375},\highlight{0.7391}] - f(0.7383)\lt 0 - - - 9 - [0.7383,\highlight{0.7391}] - f(0.7387)\lt 0 - - - 10 - [0.7387,\highlight{0.7391}] - f(0.7389)\lt 0 - - - 11 - [0.7389,\highlight{0.7391}] - f(0.7390)\lt 0 - - - 12 - [0.7390,\highlight{0.7391}] - - - - - - - -
    - -

    - Notice that in the 12th iteration we have the endpoints of the interval each starting with 0.739. - Thus we have narrowed the zero down to an accuracy of the first three places after the decimal. - Using a computer, we have - - f(0.7390) = -0.00014, f(0.7391) = 0.000024 - . -

    - -

    - Either endpoint of the interval gives a good approximation of where f is 0. - The states that the actual zero is still within this interval. - While we do not know its exact value, - we know it starts with 0.739. -

    - -

    - This type of exercise is rarely done by hand. - Rather, it is simple to program a computer to run such an algorithm and stop when the endpoints differ by a preset small amount. - One of the authors did write such a program and found the zero of f to be - 0.7390851332, accurate to 10 places after the decimal. - While it took a few minutes to write the program, - it took less than a thousandth of a second for the program to run the necessary 35 iterations. - In less than 8 hundredths of a second, - the zero was calculated to 100 decimal places - (with less than 200 iterations). -

    - - - - - - -

    - It is a simple matter to extend the Bisection Method to solve problems similar to Find x, - where f(x) = 0. For instance, - we can find x, where f(x) = 1. - It actually works very well to define a new function g where g(x) = f(x) - 1. - Then use the Bisection Method to solve g(x)=0. -

    - -

    - Similarly, given two functions f and g, - we can use the Bisection Method to solve f(x) = g(x). - Once again, create a new function h where - h(x) = f(x)-g(x) and solve h(x) = 0. -

    - -

    - In - another equation solving method will be introduced, called Newton's Method. - In many cases, Newton's Method is much faster. - It relies on more advanced mathematics, though, - so we will wait before introducing it. -

    - -

    - This section formally defined what it means to be a continuous function. - Most functions that we deal with are continuous, - so often it feels odd to have to formally define this concept. - Regardless, it is important, and forms the basis of the next chapter. -

    - - - - - - - - Terms and Concepts - - - - -

    - In your own words, describe what it means for a function to be continuous. -

    - -     - - -     - - - - -

    - In your own words, describe what the Intermediate Value Theorem states. -

    - -     - - -     - - - - -

    - What is a root of a function? -

    - -     - - - -

    - A root of a function f is a value c such that f(c)=0. -

    -     - -     - - - - -

    - Given functions f and g on an interval I, - how can the Bisection Method be used to find a value c where f(c) = g(c)? -

    - -     - - - -

    - Consider the function h(x) = g(x) - f(x), - and use the Bisection Method to find a root of h. -

    -     - -     - - - - - -

    - - If f is defined on an open interval containing c, - and \lim\limits_{x\to c}f(x) exists, - then f is continuous at c. -

    -     - -

    - There is more to the definition of continuity than existence of the limit! - What else needs to be true? -

    -     - -     - - - - - -

    - - If f is defined on an open interval containing c, and - f is continuous at c, - then \lim\limits_{x\to c}f(x) exists. -

    -     - -

    - The definition of continuity requires that \lim_{x\to c}f(x)=f(c). - This is not possible if the limit does not exist! -

    -     - -     - - - - - -

    - - If f is defined on an open interval containing c, - and f is continuous at c, - then \lim\limits_{x\to c^+}f(x) = f(c). -

    -     - -

    - If f is continuous at c, - then the limit of f at c must exist. - If the limit exists, then both one-sided limits must also exist. -

    -     - -     - - - - - -

    - - If f is continuous on [a,b], - then \lim\limits_{x\to a^-}f(x) = f(a). -

    -     - -

    - The left-hand limit at a involves points outside the interval [a,b], - and continuity on [a,b] only tells us about points inside [a,b]. - In particular, it is the right-hand limit at a that must equal f(a). -

    -     - -     - - - - - -

    - - If f is continuous on [0,1) and [1,2), - then f is continuous on [0,2). -

    -     - -

    - Review the discussion following . -

    -     - -     - - - - - -

    - - The sum of continuous functions is also continuous. -

    -     - -

    - See . -

    -     - -     -     - - - Problems - - - -

    - Use the graph to determine if the function is continuous at the given point. -

    -     - - - - parserPopUp.pl - - - $a=1; - @b=(1,2,1,2,0.5,0); - $left=Formula("(($b[1]-$b[0])/(1-0))*(x-0)+$b[0]"); - $right=Formula("(x-1)*(x-1.5)*$b[5]/((2-1)*(2-1.5))+(x-1)*(x-2)*$b[4]/((1.5-1)*(1.5-2))+(x-1.5)*(x-2)*$b[3]/((1-1.5)*(1-2))"); - $answer=DropDown(['Yes.','No.'],1,showInStatic=>0); - $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; - - -

    - Is f in the graph below continuous at ? -

    - - -

    - The graph of a piecewise-defined function defined on [0,2], consisting of two parts. - The first part is a straight line with intercept (0,1) and a positive - slope. It ends at (1,2), but this point is not part of the graph. - The second part begins at (1,2), but does not include this point, and curves downward, ending at (2,0). - There is also a solid dot at (1,1), indicating that f(1)=1. -

    -     - - The graph used for the current exercise. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-0.5, - xmax=2.5, - ymin=-0.5, - ymax=2.5 - ] - \addplot[firstcurvestyle, domain=0:1] {$left}; - \addplot[firstcurvestyle, domain=1:2] {$right}; - \addplot[soliddot] coordinates {(0,$b[0]) (1, $b[2]) (2,$b[5])}; - \addplot[hollowdot] coordinates {(1,$b[1]) (1,$b[3])}; - \end{axis} - \end{tikzpicture} - - -

    - -

    -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $a=1; - @b=(0,1,2,2,1,0); - $left=Formula("(($b[1]-$b[0])/(1-0))*(x-0)+$b[0]"); - $right=Formula("(($b[5]-$b[3])/(2-1))*(x-1)+$b[3]"); - $answer=DropDown(['Yes.','No.'],1,showInStatic=>0); - $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; - - -

    - Is f in the graph below continuous at ? -

    - - -

    - Graph of a piecewise-defined function with the domain [0,2]. The graph has - two straight lines, the first is defined by the solid point (0 ,0) - and the hollo point (1, 1), and has a positive slope. - The second line is defined by the solid points (1, 2) - and (2, 2), and has a negative slope. -

    -     - - The graph used for the current exercise. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-0.5, - xmax=2.5, - ymin=-0.5, - ymax=2.5 - ] - \addplot[firstcurvestyle, domain=0:1] {$left}; - \addplot[firstcurvestyle, domain=1:2] {$right}; - \addplot[hollowdot] coordinates {(1,$b[1]) (1,$b[3])}; - \addplot[soliddot] coordinates {(0,$b[0]) (1, $b[2]) (2,$b[5])}; - \end{axis} - \end{tikzpicture} - - -

    - -

    -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $a=1; - @b=(0,Compute("INF"),Compute("DNE"),Compute("INF"),1,0); - $left=Formula("sec(x*90)-1"); - $right=Formula("-sec(x*90)-1"); - $lowmidx = acos(1/3.5)*2/pi; - $highmidx = 2 - acos(1/3.5)*2/pi; - $answer=DropDown(['Yes.','No.'],1,showInStatic=>0); - $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; - - -

    - Is f in the graph below continuous at ? -

    - - -

    - The image shows the graph of a function that is defined on the interval [0,2], - except at x=, where it has a vertical asymptote. - On either side of the vertical asymptote, as x gets close to , - the y value increases without bound. -

    - -     - - The graph used for the current exercise. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-0.5, - xmax=2.5, - ymin=-0.5, - ymax=2.5 - ] - \addplot[firstcurvestyle, infiniteright, domain=0:$lowmidx] {$left}; - \addplot[firstcurvestyle, infiniteleft, domain=$highmidx:2] {$right}; - \addplot[asymptote, -] coordinates {(1,-0.5) (1,2.5)}; - \addplot[soliddot] coordinates {(0,$b[0]) (2,$b[5])}; - \end{axis} - \end{tikzpicture} - - -

    - -

    -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $a=0; - ($b[1],$b[2],$b[3],$b[4]) = random_subset(4,0,0.5,1,1.5,2); - for my $i(0,5) {$b[$i]=random(0,2,0.5)}; - $left=Formula("(($b[1]-$b[0])/(1-0))*(x-0)+$b[0]"); - $right=Formula("(x-1)*(x-1.5)*$b[5]/((2-1)*(2-1.5))+(x-1)*(x-2)*$b[4]/((1.5-1)*(1.5-2))+(x-1.5)*(x-2)*$b[3]/((1-1.5)*(1-2))"); - $answer=DropDown(['Yes.','No.'],0,showInStatic=>0); - $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; - - -

    - Is f in the graph below continuous at ? -

    - - -

    - Graph of a piecewise-defined function with domain [0,2]. The graph has - two lines; the first is straight and defined by the solid - point (0, ) and hollow point - (1, ). The second line - is defined by the hollow point (1, - ) and solid point - (2, ). When - x = 1 f(x) = -

    -     - - The graph used for the current exercise. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-0.5, - xmax=2.5, - ymin=-0.5, - ymax=2.5 - ] - \addplot[firstcurvestyle, domain=0:1] {$left}; - \addplot[firstcurvestyle, domain=1:2] {$right}; - \addplot[hollowdot] coordinates {(1,$b[1]) (1,$b[3])}; - \addplot[soliddot] coordinates {(0,$b[0]) (1,$b[2]) (2,$b[5])}; - \end{axis} - \end{tikzpicture} - - -

    - -

    -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $a=1; - for my$i(0..5){$b[$i]=random(0,2,0.5)}; - $b[2] = $b[1]; - $b[3] = $b[1]; - $right=Formula("(($b[1]-$b[0])/(1-2))*(x-2)+$b[0]"); - $left=Formula("(x-0.5)*(x-1)*$b[5]/((0-0.5)*(0-1))+(x-0)*(x-1)*$b[4]/((0.5-0)*(0.5-1))+(x-0)*(x-0.5)*$b[3]/((1-0)*(1-0.5))"); - $answer=DropDown(['Yes.','No.'],0,showInStatic=>0); - $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; - - -

    - Is f in the graph below continuous at ? -

    - - -

    - The graph - changes at x = 1 and is defined by the solid dots - (0, ) and - (2, ). - At the interval 0 \leq x \lt 1 the graph is curved - and on the interval 1 \lt x \leq 2 the line is straight. -

    -     - - The graph used for the current exercise. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-0.5, - xmax=2.5, - ymin=-0.5, - ymax=2.5 - ] - \addplot[firstcurvestyle, domain=0:1] {$left}; - \addplot[firstcurvestyle, domain=1:2] {$right}; - \addplot[soliddot] coordinates {(0,$b[5]) (2,$b[0])}; - \end{axis} - \end{tikzpicture} - - -

    - -

    -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $a=4; - ($b[1],$b[3]) = random_subset(2,-4..4); - for my $i (0,4) {$b[$i]=random(-4,4,1)}; - $answer=DropDown(['Yes.','No.'],0,showInStatic=>0); - $ymin=min($b[0],$b[1]-$b[0],$b[3],$b[4]-$b[3])-1; - $ymax=max($b[0],$b[1]-$b[0],$b[3],$b[4]-$b[3])+1; - $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; - - -

    - Is f in the graph below continuous at ? -

    - - -

    - The graph is made up of two curves. The first is defined - by the solid points (-4, ) and - (0, ). - The second is defined by the hollow point - (0, ) and - solid point (4, ). -

    -     - - The graph used for the current exercise. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-5, - xmax=5, - ymin=$ymin, - ymax=$ymax - ] - \addplot[firstcurvestyle, domain=-4:0] {($b[0] + $b[1] + sin((x+2)*45) * ($b[1]-$b[0]))/2}; - \addplot[firstcurvestyle, domain=0:4] {($b[3] + $b[4] - sin((x+2)*45) * ($b[4]-$b[3]))/2}; - \addplot[hollowdot] coordinates {(0,$b[3])}; - \addplot[soliddot] coordinates {(-4,$b[0]) (0,$b[1]) (4,$b[4])}; - \end{axis} - \end{tikzpicture} - - -

    - -

    -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - @b=(0,2,0,2,0,0,0,2,Compute("DNE"),2,0); - for my$i(0..3){$f[$i]=Formula("($b[3*$i+1]-$b[3*$i])/2*(x-(2*$i-4))+$b[3*$i]");} - @choices=('Yes.','No.'); - for my$i(0..2){if($b[3*$i+2]==Compute("DNE")){$correct[$i]=1;} - elsif($f[$i]->eval(x=>(2*$i-2))!=$f[$i+1]->eval(x=>(2*$i-2))){$correct[$i]=1;} - elsif($f[$i]->eval(x=>(2*$i-2))!=$b[3*$i+2]){$correct[$i]=1;} - else{$correct[$i]=0;}; - $answer[$i]=DropDown(~~@choices,$correct[$i],showInStatic=>0); - }; - $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; - - -

    - Is f in the graph below continuous at -2, - 0, and 2? -

    - - -

    - There are four lines that make up this graph. The first - going from left to right is defined by the solid point - (-4, 0) and hollow point (-2. 2), and has - a positive slope. The second line is defined by the points - (-2, 2), 0, 0 with a negative slope. The - third line by (0, 0), 2, 2 with a positive slope. - And the last line is defined by the hollow point (2, 2) - and solid point (4, 0), with a negative slope. -

    -

    - For x = -2, f(x) = 0, so there is a solid point at - (-2, 0). -

    -     - - The graph used for the current exercise. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = -5, - xmax = 5, - ymin = -4, - ymax = 4, - ] - \addplot[firstcurvestyle, domain=-4:-2] {x+4}; - \addplot[firstcurvestyle, domain=-2:0] {-x}; - \addplot[firstcurvestyle, domain=0:2] {x}; - \addplot[firstcurvestyle, domain=2:4] {4-x}; - \addplot[soliddot] coordinates {(-4,0) (-2,0) (4,0)}; - \addplot[hollowdot] coordinates {(-2,2) (2,2)}; - \end{axis} - \end{tikzpicture} - - -

    - At -2: -

    -

    - -

    -

    - At 0: -

    -

    - -

    -

    - At 2: -

    -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - Context()->flags->set(reduceConstants=>0); - $a=Formula("3pi/2"); - $g=Formula("1+sin(x*180/pi)"); - $answer=DropDown(['Yes.','No.'],0,showInStatic=>0); - $showwork = '[@ explanation_box(message => "If not, state why it is not.") @]*'; - - -

    - Is f in the graph below continuous at ? -

    - - -

    - The graph has a downward curve on the - interval 0 \leq x \leq \pi and an upward curve on - the interval \pi \leq x \leq 2\pi. The peak of the - graph is at x = \frac{\pi}{2} and the lowest point - is at x = \frac{3\pi}{2}. There are also - no breaks in the graph. -

    -     - - The graph used for the current exercise. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = -pi/2, - xmax = 2.5*pi, - ymin = -0.5, - ymax = 2.5, - xtick={ - -2*pi, -(3*pi)/2, -pi, -pi/2, - pi/2, pi, (3*pi)/2, 2*pi - }, - xticklabels={ - \(-2\pi\), \(-\frac{3\pi}{2}\), \(-\pi\), \(-\frac{\pi}{2}\), - \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\), \(2\pi\) - } - ] - \addplot[firstcurvestyle, domain=0:2*pi, smooth] {$g}; - \end{axis} - \end{tikzpicture} - - -

    - -

    -

    - -

    -     -     -     -     - - - -

    - Determine if f is continuous at the indicated values. -

    -     - - - - - parserPopUp.pl - - - Context("PiecewiseFunction"); - @choices=('Yes.', - 'No.' - ); - @answer=(DropDown(~~@choices,0,showInStatic=>0),DropDown(~~@choices,0,showInStatic=>0)); - $showwork = '[@ explanation_box(message => "If not, explain why not.") @]*'; - - -

    - f(x)=\begin{cases}1\amp x=0\\\frac{\sin(x)}{x}\amp x\neq0\end{cases} -

    -     - - - -

    - Is f is continuous at 0? -

    -

    - -

    -

    - -

    -     -     - - - -

    - Is f is continuous at \pi? -

    -

    - -

    -

    - -

    -     -     -     -     - - - - - parserPopUp.pl - - - Context("PiecewiseFunction"); - #for my$i(0..3){$a[$i]=non_zero_random(-2,2,1)}; - #if($envir{problemSeed}==1){@a=(1,-1,1,-2)}; - @a=(1,-1,1,-2); - $left=Formula("$a[0]x^3+$a[1]x^2")->reduce; - $right=Formula("$a[2]x+$a[3]")->reduce; - $f=PiecewiseFunction("x<1" => "$left","x>=1"=>"$right"); - @choices=('Yes.', - 'No.' - ); - $correct[1]=($left->eval(x=>1)==$right->eval(x=>1))?0:1; - @answer=(DropDown(~~@choices,0,showInStatic=>0),DropDown(~~@choices,$correct[1],showInStatic=>0)); - $showwork = '[@ explanation_box(message => "If not, explain why not.") @]*'; - - -

    - f(x)= -

    -     - - - -

    - Is f is continuous at 0? -

    -

    - -

    -

    - -

    -     -     - - - -

    - Is f is continuous at 1? -

    -

    - -

    -

    - -

    -     -     -     -     - - - - - parserPopUp.pl - - - Context("PiecewiseFunction"); - #if($envir{problemSeed}==1){@a=(-1,-4,-2,10)}; - @a=(-1,-4,-2,10); - $f=PiecewiseFunction("x!=$a[0]" => Formula("(x^2-($a[0]+$a[1])x+($a[0]*$a[1]))/(x^2-($a[0]+$a[2])x+($a[0]*$a[2]))")->reduce,"x=$a[0]"=>Real(($a[0]-$a[1])/($a[0]-$a[2]))); - @choices=('Yes.', - 'No.' - ); - @answer=(DropDown(~~@choices,0,showInStatic=>0),DropDown(~~@choices,0,showInStatic=>0)); - $showwork = '[@ explanation_box(message => "If not, explain why not.") @]*'; - - -

    - f(x)= -

    -     - - - -

    - Is f is continuous at ? -

    -

    - -

    -

    - -

    -     -     - - - -

    - Is f is continuous at ? -

    -

    - -

    -

    - -

    -     -     -     -     - - - - - parserPopUp.pl - - - Context("PiecewiseFunction"); - #if($envir{problemSeed}==1){@a=(0,8,-8,3,5,0)}; - @a=(0,8,-8,3,5,0); - $f=PiecewiseFunction("x!=$a[1]" => Formula("(x^2-($a[1]+$a[2])x+($a[1]*$a[2]))/(x^2-($a[1]+$a[3])x+($a[1]*$a[3]))")->reduce,"x=$a[1]"=>$a[4]); - @choices=('Yes.', - 'No.' - ); - $correct[1]=($a[4]==($a[1]-$a[2])/($a[1]-$a[3]))?0:1; - @answer=(DropDown(~~@choices,0,showInStatic=>0),DropDown(~~@choices,$correct[1],showInStatic=>0)); - $showwork = '[@ explanation_box(message => "If not, explain why not.") @]*'; - - -

    - f(x)= -

    -     - - - -

    - Is f is continuous at ? -

    -

    - -

    -

    - -

    -     -     - - - -

    - Is f is continuous at ? -

    -

    - -

    -

    - -

    -     -     -     -     -     - - - -

    - Give the intervals on which the function is continuous. -

    -     - - - - - Context("Interval"); - $a=random(-9,9,1); - $b=non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=-3;$b=9;}; - $f=Formula("x^2+$a x+$b")->reduce; - $interval=List(Interval("(-inf,inf)")); - - -

    - f(x)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     - -

    - Since f is a polynomial function, - it is continuous on (-\infty,\infty). -

    -     -     -     - - - - - Context("Interval"); - $a=random(1,5,1); - if($envir{problemSeed}==1){$a=2;}; - $f=Formula("sqrt(x^2-$a**2)")->reduce; - $interval=List("(-inf,-$a], [$a,inf)"); - - -

    - f(x)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     - -

    - The domain of f is , - and since f is a composition of continuous functions on that domain, - it is continuous on . -

    -     -     -     - - - - - Context("Interval"); - $a=random(1,5,1); - if($envir{problemSeed}==1){$a=2;}; - $f=Formula("sqrt($a**2-x^2)")->reduce; - $interval=List("[-$a,$a]"); - - -

    - f(x)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     - -

    - The domain of f is , - and since f is a composition of continuous functions on that domain, - it is continuous on . -

    -     -     -     - - - - - Context("Interval"); - $a=random(1,5,1); - $b=random(1,5,1); - if($envir{problemSeed}==1){$a=1;$b=1}; - $f1=Formula("sqrt($a-x)"); - $f2=Formula("sqrt(x+$b)"); - $f=$f1+$f2; - $interval=List(Interval("[-$b,$a]")); - - -

    - f(x)=+ -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     - -

    - The domain of is (-\infty,], - and the domain of is [-,\infty). - So the domain of f is . - And since f is the sum of two compositions of continuous functions, - it is continuous on . -

    -     -     -     - - - - - Context("Interval"); - $a=random(2,5,1); - $b=random(2,9,1); - if($envir{problemSeed}==1){$a=5;$b=6}; - Context()->variables->are(t=>'Real'); - $f=Formula("sqrt($a t^2-$a*$b)")->reduce; - $interval=List("(-inf,-sqrt($b)], [sqrt($b),inf)"); - - -

    - f(t)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     -     -     - - - - - Context("Interval"); - $a=random(1,9,1); - if($envir{problemSeed}==1){$a=1}; - Context()->variables->are(t=>'Real'); - $f=Formula("1/sqrt($a**2-t^2)")->reduce; - $interval=List(Interval("(-$a,$a)")); - - -

    - g(t)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     -     -     - - - - - Context("Interval"); - $a=random(1,9,1); - $b=random(1,9,1); - if($envir{problemSeed}==1){$a=1;$b=1;}; - Context()->variables->are(t=>'Real'); - $f=Formula("1/($a+$b*t^2)")->reduce; - $interval=Compute("(-inf,inf)"); - - -

    - g(t)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     -     -     - - - - - Context("Interval"); - $b=list_random(2..9,Formula("e"),Formula("pi")); - if($envir{problemSeed}==1){$b=Formula("e");}; - $f=Formula("$b^x")->reduce; - $interval=List(Interval("(-inf,inf)")); - - -

    - f(x)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     -     -     - - - - - Context("Interval"); - Context()->variables->are(s=>'Real'); - $b=list_random(2..9,Formula("e"),Formula("pi")); - if($envir{problemSeed}==1){$b=Formula("e");}; - $f=($b==Formula("e"))?'\ln(s)':"\log_{$b}(s)"; - $interval=List(Interval("(0,inf)")); - - -

    - g(s)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     -     -     - - - - - Context("Interval"); - Context()->variables->are(t=>'Real'); - $f=list_random(Formula("cos(t)"),Formula("sin(t)")); - if($envir{problemSeed}==1){$f=Formula("cos(t)");}; - $interval=List(Interval("(-inf,inf)")); - - -

    - h(t)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     -     -     - - - - - Context("Interval"); - $a=random(1,9,1); - if($envir{problemSeed}==1){$a=1}; - Context()->variables->are(k=>'Real'); - $f=Formula("sqrt($a-e^k)"); - $interval=($a==1)?List(Interval("(-inf,0]")):List(Interval("(-inf,ln($a)]")); - - -

    - f(k)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     -     -     - - - - - Context("Interval"); - $a=list_random('sin','cos'); - $b=list_random(2,3,'e'); - $n=random(2,5,1); - if($envir{problemSeed}==1){$a='sin';$b='e';$n=2;}; - $f=Formula("$a($b^x+x^($n))"); - $interval=List(Interval("(-inf,inf)")); - - -

    - f(x)= -

    - - Use interval notation. If the answer is a list of more than one interval, use commas to separate them. - -

    - -

    -     -     -     -     - - - -

    - Test your understanding of the Intermediate Value Theorem. -

    -     - - - - -

    - Let f be continuous on [1,5] where - f(1) = -2 and f(5) = -10. - Does a value 1\lt c\lt 5 exist such that f(c) = -9? - Why/why not? -

    - -     - - - -

    - Yes, by the Intermediate Value Theorem. -

    -     - -     - - - - -

    - Let g be continuous on [-3,7] where - g(0) = 0 and g(2) = 25. - Does a value -3\lt c\lt 7 exist such that g(c) = 15? - Why/why not? -

    - -     - - - -

    - Yes, by the Intermediate Value Theorem. - In fact, we can be more specific and state such a value c - exists in (0,2), not just in (-3,7). -

    -     - -     - - - - -

    - Let f be continuous on [-1,1] where - f(-1) = -10 and f(1) = 10. - Does a value -1\lt c\lt 1 exist such that f(c) = 11? - Why/why not? -

    - -     - - - -

    - We cannot say; - the Intermediate Value Theorem only applies to function values between -10 and 10; - as 11 is outside this range, we do not know. -

    -     - -     - - - - -

    - Let h be a function on [-1,1] where - h(-1) = -10 and h(1) = 10. - Does a value -1\lt c\lt 1 exist such that h(c) = 0? - Why/why not? -

    - -     - - - -

    - We cannot say; - the Intermediate Value Theorem only applies to continuous functions. - As we do know know if h is continuous, we cannot say. -

    -     - -     -     - - - -

    - Use the Bisection Method to approximate, - accurate to two decimal places, - the value of the root of the given function in the given interval. -

    -     - - - - - parserPopUp.pl - - - Context("Interval"); - Context()->flags->set(tolType=>'absolute',tolerance=>0.00000005); - #do{$b=non_zero_random(-9,9,1);$c=non_zero_random(-9,9,1);}until($b**2>4*$c and int(sqrt(abs($b**2-4*$c)))!=sqrt(abs($b**2-4*$c))); - #if($envir{problemSeed}==1){$b=2;$c=-4;}; - $b=2;$c=-4; - $f=Formula("x^2+$b x+$c")->reduce; - $left=floor((-$b+sqrt($b**2-4*$c))/2); - $right=$left+1; - $mid=($left+$right)/2; - ($l[0],$r[0])=($f->eval(x=>$mid)>0)?($left,$mid):($mid,$right); - $in[0]=Compute("[$l[0],$r[0]]"); - for my$i(1..7){$m[$i-1]=($l[$i-1]+$r[$i-1])/2; - ($l[$i],$r[$i])=($f->eval(x=>$m[$i-1])>0)?($l[$i-1],$m[$i-1]):($m[$i-1],$r[$i-1]); - $popup[$i-1]=($f->eval(x=>$m[$i-1])>0)?DropDown(['+','-'],'0',showInStatic=>0):DropDown(['+','-'],'1',showInStatic=>0); - $in[$i]=Compute("[$l[$i],$r[$i]]"); - }; - $z=Real(($l[7]+$r[7])/2)->with(tolType=>'absolute',tolerance=>0.01); - #$showwork = '[@ explanation_box(message => "Show the steps you used applying the Bisection Method.") @]*'; - - -

    - f(x)= on the interval -

    - - If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. - - - - Iteration - Interval - Midpoint Sign - - - 1 - - - - - 2 - - - - - 3 - - - - - 4 - - - - - 5 - - - - - 6 - - - - - 7 - - - - - 8 - - - - - - Enter the solution to f(x)=0, accurate to two decimal places. - -

    - -

    - -     -     -     - - - - - parserPopUp.pl - - - Context("Interval"); - Context()->flags->set(tolType=>'absolute',tolerance=>0.00000005,reduceConstants=>0); - #$b=random(2,6,1); - #if($envir{problemSeed}==1){$b=2;}; - $b=2; - $f=Compute("sin(x)-1/$b"); - $root=asin(1/$b); - ($l[0],$r[0])=(floor(20*$root)/20,floor(20*$root)/20+0.05); - $in[0]=Compute("[$l[0],$r[0]]"); - for my$i(1..7){$m[$i-1]=($l[$i-1]+$r[$i-1])/2; - ($l[$i],$r[$i])=($f->eval(x=>$m[$i-1])>0)?($l[$i-1],$m[$i-1]):($m[$i-1],$r[$i-1]); - $popup[$i-1]=($f->eval(x=>$m[$i-1])>0)?DropDown(['+','-'],'0',showInStatic=>0):DropDown(['+','-'],'1',showInStatic=>0); - $in[$i]=Compute("[$l[$i],$r[$i]]"); - }; - $z=Real(($l[7]+$r[7])/2)->with(tolType=>'absolute',tolerance=>0.01); - #$showwork = '[@ explanation_box(message => "Show the steps you used applying the Bisection Method.") @]*'; - - -

    - f(x)= on the interval -

    - - If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. - - - - Iteration - Interval - Midpoint Sign - - - 1 - - - - - 2 - - - - - 3 - - - - - 4 - - - - - 5 - - - - - 6 - - - - - 7 - - - - - 8 - - - - - - Enter the solution to f(x)=0, accurate to two decimal places. - -

    - -

    - -     -     -     - - - - - parserPopUp.pl - - - Context("Interval"); - Context()->flags->set(tolType=>'absolute',tolerance=>0.00000005); - #$b=random(1.5,3,0.1); - #if($envir{problemSeed}==1){$b=2;}; - $b=2; - $f=Compute("e^x-$b"); - $root=ln($b); - ($l[0],$r[0])=(floor(20*$root)/20,floor(20*$root)/20+0.05); - $in[0]=Compute("[$l[0],$r[0]]"); - for my$i(1..7){$m[$i-1]=($l[$i-1]+$r[$i-1])/2; - ($l[$i],$r[$i])=($f->eval(x=>$m[$i-1])>0)?($l[$i-1],$m[$i-1]):($m[$i-1],$r[$i-1]); - $popup[$i-1]=($f->eval(x=>$m[$i-1])>0)?DropDown(['+','-'],'0',showInStatic=>0):DropDown(['+','-'],'1',showInStatic=>0); - $in[$i]=Compute("[$l[$i],$r[$i]]"); - }; - $z=Real(($l[7]+$r[7])/2)->with(tolType=>'absolute',tolerance=>0.01); - #$showwork = '[@ explanation_box(message => "Show the steps you used applying the Bisection Method.") @]*'; - - -

    - f(x)= on the interval -

    - - If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. - - - - Iteration - Interval - Midpoint Sign - - - 1 - - - - - 2 - - - - - 3 - - - - - 4 - - - - - 5 - - - - - 6 - - - - - 7 - - - - - 8 - - - - - - Enter the solution to f(x)=0, accurate to two decimal places. - -

    - -

    - -     -     -     - - - - - parserPopUp.pl - - - Context("Interval"); - Context()->flags->set(tolType=>'absolute',tolerance=>0.00000005); - #$b=random(1.5,3,0.1); - #if($envir{problemSeed}==1){$b=2;}; - $b=2; - $f=Compute("cos(x)-sin(x)"); - $root=pi/4; - ($l[0],$r[0])=(floor(10*$root)/10,floor(10*$root)/10+0.1); - $in[0]=Compute("[$l[0],$r[0]]"); - for my$i(1..7){$m[$i-1]=($l[$i-1]+$r[$i-1])/2; - ($l[$i],$r[$i])=($f->eval(x=>$m[$i-1])<0)?($l[$i-1],$m[$i-1]):($m[$i-1],$r[$i-1]); - $popup[$i-1]=($f->eval(x=>$m[$i-1])>0)?DropDown(['+','-'],'0',showInStatic=>0):DropDown(['+','-'],'1',showInStatic=>0); - $in[$i]=Compute("[$l[$i],$r[$i]]"); - }; - $z=Real(($l[7]+$r[7])/2)->with(tolType=>'absolute',tolerance=>0.01); - #$showwork = '[@ explanation_box(message => "Show the steps you used applying the Bisection Method.") @]*'; - - -

    - f(x)= on the interval -

    - - If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign. - - - - Iteration - Interval - Midpoint Sign - - - 1 - - - - - 2 - - - - - 3 - - - - - 4 - - - - - 5 - - - - - 6 - - - - - 7 - - - - - 8 - - - - - - Enter the solution to f(x)=0, accurate to two decimal places. - -

    - -

    - -     -     -     -     -     -     -
    -
    - Limits Involving Infinity - -

    - In - we stated that in the equation \lim_{x\to c}f(x) = L, - both c and L were numbers. - In this section we relax that definition a bit by considering situations when it makes sense to let c and/or L be infinity. -

    - -

    - As a motivating example, consider f(x) = 1/x^2, - as shown in . - Note how, as x approaches 0, f(x) grows very, - very large in fact, it grows without bound. - It seems appropriate, and descriptive, to state that - - \lim_{x\to 0} \frac1{x^2}=\infty - . -

    - -

    - Also note that as x gets very large, - f(x) gets very, very small. - We could represent this concept with notation such as - - \lim_{x\to \infty} \frac1{x^2}=0 - . -

    - -
    - Graphing f(x) = 1/x^2 for values of x near 0 - - - -

    - Graph of f(x)=1/x^2 for x between -1 and 1. There is a vertical - asymptote at x = 0 and a horizontal asymptote at y = 0. - For x values near the left and right edges of the image, - the y value is close to 0. - For x values near 0, the graph extends to the top of the image (and presumably beyond), - suggesting that y approaches \infty. -

    -     - - Graph of 1 over x squared. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1.1, - xmax=1.1, - ymin=-.1, - ymax=110, - ] - \addplot[firstcurvestyle,infinite,domain=-1:-0.1] {1/x^2}; - \addplot[firstcurvestyle,infinite,domain=0.1:1] {1/x^2}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - We explore both types of use of \infty in turn. -

    -     - - - Infinite limits - - - Limit of Infinity, <m>\infty</m> - -

    - Let I be an open interval containing c, - and let f be a function defined on I, - except possibly at c. -

      -
    • -

      - The limit of f(x), - as x approaches c, - is infinity, denoted by - - \lim_{x\rightarrow c} f(x) = \infty - , - if given any N \gt 0, - there exists \delta \gt 0 such that for all x in I, - where x\neq c, - if \abs{x - c} \lt \delta, then f(x) \gt N. -

      -
      • -
    • -

      - The limit of f(x), - as x approaches c, - is negative infinity, denoted by - - \lim_{x\rightarrow c} f(x) = -\infty - , - if given any N \lt 0, - there exists \delta \gt 0 such that for all x in I, - where x\neq c, - if \abs{x - c} \lt \delta, then f(x) \lt N. -

      -
      • -
    - limitof infinity - -

    -     -     - - - - - - -

    - The first definition is similar to the - \varepsilon-\delta definition in - from . - In that definition, given any (small) value \varepsilon, - if we let x get close enough to c - (within \delta units of c) - then f(x) is guaranteed to be within \varepsilon of L. - Here, given any (large) value N, - if we let x get close enough to c - (within \delta units of c), - then f(x) will be at least as large as N. - In other words, if we get close enough to c, - then we can make f(x) as large as we want. -

    - -

    - It is important to note that by saying - \lim_{x\to c}f(x) = \infty we are implicitly stating that - the limit of f(x), - as x approaches c, - does not exist. A limit only exists when f(x) approaches an actual numeric value. - We use the concept of limits that approach infinity because it is helpful and descriptive. - It is one specific way - in which a limit can fail to exist. -

    - -

    - We define one-sided limits that approach infinity in a similar way. -

    - - - One-Sided Limits of Infinity - -

    -

      -
    • -

      - Let f be a function defined on (a,c) for some a\lt c. - We say the limit of f(x), - as x approaches c from the left, is infinity, or, - the left-hand limit of f at c is infinity, - denoted by - - \lim_{x\rightarrow c^-} f(x) = \infty - , - if given any N \gt 0, - there exists \delta \gt 0 such that for all - a\lt x\lt c, - if \abs{x - c} \lt \delta, then f(x) \gt N. -

      -
      • -
    • -

      - Let f be a function defined on (c,b) for some b \gt c. - We say the limit of f(x), - as x approaches c from the right, is infinity, or, - the right-hand limit of f at c is infinity, - denoted by - - \lim_{x\rightarrow c^+} f(x) = \infty - , - if given any N \gt 0, - there exists \delta \gt 0 such that for all - c\lt x\lt b, - if \abs{x - c} \lt \delta, then f(x) \gt N. -

      -
      • -
    • -

      - The term left- (or, - right-) hand limit of f at c is negative infinity - is defined in a manner similar to . -

      -
      • -
    -

    -     -     - - - Evaluating limits involving infinity - -

    - Find \lim\limits_{x\to 1}\frac1{(x-1)^2} as shown in . -

    - -
    - Observing infinite limit as x\to 1 in - - - -

    - Graph of f(x)=\frac{1}{(x-1)^2} for x between 0 and 1. - There is a vertical - asymptote at x = 1 and a horizontal asymptote at y = 0. - As x gets near 1 from either side of the vertical asymptote, - y approaches \infty. - For x values near the left and right edges of the image, - the value of y approaches 0. -

    -     - - Graph of 1 over (x-1) squared. Has a vertical asymptote at x = 1. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.1, - xmax=2.1, - ymin=-1, - ymax=110, - ] - \addplot[firstcurvestyle,infinite,domain=0:0.9] {1/(x-1)^2}; - \addplot[firstcurvestyle,infinite,domain=1.1:2] {1/(x-1)^2}; - \addplot[asymptote,rightarrow] coordinates {(1,1) (1,100)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -

    - In - of , - by inspecting values of x close to 1 we concluded that this limit does not exist. - That is, it cannot equal any real number. - But the limit could be infinite. - And in fact, - we see that the function does appear to be growing larger and larger, - as f(0.99)=10^4, f(0.999)=10^6, f(0.9999)=10^8. - A similar thing happens on the other side of 1. - From the graph and the numeric information, - we could state \lim_{x \to 1}1/(x-1)^2=\infty. - We can prove this by using -

    - -

    - In general, let a large value N be given. - Let \delta=1/\sqrt{N}. - If x is within \delta of 1, - , if \abs{x-1}\lt 1/\sqrt{N}, then: - - \abs{x-1} \amp \lt \frac{1}{\sqrt{N}} - (x-1)^2 \amp \lt \frac{1}{N} - \frac{1}{(x-1)^2} \amp \gt N - , - which is what we wanted to show. - So we may say \lim_{x\to 1}1/{(x-1)^2}=\infty. -

    -     - -     - - - Evaluating limits involving infinity - -

    - Find \lim\limits_{x\to 0}\frac1x, - as shown in . -

    - -
    - Evaluating \lim\limits_{x\to 0}\frac1x in - - - -

    - Graph of y=1/x, for x between -1 and 1. - There is a vertical asymptote at x = 0 and a horizontal asymptote at - y = 0. As x approaches 0 from the left, y - approaches -\infty and from the right, y approaches - \infty. As y approaches 0 from the bottom, x - approaches -\infty and from the top, x approaches - \infty. The graph conists of two parts; one in quadrant one and the other in - quadrant three. -

    -     - - Graph of 1 over x. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1.1, - xmax=1.1, - ymin=-55, - ymax=55, - ] - \addplot[firstcurvestyle,domain=-1:-0.125,leftarrow] {1/x}; - \addplot[firstcurvestyle,domain=-8:-50,rightarrow] ({1/x},{x}); - \addplot[firstcurvestyle,domain=1:0.125,leftarrow] {1/x}; - \addplot[firstcurvestyle,domain=8:50,rightarrow] ({1/x},{x}); - \end{axis} - \end{tikzpicture} - - - -
    -     - -

    - It is easy to see that the function grows without bound near 0, - but it does so in different ways on different sides of 0. - Since its behavior is not consistent, - we cannot say that \lim_{x\to 0}\frac{1}{x}=\infty. - Instead, we will say \lim_{x\to 0}\frac{1}{x} does not exist. - However, we can make a statement about one-sided limits. - We can state that \lim_{x\to 0^+}\frac1x=\infty and \lim_{x\to 0^-}\frac1x=-\infty. -

    -     - -     - - -     - - - Vertical asymptotes -

    - The graphs in the two previous examples demonstrate that if a function f has a limit (or, - left- or right-hand limit) of infinity at x=c, - then the graph of f looks similar to a vertical line near x=c. - This observation leads to a definition. -

    - - - Vertical Asymptote - -

    - Let I be an interval that either contains c or has c as an endpoint, - and let f be a function defined on I, - except possibly at c. -

    - -

    - If the limit of f(x) as x approaches c from either the left or right - (or both) - is \infty or -\infty, - then the line x=c is a vertical asymptote of f. - - asymptotevertical - -

    -     -     - - - - - Finding vertical asymptotes - -

    - Find the vertical asymptotes of f(x)=\frac{3x}{x^2-4}. -

    -     - -

    - Vertical asymptotes occur where the function grows without bound; - this can occur at values of c where the denominator is 0. - When x is near c, the denominator is small, - which in turn can make the function take on large values. - In the case of the given function, - the denominator is 0 at x=\pm 2. - Substituting in values of x close to 2 and -2 seems to indicate that the function tends toward \infty or -\infty at those points. - We can graphically confirm this by looking at . - Thus the vertical asymptotes are at x=\pm2. -

    - -
    - Graphing f(x) = \frac{3x}{x^2-4} - - - -

    - Graph of f(x)=\frac{3x}{x^2-4}. There are - two vertical asymptotes, one at x = -2 and the other at - x = 2. For x values less than -2, f(x) is - less than 0 and the graph is curved downward. For x - values greater than 2, f(x) is greater than 0 - and the graph is curved upward. For the interval -2 \lt x \lt 0 - the graph is curved upward, at x = 0 the graph changes and - starts to curve downward for the interval 0 \lt x \lt 2. -

    -

    - Because of the asymptote at x = -2, as x gets near - -2 from the left f(x) approaches -\infty. - But coming from the right f(x) approaches \infty. - Because of the other asymptote at x = 2, as x get near - 2 from the left f(x) approaches -\infty. But - coming from the right f(x) approaches \infty. -

    -     - - Graph of a rational function with two vertical asymptotes. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-6.1, - xmax=6.1, - ymin=-16, - ymax=16, - ] - \addplot[firstcurvestyle,domain=-6:-2.5,leftarrow] {3*x/(x^2-4)}; - \addplot[firstcurvestyle,domain=-3.333:-15.366,rightarrow] ({(3+sqrt(9+16*x^2))/(2*x)},{x}); - \addplot[firstcurvestyle,domain=14.615:2.571,leftarrow] ({(3-sqrt(9+16*x^2))/(2*x)},{x}); - \addplot[firstcurvestyle,domain=-1.5:1.5,-] {3*x/(x^2-4)}; - \addplot[firstcurvestyle,domain=-2.571:-14.615,rightarrow] ({(3-sqrt(9+16*x^2))/(2*x)},{x}); - \addplot[firstcurvestyle,domain=3.333:15.366,rightarrow] ({(3+sqrt(9+16*x^2))/(2*x)},{x}); - \addplot[firstcurvestyle,domain=2.5:6,rightarrow] {3*x/(x^2-4)}; - \addplot[asymptote] coordinates {(-2,-16) (-2,16)}; - \addplot[asymptote] coordinates {(2,-16) (2,16)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - -

    - When a rational function has a vertical asymptote at x=c, - we can conclude that the denominator is 0 at x=c. - However, just because the denominator is 0 at a certain point does not mean there is a vertical asymptote there. - For instance, - f(x)=(x^2-1)/(x-1) does not have a vertical asymptote at x=1, - as shown in . - While the denominator does get small near x=1, - the numerator gets small too, - matching the denominator step for step. - In fact, factoring the numerator, we get - - f(x)=\frac{(x-1)(x+1)}{x-1} - . -

    - -

    - Canceling the common term, we get that f(x)=x+1 for x\not=1. - So there is clearly no asymptote; - rather, a hole exists in the graph at x=1. -

    - -
    - Graphically showing that f(x) = \frac{x^2-1}{x-1} does not have an asymptote at x=1 - - - -

    - The graph is a single straight line with a positive slope. - At x = 1 there is a hollow dot indicating a removable discontinuity. - The exact position of the discontinuity is (1, 2). -

    -     - - Graph of a rational function. A zero in the denominator is a hole, not a vertical asymptote. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1.2, - xmax=2.2, - ymin=-.2, - ymax=3.2, - ] - \addplot+[infinite,domain=-1:2] {x+1}; - \addplot[hollowdot] coordinates {(1,2)}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - The above example may seem a little contrived. - Another example demonstrating this important concept is f(x)= (\sin(x) )/x. - We have considered this function several times in the previous sections. - We found that \lim_{x\to0}\frac{\sin(x) }{x}=1; - , there is no vertical asymptote. - No simple algebraic cancellation makes this fact obvious; - we used the in to prove this. -

    - -

    - If the denominator is 0 at a certain point but the numerator is not, - then there will usually be a vertical asymptote at that point. - On the other hand, - if the numerator and denominator are both zero at that point, - then there may or may not be a vertical asymptote at that point. - This case where the numerator and denominator are both zero returns us to an important topic. -

    -     - - - Indeterminate Forms -

    - limitindeterminate form - indeterminate form -

    - -

    - We have seen how the limits \lim_{x\to 0}\frac{\sin(x) }{x} and - \lim_{x\to1}\frac{x^2-1}{x-1} each return the indeterminate form 0/0 when we blindly plug in x=0 and x=1, - respectively. - However, 0/0 is not a valid arithmetical expression. - It gives no indication that the respective limits are 1 and 2. -

    - -

    - With a little cleverness, - one can come up with 0/0 expressions which have a limit of \infty, 0, or any other real number. - That is why this expression is called - indeterminate. -

    - -

    - A key concept to understand is that such limits do not really return 0/0. - Rather, keep in mind that we are taking limits. - What is really happening is that the numerator is shrinking to 0 while the denominator is also shrinking to 0. - The respective rates at which they do this are very important and determine the actual value of the limit. -

    - -

    - An indeterminate form indicates that one needs to do more work in order to compute the limit. - That work may be algebraic - (such as factoring and canceling), - it may involve using trigonometric identities or logarithm rules, - or it may require a tool such as the . - In - we will learn yet another technique called L'Hospital's Rule that provides another way to handle indeterminate forms. -

    - -

    - Some other common indeterminate forms are - \infty-\infty, \infty\cdot 0, - \infty/\infty, 0^0, - \infty^0 and 1^{\infty}. - Again, keep in mind that these are the blind - results of directly substituting c into the expression, - and each, - in and of itself, has no meaning. - The expression \infty-\infty does not really mean - subtract infinity from infinity. Rather, - it means One quantity is subtracted from the other, - but both are growing without bound. What is the result? - It is possible to get every value between -\infty and \infty. -

    - -

    - Note that 1/0 and \infty/0 are not indeterminate forms, - though they are not exactly valid mathematical expressions, either. - In each, the function is growing without bound, - indicating that the limit will be \infty, -\infty, - or simply not exist if the left- and right-hand limits do not match. -

    -     - - - Limits at Infinity and Horizontal Asymptotes -

    - At the beginning of this section we briefly considered what happens to - f(x) = 1/x^2 as x grew very large. - Graphically, it concerns the behavior of the function to the - far right of the graph. - We make this notion more explicit in the following definition. -

    - - - - - Limits at Infinity and Horizontal Asymptotes - -

    - Let L be a real number. -

      -
    1. -

      - Let f be a function defined on - (a,\infty) for some number a. - The limit of f at infinity is L, - denoted \lim_{x\to\infty} f(x)=L, - if for every \epsilon \gt 0 there exists M \gt a such that if x \gt M, - then \abs{f(x)-L}\lt \epsilon. - - limitat infinity - asymptotehorizontal - -

      -
      2. -
    2. -

      - Let f be a function defined on - (-\infty,b) for some number b. - The limit of f at negative infinity is L, - denoted \lim_{x\to-\infty} f(x)=L, - if for every \epsilon \gt 0 there exists M\lt b such that if x \lt M, - then \abs{f(x)-L}\lt \epsilon. -

      -
      3. -
    3. -

      - If \lim_{x\rightarrow\infty} f(x)=L or \lim_{x\rightarrow-\infty} f(x)=L, - we say the line y=L is a - horizontal asymptote of f. -

      -
      4. -
    -

    -     -     - -

    - We can also define limits such as - \lim_{x\to\infty}f(x)=\infty by combining this definition with - . -

    - - - Approximating horizontal asymptotes - -

    - Approximate the horizontal asymptote(s) of f(x)=\frac{x^2}{x^2+4}. -

    -     - -

    - We will approximate the horizontal asymptotes by approximating the limits - \lim_{x\to-\infty} \frac{x^2}{x^2+4} and - \lim_{x\to\infty} \frac{x^2}{x^2+4}. - (A rational function can have at most - one horizontal asymptote. - So we could get away with only taking x \to \infty). -

    - -

    - shows a sketch of f, - and the table in - gives values of f(x) for large magnitude values of x. - It seems reasonable to conclude from both of these sources that f has a horizontal asymptote at y=1. -

    - -
    - Using a graph and a table to approximate a horizontal asymptote in - -
    - - -

    - Graph of f(x)=\frac{x^2}{x^2+4} showing x values from -20 to 20. - There is a horizontal asymptote at y = 1. - As x approaches -\infty and \infty, f(x) gets near 1, - but never equals 1. The graph lies between y=0 and y=1, - and drops to the point (0,0) - as x approaches 0 from either direction. -

    -     - - Graph of a rational function showing a horizontal asymptote at y = 1. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-21, - xmax=21, - ymin=-.2, - ymax=1.1, - ] - \addplot+[infinite,domain=-84:84,samples=177] ({2*tan(x)},{-(cos(x))^2+1}); - \addplot[asymptote] coordinates {(-21,1) (21,1)}; - \end{axis} - \end{tikzpicture} - - - -
    - -
    - - - x - f(x) - - - 10 - 0.9615 - - - 100 - 0.9996 - - - 10000 - 0.999996 - - - -10 - 0.9615 - - - -100 - 0.9996 - - - -10000 - 0.999996 - - -
    -     -
    - -

    - Later, we will show how to determine this analytically. -

    -     -     - - - - - -

    - Horizontal asymptotes can take on a variety of forms. - - shows that f(x) = x/(x^2+1) has a horizontal asymptote of y=0, - where 0 is approached from both above and below. -

    - -

    - - shows that f(x) =x/\sqrt{x^2+1} has two horizontal asymptotes; - one at y=1 and the other at y=-1. -

    - -

    - - shows that f(x) = \sin(x)/x has even more interesting behavior than at just x=0; - as x approaches \pm\infty, - f(x) approaches 0, - but oscillates as it does this. -

    - - -
    - Considering different types of horizontal asymptotes - -
    - - - - -

    - As x approaches -\infty and \infty, f(x) gets near 0. - The graph dips to a minimum value just to the left of the y axis, - then crosses the x axis at (0,0), rising to a maximum value just to the right of the x axis, - before falling again toward the horizontal asymptote y=0. -

    -     - - Graph of a rational function with x axis as horizontal asymptote. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-21, - xmax=21, - ymin=-1.1, - ymax=1.1, - ] - \addplot+[infinite,domain=-87:87,samples=101] ({tan(x)},{sin(x)*cos(x)}); - \end{axis} - \end{tikzpicture} - - - -
    - -
    - - - - -

    - The graph of f(x)=\frac{x}{\sqrt{x^2+1}}, which has two horizontal asymptotes, - one at y = -1 (representing the limit as x\to -\infty) - and the other at y = 1 (representing the limit as x\to\infty). -

    -     - - Graph of a function with two horizontal asymptotes. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-21, - xmax=21, - ymin=-1.1, - ymax=1.1, - ] - \addplot+[infinite,domain=-87:87,samples=101] ({tan(x)},{sin(x)}); - \end{axis} - \end{tikzpicture} - - - -
    - -
    - - - - -

    - The graph of f(x)=\sin(x)/x, - which oscillates around the x axis. - As the absolute value of x gets bigger, the oscillations get smaller. - The limit as x\to \pm \infty is zero, so f(x) has a horizontal asymptote that it crosses an infinite number of times. -

    -     - - Graph of sin(x) over x, showing that a function can cross a horizontal asymptote infinitely many times. - - - \begin{tikzpicture}[declare function = {func(\x) = (\x != 0) * (sin(x*180/pi)/x) + (\x == 0) * (1);}] - \begin{axis}[ - xmin=-21, - xmax=21, - ymin=-.3, - ymax=1.1 - ] - \addplot+[infinite,domain=-20:20,samples=101] {func(x)}; - \end{axis} - \end{tikzpicture} - - - -
    -     -
    - -

    - We can analytically evaluate limits at infinity for rational functions once we understand \lim_{x\to\infty}\frac{1}{x}. - As x gets larger and larger, - 1/x gets smaller and smaller, approaching 0. - We can, in fact, - make 1/x as small as we want by choosing a large enough value of x. - Given \varepsilon, - we can make 1/x\lt \varepsilon by choosing x \gt 1/\varepsilon. - Thus we have \lim_{x\to\infty} 1/x=0. -

    - -

    - It is now not much of a jump to conclude the following: - - \lim_{x\to\infty}\frac1{x^n}\amp=0\amp\lim_{x\to-\infty}\frac1{x^n}\amp=0 - . -

    - - - - -

    - Now suppose we need to compute the following limit: - - \lim_{x\to\infty}\frac{x^3+2x+1}{4x^3-2x^2+9} - . -

    - -

    - A good way of approaching this is to divide through the numerator and denominator by x^3 - (hence multiplying by 1), - which is the largest power of x to appear in the denominator. - Doing this, we get - - \lim_{x\to\infty}\frac{x^3+2x+1}{4x^3-2x^2+9} \amp = - \lim_{x\to\infty}\frac{1/x^3}{1/x^3}\cdot\frac{x^3+2x+1}{4x^3-2x^2+9} - \amp =\lim_{x\to\infty}\frac{x^3/x^3+2x/x^3+1/x^3}{4x^3/x^3-2x^2/x^3+9/x^3} - \amp = \lim_{x\to\infty}\frac{1+2/x^2+1/x^3}{4-2/x+9/x^3} - . -

    - -

    - Then using the rules for limits - (which also hold for limits at infinity), - as well as the fact about limits of 1/x^n, - we see that the limit becomes - - \frac{1+0+0}{4-0+0}=\frac14 - . -

    - -

    - This procedure works for any rational function. - In fact, it gives us the following theorem. -

    - - - Limits of Rational Functions at Infinity - -

    - Let f(x) be a rational function of the following form: - - f(x)=\frac{a_nx^n + a_{n-1}x^{n-1}+\dots + a_1x + a_0}{b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0} - , - where m,n are positive integers and where any of the coefficients may be 0 except for a_n and b_m. - Then: -

      -
    1. -

      - If n=m, then - - \lim_{x\to\infty} f(x) = \lim_{x\to-\infty} f(x) = \frac{a_n}{b_m} - . -

      -
      2. -
    2. -

      - If n\lt m, then - - \lim_{x\to\infty} f(x) = \lim_{x\to-\infty} f(x) = 0 - . -

      -
      3. -
    3. -

      - If n \gt m, - then \lim_{x\to\infty} f(x) and \lim_{x\to-\infty} f(x) are both infinite. -

      -
      4. -
    -

    -     -     - -

    - We can see why this is true. - If the highest power of x is the same in both the numerator and denominator ( n=m), - we will be in a situation like the example above, - where we will divide by x^n and in the limit all the terms will approach 0 except for - a_nx^n/x^n and b_mx^m/x^n. - Since n=m, this will leave us with the limit a_n/b_m. - If n\lt m, then after dividing through by x^m, - all the terms in the numerator will approach 0 in the limit, - leaving us with 0/b_m or 0. - If n \gt m, and we try dividing through by x^m, - we end up with the denominator tending to b_m while the numerator tends to \infty. -

    - -

    - Intuitively, as x gets very large, - all the terms in the numerator are small in comparison to a_nx^n, - and likewise all the terms in the denominator are small compared to b_mx^m. - If n=m, looking only at these two important terms, - we have (a_nx^n)/(b_mx^m). - This reduces to a_n/b_m. - If n\lt m, the function behaves like a_n/(b_mx^{m-n}), - which tends toward 0. - If n \gt m, the function behaves like a_nx^{n-m}/b_m, - which will tend to either \infty or -\infty depending on the values of n, - m, - a_n, - b_m and whether you are looking for - \lim_{x\to\infty} f(x) or \lim_{x\to-\infty} f(x). -

    - - - Finding a limit of a rational function - -

    - Confirm analytically that y=1 is the horizontal asymptote of f(x) = \frac{x^2}{x^2+4}, - as approximated in . -

    -     - -

    - Before using , - let's use the technique of evaluating limits at infinity of rational functions that led to that theorem. - The largest power of x in f is 2, - so divide the numerator and denominator of f by x^2, - then take limits. - - \lim_{x\to\infty}\frac{x^2}{x^2+4} \amp = \lim_{x\to\infty}\frac{x^2/x^2}{x^2/x^2+4/x^2} - \amp =\lim_{x\to\infty}\frac{1}{1+4/x^2} - \amp =\frac{1}{1+0} - \amp = 1 - . -

    - -

    - We can also use directly; - in this case n=m so the limit is the ratio of the leading coefficients of the numerator and denominator, - , 1/1 = 1. -

    -     - -     - - - Finding limits of rational functions - -

    - Use - to evaluate each of the following limits. -

      -
    1. \lim_{x\to-\infty}\dfrac{x^2+2x-1}{x^3+1}
      2. -
    2. \lim_{x\to\infty}\dfrac{x^2+2x-1}{1-x-3x^2}
      3. -
    3. \lim_{x\to\infty}\dfrac{x^2-1}{3-x}
      4. -
    -

    -     - -

    -

      -
    1. -

      - The highest power of x is in the denominator. - Therefore, the limit is 0; - see . -

      -
      2. -
    2. -

      - The highest power of x is x^2, - which occurs in both the numerator and denominator. - The limit is therefore the ratio of the coefficients of x^2, - which is -1/3. - See . -

      -
      3. -
    3. -

      - The highest power of x is in the numerator so the limit will be \infty or -\infty. - To see which, - consider only the dominant terms from the numerator and denominator, - which are x^2 and -x. - The expression in the limit will behave like - x^2/(-x) = -x for large values of x. - Therefore, the limit is -\infty. - See . -

      -
      4. -
    -

    - -
    - Visualizing the functions in - -
    - - - - -

    - Graph of \dfrac{x^2+2x-1}{x^3+1} for x\lt 0. - The graph shows that - as x approaches -\infty, the limit of f(x) - is 0. -

    -     - - Graph that illustrates the highest power of x is in the denominator. Therefore, the limit is 0. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-41, - xmax=2, - ymin=-.6, - ymax=.6, - ] - \addplot[firstcurvestyle,domain=-40:-8,leftarrow] {(x^2+2*x-1)/(x^3+1)}; - \addplot[firstcurvestyle,domain=-8:-2,rightarrow] {(x^2+2*x-1)/(x^3+1)}; - \end{axis} - \end{tikzpicture} - - - -
    - -
    - - - - -

    - The graph of f(x)=\frac{x^2+2x-1}{1-x-3x^2} is shown for x\gt 0. - The graph has a horizontal asymptote at y = -\frac{1}{3}, - which it approaches from below. - The graph illustrates that the limit of f(x) as x\to\infty is -\frac{1}{3}. - The coefficient of the term in the numerator of f(x) with the highest - power of x is 1, - while the coefficient of the term in the dennominator with the - highest power of x is -3. - The ratio of these two coefficients gives the limit as x\to\pm \infty - when the highest power of x is the same in both the numerator and the denominator. -

    -     - - Graph that shows the limit is the ratio of the coefficients of the highest power of x. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1, - xmax=41, - ymin=-.55, - ymax=.55, - ] - \addplot[firstcurvestyle,domain=2:8,leftarrow] {(x^2+2*x-1)/(1-x-3*x^2)}; - \addplot[firstcurvestyle,domain=8:40,rightarrow] {(x^2+2*x-1)/(1-x-3*x^2)}; - \addplot[asymptote] coordinates {(-1,-0.33333) (41,-0.33333)}; - \end{axis} - \end{tikzpicture} - - - -
    - -
    - - - - -

    - The graph of f(x)=\frac{x^2-1}{3-x} is shown for x\gt 3. - Near x=3 the graph appears to be heading down a vertical asymptote, - suggesting that \lim_{x\to 3^+}f(x)=-\infty. - The graph then rises to a peak, before beginning to descend again. - Beyond x=10, the graph appears almost straight, - and continues downward at a slope close to -1, - showing that there is no horizontal asymptote in this case. -

    -

    - The graph shows that the limit of f(x) will be determined - by dominant terms from the numerator and denominator, which are - x^2 and -x. Since \frac{x^2}{-x} = -x for large values - of x, the graph of f(x) behaves approximately the same as that of y=-x. -

    -     - - Graph that shows the behaviour of a function can depend on the dominant terms of the numerator and denominator. - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-5, - xmax=41, - ymin=-50, - ymax=5 - ] - \addplot[firstcurvestyle,domain=3.5:8,leftarrow] {(x^2-1)/(3-x)}; - \addplot[firstcurvestyle,domain=8:40,rightarrow] {(x^2-1)/(3-x)}; - \end{axis} - \end{tikzpicture} - - - -
    -     -
    -     -     - - - -

    - With care, we can quickly evaluate limits at infinity for a large number of functions by considering the - long run behavior using dominant terms of f(x). - For instance, - consider again \lim_{x\to\pm\infty}\frac{x}{\sqrt{x^2+1}}, - graphed in . - The dominant terms are x in the numerator and \sqrt{x^2} in the denominator. - When x is very large, x^2+1 \approx x^2. - Thus - - \sqrt{x^2+1}\amp\approx \sqrt{x^2} = \abs{x} \amp \frac{x}{\sqrt{x^2+1}} \amp\approx \frac{x}{\abs{x}} - . -

    - -

    - This expression is 1 when x is positive and -1 when x is negative. - Hence we get asymptotes of y=1 and y=-1, - respectively. - We will show this more formally in the next example. -

    - - - Finding a limit using dominant terms - -

    - Confirm analytically that y=1 and y=-1 are the horizontal asymptote of \lim_{x\to\pm\infty}\frac{x}{\sqrt{x^2+1}}, - as graphed in . -

    -     - -

    - The dominating term of f in the denominator is - \sqrt{x^2}=\abs{x} so divide the numerator and denominator of f by \sqrt{x^2}, - then take limits. - - \lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}} - \amp = \lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}}\cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}} - \amp = \lim_{x\to\infty} \frac{\frac{x}{\abs{x}}}{\sqrt{\frac{x^2+1}{x^2}}} - \amp = \lim_{x\to\infty} \frac{1}{\sqrt{1+\frac{1}{x^2}}} \text{ for } x\gt 0 - \amp =\frac{1}{\sqrt{1+0}} - \amp = 1 - . -

    - -

    - As x \to -\infty, - the only thing that changes is the value of \frac{x}{\abs{x}}. - For x \lt 0, we have \frac{x}{\abs{x}}=-1, - making \lim_{x\to-\infty} \frac{x}{\sqrt{x^2+1}}=-1. - Therefore, the horizontal asymptotes are y=1 and y=-1. -

    -     -     - - - -     - - - - - - - Terms and Concepts - - - - - - -

    - - If \lim\limits_{x\to 5} f(x) = \infty, - then we are implicitly stating that the limit exists. -

    -     - -

    - When we say that a limit exists, we mean that the limit has a finite, numerical value. - An infinite limit describes a particular way in which a limit can fail to exist. -

    -     - -     - - - - - -

    - - If \lim\limits_{x\to 5} f(x) = 5, - then we are implicitly stating that the limit exists. -

    -     - -

    - In this case, the limit is a finite, numerical value; namely, 5. - This means that the limit exists. -

    -     - -     - - - - - -

    - - If \lim\limits_{x\to 1^-} f(x) = -\infty, - then \lim\limits_{x\to 1^+} f(x) = \infty. -

    -     - -

    - The sign of a function does not necessarily change at every vertical asymptote. - Consider, for example, f(x)=1/x versus f(x)=1/x^2. - We always need to carefully determine the sign of f(x) on both sides of each - vertical asymptote. -

    - -

    - It could also happen that the limit from one side is infinite, - while the limit from the other side is finite, or fails to exist in some other way. - (Can you think of an example?) -

    -     - -     - - - - - -

    - - If \lim\limits_{x\to 5} f(x) = \infty, - then f has a vertical asymptote at x=5. -

    -     - -

    - According to , - a function f has a vertical asymptote at c - if f approaches infinity from either side of c. -

    -     - -     - - - - - -

    - - \infty/0 is not an indeterminate form. -

    -     - -

    - Both the numerator approaching infinity - and the denominator approaching zero indicate an infinite limit, - so this is not indeterminate. - We may not be able to tell immediately if the result will be +\infty or -\infty, - but this is true of most infinite limits. -

    -     - -     - - - - -

    - List five indeterminate forms. -

    - -     - - -     - - - - -

    - Construct a function with a vertical asymptote at x=5 and a horizontal asymptote at y=5. -

    - -     - - -     - - - - -

    - Let \lim\limits_{x\to 7} f(x) = \infty. - Explain how we know that f is or is not continuous at x=7. -

    - -     - - - -

    - The limit of f as x approaches 7 does not exist, - hence f is not continuous. - (Note: f could be defined at 7!) -

    -     - -     -     - - - Problems - - - -

    - Evaluate the given limits using the graph of the function. -

    -     - - - - - $a=non_zero_random(-3,3,1); - $b=1; - $n=random(2,5,1); - $f=Formula("$b/(x-$a)^($n)")->reduce; - $L[0]=($f->eval(x=>$a-1)>0)?Compute("inf"):Compute("-inf"); - $L[1]=($f->eval(x=>$a+1)>0)?Compute("inf"):Compute("-inf"); - for my$i(0,1){$ev[$i]=$L[$i]->cmp()->withPostFilter(AnswerHints(Compute("DNE")=> "Technically correct, but this question is asking for more specific information."));}; - $xmin=min(-1,$a-2);$xmax=max(1,$a+2); - $e=1/50**(1/$n); - $below=$a-$e; - $above=$a+$e; - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - f(x) = has the graph: -

    - - -

    - Graph with the domain to . There - is a vertical asymptote at x = , - and f(x) grows without bound on both sides of the asymptote. - To the left of x = , f(x) is negative, - and to the right of the asymptote, f(x) is positive. -

    -     - - Graph of a reciprocal power function with a vertical asymptote at x =. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = -50, - ymax = 50, - ] - \addplot[firstcurvestyle, infiniteright, domain=$xmin:$below, samples=101] {$b/(x-$a)^($n)}; - \addplot[firstcurvestyle, infiniteleft, domain=$above:$xmax, samples=101] {$b/(x-$a)^($n)}; - \addplot[asymptote,-] coordinates {($a,-50) ($a,50)}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     -     -     - - - - - $a0 = non_zero_random(-2,3,1); - $b = random(1,2,1); - $a1 = $a0+$b; - @a = ($a0,$a1); - $f=Formula("1/((x-$a[0])(x-$a[1])^(2))")->reduce; - $L[0]=($f->eval(x=>$a[0]-0.1)>0)?Compute("inf"):Compute("-inf"); - $L[1]=($f->eval(x=>$a[0]+0.1)>0)?Compute("inf"):Compute("-inf"); - $L[3]=($f->eval(x=>$a[1]-0.1)>0)?Compute("inf"):Compute("-inf"); - $L[4]=($f->eval(x=>$a[1]+0.1)>0)?Compute("inf"):Compute("-inf"); - for my$i(2,5){$L[$i]=($L[$i-2]==$L[$i-1])?$L[$i-2]:Compute("DNE");}; - for my$i(0..5){$ev[$i]=($L[$i]==Compute("DNE"))?$L[$i]->cmp():$L[$i]->cmp()->withPostFilter(AnswerHints(Compute("DNE")=> "Technically correct, but this question is asking for more specific information."));}; - $xmin=min(0,@a)-1;$xmax=max(@a)+1; - $d = abs($a[1]-$a[0]); - @x = num_sort( - (sqrt($d*(25*$d**3+4))-5*$d**2)/(20*$d) + $a[0], - -(sqrt($d*(25*$d**3+4))-5*$d**2)/(20*$d) + $a[0], - sqrt(1/(50*$d)) + $a[1], - -sqrt(1/(50*$d)) + $a[1] - ); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - f(x) = has the graph: -

    - - -

    - There are - two vertical asymptotes, one at x = - and the other at x = . On the interval - \lt x \lt the - graph curves and as x gets near - and , f(x) - approaches \infty. For the interval - -\infty \lt x \lt as x gets near - , f(x) approaches -\infty. Lastly, - for the interval \lt x \lt \infty as x - gets near , f(x) approaches \infty. -

    -     - - Graph for problem 10. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = -50, - ymax = 50, - ] - \addplot[firstcurvestyle, infiniteright, domain=$xmin:$x[0], samples=101] {1/((x-$a[0])*(x-$a[1])^(2))}; - \addplot[firstcurvestyle, infinite, domain=$x[1]:$x[2], samples=101] {1/((x-$a[0])*(x-$a[1])^(2))}; - \addplot[firstcurvestyle, infiniteleft, domain=$x[3]:$xmax, samples=101] {1/((x-$a[0])*(x-$a[1])^(2))}; - \addplot[asymptote,-] coordinates {($a[0],-50) ($a[0],50)}; - \addplot[asymptote,-] coordinates {($a[1],-50) ($a[1],50)}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^-} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to ^+} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to } f(x) -

    -

    - -

    -     -     -     -     - - - - - $a=random(1,5,1); - $b=random(-1,1,2); - $f=Formula("$a/(e^($b*x)+1)")->reduce; - $L[0]=($b>0)?$a:0; - $L[1]=($b>0)?0:$a; - $L[2]=$f->eval(x=>0); - $L[3]=$f->eval(x=>0); - $xmin=-11;$xmax=11; - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - f(x) = has the graph: -

    - - -

    - There are two horizontal asymptotes, one at y = 0 and the other at y = . - For some versions of this question, the graph f(x)= begins with y - near when x is negative, and then descends toward 0 as x increases. -

    - -

    - For other versions of this question, f(x) will be close to 0 when x is negative, - and then will rise toward as x increases. -

    -     - - Graph for problem 11. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = -1, - ymax = $a+1, - ] - \addplot[firstcurvestyle, domain=$xmin:$xmax, smooth] {$a/(e^($b*x)+1)}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to-\infty} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to\infty} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to 0^{-}} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to 0^{+}} f(x) -

    -

    - -

    -     -     -     -     - - - - - $a=random(1,5,1); - $n=random(1,4,2); - $f=Formula("x^($n)*sin($a pi x)")->reduce; - $L[0]=Compute("DNE"); - $L[1]=Compute("DNE"); - $L[2]=0; - $L[3]=0; - $xmax = int(11/(2/$a))*(2/$a); - $xmin = -$xmax; - $samples = 1 + int(11/(2/$a))*80; - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - f(x) = has the graph: -

    - - -

    - Graph of f(x)=. - The graph is oscillating, as x approaches -\infty and \infty - the amplitude gets larger, but as x gets near 0 the amplitude - shrinks. -

    -     - - Oscillating graph with high amplitude at infinity and -infinty, but very small amplitude near 0. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ] - \addplot[firstcurvestyle, domain=$xmin:$xmax, samples=$samples] {x^($n)*sin(deg($a*pi*x))}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to-\infty} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to\infty} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to 0^{-}} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to 0^{+}} f(x) -

    -

    - -

    -     -     -     -     - - - - - $a=random(1,5,1); - $trigfun = list_random('sin','cos'); - $f=Formula("$trigfun($a x)")->reduce; - $L[0]=Compute("DNE"); - $L[1]=Compute("DNE"); - $xmax = int(14/(2*pi/$a))*(2*pi/$a); - $xmin = -$xmax; - $samples = 1 + int(14/(2*pi/$a))*80; - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - f(x) = has the graph: -

    - - -

    - The graph - is oscillating, with a peak of y = and minimum of y = -. - The graph oscillates continually between these values over the domain of the function. -

    -     - - Oscillating graph with a peak of y = 1 and minimum of y = -1. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = -1.2, - ymax = 1.2, - ] - \addplot[firstcurvestyle, domain=$xmin:$xmax, samples=$samples] {$trigfun(deg($a*x))}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to-\infty} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to\infty} f(x) -

    -

    - -

    -     -     -     -     - - - - - $b=random(1.5,2.5,0.1); - $c=non_zero_random(-12,12,1); - $n=random(-1,1,2); - $f=Formula("($b)^($n x)+$c")->reduce; - $L[0]=($n>0)?$c:Compute("inf"); - $L[1]=($n>0)?Compute("inf"):$c; - $xmin=-11;$xmax=11; - $start = $xmin; - $stop = log(20 - $c)/log($b) / $n; - if ($n < 1) { - ($start,$stop) = ($stop,$xmax); - } - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - f(x) = has the graph: -

    - - -

    - Graph with the domain to . The - line y = is a horizontal asymptote, but only in one direction. -

    - -

    - In some versions of the question, the function grows without bound as x\to \infty, - and approaches the horizontal asymptote as x\to -\infty. - In other version fo the question, these roles are reversed. -

    -     - - Graph for problem 14. - - - \begin{tikzpicture} - \begin{axis}[ - xmin = $xmin, - xmax = $xmax, - ymin = -20, - ymax = 20, - ] - \addplot[firstcurvestyle, domain=$start:$stop, smooth] {$f}; - \end{axis} - \end{tikzpicture} - - - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to-\infty} f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to\infty} f(x) -

    -

    - -

    -     -     -     -     -     - - - -

    - Numerically approximate the limits. -

    -     - - - - - @b = random_subset(4,-9..9); - if($envir{problemSeed}==1){@b=(3,1,-1,-2);}; - $a=$b[0]; - $f=Formula("(x^2-($b[1]+$b[2])*x+$b[1]*$b[2])/(x^2-($a+$b[3])*x+$a*$b[3])")->reduce; - @x=($a-0.1,$a-0.01,$a-0.001,$a+0.1,$a+0.01,$a+0.001); - @y=map{$f->eval(x=>$_)}(@x); - $L[0]=($y[2]>0)?Compute("inf"):Compute("-inf"); - $L[1]=($y[5]>0)?Compute("inf"):Compute("-inf"); - $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); - $limit=($L[2]==Compute("DNE"))?"does not exist":"is `".$L[2]->TeX."`"; - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - f(x)= -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to^{-}}f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to^{+}}f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to}f(x) -

    -

    - -

    -     -     - - -

    - Tables will vary. -

    -

    -

      -
    1. - - - x - f(x) - - - - - - - - - - - - - - -

      - It seems \lim\limits_{x\to^{-}}f(x)=. -

      -
      2. -
    2. - - - x - f(x) - - - - - - - - - - - - - - -

      - It seems \lim\limits_{x\to^{+}}f(x)=. -

      -
      3. -
    3. -

      - It seems \lim\limits_{x\to3}f(x) . -

      -
      4. -
    -

    -     -     -     - - - - - @b = random_subset(4,-9..9); - if($envir{problemSeed}==1){@b=(3,4,-9,-1);}; - $a=$b[0]; - $f=Formula("(x^2-($b[1]+$b[2])*x+$b[1]*$b[2])/(x^3-(2*$a+$b[3])*x^2+(2*$a*$b[3]+($a)^2)*x-($a)^2*$b[3])")->reduce; - @x=($a-0.1,$a-0.01,$a-0.001,$a+0.1,$a+0.01,$a+0.001); - @y=map{$f->eval(x=>$_)}(@x); - $L[0]=($y[2]>0)?Compute("inf"):Compute("-inf"); - $L[1]=($y[5]>0)?Compute("inf"):Compute("-inf"); - $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); - $limit=($L[2]==Compute("DNE"))?"does not exist":"is `".$L[2]->TeX."`"; - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - f(x)= -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to^{-}}f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to^{+}}f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to}f(x) -

    -

    - -

    -     -     - -

    - Tables will vary. -

    -

    -

      -
    1. - - - x - f(x) - - - - - - - - - - - - - - -

      - It seems \lim\limits_{x\to^{-}}f(x)=. -

      -
      2. -
    2. - - - x - f(x) - - - - - - - - - - - - - - -

      - It seems \lim\limits_{x\to^{+}}f(x)=. -

      -
      3. -
    3. -

      - It seems \lim\limits_{x\to3}f(x) . -

      -
      4. -
    -

    -     -     -     - - - - - @b = random_subset(4,-9..9); - if($envir{problemSeed}==1){@b=(3,5,6,-2);}; - $a=$b[0]; - $f=Formula("(x^2-($b[1]+$b[2])*x+$b[1]*$b[2])/(x^3-(2*$a+$b[3])*x^2+(2*$a*$b[3]+($a)^2)*x-($a)^2*$b[3])")->reduce; - @x=($a-0.1,$a-0.01,$a-0.001,$a+0.1,$a+0.01,$a+0.001); - @y=map{$f->eval(x=>$_)}(@x); - $L[0]=($y[2]>0)?Compute("inf"):Compute("-inf"); - $L[1]=($y[5]>0)?Compute("inf"):Compute("-inf"); - $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); - $limit=($L[2]==Compute("DNE"))?"does not exist":"is `".$L[2]->TeX."`"; - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - f(x)= -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to^{-}}f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to^{+}}f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to}f(x) -

    -

    - -

    -     -     - -

    - Tables will vary. -

    -

    -

      -
    1. - - - x - f(x) - - - - - - - - - - - - - - -

      - It seems \lim\limits_{x\to^{-}}f(x)=. -

      -
      2. -
    2. - - - x - f(x) - - - - - - - - - - - - - - -

      - It seems \lim\limits_{x\to^{+}}f(x)=. -

      -
      3. -
    3. -

      - It seems \lim\limits_{x\to3}f(x) . -

      -
      4. -
    -

    -     -     -     - - - - - @b = random_subset(3,-9..9); - if($envir{problemSeed}==1){@b=(3,6,-2);}; - $a=$b[0]; - $f=Formula("(x^2-($a+$b[1])*x+$a*$b[1])/(x^2-($a+$b[2])*x+$a*$b[2])")->reduce; - @x=($a-0.1,$a-0.01,$a-0.001,$a+0.1,$a+0.01,$a+0.001); - @y=map{$f->eval(x=>$_)}(@x); - $L[0]=Real(($a-$b[1])/($a-$b[2])); - $L[1]=Real(($a-$b[1])/($a-$b[2])); - $L[2]=($L[0]==$L[1])?$L[0]:Compute("DNE"); - $limit=($L[2]==Compute("DNE"))?"does not exist":"is `".$L[2]->TeX."`"; - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - f(x)= -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -     - - - -

    - \lim\limits_{x\to^{-}}f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to^{+}}f(x) -

    -

    - -

    -     -     - - - -

    - \lim\limits_{x\to}f(x) -

    -

    - -

    -     -     - -

    - Tables will vary. -

    -

    -

      -
    1. - - - x - f(x) - - - - - - - - - - - - - - -

      - It seems \lim\limits_{x\to^{-}}f(x)=. -

      -
      2. -
    2. - - - x - f(x) - - - - - - - - - - - - - - -

      - It seems \lim\limits_{x\to^{+}}f(x)=. -

      -
      3. -
    3. -

      - It seems \lim\limits_{x\to3}f(x) . -

      -
      4. -
    -

    -     -     -     -     - - - -

    - Identify the horizontal and vertical asymptotes, - if any, of the given function. -

    -     - - - - - @b = random_subset(3,-9..-1,1..9); - do {@c=(random(2,5,1),non_zero_random(-9,9,1))} until ($b[1] != $c[1]/$c[0] and $b[2] != $c[1]/$c[0]); - if($envir{problemSeed}==1){@c=(2,-2);@b=(2,4,-5)}; - $f=Formula("($c[0] x^2-($c[1]+$b[0]*$c[0])*x+$b[0]*$c[1])/(x^2-($b[1]+$b[2])*x+$b[1]*$b[2])")->reduce; - parser::Assignment->Allow; - Context()->variables->are(x=>'Real',y=>'Real'); - Context()->flags->set(reduceConstants=>0); - $asymptotes=Formula("y=$c[0],x=$b[1],x=$b[2]"); - - -

    - f(x)= -

    - - Submit your answer as a list, using commas. - Each asymptote is line of the form x=... or y=.... - If there are no such asymptotes, - you may type NONE. - -

    - -

    -     - -

    - There are horizontal/vertical asymptotes at . -

    -     -     -     - - - - - @b = random_subset(3,-9..-1,1..9); - @c=(list_random(-5..-2,2..5),non_zero_random(-9,9,1)); - if($envir{problemSeed}==1){@c=(-3,6);@b=(-1,3,5)}; - Context()->noreduce('(-x)+y','(-x)-y'); - $f=Formula("($c[0] x^2-($c[1]+$b[0]*$c[0])*x+$b[0]*$c[1])/($b[2]*x^2-$b[2]*($b[0]+$b[1])*x+$b[0]*$b[1]*$b[2])")->reduce; - parser::Assignment->Allow; - Context()->variables->are(x=>'Real',y=>'Real'); - Context()->flags->set(reduceConstants=>0); - $asymptotes=Compute("y=$c[0]/$b[2],x=$b[1]"); - - -

    - f(x)= -

    - - Submit your answer as a list, using commas. - Each asymptote is line of the form x=... or y=.... - If there are no such asymptotes, - you may type NONE. - -

    - -

    -     - -

    - There are horizontal/vertical asymptotes at . -

    -     -     -     - - - - - @b = random_subset(3,-9..-1,1..9); - @c=(random(1,5,1),non_zero_random(-9,9,1)); - if($envir{problemSeed}==1){@c=(1,-4);@b=(3,-1,7)}; - Context()->noreduce('(-x)+y','(-x)-y'); - $f=Formula("($c[0] x^2-($c[1]+$b[0]*$c[0])*x+$b[0]*$c[1])/($b[2]*x^3-$b[2]*($b[0]+$b[1])*x^2+$b[0]*$b[1]*$b[2]*x)")->reduce; - parser::Assignment->Allow; - Context()->variables->are(x=>'Real',y=>'Real'); - Context()->flags->set(reduceConstants=>0); - $asymptotestring="y=0,x=0,x=$b[1]"; - $asymptotes=Compute("$asymptotestring"); - - -

    - f(x)= -

    - - Submit your answer as a list, using commas. - Each asymptote is line of the form x=... or y=.... - If there are no such asymptotes, - you may type NONE. - -

    - -

    -     - -

    - There are horizontal/vertical asymptotes at . -

    -     -     -     - - - - - @b = random_subset(3,-9..-1,1..9); - @c=(random(1,5,1),non_zero_random(-9,9,1)); - if($envir{problemSeed}==1){@c=(1,3);@b=(-3,1,9)}; - Context()->noreduce('(-x)+y','(-x)-y'); - $f=Formula("($c[0] x^2-($c[1]+$b[0]*$c[0])*x+$b[0]*$c[1])/($b[2]*x-$b[2]*$b[1])")->reduce; - parser::Assignment->Allow; - Context()->variables->are(x=>'Real',y=>'Real'); - Context()->flags->set(reduceConstants=>0); - $asymptotestring="x=$b[1]"; - $asymptotes=Compute("$asymptotestring"); - - -

    - f(x)= -

    - - Submit your answer as a list, using commas. - Each asymptote is line of the form x=... or y=.... - If there are no such asymptotes, - you may type NONE. - -

    - -

    -     - -

    - There are horizontal/vertical asymptotes at . -

    -     -     -     - - - - - @b = random_subset(3,-9..-1,1..9); - @c=(random(1,5,1),non_zero_random(-9,9,1)); - if($envir{problemSeed}==1){@c=(1,3);@b=(-3,1,9)}; - Context()->noreduce('(-x)+y','(-x)-y'); - $f=Formula("($c[0] x^2-($c[1]+$b[0]*$c[0])*x+$b[0]*$c[1])/($b[2]*x-$b[2]*$b[0])")->reduce; - parser::Assignment->Allow; - Context()->variables->are(x=>'Real',y=>'Real'); - Context()->flags->set(reduceConstants=>0); - $asymptotes=Compute("NONE"); - - -

    - f(x)= -

    - - Submit your answer as a list, using commas. - Each asymptote is line of the form x=... or y=.... - If there are no such asymptotes, - you may type NONE. - -

    - -

    -     - -

    - There are horizontal/vertical asymptotes at . -

    -     -     -     - - - - - @b = random_subset(2,-9..-1,1..9); - $b[2]=random(-2,2,1); - $b[3]=random(1,5,1); - @c=(random(1,5,1),non_zero_random(-5,-1,1)); - if($envir{problemSeed}==1){@c=(1,-1);@b=(1,-1,0,1)}; - Context()->noreduce('(-x)+y','(-x)-y'); - $f=Formula("($c[0] x^2-($b[0]+$b[1])*$c[0]*x+$b[0]*$b[1]*$c[0])/($c[1] x^2-2*$b[2]*$c[1]*x+(($b[2])^2+$b[3])*$c[1])")->reduce; - parser::Assignment->Allow; - Context()->variables->are(x=>'Real',y=>'Real'); - Context()->flags->set(reduceConstants=>0); - $asymptotes=Compute("y=$c[0]/$c[1]"); - - -

    - f(x)= -

    - - Submit your answer as a list, using commas. - Each asymptote is line of the form x=... or y=.... - If there are no such asymptotes, - you may type NONE. - -

    - -

    -     - -

    - There are horizontal/vertical asymptotes at . -

    -     -     -     -     - - - -

    - Evaluate the given limit. -

    -     - - - - - $a=list_random(Compute("inf"),Compute("-inf")); - @b = random_subset(3,-9..-1,1..9); - @c = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$a=Compute("inf");@b=(2,0,1);@c=(1,-5)}; - $f=Formula("(x^3+$b[0]*x^2+$b[1]*x+$b[2])/($c[0]*x+$c[1])")->reduce; - $L=($c[0]>0)?Compute("inf"):Compute("-inf"); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - \lim\limits_{x\to} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - $a=list_random(Compute("inf"),Compute("-inf")); - @b = random_subset(3,-9..-1,1..9); - @c = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$a=Compute("inf");@b=(2,0,1);@c=(-1,5)}; - $f=Formula("(x^3+$b[0]*x^2+$b[1]*x+$b[2])/($c[0]*x+$c[1])")->reduce; - $L=($c[0]>0)?Compute("inf"):Compute("-inf"); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - \lim\limits_{x\to} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - $a=list_random(Compute("inf"),Compute("-inf")); - @b = random_subset(3,-9..-1,1..9); - @c = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$a=Compute("-inf");@b=(2,0,1);@c=(1,-5)}; - $f=Formula("(x^3+$b[0]*x^2+$b[1]*x+$b[2])/($c[0]*x^2+$c[1])")->reduce; - $L=($c[0]>0 and $a==Compute("inf") or $c[0]<0 and $a==Compute("-inf"))?Compute("inf"):Compute("-inf"); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - \lim\limits_{x\to} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     - - - - - $a=list_random(Compute("inf"),Compute("-inf")); - @b = random_subset(3,-9..-1,1..9); - @c = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$a=Compute("-inf");@b=(2,0,1);@c=(-1,5)}; - $f=Formula("(x^3+$b[0]*x^2+$b[1]*x+$b[2])/($c[0]*x^2+$c[1])")->reduce; - $L=($c[0]>0 and $a==Compute("inf") or $c[0]<0 and $a==Compute("-inf"))?Compute("inf"):Compute("-inf"); - Context()->strings->add('does not exist'=>{alias=>'DNE'}); - - -

    - \lim\limits_{x\to} -

    - - If you need to enter \infty, you may type infinity, or just INF. - If the limit does not exist, you may type does not exist, or just DNE. - -

    - -

    -     -     -     -     -     -     -
    - - - Chapter Summary -

    - In this chapter we: -

      -
    • -

      - defined the limit, -

      -
      • -
    • -

      - found accessible ways to approximate their values numerically and graphically, -

      -
      • -
    • -

      - developed a not-so-easy method of proving the value of a limit (\varepsilon-\delta proofs), -

      -
      • -
    • -

      - explored when limits do not exist, -

      -
      • -
    • -

      - defined continuity and explored properties of continuous functions, and -

      -
      • -
    • -

      - considered limits that involved infinity. -

      -
      • -
    -

    - -

    - Why? - Mathematics is famous for building on itself and calculus proves to be no exception. - In the next chapter we will be interested in - dividing by 0. That is, - we will want to divide a quantity by a smaller and smaller number and see what value the quotient approaches. - In other words, we will want to find a limit. - These limits will enable us to, among other things, - determine exactly how fast something is moving when we are only given position information. -

    - -

    - Later, we will want to add up an infinite list of numbers. - We will do so by first adding up a finite list of numbers, - then take a limit as the number of things we are adding approaches infinity. - Surprisingly, this sum often is finite; - that is, we can add up an infinite list of numbers and get, - for instance, - 42. -

    - -

    - These are just two quick examples of why we are interested in limits. - Many students dislike this topic when they are first introduced to it, - but over time an appreciation is often formed based on the scope of its applicability. -

    -     - - - - - - Derivatives - -

    - introduced the most fundamental of calculus topics: - the limit. - This chapter introduces the second most fundamental of calculus topics: - the derivative. - Limits describe where a function is going; - derivatives describe how fast the function is going. -

    -     - -
    - Instantaneous Rates of Change: The Derivative - - Introduction - - - -

    - A common amusement park ride lifts riders to a height then allows them to freefall a certain distance before safely stopping them. - Suppose such a ride drops riders from a height of 150 feet. - Students of physics may recall that the height - (in feet) - of the riders, t seconds after freefall - (and ignoring air resistance, etc.) - can be accurately modeled by f(t) = -16t^2+150. -

    - -

    - Using this formula, it is easy to verify that, without intervention, - the riders will hit the ground when f(t)=0 so at t=2.5\sqrt{1.5} \approx 3.06 seconds. - Suppose the designers of the ride decide to begin slowing the riders' fall after 2 seconds (corresponding to a height of f(2)= - 86). - How fast will the riders be traveling at that time? -

    - -

    - We have been given a position function, - but what we want to compute is a velocity at a specific point in time, - , we want an instantaneous velocity. - We do not currently know how to calculate this. -

    - -

    - However, we do know from common experience how to calculate an - average velocity. - (If we travel 60 miles in 2 hours, - we know we had an average velocity of - 30.) - We looked at this concept in - when we introduced the difference quotient. - We have - - \frac{\text{change in distance}}{\text{change in time}} = \frac{\text{“rise”}}{\text{“run”}} = \text{average velocity.} - -

    - -

    - We can approximate the instantaneous velocity at t=2 by considering the average velocity over some time period containing t=2. - If we make the time interval small, - we will get a good approximation. - (This fact is commonly used. - For instance, - high speed cameras are used to track fast moving objects. - Distances are measured over a fixed number of frames to generate an accurate approximation of the velocity.) -

    - -

    - Consider the interval from t=2 to t=3 - (just before the riders hit the ground). - On that interval, the average velocity is - - \frac{f(3)-f(2)}{3-2} = \frac{6-86}{1} =-80\,\text{ft/s} - , - where the minus sign indicates that the riders are moving down. - By narrowing the interval we consider, - we will likely get a better approximation of the instantaneous velocity. - On [2,2.5] we have - - \frac{f(2.5)-f(2)}{2.5-2} = \frac{50-86}{0.5} =-72\,\text{ft/s} - . -

    - - - -

    - We can do this for smaller and smaller intervals of time. - For instance, over a time span of one tenth of a second, - , on [2,2.1], we have - - \frac{f(2.1)-f(2)}{2.1-2} = \frac{79.44-86}{0.1} =-65.6\,\text{ft/s} - . -

    - -

    - Over a time span of one hundredth of a second, - on [2,2.01], the average velocity is - - \frac{f(2.01)-f(2)}{2.01-2} = \frac{85.3584-86}{0.01} =-64.16\,\text{ft/s} - . -

    - -

    - What we are really computing is the average velocity on the interval [2,2+h] for small values of h. - That is, we are computing - - \frac{f(2+h) - f(2)}{h} - - where h is small. -

    - -

    - We really want to use h=0, but this, of course, - returns the familiar 0/0 indeterminate form. - So we employ a limit, - as we did in . -

    - -

    - We can approximate the value of this limit numerically with small values of h as seen in . - It looks as though the velocity is approaching - -64. -

    - -
    - Approximating the instantaneous velocity with average velocities over a small time period h - - - h - Average Velocity () - - - 1 - \phantom{Average}{-80} - - - 0.5 - \phantom{Average}{-72} - - - 0.1 - \phantom{Average}{-65.6} - - - 0.01 - \phantom{Average}{-64.16} - - - 0.001 - \phantom{Average}{-64.016} - - -
    - -

    - Computing the limit directly gives - - \lim_{h\to 0} \frac{f(2+h)-f(2)}{h} \amp = \lim_{h\to 0}\frac{-16(2+h)^2+150 - (-16(2)^2+150)}{h} - \amp = \lim_{h\to 0}\frac{-16(4+4h+h^2)+150 - 86}{h} - \amp = \lim_{h\to 0}\frac{-64-64h-16h^2+64}{h} - \amp = \lim_{h\to 0}\frac{-64h-16h^2}{h} - \amp = \lim_{h\to 0}(-64 -16h) - \amp =-64 - . -

    - -

    - Graphically, - we can view the average velocities we computed numerically as the slopes of secant lines on the graph of f going through the points - (2,f(2)) and (2+h,f(2+h)). - In Figures, - the secant line corresponding to h=1 is shown in three contexts. - shows a zoomed out - version of f with its secant line. - In , - we zoom in around the points of intersection between f and the secant line. - Notice how well this secant line approximates f between those two points it is a common practice to approximate functions with straight lines. -

    - - - - - -
    - The function f(t) and its secant line corresponding to t=2 and t=3 - - - - A downward-curving graph of function f(t) with its secant line corresponding to t=2 and t=3. - - -

    - The graph starts at the origin and is drawn on the first quadrant. - The horizontal t axis is drawn between 0 to 3, - and the y axis is drawn between 0 to 150. - The graph has a decreasing slope, gently declining from points (0, 150) to (3,0). - A secant line is drawn on the curve from t=2 and t=3. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.5, - xmax=3.49, - ymin=-20, - ymax=180, - xlabel={$t$} - ] - \addplot+[infinite,domain=0:3.06] {-16*x^2+150}; - \addplot+[secantline,domain=1.5:3.2] {-80*(x-2)+86}; - \addplot[soliddot] coordinates {(2,86) (3,6)}; - \end{axis} - \end{tikzpicture} - - - -
    - -
    - The function f(t) and a secant line corresponding to t=2 and t=3, zoomed in near t=2 - - - - The previous graph of function f(t) that is zoomed in near t=2, with its secant line from t=2 to t=3. - - -

    - The graph appears to start at the point (1.8,0) and is drawn on the first quadrant. - The horizontal t axis is drawn between 1.8 to 3.4, and the y axis - is drawn between 0 to 120. - The curve appears to be coinciding with the secant line. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=1.8, - xmax=3.4, - ymin=0, - ymax=129, - xlabel={$t$}, - xdiscontinuity - ] - \addplot+[infinite,domain=0:3.06] {-16*x^2+150}; - \addplot+[secantline,domain=1.8:3.4] {-80*(x-2)+86}; - \addplot[soliddot] coordinates{(2,86) (3,6)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - - -
    - The function f(t) with the same secant line, zoomed in further - - - - The graph of function f(t) that is further zoomed in near t=2. - - -

    - The graph appears to start at the point (1.4,0) and is drawn on the first quadrant. - The horizontal t axis is drawn between 1.4 to 2.6, and the y axis - is drawn between 0 to 120. - The curve appears to be coinciding with the secant line that is drawn on the function at t=2. - The secant line is above the curve on the left of the point of intersection and is below the curve to its right. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=1.4, - xmax=2.6, - ymin=0, - ymax=130, - xlabel={$t$}, - xdiscontinuity - ] - \addplot+[infinite,domain=0:3.06] {-16*x^2+150}; - \addplot+[secantline,domain=1.5:2.6] {-80*(x-2)+86}; - \addplot[soliddot] coordinates{(2,86)}; - \end{axis} - \end{tikzpicture} - - - -
    - -
    - The function f(t) with its tangent line at t=2 - - - - The graph of function f(t) that is highly zoomed in near t=2. - - -

    - The graph appears to start at the point (1.4,0) and is drawn on the first quadrant. - The horizontal t axis is drawn between 1.4 to 2.6, and the y axis - is drawn between 0 to 120. - The curve appears to be coinciding with the tangent line at t=2. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=1.4, - xmax=2.6, - ymin=0, - ymax=140, - xlabel={$t$}, - xdiscontinuity - ] - \addplot+[infinite,domain=0:3.06] {-16*x^2+150}; - \addplot+[tangentline,domain=1.4:2.6] {-64*(x-2)+86}; - \addplot[soliddot] coordinates{(2,86)}; - \end{axis} - \end{tikzpicture} - - -
    -     -     - -

    - As h\to 0, these secant lines approach the - tangent line, - a line that goes through the point - (2,f(2)) with the special slope of -64. - In - and , - we zoom in around the point (2,86). - We see the secant line, which approximates f well, - but not as well the tangent line shown in . -

    - -

    - We have just introduced a number of important concepts that we will flesh out more within this section. - First, we formally define two of them. -

    - - - Derivative at a Point - -

    - Let f be a continuous function on an open interval I and let c be - in I. - - derivativeat a point - - The derivative of f at c, denoted \fp(c), is - - \lim_{h\to 0}\frac{f(c+h)-f(c)}{h} - , - provided the limit exists. - If the limit exists, we say that - f is differentiable at c; - if the limit does not exist, - then f is not differentiable at c. - If f is differentiable at every point in I, - then f is differentiable on I. - - differentiable - -

    -     -     - - - - - Tangent Line - -

    - Let f be continuous on an open interval I and differentiable at c, - for some c in I. - The line with equation \ell(x) = \fp(c)(x-c)+f(c) is the tangent line - to the graph of f at c; - that is, it is the line through - (c,f(c)) whose slope is the derivative of f at c. - - tangent line - derivativetangent line - -

    -     -     - -

    - Some examples will help us understand these definitions. -

    - - - Finding derivatives and tangent lines - -

    - Let f(x) = 3x^2+5x-7. - Find: -

      -
    1. -

      - \fp(1) -

      -
      2. -
    2. -

      - The equation of the tangent line to the graph of f at x=1. -

      -
      3. -
    3. -

      - \fp(3) -

      -
      4. -
    4. -

      - The equation of the tangent line to the graph f at x=3. -

      -
      5. -
    -

    -     - -

    -

      -
    1. -

      - We compute this directly using . - - \fp(1)\amp = \lim_{h\to 0} \frac{f(1+h)-f(1)}{h} - \amp = \lim_{h\to 0} \frac{3(1+h)^2+5(1+h)-7 - (3(1)^2+5(1)-7)}{h} - \amp = \lim_{h\to 0} \frac{3(1+2h+h^2)+5+5h-7 - 1}{h} - \amp = \lim_{h\to 0} \frac{3+6h+3h^2+5+5h-8}{h} - \amp = \lim_{h\to 0} \frac{3h^2+11h}{h} - \amp = \lim_{h\to 0} (3h+11) - \amp = 11 - . -

      -
      2. -
    2. -

      - The tangent line at x=1 has slope \fp(1) and goes through the point - (1,f(1)) = (1,1). - Thus the tangent line has equation, - in point-slope form, y = 11(x-1) + 1. - In slope-intercept form we have y = 11x-10. -

      -
      3. -
    3. -

      - Again, using the definition, - - \fp(3)\amp = \lim_{h\to 0} \frac{f(3+h)-f(3)}{h} - \amp = \lim_{h\to 0} \frac{3(3+h)^2+5(3+h)-7 - (3(3)^2+5(3)-7)}{h} - \amp = \lim_{h\to 0} \frac{3(9+6h+h^2)+15+3h-7 - 35}{h} - \amp = \lim_{h\to 0} \frac{27+18h+3h^2+15+3h-42}{h} - \amp = \lim_{h\to 0} \frac{3h^2+23h}{h} - \amp = \lim_{h\to 0} 3h+23 - \amp = 23 - . -

      -
      4. -
    4. -

      - The tangent line at x=3 has slope 23 and goes through the point - (3,f(3)) = (3,35). - Thus the tangent line has equation y=23(x-3)+35 = 23x-34. -

      -
      5. -
    -

    - -

    - A graph of f is given in - along with the tangent lines at x=1 and x=3. -

    - -
    - A graph of f(x) = 3x^2+5x-7 and its tangent lines at x=1 and x=3 - - - - A graph of function 3x^2+5x-7 and its tangent line at x =1 and x=3. - - -

    - The graph starts at origin (0,0) the x axis ends at 4 - while the y axis ends at 60. - The curve starts in the third quadrant and moves into the first quadrant with an - x intercept at 1, the curve reaches point (4,60) and continues further. - Two tangents are drawn on the curve at x=1 and x=3. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1, - xmax=4.1, - ymin=-11, - ymax=66 - ] - \addplot+[infinite,domain=-1:4]{3*x^2+5*x-7}; - \addplot+[tangentline,domain=0:4] {11*(x-1)+1}; - \addplot+[tangentline,domain=1.8:3.8] {23*(x-3)+35}; - \addplot[<->,soliddot] coordinates{(1,1) (3,35)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - - - - - - - - -

    - Another important line that can be created using information from the derivative is the - normal line. - It is perpendicular to the tangent line, - hence its slope is the negative-reciprocal of the tangent line's slope. -

    - - - Normal Line - -

    - Let f be continuous on an open interval I and differentiable at c, - for some c in I. - The normal line to the graph of f at c is the line with equation - - n(x) =\frac{-1}{\fp(c)}(x-c)+f(c) - , - when \fp(c)\neq 0. (When \fp(c)=0, - the normal line is the vertical line through \left(c,f(c)\right); - that is, x=c.) - - derivativenormal line - normal line - -

    -     -     - - - Finding equations of normal lines - -

    - Let f(x) = 3x^2+5x-7, as in . - Find the equations of the normal lines to the graph of f at x=1 and - x=3. -

    -     - -

    - In , - we found that \fp(1)=11. - Hence at x=1, the normal line will have slope -1/11. - An equation for the normal line is - - n(x) = \frac{-1}{11}(x-1)+1 - . -

    - -

    - The normal line is plotted with y=f(x) in . - Note how the line looks perpendicular to f. (A key word here is - looks. Mathematically, - we say that the normal line is - perpendicular to f at x=1 as the slope of the normal line is the negative-reciprocal of the slope of the tangent line. - However, normal lines may not always - look perpendicular. -

    - -
    - A graph of f(x)=3x^2+5x-7, along with its normal line at x=1 - - - - A graph of function 3x^2+5x-7 along with its normal line at x=1. - - -

    - The graph starts at the origin. The x axis ends at 4 in the diagram while the y axis ends at 3. - The function is a straight line that rises steeply from points (0.9,0) to point (1.1, 3.2). - The normal is drawn on the line at point (1,1). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.1, - ymax=3.5, - xmin=-0.1, - xmax=4.3, - axis equal - ] - \addplot+[infinite,domain=0.9:1.2]{3*x^2+5*x-7}; - \addplot[normalline,domain=0.1:2.5] {(-1/11)*(x-1)+1}; - \addplot[soliddot] coordinates{(1,1)}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - The aspect ratio of the picture of the graph plays a big role in this. - When using graphing software, - there is usually an option called Zoom Square that keeps the aspect ratio - 1:1 -

    - -

    - We also found that \fp(3) = 23, - so the normal line to the graph of f at x=3 will have slope -1/23. - An equation for the normal line is - - n(x) = \frac{-1}{23}(x-3)+35 - . -

    -     -     - -

    - Linear functions are easy to work with; - many functions that arise in the course of solving real problems are not easy to work with. - A common practice in mathematical problem solving is to approximate difficult functions - with not-so-difficult functions. - Lines are a common choice. - It turns out that at any given point on the graph of a differentiable function f, - the best linear approximation to f is its tangent line. - That is one reason we'll spend considerable time finding tangent lines to functions. -

    - -

    - One type of function that does not benefit from a tangent line approximation is a line; - it is rather simple to recognize that the tangent line to a line is the line itself. - We look at this in the following example. -

    - - - Finding the derivative of a linear function - -

    - Consider f(x) = 3x+5. - Find the equation of the tangent line to f at x=1 and x=7. -

    -     - -

    - We find the slope of the tangent line by using - Definition. - - \fp(1) \amp = \lim_{h\to 0}\frac{f(1+h)-f(1)}{h} - \amp = \lim_{h\to 0} \frac{3(1+h)+5 - (3+5)}{h} - \amp = \lim_{h\to 0} \frac{3h}{h} - \amp = \lim_{h\to 0} 3 - \amp = 3 - . -

    - -

    - We just found that \fp(1) = 3. - That is, we found the instantaneous rate of change - of f(x) = 3x+5 is 3. - This is not surprising; lines are characterized by being the - only functions with a - constant rate of change. - That rate of change is called the - slope of the line. - Since their rates of change are constant, - their instantaneous rates of change are always the same; - they are all the slope. -

    - -

    - So given a line f(x) = ax+b, - the derivative at any point x will be a; - that is, \fp(x) = a. -

    - -

    - It is now easy to see that the tangent line to the graph of f at x=1 - is just f, - with the same being true at x=7. -

    -     -     - -

    - We often desire to find the tangent line to the graph of a function without knowing - the actual derivative of the function. - While we will eventually be able to find derivatives of many common functions, - the algebra and limit calculations on some functions are complex. - Until we develop further techniques, - the best we may be able to do is approximate the tangent line. - We demonstrate this in the next example. -

    - - - Numerical approximation of the tangent line - -

    - Approximate the equation of the tangent line to the graph of f(x)=\sin(x) at x=0. -

    -     - -

    - In order to find the equation of the tangent line, - we need a slope and a point. - The point is given to us: (0,\sin(0)) = (0,0). - To compute the slope, we need the derivative. - This is where we will make an approximation. - Recall that - - \fp(0) \approx \frac{\sin(0+h)- \sin(0) }{h} - - for a small value of h. - We choose (somewhat arbitrarily) to let h=0.1. - Thus - - \fp(0) \approx \frac{\sin(0.1)-\sin(0) }{0.1} \approx 0.9983 - . -

    - -

    - Thus our approximation of the equation of the tangent line is - y = 0.9983(x-0) +0 = 0.9983x; - it is graphed in . - The graph seems to imply the approximation is rather good. -

    - - - -
    - f(x) = \sin(x) graphed with an approximation to its tangent line at x=0 - - - - A graph of function sin(x) with an approximation to its tangent line at x=0. - - -

    - The y axis of the graph is drawn from -1 to 1 - and the x axis is drawn from -\pi and \pi. - The sine graph has an x intercept at -\pi then enters the - third quadrant and forms an upward facing parabola - with its vertex at point (-\pi/2, -1). - It passes throgh the origin then enters the first quadrant forming a downward - facing parabola with vertex at (\pi/2, 1). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.1, - ymax=1.1, - xmin=-3.5, - xmax=3.5, - xtick={-3.14,-1.57,1.57,3.14}, - xticklabels={$-\pi$,$-\frac{\pi}2$,$\frac{\pi}2$,$\pi$} - ] - \addplot+[infinite,domain=-3.5:3.5,samples=101]{sin(deg(x))}; - \addplot+[tangentline,domain=-1:1] {sin(deg(1))*x}; - \addplot [soliddot] coordinates{(0,0)}; - \end{axis} - \end{tikzpicture} - - - -
    - -     -     - -

    - Recall from - that \lim_{x\to 0}\frac{\sin(x)}x =1, - meaning for values of x near 0, \sin(x) \approx x. - Since the slope of the line y=x is 1 at x=0, - it should seem reasonable that the slope of f(x)=\sin(x) - is near 1 at x=0. - In fact, since we approximated - the value of the slope to be 0.9983, - we might guess the actual value is 1. - We'll come back to this later. -

    - -

    - Consider again . - To find the derivative of f at x=1, - we needed to evaluate a limit. - To find the derivative of f at x=3, - we needed to again evaluate a limit. - We have this process: - - - \begin{array}{c}\text{input specific}\\\text{number }c\end{array}\longrightarrow - \begin{array}{|c|}\hline\text{do something}\\\text{to }f\text{ and }c\\\hline\end{array} - \longrightarrow\begin{array}{c}\text{return}\\\text{number }f'(c)\end{array} - -

    - -

    - This process describes a function; - given one input - (the value of c), - we return exactly one output (the value of \fp(c)). - The do something box is where the tedious work - (taking limits) - of this function occurs. -

    - -

    - Instead of applying this function repeatedly for different values of c, - let us apply it just once to the variable x. - We then take a limit just once. - The process now looks like: - - - \begin{array}{c}\text{input}\\\text{variable }x\end{array}\longrightarrow - \begin{array}{|c|}\hline\text{do something}\\\text{to }f\text{ and }x\\\hline\end{array} - \longrightarrow\begin{array}{c}\text{return}\\\text{function }f'(x)\end{array} - -

    - -

    - The output is the derivative function, \fp(x). - The \fp(x) function will take a number c as input and return the - derivative of f at c. - This calls for a definition. -

    - - - Derivative Function - -

    - Let f be a differentiable function on an open interval I. - The function - - \fp(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} - - is the derivative of f. - - derivativeas a function - derivativenotation - -

    - -

    - Let y = f(x). - The following notations all represent the derivative of f: - - \fp(x)\ =\ \yp\ =\ \frac{dy}{dx}\ =\ \frac{df}{dx}\ =\ \frac{d}{dx}(f)\ =\ - \frac{d}{dx}(y) - . -

    -     -     - - - - - -

    - Important: The notation \frac{dy}{dx} is one symbol; - it is not the fraction dy/dx. - The notation, - while somewhat confusing at first, was chosen with care. - A fraction-looking symbol was chosen because the derivative has many fraction-like properties. - Among other places, - we see these properties at work when we talk about the units of the derivative, - when we discuss the Chain Rule, - and when we learn about integration - (topics that appear in later sections and chapters). -

    - - - -

    - Examples will help us understand this definition. -

    - - - - - Finding the derivative of a function - -

    - Let f(x) = 3x^2+5x-7 as in . - Find \fp(x). -

    -     - -

    - We apply . - - \fp(x) \amp = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} - \amp = \lim_{h\to 0} \frac{3(x+h)^2+5(x+h)-7-(3x^2+5x-7)}{h} - \amp = \lim_{h\to 0} \frac{3h^2 +6xh+5h}{h} - \amp = \lim_{h\to 0} (3h+6x+5) - \amp = 6x+5 - -

    - -

    - So \fp(x) = 6x+5. - Recall earlier we found that - \fp(1) = 11 and \fp(3) = 23. - Note our new computation of \fp(x) affirms these facts. -

    -     -     - - - Finding the derivative of a function - -

    - Let f(x) = \frac{1}{x+1}. - Find \fp(x). -

    -     - -

    - We apply . - - \fp(x) \amp = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} - \amp = \lim_{h\to 0} \frac{\frac{1}{x+h+1}-\frac{1}{x+1}}{h} - Now find common denominator then subtract; pull 1/h out front to facilitate reading. - \amp = \lim_{h\to 0} \frac{1}{h}\cdot\left(\frac{x+1}{(x+1)(x+h+1)} - - \frac{x+h+1}{(x+1)(x+h+1)}\right) - Now simplify algebraically. - \amp = \lim_{h\to 0} \frac 1h\cdot\left(\frac{x+1-(x+h+1)}{(x+1)(x+h+1)}\right) - \amp = \lim_{h\to 0} \frac1h\cdot\left(\frac{-h}{(x+1)(x+h+1)}\right) - Finally, apply the limit. - \amp = \lim_{h\to 0} \frac{-1}{(x+1)(x+h+1)} - \amp = \frac{-1}{(x+1)(x+1)} - \amp = \frac{-1}{(x+1)^2} - . -

    - -

    - So \fp(x) = \frac{-1}{(x+1)^2}. - To practice using our notation, we could also state - - \frac{d}{dx}\left(\frac{1}{x+1}\right) = \frac{-1}{(x+1)^2} - . -

    -     - -     - - - - - Finding the derivative of a function - -

    - Find the derivative of f(x) = \sin(x). -

    -     - -

    - Before applying , - note that once this is found, - we can find the actual tangent line to f(x) = \sin(x) at x=0, - whereas we settled for an approximation in . - - \fp(x) \amp = \lim_{h\to 0} \frac{\sin(x+h)-\sin(x)}{h} \amp \amp - \text{Derivative definition} - \amp = \lim_{h\to 0} \frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h} \amp \amp - \text{Angle addition identity} - \amp = \lim_{h\to 0} \frac{\sin(x)(\cos(h)-1) + \cos(x)\sin(h)}{h} \amp \amp - \text{Regrouped and factored} - \amp = \lim_{h\to 0} \left(\frac{\sin(x)(\cos(h)-1)}{h} + \frac{\cos(x)\sin(h)}{h} - \right) \amp \amp \text{Split into two fractions} - \amp = \lim_{h\to 0} \sin(x) \cdot \lim_{h\to 0}\frac{\cos(h)-1}{h} - \amp\quad{}+\lim_{h\to 0}\cos(x)\cdot \lim_{h\to 0}\frac{\sin(h)}{h} \amp \amp - \text{Product/sum limit rules} - \amp = \sin(x)\cdot 0 + \cos(x) \cdot 1 \amp \amp \text{Applied } - - \amp = \cos(x). \amp \amp - -

    - -

    - We have found that when f(x) = \sin(x), \fp(x) = \cos(x). - This should be somewhat amazing; - the result of a tedious limit process on the sine function is a nice function. - Then again, perhaps this is not entirely surprising. - The sine function is periodic it repeats itself on regular intervals. - Therefore its rate of change also repeats itself on the same regular intervals. - We should have known the derivative would be periodic; - we now know exactly which periodic function it is. -

    - -

    - Thinking back to , - we can find the slope of the tangent line to - f(x)=\sin(x) at x=0 using our derivative. - We approximated the slope as 0.9983; - we now know the slope is exactly \cos(0) =1. -

    - - - - - -     - -     - - - Finding the derivative of a piecewise defined function - -

    - Find the derivative of the absolute value function, - - f(x) = \abs{x} = \begin{cases} -x \amp x\lt 0 \\ x \amp x\geq 0\end{cases} - . -

    - -

    - See . -

    - -
    - The absolute value function f(x) = \abs{x}. Notice how the slope of - the lines (and hence the tangent lines) abruptly changes at x=0. - - - - The graph of function of absolute value of x, the function forms a v shape with it's vertex at origin. - - -

    - The y axis of the graph is drawn from -0.2 to 1 - and the x axis from -1 to 1. - The function is a straight line, in the second quadrant it appears to pass through point - (-1,1) and decreases until it meets the origin. - From the origin, it steeply increases after it enters the first quadrant - and appears to pass through point (1,1). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.2, - ymax=1.1, - xmin=-1.1, - xmax=1.1, - ] - \addplot+[infinite,domain=-1:1]{abs(x)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -

    - We need to evaluate \lim_{h\to0}\frac{f(x+h)-f(x)}{h}. - As f is piecewise-defined, - we need to consider separately the limits when x\lt 0 and when x \gt 0. -

    - -

    - When x\lt 0: - - \frac{d}{dx}(-x) \amp = \lim_{h\to 0}\frac{-(x+h) - (-x)}{h} - \amp = \lim_{h\to 0}\frac{-h}{h} - \amp = \lim_{h\to 0}-1 - \amp = -1 - . -

    - -

    - When x \gt 0, - a similar computation shows that \frac{d}{dx}(x) = 1. -

    - -

    - We need to also find the derivative at x=0. - By the definition of the derivative at a point, we have - - \fp(0) = \lim_{h\to0}\frac{f(0+h)-f(0)}{h} - . -

    - -

    - Since x=0 is the point where our function's definition switches from - one piece to the other, we need to consider left and right-hand limits. - Consider the following, - where we compute the left and right hand limits side by side. -

    - - -

    - - \amp\lim_{h\to0^-}\frac{f(0+h)-f(0)}{h} - \amp= \lim_{h\to0^-}\frac{-h-0}{h} - \amp=\lim_{h\to0^-}-1 - \amp =-1 - -

    - -

    - - \amp \lim_{h\to0^+}\frac{f(0+h)-f(0)}{h} - \amp =\lim_{h\to0^+}\frac{h-0}{h} - \amp =\lim_{h\to0^+}1 - \amp =1 - -

    -     - -

    - The last lines of each column tell the story: - the left and right hand limits are not equal. - Therefore the limit does not exist at 0, - and f is not differentiable at 0. - So we have - - \fp(x) = \begin{cases} -1 \amp x\lt 0 \\ 1 \amp x \gt 0\end{cases} - . -

    - -

    - At x=0, \fp(x) does not exist; - there is a jump discontinuity at 0; - see . - So f(x) = \abs{x} is differentiable everywhere except at 0. -

    - -
    - A graph of the derivative of f(x) = \abs{x} - - - - The graph of derivative of function of absolute value of x. - - -

    - The y axis and the x axis are drawn from -1 to 1. - In the first quadrant the derivative of the function starts from the open point (0,1) - and continues as a function y=1 that moves towards right parallel to the x axis at - y=1 for all values of x, but value for x=0 does not exist. - In the third quadrant the derivative of the function starts from the open point (0,-1) - and continues as a function y=-1 that moves towards left parallel to the x axis at - y=-1 for all values of x, but value for x=0 does not exist. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.4, - ymax=1.4, - xmin=-1.1, - xmax=1.1 - ] - \addplot [firstcurvestyle,leftarrow,domain=-1:0]{-1}; - \addplot [firstcurvestyle,rightarrow,domain=0:1]{1}; - \addplot[hollowdot] coordinates{(0,-1) (0,1)}; - \end{axis} - \end{tikzpicture} - - - -
    -     -     - -

    - The point of non-differentiability came where the piecewise defined function - switched from one piece to the other. - Our next example shows that this does not always cause trouble. -

    - - - Finding the derivative of a piecewise defined function - -

    - Find the derivative of f(x), - where - - f(x) = \begin{cases}\sin(x) \amp x\leq \pi/2 \\ 1 \amp x \gt \pi/2\end{cases} - . - See . -

    - -
    - A graph of f(x) as defined in - - - - Graph of function sin(x) until pi/2 and then y = 1. - - -

    - The y axis is from 0 to 1.5 and the x axis from 0 - to \pi/2 the curve starts at 0 and increases until point (\pi/2,1) - and then moves horizontal to the x axis at y=1. -

    -     - - \begin{tikzpicture}[declare function = {func(\x) = (\x < 1.5708) * (sin(x*180/pi)) + (\x >= 1.5708) * (1);}] - \begin{axis}[ - ymin=-.4, - ymax=1.4, - xmin=-.1, - xmax=2.1, - xtick={1.57}, - xticklabels={$\frac{\pi}2$} - ] - \addplot+[infinite,domain=-.1:2,samples=101]{func(x)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -

    - Using , - we know that when x\lt \pi/2, \fp(x) = \cos(x). - It is easy to verify that when x \gt \pi/2, - \fp(x) = 0; consider: - - \lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to0}\frac{1-1}{h} = \lim_{h\to0}0 =0 - . -

    - -

    - So far we have - - \fp(x) = \begin{cases}\cos(x) \amp x\lt \pi/2\\ 0 \amp x\gt\pi/2\end{cases} - . -

    - -

    - We still need to find \fp(\pi/2). - Notice at x=\pi/2 that both pieces of \fp are 0, - meaning we can state that \fp(\pi/2)=0. -

    - -

    - Being more rigorous, - we can again evaluate the difference quotient limit at x=\pi/2, - utilizing again left- and right-hand limits. - We will begin with the left-hand limit: -

    - -

    - - \amp \lim_{h\to0^-}\frac{f(\pi/2+h)-f(\pi/2)}{h} - \amp =\lim_{h\to0^-}\frac{\sin(\pi/2+h)-\sin(\pi/2)}{h} - \amp =\lim_{h\to0^-}{ \frac{\sin(\frac{\pi}{2})\cos(h)+\sin(h)\cos(\frac{\pi}{2})-\sin(\frac{\pi}{2})}{h}} - \amp =\lim_{h\to0^-}\frac{1\cdot\cos(h)+\sin(h)\cdot 0-1}{h} - \amp =\lim_{h\to0^-}\frac{\cos(h)-1}{h} \cdot \lim_{h\to0^-}\frac{\sin(h)}{h} - - \amp=1\cdot 0 - \amp =0 - . - Notice we used to finally evaluate the limit. -

    - -

    - Now we will find the right-hand limit: - - \amp \lim_{h\to0^+}\frac{f(\pi/2+h)-f(\pi/2)}{h} - \amp =\lim_{h\to0^+}\frac{1-1}{h} - \amp =\lim_{h\to0^+}\frac{0}{h} - \amp =0 - . -

    - -

    - Since both the left and right hand limits are 0 at x=\pi/2, - the limit exists and \fp(\pi/2) exists - (and is 0). - Therefore we can fully write \fp as - - \fp(x) = \begin{cases}\cos(x) \amp x\leq\pi/2\\ 0 \amp x \gt \pi/2\end{cases} - . -

    - -

    - See - for a graph of this derivative function. -

    - -
    - A graph of \fp(x) in . - - - - - Graph of derivative of function for this example. - - -

    - Graph of the derivative function, the derivative exists at \pi/2 - and follows a curve of \cos(x) for x\leq\pi/2 and 0 after x=\pi. - There is a sharp bend at \pi/2, hence it is not differentiable at that point, - since for x=\pi there exists f(x)=0 function is continuous. -

    -     - - \begin{tikzpicture}[declare function = {func(\x) = (\x < 1.5708) * (cos(x*180/pi)) + (\x >= 1.5708) * (0);}] - \begin{axis}[ - ymin=-.4, - ymax=1.4, - xmin=-.1, - xmax=2.1, - xtick={1.57}, - xticklabels={$\frac{\pi}2$}, - ] - \addplot+[infinite,domain=-.1:2,samples=101]{func(x)}; - \end{axis} - \end{tikzpicture} - - - -
    -     -     - - - - - -

    - Recall we pseudo-defined a continuous function as one in which we could sketch its graph without lifting our pencil. - We can give a pseudo-definition for differentiability as well: - it is a continuous function that does not have any sharp corners - or a vertical tangent line. - One such sharp corner is shown in . - Even though the function f in is piecewise-defined, - the transition is smooth - hence it is differentiable. - Note how in the graph of f in - it is difficult to tell when f switches from one piece to the other; - there is no corner. -

    -     - - - Differentiability on Closed Intervals -

    - When we defined the derivative at a point in , - we specified that the interval I over which a function f was defined needed to be an open interval. - Open intervals are required so that we can take a limit at any point c in I, - meaning we want to approach c from both the left and right. -

    - -

    - differentiableon a closed interval -

    - -

    - Recall we also required open intervals in - when we defined what it meant for a function to be continuous. - Later, we used one-sided limits to extend continuity to closed intervals. - We now extend differentiability to closed intervals by again considering one-sided limits. -

    - -

    - Our motivation is three-fold. - First, we consider common sense. - In - we found that when f(x) = 3x^2+5x-7, \fp(x) = 6x+5, - and this derivative is defined for all real numbers, - hence f is differentiable everywhere. - It seems appropriate to also conclude that f is differentiable on closed intervals, - like [0,1], as well. - After all, \fp(x) is defined at both x=0 and x=1. -

    - -

    - Secondly, consider f(x) = \sqrt{x}. - The domain of f is [0,\infty). - Is f differentiable on its domain specifically, - is f differentiable at 0? (We'll consider this in the next example.) -

    - -

    - Finally, in later sections, - having the derivative defined on closed intervals will prove useful. - One such place is - where the derivative plays a role in measuring the length of a curve. -

    - -

    - After a formal definition of differentiability on a closed interval, - we explore the concept in an example. -

    - - - Differentiability on a Closed Interval - -

    - Let f be continuous on [a,b] and differentiable on (a,b). - If the one-sided limits - - \lim_{h\to0^+}\frac{f(a+h)-f(a)}{h} \amp \amp \amp \lim_{h\to0^-}\frac{f(b+h)-f(b)}{h} - - exist, then we say f is differentiable on [a,b]. -

    -     -     - -

    - For all the functions f in this text, - we can determine differentiability on [a,b] by considering the limits - \lim_{x\to a^+}\fp(x) and \lim_{x\to b^-}\fp(x). - This is often easier to evaluate than the limit of the difference quotient. -

    - - - Differentiability at an endpoint - -

    - Consider f(x) = \sqrt{x}=x^{1/2} and g(x) = \sqrt{x^3}=x^{3/2}. - The domain of each function is [0,\infty). - It can be shown that each is differentiable on (0,\infty); - determine the differentiability of each at x=0. -

    -     - -

    - We start by considering f and take the right-hand limit of the difference quotient: - - \lim_{h\to0^+}\frac{f(a+h)-f(a)}{h} \amp = \lim_{h\to0^+}\frac{\sqrt{0+h}-\sqrt{0}}h - \amp = \lim_{h\to0^+}\frac{\sqrt{h}}h - \amp = \lim_{h\to0^+}\frac{1}{h^{1/2}}\ =\ \infty - . -

    - -

    - The one-sided limit of the difference quotient does not exist at x=0 for f; - therefore f is differentiable on - (0,\infty) and not differentiable on [0,\infty). -

    - -

    - We state (without proof) that \fp(x) = 1/\big(2\sqrt{x}\big). - Note that \lim_{x\to 0^+}\fp(x) = \infty; - this limit was easier to evaluate than the limit of the difference quotient, - though it required us to already know the derivative of f. -

    - -

    - Now consider g: - - \lim_{h\to0^+}\frac{g(a+h)-g(a)}{h} \amp = \lim_{h\to0^+}\frac{\sqrt{(0+h)^3}-\sqrt{0}}h - \amp = \lim_{h\to0^+}\frac{h^{3/2}}h - \amp = \lim_{h\to0^+}h^{1/2}\ =\ 0 - . -

    - -

    - As the one-sided limit exists at x=0, - we conclude g is differentiable on its domain of [0,\infty). -

    - -

    - We state (without proof) that \gp(x) = 3\sqrt{x}/2. - Note that \lim_{x\to 0^+}\gp(x) = 0; - again, this limit is easier to evluate than the limit of the difference quotient. -

    - -
    - A graph of y=x^{1/2} and y=x^{3/2} in - - - Two curves, one of function of square root of x and the other of x to the power 3/2. - - -

    - The graph has two curves that form a leaf shape. - The y axis is drawn from 0 to 1 and the x axis is - drawn from -0.2 to 1.2. - The curve on top is of function x^(1/2) and its a downward facing curve. - The one below is of function x^(3/2) and it is an upward facing curve. - The two graphs intersect at points (0,0) and (1,1). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.2, - ymax=1.2, - xmin=-.1, - xmax=1.1, - axis equal - ] - \addplot+[domain=0:1]({\x^2},{\x^3}); - \addplot+[solid,domain=0:1]({\x^2},{\x}); - \draw (axis cs:.25,.7) node {$y=x^{1/2}$}; - \draw (axis cs:.7,.35) node {$y=x^{3/2}$}; - \end{axis} - \end{tikzpicture} - - -
    - -

    - We state (without proof) that \gp(x) = 3\sqrt{x}/2. - Note that \lim_{x\to0^+}\gp(x) = 0; - again, this limit is easier to evaluate than the limit of the difference quotient. -

    - -

    - The two functions are graphed in . - Note how f(x) = \sqrt{x} seems to go vertical - as x approaches 0, implying the slopes of its tangent lines are growing toward infinity. - Also note how the slopes of the tangent lines to - g(x)= \sqrt{x^3} approach 0 as x approaches 0. -

    -     -     - -

    - Most calculus textbooks omit this topic and simply avoid specific cases where it could be applied. - We choose in this text to not make use of the topic unless it is needed. - Many theorems in later sections require a function f to be differentiable on an - open interval I; - we could remove the word open - and just use \ldots on an interval I, - but choose to not do so in keeping with the current mathematical tradition. - Our first use of differentiability on closed intervals comes in , - where we measure the lengths of curves. -

    - -

    - This section defined the derivative; - in some sense, it answers the question of What - is the derivative? - The next section addresses the question - What does the derivative mean? -

    -     - - - - Terms and Concepts - - - - -

    - - Let f be a position function. - The average rate of change on [a,b] is the slope of the line through the points (a, f(a)) and (b,f(b)). -

    -     - -

    - Average velocity is the change in position, divided by the change in time. -

    -     - -     - - - - -

    - - The definition of the derivative of a function at a point involves taking a limit. -

    -     - -

    - A limit is necessary, since the difference quotient is undefined when h=0. -

    -     - -     - - - - -

    - In your own words, - explain the difference between the average rate of change and instantaneous rate of change. -

    - -     - - - -     - - - - -

    - In your own words, - explain the difference between Definitions - and . -

    - -     - - - -     - - - - -

    - Let y=f(x). - Give three different notations equivalent to f'(x). -

    - -     - - - -     - - - - -

    - If two lines are perpendicular, - what is true of their slopes? -

    - -     - - - -

    - The two lines have negative-reciprocal slopes. - That is, if m_1 and m_2 are the slopes, - then m_2 = -\frac{1}{m_1}. - Equivalently, the slopes of perpendicular lines satisfy m_1m_2=-1. -

    -     - -     -     - - - Problems - - - - -

    - Use the definition of the derivative to compute the derivative of the given function. -

    -     - - - - - $d = Formula('0'); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x) = 6 -

    -

    - -

    -

    - -

    -     -     -     - - - - - $d = Formula('2'); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x) = 2x -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->are(t=>'Real'); - $d = Formula('-3'); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(t) = 4-3t -

    -

    - -

    -

    - -

    -     -     -     - - - - - $d = Formula('2x'); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - g(x) = x^2 -

    -

    - -

    -

    - -

    -     -     -     - - - - - $d = Formula('3x^2'); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - h(x) = x^3 -

    -

    - -

    -

    - -

    -     -     -     - - - - - $d = Formula('6x-1'); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - f(x) = 3x^2-x+4 -

    -

    - -

    -

    - -

    -     -     -     - - - - - $d = Formula('-1/x^2'); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - r(x) = \frac{1}{x} -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->are(s=>'Real'); - $d = Formula('-1/(s-2)^2'); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - r(s) = \frac{1}{s-2} -

    -

    - -

    -

    - -

    -     -     -     -     - - - -

    - A function and an x-value are given. - (Note: these functions are the same as those given in Exercises.) - Give the equations of the tangent line and the normal line at that x-value. -

    -     - - - - - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $tangentline=Formula("y=6"); - $normalline=Formula("x=-2"); - - -

    - f(x)=6 at x=-2 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $tangentline=Compute("y=2x"); - $normalline=Compute("y=-(1/2)(x-3)+6"); - - -

    - f(x)=2x at x=3 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $tangentline=Compute("y=-3x+4"); - $normalline=Compute("y=(1/3)(x-7)-17"); - - -

    - f(x)=4-3x at x=7 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $tangentline=Compute("y=4(x-2)+4"); - $normalline=Compute("y=-(1/4)(x-2)+4"); - - -

    - g(x)=x^2 at x=2 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $tangentline=Compute("y=48(x-4)+64"); - $normalline=Compute("y=-1/48(x-4)+64"); - - -

    - h(x)=x^3 at x=4 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $tangentline=Compute("y=-7(x+1)+8"); - $normalline=Compute("y=(1/7)(x+1)+8"); - - -

    - f(x)=3x^2-x+4 at x=-1 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $tangentline=Compute("y=-(1/4)(x+2)-1/2"); - $normalline=Compute("y=4(x+2)-1/2"); - - -

    - r(x)=\frac{1}{x} at x=-2 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $tangentline=Compute("y=-(x-3)+1"); - $normalline=Compute("y=(x-3)+1"); - - -

    - r(x)=\frac{1}{x-2} at x=3 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     -     - - - -

    - A function f and an x-value a are given. - Approximate the equation of the tangent line to the graph of f at x=a by numerically approximating f'(a), - using h=0.1. -

    -     - - - - - $b=non_zero_random(-5,5,1); - $c=non_zero_random(-9,9,1); - $a=non_zero_random(-5,5,1); - if($envir{problemSeed}==1){$a=3;$b=2;$c=1;}; - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $f=Formula("x^2+$a x + $b")->reduce; - $fa=$f->eval(x=>$a); - $h=0.1; - $m=($f->eval(x=>$a+$h) - $fa)/$h; - $tangentline=Compute("y=$m(x-$a)+$fa"); - - -

    - f(x) = and a= -

    -

    - -

    -     -     -     - - - - - $b=non_zero_random(-10,10,1); - $c=non_zero_random(-9,9,1); - $a=non_zero_random(-9,9,1); - do {$a=non_zero_random(-9,9,1)} until ($a+$c != 0); - if($envir{problemSeed}==1){$a=10;$b=1;$c=9;}; - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $f=Formula("$b/(x+$c)")->reduce; - $fa=$f->eval(x=>$a); - $h=0.1; - $m=($f->eval(x=>$a+$h) - $fa)/$h; - $tangentline=Compute("y=$m(x-$a)+$fa"); - - -

    - f(x) = and a= -

    -

    - -

    -     -     -     - - - - - $a=non_zero_random(-5,5,1); - if($envir{problemSeed}==1){$a=2;}; - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $f=Formula("e^x")->reduce; - $fa=$f->eval(x=>$a); - $h=0.1; - $m=($f->eval(x=>$a+$h) - $fa)/$h; - $tangentline=Compute("y=$m(x-$a)+$fa"); - - -

    - f(x) = and a= -

    -

    - -

    -     -     -     - - - - - Context("ImplicitPlane"); - Context()->variables->are(x=>'Real',y=>'Real'); - $tangentline=Compute("y=-0.04996x+1"); - - -

    - f(x) = \cos(x) and a=0 -

    -

    - -

    -     -     -     -     - - - - - @approx=( - Compute("-2")->with(tolType=>'absolute',tolerance=>0.4), - Compute("0")->with(tolType=>'absolute',tolerance=>0.1), - Compute("4")->with(tolType=>'absolute',tolerance=>2) - ); - $approx_list=List($approx[0],$approx[1],$approx[2]); - $derivative=Formula("2x"); - $slopes=List(-2,0,4); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - The graph of f(x)=x^2-1 is shown. -

    - - - Graph of function x squared minus 1, curve is U-shaped. - - -

    - The x axis is between -2 and 2 and the y axis - is drawn between -1 and 3. - The graph is U-shaped reaching a minimum at (0,-1) and moving - further upwards after crossing point (-1,0) to the left and (1,0) to the right. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ytick={-1,1,2,3}, - ymin=-1.3,ymax=3.5,% - xmin=-2.1,xmax=2.1,% - grid=major - ] - - \addplot [firstcurvestyle,infinite,domain=-2.1:2.1] {x^2-1}; - - \end{axis} - \end{tikzpicture} - - - -     - - -

    - Use the graph to approximate the slope of the tangent line to f at (-1,0), (0,-1), and (2,3). -

    - - - Enter your answers as a comma-separated list of values. - - -

    - -

    -     -     - - - -

    - Using the definition of the derivative, find \fp(x). -

    -

    - -

    -

    - -

    -     -     - - -

    - Use the derivative to find the slope of the tangent line at the points (-1,0), (0,-1) and (2,3). -

    - - Enter your answers as a comma-separated list of values. - -

    - -

    -     -     -     -     - - - - - @approx=( - Compute("-1")->with(tolType=>'absolute',tolerance=>0.2), - Compute("-1/4")->with(tolType=>'absolute',tolerance=>0.1) - ); - $approx_list=List($approx[0],$approx[1]); - $derivative=Formula("-1/(x+1)^2"); - $slopes=List(-1,'-1/4'); - $showwork = '[@ explanation_box(message => "Show your work.") @]*'; - - -

    - The graph of f(x)=\frac{1}{x+1} is shown. -

    - - - Graph of function f(x) = 1 divided by (x+1). - - -

    - Graph of function f(x) = \frac{1}{x+1}. - The x axis is drawn from -1 - to 3 and the y axis from 0 to 5. - The curve begins in the second quadrant and steeply declines - from infinty to point (0,1) then it enters the first quadrant declining gently - and seems to come very close to the x axis on the right. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ytick={1,2,3,4,5}, - ymin=-.5,ymax=5.5,% - xmin=-1.1,xmax=3.1,% - grid=major - ] - - \addplot [firstcurvestyle,infinite,domain=-.9:3.1,samples=50] {1/(x+1)}; - - \end{axis} - \end{tikzpicture} - - -     - - - -

    - Use the graph to approximate the slope of the tangent line to f at (0,1) and (1,0.5). -

    - - Enter your answers as a comma-separated list of values. - -

    - -

    -     -     - - - -

    - Using the definition of the derivative, find \fp(x). -

    -

    - -

    -

    - -

    -     -     - - - -

    - Use the derivative to find the slope of the tangent line at the points (0,1) and (1,0.5). -

    - - Enter your answers as a comma-separated list of values. - -

    - -

    -     -     -     -     - - - - -

    - A graph of a function f(x) is given. - Using the graph, sketch \fp(x). -

    -     - - - - - - Graph of function f(x) that is a straight line. - - -

    - The y axis if drawn from -1 to 3 - and the x axis from -2 to 5. - The graph is a decreasing straight line primarily in - the first quadrant that passes through 2 on the y axis and 4 on the x axis. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-2.1, - xmax=5.1, - ymin=-1.5, - ymax=3.5 - ] - \addplot+[infinite,domain=-2.1:5.1] {-.5*x+2}; - \end{axis} - \end{tikzpicture} - - -     -     - - - - - - Graph of function is U shaped primarily in the second and third quadrant. - - -

    - The y axis is from -3 to 3 and the x axis -6 and 2. - The U-shaped curve has a minimun at point (-2,-3) - it intersects the x axis at -4.5 on the left and 0.5 on the right. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-7.1, - xmax=3.1, - ymin=-3.5, - ymax=3.5 - ] - \addplot+[infinite,domain=-5.6:1.6,samples=40] {.5*(x+2)^2-3}; - \end{axis} - \end{tikzpicture} - - -     -     - - - - - - Graph of curve that crosses the origin with a mixima at x=-1 and a minima at x=1. - - -

    - The x axis is drawn from -3 to 3 and the y axis from -5 to 5. - The graph has a maxima near point (-1,3) and a minima near point (1,-3). - It crosses the origin then passes the maxima then crosses x axis at -2 - and decreases in the third quadrant and continues to infinity on the left. - On the right side it decreases from the origin passes the minima and goes to infinity - after crossing the x axis at 2 in the first quadrant. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-3.1, - xmax=3.1, - ymin=-5.5, - ymax=5.5 - ] - \addplot+[infinite,domain=-2.45:2.45,samples=40] {(x-2)*x*(x+2)}; - \end{axis} - \end{tikzpicture} - - -     -     - - - - - - Graph of function that has maxima at even multiple of pi and minima at odd multiple of pi. - - -

    - The y axis is drawn from -1 to 1 and the x axis is drawn from - -2* \pi to 2* \pi. - At x=0 the function has a maximum of 1 on the left of the y axis the - function decreases from 1 and reaches a minima of -1 at x=-\pi after crossing -\p/2. - Then it increases to cross the x axis at x=-3*\pi/2 and increases further - to reach a maximum of 1 in the second quadrant. -

    -

    - Similarly, on the right of the y axis the function decreases from 1 and reaches - a minima of -1 at x=\pi after crossing \p/2. - Then it increases to cross the x axis at x=3*\pi/2 and increases further to reach - a maximum of 1 in the first quadrant. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-6.9, - xmax=6.9, - ymin=-1.1, - ymax=1.1, - xtick={-6.28,-3.14,3.14,6.28}, - xticklabels={$-2\pi$, $-\pi$, $\pi$, $2\pi$} - ] - \addplot+[infinite,domain=-6.8:6.8,samples=101] {cos(deg(x))}; - \end{axis} - \end{tikzpicture} - - -     -     -     - - - -

    - Use the graph of the function to answer the following questions. -

      -
    1. -

      - Where is g(x)\gt 0? -

      -
      2. -
    2. -

      - Where is g(x)\lt 0? -

      -
      3. -
    3. -

      - Where is g(x) = 0? -

      -
      4. -
    4. -

      - Where is \gp(x) \lt 0? -

      -
      5. -
    5. -

      - Where is \gp(x) \gt 0? -

      -
      6. -
    6. -

      - Where is \gp(x) = 0? -

      -
      7. -
    -

    -     - - - - Context("Interval"); - $pos=Compute("(-2,0)U(2,inf)"); - $neg=Compute("(-inf,-2)U(0,2)"); - $zer=Compute("{-2,0,2}"); - $dec=Compute("(-1,1)"); - $inc=Compute("(-inf,-1)U(1,inf)"); - $fla=Compute("{-1,1}"); - - - - - Graph of curve that crosses the origin with a mixima at x=-1 nd a minima at x=1. - - -

    - The x axis is drawn from -3 to 3 and the y axis from -5 to 5. - The graph has a maxima near point (-1,3) and a minima near point (1,-3). - It crosses the origin passes the maxima, crosses x axis at -2 - and decreases in the third quadrant and continues to infinity on the left. - On the right side it decreases from the origin passes the minima and increases - to infinity after crossing the x axis at 2 in the first quadrant. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xtick={-2,-1,1,2}, - ymin=-5.6,ymax=5.6,% - xmin=-3,xmax=3,% - ] - - \addplot [firstcurvestyle,infinite,domain=-2.5:2.5,samples=40] {(x-2)*x*(x+2)}; - - \end{axis} - \end{tikzpicture} - - - - Answer using interval notation or set notation, as appropriate. If you need to write \infty, you may type inf or infinity. - If you need the union symbol, \cup, you may type the capital letter U. - - - Enter the set on which g(x) \gt 0 using interval notation. - -

    - -

    - - Enter the set on which g(x) \lt 0 using interval notation. - -

    - -

    - - Enter the set of points where g(x) = 0 using the syntax {a,b}. - -

    - -

    - - Enter the set on which g^\prime(x) \lt 0 using interval notation. - -

    - -

    - - Enter the set on which g^\prime(x) \gt 0 using interval notation. - -

    - -

    - - Enter the set of points where g^\prime(x) = 0 using the syntax {a,b}. - -

    - -

    -     -     -     - - - - - Context("Interval"); - $pos=Compute("(-2,2)"); - $neg=Compute("(-inf,-2)U(2,inf)"); - $zer=Compute("{-2,2}"); - $dec=Compute("(-1,0)U(1,inf)"); - $inc=Compute("(-inf,-1)U(0,1)"); - $fla=Compute("{-1,0,1}"); - - - - - An M-shaped graph with two maxima. - - -

    - The y axis is drawn from -2 to 5 and the x axis - is drawn from -2 and 2. - At x=0 y has a value of 4, to the left the function increases to reach a maximum - of 4.5 in the second quadrant - and decreses and continues into the third quadrant after crossing x axis at -2. - Similarly, on the right side, the function increases to reach a maximum of 4.5 in the - first quadrant and decreses and continues into the fourth quadrant after crossing x axis at 2. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xtick={-2,-1,1,2}, - ymin=-3,ymax=6,% - xmin=-2.8,xmax=2.8,% - ] - - \addplot [firstcurvestyle,infinite,domain=-2.2:2.2,samples=50] {(-2)*(x^4/4-x^2/2)+4}; - - \end{axis} - \end{tikzpicture} - - - - Answer using interval notation or set notation, as appropriate. If you need to write \infty, you may type inf or infinity. - If you need the union symbol, \cup, you may type the capital letter U. - - - Answer using interval notation or set notation, as appropriate. If you need to write \infty, you may type inf or infinity. - If you need the union symbol, \cup, you may type the capital letter U. - - - Enter the set on which g(x) \gt 0 using interval notation. - -

    - -

    - - Enter the set on which g(x) \lt 0 using interval notation. - -

    - -

    - - Enter the set of points where g(x) = 0 using the syntax {a,b}. - -

    - -

    - - Enter the set on which g^\prime(x) \lt 0 using interval notation. - -

    - -

    - - Enter the set on which g^\prime(x) \gt 0 using interval notation. - -

    - -

    - - Enter the set of points where g^\prime(x) = 0 using the syntax {a,b}. - -

    - -

    -     -     -     -     - - - -

    - A function f(x) is given, - along with its domain and derivative. - Determine if f(x) is differentiable on its domain. -

    -     - - - - -

    - f(x) = \sqrt{x^5(1-x)}, - domain is [0,1], - \fp(x) = \frac{(5-6x)x^{3/2}}{2\sqrt{1-x}} -

    - -     - -

    - \lim_{h\to0^+}\frac{f(0+h)-f(0)}h =0; - note also that \lim_{x\to0^+}\fp(x) = 0. - So f is differentiable at x=0. -

    -

    - \lim_{h\to0^-}\frac{f(1+h)-f(1)}h =-\infty; - note also that \lim_{x\to1^-}\fp(x) = -\infty. - So f is not differentiable at x=1. -

    -

    - f is differentiable on [0,1), not its entire domain. -

    -     - -     - - - - -

    - f(x) = \cos\left(\sqrt{x}\right), - domain is [0,\infty), - \fp(x) = -\frac{\sin\left(\sqrt{x}\right)}{2\sqrt{x}} -

    - -     - -

    - The limit of the difference quotient is difficult to evaluate. - Using , - we can determine \lim_{x\to0^+}\fp(x) = -1/2. -

    -

    - Since \fp is defined on (0,\infty), - we conclude f is differentiable on [0,\infty). -

    -     - -     -     -     -     -
    -
    - Interpretations of the Derivative - -

    - - defined the derivative of a function and gave examples of how to compute it using its definition (, using limits). - The section also started with a brief motivation for this definition, that is, - finding the instantaneous velocity of a falling object given its position function. - - will give us more accessible tools for computing the derivative; - tools that are easier to use than repeated use of limits. -

    - -

    - This section falls in between the - What is the definition of the derivative? - and How do I compute the derivative? sections. - Here we are concerned with What does the derivative mean?, - or perhaps, - when read with the right emphasis, - What is the derivative? - We offer two interconnected interpretations of the derivative, - hopefully explaining why we care about it and why it is worthy of study. - derivativeinterpretation -

    -     - - - Interpretation of the Derivative as Instantaneous Rate of Change -

    - - started with an example of using the position of an object - (in this case, a falling amusement park rider) - to find the object's velocity. - This type of example is often used when introducing the derivative because we tend to readily recognize that velocity is the - instantaneous rate of change in position. In general, - if f is a function of x, - then \fp(x) measures the instantaneous rate of change of f with respect to x. - Put another way, the derivative answers - When x changes, - at what rate does f change? - Thinking back to the amusement park ride, - we asked When time changed, - at what rate did the height change? - and found the answer to be By -64 feet per second. -

    - -

    - Now imagine driving a car and looking at the speedometer, - which reads - 60 - . Five minutes later, - you wonder how far you have traveled. - Certainly, lots of things could have happened in those 5 minutes; - you could have intentionally sped up significantly, - you might have come to a complete stop, you might have slowed to - - 20 - - as you passed through construction. - But suppose that you know, as the driver, - none of these things happened. - You know you maintained a fairly consistent speed over those 5 minutes. - What is a good approximation of the distance traveled? -

    - -

    - One could argue the only good approximation, - given the information provided, - would be based on \text{distance} = \text{rate}\times\text{time.} In this case, - we assume a constant rate of - - 60 - - with a time of 5 minutes or 5/60 of an hour. - Hence we would approximate the distance traveled as 5 miles. -

    - -

    - Referring back to the falling amusement park ride, - knowing that at t=2 the velocity was -64 ft/s, we could reasonably approximate that 1 second later the riders' height would have dropped by about 64 feet. - Knowing that the riders were accelerating - as they fell would inform us that this is an - under-approximation. - If all we knew was that f(2) = 86 and \fp(2) = -64, - we'd know that we'd have to stop the riders quickly otherwise they would hit the ground. -

    - -

    - In both of these cases, - we are using the instantaneous rate of change to predict future values of the output. -

    - - - -     - - - Units of the Derivative -

    - It is useful to recognize the units - of the derivative function. - If y is a function of x, - , y=f(x) for some function f, - and y is measured in feet and x in seconds, - then the units of \yp = \fp are - feet per second, - commonly written as ft/s. In general, - if y is measured in units P and x is measured in units Q, - then \yp will be measured in units - P per Q, - or P/Q. Here we see the fraction-like behavior of the derivative in the notation: - the units of \frac{dy}{dx}are \frac{\text{units of }y}{\text{units of }x}. -

    - - - The meaning of the derivative: World Population - -

    - Let P(t) represent the world population t minutes after 12:00 a.m., January 1, 2012. - It is fairly accurate to say that P(0) = 7{,}028{,}734{,}178 - (). - It is also fairly accurate to state that P'(0) = 156; - that is, at midnight on January 1, 2012, - the population of the world was growing by about 156 - people per minute - (note the units). - Twenty days later - (or 28{,}800 minutes later) - we could reasonably assume the population grew by about 28{,}800\cdot156 = 4{,}492{,}800 people. -

    -     -     - - - The meaning of the derivative: Manufacturing - -

    - The term widget is an economic term for a generic unit of manufacturing output. - Suppose a company produces widgets and knows that the market supports a price of \$10 per widget. - Let P(n) give the profit, in dollars, - earned by manufacturing and selling n widgets. - The company likely cannot make a (positive) profit making just one widget; - the start-up costs will likely exceed \$10. - Mathematically, we would write this as P(1) \lt 0. -

    - -

    - What do P(1000) = 500 and P'(1000)=0.25 mean? - Approximate P(1100). -

    -     - -

    - The equation P(1000)=500 means that selling 1000 widgets returns a profit of \$500. - We interpret P'(1000) = 0.25 as meaning that when we are selling 1000 widgets, - the profit is increasing at rate of \$0.25 per widget - (the units are dollars per widget.) - Since we have no other information to use, - our best approximation for P(1100) is: - - P(1100) \amp \approx P(1000) + P'(1000)\times100 - \amp= \$500 + (100\text{ widgets })\cdot \$0.25/\text{widget} - \amp= \$525 - . -

    - -

    - We approximate that selling 1100 widgets returns a profit of \$525. -

    -     -     - - - - - -

    - The previous examples made use of an important approximation tool that we first used in our previous driving a car at - - 60 - example at the beginning of this section. - Five minutes after looking at the speedometer, - our best approximation for distance traveled assumed the rate of change was constant. - In Examples - and - we made similar approximations. - We were given rate of change information which we used to approximate total change. - Notationally, we would say that - - f(c+h) \approx f(c) + \fp(c)\cdot h - . -

    - -

    - This approximation is best when h is small. - Small is a relative term; - when dealing with the world population, - h=22\text{ days} = 28{,}800\text{ minutes} is small in comparison to years. - When manufacturing widgets, - 100 widgets is small when one plans to manufacture thousands. -

    -     - - - The Derivative and Motion -

    - One of the most fundamental applications of the derivative is the study of motion. - Let s(t) be a position function, - where t is time and s(t) is distance. - For instance, - s could measure the height of a projectile or the distance an object has traveled. -

    - -

    - Let s(t) measure the distance traveled, - in feet, of an object after t seconds of travel. - Then s'(t) has units feet per second, - and s'(t) measures the instantaneous rate of distance change with respect to time - it measures velocity. - velocity -

    - -

    - Now consider v(t), a velocity function. - That is, at time t, v(t) gives the velocity of an object. - The derivative of v, v'(t), - gives the instantaneous rate of velocity change with respect to time - acceleration. - (We often think of acceleration in terms of cars: - a car may go from 0 to 60 in 4.8 seconds. - This is an average acceleration, - a measurement of how quickly the velocity changed.) - If velocity is measured in feet per second, - and time is measured in seconds, - then the units of acceleration (, the units of v'(t)) are - feet per second per second, - or (ft/s)/s. - We often shorten this to feet per second squared, or - - - , but this tends to obscure the meaning of the units. - - acceleration -

    - -

    - Perhaps the most well known acceleration is that of gravity. - In this text, - we use g=32\,\text{ft}/\text{s}^2 or g=9.8\,\text{m}/\text{s}^2. - What do these numbers mean? -

    - -

    - A constant acceleration of - 32\,\frac{\text{ft}/\text{s}}{\text{s}} means that the velocity changes by 32\,\text{ft}/\text{s} each second. - For instance, - let v(t) measure the velocity of a ball thrown straight up into the air, - where v has units ft/s and t is measured in seconds. - The ball will have a positive velocity while traveling upwards and a negative velocity while falling down. - The acceleration is thus -32\,\text{ft}/\text{s}^2. - If v(1) = 20\,\text{ft}/\text{s}, - then 1 second later, - the velocity will have decreased by 32\,\text{ft}/\text{s}; - that is, v(2) = -12\,\text{ft/s}. - We can continue: v(3) = -44\,\text{ft/s}. - Working backward, - we can also figure that v(0) = 52\,\text{ft}/\text{s}. -

    - -

    - These ideas are so important we write them out as a Key Idea. -

    - - - The Derivative and Motion -

    -

      -
    1. -

      - Let s(t) be the position function of an object. - Then s'(t)=v(t) is the velocity function of the object. -

      -
      2. -
    2. -

      - Let v(t) be the velocity function of an object. - Then v'(t)=a(t) is the acceleration function of the object. -

      -
      3. -
    - derivativemotion - derivativevelocity - derivativeacceleration -

    -     -     - - - Interpretation of the Derivative as the Slope of the Tangent Line -

    - We now consider the second interpretation of the derivative given in this section. - This interpretation is not independent from the first by any means; - many of the same concepts will be stressed, - just from a slightly different perspective. -

    - -

    - Given a function y=f(x), - the difference quotient \frac{f(c+h)-f(c)}{h} gives a change in y values divided by a change in x values; - , it is a measure of the rise over run, or slope, - of the secant line that goes through two points on the graph of f: - (c, f(c)) and (c+h,f(c+h)). - As h shrinks to 0, - these two points come close together; - in the limit we find \fp(c), - the slope of a special line called the tangent line that intersects f only once near x=c. -

    - -

    - Lines have a constant rate of change, their slope. - Nonlinear functions do not have a constant rate of change, - but we can measure their instantaneous rate of change - at a given x value c by computing \fp(c). - We can get an idea of how f is behaving by looking at the slopes of its tangent lines. - We explore this idea in the following example. -

    - - - Understanding the derivative: the rate of change - -

    - Consider f(x) = x^2 as shown in . - It is clear that at x=3 the function is growing faster than at x=1, - as it is steeper at x=3. - How much faster is it growing at 3 compared to 1? -

    - -
    - A graph of f(x)=x^2 - - - A graph of f(x)=x^2 - -

    - The curve is parabolic with the focus being at the origin. - As the x values get bigger and bigger the curve gers further from the y axis - on both sides as both the ends approach infinity. Both the rate at which - it is getting further gets slower as the x values increase. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1, - xmax=4.1, - ymin=-.4, - ymax=17, - xtick={-1,0,...,4}, - minor xtick={-1,0,...,4}, - ytick={0,5,...,15}, - minor ytick={0,1,...,17}, - ] - \addplot+[infinite,domain=-.8:4] {x^2}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -

    - We can answer this exactly - (and quickly) - after , - where we learn to quickly compute derivatives. - For now, we will answer graphically, - by considering the slopes of the respective tangent lines. -

    - -
    - A graph of f(x)=x^2 and tangent lines at x=1 and x=3 - - - a graph of f(x)=x^2 - -

    - The curve is parabolic with the focus being at the origin. As the x values get bigger and bigger the curve gets - further from the y axis on both sides as both the ends approach infinity. The rate at which it is getting further - gets slower as the x values increase. The tangent lines are drawn at x=1 and x=3. -

    - -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1, - xmax=4.1, - ymin=-.4, - ymax=17, - xtick={-1,0,...,4}, - minor xtick={-1,0,...,4}, - ytick={0,5,...,15}, - minor ytick={0,1,...,17}, - grid=major, - ] - \addplot+[infinite,domain=-.8:4] {x^2}; - \addplot[tangentline,domain=0.4:3.5]{2*(x-1)+1}; - \addplot[tangentline,domain=1.6:4]{6*(x-3)+9}; - \addplot[soliddot] coordinates{(1,1) (3,9)}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - With practice, - one can fairly effectively sketch tangent lines to a curve at a particular point. - In , - we have sketched the tangent lines to f at x=1 and x=3, - along with a grid to help us measure the slopes of these lines. - At x=1, the slope is 2; - at x=3, the slope is 6. - Thus we can say not only is f growing faster at x=3 than at x=1, - it is growing three times as fast. -

    -     -     - - - Understanding the graph of the derivative - -

    - Consider the graph of f(x) and its derivative, - \fp(x), in . - Use these graphs to find the slopes of the tangent lines to the graph of f at x=1, - x=2, and x=3. -

    - -
    - Graphs of f and \fp in - - - Graphs of a function and its derivative. - -

    - The curve of f(x) intersects y axis at y=4 and slopes downwards. After intersecting the x axis at around x=0.5. - It continues sloping downwards in the negative y direction until about y=-2 and then starts sloping upwards again. - This time it intersects the x axis at x=2 and then continue upwards until y=5 then slope downwards again. - The graph of \fp is a downward parabolic curve that intersects f(x) at (2.5, 4) in the positive y direction and - y=-2 at negative y direction. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - minor y tick num=4, - minor x tick num=0, - xtick={1,2,3}, - ymin=-8.9, - ymax=5.9, - xmin=-.1, - xmax=3.49, - clip=false - ] - \addplot+[infinite,domain=0:3.5] {-x^3+7*x^2-12*x+4} node[above left] {$f(x)$}; - \addplot+[infinite,domain=0.5:3.3] {-12+14*x-3*x^2} node[pos=0.5, above left] {$f'(x)$}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -

    - To find the appropriate slopes of tangent lines to the graph of f, - we need to look at the corresponding values of \fp. -

      -
    • -

      - The slope of the tangent line to f at x=1 is \fp(1); - this looks to be about -1. -

      -
      • -
    • -

      - The slope of the tangent line to f at x=2 is \fp(2); - this looks to be about 4. -

      -
      • -
    • -

      - The slope of the tangent line to f at x=3 is \fp(3); - this looks to be about 3. -

      -
      • -
    -

    - -

    - Using these slopes, - tangent line segments to f are sketched in . - Included on the graph of \fp in this figure are points where x=1, - x=2 and x=3 to help better visualize the y value of \fp at those points. -

    - -
    - Graphs of f and \fp in - - - Graphs of function f and its derivative - -

    - The curve of f(x) starts at at y=4 and slopes downwards intersecting the x axis at around x=0.5. - It continues sloping downwards in the negative y direction until about y=-2 and then starts sloping upwards again, - this time intersecting the x axis at x=2 and then continues upwards until y=5 then slope downwards again. - The graph of \fp is a downward parabolic curve that intersects f(x) at (2.5, 4) in the positive y direction - and y=-2 at negative y direction. The tangent lines of f are drawn at (1, 2), (2, 0), (3, 4). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - minor y tick num=4, - minor x tick num=0, - xtick={1,2,3}, - ymin=-8.9, - ymax=5.9, - xmin=-.1, - xmax=3.49, - clip=false - ] - \addplot+[infinite,domain=0:3.5] {-x^3+7*x^2-12*x+4} node[above left] {$f(x)$}; - \addplot+[infinite,domain=0.5:3.3] {-12+14*x-3*x^2} node[pos=0.5, above left] {$f'(x)$}; - \addplot [tangentlineseg,domain=0.5:1.5] {-1*(x-1)-2}; - \addplot [tangentlineseg,domain=1.5:2.5] {4*(x-2)}; - \addplot [tangentlineseg,domain=2.5:3.5] {3*(x-3)+4}; - \addplot[soliddot] coordinates{(1,-2) (2,0) (3,4)}; - \addplot[soliddot] coordinates{(1,-1) (2,4) (3,3)}; - \end{axis} - \end{tikzpicture} - - - -
    -     -     - - - Approximation with the derivative - -

    - Consider again the graph of f(x) and its derivative \fp(x) in . - Use the tangent line to f at x=3 to approximate the value of f(3.1). -

    -     - -

    - - shows the graph of f along with its tangent line, - zoomed in at x=3. - Notice that near x=3, - the tangent line makes an excellent approximation of f. - Since lines are easy to deal with, - often it works well to approximate a function with its tangent line. - (This is especially true when you don't actually know much about the function at hand, - as we don't in this example.) -

    - -

    - While the tangent line to f was drawn in , - it was not explicitly computed. - Recall that the tangent line to f at x=c is y = \fp(c)(x-c)+f(c). - While f is not explicitly given, - by the graph it looks like f(3) = 4. - Recalling that \fp(3) = 3, - we can compute the tangent line to be approximately y = 3(x-3)+4. - It is often useful to leave the tangent line in point-slope form. -

    - -
    - Zooming in on f and its tangent line at x=3 for the function given in Examples and - - - Graph of a function f that is tracing the path of a projectile and its tangent line drawn at x=3. - -

    - The graph of f starts close to (0, 3). It curves upwards tracing the path of a projectile. Up to (3.1, 4.2) in the first quadrant it moves straight - upwards to the right. Then the rate at which it was moving upwards becomes slower and the graph makes a slower movement where it is still moving upwards but we can see - that the movement also has a downward component which pulling it downwards. The tangent line is drawn at (3, 4). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - minor y tick num=4, - ymin=1.9, - ymax=4.9, - xmin=2.7, - xmax=3.3, - xdiscontinuity, - ydiscontinuity, - grid=major - ] - \addplot+[infinite,domain=2.7:3.3] {-x^3+7*x^2-12*x+4} node[pos=0.85,below] {$f(x)$}; - \addplot[tangentlineseg,domain=2.7:3.3,thick] {3*(x-3)+4}; - \addplot[soliddot] coordinates{(3,4)}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - To use the tangent line to approximate f(3.1), - we simply evaluate y at 3.1 instead of f. - - f(3.1) \amp \approx y(3.1) - \amp= 3(3.1-3)+4 - \amp= 0.1\cdot3+4 - \amp = 4.3 - . -

    - -

    - We approximate f(3.1) \approx 4.3. -

    -     -     - -

    - To demonstrate the accuracy of the tangent line approximation, - we now state that in , - f(x) = -x^3+7x^2-12x+4. - We can evaluate f(3.1) = 4.279. - Had we known f all along, - certainly we could have just made this computation. - In reality, we often only know two things: -

      -
    1. -

      - what f(c) is, for some value of c, and -

      -
      2. -
    2. -

      - what \fp(c) is. -

      -
      3. -
    -

    - -

    - For instance, - we can easily observe the location of an object and its instantaneous velocity at a particular point in time. - We do not have a function f - for the location, just an observation. - This is enough to create an approximating function for f. -

    - -

    - This last example has a direct connection to our approximation method explained above after . - We stated there that - - f(c+h) \approx f(c)+\fp(c)\cdot h - . -

    - -

    - If we know f(c) and \fp(c) for some value x=c, - then computing the tangent line at (c,f(c)) is easy: - y(x) = \fp(c)(x-c)+f(c). - In , - we used the tangent line to approximate a value of f. - Let's use the tangent line at x=c to approximate a value of f near x=c; - , compute y(c+h) to approximate f(c+h), - assuming again that h is small. Note: - - y(c+h) \amp = \fp(c)\left((c+h)-c\right)+f(c) - \amp = \fp(c)\cdot h + f(c) - . -

    - -

    - This is the exact same approximation method used above! - Not only does it make intuitive sense, - as explained above, it makes analytical sense, - as this approximation method is simply using a tangent line to approximate a function's value. -

    - -

    - The importance of understanding the derivative cannot be understated. - When f is a function of x, - \fp(x) measures the instantaneous rate of change of f with respect to x and gives the slope of the tangent line to f at x. -

    -     - - - - Terms and Concepts - - - - -

    - What is the instantaneous rate of change of position called? -

    -

    - -

    -     - - - - - - - - -

    - Your answer includes the correct word but has extra text. - Unless stated otherwise, velocity is assumed to be instantaneous. -

    -     -     -     -     - - -     - - - - -

    - Given a function y=f(x), - in your own words describe how to find the units of \fp(x). -

    - -     - - - -     - - - - -

    - What functions have a constant rate of change? -

    -

    - - -

    -     - - - - - linear|linear functions? - - - - - -

    - Your answer includes the correct word but has extra text. -

    -     -     -     -     - -     -     - - Problems - - - - - $a=random(1,9,1); - $fa=random(10,20,1); - $fpa=non_zero_random(-4,4,1); - if($envir{problemSeed}==1){$a=5;$fa=10;$fpa=2;}; - $b=$a+1; - $ans=$fa+$fpa*($b-$a); - - -

    - Given f()= and \fp() = , approximate f(). -

    -

    - -

    -     -     -     - - - - - $a=random(50,150,10); - $fa=non_zero_random(-99,99,1); - $fpa=non_zero_random(-9,9,1); - $b=$a+random(5,15,5); - if($envir{problemSeed}==1){$a=100;$fa=-67;$fpa=5;$b=10;}; - $ans=$fa+$fpa*($b-$a); - - -

    - Given P()= and P'() = , approximate P(). -

    -

    - -

    -     -     -     - - - - - $a=random(10,95,5); - $fa=random(101,199,1); - $fpa=non_zero_random(-20,20,1); - $b=$a-random(5,15,5); - if($envir{problemSeed}==1){$a=25;$fa=187;$fpa=17;$b=20;}; - $ans=$fa+$fpa*($b-$a); - - -

    - Given z()= and z'() = , - approximate z(). -

    -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $popup = DropDown(['f(10.1)','f(11)','f(20)'],0,showInStatic=>0); - $showwork = '[@ explanation_box(message => "Explain your reasoning.") @]*'; - - -

    - Knowing f(10)=25 and - \fp(10) = 5 and the methods described in this section, - which approximation is likely to be most accurate? -

    -

    - -

    -     - -

    - f(10.1) is likely most accurate, - as accuracy is lost the farther from x=10 we go. -

    -     -     -     - - - - - $a=random(1,9,1); - $fa=random(11,99,1); - $fpa=non_zero_random(-10,10,1); - if($envir{problemSeed}==1){$a=7;$fa=26;$fpa=-4;}; - $b=$a+1; - $fb=$fa+$fpa*($b-$a); - - -

    - Given f()= and f() = , - approximate \fp(). -

    -

    - -

    -     -     -     - - - - - $a=random(-5,5,1); - $fa=random(11,99,1); - $fpa=non_zero_random(-10,10,1); - $b=$a+random(2,6,1); - if($envir{problemSeed}==1){$a=0;$fa=17;$fpa=6;$b=2;}; - $fb=$fa+$fpa*($b-$a); - - -

    - Given H()= and H() = , - approximate H'(). -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('decibels per customer'=>{}); - Context()->strings->add( - 'decibel per customer' => {alias=>'decibels per customer'}, - 'db per customer' => {alias=>'decibels per customer'}, - 'decibels / customer' => {alias=>'decibels per customer'}, - 'decibel / customer' => {alias=>'decibels per customer'}, - 'db / customer' => {alias=>'decibels per customer'}, - 'decibels/customer' => {alias=>'decibels per customer'}, - 'decibel/customer' => {alias=>'decibels per customer'}, - 'db/customer' => {alias=>'decibels per customer'}, - ); - $ans=Compute("decibels per customer"); - - -

    - Let V(x) measure the volume, in decibels, - measured inside a restaurant with x customers. - What are the units of V'(x)? -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('foot per second squared'=>{}); - Context()->strings->add( - 'foot per square second' => {alias=>'foot per second squared'}, - 'foot per second per second' => {alias=>'foot per second squared'}, - 'ft per second squared' => {alias=>'foot per second squared'}, - 'ft per square second' => {alias=>'foot per second squared'}, - 'ft per second per second' => {alias=>'foot per second squared'}, - 'foot per s^2' => {alias=>'foot per second squared'}, - 'foot per s per s' => {alias=>'foot per second squared'}, - 'ft per s^2' => {alias=>'foot per second squared'}, - 'ft per s per s' => {alias=>'foot per second squared'}, - 'foot / second squared' => {alias=>'foot per second squared'}, - 'foot / square second' => {alias=>'foot per second squared'}, - 'foot / second / second' => {alias=>'foot per second squared'}, - 'ft / second squared' => {alias=>'foot per second squared'}, - 'ft / square second' => {alias=>'foot per second squared'}, - 'ft / second / second' => {alias=>'foot per second squared'}, - 'foot / s^2' => {alias=>'foot per second squared'}, - 'foot / s / s' => {alias=>'foot per second squared'}, - 'ft / s^2' => {alias=>'foot per second squared'}, - 'ft / s / s' => {alias=>'foot per second squared'}, - 'foot/second squared' => {alias=>'foot per second squared'}, - 'foot/square second' => {alias=>'foot per second squared'}, - 'foot/second/second' => {alias=>'foot per second squared'}, - 'ft/second squared' => {alias=>'foot per second squared'}, - 'ft/square second' => {alias=>'foot per second squared'}, - 'ft/second/second' => {alias=>'foot per second squared'}, - 'foot/s^2' => {alias=>'foot per second squared'}, - 'foot/s/s' => {alias=>'foot per second squared'}, - 'ft/s^2' => {alias=>'foot per second squared'}, - 'ft/s/s' => {alias=>'foot per second squared'}, - 'feet per square second' => {alias=>'foot per second squared'}, - 'feet per second per second' => {alias=>'foot per second squared'}, - 'feet per s^2' => {alias=>'foot per second squared'}, - 'feet per s per s' => {alias=>'foot per second squared'}, - 'feet / second squared' => {alias=>'foot per second squared'}, - 'feet / square second' => {alias=>'foot per second squared'}, - 'feet / second / second' => {alias=>'foot per second squared'}, - 'feet / s^2' => {alias=>'foot per second squared'}, - 'feet / s / s' => {alias=>'foot per second squared'}, - 'feet/second squared' => {alias=>'foot per second squared'}, - 'feet/square second' => {alias=>'foot per second squared'}, - 'feet/second/second' => {alias=>'foot per second squared'}, - 'feet/s^2' => {alias=>'foot per second squared'}, - 'feet/s/s' => {alias=>'foot per second squared'}, - ); - $ans=Compute("foot per second squared"); - - -

    - Let v(t) measure the velocity, - in ft/s, of a car moving in a straight line t seconds after starting. - What are the units of v'(t)? -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('foot per hour'=>{}); - Context()->strings->add( - 'foot per hr' => {alias=>'foot per hour'}, - 'foot per h' => {alias=>'foot per hour'}, - 'foot / hour' => {alias=>'foot per hour'}, - 'foot / hr' => {alias=>'foot per hour'}, - 'foot / h' => {alias=>'foot per hour'}, - 'foot/hour' => {alias=>'foot per hour'}, - 'foot/hr' => {alias=>'foot per hour'}, - 'foot/h' => {alias=>'foot per hour'}, - 'feet per hour' => {alias=>'foot per hour'}, - 'feet per hr' => {alias=>'foot per hour'}, - 'feet per h' => {alias=>'foot per hour'}, - 'feet / hour' => {alias=>'foot per hour'}, - 'feet / hr' => {alias=>'foot per hour'}, - 'feet / h' => {alias=>'foot per hour'}, - 'feet/hour' => {alias=>'foot per hour'}, - 'feet/hr' => {alias=>'foot per hour'}, - 'feet/h' => {alias=>'foot per hour'}, - 'ft per hour' => {alias=>'foot per hour'}, - 'ft per hr' => {alias=>'foot per hour'}, - 'ft per h' => {alias=>'foot per hour'}, - 'ft / hour' => {alias=>'foot per hour'}, - 'ft / hr' => {alias=>'foot per hour'}, - 'ft / h' => {alias=>'foot per hour'}, - 'ft/hour' => {alias=>'foot per hour'}, - 'ft/hr' => {alias=>'foot per hour'}, - 'ft/h' => {alias=>'foot per hour'}, - ); - $ans=Compute("foot per hour"); - - -

    - The height H, in feet, - of a river is recorded t hours after midnight, April 1. - What are the units of H'(t)? -

    -

    - -

    -     -     -     - - - - -

    - P is the profit, in thousands of dollars, - of producing and selling c cars. -

    -     - - -

    - What are the units of P'(c)? -

    - -     - - - -

    - Thousands of dollars per car. -

    -     -     - - -

    - What is likely true of P(0)? -

    - -     - - - -

    - It is likely that P(0)\lt 0. - That is, negative profit for not producing any cars. -

    -     -     - -     - - - - -

    - T is the temperature in degrees Fahrenheit, - h hours after midnight on July 4 in Sidney, NE. -

    -     - - -

    - What are the units of T'(h)? -

    - -     - - - -

    - Degrees Fahrenheit per hour. -

    -     -     - - -

    - Is T'(8) likely greater than or less than 0? - Why? -

    - -     - - - -

    - It is likely that T'(8) \gt 0 since at 8 in the morning, - the temperature is likely rising. -

    -     -     - - -

    - Is T(8) likely greater than or less than 0? - Why? -

    - -     - - - -

    - It is very likely that T(8) \gt 0, - as at 8 in the morning on July 4, we would expect the temperature to be well above 0. -

    -     -     - -     - - - -

    - Graphs of functions f and g are given. - Identify which function is the derivative of the other. -

    -     - - - - - parserPopUp.pl - - - $h=non_zero_random(-2,2,1); - $k=non_zero_random(-2,2,1); - $a=non_zero_random(-1,1,0.05); - if($envir{problemSeed}==1){$a=0.25;$h=1;$k=2;}; - $f=Formula("$a*(x-$h)^2+$k"); - $fp=$f->D('x'); - if ($a>0) { - $fleft = $h - sqrt(5/$a - $k/$a); - $fright = $h + sqrt(5/$a - $k/$a); - @fpos = ('center','top'); - $fpleft = $h-5/2/$a; - $fpright = $h+5/2/$a; - @fppos = ('left','top'); - } else { - $fleft = $h - sqrt(-5/$a - $k/$a); - $fright = $h + sqrt(-5/$a - $k/$a); - @fpos = ('center','bottom'); - $fpleft = $h+5/2/$a; - $fpright = $h-5/2/$a; - @fppos = ('left','bottom'); - }; - $fleft=max($fleft,-4); - $fright=min($fright,4); - $fpleft=max($fpleft,-4); - $fpright=min($fpright,4); - $radio=DropDown([ - 'f is the derivative of g.', - 'g is the derivative of f.' - ],0,showInStatic=>0); - - - - f(x) is a line with slope 2 and x intercept . g(x) is a parabola with vertex at (,) - -

    - The graph of f is a line with slope 2a and the x intercept h. - The value of the random variable a decides what the graphs look like. - Note that the graph of f is always a line and the graph of g is always a hyperbola. - The description below is of one of the randomly generated graphs with h=-2, - a=-1 and k=1. If h=-2 and a=-1, we have a line that intersects - the negative x axis at x=-2 and the negative y axis at y=-4. - The graph of g is a hyperbola. The vertex is at (h, k). - With h=-2, a=-1 and k=1 we have the graph of g, - a downwards parabola with its vertex at (-2, 1). One of its arms intersect - the negative x axis at x=-1 and the other at x=-1. - The graph of f intersects the graph of g twice, once in the second quadrant - and once in the fourth quadrant. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-5.1,ymax=5.1,% - xmin=-4.3,xmax=4.3,% - ] - - \addplot [firstcurvestyle,domain=$fpleft:$fpright] {2*$a*(x-$h)}; - \addplot [secondcurvestyle,domain=$fleft:$fright] {$a*(x-$h)^2+$k}; - \addlegendentry{\(f(x)\)}; - \addlegendentry{\(g(x)\)}; - - \end{axis} - \end{tikzpicture} - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - ($r,$s,$t) = num_sort(random_subset(3,-3..3)); - $a=non_zero_random(-1,1,0.05); - if($envir{problemSeed}==1){$a=1;$r=0;$s=1;$t=2;}; - $f=Formula("$a*(x-$r)*(x-$s)*(x-$t)"); - $fp=$f->D('x'); - $xmin=$r-2; - $xmax=$t+2; - @cp=($xmin,$xmax); - @cv; - for $i (@cp) { - push(@cv, $f->eval(x=>$i)); - }; - $ymin=min(-5,map{Round($_)-1}(@cv)); - $ymax=max(5,map{Round($_)+1}(@cv)); - $infl=($r+$s+$t)/3; - $inflval=$fp->eval(x=>$infl); - $radio=DropDown([ - 'f is the derivative of g.', - 'g is the derivative of f.' - ],1,showInStatic=>0); - - - - f is a parabola and g is a cubic graph that has three x intercepts. - -

    - The graph of g is a parabola and the graph of f - is a graph of a cubic function that intersects the x axis three times. Note that the graph of f is always - a cubic graph and the graph of g is always a hyperbola. - The description below describes one of the randomized images for the excercise. The vertex of g is at - (-0.667,-0.367) in the third qudrant just below the negative x - and slightly to the left of the negative y axis. - From there the right arm travels upwards to the right, intersects the positive x - axis and continues travelling upwards while also moving to its right. The graph of - g is the graph of a cubic function that travels upwards from the bottom of the fourth quadrant, - intersecting the negative x axis at x=-3 it travels upwards - for a shortwhile then slopes down and again intersects the x axis at x=-2. - Next it travels downwards to the right. After crossing the negative y axis - at y=-0.9 and then it slopes upwards again. It intersects the positive x - axis for the last time at x=3 and continues upwards. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=$ymin,ymax=$ymax,% - xmin=$xmin,xmax=$xmax,% - ] - - \addplot [firstcurvestyle,domain=$xmin:$xmax,samples=101] {$a*(x-$r)*(x-$s)*(x-$t)}; - \addplot [secondcurvestyle,domain=$xmin:$xmax] {$a*(3*x^2-2*($r+$s+$t)*x+$r*$s+$r*$t+$s*$t)}; - \addlegendentry{\(f(x)\)}; - \addlegendentry{\(g(x)\)}; - - \end{axis} - \end{tikzpicture} - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $radio=DropDown([ - 'f is the derivative of g.', - 'g is the derivative of f.' - ],1,showInStatic=>0); - - - - f is a hyperbola with curves in the first and the third quadrant and g is a hyperbola with curves in the third and the fourth quadrant. - -

    - The graph of f is a hyperbola with curves in the first and the third quadrant. The curves look like infinite bows with the branches extending in - positive and negative x and y directions. The curve in the first quadrant has its vertex at (1, 1). The two branches of the curve extend - towards positive x and y axis. They continue moving in their respective directions while they get closer to the x and the y axis - without ever merging with them. The curve in the third qudrant has its vertex at (-1, -1) and the two branches of this one move towards the negative x - and y axis in the exact manner as the curve in the first quadrant creating a mirror image of the first curve. The graph of g also consists of two - curves, one in the third and one in the fourth quadrant that are mirror images of each other. The curve in the fourth quadrant is shaped like an inverted L. The - focus of the curve is located close to (1, -0.5). From there, one of branches move upwards to the right until it merges with the positive x axis and - continues along with it. The other branch moves downwards while moving slightly to its left and then contines vertically downwards parallel to the negative y - axis. The curve in the third qudrant is a mirror image of the previous curve. It follows the part of f in the third quadrant closely intersecting f - at (-1, -1). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-5.5,ymax=5.5,% - xmin=-5.5,xmax=5.5,% - ] - - \addplot [firstcurvestyle,domain=-5.5:-.182,samples=40] {1/x}; - \addplot [secondcurvestyle,domain=-5.5:-.426,samples=40] {-1/x^2}; - \addplot [firstcurvestyle,domain=.182:5.5,samples=40] {1/x}; - \addplot [secondcurvestyle,domain=.426:5.5,samples=40] {-1/x^2}; - \addlegendentry{\(f(x)\)}; - \addlegendentry{\(g(x)\)}; - - \end{axis} - \end{tikzpicture} - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - parserPopUp.pl - - - $h=random(-2,2,1); - $p=random(2,4,1); - if($envir{problemSeed}==1){$h=1;$p=2;}; - $f=Formula("cos((pi/$p)*(x-$h))"); - $fp=$f->D('x'); - $ymin=-max(int(pi/$p),1)-1; - $ymax=max(int(pi/$p),1)+1; - $radio=DropDown([ - 'f is the derivative of g.', - 'g is the derivative of f.' - ],1,showInStatic=>0); - - - - f is making a curve similar to a sine wave while g looks similar to a cosine wave both intersecting the x axis multiple times as they oscillate. - -

    - The graph of f looks like a sine wave. Starting at (0, -4) it slopes downward in the - negative y direction until (-2, -1) then starts moving upwards towards the origin. It continues - upwards through the origin until it reaches (2, 1), then starts sloping down again and eventually - intersects the x axis at (4, 0). g looks similar to a cosine curve. Starting close to (-4, -0.75) - it slopes upwards intersecting f and then intersecting the x axis at (-2, 0). It keeps moving - up and intersects the y axis close to (0, 0.75) which is where the top of the bell is. Then it starts - moving down to the right, crosses the x axis at (2, 0) then slopes down. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=$ymin,ymax=$ymax,% - xmin=-4,xmax=4,% - ] - - \addplot [firstcurvestyle,domain=-4:4,samples=101] {cos(deg((3.14/$p)*(x-$h)))}; - \addplot [secondcurvestyle,domain=-4:4,samples=40] {-(3.14/$p)*sin(deg((3.14/$p)*(x-$h)))}; - \addlegendentry{\(f(x)\)}; - \addlegendentry{\(g(x)\)}; - - \end{axis} - \end{tikzpicture} - - -

    - -

    -     -     -     -     -     -     -
    -
    - Basic Differentiation Rules -

    - The derivative is a powerful tool but is admittedly awkward given its reliance on limits. - Fortunately, one thing mathematicians are good at is - abstraction. For instance, - instead of continually finding derivatives at a point, - we abstracted and found the derivative function. -

    - -

    - Let's practice abstraction on linear functions, y=mx+b. - What is \yp? - Without limits, - recognize that linear functions are characterized by being functions with a constant rate of change - (the slope). - The derivative, \yp, gives the instantaneous rate of change; - with a linear function, this is constant, m. - Thus \yp=m. -

    - -

    - Let's abstract once more. - Let's find the derivative of the general quadratic function, - f(x) = ax^2+bx+c. - Using the definition of the derivative, we have: - - \fp(x) \amp = \lim_{h\to 0}\frac{a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h} - \amp = \lim_{h\to 0}\frac{ax^2+2ahx+ah^2+bx+bh+c-ax^2-bx-c}{h} - \amp = \lim_{h\to 0} \frac{ah^2+2ahx+bh}{h} - \amp = \lim_{h\to 0} ah+2ax+b - \amp = 2ax+b - . -

    - -

    - So if y = 6x^2+11x-13, - we can immediately compute \yp = 12x+11. -

    - -

    - In this section - (and in some sections to follow) - we will learn some of what mathematicians have already discovered about the derivatives of certain functions and how derivatives interact with arithmetic operations. - We start with a theorem. -

    - - - Derivatives of Common Functions - -

    -

      -
    1. - Constant Rule - - derivativeConstant Rule - -

      - \lzoo{x}{c} = 0, where c is a constant. -

      -
      2. - -
    2. - Power Rule - - derivativePower Rule - Power Ruledifferentiation - -

      - \lzoo{x}{x^n}= nx^{n-1}, - where n is an integer, n \gt 0. -

      -
      3. - -
    3. - -

      - \lzoo{x}{\sin(x)} = \cos(x) -

      -
      4. - -
    4. -

      - \lzoo{x}{\cos(x)} = {-\sin(x)} -

      -
      5. - -
    5. -

      - \lzoo{x}{e^x} = e^x -

      -
      6. - -
    6. -

      - \lzoo{x}{\ln(x)} = \frac{1}{x}, for x \gt 0. -

      -
      7. -
    - - derivativebasic rules - -

    -     -     - - - - - -

    - This theorem starts by stating an intuitive fact: - constant functions have zero rate of change as they are constant. - Therefore their derivative is 0 - (they change at the rate of 0). - The theorem then states some fairly amazing things. - The states that the derivatives of Power Functions - (of the form y=x^n) - are very straightforward: - multiply by the power, then subtract 1 from the power. - We see something incredible about the function y=e^x: - it is its own derivative. - We also see a new connection between the sine and cosine functions. -

    - -

    - One special case of the is when n=1, - , when f(x) = x. - What is \fp(x)? - According to the , - - \fp(x) = \lzoo{x}{x} = \lzoo{x}{x^1} = 1\cdot x^0 = 1 - . -

    - -

    - In words, we are asking At what rate does f change with respect to x? - Since f is x, - we are asking At what rate does x change with respect to x? The answer is: - 1. - They change at the same rate. - We can also interpret the derivative as the slope of the tangent line to the function at a point (c,f(c)). - Since f(x)=x is a linear function with constant slope 1, - we can say that the derivative of f(x)=x is \fp(x)=1. -

    - - - - - - -

    - Let's practice using this theorem. -

    - - - Using common derivative rules to find, and use, derivatives - -

    - Let f(x)=x^3. -

      -
    1. -

      - Find \fp(x). -

      -
      2. -
    2. -

      - Find the equation of the line tangent to the graph of f at x=-1. -

      -
      3. -
    3. -

      - Use the tangent line to approximate (-1.1)^3. -

      -
      4. -
    4. -

      - Sketch f, - \fp and the tangent line from on the same axis. -

      -
      5. -
    -

    -     - -

    -

      -
    1. -

      - The states that if f(x) = x^3, - then \fp(x) = 3x^2. -

      -
      2. -
    2. -

      - To find the equation of the line tangent to the graph of f at x=-1, - we need a point and the slope. - The point is (-1,f(-1)) = (-1, -1). - The slope is \fp(-1)= 3. - Thus the tangent line has equation y = 3(x-(-1))+(-1) = 3x+2. -

      -
      3. -
    3. -

      - We can use the tangent line to approximate - (-1.1)^3 since -1.1 is close to -1. - We have - - (-1.1)^3 \approx 3(-1.1)+2 = -1.3 - . - We can easily find the actual value: - (-1.1)^3 = -1.331. -

      -
      4. -
    4. -

      - See . -

      -
      5. -
    -

    - -
    - A graph of f(x) = x^3, along with its derivative \fp(x) = 3x^2 and its tangent line at x=-1 - - - Graph of function x^3, its derivative and a tangent line drawn at point (-1,-1). - - -

    - The y axis is drawn between -4 and 4 and the x axis - is drawn between -2 and 2. - The graph of function f(x)=x^3 is a curve that appears to conincide the - x axis from x=-0.5 to x=0.5, - then it appears to increase to infinity from x=0.5 in the first quadrant, - and curves downward from to infinity from x=-0.5 in the third quadrant. -

    -

    - The derivative \fp(x) is a parabola opening upwards. - It intersects the function f(x) at the origin and increases to infinity on - either side of the positive y axis. -

    -

    - A tangent line l(x)is drawn on the function at point (-1,-1). - It has a positive slope and and intersects the x axis at approximately - x=-0.6 and the y axis at y=2. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-2.1, - xmax=2.1, - ymin=-5.1, - ymax=5.1, - clip=false - ] - \addplot+[infinite,domain=-1.5:1.5] {x^3} node[below right] {$f(x)$}; - \addplot+[infinite,domain=-1.2:1.2] {3*x^2} node[right] {$f'(x)$}; - \addplot[tangentline,domain=-2:1] {3*(x+1)-1} node[above left] {$\ell(x)$}; - \addplot[soliddot] coordinates {(-1,-1) (-1,3)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - -

    - gives useful information, - but we will need much more. - For instance, using the theorem, - we can easily find the derivative of y=x^3, - but it does not tell how to compute the derivative of y=2x^3, - y=x^3+\sin(x) nor y=x^3\sin(x). - The following theorem helps with the first two of these examples - (the third is answered in the next section). -

    - - - - - Properties of the Derivative - -

    - Let f and g be differentiable on an open interval I and let c be a real number. - Then: -

      -
    1. - Sum/Difference Rule -

      - - \lzoo{x}{f(x) \pm g(x)} = \lzoo{x}{f(x)} \pm \lzoo{x}{g(x)} = \fp(x)\pm \gp(x) - - - derivativeSum/Difference Rule - Sum/Difference Ruleof derivatives - -

      -
      2. - -
    2. - Constant Multiple Rule -

      - - \lzoo{x}{c\cdot f(x)} = c\cdot\lzoo{x}{f(x)} = c\cdot\fp(x) - . - - derivativeConstant Multiple Rule - Constant Multiple Ruleof derivatives - -

      -
      3. -
    -

    -     -     - - - - - - - - - - - -

    - - allows us to find the derivatives of a wide variety of functions. - It can be used in conjunction with the to find the derivatives of any polynomial. - Recall in that we found, - using the limit definition, the derivative of f(x) = 3x^2+5x-7. - We can now find its derivative without expressly using limits: - - \lzoo{x}{3x^2+5x-7} \amp = 3\lzoo{x}{x^2} + 5\lzoo{x}{x} - \lzoo{x}{7} - \amp = 3\cdot 2x+5\cdot 1- 0 - \amp = 6x+5 - . -

    - -

    - We were a bit pedantic here, showing every step. - Normally we would do all the arithmetic and steps in our head and readily find \lzoo{x}{3x^2+5x+7}= 6x+5. -

    - - - Using the tangent line to approximate a function value - -

    - Let f(x) = \sin(x) + 2x+1. - Approximate f(3) using an appropriate tangent line. -

    -     - -

    - This problem is intentionally ambiguous; - we are to approximate using an - appropriate tangent line. - How good of an approximation are we seeking? - What does appropriate mean? -

    - -

    - In the real world, people solving problems deal with these issues all time. - One must make a judgment using whatever seems reasonable. - In this example, the actual answer is f(3) = \sin(3) + 7, - where the real problem spot is \sin(3). - What is \sin(3)? -

    - -

    - Since 3 is close to \pi, - we can assume \sin(3) \approx \sin(\pi) = 0. - Thus one guess is f(3) \approx 7. - Can we do better? - Let's use a tangent line as instructed and examine the results; - it seems best to find the tangent line at x=\pi. -

    - -

    - Using - we find \fp(x) = \cos(x) + 2. - The slope of the tangent line is thus \fp(\pi) = \cos(\pi) + 2 =1. - Also, f(\pi) = 2\pi+1 \approx 7.28. - So the tangent line to the graph of f at x=\pi is y=1(x-\pi)+ 2\pi+1 =x+\pi+1 \approx x+4.14. - Evaluated at x=3, our tangent line gives y=3+4.14 = 7.14. - Using the tangent line, - our final approximation is that f(3) \approx 7.14. -

    - -

    - Using a calculator, - we get an answer accurate to four places after the decimal: - f(3) = 7.1411. - Our initial guess was 7; - our tangent line approximation was more accurate, at 7.14. -

    - -

    - The point is not Here's a cool way to do some math without a calculator. Sure, - that might be handy sometime, - but your phone could probably give you the answer. - Rather, the point is to say that tangent lines are a good way of approximating, - and many scientists, - engineers and mathematicians often face problems too hard to solve directly. - So they approximate. -

    - -

    - The graphs in - shows the graph of the function f(x) along with the tangent line constructed at x=\pi. - The graph in - shows the same tangent line and function. - Once zoomed in, - you can barely distinguish the tangent line from the function. - This indicates that the tangent line is a good a approximation of the function so long as we are near the point of tangency. -

    - - -
    - A graph of f(x) = \sin(x)+2x+1 along with its tangent line approximation at x=\pi - - - Graph of the function for this example and its tangent at x=pi. - - -

    - The y axis is drawn from -1 to 10 and x axis is drawn from -1 to 5. - The graph of f(x)is drawn in the first quadrant, the curve passes the - y axis at y=1, it increases until point (5,10). - It has a slight undulation between x=2 to x=4. -

    -

    - A tangent line is drawn on the curve at approximately point (3,7) or x=\pi. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1.1, - xmax=5.1, - ymin=-1, - ymax=10, - ] - \addplot+[infinite,domain=-.5:5] {sin(deg(x))+2*x+1}; - \addplot[tangentline,domain=1:4.5] {x+4.14}; - \addplot[soliddot] coordinates {(3,7.1411) (3.14,7.28)}; - \draw (axis cs:0.5,1.2) node { $f(x)$}; - \draw (axis cs:0.5,5) node { $l(x)$}; - \end{axis} - \end{tikzpicture} - - - -
    - -
    - A graph of f(x) = \sin(x)+2x+1 along with its tangent line approximation at x=\pi, zoomed in - - - Highly zoomed in view of the previous graph and its tangent line, near x=pi. - - -

    - The y axis is drawn from 6.8 to 7.4 and x axis - is drawn from 2.6 to 3.4. - The graph of f(x) is drawn in the first quadrant, the curve and - its tangent line are aligned except near shifted origin around point (2.6, 6.8) - where they are slightly apart. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=6.7, - ymax=7.4, - xmin=2.5, - xmax=3.5, - xdiscontinuity, - ydiscontinuity - ] - \addplot+[infinite,domain=2.6:3.25] {sin(deg(x))+2*x+1}; - \addplot[tangentline,domain=2.6:3.25] {x+4.14}; - \addplot[soliddot] coordinates {(3,7.1411) (3.14,7.28)}; - %\draw (axis cs:0.5,1.2) node { $f(x)$}; - %\draw (axis cs:0.5,5) node { $l(x)$}; - \end{axis} - \end{tikzpicture} - - - -
    -     -     - -     - - - Higher Order Derivatives - -

    - The derivative of a function f is itself a function, - therefore we can take its derivative. - The following definition gives a name to this concept and introduces its notation. -

    - - - Higher Order Derivatives - -

    - Let y=f(x) be a differentiable function on I. - The following are defined, provided the corresponding limits exist. - - derivativehigher order - derivativenotation - -

      -
    1. -

      - The second derivative of f is: - - \fp'(x) = \lzoo{x}{\fp(x)} = \lzoo{x}{\lz{y}{x}} = \lzn{2}{y}{x}=\yp' - . -

      -
      2. -
    2. -

      - The third derivative of f is: - - \fp''(x) = \lzoo{x}{\fp'(x)} = \lzoo{x}{\lzn{2}{y}{x}} = \lzn{3}{y}{x}=\yp'' - . -

      -
      3. -
    3. -

      - The nth derivative of f is: - - f^{(n)}(x) = \lzoo{x}{f^{(n-1)}(x)} = \lzoo{x}{\lzn{n-1}{y}{x}} = \lzn{n}{y}{x}=y^{(n)} - . -

      -
      4. -
    -

    -     -     - - - - - -

    - In general, when finding the fourth derivative and on, - we resort to the f\,^{(4)}(x) notation, not \fp'''(x); - after a while, too many ticks is confusing. -

    - -

    - Let's practice using this new concept. -

    - - - Finding higher order derivatives - -

    - Find the first four derivatives of the following functions: -

      -
    1. f(x) = 4x^2
      2. -
    2. f(x) = \sin(x)
      3. -
    3. f(x) = 5e^x
      4. -
    -

    -     - -

    -

      -
    1. -

      - Using the Power and Constant Multiple Rules, - we have: \fp(x) = 8x. - Continuing on, we have - - \fp'(x)\amp=\lzoo{x}{8x}=8\amp\fp''(x)\amp=0\amp f^{(4)}(x)\amp=0 - . - Notice how all successive derivatives will also be 0. -

      -
      2. -
    2. -

      - We employ repeatedly. - - \fp(x)\amp=\cos(x)\amp \fp''(x)\amp= -\cos(x) - \fp'(x)\amp= -\sin(x)\amp f^{(4)}(x)\amp = \sin(x) - - Note how we have come right back to f(x) again. (Can you quickly figure what f^{(23)}(x) is?) -

      -
      3. -
    3. -

      - Employing - and the Constant Multiple Rule, we can see that - - \fp(x) = \fp'(x) = \fp''(x) = f^{(4)}(x) = 5e^x - . -

      -
      4. -
    -

    -     - -     -     - - - Interpreting Higher Order Derivatives -

    - What do higher order derivatives mean? - What is the practical interpretation? - - derivativehigher orderinterpretation -

    - -

    - Our first answer is a bit wordy, - but is technically correct and beneficial to understand. - That is, -

    - -
    -

    - The second derivative of a function f is the rate of change of the rate of change of f. -

    -
    - -

    - One way to grasp this concept is to let f describe a position function. - Then, as stated in , - \fp describes the rate of position change: velocity. - We now consider \fp', which describes the rate of velocity change. - Sports car enthusiasts talk of how fast a car can go from 0 to - - 60 - ; they are bragging about the - acceleration of the car. -

    - -

    - We started this chapter with amusement park riders free-falling with position function f(t) = -16t^2+150. - It is easy to compute \fp(t)=-32t ft/s and \fp'(t) = -32 (ft/s)/s. - We may recognize this latter constant; - it is the acceleration due to gravity. - In keeping with the unit notation introduced in the previous section, - we say the units are feet per second per second. - This is usually shortened to feet per second squared, - written as ft/s^2. -

    - -

    - It can be difficult to consider the meaning of the third, - and higher order, derivatives. - The third derivative is the rate of change of the rate of change of the rate of change of f. - That is essentially meaningless to the uninitiated. - In the context of our position/velocity/acceleration example, - the third derivative is the rate of change of acceleration, - commonly referred to as jerk. -

    - -

    - Make no mistake: - higher order derivatives have great importance even if their practical interpretations are hard - (or impossible) - to understand. - The mathematical topic of series - makes extensive use of higher order derivatives. -

    -     - - - - Terms and Concepts - - - - -

    - What is the name of the rule which states that \lzoo{x}{x^n} = nx^{n-1}, - where n \gt 0 is an integer? -

    -

    - - -

    -     - - - - - power|power rule|the power rule - - - - - -     - - - - - $ans=Formula("1/x"); - - -

    - What is \lzoo{x}{\ln(x)}? -

    -

    - -

    -     -     -     - - - - - $ans=Compute("e^x"); - $ev=$ans->cmp(checker=>sub{ - my($correct,$student,$self )=@_; - return 0 unless ($student->D('x') == $student); - return 1; - }); - - -

    - Give an example of a function f(x) where \fp(x) = f(x). -

    - - Enter the formula for the function without the f(x)=. - -

    - -

    -     - -

    - One answer is f(x) = 10e^x. - Any constant multiple of e^x will do. -

    -     -     -     - - - - - $ans=Formula("10"); - $ev=$ans->cmp(checker=>sub{my($correct,$student,$self)=@_; - return 0 unless ($student->D('x') == Formula("0")); - return 1; - }); - - -

    - Give an example of a function f(x) where \fp(x) = 0. -

    - - Enter the formula for the function without the f(x)=. - -

    - -

    -     - -

    - One answer is f(x) = 10. - Any constant function will do. -

    -     -     -     - - - - -

    - The derivative rules introduced in - explain how to compute the derivative of which of the following functions? -

    - -     - - - -

    - f(x)=3x^2-x+17 -

    -     -     - - -

    - f(x)=5\ln(x) -

    -     -     - - -

    - f(x)=\frac{3}{x^2} -

    -     -     - - -

    - f(x) = \sin(x)\cos(x) -

    -     -     - - -

    - f(x)=e^{x^2} -

    -     -     - - -

    - f(x)=\sqrt{x} -

    -     -     -     - -     - - - - -

    - Explain in your own words how to find the third derivative of a function f(x). -

    - -     - - - -     - - - - - $ans=Compute("17x-205"); - $ev=$ans->cmp(checker=>sub{my($correct,$student,$self )=@_; - return 0 if ($student->D('x') == Formula("0")); - return 0 unless ($student->D('x')->D('x') == Formula("0")); - return 1; - }); - - -

    - Give an example of a function where - \fp(x) \neq 0 and \fpp(x) = 0. -

    - - Enter the formula for the function without the f(x)=. - -

    - -

    -     - -

    - One answer is f(x) = 17x-205. - Any linear function with nonzero slope will do. -

    -     -     -     - - - - -

    - Explain in your own words what the second derivative means. -

    - -     - - - -     - - - - - Context()->strings->add('a velocity function'=>{}); - Context()->strings->add( - 'velocity'=>{alias=>'a velocity function'}, - 'velocity function'=>{alias=>'a velocity function'}, - ); - $velocity = Compute("a velocity function"); - Context()->strings->add('an acceleration function'=>{}); - Context()->strings->add( - 'acceleration'=>{alias=>'an acceleration function'}, - 'acceleration function'=>{alias=>'an acceleration function'}, - ); - $acceleration = Compute("an acceleration function"); - - -

    - If f(x) describes a position function, - then \fp(x) describes what kind of function? - What kind of function is \fpp(x)? -

    - - What kind of function is \fp(x)? - -

    - -

    - - What kind of function is \fpp(x)? - -

    - -

    -     - -

    - \fp(x) is a velocity function, - and \fpp(x) is acceleration. -

    -     -     -     - - - - - Context()->strings->add('pound per foot squared'=>{}); - Context()->strings->add( - 'pound per square foot' => {alias=>'pound per foot squared'}, - 'pound per foot per foot' => {alias=>'pound per foot squared'}, - 'lb per foot squared' => {alias=>'pound per foot squared'}, - 'lb per square foot' => {alias=>'pound per foot squared'}, - 'lb per foot per foot' => {alias=>'pound per foot squared'}, - 'pound per ft^2' => {alias=>'pound per foot squared'}, - 'pound per ft per ft' => {alias=>'pound per foot squared'}, - 'lb per ft^2' => {alias=>'pound per foot squared'}, - 'lb per ft per ft' => {alias=>'pound per foot squared'}, - 'pound / foot squared' => {alias=>'pound per foot squared'}, - 'pound / square foot' => {alias=>'pound per foot squared'}, - 'pound / foot / foot' => {alias=>'pound per foot squared'}, - 'lb / foot squared' => {alias=>'pound per foot squared'}, - 'lb / square foot' => {alias=>'pound per foot squared'}, - 'lb / foot / foot' => {alias=>'pound per foot squared'}, - 'pound / ft^2' => {alias=>'pound per foot squared'}, - 'pound / ft / ft' => {alias=>'pound per foot squared'}, - 'lb / ft^2' => {alias=>'pound per foot squared'}, - 'lb / ft / ft' => {alias=>'pound per foot squared'}, - 'pound/foot squared' => {alias=>'pound per foot squared'}, - 'pound/square foot' => {alias=>'pound per foot squared'}, - 'pound/foot/foot' => {alias=>'pound per foot squared'}, - 'lb/foot squared' => {alias=>'pound per foot squared'}, - 'lb/square foot' => {alias=>'pound per foot squared'}, - 'lb/foot/foot' => {alias=>'pound per foot squared'}, - 'pound/ft^2' => {alias=>'pound per foot squared'}, - 'pound/ft/ft' => {alias=>'pound per foot squared'}, - 'lb/ft^2' => {alias=>'pound per foot squared'}, - 'lb/ft/ft' => {alias=>'pound per foot squared'}, - 'pounds per square foot' => {alias=>'pound per foot squared'}, - 'pounds per foot per foot' => {alias=>'pound per foot squared'}, - 'pounds per ft^2' => {alias=>'pound per foot squared'}, - 'pounds per ft per ft' => {alias=>'pound per foot squared'}, - 'pounds / foot squared' => {alias=>'pound per foot squared'}, - 'pounds / square foot' => {alias=>'pound per foot squared'}, - 'pounds / foot / foot' => {alias=>'pound per foot squared'}, - 'pounds / ft^2' => {alias=>'pound per foot squared'}, - 'pounds / ft / ft' => {alias=>'pound per foot squared'}, - 'pounds/foot squared' => {alias=>'pound per foot squared'}, - 'pounds/square foot' => {alias=>'pound per foot squared'}, - 'pounds/foot/foot' => {alias=>'pound per foot squared'}, - 'pounds/ft^2' => {alias=>'pound per foot squared'}, - 'pounds/ft/ft' => {alias=>'pound per foot squared'}, - 'lbs per foot squared' => {alias=>'pound per foot squared'}, - 'lbs per square foot' => {alias=>'pound per foot squared'}, - 'lbs per foot per foot' => {alias=>'pound per foot squared'}, - 'lbs per ft^2' => {alias=>'pound per foot squared'}, - 'lbs per ft per ft' => {alias=>'pound per foot squared'}, - 'lbs / foot squared' => {alias=>'pound per foot squared'}, - 'lbs / square foot' => {alias=>'pound per foot squared'}, - 'lbs / foot / foot' => {alias=>'pound per foot squared'}, - 'lbs / ft^2' => {alias=>'pound per foot squared'}, - 'lbs / ft / ft' => {alias=>'pound per foot squared'}, - 'lbs/foot squared' => {alias=>'pound per foot squared'}, - 'lbs/square foot' => {alias=>'pound per foot squared'}, - 'lbs/foot/foot' => {alias=>'pound per foot squared'}, - 'lbs/ft^2' => {alias=>'pound per foot squared'}, - 'lbs/ft/ft' => {alias=>'pound per foot squared'}, - ); - $ans=Compute("pound per foot squared"); - - -

    - Let f(x) be a function measured in pounds (lb), - where x is measured in feet - (ft). - What are the units of \fpp(x)? -

    -

    - -

    -     -     -     -     - - Problems - - - -

    - Compute the derivative of the given function. -

    -     - - - - - $a=list_random(-9..-2,2..9); - $b=list_random(-9..-2,2..9); - $c=list_random(-9..-2,2..9); - if($envir{problemSeed}==1){$a=7;$b=-5;$c=7;}; - $f=Formula("$a x^2+$b x+$c")->reduce; - $fp=Formula("2*$a x + $b")->reduce; - - -

    - f(x) = -

    -

    - -

    -     -     -     - - - - - $a=list_random(-29..-2,2..29); - $b=list_random(-29..-2,2..29); - $c=list_random(-29..-2,2..29); - $d=list_random(-29..-2,2..29); - if($envir{problemSeed}==1){$a=14;$b=7;$c=11;$d=-9;}; - $f=Formula("$a x^3+$b x^2+$c x+ $d")->reduce; - $fp=Formula("3*$a x^2+2*$b x+$c")->reduce; - - -

    - g(x) = -

    -

    - -

    -     -     -     - - - - - $a=list_random(-9..-2,2..9); - $b=list_random(-8,-7,-5,-4,-2,2,4,5,6,8); - $c=list_random(-9..-2,2..9); - $d=list_random(-9..-2,2..9); - if($envir{problemSeed}==1){$a=9;$b=-8;$c=3;$d=-8;}; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - $f=Formula("$a t^5 + 1/$b t^3 + $c t + $d")->reduce; - $fp=Formula("5*$a t^4 + 3/$b t^2 + $c")->reduce; - - -

    - m(t) = -

    -

    - -

    -     -     -     - - - - - $a=list_random(-19..-2,2..19); - $b=list_random(-19..-2,2..19); - if($envir{problemSeed}==1){$a=9;$b=10;}; - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $f=Formula("$a sin(theta) + $b cos(theta)")->reduce; - $fp=Formula("$a cos(theta) - $b sin(theta)")->reduce; - - -

    - f(\theta) = -

    - - To enter \theta, type theta. - -

    - -

    -     -     -     - - - - - $a=list_random(-9..-2,2..9); - if($envir{problemSeed}==1){$a=6;}; - Context()->variables->are(r=>'Real'); - $f=Formula("$a e^r")->reduce; - $fp=Formula("$a e^r")->reduce; - - -

    - f(r) = -

    -

    - -

    -     -     -     - - - - - $a=non_zero_random(-19,19,1); - $n=random(2,5,1); - $b=non_zero_random(-9,9,1); - $c=non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=10;$n=4;$b=-1,$c=7;}; - Context()->variables->are(t=>'Real'); - $f=Formula("$a t^$n+$b cos(t) +$c sin(t)")->reduce; - $fp=Formula("$a*$n t^($n-1) -$b sin(t) + $c cos(t)")->reduce; - - -

    - g(t) = -

    -

    - -

    -     -     -     - - - - - $a=non_zero_random(-9,9,1); - $b=non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=2;$b=-1;}; - $f=Formula("$a ln(x) + $b x")->reduce; - $fp=Formula("$a/x + $b")->reduce; - - -

    - f(x) = -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->variables->are(s=>'Real'); - $f=Formula("1/4s^4+1/3s^3+1/2s^2+s+1")->reduce; - $fp=Formula("s^3+s^2+s+1")->reduce; - - -

    - p(s) = -

    -

    - -

    -     -     -     - - - - - $a=non_zero_random(-1,1,2); - $b=non_zero_random(-1,1,2); - $c=non_zero_random(-1,1,2); - if($envir{problemSeed}==1){$a=1;$b=-1;$c=-1;}; - Context()->variables->are(t=>'Real'); - $f=Formula("$a e^t +$b sin(t) + $c cos(t)")->reduce; - $fp=Formula("$a e^t +$b cos(t) - $c sin(t)")->reduce; - - -

    - h(t) = -

    -

    - -

    -     -     -     - - - - - $a=non_zero_random(2,9,1); - $b=non_zero_random(2,9,2); - if($envir{problemSeed}==1){$a=5;$b=3;}; - $f=Formula("ln($a x^$b)")->reduce; - $fp=Formula("$b/x")->reduce; - - -

    - f(x) = -

    -

    - -

    -     -     -     - - - - - $a=random(2,20,1); - $b=random(2,9,1); - $c=list_random('sin','cos'); - $d=list_random(2,3,4); - if($envir{problemSeed}==1){$a=17;$b=2;$c='sin';$d=2;}; - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstantFunctions=>0); - Context()->flags->set(reduceConstants=>0); - $f=Formula("ln($a)+e^$b+$c(pi/$d)"); - $fp=Formula("0"); - - -

    - f(t) = -

    -

    - -

    -     -     -     - - - - - $a=random(1,9,1); - $b=random(1,9,1); - if($envir{problemSeed}==1){$a=1;$b=3;}; - Context()->variables->are(t=>'Real'); - $f=Formula("($a+$b t)^2")->reduce; - $fp=Formula("2*($b)^2 t+2*$a*$b")->reduce; - - -

    - g(t) = -

    -

    - -

    -     -     -     - - - - - $a=random(1,4,1); - $b=non_zero_random(-4,4,1); - if($envir{problemSeed}==1){$a=2;$b=-5;}; - $f=Formula("($a x+$b)^3")->reduce; - $fp=Formula("3*($a)^3 x^2 + 6*($a)^2*$b x+3*$a*($b)^2")->reduce; - - -

    - g(x) = -

    -

    - -

    -     -     -     - - - - - $a=random(1,4,1); - $b=non_zero_random(-4,4,1); - if($envir{problemSeed}==1){$a=1;$b=-1;}; - $f=Formula("($a+$b x)^3")->reduce; - $fp=Formula("3*($b)^3 x^2 + 6*($b)^2*$a x+3*$b*($a)^2")->reduce; - - -

    - f(x) = -

    -

    - -

    -     -     -     - - - - - $a=random(1,9,1); - $b=non_zero_random(1,9,1); - if($envir{problemSeed}==1){$a=2;$b=-3;}; - $f=Formula("($a+$b x)^2")->reduce; - $fp=Formula("2*($b)^2 x+2*$a*$b")->reduce; - - -

    - f(x) = -

    -

    - -

    -     -     -     -     - - - - -

    - A property of logarithms is that \log_a(x) = \frac{\log_b(x)}{\log_b(a)}, - for all bases a,b \gt 0,\neq 1. -

    -     - - -

    - Rewrite this identity when b=e, - , using \log_e(x) =\ln(x), with a=10. -

    -     -     - - -

    - Use part (a) to find the derivative of y=\log_{10}(x). -

    -     -     - - -

    - Find the derivative of y=\log_a(x) for any a\gt0,\neq1. -

    - - -     -     - -     - - - -

    - Compute the first four derivatives of the given function. -

    -     - - - - - $n=random(5,10,1); - if($envir{problemSeed}==1){$n=6;}; - $f=Formula("x^$n")->reduce; - $d[0] = $f->D('x')->reduce; - for my $i (1..3) { - $d[$i] = $d[$i-1]->D('x')->reduce; - } - - -

    - f(x) = -

    - - Enter the first derivative. - -

    - -

    - - Enter the second derivative. - -

    - -

    - - Enter the third derivative. - -

    - -

    - - Enter the fourth derivative. - -

    - -

    -     -     -     - - - - - $a=non_zero_random(-9,9,1); - $b=list_random('sin','cos'); - if($envir{problemSeed}==1){$a=2;}; - $f=Formula("$a $b(x)")->reduce; - $d[0] = $f->D('x')->reduce; - for my $i (1..3) { - $d[$i] = $d[$i-1]->D('x')->reduce; - } - - -

    - g(x) = -

    - - Enter the first derivative. - -

    - -

    - - Enter the second derivative. - -

    - -

    - - Enter the third derivative. - -

    - -

    - - Enter the fourth derivative. - -

    - -

    -     -     -     - - - - - $a=non_zero_random(-4,4,1); - $b=random(-4,4,1); - $c=random(-1,1,2); - if($envir{problemSeed}==1){$a=1;$b=0;$c=-1;}; - Context()->variables->are(t=>'Real'); - $f=Formula("$a t^2 + $b t + $c e^t")->reduce; - $d[0] = $f->D('t')->reduce; - for my $i (1..3) { - $d[$i] = $d[$i-1]->D('t')->reduce; - } - - -

    - h(t) = -

    - - Enter the first derivative. - -

    - -

    - - Enter the second derivative. - -

    - -

    - - Enter the third derivative. - -

    - -

    - - Enter the fourth derivative. - -

    - -

    -     -     -     - - - - - ($a,$b) = random_subset(2,2..9); - $c=random(-1,1,2); - $d=random(-1,1,2); - if($envir{problemSeed}==1){$a=4;$b=3;$c=1;$d=-1;}; - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $f=Formula("$c theta^$a + $d theta^$b")->reduce; - $d[0] = $f->D('theta')->reduce; - for my $i (1..3) { - $d[$i] = $d[$i-1]->D('theta')->reduce; - } - - -

    - p(\theta) = -

    - - Enter the first derivative. - To enter \theta, type theta. - -

    - -

    - - Enter the second derivative. - To enter \theta, type theta. - -

    - -

    - - Enter the third derivative. - To enter \theta, type theta. - -

    - -

    - - Enter the fourth derivative. - To enter \theta, type theta. - -

    - -

    -     -     -     - - - - - $a=random(-1,1,2); - $b=random(-1,1,2); - if($envir{problemSeed}==1){$a=1;$b=-1;}; - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $f=Formula("$a sin(theta)+$b cos(theta)")->reduce; - $d[0] = $f->D('theta')->reduce; - for my $i (1..3) { - $d[$i] = $d[$i-1]->D('theta')->reduce; - } - - -

    - f(\theta) = -

    - - Enter the first derivative. - To enter \theta, type theta. - -

    - -

    - - Enter the second derivative. - To enter \theta, type theta. - -

    - -

    - - Enter the third derivative. - To enter \theta, type theta. - -

    - -

    - - Enter the fourth derivative. - To enter \theta, type theta. - -

    - -

    -     -     -     - - - - - $a=random(1,2000,1); - if($envir{problemSeed}==1){$a=1100}; - $f=Formula("$a"); - $d[0] = $f->D('x')->reduce; - for my $i (1..3) { - $d[$i] = $d[$i-1]->D('x')->reduce; - } - - -

    - f(x) = -

    - - Enter the first derivative. - -

    - -

    - - Enter the second derivative. - -

    - -

    - - Enter the third derivative. - -

    - -

    - - Enter the fourth derivative. - -

    - -

    -     -     -     -     - - - -

    - Find the equations of the tangent and normal lines to the graph of the function at the given point. -

    -     - - - - - $a=non_zero_random(-3,3,1); - do {$b=non_zero_random(-9,9,1)} until ($b != -3*$a**2); - if($envir{problemSeed}==1){$a=1;$b=-1}; - Context("Fraction"); - $f=Formula("x^3+$b x")->reduce; - $k=$f->eval(x=>$a); - $fp=$f->D('x'); - $m=$fp->eval(x=>$a); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=$m(x-$a)+$k"); - $n=Formula("y=-1/$m(x-$a)+$k"); - - -

    - f(x)= at x= -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - $a=random(-3,3,1); - $b=non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=0;$b=3;}; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $f=Formula("e^t+$b")->reduce; - $k=$f->substitute(t=>$a); - $fp=$f->D('t'); - $m=$fp->substitute(t=>$a); - Context()->variables->are(y=>'Real',t=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=$m(t-$a)+$k"); - $n=Formula("y=-1/$m(t-$a)+$k"); - - -

    - f(t)= at t= -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0); - Context()->variables->are(y=>'Real',x=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=x-1"); - $n=Formula("y=-(x-1)"); - - -

    - g(x)=\ln(x) at x=1 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - $a=list_random(2,3,4,6); - $b=non_zero_random(-9,9,1); - $c=list_random('sin','cos'); - if($envir{problemSeed}==1){$a=2;$b=4,$c='sin';}; - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $f=Formula("$b $c(x)")->reduce; - %trig = ( - 'sin'=>{ - 2=>Formula("1"), - 3=>Formula("sqrt(3)/2"), - 4=>Formula("sqrt(2)/2"), - 6=>Formula("1/2"), - }, - 'cos'=>{ - 2=>Formula("0"), - 3=>Formula("1/2"), - 4=>Formula("sqrt(2)/2"), - 6=>Formula("sqrt(3)/2"), - }, - ); - %trigrecip = ( - 'sin'=>{ - 2=>Formula("1"), - 3=>Formula("2*sqrt(3)/3"), - 4=>Formula("sqrt(2)"), - 6=>Formula("2"), - }, - 'cos'=>{ - 2=>Formula("DNE"), - 3=>Formula("2"), - 4=>Formula("2*sqrt(2)"), - 6=>Formula("2*sqrt(3)/3"), - }, - ); - %trigd = ( - 'sin'=>'cos', - 'cos'=>'sin', %intentional - ); - $k=Formula("$b")*$trig{$c}{$a}; - $fp=$f->D('x'); - $m=Formula("$b")*$trig{$trigd{$c}}{$a}; - $m = Formula("-$b")->reduce*$trig{$trigd{$c}}{$a} if ($c eq 'cos'); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=$m(x-pi/$a)+$k"); - $t=Formula("y=$k") if ($m == 0); - if (($c eq 'sin') and ($a==2)) - {$n= Formula("x=pi/$a");} - else { - $rm = Formula("1/$m"); - $n= Formula("y=-$rm(x-pi/$a)+$k"); - }; - - -

    - f(x)= at x=\pi/ -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - $a=list_random(2,3,4,6); - $b=non_zero_random(-9,9,1); - $c=list_random('sin','cos'); - if($envir{problemSeed}==1){$a=4;$b=-2,$c='cos';}; - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $f=Formula("$b $c(x)")->reduce; - %trig = ( - 'sin'=>{ - 2=>Formula("1"), - 3=>Formula("sqrt(3)/2"), - 4=>Formula("sqrt(2)/2"), - 6=>Formula("1/2"), - }, - 'cos'=>{ - 2=>Formula("0"), - 3=>Formula("1/2"), - 4=>Formula("sqrt(2)/2"), - 6=>Formula("sqrt(3)/2"), - }, - ); - %trigrecip = ( - 'sin'=>{ - 2=>Formula("1"), - 3=>Formula("2 sqrt(3)/3"), - 4=>Formula("sqrt(2)"), - 6=>Formula("2"), - }, - 'cos'=>{ - 2=>Formula("DNE"), - 3=>Formula("2"), - 4=>Formula("2 sqrt(2)"), - 6=>Formula("2 sqrt(3)/3"), - }, - ); - %trigd = ( - 'sin'=>'cos', - 'cos'=>'sin', %intentional - ); - $k=Formula("$b")*$trig{$c}{$a}; - $fp=$f->D('x'); - $m=Formula("$b")*$trig{$trigd{$c}}{$a}; - $m = Formula("-$b")->reduce*$trig{$trigd{$c}}{$a} if ($c eq 'cos'); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=$m(x-pi/$a)+$k"); - $t=Formula("y=$k") if ($m == 0); - if (($c eq 'sin') and ($a==2)) - {$n= Formula("x=pi/$a");} - else { - $rm = Formula("1/$b")*$trigrecip{$trigd{$c}}{$a}; - $rm = Formula("1/-$b")->reduce*$trigrecip{$trigd{$c}}{$a} if ($c eq 'cos'); - $n= Formula("y=-$rm(x-pi/$a)+$k"); - $t=Formula("y=$k") if ($m == 0); - }; - - -

    - f(x)= at x=\pi/ -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - $a=random(-9,9,1); - $b=non_zero_random(-9,9,1); - $c=non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=5;$b=2;$c=3}; - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $f=Formula("$b x + $c")->reduce; - $k=$f->eval(x=>$a); - $t=Formula("y=$f"); - $n=Formula("y=-1/$b(x-$a)+$k"); - - -

    - f(x)= at x= -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     -     -     -     -
    -
    - The Product and Quotient Rules -

    - showed that, - in some ways, derivatives behave nicely. - The and established that the derivative of f(x) = 5x^2+\sin(x) was not complicated. - We neglected computing the derivative of things like - g(x) = 5x^2\sin(x) and h(x) = \frac{5x^2}{\sin(x) } on purpose; - their derivatives are not as straightforward. - (If you had to guess what their respective derivatives are, - you would probably guess wrong.) - For these, we need the Product and Quotient Rules, respectively, - which are defined in this section. - We begin with the Product Rule. -

    - - - - - - Product Rule - -

    - Let f and g be differentiable functions on an open interval I. - Power Ruledifferentiation - derivativeProduct Rule - Then fg is a differentiable function on I, and - - \lzoo{x}{f(x)g(x)} = \fp(x)g(x) + f(x)g'(x) - . -

    -     -     - - -

    - \lzoo{x}{f(x)g(x)}\neq \fp(x)g'(x)! - While this would be simpler than the , it is wrong. -

    -     - -

    - We practice using this new rule in an example, - followed by an example that demonstrates why this theorem is true. -

    - - - Using the Product Rule - -

    - Use the Product Rule to compute the derivative of y=5x^2\sin(x). - Evaluate the derivative at x=\pi/2. -

    -     - -

    - To make our use of the explicit, - let's set f(x) = 5x^2 and g(x) = \sin(x). - We easily compute/recall that - \fp(x) = 10x and g'(x) = \cos(x). - Employing the rule, we have - - \lzoo{x}{5x^2\sin(x)} \amp= \lzoo{x}{5x^2}\sin(x) + 5x^2\lzoo{x}{\sin(x)} - \amp= 10x\sin(x) + 5x^2\cos(x) - . -

    - -

    - At x=\pi/2, we have - - \yp(\pi/2) = 10\cdot\frac{\pi}2 \sin\left(\frac{\pi}{2}\right) + 5\left(\frac{\pi}{2}\right)^2\cos\left(\frac{\pi}2\right) = 5\pi - . -

    - -

    - We graph y and its tangent line at x=\pi/2, - which has a slope of 5\pi, - in . - While this does not prove - that the Product Rule is the correct way to handle derivatives of products, - it helps validate its truth. -

    - -
    - A graph of y = 5x^2\sin(x) and its tangent line at x=\pi/2 - - - The graph of the function used in this example, along with its tangent line at pi/2 - -

    - The graph of f(x)=5x^2\sin(x) is shown, beginning at (0,0), - and ending at (\pi,0). - From (0,0) the graph moves upward with a slope that initially increases, - and then becomes steady around the point on the graph where x=\pi/2. - The slope then decreases until approximately x=3\pi/4, where the graph reaches its peak. - It then descends steeply toward the point (\pi,0). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-0.1, - xmax=3.5, - ymin=-1.1, - ymax=21, - xtick={1.57,3.14}, - xticklabels={$\frac{\pi}2$,$\pi$}, - ] - \addplot+[infinite,domain=0:3.14,samples=101] {5*(x^2)*sin(x*180/3.14)}; - \addplot [tangentline,domain=0.75:2] {15.7*(x-1.57)+12.34}; - \addplot[soliddot] coordinates{(1.57,12.34)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - - - -

    - We now investigate why the is true. -

    - - - Proof of Product Rule - - -

    - We can use the definition of the derivative to prove . -

    - -

    - By the limit definition, we have - - \lzoo{x}{f(x)g(x)} =\lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} - . -

    - -

    - We now do something a bit unexpected; - add 0 to the numerator - (so that nothing is changed) - in the form of {}-f(x)g(x+h)+f(x)g(x+h), - then do some regrouping as shown. -

    - - - -

    - - \amp\lzoo{x}{f(x)g(x)} = \lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} - \amp \text{ (now add 0 to the numerator) } - \amp = \lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h} - \amp \text{ (regroup) } - \amp = \lim_{h\to0} \frac{\left[f(x+h)g(x+h)-f(x)g(x+h)\right]+\left[f(x)g(x+h)-f(x)g(x)\right]}{h} - \amp \text{ (split fraction)} - \amp = \lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x+h)}{h}+\lim_{h\to0}\frac{f(x)g(x+h)-f(x)g(x)}{h} - \amp \text{ (factor) } - \amp = \lim_{h\to0} \left(\frac{f(x+h)-f(x)}{h}g(x+h)\right)+\lim_{h\to0}\left(f(x)\frac{g(x+h)-g(x)}{h}\right) - \amp \text{ (apply limit properties) } - \amp = \lim_{h\to0} \frac{f(x+h)-f(x)}{h}\cdot \lim_{h\to0}g(x+h)+f(x)\cdot\lim_{h\to0}\frac{g(x+h)-g(x)}{h} - \amp \text{ (apply limits) } - \amp =\fp(x)g(x) + f(x)g'(x) - \amp \text{ (by definition of the derivative)} - . - We have proven the product rule as desired. (In the last step, - we also relied on the fact that since g is differentiable, - it is also continuous, which guarantees that \lim_{h\to0}g(x+h)=g(x).) -

    -     - -

    - It is often true that we can recognize that a theorem is true through its proof yet somehow doubt its applicability to real problems. - In the following example, - we compute the derivative of a product of functions in two ways to verify that the is indeed right. -

    - - - Exploring alternate derivative methods - -

    - Let y = (x^2+3x+1)(2x^2-3x+1). - Find \yp two ways: first, - by expanding the given product and then taking the derivative, - and second, - by applying the . - Verify that both methods give the same answer. -

    -     - -

    - We first expand the expression for y; - a little algebra shows that y = 2x^4+3x^3-6x^2+1. - It is easy to compute \yp: - - \yp = 8x^3+9x^2-12x - . -

    - -

    - Instead, let's apply the to the original factored form: - - \yp \amp = \lzoo{x}{x^2+3x+1}(2x^2-3x+1)+(x^2+3x+1)\lzoo{x}{2x^2-3x+1} - \amp = (2x+3)(2x^2-3x+1)+(x^2+3x+1)(4x-3) - \amp = \left(4x^3-7x+3\right)+\left(4x^3+9x^2-5x-3\right) - \amp = 8x^3+9x^2-12x - . -

    - -

    - The uninformed usually assume that - the derivative of the product is the product of the derivatives. - Thus we are tempted to say that \yp = (2x+3)(4x-3) = 8x^2+6x-9. - Obviously this is not correct. -

    -     - -     - - - Using the Product Rule with a product of three functions - -

    - Let y = x^3\ln(x) \cos(x). - Find \yp. -

    -     - -

    - We have a product of three functions while the only specifies how to handle a product of two functions. - Our method of handling this problem is to simply group the latter two functions together, - and consider y = x^3\cdot\left[\ln(x) \cos(x) \right]. - Following the , we have - - \yp \amp = \lzoo{x}{x^3}\ln(x) \cos(x) + (x^3)\lzoo{x}{\ln(x) \cos(x)} - To evaluate \lzoo{x}{\ln(x) \cos(x)}, we apply the Product Rule again: - \yp \amp = 3x^2\left[\ln(x) \cos(x) \right] + (x^3)\left[\frac1x\cos(x) + \ln(x) (-\sin(x) )\right] - \amp = 3x^2\ln(x) \cos(x) + x^3\frac{1}{x}\cos(x) + x^3\ln(x) (-\sin(x) ) - . -

    - -

    - Recognize the pattern in our answer above: - when applying the to a product of three functions, - there are three terms added together in the final derivative. - Each term contains only one derivative of one of the original functions, - and each function's derivative shows up in only one term. - It is straightforward to extend this pattern to finding the derivative of a product of four or more functions. -

    - -

    - Ultimately though, we would simplify our final computation to: - - \yp=3x^2\ln(x)\cos(x) + x^2\cos(x) + -x^3\ln(x)\sin(x) - - If you check this answer with a CAS, - it may factor and give the answer: - - \yp=-x^2\left[x\ln(x)\sin(x) + \cos(x) + 3\ln(x)\cos(x)\right] - -

    -     - -     - - - - -

    - We consider one more example before discussing another derivative rule. -

    - - - Using the Product Rule - -

    - Find the derivatives of the following functions. -

      -
    1. f(x) = x\ln(x)
      2. -
    2. g(x) = x\ln(x) - x
      3. -
    -

    -     - -

    - Recalling that the derivative of \ln(x) is 1/x, - we use the to find our answers. -

    - -

    -

      -
    1. -

      - Applying the : - - \lzoo{x}{x\ln(x)}\amp= 1\cdot \ln(x) + x\cdot 1/x - \amp = \ln(x) + 1 - . -

      -
      2. -
    2. -

      - Using the result from above, we compute - - \lzoo{x}{x\ln(x) -x}\amp= \ln(x) + 1 - 1 - \amp= \ln(x) - . -

      -
      3. -
    -

    - -

    - This seems significant; - if the natural log function \ln(x) is an important function - (it is), - it seems worthwhile to know a function whose derivative is \ln(x). - We have found one. - (We leave it to the reader to find another; - a correct answer will be very similar to this one.) -

    -     -     - -

    - We have learned how to compute the derivatives of sums, - differences, and products of functions. - We now learn how to find the derivative of a quotient of functions. -

    - - - Quotient Rule - -

    - Let f and g be differentiable functions defined on an open interval I, - where g(x) \neq 0 on I. - derivativeQuotient Rule - Quotient Rule - Then f/g is differentiable on I, and - - \lzoo{x}{\frac{f(x)}{g(x)}} = \frac{g(x)\fp(x) - f(x)g'(x)}{g(x)^2} - . -

    -     -     - - - -

    - The is not hard to use, - although it might be a bit tricky to remember. - A useful mnemonic works as follows. - Consider a fraction's numerator and denominator as HI - and LO, respectively. - Then - - \lzoo{x}{\frac{\text{ HI } }{\text{ LO } }} = \frac{\text{ LO}\cdot\text{dHI}-\text{HI}\cdot\text{dLO }}{\text{ LOLO } } - , - read low dee high minus high dee low, - over low low. Said fast, - that phrase can roll off the tongue, - making it easy to memorize. - The dee high and dee low - parts refer to the derivatives of the numerator and denominator, - respectively. -

    - -

    - Let's practice using the Quotient Rule. -

    - - - Using the Quotient Rule - -

    - Let f(x) = \frac{5x^2}{\sin(x) }. - Find \fp(x). -

    -     - -

    - Directly applying the gives: - - \lzoo{x}{\frac{5x^2}{\sin(x) }} \amp = \frac{\sin(x) \cdot \lzoo{x}{5x^2} - 5x^2\cdot \lzoo{x}{\sin(x)} }{\sin^2(x) } - \amp = \frac{10x\sin(x) - 5x^2\cos(x) }{\sin^2(x) } - . -

    -     - -     - - - -

    - The allows us to fill in holes in our understanding of derivatives of the common trigonometric functions. - We start with finding the derivative of the tangent function. -

    - - - Using the Quotient Rule to find <m>\lzoo{x}{\tan(x)}</m> - -

    - Find the derivative of y=\tan(x). -

    -     - -

    - At first, one might feel unequipped to answer this question. - But recall that \tan(x) = \sin(x) /\cos(x), - so we can apply the . - - \lzoo{x}{\tan(x)}\amp = \lzoo{x}{\frac{\sin(x) }{\cos(x) }} - \amp = \frac{\cos(x) \lzoo{x}{\sin(x)} - \sin(x)\lzoo{x}{\cos(x)}}{\cos^2(x) } - \amp = \frac{\cos(x) \cos(x) - \sin(x) (-\sin(x) )}{\cos^2(x) } - \amp = \frac{\cos^2(x) +\sin^2(x) }{\cos^2(x) } - \amp = \frac{1}{\cos^2(x) } - \amp = \sec^2(x) - . -

    - -

    - This is a beautiful result. - To confirm its truth, - we can find the equation of the tangent line to y=\tan(x) at x=\pi/4. - The slope is \sec^2(\pi/4) = 2; - y=\tan(x), along with its tangent line, - is graphed in . -

    - -
    - A graph of y=\tan(x) along with its tangent line at x=\pi/4 - - - A graph of the tangent function between its vertical asymptotes at -pi/2 and pi/2, along with a tangent line at pi/4. - -

    - A graph of y=\tan(x) between the vertical asymptotes x=-\pi/2 and x=\pi/2. - As x approaches -\pi/2 from the right, \tan(x) approaches -\infty. - Moving right from the x=-\pi/2 asymptote, the graph initialy climbs steeply, - and then levels off, continuing to climb with a much smaller slope from -\pi/4 to \pi/4, - and passing through the origin. Past x=\pi/4, the graph again climbs steeply toward the vertical asymptote at x=\pi/2. -

    - -

    - A tangent line to the graph at the point (\pi/4,1) is also shown. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1.65, - xmax=1.7, - ymin=-11, - ymax=11, - xtick={-1.57,-.785,.785,1.57}, - xticklabels={$-\frac{\pi}{2}$,$-\frac{\pi}{4}$,$\frac{\pi}{4}$,$\frac{\pi}{2}$}, - ] - \addplot+[infinite,samples=101,domain=-11:11] ({atan(x)*pi/180},{x}); - \addplot [tangentline,domain=-1.0:1.5] {2*(x-pi/4)+1}; - \addplot [soliddot] coordinates{(0.785,1)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - -

    - We include this result in the following theorem about the derivatives of the trigonometric functions. - Recall we found the derivative of - y=\sin(x) in - and stated the derivative of the cosine function in . - The derivatives of the cotangent, - cosecant and secant functions can all be computed directly using - and the . -

    - - - Derivatives of Trigonometric Functions - derivativetrigonometric functions - -

    -

      -
    1. \lzoo{x}{\sin(x)} = \cos(x)
      2. -
    2. \lzoo{x}{\cos(x)} = -\sin(x)
      3. -
    3. \lzoo{x}{\tan(x)} = \sec^2(x)
      4. -
    4. \lzoo{x}{\cot(x)} = -\csc^2(x)
      5. -
    5. \lzoo{x}{\sec(x)} = \sec(x) \tan(x)
      6. -
    6. \lzoo{x}{\csc(x)} = -\csc(x) \cot(x)
      7. -
    -

    -     -     - -

    - To remember the above, - it may be helpful to keep in mind that the derivatives of the trigonometric functions that start with c - have a minus sign in them. -

    - - - Exploring alternate derivative methods - -

    - In - the derivative of f(x) = \frac{5x^2}{\sin(x) } was found using the . - Rewriting f as f(x) = 5x^2\csc(x), - find \fp using - and verify the two answers are the same. -

    -     - -

    - We found in - that \fp(x) = \frac{10x\sin(x) - 5x^2\cos(x) }{\sin^2(x) }. - We now find \fp using the , considering f as f(x) = 5x^2\csc(x). - - \fp(x) \amp = \lzoo{x}{5x^2\csc(x)} - \amp = 5x^2\lzoo{x}{\csc(x)} + \lzoo{x}{5x^2}\csc(x) - \amp = 5x^2\left(-\csc(x) \cot(x) \right) + 10x\csc(x) \amp\amp\text{ (now rewrite trig functions) } - \amp = 5x^2\cdot \frac{-1}{\sin(x) }\cdot \frac{\cos(x) }{\sin(x) } + \frac{10x}{\sin(x) } - \amp = \frac{-5x^2\cos(x) }{\sin^2(x) }+\frac{10x}{\sin(x) } \amp\amp\text{ (get common denominator) } - \amp = \frac{10x\sin(x) - 5x^2\cos(x) }{\sin^2(x) } - . -

    - -

    - Finding \fp using either method returned the same result. - At first, the answers looked different, - but some algebra verified they are the same. - In general, there is not one final form that we seek; - the immediate result from the Product Rule is fine. - Work to simplify your results into a form that is most readable and useful to you. -

    -     -     - -

    - The gives other useful results, - as shown in the next example. -

    - - - Using the Quotient Rule to expand the Power Rule - -

    - Find the derivatives of the following functions. -

      -
    1. -

      - f(x) = \dfrac{1}{x} -

      -
      2. -
    2. -

      - f(x)= \dfrac{1}{x^n}, where n \gt 0 is an integer. -

      -
      3. -
    -

    -     - -

    - We employ the . -

      -
    1. -

      - - \fp(x)\amp = \frac{x\cdot 0 - 1\cdot 1}{x^2} - \amp = -\frac{1}{x^2} - -

      -
      2. -
    2. -

      - - \fp(x)\amp = \frac{x^n\cdot 0 - 1\cdot nx^{n-1}}{(x^n)^2} - \amp = -\frac{nx^{n-1}}{x^{2n}} - \amp = -\frac{n}{x^{n+1}} - . -

      -
      3. -
    -

    -     - -     - -

    - The derivative of y=\frac{1}{x^n} turned out to be rather nice. - It gets better. - Consider: - - \lzoo{x}{\frac{1}{x^n}} \amp = \lzoo{x}{x^{-n}} \amp\amp\text{ (apply result from } \text{)} - \amp = -\frac{n}{x^{n+1}} \amp\amp \text{ (rewrite algebraically) } - \amp = -nx^{-(n+1)}\amp - \amp = -nx^{-n-1}\amp - . -

    - -

    - This is reminiscent of the : - multiply by the power, then subtract 1 from the power. - We now add to our previous Power Rule, - which had the restriction of n \gt 0. -

    - - - Power Rule with Integer Exponents - -

    - Let f(x) = x^n, where n\neq 0 is an integer. - derivativePower Rule - Power Ruledifferentiation - Then - - \fp(x) = n\cdot x^{n-1} - . -

    -     -     - -

    - Taking the derivative of many functions is relatively straightforward. - It is clear - (with practice) - what rules apply and in what order they should be applied. - Other functions present multiple paths; - different rules may be applied depending on how the function is treated. - One of the beautiful things about calculus is that there is not the right way; - each path, when applied correctly, - leads to the same result, the derivative. - We demonstrate this concept in an example. -

    - - - Exploring alternate derivative methods - -

    - Let f(x) = \frac{x^2-3x+1}{x}. - Find \fp(x) in each of the following ways: -

      -
    1. -

      - By applying the , -

      -
      2. -
    2. -

      - by viewing f as f(x) = \left(x^2-3x+1\right)\cdot x^{-1} and applying the and , and -

      -
      3. -
    3. -

      - by simplifying first through division. -

      -
      4. -
    - Verify that all three methods give the same result. -

    -     - -

    -

      -
    1. -

      - Applying the gives: - - \fp(x) \amp =\frac{x\cdot\lzoo{x}{x^2-3x+1}-\left(x^2-3x+1\right)\lzoo{x}{x}}{x^2} - \amp = \frac{x\cdot(2x-3)-\left(x^2-3x+1\right)\cdot 1}{x^2} - \amp = \frac{x^2-1}{x^2} - \amp = 1-\frac{1}{x^2} - . -

      -
      2. -
    2. -

      - By rewriting f, - we can apply the and as follows: - - \fp(x) \amp = \left(x^2-3x+1\right)\lzoo{x}{x^{-1}} + \lzoo{x}{x^2-3x+1} x^{-1} - \amp = \left(x^2-3x+1\right)\cdot (-1)x^{-2} + (2x-3)\cdot x^{-1} - \amp = -\frac{x^2-3x+1}{x^2}+\frac{2x-3}{x} - \amp = -\frac{x^2-3x+1}{x^2}+\frac{2x^2-3x}{x^2} - \amp = \frac{x^2-1}{x^2} = 1-\frac{1}{x^2} - , - the same result as above. -

      -
      3. -
    3. -

      - As x\neq 0, we can divide through by x first, - giving f(x) = x-3+x^{-1}. - Now apply the . - - \fp(x) = 1-\frac{1}{x^2} - , - the same result as before. -

      -
      4. -
    -

    -     - -     - -

    - - demonstrates three methods of finding \fp. - One is hard pressed to argue for a best method - as all three gave the same result without too much difficulty, - although it is clear that using the required more steps. - Ultimately, the important principle to take away from this is: - reduce the answer to a form that seems - simple and easy to interpret. - In that example, - we saw different expressions for \fp, including: - - \amp 1-\frac{1}{x^2} - \amp \frac{x\cdot(2x-3)-\left(x^2-3x+1\right)\cdot 1}{x^2} - \amp \left(x^2-3x+1\right)\cdot (-1)x^{-2} + (2x-3)\cdot x^{-1} - . -

    - -

    - They are equal; they are all correct; - only the first is simple. - Work to make answers simple. -

    - -

    - In the next section we continue to learn rules that allow us to more easily compute derivatives than using the limit definition directly. - We have to memorize the derivatives of a certain set of functions, - such as the derivative of \sin(x) is \cos(x). - The , - , - , - and show us how to find the derivatives of certain combinations of these functions. - The next section shows how to find the derivatives when we compose - these functions together. -

    - - - - Terms and Concepts - - - - -

    - - The Product Rule states that \lzoo{x}{x^2\sin(x)}= 2x\cos(x). -

    -     - -

    - The product rule results in something more complicated than the product of the derivatives. - Double-check the statement of . -

    -     - -     - - - - -

    - - The Quotient Rule states that \lzoo{x}{\frac{x^2}{\sin(x) }} = \frac{\cos(x) }{2x}. -

    -     - -

    - The quotient rule results in something more complicated than the quotient of the derivatives. - Double-check the statement of . -

    -     - -     - - - - -

    - - The derivatives of the trigonometric functions that start with c - have minus signs in them. -

    -     - -

    - We could even go so far as to say this is true of the trig functions starting with co! -

    -     - -     - - - - -

    - What derivative rule is used to extend the Power Rule to include negative integer exponents? -

    -

    - - -

    -     - - - - - quotient|quotient rule|the quotient rule - - - - - -     - - - - -

    - - Regardless of the function, - there is always exactly one right way of computing its derivative. -

    -     - -

    - Sometimes you have options! -

    -     - -     - - - - -

    - In your own words, - explain what it means to make your answers clear. -

    - -     - - - -     -     - - - Problems - - - -

    -

      -
    1. - Use the Product Rule to differentiate the function. -
      2. -
    2. - Manipulate the function algebraically and differentiate without using the Product Rule. -
      3. -
    3. - Show that the two derivatives are equivalent. -
      4. -
    -

    -     - - - - - $x = 'x'; - Context()->variables->are($x=>'Real'); - $f = Formula("$x($x^2+3$x)"); - $d = $f->D($x); - - -

    - f() = -

    -

    - -

    -     -     -     - - - - - $x = 'x'; - Context()->variables->are($x=>'Real'); - $f = Formula("2$x^2(5$x^3)"); - $d = $f->D($x); - - -

    - f() = -

    -

    - -

    -     -     -     - - - - - $x = 's'; - Context()->variables->are($x=>'Real'); - $f = Formula("(2$x-1)($x+4)"); - $d = $f->D($x); - - -

    - f() = -

    -

    - -

    -     -     -     - - - - - $x = 'x'; - Context()->variables->are($x=>'Real'); - $f = Formula("($x^2+5)(3-$x^3)"); - $d = $f->D($x); - - -

    - f() = -

    -

    - -

    -     -     -     -     - - - -

    -

      -
    1. - Use the Quotient Rule to differentiate the function. -
      2. -
    2. - Manipulate the function algebraically and differentiate without using the Quotient Rule. -
      3. -
    3. - Show that the two derivatives are equivalent. -
      4. -
    -

    -     - - - - - $x = 'x'; - Context()->variables->are($x=>'Real'); - $f = Formula("($x^2+3)/$x"); - $d = $f->D($x); - - -

    - f(x) = -

    -

    - -

    -     -     -     - - - - - $x = 'x'; - Context()->variables->are($x=>'Real'); - $f = Formula("($x^3-2$x^2)/(2$x^2)"); - $d = $f->D($x); - - -

    - f(x) = -

    -

    - -

    -     -     -     - - - - - $x = 's'; - Context()->variables->are($x=>'Real'); - $f = Formula("3/(4$x^3)"); - $d = $f->D($x); - - -

    - f(x) = -

    -

    - -

    -     -     -     - - - - - $x = 't'; - Context()->variables->are($x=>'Real'); - $f = Formula("($x^2-1)/($x+1)"); - $d = $f->D($x); - - -

    - f(x) = -

    -

    - -

    -     -     -     -     - - - -

    - Compute the derivative of the given function. -

    -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$fname='f';$x='x';$trig='sin';}; - Context()->variables->are($x => 'Real'); - $f = Formula("$x $trig($x)"); - $df = $f->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig = list_random('sin','cos'); - $n = random(2,9,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$trig='cos';$n=2;}; - Context()->variables->are($x => 'Real'); - $f = Formula("$x^$n $trig($x)"); - $df = $f->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - if($envir{problemSeed}==1){$fname='f';$x='x';}; - Context()->variables->are($x => 'Real'); - $f = Formula("e^$x ln($x)"); - $df = $f->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig = list_random('sec','csc','tan','cot'); - $n = random(2,9,1); - $a = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$fname='f';$x='t';$trig='csc';$n=2;$a=-4;}; - Context()->variables->are($x => 'Real'); - $f = Formula("1/$x^$n ($trig($x)+$a)")->reduce; - $df = $f->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - ($a,$b) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$fname='g';$x='x';$a=7;$b=-5;}; - Context()->variables->are($x => 'Real'); - $f = Formula("($x+$a)/($x+$b)")->reduce; - $df = $f->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig = list_random('sin','cos'); - $m = random(2,9,1); - $n = random(2,9,1); - $a = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$fname='g';$x='t';$trig='cos';$m=2;$n=5;$a=-2;}; - Context()->variables->are($x => 'Real'); - $f = Formula("$x^$n/($trig($x)+$a $x^$m)")->reduce; - $df = $f->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig = list_random('cot','sec','csc'); - Context()->variables->are($x => 'Real'); - $pm = list_random('+','-'); - if($envir{problemSeed}==1){$fname='h';$x='x';$trig='cot';$pm='-';}; - Context()->variables->are($x => 'Real'); - $formula = Formula("$trig($x) $pm e^$x"); - $df = $formula->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig = list_random('tan','cot','sec','csc'); - if($envir{problemSeed}==1){$fname='f';$x='x';$trig='tan';}; - Context()->variables->are($x => 'Real'); - $formula = Formula("$trig($x) ln($x)"); - $df = $formula->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $a = random(3,9,1); - $b = non_zero_random(-9,9,1); - $c = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$fname='h';$x='t';$a=7;$b=6;$c=-2;}; - Context()->variables->are($x => 'Real'); - $formula = Formula("$a $x^2+$b$x+$c")->reduce; - $df = $formula->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $a = list_random(-9..-2,2..9); - $n = random(2,5,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$a=2;$n=3;}; - Context()->variables->are($x => 'Real'); - $formula = Formula("($x^($n+1)+$a $x^($n))/($x+$a)")->reduce; - $df = Formula("$n $x^($n-1)")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - ($a,$b,$c) = random_subset(3,3..9); - if($envir{problemSeed}==1){$fname='f';$x='x';$a=3;$b=8;$c=7;}; - Context()->variables->are($x => 'Real'); - $formula = Formula("($a $x^2 + $b $x + $c) e^($x)")->reduce; - $df = Formula("($a $x^2 + ($b+2*$a) $x + ($c+$b)) e^($x)")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - ($a,$b) = random_subset(2,3,5,7,9); - $pm = list_random('+','-'); - if($envir{problemSeed}==1){$fname='g';$x='t';$a=5;$b=3;$pm='-';}; - Context()->variables->are($x => 'Real'); - $formula = Formula("($x^$a $pm $x^$b) / e^($x)")->reduce; - $df = Formula("(-$x^$a + $a $x^($a-1) - $pm $x^$b $pm $b $x^($b-1)) / e^($x)")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - ($a,$d) = random_subset(2,1..30); - ($b,$c,$e) = random_subset(3,-30..-1,1..30); - if($envir{problemSeed}==1){$fname='f';$x='x';$a=16;$b=24;$c=3;$d=7;$e=-1;}; - Context()->variables->are($x => 'Real'); - $formula = Formula("($a $x^3+$b $x^2 + $c $x) (($d $x + $e)/($a $x^3+$b $x^2 + $c $x))")->reduce; - $df = Formula("$d")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $n = random(4,6,1); - $trig = list_random('cos','sin','sec','csc','tan','cot'); - if($envir{problemSeed}==1){$fname='f';$x='t';$n=5;$trig='sec';}; - Context()->variables->are($x => 'Real'); - $formula = Formula("$x^$n($trig($x) + e^$x)"); - $df = $formula->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $a = random(2,9,1); - ($triga,$trigb) = random_subset(2,'cos','sin','sec','csc','tan','cot'); - if($envir{problemSeed}==1){$fname='f';$x='x';$a=3;$triga='sin';$trigb='cos';}; - Context()->variables->are($x => 'Real'); - $formula = Formula("$triga($x)/($trigb($x) + $a)"); - $df = $formula->D("$x")->reduce; - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $x = Formula('theta'); - $n = random(2,6,1); - $trig = list_random('cos','sin','sec','csc','tan','cot'); - if($envir{problemSeed}==1){$fname='f';$n=3;$trig='sin';}; - $formula = Formula("theta^$n $trig(theta) + $trig(theta)/theta^$n"); - $df = $formula->D("$x"); - - -

    - () = -

    - - To enter \theta, type theta. - -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - ($triga,$trigb) = random_subset(2,'cos','sin','sec','csc','tan','cot'); - if($envir{problemSeed}==1){$fname='f';$x='x';$triga='cos';$trigb='tan';}; - Context()->variables->are($x => 'Real'); - $formula = Formula("$triga($x)/$x + $x/$trigb($x)"); - $df = $formula->D("$x"); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $a = random(2,5,1); - $b = list_random(3,4,6); - $c = random(1,5,1); - $trig = list_random('\cos','\sin','\sec','\csc','\tan','\cot'); - if($envir{problemSeed}==1){$fname='g';$x='x';$a=2;$b=4;$c=1;$trig='sin';}; - Context()->variables->are($x => 'Real'); - $formula = "e^$a($trig(" . '\pi' . "/$b) - $c)"; - $df = Formula("0"); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - ($a,$b) = random_subset(2,2..9); - $trig = list_random("sin($x)*cos($x)","sec($x)*csc($x)"); - $pm = list_random('+','-'); - if($envir{problemSeed}==1){$fname='g';$x='t';$a=4;$b=3;$trig="sin($x)*cos($x)";$pm='-';}; - Context()->variables->are($x => 'Real'); - $formula = Formula("$a $x^$b e^$x $pm $trig"); - $df = $formula->D("$x"); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $n = random(2,3,1); - ($a,$b) = random_subset(2,-9..-1,1..9); - ($triga,$trigb) = random_subset(2,'sin','cos'); - if($envir{problemSeed}==1){$fname='h';$x='t';$n=2;$a=3;$b=2;$triga='sin';$trigb='cos';}; - Context()->variables->are($x => 'Real'); - $formula = Formula("($x^$n $triga($x)+$a)/($x^$n $trigb($x)+$b)")->reduce; - $df = $formula->D("$x"); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $n = random(2,5,1); - $tran = list_random("e^$x","ln($x)"); - $trig = list_random('cos','sin','sec','csc','tan','cot'); - if($envir{problemSeed}==1){$fname='f';$x='x';$n=2;$tran="e^$x";$trig='tan';}; - Context()->variables->are($x => 'Real'); - $formula = Formula("$x^$n $tran $trig($x)")->reduce; - $df = $formula->D("$x"); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $a = random(2,9,1); - $trig = list_random("sin($x) sec($x)","cos($x) csc($x)","cos($x) tan($x)","sin($x) cot($x)"); - if($envir{problemSeed}==1){$fname='g';$x='x';$a=2;$trig="sin($x) sec($x)";}; - Context()->variables->are($x => 'Real'); - $formula = Formula("$a $x $trig")->reduce; - $df = $formula->D("$x"); - - -

    - () = -

    -

    - -

    -     -     -     -     - - - -

    - Find the equations of the tangent and normal lines to the graph of g at the indicated point. -

    -     - - - - - $n = random(2,3,1); - $a = list_random(-9..-1,1..9); - if($envir{problemSeed}==1){$n=2;$a=2}; - Context("Fraction"); - $formula = Formula("e^x(x^$n+$a)")->reduce; - $x0=0; - $y0=Fraction($formula->eval(x=>$x0)); - $df=$formula->D('x'); - $m=Fraction($df->eval(x=>$x0)); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=$m x+$y0")->reduce; - if ($m == 0) { - $n=Formula("x=$x0"); - } - else { - $mn = Fraction(-1/$m); - $n=Formula("y=$mn x+$y0")->reduce; - } - - -

    - g(x) = at \left(,\right) -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0); - $trig = list_random('sin','cos','tan','cot','sec','csc'); - if ($trig eq 'sin' or $trig eq 'csc') { - $x0 = list_random(Formula("pi/6"),Formula("pi/4"),Formula("pi/3"),Formula("pi/2"),Formula("2pi/3"),Formula("3pi/4"),Formula("5pi/6"),Formula("7pi/6"),Formula("5pi/4"),Formula("4pi/3"),Formula("3pi/2"),Formula("5pi/3"),Formula("7pi/4"),Formula("11pi/6")); - } else { - $x0 = list_random(Formula("pi/6"),Formula("pi/4"),Formula("pi/3"),Formula("2pi/3"),Formula("3pi/4"),Formula("5pi/6"),Formula("7pi/6"),Formula("5pi/4"),Formula("4pi/3"),Formula("5pi/3"),Formula("7pi/4"),Formula("11pi/6")); - } - if($envir{problemSeed}==1){$trig='sin';$a=Formula("3pi/2")}; - $formula = Formula("x $trig(x)"); - $y0=$formula->eval(x=>($x0->eval(x=>0)))->value; - $sign = ($y0 > 0) ? '' : '-'; - Context("Fraction"); - $frac = Fraction(($y0/pi)**2); - ($num,$den)=$frac->value; - if ($num == 1 or $num == 4 or $num == 9 or $num == 16) {$N = sqrt($num)} else {$N = "sqrt($num)"}; - if ($den == 1 or $den == 4 or $den == 9 or $den == 16) {$D = sqrt($den)} else {$D = "sqrt($den)"}; - $y0 = Formula("($sign$N pi)/$D"); - $df=$formula->D('x'); - $m=$df->eval(x=>($x0->eval(x=>0))); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=$m (x-$x0)+$y0")->reduce; - if ($m == 0) { - $n=Formula("x=$x0"); - } - else { - $mn = Fraction(-1/$m); - $n=Formula("y=$mn (x-$x0)+$y0")->reduce; - } - - -

    - g(x) = at \left(,\right) -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - $a = list_random(-9..-1,1..9); - $x0=$a+list_random(-2,-1,1,2); - if($envir{problemSeed}==1){$a=1;$x0=2;}; - Context("Fraction"); - $formula = Formula("x^2/(x-$a)"); - $y0=Fraction($formula->eval(x=>$x0)); - $df=$formula->D('x'); - $m=Fraction($df->eval(x=>$x0)); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=$m (x-$x0)+$y0")->reduce; - if ($m == 0) { - $n=Formula("x=$x0"); - } - else { - $mn = Fraction(-1/$m); - $n=Formula("y=$mn (x-$x0)+$y0")->reduce; - } - - -

    - g(x) = at \left(,\right) -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - ($a,$b) = random_subset(2,-9..-1,1..9); - $trig = list_random('cos','sin'); - if($envir{problemSeed}==1){$a=-8;$b=1;$trig='cos'}; - Context("Fraction"); - $formula = Formula("($trig(x)+$a x)/(x+$b)")->reduce; - $x0=0; - $y0=Fraction($formula->eval(x=>$x0)); - $df=$formula->D('x'); - $m=Fraction($df->eval(x=>$x0)); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=$m (x-$x0)+$y0")->reduce; - if ($m == 0) { - $n=Formula("x=$x0"); - } - else { - $mn = Fraction(-1/$m); - $n=Formula("y=$mn (x-$x0)+$y0")->reduce; - } - - -

    - g(x) = at \left(,\right) -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     -     - - - -

    - Find the x-values where the graph of the function has a horizontal tangent line. -

    -     - - - - - Context("Fraction"); - Context()->noreduce('(-x)+y','(-x)-y'); - $a = non_zero_random(-9,9,1); - ($b,$c) = random_subset(2,-30..-1,1..30); - if($envir{problemSeed}==1){$a=6;$b=-18;$c=-24;}; - $formula = Formula("$a x^2 + $b x + $c")->reduce; - @x = (Fraction(-$b,2*$a)); - $answer = List(@x); - - -

    - f(x) = -

    - - Use commas to separate more than one x-value. - If there are no such x-values, - answer with NONE. - -

    - -

    -     -     -     - - - - - $points = List("0"); - - -

    - f(x) = x\sin(x) on [-1,1] -

    - - Use commas to separate more than one x-value. - If there are no such x-values, - answer with NONE. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->noreduce('(-x)+y','(-x)-y'); - ($a,$b,$c) = random_subset(3,-3..-1,1..3); - if($envir{problemSeed}==1){$a=1;$b=1;$c=1;}; - $formula = Formula("($a x)/($b x + $c)")->reduce; - $points = Compute("NONE"); - - -

    - f(x) = -

    - - Use commas to separate more than one x-value. - If there are no such x-values, - answer with NONE. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->noreduce('(-x)+y','(-x)-y'); - ($a,$b,$c) = random_subset(3,-3..-1,1..3); - if($envir{problemSeed}==1){$a=1;$b=1;$c=1;}; - $formula = Formula("($a x^2)/($b x + $c)")->reduce; - @x = (Fraction(0),Fraction(-(2*$c)/$b)); - $points = List(@x); - - -

    - f(x) = -

    - - Use commas to separate more than one x-value. - If there are no such x-values, - answer with NONE. - -

    - -

    -     -     -     -     - - - -

    - Find the requested higher order derivative. -

    -     - - - - - $answer=Formula("2cos(x)-x*sin(x)"); - - -

    - \fpp(x), where f(x) = x\sin(x) -

    -

    - -

    -     -     -     - - - - - $answer=Formula("-4cos(x)+x*sin(x)"); - - -

    - f^{(4)}(x), where f(x) = x\sin(x) -

    -

    - -

    -     -     -     - - - - - $trig = list_random('csc','cot','sec','tan'); - if($envir{problemSeed}==1){$trig = 'csc'}; - $formula = Formula("$trig(x)"); - $answer=$formula->D('x')->D('x'); - - -

    - f''(x), where f(x) = -

    -

    - -

    -     -     -     - - - - - ($a,$b,$c,$d) = random_subset(4,-9..-1,1..9); - $n = random(7,9,1); - if($envir{problemSeed}==1){$a=-5;$b=2;$c=1;$d=-7;$n=8;}; - $formula = Formula("(x^3+$a x+$b)(x^2+$c x+$d)")->reduce; - $answer=Formula("0"); - - -

    - f^{()}(x), where f(x) = -

    -

    - -

    -     -     -     -     -     - -     -
    -
    - The Chain Rule - -

    - We have covered almost all of the derivative rules that deal with combinations of two - (or more) - functions. - The operations of addition, subtraction, multiplication - (including by a constant) - and division led to the , - the , - the , - the and the . - To complete the list of differentiation rules, - we look at the last way two - (or more) - functions can be combined: - the process of composition ( one function - inside another). -

    - - - -

    - One example of a composition of functions is f(x) = \cos(x^2). - We currently do not know how to compute this derivative. - If forced to guess, one might guess \fp(x) = -\sin(2x), - where we recognize -\sin(x) as the derivative of \cos(x) and 2x as the derivative of x^2. - However, this is not the case; - \fp(x)\neq -\sin(2x). - One way to see this is to examine the graph of - y=\cos\mathopen{}\left(x^2\right)\mathclose{} in - and its tangent line at x=\pi/2. - Clearly the slope of the tangent line there is nonzero, - but -2\sin(2\cdot\pi/2)=0. - So it can't be correct to say that \yp=-\sin(2x). -

    - -
    - A graph of y=\cos(x^2) and a tangent line at \pi/2 - - - A cosine wave with increasing frequency with a negatively sloped tangent line. - - -

    - A cosine wave with increasing frequency. - The distance between the peaks decreases as x increases. - There is a point drawn on the curve at x= \frac{\pi}{2}, roughly at y=-0.8. - A tangent line to the point is drawn, sloping downward. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ymin=-1.5,ymax=1.5, - xmin=-.2,xmax=3.5, - clip=false, - ] - \addplot+[infinite,domain=-0.2:3.5,samples=101] {cos(deg(x^2))}; - \addplot[tangentline,domain=1:1.93] {-sin(deg((pi/2)^2))*2*pi/2*(x-pi/2)+cos(deg((pi/2)^2))}; - \addplot[soliddot] coordinates {(1.571,-0.781)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -

    - In - we'll see the correct way to compute the derivative of \sin\mathopen{}\left(x^2\right)\mathclose{}, - which employs the new rule this section introduces, the Chain Rule. -

    - -

    - Before we define this new rule, - recall the notation for composition of functions. - We write (f \circ g)(x) or f(g(x)), - read as f of g of x, - to denote composing f with g. - In shorthand, we simply write - f \circ g or f(g) and read it as - f of g. - Before giving the corresponding differentiation rule, - we note that the rule extends to multiple compositions like - f(g(h(x))) or f(g(h(j(x)))), etc. -

    - -

    - To motivate the rule, - let's look at three derivatives we can already compute. -

    - - - Exploring similar derivatives - -

    - Find the derivatives of F_1(x) = (1-x)^2, - F_2(x) = (1-x)^3, and F_3(x) = (1-x)^4. - (We'll see later why we are using subscripts for different functions and an uppercase F.) -

    -     - -

    - In order to use the rules we already have, - we must first expand each function as - - F_1(x)\amp = 1 - 2x + x^2 - F_2(x)\amp = 1 - 3x + 3x^2 - x^3 - F_3(x)\amp = 1 - 4x + 6x^2 - 4x^3 + x^4 - - It is not hard to see that: - - F_1'(x)\amp = -2 + 2x - F_2'(x)\amp = -3 + 6x - 3x^2 - F_3'(x)\amp = -4 + 12x - 12x^2 + 4x^3 - - An interesting fact is that these can be rewritten as: - - F_1'(x)\amp = -2(1-x) - F_2'(x)\amp = -3(1-x)^2 - F_3'(x)\amp = -4(1-x)^3 - -

    - -

    - A pattern might jump out at you; - note how the we end up multiplying by the old power and the new power is reduced by 1. - We also always multiply by (-1). -

    - -

    - Recognize that each of these functions is a composition, - letting g(x) = 1-x: - - F_1(x)\amp = f_1(g(x)),\amp\amp \text{ where } f_1(x) = x^2, - F_2(x)\amp = f_2(g(x)),\amp\amp \text{ where } f_2(x) = x^3, - F_3(x)\amp = f_3(g(x)),\amp\amp \text{ where } f_3(x) = x^4 - . -

    - -

    - We'll come back to this example after giving the formal statements of the Chain Rule; - for now, we are just illustrating a pattern. -

    -     -     - - - - - Chain Rule - -

    - Let g be a differentiable function on an interval I, - let the range of g be a subset of the interval J, - and let f be a differentiable function on J. - - derivativeChain Rule - Chain Rule - Then y=f(g(x)) is a differentiable function on I, and - - \yp = \fp(g(x))\cdot \gp(x) - . -

    -     -     - - - -

    - Here is the Chain Rule in words: -

    - -
    -

    - The derivative of the outside function, - evaluated at the inside function, - multiplied by the derivative of the inside function. -

    -
    - - - - - - - - - -

    - To help understand the Chain Rule, - we return to . -

    - - - Using the Chain Rule - -

    - Use the Chain Rule to find the derivatives of the functions F_1(x), - F_2(x), and F_3(x), - as given in . -

    -     - -

    - - ended with the recognition that each of the given functions was actually a composition of functions. - To avoid confusion, we ignore most of the subscripts here. -

    - -

    -

      -
    • - <m>F_1(x) = (1-x)^2</m> -

      - We found that - - y=(1-x)^2 = f(g(x)) - , - where f(x) = x^2 and g(x) = 1-x. - To find \yp, - we apply the . - We need to note that \fp(x)=2x and \gp(x)=-1. -

      - -

      - Part of the uses \fp(g(x)). - This means substitute g(x) for x in the equation for \fp(x). - That is, \fp(x) = 2(1-x). - Finishing out the we have - - \yp \amp = \fp(g(x))\cdot \gp(x) - \amp = 2(1-x)\cdot (-1) - \amp = -2(1-x) - \amp = 2x-2 - . -

      -
      • - -
    • - <m>F_2(x) = (1-x)^3</m> -

      - Let y = (1-x)^3 = f(g(x)), - where f(x) = x^3 and g(x) = (1-x). - We have \fp(x) = 3x^2, - so \fp(g(x)) = 3(1-x)^2. - The then states - - \yp \amp = \fp(g(x))\cdot \gp(x) - \amp = 3(1-x)^2\cdot (-1) - \amp = -3(1-x)^2 - . -

      -
      • - -
    • - <m>F_3(x) = (1-x)^4</m> -

      - Finally, when y = (1-x)^4, - we have f(x)= x^4 and g(x) = (1-x). - Thus \fp(x) = 4x^3 and \fp(g(x)) = 4(1-x)^3. - Thus - - \yp \amp = \fp(g(x))\cdot \gp(x) - \amp = 4(1-x)^3\cdot (-1) - \amp = -4(1-x)^3 - . -

      -
      • -
    -

    -     -     - -

    - - demonstrated a particular pattern: - when f(x)=x^n, then \yp =n\cdot (g(x))^{n-1}\cdot \gp(x). - This is called the Generalized Power Rule. -

    - - - Generalized Power Rule - -

    - Let g(x) be a differentiable function and let n\neq 0 be an integer. - - derivativeGeneralized Power Rule - Generalized Power Rule - Then - - \lzoo{x}{g(x)^n} = n\cdot\left(g(x)\right)^{n-1}\cdot \gp(x) - . -

    -     -     - -

    - This allows us to quickly find the derivative of functions like y = (3x^2-5x+7+\sin(x) )^{20}. - While it may look intimidating, - the Generalized Power Rule states that - - \yp = 20(3x^2-5x+7+\sin(x) )^{19}\cdot (6x-5+\cos(x) ) - . -

    - -

    - Treat the derivative-taking process step-by-step. - In the example just given, - first multiply by 20, then rewrite the inside of the parentheses, - raising it all to the 19th power. - Then think about the derivative of the expression inside the parentheses, - and multiply by that. -

    - -

    - We now consider more examples that employ the . -

    - - - Using the Chain Rule - -

    - Find the derivatives of the following functions: - -

      -
    1. y = \sin(2x).
      2. - -
    2. y= \ln(4x^3-2x^2).
      3. - -
    3. y = e^{-x^2}.
      4. -
    -

    -     - -

    -

      -
    1. -

      - Consider y = \sin(2x). - Recognize that this is a composition of functions, - where f(x) = \sin(x) and g(x) = 2x. - Thus - - \yp \amp = \fp(g(x))\cdot \gp(x) - \amp = \cos(2x)\cdot \lzoo{x}{2x} - \amp = \cos(2x)\cdot 2 - \amp = 2\cos(2x) - . -

      -
      2. - -
    2. -

      - Recognize that y = \ln\mathopen{}\left(4x^3-2x^2\right)\mathclose{} is the composition of - f(x) = \ln(x) and g(x) = 4x^3-2x^2. - Also, recall that - - \lzoo{x}{\ln(x)} = \frac{1}{x} - . - This leads us to: - - \yp \amp = \frac{1}{4x^3-2x^2} \cdot \lzoo{x}{4x^3-2x^2} - \amp = \frac{1}{4x^3-2x^2} \cdot \left(12x^2-4x\right) - \amp = \frac{12x^2-4x}{4x^3-2x^2} - \amp = \frac{4x(3x-1)}{2x(2x^2-x)} - \amp = \frac{2(3x-1)}{2x^2-x} - . - Note that \ln\mathopen{}\left(4x^3-2x^2\right)\mathclose{}=\ln\mathopen{}\left(4x^2(x-1/2)\right)\mathclose{} was only defined for x \gt 1/2, - so the result of \yp=\frac{2(3x-1)}{2x^2-x} is only valid for x \gt 1/2 as well. -

      -
      3. - -
    3. -

      - Recognize that y = e^{-x^2} is the composition of - f(x) = e^x and g(x) = -x^2. - Remembering that \fp(x) = e^x, we have - - \yp \amp = e^{-x^2}\cdot \lzoo{x}{-x^2} - \amp = e^{-x^2}\cdot (-2x) - \amp = -2xe^{-x^2} - . -

      -
      4. -
    -

    -     - -     - - - Using the Chain Rule to find a tangent line - -

    - Let f(x) = \cos(x^2). - Find the equation of the line tangent to the graph of f at x=1. -

    -     - -

    - The tangent line goes through the point - (1,f(1)) \approx (1,0.54) with slope \fp(1). - To find \fp, - we need the . -

    - -

    - \fp(x) = -\sin(x^2) \cdot(2x) = -2x\sin(x^2). - Evaluated at x=1, we have \fp(1) = -2\sin(1) \approx -1.68. - Thus the equation of the tangent line is approximated by - - y \approx -1.68(x-1)+0.54 - . -

    - -

    - The tangent line is sketched along with f in . -

    - -
    - f(x) = \cos(x^2) sketched along with its tangent line at x=1 - - - - A wave with increasing frequency with a negatively sloped tangent line at x=1. - - -

    - A cosine wave with increasing frequency as the graph moves further from the y-axis. - To the left, there is one peak and valley before the graph moves to the y-axis, where the graph becomes gradually horizontal. - To the right, the graph has another valley and peak, symmetrical to the left side. - At the point x=1, there is a tangent line drawn. This line has a clear negative slope. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-3.1,xmax=3.1, - ymin=-1.1,ymax=1.1,] - \addplot+[infinite,domain=-3:3,samples=200] ({x},{cos(deg(x^2))});% {}; - \addplot [tangentline,domain=.7:1.3] {-1.68*(x-1)+0.54}; - \addplot [soliddot] coordinates{(1,0.54)}; - \end{axis} - \end{tikzpicture} - - - - -
    -     -     - -

    - The is used often in taking derivatives. - Because of this, - one can become familiar with the basic process and learn patterns that facilitate finding derivatives quickly. - For instance, - - \lzoo{x}{\ln(\text{anything})} = \frac{1}{\text{anything}}\cdot\lzoo{x}{\text{anything}} = \frac{\lzoo{x}{\text{anything}}}{\text{anything}} - . -

    - -

    - A concrete example of this is - - \lzoo{x}{\ln(3x^{15}-\cos(x) +e^x)} = \frac{45x^{14}+\sin(x) +e^x}{3x^{15}-\cos(x) +e^x} - . -

    - -

    - While the derivative may look intimidating at first, look for the pattern. - The denominator is the same as what was inside the natural log function; - the numerator is simply its derivative. -

    - -

    - This pattern recognition process can be applied to lots of functions. - In general, instead of writing anything, - we use u as a generic function of x. - We then say - - \lzoo{x}{\ln(u)} = \frac{u'}{u} - . -

    - -

    - The following is a short list of how the can be quickly applied to familiar functions. - -

      -
    1. \lzoo{x}{u^n} = n\cdot u^{n-1}\cdot u'.
      2. - -
    2. \lzoo{x}{e^u} = e^u \cdot u'.
      3. - -
    3. \lzoo{x}{\sin(u)} = \cos(u) \cdot u'.
      4. - -
    4. \lzoo{x}{\cos(u)} = -\sin(u)\cdot u'.
      5. - -
    5. \lzoo{x}{\tan(u)} = \sec^2(u) \cdot u'.
      6. -
    -

    - -

    - Of course, the can be applied in conjunction with any of the other rules we have already learned. - We practice this next. -

    - - - Using the Product, Quotient and Chain Rules - -

    - Find the derivatives of the following functions. - -

      -
    1. f(x) = x^5 \sin(2x^3).
      2. - -
    2. f(x) = \dfrac{5x^3}{e^{-x^2}}.
      3. -
    -

    -     - -

    -

      -
    1. -

      - We must use the and . - Do not think that you must be able to see - the whole answer immediately; - rather, just proceed step-by-step. - - \fp(x) \amp = x^5 \cdot \lzoo{x}{\sin\mathopen{}\left(2x^3\right)\mathclose{}} + \sin\mathopen{}\left(2x^3\right)\mathclose{}\cdot \lzoo{x}{x^5} - \amp = x^5\left(\cos\mathopen{}\left(2x^3\right)\mathclose{}\cdot \lzoo{x}{2x^3} \right) + 5x^4\left(\sin\mathopen{}\left(2x^3\right)\mathclose{} \right) - \amp = x^5\big(6x^2\cos\left(2x^3\right) \big) + 5x^4\big(\sin\left(2x^3\right) \big) - \amp = 6x^7\cos\mathopen{}\left(2x^3\right)\mathclose{}+5x^4\sin\mathopen{}\left(2x^3\right)\mathclose{} - . -

      -
      2. - -
    2. -

      - We must employ the along with the . - Again, proceed step-by-step. - - \fp(x) \amp= \frac{e^{-x^2}\cdot \lzoo{x}{5x^3} - 5x^3\cdot \lzoo{x}{e^{-x^2}}}{\left(e^{-x^2}\right)^2} - \amp= \frac{e^{-x^2}\cdot 15x^2 - 5x^3\cdot e^{-x^2}\cdot \lzoo{x}{-x^2}}{\left(e^{-x^2}\right)^2} - \amp= \frac{e^{-x^2}\left(15x^2\right) - 5x^3\left((-2x)e^{-x^2}\right)}{\left(e^{-x^2}\right)^2} - \amp =\frac{e^{-x^2}\left(10x^4+15x^2\right)}{e^{-2x^2}} - \amp = e^{x^2}\left(10x^4+15x^2\right) - . -

      -
      3. -
    -

    -     - -     - -

    - A key to correctly working these problems is to break the problem down into smaller, - more manageable pieces. - For instance, - when using the and together, - just consider the first part of the at first: - f(x)\gp(x). - Just rewrite f(x), then find \gp(x). - Then move on to the \fp(x)g(x) part. - Don't attempt to figure out both parts at once. -

    - -

    - Likewise, using the , - approach the numerator in two steps and handle the denominator after completing that. - Only simplify afterward. -

    - -

    - We can also employ the itself several times, - as shown in the next example. -

    - - - Using the Chain Rule multiple times - -

    - Find the derivative of y = \tan^5(6x^3-7x). -

    -     - -

    - Recognize that we have the - g(x)=\tan\mathopen{}\left(6x^3-7x\right)\mathclose{} function inside - the f(x)=x^5 function; - that is, we have y = \left(\tan\mathopen{}\left(6x^3-7x\right)\mathclose{}\right)^5. - We begin using the ; - in this first step, we do not fully compute the derivative. - Rather, we are approaching this step-by-step. - - \yp = 5\left(\tan\mathopen{}\left(6x^3-7x\right)\mathclose{}\right)^4\cdot \gp(x) - . -

    - -

    - We now find \gp(x). - We again need the ; - - \gp(x) \amp = \sec^2\mathopen{}\left(6x^3-7x\right)\mathclose{}\cdot \lzoo{x}{6x^3-7x}. - \amp = \sec^2\mathopen{}\left(6x^3-7x\right)\mathclose{}\cdot\left(18x^2-7\right) - . -

    - -

    - Combine this with what we found above to give - - \yp \amp = 5\left(\tan\mathopen{}\left(6x^3-7x\right)\mathclose{}\right)^4\cdot\sec^2\mathopen{}\left(6x^3-7x\right)\mathclose{}\cdot\left(18x^2-7\right) - \amp = \left(90x^2-35\right)\sec^2\mathopen{}\left(6x^3-7x\right)\mathclose{}\tan^4\mathopen{}\left(6x^3-7x\right)\mathclose{} - . -

    - -

    - This function is frankly a ridiculous function, - possessing no real practical value. - It is very difficult to graph, - as the tangent function has many vertical asymptotes and 6x^3-7x grows so very fast. - The important thing to learn from this is that the derivative can be found. - In fact, it is not hard; - one can take several simple steps and should be careful to keep track of how to apply each of these steps. -

    -     - -     - -

    - It is a traditional mathematical exercise to find the derivatives of arbitrarily complicated functions just to demonstrate that it can be done. - Just break everything down into smaller pieces. -

    - - - Using the Product, Quotient and Chain Rules - -

    - Find the derivative of f(x) = \dfrac{x\cos(x^{-2})-\sin^2(e^{4x})}{\ln(x^2+5x^4)}. -

    -     - -

    - This function likely has no practical use outside of demonstrating derivative skills. - The answer is given below without simplification. - It employs the , - the , - and the three times. - - - f'(x)=\frac{\begin{matrix}\ln\left(x^2+5x^4\right)\cdot\bigg[\Big(x\cdot\left(-\sin\left(x^{-2}\right)\right)\cdot\left(-2x^{-3}\right)+1\cdot \cos\left(x^{-2}\right)\Big)\hspace{60pt}\\ - \hspace{72pt}-2\sin\left(e^{4x}\right)\cdot\cos\left(e^{4x}\right)\cdot\left(4e^{4x}\right)\bigg]\\ - \hspace{120pt}-\left(x\cos\left(x^{-2}\right)-\sin^2\left(e^{4x}\right)\right)\cdot\frac{2x+20x^3}{x^2+5x^4}\end{matrix}}{\left(\ln\left(x^2+5x^4\right)\right)^2} - . -

    - -

    - The reader is highly encouraged to look at each term and recognize why it is there. (, the is used; - in the numerator, identify the LOdHI term, - etc.) This example demonstrates that derivatives can be computed systematically, - no matter how arbitrarily complicated the function is. -

    -     -     - -

    - The also has theoretic value. - That is, it can be used to find the derivatives of functions that we have not yet learned as we do in the following example. -

    - - - The Chain Rule and exponential functions - -

    - Use the Chain Rule to find the derivative of y= 2^x. -

    -     - -

    - We only know how to find the derivative of one exponential function, - y = e^x. - We can accomplish our goal by rewriting 2 in terms of e. - Recalling that e^x and \ln(x) are inverse functions, - we can write - 2=e^{\ln 2} and so - - y=2^x = \left(e^{\ln 2}\right)^x = e^{x(\ln(2))} - , - using the power to a power property of exponents. -

    - -

    - The function is now the composition y=f(g(x)), - with f(x) = e^x and g(x) = x(\ln(2)). - Since \fp(x) = e^x and - \gp(x) = \ln(2), the gives - - \yp = e^{x (\ln(2))} \cdot \ln 2 - . -

    - -

    - Recall that the e^{x(\ln(2))} term on the right hand side is just 2^x, - our original function. - Thus, the derivative contains the original function itself. - We have - - \yp = y \cdot \ln(2) = 2^x\cdot \ln(2) - . -

    - -

    - We can extend this process to use any base a, - where a \gt 0 and a\neq 1. - All we need to do is replace each 2 - in our work with a. The Chain Rule, - coupled with the derivative rule of e^x, - allows us to find the derivatives of all exponential functions. -

    -     -     - -

    - The comment at the end of previous example is important and is restated formally as a theorem. -

    - - - Derivatives of Exponential Functions - -

    - Let f(x)=a^x, for a \gt 0, a\neq 1. - - derivativeexponential functions - - Then f is differentiable for all real numbers (, differentiable everywhere) and - - \fp(x) = \ln(a) \cdot a^x - . -

    -     -     - - - - - Alternate Chain Rule Notation -

    - It is instructive to understand what the - looks like using - \lz{y}{x} - notation instead of \yp notation. - Suppose that y=f(u) is a function of u, - where u=g(x) is a function of x, - as stated in . - Then, through the composition f \circ g, - we can think of y as a function of x, as y=f(g(x)). - Thus the derivative of y with respect to x makes sense; - we can talk about \lz{y}{x}. - This leads to an interesting progression of notation: -Chain RulenotationderivativeChain Rule - - \yp \amp = \fp(g(x))\cdot \gp(x) - \lz{y}{x} \amp = \yp(u) \cdot u'(x)\amp\amp \text{ since } y=f(u) \text{ and }u=g(x) - \lz{y}{x} \amp = \lz{y}{u} \cdot \lz{u}{x}\amp\amp \text{(using “fractional notation” for the derivative)} - -

    - -

    - Here the fractional aspect of the derivative notation stands out. - On the right hand side, it seems as though the du - terms cancel out, leaving - - \frac{dy}{dx} = \frac{dy}{dx} - . -

    - -

    - It is important to realize that we - are not canceling these terms; - the derivative notation of \lz{y}{u} is one symbol. - It is equally important to realize that this notation was chosen precisely because of this behavior. - It makes applying the easy with multiple variables. - For instance, - - \lz{y}{t} = \lz{y}{\bigcirc} \cdot \lz{\bigcirc}{\triangle} \cdot \lz{\triangle}{t} - . - where \bigcirc and \triangle are any variables you'd like to use. -

    - -

    - One of the most common ways of visualizing - the is to consider a set of gears, - as shown in . - The gears have 36, 18, - and 6 teeth, respectively. - That means for every revolution of the x gear, - the u gear revolves twice. - That is, the rate at which the u gear makes a revolution is twice as fast as the rate at which the x gear makes a revolution. -

    - - -
    - A series of gears to demonstrate the Chain Rule. Note how \lz{y}{x} = \lz{y}{u}\cdot\lz{u}{x} - - - - 3 gears of various sizes demonstrating the chain rule. - - - Three gears, connected in the order x,u,y. - x is the largest gear, having 36 teeth. It is rotating counter-clockwise. - u is connected to x, and it has 18 teeth. To the left of the connection is \frac{du}{dx} = 2. - y is connected to u, and it has 6 teeth. Below the connection is \frac{dy}{du}=3. - To the right of the gears is the expression \frac{dy}{dx} = 6. - - - - \begin{tikzpicture}[>=latex,scale=0.9] - - \begin{scope}[shift={(0,-200pt)}] - \begin{scope} - \foreach \x in {0,1,2,...,35} - {% - \draw [rotate around={{\x*10}:(0,0)}] (60pt,0)--(65pt,0) arc (0:{4.}:65pt); - \draw [rotate around={{\x*10+4.}:(0,0)}] (65pt,0) -- (60pt,0) arc (0:6:60pt); - } - \draw [->] (40pt,0) arc (0:170:40pt); - \draw (0,0) node {$x$}; - \end{scope} - - \begin{scope}[shift={(4.5pt,-99pt)}] - \foreach \x in {0,1,2,...,17} - {% - \draw [rotate around={{\x*20}:(0,0)}] (30pt,0)--(35pt,0) arc (0:{9}:35pt); - \draw [rotate around={{\x*20+9}:(0,0)}] (35pt,0) -- (30pt,0) arc (0:11:30pt); - } - \draw [->] (0,25pt) arc (90:-80:25pt); - \draw (0,0) node {$u$}; - \draw (45pt,-30pt) node {\small $\ds \frac{dy}{du} = 3$}; - \draw (-50pt,30pt) node {\small $\ds \frac{du}{dx} = 2$}; - \draw (60pt,40pt) node {\small $\ds \frac{dy}{dx} = 6$}; - \end{scope} - - \begin{scope}[shift={(53.5pt,-100pt)}] - \foreach \x in {0,1,2,...,5} - {% - \draw [rotate around={{\x*60}:(0,0)}] (10pt,0)--(15pt,0) arc (0:{29}:15pt); - \draw [rotate around={{\x*60+29}:(0,0)}] (15pt,0) -- (10pt,0) arc (0:31:10pt); - } - \draw [->] (0,-20pt) arc (-90:70:20pt); - \draw (0,0) node {$y$}; - \end{scope} - \end{scope} - \end{tikzpicture} - - - - -
    - -

    - Using the terminology of calculus, - the rate of u-change, with respect to x, - is \lz{u}{x} = 2. -

    - -

    - Likewise, every revolution of u causes 3 revolutions of y: - \lz{y}{u} = 3. - How does y change with respect to x? - For each revolution of x, - y revolves 6 times; that is, - - \frac{dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx} = 2\cdot 3 = 6 - . -

    - -

    - We can then extend the with more variables by adding more gears to the picture. -

    - -

    - It is difficult to overstate the importance of the . - So often the functions that we deal with are compositions of two or more functions, - requiring us to use this rule to compute derivatives. - It is also often used in real life when actual functions are unknown. - Through measurement, we can calculate (or, - approximate) \lz{y}{u} and \lz{u}{x}. - With our knowledge of the , - we can find \lz{y}{x}. -

    - -

    - In , - we use the to justify another differentiation technique. - There are many curves that we can draw in the plane that fail the - vertical line test. For instance, - consider x^2+y^2=1, which describes the unit circle. - We may still be interested in finding slopes of tangent lines to the circle at various points. - - shows how we can find \lz{y}{x} without first - solving for y. - While we can in this instance, - in many other instances solving for y is impossible. - In these situations, - implicit differentiation is indispensable. -

    -     - - - - Terms and Concepts - - - - -

    - - The Chain Rule describes how to evaluate the derivative of a composition of functions. -

    -     - -

    - This is precisely what it is for! -

    -     - -     - - - - -

    - - The Generalized Power Rule states that \lzoo{x}{g(x)^n} = n\left(g(x)\right)^{n-1}. -

    -     - -

    - Don't forget to multiply by the derivative of the inside function! -

    -     - -     - - - - -

    - - \lzoo{x}{\ln\mathopen{}\left(x^2\right)\mathclose{}} = \frac{1}{x^2}. -

    -     - -

    - Don't forget to multiply by the derivative of the inside function! - Note also that in this case, we can use the property \ln\mathopen{}\left(x^2\right)\mathclose{}=2\ln(x) - to compute the derivative without the chain rule. -

    -     - -     - - - - -

    - - \lzoo{x}{3^x}\approx 1.1\cdot 3^x. -

    -     - -

    - Note that \ln(3)\approx 1.1. -

    -     - -     - - - - -

    - - \lz{x}{y} = \lz{x}{t}\cdot\lz{t}{y}. -

    -     - -

    - This is one way of stating the chain rule using Leibniz notation. -

    -     - -     - - - - -

    - - Taking the derivative of f(x) = x^2\sin(5x) requires the use of both the Product and Chain Rules. -

    -     - -

    - The chain rule is needed for the argument 5x of the sine function, - and the product rule is needed since we also multiply by x^2. -

    -     - -     -     - - - Problems - - - -

    - Compute the derivative of the given function. -

    -     - - - - - - $fp=Formula("10(4x^3-x)^9(12x^2-1)"); - - -

    - f(x) = \left(4x^3-x\right)^{10} -

    -

    - -

    -     -     -     - - - - - - Context()->variables->are(t=>'Real'); - $fp=Formula("15(3t-2)^4"); - - -

    - f(t) = (3t-2)^{5} -

    -

    - -

    -     -     -     - - - - - - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $fp=Formula("3(sin(theta)+cos(theta))^2(cos(theta)-sin(theta))"); - - -

    - g(\theta)=(\sin(\theta)+\cos(\theta))^3 -

    - - For \theta, type theta. - -

    - -

    -     -     -     - - - - - - Context()->variables->are(t=>'Real'); - $fp=Formula("(6t+1)e^(3t^2+t-1)"); - - -

    - h(t) = e^{3t^2+t-1} -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $a = random(2,5,1); - $b = random(3,8,1); - $pm = list_random('+','-'); - if($envir{problemSeed}==1){$fname='f';$x='x';$a=2;$b=3;$pm='+'}; - Context()->variables->are($x=>'Real'); - $f=Formula("(ln($x)$pm $x^$a)^$b"); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $a = random(2,9,1); - $n = random(2,5,1); - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$a=2;$n=3;$b=3}; - Context()->variables->are($x=>'Real'); - $f=Formula("$a^($x^$n+$b $x)")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $pm = list_random('+','-'); - $n = random(3,6,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$pm='+';$n=4;}; - Context()->variables->are($x=>'Real'); - $f=Formula("($x$pm 1/$x)^$n"); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig= list_random('sin','cos'); - $a = random(2,9,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$trig='cos';$a=3;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$trig($a $x)"); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig= list_random('tan','cot'); - $a = random(2,9,1); - if($envir{problemSeed}==1){$fname='g';$x='x';$trig='tan';$a=5;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$trig($a $x)"); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $trig= list_random('sin','cos','tan','cot'); - $a = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$fname='h';$trig='tan';$a=4;}; - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $f=Formula("$trig(theta^2+$a)")->reduce; - $fp=$f->D('theta'); - - -

    - (\theta) = -

    - - For \theta, type theta. - -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig= list_random('sin','cos','tan','cot'); - $n = random(2,9,1); - $m = random(1,4,1); - if($envir{problemSeed}==1){$fname='g';$x='t';$trig='sin';$n=2;$m=1;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$trig($x^$n+1/$x^$m)")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig= list_random('sin','cos'); - $n = random(3,9,1); - $a = random(2,9,1); - if($envir{problemSeed}==1){$fname='h';$x='t';$trig='sin';$n=4;$a=2;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$trig^$n($a $x)")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig= list_random('sin','cos'); - $n = random(3,9,1); - ($a,$b) = random_subset(2,-9..-2,2..9); - if($envir{problemSeed}==1){$fname='p';$x='t';$trig='sin';$n=3;$a=3;$b=1;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$trig^$n($x^2+$a $x + $b)")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $trig= list_random('sin','cos','sec','csc'); - if($envir{problemSeed}==1){$fname='f';$x='x';$trig='cos';}; - Context()->variables->are($x=>'Real'); - $f=Formula("ln($trig($x))")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $n = random(2,9,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$n=2;}; - Context()->variables->are($x=>'Real'); - $f=Formula("ln($x^$n)")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $n = random(2,9,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$n=2;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$n ln($x)")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $b = random(2,9,1); - if($envir{problemSeed}==1){$fname='g';$x='r';$b=4;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$b^$x")->reduce; - Context()->flags->set(reduceConstantFunctions=>0); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - $b = random(2,9,1); - $trig= list_random('sin','cos','sec','csc','tan','cot'); - if($envir{problemSeed}==1){$fname='g';$x='t';$b=5;$trig='cos'}; - Context()->variables->are($x=>'Real'); - $f=Formula("$b^($trig($x))")->reduce; - Context()->flags->set(reduceConstantFunctions=>0); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p'); - $x = list_random('q','r','t','x','y','z'); - ($a,$b) = random_subset(2,2..20); - if($envir{problemSeed}==1){$fname='g';$x='t';$b=15;$a=2;}; - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstants=>0); - $f=Formula("$b^$a"); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - do {($a,$b) = random_subset(2,2..9);} until (gcd($a,$b) == 1); - if($envir{problemSeed}==1){$fname='m';$x='w';$a=3;$b=2;}; - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstantFunctions=>0); - $f=Formula("$a^$x/$b^$x"); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - do {($a,$b) = random_subset(2,2..9);} until (gcd($a,$b) == 1); - if($envir{problemSeed}==1){$fname='h';$x='t';$a=2;$b=3;}; - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstantFunctions=>0); - $f=Formula("($a^$x+$b)/($b^$x+$a)"); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - do {($a,$b) = random_subset(2,2..9);} until (gcd($a,$b) == 1); - $c = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$fname='m';$x='w';$a=3;$b=2;$c=1;}; - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstantFunctions=>0); - $f=Formula("($a^$x+$c)/($b^$x)")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - do {($a,$b) = random_subset(2,2..9);} until (gcd($a,$b) == 1); - $pm = list_random('+','-'); - if($envir{problemSeed}==1){$fname='f';$x='x';$a=3;$b=2;$pm='+';}; - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstantFunctions=>0); - $f=Formula("($a^($x^2)$pm$x)/($b^($x^2))"); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $n = random(2,3,1); - $trig = list_random('sin','cos','tan','cot'); - $a = random(2,9,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$n=2;$trig='sin';$a=5;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$x^$n $trig($a $x)"); - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - ($m,$n,$p) = random_subset(3,2..6); - ($a,$b,$c) = random_subset(3,1..9); - if($envir{problemSeed}==1){$fname='f';$x='x';$m=5;$n=4;$p=3;$a=1;$b=3;$c=2;}; - Context()->variables->are($x=>'Real'); - $f=Formula("($x^2+$a $x)^$m($b $x^$n+$c $x)^$p")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - ($triga,$trigb) = random_subset(2,'sin','cos'); - ($m,$n) = random_subset(2,0,2); - ($a,$b,$c,$d) = random_subset(4,1..9); - $b=$b*random(-1,1,2); - $d=$d*random(-1,1,2); - if($envir{problemSeed}==1){$fname='g';$x='t';$m=2;$n=0;$a=1;$b=3;$c=5;$d=-7;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$triga($a $x^$m + $b $x)$trigb($c $x + $d $x^$n)")->reduce->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - ($triga,$trigb) = random_subset(2,'sin','cos'); - ($a,$b,$c,$d) = random_subset(4,1..9); - $b=$b*random(-1,1,2); - $d=$d*random(-1,1,2); - if($envir{problemSeed}==1){$fname='f';$x='x';$a=3;$b=4;$c=5;$d=-2;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$triga($a $x + $b)$trigb($c + $d $x)")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $trig = list_random('sin','cos'); - $a = random(2,9,1); - $n = random(2,3,1); - if($envir{problemSeed}==1){$fname='g';$x='t';$a=5;$n=2;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$trig(1/$x)e^($a $x^$n)")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $trig = list_random('sin','cos'); - ($a,$b,$c,$d) = random_subset(4,1..9); - $b=$b*random(-1,1,2); - $d=$d*random(-1,1,2); - $n = random(2,4,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$a=4;$b=1;$c=5;$d=-9;$n=3;}; - Context()->variables->are($x=>'Real'); - $f=Formula("$trig($a $x + $b)/($c $x + $d)^$n")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $trig = list_random('sin','cos','tan','cot'); - ($a,$b,$c) = random_subset(3,1..9); - $n = random(2,4,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$a=4;$b=1;$c=5;$n=2;}; - Context()->variables->are($x=>'Real'); - $f=Formula("($a $x + $b)^$n/$trig($c $x)")->reduce; - $fp=$f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     -     - - - -

    - Find the equations of tangent and normal lines to the graph of the function at the given point. - Note: the functions here are the same as in Exercises. - -

    -     - - - - - - Context("Numeric")->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=0"); - $n=Formula("x=0"); - - -

    - f(x) = \left(4x^3-x\right)^{10} at x=0 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - - Context("Numeric")->variables->add(y=>'Real'); - parser::Assignment->Allow; - Context()->flags->set(reduceConstants=>0); - $t=Formula("y=15(x-1)+1"); - $n=Formula("y=-1/15(x-1)+1"); - - -

    - f(x) = (3x-2)^5 at x=1 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - - Context("Numeric")->variables->add(y=>'Real'); - parser::Assignment->Allow; - Context()->flags->set(reduceConstants=>0); - $t=Formula("y=-3(x - pi/2)+1"); - $n=Formula("y=1/3(x - pi/2)+1"); - - -

    - g(x) = (\sin(x)+\cos(x))^3 at x=\pi/2. -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     - - - - - - Context("Numeric")->variables->add(y=>'Real'); - parser::Assignment->Allow; - Context()->flags->set(reduceConstants=>0); - $t=Formula("y=-5e(x+1)+e"); - $n=Formula("y=1/(5e)(x+1)+e"); - - -

    - h(x) = e^{3x^2+x-1} at x=-1 -

    - - Enter the equation of the tangent line here. - -

    - -

    - - Enter the equation of the normal line here. - -

    - -

    -     -     -     -     - - - - - Context()->variables->add(k=>'Real'); - $f = Formula("ln(k x)"); - $d = Formula("1/x"); - $showwork = '[@ explanation_box(message => "Show the work for both parts.") @]*'; - - -

    - Compute \lzoo{x}{\ln(kx)} two ways. - First by using the Chain Rule. - Second, by using the logarithm rule \ln(ab) = \ln(a) + \ln(b) and then taking the derivative. -

    -

    - -

    -

    - -

    -     - -

    - In both cases the derivative is the same: 1/x. -

    -     -     -     - - - - - Context()->variables->add(k=>'Real'); - $f = Formula("ln(x^k)"); - $d = Formula("k/x"); - $showwork = '[@ explanation_box(message => "Show the work for both parts.") @]*'; - - -

    - Compute \lzoo{x}{\ln\mathopen{}\left(x^k\right)\mathclose{}} two ways. - First by using the Chain Rule. - Second, by using the logarithm rule \ln\mathopen{}\left(a^p\right)\mathclose{} = p\ln(a) - (for positive a) and then taking the derivative. -

    -

    - -

    -

    - -

    -     - -

    - In both cases the derivative is the same: k/x. -

    -     -     -     -     - -     -
    -
    - Implicit Differentiation - -

    - In the previous sections we learned to find the derivative, - \lz{y}{x}, or \yp, - when y is given explicitly - as a function of x. - That is, if we know y=f(x) for some function f, - we can find \yp. - For example, given y=3x^2-7, - we can easily find \yp=6x. - (Here we explicitly state how y depends on x. - Knowing x, we can directly find y.) -

    - - - -

    - Sometimes the relationship between y and x is not explicit; - rather, it is implicit. - For instance, we might know that x^2-y=4. - This equality defines a relationship between x and y; - if we know x, we could figure out y. - Can we still find \yp? - In this case, sure; we solve for y to get y=x^2-4 - (hence we now know y explicitly) - and then differentiate to get \yp=2x. -

    - -

    - Sometimes the implicit - relationship between x and y is complicated. - Suppose we are given \sin(y)+y^3=6-x^3. - A graph of this implicit relationship is given in . - In this case there is absolutely no way to solve for y in terms of elementary functions. - The surprising thing is, however, - that we can still find \yp via a process known as - implicit differentiation. - - implicit differentiation - derivativeimplicit - -

    - -
    - A graph of the implicit relationship \sin(y)+y^3=6-x^3 - - - - A curve beginning in the second quadrant, forming a gentle corner in the first quadrant, and decreasing into the fourth quadrant. - - -

    - The curve begins in the second quadrant. - From the left, the curve decreases as x increases. - The curve slowly flattens out, almost becoming horizontal as the curve crosses the y-axis near the point (0,1.8). - When x comes close to 0.75, the curve begins decreasing in the shape of a gentle corner. - The curve continues decreasing, becoming steepest around the point (0,1.8), at which it also crosses into the fourth quadrant. - When x is close to 2, the curve begins to decrease more gently, at around the same rate as the beginning of the curve. -

    -     - - - \begin{tikzpicture}[declare function = {cbrt(\x) = (\x < 0) * -(-\x)^(1/3) + (\x >= 0) * (\x)^(1/3);}] - \begin{axis}[,xmin=-3.2,xmax=3.2, - ymin=-3.2,ymax=3.2] - \addplot[firstcurvestyle,variable=\t,domain=-2.7:1.5,leftarrow] ({cbrt(6-sin(deg(t))-t^3)},{t}); - \addplot[firstcurvestyle,variable=\t,domain=1.5:1.8] ({cbrt(6-sin(deg(t))-t^3)},{t}); - \addplot[firstcurvestyle,variable=\t,domain=1.8:3,rightarrow] ({cbrt(6-sin(deg(t))-t^3)},{t}); - \end{axis} - \end{tikzpicture} - - - - -
    -     - - - The method of implicit differentiation -

    - Implicit differentiation is a technique based on the that is used to find a derivative when the relationship between the variables is given implicitly rather than explicitly - (solved for one variable in terms of the other). -

    - -

    - We begin by reviewing the Chain Rule. - Let f and g be functions of x. - Then - - \lzoo{x}{f(g(x))} = \fp(g(x))\cdot g'(x) - . -

    - -

    - Suppose now that y=g(x). - We can rewrite the above as - - \lzoo{x}{f(y)}=\fp(y)\cdot \yp, \quad \text{ or } \quad \lzoo{x}{f(y)}=\fp(y)\cdot\lz{y}{x} - . -

    - -

    - These equations look strange; - the key concept to learn here is that we can find \yp even if we don't exactly know how y and x relate. -

    - -

    - We demonstrate this process in the following example. -

    - - - Using Implicit Differentiation - -

    - Find \yp given that \sin(y) + y^3=6-x^3. -

    -     - -

    - We start by taking the derivative of both sides - (thus maintaining the equality.) - We have: - - \lzoo{x}{\sin(y) + y^3}=\lzoo{x}{6-x^3} - . -

    - -

    - The right hand side is easy; - it returns -3x^2. -

    - -

    - The left hand side requires more consideration. - We take the derivative term-by-term. - Using the technique derived from Equation above, - we can see that - - \lzoo{x}{\sin(y)} = \cos(y) \cdot \yp - . -

    - -

    - We apply the same process to the y^3 term. - - \lzoo{x}{y^3} = \lzoo{(y)^3} = 3(y)^2\cdot \yp - . -

    - -

    - Putting this together with the right hand side, we have - - \cos(y)\yp+3y^2\yp = -3x^2 - . -

    - -

    - Now solve for \yp. - It's important to treat \yp as an algebraically independent variable from y and x. - - \cos(y)\yp+3y^2\yp\amp = -3x^2 - \left(\cos(y) +3y^2\right)\yp \amp =-3x^2 - \yp\amp =\frac{-3x^2}{\cos(y) +3y^2} - -

    - -

    - This equation for \yp probably seems unusual for it contains both x and y terms. - How is it to be used? - We'll address that next. -

    -     - -     - - - - - -

    - Implicit functions are generally harder to deal with than explicit functions. - With an explicit function, given an x value, - we have an explicit formula for computing the corresponding y value. - With an implicit function, - one often has to find x and y values - at the same time - that satisfy the equation. - It is much easier to demonstrate that a given point satisfies the equation than to actually find such a point. -

    - -

    - For instance, we can affirm easily that the point - \left(\sqrt[3]{6},0\right) lies on the graph of the implicit function \sin(y) + y^3=6-x^3. - Plugging in 0 for y, - we see the left hand side is 0. - Setting x=\sqrt[3]6, we see the right hand side is also 0; - the equation is satisfied. - The following example finds the equation of the tangent line to this function at this point. -

    - - - Using implicit differentiation to find a tangent line - -

    - Find the equation of the line tangent to the curve of the implicitly defined function - \sin(y) + y^3=6-x^3 at the point \left(\sqrt[3]6,0\right). -

    -     - -

    - In we found that - - \yp = \frac{-3x^2}{\cos(y) +3y^2} - . -

    - -

    - We find the slope of the tangent line at the point \left(\sqrt[3]6,0\right) by substituting - \sqrt[3]6 for x and 0 for y. - Thus at the point \left(\sqrt[3]6,0\right), we have the slope as - - \yp = \frac{-3\left(\sqrt[3]{6}\right)^2}{\cos(0) + 3\cdot0^2} = \frac{-3\sqrt[3]{36}}{1} \approx -9.91 - . -

    - -

    - Therefore the equation of the tangent line to the implicitly defined function - \sin(y) + y^3=6-x^3 at the point \left(\sqrt[3]{6},0\right) is - - y = -3\sqrt[3]{36}\left(x-\sqrt[3]{6}\right)+0 \approx -9.91x+18 - . -

    - -

    - The curve and this tangent line are shown in . -

    - -
    - The function \sin(y) +y^3 = 6-x^3 and its tangent line at the point (\sqrt[3]{6},0) - - - - A decreasing curve with a negative tangent line through the negative x-axis - - -

    - The same curve as , but with a tangent line drawn at - x=\sqrt[3]{6}. The tangent line is pointing sharply downward. -

    -     - - - \begin{tikzpicture}[declare function = {cbrt(\x) = (\x < 0) * -(-\x)^(1/3) + (\x >= 0) * (\x)^(1/3);}] - \begin{axis}[,xmin=-3.2,xmax=3.2, - ymin=-3.2,ymax=3.2] - \addplot[firstcurvestyle,variable=\t,domain=-2.7:1.5,leftarrow] ({cbrt(6-sin(deg(t))-t^3)},{t}); - \addplot[firstcurvestyle,variable=\t,domain=1.5:1.8] ({cbrt(6-sin(deg(t))-t^3)},{t}); - \addplot[firstcurvestyle,variable=\t,domain=1.8:3,rightarrow] ({cbrt(6-sin(deg(t))-t^3)},{t}); - \addplot [tangentline,domain=1.617:2.017] {-9.9057*x+18}; - \end{axis} - \end{tikzpicture} - - - - -
    -     - -     - -

    - This suggests a general method for implicit differentiation. - For the steps below assume y is a function of x. -

    - -

    -

      -
    1. -

      - Take the derivative of each term in the equation. - Treat the x terms like normal. - When taking the derivatives of y terms, - the usual rules apply except that, - because of the , - we need to multiply each term by \yp. -

      -
      2. - -
    2. -

      - Get all the \yp terms on one side of the equal sign and put the remaining terms on the other side. -

      -
      3. - -
    3. -

      - Factor out \yp; solve for \yp by dividing. -

      -
      4. -
    -

    - -

    - (Practical Note: when working by hand, - it may be beneficial to use the symbol - \frac{dy}{dx} instead of \yp, - as the latter can be easily confused for y or y^1.) -

    - - - Using Implicit Differentiation - -

    - Given the implicitly defined function y^3+x^2y^4=1+2x, - find \yp. -

    -     - -

    - We will take the implicit derivatives term by term. - The derivative of y^3 is 3y^2\yp. -

    - -

    - The second term, x^2y^4, is a little tricky. - It requires the as it is the product of two functions of x: - x^2 and y^4. - Its derivative is x^2(4y^3\yp) + 2xy^4. - The first part of this expression requires a \yp because we are taking the derivative of a y term. - The second part does not require it because we are taking the derivative of x^2. -

    - -

    - The derivative of the right hand side is easily found to be 2. - In all, we get: - - 3y^2\yp + 4x^2y^3\yp + 2xy^4 = 2 - . -

    - -

    - Move terms around so that the left side consists only of the \yp terms and the right side consists of all the other terms: - - 3y^2\yp + 4x^2y^3\yp = 2-2xy^4 - . -

    - -

    - Factor out \yp from the left side and solve to get - - \yp = \frac{2-2xy^4}{3y^2+4x^2y^3} - . -

    - -

    - To confirm the validity of our work, - let's find the equation of a tangent line to this function at a point. - It is easy to confirm that the point (0,1) lies on the graph of this function. - At this point, \yp = 2/3. - So the equation of the tangent line is y = 2/3(x-0)+1. - The function and its tangent line are graphed in . -

    - -
    - A graph of the implicitly defined function y^3+x^2y^4=1+2x along with its tangent line at the point (0,1) - - - - A curve with two distinct segments and a tangent line with a positive slope - - - Two curves are drawn in the xy-plane. - The left curve stretches upwards from the left side of the y axis, curving slightly to the left. - As y approaches -2, the curve begins to widen to the left, creating a bump in the curve. - As the curve crosses the x axis, the curve moves towards the right, no longer increasing and becoming more horizontal as x increases. - At the point (0,1), a tangent line is drawn, with a moderate positive slope. - This point corresponds to the corner at which the curve begins to become horizontal. - At this point, the curve passes the vertical line test, but does not at most other points on the graph. - The second curve begins to the right of the y-axis, as a line stretching upwards from the bottom of the y-axis. - As x approaches 1, the curve also begins to become horizontal as x increases. - The entire second curve lies in the fourth quadrant. - - - - \begin{tikzpicture} - \begin{axis}[xmin=-1.5,xmax=10.5, - ymin=-10.5,ymax=2.49,] - - \addplot[firstcurvestyle,infinite,smooth] coordinates {(-0.324,-9.34) (-0.336,-8.66) (-0.351,-7.95) (-0.371,-7.13) (-0.396,-6.3) (-0.414,-5.68) (-0.451,-4.91) (-0.472,-4.48)(-0.497,-4.02) (-0.525,-3.57) (-0.56,-3.15) (-0.61,-2.63)(-0.664,-2.18) (-0.727,-1.71) (-0.762,-1.29) (-0.668,-0.805)(-0.513,-0.29) (-0.453,0.439) (-0.308,0.71) (0.0816,1.05)(0.452,1.15) (0.893,1.13) (1.33,1.08) (1.79,1.02) (2.24,0.975)(2.82,0.925) (3.4,0.881) (4.11,0.844) (4.91,0.781) (5.77,0.766)(6.76,0.74) (7.59,0.722) (8.26,0.706) (8.93,0.689) (9.69,0.671)}; - - \addplot[firstcurvestyle,infinite,smooth] coordinates {(9.6,-0.677) (9.13,-0.696) (8.66,-0.697) (8.21,-0.707) (7.72,-0.722)(7.3,-0.738) (6.79,-0.754) (5.98,-0.778) (5.1,-0.81) (4.34,-0.858)(3.84,-0.893) (3.4,-0.927) (2.77,-0.99) (2.27,-1.07) (1.82,-1.17)(1.38,-1.34) (0.97,-1.7) (0.779,-2.1) (0.678,-2.5) (0.593,-3.)(0.534,-3.66) (0.483,-4.38) (0.444,-5.04) (0.415,-5.71) (0.393,-6.38)(0.375,-7.05) (0.362,-7.59) (0.343,-8.38) (0.334,-8.9) (0.322,-9.33)(0.317,-9.83) }; - - \addplot [tangentline,domain=-1:2] {2/3*(x)+1}; - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Notice how our curve looks much different than for functions we have seen. - For one, it fails the vertical line test, - and so the complete curve is not truly representing y as a function of x. - But when we indicate we are interested in the derivative at (0,1), - we are indicating that we want the function defined by the small portion of the curve that passes through (0,1), - and that small portion does pass the vertical line test. - Such functions are important in many areas of mathematics, - so developing tools to deal with them is also important. -

    -     - -     - - - Using Implicit Differentiation - -

    - Given the implicitly defined function \sin\mathopen{}\left(x^2y^2\right)\mathclose{}+y^3=x+y, - find \yp. -

    -     - -

    - Differentiating term by term, - we find the most difficulty in the first term. - It requires both the and . - - \lzoo{x}{\sin\mathopen{}\left(x^2y^2\right)\mathclose{}} \amp = \cos\mathopen{}\left(x^2y^2\right)\mathclose{}\cdot\lzoo{x}{x^2y^2} - \amp = \cos\mathopen{}\left(x^2y^2\right)\mathclose{}\cdot\left(x^2(2y\yp)+2xy^2\right) - \amp = 2\left(x^2y\yp+xy^2\right)\cos\mathopen{}\left(x^2y^2\right)\mathclose{} - . -

    - -

    - We leave the derivatives of the other terms to the reader. - After taking the derivatives of both sides, we have - - 2\left(x^2y\yp+xy^2\right)\cos\mathopen{}\left(x^2y^2\right)\mathclose{} + 3y^2\yp = 1 + \yp - . -

    - -

    - We now have to be careful to properly solve for \yp, - particularly because of the product on the left. - It is best to multiply out the product. - Doing this, we get - - 2x^2y\cos\mathopen{}\left(x^2y^2\right)\mathclose{}\yp + 2xy^2\cos\mathopen{}\left(x^2y^2\right)\mathclose{} + 3y^2\yp = 1 + \yp - . -

    - -

    - From here we can safely move around terms to get the following: - - 2x^2y\cos\mathopen{}\left(x^2y^2\right)\mathclose{}\yp + 3y^2\yp - \yp = 1 - 2xy^2\cos\mathopen{}\left(x^2y^2\right)\mathclose{} - . -

    - -

    - Then we can solve for \yp to get - - \yp = \frac{1 - 2xy^2\cos\mathopen{}\left(x^2y^2\right)\mathclose{}}{2x^2y\cos\mathopen{}\left(x^2y^2\right)\mathclose{}+3y^2-1} - . -

    - -

    - A graph of this implicit function is given in . -

    - -
    - A graph of the implicitly defined curve \sin\mathopen{}\left(x^2y^2\right)\mathclose{}+y^3=x+y - - - - A curve beginning in the third quadrant passing through the points (0,-1), (0,0), (0,1). - - -

    - The curve begins in the third quadrant. - From there, the curve bends slightly back and increases, crossing above itself. - The curve extends to the right, increasing almost linearly as it crosses the y-axis at y = -1 into the fourth quadrant. - The curve continues to increase as such until it reaches a point close to (\frac{1}{2},-\frac{3}{4}. - The curve then bends back, increasing towards the top left linearly. - It then crosses the origin and passes into the second quadrant. - The curve quickly bends towards the right, crossing the y-axis at y = 1 into the first quadrant. - From there, the curve continues towards the right while slightly increasing. - The curves rises sharply at x = 1.5, before decreasing again. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-2.05,xmax=2.05, - ymin=-1.95,ymax=1.95,] - \addplot+[smooth,infinite] coordinates {(-2.03,-1.513)(-1.988,-1.536)(-1.972,-1.543)(-1.879,-1.586)(-1.786,-1.621)(-1.68,-1.572)(-1.707,-1.464)(-1.749,-1.357)(-1.739,-1.261)(-1.621,-1.264)(-1.5,-1.299)(-1.348,-1.357)(-1.179,-1.428)(-1.074,-1.467)(-0.9643,-1.494)(-0.8524,-1.495)(-0.6789,-1.429)(-0.5,-1.303)(-0.3182,-1.179)(-0.1408,-1.073)(0,-1.)(0.1397,-0.9317)(0.2397,-0.8826)(0.3496,-0.8214)(0.4457,-0.7329)(0.4619,-0.6071)(0.4189,-0.5)(0.276,-0.2954)(0.07143,-0.07178)(-0.07143,0.07183)(-0.2684,0.3031)(-0.3426,0.4855)(-0.3214,0.6721)(-0.2591,0.7857)(-0.1429,0.9072)(0,1.)(0.1009,1.042)(0.2857,1.083)(0.5178,1.089)(0.7793,1.065)(0.9789,1.05)(1.179,1.08)(1.248,1.145)(1.275,1.275)(1.279,1.393)(1.3,1.514)(1.393,1.582)(1.5,1.558)(1.607,1.514)(1.712,1.467)(1.804,1.426)(1.856,1.405)}; - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - It is easy to verify that the points (0,0), - (0,1) and (0,-1) all lie on the graph. - We can find the slopes of the tangent lines at each of these points using our formula for \yp. - -

      -
    • At (0,0), the slope is -1.
      • - -
    • At (0,1), the slope is 1/2.
      • - -
    • At (0,-1), the slope is also 1/2.
      • -
    - - The tangent lines have been added to the graph of the function in . -

    - -
    - A graph of the implicitly defined curve \sin\mathopen{}\left(x^2y^2\right)\mathclose{}+y^3=x+y and certain tangent lines - - - - A curve beginning in the third quadrant passing through the points (0,-1), (0,0), (0,1), with tangent lines at those points. - - -

    - The graph in , with tagent lines drawn at (0,-1), (0,0), and (0,1). - The tangent line at (0,-1) has a positive slope less than 1. - The tangent line at (0,0) has a negative slope, close to -1. - The tangent line at (0,1) has a positive slope, less than 1. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-2.05,xmax=2.05, - ymin=-1.95,ymax=1.95,] - \addplot+[infinite,smooth] coordinates {(-2.03,-1.513)(-1.988,-1.536)(-1.972,-1.543)(-1.879,-1.586)(-1.786,-1.621)(-1.68,-1.572)(-1.707,-1.464)(-1.749,-1.357)(-1.739,-1.261)(-1.621,-1.264)(-1.5,-1.299)(-1.348,-1.357)(-1.179,-1.428)(-1.074,-1.467)(-0.9643,-1.494)(-0.8524,-1.495)(-0.6789,-1.429)(-0.5,-1.303)(-0.3182,-1.179)(-0.1408,-1.073)(0,-1.)(0.1397,-0.9317)(0.2397,-0.8826)(0.3496,-0.8214)(0.4457,-0.7329)(0.4619,-0.6071)(0.4189,-0.5)(0.276,-0.2954)(0.07143,-0.07178)(-0.07143,0.07183)(-0.2684,0.3031)(-0.3426,0.4855)(-0.3214,0.6721)(-0.2591,0.7857)(-0.1429,0.9072)(0,1.)(0.1009,1.042)(0.2857,1.083)(0.5178,1.089)(0.7793,1.065)(0.9789,1.05)(1.179,1.08)(1.248,1.145)(1.275,1.275)(1.279,1.393)(1.3,1.514)(1.393,1.582)(1.5,1.558)(1.607,1.514)(1.712,1.467)(1.804,1.426)(1.856,1.405)}; - - \addplot [tangentline,domain=-.5:.5] {-x}; - \addplot [tangentline,domain=-.5:.5] {0.5*x+1}; - \addplot [tangentline,domain=-.5:.5] {0.5*(x-0)-1}; - \end{axis} - \end{tikzpicture} - - - - -
    -     - -     - -

    - Quite a few famous curves have equations that are given implicitly. - We can use implicit differentiation to find the slope at various points on those curves. - We investigate two such curves in the next examples. -

    - - - Finding slopes of tangent lines to a circle - -

    - Find the slope of the tangent line to the circle - x^2+y^2=1 at the point \left(1/2, \sqrt{3}/2\right). -

    -     - -

    - Taking derivatives, we get 2x+2y\yp=0. - Solving for \yp gives: - - \yp = \frac{-x}{y} - . -

    - -

    - This is a clever formula. - Recall that the slope of the line through the origin and the point (x,y) on the circle will be y/x. - We have found that the slope of the tangent line to the circle at that point is the opposite reciprocal of y/x, namely, - -x/y. - Hence these two lines are always perpendicular. -

    - -

    - At the point \left(1/2, \sqrt{3}/2\right), - we have the tangent line's slope as - - \yp = \frac{-1/2}{\sqrt{3}/2} = \frac{-1}{\sqrt{3}} \approx -0.577 - . -

    - -

    - A graph of the circle and its tangent line at - \left(1/2,\sqrt{3}/2\right) is given in , - along with a thin dashed line from the origin that is perpendicular to the tangent line. - (It turns out that all normal lines to a circle pass through the center of the circle.) -

    - -
    - The unit circle with its tangent line at (1/2,\sqrt{3}/2) - - - - A circle of radius 1 centered at the origin with a tangent line drawn at a point in the first quadrant. - - -

    - A circle of radius 1 centered at the origin. - A dashed line extends from the origin to a tangent line at the point (\frac{1}{2},\frac{\sqrt{3}}{2}). - At that point a tangent line is drawn with a slight negative slope. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-1.2,xmax=1.2, - ymin=-1.2,ymax=1.2, - axis equal] - \addplot+[domain=0:pi,samples=200] ({cos(deg(2*x))},{sin(deg(2*x))}); - \addplot [tangentline,domain=.1:.9] {-.577*(x-.5)+.866}; - \addplot [soliddot] coordinates {(0.5,.866)} node[above right] {$\left(1/2,\sqrt{3}/2\right)$}; - \addplot [normallineseg,domain=0:0.5]{1/0.577*x}; - \end{axis} - \end{tikzpicture} - - - - -
    -     -     - -

    - This section has shown how to find the derivatives of implicitly defined functions, - whose graphs include a wide variety of interesting and unusual shapes. - Implicit differentiation can also be used to further our understanding of - regular differentiation. -

    - -

    - One hole in our current understanding of derivatives is this: - what is the derivative of the square root function? - That is, - - \lzoo{x}{\sqrt{x}} = \lzoo{x}{x^{1/2}} = \text{?} - -

    - -

    - We allude to a possible solution, - as we can write the square root function as a power function with a rational (or, - fractional) power. - We are then tempted to apply the and obtain - - \lzoo{x}{x^{1/2}} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} - . -

    - -

    - The trouble with this is that the was initially defined only for positive integer powers, - n \gt 0. - While we did not justify this at the time, - generally the is proved using something called the Binomial Theorem, - which deals only with positive integers. - The allowed us to extend the to negative integer powers. - Implicit Differentiation allows us to extend the to rational powers, - as shown below. -

    - -

    - Let y = x^{m/n}, - where m and n are integers with no common factors - (so m=2 and n=5 is fine, - but m=2 and n=4 is not). - We can rewrite this explicit function implicitly as y^n = x^m. - Now apply implicit differentiation. - - y \amp = x^{m/n} - y^n \amp = x^m - \lzoo{x}{y^n} \amp = \lzoo{x}{x^m} - n\cdot y^{n-1}\cdot \yp \amp = m\cdot x^{m-1} - \yp\amp = \frac{m}{n} \frac{x^{m-1}}{y^{n-1}}\amp\amp\text{(now substitute }x^{m/n}\text{ for }y\text{)} - \amp = \frac{m}{n} \frac{x^{m-1}}{(x^{m/n})^{n-1}}\amp\amp\text{(apply lots of algebra)} - \amp = \frac{m}n x^{(m-n)/n} - \amp = \frac{m}n x^{m/n -1} - . -

    - -

    - The above derivation is the key to the proof extending the to rational powers. - Using limits, - we can extend this once more to include all - powers, including irrational - (even transcendental!) - powers, giving the following theorem. -

    - - - Power Rule for Differentiation - -

    - Let f(x) = x^n, where n\neq 0 is a real number. - Then f is differentiable on its domain, - except possibly at x=0, - and \fp(x) = n\cdot x^{n-1}. - derivativePower Rule - Power Ruledifferentiation -

    -     -     - -

    - This theorem allows us to say the derivative of x^\pi is \pi x^{\pi -1}. -

    - -

    - We now apply this final version of the in the next example, - the second investigation of a famous curve. -

    - - - - - Using the Power Rule - -

    - Find the slope of x^{2/3}+y^{2/3}=8 at the point (8,8). -

    -     - -

    - This is a particularly interesting curve called an astroid. - It is the shape traced out by a point on the edge of a circle that is rolling around inside of a larger circle, - as shown in . -

    - -
    - An astroid, traced out by a point on the smaller circle as it rolls inside the larger circle - - - - A four pointed star with rounded edges, surrounded by a dashed circle. - - -

    - A dashed circle of radius 20 entirely contains the curve. - In each quadrant curves connect the points on the x and y axis which also lie on the circle. - This gives the overall curve the appearence of a diamond with sides curved towards the inside. - In the third quadrant a smaller circle is drawn which touches both the outer circle and the curve. - The point on the circle touching the curve is highlighted blue. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-24,xmax=24, - ymin=-24,ymax=24, - axis equal] - \addplot+[domain=0:pi,samples=200] ({22.63*cos(deg(2*x))^3},{22.63*sin(deg(2*x))^3}); - \addplot+[domain=0:pi,samples=200] ({22.63*cos(deg(2*x))},{22.63*sin(deg(2*x))}); - \addplot+[domain=0:pi,samples=101] ({5.66*cos(deg(2*x))-12},{5.66*sin(deg(2*x))-12}); - \addplot [soliddot] coordinates {(-8,-8)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -

    - To find the slope of the astroid at the point (8,8), - we take the derivative implicitly. - - \frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}\yp\amp =0 - \frac{2}{3}y^{-1/3}\yp \amp = -\frac{2}{3}x^{-1/3} - \yp\amp =-\frac{x^{-1/3}}{y^{-1/3}} - \yp\amp =-\frac{y^{1/3}}{x^{1/3}} = -\sqrt[3]{\frac{y}{x}} - . -

    - -

    - Plugging in x=8 and y=8, - we get a slope of -1. - The astroid, with its tangent line at (8,8), - is shown in . -

    - -
    - An astroid with a tangent line - - - - A previously described astroid with a tangent line in the first quadrant. - - -

    - The curve sketched in with a tangent line at (8,8). - It has a slope of -1. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-24,xmax=24, - ymin=-24,ymax=24, - axis equal] - \addplot+[domain=0:pi,samples=200] ({22.63*cos(deg(2*x))^3},{22.63*sin(deg(2*x))^3}); - \addplot [tangentline,domain=1:15] {-1*(x-8)+8}; - \addplot [soliddot] coordinates {(8,8)} node[above right] {$(8,8)$}; - \end{axis} - \end{tikzpicture} - - - - -
    -     -     -     - - - Implicit Differentiation and the Second Derivative -

    - We can use implicit differentiation to find higher order derivatives. - In theory, this is simple: - first find \lz{y}{x}, then take its derivative with respect to x. - In practice, - it is not hard, but it often requires a bit of algebra. - We demonstrate this in an example. -

    - - - Finding the second derivative - -

    - Given x^2+y^2=1, find \lzn{2}{y}{x} = \yp'. -

    -     - -

    - We found that \yp = \lz{y}{x} = -x/y in . - To find \yp', we apply implicit differentiation to \yp. - - \yp' \amp = \lzoo{x}{\yp} - \amp = \lzoo{x}{-\frac{x}{y}}\amp\amp\text{(Now use the Quotient Rule.)} - \amp = -\frac{y\cdot1 - x(\yp)}{y^2}\amp\amp\text{replace }\yp\text{ with }-x/y\text{:} - \amp = -\frac{y-x(-x/y)}{y^2} - \amp = -\frac{y+x^2/y}{y^2} - . -

    - -

    - While this is not a particularly simple expression, it is usable. - We can see that \yp' \gt 0 when y\lt 0 and \yp'\lt 0 when y \gt 0. - In , - we will see how this relates to the shape of the graph. -

    - -

    - Also, if we remember that we are only considering points on the curve x^2+y^2=1, - then we know that x^2=1-y^2. - So we can replace the x^2 in the expression for \yp' to get - - \yp'=-\frac{y+\left(1-y^2\right)/y}{y^2}=-\frac{1}{y^3} - - which is a simpler expression. - Recognizing when simplifications like this are possible is not always easy. -

    -     - -     -     - - - Logarithmic Differentiation -

    - Consider the function y=x^x; - it is graphed in . - It is well-defined for x \gt 0 and we might be interested in finding equations of lines tangent and normal to its graph. - How do we take its derivative? - - logarithmic differentiation - derivativelogarithmic -

    - - - -
    - A plot of y=x^x - - - - An exponential curve with a discontinuity at (0,1). It decreases slightly before increasing. - - -

    - The curve is entirely contained within the first quadrant. - At the point (0,1) there is a discontinuity. - The curve begins decreasing, reaching a minimum at around (0.3,0.7). - After that point, the curve increases exponentially. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-0.1,xmax=2.2, - ymin=-.1,ymax=4.2] - \addplot+[domain=0.001:2,rightarrow,samples=50] {x^x}; - \addplot[hollowdot] coordinates{(0,1)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -

    - The function is not a power function: - it has a power of x, not a constant. - It is not an exponential function either: - it has a base of x, not a constant. -

    - -

    - A differentiation technique known as - logarithmic differentiation becomes useful here. - The basic principle is this: - take the natural log of both sides of an equation y=f(x), - then use implicit differentiation to find \yp. - We demonstrate this in the following example. -

    - - - Using Logarithmic Differentiation - -

    - Given y=x^x, use logarithmic differentiation to find \yp. -

    -     - -

    - As suggested above, - we start by taking the natural log of both sides then applying implicit differentiation. - - y \amp = x^x\amp - \ln(y) \amp = \ln(x^x)\amp\amp\text{(apply logarithm rule)} - \ln(y) \amp = x\ln(x) \amp\amp\text{(now use implicit differentiation)} - \lzoo{x}{\ln(y)} \amp = \lzoo{x}{x\ln(x)} - \frac{\yp}{y} \amp = \ln(x) + x\cdot\frac{1}{x}\amp - \frac{\yp}{y} \amp = \ln(x) + 1\amp - \yp \amp = y\left(\ln(x) +1\right) \amp\amp\text{(substitute }y=x^x\text{)} - \yp \amp = x^x\left(\ln(x) +1\right)\amp - . -

    - -

    - To test our answer, - let's use it to find the equation of the tangent line at x=1.5. - The point on the graph our tangent line must pass through is \left(1.5, 1.5^{1.5}\right) \approx (1.5, 1.837). - Using the equation for \yp, we find the slope as - - \yp = 1.5^{1.5}\left(\ln(1.5) +1\right) \approx 1.837(1.405) \approx 2.582 - . -

    - -

    - Thus the equation of the tangent line is (approximately) y \approx 2.582(x-1.5)+1.837. - - graphs y=x^x along with this tangent line. -

    - -
    - A graph of y=x^x and its tangent line at x=1.5 - - - - The previously described exponential curve with a tangent line in the increasing part of the curve. - - -

    - The graph shown in , with a tangent line at - (1.5,1.5^{1.5}). It has a slope of around 2.6. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-0.1,xmax=2.2, - ymin=-.1,ymax=4.2] - \addplot+[domain=0.001:2,rightarrow,samples=50] {x^x}; - \addplot[hollowdot] coordinates{(0,1)}; - \addplot[tangentline,domain=1:2] {2.582 * (x - 1.5) + 1.837}; - \addplot[soliddot] coordinates{(1.5,1.837)} node[below right] {$\left(1.5,1.5^{1.5}\right)$}; - \end{axis} - \end{tikzpicture} - - - - -
    -     - -     - - - - - -

    - Implicit differentiation proves to be useful as it allows us to find the instantaneous rates of change of a variety of functions. - In particular, - it extended the to rational exponents, - which we then extended to all real numbers. - In , - implicit differentiation will be used to find the derivatives of - inverse functions, - such as y=\sin^{-1}(x). -

    -     - - - - Terms and Concepts - - - - -

    - In your own words, - explain the difference between implicit functions and explicit functions. -

    - -     - - - -     - - - - -

    - Implicit differentiation is based on what other differentiation rule? -

    -

    - - -

    -     - - - - - chain|chain rule|the chain rule - - - - - -     - - - - -

    - - Implicit differentiation can be used to find the derivative of y=\sqrt{x}. -

    -     - -

    - The function \sqrt{x} is one of two solutions to the equation y^2=x. -

    -     - -     - - - - -

    - - Implicit differentiation can be used to find the derivative of y=x^{3/4}. -

    -     - -

    - x^{3/4} is one of two (real) solutions to y^4=x^3. -

    -     - -     -     - - - Problems - - -

    - Compute the derivative of the given function. -

    -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $pm = list_random('+','-'); - if($envir{problemSeed}==1){$fname='f';$x='x';$pm='+'}; - Context()->variables->are($x=>'Real'); - Context()->variables->set($x=>{limits=>[0,4]}); - $f = Formula("sqrt($x) $pm 1/sqrt($x)")->reduce; - $fp = $f->D($x); - - -

    - ()= -

    - - To enter \sqrt{x}, type sqrt(x). - -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $pm = list_random('+','-'); - $n = random(3,6,1); - if($envir{problemSeed}==1){$fname='f';$x='x';$pm='+';$n=3;}; - Context("Fraction"); - parser::Root->Enable; - Context()->variables->are($x=>'Real'); - Context()->variables->set($x=>{limits=>[0,4]}); - $f = Formula("root($n,$x) $pm $x^(($n-1)/$n)"); - $fp = $f->D($x); - - -

    - ()= -

    - - To enter \sqrt{x}, type sqrt(x). - To enter \sqrt[n]{x}, type root(n,x). - -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $pm = list_random('+','-'); - $a = random(1,4,1)**2; - $n = random(2,4,1); - if($envir{problemSeed}==1){$fname='f';$x='t';$pm='-';$a=1;$n=2;}; - Context()->variables->are($x=>'Real'); - Context()->variables->set($x=>{limits=>[-$a**(1/$n),$a**(1/$n)]}); - $f = Formula("sqrt($a $pm $x^$n)"); - $fp = $f->D($x); - - -

    - ()= -

    - - To enter \sqrt{x}, type sqrt(x). - -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $trig = list_random('sin','cos','tan'); - if($envir{problemSeed}==1){$fname='g';$x='t';$trig='sin';}; - Context()->variables->are($x=>'Real'); - Context()->variables->set($x=>{limits=>[0,pi/2]}); - $f = Formula("sqrt($x)$trig($x)"); - $fp = $f->D($x); - - -

    - ()= -

    - - To enter \sqrt{x}, type sqrt(x). - -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $n = random(1,3,1) + random(0.1,0.9,0.1); - $trig = list_random('sin','cos','tan'); - if($envir{problemSeed}==1){$fname='h';$x='x';$n=1.5;}; - Context()->variables->are($x=>'Real'); - Context()->variables->set($x=>{limits=>[0,4]}); - $f = Formula("$x^$n"); - $fp = $f->D($x); - - -

    - ()= -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $n = random(1,3,1) + random(0.1,0.9,0.1); - $b = list_random(Formula("pi"),Formula("e")); - if($envir{problemSeed}==1){$fname='f';$x='x';$n=1.9;$b=Formula("pi")}; - Context()->variables->are($x=>'Real'); - Context()->variables->set($x=>{limits=>[0,4]}); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $f = Formula("$x^$b + $x^$n + $b^$n"); - $fp = $f->D($x); - - -

    - ()= -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $a = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$fname='g';$x='x';$a=7;}; - Context()->variables->are($x=>'Real'); - Context()->variables->set($x=>{limits=>[0,4]}); - $f = Formula("($x+$a)/sqrt($x)"); - $fp = $f->D($x); - - -

    - ()= -

    - - To enter \sqrt{x}, type sqrt(x). - -

    - -

    -     -     -     - - - - - Context("Fraction"); - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $n = random(3,9,1); - $trig = list_random('sin','cos','tan','sec','cot','csc'); - if($envir{problemSeed}==1){$fname='f';$x='t';$n=5;$trig='sec';}; - Context()->variables->are($x=>'Real'); - Context()->variables->set($x=>{limits=>[0,4]}); - parser::Root->Enable; - $f = Formula("root($n,$x)($trig($x)+e^$x)"); - $fp = $f->D($x); - - -

    - ()= -

    - - To enter \sqrt{x}, type sqrt(x). - To enter \sqrt[n]{x}, type root(n,x). - -

    - -

    -     -     -     -     - - - -

    - Find \lz{y}{x} using implicit differentiation. -

    -     - - - - Context()->variables->add(y=>'Real'); - $dydx=Formula("-4x^3/(2y+1)"); - @y=(random(-2,2,0.2),random(-2,2,0.2),random(-2,2,0.2),random(-2,2,0.2),random(-2,2,0.2)); - @x=map{(7-($_)-($_)**2)**(1/4)*(-1)**(random(-1,1,2))}(@y); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - x^4+y^2+y=7 -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $dydx=Formula("-y^(3/5)/x^(3/5)"); - @y=(random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01)); - @x=map{(1-($_)**(2/5))**(5/2)}(@y); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - x^{2/5}+y^{2/5}=1 -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $dydx=Formula("sin(x)sec(y)"); - @y=(random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01)); - @x=map{acos(1-sin($_))}(@y); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - \cos(x)+\sin(y)=1 -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $dydx=Formula("y/x"); - @x=(random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1)); - @y=map{$_/10}(@x); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - \dfrac{x}{y}=10 -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $dydx=Formula("y/x"); - $a = random(5,15,1); - if($envir{problemSeed}==1){$a=10;}; - #$a=10; - @y=(random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1),random(-4.05,4.05,0.1)); - @x=map{$_/$a}(@y); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - \dfrac{y}{x}= -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - Context()->flags->set(reduceConstantFunctions=>0); - $m=list_random(2,3,5); - $b=random(2,9,1); - $c=random(4,9,1); - if($envir{problemSeed}==1){$m=2;$b=2;$c=5;}; - #$m=2;$b=2;$c=5; - $dydx=Formula("-(e^x x(x+$m) $b^(-y))/(ln($b))"); - @x=(random(-4,1,0.1),random(-4,1,0.1),random(-4,1,0.1),random(-4,1,0.1),random(-4,1,0.1)); - @y=map{ln($c-($_)**$m*exp($_))/ln($b)}(@x); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - x^e^x+^y= -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $m = random(2,9,1); - $trig = list_random('sin','cos','tan','cot'); - $c = random(1,9,1); - if($envir{problemSeed}==1){$m=2;$trig='tan';$c=50;}; - #$m=2;$trig='tan';$c=50; - if ($trig eq 'sin') {$dydx=Formula("-$m tan(y)/x");}; - if ($trig eq 'cos') {$dydx=Formula("$m cot(y)/x");}; - if ($trig eq 'tan') {$dydx=Formula("-$m sin(y) cos(y)/x");}; - if ($trig eq 'cot') {$dydx=Formula("$m sin(y) cos(y)/x");}; - # need to avoid x==0 - @x=(random(3,7,0.2),random(3,7,0.2),random(3,7,0.2),random(3,7,0.2),random(3,7,0.2)); - if ($trig eq 'sin') {@y=map{asin($c/($_)**$m)}(@x);}; - if ($trig eq 'cos') {@y=map{acos($c/($_)**$m)}(@x);}; - if ($trig eq 'tan') {@y=map{atan($c/($_)**$m)}(@x);}; - if ($trig eq 'cot') {@y=map{pi/2 - atan($c/($_)**$m)}(@x);}; - $left = Formula("x^$m $trig(y)"); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - = -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - parser::Root->Enable; - Context()->variables->add(y=>'Real'); - ($a,$b) = random_subset(2,2..5); - $c = random(1,9,1); - ($m,$n,$p) = random_subset(3,2..5); - if($envir{problemSeed}==1){$a=3;$b=2;$c=2;$m=2;$n=3;$p=4;}; - #$a=3;$b=2;$c=2;$m=2;$n=3;$p=4; - $w = ($c**(1/$p)/$a)**(1/$m); - $left = Formula("($a x^$m + $b y^$n)^$p"); - $right = Formula("$c"); - $frac = Fraction(-$a*$m,$b*$n); - $dydx=Formula("$frac x^($m-1) / y^($n-1)")->reduce; - do { - @x=(random(-$w,$w,0.01),random(-$w,$w,0.01),random(-$w,$w,0.01),random(-$w,$w,0.01),random(-$w,$w,0.01)); - if ($n % 2 == 1) { - @y=map{root($n,(root($p,$c) - $a*($_)**$m)/$b)}(@x); - } else { - @y=map{random(-1,1,2)*root($n,(root($p,$c) - $a*($_)**$m)/$b)}(@x); - } - } until (!grep( /^0$/, @y )); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - = -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - parser::Root->Enable; - Context()->variables->add(y=>'Real'); - $a = non_zero_random(-9,9,1); - $b = random(100,200,5); - $pm = list_random('+','-'); - $m = random(2,4,1); - $n = random(2,4,1); - if($envir{problemSeed}==1){$a=2;$b=200;$pm='-';$m=2;$n=2;}; - #$a=2;$b=200;$pm='-';$m=2;$n=2; - $left = Formula("(y^$m+$a y $pm x)^$n"); - $right = Formula("$b"); - $dydx=Formula("-($pm 1/($m y^($m-1)+$a))")->reduce->reduce; - @y=(random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2)); - @x=map{(root($n,$b) - $_**$m - $_) * (($pm eq '+') ? 1 : -1);}(@y); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - = -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->variables->add(y=>'Real'); - ($m,$n) = random_subset(2,1..2); - $c = random(10,20,1); - if($envir{problemSeed}==1){$m=2;$n=1;$c=17}; - #$m=2;$n=1;$c=17; - $left = Formula("(x^$m + y^$n)/(x^$n + y^$m)")->reduce; - $right = Formula("$c"); - $dydx = Formula("(($n-$m) x^2 + $n x^($n-1) y^$n - $m x^($m-1) y^$m)/($n x^$n y^($n-1) + ($n-$m) y^2 - $m x^$m y^($m-1))")->reduce; - @y=(random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2),random(-3.9,1.9,0.2)); - if ($m == 2) { - @x=map{($c+(-1)**(random(-1,1,2))*sqrt($c**2 - 4*($_)*(1 - $c*($_))))/2}(@y); - } else { - @x=map{(1+(-1)**(random(-1,1,2))*sqrt(1 - 4*$c*($_)*($c - ($_))))/(2*$c)}(@y); - } - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - = -

    -

    - -

    -

    - -

    -     - -

    - If one takes the derivative of the equation, - as shown, using the Quotient Rule, - one finds \frac{dy}{dx}=\frac{x^2+2 x y^2-y}{2 x^2 y-x+y^2}. -

    -

    - If one first clears the denominator and writes - x^2+y=17(x+y^2) then takes the derivative of both sides, - one finds \frac{dy}{dx}= \frac{2x-17}{34y-1}. -

    -

    - These expressions, by themselves, are not equal. - However, for values of x and y that satisfy the original equation - (i.e, for x and y such that \frac{x^2+y}{x+y^2}=17), - these expressions are equal. -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - ($triga,$trigb) = random_subset(2,'sin','cos'); - if($envir{problemSeed}==1){$triga='sin';$trigb='cos';}; - #$triga='sin';$trigb='cos'; - $left = Formula("($triga(x)+y)/($trigb(y)+x)"); - $right = Formula("1"); - if ($triga eq 'sin') { - $dydx = Formula("(1-cos(x))/(sin(y)+1)")->reduce; - $dydx->{test_points}=[[-3.1,-3.83041],[-0.41,0.732269],[0.59,0.759097],[2.6,1.82908],[5.3,6.93017]]; - } else { - $dydx = Formula("(sin(x)+1)/(1-cos(y))")->reduce; - $dydx->{test_points}=[[0.831958,1],[1.329583,2],[-0.0952339,-2],[-3.70455,-3],[0.7390851,0]]; - } - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - = -

    -

    - -

    -

    - -

    -     - -

    - If one takes the derivative of the equation, - as shown, using the Quotient Rule, - one finds \frac{dy}{dx} = \frac{-\cos (x) (x+\cos (y))+\sin (x)+y}{\sin (y) (\sin (x)+y)+x+\cos (y)}. -

    -

    - If one first clears the denominator and writes - \sin(x)+y = \cos(y)+x then takes the derivative of both sides, - one finds \frac{dy}{dx} = \frac{1-\cos(x)}{1+\sin(y)}. -

    -

    - These expressions, by themselves, are not equal. - However, for values of x and y that satisfy the original equation - (i.e, for x and y such that \frac{\sin(x)+y}{\cos(y)+x}=1), - these expressions are equal. -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $dydx=Formula("-x/y"); - @t=(random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01)); - @x=map{exp(exp(1)/2)*cos($_)}(@t); - @y=map{exp(exp(1)/2)*sin($_)}(@t); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - \ln\mathopen{}\left(x^2+y^2\right)\mathclose{}=e -

    -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $dydx=Formula("-(2x+y)/(2y+x)"); - @t=(random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01),random(0,6.28,0.01)); - @u=map{sqrt(2/3*exp(1))*cos($_)}(@t); - @v=map{sqrt(2*exp(1))*sin($_)}(@t); - @x=map{($u[$_]-$v[$_])/sqrt(2)}(0..4); - @y=map{($u[$_]+$v[$_])/sqrt(2)}(0..4); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $dydx->{test_points}=~~@xy; - $showwork = '[@ explanation_box(message => "Show your work using implicit differentiation.") @]*'; - - -

    - \ln\mathopen{}\left(x^2+xy+y^2\right)\mathclose{} = 1 -

    -

    - -

    -

    - -

    -     -     -     -     - - - -

    - Show that \lz{y}{x} is the same for each of the following implicitly defined functions. -

    -     - - - -

    - xy=1 -

    -     -     - - -

    - x^2y^2=1 -

    -     -     - - -

    - \sin(xy) = 1 -

    -     -     - - -

    - \ln(xy) =1 -

    -     -     -     - - - - -

    - Find the equation of the tangent line to the graph of the implicitly defined function at the indicated points. - As a visual aid, the function is graphed. -

    -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - @eq=(Formula("y=0"),Formula("y=-1.859(x-0.1)+0.2811")); - - -

    - On the curve x^{2/5}+y^{2/5} = 1. -

    - - - An astroid of radius 1 with a point drawn at (0.1,0.281). - - -

    - An astroid of radius 1. - A point is drawn on the curve, at (0.1,0.281). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.1,ymax=1.1,% - xmin=-1.1,xmax=1.1,% - ] - - \addplot [firstcurvestyle,domain=0:360,samples=101] ({(cos(x))^5},{(sin(x))^5}); - \filldraw [] (axis cs:.1,.28) node [above right] {\((0.1,0.281)\)} circle (1pt); - \end{axis} - \end{tikzpicture} - - -     - - -

    - At (1,0). -

    -

    - -

    -     -     - - -

    - At (0.1,0.2811) (which does not exactly - lie on the curve, but is very close). -

    -

    - -

    -     -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - @eq=(Formula("x=1"),Formula("y=-3sqrt(3)/8(x-sqrt(0.6))+sqrt(0.8)"),Formula("y=1")); - - -

    - On the curve x^{4}+y^{4} = 1. -

    - - - A square with rounded corners and edges with a point in the first quadrant. - - - A curve that lies in all 4 quadrants. - It has the appearance of a square with rounded sides and corners. - A point is drawn at (\sqrt{0.6},\sqrt{0.8}). - - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.1,ymax=1.1,% - xmin=-1.1,xmax=1.1,% - ] - - \addplot [firstcurvestyle] coordinates {(1.,0) (0.98901,0.45597) (0.9558,0.63776) (0.89945,0.76667) (0.818,0.86206) (0.70711,0.9306) (0.55589,0.97522) (0.32331,0.99726) (-0.32331,0.99726) (-0.55589,0.97522) (-0.70711,0.9306) (-0.818,0.86206) (-0.89945,0.76667) (-0.9558,0.63776) (-0.98901,0.45597) (-1.,0) (-1.,0) (-0.98901,-0.45597) (-0.9558,-0.63776) (-0.89945,-0.76667) (-0.818,-0.86206)(-0.70711,-0.9306) (-0.55589,-0.97522) (-0.32331,-0.99726)(0.32331,-0.99726) (0.55589,-0.97522) (0.70711,-0.9306)(0.818,-0.86206) (0.89945,-0.76667) (0.9558,-0.63776)(0.98901,-0.45597) (1.,0) }; - \filldraw [] (axis cs:.775,.894) node [below left] {\((\sqrt{0.6},\sqrt{0.8})\)} circle (1pt); - - \end{axis} - \end{tikzpicture} - - -     - - -

    - At (1,0). -

    -

    - -

    -     -     - - -

    - At \left(\sqrt{0.6},\sqrt{0.8}\right). -

    -

    - -

    -     -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - @eq=(Formula("y=4"),Formula("y=3/108^(1/4)(x-2)-108^(1/4)")); - - -

    - On the curve (x^2+y^2-4)^3 = 108y^2. -

    - - - A combination of 2 circles, with cusps at (-2,0) and (2,0) with a tangent line in the fourth quadrant. - - -

    - A curve which seems like two circles combined. - Beginning at the top of the curve, at the point (0,4), the curve decreases towards the right. - It continues to decreases, bending towards the y-axis as the curve nears the x-axis. - The curve passes through the x-axis at a point close to (2.1,0), forming a cusp at that point. - In the fourth quadrant, the curve bends outwards slightly, before curving wide towards the y-axis. - The curve passes through the y-axis at the point (0,-4). - The left side of the curve is symmetrical to the right, again curving inwards and forming a cusp at (-2,0). - From there the curve bends out, before curving in and passing through the point (0,4). - A point is drawn in the lower left of the curve at (2,-\sqrt[4]{108}). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-4.5,ymax=4.5,% - xmin=-4.5,xmax=4.5,% - ] - - \addplot [firstcurvestyle] coordinates {(-0.22282,-3.9915) (-0.57143,-3.9441) (-1.,-3.8244) (-1.2959,-3.7041)(-1.5714,-3.551) (-1.8453,-3.3571) (-2.0412,-3.1841) (-2.2149,-3.)(-2.3706,-2.7991) (-2.4727,-2.6429) (-2.5832,-2.4403)(-2.7019,-2.1552) (-2.783,-1.8571) (-2.8275,-1.4582)(-2.8002,-1.0859) (-2.7143,-0.74808) (-2.5958,-0.5)(-2.4708,-0.31491) (-2.3571,-0.19279) (-2.2143,-0.081938)(-2.0659,-0.005515) (-2.2078,0.077938) (-2.4036,0.23922) - (-2.5608,0.4392) (-2.6877,0.68771) (-2.786,1.) (-2.8262,1.3571)(-2.8065,1.7143) (-2.7075,2.136) (-2.5114,2.5714) (-2.2471,2.9614)(-2.0396,3.1825) (-1.6513,3.5) (-1.2857,3.7088) (-1.0714,3.8007)(-0.78571,3.8938) (-0.52476,3.9533) (-0.28571,3.9855) - (-0.070563,3.9991) (0.27289,3.9872) (0.64411,3.9298) (1.082,3.7963)(1.4286,3.6354) (1.7143,3.4554) (1.9675,3.2532) (2.1511,3.0714)(2.3284,2.8571) (2.4288,2.7143) (2.5351,2.5351) (2.6283,2.3426)(2.7079,2.1365) (2.7939,1.7939) (2.8255,1.3969) (2.8017,1.0874)(2.7481,0.85714) (2.6429,0.58296) (2.552,0.42857) (2.438,0.27631) - (2.3214,0.16103) (2.1786,0.06028) (2.1143,-0.028574)(2.2857,-0.13046) (2.4439,-0.28571) (2.5958,-0.5) (2.7143,-0.74808)(2.8002,-1.0859) (2.827,-1.4286) (2.7984,-1.773) (2.7079,-2.1365)(2.6052,-2.3948) (2.4292,-2.7137) (2.2453,-2.9596) (2.,-3.2231)(1.7143,-3.4554) (1.4682,-3.611) (1.1839,-3.7553) (0.86968,-3.8697) - (0.57143,-3.9441) (0.33765,-3.9805) (0.072336,-3.9991) (-0.22282,-3.9915)}; - - \filldraw [] (axis cs:2,-3.22) node [shift={(15pt,-10pt)}] {\((2,-\sqrt[4]{108})\)} circle (1pt); - - \end{axis} - \end{tikzpicture} - - -     - - -

    - At (0,4). -

    -

    - -

    -     -     - - -

    - At \left(2,-\sqrt[4]{108}\right). -

    -

    - -

    -     -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - @eq=(Formula("y=-x+1"),Formula("y=3sqrt(3)/4")); - - -

    - On the curve (x^2+y^2+x)^2 = x^2+y^2. -

    - - - An oval with a cusp on the right side. - - - An oval with a cusp on the right side. - A majority of the curve lies to the left of the y-axis. - From the top of the curve, the curve decreases towards the right. - It enters the first quadrant through the point (0,1). - As the curve nears the x-axis, it bends back toward the y-axis, forming a cusp at the origin. - The curve then bends outwards into the fourth quadrant. - The curve continues downwards and to the left, passing the y-axis through the point (0,-1). - In the third quadrant the curve bends upwards, passing vertically into the second quadrant through the point (-2,0). - The curve bends upwards and to the right, once again meeting the top of the curve. - The point at the top of the curve is drawn at (-\frac{3}{4},\frac{3\sqrt{3}}{4}). - - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.5,ymax=1.5,% - xmin=-2.5,xmax=0.5,% - ] - - \addplot [firstcurvestyle,samples=150] coordinates {(-2.,0) (-1.9547,-0.34466) (-1.8227,-0.66341) (-1.616,-0.93301)(-1.3529,-1.1352) (-1.056,-1.2584) (-0.75,-1.299) (-0.459,-1.2611)(-0.2038,-1.1558) (0,-1.) (0.14349,-0.8138) (0.22504,-0.6183)(0.25,-0.43301) (0.22961,-0.27364) (0.17922,-0.15038)(0.11603,-0.066987) (0.05667,-0.020626) (0.014961,-0.0026381) (0,0) - (0.014961,0.0026381) (0.05667,0.020626) (0.11603,0.066987)(0.17922,0.15038) (0.22961,0.27364) (0.25,0.43301) (0.22504,0.6183)(0.14349,0.8138) (0,1.) (-0.2038,1.1558) (-0.459,1.2611)(-0.75,1.299) (-1.056,1.2584) (-1.3529,1.1352) (-1.616,0.93301)(-1.8227,0.66341) (-1.9547,0.34466) (-2.,0)}; - - \filldraw [] (axis cs:-.75,1.3) node [shift={(0pt,-12pt)}] {\(\left(-\frac{3}{4},\frac{3\sqrt{3}}{4}\right)\)} circle (1pt); - - \end{axis} - \end{tikzpicture} - - -     - - -

    - At (0,1). -

    -

    - -

    -     -     - - -

    - At \left(-\frac{3}{4}, \frac{3\sqrt{3}}{4}\right). -

    -

    - -

    -     -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - @eq=(Formula("y=-1/sqrt(3)(x-7/2)+(6+3sqrt(3))/2"),Formula("y=sqrt(3)(x-(4+3sqrt(3)))/2+3/2")); - - -

    - On the curve (x-2)^2+(y-3)^2=9. -

    - - - A circle of radius 3 centered at (2,3). Two points are drawn on the circle. - - -

    - A circle of radius 3 centered at (2,3). - Two points are drawn, one at (3.5,\frac{6+3\sqrt{3}}{2}), - and the other at (\frac{4+3\sqrt{3}}{2},1.5). - The first point is in the upper right side of the circle. - The second point is in the lower right side of the circle. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.5,ymax=6.5,% - xmin=-1.5,xmax=6.5,% - ] - - \addplot [firstcurvestyle,samples=101,domain=0:360] ({(2+3*cos(x))},{(3+3*sin(x))}); - \filldraw [] (axis cs:4.6,1.5) node [left] {\(\left(\frac{4+3\sqrt{3}}{2},1.5\right)\)} circle (1pt); - \filldraw [] (axis cs:3.5,5.6) node [below left] {\(\left(3.5,\frac{6+3\sqrt{3}}{2}\right)\)} circle (1pt); - - \end{axis} - \end{tikzpicture} - - -     - - -

    - At \left(\frac{7}{2},\frac{6+3\sqrt{3}}{2}\right). -

    -

    - -

    -     -     - - -

    - At \left(\frac{4+3\sqrt{3}}{2},\frac{3}{2}\right). -

    -

    - -

    -     -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - @eq=(Formula("y=1"),Formula("y=-2/sqrt(5)(x+1)+1/2(-1+sqrt(5))"),Formula("y=2/sqrt(5)(x+1)+1/2(-1-sqrt(5))")); - - -

    - On the curve x^2+y^3+2xy=0. -

    - - - A curve that begins in the third quadrant, forms a loop in the third quadrant, and decreases in the fourth quadrant. - - -

    - The curve begins in the third quadrant. - The curve increases to the right, passing through the origin almost vertically. - From there, the curve forms a loop in the third quadrant, then decreasing and passing through the origin again. - The curve passes through the origin into the fourth qudrant, where the curve gently decreases. - Three points are drawn on the curve. - The first point is in the third quadrant, at the point (-1,\frac{-1-\sqrt{5}}{2}). - The second point is in the second quadrant, at the point (-1,1). - The third point also lies in the second quadrant, at the point (-1,\frac{-1+\sqrt{5}}{2}). - The second point lies on the top of the loop, while the third point lies on the bottom of the loop. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-2.5,ymax=2.5,% - xmin=-2.5,xmax=2.5,% - ] - \addplot [firstcurvestyle,samples=101,domain=45:157.5] ({((-sin(2*x)-(cos(x))^2)*cos(x)/((sin(x))^3))},{((-sin(2*x)-(cos(x))^2)/((sin(x))^2))}); - \filldraw [] (axis cs:-1,1) node [above] {\((-1,1)\)} circle (1pt); - \filldraw [thin,->,>=latex] (axis cs:1.25,-1.5) node [fill=white] {\(\left(-1,\frac{-1-\sqrt{5}}2\right)\)} -- (axis cs: -.9,-1.62); - \filldraw [] (axis cs: -1,-1.62) circle (1pt); - \filldraw [->,thin,>=latex] (axis cs:1.25,0.5) node [ fill=white] {\(\left(-1,\frac{-1+\sqrt{5}}2\right)\)} -- (axis cs: -.9,0.62); - \filldraw [] (axis cs:-1,0.62) circle (1pt); - - \end{axis} - \end{tikzpicture} - - -     - - -

    - At (-1,1). -

    -

    - -

    -     -     - - -

    - At \left(-1,\frac12(-1+\sqrt{5})\right). -

    -

    - -

    -     -     - - -

    - At \left(-1,\frac12(-1-\sqrt{5})\right). -

    -

    - -

    -     -     -     -     -     - - - -

    - An implicitly defined function is given. - Find \lzn{2}{y}{x}. - Note: these are the same functions used in Exercises through . -

    -     - - - - Context()->variables->add(y=>'Real'); - $ddyddx=Formula("-((2y+1)(12x^2)-4x^3(2*-(4x^3)/(2y+1)))/(2y+1)^2"); - @y=(random(-2,2,0.1),random(-2,2,0.1),random(-2,2,0.1),random(-2,2,0.1),random(-2,2,0.1)); - @x=map{(7-($_)-($_)**2)**(1/4)*(-1)**(random(-1,1,2))}(@y); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $ddyddx->{test_points}=~~@xy; - - -

    - x^4+y^2+y=7 -

    -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $ddyddx=Formula("-(x^(3/5)*3/5y^(-2/5)(-y^(3/5)/x^(3/5))-y^(3/5)*3/5x^(-2/5))/x^(6/5)"); - @y=(random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01),random(0.01,0.99,0.01)); - @x=map{(1-($_)**(2/5))**(5/2)}(@y); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $ddyddx->{test_points}=~~@xy; - - -

    - x^{2/5}+y^{2/5}=1 -

    -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $ddyddx=Formula("sin^2(x)sec^2(y)tan(y)+cos(x)sec(y)"); - @y=(random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01),random(0,3.14,0.01)); - @x=map{acos(1-sin($_))}(@y); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $ddyddx->{test_points}=~~@xy; - - -

    - \cos(x)+\sin(y)=1 -

    -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $ddyddx=Formula("0"); - @x=(random(-4,4,0.1),random(-4,4,0.1),random(-4,4,0.1),random(-4,4,0.1),random(-4,4,0.1)); - @y=map{$_/10}(@x); - @xy=map{[$x[$_],$y[$_]]}(0..4); - $ddyddx->{test_points}=~~@xy; - - -

    - \dfrac{x}{y}=10 -

    -

    - -

    -     -     -     -     - - - -

    - Use logarithmic differentiation to find \lz{y}{x}, - then find the equation of the tangent line at the indicated x-value. -

    -     - - - - Context()->variables->set(x=>{limits=>[-0.9,4]}); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $dydx=Formula("(1+x)^(1/x)(1/(x(x+1))-ln(1+x)/x^2)"); - Context("Numeric"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=(1-2ln(2))(x-1)+2"); - $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; - - -

    - y=(1+x)^{1/x} at x=1 -

    - - Enter \lz{y}{x} here. - -

    - -

    - - Enter the equation of the tangent line here. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->set(x=>{limits=>[0.01,4]}); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $dydx=Formula("(2x)^(x^2)(2x ln(2x)+x)"); - Context("Numeric"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=(2+4ln(2))(x-1)+2"); - $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; - - -

    - y=(2x)^{x^2} at x=1 -

    - - Enter \lz{y}{x} here. - -

    - -

    - - Enter the equation of the tangent line here. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->set(x=>{limits=>[0.01,4]}); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $dydx=Formula("x^x/(x+1)(ln(x)+1-1/(x+1))"); - Context("Numeric"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=1/4*(x-1)+1/2"); - $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; - - -

    - y=\dfrac{x^{x}}{x+1} at x=1 -

    - - Enter \lz{y}{x} here. - -

    - -

    - - Enter the equation of the tangent line here. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->variables->set(x=>{limits=>[0.01,4]}); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $dydx=Formula("x^(sin(x)+2)(cos(x)*ln(x)+(sin(x)+2)/x)"); - Context("Numeric"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=(3pi^2)/4*(x-pi/2)+(pi/2)^3"); - $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; - - -

    - y=x^{\sin(x)+2} at x=\pi/2 -

    - - Enter \lz{y}{x} here. - -

    - -

    - - Enter the equation of the tangent line here. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $dydx=Formula("(x+1)/(x+2)(1/(x+1)-1/(x+2))"); - Context("Numeric"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=1/9(x-1)+2/3"); - $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; - - -

    - y=\dfrac{x+1}{x+2} at x=1 -

    - - Enter \lz{y}{x} here. - -

    - -

    - - Enter the equation of the tangent line here. - -

    - -

    -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $dydx=Formula("((x+1)(x+2))/((x+3)(x+4))(1/(x+1)+1/(x+2)-1/(x+3)-1/(x+4))"); - Context("Numeric"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $t=Formula("y=11/72x+1/6"); - $showwork = '[@ explanation_box(message => "Show your work using logarithmic differentiation.") @]*'; - - -

    - y=\dfrac{(x+1)(x+2)}{(x+3)(x+4)} at x=0 -

    - - Enter \lz{y}{x} here. - -

    - -

    - - Enter the equation of the tangent line here. - -

    - -

    -

    - -

    -     -     -     -     -     -     -
    -
    - Derivatives of Inverse Functions - - - -

    - Recall that a function y=f(x) is said to be one-to-one - if it passes the horizontal line test; - that is, for two different x values x_1 and x_2, - we do not have f\mathopen{}\left(x_1\right)\mathclose{}=f\mathopen{}\left(x_2\right)\mathclose{}. - In some cases the domain of f must be restricted so that it is one-to-one. - For instance, consider f(x)=x^2. - Clearly, f(-1)= f(1), - so f is not one-to-one on its regular domain, - but by restricting f to - (0,\infty), f is one-to-one. - - derivativeinverse function -

    - -

    - Now recall that one-to-one functions have inverses. - That is, if f is one-to-one, - it has an inverse function, denoted by f^{-1}, - such that if f(a)=b, then f^{-1}(b) = a. - The domain of f^{-1} is the range of f, and vice-versa. - For ease of notation, we set - g=f^{-1} and treat g as a function of x. -

    - -

    - Since f(a)=b implies g(b)=a, - when we compose f and g we get a nice result: - - f\big(g(b)\big) = f(a) = b - . -

    - -

    - In general, f\big(g(x)\big) = x and g\big(f(x)\big) = x. - This gives us a convenient way to check if two functions are inverses of each other: - compose them and if the result is x - (on the appropriate domains), - then they are inverses. -

    - - - -

    - When the point (a,b) lies on the graph of f, - the point (b,a) lies on the graph of g. - This leads us to discover that the graph of g is the reflection of f across the line y=x. - In - we see a function graphed along with its inverse. - See how the point (1,1.5) lies on one graph, - whereas (1.5,1) lies on the other. - Because of this relationship, - whatever we know about f can quickly be transferred into knowledge about g. -

    - -
    - A function f along with its inverse f^{-1}. (Note how it does not matter which function we refer to as f; the other is f^{-1}.) - - - A function f along with its inverse. Since they are inverses of each other It doesn't matter which one we call f. - -

    - The graph of f starts at (-1,-0.5) below the negative x axis then slopes upwards while moving to the right. - A little after passing throught he point (-0.5,0.375) it slopes further and faster to the right and starts moving - horizontally parallel to the x axis. Then it starts to move upagain continuuing through (1, 1.5). The graph of g is a - similar curve starting below the x axis close to the y axis in the third quadrant it moves to the right while moving up - and then after passing through (0.375,-0.5) it moves upwards vertically for a short while and then starts moving to - the right while also moving upwards passing through (1.5,1). -

    -     - - - \begin{tikzpicture} - \begin{axis}[axis equal image, - ymin=-1.3,ymax=2, - xmin=-1.3,xmax=2, - clip=false] - \addplot+[infinite,domain=-1:1.14] ({x},{x^3+.5}); - \addplot+[infinite,domain=-1:1.14] ({x^3+.5},{x}); - \addplot [symmetryline] coordinates {(-1.3,-1.3) (1.7,1.7)}; - \addplot [soliddot] coordinates{(-.5,.375)} node[above left] {$(-0.5,0.375)$}; - \addplot [soliddot] coordinates{(.375,-.5)} node[below right] {$(0.375,-0.5)$}; - \addplot [soliddot] coordinates{(1,1.5)} node[above left] {$(1,1.5)$}; - \addplot [soliddot] coordinates{(1.5,1)} node[below right] {$(1.5,1)$}; - \end{axis} - \end{tikzpicture} - - - - -
    - -

    - For example, consider - where the tangent line to f at the point (1,1.5) is drawn. - That line has slope 3. - Through reflection across y=x, - we can see that the tangent line to g at the point (1.5,1) has slope 1/3. - Their slopes are reciprocals. - This should make sense since reflecting a line - (such as a tangent line) - across the line y=x switches the x and y values. - Also consider the point (0,0.5) on the graph of f, - where the tangent line is horizontal. - At the point (0.5,0) on g, - the tangent line is vertical. -

    - -

    - More generally, - consider the tangent line to f at the point (a,b). - That line has slope \fp(a). - Through reflection across y=x, - we can extend our above observation to say that the tangent line to g at the point (b,a) should have slope 1/\fp(a). - This then tells us that \gp(b)=1/\fp(a). -

    - -
    - Corresponding tangent lines drawn to f and f^{-1} - - - Corresponding tangent lines drawn to f and its inverse - -

    - The graph of f starts close to (-1,-0.5) below the negative x axis and slopes upwards and to the right. - After passing through (-0.5,0.375) it starts moving to the right. After intersecting the y axis at - (0,0.5) it continues to move horizontally for a short while then starts moving upwards to the right - and continues upwards through the point (1,1.5). The tangent line of f is drawn at (1,1.5), - it is a straight line that slightly touches the the curve at (1,1.5) and keeps moving straight. g starts - close to the negative y axis and starts moving to the right while sloping upwards. Once it moves through the point - (0.375,-0.5) then starts moving vertically upward and eventually starts moving again upwards and to the right. - The tangent line of the curve is drawn at (1.5,1). -

    -     - - - \begin{tikzpicture} - \begin{axis}[axis equal image, - ymin=-1.3,ymax=2, - xmin=-1.3,xmax=2, - clip=false, - %grid=both, - xtick={-1,1}, - ytick={-1,1}, - minor xtick={-1.25,-1,...,2}, - minor ytick={-1.25,-1,...,2},] - \addplot+[infinite,domain=-1:1.14] ({x},{x^3+.5}); - \addplot+[infinite,domain=-1:1.14] ({x^3+.5},{x}); - \addplot [symmetryline] coordinates {(-1.3,-1.3) (1.7,1.7)}; - \addplot [tangentlineseg,domain=0.75:1.25]{3*(x-1)+1.5}; - \addplot [tangentlineseg,domain=0.75:2]{1/3*(x-1.5)+1}; - \addplot [soliddot] coordinates{(-.5,.375)} node[above left] {$(-0.5,0.375)$}; - \addplot [soliddot] coordinates{(.375,-.5)} node[below right] {$(0.375,-0.5)$}; - \addplot [soliddot] coordinates{(1,1.5)} node[above left] {$(1,1.5)$}; - \addplot [soliddot] coordinates{(1.5,1)} node[below right] {$(1.5,1)$}; - \end{axis} - \end{tikzpicture} - - - - -
    - -

    - The information from these two graphs is summarized in below: -

    - - - - <tabular> - <row bottom="medium"> - <cell>Information about <m>f</m></cell> - <cell>Information about <m>g=f^{-1}</m></cell> - </row> - <row> - <cell><m>(1,1.5)</m> lies on <m>f</m></cell> - <cell><m>(1.5,1)</m> lies on <m>g</m></cell> - </row> - <row> - <cell> - <!-- <p> --> - Slope of tangent line to <m>f</m> at <m>x=1</m> is <m>3</m> - <!-- </p> --> - </cell> - <cell> - <!-- <p> --> - Slope of tangent line to <m>g</m> at <m>x=1.5</m> is <m>1/3</m> - <!-- </p> --> - </cell> - </row> - <row> - <cell><m>\fp(1) = 3</m></cell> - <cell><m>\gp(1.5) = 1/3</m></cell> - </row> - </tabular> - - </table> - - - - <p> - We have discovered a relationship between <m>\fp</m> and <m>\gp</m> in a mostly graphical way. - We can realize this relationship analytically as well. - Let <m>y = g(x)</m>, where again <m>g = f^{-1}</m>. - We want to find <m>\yp</m>. - Since <m>y = g(x)</m>, we know that <m>f(y) = x</m>. - Using the <xref ref="thm_chain_rule" text="title"/> and Implicit Differentiation, - take the derivative of both sides of this last equality. - <md> - <mrow>\lzoo{x}{f(y)} \amp = \lzoo{x}{x}</mrow> - <mrow>\fp(y)\cdot \yp \amp = 1</mrow> - <mrow>\yp \amp = \frac{1}{\fp(y)}</mrow> - <mrow>\yp \amp = \frac{1}{\fp(g(x))}</mrow> - </md>. - </p> - - <p> - This leads us to the following theorem. - </p> - - <theorem xml:id="thm_deriv_inverse_functions"> - <title>Derivatives of Inverse Functions - -

    - Let f be differentiable and one-to-one on an open interval I, - where \fp(x) \neq 0 for all x in I, - let J be the range of f on I, - let g be the inverse function of f, - and let f(a) = b for some a in I. - Then g is a differentiable function on J, - and in particular, - -

      -
    1. - \left(f^{-1}\right)'(b)=\gp(b) = \dfrac{1}{\fp(a)} -
      2. - -
    2. - \left(f^{-1}\right)'(x)=\gp(x) = \dfrac{1}{\fp(g(x))} -
      3. -
    -

    -     - - - - -

    - The results of are not trivial; - the notation may seem confusing at first. - Careful consideration, along with examples, - should earn understanding. -

    - - - -

    - In the next example we apply to the arcsine function. -

    - - - - - - - - Finding the derivative of an inverse trigonometric function - -

    - Let y = \arcsin(x) = \sin^{-1}(x). - Find \yp using . -

    -     - - -

    - Adopting our previously defined notation, - let g(x) = \arcsin(x) and f(x) = \sin(x). - Thus \fp(x) = \cos(x). - Applying the theorem, we have - - \gp(x) \amp = \frac{1}{\fp(g(x))} - \amp = \frac{1}{\cos(\arcsin(x))} - . -

    - -

    - This last expression is not immediately illuminating. - Drawing a figure will help, - as shown in . - Recall that the sine function can be viewed as taking in an angle and returning a ratio of sides of a right triangle, - specifically, - the ratio opposite over hypotenuse. - This means that the arcsine function takes as input a ratio of sides and returns an angle. - The equation y=\arcsin(x) can be rewritten as y=\arcsin(x/1); - that is, consider a right triangle where the hypotenuse has length 1 and the side opposite of the angle with measure y has length x. - This means the final side has length \sqrt{1-x^2}, - using the Pythagorean Theorem. -

    - -
    -
    - - - The right triangle defined by the equation sin(y)=x/1 - -

    - The right angle triangle is defined by y=\sin^{-1}(x/1). - The length of the base is \sqrt{1-x^2} and the length of the perpendicular is x. The length of the - hypotenuse is 1. The angle between the base and the hypotenuse is y. -

    -     - - - \begin{tikzpicture}[scale=1.06] - \coordinate (O) at (0,0); - \coordinate (A) at (5,0); - \coordinate (B) at (5,3.75); - \draw (O)--(A)--(B)--cycle; - - \tkzLabelSegment[below=0pt](O,A){$\sqrt{1-x^2}$} - \tkzLabelSegment[above left=0pt](O,B){$1$} - \tkzLabelSegment[right=0pt](A,B){$x$} - - \tkzMarkAngle[fill= orange, - size=1, - opacity=.4](A,O,B) - \tkzLabelAngle[pos = 0.75](A,O,B){$y$} - \end{tikzpicture} - - - - - - -

    - Therefore - - \cos\mathopen{}\left(\sin^{-1}(x)\right)\mathclose{} \amp= \cos(y) - \amp = \frac{\sqrt{1-x^2}}{1} - \amp = \sqrt{1-x^2} - , - resulting in - - \lzoo{x}{\arcsin(x)} = \frac{1}{\sqrt{1-x^2}} - . -

    - - - - -

    - Remember that the input x of the arcsine function is a ratio of a side of a right triangle to its hypotenuse; - the absolute value of this ratio will never be greater than 1. - Therefore the inside of the square root will never be negative. -

    - -

    - In order to make y=\sin(x) one-to-one, - we restrict its domain to [-\pi/2,\pi/2]; - on this domain, the range is [-1,1]. - Therefore the domain of y=\arcsin(x) is [-1,1] and the range is [-\pi/2,\pi/2]. - When x=\pm 1, - note how the derivative of the arcsine function is undefined; - this corresponds to the fact that as x\to \pm1, - the tangent lines to arcsine approach vertical lines with undefined slopes. -

    - -
    -
    - - - - The graph of sin(x) and arcsin(x) with their corresponding tangent line - -

    - The graph of \sin(x) starts at (-\pi/2,-1). It first curves upwards to the right - and then moves towards the origin making a slope. Once the slope passes through the origin it continues - upwards to the right in the first quadrant. Once it reaches the point (\pi/2,1), the graph starts - to move downwards again. The tangent line of the graph is drawn at (\pi/3,\sqrt{3/2}). -

    -     - - - \begin{tikzpicture} - \begin{axis}[ymin=-1.7,ymax=1.7, - xmin=-1.7,xmax=1.7, - xtick={-1.57,-.79,.79,1.57}, - xticklabels={$-\frac{\pi}{2}$,$-\frac{\pi}{4}$,$\frac{\pi}{4}$,$\frac{\pi}{2}$}, - axis equal] - - \addplot+[domain=-1.57:1.57] (x,{sin(deg(x))}) node [pos=0.5,above,sloped]{$y=\sin(x)$}; - \addplot [tangentline,domain=.42:1.7] {.5*(x-pi/3)+sqrt(3)/2}; - \addplot [soliddot] coordinates {(1.06,.866)} node [above left] {$\left(\frac{\pi}{3},\frac{\sqrt{3}}{2}\right)$}; - \end{axis} - \end{tikzpicture} - - - - - - Graphs of sin(x) and arcsin (x) with corresponding tangent line - -

    - Starting at the third quadrant at (-1,-\pi/2) the graph of \arcsin(x) - begins with a nearly vertical slope. The slope decreases to 1 as the graph passes through the origin. - The graph then moves toward its end at (1, \pi/2) with increasing slope. The tangent line is drawn at - (\sqrt{3/2},\pi/3). The tangent line of \sin^{-1}(x) is steeper than the tangent line of \sin(x). -

    -     - - - \begin{tikzpicture} - \begin{axis}[ymin=-1.7,ymax=1.7, - xmin=-1.7,xmax=1.7, - ytick={-1.57,-.79,.79,1.57}, - yticklabels={$-\frac{\pi}{2}$,$-\frac{\pi}{4}$,$\frac{\pi}{4}$,$\frac{\pi}{2}$}, - axis equal] - \addplot+[domain=-1.57:1.57] ({sin(deg(x))},x) node [pos=0.5,above,sloped]{$y=\sin^{-1}(x)$}; - \addplot [tangentline,domain=.42:1.7] ({.5*(x-pi/3)+sqrt(3)/2},{x}); - \addplot [soliddot] coordinates {(0.866,1.06)} node [below right] {$\left(\frac{\sqrt{3}}{2},\frac{\pi}{3}\right)$}; - \end{axis} - \end{tikzpicture} - - - -     - - -

    - In - we see f(x) = \sin(x) and - f^{-1}(x) = \sin^{-1}(x) graphed on their respective domains. - The line tangent to \sin(x) at the point - \left(\pi/3, \sqrt{3}/2\right) has slope \cos(\pi)/3 = 1/2. - The slope of the corresponding point on \sin^{-1}(x), - the point \left(\sqrt{3}/2,\pi/3\right), is - - \frac{1}{\sqrt{1-\left(\sqrt{3}/2\right)^2}} \amp = \frac{1}{\sqrt{1-3/4}} - \amp = \frac{1}{\sqrt{1/4}} - \amp = \frac{1}{1/2}=2 - , - verifying yet again that at corresponding points, - a function and its inverse have reciprocal slopes. -

    - -

    - Using similar techniques, - we can find the derivatives of all the inverse trigonometric functions. - In - we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible. -

    - -
    A right triangle defined by y=\sin^{-1}(x/1) with the length of the third leg found using the Pythagorean TheoremGraphs of \sin(x) and \sin^{-1}(x) along with corresponding tangent lines
    - Domains and ranges of the trigonometric and inverse trigonometric functions - - - Function - Domain - Range - - - \sin(x) - [-\pi/2, \pi/2] - [-1,1] - - - \sin^{-1}(x) - [-1,1] - [-\pi/2, \pi/2] - - - \cos(x) - [0,\pi] - [-1,1] - - - \cos^{-1}(x) - [-1,1] - [0,\pi] - - - \tan(x) - (-\pi/2,\pi/2) - (-\infty,\infty) - - - \tan^{-1}(x) - (-\infty,\infty) - (-\pi/2,\pi/2) - - - \csc(x) - [-\pi/2,0)\cup (0, \pi/2] - (-\infty,-1]\cup [1,\infty) - - - \csc^{-1}(x) - (-\infty,-1]\cup [1,\infty) - [-\pi/2,0)\cup (0, \pi/2] - - - \sec(x) - [0,\pi/2)\cup (\pi/2,\pi] - (-\infty,-1]\cup [1,\infty) - - - \sec^{-1}(x) - (-\infty,-1]\cup [1,\infty) - [0,\pi/2)\cup (\pi/2,\pi] - - - \cot(x) - (0,\pi) - (-\infty,\infty) - - - \cot^{-1}(x) - (-\infty,\infty) - (0,\pi) - - - -
    - - - Derivatives of Inverse Trigonometric Functions - -

    - The inverse trigonometric functions are differentiable on all open sets contained in their domains - (as listed in ) - and their derivatives are as follows: - -

      -
    1. \lzoo{x}{\sin^{-1}(x)} = \frac{1}{\sqrt{1-x^2}}
      2. - -
    2. \lzoo{x}{\cos^{-1}(x)} = -\frac{1}{\sqrt{1-x^2}}
      3. - -
    3. \lzoo{x}{\tan^{-1}(x)} = \frac{1}{1+x^2}
      4. - -
    4. \lzoo{x}{\csc^{-1}(x)} = -\frac{1}{\abs{x}\sqrt{x^2-1}}
      5. - -
    5. \lzoo{x}{\sec^{-1}(x)} = \frac{1}{\abs{x}\sqrt{x^2-1}}
      6. - -
    6. \lzoo{x}{\cot^{-1}(x)} = -\frac{1}{1+x^2}
      7. -
    - - derivativeinverse trig. - -

    -     -     - -

    - Note how each derivative is the negative of the derivative of its co function. - Because of this, derivatives of - \sin^{-1}(x), \tan^{-1}(x), - and \sec^{-1}(x) are used almost exclusively throughout this text. -

    - - - - -

    - In , - we stated without proof or explanation that \lzoo{x}{\ln(x)}=\frac{1}{x}. - We can justify that now using , - as shown in the example. -

    - - - Finding the derivative of <m>y=\ln(x)</m> - -

    - Use - to compute \lzoo{x}{\ln(x)}. -

    -     - -

    - View y= \ln(x) as the inverse of y = e^x. - Therefore, using our standard notation, - let f(x) = e^x and g(x) = \ln(x). - We wish to find \gp(x). - gives: - - \gp(x) \amp = \frac{1}{\fp(g(x))} - \amp = \frac{1}{e^{\ln(x) }} - \amp = \frac{1}{x} - . -

    -     -     - -

    - In this chapter we have defined the derivative, - given rules to facilitate its computation, - and given the derivatives of a number of standard functions. - We restate the most important of these in the following theorem, - intended to be a reference for further work. -

    - -

    - - - Glossary of Derivatives of Elementary Functions - -

    - Let f and g be differentiable functions, - and let a, - c and n be real numbers, - a \gt 0, n\neq 0. -

    - -

    -

      -
    1. \lzoo{x}{c} = 0
      2. - -
    2. \lzoo{x}{x} = 1
      3. - -
    3. \lzoo{x}{x^n} = nx^{n-1}
      4. - -
    4. \lzoo{x}{f(x)\pm g(x)} = f'(x) \pm \gp(x)
      5. - -
    5. \lzoo{x}{c\cdot f(x)} = c\cdot f'(x)
      6. - -
    6. \lzoo{x}{f(x)\cdot g(x)} = f'(x)\cdot g(x)+f(x)\cdot \gp(x)
      7. - -
    7. \lzoo{x}{f(g(x))} = f'(g(x)) \cdot \gp(x)
      8. - -
    8. \lzoo{x}{\frac{f(x)}{g(x)}} = \frac{f'(x)\cdot g(x)-f(x) \cdot \gp(x)}{(g(x))^2}
      9. - -
    9. \lzoo{x}{e^x} = e^x
      10. - -
    10. \lzoo{x}{\ln(x)} = \frac{1}{x}
      11. - -
    11. \lzoo{x}{a^x} = \ln(a)\cdot a^x
      12. - -
    12. \lzoo{x}{\log_a x} = \frac{1}{\ln(a)}\cdot\frac{1}{x}
      13. - -
    13. \lzoo{x}{\sin(x)} = \cos(x)
      14. - -
    14. \lzoo{x}{\cos(x)} = -\sin(x)
      15. - -
    15. \lzoo{x}{\tan(x)} = \sec^2(x)
      16. - -
    16. \lzoo{x}{\csc(x)} = -\csc(x)\cot(x)
      17. - -
    17. \lzoo{x}{\sec(x)} = \sec(x)\tan(x)
      18. - -
    18. \lzoo{x}{\cot(x)} = -\csc^2(x)
      19. - -
    19. \lzoo{x}{\sin^{-1}(x)} = \frac{1}{\sqrt{1-x^2}}
      20. - -
    20. \lzoo{x}{\cos^{-1}(x)} = -\frac{1}{\sqrt{1-x^2}}
      21. - -
    21. \lzoo{x}{\tan^{-1}(x)} = \frac{1}{1+x^2}
      22. - -
    22. \lzoo{x}{\csc^{-1}(x)} = -\frac{1}{\abs{x}\sqrt{x^2-1}}
      23. - -
    23. \lzoo{x}{\sec^{-1}(x)} = \frac{1}{\abs{x}\sqrt{x^2-1}}
      24. - -
    24. \lzoo{x}{\cot^{-1}(x)} = -\frac{1}{1+x^2}
      25. -
    -

    - - - - - - Terms and Concepts - - - - -

    - - Every function has an inverse. -

    -     - -

    - A necessary condition is that the function is one-to-one. -

    -     - -     - - - - -

    - In your own words explain what it means for a function to be one-to-one. -

    - -     - - - -     - - - - -

    - If (1,10) lies on the graph of y=f(x), - what can be said about the graph of y=f^{-1}(x)? -

    - -     - - - -

    - The point (10,1) lies on the graph of y=f^{-1}(x) - (assuming f is invertible). -

    -     - -     - - - - -

    - If (1,10) lies on the graph of y=f(x) and \fp(1) = 5, - what can be said about y=f^{-1}(x)? -

    - -     - - - -

    - The point (10,1) lies on the graph of y=f^{-1}(x) - (assuming f is invertible) - and (f^{-1})'(10) = 1/5. -

    -     - -     -     - - Problems - - -

    - Verify that the given functions are inverses. -

    -     - - - - -

    - f(x) = 2x+6 and - g(x) = \frac{1}{2}x-3 -

    - -     - - - -

    - Compose f(g(x)) and g(f(x)) to confirm that each equals x. -

    -     - -     - - - - -

    - f(x) = x^2+6x+11, x\geq 3 and - g(x) = \sqrt{x-2}-3, x\geq 2 -

    - -     - - - -

    - Compose f(g(x)) and g(f(x)) to confirm that each equals x. -

    -     - -     - - - - -

    - f(x) = \frac{3}{x-5}, x\neq 5 and - g(x) = \frac{3+5x}{x}, x\neq 0 -

    - -     - - - -

    - Compose f(g(x)) and g(f(x)) to confirm that each equals x. -

    -     - -     - - - - -

    - f(x) = \frac{x+1}{x-1}, x\neq 1 and - g(x) = f(x) -

    - -     - - - -

    - Compose f(g(x)) and g(f(x)) to confirm that each equals x. -

    -     - -     -     - - - -

    - An invertible function f(x) is given along with a point that lies on its graph. - Using , - evaluate \left(f^{-1}\right)'(x) at the indicated value. -

    -     - - - - - Context("Fraction"); - ($m,$b,$x0) = random_subset(3,2..10); - if($envir{problemSeed}==1){$m=5; $b=10; $x0=2;}; - $f = Formula("$m x + $b"); - $y0 = $f->eval(x=>$x0); - $point = Point($x0,$y0); - $fip = Fraction(1,$m); - - -

    - The point is on the graph of f(x) = . - Find \left(f^{-1}\right)'(). -

    -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)-y', '(-x)+y'); - ($h,$c,$x0) = random_subset(3, -9 .. -1, 1 .. 9); - if($envir{problemSeed}==1){$h=1; $c=4; $x0=3;}; - $f = Formula("x^2 - 2*$h x + $c")->reduce; - $y0 = $f->eval(x => $x0); - $point = Point($x0,$y0); - $fp = $f->D('x'); - $fpx0 = $fp->eval(x => $x0); - $fipy0 = Fraction(1/$fpx0); - $sign = ($x0 > $h) ? '\geq' : '\leq'; - - -

    - The point is on the graph of f(x) = , x . - Find \left(f^{-1}\right)'(). -

    -

    - -

    -     -     -     - - - - - Context("Fraction")->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $n = random(2,4,1); - $d = list_random(6*$n,4*$n,3*$n); - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$n=2; $d=6; $trig='sin';}; - $x0 = Formula("pi/$d"); - $f = Formula("$trig($n x)"); - ($num2,$den2) = Fraction($f->eval(x=>$x0->eval(x=>0))**2)->value; - if ($num2 == 3) { - $y0 = Formula("sqrt(3)/2"); - } elsif ($den2 == 2) { - $y0 = Formula("sqrt(2)/2"); - } else { - $y0 = Formula("1/2"); - }; - $fp = $f->D('x'); - $fpx0 = $fp->eval(x=>$x0->eval(x=>0)); - $fipy0 = 1/$fpx0; - if ($fipy0 == 1/sqrt(3)) {$fipy0 = Formula("sqrt(3)/3");} - elsif ($fipy0 == 2/(3*sqrt(3))) {$fipy0 = Formula("2 sqrt(3)/3");} - elsif ($fipy0 == 1/(2*sqrt(3))) {$fipy0 = Formula("sqrt(3)/6");} - elsif ($fipy0 == 1/sqrt(2)) {$fipy0 = Formula("sqrt(2)/2");} - elsif ($fipy0 == sqrt(2)/3) {$fipy0 = Formula("sqrt(2)/3");} - elsif ($fipy0 == 1/(2*sqrt(2))) {$fipy0 = Formula("sqrt(2)/4");} - elsif ($fipy0 == 1) {$fipy0 = Formula("1");} - elsif ($fipy0 == 2/3) {$fipy0 = Formula("2/3");} - elsif ($fipy0 == 1/2) {$fipy0 = Formula("1/2");}; - $d2 = 2*$n; - if ($trig eq 'sin') { - $low = Formula("-pi/$d2"); - $high = Formula("pi/$d2"); - } else { - $low = Formula("0"); - $high = Formula("pi/$n"); - } - - -

    - The point \left(,\right) is on the graph of f(x) = , - \leq x\leq . - Find \left(f^{-1}\right)'\left(\right). -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - ($h,$k,$c) = random_subset(3,-9..-1,1..9); - $x0 = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$h=2; $k=1; $c=-2; $x0=1;}; - $f = Formula("x^3 - (3*$h) x^2 + 3*(($h)^2+$k) x + $c")->reduce; - $y0 = $f->eval(x=>$x0); - $point = Point($x0,$y0); - $fp = $f->D('x'); - $fpx0 = $fp->eval(x=>$x0); - $fipy0 = Fraction(1/$fpx0); - - -

    - The point is on the graph of f(x) = . - Find \left(f^{-1}\right)'(). -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - $n = random(2,5,1); - $x0 = random(1,2,1); - if($envir{problemSeed}==1){$n=2; $x0=1;}; - $f = Formula("1/(1+x^$n)")->reduce; - $y0 = Fraction($f->eval(x=>$x0)); - $point = Point($x0,$y0); - $fp = $f->D('x'); - $fpx0 = $fp->eval(x=>$x0); - $fipy0 = Fraction(1/$fpx0); - - -

    - The point is on the graph of f(x) = , x\geq 0. - Find \left(f^{-1}\right)'\left(\right). -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - ($a,$k) = random_subset(2,2..9); - if($envir{problemSeed}==1){$a=6; $k=3;}; - $x0 = 0; - $f = Formula("$a e^($k x)"); - $y0 = $f->eval(x=>$x0); - $point = Point($x0,$y0); - $fp = $f->D('x'); - $fpx0 = $fp->eval(x=>$x0); - $fipy0 = Fraction(1/$fpx0); - - -

    - The point is on the graph of f(x) = . - Find \left(f^{-1}\right)'(). -

    -

    - -

    -     -     -     -     - - - -

    - Compute the derivative of the given function. -

    -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $k = random(2,9,1); - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$fname='h'; $x='t'; $k = 2; $trig = 'sin'}; - Context("Fraction"); - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->set($x=>{limits=>[-1/$k,1/$k]}); - $f = Formula("$trig^-1($k $x)"); - $fp = $f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $k = random(2,9,1); - $trig = list_random('sec','csc'); - if($envir{problemSeed}==1){$fname='f'; $x='t'; $k = 2; $trig = 'sec'}; - Context("Fraction"); - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->set($x=>{limits=>[1/$k,10/$k]}); - $f = Formula("$trig^-1($k $x)"); - $fp = $f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $k = random(2,9,1); - $trig = list_random('tan','cot'); - if($envir{problemSeed}==1){$fname='g'; $x='x'; $k = 2; $trig = 'tan'}; - Context("Fraction"); - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $f = Formula("$trig^-1($k $x)"); - $fp = $f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$fname='f'; $x='x'; $trig = 'sin'}; - Context("Fraction"); - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->set($x=>{limits=>[-1,1]}); - $f = Formula("$x $trig^-1($x)"); - $fp = $f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - ($triga,$trigb) = random_subset(2,'sin','cos','tan'); - if($envir{problemSeed}==1){$fname='g'; $x='t'; $triga = 'sin'; $trigb ='cos';}; - Context("Fraction"); - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->set($x=>{limits=>[-1,1]}); - $f = Formula("$triga($x) $trigb^-1($x)"); - $fp = $f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->set(t=>{limits=>[1,5]}); - $fp=Formula("e^t/t+ln(t)e^t"); - - -

    - f(t) = \ln(t) e^t -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - ($triga,$trigb) = random_subset(2,'sin','cos','tan'); - if($envir{problemSeed}==1){$fname='h'; $x='x'; $triga = 'sin'; $trigb ='cos';}; - Context("Fraction"); - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->set($x=>{limits=>[-1,1]}); - $f = Formula("$triga^-1($x)/$trigb^-1($x)"); - $fp = $f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $trig = list_random('tan','cot'); - $n = random(2,4,1); - if($envir{problemSeed}==1){$fname='g'; $x='x'; $trig = 'tan'; $n = 2}; - Context("Fraction"); - parser::Root->Enable; - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->set($x=>{limits=>[0,pi/2]}); - $f = Formula("$trig(root($n,$x))"); - $f = Formula("$trig(sqrt($x))") if ($n == 2); - $fp = $f->D($x); - - -

    - () = -

    - - To enter \sqrt{x}, type sqrt(x). - To enter \sqrt[n]{x}, type root(n,x). - -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $trig = list_random('sec','csc'); - $n = random(1,4,1); - if($envir{problemSeed}==1){$fname='f'; $x='x'; $trig = 'sec'; $n = 1}; - Context("Fraction"); - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->set($x=>{limits=>[0,pi/2]}); - $f = Formula("$trig(1/$x^$n)")->reduce; - $fp = $f->D($x); - - -

    - () = -

    -

    - -

    -     -     -     - - - - - $fname = list_random('f','g','h','j','k','p','m'); - $x = list_random('q','r','t','x','y','z','w'); - $trig = list_random('sin','cos','tan','cot','sec','csc'); - if($envir{problemSeed}==1){$fname='f'; $x='x'; $trig = 'sin'}; - Context()->variables->are($x=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->set($x=>{limits=>[-1,1]}); - $f = Formula("$trig($trig^-1($x))"); - $fp = Formula("1"); - - -

    - () = -

    -

    - -

    -     - -

    - () = x, so \fp(x) = 1. -

    -     -     -     -     - - - -

    - Compute the derivative of the given function in two ways: - -

      -
    1. -

      - By simplifying first, then taking the derivative, and -

      -
      2. -
    2. -

      - by using the Chain Rule first then simplifying. -

      -
      3. -
    - Verify that the two answers are the same. -

    -     - - - - -

    - f(x)=\sin(\sin^{-1}(x)) -

    - -     - - - -

    -

      -
    1. -

      - f(x) = x, so \fp(x) = 1. -

      -
      2. -
    2. -

      - \fp(x) = \cos(\sin^{-1}(x) )\frac{1}{\sqrt{1-x^2}} = 1. -

      -
      3. -
    -

    -     - -     - - - - -

    - f(x)=\tan^{-1}(\tan(x)) -

    - -     - - - -

    -

      -
    1. -

      - Recall that the inverse tangent function is defined to have range (-\pi/2, \pi/2). - If -\pi/2 \lt x\lt \pi/2, then f(x) = x. - If x is in the interval (-\pi/2+k\pi,\pi/2+k\pi) for some integer k, - then f(x)=x-k\pi. On any interval where f(x) is defined, - f(x)=x+C for some constant C, so \fp(x) = 1. -

      -
      2. -
    2. -

      - \fp(x) = \frac{1}{1+\tan^2(x) }\sec^2(x) = 1. -

      -
      3. -
    -

    -     - -     - - - - -

    - f(x)=\sin(\cos^{-1}(x)) -

    - -     - - - -

    -

      -
    1. -

      - f(x) = \sqrt{1-x^2}, so \fp(x) = \frac{-x}{\sqrt{1-x^2}}. -

      -
      2. -
    2. -

      - \fp(x) = \cos(\cos^{-1}(x) ) \frac{1}{\sqrt{1-x^2}} =\frac{-x}{\sqrt{1-x^2}}. -

      -
      3. -
    -

    -     - -     - - - - -

    - f(x)=\sin(2\sin^{-1}(x)) -

    - -     - - - -

    -

      -
    1. -

      - f(x) = 2\sin(\sin^{-1}(x))\cos(\sin^{-1}(x))=2x\sqrt{1-x^2}, - so \fp(x) = 2\sqrt{1-x^2} -\frac{2x^2}{\sqrt{1-x^2}}. -

      -
      2. -
    2. -

      - \fp(x) = \cos(2\sin^{-1}(x) ) \frac{2}{\sqrt{1-x^2}}, - but to compare, we need to simplify \cos(2\sin^{-1}(x)). - Using \cos(2\theta)=1-2\sin^2(\theta), - we get \fp(x) = (1-2x^2)\frac{2}{\sqrt{1-x^2}}. -

      -

      - This is the same as the first answer, since - - 2\sqrt{1-x^2} -\frac{2x^2}{\sqrt{1-x^2}} \amp = \frac{2(1-x^2)-2x^2}{\sqrt{1-x^2}} - \amp =\frac{2(1-2x^2)}{\sqrt{1-x^2}} - . -

      -
      3. -
    -

    -     - -     -     - - - -

    - Find the equation of the line tangent to the graph of f at the indicated x value. -

    -     - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $x0 = list_random('1/2','sqrt(2)/2','sqrt(3)/2'); - $sign = list_random('-',''); - if($envir{problemSeed}==1){$x0='sqrt(2)/2';$sign=''}; - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $f = Formula("sin^-1(x)"); - if ($x0 eq '1/2') { - $m = Formula("2sqrt(3)/3"); - $y0 = Formula("pi/6"); - } - elsif ($x0 eq 'sqrt(2)/2') { - $m = Formula("sqrt(2)"); - $y0 = Formula("pi/4"); - } - elsif ($x0 eq 'sqrt(3)/2') { - $m = Formula("2"); - $y0 = Formula("pi/3"); - }; - $x0 = Formula("$sign $x0"); - $y0 = Formula("$sign $y0"); - $t=Formula("y=$m(x-$x0)+$y0"); - - -

    - f(x)=\sin^{-1}(x) at x= -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $x0 = list_random('1/4','sqrt(2)/4','sqrt(3)/4'); - $sign = list_random('-',''); - if($envir{problemSeed}==1){$x0='sqrt(3)/4';$sign=''}; - Context()->variables->add(y=>'Real'); - parser::Assignment->Allow; - $f = Formula("cos^-1(2x)"); - if ($x0 eq '1/4') { - $m = Formula("-4sqrt(3)/3"); - if ($sign eq '') {$y0 = Formula("pi/3")} - else {$y0 = Formula("2pi/3")}; - } - elsif ($x0 eq 'sqrt(2)/4') { - $m = Formula("-2sqrt(2)"); - if ($sign eq '') {$y0 = Formula("pi/4")} - else {$y0 = Formula("3pi/4")}; - } - elsif ($x0 eq 'sqrt(3)/4') { - $m = Formula("-4"); - if ($sign eq '') {$y0 = Formula("pi/6")} - else {$y0 = Formula("5pi/6")}; - }; - $x0 = Formula("$sign $x0"); - $t=Formula("y=$m(x-$x0)+$y0"); - - -

    - f(x)=\cos^{-1}(2x) at x= -

    -

    - -

    -     -     -     -     -     - -     -
    - - -     - - - The Graphical Behavior of Functions - -

    - Our study of limits led to continuous functions, - a certain class of functions that behave in a particularly nice way. - Limits then gave us an even nicer class of functions, - functions that are differentiable. -

    - -

    - This chapter explores many of the ways we can take advantage of the information that continuous and differentiable functions provide. -

    -     - -
    - Extreme Values -

    - Given any quantity described by a function, - we are often interested in the largest and/or smallest values that quantity attains. - For instance, if a function describes the speed of an object, - it seems reasonable to want to know the fastest/slowest the object traveled. - If a function describes the value of a stock, - we might want to know the highest/lowest values the stock attained over the past year. - We call such values extreme values. -

    - - - Extreme Values - -

    - Let f be defined on an interval I containing c. - - extreme values - absolute minimum - minimumabsolute - absolute maximum - maximumabsolute - -

      -
    1. -

      - f(c) is the minimum (also, - absolute minimum) of f on I if - f(c) \leq f(x) for all x in I. -

      -
      2. - -
    2. -

      - f(c) is the maximum (also, - absolute maximum) of f on I if - f(c) \geq f(x) for all x in I. -

      -
      3. -
    -

    - -

    - The maximum and minimum values are the - extreme values, - or extrema, of f on I. - extremaabsolute -

    -     -     - - - - - -

    - Consider . - The function displayed in has a maximum, - but no minimum, - as the interval over which the function is defined is open. - In , the function has a minimum, - but no maximum; - there is a discontinuity in the natural - place for the maximum to occur. - Finally, the function shown in has both a maximum and a minimum; - note that the function is continuous and the interval on which it is defined is closed. -

    - - - -
    - Graphs of functions with and without extreme values - -
    - - - - A parabolic graph, opening downward, from -2 to 2, not including the endpoints - -

    - The image shows the graph of a function with the appearance of a parabola that opens downward, - with its vertex at (0,5). - The domain for the function is (-2,2), so the graph approaches, but does not reach, - the points (-2,1) and (2,1). These points are shown with hollow dots, - to illustrate the fact that they are not part of the graph. -

    - -

    - The maximum value f(0)=5 is reached, but there is no minimum value, - since the graph approaches, but does not reach, its end points. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ ymin=-.9,ymax=5.5,xmin=-2.2,xmax=2.4,] - \addplot+[open,domain=-2:2] {-x^2+5}; - \addplot[openinterval] coordinates {(-2,0) (2,0)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - A parabolic graph, opening downward, from -2 to 2. The vertex at (0,5) is missing. - -

    - The image shows the graph of a function with the appearance of a parabola that opens downward, - with its vertex at (0,5). - The domain for the function is [-2,2], this time, - the points (-2,1) and (2,1) at the ends of the graph are included. -

    - -

    - However, there is a hollow dot at (0,5), indicating a hole at the vertex. - A solid dot at (0,3) suggests that this is the graph of a function with f(0)=3. -

    - -

    - This graph illustrates a function with a minimum value f(-2)=f(2)=1, - but there is no maximum value, since the parabola approaches, but does not reach, its vertex. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ ymin=-.9,ymax=5.5,xmin=-2.2,xmax=2.4,] - \addplot+[domain=-2:2,closed] {-x^2+5}; - \addplot[soliddot] coordinates {(0,3)}; - \addplot[closedinterval] coordinates {(-2,0) (2,0)}; - \addplot[hollowdot] coordinates {(0,5)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - The graph of a parabola opening downward from a vertex at (0,5), for x from -2 to 2 - -

    - This graph shows a parabola on the interval [-2,2], - opening downward from its vertex at (0,5). -

    - -

    - There are no holes in the graph, and the endpoints are included. -

    - -

    - The graph illustrates a maximum value f(0)=5 and a minimum value of 1, - which is attained at both x=2 and x=-2. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ ymin=-.9,ymax=5.5,xmin=-2.2,xmax=2.4,] - \addplot+[domain=-2:2,closed] {-x^2+5}; - \addplot[closedinterval] coordinates {(-2,0) (2,0)}; - \end{axis} - \end{tikzpicture} - - - - -
    -     -
    - -

    - It is possible for discontinuous functions defined on an open interval to have both a maximum and minimum value, - but we have just seen examples where they did not. - On the other hand, - continuous functions on a closed interval always - have a maximum and minimum value. -

    - - - The Extreme Value Theorem - -

    - Let f be a continuous function defined on a closed interval I=[a,b]. - Then f has both a maximum and minimum value on I. - - Extreme Value Theorem - -

    -     -     - - - - - -

    - This theorem states that f has extreme values, - but it does not offer any advice about how/where to find these values. - The process can seem to be fairly easy, - as the next example illustrates. - After the example, - we will draw on lessons learned to form a more general and powerful method for finding extreme values. -

    - - - Approximating extreme values - -

    - Consider f(x) = 2x^3-9x^2 on I=[-1,5], - as graphed in . - Approximate the extreme values of f. -

    - -
    - A graph of f(x) = 2x^3-9x^2 as in - - - - The graph of a cubic function on the interval [-1,5], with maximum and minimum values - -

    - The image shows the graph of f(x)=2x^3-9x^2 on the interval [-1,5]. - Beginning at the point (-1,-11), the graph rises to a peak at (0,0), - before falling to its minimum value at (3,-27). - The graph then climbs to the endpoint (5,25), where it reaches its maximum value. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-2,xmax=5.7, - ymin=-33,ymax=31,] - \addplot+[domain=-1:5, samples=40] {2*x^3-9*x^2}; - \addplot[soliddot] coordinates{(-1,-11)} node[below] {$(-1,-11)$}; - \addplot[soliddot] coordinates{(0,0)} node[above right] {$(0,0)$}; - \addplot[soliddot] coordinates{(3,-27)} node[below] {$(3,-27)$}; - \addplot[soliddot] coordinates{(5,25)} node[above] {$(5,25)$}; - \end{axis} - \end{tikzpicture} - - - - -
    - -     - -

    - The graph is drawn in such a way to draw attention to certain points. - It certainly seems that the smallest y-value is -27, - found when x=3. - It also seems that the largest y-value is 25, - found at the endpoint of I, x=5. - We use the word seems, - for by the graph alone we cannot be sure the smallest value is not less than -27. - Since the problem asks for an approximation, - we approximate the extreme values to be 25 and -27. -

    -     -     - -

    - Notice how the minimum value came at - the bottom of a hill, - and the maximum value came at an endpoint. - Also note that while 0 is not an extreme value, - it would be if we narrowed our interval to [-1,4]. - The idea that the point (0,0) is the location of an extreme value for some interval is important, - leading us to a definition of a - relative maximum. - In short, a relative max - is a y-value that's the largest y-value nearby. -

    - - - - - - - Relative Minimum and Relative Maximum - -

    - Let f be defined on an interval I containing c. - -

      -
    1. -

      - If there is a \delta \gt 0 such that - f(c) \leq f(x) for all x in I where \abs{x-c}\lt \delta, - then f(c) is a relative minimum of f. - We also say that f has a relative minimum at (c,f(c)). - minimumrelative/local - extremarelative -

      -
      2. - -
    2. -

      - If there is a \delta \gt 0 such that - f(c) \geq f(x) for all x in I where \abs{x-c}\lt \delta, - then f(c) is a relative maximum of f. - We also say that f has a relative maximum at (c,f(c)). - maximumrelative/local - extremarelative -

      -
      3. -
    -

    - -

    - The relative maximum and minimum values comprise the - relative extrema of f. -

    -     -     - -

    - We briefly practice using these definitions. -

    - - - Approximating relative extrema - -

    - Consider f(x) = (3x^4-4x^3-12x^2+5)/5, - as shown in . - Approximate the relative extrema of f. - At each of these points, evaluate \fp. -

    - -
    - A graph of f(x) = (3x^4-4x^3-12x^2+5)/5 as in - - - The graph of a degree 4 function with several critical points. - -

    - The graph of f(x)=\frac15(3x^4-4x^3-12x^2+5) is shown, for x between -2 and 3. - Arrows at either end of the graph indicate that f(x) will approach \infty in either direction beyond the interval shown. - The graph shows three relative extrema: there are minima when x=-1 and x=2, and a maximum at x=0. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-2.2,xmax=3.2, - ymin=-6.5,ymax=6.9,] - \addplot+[infinite,domain=-1.8:3, samples=50] {(3*x^4-4*x^3-12*x^2+5)/5}; - \end{axis} - \end{tikzpicture} - - - - -
    -     - -

    - We still do not have the tools to exactly find the relative extrema, - but the graph does allow us to make reasonable approximations. - It seems f has relative minima at x=-1 and x=2, - with values of f(-1)=0 and f(2) = -5.4. - It also seems that f has a relative maximum at the point (0,1). -

    - -

    - We approximate the relative minima to be 0 and -5.4; - we approximate the relative maximum to be 1. -

    - -

    - It is straightforward to evaluate - \fp(x) =\frac{1}{5}\left(12x^3-12x^2-24x\right) at x=0, 1 and 2. - In each case, \fp(x) = 0. -

    -     -     - - - Approximating relative extrema - -

    - Approximate the relative extrema of f(x) = (x-1)^{2/3}+2, - shown in . - At each of these points, evaluate \fp. -

    - -
    - A graph of f(x) = (x-1)^{2/3}+2 as in - - - A graph in the shape of a curved V, with a sharp point at (1,2). - -

    - The graph of f(x)=(x-1)^{2/3}+2 resembles the outstretched wings of a bird. - It is in the shape of a wide, curved V, with a vertex at (1,2). -

    - -

    - This graph illustrates the case of a local minimum that occurs at a point of non-differentiability. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-.6,xmax=2.6, - ymin=-.5,ymax=3.5] - \addplot+[rightarrow,domain=2:3.2] ({(x-2)^1.5+1},{x}); - \addplot[firstcurvestyle,rightarrow,domain=2:3.2] ({-(x-2)^1.5+1},{x}); - \end{axis} - \end{tikzpicture} - - - - -
    -     - -

    - The figure implies that f does not have any relative maxima, - but has a relative minimum at (1,2). - In fact, the graph suggests that not only is this point a relative minimum, - y=f(1)=2 is the minimum value of the function. -

    - -

    - We compute \fp(x) = \frac23(x-1)^{-1/3}. - When x=1, \fp is undefined. -

    -     -     - - - -

    - What can we learn from the previous two examples? - We were able to visually approximate relative extrema, - and at each such point, - the derivative was either 0 or it was not defined. - This observation holds for all functions, - leading to a definition and a theorem. -

    - - - Critical Numbers and Critical Points - -

    - Let f be defined at c. - The value c is a critical number - (or critical value) - of f if \fp(c)=0 or \fp(c) is not defined. - - critical number - critical point - -

    - -

    - If c is a critical number of f, - then the point (c,f(c)) is a - critical point of f. -

    -     -     - - - - - Relative Extrema and Critical Points - -

    - Let a function f be defined on an open interval I containing c, - and let f have a relative extremum at the point (c,f(c)). - Then c is a critical number of f. -

    -     -     - - - - - -

    - Be careful to understand that this theorem states - Relative extrema on open intervals occur at critical points. - It does not say All critical numbers produce relative extrema. For instance, - consider f(x) = x^3. - Since \fp(x) = 3x^2, - it is straightforward to determine that x=0 is a critical number of f. - However, f has no relative extrema, - as illustrated in . -

    - - -
    - A graph of f(x)=x^3 which has a critical value of x=0, but no relative extrema - - - A graph of the cubic function, which has a horizontal tangent at the origin, but no maximum or minimum - -

    - The graph of f(x)=x^3 rises steadily as x increases. - It briefly levels off at (0,0), where there is a horizontal tangent. - The horizontal tangent corresponds to a critical point, but it is not a relative maximum nor a relative minimum. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-1.1,xmax=1.1, - ymin=-1.1,ymax=1.1] - \addplot+[infinite,domain=-1:1] {x^3}; - \end{axis} - \end{tikzpicture} - - - - -
    - -

    - - states that a continuous function on a closed interval will have both an absolute maximum and an absolute minimum. - Common sense tells us extrema occur either at the endpoints or somewhere in between. - It is easy to check for extrema at endpoints, - but there are infinitely many points to check that are in between. - tells us we need only check at the critical points that are in between the endpoints. - We combine these concepts to offer a strategy for finding extrema. -

    - - - Finding Extrema on a Closed Interval -

    - Let f be a continuous function defined on a closed interval [a,b]. - To find the maximum and minimum values of f on [a,b]: - extremafinding - -

      -
    1. -

      - Evaluate f at the endpoints a and b of the interval. -

      -
      2. - -
    2. -

      - Find the critical numbers of f in [a,b]. -

      -
      3. - -
    3. -

      - Evaluate f at each critical number. -

      -
      4. - -
    4. -

      - The absolute maximum of f is the largest of these values, - and the absolute minimum of f is the least of these values. -

      -
      5. -
    -

    -     - -

    - We practice these ideas in the next examples. -

    - - - Finding extreme values - -

    - Find the extreme values of f(x) = 2x^3+3x^2-12x on [0,3], - graphed in . -

    - -
    - A graph of f(x) = 2x^3+3x^2-12x on [0,3] as in - - - A graph of a cubic function on the interval [0,3]. - -

    - The image shows the graph of f(x)=2x^3+3x^2-12x on the interval [0,3]. - There appears to be a relative minimum near x=1, and no other critical points. - The absolute maximum value of the function appears to occur at the right endpoint of the interval. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-.4,xmax=3.3, - ymin=-7.5,ymax=45.5] - \addplot+[closed,domain=0:3] {2*x^3+3*x^2-12*x}; - \end{axis} - \end{tikzpicture} - - - - -
    -     - -

    - We follow the steps outlined in . - We first evaluate f at the endpoints: - - f(0)\amp=0\amp f(3)\amp=45 - . -

    - -

    - Next, we find the critical values of f on [0,3]. - \fp(x) = 6x^2+6x-12 = 6(x+2)(x-1); - therefore the critical values of f are x=-2 and x=1. - Since x=-2 does not lie in the interval [0,3], - we ignore it. - Evaluating f at the only critical number in our interval gives: - f(1) = -7. -

    - -

    - - gives f evaluated at the important - x values in [0,3]. - We can easily see the maximum and minimum values of f: - the maximum value is 45 and the minimum value is -7. -

    - -
    - Finding the extreme values of f(x)= 2x^3+3x^2-12x in - - - x - f(x) - - - 0 - 0 - - - 1 - -7 - - - 3 - 45 - - -
    - -     - -     - -

    - Note that all this was done without the aid of a graph; - this work followed an analytic algorithm and did not depend on any visualization. - - shows f and we can confirm our answer, - but it is important to understand that these answers can be found without graphical assistance. -

    - -

    - We practice again. -

    - - - Finding extreme values - -

    - Find the maximum and minimum values of f on [-4,2], where - - f(x) = \begin{cases}(x-1)^2 \amp x\leq 0 \\ x+1 \amp x \gt 0\end{cases} - . - -

    -     - -

    - Here f is piecewise-defined, - but we can still apply - as it is continuous on [-4,2] - (one should check to verify that \lim\limits_{x\to 0}f(x) =f(0)). -

    - -

    - Evaluating f at the endpoints gives: - - f(-4)\amp=25\amp f(2)\amp=3 - . -

    - -

    - We now find the critical numbers of f. - We have to define \fp in a piecewise manner; it is - - \fp(x) =\begin{cases}2(x-1) \amp x \lt 0 \\ 1 \amp x \gt 0\end{cases} - . -

    - -

    - Note that while f is defined for all of [-4,2], - \fp is not, - as the derivative of f does not exist when x=0. - (From the left, the derivative approaches -2; - from the right the derivative is 1.) - Thus one critical number of f is x=0. -

    - -

    - We now set \fp(x) = 0. - When x \gt 0, \fp(x) is never 0. - When x\lt 0, - \fp(x) is also never 0, so we find no critical values from setting \fp(x)=0. -

    - -

    - So we have three important x-values to consider: - x= -4, 2 and 0. - Evaluating f at each gives, respectively, 25, - 3 and 1, shown in . - - Thus the absolute minimum of f is 1, - the absolute maximum of f is 25. - Our answer is confirmed by the graph of f in . -

    - - - -
    - Finding the extreme values of a piecewise-defined function in - - - x - f(x) - - - -4 - 25 - - - 0 - 1 - - - 2 - 3 - - -
    - -
    - A graph of f(x) on [-4,2] as in - - - The graph of a piecewise-defined function on the interval [-4,2] - -

    - The image shows the graph of a piecewise-defined function on the interval [-4,2]. - For -4\leq x\leq 0, the graph appears to be part of a parabola, - descending from (-4,25) to (0,1). - For 0\leq x\leq 2, the graph is a straight line with positive slope, from (0,1) to (2,3). -

    - -

    - The point (0,1) is a relative minimum, and a point of non-differentiability. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-4.5,xmax=2.5, - ymin=-.9,ymax=25.5] - \addplot+[domain=-4:0] {(x-1)^2}; - \addplot[firstcurvestyle,domain=0:2] {x+1}; - \addplot[soliddot] coordinates {(-4,25) (2,3)}; - \end{axis} - \end{tikzpicture} - - - - -
    -     -     - -     - - - Finding extreme values - -

    - Find the extrema of f(x) = \cos\mathopen{}\left(x^2\right)\mathclose{} on [-2,2]. -

    -     - -

    - We again use . - Evaluating f at the endpoints of the interval gives: - f(-2) = f(2) = \cos(4) \approx -0.6536. - We now find the critical values of f. -

    - -

    - Applying the , - we find \fp(x) = -2x\sin\mathopen{}\left(x^2\right)\mathclose{}. - Set \fp(x) = 0 and solve for x to find the critical values of f. -

    - -

    - We have \fp(x) = 0 when x = 0 and when \sin\mathopen{}\left(x^2\right)\mathclose{}. - In general, \sin(t) = 0 when - t = \ldots -2\pi, -\pi, 0, \pi, \ldots Thus \sin\mathopen{}\left(x^2\right)\mathclose{} = 0 when - x^2 = 0, \pi, 2\pi, \ldots (x^2 is always nonnegative so we ignore -\pi, - etc.) So \sin\mathopen{}\left(x^2\right)\mathclose{}=0 when x= 0, \pm \sqrt{\pi}, \pm\sqrt{2\pi}, \ldots. - The only values to fall in the given interval of [-2,2] are 0 and \pm\sqrt{\pi}, - where \sqrt{\pi} \approx 1.77. -

    - -

    - We again construct a table of important values in . - In this example we have five values to consider: - x= 0, \pm 2, \pm\sqrt{\pi}. - From the table it is clear that the maximum value of f on [-2,2] is 1; - the minimum value is -1. - The graph in confirms our results. -

    - - - -
    - Finding the extrema of f(x)= \cos\mathopen{}\left(x^2\right)\mathclose{} in - - - x - f(x) - - - -2 - -0.65 - - - -\sqrt{\pi} - -1 - - - 0 - 1 - - - \sqrt{\pi} - -1 - - - 2 - -0.65 - - -
    - -
    - A graph of f(x)=\cos\mathopen{}\left(x^2\right)\mathclose{} on [-2,2] as in - - - A graph that resembles a flattened cosine wave for small values of x - -

    - The graph of f(x)=\cos(x^2) resembles a cosine wave, except that it is much flatter when x is near 0. - From an apparent relative maximum at (0,1), the graph descends in both directions toward relative minimum values close to the endpoints, - at x=2 and x=-2. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-2.2,xmax=2.2, - ymin=-1.1,ymax=1.1] - \addplot+[closed,samples=40,smooth, domain=-2:2] {cos(deg(x^2))}; - \end{axis} - \end{tikzpicture} - - - - -
    -     -     - -     - -

    - We consider one more example. -

    - - - Finding extreme values - -

    - Find the extreme values of f(x) = \sqrt{1-x^2}. -

    -     - -

    - A closed interval is not given, - so we find the extreme values of f on its domain. - f is defined whenever 1-x^2\geq 0; - thus the domain of f is [-1,1]. - Evaluating f at either endpoint returns 0. -

    - - -
    - A graph of f(x)=\sqrt{1-x^2} on [-1,1] as in - - - A semi-circle equivalent to the top half of the unit circle - -

    - The graph y=\sqrt{1-x^2} is the half of the unit circle x^2+y^2=1, where y\geq 0. - It has absolute minimum values at its endpoints, which are (-1,0) and (1,0). - It reaches its (absolute and relative) maximum at the top of the circle, which is the point (0,1). -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-1.2,xmax=1.2, - ymin=-1.2,ymax=1.2, - axis equal] - \addplot+[closed,domain=0:180] ({cos(x)},{sin(x)}); - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - Finding the extrema of the half-circle in - - - x - f(x) - - - -1 - 0 - - - 0 - 1 - - - 1 - 0 - - -
    - -     - -

    - Using the , we find \fp(x) = -x\big/\sqrt{1-x^2}. - The critical points of f are found when - \fp(x) = 0 or when \fp is undefined. - It is straightforward to find that \fp(x) = 0 when x=0, - - and \fp is undefined when x=\pm 1, - the endpoints of the interval - (which are in the domain of f.) - The table of important values is given in . - The maximum value is 1, - and the minimum value is 0. -

    -     -     - - - - - -

    - We have seen that continuous functions on closed intervals always have a maximum and minimum value, - and we have also developed a technique to find these values. - In , - we further our study of the information we can glean from nice - functions with the Mean Value Theorem. - On a closed interval, we can find the - average rate of change of a function - (as we did at the beginning of ). - We will see that differentiable functions always have a point at which their instantaneous - rate of change is same as the average rate of change. - This is surprisingly useful, as we'll see. -

    - - - - Terms and Concepts - - - - -

    - Describe what an extreme value - of a function is in your own words. -

    - -     - - - -     - - - -

    - Sketch the graph of a function f on (-1,1) that has both a maximum and minimum value. -

    -     - - - -     - - - - -

    - Describe the difference between absolute and relative maxima in your own words. -

    - -     - - - -     - - - -

    - Sketch the graph of a function f where f has a relative maximum at x=1 and \fp(1) is undefined. -

    -     - - - -     - - - - -

    - - If c is a critical value of a function f, - then f has either a relative maximum or relative minimum at x=c. -

    -     - -     - - - - - $zero = Real(0); - Context()->strings->add('undefined'=>{}); - Context()->strings->add('not defined'=>{alias=>'undefined'}); - $undefined = String("undefined"); - - -

    - Fill in the blanks: - The critical points of a function f are found where \fp(x) is equal to - or where \fp(x) is . -

    -     -     -     - -     - - - Problems - - - -

    - Identify each of the marked points as being an absolute maximum or minimum, - a relative maximum or minimum, or none of the above. -

    -     - - - - - for my $A ('A'..'G') {Context()->strings->add($A=>{})}; - $absmax = List("B"); - $absmin = List("None"); - $relmax = List("B", "G"); - $relmin = List("C","F"); - - - - A graph with several possible relative maximum and minimum values, some of which are omitted - -

    - The image shows a graph with labeled points A,B,C,D,E,F,G, going from left to right. - Each point is to be considered as potentially a relative or absolute maximum or minimum. -

    - -

    - The point A is an endpoint of the graph, but it is a hollow dot, indicating that it is not part of the graph. - Points on the graph near A are lower than anywhere else on the graph. -

    - -

    - The graph goes up until it reachs the point B, a solid dot, before going back down to the point C. - The y value at B is higher than at any other point on the graph. - At C the graph turns sharply: there is a cusp. To the right of C the graph goes up again, - until the point D. The point D is a hollow dot, indicating that it is not part of the graph. -

    - -

    - From D the graph goes down until it reaches E, which is a solid dot. - To the right of E, the graph continues to go down, - until it reaches F. - The point F is also a solid dot. To the right of F the graph goes back up, until ending at the point G. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-2.5,ymax=3.5,% - xmin=-.5,xmax=6.5,% - ] - - \draw [firstcolor] (axis cs:.1,-2) parabola [bend at end] (axis cs:1,3) parabola (axis cs:2,1) parabola [bend at end] (axis cs:3,2) parabola (axis cs:4,1) parabola [bend at end] (axis cs:5,-1) parabola (axis cs:6,2.5); - - \draw [firstcolor,fill=white] (axis cs: .1,-2) node [black,right] {\(A\)} circle (2.4pt) - (axis cs:3,2) node [above,black] {\(D\)} circle (2.4pt); - - \draw [firstcolor,fill=firstcolor] (axis cs: 1,3) node [above,black] {\(B\)} circle (2.4pt) - (axis cs:2,1) node [below,black] {\(C\)} circle (2.4pt) - (axis cs:4,1) node [right,black] {\(E\)} circle (2.4pt) - (axis cs:5,-1) node [below,black] {\(F\)} circle (2.4pt) - (axis cs:6,2.5) node [left,black] {\(G\)} circle (2.4pt); - - \end{axis} - \end{tikzpicture} - - - - Enter the absolute maxima points here, using commas to separate them if there is more than one. If there are none, enter NONE. - -

    - -

    - - Enter the absolute minima points here, using commas to separate them if there is more than one. If there are none, enter NONE. - -

    - -

    - - Enter the relative maxima points here, using commas to separate them if there is more than one. If there are none, enter NONE. - -

    - -

    - - Enter the relative minima points here, using commas to separate them if there is more than one. If there are none, enter NONE. - -

    - -

    -     -     -     - - - - - for my $A ('A'..'E') {Context()->strings->add($A=>{})}; - $absmax = List("C"); - $absmin = List("A"); - $relmax = List("C"); - $relmin = List("A","E"); - - - - A graph on the interval [1,5], with 5 marked points. - -

    - The image shows a graph on the intervl [1,5], with marked points A,B,C,D,E. - All five points are part of the graph. - The point A is the left endpoint of the graph, and its y value is below any other point on the graph. - The graph goes up toward the point B. - At B there appears to be a point of non-differentiability, - but to the right of B the graph continues to go up toward the point C. -

    - -

    - At C there is another cusp. To the right of C, the graph begins to go down, toward the point D. - There is a cusp at D as well, but to the right of D, the graph again goes down, until it reaches the endpoint at E. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-2.5,ymax=2.5,% - xmin=-.5,xmax=5.5,% - ] - - \draw [firstcolor] (axis cs:1,-2) parabola (axis cs:2,1) - -- (axis cs:3,2) - parabola [bend at end] (axis cs:4,1) - -- (axis cs:5,-1); - - \draw [firstcolor,fill=firstcolor] (axis cs: 1,-2) node [above,black] {\(A\)} circle (2.4pt) - (axis cs:2,1) node [left,black] {\(B\)} circle (2.4pt) - (axis cs:3,2) node [above,black] {\(C\)} circle (2.4pt) - (axis cs:4,1) node [right,black] {\(D\)} circle (2.4pt) - (axis cs:5,-1) node [below,black] {\(E\)} circle (2.4pt); - - \end{axis} - \end{tikzpicture} - - - - Enter the absolute maxima points here, using commas to separate them if there is more than one. If there are none, enter NONE. - -

    - -

    - - Enter the absolute minima points here, using commas to separate them if there is more than one. If there are none, enter NONE. - -

    - -

    - - Enter the relative maxima points here, using commas to separate them if there is more than one. If there are none, enter NONE. - -

    - -

    - - Enter the relative minima points here, using commas to separate them if there is more than one. If there are none, enter NONE. - -

    - -

    -     -     -     -     - - - -

    - Evaluate \fp(x) at the points indicated in the graph. -

    -     - - - - - $f=Formula("2/(x^2+1)"); - $zero=Real(0); - $dne = String("DNE"); - - -

    - f(x) = -

    - - A bell-shaped curve with a relative maximum at the marked point (0,2). - -

    - The graph of f(x)=\frac{2}{x^2+1} is bell-shaped, with the peak of the bell at the point (0,2). - The point (0,2) appears to be a relative maximum, and it is also the indicated point, - meaning that the goal of the question is to compute f'(0). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.5,ymax=2.5,% - xmin=-5.5,xmax=5.5,% - ] - - \addplot [firstcurvestyle,samples=101] {2/((x^2+1))}; - - \draw [firstcolor,fill=firstcolor] (axis cs: 0,2) node [above right,black] {\((0,2)\)} circle (2.4pt); - - \end{axis} - \end{tikzpicture} - - - - Enter f^{\prime}(0). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    -     -     -     - - - - - $f=Formula("x^2*sqrt(6-x^2)"); - $zero=Real(0); - $dne = String("DNE"); - - -

    - f(x) = -

    - - A graph consisting of a pair of arches, resembling a curved letter M. The points (0,0) and (2,4*sqrt(2)) are marked. - -

    - The graph of f(x)=x^2\sqrt{6-x^2} resembles a curved letter M, - with a pair of arches, one on either side of the y axis. -

    - -

    - The marked points are (0,0), which appears to be at a relative minimum between the two arches, - and (2,4\sqrt{2}), which appears to be at a relative maximum at the top of the right arch. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.9,ymax=6.5,% - xmin=-3.5,xmax=3.5,% - ] - - \addplot [firstcurvestyle,domain=-2.449489:2.449489,samples=104] {x^2*sqrt(6-x^2)}; - - \draw [firstcolor,fill=firstcolor] (axis cs: 0,0) node [below right,black] {\((0,0)\)} circle (2.4pt); - \draw [firstcolor,fill=firstcolor] (axis cs: 2,5.66) node [left,black] {\((2,4\sqrt{2})\)} circle (2.4pt); - \addplot[soliddot] coordinates {(2.449489,0) (-2.449489,0)}; - - \end{axis} - \end{tikzpicture} - - - - Enter f^{\prime}(0). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    - - Enter f^{\prime}(2). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    -     -     -     - - - - - $f=Formula("sin(x)"); - $zero=Real(0); - $dne = String("DNE"); - - -

    - f(x) = -

    - - Graph of a sine wave from 0 to 2 pi. The marked points are (pi/2, 1) and (3 pi/2,-1) - -

    - The graph of f(x)=\sin(x) from 0 to \pi. - There are two points marked on the graph: (\pi/2,1), at which f(x) appears to have a relative maximum, - and (3\pi/2,-1), at which f(x) appears to have a relative minimum. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.5,ymax=1.5,% - xmin=-.5,xmax=6.5,% - ] - - \addplot [firstcurvestyle,domain=0:6.28] {sin(deg(x))}; - - \draw [firstcolor,fill=firstcolor] (axis cs: 1.57,1) node [above,black] {\((\pi/2,1)\)} circle (2.4pt); - \draw [firstcolor,fill=firstcolor] (axis cs: 4.71,-1) node [below,black] {\((3\pi/2,-1)\)} circle (2.4pt); - \addplot[soliddot] coordinates {(0,0) (6.28,0)}; - - \end{axis} - \end{tikzpicture} - - - - Enter f^{\prime}(\pi/2). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    - - Enter f^{\prime}(3\pi/2). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    -     -     -     - - - - - $f=Formula("x^2*sqrt(4-x)"); - $zero=Real(0); - $dne = String("DNE"); - - -

    - f(x) = -

    - - A curved graph with a relative minimum at (0,0), and a relative maximum when x=16/5 - -

    - The graph of f(x)=x^2\sqrt{4-x} is plotted for x between -2 and 4. - There are three marked points: a relative minimum at (0,0), - a relative maximum at \left(\frac{16}{5},\frac{512}{25\sqrt{5}}\right), - and the point (4,0), which is an endpoint, since the domain of f(x) is (-\infty, 4]. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.8,ymax=11,% - xmin=-2.5,xmax=4.5,% - ] - - \addplot [firstcurvestyle,infiniteleft,domain=-2:4,samples=101] {x^2*sqrt(4-x)}; - - \draw [firstcolor,fill=firstcolor] (axis cs: 0,0) node [below right,black] {\((0,0)\)} circle (2.4pt); - \draw [firstcolor,fill=firstcolor] (axis cs: 3.2,9.16) node [above,black] {\(\left(\frac{16}{5},\frac{512}{25\sqrt{5}}\right)\)} circle (2.4pt); - \draw [firstcolor,fill=firstcolor] (axis cs: 4,0) node [above left,black] {\((4,0)\)} circle (2.4pt); - - \end{axis} - \end{tikzpicture} - - - - Enter f^{\prime}(0). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    - - Enter f^{\prime}(16/5). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    - - Enter f^{\prime}(4). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0); - $f=Formula("1+(x-2)^(2/3)/x"); - $zero=Real(0); - $dne = String("DNE"); - - -

    - f(x) = -

    - - A curve that descends steeply to a cusp at (2,1), and that is relatively flat for x greater than 1. - -

    - The graph of f(x)=1+\frac{(x-2)^{2/3}}{x} has a vertical asymptote at x=0. - It descends toward a cusp at (2,1), which is a relative minimum, and one of the marked points. - For x\gt 1 the graph rises slowly toward the point \left(6,1+\frac{\sqrt[3]{2}}{3}\right), which is also marked. - The graph is quite flat near this point, and it is hard to tell whether or not it could be a relative maximum. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.5,ymax=6.5,% - xmin=-1,xmax=11,% - ] - - \addplot [firstcurvestyle,infinite,domain=.2:11,samples=101] {(((x-2)^2)^(1/3))/x+1}; - - \draw [firstcolor,fill=firstcolor] (axis cs: 2,1) node [below,black] {\((2,1)\)} circle (2.4pt); - \draw [firstcolor,fill=firstcolor] (axis cs: 6,1.42) node [above,black] {\(\left(6,1+\frac{\sqrt[3]{2}}{3}\right)\)} circle (2.4pt); - - \end{axis} - \end{tikzpicture} - - - - - Enter f^{\prime}(2). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    - - Enter f^{\prime}(6). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0); - parser::Root->Enable; - $f=Formula("root(3,x^4-2x^2+1)"); - $zero=Real(0); - $dne = String("DNE"); - - -

    - f(x) = -

    - - A graph in the shape of a curved letter W. There are two cusps, both relative minima, at marked points (-1,0) and (1,0). - -

    - The graph of f(x)=\sqrt[3]{x^4-2x^2+1} has two cusps, at the points (-1,0) and (1,0), - which are the indicated points for this problem. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xtick={-2,-1,1,2}, - ymin=-.5,ymax=3.5,% - xmin=-2.5,xmax=2.5,% - ] - - \addplot [firstcurvestyle,infiniteleft,domain=-2:-1,samples=25] {(x^4-2*x^2+1)^(1/3)}; - \addplot [firstcurvestyle,domain=-1:1,samples=50] {(x^4-2*x^2+1)^(1/3)}; - \addplot [firstcurvestyle,infiniteright,domain=1:2,samples=25] {(x^4-2*x^2+1)^(1/3)}; - - \draw [firstcolor,fill=firstcolor] (axis cs: 1,0) node [above right,black] {\((1,0)\)} circle (2.4pt); - \draw [firstcolor,fill=firstcolor] (axis cs: -1,0) node [above left,black] {\((-1,0)\)} circle (2.4pt); - - \end{axis} - \end{tikzpicture} - - - - - Enter f^{\prime}(-1). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    - - Enter f^{\prime}(1). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    -     -     -     - - - - - $zero=Real(0); - $dne = String("DNE"); - - -

    - f(x) = \begin{cases}x^2,\amp x\leq0\\x^5,\amp x\gt0\end{cases} -

    - - The graph of a piecewise-defined function with a relative minimum at (0,0). - -

    - The piecewise-defined function for this problem is equal to x^2 when x is negative, - and x^5 when x is positive. -

    - -

    - The indicated point is (0,0), where there is a relative minimum. - Since the slope of the tangent line approaches 0 as x\to 0 for both x^2 and x^5, - it appears as though the graph will have a horizontal tangent line at the origin. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.5,ymax=1.1,% - xmin=-1.1,xmax=1.1,% - ] - - \addplot [firstcurvestyle,infiniteleft,domain=-1:0] {x^2}; - \addplot [firstcurvestyle,infiniteright,domain=0:1] {x^5}; - - \draw [firstcolor,fill=firstcolor] (axis cs: 0,0) node [below right,black] {\((0,0)\)} circle (2.4pt); - - \end{axis} - \end{tikzpicture} - - - - Enter f^{\prime}(0). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    -     -     -     - - - - - $zero=Real(0); - $dne = String("DNE"); - - -

    - f(x) = \begin{cases}x^2,\amp x\leq0\\x,\amp x\gt0\end{cases} -

    - - The graph of a piecewise-defined function with a relative minimum at the origin. - -

    - For this piecewise-defined function, there is a relative minimum at (0,0), which is the indicated point. - For x\gt 0, f(x)=x, so the slope of the tangent line for positive values of x close to 0 is 1. - For x\lt 0, f(x)=x^2, and the slope of the tangent line approaches 0 as x\to 0^-. - This suggests that f'(0) will be undefined. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-.5,ymax=1.1,% - xmin=-1.1,xmax=1.1,% - ] - - \addplot [firstcurvestyle,infiniteleft,domain=-1:0] {x^2}; - \addplot [firstcurvestyle,infiniteright,domain=0:1] {x}; - - \draw [firstcolor,fill=firstcolor] (axis cs: 0,0) node [below right,black] {\((0,0)\)} circle (2.4pt); - - \end{axis} - \end{tikzpicture} - - - - Enter f^{\prime}(0). If the derivative does not exist at the indicated point, enter DNE for does not exist. - -

    - -

    -     -     -     -     - - - -

    - Find the extreme values of the function on the given interval. - -

    -     - - - - ($b,$c) = random_subset(2,-9..-1,1..9); - ($l,$h) = random_subset(2,1..4); - if($envir{problemSeed}==1){$b=1;$c=4;$l=1;$h=2}; - $low = int(-$b/2) - $l; - $high = int(-$b/2) + $h; - $f = Formula("x^2+$b x+$c")->reduce; - Context("Fraction-NoDecimals"); - $max = Fraction(max($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>-$b/2))); - $min = Fraction(min($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>-$b/2))); - - -

    - f(x) = on [,] -

    - - Enter the maximimum value here, or enter NONE if there isn't one. - -

    - -

    - - Enter the minimimum value here, or enter NONE if there isn't one. - -

    - -

    -     -     -     - - - - - Context("Fraction-NoDecimals"); - $r = random(-9,-1,1); - $s = random(1,9,1); - $d = non_zero_random(-9,9,1); - $high = $s+random(1,2,1); - if($envir{problemSeed}==1){$r=-2;$s=5;$d=3;$high=6}; - $low = Real(0); - $b = Fraction(-3*($r+$s)/2); - $c = 3*$r*$s; - $f = Formula("x^3+$b x^2+$c x+$d")->reduce; - $max = Fraction(max($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>$s))); - $min = Fraction(min($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>$s))); - - -

    - f(x) = on [,] -

    - - Enter the maximimum value here, or enter NONE if there isn't one. - -

    - -

    - - Enter the minimimum value here, or enter NONE if there isn't one. - -

    - -

    -     -     -     - - - - - Context("Fraction-NoDecimals"); - $c = random(2,4,1); - $trig = list_random('sin','cos'); - ($lowd,$highd) = random_subset(2,3,4,6); - if ($trig eq 'sin') { - if (list_random(0,1)) {$low = Formula("pi/$lowd"); $high = Formula("(($highd-1)pi)/$highd")} - else {$low = Formula("(($lowd+1)pi)/$lowd"); $high = Formula("((2*$highd-1)pi)/$highd")} - } else { - if (list_random(0,1)) {$low = Formula("-pi/$lowd"); $high = Formula("pi/$highd")} - else {$low = Formula("(($lowd-1)pi)/$lowd"); $high = Formula("(($highd+1)pi)/$highd")} - } - if($envir{problemSeed}==1){$c=3;$trig='sin';$low=Formula("pi/4");$high=Formula("(2pi)/3")}; - $f = Formula("$c $trig(x)"); - $max = max($f->eval(x=>$low->eval(x=>0)),$f->eval(x=>$high->eval(x=>0)),$c * $f->eval(x=>$low->eval(x=>0)) / abs($f->eval(x=>$low->eval(x=>0)))); - $min = min($f->eval(x=>$low->eval(x=>0)),$f->eval(x=>$high->eval(x=>0)),$c * $f->eval(x=>$low->eval(x=>0)) / abs($f->eval(x=>$low->eval(x=>0)))); - - -

    - f(x) = on \left[,\right] -

    - - Enter the maximimum value here, or enter NONE if there isn't one. - -

    - -

    - - Enter the minimimum value here, or enter NONE if there isn't one. - -

    - -

    -     -     -     - - - - - Context("Fraction-NoDecimals"); - $n = list_random(2,4,6); - $r = random(1,4,1); - if($envir{problemSeed}==1){$n=2;$r=2;}; - $low = -$r; - $high = $r; - $f = Formula("x^$n sqrt($r^2-x^2)"); - $a = -sqrt($n*$r**2/($n+1)); - $b = sqrt($n*$r**2/($n+1)); - $max = max($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>$a),$f->eval(x=>$b),$f->eval(x=>0)); - $min = min($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>$a),$f->eval(x=>$b),$f->eval(x=>0)); - - -

    - f(x) = on [,] -

    - - Enter the maximimum value here, or enter NONE if there isn't one. - -

    - -

    - - Enter the minimimum value here, or enter NONE if there isn't one. - -

    - -

    -     -     -     - - - - - Context("Fraction-NoDecimals"); - $a = list_random(2,3,5,6,7); - $low = random(1,int(sqrt($a)),1); - $high = random(int(sqrt($a))+1,int(sqrt($a))+6); - if($envir{problemSeed}==1){$a=3;$low=1;$high=5;}; - $f = Formula("x+$a/x"); - $max = Fraction(max($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>sqrt($a)))); - $min = min($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>sqrt($a))); - - -

    - f(x) = on [,] -

    - - Enter the maximimum value here, or enter NONE if there isn't one. - -

    - -

    - - Enter the minimimum value here, or enter NONE if there isn't one. - -

    - -

    -     -     -     - - - - - Context("Fraction-NoDecimals"); - $n = random(2,4,2); - $a = random(1,9,1); - $low = random(-9,-1,1); - $high = random(1,9,1); - if($envir{problemSeed}==1){$n=2;$a=5;$low=-3;$high=5;}; - $f = Formula("x^$n/(x^$n+$a)"); - $max = Fraction(max($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>0))); - $min = Fraction(min($f->eval(x=>$low),$f->eval(x=>$high),$f->eval(x=>0))); - - -

    - f(x) = on [,] -

    - - Enter the maximimum value here, or enter NONE if there isn't one. - -

    - -

    - - Enter the minimimum value here, or enter NONE if there isn't one. - -

    - -

    -     -     -     - - - - - Context("Numeric"); - $trig='cos'; - $low = Formula("0"); - $high = Formula("pi"); - $f = Formula("e^x $trig(x)"); - $a = Formula("pi/4"); - $max = Compute("e^($a)/sqrt(2)"); - $min = Compute("-e^(pi)"); - - -

    - f(x) = on [,] -

    - - Enter the maximimum value here, or enter NONE if there isn't one. - -

    - -

    - - Enter the minimimum value here, or enter NONE if there isn't one. - -

    - -

    -     -     -     - - - - - Context("Numeric"); - $trig='sin'; - $low = Formula("0"); - $high = Formula("pi"); - $f = Formula("e^x $trig(x)"); - $a = Formula("3pi/4"); - $max = Compute("e^($a)/sqrt(2)"); - $min = Compute("0"); - - -

    - f(x) = on [,] -

    - - Enter the maximimum value here, or enter NONE if there isn't one. - -

    - -

    - - Enter the minimimum value here, or enter NONE if there isn't one. - -

    - -

    -     -     -     - - - - - Context("Numeric"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $n = random(1,4,1); - $high = random(3,9,1); - if($envir{problemSeed}==1){$n=1;$high=4;}; - $low = Compute("1"); - $a = exp(1/$n); - $f = Formula("ln(x)/x^$n")->reduce; - $max = Compute("1/($n e)"); - $min = Real(0); - - -

    - f(x) = on [,] -

    - - Enter the maximimum value here, or enter NONE if there isn't one. - -

    - -

    - - Enter the minimimum value here, or enter NONE if there isn't one. - -

    - -

    -     -     -     - - - - - Context("Fraction-NoDecimals"); - $d = random(2,5,1); - do {$n = random(1,2*$d-1,1)} until (gcd($n,$d) == 1); - $m = random(int($n/$d)+1,int($n/$d)+5,1); - $high = random(2,4,1); - if($envir{problemSeed}==1){$d=3;$n=2;$m=1;$high=2;}; - $low = Compute("0"); - $md = $m*$d; - $mdn = $md - $n; - $f = Formula("x^($n/$d) - x^$m")->reduce; - $max = max($f->substitute(x=>$low),$f->substitute(x=>$high),Formula("($mdn/$md) ($n/$md)^($n/$mdn)")); - $min = min($f->substitute(x=>$low),$f->substitute(x=>$high),Formula("($mdn/$md) ($n/$md)^($n/$mdn)")); - - -

    - f(x) = on [,] -

    - - Enter the maximimum value here, or enter NONE if there isn't one. - -

    - -

    - - Enter the minimimum value here, or enter NONE if there isn't one. - -

    - -

    -     -     -     -     -     -     -
    -
    - The Mean Value Theorem -

    - We motivate this section with the following question: Suppose you leave your house and drive to your friend's house in a city 100 miles away, - completing the trip in two hours. - At any point during the trip do you necessarily have to be going 50 miles per hour? -

    - -

    - In answering this question, - it is clear that the average - speed for the entire trip is - - 50 - - ( 100 miles in 2 hours), - but the question is whether or not your - instantaneous speed is ever exactly - - 50 - . - More simply, does your speedometer ever read exactly - - 50 - ? The answer, - under some very reasonable assumptions, is yes. -

    - - - -

    - Let's now see why this situation is in a calculus text by translating it into mathematical symbols. -

    - -

    - First assume that the function y = f(t) gives the distance - (in miles) - traveled from your home at time t - (in hours) - where 0\le t\le 2. - In particular, this gives f(0)=0 and f(2)=100. - The slope of the secant line connecting the starting and ending points - (0,f(0)) and (2,f(2)) is therefore - - \frac{\Delta f}{\Delta t} \amp = \frac{f(2)-f(0)}{2-0} - \amp = \frac{100-0}{2} - \amp = 50 \text{ mph} - . -

    - -

    - The slope at any point on the graph itself is given by the derivative \fp(t). - So, since the answer to the question above is yes, - this means that at some time during the trip, - the derivative takes on the value of - - 50 - . - Symbolically, - - \fp(c) = \frac{f(2)-f(0)}{2-0} = 50 - - for some time 0\le c \le 2. -

    - -

    - How about more generally? - Given any function y=f(x) and a range - a\le x\le b does the value of the derivative at some point between a and b have to match the slope of the secant line connecting the points (a,f(a)) and (b,f(b))? - Or equivalently, does the equation - \fp(c) = \frac{f(b)-f(a)}{b-a} have to hold for some a \lt c \lt b? -

    - -

    - Let's look at two functions in an example. -

    - - - Comparing average and instantaneous rates of change - -

    - Consider functions - - f_1(x)\amp=\frac{1}{x^2}\amp f_2(x)\amp= \abs{x} - - with a=-1 and b=1 as shown in . - Both functions have a value of 1 at a and b. - Therefore the slope of the secant line connecting the end points is 0 in each case. - But if you look at the plots of each, - you can see that there are no points on either graph where the tangent lines have slope zero. - Therefore we have found that there is no c in [-1,1] such that - - \fp(c) = \frac{f(1)-f(-1)}{1-(-1)} = 0 - . -

    - -
    - Graphs of two misbehaving functions - -
    - A graph of f_1(x) = 1/x^2 - - - - A graph of a Mean Value Theorem counterexample function. - - -

    - The graph illustrates the function f_1(x) = \frac{1}{x^2}, showcasing a scenario where the Mean Value Theorem is not applicable. - As x approaches zero, the function demonstrates a vertical asymptote, highlighting its discontinuity and the tendency of the values to become infinite. -

    -

    - Within the interval [-1, 1], although the function's value is the same at both endpoints, there is no point where the function's derivative, which represents the slope of the tangent, is zero. - This absence violates the Mean Value Theorem's requirement for the function to be both continuous on the closed interval and differentiable on the open interval, specifically at x = 0. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-1.95,xmax=1.95, - ymin=-.6,ymax=3.5,] - \addplot+[infinite,domain=-1.8:-0.58]{1/x^2}; - \addplot[firstcurvestyle,infinite,domain=.58:1.8]{1/x^2}; - \addplot[tangentlineseg, domain=-1:1]{1}; - \addplot[soliddot] coordinates {(-1,1) (1,1)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - A graph of f_2(x) = \abs{x} - - - - A graph of the absolute value function as a Mean Value Theorem counterexample. - - -

    - Displayed is the function f_2(x) = \abs{x}, chosen to highlight a case where the Mean Value Theorem cannot be applied. - The graph forms a V shape, characteristic of the absolute value function, and is continuous over the interval [-1, 1]. -

    -

    - At the endpoints of the interval, the function attains the same value, yielding a secant line with a slope of zero. - However, due to the sharp corner at the origin x = 0, the function is not differentiable at this point. - This lack of differentiability means that there does not exist a point in the interval where the slope of the tangent line is equal to the slope of the secant line, as required by the Mean Value Theorem. - Thus, the function f_2(x) serves as an example of when the Mean Value Theorem's conditions are not fulfilled. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-1.95,xmax=1.95, - ymin=-.6,ymax=3.5,] - \addplot+[infinite,domain=-1.8:1.8]{abs(x)}; - \addplot[tangentlineseg, domain=-1:1]{1}; - \addplot[soliddot] coordinates {(-1,1) (1,1)}; - \end{axis} - \end{tikzpicture} - - - - -
    -     -
    -     -     - -

    - So what went wrong? - It may not be surprising to find that the discontinuity of f_1 and the corner of f_2 play a role. - If our functions had been continuous and differentiable, - would we have been able to find that special value c? - This is our motivation for the following theorem. -

    - - - The Mean Value Theorem of Differentiation - -

    - Let y=f(x) be a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b). - There exists a value c, a \lt c \lt b, - such that Mean Value Theoremof differentiationderivativeMean Value Theorem - - \fp(c) = \frac{f(b)-f(a)}{b-a} - . -

    - -

    - That is, there is a value c in (a,b) where the instantaneous rate of change of f at c is equal to the average rate of change of f on [a,b]. -

    -     -     - -

    - Note that the reasons that the functions in - fail are indeed that f_1 has a discontinuity on the interval [-1,1] and f_2 is not differentiable at the origin. -

    - - - - -

    - We will give a proof of the Mean Value Theorem below. - To do so, we use a fact, called Rolle's Theorem, stated here. -

    - - - Rolle's Theorem - -

    - Let f be continuous on [a,b] and differentiable on (a,b), - where f(a) = f(b). - There is some c in (a,b) such that \fp(c) = 0. - Rolle's Theorem -

    -     -     - - - -

    - Consider - where the graph of a function f is given, where f(a) = f(b). - It should make intuitive sense that if f is differentiable - (and hence, continuous) - that there would be a value c in (a,b) where \fp(c)=0; - that is, there would be a relative maximum or minimum of f in (a,b). - Rolle's Theorem guarantees at least one; there may be more. -

    - -
    - A graph of f(x) = x^3-5x^2+3x+5, where f(a) = f(b). Note the existence of c, where a\lt c\lt b, where \fp(c)=0. - - - - A visual representation of Rolle's Theorem with a corresponding graph. - - -

    - The image showcases a graph of the polynomial function f(x) = x^3 - 5x^2 + 3x + 5, which is used to illustrate Rolle's Theorem. The function is plotted to show the existence of at least one point c in the open interval (a, b) where the derivative \fp(c) is zero, indicated by the horizontal tangent at the peak of the curve. - The points labeled a and b on the x-axis denote the interval on which the function's endpoints have equal values, satisfying the precondition for Rolle's Theorem. -

    -

    - Rolle's Theorem is stated, asserting that for a function continuous on the closed interval [a, b] and differentiable on the open interval (a, b), where the function values at the endpoints are equal (f(a) = f(b)), there exists at least one point c within (a, b) where the derivative f'(c) is zero. This theorem is a specific case of the Mean Value Theorem. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-1.5,xmax=2.5, - ymin=-5.2,ymax=8, - extra x ticks={-.39,1.22,.333}, - extra x tick labels={$a$,$b$,$c$},] - \addplot+[infinite,domain=-1:2.2, samples=40] {x^3-5*x^2+3*x+5}; - \addplot [secantlineseg,domain=-1:2] {3}; - \addplot [soliddot] coordinates {(-.39,3) (1.22,3) (.3333,5.48)}; - \addplot [tangentlineseg,domain=-.75:1.5] {5.48}; - \end{axis} - \end{tikzpicture} - - - - -
    - - - - - -

    - Rolle's Theorem is really just a special case of the Mean Value Theorem. - If f(a) = f(b), then the average - rate of change on (a,b) is 0, - and the theorem guarantees some c where \fp(c)=0. - We will prove Rolle's Theorem, - then use it to prove the Mean Value Theorem. -

    - - - Proof of Rolle's Theorem -

    - Let f be differentiable on (a,b) where f(a)=f(b). - We consider two cases. -

    - Case 1 -

    - Consider the case when f is constant on [a,b]; - that is, f(x) = f(a) = f(b) for all x in [a,b]. - Then \fp(x) = 0 for all x in [a,b], - showing there is at least one value c in (a,b) where \fp(c)=0. -

    -     - Case 2 -

    - Now assume that f is not constant on [a,b]. - The Extreme Value Theorem guarantees that f has a maximal and minimal value on [a,b], - found either at the endpoints or at a critical value in (a,b). - Since f(a)=f(b) and f is not constant, - it is clear that the maximum and minimum cannot both - be found at the endpoints. - Assume, without loss of generality, - that the maximum of f is not found at the endpoints. - Therefore there is a c in (a,b) such that f(c) is the maximum value of f. - By , - c must be a critical number of f; - since f is differentiable, - we have that \fp(c) = 0, - completing the proof of the theorem. -

    -     -     - -

    - We can now prove the Mean Value Theorem. -

    - - - Proof of the Mean Value Theorem - - -

    - Define the function - - g(x) = f(x) - \frac{f(b)-f(a)}{b-a}x - . -

    - -

    - We know g is differentiable on (a,b) and continuous on [a,b] since f is. - We can show g(a)=g(b) (it is actually easier to show g(b)-g(a)=0, - which suffices). - We can then apply Rolle's theorem to guarantee the existence of - c in (a,b) such that \gp(c) = 0. - But note that - - 0= \gp(c) = \fp(c) - \frac{f(b)-f(a)}{b-a} - ; - hence - - \fp(c) = \frac{f(b)-f(a)}{b-a} - , - which is what we sought to prove. -

    -     - -

    - Going back to the very beginning of the section, - we see that the only assumption we would need about our distance function f(t) is that it be continuous and differentiable for t from 0 to 2 hours - (both reasonable assumptions). - By the Mean Value Theorem, - we are guaranteed a time during the trip where our instantaneous speed is - - 50 - . - This fact is used in practice. - Some law enforcement agencies monitor traffic speeds while in aircraft. - They do not measure speed with radar, - but rather by timing individual cars as they pass over lines painted on the highway whose distances apart are known. - The officer is able to measure the average - speed of a car between the painted lines; - if that average speed is greater than the posted speed limit, - the officer is assured that the driver exceeded the speed limit at some time. -

    - -

    - Note that - is an existence theorem. - It states that a special value c exists, - but it does not give any indication about how to find it. - It turns out that when we need the Mean Value Theorem, - existence is all we need. -

    - - - Using the Mean Value Theorem - -

    - Consider f(x) = x^3+5x+5 on [-3,3]. - Find c in [-3,3] that satisfies . -

    -     - -

    - The average rate of change of f on [-3,3] is: - - \frac{f(3)-f(-3)}{3-(-3)} \amp =\frac{47-(-37)}{6} - \amp =\frac{84}{6} - \amp = 14 - . -

    - -

    - We want to find c such that \fp(c) = 14. - We find \fp(x) = 3x^2+5. - We set this equal to 14 and solve for x. - - \fp(x) \amp = 14 - 3x^2 +5 \amp = 14 - x^2 \amp = 3 - x \amp = \pm \sqrt{3} \approx \pm 1.732 - -

    - -

    - We have found two values c in [-3,3] where the instantaneous rate of change is equal to the average rate of change; - the Mean Value Theorem guaranteed at least one. - In , - f is graphed with a line representing the average rate of change; - the lines tangent to f at x=\pm \sqrt{3} are also given. - Note how these lines are parallel (, have the same slope) to the secant line. -

    - -
    - Demonstrating the Mean Value Theorem in - - - - Graphical illustration of the Mean Value Theorem in action. - - -

    - This graph depicts the function f(x) = x^3 + 5x + 5 over the interval [-3, 3]. - The function is shown as a solid blue curve, and there is a black dashed line that represents the average rate of change of the function over the given interval. -

    -

    - Two red dashed lines are drawn on the graph, indicating the tangent lines to the curve at the points where the function's derivative equals the average rate of change. - These points, marked on the x-axis, are where the slope of the tangent is the same as the slope of the secant line (the black dashed line), which is the essence of the Mean Value Theorem. - The exact points on the x-axis where these tangents touch the curve correspond to the solutions where f'(x) = 3x^2 + 5 is set equal to 14, the average rate of change. -

    -     - - - \begin{tikzpicture} - \begin{axis}[xmin=-3.2,xmax=3.2, - ymin=-45,ymax=51,] - \addplot+[domain=-3:3] {x^3+5*x+5}; - \addplot [secantlineseg, domain=-3:3]{14*(x-3)+47}; - \addplot [soliddot] coordinates {(3,47) (-3,-37) (-1.732,-8.86) (1.732, 18.86)}; - \addplot [tangentlineseg,domain=-2.8:-.1] {14*(x+1.732)-8.86}; - \addplot [tangentlineseg,domain=-.1:3] {14*(x-1.732)+18.86}; - \end{axis} - \end{tikzpicture} - - - - -
    -     - -     - - - -

    - While has practical use - (for instance, - the speed monitoring application mentioned before), - it is mostly used to advance other theory. - We will use it in the next section to relate the shape of a graph to its derivative. -

    - -

    - Before ending this section, we give two important consequences of the Mean - Value Theorem. Each of these consequences has important applications to - mathematical theory, and can be easily understood in the context of the - position and velocity of objects in motion. -

    - -

    - First, we recall that the derivative of any constant function is zero. - Is the converse true? That is, are constant functions the only ones whose derivative is zero? - The Mean Value Theorem says yes. - This officially - establishes our intuition about objects in (or, actually, not in) - motion: if the velocity of an object is 0, then the object's position is - unchanged; it is constant. - Second, if two functions f and g have the same derivative, - what does this tell us about f and g? - The Mean Value Theorem implies that these functions must only differ by a constant; - that is, f(x)=g(x)+C, for some constant C. -

    - -

    - This has an application to motion that is not intuitive to some. Suppose - two objects start moving while 5 apart, - and always move with the same velocity. - Then the two objects will always be 5 apart. - (If two pennies are dropped from the 30th and 31st stories of a tall building at - the same time, they will always be 1 story apart as they fall.) -

    - - - Consequences of the Mean Value Theorem - -

    - Let f, g, and h be differentiable - (and therefore continuous) functions on an in terval I. -

      -
    1. -

      - If \fp(x)=0 for all x in the interval I, - then f is a constant function on I. -

      -
      2. -
    2. -

      - If \gp(x)=h'(x) for all x in I, - then there is a constant C such that g(x)=h(x)+C - for all x in I. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - Choose any two points a and b in the interval I. - By the Mean Value Theorem, - we must have - - \fp(c) = \frac{f(b)-f(a)}{b-a} - - for some c between a and b. - But \fp(c)=0, so f(b)-f(a)=0, or f(a)=f(b). - Since a and b were any two points, - this tells us that f must have the same value at every point; - that is, f must be constant. -

      -
      2. -
    2. -

      - Suppose \gp(x)=h'(x) for each point x in I, - and consider the function f(x)=g(x)-h(x). - By the difference rule for derivatives, we have - - f'(x) = \gp(x)-h'(x)=0 - , - since \gp(x)=h'(x). -

      - -

      - By the previous result, this means that f(x) is a constant function. - That is, f(x)=C for each x in I, giving us - g(x)-h(x)=C, or g(x)=h(x)+C. -

      -
      3. -
    -

    -     -     - - - - - - - - - - - - - - Terms and Concepts - - - - -

    - Explain in your own words what the Mean Value Theorem states. -

    - -     - - - -     - - - - -

    - Explain in your own words what Rolle's Theorem states. -

    - -     - - - -     -     - - - Problems - - -

    - A function f(x) and interval [a,b] are given. - Check if can be applied to f on [a,b]; - if so, find c in (a,b) such that \fp(c)=0. -

    -     - - - - Context("Interval"); - Context()->strings->add('does not apply'=>{}); - $c=Compute("(-1,1)"); - $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; - - -

    - f(x) = 6 on [-1,1] -

    -

    - -

    - - If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. - You may either list individual numbers separated by commas, or use interval notation. - If Rolle's Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Interval"); - Context()->strings->add('does not apply'=>{}); - $c=Compute("does not apply"); - $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; - - -

    - f(x) = 6x on [-1,1] -

    -

    - -

    - - If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. - You may either list individual numbers separated by commas, or use interval notation. - If Rolle's Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->strings->add('does not apply'=>{}); - $c=Fraction("-1/2"); - $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; - - -

    - f(x) = x^2+x-6 on [-3,2] -

    -

    - -

    - - If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. - You may either list individual numbers separated by commas, or use interval notation. - If Rolle's Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->strings->add('does not apply'=>{}); - $c=Fraction("-1/2"); - $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; - - -

    - f(x) = x^2+x-2 on [-3,2] -

    -

    - -

    - - If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. - You may either list individual numbers separated by commas, or use interval notation. - If Rolle's Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Interval"); - Context()->strings->add('does not apply'=>{}); - $c=Compute("does not apply"); - $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; - - -

    - f(x) = x^2+x on [-2,2] -

    -

    - -

    - - If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. - You may either list individual numbers separated by commas, or use interval notation. - If Rolle's Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Interval"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Formula("pi/2"); - $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; - - -

    - f(x) = \sin(x) on [\pi/6,5\pi/6] -

    -

    - -

    - - If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. - You may either list individual numbers separated by commas, or use interval notation. - If Rolle's Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Interval"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Compute("does not apply"); - $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; - - -

    - f(x) = \cos(x) on [0,\pi] -

    -

    - -

    - - If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. - You may either list individual numbers separated by commas, or use interval notation. - If Rolle's Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Interval"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Compute("does not apply"); - $showwork = q![@ explanation_box(message => "Explain why Rolle's Theorem can or cannot be applied.") @]*!; - - -

    - f(x) = \frac{1}{x^2-2x+1} on [0,2] -

    -

    - -

    - - If Rolle's Theorem applies, enter the values of c in (a,b) such that \fp(c)=0. - You may either list individual numbers separated by commas, or use interval notation. - If Rolle's Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     -     - - - -

    - A function f(x) and interval [a,b] are given. - Check if can be applied to f on [a,b]; - if so, find c in (a,b) guaranteed by the Mean Value Theorem. -

    -     - - - - Context("Interval"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Formula("0"); - $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; - - -

    - f(x) = x^2+3x-1 on [-2,2] -

    -

    - -

    - - If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. - You may either list individual numbers separated by commas, or use interval notation. - If the Mean Value Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Fraction("5/2"); - $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; - - -

    - f(x) = 5x^2-6x+8 on [0,5] -

    -

    - -

    - - If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. - You may either list individual numbers separated by commas, or use interval notation. - If the Mean Value Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Formula("3sqrt(2)/2"); - $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; - - -

    - f(x) = \sqrt{9-x^2} on [0,3] -

    -

    - -

    - - If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. - You may either list individual numbers separated by commas, or use interval notation. - If the Mean Value Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Fraction("19/4"); - $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; - - -

    - f(x) = \sqrt{25-x} on [0,9] -

    -

    - -

    - - If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. - You may either list individual numbers separated by commas, or use interval notation. - If the Mean Value Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Compute("does not apply"); - $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; - - -

    - f(x) = \frac{x^2-9}{x^2-1} on [0,2] -

    -

    - -

    - - If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. - You may either list individual numbers separated by commas, or use interval notation. - If the Mean Value Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Formula("4/ln(5)"); - $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; - - -

    - f(x) = \ln(x) on [1,5] -

    -

    - -

    - - If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. - You may either list individual numbers separated by commas, or use interval notation. - If the Mean Value Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Compute("-asec(2/sqrt(pi)), asec(2/sqrt(pi))"); - $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; - - -

    - f(x) = \tan(x) on [-\pi/4,\pi/4] -

    -

    - -

    - - If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. - You may either list individual numbers separated by commas, or use interval notation. - If the Mean Value Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Fraction("-2/3"); - $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; - - -

    - f(x) = x^3-2x^2+x+1 on [-2,2] -

    -

    - -

    - - If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. - You may either list individual numbers separated by commas, or use interval notation. - If the Mean Value Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Compute("5+7sqrt(7)/6, 5-7sqrt(7)/6"); - $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; - - -

    - f(x) = 2x^3-5x^2+6x+1 on [-5,2] -

    -

    - -

    - - If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. - You may either list individual numbers separated by commas, or use interval notation. - If the Mean Value Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->strings->add('does not apply'=>{}); - $c=Compute("sqrt(pi^2-4)/pi, -sqrt(pi^2-4)/pi"); - $showwork = '[@ explanation_box(message => "Explain why the Mean Value Theorem can or cannot be applied.") @]*'; - - -

    - f(x) = \sin^{-1}(x) on [-1,1] -

    -

    - -

    - - If the Mean Value Theorem applies, enter the values of c in (a,b) guaranteed by the Mean Value Theorem. - You may either list individual numbers separated by commas, or use interval notation. - If the Mean Value Theorem does not apply, enter does not apply. - -

    - -

    -     -     -     -     -     -     -
    -
    - Increasing and Decreasing Functions -

    - Our study of nice functions f in this chapter has so far focused on individual points: - points where f is maximal/minimal, - points where \fp(x) = 0 or \fp does not exist, - and points c where \fp(c) is the average rate of change of f on some interval. -

    - -

    - In this section we begin to study how functions behave - between special points; - we begin studying in more detail the shape of their graphs. -

    - -

    - We start with an intuitive concept. - Given the graph in , - where would you say the function is increasing? - Decreasing? - Even though we have not defined these terms mathematically, - one likely answered that f is increasing when x \gt 1 and decreasing when x\lt 1. - We formally define these terms here. -

    - -
    - A graph of a function f used to illustrate the concepts of increasing and decreasing - - - The graph of a function with a relative minimum at x=1 - -

    - The image shows the graph of a function that has a relative minimum at x=1. - At the left edge, the graph is shown with y values close to 4 when x\lt 0. - The y values then get smaller until the minimum is reached at x=1. - Continuing to the right of x=1, the y values start to get bigger again, - reaching values above y=4 when x\gt 2. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.9, - xmax=3.2, - ymin=-.5, - ymax=5.1, - ] - \addplot+[domain=-.5:2.5] {(x-1)^2+2}; - \end{axis} - \end{tikzpicture} - - - -
    - - - - Increasing and Decreasing Functions - -

    - Let f be a function defined on an interval I. - increasing function - decreasing function - - - -

      -
    1. -

      - f is increasing - on I if for every a\lt b in I, f(a) \lt f(b). -

      -
      2. -
    2. -

      - f is decreasing - on I if for every a\lt b in I, f(a) \gt f(b). -

      -
      3. -
    -

    -     -     - - - -

    - Informally, a function is increasing if as x gets larger (, looking left to right) f(x) gets larger. -

    - - - - - - -

    - Our interest lies in finding intervals in the domain of f on which f is either increasing or decreasing. - Such information should seem useful. - For instance, if f describes the speed of an object, - we might want to know when the speed was increasing or decreasing (, when the object was accelerating vs. decelerating). - If f describes the population of a city, - we should be interested in when the population is growing or declining. -

    - -

    - To find such intervals, we again consider secant lines. - Let f be an increasing, - differentiable function on an open interval I, - such as the one shown in , - and let a\lt b be given in I. - The secant line on the graph of f from x=a to x=b is drawn; - it has a slope of (f(b)-f(a))/(b-a). -

    - -

    - But note, since b \gt a and f is increasing, - f(b) \gt f(a). - And these facts imply b-a \gt 0 and f(b)-f(a) \gt 0. - Therefore: -

    - -
    - Examining the secant line of an increasing function - -

    - - \amp\frac{f(b)-f(a)}{b-a} \gt 0 - \implies\amp \text{slope of the secant line} \gt 0 - \implies\amp \text{Average rate of change of }f - \amp\text{ on }[a,b]\text{ is } \gt 0 - . -

    - - - - The graph of a function is shown, along with the secant line between two points on the graph. - -

    - The graph of some function is shown, for x and y between 0 and 2. - Two points (a,f(a)) and (b,f(b)) are marked on the graph, - with b\gt a and f(b)\gt f(a). -

    - -

    - A secant line (dashed) is drawn between these points. - The slope of the secant line is positive, indicating a positive average rate of change for the function on the interval [a,b]. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.2, - xmax=2.2, - ymin=-.5, - ymax=2.2, - extra x ticks={.218,1.67}, - extra x tick labels={$a$,$b$} - ] - \addplot+[domain=-80:-5] ({2*cos(x)-.3},{2*sin(x)+2.3}); - \addplot [secantlineseg] coordinates { ({2*cos(-75)-.3},{2*sin(-75)+2.3}) ({2*cos(-10)-.3},{2*sin(-10)+2.3})}; - \addplot [soliddot] coordinates {(.218,.368)} node [below]{$(a,f(a))$}; - \addplot [soliddot] coordinates {(1.67,1.95)} node [above left]{$(b,f(b))$}; - \end{axis} - \end{tikzpicture} - - - -     -
    - -

    - We have shown mathematically what may have already been obvious: - when f is increasing, its secant lines will have a positive slope. - Now recall that the Mean Value Theorem - guarantees that there is a number c, - where a\lt c\lt b, such that - - \fp(c) = \frac{f(b)-f(a)}{b-a} \gt 0 - . -

    - -

    - By considering all such secant lines in I, - we strongly imply that \fp(x) \gt 0 on I. - A similar statement can be made for decreasing functions. -

    - -

    - Our above logic can be summarized as - If f is increasing, - then \fp is probably positive. - - below turns this around by stating - If \fp is positive, - then f is increasing. - This leads us to a method for finding when functions are increasing and decreasing. -

    - - - Test For Increasing/Decreasing Functions - -

    - Let f be a continuous function on [a,b] and differentiable on (a,b). -

    - -

    -

      -
    1. -

      - If \fp(c) \gt 0 for all c in (a,b), - then f is increasing on [a,b]. -

      -
      2. -
    2. -

      - If \fp(c) \lt 0 for all c in (a,b), - then f is decreasing on [a,b]. -

      -
      3. -
    3. -

      - If \fp(c) =0 for all c in (a,b), - then f is constant on [a,b]. -

      -
      4. -
    -

    - -

    - The conclusions of and also hold if - \fp(c) = 0 for a finite number of nonadjacent values of c in I. -

    -     -     - - - - - -

    - Let f be differentiable on an interval I and let a and b be in I where - \fp(a) \gt 0 and \fp(b)\lt 0. - If \fp is continuous on [a,b], - it follows from the Intermediate Value Theorem that there must be some value - c between a and b where \fp(c) = 0. - (It turns out that this is still true even if \fp is not continuous on [a,b].) - This leads us to the following method for finding intervals on which a function is increasing or decreasing. -

    - - - Finding Intervals on Which <m>f</m> is Increasing or Decreasing -

    - Let f be a continuous function on an interval I. - To find intervals on which f is increasing and decreasing: - increasing functionfinding intervals - decreasing functionfinding intervals -

    - -

    -

      -
    1. -

      - If not stated, find the domain of f, D. Begin a number line that only includes D. -

      -
      2. -
    2. -

      - Find the critical values of f. - That is, find all c in the domain of f where - \fp(c) = 0 or \fp is not defined. - (Note: Any values of c not - in the domain of f where \fp(c) is undefined should already be marked on your number line from Step). -

      -
      3. -
    3. -

      - Use the critical values to divide D into subintervals. -

      -
      4. -
    4. -

      - Pick any point p in each subinterval, - and find the sign of \fp(p). -

      - -

      -

        -
      1. -

        - If \fp(p) \gt 0, then f is increasing on that subinterval. -

        -
        2. -
      2. -

        - If \fp(p)\lt 0, then f is decreasing on that subinterval. -

        -
        3. -
      -

      -
      5. -
    -

    - -

    - Note that although allows us to use determine that a function is increasing or decreasing on a closed interval, - it is conventional to state the intervals of increase and decrease as open intervals. - We will follow this convention in the examples that follow, - but it is also acceptable to answer using closed intervals. -

    - -

    - In particular, one should note the following: -

      -
    • -

      - If \fp(x)\gt 0 on (a,b) and on (b,c), with \fp(b)=0, - then we should say that f is increasing on (a,c) (or on [a,c]) - the zero of the derivative should be included. -

      -
      • -
    • -

      - If \fp(x)\gt 0 on (a,b) and on (b,c), - but f(b) is undefined (or f is discontinuous at b), - then we should not include the point b in our interval. - Instead, we say that f is increasing on (a,b) and (b,c), - or on [a,b) and (b,c]. -

      -
      • -
    -

    -     - - - -

    - We demonstrate using this process in the following example. -

    - - - Finding intervals of increasing/decreasing - -

    - Let f(x) = x^3+x^2-x+1. - Find intervals on which f is increasing or decreasing. -

    -     - -

    - Since an interval was not specified for us to consider, using , - the domain of f is \mathbb{R} or (-\infty,\infty). - Next, we find the critical values of f. - We have \fp(x) = 3x^2+2x-1 = (3x-1)(x+1), - so \fp(x) = 0 when x=-1 and when x=1/3. - \fp is never undefined. -

    - -

    - We thus break the domain - (in this case the (-\infty, \infty)) - into three subintervals based on the two critical values we just found: - (-\infty,-1), (-1,1/3) and (1/3,\infty). - This is shown in . -

    - -
    - Number line for f in - - - Number line showing the critical points of f - -

    - A number line is shown with two marked points. - The points are labeled with the values -1 and 1/3, - which are the critical points of the function f(x)=x^3+x^2-x+1. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - numberline, - xmin=-2, - xmax=1, - extra x ticks={-1, 0.3333}, - extra x tick labels={$-1$,$1/3$}, - ] - \addplot[guideline] coordinates {(-1,0) (-1,2)}; - \addplot[guideline] coordinates {(0.3333,0) (0.3333,2)}; - \end{axis} - \end{tikzpicture} - - -
    - -

    - We now pick a value p in each subinterval and find the sign of \fp(p). - All we care about is the sign, - so we do not actually have to fully compute \fp(p); - pick nice values that make this simple. -

    - -

    -

    -
  • - Subinterval 1: <m>(-\infty,-1)</m> -

    - We (arbitrarily) pick p=-2. - We can compute \fp(-2) directly: - \fp(-2) = 3(-2)^2+2(-2)-1=7\gt 0. - We conclude that f is increasing on (-\infty,-1). -

    - -

    - Note we can arrive at the same conclusion without computation. - For instance, we could choose p=-100. - The first term in \fp(-100), - , 3(-100)^2 is clearly positive and very large. - The other terms are small in comparison, - so we know \fp(-100)\gt 0. - All we need is the sign. -

    -
    • -
  • - Subinterval 2: <m>(-1,1/3)</m> -

    - We pick p=0 since that value seems easy to deal with. - \fp(0) = -1\lt 0. - We conclude f is decreasing on (-1,1/3). -

    -
    • -
  • - Subinterval 3: <m>(1/3,\infty)</m> -

    - Pick an arbitrarily large value for p\gt 1/3 and note that \fp(p) =3p^2+2p-1 \gt 0. - We conclude that f is increasing on (1/3,\infty). -

    -
    • -
    -

    - -

    - summarizes our work. -

    - -
    - Completed number line for f in - - - - -

    - A number line is shown with two marked points. - The points are labeled with the values -1 and 1/3, - which are the critical points of the function f(x)=x^3+x^2-x+1. -

    - -

    - The marked points divide the number line into three intervals. - Above each interval, there is text indicating the sign of f', - and whether the function f is increasing or decreasing. -

    - -

    - The number line indicates the following: -

      -
    • -

      - For x\lt -1, \fp(x)\gt 0, - and f is increasing. -

      -
      • - -
    • -

      - For -1\lt x\lt 1/3, \fp(x)\lt 0, - and f is decreasing. -

      -
      • - -
    • -

      - For x\gt 1/3, \fp(x)\gt 0, - and f is increasing. -

      -
      • -
    -

    -     - - \begin{tikzpicture} - \begin{axis}[ - numberline, - xmin=-2, - xmax=1, - extra x ticks={-1, 0.3333}, - extra x tick labels={$-1$,$1/3$}, - ] - \addplot[guideline] coordinates {(-1,0) (-1,2)}; - \addplot[guideline] coordinates {(0.3333,0) (0.3333,2)}; - \addplot[mark=none] coordinates {(-1.6666,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; - \addplot[mark=none] coordinates {(-0.3333,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }}; - \addplot[mark=none] coordinates {(1,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - We can verify our calculations by considering , - where f is graphed. - The graph also presents \fp; - note how \fp\gt 0 when f is increasing and - \fp\lt 0 when f is decreasing. -

    - -
    - A graph of f(x) in , showing where f is increasing and decreasing - - - The graphs of a function and its derivative. - -

    - In the image, we see the graph of f(x)=x^3+x^2-x+1 (solid, in blue), - along with the graph of its derivative f'(x)=3x^2+2x-1 (dash-dotted, in red). -

    - -

    - In the regions where f'(x) is positive (above the x axis), f(x) is increasing. - Where f'(x) is negative (below the x axis), f(x) is decreasing. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-2.2, - xmax=2.2, - ymin=-3, - ymax=11, - extra x ticks={.33}, - extra x tick labels={$1/3$}, - ] - \addplot+[domain=-2:2] {x^3+x^2-x+1} node[pos=0.6, right]{$f(x)$}; - \addplot+[domain=-1.8:1.5] {3*x^2+2*x-1} node[pos=0.1, above right] {$\fp(x)$}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - -

    - One is justified in wondering why so much work is done when the graph seems to make the intervals very clear. - We give three reasons why the above work is worthwhile. -

    - -

    - First, the points at which f switches from increasing to decreasing are not precisely known given a graph. - The graph shows us something significant happens near x=-1 and x=0.3, - but we cannot determine exactly where from the graph. -

    - -

    - One could argue that just finding critical values is important; - once we know the significant points are x=-1 and x=1/3, - the graph shows the increasing/decreasing traits just fine. - That is true. - However, the technique prescribed here helps reinforce the relationship between increasing/decreasing and the sign of \fp. - Once mastery of this concept - (and several others) - is obtained, - one finds that either (a) just the critical points are computed and the graph shows all else that is desired, - or (b) a graph is never produced, - because determining increasing/decreasing using \fp is straightforward and the graph is unnecessary. - So our second reason why the above work is worthwhile is this: - once mastery of a subject is gained, - one has options for finding needed information. - We are working to develop mastery. -

    - -

    - Finally, our third reason: - many problems we face in the real world are very complex. - Solutions are tractable only through the use of computers to do many calculations for us. - Computers do not solve problems - on their own, however; - they need to be taught (, programmed) to do the right things. - It would be beneficial to give a function to a computer and have it return maximum and minimum values, - intervals on which the function is increasing and decreasing, - the locations of relative maxima, etc. - The work that we are doing here is easily programmable. - It is hard to teach a computer to - look at the graph and see if it is going up or down. - It is easy to teach a computer to - determine if a number is greater than or less than 0. -

    - -

    - In - we learned the definition of relative maxima and minima and found that they occur at critical points. - We are now learning that functions can switch from increasing to decreasing - (and vice-versa) - at critical points. - This new understanding of increasing and decreasing creates a great method of determining whether a critical point corresponds to a maximum, minimum, - or neither. - Imagine a function increasing until a critical point at x=c, - after which it decreases. - A quick sketch helps confirm that f(c) must be a relative maximum. - A similar statement can be made for relative minimums. - We formalize this concept in a theorem. -

    - - - First Derivative Test - -

    - Let f be continuous on an interval I, - and differentiable on I, - except possibly at c, where c is a critical number in I. - - First Derivative Test - derivativeFirst Deriv. Test - extremaand First Deriv. Test - maximumand First Deriv. Test - minimumand First Deriv. Test - -

      -
    1. -

      - If the sign of \fp switches from positive to negative at c, - then f(c) is a relative maximum of f. -

      -
      2. -
    2. -

      - If the sign of \fp switches from negative to positive at c, - then f(c) is a relative minimum of f. -

      -
      3. -
    3. -

      - If \fp is positive (or, - negative) before and after c, - then f(c) is not a relative extrema of f. -

      -
      4. -
    -

    -     -     - - - - - - Importance of Continuity -

    - The continuity of f when using the first derivative test is very important. - Without continuity, almost anything can happen at a critical number. - For example, - we can construct a piecewise function where the sign of \fp switches to positive to negative at c and f(c) is - not a local maximum. - This is shown in . -

    - -
    - A discontinuous function where \fp changes sign at 1, but f(1) is not a local maximum - - The graph of a piecewise-linear function with a discontinuity at x=1. - -

    - The image is the graph of a piecewise-linear function. - For x\lt 1, the graph is a straight line with positive slope, - ending at the point (1,2), where there is a hollow dot, indicating that this point is not part of the graph. - Beginning at the point (1,1) (a solid dot, which is included), - we have another line, with negative slope, for x\geq 1. -

    - -

    - The graph gives an example of a function whose derivative changes sign at x=1. - But since f is not continuous at 1, there is neither a maximum nor a minimum at that point. - The first derivative test can only be applied to continuous functions. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1, - xmax=3, - ymin=-3, - ymax=4, - ] - \addplot+[leftarrow,domain=-.9:1] {x+1}; - \addplot+[firstcurvestyle,rightarrow,domain=1:2.5] {-1*(x-1)+1}; - \addplot[soliddot] coordinates{(1,1)}; - \addplot[hollowdot] coordinates{(1,2)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - - - Using the First Derivative Test - -

    - Find the intervals on which f is increasing and decreasing, - and use the First Derivative Test - to determine the relative extrema of f, where - - f(x) = \frac{x^2+3}{x-1} - . -

    -     - -

    - We start by noting the domain of f: - (-\infty,1)\cup(1,\infty). -

    - -

    - Since f is not defined at x=1 - (it has a vertical asymptote), - the increasing/decreasing nature of f could switch at this value. - We know that \fp(1) will be undefined since f is discontinuous at 1. - We do not formally consider x=1 to be a critical value of f, - but we will use 1 to subdivide the real number line. -

    - -

    - Using the , we find - - \fp(x) = \frac{x^2-2x-3}{(x-1)^2} - . -

    - -

    - We need to find the critical values of f; - we want to know when \fp(x)=0 and when \fp is not defined. - That latter is straightforward: - when the denominator of \fp(x) is 0, \fp is undefined. - That occurs when x=1, - which we've already recognized as an important value, - but not a critical number. -

    - -

    - \fp(x)=0 when the numerator of \fp(x) is 0. - That occurs when x^2-2x-3 = (x-3)(x+1) = 0; - , when x=-1,3. -

    - -

    - We have found that f has two critical numbers, x=-1,3, - and at x=1 something important might also happen. - These three numbers divide the real number line into four subintervals: - - (-\infty,-1), (-1, 1), (1,3), \text{ and } (3,\infty) - . -

    - -

    - Pick a number p from each subinterval and test the sign of \fp at p to determine whether f is increasing or decreasing on that interval. - Again, we do well to avoid complicated computations; - notice that the denominator of \fp is always - positive so we can ignore it during our work. -

    - -

    -

    -
  • - Interval 1: <m>(-\infty,-1)</m> -

    - Choosing a very small number (, a negative number with a large magnitude) p returns - p^2-2p-3 in the numerator of \fp; - that will be positive. - Hence f is increasing on (-\infty,-1). -

    -
    • -
  • - Interval 2: <m>(-1,1)</m> -

    - Choosing 0 seems simple: - \fp(0)=-3\lt 0. - We conclude f is decreasing on (-1,1). -

    -
    • -
  • - Interval 3: <m>(1,3)</m> -

    - Choosing 2 seems simple: - \fp(2) = -3\lt 0. - Again, f is decreasing. -

    -
    • -
  • - Interval 4: <m>(3,\infty)</m> -

    - Choosing an very large number p from this subinterval will give a positive numerator and - (of course) - a positive denominator. - So f is increasing on (3,\infty). -

    -
    • -
    -

    - -

    - In summary, f is increasing on the intervals (-\infty,-1) and - (3,\infty) and is decreasing on the intervals (-1,1) and (1,3). - Since at x=-1, - the sign of \fp switched from positive to negative, - - states that f(-1) is a relative maximum of f. - At x=3, - the sign of \fp switched from negative to positive, - meaning f(3) is a relative minimum. - At x=1, f is not defined, - so there is no relative extremum at x=1. - As previously stated, x=1 is a vertical asymptote of f. -

    - -
    - Number line for f in - - - - Number line for this example, showing critical points of f and intervals of increase and decrease - -

    - On a number line, three points are marked. - The first point is labeled below with -1, and above with rel max. - This indicates that f has a relative maximum at the critical point x=-1. -

    - -

    - The next point is labeled below with 1 and above with VA, - indicating that f has a vertical asymptote at x=1. -

    - -

    - The last point is labeled below with 3 and above with rel min, - indicating that f has a relative minimum at the critical point x=3. -

    - -

    - In between these points there is text indicating the sign of \fp(x), - and whether f is increasing or decreasing, as follows: -

      -
    • -

      - For x\lt -1, f'\gt 0 and f is increasing -

      -
      • - -
    • -

      - For -1\lt x\lt 1, \fp\lt 0 and f is decreasing -

      -
      • - -
    • -

      - For 1\lt x\lt 3, \fp\lt 0 and f is decreasing -

      -
      • - -
    • -

      - For x\gt 3, \fp\gt 0 and f is increasing -

      -
      • -
    -

    -     - - \begin{tikzpicture} - \begin{axis}[ - numberline, - xmin=-3, - xmax=4.9, - extra x ticks={-1, 1, 3}, - extra x tick labels={$-1$,$1$, $3$}, - ] - \addplot[guideline] coordinates {(-1,0) (-1,2)}; - \addplot[guideline] coordinates {(1,0) (1,2)}; - \addplot[guideline] coordinates {(3,0) (3,2)}; - \addplot[mark=none] coordinates {(-2,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; - \addplot[mark=none] coordinates {(0,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }}; - \addplot[mark=none] coordinates {(2,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }}; - \addplot[mark=none] coordinates {(4,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; - \addplot[mark=none] coordinates {(-1,2)} node[above] {\parbox{3em}{\centering rel\\max}}; - \addplot[mark=none] coordinates {(1,2)} node[above] {\parbox{3em}{\centering VA}}; - \addplot[mark=none] coordinates {(3,2)} node[above] {\parbox{3em}{\centering rel\\min}}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - This is summarized in the number line shown in . - Also, shows a graph of f, - confirming our calculations. - This figure also shows \fp, - again demonstrating that f is increasing when \fp\gt 0 and decreasing when \fp\lt 0. -

    - -
    - A graph of f(x) in , showing where f is increasing and decreasing - - - The graphs of a function and its derivative for the current example. - -

    - The graphs of f(x)=x^{8/3}-4x^{2/3} (solid, in blue) and its derivative (dash-dotted, in red) are shown. - The graph of f resembles a large letter W. - It has two relative minima, at x=1 and x=-1, both with horizontal tangents, - and a relative maximum at x=0, where there is a cusp. -

    - -

    - The graph of f'(x) is also shown. The y axis is a vertical asymptote for f'(x), - so the graph appears as two pieces, one on either side of the axis. - We can see that the x intercepts for f'(x) correspond to the two relative minima of f(x). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-4.2, - xmax=4.2, - ymin=-21, - ymax=21, - ] - \addplot[firstcurvestyle,domain=-4:.8,samples=40] {(x^2+3)/(x-1)}; - \addplot[firstcurvestyle,domain=1.25:4] {(x^2+3)/(x-1)} node[pos=0.2,right]{$f(x)$}; - \addplot[secondcurvestyle,domain=-4:.5,samples=40] {(x^2-2*x-3)/((x-1)^2)} node[pos=0.05,above]{$\fp(x)$}; - \addplot[secondcurvestyle,domain=1.5:4] {(x^2-2*x-3)/((x-1)^2)} ; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - -

    - One is often tempted to think that functions always alternate - increasing, decreasing, increasing, - decreasing, around critical values. - Our previous example demonstrated that this is not always the case. - While x=1 was not technically a critical value, - it was an important value we needed to consider. - We found that f was decreasing on - both sides of x=1. -

    - -

    - We examine one more example. -

    - - - Using the First Derivative Test - -

    - Find the intervals on which - f(x) = x^{8/3}-4x^{2/3} is increasing and decreasing and identify the relative extrema. -

    -     - -

    - The domain of f is \mathbb{R} - (you can take the odd root of both positive and negative nubmers). - Next, we take the first derivative. - Since we know we want to solve \fp(x) = 0, - we will do some algebra after taking the derivative. - - f(x) \amp = x^{\frac{8}{3}}-4x^{\frac{2}{3}} - \fp(x) \amp = \frac{8}{3} x^{\frac{5}{3}} - \frac{8}{3}x^{-\frac{1}{3}} - \amp = \frac{8}{3}x^{-\frac{1}{3}}\left(x^{\frac{6}{3}}-1\right) - \amp =\frac{8}{3}x^{-\frac{1}{3}}\left(x^2-1\right) - \amp =\frac{8}{3}x^{-\frac{1}{3}}(x-1)(x+1) - . -

    - -

    - This derivation of \fp shows that - \fp(x) = 0 when x=\pm 1 and \fpis not defined when x=0. - Thus we have three critical values, - breaking the number line into four subintervals as shown in . -

    - -
    - Number line for f in - - - - Number line showing critical points and intervals of increase and decrease - -

    - On a number line, three points are marked. - The first point is labeled below with -1, and above with rel min, - indicating that f has a relative minimum at the critical point x=-1. -

    - -

    - The next point is labeled below with 0 and above with rel max, - indicating that f has a relative maximum at the critical point x=0. -

    - -

    - The last point is labeled below with 1 and above with rel min, - indicating that f has a relative minimum at the critical point x=1. -

    - -

    - In between these points there is text indicating the sign of \fp(x), - and whether f is increasing or decreasing, as follows: -

      -
    • -

      - For x\lt -1, f'\tt 0 and f is decreasing -

      -
      • - -
    • -

      - For -1\lt x\lt 0, \fp\gt 0 and f is increasing -

      -
      • - -
    • -

      - For 0\lt x\lt 1, \fp\lt 0 and f is decreasing -

      -
      • - -
    • -

      - For x\gt 1, \fp\gt 0 and f is increasing -

      -
      • -
    -

    -     - - \begin{tikzpicture} - \begin{axis}[ - numberline, - xmin=-2, - xmax=2, - extra x ticks={-1, 0, 1}, - extra x tick labels={$-1$,$0$, $1$}, - ] - \addplot[guideline] coordinates {(-1,0) (-1,2)}; - \addplot[guideline] coordinates {(0,0) (0,2)}; - \addplot[guideline] coordinates {(1,0) (1,2)}; - \addplot[mark=none] coordinates {(-1.5,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }}; - \addplot[mark=none] coordinates {(-0.5,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; - \addplot[mark=none] coordinates {(0.5,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }}; - \addplot[mark=none] coordinates {(1.5,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }}; - \addplot[mark=none] coordinates {(-1,2)} node[above] {\parbox{3em}{\centering rel\\min}}; - \addplot[mark=none] coordinates {(0,2)} node[above] {\parbox{3em}{\centering rel\\max}}; - \addplot[mark=none] coordinates {(1,2)} node[above] {\parbox{3em}{\centering rel\\min}}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    -

    -
  • - Interval 1: <m>(\infty,-1)</m> -

    - We choose p=-2; we can easily verify that \fp(-2)\lt 0. - So f is decreasing on (-\infty,-1). -

    -
    • -
  • - Interval 2: <m>(-1,0)</m> -

    - Choose p=-1/2. - Once more we practice finding the sign of \fp(p) without computing an actual value. - We have \fp(p) = (8/3)p^{-1/3}(p-1)(p+1); - find the sign of each of the three terms at the chosen value of p. - - \fp(p) = \frac{8}{3} \cdot \underbrace{p^{-\frac{1}{3}}}_{\lt 0}\cdot \underbrace{(p-1)}_{\lt 0}\underbrace{(p+1)}_{\gt 0} - . -

    - -

    - We have a negative negative positive - giving a positive number; - f is increasing on (-1,0). -

    -
    • -
  • - Interval 3: <m>(0,1)</m> -

    - We do a similar sign analysis as before, - using p in (0,1). - - \fp(p) = \frac{8}{3} \cdot \underbrace{p^{-\frac{1}{3}}}_{\gt 0}\cdot \underbrace{(p-1)}_{\lt 0}\underbrace{(p+1)}_{\gt 0} - . -

    - -

    - We have two positive factors and one negative factor; - \fp(p)\lt 0 and so f is decreasing on (0,1). -

    -
    • -
  • - Interval 4: <m>(1,\infty)</m> -

    - Similar work to that done for the other three intervals shows that - \fp(x)\gt 0 on (1,\infty), - so f is increasing on this interval. -

    -
    • -
    -

    - -

    - We conclude by stating that f is increasing on the intervals (-1,0) and - (1,\infty) and decreasing on the intervals (-\infty,-1) and (0,1). - The sign of \fpchanges from negative to positive around x=-1 and x=1, - meaning by - that f(-1) and f(1) are relative minima of f. - As the sign of \fp changes from positive to negative at x=0, - we have a relative maximum at f(0). - - shows a graph of f, confirming our result. - We also graph \fp, - highlighting once more that f is increasing when \fp\gt 0 and is decreasing when \fp\lt 0. -

    - -
    - A graph of f(x) in , showing where f is increasing and decreasing - - - - - \begin{tikzpicture} - \begin{axis}[ - xmin=-3.2, - xmax=3.2, - ymin=-5, - ymax=11, - ] - \addplot+[rightarrow,domain=0.0001:3,samples=50] {x^(8/3)-4*x^(2/3)}; - \addplot[firstcurvestyle,leftarrow,domain=-3:-0.001,samples=50] {(-x)^(8/3)-4*(-x)^(2/3)} node[pos=0.1, above right]{$f(x)$}; - \addplot[secondcurvestyle,domain=0.2:2,samples=50] {(8*(x^2 - 1))/(3 *x^(1/3))} node[pos=0.7,left] {$\fp(x)$}; - \addplot[secondcurvestyle,domain=-2:-0.2,samples=50] {(8*((-x)^2 - 1))/(3 *(-x)^(1/3))}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - -

    - We have seen how the first derivative of a function helps determine when the graph of a function is going up - or down. In the next section, - we will see how the second derivative helps determine how the graph of a function curves. -

    - - - - Terms and Concepts - - - - -

    - In your own words describe what it means for a function to be increasing. -

    - -     - - - -     - - - - -

    - What does a decreasing function look like? -

    - -     - - - -

    - Answers will vary, but should indicate that - the graph slopes downward from left to right. -

    -     - -     - - - -

    - Sketch a graph of a function on [0,2] that is increasing, - where it is increasing quickly - near x=0 and increasing slowly near x=2. -

    -     - - - -

    - Answers will vary; - graphs should be steeper near x=0 than near x=2. -

    -     -     - - - - -

    - Give an example of a function describing a situation where it is bad - to be increasing and good to be decreasing. -

    - -     - - - -     - - - - -

    - - Functions always switch from increasing to decreasing, - or decreasing to increasing, at critical points. -

    -     - -

    - False; for instance, - y=x^3 is always increasing though it has a critical point at x=0. -

    -     - -     - - - - -

    - A function f has derivative \fp(x) = (\sin(x) +2)e^{x^2+1}, - where \fp(x) \gt 1 for all x. - Is f increasing, decreasing, - or can we not tell from the given information? Why or why not? -

    - -     - - - -

    - The function is increasing. - Since \fp(x) \gt 1, we know in - particular that \fp(x) \gt 0, - so f is increasing by - . -

    -     - -     -     - - - Problems - - - -

    - A function f(x) is given. - Graph f and \fp on the same axes - (using technology is permitted) - and verify . -

    -     - - - -

    - f(x) = 2x+3 -

    -     -     - - - -

    - f(x) = x^2-3x+5 -

    -     -     - - - -

    - f(x) = \cos(x) -

    -     -     - - - -

    - f(x) = \tan(x) -

    -     -     - - - -

    - f(x) = x^3-5x^2+7x-1 -

    -     -     - - - -

    - f(x) = 2x^3-x^2+x-1 -

    -     -     - - - -

    - f(x) = x^4-5x^2+4 -

    -     -     - - - -

    - f(x) = \frac{1}{x^2+1} -

    -     -     -     - - - -

    - A function f(x) is given. -

      -
    1. -

      - Give the domain of f. -

      -
      2. -
    2. -

      - Find the critical numbers of f. -

      -
      3. -
    3. -

      - Find the intervals on which f is increasing. -

      -
      4. -
    4. -

      - Find the intervals on which f is decreasing. -

      -
      5. -
    5. -

      - Use the First Derivative Test to determine which critical numbers are a relative maximum. -

      -
      6. -
    6. -

      - Use the First Derivative Test to determine which critical numbers are a relative minimum. -

      -
      7. -
    -

    -     - - - - - Context("Fraction"); - ($r,$s) = random_subset(2,-9..9); - $m = ($r+$s)/2; - $crit=List("$m"); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $f = Formula("x^2-($r+$s)x+$r*$s")->reduce; - $domain = Compute("(-inf,inf)"); - $inc = List(Interval("[$m,inf)")); - $dec = List(Interval("(-inf,$m]")); - $max = List("none"); - $min = List("$m"); - - -

    - f(x)= -

    - - Give the domain of f using interval notation. - Use U for union if the domain consists of more than one interval. - -

    - -

    - - Enter the critical numbers of f, - separating with commas if needed. - If there are no critical numbers, - enter none. - -

    - -

    - - List the maximal intervals where f is increasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is decreasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the critical numbers at which there is a relative maximum. - If there are none, enter none. - -

    - -

    - - List the critical numbers at which there is a relative minimum. - If there are none, enter none. - -

    - -

    -     - - -     -     - - - - - Context("Fraction"); - ($a,$b) = random_subset(2,-9..-1,1..9); - $m = Fraction(-2*$a,3); - ($m,$n) = num_sort(0,$m); - $crit=List($m,$n); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $f = Formula("x^3 + $a x^2 + $b")->reduce; - $domain = Compute("(-inf,inf)"); - $inc = List(Interval("(-inf,$m]"),Interval("[$n,inf)")); - $dec = List(Interval("[$m,$n]")); - $max = List("$m"); - $min = List("$n"); - - -

    - f(x)= -

    - - Give the domain of f using interval notation. - Use U for union if the domain consists of more than one interval. - -

    - -

    - - Enter the critical numbers of f, - separating with commas if needed. - If there are no critical numbers, - enter none. - -

    - -

    - - List the maximal intervals where f is increasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is decreasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the critical numbers at which there is a relative maximum. - If there are none, enter none. - -

    - -

    - - List the critical numbers at which there is a relative minimum. - If there are none, enter none. - -

    - -

    -     - - -     -     - - - - - Context("Fraction"); - #roots of derivative - do { - $s = Fraction(list_random(1,2,4,5,7,8)*list_random(-1,1), list_random(3,6,9)); # make b div by 3 - ($a,$b) = $s->value; - if ($a % 2 == 0) {$t = Fraction(non_zero_random(-8,8,2),random(1,9,2));} - elsif ($b % 2 == 0) {$t = Fraction(random(-9,9,2),random(2,8,2));} - else {$t = Fraction(random(-9,9,2),random(1,9,2));} - } until ($s != $t); - ($c,$d) = $t->value; - # f'(x) = (bx - a)(dx - c) = bd x^2 - (bc+ad) x + ac - ($m,$n) = num_sort($s,$t); - $crit=List($m,$n); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $A = $b*$d/3; - $B = -($b*$c+$a*$d)/2; - $C = $a*$c; - $D = non_zero_random(-9,9,1); - $f = Formula("$A x^3 + $B x^2 + $C x + $D")->reduce; - $domain = Compute("(-inf,inf)"); - $inc = $inc = List(Interval("(-inf,$m]"),Interval("[$n,inf)")); - $dec = List(Interval("[$m,$n]")); - $max = List($m); - $min = List($n); - - -

    - f(x)= -

    - - Give the domain of f using interval notation. - Use U for union if the domain consists of more than one interval. - -

    - -

    - - Enter the critical numbers of f, - separating with commas if needed. - If there are no critical numbers, - enter none. - -

    - -

    - - List the maximal intervals where f is increasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is decreasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the critical numbers at which there is a relative maximum. - If there are none, enter none. - -

    - -

    - - List the critical numbers at which there is a relative minimum. - If there are none, enter none. - -

    - -

    -     - - -     -     - - - - - $r = non_zero_random(-5,5,1); - $crit=List($r); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $A = 1; - $B = -3*$r; - $C = 3*($r)**2; - $D = -($r)**3; - $f = Formula("$A x^3 + $B x^2 + $C x + $D")->reduce; - $domain = Compute("(-inf,inf)"); - $inc = List(Interval("(-inf,inf)")); - $dec = List(Compute("None")); - $max = List(Compute("None")); - $min = List(Compute("None")); - - -

    - f(x)= -

    - - Give the domain of f using interval notation. - Use U for union if the domain consists of more than one interval. - -

    - -

    - - Enter the critical numbers of f, - separating with commas if needed. - If there are no critical numbers, - enter none. - -

    - -

    - - List the maximal intervals where f is increasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is decreasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the critical numbers at which there is a relative maximum. - If there are none, enter none. - -

    - -

    - - List the critical numbers at which there is a relative minimum. - If there are none, enter none. - -

    - -

    -     - - -     -     - - - - - $r = non_zero_random(-5,5,1); - $crit=List($r); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $A = 1; - $B = -2*$r; - $C = ($r)**2+random(1,9,1); - $f = Formula("1/($A x^2 + $B x + $C)")->reduce; - $domain = Compute("(-inf,inf)"); - $inc = List(Interval("(-inf,$r]")); - $dec = List(Interval("[$r,inf)")); - $max = List(Compute($r)); - $min = List(Compute("None")); - - -

    - f(x)= -

    - - Give the domain of f using interval notation. - Use U for union if the domain consists of more than one interval. - -

    - -

    - - Enter the critical numbers of f, - separating with commas if needed. - If there are no critical numbers, - enter none. - -

    - -

    - - List the maximal intervals where f is increasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is decreasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the critical numbers at which there is a relative maximum. - If there are none, enter none. - -

    - -

    - - List the critical numbers at which there is a relative minimum. - If there are none, enter none. - -

    - -

    -     - - -     -     - - - - - ($r,$t) = random_subset(2,1..9); - $R = ($r)**2; - $T = ($t)**2; - $crit=List(0); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $f = Formula("(x^2-$R)/(x^2-$T)")->reduce; - $domain = Compute("(-inf,-$t)U(-$t,$t)U($t,inf)"); - $inc = List(Interval("[0,$t)"),Interval("($t,inf)")); - $dec = List(Interval("(-inf,-$t)"),Interval("(-$t,0]")); - $max = List(Compute("None")); - $min = List(0); - if ($r < $t) { - ($inc,$dec) = ($dec,$inc); - ($max,$min) = ($min,$max); - } - - -

    - f(x)= -

    - - Give the domain of f using interval notation. - Use U for union if the domain consists of more than one interval. - -

    - -

    - - Enter the critical numbers of f, - separating with commas if needed. - If there are no critical numbers, - enter none. - -

    - -

    - - List the maximal intervals where f is increasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is decreasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the critical numbers at which there is a relative maximum. - If there are none, enter none. - -

    - -

    - - List the critical numbers at which there is a relative minimum. - If there are none, enter none. - -

    - -

    -     - - -     -     - - - - - ($r,$s) = num_sort(random_subset(2,-9..-1,1..9)); - $rs = $r*$s; - $crit = ($rs > 0) ? List(Compute(-sqrt($rs)),-Compute(-sqrt($rs))) : List(Compute("None")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $f = Formula("x/(x^2-($r+$s)x+$rs)")->reduce; - $domain = Compute("(-inf,$r)U($r,$s)U($s,inf)"); - if ($r > 0) { - $inc = List(Interval("[-sqrt($rs),$r)"),Interval("($r,sqrt($rs)]")); - $dec = List(Interval("(-inf,-sqrt($rs)]"),Interval("[sqrt($rs),$s)"),Interval("($s,inf)")); - $max = List(Compute("sqrt($rs)")); - $min = List(Compute("-sqrt($rs)")); - } - elsif ($s > 0) { - $inc = List(Compute("none")); - $dec = List(Interval("(-inf,$r)"),Interval("($r,$s)"),Interval("($s,inf)")); - $max = List(Compute("none")); - $min = List(Compute("none")); - } - else { - $inc = List(Interval("[-sqrt($rs),$s)"),Interval("($s,sqrt($rs)]")); - $dec = List(Interval("(-inf,$r)"),Interval("($r,-sqrt($rs)]"),Interval("[sqrt($rs),inf)")); - $max = List(Compute("sqrt($rs)")); - $min = List(Compute("-sqrt($rs)")); - } - - -

    - f(x)= -

    - - Give the domain of f using interval notation. - Use U for union if the domain consists of more than one interval. - -

    - -

    - - Enter the critical numbers of f, - separating with commas if needed. - If there are no critical numbers, - enter none. - -

    - -

    - - List the maximal intervals where f is increasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is decreasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the critical numbers at which there is a relative maximum. - If there are none, enter none. - -

    - -

    - - List the critical numbers at which there is a relative minimum. - If there are none, enter none. - -

    - -

    -     - - -     -     - - - - - $r = non_zero_random(-9,9,1); - $crit = List($r,3*$r); - Context("Interval")->flags->set(reduceConstants=>0); - Context()->flags->set(ignoreEndpointTypes => 1); - $f = Formula("(x-$r)^(2/3)/x"); - $domain = Compute("(-inf,0)U(0,inf)"); - if ($r > 0) { - $inc = List(Interval("[$r,3*$r]")); - $dec = List(Interval("(-inf,0)"),Interval("(0,$r]"),Interval("[3*$r,inf)")); - $max = List(3*$r); - $min = List($r); - } else { - $inc = List(Interval("[3*$r,$r]")); - $dec = List(Interval("(-inf,3*$r]"),Interval("[$r,0)"),Interval("(0,inf)")); - $max = List($r); - $min = List(3*$r); - } - - -

    - f(x)= -

    - - Give the domain of f using interval notation. - Use U for union if the domain consists of more than one interval. - -

    - -

    - - Enter the critical numbers of f, - separating with commas if needed. - If there are no critical numbers, - enter none. - -

    - -

    - - List the maximal intervals where f is increasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is decreasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the critical numbers at which there is a relative maximum. - If there are none, enter none. - -

    - -

    - - List the critical numbers at which there is a relative minimum. - If there are none, enter none. - -

    - -

    -     - - -     -     - - - - - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $domain = Compute("(-pi,pi)"); - $crit = List("-3pi/4,-pi/4,pi/4,3pi/4"); - $inc = List(Interval("(-pi,-3pi/4)"),Interval("(-pi/4,pi/4)"),Interval("(3pi/4,pi)")); - $dec = List(Interval("(-3pi/4,-pi/4)"),Interval("(pi/4,3pi/4)")); - $max = List("-3pi/4,pi/4"); - $min = List("-pi/4,3pi/4"); - - -

    - f(x)=\sin(x)\cos(x) on (-\pi,\pi) -

    - - Give the domain of f using interval notation. - Use U for union if the domain consists of more than one interval. - -

    - -

    - - Enter the critical numbers of f, - separating with commas if needed. - If there are no critical numbers, - enter none. - -

    - -

    - - List the maximal intervals where f is increasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is decreasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the critical numbers at which there is a relative maximum. - If there are none, enter none. - -

    - -

    - - List the critical numbers at which there is a relative minimum. - If there are none, enter none. - -

    - -

    -     - - -     -     - - - - - $n = random(3,9,1); - if ($n == 3) { - $b = non_zero_random(-9,9,1); - } - elsif ($n == 4) { - $b = non_zero_random(-5,5,1); - } - elsif ($n == 5) { - $b = non_zero_random(-3,3,1); - } - elsif ($n >= 6 and $n <= 8) { - $b = non_zero_random(-2,2,1); - } - else { - $b = non_zero_random(-1,1,1); - } - $a = ($b)**($n-1); - Context("Interval")->flags->set(reduceConstants=>0); - Context()->flags->set(ignoreEndpointTypes => 1); - $domain = Compute("(-inf,inf)"); - $f = Formula("x^$n - $a*$n x")->reduce; - if ($n % 2 == 0) { - $crit = List($b); - $inc = List(Interval("[$b,inf)")); - $dec = List(Interval("(-inf,$b]")); - $max = List(Compute("none")); - $min = List($b); - } else { - $crit = List($b,-$b); - $inc = List(Interval("(-inf,-abs($b)]"),Interval("[abs($b),inf)")); - $dec = List(Interval("[-abs($b),abs($b)]")); - $max = List(-abs($b)); - $min = List(abs($b)); - } - - -

    - f(x)= -

    - - Give the domain of f using interval notation. - Use U for union if the domain consists of more than one interval. - -

    - -

    - - Enter the critical numbers of f, - separating with commas if needed. - If there are no critical numbers, - enter none. - -

    - -

    - - List the maximal intervals where f is increasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is decreasing, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the critical numbers at which there is a relative maximum. - If there are none, enter none. - -

    - -

    - - List the critical numbers at which there is a relative minimum. - If there are none, enter none. - -

    - -

    -     - - -     -     -     -     -     -
    -
    - Concavity and the Second Derivative - - - -

    - Our study of nice functions continues. - The previous section showed how the first derivative of a function, - \fp, can relay important information about f. - We now apply the same technique to \fp itself, - and learn what this tells us about f. -

    - -

    - The key to studying \fp is to consider its derivative, - namely \fp', which is the second derivative of f. - When \fp'\gt 0, \fp is increasing. - When \fp'\lt 0, \fp is decreasing. - \fp has relative maxima and minima where \fp'=0 or is undefined. -

    - -

    - This section explores how knowing information about \fpp gives information about f. -

    -     - - - Concavity -

    - We begin with a definition, then explore its meaning. -

    - - - Concave Up and Concave Down - -

    - Let f be continuous on an interval I. - The graph of f is concave up - on I if for any a\lt b in I, - - f\left(\frac{a+b}{2}\right) \lt \frac{f(a)+f(b)}{2} - . - The graph of f is concave down - on I if for any a\lt b in I, - - f\left(\frac{a+b}{2}\right) \gt \frac{f(a)+f(b)}{2} - . - concavity - concave up - concave down -

    -     -     - - - - - -

    - Geometrically, the condition in Equation - states that a graph is concave up if the midpoint of the secant line from (a,f(a)) - to (b,f(b)) (and hence, the secant line itself) is above the graph y=f(x). - Similarly, Equation states that the secant line lies below the graph. -

    -

    - In order for equality to hold instead of Equation or Equation, - the function would have to be of the form f(x)=mx+c, - in which case the graph is a straight line. - Straight lines are considered to have no concavity. -

    - -
    - Illustrating the nature of concave up and concave down - -
    - A graph that is concave up. Notice how the secant line lies above the graph. - - - A graph of a function with concave up curvature. - - -

    - The image shows a graph where the curve is shaped upwards, resembling a bowl. - There is a secant line from x=-1 to x=2 that lies above the curve. - The points on the curve are all below this secant line, implying that the slope of the function is increasing as you move from left to right. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1.5,xmax=2.5, - ymin=-0.5,ymax=7, - ] - \addplot+[infinite,domain=-1.4:2.4] {x^2}; - \addplot+[secantline,-,domain=-1:2] {x+2}; - \addplot[soliddot] coordinates {(-1,1) (2,4)}; - \end{axis} - \end{tikzpicture} - - -
    -
    - A graph that is concave down. Notice how the secant line lies below the graph. - - - A graph representing a function with a concave down shape. - - -

    - This image illustrates a graph with a curve that opens downwards, similar to an inverted bowl, which characterizes a concave down function. - There's a secant line from x=-1 to x=2 that lies below the curve. - The points on the curve are all above this secant line, implying that the function's slope is decreasing as you move from left to right. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1.5,xmax=2.5, - ymin=-0.5,ymax=7, - ] - \addplot+[infinite,domain=-1.4:2.3] {6-x^2}; - \addplot+[secantline,-,domain=-1:2] {-x+4}; - \addplot[soliddot] coordinates {(-1,5) (2,2)}; - \end{axis} - \end{tikzpicture} - - -
    -     -
    - - - -

    - Consider a function f such that f is continuous on [a,b] and differentiable on (a,b). - Note that \frac{a+b}{2} is the midpoint of the interval [a,b]. - By the , - there must be a point c_1 in \left[a,\frac{a+b}{2}\right] such that - - \fp(c_1) = \frac{f\left(\frac{a+b}{2}\right)-f(a)}{\frac{a+b}{2}-a}=\frac{2}{b-a}\left(f\left(\frac{a+b}{2}\right)-f(a)\right) - . - Similarly, there must be a point c_2 in \left[\frac{a+b}{2},b\right] such that - - \fp(c_2) = \frac{f(b)-f\left(\frac{a+b}{2}\right)}{b-\frac{a+b}{2}} = \frac{2}{b-a}\left(f(b)-f\left(\frac{a+b}{2}\right)\right) - . - But then we have - - \fp(c_2)-\fp(c_1) \amp = \frac{2}{b-a}\left(f(b)-f\left(\frac{a+b}{2}\right)-f\left(\frac{a+b}{2}\right)+f(a)\right) - \amp = \frac{4}{b-a}\left(\frac{f(a)+f(b)}{2}-f\left(\frac{a+b}{2}\right)\right) - . -

    - -

    - Now, let us suppose that \fp(x) is an increasing function on (a,b). - In that case, \fp(c_2)-\fp(c_1)\gt 0, and since b-a\gt 0, - this implies that - - \frac{f(a)+f(b)}{2}-f\left(\frac{a+b}{2}\right)\gt 0 - , - which, by means that the graph of f is concave up. -

    - -

    - Similarly, if \fp(x) is a decreasing function on (a,b), - then the graph of f will be concave down. - Using , we arrive at the following theorem. -

    - - - -

    - Let f be a continuous function on [a,b] and differentiable on (a,b). -

    - -

    -

      -
    1. -

      - If \fpp(c) \gt 0 for all c in (a,b), - then f is concave up on [a,b]. -

      -
      2. -
    2. -

      - If \fpp(c) \lt 0 for all c in (a,b), - then f is concave down on [a,b]. -

      -
      3. -
    3. -

      - If \fpp(c) =0 for all c in (a,b), - then f is linear on [a,b]. -

      -
      4. -
    -

    -     -     - - - -

    - The graph of a function f is concave up - when \fp is increasing. - That means as one looks at a concave up graph from left to right, - the slopes of the tangent lines will be increasing. - Consider , - where a concave up graph is shown along with some tangent lines. - Notice how the tangent line on the left is steep, - downward, corresponding to a lesser - (large negative) - value of \fp. - On the right, the tangent line is steep, - upward, corresponding to a greater - (large positive) - value of \fp. -

    - -
    - A function f with a concave up graph. Notice how the slopes of the tangent lines, when looking from left to right, are increasing. (The slope values pictured are -12, -6, 6 and 12). - - - - A graph of a function with concave up curvature and tangent lines. - - -

    - The image displays a graph of a concave up function f with tangent lines drawn at various points along the curve. - The function is shown as a continuous curve that opens upwards, like a parabola. - Tangent lines at points with x-values of (-2, -1, 1, 2) are depicted, illustrating how the slope of these lines increases as we move from left to right. - On the left side of the graph, the tangent lines slope downwards, indicating negative slopes, while on the right side, the tangent lines slope upwards, indicating positive slopes. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-3.5, - xmax=3.5, - ymin=-1, - ymax=35, - ] - \addplot+[infinite,domain=-3:3] {3*x^2+5}; - \addplot [tangentlineseg,domain =-3:-1] {-12*(x+2)+17}; - \addplot [tangentlineseg,domain =1:3] {12*(x-2)+17} ; - \addplot [tangentlineseg,domain =-2.5:.5] {-6*(x+1)+8}; - \addplot [tangentlineseg,domain =-.5:2.5] {6*(x-1)+8} ; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - If a function is decreasing and concave up, - then its rate of decrease is slowing; - it is leveling off. You can see this in the left side of . - If the function is increasing and concave up, - then the rate of increase is increasing. - The function is increasing at a faster and faster rate. - You can see this in the right side of . -

    - -

    - Now consider a function which is concave down. - We essentially repeat the above paragraphs with slight variation. -

    - -

    - The graph of a function f is concave down - when \fp is decreasing. - That means as one looks at a concave down graph from left to right, - the slopes of the tangent lines will be decreasing. - Consider , - where a concave down graph is shown along with some tangent lines. - Notice how the tangent line on the left is steep, - upward, corresponding to a greater - (large positive) - value of \fp. - On the right, the tangent line is steep, - downward, corresponding to a lesser - (large negative) - value of \fp. -

    - -
    - A function f with a concave down graph. Notice how the slopes of the tangent lines, when looking from left to right, are decreasing. - - - - A graph of a function with concave down curvature and tangent lines. - - -

    - The image shows a graph of a concave down function f with tangent lines drawn at various points along the curve. - The function forms a continuous curve that opens downwards. - Tangent lines at points with x-values of (-2, 1, 1, 2) are shown, demonstrating how the slope of these lines decreases as we move from left to right across the graph. - On the left side of the graph, the tangent lines slope upwards, indicating positive slopes, while on the right side, the tangent lines slope downwards, indicating negative slopes. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-3.5, - xmax=3.5, - ymin=-1, - ymax=35, - ] - \addplot+[infinite,domain=-3:3] {-3*x^2+31}; - \addplot [tangentlineseg, domain =-3:-1] {12*(x+2)+19}; - \addplot [tangentlineseg, domain =1:3] {-12*(x-2)+19} ; - \addplot [tangentlineseg, domain =-2.5:-.2] {6*(x+1)+28}; - \addplot [tangentlineseg, domain =.2:2.5] {-6*(x-1)+28} ; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - If a function is increasing and concave down, - then its rate of increase is slowing; - it is leveling off. If the function is decreasing and concave down, - then the rate of decrease is decreasing. - The function is decreasing at a faster and faster rate. -

    - -

    - Our definition of concave up and concave down is given in terms of when the first derivative is increasing or decreasing. - We can apply the results of the previous section to find intervals on which a graph is concave up or down. - That is, we recognize that \fp is increasing when \fpp \gt 0, etc. -

    - - - Test for Concavity - -

    - Let f be twice differentiable on an interval I. - The graph of f is concave up if \fpp \gt 0 on I, - and is concave down if \fpp\lt 0 on I. - concavitytest for - concavityinflection point -

    -     -     - - -
    - Demonstrating the four ways that concavity interacts with increasing/decreasing, along with the relationships with the first and second derivatives - - -
    - \fp\gt 0, f increasing; \fpp\lt0, f is concave down - - - A graph illustrating a function that is increasing and concave down. - - -

    - Diagram a shows a portion of a curve where the function f is increasing as the first derivative \fp\gt 0 suggests. - Despite the increase, the curve is concave down, indicated by the second derivative \fpp\lt0, meaning the rate of increase is slowing down. -

    -     - - \begin{tikzpicture} - \draw [thick] (0,0) arc (180:90:2); - \end{tikzpicture} - - -
    - -
    - \fp\lt0, f decreasing; \fpp\lt0, f is concave down - - - A graph illustrating a function that is decreasing and concave down. - - -

    - Diagram b depicts a curve that is decreasing, with the first derivative \fp\lt0, and is also concave down, as shown by the second derivative \fpp\lt0. - This represents a function f that is decreasing at an increasing rate. -

    -     - - \begin{tikzpicture} - \draw [thick] (0,0) arc (90:0:2); - \end{tikzpicture} - - -
    -     - - -
    - \fp\lt0, f decreasing; \fpp\gt 0, f is concave up - - - A graph illustrating a function that is decreasing and concave up. - - -

    - In diagram c, the curve represents a function f that is decreasing since \fp\lt0, but the concavity is up, which is indicated by the second derivative \fpp\gt 0. - This means the function is decreasing but the rate of decrease is slowing. -

    -     - - \begin{tikzpicture} - \draw [thick] (0,0) arc (180:270:2); - \end{tikzpicture} - - -
    - -
    - \fp\gt 0, f increasing; \fpp\gt 0, f is concave up - - - A graph illustrating a function that is increasing and concave up. - - -

    - Diagram d displays a curve that is increasing as the first derivative \fp\gt 0 indicates, and it is concave up as the second derivative \fpp\gt 0 confirms. - This represents a function f that is increasing at an increasing rate. -

    -     - - \begin{tikzpicture} - \draw [thick] (0,0) arc (0:-90:2); - \end{tikzpicture} - - -
    -     -     -
    - -

    - If knowing where a graph is concave up/down is important, - it makes sense that the places where the graph changes from one to the other is also important. - This leads us to a definition. -

    - - - Point of Inflection - -

    - A point of inflection - is a point on the graph of f at which the concavity of f changes. - point of inflection - inflection point -

    -     -     - -

    - - shows a graph of a function with inflection points labeled. -

    - -
    - A graph of a function with its inflection points marked. The intervals where concave up/down are also indicated. - - - - A graph depicting a function with labeled inflection points and concavity indications. - - -

    - The image shows the graph of a function featuring distinct inflection points at x=2 and x=3. - The section of the function bounded by x=0 and x=2 is concave up with \fpp\gt 0, while the section between x=2 and x=3 is concave down with \fpp\lt 0. - The section bouned by x=3 and x=4 is concave up with \fpp\gt 0. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.5, - xmax=4.5, - ymin=-1, - ymax=16, - ] - \addplot+[infinite,domain=.7:4.2, samples=40] {2*(x-1)*(x-2)*(x-3)*(x-4)+5}; - \addplot[soliddot] coordinates {(1.85,4.39) (3.15, 4.39)}; - \addplot[guideline] coordinates {(1.85,-1) (1.85,16)} node[left, anchor=north east] {\parbox{4em}{\flushright$\fpp>0$, $f$ is concave up}}; - \addplot[guideline] coordinates {(3.15,-1) (3.15,16)} node[right, anchor=north west] {\parbox{4em}{\flushleft$\fpp>0$, $f$ is concave up}}; - \draw (axis cs:2.5,16) node[anchor=north] {\parbox{4em}{\centering$\fpp<0$, $f$ is concave down}}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - If the concavity of f changes at a point (c,f(c)), - then \fp is changing from increasing to decreasing (or, - decreasing to increasing) at x=c. - That means that the sign of \fppis changing from positive to negative (or, - negative to positive) at x=c. - A sign change may occur when \fpp=0 or \fpp is undefined. - This leads to the following theorem. -

    - - - Points of Inflection - -

    - If (c,f(c)) is a point of inflection on the graph of f, - then either \fpp(c)=0 or \fpp is not defined at c. -

    -     -     - -

    - We have identified the concepts of concavity and points of inflection. - It is now time to practice using these concepts; - given a function, - we should be able to find its points of inflection and identify intervals on which it is concave up or down. - We do so in the following examples. -

    - - - Finding intervals of concave up/down, inflection points - -

    - Let f(x)=x^3-3x+1. - Find the inflection points of f and the intervals on which it is concave up/down. -

    -     - -

    - We start by finding \fp(x)=3x^2-3 and \fpp(x)=6x. - To find the inflection points, - we use - and find where \fpp(x)=0 or where \fpp is undefined. - We find \fpp is always defined, - and is 0 only when x=0. - So the point (0,f(0))=(0,1) is the only possible point of inflection. -

    - -

    - This possible inflection point divides the real line into two intervals, - (-\infty,0) and (0,\infty). - We use a process similar to the one used in the previous section to determine increasing/decreasing. - Pick any c\lt 0; - \fpp(c)\lt 0 so f is concave down on (-\infty,0). - Pick any c\gt 0; - \fpp(c)\gt 0 so f is concave up on (0,\infty). - Since the concavity changes at x=0, - the point (0,1) is an inflection point. -

    - -

    - The number line in - illustrates the process of determining concavity; - - shows a graph of f and \fpp, confirming our results. - Notice how f is concave down precisely when - \fpp(x)\lt 0 and concave up when \fpp(x)\gt 0. -

    - -
    - A number line determining the concavity of f in - - - - - A number line indicating the concavity of a function based on the second derivative. - - -

    - The image illustrates a number line used to determine the concavity of a function f. - The number line is divided at zero, with the left side showing \fpp\lt 0, indicating that the function is concave down in this region. - Conversely, the right side of the number line where \fpp\gt 0 indicates that the function is concave up. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - numberline, - xmin=-0.9, - xmax=0.9, - after end axis/.code={ - \path (axis cs:0,0) - node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; - }, - ] - \addplot[guideline] coordinates {(0,0) (0,2)}; - \addplot[mark=none] coordinates {(-0.5,1)} node {\parbox{8em}{\centering \small $\fpp<0$\\$f$ is concave down}}; - \addplot[mark=none] coordinates {(0.5,1)} node {\parbox{8em}{\centering \small $\fpp>0$\\$f$ is concave up}}; - \end{axis} - \end{tikzpicture} - - - -
    - -
    - A graph of f(x) used in - - - - A graph showing the a function and its second derivative. - - -

    - The graph shows the function f(x) and its second derivative \fpp. - The function f(x) exhibits both concave up and concave down behavior within different intervals. - The second derivative \fpp crosses the x-axis at points where the concavity of f(x) changes, indicating the locations of possible inflection points. - Above the x-axis, where \fpp is positive, the function f(x) is concave up. Below the x-axis, where \fpp is negative, the function f(x) is concave down. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-2.2, - xmax=2.2, - ymin=-3.5, - ymax=3.5, - ] - \addplot+[infinite,domain=-2.1:2,samples=60] {x^3-x*3+1} node[below right, pos=0.2] {$f(x)$} ; - \addplot+[infinite,domain=-.5:.5] {6*x} node[below right]{$\fpp(x)$}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - - - - - Finding intervals of concave up/down, inflection points - -

    - Let f(x)=x/(x^2-1). - Find the inflection points of f and the intervals on which it is concave up/down. -

    -     - -

    - We need to find \fpand \fpp. - Using the and simplifying, we find - - \fp(x)\amp=\frac{-(1+x^2)}{(x^2-1)^2}\amp\fpp(x)\amp= \frac{2x(x^2+3)}{(x^2-1)^3} - . -

    - -

    - To find the possible points of inflection, - we seek to find where \fpp(x)=0 and where \fpp is not defined. - Solving \fpp(x)=0 reduces to solving 2x(x^2+3)=0; - we find x=0. - We find that \fpp is not defined when x=\pm 1, - for then the denominator of \fpp is 0. - We also note that f itself is not defined at x=\pm1, - having a domain of (-\infty,-1)\cup(-1,1)\cup(1,\infty). - Since the domain of f is the union of three intervals, - it makes sense that the concavity of f could switch across intervals. - We technically cannot say that f has a point of inflection at x=\pm1 as they are not part of the domain, - but we must still consider these x-values to be important and will include them in our number line. -

    - -

    - The important x-values at which concavity might switch are x=-1, - x=0 and x=1, - which split the number line into four intervals as shown in . - We determine the concavity on each. - Keep in mind that all we are concerned with is the sign - of \fpp on the interval. -

    - -

    -

    -
  • - Interval 1: <m>(-\infty,-1)</m> -

    - Select a number c in this interval with a large magnitude - (for instance, c=-100). - The denominator of \fp'(x) will be positive. - In the numerator, - the \left(c^2+3\right) factor will be positive and the 2c factor will be negative. - Thus the numerator is negative and \fpp(c) is negative. - We conclude f is concave down on (-\infty,-1). -

    -
    • - -
  • - Interval 2: <m>(-1,0)</m> -

    - For any number c in this interval, - the factor 2c in the numerator will be negative, - the factor \left(c^2+3\right) in the numerator will be positive, - and the factor \left(c^2-1\right)^3 in the denominator will be negative. - Thus \fpp(c)\gt 0 and f is concave up on this interval. -

    -
    • - -
  • - Interval 3: <m>(0,1)</m> -

    - Any number c in this interval will be positive and small. - Thus the numerator is positive while the denominator is negative. - Thus \fpp(c)\lt 0 and f is concave down on this interval. -

    -
    • - -
  • - Interval 4: <m>(1,\infty)</m> -

    - Choose a large value for c. - It is evident that \fpp(c)\gt 0, - so we conclude that f is concave up on (1,\infty). -

    -
    • -
    -

    - -
    - Number line for f in - - - - - A number line indicating intervals of concavity for a function. - - -

    - The image shows a number line that is used to determine the concavity of the function f. - The line is divided into four intervals by the points -1, 0 and 1, - each of which is an inflection point. -

    - -

    - The intervals are marked as follows: -

      -
    • -

      - For x\lt -1, \fpp\lt 0, and f is concave down. -

      -
      • - -
    • -

      - For -1\lt x\lt 0, \fpp\gt 0, and f is concave up. -

      -
      • - -
    • -

      - For 0\lt x\lt 1, \fpp\lt 0, and f is concave down. -

      -
      • - -
    • -

      - For x\gt 1, \fpp\gt 0, and f is concave up. -

      -
      • -
    -

    -     - - \begin{tikzpicture} - \begin{axis}[ - numberline, - xmin=-2, - xmax=2, - extra x ticks={-1, 1, 0}, - extra x tick labels={$-1$,$1$, $0$}, - ] - \addplot[guideline] coordinates {(-1,0) (-1,2)}; - \addplot[guideline] coordinates {(0,0) (0,2)}; - \addplot[guideline] coordinates {(1,0) (1,2)}; - \addplot[mark=none] coordinates {(-1.5,1.5)} node {\parbox{3em}{\centering \small $\fpp\lt0$\\$f$ conc down}}; - \addplot[mark=none] coordinates {(-0.5,1.5)} node {\parbox{3em}{\centering \small $\fpp>0$\\$f$ conc up}}; - \addplot[mark=none] coordinates {(0.5,1.5)} node {\parbox{3em}{\centering \small $\fpp\lt0$\\$f$ conc down}}; - \addplot[mark=none] coordinates {(1.5,1.5)} node {\parbox{3em}{\centering \small $\fpp>0$\\$f$ conc up}}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - We conclude that f is concave up on (-1,0) and - (1,\infty) and concave down on (-\infty,-1) and (0,1). - There is only one point of inflection, - (0,0), as f is not defined at x=\pm 1. - Our work is confirmed by the graph of f in . - Notice how f is concave up whenever \fppis positive, - and concave down when \fppis negative. - The inflection in f occurs where \fpp changes sign. -

    - -
    - A graph of f(x) and \fpp(x) in - - - - A graph comparing a function and its second derivative. - - -

    - The graph displays two curves on the same set of axes: one for the function f(x) in blue and another for its second derivative \fpp(x) in red. - The function f(x) shows a vertical asymptote, where the curve approaches but never reaches the line x=0. - The red curve of the second derivative \fpp(x) also approaches this vertical asymptote. - Both curves demonstrate that as x gets closer to zero from either side, the values of f(x) and \fpp(x) increase or decrease without bound, indicating a point of inflection at x=0 for the function f(x). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-3.2, - xmax=3.2, - ymin=-11, - ymax=11, - ] - \addplot[firstcurvestyle, infinite,samples=25, domain=-3.1:-1.05] {x/(x^2-1)}; - \addplot[firstcurvestyle, infinite,samples=25, domain=-.95:.95] {x/(x^2-1)} node[left, pos=0.1]{$f(x)$}; - \addplot[firstcurvestyle, infinite,samples=25, domain=1.05:3.1] {x/(x^2-1)}; - \addplot[secondcurvestyle,infinite, samples=25, domain=-3.1:-1.5] {(2*x*(x^2+3))/((x^2-1)^3)}; - \addplot[secondcurvestyle,infinite, samples=25, domain=-.5:.5] {(2*x*(x^2+3))/((x^2-1)^3)}; - \addplot[secondcurvestyle, infinite,domain=1.5:3.1] {(2*x*(x^2+3))/((x^2-1)^3)} node[right, pos=0.1] {$\fpp(x)$}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - -

    - Recall that relative maxima and minima of f are found at critical points of f; - that is, they are found when - \fp(x)=0 or when \fp is undefined. - Likewise, the relative maxima and minima of \fp are found when - \fpp(x)=0 or when \fpp is undefined; - note that these are the inflection points of f. -

    - -

    - What does a relative maximum of \fp mean? - The derivative measures the rate of change of f; - maximizing \fp means finding where f is increasing the most where f has the steepest tangent line. - A similar statement can be made for minimizing \fp; it corresponds to where f has the steepest negatively-sloped tangent line. -

    - -

    - We utilize this concept in the next example. -

    - - - Understanding inflection points - -

    - The sales of a certain product over a three-year span are modeled by S(t)= t^4-8t^2+20, - where t is the time in years, - shown in . -

    - -
    - A graph of S(t) in , modeling the sale of a product over time - - - - A graph representing the sales model of a product over time. - - -

    - The figure illustrates a graph of the function S(t), which models the sales of a product over a three-year period. - The sales are modeled by the equation S(t)= t^4-8t^2+20, where t is the time in years. - The graph shows an initial decline in sales, reaching a minimum before rising again. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.2, - xmax=3.2, - ymin=-1, - ymax=21, - xlabel={$t$}, - ] - \addplot+[rightarrow, domain=0:2.8, samples=40] {x^4-8*x^2+20} node[above right, pos=0.3]{$S(t)$}; - \end{axis} - \end{tikzpicture} - - - -
    - -

    - Over the first two years, sales are decreasing. - Find the point at which sales are decreasing at their greatest rate. -

    -     - -

    - We want to maximize the rate of decrease, - which is to say, we want to find where S' has a minimum. - To do this, we find where S'' is 0 and S'' changes from negative to positive. - We find S'(t)=4t^3-16t and S''(t)=12t^2-16. - Setting S''(t)=0 and solving, - we get t=\sqrt{4/3}\approx 1.16 - (we ignore the negative solution for t since it does not lie in the domain of our function S). -

    - -

    - Since S''(1)=-4\lt 0 and S''(2)=32\gt 0, - we can say S'(\sqrt{4/3}) is a local minimum of S'. - This is both the inflection point and the point of maximum decrease. - This is the point at which things first start looking up for the company. - After the inflection point, sales are still decreasing, - but not decreasing quite as quickly as they had been. -

    - -

    - A graph of S(t) and S'(t) is given in . - When S'(t)\lt 0, sales are decreasing; - note how at t\approx 1.16, - S'(t) is minimized. - That is, sales are decreasing at the fastest rate at t\approx 1.16. - On the interval of (1.16,2), - S is decreasing but concave up, - so the decline in sales is leveling off. -

    - -
    - A graph of S(t) in , along with S'(t) - - - - A graph showing both the sales function and its rate of change over time. - - -

    - The graph depicts two curves: the blue curve represents the sales function S(t), and the red curve represents its rate of change, S'(t), over a time period t. - The sales function curve shows a parabolic shape, indicating a period of declining sales followed by an upturn. - The rate of change curve S'(t) crosses the t-axis at its lowest point, signifying the moment when the sales rate is changing from decreasing to increasing. - The point where S'(t) is most negative corresponds to the fastest rate of decrease in sales, which is the primary point of interest in this example. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-.2, - xmax=3.2, - ymin=-16, - ymax=21, - xlabel={$t$}, - ] - \addplot+[rightarrow,domain=0:2.8, samples=40] {x^4-8*x^2+20} node[above right, pos=0.3]{$S(t)$}; - \addplot+[domain=0:3, samples=40] {4*x^3-16*x} node[below right, pos=0.2]{$S'(t)$}; - \addplot[soliddot] coordinates {(1.16,11.04) (1.16,-12.32)}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - -

    - Not every critical point corresponds to a relative extrema; - f(x)=x^3 has a critical point at (0,0) but no relative maximum or minimum. - Likewise, just because \fpp(x)=0 we cannot conclude concavity changes at that point. - We were careful before to use terminology - possible point of inflection - since we needed to check to see if the concavity changed. - The canonical example of \fpp(x)=0 without - concavity changing is f(x)=x^4. - At x=0, - \fpp(x)=0 but f is always concave up, - as shown in . -

    - -
    - A graph of f(x) = x^4. Clearly f is always concave up, despite the fact that \fpp(x) = 0 when x=0. In this example, the possible point of inflection (0,0) is not a point of inflection. - - - - A graph depicting a U-shaped curve representing a function raised to the fourth power. - - -

    - The graph of f(x) = x^4 illustrates a U-shaped curve that is concave up across its domain. Despite the second derivative \fpp(x) equalling zero when x = 0, indicating a potential inflection point, the function remains concave up on both sides of the y-axis. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-1.1, - xmax=1.1, - ymin=-.1, - ymax=1.1, - ] - \addplot+[infinite,domain=-1:1] {x^4}; - \end{axis} - \end{tikzpicture} - - - -
    -     - - - The Second Derivative Test -

    - The first derivative of a function gave us a test to find if a critical value corresponded to a relative maximum, minimum, - or neither. - The second derivative gives us another way to test if a critical point is a local maximum or minimum. - The following theorem officially states something that is intuitive: - if a critical value occurs in a region where a function f is concave up, - then that critical value must correspond to a relative minimum of f, etc. - See for a visualization of this. -

    - - - The Second Derivative Test - -

    - Let c be a critical value of f where \fpp(c) is defined. - derivativeSecond Deriv. - Test - Second Derivative Test - extremaand Second Deriv. Test - maximumand Second Deriv. Test - minimumand First Deriv. Test - -

      -
    1. -

      - If \fpp(c)\gt 0, - then f has a local minimum at (c,f(c)). -

      -
      2. - -
    2. -

      - If \fpp(c)\lt 0, - then f has a local maximum at (c,f(c)). -

      -
      3. -
    -

    -     -     - -
    - Demonstrating the fact that relative maxima occur when the graph is concave down and relative minima occur when the graph is concave up - - - - A graph depicting points of maximum and minimum on a curve with changing concavity. - - -

    - The figure showcases a curve that illustrates the concepts of relative maxima and minima in relation to concavity. - The curve rises upwards, dips downward and then rises upward again, forming a valley-like shape. - At the peak of the first curve, where the shape is bending downwards indicating concave down, a point is marked at (-1, 10) signifying a relative maximum. - Similarly, at the lowest point in the valley at (1,-10), where the curve starts to bend upwards denoted as concave up, a point signifies a relative minimum. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-2.1, - xmax=2.1, - ymin=-11, - ymax=11, - ] - \addplot+[infinite,domain=-2:2, samples=50] {5*x^3-15*x}; - \addplot[soliddot] coordinates {(-1,10) (1,-10)}; - \addplot[mark=none] coordinates {(-1,7.5)} node {\parbox{6em}{\centering concave down\\$\implies$ rel max}}; - \addplot[mark=none] coordinates {(1,-7.5)} node {\parbox{6em}{\centering concave up\\$\implies$ rel min}}; - \end{axis} - \end{tikzpicture} - - - -
    - - - -

    - The Second Derivative Test relates to the in the following way. - If \fpp(c)\gt 0, - then the graph is concave up at a critical point c and \fp itself is growing. - Since \fp(c)=0 and \fp is growing at c, - then it must go from negative to positive at c. - This means the function goes from decreasing to increasing, - indicating a local minimum at c. -

    - - - Using the Second Derivative Test - -

    - Let f(x)=100/x + x. - Find the critical points of f and use the to label them as relative maxima or minima. -

    -     - -

    - We find \fp(x)=-100/x^2+1 and \fpp(x) = 200/x^3. - We set \fp(x)=0 and solve for x to find the critical values - (note that \fp is not defined at x=0, - but neither is f so this is not a critical value.) - We find the critical values are x=\pm 10. - We now evaluate the second derivative at these critical numbers. - Evaluating \fpp(10)=0.1\gt 0, - so there is a local minimum at x=10. - Evaluating \fpp(-10)=-0.1\lt 0, - determining a relative maximum at x=-10. - These results are confirmed in . -

    - -
    - A graph of f(x) in . The second derivative is evaluated at each critical point. When the graph is concave up, the critical point represents a local minimum; when the graph is concave down, the critical point represents a local maximum. - - - - A graph illustrating the evaluation of the second derivative at critical points. - - -

    - This graph depicts a function f(x) with two marked critical points where the second derivative \fpp(x) has been evaluated. - On the left, a point at x = -10 is marked with \fpp(-10)\lt 0, indicating that the graph is concave down at this point, which typically corresponds to a local maximum. - On the right, a point at x = 10 is marked with \fpp(10)\gt 0, signifying that the graph is concave up at this point, suggesting the presence of a local minimum. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xmin=-21, - xmax=21, - ymin=-51, - ymax=51, - ] - \addplot+[infinite,domain=-20:-2.2] {100/x+x}; - \addplot[firstcurvestyle,domain=2.2:20] {100/x+x}; - \addplot[soliddot] coordinates{(10,20)} node [below] {$\fpp(10)>0$}; - \addplot[soliddot] coordinates{(-10,-20)} node [above] {$\fpp(-10)\lt0$}; - \end{axis} - \end{tikzpicture} - - - -
    -     - -     - -

    - The second derivative test can only be used on a function that is twice differentiable at c. - For functions that are not twice differentiable at c, - you will need to use the . - If you've already determined the sign diagram for \fp, - the is usually - easier to apply, and it applies in cases when - does not. -

    - -

    - We have been learning how the first and second derivatives of a function relate information about the graph of that function. - We have found intervals of increasing and decreasing, - intervals where the graph is concave up and down, - along with the locations of relative extrema and inflection points. - In - we saw how limits explained asymptotic behavior. - In the next section we combine all of this information to produce accurate sketches of functions. -

    -     - - - - Terms and Concepts - - - -

    - Sketch a graph of a function f(x) that is concave up on (0,1) and is concave down on (1,2). -

    -     - - - -     - - - -

    - Sketch a graph of a function f(x) that is: -

    - -

    -

      -
    • - increasing, concave up on (0,1), -
      • - -
    • - increasing, concave down on (1,2), -
      • - -
    • - decreasing, concave down on (2,3), and -
      • - -
    • - increasing, concave down on (3,4). -
      • -
    -

    -     - - - -     - - - -

    - Is is possible for a function to be increasing and concave down on - (0,\infty) with a horizontal asymptote of y=1? - If so, give a sketch of such a function. -

    -     - - - -     - - - -

    - Is is possible for a function to be increasing and concave up on - (0,\infty) with a horizontal asymptote of y=1? - If so, give a sketch of such a function. -

    -     - - - -     -     - - - Problems - - - - -

    - A function f(x) is given. - Graph f and \fpp on the same axes - (using technology is permitted) - and verify . -

    -     - - - -

    - f(x) = -7x+3 -

    -     -     - - - -

    - f(x) = -4x^2+3x-8 -

    -     -     - - - -

    - f(x) = 4x^2+3x-8 -

    -     -     - - - -

    - f(x) = x^3-3x^2+x-1 -

    -     -     - - - -

    - f(x) = -x^3+x^2-2x+5 -

    -     -     - - - - - -

    - f(x) = \sin(x) -

    -     -     - - - -

    - f(x) = \tan(x) -

    -     -     - - - -

    - f(x) = \dfrac{1}{x^2+1} -

    -     -     - - - -

    - f(x) = \frac{1}{x} -

    -     -     - - - -

    - f(x) = \frac{1}{x^2} -

    -     -     -     - - - -

    - A function f(x) is given. -

      -
    1. -

      - Find the possible points of inflection of f. -

      -
      2. -
    2. -

      - Find the intervals on which the graph of f is concave up. -

      -
      3. -
    3. -

      - Find the intervals on which the graph of f is concave down. -

      -
      4. -
    -

    -     - - - - - $r = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$r=1;}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("x^2 - 2*$r x + ($r)^2")->reduce; - $inflecs = List(Compute("None")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,inf)")); - $dn = List(Compute("None")); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - ($a,$b) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$a=-5; $b=7}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("-x^2 + $a x + $b")->reduce; - $inflecs = List(Compute("None")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Compute("None")); - $dn = List(Interval("(-inf,inf)")); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - $s = random(1,9,1); - $c = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$s=1; $c=1;}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("x^3 - $s x + $c")->reduce; - $inflecs = List(0); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("[0,inf)")); - $dn = List(Interval("(-inf,0]")); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - $a = non_zero_random(-9,9,1); - $b = non_zero_random(-9,9,1); - $c = ($a > 0) ? random(int(($b)**2/(3*$a))+1,int(($b)**2/(3*$a))+9,1) : random(int(($b)**2/(3*$a))-9,int(($b)**2/(3*$a))-1,1); - $d = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=2; $b=-3; $c=9; $d=5}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("$a x^3 + $b x^2 + $c x + $d")->reduce; - Context("Fraction"); - $i = Fraction(-2*$b,6*$a); - $inflecs = List($i); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("[$i,inf)")); - $dn = List(Interval("(-inf,$i]")); - ($up,$dn) = ($dn,$up) if ($a < 0); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - $s = non_zero_random(-9,9,1); - $k = ($s % 2 == 0) ? non_zero_random(2,8,2) : non_zero_random(1,9,2); - $d = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$s=-1; $k=-1; $d=3;}; - $r = ($s + $k)/-2; - $t = $k*$s; - Context("Fraction"); - $b = Fraction(-($r+2*$s),3); - ($bn,$bd) = $b->value; - $c = -$r*(($s)**2+$t); - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("x^4/4 + $bn x^3/$bd + $c x + $d")->reduce; - $i = Fraction(3*$s-$k,3); - ($i,$j) = num_sort($i,0); - $inflecs = List($i,$j); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); - $dn = List(Interval("[$i,$j]")); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - ($r,$s,$t) = random_subset(3,-5..-1,1..5); - $k = list_random(-12,-8,-4,4,8,12); - if ($k % 3 != 0) { - do {$t = non_zero_random(-($r+$s) % 3 - 6, -($r+$s) % 3 + 6, 3)} until ($t != $r and $t != $s); - } - ($r,$s,$t) = num_sort($r,$s,$t); - $e = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$r=-1; $s=1; $t=2; $k=12;}; - $a = $k/4; - $b = $k*($r+$s+$t)/3; - $c = $k*($s*$t+$r*$s+$r*$t)/2; - $d = $k*$r*$s*$t; - Context("Fraction"); - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; - $disc = 9*($b)**2 - 24*$a*$c; - ($discsquare,$discnonsquare) = (1,$disc); - for my $p (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251) { - while ($discnonsquare % ($p**2) == 0) { - ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); - } - } - $G = gcd(gcd(3*$b,sqrt($discsquare)),12*$a); - $A = -3*$b/$G; - $B = sqrt($discsquare)/$G; - $C = 12*$a/$G; - if ($discnonsquare == 1) { - $i = Fraction($A+$B,$C); - $j = Fraction($A-$B,$C); - } else { - $i = Formula("($A+$B sqrt($discnonsquare))/$C"); - $j = Formula("($A-$B sqrt($discnonsquare))/$C"); - } - ($i,$j) = num_sort($i,$j); - $inflecs = List($i,$j); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); - $dn = List(Interval("[$i,$j]")); - ($up,$dn) = ($dn,$up) if ($a < 0); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - $r = non_zero_random(-4,4,1); - if($envir{problemSeed}==1){$r=1;}; - $a = 1; - $b = -4*$r; - $c = 6*($r**2); - $d = -4*($r**3); - $e = ($r)**4; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; - $inflecs = List($r); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,inf)")); - $dn = List(Compute("none")); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("sec(x)"); - $inflecs = List(Compute("none")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-pi/2,pi/2)")); - $dn = List(Interval("(-3pi/2,-pi/2)"),Interval("(pi/2,3pi/2)")); - - -

    - f(x)= on (-3\pi/2,3\pi/2) -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - $r = random(-9,9,1); - $s = random(1,9,1); - if($envir{problemSeed}==1){$r=0; $s = 1;}; - $b = -2*$r; - $c = ($r**2) + $s; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("1/(x^2 + $b x + $c)")->reduce; - $disc = 3*$s; - ($discsquare,$discnonsquare) = (1,$disc); - for my $p (2,3,5) { - while ($discnonsquare % ($p**2) == 0) { - ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); - } - } - $G = gcd(gcd(3*$3,sqrt($discsquare)),3); - $A = 3*$r/$G; - $B = sqrt($discsquare)/$G; - $C = 3/$G; - Context("Fraction"); - if ($discnonsquare == 1) { - $i = Fraction($A+$B,$C); - $j = Fraction($A-$B,$C); - } else { - $i = Formula("($A+$B sqrt($discnonsquare))/$C"); - $j = Formula("($A-$B sqrt($discnonsquare))/$C"); - } - ($i,$j) = num_sort($i,$j); - $inflecs = List($i,$j); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); - $dn = List(Interval("[$i,$j]")); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - ($r,$s) = num_sort(random_subset(2,-9..-1,1..9)); - if($envir{problemSeed}==1){$r=-1; $s = 1;}; - $b = -$r - $s; - $c = $r*$s; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("1/(x^2 + $b x + $c)")->reduce; - $inflecs = List(Compute("none")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$r)"),Interval("($s,inf)")); - $dn = List(Interval("($r,$s)")); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("sin(x) + cos(x)"); - $inflecs = List(Compute("-pi/4"),Compute("3pi/4")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-pi,-pi/4]"),Interval("[3pi/4,pi)")); - $dn = List(Interval("[-pi/4,3pi/4]")); - - -

    - f(x)= on (-\pi,\pi) -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("x^2e^x"); - $inflecs = List(Compute("-2+sqrt(2)"),Compute("-2-sqrt(2)")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,-2-sqrt(2)]"),Interval("[-2+sqrt(2),inf)")); - $dn = List(Interval("[-2-sqrt(2),-2+sqrt(2)]")); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("x^2ln(x)"); - $inflecs = List(Compute("1/e^(3/2)")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("[1/e^(3/2),inf)")); - $dn = List(Interval("(0,1/e^(3/2)]")); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("e^(-x^2)"); - $inflecs = List(Compute("1/sqrt(2)"),Compute("-1/sqrt(2)")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,-1/sqrt(2)]"),Interval("[1/sqrt(2),inf)")); - $dn = List(Interval("[-1/sqrt(2),1/sqrt(2)]")); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the maximal intervals where f is concave up, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    - - List the maximal intervals where f is concave down, - using interval notation, - and separating with commas if needed. - If there are no such intervals, - enter none. - -

    - -

    -     - - -     -     -     - - - -

    - A function f(x) is given. - Find the critical points of f and use the Second Derivative Test, - when possible, - to determine the relative extrema. - (Note: these are the same functions as in .) -

    -     - - - - - $r = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$r=1;}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("x^2 - 2*$r x + ($r)^2")->reduce; - $inflecs = List(Compute("None")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,inf)")); - $dn = List(Compute("None")); - Context("Fraction"); - $crit = List($r); - $max = List(Compute("none")); - $min = List($r); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - ($a,$b) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$a=-5; $b=7}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("-x^2 + $a x + $b")->reduce; - $inflecs = List(Compute("None")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Compute("None")); - $dn = List(Interval("(-inf,inf)")); - Context("Fraction"); - $m = Fraction($a,2); - $crit = List($m); - $max = List($m); - $min = List(Compute("none")); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $s = random(1,9,1); - $c = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$s=1; $c=1;}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("x^3 - $s x + $c")->reduce; - $inflecs = List(0); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("[0,inf)")); - $dn = List(Interval("(-inf,0]")); - Context("Fraction"); - $rad = 3*$s; - ($radsquare,$radnonsquare) = (1,$rad); - for my $p (2,3,5) { - while ($radnonsquare % ($p**2) == 0) { - ($radsquare,$radnonsquare) = ($radsquare * $p**2, $radnonsquare / $p**2); - } - } - $G = gcd(sqrt($radsquare),3); - $B = sqrt($radsquare)/$G; - $C = 3/$G; - if ($discnonsquare == 1) { - $m = Fraction($B,$C); - $n = Fraction(-$B,$C); - } else { - $m = Formula("($B sqrt($radnonsquare))/$C"); - $n = Formula("(-$B sqrt($radnonsquare))/$C"); - } - ($m,$n) = num_sort($m,$n); - $crit = List($m,$n); - $max = List($m); - $min = List($n); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $a = non_zero_random(-9,9,1); - $b = non_zero_random(-9,9,1); - $c = ($a > 0) ? random(int(($b)**2/(3*$a))+1,int(($b)**2/(3*$a))+9,1) : random(int(($b)**2/(3*$a))-9,int(($b)**2/(3*$a))-1,1); - $d = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=2; $b=-3; $c=9; $d=5}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("$a x^3 + $b x^2 + $c x + $d")->reduce; - Context("Fraction"); - $i = Fraction(-2*$b,6*$a); - $inflecs = List($i); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("[$i,inf)")); - $dn = List(Interval("(-inf,$i]")); - ($up,$dn) = ($dn,$up) if ($a < 0); - $crit = List(Compute("none")); - $max = List(Compute("none")); - $min = List(Compute("none")); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $s = non_zero_random(-9,9,1); - $k = ($s % 2 == 0) ? non_zero_random(2,8,2) : non_zero_random(1,9,2); - $d = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$s=-1; $k=-1; $d=3;}; - $r = ($s + $k)/-2; - $t = $k*$s; - Context("Fraction"); - $b = Fraction(-($r+2*$s),3); - ($bn,$bd) = $b->value; - $c = -$r*(($s)**2+$t); - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("x^4/4 + $bn x^3/$bd + $c x + $d")->reduce; - $i = Fraction(3*$s-$k,3); - ($i,$j) = num_sort($i,0); - $inflecs = List($i,$j); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); - $dn = List(Interval("[$i,$j]")); - $crit = List($r); - $max = ($i <= $r and $r <= $j) ? List($r) : List(Compute("none")); - $min = ($i <= $r and $r <= $j) ? List(Compute("none")) : List($r); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - ($r,$s,$t) = random_subset(3,-5..-1,1..5); - $k = list_random(-12,-8,-4,4,8,12); - if ($k % 3 != 0) { - do {$t = non_zero_random(-($r+$s) % 3 - 6, -($r+$s) % 3 + 6, 3)} until ($t != $r and $t != $s); - } - ($r,$s,$t) = num_sort($r,$s,$t); - $e = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$r=-1; $s=1; $t=2; $k=12;}; - $a = $k/4; - $b = $k*($r+$s+$t)/3; - $c = $k*($s*$t+$r*$s+$r*$t)/2; - $d = $k*$r*$s*$t; - Context("Fraction"); - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; - $disc = 9*($b)**2 - 24*$a*$c; - ($discsquare,$discnonsquare) = (1,$disc); - for my $p (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251) { - while ($discnonsquare % ($p**2) == 0) { - ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); - } - } - $G = gcd(gcd(3*$b,sqrt($discsquare)),12*$a); - $A = -3*$b/$G; - $B = sqrt($discsquare)/$G; - $C = 12*$a/$G; - if ($discnonsquare == 1) { - $i = Fraction($A+$B,$C); - $j = Fraction($A-$B,$C); - } else { - $i = Formula("($A+$B sqrt($discnonsquare))/$C"); - $j = Formula("($A-$B sqrt($discnonsquare))/$C"); - } - ($i,$j) = num_sort($i,$j); - $inflecs = List($i,$j); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); - $dn = List(Interval("[$i,$j]")); - ($up,$dn) = ($dn,$up) if ($a < 0); - $crit = List($r,$s,$t); - $max = ($a > 0) ? List($s) : List($r,$t); - $min = ($a > 0) ? List($r,$t): List($s); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $r = non_zero_random(-4,4,1); - if($envir{problemSeed}==1){$r=1;}; - $a = 1; - $b = -4*$r; - $c = 6*($r**2); - $d = -4*($r**3); - $e = ($r)**4; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; - $inflecs = List($r); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,inf)")); - $dn = List(Compute("none")); - $crit = List($r); - $max = List(Compute("none")); - $min = List(Compute("none")); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("sec(x)"); - $inflecs = List(Compute("none")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-pi/2,pi/2)")); - $dn = List(Interval("(-3pi/2,-pi/2)"),Interval("(pi/2,3pi/2)")); - $crit = List(Compute("-pi"),0,Compute("pi")); - $max = List(Compute("-pi"),Compute("pi")); - $min = List(0); - - -

    - f(x)= on (-3\pi/2,3\pi/2) -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $r = random(-9,9,1); - $s = random(1,9,1); - if($envir{problemSeed}==1){$r=0; $s = 1;}; - $b = -2*$r; - $c = ($r**2) + $s; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("1/(x^2 + $b x + $c)")->reduce; - $disc = 3*$s; - ($discsquare,$discnonsquare) = (1,$disc); - for my $p (2,3,5) { - while ($discnonsquare % ($p**2) == 0) { - ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); - } - } - $G = gcd(gcd(3*$3,sqrt($discsquare)),3); - $A = 3*$r/$G; - $B = sqrt($discsquare)/$G; - $C = 3/$G; - Context("Fraction"); - if ($discnonsquare == 1) { - $i = Fraction($A+$B,$C); - $j = Fraction($A-$B,$C); - } else { - $i = Formula("($A+$B sqrt($discnonsquare))/$C"); - $j = Formula("($A-$B sqrt($discnonsquare))/$C"); - } - ($i,$j) = num_sort($i,$j); - $inflecs = List($i,$j); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); - $dn = List(Interval("[$i,$j]")); - $crit = List($r); - $max = List($r); - $min = List("none"); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - ($r,$s) = num_sort(random_subset(2,-9..-1,1..9)); - if($envir{problemSeed}==1){$r=-1; $s = 1;}; - $b = -$r - $s; - $c = $r*$s; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("1/(x^2 + $b x + $c)")->reduce; - $inflecs = List(Compute("none")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$r)"),Interval("($s,inf)")); - $dn = List(Interval("($r,$s)")); - Context("Fraction"); - $m = Fraction($r+$s,2); - $crit = List($m); - $max = List($m); - $min = List("none"); - - -

    - f(x)= -

    - - Enter the possible points of inflection, - separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimum. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("sin(x) + cos(x)"); - $inflecs = List(Compute("-pi/4"),Compute("3pi/4")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-pi,-pi/4]"),Interval("[3pi/4,pi)")); - $dn = List(Interval("[-pi/4,3pi/4]")); - $crit = List(Compute("-3pi/4"),Compute("pi/4")); - $max = List(Compute("pi/4")); - $min = List(Compute("-3pi/4")); - - -

    - f(x)= on (-\pi,\pi) -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("x^2e^x"); - $inflecs = List(Compute("-2+sqrt(2)"),Compute("-2-sqrt(2)")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,-2-sqrt(2)]"),Interval("[-2+sqrt(2),inf)")); - $dn = List(Interval("[-2-sqrt(2),-2+sqrt(2)]")); - $crit = List(-2,0); - $max = List(-2); - $min = List(0); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("x^2ln(x)"); - $inflecs = List(Compute("1/e^(3/2)")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("[1/e^(3/2),inf)")); - $dn = List(Interval("(0,1/e^(3/2)]")); - $crit = List(Compute("e^(-1/2)")); - $max = List(Compute("none")); - $min = List(Compute("1/sqrt(e)")); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("e^(-x^2)"); - $inflecs = List(Compute("1/sqrt(2)"),Compute("-1/sqrt(2)")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,-1/sqrt(2)]"),Interval("[1/sqrt(2),inf)")); - $dn = List(Interval("[-1/sqrt(2),1/sqrt(2)]")); - $crit = List(0); - $max = List(0); - $min = List(Compute("none")); - - -

    - f(x)= -

    - - Enter the critical points, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative maximum. - If there are none, - enter none. - -

    - -

    - - List the critical points for which the Second Derivative Test indicates the point is a relative minimumm. - If there are none, - enter none. - -

    - -

    -     - - -     -     -     - - - -

    - A function f(x) is given. - Find the x values where \fp(x) has a relative maximum or minimum. - (Note: these are the same functions as in .) -

    -     - - - - - $r = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$r=1;}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("x^2 - 2*$r x + ($r)^2")->reduce; - $inflecs = List(Compute("None")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,inf)")); - $dn = List(Compute("None")); - Context("Fraction"); - $crit = List($r); - $max = List(Compute("none")); - $min = List($r); - $fpmax = List(Compute("none")); - $fpmin = List(Compute("none")); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - ($a,$b) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$a=-5; $b=7}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("-x^2 + $a x + $b")->reduce; - $inflecs = List(Compute("None")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Compute("None")); - $dn = List(Interval("(-inf,inf)")); - Context("Fraction"); - $m = Fraction($a,2); - $crit = List($m); - $max = List($m); - $min = List(Compute("none")); - $fpmax = List(Compute("none")); - $fpmin = List(Compute("none")); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $s = random(1,9,1); - $c = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$s=1; $c=1;}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("x^3 - $s x + $c")->reduce; - $inflecs = List(0); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("[0,inf)")); - $dn = List(Interval("(-inf,0]")); - Context("Fraction"); - $rad = 3*$s; - ($radsquare,$radnonsquare) = (1,$rad); - for my $p (2,3,5) { - while ($radnonsquare % ($p**2) == 0) { - ($radsquare,$radnonsquare) = ($radsquare * $p**2, $radnonsquare / $p**2); - } - } - $G = gcd(sqrt($radsquare),3); - $B = sqrt($radsquare)/$G; - $C = 3/$G; - if ($discnonsquare == 1) { - $m = Fraction($B,$C); - $n = Fraction(-$B,$C); - } else { - $m = Formula("($B sqrt($radnonsquare))/$C"); - $n = Formula("(-$B sqrt($radnonsquare))/$C"); - } - ($m,$n) = num_sort($m,$n); - $crit = List($m,$n); - $max = List($m); - $min = List($n); - $fpmax = List(Compute("none")); - $fpmin = List(0); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $a = non_zero_random(-9,9,1); - $b = non_zero_random(-9,9,1); - $c = ($a > 0) ? random(int(($b)**2/(3*$a))+1,int(($b)**2/(3*$a))+9,1) : random(int(($b)**2/(3*$a))-9,int(($b)**2/(3*$a))-1,1); - $d = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=2; $b=-3; $c=9; $d=5}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("$a x^3 + $b x^2 + $c x + $d")->reduce; - Context("Fraction"); - $i = Fraction(-2*$b,6*$a); - $inflecs = List($i); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("[$i,inf)")); - $dn = List(Interval("(-inf,$i]")); - ($up,$dn) = ($dn,$up) if ($a < 0); - $crit = List(Compute("none")); - $max = List(Compute("none")); - $min = List(Compute("none")); - $fpmax = List(Compute("none")); - $fpmin = List($i); - ($fpmax,$fpmin) = ($fpmin,$fpmax) if ($a < 0); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $s = non_zero_random(-9,9,1); - $k = ($s % 2 == 0) ? non_zero_random(2,8,2) : non_zero_random(1,9,2); - $d = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$s=-1; $k=-1; $d=3;}; - $r = ($s + $k)/-2; - $t = $k*$s; - Context("Fraction"); - $b = Fraction(-($r+2*$s),3); - ($bn,$bd) = $b->value; - $c = -$r*(($s)**2+$t); - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("x^4/4 + $bn x^3/$bd + $c x + $d")->reduce; - $i = Fraction(3*$s-$k,3); - ($i,$j) = num_sort($i,0); - $inflecs = List($i,$j); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); - $dn = List(Interval("[$i,$j]")); - $crit = List($r); - $max = ($i <= $r and $r <= $j) ? List($r) : List(Compute("none")); - $min = ($i <= $r and $r <= $j) ? List(Compute("none")) : List($r); - $fpmax = List($i); - $fpmin = List($j); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - ($r,$s,$t) = random_subset(3,-5..-1,1..5); - $k = list_random(-12,-8,-4,4,8,12); - if ($k % 3 != 0) { - do {$t = non_zero_random(-($r+$s) % 3 - 6, -($r+$s) % 3 + 6, 3)} until ($t != $r and $t != $s); - } - ($r,$s,$t) = num_sort($r,$s,$t); - $e = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$r=-1; $s=1; $t=2; $k=12;}; - $a = $k/4; - $b = $k*($r+$s+$t)/3; - $c = $k*($s*$t+$r*$s+$r*$t)/2; - $d = $k*$r*$s*$t; - Context("Fraction"); - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; - $disc = 9*($b)**2 - 24*$a*$c; - ($discsquare,$discnonsquare) = (1,$disc); - for my $p (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251) { - while ($discnonsquare % ($p**2) == 0) { - ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); - } - } - $G = gcd(gcd(3*$b,sqrt($discsquare)),12*$a); - $A = -3*$b/$G; - $B = sqrt($discsquare)/$G; - $C = 12*$a/$G; - if ($discnonsquare == 1) { - $i = Fraction($A+$B,$C); - $j = Fraction($A-$B,$C); - } else { - $i = Formula("($A+$B sqrt($discnonsquare))/$C"); - $j = Formula("($A-$B sqrt($discnonsquare))/$C"); - } - ($i,$j) = num_sort($i,$j); - $inflecs = List($i,$j); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); - $dn = List(Interval("[$i,$j]")); - ($up,$dn) = ($dn,$up) if ($a < 0); - $crit = List($r,$s,$t); - $max = ($a > 0) ? List($s) : List($r,$t); - $min = ($a > 0) ? List($r,$t): List($s); - $fpmax = List($i); - $fpmin = List($j); - ($fpmax,$fpmin) = ($fpmin,$fpmax) if ($a < 0); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $r = non_zero_random(-4,4,1); - if($envir{problemSeed}==1){$r=1;}; - $a = 1; - $b = -4*$r; - $c = 6*($r**2); - $d = -4*($r**3); - $e = ($r)**4; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("$a x^4 + $b x^3 + $c x^2 + $d x + $e")->reduce; - $inflecs = List($r); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,inf)")); - $dn = List(Compute("none")); - $crit = List($r); - $max = List(Compute("none")); - $min = List(Compute("none")); - $fpmax = List(Compute("none")); - $fpmin = List(Compute("none")); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("sec(x)"); - $inflecs = List(Compute("none")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-pi/2,pi/2)")); - $dn = List(Interval("(-3pi/2,-pi/2)"),Interval("(pi/2,3pi/2)")); - $crit = List(Compute("-pi"),0,Compute("pi")); - $max = List(Compute("-pi"),Compute("pi")); - $min = List(0); - $fpmax = List(Compute("none")); - $fpmin = List(Compute("none")); - - -

    - f(x)= on (-3\pi/2,3\pi/2) -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $r = random(-9,9,1); - $s = random(1,9,1); - if($envir{problemSeed}==1){$r=0; $s = 1;}; - $b = -2*$r; - $c = ($r**2) + $s; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("1/(x^2 + $b x + $c)")->reduce; - $disc = 3*$s; - ($discsquare,$discnonsquare) = (1,$disc); - for my $p (2,3,5) { - while ($discnonsquare % ($p**2) == 0) { - ($discsquare,$discnonsquare) = ($discsquare * $p**2, $discnonsquare / $p**2); - } - } - $G = gcd(gcd(3*$3,sqrt($discsquare)),3); - $A = 3*$r/$G; - $B = sqrt($discsquare)/$G; - $C = 3/$G; - Context("Fraction"); - if ($discnonsquare == 1) { - $i = Fraction($A+$B,$C); - $j = Fraction($A-$B,$C); - } else { - $i = Formula("($A+$B sqrt($discnonsquare))/$C"); - $j = Formula("($A-$B sqrt($discnonsquare))/$C"); - } - ($i,$j) = num_sort($i,$j); - $inflecs = List($i,$j); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$i]"),Interval("[$j,inf)")); - $dn = List(Interval("[$i,$j]")); - $crit = List($r); - $max = List($r); - $min = List("none"); - $fpmax = List($i); - $fpmin = List($j); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - ($r,$s) = num_sort(random_subset(2,-9..-1,1..9)); - if($envir{problemSeed}==1){$r=-1; $s = 1;}; - $b = -$r - $s; - $c = $r*$s; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("1/(x^2 + $b x + $c)")->reduce; - $inflecs = List(Compute("none")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,$r)"),Interval("($s,inf)")); - $dn = List(Interval("($r,$s)")); - Context("Fraction"); - $m = Fraction($r+$s,2); - $crit = List($m); - $max = List($m); - $min = List("none"); - $fpmax = List(Compute("none")); - $fpmin = List(Compute("none")); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("sin(x) + cos(x)"); - $inflecs = List(Compute("-pi/4"),Compute("3pi/4")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-pi,-pi/4]"),Interval("[3pi/4,pi)")); - $dn = List(Interval("[-pi/4,3pi/4]")); - $crit = List(Compute("-3pi/4"),Compute("pi/4")); - $max = List(Compute("pi/4")); - $min = List(Compute("-3pi/4")); - $fpmax = List(Compute("-pi/4")); - $fpmin = List(Compute("3pi/4")); - - -

    - f(x)= on (-\pi,\pi) -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("x^2e^x"); - $inflecs = List(Compute("-2+sqrt(2)"),Compute("-2-sqrt(2)")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,-2-sqrt(2)]"),Interval("[-2+sqrt(2),inf)")); - $dn = List(Interval("[-2-sqrt(2),-2+sqrt(2)]")); - $crit = List(-2,0); - $max = List(-2); - $min = List(0); - $fpmax = List(Compute("-2-sqrt(2)")); - $fpmin = List(Compute("-2+sqrt(2)")); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("x^2ln(x)"); - $inflecs = List(Compute("1/e^(3/2)")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("[1/e^(3/2),inf)")); - $dn = List(Interval("(0,1/e^(3/2)]")); - $crit = List(Compute("e^(-1/2)")); - $max = List(Compute("none")); - $min = List(Compute("1/sqrt(e)")); - $fpmax = List(Compute("none")); - $fpmin = List(Compute("1/e^(3/2)")); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     - - - - - $f = Formula("e^(-x^2)"); - $inflecs = List(Compute("1/sqrt(2)"),Compute("-1/sqrt(2)")); - Context("Interval"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up = List(Interval("(-inf,-1/sqrt(2)]"),Interval("[1/sqrt(2),inf)")); - $dn = List(Interval("[-1/sqrt(2),1/sqrt(2)]")); - $crit = List(0); - $max = List(0); - $min = List(Compute("none")); - $fpmax = List(Compute("-1/sqrt(2)")); - $fpmin = List(Compute("1/sqrt(2)")); - - -

    - f(x)= -

    - - Enter the points where \fp(x) has a relative maximum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    - - Enter the points where \fp(x) has a relative minimum, separating with commas if needed. - If there are none, - enter none. - -

    - -

    -     - - -     -     -     -     -     -
    -
    - Curve Sketching -

    - We have been learning how we can understand the behavior of a function based on its first and second derivatives. - While we have been treating the properties of a function separately - (increasing and decreasing, concave up and concave down, etc.), - we combine them here to produce an accurate graph of the function without plotting lots of extraneous points. -

    - -

    - Why bother? - Graphing utilities are very accessible, - whether on a computer, a hand-held calculator, or a smartphone. - These resources are usually very fast and accurate. - We will see that our method is not particularly fast it will require time - (but it is not hard). - So again: why bother? -

    - -

    - We are attempting to understand the behavior of a function f based on the information given by its derivatives. - While all of a function's derivatives relay information about it, - it turns out that most of the behavior we care about is explained by \fpand \fpp. - Understanding the interactions between the graph of f and \fpand \fppis important. - To gain this understanding, - one might argue that all that is needed is to look at lots of graphs. - This is true to a point, - but is somewhat similar to stating that one understands how an engine works after looking only at pictures. - It is true that the basic ideas will be conveyed, - but hands-on access increases understanding. -

    - - - -

    - - summarizes what we have learned so far that is applicable to sketching graphs of functions and gives a framework for putting that information together. - It is followed by several examples. -

    - - - Curve Sketching -

    - To produce an accurate sketch a given function f, - consider the following steps. - curve sketching -

      -
    1. - -

      - Find the domain of f. - Generally, we assume that the domain is the entire real line then find restrictions, - such as where a denominator is 0 or where negatives appear under the radical. -

      -
      2. - -
    2. -

      - Find the critical values of f. -

      -
      3. - -
    3. -

      - Find the possible points of inflection of f. -

      -
      4. - -
    4. -

      - Find the location of any vertical asymptotes of f - (usually done in conjunction with ). -

      -
      5. - -
    5. -

      - Consider the limits \lim\limits_{x\to-\infty}f(x) and - \lim\limits_{x\to\infty}f(x) to determine the end behavior of the function. -

      -
      6. - -
    6. -

      - Create a number line that includes all critical points, - possible points of inflection, - and locations of vertical asymptotes. - For each interval created, - determine whether f is increasing or decreasing, - concave up or down. -

      -
      7. - -
    7. -

      - Evaluate f at each critical point and possible point of inflection. - Plot these points on a set of axes. - Connect these points with curves exhibiting the proper concavity. - Sketch asymptotes and x and y intercepts where applicable. -

      -
      8. -
    -

    -     - - - - - Curve sketching - -

    - Use - to sketch f(x) = 3x^3-10x^2+7x+5. -

    -     - -

    - We follow the steps outlined in . - -

      -
    1. -

      - The domain of f is the entire real line; - there are no values x for which f(x) is not defined. -

      -
      2. - -
    2. -

      - Find the critical values of f. - We compute \fp(x) = 9x^2-20x+7. - Use the Quadratic Formula to find the roots of \fp: - - x \amp= \frac{20\pm \sqrt{(-20)^2-4(9)(7)}}{2(9)} - \amp = \frac19\left(10\pm\sqrt{37}\right) - x \amp \approx 0.435, 1.787 - . -

      -
      3. - -
    3. -

      - Find the possible points of inflection of f. - Compute \fpp(x) = 18x-20. - We have - - \fpp(x) \amp = 0 - 18x-20 \amp =0 - x \amp = 10/9 - \amp \approx 1.111 - . -

      -
      4. - -
    4. -

      - There are no vertical asymptotes. -

      -
      5. - -
    5. -

      - We determine the end behavior using limits as x approaches \pm \infty . - - \lim_{x\to -\infty} f(x) = -\infty \qquad \lim_{x\to \infty}f(x) = \infty - . - We do not have any horizontal asymptotes. -

      -
      6. - -
    6. - -

      - We place the values x=(10\pm\sqrt{37})/9 and x=10/9 on a number line, - as shown in . - We mark each subinterval as increasing or decreasing, concave up or down, - using the techniques used in Sections. -

      -
      7. - -
    7. -

      - Evaluate f at each critical number and possible inflection point. - - f(0.435)\amp\approx6.400\amp f(1.111)\amp\approx4.547\amp f(1.787)\amp\approx 2.695 - -

      - -

      - We plot the appropriate points on axes as shown in - and connect the points with straight lines - (to show increasing/decreasig behavior). - In - we adjust these lines to demonstrate the proper concavity. - In - we show a graph of f drawn with a computer program, - verifying the accuracy of our sketch. -

      -
      8. -
    -

    - -
    - Number line for f in - - - Number line showing intervals where f is increasing/decreasing and concave up/down - -

    - A number line is shown, on which three points are marked. - The first point is marked as \frac19(10-\sqrt{37}), or approximately 0.435; - it is a critical point of f, and a relative maximum. -

    - -

    - The second point is marked as \frac{10}{9}, or approximately 1.111; - it is an inflection point of f. -

    - -

    - The last point is marked as \frac19(10+\sqrt{37}), or approximately 1.787; - it is a critical point of f, and a relative minimum. -

    - -

    - These points divide the number line into four intervals. - Above each interval, the signs of both \fp and \fpp are given, - along with whether f is increasing or decreasing, and concave up or down. -

    - -

    - This information is as follows: -

      -
    • -

      - For x\lt 0.435, \fp\gt 0, so f is increasing, - and \fpp\lt 0, so f is concave down. -

      -
      • - -
    • -

      - For 0.435\lt x\lt 1.111, \fp\lt 0, so f is decreasing, - and \fpp\lt 0, so f is concave down. -

      -
      • - -
    • -

      - For 1.111\lt x\lt 1.787, \fp\lt 0, so f is decreasing, - and \fpp\gt 0, so f is concave up. -

      -
      • - -
    • -

      - For x\gt 1.787, \ft\gt 0, so f is increasing, - and \fpp\gt 0, so f is concave up. -

      -
      • -
    -

    -     - - \begin{tikzpicture} - \begin{axis}[numberline, - xmin=0,xmax=2.222, - xtick={10}, - minor xtick={{}}, - extra x ticks={0.435, 1.111, 1.787}, - extra x tick labels={\parbox{6em}{\centering \scriptsize $\frac{1}{9}\left(10-\sqrt{37}\right)$ $\approx0.435$}, - \parbox{2em}{\scriptsize $\frac{10}{9}\approx1.111$}, \parbox{6em}{\centering \scriptsize $\frac{1}{9}\left(10+\sqrt{37}\right)$ $\approx1.787$}},] - \addplot[guideline] coordinates {(0.435,0) (0.435,2)}; - \addplot[guideline] coordinates {(1.111,0) (1.111,2)}; - \addplot[guideline] coordinates {(1.787,0) (1.787,2)}; - \addplot[mark=none] coordinates {(0.18,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\gt 0$,\\ $f$ incr\\$\fpp\lt 0$,\\ $f$ c.\ down }}; - \addplot[mark=none] coordinates {(0.77,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\lt 0$,\\ $f$ decr\\$\fpp\lt 0$,\\ $f$ c.\ down }}; - \addplot[mark=none] coordinates {(1.45,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\lt 0$,\\ $f$ decr\\$\fpp\gt 0$,\\ $f$ c.\ up }}; - \addplot[mark=none] coordinates {(2.1,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\gt 0$,\\ $f$ incr\\$\fpp\gt 0$,\\ $f$ c.\ up }}; - \end{axis} - \end{tikzpicture} - - -
    - -
    - Sketching f in - -
    - - - - A hand-drawn graph based on the critical points from the number line. - -

    - The graph is hand-drawn with plotted significant points from the number line. - It connects these points with straight lines to give a general impression of the graph's shape. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ymin=-5.5,ymax=10.9,xmin=-1.2,xmax=3.2,] - \addplot[firstcurvestyle,leftarrow,domain=-1:0.435] {(5-6.4)/(0-0.435)*(x-0.435)+6.4}; - \addplot[firstcurvestyle] coordinates {(0.435,6.4) (1.79,2.69)}; - \addplot[firstcurvestyle,rightarrow,domain=1.79:2.7] {(5-6.4)/(0-0.435)*(x-1.79)+2.69}; - \addplot[soliddot] coordinates{(0.435,6.4) (1.11,4.55) (1.79,2.69)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - Adjusted graph for proper concavity. - -

    - The image shows an adjusted graph of the piecewise linear function. - The function is now a smooth, continuous curve that crosses the y-axis at y=5. - This adjustment demonstrates the proper concavity of the function. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ymin=-5.5,ymax=10.9,xmin=-1.2,xmax=3.2,] - \addplot[firstcurvestyle,leftarrow,domain=-0.8:0.435] {(x-0.435)^2/(0-0.435)^2*(5-6.4) + 6.4}; - \addplot[firstcurvestyle,domain=0.435:1.79] {(6.4-2.69)*6/(1.79-0.435)^3*((x-0.435)*(x-1.79)^2/2 - (x-1.79)^3/6)+2.69}; - \addplot[firstcurvestyle,rightarrow,domain=1.79:2.8] {-(x-1.79)^2/(0-0.435)^2*(5-6.4) + 2.69}; - \addplot[soliddot] coordinates{(0.435,6.4) (1.11,4.55) (1.79,2.69)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - Computer-generated graph for accuracy verification. - -

    - The image shows a computer-generated graph of the function. The graph verifies the accuracy of our sketch: - there is very little difference between this graph and the previous graph, - showing that accounting for concavity helps to ensure an accurate sketch. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ymin=-5.5,ymax=10.9,xmin=-1.2,xmax=3.2,] - \addplot [firstcurvestyle,infinite,domain=-.6:2.8] {3*x^3-10*x^2+7*x+5}; - \addplot[soliddot] coordinates{(0.435,6.4) (1.11,4.55) (1.79,2.69)}; - \end{axis} - \end{tikzpicture} - - - - - -
    -     -
    -     - -     - - - Curve sketching - -

    - Sketch \ds f(x) = \frac{x^2-x-2}{x^2-x-6}. -

    -     - -

    - We again follow the steps outlined in . -

    - -

    -

      -
    1. -

      - In determining the domain, - we assume it is all real numbers and look for restrictions. - We find that at x=-2 and x=3, - f(x) is not defined. - So the domain of f is D = \{ x \mid x\neq -2,3\}. -

      -
      2. - -
    2. -

      - To find the critical values of f, - we first find \fp(x). - Using the , we find - - \fp(x) = \frac{-8x+4}{(x^2+x-6)^2} = \frac{-8x+4}{(x-3)^2(x+2)^2} - . - We get \fp(x) = 0 when x = 1/2, - and \fp is undefined when x=-2,3. - Since \fpis undefined only when f is also undefined, - these are not critical values. - The only critical value is x=1/2. -

      -
      3. - -
    3. -

      - To find the possible points of inflection, we find \fpp(x), - again employing the : - - \fpp(x) = \frac{24x^2-24x+56}{(x-3)^3(x+2)^3} - . - We find that \fpp(x) is never 0 - (setting the numerator equal to 0 and solving for x, - we find the only roots to this quadratic are not real numbers) - and \fppis undefined when x=-2,3. - Thus concavity will possibly only change at x=-2 and x=3 - (which are not in the domain of f, - so these won't be inflection points). -

      -
      4. - -
    4. -

      - The vertical asymptotes of f are at x=-2 and x=3, - the places where f is undefined. -

      -
      5. - -
    5. -

      - There is a horizontal asymptote of y=1, - as \lim\limits_{x\to -\infty}f(x) = 1 and \lim\limits_{x\to\infty}f(x) =1. -

      -
      6. - -
    6. -

      - We place the values x=1/2, - x=-2 and x=3 on a number line as shown in . - We mark in each interval whether f is increasing or decreasing, - concave up or down. - We see that f has a relative maximum at x=1/2; - concavity changes only at the vertical asymptotes. -

      -
      7. - -
    7. -

      - Evaluate f at each critical number. - - f(0)\amp=1/3\amp f(1/2)\amp=9/25 - -

      - -

      - In , - we plot the points from the number line on a set of axes and connect the points with straight lines to get a general idea of what the function looks like - (these lines effectively only convey increasing/decreasing information). - In , - we adjust the graph with the appropriate concavity. - We also show f crossing the x-axis at x=-1 and x=2 and crossing the y-axis at y=1/3. - Finally, - shows a computer generated graph of f, - which verifies the accuracy of our sketch. -

      -
      8. -
    -

    - -
    - Number line for f in - - Number line showing intervals where f is increasing/decreasing and concave up/down - -

    - On a number line there are three marked points: -2, \frac12, and 3. - At x=-2 and x=3, the graph of f has a vertical asymptote. - The point x=\frac12 is a critical point of f, and a relative maximum. -

    - -

    - These three points divide the number line into four intervals. - Above each interval, the following information is indicated: -

      -
    • -

      - For x\lt -2, \fp\gt 0, so f is increasing, - and \fpp\gt 0, so f is concave up. -

      -
      • - -
    • -

      - For -2\lt x\lt \frac12, \fp\gt 0, so f is increasing, - and \fpp\lt 0, so f is concave down. -

      -
      • - -
    • -

      - For \frac12\lt x\lt 3, \fp\lt 0, so f is decreasing, - and \fpp\lt 0, so f is concave down. -

      -
      • - -
    • -

      - For x\gt 3, \fp\lt 0, so f is decreasing, - and \fpp\gt 0, so f is concave up. -

      -
      • -
    -

    -     - - \begin{tikzpicture} - \begin{axis}[numberline, - xmin=0,xmax=2.222, - minor xtick={{}}, - xtick={10}, - extra x ticks={0.435, 1.111, 1.787}, - extra x tick labels={\scriptsize $-2$, \scriptsize $\frac{1}{2}$, \scriptsize $3$},] - \addplot[guideline] coordinates {(0.435,0) (0.435,2)}; - \addplot[guideline] coordinates {(1.111,0) (1.111,2)}; - \addplot[guideline] coordinates {(1.787,0) (1.787,2)}; - \addplot[mark=none] coordinates {(0.18,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\gt 0$,\\ $f$ incr\\$\fpp\gt 0$,\\ $f$ c.\ up }}; - \addplot[mark=none] coordinates {(0.77,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\gt 0$,\\ $f$ incr\\$\fpp\lt 0$,\\ $f$ c.\ down }}; - \addplot[mark=none] coordinates {(1.45,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\lt 0$,\\ $f$ decr\\$\fpp\lt 0$,\\ $f$ c.\ down }}; - \addplot[mark=none] coordinates {(2.1,1.5)} node {\parbox{8em}{\centering \scriptsize $\fp\lt 0$,\\ $f$ incr\\$\fpp\lt 0$,\\ $f$ c.\ down }}; - \end{axis} - \end{tikzpicture} - - -
    - -
    - Sketching f in - -
    - - - - A linear piecewise graph showing changes in concavity. - -

    - The figure shows a linear piecewise graph on a Cartesian plane with the x-axis ranging from -5 to 5, - and the y-axis from -5 to 5. - The graph consists of two line segments forming a V shape, with the tip intersecting the x-axis at x = 0.5. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ymin=-5.2,ymax=5.2,xmin=-5.1,xmax=5.1,] - \addplot[firstcurvestyle,domain=-4.5:-2.5] {(x+3)+3}; - \addplot[firstcurvestyle,leftarrow,domain=-1.7:0.5] {(0-0.36)/(-1-0.5)*(x-0.5)+0.36}; - \addplot[firstcurvestyle,rightarrow,domain=0.5:2.7] {(0-0.36)/(2-0.5)*(x-0.5)+0.36}; - \addplot[firstcurvestyle,domain=3.5:5] {-(x-4)+3}; - \addplot[asymptote] coordinates {(-2,-5) (-2,5)}; - \addplot[asymptote] coordinates {(3,-5) (3,5)}; - \addplot[asymptote] {1}; - \addplot[soliddot] coordinates{(0.5,0.36)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - A concave adjusted graph. - -

    - The image displays an adjusted version of the piecewise graph, now with concavity. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ymin=-5.2,ymax=5.2,xmin=-5.1,xmax=5.1,] - \addplot[firstcurvestyle,domain=-5.1:-2.3] {-1/(x+2)+1}; - \addplot[firstcurvestyle,domain=-1.8:2.8] {(x^2-x-2)/(x^2-x-6)}; - \addplot[firstcurvestyle,domain=3.3:5.1] {1/(x-3)+1}; - \addplot[asymptote] coordinates {(-2,-5) (-2,5)}; - \addplot[asymptote] coordinates {(3,-5) (3,5)}; - \addplot[asymptote] {1}; - \addplot[soliddot] coordinates{(0.5,0.36) (-1,0) (2,0)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - Computer generated graph for accuracy verification. - -

    - The image shows a computer-generated graph of the function. The graph verifies the accuracy of our sketch: - there is very little difference between this graph and the previous graph, - showing that accounting for concavity helps to ensure an accurate sketch. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ymin=-5.5,ymax=5.5,xmin=-5.2,xmax=5.2,] - \addplot [firstcurvestyle,infinite,domain=-5:-2.1] {((x-2)*(x+1))/((x-3)*(x+2))}; - \addplot [firstcurvestyle,infinite,domain=-1.9:2.9] {((x-2)*(x+1))/((x-3)*(x+2))}; - \addplot [firstcurvestyle,infinite,domain=3.1:5] {((x-2)*(x+1))/((x-3)*(x+2))}; - \addplot [asymptote] coordinates {(-2,5.5) (-2,-5.5)}; - \addplot [asymptote] coordinates {(3,5.5) (3,-5.5)}; - \addplot [asymptote] coordinates {(-5.5,1) (5.5,1)}; - \addplot [soliddot] coordinates{(0.5,.36)}; - \end{axis} - \end{tikzpicture} - - - - - - -
    -     -
    -     - -     - - - Curve sketching - -

    - Sketch f(x) = \dfrac{5(x-2)(x+1)}{x^2+2x+4}. -

    -     - -

    - We again follow . -

    - -

    -

      -
    1. -

      - We assume that the domain of f is all real numbers and consider restrictions. - The only restrictions could come when the denominator is 0, - but this never occurs because the denominator is a quadratic polynomial with no real roots. - Therefore the domain of f is all real numbers, \mathbb{R}. -

      -
      2. - -
    2. -

      - We find the critical values of f by setting - \fp(x)=0 and solving for x. - We find - - \fp(x) \amp = \frac{15x(x+4)}{(x^2+2x+4)^2} - 0 \amp =\frac{15x(x+4)}{(x^2+2x+4)^2} - x \amp =-4,0 - . - Since the denominator of \fp is just the square of the denominator of f, - there are no values of x for which \fp is undefined. -

      -
      3. - -
    3. -

      - We find the possible points of inflection by solving \fpp(x) = 0 for x (again, - there are no values of x for which \fpp is undefined.) We find - - \fpp(x) = -\frac{30x^3+180x^2-240}{(x^2+2x+4)^3} - . - The cubic in the numerator does not factor very nicely. - We instead approximate the roots - (using a CAS) - at x= -5.759, x=-1.305 and x=1.064. -

      -
      4. - -
    4. -

      - There are no vertical asymptotes as the denominator never equals zero. -

      -
      5. - -
    5. -

      - We have a horizontal asymptote of y=5, - as \lim\limits_{x\to-\infty}f(x) = \lim\limits_{x\to\infty}f(x) = 5. -

      -
      6. - -
    6. -

      - We place the critical points and possible points on a number line as shown in - and mark each interval as increasing/decreasing, - concave up/down appropriately. -

      -
      7. - -
    7. -

      - Evaluate f at each critical number, - possible inflection point. - - f(-5.759)\amp\approx7.200\amp f(-4)\amp=7.5 - f(-1.305)\amp\approx1.630\amp f(0)\amp=2.5 - f(1.064)\amp\approx-1.331 - -

      - -

      - In - we plot the significant points from the number line as well as the x- and y-intercepts, - and connect the points with straight lines to get a general impression about the graph - (this graph only includes increasing/decreasing information). - In , - we add concavity, drawing the function so that it is smooth - (since f is differentiable everywhere, - there should be no kinks or corners). - - shows a computer generated graph of f, affirming our results. -

      -
      8. -
    -

    - -
    - Number line for f in Example - - Number line showing intervals where f is increasing/decreasing and concave up/down - -

    - A number line is shown, on which five points are marked. - The points are labeled, from left to right, as -5.579, - -4, -1.305, 0, and 1.064. - The points x=-4 and x=0 are the critical points of f. -

    - -

    - The three decimal values are the possible inflection points of f, - which had to be approximated using software. - These points divide the number line into six intervals; - above each interval, the following information is indicated: -

    - -

    -

      -
    • -

      - For x\lt -5.579, \fp\gt 0, so f is increasing, - and \fpp \gt 0, so f is concave up. -

      -
      • - -
    • -

      - For -5.579\lt x\lt -4, \fp\gt 0, so f is increasing, - and \fpp \lt 0, so f is concave down. -

      -
      • - -
    • -

      - For -4\lt x\lt -1.305, \fp\lt 0, so f is decreasing, - and \fpp \lt 0, so f is concave down. -

      -
      • - -
    • -

      - For -1.305\lt x\lt 0, \fp\lt 0, so f is decreasing, - and \fpp \gt 0, so f is concave up. -

      -
      • - -
    • -

      - For 0\lt x\lt 1.064, \fp\gt 0, so f is increasing, - and \fpp \gt 0, so f is concave up. -

      -
      • - -
    • -

      - For x\gt 1.064, \fp\gt 0, so f is increasing, - and \fpp \lt 0, so f is concave down. -

      -
      • -
    -

    -     - - \begin{tikzpicture} - \begin{axis}[axis y line=none, - x=18pt, - y=20pt, - xtick={10}, - xlabel={}, - ymax=2.5, - xmin=-7,xmax=6, - minor xtick={{}}, - extra x ticks={-5,-2.8,-0.6,1.3,3.2}, - extra x tick labels={\scriptsize $-5.579$,\scriptsize $-4$,\scriptsize $-1.305$,\scriptsize $0$,\scriptsize $1.064$},] - \addplot[guideline] coordinates {(-5,0) (-5,2)}; - \addplot[guideline] coordinates {(-2.8,0) (-2.8,2)}; - \addplot[guideline] coordinates {(-0.6,0) (-0.6,2)}; - \addplot[guideline] coordinates {(1.3,0) (1.3,2)}; - \addplot[guideline] coordinates {(3.2,0) (3.2,2)}; - \addplot[mark=none] coordinates {(-6,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\gt0$,\\ $f$ incr\\$\fpp\gt0$,\\ $f$ c.\ up }}; - \addplot[mark=none] coordinates {(-3.9,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\gt0$,\\ $f$ incr\\$\fpp\lt0$,\\ $f$ c.\ down }}; - \addplot[mark=none] coordinates {(-1.7,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\lt0$,\\ $f$ decr\\$\fpp\lt0$,\\ $f$ c.\ down }}; - \addplot[mark=none] coordinates {(0.4,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\lt0$,\\ $f$ decr\\$\fpp\gt0$,\\ $f$ c.\ up }}; - \addplot[mark=none] coordinates {(2.3,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\gt0$,\\ $f$ incr\\$\fpp\gt0$,\\ $f$ c.\ up }}; - \addplot[mark=none] coordinates {(4.4,1.2)} node {\parbox{8em}{\centering \scriptsize $\fp\gt0$,\\ $f$ incr\\$\fpp\lt0$,\\ $f$ c.\ down }}; - \end{axis} - \end{tikzpicture} - - -
    - - -
    - Sketching f in - -
    - - - - A hand-drawn graph based on the critical points from the number line. - -

    - The graph is hand-drawn with plotted significant points from the number line. - It connects these points with straight lines to give a general impression of the graph's shape. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ ymin=-3.1,ymax=8.1,xmin=-7.5,xmax=6.5,] - \addplot[firstcurvestyle,leftarrow,domain=-7:-4] {(7.2-7.5)/(-5.579+4)*(x+4)+7.5}; - \addplot[firstcurvestyle,domain=-4:-1.305] {(1.630-7.5)/(-1.305+4)*(x+4)+7.5}; - \addplot[firstcurvestyle,domain=-1.305:-1] {(0-1.630)/(-1+1.305)*(x+1)}; - \addplot[firstcurvestyle,domain=-1:0] {(-2.5-0)/(0+1)*(x+1)}; - \addplot[firstcurvestyle,domain=0:1.064] {(-1.331+2.5)/(1.064-0)*(x-1.064)-1.331}; - \addplot[firstcurvestyle,rightarrow,domain=1.064:5] {(0+1.331)/(2-1.064)*(x-2)}; - \addplot[asymptote,domain=-7.5:6.5] {5}; - \addplot[soliddot] coordinates{(0,-2.5) (-1,0) (2,0) (-5.579,7.2) (-1.305,1.630) (1.064,-1.331) (-4,7.5)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - A hand-drawn graph with concavity. - -

    - The graph is hand-drawn with plotted significant points from the number line. - It connects these points with smooth curves to give a general impression of the graph's shape. -

    -     - - - - \begin{tikzpicture} - \begin{axis}[ ymin=-3.1,ymax=8.1,xmin=-7.5,xmax=6.5,] - \addplot[firstcurvestyle,leftarrow,domain=-7.5:-4] {7.5-0.3/(exp(-1/2/(-5.579+4)^2*(-5.579+4)^2) - 1)*(-1+exp(-1/2/(-5.579+4)^2*(x+4)^2))}; - \addplot[firstcurvestyle,domain=-4:-1.305] {(1.630-7.5)/(-1.305+4)^2*(x+4)^2+7.5}; - \addplot[firstcurvestyle,domain=-1.305:-1] {(0-1.630)/(-1+1.305)*(x+1)+0}; - \addplot[firstcurvestyle,domain=-1:0] {2.5/(-1)^2*x^2-2.5}; - \addplot[firstcurvestyle,domain=0:1.064] {(-1.331+2.5)/(1.064)^2*x^2-2.5}; - \addplot[firstcurvestyle,rightarrow,domain=1.064:6] {4*(-1.331+2.5)/1.064 * sqrt(1.064+((1.331*1.064/4/(-1.331+2.5)+1.064)^2 - 1.064*2) / (1.064+2 - 2*(1.331*1.064/4/(-1.331+2.5)+1.064)))*sqrt(x+((1.331*1.064/4/(-1.331+2.5)+1.064)^2 - 1.064*2) / (1.064+2 - 2*(1.331*1.064/4/(-1.331+2.5)+1.064)))-(4*(-1.331+2.5)/1.064 * sqrt(1.064+((1.331*1.064/4/(-1.331+2.5)+1.064)^2 - 1.064*2) / (1.064+2 - 2*(1.331*1.064/4/(-1.331+2.5)+1.064))))*sqrt(2+((1.331*1.064/4/(-1.331+2.5)+1.064)^2 - 1.064*2) / (1.064+2 - 2*(1.331*1.064/4/(-1.331+2.5)+1.064)))}; - \addplot[asymptote,domain=-7.5:6.5] {5}; - \addplot[soliddot] coordinates{(0,-2.5) (-1,0) (2,0) (-5.579,7.2) (-1.305,1.630) (1.064,-1.331) (-4,7.5)}; - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - - A computer generated graph of the function. - - -

    - The graph is a computer generated graph of the function, - showing the same critical points as the hand-drawn graph. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ymin=-3.1,ymax=8.1,xmin=-7.5,xmax=6.5,] - \addplot [asymptote] coordinates {(-7.5,5) (6.5,5)}; - \addplot [firstcurvestyle,infinite,domain=-7.5:6.5] {(5*(x-2)*(x+1))/(x^2+2*x+4)}; - \addplot [soliddot] coordinates {(-5.75,7.2) (-4., 7.5) (-1.31, 1.63) (-1,0) (0,-2.5) (1.064, -1.33) (2,0)}; - \end{axis} - \end{tikzpicture} - - - - - - -
    -     -
    -     -     - - - - - - - - - -

    - In each of our examples, - we found a few significant points on the graph of f that corresponded to changes in increasing/decreasing or concavity. - We connected these points with straight lines, - then adjusted for concavity, - and finished by showing a very accurate, computer generated graph. -

    - - - -

    - Why are computer graphics so good? - It is not because computers are - smarter than we are. - Rather, it is largely because computers are much faster at computing than we are. - In general, computers graph functions much like most students do when first learning to draw graphs: - they plot equally spaced points, - then connect the dots using lines. - By using lots of points, - the connecting lines are short and the graph looks smooth. -

    - -

    - This does a fine job of graphing in most cases - (in fact, this is the method used for many graphs in this text). - However, in regions where the graph is very curvy, - this can generate noticeable sharp edges on the graph unless a large number of points are used. - High quality computer algebra systems, - such as Mathematica and Sage, - use special algorithms to plot lots of points only where the graph is curvy. -

    - -

    - In , - two graph of y=\sin(x) is given, - generated by Sage and Mathematica. - The small points represent each of the places where each CAS sampled the function. - Notice how at the bends - of \sin(x), lots of points are used; - where \sin(x) is relatively straight, fewer points are used. - (In the Mathematica plot, - many points are also used at the endpoints to ensure the - end behavior is accurate.) -

    - -
    - CAS plots of y=\sin(x) illustrating the sample points - -
    - Sage output - - A graph of the sine function generated by Sage. - -

    - The plot features a solid blue curve representing the sine wave. - The x-axis is labeled from 0 to 6, - and the y-axis ranges from -1 to 1. - Sample points are marked along the curve with blue dots, - indicating the places where the function was sampled. -

    -     - - - - - -
    - -
    - Mathematica output - - Mathematica plot of the sine graph - -

    - The same plot of the sine function, showing the location of the sample points. - This version comes from the software Mathematica. - In addition to more frequent sampling in areas with more curvature, - Mathematica also uses extra sample points at the ends of the domain. -

    -     - -
    -     -
    - -

    - How does Sage know where the graph is curvy? - Calculus. - When we study curvature in a later chapter, - we will see how the first and second derivatives of a function work together to provide a measurement of curviness. - Sage employs algorithms to determine regions of high curvature - and plots extra points there. -

    - -

    - Again, the goal of this section is not - How to graph a function when there is no computer to help. Rather, - the goal is Understand that the shape of the graph of a function is largely determined by understanding the behavior of the function at a few key places. - In , - we were able to accurately sketch a complicated graph using only five points and knowledge of asymptotes! -

    - -

    - There are many applications of our understanding of derivatives beyond curve sketching. - The next chapter explores some of these applications, - demonstrating just a few kinds of problems that can be solved with a basic knowledge of differentiation. -

    - - - - Terms and Concepts - - - - - -

    - Why is sketching curves by hand beneficial even though technology is ubiquitous? -

    - - -     - - - -     - - - - - - - - -

    - When sketching graphs of functions, - it is useful to find the critical points. - -

    -     - -     - - - - - -

    - When sketching graphs of functions, - it is useful to find the possible points of inflection. - -

    -     - -     - - - - - -

    - When sketching graphs of functions, - it is useful to find the horizontal and vertical asymptotes. - -

    -     - -     - -     - - - Problems - - - -

    - Practice using - by applying the principles to the given functions with familiar graphs. -

    -     - - - -

    - Use - to sketch a graph of f(x) = 2x+4 -

    -     - -

    - A good sketch will include the x and y intercepts and draw the appropriate line. -

    -     -     - - - -

    - Use - to sketch a graph of f(x) = -x^2+1 -

    -     - -

    - A good sketch will include the x and y intercepts.. -

    -     -     - - - -

    - Use - to sketch a graph of f(x) = \sin(x) -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of f(x) = e^x -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = \frac{1}{x} -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = \frac{1}{x^2} -

    -     - -

    - Use technology to verify sketch. -

    -     -     - -     - - - -

    - Sketch a graph of the given function using . - Show all work; check your answer with technology. -

    -     - - - -

    - Use - to sketch a graph of \ds f(x) = x^3-2x^2+4x+1 -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = -x^3+5x^2-3x+2 -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = x^3+3x^2+3x+1 -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = x^3-x^2-x+1 -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = (x-2)\ln(x-2) -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = (x-2)^2\ln(x-2) -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = \frac{x^2-4}{x^2} -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = \frac{x^2-4x+3}{x^2-6x+8} -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = \frac{x^2-2x+1}{x^2-6x+8} -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = x\sqrt{x+1} -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = x^2e^x -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = \sin(x) \cos(x) on [-\pi,\pi] -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = (x-3)^{2/3} + 2 -

    -     - -

    - Use technology to verify sketch. -

    -     -     - - - -

    - Use - to sketch a graph of \ds f(x) = \frac{(x-1)^{2/3}}{x} -

    -     - -

    - Use technology to verify sketch. -

    -     -     - -     - - - -

    - A function with the parameters a and b are given. - Describe the critical points and possible points of inflection of f in terms of a and b. -

    -     - - - - -

    - \ds f(x) = \frac{a}{x^2+b^2} -

    - -

    -

      -
    1. -

      - Find the critical points of f. -

      - -
      2. - -
    2. -

      - Find the inflection points of f. -

      - -
      3. -
    -

    -     - -

    - Critical point: x=0 - Points of inflection: \pm b/\sqrt{3} -

    -     - -     - - - - -

    - \ds f(x) = \sin(ax+b) -

    - -

    -

      -
    1. -

      - Find the critical points of f. -

      - -
      2. - -
    2. -

      - Find the inflection points of f. -

      - -
      3. -
    -

    -     - -

    - Critical points: x=\frac{n\pi/2-b}{a}, - where n is an odd integer - Points of inflection: (n\pi-b)/a, - where n is an integer. -

    -     - -     - - - - -

    - \ds f(x) = (x-a)(x-b) -

    - -

    -

      -
    1. -

      - Find the critical points of f. -

      - -
      2. - -
    2. -

      - Find the inflection points of f. -

      - -
      3. -
    -

    -     - -

    - Critical point: x=(a+b)/2 - Points of inflection: none -

    -     - -     - - - - -

    - Given x^2+y^2=1, use implicit differentiation to find - \frac{dy}{dx} and \frac{d^2y}{dx^2}. - Use this information to justify the sketch of the unit circle. -

    -     - -

    - \frac{dy}{dx} = -x/y, - so the function is increasing in second and fourth quadrants, - decreasing in the first and third quadrants. -

    - -

    - \frac{d^2y}{dx^2} = -1/y - x^2/y^3, - which is positive when y\lt 0 and is negative when y\gt0. - Hence the function is concave down in the first and second quadrants and concave up in the third and fourth quadrants. -

    -     - -     - -     -     -     -
    - - -     - - - Applications of the Derivative - -

    - In , - we learned how the first and second derivatives of a function influence its graph. - In this chapter we explore other applications of the derivative. -

    -     - -
    - Newton's Method -

    - Solving equations is one of the most important things we do in mathematics, - yet we are surprisingly limited in what we can solve analytically. - For instance, equations as simple as x^5+x+1=0 or - \cos(x) =x cannot be solved by algebraic methods in terms of familiar functions. - Fortunately, - there are methods that can give us approximate - solutions to equations like these. - These methods can usually give an approximation correct to as many decimal places as we like. - In - we learned about the Bisection Method. - This section focuses on another technique - (which generally works faster), - called Newton's Method. -

    - -

    - Newton's Method is built around tangent lines. - The main idea is that if x is sufficiently close to a root of f(x), - then the tangent line to the graph at - (x,f(x)) will cross the x-axis at a point closer to the root than x. -

    - -

    - We start Newton's Method with an initial guess about roughly where the root is. - Call this x_0. - (See .) - Draw the tangent line to the graph at - (x_0,f(x_0)) and see where it meets the x-axis. - Call this point x_1. - Then repeat the process draw the tangent line to the graph at - (x_1, f(x_1)) and see where it meets the x-axis. - (See .) - Call this point x_2. - Repeat the process again to get x_3, x_4, etc. - This sequence of points will often converge rather quickly to a root of f. -

    - -
    - Demonstrating the geometric concept behind Newton's Method - -
    - - - - A tangent line is drawn to the graph at (x_0,f(x_0)) and and meets on the x-axis at point x_1. - - -

    - This graph depicts the iterations of Newton's method for finding the root of a function. - The y axis represents function values (1,0.5,0,-0.5,-1), - while the x axis represents iterations (x_0, x_1). - The graph shows the initial guess x_0 and the subsequent approximation x_1. -

    -

    - A tangent line is drawn from the point (x_0,f(x_0)) to intersect the x axis at x_1, - providing a refined approximation of the function’s root. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xtick=\empty,% - extra x ticks={.45,2.52}, - extra x tick labels={$x_0$,$x_1$}, - ymin=-1.1,ymax=1.1, - xmin=-.3,xmax=2.7 - ] - \addplot+ [infinite,domain=-.3:2.7] {cos(deg(x))}; - \addplot [tangentlineseg,domain=.45:2.52] {-0.434*(x-.45)+.9}; - \addplot [guideline] coordinates {(0.45,0) (0.45,0.9)}; - \addplot [soliddot] coordinates {(0.45,0.9)}; - \end{axis} - \end{tikzpicture} - - -
    - -
    - - - A tangent line is drawn to the graph at (x_1,f(x_1)) and and meets on the x-axis at point x_2. - -

    - The second graph has values (1, 0.5, 0, -0.5, -1) on the y axis,while the x axis represents iterations (x_0, x_2, x_1). - A tangent line is drawn from the point (x_1) to intersect the x axis and the new point is called x_2. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xtick=\empty, - extra x ticks={.5,2.52,1.12}, - extra x tick labels={$x_0$,$x_1$,$x_2$}, - ymin=-1.1,ymax=1.1,% - xmin=-.3,xmax=2.7% - ] - \addplot+ [infinite,domain=-.3:2.7] {cos(deg(x))}; - \addplot [tangentlineseg,domain=1.12:2.52] {-0.58*(x-2.52)-.813}; - \addplot [guideline] coordinates {(2.52,0) (2.52,-0.813)}; - \addplot [soliddot] coordinates {(2.52,-.813)}; - \end{axis} - \end{tikzpicture} - - -
    - -
    - - - A tangent line is drawn to the graph at (x_2,f(x_2)) and and meets on the x-axis at point x_3. - -

    - On the third graph, the y axis has values (1,0.5, 0, -0.5, -1), - while the x axis represents iterations (x_0,x_1,x_2,x_3). - A tangent line is drawn from the point (x_2) to intersect the x-axis and the new point is called x_3. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xtick=\empty,% - extra x ticks={.5,2.52,1.12,1.6}, - extra x tick labels={$x_0$,$x_1$,$x_2$,$x_3$}, - ymin=-1.1,ymax=1.1,% - xmin=-.3,xmax=2.7% - ] - \addplot+ [infinite,domain=-.3:2.7] {cos(deg(x))}; - \addplot [tangentlineseg,domain=1.12:1.6] {-.9*(x-1.12)+.432}; - \addplot [guideline] coordinates {(1.12,0) (1.12,0.432)}; - \addplot [soliddot] coordinates {(1.12,.432)}; - \end{axis} - \end{tikzpicture} - - -
    -     -
    - - - -

    - We can use this geometric - process to create an algebraic process. - Let's look at how we found x_1. - We started with the tangent line to the graph at (x_0,f(x_0)). - The slope of this tangent line is - \fp(x_0) and the equation of the line is - - y=\fp(x_0)(x-x_0)+f(x_0) - . -

    - -

    - This line crosses the x-axis when y=0, - and the x-value where it crosses is what we called x_1. - So let y=0 and replace x with x_1, - giving the equation: - - 0 = \fp(x_0)(x_1-x_0)+f(x_0) - . -

    - -

    - Now solve for x_1: - - x_1=x_0-\frac{f(x_0)}{\fp(x_0)} - . -

    - -

    - Since we repeat the same geometric process to find x_2 from x_1, we have - - x_2=x_1-\frac{f(x_1)}{\fp(x_1)} - . -

    - -

    - In general, given an approximation x_n, - we can find the next approximation, - x_{n+1} as follows: - - x_{n+1} = x_{n} - \frac{f(x_{n})}{\fp(x_{n})} - . -

    - -

    - We summarize this process as follows. -

    - - - Newton's Method -

    - Let f be a differentiable function on an interval I with a root in I. - To approximate the value of the root, - accurate to d decimal places: - Newton's Method -

    - -

    -

      -
    1. -

      - Choose a value x_0 as an initial approximation of the root. - (This is often done by looking at a graph of f.) -

      -
      2. - -
    2. -

      - Create successive approximations iteratively; - given an approximation x_n, - compute the next approximation x_{n+1} as - - x_{n+1} = x_n - \frac{f(x_n)}{\fp(x_n)} - . -

      -
      3. - -
    3. -

      - Stop the iterations when successive approximations do not differ in the first d places after the decimal point. -

      -
      4. -
    -

    -     - - - -

    - Let's practice Newton's Method with a concrete example. -

    - - - Using Newton's Method - -

    - Approximate the real root of x^3-x^2-1=0, - accurate to the first three places after the decimal, - using Newton's Method and an initial approximation of x_0=1. -

    -     - -

    - To begin, we compute \fp(x)=3x^2-2x. - Then we apply the Newton's Method algorithm, - outlined in . -

      -
    • -

      - \begin{aligned} - x_1\amp=1-\frac{f(1)}{\fp(1)}\\ - \amp=1-\frac{1^3-1^2-1}{3\cdot 1^2-2\cdot 1}\\ - \amp =2\end{aligned} - -

      -
      • -
    • -

      - \begin{aligned} - x_2\amp =2-\frac{f(2)}{\fp(2)}\\ - \amp =2-\frac{2^3-2^2-1}{3\cdot 2^2-2\cdot 2}\\ - \amp=1.625\end{aligned} - -

      -
      • -
    • -

      - \begin{aligned} - x_3\amp =1.625-\frac{f(1.625)}{\fp(1.625)}\\ - \amp = 1.625-\frac{1.625^3-1.625^2-1}{3\cdot 1.625^2-2\cdot 1.625}\\ - \amp \approx 1.48579\end{aligned} - -

      -
      • -
    • -

      - \begin{aligned} - x_4 \amp =1.48579-\frac{f(1.48579)}{\fp(1.48579)}\\ - \amp \approx 1.46596\end{aligned} - -

      -
      • -
    • -

      - \begin{aligned} - x_5 \amp = 1.46596 - \frac{f(1.46596)}{\fp(1.46596)}\\ - \amp \approx 1.46557\end{aligned} - -

      -
      • -
    -

    - -

    - We performed five iterations of Newton's Method to find a root accurate to the first three places after the decimal; - our final approximation is 1.465. - The exact value of the root, to six decimal places, - is 1.465571; It turns out that our x_5 is accurate to more than just three decimal places. -

    - -

    - A graph of f(x) is given in . - We can see from the graph that our initial approximation of x_0=1 was not particularly accurate; - a closer guess would have been x_0=1.5. - Our choice was based on ease of initial calculation, - and shows that Newton's Method can be robust enough that we do not have to make a very accurate initial approximation. -

    - -
    - A graph of f(x) = x^3-x^2-1 in - - - A cubic graph features a curve with two turning points. - - -

    - The y axis represents function values (0.5, 0, -0.5, -1,-1.5), while the x axis represents values of (0,0.5,1,1.5). - The graph intersects the y axis at -1 and 1.5 on x axis and forms a curve. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.5,ymax=.6,% - xmin=-0.1,xmax=1.6% - ] - \addplot+ [infinite,domain=-.05:1.58] {x^3-x^2-1}; - \end{axis} - \end{tikzpicture} - - -
    -     -     - -

    - We can automate this process on a calculator that has an ANS key that returns the result of the previous calculation. - Start by pressing 1 and then Enter. (We have just entered our initial guess, - x_0=1.) Now compute - - \texttt{ ANS } - \frac{f(\texttt{ ANS } )}{\fp(\texttt{ ANS } )} - - by entering the following and repeatedly press the Enter key. - - ANS-(ANS^3-ANS^2-1)/(3*ANS^2-2*ANS) - -

    - -

    - Each time we press the Enter key, - we are finding the successive approximations, x_1, - x_2, - , and each one is getting closer to the root. - In fact, once we get past around x_7 or so, - the approximations don't appear to be changing. - They actually are changing, - but the change is far enough to the right of the decimal point that it doesn't show up on the calculator's display. - When this happens, - we can be pretty confident that we have found an accurate approximation. -

    - -

    - Using a calculator in this manner makes the calculations simple; - many iterations can be computed very quickly. -

    - - - Using Newton's Method to find where functions intersect - -

    - Use Newton's Method to approximate a solution to \cos(x) = x, - accurate to five places after the decimal. -

    -     - -

    - Newton's Method provides a method of solving f(x) = 0; - it is not (directly) a method for solving equations like f(x) = g(x). - However, this is not a problem; - we can rewrite the latter equation as - f(x) - g(x)=0 and then use Newton's Method. -

    - -

    - So we rewrite \cos(x) =x as \cos(x)-x=0. - Written this way, we are finding a root of f(x)=\cos(x) -x. - We compute \fp(x)=-\sin(x) - 1. - Next we need a starting value, x_0. - Consider , - where f(x) = \cos(x) -x is graphed. - It seems that x_0=0.75 is pretty close to the root, - so we will use that as our x_0. (The figure also shows the graphs of y=\cos(x) and y=x. - Note how they intersect at the same x value as when f(x) = 0.) -

    - -
    - A graph of f(x)=\cos(x) -x used to find an initial approximation of its root - - - A graph of f(x) = cos x - x used to find an initial approximation of its root. - - -

    - The y axis and x axis both have values of (-1,0.5,0,0.5,1). - A broken line with equation y=x crosses the origin. - A second broken line depicts a downward parabola of equation y=\cos(x), the parabola intersects the first line y=x. - A solid line from the point where y=1 on the y axis to approximately x=0.75 on the x axis demonstrates an approach to estimating the root of the equation f(x) = \cos(x) - x. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.1,ymax=1.1,% - xmin=-1.1,xmax=1.1% - ] - \addplot+ [infinite,domain=-.1:1] {cos(deg(x))-x}; - \addplot+ [infinite,domain=-1:1] {cos(deg(x))}; - \addplot+ [infinite,domain=-1:1] {x}; - \addplot [guideline] coordinates {(0.739,0) (0.739,0.739)}; - \end{axis} - \end{tikzpicture} - - -
    - -

    - We now compute x_1, x_2, etc. - The formula for x_1 is - - x_1 \amp= 0.75 - \frac{\cos(0.75)-0.75}{-\sin(0.75)-1} - \amp \approx 0.7391111388 - . -

    - -

    - Apply Newton's Method again to find x_2: - - x_2 \amp= 0.7391111388 - \frac{\cos(0.7391111388)-0.7391111388}{-\sin(0.7391111388)-1} - \amp \approx 0.7390851334 - . -

    - -

    - We can continue this way, but it is really best to automate this process. - On a calculator with an ANS key, - we would start by entering 0.75, - then Enter, - inputting our initial approximation. - We then enter: - - ANS - (cos(ANS)-ANS)/(-sin(ANS)-1) - -

    - -

    - Repeatedly pressing the Enter key gives successive approximations. - We quickly find: - - x_3 \amp = 0.7390851332 - x_4 \amp = 0.7390851332 - . -

    - -

    - Our approximations x_2 and x_3 did not differ for at least the first five places after the decimal, - so we could have stopped. - However, using our calculator in the manner described is easy, - so finding x_4 was not hard. - It is interesting to see how we found an approximation, - accurate to as many decimal places as our calculator displays, - in just four iterations. -

    -     -     - - - -

    - If you know how to program, - you can translate the following pseudocode into your favorite language to perform the computation in this problem. -

    - - 
    -  x = 0.75
-  while true
-      oldx = x
-      x = x - (cos(x)-x)/(-sin(x)-1)
-      print x
-      if abs(x-oldx) < 0.0000000001
-          break
-
    - -

    - This code calculates x_1, x_2, - etc., storing each result in the variable x. - The previous approximation is stored in the variable oldx. - We continue looping until the difference between two successive approximations, abs(x-oldx), - is less than some small tolerance, - in this case, 0.0000000001. -

    - - - Convergence of Newton's Method -

    - What should one use for the initial guess, x_0? - Generally, the closer to the actual root the initial guess is, - the better. - However, some initial guesses should be avoided. - For instance, consider - where we sought the root to f(x) = x^3-x^2-1. - Choosing x_0=0 would have been a particularly poor choice. - Consider , - where f(x) is graphed along with its tangent line at x=0. - Since \fp(0)=0, - the tangent line is horizontal and does not intersect the x-axis. - Graphically, we see that Newton's Method fails. -

    - -
    - A graph of f(x) = x^3-x^2-1, showing why an initial approximation of x_0=0 with Newton's Method fails - - - A cubic graph features an "S" shaped curve with two turning points. - - -

    - The y axis represents function values (0.5, 0, -0.5, -1,-1.5), while the x axis represents values of (-0.5,0,0.5,1,1.5). - The graph intersects the y axis at -1 and 1.5 on the x axis, and forms an S shaped curve. -

    -

    - A horizontal tangent line is drawn at -1 on the y axis and 0 on the x axis -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.5,ymax=.6, - xmin=-0.6,xmax=1.6 - ] - - \addplot+ [infinite,domain=-.5:1.55] {-1}; - \addplot [tangentline,domain=-.5:1.55] {x^3-x^2-1}; - - \end{axis} - \end{tikzpicture} - - - -
    - -

    - We can also see analytically that it fails. - Since - - x_1 = 0 -\frac{f(0)}{\fp(0)} - - and \fp(0)=0, we see that x_1 is not well defined. -

    - -

    - This problem can also occur if, for instance, - it turns out that \fp(x_5)=0. - Adjusting the initial approximation x_0 by a very small amount will likely fix the problem. -

    - -

    - It is also possible for Newton's Method to not converge while each successive approximation is well defined. - Consider f(x) = x^{1/3}, - as shown in . - It is clear that the root is x=0, - but let's approximate this with x_0=0.1. - - shows graphically the calculation of x_1; - notice how it is farther from the root than x_0. - - and - show the calculation of x_2 and x_3, - which are even farther away; - our successive approximations are getting worse. - (It turns out that in this particular example, - each successive approximation is twice as far from the true answer as the previous approximation.) -

    - -
    - Newton's Method fails to find a root of f(x) = x^{1/3}, regardless of the choice of x_0. - -
    - - - - The graph illustrates the cube root function with initial guess x_0 as 0.1 - - -

    - The y axis represents function values (-1,0,1), while the x axis represents values of (-1,0,1). - The graph illustrates the cube root function, f(x) = x^{1/3}. It exhibits a curve that passes through the origin (0,0) - and extends into the positive and negative regions of the x axis. -

    -

    - It starts with an initial guess x_0=0.1. A tangent line is drawn at this point and intersects the x axis at the next - approximation, bringing it further from the root x=0. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - xtick={-1,1}, - extra x ticks={.1,-.2}, - extra x tick labels={$x_0$,$x_1$}, - ymin=-1.6,ymax=1.6,% - xmin=-1.6,xmax=1.6% - ] - - \addplot+ [infinite,domain=-1.16:1.16] ({x^3},{x}); - \addplot [tangentlineseg,domain = -.2:.1] {1.547*(x-.1)+.464}; - \addplot [guideline] coordinates {(0.1,0) (0.1,0.464)}; - \addplot [soliddot] coordinates {(.1,.464)}; - - \end{axis} - \end{tikzpicture} - - - - -
    -
    - - - - The graph illustrates the cube root function with point x_1 - - -

    - The graph illustrates the cube root function, f(x) = x^{1/3}. It exhibits a curve that passes through the origin (0,0) - and extends into the positive and negative regions of the x axis. -

    -

    - Here x_1 is twice as much as the initial guess bringing it more further from the root. - A tangent line is drawn at this point and intersects the x axis at the next - approximation, x_2. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - xtick={-1,1}, - extra x ticks={.1,-.2,.4}, - extra x tick labels={$x_0$, $x_1$,$x_2$}, - ymin=-1.6,ymax=1.6,% - xmin=-1.6,xmax=1.6% - ] - - \addplot+ [infinite,domain=-1.16:1.16]({x^3},{x}); - \addplot [tangentlineseg,domain = -.2:.4] {.975*(x+.2)-.585}; - \addplot [guideline] coordinates {(-0.2,0) (-0.2,-0.585)}; - \addplot [soliddot] coordinates {(-.2,-.585)}; - - \end{axis} - \end{tikzpicture} - - - -
    - -
    - - - - The graph illustrates the cube root function with point x_2 - - -

    - The graph illustrates the cube root function, f(x) = x^{1/3}. It exhibits a curve that passes through the origin (0,0) - and extends into the positive and negative regions of the x axis. -

    -

    - Here x_2 is twice as much as the x_1bringing it even more further from the root. - A tangent line is drawn at this point and intersects the x axis at the next approximation,x_3. -

    - -

    - Each successive approximation is twice as far from the true answer as the previous approximation. - This shows that Newton's Method doesn't work every time. -

    - -     - - - \begin{tikzpicture} - \begin{axis}[ - xtick={-1,1}, - extra x ticks={.1,-.2,.4,-.8}, - extra x tick labels={$x_0$,$x_1$,$x_2$,$x_3$}, - ymin=-1.6,ymax=1.6,% - xmin=-1.6,xmax=1.6% - ] - - \addplot+ [infinite,domain=-1.16:1.16] ({x^3},{x}); - \addplot [tangentlineseg,domain = -.8:.4] {.614*(x-.4)+.737}; - \addplot [guideline] coordinates {(0.4,0) (0.4,0.737)}; - \addplot [soliddot] coordinates {(.4,.737)}; - - \end{axis} - \end{tikzpicture} - - - -
    -     -
    - -

    - There is no fix to this problem; Newton's Method simply will not work and another method must be used. (In this case the particular reason Newton's Method fails is that the tangent line is vertical at the root). -

    - - - -

    - While Newton's Method does not always work, - it does work most of the time, - and it is generally very fast. - Once the approximations get close to the root, Newton's Method can as much as double the number of correct decimal places with each successive approximation. - A course in Numerical Analysis will introduce the reader to more iterative root finding methods, - as well as give greater detail about the strengths and weaknesses of Newton's Method. -

    -     - - - - Terms and Concepts - - - - -

    - - Given a function f(x), Newton's Method produces an exact solution to f(x) = 0. -

    -     - -     - - - - -

    - - In order to get a solution to f(x)=0 accurate to d places after the decimal, - at least d+1 iterations of Newton's Method must be used. -

    -     - -     -     - - - - Problems - - - -

    - The roots of the function f(x) are known or are easily found. - Use five iterations of Newton's Method with the given initial approximation - to approximate the root. - Compare it to the known value of the root. -

    -     - - - - - $f=Compute('cos(x)'); - Context("LimitedNumeric"); - $x[0]=1.5; - $df=$f->D('x'); - for my $i (1..5) { - $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); - }; - $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; - - -

    - f(x) = , x_0= -

    - - Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. - -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -     -     -     - - - - - $f=Compute('sin(x)'); - Context("LimitedNumeric"); - $x[0]=1; - $df=$f->D('x'); - for my $i (1..5) { - $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); - }; - $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; - - -

    - f(x) = , x_0= -

    - - Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. - -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -     -     -     - - - - - $f=Compute('x^2+x-2'); - Context("LimitedNumeric"); - $x[0]=0; - $df=$f->D('x'); - for my $i (1..5) { - $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); - }; - $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; - - -

    - f(x) = , x_0= -

    - - Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. - -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -     -     -     - - - - - $f=Compute('x^2-2'); - Context("LimitedNumeric"); - $x[0]=1.5; - $df=$f->D('x'); - for my $i (1..5) { - $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); - }; - $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; - - -

    - f(x) = , x_0= -

    - - Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. - -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -     -     -     - - - - - $f=Compute('ln(x)'); - Context("LimitedNumeric"); - $x[0]=2; - $df=$f->D('x'); - for my $i (1..5) { - $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); - }; - $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; - - -

    - f(x) = , x_0= -

    - - Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. - -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -     -     -     - - - - - $f=Compute('x^3-x^2+x-1'); - Context("LimitedNumeric"); - $x[0]=2; - $df=$f->D('x'); - for my $i (1..5) { - $x[$i] = Compute($x[$i-1] - ($f->eval(x=>$x[$i-1]))/($df->eval(x=>$x[$i-1]))); - }; - $showwork = q![@ explanation_box(message => "Compare ".math_ev3('x_5')." to the known value of the root.") @]*!; - - -

    - f(x) = , x_0= -

    - - Enter x_1, x_2, x_3, x_4, x_5, into the given answer blanks. - -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -

    - -

    -     -     -     -     - - - - -

    - Use Newton's Method to approximate all roots of the given function accurate to three places after the decimal. - If an interval is given, find only the roots that lie within that interval. - Use technology to obtain good initial approximations. -

    -     - - - - - $f = Compute('x^3+5x^2-x-1'); - Context("FiniteSolutionSets"); - Context()->flags->remove("NumberCheck"); - Context()->flags->set(preferSetNotation=>0); - Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); - $L = Formula("{-5.15632517465866,-0.369102386184855,0.525427560843517}"); - $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; - - -

    - f(x)= -

    - - Use commas to separate roots. - -

    - -

    -

    - -

    -     -     -     - - - - - $f = Compute('x^4+2x^3-7x^2-x+5'); - Context("FiniteSolutionSets"); - Context()->flags->remove("NumberCheck"); - Context()->flags->set(preferSetNotation=>0); - Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); - $L = Formula("{-3.71447874438776,-0.856722678160357,1,1.57120142254812}"); - $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; - - -

    - f(x)= -

    - - Use commas to separate roots. - -

    - -

    -

    - -

    -     -     -     - - - - - $f = Compute('x^17-2x^13-10x^8+10'); - Context("FiniteSolutionSets"); - Context()->flags->remove("NumberCheck"); - Context()->flags->set(preferSetNotation=>0); - Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); - $L = Formula("{-1.0134032173648748818979878,0.98831169481331833285038046,1.3934095863189322104023657}"); - $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; - - -

    - f(x)= on (-2,2) -

    - - Use commas to separate roots. - -

    - -

    -

    - -

    -     -     -     - - - - - $f = Compute('x^2cos(x) + (x-1)sin(x)'); - Context("FiniteSolutionSets"); - Context()->flags->remove("NumberCheck"); - Context()->flags->set(preferSetNotation=>0); - Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); - $L = Formula("{-2.16477423608072,0,0.524501487448536,1.81327843509828}"); - $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; - - -

    - f(x)= on (-3,3) -

    - - Use commas to separate roots. - -

    - -

    -

    - -

    -     -     -     -     - - - - -

    - Use Newton's Method to approximate when the given functions are equal, - accurate to 3 places after the decimal. - Use technology to obtain good initial approximations. -

    -     - - - - - - $f = Compute('x^2'); - $g = Compute('cos(x)'); - Context("FiniteSolutionSets"); - Context()->flags->remove("NumberCheck"); - Context()->flags->set(preferSetNotation=>0); - Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); - $L = Formula("{-0.82413231230252242296,0.82413231230252242296}"); - $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; - - -

    - f(x)=, g(x)= -

    - - Use commas to separate solutions. - -

    - -

    -

    - -

    -     -     -     - - - - - - $f = Compute('x^2-1'); - $g = Compute('sin(x)'); - Context("FiniteSolutionSets"); - Context()->flags->remove("NumberCheck"); - Context()->flags->set(preferSetNotation=>0); - Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); - $L = Formula("{-0.63673265080528201089,1.4096240040025962492}"); - $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; - - -

    - f(x)=, g(x)= -

    - - Use commas to separate solutions. - -

    - -

    -

    - -

    -     -     -     - - - - - - $f = Compute('e^(x^2)'); - $g = Compute('cos(x)'); - Context("FiniteSolutionSets"); - Context()->flags->remove("NumberCheck"); - Context()->flags->set(preferSetNotation=>0); - Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); - $L = Formula("{0}"); - $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; - - -

    - f(x)=, g(x)= -

    - - Use commas to separate solutions. - -

    - -

    -

    - -

    -     -     -     - - - - - - $f = Compute('x'); - $g = Compute('tan(x)'); - Context("FiniteSolutionSets"); - Context()->flags->remove("NumberCheck"); - Context()->flags->set(preferSetNotation=>0); - Context()->flags->set(tolType=>'absolute',tolerance=>0.0005); - $L = Formula("{-4.49340945790906,0,4.49340945790906}"); - $showwork = q![@ explanation_box(message => "Show the steps you took using Newton's Method.") @]*!; - - -

    - f(x)=, g(x)= on [-6,6] -

    - - Use commas to separate solutions. - -

    - -

    -

    - -

    -     -     -     -     - - - - -

    - Why does Newton's Method fail in finding a root of f(x) = x^3-3x^2+x+3 when x_0=1? -

    - -     - - - -

    - The approximations alternate between x=1 and x=2. -

    -     - -     - - - - -

    - Why does Newton's Method fail in finding a root of f(x) = -17x^4+130x^3-301x^2+156x+156 when x_0=1? -

    - -     - - - -

    - The approximations alternate between x=1, - x=2 and x=3. -

    -     - -     -     -     -
    -
    - Related Rates - - - -

    - When two quantities are related by an equation, - knowing the value of one quantity can determine the value of the other. - For instance, - the circumference and radius of a circle are related by C=2\pi r; - knowing that C is - - 6\pi - - determines the radius must be - - 3 - . -

    - -

    - But what if both variables are changing with time? - If we know how two variables are related and we know how one of them changes with time, - can we find how the other variable changes with time? -

    - -

    - The topic of related rates - allows us to answer this question: - knowing the rate at which one quantity is changing can determine the - rate at which another changes. - - related rates - -

    - - - -

    - We demonstrate the concepts of related rates through examples. -

    - - - Understanding related rates - -

    - The radius of a circle is growing at a rate of - - 5 - . - At what rate is the circumference growing? -

    -     - -

    - The circumference and radius of a circle are related by C = 2\pi r. - We are given information about how the length of r changes with respect to time; - that is, we are told \lz{r}{t} is - - 5 - . - We want to know how the length of C changes with respect to time, - , we want to know \lz{C}{t}. -

    - -

    - Implicitly differentiate both sides of - C = 2\pi r with respect to t: - - C \amp = 2\pi r - \lzoo{t}{C} \amp = \lzoo{t}{2\pi r} - \lz{C}{t} \amp =2\pi \lz{r}{t} - . -

    - -

    - As we know \lz{r}{t} is - - 5 - , we know - - \lz{C}{t} = 2\pi 5 = 10\pi \approx 31.4\text{ in/hr } - . -

    - - - - - -     - -     - -

    - In related rates problems, - we will be presented with an application problem that involves two or more variables and one or more rate. - It is the job of the reader to construct the appropriate model that can be used to answer the posed question. - - outlines the basic steps for solving a related rates problem. -

    - - - Related Rates - Related Rates - -

    -

      -
    1. -

      - Understand the problem. - Clearly identify the quantity whose rate of change you need to determine. - Make sketch, if helpful. -

      -
      2. - -
    2. -

      - Identify other quantities relevant to the context of the problem - and create an equation that relates them to the quantity identified in Step. - If values for certain quantities are already known, do not substitute these values into your equation yet. - These instantaneous values will be used in Step. -

      -
      3. - -
    3. -

      - Implicitly differentiate both sides of the equation found in Step - with respect to t. -

      -
      4. - -
    4. -

      - Substitute in the known values of rates and known instantaneous values of the variables. -

      -
      5. - -
    5. -

      - Solve for the unknown rate identified in Step. -

      -
      6. - -
    6. -

      - Write a full sentence conclusion. -

      -
      7. -
    -

    -     - -

    - Consider another, similar example. -

    - - - Finding related rates - -

    - Water streams out of a faucet at a rate of - - 2 - - onto a flat surface at a constant rate, - forming a circular puddle that is - - 1/8 - - deep. -

    - -

    -

      -
    1. - -

      - At what rate is the area of the puddle growing? -

      -
      2. - -
    2. - -

      - At what rate is the radius of the circle growing? -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - We can answer this question two ways: - using common sense or related rates. - The common sense method states that the volume of the puddle is - growing by - - 2 - , where - - \text{volume of puddle}=\text{area of circle}\times\text{depth} - . - Since the depth is constant at - - 1/8 - , the area must be growing by - - 16 - - since 16\cdot \frac{1}{8}=2. - This approach reveals the underlying related rates principle. -

      - -

      - Now let's solve the problem using - . - Based on the problem description, - the quantities that change with time are the volume of water - (the volume of the puddle), - the area of the circular puddle and the radius of the circle. - We don't need a diagram for this problem. - The important variables for this part of the problem are the - volume and area. -

      - -

      - Let V and A represent the Volume and Area of the puddle. - We know V= A\times \frac{1}{8}. - Take the derivative of both sides with respect to t, - employing implicit differentiation. - - V \amp = \frac{1}{8}A - \lzoo{t}{V}\amp = \lzoo{t}{\frac{1}{8}A} - \lz{V}{t} \amp = \frac{1}{8}\lz{A}{t} - - We know the change in volume, \lz{V}{t} = 2, - so we substitute this value into our related rates equation: - 2 = \frac{1}{8}\lz{A}{t}, - and hence \lz{A}{t} = 16. - Thus the area is growing by - - 16 - . -

      -
      2. - -
    2. -

      - We already identified the quantities that are changing in - Part. - The variables of interest in this problem are the radius and the volume. - We need an equation that relates the volume of the circle to the radius. - Since the puddle is a right circular cylinder, - we will use a known volume formula, - V=\pi r^2 h where V is the volume of the puddle (in - - - , r is the radius - (in inches) - and h is the height ( depth) of the puddle in inches. - (Notice that this formula is equivalent to - V=\text{area} \times \text{depth}.) - We know that the height (depth) is a constant 1/8 inch. - Since this quantity does not change in the problem, - we can safely substitute this value now. -

      - -

      - Implicitly derive both sides of - V=\pi r^2 \frac{1}{8} with respect to t: - - V \amp = \frac{1}{8} \pi r^2 - \frac{d}{dt}\big(V\big) \amp = \frac{d}{dt}\left(\frac{1}{8}\pi r^2\right) - \lz{V}{t} \amp = \frac{1}{8} 2\pi r\lz{r}{t} - \lz{V}{t} \amp = \frac{1}{4} \pi r\lz{r}{t} - - We know that \lz{V}{t} is - - 2 - . - So we have: - - 2 = \frac14 \pi r\lz{r}{t} - - Solving for \lz{r}{t}, we have - - \lz{r}{t} = \frac{8}{\pi r} - . - Note how our answer is not a number, - but rather a function of r. - In other words, - the rate at which the radius is growing depends on how big - the circle already is. - If the circle is very large, adding - - 2 - - of water will not make the circle much bigger at all. - If the circle is dime-sized, - adding the same amount of water will make a radical change in the - radius of the circle. -

      - -

      - In some ways, our problem was (intentionally) ill-posed. - We need to specify a current (instantaneous) value of the radius - in order to know a rate of change. - When the puddle has a radius of - - 10 - , the radius is growing at a rate of - - \lz{r}{t} = \frac{8}{10\pi} = \frac{4}{5\pi} \approx 0.25\text{ in/s } - . -

      -
      3. -
    -

    -     - -     - - - - - - - Studying related rates - -

    - Radar guns measure the rate of distance change between the gun and the - object it is measuring. - For instance, a reading of - 55 - means the object is moving away from the gun at a rate - of 55 miles per hour, - whereas a measurement of - -25 - would mean that the object is approaching the gun at a - rate of 25 miles per hour. -

    - -

    - If the radar gun is moving (say, - attached to a police car) then radar readouts are only immediately - understandable if the gun and the object are moving along the same line. - If a police officer is traveling - - 60 - - and gets a readout of - - 15 - , he knows that the car ahead of him is moving away at a rate - of 15 miles an hour, - meaning the car is traveling - - 75 - . (This straight-line principle is one reason officers park - on the side of the highway and try to shoot straight back down the road. - It gives the most accurate reading.) -

    - -

    - Suppose an officer is driving due north at - - 30 - - and sees a car moving due east, - as shown in . - Using his radar gun, he measures a reading of - - 20 - . - By using landmarks, - he believes both he and the other car are about 1/2 mile from the - intersection of their two roads. -

    - -
    - A sketch of a police car (at bottom) attempting to measure the - speed of a car (at right) in - - - A passenger car is shown moving east, while a police car is moving north. Their roads are set to intersect. - - -

    - The passenger car is on an eastward road, while the police car is on a northward road, and these roads are perpendicular to each other. The police car is 1/2 units south of the intersection point, and the passenger car is 1/2 units east of the same intersection. -

    -

    - A straight line, labeled as C, connects the current positions of both the police car and the passenger car. -

    -     - - - - \begin{tikzpicture}[>=latex,scale=1.32] - \draw(0,0) -- (3,0) node [pos=.5,above] { $B=1/2$} -- (0,-3) node [pos=.4,below] { $C$}-- (0,0) node [pos=.5,rotate=90,shift={(.1,.2)}] { $A=1/2$}; - - \draw [thick] (0,-3.5) -- (0,.5) node [above] { $N$}; - \draw [thick] (-.5,0) -- (3.5,0) node [right] { $E$}; - - \filldraw [fill=firstcolor,draw=secondcolor] (0,-3) circle (2pt) node [right] { Officer}; - \filldraw [fill=black,draw=black] (3,0) circle (2pt) node [below ] { \quad Car}; - - \draw[->,thick] (-.2,-3.3) -- (-.2,-2.7); - \draw[->,thick] (2.7,.2) -- (3.3,.2); - - \end{tikzpicture} - - - -
    - -

    - If the speed limit on the other road is - - 55 - , is the other driver speeding? -

    -     - -

    - The important quantities that are changing are: - the distance of the officer to the intersection, - the distance of the car to the intersection, - and the distance of the officer to the car. - (There are other quantities that are changing as well such as the - angles and area of the triangle, - but these are not important to this problem.) -

    - -

    - Using the diagram in , - let's label what we know about the situation. - As both the police officer and other driver are 1/2 mile from - the intersection, - we have A = 1/2, B = 1/2, - and through the Pythagorean Theorem, - C = 1/\sqrt{2}\approx 0.707. - These values are instantaneous - values for our variables, - so we won't use them until the end of the problem. - Instead, we will use the variables A, B, - and C. -

    - -

    - We need an equation that relates A, B, and C. - The Pythagorean Theorem is a good choice: - A^2+B^2 = C^2. - Differentiate both sides with respect to t: - - A^2 + B^2 \amp = C^2 - \lzoo{t}{A^2+B^2} \amp = \lzoo{t}{C^2} - 2A\lz{A}{t} + 2B\lz{B}{t} \amp = 2C\lz{C}{t} - -

    - -

    - We know the police officer is traveling at - - 30 - ; that is, \lz{A}{t} = -30. - The reason this rate of change is negative is that A is getting - smaller; - the distance between the officer and the intersection is shrinking. - The radar measurement is \lz{C}{t} = 20. - We want to find \lz{B}{t}. -

    - - - -

    - We have values for everything except \lz{B}{t}. - Solving for this we have: - - \lz{B}{t} = \frac{C\lz{C}{t}- A\lz{A}{t}}{B} - . - Now we substitue in our known rates and instantaneous values of our - variables: - - \lz{B}{t} \amp \approx \frac{0.707(20)- 0.5(-30)}{(0.5)} - \amp= 58.28 \text{ mph } - . -

    - -

    - The other driver appears to be speeding slightly. -

    -     - -     - - - Studying related rates - -

    - A camera is placed on a tripod - - 10 - - from the side of a road. - The camera is to turn to track a car that is to drive by at - - 100 - - for a promotional video. - The video's planners want to know what kind of motor the tripod should - be equipped with in order to properly track the car as it passes by. - shows the proposed setup. -

    - -
    - Tracking a speeding car (at left) with a rotating camera - - - A tripod with a rotating camera is shown 10ft from the side of the road, tracking a speeding car. - - -

    - A car is shown to the left of the image, driving at a 100 mph east, - with a tripod placed 10 from the side of the road, set to track the car. -

    -

    - A right triangle is drawn, with the hypotenuse connecting the car to the tripod, and the other two sides parallel to the road and perpendicular to the road. - The side parallel to the road has length x, and the angle between the hypotenuse and the side perpendicular to the road is labeled as \theta. The side perpendicular to the road has a length of 10. -

    -     - - - \begin{tikzpicture}[>=latex,scale=1.32] - - \draw (-3,0) -- (0,0) -- (0,-2) node [shift={(-5pt,10pt)}] {\small $\theta$} node [right,pos=.5] { 10ft} -- cycle; - \draw [<->, thick] (-3.5,0) -- node [below,pos=.5] { $x$} (.5,0); - \draw [fill=firstcolor,draw=secondcolor] (-3,0) circle (2pt); - \draw [fill=black] (0,-2) circle (2pt); - \draw [->] (-3,.2) -- (-2,.2) node [above,pos=.5] { 100mph}; - - \end{tikzpicture} - - - -
    - -

    - How fast must the camera be able to turn to track the car? -

    -     - -

    - The quantities that changing are x and \theta as drawn on - . (The hypotenuse of the triangle is - also changing, but this isn't important to the problem). - We seek information about how fast the camera is to turn; - therefore, we need an equation that will relate an angle \theta - to the position of the camera and the speed and position of the car. -

    - -

    - - suggests we use a trigonometric equation. - Letting x represent the distance the car is from the point on the - road directly in front of the camera, we have - - \tan(\theta) = \frac{x}{10} - . -

    - -

    - Now take the derivative of both sides of Equation - using implicit differentiation: - - \tan(\theta) \amp = \frac{x}{10} - \lzoo{t}{\tan(\theta)} \amp = \lzoo{t}{\frac{x}{10}} - \sec^2(\theta)\lz{\theta}{t} \amp = \frac{1}{10}\lz{x}{t} - - Now we solve for \lz{\theta}{t}: - - \lz{\theta}{t} = \frac{\cos^2(\theta)}{10}\lz{x}{t} - -

    - -

    - As the car is moving at - - 100 - , we have that \lz{x}{t} is - - -100 - - (as in the last example, - since x is getting smaller as the car travels, - \lz{x}{t} is negative). - We need to convert the measurements so they use the same units - (we chose - - - ); - rewrite - - -100 - - in terms of - - - : - - \lz{x}{t} \amp = -100\frac{\text{mi} }{\text{ hr } } - \amp = -100\frac{\text{mi} }{\text{ hr } }\cdot5280\frac{\text{ ft } }{\text{mi} }\cdot\frac{1}{3600}\frac{\text{ hr } }{\text{s} } - \amp =-146.\overline{6}\text{ ft/s } - . -

    - -

    - We want to know the fastest the camera has to turn. - Common sense tells us this is when the car is directly in front of the - camera (, when \theta = 0). - Our mathematics bears this out. - In Equation - we see this is when \cos^2(\theta) is largest; - this is when \cos(\theta) = 1, - or when \theta = 0. - We also know that we should get an answer that is in - - - . - Since \cos(\theta) is a dimensionless measure, - it won't contribute to the units. - However, radians are also dimensionless. - This means we can write - (or erase) - the word radian without any unit consequences. (The same is not - true of degrees always convert degress to radians). -

    - -

    - With \lz{x}{t} approximately - - -146.7 - , we have - - \lz{\theta}{t} \amp \approx -\frac{1}{10\text{ ft } }146.67\text{ ft/s } - \amp = -14.667\text{ radians/s } - -

    - -

    - We find that \lz{\theta}{t} is negative; - this matches our diagram in - for \theta is getting smaller as the car approaches the camera. -

    - -

    - What is the practical meaning of - - -14.667 - ? - Recall that 1 circular revolution goes through 2\pi - radians, thus - - 14.667 - - means 14.667/(2\pi)\approx 2.33 revolutions per second. - The negative sign indicates the camera is rotating in a clockwise fashion. -

    -     - -     - - - - - -

    - We introduced the derivative as a function that gives the slopes of tangent - lines of functions. - This chapter emphasizes using the derivative in other ways. - Newton's Method uses the derivative to approximate roots of functions; - this section stresses the rate of change - aspect of the derivative to find a relationship between the rates of change - of two related quantities. -

    - -

    - In the next section we use Extreme Value concepts to - optimize quantities. -

    - - - - Terms and Concepts - - - - -

    - - Implicit differentiation is often used when solving - related rates type problems. -

    -     - -     - - - - -

    - - A study of related rates is part of the standard police officer training. -

    -     - -     -     - - - Problems - - - - - ($rate,$depth) = random_subset(2,4..10); - $a = random(1,4,1); - if($envir{problemSeed}==1){$rate=5; $depth=10; $a=1;}; - Context()->flags->set(reduceConstants=>0); - $rateU = NumberWithUnits("$rate cm^3/s"); - $depthU = NumberWithUnits("$depth mm"); - @r=($a,$a*10,$a*100); - @rU = map{NumberWithUnits("$_ cm")}(@r); - Context("Fraction"); - @f = map{Fraction(5*$rate,$_*$depth)}(@r); - @num = map{($_->value)[0]}(@f); - @den = map{($_->value)[1]}(@f); - @sU = map{NumberWithUnits("$num[$_]/($den[$_] pi) cm/s")}(0,1,2); - - -

    - Water flows onto a flat surface at a rate of - - forming a circular puddle deep. - How fast is the radius growing when the radius is: -

    -     - - - -

    - -

    -

    - -

    -     -     - - - -

    - -

    -

    - -

    -     -     - - - -

    - -

    -

    - -

    -     -     -     -     - - - - - $rate = random(4,12,1); - $a = random(1,4,1); - if($envir{problemSeed}==1){$rate=10; $a=1;}; - Context()->flags->set(reduceConstants=>0); - $rateU = NumberWithUnits("$rate cm^3/s"); - @r=($a,$a*10,$a*100); - @rU = map{NumberWithUnits("$_ cm")}(@r); - Context("Fraction"); - @f = map{Fraction($rate,4*$_**2)}(@r); - @num = map{($_->value)[0]}(@f); - @den = map{($_->value)[1]}(@f); - @sU = map{NumberWithUnits("$num[$_]/($den[$_] pi) cm/s")}(0,1,2); - - -

    - A spherical balloon is inflated with air flowing at a rate of - . - How fast is the radius of the balloon increasing when the radius is: -

    -     - - - -

    - -

    -

    - -

    -     -     - - - -

    - -

    -

    - -

    -     -     - - - -

    - -

    -

    - -

    -     -     -     -     - - - - - Context("Fraction"); - $f = list_random(Fraction(1,2),Fraction(3,4),Fraction(2,3),Fraction(3,5),Fraction(4,5)); - $speed = random(45,55,5); - $rate = random(70,90,5); - if($envir{problemSeed}==1){$f=Fraction(1,2);$speed=50;$rate=80;}; - Context("Numeric"); - Context()->flags->set(reduceConstants=>0); - $a = Compute("$rate*sqrt(2)-$speed"); - $a = NumberWithUnits("$a mi/h"); - - -

    - Consider the traffic situation introduced in . - How fast is the other car traveling if the officer and the other car are each - mile from the intersection, the other car is traveling due west, - the officer is traveling north at \,\text{mph}, - and the radar reading is -\,\text{mph}? -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - $f = list_random(Fraction(1,2),Fraction(3,4),Fraction(2,3),Fraction(3,5),Fraction(4,5)); - $speed = random(45,55,5); - $rate = random(70,90,5); - if($envir{problemSeed}==1){$f=Fraction(1,2);$speed=50;$rate=80;}; - Context("Numeric"); - Context()->flags->set(reduceConstants=>0); - $f2 = 1+$f**2; - $f3 = $speed*$f; - $a = Compute("$rate*sqrt($f2)-$f3"); - $a = NumberWithUnits("$a mi/h"); - $f4 = $rate/$f; - $f5 = $speed/$f; - $b = Compute("$f4 sqrt($f2)-$f5"); - $b = NumberWithUnits("$b mi/h"); - - -

    - Consider the traffic situation introduced in . - Calculate how fast the other car is traveling in each of the following situations. -

    -     - - - -

    - The officer is traveling due north at \,\text{mph} - and is mile from the intersection, - while the other car is 1 mile from the intersection - traveling west and the radar reading is -\,\text{mph}? -

    -

    - -

    -     -     - - - -

    - The officer is traveling due north at \,\text{mph} - and is 1 mile from the intersection, - while the other car is mile from the - intersection traveling west and the radar reading is -\,\text{mph}? -

    -

    - -

    -     -     -     -     - - - - - Context()->flags->set(reduceConstants=>0); - $spd = random(300,800,10); - $elev = random(5000,15000,100); - $a = Compute("($elev*$spd*5280)/(5280^2+$elev^2)"); - $aU = NumberWithUnits("$a rad/hr"); - $b = Compute("($elev*$spd*5280)/(1056^2+$elev^2)"); - $bU = NumberWithUnits("$b rad/hr"); - $c = Compute("$spd*5280/$elev"); - $cU = NumberWithUnits("$c rad/hr"); - - -

    - An F-22 aircraft is flying at \,\text{mph} - with an elevation of \,\text{ft} on a - straight-line path that will take it directly over an - anti-aircraft gun. -

    - - - An airplane is shown on the right, over horizontal ground with a gun on the ground to the left. - - -

    - To the left of a horizontal surface, a square marks an anti-aircraft gun. - Above the ground, on the right of the image, is an airplane flying to the left. - The horizontal distance along the ground from the gun to the airplane position is marked with the variable x. - The elevation of the plane is shown as 6600. -

    -

    - There is also a dashed line from the gun to the plane. - The angle that this line makes with the ground is marked with the variable \theta. -

    -     - - \begin{tikzpicture}[>=latex,scale=1.3] - \begin{scope} - \clip (0,0) rectangle (120pt,50pt); - \draw [inner color=firstcolor,draw = white] (-10,-25pt) rectangle (180pt,70pt); - \end{scope} - \draw [top color=treestump, bottom color=white,draw=white] (0,0) rectangle (120pt,-15pt); - \draw (0,0) -- (120pt,0); - \draw [dashed] (12.5pt,0pt) node [xshift=22pt,yshift=4pt] { \(\theta\)} -- (105pt,35pt); - \draw [ultra thick] (12.5pt,0) -- (18pt,4pt); - \filldraw [fill=white] (10pt,-2.5pt) rectangle (15pt,2.5pt); - \draw [<->] (15pt,-10pt) -- (110pt,-10pt) node [below,pos=.5] { \(x\)}; - \draw [<->] (120pt,0pt) -- (120pt,35pt) node [right, pos=.5] { $elev ft}; - \begin{scope}[shift={(105pt,35pt)}] - \draw [fill=black] (0,0) rectangle (10pt,2pt); - \draw [very thick] (3pt,1pt) -- (6pt,-1.5pt) -- (5pt,0pt); - \draw [very thick] (9pt,2pt) -- (9pt,4pt) -- (7pt,2pt); - \end{scope} - \end{tikzpicture} - - -

    - How fast (in radians per second) must the gun be able to turn to accurately track the - aircraft when the plane is: -

    -     - - - -

    - 1 mile away? -

    -

    - -

    - - Use rad/s for radians per second. - -     -     - - - -

    - 1/5 mile away? -

    -

    - -

    - - Use rad/s for radians per second. - -     -     - - - -

    - Directly overhead? -

    -

    - -

    - - Use rad/s for radians per second. - -     -     - -

    - There are 5280 feet in one mile. -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0); - $speed = random(400,600,50); - $speedU = NumberWithUnits("$speed mi/h"); - @a = (random(1000,2000,100), random(100,500,50), 0); - if($envir{problemSeed}==1){$a[0]=1000,$a[1]=100;}; - @aU = map{NumberWithUnits("$_ ft")}(@a); - @b = map{100*$speed*5280/3600/(100**2 + $_**2)}(@a); - @bU = map{NumberWithUnits("$_ rad/s")}(@b); - - -

    - An F-22 aircraft is flying at - - with an elevation of 100\,\text{ft} - on a straight-line path that will take it directly over an - anti-aircraft gun as in - (note the lower elevation here). -

    -

    - How fast (in radians per second) must the gun be able to turn to accurately track the - aircraft when the plane is: -

    -     - - - -

    - away? -

    - - Use rad/s for radians per second. - -

    - -

    -     -     - - - -

    - away? -

    - - Use rad/s for radians per second. - -

    - -

    -     -     - - - -

    - Directly overhead? -

    - - Use rad/s for radians per second. - -

    - -

    -     -     -     -     - - - - - - $f = Compute("x/sqrt(24^2-x^2)"); - $a = Compute($f->eval(x=>1)); - $a = NumberWithUnits("$a ft/s"); - $b = Compute($f->eval(x=>10)); - $b = NumberWithUnits("$b ft/s"); - $c = Compute($f->eval(x=>23)); - $c = NumberWithUnits("$c ft/s"); - $d = Compute("inf"); - - -

    - A 24 ladder - is leaning against a house while the base is pulled away at a - constant rate of - - 1 - . -

    - - - A ladder is shown leaning against a house, with its base being pulled away. - - -

    - A 24 ladder is positioned against the side of a house. The base of the ladder is being pulled away from the house at a rate of 1 ft/s. The exact distance of the ladder's base from the house varies based on different scenarios provided. -

    -     - - \begin{tikzpicture}[>=latex,xscale=4,yscale=3] - \draw [top color=treestump, bottom color=treestump!50!white,draw=black] (0,0) rectangle (20pt,30pt); - \draw [fill=black] (0,30pt) -- (10pt,38pt) -- (20pt,30pt) -- cycle; - \draw [fill=white] (7pt,15pt) rectangle (13pt,25pt); - \draw (10pt,15pt) -- (10pt,25pt) (7pt,20pt) -- (13pt,20pt); - \draw [thick] (20pt,25pt) -- node [pos=.5,rotate=-67,yshift=5pt] { 24 ft} (30pt,0); - \draw [thick] (0,0) -- (35pt,0); - \draw [->,>=latex] (34pt,2pt) -- node [above right,pos=.5] { 1 ft/s} (50pt,2pt); - \end{tikzpicture} - - -

    - At what rate is the top of the ladder sliding down the side of - the house when the base is: -

    -     - - - -

    - 1 foot from the house? -

    -

    - -

    - - Use ft/s for feet per second. - -     -     - - - -

    - 10 feet from the house? -

    -

    - -

    - - Use ft/s for feet per second. - -     -     - - - -

    - 23 feet from the house? -

    - -

    - -

    - - Use ft/s for feet per second. - -     -     - - - -

    - 24 feet from the house? -

    -

    - -

    - - Use ft/s for feet per second. - -     -     -     -     - - - - - - $f = Compute("30*sqrt(10^2+x^2)/x"); - $a = Compute($f->eval(x=>50)); - $a = NumberWithUnits("$a ft/min"); - $b = Compute($f->eval(x=>15)); - $b = NumberWithUnits("$b ft/min"); - $c = Compute($f->eval(x=>1)); - $c = NumberWithUnits("$c ft/min"); - - -

    - A boat is being pulled into a dock at a constant rate of - - 30 - - by a winch located 10 - above the deck of the boat. -

    - - - A boat is being pulled towards a dock by an overhead winch. - - -

    - A boat is approaching a dock, being pulled by a winch located 10 above the deck of the boat. The rope from the winch to the boat forms a hypotenuse, creating a right-angle triangle with the water's surface and the vertical winch post. The boat is being pulled at a constant rate of 30 ft/min towards the dock. -

    -     - - \begin{tikzpicture}[>=latex,scale=1.8] - \draw [top color=firstcolor!60!white, bottom color=firstcolor!10!white,draw=firstcolor] (-5pt,5pt) rectangle (100pt,-5pt); - \draw [thick,fill=treestump] (-10pt,20pt) rectangle (8pt,22pt); - \draw [top color=treestump!30!black, bottom color=treestump!70!white,draw=black] (0,0) rectangle (5pt,30pt); - \draw [fill=white] (50pt,0pt) -- (90pt,0pt) -- (90pt, 15pt) -- (45pt,15pt) -- cycle; - \draw [fill=black] (65pt,15pt) rectangle (67pt,20pt); - \draw [fill=yellowcolfill] (55pt,20pt) -- (80pt,20pt) -- (65pt,40pt) -- cycle; - \draw [fill=black] (8pt,35pt) circle (2pt); - \draw (45pt,15pt) [dashed] -- (8pt,15pt) -- (8pt,35pt) node [pos=.4,xshift=12pt] { 10 ft}; - \draw [-,thick] (8pt,35pt) -- (45pt,15pt); - \end{tikzpicture} - - -

    - At what rate is the boat approaching the dock when the boat is: -

    -     - - - -

    - 50 feet out? -

    -

    - -

    - - Use ft/min for feet per minute. - -     -     - - - -

    - 15 feet out? -

    -

    - -

    - - Use ft/min for feet per minute. - -     -     - - - -

    - 1 foot from the dock? -

    -

    - -

    - - Use ft/min for feet per minute. - -     -     - - - -

    - What happens when the length of rope pulling in the boat is less than 10 feet long? -

    - -

    - -

    -     -     -     -     - - - - - contextUnits.pl - - - # make $depth and $diameter two distinct numbers - ($depth, $diameter) = random_subset(2, 10 .. 30); - # the rate water is entering the tank - $rate = random(8, 12); - # the depths that water will be at when we ask how fast the water is rising - # the first depth will just be 1, and the other two should be no more than $depth - @a = (1, num_sort(random_subset(2, 2 .. $depth-1))); - Context("Units")->withUnitsFor('length', 'time'); - $depthU = Compute("$depth ft"); - $diameterU = Compute("$diameter ft"); - $rateU = Compute("$rate ft^3/s"); - # push the depths through the function that produces the rates - @b = map{$rate * 4 / pi * $depth**2 / $diameter**2 / $_**2} (@a); - @bU = map{Compute("$_ ft/s")} (@b); - $T = pi/12 * $diameter**2 * $depth / $rate; - $TU = Compute("$T s"); - - -

    - An inverted cylindrical cone, - deep and - across - at the top, is being filled with water at a rate of - . - At what rate is the water rising in the tank when the depth of - the water is: -

    -     - - - -

    - foot? -

    -

    - -

    -     -     - - - -

    - feet? -

    -

    - -

    -     -     - - - -

    - feet? -

    -

    - -

    -     -     - - - -

    - How long will the tank take to fill when starting at empty? -

    -

    - -

    -     -     -     -     - - - - - - $f = Compute("2x/sqrt(x^2+30^2)"); - $a = Compute($f->eval(x=>10)); - $a = NumberWithUnits("$a ft/s"); - $b = Compute($f->eval(x=>40)); - $b = NumberWithUnits("$b ft/s"); - $c = Compute("sqrt(60^2-30^2)"); - $c = NumberWithUnits("$c ft"); - - -

    - A rope, attached to a weight, - goes up through a pulley at the ceiling and back down to a worker. - The man holds the rope at the same height as the connection - point between rope and weight. -

    - - - A worker pulls a rope attached to a weight through a ceiling pulley. - - -

    - The diagram illustrates a weight connected to a 60 long rope, with the other end looped through a pulley affixed to the ceiling, 30 above the weight's resting position. Adjacent to the weight, a worker is depicted holding the rope, specifically at its attachment point to the weight. As the worker moves away at 2 ft/s, pulling the rope with him, the weight begins its ascent. -

    -     - - \begin{tikzpicture}[>=latex,x=1pt,y=1pt,scale=2.4] - \draw [thick](0,40) -- (55,40); - \draw [thick] (0,0) -- (55,0); - \draw (2,0) -- (12,0) -- (9,5) -- (5,5)--cycle; - \draw (7,5) -- (7,35) -- (33,5); - \draw (7,35) -- (7,40); - \draw [fill=gray] (7,35) circle (2pt); - \draw (35,10) circle (2pt); - \draw (35,8) -- (35,4) -- (33,0) (35,4) -- (37,0) (33,6) -- (37,6); - \draw (-4,5) -- (-1,5) (-2.5,5) -- (-2.5,35) node [pos=.5,draw,fill=white,draw=white,rotate=90] { 30 ft} - (-4,35) -- (-1,35); - \draw [thick,->](40,5) --(55,5) node[pos=.5,above] { 2 ft/s}; - \end{tikzpicture} - - -

    - Suppose the man stands directly next to the weight - (, a total rope length of 60 feet) and begins to - walk away at a rate of - - 2 - . - How fast is the weight rising when the man has walked: -

    -     - - - -

    - 10 feet? -

    -

    - -

    - - Use ft/s for feet per second. - -     -     - - - -

    - 40 feet? -

    -

    - -

    - - Use ft/s for feet per second. - -     -     - - - -

    - How far must the man walk to raise the weight all the way to - the pulley? -

    - -

    - -

    - - Use ft for feet. - -     -     -     -     - - - - - # Height is fixed at 30 ft - # Need an initial distance that makes a Pythagorean triple with 30 - $b = random(40,72,224); - $speed = random(2,4,1); - if($envir{problemSeed}==1){$b=40;$speed=2;}; - $bU = NumberWithUnits("$b ft"); - $speedU = NumberWithUnits("$speed ft/s"); - $c = sqrt(30**2 + $b**2); - $length = 30 + $c; - $lengthU = NumberWithUnits("$length ft"); - $newb = $b+10; - $newc = sqrt(30**2 + $newb**2); - $newrate = $newb/$newc * $speed; - $newrateU = NumberWithUnits("$newrate ft/s"); - $newnewb = $b+30; - $newnewc = sqrt(30**2 + $newnewb**2); - $newnewrate = $newnewb/$newnewc * $speed; - $newnewrateU = NumberWithUnits("$newnewrate ft/s"); - $far = sqrt($length**2 - 30**2); - $farU = NumberWithUnits("$far ft"); - - -

    - Consider the situation described in - . - Suppose the man starts from the weight and begins - to walk away at a rate of . -

    -     - - - -

    - How long is the rope? -

    -

    - -

    -     -     - - - -

    - How fast is the weight rising after the man has walked - 10 feet? -

    -

    - -

    -     -     - - - -

    - How fast is the weight rising after the man has walked - 30 feet? -

    -

    - -

    -     -     - - - -

    - How far must the man walk to raise the weight all the way - to the pulley? -

    -

    - -

    -     -     -     -     - - - - - $distance = random(80,120,5); - $height = random(5,6,1); - $rate = random(3,8,1); - if($envir{problemSeed}==1){$distance=100;$height=5;}; - $heightU = NumberWithUnits("$height ft"); - $elev = $distance + $height; - $elevU = NumberWithUnits("$elev ft"); - $speed = 2*$distance*$rate*pi/180; - $speedU = NumberWithUnits("$speed ft/min"); - - -

    - A hot air balloon lifts off from ground rising vertically. - From feet away, a - tall woman - tracks the path of the balloon. - When her sightline with the balloon makes a - 45^\circ angle with the horizontal, she notes the angle - is increasing at about ^\circ per minute. -

    -     - - - -

    - What is the elevation of the balloon? -

    -

    - -

    -     -     - - - -

    - How fast is it rising? -

    -

    - -

    -     -     -     -     - - - - - $rate = random(3,8,1); - $nd = list_random([2,3],[4,5],[4,7],[5,6],[5,7],[5,8]); - $height = random(20,40,5); - if($envir{problemSeed}==1){$rate=5;$nd=[2,3];$height=30;}; - $rateU = NumberWithUnits("$rate ft^3/s"); - Context("Fraction"); - $frac = Fraction($nd->[0],$nd->[1]); - Context("Numeric"); - $speed = 4*$frac**2/pi/$height**2*$rate; - $speedU = NumberWithUnits("$speed ft/s"); - - -

    - A company that produces landscaping materials is dumping sand - into a conical pile. - The sand is being poured at a rate of - . - The physical properties of the sand, - in conjunction with gravity, - ensure that the cone's height is roughly the length - of the diameter of the circular base. -

    -

    - How fast is the cone rising when it has a height of feet? -

    -

    - -

    -     -     -     -     -     -
    -
    - Optimization - - - - -

    - In - we learned about extreme values the largest and smallest values a - function attains on an interval. - We motivated our interest in such values by discussing how it made sense to - want to know the highest/lowest values of a stock, - or the fastest/slowest an object was moving. - In this section we apply the concepts of extreme values to solve - word problems, - , problems stated in terms of situations that require us to create the - appropriate mathematical framework in which to solve the problem. - optimization -

    - -

    - We start with a classic example which is followed by a discussion of the - topic of optimization. -

    - - - Optimization: perimeter and area - -

    - A man has 100 feet of fencing, - a large yard, and a small dog. - He wants to create a rectangular enclosure for his dog with the fencing - that provides the maximal area. - What dimensions provide the maximal area? -

    -     - -

    - One can likely guess the correct answer that is great. - We will proceed to show how calculus can provide this answer in a context - that proves this answer is correct. -

    - -

    - It helps to make a sketch of the situation. - Our enclosure is sketched twice in , - either with treetop grass and nice fence boards or as a simple rectangle. - Either way, drawing a rectangle forces us to realize that we need to - know the dimensions of this rectangle so we can create an area function - after all, - we are trying to maximize the area. -

    - -
    - A sketch of the enclosure in . - - - - - A rectangular enclosure with green grass and a fence with x and y dimensions. - - -

    - A 3-D rectangular enclosure is drawn with green grass and nice fence boards on the x and x axis. -

    -     - - - - \begin{tikzpicture}[>=latex,scale=1.2] - - \shadedraw [top color = treetop,bottom color = treetop!50!black] (0,0) -- (33.6pt,35.28pt) -- (95.2pt,35.28pt) -- (61.6pt,0) -- cycle; - - \begin{scope}[cm={1,1.05,0,1,(0,0)}] - \foreach \x in {0,...,7} - { - \shadedraw [xscale=.3,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++ (0,20pt) -- ++ (4pt,4pt) -- ++ (6pt,0pt) -- ++ (4pt,-4pt) -- ++ (0pt,-20pt) -- cycle; - } - \end{scope} - - \begin{scope}[shift={(33.6pt,35.28pt)}] - \foreach \x in {0,...,10} - { - \shadedraw [xscale=.4,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++ (0,20pt) -- ++ (4pt,4pt) -- ++ (6pt,0pt) -- ++ (4pt,-4pt) -- ++ (0pt,-20pt) -- cycle; - } - \end{scope} - - \foreach \x in {0,...,10} - { - \shadedraw [xscale=.4,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++ (0,20pt) -- ++ (4pt,4pt) -- ++ (6pt,0pt) -- ++ (4pt,-4pt) -- ++ (0pt,-20pt) -- cycle; - } - - \begin{scope}[cm={1,1.05,0,1,(61.6pt,0)}] - \foreach \x in {0,...,7} - { - \shadedraw [xscale=.3,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++ (0,20pt) -- ++ (4pt,4pt) -- ++ (6pt,0pt) -- ++ (4pt,-4pt) -- ++ (0pt,-20pt) -- cycle; - } - \end{scope} - - \draw [<->] (0,-5pt) -- (61.6pt,-5pt) node [below,pos=.5] {$x$}; - \draw [<->] (100.2pt,35.28pt) -- (66.6pt,0pt) node [right,pos=.5] {$y$}; - - \end{tikzpicture} - - - - - - - A rectangle shape is drawn with sides labeled x and y. - - -

    - The 2-D plane rectangle has sides labeled x and y. The dimensions are both unknown. -

    -     - - - \begin{tikzpicture}[>=latex,scale=0.66] - \draw (0,0) -- (8,0) node [pos=.5,below] {$x$} -- (8,5) node [pos=.5,right] {$y$} -- (0,5) -- cycle; - \end{tikzpicture} - - - - -     -
    -

    - We let x and y denote the lengths of the sides of the - rectangle. Clearly, - - \text{ Area } =xy - . -

    - -

    - We do not yet know how to handle functions with two variables; - we need to reduce this down to a single variable. - We know more about the situation: - the man has 100 feet of fencing. - By knowing the perimeter of the rectangle must be 100, - we can create another equation: - - \text{ Perimeter } = 100 = 2x+2y - . -

    - -

    - We now have two equations and two unknowns. - In the latter equation, we solve for y: - - y = 50-x - . -

    - -

    - Now substitute this expression for y in the area equation: - - \text{ Area } = A(x) = x(50-x) - . -

    - -

    - Note we now have an equation of one variable; - we can truly call the Area a function of x. -

    - -

    - This function only makes sense when 0\leq x \leq 50, - otherwise we get negative values of area. - So we find the extreme values of A(x) on the interval - [0,50] using . -

    - -

    - To find the critical points, - we take the derivative of A(x) and set it equal to 0, - then solve for x. - - A(x) \amp = x(50-x) - \amp = 50x-x^2 - A'(x) \amp = 50-2x - -

    - -

    - We solve 50-2x=0 to find x=25; - this is the only critical point. - We evaluate A(x) at the endpoints of our interval and at this - critical point to find the extreme values; - in this case, all we care about is the maximum. -

    - -

    - Clearly A(0)=0 and A(50)=0, - whereas A(25) = 625 - - . - This is the maximum. - Since we earlier found y = 50-x, - we find that y is also 25. - Thus the dimensions of the rectangular enclosure with perimeter of - - 100 - . with maximum area is a square, with sides of length - - 25 - . -

    -     -     - -

    - This example is very simplistic and a bit contrived. - (After all, most people create a design then buy fencing to meet their needs, - and not buy fencing and plan later.) - But it models well the necessary process: - create equations that describe a situation, - reduce an equation to a single variable, - then find the needed extreme value. -

    - -

    - In real life, problems are much more complex. - The equations are often not - reducible to a single variable - (hence multi-variable calculus is needed) - and the equations themselves may be difficult to form. - Understanding the principles here will provide a good foundation for the - mathematics you will likely encounter later. -

    - -

    - We outline here the basic process of solving these optimization problems. -

    - - - Solving Optimization Problems -

    -

      -
    1. -

      - Understand the problem. - Clearly identify what quantity is to be maximized or minimized. - Make a sketch if helpful. -

      -
      2. - -
    2. -

      - Create equations relevant to the context of the problem, - using the information given. One of these should describe the - quantity to be optimized. - We'll call this the fundamental equation. -

      -
      3. - -
    3. -

      - If the fundamental equation defines the quantity to be optimized as - a function of more than one variable, - reduce it to a single variable function using substitutions derived - from the other equations. We call these other equations constraint equations. -

      -
      4. - -
    4. -

      - Identify the domain of this function, - keeping in mind the context of the problem. -

      -
      5. - -
    5. -

      - Find the extreme values of this function on the determined domain. -

      -
      6. - -
    6. -

      - Identify the values of all relevant quantities of the problem and write a full sentence conclusion. -

      -
      7. -
    -

    -     - -

    - We will use in a variety of - examples. -

    - - - - - Optimization: perimeter and area - -

    - Here is another classic calculus problem: A woman has a 100 feet of - fencing, a small dog, and a large yard that contains a stream - (that is mostly straight). - She wants to create a rectangular enclosure with maximal area that uses - the stream as one side. - (Apparently her dog won't swim away.) - What dimensions provide the maximal area? -

    -     - -

    - We will follow the steps outlined by - . -

    - -

    -

      -
    1. -

      - We are maximizing area. - A sketch of the region will help; - - gives two sketches of the proposed enclosed area. - A key feature of the sketches is to acknowledge that one side is - not fenced. -

      - -
      - A sketch of the enclosure in - - - - A rectangular fenced yard with one side not fenced. - - -

      - A rectangular fenced yard with one side open for a stream. - The dimensions are labeled x and y and are not provided. -

      -       - - - \begin{tikzpicture}[>=latex,scale=1.2] - - \shadedraw [top color = treetop,bottom color = treetop!50!black] (0,0) -- (33.6pt,35.28pt) -- (95.2pt,35.28pt) -- (61.6pt,0) -- cycle; - - \begin{scope}[cm={1,1.05,0,1,(0,0)}] - \foreach \x in {0,...,7} - { - \shadedraw [xscale=.3,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++(0,20pt) -- ++(4pt,4pt) -- ++(6pt,0pt) -- ++(4pt,-4pt)-- ++(0pt,-20pt) -- cycle; - } - \end{scope} - - \begin{scope}[shift={(33.6pt,35.28pt)}] - \foreach \x in {0,...,10} - { - \shadedraw [xscale=.4,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++(0,20pt) -- ++(4pt,4pt) -- ++(6pt,0pt) -- ++(4pt,-4pt)-- ++(0pt,-20pt) -- cycle; - } - \end{scope} - - \begin{scope}[cm={1,1.05,0,1,(61.6pt,0)}] - \foreach \x in {0,...,7} - { - \shadedraw [xscale=.3,shift={(\x*14 pt,0)},left color=treestump!40!white,right color = treestump!99!white] (0,0) -- ++(0,20pt) -- ++(4pt,4pt) -- ++(6pt,0pt) -- ++(4pt,-4pt)-- ++(0pt,-20pt) -- cycle; - } - \end{scope} - - \shadedraw [top color=firstcolor, bottom color=firstcolor!50!white] (-10pt,0) sin (30pt,2pt) cos (70pt,0pt) -- (70pt,-10pt) sin (35pt,-8pt) cos (-10pt,-10pt) --cycle; - - \draw [<->] (0,-15pt) -- (61.6pt,-15pt) node [below,pos=.5] {$x$}; - \draw [<->] (100.2pt,35.28pt) -- (66.6pt,0pt) node [right,pos=.5] {$y$}; - - \end{tikzpicture} - - - - - - - A rectangle shape is drawn below also not closed on one side. - - -

      - A 2-D sketch is drawn below showing a rectangle shape with one side opened and the sides labeled x and y. - The dimensions are unknown. -

      -       - - - \begin{tikzpicture}[scale=0.66] - - \draw (0,0) -- (0,5) -- (8,5) node [pos=.5,below] {$x$} -- (8,0) node [pos=.5,right] {$y$}; - \draw [firstcolor,very thick] (0,0) sin (4,.2) cos (8,0); - - \end{tikzpicture} - - - - -       -
      - -
      2. - -
    2. -

      - We want to maximize the area; - as in the example before, - - \text{ Area } = xy - . - This is our fundamental equation. - This defines area as a function of two variables, - so we need another equation to reduce it to one variable. -

      - -

      - We again appeal to the perimeter; - here the perimeter is - - \text{ Perimeter } = 100 = x+2y - . - The perimeter is our constraint equation. - Note how this is a different equation for perimeter than in - , - since one of the sides does not need to be fenced. -

      -
      3. - -
    3. -

      - We now reduce the fundamental equation to a single variable using - our constraint equation. - In the perimeter equation, solve for y: y = 50 - x/2. - We can now write Area as - - \text{ Area } = A(x) \amp= x(50-x/2) - \amp = 50x - \frac12x^2 - . - Area is now defined as a function of one variable. -

      -
      4. - -
    4. -

      - We want the area to be non-negative. - Since A(x) = x(50-x/2), - we want x\geq 0 and 50-x/2\geq 0. - The latter inequality implies that x\leq100, - so 0\leq x\leq 100. -

      -
      5. - -
    5. -

      - We now find the extreme values. - At the endpoints, the minimum is found, - giving an area of 0. -

      - -

      - Find the critical points. - We have A'(x) = 50-x; - setting this equal to 0 and solving for x returns x=50. - This gives an area of - - A(50) = 50(25) = 1250 - . -

      -
      6. - -
    6. -

      - We earlier set y = 50-x/2; thus y = 25. - Thus our rectangle will have two sides of length 25 and one - side of length 50, - with a total area of - - 1250 - . -

      -
      7. -
    -

    -     - -     - -

    - Keep in mind as we do these problems that we are practicing a process; - that is, we are learning to turn a situation into a system of equations. - These equations allow us to write a certain quantity as a function of one - variable, which we then optimize. -

    - - - - - - Optimization: minimizing cost - -

    - A power line needs to be run from a power station located on the beach - to an offshore facility. - - shows the distances between the power station to the facility. -

    - -

    - It costs \$50/\text{ ft } to run a - power line along the land, and - \$130/\text{ ft } to run a power - line under water. - How much of the power line should be run along the land to minimize the - overall cost? What is the minimal cost? -

    - -
    - Running a power line from the power station to an offshore - facility with minimal cost in - - - a power line is ran from the power station to an offshore facility. - - -

    - A power line is run from a power station to an offshore facility 5000 underwater. - It is also run 1000 along the offshore facility forming a right angle triangle with the under water powerline. -

    -

    - The hypotenuse of the triangle shows the distance between the station and offshore facility. -

    -     - - - \begin{tikzpicture}[>=latex] - - \begin{scope} - \clip (0,0) rectangle (140pt,50pt); - \draw [inner color=firstcolor,draw = white] (-10,-25pt) rectangle (180pt,70pt); - - \end{scope} - \draw [top color=treestump, bottom color=white,draw=white] (0,0) rectangle (140pt,-15pt); - \draw (0,0) -- (150pt,0); - \filldraw [fill=white] (10pt,-5pt) rectangle (20pt,5pt); - %\filldraw [fill=white] (130pt,35pt) circle (7pt); - - \draw [<->] (15pt,-10pt) -- (130pt,-10pt) node [below,pos=.5] { 5000 ft}; - \draw [<->] (140pt,0pt) -- (140pt,35pt) node [right, pos=.5] { 1000 ft}; - - \draw [ultra thick] (20pt,0) -- (70pt,0); - \filldraw (70pt,0) circle (2pt); - \draw [dashed,thick] (70pt,0) -- (130pt,35pt); - \filldraw [fill=white] (130pt,35pt) circle (7pt); - - \end{tikzpicture} - - - -
    -     - -

    - We will follow the strategy of - implicitly, - without specifically numbering steps. -

    - -

    - There are two immediate solutions that we could consider, - each of which we will reject through - common sense. First, - we could minimize the distance by directly connecting the two - locations with a straight line. - However, this requires that all the wire be laid underwater, - the most costly option. - Second, we could minimize the underwater length by running a wire all - - 5000 - - along the beach, directly across from the offshore facility. - This has the undesired effect of having the longest distance of all, - probably ensuring a non-minimal cost. -

    - -

    - The optimal solution likely has the line being run along the ground for - a while, then underwater, as the figure implies. - We need to label our unknown distances the distance run along - the ground and the distance run underwater. - Recognizing that the underwater distance can be measured as the - hypotenuse of a right triangle, - we choose to label the distances as shown in - . -

    - -
    - Labeling unknown distances in - - - A power line is ran from the power station to an offshore facility with the distance run along - the ground and the distance run underwater unknown. - - -

    - A power line is run from a power station to an offshore facility 5000 underwater. Here, the distance run along - the ground and the distance run underwater is unknown and thereby labeled as x. - It is also run 1000 along the offshore facility forming a right angle triangle with the power line. -

    -

    - The horizontal line of the image is labelled 5000-x which is the distance of the power line laid along land,and - x from the middle to the pont of the offshore facility which is the distance the power line is not laid. - The hypotenuse of the right angle triangle measures underwater distance with the equation \sqrt{x^2-1000^2}. -

    -     - - - \begin{tikzpicture}[>=latex] - - \begin{scope} - \clip (0,0) rectangle (140pt,50pt); - \draw [inner color=firstcolor,draw = white] (-10,-25pt) rectangle (180pt,70pt); - - \end{scope} - \draw [top color=treestump, bottom color=white,draw=white] (0,0) rectangle (140pt,-15pt); - \draw (0,0) -- (150pt,0); - \filldraw [fill=white] (10pt,-5pt) rectangle (20pt,5pt); - %\filldraw [fill=white] (130pt,35pt) circle (7pt); - - \draw [<->] (15pt,-10pt) -- (70pt,-10pt) node [below,pos=.5] { $5000 -x$}; - \draw [<->] (70pt,-10pt) -- (130pt,-10pt) node [below,pos=.5] { $x$}; - \draw [<->] (140pt,0pt) -- (140pt,35pt) node [right, pos=.5] { 1000 ft}; - - \draw [ultra thick] (20pt,0) -- (70pt,0); - \filldraw (70pt,0) circle (2pt); - \draw [dashed,thick] (70pt,0) -- (130pt,35pt) ; - \draw (100pt,23pt) node [rotate=30] { $\sqrt{x^2+1000^2}$}; - \filldraw [fill=white] (130pt,35pt) circle (7pt); - - \end{tikzpicture} - - - -
    - -

    - By choosing x as we did - (instead of letting x be the distance along the land), - we make the expression under the square root simple. - We now create the cost function. - - \text{Cost} \amp ={}\amp\amp \text{land cost} \amp\amp +\text{water cost} - \amp\amp\amp \$50\times \text{land distance} \amp\amp +\$130 \times \text{water distance} - \amp\amp\amp 50(5000-x) \amp\amp +130\sqrt{x^2+1000^2} - . -

    - -

    - So we have c(x) = 50(5000-x)+ 130\sqrt{x^2+1000^2}. - This function only makes sense on the interval [0,5000]. - While we are fairly certain the endpoints will not give a minimal cost, - we still evaluate c(x) at each to verify. - - c(0) \amp = 380{,}000\amp c(5000) \amp \approx 662{,}873 - . - (Notice that if x=0, the line is run the full - - 5000 - - along land and a full - - 1000 - - under water. - If x=5000, the line is run the maximum distance underwater.) -

    - -

    - We now find the critical values of c(x). - We compute c'(x) as - - c'(x) = -50+\frac{130x}{\sqrt{x^2+1000^2}} - . -

    - -

    - Recognize that this is never undefined. - Setting c'(x)=0 and solving for x, we have: - - -50+\frac{130x}{\sqrt{x^2+1000^2}} \amp = 0 - \frac{130x}{\sqrt{x^2+1000^2}} \amp = 50 - \frac{130^2x^2}{x^2+1000^2} \amp = 50^2 - 130^2x^2 \amp = 50^2(x^2+1000^2) - 130^2x^2-50^2x^2 \amp = 50^2\cdot1000^2 - (130^2-50^2)x^2 \amp = 50,000^2 - x^2 \amp = \frac{50,000^2}{130^2-50^2} - x \amp = \frac{50,000}{\sqrt{130^2-50^2}} - x \amp = \frac{50,000}{120} =\frac{1250}3\approx 416.67 - . -

    - -

    - Evaluating c(x) at x=416.67 gives a minimal cost of about - \$370{,}000. - The distance the power line is laid along land is - 5000-416.67 = 4583.33 ft., and the underwater distance is - \sqrt{416.67^2+1000^2} \approx 1083 ft. -

    -     - -     - - - -

    - In the exercises you will see a variety of situations that require you to - combine problem-solving skills with calculus. - Focus on the process; - learn how to form equations from situations that can be manipulated into - what you need. - Eschew memorizing how to do this kind of problem - as opposed to that kind of problem. - Learning a process will benefit one far longer than memorizing a specific - technique. -

    - - - - - - -

    - - introduces our final application of the derivative: - differentials. - Given y=f(x), - they offer a method of approximating the change in y after x - changes by a small amount. -

    - - - - Terms and Concepts - - - - -

    - - An optimization problem - is essentially an extreme values - problem in a story problem setting. -

    -     - -     - - - - -

    - - This section teaches one to find the extreme values of a - function that has more than one variable. -

    -     - -     -     - - - Problems - - - - - $sum = random(50,200,2); - if($envir{problemSeed}==1){$sum=100;}; - $maxprod = ($sum/2)**2; - - -

    - Find the maximum product of two numbers - (not necessarily integers) - that have a sum of . -

    - - If there is no maximum product, enter DNE. - -

    - -

    -     -     -     - - - - - $prod = random(400,600,10); - if($envir{problemSeed}==1){$prod=500;}; - $minsum = Compute("2 sqrt($prod)"); - - -

    - Find the minimum sum of two positive numbers - whose product is . -

    - - If there is no minimum sum, enter DNE. - -

    - -

    -     -     -     - - - - - $prod = random(400,600,10); - if($envir{problemSeed}==1){$prod=500;}; - $maxsum = Compute("DNE"); - - -

    - Find the maximum sum of two positive numbers - whose product is . -

    - - If there is no maximum sum, enter DNE. - -

    - -

    -     - -

    - There is no maximum sum; - the fundamental equation has only 1 critical value that - corresponds to a minimum. -

    -     -     -     - - - - - $prod = random(400,600,10); - $upper = random(200,500,10); - if($envir{problemSeed}==1){$prod=500;$upper=300;}; - Context("Fraction-NoDecimals"); - $maxsum = Fraction("$upper + $prod/$upper"); - - -

    - Find the maximum sum of two numbers, - each of which is less than or equal to , - whose product is . -

    - - If there is no maximum sum, enter DNE. - -

    - -

    -     -     -     - - - - - $hyp = random(1,9,1); - if($envir{problemSeed}==1){$hyp=1;}; - $maxarea=Real($hyp**2/4); - - -

    - Find the maximal area of a right triangle with hypotenuse of - length . -

    -

    - -

    -     -     -     - - - - - contextUnits.pl - - - Context(context::Fraction::extending(Context('Units')->withUnitsFor('length'))); - - $total = random(800, 1500, 100); - if ($envir{problemSeed} == 1) { $total = 1000; } - - $length = Fraction($total, 6); - $width = Fraction($total, 8); - $l = FormulaWithUnits($length, 'ft'); - $w = FormulaWithUnits($width, 'ft'); - $multians = MultiAnswer($l, $w)->with( - singleResult => 1, - checker => sub { - my ($correct, $student, $self) = @_; - my ($stu1, $stu2) = @{$student}; - my ($cor1, $cor2) = @{$correct}; - if (($cor1 == $stu1 && $cor2 == $stu2) - || ($cor1 == $stu2 && $cor2 == $stu1)) - { - return 1; - } else { - return 0; - } - } - ); - - -

    - A rancher has feet of fencing in which to construct - adjacent, equally sized rectangular pens, as shown below. - What dimensions should these pens have to maximize the enclosed area? -

    - - - A rectangle divided into two parts. - - - A large rectangle divided into two equal rectangles. - - - \begin{tikzpicture} - \draw [thick] (0,0) rectangle (3,2); - \draw [thick] (1.5,0) -- (1.5,2); - \end{tikzpicture} - - - - Enter the two dimensions, one in each blank, for one of the two pens here. Use ft for feet. - -

    - -

    -

    - -

    -     -     -     - - - - - $V = NumberWithUnits("355 cm^3"); - parser::Root->Enable; - $r = Compute("root(3,355/(2pi))"); - $h = Compute("root(3,1420/pi)"); - $rU = NumberWithUnits("$r cm"); - $hU = NumberWithUnits("$h cm"); - $showwork = '[@ explanation_box(message => "Discuss whether or not your calculation suggests that a real world soda can is designed to minimize the materials cost.") @]*'; - - -

    - A standard soda can is roughly cylindrical and holds - of liquid. - What dimensions should the cylinder have to minimize the material - needed to produce the can? - Based on your dimensions, - determine whether or not the standard can is produced to - minimize the material costs. -

    - - Enter its radius here. - -

    - -

    - - Enter its height here. - -

    - -

    -

    - -

    -     -     -     - - - - - $V = NumberWithUnits("206 in^3"); - parser::Root->Enable; - $r = Compute("root(3,206/(2pi))"); - $h = Compute("root(3,824/pi)"); - $rU = NumberWithUnits("$r in"); - $hU = NumberWithUnits("$h in"); - $showwork = '[@ explanation_box(message => "Discuss whether or not your calculation suggests that a #10 can is designed to minimize the materials cost.") @]*'; - - -

    - Find the dimensions of a cylindrical can with a volume of - that minimizes the surface area. -

    - - Enter its radius here. - -

    - -

    - - Enter its height here. - -

    - -

    -

    - The #10 can is a standard sized can used by the restaurant - industry that holds about - with a diameter of - 6\,\frac{3}{16}\,\text{in} - and height of - 7\,\text{in}. - Does it seem these dimensions where chosen with minimization in mind? -

    -

    - -

    -     -     -     - - - - - $V = NumberWithUnits("355 cm^3"); - parser::Root->Enable; - $r = Compute("root(3,355/(4pi))"); - $h = Compute("root(3,5680/pi)"); - $rU = NumberWithUnits("$r cm"); - $hU = NumberWithUnits("$h cm"); - $showwork = '[@ explanation_box(message => "Discuss whether or not your calculation suggests that a real world soda can is designed to minimize the materials cost.") @]*'; - - -

    - A standard soda can is roughly cylindrical and holds - of liquid. - A real-world soda can has material on the top and bottom that is thicker - than the material around the side. - Assume that the top/bottom material is twice as thick as the material around the side. - What dimensions should the cylinder have to minimize the material - needed to produce the can? - Based on your dimensions and the assumption about material thickness, - determine whether or not the standard can is produced to - minimize the material costs. -

    - - Enter its radius here. - -

    - -

    - - Enter its height here. - -

    - -

    -

    - -

    -     -     -     - - - - - $V = NumberWithUnits("36*18**2 in^3"); - - -

    - The United States Postal Service charges more for boxes whose - combined length and girth exceeds 108 inches. - (The length of a package is the length of its longest side; - the girth is the perimeter of the cross section, i.e., 2w+2h). -

    -

    - What is the maximum volume of a package with a square cross - section (w=h) that does not exceed the - 108 inch standard? -

    -

    - -

    -     -     -     - - - - - $D = random(10,20,1); - if($envir{problemSeed}==1){$D=12;}; - $w = Compute("$D/sqrt(3)"); - $h = Compute("sqrt($D^2 - $w^2)"); - $wU = NumberWithUnits("$w in"); - $hU = NumberWithUnits("$h in"); - $multians = MultiAnswer($wU, $hU)->with( - singleResult => 1, - checker => sub { - my ( $correct, $student, $self ) = @_; - my ( $stu1, $stu2 ) = @{$student}; - my ( $cor1, $cor2 ) = @{$correct}; - if ( ($cor1 == $stu1 && $cor2 == $stu2) || - ($cor1 == $stu2 && $cor2 == $stu1) ) { - return 1; - } else { - return 0; - } - } - ); - - -

    - The strength S of a wooden beam is directly proportional - to its cross sectional width w and the square of its - height h. - that is, S = kwh^2 for some constant k. -

    - - - A circle with a rectangle divided diagonally inscribed inside it. - - -

    - A circular log with diameter of 12 has rectangle inscribed inside. The rectangle is divided diagonally forming - two equal right-angle triangles. The line that divides it is the diameter. -

    -

    - The shorter side of the rectangle is labeled w which serves as the cross sectional width and the - longer side is labelled h which is the height. -

    -     - - \begin{tikzpicture}[scale=1.8] - \draw [thick] (0,0) circle (1cm); - \draw [thick] (-.6,.8) -- node [pos=.5,right] { \(12\)} (.6,-.8); - \draw [thick] (-.6,-.8) rectangle (.6,.8); - \draw (.7,0) node { \(h\)} (0,.7) node { \(w\)}; - \end{tikzpicture} - - -

    - Given a circular log with diameter of inches, - what sized beam can be cut from the log with maximum strength? -

    - - Enter the cross-sectional dimensions here. - -

    - -

    -

    - -

    -     -     -     - - - - - do { - ($x,$y) = random_subset(2,2..6); - $ycost=random(30000,50000,5000); - $xcost=$ycost + random(30000,50000,5000); - } until ($y - $ycost*$x/sqrt($xcost**2 - $ycost**2) > 0); - if($envir{problemSeed}==1){$x=2;$y=5;$xcost=80000;$ycost=50000}; - Context("Currency")->flags->set(trimTrailingZeros=>1,tolerance=>1); - $xC = Currency($xcost); - $yC = Currency($ycost); - $g = Compute("50000"); - $s = Compute("80000"); - $a = Compute("5"); - $b = Compute("2"); - $z = max(0,$y - $ycost*$x/sqrt($xcost**2 - $ycost**2)); - $C = Currency($z*$ycost + $xcost*sqrt($x**2+($y - $z)**2)); - Context("Numeric"); - $zU = NumberWithUnits("$z mi"); - - -

    - A power line is to be run to an offshore facility in the manner - described in . - The offshore facility is miles at sea and - miles along the shoreline from the power plant. - It costs per mile to lay a power line underground and - to run the line underwater. -

    -

    - How much of the power line should be run underground? - What is the minimum overall cost? -

    - - Enter the length of line under ground here. - -

    - -

    - - Enter the minimum overall cost here. - -

    - -

    -     -     -     - - - - - do { - ($x,$y) = random_subset(2,2..6); - $ycost=random(30000,50000,5000); - $xcost=$ycost + random(30000,50000,5000); - } until ($y - $ycost*$x/sqrt($xcost**2 - $ycost**2) < 0); - if($envir{problemSeed}==1){$x=5;$y=2;$xcost=80000;$ycost=50000}; - Context("Currency")->flags->set(trimTrailingZeros=>1,tolerance=>1); - $xC = Currency($xcost); - $yC = Currency($ycost); - $g = Compute("50000"); - $s = Compute("80000"); - $a = Compute("5"); - $b = Compute("2"); - $z = max(0,$y - $ycost*$x/sqrt($xcost**2 - $ycost**2)); - $C = Currency($z*$ycost + $xcost*sqrt($x**2+($y - $z)**2)); - Context("Numeric"); - $zU = NumberWithUnits("$z mi"); - - -

    - A power line is to be run to an offshore facility in the manner - described in . - The offshore facility is miles at sea and - miles along the shoreline from the power plant. - It costs per mile to lay a power line underground and - to run the line underwater. -

    -

    - How much of the power line should be run underground? - What is the minimum overall cost? -

    - - Enter the length of line under ground here. - -

    - -

    - - Enter the minimum overall cost here. - -

    - -

    -     -     -     - - - - - ($x,$y) = random_subset(2,10..40); - $run = random(18,25,1); - $swim = random(1.2,2.4,0.1); - if($envir{problemSeed}==1){$x=15;$y=20;$run=22;$swim=1.5;}; - $runU = NumberWithUnits("$run ft/s"); - $swimU = NumberWithUnits("$swim ft/s"); - $z = max(0,$y - $swim*$x/sqrt($run**2 - $swim**2)); - $zU = NumberWithUnits("$z ft"); - - -

    - A woman throws a stick into a lake for her dog to fetch; - the stick is feet down the shore line - and feet into the water from there. - The dog may jump directly into the water and swim, - or run along the shore line to get closer to the stick before - swimming. The dog runs about - and swims about . -

    -

    - How far along the shore should the dog run to minimize the - time it takes to get to the stick? (Hint: the figure from - can be useful.) -

    -

    - -

    -     -     -     - - - - - ($x,$y) = random_subset(2,10..40); - $run = random(18,25,1); - $swim = random(1.2,2.4,0.1); - if($envir{problemSeed}==1){$x=30;$y=15;$run=22;$swim=1.5;}; - $runU = NumberWithUnits("$run ft/s"); - $swimU = NumberWithUnits("$swim ft/s"); - $z = max(0,$y - $swim*$x/sqrt($run**2 - $swim**2)); - $zU = NumberWithUnits("$z ft"); - - -

    - A woman throws a stick into a lake for her dog to fetch; - the stick is feet down the shore line - and feet into the water from there. - The dog may jump directly into the water and swim, - or run along the shore line to get closer to the stick before - swimming. The dog runs about - and swims about . -

    -

    - How far along the shore should the dog run to minimize the - time it takes to get to the stick? - (Google calculus dog to learn more about a dog's ability to minimize times.) -

    -

    - -

    -     -     -     - - - - - $w = Compute("sqrt(2)"); - $multians = MultiAnswer($w,$w)->with( - singleResult => 1, - checker => sub { - my ( $correct, $student, $self ) = @_; - my ( $stu1, $stu2 ) = @{$student}; - my ( $cor1, $cor2 ) = @{$correct}; - if ( ($cor1 == $stu1 && $cor2 == $stu2) || - ($cor1 == $stu2 && $cor2 == $stu1) ) { - return 1; - } else { - return 0; - } - } - ); - - -

    - What are the dimensions of the rectangle with largest area that - can be drawn inside the unit circle? -

    -

    - -

    -

    - -

    -     -     -     -     -     -
    -
    - Differentials - - - -

    - In - we explored the meaning and use of the derivative. - This section starts by revisiting some of those ideas. -

    - -

    - Recall that the derivative of a function f can be used to find the slopes of lines tangent to the graph of f. - At x=c, - the tangent line to the graph of f has equation - - y = \fp(c)(x-c)+f(c) - . -

    - -

    - The tangent line can be used to find good approximations of f(x) for values of x near c. -

    - -

    - For instance, we can approximate - \sin(1.1) using the tangent line to the graph of - f(x)=\sin(x) at x=\pi/3 \approx 1.05. - Recall that \sin(\pi/3) = \sqrt{3}/2 \approx 0.866, - and \fp(\pi/3)=\cos(\pi/3) = 1/2. - Thus the tangent line to f(x) = \sin(x) at x=\pi/3 is: - - \ell(x) = \frac12(x-\pi/3)+0.866 - . -

    - -
    - Graphing f(x) = \sin(x) and its tangent line at x=\pi/3 in order to estimate \sin(1.1) - -
    - - - - A graph showcasing the sine function and a tangent line at a point. - - -

    - The graph portrays f(x) = \sin(x) in its typical wave-like form, beginning from x=0. A tangent line intersects the curve at x=\pi/3, giving us an insight into the function's behavior around this point. A small rectangular region on the graph suggests a section of interest, likely to be examined more closely in a subsequent depiction. The overall visualization offers a comprehensive understanding of the sine function's progression and how the tangent line serves as a useful approximation around x=\pi/3. -

    -     - - - - \begin{tikzpicture} - \begin{axis}[ - xtick=\empty, - extra x ticks={1.047}, - extra x tick labels={$\frac{\pi}{3}$}, - ytick = {0,0.5,1}, - extra y ticks={0.866}, - extra y tick labels={$\frac{\sqrt{3}}{2}$}, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=2.36, - ] - \addplot+[domain=0:2.1, rightarrow] {sin(deg(x))}; - \addplot [tangentline, domain=-.1:1.5] {.5*(x-1.047197551)+0.866025}; - \addplot [soliddot] coordinates {(1.047,0.866)} node [below right] {\small $(\pi/3,\sqrt{3}/3)$}; - \draw (axis cs:1.04,.86) rectangle (axis cs:1.122,.9); - \end{axis} - \end{tikzpicture} - - - -
    - -
    - - - - A close-up view of the sine function and its tangent line around a specific point. - - -

    - Zooming in on the small rectangle from the prior graph, - we get a magnified view around the point x=\pi/3 for f(x) = \sin(x) and its tangent line. - This zoomed-in perspective demonstrates the precision with which the tangent line estimates \sin(1.1) at x=1.1. - We can see how points on the tangent line lie quite close to the graph of f(x), - illustrating the fact that the tangent line gives a good approximation to the original graph, - near the point where it meets the graph. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - xtick=\empty,% - extra x ticks={1.047,1.1}, - extra x tick labels={$\frac{\pi}{3}$,$1.1$}, - extra y ticks={0.866}, - extra y tick labels={$\frac{\sqrt{3}}{2}$}, - ytick={0.86,0.87,0.88,0.89}, - %extra y ticks={0.866}, - %extra y tick labels={$\frac{\sqrt{3}}{2}$}, - xdiscontinuity, - ydiscontinuity, - ymin=.86,ymax=.9,% - xmin=1.04,xmax=1.122, - ] - \addplot+[infinite,domain=1.04:1.12] {sin(deg(x))}; - \addplot [tangentline,domain=1.04:1.115] {.5*(x-1.0471)+0.866025}; - \addplot [soliddot] coordinates{(1.047,0.866)} node [below right] {\small $\left(\pi/3,\sqrt{3}/3\right)$}; - \addplot [soliddot,guideline] coordinates {(1.1,.8925)} node [pin={180:{$\ell(1.1)\approx \sin(1.1)$}}] {}; - \addplot [soliddot] coordinates {(1.1,.8912)} node [pin={0:{$\sin(1.1)$}}] {}; - - \end{axis} - \end{tikzpicture} - - - -
    -     -
    - -

    - In , - we see a graph of f(x) = \sin(x) graphed along with its tangent line at x=\pi/3. - The small rectangle shows the region that is displayed in . - In this figure, we see how we are approximating - \sin(1.1) with the tangent line, evaluated at 1.1. - Together, the two figures show how close these values are. -

    - -

    - Using this line to approximate \sin(1.1), we have: - - \ell(1.1) \amp = \frac{1}{2}(1.1-\pi/3)+0.866 - \amp = \frac12(0.053)+0.866 = 0.8925 - . -

    - -

    - (We leave it to the reader to see how good of an approximation this is.) -

    - - - - - - - - -

    - We now generalize this concept. - Given f(x) and an x-value c, - the tangent line is y=\ell(x), where \ell(x) = \fp(c)(x-c)+f(c). - Clearly, f(c) = \ell(c). - Let \dx be a small number, - representing a small change in the x-value. - We assert that: - - f(c+\dx) \approx \ell(c+\dx) - , - since the tangent line to a function approximates well the values of that function near x=c. - In fact, the tangent line is the graph of the linear function that best approximates the value of - f(x) for x near c. - Because of its value in applications, we give it a name. -

    - - - -

    - Let f be differentiable on an open inverval I containing c. - The function \ell(x) = \fp(c)(x-c)+f(c) is called the linearization, - or linear approximation of f at c. - - linearization - approximationtangent line - approximationlinear -

    -     -     - -

    - As the x-value changes from c to c+\dx, - the y-value of f changes from f(c) to f(c+\dx). - We call this change of y-value \dy. - That is: - - \dy = f(c+\dx)-f(c) - . -

    - -

    - Replacing f(c+\dx) with its tangent line approximation, we have - - \dy \amp \approx \ell(c+\dx) - f(c) - \amp = \fp(c)\big((c+\dx)-c\big)+f(c) - f(c) - \amp =\fp(c)\dx - . -

    - -

    - This final equation is important; - it becomes the basis of and . - In short, it says that when the x-value changes from c to c+\Delta x, - the y value of a function f changes by about \fp(c)\Delta x. -

    - - - -

    - We introduce two new variables, - dx and dy in the context of a formal definition. -

    - - - Differentials of <m>x</m> and <m>y</m> - -

    - Let y=f(x) be differentiable. - The differential of x, - denoted dx, is any nonzero real number - (usually taken to be a small number). - - differential - - derivativedifferential - - The differential of y, denoted dy, is - - dy = \fp(x)dx - . -

    -     -     - - - - -

    - We can solve for \fp(x) in the above equation: - \fp(x) = dy/dx. - This states that the derivative of f with respect to x is the differential of y divided by the differential of x; - this is not the alternate notation for the derivative, - \lz{y}{x}. - This latter notation was chosen because of the fraction-like qualities of the derivative, - but again, - it is one symbol and not a fraction. -

    - - - -

    - It is helpful to organize our new concepts and notations in one place. -

    - - - - - Differential Notation -

    - Let y = f(x) be a differentiable function. - - differentialnotation - -

    - -

    -

      -
    1. -

      - Let \dx represent a small, - nonzero change in x value. -

      -
      2. - -
    2. -

      - Let dx represent a small, - nonzero change in x value (, \dx = dx). -

      -
      3. - -
    3. -

      - Let \dy be the change in y value as x changes by \dx; hence - - \dy = f(x+\dx)-f(x) - . -

      -
      4. - -
    4. -

      - Let dy = \fp(x)dx which, - by Equation, - is an approximation of the change in y-value as x changes by \dx; - dy \approx \dy. -

      -
      5. -
    -

    -     - - - -

    - What is the value of differentials? - Like many mathematical concepts, - differentials provide both practical and theoretical benefits. - We explore both here. -

    - - - Finding and using differentials - -

    - Consider f(x) = x^2. - Knowing f(3) = 9, approximate f(3.1). -

    -     - -

    - The x-value is changing from x=3 to x=3.1; - therefore, we see that dx=0.1. - If we know how much the y-value changes from f(3) to f(3.1) (, if we know \dy), - we will know exactly what f(3.1) is (since we already know f(3)). - We can approximate \dy with dy. - - \dy \amp \approx dy - \amp = \fp(3)dx - \amp = 2\cdot 3\cdot 0.1 = 0.6 - . -

    - -

    - We expect the y-value to change by about 0.6, - so we approximate f(3.1) \approx 9.6. -

    - -

    - We leave it to the reader to verify this, - but the preceding discussion links the differential to the tangent line of f(x) at x=3. - One can verify that the tangent line, - evaluated at x=3.1, also gives y=9.6. -

    -     - -     - - - -

    - Of course, it is easy to compute the actual answer (by hand or with a calculator): - 3.1^2 = 9.61. (Before we get too cynical and say - Then why bother?, - note our approximation is really good!) -

    - -

    - So why bother? -

    - -

    - In most real life situations, - we do not know the function that describes a particular behavior. - Instead, we can only take measurements of how things change measurements of the derivative. -

    - -

    - Imagine water flowing down a winding channel. - It is easy to measure the speed and direction (, the - velocity) of water at any location. - It is very hard to create a function that describes the overall flow, - hence it is hard to predict where a floating object placed at the beginning of the channel will end up. - However, we can approximate - the path of an object using differentials. - Over small intervals, - the path taken by a floating object is essentially linear. - Differentials allow us to approximate the true path by piecing together lots of short, - linear paths. - This technique is called Euler's Method, - studied in introductory Differential Equations courses. -

    - - - -

    - We use differentials once more to approximate the value of a function. - Even though calculators are very accessible, - it is neat to see how these techniques can sometimes be used to easily compute something that looks rather hard. -

    - - - Using differentials to approximate a function value - -

    - Approximate \sqrt{4.5}. -

    -     - -

    - We expect \sqrt{4.5} \approx 2, yet we can do better. - Let f(x) = \sqrt{x}, and let c=4. - Thus f(4) = 2. - We can compute \fp(x) = 1/(2\sqrt{x}), so \fp(4) = 1/4. -

    - -

    - We approximate the difference between f(4.5) and f(4) using differentials, - with dx = 0.5: - - f(4.5)-f(4) = \dy \approx dy = \fp(4)\cdot dx = \frac14 \cdot \frac12 = \frac18 = 0.125 - . -

    - -

    - The approximate change in f from x=4 to x=4.5 is 0.125, - so we approximate \sqrt{4.5} \approx 2.125. -

    -     - -     - -

    - Differentials are important when we discuss integration. - When we study that topic, we will use notation such as - - \int f(x)\,dx - - quite often. - While we don't discuss here what all of that notation means, - note the existence of the differential dx. - Proper handling of integrals - comes with proper handling of differentials. -

    - -

    - In light of that, we practice finding differentials in general. -

    - - - Finding differentials - -

    - In each of the following, find the differential dy. -

    - -

    -

      -

    1. y = \sin(x)

      2. - -

    2. y = e^x\left(x^2+2\right)

      3. - -

    3. y = \sqrt{x^2+3x-1}

      4. -
    -

    -     - -

    -

      -
    1. -

      - y = \sin(x): As f(x) = \sin(x), \fp(x) = \cos(x). - Thus - - dy = \cos(x)dx - . -

      -
      2. - -
    2. -

      - y = e^x\left(x^2+2\right): Let f(x) = e^x\left(x^2+2\right). - We need \fp(x), - requiring the . -

      - -

      - We have \fp(x) = e^x\left(x^2+2\right) + 2xe^x, so - - dy = \left(e^x\left(x^2+2\right) + 2xe^x\right)dx - . -

      -
      3. - -
    3. -

      - y = \sqrt{x^2+3x-1}: Let f(x) = \sqrt{x^2+3x-1}; - we need \fp(x), - requiring the . -

      - -

      - We have \fp(x) = \frac{1}{2}\left(x^2+3x-1\right)^{-\frac{1}{2}}(2x+3) = \frac{2x+3}{2\sqrt{x^2+3x-1}}. - Thus - - dy = \frac{(2x+3)dx}{2\sqrt{x^2+3x-1}} - . -

      -
      4. -
    -

    -     -     - -

    - Finding the differential dy of y=f(x) is really no harder than finding the derivative of f; - we just multiply \fp(x) by dx. - It is important to remember that we are not simply adding the symbol - dx at the end. -

    - -

    - We have seen a practical use of differentials as they offer a good method of making certain approximations. - Another use is error propagation. - Suppose a length is measured to be x, - although the actual value is x+\dx - (where \dx is the error, which we hope is small). - This measurement of x may be used to compute some other value; - we can think of this latter value as f(x) for some function f. - As the true length is x+\dx, - one really should have computed f(x+\dx). - The difference between f(x) and - f(x+\dx) is the propagated error. -

    - -

    - How close are f(x) and f(x+\dx)? - This is a difference in y values: - - f(x+\dx)-f(x) = \dy \approx dy - . -

    - -

    - We can approximate the propagated error using differentials. -

    - - - Using differentials to approximate propagated error - -

    - A steel ball bearing is to be manufactured with a diameter of - - 2 - . - The manufacturing process has a tolerance of \pm 0.1 - - - in the diameter. - Given that the density of steel is about - - 7.85 - , estimate the propagated error in the mass of the ball bearing. -

    -     - - -

    - The mass of a ball bearing is found using the equation - mass = volume density. - In this situation the mass function is a product of the radius of the ball bearing, - hence it is m = 7.85\frac43\pi r^3. - The differential of the mass is - - dm = 31.4\pi r^2 dr - . -

    - -

    - The radius is to be - - 1 - ; the manufacturing tolerance in the radius is \pm 0.05 - - , or \pm 0.005 - - . - The propagated error is approximately: - - \Delta m \amp \approx dm - \amp = 31.4\pi (1)^2 (\pm 0.005) - \amp = \pm 0.493\text{g} - -

    - -

    - Is this error significant? - It certainly depends on the application, - but we can get an idea by computing the - relative error. - The ratio between amount of error to the total mass is - - \frac{dm}{m} \amp = \pm \frac{0.493}{7.85\frac43\pi} - \amp =\pm \frac{0.493}{32.88} - \amp =\pm 0.015 - , - or \pm 1.5\%. -

    - -

    - We leave it to the reader to confirm this, - but if the diameter of the ball was supposed to be - - 10 - , the same manufacturing tolerance would give a propagated error in mass of \pm12.33 - - , which corresponds to a - percent error of \pm0.188\%. - While the amount of error is much greater (12.33 \gt 0.493), - the percent error is much lower. -

    -     - -     - - - - Terms and Concepts - - - - -

    - Given a differentiable function y=f(x), - we are generally free to choose a value for dx, - which then determines the value of dy. -

    -     -     - - - - - -

    - - The symbols dx and \dx - represent the same concept. -

    -     - -     - - - - -

    - - The symbols dy and \dy - represent the same concept. -

    -     - -     - - - - -

    - - Differentials are important in the study of integration. -

    -     - -     - - - - -

    - How are differentials and tangent lines related? -

    - -     - - - -     - - - - -

    - - In real life, - differentials are used to approximate function values - when the function itself is not known. -

    -     - -     -     - - - Problems - - - -

    - Use differentials to approximate the given value by hand. -

    -     - - - - - $x0 = random(2,9,1); - $dx = random(0.03,0.09,0.01); - if($envir{problemSeed}==1){$x0=2;$dx=0.05}; - Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); - $f = Compute("x^2"); - $df = $f->D('x'); - $x1 = $x0 + $dx; - $dy = $dx * $df->eval(x=>$x0); - $y = $f->eval(x=>$x0); - $r = $y+$dy; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $given = $f->substitute(x=>Formula("$x1")); - - -

    - -

    -

    - -

    -     - -

    - Use y = ; - dy = \cdot dx with x= - and dx = . - Thus dy = ; knowing - =, - we have \approx . -

    -     -     -     - - - - $x0 = random(2,9,1); - $dx = -random(0.03,0.09,0.01); - if($envir{problemSeed}==1){$x0=6;$dx=-0.07}; - Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); - $f = Compute("x^2"); - $df = $f->D('x'); - $x1 = $x0 + $dx; - $dy = $dx * $df->eval(x=>$x0); - $y = $f->eval(x=>$x0); - $r = $y+$dy; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $given = $f->substitute(x=>Formula("$x1")); - - -

    - -

    -

    - -

    -     - -

    - Use y = ; - dy = \cdot dx with x= - and dx = . - Thus dy = ; knowing - =, - we have \approx . -

    -     -     -     - - - - - $x0 = random(2,9,1); - $dx = random(0.1,0.4,0.1); - if($envir{problemSeed}==1){$x0=5;$dx=0.1}; - Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); - $f = Compute("x^3"); - $df = $f->D('x'); - $x1 = $x0 + $dx; - $dy = $dx * $df->eval(x=>$x0); - $y = $f->eval(x=>$x0); - $r = $y+$dy; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $given = $f->substitute(x=>Formula("$x1")); - - -

    - -

    -

    - -

    -     - -

    - Use y = ; - dy = \cdot dx with x= - and dx = . - Thus dy = ; knowing - =, - we have \approx . -

    -     -     -     - - - - $x0 = random(2,9,1); - $dx = -random(0.1,0.4,0.1); - if($envir{problemSeed}==1){$x0=7;$dx=-0.2}; - Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); - $f = Compute("x^3"); - $df = $f->D('x'); - $x1 = $x0 + $dx; - $dy = $dx * $df->eval(x=>$x0); - $y = $f->eval(x=>$x0); - $r = $y+$dy; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $given = $f->substitute(x=>Formula("$x1")); - - -

    - -

    -

    - -

    -     - -

    - Use y = ; - dy = \cdot dx with x= - and dx = . - Thus dy = ; knowing - =, - we have \approx . -

    -     -     -     - - - - - $x0 = list_random(9,16,25,36,49); - $dx = random(0.3,0.9,0.1); - if($envir{problemSeed}==1){$x0=16;$dx=0.5}; - Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); - $f = Compute("sqrt(x)"); - $df = $f->D('x'); - $x1 = $x0 + $dx; - $dy = $dx * $df->eval(x=>$x0); - $y = $f->eval(x=>$x0); - $r = $y+$dy; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $given = $f->substitute(x=>Formula("$x1")); - - -

    - -

    -

    - -

    -     - -

    - Use y = ; - dy = \cdot dx with x= - and dx = . - Thus dy = ; knowing - =, - we have \approx . -

    -     -     -     - - - - $x0 = list_random(9,16,25,36,49); - $dx = -random(0.3,1.5,0.1); - if($envir{problemSeed}==1){$x0=25;$dx=-1}; - Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); - $f = Compute("sqrt(x)"); - $df = $f->D('x'); - $x1 = $x0 + $dx; - $dy = $dx * $df->eval(x=>$x0); - $y = $f->eval(x=>$x0); - $r = $y+$dy; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $given = $f->substitute(x=>Formula("$x1")); - - -

    - -

    -

    - -

    -     - -

    - Use y = ; - dy = \cdot dx with x= - and dx = . - Thus dy = ; knowing - =, - we have \approx . -

    -     -     -     - - - - - $x0 = list_random(8,27,64,125,216); - $dx = -random(0.3,1.5,0.1); - if($envir{problemSeed}==1){$x0=64;$dx=-1}; - Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); - parser::Root->Enable; - $f = Compute("root(3,x)"); - $df = $f->D('x'); - $x1 = $x0 + $dx; - $dy = $dx * $df->eval(x=>$x0); - $y = $f->eval(x=>$x0); - $r = $y+$dy; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $given = $f->substitute(x=>Formula("$x1")); - - -

    - -

    -

    - -

    -     - -

    - Use y = ; - dy = \cdot dx with x= - and dx = . - Thus dy = ; knowing - =, - we have \approx . -

    -     -     -     - - - - $x0 = list_random(8,27,64,125,216); - $dx = random(0.3,1.5,0.1); - if($envir{problemSeed}==1){$x0=8;$dx=0.5}; - Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); - parser::Root->Enable; - $f = Compute("root(3,x)"); - $df = $f->D('x'); - $x1 = $x0 + $dx; - $dy = $dx * $df->eval(x=>$x0); - $y = $f->eval(x=>$x0); - $r = $y+$dy; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $given = $f->substitute(x=>Formula("$x1")); - - -

    - -

    -

    - -

    -     - -

    - Use y = ; - dy = \cdot dx with x= - and dx = . - Thus dy = ; knowing - =, - we have \approx . -

    -     -     -     - - - - - $x0 = Compute("pi"); - $dx = 3-$x0; - Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); - $f = Compute("sin(x)"); - $df = $f->D('x'); - $x1 = 3; - $dy = $dx * $df->eval(x=>$x0); - $y = $f->eval(x=>$x0); - $r = $y+$dy; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $given = $f->substitute(x=>Formula("$x1")); - - -

    - -

    -

    - -

    -     - -

    - Use y = ; - dy = \cdot dx with x= - and dx = . - Thus dy = ; knowing - =, - we have \approx . -

    -     -     -     - - - - - $x0 = 0; - $dx = random(0.1,0.5,0.1); - if($envir{problemSeed}==1){$dx=0.1}; - Context()->flags->set(tolType=>'absolute',tolerance=>10**-8); - parser::Root->Enable; - $f = Compute("e^x"); - $df = $f->D('x'); - $x1 = $x0 + $dx; - $dy = $dx * $df->eval(x=>$x0); - $y = $f->eval(x=>$x0); - $r = $y+$dy; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $given = $f->substitute(x=>Formula("$x1")); - - -

    - -

    -

    - -

    -     - -

    - Use y = ; - dy = \cdot dx with x= - and dx = . - Thus dy = ; knowing - =, - we have \approx . -

    -     -     -     -     - - - - -

    - Compute the differential dy. -

    -     - - - - $b = non_zero_random(-9,9,1); - $c = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b=3;$c=-5;}; - Context()->variables->add(dx=>"Real"); - $f = Compute("x^2+$b x + $c")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y= -

    -

    - -

    -     -     -     - - - - - ($m,$n) = random_subset(2,3..9); - $c = list_random(1,-1); - if($envir{problemSeed}==1){$m=7;$n=5;$c=-1}; - Context()->variables->add(dx=>"Real"); - $f = Compute("x^$m+$c x^$n")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y= -

    -

    - -

    -     -     -     - - - - - ($m,$n) = random_subset(2,1..9); - if($envir{problemSeed}==1){$m=4;$n=2;}; - Context()->variables->add(dx=>"Real"); - $f = Compute("1/($m x^$n)")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y=\displaystyle -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $n = random(2,3,1); - if($envir{problemSeed}==1){$a=2;$n=2;}; - Context()->variables->add(dx=>"Real"); - $f = Compute("($a x + sin(x))^$n")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y= -

    -

    - -

    -     -     -     - - - - - ($m,$n) = random_subset(2,2..9); - if($envir{problemSeed}==1){$m=2;$n=3;}; - Context()->variables->add(dx=>"Real"); - $f = Compute("x^$m + e^($n x)")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y= -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $b = random(2,9,1); - if($envir{problemSeed}==1){$a=4;$b=4;}; - Context()->variables->add(dx=>"Real"); - $f = Compute("$a/x^$b")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y=\displaystyle -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $b = random(1,9,1); - if($envir{problemSeed}==1){$a=2;$b=1;}; - Context()->variables->add(dx=>"Real"); - $f = Compute("($a x)/(tan(x) + $b)")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y=\displaystyle -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - if($envir{problemSeed}==1){$a=5;}; - Context()->variables->add(dx=>"Real"); - $f = Compute("ln($a x)")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y= -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','cos','tan','sec','csc','cot'); - if($envir{problemSeed}==1){$trig='sin';}; - Context()->variables->add(dx=>"Real"); - $f = Compute("e^x $trig(x)")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y= -

    -

    - -

    -     -     -     - - - - - ($trig1,$trig2) = random_subset(2,'sin','cos','tan','sec','csc','cot'); - if($envir{problemSeed}==1){$trig1='cos';$trig2='sin';}; - Context()->variables->add(dx=>"Real"); - $f = Compute("$trig1($trig2(x))")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y= -

    -

    - -

    -     -     -     - - - - - ($a,$b) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$a=1;$b=2;}; - Context()->variables->add(dx=>"Real"); - $f = Compute("(x+$a)/(x+$b)")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y=\displaystyle -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - if($envir{problemSeed}==1){$a=3;}; - Context()->variables->add(dx=>"Real"); - $f = Compute("$a^x ln(x)")->reduce; - $df = $f->D('x')->reduce; - $A = Formula("$df dx"); - - -

    - y= -

    -

    - -

    -     -     -     - - - - - $fstring = list_random('x ln(x) - x','x arctan(x) - (1/2) ln(1+x^2)'); - if($envir{problemSeed}==1){$fstring='x ln(x) - x';}; - Context()->variables->add(dx=>"Real"); - $f = Compute("$fstring")->reduce; - $df = ($fstring eq 'x ln(x) - x') ? Compute("ln(x)") : Compute("arctan(x)"); - $A = Formula("$df dx"); - - -

    - y= -

    -

    - -

    -     -     -     - - - - - $fstring = list_random('ln(sec(x))','ln(sin(x))',); - if($envir{problemSeed}==1){$fstring='ln(sec(x))';}; - Context()->variables->add(dx=>"Real"); - $f = Compute("$fstring")->reduce; - $df = ($fstring eq 'ln(sec(x))') ? Compute("tan(x)") : Compute("cot(x)"); - $A = Formula("$df dx"); - - -

    - y= -

    -

    - -

    -     -     -     -     - - - - - - $diameter = random(1,5,1); - $dx = random(1,$diameter,1); - if($envir{problemSeed}==1){$diamter=1;$dx=1;}; - $diameterU = NumberWithUnits("$diameter cm"); - $dxU = NumberWithUnits("$dx mm"); - $f = Compute("1/6*pi*x^3"); - $df = $f->D('x'); - $err = $df->eval(x=>$diameter) * $dx/10; - $errU = NumberWithUnits("$err cm^3"); - - -

    - A set of plastic spheres are to be made with a diameter of . - If the manufacturing process is accurate to , - what is the propagated error in volume of the spheres? -

    -

    - -

    -     -     -     - - - - - $dx = random(1,4,1); - $a = random(2,4,1); - $b = random(5,8,1); - if($envir{problemSeed}==1){$dx=2;$a=2;$b=5;}; - $f = Compute("16*x^2"); - $df = $f->D('x'); - $err_a = $df->eval(x=>$a) * $dx/10; - $err_aU = NumberWithUnits("$err_a ft"); - $err_b = $df->eval(x=>$b) * $dx/10; - $err_bU = NumberWithUnits("$err_b ft"); - - -

    - The distance, in feet, - a stone drops in t seconds is given by d(t) = 16t^2. - The depth of a hole is to be approximated by dropping a rock and listening for it to hit the bottom. - What is the propagated error if the time measurement is accurate to - /10 of a second and the measured time is: -

    -     - - -

    - seconds? -

    -

    - -

    -     -     - - -

    - seconds? -

    -

    - -

    -     -     -     -     - - - - - $diameter = random(12,20,1); - $dx = list_random(2,4,8,16); - if($envir{problemSeed}==1){$diameter=15;$dx=4}; - $f = Compute("pi/4*x^2"); - $df = $f->D('x'); - $err = $df->eval(x=>$diameter) * 1/$dx; - $errU = NumberWithUnits("$err in^2"); - - -

    - What is the propagated error in the measurement of the cross sectional area - of a circular log if the diameter is measured at '', - accurate to 1/''? -

    -

    - -

    -     -     -     - - - - - $h = random(8,10,1); - $lft = random(10,20,1); - $lin = random(1,11); - $dx = list-random(2,4,8,16); - if($envir{problemSeed}==1){$h=8;$lft=10;$lin=7;$dx=2}; - $f = Compute("$h*x"); - $df = $f->D('x'); - $x = $lft + $lin/12; - $err = $df->eval(x=>$x) * 1/$dx; - $errU = NumberWithUnits("$err ft^2"); - - -

    - A wall is to be painted that is ' high and is measured - to be ',\,'' long. - Find the propagated error in the measurement of the wall's surface area - if the measurement is accurate to 1/''. -

    -

    - -

    -     -     -     - - - -

    - The following exercises explore some issues related to surveying in which - distances are approximated using other measured distances and measured angles. - (Hint: Convert all angles to radians before computing.) -

    -     - - - - - $theta = 85.2; - $d = 25; - $f = Compute("$d*tan(x)"); - $l = $f->eval(x=>$theta*pi/180); - $lU = NumberWithUnits("$l ft"); - $df = $f->D('x'); - $err = $df->eval(x=>$theta*pi/180) * pi/180; - $errU = NumberWithUnits("$err ft"); - Context("Percent"); - $err_pct = Percent("$err/$l"); - - -

    - The length L of a long wall is to be approximated. - The angle \theta, as shown in the diagram - (not to scale), - is measured at a distance of 25 feet from the wall, and found to be ^\circ, - accurate to 1^\circ. - Assume that the triangle formed is a right triangle. -

    - - - A right-angled triangle representing an approximation challenge for a wall's length. - - -

    - The diagram presents a right-angled triangle utilized to navigate an approximation problem concerning the length, l, of a long wall. - The angle labeled as \theta is opposite the wall, - and the adjacent side is marked as 25. - The opposite side, representing the wall's length, is denoted by the variable l. - The triangle's hypotenuse is not specified in the given representation. -

    -     - - - \begin{tikzpicture}[scale=2.1] - \draw [ultra thick] (1,-1) -- node [pos=.5,right] {\(l=\)?}(1,1); - \draw [dashed] (1,1) -- (-1,-1) node [xshift=10pt,yshift=5pt] {\(\theta\)} -- node [pos=.5,below] {\(25'\)} (1,-1); - \end{tikzpicture} - - -     - - -

    - What is the measured length L of the wall? -

    -

    - -

    -     -     - - -

    - What is the propagated error? -

    -

    - -

    -     -     - - -

    - What is the percent error? -

    -

    - -

    -     -     -     -     - - - - - $theta = 71.5; - $d = 100; - $f = Compute("$d*tan(x)"); - $l = $f->eval(x=>$theta*pi/180); - $lU = NumberWithUnits("$l ft"); - $df = $f->D('x'); - $err = $df->eval(x=>$theta*pi/180) * pi/180; - $errU = NumberWithUnits("$err ft"); - Context("Percent"); - $err_pct = Percent("$err/$l"); - - -

    - The length L of a long wall is to be approximated. - The angle \theta, as shown in the diagram - (not to scale), - is measured at a distance of 100 feet from the wall, and found to be ^\circ, - accurate to 1^\circ. - Assume that the triangle formed is a right triangle. -

    - - - A right-angled triangle illustrating an approximation problem related to a long wall. - - -

    - The diagram showcases a right-angled triangle, - which aids in determining an approximation for the length, l, of a long wall. - The angle labeled as \theta is opposite the wall, - and the adjacent side is marked as 25. - The opposite side, representing the wall's length, is denoted by the variable l. - The triangle's hypotenuse is not specified in the given representation. -

    -     - - \begin{tikzpicture}[scale=2.1] - \draw [ultra thick] (1,-1) -- node [pos=.5,right] {\(l=\)?}(1,1); - \draw [dashed] (1,1) -- (-1,-1) node [xshift=10pt,yshift=5pt] {\(\theta\)} -- node [pos=.5,below] {\(100'\)} (1,-1); - \end{tikzpicture} - - -     - - -

    - What is the measured length L of the wall? -

    -

    - -

    -     -     - - -

    - What is the propagated error? -

    -

    - -

    -     -     - - -

    - What is the percent error? -

    -

    - -

    -     -     -     -     - - - - - $theta = 143; - $d = 50; - $f = Compute("2*$d*tan(x/2)"); - $l = $f->eval(x=>$theta*pi/180); - $lU = NumberWithUnits("$l ft"); - $df = $f->D('x'); - $err = $df->eval(x=>$theta*pi/180) * pi/180; - $errU = NumberWithUnits("$err ft"); - Context("Percent"); - $err_pct = Percent("$err/$l"); - - -

    - The length L of a long wall is to be calculated - by measuring the angle \theta shown in the diagram - (not to scale) at a distance of 50 feet from the wall. - Assume the formed triangle is an isosceles triangle. - The measured angle is ^\circ, - accurate to 1^\circ. -

    - - - An isosceles triangle used for calculating a wall's length, l. - - -

    - The diagram illustrates an isosceles triangle to facilitate a calculation related to determining the length, l, of a long wall. - The triangle's height, perpendicular to the base representing the wall, - is given as 50. The length of the wall is denoted as l. - The angle opposite the wall is labeled as \theta. -

    -     - - - \begin{tikzpicture}[scale=2.1] - \draw [ultra thick] (1,-1) -- node [pos=.5,right] {\(l=\)?}(1,1); - \draw [dashed] (1,1) -- (-1,0) node [xshift=10pt] {\(\theta\)} -- (1,-1); - \draw (-.5,0) -- node [pos=.5,draw=white,fill=white] {\(50'\)} (1,0); - \end{tikzpicture} - - -     - - -

    - What is the measured length L of the wall? -

    -

    - -

    -     -     - - -

    - What is the propagated error? -

    -

    - -

    -     -     - - -

    - What is the percent error? -

    -

    - -

    -     -     -     -     - - - - - - $theta = 143; - $d = 50; - $f = Compute("2*x*tan( ($theta*pi/180) /2)"); - $l = $f->eval(x=>$d); - $lU = NumberWithUnits("$l ft"); - $df = $f->D('x'); - $err = $df->eval(x=>$d) * 0.5; - $errU = NumberWithUnits("$err ft"); - Context("Percent"); - $err_pct = Percent("$err/$l"); - - -

    - Consider the setup in . - This time, assume the angle measurement of - 143^\circ is exact but the measured 50' from the wall is accurate to 6''. -

    - - - An isosceles triangle used for calculating a wall's length, l. - - -

    - The diagram illustrates an isosceles triangle to facilitate a calculation related to determining the length, l, of a long wall. - The triangle's height, perpendicular to the base representing the wall, - is given as 50. The length of the wall is denoted as l. - The angle opposite the wall is labeled as \theta. -

    -     - - \begin{tikzpicture} - \draw [ultra thick] (1,-1) -- node [pos=.5,right] {\scriptsize \(l=\)?}(1,1); - \draw [dashed] (1,1) -- (-1,0) node [xshift=10pt] {\scriptsize \(\theta\)} -- (1,-1); - \draw (-.5,0) -- node [pos=.5,draw=white,fill=white] {\scriptsize \(50'\)} (1,0); - \end{tikzpicture} - - -

    - What is the approximate percent error? -

    -

    - -

    -     -     -     - - - - - - parserPopUp.pl - - - $buttons = DropDown( - [ - 'Right triangle at 25 feet', - 'Right triangle at 100 feet', - 'Isosceles triangle at 50 feet', - ],2,showInStatic=>0); - - -

    - The length of the walls in - - are essentially the same. - Which setup gives the most accurate result? -

    -

    - -

    -     -     -     -     -     -     -
    - - -

    - We first learned of the derivative in the context of instantaneous rates of change and slopes of tangent lines. - We furthered our understanding of the power of the derivative by studying how it relates to the graph of a function - (leading to ideas of increasing/decreasing and concavity). - This chapter has put the derivative to yet more uses: -

    - -

    -

      -
    • -

      - Equation solving (Newton's Method), -

      -
      • - -
    • -

      - Related Rates (furthering our use of the derivative to find instantaneous rates of change), -

      -
      • - -
    • -

      - Optimization - (applied extreme values), - and -

      -
      • - -
    • -

      - Differentials - (useful for various approximations and for something called integration). -

      -
      • - - - - - - - -
    -

    - -

    - In the next chapters, we will consider the reverse - problem to computing the derivative: - given a function f, - can we find a function whose derivative is f? - Being able to do so opens up an incredible world of mathematics and applications. -

    -     - -     - - - Integration - -

    - We have spent considerable time considering the derivatives of a function and their applications. - In the following chapters, - we are going to starting thinking in - the other direction. That is, - given a function f(x), - we are going to consider functions F(x) such that F'(x) = f(x). - There are numerous reasons this will prove to be useful: - these functions will help us compute area, - volume, mass, force, pressure, work, and much more. -

    -     - -
    - Antiderivatives and Indefinite Integration - - - -

    - Given a function y=f(x), - a differential equation - is an equation that incorporates y, x, - and the derivatives of y. - For instance, a simple differential equation is: - - \yp = 2x - . -

    -

    - Solving a differential equation amounts to finding a function y - that satisfies the given equation. - Take a moment and consider that equation; - can you find a function y such that \yp = 2x? -

    -

    - Can you find another? -

    -

    - And yet another? -

    -

    - Hopefully you were able to come up with at least one solution: - y = x^2. - Finding another may have seemed impossible until one realizes that - a function like y=x^2+1 also has a derivative of 2x. - Once that discovery is made, finding - yet another is not difficult; - the function y = x^2 + 123{,}456{,}789 also has a derivative of - 2x. - The differential equation \yp = 2x has many solutions. - This leads us to some definitions. -

    - - - Antiderivatives and Indefinite Integrals - -

    - Let a function f(x) be given. - An antiderivative of f(x) is a function F(x) - such that \Fp(x) = f(x). - antiderivative - indefinite integral - integrationindefinite -

    -

    - The set of all antiderivatives of f(x) is the - indefinite integral of f, denoted by - - \int f(x) \,dx - . -

    -     -     - - - - - - -

    - Make a note about our definition: - we refer to an antiderivative of f, - as opposed to the antiderivative of f, - since there is always an infinite number of them. - We often use upper-case letters to denote antiderivatives. -

    -

    - When f is continuous, - knowing one antiderivative of f allows us to find infinitely more, - simply by adding a constant. - Not only does this give us more antiderivatives, - it gives us all of them. -

    - - - Antiderivative Forms - -

    - Let F(x) and G(x) be antiderivatives of a continuous - function f(x) on an interval I. - Then there exists a constant C such that, on I, - - G(x) = F(x) + C - . -

    -     -     - -

    - Given a continuous function f defined on an interval I and - one of its antiderivatives F, - we know all antiderivatives of f on I have the form - F(x) + C for some constant C. - Using , we can say that - - \int f(x) \,dx = F(x) + C - . -

    -

    - Note that we are abusing notation somewhat: - when we write F(x)+C on the right-hand side, - we really mean the set of all such functions, - for each real number value of C. - Let's analyze this indefinite integral notation. - integrationnotation -

    - -
    - Antiderivative notation - - - - - A labeled formula for the indefinite integral/ - -

    - The equation \int f(x) \cdot dx = F(x) + C, with labels for each part. - \int is labeled as the integral symbol. - f(x) is labeled as the integrand function. - dx is labeled as the differential of x. - F(x) is labeled as any "antiderivative of f." - C is labeled as the constant of integration. -

    -     - - - \begin{tikzpicture}[>=latex, node distance=6pt,scale=0.9] - - \node[] (integral) - {$\displaystyle\int$}; - \node[above left=of integral, anchor=south east] (l1) {Integral symbol}; - \draw[->] (l1) edge [out=-90, in=180] (integral); - \node[right=of integral] (integrand) - {$\displaystyle f(x)$}; - \node[below=of integrand, anchor=north east] (l2) {Integrand function}; - \draw[->] (l2) edge [out=0, in=-90] (integrand); - \node[right=of integrand] (dot) {$\cdot$}; - \node[right=of dot] (differential) - {$\displaystyle dx$}; - \node[above=of differential, anchor=south west] (l3) {Differential of $x$}; - \draw[->] (l3) edge [out=-90, in=90] (differential); - \node[right=of differential] (equals) - {$\displaystyle =$}; - \node[right=of equals] (antiderivative) - {$\displaystyle F(x)$}; - \node[below=of antiderivative, anchor=north west] (l4) {Any antiderivative of $f$}; - \draw[->] (l4) edge [out=180, in=-90] (antiderivative); - \node[right=of antiderivative] (plus) - {$\displaystyle {+}$}; - \node[right=of plus] (constant) - {$\displaystyle C$}; - \node[above right=of constant, anchor=south west] (l5) {Constant of integration}; - \draw[->] (l5) edge [out=-90, in=0] (constant); - - \end{tikzpicture} - - - - -
    - -

    - - shows the typical notation of the indefinite integral. - The integration symbol, \int, - is in reality an elongated S, - representing take the sum. - We will later see how sums - and antiderivatives are related. -

    -

    - The function we want to find an antiderivative of is called the - integrand. - It contains the differential of the variable we are integrating with respect to. - The \int symbol and the differential dx are not bookends - with a function sandwiched in between; - rather, the symbol \int means - find all antiderivatives of what follows, - and the function f(x) and dx are multiplied together; - the dx does not just sit there. -

    -

    - Another way of looking at the notation is that it tells us that - f(x)\,dx is the differential of F(x): dF(x) = f(x)\,dx, - confirming that F'(x)=f(x), as required of an antiderivative. - The integral symbol can then be viewed as an instruction to undo - the differential and recover the antiderivative F(x). -

    -

    - Another important aspect of the dx is that it tells us which - variable we're taking the antiderivative with respect to, - much like how \lzo{x} would mean to take the derivative with - respect to x, - while \lzo{t} would be the derivative with respect to t. -

    -

    - Let's practice using this notation. -

    - - - Evaluating indefinite integrals - -

    - Evaluate \int \sin(x) \,dx. -

    -     - -

    - We are asked to find all functions F(x) such that - \Fp(x) = \sin(x). - Some thought will lead us to one solution: - F(x) = -\cos(x), because \lzoo{x}{-\cos(x)} = \sin(x). -

    -

    - The indefinite integral of \sin(x) is thus -\cos(x), - plus a constant of integration. - So: - - \int \sin(x) \,dx = -\cos(x) + C - . -

    -     - -     - -

    - A commonly asked question is What happened to the dx? - The unenlightened response is Don't worry about it. - It just goes away. A full understanding includes the following. -

    -

    - This process of antidifferentiation - is really solving a differential question. - The integral - - \int \sin(x) \,dx - - presents us with a differential, dy = \sin(x) \, dx. - It is asking: What is y? - We found lots of solutions, - all of the form y = -\cos(x) +C. -

    -

    - Letting dy = \sin(x)\,dx, rewrite - - \int \sin(x) \,dx \text{ as } \int\,dy - . -

    -

    - This is asking: - What functions have a differential of the form dy? - The answer is Functions of the form y+C, - where C is a constant. What is y? - We have lots of choices, all differing by a constant; - the simplest choice is y = -\cos(x). -

    -

    - Understanding all of this is more important later as we try to find - antiderivatives of more complicated functions. - In this section, - we will simply explore the rules of indefinite integration, - and one can succeed for now with answering - What happened to the dx? - with It went away. -

    -

    - Let's practice once more before stating integration rules. -

    - - - Evaluating indefinite integrals - -

    - Evaluate \int\left(3x^2 + 4x+5\right)\,dx. -

    -     - -

    - We seek a function F(x) whose derivative is 3x^2+4x+5. - When taking derivatives, - we can consider functions term-by-term, - so we can likely do that here. -

    -

    - What functions have a derivative of 3x^2? - Some thought will lead us to a cubic, - specifically x^3+C_1, where C_1 is a constant. -

    -

    - What functions have a derivative of 4x? - Here the x term is raised to the first power, - so we likely seek a quadratic. - Some thought should lead us to 2x^2+C_2, - where C_2 is a constant. -

    -

    - Finally, what functions have a derivative of 5? - Functions of the form 5x+C_3, - where C_3 is a constant. -

    -

    - Our answer appears to be - - \int\left(3x^2+4x+5\right)\,dx = x^3+C_1+2x^2+C_2+5x+C_3 - . -

    -

    - We do not need three separate constants of integration; - combine them as one constant, - giving the final answer of - - \int\left(3x^2+4x+5\right)\,dx = x^3+2x^2+5x+C - . -

    -

    - It is easy to verify our answer; - take the derivative of x^3+2x^2+5x+C and see we indeed get - 3x^2+4x+5. -

    -     - -     - -

    - This final step of verifying our answer - is important both practically and theoretically. - In general, taking derivatives is easier than finding antiderivatives so - checking our work is easy and vital as we learn. -

    -

    - We also see that taking the derivative of our answer returns the function - in the integrand. - Thus we can say that: - - \lzoo{x}{\int f(x)\,dx} = f(x) - . -

    -

    - Differentiation undoes the work done by antidifferentiation. -

    -

    - - gave a list of the derivatives of common functions we had learned at that point. - We restate part of that list here to stress the relationship between - derivatives and antiderivatives. - This list will also be useful as a glossary of common antiderivatives - as we learn. -

    - - - Derivatives and Antiderivatives - -

    - Here are the Common Differentiation Rules and their Common Indefinite - Integral Rule counterparts. -

    -

    - - \amp\lzoo{x}{cf(x)}=c\cdot\fp(x)\amp\amp\int c\cdot f(x)\,dx= c\cdot \int f(x)\,dx\quad(c\neq 0) - \amp\lzoo{x}{f(x)\pm g(x)} =\fp(x)\pm \gp(x)\amp\amp \int \big(f(x)\pm g(x)\big)\,dx =\int f(x)\,dx\pm \int g(x)\,dx - \amp\lzoo{x}{C} = 0\amp\amp \int 0\,dx = C - \amp\lzoo{x}{x} = 1\amp\amp \int 1\,dx = \int dx = x+C - \amp\lzoo{x}{x^n} = n\cdot x^{n-1}\amp\amp \int x^n\,dx =\frac{1}{n+1}x^{n+1}+ C\quad(n\neq -1) - \amp\lzoo{x}{\sin(x)} = \cos(x)\amp\amp \int \cos(x) \,dx = \sin(x) +C - \amp\lzoo{x}{\cos(x)} = -\sin(x)\amp\amp \int \sin(x) \,dx = -\cos(x) +C - \amp\lzoo{x}{\tan(x)} = \sec^2(x)\amp\amp \int \sec^2(x) \,dx = \tan(x) +C - \amp\lzoo{x}{\csc(x)} = -\csc(x) \cot(x)\amp\amp \int \csc(x) \cot(x) \,dx = -\csc(x) +C - \amp\lzoo{x}{\sec(x)} = \sec(x) \tan(x)\amp\amp \int \sec(x) \tan(x) \,dx = \sec(x) +C - \amp\lzoo{x}{\cot(x)} = -\csc^2(x)\amp\amp \int \csc^2(x) \,dx = -\cot(x) +C - \amp\lzoo{x}{e^ x} = e^x\amp\amp \int e^x\,dx = e^x+C - \amp\lzoo{x}{a^x} = \ln(a) \cdot a^x\amp\amp \int a^x\,dx = \frac{1}{\ln(a) }\cdot a^x+C - \amp\lzoo{x}{\ln(x)} = \frac1 x,\ x \gt 0\amp\amp \int \frac{1}x\,dx = \ln\abs{x}+C - -

    -     -     - - - -

    - We highlight a few important points from . -

    -

    -

      -
    • -

      - - \int c\cdot f(x)\,dx = c\cdot \int f(x)\,dx - -

      -

      - This is the Constant Multiple Rule: - - Constant Multiple Ruleof integration - - we can temporarily ignore constants when finding antiderivatives, - just as we did when computing derivatives (, - \lzoo{x}{3x^2} is just as easy to compute as - \lzoo{x}{x^2}). An example: - - \int 5\cos(x) \,dx = 5\cdot\int \cos(x) \,dx = 5\cdot (\sin(x) +C) = 5\sin(x) + C - . - In the last step we can consider the constant as also being - multiplied by 5, but - 5 times a constant is still a constant, - so we just write C . -

      -
      • - -
    • -

      - - \int \big(f(x)\pm g(x)\big)\,dx =\int f(x)\,dx\pm \int g(x)\,dx - -

      -

      - This is the Sum/Difference Rule: - - integrationSum/Difference Rule - - Sum/Difference Ruleof integration - - we can split integrals apart when the integrand contains terms - that are added/subtracted, - as we did in . - So: - - \int(3x^2+4x+5)\, dx \amp = \int 3x^2\, dx + \int 4x\,dx + \int 5\,dx - \amp = 3\int x^2\, dx + 4\int x\,dx + \int 5 \,dx - \amp = 3\cdot \frac13x^3 + 4\cdot \frac12x^2+5x+C - \amp = x^3+2x^2+5x+C - - In practice we generally do not write out all these steps, - but we demonstrate them here for completeness. -

      -
      • - -
    • -

      - - \int x^n\,dx =\frac{1}{n+1}x^{n+1}+ C\quad(n\neq -1) - -

      -

      - This is the Power Rule of indefinite integration. - - integrationPower Rule - - Power Ruleintegration - - There are two important things to keep in mind: - -

        -
      1. -

        - Notice the restriction that n\neq -1. - This is important: \int \frac{1}{x}\,dx \neq - \frac{1}{0}x^0+C; - rather, see the last rule from the list. -

        -
        2. - -
      2. -

        - We are presenting antidifferentiation as the - inverse operation of differentiation. - Here is a useful quote to remember: -

        - -
        -

        - Inverse operations do the opposite things in the opposite - order. -

        -
        - -

        - When taking a derivative using the Power Rule, - we first multiply by the power, - then second subtract 1 from the power. - To find the antiderivative, do the opposite things in the - opposite order: - first add 1 to the power, - then second divide by the power. -

        -
        3. -
      -

      -
      • - -
    • -

      - - \int \frac{1}x\,dx = \ln\abs{x}+C - -

      -

      - Note that this rule uses the absolute value of x. - The exercises will work the reader through why this is the case; - for now, know the absolute value is important and cannot be ignored. -

      -
      • -
    -

    - - - - Initial Value Problems - -

    - In - we saw that the derivative of a position function gave a velocity function, - and the derivative of a velocity function describes acceleration. - initial value problem - We can now go the other way: - the antiderivative of an acceleration function gives a velocity function, - . While there is just one derivative of a given function, - there are infinitely many antiderivatives. - Therefore we cannot ask What is the - velocity of an object whose acceleration is - - -32 - ?, since there is more than one answer. -

    - - - -

    - We can find the answer if we provide more information with the - question, as done in the following example. - Often the additional information comes in the form of an initial value, - a value of the function that one knows beforehand. -

    - - - Solving initial value problems - -

    - The acceleration due to gravity of a falling object is - - -32 - . - At time t=3, a falling object had a velocity of - - -10 - . - Find the equation of the object's velocity. -

    -     - -

    - We want to know a velocity function, v(t). - We know two things: - -

      -
    • -

      - The acceleration, , v'(t)= -32, and -

      -
      • - -
    • -

      - the velocity at a specific time, , v(3) = -10. -

      -
      • -
    -

    -

    - Using the first piece of information, - we know that v(t) is an antiderivative of v'(t)=-32. - So we begin by finding the indefinite integral of -32: - - \int (-32)\,dt = -32t+C=v(t) - . -

    -

    - Now we use the fact that v(3)=-10 to find C: - - v(t) \amp = -32t+C - v(3) \amp = -10 - -32(3)+C \amp = -10 - C \amp = 86 - -

    -

    - Thus v(t)= -32t+86. - We can use this equation to understand the motion of the object: - when t=0, the object had a velocity of v(0) = 86 - - . - Since the velocity is positive, - the object was moving upward. -

    -

    - When did the object begin moving down? - Immediately after v(t) = 0: - - -32t+86 = 0 \implies t = \frac{43}{16} \approx 2.69\text{s} - . -

    -

    - Recognize that we are able to determine quite a bit about the path of - the object knowing just its acceleration and its velocity at a single - point in time. -

    -     - -     - - - Solving initial value problems - -

    - Find f(t), given that \fpp(t) = \cos(t), - \fp(0) = 3 and f(0) = 5. -

    -     - -

    - We start by finding \fp(t), - which is an antiderivative of \fpp(t): - - \int \fpp(t)\,dt \amp = \int \cos(t) \,dt - \amp = \sin(t) + C - \amp = \fp(t) - . -

    -

    - So \fp(t) = \sin(t) +C for the correct value of C. - We are given that \fp(0) = 3, so: - - \sin(0) +C \amp = 3 - C \amp = 3 - . -

    -

    - Using the initial value, we have found \fp(t) = \sin(t) + 3. - We now find f(t) by integrating again. - We will use a different integration constant since we have already - defined C to equal 3 above. - - f(t)=\int \fp(t) \,dt = \int (\sin(t) +3)\,dt = -\cos(t) + 3t + D - . -

    -

    - We are given that f(0) = 5, so - - -\cos(0) + 3(0) + D \amp = 5 - -1 + C \amp = 5 - C \amp = 6 - -

    -

    - Thus f(t) = -\cos(t) + 3t + 6. -

    -     - -     - - - -

    - This section introduced antiderivatives and the indefinite integral. - We found they are needed when finding a function given information about - its derivative(s). For instance, - we found a velocity function given an acceleration function. -

    -

    - In the next section, - we will see how position and velocity are unexpectedly related by the - areas of certain regions on a graph of the velocity function. - Then, in , - we will see how areas and antiderivatives are closely tied together. - This connection is incredibly important, - as indicated by the name of the theorem that describes it: - The Fundamental Theorem of Calculus. -

    -     - - - - Terms and Concepts - - - - -

    - Define the term antiderivative in your own words. -

    - -     - - - -     - - - - -

    - Is it more accurate to refer to the - antiderivative of f(x) or an - antiderivative of f(x)? -

    - -     - - - -

    - An antiderivative -

    -     -     - - -

    - The antiderivative -

    -     -     - -     - -     - - - - -

    - Use your own words to define the indefinite integral of f(x). -

    - -     - - - -     - - - - -

    - Fill in the blanks: - Inverse operations do the - things in the order. -

    -     - - - - opposite|reverse - - - - - opposite|reverse - - - - -     - - - - -

    - What is an initial value problem? -

    - -     - - - -     - - - - -

    - The derivative of a position function is a/an function. -

    -     - - - - - - - - -     - - - - -

    - An antiderivative of an acceleration function is a/an - function. -

    -     - - - - - - - - -     - - - - - parserFunction("F" => "exp(x)*sin(x)/sqrt(abs(x))"); - parserFunction("G" => "ln(abs(x))*cos(x)*sqrt(abs(x))"); - $answer = Formula('F(x)+G(x)'); - $evaluator = $answer->cmp(upToConstant=>1); - - -

    - If F(x) is an antiderivative of f(x), - and G(x) is an antiderivative of g(x), - give an antiderivative of f(x)+g(x). -

    -

    - -

    -     -     -     -     - - - Problems - - -

    - Evaluate the indefinite integral. - Don't forget your constant of integration! -

    -     - - - - - $a = random(2,9,1); - $n = random(2,9,1); - if($envir{problemSeed}==1){$a=3;$n=3}; - $f = Formula("$a x^$n"); - Context("Fraction"); - $b = Fraction($a,$n+1); - $F = FormulaUpToConstant("$b x^($n+1)"); - - -

    - \int \, dx -

    -

    - -

    -     -     -     - - - - - $n = random(2,9,1); - if($envir{problemSeed}==1){$n=9}; - $f = Formula("x^$n"); - Context("Fraction"); - $F = FormulaUpToConstant("1/($n+1) x^($n+1)"); - - -

    - \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(5,15,1); - $n = random(2,9,1); - $c = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=10;$n=2;$c=-2}; - $f = Formula("$a x^$n + $c")->reduce; - Context("Fraction"); - $b = Fraction($a,$n+1); - $F = FormulaUpToConstant("$b x^($n+1) + $c x")->reduce; - - -

    - \int \left(\right)\, dx -

    -

    - -

    -     -     -     - - - - - Context()->variables->are(t=>"Real"); - $F = FormulaUpToConstant("t"); - - -

    - \int dt -

    -

    - -

    -     -     -     - - - - - Context()->variables->are(s=>"Real"); - $F = FormulaUpToConstant("s"); - - -

    - \int 1\, ds -

    -

    - -

    -     -     -     - - - - - ($a,$n) = random_subset(2,2..9); - if($envir{problemSeed}==1){$a=3;$n=2}; - Context("Fraction"); - Context()->variables->are(t=>"Real"); - $f = Formula("1/($a t^$n)")->reduce; - $F = FormulaUpToConstant("-1/($a*($n-1) t^($n-1))")->reduce; - - -

    - \int \, dt -

    -

    - -

    -     -     -     - - - - - ($a,$n) = random_subset(2,2..9); - if($envir{problemSeed}==1){$a=3;$n=2}; - Context("Fraction"); - Context()->variables->are(t=>"Real"); - $f = Formula("$a/(t^$n)")->reduce; - ($num,$den) = Fraction($a,$n-1)->value; - $F = FormulaUpToConstant("-$num/($den t^($n-1))")->reduce; - - -

    - \int \, dt -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("2*sqrt(x)"); - - -

    - \int \frac{1}{\sqrt{x}}\, dx -

    -

    - -

    -     -     -     - - - - - $F = list_random('tan','cot','sec','csc'); - if($envir{problemSeed}==1){$F='tan';}; - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $F = Formula("$F(theta)"); - $f = $F->D('theta')->reduce; - $F = FormulaUpToConstant("$F"); - - -

    - \int \, d\theta -

    - - To enter \theta, type theta. - -

    - -

    -     -     -     - - - - - $F = list_random('-sin','-cos'); - if($envir{problemSeed}==1){$F='-cos';}; - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $F = Formula("$F(theta)"); - $f = $F->D('theta')->reduce; - $F = FormulaUpToConstant("$F"); - - -

    - \int \, d\theta -

    - - To enter \theta, type theta. - -

    - -

    -     -     -     - - - - - ($G,$H) = ('sec','csc','tan','cot'); - $c = list_random(-1,1); - if($envir{problemSeed}==1){$G='sec';$H=csc;$c=-1}; - $F = Formula("$G(x) + $c*$H(x)")->reduce; - $f = $F->D('x')->reduce; - $F = FormulaUpToConstant("$F"); - - -

    - \int \left(\right)\, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - if($envir{problemSeed}==1){$a=5;}; - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $F = Formula("$a e^(theta)"); - $f = $F->D('theta')->reduce; - $F = FormulaUpToConstant("$F"); - - -

    - \int \, d\theta -

    - - To enter \theta, type theta. - -

    - -

    -     -     -     - - - - - $b = random(2,9,1); - if($envir{problemSeed}==1){$b=3;}; - Context()->variables->are(t=>'Real'); - $f = Formula("$b^t"); - Context()->flags->set(reduceConstantFunctions=>0); - $F = FormulaUpToConstant("$b^t/ln($b)"); - - -

    - \int \, dt -

    -

    - -

    -     -     -     - - - - - ($b,$c) = random_subset(2,2..9); - if($envir{problemSeed}==1){$b=5;$c=2}; - Context()->variables->are(t=>'Real'); - $f = Formula("$b^t/$c"); - Context()->flags->set(reduceConstantFunctions=>0); - $F = FormulaUpToConstant("$b^t/($c ln($b))"); - - -

    - \int \, dt -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=2;$b=3}; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - $f = Formula("($a t + $b)^2")->reduce; - $A = Fraction($a**3,3*$a); - $B = Fraction(3*$a**2*$b,3*$a); - $C = Fraction(3*$a*$b**2,3*$a); - $D = Fraction($b**3,3*$a); - $F = FormulaUpToConstant("$A t^3 + $B t^2 + $C t + $D")->reduce; - - -

    - \int \, dt -

    -

    - -

    -     -     -     - - - - - $m = random(2,5,1); - ($a,$b) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$m=2;$a=3;$b=-2}; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - $n = $m+1; - $f = Formula("(t^$m+$a)(t^$n+$b t)")->reduce; - ($num2,$den2) = Fraction($b/($m+2) + $a/($n+1))->value; - ($num3,$den3) = Fraction($a*$b,2)->value; - $F = FormulaUpToConstant("t^($m+$n+1)/($m+$n+1) + $num2 t^($n+1)/$den2 + $num3 t^2/$den3")->reduce; - - -

    - \int \, dt -

    -

    - -

    -     -     -     - - - - - ($m,$n) = random_subset(2,2..9); - if($envir{problemSeed}==1){$m=2;$n=3}; - $f = Formula("x^$m x^$n")->reduce; - $F = FormulaUpToConstant("x^($m+$n+1)/($m+$n+1)")->reduce; - - -

    - \int \, dx -

    -

    - -

    -     -     -     - - - - - ($b,$p) = random_subset(2,'pi','e','sqrt(2)'); - if($envir{problemSeed}==1){$b='e';$p='pi'}; - Context()->flags->set(reduceConstants=>0); - $f = Formula("$b^$p")->reduce; - $F = FormulaUpToConstant("$b^$p x")->reduce; - - -

    - \int \, dx -

    -

    - -

    -     -     -     - - - - - $c = list_random('a'..'d','k','m','n','p'..'s'); - if($envir{problemSeed}==1){$c='a'}; - Context()->flags->set(reduceConstants=>0); - Context()->constants->add($c=>Real(exp(pi/sqrt(2)+sin(1)))); - $f = Formula("$c")->reduce; - $F = FormulaUpToConstant("$c x")->reduce; - - -

    - \int \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Consider the two integrals, \ds \int s^n\,ds and - \ds \int s^n\,dn. -

    -     - - -

    - What is the difference between these two indefinite integrals? -

    -     - - - -

    - The indefinite integrals have different differentials, - , different variables of integration. -

    -     -     - - -

    - Evaluate \ds \int s^n\,ds. -

    -     - - - -

    - \frac{s^{n+1}}{n+1}+C, n\neq -1 -

    -     -     - - -

    - Evaluate \ds \int s^n\,dn. -

    -     - - - -

    - \frac{s^{n}}{\ln s}+C, s>0 -

    -     -     -     - - - - -

    - This problem investigates why - states that \ds \int \frac1x\,dx = \ln\abs{x} + C. -

    -     - - -

    - What is the domain of y = \ln(x)? -

    -     -     - - -

    - Find \lzoo{x}{\ln(x) }. -

    -     -     - - -

    - What is the domain of y = \ln(-x)? -

    -     -     - - -

    - Find \lzoo{x}{\ln(-x)}. -

    -     -     - - -

    - You should find that 1/x has two types of antiderivatives, - depending on whether x \gt 0 or x\lt 0. - In one expression, give a formula for - \ds \int \frac{1}{x}\, dx that takes these different domains into account, - and explain your answer. -

    - - -     -     - -     - - - -

    - Find the function determined by the given initial value problem. -

    -     - - - - - $trig = list_random('-sin','-cos'); - $f0 = list_random(-9..-1,2..9); - if($envir{problemSeed}==1){$trig='-cos';$f0=2;}; - $F = Formula("$trig(x)"); - $f = $F->D('x'); - $F0 = $F->eval(x=>0); - $c = $f0 - $F0; - $F = Formula($F + $c)->reduce; - - -

    - \fp(x) = and f(0)= -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $b = random(2,9,1); - if($envir{problemSeed}==1){$a=5;$b=5;}; - $f0 = $a+$b; - $f = Formula("$a e^x"); - $F = Formula("$a e^x + $b"); - - -

    - \fp(x) = and f(0)= -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $b = non_zero_random(-9,9,1); - $m = random(1,2,1); - $x0 = non_zero_random(-2,2,1); - $f0 = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=4;$b=-3;$m=2;$x0=-1;$f0=9}; - $f = Formula("$a x^3 + $b x^$m")->reduce; - Context("Fraction"); - ($num1,$den1) = Fraction($a,4)->value; - ($num2,$den2) = Fraction($b,$m+1)->value; - $F = Formula("$num1 x^4/$den1 + $num2 x^($m+1)/$den2"); - $c = Fraction($f0 - $F->eval(x=>$x0)); - $F = Formula("$num1 x^4/$den1 + $num2 x^($m+1)/$den2 + $c")->reduce; - - -

    - \fp(x) = and f()= -

    -

    - -

    -     -     -     - - - - - $trig = list_random(['tan','pi/4'],['sec','pi/3'],['cot','pi/4'],['csc','pi/6']); - $f0 = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$trig=['tan','pi/4'];$f0=5;}; - $trf = $trig->[0]; - $x0 = $trig->[1]; - Context()->flags->set(reduceConstants=>0); - $x0 = Formula("$x0"); - $F = Formula("$trf(x)"); - $f = $F->D('x'); - $c = $f0 - $F->eval(x=>Real("$x0")); - $F = Formula("$F + $c"); - - -

    - \fp(x) = and f\mathopen{}\left(\right)\mathclose{}= -

    -

    - -

    -     -     -     - - - - - $b = random(2,9,1); - $f0 = random(1,9,1); - if($envir{problemSeed}==1){$b=7;$f0=1;}; - $f = Formula("$b^x"); - Context()->flags->set(reduceConstantFunctions=>0); - $F = Formula("$b^x/ln($b) - $b^2/ln($b) + $f0"); - - -

    - \fp(x) = and f(2) = -

    -

    - -

    -     -     -     - - - - - ($fpp,$fp0,$f0) = random_subset(3,2..9); - if($envir{problemSeed}==1){$fpp=5;$fp0=7;$f0=3;}; - Context("Fraction"); - $a = Fraction($fpp,2); - $F = Formula("$a x^2 + $fp0 x + $f0"); - - -

    - \fpp(x)= and \fp(0)=, f(0)= -

    -

    - -

    -     -     -     - - - - - ($fpp,$fp0,$f0) = random_subset(3,-10..-2,2..10); - if($envir{problemSeed}==1){$fpp=7;$fp0=-1;$f0=10;}; - Context("Fraction"); - $a = Fraction($fpp,6); - $a2 = Fraction($fpp,2); - $a3 = Fraction($fpp,3); - $b = Fraction($fp0 - $a2); - $c = Fraction($f0 + $a3 - $fp0); - $F = Formula("$a x^3 + $b x + $c"); - - -

    - \fpp(x)=x and \fp(1)=, f(1)= -

    -

    - -

    -     -     -     - - - - - $fpp = random(2,9,1); - ($fp0,$f0) = random_subset(2,-10..-2,2..10); - if($envir{problemSeed}==1){$fpp=5;$fp0=3;$f0=5;}; - $f = Formula("$fpp e^x"); - $F = Formula("$f + ($fp0-$fpp)x + ($f0-$fpp)")->reduce; - - -

    - \fpp(x)= and \fp(0)=, f(0)= -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','cos'); - $x0 = list_random(0,'pi/2','pi'); - ($fp0,$f0) = random_subset(2,2..9); - if($envir{problemSeed}==1){$trig='sin';$x0='pi';$fp0=2;$f0=4;}; - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $x0 = Formula("$x0"); - $f = Formula("$trig(theta)"); - $F = Formula("-$trig(theta)"); - $m = $fp0 - $F->D('theta')->eval(theta=>Real("$x0")); - $F = Formula("-$trig(theta) + $m theta + ($f0 + $trig($x0) - $m $x0 ) ")->reduce; - - -

    - \fpp(\theta)= and \fp\left(\right)=, f\left(\right)= -

    - - To enter \theta, type theta. - -

    - -

    -     -     -     - - - - - $a = random(10,30,1); - $m = random(2,4,1); - $b = random(2,9,1); - $trig = list_random('sin','cos'); - $c = list_random(1,-1); - ($fp0,$f0) = random_subset(2,-9..9); - if($envir{problemSeed}==1){$a=24;$m=2;$b=2;$c=-1;$trig='cos';$fp0=5;$f0=0;}; - Context("Fraction"); - Context()->flags->set(reduceConstants=>0); - ($num,$den) = Fraction($a,($m+1)*($m+2))->value; - $f = Formula("$a x^$m + $b^x + $c $trig(x)")->reduce; - $F = Formula("$num x^($m+2)/$den + $b^x/(ln($b))^2 - $c $trig(x)")->reduce; - $m = $fp0 - $F->D('x')->eval(x=>0); - $F = Formula("$F + $m x + ($f0 + $c*$trig(0)- 1/(ln($b))^2)")->reduce; - - -

    - \fpp(x)= and \fp(0)=, f(0)= -

    -

    - -

    -     -     -     - - - - - $x0 = non_zero_random(-5,5,1); - ($fp0,$f0) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$x0=1;$fp0=5;$f0=0;}; - $F = Formula("$fp0 x + ($f0 - $fp0*$x0)")->reduce; - - -

    - \fpp(x)=0 and \fp()=, f()= -

    -

    - -

    -     -     -     -     -     -     -
    -
    - The Definite Integral - - - - -

    - We start with an easy problem. - An object travels in a straight line at a constant velocity of - 5 - for 10 seconds. - How far away from its starting point is the object? -

    - -

    - We approach this problem with the familiar - \text{Distance } = \text{ Rate } \times \text{ Time} equation. - In this case, the distance traveled is - 5 - - 10 - = 50 feet. -

    - -

    - It is interesting to note that this solution of 50 feet can be represented - graphically. - Consider , - where the constant velocity of - 5 - is graphed on the axes. - Shading the area under the line from t=0 to t=10 gives a - rectangle with an area of 50 square units; - when one considers the units of the axes, - we can say this area represents - 50. -

    - - - -
    - The area under a constant velocity function corresponds to distance traveled - - - Graph of a linear function, the area under the curve is rectangular in shape. - - -

    - The y axis is drawn from 0 to 5 represents velocity in feets per second - and the x axis is drawn from 0 to 10 represents time in second. - The function y=5 for all values of x until 10, the function - is a straight line parallel to the x axis. - The area under the line to the x axis is shaded, and is drawn between x=0 and x=10. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - x label style={at={(axis description cs:0.85,0.05)},anchor=north}, - y label style={at={(axis description cs:0,.85)},rotate=90,anchor=south}, - xlabel={$t$ (s)}, - ylabel={$v$ (ft/s)}, - ytick={5}, - ymin=-.9,ymax=5.9, - xmin=-.9,xmax=10.9 - ] - \addplot [firstcurvestyle,areastyle,domain=0:10] {5} \closedcycle; - \addplot [firstcurvestyle,domain=0:10] {5}; - \end{axis} - \end{tikzpicture} - - -
    - -

    - Now consider a slightly harder situation (and not particularly realistic): - an object travels in a straight line with a constant velocity of - 5 - for 10 seconds, - then instantly reverses course at a rate of - 2 - for 4 seconds. - (Since the object is traveling in the opposite direction when reversing course, - we say the velocity is a constant - -2.) - How far away from the starting point is the object what is its - displacement? -

    - -

    - Here we use \text{Distance } = \text{ Rate}_1\, \times \text{ Time}_1\, + \text{ Rate}_2\, \times \text{ Time}_2, which is - - \text{ Distance } = 5\cdot10 + (-2)\cdot 4 = 42\text{ ft. } - -

    - -

    - Hence the object is 42 feet from its starting location. -

    - - - - - -

    - We can again depict this situation graphically. - In - we have the velocities graphed as straight lines on [0,10] and - [10,14], - respectively. - The displacement of the object is -

    - -
    -

    Area above the t-axis -Area below the t-axis,

    -
    - -

    - which is easy to calculate as 50-8=42 feet. -

    - -
    - The total displacement is the area above the t-axis minus the area below the t-axis - - - The graph has two areas shaded that are rectangular in shape. - - -

    - The y axis is drawn from 0 to 5 represents velocity in feets per second - and the x axis is drawn from 0 to 14 represents time in seconds. - There are two areas in the graph both rectangular in shape, - the bigger one lies in the first quadrant and the smaller one lies in the fourth quadrant. - The function y=5 is a straight line parallel to the x axis and creates the bigger area under graph, - and is drawn between x=0 and x=10. - The function y=-2, is a straight line parallel to the x axis and creates the smaller area, - and is drawn between x=10 and x = 14. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - x label style={at={(axis description cs:0.85,0.33)},anchor=south}, - y label style={at={(axis description cs:0.05,.75)},rotate=90,anchor=south}, - xlabel={$t$ (s)}, - ylabel={$v$ (ft/s)}, - ytick={-2,5}, - minor x tick num=4, - ymin=-2.9,ymax=5.9, - xmin=-.9,xmax=15.9 - ] - \addplot [firstcurvestyle,areastyle,domain=0:10] {5} \closedcycle; - \addplot [firstcurvestyle,domain=0:10] {5}; - \addplot [secondcurvestyle,areastyle,domain=10:14] {-2} \closedcycle; - \addplot [secondcurvestyle,domain=10:14] {-2}; - \end{axis} - \end{tikzpicture} - - -
    - -

    - Now consider a more difficult problem. -

    - - - Finding position using velocity - -

    - The velocity of an object moving straight up/down under the acceleration - of gravity is given as v(t) = -32t+48, - where time t is given in seconds and velocity is in - . - When t=0, the object had a height of - 0. -

    - -

    -

      -
    1. -

      - What was the initial velocity of the object? -

      -
      2. - -
    2. -

      - What was the maximum height of the object? -

      -
      3. - -
    3. -

      - What was the height of the object at time t=2? -

      -
      4. -
    -

    -     - -

    - It is straightforward to find the initial velocity; at time t=0, - - v(0) \amp =-32\cdot 0+48 - \amp = 48 - - The initial velocity was - 48. -

    - -

    - To answer questions about the height of the object, - we need to find the object's position function s(t). - This is an initial value problem, - which we studied in the previous section. - We are told the initial height is 0, , s(0) = 0. - We know s'(t) = v(t) = -32t+48. - To find s, we find the indefinite integral of v(t): - - s(t) \amp =\int v(t)\, dt - \amp = \int (-32t+48)\, dt - \amp = -16t^2+48t+C - . -

    - -

    - Since s(0) = 0, - we conclude that C=0 and s(t) = -16t^2+48t. -

    - -

    - To find the maximum height of the object, - we need to find the maximum of s. - Recalling our work finding extreme values, - we find the critical points of s by setting its derivative - (the velocity function) - equal to 0 and solving for t: - - 0 \amp = -32t+48 - t \amp =48/32 - \amp = 1.5\text{ s } - . -

    - -

    - (Notice how we ended up just finding when the velocity was 0ft/s!) - The first derivative test shows this is a maximum, - so the maximum height of the object is found at - - s(1.5) = -16(1.5)^2+48(1.5)=36\text{ ft } - . -

    - -

    - The height at time t=2 is now straightforward to compute: - - s(2) \amp =-16(2)^2+48(2) - \amp = 32 - . - The height is 32 - after 2 seconds. -

    - -

    - While we have answered all three questions - (using derivatives and antiderivatives), - let's look at them again graphically, - using the concepts of area that we explored earlier. -

    - -

    - - shows a graph of v(t) on axes from t=0 to t=3. - It is again straightforward to find v(0). - How can we use the graph to find the maximum height of the object? -

    - -
    - A graph of v(t)=-32t+48; the shaded areas help determine displacement - - - The graph of the function is a straight line that intersects the x axis. There are two triangular areas. - - -

    - The y axis is drawn from -40 to 40 and the x axis is drawn from 0 to 3. - The line starts from point (0, 48) and crosses the x axis at x=1.5, - the area under the curve forms a right angled triangle in the first quadrant with the positive x and y axes. - After crossing the x axis at x=1.5 the line goes down to point (3,-48). - Another right angle is formed on the x axis but in the fourth quadrant with the x axis and line x=3. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - ymin=-55,ymax=55, - xmin=-.5,xmax=3.2, - x label style={at={(axis description cs:0.85,0.3)},anchor=south}, - y label style={at={(axis description cs:0,.85)},rotate=90,anchor=south}, - xlabel={$t$ (s)}, - ylabel={$v$ (ft/s)}, - ] - \addplot [firstcurvestyle,areastyle,domain=0:3] {-32*x+48} \closedcycle; - \addplot [firstcurvestyle,domain=0:3] {-32*x+48}; - \end{axis} - \end{tikzpicture} - - -
    - -

    - Recall how in our previous work that the displacement of the object - (in this case, its height) - was found as the area under the velocity curve, as shaded in the figure. - Moreover, the area between the curve and the t-axis that is - below the t-axis counted as negative area. - That is, it represents the object coming back toward its starting position. - So to find the maximum distance from the starting point the - maximum height we find the area under the velocity line that - is above the t-axis, - , from t=0 to t=1.5. - This region is a triangle; its area is - - \text{ Area } \amp = \frac12\text{ Base } \times \text{ Height } - \amp =\frac12\times 1.5\text{ s } \times 48\text{ ft/s } - \amp = 36\text{ ft } - - which matches our previous calculation of the maximum height. -

    - -

    - Finally, to find the height of the object at time t=2 we - calculate the total signed area - (where some area is negative) - under the velocity function from t=0 to t=2. - This signed area is equal to s(2), - the displacement (, signed distance) from the starting position at - t=0 to the position at time t=2. - That is, -

    - -

    - Displacement = Area above the t-axis - Area below - t-axis. -

    - -

    - The regions are triangles, and we find - - \text{ Displacement } \amp = \frac12(1.5\text{s} )(48\text{ ft/s } ) - \frac12(0.5\text{s} )(16\text{ ft/s } ) - \amp = 32\text{ ft } - . -

    - -

    - This also matches our previous calculation of the height at t=2. -

    - -

    - Notice how we answered each question in this example in two ways. - Our first method was to manipulate equations using our understanding of - antiderivatives and derivatives. - Our second method was geometric: - we answered questions looking at a graph and finding the areas of - certain regions of this graph. -

    -     - -     - -

    - The above example does not prove - a relationship between area under a velocity function and displacement, - but it does imply a relationship exists. - - will fully establish fact that the area under a velocity function is - displacement. -

    - -

    - Given a graph of a function y=f(x), - we will find that there is great use in computing the area between the - curve y=f(x) and the x-axis. - Because of this, we need to define some terms. -

    - - - The Definite Integral, Total Signed Area - -

    - Let y=f(x) be defined on a closed interval [a,b]. - The total signed area from x=a to x=b under - f is: - integrationdefinite - definite integral - signed area - total signed area - integrationnotation - integrationarea -

    - -

    - (area under y=f(x) and above the x-axis on [a,b]) - - (area above y=f(x) and under the x-axis on [a,b]). -

    - -

    - The definite integral of f on [a,b] - is the total signed area of f on [a,b], denoted - - \int_a^b f(x)\, dx - , - where a and b are the - bounds of integration. -

    -     -     - - - - -

    - By our definition, the definite integral gives the - signed area under f. - We usually drop the word signed - when talking about the definite integral, - and simply say the definite integral gives - the area under f or, more commonly, - the area under the curve. -

    - -

    - The previous section introduced the indefinite integral, - which related to antiderivatives. - We have now defined the definite integral, - which relates to areas under a function. - The two are very much related, - as we'll see when we learn the Fundamental Theorem of Calculus in - . - Recall that earlier we said that the - \int symbol was an elongated S - that represented finding a sum. - In the context of the definite integral, - this notation makes a bit more sense, - as we are adding up areas under the function f. -

    - -

    - We practice using this notation. -

    - - - Evaluating definite integrals - -

    - Consider the function f given in . -

    - -
    - A graph of f(x) in - - - The graph of f(x) that shows the areas it forms with the x axis. - - -

    - The y axis is drawn between -1 to 1 and the x axis is drawn between 0 to 5. - There is a triangle in the first quadrant drawn on the x axis - with its base from x=0 to x=3 the peak of the triangle is at point (1,1). - The area in the first quadrant is the shaded portion inside the triangle. -

    -

    - The second area is a right angle triangle and is drawn on the x axis between x=3 and x=5. - The triangle lies in the fourth quadrant with its peak at point (5,-1) and the function forming the hypotenuse. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - ymin=-1.1,ymax=1.1, - xmin=-.5,xmax=5.5 - ] - \addplot [firstcurvestyle,areastyle] coordinates {(0,0) (1,1) (5,-1)} \closedcycle; - \addplot [firstcurvestyle,-] coordinates {(0,0) (1,1) (5,-1)}; - \end{axis} - \end{tikzpicture} - - -
    - -

    - Find: -

    - -

    -

      -
    1. \int_0^3 f(x)\, dx
      2. - -
    2. \int_3^5 f(x)\, dx
      3. - -
    3. \int_0^5 f(x)\, dx
      4. - -
    4. \int_0^3 5f(x)\, dx
      5. - -
    5. \int_1^1 f(x) \, dx
      6. -
    -

    -     - -

    -

      -
    1. -

      - \int_0^3 f(x)\, dx is the area under f on the - interval [0,3]. - This region is a triangle, - so the area is \int_0^3 f(x)\, dx=\frac12(3)(1) = 1.5. -

      -
      2. - -
    2. -

      - \int_3^5 f(x)\, dx represents the area of the triangle - found under the x-axis on [3,5]. - The area is \frac12(2)(1) = 1; - since it is found under - the x-axis, - this is negative area. Therefore \int_3^5 f(x)\, dx = -1. -

      -
      3. - -
    3. -

      - \int_0^5f(x)\, dx is the total signed area under f - on [0,5]. - This is 1.5 + (-1) = 0.5. -

      -
      4. - -
    4. -

      - \int_0^35f(x)\, dx is the area under 5f on [0,3]. - This is sketched in . - Again, the region is a triangle, - with height 5 times that of the height of the original triangle. - Thus the area is \int_0^35f(x)\, dx = \frac12(15)(1) = 7.5. -

      -
      5. - -
    5. -

      - \int_1^1f(x)\, dx is the area under f on the - interval [1,1]. - This describes a line segment, - not a region; it has no width. - Therefore the area is 0. -

      -
      6. -
    -

    -
    - - A graph of 5f in . - (Yes, it looks just like the graph of f in , just with a different y-scale.) - - - - The graph of f(x) that shows the areas it forms with the x axis. - - -

    - The y axis is drawn between -5 to 5 and the x axis is drawn between 0 to 5. - There is a triangle in the first quadrant drawn on the x axis with - its base from x=0 to x=3 the peak of the triangle is at point (1,5). - The area in the first quadrant is the shaded portion inside the triangle. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - ymin=-5.5,ymax=5.5, - xmin=-.5,xmax=5.5 - ] - \addplot [firstcurvestyle,areastyle] coordinates {(0,0) (1,5) (5,-5)} \closedcycle; - \addplot [firstcurvestyle,-] coordinates {(0,0) (1,5) (5,-5)}; - \end{axis} - \end{tikzpicture} - - -
    -     - -     - -

    - This example illustrates some of the properties of the definite integral, - given here. -

    - - - Properties of the Definite Integral - -

    - Let f and g be defined on a closed interval I that - contains the values a, - b and c, and let k be a constant. - The following hold: - integrationdefinite!properties - definite integralproperties -

    - -

    -

      -
    1. \int_a^a f(x)\, dx = 0
      2. - -
    2. \int_a^b f(x)\, dx + \int_b^c f(x)\, dx = \int_a^cf(x)\, dx
      3. - -
    3. \int_a^bf(x)\, dx = -\int_b^a f(x)\, dx
      4. - -
    4. \int_a^b\big(f(x)\pm g(x)\big)\, dx = \int_a^bf(x)\, dx \pm \int_a^bg(x)\, dx
      5. - -
    5. \int_a^bk\cdot f(x)\, dx = k\cdot\int_a^bf(x)\, dx
      6. -
    -

    -     -     - - - - - - -

    - We give a brief justification of here. -

    - -

    -

    -
  • - 1. -

    - As demonstrated in , - there is no area under the curve - when the region has no width; - hence this definite integral is 0. -

    -
    • - -
  • 2. -

    - This states that total area is the sum of the areas of subregions. - It is easily considered when we let a\lt b\lt c. - We can break the interval [a,c] into two subintervals, - [a,b] and [b,c]. - The total area over [a,c] is the area over [a,b] plus - the area over [b,c]. - It is important to note that this still holds true even if - a\lt b\lt c is not true. - We discuss this in the next point. -

    -
    • - -
  • 3. -

    - This property can be viewed a merely a convention to make other - properties work well. - (Later we will see how this property has a justification all its own, - not necessarily in support of other properties.) - Suppose b\lt a\lt c. - The discussion from the previous point clearly justifies - - \int_b^a f(x)\, dx + \int_a^c f(x)\, dx = \int_b^c f(x)\, dx - . - However, we still claim that, as originally stated, - - \int_a^b f(x)\, dx + \int_b^c f(x)\, dx = \int_a^c f(x)\, dx - . - How do Equations and - relate? - Start with Equation: - - \int_b^a f(x)\, dx + \int_a^c f(x)\, dx \amp = \int_b^c f(x)\, dx - \int_a^c f(x)\, dx \amp = -\int_b^a f(x)\, dx + \int_b^c f(x)\, dx - - Property (3) justifies changing the sign and switching the - bounds of integration on the \ds -\int_b^a f(x)\, dx term; - when this is done, Equations and - are equivalent. - The conclusion is this: - by adopting the convention of Property (3), Property (2) holds no - matter the order of a, - b and c. - Again, in the next section we will see another justification for this - property. -

    -
    • - -
  • - 4,5. -

    - Each of these may be non-intuitive. - Property (5) states that when one scales a function by, - for instance, 7, the area of the enclosed region also is scaled by a - factor of 7. - Both Properties (4) and (5) can be proved using geometry. - The details are not complicated but are not discussed here. -

    -
    • -
    -

    - - - Evaluating definite integrals using <xref ref="thm_defintprop"/> - -

    - Consider the graph of a function f(x) shown in - . -

    - -
    - A graph of a function in - - - The graph of function in this example. - - -

    - The x and the y axes are uncalibrated, there are three positions a, - b and c in order on the x axis where the inflection changes. - The curve in the first quadrant is smaller, the area under the curve forms a bell jar shape - it extends from x=a to x=b. - The other curve is twice in height as the first and lies in the fourth quadrant. - It is also bell jar shaped, but is inverted on the positive x axis and extends between x=b to x=c. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={.5,1.5,3}, - extra x tick labels={$a$,$b$,$\ c$}, - ytick=\empty, - ymin=-1.1,ymax=1.1, - xmin=-.2,xmax=3.5 - ] - \addplot [firstcurvestyle,areastyle,domain=0.5:3] {(x-.5)*(x-1.5)*(x-3)} \closedcycle; - \addplot [firstcurvestyle,domain=0.5:3,samples=40] {(x-.5)*(x-1.5)*(x-3)}; - \end{axis} - \end{tikzpicture} - - -
    - -

    - Answer the following: -

    - -

    -

      -
    1. -

      - Which value is greater: - \ds \int_a^b f(x)\, dx or \ds \int_b^c f(x)\, dx? -

      -
      2. - -
    2. -

      - Is \ds \int_a^c f(x)\, dx greater or less than 0? -

      -
      3. - -
    3. -

      - Which value is greater: - \ds \int_a^b f(x)\, dx or \ds \int_c^b f(x)\, dx? -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - \int_a^b f(x)\, dx has a positive value - (since the area is above the x-axis) - whereas \int_b^c f(x)\, dx has a negative value. - Hence \int_a^b f(x)\, dx is bigger. -

      -
      2. - -
    2. -

      - \int_a^c f(x)\, dx is the total signed area under f - between x=a and x=c. - Since the region below the x-axis looks to be larger than - the region above, - we conclude that the definite integral has a value less than 0. -

      -
      3. - -
    3. -

      - Note how the second integral has the bounds reversed. - Therefore \int_c^b f(x)\, dx=-\int_b^c f(x)\, dx represents - a positive number, - greater than the area described by the first definite integral. - Hence \int_c^b f(x)\, dx is greater. -

      -
      4. -
    -

    -     -     - -

    - The area definition of the definite integral allows us to use geometry to - compute the definite integral of some simple functions. -

    - - - Evaluating definite integrals using geometry - -

    - Evaluate the following definite integrals: - - 1. \, \int_{-2}^5 (2x-4)\, dx \qquad 2.\, \int_{-3}^3 \sqrt{9-x^2}\, dx - . -

    -     - -

    -

      -
    1. -

      - It is useful to sketch the function in the integrand, - as shown in . - We see we need to compute the areas of two regions, - which we have labeled R_1 and R_2. - Both are triangles, so the area computation is straightforward: - - R_1: \frac12(4)(8) = 16 \qquad R_2: \frac12(3)6 = 9 - . - Region R_1 lies under the x-axis, - hence it is counted as negative area - (we can think of the triangle's height as being -8), - so - - \int_{-2}^5(2x-4)\, dx = -16+9 = -7 - . -

      -
      2. - -
    2. -

      - Recognize that the integrand of this definite integral describes - a half circle, - as sketched in , - with radius 3. - Thus the area is: - - \int_{-3}^3 \sqrt{9-x^2}\, dx = \frac12\pi r^2 = \frac 92\pi - . -

      -
      3. -
    -

    - - -
    - f(x) = 2x-4 - - - Graph of function of a positive sloped straight line along with the areas it forms on the x axis. - - -

    - The y axis is drawn from -10 to 10 and the - x axis is drawn from -3 to 5. - The function is a straight line that starts from point (-2,-8) and ends at point (5,6). - The function intersects the x axis at x=2. -

    -

    - The function, below the x axis creates a right angled triangle - with the x axis and line x=-2 from y=0 to y=-8, - with its base from x=-2 to x=2. This area is named R1. -

    -

    - The function above the x axis forms a right angled triangle with the x axis and line x=5 - from y=0 to y=6, with its base from x=2 and x=5. This area is named R2. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - ymin=-11,ymax=11, - xmin=-3.5,xmax=5.5 - ] - \addplot [firstcurvestyle,areastyle,domain=-2:5] {2*x-4} \closedcycle; - \addplot [firstcurvestyle,domain=-2:5] {2*x-4}; - \addplot [soliddot] coordinates {(-2,-8)} node[right] {\small $(-2,-8)$}; - \addplot [soliddot] coordinates {(5,6)} node[left] {\small $(5,6)$}; - \draw (axis cs:-1,-2) node { $R_1$}; - \draw (axis cs:4.2,2) node { $R_2$}; - \end{axis} - \end{tikzpicture} - - -
    -
    - f(x) = \sqrt{9-x^2} - - - The graph area under the curve is a semicircle with radius 3 on the x axis with centre at origin. - - - The graph shows the area under the curve that is a semicircle with radius 3 on the x axis - with centre at origin. It lies on the first and the second quadrant. - - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - xtick={-3,3}, - extra y ticks={1,2,3,4}, - extra y tick labels=\empty, - ytick={5}, - ymin=-.5,ymax=5.5, - xmin=-3.5,xmax=3.5] - \addplot [firstcurvestyle,areastyle,domain=0:180] ({3*cos (x)},{3*sin(x)}) \closedcycle; - \addplot [firstcurvestyle,domain=0:180,samples=40] ({3*cos (x)},{3*sin(x)}); - \end{axis} - \end{tikzpicture} - - -
    -     -     - -     - - - Understanding motion given velocity - -

    - Consider the graph of a velocity function of an object moving in a - straight line, - given in , - where the numbers in the given regions gives the area of that region. - Assume that the definite integral of a velocity function gives displacement. - Find the maximum speed of the object and its maximum displacement from - its starting position. -

    - -
    - A graph of a velocity in - - - The graph of velocity for this example. - - -

    - The y axis is drawn from -10 to 15 and - it represents velocity in feets per second, the x axis is uncalibrated and represents time in seconds. - There are three points on the x axis marked a, b and c in that order. -

    -

    - In the fourth quadrant, on the x axis from 0 to a, a parabola is drawn with its peak at - y=-10. The number 11 is written inside the shaded portion. -

    -

    - From a to b there is a second parabola in the first quadrant with its peak at y= 15, - the number 38 is written inside it. -

    -

    - From b to c the third parabola is located in the fourth quadrant with its peak at x=-10, - the number 11 is written inside it, indicating same size as the first parabola. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - x label style={at={(axis description cs:0.85,0.35)},anchor=south}, - y label style={at={(axis description cs:-.05,.85)},rotate=90,anchor=south}, - xlabel={$t$ (s)}, - ylabel={$v$ (ft/s)}, - xtick=\empty, - extra x ticks={2,6,8}, - extra x tick labels={$\ a$,$b\ $,$\ c$}, - ytick={-5,5,10,15}, - ymin=-9,ymax=17, - xmin=-.5,xmax=8.5 - ] - \addplot [firstcurvestyle,areastyle,domain=0:8,samples=40] {15/64*x*(x-2)*(x-6)*(x-8)} \closedcycle; - \addplot [firstcurvestyle,domain=0:8,samples=70] {15/64*x*(x-2)*(x-6)*(x-8)}; - \draw (axis cs:.9,-4) node { $11$}; - \draw (axis cs:7.1,-4) node { $11$}; - \draw (axis cs:4,8) node { $38$}; - \end{axis} - \end{tikzpicture} - - -
    -     - -

    - Since the graph gives velocity, - finding the maximum speed is simple: - it looks to be 15ft/s. -

    - -

    - At time t=0, the displacement is 0; - the object is at its starting position. - At time t=a, the object has moved backward 11 feet. - Between times t=a and t=b, - the object moves forward 38 feet, - bringing it into a position 27 feet forward of its starting position. - From t=b to t=c the object is moving backwards again, - hence its maximum displacement is 27 feet from its starting position. -

    -     - -     - -

    - In our examples, - we have either found the areas of regions that have nice geometric shapes - (such as rectangles, triangles and circles) - or the areas were given to us. - Consider , - where a region below y=x^2 is shaded. - What is its area? - The function y=x^2 is relatively simple, - yet the shape it defines has an area that is not simple to find geometrically. -

    - -
    - What is the area below y=x^2 on [0,3]? The region is not a usual geometric shape. - - - Graph of function described above between x=0 and x=3. - - -

    - The y axis is drawn from 0 to 10 and the x axis is drawn from 0 to 3. - The curve is drawn in the first quadrant, it starts at the origin and - increases gently until x=1 and steeply from x=1 to x=3 and ends at point (3,9). -

    -

    - The area under the curve roughly looks like a right angled triangle. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - xtick={1,2,3}, - ymin=-1,ymax=11, - xmin=-1,xmax=3.5 - ] - \addplot [firstcurvestyle,areastyle,domain=0:3] {x^2} \closedcycle; - \addplot [firstcurvestyle,domain=0:3] {x^2}; - \end{axis} - \end{tikzpicture} - - -
    - -

    - In - we will explore how to find the areas of such regions. -

    - - - - Terms and Concepts - - - - -

    - What is total signed area? -

    - -     - - - -     - - - - -

    - What is displacement? -

    - -     - - - -     - - - - -

    - What is \ds \int_3^3 \sin(x) \, dx? -

    -

    - -

    -     - - - - - - - -

    - Notice that the upper and lower bounds of the integral are equal. -

    -     -     -     -     - -     - - - -

    - Give a single definite integral that has the same value as - I = \int_0^1(2x+3)\, dx + \int_1^2 (2x+3)\, dx. -

    -     - - - -

    - \int 0^2 (2x+3)\, dx -

    -     -     -     - - - Problems - - -

    - A graph of a function f(x) is given. - Using the geometry of the graph, - evaluate the definite integrals. -

    -     - - - - - @a = (3,4,3,0,-4,9); - - - - - - The graph of the function is a decreasing straight line drawn in the first and fourth quadrant. - - -

    - The y axis is drawn from -4 to 4 and the x axis is drawn from 0 to 4. - The function y=-2x+4 is a line that decreases from point (0,4) to point (4,-4). - The function is linear hence the graph is a straight line between the two points. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - ymin=-4.5, - ymax=4.5, - xmin=-.9, - xmax=4.5 - ] - \addplot+ [domain=0:4] {-2*x+4}; - \draw (axis cs:2,2.5) node { \(y = -2x+4\)}; - \end{axis} - \end{tikzpicture} - - -     - - - -

    - \int_0^1 (-2x+4)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^2 (-2x+4)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^3 (-2x+4)\, dx -

    -

    - -

    -     -     - - -

    - \int_1^3 (-2x+4)\, dx -

    -

    - -

    -     -     - - -

    - \int_2^4 (-2x+4)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^1 (-6x+12)\, d -

    -

    - -

    -     -     -     -     - - - - - @a = (-4,-5,-3,1,-2,10); - - - - - - Graph of function f(x). - - -

    - The y axis is drawn from -2 to 2 and the x axis is drawn from 0 to 5. - The graph of function is drawn in the fourth and the first quadrants. In the fourth quadrant, - the function is a straight line y=-2, from x=0 to x=2. - Then the function increases steeply and meets the x axis at x=3. - After intersecting the x axis at 3, - the function assumes a gentler slope and increases until point (5,2). -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - xtick={1,2,3,4,5}, - ymin=-2.2, - ymax=2.2, - xmin=-.9, - xmax=5.5 - ] - \addplot+ [domain=0:2] {-2}; - \addplot [firstcurvestyle,domain=2:3] {2*x-6}; - \addplot [firstcurvestyle,domain=3:5] {x-3}; - \draw (axis cs:3.2,-1.5) node { \(y = f(x)\)}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - \int_0^2 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^3 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^5 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_2^5 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_5^3 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^3 -2f(x)\, dx -

    -

    - -

    -     -     -     -     - - - - - @a = (4,2,4,2,1,2); - - - - - - The graph has two triangles from 0 to 2 and from 2 to 4, - the first one is twice the height of the second. - - -

    - The x axis is drawn from 0 to 4 and the y axis - is drawn from 0 to 4. - The function increases from the origin to peak at (1,4) - then it decreases sharply to point (2,0). - Then it increases again, gently to the second peak at (3,2) - after which it decreases to point (4,0). -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - xtick={1,2,3,4,5}, - ymin=-.5, - ymax=4.5, - xmin=-.9, - xmax=4.5 - ] - \addplot+ [domain=0:1] {4*x}; - \addplot [firstcurvestyle,domain=1:2] {-4*(x-2)}; - \addplot [firstcurvestyle,domain=2:3] {2*(x-2)}; - \addplot [firstcurvestyle,domain=3:4] {-2*(x-4)}; - \draw (axis cs:3,2.5) node { \(y = f(x)\)}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - \int_0^2 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_2^4 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_2^4 2f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^1 4x\, dx -

    -

    - -

    -     -     - - -

    - \int_2^3 (2x-4)\, dx -

    -

    - -

    -     -     - - -

    - \int_2^3 (4x-8)\, dx -

    -

    - -

    -     -     -     -     - - - - - Context("Fraction"); - @a = (Fraction(-1/2),Fraction(0),Fraction(3/2),Fraction(3/2),Fraction(9/2),Fraction(15/2)); - - - - - - The graph of the function is a straight line with a positive slope. - - -

    - The y axis is drawn from -1 to 3 and the x axis - is drawn from 0 to 4. - The graph of function y=x-1 is a straight line that - increases from point (0,-1) to point (4,3), - and it crosses the x axis at x=1. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - xtick={1,2,3,4,5}, - ytick={-1,1,2,3}, - ymin=-1.5, - ymax=3.5, - xmin=-.9, - xmax=4.5 - ] - \addplot+ [domain=0:4] {x-1}; - \draw (axis cs:2,2.5) node { \(y = x-1\)}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - \int_0^1 (x-1)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^2 (x-1)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^3 (x-1)\, dx -

    -

    - -

    -     -     - - -

    - \int_2^3 (x-1)\, dx -

    -

    - -

    -     -     - - -

    - \int_1^4 (x-1)\, dx -

    -

    - -

    -     -     - - -

    - \int_1^4 \big((x-1)+1\big)\, dx -

    -

    - -

    -     -     -     -     - - - - - @a = (Formula("pi"),Formula("pi"),Formula("2 pi"),Formula("10 pi")); - - - - - - Graph of function that is a semicircle on the x axis. - - -

    - The y axis is drawn from 0 to 3 and the x axis is - drawn from 0 to 4. - The graph of function f(x) = \sqrt{4-(x-2)^2} - is a semi circle drawn on the x axis with centre at point - (2,0) and radius of 2. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - xtick={1,2,3,4}, - ytick={1,2,3}, - ymin=-.5, - ymax=3.5, - xmin=-.9, - xmax=4.5 - ] - \addplot+ [samples=40,domain=0:3.14159] ({2*cos(deg(x))+2}, {2*sin(deg(x))}); - \draw (axis cs:3,2.4) node { \(f(x) = \sqrt{4-(x-2)^2}\)}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - \int_0^2 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_2^4 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^4 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^4 5f(x)\, dx -

    -

    - -

    -     -     -     -     - - - - - Context()->variables->add(a=>'Real',b=>'Real'); - @a = (15,12,0,Formula("3(b-a)")); - - - - - - The graph of a function is a straight line parallel to the x axis. - - -

    - The y axis is drawn from 0 to 4 and - the x axis is drawn from 0 to 10. - The graph of function f(x) = 3 is a straight - line parallel to the x axis and goes from x=0 to x=10. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - ytick={1,2,3}, - ymin=-.5, - ymax=5, - xmin=-0.5, - xmax=10.5 - ] - \addplot+[domain=0:10] ({x}, {3}); - \draw (axis cs:5,3.4) node[] { \(f(x) = 3\)}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - \int_0^5 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_3^7 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^0 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \ds \int_a^b f(x)\, dx, where - 0\leq a\leq b\leq 10 -

    -

    - -

    -     -     -     -     -     - - - -

    - A graph of a function f(x) is given; - the numbers inside the shaded regions give the area of that region. - Evaluate the definite integrals using this area information. -

    -     - - - - - @a = (-59,-48,-27,-33); - - - - - Graph of function f(x) for this exercise. - - -

    - The y axis is drawn from -100 to 50 and the - x axis is drawn from 0 to 3. The function - forms three parabolic shapes. The areas inside the parabolas - until the x axis are shaded. The biggest one lies in the - fourth quadrant and the other two in the first. -

    -

    - Between x=0 to x=1 the highest parabola is formed - with peak at y=-100, it is labeled with number 59. -

    -

    - Between x=1 to x=2 the smallest parabola is formed - with the peak at nearly y=20, it is labeled with number 11. -

    -

    - Between x=2 to x=3 the third parabola is formed with - the peak at nearly y=45, it is labeled with number 21. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - xtick={1,2,3}, - ymin=-105, - ymax=55, - xmin=-.9, - xmax=3.5 - ] - \addplot [firstcurvestyle,areastyle,domain=0:3,samples=40] {60*x*(1-x)*(x-2)*(x-2)*(x-3)} \closedcycle; - \addplot [firstcurvestyle,domain=0:3,samples=90] {60*x*(1-x)*(x-2)*(x-2)*(x-3)}; - \draw (axis cs:1.5,30) node { \(y=f(x)\)}; - \draw (axis cs:.45,-10) node { \(59\)}; - \draw (axis cs:1.4,10) node { \(11\)}; - \draw (axis cs:2.6,10) node { \(21\)}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - \int_0^1 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^2 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^3 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_1^2 -3f(x)\, dx -

    -

    - -

    -     -     -     -     - - - - - @a = (Formula("4/pi"),Formula("-4/pi"),Formula("0"),Formula("2/pi")); - - - - - - Graph of sine function with angle as multiple of pi/2. - - -

    - The y axis is drawn from -1 to 1 and - the x axis is drawn from 0 to 4. The function - crosses the x axis at 0, 2 and 4. -

    -

    - There is a downward facing parabola between x=0 and x=2 with - peak at point (1,1) and it lies in the first quadrant. - The shaded portion inside this downward facing parabola until the x axis is labeled 4/\pi. - From x=2 and x=4 the secong prabola is present - in the fourth quadrant. - The shaded portion inside this parabola until the x axis is labeled 4/\pi. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ymin=-1.5, - ymax=1.5, - xmin=-.9, - xmax=4.5 - ] - \addplot [firstcurvestyle,areastyle,domain=0:4] {sin(deg(3.14159*x/2))} \closedcycle; - \addplot [firstcurvestyle,domain=0:4,samples=50] {sin(deg(3.14159*x/2))}; - \draw (axis cs:3,0.75) node { \(f(x)=\sin(\pi x/2)\)}; - \draw (axis cs:1,.6) node { \(4/\pi\)}; - \draw (axis cs:3,-.6) node { \(4/\pi\)}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - \int_0^2 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_2^4 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^4 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^1 f(x)\, dx -

    -

    - -

    -     -     -     -     - - - - - @a = (4,4,-4,-2); - - - - - - Graph of function for this exercise that represents a parabola. - - -

    - The y axis is drawn from -5 to 10 and the x axis - is drawn from -2 to 2. - The function f(x) = 3x^2 -3 is parabolic, with vertex at point (0,-3). - It has x intercepts at x=-1 and x=1. -

    -

    - There are three distinct shaded regions, from x=-1 to x=-2 and from - x=1 to x=2 the area under the arms of the parabola to the x - axis are shaded and both are labeled with number 4. -

    -

    - From x=-1 to x=1, the area from above the parabola to the x axis - is shaded and labeled as -4. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - xtick={-2,-1,1,2}, - ymin=-5.5, - ymax=11, - xmin=-2.5, - xmax=2.5 - ] - \addplot [firstcurvestyle,areastyle,domain=-2:2] {3*x^2-3} \closedcycle; - \addplot [firstcurvestyle,domain=-2:2,samples=40] {3*x^2-3}; - \draw (axis cs:1.5,10) node { \(f(x)=3x^2-3\)}; - \draw (axis cs:-1.5,1) node { \(4\)}; - \draw (axis cs:1.5,1) node { \(4\)}; - \draw (axis cs:.3,-1) node { \(-4\)}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - \int_{-2}^{-1} f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_1^2 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_{-1}^1 f(x)\, dx -

    -

    - -

    -     -     - - -

    - \int_0^1 f(x)\, dx -

    -

    - -

    -     -     -     -     - - - - - Context("Fraction"); - @a = (Fraction("40/3"),Fraction("26/3"),Fraction("8/3"),Fraction("38/3")); - - - - - - Graph of function is a curve with a positive slope. - - -

    - The y axis is drawn from 0 to - 4 and the x axis is drawn from 0 to 2. - The function f(x)=x^2 is drawn in the first quadrant, - it starts at the origin and gently curves up to point (1,1) - then steeply to point (2,4). The area under the curve is - shaded from x=0 to x=2. It forms a roughly right-angled - triangle shape. -

    -

    - From x=0 to x=1, the number 1/3 is written - inside the shaded region, and from x=1 to x=2, - the number 7/3 is written. -

    - -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - xtick={1,2}, - ymin=-.1, - ymax=4.2, - xmin=-.2, - xmax=2.1 - ] - \addplot [firstcurvestyle,areastyle,domain=0:2] {x^2} \closedcycle; - \addplot [firstcurvestyle,domain=0:2] {x^2}; - \draw [thick,firstcolor,dashed] (axis cs:1,0) -- (axis cs:1,1); - \draw (axis cs:1,3) node { \(f(x)=x^2\)}; - \draw (axis cs:.7,.2) node { \(1/3\)}; - \draw (axis cs:1.5,.2) node { \(7/3\)}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - \int_{0}^{2} 5x^2\, dx -

    -

    - -

    -     -     - - -

    - \int_0^2 (x^2+3)\, dx -

    -

    - -

    -     -     - - -

    - \int_{1}^3 (x-1)^2\, dx -

    -

    - -

    -     -     - - -

    - \int_2^4 \big((x-2)^2+5\big)\, dx -

    -

    - -

    -     -     -     -     -     - - - -

    - A graph is given of the velocity function of an object moving in a straight line. - Answer the questions based on the graph. -

    -     - - - - - $mv = NumberWithUnits("2 ft/s"); - $md = NumberWithUnits("2 ft"); - $td = NumberWithUnits("1.5 ft"); - - - - - - Graph of function for this exercise. It is a straight line with a negative slope. - - -

    - The y axis is drawn from -1 to 3 and represents velocity in - feets per second and the x axis is drawn from 0 to 3 and - it represents time in seconds. - The graph of the function is an inclined line. It starts from point (0,2) - then decreases in the first quadrant, crosses the x axis at x=2 - and continues decreasing in the fourth quadrant. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - xtick={1,2,3}, - ymin=-1.1, - ymax=3, - xmin=-.2, - xmax=3.1, - xlabel={\(t\) (s)}, - ylabel={\(v\) (ft/s)} - ] - \addplot+ [domain=0:3] {-x+2}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - What is the object's maximum velocity? -

    -

    - -

    -     -     - - -

    - What is the object's maximum displacement? -

    -

    - -

    -     -     - - -

    - What is the object's total displacement on [0,3]? -

    -

    - -

    -     -     -     -     - - - - - $mv = NumberWithUnits("3 ft/s"); - $md = NumberWithUnits("9.5 ft"); - $td = NumberWithUnits("9.5 ft"); - - - - - - The graph of function for this example. - - -

    - The y axis is drawn from 0 to 4 and represents velocity - in feets per second and the x axis is drawn from 0 to 5 - and it represents time in seconds. - The graph is a straight line y=3 from x=0 to x=1 then - it decreases to point (2,0). The function then increases to (3,2) - then it becomes a straight line y=2. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - xtick={1,2,3,4,5}, - ymin=-.5, - ymax=4, - xmin=-.2, - xmax=5.3, - xlabel={\(t\) (s)}, - ylabel={\(v\) (ft/s)} - ] - \addplot+ [domain=0:1] {3}; - \addplot [firstcurvestyle,domain=1:2] {-3*(x-2)}; - \addplot [firstcurvestyle,domain=2:3] {2*(x-2)}; - \addplot [firstcurvestyle,domain=3:5] {2}; - \end{axis} - \end{tikzpicture} - - -     - - -

    - What is the object's maximum velocity? -

    -

    - -

    -     -     - - -

    - What is the object's maximum displacement? -

    -

    - -

    -     -     - - -

    - What is the object's total displacement on [0,5]? -

    -

    - -

    -     -     -     -     -     - - - - - $mv = NumberWithUnits("64 ft/s"); - $md = NumberWithUnits("64 ft"); - $md_t = NumberWithUnits("2 s"); - $h0_t = FormulaWithUnits("2+sqrt(7) s"); - - -

    - An object is thrown straight up with a velocity, - in ft/s, given by v(t) = -32t+64, - where t is in seconds, from a height of 48 feet. -

    -     - - -

    - What is the object's maximum velocity? -

    -

    - -

    -     -     - - -

    - What is the object's maximum displacement? -

    -

    - -

    -     -     - - -

    - When does the maximum displacement occur? -

    -

    - -

    -     -     - - -

    - When will the object reach a height of 0? (Hint: - find when the displacement is -48ft.) -

    -

    - -

    -     -     -     -     - - - - - $v0 = NumberWithUnits("96 ft/s"); - $d = NumberWithUnits("6 s"); - $h0_2 = NumberWithUnits("6 s"); - $hmax = NumberWithUnits("-16*3^2+96*3+64 ft"); - - -

    - An object is thrown straight up with a velocity, - in ft/s, given by v(t) = -32t+96, - where t is in seconds, from a height of 64 feet. -

    -     - - -

    - What is the object's initial velocity? -

    -

    - -

    -     -     - - -

    - When is the object's displacement 0? -

    -

    - -

    -     -     - - -

    - How long does it take for the object to return to its initial height? -

    -

    - -

    -     -     - - -

    - What is the maximum height the object reaches? -

    -

    - -

    -     -     -     -     - - - -

    - The values of several definite integrals are given as follows: - \int_0^2f(x)\,dx = 5 \quad \int_0^3f(x)\,dx = 7 \quad \int_0^2g(x)\,dx = -3 \quad \int_2^3g(x)\,dx = 5 - Use these values and properties of definite integrals to evaluate the indicated definite integral. -

    -     - - - - -

    - \int_0^2 \big(f(x)+g(x)\big) \, dx -

    -

    - -

    -     -     -     - - - - -

    - \int_0^3 \big(f(x)-g(x)\big) \, dx -

    -

    - -

    -     -     -     - - - - -

    - \int_2^3 \big(3f(x)+2g(x)\big) \, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->variables->set(a=>"Real",b=>"Real"); - parser::Assignment->Allow; - $a = Formula("a = -2/7*b"); - - -

    - Find a formula for a in terms of b such that - \ds \int_0^3 \big(af(x)+bg(x)\big) \, dx=0. -

    -

    - -

    -     -     -     -     - - - -

    - The values of several definite integrals are given as follows: - \int_0^3s(t)\,dt = 10 \quad \int_3^5s(t)\,dt = 8 \quad \int_3^5r(t)\,dt = -1 \quad \int_0^5r(t)\,dt = 11 - Use these values and properties of definite integrals to evaluate the indicated definite integral. -

    -     - - - - -

    - \int_0^3 \big(s(t) + r(t)\big)\, dt -

    -

    - -

    -     -     -     - - - - -

    - \int_5^0 \big(s(t) - r(t)\big)\, dt -

    -

    - -

    -     -     -     - - - - -

    - \int_3^3 \big(\pi s(t) - 7r(t)\big)\, dt -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->variables->set(a=>"Real",b=>"Real"); - parser::Assignment->Allow; - $a = Formula("a = -18/11*b"); - - -

    - Find a formula for a in terms of b such that - \ds \int_0^5 \big(ar(t)+bs(t)\big) \, dt=0. -

    -

    - -

    -     -     -     -     -     -     -
    -
    - Riemann Sums - - - -

    - In the previous section we defined the definite integral of a function on [a,b] to be the signed area between the curve and the x-axis. - Some areas were simple to compute; - we ended the section with a region whose area was not simple to compute. - In this section we develop a technique to find such areas. - Riemann Sum -

    - -

    - A fundamental calculus technique is to first answer a given problem with an approximation, - then refine that approximation to make it better, - then use limits in the refining process to find the exact answer. - That is what we will do here. -

    -     - - - Approximating area with rectangles - -

    - Consider the region given in , - which is the area under y=4x-x^2 on [0,4]. - What is the signed area of this region , what is \int_0^4(4x-x^2)\, dx? -

    - -
    - A graph of f(x) = 4x-x^2. What is the area of the shaded region? - - - A shaded downward opening parabola in the first quadrant. - -

    - A parabola with the equation y = 4x -x^2. - It is a downward opening parabola lying in the first quadrant. - It has a maximum of (2,4), and crosses the x-axis at x=0 and x = 4. - The area between the curve and the x-axis is shaded on the interval [0,4]. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4}, - ymin=-1,ymax=5, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; - \addplot [firstcurvestyle,domain=-0.2:4.2,samples=45] {4*x-x^2}; - - \end{axis} - \end{tikzpicture} - - - - -
    - -

    - We start by approximating. - We can surround the region with a rectangle with height and width of 4 and find the area is approximately 16 square units. - This is obviously an over-approximation; - we are including area in the rectangle that is not under the parabola. -

    - -
    - Approximating area under a curve with one rectangle - - The previously described parabola with a large rectangle containing the shaded area. - -

    - The curve shown in , with a rectangle entirely containing the shaded area. - The rectangle has a height and width of 4. - On the left and right sides of the parabola, area is included in the rectangle which is not shaded. - The rectangle clearly contains a greater area than is under the parabola. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4}, - ymin=-1.2,ymax=5, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2} \closedcycle; - \addplot [firstcurvestyle,domain=-.2:4.2,samples=40] {4*x-x^2} node [pos=1, left] { $y=4x-x^2$}; - - \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:4,4); - - \end{axis} - \end{tikzpicture} - - - - -
    - -

    - We have an approximation of the area, using one rectangle. - How can we refine our approximation to make it better? - The key to this section is this answer: - use more rectangles. -

    - -

    - Let's use four rectangles with an equal width of 1. - This partitions the interval [0,4] into 4 subintervals, - [0,1], [1,2], [2,3] and [3,4]. - On each subinterval we will draw a rectangle. -

    - -

    - There are three common ways to determine the height of these rectangles: - the Left Hand Rule, - the Right Hand Rule, - and the Midpoint Rule. - The Left Hand Rule says to evaluate the function at the left-hand endpoint of the subinterval and make the rectangle that height. - In , - the rectangle drawn on the interval [2,3] has height determined by the Left Hand Rule; - it has a height of f(2). (The rectangle is labeled LHR.) - Left Hand Rule - Right Hand Rule - Midpoint Rule -

    - -
    - Approximating \int_0^4(4x-x^2)\, dx using rectangles. The heights of the rectangles are determined using different rules. - - - The area of a downward opening parabola being approximated by 4 rectangles. - -

    - The parabola shown in . The shaded are is being approximated by 4 rectangles of width 1. - The left most rectangle has a height of 3 and is labeled "RHR," for "Right Hand Rule." - The height of the rectangle is equal to the height of the parabola om the rightmost side, where x = 1. - The second rectangle has a height of 3.75, and is labeled "MPR," for "Midpoint Rule." - The height of the rectangle is equal to the height of the parabola at the center of the rectangle, where x=1.5. - The third rectangle had a height of 4, and is labeled "LHR," for "Left Hand Rule." - The height of the rectangle is given by the height of the parabola on the leftmost side, where x = 2. - The rightmost rectangle has a height of 2.6, and is labeled "other". - The height is given by the height of the parabola at a point slightly to the right of the center of the rectangle. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4}, - ymin=-1,ymax=5, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; - \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; - - \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:1,3); - \draw [thick,secondcolor] (axis cs:1,0) rectangle (axis cs:2,3.75); - \draw [thick,secondcolor] (axis cs:2,0) rectangle (axis cs:3,4); - \draw [thick,secondcolor] (axis cs:3,0) rectangle (axis cs:4,1.666); - - \draw [fill=black] (axis cs:1,3) circle (1pt); - \draw [fill=black] (axis cs:1.5,3.75) circle (1pt); - \draw [fill=black] (axis cs:2,4) circle (1pt); - \draw [fill=black] (axis cs:3.53,1.666) circle (1pt); - - \draw (axis cs:.5,.3) node { RHR}; - \draw (axis cs:1.5,.3) node { MPR}; - \draw (axis cs:2.5,.3) node { LHR}; - \draw (axis cs:3.5,.3) node { other}; - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - The Right Hand Rule says the opposite: - on each subinterval, - evaluate the function at the right endpoint and make the rectangle that height. - In the figure, - the rectangle drawn on [0,1] is drawn using f(1) as its height; - this rectangle is labeled RHR.. -

    - -

    - The Midpoint Rule says that on each subinterval, - evaluate the function at the midpoint and make the rectangle that height. - The rectangle drawn on [1,2] was made using the Midpoint Rule, - with a height of f(1.5). - That rectangle is labeled MPR. -

    - -

    - These are the three most common rules for determining the heights of approximating rectangles, - but one is not forced to use one of these three methods. - The rectangle on [3,4] has a height of approximately f(3.53), - very close to the Midpoint Rule. - It was chosen so that the area of the rectangle is exactly - the area of the region under f on [3,4]. - (Later you'll be able to figure how to do this, too.) -

    - -

    - The following example will approximate the value of \int_0^4 (4x-x^2)\, dx using these rules. -

    - - - Using the Left Hand, Right Hand and Midpoint Rules - -

    - Approximate the value of \int_0^4 (4x-x^2)\, dx using the Left Hand Rule, - the Right Hand Rule, - and the Midpoint Rule, using 4 equally spaced subintervals. -

    -     - -

    - We break the interval [0,4] into four subintervals as before. - In - we see 4 rectangles drawn on - f(x) = 4x-x^2 using the Left Hand Rule. - (The areas of the rectangles are given in each figure.) -

    - -

    - Note how in the first subinterval, - [0,1], the rectangle has height f(0)=0. - We add up the areas of each rectangle (height width) for our Left Hand Rule approximation: - - \amp f(0)\cdot 1 + f(1)\cdot 1+ f(2)\cdot 1+f(3)\cdot 1 - =\amp 0+3+4+3 = 10 - . -

    - -

    - - shows 4 rectangles drawn under f using the Right Hand Rule; - note how the [3,4] subinterval has a rectangle of height 0. -

    - -

    - In this example, - these rectangles seem to be the mirror image of those found in . - This is because of the symmetry of our shaded region. - Our approximation gives the same answer as before, - though calculated a different way: - - \amp f(1)\cdot 1 + f(2)\cdot 1+ f(3)\cdot 1+f(4)\cdot 1 - \amp =3+4+3+0= 10 - . -

    - -

    - - shows 4 rectangles drawn under f using the Midpoint Rule. -

    - -

    - This gives an approximation of \int_0^4(4x-x^2)\, dx as: - - f(0.5)\cdot 1 + f(1.5)\cdot 1+ f(2.5)\cdot 1 \amp +f(3.5)\cdot 1 - \amp =1.75+3.75+3.75+1.75 = 11 - . -

    - -

    - Our three methods provide two approximations of \int_0^4(4x-x^2)\, dx: 10 and 11. -

    -
    - Approximating \int_0^4(4x-x^2)\, dx in - -
    - using the Left Hand Rule - - - A parabola approximated using the Left Hand Rule. - -

    - The parabola in approximated using 4 strips of width 1. - The height of each strip is fiven by the left hand rule. - The first strip has a height of 0, and an area of 0. - The second strip has a height of 3, and an area of 3. - The third strip has a height of 4, and an area of 4. - The fourth strip has a height of 3, and an area of 3. - Each rectangle touches the curve in the top left corner. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4}, - ymin=-1,ymax=5, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; - \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; - - \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:1,0); - \draw [thick,secondcolor] (axis cs:1,0) rectangle (axis cs:2,3); - \draw [thick,secondcolor] (axis cs:2,0) rectangle (axis cs:3,4); - \draw [thick,secondcolor] (axis cs:3,0) rectangle (axis cs:4,3); - - \draw [fill=black] (axis cs:0,0) circle (1pt); - \draw [fill=black] (axis cs:1,3) circle (1pt); - \draw [fill=black] (axis cs:2,4) circle (1pt); - \draw [fill=black] (axis cs:3,3) circle (1pt); - - \draw (axis cs:.5,.3) node {$0$}; - \draw (axis cs:1.5,1.5) node {$3$}; - \draw (axis cs:2.5,2) node {$4$}; - \draw (axis cs:3.5,1.5) node {$3$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - using the Right Hand Rule - - - A parabola approximated using the Right Hand Rule. - -

    - The parabola in approximated using 4 strips of width 1. - The height of each strip is fiven by the right hand rule. - The first strip has height of 3, and an area of 3. - The second strip has height of 4, and an area of 4. - The third strip has height of 3, and an area of 3. - The fourth strip has height of 0, and an area of 0. - Each rectangle touches the curve in the top right corner. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4}, - ymin=-1,ymax=5, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; - \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; - - \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:1,3); - \draw [thick,secondcolor] (axis cs:1,0) rectangle (axis cs:2,4); - \draw [thick,secondcolor] (axis cs:2,0) rectangle (axis cs:3,3); - \draw [thick,secondcolor] (axis cs:3,0) rectangle (axis cs:4,0); - - \draw [fill=black] (axis cs:4,0) circle (1pt); - \draw [fill=black] (axis cs:1,3) circle (1pt); - \draw [fill=black] (axis cs:2,4) circle (1pt); - \draw [fill=black] (axis cs:3,3) circle (1pt); - - \draw (axis cs:.5,1.5) node {$3$}; - \draw (axis cs:1.5,2) node {$4$}; - \draw (axis cs:2.5,1.5) node {$3$}; - \draw (axis cs:3.5,.3) node {$0$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - using the Midpoint Rule - - - A parabola approximated using the Midpoint Rule - -

    - The parabola in approximated using 4 strips of width 1. - The height of each strip is fiven by the Midpoint Rule. - The first strip has a height and area of 1.75. - The second strip has a height and area of 3.75. - The third strip has a height and area of 3.75. - The fourth strip has a height and area of 1.75. - Each rectangle touches the curve in the top center of the rectangle. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4}, - ymin=-1,ymax=5, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; - \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; - - \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:1,1.75); - \draw [thick,secondcolor] (axis cs:1,0) rectangle (axis cs:2,3.75); - \draw [thick,secondcolor] (axis cs:2,0) rectangle (axis cs:3,3.75); - \draw [thick,secondcolor] (axis cs:3,0) rectangle (axis cs:4,1.75); - - \draw [fill=black] (axis cs:.5,1.75) circle (1pt); - \draw [fill=black] (axis cs:1.5,3.75) circle (1pt); - \draw [fill=black] (axis cs:2.5,3.75) circle (1pt); - \draw [fill=black] (axis cs:3.5,1.75) circle (1pt); - - \draw (axis cs:.5,.875) node {\small $1.75$}; - \draw (axis cs:1.5,1.875) node {\small $3.75$}; - \draw (axis cs:2.5,1.875) node {\small $3.75$}; - \draw (axis cs:3.5,.875) node {\small $1.75$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - -     -
    -     - -     - - -     - - - Summation Notation -

    - It is hard to tell at this moment which is a better approximation: 10 or 11? - We can continue to refine our approximation by using more rectangles. - The notation can become unwieldy, - though, as we add up longer and longer lists of numbers. - We introduce summation notation - to ameliorate this problem. - summationnotation -

    - - - - -

    - Suppose we wish to add up a list of numbers a_1, - a_2, - a_3, , a_9. - Instead of writing - - a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8+a_9 - , - we use summation notation and write \sum_{i=1}^9 a_i. - - The upper case sigma, - Sigma represents the term sum. The index (counter) of summation in this example is i; any symbol can be used. - By convention, the index takes on only the integer values between - (and including) - the lower and upper bounds. - To the right of \Sigma, - the expression a_i is called the summand. - It tells us what we are summing. - This is summarized in Equation. - - \sum_{\underbrace{i=1}_{i \text{-index of summation}}}^{\overbrace{9}^\text{upper bound}} \underbrace{a_i}_\text{summand} - -

    - -

    - Let's practice using this notation. -

    - - - Using summation notation - -

    - Let the numbers \{a_i\} be defined as - a_i = 2i-1 for integers i, where i\geq 1. - So a_1 = 1, a_2 = 3, - a_3 = 5, etc. (The output is the positive odd integers). - Evaluate the following summations: - -

      -
    1. \sum_{i=1}^6 a_i
      2. - -
    2. \sum_{i=3}^7 (3a_i-4)
      3. - -
    3. \sum_{i=1}^4 (a_i)^2
      4. -
    -

    -     - -

    -

      -
    1. -

      - - \sum_{i=1}^6 a_i \amp = a_1+a_2+a_3+a_4+a_5+a_6 - \amp = 1+3+5+7+9+11 - \amp = 36 - . -

      -
      2. - -
    2. -

      - Note the starting value is different than 1: - - \sum_{i=3}^7 (3a_i-4) \amp = (3a_3-4)+(3a_4-4)+(3a_5-4)+(3a_6-4)+(3a_7-4) - \amp = 11+17+23+29+35 - \amp = 115 - . -

      -
      3. - -
    3. -

      - - \sum_{i=1}^4 (a_i)^2 \amp = (a_1)^2+(a_2)^2+(a_3)^2+(a_4)^2 - \amp = 1^2+3^2+5^2+7^2 - \amp = 84 - . -

      -
      4. -
    -

    -     - -     - -

    - It might seem odd to stress a new, - concise way of writing summations only to write each term out as we add them up. - It is. - The following theorem gives some of the properties of summations that allow us to work with them without writing individual terms. - Examples will follow. -

    - - - Properties of Summations - -

    - summationproperties - -

      -
    1. -

      - \ds \sum_{i=1}^n c = c\cdot n, - where c is a constant. -

      -
      2. - -
    2. \sum_{i=m}^n (a_i\pm b_i) = \sum_{i=m}^n a_i \pm \sum_{i=m}^n b_i
      3. - -
    3. \sum_{i=m}^n c\cdot a_i = c\cdot\sum_{i=m}^n a_i
      4. - -
    4. \sum_{i=m}^j a_i + \sum_{i=j+1}^n a_i = \sum_{i=m}^n a_i
      5. - -
    5. \sum_{i=1}^n i = \frac{n(n+1)}2
      6. - -
    6. \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}6
      7. - -
    7. \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}2\right)^2
      8. -
    -

    -     -     - - - - - - Evaluating summations using <xref ref="thm_summation"/> - -

    - Revisit and, - using , evaluate - - \sum_{i=1}^6 a_i = \sum_{i=1}^6 (2i-1) - . -

    -     - -

    - - \sum_{i=1}^6 (2i-1) \amp = \sum_{i=1}^6 2i - \sum_{i=1}^6 (1) - \amp = \left(2\sum_{i=1}^6 i \right)- 6 - \amp = 2\frac{6(6+1)}{2} - 6 - \amp = 42-6 = 36 - -

    -

    - We obtained the same answer without writing out all six terms. - When dealing with small sizes of n, - it may be faster to write the terms out by hand. - However, - is incredibly important when dealing with large sums as we'll soon see. -

    -     -     -     - - - Riemann Sums -

    - Consider again \int_0^4(4x-x^2)\, dx. - We will approximate this definite integral using 16 equally spaced subintervals and the Right Hand Rule in . - Before doing so, it will pay to do some careful preparation. - Riemann Sum -

    - -
    - Dividing [0,4] into 16 equally spaced subintervals - - A number line between 0 and 4 divided into 16 sections - -

    - A number line ranging between 0 and 4. - The space between each number is divided into 4 equal sections, giving a total of 16 equal sections. - x_0 is written under 0. - x_4 is written under 1. - x_8 is written under 2. - x_12 is written under 3. - x_16 is written under 4. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-.5,0) -- (4.5,0); - - \foreach \x in {0,.25,...,4} { - \draw (\x,-.1) -- (\x,.1); - } - \draw (0,-.25) node {\scriptsize $0$}; - \draw (1,-.25) node {\scriptsize $1$}; - \draw (2,-.25) node {\scriptsize $2$}; - \draw (3,-.25) node {\scriptsize $3$}; - \draw (4,-.25) node {\scriptsize $4$}; - - \draw (0,-.55) node {\scriptsize $x_0$}; - \draw (1,-.55) node {\scriptsize $x_4$}; - \draw (2,-.55) node {\scriptsize $x_8$}; - \draw (3,-.55) node {\scriptsize $x_{12}$}; - \draw (4,-.55) node {\scriptsize $x_{16}$}; - - \end{tikzpicture} - - - -
    - -

    - - shows a number line of [0,4] divided, - or partitioned, into 16 equally spaced subintervals. - We denote 0 as x_0; - we have marked the values of x_4, x_8, - x_{12} and x_{16}. - We could mark them all, but the figure would get crowded. - While it is easy to figure that x_{9} = 2.25, in general, - we want a method of determining the value of x_i without consulting the figure. - Consider: -

    - - - A labeled equation for caclulating values at subintervals - -

    - The equation x_i = x_0 + i\Delta x, with labels for each component. - x_0 is labeled as the starting value. - i is labeled as the numer of subintervals between x_0 and x_i. - \Delta x is labeled as the subinterval size. -

    -     - - - \begin{tikzpicture} - \draw (1.5,1) node {$\displaystyle x_i = x_0 + i\Delta x$}; - - \draw [firstcolor] (1,0) node [text width=32pt,align=center] (a) { \centering starting\\[-5pt] value}; - \draw [firstcolor,->] (a) -- (.95,.75); - - \draw [firstcolor] (2,2.5) node [text width=120pt,align=center] (b) { \centering number of subintervals\\[-5pt] between $x_0$ and $x_i$}; - \draw [firstcolor,->] (b) -- (1.95,1.25); - - \draw [firstcolor] (3,0) node [text width=60pt,align=center] (c) { \centering subinterval\\[-5pt] size}; - \draw [firstcolor,->] (c) -- (2.75,.75); - - \end{tikzpicture} - - - - -

    - So x_{9} = x_0 + 9(4/16) = 2.25. -

    - -

    - If we had partitioned [0,4] into 100 equally spaced subintervals, - each subinterval would have length \Delta x=4/100 = 0.04. - We could compute x_{31} as - - x_{31} = x_0 + 31(4/100) = 1.24 - . -

    - -

    - (That was far faster than creating a sketch first.) -

    - -

    - Given any subdivision of [0,4], - the first subinterval is [x_0,x_1]; - the second is [x_1,x_2]; - the ith subinterval is [x_{i-1},x_{i}]. -

    - -

    - When using the Left Hand Rule, - the height of the ith rectangle will be f(x_{i-1}). -

    - -

    - When using the Right Hand Rule, - the height of the ith rectangle will be f(x_{i}). -

    - -

    - When using the Midpoint Rule, - the height of the ith rectangle will be \ds f\left(\frac{x_{i-1}+x_{i}}2\right). -

    - -

    - Thus approximating \int_0^4(4x-x^2)\, dx with 16 equally spaced subintervals can be expressed as follows, - where \Delta x = 4/16 = 1/4: -

    - -

    -

    -
  • - Left Hand Rule -

    - \ds \sum_{i=1}^{16} f(x_{i-1})\Delta x -

    -
    • - -
  • - Right Hand Rule -

    - \ds \sum_{i=1}^{16} f(x_{i})\Delta x -

    -
    • - -
  • - Midpoint Rule -

    - \ds \sum_{i=1}^{16} f\left(\frac{x_{i-1}+x_{i}}2\right)\Delta x -

    -
    • -
    - Left Hand Rule - Right Hand Rule - Midpoint Rule -

    - -

    - We use these formulas in the next two examples. - The following example lets us practice using the Right Hand Rule and the summation formulas introduced in . -

    - - - Approximating definite integrals using sums - -

    - Approximate \int_0^4(4x-x^2)\, dx using the Right Hand Rule and summation formulas with 16 and 1000 equally spaced intervals. -

    -     - -

    - Using the formula derived before, - using 16 equally spaced intervals and the Right Hand Rule, - we can approximate the definite integral as - - \sum_{i=1}^{16}f(x_{i})\Delta x - . -

    - -

    - We have \Delta x = 4/16 = 0.25. - Since x_i = 0+i\Delta x, we have - - x_{i} = 0 + i\Delta x = i\Delta x - . -

    - -

    - Using the summation formulas, consider: - - \int_0^4 (4x-x^2)\, dx \amp \approx \sum_{i=1}^{16} f(x_{i})\Delta x - \amp = \sum_{i=1}^{16} f(i\Delta x) \Delta x - \amp = \sum_{i=1}^{16} \big(4i\Delta x - (i\Delta x)^2\big)\Delta x - \amp = \sum_{i=1}^{16} (4i\Delta x^2 - i^2\Delta x^3) - \amp = (4\Delta x^2)\sum_{i=1}^{16} i - \Delta x^3 \sum_{i=1}^{16} i^2 - \amp = (4\Delta x^2)\frac{16\cdot 17}{2} - \Delta x^3 \frac{16(17)(33)}6 - \amp = 4\cdot 0.25^2\cdot 136-0.25^3\cdot 1496 - \amp =10.625 - -

    - -

    - We were able to sum up the areas of 16 rectangles with very little computation. - In - the function and the 16 rectangles are graphed. - While some rectangles over-approximate the area, - other under-approximate the area - (by about the same amount). - Thus our approximate area of 10.625 is likely a fairly good approximation. -

    - -

    - Notice Equation; - by replacing 16 by 1,000 - (and appropriately changing the value of \Delta x), - we can use that equation to sum up 1000 rectangles! -

    - -
    - Approximating \int_0^4(4x-x^2)\, dx with the Right Hand Rule and 16 evenly spaced subintervals - - - A parabola approximated using 16 equal subintervals. - -

    - The parabola in approximated using 16 strips of equal width. - The height of each strips are given by the Right Hand Rule. - On the left side the area of the rectangles is slightly greater than the area of the parabola. - On the right side the area of the rectangles is slightly less than the area of the parabola. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4}, - ymin=-1,ymax=5, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; - \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; - - \foreach \x / \y in {0.25 / 0.9375, 0.5 / 1.75, 0.75 / 2.4375, 1. / 3., 1.25 / 3.4375, - 1.5 / 3.75, 1.75 / 3.9375, 2. / 4., 2.25 / 3.9375, 2.5 / 3.75, - 2.75 /3.4375, 3. / 3., 3.25 / 2.4375, 3.5 / 1.75, 3.75 / 0.9375, 4. / 0.} - {\addplot [thick,secondcolor] coordinates {(\x-.25,0) (\x-.25,\y) (\x,\y) (\x,0) (\x-.25,0)}; - } - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - We do so here, - skipping from the original summand to the equivalent of Equation to save space. - Note that \Delta x = 4/1000 = 0.004. - - \int_0^4 (4x-x^2)\, dx \amp \approx \sum_{i=1}^{1000} f(x_{i})\Delta x - \amp = (4\Delta x^2)\sum_{i=1}^{1000} i - \Delta x^3 \sum_{i=1}^{1000} i^2 - \amp = (4\Delta x^2)\frac{1000\cdot 1001}{2} - \Delta x^3 \frac{1000(1001)(2001)}6 - \amp =10.666656 - -

    - -

    - Using many, many rectangles, we have a likely good approximation of - \int_0^4 (4x-x^2)\dx. - That is, - - \int_0^4(4x-x^2)\, dx \approx 10.666656 - . -

    -     - -     - -

    - Before the above example, - we stated what the summations for the Left Hand, Right Hand and Midpoint Rules looked like. - Each had the same basic structure, which was: -

    - -

    -

      -
    1. -

      - each rectangle has the same width, - which we referred to as \Delta x, and -

      -
      2. - -
    2. -

      - each rectangle's height is determined by evaluating f at a particular point in each subinterval. - For instance, - the Left Hand Rule states that each rectangle's height is determined by evaluating f at the left hand endpoint of the subinterval the rectangle lives on. -

      -
      3. -
    -

    - -

    - One could partition an interval [a,b] with subintervals that do not have the same size. - We refer to the length of the - ith subinterval as \Delta x_i. - Also, one could determine each rectangle's height by evaluating f at any - point c_i in the ith subinterval. - Thus the height of the ith subinterval would be f(c_i), - and the area of the ith rectangle would be f(c_i)\Delta x_i. - These ideas are formally defined below. -

    - - - Partition - -

    - A partition \Delta x of a closed interval [a,b] is a set of numbers x_0, - x_1, - \ldots x_{n} where - - a=x_0 \lt x_1 \lt \ldots \lt x_{n-1} \lt x_{n}=b - . -

    - -

    - The length of the ith subinterval, - [x_{i-1},x_{i}], is \Delta x_i = x_{i}-x_{i-1}. - If [a,b] is partitioned into subintervals of equal length, - we let \Delta x represent the length of each subinterval. -

    - -

    - The size of the partition, - denoted \norm{\Delta x}, - is the length of the largest subinterval of the partition. - partition - partitionsize of -

    -     -     - - - - -

    - Summations of rectangles with area - f(c_i)\Delta x_i are named after mathematician Georg Friedrich Bernhard Riemann, - as given in the following definition. -

    - - - Riemann Sum - -

    - Let f be defined on a closed interval [a,b], - let \Delta x be a partition of [a,b] - as given in , - Riemann Sum - and let c_i denote any value in the ith subinterval. -

    - -

    - The sum - - \sum_{i=1}^n f(c_i)\Delta x_i - - is a Riemann sum of f on [a,b]. -

    -     -     - - - - -

    - - shows the approximating rectangles of a Riemann sum of \int_0^4(4x-x^2)\, dx. - While the rectangles in this example do not approximate well the shaded area, - they demonstrate that the subinterval widths may vary and the heights of the rectangles can be determined without following a particular rule. -

    - -
    - An example of a general Riemann sum to approximate \int_0^4(4x-x^2)\, dx - - - A parabola approximated by 3 rectangles of different width - -

    - The parabola in approximated using 3 rectangles of different width. - The height of each rectangle is not given by any particular rule. - The left most strip has a width of 2, and a height of around 0.95. This height is given by a point at around x=0.3. - The center strip has a width of 2.5. It has a height of around 2.8, given by a point at around x=3.2. - The right strip has a width of 0.5. It has a height of around 0.8, given by a point around x=3.8. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4}, - ymin=-1,ymax=5, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:4] {4*x-x^2}; - \addplot [firstcurvestyle,domain=0:4,samples=40] {4*x-x^2}; - - \draw [thick,secondcolor] (axis cs:0,0) rectangle (axis cs:2,.9375); - \draw [thick,secondcolor] (axis cs:2,0) rectangle (axis cs:3.5,2.79); - \draw [thick,secondcolor] (axis cs:3.5,0) rectangle (axis cs:4,.76); - \draw [fill=black] (axis cs:.25,.9375) circle (1pt); - \draw [fill=black] (axis cs:3.1,2.79) circle (1pt); - \draw [fill=black] (axis cs:3.8,.76) circle (1pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Usually Riemann sums are calculated using one of the three methods we have introduced. - The uniformity of construction makes computations easier. - Before working another example, - let's summarize some of what we have learned in a convenient way. -

    - - - Riemann Sum Concepts -

    - Consider \ds \int_a^b f(x) \, dx \approx \sum_{i=1}^n f(c_i)\Delta x_i. -

    - -

    -

      -
    1. -

      - When the n subintervals have equal length, - \ds \Delta x_i = \Delta x = \frac{b-a}n. -

      -
      2. - -
    2. -

      - The ith term of an equally spaced partition is x_i = a + i\Delta x. - (Thus x_0=a and x_{n} = b.) -

      -
      3. - -
    3. -

      - The Left Hand Rule summation is: - \ds \sum_{i=1}^n f(x_{i-1})\Delta x. -

      -
      4. - -
    4. -

      - The Right Hand Rule summation is: - \ds \sum_{i=1}^n f(x_{i})\Delta x. -

      -
      5. - -
    5. -

      - The Midpoint Rule summation is: - \ds \sum_{i=1}^n f\left(\frac{x_{i-1}+x_{i}}{2}\right)\Delta x. -

      -
      6. -
    -

    -     - -

    - Let's do another example. -

    - - - Approximating definite integrals with sums - -

    - Approximate \int_{-2}^3 (5x+2)\, dx using the Midpoint Rule and 10 equally spaced intervals. -

    -     - -

    - Following , we have - - \Delta x = \frac{3 - (-2)}{10} = 1/2 \text{ and } x_i = (-2) + (1/2)(i) = i/2-2 - . -

    - -

    - As we are using the Midpoint Rule, - we will also need x_{i-1} and \ds \frac{x_{i-1}+x_{i}}2. - Since x_i = i/2-2,x_{i-1} = (i-1)/2 - 2 = i/2 -5/2. - This gives - - \frac{x_{i-1}+x_{i}}2 = \frac{(i/2-5/2) + (i/2-2)}{2} = \frac{i-9/2}{2} = i/2 - 9/4 - . -

    - -

    - We now construct the Riemann sum and compute its value using summation formulas. - - \int_{-2}^3 (5x+2)\, dx \amp \approx \sum_{i=1}^{10} f\left(\frac{x_{i-1}+x_{i}}{2}\right)\Delta x - \amp = \sum_{i=1}^{10} f(i/2 - 9/4)\Delta x - \amp = \sum_{i=1}^{10} \big(5(i/2-9/4) + 2\big)\Delta x - \amp = \Delta x\sum_{i=1}^{10}\left[\left(\frac{5}{2}\right)i - \frac{37}{4}\right] - \amp = \Delta x\left(\frac{5}2\sum_{i=1}^{10} (i) - \sum_{i=1}^{10}\left(\frac{37}{4}\right)\right) - \amp = \frac12\left(\frac52\cdot\frac{10(11)}{2} - 10\cdot\frac{37}4\right) - \amp = \frac{45}2 = 22.5 - -

    - -
    - Approximating \int_{-2}^3 (5x+2)\, dx using the Midpoint Rule and 10 evenly spaced subintervals in - - - The area under a line approximated with 10 even subintervals. - -

    - The graph 5x + 2, plotted from x = -2 to x = 3. - The line begins in the third quadrant and crosses the x-axis at x = -\frac{2}{5}. - The line crosses the y-axis at y=2. - The area between the line and the x-axis is shaded. - 10 rectangles of equal width are approximating the shaded area. - The height of each rectangle is given by the Midpoint Rule. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-2,-1,1,2,3}, - ytick={10,17,-8}, - ymin=-9,ymax=19, - xmin=-2.5,xmax=3.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=-2:3] {5*x+2} \closedcycle; - \addplot [firstcurvestyle,domain=-2:3] {5*x+2}; - - \foreach \x / \y in {-1.75 / -6.75, -1.25 / -4.25, -0.75 / -1.75, -0.25 / 0.75, - 0.25 /3.25, 0.75 / 5.75, 1.25 / 8.25, 1.75 / 10.75, 2.25 / 13.25, - 2.75 /15.75} - {\addplot [thick,secondcolor] coordinates {(\x-.25,0) (\x-.25,\y) (\x+.25,\y) (\x+.25,0) (\x-.25,0)}; - } - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Note the graph of f(x) = 5x+2 in . - The regions whose area is computed by the definite integral are triangles, - meaning we can find the exact answer without summation techniques. - We find that the exact answer is indeed 22.5. - One of the strengths of the Midpoint Rule is that often each rectangle includes area that should not be counted, - but misses other area that should. - When the partition size is small, - these two amounts are about equal and these errors almost - cancel each other out. In this example, - since our function is a line, - these errors are exactly equal and they do cancel each other out, - giving us the exact answer. -

    - -

    - Note too that when the function is negative, - the rectangles have a negative height. - When we compute the area of the rectangle, - we use f(c_i)\Delta x; - when f is negative, the area is counted as negative. -

    -     - -     - -

    - Notice in the previous example that while we used 10 equally spaced intervals, - the number 10 didn't play a big role in the calculations until the very end. - Mathematicians love to abstract ideas; - let's approximate the area of another region using n subintervals, - where we do not specify a value of n until the very end. -

    - - - Approximating definite integrals with a formula, using sums - -

    - Revisit \int_0^4(4x-x^2)\, dx yet again. - Approximate this definite integral using the Right Hand Rule with n equally spaced subintervals. -

    -     - -

    - Using , - we know \Delta x = \frac{4-0}{n} = 4/n. - We also find x_i = 0 + i\Delta x = 4i/n. -

    - -

    - We construct the Right Hand Rule Riemann sum as follows. - Be sure to follow each step carefully. - If you get stuck, - and do not understand how one line proceeds to the next, - you may skip to the result and consider how this result is used. - You should come back, though, - and work through each step for full understanding. - - \int_0^4(4x-x^2)\, dx \amp \approx \sum_{i=1}^n f(x_{i})\Delta x - \amp = \sum_{i=1}^n f\left(\frac{4i}{n}\right) \Delta x - \amp = \sum_{i=1}^n \left[4\frac{4i}n-\left(\frac{4i}n\right)^2\right]\Delta x - \amp = \sum_{i=1}^n \left(\frac{16\Delta x}{n}\right)i - \sum_{i=1}^n \left(\frac{16\Delta x}{n^2}\right)i^2 - \amp = \left(\frac{16\Delta x}{n}\right)\sum_{i=1}^n i - \left(\frac{16\Delta x}{n^2}\right)\sum_{i=1}^n i^2 - \amp = \left(\frac{16\Delta x}{n}\right)\cdot \frac{n(n+1)}{2} - \left(\frac{16\Delta x}{n^2}\right)\frac{n(n+1)(2n+1)}{6} - \amp =\frac{32(n+1)}{n} - \frac{32(n+1)(2n+1)}{3n^2} (\text{ recall } \Delta x = 4/n) - \amp = \frac{32}{3}\left(1-\frac{1}{n^2}\right) \text{ (after simplifying) } - -

    - -

    - The result is an amazing, easy to use formula. - To approximate the definite integral with 10 equally spaced subintervals and the Right Hand Rule, - set n=10 and compute - - \int_0^4 (4x-x^2)\, dx \approx \frac{32}{3}\left(1-\frac{1}{10^2}\right) = 10.56 - . -

    - -

    - Recall how earlier we approximated the definite integral with 4 subintervals; - with n=4, the formula gives 10, our answer as before. -

    - -

    - It is now easy to approximate the integral with 1,000,000 subintervals! - Hand-held calculators will round off the answer a bit prematurely giving an answer of 10.66666667. - (The actual answer is 10.666666666656.) -

    - -

    - We now take an important leap. - Up to this point, - our mathematics has been limited to geometry and algebra - (finding areas and manipulating expressions). - Now we apply calculus. - For any finite n, we know that - - \int_0^4 (4x-x^2)\, dx \approx \frac{32}{3}\left(1-\frac{1}{n^2}\right) - . -

    - -

    - Both common sense and high-level mathematics tell us that as n gets large, - the approximation gets better. - In fact, if we take the limit - as n\rightarrow \infty, - we get the exact area described by \int_0^4 (4x-x^2)\, dx. - That is, - - \int_0^4 (4x-x^2)\, dx \amp = \lim_{n\rightarrow \infty} \frac{32}{3}\left(1-\frac{1}{n^2}\right) - \amp = \frac{32}{3}\left(1-0\right) - \amp = \frac{32}{3} = 10.\overline{6} - -

    - -

    - This is a fantastic result. - By considering n equally-spaced subintervals, - we obtained a formula for an approximation of the definite integral that involved our variable n. - As n grows large without bound the error shrinks to zero and we obtain the exact area. -

    -     - -     - -

    - This section started with a fundamental calculus technique: - make an approximation, refine the approximation to make it better, - then use limits in the refining process to get an exact answer. - That is precisely what we just did. -

    - -

    - Let's practice this again. -

    - - - Approximating definite integrals with a formula, using sums - -

    - Find a formula that approximates - \int_{-1}^5 x^3\, dx using the Right Hand Rule and n equally spaced subintervals, - then take the limit as n\to\infty to find the exact area. -

    -     - -

    - Following , - we have \Delta x = \frac{5-(-1)}{n} = 6/n. - We have x_i = (-1) + i\Delta x, - which is the right endpoint of the ith subinterval. -

    - -

    - The Riemann sum corresponding to the Right Hand Rule is (followed by simplifications): - - \int_{-1}^5 x^3\, dx \amp \approx \sum_{i=1}^n f(x_{i})\Delta x - \amp = \sum_{i=1}^n f(-1+i\Delta x)\Delta x - \amp = \sum_{i=1}^n (-1+i\Delta x)^3\Delta x - \amp = \sum_{i=1}^n \big((i\Delta x)^3 -3(i\Delta x)^2 + 3i\Delta x -1\big)\Delta x \text{ (now distribute \(\Delta x\)) } - \amp = \sum_{i=1}^n \big(i^3\Delta x^4 - 3i^2\Delta x^3 + 3i\Delta x^2 -\Delta x\big) \text{ (now split up summation) } - \amp = \Delta x^4 \sum_{i=1}^ni^3 -3\Delta x^3 \sum_{i=1}^n i^2+ 3\Delta x^2 \sum_{i=1}^n i - \sum_{i=1}^n \Delta x - \amp = \Delta x^4 \left(\frac{n(n+1)}{2}\right)^2 -3\Delta x^3 \frac{n(n+1)(2n+1)}{6}+ 3\Delta x^2 \frac{n(n+1)}{2} - n\Delta x - (use \Delta x = 6/n) - \amp = \frac{1296}{n^4}\cdot\frac{n^2(n+1)^2}{4} - 3\frac{216}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} + 3\frac{36}{n^2}\frac{n(n+1)}2 -6 - (now do a sizable amount of algebra to simplify) - \amp =156 + \frac{378}n + \frac{216}{n^2} - -

    - -

    - Once again, we have found a compact formula for approximating the definite integral with n equally spaced subintervals and the Right Hand Rule. - Using 10 subintervals, we have an approximation of 195.96 - (these rectangles are shown in ). - Using n=100 gives an approximation of 159.802. -

    - -
    - Approximating \int_{-1}^5 x^3\, dx using the Right Hand Rule and 10 evenly spaced subintervals - - - A cubic graph approximated using 10 subintervals and the Right Hand Rule. - -

    - The curve begins at x=-1, at which point the curve is close to the x-axis. - The curve increases to the right, ending at x=5, at which point the curve has a value of 125. - The area under the curve is approsimated by 10 rectangles of equal width. - The height of each rectangle is given by the Right Hand Rule. - The left of each rectangle includes some unshaded area, so each rectangle overestimates the area under the curve. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,1,2,3,4,5}, - ymin=-2,ymax=130, - xmin=-1.5,xmax=5.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=-1:5] {x^3} \closedcycle; - \addplot [firstcurvestyle,domain=-1:5] {x^3}; - - \foreach \x / \y in {-0.4 / -0.064, 0.2 / 0.008, 0.8 / 0.512, 1.4 / 2.744, 2. / 8., 2.6 /17.576, 3.2 / 32.768, 3.8 / 54.872, 4.4 / 85.184, 5. / 125.} - {\addplot [thick,secondcolor] coordinates {(\x-.6,0) (\x-.6,\y) (\x,\y) (\x,0) (\x-.6,0)}; - } - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Now find the exact answer using a limit: - - \int_{-1}^5 x^3\, dx = \lim_{n\to\infty} \left(156 + \frac{378}n + \frac{216}{n^2}\right) = 156 - . -

    -     - -     -     - - - Limits of Riemann Sums -

    - We have used limits to evaluate given definite integrals. - Will this always work? - It can be shown, - given not-very-restrictive conditions, - that yes, it will always work this is the content of - below. -

    - -

    - The previous two examples demonstrated how an expression such as - - \sum_{i=1}^n f(x_{i})\Delta x - - can be rewritten as an expression explicitly involving n, - such as 32/3(1-1/n^2). -

    - -

    - Viewed in this manner, - we can think of the summation as a function of n. - An n value is given - (where n is a positive integer), - and the sum of areas of n equally spaced rectangles is returned, - using the Left Hand, Right Hand, or Midpoint Rules. -

    - -

    - Given a definite integral \int_a^b f(x)\, dx, let: -

    - -

    -

      -
    • -

      - \ds S_L(n) = \sum_{i=1}^n f(x_{i-1})\Delta x, - the sum of equally spaced rectangles formed using the Left Hand Rule, -

      -
      • - -
    • -

      - \ds S_R(n) = \sum_{i=1}^n f(x_{i})\Delta x, - the sum of equally spaced rectangles formed using the Right Hand Rule, and -

      -
      • - -
    • -

      - \ds S_M(n) = \sum_{i=1}^n f\left(\frac{x_{i-1}+x_{i}}{2}\right)\Delta x, - the sum of equally spaced rectangles formed using the Midpoint Rule. -

      -
      • -
    -

    - -

    - Recall the definition of a limit as n\to\infty: - \lim\limits_{n\to\infty}S_L(n) = K if, given any \varepsilon \gt 0, - there exists N \gt 0 such that - - \abs{S_L(n)-K} \lt \varepsilon \text{ when } n\geq N - . -

    - -

    - The following theorem states that we can use any of our three rules to find the exact value of a definite integral \int_a^b f(x)\, dx. - It also goes two steps further. - The theorem states that the height of each rectangle doesn't have to be determined following a specific rule, - but could be f(c_i), - where c_i is any point in the ith subinterval, - as discussed before Riemann Sums were defined in . -

    - -

    - The theorem goes on to state that the rectangles do not need to be of the same width. - Using the notation of , - let \Delta x_i denote the length of the - ith subinterval in a partition of [a,b] and let - \norm{\Delta x} represent the length of the largest subinterval in the partition: - that is, \norm{\Delta x} is the largest of all the \Delta x_i. - If \norm{\Delta x} is small, - then [a,b] must be partitioned into many subintervals, - since all subintervals must have small lengths. - Taking the limit as \norm{\Delta x} goes to zero - implies that the number n of subintervals in the partition is growing to infinity, - as the largest subinterval length is becoming arbitrarily small. - We then interpret the expression - - \lim_{\norm{\Delta x}\to 0}\sum_{i=1}^nf(c_i)\Delta x_i - - as the limit of the sum of the areas of rectangles, - where the width of each rectangle can be different but getting small, - and the height of each rectangle is not necessarily determined by a particular rule. - The theorem states that this Riemann Sum also gives the value of the definite integral of f over [a,b]. -

    - - - Definite Integrals and the Limit of Riemann Sums - -

    - Let f be continuous on the closed interval [a,b] and let S_L(n), - S_R(n), - S_M(n), \Delta x, - \Delta x_i and c_i be defined as before. - Then: - - Riemann Sumand definite integral - - integrationdefinite!Riemann Sums - -

    - -

    -

      -
    1. \begin{aligned}\lim_{n\to\infty} S_L(n) \amp = \lim_{n\to\infty} S_R(n)\\ - \amp = \lim_{n\to\infty} S_M(n)\\ - \amp = \lim_{n\to\infty}\sum_{i=1}^n f(c_i)\Delta x\end{aligned}
      2. - -
    2. \ds \lim_{n\to\infty}\sum_{i=1}^n f(c_i)\Delta x = \int_a^b f(x)\, dx
      3. - -
    3. \lim_{\norm{\Delta x}\to 0} \sum_{i=1}^n f(c_i)\Delta x_i = \int_a^b f(x)\, dx
      4. -
    -

    -     -     - - - - - - -

    - We summarize what we have learned over the past few sections here. -

    - -

    -

      -
    • -

      - Knowing the area under the curve can be useful. - One common example: the area under a velocity curve is displacement. -

      -
      • - -
    • -

      - We have defined the definite integral, \int_a^b f(x)\, dx, - to be the signed area under f on the interval [a,b]. -

      -
      • - -
    • -

      - While we can approximate a definite integral many ways, - we have focused on using rectangles whose heights can be determined using the Left Hand Rule, - the Right Hand Rule and the Midpoint Rule. -

      -
      • - -
    • -

      - Sums of rectangles of this type are called Riemann sums. -

      -
      • - -
    • -

      - The exact value of the definite integral can be computed using the limit of a Riemann sum. - We generally use one of the above methods as it makes the algebra simpler. -

      -
      • -
    -

    - -

    - We first learned of derivatives through limits then learned rules that made the process simpler. - We know of a way to evaluate a definite integral using limits; - in the next section we will see how the Fundamental Theorem of Calculus makes the process simpler. - The key feature of this theorem is its connection between the indefinite integral and the definite integral. -

    -     - - - - Terms and Concepts - - - - -

    - A fundamental calculus technique is to use - to refine approximations to get an exact answer. -

    -     - - - - limits? - - - - -     - - - - - $lb = random(2,9,1); - $ub = random(12,20,1); - $m = random(30,60,1); - $b = non_zero_random(-300,300,1); - if($envir{problemSeed}==1){$lb=7;$ub=14;$m=48;$b=-201;}; - Context()->variables->are(i=>'Real'); - $f = Formula("$m i + $b")->reduce; - - -

    - What is the upper bound in the summation - \sum\limits_{i=}^{} ()? -

    -

    - -

    -     -     -     - - - - - Context()->strings->add('rectangles'=>{}); - Context()->strings->add('rectangle'=>{alias=>'rectangles'}); - - -

    - This section approximates definite integrals using what geometric shape? -

    -

    - -

    -     -     -     - - - - -

    - - A sum using the Right Hand Rule is an example of a Riemann Sum. -

    -     - -     -     - - - Problems - - -

    - Write out each term of the summation and compute the sum. -

    -     - - - - - $lb = random(2,5,1); - $ub = $lb + random(2,3,1); - if($envir{problemSeed}==1){$lb=2;$ub=4;}; - @i = ($lb..$ub); - @fi = map{$_**2}(@i); - $string = join('+',@fi); - $sum = Real($string); - Context("Form"); - Context()->operators->set( - '+' => {class => 'bizarro::BOP::add', isCommand => 1}, - '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, - ); - $terms = Formula("$string"); - - -

    - \sum\limits_{i=}^{} i^2 -

    - - Enter the sum of terms here. For example, 1+2+3. - -

    - -

    - - Enter the actual sum here. - -

    - -

    -     -     -     - - - - - $lb = random(-3,-1,1); - $ub = random(1,3,1); - $m = random(2,5,1); - $b = non_zero_random(-3,3,1); - if($envir{problemSeed}==1){$lb=-1;$ub=3;$m=4;$b=-2;}; - Context()->variables->are(i=>'Real'); - $f = Formula("$m i + $b")->reduce; - @i = ($lb..$ub); - @fi = map{$m*$_ + $b}(@i); - $string = join('+',@fi); - $sum = Real($string); - Context("Form"); - Context()->operators->set( - '+' => {class => 'bizarro::BOP::add', isCommand => 1}, - '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, - ); - $terms = Formula("$string"); - - -

    - \sum\limits_{i=}^{} \left(\right) -

    - - Enter the sum of terms here. For example, 1+2+3. - -

    - -

    - - Enter the actual sum here. - -

    - -

    -     -     -     - - - - - $lb = random(-3,-1,1); - $ub = random(1,3,1); - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$lb=-2;$ub=2;$trig='sin';}; - Context()->variables->are(i=>'Real'); - $f = Formula("sin(pi i /2)")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - Context("Form"); - Context()->operators->set( - '+' => {class => 'bizarro::BOP::add', isCommand => 1}, - '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, - ); - $terms = Formula("$string"); - - -

    - \sum\limits_{i=}^{} -

    - - Enter the sum of terms here. For example, 1+2+3. - -

    - -

    - - Enter the actual sum here. - -

    - -

    -     -     -     - - - - - $lb = 1; - ($ub,$c) = random_subset(2,5..10); - if($envir{problemSeed}==1){$ub=10;$c=5;}; - Context()->variables->are(i=>'Real'); - $f = Formula("$c")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - Context("Form"); - Context()->operators->set( - '+' => {class => 'bizarro::BOP::add', isCommand => 1}, - '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, - ); - $terms = Formula("$string"); - - -

    - \sum\limits_{i=}^{} -

    - - Enter the sum of terms here. For example, 1+2+3. - -

    - -

    - - Enter the actual sum here. - -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(4,6,1); - if($envir{problemSeed}==1){$ub=5;}; - Context("Fraction"); - Context()->variables->are(i=>'Real'); - $f = Formula("1/i")->reduce; - @i = ($lb..$ub); - @fi = map{Fraction($f->eval(i=>$_))}(@i); - $string = join('+',@fi); - $sum = Fraction($string); - Context("Form"); - Context()->operators->set( - '+' => {class => 'bizarro::BOP::add', isCommand => 1}, - '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, - ); - $terms = Formula("$string"); - - -

    - \sum\limits_{i=}^{} -

    - - Enter the sum of terms here. For example, 1+2+3. - -

    - -

    - - Enter the actual sum here. - -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(4,8,1); - if($envir{problemSeed}==1){$ub=6;}; - Context()->variables->are(i=>'Real'); - $f = Formula("(-1)^i i")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - Context("Form"); - Context()->operators->set( - '+' => {class => 'bizarro::BOP::add', isCommand => 1}, - '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, - ); - $terms = Formula("$string"); - - -

    - \sum\limits_{i=}^{} -

    - - Enter the sum of terms here. For example, 1+2+3. - -

    - -

    - - Enter the actual sum here. - -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(3,5,1); - if($envir{problemSeed}==1){$ub=4;}; - Context()->variables->are(i=>'Real'); - $f = Formula("1/i - 1/(i+1)")->reduce; - @i = ($lb..$ub); - Context("Fraction"); - @fi = map{Fraction($f->eval(i=>$_))}(@i); - $string = join('+',@fi); - $sum = Fraction($string); - Context("Form"); - Context()->operators->set( - '+' => {class => 'bizarro::BOP::add', isCommand => 1}, - '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, - ); - $terms = Formula("$string"); - - -

    - \sum\limits_{i=}^{} \left(\right) -

    - - Enter the sum of terms here. For example, 1+2+3. - -

    - -

    - - Enter the actual sum here. - -

    - -

    -     -     -     - - - - - $lb = random(0,1,1); - $ub = random(4,6,1); - if($envir{problemSeed}==1){$lb=0;$ub=5;}; - Context()->variables->are(i=>'Real'); - $f = Formula("(-1)^i cos(pi i)")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - Context("Form"); - Context()->operators->set( - '+' => {class => 'bizarro::BOP::add', isCommand => 1}, - '-' => {class => 'bizarro::BOP::subtract', isCommand => 1}, - ); - $terms = Formula("$string"); - - -

    - \sum\limits_{i=}^{} -

    - - Enter the sum of terms here. For example, 1+2+3. - -

    - -

    - - Enter the actual sum here. - -

    - -

    -     -     -     -     - - - -

    - Write the sum in summation notation. -

    -     - - - - - $lb = 1; - $ub = random(4,6,1); - $m = random(2,9,1); - if($envir{problemSeed}==1){$ub=5;$m=3;}; - Context()->variables->are(i=>'Real'); - $f = Formula("$m i")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - Context()->flags->set(reduceConstants=>0); - $terms = Formula("$string"); - $multians = MultiAnswer($lb, $ub, $f)->with( - singleResult => 1, - checker => sub { - my ( $correct, $student, $self ) = @_; - my ( $stulb, $stuub, $stuf ) = @{$student}; - my ( $corlb, $corub, $corf ) = @{$correct}; - if ($stuub - $stulb != $corub - $corlb) {Value::Error("Your sum does not have the correct number of terms");}; - @stui = ($stulb..$stuub); - @stufi = map{$stuf->eval(i=>$_)}(@stui); - @cori = ($corlb..$corub); - @corfi = map{$corf->eval(i=>$_)}(@cori); - $score = 1; - for $j (0..$#corfi) { - if ($stufi[$j] != $corfi[$j]) { - $score = 0; - last; - } - } - return $score; - } - ); - - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the start value here. - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the end value here. - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the summand expression here. - -

    - -

    -     -     -     - - - - - $lb = 0; - $ub = random(5,9,1); - $b = non_zero_random(-2,2,1); - if($envir{problemSeed}==1){$ub=8;$b=-1;}; - Context()->variables->are(i=>'Real'); - $f = Formula("i^2 + $b")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - Context()->flags->set(reduceConstants=>0); - $terms = Formula("$string"); - $multians = MultiAnswer($lb, $ub, $f)->with( - singleResult => 1, - checker => sub { - my ( $correct, $student, $self ) = @_; - my ( $stulb, $stuub, $stuf ) = @{$student}; - my ( $corlb, $corub, $corf ) = @{$correct}; - if ($stuub - $stulb != $corub - $corlb) {Value::Error("Your sum does not have the correct number of terms");}; - @stui = ($stulb..$stuub); - @stufi = map{$stuf->eval(i=>$_)}(@stui); - @cori = ($corlb..$corub); - @corfi = map{$corf->eval(i=>$_)}(@cori); - $score = 1; - for $j (0..$#corfi) { - if ($stufi[$j] != $corfi[$j]) { - $score = 0; - last; - } - } - return $score; - } - ); - - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the start value here. - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the end value here. - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the summand expression here. - -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(4,6,1); - $b = random(1,5,1); - if($envir{problemSeed}==1){$ub=4;$b=1;}; - Context("Fraction"); - Context()->flags->set(reduceFractions=>0); - Context()->variables->are(i=>'Real'); - $f = Formula("i/(i+$b)")->reduce; - @i = ($lb..$ub); - @fi = map{Fraction($_,$_+$b)}(@i); - $string = join('+',@fi); - Context()->flags->set(reduceConstants=>0); - $terms = Formula("$string"); - $multians = MultiAnswer($lb, $ub, $f)->with( - singleResult => 1, - checker => sub { - my ( $correct, $student, $self ) = @_; - my ( $stulb, $stuub, $stuf ) = @{$student}; - my ( $corlb, $corub, $corf ) = @{$correct}; - if ($stuub - $stulb != $corub - $corlb) {Value::Error("Your sum does not have the correct number of terms");}; - @stui = ($stulb..$stuub); - @stufi = map{$stuf->eval(i=>$_)}(@stui); - @cori = ($corlb..$corub); - @corfi = map{$corf->eval(i=>$_)}(@cori); - $score = 1; - for $j (0..$#corfi) { - if ($stufi[$j] != $corfi[$j]) { - $score = 0; - last; - } - } - return $score; - } - ); - - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the start value here. - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the end value here. - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the summand expression here. - -

    - -

    -     -     -     - - - - - $lb = list_random(0,1); - $ub = $lb + 4; - if($envir{problemSeed}==1){$lb=0;}; - Context()->variables->are(i=>'Real'); - $f = ($lb == 0) ? Formula("(-1)^i") : Formula("(-1)^(i+1)"); - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_) * Formula("e^$_")}(@i); - $f = ($lb == 0) ? Formula("(-e)^i") : Formula("-(-e)^i"); - $string = join('+',@fi); - Context()->flags->set(reduceConstants=>0); - $terms = Formula("$string")->reduce; - $multians = MultiAnswer($lb, $ub, $f)->with( - singleResult => 1, - checker => sub { - my ( $correct, $student, $self ) = @_; - my ( $stulb, $stuub, $stuf ) = @{$student}; - my ( $corlb, $corub, $corf ) = @{$correct}; - if ($stuub - $stulb != $corub - $corlb) {Value::Error("Your sum does not have the correct number of terms");}; - @stui = ($stulb..$stuub); - @stufi = map{$stuf->eval(i=>$_)}(@stui); - @cori = ($corlb..$corub); - @corfi = map{$corf->eval(i=>$_)}(@cori); - $score = 1; - for $j (0..$#corfi) { - if ($stufi[$j] != $corfi[$j]) { - $score = 0; - last; - } - } - return $score; - } - ); - - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the start value here. - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the end value here. - -

    - -

    - - If the sum were in sigma notation, \sum\limits_{i=\text{start}}^{\text{end}}(\text{summand expression}), enter the summand expression here. - -

    - -

    -     -     -     -     - - - -

    - Evaluate the summation using . -

    -     - - - - - $lb = 1; - ($ub,$c) = random_subset(2,5..10); - if($envir{problemSeed}==1){$ub=10;$c=5;}; - Context()->variables->are(i=>'Real'); - $f = Formula("$c")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - - -

    - \sum\limits_{i=}^{} -

    -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(20,30,1); - if($envir{problemSeed}==1){$ub=25;}; - Context()->variables->are(i=>'Real'); - $f = Formula("i")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - - -

    - \sum\limits_{i=}^{} -

    -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(8,12,1); - $a = random(2,6,1); - $b = non_zero_random(-4,4,1); - if($envir{problemSeed}==1){$ub=10;$a=3;$b=-2;}; - Context()->variables->are(i=>'Real'); - $f = Formula("$a i^2 + $b i")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - - -

    - \sum\limits_{i=}^{} \left(\right) -

    -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(12,20,1); - $a = random(2,6,1); - $b = non_zero_random(-10,10,1); - if($envir{problemSeed}==1){$ub=15;$a=2;$b=-10;}; - Context()->variables->are(i=>'Real'); - $f = Formula("$a i^3 + $b")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - - -

    - \sum\limits_{i=}^{} \left(\right) -

    -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(8,12,1); - ($a,$b,$c,$d) = random_subset(4,-12..-1,1..12); - if($envir{problemSeed}==1){$ub=10;$a=-4;$b=10;$c=-7;$d=11;}; - Context()->variables->are(i=>'Real'); - Context()->noreduce('(-x)+y','(-x)-y'); - $f = Formula("$a i^3 + $b i^2 + $c i + $d")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - - -

    - \sum\limits_{i=}^{} \left(\right) -

    -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(8,12,1); - ($b,$c,$d) = random_subset(3,-9..-1,1..9); - if($envir{problemSeed}==1){$ub=10;$b=-3;$c=2;$d=7;}; - Context()->variables->are(i=>'Real'); - Context()->noreduce('(-x)+y','(-x)-y'); - $f = Formula("i^3 + $b i^2 + $c i + $d")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - - -

    - \sum\limits_{i=}^{} \left(\right) -

    -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(80,120,5); - if($envir{problemSeed}==1){$ub=100;}; - Context()->variables->are(i=>'Real'); - $f = Formula("i")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - $penultimate = $ub - 1; - - -

    - 1+2+3+\cdots++ -

    -

    - -

    -     -     -     - - - - - $lb = 1; - $ub = random(16,30,1); - if($envir{problemSeed}==1){$ub=20;}; - Context()->variables->are(i=>'Real'); - $f = Formula("i^2")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - $last = $ub**2; - $penultimate = ($ub - 1)**2; - - -

    - 1+4+9+\cdots++ -

    -

    - -

    -     -     -     -     - - - -

    - states - \sum\limits_{i=1}^na_i = \sum\limits_{i=1}^k a_i + \sum\limits_{i=k+1}^n a_i, - so - \sum\limits_{i=k+1}^na_i = \sum\limits_{i=1}^n a_i - \sum\limits_{i=1}^k a_i. - Use this fact, - along with other parts of , - to evaluate the summation. -

    -     - - - - - $lb = random(7,13,1); - $ub = $lb + random(8,12,1); - if($envir{problemSeed}==1){$lb=11;$ub=20;}; - Context()->variables->are(i=>'Real'); - $f = Formula("i")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - - -

    - \sum\limits_{i=}^{} -

    -

    - -

    -     -     -     - - - - - $lb = random(14,18,1); - $ub = $lb + random(8,12,1); - if($envir{problemSeed}==1){$lb=16;$ub=25;}; - Context()->variables->are(i=>'Real'); - $f = Formula("i**3")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - - -

    - \sum\limits_{i=}^{} -

    -

    - -

    -     -     -     - - - - - $lb = random(5,8,1); - $ub = $lb + random(5,8,1); - $c = random(3,8,1); - if($envir{problemSeed}==1){$lb=7;$ub=12;$c=4;}; - Context()->variables->are(i=>'Real'); - $f = Formula("$c")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - - -

    - \sum\limits_{i=}^{} -

    -

    - -

    -     -     -     - - - - - $lb = random(5,8,1); - $ub = $lb + random(5,8,1); - $c = random(3,8,1); - if($envir{problemSeed}==1){$lb=5;$ub=10;$c=4;}; - Context()->variables->are(i=>'Real'); - $f = Formula("$c i**3")->reduce; - @i = ($lb..$ub); - @fi = map{$f->eval(i=>$_)}(@i); - $string = join('+',@fi); - $sum = Real($string); - - -

    - \sum\limits_{i=}^{} -

    -

    - -

    -     -     -     -     - - - -

    - A definite integral \ds \int_a^b f(x) \, dx is given. -

      -
    1. -

      - Graph f(x) on [a,b]. -

      -
      2. -
    2. -

      - Add to the sketch rectangles using the provided rule. -

      -
      3. -
    3. -

      - Approximate \ds \int_a^b f(x) \, dx by summing the areas of the rectangles. -

      -
      4. -
    -

    -     - - - -

    - \ds \int_{-3}^3 x^2\, dx, - with 6 rectangles using the Left Hand Rule. -

    -     -     - - - -

    - \ds \int_{0}^2 (5-x^2)\, dx, - with 4 rectangles using the Midpoint Rule. -

    -     -     - - - -

    - \ds \int_{0}^\pi \sin(x) \, dx, - with 6 rectangles using the Right Hand Rule. -

    -     -     - - - -

    - \ds \int_{0}^3 2^x\, dx, - with 5 rectangles using the Left Hand Rule. -

    -     -     - - - -

    - \ds \int_{1}^2 \ln(x) \, dx, - with 3 rectangles using the Midpoint Rule. -

    -     -     - - - -

    - \ds \int_{1}^9 \frac1x\, dx, - with 4 rectangles using the Right Hand Rule. -

    -     -     -     - - - -

    - A definite integral is given below. - As demonstrated in Examples - and , do the following: -

      -
    1. -

      - Find a formula to approximate the definite integral - using n subintervals and the provided rule. -

      -
      2. -
    2. -

      - Evaluate the formula using n=10, - 100, and 1000. -

      -
      3. -
    3. -

      - Find the limit of the formula, as n\to \infty, - to find the exact value of the definite integral. -

      -
      4. -
    -

    -     - - - - - $rule = list_random('Left Hand Rule','Right Hand Rule'); - if($envir{problemSeed}==1){$rule = 'Right Hand Rule';}; - Context()->variables->are(n=>"Real"); - if ($rule eq 'Left Hand Rule') { - $sum = Formula("(n-1)^2/(4n^2)"); - } - elsif ($rule eq 'Right Hand Rule') { - $sum = Formula("(n+1)^2/(4n^2)"); - }; - $sum10 = $sum->eval(n=>10); - $sum100 = $sum->eval(n=>100); - $sum1000 = $sum->eval(n=>1000); - Context("Fraction"); - $exact = Fraction(1,4); - - -

    - \ds \int_{0}^1 x^3\, dx, using the . -

    - - Enter the formula in terms of n here. - -

    - -

    - - Evaluate the formula using n=10 here. - -

    - -

    - - Evaluate the formula using n=100 here. - -

    - -

    - - Evaluate the formula using n=1000 here. - -

    - -

    - - Enter the limit here. - -

    - -

    -     -     -     - - - - - $rule = list_random('Left Hand Rule','Right Hand Rule'); - $lb = random(-2,-1,1); - $ub = random(1,2,1); - $c = random(2,5,1); - if($envir{problemSeed}==1){$rule = 'Left Hand Rule';$lb=-1;$ub=1;$c=3;}; - Context("Fraction"); - Context()->variables->are(n=>"Real"); - $A = Fraction($c*($ub - $lb)*(6*$lb*$ub + 2*($ub - $lb)**2),6); - $B = Fraction($c*($ub - $lb)*(6*$lb*($ub - $lb) + 3*($ub - $lb)**2),6); - if ($rule eq 'Left Hand Rule') {$B = -$B;}; - $C = Fraction($c*($ub - $lb)**3,6); - ($Bn,$Bd)=$B->value; - ($Cn,$Cd)=$C->value; - $sum = Formula("$A + $Bn/($Bd n) + $Cn/($Cd n^2)"); - $sum10 = $sum->eval(n=>10); - $sum100 = $sum->eval(n=>100); - $sum1000 = $sum->eval(n=>1000); - $exact = $A; - - -

    - \ds \int_{}^{} x^2\, dx, using the . -

    - - Enter the formula in terms of n here. - -

    - -

    - - Evaluate the formula using n=10 here. - -

    - -

    - - Evaluate the formula using n=100 here. - -

    - -

    - - Evaluate the formula using n=1000 here. - -

    - -

    - - Enter the limit here. - -

    - -

    -     -     -     - - - - - $lb = random(-2,-1,1); - $ub = random(1,4,1); - $m = random(2,5,1); - $b = non_zero_random(-2,2,1); - if($envir{problemSeed}==1){$lb=-1;$ub=3;$m=3;$b=-1;}; - $f = Formula("$m x + $b")->reduce; - Context("Fraction"); - Context()->variables->are(n=>"Real"); - $sum = Fraction(($ub-$lb)*($m*($ub+$lb)+2*$b),2); - $sum = Formula("$sum"); - $sum10 = $sum->eval(n=>10); - $sum100 = $sum->eval(n=>100); - $sum1000 = $sum->eval(n=>1000); - $exact = Fraction("$sum"); - - -

    - \ds \int_{}^{} \left(\right)\, dx, using the Midpoint Rule. -

    - - Enter the formula in terms of n here. - -

    - -

    - - Evaluate the formula using n=10 here. - -

    - -

    - - Evaluate the formula using n=100 here. - -

    - -

    - - Evaluate the formula using n=1000 here. - -

    - -

    - - Enter the limit here. - -

    - -

    -     -     -     - - - - - $rule = list_random('Left Hand Rule','Right Hand Rule'); - $lb = random(1,2,1); - $ub = $lb + random(2,4,1); - $m = random(2,5,1); - $b = non_zero_random(-4,4,1); - if($envir{problemSeed}==1){$rule='Left Hand Rule';$lb=1;$ub=4;$m=2;$b=-3;}; - $f = Formula("$m x^2 + $b")->reduce; - Context("Fraction"); - $A = Fraction(($ub - $lb)*($m*$ub**2 + $m*$ub*$lb + $m*$lb**2 + 3*$b),3); - $B = Fraction($m*($ub - $lb)**2*($ub + $lb),2); - if ($rule eq 'Left Hand Rule') {$B = -$B;}; - $C = Fraction($m*($ub - $lb)**3,6); - ($Bn,$Bd)=$B->value; - ($Cn,$Cd)=$C->value; - Context()->variables->are(n=>"Real"); - $sum = Formula("$A + $Bn/($Bd n) + $Cn/($Cd n^2)"); - $sum10 = $sum->eval(n=>10); - $sum100 = $sum->eval(n=>100); - $sum1000 = $sum->eval(n=>1000); - $exact = $A; - - -

    - \ds \int_{}^{} \left(\right)\, dx, using the . -

    - - Enter the formula in terms of n here. - -

    - -

    - - Evaluate the formula using n=10 here. - -

    - -

    - - Evaluate the formula using n=100 here. - -

    - -

    - - Evaluate the formula using n=1000 here. - -

    - -

    - - Enter the limit here. - -

    - -

    -     -     -     - - - - - $rule = list_random('Left Hand Rule','Right Hand Rule'); - $lb = random(-12,-7,1); - $ub = -$lb; - $b = non_zero_random(3,8,1); - if($envir{problemSeed}==1){$rule='Right Hand Rule';$lb=-10;$ub=10;$b=5;}; - $f = Formula("$b - x")->reduce; - Context()->variables->are(n=>"Real"); - $sum = Formula("2*$ub*$b + (2*$ub**2)/n")->reduce; - if ($rule eq 'Left Hand Rule') { - $sum = Formula("2*$ub*$b - (2*$ub**2)/n")->reduce; - }; - $sum10 = $sum->eval(n=>10); - $sum100 = $sum->eval(n=>100); - $sum1000 = $sum->eval(n=>1000); - $exact = Real(2*$ub*$b); - - -

    - \ds \int_{}^{} \left(\right)\, dx, using the . -

    - - Enter the formula in terms of n here. - -

    - -

    - - Evaluate the formula using n=10 here. - -

    - -

    - - Evaluate the formula using n=100 here. - -

    - -

    - - Evaluate the formula using n=1000 here. - -

    - -

    - - Enter the limit here. - -

    - -

    -     -     -     - - - - - $rule = list_random('Left Hand Rule','Right Hand Rule'); - if($envir{problemSeed}==1){$rule='Right Hand Rule';}; - $f = Formula("x^3 - x^2")->reduce; - Context("Fraction"); - Context()->variables->are(n=>"Real"); - $sum = Formula("-1/12+1/(12n^2)"); - $sum10 = $sum->eval(n=>10); - $sum100 = $sum->eval(n=>100); - $sum1000 = $sum->eval(n=>1000); - $exact = Fraction(-1,12); - - -

    - \ds \int_{0}^{1} \left(\right)\, dx, using the . -

    - - Enter the formula in terms of n here. - -

    - -

    - - Evaluate the formula using n=10 here. - -

    - -

    - - Evaluate the formula using n=100 here. - -

    - -

    - - Evaluate the formula using n=1000 here. - -

    - -

    - - Enter the limit here. - -

    - -

    -     -     -     -     -     -     -
    -
    - The Fundamental Theorem of Calculus - - - - -

    - Let f(t) be a continuous function defined on [a,b]. - The definite integral \int_a^b f(x)\, dx is the - area under f on [a,b]. - We can turn this concept into a function by letting the upper - (or lower) - bound vary. -

    - -

    - Let F(x) = \int_a^x f(t)\, dt. - It computes the area under f on [a,x] as illustrated in . - We can study this function using our knowledge of the definite integral. - For instance, F(a)=0 since \int_a^af(t)\, dt=0. -

    - -
    - The area of the shaded region is F(x) = \int_a^x f(t)\, dt - - - A concave-down graph in the first quadrant, with the area underneath shaded from a to x. - - -

    - The t and the y axes are uncalibrated. - There are three distinct positions on the t axis, - a, x and b in the order from left to right. - The function f(t) is a curve facing downward, - the areas under the curve between a and x - are shaded. b lies to the right of x - outside of the shaded portion. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={1,2.5,3}, - extra x tick labels={$a$,$x$,$b$}, - ytick=\empty, - xlabel=$t$, - ymin=-.5,ymax=2, - xmin=-.5,xmax=3.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:2.5] {.5*sin(deg(x))+1} \closedcycle; - \addplot [firstcurvestyle,domain=0:3.25,samples=30] {.5*sin(deg(x))+1}; - - \end{axis} - \end{tikzpicture} - - - -
    - - - Exploring the <q>Area so far</q> function - -

    - Consider f(t)=2t pictured in Figure - and its associated area so far function, - F(x)=\int_1^x 2t\, dt. - Using the graph of f and geometry, - find an explicit formula for F. -

    - -
    - The area of the shaded region is F(x) = \int_1^x 2t\, dt - - - The graph y=2t, and below it, the trapezoidal region from t=1 to t=x is shaded. - - -

    - The y axis is drawn from -2 to 8 while the - t axis is drawn without calibration but it has - two distinct points 1 and x. - The graph is a straight line passing through the origin and - has a positive slope. The shaded portion is drawn under the - graph between 1 and x. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={1,2.5}, - extra x tick labels={$1$,$x$}, - %ytick=\empty, - xlabel=$t$, - ymin=-2,ymax=8, - xmin=-1,xmax=3.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:2.5] {2*x} \closedcycle; - \addplot [firstcurvestyle,domain=-1:3.25] {2*x}; - - \end{axis} - - \end{tikzpicture} - - - -
    -     - -

    - We can see from that for x \geq 1, - the area under the curve can be found by subtracting the area of two triangles. - The larger triangle will have a base of x and a height of f(x)=2x, - while the smaller triangle will have a base of 1 and a height of 2. - Therefore, the area under the curve for - x \geq 1 is given by A(x)=\frac12 (x)(2x)-\frac12 (1)(2)=x^2-1. -

    - -
    - The area of the shaded region is F(x) = \int_1^x 2t\, dt - - - The previous trapezoidal region, with heights 2 and 2x shown. - - -

    - Graph of function same as the one used for the previous example. -

    -

    - The y axis is drawn from -2 to 8 while the - t axis is drawn without calibration but it has two - distinct positions 1 and x. The graph is a straight - line passing through the origin and has a positive slope. - The shaded portion is drawn under the graph between 1 and x. -

    -

    - The distance from the origin to point x on the t - axis is marked x. The distance from origin to 1 - is labeled 1. The left boundary of the shaded region the - y value is labeled 2 and on the right boundaries - the y value is labeled 2x. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={1,2.5}, - extra x tick labels={$1$,$x$}, - xlabel=$t$, - ymin=-2,ymax=8, - xmin=-1,xmax=3.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:2.5] {2*x} \closedcycle; - \addplot [firstcurvestyle,domain=-1:3.25] {2*x}; - - \addplot [lineseg,domain=0:2]({1},{x}) node [pos=0.5, right] {$2$}; - \addplot [lineseg,domain=0:1]({x},{0}) node [pos=0.5, above] {$1$}; - \addplot [lineseg,<->, domain=0:2.5] ({x},{-1}) node [pos=0.5, below] {$x$}; - \addplot [lineseg,domain=0:5]({2.5},{x}) node [pos=0.5, right] {$2x$}; - - \end{axis} - - \end{tikzpicture} - - - -
    - -

    - Note that this same formula holds for x\lt 1. - If x \lt 1, - then F(x) = \int_1^x 2t\, dt=-\int_x^1 2t\, dt. - The areas to the left of x=1 will have opposite signs - (since they areas are accumulated - before x=1). - For example, when x=0, - F(0) = -\int_0^1 2t\, dt=-\frac12 (1)(2)=-1. - This is the same value we get from evaluating x^2-1 for x=0. - Also notice that F(-1)=\int_1^{-1} 2t \, dt=-\int_{-1}^1 2t\, dt. - This integral is clearly 0 since the areas over [-1,0] and [0,1] will sum to zero. - Again, this is the same answer obtained by evaluating x^2-1 for x=-1. -

    - -

    - Therefore, we can reasonably say that F(x)=x^2-1. - A plot of both f(x)=2x and - F(x)=x^2-1 are given in Figure . - You should notice a familiar relationship between these two functions. - This relationship is formally stated in . -

    - -
    - Graphs of f(x)=2x and F(x)=x^2-1 - - - Graph of f(x) = 2*x and F(x)= x^2-1. - - -

    - The y axis is drawn from -2 to 8 and - the x axis is drawn from -1 to 3. - The graphs of two functions are shown. - The first is a line through the origin with slope 2, - and the second is a parabola that opens upward, with its vertex at (0,-1). - Both functions intersect at approximately x=-0.4 - and x=2.4, it is also between these points that - the line is above the curve beyond which the curve is - above the line. -

    -

    - The straight line that passes through the origin and has - a positive slope. It lies in the first and the third - quadrant. -

    -

    - The curve appears to start from x intercept - 1. It has a y intercept at -1 and - the second x intercept at 1. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-2,ymax=8, - xmin=-1,xmax=3.5 - ] - - \addplot+ [domain=-1:3.25] {2*x}; - \addplot+ [domain=-1:3.25] {x^2-1}; - - \end{axis} - - \end{tikzpicture} - - - -
    -     - -     -     - - - Fundamental Theorem of Calculus, Parts 1 and 2 -

    - As hinted, - we can apply calculus ideas to F(x); - in particular, we can compute its derivative. - In , - F(x)=x^2-1, so F'(x)=2x=f(x). - While this may seem like an innocuous thing to do, - it has far-reaching implications, - as demonstrated by the fact that the result is given as an important theorem. -

    - - - The Fundamental Theorem of Calculus, Part 1 - -

    - Let f be continuous on [a,b] and let F(x) = \int_a^x f(t)\, dt. - Then F is continuous on [a,b], differentiable on (a,b), and - - Fundamental Theorem of Calculus - - integrationFun. Thm. of Calc. - - - \Fp(x)=f(x) - . - In other words: - - \lzoo{x}{\int_a^x f(t)\, dt}=f(x) - . -

    -     -     - - - - -

    - Initially this seems simple, as demonstrated in the following example. -

    - - - Using the Fundamental Theorem of Calculus, Part 1 - -

    - Let \ds F(x) = \int_{-5}^x (t^2+\sin(t) )\, dt. - What is \Fp(x)? -

    -     - -

    - Using the Fundamental Theorem of Calculus, - we have \Fp(x) = x^2+\sin(x). - That is, the derivative of the area so far function, - is simply the integrand replacing x with t. -

    - -

    - This simple example reveals something incredible: - F(x) is an antiderivative of x^2+\sin(x)! - Therefore, F(x) = \frac13x^3-\cos(x) +C for some value of C. (We can find C, - but generally we do not care. - We know that F(-5)=0, which allows us to compute C. - In this case, C=\cos(-5)+\frac{125}3.) -

    -     - -     - - - - - - - - -

    - What we have done in - was more than finding a complicated way of computing an antiderivative. - Consider a function f defined on an open interval containing a, - b and c. - Suppose we want to compute \int_a^b f(t)\, dt. - First, let - - F(x) = \int_c^x f(t)\, dt - . - Using the properties of the definite integral found in , we know - - \int_a^b f(t)\, dt \amp = \int_a^c f(t)\, dt + \int_c^b f(t)\, dt - \amp = -\int_c^a f(t)\, dt + \int_c^b f(t)\, dt - - Using Equation, - let x=a in the first integral and x=b in the second integral so that - \int_c^a f(t)\, dt =F(a) and \int_c^b f(t)\, dt =F(b). - Therefore: - - \int_a^b f(t)\, dt \amp =-F(a) + F(b) - \amp = F(b) - F(a) - . -

    - -

    - We now see how indefinite integrals and definite integrals are related: - we can evaluate a definite integral using antiderivatives! - In fact, this is exactly what we noticed in . - The area so far function was indeed an anti-derivative of the integrand. - This is the second part of the Fundamental Theorem of Calculus. -

    - - - Fundamental Theorem of Calculus, Part 2 - -

    - Let f be continuous on [a,b] and let F be any - antiderivative of f. - Then - - Fundamental Theorem of Calculus - - integrationFun. Thm. of Calc. - - - \int_a^b f(x)\, dx = F(b) - F(a) - . -

    -     -     - - - - - - - - - - Using the Fundamental Theorem of Calculus, Part 2 - -

    - We spent a great deal of time in the previous section studying \int_0^4(4x-x^2)\, dx. - Using the Fundamental Theorem of Calculus, - evaluate this definite integral. -

    -     - -

    - We need an antiderivative of f(x)=4x-x^2. - All antiderivatives of f have the form F(x) = 2x^2-\frac13x^3+C; - for simplicity, choose C=0. -

    - -

    - The Fundamental Theorem of Calculus states - - \int_0^4(4x-x^2)\, dx \amp= F(4)-F(0) - \amp = \big(2(4)^2-\frac134^3\big)-\big(0-0\big) - \amp = 32-\frac{64}3 = 32/3 - . -

    - -

    - This is the same answer we obtained using limits in the previous section, - just with much less work. -

    -     -     - -

    - Notation: - - integrationnotation - - A special notation is often used in the process of evaluating definite integrals using the Fundamental Theorem of Calculus. - Instead of explicitly writing F(b)-F(a), - the notation F(x)\Big|_a^b is used. - Thus the solution to would be written as: - - \int_0^4(4x-x^2)\, dx \amp = \left.\left(2x^2-\frac13x^3\right)\right|_0^4 - \amp = \big(2(4)^2-\frac134^3\big)-\big(0-0\big) = 32/3 - . -

    - -

    - The Constant C: Any - antiderivative F(x) can be chosen when using the Fundamental Theorem of Calculus to evaluate a definite integral, - meaning any value of C can be picked. - The constant always cancels out of the expression when evaluating F(b)-F(a), - so it does not matter what value is picked. - This being the case, we might as well let C=0. -

    - - - Using the Fundamental Theorem of Calculus, Part 2 - -

    - Evaluate the following definite integrals. - -

      -
    1. \int_{-2}^2 x^3\, dx
      2. - -
    2. \int_0^\pi \sin(x) \, dx
      3. - -
    3. \int_0^5 e^t\, dt
      4. - -
    4. \int_4^9 \sqrt{u}\, du
      5. - -
    5. \int_1^5 2\, dx
      6. -
    -

    -     - -

    -

      -
    1. -

      - - \int_{-2}^2 x^3\, dx \amp = \left.\frac14x^4\right|_{-2}^2 \amp - \amp = \left(\frac142^4\right) - \left(\frac14(-2)^4\right) - \amp = 0 - . -

      -
      2. - -
    2. -

      - - \int_0^\pi \sin(x) \, dx \amp = -\cos(x) \Big|_0^\pi - \amp = -\cos(\pi) - \big(-\cos(0) \big) - \amp = 1+1=2 - . - (This is interesting; it says that the area under one - hump of a sine curve is 2.) -

      -
      3. - -
    3. -

      - - \int_0^5e^t\, dt \amp = e^t\Big|_0^5 - \amp = e^5 - e^0 - \amp= e^5-1 \approx 147.41 - . -

      -
      4. - -
    4. -

      - - \int_4^9 \sqrt{u}\, du \amp= \int_4^9 u^\frac12\, du - \amp = \frac23u^\frac32\Big|_4^9 - \amp = \frac23\left(9^\frac32-4^\frac32\right) - \amp = \frac23\big(27-8\big) =\frac{38}3 - . -

      -
      5. - -
    5. -

      - - \int_1^5 2\, dx \amp = 2x\Big|_1^5 - \amp = 2(5)-2 - \amp =2(5-1)=8 - . - This integral is interesting; - the integrand is a constant function, - hence we are finding the area of a rectangle with width (5-1)=4 and height 2. - Notice how the evaluation of the definite integral led to 2(4)=8. - - In general, if c is a constant, - then \int_a^b c\, dx = c(b-a). -

      -
      6. -
    -

    -     - -     -     - - - Understanding Motion with the Fundamental Theorem of Calculus -

    - We established, - starting with , - that the derivative of a position function is a velocity function, - and the derivative of a velocity function is an acceleration function. - Now consider definite integrals of velocity and acceleration functions. - Specifically, if v(t) is a velocity function, - what does \ds \int_a^b v(t)\, dt mean? - - integrationdisplacement - - displacement - -

    - -

    - The Fundamental Theorem of Calculus states that - - \int_a^b v(t)\, dt = V(b) - V(a) - , - where V(t) is any antiderivative of v(t). - Since v(t) is a velocity function, - V(t) must be a position function, - and V(b) - V(a) measures a change in position, or displacement. -

    - - - Finding displacement and distance - -

    - A ball is thrown straight up with velocity given by - v(t) = -32t+20ft/s, where t is measured in seconds. - Find, and interpret, - \int_0^1 v(t)\, dt and \int_0^1 \abs{v(t)}\, dt. -

    -     - -

    - Using the Fundamental Theorem of Calculus, we have - - \int_0^1 v(t)\, dt \amp = \int_0^1 (-32t+20)\, dt - \amp = \left(-16t^2 + 20t\right)\Big|_0^1 - \amp = 4 - . -

    - -

    - Thus if a ball is thrown straight up into the air with velocity v(t) = -32t+20, - the height of the ball, 1 second later, - will be 4 feet above the initial height. -

    - -

    - Note that the ball has traveled much farther. - It has gone up to its peak and is falling down, - but the difference between its height at t=0 and t=1 is 4ft. -

    - -

    - If we wish to find the total distance traveled, - we must evaluate \int_0^1 \abs{v(t)}\, dt - (noting that negative velocities will reduce the diplacement, - but we want distance, not displacement). - In this case, - we know that the velocity changes sign once when v(t)=0,so t=20/32=5/8 seconds. - The velocity is positive over [0,5/8] and negative over [5/8,1]. - Therefore - - \int_0^1 \abs{v(t)}\, dt \amp = \int_0^{5/8}v(t)\,dt + \int_{5/8}^1 -v(t)\,dt - \amp =\int_0^{5/8}(-32t+20)\, dt-\int_{5/8}^1 (-32t+20)\, dt - \amp =\left(-16t^2 + 20t\right)\Big|_0^{5/8}-\left(-16t^2 + 20t\right)\Big|_0^1 - \amp = \frac{25}4-\left(-\frac94\right)=9 - . - So the total distance traveled over [0,1] is \int_0^1 \abs{-32t+20}\, dt=9 \text{ feet }. -

    - -

    - As we can see in , - the positive area between v(t) and the t-axis, - A_1=25/4, while the negative area, A_2=-9/4. - When we add these two areas, we get the displacement of 4 ft. - But when we add the absolute value of both of these areas - (as in ), - we get the total distance of 9 ft. -

    - - -
    - The area between v(t) and the t-axis can be used to represent displacement - - - Graph showing area between v(t) and the t axis can be used to represent displacement. - - -

    - The y axis is drawn from -10 to 20 - and the t axis is drawn from 0 to 1. - The function is a straight line with a negative slope. - The function has an y intercept of 20 and - an t intercept of 5/8. -

    -

    - The line forms two shaded regions with the t axis. - Between 0 to 5/8 on the t axis the region - under the graph is shaded and labeled A1. - The second area is between 5/8 and 1 that lies - in the fourth quadrant and the area is labeled A2. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={0.625,1}, - extra x tick labels={$5/8$,$1$}, - xlabel=$t$, - ymin=-15,ymax=25, - xmin=-1,xmax=1.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1] {-32*x+20} \closedcycle; - \addplot [firstcurvestyle,domain=-.5:1.5] {-32*x+20}; - - \node at (axis cs:.3125,5) { $A_1$}; - \node at (axis cs:.8125,-3) { $A_2$}; - - \end{axis} - - \end{tikzpicture} - - - -
    - -
    - The area between \abs{v(t)} and the t-axis can be used to represent distance - - - Graph showing area between absolute value of v(t) and the t axis can be used to represent displacement. - - -

    - The y axis is drawn from -10 to 20 and the - t axis is drawn from 0 to 1. - The function is a straight line with a negative slope. - The function has an y intercept of 20 and an - t intercept of 5/8. -

    -

    - The line forms two shaded regions with the t axis. - Between 0 to 5/8 on the t axis the region - under the graph is shaded and labeled A1. - Between 5/8 and 1 lies the second shaded region - labeled A3. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={0.625,1}, - extra x tick labels={$5/8$,$1$}, - xlabel=$t$, - ymin=-15,ymax=25, - xmin=-1,xmax=1.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:5/8] {-32*x+20} \closedcycle; - \addplot [firstcurvestyle,areastyle,domain=5/8:1] {32*x-20} \closedcycle; - - \addplot [firstcurvestyle,domain=-.5:5/8] {-32*x+20}; - \addplot [firstcurvestyle,domain=5/8:1.5] {32*x-20}; - - \node at (axis cs:.3125,5) { $A_1$}; - \node at (axis cs:.8125,3) { $A_3$}; - - \end{axis} - - \end{tikzpicture} - - - -
    -     -     -     - -

    - Integrating a rate of change function gives total change. - Velocity is the rate of position change; - integrating velocity gives the total change of position, - , displacement. -

    - -

    - Integrating a speed function gives a similar, though different, result. - Speed is also the rate of position change, - but does not account for direction. - That is, the speed an object is the absolute value of its velocity. - This is what we saw in - when we evaluated \int_0^1 \abs{v(t)}\, dt. - So integrating a speed function gives total change of position, - without the possibility of negative position change. - Hence the integral of a speed function gives - distance traveled. -

    - -

    - As acceleration is the rate of velocity change, - integrating an acceleration function gives total change in velocity. - We do not have a simple term for this analogous to displacement. - If a(t) = 5miles/h^2 and t is measured in hours, then - - \int_0^3 a(t)\, dt = 15 - - means the velocity has increased by 15m/h from t=0 to t=3. -

    -     - - - The Fundamental Theorem of Calculus and the Chain Rule -

    - Part 1 of the Fundamental Theorem of Calculus (FTC) states that given - - F(x) = \int_a^x f(t)\, dt - , - we have \Fp(x) = f(x). Using other notation, - - \lzoo{x}{F(x)}=\lzoo{x}{\int_a^x f(t)\, dt} = f(x) - . - While we have just practiced evaluating definite integrals, - sometimes finding antiderivatives is impossible and we need to rely on other techniques to approximate the value of a definite integral. - Functions written as F(x) = \int_a^x f(t)\, dt are useful in such situations. -

    - -

    - It may be of further use to compose such a function with another. - As an example, we may compose F(x) with g(x) to get - - F\big(g(x)\big) = \int_a^{g(x)} f(t)\, dt - . -

    - -

    - What is the derivative of such a function? - Fundamental Theorem of Calculusand Chain Rule - The Chain Rule can be employed to state - - \frac{d}{dx}\Big(F\big(g(x)\big)\Big) = \Fp\big(g(x)\big)\gp(x) = f\big(g(x)\big)\gp(x) - . -

    - - - -

    - An example will help us understand this. -

    - - - The FTC, Part 1, and the Chain Rule - -

    - Find the derivative of \ds F(x) = \int_2^{x^2} \ln(t) \, dt. -

    -     - -

    - We can view F(x) as being the function - G(x) = \int_2^x \ln(t) \, dt composed with g(x) = x^2; - that is, F(x) = G\big(g(x)\big). - The Fundamental Theorem of Calculus states that G'(x) = \ln(x). - The Chain Rule gives us - - F'(x) \amp = G'\big(g(x)\big) \gp(x) - \amp = \ln(g(x)) \gp(x) - \amp = \ln(x^2) 2x - \amp =2x\ln(x^2) - -

    - -

    - Normally, the steps defining G(x) and g(x) are skipped. -

    -     -     - - - -

    - Let's practice this once more. -

    - - - - - The FTC, Part 1, and the Chain Rule - -

    - Find the derivative of \ds F(x) = \int_{\cos(x) }^5 t^3\, dt. -

    -     - -

    - Note that \ds F(x) = -\int_5^{\cos(x) } t^3\, dt. - Viewed this way, the derivative of F is straightforward: - - F'(x) \amp= -\cos^3(x)\left(-\sin(x)\right) - \amp= \cos^3(x)\sin(x) - . -

    -     - -     -     - - - Area Between Curves - - -

    - Consider continuous functions f(x) and g(x) defined on [a,b], - where f(x) \geq g(x) for all x in [a,b], - as demonstrated in . - What is the area of the shaded region bounded by the two curves over [a,b]? - integrationarea between curves -

    - -
    - Finding the area bounded by two functions on an interval by subtracting the area under g from the area under f - -
    - - - - Graph showing area bounded by two functions f(x) and g(x) between x=a and x=b. - - -

    - The x and the y axes are uncalibrated but there - are two positions a and b marked on the x axis. - There are two functions f(x) and g(x) graphed. - The graph shows the area that f(x) forms on g(x) - between positions a and b. - The two functions intersect at one point. -

    -

    - The function f(x) is above g(x), - it first curves up and reaches a maxima after which it dips - down for a minima and then continues to increase. - The function g(x) has a dip between a and - b then it increases. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={.5,3}, - extra x tick labels={$a$,$b$}, - ytick=\empty, - ymin=-.5,ymax=2, - xmin=-.5,xmax=3.5 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.25:3.25,samples=40] {.25*sin(deg(2*x))+1.25} node [shift={(5pt,7pt)},black] { $f(x)$}; - \addplot [name path=B,firstcurvestyle,domain=-.25:3.25] {.25*(x-2)^2+.2} node [shift={(5pt,7pt)},black] { $g(x)$}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=.5:3}]; - - \end{axis} - - \end{tikzpicture} - - - -
    - -
    - - - - Graph showing area bounded by two functions f(x) and g(x) between x=a and x=b. - - -

    - Graph same as previous with additional details. - The x and the y axes are uncalibrated but there - are two positions a and b marked on the x axis. - There are two functions f(x) and g(x) graphed. - The graph shows the area that f(x) forms on g(x) - between positions a and b. - The two functions intersect at one point. -

    -

    - The function f(x) is above g(x), it first curves - up and reaches a maxima after which it dips down for a minima - and then continues to increase.The function g(x) has a - dip between a and b then it increases. - The graph shows areas under both curves as different shaded regions. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={.5,3}, - extra x tick labels={$a$,$b$}, - ytick=\empty, - ymin=-.5,ymax=2, - xmin=-.5,xmax=3.5 - ] - - \addplot [firstcurvestyle,areastyle,area style,domain=.5:3] {.25*sin(deg(2*x))+1.25} \closedcycle; - \addplot [secondcurvestyle,areastyle,domain=.5:3] {.25*(x-2)^2+.2} \closedcycle; - - \addplot [firstcurvestyle,domain=-.25:3.25,samples=40] {.25*sin(deg(2*x))+1.25} node [shift={(5pt,7pt)},black] { $f(x)$}; - \addplot [secondcurvestyle,domain=-.25:3.25] {.25*(x-2)^2+.2}node [shift={(5pt,7pt)},black] { $g(x)$}; - - \end{axis} - - \end{tikzpicture} - - - -
    -     -
    - -

    - The area can be found by recognizing that this area is - the area under f - the area under g. - Using mathematical notation, the area is - - \int_a^b f(x)\, dx - \int_a^b g(x)\, dx - . -

    - -

    - Properties of the definite integral allow us to simplify this expression to - - \int_a^b\big(f(x) - g(x)\big)\, dx - . -

    - - - Area Between Curves - -

    - Let f(x) and g(x) be continuous functions defined on [a,b] where - f(x)\geq g(x) for all x in [a,b]. - The area of the region bounded by the curves y=f(x), - y=g(x) and the lines x=a and x=b is - - \int_a^b \big(f(x)-g(x)\big)\, dx - . -

    -     -     - - - Finding area between curves - -

    - Find the area of the region enclosed by y=x^2+x-5 and y=3x-2. -

    -     - -

    - It will help to sketch these two functions, - as done in . -

    - -
    - Sketching the region enclosed by y=x^2+x-5 and y=3x-2 in - - - Graph of region enclosed by y=x^2+x-5 and y=3*x-2. - - -

    - The y axis is drawn from -5 to 15 and - the x axis is drawn from -2 to 4. - The first function x^2 +x-5is a curve that has a - y intercept at -5 and the x intercept - near 1.8. The second function 3*x-2 is a - straight line. The two lines intersect at points (-1,-5) - and (3,7). The area bounded by the two functions is shaded. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-2,-1,1,2,3,4}, - ytick={5,10,15}, - ymin=-10,ymax=16, - xmin=-2.5,xmax=4.5 - ] - - \addplot+ [name path=F,domain={-2:4},samples=40] {x^2+x-5} node [pos=1, left] { $y=x^2+x-5$}; - \addplot+ [name path=G,domain={-2:4}] {3*x-2} node [pos=0.7, above left] { $y=3x-2$}; - \addplot [gray!50] fill between [of=F and G, soft clip={domain=-1:3}]; - - \end{axis} - - \end{tikzpicture} - - - -
    - -

    - The region whose area we seek is completely bounded by these two functions; - they seem to intersect at x=-1 and x=3. - To check, set x^2+x-5=3x-2 and solve for x: - - x^2+x-5 \amp = 3x-2 - (x^2+x-5) - (3x-2) \amp = 0 - x^2-2x-3 \amp = 0 - (x-3)(x+1) \amp = 0 - x\amp =-1,\,3 - . -

    - -

    - Following , - the area is - - \int_{-1}^3\big(3x-2 -(x^2+x-5)\big)\, dx \amp = \int_{-1}^3 (-x^2+2x+3)\, dx - \amp =\left.\left(-\frac13x^3+x^2+3x\right)\right|_{-1}^3 - \amp =-\frac13(27)+9+9-\left(\frac13+1-3\right) - \amp = 10\frac23 = 10.\overline{6} - -

    -     - -     - - - - - -     - - - The Mean Value Theorem and Average Value -
    - A graph of a function f to introduce the Mean Value Theorem - - - A graph of a function to introduce the Mean Value Theorem. - - -

    - The y axis is uncalibrated and the x axis is - drawn from 0 to 4. - The curve first decreases from 0 to 3 then - it increases until x=4. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick=\empty, - ymin=-.5,ymax=1.75, - xmin=-.5,xmax=4.5 - ] - - \addplot+ [domain=-.5:4.5,samples=40] {.3*cos(deg(x))+1}; - - \end{axis} - - \end{tikzpicture} - - - -
    - -

    - Consider the graph of a function f in - and the area defined by \int_1^4 f(x)\, dx. - Three rectangles are drawn in ; - in , - the height of the rectangle is greater than f on [1,4], - hence the area of this rectangle is is greater than \int_1^4 f(x)\, dx. -

    - -

    - In , - the height of the rectangle is smaller than f on [1,4], - hence the area of this rectangle is less than \int_1^4 f(x)\, dx. -

    - -

    - Finally, in - the height of the rectangle is such that the area of the rectangle is exactly - that of \int_1^4 f(x)\, dx. - Since rectangles that are too big, - as in (a), and rectangles that are - too little, as in (b), - give areas greater/lesser than \int_1^4 f(x)\, dx, - it makes sense that there is a rectangle, - whose top intersects f(x) somewhere on [1,4], - whose area is exactly that of the definite integral. -

    - -
    - Differently sized rectangles give upper and lower bounds on \int_1^4 f(x)\, dx; the last rectangle matches the area exactly - -
    - - - - Graph of function with rectangle to show upper bound on function. - - -

    - The y axis is uncalibrated and the x axis is drawn - from 0 to 4. The curve first decreases from 0 - to 3 then it increases until x=4. - The area under the curve between x=1 to x=4 is shaded. - A rectangular boundary is drawn on the x axis with height - above the curve. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick=\empty, - ymin=-.5,ymax=1.75, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:4] {.3*cos(deg(x))+1} \closedcycle; - \addplot [firstcurvestyle,domain=-.5:4.5,samples=40] {.3*cos(deg(x))+1}; - - \draw [thick,secondcolor] (axis cs: 1,0) rectangle (axis cs:4,1.35); - - \end{axis} - - \end{tikzpicture} - - - -
    - -
    - - - - Graph of function with rectangle to show lower bound on function. - - -

    - The y axis is uncalibrated and the x axis is drawn - from 0 to 4. The curve first decreases from 0 - to 3 then it increases until x=4. - The area under the curve between x=1 to x=4 is shaded. - A rectangular boundary is drawn on the x axis with height - a little below the curve. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick=\empty, - ymin=-.5,ymax=1.75, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:4] {.3*cos(deg(x))+1} \closedcycle; - \addplot [firstcurvestyle,domain=-.5:4.5,samples=40] {.3*cos(deg(x))+1}; - - \draw [thick,secondcolor] (axis cs: 1,0) rectangle (axis cs:4,.65); - - \end{axis} - - \end{tikzpicture} - - - -
    - -
    - - - - Graph of function with rectangle that matches area of function exactly. - - -

    - The y axis is uncalibrated and the x axis is drawn from - 0 to 4. The curve first decreases from 0 to 3 - then it increases until x=4. - The area under the curve between x=1 to x=4 is shaded. - A rectangular boundary is drawn on the x axis such that from - 1 to 2 the curve is above the rectangle and from 2 - to 4 the curve is below the rectangle. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick=\empty, - ymin=-.5,ymax=1.75, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:4] {.3*cos(deg(x))+1} \closedcycle; - \addplot [firstcurvestyle,domain=-.5:4.5,samples=50] {.3*cos(deg(x))+1}; - - \draw [thick,secondcolor] (axis cs: 1,0) rectangle (axis cs:4,.84); - - \end{axis} - - \end{tikzpicture} - - - -
    -     -
    - -

    - We state this idea formally in a theorem. -

    - - - The Mean Value Theorem of Integration - -

    - Let f be continuous on [a,b]. - Mean Value Theoremof integration - integrationMean Value Theorem - There exists a value c in [a,b] such that - - \int_a^bf(x)\, dx = f(c)(b-a) - . -

    -     -     - - - - -

    - This is an existential statement; - c exists, but we do not provide a method of finding it. - - is directly connected to the Mean Value Theorem of Differentiation, given - as ; - we leave it to the reader to see how. -

    - - - -

    - We demonstrate the principles involved in this version of the Mean Value Theorem in the following example. -

    - - - Using the Mean Value Theorem - -

    - Consider \int_0^\pi \sin(x) \, dx. - Find a value c guaranteed by the Mean Value Theorem. -

    -     - -

    - We first need to evaluate \int_0^\pi \sin(x) \, dx. - (This was previously done in .) - - \int_0^\pi\sin(x) \, dx = -\cos(x) \Big|_0^\pi = 2 - . -

    - -

    - Thus we seek a value c in [0,\pi] such that \pi\sin(c) =2. - - \pi\sin(c) = 2\,\,\Rightarrow\,\,\sin(c) = 2/\pi\,\,\Rightarrow\,\,c = \arcsin(2/\pi) \approx 0.69 - . -

    - -
    - A graph of y=\sin(x) on [0,\pi] and the rectangle guaranteed by the Mean Value Theorem - - - Graph of function y=sin(x) and the rectangle guaranteed by the Mean Value Theorem. - - -

    - The y axis has two positions marked sin(0.69) and - 1, the x axis has c, 1, 2 and - \pi marked in order. The function is an inverted parabola - of height 1 and has x intercepts at x=0 and - x=\pi. The area under the parabola until the x axis - is shaded. - A rectangular boundary is drawn with height sin(0.69) and - width \pi -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2}, - extra x ticks={.69,3.14}, - extra x tick labels={$c$,$\pi$}, - ytick={1}, - extra y ticks={.637}, - extra y tick labels={$\sin(0.69) $}, - ymin=-.25,ymax=1.25, - xmin=-.5,xmax=3.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:3.14] {sin(deg(x))}; - \addplot [firstcurvestyle,domain=-.5:3.5,samples=40] {sin(deg(x))}; - - \draw [thick,secondcolor] (axis cs: 0,0) rectangle (axis cs:3.14,.637); - - \end{axis} - - \end{tikzpicture} - - - -
    - -

    - In - \sin(x) is sketched along with a rectangle with height \sin(0.69). - The area of the rectangle is the same as the area under \sin(x) on [0,\pi]. -

    -     -     - -

    - We now turn our attention to a related topic average value. - Let f be a function on [a,b] with c such that f(c)(b-a) = \int_a^bf(x)\, dx. - Consider \int_a^b\big(f(x)-f(c)\big)\, dx: - - \int_a^b\big(f(x)-f(c)\big)\, dx \amp = \int_a^b f(x) - \int_a^b f(c)\, dx - \amp = f(c)(b-a) - f(c)(b-a) - \amp = 0 - . -

    - -

    - When f(x) is shifted by -f(c), - the amount of area under f above the x-axis on [a,b] is the same as the amount of area below the x-axis above f; - see for an illustration of this. - In this sense, - we can say that f(c) is the average value - of f on [a,b]. -

    - - -
    - On the left, a graph of y=f(x) and the rectangle guaranteed by the Mean Value Theorem. On the right, y=f(x) is shifted down by f(c); the resulting area under the curve is 0 - - - - A graph of y = f(x) and the rectangle guaranteed by the Mean Value Theorem. - - -

    - The x and the y axes are uncalibrated. - The y axis is labeled as f(x). - The x axis has three distinct positions marked a, - c and b in the order. -

    -

    - The function represents a parabola and the graph shows - the area under the parabola. a and b are the - two ends for the area. c is to the left of the vertex - of the parabola almost at half the distance from a. -

    -

    - The rectangle guaranteed by the Mean Value Theorem lies above - the vertex of the parabola. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={1,4,1.63}, - extra x tick labels={$a$,$b$,$c$}, - ytick=\empty, - extra y ticks={1.75}, - extra y tick labels={$f(c)$}, - ymin=-1,ymax=3.5, - xmin=-.5,xmax=4.25 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:4] {(x-2.5)^2+1} \closedcycle; - \addplot [firstcurvestyle,domain=-.5:4.25,samples=40] {(x-2.5)^2+1}; - - \draw [thick,secondcolor] (axis cs: 1,0) rectangle (axis cs:4,1.75); - \draw (axis cs:3.35,3) node { $y=f(x)$}; - \end{axis} - - \end{tikzpicture} - - - - - - - Graph of f(x) is shifted down by f(c), the resulting “area under the curve” is 0. - - -

    - The x and the y axes are uncalibrated. - The y axis is labeled as f(x). - The x axis has three distinct positions marked - a, c and b in the order. -

    -

    - The function represents a parabola and the graph shows - the area under the parabola. a and b are the - two ends for the area. c is to the left of the - vertex of the parabola almost at half the distance from - a. -

    -

    - Sine the function is shifted by f(c), there are three - parts to the area, two are positive and are above the - x axis, from a to c and from a little - before b to b. From a little before b - to b the area is drawn in the fourth quadrant from - above the parabola on the x axis. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={1,4,1.63}, - extra x tick labels={$a$,$b$,$c$}, - ytick=\empty, - extra y ticks={1.75}, - extra y tick labels={$f(c)$}, - ymin=-1,ymax=3.5, - xmin=-.5,xmax=4.25 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:4] {(x-2.5)^2-.75} \closedcycle; - \addplot [firstcurvestyle,domain=.75:4.25,samples=40] {(x-2.5)^2-.75}; - - \draw (axis cs:3.2,2.5) node { $y=f(x)-f(c)$}; - \end{axis} - - \end{tikzpicture} - - - -     -
    - -

    - The value f(c) is the average value in another sense. - First, recognize that the Mean Value Theorem can be rewritten as - - f(c) = \frac{1}{b-a}\int_a^b f(x)\, dx - , - for some value of c in [a,b]. - - - Replacing the integral with the limit of a Riemann sum (as in ): - - f(c) \amp = \frac{1}{b-a}\int_a^b f(x)\, dx \amp - \amp = \frac{1}{b-a} \lim\limits_{n \to \infty}\sum_{i=1}^n f(c_i)\,\Delta x \amp \text{Using } - \amp =\frac{1}{b-a} \lim\limits_{n \to \infty}\sum_{i=1}^n f(c_i)\,\frac{b-a}{n}\amp \Delta x =\frac{b-a}{n} - \amp =\lim\limits_{n \to \infty}\sum_{i=1}^n f(c_i)\frac{1}{n} \amp \text{Cancelling the common factor of } b-a - . -

    - -

    - Examining this last line closely, - the expression \sum_{i=1}^n f(c_i)\frac{1}{n} represents adding up n sample values of f(x)and then dividing by n. - This is exactly what we do when we calculate the average of a set of n numbers. - Now when we consider taking the limit as n goes to \infty, - \lim\limits_{n \to \infty}\sum_{i=1}^n f(c_i)\frac{1}{n}, - we are adding up all of the function's output values over [a,b] and dividing by the - number of numbers. - In a sense, we are adding up an infinite number of output values and then dividing by the number of terms we summed - (which is again infinite). -

    - -

    - This leads us to a definition. -

    - - - The Average Value of <m>f</m> on <m>[a,b]</m> - -

    - Let f be continuous on [a,b]. - The average value of f on [a,b] is f(c), - where c is a value in [a,b] guaranteed by the Mean Value Theorem. - integrationaverage value - average value of function - , - - \text{ Average Value of \(f\) on \([a,b]\) } = \frac{1}{b-a}\int_a^b f(x)\, dx - . -

    -     -     - - - -

    - An application of this definition is given in the following example. -

    - - - Finding the average value of a function - -

    - An object moves back and forth along a straight line with a velocity given by v(t) = (t-1)^2 on [0,3], - where t is measured in seconds and v(t) is measured in ft/s. -

    - -

    - What is the average velocity of the object? -

    -     - -

    - By our definition, the average velocity is: - - \frac{1}{3-0}\int_0^3 (t-1)^2\, dt \amp =\frac13 \int_0^3 \big(t^2-2t+1\big)\, dt - \amp = \left.\frac13\left(\frac13t^3-t^2+t\right)\right|_0^3 - \amp =\frac13\left[\left(\frac13(3)^3-(3)^2+(3)\right)-\left(\frac13(0)^3-(0)^2+(0)\right)\right] - \amp = 1\text{ ft/s } - . -

    -     -     - -

    - We can understand the above example through a simpler situation. - Suppose you drove 100 miles in 2 hours. - What was your average speed? - The answer is simple: displacement/time = 100 miles/2 hours = 50 mph. -

    - -

    - What was the displacement of the object in ? - We calculate this by integrating its velocity function: - \int_0^3 (t-1)^2\, dt = 3 ft. - Its final position was 3 feet from its initial position after 3 seconds: - its average velocity was 1 ft/s. -

    - -

    - This section has laid the groundwork for a lot of great mathematics to follow. - The most important lesson is this: - definite integrals can be evaluated using antiderivatives. - Since - established that definite integrals are the limit of Riemann sums, - we can later create Riemann sums to approximate values other than - area under the curve, - convert the sums to definite integrals, - then evaluate these using the . - This will allow us to compute the work done by a variable force, - the volume of certain solids, - the arc length of curves, and more. -

    - -

    - The downside is this: generally speaking, - computing antiderivatives is much more difficult than computing derivatives. - - is devoted to techniques of finding antiderivatives so that a wide variety of definite integrals can be evaluated. - Before that, - - explores techniques of approximating the value of definite integrals beyond using the Left Hand, Right Hand and Midpoint Rules. - These techniques are invaluable when antiderivatives cannot be computed, - or when the actual function f is unknown and all we know is the value of f at certain x-values. -

    -     - - - - Terms and Concepts - - - - -

    - How are definite and indefinite integrals related? -

    - -     - - - -     - - - - -

    - What constant of integration is most commonly used when evaluating definite integrals? -

    -

    - -

    -     -     -     - - - - -

    - - If f is a continuous function, - then \ds F(x) = \int_a^x f(t)\, dt is also a continuous function. -

    -     - -     - - - - -

    - The definite integral can be used to find - the area under a curve. - Give two other uses for definite integrals. -

    - -     - - - -     -     - - - Problems - - -

    - Evaluate the definite integral. -

    -     - - - - - $b = list_random(-1,1); - $c = non_zero_random(-9,9,1); - $low = random(1,2,1); - $high = $low + random(1,3,1); - if($envir{problemSeed}==1){$b=-1;$c=1;$low=1;$high=3}; - Context()->noreduce('(-x)-y','(-x)+y'); - $F = Formula("x^3 + $b x^2 + $c x")->reduce; - $f = $F->D('x')->reduce; - $r = $F->eval(x=>$high) - $F->eval(x=>$low); - - -

    - \ds\int_{}^{} \left(\right)\, dx -

    -

    - -

    -     -     -     - - - - - $c = non_zero_random(-9,9,1); - $low = 0; - $high = random(2,6,1); - if($envir{problemSeed}==1){$c=-1;$high=4;}; - Context("Fraction"); - Context()->noreduce('(-x)-y','(-x)+y'); - $F = Formula("1/3(x + $c)^3")->reduce; - $f = Formula("(x + $c)^2")->reduce; - $r = Fraction($F->eval(x=>$high) - $F->eval(x=>$low)); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random(-1,1); - $low = random(-4,-1,1); - ($m,$n) = random_subset(2,3,5,7); - if($envir{problemSeed}==1){$b=-1;$low=-1;}; - $high = -$low; - Context("Fraction"); - Context()->noreduce('(-x)-y','(-x)+y'); - $F = Formula("x^($m+1)/($m+1) + $b x^($n+1)/($n+1)")->reduce; - $f = Formula("x^($m) + $b x^($n)")->reduce; - $r = $F->eval(x=>$high) - $F->eval(x=>$low); - - -

    - \ds\int_{}^{} \left(\right)\, dx -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','cos'); - $lh = list_random(['0','pi/2'],['0','pi'],['pi/2','pi']); - if($envir{problemSeed}==1){$trig='cos';$lh=['pi/2','pi'];}; - $low = Formula($lh->[0]); - $high = Formula($lh->[1]); - $f = Formula("$trig(x)"); - $F = -$f->D('x')->reduce; - $r = $F->eval(x=>Real("$high")) - $F->eval(x=>Real("$low")); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $trig = list_random('tan','sec'); - $lh = list_random(['0','pi/4'],['0','pi/3'],['pi/4','pi/3']); - if($envir{problemSeed}==1){$trig='tan';$lh=['0','pi/4'];}; - $low = Formula($lh->[0]); - $high = Formula($lh->[1]); - $F = Formula("$trig(x)"); - $f = $F->D('x')->reduce; - $r = $F->eval(x=>Real("$high")) - $F->eval(x=>Real("$low")); - Context()->flags->set(reduceConstantFunctions=>0); - if ($r == 1) {$r = 1;} - elsif ($r == sqrt(3)) {$r = Formula("sqrt(3)");} - elsif ($r == sqrt(3) - 1) {$r = Formula("sqrt(3) - 1");} - elsif ($r == sqrt(2) - 1) {$r = Formula("sqrt(2) - 1");} - elsif ($r == 2 - sqrt(2)) {$r = Formula("2 - sqrt(2)");}; - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $n = random(1,9,1); - if($envir{problemSeed}==1){$n=1;}; - $low = 1; - $high = Formula("e^$n")->reduce; - $F = Formula("ln(x)")->reduce; - $f = Formula("1/x")->reduce; - $r = $n; - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $b = random(2,9,1); - $low = random(-3,-1,1); - $high = random(1,3,1); - if($envir{problemSeed}==1){$b=5;$low=-1;$high=1;}; - $f = Formula("$b^x")->reduce; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $r = Formula("($b**$high - 1/$b**(-$low))/ln($b)"); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - ($low,$high) = num_sort(random_subset(2,-4..-1)); - ($a,$b) = random_subset(2,-9..-1,1..9); - $m = random(2,5,1); - if($envir{problemSeed}==1){$low=-2;$high=-1;$a=4,$b=-2;$m=3;}; - $f = Formula("$a + $b x^$m")->reduce; - Context("Fraction"); - $r = Fraction(($a*$high + $b/($m+1)*$high**($m+1)) - ($a*$low + $b/($m+1)*$low**($m+1))); - - -

    - \ds\int_{}^{} \left(\right)\, dx -

    -

    - -

    -     -     -     - - - - - ($a,$b) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$a=2;$b=-2;}; - $f = Formula("$a cos(x) + $b sin(x)")->reduce; - $r = 2*$b; - - -

    - \ds\int_{0}^{\pi} \left(\right)\, dx -

    -

    - -

    -     -     -     - - - - - ($low,$high) = num_sort(random_subset(2,1..4)); - if($envir{problemSeed}==1){$low=1;$high=4;}; - $f = Formula("e^x")->reduce; - $r = Formula("e^$high - e^$low"); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - ($low,$high) = num_sort(random_subset(2,0,1,4,9,16,25)); - if($envir{problemSeed}==1){$low=0;$high=4;}; - Context()->variables->are(t=>'Real'); - $f = Formula("sqrt(t)")->reduce; - Context("Fraction"); - $r = Fraction(2/3 * ($high**(3/2) - $low**(3/2))); - - -

    - \ds\int_{}^{} \, dt -

    -

    - -

    -     -     -     - - - - - ($low,$high) = num_sort(random_subset(2,1,4,9,16,25)); - if($envir{problemSeed}==1){$low=9;$high=25;}; - Context()->variables->are(t=>'Real'); - $f = Formula("1/sqrt(t)")->reduce; - Context("Fraction"); - $r = Fraction(2 * ($high**(1/2) - $low**(1/2))); - - -

    - \ds\int_{}^{} \, dt -

    -

    - -

    -     -     -     - - - - - $n = random(3,5,1); - ($low,$high) = num_sort(random_subset(2,0**$n,1**$n,2**$n,3**$n,4**$n)); - if($envir{problemSeed}==1){$n=3;$low=1;$high=8;}; - parser::Root->Enable; - $f = Formula("root($n,x)")->reduce; - Context("Fraction"); - $r = Fraction($high**(1/$n + 1)/(1/$n + 1) - $low**(1/$n + 1)/(1/$n + 1)); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $low = 1; - $high = random(2,9,1); - if($envir{problemSeed}==1){$high=2;}; - $f = Formula("1/x")->reduce; - Context()->flags->set(reduceConstantFunctions=>0); - $r = Formula("ln($high)"); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $low = 1; - $high = random(2,9,1); - if($envir{problemSeed}==1){$high=2;}; - $f = Formula("1/x^2")->reduce; - Context("Fraction"); - $r = Fraction(1/$low - 1/$high); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $n = random(3,6,1); - $low = 1; - $high = random(2,9,1); - if($envir{problemSeed}==1){$n=3;$high=2;}; - $f = Formula("1/x^$n")->reduce; - Context("Fraction"); - $r = Fraction($high**(-$n+1)/(-$n+1) - $low**(-$n+1)/(-$n+1)); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = 1; - $f = Formula("x")->reduce; - Context("Fraction"); - $r = Fraction(1/2); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = 1; - $f = Formula("x^2")->reduce; - Context("Fraction"); - $r = Fraction(1/3); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = 1; - $f = Formula("x^3")->reduce; - Context("Fraction"); - $r = Fraction(1/4); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $n = random(80,120,1); - if($envir{problemSeed}==1){$n=100;}; - $low = 0; - $high = 1; - $f = Formula("x^$n")->reduce; - Context("Fraction"); - $r = Fraction(1/($n+1)); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $low = random(-9,-2,1); - if($envir{problemSeed}==1){$low=-4;}; - $high = -$low; - $r = Real(2*$high); - - -

    - \ds\int_{}^{} dx -

    -

    - -

    -     -     -     - - - - - ($low,$high) = num_sort(random_subset(2,-9..-1)); - $c = random(2,9,1); - if($envir{problemSeed}==1){$low=-10;$high=-5;$c=3;}; - $f = Formula("$c")->reduce; - $r = Real($c*($high - $low)); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $low = random(-9,-2,1); - if($envir{problemSeed}==1){$low=-2;}; - $high = -$low; - $f = Formula("0")->reduce; - $r = Real(0); - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $trig = list_random('-cot','-csc'); - $lh = list_random(['pi/6','pi/4'],['pi/6','pi/3'],['pi/6','pi/2'],['pi/4','pi/3'],['pi/4','pi/2'],['pi/3','pi/2']); - if($envir{problemSeed}==1){$trig='-csc';$lh=['pi/6','pi/3'];}; - $low = Formula($lh->[0]); - $high = Formula($lh->[1]); - $F = Formula("$trig(x)"); - $f = $F->D('x')->reduce; - $r = $F->eval(x=>Real("$high")) - $F->eval(x=>Real("$low")); - Context()->flags->set(reduceConstantFunctions=>0); - if ($r == 1) {$r = 1;} - elsif ($r == sqrt(3) - 1) {$r = Formula("sqrt(3) - 1");} - elsif ($r == sqrt(3) - 1/sqrt(3)) {$r = Formula("sqrt(3) - 1/sqrt(3)");} - elsif ($r == sqrt(3)) {$r = Formula("sqrt(3)");} - elsif ($r == 1 - 1/sqrt(3)) {$r = Formula("1 - 1/sqrt(3)");} - elsif ($r == 1/sqrt(3)) {$r = Formula("1/sqrt(3)");} - elsif ($r == 2 - sqrt(2)) {$r = Formula("2 - sqrt(2)");} - elsif ($r == 2 - 2/sqrt(3)) {$r = Formula("2 - 2/sqrt(3)");} - elsif ($r == sqrt(2) - 2/sqrt(3)) {$r = Formula("sqrt(2) - 2/sqrt(3)");} - elsif ($r == sqrt(2) - 1) {$r = Formula("sqrt(2) - 1");} - elsif ($r == 2/sqrt(3) - 1) {$r = Formula("2/sqrt(3) - 1");} - - -

    - \ds\int_{}^{} \, dx -

    -

    - -

    -     -     -     -     - - - - - - -

    - Explain why \ds\int_{-1}^1 x^n\, dx=0 - when n is a positive, odd integer. -

    - - - -     - - -     - - - -

    - Explain why \ds\int_{-1}^1 x^n\, dx = 2\int_{0}^1 x^n\, dx when n is a positive, - even integer. -

    - - -     - - -     - -     - - - - -

    - Explain why \ds\int_{a}^{a+2\pi} \sin(t)\, dt = 0 for all values of a. -

    - -     - - - -     - - - -

    - Find all values c such that \int_a^b f(x)\, dx = f(c)(b-a), - as guaranteed by the . -

    -     - - - - - $low = 0; - $high = random(2,4,1); - if($envir{problemSeed}==1){$high=2;}; - $f = Formula("x^2"); - $D = $high - $low; - $F = Formula("x^3/3"); - $f_inv = Formula("sqrt(x)"); - $f_c = ($F->eval(x=>$high) - $F->eval(x=>$low))/$D; - $c = List($f_inv->eval(x=>$f_c)); - - -

    - \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $low = random(-9,-1,1); - if($envir{problemSeed}==1){$low=-2;}; - $high = -$low; - $f = Formula("x^2"); - $D = $high - $low; - $F = Formula("x^3/3"); - $f_inv = Formula("sqrt(x)"); - $f_c = ($F->eval(x=>$high) - $F->eval(x=>$low))/$D; - $c = $f_inv->eval(x=>$f_c); - $c = List(-$c,$c); - - -

    - \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = 1; - $f = Formula("e^x"); - $D = $high - $low; - $F = Formula("e^x"); - $f_inv = Formula("ln(x)"); - $f_c = ($F->eval(x=>$high) - $F->eval(x=>$low))/$D; - $c = $f_inv->eval(x=>$f_c); - $c = List($c); - - -

    - \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = list_random(1,4,9,16,25,36,49); - if($envir{problemSeed}==1){$high=16;}; - $f = Formula("sqrt(x)"); - $D = $high - $low; - $F = Formula("2/3 x^(3/2)"); - $f_inv = Formula("x^2"); - $f_c = ($F->eval(x=>$high) - $F->eval(x=>$low))/$D; - $c = $f_inv->eval(x=>$f_c); - $c = List($c); - - -

    - \int_{}^{} \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Find the average value of the function on the given interval. -

    -     - - - - - $trig = list_random('sin','cos'); - $lh = list_random(['0','pi/2'],['0','pi'],['pi/2','pi']); - if($envir{problemSeed}==1){$trig='sin';$lh=['0','pi/2'];}; - $low = Formula($lh->[0]); - $high = Formula($lh->[1]); - $D = $high - $low; - $f = Formula("$trig(x)"); - $F = -$f->D('x')->reduce; - $A = ($F->eval(x=>Real("$high")) - $F->eval(x=>Real("$low")))/$D; - $num = Real($A*pi); - $A = Formula("$num/pi"); - - -

    - f(x)= on \left[,\right] -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','cos'); - $lh = list_random(['0','pi/2'],['0','pi'],['pi/2','pi']); - if($envir{problemSeed}==1){$trig='sin';$lh=['0','pi'];}; - $low = Formula($lh->[0]); - $high = Formula($lh->[1]); - $D = $high - $low; - $f = Formula("$trig(x)"); - $F = -$f->D('x')->reduce; - $A = ($F->eval(x=>Real("$high")) - $F->eval(x=>Real("$low")))/$D; - $num = Real($A*pi); - $A = Formula("$num/pi"); - - -

    - y= on [,] -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = random(1,9,1); - if($envir{problemSeed}==1){$high=4;}; - $D = $high - $low; - $f = Formula("x"); - $F = Formula("x^2/2"); - Context("Fraction"); - $A = Fraction(($F->eval(x=>Real($high)) - $F->eval(x=>Real($low)))/$D); - - -

    - y= on [,] -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = random(1,9,1); - if($envir{problemSeed}==1){$high=4;}; - $D = $high - $low; - $f = Formula("x^2"); - $F = Formula("x^3/3"); - Context("Fraction"); - $A = Fraction(($F->eval(x=>Real($high)) - $F->eval(x=>Real($low)))/$D); - - -

    - y= on [,] -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = random(1,9,1); - if($envir{problemSeed}==1){$high=4;}; - $D = $high - $low; - $f = Formula("x^3"); - $F = Formula("x^4/4"); - Context("Fraction"); - $A = Fraction(($F->eval(x=>Real($high)) - $F->eval(x=>Real($low)))/$D); - - -

    - y= on [,] -

    -

    - -

    -     -     -     - - - - - $low = 1; - $m = random(1,9,1); - if($envir{problemSeed}==1){$m=1;}; - $high = Formula("e^$m")->reduce; - $D = $high - $low; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - $f = Formula("1/t"); - $F = Formula("ln(t)"); - $A = Formula("$m/(e^$m - 1)"); - - -

    - y=\ds on \left[,\right] -

    -

    - -

    -     -     -     -     - - - -

    - A velocity function is given for an object moving along a straight line. - Find the displacement of the object over the given time interval. -

    -     - - - - - $low = 0; - $high = random(4,8,1); - $v0 = random(10,30,2); - if($envir{problemSeed}==1){$high=5;$v0=20}; - $a = -32; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - $f = FormulaWithUnits("$a t + $v0 ft/s"); - $F = Formula("$a/2 t^2 + $v0 t"); - $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft'); - - -

    - v(t)= on [,] -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = random(7,15,1); - $v0 = random(100,300,20); - if($envir{problemSeed}==1){$high=10;$v0=200}; - $a = -32; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - $f = FormulaWithUnits("$a t + $v0 ft/s"); - $F = Formula("$a/2 t^2 + $v0 t"); - $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft'); - - -

    - v(t)= on [,] -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = random(2,9,1); - $v0 = random(5,20,1); - if($envir{problemSeed}==1){$high=3;$v0=10}; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - $f = FormulaWithUnits("$v0 ft/s"); - $F = Formula("$v0 t"); - $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft'); - - -

    - v(t)= on [,] -

    -

    - -

    -     -     -     - - - - - $low = random(-3,-1,1); - $high = random(1,3,1); - $b = random(2,9,1); - if($envir{problemSeed}==1){$low=-1;$high=1;$b=2;}; - Context()->variables->are(t=>'Real'); - $f = FormulaWithUnits("$b^t mph"); - $F = Formula("$b^t/ln($b)"); - $D = FormulaWithUnits($F->eval(t=>Real($high)) - $F->eval(t=>Real($low)), 'mi'); - - -

    - v(t)= on [,] -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','cos'); - $lh = list_random(['0','pi/2'],['0','pi'],['0','3*pi/2'],['pi/2','pi'],['pi/2','3*pi/2'],['pi','3*pi/2']); - if($envir{problemSeed}==1){$trig='cos';$lh=['0','3*pi/2'];}; - $low = Formula($lh->[0]); - $high = Formula($lh->[1]); - Context()->variables->are(t=>'Real'); - $f = FormulaWithUnits("$trig(t) ft/s"); - $F = Formula("-$trig(t)")->D('t')->reduce; - $D = FormulaWithUnits($F->eval(t=>Real("$high")) - $F->eval(t=>Real("$low")), 'ft'); - - -

    - v(t)= on \left[,\right] -

    -

    - -

    -     -     -     - - - - - $low = 0; - $n = random(3,5,1); - $high = list_random(2**$n,3**$n,4**$n); - if($envir{problemSeed}==1){$high=1;$n=4;$high=16;}; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - parser::Root->Enable; - $f = FormulaWithUnits("root($n,t) ft/s"); - $F = Formula("t^(1/$n+1)/(1/$n+1)"); - $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft'); - - -

    - v(t)= on [,] -

    -

    - -

    -     -     -     -     - - - -

    - An acceleration function of an object moving along a straight line is given. - Find the change of the object's velocity over the given time interval. -

    -     - - - - - $low = 0; - $high = random(1,9,1); - if($envir{problemSeed}==1){$high=2;}; - $a = -32; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - $f = FormulaWithUnits("$a ft/s^2"); - $F = Formula("$a t"); - $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft/s'); - - -

    - a(t)= on [,] -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = random(1,9,1); - $a = random(5,20,1); - if($envir{problemSeed}==1){$high=5;$a=10;}; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - $f = FormulaWithUnits("$a ft/s^2"); - $F = Formula("$a t"); - $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft/s'); - - -

    - a(t)= on [,] -

    -

    - -

    -     -     -     - - - - - $low = 0; - $high = random(1,9,1); - if($envir{problemSeed}==1){$high=2;}; - Context("Fraction"); - Context()->variables->are(t=>'Real'); - $f = FormulaWithUnits("t ft/s^2"); - $F = Formula("t^2/2"); - $D = FormulaWithUnits(Fraction($F->eval(t=>Real($high)) - $F->eval(t=>Real($low))), 'ft/s'); - - -

    - a(t)= on [,] -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','cos'); - $lh = list_random(['0','pi/2'],['0','pi'],['0','3*pi/2'],['pi/2','pi'],['pi/2','3*pi/2'],['pi','3*pi/2']); - if($envir{problemSeed}==1){$trig='cos';$lh=['0','pi/2'];}; - $low = Formula($lh->[0]); - $high = Formula($lh->[1]); - Context()->variables->are(t=>'Real'); - $f = FormulaWithUnits("$trig(t) ft/s^2"); - $F = Formula("-$trig(t)")->D('t')->reduce; - $D = FormulaWithUnits($F->eval(t=>Real("$high")) - $F->eval(t=>Real("$low")), 'ft/s'); - - -

    - a(t)= on \left[,\right] -

    -

    - -

    -     -     -     -     - - - -

    - Sketch the given relations and find the area of the enclosed region. -

    -     - - - -

    - y=2x, y=5x, and x= 3 -

    -     -     - - - -

    - y=-x+1, y=3x+6, - x=2 and x= -1 -

    -     -     - - - -

    - y=x^2-2x+5, y=5x-5 -

    -     -     - - - -

    - y=2x^2+2x-5, y=x^2+3x+7, -

    -     -     -     - - - -

    - Find F'(x). -

    -     - - - - - $m = random(3,5,1); - $b = non_zero_random(-9,9,1); - $low = random(1,9,1); - if($envir{problemSeed}==1){$m=3;$b=1;$low=2;}; - Context()->variables->add(t=>'Real'); - $high = Formula("x^$m + $b x")->reduce; - $f = Formula("1/t"); - $D = $high->D('x')->reduce / $high; - - -

    - F(x) = \ds\int_{}^{} \, dt -

    -

    - -

    -     -     -     - - - - - $m = random(2,5,1); - $high = random(-9,9,1); - if($envir{problemSeed}==1){$m=3;$high=0;}; - Context()->variables->add(t=>'Real'); - $low = Formula("x^$m"); - $f = Formula("t^$m"); - $D = Formula("-$m x^($m**2+$m-1)")->reduce; - - -

    - F(x) = \ds\int_{}^{} \, dt -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - $m = random(2,4,1); - if($envir{problemSeed}==1){$b=1;$m=2}; - Context()->variables->add(t=>'Real'); - $low = Formula("x"); - $high = Formula("x^$m"); - $f = Formula("t+$b")->reduce; - $D = Formula("$m x^($m-1)(x^$m+$b) - (x+$b)")->reduce; - - -

    - F(x) = \ds\int_{}^{} ()\, dt -

    -

    - -

    -     -     -     - - - - - ($low,$high,$f) = random_subset(3,'e^','ln','sin','cos','sqrt'); - if($envir{problemSeed}==1){$low='ln';$high='e^';$f='sin';}; - Context()->variables->add(t=>'Real'); - # interval where all these functions have positive output - Context()->variables->set(x=>{limits=>[1,pi/2]}); - $low = Formula("$low(x)"); - $high = Formula("$high(x)"); - $f = Formula("$f(t)"); - $D = $high->D('x') * $f->substitute(t=>$high) - $low->D('x') * $f->substitute(t=>$low); - $D = $D->reduce; - - -

    - F(x) = \ds\int_{}^{} \, dt -

    -

    - -

    -     -     -     - - - - $b = non_zero_random(-9,9,1); - $m = random(2,4,1); - Context()->variables->add(t=>'Real'); - $low = random(1,9,1); - $high = Formula("x^$m"); - $f = Formula("sin($b t^2)")->reduce; - $D = Formula("$m x^($m-1)sin($b x^(2*$m))")->reduce; - - -

    - F(x) = \ds\int_{}^{} ()\, dt -

    -

    - -

    -     -     -     - - - - $b = non_zero_random(-9,9,1); - ($low,$high) = random_subset(2,'e^','ln','sin','cos','sqrt'); - Context()->variables->add(t=>'Real'); - $low = Formula("$low(x)"); - $high = Formula("$high(x)"); - $f = Formula("sqrt(t^4+$b t^2)")->reduce; - $D = $high->D('x') * $f->substitute(t=>$high) - $low->D('x') * $f->substitute(t=>$low); - $D = $D->reduce; - - -

    - F(x) = \ds\int_{}^{} ()\, dt -

    -

    - -

    -     -     -     -     -     -     -
    -
    - Numerical Integration - -

    - The Fundamental Theorem of Calculus gives a concrete technique for finding the exact value of a definite integral. - That technique is based on computing antiderivatives. - Despite the power of this theorem, - there are still situations where we must approximate - the value of the definite integral instead of finding its exact value. - The first situation we explore is where we cannot - compute the antiderivative of the integrand. - The second case is when we actually do not know the function in the integrand, - but only its value when evaluated at certain points. - integrationnumerical - numerical integration -

    - -

    - An elementary function - elementary function - is any function that is a combination of polynomial, - nth root, rational, - exponential, logarithmic and trigonometric functions. - We can compute the derivative of any elementary function, - but there are many elementary functions of which we cannot compute an antiderivative. - For example, - the following functions do not have antiderivatives that we can express with elementary functions: -

      -
    • e^{x^2},
      • -
    • \sin(x^3),
      • -
    • \frac{\sin(x)}{x}.
      • -
    -

    - -

    - The simplest way to refer to the antiderivatives of - e^{-x^2} is to simply write \int e^{-x^2}\, dx. -

    - -

    - This section outlines three common methods of approximating the value of definite integrals. - We describe each as a systematic method of approximating area under a curve. - By approximating this area accurately, - we find an accurate approximation of the corresponding definite integral. -

    - -

    - We will apply the methods we learn in this section to the following definite integrals: -

      -
    • \int_0^1 e^{-x^2} \, dx,
      • -
    • \int_{-\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(x^3) \, dx,
      • -
    • \int_{0.5}^{4\pi} \frac{\sin(x)}{x} \, dx,
      • -
    - as pictured in . -

    - -
    - Graphically representing three definite integrals that cannot be evaluated using antiderivatives - - -
    - - - - Integral of e^{-x^2} from 0 to 1 - -

    - This is a graph of f(x)=e^{-x^2}.starting at (0, 1) the line slopes down towards the x axis. The line goes from (0, 1) to close - to (1, 0.75). We consider the area between (0, 1), (1, 0.75),(1, 0) and (0, 0). - -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.2,ymax=1.2, - xmin=-.2,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1] {e^-x^2} \closedcycle; - \addplot [firstcurvestyle,domain=0:1] {e^-x^2}; - - \draw (axis cs:.75,1) node { $y=e^{-x^2}$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - - -
    - - - - Graph of the function sin(x^3) - -

    - The graph of \sin(x^3). Starting at (-0.75, -0.5) the graph moves upwards to the right. It merges to the x axis close to (-0.25, 0) and - continues along the x axis. It goes through and continues along the positive x axis for a short while then rises up until it reaches (1.25, 1) - then sharply descends back towards the x axis. It intersects the y axis close to (1.5, 0) and continues downwards. We consider the area above the part of the - curve that is in the fourth quadrant and below the negative x axis. Then below the curve that rises up from the positive x axis and comes down. And the area above the - part of the curve that continues downwards after intersecting the x axis. - -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.6,ymax=1.2, - xmin=-1,xmax=1.7 - ] - - \addplot [firstcurvestyle,areastyle,domain=-.7854:1.5708,samples=40] {sin(deg(x^3))} \closedcycle; - \addplot [firstcurvestyle,domain=-.7854:1.5708,samples=80] {sin(deg(x^3))}; - - \draw (axis cs:.6,.9) node { $y=\sin(x^3)$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - - -
    - - - - The graph of y=sin(x)/x and its integral from .5 to 4 pi - -

    - We consider the area under the graph of \sin(x)/x starting close to (0.5, 0.9) to (12.5, 0). The structure of the graph is similar to that - of a wave. The distance between the highest point of the graph in the positive y axis and the lowest point in the negative y axis shows a - decreasing trend as it continues in the positive x direction but the wavelength increases in the positive x direction.The graph starts at (0.5, 0.9) - close to the x axis and curves downwards. It crosses the x axis close to (3, 0) while travelling downwards. It continues downwards for - a short amount and the curves upwards and this time crosses the x axis at (6.25, 0). From (6.25, 0) it travels another wavelength crossing - the x axis once close to (10,0) while traveling upwards and at (12.5, 0) where the graph ends. We consider the area under the graph when - it is above the x axis and above it when it is below the x axis. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={5,10,15}, - ymin=-.3,ymax=1.1, - xmin=-.9,xmax=15.9 - ] - - \addplot [firstcurvestyle,areastyle,domain=0.5:12.57] {sin(deg(x))/x} \closedcycle; - \addplot [firstcurvestyle,domain=0.5:12.57,samples=60] {sin(deg(x))/x}; - - \draw (axis cs:10,.7) node { $\displaystyle y=\frac{\sin(x) }{x}$}; - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    -     - - - The Left and Right Hand Rule Methods -

    - In - we addressed the problem of evaluating definite integrals by approximating the area under the curve using rectangles. - We revisit those ideas here before introducing other methods of approximating definite integrals. - numerical integrationLeft/Right Hand Rule - Right Hand Rule - Left Hand Rule - integrationnumerical!Left/Right Hand Rule -

    - -

    - We start with a review of notation. - Let f be a continuous function on the interval [a,b]. - We wish to approximate \ds \int_a^b f(x)\, dx. - We partition [a,b] into n equally spaced subintervals, - each of length \ds\dx = \frac{b-a}{n}. - The endpoints of these subintervals are labeled as - - x_0=a,\,x_1 = a+\dx,\,x_2 = a+ 2\dx,\,\ldots,\,x_i = a+i\dx,\,\ldots,\,x_{n} = b - . -

    - -

    - - states that to use the Left Hand Rule we use the summation - \ds \sum_{i=1}^n f(x_{i-1})\dx and to use the Right Hand Rule we use \ds \sum_{i=1}^n f(x_{i})\dx. - We review the use of these rules in the context of examples. -

    - - - Approximating definite integrals with rectangles - -

    - Approximate \ds \int_0^1e^{-x^2}\, dx using the Left and Right Hand Rules with 5 equally spaced subintervals. -

    -     - -

    - We begin by partitioning the interval [0,1] into 5 equally spaced intervals. - We have \dx = \frac{1-0}5 = 1/5=0.2, so - - x_0 = 0,\,x_1 = 0.2,\,x_2 = 0.4,\,x_3 = 0.6,\,x_4 = 0.8,\,\text{ and } \,x_5 = 1 - . -

    - -

    - Using the Left Hand Rule, we have: - - \sum_{i=1}^n f(x_{i-1})\dx \amp = \big(f(x_0)+f(x_1) + f(x_2) + f(x_3) + f(x_4)\big)\dx - \amp = \big(f(0) + f(0.2) + f(0.4) + f(0.6) + f(0.8)\big)\dx - \amp \approx (1+0.9608 +0.8521 + 0.6977 + 0.5273)(0.2) - \amp \approx 0.8076 - . -

    - -

    - Using the Right Hand Rule, we have: - - \sum_{i=1}^n f(x_{i})\dx \amp = \big(f(x_1) + f(x_2) + f(x_3) + f(x_4)+f(x_5)\big)\dx - \amp = \big(f(0.2) + f(0.4) + f(0.6) + f(0.8)+f(1)\big)\dx - \amp \approx (0.9608 +0.8521 + 0.6977 + 0.5273 + 0.3678)(0.2) - \amp \approx 0.6812 - . -

    - -
    - Approximating \int_0^1e^{-x^2}\, dx in - - - -
    - Using the Left Hand Rule - - Using the Left Hand rule - -

    The graph of e^{-x^2} starts at (0, 1) and slowly slopes downwards to the right as it slowly gets closer to the positive x axis. - To use the left hand rule we divide the area between [0, 1] five rectangles of equal widhth and decreasing height. As the curve moves closer to the x axis, - the heights of the rectangles gets shorter even thought the widths stay the same. Because the curve is downwards sloping to the right, the top right corners - of the rectangles contain a little region which is above the curve. As the curve moves downwards and the height of the rectangles become shorter, the area - that is contained by the top right corners of the rectangles and the top of the curve gets larger. We calculate the sum of the areas of these rectangles. -

    - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={.2,.4,.6,.8,1}, - ymin=-.2,ymax=1.2, - xmin=-.2,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1] {e^-x^2} \closedcycle; - \addplot [firstcurvestyle,domain=0:1] {e^-x^2}; - - \foreach \x / \y in {0 / 1, 0.2 / 0.96, 0.4 / .85, 0.6 / .7, .8 / .53} - {\addplot [thick,secondcolor] coordinates {(\x+.2,0) (\x+.2,\y) (\x,\y) (\x,0) (\x+.2,0)}; - } - - \draw (axis cs:.75,1) node { $y=e^{-x^2}$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - - -
    - Using the Right Hand Rule - - Using the Right Hand rule - -

    For this image we consider the same downwards sloping curve of e^{-x^2}. We divide the area between [0,1] into five - rectangles of equal width and decreasing height. For the right hand rule the rectangles are drawn in such a way that the top right corners of the rectangles - touch the bottom of the curve. There is a small area between the top left of the triangles and the bottom of the curve. This area gets larger as the - curve moves downwards and the lengths of the rectangles decrease. For the left hand rule we add up the areas of these rectangles. -

    - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={.2,.4,.6,.8,1}, - ymin=-.2,ymax=1.2, - xmin=-.2,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1] {e^-x^2} \closedcycle; - \addplot [firstcurvestyle,domain=0:1] {e^-x^2}; - - \foreach \x / \y in {0.2 / 0.96, 0.4 / .85, 0.6 / .7, .8 / .53, 1/ .37} - {\addplot [thick,secondcolor] coordinates {(\x-.2,0) (\x-.2,\y) (\x,\y) (\x,0) (\x-.2,0)}; - } - - \draw (axis cs:.75,1) node { $y=e^{-x^2}$}; - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    - -

    - - shows the rectangles used in each method to approximate the definite integral. - These graphs show that in this particular case, - the Left Hand Rule is an over approximation and the Right Hand Rule is an under approximation. - To get a better approximation, - we could use more rectangles, - as we did in . - We could also average the Left and Right Hand Rule results together, giving - - \frac{0.8076 + 0.6812}{2} = 0.7444 - . -

    - -

    - The actual answer, accurate to 4 places after the decimal, - is 0.7468, showing our average is a good approximation. -

    -     -     - - - Approximating definite integrals with rectangles - -

    - Approximate \ds\int_{-\frac{\pi}4}^{\frac{\pi}2} \sin(x^3)\, dx using the Left and Right Hand Rules with 10 equally spaced subintervals. -

    -     - -

    - We begin by finding \dx: - - \frac{b-a}{n} = \frac{\pi/2 - (-\pi/4)}{10} = \frac{3\pi}{40}\approx 0.2356 - . -

    - -

    - It is useful to write out the endpoints of the subintervals in a table; - in , - we give the exact values of the endpoints, - their decimal approximations, - and decimal approximations of - \sin(x^3) evaluated at these points. -

    - -
    - Values used to approximate \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in - - - x_i - Exact - Approx. - \sin(x_i^3) - - - - - - - - - x_0 - -\pi/4 - -0.7854 - -0.4657 - - - x_1 - -7 \pi/40 - -0.5498 - -0.1654 - - - x_2 - -{\pi }/{10} - -0.3142 - -0.0310 - - - x_3 - -{\pi }/{40} - -0.0785 - -0.0005 - - - x_4 - {\pi }/{20} - 0.1571 - 0.0039 - - - x_5 - {\pi }/{8} - 0.3927 - 0.0605 - - - x_6 - {\pi }/{5} - 0.6283 - 0.2455 - - - x_7 - {11 \pi }/{40} - 0.8639 - 0.6011 - - - x_8 - {7 \pi }/{20} - 1.0996 - 0.9710 - - - x_{9} - {17 \pi }/{40} - 1.3352 - 0.6899 - - - x_{10} - {\pi }/{2} - 1.5708 - -0.6700 - - -
    - -

    - Once this table is created, - it is straightforward to approximate the definite integral using the Left and Right Hand Rules. - (Note: the table itself is easy to create, - especially with a standard spreadsheet program on a computer. - The last two columns are all that are needed.) - The Left Hand Rule sums the first 10 values of - \sin(x_i^3) and multiplies the sum by \dx; - the Right Hand Rule sums the last 10 values of - \sin(x_i^3) and multiplies by \dx. - Therefore we have: -

    - -

    - Left Hand Rule: - \ds \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx \approx (1.9093)(0.2356) \approx 0.4498. -

    - -

    - Right Hand Rule: - \ds \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx \approx (1.705)(0.2356) \approx 0.4017. -

    - -

    - Average of the Left and Right Hand Rules: 0.4258. -

    - -
    - Approximating \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in - - - -
    - - - - Using the Left Hand rule - -

    We calculate \int_{-\frac{\pi}4}^{\frac{\pi}/2}\sin(x^3)\, dx using the left hand rule. The graph of \sin(x^3) starts at (-0.75, -0.5) - in the fourth quadrant and curves upwards to the right. It merges with the negative x axis close to (-0.25, 0) and continues along the x axis - until (0.25, 0) and then slopes upwards to the right. The graph continues upwards until (1.25, 1) in the first quadrant and then descends sharply. It - crosses the positive x axis close to (1.5, 0) and continues downwards. We divide the areas under the curve into boxes of equal width but varying - height. The far left side of the graph is divided into three boxes of decreasing length that are below the negative x axis. THe rectangle farthest to the - left is the tallest and the other two get get shorter as the graph moves upwards towards the negative x axis. The region roughly from (-0.25, 0) to - (0.25, 0) where the curve moves along the x axis there are no boxes since the are of that part is zero. Then the part of the graph in the first - quadrant is divided into five rectangles. As the graph rises up, the heights of the rectangles increase. The first one being the shortest and the fourth one - being the tallest. The fifth rectangle is shorter than the fourth one but taller than the first three. We add up the areas of the rectangles to approximate the - integral.

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.7,ymax=1.2, - xmin=-1,xmax=1.7 - ] - - \addplot [firstcurvestyle,areastyle,domain=-.7854:1.5708,samples=40] {sin(deg(x^3))} \closedcycle; - \addplot [firstcurvestyle,domain=-.7854:1.5708,samples=80] {sin(deg(x^3))}; - - \foreach \x / \y in {-0.785/ -0.466, -0.550/-0.165, -0.314/ -0.0310, - -0.0785/-0.000484, 0.157/ 0.00388, 0.393/ 0.0605, - 0.628/ 0.246, 0.864/ 0.601, 1.10/ 0.971, 1.34/ 0.690} - {\addplot [thick,secondcolor] coordinates {(\x+.2356,0) (\x+.2356,\y) (\x,\y) (\x,0) (\x+.2356,0)}; - } - - \draw (axis cs:.6,.9) node { $y=\sin(x^3)$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - - -
    - - - - Using the Right Hand rule - -

    We calculate \int_{-\frac{\pi}4}^{\frac{\pi}/2}\sin(x^3)\, dx using the Right hand rule. The far left part of the graph is divided into two - rectangles of decreasing height. The region from (-0.25, 0) to (0.25, 0) on the x axis does not contribute to the area since the curve - here continues along the axis and therefore the area is zero. The part in the first quadrant is divided into five adjacent rectangles of increasing height. - The upper left side of the first four rectangles contain some area outside of the curve on the left. As the graph moves up, the heights of the rectangles increase, - the fourth one being the tallest. The fifth rectangle is below the graph with its upper right corner touching the inside of the graph. The sixth and the last rectangle - is below the positive x axis. It captures the area adjacent to the part of the graph that is below the positive x axis.

     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.7,ymax=1.2, - xmin=-1,xmax=1.7 - ] - - \addplot [firstcurvestyle,areastyle,domain=-.7854:1.5708,samples=40] {sin(deg(x^3))} \closedcycle; - \addplot [firstcurvestyle,domain=-.7854:1.5708,samples=80] {sin(deg(x^3))}; - - \foreach \x / \y in {-0.550/-0.165, -0.314/ -0.0310, - -0.0785/-0.000484, 0.157/ 0.00388, 0.393/ 0.0605, - 0.628/ 0.246, 0.864/ 0.601, 1.10/ 0.971, 1.34/ 0.690, 1.57/ -0.670} - {\addplot [thick,secondcolor] coordinates {(\x-.2356,0) (\x-.2356,\y) (\x,\y) (\x,0) (\x-.2356,0)}; - } - - \draw (axis cs:.6,1.1) node { $y=\sin(x^3)$}; - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    - -

    - The actual answer, accurate to 4 places after the decimal, - is 0.4609. - Our approximations were once again fairly good. - The rectangles used in each approximation are shown in . - It is clear from the graphs that using more rectangles - (and hence, narrower rectangles) - should result in a more accurate approximation. -

    -     -     -     - - - The Trapezoidal Rule -

    - In - we approximated the value of - \ds \int_0^1 e^{-x^2}\, dx with 5 rectangles of equal width. - - shows the rectangles used in the Left and Right Hand Rules. - These graphs clearly show that rectangles do not match the shape of the graph all that well, - and that accurate approximations will only come by using lots of rectangles. - Trapezoidal Rule - numerical integrationTrapezoidal Rule - integrationnumerical!Trapezoidal Rule -

    - -

    - Instead of using rectangles to approximate the area, - we can instead use trapezoids. - In , - we show the region under f(x) = e^{-x^2} on [0,1] approximated - with 5 trapezoids of equal width; - the top corners of each trapezoid lies on the graph of f(x). - It is clear from this figure that these trapezoids more accurately approximate - the area under f and hence should give a better approximation of - \int_0^1 e^{-x^2}\, dx. - (In fact, these trapezoids seem to give a great - approximation of the area!) -

    - -
    - Approximating \int_0^1 e^{-x^2}\, dx using 5 trapezoids of equal widths - - - Approximating the integral of e to the power of negative x squared from 0 to 1 using five trapezoids - -

    The graph of e^{-x^2} starts from (0,1) and slopes downwards to the right in the positive x direction. We consider the are under the curve from - (0,1) to (1, 0.25) in the first quadrant. The area is approximated with five adjacent trapezoids of equal width but decreasing heights. The upper arms - of the trapezoids are downwar sloping and they follow the graph of e^{-x^2} very closely which allows for more precise approximation. Both of the upper corners - of the trapezoids lie on the curve. The first trapezoid is the tallest one and the fifth one is the shortest. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={.2,.4,.6,.8,1}, - ymin=-.2,ymax=1.2, - xmin=-.2,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1] {e^-x^2} \closedcycle; - \addplot [firstcurvestyle,domain=0:1] {e^-x^2}; - - \foreach \x / \y / \z in {0/1 /0.96 , 0.2/0.96 / .85, 0.4 /.85 / .7, .6 / .7/ .53, .8 / .53 / .37} - {\addplot [thick,secondcolor] coordinates {(\x,0) (\x,\y) (\x+.2,\z) (\x+.2,0) (\x,0)}; - } - - \draw (axis cs:.75,1) node { $y=e^{-x^2}$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - The formula for the area of a trapezoid is given in . - We approximate \int_0^1 e^{-x^2}\, dx with these trapezoids in the following example. -

    - -
    - The area of a trapezoid - - - The area of a trapezoid. - -

    This is a picture of a trapezoid. The height is h. The left arm of the trapezoid is shorter than the right arm. The arm connecting the left and the right - arm is thus upward sloping. The left arm is named a and the right arm is named b. -

    - - - \begin{tikzpicture}[scale=1.3] - - \draw (0,0) -- node [pos=.5,left] {\small $a$} (0,1) -- (1,1.5) -- node [pos=.5,right] {\small $b$} (1,0) -- node [pos=.5,below] {\small $h$} (0,0); - - \draw (0,.1) -- (.1,.1) -- (.1,0); - \draw (.9,0) -- (.9,.1) -- (1,.1); - - \draw (3,.75) node {Area = $\frac{a+b}2h$}; - - \end{tikzpicture} - - - - -
    - - - Approximating definite integrals using trapezoids - -

    - Use 5 trapezoids of equal width to approximate \ds \int_0^1e^{-x^2}\, dx. -

    -     - -

    - To compute the areas of the 5 trapezoids in , - it will again be useful to create a table of values as shown in . -

    - -
    - A table of values of e^{-x^2} - - - x_i - e^{-x_i^2} - - - - - - - 0 - 1 - - - 0.2 - 0.9608 - - - 0.4 - 0.8521 - - - 0.6 - 0.6977 - - - 0.8 - 0.5273 - - - 1 - 0.3679 - - -
    - -

    - The leftmost trapezoid has legs of length 1 and 0.9607 and a height of 0.2. - Thus, by our formula, the area of the leftmost trapezoid is: - - \frac{1+0.9608}{2}(0.2) = 0.1961 - . -

    - -

    - Moving right, - the next trapezoid has legs of length 0.9607 and 0.8521 and a height of 0.2. - Thus its area is: - - \frac{0.9608+0.8521}2(0.2) = 0.1813 - . -

    - -

    - The sum of the areas of all 5 trapezoids is: - - \frac{1+0.9608}{2}(0.2) + \frac{0.9608+0.8521}2(0.2)+\frac{0.8521+0.6977}2(0.2)\amp + - \frac{0.6977+0.5273}2(0.2)+\frac{0.5273+0.3679}2(0.2)\amp = 0.7444 - . -

    - -

    - We approximate \int_0^1 e^{-x^2}\, dx \approx 0.7444. -

    -     -     - -

    - There are many things to observe in this example. - Note how each term in the final summation was multiplied by both 1/2 and by \dx = 0.2. - We can factor these coefficients out, - leaving a more concise summation as: - - \frac12(0.2)\amp\Big[(1+0.9608) + (0.9608+0.8521) + (0.8521+0.6977) - \amp\quad + ( 0.6977+ 0.5273) +(0.5273 + 0.3679)\Big] - . -

    - -

    - Now notice that all numbers except for the first and the last are added twice. - Therefore we can write the summation even more concisely as - - \frac{0.2}{2}\Big[1 + 2(0.9608+0.8521+0.6977+0.5273) + 0.3679\Big] - . -

    - -

    - This is the heart of the Trapezoidal Rule, - wherein a definite integral - \int_a^b f(x) \, dx is approximated by using trapezoids of equal widths - to approximate the corresponding area under f. - Using n equally spaced subintervals with endpoints x_0, - x_1, - \ldots, x_{n}, - we again have \ds \dx = \frac{b-a}n. - Thus: - - \int_a^b f(x)\, dx \amp \approx \sum_{i=1}^n \frac{f(x_{i-1})+f(x_{i})}2\dx - \amp = \frac{\dx}2 \sum_{i=1}^n \big(f(x_{i-1})+f(x_{i})\big) - \amp = \frac{\dx}2\Big[f(x_0)+ \left(2\sum_{i=1}^{n-1} f(x_i)\right) - + f(x_{n})\Big] - . -

    - - - Using the Trapezoidal Rule - -

    - Revisit - and approximate \ds\int_{-\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(x^3)\, dx using the Trapezoidal Rule and 10 equally spaced subintervals. -

    -     - -

    - We refer back to - for the table of values of \sin(x^3). - Recall that \dx = 3\pi/40 \approx 0.236. - Thus we have: - - \amp \int_{-\frac{\pi}4}^{\frac{\pi}2} \sin(x^3)\, dx - \amp \approx \frac{0.236}{2}\Big[-0.4657 + 2\Big(-0.1654+(-0.031)+\ldots+ - 0.68999\Big)+(-0.67)\Big] - \amp = 0.4258 - . - The actual answer, accurate to 4 decimal places is 0.4609. - So the Trapezoidal Rule with 10 subintervals is an under-approximation - by about 0.0351. -

    -     -     - -

    - Notice how quickly the Trapezoidal Rule can be implemented once the - table of values is created. - This is true for all the methods explored in this section; - the real work is creating a table of x_i and f(x_i) values. - Once this is completed, - approximating the definite integral is not difficult. - Again, using technology is wise. - Spreadsheets can make quick work of these computations and make - using lots of subintervals easy. -

    - -

    - Also notice the approximations the Trapezoidal Rule gives. - It is the average of the approximations given by the Left and Right Hand Rules! - This effectively renders the Left and Right Hand Rules obsolete. - They are useful when first learning about definite integrals, - but if a real approximation is needed, - one is generally better off using the Trapezoidal Rule instead of either - the Left or Right Hand Rule. - However, there are two other methods that are also generally more accurate - than the Left or Right Hand Rule. -

    -     - - - The Midpoint Rule -

    - Another method that can be more accurate than the Trapezoidal Rule is the Midpoint Rule: - - S_M(n)\amp =\sum_{i=1}^n f\left(\frac{x_{i-1}+x_{i}}{2}\right)\Delta x - \amp = \sum_{i=1}^n f\left(\overline{x_i}\right)\Delta x - \amp \text{ where } \overline{x_i} \text{ is the midpoint of each subinterval,} - \amp \overline{x_i}=a+\dx\left(i-\frac12\right) - -

    - - - Using the Midpoint Rule - -

    - Use the Midpoint Rule with n=5 to approximate \ds \int_0^1e^{-x^2}\, dx. -

    -     - -

    - We cannot use the table in - that we used for the Trapezoidal, Right and Left Hand Rules when using the Midpoint Rule. - The Trapezoidal rule averages the outputs - of the function to obtain a more accurate estimate of the definite integral. - The Midpoint Rule averages the inputs - of each subinterval to create a rectangle with height f\left(\frac{x_{i-1}+x_{i}}{2}\right). - Generally f\left(\frac{x_{i-1}+x_{i}}{2}\right)\neq \frac{f(x_{i-1})+f(x_{i})}{2}. -

    - -

    - So we will create a new table of values as shown in . - We have \dx=(1-0)/5=0.2. - The midpoint of the first subinteval is at - 0+0.2(1/2)=0.1 and each successive midpoint is 0.2 from the last. -

    - -
    - A table of values of e^{-x^2} - - - x_i - e^{-x_i^2} - - - - - - - 0.1 - 0.9900 - - - 0.3 - 0.9139 - - - 0.5 - 0.7788 - - - 0.7 - 0.6126 - - - 0.9 - 0.4449 - - -
    - -

    - So we have - - \int_0^1 e^{-x^2}\, dx \amp \approx 0.2(0.99+0.9139+0.7788+0.6126+0.4449) - \amp \approx 0.7480 - -

    - -

    - We approximate \ds \int_0^1 e^{-x^2}\, dx \approx 0.7480. -

    -     -     - - - Using the Midpoint Rule - -

    - Revisit - and approximate \ds\int_{-\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(x^3)\, dx using the Midpoint Rule and 10 equally spaced subintervals. -

    -     - -

    - Again, a table will be useful. - Recall that \dx = 3\pi/40 \approx 0.2356. - The midpoint of the first subinterval is \overline{x_1}=a+\dx/2=-\pi/4+3\pi/40(1/2)=-17\pi/80 - (notice that \overline{x_1} is half of a subinterval width to the right of a). - Each successive midpoint is - \dx=3 \pi/40=6\pi/80 to the right of the last. - So we have: -

    - -
    - Values used to approximate \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in - - - \overline{x_i} - Exact - Approx. - \sin(x_i^3) - - - - - - - - - \overline{x_1} - -\frac{17\pi}{80} - -0.6676 - -0.2932 - - - \overline{x_2} - -\frac{11 \pi}{80} - -0.4320 - -0.0805 - - - \overline{x_3} - -\frac{5\pi }{80} - -0.1963 - -0.0076 - - - \overline{x_4} - \frac{\pi }{80} - -0.0393 - 0.0001 - - - \overline{x_5} - \frac{7\pi }{80} - 0.2749 - 0.0208 - - - \overline{x_6} - \frac{13\pi }{80} - 0.5105 - 0.1327 - - - \overline{x_7} - \frac{19\pi}{80} - 0.7461 - 0.4035 - - - \overline{x_8} - \frac{25 \pi}{80} - 0.9817 - 0.8112 - - - \overline{x_9} - \frac{31 \pi}{80} - 1.2174 - 0.9729 - - - \overline{x_{10}} - \frac{37 \pi}{80} - 1.4530 - 0.0740 - - - - - - -
    - -

    - Thus we have: - - \amp \int_{-\frac{\pi}4}^{\frac{\pi}2} \sin(x^3)\, dx - \amp \approx 0.2356\Big[-0.2932+(-0.0805)+(-0.0076)+\dots+0.9729+0.0740\Big] - \amp=0.2356\cdot 2.0339 - \amp \approx 0.4792 - . - The actual answer, accurate to 4 decimal places is 0.4609. - So the Midpoint Rule with 10 subintervals is an overrapproximation by about 0.0183 . - Notice that this error is about half of the error in using the Trapezoidal Rule. -

    -     -     - - -

    - In many cases, - the Midpoint Rule will more accurate than the Trapezoidal Rule. - You may wonder though, - how can we improve on the Trapezoidal and Midpoint Rules, - apart from using more and more subintervals? - The answer is clear once we look back and consider what we have - really done so far. - The Left Hand Rule, Right Hand Rule and Midpoint Rules are not really - about using rectangles to approximate area. - Instead, they approximate a function f with constant functions on small - subintervals and then compute the definite integral of these constant functions. - The Trapezoidal Rule is really approximating a function f with a linear - function on a small subinterval, - then computing the definite integral of this linear function. - In all of these cases the definite integrals are easy to compute in geometric terms. -

    - -

    - So we have a progression: - we start by approximating f with a constant function and then with a linear function. - What is next? - A quadratic function. - By approximating the curve of a function with lots of parabolas, - we generally get an even better approximation of the definite integral. - We call this process Simpson's Rule, - named after Thomas Simpson (1710-1761), - even though others had used this rule as much as 100 years prior. -

    -     - - - Simpson's Rule -

    - Given one point, - we can create a constant function that goes through that point. - Given two points, - we can create a linear function that goes through those points. - Given three points, - we can create a quadratic function that goes through those three points - (given that no two have the same x-value). - Simpson's Rule - numerical integrationSimpson's Rule - integrationnumerical!Simpson's Rule -

    - -

    - Consider three points (x_0,y_0), - (x_1,y_1) and (x_2,y_2) whose x-values are equally spaced and x_0\lt x_1\lt x_2. - Let f be the quadratic function that goes through these three points. - It is not hard to show that - - \int_{x_0}^{x_2} f(x)\, dx = \frac{x_2-x_0}{6}\big(y_0+4y_1+y_2\big) - . -

    - - - - - -

    - Consider . - A function f goes through the 3 points shown and the parabola g that also goes through those points is graphed with a dashed line. - Using our equation from above, we know exactly that - - \int_1^3 g(x) \, dx = \frac{3-1}{6}\big(3+4(1)+2\big)= 3 - . -

    - -

    - Since g is a good approximation for f on [1,3], - we can state that - - \int_1^3 f(x)\, dx \approx 3 - . -

    - -
    - A graph of a function f and a parabola that approximates it well on [1,3] - - - A parabola approximating the function f on [1, 3]. - -

    The graph of the function f starts close to (1, 4) in the first quadrant and then slopes downwards to the right towards the x axis. Once - it reaches the point (2, 1), the graph starts sloping upwards to the right and continues upwards. It ends close to (3, 2). The shape of the graph - resembles the shape of a parabola. We approximate the graph of f using a parabola whose path follows f very closely. The parabola intersects f - three times, once at (1, 3) then at (2, 1) and at (3, 2). -

    - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3}, - ytick={1,2,3}, - ymin=-.2,ymax=4, - xmin=.5,xmax=3.5 - ] - - \addplot+ [domain=.9:3.1,samples=30] {11-12*x+4.5*x^2-.5*x^3}; - \addplot+ [domain=.9:3.1] {8-6.5*x+1.5*x^2}; - - \filldraw (axis cs:1,3) circle (1pt); - \filldraw (axis cs:2,1) circle (1pt); - \filldraw (axis cs:3,2) circle (1pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Notice how the interval [1,3] was split into two subintervals as we needed 3 points. - Because of this, whenever we use Simpson's Rule, - we need to break the interval into an even number of subintervals. -

    - -

    - In general, to approximate - \ds \int_a^b f(x)\, dx using Simpson's Rule, - subdivide [a,b] into n subintervals, - where n is even and each subinterval has width \dx = (b-a)/n. - We approximate f with n/2 parabolic curves, - using Equation to compute the area under these parabolas. - Adding up these areas gives the formula: - - \int_a^b f(x) \, dx \approx \frac{\dx}3\Big[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+\ldots+2f(x_{n-2})+4f(x_{n-1})+f(x_{n})\Big] - . -

    - -

    - Note how the coefficients of the terms in the summation have the pattern 1, 4, 2, 4, 2, 4, \ldots, 2, 4, 1. -

    - -

    - - illustrates how the area calculated by Simpson's Rule approximates - \int_0^5 f(x)\, dx for the function f(x)=\sin(\pi x). - In this case, 8 subintervals were used, - resulting in 4 quadratic curves - (dashed lines) - being fitted to each pair of subintervals. - The actual answer - (accurate to 4 decimal places) - is about 10.6366, while Simpson's rule gives 10.7294. - Of course more subintervals would result in better accuracy. - However 8 intervals were chosen specifically so that you could see how the parabolas compare to the original function. - With larger values of n, - it becomes difficult to distinguish the function and its quadratic approximations on each subinterval. -

    - -
    - An illustration of Simpson's rule on f(x)=\sin(\pi x)+2 over [0,5] using 8 subintervals, resulting in 4 quadratic approximations - - Approximating f(x)=sin(pi x)+2 using simpson's rule. - -

    The graph of f(x)=\sin(\pi x)+2 is shaped like wave that oscillates between y=3 and y=1 as it moves in the positive x direction. We consider - the area under the curve from x=0 to x=5. THe area is subdivided into eight intervals and each interval is approximated by quadratic curves. The first - quadratic curve is a downward parabola with its vertex at (3,0.5) which approximates the first two subintervals. The second quadratic curve is also a parabola - whose vertex is located at (1.5,1). This parabola approximates the second two subintervals between x=1.25 and x=2.5. The next two subintervals are - are approximated by the left half of a parabola whose vertex is at (3.75,1.25). The area under the last two subintervals are approximated by a downwards parabola - whose vertex is at (3,4.5). -

    - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.2,ymax=3.3, - xmin=-1,xmax=5.8, - ] - - \addplot [firstcurvestyle,name path=F,domain={-0.9:5.2},samples=120] {sin(deg(3.14159*x))+2} node [pos=.87, above] { $y=\sin(\pi x)+2$}; - \addplot [name path=G,domain={0:5}]{0}; - \addplot [color=firstcolor!50] fill between [of=F and G, soft clip={domain=0:5}]; - - \addplot [secondcurvestyle,name path=H,domain=0:1.25] {-3.27*x^2+3.52*x+2}; - \addplot [color=gray!30] fill between [of=H and G, soft clip={domain=0:1.25}]; - - \addplot [secondcurvestyle,name path=I,domain=1.25:2.5] {1.35*x^2-3.71*x+3.82}; - \addplot [color=gray!30] fill between [of=I and G, soft clip={domain=1.25:2.5}]; - - \addplot [secondcurvestyle,name path=J,domain=2.5:3.75] {1.35*x^2-9.83*x+19.11}; - \addplot [color=gray!30] fill between [of=J and G, soft clip={domain=2.5:3.75}]; - - \addplot [secondcurvestyle,name path=K,domain=3.75:5] {-3.27*x^2+29.18*x-62.145}; - \addplot [color=gray!30] fill between [of=K and G, soft clip={domain=3.75:5}]; - - \addplot [-,color=gray!70,dashed,thin,domain=0:2.924] ({.625},{x}); - \addplot [-,color=gray,dashed,thin,domain=0:1.293] ({1.25},{x}); - \addplot [-,color=gray!70,dashed,thin,domain=0:1.617] ({1.875},{x}); - \addplot [-,color=gray,dashed,thin,domain=0:3] ({2.5},{x}); - \addplot [-,color=gray!70,dashed,thin,domain=0:1.617] ({3.125},{x}); - \addplot [-,color=gray,dashed,thin,domain=0:1.293] ({3.75},{x}); - \addplot [-,color=gray!70,dashed,thin,domain=0:2.924] ({4.375},{x}); - \addplot [-,color=gray,dashed,thin,domain=0:3] ({5},{x}); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Let's demonstrate Simpson's Rule with a concrete example. -

    - - - Using Simpson's Rule - -

    - Approximate \ds\int_0^1 e^{-x^2}\, dx using Simpson's Rule and 4 equally spaced subintervals. -

    -     - -

    - We begin by making a table of values as we have in the past, - as shown in . -

    - -
    - A table of values to approximate \int_0^1e^{-x^2}\, dx, along with a graph of the function - - -
    - - - x_i - e^{-x_i^2} - - - - - - - 0 - 1 - - - 0.25 - 0.939 - - - 0.5 - 0.779 - - - 0.75 - 0.570 - - - 1 - 0.368 - - - - - -
    - -
    - - - - - Approximating the integral of e^{-x^2} using Simpson's rule - -

    The graph of e^{-x^2} starts from (0, 1) and slopes downwards to the right in the positive x direction. We consider the are under the curve from - (0, 1) to (1, 0.25) in the first quadrant. The area is subdivided into four equally spaced intervals. The top of the intervals are parts of parabolas that - track the curve of e^{-x^2}. These approximating curves are almost identical to the parts of the graph they are tracking thus resulting in a much more accurate - approximation. -

    - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={.25,.5,.75,1}, - ymin=-.2,ymax=1.2, - xmin=-.2,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1] {e^-x^2} \closedcycle; - \addplot [firstcurvestyle,domain=0:1] {e^-x^2}; - - \addplot [thick,secondcolor,domain=0:.5] {1-0.042*x-0.8*x^2}; - \addplot [thick,secondcolor,domain=.5:1] {1.218-0.907*x+0.0569*x^2}; - \addplot [thick,secondcolor] coordinates {(0,1) (0,0) (.5,0) (.5,.779)}; - \addplot [thick,secondcolor] coordinates {(.5,0) (1,0) (1,.368)}; - - \draw (axis cs:.75,1) node { $y=e^{-x^2}$}; - - \filldraw (axis cs:0,1) circle (1pt) - (axis cs:.25,.939) circle (1pt) - (axis cs:.5,.779) circle (1pt) - (axis cs:.75,.57) circle (1pt) - (axis cs:1,.368) circle (1pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    - -

    - Simpson's Rule states that - - \int_0^1e^{-x^2}\, dx \approx \frac{0.25}{3}\Big[1+4(0.939)+2(0.779)+4(0.570) + 0.368\Big] = 0.7468\overline{3} - . -

    - -

    - Recall in - we stated that the correct answer, - accurate to 4 places after the decimal, was 0.7468. - Our approximation with Simpson's Rule, with 4 subintervals, - is better than our approximation with the Trapezoidal Rule using 5! -

    - -

    - - shows f(x) = e^{-x^2} along with its approximating parabolas, - demonstrating how good our approximation is. - The approximating curves are nearly indistinguishable from the actual function. -

    -     -     - - - Using Simpson's Rule - -

    - Approximate \ds\int_{-\frac{\pi}4}^{\frac{\pi}2} \sin(x^3)\, dx using Simpson's Rule and 10 equally spaced intervals. -

    -     - -

    - - shows the table of values that we used in the past for this problem, - shown here again for convenience. - Again, \dx = (\pi/2+\pi/4)/10 \approx 0.236. -

    - -
    - Values used to approximate \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in - - - x_i - \sin(x_i^3) - - - -0.7854 - -0.4657 - - - -0.5498 - -0.1654 - - - -0.3142 - -0.0310 - - - -0.0785 - -0.0005 - - - 0.1571 - 0.0039 - - - 0.3927 - 0.0605 - - - 0.6283 - 0.2455 - - - 0.8639 - 0.6011 - - - 1.0996 - 0.9710 - - - 1.3352 - 0.6899 - - - 1.5708 - -0.6700 - - -
    - -

    - Simpson's Rule states that - - \int_{-\frac{\pi}4}^{\frac{\pi}2} \sin(x^3)\, dx \amp \approx \frac{0.2356}3\Big[(-0.4657)+4(-0.1654)+2(-0.0310) + \ldots - \amp \ldots + 2(0.9710) + 4(0.6899) + (-0.6700)\big] - \amp \approx 0.4701 - -

    - -
    - Approximating \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in with Simpson's Rule and 10 equally spaced intervals - - - Approximating the integral of sin(x^3) using Simpson's rule - -

    The area under the graph of \sin(x^3) is subdivided into ten equally spaced subintervals. The leftmost part of the graph that is below the negative - x axis is approximated by a straight line and the upper left part of a downward parabola on its right. The part of the graph on the first quadrant is - subdivided into six subintervals. These are approximated by parabolas that are almost identical to the parts of the graphs they are approximating. The last part - of the graph that is below the positive x axis, is approximated with the descending part of a downward parabola and a straight line. -

    - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.7,ymax=1.2, - xmin=-1,xmax=1.7 - ] - - \addplot [firstcurvestyle,areastyle,domain=-.7854:1.5708,samples=40] {sin(deg(x^3))} \closedcycle; - \addplot [firstcurvestyle,domain=-.7854:1.5708,samples=80] {sin(deg(x^3))}; - - \addplot [thick,secondcolor,domain=-.785:-.314] {-.11-.721*x-1.49*x^2}; - \addplot [thick,secondcolor] coordinates {(-.785,-.466) (-.785,0) (-.314,0) (-.314,-.03)}; - - \addplot [thick,secondcolor,domain=-.314:.157] {.004+.037*x-.236*x^2}; - \addplot [thick,secondcolor] coordinates {(-.314,0) (.157,0) (.157,.004)}; - - \addplot [thick,secondcolor,domain=.157:.628] {.037-.395*x+1.156*x^2}; - \addplot [thick,secondcolor] coordinates {(.157,0) (.628,0) (.628,.246)}; - - \addplot [thick,secondcolor,domain=.628:1.1] {-.632+1.32*x+.13*x^2}; - \addplot [thick,secondcolor] coordinates {(.628,0) (1.1,0) (1.1,.971)}; - - \addplot [thick,secondcolor,domain=1.1:1.57] {-11.98+22.46*x-9.72*x^2}; - \addplot [thick,secondcolor] coordinates {(1.1,0) (1.57,0) (1.57,-.67)}; - - \filldraw (axis cs:-.785,-.466 ) circle (1pt) - (axis cs:-.55,-.165 ) circle (1pt) - (axis cs:-.314,-.031 ) circle (1pt) - (axis cs:-.0785,0 ) circle (1pt) - (axis cs:.157,.004 ) circle (1pt) - (axis cs:.393,.061 ) circle (1pt) - (axis cs:.628,.246 ) circle (1pt) - (axis cs:.864,.601 ) circle (1pt) - (axis cs:1.1,.971 ) circle (1pt) - (axis cs:1.34,.69 ) circle (1pt) - (axis cs:1.57,-.67 ) circle (1pt); - - \draw (axis cs:.6,.9) node { $y=\sin(x^3)$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Recall that the actual value, - accurate to 3 decimal places, is 0.4609. - Our approximation is within one 1/100th of the correct value. - The graph in - shows how closely the parabolas match the shape of the graph. -

    -     -     -     - - - Summary and Error Analysis -

    - We summarize the key concepts of this section thus far in the following Key Idea. -

    - - - Numerical Integration -

    - Let f be a continuous function on [a,b], - let n be a positive integer, - and let \ds\dx = \frac{b-a}{n}. - integrationnumerical!Left/Right Hand Rule - integrationnumerical!Trapezoidal Rule - integrationnumerical!Simpson's Rule - Left/Right Hand Rule - Trapezoidal Rule - Simpson's Rule - numerical integrationLeft/Right Hand Rule - numerical integrationTrapezoidal Rule - numerical integrationSimpson's Rule -

    - -

    - Set x_0=a, x_1 = a+\dx, - \ldots, x_i = a+i\dx, x_{n}=b. -

    - -

    - Consider \ds\int_a^b f(x)\, dx. -

    - -

    - Left Hand Rule: - \ds\int_a^b f(x)\, dx \approx \dx\big[f(x_0) + f(x_1) + \ldots + f(x_{n-1})\big]. -

    - -

    - Right Hand Rule: - \ds\int_a^b f(x)\, dx \approx \dx\big[f(x_1) + f(x_2) + \ldots + f(x_{n})\big]. -

    - -

    - Trapezoidal Rule: - \ds\int_a^b f(x)\, dx \approx \frac{\dx}2\big[f(x_0) + 2f(x_1) + 2f(x_2) +\ldots + 2f(x_{n-1})+ f(x_{n})\big]. -

    - -

    - Midpoint Rule: - \ds \int_a^b f(x)\, dx \approx \sum_{i=1}^n f\left(\frac{x_{i-1}+x_{i}}{2}\right)\Delta x. -

    - -

    - Simpson's Rule: - \ds\int_a^b f(x)\, dx \approx \frac{\dx}3\big[f(x_0) + 4f(x_1) + 2f(x_2) +\ldots + 4f(x_{n-1})+ f(x_{n})\big] \text{ for } n \text{ even}. -

    -     - -

    - In our examples, - we approximated the value of a definite integral using a given method then compared it to the right answer. - This should have raised several questions in the reader's mind, such as: -

    - -

    -

      -
    1. -

      - How was the right answer computed? -

      -
      2. - -
    2. -

      - If the right answer can be found, - what is the point of approximating? -

      -
      3. - -
    3. -

      - If there is value to approximating, - how are we supposed to know if the approximation is any good? -

      -
      4. -
    -

    - -

    - These are good questions, and their answers are educational. - In the examples, the right answer was never computed. - Rather, an approximation accurate to a certain number of places after the decimal was given. - In , - we do not know the exact answer, - but we know it starts with 0.7468. - These more accurate approximations were computed using numerical integration but with more precision (, more subintervals and the help of a computer). -

    - -

    - Since the exact answer cannot be found, - approximation still has its place. - How are we to tell if the approximation is any good? -

    - -

    - Trial and error provides one way. - Using technology, make an approximation with, - say, 10, 100, and 200 subintervals. - This likely will not take much time at all, - and a trend should emerge. - If a trend does not emerge, try using yet more subintervals. - Keep in mind that trial and error is never foolproof; - you might stumble upon a problem in which a trend will not emerge. -

    - -

    - A second method is to use Error Analysis. - While the details are beyond the scope of this text, - there are some formulas that give bounds - for how good your approximation will be. - For instance, - the formula might state that the approximation is within 0.1 of the correct answer. - If the approximation is 1.58, - then one knows that the correct answer is between 1.48 and 1.68. - By using lots of subintervals, - one can get an approximation as accurate as one likes. - states what these bounds are. -

    - - - Error Bounds in the Trapezoidal Rule and Simpson's Rule - -

    - integrationnumerical!Trapezoidal Rule - integrationnumerical!Simpson's Rule - Trapezoidal Ruleerror bounds - Simpson's Ruleerror bounds - numerical integrationTrapezoidal Rule!error bounds - numerical integrationSimpson's Rule!error bounds -

      -
    1. -

      - Let E_T and E_Mbe the error in approximating - \ds \int_a^b f(x)\, dx using the Trapezoidal and Midpoint Rules respectively, with n subintervals. - - If f has a continuous second derivative on [a,b] and K is any upper bound of - \abs{\fpp(x)} on [a,b], then - - \ds E_T \leq \frac{(b-a)^3}{12n^2}K - . - and - - \ds E_M \leq \frac{(b-a)^3}{24n^2}K - . -

      -
      2. - -
    2. -

      - Let E_S be the error in approximating - \ds \int_a^b f(x)\, dx using Simpson's Rule with n subintervals.. - - If f has a continuous 4th derivative on [a,b] and K is any upper bound of - \abs{f^{(4)}(x)} on [a,b], then - - E_S \leq \frac{(b-a)^5}{180n^4}K - . -

      -
      3. -
    -

    -     -     - -

    - There are some key things to note about this theorem. - -

      -
    1. -

      - The larger the interval, the larger the error. - This should make sense intuitively. -

      -
      2. - -
    2. -

      - The error shrinks as more subintervals are used (, as n gets larger). -

      -
      3. - -
    3. -

      - The maximum error in the Midpoint Rule is half of the maximum error in the Trapezoidal Rule. - (Usually the errors in these two rules have opposite signs as well, - that is one will be an under approximation and the other will be an over approximation). -

      -
      4. - -
    4. -

      - The error in Simpson's Rule has a term relating to the 4th derivative of f. - Consider a cubic polynomial: - its 4th derivative is 0. - Therefore, the error in approximating the definite integral of a cubic polynomial with Simpson's Rule is 0 Simpson's Rule computes the exact answer! -

      -
      5. -
    -

    - -

    - We revisit Examples - and - and compute the error bounds using in the following example. -

    - - - Computing error bounds - -

    - Find the error bounds when approximating - \ds \int_0^1 e^{-x^2}\, dx using the Trapezoidal and Midpoint Rules and 5 subintervals, - and using Simpson's Rule with 4 subintervals. -

    -     - -

    - Trapezoidal and Midpoints Rules with n=5: -

    - -

    - We start by computing the 2nd derivative of f(x) = e^{-x^2}: - - \fpp(x) = e^{-x^2}(4x^2-2) - . -

    - -

    - - shows a graph of \fpp(x) on [0,1]. - It is clear that the largest value of \fpp, in absolute value, is 2. -

    - -
    - Graphing \fpp(x) in to help establish error bounds - - - Shows the double derivative of f. - -

    The graph of \fpp(x) starts at (0, -2) and curves upwards to the left towards the positive x axis. It intersects the x axis at - (0.7, 0). Then it moves futher to the right in the positive x direction while moving upwards and ends at (1, 0.5). -

    - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-2.1,ymax=.7, - xmin=-.1,xmax=1.1 - ] - - \addplot+ [domain=0:1,samples=30] {(e^-x^2)*(4*x^2-2)}; - - \draw (axis cs:.8,-1.5) node { $y=e^{-x^2}(4x^2-2)$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Thus we let K=2 and apply the error formula from . - - E_T \leq \frac{(1-0)^3}{12\cdot 5^2}\cdot 2 = 0.00\overline{6} - . - Since the maximum error in the Midpoint rule is half the error in the Trapezoidal Rule, - we can say: - E_M \leq 0.00\overline{3} -

    - - - -

    - Our error estimation formula states that our approximation of 0.7444 found in - is within 0.0067 of the correct answer. - Hence we know that the actual value is within - [0.7444-0.0067, 0.7444+0.0067]=[0.7377,0.7511]. So: - - 0.7377 \leq \int_0^1e^{-x^2}\, dx \leq 0.7511 - - But we can do better than this with the Midpoint Rule since its error is at most half of the error of the Trapezoidal Rule. - Our error estimate formula state that our approximate of 0.7480 found in - is within 0.0034 of the correct answer. - Hence Hence we know that the actual value is within [0.7480-0.0034, 0.7480+0.0033]=[0.7447,0.7513]. -

    - -

    - We had earlier stated the actual answer, - correct to 4 decimal places, to be 0.7468, - affirming the validity of . -

    - -

    - Simpson's Rule with n=4: -

    - -

    - We start by computing the 4th derivative of f(x) = e^{-x^2}: - - f^{(4)}(x) = e^{-x^2}(16x^4-48x^2+12) - . -

    - -

    - - shows a graph of f^{(4)}(x) on [0,1]. - It is clear that the largest value of f^{(4)}, in absolute value, is 12. - Thus we let K=12 and apply the error formula from . - - E_s =\leq \frac{(1-0)^5}{180\cdot 4^4}\cdot 12 = 0.00026 - . -

    - -
    - Graphing f^{(4)}(x) in to help establish error bounds - - - Graphing the fourth derivative of f to help establish error bounds. - -

    The graph of f^{(4)}(x) starts from (12,0) in the positive y-axis and curves downwards to the right towards the positive x axis. - It intersects the x axis close to x=0.5 and continues moving downwards while also moving to the right in the positive x direction. It stops at the point - (1, -8). -

    - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - minor y tick num=4, - ymin=-8,ymax=13.5, - xmin=-.1,xmax=1.1 - ] - - \addplot+ [domain=0:1,samples=30] {(e^-x^2)*(16*x^4-48*x^2+12)}; - - \draw (axis cs:.7,11) node { $y=e^{-x^2}(16x^4-48x^2+12)$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Our error estimation formula states that our approximation of - 0.7468\overline{3} found in - is within 0.00026 of the correct answer, - hence we know that the correct answer is in the interval [0.74683-0.00026 , 0.74683 + 0.00026]=[0.74657,0.74709]. So: - - 0.74657 \leq \int_0^1e^{-x^2}\, dx \leq 0.74709 - . -

    - -

    - Once again we affirm the validity of - since the answer to 4 decimal places is actually 0.7468. -

    -     -     - -

    - At the beginning of this section we mentioned two main situations where numerical integration was desirable. - We have considered the case where an antiderivative of the integrand cannot be computed. - We now investigate the situation where the integrand is not known. - This is, in fact, - the most widely used application of Numerical Integration methods. - Most of the time we observe behavior but do not know the - function that describes it. - We instead collect data about the behavior and make approximations based on this data. - We demonstrate this in an example. -

    - - - Approximating distance traveled - -

    - One of the authors drove his daughter home from school while she recorded their speed every 30 seconds. - The data is given in . - Approximate the distance they traveled. -

    -
    - Speed data collected at 30 second intervals for - - - TimeSpeed - - - (min)() - - - 0 - 0 - - - 1 - 25 - - - 2 - 22 - - - 3 - 19 - - - 4 - 39 - - - 5 - 0 - - - 6 - 43 - - - 7 - 59 - - - 8 - 54 - - - 9 - 51 - - - 10 - 43 - - - 11 - 35 - - - 12 - 40 - - - 13 - 43 - - - 14 - 30 - - - 15 - 0 - - - 16 - 0 - - - 17 - 28 - - - 18 - 40 - - - 19 - 42 - - - 20 - 40 - - - 21 - 39 - - - 22 - 40 - - - 23 - 23 - - - 24 - 0 - - -
    -     - -

    - Recall that by integrating a speed function we get distance traveled. - We have information about v(t); - we will use Simpson's Rule to approximate \ds \int_a^b v(t)\, dt. -

    - -

    - The most difficult aspect of this problem is converting the given data into the form we need it to be in. - The speed is measured in miles per hour, - whereas the time is measured in minutes. -

    - -

    - We need to compute \dx = (b-a)/n. - With 25 data points collected, - there are n=24 subintervals. - What are a and b? - Since we start at time t=0, we have a=0. - The final recorded time was t=12 minutes, which is 1/5 of an hour. - Thus we have - - \dx = \frac{b-a}{n} = \frac{1/5-0}{24} = \frac1{120}; \frac{\dx}{3} = \frac{1}{360} - . -

    - -

    - Thus the distance traveled is approximately: - - \int_0^{0.2}v(t)\, dt \amp \approx \frac{1}{360}\Big[f(x_0)+4f(x_1) + 2f(x_2) + \cdots + 4f(x_{n-1})+f(x_{n})\Big] - \amp = \frac{1}{360}\Big[0+4\cdot25+2\cdot 22 + \cdots + 2\cdot40+4\cdot 23 + 0\Big] - \amp \approx 6.2167 \,\text{ miles. } - -

    - -

    - We approximate the author drove 6.2 miles. - (Because we are sure the reader wants to know, - the author's odometer recorded the distance as about 6.05 miles.) -

    -     -     - -     - - - - Terms and Concepts - - - - - -

    - Simpson's Rule is a method of approximating antiderivatives. - -

    -     - -     - - - - - -

    - What are the two basic situations where approximating the value of a definite integral is necessary? -

    - - -     - - - -     - - - - - -

    - Why are the Left and Right Hand Rules rarely used? -

    -     - - - -

    - They are superseded by the Trapezoidal Rule; - it takes an equal amount of work and is generally more accurate. -

    -     - -     - - - -

    - Simpson's Rule is based on approximating portions of a function with what type of function? -

    -

    - -

    -     - - - - quadratic|quadratic function|parabola|parabolic - - - -     -     - - Problems - - - -

    - Approximate the definite integral with the Trapezoidal Rule and Simpson's Rule, - with n=4. Then find the exact value. -

    -     - - - - - $a = -1; - $b = 1; - $n = 4; - $f = Formula("x^2"); - $F = Formula("x^3/3"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - $exact = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - For the integral \int_{-1}^1 x^2\, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     - - - -

    - Find the exact value. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = 10; - $n = 4; - $f = Formula("5*x"); - $F = Formula("5*x^2/2"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - $exact = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - For the integral \int_{0}^{10} 5x\, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     - - - -

    - Find the exact value. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = pi; - $n = 4; - $f = Formula("sin(x)"); - $F = Formula("-cos(x)"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - $exact = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - For the integral \int_{0}^{\pi} \sin(x) \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     - - - -

    - Find the exact value. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = 4; - $n = 4; - $f = Formula("sqrt(x)"); - $F = Formula("2/3*x^(3/2)"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - $exact = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - For the integral \int_{0}^{4} \sqrt x\, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     - - - -

    - Find the exact value. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = 3; - $n = 4; - $f = Formula("x^3+2x^2-5x+7"); - $F = Formula("x^4/4+2x^3/3-5x^2/2+7x"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - $exact = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - For the integral \int_{0}^{3} (x^3+2x^2-5x+7)\, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     - - - -

    - Find the exact value. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = 1; - $n = 4; - $f = Formula("x^4"); - $F = Formula("x^5/5"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - $exact = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - For the integral \int_{0}^{1} x^4\, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     - - - -

    - Find the exact value. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = 2*pi; - $n = 4; - $f = Formula("cos(x)"); - $F = Formula("sin(x)"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - $exact = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - For the integral \int_{0}^{2\pi} \cos(x) \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     - - - -

    - Find the exact value. -

    - -

    - -

    -     -     -     -     - - - - - $a = -3; - $b = 3; - $n = 4; - $f = Formula("sqrt(9-x^2)"); - $F = Formula("1/2*(sqrt(9-x^2)*x + 9 *asin(x/3))"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - $exact = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - For the integral \int_{-3}^{3} \sqrt{9-x^2} \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     - - - -

    - Find the exact value. -

    - -

    - -

    -     -     -     -     - -     - - - -

    - Approximate the definite integral with the Trapezoidal Rule and Simpson's Rule, - with n=6. -

    -     - - - - - $a = 0; - $b = 1; - $n = 6; - $f = Formula("cos(x^2)"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - - - -

    - For the integral \int_{0}^{1} \cos\big(x^2\big) \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     -     -     - - - - - $a = -1; - $b = 1; - $n = 6; - $f = Formula("e^(x^2)"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - - - -

    - For the integral \int_{-1}^{1} e^{x^2} \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = 5; - $n = 6; - $f = Formula("sqrt(x^2+1)"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - - - -

    - For the integral \int_{0}^{5} \sqrt{x^2+1} \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = pi; - $n = 6; - $f = Formula("x*sin(x)"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - - - -

    - For the integral \int_{0}^{\pi} x\sin(x) \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = pi/2; - $n = 6; - $f = Formula("sqrt(cos(x))"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - - - -

    - For the integral \int_{0}^{\pi/2} \sqrt{\cos(x) } \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     -     -     - - - - - $a = 1; - $b = 4; - $n = 6; - $f = Formula("ln(x)"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - - - -

    - For the integral \int_{1}^{4} \ln(x) \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     -     -     - - - - - $a = -1; - $b = 1; - $n = 6; - $f = Formula("1/(sin(x)+2)"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - - - -

    - For the integral \int_{-1}^{1} \frac{1}{\sin(x) +2} \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = 6; - $n = 6; - $f = Formula("1/(sin(x)+2)"); - $delta_x = ($b - $a)/$n; - @l = (); - $trap = 0; - $simp = 0; - foreach $i (0..$n-1) { - push @l, $i*$delta_x + $a; - } - $flip = 1; - foreach $i (@l) { - $trap += $f->eval(x=>$i) + $f->eval(x=>($i + $delta_x)); - if ($flip == 1) { - $simp += $f->eval(x=>$i) + 4*$f->eval(x=>($i + $delta_x)) + $f->eval(x=>($i+2*$delta_x)); - $flip = 0; - } else { - $flip = 1; - } - } - $trap = $trap*$delta_x/2; - $simp = $simp*$delta_x/3; - - - -

    - For the integral \int_{0}^{6} \frac{1}{\sin(x) +2} \, dx: -

    -     - - - -

    - Approximate using the trapezoidal rule. -

    - -

    - -

    -     -     - - - -

    - Approximate using Simpson's rule. -

    - -

    - -

    -     -     -     -     - -     - - - -

    - Find n such that the error in approximating the given definite integral is less than 0.0001 - when using the Trapezoidal Rule and Simpson's Rule. -

    -     - - - - - $a = 0; - $b = pi; - $err = 0.0001; - $Mt = 1; - $Ms = 1; - $nt = sqrt(($b-$a)**3*$Mt/(12*$err)); - $nt = int($nt + 0.5); - $ns = (($b-$a)**5*$Ms/(180*$err))**(1./4); - $ns = int($ns + .99999); - if ($ns%2==1) { - $ns++; - } - - - -

    - For the integral \int_{0}^{\pi} \sin(x) \, dx: -

    -     - - - -

    - Using the trapezoid rule. -

    - -

    - -

    -     - -

    - n=161 (using \max\big(\fpp(x)\big)=1) -

    -     -     - - - -

    - Using Simpson's rule. -

    - -

    - -

    -     - -

    - n=12 (using \max\big(f^{(4)}(x)\big)=1) -

    -     -     -     -     - - - - - $a = 1; - $b = 4; - $err = 0.0001; - $Mt = 0.75; - $Ms = 105/16; - $nt = sqrt(($b-$a)**3*$Mt/(12*$err)); - $nt = int($nt + 0.99999); - $ns = (($b-$a)**5*$Ms/(180*$err))**(1./4); - $ns = int($ns + .99999); - if ($ns%2==1) { - $ns++; - } - - - -

    - For the integral \int_{1}^{4} \frac{1}{\sqrt x} \, dx: -

    -     - - - -

    - Using the trapezoid rule. -

    - -

    - -

    -     -     - - - -

    - Using Simpson's rule. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = pi; - $err = 0.0001; - $Mt = 38.1865; - $Ms = 804.2; - $nt = sqrt(($b-$a)**3*$Mt/(12*$err)); - $nt = int($nt + 0.99999); - $ns = (($b-$a)**5*$Ms/(180*$err))**(1./4); - $ns = int($ns + .99999); - if ($ns%2==1) { - $ns++; - } - - - -

    - For the integral \int_{0}^{\pi} \cos\big(x^2\big) \, dx: -

    -     - - - -

    - Using the trapezoid rule. -

    - -

    - -

    -     -     - - - -

    - Using Simpson's rule. -

    - -

    - -

    -     -     -     -     - - - - - $a = 0; - $b = 5; - $err = 0.0001; - $Mt = 300; - $Ms = 24; - $nt = sqrt(($b-$a)**3*$Mt/(12*$err)); - $nt = int($nt + 0.99999); - $ns = (($b-$a)**5*$Ms/(180*$err))**(1./4); - $ns = int($ns + .99999); - if ($ns%2==1) { - $ns++; - } - - - -

    - For the integral \int_{0}^{5} x^4 \, dx: -

    -     - - - -

    - Using the trapezoid rule. -

    - -

    - -

    -     -     - - - -

    - Using Simpson's rule. -

    - -

    - -

    -     -     -     -     - -     - - - -

    - A region is given. - Find the area of the region using Simpson's Rule: -

    - -

    -

      -
    1. -

      - where the measurements are in centimeters, - taken in 1 cm increments, and -

      -
      2. - -
    2. -

      - where the measurements are in hundreds of feet, - taken in 100 ft increments. -

      -
      3. -
    -

    -     - - - - - $a = 0; - $b = 6; - $n = 6; - $delta_x = ($b - $a)/$n; - @y = (0, 4.7, 6.3, 6.9, 6.6, 5.1, 0); - @coeffs = (1, 4, 2, 4, 2, 4, 1); - $area = 0; - $size = scalar @y; - foreach $i (0..$size) { - $area += $coeffs[$i] * $y[$i]; - } - $area = $area*$delta_x/3; - $area_ft = $area * 100 * 100; - $area = NumberWithUnits("$area cm^2"); - $area_ft = NumberWithUnits("$area_ft ft^2"); - - - - - - Finding area using Simpson's rule. - -

    We have an area subdivided into six subintervals. The heights of the subintervals are 4.7,6.3,6.9,6.6 and 5.1. -

    - - - \begin{tikzpicture}[xscale=1.19,yscale=.7] - - \draw [firstcolor,thick,fill=firstcolor!30,smooth] plot coordinates { - (0,2.)(0.06639,2.527)(0.2407,2.839)(0.4857,3.009)(0.7641,3.112)(1.039, - 3.221)(1.294,3.367)(1.532,3.53)(1.759,3.689)(1.98,3.823)(2.2,3.915)(2. - 42,3.967)(2.64,3.992)(2.86,4.)(3.08,4.)(3.305,3.987)(3.543,3.93)(3. - 808,3.797)(4.107,3.558)(4.45,3.185)(4.815,2.703)(5.17,2.152)(5.483,1. - 577)(5.723,1.018)(5.872,0.5022)(5.938,0.02083)(5.93,-0.4363)(5.861,-0. - 88)(5.741,-1.319)(5.58,-1.742)(5.389,-2.128)(5.179,-2.456)(4.96,-2. - 705)(4.737,-2.865)(4.502,-2.953)(4.243,-2.991)(3.95,-3.)(3.614,-2.999) - (3.244,-2.984)(2.86,-2.935)(2.484,-2.829)(2.137,-2.646)(1.832,-2.378)( - 1.558,-2.06)(1.299,-1.734)(1.039,-1.443)(0.7641,-1.222)(0.4857,-0. - 9779)(0.2407,-0.5015)(0.06639,0.4201)(0,2.)}; - - \draw (1,3.22) -- (1,-1.443) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.7}; - - \draw (2,3.823) -- (2,-2.5) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.3}; - - \draw (3,4) -- (3,-2.95) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.9}; - - \draw (4,3.62) -- (4,-3) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.6}; - - \draw (5,2.4) -- (5,-2.7) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 5.1}; - - \end{tikzpicture} - - - - - - Enter your answer using the metric measurements. - -

    - -

    - - - Enter your answer using the imperial measurements. - -

    - -

    - -     -     -     - - - - - $a = 0; - $b = 6; - $n = 6; - $delta_x = ($b - $a)/$n; - @y = (0, 3.6, 3.6, 4.5, 6.6, 5.6, 0); - @coeffs = (1, 4, 2, 4, 2, 4, 1); - $area = 0; - $size = scalar @y; - foreach $i (0..$size) { - $area += $coeffs[$i] * $y[$i]; - } - $area = $area*$delta_x/3; - $area_ft = $area * 100 * 100; - $area = NumberWithUnits("$area cm^2"); - $area_ft = NumberWithUnits("$area_ft ft^2"); - - - - - - Finding area using Simpson's rule. - -

    We have an area subdivided into six subintervals. The heights of the subintervals are 3.6,3.6,4.5,6.6 and 5.6. -

    - - - \begin{tikzpicture}[xscale=1.19,yscale=.7] - - \draw [firstcolor,thick,fill=firstcolor!30,smooth] plot coordinates { - (0,1.)(0.05533,1.541)(0.2027,1.965)(0.414,2.278)(0.6613,2.483)(0.9167, - 2.583)(1.157,2.588)(1.382,2.521)(1.595,2.409)(1.799,2.282)(2.,2.167)( - 2.2,2.089)(2.4,2.068)(2.6,2.119)(2.8,2.257)(3.,2.5)(3.201,2.847)(3. - 411,3.233)(3.636,3.58)(3.885,3.807)(4.167,3.833)(4.483,3.604)(4.815,3. - 159)(5.139,2.561)(5.431,1.876)(5.667,1.167)(5.828,0.488)(5.917,-0. - 1427)(5.943,-0.7173)(5.912,-1.228)(5.833,-1.667)(5.715,-2.028)(5.564,- - 2.317)(5.389,-2.543)(5.199,-2.712)(5.,-2.833)(4.799,-2.912)(4.589,-2. - 943)(4.364,-2.917)(4.115,-2.828)(3.833,-2.667)(3.513,-2.432)(3.153,-2. - 149)(2.753,-1.851)(2.313,-1.568)(1.833,-1.333)(1.323,-1.155)(0.828,-0. - 944)(0.4053,-0.5893)(0.1107,0.02133)(0,1)}; - - \draw (1,2.6) -- (1,-1.) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 3.6}; - - \draw (2,2.167) -- (2,-1.43) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 3.6}; - - \draw (3,2.5) -- (3,-2) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.5}; - - \draw (4,3.8) -- (4,-2.75) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.6}; - - \draw (5,2.8) -- (5,-2.83) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 5.6}; - - \end{tikzpicture} - - - - - - Enter your answer using the metric measurements. - -

    - -

    - - - Enter your answer using the imperial measurements. - -

    - -

    - -     -     -     - -     -     -     -
    - - -

    - We started this chapter learning about antiderivatives and indefinite integrals. - We then seemed to change focus by looking at areas between the graph of a function and the x-axis. - We defined these areas as the definite integral of the function, - using a notation very similar to the notation of the indefinite integral. - The Fundamental Theorem of Calculus tied these two seemingly separate concepts together: - we can find areas under a curve, - , we can evaluate a definite integral, using antiderivatives. -

    - -

    - We ended the chapter by noting that antiderivatives are sometimes more than difficult to find: - they are impossible. - Therefore we developed numerical techniques that gave us good approximations of definite integrals. -

    - -

    - We used the definite integral to compute areas, - and also to compute displacements and distances traveled. - There is far more we can do than that. - In - we'll see more applications of the definite integral. - Before that, in - we'll learn advanced techniques of integration, - analogous to learning rules like the Product, Quotient and Chain Rules of differentiation. -

    -     - - -     - - - Techniques of Antidifferentiation - -

    - The previous chapter introduced the antiderivative and connected it to signed areas under a curve through the Fundamental Theorem of Calculus. - The next chapter explores more applications of definite integrals than just area. - As evaluating definite integrals will become important, - we will want to find antiderivatives of a variety of functions. -

    - -

    - This chapter is devoted to exploring techniques of antidifferentiation. - While not every function has an antiderivative in terms of elementary functions (a concept introduced in the section on Numerical Integration), - we can still find antiderivatives of a wide variety of functions. -

    -     - -
    - Substitution - - - - -

    - We motivate this section with an example. - Let f(x) = (x^2+3x-5)^{10}. - We can compute \fp(x) using the Chain Rule. - It is: - - \fp(x) \amp = 10(x^2+3x-5)^9\cdot(2x+3) - \amp = (20x+30)(x^2+3x-5)^9 - . -

    - -

    - Now consider this: What is \int (20x+30)(x^2+3x-5)^9\, dx? - We have the answer in front of us; - - \int (20x+30)(x^2+3x-5)^9\, dx = (x^2+3x-5)^{10}+C - . -

    - -

    - How would we have evaluated this indefinite integral without starting with f(x) as we did? -

    - -

    - This section explores integration by substitution. - It allows us to undo the Chain Rule. - Substitution allows us to evaluate the above integral without knowing the original function first. -

    - -

    - The underlying principle is to rewrite a complicated - integral of the form \int f(x)\, dx as a not-so-complicated integral \int h(u)\, du. - We'll formally establish later how this is done. - First, consider again our introductory indefinite integral, - \int (20x+30)(x^2+3x-5)^9\, dx. - Arguably the most complicated - part of the integrand is (x^2+3x-5)^9. - We wish to make this simpler; - we do so through a substitution. - Let u=x^2+3x-5. - Thus - - (x^2+3x-5)^9 = u^9 - . -

    - -

    - We have established u as a function of x, - so now consider the differential of u: - - du = (2x+3)dx - . -

    - -

    - Keep in mind that (2x+3) and dx are multiplied; - the dx is not just sitting there. -

    - -

    - Return to the original integral and do some substitutions through algebra: - - \int (20x+30)(x^2+3x-5)^9\, dx \amp = \int 10(2x+3)(x^2+3x-5)^9\, dx - \amp =\int 10(\underbrace{x^2+3x-5}_u)^9\underbrace{(2x+3)\, dx}_{du} - \amp =\int 10u^9\, du - \amp = u^{10} + C \quad \text{(replace \(u\) with \(x^2+3x-5\)) } - \amp = (x^2+3x-5)^{10} +C - -

    - -

    - One might well look at this and think I - (sort of) - followed how that worked, - but I could never come up with that on my own, - but the process is learnable. - This section contains numerous examples through which the reader will gain understanding and mathematical maturity enabling them to regard substitution as a natural tool when evaluating integrals. -

    - -

    - We stated before that integration by substitution - undoes the Chain Rule. - Specifically, - let F(x) and g(x) be differentiable functions and consider the derivative of their composition: - - \frac{d}{dx}\Big(F\big(g(x)\big)\Big) = \Fp(g(x))\gp(x) - . -

    - -

    - Thus - - \int \Fp(g(x))\gp(x)\, dx = F(g(x)) + C - . -

    -     - - - - Substitution for indefinite integrals - -

    - Integration by substitution works by recognizing the inside - function g(x) and replacing it with a variable. - By setting u=g(x), we can rewrite the derivative as - - \frac{d}{dx}\Big(F\big(u\big)\Big) = \Fp(u)u' - . -

    - -

    - Since du = \gp(x)dx, we can rewrite the above integral as - - \int \Fp(g(x))\gp(x)\, dx = \int \Fp(u) du = F(u)+C = F(g(x))+ C - . -

    - -

    - This concept is important so we restate it in the context of a theorem. -

    - - - Integration by Substitution - -

    - Let F and g be differentiable functions, - where the range of g is an interval I contained in the domain of F. - Then integrationby substitution - - \int \Fp(g(x))\gp(x)\, dx = F(g(x)) + C - . -

    - -

    - If u = g(x), then du = \gp(x)dx and - - \int \Fp(g(x))\gp(x)\, dx = \int \Fp(u)\, du = F(u)+C = F(g(x))+C - . -

    -     -     - - - -

    - The point of substitution is to make the integration step easy. - Indeed, the step \int \Fp(u)\, du = F(u) + C looks easy, - as the antiderivative of the derivative of F is just F, - plus a constant. - The work involved is making the proper substitution. - There is not a step-by-step process that one can memorize; - rather, experience will be one's guide. - To gain experience, we now embark on many examples. -

    - - - Integrating by substitution - -

    - Evaluate \ds \int x\sin(x^2+5)\, dx. -

    -     - -

    - Knowing that substitution is related to the Chain Rule, - we choose to let u be the inside - function of \sin(x^2+5). - (This is not always a good choice, - but it is often the best place to start.) -

    - -

    - Let u = x^2+5, hence du = 2x\,dx. - The integrand has an x\,dx term, - but not a 2x\,dx term. - (Recall that multiplication is commutative, - so the x does not physically have to be next to dx for there to be an x\,dx term.) - We can divide both sides of the du expression by 2: - - du = 2x\,dx \quad \Rightarrow \quad \frac12du = x\,dx - . -

    - -

    - We can now substitute. - - \int x\sin(x^2+5)\, dx \amp = \int \sin(\underbrace{x^2+5}_u) \underbrace{x\, dx}_{\frac12du} - \amp = \int \frac12\sin(u) \, du - \amp = -\frac12\cos(u) + C \quad \text{ (now replace \(u\) with \(x^2+5\)) } - \amp =-\frac12\cos(x^2+5) + C - . -

    - -

    - Thus \int x\sin(x^2+5)\, dx = -\frac12\cos(x^2+5)+C. - We can check our work by evaluating the derivative of the right hand side. -

    -     -     - - - Integrating by substitution - -

    - Evaluate \ds \int \cos(5x)\, dx. -

    -     - -

    - Again let u replace the inside function. - Letting u = 5x, we have du = 5\, dx. - Since our integrand does not have a 5\, dx term, - we can divide the previous equation by 5 to obtain \frac15du = dx. - We can now substitute. - - \int \cos(5x)\, dx \amp = \int \cos(\underbrace{5x}_u) \underbrace{dx}_{\frac15du} - \amp = \int \frac15\cos(u) \, du - \amp = \frac15\sin(u) + C - \amp = \frac15\sin(5x)+C - . -

    - -

    - We can again check our work through differentiation. -

    -     -     - -

    - The previous example exhibited a common, - and simple, type of substitution. - The inside function was a linear function - (in this case, y = 5x). - When the inside function is linear, - the resulting integration is very predictable, outlined here. -

    - - - Substitution With A Linear Function -

    - Consider \int \Fp(ax+b)\, dx, - where a\neq 0 and b are constants. - Letting u = ax+b gives - du = a\cdot dx, leading to the result - - \int \Fp(ax+b)\, dx = \frac{1}{a}F(ax+b) + C - . -

    -     - -

    - Thus \int \sin(7x-4)\, dx = -\frac17\cos(7x-4)+C. - Our next example can use , - but we will only employ it after going through all of the steps. -

    - - - Integrating by substituting a linear function - -

    - Evaluate \ds \int \frac{7}{-3x+1}\, dx. -

    -     - -

    - View the integrand as the composition of functions f(g(x)), - where f(x) = 7/x and g(x) = -3x+1. - Employing our understanding of substitution, - we let u = -3x+1, the inside function. - Thus du = -3\, dx. - The integrand lacks a -3; - hence divide the previous equation by -3 to obtain -du/3 = dx. - We can now evaluate the integral through substitution. - - \int \frac{7}{-3x+1}\, dx \amp = \int \frac{7}{u}\frac{du}{-3} - \amp = \frac{-7}3\int \frac{du}{u} - \amp = \frac{-7}3\ln\abs{u} + C - \amp =-\frac73\ln\abs{-3x+1} + C - . -

    - -

    - Using is faster, - recognizing that u is linear and a = -3. - One may want to continue writing out all the steps until they are comfortable with this particular shortcut. -

    -     -     - - - - - -

    - Not all integrals that benefit from substitution have a clear - inside function. - Several of the following examples will demonstrate ways in which this occurs. -

    - - - Integrating by substitution - -

    - Evaluate \ds \int \sin(x) \cos(x) \, dx. -

    -     - -

    - There is not a composition of functions here to exploit; - rather, just a product of functions. - Do not be afraid to experiment; - when given an integral to evaluate, - it is often beneficial to think - If I let u be this, - then du must be that - and see if this helps simplify the integral at all. -

    - -

    - In this example, let's set u = \sin(x). - Then du = \cos(x) \, dx, - which we have as part of the integrand! - The substitution becomes very straightforward: - - \int \sin(x) \cos(x) \, dx \amp = \int u\, du - \amp = \frac12u^2+ C - \amp = \frac12\sin^2(x) + C - . -

    - -

    - One would do well to ask What would happen if we let u = \cos(x)? - The result is just as easy to find, yet looks very different. - The challenge to the reader is to evaluate the integral letting - u = \cos(x) and discover why the answer is the same, - yet looks different. -

    -     - -     - -

    - Our examples so far have required - basic substitution. - The next example demonstrates how substitutions can be made that often strike the new learner as being nonstandard. -

    - - - Integrating by substitution - -

    - Evaluate \ds\int x\sqrt{x+3}\, dx. -

    -     - -

    - Recognizing the composition of functions, set u = x+3. - Then du = dx, - giving what seems initially to be a simple substitution. - But at this stage, we have: - - \int x\sqrt{x+3}\, dx = \int x\sqrt{u}\, du - . -

    - -

    - We cannot evaluate an integral that has both an x and an u in it. - We need to convert the x to an expression involving just u. -

    - -

    - Since we set u = x+3, we can also state that u-3 = x. - Thus we can replace x in the integrand with u-3. - It will also be helpful to rewrite \sqrt{u} as u^\frac12. - - \int x\sqrt{x+3} \, dx \amp = \int (u-3)u^\frac12\, du - \amp = \int \big(u^\frac32 - 3u^\frac12\big) \, du - \amp = \frac25u^\frac52 - 2u^\frac32 + C - \amp = \frac25(x+3)^\frac52 - 2(x+3)^\frac32 + C - . -

    - -

    - Checking your work is always a good idea. - In this particular case, - some algebra will be needed to make one's answer match the integrand in the original problem. -

    -     -     - - - Integrating by substitution - -

    - Evaluate \ds \int \frac{1}{x\ln(x) }\, dx. -

    -     - -

    - This is another example where there does not seem to be an obvious composition of functions. - The line of thinking used in is useful here: - choose something for u and consider what this implies du must be. - If u can be chosen such that du also appears in the integrand, - then we have chosen well. -

    - -

    - Choosing u = 1/x makes du = -1/x^2\, dx; - that does not seem helpful. - However, setting u = \ln(x) makes du = 1/x\, dx, - which is part of the integrand. - Thus: - - \int \frac1{x\ln(x)}\, dx \amp = \int \frac{1}{\underbrace{\ln(x)}_{u}}\underbrace{\frac1x\, dx}_{du} - \amp = \int \frac1u\, du - \amp = \ln\abs{u} + C - \amp = \ln\abs{ \ln(x) } + C - . -

    - -

    - The final answer is interesting; - the natural log of the natural log. - Take the derivative to confirm this answer is indeed correct. -

    -     -     - - - -     - - - Integrals Involving Trigonometric Functions -

    - - delves deeper into integrals of a variety of trigonometric functions; - here we use substitution to establish a foundation that we will build upon. -

    - -

    - The next three examples will help fill in some missing pieces of our antiderivative knowledge. - We know the antiderivatives of the sine and cosine functions; - what about the other standard functions tangent, - cotangent, secant and cosecant? - We discover these next. -

    - - - Integrating by substitution: the antiderivative of <m>\tan(x)</m> - -

    - Evaluate \int \tan(x) \, dx. -

    -     - -

    - The previous paragraph established that we did not know the antiderivatives of tangent, - hence we must assume that we have learned something in this section that can help us evaluate this indefinite integral. -

    - -

    - Rewrite \tan(x) as \sin(x) /\cos(x). - While the presence of a composition of functions may not be immediately obvious, - recognize that \cos(x) is - inside the 1/x function. - Therefore, we see if setting - u = \cos(x) returns usable results. - We have that du = -\sin(x) \, dx, - hence -du = \sin(x) \, dx. - We can integrate: - - \int \tan(x) \, dx \amp = \int \frac{\sin(x) }{\cos(x) }\, dx - \amp = \int \frac1{\underbrace{\cos(x) }_u}\underbrace{\sin(x) \, dx}_{-du} - \amp = \int \frac {-1}u \, du - \amp = -\ln\abs{u} + C - \amp = -\ln\abs{\cos(x) } + C - . -

    - -

    - Some texts prefer to bring the -1 inside the logarithm as a power of \cos(x), as in: - - -\ln\abs{\cos(x) } + C \amp = \ln\abs{(\cos(x) )^{-1}} + C - \amp = \ln\abs{ \frac{1}{\cos(x) }} + C - \amp = \ln\abs{\sec(x) } + C - . -

    - -

    - Thus the result they give is \int \tan(x) \, dx = \ln\abs{\sec(x) } + C. - These two answers are equivalent. -

    -     - -     - - - Integrating by substitution: the antiderivative of <m>\sec(x)</m> - -

    - Evaluate \int \sec(x) \, dx. -

    -     - -

    - This example employs a wonderful trick: - multiply the integrand by 1 - so that we see how to integrate more clearly. - In this case, we write 1 as - - 1 = \frac{\sec(x) + \tan(x) }{\sec(x) + \tan(x) } - . -

    - -

    - This may seem like it came out of left field, but it works beautifully. - Consider: - - \int \sec(x) \, dx \amp = \int \sec(x) \cdot \frac{\sec(x) + \tan(x) }{\sec(x) + \tan(x) }\, dx - \amp = \int \frac{\sec^2(x) + \sec(x) \tan(x) }{\sec(x) + \tan(x) }\, dx. - Now let u = \sec(x) +\tan(x); this means du = (\sec(x) \tan(x) + \sec^2(x) )\, dx, which is our numerator. Thus: - \amp = \int \frac{du}{u} - \amp = \ln\abs{u} + C - \amp = \ln\abs{\sec(x) +\tan(x) } + C - . -

    -     - -     - - - - -

    - We can use similar techniques to those used in Examples - and - to find antiderivatives of \cot(x) and \csc(x) - (which the reader can explore in the exercises.) - We summarize our results here. -

    - - - Antiderivatives of Trigonometric Functions - - -

    - integrationof trig. functions -

    - - - - -

    -

      -
    1. \int \sin(x) \, dx = -\cos(x) +C,
      2. - -
    2. \int \cos(x) \, dx = \sin(x) + C,
      3. - -
    3. \int \tan(x) \, dx = -\ln\abs{\cos(x) }+C,
      4. - -
    4. \int \csc(x) \, dx = -\ln\abs{\csc(x) +\cot(x) } +C,
      5. - -
    5. \int \sec(x) \, dx = \ln\abs{\sec(x) +\tan(x) } + C,
      6. - -
    6. \int \cot(x) \, dx = \ln\abs{\sin(x) }+C,
      7. -
    -

    -     -     - -

    - We explore one more common trigonometric integral. -

    - - - Integration by substitution: powers of <m>\cos(x)</m> and <m>\sin(x)</m> - -

    - Evaluate \int \cos^2(x) \, dx. -

    -     - -

    - We have a composition of functions as \cos^2(x) = \big(\cos(x) \big)^2. - However, setting u = \cos(x) means du = -\sin(x) \, dx, - which we do not have in the integral. - Another technique is needed. -

    - - - -

    - The process we'll employ is to use a Power Reducing formula for \cos^2(x), - which states - - \cos^2(x) = \frac{1+\cos(2x)}{2} - . -

    - -

    - The right hand side of this equation is not difficult to integrate. - We have: - - \int \cos^2(x) \, dx \amp = \int \frac{1+\cos(2x)}2\, dx - \amp = \int \left( \frac12 + \frac12\cos(2x)\right)\, dx - \amp = \frac12x + \frac12\frac{\sin(2x)}{2} + C - \amp = \frac12x + \frac{\sin(2x)}4 + C - , - where we used for the antiderivative of \cos(2x). -

    - -

    - We'll make significant use of this power-reducing technique in future sections. -

    -     -     - - -     - - - Simplifying the Integrand -

    - It is common to be reluctant to manipulate the integrand of an integral; - at first, our grasp of integration is tenuous and one may think that working with the integrand will improperly change the results. - Integration by substitution works using a different logic: - as long as equality is maintained, - the integrand can be manipulated so that its - form is easier to deal with. - The next two examples demonstrate common ways in which using algebra first makes the integration easier to perform. -

    - - - Integration by substitution: simplifying first - -

    - Evaluate \ds\int \frac{x^3+4x^2+8x+5}{x^2+2x+1}\, dx. -

    -     - -

    - One may try to start by setting u equal to either the numerator or denominator; - in each instance, the result is not workable. -

    - -

    - When dealing with rational functions (, quotients made up of polynomial functions), - it is an almost universal rule that everything works better when the degree of the numerator is less than the degree of the denominator. - Hence we use polynomial division. -

    - -

    - We skip the specifics of the steps, - but note that when x^2+2x+1 is divided into x^3+4x^2+8x+5, - it goes in x+2 times with a remainder of 3x+3. - Thus - - \frac{x^3+4x^2+8x+5}{x^2+2x+1} = x+2 + \frac{3x+3}{x^2+2x+1} - . -

    - -

    - Integrating x+2 is simple. - The fraction can be integrated by setting u = x^2+2x+1, - giving du = (2x+2)\, dx. - This is very similar to the numerator. - Note that du/2 = (x+1)\, dx and then consider the following: - - \int \frac{x^3+4x^2+8x+5}{x^2+2x+1}\, dx \amp = \int \left(x+2 + \frac{3x+3}{x^2+2x+1}\right)\, dx - \amp = \int (x+2)\, dx + \int \frac{3(x+1)}{x^2+2x+1}\, dx - \amp = \frac12x^2+2x+C_1 + \int \frac{3}{u}\frac{du}{2} - \amp = \frac12x^2+2x+C_1 + \frac32\ln\abs{u} + C_2 - \amp = \frac12x^2+2x+\frac32\ln\abs{x^2+2x+1} + C - . -

    - -

    - In some ways, we lucked out in that after dividing, - substitution was able to be done. - In later sections we'll develop techniques for handling rational functions where substitution is not directly feasible. -

    -     - -     - - - Integration by alternate methods - -

    - Evaluate \ds\int \frac{x^2+2x+3}{\sqrt{x}}\, dx with, - and without, substitution. -

    -     - -

    - We already know how to integrate this particular example. - Rewrite \sqrt{x} as - x^\frac12 and simplify the fraction: - - \frac{x^2+2x+3}{x^{1/2}} = x^\frac32 + 2x^\frac12 + 3x^{-\frac12} - . -

    - -

    - We can now integrate using the Power Rule: - - \int \frac{x^2+2x+3}{x^{1/2}}\, dx \amp = \int\left(x^\frac32 + 2x^\frac12 + 3x^{-\frac12}\right)\, dx - \amp = \frac25x^\frac52 + \frac43x^\frac32 + 6x^\frac12 + C - -

    - -

    - This is a perfectly fine approach. - We demonstrate how this can also be solved using substitution as its implementation is rather clever. -

    - -

    - Let u = \sqrt{x} = x^\frac12; therefore - - du = \frac{1}{2\sqrt{x}}\, dx \quad \Rightarrow \quad 2du = \frac{1}{\sqrt{x}}\, dx - . -

    - -

    - This gives us \ds \int \frac{x^2+2x+3}{\sqrt{x}}\, dx = \int (x^2+2x+3)\cdot2\, du. - What are we to do with the other x terms? - Since u = x^\frac12, u^2 = x, etc. - We can then replace x^2 and x with appropriate powers of u. - We thus have - - \int \frac{x^2+2x+3}{\sqrt{x}}\, dx \amp = \int (x^2+2x+3)\cdot2\, du - \amp = \int 2(u^4 + 2u^2 + 3)\, du - \amp = \frac25u^5 + \frac43u^3 + 6u + C - \amp = \frac25x^\frac52 + \frac43x^\frac32 + 6x^\frac12+C - , - which is obviously the same answer we obtained before. - In this situation, - substitution is arguably more work than our other method. - The fantastic thing is that it works. - It demonstrates how flexible integration is. -

    -     -     -     - - - Substitution and Inverse Trigonometric Functions -

    - When studying derivatives of inverse functions, we learned that - - \frac{d}{dx}\big(\tan^{-1}(x) \big) = \frac{1}{1+x^2} - . -

    - -

    - Applying the Chain Rule to this is not difficult; for instance, - - \frac{d}{dx}\big(\tan^{-1}(5x) \big) = \frac{5}{1+25x^2} - . -

    - -

    - We now explore how Substitution can be used to undo - certain derivatives that are the result of the Chain Rule applied to Inverse Trigonometric functions. - We begin with an example. -

    - - - Integrating by substitution: inverse trigonometric functions - -

    - Evaluate \ds \int \frac{1}{25+x^2}\, dx. -

    -     - -

    - The integrand looks similar to the derivative of the arctangent function. - Note: - - \frac{1}{25+x^2} \amp = \frac{1}{25\left(1+\frac{x^2}{25}\right)} - \amp = \frac{1}{25(1+\left(\frac{x}{5}\right)^2)} - \amp = \frac{1}{25}\frac{1}{1+\left(\frac{x}{5}\right)^2} - . -

    - -

    - Thus - - \int\frac{1}{25+x^2}\, dx = \frac{1}{25}\int \frac{1}{1+\left(\frac{x}{5}\right)^2}\, dx - . -

    - -

    - This can be integrated using Substitution. - Set u = x/5, hence du = dx/5 or dx=5\, du. - Thus - - \int\frac{1}{25+x^2}\, dx \amp = \frac{1}{25}\int \frac{1}{1+\left(\frac{x}{5}\right)^2}\, dx - \amp = \frac15\int \frac{1}{1+u^2}\, du - \amp = \frac15\tan^{-1}(u) + C - \amp = \frac15\tan^{-1}\left(\frac x5\right)+C - -

    -     - -     - -

    - - demonstrates a general technique that can be applied to other integrands that result in inverse trigonometric functions. - The results are summarized here. -

    - - - Integrals Involving Inverse Trigonometric Functions - -

    - Let a \gt 0. -

    - -

    -

      -
    1. \int \frac{1}{a^2+x^2}\, dx = \frac1a\tan^{-1}\left(\frac{x}{a}\right) + C
      2. - -
    2. \int \frac{1}{\sqrt{a^2-x^2}}\, dx = \sin^{-1}\left(\frac{x}{a}\right)+C
      3. - -
    3. \int \frac{1}{x\sqrt{x^2-a^2}}\, dx = \frac1a\sec^{-1}\left(\frac{\abs{x}}{a}\right)+C
      4. -
    -

    -     -     - -

    - Let's practice using . -

    - - - Integrating by substitution: inverse trigonometric functions - -

    - Evaluate the given indefinite integrals: -

      -
    1. \int \frac{1}{9+x^2}\, dx
      2. - -
    2. \int \frac{1}{\sqrt{5-x^2}}\, dx
      3. - -
    3. \int \frac{1}{x\sqrt{x^2-\frac{1}{100}}}\, dx
      4. -
    -

    -     - -

    - Each can be answered using a straightforward application of . - -

      -
    1. -

      - \ds \int \frac{1}{9+x^2}\, dx = \frac13\tan^{-1}\left(\frac x3\right) + C, - as a = 3. -

      -
      2. - -
    2. -

      - \ds \int \frac{1}{\sqrt{5-x^2}} = \sin^{-1}\left(\frac{x}{\sqrt{5}}\right)+C, - as a = \sqrt{5}. -

      -
      3. - -
    3. -

      - \ds \int \frac{1}{x\sqrt{x^2-\frac{1}{100}}}\, dx = 10\sec^{-1}(10x) + C, - as a = \frac1{10}. -

      -
      4. -
    -

    -     -     - -

    - Most applications of are not as straightforward. - The next examples show some common integrals that can still be approached with this theorem. -

    - - - Integrating by substitution: completing the square - -

    - Evaluate \ds \int\frac{1}{x^2-4x+13}\, dx. -

    -     - -

    - Initially, this integral seems to have nothing in common with the integrals in . - As it lacks a square root, - it almost certainly is not related to arcsine or arcsecant. - It is, however, related to the arctangent function. -

    - -

    - We see this by completing the square in the denominator. - We give a brief reminder of the process here. -

    - -

    - Start with a quadratic with a leading coefficient of 1. - It will have the form of x^2 + bx + c. - Take 1/2 of b, square it, - and add/subtract it back into the expression. - , - - x^2+bx+ c \amp = \underbrace{x^2 + bx + \frac{b^2}4}_{(x+b/2)^2} - \frac{b^2}4 + c - \amp = \left(x+\frac b2\right)^2 + c-\frac{b^2}4 - -

    - -

    - In our example, - we take half of -4 and square it, getting 4. - We add/subtract it into the denominator as follows: - - \frac{1}{x^2-4x+13} \amp = \frac{1}{\underbrace{x^2-4x+4}_{(x-2)^2}-4+13} - \amp =\frac{1}{(x-2)^2 + 9} - -

    - -

    - We can now integrate this using the arctangent rule. - Technically, we need to substitute first with u=x-2, - but we can employ instead. - Thus we have - - \int \frac{1}{x^2-4x+13}\, dx \amp = \int \frac{1}{(x-2)^2+9}\, dx - \amp = \frac13\tan^{-1}\left(\frac{x-2}{3}\right)+C - . -

    -     - -     - - - Integrals requiring multiple methods - -

    - Evaluate \ds \int \frac{4-x}{\sqrt{16-x^2}}\, dx. -

    -     - -

    - This integral requires two different methods to evaluate it. - We get to those methods by splitting up the integral into two terms: - - \int \frac{4-x}{\sqrt{16-x^2}}\, dx = \int \frac{4}{\sqrt{16-x^2}}\, dx - \int \frac{x}{\sqrt{16-x^2}}\, dx - . -

    - -

    - We handle each separately. - The first integral is handled using a straightforward application of : -

    - -

    - \ds \int \frac{4}{\sqrt{16-x^2}}\, dx = 4\sin^{-1}\left(\frac{x}{4}\right) + C. -

    - -

    - The second integral is handled by substitution, with u = 16-x^2. - \ds \int\frac{x}{\sqrt{16-x^2}}\, dx: Set u = 16-x^2, - so du = -2x\, dx and x\, dx = -du/2. - We have - - \int\frac{x}{\sqrt{16-x^2}}\, dx \amp = \int\frac{-du/2}{\sqrt{u}} - \amp = -\frac12\int \frac{1}{\sqrt{u}}\, du - \amp = - \sqrt{u} + C - \amp = -\sqrt{16-x^2} + C - . -

    - -

    - Combining these together, we have - - \int \frac{4-x}{\sqrt{16-x^2}}\, dx = 4\sin^{-1}\left(\frac{x}{4}\right) + \sqrt{16-x^2}+C - . - As with all definite integrals, - you can check your work by differentiation. -

    -     - -     -     - - - Substitution and Definite Integration - - - -

    - This section has focused on evaluating indefinite integrals as we are learning a new technique for finding antiderivatives. - However, much of the time integration is used in the context of a definite integral. - Definite integrals that require substitution can be calculated using the following workflow: -

    - -

    -

      -
    1. -

      - Start with a definite integral - \ds \int_a^b f(x)\, dx that requires substitution. -

      -
      2. - -
    2. -

      - Ignore the bounds; use substitution to evaluate - \ds \int f(x)\, dx and find an antiderivative F(x). -

      -
      3. - -
    3. -

      - Evaluate F(x) at the bounds; - that is, evaluate F(x)\Big|_a^b = F(b) - F(a). -

      -
      4. -
    -

    - -

    - This workflow works fine, - but substitution offers an alternative that is powerful and amazing - (and a little time saving). -

    - -

    - At its heart, - (using the notation of ) - substitution converts integrals of the form - \int \Fp(g(x))\gp(x)\, dx into an integral of the form - \int \Fp(u)\, du with the substitution of u = g(x). - The following theorem states how the bounds of a definite integral can be changed as the substitution is performed. -

    - - - Substitution with Definite Integrals - -

    - Let F and g be differentiable functions, - where the range of g is an interval I that is contained in the domain of F and u=g(x). - Then - - integrationdefinite!and substitution - - definite integraland substitution - - - \int_a^b \Fp\big(g(x)\big)\gp(x)\, dx = \int_{g(a)}^{g(b)} \Fp(u)\, du - . -

    -     -     - - - -

    - In effect, - states that once you convert to integrating with respect to u, - you do not need to switch back to evaluating with respect to x. - A few examples will help one understand. -

    - - - Definite integrals and substitution: changing the bounds - -

    - Evaluate \ds\int_0^2 \cos(3x-1)\, dx using . -

    -     - -

    - Observing the composition of functions, - let u=3x-1, hence du = 3\, dx. - As 3\, dx does not appear in the integrand, - divide the latter equation by 3 to get du/3 = dx. -

    - -

    - By setting u = 3x-1, we are implicitly stating that g(x) = 3x-1. - - states that the new lower bound is g(0) = -1; - the new upper bound is g(2) = 5. - We now evaluate the definite integral: - - \int_0^2 \cos(3x-1) \, dx \amp = \int_{-1}^5 \cos(u) \frac{du}{3} - \amp = \frac{1}{3} \sin(u) \Big|_{-1}^5 - \amp = \frac{1}{3}\big(\sin(5) - \sin(-1)\big) - \amp \approx -0.039 - . -

    - -

    - Notice how once we converted the integral to be in terms of u, - we never went back to using x. -

    - -
    - Graphing the areas defined by the definite integrals of - -
    - - - - - Graph of function cos(3x-1) with integrals shown as shaded portions under the curve between 0 and 2. - - -

    - The y axis is drawn from -1 to 1 and the x axis is drawn from - -1 to 5. - The function y=\cos(3x-1) has several maxima of 1 and minima of -1. - The function has several complete waves. At x intercept -0.9 the first wave - starts it forms a y intercept of 0.5 after which it reaches the maxima, it - decreases after and crosses the x axis at 0.9 then it decreases further to - reach the minima then it gets a positive slope and meets the x axis again at 2. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,1,2,3,4,5}, - ymin=-1.1,ymax=1.1, - xmin=-1.5,xmax=5.5] - - \addplot [firstcurvestyle,areastyle,domain=0:2] {cos(deg(3*x-1))} \closedcycle; - \addplot [firstcurvestyle,domain=-1.5:5.5,samples=140] {cos(deg(3*x-1))} node [pos=0.54, above right]{ $y=\cos(3x-1)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - Graph of a cosine wave with amplitude 1/3. - - -

    - The y axis is drawn from -1 to 1 - and the u axis is drawn from -1 to 5. - The area that the curve forms with the u axis is - shaded between -1 and 5. The function has - two u intercepts at u = 1.5 and u = 4.6. There is a parabolic curve above the u axis the major part of which is shaded from -1.4 to 1.5, there is a second shaded inverted parabolic curve of the same size as the first below the u axis, from 1.5 to 4.6. A third parabola is shaded from 4.6 to 5. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,1,2,3,4,5}, - ymin=-1.1,ymax=1.1, - xmin=-1.5,xmax=5.5, - xlabel={$u$}, - ] - - \addplot [firstcurvestyle,areastyle,domain=-1:5] {cos(deg(x))/3} \closedcycle; - \addplot [firstcurvestyle,domain=-1.5:5.5,samples=40] {cos(deg(x))/3} node [pos=0.45, above right] { $y=\frac13\cos(u)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    - -

    - The graphs in tell more of the story. - In - the area defined by the original integrand is shaded, - whereas in - the area defined by the new integrand is shaded. - In this particular situation, - the areas look very similar; - the new region is shorter - but wider, giving the same area. -

    -     -     - - - Definite integrals and substitution: changing the bounds - -

    - Evaluate \ds \int_0^{\pi/2} \sin(x) \cos(x) \, dx using . -

    -     - -

    - We saw the corresponding indefinite integral in . - In that example we set u = \sin(x) but stated that we could have let u = \cos(x). - For variety, we do the latter here. -

    - -

    - Let u = g(x) = \cos(x), - giving du = -\sin(x) \, dx and hence \sin(x) \, dx = -du. - The new upper bound is g(\pi/2) = 0; - the new lower bound is g(0) = 1. - Note how the lower bound is actually larger than the upper bound now. - We have - - \int_0^{\pi/2} \sin(x) \cos(x) \, dx \amp = \int_1^0 -u\, du \quad \text{(switch bounds and change sign) } - \amp = \int_0^1 u\, du - \amp = \frac12u^2\Big|_0^1= 1/2 - . -

    - -

    - In - we have again graphed the two regions defined by our definite integrals. - Unlike the previous example, they bear no resemblance to each other. - However, - guarantees that they have the same area. -

    - -
    - Graphing the areas defined by the definite integrals of - -
    - - - - - Graph of function showing area under the curve as the definite integrals. - - -

    - The y axis is drawn from -0.5 to 1 and the x - axis is drawn from a little before 0 to pi/2. - The function y=\sin(x)*\cos(x) has a parabolic shape shown in the - graph; it is drawn in the first quadrant. The function has two x - intercepts at x=0 and x = \pi/2. The area under the curve - between the x intercepts until the x axis is shaded. The - curve continues from the x intercepts under the x axis. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={1.57}, - extra x tick labels={$\frac{\pi}{2}$}, - ymin=-.6,ymax=1.1, - xmin=-.5,xmax=2 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1.57] {cos(deg(x))*sin(deg(x)} \closedcycle; - \addplot [firstcurvestyle,domain=-.5:2,samples=40] {cos(deg(x))*sin(deg(x)} node [pos=0.5, above right] { $y=\sin(x) \cos(x)$}; - - \end{axis} - \end{tikzpicture} - - - - - -
    - -
    - - - - - Graph of function y=u, showing shaded portion from 0 to 1. - - -

    - The y axis is drawn from -0.5 to 1 and the - u axis is drawn from 0 to \pi/2. The graph - of function y=u is a straight line with a positive slope - and passes through the origin. The area under the line is shaded - from 0 to 1 indicating the limit for the integral. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1}, - extra x ticks={1.57}, - extra x tick labels={$\frac{\pi}{2}$},ymin=-.6,ymax=1.1, - xmin=-.5,xmax=2, - xlabel={$u$} - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1] {x} \closedcycle; - \addplot [firstcurvestyle,domain=-.5:1.1] {x} node [pos=0.9, right] { $y=u$}; - - \end{axis} - \end{tikzpicture} - - - - -
    -     -
    -     - -     - -

    - Integration by substitution is a powerful and useful integration technique. - The next section introduces another technique, - called Integration by Parts. - As substitution undoes the Chain Rule, - integration by parts undoes the Product Rule. - Together, these two techniques provide a strong foundation on which most other integration techniques are based. -

    -     - - - - Terms and Concepts - - - -

    - Substitution undoes what derivative rule? -

    -

    - - -

    -     - - - - - chain|chain rule|the chain rule - - - - - -     - - - - -

    - - One can use algebra to rewrite the integrand of an integral to make it easier to evaluate. -

    -     - -     -     - - - Problems - - -

    - Evaluate the indefinite integral to develop an understanding of Substitution. -

    -     - - - - - $m = random(3,4,1); - $b = non_zero_random(-9,9,1); - $n = random(5,9,1); - if($envir{problemSeed}==1){$m=3;$b=-5;$n=7;}; - Context("Fraction"); - $u = Formula("x^$m+$b")->reduce; - $du = $u->D('x'); - $f = $du * Formula("($u)^$n"); - $F = FormulaUpToConstant("1/($n+1)*($u)^($n+1)"); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($m,$b) = random_subset(2,-9..-1,1..9); - $n = random(3,9,1); - if($envir{problemSeed}==1){$m=-5;$b=7;$n=3;}; - Context("Fraction"); - $u = Formula("x^2+$m x+$b")->reduce; - $du = $u->D('x'); - $f = $du * Formula("($u)^$n"); - $F = FormulaUpToConstant("1/($n+1)*($u)^($n+1)"); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - $n = random(3,9,1); - if($envir{problemSeed}==1){$b=1;$n=8;}; - Context("Fraction"); - $u = Formula("x^2+$b")->reduce; - $f = Formula("x ($u)^$n"); - $F = FormulaUpToConstant("1/(2*($n+1))*($u)^($n+1)"); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($a,$b,$c) = random_subset(3,-9..-1,1..9); - $n = random(3,9,1); - $s = random(2,4,1); - if($envir{problemSeed}==1){$a=3;$b=7;$c=-1;$n=5;$s=2;}; - Context("Fraction"); - $u = Formula("$a x^2 + $b x + $c")->reduce; - $f = Formula("(2*$a*$s x + $b*$s) ($u)^$n")->reduce; - $frac = Fraction($s,$n+1); - $F = FormulaUpToConstant("$frac*($u)^($n+1)"); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($m,$b) = random_subset(2,2..9); - if($envir{problemSeed}==1){$m=2;$b=7;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [-$b/$m-2,-$b/$m+2]}); - $f = Formula("1/($m x + $b)")->reduce; - $F = FormulaUpToConstant("1/$m ln(abs($m x + $b))")->reduce; - $F->{test_at} = [[-$b/$m-2],[-$b/$m+2]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($m,$b) = random_subset(2,2..9); - if($envir{problemSeed}==1){$m=2;$b=3;}; - Context("Fraction"); - $f = Formula("1/sqrt($m x + $b)")->reduce; - $frac = Fraction(2,$m); - $F = FormulaUpToConstant("$frac sqrt($m x + $b)")->reduce; - $F->{limits} = [-$b/$m,-$b/$m+4]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b=3;}; - Context("Fraction"); - $f = Formula("x/sqrt(x + $b)")->reduce; - $F = FormulaUpToConstant("2/3 (x - 2*$b) sqrt(x + $b)")->reduce; - $F->{limits} = [-$b,-$b+4]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - $m = random(3,5,1); - if($envir{problemSeed}==1){$m=3;$b=-1;}; - Context("Fraction"); - $f = Formula("(x^$m+$b x)/sqrt(x)")->reduce; - $frac = Fraction(2*$b,3); - $F = FormulaUpToConstant("x^(3/2) (2/(2*$m+1) x^($m-1) + $frac )")->reduce; - $F->{limits} = [0,4]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("2*e^sqrt(x)"); - $F->{limits} = [0,4]; - - -

    - \ds \int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - $m = random(4,8,1); - if($envir{problemSeed}==1){$m=5;$b=1;}; - Context("Fraction"); - $f = Formula("x^($m-1)/sqrt(x^$m+$b)")->reduce; - $frac = Fraction(2,$m); - $F = FormulaUpToConstant("$frac sqrt(x^$m+$b)")->reduce; - if ($b < 0) { - $absb = abs($b); - $F->{limits} = [$absb**(1/$m)+1,$absb**(1/$m)+4]; - } - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - $m = random(1,3,1); - if($envir{problemSeed}==1){$m=1;$b=1;}; - Context("Fraction"); - $f = Formula("(1/x^$m + $b)/x^($m+1)")->reduce; - $frac = Fraction(2,$m); - $F = FormulaUpToConstant("-1/(2*$m) (1/x^$m + $b)^2")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("ln(x)^2/2"); - - -

    - \ds \int \frac{\ln(x)}{x} dx -

    -

    - -

    -     -     -     -     - - - -

    - Use Substitution to evaluate the indefinite integral involving trigonometric functions. -

    -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=2;}; - Context("Fraction"); - $f = Formula("sin^$m(x) cos(x)")->reduce; - $F = FormulaUpToConstant("sin^($m+1)(x)/($m+1)")->reduce; - - -

    - \ds \int \sin^{}(x) \cos(x) \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=3;}; - Context("Fraction"); - $f = Formula("cos^$m(x) sin(x)")->reduce; - $F = FormulaUpToConstant("-cos^($m+1)(x)/($m+1)")->reduce; - - -

    - \ds \int \cos^{}(x) \sin(x) \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(-9,-1,1); - $b = random(1,9,1); - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$m=-6;$b=3;$trig='cos'}; - Context("Fraction"); - $f = Formula("$trig($b+$m x)")->reduce; - $trigAD =Formula("-$trig(x)")->D('x'); - $F = $trigAD->substitute(x=>Formula("$b+$m x")); - $F = FormulaUpToConstant("$F/$m")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(-9,-1,1); - $b = random(1,9,1); - if($envir{problemSeed}==1){$m=-1;$b=4;}; - Context("Fraction"); - $g = Formula("$b + $m x")->reduce; - $f = Formula("sec^2($b + $m x)")->reduce; - $F = FormulaUpToConstant("tan($b + $m x)/$m")->reduce; - - -

    - \ds \int \sec^2() \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=2;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [0,2*pi/$m]}); - $f = Formula("sec($m x)")->reduce; - $F = FormulaUpToConstant("1/$m ln(abs(sec($m x) + tan($m x)))")->reduce; - $F->{test_at} = [[pi/4/$m],[3*pi/4/$m],[5*pi/4/$m],[7*pi/4/$m]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=2;}; - Context("Fraction"); - $f = Formula("tan^$m(x) sec^2(x)")->reduce; - $F = FormulaUpToConstant("1/($m+1) tan^($m+1)(x)")->reduce; - - -

    - \ds \int \tan^{}(x)\sec^2(x) \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$m=2;$trig='cos'}; - Context("Fraction"); - $f = Formula("x^($m-1) $trig(x^$m)")->reduce; - $trigAD = Formula("-$trig(x)")->D('x'); - $F = $trigAD->substitute(x=>Formula("x^$m")); - $F = FormulaUpToConstant("1/($m) $F")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("tan(x)-x"); - - -

    - \ds \int \tan^2(x) dx -

    -

    - -

    -     -     -     - - - - - Context()->variables->set(x => {limits => [0,2*pi]}); - $F = FormulaUpToConstant("ln(abs(sin(x)))"); - $F->{test_at} = [[pi/4],[3*pi/4],[5*pi/4],[7*pi/4]]; - - -

    - \ds \int \cot(x) \, dx -

    -

    - Do not just refer to for the answer; - justify it through Substitution. -

    -

    - -

    -     -     -     - - - - - Context()->variables->set(x => {limits => [0,2*pi]}); - $F = FormulaUpToConstant("-ln(abs(csc(x) + cot(x)))"); - $F->{test_at} = [[pi/4],[3*pi/4],[5*pi/4],[7*pi/4]]; - - -

    - \ds \int \csc(x) \, dx -

    -

    - Do not just refer to for the answer; - justify it through Substitution. -

    -

    - -

    -     -     -     -     - - - -

    - Use Substitution to evaluate the indefinite integral involving exponential functions. -

    -     - - - - - $m = random(2,9,1); - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$m=3;$b=-1}; - Context("Fraction"); - $f = Formula("e^($m x+$b)")->reduce; - $F = FormulaUpToConstant("1/$m e^($m x+$b)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=3;}; - Context("Fraction"); - $f = Formula("e^(x^$m)x^($m-1)")->reduce; - $F = FormulaUpToConstant("1/$m e^(x^$m)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b=-1}; - Context("Fraction"); - Context()->variables->set(x => {limits => [-$b-2,-$b+2]}); - $f = Formula("e^(x^2 + 2*$b x + ($b)^2) (x + $b)")->reduce; - $F = FormulaUpToConstant("1/2 e^((x+$b)^2)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b=1}; - Context("Fraction"); - $f = Formula("(e^x+$b)/e^x")->reduce; - $F = FormulaUpToConstant("x - $b e^(-x)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(1,9,1); - if($envir{problemSeed}==1){$b=1}; - Context("Fraction"); - $f = Formula("e^x/(e^x+$b)")->reduce; - $F = FormulaUpToConstant("ln(e^x+$b)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $c = list_random(-1,1); - $m = random(2,5,1); - if($envir{problemSeed}==1){$c=-1;$m=2;}; - Context("Fraction"); - $f = Formula("(e^x + $c e^(-x))/e^($m x)")->reduce; - $F = FormulaUpToConstant("1/(1-$m) e^((1-$m)x) + $c/(-1-$m) e^((-1-$m)x)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = random(2,9,1); - if($envir{problemSeed}==1){$b=3;}; - Context("Fraction"); - $f = Formula("$b^($b x)")->reduce; - Context()->flags->set(reduceConstantFunctions=>0); - $F = FormulaUpToConstant("$b^($b x)/($b ln($b))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($b,$c) = random_subset(2,2..9); - if($envir{problemSeed}==1){$b=4;$c=2;}; - Context("Fraction"); - $f = Formula("$b^($c x)")->reduce; - Context()->flags->set(reduceConstantFunctions=>0); - $F = FormulaUpToConstant("$b^($c x)/($c ln($b))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Use Substitution to evaluate the indefinite integral involving logarithmic functions. -

    -     - - - - - - $F = FormulaUpToConstant("ln(x)^2/2"); - - -

    - \ds \int \frac{\ln(x)}{x} dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=2;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [0.1,10]}); - $f = Formula("(ln(x))^$m/x")->reduce; - $F = FormulaUpToConstant("(ln(x))^($m+1)/($m+1)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(3,9,2); - if($envir{problemSeed}==1){$m=3;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [0.1,10]}); - $f = Formula("ln(x^$m)/x")->reduce; - $frac = Fraction($m,2); - $F = FormulaUpToConstant("$frac (ln(x))^2")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,8,2); - if($envir{problemSeed}==1){$m=2;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [0.1,10]}); - $f = Formula("1/(x ln(x^$m))")->reduce; - $F = FormulaUpToConstant("1/$m ln(abs(ln(x^$m)))")->reduce; - $F->{test_at} = [[0.5],[2]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Use Substitution to evaluate the indefinite integral involving rational functions. -

    -     - - - - - ($a,$b) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$a=3;$b=1;}; - Context("Fraction"); - $f = Formula("(x^2 + $a x + $b)/x")->reduce; - $F = FormulaUpToConstant("x^2/2 + $a x + $b ln(abs(x))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - $f = Formula("(x^3 + x^2 + x + 1)/x")->reduce; - $F = FormulaUpToConstant("x^3/3 + x^2/2 + x + ln(abs(x))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $c = list_random(-1,1); - $b = non_zero_random(-9,9,1); - if ($b == $c) {$b = list_random(-9..-2,2..9);} - if($envir{problemSeed}==1){$c=1;$b=-1}; - Context("Fraction"); - Context()->variables->set(x => {limits => [-$c-2,-$c+2]}); - $f = Formula("(x^3 + $b)/(x + $c)")->reduce; - $frac = Fraction(3*$c,2); - $F = FormulaUpToConstant("1/3 (x+$c)^3 + $frac (x+$c)^2 + 3*($c)^2 (x+$c) + (($c)^3 + $b) ln(abs(x+$c))")->reduce; - $F->{test_at} = [[-$c-2],[-$c+2]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - do {($a,$b,$c) = random_subset(3,-9..-1,1..9);} until ((-$c)^2 + $a*(-$c) + $b != 0); - if($envir{problemSeed}==1){$a=2;$b=-5;$c=-3;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [-$c-2,-$c+2]}); - $f = Formula("(x^2 + $a x + $b)/(x + $c)")->reduce; - $F = FormulaUpToConstant("(x+$c)^2/2 + ($a-2*$c)(x+$c) + (($c)^2 - $a*$c + $b)ln(abs(x+$c))")->reduce; - $F->{test_at} = [[-$c-2],[-$c+2]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - do {($a,$b,$c,$d) = random_subset(4,-9..-1,1..9);} until ($a*(-$d)^2 + $b*(-$d) + $c != 0); - if($envir{problemSeed}==1){$a=3;$b=-5;$c=7;$d=1;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [-$d-2,-$d+2]}); - $f = Formula("($a x^2 + $b x + $c)/(x + $d)")->reduce; - $frac = Fraction($a,2); - $F = FormulaUpToConstant("$frac (x+$d)^2 + ($b - 2*$a*$d)(x+$d) + ($a*($d)^2 - $b*$d + $c)ln(abs(x+$d))")->reduce; - $F->{test_at} = [[-$d-2],[-$d+2]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($a,$b,$c) = random_subset(3,-9,-6,-3,0,3,6,9); - if($envir{problemSeed}==1){$a=3;$b=3;$c=0;}; - Context("Fraction"); - $denom = Formula("x^3 + $a x^2 + $b x + $c")->reduce; - $A = 2*$a/3; - $B = $b/3; - $numer = Formula("x^2 + $A x + $B")->reduce; - $f = Formula("($numer)/($denom)")->reduce; - $F = FormulaUpToConstant("1/3 ln(abs($denom))")->reduce; - $bound = abs($c)+abs($b)+abs($a); # be sure to test where $denom is pos and neg - $F->{test_at} = [[-1],[1],[-$bound-1],[$bound+1]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Use Substitution to evaluate the indefinite integral involving inverse trigonometric functions. -

    -     - - - - - $a = list_random(2,3,5,6,7,11,13,14,15,17); - if($envir{problemSeed}==1){$a=7;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$a/(x^2+$a)")->reduce; - $F = FormulaUpToConstant("sqrt($a) atan(x/sqrt($a))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - if($envir{problemSeed}==1){$a=3;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [-$a,$a]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$a/sqrt($a^2-x^2)")->reduce; - $F = FormulaUpToConstant("$a asin(x/$a)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = list_random(2,3,5,6,7,10,11,13,14,15); - $b = random(2,20,1); - if($envir{problemSeed}==1){$a=5;$b=14;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [-sqrt($a),sqrt($a)]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$b/sqrt($a-x^2)")->reduce; - $F = FormulaUpToConstant("$b asin(x/sqrt($a))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($a,$b) = random_subset(2,2..9); - if($envir{problemSeed}==1){$a=3;$b=2;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [$a,$a+10]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$b/(x sqrt(x^2-$a^2))")->reduce; - $frac = Fraction($b,$a); - $F = FormulaUpToConstant("$frac asec(abs(x)/$a)")->reduce; - $F->{test_at} = [[-$a-1],[-$a-2],[-$a-3],[-$a-4]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($a,$b) = random_subset(2,2..9); - if($envir{problemSeed}==1){$a=4;$b=5;}; - Context("Fraction"); - Context()->variables->set(x => {limits => [$a,$a+10]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("($b x)/sqrt(x^6 - ($a)^2 x^4)")->reduce; - $frac = Fraction($b,$a); - $F = FormulaUpToConstant("$frac asec(abs(x)/$a)")->reduce; - $F->{test_at} = [[-$a-1],[-$a-2],[-$a-3],[-$a-4]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - Context()->variables->set(x => {limits => [-1,1]}); - $F = FormulaUpToConstant("1/2*asin(x^2)"); - - -

    - \ds \int \frac{x}{\sqrt{1-x^4}} dx -

    -

    - -

    -     -     -     - - - - - $a = non_zero_random(-9,9,1); - $b = list_random(2,3,5,6,7,10,11,13,14,15); - if($envir{problemSeed}==1){$a=-1;$b=7;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("1/(x^2 + (2*$a) x + (($a)^2+$b))")->reduce; - $F = FormulaUpToConstant("1/sqrt($b) atan((x+$a)/$b)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = non_zero_random(-9,9,1); - ($b,$c) = random_subset(2,2..9); - if($envir{problemSeed}==1){$a=-3;$b=4;$c=2}; - Context("Fraction"); - Context()->noreduce('(-x)-y','(-x)+y'); - Context()->variables->set(x => {limits => [-$a-$b,-$a+$b]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$c/sqrt(-x^2 - (2*$a) x + (($b)^2-($a)^2))")->reduce; - $F = FormulaUpToConstant("$c asin((x+$a)/$b)")->reduce; - $F->{limits} = [-$b-$a,$b-$a]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = non_zero_random(-9,9,1); - ($b,$c) = random_subset(2,2..9); - if($envir{problemSeed}==1){$a=-4;$b=5;$c=3}; - Context("Fraction"); - Context()->noreduce('(-x)-y','(-x)+y'); - Context()->variables->set(x => {limits => [-$a-$b,-$a+$b]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$c/sqrt(-x^2 - (2*$a) x + (($b)^2-($a)^2))")->reduce; - $F = FormulaUpToConstant("$c asin((x+$a)/$b)")->reduce; - $F->{limits} = [-$b-$a,$b-$a]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = non_zero_random(-9,9,1); - $b = random(2,9,1); - if($envir{problemSeed}==1){$a=3;$b=5}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$b/(x^2+(2*$a)x+(($a)^2+($b)^2))")->reduce; - $F = FormulaUpToConstant("atan((x+$a)/$b)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Evaluate the indefinite integral. -

    -     - - - - - $m = random(3,9,1); - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$m=3;$b=3}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("x^($m-1)/(x^$m + $b)^2")->reduce; - $F = FormulaUpToConstant("-1/($m(x^$m + $b))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(1,9,1); - ($b,$c) = random_subset(2,-9..-1,1..9); - ($n,$m) = num_sort(random_subset(2,2..6)); - $p = random(4,9,1); - if($envir{problemSeed}==1){$a=5;$b=5;$c=2;$m=3;$n=2;$p=8;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $u = Formula("$a x^$m + $b x^$n + $c")->reduce; - $g = gcd($a*$m,$b*$n); - $A = $a*$m/$g; - $B = $b*$n/$g; - $f = Formula("($A x^($m-1) + $B x^($n-1))($u)^$p")->reduce; - $F = FormulaUpToConstant("1/(($p+1)*$g) ($u)^($p+1)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(1,9,1); - $c = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$a=1;$c=-1;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("x/sqrt($a + $c x^2)")->reduce; - $frac = Fraction(1,$c); - $F = FormulaUpToConstant("$frac sqrt($a + $c x^2)")->reduce; - if ($c < 0) { - $F->{limits} = [-sqrt(-$a/$c),sqrt(-$a/$c)]; - }; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(3,9,1); - $c = non_zero_random(-9,9,1); - $trig = list_random('tan','-cot'); - if($envir{problemSeed}==1){$m=3;$c=1;$trig='-cot'}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $dt = ($trig eq 'tan') ? '\sec^2' : '\csc^2'; - $h = Formula("x^$m+$c")->reduce; - $g = Formula("$trig(x)")->D('x')->reduce->substitute(x=>$h)->reduce; - $f = Formula("x^($m-1) $g")->reduce; - $F = FormulaUpToConstant("$trig(x^$m+$c)")->reduce; - - -

    - \ds \int x^{} \left(\right) \, dx -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','-cos'); - if($envir{problemSeed}==1){$trig='-cos'}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $g = Formula("$trig(x)")->D('x')->reduce; - $f = ($trig eq 'sin') ? Formula("cos(x) sqrt(sin(x))") : Formula("sin(x) sqrt(cos(x))"); - $F = FormulaUpToConstant("2/3*$trig(x)^(3/2)")->reduce; - $F->{limits} = [0,pi/2]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - $b = non_zero_random(-9,9,1); - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$trig='sin';$m=5;$b=1;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$trig($m x + $b)")->reduce; - $F = Formula("-$trig(x)")->D('x')->reduce->substitute(x=>Formula("$m x + $b")); - $F = FormulaUpToConstant("1/$m $F")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b=-5}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - Context()->variables->set(x => {limits => [-$b-2,-$b+2]}); - $f = Formula("1/(x+$b)")->reduce; - $F = FormulaUpToConstant("ln(abs(x+$b))")->reduce; - $F->{test_at} = [[-$b-2],[-$b+2]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($a,$b,$c) = random_subset(3,1..9); - if($envir{problemSeed}==1){$a=7;$b=3;$c=2;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - Context()->variables->set(x => {limits => [-$c/$b-2,-$c/$b+2]}); - $f = Formula("$a/($b x+$c)")->reduce; - $frac = Fraction($a,$b); - $F = FormulaUpToConstant("$frac ln(abs($b x+$c))")->reduce; - $F->{test_at} = [[-$c/$b-2],[-$c/$b+2]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - do {($b,$c) = random_subset(2,-9..-1,1..9)} until (($b)**2 != 4*$c); - $m = random(2,4,1); - $k = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b=3;$c=5;$m=3;$k=-5;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - if (($b)**2 > 4*$c) { - ($min, $max) = num_sort((-$b - sqrt(($b)**2 - 4*$c))/2,(-$b + sqrt(($b)**2 - 4*$c))/2); - $D = sqrt(($b)**2 - 4*$c); - Context()->variables->set(x => {limits => [$min-$D,$max+$D]}); - }; - $denom = Formula("x^2 + $b x + $c")->reduce; - $numer = Formula("$m x^3 + ($m*$b + $k)x^2 + ($m*$c + $b*$k + 2) x + ($c*$k + $b)")->reduce; - $f = Formula("($numer)/($denom)"); - $frac = Fraction($m,2); - $F = FormulaUpToConstant("$frac x^2 + $k x + ln(abs($denom))")->reduce; - if (($b)**2 > 4*$c) { - $F->{test_at} = [[$min-$D],[$max+$D]]; - }; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - do {($b,$c) = random_subset(2,-9..-1,1..9)} until (($b)**2 != 4*$c); - if($envir{problemSeed}==1){$b=7;$c=3;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - if (($b)**2 > 4*$c) { - ($min, $max) = num_sort((-$b - sqrt(($b)**2 - 4*$c))/2,(-$b + sqrt(($b)**2 - 4*$c))/2); - $D = sqrt(($b)**2 - 4*$c); - Context()->variables->set(x => {limits => [$min-$D,$max+$D]}); - }; - $denom = Formula("x^2 + $b x + $c")->reduce; - $numer = Formula("2x + $b")->reduce; - $f = Formula("($numer)/($denom)"); - $F = FormulaUpToConstant("ln(abs($denom))")->reduce; - if (($b)**2 > 4*$c) { - $F->{test_at} = [[$min-$D],[$max+$D]]; - }; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - do {($a,$b,$c) = random_subset(3,-9..-1,1..9)} until (($b)**2 != 4*$a*$c); - $k = random(2,5,1); - if($envir{problemSeed}==1){$a=3;$b=9;$c=7;$k=3;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - if (($b)**2 > 4*$a*$c) { - ($min, $max) = num_sort((-$b - sqrt(($b)**2 - 4*$a*$c))/2,(-$b + sqrt(($b)**2 - 4*$a*$c))/2); - $D = sqrt(($b)**2 - 4*$a*$c); - Context()->variables->set(x => {limits => [$min-$D,$max+$D]}); - }; - $denom = Formula("$a x^2 + $b x + $c")->reduce; - $g = gcd(2*$a,$b); - $A = 2*$a/$g; - $B = $b/$g; - $numer = Formula("$g*$k ($A x + $B)")->reduce; - $f = Formula("($numer)/($denom)"); - $F = FormulaUpToConstant("$k ln(abs($denom))")->reduce; - if (($b)**2 > 4*$c) { - $F->{test_at} = [[$min-$D],[$max+$D]]; - }; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - do {($b,$c) = random_subset(2,-9..-1,1..9)} until (($b)**2 != 4*$c); - $k = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b=-7;$c=1;$k=7;}; - $m = -1; - Context("Fraction"); - Context()->noreduce('(-x)-y','(-x)+y'); - Context()->flags->set(reduceConstantFunctions=>0); - if (($b)**2 > 4*$c) { - ($min, $max) = num_sort((-$b - sqrt(($b)**2 - 4*$c))/2,(-$b + sqrt(($b)**2 - 4*$c))/2); - $D = sqrt(($b)**2 - 4*$c); - Context()->variables->set(x => {limits => [$min-$D,$max+$D]}); - }; - $denom = Formula("x^2 + $b x + $c")->reduce; - $numer = Formula("$m x^3 + ($m*$b + $k)x^2 + ($m*$c + $b*$k + 2) x + ($c*$k + $b)")->reduce; - $f = Formula("($numer)/($denom)"); - $frac = Fraction($m,2); - $F = FormulaUpToConstant("$frac x^2 + $k x + ln(abs($denom))")->reduce; - if (($b)**2 > 4*$c) { - $F->{test_at} = [[$min-$D],[$max+$D]]; - }; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - if($envir{problemSeed}==1){$a=9;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("x/(x^4+$a^2)")->reduce; - $F = FormulaUpToConstant("1/(2*$a) atan(x^2/$a)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - if($envir{problemSeed}==1){$a=2;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$a/($a^2x^2+1)")->reduce; - $F = FormulaUpToConstant("atan($a x)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - if($envir{problemSeed}==1){$a=2;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - Context()->variables->set(x => {limits => [1/$a,1/$a+4]}); - $f = Formula("1/(x sqrt($a^2x^2-1))")->reduce; - $F = FormulaUpToConstant("asec(abs($a x))")->reduce; - $F->{test_at} = [[-1/$a-1],[-1/$a-2],[-1/$a-3],[-1/$a-4]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($a,$b) = random_subset(2,2..9); - if($envir{problemSeed}==1){$a=4;$b=3;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - Context()->variables->set(x => {limits => [-$a/$b,$a/$b]}); - $f = Formula("1/sqrt($a^2 - $b^2 x^2)")->reduce; - $F = FormulaUpToConstant("1/$b asin($b x/$a)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = non_zero_random(-9,9,1); - $b = random(2,9,1); - $c = non_zero_random(-9,9,1); - $k = random(2,9,1); - if($envir{problemSeed}==1){$a=-1;$b=3;$c=1;$k = 3;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $denom = Formula("x^2 + 2*$a x + (($a)^2 + $b^2)")->reduce; - $numer = Formula("$k x + ($k*$a + $c)")->reduce; - $f = Formula("($numer)/($denom)")->reduce; - $frac = Fraction($k,2); - $frac2 = Fraction($c,$b); - $F = FormulaUpToConstant("$frac ln(abs($denom)) + $frac2 atan((x+$a)/$b)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = non_zero_random(-9,9,1); - $b = random(2,9,1); - $c = non_zero_random(1,19,1); - $k = random(2,9,1); - if($envir{problemSeed}==1){$a=6;$b=5;$c=19;$k=-2;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $denom = Formula("x^2 + 2*$a x + (($a)^2 + $b^2)")->reduce; - $numer = Formula("$k x + ($k*$a + $c)")->reduce; - $f = Formula("($numer)/($denom)")->reduce; - $frac = Fraction($k,2); - $frac2 = Fraction($c,$b); - $F = FormulaUpToConstant("$frac2 atan((x+$a)/$b) + $frac ln(abs($denom))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = non_zero_random(-9,9,1); - $b = list_random(2,3,5,6,7,10,11,13,14,15); - $c = non_zero_random(1,50,1); - $k = random(2,20,1); - if($envir{problemSeed}==1){$a=-5;$b=7;$c=41;$k=15;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $denom = Formula("x^2 + 2*$a x + (($a)^2 + $b)")->reduce; - $numer = Formula("x^2 + (2*$a+$k) x + (($a)^2 + $b + $k*$a + $c)")->reduce; - $f = Formula("($numer)/($denom)")->reduce; - $frac = Fraction($k,2); - $F = FormulaUpToConstant("x+$c/sqrt($b) atan((x+$a)/sqrt($b)) + $frac ln(abs($denom))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = random(2,9,1); - if($envir{problemSeed}==1){$b=3;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("x^3/(x^2 + $b^2)")->reduce; - $frac = Fraction($b**2,2); - $F = FormulaUpToConstant("x^2/2 - $frac ln(abs(x^2+$b^2))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = non_zero_random(-9,9,1); - $b = list_random(2,3,5,6,7,10,11,13,14,15); - $c = non_zero_random(1,50,1); - $k = random(2,20,1); - if($envir{problemSeed}==1){$a=2;$b=5;$c=24;$k=6;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $denom = Formula("x^2 + 2*$a x + (($a)^2 + $b)")->reduce; - $numer = Formula("x^3 + ($b - 3*($a)^2 + $k)x + ((($a)^2+$b) (-2*$a) + $a*$k + $c)")->reduce; - $f = Formula("($numer)/($denom)")->reduce; - $frac = Fraction($k,2); - $F = FormulaUpToConstant("1/2 x^2 - (2*$a) x + $frac ln(abs($denom)) + $c/sqrt($b) atan((x+$a)/sqrt($b))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("-atan(cos(x))"); - - -

    - \ds \int \frac{\sin(x)}{\cos^2(x)+1} dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("atan(sin(x))"); - - -

    - \ds \int \frac{\cos(x)}{\sin^2(x)+1} dx -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$trig='sin'}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $num = ($trig eq 'cos') ? 'sin' : 'cos'; - $numtex = '\'.$num; - $trigtex = '\'.$trig; - $f = Formula("$num(x)/(1 - $trig^2(x))")->reduce; - $F = FormulaUpToConstant("ln(abs(sec(x) + tan(x)))")->reduce; - if ($trig eq 'cos') { - $F = FormulaUpToConstant("-ln(abs(csc(x) + cot(x)))")->reduce; - }; - $F->{test_at} = [[pi/4],[3*pi/4],[5*pi/4],[7*pi/4]]; - - -

    - \ds \int \frac{(x)}{1-^2(x)} \, dx -

    -

    - -

    -     -     -     - - - - - $h = non_zero_random(-9,9,1); - $k = random(-9,-1,1); - $c = random(3,9,1); - if($envir{problemSeed}==1){$h=1;$k=-7;$c=3;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - Context()->variables->set(x => {limits => [$h-sqrt(-$k),$h+sqrt(-$k)]}); - $numer = Formula("$c x - $c*$h")->reduce; - $quad = Formula("x^2 - (2*$h) x + (($h)^2+$k)")->reduce; - $f = Formula("($numer)/(sqrt($quad))")->reduce; - $F = FormulaUpToConstant("$c sqrt($quad)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $h = non_zero_random(-9,9,1); - $k = random(-9,-1,1); - if($envir{problemSeed}==1){$h=3;$k=-1;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - Context()->variables->set(x => {limits => [$h-sqrt(-$k),$h+sqrt(-$k)]}); - $numer = Formula("x - $h")->reduce; - $quad = Formula("x^2 - (2*$h) x + (($h)^2+$k)")->reduce; - $f = Formula("($numer)/(sqrt($quad))")->reduce; - Context()->variables->set(x => {limits => [sqrt(-$k)+$h,sqrt(-$k)+$h+4]}); - $F = FormulaUpToConstant("sqrt($quad)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Evaluate the definite integral. -

    -     - - - - - $b = non_zero_random(-9,9,1); - ($lb,$ub) = num_sort(random_subset(2,-$b-9..-$b-1)); - if($envir{problemSeed}==1){$b=-5;$lb=1;$ub=3}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("1/(x+$b)")->reduce; - $frac = Fraction($ub+$b,$lb+$b); - $A = Formula("ln($frac)"); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - ($lb,$ub) = num_sort(random_subset(2,0,1,4,9,16,25,36,49,64,81)); - $lb = $lb - $b; - $ub = $ub - $b; - if($envir{problemSeed}==1){$b=-2;$lb=2;$ub=6}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("x sqrt(x+$b)")->reduce; - $fracu = Fraction(2*(($ub+$b)/5 - $b/3)); - $fracl = Fraction(2*(($lb+$b)/5 - $b/3)); - $A = Fraction($fracu*($ub + $b)**(3/2) - $fracl*($lb + $b)**(3/2)); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','cos'); - $lb = list_random('-pi','-pi/2'); - $ub = list_random('pi','pi/2'); - $n = random(2,5,1); - if($envir{problemSeed}==1){$trig='sin';$lb='-pi/2';$ub='pi/2';$n=2;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("sin^$n(x) cos(x)"); - $ftex = '\sin^' . $n . '(x)\cos(x)'; - $A = Fraction(1/($n+1)*sin(Real($ub))**($n+1) - 1/($n+1)*sin(Real($lb))**($n+1)); - if ($trig eq 'cos') { - $f = Formula("cos^$n(x) sin(x)"); - $ftex = '\cos^' . $n . '(x)\sin(x)'; - $A = Fraction(-1/($n+1)*cos(Real($ub))**($n+1) + 1/($n+1)*cos(Real($lb))**($n+1)); - } - $lb = Formula("$lb"); - $ub = Formula("$ub"); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $n = random(3,9,1); - if($envir{problemSeed}==1){$n=4;}; - $lb = 0; - $ub = 1; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("2x (1 - x^2)^$n"); - $A = Fraction(-1/($n+1)*(1 - $ub**2)**($n+1) + 1/($n+1)*(1 - $lb**2)**($n+1)); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - ($lb,$ub) = num_sort(random_subset(2,-$b-2..-$b+2)); - if($envir{problemSeed}==1){$b=1;$lb=-2;$ub=-1}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("(x+$b) e^(x^2 + 2*$b x + ($b)^2)")->reduce; - $A = Formula("1/2 (e^(($ub + $b)^2) - e^(($lb + $b)^2))")->reduce; - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $A = Formula("pi/2")->reduce; - - -

    - \ds \int_{-1}^{1} \frac{1}{1+x^2} \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b=-3;}; - $lb = -$b-1; - $ub = -$b+1; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("1/(x^2 + 2*$b x + (($b)^2+1))")->reduce; - $A = Formula("pi/2")->reduce; - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - ($lbindex,$ubindex) = num_sort(random_subset(2,0..8)); - if($envir{problemSeed}==1){$lbindex=5;$ubindex=7;}; - ($lb,$ub) = (-2,'-sqrt(3)','-sqrt(2)',-1,0,1,'sqrt(2)','sqrt(3)',2)[$lbindex,$ubindex]; - Context()->flags->set(reduceConstantFunctions=>0); - $lb = Formula("$lb"); - $ub = Formula("$ub"); - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("1/sqrt(4 - x^2)")->reduce; - $A = Real("arcsin($ub/2) - arcsin($lb/2)"); - $frac = Fraction($A/pi); - $A = Formula("$frac pi")->reduce; - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     -     -     -     -
    -
    - Integration by Parts -

    - Here's a simple integral that we can't yet evaluate: - - \int x\cos(x) \,dx - . -

    - -

    - It's a simple matter to take the derivative of the integrand using the Product Rule, - but there is no Product Rule for integrals. - However, this section introduces - Integration by Parts, - a method of integration that is based on the Product Rule for derivatives. - It will enable us to evaluate this integral. -

    - - - -

    - The Product Rule says that if u and v are functions of x, - then (uv)' = u'v + uv'. - For simplicity, - we've written u for u(x) and v for v(x). - Suppose we integrate both sides with respect to x. - This gives - - \int (uv)'\,dx = \int (u'v+uv')\,dx - . -

    - -

    - By the Fundamental Theorem of Calculus, - the left side integrates to uv. - The right side can be broken up into two integrals, and we have - - uv = \int u'v\,dx + \int uv'\,dx - . -

    - -

    - Solving for the second integral we have - - \int uv'\,dx = uv - \int u'v\,dx - . -

    - -

    - Using differential notation, - we can write du = u'(x)dx and - dv=v'(x)dx and the expression above can be written as follows: - - \int u\,dv = uv - \int v\,du - . -

    - -

    - This is the Integration by Parts formula. - For reference purposes, we state this in a theorem. -

    - - - Integration by Parts - -

    - Let u and v be differentiable functions of - x on an interval I containing a and b. - Then - - \int u\, dv = uv - \int v\, du - , - and integrationby parts - - \int_{x=a}^{x=b} u\, dv = uv\Big|_a^b - \int_{x=a}^{x=b}v\, du - . -

    -     -     - - - - - -

    - Let's try an example to understand our new technique. -

    - - - Integrating using Integration by Parts - -

    - Evaluate \ds\int x\cos(x) \, dx. -

    -     - -

    - The key to Integration by Parts is to identify part of the integrand as - u and part as dv. - Regular practice will help one make good identifications, - and later we will introduce some principles that help. - For now, let u=x and dv=\cos(x) \, dx. -

    - -

    - It is generally useful to make a small table of these values as done below. - Right now we only know u and dv as shown on the left of - ; - on the right we fill in the rest of what we need. - If u = x, then du = dx. - Since dv = \cos(x)\, dx, - v is an antiderivative of \cos(x). - We choose v = \sin(x). -

    - -
    - Setting up Integration by Parts - -

    - - u\amp = x \amp v\amp =\mathord{?} - du\amp = \mathord{?} \amp dv\amp =\cos(x) \, dx - -

    - -

    - \implies -

    - -

    - - u\amp = x \amp v\amp =\sin(x) - du\amp = dx \amp dv\amp =\cos(x) \, dx - -

    -     -
    - -

    - Now substitute all of this into the Integration by Parts formula, giving - - \int x\cos(x)\,dx = x\sin(x) - \int \sin(x) \,dx - . -

    - -

    - We can then integrate \sin(x) to get - -\cos(x) + C and overall our answer is - - \int x\cos(x)\, dx = x\sin(x) + \cos(x) + C - . -

    - -

    - Note how the antiderivative contains a product, x\sin(x). - This product is what makes Integration by Parts necessary. -

    - -

    - We can check our work by taking the derivative: - - \lzoo{x}{x\sin(x) + \cos(x) + C} \amp =x\cos(x)+\sin(x)-\sin(x)+0 - \amp = x\cos(x) - . -

    -     - -     - -

    - You may wonder what would have happened in - if we had chosen our u and dv differently. - If we had chosen u=\cos(x) and - dv=x \, dx then du=-\sin(x)\, dx and v=x^2/2. - Our second integral is not simpler than the first; - we would have - - \int x\cos(x)\,dx=\cos(x)\frac{x^2}{2}-\int \frac{x^2}{2}\left(-\sin(x)\right)\,dx - . - The only way to approach this second integral would be yet another integration by parts. -

    - -

    - - demonstrates how Integration by Parts works in general. - We try to identify u and dv in the integral we are given, - and the key is that we usually want to choose u and dv - so that du is simpler than u and v - is hopefully not too much more complicated than dv. - This will mean that the integral on the right side of the Integration by Parts formula, - \int v\,du will be simpler to integrate than the original integral \int u\,dv. -

    - -

    - In the example above, we chose u=x and dv=\cos(x)\,dx. - Then du=dx was simpler than u and - v=\sin(x) is no more complicated than dv. - Therefore, instead of integrating x\cos(x) \,dx, - we could integrate \sin(x)\,dx, which we knew how to do. -

    - -

    - A useful mnemonic for helping to determine u is LIATE, where -

    - -

    - L = Logarithmic, I = Inverse Trig., A = - Algebraic (polynomials, roots, - power functions), T = Trigonometric, - and E = Exponential. -

    - -

    - If the integrand contains both a logarithmic and an algebraic term, - in general letting u be the logarithmic term works best, - as indicated by L coming before A in LIATE. -

    - - - -

    - We now consider another example. -

    - - - Integrating using Integration by Parts - -

    - Evaluate \displaystyle \int x e^x\,dx. -

    -     - -

    - The integrand contains an Algebraic term (x) and an - Exponential term - (e^x). - Our mnemonic suggests letting u be the algebraic term, - so we choose u=x and dv=e^x\,dx. - Then du=dx and v=e^x as indicated by the tables below. -

    - -
    - Setting up Integration by Parts - -

    - - u\amp = x \amp v\amp =\mathord{?} - du\amp = \mathord{?} \amp dv\amp =e^x \, dx - -

    - -

    - \implies -

    - -

    - - u\amp = x \amp v\amp =e^x - du\amp = dx \amp dv\amp =e^x \, dx - -

    -     -
    - -

    - We see du is simpler than u, - while there is no change in going from dv to v. - This is good. - The Integration by Parts formula gives - - \int x e^x\,dx = xe^x - \int e^x\,dx - . -

    - -

    - The integral on the right is simple; our final answer is - - \int xe^x\, dx = xe^x - e^x + C - . -

    - -

    - Note again how the antiderivatives contain a product term. -

    -     -     - - - Integrating using Integration by Parts - -

    - Evaluate \displaystyle \int x^2\cos(x) \,dx. -

    -     - -

    - The mnemonic suggests letting u=x^2 instead of the trigonometric function, - hence dv=\cos(x)\,dx. - Then du=2x\,dx and v=\sin(x) as shown below. -

    - -
    - Setting up Integration by Parts - -

    - - u\amp = x^2 \amp v\amp =\mathord{?} - du\amp = \mathord{?} \amp dv\amp =\cos(x) \, dx - -

    - -

    - \implies -

    - -

    - - u\amp = x^2 \amp v\amp =\sin(x) - du\amp = 2x\, dx \amp dv\amp =\cos(x) \, dx - -

    -     -
    - -

    - The Integration by Parts formula gives - - \int x^2\cos(x)\,dx = x^2\sin(x) - \int 2x\sin(x)\,dx - . -

    - -

    - At this point, - the integral on the right is indeed simpler than the one we started with, - but to evaluate it, we need to do Integration by Parts again. - Here we choose r=2x and - ds=\sin(x) and fill in the rest below. - (We are choosing new names since we have already used u and v. - Our integration by parts formula is now \int r\,ds=rs-\int s\,dr.) -

    - -
    - Setting up Integration by Parts (again) - -

    - - u\amp = 2x \amp v\amp =\text{?} - du\amp = \text{?} \amp dv\amp =\sin(x)\, dx - -

    - -

    - \Rightarrow -

    - -

    - - u\amp = 2x \amp v\amp =-\cos(x) - du\amp = 2\, dx \amp dv\amp =\sin(x)\, dx - -

    -     -
    - -

    - - \int x^2\cos(x)\,dx = x^2\sin(x) - \left(-2x\cos(x) - \int -2\cos(x)\,dx\right) - . -

    - -

    - The integral all the way on the right is now something we can evaluate. - It evaluates to -2\sin(x). - Then going through and simplifying, - being careful to keep all the signs straight, our answer is - - \int x^2\cos(x)\, dx = x^2\sin(x) + 2x\cos(x) - 2\sin(x) + C - . -

    -     - -     - - - Integrating using Integration by Parts - -

    - Evaluate \displaystyle \int e^x\cos(x) \,dx. -

    -     - -

    - This is a classic problem. - Our mnemonic suggests letting u - be the trigonometric function instead of the exponential. - In this particular example, - one can let u be either \cos(x) or e^x; - to demonstrate that we do not have to follow LIATE, we choose - u=e^x and hence dv = \cos(x)\,dx. - Then du=e^x\,dx and v=\sin(x) as shown below. -

    - -
    - Setting up Integration by Parts - -

    - - u\amp = e^x \amp v\amp =\mathord{?} - du\amp = \mathord{?} \amp dv\amp =\cos(x) \, dx - -

    - -

    - \implies -

    - -

    - - u\amp = e^x \amp v\amp =\sin(x) - du\amp = e^x \amp dv\amp =\cos(x) \, dx - -

    -     -
    - -

    - Notice that du is no simpler than u, - going against our general rule - (but bear with us). - The Integration by Parts formula yields - - \int e^x\cos(x)\, dx = e^x\sin(x) - \int e^x\sin(x)\,dx - . -

    - -

    - The integral on the right is not much different than the one we started with, - so it seems like we have gotten nowhere. - Let's keep working and apply Integration by Parts to the new integral, - using u=e^x and dv = \sin(x)\,dx. - This leads us to the following: -

    - -
    - Setting up Integration by Parts (again) - -

    - - r\amp = e^x \amp s\amp =\mathord{?} - dr\amp = \mathord{?} \amp ds\amp =\sin(x) \, dx - -

    - -

    - \implies -

    - -

    - - r\amp = e^x \amp s\amp =-\cos(x) - dr\amp = e^x\, dx \amp ds\amp =\sin(x) \, dx - -

    -     -
    - -

    - The Integration by Parts formula then gives: - - \int e^x\cos(x)\,dx \amp = e^x\sin(x) - \left(-e^x\cos(x) - \int -e^x\cos(x)\,dx\right) - \amp = e^x\sin(x)+ e^x\cos(x) - \int e^x\cos(x)\, dx - . -

    - -

    - It seems we are back right where we started, - as the right hand side contains \int e^x\cos(x)\,dx. - But this is actually a good thing. -

    - -

    - Add \ds\int e^x\cos(x)\, dx to both sides. - This gives - - 2\int e^x\cos(x) \, dx \amp = e^x\sin(x) + e^x\cos(x) - Now divide both sides by 2 and then add the integration constant: - \int e^x\cos(x) \, dx \amp = \frac{1}{2}\big(e^x\sin(x) + e^x\cos(x) \big)+C - . -

    - -

    - Simplifying a little, our answer is thus - - \int e^x\cos(x)\, dx = \frac12e^x\left(\sin(x) + \cos(x)\right)+C - . -

    -     - -     - - - Integrating using Integration by Parts: antiderivative of <m>\ln(x)</m> - -

    - Evaluate \int \ln(x)\,dx. -

    -     - -

    - One may have noticed that we have rules for integrating the familiar trigonometric functions and e^x, - but we have not yet given a rule for integrating \ln(x). - That is because \ln(x) can't easily be integrated with any of the rules we have learned up to this point. - But we can find its antiderivative by a clever application of Integration by Parts. - Set u=\ln(x) and dv=dx. - This is a good, - sneaky trick to learn as it can help in other situations. - This determines du=(1/x)\,dx and v=x as shown below. -

    - -
    - Setting up Integration by Parts - -

    - - u\amp = \ln(x) \amp v\amp =\mathord{?} - du\amp = \mathord{?} \amp dv\amp =1 \, dx - -

    - -

    - \implies -

    - -

    - - u\amp = \ln(x) \amp v\amp =x - du\amp = 1/x\, dx \amp dv\amp =1 \, dx - -

    -     -
    - -

    - Putting this all together in the Integration by Parts formula, - things work out very nicely: - - \int \ln(x)\,dx = x\ln(x) - \int x\,\frac1x\,dx - . -

    - -

    - The new integral simplifies to \int 1\,dx, - which is about as simple as things get. - Its integral is x+C and our answer is - - \int \ln(x) \, dx = x\ln(x) - x + C - . -

    -     - -     - - - Integrating using Int. by Parts: antiderivative of <m>\arctan x</m> - -

    - Evaluate \displaystyle \int \arctan x \,dx. -

    -     - -

    - The same sneaky trick we used above works here. - Let u=\arctan x and dv=dx. - Then du=1/(1+x^2)\,dx and v=x. - The Integration by Parts formula gives - - \int \arctan x \,dx = x\arctan x - \int \frac x{1+x^2}\,dx - . -

    - -

    - The integral on the right can be solved by substitution. - Taking w=1+x^2, we get dw=2x\,dx. - The integral then becomes - - \int \arctan x \,dx = x\arctan x - \frac12\int \frac 1{w}\,dw - . -

    - -

    - The integral on the right evaluates to \ln\abs{w}+C, - which becomes \ln(1+x^2)+C - (we can drop the absolute values as 1+x^2 is always positive). - Therefore, the answer is - - \int \arctan x\, dx = x\arctan x - \frac12\ln(1+x^2) + C - . -

    -     - -     - - - - - Substitution Before Integration -

    - When taking derivatives, it was common to employ multiple rules - (such as using both the Quotient and the Chain Rules). - It should then come as no surprise that some integrals are best evaluated by combining integration techniques. - In particular, here we illustrate making an unusual - substitution first before using Integration by Parts. -

    - - - Integration by Parts after substitution - -

    - Evaluate \ds \int \cos(\ln(x))\, dx. -

    -     - -

    - The integrand contains a composition of functions, - leading us to think Substitution would be beneficial. - Letting u=\ln(x), we have du = 1/x\, dx. - This seems problematic, as we do not have a 1/x in the integrand. - But consider: - - du = \frac 1x\, dx \quad \Rightarrow \quad x\cdot du = dx - . -

    - -

    - Since u = \ln(x), - we can use inverse functions and conclude that x = e^u. - Therefore we have that - - dx \amp = x\cdot du - \amp = e^u\, du - . -

    - -

    - We can thus replace \ln(x) with u and dx with e^u\, du. - Thus we rewrite our integral as - - \int \cos(\ln(x))\, dx = \int e^u\cos u \, du - . -

    - -

    - We evaluated this integral on the right in . (This integral can also be found in a table of integrals). - Using the result there, we have: - - \int \cos(\ln(x) )\, dx \amp = \int e^u\cos(u) \, du - \amp = \frac12e^u\big(\sin(u) + \cos(u) \big) + C - \amp = \frac12e^{\ln(x) } \big(\sin(\ln(x) ) + \cos(\ln(x) )\big)+C - \amp = \frac12x \big(\sin(\ln(x) ) + \cos(\ln(x) )\big)+C - . -

    -     - -     -     - - - Definite Integrals and Integration By Parts -

    - So far we have focused only on evaluating indefinite integrals. - Of course, we can use Integration by Parts to evaluate definite integrals as well, - as states. - We do so in the next example. -

    - - - Definite integration using Integration by Parts - -

    - Evaluate \displaystyle \int_1^2 x^2 \ln(x) \,dx. -

    -     - -

    - Our mnemonic suggests letting u=\ln(x), hence dv =x^2\,dx. - We then get du = (1/x)\,dx and v=x^3/3 as shown below. -

    - -
    - Setting up Integration by Parts - -

    - - u\amp = \ln(x) \amp v\amp =\text{?} - du\amp = \text{?} \amp dv\amp =x^2\, dx - -

    - -

    - \Rightarrow -

    - -

    - - u\amp = \ln(x)\amp v\amp =x^3/3 - du\amp = 1/x\, dx \amp dv\amp =x^2\, dx - -

    -     -
    - -

    - The Integration by Parts formula then gives - - \int_1^2 x^2 \ln(x)\,dx \amp = \left.\frac{x^3}3\ln(x)\right|_1^2 - \int_1^2 \frac{x^3}{3}\,\frac 1x\,dx - \amp = \left.\frac{x^3}3\ln(x)\right|_1^2 - \int_1^2 \frac{x^2}{3}\,dx - \amp = \left.\frac{x^3}3\ln(x)\right|_1^2 - \left.\frac{x^3}{9}\right|_1^2 - \amp = \left.\left(\frac{x^3}3\ln(x) - \frac{x^3}{9}\right)\right|_1^2 - \amp = \left(\frac83\ln(2) - \frac89\right)-\left(\frac13\ln(1) - \frac19\right) - \amp = \frac83\ln(2) - \frac79 - \amp \approx 1.07 - . -

    -     - -     - -

    - In general, Integration by Parts is useful for integrating certain products of functions, - like \int x e^x\,dx or \int x^3\sin(x)\,dx. - It is also useful for integrals involving logarithms and inverse trigonometric functions. -

    - -

    - As stated before, - integration is generally more difficult than derivation. - We are developing tools for handling a large array of integrals, - and experience will tell us when one tool is preferable/necessary over another. - For instance, - consider the three similar-looking integrals - - \int xe^x\,dx, \qquad \int x e^{x^2}\,dx \qquad \text{ and } \qquad \int xe^{x^3}\,dx - . -

    - -

    - While the first is calculated easily with Integration by Parts, - the second is best approached with Substitution. - Taking things one step further, - the third integral has no answer in terms of elementary functions, - so none of the methods we learn in calculus will get us the exact answer. -

    - -

    - Integration by Parts is a very useful method, - second only to Substitution. - In the following sections of this chapter, - we continue to learn other integration techniques. - - focuses on handling integrals containing trigonometric functions. -

    -     - - - - Terms and Concepts - - - - -

    - - Integration by Parts is useful in evaluating integrands that contain products of functions. -

    -     - -     - - - - -

    - - Integration by Parts can be thought of as the - opposite of the Chain Rule. -

    -     - -     - - - - -

    - For what is LIATE useful? -

    - -     - - - -

    - Determining which functions in the integrand to set equal to u - and which to set equal to dv. -

    -     - -     - - - - -

    - - If the integral that results from Integration by Parts appears to also need Integration by Parts, - then a mistake was made in the original choice of u. -

    -     - -

    - False; it is not uncommon to need to use Integration by Parts several times to fully evaluate an integral. -

    -     - -     -     - - - Problems - - - -

    - Evaluate the given indefinite integral. -

    -     - - - - - $F = FormulaUpToConstant("sin(x)-x*cos(x)"); - - -

    - \ds \int x\sin(x) \, dx -

    -

    - -

    -     - -

    - \sin(x) - x\cos(x) +C -

    -     -     -     - - - - - $F = FormulaUpToConstant("-e^(-x)*(x+1)"); - - -

    - \ds \int xe^{-x}\, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("-x^2*cos(x)+2x*sin(x)+2*cos(x)"); - - -

    - \ds \int x^2\sin(x) \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("-x^3*cos(x)+3x^2*sin(x)+6x*cos(x)-6*sin(x)"); - - -

    - \ds \int x^3\sin(x) \, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - $F = FormulaUpToConstant("1/2*e^(x^2)"); - - -

    - \ds \int xe^{x^2}\, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("e^x*(x^3-3x^2+6x-6)"); - - -

    - \ds \int x^3e^{x}\, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - $F = FormulaUpToConstant("-1/2*x*e^(-2x)-e^(-2x)/4"); - - -

    - \ds \int xe^{-2x}\, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - $F = FormulaUpToConstant("1/2*e^x*(sin(x)-cos(x))"); - - -

    - \ds \int e^x\sin(x) \, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - $F = FormulaUpToConstant("1/5*e^(2x)*(sin(x)+2*cos(x))"); - - -

    - \ds \int e^{2x}\cos(x) \, dx -

    -

    - -

    -     -     -     - - - - - ($m,$n) = random_subset(2,2..9); - if($envir{problemSeed}==1){$m=2;$n=3}; - $g = gcd($m,$n); - $M = $m/$g; - $N = $n/$g; - $f = Formula("e^($m x) sin($n x)"); - Context("Fraction"); - $frac = Fraction($g,$m**2+$n**2); - $F = FormulaUpToConstant("$frac e^($m x)*($M sin($n x) - $N cos($n x))"); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=5;}; - $f = Formula("e^($m x) cos($m x)"); - Context("Fraction"); - $frac = Fraction(1,2*$m); - $F = FormulaUpToConstant("$frac e^($m x) (sin($m x) + cos($m x))"); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("1/2*sin(x)^2"); - - -

    - \ds \int \sin(x) \cos(x) \, dx -

    -

    - -

    -     -     -     - - - - - $f = Formula("sin^-1(x)"); - $F = FormulaUpToConstant("sqrt(1-x^2)+x*asin(x)"); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m = 2;}; - $f = Formula("tan^-1($m x)"); - $F = FormulaUpToConstant("x*tan^-1($m x)-1/(2*$m)*ln($m^2x^2+1)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("1/2*x^2*atan(x)-x/2+1/2*atan(x)"); - - -

    - \ds \int x\tan^{-1}(x) \, dx -

    -

    - -

    -     -     -     - - - - - $f = Formula("cos^-1(x)"); - $F = FormulaUpToConstant("-sqrt(1-x^2)+x*acos(x)"); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - Context()->variables->set(x=>{limits=>[0,4]}); - $F = FormulaUpToConstant("1/2*x^2*ln(x)-x^2/4"); - - -

    - \ds \int x\ln(x) \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b = -2;}; - $f = Formula("(x+$b)ln(x)")->reduce; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[0,4]}); - $F = FormulaUpToConstant("1/2*x^2*ln(x)-x^2/4 + $b x ln(x) - $b x")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b = 1;}; - $f = Formula("x ln(x-$b)")->reduce; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[$b,$b+4]}); - $frac = Fraction(($b)**2,2); - $F = FormulaUpToConstant("1/2 x^2 ln(x-$b) - 1/4 (x-$b)^2 - $b x - $frac ln(x-$b)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("1/2*x^2*ln(x^2) - x^2/2"); - - -

    - \ds \int x\ln(x^2)\, dx -

    -

    - -

    -     -     -     - - - - - Context()->variables->set(x => {limits => [0,4]}); - $F = FormulaUpToConstant("1/3*x^3*ln(x) - x^3/9"); - - -

    - \ds \int x^2\ln(x) \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("2x+x*(ln(x))^2-2x*ln(x)"); - $F->{limits} = [1,5]; - - -

    - \ds \int \left(\ln(x) \right)^2\, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b = 1;}; - $f = Formula("(ln(x+$b))^2")->reduce; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[-$b,-$b+4]}); - $F = FormulaUpToConstant("2(x+$b)+(x+$b)*ln(x+$b)^2-2*(x+$b)*ln(x+$b)")->reduce; - $F->{limits} = [-$b+1,-$b+5]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("x*tan(x)+ln(abs(cos(x)))"); - $F->{test_at} = [[pi/4],[3*pi/4],[5*pi/4],[7*pi/4]]; - - -

    - \ds \int x\sec^2(x) \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("ln(abs(sin(x)))-x*cot(x)"); - $F->{test_at} = [[pi/4],[3*pi/4],[5*pi/4],[7*pi/4]]; - - -

    - \ds \int x\csc^2(x) \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$b = -2;}; - $f = Formula("x sqrt(x+$b)")->reduce; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[-$b,-$b+4]}); - $frac = Fraction(2*$b,3); - $F = FormulaUpToConstant("(2/5 (x+$b)^2 - $frac (x+$b))sqrt(x+$b)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = non_zero_random(2,9,1); - if($envir{problemSeed}==1){$b = 2;}; - $f = Formula("x sqrt(x^2-$b)")->reduce; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[sqrt($b),sqrt($b)+4]}); - $F = FormulaUpToConstant("1/3(x^2-$b)^(3/2)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("sec(x)"); - - -

    - \ds \int \sec(x) \tan(x) \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("x*sec(x)-ln(abs(sec(x)+tan(x)))"); - - -

    - \ds \int x\sec(x) \tan(x) \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("-x*csc(x)-ln(abs(csc(x)+cot(x)))"); - - -

    - \ds \int x\csc(x) \cot(x) \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Evaluate the indefinite integral after first making a substitution. -

    -     - - - - - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$trig = 'sin';}; - $f = Formula("$trig(ln(x))"); - Context()->variables->set(x=>{limits=>[0,4]}); - $F = FormulaUpToConstant("x/2*(sin(ln(x)) - cos(ln(x)))"); - $F = FormulaUpToConstant("x/2*(sin(ln(x)) + cos(ln(x)))") if ($trig eq 'cos'); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$trig = 'cos';}; - $f = Formula("e^(2x) $trig(e^x)"); - $F = FormulaUpToConstant("cos(e^x)+(e^x)*sin(e^x)"); - $F = FormulaUpToConstant("sin(e^x)-(e^x)*cos(e^x)") if ($trig eq 'sin'); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $trig = list_random('sin','cos'); - if($envir{problemSeed}==1){$trig = 'sin';}; - $f = Formula("$trig(sqrt(x))"); - Context()->variables->set(x=>{limits=>[0,4]}); - $F = FormulaUpToConstant("2*sin(sqrt(x)) - 2*sqrt(x)*cos(sqrt(x))"); - $F = FormulaUpToConstant("2*cos(sqrt(x)) + 2*sqrt(x)*sin(sqrt(x))") if ($trig eq 'cos'); - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("x ln(sqrt(x)) - x/2"); - Context()->variables->set(x=>{limits=>[0,4]}); - - -

    - \ds \int \ln(\sqrt{x})\, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("2*sqrt(x)*e^sqrt(x)-2*e^sqrt(x)"); - Context()->variables->set(x=>{limits=>[0,4]}); - - -

    - \ds \int e^{\sqrt{x}}\, dx -

    -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("x^2/2"); - Context()->variables->set(x=>{limits=>[0,4]}); - - -

    - \ds \int e^{\ln(x) }\, dx -

    -

    - -

    -     -     -     -     - - - -

    - Evaluate the definite integral. - Note: the corresponding indefinite integral appears in - Exercises. -

    -     - - - - - $b = list_random('\pi/2','\pi','3\pi/2','2\pi'); - if($envir{problemSeed}==1){$b = '\pi';}; - $F = Formula("sin(x)-x*cos(x)"); - %r = ('\pi/2'=>'1','\pi'=>'pi','3\pi/2'=>'-1','2\pi'=>'-2pi'); - $r = Compute($r{$b}); - - -

    - \ds \int_0^{} x\sin(x) \, dx -

    -

    - -

    -     -     -     - - - - - ($a,$b) = num_sort(random_subset(2,-2,-1,1,2)); - if($envir{problemSeed}==1){$a=-1;$b=1;}; - $f = Formula("x e^(-x)"); - $F = Formula("-e^(-x)*(x+1)"); - $r = Formula("-($b+1)e^(-$b)+($a+1)e^(-$a)")->reduce; - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random('\pi/6','\pi/4','\pi/3','\pi/2'); - if($envir{problemSeed}==1){$b='\pi/4';}; - $a = '-'.$b; - $f = Formula("x^2 sin(x)"); - $F = Formula("-x^2*cos(x)+2x*sin(x)+2*cos(x)"); - $r = Formula("0")->reduce; - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random('\pi/6','\pi/4','\pi/3','\pi/2'); - if($envir{problemSeed}==1){$b='\pi/2';}; - $a = '-'.$b; - $f = Formula("x^3 sin(x)"); - $F = Formula("-x^3*cos(x)+3x^2*sin(x)+6x*cos(x)-6*sin(x)"); - $r = Formula("-pi^3/(32sqrt(2))+3pi^2/(8sqrt(2))+3pi/sqrt(2)-12/sqrt(2)")->reduce; - $r = Formula("-pi^3sqrt(3)/216+pi^2/12+pi sqrt(3)-6")->reduce if ($b eq '\pi/6'); - $r = Formula("-pi^3/27+pi^2sqrt(3)/3+2pi-6*sqrt(3)")->reduce if ($b eq '\pi/3'); - $r = Formula("3pi^2/2-12")->reduce if ($b eq '\pi/2'); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $c = random(2,9,1); - if($envir{problemSeed}==1){$c=2;}; - $a = 0; - Context()->flags->set(reduceConstantFunctions=>0); - $b = Formula("sqrt(ln(2))"); - $f = Formula("x e^(x^2)"); - $F = Formula("1/2*e^(x^2)"); - Context("Fraction"); - $r = Fraction($c/2-1/2); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $a = 0; - $b = 1; - $F = Formula("x^3*e^x-3*x^2*e^x+6*x*e^x-6*e^x"); - $r = $F->eval(x=>$b) - $F->eval(x=>$a); - - -

    - \ds \int_0^1 x^3e^{x}\, dx -

    -

    - -

    -     -     -     - - - - - ($a,$b) = num_sort(random_subset(2,1..4)); - if($envir{problemSeed}==1){$a=1;$b=2}; - $f = Formula("x e^(-2x)"); - $F = Formula("(-x/2-1/4) e^(-2x)"); - Context("Fraction"); - $fracb = Fraction(-$b/2-1/4); - $fraca = Fraction(-$a/2-1/4); - $b2 = 2*$b; - $a2 = 2*$a; - $r = Formula("$fracb e^(-$b2) - $fraca e^(-$a2)"); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random('\pi/2','\pi','3\pi/2','2\pi'); - if($envir{problemSeed}==1){$b='\pi'}; - $a=0; - $f = Formula("e^x sin(x)"); - $F = Formula("1/2*e^x*(sin(x)-cos(x))"); - $r = Formula("1/2*e^(pi/2) + 1/2"); - $r = Formula("1/2*e^pi + 1/2") if ($b eq '\pi'); - $r = Formula("-1/2*e^(3pi/2) + 1/2") if ($b eq '3\pi/2'); - $r = Formula("-1/2*e^(2pi) + 1/2") if ($b eq '2\pi'); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random('\pi/2','\pi','3\pi/2','2\pi'); - if($envir{problemSeed}==1){$b='\pi/2'}; - $a='-'.$b; - $f = Formula("e^(2x) cos(x)"); - $F = Formula("1/5*e^(2x)*(sin(x)+2*cos(x))"); - $r = Formula("1/5 (e^(pi) + e^(-pi))"); - $r = Formula("2/5 (-e^(2pi) + e^(-2pi))") if ($b eq '\pi'); - $r = Formula("1/5 (-e^(3pi) - e^(-3pi))") if ($b eq '3\pi/2'); - $r = Formula("2/5 (e^(4pi) - e^(-4pi))") if ($b eq '2\pi'); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     -     -     -     -
    -
    - Trigonometric Integrals - -

    - Functions involving trigonometric functions are useful as they are good at describing periodic behavior. - This section describes several techniques for finding antiderivatives of certain combinations of trigonometric functions. -

    -     - - - Integrals of the form <m>\int \sin^m(x) \cos^n(x) \, dx</m> -

    - In learning the technique of Substitution, - we saw the integral \int \sin(x) \cos(x) \, dx in . - The integration was not difficult, - and one could easily evaluate the indefinite integral by letting - u=\sin(x) or by letting u = \cos(x). - This integral is easy since the power of both sine and cosine is 1. -

    - -

    - We generalize this integral and consider integrals of the form - \int \sin^m(x) \cos^n(x) \, dx, - where m,n are nonnegative integers. - Our strategy for evaluating these integrals is to use the identity - \cos^2(x) +\sin^2(x) =1 to convert high powers of one trigonometric function into the other, - leaving a single sine or cosine term in the integrand. - Let's see an example of how this technique works. -

    - - - Integrating powers of sine and cosine - -

    - Evaluate \ds\int\sin^3(x) \cos(x) \, dx. -

    -     - -

    - We have used substitution on problems similar to this problem in - . - If we let u=\sin(x), - then du=\cos(x)\,dx, and - - \int \sin^3(x)\cos(x)\,dx = \int u^3 \,du = \frac{u^4}{4}+C = \frac14 \sin^4(x)+C - . - But what if, for some reason, we wanted to let u=\cos(x) instead? - Unfortunately, - we have \sin^3(x) as part of our integrand, not just \sin(x). - The solution to this problem is to replace some of our powers of sine - (two of them to be exact) - with expressions that involve cosine. - We will use the Pythagorean Identity \sin^2(x)=1-\cos^2(x). - - \int\sin^3(x) \cos(x) \, dx \amp =\int\sin(x)\cdot \sin^2(x) \cos(x) \, dx - \amp =\int\sin(x)\left(1-\cos^2(x)\right) \cos(x) \, dx - . -

    - -

    - Now we let u=\cos(x) so that -du=\sin(x)\,dx. - - \int\sin^3(x) \cos(x) \, dx \amp =\int\sin(x)\left(1-\cos^2(x)\right) \cos(x) \, dx - \amp =\int -\left(1-u^2\right)u\,du - \amp =\int -\left(u-u^3\right)\,du - \amp =-\frac{u^2}{2}+\frac{u^4}{4}+C - \amp =-\frac{\cos^2(x)}{2}+\frac{\cos^4(x)}{4}+C - . -

    - -

    - This looks like a very different answer, so you might wonder if we went wrong somewhere. - But in fact, the two answers are equivalent, in the sense that they differ by a constant! - (So the +C is different in each case, if you like.) - Notice that - - \frac14\sin^4(x) \amp = \frac14(1-\cos^2(x))^2 - \amp = \frac14-\frac12\cos^2(x)+\frac14\cos^4(x) - , - so the difference between the two answers is the constant \frac14 . -

    -     - -     - -

    - We summarize the general technique in the following Key Idea. -

    - - - Integrals Involving Powers of Sine and Cosine -

    - Consider \ds \int \sin^m(x) \cos^n(x) \, dx, - where m,n are nonnegative integers. - integrationof trig. powers -

    - -

    -

      -
    1. -

      - If m is odd, then m=2k+1 for some integer k. - Rewrite - - \sin^m(x) \amp = \sin^{2k+1}(x) - \amp = \sin^{2k}(x) \sin(x) - \amp = (\sin^2(x) )^k\sin(x) - \amp = (1-\cos^2(x) )^k\sin(x) - . - Then - - \int \sin^m(x) \cos^n(x) \, dx \amp = \int (1-\cos^2(x) )^k\sin(x) \cos^n(x) \, dx - \amp = -\int (1-u^2)^ku^n\,du - , - where u = \cos(x) and du = -\sin(x) \, dx. -

      -
      2. - -
    2. -

      - If n is odd, - then using substitutions similar to that outlined above - (replacing all of the even powers of cosine using a Pythagorean identity) - we have: - - \int \sin^m(x) \cos^n(x) \, dx = \int u^m(1-u^2)^k\,du - , - where u = \sin(x) and du = \cos(x) \, dx. -

      -
      3. - -
    3. -

      - If both m and n are even, - use the power-reducing identities: - - \cos^2(x) = \frac{1+\cos(2x)}{2} \text{ and } \sin^2(x) = \frac{1-\cos(2x)}2 - - to reduce the degree of the integrand. - Expand the result and apply the principles of this Key Idea again. -

      -
      4. -
    -

    -     - -

    - We practice applying in the next examples. -

    - - - Integrating powers of sine and cosine - -

    - Evaluate \ds\int\sin^5(x) \cos^8(x) \, dx. -

    -     - -

    - The power of the sine term is odd, - so we rewrite \sin^5(x) as - - \sin^5(x) \amp = \sin^4(x) \sin(x) - \amp = (\sin^2(x) )^2\sin(x) - \amp = (1-\cos^2(x) )^2\sin(x) - . -

    - -

    - Our integral is now \ds \int (1-\cos^2(x) )^2\cos^8(x) \sin(x) \, dx. - Let u = \cos(x), hence du = -\sin(x) \, dx. - Making the substitution and expanding the integrand gives - - \int (1-\cos^2)^2\cos^8(x) \sin(x) \, dx \amp = -\int (1-u^2)^2u^8\,du - \amp = -\int \big(1-2u^2+u^4\big)u^8\,du - \amp = -\int \big(u^8-2u^{10}+u^{12}\big)\,du - . -

    - -

    - This final integral is not difficult to evaluate, giving - - -\int \big(u^8-2u^{10}+u^{12}\big)\, du \amp = -\frac19u^9 + \frac2{11}u^{11} - \frac1{13}u^{13} + C - \amp =-\frac19\cos^9(x) + \frac2{11}\cos^{11}(x) - \frac1{13}\cos^{13}(x) + C - . -

    -     - -     - - - Integrating powers of sine and cosine - -

    - Evaluate \ds \int\sin^5(x) \cos^9(x) \, dx. -

    -     - -

    - The powers of both the sine and cosine terms are odd, - therefore we can apply the techniques of - to either power. - We choose to work with the power of the cosine term since the previous example used the sine term's power. -

    - -

    - We rewrite \cos^9(x) as - - \cos^9(x) \amp = \cos^8(x) \cos(x) - \amp = \left(\cos^2(x)\right)^4\cos(x) - \amp = \left(1-\sin^2(x)\right)^4\cos(x) - . -

    - -

    - We rewrite the integral as - - \int\sin^5(x) \cos^9(x) \, dx = \int\sin^5(x) \left(1-\sin^2(x)\right)^4\cos(x) \, dx - . -

    - -

    - Now substitute and integrate, - using u = \sin(x) and du = \cos(x) \, dx. - Expand the binomial using algebra. -

    - -

    - - \amp \int u^5(1-u^2)^4\,du - \amp = \int u^5(1-4u^2+6u^4-4u^6+u^8)\,du - \amp = \int\big(u^5-4u^7+6u^9-4u^{11}+u^{13}\big)\,du - \amp = \frac16u^6-\frac12u^8+\frac35u^{10}-\frac13u^{12}+\frac{1}{14}u^{14}+C - \amp = \frac16\sin^6(x) -\frac12\sin^8(x) +\frac35\sin^{10}(x)-\frac13\sin^{12}(x) +\frac{1}{14}\sin^{14}(x) +C - . -

    -     -     - -

    - Technology Note: The work we are doing here can be a bit tedious, - but the skills developed - (problem solving, algebraic manipulation, etc.) - are important. - Nowadays problems of this sort are often solved using a computer algebra system. - The powerful program Mathematica integrates - \int \sin^5(x) \cos^9(x) \, dx as - - f(x) =\amp -\frac{45 \cos(2 x)}{16384}-\frac{5 \cos(4 x)}{8192}+\frac{19 \cos(6x)}{49152} - \amp+\frac{\cos(8 x)}{4096}-\frac{\cos(10 x)}{81920}-\frac{\cos(12x)}{24576}-\frac{\cos(14 x)}{114688} - , - which clearly has a different form than our answer in - , which is - - g(x)=\frac16\sin^6(x) -\frac12\sin^8(x) +\frac35\sin^{10}(x) -\frac13\sin^{12}(x) +\frac{1}{14}\sin^{14}(x) - . -

    - -

    - - shows a graph of f and g; - they are clearly not equal, but they differ - only by a constant. - That is g(x) = f(x) + C for some constant C. - So we have two different antiderivatives of the same function, - meaning both answers are correct. -

    - -
    - A plot of f(x) and g(x) from and the Technology Note - - - - Graph of f(x) and g(x) that are different only by a constant. - - -

    - The y axis is drawn from -0.002 and 0.004 and the x axis - is drawn from 0 to 3. The graph has two functions drawn. Both of them - are plateau shaped. -

    -

    - The first function g(x) is drawn on the x axis, the function starts at the - origin and rises up at x=0.25 and keeps rising steeply until x=1.75 then - it runs parallel to the x axis and starts declining at x=2.25, it declines - sharply till x=2.75 then it merged with the x axis. -

    -

    - The second function f(x) is the same as the first but the line on which it is placed - is x=0.0025 instead of x=0. -

    -     - - - \begin{tikzpicture} - \begin{axis}[% - axis on top,% - ytick={-.002,.002,.004}, - yticklabels={$-0.002$,$0.002$,$0.004$}, - ymin=-.003,ymax=0.005,% - xmin=-.1,xmax=3.15,% - scaled ticks=false - ] - - \addplot [secondcurvestyle] coordinates {(0,-0.0027879) (0.15708,-0.0027856) (0.31416,-0.0026798) (0.47124,-0.0020322) (0.62832,-0.00060997) (0.7854,0.00089518)(0.94248,0.0017233) (1.0996,0.0019485) (1.2566,0.0019734) (1.4137,0.0019741) (1.5708,0.0019741) (1.7279,0.0019741) (1.885,0.0019734) (2.042,0.0019485) (2.1991,0.0017233) (2.3562,0.00089518) (2.5133,-0.00060997) (2.6704,-0.0020322) (2.8274,-0.0026798) (2.9845,-0.0027856) (3.1416,-0.0027879)}; - - \draw (axis cs:2.6,0.004) node { $g(x)$}; - - \addplot [firstcurvestyle] coordinates {(0,0) (0.15708,0) (0.31416,0.00010807) (0.47124,0.0007557) (0.62832,0.0021779) (0.7854,0.003683) (0.94248,0.0045111) - (1.0996,0.0047364) (1.2566,0.0047612) (1.4137,0.0047619) - (1.5708,0.0047619) (1.7279,0.0047619) (1.885,0.0047612) (2.042,0.0047364) (2.1991,0.0045111) (2.3562,0.003683) (2.5133,0.0021779) (2.6704,0.0007557) (2.8274,0.00010807) (2.9845,0) (3.1416,0)}; - - \draw (axis cs:2.4,-0.002) node { $f(x)$}; - \end{axis} - - \end{tikzpicture} - - - - -
    - - - Integrating powers of sine and cosine - -

    - Evaluate \ds\int\cos^4(x) \sin^2(x) \, dx. -

    -     - -

    - The powers of sine and cosine are both even, - so we employ the power-reducing formulas and algebra as follows. - - \amp \int \cos^4(x) \sin^2(x) \, dx = \int\left(\frac{1+\cos(2x)}{2}\right)^2\left(\frac{1-\cos(2x)}2\right)\, dx - \amp = \int\frac{1+2\cos(2x)+\cos^2(2x)}4\cdot\frac{1-\cos(2x)}2\, dx - \amp = \int \frac18\big(1+\cos(2x)+\cos^2(2x)-\cos^3(2x)\big)\, dx - \amp = \frac18 \left(\underbrace{\int 1 \, dx}_{a}+\underbrace{\int \cos(2x)\, dx}_{b}-\underbrace{\int \cos^2(2x)\, dx}_{c}-\underbrace{\int \cos^3(2x)\, dx}_{d}\right) - -

    - -

    - The first integral labeled a is easy to integrate. - The \cos(2x) term is also easy to integrate, - especially with . - The \cos^2(2x) term is another trigonometric integral with an even power, - requiring the power-reducing formula again. - The \cos^3(2x) term is a cosine function with an odd power, - requiring a substitution as done before. - We integrate each in turn below. - - \underbrace{\int\cos(2x)\, dx}_{b} \amp = \frac12\sin(2x)+C - \underbrace{\int\cos^2(2x)\, dx}_{c} \amp = \int \frac{1+\cos(4x)}2\, dx - \amp = \frac12\big(x+\frac14\sin(4x)\big)+C - . -

    - -

    - Finally, we rewrite \cos^3(2x) as - - \cos^3(2x) \amp = \cos^2(2x)\cos(2x) - \amp = \big(1-\sin^2(2x)\big)\cos(2x) - . -

    - -

    - Letting u=\sin(2x), we have du = 2\cos(2x)\, dx, hence - - \underbrace{\int \cos^3(2x)\, dx}_{d} \amp = \int\big(1-\sin^2(2x)\big)\cos(2x)\, dx - \amp = \int \frac12(1-u^2)\,du - \amp = \frac12\Big(u-\frac13u^3\Big)+C - \amp = \frac12\Big(\sin(2x)-\frac13\sin^3(2x)\Big)+C - . -

    - -

    - Putting all the pieces together, we have - - \amp \int \cos^4(x) \sin^2(x) \, dx - \amp =\int \frac18\big(1+\cos(2x)-\cos^2(2x)-\cos^3(2x)\big)\, dx - \amp = \frac18\Big[x+\frac12\sin(2x)-\frac12\big(x+\frac14\sin(4x)\big)-\frac12\Big(\sin(2x)-\frac13\sin^3(2x)\Big)\Big]+C - \amp =\frac18\Big[\frac12x-\frac18\sin(4x)+\frac16\sin^3(2x)\Big]+C - . -

    -     - -     - -

    - The process above was a bit long and tedious, - but being able to work a problem such as this from start to finish is important. -

    - - - - -     - - - Integrals of the form <m>\int\sin(mx)\sin(nx)\, dx</m>, <m>\int \cos(mx)\cos(nx)\, dx</m>, and <m>\int \sin(mx)\cos(nx)\, dx</m> -

    - Functions that contain products of sines and cosines of differing periods are important in many applications including the analysis of sound waves. - Integrals of the form - - \int\sin(mx)\sin(nx)\, dx, \int \cos(mx)\cos(nx)\, dx \text{ and } \int \sin(mx)\cos(nx)\, dx - - are best approached by first applying the Product to Sum Formulas found in the back cover of this text, namely - - \sin(mx)\sin(nx) \amp = \frac12\Big[\cos\big((m-n)x\big)-\cos\big((m+n)x\big)\Big] - \cos(mx)\cos(nx) \amp = \frac12\Big[\cos\big((m-n)x\big)+\cos\big((m+n)x\big)\Big] - \sin(mx)\cos(nx) \amp = \frac12\Big[\sin\big((m-n)x\big)+\sin\big((m+n)x\big)\Big] - . -

    - - - Integrating products of <m>\sin(mx)</m> and <m>\cos(nx)</m> - -

    - Evaluate \ds\int\sin(5x)\cos(2x)\, dx. -

    -     - -

    - The application of the formula and subsequent integration are straightforward: - - \int\sin(5x)\cos(2x)\, dx \amp = \int \frac12\Big[\sin((5-2)x)+\sin((5+2)x)\Big]\, dx - \amp= \int \frac12\Big[\sin(3x)+\sin(7x)\Big]\, dx - \amp = -\frac16\cos(3x) - \frac1{14}\cos(7x) + C - -

    -     - -     -     - - - Integrals of the form <m>\int\tan^m(x) \sec^n(x) \, dx</m> -

    - When evaluating integrals of the form \int \sin^m(x) \cos^n(x) \, dx, - the Pythagorean Theorem allowed us to convert even powers of sine into even powers of cosine, - and vise-versa. - If, for instance, the power of sine was odd, - we pulled out one \sin(x) and converted the remaining even power of \sin(x) into a function using powers of \cos(x), - leading to an easy substitution. -

    - -

    - The same basic strategy applies to integrals of the form \int \tan^m(x) \sec^n(x) \, dx, - albeit a bit more nuanced. - The following three facts will prove useful: -

    - -

    -

      -
    • \frac{d}{dx}(\tan(x) ) = \sec^2(x),
      • - -
    • \frac{d}{dx}(\sec(x) ) = \sec(x) \tan(x),
      • - -
    • -

      - 1+\tan^2(x) = \sec^2(x) - (the Pythagorean Theorem). -

      -
      • -
    -

    - -

    - If the integrand can be manipulated to separate a - \sec^2(x) term with the remaining secant power even, - or if a \sec(x) \tan(x) term can be separated with the remaining \tan(x) power even, - the Pythagorean Theorem can be employed, - leading to a simple substitution. - This strategy is outlined in the following Key Idea. -

    - - - Integrals Involving Powers of Tangent and Secant -

    - Consider \ds\int\tan^m(x) \sec^n(x) \, dx, - where m,n are nonnegative integers. - integrationof trig. powers - -

    - -

    -

      -
    1. - -

      - If n is even, then n=2k for some integer k. - Rewrite \sec^n(x) as - - \sec^n(x) \amp = \sec^{2k}(x) - \amp = \sec^{2k-2}(x) \sec^2(x) - \amp = (1+\tan^2(x) )^{k-1}\sec^2(x) - . - Then - - \int\tan^m(x) \sec^n(x) \, dx \amp =\int\tan^m(x) (1+\tan^2(x) )^{k-1}\sec^2(x) \, dx - \amp =\int u^m(1+u^2)^{k-1}\,du - , - where u = \tan(x) and du = \sec^2(x) \, dx. -

      -
      2. - -
    2. - -

      - If m is odd, then m=2k+1 for some integer k. - Rewrite \tan^m(x) \sec^n(x) as - - \tan^m(x) \sec^n(x) \amp = \tan^{2k+1}(x) \sec^n(x) - \amp = \tan^{2k}(x) \sec^{n-1}(x) \sec(x) \tan(x) - \amp = (\sec^2(x) -1)^k\sec^{n-1}(x) \sec(x) \tan(x) - . - Then - - \int\tan^m(x) \sec^n(x) \, dx \amp =\int(\sec^2(x) -1)^k\sec^{n-1}(x) \sec(x) \tan(x) \, dx - \amp = \int(u^2-1)^ku^{n-1}\,du - , - where u = \sec(x) and du = \sec(x) \tan(x) \, dx. -

      -
      3. - -
    3. - -

      - If n is odd and m is even, - then m=2k for some integer k. - Convert \tan^m(x) to (\sec^2(x) -1)^k. - Expand the new integrand and use Integration By Parts, - with dv = \sec^2(x) \, dx. -

      -
      4. - -
    4. - -

      - If m is even and n=0, - rewrite \tan^m(x) as - - \tan^m(x) \amp = \tan^{m-2}(x) \tan^2(x) - \amp = \tan^{m-2}(x) (\sec^2(x) -1) - \amp = \tan^{m-2}\sec^2(x) -\tan^{m-2}(x) - . - So - - \int\tan^m(x) \, dx = \underbrace{\int\tan^{m-2}\sec^2(x) \, dx}_{\text{ apply rule 1 } } - \underbrace{\int\tan^{m-2}(x) \, dx}_{\text{ apply rule 4 again } } - . -

      -
      5. -
    -

    -     - -

    - The techniques described in Item - and Item - of - are relatively straightforward, - but the techniques in Item - and Item can be rather tedious. - A few examples will help with these methods. -

    - - - Integrating powers of tangent and secant - -

    - Evaluate \ds\int \tan^2(x) \sec^6(x) \, dx. -

    -     - -

    - Since the power of secant is even, - we use Rulefrom - and pull out a \sec^2(x) in the integrand. - We convert the remaining powers of secant into powers of tangent. - - \int \tan^2(x) \sec^6(x) \, dx \amp = \int\tan^2(x) \sec^4(x) \sec^2(x) \, dx - \amp = \int \tan^2(x) \big(1+\tan^2(x) \big)^2\sec^2(x) \, dx - Now substitute, with u=\tan(x), with du = \sec^2(x) \, dx. - \amp =\int u^2\big(1+u^2\big)^2\,du - We leave the integration and subsequent substitution to the reader. The final answer is - \amp =\frac13\tan^3(x) +\frac25\tan^5(x) +\frac17\tan^7(x) +C - . -

    -     - -     - - - - - - - Integrating powers of tangent and secant - -

    - Evaluate \ds\int \sec^3(x) \, dx. -

    -     - -

    - We apply Rule - from - as the power of secant is odd and the power of tangent is even (0 is an even number). - We use Integration by Parts; - the rule suggests letting dv = \sec^2(x) \, dx, - meaning that u = \sec(x). -

    - - -
    - Setting up Integration by Parts - -

    - - u\amp = \sec(x) \amp v\amp =\mathord{?} - du\amp = \mathord{?} \amp dv\amp =\sec^2(x) \, dx - -

    - -

    - \implies -

    - -

    - - u\amp = \sec(x) \amp v\amp =\tan(x) - du\amp = \sec(x) \tan(x) \, dx \amp dv\amp =\sec^2(x) \, dx - -

    -     -
    - -

    - Employing Integration by Parts, we have - - \int \sec^3(x) \, dx \amp = \int \underbrace{\sec(x) }_u\cdot\underbrace{\sec^2(x) \, dx}_{dv} - \amp = \sec(x) \tan(x) - \int \sec(x) \tan^2(x) \, dx. - This new integral also requires applying Rule - of : - \int \sec^3(x)\, dx\amp = \sec(x) \tan(x) - \int \sec(x) \big(\sec^2(x) -1\big)\, dx - \amp = \sec(x) \tan(x) - \int \sec^3(x) \, dx + \int \sec(x) \, dx - \amp = \sec(x) \tan(x) -\int \sec^3(x) \, dx + \ln\abs{\sec(x) +\tan(x) } - In previous applications of Integration by Parts, we have seen where the original integral has reappeared in our work. We resolve this by adding \int \sec^3(x) \, dx to both sides, giving: - 2\int \sec^3(x) \, dx \amp = \sec(x) \tan(x) + \ln\abs{\sec(x) +\tan(x) } - \int \sec^3(x) \, dx \amp = \frac12\Big(\sec(x) \tan(x) + \ln\abs{\sec(x) +\tan(x) }\Big)+C - -

    -     - -     - - - - - -

    - We give one more example. -

    - - - - Integrating powers of tangent and secant - -

    - Evaluate \ds\int\tan^6(x) \, dx. -

    -     - -

    - We employ Rule of - . - - \int \tan^6(x) \, dx \amp = \int \tan^4(x) \tan^2(x) \, dx - \amp = \int\tan^4(x) \big(\sec^2(x) -1\big)\, dx - \amp = \int\tan^4(x) \sec^2(x) \, dx - \int\tan^4(x) \, dx - Integrate the first integral with substitution, u=\tan(x); integrate the second by employing rule - Rule again. - \amp = \frac15\tan^5(x) -\int\tan^2(x) \tan^2(x) \, dx - \amp = \frac15\tan^5(x) -\int\tan^2(x) \big(\sec^2(x) -1\big)\, dx - \amp = \frac15\tan^5(x) -\underbrace{\int\tan^2(x) \sec^2(x) \, dx}_{a} + \underbrace{\int\tan^2(x) \, dx}_b - Again, use substitution (u=\tan(x)) for the first integral (a) and - Rule for the second (b). - \amp = \frac15\tan^5(x) -\frac13\tan^3(x) +\int\big(\sec^2(x) -1\big)\, dx - \int \tan^6(x)\, dx\amp = \frac15\tan^5(x) -\frac13\tan^3(x) +\tan(x) - x+C - . -

    -     - -     - - - -

    - These latter examples were admittedly long, - with repeated applications of the same rule. - Try to not be overwhelmed by the length of the problem, - but rather admire how robust this solution method is. - A trigonometric function of a high power can be systematically reduced to trigonometric functions of lower powers until all antiderivatives can be computed. -

    - -

    - - introduces an integration technique known as Trigonometric Substitution, - a clever combination of Substitution and the Pythagorean Theorem. -

    -     - - - - Terms and Concepts - - - - -

    - - \ds \int \sin^2(x) \cos^2(x) \, dx cannot be evaluated using the techniques described in this section - since both powers of \sin(x) and \cos(x) are even. -

    -     - -     - - - - -

    - - \ds \int \sin^3(x) \cos^3(x) \, dx cannot be evaluated using the techniques described in this section - since both powers of \sin(x) and \cos(x) are odd. -

    -     - -     - - - - -

    - - This section addresses how to evaluate indefinite integrals such as - \ds \int \sin^5(x) \tan^3(x) \, dx. -

    -     - -     - - - - -

    - - Sometimes computer programs evaluate integrals involving trigonometric functions differently - than one would using the techniques of this section. - When this is the case, - the techniques of this section have failed and one should only trust the answer given by the computer. -

    -     - -     -     - - - Problems - - -

    - Evaluate the indefinite integral. -

    -     - - - - - - $F = FormulaUpToConstant("-1/5*cos(x)^5"); - - -

    - \ds \int \sin(x) \cos^4(x) \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("1/4*sin(x)^4"); - - -

    - \ds \int \sin^3(x) \cos(x) \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=2;}; - Context("Fraction"); - $f = Formula("sin^3(x) cos^$m(x)"); - $F = FormulaUpToConstant("1/($m+3)*cos(x)^($m+3)-1/($m+1)*cos(x)^($m+1)"); - - -

    - \ds \int \sin^3(x)\cos^{}(x) \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - $n = 3; - if (list_random(-1,1)) { - ($m,$n) = ($n,$m); - } - if($envir{problemSeed}==1){$m=3;$n=3;}; - Context("Fraction"); - $f = Formula("sin^$m(x) cos^$n(x)"); - $F = FormulaUpToConstant("1/($m+1)*sin(x)^($m+1)-1/($m+3)*sin(x)^($m+3)"); - if ($n != 3) { - $F = FormulaUpToConstant("1/($n+3)*cos(x)^($n+3)-1/($n+1)*cos(x)^($n+1)"); - } - - -

    - \ds \int \sin^{}(x)\cos^{}(x) \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=6;}; - Context("Fraction"); - $f = Formula("sin^$m(x) cos^5(x)"); - $F = FormulaUpToConstant("1/($m+5)*sin(x)^($m+5)-2/($m+3)*sin(x)^($m+3)+1/($m+1)*sin(x)^($m+1)"); - - -

    - \ds \int \sin^{}(x)\cos^{5}(x) \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("-1/9*sin(x)^9+3/7*sin(x)^7-3/5*sin(x)^5+1/3*sin(x)^3"); - - -

    - \ds \int \sin^2(x) \cos^7(x) \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("x/8-1/32*sin(4x)"); - - -

    - \ds \int \sin^2(x) \cos^2(x) \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("1/2*(-1/8*cos(8x)-1/2*cos(2x))"); - - -

    - \ds \int \sin(5x)\cos(3x)\, dx -

    -

    - -

    -     -     -     - - - - - ($m,$n) = random_subset(2,2..9); - if (list_random(-1,1)) {$m = 1;} else {$n = 1;}; - if($envir{problemSeed}==1){$m=1;$n=2}; - Context("Fraction"); - $f = Formula("sin($m x) cos($n x)")->reduce; - $a = abs($m-$n); - $b = abs($m+$n); - $A = Fraction(-1,2*$a); - $B = Fraction(-1,2*$b); - $F = FormulaUpToConstant("$A cos($a x) + $B cos($b x)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($m,$n) = random_subset(2,2..9); - if($envir{problemSeed}==1){$m=3;$n=7}; - Context("Fraction"); - $f = Formula("sin($m x) sin($n x)")->reduce; - $a = abs($m-$n); - $b = abs($m+$n); - $A = Fraction(1,2*$a) * (abs($a)/$a); - $B = Fraction(1,2*$b) * (abs($b)/$b); - $F = FormulaUpToConstant("$A sin($a x) - $B sin($b x)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($m,$n) = random_subset(2,2..9); - if (list_random(-1,1)) {$m = 1;} else {$n = 1;}; - if($envir{problemSeed}==1){$m=1;$n=2}; - Context("Fraction"); - $f = Formula("sin($m pi x) sin($n pi x)")->reduce; - $a = abs($m-$n); - $b = abs($m+$n); - $A = Fraction(1,2*$a) * (abs($a)/$a); - $B = Fraction(1,2*$b) * (abs($b)/$b); - ($Anum,$Aden) = $A->value; - ($Bnum,$Bden) = $B->value; - $A = Formula("$Anum/($Aden pi)"); - $B = Formula("$Bnum/($Bden pi)"); - $F = FormulaUpToConstant("$A sin($a pi x) - $B sin($b pi x)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("1/2*(sin(x)+1/3*sin(3x))"); - - -

    - \ds \int \cos(x)\cos(2x)\, dx -

    -

    - -

    -     -     -     - - - - - ($m,$n) = random_subset(2,'pi/6','pi/4','pi/3','pi/2','pi'); - if($envir{problemSeed}==1){$m='pi/2';$n='pi'}; - Context("Fraction"); - $f = Formula("cos($m x) cos($n x)")->reduce; - $a = Fraction(abs((Real($m)-Real($n))/pi)); - $b = Fraction(abs((Real($m)+Real($n))/pi)); - $A = Fraction(1/(2*$a)); - $B = Fraction(1/(2*$b)); - ($anum,$aden) = $a->value; - ($bnum,$bden) = $b->value; - $a = Formula("$anum pi/$aden"); - $b = Formula("$bnum pi/$bden"); - ($Anum,$Aden) = $A->value; - ($Bnum,$Bden) = $B->value; - $A = Formula("$Anum/($Aden pi)"); - $B = Formula("$Bnum/($Bden pi)"); - $F = FormulaUpToConstant("$A cos($a pi x) + $B cos($b pi x)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("tan(x)^5/5"); - - -

    - \ds \int \tan^4(x) \sec^2(x) \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("tan(x)^5/5+tan(x)^3/3"); - - -

    - \ds \int \tan^2(x) \sec^4(x) \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=3;}; - Context("Fraction"); - $f = Formula("tan^$m(x) sec^4(x)"); - $F = FormulaUpToConstant("1/($m+3)*tan(x)^($m+3)+1/($m+1)*tan(x)^($m+1)"); - - -

    - \ds \int \tan^{}(x)\sec^{4}(x) \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=3;}; - Context("Fraction"); - $f = Formula("tan^$m(x) sec^2(x)"); - $F = FormulaUpToConstant("1/($m+1)*tan(x)^($m+1)"); - - -

    - \ds \int \tan^{}(x)\sec^{2}(x) \, dx -

    -

    - -

    -     -     -     - - - - - $n = random(2,9,1); - if($envir{problemSeed}==1){$n=3;}; - Context("Fraction"); - $f = Formula("tan^3(x) sec^$n(x)"); - $F = FormulaUpToConstant("1/($n+2)*sec(x)^($n+2)-1/($n)*sec(x)^($n)"); - - -

    - \ds \int \tan^{3}(x)\sec^{}(x) \, dx -

    -

    - -

    -     -     -     - - - - - $n = random(2,9,1); - if($envir{problemSeed}==1){$n=5;}; - Context("Fraction"); - Context()->flags->set(limits=>[-1,1]); - $f = Formula("tan^5(x) sec^$n(x)"); - $F = FormulaUpToConstant("1/($n+4)*sec(x)^($n+4)-2/($n+2)*sec(x)^($n+2) + 1/$n sec(x)^($n)"); - - -

    - \ds \int \tan^{5}(x)\sec^{}(x) \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("tan(x)^3/3-tan(x)+x"); - - -

    - \ds \int \tan^4(x) \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("1/4*tan(x)*sec(x)^3+3/8*(sec(x)*tan(x)+ln(abs(sec(x) + tan(x))))"); - - -

    - \ds \int \sec^5(x) \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("1/2*(sec(x)*tan(x)-ln(abs(sec(x)+tan(x))))"); - - -

    - \ds \int \tan^2(x) \sec(x) \, dx -

    -

    - -

    -     -     -     - - - - - - $F = FormulaUpToConstant("1/4*tan(x)*sec(x)^3-1/8*(sec(x)*tan(x)+ln(abs(sec(x)+tan(x))))"); - - -

    - \ds \int \tan^2(x) \sec^3(x) \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Evaluate the definite integral. - Note: the corresponding indefinite integrals appear in - Exercises. -

    -     - - - - - $b = list_random(pi/2,pi,3*pi/2,2*pi); - if($envir{problemSeed}==1){$b=pi}; - $a=0; - Context("Fraction"); - $F = Formula("-1/5*cos(x)^5"); - $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); - $a = Fraction($a/pi); - $b = Fraction($b/pi); - ($anum,$aden) = $a->value; - ($bnum,$bden) = $b->value; - $a = Formula("($anum pi)/$aden")->reduce; - $b = Formula("($bnum pi)/$bden")->reduce; - - -

    - \ds \int_{}^{} \sin(x) \cos^4(x) \, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random(pi/2,pi,3*pi/2,2*pi); - if($envir{problemSeed}==1){$b=pi}; - $a = -$b; - Context("Fraction"); - $F = Formula("1/4*sin(x)^4"); - $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); - $a = Fraction($a/pi); - $b = Fraction($b/pi); - ($anum,$aden) = $a->value; - ($bnum,$bden) = $b->value; - $a = Formula("($anum pi)/$aden")->reduce; - $b = Formula("($bnum pi)/$bden")->reduce; - - -

    - \ds \int_{}^{} \sin^3(x) \cos(x) \, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random(pi/2,pi,3*pi/2,2*pi); - if($envir{problemSeed}==1){$b=pi/2}; - $a = -$b; - Context("Fraction"); - $F = Formula("-1/9*sin(x)^9+3/7*sin(x)^7-3/5*sin(x)^5+1/3*sin(x)^3"); - $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); - $a = Fraction($a/pi); - $b = Fraction($b/pi); - ($anum,$aden) = $a->value; - ($bnum,$bden) = $b->value; - $a = Formula("($anum pi)/$aden")->reduce; - $b = Formula("($bnum pi)/$bden")->reduce; - - -

    - \ds \int_{}^{} \sin^2(x) \cos^7(x) \, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random(pi/4,pi/2,3*pi/4,pi,3*pi/2,2*pi); - if($envir{problemSeed}==1){$b=pi/2}; - $a = 0; - Context("Fraction"); - $F = Formula("1/2*(-1/8*cos(8x)-1/2*cos(2x))"); - $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); - $a = Fraction($a/pi); - $b = Fraction($b/pi); - ($anum,$aden) = $a->value; - ($bnum,$bden) = $b->value; - $a = Formula("($anum pi)/$aden")->reduce; - $b = Formula("($bnum pi)/$bden")->reduce; - - -

    - \ds \int_{}^{} \sin(5 x)\cos(3x)\, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random(pi/2,pi,3*pi/2,2*pi); - if($envir{problemSeed}==1){$b=pi/2}; - $a = -$b; - Context("Fraction"); - $F = Formula("1/2*(sin(x)+1/3*sin(3x))"); - $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); - $a = Fraction($a/pi); - $b = Fraction($b/pi); - ($anum,$aden) = $a->value; - ($bnum,$bden) = $b->value; - $a = Formula("($anum pi)/$aden")->reduce; - $b = Formula("($bnum pi)/$bden")->reduce; - - -

    - \ds \int_{}^{} \cos(x)\cos(2x)\, dx -

    -

    - -

    -     -     -     - - - - - $a = list_random(0,-pi/4); - if($envir{problemSeed}==1){$a=0}; - $b = pi/4; - Context("Fraction"); - $F = Formula("tan(x)^5/5"); - $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); - $a = Fraction($a/pi); - $b = Fraction($b/pi); - ($anum,$aden) = $a->value; - ($bnum,$bden) = $b->value; - $a = Formula("($anum pi)/$aden")->reduce; - $b = Formula("($bnum pi)/$bden")->reduce; - - -

    - \ds \int_{}^{} \tan^4(x) \sec^2(x) \, dx -

    -

    - -

    -     -     -     - - - - - $a = list_random(0,-pi/4); - if($envir{problemSeed}==1){$a=-pi/4}; - $b = pi/4; - Context("Fraction"); - $F = Formula("tan(x)^5/5+tan(x)^3/3"); - $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); - $a = Fraction($a/pi); - $b = Fraction($b/pi); - ($anum,$aden) = $a->value; - ($bnum,$bden) = $b->value; - $a = Formula("($anum pi)/$aden")->reduce; - $b = Formula("($bnum pi)/$bden")->reduce; - - -

    - \ds \int_{-\pi/4}^{\pi/4} \tan^2(x) \sec^4(x) \, dx -

    -

    - -

    -     -     -     -     -     -     -
    -
    - Trigonometric Substitution -

    - In - we defined the definite integral as the - signed area under the curve. - In that section we had not yet learned the Fundamental Theorem of Calculus, - so we only evaluated special definite integrals which described nice, - geometric shapes. - For instance, we were able to evaluate - - \int_{-3}^3\sqrt{9-x^2}\, dx = \frac{9\pi}{2} - - as we recognized that f(x) = \sqrt{9-x^2} described the upper half of a circle with radius 3. -

    - -

    - We have since learned a number of integration techniques, - including Substitution and Integration by Parts, - yet we are still unable to evaluate the above integral without resorting to a geometric interpretation. - This section introduces Trigonometric Substitution, - a method of integration that fills this gap in our integration skill. - This technique works on the same principle as Substitution as found in , - though it can feel backward. - In , we set u=f(x), - for some function f, and replaced f(x) with u. - In this section, we will set x=f(\theta), - where f is a trigonometric function, - then replace x with f(\theta). -

    - - - -

    - We start by demonstrating this method in evaluating the integral in Equation. - After the example, - we will generalize the method and give more examples. -

    - - - Using Trigonometric Substitution - -

    - Evaluate \ds \int_{-3}^3\sqrt{9-x^2}\, dx. -

    -     - -

    - We begin by noting that 9\left(\sin^2(\theta) + \cos^2(\theta)\right) = 9, - and hence 9\cos^2(\theta) = 9-9\sin^2(\theta). - If we let x=3\sin(\theta), - then 9-x^2 = 9-9\sin^2(\theta) = 9\cos^2(\theta). -

    - -

    - Setting x=3\sin(\theta) gives dx = 3\cos(\theta) \, d\theta. - We are almost ready to substitute. - We also wish to change our bounds of integration. - The bound x=-3 corresponds to - \theta = -\pi/2 (for when \theta = -\pi/2, - x=3\sin(\theta) = -3). - Likewise, the bound of x=3 is replaced by the bound \theta = \pi/2. - Thus - - \int_{-3}^3\sqrt{9-x^2}\, dx \amp = \int_{-\pi/2}^{\pi/2} \sqrt{9-9\sin^2(\theta) }\,(3\cos(\theta) )\, d\theta - \amp = \int_{-\pi/2}^{\pi/2} 3\sqrt{9\cos^2(\theta) } \cos(\theta) \, d\theta - \amp =\int_{-\pi/2}^{\pi/2} 3\abs{3\cos(\theta) } \cos(\theta) \, d\theta - . - On [-\pi/2,\pi/2], \cos(\theta) is always positive, so we can drop the absolute value bars, then employ a power-reducing formula: - - \int_{-3}^3\sqrt{9-x^2}\, dx \amp = \int_{-\pi/2}^{\pi/2} 9\cos^2(\theta) \, d\theta - \amp = \int_{-\pi/2}^{\pi/2} \frac{9}{2}\big(1+\cos(2\theta)\big)\, d\theta - \amp = \left.\frac92 \big(\theta +\frac12\sin(2\theta)\big)\right|_{-\pi/2}^{\pi/2} - \amp = \frac92\pi - . -

    - -

    - This matches our answer from before. -

    -     - -     - -

    - We now describe in detail Trigonometric Substitution. - This method excels when dealing with integrands that contain \sqrt{a^2-x^2}, - \sqrt{x^2-a^2} and \sqrt{x^2+a^2}. - The following Key Idea outlines the procedure for each case, - followed by more examples. - Each right triangle acts as a reference to help us understand the relationships between x and \theta. -

    - - - - Trigonometric Substitution -

    -

      -
    1. - Integrands containing <m>\sqrt{a^2-x^2}</m> - - -

      - integrationtrig. subst. - - Let x=a\sin(\theta), dx = a\cos(\theta) \, d\theta. - Thus \theta = \sin^{-1}(x/a), - for -\pi/2\leq \theta\leq \pi/2. - On this interval, \cos(\theta) \geq 0, - so \sqrt{a^2-x^2} = a\cos(\theta). -

      - -
      - - - - - Diagram showing trigonometric substitution with integrands containing square root of a^2-x^2. - - -

      - The diagram is of a right angled triangle. The perpendicular is marked x - the base as \sqrt{a^2 -x^2}. The hypotenuse is marked a. The angle - opposite to the perpendicular is marked as \theta. -

      -       - - - \begin{tikzpicture}[scale=1.5] - - \draw [very thick] (0,0) -- node [below,pos=.5] { $\sqrt{a^2-x^2}$} (3,0) -- node [right,pos=.5] { $x$} (3,2) -- node [pos=.5,above] { $a$} (0,0); - \draw [thick] (2.7,0) -- (2.7,.3) -- (3,.3); - \draw (.75,.25) node {$\theta$}; - - \end{tikzpicture} - - - - -
      -       -
      2. - -
    2. - Integrands containing <m>\sqrt{x^2+a^2}</m> - - -

      - Let x=a\tan(\theta), dx = a\sec^2(\theta) \, d\theta. - Thus \theta = \tan^{-1}(x/a), - for -\pi/2 \lt \theta \lt \pi/2. - On this interval, \sec(\theta) \gt 0, - so \sqrt{x^2+a^2} = a\sec(\theta). -

      - -
      - - - - - Diagram showing trigonometric substitution with integrands containing square root of x^2+a^2. - - -

      - The diagram is of a right angled triangle. The perpendicular is marked x - the base as a. The hypotenuse is marked \sqrt{x^2+a^2}. The angle - opposite to the perpendicular is marked as \theta. -

      -       - - - \begin{tikzpicture}[scale=1.5] - - \draw [very thick] (0,0) -- node [below,pos=.5] { $a$} (3,0) -- node [right,pos=.5] { $x$} (3,2) -- node [pos=.5,above,sloped] { $\sqrt{x^2+a^2}$} (0,0); - \draw [thick] (2.7,0) -- (2.7,.3) -- (3,.3); - \draw (.75,.25) node {$\theta$}; - - \end{tikzpicture} - - - - -
      -       -
      3. - -
    3. - Integrands containing <m>\sqrt{x^2-a^2}</m> - - -

      - Let x=a\sec(\theta), - dx = a\sec(\theta) \tan(\theta) \, d\theta. - Thus \theta = \sec^{-1}(x/a). - If x/a\geq 1, then 0\leq\theta\lt \pi/2; - if x/a \leq -1, then \pi/2\lt \theta\leq \pi. - We restrict our work to where x\geq a, - so x/a\geq 1, and 0\leq\theta\lt \pi/2. - On this interval, \tan(\theta) \geq 0, - so \sqrt{x^2-a^2} = a\tan(\theta). -

      - -
      - - - - - Diagram showing trigonometric substitution with integrands containing square root of x^2-a^2. - - -

      - The diagram is of a right angled triangle. The perpendicular is marked \sqrt{x^2-a^2} - the base as a. The hypotenuse is marked x. The angle opposite to the perpendicular - is marked as \theta. -

      -       - - - \begin{tikzpicture}[scale=1.5] - - \draw [very thick] (0,0) -- node [below,pos=.5] { $a$} (3,0) -- node [right,xshift=3mm,pos=.9,rotate=-90] { $\sqrt{x^2-a^2}$} (3,2) -- node [pos=.5,above] { $x$} (0,0); - \draw [thick] (2.7,0) -- (2.7,.3) -- (3,.3); - \draw (.75,.25) node {$\theta$}; - - \end{tikzpicture} - - - - -
      -       -
      4. -
    -

    -     - - - - - - - Using Trigonometric Substitution - -

    - Evaluate \ds \int \frac{1}{\sqrt{5+x^2}}\, dx. -

    -     - -

    - Using Item in , - we recognize a=\sqrt{5} and set x= \sqrt{5}\tan(\theta). - This makes dx = \sqrt{5}\sec^2(\theta) \, d\theta. - We will use the fact that \sqrt{5+x^2} = \sqrt{5+5\tan^2(\theta) } = \sqrt{5\sec^2(\theta) } = \sqrt{5}\sec(\theta). - Substituting, we have: - - \int \frac{1}{\sqrt{5+x^2}}\, dx \amp = \int \frac{1}{\sqrt{5+5\tan^2(\theta) }}\sqrt{5}\sec^2(\theta) \, d\theta - \amp = \int \frac{\sqrt{5}\sec^2(\theta) }{\sqrt{5}\sec(\theta) } \, d\theta - \amp = \int \sec(\theta) \, d\theta - \amp = \ln\abs{\sec(\theta) +\tan(\theta) }+C - . -

    - -

    - While the integration steps are over, we are not yet done. - The original problem was stated in terms of x, - whereas our answer is given in terms of \theta. - We must convert back to x. -

    - -

    - The reference triangle given in helps. - With x=\sqrt{5}\tan(\theta), we have - - \tan(\theta) = \frac x{\sqrt{5}} \text{ and } \sec(\theta) = \frac{\sqrt{x^2+5}}{\sqrt{5}} - . -

    - -

    - This gives - - \int \frac{1}{\sqrt{5+x^2}}\, dx \amp = \ln\abs{\sec(\theta) +\tan(\theta) }+C - \amp = \ln\abs{\frac{\sqrt{x^2+5}}{\sqrt{5}}+ \frac x{\sqrt{5}}}+C - . -

    - -

    - We can leave this answer as is, - or we can use a logarithmic identity to simplify it. - Note: - - \ln\abs{\frac{\sqrt{x^2+5}}{\sqrt{5}}+ \frac x{\sqrt{5}}}+C \amp = \ln\abs{\frac{1}{\sqrt{5}}\big(\sqrt{x^2+5}+ x\big)}+C - \amp = \ln\abs{\frac{1}{\sqrt{5}}} + \ln\abs{\sqrt{x^2+5}+ x}+C - \amp = \ln\abs{\sqrt{x^2+5}+ x}+C - , - where the \ln\big(1/\sqrt{5}\big) term is absorbed into the constant C. - (In - we will learn another way of approaching this problem.) -

    -     - -     - - - Using Trigonometric Substitution - -

    - Evaluate \ds \int \sqrt{4x^2-1}\, dx. -

    -     - -

    - We start by rewriting the integrand so that it looks like - \sqrt{x^2-a^2} for some value of a: - - \sqrt{4x^2-1} \amp = \sqrt{4\left(x^2-\frac14\right)} - \amp = 2\sqrt{x^2-\left(\frac12\right)^2} - . -

    - -

    - So we have a=1/2, - and following Part of , - we set x= \frac12\sec(\theta), - and hence dx = \frac12\sec(\theta) \tan(\theta) \, d\theta. - We now rewrite the integral with these substitutions: - - \int \sqrt{4x^2-1}\, dx \amp = \int 2\sqrt{x^2-\left(\frac12\right)^2}\, dx - \amp = \int 2\sqrt{\frac14\sec^2(\theta) - \frac14}\left(\frac12\sec(\theta) \tan(\theta) \right)\, d\theta - \amp =\int \sqrt{\frac14(\sec^2(\theta) -1)}\Big(\sec(\theta) \tan(\theta) \Big)\, d\theta - \amp =\int\sqrt{\frac14\tan^2(\theta) }\Big(\sec(\theta) \tan(\theta) \Big)\, d\theta - \amp =\int \frac12\tan^2(\theta) \sec(\theta) \, d\theta - \amp =\frac12\int \Big(\sec^2(\theta) -1\Big)\sec(\theta) \, d\theta - \amp =\frac12\int \big(\sec^3(\theta) - \sec(\theta) \big)\, d\theta - . -

    - -

    - We integrated \sec^3(\theta) in , - finding its antiderivatives to be - - \int \sec^3(\theta) \, d\theta = \frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C - . -

    - -

    - Thus - - \amp \int \sqrt{4x^2-1}\, dx =\frac12\int \big(\sec^3(\theta) - \sec(\theta) \big)\, d\theta - \amp = \frac12\left(\frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big) -\ln\abs{\sec(\theta) + \tan(\theta) }\right) + C - \amp = \frac14\left(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\right)+C - . -

    - -

    - We are not yet done. - Our original integral is given in terms of x, - whereas our final answer, as given, - is in terms of \theta. - We need to rewrite our answer in terms of x. - With a=1/2, and x=\frac12\sec(\theta), - the reference triangle in shows that - - \tan(\theta) = \sqrt{x^2-1/4}\Big/(1/2) = 2\sqrt{x^2-1/4} \text{ and } \sec(\theta) = 2x - . -

    - -

    - Thus - - \amp \frac14\Big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C - \amp\quad = \frac14\Big(2x\cdot 2\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C - \amp\quad = \frac14\Big(4x\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C - . -

    - -

    - The final answer is given in the last line above, repeated here: - - \int \sqrt{4x^2-1}\, dx = \frac14\Big(4x\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C - . -

    -     - -     - - - Using Trigonometric Substitution - -

    - Evaluate \ds \int \frac{\sqrt{4-x^2}}{x^2}\, dx. -

    -     - -

    - We use Part of with a=2, - x=2\sin(\theta), - dx = 2\cos(\theta) and hence \sqrt{4-x^2} = 2\cos(\theta). - This gives - - \int \frac{\sqrt{4-x^2}}{x^2}\, dx \amp = \int \frac{2\cos(\theta) }{4\sin^2(\theta) }(2\cos(\theta) )\, d\theta - \amp = \int \cot^2(\theta) \, d\theta - \amp = \int (\csc^2(\theta) -1)\, d\theta - \amp = -\cot(\theta) -\theta + C - . -

    - -

    - We need to rewrite our answer in terms of x. - Using the reference triangle found in , - we have \cot(\theta) = \sqrt{4-x^2}/x and \theta = \sin^{-1}(x/2). - Thus - - \int \frac{\sqrt{4-x^2}}{x^2}\, dx = -\frac{\sqrt{4-x^2}}x-\sin^{-1}\left(\frac x2\right) + C - . -

    -     - -     - -

    - Trigonometric Substitution can be applied in many situations, - even those not of the form \sqrt{a^2-x^2}, - \sqrt{x^2-a^2} or \sqrt{x^2+a^2}. - In the following example, - we apply it to an integral we already know how to handle. -

    - - - Using Trigonometric Substitution - -

    - Evaluate \ds \int\frac1{x^2+1}\, dx. -

    -     - -

    - We know the answer already as \tan^{-1}(x) +C. - We apply Trigonometric Substitution here to show that we get the same answer without inherently relying on knowledge of the derivative of the arctangent function. -

    - -

    - Using Part of , - let x=\tan(\theta), - dx=\sec^2(\theta) \, d\theta and note that x^2+1 = \tan^2(\theta) +1 = \sec^2(\theta). - Thus - - \int \frac1{x^2+1}\, dx \amp = \int \frac{1}{\sec^2(\theta) }\sec^2(\theta) \, d\theta - \amp = \int 1\, d\theta - \amp = \theta + C - . -

    - -

    - Since x=\tan(\theta), \theta = \tan^{-1}(x), - and we conclude that \ds \int\frac1{x^2+1}\, dx = \tan^{-1}(x) +C. -

    -     -     - -

    - The next example is similar to the previous one in that it does not involve a square-root. - It shows how several techniques and identities can be combined to obtain a solution. -

    - - - Using Trigonometric Substitution - -

    - Evaluate \ds\int\frac1{(x^2+6x+10)^2}\, dx. -

    -     - -

    - We start by completing the square, - then make the substitution u=x+3, - followed by the trigonometric substitution of u=\tan(\theta): - - \int \frac1{(x^2+6x+10)^2}\, dx \amp =\int \frac1{\big((x+3)^2+1\big)^2}\, dx = \int \frac1{(u^2+1)^2}\,du. - Now make the substitution u=\tan(\theta), du=\sec^2(\theta) \, d\theta: - \amp = \int \frac1{(\tan^2(\theta) +1)^2}\sec^2(\theta) \, d\theta - \amp = \int\frac 1{(\sec^2(\theta) )^2}\sec^2(\theta) \, d\theta - \amp = \int \cos^2(\theta) \, d\theta. - Applying a power reducing formula, we have - \amp = \int \left(\frac12 +\frac12\cos(2\theta)\right)\, d\theta - \amp = \frac12\theta + \frac14\sin(2\theta) + C - . -

    - -

    - We need to return to the variable x. - As u=\tan(\theta), \theta = \tan^{-1}(u). - Using the identity \sin(2\theta) = 2\sin(\theta) \cos(\theta) and using the reference triangle found in - , we have - - \frac14\sin(2\theta) = \frac12\frac u{\sqrt{u^2+1}}\cdot\frac 1{\sqrt{u^2+1}} = \frac12\frac u{u^2+1} - . -

    - -

    - Finally, we return to x with the substitution u=x+3. - We start with the expression in Equation: - - \frac12\theta + \frac14\sin(2\theta) + C \amp = \frac12\tan^{-1}(u) + \frac12\frac{u}{u^2+1}+C - \amp = \frac12\tan^{-1}(x+3) + \frac{x+3}{2(x^2+6x+10)}+C - . -

    - -

    - Stating our final result in one line, - - \int\frac1{(x^2+6x+10)^2}\, dx=\frac12\tan^{-1}(x+3) + \frac{x+3}{2(x^2+6x+10)}+C - . -

    -     - -     - - - -

    - Our last example returns us to definite integrals, - as seen in our first example. - Given a definite integral that can be evaluated using Trigonometric Substitution, - we could first evaluate the corresponding indefinite integral - (by changing from an integral in terms of x to one in terms of \theta, - then converting back to x) - and then evaluate using the original bounds. - It is much more straightforward, though, - to change the bounds as we substitute. -

    - - - Definite integration and Trigonometric Substitution - -

    - Evaluate \ds\int_0^5\frac{x^2}{\sqrt{x^2+25}}\, dx. -

    -     - -

    - Using Part of , - we set x=5\tan(\theta), - dx = 5\sec^2(\theta) \, d\theta, - and note that \sqrt{x^2+25} = 5\sec(\theta). - As we substitute, we can also change the bounds of integration. -

    - -

    - The lower bound of the original integral is x=0. - As x=5\tan(\theta), - we solve for \theta and find \theta = \tan^{-1}(x/5). - Thus the new lower bound is \theta = \tan^{-1}(0) = 0. - The original upper bound is x=5, - thus the new upper bound is \theta = \tan^{-1}(5/5) = \pi/4. -

    - -

    - Thus we have - - \int_0^5\frac{x^2}{\sqrt{x^2+25}}\, dx \amp = \int_0^{\pi/4} \frac{25\tan^2(\theta) }{5\sec(\theta) }5\sec^2(\theta) \, d\theta - \amp = 25\int_0^{\pi/4} \tan^2(\theta) \sec(\theta) \, d\theta - . -

    - -

    - We encountered this indefinite integral in where we found - - \int \tan^2(\theta) \sec(\theta) \, d\theta = \frac12\big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\big) - . -

    - -

    - So - - 25\int_0^{\pi/4} \tan^2(\theta) \sec(\theta) \, d\theta \amp = \left.\frac{25}2\big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\big)\right|_0^{\pi/4} - \amp = \frac{25}2\big(\sqrt2-\ln(\sqrt2+1)\big) - \amp \approx 6.661 - . -

    -     - -     - -

    - The following equalities are very useful when evaluating integrals using Trigonometric Substitution. -

    - - - Useful Equalities with Trigonometric Substitution -

    -

      -
    1. -

      - \sin(2\theta) = 2\sin(\theta) \cos(\theta) -

      -
      2. - -
    2. -

      - \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) -1 = 1-2\sin^2(\theta) -

      -
      3. - -
    3. -

      - \ds \int \sec^3(\theta) \, d\theta = \frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C -

      -
      4. - -
    4. -

      - \ds \int \cos^2(\theta) \, d\theta = \int \frac12\big(1+\cos(2\theta)\big)\, d\theta = \frac12\big(\theta+\sin(\theta) \cos(\theta) \big)+C. -

      -
      5. -
    -

    -     - -

    - The next section introduces Partial Fraction Decomposition, - which is an algebraic technique that turns complicated - fractions into sums of simpler fractions, - making integration easier. -

    - - - - Terms and Concepts - - - - -

    - Trigonometric Substitution works on the same principles as Integration by Substitution, - though it can feel . -

    -     - - - - backwards? - - - - -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=5;}; - Context()->variables->add(theta=>['Real',TeX=>'\theta']); - $integrand = Formula("sqrt($m^2-x^2)"); - $answer = OneOf(Formula("$m*sin(theta)"),Formula("$m*cos(theta)")); - - -

    - If one uses Trigonometric Substitution on an integrand containing , - then one should set x equal to what? -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=9;}; - Context()->variables->add(theta=>['Real',TeX=>'\theta']); - # Redefine tan() so the following equtaion is not an identity - package my::Function::numeric; - our @ISA = ('Parser::Function::numeric'); - sub tan { - shift; my $x = shift; - return CORE::sin($x+2)/CORE::cos($x+2); - } - package main; - Context()->functions->set(tan => {class => 'my::Function::numeric'}); - $I = ImplicitEquation("tan(theta)^2+1=sec(theta)^2"); - $S = Formula("$m sec(theta)^2"); - - -

    - Consider the Pythagorean Identity \sin^2(\theta) +\cos^2(\theta) = 1. -

    -

    -

      -
    1. -

      - What identity is obtained when both sides are divided by \cos^2(\theta)? -

      -

      - -

      -
      2. -
    2. -

      - Use the new identity to simplify 9\tan^2(\theta) + 9. -

      -

      - -

      -
      3. -
    -

    -     -     -     - - - - -

    - Why does Part of - state that \sqrt{a^2-x^2} = a\cos(\theta), - and not \abs{a\cos(\theta) }? -

    - -     - - - -

    - Because we are considering a>0 and x=a\sin(\theta), - which means \theta = \sin^{-1}(x/a). - The arcsine function has a domain of -\pi/2\leq \theta \leq \pi/2; - on this domain, \cos(\theta) \geq 0, - so a\cos(\theta) is always non-negative, - allowing us to drop the absolute value signs. -

    -     - -     -     - - - Problems - - -

    - Apply Trigonometric Substitution to evaluate the indefinite integral. -

    -     - - - - - Context("Fraction"); - $F = FormulaUpToConstant("1/2 (x sqrt(x^2 + 1) + ln(sqrt(x^2 + 1) + x))"); - - -

    - \ds \int \sqrt{x^2+1}\, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - $F = FormulaUpToConstant("x/2 sqrt(x^2 + 4) + 2 ln(sqrt(x^2 + 4)/2 + x/2)"); - - -

    - \ds \int \sqrt{x^2+4}\, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->variables->set(x=>{limits=>[-1,1]}); - $F = FormulaUpToConstant("1/2 asin(x) + x/2 sqrt(1 - x^2)"); - - -

    - \ds \int \sqrt{1-x^2}\, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->variables->set(x=>{limits=>[-3,3]}); - $F = FormulaUpToConstant("9/2 asin(x/3) + x/2 sqrt(9 - x^2)"); - - -

    - \ds \int \sqrt{9-x^2}\, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->variables->set(x=>{limits=>[1,5]}); - $F = FormulaUpToConstant("1/2 x sqrt(x^2 - 1) - 1/2 ln(abs(x+sqrt(x^2 - 1)))"); - $F->{test_at} = [-2,-3,-4]; - - -

    - \ds \int \sqrt{x^2-1}\, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->variables->set(x=>{limits=>[4,9]}); - $F = FormulaUpToConstant("1/2 x sqrt(x^2 - 16) - 8 ln(abs(x/4 + sqrt(x^2 - 16)/4))"); - $F->{test_at} = [-5,-6,-7]; - - -

    - \ds \int \sqrt{x^2-16}\, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=2;}; - Context("Fraction"); - $f = Formula("sqrt($m^2 x^2+1)"); - $F = FormulaUpToConstant("x/2 sqrt($m^2 x^2+1)+1/(2*$m)*ln($m x + sqrt($m^2 x^2+1))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=3;}; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[-1/$m,1/$m]}); - $f = Formula("sqrt(1 - $m^2 x^2)"); - $F = FormulaUpToConstant("x/2 sqrt(1 - $m^2 x^2) + 1/(2*$m) asin($m x)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=4;}; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[1/$m,1/$m+4]}); - $f = Formula("sqrt($m^2 x^2 - 1)"); - $F = FormulaUpToConstant("x/2 sqrt($m^2 x^2 - 1) - 1/(2*$m) ln(abs($m x + sqrt($m^2 x^2 - 1)))")->reduce; - $F->{test_at} = [-1/$m-1,-1/$m-2,-1/$m-3]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $b = list_random(2,3,5,6,7,10,11,13,14,15); - if($envir{problemSeed}==1){$a=8;$b=2}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$a/sqrt(x^2 + $b)"); - $F = FormulaUpToConstant("$a ln(x/sqrt($b) + sqrt(x^2/$b + 1))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $b = list_random(2,3,5,6,7,10,11,13,14,15); - if($envir{problemSeed}==1){$a=3;$b=7}; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[-sqrt($b),sqrt($b)]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$a/sqrt($b - x^2)"); - $F = FormulaUpToConstant("$a asin(x/sqrt($b))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $b = list_random(2,3,5,6,7,10,11,13,14,15); - if($envir{problemSeed}==1){$a=5;$b=6}; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[sqrt($b),sqrt($b)+4]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("$a/sqrt(x^2-$b)"); - $F = FormulaUpToConstant("$a ln(abs(x/sqrt($b) + sqrt(x^2/$b - 1)))")->reduce; - $F->{test_at} = [-sqrt($b)-1,-sqrt($b)-2,-sqrt($b)-3]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Evaluate the indefinite integral. - Trigonometric Substitution may not be required. -

    -     - - - - - $b = list_random(2,3,5,6,7,10,11,13,14,15); - if($envir{problemSeed}==1){$b=11}; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[sqrt($b),sqrt($b)+4]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("sqrt(x^2-$b)/x"); - $F = FormulaUpToConstant("sqrt(x^2 - $b) - sqrt($b) asec(x/sqrt($b))")->reduce; - $F->{test_at} = [-sqrt($b)-1,-sqrt($b)-2,-sqrt($b)-3]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - $F = FormulaUpToConstant("1/2 atan(x) + x/[2(x^2+1)]"); - - -

    - \ds \int \frac {1}{(x^2+1)^2}\, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random(2,3,5,6,7,10,11,13,14,15); - if($envir{problemSeed}==1){$b=3}; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[sqrt($b),sqrt($b)+4]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("x/sqrt(x^2-$b)"); - $F = FormulaUpToConstant("sqrt(x^2 - $b)")->reduce; - $F->{test_at} = [-sqrt($b)-1,-sqrt($b)-2,-sqrt($b)-3]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->variables->set(x=>{limits=>[-1,1]}); - $F = FormulaUpToConstant("1/8 asin(x) + x/8 sqrt(1-x^2) (2x^2 - 1)"); - - -

    - \ds \int x^2\sqrt{1-x^2}\, dx -

    -

    - -

    -     -     -     - - - - - $m = random(2,9,1); - if($envir{problemSeed}==1){$m=3}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("x/(x^2 + $m^2)^(3/2)"); - $F = FormulaUpToConstant("-1/sqrt(x^2 + $m^2)")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $b = list_random(2,3,5,6,7,10,11,13,14,15); - if($envir{problemSeed}==1){$a=5;$b=10;}; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[sqrt($b),sqrt($b)+4]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("($a x^2)/sqrt(x^2 - $b)"); - $frac = Fraction($a,2); - ($num,$den) = $frac->value; - $frac2 = Fraction($a*$b,2); - $F = FormulaUpToConstant("($num x)/$den sqrt(x^2 - $b) + $frac2 ln(abs(x/sqrt($b) + sqrt(x^2/$b - 1)))")->reduce; - $F->{test_at} = [-sqrt($b)-1,-sqrt($b)-2,-sqrt($b)-3]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $h = non_zero_random(-9,9,1); - $m = random(1,9,1); - if($envir{problemSeed}==1){$h=-2;$m=3;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("1/(x^2 - (2*$h) x + (($h)^2+$m^2))^2")->reduce; - $frac = Fraction(1,2*$m**2); - $frac2 = Fraction(1,2*$m**3); - $F = FormulaUpToConstant("$frac (x-$h)/(x^2 - (2*$h) x + (($h)^2+$m^2)) + $frac2 atan((x-$h)/$m) ")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - Context("Fraction"); - Context()->variables->set(x=>{limits=>[-1,1]}); - $F = FormulaUpToConstant("x/sqrt(1-x^2)-asin(x)"); - - -

    - \ds \int x^2(1-x^2)^{-3/2}\, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - $b = list_random(2,3,5,6,7,10,11,13,14,15); - if($envir{problemSeed}==1){$a=7;$b=5;}; - Context("Fraction"); - Context()->variables->set(x=>{limits=>[-sqrt($b),sqrt($b)]}); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("sqrt($b - x^2)/($a x^2)"); - $F = FormulaUpToConstant("-sqrt($b - x^2)/($a x) - 1/$a asin(x/sqrt($b))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $b = list_random(2,3,5,6,7,10,11,13,14,15); - if($envir{problemSeed}==1){$b=3;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("x^2/sqrt(x^2 + $b)"); - $frac = Fraction($b,2); - $F = FormulaUpToConstant("x/2 sqrt(x^2 + $b) - $frac ln(x/sqrt(3) + sqrt(x^2/$b + 1))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Evaluate the definite integral by making the proper trigonometric substitution and - changing the bounds of integration. - (Note: the corresponding indefinite integrals appeared previously in the exercises.) -

    -     - - - - - $answer = Compute("pi/2"); - - -

    - \ds \int_{-1}^1 \sqrt{1-x^2}\, dx -

    -

    - -

    -     -     -     - - - - - $b = random(5,10,1); - if($envir{problemSeed}==1){$b=8;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $arg = $b**2 - 16; - ($sq,$nsq) = (1,$arg); - while ($nsq % 4 == 0) { - $sq *= 2; - $nsq /= 4; - } - while ($nsq % 9 == 0) { - $sq *= 3; - $nsq /= 9; - } - $frac = Fraction($b*$sq,2); - $frac2 = Fraction($b,4); - $frac3 = Fraction($sq,4); - $answer = Formula("$frac sqrt($nsq) - 8 ln(abs($frac2 + $frac3 sqrt($nsq)))"); - - -

    - \ds \int_{4}^{} \sqrt{x^2-16}\, dx -

    -

    - -

    -     -     -     - - - - - $b = random(1,10,1); - if($envir{problemSeed}==1){$b=2;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $arg = $b**2 + 4; - ($sq,$nsq) = (1,$arg); - while ($nsq % 4 == 0) { - $sq *= 2; - $nsq /= 4; - } - while ($nsq % 9 == 0) { - $sq *= 3; - $nsq /= 9; - } - $frac = Fraction($b*$sq,2); - $frac2 = Fraction($b,2); - $frac3 = Fraction($sq,2); - $answer = Formula("$frac sqrt($nsq) + 2 ln($frac2 + $frac3 sqrt($nsq))"); - - -

    - \ds \int_{0}^{} \sqrt{x^2+4}\, dx -

    -

    - -

    -     -     -     - - - - - $b = random(1,10,1); - if($envir{problemSeed}==1){$b=1;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $frac = Fraction($b,$b**2 + 1); - $answer = Formula("atan($b) + $frac"); - - -

    - \ds \int_{-}^{} \frac1{(x^2+1)^2}\, dx -

    -

    - -

    -     -     -     - - - - - $b = random(1,2,1); - if($envir{problemSeed}==1){$b=1;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $arg = 9 - $b**2; - ($sq,$nsq) = (1,$arg); - while ($nsq % 4 == 0) { - $sq *= 2; - $nsq /= 4; - } - while ($nsq % 9 == 0) { - $sq *= 3; - $nsq /= 9; - } - $frac = Fraction($b,3); - $frac2 = Fraction($b*$sq,1); - $answer = Formula("9 asin($frac) + $frac2 sqrt($nsq)"); - - -

    - \ds \int_{-}^{} \sqrt{9-x^2}\, dx -

    -

    - -

    -     -     -     - - - - - $answer = Formula("pi/8"); - - -

    - \ds \int_{-1}^1 x^2\sqrt{1-x^2}\, dx -

    -

    - -

    -     -     -     -     -     -     -
    -
    - Partial Fraction Decomposition -

    - In this section we investigate the antiderivatives of rational functions. - Recall that rational functions are functions of the form f(x)= \frac{p(x)}{q(x)}, - where p(x) and q(x) are polynomials and q(x)\neq 0. - Such functions arise in many contexts, - one of which is the solving of certain fundamental differential equations. -

    - - - -

    - We begin with an example that demonstrates the motivation behind this section. - Consider the integral \ds\int \frac{1}{x^2-1}\, dx. - We do not have a simple formula for this - (if the denominator were x^2+1, - we would recognize the antiderivative as being the arctangent function). - It can be solved using Trigonometric Substitution, - but note how the integral is easy to evaluate once we realize: - - \frac{1}{x^2-1} = \frac{1/2}{x-1} - \frac{1/2}{x+1} - . -

    - -

    - Thus - - \int\frac{1}{x^2-1}\, dx \amp = \int\frac{1/2}{x-1}\, dx - \int\frac{1/2}{x+1}\, dx - \amp = \frac12\ln\abs{x-1} - \frac12\ln\abs{x+1} + C - . -

    - -

    - This section teaches how to decompose - - \frac{1}{x^2-1} \text{ into } \frac{1/2}{x-1}-\frac{1/2}{x+1} - . -

    - -

    - We start with a rational function f(x)=\frac{p(x)}{q(x)}, - where p and q do not have any common factors and the degree of p is less than the degree of q. - It can be shown that any polynomial, and hence q, - can be factored into a product of linear and irreducible quadratic terms. - The following Key Idea states how to decompose a rational function into a sum of rational functions whose denominators are all of lower degree than q. -

    - - - Partial Fraction Decomposition -

    - Let \ds \frac{p(x)}{q(x)} be a rational function, - where the degree of p is less than the degree of q. - integrationpartial fraction decomp. - -

    - -

    -

      -
    1. -

      - Linear Terms: Let (x-a) divide q(x), - where (x-a)^n is the highest power of (x-a) that divides q(x). - Then the decomposition of \frac{p(x)}{q(x)} will contain the sum - - \frac{A_1}{(x-a)} + \frac{A_2}{(x-a)^2} + \cdots +\frac{A_n}{(x-a)^n} - . -

      -
      2. - -
    2. -

      - Quadratic Terms: Let - x^2+bx+c be an irreducible quadratic that divides q(x), - where (x^2+bx+c)^n is the highest power of - x^2+bx+c that divides q(x). - Then the decomposition of \frac{p(x)}{q(x)} will contain the sum - - \frac{B_1x+C_1}{x^2+bx+c}+\frac{B_2x+C_2}{(x^2+bx+c)^2}+\cdots+\frac{B_nx+C_n}{(x^2+bx+c)^n} - . -

      -
      3. -
    -

    - - - -

    - To find the coefficients A_i, - B_i and C_i: -

    - -

    -

      -
    1. -

      - Multiply all fractions by q(x), - clearing the denominators. - Collect like terms. -

      -
      2. - -
    2. -

      - Equate the resulting coefficients of the powers of x and solve the resulting system of linear equations. -

      -
      3. -
    -

    -     - - -

    - The following examples will demonstrate how to put this Key Idea into practice. - - stresses the decomposition aspect of the Key Idea. -

    - - - Decomposing into partial fractions - -

    - Decompose \ds f(x)=\frac{1}{(x+5)(x-2)^3(x^2+x+2)(x^2+x+7)^2} without solving for the resulting coefficients. -

    -     - -

    - The denominator is already factored, - as both x^2+x+2 and x^2+x+7 cannot be factored further. - We need to decompose f(x) properly. - Since (x+5) is a linear term that divides the denominator, - there will be a - - \frac{A}{x+5} - - term in the decomposition. -

    - -

    - As (x-2)^3 divides the denominator, - we will have the following terms in the decomposition: - - \frac{B}{x-2}, \frac{C}{(x-2)^2} \text{ and } \frac{D}{(x-2)^3} - . -

    - -

    - The x^2+x+2 term in the denominator results in a \ds\frac{Ex+F}{x^2+x+2} term. -

    - -

    - Finally, the (x^2+x+7)^2 term results in the terms - - \frac{Gx+H}{x^2+x+7} \text{ and } \frac{Ix+J}{(x^2+x+7)^2} - . -

    - -

    - All together, we have - - \amp\frac{1}{(x+5)(x-2)^3(x^2+x+2)(x^2+x+7)^2} - \amp \quad\quad = \frac{A}{x+5} + \frac{B}{x-2}+ \frac{C}{(x-2)^2}+\frac{D}{(x-2)^3} - \amp \quad\quad\quad\quad + \frac{Ex+F}{x^2+x+2}+\frac{Gx+H}{x^2+x+7}+\frac{Ix+J}{(x^2+x+7)^2} - -

    - -

    - Solving for the coefficients A, - B \ldots J would be a bit tedious but not hard. -

    -     - -     - - - Decomposing into partial fractions - -

    - Perform the partial fraction decomposition of \ds \frac{1}{x^2-1}. -

    -     - -

    - The denominator factors into two linear terms: - x^2-1 = (x-1)(x+1). - Thus - - \frac{1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1} - . -

    - -

    - To solve for A and B, - first multiply through by x^2-1 = (x-1)(x+1): - - 1 \amp = \frac{A(x-1)(x+1)}{x-1}+\frac{B(x-1)(x+1)}{x+1} - \amp = A(x+1) + B(x-1) - \amp = Ax+A + Bx-B - - \amp = (A+B)x + (A-B) - , - by collecting like terms. -

    - -

    - The next step is key. - Note the equality we have: - - 1 = (A+B)x+(A-B) - . -

    - -

    - For clarity's sake, rewrite the left hand side as - - 0x+1 = (A+B)x+(A-B) - . -

    - -

    - On the left, the coefficient of the x term is 0; - on the right, it is (A+B). - Since both sides are equal, we must have that 0=A+B. -

    - -

    - Likewise, on the left, we have a constant term of 1; - on the right, the constant term is (A-B). - Therefore we have 1=A-B. -

    - -

    - We have two linear equations with two unknowns. - This one is easy to solve by hand, leading to - - A+B \amp = 0 - A-B \amp = 1 - - If we add these two equations, - we get 2A=1 \Rightarrow A=1/2. - Substitution into the first equation gives B=-1/2. -

    - -

    - Thus - - \frac{1}{x^2-1} = \frac{1/2}{x-1}-\frac{1/2}{x+1} - . -

    -     - -     - - - -

    - There is another method for finding the partial fraction decomposition called the - Heaviside method, - named after Oliver Heaviside. - We show a variation of this process using the same example as in . -

    - - - Decomposing into partial fractions using the Heaviside method - -

    - Perform the partial fraction decomposition of \ds \frac{1}{x^2-1}. -

    -     - -

    - As we saw in , - - \frac{1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1} - . -

    - -

    - To solve for A and B using the Heaviside method, - we will build to a common denominator: - - \frac{1}{x^2-1} \amp = \frac{A(x+1)}{(x-1)(x+1)}+\frac{B(x-1)}{(x+1)(x-1)} - \amp = \frac{A(x+1)+B(x-1)}{(x-1)(x+1)} - -

    - -

    - Now since the denomiators match, - we will only consider the numerator equation (essentially if we multiply both sides of the equation by (x-1)(x+1), - we will clear the denomiators): - - 1=A(x+1)+B(x-1) - - Now we substitute in convenient values of x. - When x=1, we get 1=2A \Rightarrow A=1/2. - When x=-1, we get 1=-2B \Rightarrow B=-1/2. -

    - -

    - You may note that x=1 and x=-1 were not in the domain of the original fraction. - However, - - \frac{1}{x^2-1}=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)} - - is an identity, meaning it is true for all values of x, - even those for which the equation is undefined. - We could have chosen any values of x to substitute. - Whenever possible, - we choose values of x that will make one of the factors zero. - In this way, we can avoid solving a system of equations. -

    - -

    - Thus as in , we get - - \frac{1}{x^2-1} = \frac{1/2}{x-1}-\frac{1/2}{x+1} - . -

    -     - -     - -

    - For the remaining examples, we will use a combination of systems of equations and the Heaviside method to get partial fraction decompositions. -

    - - - Integrating using partial fractions - -

    - Use partial fraction decomposition to integrate \ds\int\frac{1}{(x-1)(x+2)^2}\, dx. -

    -     - -

    - We decompose the integrand as follows, - as described by : - - \frac{1}{(x-1)(x+2)^2} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} - . -

    - -

    - To solve for A, B and C, - we multiply both sides by (x-1)(x+2)^2: - - 1 = A(x+2)^2 + B(x-1)(x+2) + C(x-1) - -

    - -

    - Now we collect like terms: - - 1 \amp = A(x+2)^2 + B(x-1)(x+2) + C(x-1) - \amp = Ax^2+4Ax+4A + Bx^2 + Bx-2B + Cx-C - \amp = (A+B)x^2 + (4A+B+C)x + (4A-2B-C) - -

    - - - -

    - We have - - 0x^2+0x+ 1 = (A+B)x^2 + (4A+B+C)x + (4A-2B-C) - - leading to the equations - - A+B = 0, 4A+B+C = 0 \text{ and } 4A-2B-C = 1 - . -

    - -

    - These three equations of three unknowns lead to a unique solution: - - A = 1/9, B = -1/9 \text{ and } C = -1/3 - . -

    - -

    - Thus - - \int\frac{1}{(x-1)(x+2)^2}\, dx = \int \frac{1/9}{x-1}\, dx + \int \frac{-1/9}{x+2}\, dx + \int \frac{-1/3}{(x+2)^2}\, dx - . -

    - -

    - Each can be integrated with a simple substitution with u=x-1 or u=x+2 - (or by directly applying - as the denominators are linear functions). - The end result is - - \int\frac{1}{(x-1)(x+2)^2}\, dx = \frac19\ln\abs{x-1} -\frac19\ln\abs{x+2} +\frac1{3(x+2)}+C - . -

    -     - -     - - - - - - - Integrating using partial fractions - -

    - Use partial fraction decomposition to integrate \ds \int \frac{x^3}{(x-5)(x+3)}\, dx. -

    -     - -

    - - presumes that the degree of the numerator is less than the degree of the denominator. - Since this is not the case here, - we begin by using polynomial division to reduce the degree of the numerator. - We omit the steps, but encourage the reader to verify that - - \frac{x^3}{(x-5)(x+3)} = x+2+\frac{19x+30}{(x-5)(x+3)} - . -

    - -

    - Using , - we can rewrite the new rational function as: - - \frac{19x+30}{(x-5)(x+3)} = \frac{A}{x-5} + \frac{B}{x+3} - - for appropriate values of A and B. - Clearing denominators, we have -

    - - -

    - - 19x+30 \amp = A(x+3) + B(x-5) - \amp = (A+B)x + (3A-5B). - This implies that: - 19\amp = A+B - 30\amp = 3A-5B. - Solving this system of linear equations gives - 125/8 \amp =A - 27/8 \amp =B - . -

    -

    - We can now integrate. - - \int \frac{x^3}{(x-5)(x+3)}\, dx \amp = \int\left(x+2+\frac{125/8}{x-5}+\frac{27/8}{x+3}\right)\, dx - \amp = \frac{x^2}2 + 2x + \frac{125}{8}\ln\abs{x-5} + \frac{27}8\ln\abs{x+3} + C - . -

    -     - -     - - - Integrating using partial fractions - -

    - Use partial fraction decomposition to evaluate \ds \int\frac{7x^2+31x+54}{(x+1)(x^2+6x+11)}\, dx. -

    -     - -

    - The degree of the numerator is less than the degree of the denominator so we begin by applying . - We have: - - - \frac{7x^2+31x+54}{(x+1)(x^2+6x+11)} \amp = \frac{A}{x+1} + \frac{Bx+C}{x^2+6x+11}. - Now clear the denominators. - 7x^2+31x+54 \amp = A(x^2+6x+11) + (Bx+C)(x+1) - - Now, letting x=-1 we have 30 = 6A \Rightarrow A=5. - When x=0, 54 = 11A+C. - But we know that A=5, - so 54 =55+C \Rightarrow C=-1 Finally, we choose x=1 - (with A=5, C=-1) - we have 92=90+(B-1)(2)\Rightarrow B=2. -

    - -

    - - Thus - - \int\frac{7x^2+31x+54}{(x+1)(x^2+6x+11)}\, dx = \int\left(\frac{5}{x+1} + \frac{2x-1}{x^2+6x+11}\right)\, dx - . -

    - -

    - The first term of this new integrand is easy to evaluate; - it leads to a 5\ln\abs{x+1} term. - The second term is not hard, - but takes several steps and uses substitution techniques. -

    - -

    - The integrand \ds \frac{2x-1}{x^2+6x+11} has a quadratic in the denominator and a linear term in the numerator. - This leads us to try substitution. - Let u = x^2+6x+11, so du = (2x+6)\, dx. - The numerator is 2x-1, not 2x+6, - but we can get a 2x+6 term in the numerator by adding 0 in the form of 7-7. - - \frac{2x-1}{x^2+6x+11} \amp = \frac{2x-1+7-7}{x^2+6x+11} - \amp = \frac{2x+6}{x^2+6x+11} - \frac{7}{x^2+6x+11} - . -

    - -

    - We can now integrate the first term with substitution, - leading to a \ln\abs{x^2+6x+11} term. - The final term can be integrated using arctangent. (We can tell there is no further factoring for this quadratic since the denominator has no real solutions). - First, complete the square in the denominator: - - \frac{7}{x^2+6x+11} = \frac{7}{(x+3)^2+2} - . -

    - -

    - An antiderivative of the latter term can be found using and substitution: - - \int \frac{7}{x^2+6x+11}\, dx = \frac{7}{\sqrt{2}}\tan^{-1}\left(\frac{x+3}{\sqrt{2}}\right)+C - . -

    - -

    - Let's start at the beginning and put all of the steps together. - - \amp \int\frac{7x^2+31x+54}{(x+1)(x^2+6x+11)}\, dx - \amp = \int\left(\frac{5}{x+1} + \frac{2x-1}{x^2+6x+11}\right)\, dx - \amp = \int\frac{5}{x+1}\, dx + \int\frac{2x+6}{x^2+6x+11}\, dx -\int\frac{7}{(x+3)^2+2}\, dx - \amp = 5\ln\abs{x+1}+ \ln\abs{x^2+6x+11} -\frac{7}{\sqrt{2}}\tan^{-1}\left(\frac{x+3}{\sqrt{2}}\right)+C - . -

    - -

    - As with many other problems in calculus, - it is important to remember that one is not expected to see - the final answer immediately after seeing the problem. - Rather, given the initial problem, - we break it down into smaller problems that are easier to solve. - The final answer is a combination of the answers of the smaller problems. -

    -     - -     - - - -

    - Partial Fraction Decomposition is an important tool when dealing with rational functions. - Note that at its heart, it is a technique of algebra, - not calculus, as we are rewriting a fraction in a new form. - Regardless, it is very useful in the realm of calculus as it lets us evaluate a certain set of - complicated integrals. -

    - -

    - introduces new functions, - called the Hyperbolic Functions. - They will allow us to make substitutions similar to those found when studying Trigonometric Substitution, - allowing us to approach even more integration problems. -

    - - - - Terms and Concepts - - - -

    - Partial Fraction Decomposition is a method of rewriting functions. -

    -     - - - - - rational - - - - - -     - - - - -

    - - It is sometimes necessary to use polynomial division before using Partial Fraction Decomposition. -

    -     - -     - - -

    - Decompose without solving for the coefficients, - as done in . -

    -     - - - - - $r = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$r=-3}; - Context("Form")->variables->add(s=>"Real",t=>"Real",A=>"Real",B=>"Real",C=>"Real"); - Context()->reductions->set("(-x)+y" => 1); - $f = Formula("1/(x^2 + $r x)")->reduce; - $template = Formula("s/x + t/(x+$r)")->reduce, - @answer = (); - for my $s (A..C) { - for my $t (A..C) { - if ($s ne $t) { - for my $ss ("","-") { - for my $st ("","-") { - push(@answer,$template->substitute(s=>Formula("$ss $s"),t=>Formula("$st $t"))->reduce); - } - } - } - } - } - $evaluator = $answer[0]->cmp( - bypass_equivalence_test => 1, - checker=> sub { - my ($correct,$student,$ansHash) = @_; - for my $ans (@answer) { - if ($ans->cmp->evaluate($student)->{score} == 1) {return 1}; - } - Value->Error("Your answer should use paramaters A, B, and/or C.") unless ($student->usesOneOf(A..C)); - return 0; - }); - - -

    - \ds -

    - - Use A, B, and C as needed. - Do not actually solve for these coefficients. - -

    - -

    -     -     -     - - - - - $r = random(1,9,1); - ($a,$b) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$r=3;$a=-1;$b=7}; - Context("Form")->variables->add(s=>"Real",t=>"Real",A=>"Real",B=>"Real",C=>"Real"); - Context()->reductions->set("(-x)+y" => 1); - $f = Formula("($a x+$b)/(x^2 - $r^2)")->reduce; - $template = Formula("s/(x-$r) + t/(x+$r)")->reduce, - @answer = (); - for my $s (A..C) { - for my $t (A..C) { - if ($s ne $t) { - for my $ss ("","-") { - for my $st ("","-") { - push(@answer,$template->substitute(s=>Formula("$ss $s"),t=>Formula("$st $t"))->reduce); - } - } - } - } - } - $evaluator = $answer[0]->cmp( - bypass_equivalence_test => 1, - checker=> sub { - my ($correct,$student,$ansHash) = @_; - for my $ans (@answer) { - if ($ans->cmp->evaluate($student)->{score} == 1) {return 1}; - } - Value->Error("Your answer should use paramaters A, B, and/or C.") unless ($student->usesOneOf(A..C)); - return 0; - }); - - -

    - \ds -

    - - Use A, B, and C as needed. - Do not actually solve for these coefficients. - -

    - -

    -     -     -     - - - - - $r = list_random(2,3,5,6,7,10,11,13,14,15); - $b = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$r=7;$b=-3}; - Context("Form")->variables->add(s=>"Real",t=>"Real",A=>"Real",B=>"Real",C=>"Real"); - Context()->reductions->set("(-x)+y" => 1); - $f = Formula("(x+$b)/(x^2 - $r)")->reduce; - $template = Formula("s/(x-sqrt($r)) + t/(x+sqrt($r))"); - @answer = (); - for my $s (A..C) { - for my $t (A..C) { - if ($s ne $t) { - for my $ss ("","-") { - for my $st ("","-") { - push(@answer,$template->substitute(s=>Formula("$ss $s"),t=>Formula("$st $t"))); - } - } - } - } - } - $evaluator = $answer[0]->cmp( - bypass_equivalence_test => 1, - checker=> sub { - my ($correct,$student,$ansHash) = @_; - for my $ans (@answer) { - if ($ans->cmp->evaluate($student)->{score} == 1) {return 1}; - } - Value->Error("Your answer should use paramaters A, B, and/or C.") unless ($student->usesOneOf(A..C)); - return 0; - }); - - -

    - \ds -

    - - Use A, B, and C as needed. - Do not actually solve for these coefficients. - -

    - -

    -     -     -     - - - - - $r = random(2,9,1); - ($a,$b) = random_subset(2,1..9); - if($envir{problemSeed}==1){$r=7;$a=2;$b=5}; - Context("Form")->variables->add(r=>"Real",s=>"Real",t=>"Real",A=>"Real",B=>"Real",C=>"Real"); - Context()->reductions->set("(-x)+y" => 1); - $f = Formula("($a x + $b)/(x^3 + $r x)")->reduce; - $template = Formula("r/x + (s x + t)/(x^2 + $r)")->reduce, - @answer = (); - for my $r (A..C) { - for my $s (A..C) { - if ($r ne $s) { - $t = 'C' if ($r.$s eq 'AB' or $r.$s eq 'BA'); - $t = 'B' if ($r.$s eq 'AC' or $r.$s eq 'CA'); - $t = 'A' if ($r.$s eq 'BC' or $r.$s eq 'CB'); - for my $sr ("","-") { - for my $ss ("","-") { - for my $st ("","-") { - push(@answer,$template->substitute(r=>Formula("$sr $r"), s=>Formula("$ss $s"),t=>Formula("$st $t"))->reduce); - } - } - } - } - } - } - $evaluator = $answer[0]->cmp( - bypass_equivalence_test => 1, - checker=> sub { - my ($correct,$student,$ansHash) = @_; - for my $ans (@answer) { - if ($ans->cmp->evaluate($student)->{score} == 1) {return 1}; - } - Value->Error("Your answer should use paramaters A, B, and/or C.") unless ($student->usesOneOf(A..C)); - return 0; - }); - - -

    - \ds -

    - - Use A, B, and C as needed. - Do not actually solve for these coefficients. - -

    - -

    -     -     -     -     -     - - - Problems - - -

    - Evaluate the indefinite integral. -

    -     - - - - - ($r,$s) = num_sort(random_subset(2,-9..-1,1..9)); - ($A,$B) = random_subset(2,1..9); - if($envir{problemSeed}==1){$r=-5;$s=2;$A=4;$B=3;}; - $f = Formula("(($A+$B)x-($A*$s+$B*$r))/(x^2 - ($r+$s)x + $r*$s)")->reduce; - $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(abs(x-$s))")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $r = non_zero_random(-9,9,1); - ($A,$B) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$r=-1;$A=9;$B=-2;}; - $s = 0; - $f = Formula("(($A+$B)x-($A*$s+$B*$r))/(x^2 - ($r+$s)x + $r*$s)")->reduce; - $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(abs(x-$s))")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $r = random(-9,-1,1); - $c = random(2,5,1); - $A = non_zero_random(-9,9,1); - if($envir{problemSeed}==1){$r=-2;$c=3;$A=1;}; - $B = -$A; - $s = -$r; - Context()->noreduce('(-x)/y'); - $f = Formula("($A($r-$s))/($c x^2 - $c*($r+$s)x + $c*$r*$s)")->reduce; - Context("Fraction"); - $Afrac = Fraction($A,$c); - $Bfrac = Fraction($B,$c); - $F = FormulaUpToConstant("$Afrac ln(abs(x-$r)) + $Bfrac ln(abs(x-$s))")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $r = non_zero_random(-9,9,1); - $m = list_random(-9..-2,2..9); - if($envir{problemSeed}==1){$r=-1;$m=3;}; - $f = Formula("((2*$m)x+(1-$m*$r))/($m x^2 + (1 - $r*$m)x - $r)")->reduce; - Context("Fraction"); - $F = FormulaUpToConstant("ln(abs(x-$r)) + ln(abs($m x + 1))")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $r = non_zero_random(-9,9,1); - ($A,$B) = random_subset(2,1..9); - if($envir{problemSeed}==1){$r=-5;$A=1;$B=2;}; - $f = Formula("($A x + ($B-$A*$r))/(x-$r)^2")->reduce; - Context("Fraction"); - $F = FormulaUpToConstant("$A ln(abs(x-$r)) - $B/(x - $r)")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $r = non_zero_random(-9,9,1); - ($A,$B) = random_subset(2,1..9); - if($envir{problemSeed}==1){$r=-8;$A=-3;$B=4;}; - Context()->noreduce('(-x)-y'); - $f = Formula("($A x + ($B-$A*$r))/(x-$r)^2")->reduce; - Context("Fraction"); - $F = FormulaUpToConstant("$A ln(abs(x-$r)) - $B/(x - $r)")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $r = non_zero_random(-9,9,1); - ($A,$B,$C) = random_subset(3,1..9); - if($envir{problemSeed}==1){$r=-1;$A=7;$B=2;$C=5}; - $f = Formula("(($A+$B)x^2 + (-2*$A*$r-$B*$r-$C)x + $A*($r)^2)/(x (x-$r)^2)")->reduce; - Context("Fraction"); - $F = FormulaUpToConstant("$A ln(abs(x)) + $B ln(abs(x-$r)) + $C/(x-$r)")->reduce; - $F->{test_at} = [[-0.25],[0.25],[$r-0.25],[$r+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($r,$s) = random_subset(2,-9..-1,1..9); - ($m,$n) = random_subset(2,-9..-2,2..9); - ($A,$B,$C) = random_subset(3,-9..-1,1..9); - if($envir{problemSeed}==1){$r=1;$s=-3;$m=-2;$n=3;$A=5;$B=2;$C=-1}; - Context()->noreduce('(-x)-y','(-x)+y'); - $f = Formula("(($A+$B*$m+$C*$m)x^2 + ($A*$m*(-$r-$s)+$B*($n-$m*$s)+$C*($n-$r*$m))x + ($A*$m*$r*$s-$B*$n*$s-$C*$r*$n))/((x-$r)(x-$s)($n + $m x))")->reduce; - Context("Fraction"); - $F = FormulaUpToConstant("$A ln(abs($m*x+$n)) + $B ln(abs(x-$r)) + $C ln(abs(x-$s))")->reduce; - $F->{test_at} = [[$s-0.1],[$s+0.1],[$r-0.1],[$r+0.1],[-$n/$m+0.1],[-$n/$m-0.1]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($m,$n,$p) = random_subset(3,2..9); - ($A,$B,$C) = random_subset(3,1..3); - $A = $A*random(-1,1,2); - $B = $B*random(-1,1,2); - $C = $C*random(-1,1,2); - $pm1 = list_random(-1,1); - do {$r = non_zero_random(-3,3,1);} until (($pm1 - $B*$r) % $A == 0 and ($pm1 - $B*$r) != 0); - $s = ($pm1 - $B*$r)/$A; - $t = -$C*$r*$s/$pm1; - if($envir{problemSeed}==1){$m=5;$n=3;$p=7;$r=-1;$s=-1;$t=3;$A=-1;$B=2;$C=3}; - $f = Formula("(($A*$n*$p+$B*$m*$p+$C*$m*$n)x^2 + ($A*($n*$t+$p*$s)+$B*($m*$t+$p*$r)+$C*($m*$s+$n*$r))x + ($A*$s*$t+$B*$r*$t+$C*$r*$s))/(($p x+$t)($m x+$r)($n x+$s))")->reduce; - Context("Fraction"); - $a =Fraction($A,$m); - $b =Fraction($B,$n); - $c =Fraction($C,$p); - $F = FormulaUpToConstant("$a ln(abs($m x+$r)) + $b ln(abs($n x+$s)) + $c/($p x+$t)")->reduce; - $F->{test_at} = [[-$r/$m-0.1],[-$r/$m+0.1],[-$s/$n-0.1],[-$s/$n+0.1],[-$t/$p-0.1],[-$t/$p+0.1]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($r,$s) = random_subset(2,-9..-1,1..9); - ($A,$B) = random_subset(2,-3..-1,1..3); - if($envir{problemSeed}==1){$r=1;$s=-2;$A=1;$B=-1;}; - $f = Formula("(x^2+($A+$B-$r-$s)x+($r*$s-$A*$s-$B*$r))/(x^2-($r+$s)x+$r*$s)")->reduce; - Context("Fraction"); - $F = FormulaUpToConstant("x + $A ln(abs(x-$r)) + $B ln(abs(x-$s))")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - ($r,$s) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$r=-4;$s=5;}; - $f = Formula("x^3/(x^2-($r+$s)x+$r*$s)")->reduce; - Context("Fraction"); - $A = Fraction($r*($r**2+$s**2+$r*$s) - ($r+$s)*$r*$s, $r-$s); - $B = Fraction(-$s*($r**2+$s**2+$r*$s) + ($r+$s)*$r*$s, $r-$s); - $F = FormulaUpToConstant("1/2 x^2 + ($r+$s) x + $A ln(abs(x-$r)) + $B ln(abs(x-$s))")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25],[$s-0.25],[$s+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $h = non_zero_random(-4,4,1); - $k = random(1,5,1); - $c = random(2,4,1); - if($envir{problemSeed}==1){$h=1;$k=2;$c=2;}; - $f = Formula("($c x^2+(-2*$h*$c) x+$c*(($h)^2+$k))/(x^2+(-2*$h) x+(($h)^2+$k))")->reduce; - Context("Fraction"); - $F = FormulaUpToConstant("$c x")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $h = non_zero_random(-4,4,1); - $k = list_random(2,3,5); - if($envir{problemSeed}==1){$h=-1;$k=2;}; - $f = Formula("1/(x^3+(-2*$h) x^2+(($h)^2+$k)x)")->reduce; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $A = Fraction(1,($h)**2+$k); - $B = -$A/2; - $C = 2*$h*($A+$B); - $F = FormulaUpToConstant("$A ln(abs(x)) + $B ln(x^2+(-2*$h) x+(($h)^2+$k)) + $C/sqrt($k) atan((x-$h)/sqrt($k))")->reduce; - $F->{test_at} = [[-0.25],[0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $h = non_zero_random(-4,4,1); - $k = list_random(2,3,5,6); - ($m,$b) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$h=-2;$k=6;$m=-3;$b=-5;}; - $f = Formula("(x^2+(-2*$h+$m) x+(($h)^2+$k+$b))/(x^2+(-2*$h) x+(($h)^2+$k))")->reduce; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $A = Fraction($m,2); - $B = $b + 2*$A*$h; - $F = FormulaUpToConstant("x + $A ln(x^2+(-2*$h) x+(($h)^2+$k)) + $B/sqrt($k) atan((x-$h)/sqrt($k))")->reduce; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $a = random(2,9,1); - ($b,$c) = random_subset(2,-9..-1,1..9); - do {$r = non_zero_random(-9,9,1);} until ($a*($r)**2 + $b*$r + $c != 0); - $B = non_zero_random(-5,5,1); - if($envir{problemSeed}==1){$a=3;$b=5;$c=-1;$r=-1;$B=2;}; - $f = Formula("($a*(2+$B)x^2 + ($b-2*$a*$r+$B*$b)x + ($B*$c-$b*$r))/((x - $r)($a x^2 + $b x + $c))")->reduce; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $F = FormulaUpToConstant("ln(abs($a x^2 + $b x + $c)) + $B ln(abs(x-$r))")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25],[-$b/(2*$a)],[6]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $h = non_zero_random(-4,4,1); - $r = non_zero_random(-9,9,1); - $A = random(2,5,1); - $B = random(2,5,1); - $C = non_zero_random(-5,5,1); - if($envir{problemSeed}==1){$h=-3;$r=3;$A=2;$B=2;$C=-4;}; - $f = Formula("(($A+2*$B)x^2 + (-2*$A*$h - 2*$B*$r + $C - 2*$B*$h)x + ($A*($h)**2 + $A - $r*$C + 2*$r*$B*$h))/((x-$r)(x^2+(-2*$h) x+(($h)^2+1)))")->reduce; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(x^2+(-2*$h) x+(($h)^2+1)) + $C atan(x-$h)")->reduce; - $F->{test_at} = [[-0.25],[0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $k = list_random(4,9,16,25); - $r = non_zero_random(-9,9,1); - $a = random(2,9,1); - ($b,$c) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$k=9;$r=-1;$a=2;$b=1;$c=1;}; - $sqrtk = sqrt($k); - Context("Fraction"); - $C = Fraction(($c*$r - $k*$a*$r - $b*$k)/-$sqrtk/(($r)**2+$k)); - $B = ($b*$k - $C*$k*$sqrtk)/(-2*$r*$k); - $A = $a - 2*$B; - $f = Formula("(($A+2*$B)x^2 + (-2*$B*$r + $C*$sqrtk)x + ($A*$k - $r*$C*$sqrtk))/((x-$r)(x^2+$k))")->reduce; - Context()->flags->set(reduceConstantFunctions=>0); - $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(x^2+$k) + $C atan(x/$sqrtk)")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $h = non_zero_random(-4,4,1); - $k = list_random(4,9,16,25,36); - $r = non_zero_random(-9,9,1); - $A = non_zero_random(-5,5,1); - $C = non_zero_random(-5,5,1); - if($envir{problemSeed}==1){$h=-1;$k=16;$r=7;$A=-2;$C=1;}; - $sqrtk = sqrt($k); - Context("Fraction"); - $B = Fraction(1-$A,2); - $C = Fraction($C,2); - $f = Formula("(($A+2*$B)x^2 + (-2*$A*$h - 2*$B*$r + $C*$sqrtk - 2*$B*$h)x + ($A*($h)**2 + $A*$k - $r*$C*$sqrtk + 2*$r*$B*$h))/((x-$r)(x^2+(-2*$h) x+(($h)^2+$k)))")->reduce; - Context()->flags->set(reduceConstantFunctions=>0); - $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(x^2+(-2*$h) x+(($h)^2+$k)) + $C atan((x-$h)/$sqrtk)")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $h = non_zero_random(-4,4,1); - $k = list_random(2,3,5,6,7,10,11,13,14,15); - $r = non_zero_random(-9,9,1); - $A = non_zero_random(-5,5,1); - $B = non_zero_random(-5,5,1); - $C = non_zero_random(-5,5,1); - if($envir{problemSeed}==1){$h=1;$k=10;$r=9;$A=3;$B=3;$C=6;}; - Context("Fraction"); - $f = Formula("(($A+2*$B)x^2 + (-2*$A*$h - 2*$B*$r + $C - 2*$B*$h)x + ($A*($h)**2 + $A*$k - $r*$C + 2*$r*$B*$h))/((x-$r)(x^2+(-2*$h) x+(($h)^2+$k)))")->reduce; - Context()->flags->set(reduceConstantFunctions=>0); - $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(x^2+(-2*$h) x+(($h)^2+$k)) + $C/sqrt($k) atan((x-$h)/sqrt($k))")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     - - - - - $h = non_zero_random(-9,9,1); - $k = list_random(2,3,5); - $r = non_zero_random(-9,9,1); - $A = non_zero_random(-5,5,1); - $B = non_zero_random(-5,5,2); - $C = non_zero_random(-15,15,1); - if($envir{problemSeed}==1){$h=-5;$k=2;$r=-2;$A=5;$B=1;$C=-12;}; - Context("Fraction"); - $B = Fraction($B,2); - $f = Formula("(($A+2*$B)x^2 + (-2*$A*$h - 2*$B*$r + $C - 2*$B*$h)x + ($A*($h)**2 + $A*$k - $r*$C + 2*$r*$B*$h))/((x-$r)(x^2+(-2*$h) x+(($h)^2+$k)))")->reduce; - Context()->flags->set(reduceConstantFunctions=>0); - $F = FormulaUpToConstant("$A ln(abs(x-$r)) + $B ln(x^2+(-2*$h) x+(($h)^2+$k)) + $C/sqrt($k) atan((x-$h)/sqrt($k))")->reduce; - $F->{test_at} = [[$r-0.25],[$r+0.25]]; - - -

    - \ds \int \, dx -

    -

    - -

    -     -     -     -     - - - -

    - Evaluate the definite integral. -

    -     - - - - - $a = 1; - $b = 2; - ($r,$s) = num_sort(random_subset(2,-9..-1,3..9)); - ($A,$B) = random_subset(2,1..9); - if($envir{problemSeed}==1){$r=-3;$s=-2;$A=3;$B=5;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("(($A+$B)x-($A*$s+$B*$r))/((x-$r)(x-$s))")->reduce; - $arg = Fraction(abs($b-$r)**$A * abs($b-$s)**$B * abs($a-$r)**(-$A) * abs($a-$s)**(-$B)); - $answer = Formula("ln($arg)"); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $a = 0; - $b = random(2,9,1); - $r = random(-9,-1,1); - ($m,$n) = random_subset(2,2..5); - ($A,$B) = random_subset(2,-9..-1,1..9); - if($envir{problemSeed}==1){$b=5;$r=-4;$m=3;$n=2;$A=5;$B=-1;}; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $f = Formula("(($A*$m+$B)x+($A*$n-$B*$r))/(($m x + $n)(x-$r))")->reduce; - $F = Formula("$A ln(abs(x-$r)) + $B/$m ln(abs($m x+$n))")->reduce; - $answer = $F->eval(x=>$b) - $F->eval(x=>$a); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - $a = -1; - $b = 1; - $h = non_zero_random(-4,4,1); - $r = list_random(-10..-2,2..10); - $A = random(2,5,1); - $B = random(2,5,1); - $C = non_zero_random(-5,5,1); - if($envir{problemSeed}==1){$h=-2;$r=10;}; - $f = Formula("(x^2 + (1-2*$h)x + (($h)^2+1-$r))/((x-$r)(x^2+(-2*$h) x+(($h)^2+1)))")->reduce; - Context("Fraction"); - Context()->flags->set(reduceConstantFunctions=>0); - $F = Formula("ln(abs(x-$r)) + atan(x-$h)")->reduce; - $frac=Fraction($b-$r,$a-$r); - $Fb = Formula("atan($b-$h)"); - if ($b-$h == 0) {$Fb = 0;}; - if ($b-$h == 1) {$Fb = Formula("pi/4");}; - if ($b-$h == -1) {$Fb = Formula("-pi/4");}; - $Fa = Formula("atan($a-$h)"); - if ($a-$h == 0) {$Fa = 0;}; - if ($a-$h == 1) {$Fa = Formula("pi/4");}; - if ($a-$h == -1) {$Fa = Formula("-pi/4");}; - $Fba = $Fb - $Fa; - $Fba = $Fb if ($Fa == 0); - $Fba = -$Fa if ($Fb == 0); - $answer = Formula("ln($frac) + $Fba"); - - -

    - \ds \int_{}^{} \, dx -

    -

    - -

    -     -     -     - - - - - - $a = 0; - $b = 1; - $F = Formula("-(2x+1)/(2*(x+1)^2)"); - Context("Fraction"); - $r = Fraction($F->eval(x=>$b) - $F->eval(x=>$a)); - - -

    - \ds \int_{0}^1 \frac{x}{(x+1)(x^2+2x+1)}\, dx -

    -

    - -

    -     -     -     -     -     -     -
    -
    - Hyperbolic Functions - -

    - The hyperbolic functions - are a set of functions that have many applications to mathematics, physics, - and engineering. - Among many other applications, - they are used to describe the formation of satellite rings around planets, - to describe the shape of a rope hanging from two points, - and have application to the theory of special relativity. - This section defines the hyperbolic functions and describes many of their properties, - especially their usefulness to calculus. -

    - -

    - These functions are sometimes referred to as the - hyperbolic trigonometric functions as there are many, - many connections between them and the standard trigonometric functions. - - demonstrates one such connection. - Just as cosine and sine are used to define points on the circle defined by x^2+y^2=1, - the functions hyperbolic cosine and - hyperbolic sine - are used to define points on the hyperbola x^2-y^2=1. -

    - - - -
    - Using trigonometric functions to define points on a circle and hyperbolic functions to define points on a hyperbola. The area of the shaded regions are included in them. - - -
    - - - - - Graph showing cosine and sine function used to define points on a circle. - - -

    - The y and the x axes are drawn from -1 to - 1. The function x^2+y^2=1 represents a circle of - radius 1 and center at origin. -

    -

    - A sector in the circle is shaded, it is present in the first - quadrant and is drawn with one side on the x axis. - It has an angle \theta/2 drawn from the x axis - and is marked inside the sector. The points between which the - sector is drawn on the circumference are (1,0) and - (\cos(\theta), \sin(\theta)). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis equal image, - axis on top, - extra x tick labels={$a$,$b$}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.4 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:45] ({cos(x)},{sin(x)}) -- (axis cs:0,0) -- cycle; - \addplot [firstcurvestyle,domain=0:360,samples=80] ({cos(x)},{sin(x)}); - - \filldraw (axis cs:.707,.707) circle (1pt) node [shift={(20pt,12pt)}] { ($\cos(\theta) $,$\sin(\theta) $)}; - - \draw (axis cs:.6,.25) node { $\ds\frac{\theta}{2}$}; - \draw (axis cs:-.75,1) node { $x^2+y^2=1$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - Graph showing hyperbolic cosine and hyperbolic sine function used to define points on a hyperbola. - - -

    - The y and the x axes are both drawn from -2 to 2. - The graph of function x^2 -y^2=1 has two x intercepts at x=1 - and x=-1. The function represents a hyperbola and has two conic sections - facing opposite to each other, opening along the positive and negative x - axis with vertices at the x intercepts. An angle of \theta/2 is - marked at the origin starting from the x axis, it is drawn from point - (1,0) to (\cos(\theta), \sin(\theta)) on the conic section to the - right of the y axis. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis equal image, - axis on top, - ymin=-3.1,ymax=3.1, - xmin=-3.1,xmax=3.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1.6] ({cosh(x)},{sinh(x)}) -- (axis cs:0,0) -- cycle; - - \addplot [firstcurvestyle,domain=-2:2,samples=40] ({cosh(x)},{sinh(x)}); - \addplot [firstcurvestyle,domain=-2:2,samples=40] ({-cosh(x)},{sinh(x)}); - - \filldraw (axis cs:2.577,2.376) circle (1pt) node [left] { ($\cosh(\theta) $,$\sinh(\theta) $)}; - - \draw (axis cs:.73,.32) node { $\ds\frac{\theta}{2}$}; - \draw (axis cs:-1.75,2.75) node { $x^2-y^2=1$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    -     - - - The Hyperbolic Functions and their Properties -

    - We begin with their definition. -

    - - - Hyperbolic Functions - -

    -

      -
    1. -

      - \ds \cosh(x) = \frac{e^x+e^{-x}}2 - - hyperbolic functiondefinition - -

      -
      2. - -
    2. \sinh(x) = \frac{e^x-e^{-x}}2
      3. - -
    3. \tanh(x) = \frac{\sinh(x) }{\cosh(x) }
      4. - -
    4. \sech(x) = \frac{1}{\cosh(x) }
      5. - -
    5. \csch(x) = \frac{1}{\sinh(x) }
      6. - -
    6. \coth(x) = \frac{\cosh(x) }{\sinh(x) }
      7. -
    -

    -     -     - - -

    - These hyperbolic functions are graphed in - and . -

    - -

    - In the graph of \cosh(x) in , - the graphs of e^x/2 and - e^{-x}/2 are included with dashed lines. - In the graph of \sinh(x) in , - the graphs of e^x/2 and - - - e^{-x}/2 are included with dashed lines. - As x gets large, - \cosh(x) and \sinh(x) each act like e^x/2; - when x is a large negative number, - \cosh(x) acts like e^{-x}/2 whereas - \sinh(x) acts like -e^{-x}/2. -

    - -
    - Graphs of \sinh(x) and \cosh(x) - - - -
    - - - - - Graph of hyperbolic cosine function. - - -

    - The y axis is drawn from -10 to 10 and the x - axis is drawn from -3 to 3. The function f(x)=\cosh(x) - is shown as a U shaped curve that opens upwards along the positive y - axis, it is symmetrical about the y axis. The graphs of e^{x}/2 - and e^{-x}/2 are also included. -

    -

    - For large values of x the function f(x)=\cosh(x) is approximately equal - to e^{x}/2. From left to right, the function e^{x}/2 appears to - coincide with the x axis in the fourth quadrant, it has a positive slope, - it is some distance apart from f(x) then it rises to coincide with f(x) - after x=1. -

    -

    - For large negative values of x, f(x) becomes equal to the function - e^{-x}/2. From right to left, the function e^{-x}/2 appears to start - from the first quadrant and enters the second quadrant with a positive slope. It - coincides with f(x) after approximately x= -1. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-11,ymax=11, - xmin=-3.5,xmax=3.5, - scaled ticks=false - ] - - \addplot+ [domain=-3.1:3.1] {cosh(x)} node [pos=0.9, above left] { $f(x)=\cosh(x)$}; - \addplot+ [domain=-3:3] {e^x/2} node [pos=0, above right] { $e^x/2$}; - \addplot [secondcurvestyle,domain=-3:3] {e^(-x)/2} node [pos=0.9, above right] { $e^{-x}/2$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - - -
    - - - - - Graph of hyperbolic sine function. - - -

    - The y axis is drawn from -10 to 10 and the x axis is drawn - from -3 to 3. The function f(x) = \sinh(x), e^{x}/2 - and e^-{x}/2 is also shown in the graph. -

    -

    - From left to right, the function e^{x}/2 starts in the second quadrant - and gets close to the x axis, it gains the positive slope into the first - quadrant. The function f(x) in the first quadrant, starts at the origin and - rises with a positive slope after a bend.It is separated by a small distance from - e^{x}/2, after approximately x=1 the function e^{x}/2 coincides - with f(x). -

    -

    - From right to left, the function e^{-x}/2 starts in the fourth quadrant and - coincides with the x axis, it moves downward after entering the third quadrant. - The function f(x) in the third quadrant, starts at the origin and moves downwards - from right to left. It is separated by a small distance from e^{-x}/2 after - approximately x=-1 the function e^{-x}/2 coincides with f(x). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-11,ymax=11, - xmin=-3.5,xmax=3.5, - scaled ticks=false - ] - - \addplot+ [domain=-3.1:3.1] {0.5*(e^x-e^(-x))} node [pos=0.9, above left] { $f(x)=\sinh(x)$}; - \addplot+ [domain=-3:3] {0.5*(e^x)} node [pos=0, above right] { $e^x/2$}; - \addplot+ [secondcurvestyle,domain=-3:3] {-0.5*(e^(-x)} node [pos=0.9, above right] { $-e^{-x}/2$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    - - - -

    - In Figure , - notice the domains of \tanh(x) and - \sech(x) are (-\infty,\infty), - whereas both \coth(x) and - \csch(x) have vertical asymptotes at x=0. - Also note the ranges of these functions, - especially \tanh(x): - as x\to\infty, both \sinh(x) and - \cosh(x) approach e^{-x}/2, - hence \tanh(x) approaches 1. -

    - -
    - Graphs of \tanh(x), \coth(x), \csch(x) and \cosh(x) - - -
    - - - - - Graph of hyperbolic tangent and hyperbolic cotangent functions. - - -

    - The y axis is drawn from -2 to 2 and the x axis - is drawn from -3 to 3. The functions \tanh(x) and - \coth(x) are shown. There are two lines drawn at y=-1 and y=1. -

    -

    - The \tanh(x) function is drawn in the third and the first quadrant. - In the first quadrant the function starts at the origin and gets a positive - slope then after x=2 it becomes parallel to the x axis at - y=1. In the third quadrant the function starts at the origin and - decreases until x=-2 after which it becomes parallel to the x - axis as y=-1. -

    -

    - The \coth(x) function is drawn in the first and the third quadrants. - It is hyperbolic in shape with the two parts being symmetrical about the - axis y=-x. It has a horizontal asymptote at x=0. This function - coincides with the \tanh(x) curve after x=2 and extends along - the positive x axis and x=-2 and extends further along the - negative x axis. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-3.5,ymax=3.5, - xmin=-3.5,xmax=3.5, - scaled ticks=false - ] - - \addplot+ [domain=-3:3] {(e^x-e^(-x))/(e^x+e^(-x))} node [pos=0.60,pin={120:$ \tanh(x)$},inner sep=0pt] {}; - - \addplot+ [solid,domain=-3:-.33] {(e^x+e^(-x))/(e^x-e^(-x))}; - \addplot [secondcurvestyle,solid,domain=.33:3,samples=30] {(e^x+e^(-x))/(e^x-e^(-x))} node [pos=0.5,pin={45:$ \coth(x)$},inner sep=0pt] {}; - \addplot+ [lineseg,dotted,domain=-3:3] {1}; - \addplot+ [lineseg,dotted,domain=-3:3] {-1}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - - -
    - - - - - Graph of hyperbolic secant and cosecant functions. - - -

    - The y axis is drawn from -2 to 2 and the x - axis is drawn from -3 to 3. The functions \sech(x) - and \csch(x) are shown. -

    -

    - The \sech(x) is drawn in the second and the first quadrant. - From point (1,0) the function slowly decreases moving left to - right, almost touching the x axis at x=3. It is symmetrical - about the x axis and in the third quadrant it decreases from - (1,0), moving from right to left, and almost touches the x axis at - x=-3. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-3.5,ymax=3.5, - xmin=-3.5,xmax=3.5, - scaled ticks=false - ] - - \addplot+ [domain=-3:3] {(2/(e^x+e^(-x))} node [pos=0.50,pin={120:$ \sech(x)$},inner sep=0pt] {}; - \addplot+ [solid,domain=-3:-.33] {2/(e^x-e^(-x))}; - \addplot+ [secondcurvestyle,solid,domain=.33:3,samples=30] {2/(e^x-e^(-x))} node [pos=0.5,pin={45:$ \csch(x)$},inner sep=0pt] {}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -     -
    - -

    - The following example explores some of the properties of these functions that bear remarkable resemblance to the properties of their trigonometric counterparts. -

    - - - Exploring properties of hyperbolic functions - -

    - Use - to rewrite the following expressions. -

    - -

    -

      -
    1. \cosh^2(x) -\sinh^2(x)
      2. - -
    2. \tanh^2(x) +\sech^2(x)
      3. - -
    3. 2\cosh(x) \sinh(x)
      4. - -
    4. \frac{d}{dx}\big(\cosh(x) \big)
      5. - -
    5. \frac{d}{dx}\big(\sinh(x) \big)
      6. - -
    6. \frac{d}{dx}\big(\tanh(x) \big)
      7. -
    -

    -     - -

    -

      -
    1. - -

      - By - - \cosh^2(x) -\sinh^2(x) \amp = \left(\frac{e^x+e^{-x}}2\right)^2 -\left(\frac{e^x-e^{-x}}2\right)^2 - \amp = \frac{e^{2x}+2e^xe^{-x} + e^{-2x}}4 - \frac{e^{2x}-2e^xe^{-x} + e^{-2x}}4 - \amp = \frac44=1 - . - So \cosh^2(x) -\sinh^2(x) =1. -

      -
      2. - -
    2. -

      - Again, use - - \tanh^2(x) +\sech^2(x) \amp =\frac{\sinh^2(x) }{\cosh^2(x) } + \frac{1}{\cosh^2(x) } - \amp= \frac{\sinh^2(x) +1}{\cosh^2(x) }\qquad \text{ Now use identity from } Part - \amp = \frac{\cosh^2(x) }{\cosh^2(x) } = 1 - . - So \tanh^2(x) +\sech^2(x) =1. -

      -
      3. - -
    3. -

      - Again, use - - 2\cosh(x) \sinh(x) \amp = 2\left(\frac{e^x+e^{-x}}2\right)\left(\frac{e^x-e^{-x}}2\right) - \amp = 2 \cdot\frac{e^{2x} - e^{-2x}}4 - \amp = \frac{e^{2x} - e^{-2x}}2 = \sinh(2x) - . - Thus 2\cosh(x) \sinh(x) = \sinh(2x). -

      -
      4. - -
    4. -

      - Again, use - - \frac{d}{dx}\big(\cosh(x) \big) \amp = \frac{d}{dx}\left(\frac{e^x+e^{-x}}2\right) - \amp = \frac{e^x-e^{-x}}2 - \amp = \sinh(x) - - So \frac{d}{dx}\big(\cosh(x) \big) = \sinh(x). -

      -
      5. - -
    5. -

      - Apply derivatives to : - - \frac{d}{dx}\big(\sinh(x) \big) \amp = \frac{d}{dx}\left(\frac{e^x-e^{-x}}2\right) - \amp = \frac{e^x+e^{-x}}2 - \amp = \cosh(x) - . - So \frac{d}{dx}\big(\sinh(x) \big) = \cosh(x). -

      -
      6. - -
    6. -

      - Apply derivatives to : - - \frac{d}{dx}\big(\tanh(x) \big) \amp = \frac{d}{dx}\left(\frac{\sinh(x) }{\cosh(x) }\right) - \amp = \frac{\cosh(x) \cosh(x) - \sinh(x) \sinh(x) }{\cosh^2(x) } - \amp = \frac{1}{\cosh^2(x) } - \amp = \sech^2(x) - . - So \frac{d}{dx}\big(\tanh(x) \big) = \sech^2(x). -

      -
      7. -
    -

    -     - -     - -

    - The following Key Idea summarizes many of the important identities relating to hyperbolic functions. - Each can be verified by referring back to . -

    - - - Useful Hyperbolic Function Properties - - - - - Basic Identities -
      -
    1. -

      - \cosh^2(x) -\sinh^2(x) =1 - - hyperbolic functionidentities - - hyperbolic functionderivatives - - hyperbolic functionintegrals - - derivativehyperbolic funct. - - integrationhyperbolic funct. - -

      -
      2. - -
    2. \tanh^2(x) +\sech^2(x) =1
      3. - -
    3. \coth^2(x) -\csch^2(x) = 1
      4. - -
    4. \cosh(2x) =\cosh^2(x) +\sinh^2(x)
      5. - -
    5. \sinh(2x) = 2\sinh(x) \cosh(x)
      6. - -
    6. \cosh^2(x) = \frac{\cosh(2x) +1}{2}
      7. - -
    7. \sinh^2(x) =\frac{\cosh(2x) -1}{2}
      8. -
    -     - - - Derivatives -
      -
    1. \frac{d}{dx}\big(\cosh(x) \big) = \sinh(x)
      2. - -
    2. \frac{d}{dx}\big(\sinh(x) \big) = \cosh(x)
      3. - -
    3. \frac{d}{dx}\big(\tanh(x) \big) = \sech^2(x)
      4. - -
    4. \frac{d}{dx}\big(\sech(x) \big) = -\sech(x) \tanh(x)
      5. - -
    5. \frac{d}{dx}\big(\csch(x) \big) = -\csch(x) \coth(x)
      6. - -
    6. \frac{d}{dx}\big(\coth(x) \big) = -\csch^2(x)
      7. -
    -     - - - Integrals -
      -
    1. \int \cosh(x) \, dx = \sinh(x) +C
      2. - -
    2. \int \sinh(x) \, dx = \cosh(x) +C
      3. - -
    3. \int \tanh(x) \, dx = \ln(\cosh(x) ) +C
      4. - -
    4. \int \coth(x) \, dx = \ln\abs{\sinh(x) \,}+C
      5. -
    -     - -     - - - -

    - We practice using . -

    - - - Derivatives and integrals of hyperbolic functions - -

    - Evaluate the following derivatives and integrals. -

    - -

    -

      -
    1. \frac{d}{dx}\big(\cosh(2x) \big)
      2. - -
    2. \int \sech^2(7t-3)\,dt
      3. - -
    3. \int_0^{\ln(2) } \cosh(x) \, dx
      4. -
    -

    -     - -

    -

      -
    1. -

      - Using the Chain Rule directly, - we have \frac{d}{dx} \big(\cosh(2x) \big) = 2\sinh(2x). - - Just to demonstrate that it works, - let's also use the Basic Identity found in : - \cosh(2x) = \cosh^2(x) +\sinh^2(x). - - \frac{d}{dx}\big(\cosh(2x) \big) \amp = \frac{d}{dx}\big(\cosh^2(x) +\sinh^2(x) \big) - \amp = 2\cosh(x) \sinh(x) + 2\sinh(x) \cosh(x) - \amp = 4\cosh(x) \sinh(x) - . - Using another Basic Identity, - we can see that 4\cosh(x) \sinh(x) = 2\sinh(2x). - We get the same answer either way. -

      -
      2. - -
    2. -

      - We employ substitution, with u = 7t-3 and du = 7dt. - Applying Key Ideas - and we have: - - \int \sech^2(7t-3)\,dt = \frac17\tanh(7t-3) + C - . -

      -
      3. - -
    3. -

      - - \int_0^{\ln(2) } \cosh(x) \, dx \amp = \sinh(x) \Big|_0^{\ln(2) } - \amp = \sinh(\ln(2) ) - \sinh(0) - \amp = \sinh(\ln(2) ) - . - We can simplify this last expression as - \sinh(x) is based on exponentials: - - \sinh(\ln(2) ) \amp = \frac{e^{\ln(2) }-e^{-\ln(2) }}2 - \amp = \frac{2-1/2}{2} - \amp = \frac34 - . -

      -
      4. -
    -

    -     - -     -     - - - Inverse Hyperbolic Functions -

    - Just as the inverse trigonometric functions are useful in certain applications, - the inverse hyperbolic functions are useful with others. - - shows restriction on the domain of - \cosh(x) to make the function one-to-one and the resulting domain and range of its inverse function. - Since \sinh(x) is already one-to-one, - no domain restriction is needed as shown in . - Since \sech(x) is not one to one, - it also needs a restricted domain in order to be invertible. - - shows the graph of \sech^{-1}(x). - You should carefully compare the graph of this function to the graph given in - to see how this inverse was constructed. - The rest of the hyperbolic functions area already one-to-one and need no domain restrictions. - Their graphs are also shown in . - - hyperbolic functioninverse - -

    - -

    - Because the hyperbolic functions are defined in terms of exponential functions, - their inverses can be expressed in terms of logarithms as shown in . - It is often more convenient to refer to - \sinh^{-1}(x) than to \ln\big(x+\sqrt{x^2+1}\big), - especially when one is working on theory and does not need to compute actual values. - On the other hand, when computations are needed, - technology is often helpful but many hand-held calculators lack a convenient - \sinh^{-1}(x) button. - (Often it can be accessed under a menu system, but not conveniently.) - In such a situation, the logarithmic representation is useful. - The reader is not encouraged to memorize these, - but rather know they exist and know how to use them when needed. -

    - - - - - Domains and ranges of the hyperbolic and inverse hyperbolic functions - - - - Function - Domain - Range - Function - Domain - Range - - - \cosh(x) - [0,\infty) - [1,\infty) - \cosh^{-1}(x) - [1,\infty) - [0,\infty) - - - \sinh(x) - (-\infty,\infty) - (-\infty,\infty) - \sinh^{-1}(x) - (-\infty,\infty) - (-\infty,\infty) - - - \tanh(x) - (-\infty,\infty) - (-1,1) - \tanh^{-1}(x) - (-1,1) - (-\infty,\infty) - - - \sech(x) - [0,\infty) - (0,1] - \sech^{-1}(x) - (0,1] - [0,\infty) - - - \csch(x) - (-\infty,0) \cup (0,\infty) - (-\infty,0) \cup (0,\infty) - \csch^{-1}(x) - (-\infty,0) \cup (0,\infty) - (-\infty,0) \cup (0,\infty) - - - \coth(x) - (-\infty,0) \cup (0,\infty) - (-\infty,-1) \cup (1,\infty) - \coth^{-1}(x) - (-\infty,-1) \cup (1,\infty) - (-\infty,0) \cup (0,\infty) - - - -
    - -
    - Graphs of the hyperbolic functions (with restricted domains) and their inverses - - - -
    - - - - - Graph of hyperbolic cosine function and its inverse. - - -

    - The y and the x axes are drawn from 0 to 10. - The functions y=\cosh(x) and y=\cosh^{-1}(x) are shown. - They are symmetrical about the axis y=x. -

    -

    - From left to right, the function y=\cosh(x) starts at point - (0,1) then slowly rises from (1,1) then it rises up - steeply and it appears to run almost parallel to the y axis. - The function \cosh^{-1}(x) starts at point (1,0) and - curves up steeply until (2,1) then it rises very slowly and - appears to almost run parallel to the x axis. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.9,ymax=10.9, - xmin=-.9,xmax=10.9, - scaled ticks=false - ] - - \addplot+ [domain=0:3] {cosh(x)}; - \addplot+ [domain=0:3] (cosh(x),x); - - \addplot [dashed,domain=-.5:10] {x}; - - \draw (axis cs:8,4) node { $y=\cosh^{-1}(x) $}; - \draw (axis cs:5.6,10) node { $y=\cosh(x) $}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - - -
    - - - - - Graph of hyperbolic sine function and its inverse. - - -

    - The y and the x axes are drawn from -10 to - 10. The functions y=\sinh(x) and y=\sinh^{-1}(x). - The axis y=x is shown. -

    -

    - From left to right, the \sinh(x) function starts in the third - quadrant and it rises steeply, very closely to the y axis. It - crosses the origin along the y=x line, has a dip then increases - very steeply and closely to the y axis in the first quadrant. -

    -

    - From left to right, in the third quadrant, the \sinh{-1}(x) - function runs very closely to the x axis, it crosses the origin - along the y=x line and bends to move very closely to the x - axis in the first quadrant. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-10.9,ymax=10.9, - xmin=-10.9,xmax=10.9, - scaled ticks=false - ] - - \addplot+ [domain=-3:3] {sinh(x)}; - \addplot+ [domain=-3:3] (sinh(x),x); - - \addplot [dashed,domain=-10:10] {x}; - - \draw (axis cs:-6,7) node { $y=\sinh(x) $}; - \draw (axis cs:6,-5) node { $y=\sinh^{-1}(x) $}; - \draw[->,>=stealth] (axis cs:-3,7) -- (axis cs:2,7); - \draw[->,>=stealth] (axis cs:7,-3) -- (axis cs:7,2); - - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - - - -
    - - - - - Graph of inverse of hyperbolic tangent and inverse of hyperbolic cotangent functions. - - -

    - The y and the x axes are drawn from -3 to 3. - There are two functions drawn, \coth^{-1}(x) and \tanh{-1}(x) - along with two dashed lines x=-1 and x=1. -

    -

    - The \tanh{-1}(x) function is drawn in the third and the first quadrant. - From left to right, in the third quadrant the function is aligned with the line - x=-1 at around y=-2 it diverges to the right side of the line, it - crosses the origin then bends and merges with the line x=1 from its left - in the first quadrant. -

    -

    - The coth^{-1}(x) is also drawn in the third and the first quadrant. From - left to right, in the third quadrant, the function appears to be parallel to the - x axis; it diverges and bends down to join the line x=-1. In the - first quadrant, from left to right the function is along the line x=1, it - decreases and diverges from the line, there is a bend after x=2 after which - it becomes parallel to the x axis. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ytick={-2,2}, - ymin=-3.2,ymax=3.2, - xmin=-3.2,xmax=3.2, - scaled ticks=false - ] - - \addplot+ [domain=-3:3] (tanh(x),x); - \addplot+ [solid,smooth] coordinates {(1.005,2.997)(1.01,2.65)(1.02,2.31)(1.03,2.11)(1.04,1.97)(1.05,1.86)(1.06,1.77)(1.07,1.69)(1.08,1.63)(1.09,1.57)(1.1,1.52)(1.2,1.2)(1.3,1.02)(1.4,0.896)(1.5,0.805)(1.6,0.733)(1.7,0.675)(1.8,0.626)(1.9,0.585)(2.,0.549)(2.1,0.518)(2.2,0.49)(2.3,0.466)(2.4,0.444)(2.5,0.424)(2.6,0.405)(2.7,0.389)(2.8,0.374)(2.9,0.36)(3.,0.347)}; - \addplot [secondcurvestyle,solid,smooth] coordinates {(-3.,-0.347)(-2.9,-0.36)(-2.8,-0.374)(-2.7,-0.389)(-2.6,-0.405)(-2.5,-0.424)(-2.4,-0.444)(-2.3,-0.466)(-2.2,-0.49)(-2.1,-0.518)(-2.,-0.549)(-1.9,-0.585)(-1.8,-0.626)(-1.7,-0.675)(-1.6,-0.733)(-1.5,-0.805)(-1.4,-0.896)(-1.3,-1.02)(-1.2,-1.2)(-1.1,-1.52)(-1.09,-1.57)(-1.08,-1.63)(-1.07,-1.69)(-1.06,-1.77)(-1.05,-1.86)(-1.04,-1.97)(-1.03,-2.11)(-1.02,-2.31)(-1.01,-2.65)(-1.005,-2.997)}; - - \draw (axis cs:2.2,1.5) node {\tiny $y=\coth^{-1}(x) $}; - \draw (axis cs:2.2,-1.5) node {\tiny $y=\tanh^{-1}(x) $}; - \draw [->,>=stealth] (axis cs:1.8,-1.3) -- (axis cs:.2,-.2); - - \addplot [asymptote] coordinates {(-1,-5) (-1,5)}; - \addplot [asymptote] coordinates {(1,-5) (1,5)}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - Graph of inverse of hyperbolic cosecant and inverse of hyperbolic secant functions. - - -

    - The sec{-1}(x) is drawn only in the first quadrant.From left to right, - it starts very close to the y axis without touching it, at around - y=3. It moves away from the y axis while declining then gets - a small bend before making an x intercept at x=1. -

    -

    - The \csch{-1}(x) is drawn in the third and the first quadrant. In the - third quadrant from left to right the function appears to be parallel to the - x axis. It bends toward the negative y axis and comes very close - to it at y=-3. In the first quadrant from left to right, the function - appears to start very close to the y axis coinciding with the sech{-1}(x) - function. It has a negative slope and it moves down, gets a bend and runs parallel - to the x axis. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ytick={-3,-2,-1,1,2,3}, - ymin=-3.5,ymax=3.5, - xmin=-3.5,xmax=3.5, - scaled ticks=false - ] - - \addplot+ [smooth] coordinates {(0.1,2.993)(0.2,2.292)(0.3,1.874)(0.4,1.567)(0.5,1.317) - (0.55,1.205)(0.6,1.099)(0.65,0.9961)(0.7,0.8956)(0.75,0.7954)(0.8,0.6931)(0.85,0.5857)(0.9,0.4671)(0.95,0.323)(1.,0)}; - - \addplot+ [solid,smooth] coordinates {(-3.,-0.3275)(-2.9,-0.3383)(-2.8,-0.35)(-2.7,-0.3624)(-2.6,-0.3757)(-2.5,-0.39)(-2.4,-0.4055)(-2.3,-0.4221)(-2.2,-0.4402)(-2.1,-0.4598)(-2.,-0.4812)(-1.9,-0.5046)(-1.8,-0.5303)(-1.7,-0.5587)(-1.6,-0.5901)(-1.5,-0.6251)(-1.4,-0.6643)(-1.3,-0.7085)(-1.2,-0.7585)(-1.1,-0.8156)(-1.,-0.8814)(-0.9,-0.9578)(-0.8,-1.048)(-0.7,-1.154)(-0.6,-1.284)(-0.5,-1.444)(-0.4,-1.647)(-0.3,-1.919)(-0.2,-2.312)(-0.1,-2.998)}; - \addplot+ [secondcurvestyle,solid,smooth] coordinates {(0.1,2.998)(0.2,2.312)(0.3,1.919)(0.4,1.647)(0.5,1.444)(0.6,1.284)(0.7,1.154)(0.8,1.048)(0.9,0.9578)(1.,0.8814)(1.1,0.8156)(1.2,0.7585)(1.3,0.7085)(1.4,0.6643)(1.5,0.6251)(1.6,0.5901)(1.7,0.5587)(1.8,0.5303)(1.9,0.5046)(2.,0.4812)(2.1,0.4598)(2.2,0.4402)(2.3,0.4221)(2.4,0.4055)(2.5,0.39)(2.6,0.3757)(2.7,0.3624)(2.8,0.35)(2.9,0.3383)(3.,0.3275)}; - - \draw (axis cs:-1.2,.5) node {\tiny $y=\sech^{-1}(x) $}; - \draw (axis cs:-2.2,-1.6) node {\tiny $y=\csch^{-1}(x) $}; - \draw [->,>=latex] (axis cs:-.2,.5) -- (axis cs:0.7,.5); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -     -     -
    - - - Logarithmic definitions of Inverse Hyperbolic Functions -

    -

      -
    1. -

      - \ds\cosh^{-1}(x) =\ln\big(x+\sqrt{x^2-1}\big);\, x\geq1 - - hyperbolic functioninverse!logarithmic def. - -

      -
      2. - -
    2. \tanh^{-1}(x) = \frac12\ln\left(\frac{1+x}{1-x}\right);\, \abs{x}\lt 1
      3. - -
    3. \sech^{-1}(x) = \ln\left(\frac{1+\sqrt{1-x^2}}x\right);\, 0\lt x\leq1
      4. - -
    4. \sinh^{-1}(x) = \ln\big(x+\sqrt{x^2+1}\big)
      5. - -
    5. \coth^{-1}(x) = \frac12\ln\left(\frac{x+1}{x-1}\right);\, \abs{x} \gt 1
      6. - -
    6. \csch^{-1}(x) = \ln\left(\frac1x+\frac{\sqrt{1+x^2}}{\abs{x}}\right);\, x\neq0
      7. -
    -

    -     - -

    - The following Key Ideas give the derivatives and integrals relating to the inverse hyperbolic functions. - In , - both the inverse hyperbolic and logarithmic function representations of the antiderivative are given, - based on . - Again, these latter functions are often more useful than the former. - Note how inverse hyperbolic functions can be used to solve integrals we used Trigonometric Substitution to solve in . -

    - - - - - Derivatives Involving Inverse Hyperbolic Functions -

    -

      -
    1. -

      - \ds\frac{d}{dx}\big(\cosh^{-1}(x) \big) = \frac{1}{\sqrt{x^2-1}};\\ x \gt 1 - - derivativeinverse hyper. - - hyperbolic functioninverse!derivative - -

      -
      2. - -
    2. \frac{d}{dx}\big(\sinh^{-1}(x) \big) = \frac{1}{\sqrt{x^2+1}}
      3. - -
    3. \frac{d}{dx}\big(\tanh^{-1}(x) \big) = \frac{1}{1-x^2};\\ \abs{x}\lt 1
      4. - -
    4. \frac{d}{dx}\big(\sech^{-1}(x) \big) = \frac{-1}{x\sqrt{1-x^2}};\\ 0\lt x\lt 1
      5. - -
    5. \frac{d}{dx}\big(\csch^{-1}(x) \big) = \frac{-1}{\abs{x}\sqrt{1+x^2}};\\ x\neq0
      6. - -
    6. \frac{d}{dx}\big(\coth^{-1}(x) \big) = \frac{1}{1-x^2};\\ \abs{x} \gt 1
      7. -
    -

    -     - - - - - Integrals Involving Inverse Hyperbolic Functions -

    - Assume a\gt 0. -

      -
    1. -

      - - \int \frac{1}{\sqrt{x^2-a^2}}\, dx \amp = \ln\abs{x+\sqrt{x^2-a^2}}+C - (\text{for } 0\lt x\lt a)\, \amp = \cosh^{-1}\left(\frac xa\right)+C - -

      -
      2. - -
    2. -

      - - \int \frac{1}{\sqrt{x^2+a^2}}\, dx \amp =\ln\abs{x+\sqrt{x^2+a^2}}+C - \amp = \sinh^{-1}\left(\frac xa\right)+C - -

      -
      3. - -
    3. -

      - - \int \frac{1}{a^2-x^2}\, dx \amp = \frac1{2a}\ln\abs{\frac{a+x}{a-x}}+C - \amp = \begin{cases} - \frac1a\tanh^{-1}\left(\frac xa\right)+C \amp x^2\lt a^2\\ - \frac1a\coth^{-1}\left(\frac xa\right)+C \amp a^2\lt x^2 - \end{cases} - -

      -
      4. - -
    4. -

      - - \int \frac{1}{x\sqrt{a^2-x^2}}\, dx \amp =\frac1a \ln\left(\frac{x}{a+\sqrt{a^2-x^2}}\right)+C - (\text{for } 0\lt x\lt a)\, \amp = -\frac1a\sech^{-1}\left(\frac xa\right)+C - -

      -
      5. - -
    5. -

      - - \int \frac{1}{x\sqrt{x^2+a^2}}\, dx \amp = \frac1a \ln\abs{\frac{x}{a+\sqrt{a^2+x^2}}}+C - \amp = -\frac1a\csch^{-1}\abs{\frac xa} + C - -

      -
      6. -
    - integrationinverse hyperbolic - - hyperbolic functioninverse!integration - -

    -     - - - - - -

    - We practice using the derivative and integral formulas in the following example. -

    - - - Derivatives and integrals involving inverse hyperbolic functions - -

    - Evaluate the following. -

    - -

    -

      -
    1. \frac{d}{dx}\left[\cosh^{-1}\left(\frac{3x-2}{5}\right)\right]
      2. - -
    2. \int\frac{1}{x^2-1}\, dx
      3. - -
    3. \int \frac{1}{\sqrt{9x^2+10}}\, dx
      4. -
    -

    -     - -

    -

      -
    1. -

      - Applying with the Chain Rule gives: - - \frac{d}{dx}\left[\cosh^{-1}\left(\frac{3x-2}5\right)\right] = \frac{1}{\sqrt{\left(\frac{3x-2}5\right)^2-1}}\cdot\frac35 - . -

      -
      2. - -
    2. -

      - Multiplying the numerator and denominator by (-1) gives: - \ds \int \frac{1}{x^2-1}\, dx = \int \frac{-1}{1-x^2}\, dx. - The second integral can be solved with a direct application of item #3 from , - with a=1. - Thus - - \int \frac{1}{x^2-1}\, dx \amp = -\int \frac{1}{1-x^2}\, dx - \amp = \left\{\begin{array}{ccc} -\tanh^{-1}\left(x\right)+C \amp \amp x^2\lt 1 \\ \\ - -\coth^{-1}\left(x\right)+C \amp \amp 1\lt x^2 - \end{array} \right. - \amp =-\frac12\ln\abs{\frac{x+1}{x-1}}+C - \amp =\frac12\ln\abs{\frac{x-1}{x+1}}+C - . - We should note that this exact problem was solved at the beginning of . - In that example the answer was given as \frac12\ln\abs{x-1}-\frac12\ln\abs{x+1}+C. - Note that this is equivalent to the answer given in Equation, - as \ln(a/b) = \ln(a) - \ln(b). -

      -
      3. - -
    3. -

      - This requires a substitution, - then item #2 of can be applied. - - Let u = 3x, hence du = 3dx. - We have - - \int \frac{1}{\sqrt{9x^2+10}}\, dx \amp = \frac13\int\frac{1}{\sqrt{u^2+10}}\,du. - Note a^2=10, hence a = \sqrt{10}. Now apply the integral rule. - \amp = \frac13 \sinh^{-1}\left(\frac{3x}{\sqrt{10}}\right) + C - \amp = \frac13 \ln\abs{3x+\sqrt{9x^2+10}}+C - . -

      -
      4. -
    -

    -     - -     - -

    - This section covers a lot of ground. - New functions were introduced, - along with some of their fundamental identities, - their derivatives and antiderivatives, their inverses, - and the derivatives and antiderivatives of these inverses. - Four Key Ideas were presented, - each including quite a bit of information. -

    - -

    - Do not view this section as containing a source of information to be memorized, - but rather as a reference for future problem solving. - - contains perhaps the most useful information. - Know the integration forms it helps evaluate and understand how to use the inverse hyperbolic answer and the logarithmic answer. -

    - -

    - The next section takes a brief break from demonstrating new integration techniques. - It instead demonstrates a technique of evaluating limits that return indeterminate forms. - This technique will be useful in , - where limits will arise in the evaluation of certain definite integrals. -

    -     - - - - Terms and Concepts - - - - - -

    - In , - the equation \ds \int \tanh(x) \, dx = \ln(\cosh(x) )+C is given. - Why is \ln\abs{\cosh(x) } - not used , why are absolute values not necessary? -

    - - -     - - - -

    - Because \cosh(x) is always positive. -

    -     - -     - - - - - -

    - The hyperbolic functions are used to define points on the right hand portion of the hyperbola x^2-y^2=1, - as shown in . - How can we use the hyperbolic functions to define points on the left hand portion of the hyperbola? -

    - - -     - - - -

    - The points on the left hand side can be defined as (-\cosh(x) , \sinh(x) ). -

    -     - -     -     - - Problems - - - -

    - Verify the given identity using , - as done in . -

    -     - - - - - -

    - Verify the identity \coth^2(x) -\csch^2(x) =1 using the definitions of the hyperbolic functions. -

    - - -     - - - -

    - \begin{aligned}\coth^2(x) -\csch^2(x) \amp = \left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right)^2 - \left(\frac{2}{e^x-e^{-x}}\right)^2 \\ - \amp = \frac{(e^{2x} + 2 + e^{-2x}) - (4)}{e^{2x} - 2 + e^{-2x}}\\ - \amp = \frac{e^{2x} - 2 + e^{-2x}}{e^{2x} - 2 + e^{-2x}}\\ - \amp = 1 - \end{aligned} -

    -     - -     - - - - - -

    - Verify the identity \cosh(2x) = \cosh^2(x) +\sinh^2(x) using the definitions of the hyperbolic functions. -

    - - -     - - - -

    - \begin{aligned}\cosh^2(x) +\sinh^2(x) \amp = \left(\frac{e^x+e^{-x}}{2}\right)^2 + \left(\frac{e^x-e^{-x}}{2}\right)^2 \\ - \amp = \frac{e^{2x} + 2 + e^{-2x}}{4} + \frac{e^{2x} - 2 + e^{-2x}}{4}\\ - \amp = \frac{2e^{2x} + 2e^{-2x}}{4}\\ - \amp = \frac{e^{2x} + e^{-2x}}{2} \\ - \amp = \cosh(2x) . - \end{aligned} -

    -     - -     - - - - - -

    - Verify the identity \ds\cosh^2(x) = \frac{\cosh(2x) +1}{2} using the definitions of the hyperbolic functions. -

    - - -     - - - -

    - \begin{aligned}\cosh^2(x) \amp = \left(\frac{e^x+e^{-x}}{2}\right)^2 \\ - \amp = \frac{e^{2x} + 2 + e^{-2x}}{4} \\ - \amp = \frac12\frac{(e^{2x} + e^{-2x})+2}{2}\\ - \amp = \frac12\left(\frac{e^{2x} + e^{-2x}}{2}+1\right)\\ - \amp = \frac{\cosh(2x) +1}{2}. - \end{aligned} -

    -     - -     - - - - - -

    - Verify the identity \ds\sinh^2(x) = \frac{\cosh(2x) -1}{2} using the definitions of the hyperbolic functions. -

    - - -     - - - -

    - \begin{aligned}\sinh^2(x) \amp = \left(\frac{e^x-e^{-x}}{2}\right)^2 \\ - \amp = \frac{e^{2x} - 2 + e^{-2x}}{4} \\ - \amp = \frac12\frac{(e^{2x} + e^{-2x})-2}{2}\\ - \amp = \frac12\left(\frac{e^{2x} + e^{-2x}}{2}-1\right)\\ - \amp = \frac{\cosh(2x) -1}{2}. - \end{aligned} -

    -     - -     - - - - - -

    - Verify the identity \ds\frac{d}{dx}\left[\sech(x) \right] = -\sech(x) \tanh(x) using the definitions of the hyperbolic functions. -

    - - -     - - - -

    - \begin{aligned}\frac{d}{dx}\left[\sech(x) \right] \amp = \frac{d}{dx}\left[\frac{2}{e^x+e^{-x}}\right] \\ - \amp = \frac{-2(e^x-e^{-x})}{(e^x+e^{-x})^2} \\ - \amp = -\frac{2(e^x-e^{-x})}{(e^x+e^{-x})(e^x+e^{-x})} \\ - \amp = -\frac{2}{e^x+e^{-x}}\cdot \frac{e^x-e^{-x}}{e^x+e^{-x}}\\ - \amp = -\sech(x) \tanh(x) \end{aligned} -

    -     - -     - - - - - -

    - Verify the identity \ds\frac{d}{dx}\left[\coth(x) \right] = -\csch^2(x) using the definitions of the hyperbolic functions. -

    - - -     - - - -

    - \begin{aligned}\frac{d}{dx}\left[\coth(x) \right] \amp = \frac{d}{dx}\left[\frac{e^x+e^{-x}}{e^x-e^{-x}}\right] \\ - \amp = \frac{(e^x-e^{-x})(e^x-e^{-x})-(e^x+e^{-x})(e^x+e^{-x})}{(e^x-e^{-x})^2}\\ - \amp = \frac{e^{2x}+e^{-2x} - 2 - (e^{2x}+e^{-2x}+2)}{(e^x-e^{-x})^2}\\ - \amp = -\frac{4}{(e^x-e^{-x})^2}\\ - \amp =-\csch^2(x) \end{aligned} -

    -     - -     - - - - - -

    - Verify the identity \ds \int \tanh(x) \, dx = \ln(\cosh(x) ) + C using the definitions of the hyperbolic functions. -

    - - -     - - - -

    - \ds \int \tanh(x) \, dx = \int \frac{\sinh(x) }{\cosh(x) }\, dx -

    - -

    - Let u = \cosh(x); du = (\sinh(x) ) dx -

    - -

    - \begin{aligned}\amp = \int \frac{1}{u}\,du \\ - \amp = \ln\abs{u} + C \\ - \amp = \ln(\cosh(x) ) + C. - \end{aligned} -

    -     - -     - - - - - -

    - Verify the identity \ds \int \coth(x) \, dx = \ln\abs{\sinh(x) } + C using the definitions of the hyperbolic functions. -

    - - -     - - - -

    - \begin{aligned}\int \coth(x) \, dx \amp = \int \frac{\cosh(x) }{\sinh(x) }\, dx\\ - \text{ Let \(u = \sinh(x) \); \(du = (\cosh(x) ) dx\) } \\ - \amp = \int \frac{1}{u}\,du \\ - \amp = \ln\abs{u} + C \\ - \amp = \ln\abs{\sinh(x) } + C. - \end{aligned} -

    -     - -     - -     - - - -

    - Find the derivative of the given function. -

    -     - - - - - $f = Formula("sinh(2x)"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = \sinh(2x). -

    - -

    - -

    -     -     -     - - - - - $f = Formula("(cosh(x))^2"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = \cosh ^2x. -

    - -

    - -

    -     - -

    - Taking the derivative of (\cosh x)^2 directly, - one gets 2\cosh x\sinh x; - using the identity \cosh^2x=\frac12(\cosh2x+1) first, - one gets \sinh 2x; - by known hyperbolic identities, - these are equal. -

    -     -     -     - - - - - - $f = Formula("tanh(x^2)"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = \tanh(x^2). -

    - -

    - -

    -     -     -     - - - - - $f = Formula("ln(sinh(x))"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = \ln(\sinh(x) ). -

    - -

    - -

    -     -     -     - - - - - $f = Formula("sinh(x)*cosh(x)"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = \sinh(x) \cosh(x). -

    - -

    - -

    -     -     -     - - - - - $f = Formula("x*sinh(x)-cosh(x)"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = x\sinh(x) -\cosh(x). -

    - -

    - -

    -     -     -     - - - - - $f = Formula("asech(x^2)"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = \sech^{-1}(x^2). -

    - -

    - -

    -     -     -     - - - - - $f = Formula("asinh(3x)"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = \sinh^{-1}(3x). -

    - -

    - -

    -     -     -     - - - - - $f = Formula("acosh(2x^2)"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = \cosh^{-1}(2x^2). -

    - -

    - -

    -     -     -     - - - - - $f = Formula("atanh(x+5)"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = \tanh^{-1}(x+5). -

    - -

    - -

    -     -     -     - - - - - $f = Formula("atanh(cos(x))"); - $df = $f->D('x'); - - - -

    - Find the derivative of f(x) = \tanh^{-1}(\cos(x) ). -

    - -

    - -

    -     -     -     - - - - - $f = Formula("acosh(sec(x))"); - $df = $f->D('x'); - - - - -

    - Find the derivative of f(x) = \cosh^{-1}(\sec(x) ). -

    - -

    - -

    -     -     -     - -     - - - -

    - Find the equation of the line tangent to the function at the given x-value. -

    -     - - - - - $x0 = 0; - $f = Formula("sinh(x)"); - $df = $f->D('x'); - $m = $df->eval(x=>$x0); - $y0 = $f->eval(x=>$x0); - $line = Formula("$m*(x - $x0) + $y0"); - - - -

    - Find the equation of the tangent line to y=f(x) at x=0, where f(x) = \sinh(x). -

    - -

    - y = -

    -     - -

    - y=x -

    - -     -     -     - - - - - $x0 = ln(2); - $f = Formula("cosh(x)"); - $df = $f->D('x'); - $m = $df->eval(x=>$x0); - $y0 = $f->eval(x=>$x0); - $line = Formula("$m*(x - $x0) + $y0"); - - - -

    - Find the equation of the tangent line to y=f(x) at x=\ln(2), where f(x) = \cosh(x). -

    - -

    - y = -

    -     - -

    - y=3/4(x-\ln(2) )+5/4 -

    - -     -     -     - - - - - $x0 = -ln(3); - $f = Formula("tanh(x)"); - $df = $f->D('x'); - $m = $df->eval(x=>$x0); - $y0 = $f->eval(x=>$x0); - $line = Formula("$m*(x - $x0) + $y0"); - - - - -

    - Find the equation of the tangent line to y=f(x) at x=-\ln(3), where f(x) = \tanh(x). -

    - -

    - y = -

    -     - -

    - y=\frac9{25}(x+\ln(3))-\frac45 -

    - -     -     -     - - - - - - $x0 = ln(3); - $f = Formula("sech(x)^2"); - $df = $f->D('x'); - $m = $df->eval(x=>$x0); - $y0 = $f->eval(x=>$x0); - $line = Formula("$m*(x - $x0) + $y0"); - - - -

    - Find the equation of the tangent line to y=f(x) at x=\ln(3), where f(x) = \sech^2(x). -

    - -

    - y = -

    -     - -

    - y=-72/125(x-\ln(3) )+9/25 -

    - -     -     -     - - - - - $x0 = 0; - $f = Formula("asinh(x)"); - $df = $f->D('x'); - $m = $df->eval(x=>$x0); - $y0 = $f->eval(x=>$x0); - $line = Formula("$m*(x - $x0) + $y0"); - - - -

    - Find the equation of the tangent line to y=f(x) at x=0, where f(x) = \sinh^{-1}(x). -

    - -

    - y = -

    -     - -

    - y=x -

    - -     -     -     - - - - - $x0 = sqrt(2); - $f = Formula("acosh(x)"); - $df = $f->D('x'); - $m = $df->eval(x=>$x0); - $y0 = $f->eval(x=>$x0); - $line = Formula("$m*(x - $x0) + $y0"); - - - -

    - Find the equation of the tangent line to y=f(x) at x=\sqrt 2, where f(x) = \cosh^{-1}(x). -

    - -

    - y = -

    -     - -

    - y=(x-\sqrt{2})+\cosh^{-1}(\sqrt{2}) \approx (x-1.414)+0.881 -

    - -     -     -     - -     - - - -

    - Evaluate the given indefinite integral. -

    -     - - - - - $F = FormulaUpToConstant("1/2*ln(cosh(2x))"); - - - -

    - Evaluate the indefinite integral \ds \int \tanh(2x)\, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("1/3*sinh(3x-7)"); - - - -

    - Evaluate the indefinite integral \ds \int \cosh(3x-7)\, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("1/2*sinh(x)^2"); - - - -

    - Evaluate the indefinite integral \ds \int \sinh(x) \cosh(x) \, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("x*sinh(x)-cosh(x)"); - - - -

    - Evaluate the indefinite integral \ds \int x\cosh(x) \, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("x*cosh(x)-sinh(x)"); - - - -

    - Evaluate the indefinite integral \ds \int x\sinh(x) \, dx. -

    - -

    - -

    -     -     -     - - - -

    - Evaluate the indefinite integral \ds\int \frac1{\sqrt{x^2+1}}\, dx. -

    -     - -

    - \sinh^{-1} x +C=\ln\big(x+\sqrt{x^2+1}\big)+C -

    -     -     - - - -

    - Evaluate the indefinite integral \ds\int \frac1{\sqrt{x^2-9}}\, dx. -

    -     - -

    - \cosh^{-1} x/3 +C=\ln\big(x+\sqrt{x^2-9}\big)+C -

    -     -     - - - - - - $F = FormulaUpToConstant("1/2*ln(abs(x+1))-1/2*ln(abs(x-1))"); - - - -

    - Evaluate the indefinite integral \ds \int \frac{1}{9-x^2}\, dx. -

    - -

    - -

    -     - -

    - \left\{\begin{array}{ccc} \frac13\tanh^{-1}\left(\frac x3\right)+C \amp \amp x^2\lt 9 \\ \\ - \frac13\coth^{-1}\left(\frac x3\right)+C \amp \amp 9\lt x^2 - \end{array} \right. = \frac12\ln\abs{x+1} - \frac12\ln\abs{x-1}+C -

    - -     -     -     - - - - - $F = FormulaUpToConstant("acosh(x^2/2)"); - - - -

    - Evaluate the indefinite integral \ds \int \frac{2x}{\sqrt{x^4-4}}\, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("2/3*asinh(x^(3/2))"); - - - -

    - Evaluate the indefinite integral \ds \int \frac{\sqrt{x}}{\sqrt{1+x^3}}\, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("-1/16*atan(x/2)+1/32*ln(abs(x-2))-1/32*ln(abs(x+2))"); - - - -

    - Evaluate the indefinite integral \ds \int \frac{1}{x^4-16}\, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("ln(x)-ln(abs(x+1))"); - - - -

    - Evaluate the indefinite integral \ds \int \frac{1}{x^2+x}\, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("atan(e^x)"); - - - -

    - Evaluate the indefinite integral \ds \int \frac{e^x}{e^{2x}+1}\, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("x*asinh(x)-sqrt(x^2+1)"); - - - -

    - Evaluate the indefinite integral \ds \int \sinh^{-1}(x) \, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("x*atanh(x)+1/2*ln(abs(x^2-1))"); - - - -

    - Evaluate the indefinite integral \ds \int \tanh^{-1}(x) \, dx. -

    - -

    - -

    -     -     -     - - - - - $F = FormulaUpToConstant("atan(sinh(x))"); - - - -

    - Evaluate the indefinite integral \ds \int \sech(x) \, dx. -

    - -

    - (Hint: mutiply by \frac{\cosh(x) }{\cosh(x) }; - set u = \sinh(x).) -

    - -

    - -

    -     -     -     - -     - - - -

    - Evaluate the given definite integral. -

    -     - - - - - $a = -1; - $b = 1; - $F = Formula("cosh(x)"); - $r = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - Evaluate the definite integral \ds \int_{-1}^1 \sinh(x) \, dx. -

    - -

    - -

    - -     -     -     - - - - - $a = -ln(2); - $b = ln(2); - $F = Formula("sinh(x)"); - $r = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - Evaluate the definite integral \ds \int_{-\ln(2) }^{\ln(2) } \cosh(x) \, dx. -

    - -

    - -

    - -     -     -     - - - - - $a = 0; - $b = 1; - $F = Formula("tanh(x)"); - $r = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - Evaluate the definite integral \ds \int_{0}^{1} \sech^{2}(x) \, dx. -

    - -

    - -

    - -     -     -     - - - - - $a = 0; - $b = 2; - $F = Formula("asinh(x)"); - $r = $F->eval(x=>$b) - $F->eval(x=>$a); - - - -

    - Evaluate the definite integral \ds \int_{0}^{2}\frac1{\sqrt{x^2+1}}\, dx. -

    - -

    - -

    -     - -     -     -     -     -     -
    -
    - L'Hospital's Rule - -

    - While this chapter is devoted to learning techniques of integration, - this section is not about integration. - Rather, it is concerned with a technique of evaluating certain limits that will be useful in the following section, - where integration is once more discussed. -

    - -

    - Our treatment of limits exposed us to the notion of - 0/0, an indeterminate form. - If \lim\limits_{x\to c}f(x)=0 and \lim\limits_{x\to c} g(x) =0, - we do not conclude that \lim\limits_{x\to c} f(x)/g(x) is 0/0; - rather, we use 0/0 as notation to describe the fact that both the numerator and denominator approach 0. - The expression 0/0 has no numeric value; - other work must be done to evaluate the limit. -

    - -

    - Other indeterminate forms exist; they are: - \infty/\infty, 0\cdot\infty, - \infty-\infty, 0^0, - 1^\infty and \infty^0. - Just as 0/0 does not mean divide 0 by 0, - the expression \infty/\infty - does not mean divide infinity by infinity. Instead, - it means a quantity is growing without bound and is being divided by another quantity that is growing without bound. - We cannot determine from such a statement what value, - if any, results in the limit. - Likewise, 0\cdot \infty - does not mean multiply zero by infinity. Instead, - it means one quantity is shrinking to zero, - and is being multiplied by a quantity that is growing without bound. - We cannot determine from such a description what the result of such a limit will be. -

    - - - -

    - This section introduces l'Hospital's Rule, - a method of resolving limits that produce the indeterminate forms 0/0 and \infty/\infty. - We'll also show how algebraic manipulation can be used to convert other indeterminate expressions into one of these two forms so that our new rule can be applied. -

    -     - - - L'Hospital's Rule with Indeterminate Forms <m>0/0</m> and <m>\infty/\infty</m> - - L'Hospital's Rule, Part 1 - -

    - Let \lim\limits_{x\to c}f(x) = 0 and \lim\limits_{x\to c}g(x)=0, - where f and g are differentiable functions on an open interval I containing c, - and \gp(x)\neq 0 on I except possibly at c. - If limitl'Hospital's Rule - l'Hospital's Rulezero over zero - - \lim_{x\to c} \frac{\fp(x)}{\gp(x)}=L - , - then - - \lim_{x\to c} \frac{f(x)}{g(x)}=L - , - where L is a real number, or L=\pm \infty. - The result applies to one-sided limits as well. -

    -     -     - - - -

    - We demonstrate the use of l'Hospital's Rule in the following examples; - we will often use LHR as an abbreviation of - l'Hospital's Rule. -

    - - - Using l'Hospital's Rule - -

    - Evaluate the following limits, - using l'Hospital's Rule as needed. -

    - -

    -

      -
    1. - \lim\limits_{x\to0}\frac{\sin(x) }x -
      2. - -
    2. - \lim\limits_{x\to 1}\frac{\sqrt{x+3}-2}{1-x} -
      3. - -
    3. - \lim\limits_{x\to0}\frac{x^2}{1-\cos(x) } -
      4. - -
    4. - \lim\limits_{x\to 2}\frac{x^2+x-6}{x^2-3x+2} -
      5. -
    -

    -

    - - -

    -

      -
    1. -

      - We proved this limit is 1 in using the Squeeze Theorem. - Here we use l'Hospital's Rule to show its power. - - \lim_{x\to0}\frac{\sin(x) }x \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to0} \frac{\cos(x) }{1}=1 - . -

      -
      2. - -
    2. - - \lim\limits_{x\to 1}\frac{\sqrt{x+3}-2}{1-x} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x \to 1} \frac{\frac12(x+3)^{-1/2}}{-1} =-\frac 14 - . -
      3. - -
    3. -

      - \lim\limits_{x\to 0}\frac{x^2}{1-\cos(x) } \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to 0} \frac{2x}{\sin(x) }. - - This latter limit also evaluates to the 0/0 indeterminate form. - To evaluate it, we apply l'Hospital's Rule again. - - \lim\limits_{x\to 0} \frac{2x}{\sin(x) } \stackrel{\,\text{ by LHR } \,}{=} \frac{2}{\cos(x) } = 2. - - Thus \lim\limits_{x\to0}\frac{x^2}{1-\cos(x) }=2. -

      -
      4. - -
    4. -

      - We already know how to evaluate this limit; - first factor the numerator and denominator. - We then have: - - \lim_{x\to 2}\frac{x^2+x-6}{x^2-3x+2} = \lim_{x\to 2}\frac{(x-2)(x+3)}{(x-2)(x-1)} = \lim_{x\to 2}\frac{x+3}{x-1} = 5 - . - We now show how to solve this using l'Hospital's Rule. - - \lim_{x\to 2}\frac{x^2+x-6}{x^2-3x+2}\stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to 2}\frac{2x+1}{2x-3} = 5 - . -

      -
      5. -
    -

    - - -     - -

    - Note that at each step where l'Hospital's Rule was applied, - it was needed: - the initial limit returned the indeterminate form of 0/0. - If the initial limit returns, - for example, 1/2, then l'Hospital's Rule does not apply. -

    - -

    - The following theorem extends our initial version of l'Hospital's Rule in two ways. - It allows the technique to be applied to the indeterminate form - \infty/\infty and to limits where x approaches \pm\infty. -

    - - - L'Hospital's Rule, Part 2 - -

    -

      -
    1. -

      - Let \lim\limits_{x\to a}f(x) = \pm\infty and \lim\limits_{x\to a}g(x)=\pm \infty, - where f and g are differentiable on an open interval I containing a. - If - - limitl'Hospital's Rule - l'Hospital's Ruleinfinity over infinity - - - \lim_{x\to a}\frac{\fp(x)}{\gp(x)}=L - , - then - - \lim_{x\to a}\frac{f(x)}{g(x)}=L - , - where L is a real number, or L=\pm\infty. - The result applies to one-sided limits as well. -

      -
      2. - -
    2. -

      - Let f and g be differentiable functions on the open interval - (a,\infty) for some value a, - where \gp(x)\neq 0 on (a,\infty) and - \lim\limits_{x\to\infty} f(x)/g(x) returns either 0/0 or \infty/\infty. - If - - \lim_{x\to \infty}\frac{\fp(x)}{\gp(x)}=L - , - then - - \lim_{x\to \infty}\frac{f(x)}{g(x)}=L - , - where L is a real number, or L=\pm \infty. - A similar statement can be made for limits where x approaches -\infty. -

      -
      3. -
    -

    -     -     - - - Using l'Hospital's Rule with limits involving <m>\infty</m> - -

    - Evaluate the following limits. -

    - -

    - \ds 1.\,\lim_{x\to\infty} \frac{3x^2-100x+2}{4x^2+5x-1000} \qquad\qquad 2. \,\lim_{x\to \infty}\frac{e^x}{x^3}. -

    -     - -

    -

      -
    1. -

      - We can evaluate this limit already using ; - the answer is 3/4. - We apply l'Hospital's Rule to demonstrate its applicability. - - \lim_{x\to\infty} \frac{3x^2-100x+2}{4x^2+5x-1000}\stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{6x-100}{8x+5} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac68 = \frac34 - . -

      -
      2. - -
    2. -

      - - \lim\limits_{x\to \infty}\frac{e^x}{x^3} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{e^x}{3x^2} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{e^x}{6x} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{e^x}{6} = \infty - . - - Recall that this means that the limit does not exist; - as x approaches \infty, - the expression e^x/x^3 grows without bound. - We can infer from this that e^x grows - faster than x^3; - as x gets large, e^x is far larger than x^3. - (This has important implications in computing when considering efficiency of algorithms.) -

      -
      3. -
    -

    -     - -     - - -     - - - Indeterminate Forms <m>0\cdot\infty</m> and <m>\infty-\infty</m> -

    - L'Hospital's Rule can only be applied to ratios of functions. - When faced with an indeterminate form such as - 0\cdot\infty or \infty-\infty, - we can sometimes apply algebra to rewrite the limit so that l'Hospital's Rule can be applied. - We demonstrate the general idea in the next example. - limitindeterminate form - indeterminate form -

    - - - Applying l'Hospital's Rule to other indeterminate forms - -

    - Evaluate the following limits. -

    - -

    -

      -
    1. \lim\limits_{x\to0^+} x\cdot e^{1/x}
      2. - -
    2. \lim\limits_{x\to0^-} x\cdot e^{1/x}
      3. - -
    3. \lim\limits_{x\to\infty} \ln(x+1)-\ln(x)
      4. - -
    4. \lim\limits_{x\to\infty} x^2-e^x
      5. -
    -

    -     - -

    -

      -
    1. -

      - As x\rightarrow 0^+, - x\rightarrow 0 and e^{1/x}\rightarrow \infty. - Thus we have the indeterminate form 0\cdot\infty. - We rewrite the expression x\cdot e^{1/x} as \ds\frac{e^{1/x}}{1/x}; - now, as x\rightarrow 0^+, - we get the indeterminate form - \infty/\infty to which l'Hospital's Rule can be applied. - - \lim_{x\to0^+} x\cdot e^{1/x} = \lim_{x\to 0^+} \frac{e^{1/x}}{1/x} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to 0^+}\frac{(-1/x^2)e^{1/x}}{-1/x^2} =\lim_{x\to 0^+}e^{1/x} =\infty - . - Interpretation: e^{1/x} grows faster - than x shrinks to zero, - meaning their product grows without bound. -

      -
      2. - -
    2. -

      - As x\rightarrow 0^-, - x\rightarrow 0 and e^{1/x}\rightarrow e^{-\infty}\rightarrow 0. - The the limit evaluates to - 0\cdot 0 which is not an indeterminate form. - We conclude then that - - \lim_{x\to 0^-}x\cdot e^{1/x} = 0 - . -

      -
      3. - -
    3. -

      - This limit initially evaluates to the indeterminate form \infty-\infty. - By applying a logarithmic rule, - we can rewrite the limit as - - \lim_{x\to\infty} \ln(x+1)-\ln(x) = \lim_{x\to \infty} \ln\left(\frac{x+1}x\right) - . - As x\rightarrow \infty, - the argument of the \ln term approaches \infty/\infty, - to which we can apply l'Hospital's Rule. - - \lim_{x\to\infty} \frac{x+1}x \stackrel{\,\text{ by LHR } \,}{=} \frac11=1 - . - Since x\rightarrow \infty implies \ds\frac{x+1}x\rightarrow 1, - it follows that - - x\rightarrow \infty \text{ implies } \ln\left(\frac{x+1}x\right)\rightarrow \ln(1) =0 - . - Thus - - \lim_{x\to\infty} \ln(x+1)-\ln(x) = \lim_{x\to \infty} \ln\left(\frac{x+1}x\right)=0 - . - Interpretation: - since this limit evaluates to 0, it means that for large x, - there is essentially no difference between \ln(x+1) and \ln(x); - their difference is essentially 0. -

      -
      4. - -
    4. -

      - The limit \lim\limits_{x\to\infty} x^2-e^x initially returns the indeterminate form \infty-\infty. - We can rewrite the expression by factoring out x^2; - \ds x^2-e^x = x^2\left(1-\frac{e^x}{x^2}\right). - We need to evaluate how e^x/x^2 behaves as x\rightarrow \infty: - - \lim_{x\to\infty}\frac{e^x}{x^2} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{e^x}{2x} \stackrel{\,\text{ by LHR } \,}{=} \lim_{x\to\infty} \frac{e^x}{2} = \infty - . - Thus \lim_{x\to\infty}x^2(1-e^x/x^2) evaluates to \infty\cdot(-\infty), - which is not an indeterminate form; - rather, \infty\cdot(-\infty) evaluates to -\infty. - We conclude that - \lim\limits_{x\to\infty} x^2-e^x = -\infty. - - Interpretation: as x gets large, - the difference between x^2 and e^x grows very large. -

      -
      5. -
    -

    -     - -     -     - - - Indeterminate Forms<nbsp/> <m>0^0</m>, <m>1^\infty</m> and <m>\infty^0</m> -

    - When faced with an indeterminate form that involves a power, - it often helps to employ the natural logarithmic function. - The following Key Idea expresses the concept, - which is followed by an example that demonstrates its use. -

    - - - Evaluating Limits Involving Indeterminate Forms <m>0^0</m>, <m>1^\infty</m> and <m>\infty^0</m> -

    - If \lim\limits_{x\to c} \ln\big(f(x)\big) = L, then - \lim\limits_{x\to c} f(x) = \lim_{x\to c} e^{\ln(f(x))} = e^L. - - limitindeterminate form - indeterminate form -

    -     - - - Using l'Hospital's Rule with indeterminate forms involving exponents - -

    - Evaluate the following limits. -

    - -

    -

      -
    1. \lim_{x\to\infty} \left(1+\frac1x\right)^x
      2. -
    2. \lim_{x\to0^+} x^x
      3. -
    -

    -     - -

    -

      -
    1. -

      - This is equivalent to a special limit given in ; - these limits have important applications within mathematics and finance. - Note that the exponent approaches \infty while the base approaches 1, leading to the indeterminate form 1^\infty. - Let f(x) = (1+1/x)^x; - the problem asks to evaluate \lim\limits_{x\to\infty}f(x). - Let's first evaluate \lim\limits_{x\to\infty}\ln\big(f(x)\big). - - \lim_{x\to\infty}\ln\big(f(x)\big) \amp = \lim_{x\to\infty} \ln\left(1+\frac1x\right)^x - \amp = \lim_{x\to\infty} x\ln\left(1+\frac1x\right) - \amp = \lim_{x\to\infty} \frac{\ln\left(1+\frac1x\right)}{1/x} - This produces the indeterminate form 0/0, so we apply l'Hospital's Rule. - \amp = \lim_{x\to\infty} \frac{\frac{1}{1+1/x}\cdot(-1/x^2)}{(-1/x^2)} - \amp = \lim_{x\to\infty}\frac{1}{1+1/x} - \amp = 1 - . - Thus \lim\limits_{x\to\infty} \ln\big(f(x)\big) = 1. - We return to the original limit and apply . - - \lim_{x\to\infty}\left(1+\frac1x\right)^x = \lim_{x\to\infty} f(x) = \lim_{x\to\infty}e^{\ln(f(x))} = e^1 = e - . -

      -
      2. - -
    2. -

      - This limit leads to the indeterminate form 0^0. - Let f(x) = x^x and consider first \lim\limits_{x\to0^+} \ln\big(f(x)\big). - - \lim_{x\to0^+} \ln\big(f(x)\big) \amp = \lim_{x\to0^+} \ln\left(x^x\right) - \amp = \lim_{x\to0^+} x\ln(x) - \amp = \lim_{x\to0^+} \frac{\ln(x) }{1/x}. - This produces the indeterminate form -\infty/\infty so we apply l'Hospital's Rule. - \amp = \lim_{x\to0^+} \frac{1/x}{-1/x^2} - \amp = \lim_{x\to0^+} -x - \amp = 0 - . - Thus \lim\limits_{x\to0^+} \ln\big(f(x)\big) =0. - We return to the original limit and apply . - - \lim_{x\to0^+} x^x = \lim_{x\to0^+} f(x) = \lim_{x\to0^+} e^{\ln(f(x))} = e^0 = 1 - . - This result is supported by the graph of - f(x)=x^x given in . -

      -
      - A graph of f(x)=x^x supporting the fact that as x\to 0^+, f(x)\to 1 - - - - Graph of function x^x, used in this the example. - - -

      - The y axis is drawn from 0 to 4 and the - x axis is drawn from 0 to 2. The - function f(x) = x^x is drawn as a curve opening - towards the positive y axis with arrows towards the - ends. The function is drawn from point (0,1) from - where it dips gently then rises up slowly. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ytick={1,2,3,4}, - ymin=-.4,ymax=4.5, - xmin=-.1,xmax=2.2, - scaled ticks=false - ] - - \addplot+ [infinite,domain=0:2,samples=40] {x^x}; - - \draw (axis cs:1,1) node [below right] { $f(x)=x^x$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      3. -
    -

    -     - -     - - - - - -

    - Our brief revisit of limits will be rewarded in the next section where we consider - improper integration. So far, - we have only considered definite integrals where the bounds are finite numbers, - such as \ds \int_0^1 f(x)\, dx. - Improper integration considers integrals where one, or both, - of the bounds are infinity. - Such integrals have many uses and applications, - in addition to generating ideas that are enlightening. -

    -     - - - - Terms and Concepts - - - - -

    - List the different indeterminate forms described in this section. -

    - - -     - - - -

    - 0/0, \infty/\infty, 0\cdot\infty,\infty-\infty,0^0,1^\infty,\infty^0 -

    - -     - -     - - - - -

    - L'Hospital's Rule provides a faster method of computing derivatives. - -

    -     - -     - - - - -

    - L'Hospital's Rule states that - \ds \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\fp(x)}{\gp(x)}. - -

    -     - -     - - - - -

    - Explain what the indeterminate form - 1^\infty means. -

    - - -     - - - -

    - The base of an expression is approaching 1 while its power is growing without bound. -

    - -     - -     - - - - -

    - Fill in the blanks: The Quotient Rule is applied to - \ds \frac{f(x)}{g(x)} when taking ; - l'Hospital's Rule is applied when taking certain . -

    -     - - - - a derivative|derivatives? - - - - -

    - Your answer includes the correct word but has extra text. -

    -     -     -     - - - limits? - - - - -

    - Your answer includes the correct word but has extra text. -

    -     -     -     -     - -     - - - - -

    - Create (but do not evaluate!) a limit that returns - \infty^0. -

    - -     - - - -     - - - - -

    - Create a function f(x) such that - \lim\limits_{x\to 1}f(x) returns 0^0. -

    - - -     - - - -     - - - - -

    - Create a function f(x) such that \ds \lim_{x\to \infty}f(x) returns - 0\cdot\infty. -

    - - - -     - - - -     -     - - Problems - - - -

    - Evaluate the given limit using l'Hospital's rule. -

    -     - - - - - - $xlim = 1; - $fu = Formula("x^2+x-2"); - $fl = Formula("x-1"); - $dfu = $fu->D('x'); - $dfl = $fl->D('x'); - if ($dfl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to 1} \frac{ }{} -

    - -

    - -

    -     -     -     - - - - - - $xlim = 2; - $fu = Formula("x^2+x-6"); - $fl = Formula("x^2-7x+10"); - $dfu = $fu->D('x'); - $dfl = $fl->D('x'); - if ($dfl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to 2} \frac{ }{} -

    - -

    - -

    - -     -     -     - - - - - - $xlim = pi; - $fu = Formula("sin(x)"); - $fl = Formula("x-pi"); - $dfu = $fu->D('x'); - $dfl = $fl->D('x'); - if ($dfl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to \pi} \frac{ }{} -

    - -

    - -

    - -     -     -     - - - - - - $xlim = pi/4; - $fu = Formula("sin(x)-cos(x)"); - $fl = Formula("cos(2x)"); - $dfu = $fu->D('x'); - $dfl = $fl->D('x'); - if ($dfl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to \pi/4} \frac{ }{} -

    - -

    - -

    - -     -     -     - - - - - - $xlim = 0; - $fu = Formula("sin(5x)"); - $fl = Formula("x"); - $dfu = $fu->D('x'); - $dfl = $fl->D('x'); - if ($dfl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to 0} \frac{ }{} -

    - -

    - -

    - -     -     -     - - - - - - $xlim = 0; - $fu = Formula("sin(2x)"); - $fl = Formula("x+2"); - $l = $fu->eval(x=>$xlim) / $fl->eval(x=>$xlim); - - -

    - \lim\limits_{x\to 0} \frac{ }{} -

    - -

    - -

    - -     -     -     - - - - - - $xlim = 0; - $fu = Formula("sin(2x)"); - $fl = Formula("sin(3x)"); - $dfu = $fu->D('x'); - $dfl = $fl->D('x'); - if ($dfl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to 0} \frac{ }{} -

    - -

    - -

    - -     -     -     - - - - - - - Context()->variables->add(a=>"Real",b=>"Real"); - $xlim = 0; - $fu = Formula("sin(a*x)"); - $fl = Formula("sin(b*x)"); - $dfu = $fu->D('x'); - $dfl = $fl->D('x'); - if ($dfl->substitute(x=>$xlim)==Formula("0")) { - $l = Compute("inf"); - } - else { - $l = $dfu->substitute(x=>$xlim) / $dfl->substitute(x=>$xlim); - } - - -

    - \lim\limits_{x\to 0} \frac{ }{} -

    - -

    - -

    -     -     -     - - - - - - $xlim = 0; - $fu = Formula("e^x-1"); - $fl = Formula("x^2"); - $dfu = $fu->D('x'); - $dfl = $fl->D('x'); - if ($dfl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to 0^+} \frac{ }{} -

    - -

    - -

    - -     -     -     - - - - - - $xlim = 0; - $fu = Formula("e^x-x-1"); - $fl = Formula("x^2"); - $d2fu = $fu->D('x')->D('x'); - $d2fl = $fl->D('x')->D('x'); - if ($d2fl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $d2fu->eval(x=>$xlim) / $d2fl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to 0^+} \frac{ }{} -

    - -

    - -

    - -     -     -     - - - - - - $xlim = 0; - $fu = Formula("x-sin(x)"); - $fl = Formula("x^3-x^2"); - $d2fu = $fu->D('x')->D('x'); - $d2fl = $fl->D('x')->D('x'); - if ($d2fl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $d2fu->eval(x=>$xlim) / $d2fl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to 0^+} \frac{ }{} -

    - -

    - -

    - -     -     -     - - - - - - - $l = Compute("0"); - $fu = Formula("x^4"); - $fl = Formula("e^x"); - - -

    - \lim\limits_{x\to \infty} \frac{}{} -

    - -

    - -

    -     -     -     - - - - - - $l = Compute("0"); - $fu = Formula("sqrt(x)"); - $fl = Formula("e^x"); - - -

    - \lim\limits_{x\to \infty} \frac{}{} -

    - -

    - -

    -     -     -     - - - - - - $l = Compute("infinity"); - $fu = Formula("e^x"); - $fl = Formula("x^2"); - - -

    - \lim\limits_{x\to\infty} \frac{}{} -

    -     -     -     - - - - - - $fu = Formula("e^x"); - $fl = Formula("sqrt(x)"); - $l = Compute("infinity"); - - -

    - \lim\limits_{x\to \infty} \frac{}{} -

    - -

    - -

    -     -     -     - - - - - - $fu = Formula("e^x"); - $fl = Formula("2^x"); - $l = Compute("infinity"); - - -

    - \lim\limits_{x\to \infty} \frac{}{} -

    - -

    - -

    -     -     -     - - - - - - $fu = Formula("e^x"); - $fl = Formula("3^x"); - $l = Compute("0"); - - -

    - \lim\limits_{x\to \infty} \frac{}{} -

    - -

    - -

    -     -     -     - - - - - - $xlim = 3; - $fu = Formula("x^3-5x^2+3x+9"); - $fl = Formula("x^3-7x^2+15x-9"); - $d2fu = $fu->D('x')->D('x'); - $d2fl = $fl->D('x')->D('x'); - if ($d2fl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $d2fu->eval(x=>$xlim) / $d2fl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to 3} \frac{}{} -

    - -

    - -

    - -     -     -     - - - - - - $xlim = -2; - $fu = Formula("x^3+4x^2+4x"); - $fl = Formula("x^3+7x^2+16x+12"); - $d2fu = $fu->D('x')->D('x'); - $d2fl = $fl->D('x')->D('x'); - if ($d2fl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $d2fu->eval(x=>$xlim) / $d2fl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to -2} \frac{}{} -

    - -

    - -

    - -     -     -     - - - - - - $fu = Formula("ln(x)"); - $fl = Formula("x"); - $l = Compute("0"); - - -

    - \lim\limits_{x\to \infty} \frac{}{} -

    - -

    - -

    - -     -     -     - - - - - - $fu = Formula("ln(x^2)"); - $fl = Formula("x"); - $l = Compute("0"); - - - - -

    - \lim\limits_{x\to \infty} \frac{}{} -

    - -

    - -

    - -     -     -     - - - - - - $fu = Formula("ln(x)^2"); - $fl = Formula("x"); - $l = Compute("0"); - - - - -

    - \lim\limits_{x\to \infty} \frac{}{} -

    - -

    - -

    - -     -     -     - - - - - - $f1 = Formula("x"); - $f2 = Formula("ln(x)"); - $l = Compute("0"); - - -

    - \lim\limits_{x\to 0^+} \cdot -

    - -

    - -

    -     -     -     - - - - - - $f1 = Formula("sqrt(x)"); - $f2 = Formula("ln(x)"); - $l = Compute("0"); - - -

    - \lim\limits_{x\to 0^+} \cdot -

    - -

    - -

    - -     -     -     - - - - - - - $f1 = Formula("x"); - $f2 = Formula("e^(1/x)"); - $l = Compute("inf"); - - -

    - \lim\limits_{x\to 0^+} \cdot -

    - -

    - -

    -     -     -     - - - - - - $f1 = Formula("x^3"); - $f2 = Formula("x^2"); - $l = Compute("inf"); - - -

    - \lim\limits_{x\to \infty} - -

    - -

    - -

    - -     -     -     - - - - - - $f1 = Formula("sqrt(x)"); - $f2 = Formula("ln(x)"); - $l = Compute("inf"); - - -

    - \lim\limits_{x\to \infty} - -

    - -

    - -

    -     -     -     - - - - - - $f1 = Formula("x"); - $f2 = Formula("e^x"); - $l = Compute("0"); - - -

    - \lim\limits_{x\to -\infty} \cdot -

    - -

    - -

    -     -     -     - - - - - - $f1 = Formula("1/x^2"); - $f2 = Formula("e^(-1/x)"); - $l = Compute("0"); - - -

    - \lim\limits_{x\to 0^+} \cdot -

    - -

    - -

    -     -     -     - - - - - - $fb = Formula("(1+x)"); - $fp = Formula("1/x"); - $l = Compute("e"); - - -

    - \lim\limits_{x\to 0^+} ()^ -

    - -

    - -

    -     -     -     - - - - - - $fb = Formula("2x"); - $fp = Formula("x"); - $l = Compute("1"); - - -

    - \lim\limits_{x\to 0^+} ()^{} -

    - -

    - -

    -     -     -     - - - - - - $fb = Formula("2/x"); - $fp = Formula("x"); - $l = Compute("1"); - - -

    - \lim\limits_{x\to 0^+} ()^{} -

    - -

    - -

    -     -     -     - - - - - - $fb = Formula("sin(x)"); - $fp = Formula("x"); - $l = Compute("1"); - - -

    - \lim\limits_{x\to 0^+} ()^{} - Hint: use the Squeeze Theorem. -

    - -

    - -

    -     -     -     - - - - - - $fb = Formula("1-x"); - $fp = Formula("1-x"); - $l = Compute("1"); - - -

    - \lim\limits_{x\to 1^-} ()^{} -

    - -

    - -

    -     -     -     - - - - - - $fb = Formula("x"); - $fp = Formula("1/x"); - $l = Compute("1"); - - -

    - \lim\limits_{x\to \infty} ()^{} -

    - -

    - -

    -     -     -     - - - - - - $fb = Formula("1/x"); - $fp = Formula("x"); - $l = Compute("0"); - - -

    - \lim\limits_{x\to \infty} ()^{} -

    - -

    - -

    -     -     -     - - - - - - $fb = Formula("ln(x)"); - $fp = Formula("1-x"); - $l = Compute("1"); - - -

    - \lim\limits_{x\to 1^+} ()^{} -

    - -

    - -

    -     -     -     - - - - - - $fb = Formula("1+x"); - $fp = Formula("1/x"); - $l = Compute("1"); - - -

    - \lim\limits_{x\to \infty} ()^{} -

    - -

    - -

    -     -     -     - - - - - - $fb = Formula("1+x^2"); - $fp = Formula("1/x"); - $l = Compute("1"); - - -

    - \lim\limits_{x\to \infty} ()^{} -

    - -

    - -

    -     -     -     - - - - - - $f1 = Formula("tan(x)"); - $f2 = Formula("cos(x)"); - $l = Compute("1"); - - -

    - \lim\limits_{x\to \pi/2} -

    - -

    - -

    -     -     -     - - - - - - $f1 = Formula("tan(x)"); - $f2 = Formula("sin(2x)"); - $l = Compute("2"); - - -

    - \lim\limits_{x\to \pi/2} -

    - -

    - -

    -     -     -     - - - - - - $f1 = Formula("1/ln(x)"); - $f2 = Formula("1/(x-1)"); - $l = Compute("1/2"); - - -

    - \lim\limits_{x\to 1^+} - -

    - -

    - -

    -     -     -     - - - - - - $f1 = Formula("5/(x^2-9)"); - $f2 = Formula("x/(x-3)"); - $l = Compute("-inf"); - - -

    - \lim\limits_{x\to 3^+} - -

    - -

    - -

    -     -     -     - - - - - - $f1 = Formula("x"); - $f2 = Formula("tan(1/x)"); - $l = Compute("1"); - - -

    - \lim\limits_{x\to \infty} -

    - -

    - -

    -     -     -     - - - - - - $fu = Formula("ln(x)^3"); - $fl = Formula("x"); - $l = Compute("0"); - - -

    - \lim\limits_{x\to \infty} \frac{}{} -

    - -

    - -

    -     -     -     - - - - - - $xlim = 1; - $fu = Formula("x^2+x-2"); - $fl = Formula("ln(x)"); - $dfu = $fu->D('x'); - $dfl = $fl->D('x'); - if ($dfl->eval(x=>$xlim)==0) { - $l = Compute("inf"); - } - else { - $l = $dfu->eval(x=>$xlim) / $dfl->eval(x=>$xlim); - } - - -

    - \lim\limits_{x\to 1} \frac{}{ } -

    - -

    - -

    - -     -     -     - -     -     -     -
    -
    - Improper Integration - -

    - We begin this section by considering the following definite integrals: -

    - -

    -

      -
    • \int_0^{100}\frac1{1+x^2}\, dx \approx 1.5608
      • - -
    • \int_0^{1000}\frac1{1+x^2}\, dx \approx 1.5698
      • - -
    • \int_0^{10,000}\frac1{1+x^2}\, dx \approx 1.5707
      • -
    -

    - -

    - Notice how the integrand is 1/(1+x^2) in each integral - (which is sketched in ). - As the upper bound gets larger, - one would expect the area under the curve would also grow. - While the definite integrals do increase in value as the upper bound grows, - they are not increasing by much. - In fact, consider: - - \int_0^b \frac{1}{1+x^2}\, dx = \tan^{-1}(x) \Big|_0^b = \tan^{-1}(b) -\tan^{-1}(0) = \tan^{-1}(b) - . -

    - -

    - As b\rightarrow \infty, - \tan^{-1}(b) \rightarrow \pi/2. - Therefore it seems that as the upper bound b grows, - the value of the definite integral - \ds \int_0^b\frac{1}{1+x^2}\, dx approaches \pi/2\approx 1.5708. - This should strike the reader as being a bit amazing: - even though the curve extends to infinity, - it has a finite amount of area underneath it. -

    - -
    - Graphing \ds f(x)=\frac{1}{1+x^2} - - - - Graph of function 1/(1+x^2) showing finite area under curves that extends to infinity. - - -

    - The y axis is drawn from 0 to 1 and the x axis is drawn - from 0 to 10. - The function f(x)=\frac{1}{1+x^2} starts at point (0,1), then decreases - sharply, has a dip at (2,0.2) then decreases gently and almost coincides with - the x axis after x=6. The area under the curve is shaded. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-1,xmax=11 - ] - - \addplot [firstcurvestyle,areastyle,domain=-0:1.5] {1/(1+x^2)} \closedcycle; - \addplot [firstcurvestyle,areastyle,domain=1.5:10.5] {1/(1+x^2)} \closedcycle; - - \addplot [firstcurvestyle,domain=0:1.5] {1/(1+x^2)}; - \addplot [firstcurvestyle,domain=1.5:10.5,samples=40] {1/(1+x^2)}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - - - -

    - When we defined the definite integral - \ds\int_a^b f(x)\, dx, we made two stipulations: -

    - -

    -

      -
    1. -

      - The interval over which we integrated, - [a,b], was a finite interval, and -

      -
      2. - -
    2. -

      - The function f(x) was continuous on [a,b] - (ensuring that the range of f was finite). -

      -
      3. -
    -

    - -

    - In this section we consider integrals where one or both of the above conditions do not hold. - Such integrals are called improper integrals. -

    -     - - - Improper Integrals with Infinite Bounds - - - Improper Integrals with Infinite Bounds; Converge, Diverge - -

    -

      -
    1. -

      - Let f be a continuous function on [a,\infty). - Define - - integrationimproper - improper integration - convergenceof improper int. - divergenceof improper int. - - - \int_a^\infty f(x)\, dx \text{ to be } \lim_{b\to\infty}\int_a^b f(x)\, dx - . -

      -
      2. - -
    2. -

      - Let f be a continuous function on (-\infty,b]. - Define - - \int_{-\infty}^b f(x)\, dx \text{ to be } \lim_{a\to-\infty}\int_a^b f(x)\, dx - . -

      -
      3. - -
    3. -

      - Let f be a continuous function on (-\infty,\infty). - Let c be any real number; define - - \int_{-\infty}^\infty f(x)\, dx \text{ to be } \lim_{a\to-\infty}\int_a^c f(x)\, dx\,+\,\lim_{b\to\infty}\int_c^b f(x)\, dx - . -

      -
      4. -
    -

    - -

    - An improper integral is said to converge - if its corresponding limit exists; - otherwise, it diverges. - The improper integral in part 3 converges if and only if both of its limits exist. -

    -     -     - - - Evaluating improper integrals - -

    - Evaluate the following improper integrals. -

    - -

    -

      -
    1. \int_1^\infty \frac1{x^2}\, dx
      2. - -
    2. \int_1^\infty \frac1x\, dx
      3. - -
    3. \int_{-\infty}^0 e^x\, dx
      4. - -
    4. \int_{-\infty}^\infty \frac1{1+x^2}\, dx
      5. -
    -

    -     - -

    -

      -
    1. -

      - - \int_1^\infty \frac{1}{x^2}\, dx = \lim_{b\to\infty} \int_1^b\frac1{x^2}\, dx \amp = \lim_{b\to\infty} \frac{-1}{x}\Big|_1^b - \amp = \lim_{b\to\infty} \frac{-1}{b} + 1 - \amp = 1 - . - - A graph of the area defined by this integral is given in . -

      - -
      - A graph of f(x) = \frac{1}{x^2} in - - - - Graph of the area under the curve 1/x^2 from 1 to infinity. - - -

      - The y axis is drawn from 0 to 1 and - the x axis is drawn from 0 to 10. The - function f(x) = \frac{1}{x^2} starts from point - (1,1), the curve drops sharply then bends before - x=5 after which it runs very close to the x - axis. The area under the curve is shaded from x=1 to - x=10. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,5,10}, - ymin=-.1,ymax=1.1, - xmin=-1,xmax=11 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:10.5] {1/(x^2)} \closedcycle; - \addplot [firstcurvestyle,domain=1:10.5,samples=50] {1/(x^2)}; - - \draw (axis cs:5,.75) node { $\ds f(x)=\frac{1}{x^2}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      2. - -
    2. -

      - - \int_1^\infty \frac1x\, dx \amp = \lim_{b\to\infty}\int_1^b\frac1x\, dx - \amp = \lim_{b\to\infty} \ln\abs{x}\Big|_1^b - \amp = \lim_{b\to\infty} \ln(b) - \amp = \infty - . - - The limit does not exist, - hence the improper integral \ds\int_1^\infty\frac1x\, dx diverges. - Compare the graphs in Figures - and ; - notice how the graph of f(x) = 1/x is noticeably larger. - This difference is enough to cause the improper integral to diverge. -

      - -
      - A graph of f(x) = \frac{1}{x} in - - - - Graph of area of the curve 1/x from 1 to 10. - - -

      - The y axis is drawn from 0 to 1 and the - x axis is drawn from 0 to 10. The function - f(x) = \frac{1}{x} starts from point (1,1), the - curve drops sharply then bends before x=5 after which it - runs almost parallel to the x axis about line y=0.2. - The area under the curve is shaded from x=1 to x=10. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,5,10}, - ymin=-.1,ymax=1.1, - xmin=-1,xmax=11 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:10.5] {1/x} \closedcycle; - \addplot [firstcurvestyle,domain=1:10.5,samples=40] {1/x}; - - \draw (axis cs:5,.75) node { $\ds f(x)=\frac{1}{x}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      3. - -
    3. -

      - - \int_{-\infty}^0 e^x \, dx \amp = \lim_{a\to-\infty} \int_a^0e^x\, dx - \amp = \lim_{a\to-\infty} e^x\Big|_a^0 - \amp = \lim_{a\to-\infty} e^0-e^a - \amp = 1 - . - - A graph of the area defined by this integral is given in . -

      - -
      - A graph of f(x) = e^x in - - - - Graph of area under the curve f(x)=e^x from -10 to 0. - - -

      - The graph of function f(x)= e^x is drawn in the - second quadrant.The y axis is drawn from 0 - to 1 and the x axis is drawn from -10 - to 0. From left to right the function appears to - coincide with the x axis, at about x=5 it gets - a positive slope and rises sharply until it reaches point (0,1). -

      -       - - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,-5,-10}, - ytick={1}, - ymin=-.1,ymax=1.1, - xmin=-11,xmax=1 - ] - - \addplot [firstcurvestyle,areastyle,domain=-10.5:0] {e^x} \closedcycle; - \addplot [firstcurvestyle,domain=-10.5:0,samples=50] {e^x}; - - \draw (axis cs:-5,.75) node { $\ds f(x)=e^x$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      4. - -
    4. -

      - We will need to break this into two improper integrals and choose a value of c as in part 3 of . - Any value of c is fine; - we choose c=0. - - \int_{-\infty}^\infty \frac1{1+x^2}\, dx \amp = \lim_{a\to-\infty} \int_a^0\frac{1}{1+x^2}\, dx + \lim_{b\to\infty} \int_0^b\frac{1}{1+x^2}\, dx - \amp = \lim_{a\to-\infty} \tan^{-1}(x) \Big|_a^0 + \lim_{b\to\infty} \tan^{-1}(x) \Big|_0^b - \amp = \lim_{a\to-\infty} \left(\tan^{-1}(0) -\tan^{-1}(a) \right) - \amp \quad\quad + \lim_{b\to\infty} \left(\tan^{-1}(b) -\tan^{-1}(0) \right) - \amp = \left(0-\frac{-\pi}2\right) + \left(\frac{\pi}2-0\right). - Each limit exists, hence the original integral converges and has value: - \amp = \pi - . - A graph of the area defined by this integral is given in . -

      - -
      - A graph of f(x) = \frac{1}{1+x^2} in - - - - Graph of function 1/(1+x^2). - - -

      - The y axis is drawn from 0 to 1 and the x - axis is drawn from -10 to 10.The function f(x)= \frac{1}{1+x^2} - is symmetrical about the y axis. From left to right the function - moves very closely to the x axis after x=-5 it bends and sharply - increases to point (0,1), from this point it decreases sharply and then - gets a dip then after x=5 runs very closely to the x axis. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ytick={1}, - ymin=-.1,ymax=1.1, - xmin=-11,xmax=11 - ] - - \addplot [firstcurvestyle,areastyle,domain=-10.5:-.1] {1/(1+x^2)} \closedcycle; - \addplot [firstcurvestyle,areastyle,domain=-.1:.1] {1/(1+x^2)} \closedcycle; - \addplot [firstcurvestyle,areastyle,domain=.1:10.5] {1/(1+x^2)} \closedcycle; - - \addplot [firstcurvestyle,domain=-10.5:-1,samples=40] {1/(1+x^2)}; - \addplot [firstcurvestyle,domain=-1:1,samples=30] {1/(1+x^2)}; - \addplot [firstcurvestyle,domain=1:10.5,samples=40] {1/(1+x^2)}; - - \draw (axis cs:6,.75) node { $\ds f(x)=\frac{1}{1+x^2}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      5. -
    -

    -     - -     - -

    - introduced L'Hospital's Rule, - a method of evaluating limits that return indeterminate forms. - It is not uncommon for the limits resulting from improper integrals to need this rule as demonstrated next. -

    - - - Improper integration and L'Hospital's Rule - -

    - Evaluate the improper integral \ds \int_1^\infty \frac{\ln(x) }{x^2}\, dx. -

    -     - -

    - This integral will require the use of Integration by Parts. - Let u = \ln(x) and dv = 1/x^2\, dx. - Then -

    - -
    - A graph of f(x) = \frac{\ln(x) }{x^2} in - - - - Graph of lx(x)/x^2. - - -

    - The y axis is drawn from 0 to 0.4 and the x - axis is drawn from 0 to 10. From left to right, the function - f(x) = \frac{\ln(x) }{x^2} starts at point (1,0) it rises sharply, - reaches a height of 0.2 near x= 1.5 then it decreases gently until - x=10 but stays a bit above the x axis. The area under the curve - until x=10 is shaded. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,5,10}, - ymin=-.1,ymax=.5, - xmin=-1,xmax=11 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:2] {ln(x)/x^2} \closedcycle; - \addplot [firstcurvestyle,areastyle,domain=2:10.5] {ln(x)/x^2} \closedcycle; - - \addplot [firstcurvestyle,domain=1:2] {ln(x)/x^2}; - \addplot [firstcurvestyle,domain=2:10.5] {ln(x)/x^2}; - - \draw (axis cs:5,.3) node { $\ds f(x)=\frac{\ln(x) }{x^2}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -

    - - \int_1^\infty\frac{\ln(x) }{x^2}\, dx \amp = \lim_{b\to\infty}\int_1^b\frac{\ln(x) }{x^2}\, dx - \amp = \lim_{b\to\infty}\left(-\frac{\ln(x) }{x}\Big|_1^b +\int_1^b \frac{1}{x^2} \, dx \right) - \amp = \lim_{b\to\infty} \left.\left(-\frac{\ln(x) }{x} -\frac1x\right)\right|_1^b - \amp = \lim_{b\to\infty} \left(-\frac{\ln(b) }{b}-\frac1b - \left(-\ln(1) -1\right)\right). - The 1/b and \ln(1) terms go to 0, leaving \lim\limits_{b\to\infty} -\frac{\ln(b) }b + 1. We need to evaluate \lim\limits_{b\to\infty} \frac{\ln(b) }{b} with l'Hospital's Rule. We have: - \lim_{b\to\infty}\frac{\ln(b) }b \amp \stackrel{\,\text{ by LHR } \,}{=} \lim_{b\to\infty} \frac{1/b}{1} - \amp = 0. - Thus the improper integral evaluates as: - \int_1^\infty\frac{\ln(x) }{x^2}\, dx \amp = 1 - . -

    -     - -     -     - - - Improper Integrals with Infinite Range - -

    - We have just considered definite integrals where the interval of integration was infinite. - We now consider another type of improper integration, - where the range of the integrand is infinite. -

    - - - Improper Integration with Infinite Range - -

    - Let f(x) be a continuous function on [a,b] except at c, - a\leq c\leq b, - where x=c is a vertical asymptote of f. - Define - - integrationimproper - improper integration - - - \int_a^b f(x)\, dx = \lim_{t\to c^-}\int_a^t f(x)\, dx + \lim_{t\to c^+}\int_t^b f(x)\, dx - . -

    -     -     - - - Improper integration of functions with infinite range - -

    - Evaluate the following improper integrals: -

    - -

    -

      -
    1. \int_0^1\frac1{\sqrt{x}}\, dx
      2. -
    2. \int_{-1}^1\frac{1}{x^2}\, dx
      3. -
    -

    -     - -

    -

      -
    1. -

      - A graph of f(x) = 1/\sqrt{x} is given in . - Notice that f has a vertical asymptote at x=0; - in some sense, - we are trying to compute the area of a region that has no top. - Could this have a finite value? - - \int_0^1 \frac{1}{\sqrt{x}}\, dx \amp = \lim_{a\to0^+}\int_a^1 \frac1{\sqrt{x}}\, dx - \amp = \lim_{a\to0^+} 2\sqrt{x}\Big|_a^1 - \amp = \lim_{a\to0^+} 2\left(\sqrt{1}-\sqrt{a}\right) - \amp = 2 - . - It turns out that the region does have a finite area even though it has no upper bound - (strange things can occur in mathematics when considering the infinite). -

      - - - - -
      - A graph of f(x)=\frac{1}{\sqrt{x}} in - - - - Graph of function 1 over square root of x. - - -

      - The y axis is drawn from 0 to 10 and the x - axis is drawn from 0 to 2. The function f(x)=\frac{1}{\sqrt{x}} - starts at (0,10) then it decreases sharply while being very close to the - y axis at about y=4 it starts diverging away, it moves away from - the y axis and appears to move almost parallel to the x axis until x=1. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=11, - xmin=-.1,xmax=2.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0.01:.2] {1/sqrt(x)} \closedcycle; - \addplot [firstcurvestyle,areastyle,domain=.2:1] {1/sqrt(x)} \closedcycle; - - \addplot [firstcurvestyle,domain=0.01:.2] {1/sqrt(x)}; - \addplot [firstcurvestyle,domain=.2:1] {1/sqrt(x)}; - - \draw (axis cs:.5,5) node { $\ds f(x)=\frac{1}{\sqrt{x}}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      2. - -
    2. -

      - The function f(x) = 1/x^2 has a vertical asymptote at x=0, - as shown in , - so this integral is an improper integral. - Let's eschew using limits for a moment and proceed without recognizing the improper nature of the integral. - This leads to: - - \int_{-1}^1\frac1{x^2}\, dx \amp = -\frac1x\Big|_{-1}^1 - \amp = -1 - (1) - \amp =-2. \,(!) - -

      - -
      - A graph of f(x)=\frac{1}{x^2} in - - - - Graph of function f(x) = 1/x^2. - - -

      - The y axis is drawn from -1 to 1 and the x axis is drawn - from 0 to 20. The graph of function f(x)=\frac{1}{x^2} is drawn - in the first and the second quadrants. From left to right, in the second quadrant the - function starts at x=-1 a little above the x axis then it curves up sharply - and runs parallel to the y axis, at x=-0.4. In the first quadrant, from - left to right the function runs parallel to the y axis and declines sharply, it - then bends and moves along the x axis until x=1. Between x=-1 and - x=1 and from y=0 to y=20 the area within the curves is shaded. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=21, - xmin=-1.1,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0.1:1,samples=30] {1/x^2} \closedcycle; - \addplot [firstcurvestyle,domain=0.1:1,samples=50] {1/x^2}; - \addplot [firstcurvestyle,areastyle,domain=-0.1:0.1] {22} \closedcycle; - \addplot [firstcurvestyle,areastyle,domain=-1:-0.1,samples=30] {1/x^2} \closedcycle; - \addplot [firstcurvestyle,domain=-1:-0.1,samples=50] {1/x^2}; - - \draw (axis cs:.7,7) node { $\ds f(x)=\frac{1}{x^2}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      - -

      - Clearly the area in question is above the x-axis, - yet the area is supposedly negative! - Why does our answer not match our intuition? - To answer this, - evaluate the integral using . - - \int_{-1}^1\frac1{x^2}\, dx \amp = \lim_{t\to0^-}\int_{-1}^t \frac1{x^2}\, dx + \lim_{t\to0^+}\int_t^1\frac1{x^2}\, dx - \amp = \lim_{t\to0^-}-\frac1x\Big|_{-1}^t + \lim_{t\to0^+}-\frac1x\Big|_t^1 - \amp = \lim_{t\to0^-}-\frac1t-1 + \lim_{t\to0^+} -1+\frac1t - \amp \Rightarrow \Big(\infty-1\Big)\,+ \,\Big(- 1+\infty\Big) - . - Neither limit converges hence the original improper integral diverges. - The nonsensical answer we obtained by ignoring the improper nature of the integral is just that: - nonsensical. -

      -
      3. -
    -

    -     - -     -     - - - Understanding Convergence and Divergence -

    - Oftentimes we are interested in knowing simply whether or not an improper integral converges, - and not necessarily the value of a convergent integral. - We provide here several tools that help determine the convergence or divergence of improper integrals without integrating. -

    - - - -

    - Our first tool is to understand the behavior of functions of the form \ds \frac1{x^p}. -

    - - - Improper integration of <m>1/x^p</m> - -

    - Determine the values of p for which \ds \int_1^\infty \frac1{x^p}\, dx converges. -

    -     - -

    - We begin by integrating and then evaluating the limit. - - \int_1^\infty \frac1{x^p}\, dx \amp = \lim_{b\to\infty}\int_1^b\frac1{x^p}\, dx - \amp = \lim_{b\to\infty}\int_1^b x^{-p}\, dx \qquad \text{ (assume \(p\neq 1\)) } - \amp = \lim_{b\to\infty} \frac{1}{-p+1}x^{-p+1}\Big|_1^b - \amp = \lim_{b\to\infty} \frac{1}{1-p}\big(b^{1-p}-1^{1-p}\big) - . -

    - -

    - When does this limit converge , when is this limit - not \infty? - This limit converges precisely when the power of b is less than 0: - when 1-p\lt 0 \Rightarrow 1\lt p. -

    - -
    - Plotting functions of the form 1/x^p in - - - - Graph of two functions 1/x^p and 1/x^q with p <1 <q. - - -

    - The y and the x axes are uncalibrated except that 1 - is marked in the middle of the x axis. There are two functions - f(x) = 1/x^p and 1/x^q shown in the graph. The two functions - intersect when x=1. Before x=1 both functions have negative - slopes and decline sharply along the y axis then gently as they approach - x=1, p<q hence 1/x^p graph is below 1/x^q. After - x=1, 1/x^p is above 1/x^q but they are only slightly apart. - There is a dashed line between the two functions. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1}, - ytick=\empty, - ymin=-.1,ymax=6, - xmin=-.1,xmax=2.1 - ] - - \addplot+ [smooth] coordinates {(0.1,31.62)(0.15,17.21)(0.2,11.18)(0.25,8.)(0.3,6.086)(0.35,4.829)(0.4,3.953)(0.45,3.313)(0.5,2.828)(0.55,2.452)(0.6,2.152)(0.65,1.908)(0.7,1.707)(0.75,1.54)(0.8,1.398)(0.85,1.276)(0.9,1.171)(0.95,1.08)(1.,1.)(1.05,0.9294)(1.1,0.8668)(1.15,0.8109)(1.2,0.7607)(1.25,0.7155)(1.3,0.6747)(1.35,0.6375)(1.4,0.6037)(1.45,0.5727)(1.5,0.5443)(1.55,0.5182)(1.6,0.4941)(1.65,0.4718)(1.7,0.4512)(1.75,0.432)(1.8,0.4141)(1.85,0.3974)(1.9,0.3818)(1.95,0.3672)(2.,0.3536)}; - \addplot+ [solid,smooth] coordinates {(.05,4.47)(0.1,3.162)(0.15,2.582)(0.2,2.236)(0.25,2.)(0.3,1.826)(0.35,1.69)(0.4,1.581)(0.45,1.491)(0.5,1.414)(0.55,1.348)(0.6,1.291)(0.65,1.24)(0.7,1.195)(0.75,1.155)(0.8,1.118)(0.85,1.085)(0.9,1.054)(0.95,1.026)(1.,1.)(1.05,0.9759)(1.1,0.9535)(1.15,0.9325)(1.2,0.9129)(1.25,0.8944)(1.3,0.8771)(1.35,0.8607)(1.4,0.8452)(1.45,0.8305)(1.5,0.8165)(1.55,0.8032)(1.6,0.7906)(1.65,0.7785)(1.7,0.767)(1.75,0.7559)(1.8,0.7454)(1.85,0.7352)(1.9,0.7255)(1.95,0.7161)(2.,0.7071)}; - \addplot [lineseg,dashed,thick,smooth] coordinates {(0.1,10.)(0.15,6.667)(0.2,5.)(0.25,4.)(0.3,3.333)(0.35,2.857)(0.4,2.5)(0.45,2.222)(0.5,2.)(0.55,1.818)(0.6,1.667)(0.65,1.538)(0.7,1.429)(0.75,1.333)(0.8,1.25)(0.85,1.176)(0.9,1.111)(0.95,1.053)(1.,1.)(1.05,0.9524)(1.1,0.9091)(1.15,0.8696)(1.2,0.8333)(1.25,0.8)(1.3,0.7692)(1.35,0.7407)(1.4,0.7143)(1.45,0.6897)(1.5,0.6667)(1.55,0.6452)(1.6,0.625)(1.65,0.6061)(1.7,0.5882)(1.75,0.5714)(1.8,0.5556)(1.85,0.5405)(1.9,0.5263)(1.95,0.5128)(2.,0.5)}; - - \draw (axis cs:.7,5) node { $\ds f(x)=\frac{1}{x^q}$}; - \draw (axis cs:.35,.8) node { $\ds f(x)=\frac{1}{x^p}$}; - \draw (axis cs:1.5,3) node { $p<1<q$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Our analysis shows that if p \gt 1, - then \ds\int_1^\infty \frac1{x^p}\, dx converges. - When p\lt 1 the improper integral diverges; - we showed in - that when p=1 the integral also diverges. -

    - -

    - - graphs y=1/x with a dashed line, - along with graphs of y=1/x^p, p\lt 1, - and y=1/x^q, q \gt 1. - Somehow the dashed line forms a dividing line between convergence and divergence. -

    -     - -     - -

    - The result of - provides an important tool in determining the convergence of other integrals. - A similar result is proved in the exercises about improper integrals of the form \ds \int_0^1\frac1{x^p}\, dx. - These results are summarized in the following Key Idea. -

    - - - Convergence of Improper Integrals involving <m>1/x^p</m> -

    -

      -
    1. -

      - The improper integral \ds \int_1^\infty\frac1{x^p}\, dx converges when p\gt 1 and diverges when p\leq 1. - - convergenceof improper int. - divergenceof improper int. -

      -
      2. - -
    2. -

      - The improper integral \ds \int_0^1\frac1{x^p}\, dx converges when p\lt 1 and diverges when p\geq 1. -

      -
      3. -
    -

    -     - -

    - A basic technique in determining convergence of improper integrals is to compare an integrand whose convergence is unknown to an integrand whose convergence is known. - We often use integrands of the form 1/x^p to compare to as their convergence on certain intervals is known. - This is described in the following theorem. -

    - - - - - Direct Comparison Test for Improper Integrals - -

    - Let f and g be continuous on [a,\infty) where - 0\leq f(x)\leq g(x) for all x in [a,\infty). -

    - -

    -

      -
    1. -

      - If \ds \int_a^\infty g(x)\, dx converges, - then \ds \int_a^\infty f(x)\, dx converges. - integrationimproper - convergenceDirect Comparison Test!for integration - divergenceDirect Comparison Test!for integration - Direct Comparison Testfor integration - convergenceof improper int. - divergenceof improper int. -

      -
      2. - -
    2. -

      - If \ds \int_a^\infty f(x)\, dx diverges, - then \ds \int_a^\infty g(x)\, dx diverges. -

      -
      3. -
    -

    -     -     - - - Determining convergence of improper integrals - -

    - Determine the convergence of the following improper integrals. -

    - -

    -

      -
    1. \int_1^\infty e^{-x^2}\, dx
      2. -
    2. \int_3^\infty \frac{1}{\sqrt{x^2-x}}\, dx
      3. -
    -

    -     - -

    -

      -
    1. -

      - The function f(x) = e^{-x^2} does not have an antiderivative expressible in terms of elementary functions, - so we cannot integrate directly. - It is comparable to g(x)=1/x^2, - and as demonstrated in , - e^{-x^2} \lt 1/x^2 on [1,\infty). - We know from - that \ds \int_1^\infty \frac{1}{x^2}\, dx converges, - hence \ds\int_1^\infty e^{-x^2}\, dx also converges. -

      - -
      - Graphs of f(x) = e^{-x^2} and f(x)= 1/x^2 in - - - - Graph of two functions e to the power negative x^2 and 1/x^2. - - -

      - The y axis is drawn from 0 to 1 and the x axis - is drawn from 0 to 4. There are two functions drawn - f(x) = e^{-x^2} and f(x)= 1/x^2. The function e appears - to start from (1,0.4) then decreases and merges with the x axis - after x=2. The function 1/x^2 is above the function e^{-x^2}, - it starts from point (1,1) and curves down and appears to be parallel to the - x axis after x=3. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=4.1 - ] - - \addplot+ [domain=1:4,samples=40] {e^-x^2}; - \addplot+ [solid,domain=1:4,samples=40] {1/x^2}; - - \draw (axis cs:.7,.45) node { $\ds f(x)=e^{-x^2}$}; - \draw (axis cs:1.9,.8) node { $\ds f(x)=\frac{1}{x^2}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      2. - -
    2. -

      - Note that for large values of x, - \ds \frac{1}{\sqrt{x^2-x}} \approx \frac{1}{\sqrt{x^2}} =\frac{1}{x}. - We know from - and the subsequent note that \ds \int_3^\infty \frac1x\, dx diverges, - so we seek to compare the original integrand to 1/x. - - It is easy to see that when x \gt 0, - we have x = \sqrt{x^2} \gt \sqrt{x^2-x}. - Taking reciprocals reverses the inequality, giving - - \frac1x \lt \frac1{\sqrt{x^2-x}} - . - Using , - we conclude that since \ds\int_3^\infty\frac1x\, dx diverges, - \ds\int_3^\infty\frac1{\sqrt{x^2-x}}\, dx diverges as well. - illustrates this. -

      - -
      - Graphs of f(x) = 1/\sqrt{x^2-x} and f(x)= 1/x in - - - - Graph of the two functions used in this example. - - -

      - The y axis is drawn from 0 to 0.4 and the x axis is - drawn from 0 to 6. Two functions f(x)=1/x and - f(x) = 1/\sqrt{x^2-x} are drawn. At x=3, the function 1/x - starts at y=3.33 then curves down with a negative slope. The function - f(x) = 1/\sqrt{x^2-x} at x=3 starts at y=0.41 and also - curves down with a negative slope. The function 1/x is below function - f(x) = 1/\sqrt{x^2-x} they are comparatively more apart at x=3 - than at x=6. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=.5, - xmin=-.1,xmax=6.2 - ] - - \addplot+ [domain=3:6] {1/sqrt((x^2)-x)}; - \addplot+ [solid,domain=3:6] {1/x}; - - \draw (axis cs:4,.45) node { $\ds f(x)=\frac{1}{\sqrt{x^2-x}}$}; - \draw (axis cs:2.5,.2) node { $\ds f(x)=\frac{1}{x}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      3. -
    -

    -     - -     - -

    - Being able to compare unknown - integrals to known - integrals is very useful in determining convergence. - However, some of our examples were a little - too nice. For instance, - it was convenient that \ds \frac{1}x \lt \frac{1}{\sqrt{x^2-x}}, - but what if the -x - were replaced with a +2x+5? - That is, what can we say about the convergence of \ds \int_3^\infty\frac{1}{\sqrt{x^2+2x+5}}\, dx? - We have \ds \frac{1}{x} \gt \frac1{\sqrt{x^2+2x+5}}, - so we cannot use . -

    - -

    - In cases like this - (and many more) - it is useful to employ the following theorem. -

    - - - Limit Comparison Test for Improper Integrals - -

    - Let f and g be continuous functions on - [a,\infty) where f(x) \gt 0 and g(x) \gt 0 for all x. - If - - \lim_{x\to\infty} \frac{f(x)}{g(x)} = L,\qquad 0\lt L\lt \infty - , - then - - \int_a^\infty f(x)\, dx \text{ and } \int_a^\infty g(x)\, dx - - either both converge or both diverge. - integrationimproper - convergenceLimit Comparison Test!for integration - divergenceLimit Comparison Test!for integration - Limit Comparison Testfor integration - convergenceof improper int. - divergenceof improper int. -

    -     -     - - - Determining convergence of improper integrals - -

    - Determine the convergence of \ds \int_3^{\infty} \frac{1}{\sqrt{x^2+2x+5}}\, dx. -

    -     - -

    - As x gets large, - the denominator of the integrand will begin to behave much like y=x. - So we compare \ds\frac{1}{\sqrt{x^2+2x+5}}to \ds\frac1xwith the Limit Comparison Test: - - \lim_{x\to\infty} \frac{1/\sqrt{x^2+2x+5}}{1/x} = \lim_{x\to\infty}\frac{x}{\sqrt{x^2+2x+5}} - . -

    - -

    - The immediate evaluation of this limit returns - \infty/\infty, an indeterminate form. - Using L'Hospital's Rule seems appropriate, - but in this situation, it does not lead to useful results. - (We encourage the reader to employ L'Hospital's Rule at least once to verify this.) -

    - -

    - The trouble is the square root function. - To get rid of it, - we employ the following fact: If \lim\limits_{x\to c} f(x) = L, - then \lim\limits_{x\to c} f(x)^2 = L^2. - (This is true when either c or L is \infty.) - So we consider now the limit - - \lim_{x\to\infty} \frac{x^2}{x^2+2x+5} - . -

    - -

    - This converges to 1, meaning the original limit also converged to 1. - As x gets very large, the function - \ds\frac{1}{\sqrt{x^2+2x+5}}looks very much like \ds\frac1x. - Since we know that \ds\int_3^{\infty} \frac1x\, dxdiverges, - by the Limit Comparison Test we know that \ds\int_3^\infty\frac{1}{\sqrt{x^2+2x+5}}\, dxalso diverges. - - graphs f(x)=1/\sqrt{x^2+2x+5} and f(x)=1/x, - illustrating that as x gets large, - the functions become indistinguishable. -

    - -
    - Graphing f(x)=\frac{1}{\sqrt{x^2+2x+5}} and f(x)=\frac1x in - - - - Graph of the two functions used in this example. - - -

    - The y axis is drawn from -0.1 to 0.3 and the x axis - is drawn from 0 to 20. Two functions are drawn f(x)= 1/x - and f(x)=\frac{1}{\sqrt{x^2+2x+5}}. The function f(x)=\frac{1}{\sqrt{x^2+2x+5}} - is drawn below 1/x. Both functions have negative slopes, they are apart by a small - distance around x=4, after x=10 they come very close and almost coincide. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=.35, - xmin=-.1,xmax=21 - ] - - \addplot+ [domain=3:20,samples=40] {1/sqrt(x^2+2*x+5)}; - \addplot+ [solid,domain=3:20,samples=50] {1/x}; - - \draw (axis cs:4.5,.08) node { $\ds f(x)=\frac{1}{\sqrt{x^2+2x+5}}$}; - \draw (axis cs:6,.3) node { $\ds f(x)=\frac{1}{x}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -     - - - -

    - Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. - There are versions that apply to improper integrals with an infinite range, - but as they are a bit wordy and a little more difficult to employ, - they are omitted from this text. -

    - - - -

    - This chapter has explored many integration techniques. - We learned Substitution, - which undoes the Chain Rule of differentiation, - as well as Integration by Parts, - which undoes the Product Rule. - We learned specialized techniques for handling trigonometric functions and introduced the hyperbolic functions, - which are closely related to the trigonometric functions. - All techniques effectively have this goal in common: - rewrite the integrand in a new way so that the integration step is easier to see and implement. -

    - -

    - As stated before, integration is, in general, hard. - It is easy to write a function whose antiderivative is impossible to write in terms of elementary functions, - and even when a function does have an antiderivative expressible by elementary functions, - it may be really hard to discover what it is. - The powerful computer algebra system - Mathematica has approximately 1,000 pages of code dedicated to integration. -

    - -

    - Do not let this difficulty discourage you. - There is great value in learning integration techniques, - as they allow one to manipulate an integral in ways that can illuminate a concept for greater understanding. - There is also great value in understanding the need for good numerical techniques: - the Trapezoidal and Simpson's Rules are just the beginning of powerful techniques for approximating the value of integration. -

    - -

    - The next chapter stresses the uses of integration. - We generally do not find antiderivatives for antiderivative's sake, - but rather because they provide the solution to some type of problem. - The following chapter introduces us to a number of different problems whose solution is provided by integration. -

    -     - - - - Terms and Concepts - - - - -

    - The definite integral was defined with what two stipulations? -

    - - -     - - - -

    - The interval of integration is finite, - and the integrand is continuous on that interval. -

    - -     - -     - - - - -

    - If \lim\limits_{b\to \infty} \int_0^b f(x)\, dx exists, - then the integral \ds \int_0^\infty f(x)\, dx is said to . -

    -     - - - - - - - - -

    - converge -

    -     - -     - - - - -

    - If \ds \int_1^\infty f(x)\, dx=10, - and 0\leq g(x)\leq f(x) for all x, - then we know that \ds \int_1^\infty g(x)\, dx . -

    -     - - - - converges|<10|< 10 - - - - -

    - converges; could also state \lt 10. -

    -     - -     - - - -

    - For what values of p will \ds \int_1^\infty \frac1{x^p}\, dx converge? -

    -     - - - -

    - p\lt 1 -

    -     -     - - -

    - p\leq 1 -

    -     -     - - -

    - p\gt 1 -

    -     -     - - -

    - p\geq 1 -

    -     -     -     -     - - - -

    - For what values of p will \ds \int_{10}^\infty \frac1{x^p}\, dx converge? -

    -     - - - -

    - p\lt 1 -

    -     -     - - -

    - p\leq 1 -

    -     -     - - -

    - p\gt 1 -

    -     -     - - -

    - p\geq 1 -

    -     -     -     -     - - - -

    - For what values of p will \ds \int_{0}^1 \frac1{x^p}\, dx converge? -

    -     - - - -

    - p\lt 1 -

    -     -     - - -

    - p\leq 1 -

    -     -     - - -

    - p\gt 1 -

    -     -     - - -

    - p\geq 1 -

    -     -     -     -     - -     - - Problems - - - -

    - Evaluate the given improper integral. -

    -     - - - - - $f = Formula("e^(5-2x)"); - $l = Compute("e^5/2"); - - -

    - \ds \int_0^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("1/x^3"); - $l = Compute("1/2"); - - -

    - \ds \int_1^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("x^-4"); - $l = Compute("1/3"); - - -

    - \ds \int_1^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("1/(x^2+9)"); - $l = Compute("pi/3"); - - -

    - \ds \int_{-\infty}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("2^x"); - $l = Compute("1/ln(2)"); - - -

    - \ds \int_{-\infty}^0 \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("(1/2)^x"); - $l = Compute("inf"); - - -

    - \ds \int_{-\infty}^0 \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("x/(x^2+1)"); - $l = Compute("inf"); - - -

    - \ds \int_{-\infty}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("x/(x^2-4)"); - $l = Compute("inf"); - - -

    - \ds \int_{3}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("1/(x-1)^2"); - $l = Compute("1"); - - -

    - \ds \int_{2}^\infty\, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("1/(x-1)^2"); - $l = Compute("inf"); - - -

    - \ds \int_{1}^2\, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("1/(x-1)"); - $l = Compute("inf"); - - -

    - \ds \int_{2}^\infty\, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("1/(x-1)"); - $l = Compute("inf"); - - -

    - \ds \int_{1}^2\, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("1/x"); - $l = Compute("inf"); - - -

    - \ds \int_{-1}^1 \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("1/(x-2)"); - $l = Compute("inf"); - - -

    - \ds \int_{1}^3\, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("sec(x)^2"); - $l = Compute("inf"); - - -

    - \ds \int_{0}^\pi \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("1/sqrt(abs(x))"); - $l = Compute("2+2*sqrt(2)"); - - -

    - \ds \int_{-2}^1 \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("x*e^-x"); - $l = Compute("1"); - - -

    - \ds \int_{0}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("x*e^(-x^2)"); - $l = Compute("1/2"); - - -

    - \ds \int_{0}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("x*e^(-x^2)"); - $l = Compute("0"); - - -

    - \ds \int_{-\infty}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("1/(e^x+e^-x)"); - $l = Compute("pi/2"); - - -

    - \ds \int_{-\infty}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("x*ln(x)"); - $l = Compute("-1/4"); - - -

    - \ds \int_{0}^1 \, dx -

    - -

    - -

    -     -     -     - - - - - - $f = Formula("x^2*ln(x)"); - $l = Compute("-1/9"); - - -

    - \ds \int_{0}^1 \, dx -

    - -

    - -

    -     -     -     - - - - - - $f = Formula("ln(x)/x"); - $l = Compute("inf"); - - -

    - \ds \int_{1}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("ln(x)"); - $l = Compute("-1"); - - -

    - \ds \int_{0}^1 \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("ln(x)/x^2"); - $l = Compute("1"); - - -

    - \ds \int_{1}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("ln(x)/sqrt(x)"); - $l = Compute("inf"); - - -

    - \ds \int_{1}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("e^-x*sin(x)"); - $l = Compute("1/2"); - - -

    - \ds \int_{0}^\infty \, dx -

    - -

    - -

    -     -     -     - - - - - $f = Formula("e^-x*cos(x)"); - $l = Compute("1/2"); - - -

    - \ds \int_{0}^\infty \, dx -

    - -

    - -

    -     -     -     - -     - - - -

    - Use the Direct Comparison Test or the Limit Comparison Test to determine whether the given definite integral converges or diverges. - Clearly state what test is being used and what function the integrand is being compared to. -

    -     - - - - - parserPopUp.pl - - - $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); - $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); - $converges = DropDown(['converges','diverges'],0,showInStatic=>0); - $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); - $comp_f = Formula("1/x"); - - -

    - \ds \int_{10}^\infty \frac{3}{\sqrt{3x^2+2x-5}} \, dx -

    - - - Select the test to be used. - - -

    - -

    - - - Determine whether the integral converges or diverges. - - -

    - -

    - - - Enter the function used for comparison. - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); - $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); - $converges = DropDown(['converges','diverges'],0,showInStatic=>0); - $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); - $comp_f = Formula("1/x^(3/2)"); - - -

    - \ds \int_{2}^\infty \frac{4}{\sqrt{7x^3-x}} \, dx -

    - - - Select the test to be used. - - -

    - -

    - - - Determine whether the integral converges or diverges. - - -

    - -

    - - - Enter the function used for comparison. - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); - $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); - $converges = DropDown(['converges','diverges'],0,showInStatic=>0); - $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); - $comp_f = Formula("1/x"); - - -

    - \ds \int_{0}^\infty \frac{\sqrt{x+3}}{\sqrt{x^3-x^2+x+1}} \, dx -

    - - - Select the test to be used. - - -

    - -

    - - - Determine whether the integral converges or diverges. - - -

    - -

    - - - Enter the function used for comparison. - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); - $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); - $converges = DropDown(['converges','diverges'],0,showInStatic=>0); - $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); - $comp_f = Formula("x*e^-x"); - - -

    - \ds \int_{1}^\infty e^{-x}\ln(x) \, dx -

    - - - Select the test to be used. - - -

    - -

    - - - Determine whether the integral converges or diverges. - - -

    - -

    - - - Enter the function used for comparison. - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); - $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); - $converges = DropDown(['converges','diverges'],0,showInStatic=>0); - $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); - $comp_f = Formula("e^-x"); - - -

    - \ds \int_{5}^\infty e^{-x^2+3x+1} \, dx -

    - - - Select the test to be used. - - -

    - -

    - - - Determine whether the integral converges or diverges. - - -

    - -

    - - - Enter the function used for comparison. - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); - $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); - $converges = DropDown(['converges','diverges'],0,showInStatic=>0); - $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); - $comp_f = Formula("x*e^-x"); - - -

    - \ds \int_{0}^\infty \frac{\sqrt{x}}{e^x} \, dx -

    - - - Select the test to be used. - - -

    - -

    - - - Determine whether the integral converges or diverges. - - -

    - -

    - - - Enter the function used for comparison. - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); - $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); - $converges = DropDown(['converges','diverges'],0,showInStatic=>0); - $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); - $comp_f = Formula("1/(x^2-1)"); - - -

    - \ds \int_{2}^\infty \frac{1}{x^2+\sin(x) } \, dx -

    - - - Select the test to be used. - - -

    - -

    - - - Determine whether the integral converges or diverges. - - -

    - -

    - - - Enter the function used for comparison. - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); - $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); - $converges = DropDown(['converges','diverges'],0,showInStatic=>0); - $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); - $comp_f = Formula("x/(x^2+1)"); - - - -

    - \ds \int_{0}^\infty \frac{x}{x^2+\cos(x) } \, dx -

    - - - Select the test to be used. - - -

    - -

    - - - Determine whether the integral converges or diverges. - - -

    - -

    - - - Enter the function used for comparison. - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); - $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); - $converges = DropDown(['converges','diverges'],0,showInStatic=>0); - $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); - $comp_f = Formula("1/e^x"); - - -

    - \ds \int_{0}^\infty \frac{1}{x+e^x} \, dx -

    - - - Select the test to be used. - - -

    - -

    - - - Determine whether the integral converges or diverges. - - -

    - -

    - - - Enter the function used for comparison. - - -

    - -

    -     -     -     - - - - - parserPopUp.pl - - - $DCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],0,showInStatic=>0); - $LCT = DropDown(['Direct Comparison Test','Limit Comparison Test'],1,showInStatic=>0); - $converges = DropDown(['converges','diverges'],0,showInStatic=>0); - $diverges = DropDown(['converges','diverges'],1,showInStatic=>0); - $comp_f = Formula("1/e^x"); - - -

    - \ds \int_{0}^\infty \frac{1}{e^x-x} \, dx -

    - - - Select the test to be used. - - -

    - -

    - - - Determine whether the integral converges or diverges. - - -

    - -

    - - - Enter the function used for comparison. - - -

    - -

    -     -     -     - -     -     -     -
    -     - - - Applications of Integration - -

    - We begin this chapter with a reminder of a few key concepts from . - Let f be a continuous function on [a,b] which is partitioned into n equally spaced subintervals as - - a=x_0 \lt x_1 \lt \cdots \lt x_n\lt x_{n}=b - . -

    - -

    - Let \dx=(b-a)/n denote the length of the subintervals, - and let c_i be any x-value in the ith subinterval. - states that the sum - - \sum_{i=1}^n f(c_i)\dx - - is a Riemann Sum. Riemann Sums are often used to approximate some quantity - (area, volume, work, pressure, etc.). - The approximation becomes - exact by taking the limit - - \lim_{n\to\infty} \sum_{i=1}^n f(c_i)\dx - . -

    - -

    - - connects limits of Riemann Sums to definite integrals: - - \lim_{n\to\infty} \sum_{i=1}^n f(c_i)\dx = \int_a^b f(x)\, dx - . -

    - -

    - Finally, the Fundamental Theorem of Calculus states how definite integrals can be evaluated using antiderivatives. -

    - -

    - This chapter employs the following technique to a variety of applications. - Suppose the value Q of a quantity is to be calculated. - We first approximate the value of Q using a Riemann Sum, - then find the exact value via a definite integral. - We spell out this technique in the following Key Idea. -

    - - - Application of Definite Integrals Strategy -

    - Let a quantity be given whose value Q is to be computed. - integrationgeneral application technique -

      -
    1. -

      - Divide the quantity into n smaller - subquantities of value Q_i. -

      -
      2. - -
    2. -

      - Identify a variable x and function f(x) such that each subquantity can be approximated with the product f(c_i)\dx, - where \dx represents a small change in x. - Thus Q_i \approx f(c_i)\dx. - A sample approximation f(c_i)\dx of Q_i is called a - differential element. -

      -
      3. - -
    3. -

      - Recognize that \ds Q= \sum_{i=1}^n Q_i \approx \sum_{i=1}^n f(c_i)\dx, - which is a Riemann Sum. -

      -
      4. - -
    4. -

      - Taking the appropriate limit gives \ds Q = \int_a^b f(x)\, dx -

      -
      5. -
    -

    -     - -

    - This Key Idea will make more sense after we have had a chance to use it several times. - We begin with Area Between Curves, - which we addressed briefly in . -

    -     - -
    - Area Between Curves -

    - We are often interested in knowing the area of a region. - Forget momentarily that we addressed this already in - and approach it instead using the technique described in . -

    - - - -

    - Let Q be the area of a region bounded by continuous functions f and g. - If we break the region into many subregions, - we have an obvious equation: -

    - -

    - Total Area = sum of the areas of the subregions. -

    - -

    - The issue to address next is how to systematically break a region into subregions. - A graph will help. - Consider - where a region between two curves is shaded. - While there are many ways to break this into subregions, - one particularly efficient way is to - slice it vertically, - as shown in , - into n equally spaced slices. -

    - -

    - We now approximate the area of a slice. - Again, we have many options, but using a rectangle seems simplest. - Picking any x-value c_i in the ith slice, - we set the height of the rectangle to be f(c_i)-g(c_i), - the difference of the corresponding y-values. - The width of the rectangle is a small difference in x-values, - which we represent with \dx. - - shows sample points c_i chosen in each subinterval and appropriate rectangles drawn. - (Each of these rectangles represents a differential element.) - Each slice has an area approximately equal to \big(f(c_i)-g(c_i)\big)\dx; - hence, the total area is approximately the Riemann Sum - - Q = \sum_{i=1}^n \big(f(c_i)-g(c_i)\big)\dx - . -

    - -

    - Taking the limit as n\to \infty gives the exact area as \int_a^b \big(f(x)-g(x)\big)\, dx. -

    - -
    - Subdividing a region into vertical slices and approximating the areas with rectangles - -
    - - - - -

    - Graph of two functions f(x) and g(x) lying on the xy plane. - The x-axis contains two marked points, a and b, which are both plotted on the positive x-axis. - The functions f(x) and g(x) cross at the y-axis before diverging. - The function f(x) lies above the function g(x) and resembles a sine wave. - The function g(x) is a quadratic function which lies below f(x) on the interval (a,b). - The area lying between the marked points a and b and below f(x) and above g(x) is shaded, which gives the exact area of this region. -

    -     - Graph of two curves f and g with a shaded region between two points on the x-axis and the two curves. - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={.5,3}, - extra x tick labels={$a$,$b$}, - ytick=\empty, - ymin=-.2,ymax=2, - xmin=-.5,xmax=3.5 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.25:3.25,samples=40] {.25*sin(deg(2*x))+1.25} node [shift={(5pt,7pt)} ,black] { $f(x)$}; - \addplot [name path=B,firstcurvestyle,domain=-.25:3.25] {.25*(x-2)^2+.2}node [shift={(5pt,7pt)} ,black] { $g(x)$}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0.5:3}]; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - -

    - Graph of the same two functions f(x) and g(x) as in the previous image. - Now, the area lying between the marked points a and b and below f(x) and above g(x) is subdivided into 10 exact vertical slices, whose areas added together give the exact area of this region. -

    -     - Graph of the shaded region between the points a and b and the two curves is subdivided into 10 vertical slices. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={.5,3}, - extra x tick labels={$a$,$b$}, - ytick=\empty, - ymin=-.2,ymax=2, - xmin=-.5,xmax=3.5 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.25:3.25,samples=40] {.25*sin(deg(2*x))+1.25} node [shift={(5pt,7pt)} ,black] { $f(x)$}; - \addplot [name path=B,firstcurvestyle,domain=-.25:3.25] {.25*(x-2)^2+.2}node [shift={(5pt,7pt)} ,black] { $g(x)$}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0.5:3}]; - - \draw [thick] (axis cs:0.5,0.7625)--(axis cs:0.5,1.46) (axis cs:1.,0.45)--(axis cs:1.,1.477) (axis cs:1.5,0.2625)--(axis cs:1.5,1.285) (axis cs:2.,0.2)--(axis cs:2.,1.061) (axis cs:2.5,0.2625)--(axis cs:2.5,1.01) (axis cs:3.,1.18)--(axis cs:3.,0.45) (axis cs:0.75,1.499)--(axis cs:0.75,0.5906) (axis cs:1.25,0.3406)--(axis cs:1.25,1.4) (axis cs:1.75,1.162)--(axis cs:1.75,0.2156) (axis cs:2.25,0.2156)--(axis cs:2.25,1.006) (axis cs:2.75,1.074)--(axis cs:2.75,0.3406); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - -

    - Graph of the same two functions f(x) and g(x) as in the previous two images. - Now, the area lying between the marked points a and b and below f(x) and above g(x) is subdivided into 10 approximate vertical slices. - These slices showcase the slight inaccuracy of approximation using Riemann sums for small n , while the first image shows us that taking the limit as n\to \infty gives the exact area of this region between the two functions. -

    -     - Graph of the shaded region between the two curves is approximated using 10 rectangles. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={.5,3}, - extra x tick labels={$a$,$b$}, - ytick=\empty, - ymin=-.2,ymax=2, - xmin=-.5,xmax=3.5 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.25:3.25,samples=40] {.25*sin(deg(2*x))+1.25} node [shift={(5pt,7pt)} ,black] { $f(x)$}; - \addplot [name path=B,firstcurvestyle,domain=-.25:3.25] {.25*(x-2)^2+.2}node [shift={(5pt,7pt)} ,black] { $g(x)$}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0.5:3}]; - - \draw [thick] (axis cs:0.5,0.6225) rectangle (axis cs:0.75,1.5) (axis cs:1.,1.487) rectangle (axis cs:0.75,0.4756) (axis cs:1.,0.4025) rectangle (axis cs:1.25,1.452) (axis cs:1.5,1.334) rectangle (axis cs:1.25,0.29) (axis cs:1.5,0.24) rectangle (axis cs:1.75,1.235) (axis cs:2.,1.097) rectangle (axis cs:1.75,0.2025) (axis cs:2.,0.21) rectangle (axis cs:2.25,1.012) (axis cs:2.5,1.002) rectangle (axis cs:2.25,0.2225) (axis cs:2.5,0.29) rectangle (axis cs:2.75,1.029)(axis cs:3.,1.134) rectangle (axis cs:2.75,0.4025); - - \filldraw (axis cs:.7,.6225) circle (1.3pt) - (axis cs:0.95,0.4756) circle (1.3pt) - (axis cs:1.1,0.4025) circle (1.3pt) - (axis cs:1.4,0.29) circle (1.3pt) - (axis cs:1.6,0.24) circle (1.3pt) - (axis cs:1.9,0.2025) circle (1.3pt) - (axis cs:2.2,0.21) circle (1.3pt) - (axis cs:2.3,0.2225) circle (1.3pt) - (axis cs:2.6,0.29) circle (1.3pt) - (axis cs:2.9,0.4025) circle (1.3pt) - (axis cs:0.7,1.496) circle (1.3pt) - (axis cs:0.95,1.487) circle (1.3pt) - (axis cs:1.1,1.452) circle (1.3pt) - (axis cs:1.4,1.334) circle (1.3pt) - (axis cs:1.6,1.235) circle (1.3pt) - (axis cs:1.9,1.097) circle (1.3pt) - (axis cs:2.2,1.012) circle (1.3pt) - (axis cs:2.3,1.002) circle (1.3pt) - (axis cs:2.6,1.029) circle (1.3pt) - (axis cs:2.9,1.134) circle (1.3pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - - - Area Between Curves (restatement of <xref ref="thm_areabtwncurves"/>) - -

    - Let f(x) and g(x) be continuous functions defined on [a,b] where - f(x)\geq g(x) for all x in [a,b]. - The area of the region bounded by the curves y=f(x), - y=g(x) and the lines x=a and x=b is - integrationarea between curves - - \int_a^b \big(f(x)-g(x)\big)\, dx - . -

    -     -     - - - - - Finding area enclosed by curves - -

    - Find the area of the region bounded by f(x) = \sin(x) +2, - g(x) = \frac12\cos(2x)-1, - x=0 and x=4\pi, - as shown in . -

    - -
    - Graphing an enclosed region in - - - -

    - Graph showing the area of the region bounded by f(x) = \sin(x) +2, g(x) = \frac12\cos(2x)-1, x=0 and x=4\pi. - The function f(x) is drawn starting from the y-axis and ends at the point x=4 \pi . - The function g(x) is also drawn starting from the y-axis and ending at the point x=4 \pi . - For the duration of the region between x=0 and x=4 \pi , the curve f(x) = \sin(x) +2 lies above the curve g(x) = \frac12\cos(2x)-1. -

    -     - Graph of the shaded region between two points on the x-axis and the functions f and g. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick = {2,4,6,8,10}, - extra x ticks={12.57}, - extra x tick labels={$4\pi$}, - ymin=-2,ymax=3.5, - xmin=-.5,xmax=13.5 - ] - - \addplot [name path=A,firstcurvestyle,domain=0:12.57,samples=70] {sin(deg(x))+2} node [shift={(-15pt,7pt)},black] { $f(x)$}; - \addplot [name path=B,firstcurvestyle,domain=0:12.57,samples=90] {.5*cos(deg(2*x))-1.2} node [shift={(-20pt,-16pt)},black] { $g(x)$}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B]; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - The graph verifies that the upper boundary of the region is given by f and the lower bound is given by g. - Therefore the area of the region is the value of the integral - - \int_0^{4\pi} \big(f(x)- g(x)\big)\, dx \amp = \int_0^{4\pi} \Big(\sin(x) +2 - \big(\frac12\cos(2x)-1\big)\Big)\, dx - \amp = -\cos(x) -\frac14\sin(2x)+3x\Big|_0^{4\pi} - \amp = 12\pi \approx 37.7\,\text{units}^2 - . -

    -     - -     - - - Finding total area enclosed by curves - -

    - Find the total area of the region enclosed by the functions f(x) = -2x+5 and - g(x) = x^3-7x^2+12x-3 as shown in . -

    - -
    - Graphing a region enclosed by two functions in - - - -

    - Graph showing the area of the region bounded by f(x) = -2x+5, g(x) = x^3 -7x^2 +12x -3. - The line given by f(x) is drawn starting approximately at x=1 and ending at the point x=5 . - The cubic function given by g(x) is drawn starting from the y-axis as it comes up, before intersecting and going above the line f(x) at the point (1,3) . - The function g(x) then meets f(x) at the point (2,1) and continues below f(x) until once again intercepting at the point (4,-3) . - After this point, the function g(x) continues above the x-axis while the line f(x) slopes downwards never to meet again. -

    -

    - The area encosed by the curves f(x) and g(x) contains two parts. - The first part begins when g(x) rises above f(x) at x=1 and ends at x=2 , where g(x) falls below f(x) . - The second part begins at the point x=2 where g(x) is below f(x) and ends at x=4 , where g(x) rises above f(x) . - The first region is bounded below by f(x) , above by g(x) , x=1 and x=2 . - The second region is bounded above by f(x) , below by g(x) , x=2 and x=4 . - The two regions are shaded to showcase the total area enclosed by the two functions. -

    -     - Graph of the shaded region enclosed by the functions f and g. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ymin=-5,ymax=3.5, - xmin=-.5,xmax=4.7 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.2:4.4,samples=60] {x^3-7*x^2+12*x-3}; - \addplot [name path=B,firstcurvestyle,domain=0:4.4] {5-2*x}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=1:4}]; - - \end{axis} - - \end{tikzpicture} - - - - - -
    -     - -

    - A quick calculation shows that f=g at x=1, 2 and 4. - One can proceed thoughtlessly by computing \ds \int_1^4\big(f(x)-g(x)\big)\, dx, - but this ignores the fact that on [1,2], g(x) \gt f(x). - (In fact, the thoughtless integration returns -9/4, - hardly the expected value of an area.) - Thus we compute the total area by breaking the interval [1,4] into two subintervals, - [1,2] and [2,4] and using the proper integrand in each. - - \text{Total Area} \amp = \int_1^2 \big(g(x)-f(x)\big)\, dx + \int_2^4\big(f(x)-g(x)\big)\, dx - \amp = \int_1^2 \big(x^3-7x^2+14x-8\big) \, dx + \int_2^4\big(-x^3+7x^2-14x+8\big)\, dx - \amp = 5/12 + 8/3 - \amp = 37/12 = 3.083\,\text{units}^2 - . -

    -     - -     - -

    - The previous example makes note that we are expecting area to be positive. - When first learning about the definite integral, - we interpreted it as signed area under the curve, - allowing for negative area. That doesn't apply here; - area is to be positive. -

    - -

    - The previous example also demonstrates that we often have to break a given region into subregions before applying . - The following example shows another situation where this is applicable, - along with an alternate view of applying the Theorem. -

    - - - Finding area: integrating with respect to <m>y</m> - -

    - Find the area of the region enclosed by the functions y=\sqrt{x}+2, - y=-(x-1)^2+3 and y=2, - as shown in . -

    -
    - Graphing a region for - - - -

    - Graph showing the area of the region bounded by y=\sqrt{x}+2, y=-(x-1)^2+3 and y=2. - The curve given by y=\sqrt{x}+2 is drawn starting at the y-axis and ending at the point (1,3) . - The curve given by y=-(x-1)^2+3 is drawn starting from the end of the previous curve, at the point (1,3) . - This curve then slopes downwards before intersecting the horizontal line y=2 at the point (2,2) . - Both curves lie entirely above the line y=2. - Additionally, the curve y=\sqrt{x}+2 lies to the left of y=-(x-1)^2+3 for the entirety of the enclosed region. -

    -     - Graph of the shaded region enclosed by the two functions and the line y=2. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2}, - ymin=-.1,ymax=3.5, - xmin=-.1,xmax=2.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(0,2.)(0.02,2.141)(0.04,2.2)(0.06,2.245)(0.08,2.283)(0.1,2.316)(0.12,2.346)(0.14,2.374)(0.16,2.4)(0.18,2.424)(0.2,2.447)(0.22,2.469)(0.24,2.49)(0.26,2.51)(0.28,2.529)(0.3,2.548)(0.32,2.566)(0.34,2.583)(0.36,2.6)(0.38,2.616)(0.4,2.632)(0.42,2.648)(0.44,2.663)(0.46,2.678)(0.48,2.693)(0.5,2.707)(0.52,2.721)(0.54,2.735)(0.56,2.748)(0.58,2.762)(0.6,2.775)(0.62,2.787)(0.64,2.8)(0.66,2.812)(0.68,2.825)(0.7,2.837)(0.72,2.849)(0.74,2.86)(0.76,2.872)(0.78,2.883)(0.8,2.894)(0.82,2.906)(0.84,2.917)(0.86,2.927)(0.88,2.938)(0.9,2.949)(0.92,2.959)(0.94,2.97)(0.96,2.98)(0.98,2.99)(1.,3.)(1.04,2.998)(1.08,2.994)(1.12,2.986)(1.16,2.974)(1.2,2.96)(1.24,2.942)(1.28,2.922)(1.32,2.898)(1.36,2.87)(1.4,2.84)(1.44,2.806)(1.48,2.77)(1.52,2.73)(1.56,2.686)(1.6,2.64)(1.64,2.59)(1.68,2.538)(1.72,2.482)(1.76,2.422)(1.8,2.36)(1.84,2.294)(1.88,2.226)(1.92,2.154)(1.96,2.078)(2.,2.)(0,2)}; - - \addplot [firstcurvestyle,domain=0:.1] {sqrt(x) + 2}; - \addplot [firstcurvestyle,domain=.1:1] {sqrt(x) + 2}; - \addplot [firstcurvestyle,domain=1:2] {-(x-1)^2+3}; - \addplot [firstcurvestyle,domain=0:2] {2}; - - \draw (axis cs:.5,3.1) node { $y=\sqrt{x}+2$} (axis cs:1.8,3.1) node { $y=-(x-1)^2+3$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - We give two approaches to this problem. - In the first approach, we notice that the region's top - is defined by two different curves. - On [0,1], the top function is y=\sqrt{x}+2; - on [1,2], the top function is y=-(x-1)^2+3. -

    - -

    - Thus we compute the area as the sum of two integrals: - - \text{Total Area} \amp = \int_0^1 \Big(\big(\sqrt{x}+2\big)-2\Big)\, dx + \int_1^2 \Big(\big(-(x-1)^2+3\big)-2\Big)\, dx - \amp = 2/3 + 2/3 - \amp =4/3 - . -

    - -

    - The second approach is clever and very useful in certain situations. - We are used to viewing curves as functions of x; - we input an x-value and a y-value is returned. - Some curves can also be described as functions of y: - input a y-value and an x-value is returned. - We can rewrite the equations describing the boundary by solving for x: - - y=\sqrt{x}+2 \Rightarrow x=(y-2)^2 - - - y=-(x-1)^2+3 \Rightarrow x=\sqrt{3-y}+1 - . -

    - -
    - The region used in with boundaries relabeled as functions of y - - - - Graph of the shaded region bounded by y=\sqrt{x}+2, y=-(x-1)^2+3 and y=2 with a red horizontal rectangle slice showing integration with respect to y. - The rectangle starts near the beginning of the graph of y=\sqrt{x}+2 at a y-level of about 2.2 and spans across to meet the graph of y=-(x-1)^2+3. - - Graph of the shaded region with boundaries relabeled as functions of y. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2}, - ymin=-.1,ymax=3.5, - xmin=-.1,xmax=2.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(0,2.)(0.02,2.141)(0.04,2.2)(0.06,2.245)(0.08,2.283)(0.1,2.316)(0.12,2.346)(0.14,2.374)(0.16,2.4)(0.18,2.424)(0.2,2.447)(0.22,2.469)(0.24,2.49)(0.26,2.51)(0.28,2.529)(0.3,2.548)(0.32,2.566)(0.34,2.583)(0.36,2.6)(0.38,2.616)(0.4,2.632)(0.42,2.648)(0.44,2.663)(0.46,2.678)(0.48,2.693)(0.5,2.707)(0.52,2.721)(0.54,2.735)(0.56,2.748)(0.58,2.762)(0.6,2.775)(0.62,2.787)(0.64,2.8)(0.66,2.812)(0.68,2.825)(0.7,2.837)(0.72,2.849)(0.74,2.86)(0.76,2.872)(0.78,2.883)(0.8,2.894)(0.82,2.906)(0.84,2.917)(0.86,2.927)(0.88,2.938)(0.9,2.949)(0.92,2.959)(0.94,2.97)(0.96,2.98)(0.98,2.99)(1.,3.)(1.04,2.998)(1.08,2.994)(1.12,2.986)(1.16,2.974)(1.2,2.96)(1.24,2.942)(1.28,2.922)(1.32,2.898)(1.36,2.87)(1.4,2.84)(1.44,2.806)(1.48,2.77)(1.52,2.73)(1.56,2.686)(1.6,2.64)(1.64,2.59)(1.68,2.538)(1.72,2.482)(1.76,2.422)(1.8,2.36)(1.84,2.294)(1.88,2.226)(1.92,2.154)(1.96,2.078)(2.,2.)(0,2)}; - - \addplot [firstcurvestyle,domain=0:.1] {sqrt(x) + 2}; - \addplot [firstcurvestyle,domain=.1:1] {sqrt(x) + 2}; - \addplot [firstcurvestyle,domain=1:2] {-(x-1)^2+3}; - \addplot [firstcurvestyle,domain=0:2] {2}; - - \addplot [secondcurvestyle,solid,fill] coordinates {(0.0625,2.2)(0.0625,2.3)(1.867,2.3)(1.867,2.2)(0.0625,2.2)}; - - \draw (axis cs:.5,3.1) node { $x=(y-2)^2$} (axis cs:1.8,3.2) node { $x=\sqrt{3-y}+1$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - - shows the region with the boundaries relabeled. - A differential element, a horizontal rectangle, is also pictured. - The width of the rectangle is a small change in y: - \Delta y. - The height of the rectangle is a difference in x-values. - The top x-value is the largest value, , the rightmost. - The bottom x-value is the smaller, , the leftmost. - Therefore the height of the rectangle is - - \big(\sqrt{3-y}+1\big) - (y-2)^2 - . -

    - -

    - The area is found by integrating the above function with respect to y with the appropriate bounds. - We determine these by considering the y-values the region occupies. - It is bounded below by y=2, - and bounded above by y=3. - That is, both the top and bottom - functions exist on the y interval [2,3]. - Thus - - \text{Total Area} \amp = \int_2^3 \big(\sqrt{3-y}+1 - (y-2)^2\big)\, dy - \amp = \Big(-\frac23(3-y)^{3/2}+y-\frac13(y-2)^3\Big)\Big|_2^3 - \amp = 4/3 - . -

    -     - -     - - - -

    - This calculus-based technique of finding area can be useful even with shapes that we normally think of as easy. - - computes the area of a triangle. - While the formula \frac12\times\,\text{base}\, \times\,\text{height} is well known, - in arbitrary triangles it can be nontrivial to compute the height. - Calculus makes the problem simple. -

    - - - Finding the area of a triangle - -

    - Compute the area of the regions bounded by the lines -

    - -

    - y=x+1, y=-2x+7 and y=-\frac12x+\frac52, - as shown in . -

    - -
    - Graphing a triangular region in - - - -

    - Graph showing the enclosed area of the triangular region bounded by y=x+1, y=-2x+7 and y=-\frac12x+\frac52. - The corners of the triangle are the points (1,2) , (2,3) and (3,1) . - The line y=x+1 lies above the line y=-\frac12x+\frac52 between x=1 and x=2. - The line y=-2x+7 also lies above the line y=-\frac12x+\frac52 between x=2 and x=3. - For integrating in terms of y, the line y=x+1 lies to the left of y=-2x+7 between y=2 and y=3. - The line y=-\frac12x+\frac52 lies to the left of y=-2x+7 between y=1 and y=2. -

    -     - Graph of the triangle having corners on the points (1,2), (2,3) and (3,1). - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3}, - ymin=-.1,ymax=3.5, - xmin=-.1,xmax=3.9 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(1,2) (2,3) (3,1) (1,2)}; - - \addplot [firstcurvestyle,domain=1:2] {x+1} node [shift={(-50pt,-20pt)},black] { $y=x+1$}; - \addplot [firstcurvestyle,domain=2:3] {-2*x+7} node [shift={(0pt,60pt)},black] { $y=-2x+7$}; - \addplot [firstcurvestyle,domain=1:3] {-.5*x+(5/2)} node [shift={(-50pt,0pt)},black] { $y=-\frac12x+\frac52$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - Recognize that there are two top - functions to this region, - causing us to use two definite integrals. - - \text{Total Area} \amp = \int_1^2\big((x+1)-(-\frac12x+\frac52)\big)\, dx - \amp \quad + \int_2^3\big((-2x+7)-(-\frac12x+\frac52)\big)\, dx - \amp = 3/4+3/4 - \amp =3/2 - . -

    - -

    - We can also approach this by converting each function into a function of y. - This also requires 2 integrals, - so there isn't really any advantage to doing so. - We do it here for demonstration purposes. -

    - -

    - The top function is always x=\frac{7-y}2 while there are two - bottom functions. - Being mindful of the proper integration bounds, we have - - \text{Total Area} \amp = \int_1^2\big(\frac{7-y}2 - (5-2y)\big)\, dy + \int_2^3\big(\frac{7-y}2-(y-1)\big)\, dy - \amp = 3/4 + 3/4 - \amp = 3/2 - . -

    - -

    - Of course, the final answer is the same. - (It is interesting to note that the area of all 4 subregions used is 3/4. - This is coincidental.) -

    -     - -     - -

    - While we have focused on producing exact answers, - we are also able to make approximations using the principle of . - The integrand in the theorem is a distance (top minus bottom); - integrating this distance function gives an area. - By taking discrete measurements of distance, - we can approximate an area using numerical integration techniques developed in . - The following example demonstrates this. -

    - - - Numerically approximating area - -

    - To approximate the area of a lake, - shown in , - the length of the lake is measured at 200-foot increments, as shown in . - The lengths are given in hundreds of feet. - Approximate the area of the lake. -

    - -
    - (a) A sketch of a lake, and (b) the lake with length measurements - -
    - - - - -

    - A sketch of a lake. This image contains no included measurements and no coordinate plots. -

    -     - A sketch of a lake. - - - \begin{tikzpicture}[scale=.6] - - \draw [firstcolor,thick,fill=firstcolor!15,smooth] plot coordinates {(0,3.2)(0.5,3.417)(1.,3.637)(1.5,4.041)(2.,4.505)(2.5,5.)(3.,5.495)(3.5,5.959)(4.,6.363)(4.5,6.683)(5.,6.898)(5.5,6.995)(6.,6.968)(6.5,6.819)(7.,6.556)(7.5,6.197)(8.,5.763)(8.5,5.282)(9.,4.784)(9.5,4.298)(10.,3.857)(10.5,3.486)(11.,3.21)(11.5,2.8)(11.5,2)(11.,1.545)(10.5,1.314)(10.,1.102)(9.5,0.9154)(9.,0.7585)(8.5,0.6361)(8.,0.5514)(7.5,0.5069)(7.,0.5038)(6.5,0.5421)(6.,0.6208)(5.5,0.7378)(5.,0.8897)(4.5,1.072)(4.,1.281)(3.5,1.509)(3.,1.751)(2.5,2.)(2.,2.249)(1.5,2.491)(1.,2.719)(0.5,2.9102)(0,3.15)(0,3.2)}; - - \end{tikzpicture} - - - - -
    - -
    - - - - -

    - A graph of the previous sketch of a lake with vertically strenching length measurements. - The curve begins at the y-axis at the point (0,3). - From this point, the curve slopes upwards until it reaches a peak near the point (6,7). - After reaching a peak, the curve slopes downwards towards the point (12,3). - From the point (0,3), the downward portion of the curve reaches its minimum near (7,1), after which it goes upwards to meet at the point (12,3). -

    -

    - There are 5 vertical length measurements given which occur at all even x starting at x=2. - The first vertical length measurement is 2.25. - The second is 5.08. - The third is 6.35. - The fourth is 5.21. - The fifth is 2.76. - There is then an increment of length 2 on the x-axis between the last measurement occuring at x=10m, and the edge of the lake occuring at x=12. -

    -     - A graph of the lake with five vertical length measurements. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4,5,6,7,8,9,10,11,12}, - ytick={1,2,3,4,5,6,7,8}, - ymin=-.1,ymax=8.5, - xmin=-.1,xmax=12.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(0,3.2)(0.5,3.417)(1.,3.637)(1.5,4.041)(2.,4.505)(2.5,5.)(3.,5.495)(3.5,5.959)(4.,6.363)(4.5,6.683)(5.,6.898)(5.5,6.995)(6.,6.968)(6.5,6.819)(7.,6.556)(7.5,6.197)(8.,5.763)(8.5,5.282)(9.,4.784)(9.5,4.298)(10.,3.857)(10.5,3.486)(11.,3.21)(11.5,2.8)(11.55,2)(11.,1.545)(10.5,1.314)(10.,1.102)(9.5,0.9154)(9.,0.7585)(8.5,0.6361)(8.,0.5514)(7.5,0.5069)(7.,0.5038)(6.5,0.5421)(6.,0.6208)(5.5,0.7378)(5.,0.8897)(4.5,1.072)(4.,1.281)(3.5,1.509)(3.,1.751)(2.5,2.)(2.,2.249)(1.5,2.491)(1.,2.719)(0.5,2.9102)(0,3.15)(0,3.2)}; - \addplot [firstcurvestyle,smooth] coordinates {(0,3.2)(0.5,3.417)(1.,3.637)(1.5,4.041)(2.,4.505)(2.5,5.)(3.,5.495)(3.5,5.959)(4.,6.363)(4.5,6.683)(5.,6.898)(5.5,6.995)(6.,6.968)(6.5,6.819)(7.,6.556)(7.5,6.197)(8.,5.763)(8.5,5.282)(9.,4.784)(9.5,4.298)(10.,3.857)(10.5,3.486)(11.,3.21)(11.5,2.8)(11.5,2)(11.,1.545)(10.5,1.314)(10.,1.102)(9.5,0.9154)(9.,0.7585)(8.5,0.6361)(8.,0.5514)(7.5,0.5069)(7.,0.5038)(6.5,0.5421)(6.,0.6208)(5.5,0.7378)(5.,0.8897)(4.5,1.072)(4.,1.281)(3.5,1.509)(3.,1.751)(2.5,2.)(2.,2.249)(1.5,2.491)(1.,2.719)(0.5,2.9102)(0,3.15)(0,3.2)}; - - \draw (axis cs:2,4.5) -- (axis cs:2,2.249) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 2.25} - (axis cs:4,6.36) -- (axis cs:4,1.28) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 5.08} - (axis cs:6,6.97) -- (axis cs:6,0.62) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.35} - (axis cs:8,5.76) -- (axis cs:8,.55) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 5.21} - (axis cs:10,3.86) -- (axis cs:10,1.1) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 2.76}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    -     - -

    - The measurements of length can be viewed as measuring - top minus bottom of two functions. - The exact answer is found by integrating \ds \int_0^{12} \big(f(x)-g(x)\big)\, dx, - but of course we don't know the functions f and g. - Our discrete measurements instead allow us to approximate. -

    - -

    - We have the following data points: - - (0,0),\,(2,2.25),\,(4,5.08),\,(6,6.35),\,(8,5.21),\,(10,2.76),\,(12,0) - . -

    - -

    - We also have that \dx=\frac{b-a}{n} = 2, so Simpson's Rule gives - - \text{Area} \amp \approx \frac{2}{3}\Big(1\cdot0+4\cdot2.25+2\cdot5.08+4\cdot6.35+2\cdot5.21+4\cdot2.76+1\cdot0\Big) - \amp = 44.01\overline{3} \,\text{units}^2 - . -

    - -

    - Since the measurements are in hundreds of feet, - square units are given by (100)^2 = 10,000, - giving a total area of 440,133. - (Since we are approximating, - we'd likely say the area was about 440,000, - which is a little more than 10 acres.) -

    -     -     - -

    - In the next section we apply our applications of integration techniques to finding the volumes of certain solids. -

    - - - - Terms and Concepts - - - -

    - The area between curves is always positive. - -

    -     - -     - - - - -

    - Calculus can be used to find the area of basic geometric shapes. - -

    -     - -     - - - - -

    - In your own words, - describe how to find the total area enclosed by y=f(x) and y=g(x). -

    - - -     - - - -     - - - - -

    - Describe a situation where it is advantageous to find an area enclosed by curves through integration with respect to y instead of x. -

    - - -     - - - -

    - Answers may vary; - one common answer is when the region has two or more - top or bottom - functions when viewing the region with respect to x, - but has only 1 top function and 1 bottom - function when viewed with respect to y. - The former area requires multiple integrals to compute, - whereas the latter area requires one. -

    -     - -     -     - - Problems - - - -

    - Find the area of the shaded region in the given graph. -

    -     - - - - - - $x0 = 0; - $x1 = Compute("2*pi"); - $f = Formula("x/2+3"); - $g = Formula("1/2*cos(x)+1"); - $F = Formula("x^2/4+3x"); - $G = Formula("1/2*sin(x)+x"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - -

    - Between y=\frac12 x +3 and y=\frac12\cos(x)+1, - for 0\leq x\leq 2\pi. -

    - -

    - -

    - - - - Graph of the region enclosed by the functions y=\frac12 x +3 and y=\frac12\cos(x)+1 between x = 0 and x = 2\pi. - The function y=\frac12 x +3 lies above the function y=\frac12\cos(x)+1 for the entirety of the region between x = 0 and x = 2\pi. - - Graph of the region between two functions between x=0 and x=2pi. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={3.14,6.28}, - extra x tick labels={\(\pi\),\(2\pi\)}, - ymin=-.2,ymax=6.5, - xmin=-1,xmax=7 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.9:6.5,samples=40] {.5*cos(deg(x))+1} node [shift={(-50pt,7pt)} ,black] { \(y=\frac12\cos(x) +1\)}; - \addplot [name path=B,firstcurvestyle,domain=-.9:6.5] {.5*x+3}node [shift={(-50pt,-5pt)} ,black] { \(y=\frac12x+3\)}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0:2*pi}]; - - \end{axis} - - \end{tikzpicture} - - - - - - - -     - -

    - 4\pi+\pi^2\approx 22.436 -

    -     -     -     - - - - - $x0 = Compute("-1"); - $x1 = Compute("1"); - $f = Formula("-3x^3+2x+2"); - $g = Formula("x^2+x-1"); - $F = Formula("-x^4/3+3x^2/2+2x"); - $G = Formula("x^3/3+x^2/2-x"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - -

    - Between y=-3x^3+3x+2 and y=x^2+x-1, - for -1\leq x\leq 1. -

    - -

    - -

    - - - - Graph of the region enclosed by the functions y=-3x^3+3x+2 and y=x^2+x-1 between x = -1 and x = 1. - The function y=-3x^3+3x+2 lies above the function y=x^2+x-1 for the entirety of the region between x = -1 and x = 1. - - Graph of the region enclosed by the two functions between x=0 and x=2pi. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,1}, - ytick={-1,1,2,3}, - ymin=-1.5,ymax=3.5, - xmin=-1.1,xmax=1.1 - ] - - \addplot [name path=A,firstcurvestyle,domain=-1.05:1.05] {x^2+x-1} node [shift={(-40pt,-65pt)},black] { \(y=x^2+x-1\)}; - \addplot [name path=B,firstcurvestyle,domain=-1.05:1.05,samples=60] {-3*x^3+3*x+2} node [shift={(-140pt,30pt)},black] { \(y=-3x^3+3x+2\)}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=-1:1}]; - - \end{axis} - - \end{tikzpicture} - - - - - - - -     - -

    - 10/3 -

    -     -     -     - - - - - $x0 = Compute("0"); - $x1 = Compute("pi"); - $f = Formula("2"); - $g = Formula("1"); - $F = Formula("2x"); - $G = Formula("x"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - -

    - Between y=1 and y=2, for 0\leq x\leq \pi. -

    - -

    - -

    - - - - Graph of the region enclosed by horizontal lines y=0 and y=1 between x = 0 and x = \pi. - - Graph of the rectangular region between y=1, y=2, and x=0, x=pi. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={3.14,1.57}, - extra x tick labels={\(\pi\),\(\pi/2\)}, - ytick={-1,1,2,3}, - ymin=-.5,ymax=2.4, - xmin=-.1,xmax=3.5 - ] - - \addplot [name path=A,firstcurvestyle,-] {1} node [shift={(-100pt,-10pt)},black] { \(y=1\)}; - \addplot [name path=B,firstcurvestyle,-] {2} node [shift={(-100pt,10pt)},black] { \(y=2\)}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0:pi}]; - - \end{axis} - - \end{tikzpicture} - - - - - - - -     - -

    - \pi -

    -     -     -     - - - - - $x0 = Compute("0"); - $x1 = Compute("pi"); - $f = Formula("sin(x)+1"); - $g = Formula("sin(x)"); - $F = Formula("-cos(x)+x"); - $G = Formula("-cos(x)"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - -

    - Between y=\sin(x)+1 and y=\sin(x), - for 0\leq x\leq \pi. -

    - -

    - -

    - - - - Graph of the region enclosed by the functions y=\sin(x)+1 and y=\sin(x) between x = 0 and x = \pi. - The function y=\sin(x)+1 lies above the function y=\sin(x) for the entirety of the region between x = 0 and x = \pi. - - Graph of the region between the two sine graphs between x=0 and x=pi. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={3.14,1.57}, - extra x tick labels={\(\pi\),\(\pi/2\)}, - ytick={-1,1,2,3}, - ymin=-.5,ymax=2.2, - xmin=-.1,xmax=3.5 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.1:3.5,samples=40] {sin(deg(x))} node [shift={(-80pt,45pt)} ,black] { \(y=\sin(x) \)}; - \addplot [name path=B,firstcurvestyle,domain=-.1:3.5,samples=40] {sin(deg(x))+1} node [shift={(-45pt,85pt)} ,black] { \(y=\sin(x) +1\)}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0:pi}]; - - \end{axis} - - \end{tikzpicture} - - - - - - - -     - -

    - \pi -

    -     -     -     - - - - - $x0 = Compute("0"); - $x1 = Compute("pi/4"); - $f = Formula("sec(x)^2"); - $g = Formula("sin(4x)"); - $F = Formula("tan(x)"); - $G = Formula("-cos(4x)/4"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - - -

    - Between y=\sin(4x) and y=\sec^2(x), - for 0\leq x\leq \pi/4. -

    - -

    - -

    - - - - Graph of the region enclosed by the functions y=\sin(4x) and y=\sec^2(x) between x = 0 and x = \pi/4. - The function y=\sec^2(x) lies above the function y=\sin(4x) for the entirety of the region between x = 0 and x = \pi/4. - - Graph of the region enclosed by the two functions between x=0 and x=pi/4. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={0.785,0.392}, - extra x tick labels={\(\pi/4\),\(\pi/8\)}, - ytick={-1,1,2,3}, - ymin=-.5,ymax=2.2, - xmin=-.1,xmax=1 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.1:.9,samples=60] {sin(deg(4*x))} node [shift={(-65pt,40pt)} ,black] { \(y=\sin(4x) \)}; - \addplot [name path=B,firstcurvestyle,domain=-.1:.9,samples=60] {sec(deg(x))^2} node [shift={(-60pt,-40pt)} ,black] { \(y=\sec^2(x)\)}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0:pi/4}]; - - \end{axis} - - \end{tikzpicture} - - - - - - - -     - -

    - 1/2 -

    -     -     -     - - - - - $x0 = Compute("pi/4"); - $x1 = Compute("5*pi/4"); - $f = Formula("sin(x)"); - $g = Formula("cos(x)"); - $F = Formula("-cos(x)"); - $G = Formula("sin(x)"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - -

    - Between y=\sin(x) and y=\cos(x), - for \pi/4\leq x\leq 5\pi/4. -

    - -

    - -

    - - - - Graph of the region enclosed by the functions y=\sin(x) and y=\cos(x) between x = \pi/4 and x = 5\pi/4. - The two functions intersect at x = \pi/4 and x = 5\pi/4. - The function y=\sec^2(x) lies above the function y=\sin(4x) for the entirety of the region between x = \pi/4 and x = 5\pi/4. - - Graph of the region enclosed by the two functions between x=pi/4 and x=5 pi/4. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={.785,1.57,2.36,3.14,3.92}, - extra x tick labels={\(\pi/4\),\(\pi/2\),\(3\pi/4\),\(\pi\),\(5\pi/4\)}, - ymin=-1.1,ymax=1.1, - xmin=-.1,xmax=4.1 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.1:4.1,samples=40] {sin(deg(x))} node [shift={(-30pt,90pt)},black] { \(y=\sin(x) \)}; - \addplot [name path=B,firstcurvestyle,domain=-.1:4.1,samples=40] {cos(deg(x))} node [shift={(-125pt,-5pt)},black] { \(y=\cos(x) \)}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=pi/4:5*pi/4}]; - - \end{axis} - - \end{tikzpicture} - - - - - - - -     - -

    - 2\sqrt{2} -

    -     -     -     - - - - - $x0 = Compute("0"); - $x1 = Compute("1"); - $f = Formula("4^x"); - $g = Formula("2^x"); - $F = Formula("4^x/ln(4)"); - $G = Formula("2^x/ln(2)"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - -

    - Between y=2^x and y=4^x, - for 0\leq x\leq 1. -

    - -

    - -

    - - - - Graph of the region enclosed by the functions y=2^x and y=4^x between x = 0 and x = 1. - The two functions intersect at x = 0 before y=4^x overtakes y=2^x. - The function y=4^x lies above the function y=2^x for the entirety of the region between 0 and 1. - - Graph of the region enclosed by the two functions between x=0 and x=1. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=4.1, - xmin=-.1,xmax=1.1 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.1:1.1] {2^x} node [shift={(-25pt,-20pt)},black] { \(y=2^x\)}; - \addplot [name path=B,firstcurvestyle,domain=-.1:1.1] {4^x} node [shift={(-45pt,-30pt)},black] { \(y=4^x\)}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=0:1}]; - - \end{axis} - - \end{tikzpicture} - - - - - - - -     - -

    - 1/\ln(4) -

    -     -     -     - - - -

    - Bounded by the curves y=\sqrt{x}+1, y=\sqrt{2-x}+1, - and y=1. -

    - - Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1 and the horizontal line y=1. - The curve given by y=\sqrt{x}+1 is drawn starting at the y-axis and ending at the point (1,2) . - The curve given by y=\sqrt{2-x}+1 is drawn starting from the end of the previous curve, at the point (1,2) . - This curve then falls downwards before intersecting the horizontal line y=1 at the point (2,1) . - Both curves lie entirely above the horizontal line y=1. - The curve y=\sqrt{x}+1 also lies to the left of y=\sqrt{2-x}+1 throughout the enclosed region. - - Graph of the region enclosed by the two functions and the line y=1. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=2.5, - xmin=-.1,xmax=2.25 - ] - - \addplot [name path=A,firstcurvestyle,domain=0:1,samples=40] {1}; - \addplot [name path=B,firstcurvestyle,domain=0:1,samples=40] ({x^2},{x+1}) node [pos=.85,above left,black] { \(y=\sqrt{x}+1\)}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B]; - - \addplot [name path=C,firstcurvestyle,domain=1:2,samples=2] {1}; - \addplot [name path=D,firstcurvestyle,domain=0:1,samples=40] ({-x^2+2},{x+1}) node [pos=.9,above right,black] { \(y=\sqrt{2-x}+1\)}; - - \addplot [firstcurvestyle,areastyle] fill between [of=C and D]; - - \end{axis} - - \end{tikzpicture} - - - -     - -

    - 4/3 -

    -     -     - -     - - - -

    - Find the total area enclosed by the functions f and g. -

    -     - - - - - $x0 = Compute("-2"); - $x1 = Compute("1"); - $f = Formula("x^2+4x-1"); - $g = Formula("2x^2+5x-3"); - $F = Formula("x^3/3+2x^2-x"); - $G = Formula("2x^3/3+5x^2/2-3x"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - -

    - f(x) = 2x^2+5x-3, g(x) = x^2+4x-1 -

    - -

    - -

    - -     -     -     - - - - - $x0 = Compute("-1"); - $x1 = Compute("1"); - $f = Formula("-3x+3"); - $g = Formula("x^2-3x+2"); - $F = Formula("3x^2/2+3x"); - $G = Formula("x^3/3-3x^2+2x"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - -

    - f(x) = x^2-3x+2, g(x) = -3x+3 -

    - -

    - -

    - -     -     -     - - - - - $x0 = Compute("0"); - $x1 = Compute("pi/2"); - $f = Formula("sin(x)"); - $g = Formula("2x/pi"); - $F = Formula("-cos(x)"); - $G = Formula("x^2/pi"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - $area = 2*$area; - - -

    - f(x) = \sin(x), g(x) = 2x/\pi -

    - -

    - -

    - -     -     -     - - - - - $x0 = Compute("-1"); - $x1 = Compute("1"); - $x2 = Compute("3"); - - $f = Formula("x^3-4x^2+x-1"); - $g = Formula("-x^2+2x-4"); - $F = Formula("x^4/4-4x^3/3+x^2/2-x"); - $G = Formula("-x^3/3+x^2-4x"); - $D1 = Formula("$F-$G"); - $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); - $D2 = Formula("-$D1"); - $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); - $area = $area1 + $area2; - - -

    - f(x) = x^3-4x^2+x-1, g(x) = -x^2+2x-4 -

    - -

    - -

    - -     -     -     - - - - - $x0 = Compute("0"); - $x1 = Compute("1"); - $f = Formula("sqrt(x)"); - $g = Formula("x"); - $F = Formula("2/3*x^(3/2)"); - $G = Formula("x^2/2"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - -

    - f(x) = x, g(x) = \sqrt{x} -

    - -

    - -

    - -     -     -     - - - - - $x0 = Compute("-1"); - $x1 = Compute("1"); - $x2 = Compute("2"); - $f = Formula("3x^2+x+3"); - $g = Formula("-x^3+5x^2+2x+1"); - $F = Formula("x^3+x^2/2+3x"); - $G = Formula("-x^4/4+5x^3/3+x^2+x"); - $D1 = Formula("$F-$G"); - $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); - $D2 = Formula("-$D1"); - $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); - $area = $area1 + $area2; - - -

    - f(x) = -x^3+5x^2+2x+1, g(x) = 3x^2+x+3 -

    - -

    - -

    - -     -     -     - -     - - - -

    - The functions f(x) = \cos (x) and - g(x) = \sin(x) intersect infinitely many times, - forming an infinite number of repeated, enclosed regions. - Find the areas of these regions. -

    -     - -

    - All enclosed regions have the same area, - with regions being the reflection of adjacent regions. - One region is formed on [\pi/4,5\pi/4], - with area 2\sqrt{2}. -

    -     -     - - - - - - $x0 = Compute("pi/6"); - $x1 = Compute("5*pi/6"); - $x2 = Compute("3*pi/2"); - - $f = Formula("sin(x)"); - $g = Formula("cos(2x)"); - $F = Formula("-cos(x)"); - $G = Formula("sin(2x)/2"); - $D1 = Formula("$F-$G"); - $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); - $D2 = Formula("-$D1"); - $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); - $area = $area1 + $area2; - - -

    - The functions f(x) = \cos(2x) and - g(x) = \sin(x) intersect infinitely many times, - forming an infinite number of repeated, enclosed regions. - Find the areas of these regions. -

    - -

    - -

    - -     - -

    - On regions such as [\pi/6,5\pi/6], - the area is 3\sqrt{3}/2. - On regions such as [-\pi/2,\pi/6], - the area is 3\sqrt{3}/4. -

    -     -     -     - - - -

    - Find the area of the enclosed region in two ways: -

    - -

    -

      -
    1. -

      - by treating the boundaries as functions of x, and -

      -
      2. - -
    2. -

      - by treating the boundaries as functions of y. -

      -
      3. -
    -

    -     - - - - - $x0 = Compute("0"); - $x1 = Compute("1"); - $x2 = Compute("3"); - - $f1 = Formula("x^2+1"); - $f2 = Formula("1/4*(x-3)^2+1"); - $g = Formula("1"); - $F1 = Formula("x^3/3+x"); - $F2 = Formula("1/12*(x-3)^3+x"); - $G = Formula("x"); - $D1 = Formula("$F1-$G"); - $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); - $D2 = Formula("$F2-$G"); - $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); - $area = $area1 + $area2; - - -

    - Bounded by y=x^2+1, y=\frac14(x-3)^2+1, and y=1. -

    - - - - Graph of the region enclosed by the functions y=x^2+1, y=\frac14(x-3)^2+1 and the horizontal line y=1. - The curve given by y=x^2+1 is drawn starting at the y-axis and ending at the point (1,2) . - The curve given by y=\frac14(x-3)^2+1 is drawn starting from the end of the previous curve at the point (1,2) . - This curve then falls downwards before intersecting the horizontal line y=1 at the point (3,1) . - Both curves lie above the line y=1 for the entirety of the enclosed region. - The curve y=\frac14(x-3)^2+1 also lies to the right of the curve y=x^2+1 throughout the enclosed region. - - Graph of the region enclosed by the two functions and the line y=1. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=2.2, - xmin=-.5,xmax=3.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(0,1.)(0.1,1.01)(0.2,1.04)(0.3,1.09)(0.4,1.16)(0.5,1.25)(0.6,1.36)(0.7,1.49)(0.8,1.64)(0.9,1.81)(1.,2.)(1.1,1.903)(1.2,1.81)(1.3,1.723)(1.4,1.64)(1.5,1.563)(1.6,1.49)(1.7,1.423)(1.8,1.36)(1.9,1.303)(2.,1.25)(2.1,1.203)(2.2,1.16)(2.3,1.123)(2.4,1.09)(2.5,1.063)(2.6,1.04)(2.7,1.023)(2.8,1.01)(2.9,1.003)(3.,1.)(0,1)}; - - \addplot [firstcurvestyle,domain=1:3] {.25*(x-3)^2+1} node [shift={(-20pt,40pt)},black] { \(y=\frac14(x-3)^2+1\)}; - \addplot [firstcurvestyle,domain=0:1] {x^2+1} node [shift={(-40pt,-15pt)},black] { \(y=x^2+1\)}; - \addplot [firstcurvestyle,domain=0:3] {1} node [pos=.5,below,black] { \(y=1\)} (axis cs:3,1); - - \end{axis} - - \end{tikzpicture} - - - - - -

    - -

    - -     - -

    - 1 -

    -     -     -     - - - - - $x0 = Compute("0"); - $x1 = Compute("1"); - $x2 = Compute("2"); - - $f1 = Formula("sqrt(x)"); - $f2 = Formula("-2x+3"); - $g = Formula("-1/2*x"); - $F1 = Formula("2/3*x^(3/2)"); - $F2 = Formula("-x^2+3x"); - $G = Formula("-x^2/4"); - $D1 = Formula("$F1-$G"); - $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); - $D2 = Formula("$F2-$G"); - $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); - $area = $area1 + $area2; - - -

    - Bounded by y=\sqrt{x}, y=-2x+3, and y=-\frac12 x. -

    - - - - Graph of the region enclosed by the functions y=\sqrt{x}, y=-2x+3 and y=-\frac12 x. - The curve given by y=\sqrt{x} is drawn starting at the origin, from which it goes upwards and ends at the point (1,1) . - The curve given by y=-2x+3 is drawn starting from the end of the previous curve at the point (1,1) . - This curve then goes downwards before ending at the point (2,-1) . - The curve given by y=-\frac12 x is drawn starting from the origin, from which it goes downwards until it meets the previous curve at the point (2,-1) . - The curve y=\sqrt{x} graphed between x=0 and x=1 and the line y=-2x+3 graphed between x=1 and x=2 lie entirely above the line y=-\frac12 x. - The curve y=\sqrt{x} between y=0 and y=1 and the line y=-\frac12 x between y=-1 and y=0 lie to the left of the curve y=-2x+3. - - Graph of the region enclosed by the three functions. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-1.1,ymax=1.1, - xmin=-.5,xmax=2.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates { (0,0)(0.02,0.1414)(0.04,0.2)(0.06,0.2449)(0.08,0.2828)(0.1,0.3162)(0.2,0.4472)(0.3,0.5477)(0.4,0.6325)(0.5,0.7071)(0.6,0.7746)(0.7,0.8367)(0.8,0.8944)(0.9,0.9487)(1.,1.)(2,-1)(0,0)}; - - \addplot [firstcurvestyle,domain=0:1,samples=60] {sqrt(x)} node [shift={(-40pt,-10pt)} ,black] { \(y=\sqrt{x}\)}; - \addplot [firstcurvestyle,domain=0:2] {-.5*x} node [pos=.3,shift={(5pt,-25pt)},black] { \(y=-\frac12x\)}; - \addplot [firstcurvestyle,domain=1:2] {-2*x+3} node [pos=.5,shift={(25pt,25pt)},black] { \(y=-2x+3\)}; - - \end{axis} - - \end{tikzpicture} - - - - - -

    - -

    - -     - -

    - 5/3 -

    -     -     -     - - - - - $x0 = Compute("-1"); - $x1 = Compute("2"); - $f = Formula("x+2"); - $g = Formula("x^2"); - $F = Formula("x^2/2+2x"); - $G = Formula("x^3/3"); - $D = Formula("$F-$G"); - $area = $D->eval(x=>$x1) - $D->eval(x=>$x0); - - -

    - Between the curves y=x+2 and y=x^2. -

    - - - - Graph of the region enclosed by the functions y=x+2 and y=x^2. - The line y=x+2 is drawn starting at the point (-1,1), from which it goes upwards and ends at the point (2,4) . - The parabola given by y=x^2 is also drawn starting from the point (-1,1), from which it slopes downwards until the point (0,0) . - The parabola then goes upwards before meeting the line at the point (2,4) . - The line y=x+2 graphed between x=-1 and x=2 lies entirely above the parabola y=x^2 graphed on the same bounds, only interesecting at the end points. - The line y=\sqrt{x} lies to the left of the parabola y=x^2 between y=1 and y=4. - Between y=0 and y=1, the enclosed region spans from the left side of y=x^2 to the right side of the y=x^2. - - Graph of the region enclosed by the line and parabola. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=4.5, - xmin=-1.5,xmax=2.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates { (-1.,1.)(-0.9,0.81)(-0.8,0.64)(-0.7,0.49)(-0.6,0.36)(-0.5,0.25)(-0.4,0.16)(-0.3,0.09)(-0.2,0.04)(-0.1,0.01)(0,0)(0.1,0.01)(0.2,0.04)(0.3,0.09)(0.4,0.16)(0.5,0.25)(0.6,0.36)(0.7,0.49)(0.8,0.64)(0.9,0.81)(1.,1.)(1.1,1.21)(1.2,1.44)(1.3,1.69)(1.4,1.96)(1.5,2.25)(1.6,2.56)(1.7,2.89)(1.8,3.24)(1.9,3.61)(2.,4.)(-1,1)}; - - \addplot [firstcurvestyle,domain=-1:2] {x^2} node [shift={(0pt,-60pt)} ,black] { \(y=x^2\)}; - \addplot [firstcurvestyle,domain=-1:2] {x+2} node [pos=.5,shift={(10pt,35pt)},black] { \(y=x+2\)} (axis cs:-1,1); - - \end{axis} - - \end{tikzpicture} - - - - - -

    - -

    - -     - -

    - 9/2 -

    -     -     -     - - - - - $x0 = Compute("0"); - $x1 = Compute("1/2"); - $x2 = Compute("2"); - - $f1 = Formula("sqrt(2*x)"); - $f2 = Formula("-2*(x-1)"); - $g = Formula("-sqrt(2*x)"); - $F1 = Formula("sqrt(2)*2/3*x^(3/2)"); - $F2 = Formula("-(x-1)^2"); - $G = Formula("-sqrt(2)*2/3*x^(3/2)"); - $D1 = Formula("$F1-$G"); - $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); - $D2 = Formula("$F2-$G"); - $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); - $area = $area1 + $area2; - - -

    - Between the curves x=-\frac12 y+1 and x=\frac12 y^2. -

    - - - - Graph of the region enclosed by the functions x=-\frac12 y+1 and x=\frac12 y^2. - The line y=x+2 is drawn starting at the point (0.5,1), from which it goes downwards and ends at the point (2,-2) . - The sideways parabola given by x=\frac12 y^2 is also drawn starting from the point (0.5,1), from which it goes backwards until the point (0,0). - The parabola then heads to the right before meeting the line at the point (2,-2) . - The line x=-\frac12 y+1 graphed between x=-0.5 and x=2 lies entirely above the parabola x=\frac12 y^2 between the same bounds, only interesecting at the end points. - However, between x=0 and x=0.5 , both the top and bottom of the enclosed region are made up by the parabola. - The sideways parabola x=\frac12 y^2 lies to the left of the line x=-\frac12 y+1 for the entirety of the region between y=-2 and y=1 . - - Graph of the region enclosed by the line and parabola. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-2.5,ymax=1.5, - xmin=-.5,xmax=2.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(2.,-2.)(1.805,-1.9)(1.62,-1.8)(1.445,-1.7)(1.28,-1.6)(1.125,-1.5)(0.98,-1.4)(0.845,-1.3)(0.72,-1.2)(0.605,-1.1)(0.5,-1.)(0.405,-0.9)(0.32,-0.8)(0.245,-0.7)(0.18,-0.6)(0.125,-0.5)(0.08,-0.4)(0.045,-0.3)(0.02,-0.2)(0.005,-0.1)(0,0)(0.005,0.1)(0.02,0.2)(0.045,0.3)(0.08,0.4)(0.125,0.5)(0.18,0.6)(0.245,0.7)(0.32,0.8)(0.405,0.9)(0.5,1.)(2,-2)}; - - \addplot [firstcurvestyle,domain=-2:1] ({.5*x^2},x) node [shift={(5pt,-110pt)},black] { \(x=\frac12y^2\)}; - \addplot [firstcurvestyle,domain=-2:1] ({-.5*x+1},x) node [pos=.85,shift={(40pt,0pt)},black] { \(x=-\frac12y+1\)}; - - \end{axis} - - \end{tikzpicture} - - - - - -

    - -

    - -     - -

    - 9/4 -

    -     -     -     - - - - - $x0 = Compute("0"); - $x1 = Compute("1/2"); - $x2 = Compute("1"); - - $f1 = Formula("x^(1/3)"); - $f2 = Formula("sqrt(x-1/2)"); - $g = Formula("0"); - $F1 = Formula("3/4*x^(4/3)"); - $F2 = Formula("2/3*(x-1/2)^(3/2)"); - $G = Formula("0"); - $D1 = Formula("$F1-$G"); - $area1 = $D1->eval(x=>$x1) - $D1->eval(x=>$x0); - $D2 = Formula("$F1-$F2"); - $area2 = $D2->eval(x=>$x2) - $D2->eval(x=>$x1); - $area = $area1 + $area2; - - -

    - Bounded by y=x^{1/3}, y=\sqrt{x-1/2}, - y=0, and x=1. -

    - - - - Graph of the region enclosed by the functions y=x^{1/3}, y=\sqrt{x-1/2} and the lines y=0, x=1. - The curve y=x^{1/3} is drawn starting at the origin, from which it goes upwards and ends at the point (1,1) . - The curve y=\sqrt{x-1/2} is drawn starting from the point (0.5,0), from which it goes upwards until ending at the point (1,sqrt{\frac12}). - The curve y=x^{1/3} graphed between x=0 and x=0.5 lies above x-axis. - Between x=0.5 and x=1 , the curve y=x^{1/3} lies above the curve y=\sqrt{x-1/2}. - The curve y=x^{1/3} lies to the left of the curve y=\sqrt{x-1/2} for the entirety of the length of the curve y=\sqrt{x-1/2}. - After this point, the curve y=x^{1/3} lies to the left of the boundary line occuring at x=1. - - Graph of the region enclosed by the two functions and y=0 and x=1. - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle] coordinates { (0,0)(0.01,0.2154)(0.02,0.2714)(0.03,0.3107)(0.04,0.342)(0.05,0.3684)(0.06,0.3915)(0.07,0.4121)(0.08,0.4309)(0.09,0.4481)(0.1,0.4642)(0.12,0.4932)(0.14,0.5192)(0.16,0.5429)(0.18,0.5646)(0.2,0.5848)(0.22,0.6037)(0.24,0.6214)(0.26,0.6383)(0.28,0.6542)(0.3,0.6694)(0.4,0.7368)(0.5,0.7937)(0.6,0.8434)(0.7,0.8879)(0.8,0.9283)(0.9,0.9655)(1.,1.)(1.,0.7071)(0.9,0.6325)(0.8,0.5477)(0.7,0.4472)(0.6,0.3162)(0.59,0.3)(0.58,0.2828)(0.57,0.2646)(0.56,0.2449)(0.55,0.2236)(0.54,0.2)(0.53,0.1732)(0.52,0.1414)(0.51,0.1)(0.5,0)(0,0)}; - - \addplot [firstcurvestyle,domain=0:1] (x^3,x) node [shift={(-40pt,0pt)},black] { \(y=x^{1/3}\)}; - \addplot [firstcurvestyle,domain=0:.7071] ({(x)^2+.5},x) node [shift={(-20pt,-50pt)},black] { \(y=\sqrt{x-1/2}\)}; - - \end{axis} - - \end{tikzpicture} - - - - - -

    - -

    - -     - -

    - \frac{1}{12}(9-2\sqrt{2})\approx 0.514 -

    -     -     -     - - - -

    - Bounded by the curves y=\sqrt{x}+1, y=\sqrt{2-x}+1, - and y=1. -

    - - Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1 and the line y=1. - The curve y=\sqrt{x}+1 is drawn starting at the point (0,1), from which it goes upwards and ends at the point (1,2) . - The curve y=\sqrt{2-x}+1 is drawn starting from the end of the previous curve, at the point (1,2), from which it goes downward until ending at the point (2,1). - Both of these curves lie entirely above the horizontal line y=1. - The curve y=\sqrt{x}+1 lies to the left of the curve y=\sqrt{2-x}+1 for the entirety of the enclosed region. - - Graph of the region enclosed by the two functions and the line y=1. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=2.5, - xmin=-.1,xmax=2.25 - ] - - \addplot [name path=A,firstcurvestyle,domain=0:1,samples=40] {1}; - \addplot [name path=B,firstcurvestyle,domain=0:1,samples=40] ({x^2},{x+1}) node [pos=.85,above left,black] { $y=\sqrt{x}+1$}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B]; - - \addplot [name path=C,firstcurvestyle,domain=1:2,samples=2] {1}; - \addplot [name path=D,firstcurvestyle,domain=0:1,samples=40] ({-x^2+2},{x+1}) node [pos=.9,above right,black] { $y=\sqrt{2-x}+1$}; - - \addplot [firstcurvestyle,areastyle] fill between [of=C and D]; - - \end{axis} - - \end{tikzpicture} - - - -     - -

    - 4/3 -

    -     -     - -     - - - -

    - Find the area of the triangle formed by the given three points. -

    -     - - - - - @p1 = (1,1); - @p2 = (2,3); - @p3 = (3,3); - $area = 1/2*($p1[1]*($p2[0]-$p3[0])+$p2[1]*($p3[0]-$p1[0]) + $p3[1]*($p1[0] - $p2[0])); - $area = Formula(abs($area)); - $s1 = join(", ", @p1); - $s2 = join(", ", @p2); - $s3 = join(", ", @p3); - - -

    - (),(), and () -

    - -

    - -

    - -     -     -     - - - - - @p1 = (-1,1); - @p2 = (1,3); - @p3 = (2,-1); - $area = 1/2*($p1[1]*($p2[0]-$p3[0])+$p2[1]*($p3[0]-$p1[0]) + $p3[1]*($p1[0] - $p2[0])); - $area = Formula(abs($area)); - $s1 = join(", ", @p1); - $s2 = join(", ", @p2); - $s3 = join(", ", @p3); - - -

    - (),(), and () -

    - -

    - -

    - -     -     -     - - - - - @p1 = (1,1); - @p2 = (3,3); - @p3 = (0,4); - $area = 1/2*($p1[1]*($p2[0]-$p3[0])+$p2[1]*($p3[0]-$p1[0]) + $p3[1]*($p1[0] - $p2[0])); - $area = Formula(abs($area)); - $s1 = join(", ", @p1); - $s2 = join(", ", @p2); - $s3 = join(", ", @p3); - - -

    - (),(), and () -

    - -

    - -

    - -     -     -     - - - - - @p1 = (0,0); - @p2 = (2,5); - @p3 = (5,2); - $area = 1/2*($p1[1]*($p2[0]-$p3[0])+$p2[1]*($p3[0]-$p1[0]) + $p3[1]*($p1[0] - $p2[0])); - $area = Formula(abs($area)); - $s1 = join(", ", @p1); - $s2 = join(", ", @p2); - $s3 = join(", ", @p3); - - -

    - (),(), and () -

    - -

    - -

    - -     -     -     - -     - - - - - - $a = 0; - $b = 6; - $n = 5; - $delta_x = ($b - $a)/$n; - @y = (0, 4.9, 5.2, 7.3, 4.5, 0); - @coeffs = (1, 2, 2, 2, 2, 1); - $area = 0; - $size = scalar @y; - foreach $i (0..$size) { - $area += $coeffs[$i] * $y[$i]; - } - $area = $area*$delta_x/2; - $area_ft = $area * 100 * 100; - $area = NumberWithUnits("$area cm^2"); - $area_ft = NumberWithUnits("$area_ft ft^2"); - - -

    - Use the Trapezoidal Rule to approximate the area of the pictured lake whose lengths, - in hundreds of feet, - are measured in 100-foot increments. -

    - - - - -

    - A sketch of a lake with four vertically strenching length measurements measured with lengths in hundreds of feet. - Each length measurement is given in 100-foot increments from the previous measurement, starting 100 feet from the leftmost point of the lake. - The first vertical length measurement is 4.9. - The second is 5.2. - The third is 7.3. - The fourth is 4.5. - There is then a 100-foot distance from the last measurement to the end of the lake. -

    -     - A sketch of a lake with four vertical length measurements. - - - \begin{tikzpicture}[xscal=1.19,yscale=.6] - - \draw [firstcolor,thick,fill=firstcolor!15,smooth] plot coordinates {(0,1.)(0.1013,1.899)(0.3567,2.785)(0.6934,3.42)(1.039,3.569)(1.355,3.224)(1.646,2.705)(1.925,2.355)(2.2,2.473)(2.475,2.968)(2.75,3.529)(3.025,3.844)(3.3,3.709)(3.575,3.249)(3.85,2.651)(4.125,2.098)(4.389,1.661)(4.624,1.285)(4.807,0.9071)(4.921,0.4743)(4.958,0)(4.921,-0.4743)(4.807,-0.9071)(4.624,-1.279)(4.389,-1.625)(4.125,-1.986)(3.85,-2.401)(3.575,-2.843)(3.3,-3.233)(3.025,-3.487)(2.75,-3.536)(2.475,-3.397)(2.2,-3.111)(1.925,-2.718)(1.646,-2.263)(1.355,-1.792)(1.039,-1.352)(0.6934,-0.9844)(0.3567,-0.6744)(0.1013,-0.3653)(0,0)(0,1)}; - - \draw (1,3.56) -- (1,-1.35) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.9}; - \draw (2,2.37) -- (2,-2.8) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 5.2}; - \draw (3,3.8) -- (3,-3.5) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 7.3}; - \draw (4,2.35) -- (4,-2.15) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.5}; - - \end{tikzpicture} - - - - - -

    - -

    - -     -     -     - - - - - - $a = 0; - $b = 6; - $n = 6; - $delta_x = ($b - $a)/$n; - @y = (0, 4.25, 6.6, 7.7, 6.45, 4.9, 0); - @coeffs = (1, 4, 2, 4, 2, 4, 1); - $area = 0; - $size = scalar @y; - foreach $i (0..$size) { - $area += $coeffs[$i] * $y[$i]; - } - $area = $area*$delta_x/3; - $area_ft = $area * 100 * 200; - $area = NumberWithUnits("$area cm^2"); - $area_ft = NumberWithUnits("$area_ft ft^2"); - - -

    - Use Simpson's Rule to approximate the area of the pictured lake whose lengths, - in hundreds of feet, - are measured in 200-foot increments. -

    - - - - - -

    - A sketch of a lake with five vertically strenching length measurements with lengths in hundreds of feet. - Each length measurement is given in 200-foot increments from the previous measurement, starting 200 feet from the leftmost point of the lake. - The first vertical length measurement is 4.25. - The second is 6.6. - The third is 7.7. - The fourth is 6.45. - The fifth is 4.9. - There is then a 200-foot distance from the last measurement to the end of the lake. -

    -     - A sketch of a lake with five vertical length measurements. - - - \begin{tikzpicture}[xscale=1.19,yscale=.6] - - \draw [firstcolor,thick,fill=firstcolor!15,smooth] plot coordinates {(0,2.)(0.1013,2.717)(0.3567,3.238)(0.6934,3.594)(1.039,3.818)(1.355,3.948)(1.646,4.035)(1.925,4.132)(2.2,4.28)(2.475,4.428)(2.75,4.471)(3.025,4.307)(3.305,3.88)(3.607,3.287)(3.952,2.652)(4.361,2.098)(4.815,1.661)(5.253,1.285)(5.614,0.9071)(5.843,0.4705)(5.938,-0.04167)(5.919,-0.6295)(5.807,-1.293)(5.624,-2.009)(5.389,-2.703)(5.125,-3.29)(4.849,-3.692)(4.562,-3.892)(4.243,-3.929)(3.871,-3.846)(3.432,-3.674)(2.956,-3.409)(2.484,-3.028)(2.057,-2.511)(1.692,-1.869)(1.363,-1.168)(1.039,-0.4819)(0.6934,0.1248)(0.3567,0.679)(0.1013,1.273)(0,2.)}; - - \draw (1,3.8) -- (1,-.45) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.25}; - \draw (2,4.15) -- (2,-2.45) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.6}; - \draw (3,4.3) -- (3,-3.4) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 7.7}; - \draw (4,2.6) -- (4,-3.85) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 6.45}; - \draw (5,1.45) -- (5,-3.45) node [shift={(-3pt,0pt)},rotate=90,pos=.5] { 4.9}; - - \end{tikzpicture} - - - - - -

    - -

    - -     -     -     -     -     -
    -
    - Volume by Cross-Sectional Area; Disk and Washer Methods -

    - The volume of a general right cylinder, - as shown in , is -

    - -
    -

    - Area of the base height. -

    -
    - -
    - The volume of a general right cylinder - - - - -

    - An image of a general right cylinder. - The area of the base of the cylinder is given to be A. - The base of the cylinder resemembles a parallelogram with curved edges, with each side slightly bowing in at the half way point between two corners. - The height of the cylinder is h. - The base of the general cylinder is identical to the top of the general cylinder, with the two faces being parallel. - These faces also coincide on top of eachother, which leads to right angles when the top and base are connected to form the general right cylinder. - The volume of the general right cylinder is V=A\cdot h. -

    -     - An image of a general right cylider with a base area of A and heigh h. - - - - - - //ASY file for figcross1.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(3.4,5.4,.9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={1,2,3}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,2.5); - pair ybounds=(-0.25,2.5); - pair zbounds=(-0.25,1.5); - - //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple f(pair t) { - return (t.y*(cos(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))+.5*sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))),t.y*(sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))),0); - } - surface s=surface(f,(0,0),(2*pi,1),16,2,Spline); - pen p=apexmeshpen; - draw(s,darksurfacepen,meshpen=p); - - triple f(pair t) { - return (cos(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))+.5*sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x)),sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x)),t.y); - } - surface s=surface(f,(0,0),(2*pi,1),16,2,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (t.y*(cos(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))+.5*sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))),t.y*(sin(t.x)*(1.5+.5*cos(t.x)*sin(3*t.x))),1); - } - surface s=surface(f,(0,0),(2*pi,1),16,2,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return (cos(t)*(1.5+.5*cos(t)*sin(3*t))+.5*sin(t)*(1.5+.5*cos(t)*sin(3*t)),sin(t)*(1.5+.5*cos(t)*sin(3*t)),1);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (cos(t)*(1.5+.5*cos(t)*sin(3*t))+.5*sin(t)*(1.5+.5*cos(t)*sin(3*t)),sin(t)*(1.5+.5*cos(t)*sin(3*t)),0);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - label("base area $= A$",(2,2,-.1)); - label("Volume $=A\cdot h$",(2,2,-.4)); - label("$h$",(-1.9,0,.5)); - - - - -
    - -

    - We can use this fact as the building block in finding volumes of a variety of shapes. -

    - - - -

    - Given an arbitrary solid, - we can approximate its volume by cutting it into n thin slices. - When the slices are thin, - each slice can be approximated well by a general right cylinder. - Thus the volume of each slice is approximately its cross-sectional area thickness. - (These slices are the differential elements.) -

    - -

    - By orienting a solid along the x-axis, - we can let A(x_i) represent the cross-sectional area of the ith slice, - and let \dx_i represent the thickness of this slice - (the thickness is a small change in x). - The total volume of the solid is approximately: - - \text{Volume} \amp \approx \sum_{i=1}^n \Big[\text{Area} \,\times\,\text{thickness} \Big] - \amp = \sum_{i=1}^n A(x_i)\dx_i - . -

    - -

    - Recognize that this is a Riemann Sum. - By taking a limit - (as the thickness of the slices goes to 0) - we can find the volume exactly. -

    - - - Volume By Cross-Sectional Area - -

    - The volume V of a solid, - oriented along the x-axis with cross-sectional area A(x) from x=a to x=b, is - integrationvolume!cross-sectional area - - V = \int_a^b A(x)\, dx - . -

    -     -     - - - Finding the volume of a solid - -

    - Find the volume of a pyramid with a square base of side length 10 - and a height of 5. -

    -     - -

    - There are many ways to orient - the pyramid along the x-axis; - gives one such way, - with the pointed top of the pyramid at the origin and the x-axis going through the center of the base. -

    - -
    - Orienting a pyramid along the x-axis in - - - - -

    - Three-dimensional plot of a pyramid with a square base of side length 10 and height 5. - The peak of the pyramid is placed on the origin, from which it expands towards the positive x-axis. - The square base of the pyramid is centered at the point (5,0). - Additionally, the plot contains an arbitrarily chosen point labeled x, lying on the x-axis. - Slicing the pyramid parallel to the base at this point x results in side lengths of the square slice to be 2x. -

    -     - A three-dimensional graph of a pyramid with a square base of side length 10 and height 5. - - - - - //ASY file for fig13_06_ex_143D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((8.1,15.5,23.4),(0,1,0),(0,0,0),1,(-0.13,0.0046)); - defaultrender.merge=true; - - //(-0.01,0.061,-0.029) - - // setup and draw the axes - real[] myxchoice={5}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,8); - pair ybounds=(-8,8); - pair zbounds=(-8,8); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - path3 p=(5,5,5)--(5,-5,5)--(5,-5,-5)--(5,5,-5); - draw(surface(p -- cycle), simplesurfacepen); - draw(p--cycle,bluepen+.3mm); - - path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); - draw(surface(p1 -- cycle), simplesurfacepen); - draw(p1--cycle,bluepen+.3mm); - - path3 p2=(0,0,0)--(5,5,5)--(5,5,-5); - draw(surface(p2 -- cycle), simplesurfacepen); - draw(p2--cycle,bluepen+.3mm); - - path3 p3=(0,0,0)--(5,5,-5)--(5,-5,-5); - draw(surface(p3 -- cycle), simplesurfacepen); - draw(p3--cycle,bluepen+.3mm); - - path3 p4=(0,0,0)--(5,-5,5)--(5,-5,-5); - draw(surface(p4 -- cycle), simplesurfacepen); - draw(p4--cycle,bluepen+.3mm); - - draw((3,-3,3) -- (3,-3,-3) -- (3,3,-3)-- (3,3,3)--cycle,black+.5mm); - - dotfactor=3; - dot((3,0,0)); - label("$x$",(3,0,0),S); - - label("$2x$",(3,0,3),W); - label("$2x$",(3,3,0),N); - label("$10$",(5,-5,0),S); - label("$10$",(5,0,-5),E); - - - - -
    - -

    - Each cross section of the pyramid is a square; - this is a sample differential element. - To determine its area A(x), - we need to determine the side lengths of the square. -

    - -

    - When x=5, the square has side length 10; - when x=0, the square has side length 0. - Since the edges of the pyramid are lines, - it is easy to figure that each cross-sectional square has side length 2x, - giving A(x) = (2x)^2=4x^2. -

    - -

    - If one were to cut a slice out of the pyramid at x=3, - as shown in , - one would have a shape with square bottom and top with sloped sides. - If the slice were thin, - both the bottom and top squares would have sides lengths of about 6, and thus the cross-sectional area of the bottom and top would be about - 36. - Letting \Delta x_i represent the thickness of the slice, - the volume of this slice would then be about 36\Delta x_i . -

    - -
    - Cutting a slice in the pyramid in at x=3 - - - - -

    - Three-dimensional plot of the same pyramid with a square base of side length 10 and height 5 as in the previous image. - This time, the plot contains a thin slice of the pyramid at x=3 which is parallel to the base of the pyramid. - The side lengths of the resulting square slice are 6, giving the square a surface area of 36 with an arbitrarily small thickness \Delta x. -

    -     - A three-dimensional plot of the same pyramid showing a small slice being used to approximate its volume. - - - - - //ASY file for fig13_06_ex_143D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((8.1,15.5,23.4),(0,1,0),(0,0,0),1,(-0.13,0.0046)); - defaultrender.merge=true; - - //(-0.01,0.061,-0.029) - - // setup and draw the axes - real[] myxchoice={3,5}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,8); - pair ybounds=(-8,8); - pair zbounds=(-8,8); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple f1(real t) { - return (t,t,t); - } - triple f2(real t) { - return (t,-t,t); - } - triple f3(real t) { - return (t,t,-t); - } - triple f4(real t) { - return (t,-t,-t); - } - //path3 mypath1=graph(f1,2.8,3.2,operator ..); - //draw(mypath1,bluepen+.3mm); - - real x1=2.8; - real x2=3.2; - - path3 p1=f1(x1)--f1(x2)--f2(x2)--f2(x1); - draw(surface(p1 -- cycle), simplesurfacepen); - draw(p1--cycle,bluepen+.3mm); - - path3 p2=f1(x1)--f1(x2)--f3(x2)--f3(x1); - draw(surface(p2 -- cycle), simplesurfacepen); - draw(p2--cycle,bluepen+.3mm); - - path3 p3=f4(x1)--f4(x2)--f3(x2)--f3(x1); - draw(surface(p3 -- cycle), simplesurfacepen); - draw(p3--cycle,bluepen+.3mm); - - path3 p4=f4(x1)--f4(x2)--f2(x2)--f2(x1); - draw(surface(p4 -- cycle), simplesurfacepen); - draw(p4--cycle,bluepen+.3mm); - - path3 p5=f1(x1)--f3(x1)--f4(x1)--f2(x1); - draw(surface(p5 -- cycle), simplesurfacepen); - draw(p5--cycle,bluepen+.3mm); - - path3 p6=f1(x2)--f3(x2)--f4(x2)--f2(x2); - draw(surface(p6 -- cycle), simplesurfacepen); - draw(p6--cycle,bluepen+.3mm); - - label("$\Delta x$",f3(3),N); - - path3 p=(5,5,5)--(5,-5,5)--(5,-5,-5)--(5,5,-5); - draw(p--cycle,bluepen+.2mm+dashed); - - path3 p1=(0,0,0)--(5,5,5); - draw(p1,bluepen+.2mm+dashed); - - path3 p2=(0,0,0)--(5,-5,-5); - draw(p2,bluepen+.2mm+dashed); - - path3 p3=(0,0,0)--(5,5,-5); - draw(p3,bluepen+.2mm+dashed); - - path3 p4=(0,0,0)--(5,-5,5); - draw(p4,bluepen+.2mm+dashed); - - //draw((3,-3,3) -- (3,-3,-3) -- (3,3,-3)-- (3,3,3)--cycle,black+.5mm); - - //dotfactor=3; - //dot((3,0,0)); - //label("$x$",(3,0,0),S); - - //label("$2x$",(3,0,3),W); - //label("$2x$",(3,3,0),N); - //label("$10$",(5,-5,0),S); - //label("$10$",(5,0,-5),E); - - - - -
    - -

    - Cutting the pyramid into n slices divides the total volume into n equally-spaced smaller pieces, - each with volume (2x_i)^2\Delta x, - where x_i is the approximate location of the slice along the x-axis and - \Delta x represents the thickness of each slice. - One can approximate total volume of the pyramid by summing up the volumes of these slices: - - \text{Approximate volume} = \sum_{i=1}^n (2x_i)^2\Delta x - . -

    - -

    - Taking the limit as n\to\infty gives the actual volume of the pyramid; - recoginizing this sum as a Riemann Sum allows us to find the exact answer using a definite integral, - matching the definite integral given by . -

    - -

    - We have - - V \amp = \lim_{n\to\infty} \sum_{i=1}^n (2x_i)^2\Delta x - \amp = \int_0^5 4x^2\, dx - \amp = \frac43x^3\Big|_0^5 - \amp =\frac{500}{3}\,\text{in}^3 \approx 166.67\,\text{in}^3 - . -

    - -

    - We can check our work by consulting the general equation for the volume of a pyramid - (a pyramid is a special type of cone see in the back matter under - Volume of A General Cone): -

    - -

    - \frac13\times \,\text{area of base}\, \times \,\text{height}. -

    - -

    - Certainly, using this formula from geometry is faster than our new method, - but the calculus-based method can be applied to much more than just cones. -

    -     - -     - -

    - An important special case of - is when the solid is a solid of revolution, that is, - when the solid is formed by rotating a shape around an axis. -

    - -

    - Start with a function y=f(x) from x=a to x=b. - Revolving this curve about a horizontal axis creates a three-dimensional solid whose cross sections are disks - (thin circles). - Let R(x) represent the radius of the cross-sectional disk at x; - the area of this disk is \pi R(x)^2. - Applying gives the Disk Method. -

    - - - The Disk Method -

    - Let a solid be formed by revolving the curve y=f(x) from x=a to x=b around a horizontal axis, - and let R(x) be the radius of the cross-sectional disk at x. - The volume of the solid is - integrationvolume!Disk Method - Disk Method - - V = \pi \int_a^b R(x)^2\, dx - . -

    -     - - - - - Finding volume using the Disk Method - -

    - Find the volume of the solid formed by revolving the curve y=1/x, - from x=1 to x=2, around the x-axis. -

    -     - -

    - A sketch can help us understand this problem. - In , the curve y=1/x is sketched along with the differential element a disk at x with radius R(x)=1/x. - In the whole solid is pictured, along with the differential element. -

    - -

    - The volume of the differential element shown in is approximately \pi R(x_i)^2\Delta x, - where R(x_i) is the radius of the disk shown and - \Delta x is the thickness of that slice. - The radius R(x_i) is the distance from the x-axis to the curve, - hence R(x_i) = 1/x_i. -

    - -
    - Sketching a solid in - -
    - - - - - -

    - Three-dimensional plot the curve y=1/x between x=1 and x=2 lying in the xy plane. - The volume of the inside of the rotated curve is then approximated using thin circular disks. - Taking an arbitrary circular disk, it would have a radius which whose length is equal to the distance between the x-axis and the curve y=1/x. - The image depicts a circular disk centered at some arbitrary value of x between x=1 and x=2, with the radius given by the function R(x)=1/x. -

    -     - A three-dimensional plot of the curve from the example showing a thin vertical slice being used to approximate its volume. - - - - - //ASY file for figdisk1.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((18,7.7,21.1),(0,1,0),(0,0,0),1,(-0.088,.0039)); - defaultrender.merge=true; - - - //(-0.01,0.061,-0.029) - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={.5,1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-0.1,2.2); - pair ybounds=(-1.2,1.2); - pair zbounds=(-1.2,1.2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - path3 p=(3/2,2/3,0)..(3/2,0,2/3)..(3/2,-2/3,0)..(3/2,0,-2/3); - draw(surface(p .. cycle), simplesurfacepen2); - draw(p..cycle,redpen+.4mm); - - triple g(real t) {return (t,1/t,0);} - path3 mypath=graph(g,1,2,operator ..); - draw(mypath,bluepen+linewidth(2)); - - draw((1.5,0,0)--(1.5,2/3,0),black+.3mm); - - triple pt=(1.3,Sin(-20),Cos(-20)); - - draw(pt--(1.48,.3,.05),linewidth(.75),Arrow3); - label("$R(x)=1/x$",pt,S); - - label("$y=1/x$",(1,1,0),N); - - //triple f(pair t) {return (t.x,1/t.x*cos(t.y),1/t.x*sin(t.y));} - //surface s=surface(f,(1,0),(2,2*pi),5,16,Spline); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - //path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); - //draw(surface(p1 -- cycle), simplesurfacepen); - //draw(p1--cycle,bluepen+.3mm); - - - - -
    - -
    - - - - - -

    - Three-dimensional plot the curve y=1/x between x=1 and x=2 lying in the xy plane. - The curve is then rotated in a full circle about the x-axis. - The solid is then bounded on both sides by the planes x=1/x and x=2/x which close off the solid with cirles of radius 1 and \frac12 respectively. - The volume of the inside of the curve is then approximated using thin circular disks which lie completely inside the solid as described in the previous image. -

    -     - A three-dimensional plot of the solid which comes from rotating the curve from the example. - - - - - - //ASY file for figdisk1b.asy in Chapter 7 - - size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((18,7.7,21.1),(0,1,0),(0,0,0),1,(-0.088,.0039)); - defaultrender.merge=true; - - //(-0.01,0.061,-0.029) - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={.5,1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-0.1,2.2); - pair ybounds=(-1.2,1.2); - pair zbounds=(-1.2,1.2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - path3 p2=(1,1,0)..(1,0,1)..(1,-1,0)..(1,0,-1); - draw(surface(p2 .. cycle), simplesurfacepen); - draw(p2..cycle,bluepen+.4mm); - - triple f(pair t) {return (t.x,1/t.x*cos(t.y),1/t.x*sin(t.y));} - surface s=surface(f,(1,0),(2,2*pi),5,16,Spline); - pen p=apexmeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - path3 p1=(2,.5,0)..(2,0,.5)..(2,-.5,0)..(2,0,-.5); - draw(surface(p1 .. cycle), simplesurfacepen); - draw(p1..cycle,bluepen+.4mm); - - path3 p=(3/2,2/3,0)..(3/2,0,2/3)..(3/2,-2/3,0)..(3/2,0,-2/3); - draw(surface(p .. cycle), simplesurfacepen2); - draw(p..cycle,redpen+.4mm); - - triple g(real t) {return (t,1/t,0);} - path3 mypath=graph(g,1.01,1.99,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); - //draw(surface(p1 -- cycle), simplesurfacepen); - //draw(p1--cycle,bluepen+.3mm); - - - - -
    -     - -
    - -

    - Slicing the solid into n equally-spaced slices, - we can approximate the total volume by adding up the approximate volume of each slice: - - \text{Approximate volume} = \sum_{i=1}^n \pi \left(\frac1{x_i}\right)^2\Delta x - . -

    - -

    - Taking the limit of the above sum as - n\to\infty gives the actual volume; - recognizing this sum as a Riemann sum allows us to evaluate the limit with a definite integral, - which matches the formula given in : - - V \amp = \lim_{n\to\infty}\sum_{i=1}^n \pi \left(\frac1{x_i}\right)^2\Delta x - \amp = \pi\int_1^2 \left(\frac1x\right)^2\, dx - \amp = \pi\int_1^2 \frac1{x^2}\, dx - - - \amp = \pi\left[-\frac1x\right]\Big|_1^2 - \amp = \pi \left[-\frac12 - \left(-1\right)\right] - \amp = \frac{\pi}{2}\,\text{units}^3 - . -

    -     - -     - - - - - - - - - -

    - While - is given in terms of functions of x, - the principle involved can be applied to functions of y when the axis of rotation is vertical, - not horizontal. - We demonstrate this in the next example. -

    - - - Finding volume using the Disk Method - -

    - Find the volume of the solid formed by revolving the curve y=1/x, - from x=1 to x=2, about the y-axis. -

    -     - -

    - Since the axis of rotation is vertical, - we need to convert the function into a function of y and convert the x-bounds to y-bounds. - Since y=1/x defines the curve, - we rewrite it as x=1/y. - The bound x=1 corresponds to the y-bound y=1, - and the bound x=2 corresponds to the y-bound y=1/2. -

    - -

    - Thus we are rotating the curve x=1/y, - from y=1/2 to y=1 about the y-axis to form a solid. - The curve and sample differential element are sketched in , - with a full sketch of the solid in . -

    - -
    - Sketching a solid in - -
    - - - - - -

    - Three-dimensional plot the curve y=1/x between x=1 and x=2 lying in the xy plane. - The volume of the inside of the rotated curve is approximated thin horizontal circular disks. - Taking an arbitrary circular disk, it would have a radius which whose length is equal to the distance between the y-axis and the curve x=1/y. - The image depicts a circular disk centered at some arbitrary value of y between y=\frac12 and y=1, with the radius given by the function R(y)=1/y. -

    -     - A three-dimensional plot of the curve from the example showing a thin horizontal slice being used to approximate its volume. - - - - - - //ASY file for figdisk1a.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((24.2,2.9,24.5),(0,1,0),(0,0,0),.95,(0.011,0.0033)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,-1,1,2}; - real[] myychoice={1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-2.3,2.3); - pair ybounds=(-.2,1.2); - pair zbounds=(-2.3,2.3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); - draw(surface(p .. cycle), simplesurfacepen2); - draw(p..cycle,redpen+.4mm); - - triple g(real t) {return (t,1/t,0);} - path3 mypath=graph(g,1,2,operator ..); - draw(mypath,bluepen+linewidth(2)); - - draw((0,2/3,0)--(1.5,2/3,0),black+.3mm); - - label("$x=1/y$",(1,1,0),N); - - triple pt=(1,.2,0); - draw(pt--(.75,.65,0),linewidth(.75),Arrow3); - label("$R(y)=1/y$",pt,S); - - triple f(pair t) {return (1/t.x*cos(t.y),t.x,1/t.x*sin(t.y));} - surface s=surface(f,(.5,0),(1,2*pi),5,16,Spline); - pen p=apexmeshpen+.1mm; - draw(s,invisible); - - //triple f(pair t) {return (t.x,1/t.x*cos(t.y),1/t.x*sin(t.y));} - //surface s=surface(f,(1,0),(2,2*pi),5,16,Spline); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - //path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); - //draw(surface(p1 -- cycle), simplesurfacepen); - //draw(p1--cycle,bluepen+.3mm); - - - - -
    - -
    - - - - - -

    - Three-dimensional plot the curve y=1/x from x=1 and x=2 lying in the xy plane. - The curve is then rotated in a full circle about the y-axis. - The solid is then bounded on both the top and bottom by the planes y=1 and y=\frac12. - The plane y=1 closes off the top of the solid with a circle of radius 1, and the plane y=\frac12 closes off the bottom of the solid with a circle of radius 2. - The volume of the inside of the curve is then approximated using thin circular disks which are contained completely inside the solid as described in the previous image. -

    -     - A three-dimensional plot of the solid which comes from rotating the curve from the example. - - - - - - //ASY file for figdisk2a.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((24.2,2.9,24.5),(0,1,0),(0,0,0),.95,(0.011,0.0033)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,-1,1,2}; - real[] myychoice={1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-2.3,2.3); - pair ybounds=(-.2,1.2); - pair zbounds=(-2.3,2.3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple g(real t) {return (t,1/t,0);} - path3 mypath=graph(g,1.005,1.99,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple f(pair t) {return (1/t.x*cos(t.y),t.x,1/t.x*sin(t.y));} - surface s=surface(f,(.5,0),(1,2*pi),5,16,Spline); - pen p=apexmeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - path3 p1=(1,1,0)..(0,1,1)..(-1,1,0)..(0,1,-1); - draw(surface(p1 .. cycle), simplesurfacepen); - draw(p1..cycle,bluepen+.4mm); - - path3 p2=(2,.5,0)..(0,.5,2)..(-2,.5,0)..(0,.5,-2); - draw(surface(p2 .. cycle), simplesurfacepen); - draw(p2..cycle,bluepen+.4mm); - - path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); - draw(surface(p .. cycle), simplesurfacepen2); - draw(p..cycle,redpen+.4mm); - - //triple f(pair t) {return (t.x,1/t.x*cos(t.y),1/t.x*sin(t.y));} - //surface s=surface(f,(1,0),(2,2*pi),5,16,Spline); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - //path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); - //draw(surface(p1 -- cycle), simplesurfacepen); - //draw(p1--cycle,bluepen+.3mm); - - - - -
    -     - -
    - -

    - We integrate to find the volume: - - V \amp = \pi\int_{1/2}^1 \frac{1}{y^2}\,dy - \amp = -\frac{\pi}y\Big|_{1/2}^1 - \amp = \pi\,\text{units}^3 - . -

    -     - -     - - - -

    - We can also compute the volume of solids of revolution that have a hole in the center. - The general principle is simple: - compute the volume of the solid irrespective of the hole, - then subtract the volume of the hole. - If the outside radius of the solid is R(x) and the inside radius - (defining the hole) - is r(x), then the volume is - - V = \pi\int_a^b R(x)^2 \,dx - \pi\int_a^b r(x)^2\,dx = \pi\int_a^b \left(R(x)^2-r(x)^2\right)\,dx - . -

    - -
    - Establishing the Washer Method; see also - -
    - - - - - -

    - Graph of two curves lying in the xy plane. - One of the curves is blue and resembles the peak of a wave. - The other curve is red, and resembles the trough of a wave. - Both curves are plotted between the points a and b on the x-axis. - For the interval [a,b] the blue curve lies entirely above the red curve. - Once the region between the blue and red curve is rotated about the y-axis, the blue curve which has a radius given by the function R(x) will be on the outside of the solid. - The red curve will be on the inside of the solid, which will have a radius given by the function r(x). - Both the functions R(x) and r(x) give the distance of the curve from the axis of rotation, which in this case is some arbitrarily chosen line paralell and above the x-axis. - The space on the inside of the red curve will then be completely empty once the region is rotated about the chosen axis of rotation. -

    -     - Graph of two curves which will be used to showcase the Washer Method. - - - - - //ASY file for figwasher_idea_a_3D.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.2,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-3.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*(-.5*(x-2)^2+3-((x-2)^2+1))+(x-2)^2+1}); - - //path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); - //draw(surface(p .. cycle), simplesurfacepen2); - //draw(p..cycle,redpen+.4mm); - - triple g1(real t) {return (t,((t-2)^2+1),0);} - path3 p1=graph(g1,1,3,operator ..); - draw(p1,redpen+.4mm); - - triple g2(real t) {return (t,(-.5*(t-2)^2+3),0);} - path3 p2=graph(g2,3,1,operator ..); - draw(p2,bluepen+.4mm); - - path3 p3=p1--p2; - draw(surface(p3 -- cycle), simplesurfacepen); - - draw((-.2,.5,0)--(3.5,.5,0),dashed+.2mm); - - draw((1,0,0)--(1,-.1,0)); - draw((3,0,0)--(3,-.1,0)); - label("$a$",(1,-.1,0),S); - label("$b$",(3,-.1,0),S); - - //\draw (axis cs:1.2,0,2.68) -- node [pos=.6,left] { $R(x)$} (axis cs:1.2,0,.5); - //\draw (axis cs: 2.8,0,1.64) -- node [pos=.5,right] { $r(x)$} (axis cs: 2.8,0,.5); - - triple dot1=(1.2,2.68,0); - triple dot2=(1.2,.5,0); - triple dot3=(2.8,1.64,0); - triple dot4=(2.8,.5,0); - - draw(dot1--dot2,black+.2mm); - draw(dot3--dot4,black+.2mm); - - dotfactor=2; - dot(dot1); - dot(dot2); - dot(dot3); - dot(dot4); - - label("$R(x)$",dot2+.3*(dot1-dot2),W); - label("$r(x)$",dot4+.4*(dot3-dot4),E); - - - - -
    - -
    - - - - - -

    - Three dimensional graph of the region between the blue and red curves on the interval [a,b] rotated about some line parallel and above x-axis. - The outside of the solid is coloured blue, which resembles the blue curve which has a radius given by the function R(x) being rotated about the axis of rotation. - The inside of the solid is coloured red, which resembles the red curve which has a radius by the function r(x) being rotated about the axis of rotation. - The space on the inside of the red region is completely empty, which makes us subtract the outside radius R(x) from the inside radius r(x) to create washers which will approximate the volume of the solid. -

    -     - Three-dimensional plot of solid coming from rotating the region between the two curves. - - - - - //ASY file for figwasher_idea_a_3D.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.2,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-3.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*(-.5*(x-2)^2+3-((x-2)^2+1))+(x-2)^2+1}); - - //path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); - //draw(surface(p .. cycle), simplesurfacepen2); - //draw(p..cycle,redpen+.4mm); - - pen p=apexmeshpen+.1mm; - - triple g1(real t) {return (t,((t-2)^2+1),0);} - path3 p1=graph(g1,1,3,operator ..); - draw(p1,redpen+.5mm); - - triple g2(real t) {return (t,(-.5*(t-2)^2+3),0);} - path3 p2=graph(g2,3,1,operator ..); - draw(p2,bluepen+.5mm); - - triple f2(pair t) {return (1,cos(t.x)*t.y+.5,sin(t.x)*t.y);} - surface s2=surface(f2,(0,1.5),(2*pi,2),16,1,Spline); - draw(s2,simplesurfacepen,meshpen=p); - - triple f3(pair t) {return (3,cos(t.x)*t.y+.5,sin(t.x)*t.y);} - surface s3=surface(f3,(0,1.5),(2*pi,2),16,1,Spline); - draw(s3,simplesurfacepen,meshpen=p); - - triple g3(real t) {return (1,1.5*cos(t)+.5,1.5*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,redpen+.3mm); - - triple g3(real t) {return (3,1.5*cos(t)+.5,1.5*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,redpen+.3mm); - - triple g3(real t) {return (1,2*cos(t)+.5,2*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.3mm); - - triple g3(real t) {return (3,2*cos(t)+.5,2*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.3mm); - - //path3 p3=p1--p2; - //draw(surface(p3 -- cycle), simplesurfacepen); - - draw((-.2,.5,0)--(3.5,.5,0),dashed+.2mm); - - triple f(pair t) {return (t.x,(-.5*(t.x-2)^2+3-.5)*cos(t.y)+.5,(-.5*(t.x-2)^2+3-.5)*sin(t.y));} - surface s=surface(f,(1,0),(3,2*pi),5,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - triple f1(pair t) {return (t.x,((t.x-2)^2+1-.5)*cos(t.y)+.5,((t.x-2)^2+1-.5)*sin(t.y));} - surface s1=surface(f1,(1,0),(3,2*pi),5,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s1,simplesurfacepen2,meshpen=p); - - //draw(p..cycle,redpen+.4mm); - - //label("$x=1/y$",(1,1,0),N); - - //triple pt=(1,.2,0); - //draw(pt--(.75,.65,0),linewidth(.75),Arrow3); - //label("$R(y)=1/y$",pt,S); - - //triple f(pair t) {return (1/t.x*cos(t.y),t.x,1/t.x*sin(t.y));} - //surface s=surface(f,(.5,0),(1,2*pi),5,16,Spline); - //pen p=apexmeshpen+.1mm; - //draw(s,invisible); - - //triple f(pair t) {return (t.x,1/t.x*cos(t.y),1/t.x*sin(t.y));} - //surface s=surface(f,(1,0),(2,2*pi),5,16,Spline); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - //path3 p1=(0,0,0)--(5,5,5)--(5,-5,5); - //draw(surface(p1 -- cycle), simplesurfacepen); - //draw(p1--cycle,bluepen+.3mm); - - - - -
    -     - -
    - -

    - One can generate a solid of revolution with a hole in the middle by revolving a region about an axis. - Consider , - where a region is sketched along with a dashed, - horizontal axis of rotation. - By rotating the region about the axis, - a solid is formed as sketched in . - The outside of the solid has radius R(x), - whereas the inside has radius r(x). - Each cross section of this solid will be a washer (a disk with a hole in the center) as sketched in . This leads us to the Washer Method. -

    - -
    - Establishing the Washer Method; see also - - - - -

    - Three dimensional graph of the region between the blue and red curves on the interval [a,b]. - The graph also cotains a washer plotted on an arbitrary value of x in the interval [a,b], which will lie entirely inside the area between the blue and red curve. - The outside of the washer has a radius of R(x), while the inside of the washer has a radius of r(x). - In other words, the washer is a circle of radius R(x) with a circular hole of radius r(x) in the center. - Taking these washers to be arbitrarily thin, we can calculate the volume of the solid which comes from rotating the region. -

    -     - Three-dimensional plot of solid coming from rotating the region between the two curves. - - - - - //ASY file for figwasher_idea_a_3D.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.2,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-3.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*(-.5*(x-2)^2+3-((x-2)^2+1))+(x-2)^2+1}); - - //path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); - //draw(surface(p .. cycle), simplesurfacepen2); - //draw(p..cycle,redpen+.4mm); - - triple g1(real t) {return (t,((t-2)^2+1),0);} - path3 p1=graph(g1,1,3,operator ..); - draw(p1,redpen+.4mm); - - triple g2(real t) {return (t,(-.5*(t-2)^2+3),0);} - path3 p2=graph(g2,3,1,operator ..); - draw(p2,bluepen+.4mm); - - path3 p3=p1--p2; - draw(surface(p3 -- cycle), simplesurfacepen); - - draw((-.2,.5,0)--(3.5,.5,0),dashed+.2mm); - - draw((1,0,0)--(1,-.1,0)); - draw((3,0,0)--(3,-.1,0)); - label("$a$",(1,-.1,0),S); - label("$b$",(3,-.1,0),S); - - //\draw (axis cs:1.2,0,2.68) -- node [pos=.6,left] { $R(x)$} (axis cs:1.2,0,.5); - //\draw (axis cs: 2.8,0,1.64) -- node [pos=.5,right] { $r(x)$} (axis cs: 2.8,0,.5); - - triple dot1=(1.2,2.68,0); - triple dot2=(1.2,.5,0); - triple dot3=(2.8,1.64,0); - triple dot4=(2.8,.5,0); - - draw(dot1--dot2,black+.2mm); - draw(dot3--dot4,black+.2mm); - - dotfactor=2; - dot(dot1); - dot(dot2); - dot(dot3); - dot(dot4); - - label("$R(x)$",dot2+.3*(dot1-dot2),W); - label("$r(x)$",dot4+.4*(dot3-dot4),E); - - triple f2(pair t) {return (2,cos(t.x)*t.y+.5,sin(t.x)*t.y);} - surface s2=surface(f2,(0,.5),(2*pi,2.5),16,1,Spline); - draw(s2,simplesurfacepen2); - - triple g3(real t) {return (2,.5*cos(t)+.5,.5*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,redpen+.4mm); - - triple g3(real t) {return (2,2.5*cos(t)+.5,2.5*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.4mm); - - draw((2,1,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); - - - - -
    - - - The Washer Method -

    - Let a region bounded by y=f(x), y=g(x), - x=a and x=b be rotated about a horizontal axis that does not intersect the region, - forming a solid. - Each cross section at x will be a washer with outside radius R(x) and inside radius r(x). - The volume of the solid is - integrationvolume!Washer Method - Washer Method - - V = \pi\int_a^b \Big(R(x)^2-r(x)^2\Big)\, dx - . -

    -     - -

    - Even though we introduced it first, - the Disk Method is just a special case of the Washer Method with an inside radius of r(x)=0. -

    - - - Finding volume with the Washer Method - -

    - Find the volume of the solid formed by rotating the region bounded by - y=x^2-2x+2 and y=2x-1 about the x-axis. -

    -     - -

    - A sketch of the region will help, - as given in . - Rotating about the x-axis will produce cross sections in the shape of washers, - as shown in ; - the complete solid is shown in . - The outside radius of this washer is R(x) = 2x-1; - the inside radius is r(x) = x^2-2x+2. - As the region is bounded from x=1 to x=3, - we integrate as follows to compute the volume. - - V \amp = \pi\int_1^3 \Big((2x-1)^2-(x^2-2x+2)^2\Big)\, dx - \amp = \pi\int_1^3 \big(-x^4+4x^3-4x^2+4x-3\big)\, dx - \amp = \pi\Big[-\frac{1}{5}x^5+x^4-\frac43x^3+2x^2-3x\Big]\Big|_1^3 - \amp =\frac{104}{15}\pi \approx 21.78\,\text{units}^3 - . -

    - -
    - Sketching the differential element and solid in - -
    - - - - - -

    - Graph of the region enclosed by the curves y=x^2-2x+2 and y=2x-1. - The curves which create an enclosed region both start at the point (1,1) and end at the point (3,5). - The line y=2x-1 lies entirely above the curve y=x^2-2x+2 for the entirety of the enclosed region. - The radius of the outside of the washer is then given as distance between the x-axis and the line y=2x-1, and is given by the function R(x)=2x-1. - The inside radius of the washer is given as the distance between the x-axis and the inside curve y=x^2-2x+2, and is given by the function r(x)=x^2-2x+2. -

    -     - Graph of the region bounded by the two curves lying in the xy plane. - - - - - //ASY file for figwash1_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3}; - real[] myychoice={-5,5}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.2,3.3); - pair ybounds=(-5.6,5.6); - pair zbounds=(-5.6,5.6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); - - triple g1(real t) {return (t,(2*t-1),0);} - path3 p1=graph(g1,1,3,operator ..); - draw(p1,bluepen+.4mm); - - triple g2(real t) {return (t,(t^2-2*t+2),0);} - path3 p2=graph(g2,3,1,operator ..); - draw(p2,redpen+.4mm); - - path3 p3=p1--p2; - draw(surface(p3 -- cycle), simplesurfacepen); - - - - -
    - -
    - - - - - -

    - Graph of the region enclosed by the curves y=x^2-2x+2 and y=2x-1 with a washer centered at x=2. - The washer lies parallel to the y-axis, and has an outside radius of R(2)=3 and an inside radius of r(2)=2. - In the original graph of the functions, R(2)=3 is the distance between the line y=2x-1 and the x-axis, and r(2)=2 is the distance between the curve y=x^2-2x+2 and the x-axis. - The washer will lie entirely in the space enclosed by rotating the region between the two curves. -

    -     - Graph of the region bounded by the two curves with an arbitrary washer used to calculate the volume of the solid. - - - - - //ASY file for figwash1_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3}; - real[] myychoice={-5,5}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.2,3.3); - pair ybounds=(-5.6,5.6); - pair zbounds=(-5.6,5.6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); - - triple g1(real t) {return (t,(2*t-1),0);} - path3 p1=graph(g1,1,3,operator ..); - draw(p1,bluepen+.4mm); - - triple g2(real t) {return (t,(t^2-2*t+2),0);} - path3 p2=graph(g2,3,1,operator ..); - draw(p2,redpen+.4mm); - - path3 p3=p1--p2; - draw(surface(p3 -- cycle), simplesurfacepen); - - triple f2(pair t) {return (2,cos(t.x)*t.y,sin(t.x)*t.y);} - surface s2=surface(f2,(0,2),(2*pi,3),16,1,Spline); - draw(s2,simplesurfacepen2); - - triple g3(real t) {return (2,2*cos(t),2*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,redpen+.4mm); - - triple g3(real t) {return (2,3*cos(t),3*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.4mm); - - draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); - - - - -
    - -
    - - - - - -

    - Three-dimensional graph of the space enclosed by rotating the curves y=x^2-2x+2 and y=2x-1 about the x-axis. - The outside line y=2x-1 forms the outer border of the solid once rotated about the x-axis, whose surface is coloured in blue. - The inside curve y=x^2-2x+2 forms the inner border of the solid once rotated about the x-axis, whose surface is coloured in red. - The washers having outside radius R(x)=2x-1 and inner radius r(x)=x^2-2x+2 will lie entirely in the space enclosed by rotating the region between the two curves. -

    -     - Three-dimensional graph of the solid formed by rotating the region bounded by the two curves about the x axis. - - - - - //ASY file for figwash1_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((10.17,5.3,46),(0,1,0),(0,0,0),1,(-0.0999,0.0072)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3}; - real[] myychoice={-5,5}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.2,3.3); - pair ybounds=(-5.6,5.6); - pair zbounds=(-5.6,5.6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); - - pen p=apexmeshpen+.1mm; - pen q=redmeshpen+.1mm; - - triple g1(real t) {return (t,(2*t-1),0);} - path3 p1=graph(g1,1,3,operator ..); - draw(p1,bluepen+.4mm); - - triple g2(real t) {return (t,(t^2-2*t+2),0);} - path3 p2=graph(g2,3,1,operator ..); - draw(p2,redpen+.4mm); - - path3 p3=p1--p2; - draw(surface(p3 -- cycle), simplesurfacepen); - - triple f2(pair t) {return (t.x,cos(t.y)*(2*t.x-1),sin(t.y)*(2*t.x-1));} - surface s2=surface(f2,(1,0),(3,2*pi),4,16,Spline); - draw(s2,simplesurfacepen,meshpen=p); - - triple f2(pair t) {return (t.x,cos(t.y)*(t.x^2-2*t.x+2),sin(t.y)*(t.x^2-2*t.x+2));} - surface s2=surface(f2,(1,0),(3,2*pi),4,16,Spline); - draw(s2,simplesurfacepen2,meshpen=q); - - triple g3(real t) {return (1,cos(t),sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,black+.4mm); - - triple g3(real t) {return (3,5*cos(t),5*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,black+.4mm); - - - - -
    -     - -
    -     - -     - - - -

    - When rotating about a vertical axis, - the outside and inside radius functions must be functions of y. -

    - - - Finding volume with the Washer Method - -

    - Find the volume of the solid formed by rotating the triangular region with vertices at (1,1), - (2,1) and (2,3) about the y-axis. -

    -     - -

    - The triangular region is sketched in ; - the differential element is sketched in and the full solid is drawn in . - They help us establish the outside and inside radii. - Since the axis of rotation is vertical, - each radius is a function of y. -

    - -

    - The outside radius R(y) is formed by the line connecting (2,1) and (2,3); - it is a constant function, - as regardless of the y-value the distance from the line to the axis of rotation is 2. - Thus R(y)=2. -

    - -
    - Sketching the solid in - -
    - - - - - -

    - Graph of the triangular region with vertices at (1,1), (2,1) and (2,3). - As we are rotating about the y-axis, the outer radius is the outermost edge of the triangle, which is a horizontal line at x=2, giving us a constant R(y)=2. - The inner radius is formed by the line between (1,1) and (2,3), which is the line y=2x-1. - This gives us an inner radius r(y)=\frac12(y+1). - The triangular region which we will rotate about the y-axis then consists of the area to the right of the line y=2x-1 and to the left of the horizontal line x=2 for 1 \leq y \leq 3. -

    -     - Graph of the triangular region showing the inner and outer radius functions. - - - - - //ASY file for figwasher_idea_a_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0.029,-0.015)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,-2,1,2}; - real[] myychoice={1,2,3}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-2.2,2.2); - pair ybounds=(-.2,3.2); - pair zbounds=(-2.2,2.2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (axis cs:1,0,1) -- (axis cs:2,0,1)--(axis cs:2,0,3) -- cycle; - - //path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); - //draw(surface(p .. cycle), simplesurfacepen2); - //draw(p..cycle,redpen+.4mm); - - path3 p1=(1,1,0)--(2,1,0)--(2,3,0); - draw(surface(p1 -- cycle), simplesurfacepen); - draw(p1--cycle,bluepen+.4mm); - - //\draw (axis cs: 0,0,1.2) -- (axis cs:2,0,1.2); - //\draw (axis cs: .8,0,.5) node { $R(y)=2$}; - //\filldraw (axis cs:0,0,1.2) circle (1pt) - // (axis cs:2,0,1.2) circle (1pt) - // (axis cs:0,0,2.6) circle (1pt) - // (axis cs:1.8,0,2.6) circle (1pt); - //\draw (axis cs: 0,0,2.6) -- (axis cs:1.8,0,2.6); - //\draw (axis cs: 1.5,0,3.2) node { $r(y)=\frac12(y+1)$}; - - triple dot1=(0,1.2,0); - triple dot2=(2,1.2,0); - triple dot3=(0,2.6,0); - triple dot4=(1.8,2.6,0); - - draw(dot1--dot2,black+.2mm); - draw(dot3--dot4,black+.2mm); - - dotfactor=2; - dot(dot1); - dot(dot2); - dot(dot3); - dot(dot4); - - label("$R(y)$",dot2+.7*(dot1-dot2),S); - label("$r(y)$",dot4+.7*(dot3-dot4),N); - - - -
    - - - -
    - - - -

    - Graph of the triangular region with vertices at (1,1), (2,1) and (2,3) with a washer centered at y=2. - The washer lies parallel to the x-axis, and has an outside radius of R(2)=2 and an inside radius of r(2)=\frac32. - In the original graph of the functions, R(2)=2 is the distance between the rightmost edge of the triangular region and the y-axis. - Similarly, r(2)=\frac32 is the distance betweeen the leftmost edge of the triangular region given by the line y=2x-1, and the y-axis - The washer will lie entirely in the space enclosed by rotating the region between the two curves as described in the following image. -

    -     - Graph of the triangular region with an arbitrary washer which will be used to calculate the volume of the solid. - - - - - //ASY file for figwasher_idea_a_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0.029,-0.015)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,-2,1,2}; - real[] myychoice={1,2,3}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-2.2,2.2); - pair ybounds=(-.2,3.2); - pair zbounds=(-2.2,2.2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (axis cs:1,0,1) -- (axis cs:2,0,1)--(axis cs:2,0,3) -- cycle; - - //path3 p=(3/2,2/3,0)..(0,2/3,3/2)..(-3/2,2/3,0)..(0,2/3,-3/2); - //draw(surface(p .. cycle), simplesurfacepen2); - //draw(p..cycle,redpen+.4mm); - - path3 p1=(1,1,0)--(2,1,0)--(2,3,0); - draw(surface(p1 -- cycle), simplesurfacepen); - draw(p1--cycle,bluepen+.4mm); - - //\draw (axis cs: 0,0,1.2) -- (axis cs:2,0,1.2); - //\draw (axis cs: .8,0,.5) node { $R(y)=2$}; - //\filldraw (axis cs:0,0,1.2) circle (1pt) - // (axis cs:2,0,1.2) circle (1pt) - // (axis cs:0,0,2.6) circle (1pt) - // (axis cs:1.8,0,2.6) circle (1pt); - //\draw (axis cs: 0,0,2.6) -- (axis cs:1.8,0,2.6); - //\draw (axis cs: 1.5,0,3.2) node { $r(y)=\frac12(y+1)$}; - - triple g1(real t) {return (1.5*cos(t),2,1.5*sin(t));} - path3 p1=graph(g1,0,2*pi,operator ..); - draw(p1,redpen+.4mm); - - triple g1(real t) {return (2*cos(t),2,2*sin(t));} - path3 p1=graph(g1,0,2*pi,operator ..); - draw(p1,redpen+.4mm); - - triple f2(pair t) {return (cos(t.x)*t.y,2,sin(t.x)*t.y);} - surface s2=surface(f2,(0,1.5),(2*pi,2),16,1,Spline); - draw(s2,simplesurfacepen2); - - draw((1.5,2,0)--(2,2,0),purple+linetype(new real[] {4,4})+.3mm); - - - -
    - - - -
    - - - -

    - Three-dimensional graph of the space enclosed by rotating the triangular region about the y-axis. - The outside line x=2 forms the outer border of the solid once rotated about the y-axis, whose surface is coloured in blue. - The inside curve y=2x-1, or written in terms of y as x=\frac12(y+1) forms the inner border of the solid once rotated about the y-axis, whose surface is coloured in red. - The washers having outside radius R(y)=2 and inner radius r(y)=\frac12(y+1) will lie entirely in the space enclosed by rotating the region between the two curves and will allow us to calculate the volume of this solid. -

    -     - Graph of the triangular region with an arbitrary washer which will be used to calculate the volume of the solid. - - - - - //ASY file for figwasher_idea_a_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0.029,-0.015)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,-2,1,2}; - real[] myychoice={1,2,3}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-2.2,2.2); - pair ybounds=(-.2,3.2); - pair zbounds=(-2.2,2.2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (axis cs:1,0,1) -- (axis cs:2,0,1)--(axis cs:2,0,3) -- cycle; - - pen p=apexmeshpen+.1mm; - pen q=redmeshpen+.1mm; - - path3 p1=(1,1,0)--(2,1,0)--(2,3,0); - draw(surface(p1 -- cycle), simplesurfacepen); - draw(p1--cycle,bluepen+.4mm); - - //\draw (axis cs: 0,0,1.2) -- (axis cs:2,0,1.2); - //\draw (axis cs: .8,0,.5) node { $R(y)=2$}; - //\filldraw (axis cs:0,0,1.2) circle (1pt) - // (axis cs:2,0,1.2) circle (1pt) - // (axis cs:0,0,2.6) circle (1pt) - // (axis cs:1.8,0,2.6) circle (1pt); - //\draw (axis cs: 0,0,2.6) -- (axis cs:1.8,0,2.6); - //\draw (axis cs: 1.5,0,3.2) node { $r(y)=\frac12(y+1)$}; - - triple g1(real t) {return (2*cos(t),3,2*sin(t));} - path3 p1=graph(g1,0,2*pi,operator ..); - draw(p1,bluepen+.4mm); - - triple g1(real t) {return (2*cos(t),1,2*sin(t));} - path3 p1=graph(g1,0,2*pi,operator ..); - draw(p1,bluepen+.4mm); - - triple g1(real t) {return (1*cos(t),1,1*sin(t));} - path3 p1=graph(g1,0,2*pi,operator ..); - draw(p1,redpen+.4mm); - - triple f2(pair t) {return (cos(t.x)*t.y,1,sin(t.x)*t.y);} - surface s2=surface(f2,(0,1),(2*pi,1),16,1,Spline); - draw(s2,simplesurfacepen); - - triple f2(pair t) {return (cos(t.x)*(2),t.y,sin(t.x)*2);} - surface s2=surface(f2,(0,1),(2*pi,3),16,2,Spline); - draw(s2,simplesurfacepen,meshpen=p); - - triple f2(pair t) {return (cos(t.x)*(t.y/2+1/2),t.y,sin(t.x)*(t.y/2+1/2));} - surface s2=surface(f2,(0,1),(2*pi,3),16,2,Spline); - draw(s2,simplesurfacepen2,meshpen=q); - - - -
    - -     -
    - -

    - The inside radius is formed by the line connecting (1,1) and (2,3). - The equation of this line is y=2x-1, - but we need to refer to it as a function of y. - Solving for x gives r(y) = \frac12(y+1). -

    - -

    - We integrate over the y-bounds of y=1 to y=3. - Thus the volume is - - V \amp = \pi\int_1^3\Big(2^2 - \big(\frac12(y+1)\big)^2\Big)\, dy - \amp = \pi\int_1^3\Big(-\frac14y^2-\frac12y+\frac{15}4\Big)\, dy - \amp = \pi\Big[-\frac1{12}y^3-\frac14y^2+\frac{15}4y\Big]\Big|_1^3 - \amp = \frac{10}3\pi \approx 10.47\,\text{units}^3 - . -

    -     - -     - -

    - This section introduced a new application of the definite integral. - Our default view of the definite integral is that it gives - the area under the curve. However, - we can establish definite integrals that represent other quantities; - in this section, we computed volume. -

    - -

    - The ultimate goal of this section is not to compute volumes of solids. - That can be useful, - but what is more useful is the understanding of this basic principle of integral calculus, - outlined in : - to find the exact value of some quantity, -

    - -

    -

      -
    • -

      - we start with an approximation (in this section, - slice the solid and approximate the volume of each slice), -

      -
      • - -
    • -

      - then make the approximation better by refining our original approximation (, use more slices), -

      -
      • - -
    • -

      - then use limits to establish a definite integral which gives the exact value. -

      -
      • -
    -

    - -

    - We practice this principle in the next section where we find volumes by slicing solids in a different way. -

    - - - - Terms and Concepts - - - - -

    - A solid of revolution is formed by revolving a shape around an axis. -

    -     - -     - - - - -

    - In your own words, - explain how the Disk and Washer Methods are related. -

    -     - - - -     - - - - -

    - Explain the how the units of volume are found in the integral of : - if A(x) has units of \text{in}^2, - how does \int A(x)\,dx have units of \text{in}^3? -

    -     - - - -

    - Recall that dx does not just sit there; - it is multiplied by A(x) and represents the thickness of a small slice of the solid. - Therefore dx has units of in, - giving A(x)\,dx the units of in^3. -

    -     - -     -     - - Problems - - - -

    - Use the Disk/Washer Method to find the volume of the solid of revolution - formed by revolving the given region about the x-axis. -

    -     - - - - -

    - The region between y=3-x^2 and the x axis: -

    - - - -

    - Graph of the region bounded by the curve y=3-x^2 and the x-axis. - The curve y=3-x^2 begins at x-axis at the point (-\sqrt{3},0). - From this point the curve rises until reaching the y-axis at the point (0,3). - After reaching the y-axis, the curve slopes down until reaching the x-axis at the point (\sqrt{3},0). - The region then contains the entire area below this curve, and above the x-axis. -

    -     - Graph of the region bounded by the curve and the x axis. - - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=3.5, - xmin=-2.1,xmax=2.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=-1.73:1.73] {3-x^2}; - \addplot [firstcurvestyle,domain=-1.73:1.73,samples=50] {3-x^2} node [shift={(-15pt,90pt)} ,black] { $y=3-x^2$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - 48\pi\sqrt{3}/5 units^3 -

    -     - -     - - - - -

    - The region between y=5x and the x axis, - for 1\leq x\leq 2: -

    - - - - -

    - Graph of the region between y=5x and the x-axis, for 1\leq x\leq 2. - The line y=5x begins at the point (1,5) from which it linearly increases until reaching the rightmost bound which is at the point (2,10). - The region then contains the entire area below the line y=5x and above the x-axis for 1\leq x\leq 2. -

    -     - Graph of the region bounded by the the line y=5x and the x-axis for x between 1 and 2. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=11, - xmin=-.1,xmax=2.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:2] {5*x} \closedcycle; - \addplot [firstcurvestyle,domain=0:2.1] {5*x} node [shift={(-30pt,-5pt)},black] { $y=5x$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - 175\pi/3 units^3 -

    -     - -     - - - - -

    - The region between y=\cos(x) and the x axis, - for 0\leq x\leq \pi/2: -

    - - - - -

    - Graph of the region between y=\cos(x) and the x-axis, for 0\leq x\leq \pi/2. - The curve y=\cos(x) begins at the point (0,1) from which it slopes downwards until ending after reaching the x-axis at the point (\pi/2,0). - The region then contains the entire area below the curve y=\cos(x) and above the x-axis for 0\leq x\leq \pi/2. -

    -     - Graph of the region bounded by the the cosine function and the two coordinate axes. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.7 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1.57] {cos(deg(x))} \closedcycle; - \addplot [firstcurvestyle,domain=0:1.57] {cos(deg(x))} node [shift={(-20pt,65pt)},black] { $y=\cos(x) $}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \pi^2/4 units^3 -

    -     - -     - - - - -

    - The region between the curves y=x and y=\sqrt{x}: -

    - - - - -

    - Graph of the region between the curves y=x and y=\sqrt{x}. - The curves y=x and y=\sqrt{x} both begin at the origin. - From this point the curve y=\sqrt{x} rises above the line y=x until reaching the point (1,1), where the curve once again intersects the line. - The region then contains the area below the curve y=\sqrt{x} and above the line y=x. -

    -     - Graph of the region bounded by the two curves. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1,samples=60] {sqrt(x)}; - - \addplot [firstcurvestyle,domain=0:.05] {sqrt(x)}; - \addplot [firstcurvestyle,domain=.05:1,samples=30] {sqrt(x)} node [shift={(-55pt,-5pt)} ,black] { $y=\sqrt{x}$}; - - \addplot [firstcurvestyle,domain=0:1] {x} node [shift={(-30pt,-45pt)},black] { $y=x$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \pi/6 units^3 -

    -     - -     - -     - - - -

    - Use the Disk/Washer Method to find the volume of the solid of revolution - formed by revolving the given region about the y-axis. -

    -     - - - - -

    - The region bounded by the curve y=3-x^2, - the x axis, and the y axis: -

    - - - - -

    - Graph of the region bounded by the curve y=3-x^2, the x-axis, and the y-axis. - Note that this region can reference both the regions to the left or right of the y-axis, but we will consider the region to the right of the y-axis. - The curve y=3-x^2 begins at the point (0,3) from which it slopes down until reaching the x-axis. - The region then contains the area to the right of the y-axis, and to the left of the curve y=3-x^2 for y values between 0 and 3. -

    -     - Graph of the region lying in the first quadrant bounded by the curve and the two coordinate axes. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=3.5, - xmin=-2.1,xmax=2.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1.73] {3-x^2} \closedcycle; - \addplot [firstcurvestyle,domain=-1.73:1.73,samples=50] {3-x^2} node [shift={(-15pt,90pt)},black] { $y=3-x^2$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - 9\pi/2 units^3 -

    -     - -     - - - - -

    - The region between y=5x and the y axis, - for 5\leq y\leq 10: -

    - - - - -

    - Graph of the region between y=5x and the y axis, for 5\leq y\leq 10. - The line y=5x begins at the point (1,5), from which it linearly increases until ending upper bound for y at the point (2,10). - The region then contains the entire area to the right of x=0 and to the left of the line y=5x for y between 5 and 10. -

    -     - Graph of the region bounded by the line and the y axis for y values between 5 and 10. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=11, - xmin=-.1,xmax=2.1 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(1,5) (2,10) (0,10) (0,5) (1,5)}; - \addplot [firstcurvestyle,domain=0:2.1] {5*x} node [shift={(-20pt,-40pt)},black] { $y=5x$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - 35\pi/3 units^3 -

    -     - -     - - - - -

    - The region between y=\cos(x) and the x axis, - for 0\leq x\leq \pi/2: -

    - -

    - (Hint: Integration By Parts will be necessary, twice. - First let u = \arccos^2x, - then let u=\arccos x.) -

    - - - -

    - Graph of the region between y=\cos(x) and the x axis, for 0\leq x\leq \pi/2. - The curve y=\cos(x) begins at the point (0,1) from which it slopes downwards until ending after reaching the x-axis at the point (\pi/2,0). - The region then contains the entire area to the right of x=0 and to the left of of the curve y=\cos(x) for y values between 0 and 1. -

    -     - Graph of the region bounded by the cosine curve and the coordinate axes. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.7 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1.57] {cos(deg(x))} \closedcycle; - \addplot [firstcurvestyle,domain=0:1.57] {cos(deg(x))} node [shift={(-20pt,65pt)},black] { $y=\cos(x) $}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \pi^2-2\pi units^3 -

    -     - -     - - - - -

    - The region between the curves y=x and y=\sqrt{x}: -

    - - - - -

    - Graph of the region between the curves y=x and y=\sqrt{x}. - The curves y=x and y=\sqrt{x} both begin at the origin. - From this point the curve y=\sqrt{x} rises above the line y=x until reaching the point (1,1), where the curve once again intersects the line. - The region consists of the area to the right of the curve y=\sqrt{x} and the to the left line y=x for y values between 0 and 1. -

    -     - Graph of the region bounded by the two curves. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1,samples=60] {sqrt(x)}; - - \addplot [firstcurvestyle,domain=0:.05] {sqrt(x)}; - \addplot [firstcurvestyle,domain=.05:1,samples=30] {sqrt(x)} node [shift={(-60pt,-10pt)} ,black] { $y=\sqrt{x}$}; - - \addplot [firstcurvestyle,domain=0:1] {x} node [shift={(-30pt,-45pt)},black] { $y=x$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - 2\pi/15 units^3 -

    -     - -     - -     - - - -

    - Use the Disk/Washer Method to find the volume of the solid of revolution - formed by rotating the given region about each of the given axes. -

    -     - - - - -

    - Region bounded by: y=\sqrt{x}, - y=0 and x=1. -

    -     - - - -

    - Rotate about the x axis. -

    -     - -

    - \pi/2 -

    -     -     - - - -

    - Rotate about y=1. -

    -     - -

    - 5\pi/6 -

    -     -     - - - -

    - Rotate about the y axis. -

    -     - -

    - 4\pi/5 -

    -     -     - - - -

    - Rotate about x=1. -

    -     - -

    - 8\pi/15 -

    -     -     - -     - - - - -

    - Region bounded by: y=4-x^2 and y=0. -

    -     - - - -

    - Rotate about the x axis. -

    -     - -

    - 512\pi/15 -

    -     -     - - - -

    - Rotate about y=4. -

    -     - -

    - 256\pi/5 -

    -     -     - - - -

    - Rotate about y=-1. -

    -     - -

    - 832\pi/15 -

    -     -     - - - -

    - Rotate about x=2. -

    -     - -

    - 128\pi/3 -

    -     -     - -     - - - - -

    - The triangle with vertices (1,1), - (1,2) and (2,1). -

    -     - - - -

    - Roate about the x axis. -

    -     - -

    - 4\pi/3 -

    -     -     - - - -

    - Roate about y=2. -

    -     - -

    - 2\pi/3 -

    -     -     - - - -

    - Rotate about the y axis. -

    -     - -

    - 4\pi/3 -

    -     -     - - - -

    - Rotate about x=1. -

    -     - -

    - \pi/3 -

    -     -     - -     - - - - -

    - Region bounded by y=x^2-2x+2 and y=2x-1. -

    -     - - - -

    - Rotate about the x axis. -

    -     - -

    - 104\pi/15 -

    -     -     - - - -

    - Rotate about y=1. -

    -     - -

    - 64\pi/15 -

    -     -     - - - -

    - Rotate about y=5. -

    -     - -

    - 32\pi/5 -

    -     -     - -     - - - - -

    - Region bounded by y=1/\sqrt{x^2+1}, x=-1, - x=1 and the x-axis. -

    -     - - - -

    - Rotate about the x axis. -

    -     - -

    - \pi^2/2 -

    -     -     - - - -

    - Rotate about y=1. -

    -     - -

    - \pi^2/2-4\pi\sinh^{-1}(1) -

    -     -     - - - -

    - Rotate about y=-1. -

    -     - -

    - \pi^2/2+4\pi\sinh^{-1}(1) -

    -     -     - -     - - - - -

    - Region bounded by y=2x, - y=x and x=2. -

    -     - - - -

    - Rotate about the x axis. -

    -     - -

    - 8\pi -

    -     -     - - - -

    - Rotate about y=4. -

    -     - -

    - 8\pi -

    -     -     - - - -

    - Rotate about the y axis. -

    -     - -

    - 16\pi/3 -

    -     -     - - - -

    - Rotate about x=2. -

    -     - -

    - 8\pi/3 -

    -     -     - -     - -     - - - -

    - Orient the given solid along the x-axis such that a cross-sectional area function A(x) can be obtained, - then apply - to find the volume of the solid. -

    -     - - - - -

    - A right circular cone with height of 10 and base radius of 5. -

    - - - - -

    - Image of a right circular cone with height of 10 and base radius of 5. - The base of the cone is a circle of radius 5. - From the center of the circle to the peak of the cone we measure the height to be 10. -

    -     - An image of a right circular cone with height of 10 and base radius of 5. - - - \begin{tikzpicture}[scale=.8] - - \begin{scope}[xscale=2] - - \draw [thick] (-1,0) arc (180:360:1); - \draw [thick,dashed] (1,0) arc (0:180:1); - - \end{scope} - - \draw [fill=black] (0,0) circle (1pt) -- node [pos=.5,above] {5} (2,0); - \draw (0,0) -- node [pos=.5,rotate=90,above] {10} (0,4); - \draw [thick] (-2,0) -- (0,4)-- (2,0); - - \end{tikzpicture} - - - - - -     - -

    - 250\pi/3 -

    -     - -

    - Placing the tip of the cone at the origin such that the x-axis runs through the center of the circular base, - we have A(x)=\pi x^2/4. - Thus the volume is 250\pi/3 units^3. -

    -     - -     - - - - -

    - A skew right circular cone with height of 10 and base radius of 5. (Hint: - all cross-sections are circles.) -

    - - - - -

    - Image of a skew right circular cone with height of 10 and base radius of 5. - The base of the cone is a circle of radius 5. - From the rightmost edge of the circle to the peak of the cone we measure the height to be 10. -

    -     - An image of a skew right circular cone with height of 10 and base radius of 5. - - - \begin{tikzpicture}[scale=.8] - - \begin{scope}[xscale=2] - - \draw [thick] (-1,0) arc (180:360:1); - \draw [thick,dashed] (1,0) arc (0:180:1); - \draw [dashed] (.4,1.7) circle (.6); - - \end{scope} - - \draw [fill=black] (0,0) circle (1pt) -- node [pos=.5,above] { 5} (2,0); - \draw [thick] (-2,0.2) -- (2,4)-- node [pos=.5,rotate=-90,above] { 10}(2,0); - - \end{tikzpicture} - - - - - -     - -

    - 250\pi/3 -

    -     - -

    - The cross-sections of this cone are the same as the cone in . - Thus they have the same volume of 250\pi/3 units^3. -

    -     - -     - - - - -

    - A right triangular cone with height of 10 and whose base is a right, - isosceles triangle with side length 4. -

    - - - -

    - Image of a right triangular cone with height of 10 and whose base is a right isosceles triangle with side length 4. - The base of the cone is a right isosceles triangle having two side lengths of 4. - The two sides of length 4 are connected by a right angle. - Additionally, the distance from the right angle occuring at the connection of the two sides of length 4 from the base of the triangle to the peak of the cone is measured to be 10. -

    -     - Image of a right triangular cone whose base is a right isosceles triangle. - - - \begin{tikzpicture}[scale=1.58] - - \draw [thick](-1,0) -- (1,0) -- (0,3)--cycle; - \draw [thick,dashed] (-1,0) -- node [pos=.5,above] {4} (0,.5) -- node [pos=.5,above] {4} (1,0) - (0,.5) -- node [pos=.3,above,rotate=90] {10} (0,3); - \draw (-.2,.4) -- (0,.3) -- (.2,.4); - - \end{tikzpicture} - - - -     - -

    - 80/3 -

    -     - -

    - Orient the cone such that the tip is at the origin and the x-axis is perpendicular to the base. - The cross-sections of this cone are right, - isosceles triangles with side length 2x/5; - thus the cross-sectional areas are A(x) = 2x^2/25, - giving a volume of 80/3 units^3. -

    -     - -     - - - - -

    - A solid with length 10 with a rectangular base and triangular top, - wherein one end is a square with side length 5 and the other end is a triangle with base and height of 5. -

    - - - -

    - Image of a solid with length 10 with a rectangular base and triangular top, wherein one end is a square with side length 5 and the other end is a triangle with base and height of 5.. - The rectangular base measures a length of 10 and a depth of 5. - The square of side length 5 connects to the rectangle on the left side of the solid. - The triangle with a base length and height of 5 connects to the rectangle on the right side of the solid. - The square and triangle are then connected by two slanted trapezoidal regions on the front and back of the solid. - On top of the solid is a triangle whose base of length 5 is connected to the square, having a height of 10. -

    -     - Image solid with a rectangular base, and a square and triangular sides which are connected by two trapezoids and a triangle. - - - \begin{tikzpicture}[x={(1,0)},z={(0,1)},y={(.5,.87)},scale=.35] - - \draw [thick] (0,0,0) -- node[pos=.5,below] {10} (10,0,0) -- (10,2.5,5) -- (10,5,0) -- node [pos=.5,below right] { 5} (10,0,0) - (0,0,0) -- node [pos=.5,left] {5} (0,0,5) -- (10,2.5,5) -- (0,5,5) -- node [pos=.5,left] { 5} (0,0,5); - - \draw [thick,dashed] (0,0,0) -- (0,5,0) -- (0,5,5) - (0,5,0) -- (10,5,0); - - \end{tikzpicture} - - - -     - -

    - 187.5 -

    -     - -

    - Orient the solid so that the x-axis is parallel to long side of the base. - All cross-sections are trapezoids - (at the far left, the trapezoid is a square; - at the far right, - the trapezoid has a top length of 0, making it a triangle). - The area of the trapezoid at x is A(x) = 1/2(-1/2x+5+5)(5) = -5/4x+25. - The volume is 187.5 units^3. -

    -     - -     - -     -     -     -
    -
    - The Shell Method -

    - Often a given problem can be solved in more than one way. - A particular method may be chosen out of convenience, - personal preference, or perhaps necessity. - Ultimately, it is good to have options. -

    - -

    - The previous section introduced the Disk and Washer Methods, - which computed the volume of solids of revolution by integrating the cross-sectional area of the solid. - This section develops another method of computing volume, - the Shell Method. Instead of slicing the solid perpendicular to the axis of rotation creating cross-sections, - we now slice it parallel to the axis of rotation, creating shells. -

    - - - -

    - Consider , - where the region shown in - is rotated around the y-axis forming the solid shown in - . - A small slice of the region is drawn in , - parallel to the axis of rotation. - When the region is rotated, - this thin slice forms a cylindrical shell, - as pictured in . - The previous section approximated a solid with lots of thin disks (or washers); - we now approximate a solid with many thin cylindrical shells. -

    - -
    - Introducing the Shell Method - -
    - - - - - -

    - Graph of the function y=\frac{1}{1+x^2} . - The curve begins at the point (0,1). - The curve then decreases until it meets the horizontal line x=1 where it stops at the point (1,\frac12) . - The entire area enclosed by y=\frac{1}{1+x^2} , x=0 , x=1 and y=0 is shaded. - In other words, the region consists of the area lying below the curve and above the x-axis - A vertical red rectangular slice with a base having an approximate length of 0.5 coming up from approximately the point (0.5,0) reaches up to meet the curve. - This rectangle when rotated about the y-axis will form a cylindrical shell in the following figures. -

    -     - A graph of a function drawn in the first quadrant in the xy plane. - - - - - //ASY file for figwasher_shell_intro_b_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0,0)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-1.2,1.2); - pair ybounds=(-.2,1.2); - pair zbounds=(-1.2,1.2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple g1(real t) {return (t,1/(1+t^2),0);} - path3 p1=graph(g1,0,1,operator ..); - draw(p1--(1,0,0),bluepen+.4mm); - draw(surface(p1 -- (1,0,0)--(0,0,0)--cycle), simplesurfacepen); - - // \filldraw [bottom color=colormaptwobottom,top color=colormaptwotop] - // (axis cs:.6,0,0) -- (axis cs:.7,0,0) -- (axis cs:.7,0,.7) -- (axis cs:.6,0,.7) -- cycle; - - path3 p2=(.6,0,0)--(.7,0,0)--(.7,.7,0)--(.6,.7,0)--cycle; - draw(surface(p2),simplesurfacepen2); - draw(p2,redpen+.4mm); - - label("$\displaystyle y=\frac{1}{1+x^2}$",(.75,1,0)); - - - - -
    - -
    - - - - - -

    - A three dimensional diagram of the area enclosed by the curve y=\frac{1}{1+x^2} , and the lines x=0 , x=1 and y=0 rotated about the y-axis. - The base of this shape is a unit circle lying on the xz plane. - This unit circle comes up from y=0 to y=0.5 creating a solid cylinder. - The region above y=0.5 forms a shape which looks like a perfectly symmetrical mountain whose peak occurs at the point (0,1,0). - The space below this region above the y-axis is contained in the shape which comes from rotating the curve. -

    -     - A 3d graph of the function in the first quadrant rotated fully about the y axis. - - - - - //ASY file for figshell_intro_a_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0,0)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-1.2,1.2); - pair ybounds=(-.2,1.2); - pair zbounds=(-1.2,1.2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple g1(real t) {return (t,1/(1+t^2),0);} - path3 p1=graph(g1,0,1,operator ..); - draw(p1--(1,0,0),bluepen+.4mm); - //draw(surface(p1 -- (1,0,0)--(0,0,0)--cycle), simplesurfacepen); - - // \filldraw [bottom color=colormaptwobottom,top color=colormaptwotop] - // (axis cs:.6,0,0) -- (axis cs:.7,0,0) -- (axis cs:.7,0,.7) -- (axis cs:.6,0,.7) -- cycle; - - pen p=apexmeshpen; - - triple f(pair t) { - return (cos(t.x),t.y,sin(t.x));} - surface s=surface(f,(0,0),(2*pi,.5),16,2,Spline); - draw(s,simplesurfacepen,meshpen=p); - - triple f(pair t) { - return (cos(t.x)*t.y,1/(1+t.y^2),t.y*sin(t.x));} - surface s=surface(f,(0,0),(2*pi,1),16,16,Spline); - draw(s,simplesurfacepen,meshpen=p); - - - - -
    - -
    - - - - - -

    - The red rectangle from figure (a) is rotated 360 degrees about the y-axis, creating a cylindrical shell. - Two circles, one of radius 0.5 and one of radius 0.6 are drawn on the xz-plane. - The area between the circles is shaded, creating a circular disc in the xz-plane. - The disc then goes up from y=0 until it meets the curve y=\frac{1}{1+x^2} , which creates the cylindrical shell. - Adding up many of these cylindrical shells will then allow us to approximate the volume of the solid described in the previous images. -

    -     - A three dimensional graph of the cylindrical shell which comes from rotating the rectangle in the first image about the y axis. - - - - - //ASY file for figshell_intro_d_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((13.6,14.55,43),(0,1,0),(0,0,0),1,(0,0)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-1.2,1.2); - pair ybounds=(-.2,1.2); - pair zbounds=(-1.2,1.2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple g1(real t) {return (t,1/(1+t^2),0);} - path3 p1=graph(g1,0,1,operator ..); - draw(p1--(1,0,0),bluepen+.4mm); - //draw(surface(p1 -- (1,0,0)--(0,0,0)--cycle), simplesurfacepen); - - // \filldraw [bottom color=colormaptwobottom,top color=colormaptwotop] - // (axis cs:.6,0,0) -- (axis cs:.7,0,0) -- (axis cs:.7,0,.7) -- (axis cs:.6,0,.7) -- cycle; - - pen p=apexmeshpen; - pen q=redmeshpen+.1mm; - - triple f(pair t) { - return (cos(t.x),t.y,sin(t.x));} - surface s=surface(f,(0,0),(2*pi,.5),8,2,Spline); - //draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (cos(t.x)*t.y,1/(1+t.y^2),t.y*sin(t.x));} - surface s=surface(f,(0,0),(2*pi,1),8,16,Spline); - //draw(s,surfacepen,meshpen=p); - - path3 p2=(.6,0,0)--(.7,0,0)--(.7,.7,0)--(.6,.7,0)--cycle; - //draw(surface(p2),simplesurfacepen2); - draw(p2,redpen+.4mm); - - triple f(pair t) { - return (.6*cos(t.x),t.y,.6*sin(t.x));} - surface s=surface(f,(0,0),(2*pi,.7),8,2,Spline); - draw(s,simplesurfacepen2,meshpen=q); - - triple f(pair t) { - return (.7*cos(t.x),t.y,.7*sin(t.x));} - surface s=surface(f,(0,0),(2*pi,.7),8,2,Spline); - draw(s,simplesurfacepen2,meshpen=q); - - triple g1(real t) {return (.6*cos(t),.7,.6*sin(t));} - path3 p1=graph(g1,0,2*pi,operator ..); - draw(p1,redpen+.3mm); - - triple g1(real t) {return (.7*cos(t),.7,.7*sin(t));} - path3 p1=graph(g1,0,2*pi,operator ..); - draw(p1,redpen+.3mm); - - triple g1(real t) {return (.6*cos(t),0,.6*sin(t));} - path3 p1=graph(g1,0,2*pi,operator ..); - draw(p1,redpen+.3mm); - - triple g1(real t) {return (.7*cos(t),0,.7*sin(t));} - path3 p1=graph(g1,0,2*pi,operator ..); - draw(p1,redpen+.3mm); - - triple g3(real t) {return (.7*cos(t),1/(1+.7^2),.7*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,purple+.3mm); - - triple g4(real t) {return (.655*cos(t),.7,.655*sin(t));} - path3 p4=graph(g4,0,2*pi,operator ..); - //draw(p4,purple+.3mm); - - triple f(pair t) { - return (t.y*cos(t.x),.7,t.y*sin(t.x));} - surface s=surface(f,(0,.6),(2*pi,.7),8,1,Spline); - draw(s,simplesurfacepen2,meshpen=q); - - - - -
    -     - -
    - -

    - To compute the volume of one shell, - first consider the paper label on a soup can with radius r and height h. - What is the area of this label? - A simple way of determining this is to cut the label and lay it out flat, - forming a rectangle with height h and length 2\pi r. - Thus the area is A = 2\pi rh; - see . -

    - -

    - Do a similar process with a cylindrical shell, with height h, - thickness \Delta x, and approximate radius r. - Cutting the shell and laying it flat forms a rectangular solid with length 2\pi r, - height h and depth \dx. - Thus the volume is V \approx 2\pi rh\dx; - see . - (We say approximately since our radius was an approximation.) -

    - -

    - By breaking the solid into n cylindrical shells, - we can approximate the volume of the solid as - - V \approx \sum_{i=1}^n 2\pi r_ih_i\dx_i - , - where r_i, h_i and \dx_i are the radius, - height and thickness of the ith shell, - respectively. -

    - -

    - This is a Riemann Sum. - Taking a limit as the thickness of the shells approaches 0 leads to a definite integral. -

    - -
    - Determining the volume of a thin cylindrical shell - - -
    - - - -

    - The left side of the image contains a cylinder, and the right side of the image shows a rectangle which comes from unraveling the side of the cylinder. - The cylinder has a height h , radius r and is pictured lying on its circular base. - Both the circular base, and the top of the cylinder are shaded, while the curved surface is unshaded, but contains a cut here instruction. - Once the curved surface of the cylinder is cut, it is unraveled into a rectangular region as seen on the right side of the image. - This rectangular region has a height h and a base 2\pi r . - The area of the rectangle is A=2\pi r h . -

    -     - A visualization of how we can unravel the outside shell of a cylinder into a rectangle. - - - \begin{tikzpicture}[scale=.75] - - \begin{scope}[xscale=1.5] - - \draw [left color=black!60,right color=black!20,thick] (0,0) circle (1cm); - \draw [left color=black!60,right color=black!20,thick] (-1,0) -- (-1,-3.5) arc (180:360:1) -- (1,0) arc (360:180:1); - \draw [fill=white](-1,-.2) -- node [pos=.5,left] { $h$} (-1,-3.3) arc (180:360:1) -- (1,-.2) arc (360:180:1); - \draw [dashed] (.71,-.91) -- node [pos=.5, above,rotate=90] { cut here} (.71,-4.01); - \draw (0,0) -- node [pos=.5,above] { $r$} (1,0); - - \end{scope} - - \draw [->,>=latex,ultra thick] (2,-2)--(3,-2); - - \begin{scope}[shift={(3.5,-.3)}] - - \draw (0,-.2) -- node [pos=.5,above] { $2\pi r$} (5,-.2) -- node [pos=.5,right] { $h$} (5,-3.3)--(0,-3.3) -- cycle; - \draw (2.5,-1.75) node { $A = 2\pi r h$}; - - \end{scope} - - \end{tikzpicture} - - - -
    - - - -
    - - - -

    - The left side of the image contains a cylindrical shell, which is just a cylinder with the middle part cut out. - The right side of the image shows a rectangular box which comes from unraveling the cylindrical shell. - The cylindrical shell has a height h , radius r which spans from the cut out center to the edge of the shell, and the shell has thickness \dx and is pictured lying on its circular base. - The entire cylindrical shell is shaded, and contains a cut here instruction along the curved surface as in the previous image. - Once the curved surface of the cylinderical shell is cut, the entire shell is unraveled into a rectangular box as seen on the right side of the image. - This rectangular box has a height h and a base length of 2\pi r and a thickness of \dx . - The volume of the unraveled cylindrical shell is approximated the volume of the box given by V=2\pi r h \dx . -

    -     - Unraveling the shell of a cylinder, but this time into a rectangular box with an arbitrary thickness. - - - \begin{tikzpicture}[scale=.75] - - \begin{scope}[xscale=1.5] - - \draw [left color=black!60,right color=black!20,thick] (0,0) circle (1cm); - \draw [left color=black!60,right color=black!20,thick] (-1,0) -- (-1,-3.5) arc (180:360:1) -- (1,0) arc (360:180:1); - \draw [left color=black!20,right color=black!60,thick] (0,0) circle (.8cm); - \draw (-1,-1.75) node [left] { $h$}; - \draw [dashed] (.565,-.565) -- (.71,-.71) -- node [pos=.5, above,rotate=90] { cut here} (.71,-4.2); - \draw (0,0) -- node [pos=.5,above] { $r$} (.9,0); - \draw [->,>=stealth](-1.2,.3) node [above] { $\dx$}-- (-.87,0); - - \end{scope} - - \draw [->,>=latex,ultra thick] (2,-2)--(3,-2); - - \begin{scope}[shift={(3.5,-.3)}] - - \draw [left color=black!60,right color=black!20,thick] (0,0) -- (.2,.2) -- node [pos=.5,above] { $2\pi r$} (5.2,.2) -- (5,0); - \draw [left color=black!60,right color=black!20,thick] (5,0) -- (5.2,.2) -- node [pos=.5,right] { $h$} (5.2,-3.3) -- (5,-3.5); - \draw [left color=black!60,right color=black!20,thick] (0,0) -- (5,0) -- (5,-3.5)--(0,-3.5) -- cycle; - \draw [>=stealth,->] (.5,0.3) node [above] { $\dx$} --(0.3,.1); - \draw (2.5,-1.75) node { $V\approx 2\pi r h\dx$}; - - \end{scope} - - \end{tikzpicture} - - - - -
    -     -
    - - - The Shell Method -

    - Let a solid be formed by revolving a region R, - bounded by x=a and x=b, around a vertical axis. - Let r(x) represent the distance from the axis of rotation to x (, the radius of a sample shell) and let h(x) represent the height of the solid at x (, the height of the shell). - The volume of the solid is - integrationvolume!Shell Method - Shell Method - - V = 2\pi\int_a^b r(x)h(x)\, dx - . -

    -     - -

    - Special Cases: -

    - -

    -

      -
    1. -

      - When the region R is bounded above by y=f(x) and below by y=g(x), - then h(x) = f(x)-g(x). -

      -
      2. - -
    2. -

      - When the axis of rotation is the y-axis (, x=0) then r(x) = x. -

      -
      3. -
    -

    - -

    - Let's practice using the Shell Method. -

    - - - Finding volume using the Shell Method - -

    - Find the volume of the solid formed by rotating the region bounded by y=0, - y=1/(1+x^2), x=0 and x=1 about the y-axis. -

    -     - -

    - This is the region used to introduce the Shell Method in , - but is sketched again in for closer reference. - A line is drawn in the region parallel to the axis of rotation representing a shell that will be carved out as the region is rotated about the y-axis. - (This is the differential element.) -

    - -
    - Graphing a region in - - - -

    - Graph of the region bounded by y=0, y=1/(1+x^2), x=0 and x=1. - The curve y=1/(1+x^2) begins at the y-axis at point (0,1) and slopes downwards before ending at the point (1,\frac12) . - The region contains the entire area below the curve, lying above the horizontal line y=0 . - On the x-axis, the graph contains an arbitrarily chosen point x . - The distance between the origin and the point x is given as the function r(x) , which will give us the radius of the cylindrical shell once the region is rotated about the y-axis. - Additionally, coming up from the point x is a red vertical line, which ends after meeting the curve y=1/(1+x^2). - This vertical line is labeled h(x), and will give the us the height of the cylindrical shell. -

    -     - Two dimensional graph of the region from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1}, - extra x ticks={.4}, - extra x tick labels={$x$}, - ytick={1}, - ymin=-.2,ymax=1.25, - xmin=-.15,xmax=1.1, - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1] {(1/(1+x^2))} \closedcycle; - \addplot [firstcurvestyle,domain=0:1] {(1/(1+x^2))}; - - \draw [thick,secondcolor] (axis cs:.4,0) -- (axis cs:.4,.86); - - \draw (axis cs:.4,.45) node [left] {$ h(x)\left\{\rule{0pt}{33pt}\right.$} - (axis cs:.2,-.12) node { $\underbrace{\rule{35pt}{0pt}}_{r(x)}$} - (axis cs: .7,1) node { $\ds y=\frac{1}{1+x^2}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - The distance this line is from the axis of rotation determines r(x); - as the distance from x to the y-axis is x, - we have r(x)=x. - The height of this line determines h(x); - the top of the line is at y=1/(1+x^2), - whereas the bottom of the line is at y=0. - Thus h(x) = 1/(1+x^2)-0 = 1/(1+x^2). - The region is bounded from x=0 to x=1, so the volume is - - V \amp = 2\pi\int_0^1 \frac{x}{1+x^2}\, dx. - This requires substitution. Let u=1+x^2, so du = 2x\, dx. We also change the bounds: u(0) = 1 and u(1) = 2. Thus we have: - \amp = \pi\int_1^2 \frac{1}{u}\, du - \amp = \pi\ln(u)\Big|_1^2 - \amp = \pi\ln(2) \approx 2.178 \,\text{units}^3 - . -

    - -

    - Note: in order to find this volume using the Disk Method, - two integrals would be needed to account for the regions above and below y=1/2. -

    -     - -     - -

    - With the Shell Method, - nothing special needs to be accounted for to compute the volume of a solid that has a hole in the middle, - as demonstrated next. -

    - - - Finding volume using the Shell Method - -

    - Find the volume of the solid formed by rotating the triangular region determined by the points (0,1), - (1,1) and (1,3) about the line x=3. -

    -     - -

    - The region is sketched in along with the differential element, - a line within the region parallel to the axis of rotation. - In , - we see the shell traced out by the differential element, - and in - the whole solid is shown. -

    - -
    - Graphing a region in - -
    - - - - -

    - Graph of the triangular region with edges on the points (0,1), (1,1) and (1,3). - The leftmost point of the triangular region begins at the y-axis at point (0,1) and linearly increases until the point (1,3) . - The equation of the line between the points (0,1) and (1,3) is given to be y=2x+1 . - The region contains the entire area below the curve, lying above the horizontal line y=1 . - Above the x-axis, the graph showcases the distance between an arbitrarily chosen value on the x-axis, labeled x , which is between 0 and 1, and the axis of rotation at x=3 . - The distance between these two x values is given as the function r(x) , which will give us the radius of the cylindrical shell once the region is rotated about the vertical line x=3. - Additionally, coming up from the point (x,1) is a red vertical line, which meets the upper bound of the triangular region, which is given by the line y=2x+1 . - This vertical line is labeled h(x), and will give the us the height of the cylindrical shell. -

    -     - Graph of the triangular region described in the example. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3}, - extra x ticks={.4}, - extra x tick labels={$x$}, - ytick={1,2,3}, - ymin=-.2,ymax=3.25, - xmin=-.15,xmax=3.2, - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(0,1)(1,1)(1,3)}; - \addplot [firstcurvestyle,domain=0:1] {2*x+1} node [rotate=57,pos=.5,above,black] { $y=2x+1$}; - \addplot [firstcurvestyle] coordinates {(0,1)(1,1)(1,3)}; - - \draw [thick,secondcolor] (axis cs: .4,1) -- (axis cs:.4,1.8); - - \draw [dashed] (axis cs:3,0) -- (axis cs:3,3.2); - - \draw (axis cs:1,1.4) node [right] {$ \left.\rule{0pt}{14pt}\right\} h(x).$} - (axis cs:1.7,.7) node { $\underbrace{\rule{100pt}{0pt}}_{r(x)}$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - -

    - A three dimensional graph of the same triangular region with edges on the points (0,1), (1,1) and (1,3) and an arbitrary cylindrical shell which lies inside the triangular region rotated about x=3. - The cylindrical shell is centered at the line x=3 and radius r(x) , which spans from arbitrarily chosen value on the x-axis between x=0 and x=1, and the axis of rotation at x=3 . - The height of the cylindrical shell is h(x) , which is the distance between the point (x,1) and upper bound of the triangular region at the particular value of x. -

    -     - Three dimensional depiction of an arbitrary cylindrical shell from rotating the region described in the example. - - - - - //ASY file for figshell2b_3D.asy in Chapter 7 - - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - //currentprojection=orthographic((6.1,13,42.3),(0,1,0),(0,0,0),.95,(.005,-.024)); - currentprojection=orthographic((2.6,4.6,47),(0,1,0),(0,0,0),.95,(0.004,0.004)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3,4,5,6}; - real[] myychoice={1,2,3}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.2,6.6); - pair ybounds=(-.2,3.4); - pair zbounds=(-3.4,3.4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); - - pen p=apexmeshpen; - - triple f2(pair t) {return ((3-t.x)*cos(t.y)+3,2*t.x+1,(3-t.x)*sin(t.y));} - surface s2=surface(f2,(0,0),(1,2*pi),2,16,Spline); - //draw(s2,simplesurfacepen); - - triple f3(pair t) {return (2*cos(t.y)+3,t.x,2*sin(t.y));} - surface s3=surface(f3,(1,0),(3,2*pi),2,16,Spline); - //draw(s3,simplesurfacepen); - - triple f4(pair t) {return (2.6*cos(t.y)+3,t.x,2.6*sin(t.y));} - surface s4=surface(f4,(1,0),(1.8,2*pi),2,16,Spline); - draw(s4,simplesurfacepen2); - - triple g3(real t) {return (2.6*cos(t)+3,1.8,2.6*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,redpen+.4mm); - - triple g3(real t) {return (2.6*cos(t)+3,1,2.6*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,redpen+.4mm); - - draw((0,1,0)--(1,3,0)--(1,1,0)--cycle,bluepen+.6mm); - - triple g3(real t) {return (2*cos(t)+3,3,2*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - draw((3,0,0)--(3,3.1,0),dashed+.2mm); - - draw((.4,1,0)--(.4,1.8,0),redpen+.4mm); - - triple g3(real t) {return (2*cos(t)+3,1,2*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - triple g3(real t) {return (3*cos(t)+3,1,3*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - //triple g3(real t) {return (2,3*cos(t),3*sin(t));} - //path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); - - - - -
    - -
    - - - -

    - A three dimensional plot showing the same triangular region rotated about x=3. - The resulting shape lies entirely above x=1 . - The leftmost point of the shape is the point (0,1) , which is the leftmost point of the original triangle. - The shape is then rotated in a full circle around the vertical line x=3 . - The resulting shape has a base of radius of 3 , whose center is an empty central cylinder of radius 2 centered at the axis of rotation. - As we move up the shape, the thickness tapers off as the line y=2x+1 reaches the point (3,1) in the original triangular region. - The point (3,1) rotated about the line x=3 then becomes the top of the resulting shape, while the base of the triangle, which is between the points (0,1) and (1,1) becomes the base. -

    -     - Three dimensional plot of the entire space which comes from rotating the triangular region. - - - - //ASY file for figshell2c_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - //currentprojection=orthographic((6.1,13,42.3),(0,1,0),(0,0,0),.95,(.005,-.024)); - currentprojection=orthographic((2.6,4.6,47),(0,1,0),(0,0,0),.95,(0.004,0.004)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3,4,5,6}; - real[] myychoice={1,2,3}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.2,6.6); - pair ybounds=(-.2,3.4); - pair zbounds=(-3.4,3.4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); - - pen p=apexmeshpen; - - triple f2(pair t) {return ((3-t.x)*cos(t.y)+3,2*t.x+1,(3-t.x)*sin(t.y));} - surface s2=surface(f2,(0,0),(1,2*pi),2,16,Spline); - draw(s2,simplesurfacepen); - - triple f3(pair t) {return (2*cos(t.y)+3,t.x,2*sin(t.y));} - surface s3=surface(f3,(1,0),(3,2*pi),2,16,Spline); - draw(s3,simplesurfacepen); - - triple f4(pair t) {return (2.6*cos(t.y)+3,t.x,2.6*sin(t.y));} - surface s4=surface(f4,(1,0),(1.8,2*pi),2,16,Spline); - //draw(s4,surfacepen2); - - triple g3(real t) {return (2.6*cos(t)+3,1.8,2.6*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,redpen+.4mm); - - draw((0,1,0)--(1,3,0)--(1,1,0)--cycle,bluepen+.6mm); - - triple g3(real t) {return (2*cos(t)+3,3,2*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.4mm); - - draw((3,0,0)--(3,3.1,0),dashed+.2mm); - - draw((.4,1,0)--(.4,1.8,0),redpen+.4mm); - - triple g3(real t) {return (2*cos(t)+3,1,2*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.4mm); - - triple g3(real t) {return (3*cos(t)+3,1,3*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.4mm); - - //triple g3(real t) {return (2,3*cos(t),3*sin(t));} - //path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); - - - -
    -     - -
    - -

    - The height of the differential element is the distance from y=1 to y=2x+1, - the line that connects the points (0,1) and (1,3). - Thus h(x) = 2x+1-1 = 2x. - The radius of the shell formed by the differential element is the distance from x to x=3; - that is, it is r(x)=3-x. - The x-bounds of the region are x=0 to x=1, giving - - V \amp = 2\pi\int_0^1 (3-x)(2x)\, dx - \amp = 2\pi\int_0^1 \big(6x-2x^2\big)\, dx - \amp = 2\pi\left(3x^2-\frac23x^3\right)\Big|_0^1 - \amp = \frac{14}{3}\pi\approx 14.66 \,\text{units}^3 - . -

    -     - -     - -

    - When revolving a region around a horizontal axis, - we must consider the radius and height functions in terms of y, - not x. -

    - - - Finding volume using the Shell Method - -

    - Find the volume of the solid formed by rotating the region given in about the x-axis. -

    -     - -

    - The region is sketched in with a sample differential element. - In the shell formed by the differential element is drawn, - and the solid is sketched in . (Note that the triangular region looks - short and wide here, - whereas in the previous example the same region looked - tall and narrow. - This is because the bounds on the graphs are different.) -

    - -

    - The height of the differential element is an x-distance, - between x=\frac12y-\frac12 and x=1. - Thus h(y) = 1-(\frac12y-\frac12) = -\frac12y+\frac32. - The radius is the distance from y to the x-axis, - so r(y) =y. - The y bounds of the region are y=1 and y=3, - leading to the integral -

    - -
    - Graphing a region in - -
    - - - - -

    - Graph of the triangular region with edges on the points (0,1), (1,1) and (1,3). - The leftmost point of the triangular region begins at the y-axis at point (0,1) and linearly increases until the point (1,3) . - The equation of the line between the points (0,1) and (1,3) is given as x=\frac12y-\frac12 . - The region contains the entire area to the right of the line x=\frac12y-\frac12 and to the left of the vertical line x=1 . - Coming up from the x-axis, the graph contains an arbitrarily chosen value on the y-axis, labeled y , which is between 1 and 3. - The graph showcases the distance between the point labeled y and the axis of rotation which is y=0 . - The distance between these two y values is given as the function r(y) , which will give us the radius of the cylindrical shell once the solid is formed by rotating about the x-axis. - Additionally, coming up from the line x=\frac12y-\frac12 is a red horizontal line, which meets the right side of the triangular region, which is given by the vertical line x=1 . - This vertical line is labeled h(y), and will give the us the height of the cylindrical shells. -

    -     - Graph of the triangular region described in the example. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3}, - ytick={1,2,3}, - extra y ticks={1.8}, - extra y tick labels={$y$}, - ymin=-.2,ymax=3.25, - xmin=-.15,xmax=1.5, - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(0,1)(1,1)(1,3)}; - \addplot [firstcurvestyle] coordinates {(0,1)(1,1)(1,3)(0,1)} node [rotate=41,shift={(75pt,0pt)},above,black] { $x=\frac12y-\frac12$}; - - \draw [thick,secondcolor] (axis cs: .4,1.8) -- node [black,pos=.5,below] {$\underbrace{\rule{42pt}{0pt}}_{h(y)}$} (axis cs:1,1.8); - - \draw (axis cs:1,.9) node [right] {$ \left.\rule{0pt}{29pt}\right\} r(y)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - -

    - A three dimensional graph of the same triangular region as the previous image and an arbitrary cylindrical shell which lies inside the triangular region rotated about the x-axis. - The cylindrical shell lies horizontally resting on its curved surface, and is centered at the line y=0 . - The cylindrical shell has radius r(y) , which spans from the y-axis to the arbitrarily chosen value on the y-axis between y=1 and y=3. - The height of the cylindrical shell is h(y) , which is the distance between the line x=\frac12y-\frac12 and the rightmost bound of the triangular region, which is the vertical line x=1. -

    -     - Three dimensional depiction of an arbitrary cylindrical shell from rotating the region described in the example. - - - - - //ASY file for figshell3b_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((3.13,8.3,49),(0,1,0),(0,0,0),.95,(0.0148,0.00673)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1,2,3}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.2,1.25); - pair ybounds=(-3.4,3.4); - pair zbounds=(-3.4,3.4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); - - pen p=apexmeshpen; - - triple f2(pair t) {return (t.x,(2*t.x+1)*cos(t.y),(2*t.x+1)*sin(t.y));} - surface s2=surface(f2,(0,0),(1,2*pi),2,16,Spline); - //draw(s2,simplesurfacepen); - - triple f3(pair t) {return (t.x,cos(t.y),sin(t.y));} - surface s3=surface(f3,(0,0),(1,2*pi),2,16,Spline); - //draw(s3,surfacepen); - - triple f4(pair t) {return (t.x,1.8*cos(t.y),1.8*sin(t.y));} - surface s4=surface(f4,(.4,0),(1,2*pi),2,16,Spline); - draw(s4,simplesurfacepen2); - - triple g3(real t) {return (.4,1.8*cos(t),1.8*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,redpen+.4mm); - - triple g3(real t) {return (1,1.8*cos(t),1.8*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,redpen+.4mm); - - draw((0,1,0)--(1,3,0)--(1,1,0)--cycle,bluepen+.6mm); - - draw((.4,1.8,0)--(1,1.8,0),redpen+.4mm); - - triple g3(real t) {return (0,cos(t),sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - triple g3(real t) {return (1,cos(t),sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - triple g3(real t) {return (1,3*cos(t),3*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - //triple g3(real t) {return (2,3*cos(t),3*sin(t));} - //path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); - - - - -
    - -
    - - - - - -

    - A three dimensional plot showing the same triangular region rotated about the x-axis. - The resulting shape lies entirely between x=0 and x=1 . - The highest point of the shape is the point (1,3) , which is the highest point of the original triangle. - The shape is then rotated in a full circle about the x-axis. - The resulting shape has a height given by the value h(y)= -\frac12y+\frac32 . - The radius of the shape is given by r(y)=y, with the y bounds between y=1 and y=3 as the center of the shape is an empty central cylinder of radius 1 centered at the axis of rotation. -

    -     - Three dimensional plot of the entire space which comes from rotating the triangular region. - - - - - //ASY file for figshell3c_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((3.13,8.3,49),(0,1,0),(0,0,0),.95,(0.0148,0.00673)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1,2,3}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.2,1.25); - pair ybounds=(-3.4,3.4); - pair zbounds=(-3.4,3.4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); - - pen p=apexmeshpen; - - triple f2(pair t) {return (t.x,(2*t.x+1)*cos(t.y),(2*t.x+1)*sin(t.y));} - surface s2=surface(f2,(0,0),(1,2*pi),2,16,Spline); - draw(s2,simplesurfacepen); - - triple f3(pair t) {return (t.x,cos(t.y),sin(t.y));} - surface s3=surface(f3,(0,0),(1,2*pi),2,16,Spline); - draw(s3,simplesurfacepen); - - triple f3(pair t) {return (1,t.x*cos(t.y),t.x*sin(t.y));} - surface s3=surface(f3,(1,0),(3,2*pi),2,16,Spline); - draw(s3,simplesurfacepen); - - triple f4(pair t) {return (t.x,1.8*cos(t.y),1.8*sin(t.y));} - surface s4=surface(f4,(.4,0),(1,2*pi),2,16,Spline); - //draw(s4,surfacepen2); - - triple g3(real t) {return (.4,1.8*cos(t),1.8*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,redpen+.4mm); - - draw((0,1,0)--(1,3,0)--(1,1,0)--cycle,bluepen+.6mm); - - draw((.4,1.8,0)--(1,1.8,0),redpen+.4mm); - - triple g3(real t) {return (0,cos(t),sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.4mm); - - triple g3(real t) {return (1,cos(t),sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.4mm); - - triple g3(real t) {return (1,3*cos(t),3*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.4mm); - - //triple g3(real t) {return (2,3*cos(t),3*sin(t));} - //path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); - - - - -
    -     - -
    - -

    - - V \amp = 2\pi\int_1^3\left[y\left(-\frac12y+\frac32\right)\right]\, dy - \amp = 2\pi\int_1^3\left[-\frac12y^2+\frac32y\right]\, dy - \amp = 2\pi\left[-\frac16y^3+\frac34y^2\right]\Big|_1^3 - \amp = 2\pi\left[\frac94-\frac7{12}\right] - \amp = \frac{10}{3}\pi \approx 10.472\,\text{units}^3 - . -

    -     - -     - -

    - At the beginning of this section it was stated that - it is good to have options. - The next example finds the volume of a solid rather easily with the Shell Method, - but using the Washer Method would be quite a chore. -

    - - - Finding volume using the Shell Method - -

    - Find the volume of the solid formed by revolving the region bounded by - y= \sin(x) and the x-axis from x=0 to x=\pi about the y-axis. -

    -     - -

    - The region and a differential element, - the shell formed by this differential element, - and the resulting solid are given in . -

    - -
    - Graphing a region in - -
    - - - - -

    - Graph of the region bounded by y= \sin(x) and the x-axis from x=0 to x=\pi. - The leftmost point of the triangular region begins at the origin after which we see a singular sine wave ending at the point (\pi,0) . - The region contains the entire area to below the curve y= \sin(x), lying above the x-axis. - The graph contains an arbitrarily chosen value on the x-axis, labeled x , which is between 0 and \pi. - The graph showcases the distance between the axis of rotation which is x=0 and the point labeled x as the function r(x) , which will be the radius of the cylindrical shells. - Additionally, coming up from the x-axis is a red vertical line, which meets the curve y= \sin(x) at the point (x,\sin(x)). - This vertical line is labeled h(x), and will give the us the height of the cylindrical shells. -

    -     - Graph of the region described in the example. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={1,1.57,3.14}, - extra x tick labels={$x$,$\frac{\pi}2$,$\pi$}, - ytick={1}, - ymin=-.2,ymax=1.25,% - xmin=-.1,xmax=3.5,% - ] - - \addplot [firstcurvestyle,areastyle,domain=0:3.14] {sin(deg(x))}; - \addplot [firstcurvestyle,domain=0:3.14,samples=50] {sin(deg(x))}; - - \draw [thick,secondcolor] (axis cs: 1,.84) -- node [black,pos=.49,right] {$ \left.\rule{0pt}{32pt}\right\} h(x)$} (axis cs:1,0); - - \draw (axis cs:.52,0) node [above] {$\overbrace{\rule{30pt}{0pt}}^{r(x)}$} ; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - -

    - A three dimensional graph of the same region as the previous image and an arbitrary cylindrical shell which will lie inside the region rotated about the x-axis. - The cylindrical shell lies vertically resting on its flat base and is centered on the y-axis. - The cylindrical shell has radius r(x)=x , which spans from the y-axis to the arbitrarily chosen value on the x-axis between x=0 and x=\pi. - The height of the cylindrical shell is h(x)=\sin(x) , which is the distance between the x-axis and the upper bound of the region, which curve y=\sin(x). -

    -     - Three dimensional depiction of an arbitrary cylindrical shell lying inside the rotated region described in the example. - - - - - //ASY file for figshell4b_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((23.8,4.1,64),(0,1,0),(0,0,0),1,(-.00343,0.004889)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-3.4,3.4); - pair ybounds=(-.2,1.2); - pair zbounds=(-3.4,3.4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); - - pen p=apexmeshpen; - - triple pt1=(pi,-.05,0); - draw((pi,0,0)--pt1,black+.2mm); - label("$\pi$",pt1,S); - - triple f2(pair t) {return (t.x*cos(t.y),sin(t.x),t.x*sin(t.y));} - surface s2=surface(f2,(0,0),(pi,2*pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - //draw(s2,surfacepen,meshpen=p); - - triple f4(pair t) {return (cos(t.y),t.x,sin(t.y));} - surface s4=surface(f4,(0,0),(.84,2*pi),2,16,Spline); - draw(s4,simplesurfacepen2); - - draw((1,0,0)--(1,.84,0),redpen+.4mm); - - triple g3(real t) {return (t,sin(t),0);} - path3 p3=graph(g3,0,pi,operator ..); - draw(p3,bluepen+.6mm); - - triple g3(real t) {return (cos(t),.84,sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,redpen+.4mm); - - triple g3(real t) {return (cos(t),0,sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,redpen+.4mm); - - triple g3(real t) {return (pi*cos(t),0,pi*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - //triple g3(real t) {return (2,3*cos(t),3*sin(t));} - //path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); - - - - -
    - -
    - - - - - -

    - A three dimensional plot showing the entire region rotated about the y-axis. - The base of the shape is a circle of radius \pi centered at the origin. - From the outside edge of the circle, the surface curves both upwards and inwards towards the positive y-axis until reaching a peak at y=1. - From here, the surface curves downwards and inwards until closing in at the origin. - The cylinderical shells lying inside the shape have a height given by h(x)= \sin(x) and radius r(x)=x. -

    -     - Three dimensional plot of the entire space which comes from rotating sine wave. - - - - - //ASY file for figshell4c_3D.asy in Chapter 7 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((23.8,4.1,64),(0,1,0),(0,0,0),1,(-.00343,0.004889)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-3.4,3.4); - pair ybounds=(-.2,1.2); - pair zbounds=(-3.4,3.4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // \addplot3[domain=1:3,y domain=0:1,surf,shader=flat,colormap={mp2}{\colormapone}] - // (x,0,{y*((2*x-1)-(x^2-2*x+2))+x^2-2*x+2}); - - pen p=apexmeshpen; - - triple pt1=(pi,-.05,0); - draw((pi,0,0)--pt1,black+.2mm); - label("$\pi$",pt1,S); - - triple f2(pair t) {return (t.x*cos(t.y),sin(t.x),t.x*sin(t.y));} - surface s2=surface(f2,(0,0),(pi,2*pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s2,simplesurfacepen,meshpen=p); - - triple f4(pair t) {return (cos(t.y),t.x,sin(t.y));} - surface s4=surface(f4,(0,0),(.84,2*pi),2,16,Spline); - //draw(s4,emissive(rgb(1,.6,.6)+opacity(.2))); - - draw((1,0,0)--(1,.84,0),redpen+.4mm); - - triple g3(real t) {return (t,sin(t),0);} - path3 p3=graph(g3,0,pi,operator ..); - draw(p3,bluepen+.6mm); - - triple g3(real t) {return (cos(t),.84,sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,redpen+.4mm); - - triple g3(real t) {return (cos(t),0,sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,redpen+.4mm); - - triple g3(real t) {return (pi*cos(t),0,pi*sin(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.4mm); - //triple g3(real t) {return (2,3*cos(t),3*sin(t));} - //path3 p3=graph(g3,0,2*pi,operator ..); - //draw(p3,bluepen+.4mm); - - //draw((2,2,0)--(2,3,0),purple+linetype(new real[] {4,4})+.3mm); - - - - -
    -     - -
    - -

    - The radius of a sample shell is r(x) = x; - the height of a sample shell is h(x) = \sin(x), - each from x=0 to x=\pi. - Thus the volume of the solid is - - V \amp = 2\pi\int_0^{\pi} x\sin(x) \, dx. - This requires Integration By Parts. Set u=x and dv=\sin(x) \, dx; we leave it to the reader to fill in the rest. We have: - \amp = 2\pi\Big[-x\cos(x) \Big|_0^{\pi} +\int_0^{\pi}\cos(x) \, dx \Big] - \amp = 2\pi\Big[\pi + \sin(x) \Big|_0^{\pi}\,\Big] - \amp = 2\pi\Big[\pi + 0 \Big] - \amp = 2\pi^2 \approx 19.74 \,\text{units}^3 - . -

    - -

    - Note that in order to use the Washer Method, - we would need to solve y=\sin(x) for x, - requiring the use of the arcsine function. - We leave it to the reader to verify that the outside radius function is - R(y) = \pi-\arcsin y and the inside radius function is r(y)=\arcsin y. - Thus the volume can be computed as - - \pi\int_0^1 \Big[ (\pi-\arcsin y)^2-(\arcsin y)^2\Big]\, dy - . -

    - -

    - This integral isn't terrible given that the \arcsin^2 y terms cancel, - but it is more onerous than the integral created by the Shell Method. -

    -     - -     - -

    - We end this section with a table summarizing the usage of the Washer and Shell Methods. -

    - - - Summary of the Washer and Shell Methods -

    - Let a region R be given with x-bounds x=a and x=b and y-bounds y=c and y=d. -

    - - - - - Washer Method - - Shell Method - - - Horizontal Axis - \ds \pi\int_a^b \big(R(x)^2-r(x)^2\big)\, dx - - \ds 2\pi\int_c^d r(y)h(y)\, dy - - - - - - Vertical Axis - \ds\pi \int_c^d\big(R(y)^2-r(y)^2\big)\, dy - - \ds 2\pi\int_a^b r(x)h(x)\, dx - - - -

    - integrationvolume!Washer Method - Washer Method - integrationvolume!Shell Method - Shell Method -

    -     - - - - - - - - - - - -

    - As in the previous section, - the real goal of this section is not to be able to compute volumes of certain solids. - Rather, it is to be able to solve a problem by first approximating, - then using limits to refine the approximation to give the exact value. - In this section, - we approximate the volume of a solid by cutting it into thin cylindrical shells. - By summing up the volumes of each shell, - we get an approximation of the volume. - By taking a limit as the number of equally spaced shells goes to infinity, - our summation can be evaluated as a definite integral, - giving the exact value. -

    - -

    - We use this same principle again in the next section, - where we find the length of curves in the plane. -

    - - - - Terms and Concepts - - - -

    - A solid of revolution is formed by revolving a shape around an axis. -

    -     - - -     - - - - -

    - The Shell Method can only be used when the Washer Method fails. -

    -     - -     - - - - -

    - The Shell Method works by integrating cross-sectional areas of a solid. -

    -     - - -     - - - - -

    - When finding the volume of a solid of revolution that was revolved around a vertical axis, - the Shell Method integrates with respect to x. -

    -     - - -     -     - - Problems - - - - -

    - Use the Shell Method to find the volume of the solid of revolution formed by - revolving the given region about the y-axis. -

    -     - - - - -

    - The region bounded by the curve y=3-x^2, - the x axis, and the y axis: -

    - - - -

    - Graph of the region bounded by the curve y=3-x^2, the x-axis, and the y-axis. - Note that this region can reference both the regions to the left or right of the y-axis, but we will consider the region to the right of the y-axis. - The curve y=3-x^2 begins at the point (0,3) from which it slopes down until reaching the x-axis. - The region then contains the entire area below this curve, and above the x-axis. -

    -     - Graph of the region lying in the first quadrant bounded by the curve and the two coordinate axes. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=3.5, - xmin=-2.1,xmax=2.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1.73] {3-x^2} \closedcycle; - \addplot [firstcurvestyle,domain=-1.73:1.73,samples=40] {3-x^2} node [shift={(-15pt,90pt)},black] { $y=3-x^2$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - 9\pi/2 units^3 -

    -     - -     - - - - -

    - The region between y=5x and the x axis, - for 1\leq x\leq 2: -

    - - - -

    - Graph of the region between y=5x and the x-axis, for 1\leq x\leq 2. - The line y=5x begins at the point (1,5) from which it linearly increases until reaching the rightmost bound which is at the point (2,10). - The region then contains the entire area below the line y=5x and above the x-axis for 1\leq x\leq 2. -

    -     - Graph of the region bounded by the the line y=5x and the x-axis for x between 1 and 2. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=11, - xmin=-.1,xmax=2.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=1:2] {5*x} \closedcycle; - \addplot [firstcurvestyle,domain=0:2.1] {5*x} node [shift={(-35pt,-10pt)},black] { $y=5x$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - 70\pi/3 units^3 -

    -     - -     - - - - -

    - The region between y=\cos(x) and the x axis, - for 0\leq x\leq \pi/2: -

    - - - -

    - Graph of the region between y=\cos(x) and the x-axis, for 0\leq x\leq \pi/2. - The curve y=\cos(x) begins at the point (0,1) from which it slopes downwards until ending after reaching the x-axis at the point (\pi/2,0). - The region then contains the entire area below the curve y=\cos(x) and above the x-axis for 0\leq x\leq \pi/2. - The bound 0\leq x\leq \pi/2 can also be interpreted as the leftmost bound of the region being the y-axis. -

    -     - Graph of the region bounded by the the cosine function and the two coordinate axes. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.7 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1.57] {cos(deg(x))} \closedcycle; - \addplot [firstcurvestyle,domain=0:1.57,samples=30] {cos(deg(x))} node [shift={(-20pt,65pt)},black] { $y=\cos(x) $}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \pi^2-2\pi units^3 -

    -     - -     - - - - -

    - The region between the curves y=x and y=\sqrt{x}: -

    - - - -

    - Graph of the region between the curves y=x and y=\sqrt{x}. - The curves y=x and y=\sqrt{x} both begin at the origin. - From this point the curve y=\sqrt{x} rises above the line y=x until reaching the point (1,1), where the curve once again intersects the line. - The region then contains the area below the curve y=\sqrt{x} and above the line y=x. -

    -     - Graph of the region bounded by the two curves. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1,samples=60] {sqrt(x)}; - - \addplot [firstcurvestyle,domain=0:.05] {sqrt(x)}; - \addplot [firstcurvestyle,domain=.05:1,samples=30] {sqrt(x)} node [shift={(-65pt,-15pt)} ,black] { $y=\sqrt{x}$}; - - \addplot [firstcurvestyle,domain=0:1] {x} node [shift={(-30pt,-45pt)},black] { $y=x$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - 2\pi/15 units^3 -

    -     - -     - -     - - - -

    - Use the Shell Method to find the volume of the solid of revolution formed by - revolving the given region about the x-axis. -

    -     - - - - -

    - The region between y=3-x^2 and the x axis: -

    - - - -

    - Graph of the region between y=3-x^2 and the x-axis. - The curve y=3-x^2 begins at x-axis at the point (-\sqrt{3},0). - From this point the curve rises until reaching the y-axis at the point (0,3). - From this point, the curve slopes downwards until once again reaching the x-axis at the point (\sqrt{3},0). - On both the left and right sides of the region, the area is bounded by the curve y=3-x^2. -

    -     - Graph of the region bounded by the curve and the x axis. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=3.5, - xmin=-2.1,xmax=2.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=-1.73:1.73] {3-x^2}; - \addplot [firstcurvestyle,domain=-1.73:1.73,samples=40] {3-x^2} node [shift={(-15pt,90pt)},black] { $y=3-x^2$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - 48\pi\sqrt{3}/5 units^3 -

    -     - -     - - - - -

    - The region between y=5x and the y axis, - for 5\leq y\leq 10: -

    - - - -

    - Graph of the region between y=5x and the y axis, for 5\leq y\leq 10. - The line y=5x begins at the point (1,5), from which it linearly increases until ending upper bound for y at the point (2,10). - The region then contains the entire area to the right of x=0 and to the left of the line y=5x for y between 5 and 10. -

    -     - Graph of the region bounded by the line and the y axis for y values between 5 and 10. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=11, - xmin=-.1,xmax=2.1 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(1,5) (2,10) (0,10) (0,5) (1,5)}; - \addplot [firstcurvestyle,domain=0:2.1] {5*x} node [shift={(-20pt,-40pt)},black] { $y=5x$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - 350\pi/3 units^3 -

    -     - -     - - - - -

    - The region between y=\cos(x) and the x axis, - for 0\leq x\leq \pi/2: -

    - - - -

    - Graph of the region between y=\cos(x) and the x axis, for 0\leq x\leq \pi/2. - The curve y=\cos(x) begins at the point (0,1) from which it slopes downwards until ending after reaching the x-axis at the point (\pi/2,0). - The region then contains the entire area to the right of x=0 and to the left of of the curve y=\cos(x) for y values between 0 and 1. -

    -     - Graph of the region bounded by the cosine curve and the coordinate axes. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.7 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1.57] {cos(deg(x))} \closedcycle; - \addplot [firstcurvestyle,domain=0:1.57,samples=30] {cos(deg(x))} node [shift={(-20pt,65pt)},black] { $y=\cos(x) $}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \pi^2/4 units^3 -

    -     - -     - - - - -

    - The region between the curves y=x and y=\sqrt{x}: -

    - - - -

    - Graph of the region between the curves y=x and y=\sqrt{x}. - The curves y=x and y=\sqrt{x} both begin at the origin. - From this point the curve y=\sqrt{x} rises above the line y=x until reaching the point (1,1), where the curve once again intersects the line. - The region consists of the area to the right of the curve y=\sqrt{x} and the to the left line y=x for y values between 0 and 1. -

    -     - Graph of the region bounded by the two curves. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:1,samples=60] {sqrt(x)}; - - \addplot [firstcurvestyle,domain=0:.05] {sqrt(x)}; - \addplot [firstcurvestyle,domain=.05:1,samples=30] {sqrt(x)} node [shift={(-65pt,-15pt)} ,black] { $y=\sqrt{x}$}; - - \addplot [firstcurvestyle,domain=0:1] {x} node [shift={(-30pt,-45pt)},black] { $y=x$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \pi/6 units^3 -

    -     - -     - -     - - - -

    - Use the Shell Method to find the volume of the solid of revolution formed - by revloving the given region about each of the given axes. -

    -     - - - - -

    - Region bounded by: y=\sqrt{x}, - y=0 and x=1. -

    -     - - - -

    - Rotate about the y axis. -

    -     - -

    - 4\pi/5 -

    -     -     - - - -

    - Rotate about x=1. -

    -     - -

    - 8\pi/15 -

    -     -     - - - -

    - Rotate about the x axis. -

    -     - -

    - \pi/2 -

    -     -     - - - -

    - Rotate about y=1. -

    -     - -

    - 5\pi/6 -

    -     -     - -     - - - - -

    - Region bounded by: y=4-x^2 and y=0. -

    -     - - - -

    - Rotate about x=2. -

    -     - -

    - 128\pi/3 -

    -     -     - - - -

    - Rotate about x=-2. -

    -     - -

    - 128\pi/3 -

    -     -     - - - -

    - Rotate about the x axis. -

    -     - -

    - 512\pi/15 -

    -     -     - - - -

    - Rotate about y=4. -

    -     - -

    - 256\pi/5 -

    -     -     - -     - - - - -

    - The triangle with vertices (1,1), - (1,2) and (2,1). -

    -     - - - -

    - Rotate about the y axis. -

    -     - -

    - 4\pi/3 -

    -     -     - - - -

    - Rotate about x=1. -

    -     - -

    - \pi/3 -

    -     -     - - - -

    - Rotate about the x axis. -

    -     - -

    - 4\pi/3 -

    -     -     - - - -

    - Rotate about y=2. -

    -     - -

    - 2\pi/3 -

    -     -     - -     - - - - -

    - Region bounded by y=x^2-2x+2 and y=2x-1. -

    -     - - - -

    - Rotate about the y axis. -

    -     - -

    - 16\pi/3 -

    -     -     - - - -

    - Rotate about x=1. -

    -     - -

    - 8\pi/3 -

    -     -     - - - -

    - Rotate about x=-1. -

    -     - -

    - 8\pi -

    -     -     - -     - - - - -

    - Region bounded by y=1/\sqrt{x^2+1}, - x=1 and the x and y axes. -

    -     - - - -

    - Rotate about the y axis. -

    -     - -

    - 2\pi(\sqrt{2}-1) -

    -     -     - - - -

    - Rotate about x=1. -

    -     - -

    - 2\pi(1-\sqrt{2}+\sinh^{-1}(1)) -

    -     -     - -     - - - - -

    - Region bounded by y=2x, - y=x and x=2. -

    -     - - - -

    - Rotate about the y axis. -

    -     - -

    - 16\pi/3 -

    -     -     - - - -

    - Rotate about x=2. -

    -     - -

    - 8\pi/3 -

    -     -     - - - -

    - Rotate about the x axis. -

    -     - -

    - 8\pi -

    -     -     - - - -

    - Rotate about y=4. -

    -     - -

    - 8\pi -

    -     -     - -     - -     -     -     -
    -
    - Arc Length and Surface Area - -

    - In previous sections we have used integration to answer the following questions: -

    - -

    -

      -
    1. -

      - Given a region, what is its area? -

      -
      2. - -
    2. -

      - Given a solid, what is its volume? -

      -
      3. -
    -

    - - - -

    - In this section, - we address two related questions: -

      -
    1. -

      - Given a curve, what is its length? - This is often referred to as arc length. -

      -
      2. - -
    2. -

      - Given a solid, what is its surface area? -

      -
      3. -
    -

    -     - - - Arc Length -

    - Consider the graph of y=\sin(x) on [0,\pi] given in . - How long is this curve? - That is, if we were to use a piece of string to exactly match the shape of this curve, - how long would the string be? -

    - -

    - As we have done in the past, we start by approximating; - later, we will refine our answer using limits to get an exact solution. -

    - -

    - The length of straight-line segments is easy to compute using the Distance Formula. - We can approximate the length of the given curve by approximating the curve with straight lines and measuring their lengths. -

    - -
    - Graphing y=\sin(x) on [0,\pi] and approximating the curve with line segments - - -
    - - - - - Graph of the function y=\sin(x) on [0,\pi]. - The curve y=\sin(x) begins at the point (0,0), from which it slopes upwards until reaching a peak at the point (\frac{\pi}{2},1). - From the point, the curve slopes downwards until reaching the x-axis at the point (\pi,0). - - Graph of the sine function for x between 0 and pi. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={.79,1.57,2.36,3.14}, - extra x tick labels={$\frac{\pi}4$,$\frac{\pi}2$,$\frac{3\pi}{4}$,$\pi$}, - ymin=-.2,ymax=1.25, - xmin=-.1,xmax=3.5, - ] - - \addplot [firstcurvestyle,closed,domain=0:3.14,samples=40] {sin(deg(x))}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - Graph of the function y=\sin(x) on [0,\pi], with four straight lines which will be used to approximate the length of this curve. - The four lines are evenly spaced out in intervals of \frac{\pi}{4} on the x-axis. - Each line begins at a point on the curve y=\sin(x), and ends at a point on the same curve after travelling a distance of \frac{\pi}{4} on the x-axis. - The first line begins at the same point (0,0) as the curve, from which it linearly increases until reaching the point (\frac{\pi}{4},\frac{sqrt2}{2}), which is also a point on the curve. - The second line begins at the same point the first line ends, given by (\frac{\pi}{4},\frac{sqrt2}{2}), from which it linearly increases until reaching the point (\frac{\pi}{2},1), which is the peak of the curve. - The third line begins at the same point the second line ends, given by (\frac{\pi}{2},1), from which it linearly decreases until reaching the point (\frac{3\pi}{4},\frac{sqrt2}{2}), which is also a point on the curve. - The fourth line begins at the same point the third line ends, given by (\frac{3\pi}{4},\frac{sqrt2}{2}), from which it linearly decreases until reaching the point (\pi,0), which is the end of the curve. - - Graph of the sine function for x between 0 and pi and four straight lines approximating this curve. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={.79,1.57,2.36,3.14}, - extra x tick labels={$\frac{\pi}4$,$\frac{\pi}2$,$\frac{3\pi}{4}$,$\pi$}, - extra y ticks={.71}, - extra y tick labels={$\frac{\sqrt{2}}2$}, - ymin=-.2,ymax=1.25, - xmin=-.1,xmax=3.5, - ] - - \addplot+ [domain=0:3.14,samples=40] {sin(deg(x))}; - - \draw [secondcolor,thick] (axis cs:0,0) -- (axis cs:.79,.71) -- (axis cs: 1.57,1) -- (axis cs:2.36,.71) -- (axis cs: 3.14,0); - - \filldraw (axis cs:0,0) circle (2pt) (axis cs:.79,.71) circle (2pt) (axis cs: 1.57,1) circle (2pt) (axis cs:2.36,.71) circle (2pt) (axis cs: 3.14,0) circle (2pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - In , - the curve y=\sin(x) has been approximated with 4 line segments - (the interval [0,\pi] has been divided into 4 subintervals of equal length). - It is clear that these four line segments approximate - y=\sin(x) very well on the first and last subinterval, - though not so well in the middle. - Regardless, the sum of the lengths of the line segments is 3.79, - so we approximate the arc length of - y=\sin(x) on [0,\pi] to be 3.79. -

    - -

    - In general, we can approximate the arc length of y=f(x) on [a,b] in the following manner. - Let a=x_0 \lt x_1 \lt \ldots \lt x_{n-1}\lt x_{n}=b be a partition of [a,b] into n subintervals. - Let \dx_i represent the length of the - ith subinterval [x_{i-1},x_{i}]. -

    - -
    - Zooming in on the ith subinterval [x_{i-1},x_{i}] of a partition of [a,b] - - - - Graph of the ith subinterval of the function y=f(x), which is graphed on the interval [x_{i-1},x_{i}]. - The graph contains a line between the start and endpoint of the curve, which will be used to approximate the length of the ith subinterval curve. - The ith subinterval of the curve y=f(x) begins at the point (x_{i-1},y_{i-1}) from which it heads upwards in a concave arc until reaching the point (x_{i},y_{i}). - The straight line then passes through the start and endpoints of the curve, (x_{i-1},y_{i-1}) and (x_{i},y_{i}) respectively. - The line lies below the curve for the entire interval that the curve is plotted on. - The graph also contains the measurements \dx_i and \dy_i, giving the respective length of the change in x and y between the start and endpoint of the subinterval of the curve. - - Graph of a portion of a curve with a line between the start and endpoint which is used to approximate the length of the curve. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={.3,1.37}, - extra x tick labels={$x_{i-1}$,$x_{i}$}, - ytick=\empty, - extra y ticks={.48,1}, - extra y tick labels={$y_{i-1}$,$y_{i}$}, - ymin=-.2,ymax=1.25,% - xmin=-.1,xmax=1.6,% - ] - - \addplot+ [domain=0.3:1.37,samples=30] {sin(deg(x+.2))}; - - \draw (axis cs: .25,1) -- (axis cs:.1,1) - (axis cs: .25,.48) -- (axis cs:.1,.48) - (axis cs: .175,.48) -- node [pos=.5,fill=white] { $\Delta y_i$} (axis cs:.175,1); - - \draw (axis cs: .3,.25) -- (axis cs: .3,.4) - (axis cs: 1.37,.25) -- (axis cs: 1.37,.4) - (axis cs: .3,.325) -- node [pos=.5,fill=white] { $\Delta x_i$} (axis cs: 1.37,.325); - - \draw [secondcolor,thick] (axis cs: .3,.48) -- (axis cs:1.37,1); - - \draw [dashed,thin] (axis cs: .3,.48) -| (axis cs:1.37,1); - - \filldraw (axis cs: .3,.48) circle (1pt) (axis cs:1.37,1) circle (1pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - - zooms in on the ith subinterval where y=f(x) is approximated by a straight line segment. - The dashed lines show that we can view this line segment as the hypotenuse of a right triangle whose sides have length \dx_i and \dy_i. - Using the Pythagorean Theorem, - the length of this line segment is - - \sqrt{\dx_i^2 + \Delta y_i^2} - . - Summing over all subintervals gives an arc length approximation - - L \approx \sum_{i=1}^n \sqrt{\dx_i^2 + \Delta y_i^2} - . -

    - -

    - As shown here, this is not a Riemann Sum. - While we could conclude that taking a limit as the subinterval length goes to zero gives the exact arc length, - we would not be able to compute the answer with a definite integral. - We need first to do a little algebra. -

    - -

    - In the above expression factor out a \dx_i^2 term: - - \sum_{i=1}^n \sqrt{\dx_i^2 + \Delta y_i^2} \amp = \sum_{i=1}^n \sqrt{\dx_i^2\left(1 + \frac{\Delta y_i^2}{\dx_i^2}\right)}. - \amp = \sum_{i=1}^n\sqrt{1 + \frac{\Delta y_i^2}{\dx_i^2}}\,\dx_i - - after pulling the \dx_i^2 term out of the square root. -

    - -

    - This is nearly a Riemann Sum. Consider the \Delta y_i^2/\dx_i^2 term. - The expression \Delta y_i/\dx_i measures the change in y/change in x, - that is, the rise over run of f on the ith subinterval. - The Mean Value Theorem of Differentiation () - states that there is a c_i in the ith subinterval where - \fp(c_i) = \Delta y_i/\dx_i. Thus we can rewrite our above expression as: - - L \approx \sum_{i=1}^n\sqrt{1+\fp(c_i)^2}\,\dx_i - . - This is a Riemann Sum. - As long as \fp is continuous, we can invoke - and conclude - - L = \int_a^b\sqrt{1+\fp(x)^2}\, dx - . -

    - - - Arc Length - -

    - Let f be differentiable on [a,b], - where \fp is also continuous on [a,b]. - Then the arc length of f from x=a to x=b is - integrationarc length - arc length - - L = \int_a^b \sqrt{1+\fp(x)^2}\, dx - . -

    -     -     - - - -

    - As the integrand contains a square root, - it is often difficult to use the formula in to find the length exactly. - When exact answers are difficult to come by, - we resort to using numerical methods of approximating definite integrals. - The following examples will demonstrate this. -

    - - - Finding arc length - -

    - Find the arc length of f(x) = x^{3/2} from x=0 to x=4. -

    -     - -

    - We find \fp(x)= \frac32x^{1/2}; - note that on [0,4], - f is differentiable and \fp is also continuous. - Using the formula, we find the arc length L as - - L \amp = \int_0^4 \sqrt{1+\left(\frac32x^{1/2}\right)^2}\, dx - \amp = \int_0^4 \sqrt{1+\frac94x} \, dx - \amp = \int_0^4 \left(1+\frac94x\right)^{1/2}\, dx - \amp = \frac23\cdot\frac49\cdot\left(1+\frac94x\right)^{3/2}\Big|_0^4 - \amp =\frac{8}{27}\left(10^{3/2}-1\right) \approx 9.07 \,\text{units} - . -

    - -
    - A graph of f(x) = x^{3/2} from - - - - Graph of the function f(x) = x^{3/2} on the interval between x=0 and x=4. - The curve f(x) = x^{3/2} begins at the point (0,0) from which it heads upwards in a convex arc until reaching the point (4,8). - A straight line plotted between the start and endpoints of the curve would lie entirely above the curve on the interval between x=0 and x=4 and would showcase the shortest distance between the two points. - - Graph of the function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.2,ymax=8.5, - xmin=-.1,xmax=4.5, - ] - - \addplot+ [closed,domain=0:4] {x^(3/2)}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - A graph of f is given in . -

    -     - -     - - - Finding arc length - -

    - Find the arc length of \ds f(x) =\frac18x^2-\ln(x) from x=1 to x=2. -

    -     - -

    - This function was chosen specifically because the resulting integral can be evaluated exactly. - We begin by finding \fp(x) = x/4-1/x. - The arc length is - - L \amp = \int_1^2 \sqrt{1+ \left(\frac x4-\frac1x\right)^2}\, dx - \amp = \int_1^2 \sqrt{1 + \frac{x^2}{16} -\frac12 + \frac1{x^2} } \, dx - \amp = \int_1^2 \sqrt{\frac{x^2}{16} +\frac12 + \frac1{x^2} } \, dx - \amp = \int_1^2 \sqrt{ \left(\frac x4 + \frac1x\right)^2}\, dx - - - \amp = \int_1^2 \left(\frac x4 + \frac1x\right) \, dx - \amp = \left.\left(\frac{x^2}8 + \ln(x)\right)\right|_1^2 - \amp = \frac38+\ln(2) \approx 1.07 \,\text{units} - . -

    - -
    - A graph of f(x) =\frac18x^2-\ln(x) from - - - - Graph of the function f(x) =\frac18x^2-\ln(x). - The curve is highlighted on the interval between x=1 and x=2. - The curve f(x) =\frac18x^2-\ln(x) begins near the point (0.4,1) from which it heads downwards in a convex arc until crossing the x-axis at approximately x=1.25. - The curve then continues in the convex arc, until it once again reaches the x-axis at approximately x=3. - - Graph of the function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.26,ymax=1.2, - xmin=-.1,xmax=3.1, - ] - - \addplot [firstcurvestyle,closed,semithick,domain=.2:3,samples=50] {(1/8)*x^2-ln(x)}; - \addplot [firstcurvestyle,-,very thick,domain=1:2] {(1/8)*x^2-ln(x)}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - A graph of f is given in ; - the portion of the curve measured in this problem is in bold. -

    -     - -     - - - -

    - The previous examples found the arc length exactly through careful choice of the functions. - In general, exact answers are much more difficult to come by and numerical approximations are necessary. -

    - - - Approximating arc length numerically - -

    - Find the length of the sine curve from x=0 to x=\pi. -

    -     - -

    - This is somewhat of a mathematical curiosity; - in - we found the area under one hump - of the sine curve is 2 square units; - now we are measuring its arc length. -

    - -

    - The setup is straightforward: - f(x) = \sin(x) and \fp(x) = \cos(x). - Thus - - L = \int_0^\pi \sqrt{1+\cos^2(x)}\, dx - . -

    - -

    - This integral cannot be evaluated in terms of elementary functions so we will approximate it with Simpson's Method with n=4. -

    - -
    - A table of values of y=\sqrt{1+\cos^2(x) } to evaluate a definite integral in - - - x\sqrt{1+\cos^2(x) } - - - 0\sqrt{2} - - - \pi/4\sqrt{3/2} - - - \pi/21 - - - 3 \pi/4\sqrt{3/2} - - - \pi\sqrt{2} - - -
    - -

    - - gives \sqrt{1+\cos^2(x) } evaluated at 5 evenly spaced points in [0,\pi]. - Simpson's Rule then states that - - \int_0^\pi \sqrt{1+\cos^2(x)}\, dx \amp \approx \frac{\pi-0}{4\cdot 3}\left(\sqrt{2}+4\sqrt{3/2}+2(1)+4\sqrt{3/2}+\sqrt{2}\right) - \amp =3.82918 - . -

    - -

    - Using a computer with n=100 the approximation is L\approx 3.8202; - our approximation with n=4 is quite good. -

    -     -     -     - - - Surface Area of Solids of Revolution -

    - We have already seen how a curve y=f(x) on [a,b] can be revolved around an axis to form a solid. - Instead of computing its volume, - we now consider its surface area. -

    - - - -
    - Establishing the formula for surface area - -
    - - - - - Graph of an arbitrary function y=f(x) on the interval [a,b]. - The curve is a concave arc starting at x=a at some arbitrary y value from which it slopes upwards until ending at x=b at some slightly higher y value. - The plot of the graph also contains a subinterval on the x-axis, given by [x_{i-1},x_{i}]. - A line is drawn through the points (x_{i-1},f(x_{i-1})) and (x_{i},f(x_{i})), which approximates the length of the curve y=f(x) on the interval [x_{i-1},x_{i}]. - - Graph of an arbitrary function on the interval from a to b, with a line approximating a small portion of the curve. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - axis y line=none, - axis lines=center, - y dir=reverse, - view={10}{30}, - xtick=\empty, - ztick=\empty, - extra x ticks={.5,.9,1.4,1.9}, - extra x tick labels={$a$,$x_{i-1}$,$x_{i}$,$b$}, - ymin=-1,ymax=1, - xmin=.25,xmax=2.1, - zmin=-.2,zmax=1.2 - ] - - \addplot3 [firstcurvestyle,closed,domain=.5:1.9,samples y=0] ({x},{0},{(sin(deg(x)))}); - - \draw [secondcolor,thick] (axis cs:.9,0,.78) -- (axis cs:1.4,0,.985); - \draw [fill=black] (axis cs:.9,.0,.78) circle (1pt) (axis cs:1.4,0,.985) circle (1pt); - - \end{axis} - - %\node [right] at (myplot.right of origin)[shift={(-10pt,-6pt)}] { $x$}; - %\node [above] at (myplot.above origin) [shift={(0,-17pt)}] { $y$}; - - \end{tikzpicture} - - - - -
    - -
    - - - - - - Graph of an arbitrary function y=f(x) on the interval [a,b]. - The line drawn through the points (x_{i-1},f(x_{i-1})) and (x_{i},f(x_{i})) is then rotated about the x-axis. - The resulting shape resembles a part of a cone which is lying horizontally, and can be used to approximate the surface area of the function y=f(x) being rotated about the x-axis on the interval [x_{i-1},x_{i}]. - The plot also contains two vertical measurements. - The first vertical measurement is r, which gives the radius of the cone at x=x_{i-1} and the second is R, which gives the radius of the cone at x=x_{i} - The plot also contains an additional measurement L which gives the length of the line connecting f(x_{i-1}) and f(x_{i}). - The measurement L is also the length of the part of the resulting part of a cone that comes from rotating the line about the x-axis. - - Graph of a function with the line approximating the length of a part of the curve being rotated about the x axis. - - - - - //ASY file for figarc4_3D.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((16.7,13.2,46),(0,1,0),(0,0,0),1,(-.0323,0.0012)); - //currentprojection=orthographic((16.7,13.2,46),(0,1,0),(0,0,0),1,(0.0148,0.00673)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.25,2.1); - pair ybounds=(-1.1,1.1); - pair zbounds=(-1.1,1.1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //\addplot3[domain=.9:1.4,y domain=-210:150,samples y=36,surf,shader=flat,colormap={mp2}{\colormapone}] - // ({x},{(.41*(x-.9)+.78)*sin(y)},{(.41*(x-.9)+.78)*cos(y)}); - - //\addplot3[domain=.5:1.9,,samples y=0,{\colorone},] ({x},{0},{(sin(deg(x)))}); - - //\addplot3[domain=0:360,,samples y=0,black,smooth] ({1.4},{.98*cos(x)},{.98*sin(x)}); - - //\addplot3[domain=130:330,,samples y=0,black,dashed,smooth] ({.9},{.78*cos(x)},{.78*sin(x)}); - - pen p=apexmeshpen; - - triple f2(pair t) {return (t.x,(.41*(t.x-.9)+.78)*sin(t.y),(.41*(t.x-.9)+.78)*cos(t.y));} - surface s2=surface(f2,(.9,0),(1.4,2*pi),2,16,Spline); - draw(s2,simplesurfacepen,meshpen=p); - - triple g3(real t) {return (t,sin(t),0);} - path3 p3=graph(g3,.5,1.9,operator ..); - draw(p3,bluepen+.4mm); - - triple g3(real t) {return (.9,.78*sin(t),.78*cos(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.2mm); - - triple g3(real t) {return (1.4,.985*sin(t),.985*cos(t));} - path3 p3=graph(g3,0,2*pi,operator ..); - draw(p3,bluepen+.2mm); - - draw((.9,.78,0)--(1.4,.985,0),redpen+.6mm); - - draw((.5,.05,0)--(.5,-.05,0),black+.2mm); - label("$a$",(.5,-.05,0),S); - - draw((1.9,.05,0)--(1.9,-.05,0),black+.2mm); - label("$b$",(1.9,-.05,0),S); - - draw((.9,.05,0)--(.9,-.05,0),black+.2mm); - label("$x_{i-1}$",(.9,-.05,0),S); - - draw((1.4,.05,0)--(1.4,-.05,0),black+.2mm); - label("$x_{i}$",(1.4,-.05,0),S); - - label(XY*"$\left.\rule{0pt}{40pt}\right\}\ R$",(1.4,.49,0),E,Embedded); - label(XY*"$r\left\{\rule{0pt}{30pt}\right.$",(.85,.37,0),W,Embedded); - label(rotate(20,(0,0,1))*"$\overbrace{\rule{41pt}{0pt}}^{\text{\normalsize \textit{L}}}$",(1.1,1.03,0),Embedded); - - - - -
    -     - -
    - -

    - We begin as we have in the previous sections: - we partition the interval [a,b] with n subintervals, - where the ith subinterval is [x_{i-1},x_{i}]. - On each subinterval, - we can approximate the curve y=f(x) with a straight line that connects f(x_{i-1}) and - f(x_{i}) as shown in . - Revolving this line segment about the x-axis creates part of a cone - (called a frustum of a cone) - as shown in . - The surface area of a frustum of a cone is - - 2\pi\cdot\,\text{length} \,\cdot\,\text{average of the two radii \(R\) and \(r\)} - . -

    - -

    - The length is given by L; - we use the material just covered by arc length to state that - - L\approx \sqrt{1+\fp(c_i)^2}\dx_i - - for some c_i in the ith subinterval. - The radii are just the function evaluated at the endpoints of the interval. - That is, - - R = f(x_{i}) \text{ and } r = f(x_{i-1}) - . -

    - -

    - Thus the surface area of this sample frustum of the cone is approximately - - 2\pi\frac{f(x_{i-1})+f(x_{i})}2\sqrt{1+\fp(c_i)^2}\dx_i - . -

    - -

    - Since f is a continuous function, - the Intermediate Value Theorem states there is some d_i in - [x_{i-1},x_{i}] such that \ds f(d_i) = \frac{f(x_{i-1})+f(x_{i})}2; - we can use this to rewrite the above equation as - - 2\pi f(d_i)\sqrt{1+\fp(c_i)^2}\dx_i - . -

    - -

    - Summing over all the subintervals we get the total surface area to be approximately - - \text{Surface Area}\, \approx \sum_{i=1}^n 2\pi f(d_i)\sqrt{1+\fp(c_i)^2}\dx_i - , - which is a Riemann Sum. - Taking the limit as the subinterval lengths go to zero gives us the exact surface area, - given in the following theorem. -

    - - - Surface Area of a Solid of Revolution - -

    - Let f be differentiable on [a,b], - where \fp is also continuous on [a,b]. - integrationsurface area - surface areasolid of revolution -

    - -

    -

      -
    1. -

      - The surface area of the solid formed by revolving the graph of y=f(x), - where f(x)\geq0, about the x-axis is - - \text{Surface Area}\, = 2\pi\int_a^b f(x)\sqrt{1+\fp(x)^2}\, dx - . -

      -
      2. - -
    2. -

      - The surface area of the solid formed by revolving the graph of y=f(x) about the y-axis, - where a,b\geq0, is - - \text{Surface Area}\, = 2\pi\int_a^b x\sqrt{1+\fp(x)^2}\, dx - . -

      -
      3. -
    -

    -     -     - -

    - (When revolving y=f(x) about the y-axis, - the radii of the resulting frustum are x_{i-1} and x_{i}; - their average value is simply the midpoint of the interval. - In the limit, this midpoint is just x. - This gives the second part of .) -

    - - - - - Finding surface area of a solid of revolution - -

    - Find the surface area of the solid formed by revolving - y=\sin(x) on [0,\pi] around the x-axis, - as shown in . -

    - -
    - Revolving y=\sin(x) on [0,\pi] about the x-axis - - - - - Three dimensional graph of the shape coming from revolving y=\sin(x) on [0,\pi] about the x-axis. - The curve y=\sin(x) is drawn on the interval [0,\pi]. - This concave curve begins at the point (0,0), from which it increases until reaching a maximum at the point (\frac{\pi}{2},1). - The curve then decreases until ending at the x-axis at the point (\pi,0). - The curve is then rotated about the x-axis, which creates the solid of revolution. - This solid has the largest diameter and is symmetric about x=\frac{pi}{2}, from which it shrinks down until closing in on itself at x=0 and x=\pi. - - Three dimensional graph of the sine curve being rotated about the x axis. - - - - - //ASY file for figarc4_3D.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((16.7,13.2,46),(0,1,0),(0,0,0),1,(-.0323,0.0012)); - //currentprojection=orthographic((16.7,13.2,46),(0,1,0),(0,0,0),1,(0.0148,0.00673)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={-1,1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.1,3.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //\addplot3[domain=.9:1.4,y domain=-210:150,samples y=36,surf,shader=flat,colormap={mp2}{\colormapone}] - // ({x},{(.41*(x-.9)+.78)*sin(y)},{(.41*(x-.9)+.78)*cos(y)}); - - //\addplot3[domain=.5:1.9,,samples y=0,{\colorone},] ({x},{0},{(sin(deg(x)))}); - - //\addplot3[domain=0:360,,samples y=0,black,smooth] ({1.4},{.98*cos(x)},{.98*sin(x)}); - - //\addplot3[domain=130:330,,samples y=0,black,dashed,smooth] ({.9},{.78*cos(x)},{.78*sin(x)}); - - pen p=apexmeshpen+.1mm; - - triple f2(pair t) {return (t.x,sin(t.x)*sin(t.y),sin(t.x)*cos(t.y));} - surface s2=surface(f2,(0,0),(pi,2*pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - draw(s2,simplesurfacepen,meshpen=p); - - triple g3(real t) {return (t,sin(t),0);} - path3 p3=graph(g3,0,pi,32,operator ..); - draw(p3,bluepen+.6mm); - - draw((pi,.05,0)--(pi,-.05,0),black+.2mm); - label("$\pi$",(pi,-.05,0),S); - - - - -
    -     - -

    - The setup is relatively straightforward. - Using , - we have the surface area SA is: - - SA \amp = 2\pi\int_0^\pi \sin(x) \sqrt{1+\cos^2(x) }\, dx - \amp = -2\pi\frac12\left.\left(\sinh^{-1}(\cos(x) )+\cos(x) \sqrt{1+\cos^2(x) }\right)\right|_0^\pi - \amp = 2\pi\left(\sqrt{2}+\sinh^{-1}(1) \right) - \amp \approx 14.42\,\text{units}^2 - . -

    - -

    - The integration step above is nontrivial, - utilizing the integration method of Trigonometric Substitution from . -

    - -

    - It is interesting to see that the surface area of a solid, - whose shape is defined by a trigonometric function, - involves both a square root and an inverse hyperbolic trigonometric function. -

    -     - -     - - - Finding surface area of a solid of revolution - -

    - Find the surface area of the solid formed by revolving the curve y=x^2 on [0,1] about: -

    - -

    -

      -
    1. -

      - the x-axis -

      -
      2. - -
    2. -

      - the y-axis. -

      -
      3. -
    -

    - -
    - The solids used in - - -
    - - - - - Three dimensional graph of the shape coming from revolving y=x^2 on [0,1] about the x-axis. - The quadratic function y=x^2 is drawn on the interval [0,1]. - This curve begins at the point (0,0), from which it quadratically increases until reaching a maximum at the point (1,1). - The curve is then rotated about the x-axis, creating a solid of revolution. - This shape has a circular vertical cross-section which has a radius of r(x)=x^2 and is hollow on the inside. - The shape also is not closed off on its rightmost boundary at x=1. - - Three dimensional graph of the quadratic function being rotated about the x axis. - - - - - //ASY file for figarc4_3D.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((7.4,20,47),(0,1,0),(0,0,0),1,(-.004432,-0.00537)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={-1,1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.1,1.5); - pair ybounds=(-1.1,1.1); - pair zbounds=(-1.1,1.1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - pen p=apexmeshpen+.1mm; - - triple f2(pair t) {return (t.x,t.x^2*sin(t.y),t.x^2*cos(t.y));} - surface s2=surface(f2,(0,0),(1,2*pi),8,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - draw(s2,simplesurfacepen,meshpen=p); - - triple g3(real t) {return (t,t^2,0);} - path3 p3=graph(g3,0,1,operator ..); - draw(p3,bluepen+.6mm); - - - - -
    - - -
    - - - - - Three dimensional graph of the shape coming from revolving y=x^2 on [0,1] about the y-axis. - The quadratic function y=x^2 is drawn on the interval [0,1]. - The curve is then rotated about the y-axis, creating a solid of revolution. - This shape has a circular horizontal cross-section, which has a radius of r(x)=x and is hollow on the inside. - The shape also is not closed off on its top boundary at y=1. - - Three dimensional graph of the quadratic function being rotated about the y axis. - - - - - //ASY file for figarc4_3D.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((10.4,3.7,64),(0,1,0),(0,0,0),.95,(0.00404,0.00136)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,-1}; - real[] myychoice={-1,1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(-.1,1.1); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - pen p=apexmeshpen+.1mm; - - triple f2(pair t) {return (t.x*sin(t.y),t.x^2,t.x*cos(t.y));} - surface s2=surface(f2,(0,0),(1,2*pi),8,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - draw(s2,simplesurfacepen,meshpen=p); - - triple g3(real t) {return (t,t^2,0);} - path3 p3=graph(g3,0,1,operator ..); - draw(p3,bluepen+.6mm); - - - - -
    -     -
    -     - -

    -

      -
    1. -

      - The integral is straightforward to setup: - - SA \amp = 2\pi\int_0^1 x^2\sqrt{1+(2x)^2}\, dx. - Like the integral in , this requires Trigonometric Substitution. - \amp = \left.\frac{\pi}{32}\left(2(8x^3+x)\sqrt{1+4x^2}-\sinh^{-1}(2x)\right)\right|_0^1 - \amp =\frac{\pi}{32}\left(18\sqrt{5}-\sinh^{-1}(2) \right) - \amp \approx 3.81\,\text{units}^2 - . - The solid formed by revolving y=x^2 around the x-axis is graphed in . -

      -
      2. - -
    2. -

      - Since we are revolving around the y-axis, - the radius of the solid is not f(x) but rather x. - Thus the integral to compute the surface area is: - - SA \amp = 2\pi\int_0^1x\sqrt{1+(2x)^2}\, dx. - This integral can be solved using substitution. Set u=1+4x^2; the new bounds are u=1 to u=5. We then have - \amp = \frac{\pi}4\int_1^5 \sqrt{u}\, du - \amp = \left.\frac{\pi}{4}\frac23 u^{3/2}\right|_1^5 - \amp = \frac{\pi}6\left(5\sqrt{5}-1\right) - \amp \approx 5.33\,\text{units}^2 - . - The solid formed by revolving y=x^2 about the y-axis is graphed in . -

      -
      3. -
    -

    -     - -     - -

    - Our final example is a famous mathematical paradox. -

    - - - The surface area and volume of Gabriel's Horn - -

    - Consider the solid formed by revolving y=1/x about the x-axis on [1,\infty). - Find the volume and surface area of this solid. - (This shape, - as graphed in , - is known as Gabriel's Horn - since it looks like a very long horn that only a supernatural person, - such as an angel, could play.) - Gabriel's Horn -

    - -
    - A graph of Gabriel's Horn - - - - - Three dimensional graph of the shape coming from revolving y=1/x on [1,\infty) about the x-axis. - The function y=1/x is drawn on the interval [1,\infty). - The curve is then rotated about the x-axis, creating the shape called Gabriel's Horn. - This shape has a circular vertical cross-section, which has a radius of r(x)=1/x and is hollow on the inside. - The shape also is not closed off on its leftmost boundary at x=1. - For calculating the volume, we consider how much space is enclosed by x=1 and Gabriel's Horn itself. - - Three dimensional graph of Gabriel's Horn. - - - - - //ASY file for figgabriel_3D.asy in Chapter 7 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((372,29,113),(0,1,0),(0,0,0),.95,(-0.024,-0.0032)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={5,10,15}; - real[] myychoice={-1,1}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-1,17); - pair ybounds=(-1.3,1.3); - pair zbounds=(-1.3,1.3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - pen p=apexmeshpen+.1mm; - - triple f2(pair t) {return (t.x,1/t.x*sin(t.y),1/t.x*cos(t.y));} - surface s2=surface(f2,(1,0),(15,2*pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - draw(s2,simplesurfacepen,meshpen=p); - - triple g3(real t) {return (t,1/t,0);} - path3 p3=graph(g3,1,15,16,operator ..); - draw(p3,bluepen+.6mm); - - draw((0,-2,0)--(0,2,0),invisible); - - - - -
    -     - -

    - To compute the volume it is natural to use the Disk Method. - We have: - - V \amp = \pi\int_1^\infty \frac{1}{x^2}\, dx - \amp = \lim_{b\to\infty}\pi\int_1^b\frac{1}{x^2}\, dx - \amp = \lim_{b\to\infty} \left.\pi\left(\frac{-1}{x}\right)\right|_1^b - \amp = \lim_{b\to\infty} \pi\left(1-\frac1b\right) - \amp = \pi \,\text{units}^3 - . -

    - -

    - Gabriel's Horn has a finite volume of \pi cubic units. - Since we have already seen that regions with infinite length can have a finite area, - this is not too difficult to accept. -

    - -

    - We now consider its surface area. - The integral is straightforward to setup: - - SA \amp = 2\pi\int_1^\infty \frac{1}{x}\sqrt{1+1/x^4}\, dx. - Integrating this expression is not trivial. We can, however, compare it to other improper integrals. Since 1\lt \sqrt{1+1/x^4} on [1,\infty), we can state that - 2\pi\int_1^\infty \frac{1}{x}\, dx \amp \lt 2\pi\int_1^\infty \frac{1}{x}\sqrt{1+1/x^4}\, dx - . -

    - -

    - By , - the improper integral on the left diverges. - Since the integral on the right is larger, - we conclude it also diverges, - meaning Gabriel's Horn has infinite surface area. -

    - -

    - Hence the paradox: - we can fill Gabriel's Horn with a finite amount of paint, - but since it has infinite surface area, we can never paint it. -

    - -

    - Somehow this paradox is striking when we think about it in terms of volume and area. - However, we have seen a similar paradox before, as referenced above. - We know that the area under the curve y=1/x^2 on [1,\infty) is finite, - yet the shape has an infinite perimeter. - Strange things can occur when we deal with the infinite. -

    -     - -     - -

    - A standard equation from physics is - Work = force distance, - when the force applied is constant. - In we learn how to compute work when the force applied is variable. -

    -     - - - - Terms and Concepts - - -

    - The integral formula for computing arc length was found by first approximating arc length with straight line segments. -

    -     - -     - - - -

    - The integral formula for computing arc length includes a square root, - meaning the integration is probably easy. -

    -     - -     -     - - Problems - - - -

    - Find the arc length of the function on the given interval. -

    -     - - - - - $f = Formula("x"); - package my::Function::numeric; - our @ISA = ('Parser::Function::numeric'); - sub sqrt { - my $self = shift; - my $value = $self->context->flag("setSqrt"); - return $value if $value; - return $self->SUPER::sqrt(@_); - } - package main; - Context("Numeric"); - Context()->functions->set(sqrt=>{class=>'my::Function::numeric'}); - Context()->flags->set(reduceConstantFunctions=>0); - - $ans = Compute("sqrt(2)"); - - -

    - \ds f(x) = on [0, 1]. -

    - -

    - -

    -     - -

    - \sqrt{2} -

    -     -     -     - - - - - - -

    - \ds f(x) = \sqrt{8}x on [-1, 1]. -

    -

    - -

    -     -     -     - - - - - - -

    - \ds f(x) = \frac13x^{3/2}-x^{1/2} on [0,1]. -

    - -

    - -

    -     -     -     - - - - - - -

    - \ds f(x) = \frac1{12}x^{3}+\frac1x on [1,4]. -

    - -

    - -

    -     -     -     - - - - - - -

    - \ds f(x) = 2x^{3/2}-\frac16\sqrt{x} on [1,4]. -

    -

    - -

    -     -     -     - - - - - - -

    - \ds f(x) = \cosh(x) on [-\ln(2) , \ln(2) ]. -

    - -

    - -

    -     -     -     - - - - - - -

    - \ds f(x) = \frac12\big(e^x+e^{-x}\big) on [0, \ln(5) ]. -

    - -

    - -

    -     -     -     - - - - - - -

    - \ds f(x) = \frac1{12}x^5+\frac1{5x^3} on [0.1, 1]. -

    - - -     -     -     - - - - -

    - \ds f(x) = \ln\big(\sin(x) \big) on [\pi/6, \pi/2]. -

    -     - -

    - -\ln(2-\sqrt{3}) \approx 1.31696 -

    -     - -     - - - - -

    - \ds f(x) = \ln\big(\cos(x) \big) on [0, \pi/4]. -

    -     - -

    - \sinh^{-1}(1) -

    -     - -     - -     - - - -

    - Set up the integral to compute the arc length of the function on the given interval. - Do not evaluate the integral. -

    -     - - - - -

    - \ds f(x) = x^2 on [0, 1]. -

    -     - -

    - \int_0^1 \sqrt{1+4x^2}\, dx -

    -     - -     - - - - -

    - \ds f(x) = x^{10} on [0, 1]. -

    -     - -

    - \int_0^1 \sqrt{1+100x^{18}}\, dx -

    -     - -     - - - - -

    - \ds f(x) = \ln(x) on [1, e]. -

    -     - -

    - \int_1^e \sqrt{1+\frac1{x^2}}\, dx -

    -     - -     - - - - -

    - \ds f(x) = \frac1x on [1,2]. -

    -     - -

    - \int_{1}^2 \sqrt{1+\frac1{x^4}}\, dx -

    -     - -     - - - - -

    - \ds f(x) = cos(x) on [0,\pi/2]. -

    -     - -

    - \int_0^{\pi/2}\sqrt{1+\sin^2(x)}\,dx -

    -     -     - - - - -

    - \ds f(x) = \sec(x) on [-\pi/4,\pi/4]. -

    -     - -

    - \int_{-\pi/4}^{\pi/4} \sqrt{1+\sec^2(x) \tan^2(x) }\, dx -

    -     - -     -     - - - -

    - Use Simpson's Rule, with n=4, - to approximate the arc length of the function on the given interval. - Note: these are the same problems as in Exercises. -

    -     - - - - -

    - \ds f(x) = x^2 on [0, 1]. -

    -     - -

    - 1.4790 -

    -     - -     - - - - -

    - \ds f(x) = x^{10} on [0, 1]. -

    -     - -

    - 1.8377 -

    -     - -     - - - - -

    - \ds f(x) = \ln(x) on [1, e]. -

    -     - -

    - 2.1300 -

    -     - -     - - - - -

    - \ds f(x) = \frac1x on [1,2]. -

    -     - -

    - 1.3254 -

    -     - -     - - - - -

    - \ds f(x) = \cos(x) on [0, \pi/2]. -

    -     - -

    - 1.00013 -

    -     - -     - - - - -

    - \ds f(x) = \sec(x) on [-\pi/4,\pi/4]. -

    -     - -

    - 1.7625 -

    -     - -     - -     - - - -

    - Find the surface area of the described solid of revolution. -

    -     - - - - -

    - The solid formed by revolving y=2x on [0,1] about the x-axis. -

    -     - -

    - 2\pi\int_0^1 2x\sqrt{5}\, dx = 2\pi\sqrt{5} -

    -     - -     - - - - -

    - The solid formed by revolving y=2x on [0,1] about the y-axis. -

    -     - -

    - 2\pi\int_0^1 x\sqrt{5}\, dx = \pi\sqrt{5} -

    -     - -     - - - - -

    - The solid formed by revolving y=x^2 on [0,1] about the y-axis. -

    -     - -

    - 2\pi\int_0^1 x\sqrt{1+4x^2}\, dx = \pi/6(5\sqrt{5}-1) -

    -     - -     - - - - -

    - The solid formed by revolving y=x^3 on [0,1] about the x-axis. -

    -     - -

    - 2\pi\int_0^1 x^3\sqrt{1+9x^4}\, dx = \pi/27(10\sqrt{10}-1) -

    -     - -     -     - - - -

    - The following arc length and surface area problems lead to improper integrals. - Although the hypotheses of and - are not satisfied, the improper integrals converge, - and formulas for arc length and surface area still give the correct result. -

    -     - - - - -

    - Find the length of the curve \ds f(x) = \sqrt{x} on [0, 1]. - (Note: this is the same as the length of f(x)=x^2 on [0,1]. Why?) -

    -     - -

    - \int_0^1 \sqrt{1+\frac{1}{4x}}\, dx -

    -     - -     - - - - -

    - Find the length of the curve - \ds f(x) = \sqrt{1-x^2} on [-1, 1]. - (Note: this describes the top half of a circle with radius 1.) -

    -     - -

    - \int_{-1}^1 \sqrt{1+\frac{x^2}{1-x^2}}\, dx -

    -     - -     - - - - -

    - Find the length of the curve \ds f(x) = \sqrt{1-x^2/9} on [-3, 3]. - (Note: this describes the top half of an ellipse with a major axis of length 6 and a minor axis of length 2.) -

    -     - -

    - \int_{-3}^3 \sqrt{1+\frac{x^2}{81-9x^2}}\, dx -

    -     - -     - - - - -

    - Find the surface area of the solid formed by revolving - y=\sqrt{x} on [0,1] about the x-axis. -

    -     - -

    - 2\pi\int_0^1 \sqrt{x}\sqrt{1+1/(4x)}\, dx = \pi/6(5\sqrt{5}-1) -

    -     - -     - - - - -

    - Find the surface area of the sphere formed by revolving - y=\sqrt{1-x^2} on [-1,1] about the x-axis. -

    -     - -

    - 2\pi\int_0^1 \sqrt{1-x^2}\sqrt{1+x/(1-x^2)}\, dx = 4\pi -

    -     - -     -     -     -     -
    -
    - Work - -

    - Work is the scientific term used to describe the action of a force which moves an object. - When a constant force F is applied to move an object a distance d, - the amount of work performed is W=F\cdot d. -

    - -

    - The SI unit of force is the newton; one newton is equal to one - , - and the SI unit of distance is a meter (m). - The fundamental unit of work is one newtonmeter, - or a joule (J). - That is, applying a force of one newton for one meter performs one joule of work. - In Imperial units - (as used in the United States), - force is measured in pounds (lb) and distance is measured in feet (ft), - hence work is measured in ftlb. -

    - -

    - When force is constant, the measurement of work is straightforward. - For instance, lifting a 200 - object 5 performs - 200\cdot 5 = 1000 ftlb of work. -

    - -

    - What if the force applied is variable? - For instance, - imagine a climber pulling a 200 rope up a vertical face. - The rope becomes lighter as more is pulled in, - requiring less force and hence the climber performs less work. -

    -     - - - Work Done by a Variable Force -

    - In general, let F(x) be a force function on an interval [a,b]. - We want to measure the amount of work done applying the force F from x=a to x=b. - We can approximate the amount of work being done by partitioning [a,b] into subintervals - a=x_0\lt x_1 \lt \cdots \lt x_{n}=b and assuming that F is constant on each subinterval. - Let c_i be a value in the - ith subinterval [x_{i-1},x_{i}]. - Then the work done on this interval is approximately W_i\approx F(c_i)\cdot(x_{i}-x_{i-1}) = F(c_i)\dx_i, - a constant force the distance over which it is applied. - The total work is - - W = \sum_{i=1}^n W_i \approx \sum_{i=1}^n F(c_i)\dx_i - . -

    - -

    - This, of course, is a Riemann sum. - Taking a limit as the subinterval lengths go to zero gives an exact value of work which can be evaluated through a definite integral. -

    - - - Work -

    - Let F(x) be a continuous function on [a,b] describing the amount of force being applied to an object in the direction of travel from distance x=a to distance x=b. - The total work W done on [a,b] is - integrationwork - work - - W = \int_a^b F(x)\, dx - . -

    -     - - - Computing work performed: applying variable force - -

    - A 60 climbing rope is hanging over the side of a tall cliff. - How much work is performed in pulling the rope up to the top, - where the rope has a linear mass density of - 66? -

    -     - -

    - We need to create a force function F(x) on the interval [0,60]. - To do so, we must first decide what x is measuring: - is it the length of the rope still hanging or is it the amount of rope pulled in? - As long as we are consistent, either approach is fine. - We adopt for this example the convention that x is the amount of rope pulled in. - This seems to match intuition better; - pulling up the first 10 meters of rope involves x=0 to x=10 instead of x=60 to x=50. -

    - -

    - As x is the amount of rope pulled in, - the amount of rope still hanging is 60-x. - This length of rope has a mass of 66 or - 0.066. - The mass of the rope still hanging is 0.066(60-x) ; - multiplying this mass by the acceleration of gravity, - 9.8, - gives our variable force function - - F(x) = (9.8)(0.066)(60-x) = 0.6468(60-x) - . -

    - -

    - Thus the total work performed in pulling up the rope is - - W = \int_0^{60} 0.6468(60-x)\, dx = 1,164.24 \,\text{J} - . -

    - -

    - By comparison, - consider the work done in lifting the entire rope 60 meters. - The rope weighs 60\times 0.066 \times 9.8 = 38.808 N, so the work applying this force for 60 meters is - 60\times 38.808 = 2,328.48 J. This is exactly twice the work calculated before (and we leave it to the reader to understand why.) -

    -     -     - - - Computing work performed: applying variable force - -

    - Consider again pulling a 60 rope up a cliff face, - where the rope has a mass of 66. - At what point is exactly half the work performed? -

    -     - -

    - From - we know the total work performed is 1,164.24 J. - We want to find a height h such that the work in pulling the rope from a height of x=0 - to a height of x=h is 582.12, or half the total work. - Thus we want to solve the equation - - \int_0^h 0.6468(60-x)\, dx = 582.12 - - for h. - - \int_0^h 0.6468(60-x)\, dx \amp = 582.12 - \left(38.808x-0.3234x^2\right)\Big|_0^h \amp =582.12 - 38.808h-0.3234h^2 \amp =582.12 - -0.3234h^2+38.808h-582.12 \amp =0 . - Apply the Quadratic Formula: - h\amp =17.57 \,\text{ and } \,102.43 - -

    - -

    - As the rope is only 60 long, - the only sensible answer is h=17.57. - Thus about half the work is done pulling up the first - 17.57; - the other half of the work is done pulling up the remaining - 42.43. -

    - - -     -     - - - Computing work performed: applying variable force - -

    - A box of 100 of sand is being pulled up at a uniform rate a distance of - 50 over 1 minute. - The sand is leaking from the box at a rate of 1. - The box itself weighs 5 and is pulled by a rope weighing - 0.2. -

    - -

    -

      -
    1. -

      - How much work is done lifting just the rope? -

      -
      2. - -
    2. -

      - How much work is done lifting just the box and sand? -

      -
      3. - -
    3. -

      - What is the total amount of work performed? -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - We start by forming the force function F_r(x) for the rope - (where the subscript denotes we are considering the rope). - As in the previous example, - let x denote the amount of rope, in feet, pulled in. - (This is the same as saying x denotes the height of the box.) - The weight of the rope with x feet pulled in is F_r(x) = 0.2(50-x) = 10-0.2x. - (Note that we do not have to include the acceleration of gravity here, - for the weight of the rope per foot is given, - not its mass per meter as before.) The work performed lifting the rope is - - W_r = \int_0^{50} (10-0.2x)\, dx = 250\,\text{ft--lb} - . -

      -
      2. - -
    2. -

      - The sand is leaving the box at a rate of - 1. - As the vertical trip is to take one minute, - we know that 60 - will have left when the box reaches its final height of - 50. - Again letting x represent the height of the box, - we have two points on the line that describes the weight of the sand: - when x=0, the sand weight is - 100, - producing the point (0,100); - when x=50, the sand in the box weighs - 40, - producing the point (50,40). - The slope of this line is \frac{100-40}{0-50} = -1.2, - giving the equation of the weight of the sand at height x as w(x) = -1.2x+100. - The box itself weighs a constant - 5, - so the total force function is F_b(x) = -1.2x+105. - Integrating from x=0 to x=50 gives the work performed in lifting box and sand: - - W_b = \int_0^{50} (-1.2x+105)\, dx = 3750\,\text{ft--lb} - . -

      -
      3. - -
    3. -

      - The total work is the sum of W_r and W_b: - 250+3750=4000 ftlb. - We can also arrive at this via integration: - - W \amp = \int_0^{50} (F_r(x)+F_b(x))\, dx - \amp = \int_0^{50} (10-0.2x-1.2x+105)\, dx - \amp = \int_0^{50} (-1.4x+115) \, dx - \amp = 4000 \,\text{ft--lb} - . -

      -
      4. -
    -

    -     -     -     - - - Hooke's Law and Springs - -

    - Hooke's Law states that the force required to compress or stretch a spring x - units from its natural length is proportional to x; - that is, this force is F(x) = kx for some constant k. - For example, if a force of 1 - stretches a given spring 2, - then a force of 5 - will stretch the spring 10. - Converting the distances to meters, - we have that stretching this spring 0.02 - requires a force of F(0.02) = k(0.02) = 1 - , - hence k = 1/0.02 = 50 - . - Hooke's Law -

    - - - Computing work performed: stretching a spring - -

    - A force of 20 - stretches a spring from a natural length of 7 inches to a length of 12 inches. - How much work was performed in stretching the spring to this length? -

    -     - -

    - In many ways, - we are not at all concerned with the actual length of the spring, - only with the amount of its change. - Hence, we do not care that 20 - of force stretches the spring to a length of 12 inches, - but rather that a force of - 20 - stretches the spring by 5 inches. - This is illustrated in ; - we only measure the change in the spring's length, - not the overall length of the spring. -

    - - -
    - Illustrating the important aspects of stretching a spring in computing work in - - - Image of a spring showing the spring both before and after streching. - In the first illustration, we see the unstreched spring with a block attached to it on its right side, with a rightward facing force vector labeled F, showing the direction of the force that is applied by the spring. - On the x-axis, the block that the spring is going to move is pictured to be centered at x=1. - In the second illustration, we see the streched spring, with the same block still attached on the right side. - This time, the block is centered at x=6, showing the spring moved the block a total of 5 inches to the right. - - Image showing the important aspects of strenching a spring which are used in computing work. - - - \begin{tikzpicture} - - \begin{scope}[xscale=.2,yscale=.5,shift={(.5,0)}] - \draw [thick,firstcolor] (-1,.5) -- (-.5,.5) -- (0,0); - - \foreach \x in {0,1,...,5} - { - \begin{scope}[shift={(\x*2,0)}] - \draw [thick,firstcolor] (0,0) -- (1,1) -- (2,0); - \end{scope} - } - - \begin{scope}[shift={(12,0)}] - \draw [thick,firstcolor] (0,0) -- (1,1) -- (1.5,.5) -- (2,.5); - \end{scope} - - \end{scope} - - \draw [shift={(.9,0)},thick,left color=firstcolor!70,right color=firstcolor!15,firstcolor] (2,0) rectangle (2.5,.5); - \draw [shift={(.9,0)},->,>=stealth] (2.75,.15)-= node [pos=.5,above] {$F$} (3.25,.15); - - \begin{scope}[shift={(0,-.25)}] - \draw [thick] (-.5,0) -- (5.5,0); - - \foreach \x in {0,...,6} - { - \draw (\x*.4+2.8,.1) -- (\x*.4+2.8,-.1) node [below] { $\x$}; - } - \end{scope} - - \begin{scope}[shift={(0,-1.75)}] - - \begin{scope}[xscale=.343,yscale=.5,shift={(.5,0)}] - \draw [thick,firstcolor] (-1,.5) -- (-.5,.5) -- (0,0); - \foreach \x in {0,1,...,5} - { - \begin{scope}[shift={(\x*2,0)}] - \draw [thick,firstcolor] (0,0) -- (1,1) -- (2,0); - \end{scope} - } - - \begin{scope}[shift={(12,0)}] - \draw [thick,firstcolor] (0,0) -- (1,1) -- (1.5,.5) -- (2,.5); - \end{scope} - \end{scope} - - \begin{scope}[shift={(0,-.25)}] - \draw [thick] (-.5,0) -- (5.5,0); - - \foreach \x in {0,...,6} - { - \draw (\x*.4+2.8,.1) -- (\x*.4+2.8,-.1) node [below] { $\x$}; - } - \end{scope} - - \draw [shift={(2.9,0)},thick,left color=firstcolor!70,right color=firstcolor!15,firstcolor] (2,0) rectangle (2.5,.5); - - \end{scope} - - \end{tikzpicture} - - - - -
    - -

    - Converting the units of length to feet, we have - - F(5/12) = 5/12k = 20\,\text{lb} - . -

    - -

    - Thus k = 48 and F(x) = 48x. -

    - -

    - We compute the total work performed by integrating F(x) from x=0 to x=5/12: - - W \amp = \int_0^{5/12} 48x \, dx - \amp = 24x^2\Big|_0^{5/12} - \amp = 25/6 \approx 4.1667\,\text{ft--lb} - . -

    -     -     -     - - - Pumping Fluids -

    - Another useful example of the application of integration to compute work comes in the pumping of fluids, - often illustrated in the context of emptying a storage tank by pumping the fluid out the top. - This situation is different than our previous examples for the forces involved are constant. - After all, the force required to move one cubic foot of water (about 62.4 ) - is the same regardless of its location in the tank. - What is variable is the distance that cubic foot of water has to travel; - water closer to the top travels less distance than water at the bottom, - producing less work. -

    - - - Weight and Mass densities - - - Fluid - lb/ft^3 - kg/m^3 - - - - - - - - Concrete - 150 - 2400 - - - Fuel Oil - 55.46 - 890.13 - - - Gasoline - 45.93 - 737.22 - - - Iodine - 307 - 4927 - - - Methanol - 49.3 - 791.3 - - - Mercury - 844 - 13546 - - - Milk - 63.665.4 - 10201050 - - - Water - 62.4 - 1000 - - - -
    - -

    - We demonstrate how to compute the total work done in pumping a fluid out of the top of a tank in the next two examples. -

    - - - Computing work performed: pumping fluids - -

    - A cylindrical storage tank with a radius of - 10 - and a height of 30 - is filled with water, which weighs approximately - 62.4. - Compute the amount of work performed by pumping the water up to a point 5 feet above the top of the tank. -

    -     - -

    - We will refer often to - which illustrates the salient aspects of this problem. -

    - -
    - Illustrating a water tank in order to compute the work required to empty it in - - - - Image of a cylindrical storage tank with a radius of 10 and a height of 30ft. - On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank. - The image also contains the ith subinterval of y, which corresponds to the shaded region between y_{i-1} and y_i on the y-axis. - The height of this subinterval of y given by y_{i}-y_{i-1} is also labeled \Delta y_i. - Additionally, the distance from the top of this subinterval, occuring at y_{i} and the point 5 feet above the tank is given on the y-axis as 35-y_i. - - Illustration of a cylindrical water tank with measurements used to compute the work required to empty it. - - - \begin{tikzpicture}[x=.1cm,y=.125cm,>=stealth] - - \draw [->] (0,-2) -- (0,37) node [above] { $y$}; - - \draw (-1,0) node [left] { 0} -- (1,0) - (-1,30) node [left] { 30} -- (1,30) - (-1,35) node [left] { 35} -- (1,35); - - \draw (2,12) -- (4,12) (2,35) -- (4,35) - (3,12) -- node [pos=2,rotate=90] { $35-y_i$} (3,18) - (3,29) -- (3,35); - - \begin{scope}[xscale=4,shift={(4.5,0)}] - - \draw (0,30) -- node [pos=.5,above,shift={(-4pt,-1pt)}] { $10$} (2.9,29.22); - \draw [thick] (3,30) --(3,0) arc (0:-180:3) -- (-3,30); - \draw [thick](0,30) circle (3); - \draw [thick,dashed] (3,0) arc (0:180:3); - - \foreach \y/\x in {12/{$y_{i-1}$},16/{$y_{i}$}} - { - \draw (3,\y) node [right] {\x} arc (0:-180:3); - \draw [dashed] (3,\y) arc (0:180:3); - } - - \draw (5.9,14) node { $\left.\rule{0pt}{.3cm}\right\}\Delta y_i$}; - \draw [left color=firstcolor,right color=firstcolor!15,thick] (0,16) circle (3); - \draw [left color=firstcolor,right color=firstcolor!15,thick] (-3,16) -- (-3,12) arc (180:360:3) -- (3,16) arc (360:180:3); - - \end{scope} - - \end{tikzpicture} - - - - -
    - -

    - We start as we often do: we partition an interval into subintervals. - We orient our tank vertically since this makes intuitive sense with the base of the tank at y=0. - Hence the top of the water is at y=30, - meaning we are interested in subdividing the y-interval [0,30] into n subintervals as - - 0 = y_0 \lt y_1 \lt \cdots \lt y_{n} = 30 - . -

    - -

    - Consider the work W_i of pumping only the water residing in the ith subinterval, - illustrated in . - The force required to move this water is equal to its weight which we calculate as volume density. - The volume of water in this subinterval is V_i = 10^2\pi \Delta y_i; - its density is 62.4. - Thus the required force is 6240\pi\Delta y_i . -

    - -

    - We approximate the distance the force is applied by using any y-value contained in the ith subinterval; - for simplicity, we arbitrarily use y_i for now - (it will not matter later on). - The water will be pumped to a point 5 feet above the top of the tank, that is, - to the height of y=35 . - Thus the distance the water at height y_i travels is 35-y_i . -

    - -

    - In all, the approximate work W_i performed in moving the water in the - ith subinterval to a point 5 feet above the tank is - - W_i \approx 6240\pi\Delta y_i(35-y_i) - . -

    - -

    - To approximate the total work performed in pumping out all the water from the tank, - we sum all the work W_i performed in pumping the water from each of the n subintervals of [0,30]: - - W \approx \sum_{i=1}^n W_i = \sum_{i=1}^n 6240\pi\Delta y_i(35-y_i) - . -

    - -

    - This is a Riemann sum. - Taking the limit as the subinterval length goes to 0 gives - - W \amp = \int_0^{30} 6240\pi(35-y)\, dy - \amp = 6240\pi\left(35y-1/2y^2\right)\Big|_0^{30} - \amp = 11,762,123 \,\text{ft--lb} - \amp \approx 1.176 \times 10^7 \,\text{ft--lb} - . -

    -     -     - -

    - We can streamline the above process a bit as we may now recognize what the important features of the problem are. - - shows the tank from - without the ith subinterval identified. -

    - -
    - A simplified illustration for computing work - - - - Image of the same cylindrical storage tank with a radius of 10 and a height of 30. - On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank. - Now instead of containing the ith subinterval of y, the image contains an arbitrarily chosen point y which is contained in the tank. - At this point y, the volume is given as V(y)=100\pi dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y. - Additionally, the distance from the arbitrary point y and the point 5 feet above the tank is given on the y-axis as 35-y. - - Illustration of a cylindrical water tank with simplified measurements used to compute the work required to empty it. - - - \begin{tikzpicture}[x=.1cm,y=.125cm,>=stealth] - - \draw [->] (0,-2) -- (0,37) node [above] { $y$}; - - \draw (-1,0) node [left] { 0} -- (1,0) - (-1,30) node [left] { 30} -- (1,30) - (-1,35) node [left] { 35} -- (1,35) - (-1,14) node [left] { $y$} -- (1,14); - - \draw (2,12) -- (4,12) (2,35) -- (4,35) - (3,12) -- node [pos=2,rotate=90] { $35-y_i$} (3,18) - (3,29) -- (3,35); - - \begin{scope}[xscale=4,shift={(4.5,0)}] - - \draw (0,30) -- node [pos=.5,above,shift={(-4pt,-1pt)}] { $10$} (2.9,29.22); - \draw [thick] (3,30) --(3,0) arc (0:-180:3) -- (-3,30); - \draw [thick] (0,30) circle (3); - \draw [thick,dashed] (3,0) arc (0:180:3); - \draw [left color=firstcolor,right color=firstcolor!15,thick,draw=firstcolor] (0,14) circle (3); - \draw [->] (4,16) node [above] { $V(y)=100\pi dy$} -- (2,14); - - \end{scope} - - \end{tikzpicture} - - - - -
    - -

    - Instead, we just draw one differential element. - This helps establish the height a small amount of water must travel along with the force required to move it - (where the force is volume density). -

    - -

    - We demonstrate the concepts again in the next examples. -

    - - - Computing work performed: pumping fluids - -

    - A conical water tank has its top at ground level and its base 10 feet below ground. - The radius of the cone at ground level is 2. - It is filled with water weighing 62.4 and is to be emptied by pumping the water to a spigot 3 feet above ground level. - Find the total amount of work performed in emptying the tank. -

    -     - -

    - The conical tank is sketched in . - We can orient the tank in a variety of ways; - we could let y=0 represent the base of the tank and y=10 represent the top of the tank, - but we choose to keep the convention of the wording given in the problem and let y=0 represent ground level and hence y=-10 represents the bottom of the tank. - The actual height of the water does not matter; - rather, we are concerned with the distance the water travels. -

    - -
    - A graph of the conical water tank in - - - - Image of a conical storage tank with its tip 10 below ground and the base at ground level with a radius of 2. - This means as we move further below ground level, the area of the circular cross-section of the conical storage tank decreases. - On the left side of the storage tank the y-axis is shown with measurements of y=3 which is 3 feet above the tank, y=0 which is the top of the tank, and y=-10 which is the tip of the cone, 10 feet below ground level. - The image contains an arbitrarily chosen point y between y=-10 and y=0 which is contained in the tank. - At this point y, the volume is given as V(y)=\pi (\frac{y}{5} + 2)^2 dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y. - Additionally, the distance from the arbitrary point y and the point 3 feet above the tank is given on the y-axis as 3-y. - - Illustration of a conical water tank with measurements used to compute the work required to empty it. - - - \begin{tikzpicture}[x=.1cm,y=.125cm,>=stealth] - - \draw [->] (0,-2) -- (0,37) node [above] { $y$}; - - \draw (-1,0) node [left] { $-10$} -- (1,0) - (-1,30) node [left] { 0} -- (1,30) - (-1,35) node [left] { 3} -- (1,35) - (-1,14) node [left] { $y$} -- (1,14); - - \draw (2,12) -- (4,12) (2,35) -- (4,35) - (3,12) -- node [pos=2,rotate=90] { $3-y_i$} (3,18) - (3,29) -- (3,35); - - \begin{scope}[xscale=4,shift={(5,0)}] - - \draw (0,30) -- node [pos=.5,above,shift={(-4pt,-1pt)}] { $2$} (2.9,29.22); - \draw [thick](3,30) -- (0,0) -- (-3,30); - \draw [thick] (0,30) circle (3); - \draw [left color=firstcolor,right color=firstcolor!15,thick,draw=firstcolor] (0,14) circle (1.4); - \draw [->] (5,6.5) node [below,shift={(10pt,0pt)}] { $V(y)=\pi(\frac y5+2)^2 dy$} -- (0,14); - - \end{scope} - - \end{tikzpicture} - - - - -
    - -

    - The figure also sketches a differential element, - a cross-sectional circle. - The radius of this circle is variable, depending on y. - When y=-10, the circle has radius 0; - when y=0, the circle has radius 2. - These two points, (-10,0) and (0,2), - allow us to find the equation of the line that gives the radius of the cross-sectional circle, - which is r(y) = 1/5y+2. - Hence the volume of water at this height is V(y)=\pi(1/5y+2)^2dy, - where dy represents a very small height of the differential element. - The force required to move the water at height y is F(y) = 62.4\times V(y). -

    - -

    - The distance the water at height y travels is given by h(y)=3-y. - Thus the total work done in pumping the water from the tank is - - W \amp = \int_{-10}^0 62.4\pi(1/5y+2)^2(3-y)\, dy - \amp = 62.4\pi\int_{-10}^0\left(-\frac1{25}y^3-\frac{17}{25}y^2-\frac85y+12\right)\, dy - \amp = 62.2\pi\cdot\frac{220}{3} \approx 14,376 \text{ft--lb} - . -

    -     -     - - - Computing work performed: pumping fluids - -

    - A rectangular swimming pool is 20 wide and has a 3 - shallow end and a 6 deep end. - It is to have its water pumped out to a point 2 above the current top of the water. - The cross-sectional dimensions of the water in the pool are given in ; - note that the dimensions are for the water, not the pool itself. - Compute the amount of work performed in draining the pool. -

    -
    - The cross-section of a swimming pool filled with water in - - - - Image of the swimming pool described in the example. - The pool is 25 long. - On the leftmost side of the image is the deep end of the pool, which is - 10 long having a depth of 6. - After the 10 mark, - the bottom of the pool slopes up until it rises to a depth of 3 - a distance of 5 away in the x dimension. - The shallow end of the pool has a depth of 3 - and length of 10. - The end of the shallow end marks the end of the pool. - - Illustration of the swimming pool from the example. - - - \begin{tikzpicture}[x=.18cm,y=.2cm,>=stealth] - - \draw [fill=firstcolor!15,draw=firstcolor!15] (0,0) -- node [below,pos=.5] { 10 ft.} (10,0) -- (15,3) -- node [below,pos=.5] { 10 ft.} (25,3)-- node [right,pos=.5] { 3 ft.} (25,6)--(0,6) -- node [left,pos=.5] { 6 ft.} (0,0); - - \draw [thick] (0,0) -- (10,0) -- (15,3) -- (25,3) -- (25,7) -- node [above,pos=.5] { 25 ft} (0,7) -- cycle; - - \end{tikzpicture} - - - - -
    -     - -

    - For the purposes of this problem we choose to set y=0 to represent the bottom of the pool, - meaning the top of the water is at y=6. -

    - -
    - Orienting the pool and showing differential elements for - - - - Image of the same swimming pool described in the example. - The image now contains the two coordinate axes x and y. - On the y axis are five markings. - The label y=0 marks the bottom of the deep end of the pool, the label y=3 marks the bottom of the shallow end of the pool. - The label y=6 marks the top of the water in the pool, and the label y=8 marks a point 2 above the top of the water. - The image also contains an arbitrary label y lying somewhere below the water level of the pool. - On the x-axis, the label x=0 marks the leftmost edge of the pool, x=10 marks the end of the deep end and after the pool slopes up to a depth of 3, the label x=15 marks the start of the shallow end. - The distance between the leftmost edge and rightmost edge of the pool when y is above y=3, which is the depth of shallow end, is constant, and is given by the length of the pool. - On the other hand, the distance between the same two edges of the pool when y is below y=3 is the length of the deep end, plus the additional distance in the x direction between the end of the deep end, x=10, and the point at which the arbitrary y level meets the sloped portion of the pool. - The sloped portion of the pool is given as the line which passes through the points (10,0) and (15,3), which are the end of the deep end, and the start of the shallow end, respectively. - - Illustration of the swimming pool from the example showing necessary differential elements. - - - \begin{tikzpicture}[x=.18cm,y=.2cm,>=stealth] - - \draw[>=stealth,->] (-2,-1) -- ( -2,10) node [above] { $y$}; - - \foreach \y/\x in {0/0,1.5/$y$,3/3,6/6,8/8} - { - \draw (-2.5,\y) node [left] { $\x$} -- (-1.5,\y); - } - - \draw [thick] (0,0) -- (10,0) -- (15,3) -- (25,3) -- (25,6) -- (0,6) -- (0,0); - - \draw [thick,firstcolor] (0,5) -- (25,5) - (0,1.5) -- (12.5,1.5); - - \draw [fill=black] (10,0) circle (2.4pt) node [shift={(15pt,-2pt)}] { $(10,0)$} - (15,3) circle (2.4pt) node [shift={(15pt,-6pt)}] { $(15,3)$}; - - \draw[>=stealth,->] (-2,-2) -- (20,-2) node [right] { $x$}; - - \foreach \y in {0,10,15} - { - \draw (\y,-2.5) node [below] { $\y$} -- (\y,-1.5); - } - - \end{tikzpicture} - - - - -
    - -

    - - shows the pool oriented with this y-axis, - along with 2 differential elements as the pool must be split into two different regions. -

    - -

    - The top region lies in the y-interval of [3,6], - where the length of the differential element is 25 as shown. - As the pool is 20 wide, - this differential element represents a thin slice of water with volume V(y) = 20\cdot25\cdot dy. - The water is to be pumped to a height of y=8, - so the height function is h(y) = 8-y. - The work done in pumping this top region of water is - - W_t = 62.4\int_3^6 500(8-y)\, dy = 327,600 \,\text{ft--lb} - . -

    - -

    - The bottom region lies in the y-interval of [0,3]; - we need to compute the length of the differential element in this interval. -

    - -

    - One end of the differential element is at x=0 and the other is along the line segment joining the points (10,0) and (15,3). - The equation of this line is y= 3/5(x-10); - as we will be integrating with respect to y, - we rewrite this equation as x=5/3y+10. - So the length of the differential element is a difference of x-values: - x=0 and x=5/3y+10, - giving a length of x=5/3y+10. -

    - -

    - Again, as the pool is 20 wide, - this differential element represents a thin slice of water with volume V(y) = 20\cdot(5/3y+10)\cdot dy; - the height function is the same as before at h(y)=8-y. - The work performed in emptying this part of the pool is - - W_b = 62.4\int_0^3 20(5/3y+10)(8-y)\, dy = 299,520\,\text{ft--lb} - . -

    - -

    - The total work in empyting the pool is - - W = W_b+W_t = 327,600+299,520 = 627,120\,\text{ft--lb} - . -

    - -

    - Notice how the emptying of the bottom of the pool performs almost as much work as emptying the top. - The top portion travels a shorter distance but has more water. - In the end, this extra water produces more work. -

    -     -     - -

    - The next section introduces one final application of the definite integral, - the calculation of fluid force on a plate. -

    -     - - - - Terms and Concepts - - -

    - What are the typical units of work? -

    -     - - - -

    - In SI units, it is one joule, - , one newtonmeter, - or - - - In Imperial Units, it is ftlb. -

    -     -     - - - -

    - If a man has a mass of - 80 - on Earth, - will his mass on the moon be bigger, smaller, or the same? -

    -     - - - -

    - Bigger -

    -     -     - - -

    - Smaller -

    -     -     - - -

    - The same -

    -     -     -     -     - - - -

    - If a woman weighs 130 on Earth, - will her weight on the moon be bigger, smaller, or the same? -

    -     - - - -

    - Bigger -

    -     -     - - -

    - Smaller -

    -     -     - - -

    - The same -

    -     -     -     -     - - - -

    - Fill in the blanks: -

    - -

    - Some integrals in this section are set up by multiplying a variable by a constant distance; - others are set up by multiplying a constant force by a variable . -

    -     - - - - - - - - - - - - -     -     - - Problems - - -

    - A 100 rope, weighing - 0.1, - hangs over the edge of a tall building. -

    -     - - - -

    - How much work is done pulling the entire rope to the top of the building? -

    -     - -

    - 500 ftlb -

    -     -     - - - -

    - How much rope is pulled in when half of the total work is done? -

    -     - -

    - 100-50\sqrt{2} \approx 29.29 ft -

    -     -     -     - - - -

    - A 50 rope, - with a mass density of 0.2, - hangs over the edge of a tall building. -

    -     - - - -

    - How much work is done pulling the entire rope to the top of the building? -

    -     - -

    - 2450 J -

    -     -     - - - -

    - How much work is done pulling in the first 20 m? -

    -     - -

    - 1568 J -

    -     -     -     - - - -

    - A rope of length \ell hangs over the edge of tall cliff. - (Assume the cliff is taller than the length of the rope.) - The rope has a weight density of d . -

    -     - - - -

    - How much work is done pulling the entire rope to the top of the cliff? -

    -     - -

    - \frac12\cdot d\cdot l^2 ftlb -

    -     -     - - - -

    - What percentage of the total work is done pulling in the first half of the rope? -

    -     - -

    - 75 % -

    -     -     - - - -

    - How much rope is pulled in when half of the total work is done? -

    -     - -

    - \ell(1-\sqrt{2}/2) \approx 0.2929\ell -

    -     -     -     - - - -

    - A 20 rope with mass density of 0.5 hangs over the edge of a 10 building. - How much work is done pulling the rope to the top? -

    -     - -

    - 735 J -

    -     -     - - - -

    - A crane lifts a 2000 - load vertically 30 - with a 1 cable weighing - 1.68. -

    -     - - - -

    - How much work is done lifting the cable alone? -

    -     - -

    - 756 ftlb -

    -     -     - - - -

    - How much work is done lifting the load alone? -

    -     - -

    - 60,000 ftlb -

    -     -     - - - -

    - Could one conclude that the work done lifting the cable is negligible compared to the work done lifting the load? -

    -     - -

    - Yes, for the cable accounts for about 1% of the total work. -

    -     -     -     - - - -

    - A100 bag of sand is lifted uniformly 120 in one minute. - Sand leaks from the bag at a rate of 1/4. - What is the total work done in lifting the bag? -

    -     - -

    - 11,100 ftlb -

    -     -     - - - -

    - A box weighing 2 lifts 10 of sand vertically 50. - A crack in the box allows the sand to leak out such that 9 of sand is in the box at the end of the trip. - Assume the sand leaked out at a uniform rate. - What is the total work done in lifting the box and sand? -

    -     - -

    - 575 ftlb -

    -     -     - - - -

    - A force of 1000 compresses a spring 3. - How much work is performed in compressing the spring? -

    -     - -

    - 125 ftlb -

    -     -     - - - -

    - A force of 2 stretches a spring 5. - How much work is performed in stretching the spring? -

    -     - -

    - 0.05 J -

    -     -     - - - -

    - A force of 50 compresses a spring from a natural length of 18 to 12. - How much work is performed in compressing the spring? -

    -     - -

    - 12.5 ftlb -

    -     -     - - - -

    - A force of 20 stretches a spring from a natural length of 6 to 8. - How much work is performed in stretching the spring? -

    -     - -

    - 5/3 ftlb -

    -     -     - - - -

    - A force of 7 stretches a spring from a natural length of 11 to 21. - How much work is performed in stretching the spring from a length of 16 to 21? -

    -     - -

    - 0.2625 = 21/80 J -

    -     -     - - - -

    - A force of f - stretches a spring d from its natural length. - How much work is performed in stretching the spring? -

    -     - -

    - f\cdot d/2 J -

    -     -     - - - -

    - A 20 weight is attached to a spring. - The weight rests on the spring, - compressing the spring from a natural length of 1 to 6. -

    - -

    - How much work is done in lifting the box 1.5 (i.e, the spring will be stretched 1 beyond its natural length)? -

    -     - -

    - 45 ftlb -

    -     -     - - - -

    - A 20 weight is attached to a spring. - The weight rests on the spring, - compressing the spring from a natural length of 1 to 6. -

    - -

    - How much work is done in lifting the box 6 (i.e, bringing the spring back to its natural length)? -

    -     - -

    - 5 ftlb -

    -     -     - - - -

    - A 5 tall cylindrical tank with radius of 2 is filled with 3 of gasoline, - with a mass density of 737.22. - Compute the total work performed in pumping all the gasoline to the top of the tank. -

    -     - -

    - 953,284 J -

    -     -     - - - -

    - A 6 cylindrical tank with a radius of 3 is filled with water, - which has a weight density of 62.4. - The water is to be pumped to a point 2 above the top of the tank. -

    -     - - - -

    - How much work is performed in pumping all the water from the tank? -

    -     - -

    - 52,929.6 ftlb -

    -     -     - - - -

    - How much work is performed in pumping 3 of water from the tank? -

    -     - -

    - 18,525.3 ftlb -

    -     -     - - - -

    - At what point is 1/2 of the total work done? -

    -     - -

    - When 3.83 of water have been pumped from the tank, - leaving about 2.17 in the tank. -

    -     -     -     - - - -

    - A gasoline tanker is filled with gasoline with a weight density of 45.93. - The dispensing valve at the base is jammed shut, - forcing the operator to empty the tank via pumping the gas to a point 1 above the top of the tank. - Assume the tank is a perfect cylinder, 20 long with a diameter of 7.5. - How much work is performed in pumping all the gasoline from the tank? -

    -     - -

    - 192,767 ftlb. - Note that the tank is oriented horizontally. - Let the origin be the center of one of the circular ends of the tank. - Since the radius is 3.75, the fluid is being pumped to y=4.75; - thus the distance the gas travels is h(y)=4.75-y. - A differential element of water is a rectangle, - with length 20 and width 2\sqrt{3.75^2-y^2}. - Thus the force required to move that slab of gas is F(y) = 40\cdot45.93\cdot\sqrt{3.75^2-y^2}dy. - Total work is \int_{-3.75}^{3.75} 40\cdot45.93\cdot(4.75-y)\sqrt{3.75^2-y^2}\, dy. - This can be evaluated without actual integration; - split the integral into \int_{-3.75}^{3.75} 40\cdot45.93\cdot(4.75)\sqrt{3.75^2-y^2}\, dy + \int_{-3.75}^{3.75} 40\cdot45.93\cdot(-y)\sqrt{3.75^2-y^2}\, dy. - The first integral can be evaluated as measuring half the area of a circle; - the latter integral can be shown to be 0 without much difficulty. - (Use substitution and realize the bounds are both 0.) -

    -     -     - - - -

    - A fuel oil storage tank is 10 deep with trapezoidal sides, 5 at the top and 2 at the bottom, - and is 15 wide - (see diagram below). - Given that fuel oil weighs 55.46, - find the work performed in pumping all the oil from the tank to a point 3 above the top of the tank. -

    - - - - Image of a fuel storage tank. - The storage tank has a base that has a 2 length and 15 width. - The forward, left and back walls of the tank are perpendicular to the base, while the right side of the tank slopes upward further away to the right side. - The top of the tank has a length of 5 due to the slope of the right wall of the fuel tank and the top same 15 width as the base. - The depth of the tank, which is the distance between the top and bottom of the tank is 10. - - Illustration of the fuel storage tank from the problem. - - - \begin{tikzpicture}[x={(1,0)},z={(0,1)},y={(.5,.87)},xscale=.38,yscale=.25] - - \draw [thick] (0,0,0) -- node [above,rotate=90,pos=.5] {10} (0,0,-10) -- node [below, pos=.5] {2} (2,0,-10) -- node [right, pos=.5] {15} (2,5,-10) -- (5,5,0) - (5,5,0) -- (5,0,0) - (5,0,0) -- (0,0,0) -- (0,5,0) -- node [above,pos=.5] { 5}(5,5,0) - (5,0,0) -- (2,0,-10); - - \draw [thick,dashed] (0,0,-10) -- (0,5,-10) -- (2,5,-10) - (0,5,-10) -- (0,5,0); - - \end{tikzpicture} - - - - -     - -

    - 212,135 ftlb -

    -     -     - - - -

    - A conical water tank is 5 deep with a top radius of 3. - (This is similar to .) - The tank is filled with pure water, - with a mass density of 1000. -

    -     - - - -

    - Find the work performed in pumping all the water to the top of the tank. -

    -     - -

    - approx. 577,000 J -

    -     -     - - - -

    - Find the work performed in pumping the top - 2.5 of water to the top of the tank. -

    -     - -

    - approx. 399,000 J -

    -     -     - - - -

    - Find the work performed in pumping the top half of the water, - by volume, to the top of the tank. -

    -     - -

    - approx 110,000 J (By volume, - half of the water is between the base of the cone and a height of 3.9685 m. - If one rounds this to 4, the work is approx 104,000 J.) -

    -     -     -     - - - -

    - A water tank has the shape of a truncated cone, - with dimensions given below, - and is filled with water with a weight density of 62.4. - Find the work performed in pumping all water to a point 1 above the top of the tank. -

    - - - - Image of a water tank in the shape of a circular truncated cone. - The water tank is laying on its base and is in the shape of a cone, from which a small portion including the top is sliced off paralell to the base. - The water tank has a circular base with a radius of 5. - From the base, the tank has a height of 10, and a circular top with a radius of 2. - - Illustration of the water tank in the shape of a truncated cone. - - - \begin{tikzpicture}[xscale=1.2,yscale=0.9] - - \draw [thick] (0,0) circle (1) - (-1,.1) -- (-1.5,-1.9) - (1,.1) -- (1.5,-1.9); - - \draw (0,0)-- node [above,pos=.5] {2 ft} (1,0) - (0,-2) -- node [above,pos=.5] {5 ft} (1.5,-2) - (1.7,0) -- (2.3,0) - (1.7,-2) -- (2.3,-2) - (2,0) -- node [pos=.5,right] {10 ft} (2,-2); - - \draw [thick] (-1.5,-2) arc (180:360:1.5); - \draw [thick,dashed] (-1.5,-2) arc (180:0:1.5); - - \end{tikzpicture} - - - - -     - -

    - 187,214 ftlb -

    -     -     - - - -

    - A water tank has the shape of an inverted pyramid, - with dimensions given below, - and is filled with water with a mass density of 1000. - Find the work performed in pumping all water to a point 5 above the top of the tank. -

    - - - - Image of the pyramidal water tank. - The water tank is in the shape of a pyramid with a square base which is laying on its tip, meaning that the square base is the top of the tank. - The top of the water tank has a square top with a side length of 2. - From the top, the tank has a depth of 7, which is the shortest distance between the square top of the tank and the peak of the pyramid, which is the bottom of the tank. - - Illustration of the water tank in the shape of an inverted pyramid. - - - \begin{tikzpicture}[scale=1.25] - - \draw [thick] (0,0) -- node [pos=.5,left]{2 m} (1,1) --node [above,pos=.5] {2 m} (3,1) -- (2,0) -- cycle - (0,0) -- (1.5,-2) - (3,1) -- (1.5,-2) - (2,0) -- (1.5,-2); - - \draw [thick,dashed] (1,1) -- (1.5,-2); - - \draw (3,.5) -- (3.6,.5) - (3,-2) -- (3.6,-2) - (3.3,.5) -- node [pos=.5,right] {7 m} (3.3,-2); - - \end{tikzpicture} - - - -     - -

    - 617,400 J -

    -     -     - - - -

    - A water tank has the shape of a truncated, - inverted pyramid, with dimensions given below, - and is filled with water with a mass density of 1000. - Find the work performed in pumping all water to a point 1 above the top of the tank. -

    - - - - Image of a water tank in the shape of a truncated pyramid. - The water tank has a square base with a side length of 2. - From the base, the pyramidal water tank expands on all sides until reaching the top of the tank. - The top of the water tank is a square with a side length of 5. - From the top, the tank has a depth of 9, which is the shortest distance between the bottom and top of the water tank. - - Illustration of the water tank in the shape of a truncated pyramid. - - - \begin{tikzpicture}[scale=1.25] - - \draw [thick] (0,0) node (A) {} -- node [pos=.5,left] {5 m} (1,1) node (B) {} -- node [above,pos=.5] {5 m} (3,1) node (C) {} -- (2,0) node (D) {} -- cycle; - - \begin{scope}[scale=.75,shift={(.5,-3)}] - \draw [thick] (0,0) node (AA) {} -- (1,1) node (BB) {} -- (3,1) node (CC) {} -- node [right,pos=.5] {2 m} (2,0) node (DD) {} -- node [pos=.5,below] {2 m} (0,0); - \end{scope} - - \draw [dashed,thick] (BB.center) -- (B.center); - - \draw [thick] (AA.center) -- (A.center) - (DD.center) -- (D.center) - (CC.center) -- (C.center); - - \draw (3.4,1) -- (4,1) - (3.4,-1.5) -- (4,-1.5) - (3.7,1) -- node [pos=.5,right] { 9 m} (3.7,-1.5); - - \end{tikzpicture} - - - - -     - -

    - 4,917,150 J -

    -     -     -     -     -
    -
    - Fluid Forces -

    - In the unfortunate situation of a car driving into a body of water, - the conventional wisdom is that the water pressure on the doors will quickly be so great that they will be effectively unopenable. - (Survival techniques suggest immediately opening the door, - rolling down or breaking the window, - or waiting until the water fills up the interior at which point the pressure is equalized and the door will open. - See Mythbusters episode #72 to watch Adam Savage test these options.) -

    - -

    - How can this be true? - How much force does it take to open the door of a submerged car? - In this section we will find the answer to this question by examining the forces exerted by fluids. -

    - -

    - We start with pressure, - which is related to force - by the following equations: - - \text{Pressure}\, = \,\frac{\text{Force}}{\text{Area}}\, \Leftrightarrow\, \text{Force}\, = \, \text{Pressure} \times\text{Area} - . -

    - -

    - In the context of fluids, we have the following definition. -

    - - - Fluid Pressure - -

    - Let w be the weight-density of a fluid. - The pressure p exerted on an object at depth d in the fluid is p = w\cdot d. - fluid pressure/force - integrationfluid force -

    -     -     - -

    - We use this definition to find the force - exerted on a horizontal sheet by considering the sheet's area. -

    - - - Computing fluid force - -

    -

      -
    1. -

      - A cylindrical storage tank has a radius of - 2 - and holds 10 - of a fluid with a weight-density of - 50. - (See .) - What is the force exerted on the base of the cylinder by the fluid? -

      - -
      - A cylindrical tank in - - - -

      - Illustration of a cylindrical storage tank with a radius of 2. - The cylindrical storage tank is then filled with water up to a marked level of 10 from the base of the tank. - The tank extends slightly above the 10 water level, but this region of the tank contains no water. -

      -       - Illustration of the cylindrical tank from the example. - - - \begin{tikzpicture}[>=stealth,scale=0.4] - - \begin{scope}[xscale=2] - - \draw [left color=firstcolor!15,right color=firstcolor!40,thick] (-2,10) -- (-2,0) arc (180:360:2) -- (2,10); - \draw [right color=firstcolor!15,left color=firstcolor!40,thick] (0,10) circle (2); - \draw [thick] (0,12) circle (2); - \draw [thick] (-2,12) -- (-2,10) - (2,12) -- (2,10); - - \draw [dashed,thick] (-2,0) arc (180:0:2); - \draw (0,0) -- node [above,pos=.5] { 2 ft} (2,0); - - \end{scope} - - \begin{scope}[shift={(5,0)}] - - \draw (-.5,0) -- (.5,0) - (-.5,10) -- (.5,10) - (0,0) -- node [pos=.5,fill=white,rotate=90] { 10 ft} (0,10); - - \end{scope} - - \end{tikzpicture} - - - - -
      -
      2. - -
    2. -

      - A rectangular tank whose base is a 5 - square has a circular hatch at the bottom with a radius of - 2. - The tank holds 10 - of a fluid with a weight-density of - 50. - (See .) - What is the force exerted on the hatch by the fluid? -

      - -
      - A rectangular tank in - - - -

      - Illustration of a rectangular storage tank with a square base. - The square base has a side length of 5. - The height of the storage tank extends slightly above the 10 mark, which is the amount of water contained in the tank. - On the square base of the tank is a centered circular hatch, having a radius of 2. -

      -       - Illustration of the rectangular storage tank with a circular hatch at the bottom. - - - \begin{tikzpicture}[x=.25cm,y=.25cm,>=stealth] - - \begin{scope}[xscale=2] - \begin{scope}[rotate=-30] - - \draw [thick] (-2.5,0) node (A) {} - (2.5,0) node (B) {} - (2.5,5) node (C) {}; - \draw [dashed,thick] (A.center) -- (-2.5,5) node (D) {} -- (C.center); - - \draw (0,2.5) node (O) {} - (-1,.77) node (OC) {}; - - \end{scope} - - \begin{scope}[shift={(0,10)}] - \begin{scope}[rotate=-30] - - \draw (-2.5,0) node (AAA) {} - (2.5,0) node (BBB) {} - (2.5,5) node (CCC) {} - (-2.5,5) node (DDD) {}; - - \end{scope} - \end{scope} - - \begin{scope}[shift={(0,12)}] - \begin{scope}[rotate=-30] - - \draw [thick] (-2.5,0) node (AA) {} - (2.5,0) node (BB) {} - (2.5,5) node (CC) {} - (-2.5,5) node (DD) {}; - - \end{scope} - \end{scope} - - \draw [left color=firstcolor!15,right color=firstcolor!40,thick] (AAA.center) -- (A.center)-- node [pos=.5,rotate=-15,below] { 5 ft} (B.center)--node [pos=.5,rotate=45,below] { 5 ft} (C.center) -- (CCC.center); - \draw [right color=firstcolor!15,left color=firstcolor!40,thick] (AAA.center) -- (BBB.center) -- (CCC.center)--(DDD.center) -- cycle; - - \draw [thick] (AA.center) -- (BB.center) -- (CC.center) -- (DD.center) -- cycle - (AA.center) -- (AAA.center) - (CC.center) -- (CCC.center) - (B.center) -- (BB.center) - (DD.center) --(DDD.center); - - \draw [thick,dashed] (DDD.center)--(D.center) -- (C.center) - (D.center) -- (A.center) - (O.center) circle (2) - (O.center) -- node [pos=.5,above,rotate=15] { 2 ft} (OC.center); - - \begin{scope}[shift={(5.5,3)}] - - \draw (-.5,0) -- (.5,0) - (-.5,10) -- (.5,10) - (0,0) -- node [pos=.5,fill=white,rotate=90] { 10 ft} (0,10); - - \end{scope} - \end{scope} - - \end{tikzpicture} - - - - -
      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - Using , - we calculate that the pressure exerted on the cylinder's base is - w\cdot d = 50 - 10=500. - The area of the base is \pi\cdot 2^2 = 4\pi . - So the force exerted by the fluid is - - F = 500\times 4\pi = 6283\,\text{lb} - . - Note that we effectively just computed the weight - of the fluid in the tank. -

      -
      2. - -
    2. -

      - The dimensions of the tank in this problem are irrelevant. - All we are concerned with are the dimensions of the hatch and the depth of the fluid. - Since the dimensions of the hatch are the same as the base of the tank in the previous part of this example, - as is the depth, we see that the fluid force is the same. - That is, F = 6283 . - A key concept to understand here is that we are effectively measuring the weight of a - 10 - column of water above the hatch. - The size of the tank holding the fluid does not matter. -

      -
      3. -
    -

    -     -     - -

    - The previous example demonstrates that computing the force exerted on a horizontally oriented plate is relatively easy to compute. - What about a vertically oriented plate? - For instance, - suppose we have a circular porthole located on the side of a submarine. - How do we compute the fluid force exerted on it? -

    - -

    - Pascal's Principle states that the pressure exerted by a fluid at a depth is equal in all directions. - Thus the pressure on any portion of a plate that is - 1 - below the surface of water is the same no matter how the plate is oriented. - (Thus a hollow cube submerged at a great depth will not simply be - crushed from above, - but the sides will also crumple in. - The fluid will exert force on all sides of the cube.) -

    - -

    - So consider a vertically oriented plate as shown in - submerged in a fluid with weight-density w. - What is the total fluid force exerted on this plate? - We find this force by first approximating the force on small horizontal strips. -

    - -
    - A thin, vertically oriented plate submerged in a fluid with weight-density w - - - -

    - Illustration of a thin, vertically oriented plate fully submerged in a fluid with weight-density w. - The plate is in the shape of an acute trapezoid. - On the plate, the image contains an arbitrarily placed small horizontal strip which stretches horizontally between the two edges of the submerged plate. - The small strip has a horizontal length measurement of \ell(c_i), and a vertical height measurement of \Delta y_i. - The vertical distance between the water level and the center of the small horizontal strip is given as d_i. -

    -     - Illustration of a thin, vertically oriented plate submerged in a fluid. - - - \begin{tikzpicture}[scale=1.32,>=stealth] - - \begin{scope} - - \draw [thick] (0,0) -- (1,2) -- (2,2) -- (4,0)--cycle; - \draw [secondcolor,fill=secondcolor!15,thick] (.5,.9) rectangle (3,1.1); - \draw (3,1) node [right] { $\left. \rule{0pt}{.2cm}\right\}\Delta y_i$}; - - \draw (.5,.5) -- (.5,.7) - (3,.5) -- (3,.7) - (.5,.6) -- node [pos=.5,below,] { $\ell(c_i)$} (3,.6); - - \foreach \x in {0.5,1.5,2.5,3.5} - { - \begin{scope}[shift={(\x*1,2.5)}] - \draw [firstcolor,thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw (.2,1) -- (.4,1) - (.2,2.5) -- (.4,2.5) - (.3,1) -- node [pos=.5,left] { $d_i$} (.3,2.5); - - \end{scope} - - \end{tikzpicture} - - - - -
    - -

    - Let the top of the plate be at depth b and let the bottom be at depth a. - (For now we assume that surface of the fluid is at depth 0, so if the bottom of the plate is - 3 under the surface, - we have a=-3. - We will come back to this later.) - We partition the interval [a,b] into n subintervals - - a = y_0 \lt y_1 \lt \cdots \lt y_{n} = b - , - with the ith subinterval having length \Delta y_i. - The force F_i exerted on the plate in the - ith subinterval is F_i = \text{Pressure} \times \text{Area}. -

    - -

    - The pressure is depth times the weight density w. - We approximate the depth of this thin strip by choosing any value d_i in [y_{i-1},y_{i}]; - the depth is approximately -d_i. - (Our convention has d_i being a negative number, - so -d_i is positive.) - For convenience, - we let d_i be an endpoint of the subinterval; - we let d_i = y_i. -

    - -

    - The area of the thin strip is approximately length width. - The width is \Delta y_i. - The length is a function of some y-value c_i in the ith subinterval. - We state the length is \ell(c_i). - Thus - - F_i \amp = \text{Pressure} \times \text{Area} - \amp = -y_i\cdot w \times \ell(c_i)\cdot\Delta y_i - . -

    - -

    - To approximate the total force, - we add up the approximate forces on each of the n thin strips: - - F = \sum_{i=1}^n F_i \approx \sum_{i=1}^n -w\cdot y_i\cdot\ell(c_i)\cdot\Delta y_i - . -

    - -

    - This is, of course, another Riemann Sum. - We can find the exact force by taking a limit as the subinterval lengths go to 0; - we evaluate this limit with a definite integral. -

    - - - Fluid Force on a Vertically Oriented Plate -

    - Let a vertically oriented plate be submerged in a fluid with weight-density w, - where the top of the plate is at y=b and the bottom is at y=a. - Let \ell(y) be the length of the plate at y. - fluid pressure/force - integrationfluid force - -

      -
    1. -

      - If y=0 corresponds to the surface of the fluid, - then the force exerted on the plate by the fluid is - - F=\int_a^b w\cdot(-y)\cdot\ell(y)\, dy - . -

      -
      2. - -
    2. -

      - In general, let d(y) represent the distance between the surface of the fluid and the plate at y. - Then the force exerted on the plate by the fluid is - - F=\int_a^b w\cdot d(y)\cdot\ell(y)\, dy - . -

      -
      3. -
    -

    -     - - - Finding fluid force - -

    - Consider a thin plate in the shape of an isosceles triangle as shown in , - submerged in water with a weight-density of - 62.4. - If the bottom of the plate is 10 - below the surface of the water, - what is the total fluid force exerted on this plate? -

    - -
    - A thin plate in the shape of an isosceles triangle in - - - -

    - Illustration of a thin plate in the shape of an isosceles triangle. - The triangle has one side having a length of 4. - The height of the triangle is also 4. - The length of the two remaining sides is equal but is not given in the image. - The triangle is oriented such that the side having a length of 4 is the top of the plate and lies parallel to the water level. - The distance between the top of the plate and the vertex lying across from the top plate is the height of the triangle, which is 4. -

    -     - Illustration of a thin plate in the shape of an isosceles triangle. - - - \begin{tikzpicture}[x=.8cm,y=.8cm,>=stealth] - - \draw[thick] (0,0) -- (2,4) -- node [pos=.5,above] { 4 ft} (-2,4) -- cycle; - - \draw (2.2,0) -- (2.6,0) - (2.2,4) -- (2.6,4) - (2.4,0) -- node [pos=.5,rotate=-90,above] { 4 ft} (2.4,4); - - \end{tikzpicture} - - - - -
    -     - -

    - We approach this problem in two different ways to illustrate the different ways can be implemented. - First we will let y=0 represent the surface of the water, - then we will consider an alternate convention. -

    - -

    -

      -
    1. -

      - We let y=0 represent the surface of the water; - therefore the bottom of the plate is at y=-10. - We center the triangle on the y-axis as shown in . - The depth of the plate at y is -y as indicated by the Key Idea. - We now consider the length of the plate at y. - - We need to find equations of the left and right edges of the plate. - The right hand side is a line that connects the points (0,-10) and (2,-6): - that line has equation x=1/2(y+10). - (Find the equation in the familiar y=mx+b format and solve for x.) - Likewise, the left hand side is described by the line x=-1/2(y+10). - The total length is the distance between these two lines: - \ell(y)=1/2(y+10) - (-1/2(y+10)) = y+10. -

      - -
      - Sketching the triangular plate in with the convention that the water level is at y=0 - - - -

      - Graph of the thin plate in the shape of an isosceles triangle with the convention that the water level is at y=0. - The triangle is oriented in the same way as described previously, with the top of the plate being the side having a length of 4 running parallel to the water level. - The graph contains the two coordinate axes. - The bottom of the plate, which is the vertex of the triangle lies 10 below the water level, which is marked as the point (0,-10). - The remaining two vertices of the triangle are the points (-2,-6) and (2,-6), with the line joining these two vertices making up the top of the plate. - The graph also contains an arbitrarily chosen thin slice of the plate, which spans the horizontal length of the triangular plate. - The red coloured thin slice occurs at the level y, which is an arbitrarily chosen value between y = -6 which marks the top of the plate and y = -10 which marks the bottom of the plate. - The distance between the water level and the thin horizontal slice is also given as d(y)=-y. -

      -       - Graph of a thin plate in the shape of an isosceles triangle. - - - \begin{tikzpicture}[scale=0.755,>=stealth] - - \draw[thick] (0,0) -- (2,4) -- (-2,4) -- cycle; - - \draw [fill=black] (2,4) circle (1pt) node [right] { $(2,-6)$} - (-2,4) circle (1pt) node [left] { $(-2,-6)$} - (0,0) circle (1pt) node [below right] { $(0,-10)$}; - - \draw [secondcolor,thick] (-1.5,3) -- (1.5,3) node [right,black] { $y$}; - \draw [->] (0,-1) -- (0,12) node [above] { $y$}; - \draw [->] (-3,10) -- (3,10) node [right] { $x$}; - - \foreach \x in {-2,-1,1,2} - { - \draw (\x,10.2) -- (\x,9.8) node [below] { $\x$}; - } - - \foreach \x in {-10,-8,-4,-2} - { - \draw (.2,{\x+10}) -- (-.2,{\x+10}) node [left] { $\x$}; - } - - \foreach \x in {-2,-1,0,1,2} - { - \begin{scope}[shift={(\x*1,10)}] - \draw [firstcolor,thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw (2,11) node { water line}; - - \draw (4.3,3) -- (4.7,3) - (4.3,10) -- (4.7,10) - (4.5,3) -- node [pos=.5,rotate=-90,above] { $d(y) = -y$} (4.5,10); - - \end{tikzpicture} - - - - -
      - -

      - The total fluid force is then: - - F \amp = \int_{-10}^{-6} 62.4(-y)(y+10)\, dy - \amp = 62.4\cdot \frac{176}{3} \approx 3660.8\,\text{lb} - . -

      -
      2. - -
    2. -

      - Sometimes it seems easier to orient the thin plate nearer the origin. - For instance, - consider the convention that the bottom of the triangular plate is at (0,0), - as shown in . - The equations of the left and right hand sides are easy to find. - They are y=2x and y=-2x, respectively, - which we rewrite as x= 1/2y and x=-1/2y. - Thus the length function is \ell(y) = 1/2y-(-1/2y) = y. -

      - -
      - Sketching the triangular plate in with the convention that the base of the triangle is at (0,0) - - - -

      - Graph of the thin plate in the shape of an isosceles triangle with the convention that the water level is at y=10. - The triangle is oriented in the same way as described previously, with the top of the plate being the side having a length of 4 running parallel to the water level. - The graph contains two the coordinate axes. - The bottom of the plate, which is the vertex of the triangle lies 10 below the water level, which is marked as the point (0,0). - The remaining two vertices of the triangle are the points (-2,4) and (2,4), with the line joining these two vertices making up the top of the plate. - The graph also contains an arbitrarily chosen thin slice of the plate, which spans the horizontal length of the triangular plate. - The red coloured thin slice occurs at the level y, which is an arbitrarily chosen value between y = 0 which marks the bottom of the plate and y = 4 which marks the top of the plate. - The distance between the water level and the thin horizontal slice is also given as d(y)=10-y. -

      -       - Graph of a thin plate in the shape of an isosceles triangle. - - - \begin{tikzpicture}[scale=0.755,>=stealth] - - \draw[thick] (0,0) -- (2,4) -- (-2,4) -- cycle; - - \draw [fill=black] (2,4) circle (1pt) node [right] { $(2,4)$} - (-2,4)circle (1pt) node [left] { $(-2,4)$} - (0,0)circle (1pt); - - \draw [secondcolor,thick] (-1.5,3) -- (1.5,3) node [right,black] { $y$}; - \draw [->] (0,-1) -- (0,12) node [above] { $y$}; - \draw [->] (-3,0) -- (3,0) node [right] { $x$}; - - \foreach \x in {-2,-1,1,2} - { - \draw (\x,0.2) -- (\x,-.2) node [below] { $\x$}; - } - - \foreach \x in {10,8,6,2} - { - \draw (.2,{\x}) -- (-.2,{\x}) node [left] { $\x$}; - } - - \foreach \x in {-2,-1,0,1,2} - { - \begin{scope}[shift={(\x*1,10)}] - \draw [firstcolor,thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw (2,11) node { water line}; - - \draw (4.1,3) -- (4.5,3) - (4.1,10) -- (4.5,10) - (4.3,3) -- node [pos=.5,rotate=-90,above] { $d(y) = 10-y$} (4.3,10); - - \end{tikzpicture} - - - - -
      - -

      - As the surface of the water is 10 above the base of the plate, - we have that the surface of the water is at y=10. - Thus the depth function is the distance between y=10 and y; - d(y) = 10-y. - We compute the total fluid force as: - - F \amp =\int_0^4 62.4(10-y)(y)\, dy - \amp \approx 3660.8\,\text{lb} - . -

      -
      3. -
    -

    - -

    - The correct answer is, of course, - independent of the placement of the plate in the coordinate plane as long as we are consistent. -

    -     -     - - - Finding fluid force - -

    - Find the total fluid force on a car door submerged up to the bottom of its window in water, - where the car door is a rectangle - 40 long and - 27 high - (based on the dimensions of a 2005 Fiat Grande Punto.) -

    -     - -

    - The car door, as a rectangle, - is drawn in . - Its length is 10/3 ft and its height is 2.25. - We adopt the convention that the top of the door is at the surface of the water, - both of which are at y=0. - Using the weight-density of water of 62.4, - we have the total force as - - F \amp = \int_{-2.25}^0 62.4(-y)10/3\, dy - \amp = \int_{-2.25}^0 -208y\, dy - \amp = -104y^2\Big|_{-2.25}^0 - \amp = 526.5 \,\text{lb} - . -

    - -

    - Most adults would find it very difficult to apply over 500 of force to a car door while seated inside, - making the door effectively impossible to open. - This is counter-intuitive as most assume that the door would be relatively easy to open. - The truth is that it is not, - hence the survival tips mentioned at the beginning of this section. -

    - -
    - Sketching a submerged car door in - - - -

    - Graph of a perfectly rectangular car door submerged in water with the convention that the water level is at y=0. - The coordinate axes are labeled in terms of feet. - The top of the rectangular door is 40 inches or 3.\overline{3} feet long and perfectly coincides with the water level at y=0. - The leftmost upper corner of the rectangular door is labeled to be the point (0,0), while the upper right corner is labeled by the point (3.\overline{3},0). - The leftmost bottom corner of the rectangular door is labeled to be the point (0,-2.25), while the upper right corner is labeled by the point (3.\overline{3},-2.25). - The graph also contains an arbitrarily chosen thin slice of the door, which spans the horizontal length of the rectangular door. - The red coloured thin slice occurs at the level y, which is an arbitrarily chosen value between y = -2.25 which marks the distance from the water level and the bottom of the door and y = 0 which marks the top of the door. - The distance between the water level and the thin horizontal slice will be given by d(y)=-y. -

    -     - Graph of a perfectly rectangular car door submerged in water. - - - \begin{tikzpicture}[scale=0.66,>=stealth] - - \draw [thick] (0,0) -- (3.33,0) -- (3.33,-2.25) -- (0,-2.25) -- cycle; - - \draw [fill=black] (3.33,0) circle (1pt) node [above right] { $(3.\overline{3},0)$} - (3.33,-2.25) circle (1pt) node [below right] { $(3.\overline{3},-2.25)$} - (0,-2.25) circle (1pt) node [below left] { $(0,-2.25)$}; - - \draw [fill=black] (0,0) circle (1pt) node [above left] { $(0,0)$}; - \draw [secondcolor,thick] (0,-1.5) -- (3.33,-1.5) node [right,black] { $y$}; - \draw [->] (0,-3) -- (0,1) node [above] { $y$}; - \draw [->] (-1,0) -- (6,0) node [right] { $x$}; - - \foreach \x in {-1,0,1,2,3,4} - { - \begin{scope}[shift={(\x*1,0)}] - \draw [firstcolor,thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \end{tikzpicture} - - - - -
    -     -     - - - Finding fluid force - -

    - An underwater observation tower is being built with circular viewing portholes enabling visitors to see underwater life. - Each vertically oriented porthole is to have a 3 - diameter whose center is to be located 50 underwater. - Find the total fluid force exerted on each porthole. - Also, compute the fluid force on a horizontally oriented porthole that is under - 50 of water. -

    - -
    - Measuring the fluid force on an underwater porthole in - - - -

    - Graph of an underwater circular porthole with the convention that the water level is at y=50. - The circular porthole having a diameter of 3 is centered at the origin. - The top of the circular porthole is at the point (0,1.5), while the bottom is at the point (0,-1.5). - The leftmost point of the circle occurs at the point (-1.5,0), whilet the rightmost point of the circle occurs at the point (-1.5,0). - The graph also contains an arbitrarily chosen thin slice of the circular porthole, which spans the horizontal length of the circle. - The red-coloured thin slice occurs at the level y, which is an arbitrarily chosen value between y = -1.5 which marks the bottom of the porthole and y = 1.5 which marks the top of the porthole. - The distance between the water level and the thin horizontal slice is given by d(y)=50-y. -

    -     - Graph of a circular underwater porthole. - - - \begin{tikzpicture}[scale=1.06,>=stealth] - - \draw [thick] (0,0) circle (1.5); - \draw [secondcolor,thick] (-1.3,.75) -- (1.3,.75) node [right,black] { $y$}; - \draw [->] (0,-2.2) -- (0,3)--(-.25,3.25) -- (0,3.5) -- (.25,3.75) -- (0,4) --(0,6) node [above] { $y$}; - \draw [->] (-2.5,0) -- (2.5,0) node [right] { $x$}; - - \foreach \x in {-2,-1,1,2} - { - \draw (\x,0.2) -- (\x,-.2) node [below] { $\x$}; - } - - \foreach \x in {-2,-1,1,2} - { - \draw (.2,{\x}) -- (-.2,{\x}) node [left] { $\x$}; - } - - \draw (.2,5) -- (-.2,5) node [left] { $50$}; - - \foreach \x in {-1.5,-0.5,0.5,1.5} - { - \begin{scope}[shift={(\x*1,5)}] - \draw [firstcolor,thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw (2,6) node { water line} - (2,-2.5) node { \textit{not to scale}}; - - \draw (2.1,.75) -- (2.5,.75) - (2.1,5) -- (2.5,5) - (2.3,.75) -- node [pos=.5,rotate=-90,above] { $d(y) = 50-y$} (2.3,5); - - \end{tikzpicture} - - - - -
    -     - -

    - We place the center of the porthole at the origin, - meaning the surface of the water is at y=50 and the depth function will be d(y)=50-y; - see -

    - -

    - The equation of a circle with a radius of 1.5 is x^2+y^2=2.25; - solving for x we have x=\pm \sqrt{2.25-y^2}, - where the positive square root corresponds to the right side of the circle and the negative square root corresponds to the left side of the circle. - Thus the length function at depth y is \ell(y) = 2\sqrt{2.25-y^2}. - Integrating on [-1.5,1.5] we have: - - F \amp = 62.4\int_{-1.5}^{1.5} 2(50-y)\sqrt{2.25-y^2}\, dy - \amp = 62.4\int_{-1.5}^{1.5} \big(100\sqrt{2.25-y^2} - 2y\sqrt{2.25-y^2}\big)\, dy - \amp = 6240\int_{-1.5}^{1.5} \big(\sqrt{2.25-y^2}\big)\, dy - 62.4\int_{-1.5}^{1.5} \big(2y\sqrt{2.25-y^2}\big)\, dy - . -

    - -

    - The second integral above can be evaluated using substitution. - Let u=2.25-y^2 with du = -2y\,dy. - The new bounds are: u(-1.5)=0 and u(1.5)=0; - the new integral will integrate from u=0 to u=0, - hence the integral is 0. -

    - -

    - The first integral above finds the area of half a circle of radius 1.5, thus the first integral evaluates to 6240\cdot\pi\cdot1.5^2/2 = 22,054. - Thus the total fluid force on a vertically oriented porthole is 22,054. -

    - -

    - Finding the force on a horizontally oriented porthole is more straightforward: - - F = \text{Pressure} \times\text{Area} = 62.4\cdot50\times \pi\cdot1.5^2 = 22,054\, \text{lb} - . -

    - -

    - That these two forces are equal is not coincidental; - it turns out that the fluid force applied to a vertically oriented circle whose center is at depth d is the same as force applied to a horizontally oriented circle at depth d. -

    -     -     - -

    - We end this chapter with a reminder of the true skills meant to be developed here. - We are not truly concerned with an ability to find fluid forces or the volumes of solids of revolution. - Work done by a variable force is important, - though measuring the work done in pulling a rope up a cliff is probably not. -

    - -

    - What we are actually concerned with is the ability to solve certain problems by first approximating the solution, - then refining the approximation, - then recognizing if/when this refining process results in a definite integral through a limit. - Knowing the formulas found inside the special boxes within this chapter is beneficial as it helps solve problems found in the exercises, - and other mathematical skills are strengthened by properly applying these formulas. - However, more importantly, - understand how each of these formulas was constructed. - Each is the result of a summation of approximations; - each summation was a Riemann sum, - allowing us to take a limit and find the exact answer through a definite integral. -

    - - - - - - Terms and Concepts - - - -

    - State in your own words Pascal's Principle. -

    -     - - -     - - - -

    - State in your own words how pressure is different from force. -

    -     - - -     -     - - Problems - - -

    - Find the fluid force exerted on the given plate, - submerged in water with a weight density of 62.4. -

    -     - - - - - -

    - Image of a submerged square plate having a side length of 2. - The top side of the square is parallel to the water level and lies 1 below the water level. -

    -     - Illustration of a submerged square plate. - - - \begin{tikzpicture}[scale=0.96] - - \draw [shift={(-1,0)},thick] (0,0) -- node [below,pos=.5] {2 ft} (2,0) -- node [right,pos=.5] {2 ft} (2,2) -- (0,2) -- cycle; - - \foreach \x in {-2,-1,0,1,2} - { - \begin{scope}[shift={(\x*1,3)}] - \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - \draw [dashed] (0,2.1) -- node [right,pos=.5] {1 ft} (0,2.9); - - \end{tikzpicture} - - - - -     - -

    - 499.2 -

    -     -     - - - - - -

    - Image of a submerged rectangular plate. - The rectangle has a horizontal length of 1 and a veritcal length of 2. - The shorter side of the rectangle is the top of the plate and is parallel to the water level and lies 1 below the water level. -

    -     - Illustration of a submerged rectangular plate. - - - \begin{tikzpicture}[scale=0.96] - - \draw [shift={(-.5,0)},thick] (0,0) -- node [below,pos=.5] {1 ft} (1,0)-- node [right,pos=.5] {2 ft} (1,2) -- (0,2) -- cycle; - - \foreach \x in {-2,-1,0,1,2} - { - \begin{scope}[shift={(\x*1,3)}] - \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw [dashed] (0,2.1) -- node [right,pos=.5] {1 ft} (0,2.9); - - \end{tikzpicture} - - - - -     - -

    - 249.6 -

    -     -     - - - - - -

    - Image of a submerged plate in the shape of an isosceles triangle. - The bottom of the triangle has a horizontal length of 4. - The distance from the bottom of the triangle to the vertex which is the nearest point on the plate to the water level is 6, which is the height of the triangle. - This vertex is the top of the plate and lies 5 below the water level. - The remaining two sides of the triangle have equal but unmarked lengths and connect the vertex to the base of the triangular plate. -

    -     - Illustration of a submerged triangular plate. - - - \begin{tikzpicture}[xscale=.8,yscale=.64] - - \draw [thick] (0,0) -- (-2,-6) -- node [below,pos=.5] {4 ft} (2,-6) -- cycle; - - \foreach \x in {-2,-1,0,1,2} - { - \begin{scope}[shift={(\x*1,3)}] - \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw [dashed] (0,0.1) -- node [right,pos=.5] {5 ft} (0,2.9); - - \draw (2.2,-6) -- (2.6,-6) - (2.2,0) -- (2.6,0) - (2.4,-6) -- node [pos=.5,right] {6 ft} (2.4,0); - - \end{tikzpicture} - - - - -     - -

    - 6739.2 -

    -     -     - - - - - -

    - Image of a submerged plate in the shape of an isosceles triangle. - The top of the triangle has a horizontal length of 4. - The distance from the top of the triangle to the vertex which is the lowest point below the water level on the plate is 6, which is the height of the triangle. - This top side of the plate having a horizontal length of 4 lies 5 below the water level. - The remaining two sides of the triangle have equal but unmarked lengths and connect the vertex to the top of the triangular plate. -

    -     - Illustration of a submerged triangular plate. - - - \begin{tikzpicture}[xscale=.8,yscale=.64] - - \draw [thick] (0,0) -- (-2,6) -- node [below,pos=.5] {4 ft} (2,6) -- cycle; - - \foreach \x in {-2,-1,0,1,2} - { - \begin{scope}[shift={(\x*1,9)}] - \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw [dashed] (0,6.1) -- node [right,pos=.5] {5 ft} (0,8.9); - - \draw (2.2,6) -- (2.6,6) - (2.2,0) -- (2.6,0) - (2.4,6) -- node [pos=.5,right] {6 ft} (2.4,0); - - \end{tikzpicture} - - - - -     - -

    - 5241.6 -

    -     -     - - - - - -

    - Image of a submerged circular plate with a radius of 2. - The center of the circular plate lies 5 below the water level. - The uppermost part of the circular plate lies 3 below the water level, and the lowest part lies 7 below the water level. -

    -     - Illustration of a submerged circular plate. - - - \begin{tikzpicture}[scale=.8] - - \draw [thick] (0,0) circle (2); - - \foreach \x in {-2,-1,0,1,2} - { - \begin{scope}[shift={(\x*1,5)}] - \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw [dashed] (0,0) -- node [above,pos=.5] {2 ft} (2,0); - - \draw (2.2,5) -- (2.6,5) - (2.2,0) -- (2.6,0) - (2.4,5) -- node [pos=.5,right] {5 ft} (2.4,0); - - \end{tikzpicture} - - - - -     - -

    - 3920.7 -

    -     -     - - - - - -

    - Image of a submerged circular plate with a radius of 4. - The center of the circular plate lies 5 below the water level. - The uppermost part of the circular plate lies 1 below the water level, and the lowest part lies 9 below the water level. -

    -     - Illustration of a submerged circular plate. - - - \begin{tikzpicture}[scale=0.5] - - \draw [thick] (0,0) circle (4); - - \foreach \x in {-3,-2,-1,0,1,2,3} - { - \begin{scope}[shift={(\x*1,5)}] - \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw [dashed] (0,0) -- node [above,pos=.5] {4 ft} (4,0); - - \draw (4.2,5) -- (4.6,5) - (4.2,0) -- (4.6,0) - (4.4,5) -- node [pos=.5,right] {5 ft} (4.4,0); - - \end{tikzpicture} - - - - -     - -

    - 15682.8 -

    -     -     - - - - - -

    - Image of a submerged rectangular plate with a horizontal length of 4 and a height of 2. - The horizontal line through the middle of the plate lies 5 below the water level. - The uppermost edge of the rectangular plate having a length of 4 lies 4 below the water level, and the lowest edge lies 6 below the water level. - Both the top and bottom edges lie parallel to the water level. -

    -     - Illustration of a submerged rectangular plate. - - - \begin{tikzpicture}[scale=0.688] - - \draw [thick] (-2,-1) -- node [below,pos=.5] {4 ft} (2,-1) -- (2,1) -- (-2,1) -- node [left,pos=.5] {2 ft} (-2,-1); - - \foreach \x in {-3,-2,-1,0,1,2,3} - { - \begin{scope}[shift={(\x*1,5)}] - \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw (2.2,5) -- (2.6,5) - (2.2,0) -- (2.6,0) - (2.4,5) -- node [pos=.5,right] {5 ft} (2.4,0); - - \end{tikzpicture} - - - - -     - -

    - 2496 -

    -     -     - - - - - -

    - Image of a submerged rectangular plate with a horizontal length of 2 and a height of 4. - The horizontal line through the middle of the plate lies 5 below the water level. - The uppermost edge of the rectangular plate having a length of 2 lies 3 below the water level, and the lowest edge lies 7 below the water level. - Both the top and bottom edges lie parallel to the water level. -

    -     - Illustration of a submerged rectangular plate. - - - \begin{tikzpicture}[scale=0.688] - - \draw [thick,rotate=90] (-2,-1) -- node [right,pos=.5] {4 ft} (2,-1) -- (2,1) -- (-2,1) -- node [below,pos=.5] {2 ft} (-2,-1); - - \foreach \x in {-3,-2,-1,0,1,2,3} - { - \begin{scope}[shift={(\x*1,5)}] - \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw (2.2,5) -- (2.6,5) - (2.2,0) -- (2.6,0) - (2.4,5) -- node [pos=.5,right] {5 ft} (2.4,0); - - \end{tikzpicture} - - - - -     - -

    - 2496 -

    -     -     - - - - - -

    - Image of a submerged rotated square plate with a side length of 2. - The square is rotated such that one of its vertices is the uppermost point of the plate, while the opposite vertex lying directly below the uppermost vertex is the lowest point of the plate. - The uppermost vertex of the square plate lies 1 below the water level. -

    -     - Illustration of a submerged rotated square plate. - - - \begin{tikzpicture}[scale=0.96] - - \draw [shift={(0,-.85)},thick,rotate=45] (0,0) -- node [shift={(8pt,-5pt)},pos=.5] {2 ft} (2,0)-- node [shift={(8pt,5pt)},pos=.5] {2 ft} (2,2) -- (0,2) -- cycle; - - \foreach \x in {-2,-1,0,1,2} - { - \begin{scope}[shift={(\x*1,3)}] - \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - - \draw [dashed] (0,2.1) -- node [right,pos=.5] {1 ft} (0,2.9); - - \end{tikzpicture} - - - - -     - -

    - 602.59 -

    -     -     - - - - - -

    - Image of a submerged plate in the shape of an equilateral right triangle. - The right angle occurs at the bottom left part of the triangle. - The two edges connected to the right angle lie parallel and perpendicular to the water level, and both have a side length of 2. - The edge that is perpendicular to the water level extends up from the right angle for 2, until reaching the uppermost vertex of the triangle. - This uppermost vertex lies 1 below the water level. - The edge that is connected to the right angle and parallel to the water level extends to the right of the right angle for a distance of 2, until reaching the rightmost vertex of the triangle. - The rightmost and uppermost vertices are then connected by an edge, which completes the equilateral right triangle. -

    -     - Illustration of a submerged plate in the shape of an equilateral right triangle. - - - \begin{tikzpicture}[scale=0.96] - - \draw [shift={(-.5,0)},thick] (0,0) -- node [below,pos=.5] {2 ft} (2,0) -- (0,2) -- node [left,pos=.5] {2 ft} (0,0); - - \foreach \x in {-2,-1,0,1,2} - { - \begin{scope}[shift={(\x*1,3)}] - \draw [thick] (-.5,.25) parabola bend (0,0) (.5,.25); - \end{scope} - } - \draw [dashed] (0,2.1) -- node [right,pos=.5] {1 ft} (0,2.9); - - \end{tikzpicture} - - - - -     - -

    - 291.2 -

    -     -     - -     - - - -

    - The side of a container is pictured. - Find the fluid force exerted on this plate when the container is full of: -

    - -

    -

      -
    1. -

      - water, with a weight density of 62.4, and -

      -
      2. - -
    2. -

      - concrete, with a weight density of 150. -

      -
      3. -
    -

    - -     - - - - - -

    - Image of a rectangular container having horizontal length of 3 and a vertical height measurement of 5. -

    -     - Illustration of a rectangular container. - - - \begin{tikzpicture}[scale=1.2] - - \draw [thick] (0,0) -- node [below,pos=.5] {3 ft} (3,0) -- node [right,pos=.5] {5 ft} (3,5) -- (0,5) -- cycle; - - \end{tikzpicture} - - - - -     - -

    -

      -
    1. -

      - 2340 -

      -
      2. - -
    2. -

      - 5625 -

      -
      3. -
    -

    -     -     - - - - - -

    - Image of a container that is the combination of the parabola y=x^2 and a horizontal line having a length of 4. - The parabola is drawn such that the two sides of the parabola end when the horizontal distance between them measures 4, from which they are connected by the horizontal line. - The height, measured from the base of the parabola to the horizontal line measures 4. - A horizontal line which stretches from the left side of the container to the right side would lie entirely between the parabola given by y=x^2. -

    -     - Illustration of a container made with a parabola and a line. - - - \begin{tikzpicture}[scale=.8] - - \draw [thick] (-2,4) parabola bend (0,0) (2,4) -- node [above,pos=.5] {4 ft} (-2,4); - - \draw (0,0) node [below] {\(y=x^2\)}; - - \draw (2.2,0) -- (2.6,0) - (2.2,4) -- (2.6,4) - (2.4,0) -- node [pos=.5,right] {4 ft} (2.4,4); - - \end{tikzpicture} - - - - -     - -

    -

      -
    1. -

      - 1064.96 -

      -
      2. - -
    2. -

      - 2560 -

      -
      3. -
    -

    -     -     - - - - - -

    - Image of a container that is the combination of the parabola y=4-x^2 and a horizontal line having a length of 4. - The upsidedown parabola is drawn such that two sides of the parabola end when the horizontal distance between them measures 4, from which they are connected by the horizontal line. - The height, measured from the top of the parabola to the horizontal line which makes up the base of the container measures 4. - A horizontal line which stretches from the left side of the container to the right side would lie entirely between the parabola given by y=4-x^2. -

    -     - Illustration of a container made with a parabola and a line. - - - \begin{tikzpicture}[scale=.8] - - \draw [thick] (-2,0) parabola bend (0,4) (2,0) -- node [below,pos=.5] {4 ft} (-2,0); - - \draw (0,4) node [above] {\(y=4-x^2\)}; - - \draw (2.2,0) -- (2.6,0) - (2.2,4) -- (2.6,4) - (2.4,0) -- node [pos=.5,right] {4 ft} (2.4,4); - - \end{tikzpicture} - - - - -     - -

    -

      -
    1. -

      - 1597.44 -

      -
      2. - -
    2. -

      - 3840 -

      -
      3. -
    -

    -     -     - - - - - -

    - Image of a container in the shape of a semicircle having a radius of 1 that is the combination of the equation y=-\sqrt{1-x^2} and a horizontal line having a length of 2. - The circular arc coming from the equation y=-\sqrt{1-x^2} makes up the base of the container and is plotted for all x in the domain of y=-\sqrt{1-x^2}. - Connecting the two vertices of the circular arc is the horizontal line having a length of 2 which makes up the top of the container. - A horizontal line which stretches from the left side of the container to the right side would lie entirely between the circular arc given by y=-\sqrt{1-x^2}. -

    -     - Illustration of a container in the shape of a semicircle. - - - \begin{tikzpicture}[scale=2.16] - - \draw [thick] (-1,0) arc (180:360:1) -- node [above,pos=.5] {2 ft} (-1,0); - \draw (0,-1) node [below] {\(y=-\sqrt{1-x^2}\)}; - - \end{tikzpicture} - - - - -     - -

    -

      -
    1. -

      - 41.6 -

      -
      2. - -
    2. -

      - 100 -

      -
      3. -
    -

    -     -     - - - - - -

    - Image of a container in the shape of a semicircle having a radius of 1 that is the combination of the equation y=\sqrt{1-x^2} and a horizontal line having a length of 2. - The circular arc coming from the equation y=\sqrt{1-x^2} makes up the top of the container and is plotted for all x in the domain of y=\sqrt{1-x^2}. - Connecting the two vertices of the circular arc is the horizontal line having a length of 2 which makes up the bottom part of the container. - A horizontal line which stretches from the left side of the container to the right side would lie entirely between the circular arc given by y=\sqrt{1-x^2}. -

    -     - Illustration of a container in the shape of a semicircle. - - - \begin{tikzpicture}[scale=2] - - \draw [thick] (-1,0) arc (180:0:1) -- node [below,pos=.5] {2 ft} (-1,0); - \draw (0,1) node [above] {\(y=\sqrt{1-x^2}\)}; - - \end{tikzpicture} - - - - -     - -

    -

      -
    1. -

      - 56.42 -

      -
      2. - -
    2. -

      - 135.62 -

      -
      3. -
    -

    -     -     - - - - - -

    - Image of a container in the shape of a semicircle having a radius of 3 that is the combination of the equation y=-\sqrt{9-x^2} and a horizontal line having a length of 6. - The circular arc coming from the equation y=-\sqrt{9-x^2} makes up the top of the container and is plotted for all x in the domain of y=-\sqrt{9-x^2}. - Connecting the two vertices of the circular arc is the horizontal line having a length of 6 which makes up the bottom part of the container. - A horizontal line which stretches from the left side of the container to the right side would lie entirely between the circular arc given by y=-\sqrt{9-x^2}. -

    -     - Illustration of a container in the shape of a semicircle. - - - \begin{tikzpicture}[scale=.67] - - \draw [thick] (-3,0) arc (180:360:3) -- node [above,pos=.5] {6 ft} (-3,0); - \draw (0,-3) node [below] {\(y=-\sqrt{9-x^2}\)}; - - \end{tikzpicture} - - - - -     - -

    -

      -
    1. -

      - 1123.2 -

      -
      2. - -
    2. -

      - 2700 -

      -
      3. -
    -

    -     -     -     - - - -

    - How deep must the center of a vertically oriented circular plate with a radius of 1 be submerged in water, - with a weight density of 62.4, - for the fluid force on the plate to reach 1,000? -

    -     - -

    - 5.1 -

    -     -     - - - -

    - How deep must the center of a vertically oriented square plate with a side length of 2 be submerged in water, - with a weight density of 62.4, - for the fluid force on the plate to reach 1,000 lb? -

    -     - -

    - 4.1 -

    -     -     -     -     -
    -     - - - Differential Equations - -

    - One of the strengths of calculus is its ability to describe real-world phenomena. - We have seen hints of this in our discussion of the applications of derivatives and integrals in the previous chapters. - The process of formulating an equation or multiple equations to describe a physical phenomenon is called - mathematical modeling. As a simple example, - populations of bacteria are often described as - growing exponentially. - Looking in a biology text, we might see P(t) = P_0e^{kt}, - where P(t) is the bacteria population at time t, - P_0 is the initial population at time t=0, - and the constant k describes how quickly the population grows. - This equation for exponential growth arises from the assumption that the population of bacteria grows at a rate proportional to its size. - Recalling that the derivative gives the rate of change of a function, - we can describe the growth assumption precisely using the equation P' = kP. - This equation is called a differential equation, - and these equations are the subject of the current chapter. -

    -     -
    - Graphical and Numerical Solutions to Differential Equations - -

    - In , - we were introduced to the idea of a differential equation. - Given a function y = f(x), - we defined a differential equation - as an equation involving y, - x, and derivatives of y. - We explored the simple differential equation \yp = 2x, - and saw that a solution - to a differential equation is simply a function that satisfies the differential equation. -

    - -     - - Introduction and Terminology - - Differential Equation - -

    - Given a function y=f(x), - a differential equation is an equation relating - x, - y, and derivatives of y. -

      -
    • -

      - The variable x is called the - independent variable. -

      -
      • -
    • -

      - The variable y is called the - dependent variable. -

      -
      • -
    • -

      - The order of the differential equation is the order of the highest derivative of y that appears in the equation. -

      -
      • -
    - differential equationdefinition - orderof a differential equation - differential equationorder of -

    -     -     -

    - Let us return to the simple differential equation - - \yp = 2x - . -

    -

    - To find a solution, - we must find a function whose derivative is 2x. - In other words, we seek an antiderivative of 2x. - The function - - y = x^2 - - is an antiderivative of 2x, - and solves the differential equation. - So do the functions - - y = x^2 + 1 - - and - - y = x^2 - 2346 - . -

    -

    - We call the function - - y = x^2 + C - , - with C an arbitrary constant of integration, - the general solution to the differential equation. - general solutionof a differential equation - differential equationgeneral solution -

    -

    - In order to specify the value of the integration constant C, - we require additional information. - For example, if we know that y(1) = 3, - it follows that C=2. - This additional information is called an - initial condition. -

    - - Initial Value Problem - -

    - A differential equation paired with an initial condition - (or initial conditions) - is called an initial value problem. - initial value problemfor differential equations - initial condition -

    -

    - The solution to an initial value problem is called a - particular solution. - differential equationparticular solution - A particular solution does not include arbitrary constants. -

    -

    - The family of solutions to a differential equation that encompasses all possible solutions is called the - general solution to the differential equation. - differential equationgeneral solution -

    -     -     - - - - - - - - - A simple first-order differential equation - -

    - Solve the differential equation \yp = 2y. -

    -     - -

    - The solution is a function y such that differentiation yields twice the original function. - Unlike our starting example, - finding the solution here does not involve computing an antiderivative. - Notice that - integrating both sides - would yield the result y = \int 2y\,dx, which is not useful. - Without knowledge of the function y, - we can't compute the indefinite integral. - Later sections will explore systematic ways to find analytic solutions to simple differential equations. - For now, a bit of thought might let us guess the solution - - y = e^{2x} - . -

    -

    - Notice that application of the chain rule yields \yp = 2e^{2x} = 2y. - Another solution is given by - - y = -3e^{2x} - . -

    -

    - In fact, - - y = Ce^{2x} - , - where C is any constant, - is the general solution - to the differential equation because \yp = 2Ce^{2x} = 2y. -

    -

    - If we are provided with a single initial condition, - say y(0) = 3/2, we can identify C=3/2 so that - - y = \frac{3}{2}e^{2x} - - is the particular solution - to the initial value problem - - \yp = 2y, \text{ with } y(0) = \frac{3}{2} - . -

    -

    - - shows various members of the general solution to the differential equation \displaystyle \yp = 2y. - Each C value yields a different member of the family, - and a different function. - We emphasize the particular solution corresponding to the initial condition y(0)=3/2. -

    -
    - A representation of some of the members of general solution to the differential equation \yp = 2y, - including the particular solution to the initial value problem with y(0)=\displaystyle 3/2, - from - - - Graph of the members of the general solution to the equation \yp =2y, including a particular solution. - - -

    - The y axis is drawn from -10 to 10 and the x axis is drawn from -2 - to 2. A group of dashed lines emerge very close to the x axis on either side of it. - From left to right, these lines diverge with increasing values of x. At about x =1 the - lines diverge greatly, forming a trumpet shape. -

    -

    - The particular solution to the initial value problem with y(0) = 3/2 from the example is shown - as a curve in the second and the third quadrant that runs along one of the groups of dashed lines that - crosses the y axis at 3/2. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-10.5,ymax=10.75, - xmin=-2,xmax=2.15% - ] - - \foreach \n in {-6,...,6} - { - \addplot[secondcurvestyle,domain=-2:2,samples=50] {\n/2*exp(2*x)}; - } - \addplot[firstcurvestyle,domain=-2:2,samples=50] {3/2*exp(2*x)}; - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -
    -     - -     - - - A second-order differential equation - -

    - Solve the differential equation \yp' + 9y = 0. -

    -     - -

    - We seek a function whose second derivative is negative 9 multiplied by the original function. - Both \sin(3x) and \cos(3x) have this feature. - The general solution to the differential equation is given by - - y = C_1\sin(3x) + C_2\cos(3x) - , - where C_1 and C_2 are arbitrary constants. - To fully specify a particular solution, - we require two additional conditions. - For example, the initial conditions y(0)=1 and - \yp(0)=3 yield C_1 = C_2 = 1. -

    -     - -     - -

    - The differential equation in - is second order, because the equation involves a second derivative. - In general, the number of initial conditions required to specify a particular solution depends on the order of the differential equation. - For the remainder of the chapter, - we restrict our attention to first order differential equations and first order initial value problems. -

    - - - Verifying a solution to the differential equation - -

    - Which of the following is a solution to the differential equation - - \yp + \frac{y}{x} - \sqrt{y} = 0 - ? -

    -

    -

      -
    1. y = C \left( 1 + \ln(x) \right)^2
      2. -
    2. y = \left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right)^2
      3. -
    3. y = C e^{-3x} + \sqrt{\sin(x)}
      4. -
    -

    -     - -

    - Verifying a solution to a differential equation is simply an exercise in differentiation and simplification. - We substitute each potential solution into the differential equation to see if it satisfies the equation. -

    -

    -

      -
    1. -

      - Testing the potential solution y = C \left( 1 + \ln(x) \right)^2: -

      -

      - Differentiating, - we have \displaystyle \yp = \frac{2C(1 + \ln(x))}{x}. - Substituting into the differential equation, - - \amp \frac{2C(1+\ln(x))}{x} + \frac{C(1+\ln(x))^2}{x} -\sqrt{C}(1+\ln(x)) - \amp = (1+\ln(x))\left( \frac{2C}{x} + \frac{C(1+\ln(x))}{x} - \sqrt{C}\right) - \amp \neq 0 - . -

      -

      - Since it doesn't satisfy the differential equation, - y = C(1 + \ln(x))^2 is not a solution. -

      -
      2. - -
    2. -

      - Testing the potential solution \displaystyle y = \left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right)^2: -

      -

      - Differentiating, - we have \displaystyle \yp = 2\left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right)\left( \frac{1}{3} - \frac{C}{2x^{3/2}}\right). - Substituting into the differential equation, - - \amp 2\left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right)\left( \frac{1}{3} - \frac{C}{2x^{3/2}}\right) + \frac{1}{x}\left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right)^2 - \left(\frac{1}{3}x + \frac{C}{\sqrt{x}}\right) - \amp = \left( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right) \left( \frac{2}{3} - \frac{C}{x^{3/2}} + \frac{1}{3} + \frac{C}{x^{3/2}} - 1 \right) - \amp = 0. \text{ (Note how the second parenthetical grouping above reduces to 0.) } - -

      -

      - Thus \displaystyle y = \left(\frac{1}{3}x + \frac{C}{\sqrt{x}} \right)^2 is - a solution to the differential equation. -

      -
      3. - -
    3. -

      - Testing the potential solution \displaystyle y = C e^{-3x} + \sqrt{\sin(x)}: -

      -

      - Differentiating, - \displaystyle \yp = -3Ce^{-3x} + \frac{\cos(x)}{2\sqrt{\sin(x)}}. - Substituting into the differential equation, - - -3Ce^{-3x} + \frac{\cos(x)}{2\sqrt{\sin(x)}} + \frac{C e^{-3x} + \sqrt{\sin(x)}}{x} - \sqrt{C e^{-3x} + \sqrt{\sin(x)}} \neq 0 - . -

      -

      - The function \displaystyle y = C e^{-3x} + \sqrt{\sin(x)} is not - a solution to the differential equation. -

      -
      4. -
    -

    -     - -     - - - - - Verifying a solution to a differential equation - -

    - Verify that x^2+y^2 = Cy is a solution to \displaystyle \yp = \frac{2xy}{x^2-y^2}. -

    -     - -

    - The solution in this example is called an - implicit solution. - That means the dependent variable y is a function of x, - but has not been explicitly solved for. - Verifying the solution still involves differentiation, - but we must take the derivatives implicitly. - Differentiating, we have - - 2x + 2y\yp = C\yp - . - differential equationimplicit soution -

    -

    - Solving for \yp, we have - - \yp = \frac{2x}{C-2y} - . -

    -

    - From the solution, - we know that \displaystyle C = \frac{x^2+y^2}{y}. - Then - - \yp \amp = \frac{2x}{\displaystyle \frac{x^2+y^2}{y} - 2y} - \amp =\frac{2xy}{x^2+y^2-2y^2} - \amp = \frac{2xy}{x^2-y^2} - . -

    -

    - We have verified that x^2+y^2 = Cy is a solution to \displaystyle \yp = \frac{2xy}{x^2-y^2}. -

    -     - -     -     - - - Graphical Solutions to Differential Equations -

    - In the examples we have explored so far, - we have found exact forms for the functions that solve the differential equations. - Solutions of this type are called - analytic solutions. - Many times a differential equation has a solution, - but it is difficult or impossible to find the solution analytically. - This is analogous to algebraic equations. - The algebraic equation x^2 + 3x - 1 = 0 has two real solutions that can be found analytically by using the quadratic formula. - The equation \cos(x) = x has one real solution, - but we can't find it analytically. - As shown in , - we can find an approximate solution graphically by plotting \cos(x) and x and observing the x-value of the intersection. - We can similarly use graphical tools to understand the qualitative behavior of solutions to a first order-differential equation. -

    - -
    - Graphically finding an approximate solution to \cos(x) = x - - - Graphically finding an approximate solution to cos(x)=x. - - -

    - The y and the x axes are drawn from 0 to 1. There is a curve and a - dashed line shown in the graph. The dashed straight line is drawn from the origin and has a - positive slope. The curve starts at point (0, 1) and curves down to about point (1, 0.5). -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=1, - %extra x ticks={.25,.75}, - ymin=-.1,ymax=1.1,% - xmin=-.1,xmax=1.1% - ] - - \addplot [firstcurvestyle,domain=0:1] {cos(deg(x))}; - \addplot [secondcurvestyle,-] coordinates {(0,0) (1,1)}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -
    -

    - Consider the first-order differential equation - - \yp = f(x,y) - . -

    -

    - The function f could be any function of the two variables x and y. - Written in this way, - we can think of the function f as providing a formula to find the slope of a solution at a given point in the xy-plane. - In other words, - suppose a solution to the differential equation passes through the point (x_0,y_0). - At the point (x_0,y_0), - the slope of the solution curve will be f(x_0,y_0). - Since this calculation of the slope is possible at any point (x,y) where the function f(x,y) is defined, - we can produce a plot called a slope field - (or direction field) - that shows the slope of a solution at any point in the xy-plane where the solution is defined. - Further, this process can be done purely by working with the differential equation itself. - In other words, - we can draw a slope field and use it to determine the qualitative behavior of solutions to a differential equation without having to solve the differential equation. - differential equationgraphical solution - direction fieldslope field -

    - - - Slope Field - -

    - A slope field - for a first-order differential equation - \yp = f(x,y) is a plot in the xy-plane made up of short line segments or arrows. - At each point (x_0,y_0) where f(x,y) is defined, - the slope of the line segment is given by f(x_0,y_0). - Plots of solutions to a differential equation are tangent to the line segments in the slope field. - slope field -

    -     -     - - - Sketching a slope field - -

    - Find a slope field for the differential equation \displaystyle \yp = x+y. -

    -     - -

    - Because the function f(x,y) = x+y is defined for all points (x,y), - every point in the xy-plane has an associated line segment. - It is not practical to draw an entire slope field by hand, - but many tools exist for drawing slope fields on a computer. - Here, we explicitly calculate a few of the line segments in the slope field. -

    - -

    -

      -
    • -

      - The slope of the line segment at (0,0) is f(0,0) = 0 + 0 = 0. -

      -
      • -
    • -

      - The slope of the line segment at (1,1) is f(1,1) = 1 + 1 = 2. -

      -
      • -
    • -

      - The slope of the line segment at (1,-1) is f(1,-1) = 1 - 1 = 0. -

      -
      • -
    • -

      - The slope of the line segment at (-2,-1) is f(-2,-1) = -2 - 1 = -3. -

      -
      • -
    -

    - -

    - Though it is possible to continue this process to sketch a slope field, - we usually use a computer to make the drawing. - Most popular computer algebra systems can draw slope fields. - There are also various online tools that can make the drawings. - The slope field for \yp = x+y is shown in . -

    -
    - Slope field for \yp = x+y from - - - Graph of slope field of \yp = y+x. - - -

    - The y and the x axis are shown, the graph shows the slope field as a group of - dashed lines. In the first quadrant, from left to right the slope fields are shown moving - from north-east to north. In the second quadrant the lines are south facing on the bottom - left and curving towards north-east to the top right. In the third quadrant all lines are - facing south-east. In the fourth quadrant the lines are facing south-east on the bottom left - and curving towards north-east to the top right, in the transition the lines are facing east. -

    -     - - \def\length{sqrt(1+(x+y)^2)} - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -2:2, - ytick={-2,-1,0,1,2}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(\length)}, v = {(x+y)/(\length)},scale arrows=.25},samples=9]{0}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - - -
    -     -     - - - A graphical solution to an initial value problem - -

    - Approximate, with a sketch, - the solution to the initial value problem - \displaystyle \yp = x+y, with y(1)=-1. -

    -     - -

    - The solution to the initial value problem should be a continuous smooth curve. - Using the slope field, - we can draw of a sketch of the solution using the following two criteria: -

      -
    1. -

      - The solution must pass through the point (1,-1). -

      -
      2. -
    2. -

      - When the solution passes through a point - (x_0,y_0) it must be tangent to the line segment at (x_0,y_0). -

      -
      3. -
    -

    -

    - Essentially, - we sketch a solution to the initial value problem by starting at the point (1,-1) and - following the lines - in either direction. - A sketch of the solution is shown in . -

    -
    - Solution to the initial value problem \yp = x+y, with y(1)=-1 from - - - Graph of slope field of \yp = y+x with initial value y(1) = -1. - - -

    - The y and the x axis are shown, the graph shows the slope field as a group - of dashed lines. In the first quadrant, from left to right the slope fields are shown - moving from north-east to north. In the second quadrant the lines are south facing on the - bottom left and curving towards north-east to the top right. In the third quadrant all - lines are facing south-east. In the fourth quadrant the lines are facing south-east on - the bottom left and curving towards north-east to the top right, in the transition the - lines are facing east. -

    -

    - The initial value at y(1) = -1 is a curve that moves down starting in the second - quadrant then crossing the third quadrant, in the fourth quadrant it curves up after passing - the point (1,-1) and moves up to touch the x axis. -

    -     - - \def\length{sqrt(1+(x+y)^2)} - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -2:2, - xtick={-2,-1,0,1,2}, - ytick={-2,-1,0,1,2}, - xmin=-2,xmax=2.3 - ] - - \addplot3 [secondcolor,quiver = {u = {1/(\length)}, v = {(x+y)/(\length)},scale arrows=.25},samples=9]{0}; - \addplot [firstcurvestyle,domain=-2.5:2.2,samples=51] {-(x+1)+exp(x-1)}; - - \fill[black,draw=black] (axis cs:1,-1) circle (2.4pt); - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -
    -     -     - - - - - Using a slope field to predict long term behavior - -

    - Use the slope field for the differential equation \yp = y(1-y), - shown in , - to predict long term behavior of solutions to the equation. -

    -
    - Slope field for the logistic differential equation \yp = y(1-y) from - - - Graph of slope field for the logistic differential equation y’=y(1-y) from the example. - - -

    - The y and the t axis are drawn. There are two positions where the slope field - runs parallel to the t axis, once along the t axis itself and again at some - positive value of y. The two horizontal lines appear to divide the slope files into - three distinct parts. Over the horizontal line about some y value, the field lines are - directed south-east. Between the line and the t axis the field lines are directed south-west. - Below the t axis the field lines are again directed south-east. -

    -     - - \def\length{sqrt(1+(y*(1-y))^2)} - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -1:3, - y domain=-.5:1.5, - xtick={-1,0,1,2,3}, - ytick={-.5,0,.5,1}, - ymin=-0.5,ymax=1.5, - xmin=-1,xmax=3, - xlabel={$t$} - ] - - \addplot3 [secondcolor,quiver = {u = {1/(\length)}, v = {(y*(1-y))/(\length)},scale arrows=.1},samples=20]{0}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $t$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -
    -     - -

    - This differential equation, - called the logistic differential equation, often appears in population biology to describe the size of a population. - For that reason, - we use t (time) as the independent variable instead of x. - We also often restrict attention to non-negative y-values because negative values correspond to a negative population. - differential equationlogistic -

    - -

    - Looking at the slope field in , - we can predict long term behavior for a given initial condition. -

      -
    • -

      - If the initial y-value is negative (y(0)\lt 0), - the solution curve must pass though the point - (0,y(0)) and follow the slope field. - We expect the solution y to become more and more negative as time increases. - Note that this result is not physically relevant when considering a population. -

      -
      • -
    • -

      - If the initial y-value is greater than 0 but less than 1, we expect the solution y to increase and level off at y=1. -

      -
      • -
    • -

      - If the initial y-value is greater than 1, we expect the solution y to decrease and level off at y=1. -

      -
      • -
    -

    - -

    - The slope field for the logistic differential equation, - along with representative solution curves, - is shown in . - Notice that any solution curve with positive initial value will tend towards the value y=1. - We call this the carrying capacity. - carrying capacity -

    -
    - Slope field for the logistic differential equation \yp = y(1-y) from - with a few representative solution curves - - - Graph of slope field for the logistic differential equation y’=y(1-y) with representative solution curves. - - -

    - The y and the t axis are drawn. There are two positions where the slope field runs - parallel to the t axis, once along the t axis itself and again at some positive - value of y. The two horizontal lines appear to divide the slope files into three distinct parts. -

    -

    - Over the horizontal line, about some y value, the field lines are directed south-east. - A curve that starts a little before the y axis in the second quadrant, enters the first - quadrant with a negative slope it bends and moves along the horizontal line. -

    -

    - Between the horizontal line and the t axis the field lines are directed south-west. - From left to right, a curve is drawn that moves with a positive slope from the second quadrant - to the first, then becomes parallel to the horizontal line. -

    -

    - Below the t axis the field lines are again directed south-east. From left to right, the - third curve starts in the third quadrant and moves parallel to the t axis then it moves away - from the t axis with a negative slope. -

    -     - - \def\length{sqrt(1+(y*(1-y))^2)} - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -1:3, - y domain=-.5:1.5, - xtick={-1,0,1,2,3}, - ytick={-.5,0,.5,1}, - xlabel={$t$}, - ymin=-0.5,ymax=1.5, - xmin=-1,xmax=3 - ] - - \addplot3 [secondcolor,quiver = {u = {1/(\length)}, v = {(y*(1-y))/(\length)},scale arrows=.1},samples=20]{0}; - \addplot [firstcurvestyle,domain = -1:2, samples=50] {1/(1-10*exp(-x))}; - \addplot [firstcurvestyle,domain = -1:3, samples=50] {1/(1+1*exp(-x))}; - \addplot [firstcurvestyle,domain = -1:3, samples=50] {1/(1-.3*exp(-x))}; - - \end{axis} - %\node [right] at (myplot.right of origin) { $t$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -
    -     -     - - -     - - - Numerical Solutions to Differential Equations: Euler's Method -

    - While the slope field is an effective way to understand the qualitative behavior of solutions to a differential equation, - it is difficult to use a slope field to make quantitative predictions. - For example, - if we have the slope field for the differential equation - \yp = x+y from - along with the initial condition y(0)=1, - we can understand the qualitative behavior of the solution to the initial value problem, - but will struggle to predict a specific value, - y(2) for example, with any degree of confidence. - The most straightforward way to predict y(2) is to find the analytic solution to the the initial value problem and evaluate it at x=2. - Unfortunately, - we have already mentioned that it is impossible to find analytic solutions to many differential equations. - In the absence of an analytic solution, - a numerical solution can serve as an effective tool to make quantitative predictions about the solution to an initial value problem. -

    -

    - There are many techniques for computing numerical solutions to initial value problems. - A course in numerical analysis will discuss various techniques along with their strengths and weaknesses. - The simplest technique is called - Euler's Method. - differential equationnumerical solution -

    - - - -

    - Consider the first-order initial value problem - - \yp = f(x,y), \text{ with } y(x_0) = y_0 - . -

    -

    - Using the definition of the derivative, - - \yp(x) = \lim_{h \to 0} \frac{y(x+h) - y(x)}{h} - . -

    -

    - This notation can be confusing at first, but - y(x) - simply means - the y-value of the solution when the x-value is x, and - y(x+h) - means - the y-value of the solution when the x-value is x+h. -

    -

    - If we remove the limit but restrict h to be - small, - we have - - \yp(x) \approx \frac{y(x+h) - y(x)}{h} - , - so that - - f(x,y) \approx \frac{y(x+h)-y(x)}{h} - , - because \yp = f(x,y) according to the differential equation. - Rearranging terms, - - y(x + h) \approx y(x) + h\,f(x,y) - . -

    -

    - This statement says that if we know the solution (y-value) to the initial value problem for some given x-value, - we can find an approximation for the solution at the value x+h by taking our y-value and adding h times the function f evaluated at the x and y values. - Euler's method uses the initial condition of an initial value problem as the starting point, - and then uses the above idea to find approximate values for the solution y at later x-values. - The algorithm is summarized in . -

    - - Euler's Method -

    - Consider the initial value problem - - \yp = f(x,y) \text{ with } y(x_0)=y_0 - . -

    -

    - Let h be a small positive number and N be an integer. -

      -
    1. -

      - For i = 0, 1, 2, \ldots, N, define - - x_i = x_0 + ih - . -

      -
      2. -
    2. -

      - The value y_0 is given by the initial condition. - For i = 0, 1, 2, \ldots, N-1, define - - y_{i+1} = y_i + hf(x_i,y_i) - . -

      -
      3. -
    -

    -

    - This process yields a sequence of N+1 points - (x_i,y_i) for i= 0,1,2,\ldots,N, - where (x_i, y_i) is an approximation for (x_i,y(x_i)). - Euler's Method -

    -     - - - -

    - Let's practice Euler's Method using a few concrete examples. -

    - - - Using Euler's Method 1 - -

    - Find an approximation at x=2 for the solution to - \yp = x + y with y(1)=-1 using Euler's Method with h=0.5. -

    -     - -

    - Our initial condition yields the starting values x_0 = 1 and y_0 = -1. - With h = 0.5, it takes N=2 steps to get to x=2. - Using steps 1 and 2 from the Euler's Method algorithm, - - x_0 \amp = 1 \amp y_0 \amp = -1 - x_1 \amp = x_0 + h \amp y_1 \amp = y_0 + hf(x_0,y_0) - \amp = 1 + 0.5 \amp \amp = -1 + 0.5(1 - 1) - \amp = 1.5 \amp \amp = -1 - x_2 \amp = x_0 + 2h \amp y_2 \amp = y_1 + hf(x_1,y_1) - \amp = 1 + 2(0.5) \amp \amp = -1 + 0.5(1.5 -1) - \amp = 2 \amp \amp = -0.75 - . -

    -

    - Using Euler's method, we find the approximation y(2) \approx -0.75. -

    - -
    - Euler's Method approximation to \yp = x + y with y(1) = -1 from , - along with the analytical solution to the initial value problem - - - Euler's Method approximation to \yp=x+y with y(1)=-1, with the analytical solution. - - -

    - The y axis is drawn from -1 to 0 and the x axis is drawn from 0 - to 2. The function \yp=x+y with y(1)=-1 is drawn in the fourth quadrant. From left to right from point (1, -1) - the function curves up till point (2, -0.25). There is a plot of three points which are joined - by straight lines; this plot starts at the same point (1,-1) as the curve. The second point is - a (1.5, -1) and the third point is at (2, -0.75). -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,1.5,2}, - ytick={-1,-.5,0}, - ymin=-1.25,ymax=0.1, - xmin=.9,xmax=2.1 - ] - - \addplot [secondcurvestyle,solid,domain=1:2,samples = 51] {-(x+1)+exp(x-1)}; - \addplot [firstcolor, mark = *] coordinates{ (1,-1) (1.5,-1) (2,-0.75)}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -
    - -

    - To help visualize the Euler's method approximation, these three points - (connected by line segments) - are plotted along with the analytical solution to the initial value problem in . -

    -     -     - -

    - This approximation doesn't appear terrific, - though it is better than merely guessing. - Let's repeat the previous example using a smaller h-value. -

    - - - Using Euler's Method 2 - -

    - Find an approximation on the interval [1,2] for the solution to - \yp = x + y with y(1)=-1 using Euler's Method with h=0.25. -

    -     - -

    - Our initial condition yields the starting values x_0 = 1 and y_0 = -1. - With h = 0.25, - we need N=4 steps on the interval [1,2] Using steps 1 and 2 from the Euler's Method algorithm - (and rounding to 4 decimal points), - we have - - x_0 \amp = 1 \amp y_0 \amp = -1 - x_1 \amp = 1.25 \amp y_1 \amp = -1 + 0.25(1 - 1) - \amp \amp \amp = -1 - x_2 \amp = 1.5 \amp y_2 \amp = -1 + 0.25(1.25-1) - \amp \amp \amp = -0.9375 - x_3 \amp = 1.75 \amp y_3 \amp = -0.9375 + 0.25(1.5-0.9375) - \amp \amp \amp = -0.7969 - x_4 \amp = 2 \amp y_4 \amp = -0.7969 + 0.25(1.75 - 0.7969) - \amp \amp \amp = -0.5586 - . -

    - -

    - Using Euler's method, we find y(2) \approx -0.5586. -

    - -

    - These five points, - along with the points from and the analytic solution, - are plotted in . -

    -
    - Euler's Method approximations to \yp = x + y with y(1) = -1 from Examples - and , along with the analytical solution - - - Euler's Method approximation to \yp=x+y with y(1)=-1 along with the analytical solution. - - -

    - The y axis is drawn from -1 to 0 and the x axis is drawn from 0 - to 2. The function is drawn in the fourth quadrant. From left to right from point (1, -1) - the function curves up till point (2, -0.25). -

    -

    - There is a plot of three points which are joined by straight lines; this plot starts at the same point - (1,-1) as the curve. The second point is a (1.5, -1) and the third point is at - (2, -0.75), this plot is marked as h=0.5. -

    -

    - A second plot is drawn with five points every 0.25 interval of the x value. It coincides - with the first plot from x=1 to x=1.25, after which the third, the fourth and the fifth - points of the second plot are all above the first plot at x=1.5, x= 1.75 and x=2. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,1.5,2}, - ytick={-1,-.5,0}, - ymin=-1.25,ymax=0.1, - xmin=.9,xmax=2.35 - ] - - \addplot [secondcolor,thick,domain=1:2,samples = 51] {-(x+1)+exp(x-1)}; - \addplot [firstcolor, mark = *] coordinates{ (1,-1) (1.5,-1) (2,-0.75)}; - \addplot [firstcolor, mark = *,dashed] coordinates{ (1,-1) (1.25,-1) (1.5,-.9375) (1.75,-.7969) (2,-.5586)}; - - \node [right] at (axis cs:2,-0.75) { $h = 0.5$}; - \node [right] at (axis cs:2,-0.5586) { $h = 0.25$}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -
    -     -     - - - -

    - Using the results from Examples - and , - we can make a few observations about Euler's method. - First, the Euler approximation generally gets worse as we get farther from the initial condition. - This is because Euler's method involves two sources of error. - The first comes from the fact that we're using a positive h-value in the derivative approximation instead of using a limit as h approaches zero. - Essentially, - we're using a linear approximation to the solution y - (similar to the process described in on Differentials.) - This error is often called the local truncation error. - The second source of error comes from the fact that every step in Euler's method uses the result of the previous step. - That means we're using an approximate y-value to approximate the next y-value. - Doing this repeatedly causes the errors to build on each other. - This second type of error is often called the propagated - or accumulated error. - accumulated errorusing Euler's method - Euler's methodaccumulated error -

    -

    - A second observation is that the Euler approximation is more accurate for smaller h-values. - This accuracy comes at a cost, though. - - is more accurate than , - but takes twice as many computations. - In general, numerical algorithms - (even when performed by a computer program) - require striking a balance between a desired level of accuracy and the amount of computational effort we are willing to undertake. -

    -

    - Let's do one final example of Euler's Method. -

    - - - Using Euler's Method 3 - -

    - Find an approximation for the solution to the logistic differential equation -

    -

    - \yp = y(1-y) with y(0) = 0.25, - for 0 \leq y \leq 4. - Use N = 10 steps. -

    -     - -

    - The logistic differential equation is what is called an - autonomous equation. - An autonomous differential equation has no explicit dependence on the independent variable - (t in this case). - This has no real effect on the application of Euler's method other than the fact that the function f(t,y) is really just a function of y. - To take steps in the y variable, we use - - y_{i+1} = y_i + hf(t_i,y_i) = y_i + hy_i(1-y_i) - . -

    -

    - Using N=10 steps requires \displaystyle h = \frac{4-0}{10} = 0.4. - Implementing Euler's Method, we have - - x_0 \amp = 0 \amp y_0 \amp = 0.25 - x_1 \amp = 0.4 \amp y_1 \amp = 0.25 + 0.4(0.25)(1-0.25) - \amp \amp \amp = 0.325 - x_2 \amp = 0.8 \amp y_2 \amp = 0.325 + 0.4(0.325)(1-0.325) - \amp \amp \amp = 0.41275 - x_3 \amp = 1.2 \amp y_3 \amp = 0.41275 + 0.4(0.41275)(1-0.41275) - \amp \amp \amp = 0.50970 - x_4 \amp = 1.6 \amp y_4 \amp = 0.50970 + 0.4(0.50970)(1 - 0.50970) - \amp \amp \amp = 0.60966 - x_5 \amp = 2.0 \amp y_5 \amp = 0.60966 + 0.4(0.60966)(1-0.60966) - \amp \amp \amp = 0.70485 - x_6 \amp = 2.4 \amp y_6 \amp = 0.70485 + 0.4(0.70485)(1 - 0.70485) - \amp \amp \amp = 0.78806 - x_7 \amp = 2.8 \amp y_7 \amp = 0.78806 + 0.4(0.78806)(1-0.78806) - \amp \amp \amp = 0.85487 - x_8 \amp = 3.2 \amp y_8 \amp = 0.85487 + 0.4(0.85487)(1-0.85487) - \amp \amp \amp = 0.90450 - x_9 \amp = 3.6 \amp y_9 \amp = 0.90450 + 0.4(0.90450)(1 - 0.90450) - \amp \amp \amp = 0.93905 - x_{10}\amp = 4.0 \amp y_{10}\amp = 0.93905 + 0.4(0.93905)(1 - 0.93905) - \amp \amp \amp = 0.96194 - . -

    -

    - These 11 points, along with the the analytic solution, - are plotted in . - Notice how well they seem to match the true solution. -

    -
    - Euler's Method approximation to \yp = y(1-y) with y(0) = 0.25 from , - along with the analytical solution - - - Euler's Method approximation to \yp = y(1-y) with y(0) = 0.25 along with its analytical solution. - - -

    - The y axis is drawn from 0 to 1 and the x axis is drawn from 0 - to 4. There is a curve and a plot of 11 points, the curve and the plot overlap indicating - the appropriateness of the approximation. - The curve starts at y intercept 0.25 and rises with a positive slope and ends in the graph at - point (4,1), the plot has the points distributed evenly along the curve, with few points slightly - away from the curve. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - xtick={0,1,2,3,4}, - ytick={0,0.5,1}, - xlabel={$t$}, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=4.1 - ] - - \addplot [firstcurvestyle,domain = 0:4, samples=50] {1/(1+3*exp(-x)}; - \addplot [firstcolor, mark = *] coordinates{(0,.25)(0.4,.325)(0.8,.41275)(1.2,.50970)(1.6,.60966)(2.0,.70485)(2.4,.78806)(2.8,.85487)(3.2,.90450)(3.6,.93905)(4.0,.96194)}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $t$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -
    -     -     - - - -

    - The study of differential equations is a natural extension of the study of derivatives and integrals. - The equations themselves involve derivatives, - and methods to find analytic solutions often involve finding antiderivatives. - In this section, - we focus on graphical and numerical techniques to understand solutions to differential equations. - We restrict our examples to relatively simple initial value problems that permit analytic solutions to the equations, - but we should remember that this is only for comparison purposes. - In reality, many differential equations, - even some that appear straightforward, - do not have solutions we can find analytically. - Even so, we can use the techniques presented in this section to understand the behavior of solutions. - In the next two sections, - we explore two techniques to find analytic solutions to two different classes of differential equations. -

    -     - - - - Terms and Concepts - - -

    - In your own words, what is an initial value problem, - and how is it different from a differential equation? -

    -     - - - -

    - An initial value problems is a differential equation that is paired with one or more initial conditions. - A differential equation is simply the equation without the initial conditions. -

    -     -     - - -

    - In your own words, - describe what it means for a function to be a solution to a differential equation. -

    -     - - -     - - -

    - How can we verify that a function is a solution to a differential equation? -

    -     - - - -

    - Substitute the proposed function into the differential equation, - and show the the statement is satisfied. -

    -     -     - - -

    - Describe the difference between a particular solution and a general solution. -

    -     - - - -

    - A particular solution is one specifica member of a family of solutions, - and has no arbitrary constants. - A general solution is a family of solutions, - includes all possible solutions to the differential equation, - and typically includes one or more arbitrary constants. -

    -     -     - - -

    - Why might we use a graphical or numerical technique to study solutions to a differential equation - instead of simply solving the differential equation to find an analytic solution? -

    -     - - - -

    - Many differential equations are impossible to solve analytically. -

    -     -     - - -

    - Describe the considerations that should be made when choosing an h value to use in a numerical method like Euler's Method. -

    -     - - - -

    - A smaller h value leads to a numerical solution that is closer to the true solution, - but decreasing the h value leads to more computational effort. -

    -     -     -     - - - Problems - - -

    - Verify that the given function is a solution to the differential equation or initial value problem. -

    -     - - - -

    - y = Ce^{-6x^2}; \yp = -12xy. -

    -     - - -

    - If y=Ce^{-6x^2}, then \yp = -12xCe^{-6x^2} = -12x(Ce^{-6x^2})=-12xy. -

    -     -     - - -

    - y = x\sin(x); - \yp - x\cos(x)= (x^2+1)\sin(x) - xy, with y(\pi) = 0. -

    -     - - -

    - Given y=x\sin(x), we find \yp = \sin(x)+x\cos(x), using the product rule. - Then, \yp-x\cos(x) = \sin(x); on the other hand, - - (x^2+1)\sin(x)-xy = x^2\sin(x)+\sin(x)-x^2\sin(x)=\sin(x) - . - Finally, we check that y(\pi) = \pi\sin(\pi)=0. -

    -     -     - - -

    - 2x^2-y^2=C; y\yp-2x = 0 -

    -     - - - -

    - Using implicit differentiation, 2x^2-y^2=C - implies that 4x - 2y\yp = 0. - Dividing by -2 gives us y\yp - 2x=0. -

    -     -     - - -

    - y=xe^x; \yp' - 2\yp + y = 0 -

    -     - - - -

    - If y=xe^x, we find \yp = e^x+xe^x, - and \yp' = 2e^x+xe^x. Therefore, - - \yp' - 2\yp + y = 2e^x+xe^x -2e^x-2xe^x + xe^x = 0 - . - -

    -     -     -     - - -

    - Verify that the given function is a solution to the differential equation and find the C - value required to make the function satisfy the initial condition. -

    -     - - -

    - y = 4e^{3x}\sin(x) + Ce^{3x}; - \yp - 3y = 4e^{3x}\cos(x), with y(0)=2 -

    -     - - - -

    - With y = 4e^{3x}\sin(x) + Ce^{3x}, we find - \yp = 12 e^{3x}\sin(x) + 4e^{3x}\cos(x) + 3Ce^{3x}. - This gives \yp - 3y = 4e^{3x}\cos(x), as required. -

    - -

    - We also find y(0) = 4(1)(0)+C(1) = C, so we must have C=2. -

    -     -     - - - -

    - y(x^2+y) = C; 2xy + (x^2+2y)\yp= 0, with y(1)=2 -

    -     - - - -

    - If y(x^2+y)=C, implicit differentiation gives us - - \yp(x^2+y)+y(2x+\yp) = 2xy + 2y\yp +x^2\yp = 0 - , - so the differential equation is satisfied. When x=1, y=2, - which gives 2(1+2)=C, or C=6. - -

    -     -     -     - - - -

    - Sketch a slope field for the given differential equation. - Let x and y range between -2 and 2. -

    -     - - - -

    - \yp = y-x -

    -     - - - - Graph showing slope field for the given differential equation. - - -

    - The x and y axes are uncalibrated.In the first quadrant in the top left, - the field lines are north-east facing and in the bottom right they are southeast facing. - In the second quadrant the field lines are all north-east facing. In the third quadrant - like in the first quadrant in the top left the field lines are northeast facing and in - the bottom right they are southeast facing. In the fourth quadrant all lines are - southeast facing. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -2:2, - ytick={-2,-1,0,1,2}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(y-x)^2))}, v = {(y-x)/(sqrt(1+(y-x)^2))},scale arrows=.25},samples=9]{0}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -     -     - - - -

    - \yp = \displaystyle \frac{x}{2y} -

    -     - - - - Graph showing slope field for the given differential equation. - - -

    - The x and y axes are uncalibrated. The field lines form concentric - ovals facing away from the origin on both positive and negative x and - y axes. The concentric shorter arcs are on either end of the x axis. - On the two ends of the y axis concentric wider arcs are drawn. - The field lines intermix to form an 'X' with centre at the origin. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -2:2, - ytick={-2,-1,0,1,2}, - ] - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(x/(2*y))^2))}, v = {(x/(2*y))/(sqrt(1+(x/(2*y))^2))},scale arrows=.2},samples=10]{0}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -     -     - - - -

    - \yp = \sin(\pi y) -

    -     - - - - Graph showing slope field for the given differential equation. - - -

    - The x and y axes are uncalibrated. There are five instances where - the field lines run parallel to the x axis. One of them is on the x - axis itself, other two pairs of such field lines are above and below the x - axis. In between the x axis and the first horizontal field line for some - positive y value, the field lines are all northeast facing. Above the - horizontal field line for some y value until another with a higher y - value, the field lines in between are southeast facing. -

    -

    - Similarly below the x axis till the first horizontal line with some negative - y value, the field lines in between are southeast facing. In between this - horizontal line and another horizontal line with a higher negative y value, - the field lines are northeast facing. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -2:2, - ytick={-2,-1,0,1,2}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(sin(deg pi*y))^2))}, v = {(sin(deg pi*y))/(sqrt(1+(sin(deg pi*y))^2))},scale arrows=.15},samples=20]{0}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -     -     - - - -

    - \yp = \frac{y}{4} -

    -     - - - - Graph showing slope field for the given differential equation. - - -

    - The x and y axes are uncalibrated. The field lines run - almost parallel to the x axis. Above the axis the field lines - are slightly facing north east. Below the x axis the lines are - directed facing southeast. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -2:2, - ytick={-2,-1,0,1,2}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(y/4)^2))}, v = {(y/4)/(sqrt(1+(y/4)^2))},scale arrows=.2},samples=10]{0}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -     -     -     - - - -

    - Match each slope field below with the appropriate differential equation. -

    - - - - - Graph of slope field used in the exercise. - - -

    - The y and the x axis are shown both uncalibrated. In the second and the third - quadrants the field lines towards larger negative values are south facing. The field lines - closer to the y axis are south-east facing. Around the y axis the field lines - are almost horizontal. In the first and the fourth quadrants the field lines closer to the - y axis are east facing but for greater values of x they are south-east facing. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -2:2, - y domain = -2:2, - ytick={-3,-2,-1,0,1,2,3}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(x*(1-x))^2))}, v = {(x*(1-x))/(sqrt(1+(x*(1-x))^2))},scale arrows=.1},samples=15]{0}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - - - - Graph of slope field used in the exercise. - - -

    - The field lines are concentric facing upwards in the second and the first quadrant along the - positive y axis. The lines furthest away from the origin are more circular and the lines - closest to the x axis are almost parallel to the x axis. Similarly in the third - and the fourth quadrant the field lines are concentric along the negative y axis and the - ones away from origin are more circular and the ones closest to the x axis are almost - parallel to the x axis. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -2:2, - y domain = -2:2, - ytick={-3,-2,-1,0,1,2,3}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(x*y)^2))}, v = {(x*y)/(sqrt(1+(x*y)^2))},scale arrows=.1},samples=15]{0}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -     - -

    - (a) -

    -

    - (b) -

    -     - - - - Graph of slope field used in the exercise. - - -

    - The field lines in the second and the first quadrant are south-east facing and appear to be east - facing when they come very close to the x axis. Similarly the field lines in the third - and the fourth quadrants are north-east facing and become east facing when they come very close - to the x axis. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -2:2, - y domain = -2:2, - ytick={-3,-2,-1,0,1,2,3}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(-y)^2))}, v = {(-y)/(sqrt(1+(-y)^2))},scale arrows=.1},samples=15]{0}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - - - - Graph of slope field used in the exercise. - - -

    - The field lines appear to be concentric dome shaped lines with peaks along the positive y - axis. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -2:2, - y domain = -2:2, - ytick={-3,-2,-1,0,1,2,3}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(-x)^2))}, v = {(-x)/(sqrt(1+(-x)^2))},scale arrows=.1},samples=15]{0}; - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -     - -

    - (c) -

    - -

    - (d) -

    -     -     -     - - - - \yp=xy - (b) - - - \yp = -y - (c) - - - \yp = -x - (d) - - - \yp = x(1-x) - (a) - - - - -     - - - -

    - Sketch the slope field for the differential equation, - and use it to draw a sketch of the solution to the initial value problem. -

    -     - - -

    - \yp = \displaystyle \frac{y}{x} - y, with y(0.5)=1. -

    -     - - - - Graph showing slope field for the given differential equation. - - -

    - The x and y axes are uncalibrated, the field lines in the first quadrant - are shown. The field lines very close to the y axis are almost north facing for - higher values of y and almost east facing for lower values of y. With - smaller values of x, the field lines, from left to right the lines first face - northeast then east and southeast after for greater values of x. -

    -

    - A curve is drawn that starts at a point for some small value of x and a high - value of y. The curve has a positive slope at first after reaching a peak it - declines almost close to the x axis. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - y domain = -.5:1.5, - domain = -.5:5, - minor x tick num=1, - extra x ticks={1,3,5}, - ymin=-.1,ymax=1.6, - xmin=-.1,xmax=5.2 - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(y/x-y)^2))}, v = {(y/x-y)/(sqrt(1+(y/x-y)^2))},scale arrows=.15},samples=15]{0}; - \addplot [firstcurvestyle,domain=0.5:5.1,samples = 51] {2*x*exp(1/2-x)}; - - \fill[black,draw=black] (axis cs:.5,1) circle (2.4pt); - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -     -     - - -

    - \yp = y\sin(x), with y(0)=1. -

    -     - - - - Graph showing slope field for the given differential equation. - - -

    - The x and y axes are uncalibrated, the field lines in the first - quadrant are shown. - Front left to right, a little away from the x axis the field lines are northeast - facing that transition to north facing. Moving further right then again become - northeast facing then transition to southeast facing, further right they become - south facing then east facing. The pattern then repeats. Very close to the x - axis the field lines are almost parallel to it. -

    -

    - A wave is drawn that starts at some y intercept above the origin. It has a high - positive slope, it reaches peak when the field lines change from northeast facing - to southeast facing, then it declines until the point the field lines are parallel - to the x axis. The curve continues to form a second wave. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - domain = -.5:10, - minor x tick num=1, - extra x ticks={2.5,7.5}, - minor y tick num=1, - extra y ticks={2.5,7.5}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(y*sin(deg x))^2))}, v = {(y*sin(deg x))/(sqrt(1+(y*sin(deg x))^2))},scale arrows=.4},samples=15]{0}; - \addplot [firstcurvestyle,domain=0:10,samples = 100] {exp(1-cos(deg x))}; - - \fill[black,draw=black] (axis cs:0,1) circle (2.4pt); - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -     -     - - -

    - \yp = y^2-3y+2, with y(0)=2. -

    -     - - - - Graph showing slope field for the given differential equation. - - -

    - The x and y axes are uncalibrated, the field lines in the first - quadrant are shown. There are two instances where the field lines are parallel - to the x axis. From under the x axis to the first such line the - field lines transition from almost north facing to northeast facing. Between - the horizontal field line for a small y value and a greater y value - the field lines are facing southeast. Above the line with a higher y value - the field lines transition from northeast facing to north facing. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - y domain = -.15:3, - domain = -.15:4, - minor x tick num=1, - extra x ticks={1,3}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(y^2-3*y+2)^2))}, v = {(y^2-3*y+2)/(sqrt(1+(y^2-3*y+2)^2))},scale arrows=.2},samples=15]{0}; - \addplot [firstcurvestyle,-] coordinates {(0,2) (4.1,2)}; - - \fill[black,draw=black] (axis cs:0,2) circle (2.4pt); - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -     -     - - -

    - \displaystyle \yp = -\frac{xy}{1+x^2}, with y(0)=1. -

    -     - - - - Graph showing slope field for the given differential equation. - - -

    - The x and y axes are uncalibrated, the field lines in the first quadrant are shown. - In the top right and the centre the field lines are southeast facing, very close to the x - and y axis the field lines are almost parallel to the x axis. - A curve is drawn that starts from a y intercept and decreases along the slope lines coming - close to the x axis. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - y domain = -.15:3, - domain = -.15:4, - minor x tick num=1, - extra x ticks={1,3}, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(-x*y/(1+x^2))^2))}, v = {(-x*y/(1+x^2))/(sqrt(1+(-x*y/(1+x^2))^2))},scale arrows=.2},samples=15]{0}; - \addplot [firstcurvestyle,domain=0:4,samples=50] {1/sqrt(1+x^2)}; - - \fill[black,draw=black] (axis cs:0,1) circle (2.4pt); - - \end{axis} - - %\node [right] at (myplot.right of origin) { $x$}; - %\node [above] at (myplot.above origin) { $y$}; - - \end{tikzpicture} - - -     -     -     - - - -

    - Use Euler's Method to make a table of values that approximates the solution to the initial value problem on the given interval. - Use the specified h or N value. -

    -     - - - -

    - \yp = x+2y -

    -

    - y(0)=1 -

    -

    - interval: [0,1] -

    -

    - h=0.25 -

    -     - -

    - - x_i \amp \quad \amp \quad \amp y_i - 0.00 \amp \quad \amp \quad \amp 1.0000 - 0.25 \amp \quad \amp \quad \amp 1.5000 - 0.50 \amp \quad \amp \quad \amp 2.3125 - 0.75 \amp \quad \amp \quad \amp 3.5938 - 1.00 \amp \quad \amp \quad \amp 5.5781 - -

    -     -     - - - -

    - \yp = xe^{-y} -

    -

    - y(0)=1 -

    -

    - interval: [0,0.5] -

    -

    - N=5 -

    -     - -

    - - x_i \amp \quad \amp \quad \amp y_i - 0.0 \amp \quad \amp \quad \amp 1.0000 - 0.1 \amp \quad \amp \quad \amp 1.0000 - 0.2 \amp \quad \amp \quad \amp 1.0037 - 0.3 \amp \quad \amp \quad \amp 1.0110 - 0.4 \amp \quad \amp \quad \amp 1.0219 - 0.5 \amp \quad \amp \quad \amp 1.0363 - -

    -     -     - - - -

    - \yp = y + \sin(x) -

    -

    - y(0)=2 -

    -

    - interval: [0,1] -

    -

    - h = 0.2 -

    -     - -

    - - x_i \amp \quad \amp \quad \amp y_i - 0.0 \amp \quad \amp \quad \amp 2.0000 - 0.2 \amp \quad \amp \quad \amp 2.4000 - 0.4 \amp \quad \amp \quad \amp 2.9197 - 0.6 \amp \quad \amp \quad \amp 3.5816 - 0.8 \amp \quad \amp \quad \amp 4.4108 - 1.0 \amp \quad \amp \quad \amp 5.4364 - -

    -     -     - - - -

    - \yp = e^{x-y} -

    -

    - y(0)=0 -

    -

    - interval: [0,2] -

    -

    - h = 0.5 -

    -     - -

    - - x_i \amp \quad \amp \quad \amp y_i - 0.0 \amp \quad \amp \quad \amp 0.0000 - 0.5 \amp \quad \amp \quad \amp 0.5000 - 1.0 \amp \quad \amp \quad \amp 1.8591 - 1.5 \amp \quad \amp \quad \amp 10.5824 - 2.0 \amp \quad \amp \quad \amp 88378.1190 - -

    -     -     -     - - - -

    - Use the provided solution y(x) and Euler's Method with the h=0.2 and h=0.1 to complete the following table. -

    - - - - - - - - - - x 0.0 0.20.4 0.6 0.8 1.0 - - - y(x) - - - h = 0.2 - - - h = 0.1 - - -     - - - -

    - \yp = xy^2 -

    -

    - y(0)=1 -

    -

    - Solution: \displaystyle y(x) = \frac{2}{1-x^2} -

    -     - - - - - - - - - - - x 0.0 0.20.4 0.6 0.8 1.0 - - - y(x)1.00001.02041.08701.21951.47062.0000 - - - h = 0.21.00001.00001.04001.12651.27881.5405 - - - h = 0.11.00001.01001.06231.16871.36011.7129 - - - -     - - -

    - \displaystyle \yp = xe^{x^2}+\frac{1}{2}xy -

    -

    - \displaystyle y(0)=\frac{1}{2} -

    -

    - Solution: \displaystyle y(x) = \frac{1}{2}(x^2+1)e^{x^2} -

    -     - - - - - - - - - - - x 0.0 0.20.4 0.6 0.8 1.0 - - - y(x)0.50000.54120.68060.97471.55512.7183 - - - h = 0.20.50000.50000.58160.76861.12501.7885 - - - h = 0.10.50000.52010.62820.86221.31322.1788 - - - -     -     -     -     -
    -
    - Separable Differential Equations - -

    - There are specific techniques that can be used to solve specific types of differential equations. - This is similar to solving algebraic equations. - In algebra, we can use the quadratic formula to solve a quadratic equation, - but not a linear or cubic equation. - In the same way, - techniques that can be used for a specific type of differential equation are often ineffective for a differential equation of a different type. - In this section, - we describe and practice a technique to solve a class of differential equations called - separable equations. -

    -     - - - Separation of Variables - - - Separable Differential Equation - -

    - A separable differential equation is one that can be written in the form - - \displaystyle n(y) \frac{dy}{dx} = m(x) - , - where n is a function that depends only on the dependent variable y, - and m is a function that depends only on the independent variable x. - differential equationseparable -

    -     -     -

    - Below, we show a few examples of separable differential equations, - along with similar looking equations that are not separable. -

    - - - Separable -
      -
    1. \displaystyle \frac{dy}{dx} = x^2y
      2. -
    2. \displaystyle y\sqrt{y^2-5} \frac{dy}{dx} - \sin(x) \cos(y) = 0
      3. -
    3. \displaystyle \frac{dy}{dx} = \frac{(x^2 + 1)e^{y}}{y}
      4. -
    -     - - Not Separable -
      -
    1. \displaystyle \frac{dy}{dx} = x^2 + y
      2. -
    2. \displaystyle y\sqrt{y^2-5} \frac{dy}{dx} - \sin(x) \cos(y) = 1
      3. -
    3. \displaystyle \frac{dy}{dx} = \frac{(xy + 1)e^{y}}{y}
      4. -
    -     -     -

    - Notice that a separable equation requires that the functions of the dependent and independent variables be multiplied, - not added - (like Item in List). - An alternate definition of a separable differential equation states that an equation is separable if it can be written in the form - - \frac{dy}{dx} = f(x)g(y) - , - for some functions f and g. -

    - - - -

    - separation of variables - Let's find a formal solution to the separable equation - - \displaystyle n(y) \frac{dy}{dx} = m(x) - . -

    -

    - Since the functions on the left and right hand sides of the equation are equal, - their antiderivatives should be equal up to an arbitrary constant of integration. - That is - - \displaystyle \int n(y) \frac{dy}{dx}\,dx = \int m(x)\, dx + C - . -

    -

    - Though the integral on the left may look a bit strange, - recall that y itself is a function of x. - Consider the substitution u = y(x). - The differential is du = \displaystyle \frac{dy}{dx}\,dx. - Using this substitution, the above equation becomes - - \int n(u)\,du = \int m(x)\,dx + C - . -

    -

    - Let N(u) and M(x) be antiderivatives of n(u) and m(x), - respectively. - Then - - N(u) = M(x) + C - . -

    -

    - Since u = y(x), this is - - N(y) = M(x) + C - . -

    -

    - This relationship between y and x is an implicit form of the solution to the differential equation. - Sometimes (but not always) it is possible to solve for y to find an explicit version of the solution. -

    -

    - Though the technique outlined above is formally correct, - what we did essentially amounts to integrating the function n with respect to its variable and integrating the function - m with respect to its variable. - The informal way to solve a separable equation is to treat the derivative - \displaystyle \frac{dy}{dx} as if it were a fraction. - The separated form of the equation is - - n(y)\,dy = m(x)\, dx - . -

    -

    - To solve, we integrate the left hand side with respect to y and the right hand side with respect to x and add a constant of integration. - As long as we are able to find the antiderivatives, - we can find an implicit form for the solution. - Sometimes we are able to solve for y in the implicit solution to find an explicit form of the solution to the differential equation. - We practice the technique by solving the three differential equations listed in the separable column above, - and conclude by revisiting and finding the general solution to the logistic differential equation from . -

    - - Solving a Separable Differential Equation - -

    - Find the general solution to the differential equation \yp = x^2y. -

    -     - -

    - Using the informal solution method outlined above, - we treat \displaystyle \frac{dy}{dx} as a fraction, - and write the separated form of the differential equation as - - \frac{dy}{y} = x^2 dx - . -

    - - - -

    - Integrating the left hand side of the equation with respect to y and the right hand side of the equation with respect to x yields - - \ln \abs{y} = \frac{1}{3}x^3 + C - . -

    -

    - This is an implicit form of the solution to the differential equation. - Solving for y yields an explicit form for the solution. - Exponentiating both sides, we have - - \abs{y} = e^{x^3/3 + C} = e^{x^3/3}e^C - . -

    - -

    - This solution is a bit problematic. - First, the absolute value makes the solution difficult to understand. - The second issue comes from our desire to find the - general solution. - Recall that a general solution includes all possible solutions to the differential equation. - In other words, for any given initial condition, - the general solution must include the solution to that specific initial value problem. - We can often satisfy any given initial condition by choosing an appropriate C value. - When solving separable equations, though, - it is possible to lose solutions that have the form y = \text{ constant}. - Notice that y=0 solves the differential equation, - but it is not possible to choose a finite C to make our solution look like y=0. - Our solution cannot solve the initial value problem - \displaystyle \frac{dy}{dx} = x^2y, with y(a) = 0 - (where a is any value). - Thus, we haven't actually found a general solution to the problem. - We can clean up the solution and recover the missing solution with a bit of clever thought. -

    - - - -

    - Recall the formal definition of the absolute value: - \abs{y} = y if y \geq 0 and \abs{y} = -y if y \lt 0. - Our solution is either y = e^C e^{\frac{x^3}{3}} or y = - e^C e^{{\frac{x^3}{3}}}. - Further, note that C is constant, - so e^C is also constant. - If we write our solution as y = Ae^{\frac{x^3}{3}}, - and allow the constant A to take on either positive or negative values, - we incorporate both cases of the absolute value. - Finally, if we allow A to be zero, - we recover the missing solution discussed above. - The best way to express the general solution to our differential equation is - - y = Ae^{\frac{x^3}{3}} - . -

    -     - -     - - - - - Solving a Separable Initial Value Problem - -

    - Solve the initial value problem - \displaystyle (y\sqrt{y^2-5}) \yp - \sin(x) \cos(x) = 0, - with y(0) = -3. -

    -     - -

    - We first put the differential equation in separated form - - y\sqrt{y^2-5}\,dy = \sin(x) \cos(x)\, dx - . -

    -

    - The indefinite integral \displaystyle \int y\sqrt{y^2-5}\,dy requires the substitution u = y^2-5. - Using this substitute yields the antiderivative \displaystyle \frac{1}{3} (y^2-5)^{3/2}. - The indefinite integral \displaystyle \int \sin(x) \cos(x)\,dx requires the substitution u = \sin(x). - Using this substitution yields the antiderivative \displaystyle \frac{1}{2} \sin^2 x. - Thus, we have an implicit form of the solution to the differential equation given by - - \frac{1}{3} (y^2-5)^{3/2} = \frac{1}{2} \sin^2 x + C - . -

    -

    - The initial condition says that y should be -3 when x is 0, or - - \frac{1}{3} ((-3)^2 - 5)^{3/2} = \frac{1}{2} \sin^2 0 + C - . -

    -

    - Evaluating the line above, we find C = 8/3, - yielding the particular solution to the initial value problem - - \frac{1}{3} (y^2-5)^{3/2} = \frac{1}{2} \sin^2 x + \frac{8}{3} - . -

    -     - -     - - - Solving a Separable Differential Equation - -

    - Find the general solution to the differential equation \displaystyle \frac{dy}{dx} = \frac{(x^2 + 1)e^{y}}{y}. -

    -     - -

    - We start by observing that there are no constant solutions to this differential equation because there are no constant y - values that make the right hand side of the equation identically zero. - Thus, we need not worry about losing solutions during the separation of variables process. - The separated form of the equation is given by - - ye^{-y}\,dy = (x^2+1)\,dx - . -

    -

    - The antiderivative of the left hand side requires Integration by Parts. - Evaluating both indefinite integrals yields the implicit solution - - -(y+1)e^{-y} = \frac{1}{3}x^3 + x + C - . -

    -

    - Since we cannot solve for y, - we cannot find an explicit form of the solution. -

    -     - -     - - - - - Solving the Logistic Differential Equation - -

    - Solve the logistic differential equation \displaystyle \frac{dy}{dt} = ky\left( 1 - \frac{y}{M}\right) -

    -     - -

    - We looked at a slope field for this equation in - in the specific case of k = M = 1. - Here, we use separation of variables to find an analytic solution to the more general equation. - Notice that the independent variable t does not explicitly appear in the differential equation. - We mentioned that an equation of this type is called autonomous. - All autonomous first order differential equations are separable. -

    -

    - We start by making the observation that both y=0 and y = M are constant solutions to the differential equation. - We must check that these solutions are not lost during the separation of variables process. - The separated form of the equation is - - \frac{1}{y \left(1-\displaystyle\frac{y}{M}\right)}\,dy = k\,dt - . -

    -

    - The antiderivative of the left hand side of the equation can be found by making use of partial fractions. - Using the techniques discussed in , we write - - \frac{1}{y\left(1-\frac{y}{M}\right)} = \frac{1}{y} + \frac{1}{M-y} - . -

    -

    - Then an implicit form of the solution is given by - - \ln \abs{y} - \ln \abs{M-y} = kt + C - . -

    -

    - Combining the logarithms, - - \ln \abs{\frac{y}{M-y}} = kt + C - . -

    -

    - Similarly to , - we can write - - \frac{y}{M-y} = Ae^{kt} - . -

    -

    - Letting A take on positive values or negative values incorporates both cases of the absolute value. - This is another implicit form of the solution. - Solving for y gives the explicit form - - y = \frac{M}{1 + be^{-kt}} - , - where b is an arbitrary constant. - Notice that b=0 recovers the constant solution y = M. - The constant solution y=0 cannot be produced with a finite b value, - and has been lost. - The general solution the logistic differential equation is the set containing \displaystyle y = \frac{M}{1 + be^{-kt}} and y=0. -

    -     - -     - - -     - - - - Problems - - -

    - Decide whether the differential equation is separable or not separable. - If the equation is separable, write it in separated form. -

    -     - - -

    - \displaystyle \yp = y^2 - y -

    -     - -

    - Separable. - \displaystyle \frac{1}{y^2-y}\,dy = dx -

    -     -     - - -

    - \displaystyle x\yp + x^2y = \frac{\sin(x)}{x-y} -

    -     - -

    - Not separable. -

    -     -     - - -

    - \displaystyle (y + 3)\yp + (\ln(x)) \yp - x\sin(y) = (y+3)\ln(x) -

    -     - -

    - Not separable. -

    -     -     - - -

    - \displaystyle \yp -x^2\cos(y) + y = \cos(y) - x^2 y -

    -     - -

    - Separable. - \displaystyle \frac{1}{\cos(y) - y}\,dy = (x^2+1)\,dx -

    -     -     -     - - -

    - Find the general solution to the separable differential equation. - Be sure to check for missing constant solutions. -

    -     - - -

    - \displaystyle \yp +1 - y^2 = 0 -

    -     - -

    - \left\{ \displaystyle y = \frac{1 + Ce^{2x}}{1 - Ce^{2x}}, y = -1\right\} -

    -     -     - - -

    - \displaystyle \yp = y-2 -

    -     - -

    - y = 2 + Ce^x -

    -     -     - - -

    - \displaystyle x \yp = 4y -

    -     - -

    - y = Cx^4 -

    -     -     - - -

    - \displaystyle y\yp = 4x -

    -     - -

    - y^2 - 4x^2 = C -

    -     -     - - -

    - \displaystyle e^xy \yp = e^{-y} + e^{-2x - y} -

    -     - -

    - \displaystyle (y-1)e^y = -e^{-x} - \frac{1}{3}e^{-3x} + C -

    -     -     - - -

    - \displaystyle (x^2 + 1) \yp = \frac{x}{y-1} -

    -     - -

    - \displaystyle (y-1)^2 = \ln(x^2+1) + C -

    -     -     - - -

    - \displaystyle \yp = \frac{x\sqrt{1-4y^2}}{x^4 + 2x^2 + 2} -

    -     - -

    - \left\{ \arcsin{2y} - \arctan(x^2+1) = C, y = \pm \displaystyle \frac{1}{2} \right\} -

    -     -     - - -

    - \displaystyle (e^x + e^{-x})\yp = y^2 -

    -     - -

    - \left\{ \displaystyle y = \frac{1}{C - \arctan x}, y = 0 \right\} -

    -     -     -     - - -

    - Find the particular solution to the separable initial value problem. -

    -     - - -

    - \displaystyle \yp = \frac{\sin(x)}{\cos(y)}, - with y(0) = \displaystyle \frac{\pi}{2} -

    -     - -

    - \sin(y) + \cos(x) = 2 -

    -     -     - - -

    - \displaystyle \yp = \frac{x^2}{1-y^2}, with y(0) = -2 -

    -     - -

    - -x^3 + 3y - y^3 = 2 -

    -     -     - - -

    - \displaystyle \yp = \frac{2x}{y+x^2y}, with y(0) = -4 -

    -     - -

    - \frac{1}{2}y^2 - \ln(1+x^2) = 8 -

    -     -     - - -

    - \displaystyle x + ye^{-x}\yp = 0, with y(0) = -2 -

    -     - -

    - y^2+2xe^x - 2e^x = 2 -

    -     -     - - -

    - \displaystyle \yp = \frac{x\ln(x^2+1)}{y-1}, with y(0) = 2 -

    -     - -

    - \displaystyle \frac{1}{2}y^2 - y = \frac{1}{2}\big ( (x^2+1)\ln(x^2+1) - (x^2 + 1)\big) + \frac{1}{2} -

    -     -     - - -

    - \displaystyle \sqrt{1-x^2}\,\yp - \frac{\arcsin x}{y\cos(y^2)}= 0, - with y(0) = \sqrt{\displaystyle\frac{7\pi}{6}} -

    -     - -

    - \sin(y^2)-(\arcsin x)^2 = -\frac{1}{2} -

    -     -     - - -

    - \displaystyle \yp = (\cos^2(x))(\cos^2 (2y)), with y(0) = 0 -

    -     - -

    - 2\tan(2y) = 2x + \sin(2x) -

    -     -     - - -

    - \displaystyle \yp = \frac{y^2\sqrt{1-y^2}}{x}, with y(0) = 1 -

    -     - -

    - x = exp \displaystyle \left( -\frac{\sqrt{1-y^2}}{y}\right) -

    -     -     -     -     -     -
    -
    - First Order Linear Differential Equations - -

    - In the previous section, - we explored a specific techique to solve a specific type of differential equation called a separable differential equation. - In this section, - we develop and practice a technique to solve a type of differential equation called a - first order linear differential equation. -

    - -

    - Recall than a linear algebraic equation in one variable is one that can be written ax + b = 0, - where a and b are real numbers. - Notice that the variable x appears to the first power. - The equations \sqrt{x}+1=0 and \sin(x)-3x = 0 are both nonlinear. - A linear differential equation is one in which the dependent variable and its derivatives appear only to the first power. - We focus on first order equations, which involve first - (but not higher order) - derivatives of the dependent variable. -

    - -     - - - Solving First Order Linear Equations - - - First Order Linear Differential Equation - -

    - A first order linear differential equation is a differential equation that can be written in the form - - \frac{dy}{dx} + p(x)y = q(x) - , - where p and q are arbitrary functions of the independent variable x. - differential equationfirst order linear -

    -     -     - - - Classifying Differential Equations - -

    - Classify each differential equation as first order linear, - separable, both, or neither. -

    -

    -

      -
    1. \displaystyle \yp = xy
      2. -
    2. \displaystyle \yp = e^y + 3x
      3. -
    3. \displaystyle \yp - (\cos(x))y = \cos(x)
      4. -
    4. \displaystyle y\yp -3xy = 4\ln(x)
      5. -
    -

    -     - -

    -

      -
    1. -

      Both. - We identify p(x) = -x and q(x) = 0. - The separated form of the equation is \displaystyle \frac{dy}{y} = x\,dx. -

      -
      2. -
    2. -

      - Neither. - The e^y term makes the equation nonlinear. - Because of the addition, - it is not possible to write the equation in separated form. -

      -
      3. -
    3. -

      - First order linear. - We identify p(x) = -\cos(x) and q(x) = \cos(x). - The equation cannot be written in separated form. -

      -
      4. -
    4. -

      - Neither. - Notice that dividing by y results in the nonlinear term \displaystyle \frac{4\ln(x)}{y}. - It is not possible to write the equation in separated form. -

      -
      5. -
    -

    -     -     - - - -

    - Notice that linearity depends on the dependent variable y, - not the independent variable x. - The functions p(x) and q(x) need not be linear, - as demonstrated in part (c) of . - Neither \cos(x) nor \sin(x) are linear functions of x, - but the differential equation is still linear. -

    - -

    - Before working out a general technique for solving first order linear differential equations, - we look at a specific example. - Consider the differential equation - - \frac{d}{dx}\bigl(xy\bigr) = \sin(x) \cos(x) - . -

    -

    - This is an easy differential equation to solve. - On the left, - the antiderivative of the derivative is simply the function xy. - Using the substitution u = \sin(x) on the right and integrating results in the implicit solution - - xy = \frac{1}{2}\sin^2 x + C - . -

    -

    - Solving for y yields the explicit solution - - y = \frac{\sin^2 x}{2x} + \frac{C}{x} - . -

    -

    - Though not obvious, - the differential equation above is actually a linear differential equation. - Using the product rule and implicit differentiation, - we can write \displaystyle \frac{d}{dx}\big(xy\big) = x\frac{dy}{dx} + y. - Our original differential equation can be written - - x\frac{dy}{dx} + y = \sin(x) \cos(x) - . -

    -

    - If we divide by x, we have - - \frac{dy}{dx} + \frac{1}{x} y = \frac{\sin(x) \cos(x)}{x} - , - which matches the form in . - Reversing our steps would lead us back to the original form our our differential equation. -

    - - -

    - As motivated by the problem we just explored, - the basic idea behind solving first order linear differential equations is to multiply both sides of the differential equation by a function, - called an integrating factor, - that makes the left hand side of the equation look like an expanded Product Rule. - We then condense the left hand side into the derivative of a product and integrate both sides. - An obvious question is, - How do you find this integrating factor? - differential equationintegrating factor -

    - - -

    - Consider the first order linear equation - - \frac{dy}{dx} + p(x)y = q(x) - . -

    -

    - Let's call the integrating factor \mu(x). - We multiply both sides of the differential equation by \mu(x) to get - - \mu(x) \left( \frac{dy}{dx} + p(x)y \right) = \mu(x)q(x) - . -

    - - - -

    - Our goal is to choose \mu(x) so that the left hand side of the differential equation looks like the result of a Product Rule. - The left hand side of the equation is - - \mu(x) \frac{dy}{dx} + \mu(x)p(x)y - . -

    -

    - Using the Product Rule and Implicit Differentiation, - - \frac{d}{dx} \bigl( \mu(x) y \bigr) = \frac{d\mu}{dx}y + \mu(x)\frac{dy}{dx} - . -

    - -

    - Equating - \frac{d}{dx} \big ( \mu(x) y \big ) and \mu(x) \left( \frac{dy}{dx} + p(x)y \right) gives - - \frac{d\mu}{dx}y + \mu(x)\frac{dy}{dx} = \mu(x) \frac{dy}{dx} + \mu(x)p(x)y - , - which is equivalent to - - \frac{d\mu}{dx} = \mu(x)p(x) - . -

    -

    - In order for the integrating factor \mu(x) to perform its job, - it must solve the differential equation above. - But that differential equation is separable, so we can solve it. - The separated form is - - \frac{d\mu}{\mu} = p(x)\,dx - . -

    -

    - Integrating, - - \ln(\mu) = \int p(x)\,dx - , - or - - \mu(x) = e^{\int p(x)\,dx} - . -

    - - - -

    - If \mu(x) is chosen this way, - after multiplying by \mu(x), - we can always write the differential equation in the form - - \frac{d}{dx} \bigl( \mu(x)y \bigr) = \mu(x)q(x) - . -

    -

    - Integrating and solving for y, the explicit solution is - - y = \frac{1}{\mu(x)}\int \bigl( \mu(x)q(x) \bigr)\,dx - . -

    -

    - Though this formula can be used to write down the solution to a first order linear equation, - we shy away from simply memorizing a formula. - The process is lost, and it's easy to forget the formula. - Rather, we always always follow the steps outlined in - when solving equations of this type. -

    - - - Solving First Order Linear Equations -

    -

      -
    1. -

      - Write the differential equation in the form - - \frac{dy}{dx} + p(x)y = q(x) - . -

      -
      2. -
    2. -

      - Compute the integrating factor - - \mu(x) = e^{\int p(x)\,dx} - . -

      -
      3. -
    3. -

      - Multiply both sides of the differential equation by \mu(x), - and condense the left hand side to get - - \frac{d}{dx}\big( \mu(x)y \big) = \mu(x)q(x) - . -

      -
      4. -
    4. -

      - Integrate both sides of the differential equation with respect to x, - taking care to remember the arbitrary constant. -

      -
      5. -
    5. -

      - Solve for y to find the explicit solution to the differential equation. -

      -
      6. -
    -

    -     - -

    - Let's practice the process by solving the two first order linear differential equations from . -

    - - - Solving a First Order Linear Equation - -

    - Find the general solution to \yp = xy. -

    -     - -

    - We solve by following the steps in . - Unlike the process for solving separable equations, - we need not worry about losing constant solutions. - The answer we find will - be the general solution to the differential equation. - We first write the equation in the form - - \frac{dy}{dx} - xy = 0 - . -

    -

    - By identifying p(x) = -x, - we can compute the integrating factor - - \mu(x) = e^{\int -x\,dx} = e^{-\frac{1}{2}x^2} - . -

    -

    - Multiplying both side of the differential equation by \mu(x), we have - - e^{-\frac{1}{2}x^2}\left( \frac{dy}{dx} - xy\right) = 0 - . -

    -

    - The left hand side of the differential equation condenses to yield - - \frac{d}{dx} \left( e^{-\frac{1}{2}x^2}y\right) = 0 - . -

    - - - -

    - We integrate both sides with respect to x to find the implicit solution - - e^{-\frac{1}{2}x^2}y = C - , - or the explicit solution - - y = Ce^{\frac{1}{2}x^2} - . -

    -     - -     - - - - - Solving a First Order Linear Equation - -

    - Find the general solution to \yp -(\cos(x))y = \cos(x). -

    -     - -

    - The differential equation is already in the correct form. - The integrating factor is given by - - \mu(x) = e^{-\int\cos(x)\,dx} = e^{-\sin(x)} - . -

    -

    - Multiplying both sides of the equation by the integrating factor and condensing, - - \frac{d}{dx}\left(e^{-\sin(x)}y \right) = (\cos(x)) e^{-\sin(x)} - -

    -

    - Using the substitution u = -\sin(x), - we can integrate to find the implicit solution - - e^{-\sin(x)} y = -e^{-\sin(x)} + C - . -

    -

    - The explicit form of the general solution is - - y = -1 + Ce^{\sin(x)} - . -

    -     - -     - -

    - We continue our practice by finding the particular solution to an initial value problem. -

    - - - Solving a First Order Linear Initial Value Problem - -

    - Solve the initial value problem \displaystyle x\yp - y = x^3\ln(x), - with y(1)=0. -

    -     - -

    - We first divide by x to get - - \frac{dy}{dx}-\frac{1}{x}y = x^2\ln(x) - . -

    -

    - The integrating factor is given by - - \mu(x) \amp = e^{\int-\frac{1}{x}\,dx} - \amp = e^{-\ln(x)} - \amp = e^{\ln(x)^{-1}} - \amp = x^{-1} - . -

    -

    - Multiplying both sides of the differential equation by the integrating factor and condensing the left hand side, we have - - \frac{d}{dx} \left( \frac{y}{x}\right) = x\ln(x) - . -

    -

    - Using Integrating by Parts to find the antiderivative of x\ln(x), - we find the implicit solution - - \frac{y}{x} = \frac{1}{2}x^2\ln(x) - \frac{1}{4}x^2 + C - . -

    -

    - Solving for y, the explicit solution is - - y = \frac{1}{2}x^3\ln(x) - \frac{1}{4}x^3 + Cx - . -

    -

    - The initial condition y(1) = 0 yields C = 1/4. - The solution to the initial value problem is - - y = \frac{1}{2}x^3\ln(x) - \frac{1}{4}x^3 + \frac{1}{4}x - . -

    -     - -     - -

    - Differential equations are a valuable tool for exploring various physical problems. - This process of using equations to describe real world situations is called mathematical modeling, - and is the topic of the next section. - The last two examples in this section begin our discussion of mathematical modeling. -

    - - - A Falling Object Without Air Resistance - -

    - Suppose an object with mass m is dropped from an airplane. - Find and solve a differential equation describing the vertical velocity of the object assuming no air resistance. -

    -     - -

    - The basic physical law at play is Newton's second law, - - \text{ mass } \times \text{ acceleration } = \text{ the sum of the forces } - . -

    - -

    - Using the fact that acceleration is the derivative of velocity, - mass acceleration can be writting mv'. - In the absence of air resistance, - the only force of interest is the force due to gravity. - This force is approximately constant, - and is given by mg, where g is the gravitational constant. - The word equation above can be written as the differential equation - - m\frac{dv}{dt} = mg - . -

    - -

    - Because g is constant, - this differential equation is simply an integration problem, - and we find - - v = gt + C - . -

    - -

    - Since v = C with t=0, - we see that the arbitrary constant here corresponds to the initial vertical velocity of the object. -

    -     -     - -

    - The process of mathematical modeling does not stop simply because we have found an answer. - We must examine the answer to see how well it can describe real world observations. - In the previous example, - the answer may be somewhat useful for short times, - but intuition tells us that something is missing. - Our answer says that a falling object's velocity will increase linearly as a function of time, - but we know that a falling object does not speed up indefinitely. - In order to more fully describe real world behavior, - our mathematical model must be revised. -

    - - - - - A Falling Object with Air Resistance - -

    - Suppose an object with mass m is dropped from an airplane. - Find and solve a differential equation describing the vertical velocity of the object, - taking air resistance into account. -

    -     - -

    - We still begin with Newon's second law, - but now we assume that the forces in the object come both from gravity and from air resistance. - The gravitational force is still given by mg. - For air resistance, - we assume the force is related to the velocity of the object. - A simple way to describe this assumption might be kv^{p}, - where k is a proportionality constant and p is a positive real number. - The value k depends on various factors such as the density of the object, - surface area of the object, and density of the air. - The value p affects how changes in the velocity affect the force. - Taken together, - a function of the form kv^{p} is often called a power law. - The differential equation for the velocity is given by - - m\frac{dv}{dt} = mg - kv^{p} - . -

    - -

    - (Notice that the force from air resistance opposes motion, - and points in the opposite direction as the force from gravity.) This differential equation is separable, - and can be written in the separated form - - \frac{m}{mg - kv^{p}}\,dv = dt - . -

    - -

    - For arbitrary positive p, - the integration is difficult, - making this problem hard to solve analytically. - In the case that p=1, the differential equation becomes linear, - and is easy to solve either using either separation of variables or integrating factor techniques. - We assume p=1, - and proceed with an integrating factor so we can continue practicing the process. - Writing - - \frac{dv}{dt} + \frac{k}{m}v = g - , - we identify the integrating factor - - \mu(t) = e^{\int \frac{k}{m}\,dt} = e^{\frac{k}{m}t} - . -

    - -

    - Then - - \frac{d}{dt}\left(e^{\frac{k}{m}t}v \right) = ge^{\frac{k}{m}t} - , - so - - e^{\frac{k}{m}t}v = \frac{mg}{k}e^{\frac{k}{m}t} + C - , - or - - v = \frac{mg}{k}+ Ce^{-\frac{k}{m}t} - . -

    -     -     - -
    - The velocity functions from Examples (dashed) and (solid) under the assumption that v(0)=0, with g=9.8, m=1, and k=1 - - - The velocity functions from the previous examples. - - -

    - The horizontal axis represents time and the vertical axis represents the velocity, - both axes are drawn from 0 to 10. The graph has assumptions that - v(0)=0 , g =9.8, m=1 and k=1. The velocity of a falling - object without air resistance is a straight line that is very close to the vertical - axis. The velocity function of a falling object with air resistance is along the - previous velocity till v=3 after which it diverges away, gets a bend at - y=10 and then runs parallel to the x axis after x=4. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - xtick={0,2,...,10}, - ymin=-.1,ymax=10.5, - xmin=-.1,xmax=10.5, - xlabel={$t$}, - ylabel={$v$} - ] - - \addplot [firstcurvestyle,domain = 0:10, samples=50] {9.8-9.8*exp(-x)}; - \addplot [secondcurvestyle,domain = 0:10, samples=50] {9.8*x}; - - \end{axis} - \end{tikzpicture} - - -
    - -

    - In the solution above, the exponential term decays as time increases, - causing the velocity to approach the constant value mg/k in the limit as t approaches infinity. - This value is called the terminal velocity. - If we assume a zero initial velocity - (the object is dropped, not thrown from the plane), - the velocities from Examples - and - are given by v = gt and v = \displaystyle \frac{mg}{k}\left(1 - e^{-\frac{k}{m}t}\right), - respectively. - These two functions are shown in , - with g = 9.8, m=1, and k=1. - Notice that the two curves agree well for short times, - but have dramatically different behaviors as t increases. - Part of the art in mathematical modeling is deciding on the level of detail required to answer the question of interest. - If we are only interested in the initial behavior of the falling object, - the simple model in may be sufficient. - If we are interested in the longer term behavior of the object, - the simple model is not sufficient, - and we should consider a more complicated model. -

    -     - - - - Problems - - -

    - Find the general solution to the first order linear differential equation. -

    -     - - -

    - \displaystyle \yp = 2y - 3 -

    -     - -

    - y = \displaystyle \frac{3}{2} + Ce^{2x} -

    -     -     - - -

    - \displaystyle x^2\yp + xy = 1 -

    -     - -

    - y = \displaystyle \frac{\ln \abs{x} + C}{x} -

    -     -     - - -

    - \displaystyle x^2\yp - xy = 1 -

    -     - -

    - y = \displaystyle -\frac{1}{2x} + Cx -

    -     -     - - -

    - \displaystyle x\yp +4y = x^3-x -

    -     - -

    - y = \displaystyle \frac{x^3}{7} - \frac{x}{5} + \frac{C}{x^4} -

    -     -     - - -

    - \displaystyle (\cos^2 x \sin(x))\yp + (\cos^3 x)y = 1 -

    -     - -

    - y = \sec x + C(\csc x) -

    -     -     - - -

    - \displaystyle \frac{\yp}{x} = 1-2y -

    -     - -

    - y = \displaystyle \frac{1}{2} + Ce^{-x^2} -

    -     -     - - -

    - \displaystyle x^3\yp-3x^3y=x^4e^{2x} -

    -     - -

    - y = \displaystyle Ce^{3x}-(x+1)e^{2x} -

    -     -     - - -

    - \displaystyle \yp + y = 5\sin(2x) -

    -     - -

    - y = sin(2x) - 2\cos(2x) + Ce^{-x} -

    -     -     -     - - -

    - Find the particular solution to the initial value problem. -

    -     - - -

    - \displaystyle \yp = y + 2xe^x, y(0) = 2 -

    -     - -

    - y = (x^2+2)e^x -

    -     -     - - -

    - \displaystyle x\yp + 2y = x^2 - x + 1, y(1) = 1 -

    -     - -

    - y = \displaystyle \frac{1}{4}x^2-\frac{1}{3}x+\frac{1}{2}+\frac{7}{12x^2} -

    -     -     - - -

    - \displaystyle x\yp + (x+2)y = x, y(1) = 0 -

    -     - -

    - y = \displaystyle 1 - \frac{2}{x} + \frac{2-e^{1-x}}{x^2} -

    -     -     - - -

    - \displaystyle \yp + 2y = 0, y(0) = 3 -

    -     - -

    - y = \displaystyle 3e^{-2x} -

    -     -     - - -

    - \displaystyle (x+1)\yp + (x+2)y = 2xe^{-x}, y(0) = 1 -

    -     - -

    - y = \displaystyle \frac{x^2+1}{x+1}e^{-x} -

    -     -     - - -

    - \displaystyle (\cos(x))\yp + (\sin(x))y = 1, y(0) = -3 -

    -     - -

    - y = \sin(x) - 3\cos(x) -

    -     -     - - -

    - \displaystyle (x^2-1)\yp + 2y = (x+1)^2, y(0) = 2 -

    -     - -

    - y = \displaystyle \frac{(x-2)(x+1)}{x-1} -

    -     -     - - -

    - \displaystyle x\yp-2y = \frac{x^3}{1+x^2}, y(1) = 0 -

    -     - -

    - y = \displaystyle x^2\left(\arctan x - \frac{\pi}{4}\right) -

    -     -     -     - - -

    - Classify the differential equation as separable, - first order linear, or both, - and solve the initial value problem using an appropriate method. -

    -     - - -

    - \displaystyle \yp = y + yx^2, y(0) = -5 -

    -     - -

    - Both; \displaystyle y = -5e^{x + \frac{1}{3}x^3} -

    -     -     - - -

    - \displaystyle xe^y \yp = x^2\sin(x), y(0) = 0 -

    -     - -

    - separable; \displaystyle e^y = \sin(x) - x\cos(x) + 1 -

    -     -     - - -

    - \displaystyle (x-1)\yp+y = x^2-1, y(0) = 2 -

    -     - -

    - linear; \displaystyle y = \frac{x^3-3x-6}{3(x-1)} -

    -     -     - - -

    - \displaystyle \yp = y^2+y-2, y(0) = 1 -

    -     - -

    - separable; \displaystyle y = 1 -

    -     -     -     - - -

    - Draw a slope field for the differential equation. - Use the slope field to predict the behavior of the solution to the initial value problem for large x values. - Solve the initial value problem, - and verify your prediction. -

    -     - - -

    - \displaystyle \yp = x-y, y(0) = 0 -

    -     - - - - Graph showing slope field for the given differential equation. - - -

    - The x and y axes are uncalibrated, the field lines - in the first quadrant are shown. - On the bottom right the field lines are facing northeast. On the - top left the field lines transition from southeast facing to east - facing moving downwards. - A curve is shown that almost represents a straight line with a - positive slope. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - y domain = -.15:4, - domain = -.15:4, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(x - y)^2))}, v = {(x-y)/(sqrt(1+(x - y)^2))},scale arrows=.2},samples=15]{0}; - \addplot [firstcurvestyle,domain=0:4.1,samples=50] {x-1+exp(-x)}; - - \end{axis} - - \end{tikzpicture} - - -

    - The solution will increase and begin to follow the line y=x-1. -

    -

    - y = x-1 + e^{-x} -

    -     -     - - -

    - \displaystyle (x+1)\yp + y = \frac{1}{x+1}, y(0) = 2 -

    -     - - - - Graph showing slope field for the given differential equation. - - -

    - The x and y axes are uncalibrated, the field lines in - the first quadrant are shown. The lines in the top are southeast facing, - for lower values of y from left to right the field lines are - northeast facing then they transition to east facing. - A downward sloping curve is shown on the field lines. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - view={0}{90}, - y domain = -.15:3, - domain = -.15:4, - ] - - \addplot3 [secondcolor,quiver = {u = {1/(sqrt(1+(-y/(x+1)+1/(x+1)^2)^2))}, v = {(-y/(x+1)+1/(x+1)^2)/(sqrt(1+(-y/(x+1)+1/(x+1)^2)^2))},scale arrows=.2},samples=15]{0}; - \addplot [firstcurvestyle,domain=0:4.15,samples=50] {(2+ln(x+1))/(x+1)}; - - \end{axis} - - \end{tikzpicture} - - -

    - The solution will decrease and approach y=0. -

    -

    - \displaystyle y = \frac{2 + \ln(x+1)}{x+1} -

    -     -     -     -     -     -
    -
    - Modeling with Differential Equations - -

    - In the first three sections of this chapter, - we focused on the basic ideas behind differential equations and the mechanics of solving certain types of differential equations. - We have only hinted at their practical use. - In this section, - we use differential equations for mathematical modeling, - the process of using equations to describe real world processes. - We explore a few different mathematical models with the goal of gaining an introduction to this large field of applied mathematics. - differential equationmodeling -

    -     - - Models Involving Proportional Change -

    - Some of the simplest differential equation models involve one quantity that changes at a rate proportional to another quantity. - In the introduction to this chapter, - we considered a population that grows at a rate proportional to the current population. - The words in this assumption can be directly translated into a differential equation as shown below. -

    - -
    - Translating words into a differential equation - - - The formula showing the rate of population is proportional to the population. - - -

    - Image shows the equation of derivative of population with respect to time, and it being - proportional to the population. -

    -     - - \begin{tikzpicture} - - \draw (1,1) node {$\displaystyle \frac{dp}{dt} = kp$}; - - \draw [firstcolor] (-1.5,0) node [text width=60pt,align=center] (a) { \centering The rate of change of the population}; - \draw [firstcolor,->] (a) -- (.3,.7); - - \draw [firstcolor] (2,.25) node [text width=60pt,align=center] (b) { \centering the population.}; - \draw [firstcolor,->] (b) -- (1.6,.8); - - \draw [firstcolor] (.5,2) node [text width=32pt,align=center] (c) { \centering is}; - \draw [firstcolor,->] (c) -- (1,1.2); - - \draw [firstcolor] (2,2) node [text width=60pt,align=center] (d) { \centering proportional to}; - \draw [firstcolor,->] (d) -- (1.4,1.2); - - \end{tikzpicture} - - -
    - -

    - There are some key ideas that can be helpful when translating words into a differential equation. - Any time we see something about rates or changes, - we should think about derivatives. - The word - is - usually corresponds to an equal sign in the equation. - The words - proportional to - mean we have a constant multiplied by something. -

    -

    - The differential equation in - is easily solved using separation of variables. - We find - - p = Ce^{kt} - . -

    -

    - Notice that we need values for both C and k before we can use this formula to predict population size. - We require information about the population at two different times in order to fully determine the population model. -

    - - Bacterial Growth - -

    - Suppose a population of e-coli - bacteria grows at a rate proportional to the current population. - If an initial popluation of 200 bacteria has grown to 1600 three hours later, - find a function for the size of the population at time t, - and use it to predict when the population size will reach 10,000. - bacterial growth -

    -     - -

    - We already know that the population at time t is given by - p = Ce^{kt} for some C and k. - The information about the initial size of the population means that p(0)=200. - Thus C=200. - Our knowledge of the population size after three hours allows us to solve for k via the equation - - 1600 = 200e^{3k} - . -

    -

    - Solving this exponential equation yields k =\ln(8)/3 \approx 0.6931. - The popluation at time t is given by - - p = 200 e^{(\ln(8)/3)t} - . -

    -

    - Solving - - 10000 = 200e^{(\ln(8)/3)t} - - yields t =(3\ln(50))/\ln(8) \approx 5.644. - The population is predicted to reach 10,000 bacteria in slightly more than five and a half hours. -

    -     - -     - -

    - Another example of porportional change is - Newton's Law of Cooling. - The laws of thermodynamics state that heat flows from areas of higher temperature to areas of lower temperature. - A simple example is a hot object that cools down when placed in a cool room. - Newton's Law of Cooling is the simple assumption that the temperature of the object changes at a rate proportional to the difference between the temperature of the object and the ambient temperature of the room. - If T is the temperature of the object and A is the constant ambient temperature, - Newton's Law of Cooling can be expressed as the differential equation - - \frac{dT}{dt} = k(A - T) - . - Newton's Law of Cooling -

    -

    - This differential equation is both linear and separable. - The separated form is - - \frac{1}{A-T}\,dT = k\,dt - . -

    -

    - Then an implicit definition of the temperature is given by - - -\ln\abs{A-T} = kt + C - . -

    -

    - If we solve for T, we find the explicit temperature - - T = A-Ce^{-kt} - . -

    -

    - Though we didn't show the steps, - the explicit solution involves the typical process of renaming the constant \pm e^{-C} as C, - and allowing C to be positive, negative, - or zero to account for both cases of the absolution value and to catch the constant solution T=A. - Notice that the temperature of the object approaches the ambient temperature in the limit as t\to\infty. -

    - - - - - Hot Coffee - -

    - A freshly brewed cup of coffee is set on the counter and has a temperature of 200^\circ Fahrenheit. - After 3 minutes, it has cooled to 190^\circ, - but is still too hot to drink. - If the room is 72^\circ and the coffee cools according to Newton's Law of Cooling, - how long will the impatient coffee drinker have to wait until the coffee has cooled to 165^\circ? -

    -     - -

    - Since we have already solved the differential equation for Newton's Law of Cooling, - we can immediately use the function - - T = A - Ce^{-kt} - . -

    -

    - Since the room is 72^\circ, we know A = 72. - The initial temperature is 200^\circ, - which means C = -128. - At this point, we have - - T = 72 + 128e^{-kt} - -

    -

    - The information about the coffee cooling to 190^\circ in 3 minutes leads to the equation - - 190 = 72 + 128e^{-3k} - . -

    -

    - Solving the exponential equation for k, we have - - k = -\frac{1}{3}\ln \left(\frac{59}{64}\right) \approx 0.0271 - . -

    -

    - Finally, we finish the problem by solving the exponential equation - - 165 = 72 + 128e^{\frac{1}{3}\ln \left(\frac{59}{64}\right)t} - . -

    -

    - The coffee drinker must wait \displaystyle t = \frac{3 \ln \left(\frac{93}{128}\right)}{\ln \left(\frac{59}{64}\right)} \approx 11.78 minutes. -

    -     - -     -

    - We finish our discussion of models of proportional change by exploring three different models of disease spread through a population. - In all of the models, - we let y denote the proportion of the population that is sick - (0 \leq y \leq 1). - We assume a proportion of 0.05 is initially sick and that a proportion of 0.1 is sick 1 week later. -

    - - Disease Spread 1 - -

    - Suppose a disease spreads through a population at a rate proportional to the number of individuals who are sick. - If 5% of the population is sick initially and 10% of the population is sick one week later, - find a formula for the proportion of the popoulation that is sick at time t. -

    -     - -

    - The assumption here seems to have some merit because it matches our intuition that a disease should spread more rapidly when more individuals are sick. - The differential equation is simply - - \frac{dy}{dt} = ky - , - with solution - - y = Ce^{kt} - . -

    -

    - The conditions y(0)=0.05 and y(1) = 0.1 lead to - C = 0.05 a and k = \ln(2), so the function is - - y = 0.05e^{(\ln(2)t} - . -

    -

    - We should point out a glaring problem with this model. - The variable y is a proportion and should take on values between 0 and 1, but the function y = 0.05e^{2t} grows without bound. - After t \approx 4.32 weeks, - y exceeds 1, and the model ceases to make physical sense. -

    -     -     - - - - Disease Spread 2 - -

    - Suppose a disease spreads through a population at a rate proportional to the number of individuals who are not sick. - If 5% of the population is sick initially and 10% of the population is sick one week later, - find a formula for the proportion of the popoulation that is sick at time t. -

    -     - -

    - The intuition behind the assumption here is that a disease can only spread if there are individuals who are susceptible to the infection. - As fewer and fewer people are able to be infected, - the disease spread should slow down. - Since y is proportion of the population that is sick, - 1-y is the proportion who are not sick, - and the differential equation is - - \frac{dy}{dt} = k(1-y) - . -

    -

    - Though the context is quite different, - the differential equation is identical to the differential equation for Newton's Law of Cooling, - with A=1. - The solution is - - y = 1 - Ce^{-kt} - . -

    -

    - The conditions y(0)=0.05 and - y(1) = 0.1 yield C = 0.95 and - k = -\ln\left(\frac{18}{19}\right) \approx 0.0541, - so the final function is - - y = 1-.95e^{\ln\left(\frac{18}{19}\right)t} - . -

    -

    - Notice that this function approaches y=1 in the limit as t \to \infty, - and does not suffer from the non-physical behavior described in . -

    -     -     - -

    - In , - we assumed disease spread depends on the number of infected individuals. - In , - we assumed disease spread depends on the number of susceptible individuals who are able to become infected. - In reality, we would expect many diseases to require the interaction of both infected and susceptible individuals in order to spread. - One of the simplest ways to model this required interaction is to assume disease spread depends on the product of the proportions of infected and uninfected individuals. - This assumption - (regularly seen in the context of chemical reactions) - is often called the law of mass action. -

    - - - - Disease Spread 3 - -

    - Suppose a disease spreads through a population at a rate proportional to the product of the number of infected and uninfected individuals. - If 5% of the population is sick initially and 10% of the population is sick one week later, - find a formula for the proportion of the population that is sick at time t. - differential equationlogistic -

    -     - -

    - The differential equation is - - \frac{dy}{dt} = ky(1-y) - . -

    -

    - This is exactly the logistic equation with M = 1. - We solved this differential equation in , - and found - - y = \frac{1}{1 + be^{-kt}} - . -

    -

    - The conditions y(0)=0.05 and - y(1) = 0.1 yield b = 19 and k = -\ln\left(\frac{9}{19}\right) \approx 0.7472. - The final function is - - y = \frac{1}{1+19e^{\ln\left(\frac{9}{19}\right)t}} - . -

    -

    - Based on the three different assumptions about the rate of disease spread explored in the last three examples, - we now have three different functions giving the proportion of a population that is sick at time t. - Each of the three functions meets the conditions - y(0)=0.05 and y(1) = 0.1. - The three functions are shown in . -

    - -

    - Notice that the logistic function mimics specific parts of the functions from Examples - and . - We see in that the logistic and exponential functions are virtually indistinguishable for small t values. - When there are few infected individuals and lots of susceptible individuals, - the spread of a disease is largely determined by the number of sick people. - The logistic curve captures this feature, and is - almost exponential - early on. -

    - -

    - In , - we see that the logistic curve leaves the exponential curve from - and approaches the curve from . - This result implies that when most of the population is sick, - the spread of the disease is largely dependent on the number of susceptible individuals. - Though there are much more sophisticated mathematical models describing the spread of infections, - we could argue that the logistic model presented in this example is the - best - of the three. -

    - -
    - Plots of the functions from (dotted), - (dashed), - and (solid) - -
    - - - - Plots of functions from the last three examples for small values of t. - - -

    - The y axis is drawn from 0 to 0.1 and the t axis is drawn from - 0 to 1.2. The first function is a straight line that is positively inclined, - it starts from the point (0, 0.05) and moves away from the t axis with increasing - values of t. The exponential and logistic functions are indistinguishable and they - also begin from the point (0,0.05), they also have a positive slope but they have a dip - and separates from the line before merging again at point (1,1) after which they are - above the line. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ytick={0,0.05,.1}, - yticklabels = {0,0.05,0.1}, - ymin=-.01,ymax=.11, - xmin=-.1,xmax=1.25, - xlabel={$t$} - ] - - \addplot [firstcurvestyle,domain = 0:10, samples=50] {1/(1+19*exp(ln(9/19)*x))}; - \addplot [firstcurvestyle,dashed,domain = 0:10, samples=50] {1-.95*exp(ln(18/19)*x)}; - \addplot [firstcurvestyle,dotted,domain = 0:2, samples=50] {.05*exp(ln(2)*x)}; - - \fill[black,draw=black](axis cs:0,.05) circle (2.4pt); - \fill[black,draw=black](axis cs:1,.1) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - -
    -
    - - - - Plots of functions from the last three examples for large values of t. - - -

    - The y axis is drawn from 0 to 1 and the x axis is drawn from - 0 to 50. The three functions in the previous examples are shown. The three - curves are aligned till about x=2 after which they diverge. The exponential curve - moves vertically, very close to the y axis. The logistic curve diverges to the left - from the exponential curve at about point (3,0.3), it acquires a sharp bend at point - (10,1) after which it runs parallel to the x axis. The third curve is below the - other two and starts close to the origin and curves up to reach point (50, 0.9). -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.01,ymax=1.1, - xmin=-.1,xmax=50.1, - xlabel={$t$} - ] - - \addplot [firstcurvestyle,domain = 0:50, samples=101] {1/(1+19*exp(ln(9/19)*x))}; - \addplot [firstcurvestyle,dashed,domain = 0:50, samples=50] {1-.95*exp(ln(18/19)*x)}; - \addplot [firstcurvestyle,dotted,domain = 0:5, samples=50] {.05*exp(ln(2)*x)}; - - \fill[black,draw=black](axis cs:0,.05) circle (2.4pt); - \fill[black,draw=black](axis cs:1,.1) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - -
    -     -
    - -     - -     -     - - Rate-in Rate-out Problems -

    - One of the classic ways to build a mathematical model involves tracking the way the amount of something can change. - We sometimes say these models are based on - conservation laws. - Consider a box with some amount of a specific type of material inside. - (Some type of chemical, for example.) - The amount of material of the specific type in the box can only change in four ways; - we can add more to the box, we can remove some from the box, - some of the material can change into material of a different type, - or some other type of material can turn into the type we're tracking. - In the examples that follow, we assume material doesn't change type, - so we only need to keep track of material coming into the box and material leaving the box. - To derive a differential equation, we track rates: - - \text{ rate of change of some quantity } = \text{ rate in } - \text{ rate out } - . -

    - - -

    - Though we stick to relatively simple examples, - this basic idea can be used to derive some very important differential equations in mathematics and physics. -

    -

    - The examples to follow involve tracking the amount of a chemical in solution. - We assume liquid containing some chemical flows into a container at some rate. - That liquid mixes instantaneously with the liquid already in the container. - Then the liquid from the container flows out at some - (potentially different) - rate. -

    - - - - - Equal Flow Rates - -

    - Suppose a 10 liter tank has 5 liters of salt solution in it. - The initial concentration of the salt solution is 1 gram per liter. - A salt solution with concentration - 3 flows into the tank at a rate of - 2. - Suppose the salt solution mixes instantaneously with the solution already in the tank, - and that the mixed solution from the tank flows out at a rate of - 2. - Find a function that gives the amount of salt in the tank at time t. -

    -     - -

    - We use the rate in - rate out setup described above. - The quantity here is the amount - (in grams) - of salt in the tank at time t. - Let y denote the amount of salt. - In words, the differential equation is given by - - \frac{dy}{dt} = \text{ rate in } - \text{ rate out } - . -

    -

    - Thinking in terms of units can help fill in the details of the differential equation. - Since y has units of grams, - the left hand side of the equation has units g/min. - Both terms on the right hand side must have these same units. - Notice that the product of a concentration - (with units g/L) - and a flow rate - (with units L/min) - results in a quantity with units g/min. - Both terms on the right hand side of the equation will include a concentration multiplied by a flow rate. -

    -

    - For the rate in, - we multiply the inflow concentration by the rate that fluid is flowing into the bucket. - This is \displaystyle \left(3 \frac{\text{g} }{\text{L} }\right)\left(2 \frac{\text{L} }{\text{ min } }\right) = 6 g/min. -

    -

    - The rate out is more complicated. - The flow rate is still 2, - meaning that the overall volume of the fluid in the bucket is the constant - 5. - The salt concentration in the bucket is not constant though, - meaning that the outflow concentration is not constant. - In particular, the outflow concentration is not - the constant 1. - This is simply the initial concentration. - To find the concentration at any time, - we need the amount of salt in the bucket at that time and the volume of liquid in the bucket at that time. - The volume of liquid is the constant 5, - and the amount of salt is given by the dependent variable y. - Thus, the outflow concentration is - \displaystyle \frac{y}{5} g/L, yielding a rate out given by - - \left(\frac{y}{5}\frac{\text{ g } }{\text{ L } }\right)\left(2 \frac{\text{ L } }{\text{ min } }\right) = \frac{2y}{5}\text{ g/min } - . -

    -

    - The differential equation we wish to solve is given by - - \frac{dy}{dt} = 6 - \frac{2y}{5} - . -

    -

    - To furnish an initial condition, - we must convert the initial salt concentration into an initial amount of salt. - This is \left(1\displaystyle \frac{\text{ g } }{\text{ L } }\right)(5 \text{ L } ) = 5 g, so - y(0) = 5 is our initial condition. -

    -

    - Our differential equation is both separable and linear. - We solve using separation of variables. - The separated form of the differential equation is - - \frac{5}{30 - 2y}\,dy = dt - . -

    -

    - Integration yields the implicit solution - - -\frac{5}{2}\ln\abs{30 - 2y} = t+C - . -

    -

    - Solving for y - (and redefining the arbitrary constant C as necessary) - yields the explicit solution - - y = 15 + Ce^{-\frac{2}{5}t} - . -

    -

    - The initial condition y(0) = 5 means that C = -10 so that - - y = 15 - 10e^{-\frac{2}{5}t} - - is the particular solution to our initial value problem. -

    -

    - This function is plotted in . - Notice that in the limit as t\to\infty, - y approaches 15. - This corresponds to a bucket concentration of - 15/5 = 3 g/L. It should not be surprising that salt concentration inside the tank will move to match the inflow salt concentration. -

    -
    - Salt concentration at time t, from - - - Graph of salt concentration at time t used in this example. - - -

    - The y axis is drawn from 0 to 15 and the x axis is drawn from - 0 to 10. The function starts at point (0, 5) and curves up with a great - positive slope then the slope decreases. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=15.5, - xmin=-.1,xmax=10.3, - xlabel={$t$} - ] - - \addplot [firstcurvestyle,domain = 0:10, samples=50] {15-10*exp(-2/5*x)}; - - \end{axis} - - \end{tikzpicture} - - -
    -     - -     - - - Unequal Flow Rates - -

    - Suppose the setup is identical to the setup in - except that now liquid flows out of the bucket at a rate of 1 L/min. - Find a function that gives the amount of salt in the bucket at time t. - What is the salt concentration when the solution ceases to be valid? -

    -     - -

    - Because the inflow and outflow rates no longer match, - the volume of liquid in the bucket is not the constant 5 L. In general, - we can find the volume of liquid via the equation - - \text{ volume } = \text{ initial volume } + \text{ (inflow rate - outflow rate) } t - . -

    -

    - In this example, - the volume at time t is 5 + t liters. - Because the total volume of the bucket is only 10 L, it follows that our solution will only be valid for 0 \leq t \leq 5. - At that point it is no longer possible to have liquid flow into a the bucket at a rate of 2 L/min and out of the bucket at a rate of 1 L/min. -

    -

    - To update the differential equation, - we must modify the rate out. - Since the volume is 5 + t, - the concentration at time t is given by - \frac{y}{5+t} g/L. Thus for rate out, - we must use \left( \frac{y}{5+t}\right)(1) g/min. - The initial value problem is - - \frac{dy}{dt} = 6 - \frac{y}{5+t}, \text{ with } y(0)=5 - . -

    -

    - Unlike , - where we had equal flow rates, - this differential equation is no longer separable. - We must proceed with an integrating factor. - Writing the differential equation in the form - - \frac{dy}{dt} + \frac{1}{5+t}y = 6 - , - we identify the integrating factor - - \mu(t) = e^{\int \frac{1}{5+t}\,dt} = e^{\ln(5+t)} = 5+t - . -

    -

    - Then - - \frac{d}{dt}\big((5+t)y\big) = 6(5+t) - , - yielding the implicit solution - - (5+t)y = 30t + 3t^2 + C - . -

    -

    - The initial condition y(0) = 5 implies C = 25, - so the explicit solution to our initial value problem is given by - - y = \frac{3t^2 + 30t + 25}{5+t} - . -

    -

    - This solution ceases to be valid at t=5. - At that time, there are 25 g of salt in the tank. - The volume of liquid is 10 L, resulting in a salt concentration of 2.5 g/L. -

    -     - -     -

    - Differential equations are powerful tools that can be used to help describe the world around us. - Though relatively simple in concept, - the ideas of proportional change and matching rates can serve as building blocks in the development of more sophisticated mathematical models. - As we saw in this section, - some simple mathematical models can be solved analytically using the techniques developed in this chapter. - Most sophisticated mathematical models don't allow for analytic solutions. - Even so, there are an array of graphical and numerical techniques that can be used to analyze the model to make predictions and infer information about real world phenomena. -

    -     - - - - Problems - - -

    - Use the tools in the section to answer the questions presented. -

    -     - - -

    - Suppose the rate of change of y with respect to x is proportional to 10 - y. - Write down and solve a differential equation for y. -

    -     - -

    - y = 10 + Ce^{-kx} -

    -     -     - - -

    - A rumor is spreading through a middle school with 250 students. - Suppose the rumor spreads at a rate proportional to the number of students who haven't heard the rumor yet. - If 1 person starts the rumor, - and 75 students have heard the rumor 3 days later, - how many days will it take until 80% of the students in the school have heard the rumor? -

    -     - -

    - 13.66 days -

    -     -     - - -

    - A rumor is spreading through a middle school with 250 students. - Suppose the rumor spreads at a rate proportional to the product of number of students who have heard the rumor and the number who haven't heard the rumor. - If 1 person starts the rumor, - and 75 students have heard the rumor 3 days later, - how many days will it take until 80% of the students in the school have heard the rumor? -

    -     - -

    - 4.43 days -

    -     -     - - -

    - A feature of radioactive decay is that the amount of a radioactive substance decreases at a rate proportional to the current amount of the substance. - The half life - half life - of a substance is the amount of time it takes for half of a given amount of substance to decay. - The half life of carbon-14 is approximately 5730 years. - If an ancient object has a carbon-14 amount that is 20% of the original amount, - how old is the object? -

    -     - -

    - 13,304.65 years old -

    -     -     - - -

    - Consider a chemical reaction where molecules of type A combine with molecules of type B to form molecules of type C. Suppose one molecule of type A combines with one molecule of type B to form one molecule of type C, and that type C is produced at a rate proportional the product of the remaining number of molecules of types A and B. Let x denote moles of molecules of type C. - Find a function giving the number of moles of type C at time t if there are originally a moles of type A, b moles of type B, and zero moles of type C. -

    -     - -

    - x = \begin{cases}\displaystyle\frac{ab(1 - e^{(a-b)kt})}{b-ae^{(a-b)kt}} \amp \text{ if } a \neq b\\ \displaystyle \frac{a^2kt}{1+akt} \amp \text{ if } a = b \end{cases} -

    -     -     - - -

    - Suppose an object with a temperature of 100^\circ is introduced into a room with an ambient temperature of 70^\circ. - Suppose the temperature of the object changes at a rate proportional to the difference between the temperature of the object and the temperature of the room (Newton's Law of Cooling). - If the object has cooled to 92^\circ in 10 minutes, - how long until the object has cooled to 84^\circ? -

    -     - -

    - 24.57 minutes -

    -     -     - - -

    - Suppose an object with a temperature of 100^\circ is introduced into a room with an ambient temperature given by 60 + 20e^{-\frac{1}{4}t} degrees. - Suppose the temperature of the object changes at a rate proportional to the difference between the temperature of the object and the temperature of the room (Newton's Law of Cooling). - If the object is 80^\circ after 20 minutes, - find a formula giving the temperature of the object at time t. - (Note: This problem requires a numerical technique to solve for the unknown constants.) -

    -     - -

    - \displaystyle y = 60 - 3.69858e^{-\frac{1}{4}t} + 43.69858e^{-0.0390169 t} -

    -     -     - - -

    - A tank contains 5 gallons of salt solution with concentration 0.5 g/gal. - Pure water flows into the tank at a rate of 1 gallon per minute. - Salt solution flows out of the tank at a rate of 1 gallon per minute. - (Assume instantaneous mixing.) - Find the concentration of the salt solution at 10 minutes. -

    -     - -

    - 0.06767 g/gal -

    -     -     - - -

    - Dead leaves accumulate on the ground at a rate of 4 grams per square centimeter per year. - The dead leaves on the ground decompose at a rate of 50% per year. - Find a formula giving grams per square centimeter on the ground if there are no leaves on the ground at time t=0. -

    -     - -

    - y = 8(1-e^{-\frac{1}{2}t}) g/cm^2 -

    -     -     - - -

    - A pond initially contains 10 million gallons of fresh water. - Water containing an undesirable chemical flows into the pond at a rate of 5 million gallons per year, - and fluid from the pond flows out at the same rate. - (Assume instantaneous mixing.) - If the concentration - (in grams per million gallons) - of the incoming chemical varies periodically according to the expression 2 + \sin(2t), - find a formula giving the amount of chemical in the pond at time t. -

    -     - -

    - y = \displaystyle 20 - \frac{10}{17}\left(4\cos(2t)- \sin(2t)\right) - \frac{300}{17}e^{-\frac{1}{2}t} g -

    -     -     - - -

    - A large tank contains 1 gallon of a salt solution with concentration 2 g/gal. - A salt solution with concentration 1 g/gal flows into the tank at a rate of 4 gal/min. - Salt solution flows out of the tank at a rate of 3 gal/min. - (Assume instantaneous mixing.) - Find the amount of salt in the tank at 10 minutes. -

    -     - -

    - 11.00075 g -

    -     -     - - -

    - A stream flows into a pond containing 2 million gallons of fresh water at a rate of 1 million gallons per day. - The stream flows out of the first pond and into a second pond containing 3 million gallons of fresh water. - The stream then flows out of the second pond. - Suppose the inflow and outflow rates are the same so that both ponds maintain their volumes. - A factory upstream of the first pond starts polluting the stream. - Directly below the factory, - pollutant has a concentration of 55 grams per million gallons, - and this concentration starts to flow into the first pond. - Find the concentration of pollutant in the first and second ponds at 5 days. -

    -     - -

    - pond 1: 50.4853 grams per million gallons -

    -

    - pond 2: 32.8649 grams per million gallons -

    -     -     -     -     -     -
    -     - - - Sequences and Series - -

    - This chapter introduces sequences and series, - important mathematical constructions that are useful when solving a large variety of mathematical problems. - The content of this chapter is considerably different from the content of the chapters before it. - While the material we learn here definitely falls under the scope of calculus, - we will make very little use of derivatives or integrals. - Limits are extremely important, though, - especially limits that involve infinity. -

    - -

    - One of the problems addressed by this chapter is this: - suppose we know information about a function and its derivatives at a point, - such as f(1) = 3, \fp(1) = 1, - \fp'(1) = -2, \fp''(1) = 7, and so on. - What can I say about f(x) itself? - Is there any reasonable approximation of the value of f(2)? - The topic of Taylor Series addresses this problem, - and allows us to make excellent approximations of functions when limited knowledge of the function is available. -

    -     - -
    - Sequences -

    - We commonly refer to a set of events that occur one after the other as a - sequence of events. - In mathematics, we use the word sequence - to refer to an ordered set of numbers, - , a set of numbers that occur one after the other. -

    - - - -

    - For instance, - the numbers 2, 4, 6, 8, , form a sequence. - The order is important; the first number is 2, the second is 4, etc. - It seems natural to seek a formula that describes a given sequence, - and often this can be done. - For instance, - the sequence above could be described by the function a(n) = 2n, - for the values of n = 1, 2, \ldots To find the 10th term in the sequence, - we would compute a(10). - This leads us to the following, - formal definition of a sequence. -

    - - - Sequence - -

    - A sequence is a function a(n) whose domain is \mathbb{N}. - The range of a sequence is the set of all distinct values of a(n). - sequencesdefinition -

    - -

    - The terms of a sequence are the values a(1), - a(2), - , which are usually denoted with subscripts as a_1, a_2, - . -

    - -

    - A sequence a(n) is often denoted as \{a_n\}. -

    -     -     - - - - - -

    - A factorial refers to the product of a descending sequence of natural numbers. - For example, the expression 4! - (read as 4 factorial) - refers to the number 4\cdot 3\cdot2\cdot1 = 24. - - factorial - aa@"! -

    - -

    - In general, n! = n\cdot (n-1)\cdot(n-2)\cdots 2\cdot1, - where n is a natural number. -

    - -

    - We define 0! = 1. - While this does not immediately make sense, - it makes many mathematical formulas work properly. -

    -     -     - - - - - Listing terms of a sequence - -

    - List the first four terms of the following sequences. -

    - -

    -

      -
    1. \{a_n\} = \left\{\frac{3^n}{n!}\right\}
      2. - -
    2. \{a_n\} = \{4+(-1)^n\}
      3. - -
    3. - \{a_n\} = \left\{\frac{(-1)^{n(n+1)/2}}{n^2}\right\} -
      4. -
    -

    -     - -

    -

      -
    1. -

      - \ds a_1=\frac{3^1}{1!} = 3;\quad \ds a_2= \frac{3^2}{2!} = \frac92;\quad - \ds a_3 = \frac{3^3}{3!} = \frac92;\quad \ds a_4 = \frac{3^4}{4!} = \frac{27}8\quad -

      - -

      - We can plot the terms of a sequence with a scatter plot. - The horizontal axis is used for the values of n, - and the values of the terms are plotted on the vertical axis. - To visualize this sequence, see . -

      -
      - Plotting the sequence in of - - - Plot of the first four terms of the sequence in part 1 of this example. - -

      - A pair of coordinate axes are given, with the horizontal axis labeled n (rather than x), - and the origin at the bottom-left of the image. - The first four terms of the sequence 3^n/n! are plotted, - at coordinates (1,3),(2,4.5),(3,4.5), and (4,3.375). - The four points lie in an inverted U shape. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4,5}, - ymin=-.1,ymax=5.5, - xmin=-.1,xmax=4.5, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt}] coordinates {(1,3)(2,4.5)(3,4.5)(4,3.375)}; - - \draw (axis cs:2.5,1.5) node { $\ds a_n = \frac{3^n}{n!}$}; - - \end{axis} - \end{tikzpicture} - - - - -
      -
      2. - -
    2. -

      - a_1= 4+(-1)^1 = 3;\quad a_2 = 4+(-1)^2 = 5;\quad - - a_3=4+(-1)^3 = 3;\quad a_4 = 4+(-1)^4 = 5\quad. -

      - -

      - Note that the range of this sequence is finite, - consisting of only the values 3 and 5. - This sequence is plotted in . -

      -
      - Plotting the sequence in of - - - A plot of the first four terms of the sequence in part 2 of this example. - -

      - A pair of coordinate axes are shown, with the horizontal axis labeled n, - and the origin at the bottom-left of the image. - The first four terms of the sequence 4+(-1)^n are plotted, - illustrating the oscillatory nature of this sequence: - the four points make a sort of zig-zag configuration, with the y value - going back and forth between 3 and 5. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4,5}, - ymin=-.1,ymax=5.5, - xmin=-.1,xmax=4.5, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:4] {4+(-1)^x}; - - \draw (axis cs:2,1) node { $\ds a_n = 4+(-1)^n$}; - - \end{axis} - \end{tikzpicture} - - - - -
      -
      3. - -
    3. -

      - \ds a_1= \frac{(-1)^{1(2)/2}}{1^2} = -1;\quad \ds a_2 = \frac{(-1)^{2(3)/2}}{2^2} =-\frac14 ;\quad - - \ds a_3 = \frac{(-1)^{3(4)/2}}{3^2} = \frac19;\quad \ds a_4 = \frac{(-1)^{4(5)/2}}{4^2} = \frac1{16};\quad; - - \ds a_5 = \frac{(-1)^{5(6)/2}}{5^2}=-\frac1{25}\quad. -

      - -

      - We gave one extra term to begin to show the pattern of signs is -, - -, +, +, - -, -, \ldots, - due to the fact that the exponent of -1 is a special quadratic. - This sequence is plotted in . -

      -
      - Plotting the sequence in of - - - A plot of the first four terms of the sequence in part 3 of this example. - -

      - Another set of coordinate axes are shown; in this case, the horizontal axis (labeled n) - is about two thirds of the way from the bottom of the image. - The first five terms of the sequence (-1)^{n(n+1)/2}/n^2 are shown. - Since the triangular numbers n(n+1)/2 follow an odd, odd, even, even pattern, - the points plotted for the first two terms have negative y value, - the next two have positive y value, and the fifth term is again negative. - As n increases, the points get closer and closer to the n axis. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4,5}, - ytick={-1}, - extra y ticks={.5,.25}, - extra y tick labels={$1/2$,$1/4$}, - ymin=-1.1,ymax=0.6, - xmin=-.1,xmax=5.5, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt}] coordinates {(1,-1)(2,-.25)(3,.111)(4,0.0625)(5,-.04)}; - - \draw (axis cs:3,-.5) node { $\ds a_n = \frac{(-1)^{n(n+1)/2}}{n^2}$}; - - \end{axis} - \end{tikzpicture} - - - - -
      - -
      4. -
    -

    -     - -     - - - Determining a formula for a sequence - -

    - Find the nth term of the following sequences, - , find a function that describes each of the given sequences. -

    - -

    -

      -
    1. -

      - \{a_n\}=\{2, 5, 8, 11, 14, \ldots\} -

      -
      2. - -
    2. -

      - \{b_n\}=\{2,-5, 10, -17, 26, -37,\ldots\} -

      -
      3. - -
    3. -

      - \{c_n\}=\{1, 1, 2, 6, 24, 120, 720, \ldots\} -

      -
      4. - -
    4. -

      - \{d_n\}=\left\{\frac52, \frac52, \frac{15}8, \frac54,\frac{25}{32},\ldots\right\} -

      -
      5. -
    -

    -     - -

    - We should first note that there is never exactly one function that describes a finite set of numbers as a sequence. - There are many sequences that start with 2, then 5, as our first example does. - We are looking for a simple formula that describes the terms given, - knowing there is possibly more than one answer. -

    - -

    -

      -
    1. -

      - Note how each term is 3 more than the previous one. - This implies a linear function would be appropriate: - a(n) = a_n = 3n + b for some appropriate value of b. - If we were to think in terms of ordered pairs, - they would be of the form (n,a(n)). - So one such ordered pair would be (1,2). - As we want a_1=2, we set b=-1. - Thus a_n = 3n-1. -

      -
      2. - -
    2. -

      - First notice how the sign changes from term to term. - This is most commonly accomplished by multiplying the terms by either (-1)^n or (-1)^{n+1}. - Using (-1)^n multiplies the odd indexed terms by (-1). - Thus the first term would be negative and the second term would be positive. - Multiplying by (-1)^{n+1} multiplies the even indexed terms by (-1). - Thus the second term would be negative and the first term would be positive. - As this sequence has negative even indexed terms, - we will multiply by (-1)^{n+1}. -

      - -

      - After this, we might feel a bit stuck as to how to proceed. - At this point, we are just looking for a pattern of some sort: - what do the numbers 2, 5, 10, 17, etc., have in common? - There are many correct answers, - but the one that we'll use here is that each is one more than a perfect square. - That is, 2=1^2+1, 5=2^2+1, 10=3^2+1, etc. - Thus our formula is b_n= (-1)^{n+1}(n^2+1). -

      -
      3. - -
    3. -

      - One who is familiar with the factorial function will readily recognize these numbers. - They are 0!, 1!, - 2!, 3!, etc. - Since our sequences start with n=1, - we cannot write c_n = n!, - for this misses the 0! term. - Instead, we shift by 1, and write c_n = (n-1)!. -

      -
      4. - -
    4. -

      - This one may appear difficult, - especially as the first two terms are the same, - but a little sleuthing will help. - Notice how the terms in the numerator are always multiples of 5, and the terms in the denominator are always powers of 2. - Does something as simple as d_n = \frac{5n}{2^n} work? -

      - -

      - When n=1, we see that we indeed get 5/2 as desired. - When n=2, we get 10/4 = 5/2. - Further checking shows that this formula indeed matches the other terms of the sequence. -

      -
      5. -
    -

    -     - -     - -

    - A common mathematical endeavor is to create a new mathematical object - (for instance, a sequence) - and then apply previously known mathematics to the new object. - We do so here. - The fundamental concept of calculus is the limit, - so we will investigate what it means to find the limit of a sequence. -

    - - - Limit of a Sequence, Convergent, Divergent - -

    - Let \{a_n\} be a sequence and let L be a real number. - Given any \varepsilon \gt 0, - if an N can be found such that - \abs{a_n-L}\lt \varepsilon for all n\gt N, - then we say the limit of \{a_n\}, - as n approaches infinity, - is L, denoted - - \lim_{n\to\infty}a_n = L - . -

    - -

    - If \lim\limits_{n\to\infty} a_n exists, - we say the sequence converges; - otherwise, the sequence diverges. - limitof sequence - sequenceslimit - convergenceof sequence - divergenceof sequence - sequencesconvergent - sequencesdivergent -

    -     -     - -

    - This definition states, informally, - that if the limit of a sequence is L, - then if you go far enough out along the sequence, - all subsequent terms will be really close to L. - Of course, the terms far enough - and really close are subjective terms, - but hopefully the intent is clear. -

    - - - -

    - This definition is reminiscent of the - \varepsilon-\delta proofs of . - In that chapter we developed other tools to evaluate limits apart from the formal definition; - we do so here as well. -

    - - - Limit of Infinity, Divergent Sequence - -

    - Let \{a_n\} be a sequence. - We say \lim\limits_{n\to\infty} a_n=\infty if for all M \gt 0, - there exists a number N such that if n\ge N, - then a_n \gt M. - In this case, we say the sequence diverges to \infty. -

    -     -     - -

    - This definition states, informally, - that if the limit of a_n is \infty, - then you can guarantee that the terms of a_n will get arbitrarily large - (larger than any value of M that you think of), - by going out far enough in the sequence. -

    - - - Limit of a Sequence - -

    - Let \{a_n\} be a sequence, - let L be a real number, - and let f(x) be a function whose domain contains the positive real numbers where - f(n) = a_n for all n in \mathbb{N}. -

    - -

    -

      -
    1. -

      - If \lim\limits_{x\to\infty} f(x) = L, - then \lim\limits_{n\to\infty} a_n = L. -

      -
      2. - -
    2. -

      - If \lim\limits_{x\to\infty} f(x) = \infty, - then \lim\limits_{n\to\infty} a_n = \infty. -

      -
      3. -
    -

    -     -     - -

    - allows us, in certain cases, - to apply the tools developed in to limits of sequences. - Note two things not stated by the theorem: -

    - -

    -

      -
    1. -

      - If \lim\limits_{x\to\infty}f(x) does not exist, - we cannot conclude that \lim\limits_{n\to\infty} a_n does not exist. - It may, or may not, exist. - For instance, - we can define a sequence \{a_n\} = \{\cos(2\pi n)\}. - Let f(x) = \cos(2\pi x). - Since the cosine function oscillates over the real numbers, - the limit \lim\limits_{x\to\infty}f(x) does not exist. - - However, for every positive integer n, \cos(2\pi n) = 1, so \lim\limits_{n\to\infty} a_n = 1. -

      -
      2. - -
    2. -

      - If we cannot find a function f(x) whose domain contains the positive real numbers where - f(n) = a_n for all n in \mathbb{N}, - we cannot conclude \lim\limits_{n\to\infty} a_n does not exist. - It may, or may not, exist. -

      -
      3. -
    -

    - - - Determining convergence/divergence of a sequence - -

    - Determine the convergence or divergence of the following sequences. -

    - -

    -

      -
    1. - \ds\{a_n\} = \left\{\frac{3n^2-2n+1}{n^2-1000}\right\} -
      2. - -
    2. \{b_n\} = \{\cos(n) \}
      3. - -
    3. - \ds\{c_n\} = \left\{\frac{(-1)^n}{n}\right\} -
      4. -
    -

    -     - -

    -

      -
    1. -

      - Using , - we can state that \lim\limits_{x\to\infty} \frac{3x^2-2x+1}{x^2-1000} = 3. - (We could have also directly applied L'Hospital's Rule.) - Thus the sequence \{a_n\} converges, and its limit is 3. - A scatter plot of every 5 values of a_n is given in . - The values of a_n vary widely near n=30, - ranging from about -73 to 125, - but as n grows, the values approach 3. -

      -
      - Scatter plot for the sequence in of - - - A scatter plot showing a representative sample of points from the first sequence in this example. - -

      - A pair of coordinate axes are shown, with the horizontal axis (labeled n) in the center of the image. - The range for n is from 0 to 100, and every fourth point in the sequence has been plotted. - The scatter plot for the sequence a_n = \frac{3n^2-2n+1}{n^2-1000} follows the graph of - f(x)=\frac{3x^2-3x+1}{x^2-1000}. - The graph of this function has a vertical asymptote at x=\sqrt{1000}\approx 31.6. - Since this is not an integer value, a_n is defined for each n, - but we can see that the largest negative value of a_n occurs when n=31 - (to the left of the asymptote), and the largest positive value of a_n occurs when n=32 - (to the right of the asymptote). As n gets large, the y coordinate of each point gets closer to 3. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={20,40,60,80,100}, - ymin=-11,ymax=11, - xmin=-.1,xmax=110, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={1.75pt},domain=1:100,samples=21] {(3*x^2 - 2*x + 1)/(x^2 - 1000)}; - - \draw (axis cs:60,-8) node { $\ds a_n = \frac{3n^2 - 2n + 1}{n^2 - 1000}$}; - - \end{axis} - \end{tikzpicture} - - - - -
      -
      2. - -
    2. -

      - The limit \lim\limits_{x\to\infty}\cos(x) does not exist, - as \cos(x) oscillates - (and takes on every value in [-1,1] infinitely many times). - Thus we cannot apply . - - The fact that the cosine function oscillates strongly hints that \cos(n), - when n is restricted to - \mathbb{N}, will also oscillate. - , - where the sequence is plotted, shows that this is true. - Because only discrete values of cosine are plotted, - it does not bear strong resemblance to the familiar cosine wave. - The proof of the following statement is beyond the scope of this text, - but it is true: - there are infinitely many integers n that are arbitrarily (, - very) close to an even multiple of \pi, - so that \cos n \approx 1. - Similarly, there are infinitely many integers m that are arbitrarily close to an odd multiple of \pi, - so that \cos m \approx -1. - As the sequence takes on values near 1 and -1 infinitely many times, - we conclude that \lim\limits_{n\to\infty}a_n does not exist. -

      -
      - Scatter plot for the sequence in of - - - A scatter plot showing a representative sample of points from the second sequence in this example. - -

      - The scatter plot for this example shows the first 100 terms in the sequence a_n=\cos(n). - The plot shows a collection of points whose y values seem to be randomly scattered - between -1 and 1. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={20,40,60,80,100}, - ymin=-1.1,ymax=1.1, - xmin=-.1,xmax=110, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={1.5pt},domain=1:100,samples=101] {cos(deg(x))}; - - \end{axis} - - - \node at (myplot.north) { $\ds a_n = \cos(n) $}; - - \end{tikzpicture} - - - - -
      -
      3. - -
    3. -

      - We cannot actually apply here, - as the function f(x) = (-1)^x/x is not well defined. (What does (-1)^{\sqrt{2}} mean? - In actuality, there is an answer, - but it involves complex analysis, - beyond the scope of this text.) Instead, - we invoke the definition of the limit of a sequence. - By looking at the plot in , - we would like to conclude that the sequence converges to L=0. -

      - -
      - Scatter plot for the sequence in - - - A scatter plot showing a representative sample of points from the third sequence in this example. - -

      - The scatter plot for the sequence a_n = (-1)^n/n shows a pair of coordinate axes, - with the horizontal axis labeled n, and positioned in the center of the plot. -

      - -

      - When n is even, the points (n,a_n) follow the graph y=1/x, - and get closer and closer to the n axis as n increases. -

      - -

      - When n is odd, the points (n,a_n) follow the graph y=-1/x, - and approach the n axis from below. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-1.1,ymax=1.1, - xmin=-.1,xmax=22, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={1.75pt},domain=1:20,samples=20] {((-1)^x)/x}; - - \draw (axis cs:10,-.7) node { $\ds a_n = \frac{(-1)^n}{n}$}; - - \end{axis} - \end{tikzpicture} - - - - -
      - -

      - Let \epsilon\gt 0 be given. - We can find a natural number m such that 1/m \lt \varepsilon. - Let n\gt m, and consider \abs{a_n - L}: - - \abs{a_n - L} \amp = \left\lvert\frac{(-1)^n}{n} - 0\right\rvert - \amp = \frac1n - \amp \lt \frac1m \text{ (since \(n\gt m\)) } - \amp \lt \varepsilon - . - We have shown that by picking m large enough, - we can ensure that a_n is arbitrarily close to our limit, - L=0, - hence by the definition of the limit of a sequence, - we can say \lim\limits_{n\to\infty}a_n = 0. -

      -
      4. -
    -

    -     - -     - -

    - In the previous example we used the definition of the limit of a sequence to determine the convergence of a sequence as we could not apply . - In general, we like to avoid invoking the definition of a limit, - and the following theorem gives us tool that we could use in that example instead. -

    - - - Absolute Value Theorem - -

    - Let \{a_n\} be a sequence. - If \lim\limits_{n\to\infty} \abs{a_n} = 0, - then \lim\limits_{n\to\infty} a_n = 0 - - Absolute Value Theorem - limitAbsolute Value Theorem - sequenceAbsolute Value Theorem -

    -     - -

    - Let \lim\limits_{n\to\infty} \abs{a_n} = 0. - We start by noting that -\abs{a_n}\leq a_n \leq \abs{a_n}. - If we apply limits to this inequality: - - \lim\limits_{n \to \infty}\left(-\abs{a_n}\right) \leq \lim\limits_{n \to \infty} a_n \leq \lim\limits_{n \to \infty} \abs{a_n} - -\lim\limits_{n \to \infty}\abs{a_n} \leq \lim\limits_{n \to \infty} a_n \leq \lim\limits_{n \to \infty} \abs{a_n} - Using the fact that \lim\limits_{n\to\infty} \abs{a_n} = 0: - 0 \leq \lim\limits_{n \to \infty} a_n \leq 0 - - We conclude that the only possible answer for \lim\limits_{n \to \infty} a_n is 0. -

    -     -     - - - - - Determining the convergence/divergence of a sequence - -

    - Determine the convergence or divergence of the following sequences. -

    - -

    -

      -
    1. - \ds \{a_n\} = \left\{\frac{(-1)^n}{n}\right\} -
      2. - -
    2. - \ds \{a_n\} = \left\{\frac{(-1)^n(n+1)}{n}\right\} -
      3. -
    -

    -     - -

    -

      -
    1. -

      - This appeared in . - We want to apply , - so consider the limit of \{\abs{a_n}\}: - - \lim_{n\to\infty} \abs{a_n} \amp = \lim_{n\to\infty} \abs{\frac{(-1)^n}{n}} - \amp = \lim_{n\to\infty} \frac{1}{n} - \amp = 0 - . - Since this limit is 0, we can apply - and state that \lim\limits_{n\to\infty} a_n=0. -

      -
      2. - -
    2. -

      - Because of the alternating nature of this sequence (, every other term is multiplied by -1), - we cannot simply look at the limit \lim\limits_{x\to\infty} \frac{(-1)^x(x+1)}{x}. - We can try to apply the techniques of : - - \lim_{n\to\infty} \abs{a_n} \amp = \lim_{n\to\infty} \abs{\frac{(-1)^n(n+1)}{n}} - \amp = \lim_{n\to\infty} \frac{n+1}{n} - \amp = 1 - . - We have concluded that when we ignore the alternating sign, - the sequence approaches 1. - This means we cannot apply ; - it states the the limit must be 0 in order to conclude anything. -

      - -
      - A plot of a sequence in , part 2 - - - Scatter plot illustrating the first 20 points in the sequence from the second part of this example. - -

      - The scatter plot for a_n=(-1)^n(n+1)/n splits into two parts. - When n is even, the points (n,a_n) follow the graph y=1+1/x, - and approach the line y=1 from above as n gets large. - When n is odd, the points (n,a_n) follow the graph y=-1-1/x, - and approach the line y=-1 from below as n gets large. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ytick={-1,-2,1,2}, - ymin=-2.5,ymax=2.5, - xmin=-.1,xmax=22, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:20,samples=20] {((-1)^x*(x+1))/x}; - - \draw (axis cs:15,1.9) node { $\ds a_n = \frac{(-1)^n(n+1)}{n}$}; - - \end{axis} - \end{tikzpicture} - - - - -
      - -

      - Since we know that the signs of the terms alternate and - we know that the limit of \abs{a_n} is 1, we know that as n approaches infinity, - the terms will alternate between values close to 1 and -1, - meaning the sequence diverges. - A plot of this sequence is given in . -

      -
      3. -
    -

    -     -     - -

    - We continue our study of the limits of sequences by considering some of the properties of these limits. -

    - - - Properties of the Limits of Sequences - -

    - Let \{a_n\} and \{b_n\} be sequences such that \lim\limits_{n\to\infty} a_n = L, - \lim\limits_{n\to\infty} b_n = K, - and let c be a real number. -

    - -

    -

      -
    1. -

      - \lim\limits_{n\to\infty} (a_n\pm b_n) = L\pm K - - sequenceslimit properties - -

      -
      2. - -
    2. -

      - \lim\limits_{n\to\infty} (a_n\cdot b_n) = L\cdot K -

      -
      3. - -
    3. -

      - \lim\limits_{n\to\infty} \left(\frac{a_n}{b_n}\right) = \frac{L}{K}, K\neq 0 -

      -
      4. - -
    4. -

      - \lim\limits_{n\to\infty} c\cdot a_n = c\cdot L -

      -
      5. - -
    5. -

      - If g is continuous at L, then \lim\limits_{n\to\infty}g(a_n)=g(L). -

      -
      6. -
    -

    -     -     - - - - - Applying properties of limits of sequences - -

    - Let the following sequences, and their limits, be given: -

    - -

    -

      -
    • -

      - \ds \{a_n\} = \left\{\frac{n+1}{n^2}\right\}, - and \lim\limits_{n\to\infty} a_n = 0; -

      -
      • - -
    • -

      - \ds \{b_n\} = \left\{\left(1+\frac1n\right)^{n}\right\}, - and \lim\limits_{n\to\infty} b_n = e; and -

      -
      • - -
    • -

      - \ds \{c_n\} = \big\{n\cdot \sin(5/n)\big\}, - and \lim\limits_{n\to\infty} c_n = 5. -

      -
      • -
    -

    - -

    - Evaluate the following limits. -

    - -

    -

      -
    1. \lim\limits_{n\to\infty} (a_n+b_n)
      2. - -
    2. \lim\limits_{n\to\infty} (b_n\cdot c_n)
      3. - -
    3. \lim\limits_{n\to\infty} (1000\cdot a_n)
      4. -
    -

    -     - -

    - We will use to answer each of these. -

    - -

    -

      -
    1. -

      - Since \lim\limits_{n\to\infty} a_n = 0 and \lim\limits_{n\to\infty} b_n = e, - we conclude that \lim\limits_{n\to\infty} (a_n+b_n) = 0+e = e. - So even though we are adding something to each term of the sequence b_n, - we are adding something so small that the final limit is the same as before. -

      -
      2. - -
    2. -

      - Since \lim\limits_{n\to\infty} b_n = e and \lim\limits_{n\to\infty} c_n = 5, - we conclude that \lim\limits_{n\to\infty} (b_n\cdot c_n) = e\cdot 5 = 5e. -

      -
      3. - -
    3. -

      - Since \lim\limits_{n\to\infty} a_n = 0, - we have \lim\limits_{n\to\infty} 1000a_n =1000\cdot 0 = 0. - It does not matter that we multiply each term by 1000; - the sequence still approaches 0. (It just takes longer to get close to 0.) -

      -
      4. -
    -

    -     - -     - -

    - There is more to learn about sequences than just their limits. - We will also study their range and the relationships terms have with the terms that follow. - We start with some definitions describing properties of the range. -

    - - - Bounded and Unbounded Sequences - -

    - A sequence \{a_n\} is said to be bounded - if there exist real numbers m and M such that - m \leq a_n \leq M for all n in \mathbb{N}. - The number m is called a lower bound for the sequence, - and the number M is called an upper bound for the sequence. -

    - -

    - A sequence \{a_n\} is said to be - unbounded if it is not bounded. -

    - -

    - A sequence \{a_n\} is said to be - bounded above - if there exists an M such that - a_n \lt M for all n in \mathbb{N}; - it is bounded below if there exists an m such that - m\lt a_n for all n in \mathbb{N}. - sequencesboundedness - bounded sequence - unbounded sequence -

    -     -     - -

    - It follows from this definition that an unbounded sequence may be bounded above or bounded below; - a sequence that is both bounded above and below is simply a bounded sequence. -

    - - - Determining boundedness of sequences - -

    - Determine the boundedness of the following sequences. -

    - -

    -

      -
    1. \ds\{a_n\} = \left\{\frac1n\right\}
      2. - -
    2. \{a_n\} = \{2^n\}
      3. -
    -

    -     - -

    -

      -
    1. -

      - The terms of this sequence are always positive but are decreasing, - so we have 0\lt a_n\lt 2 for all n. - Thus this sequence is bounded. - illustrates this. -

      -
      2. - -
    2. -

      - The terms of this sequence obviously grow without bound. - However, it is also true that these terms are all positive, - meaning 0\lt a_n. - Thus we can say the sequence is unbounded, but also bounded below. - illustrates this. -

      -
      3. -
    -

    - -
    - A plot of \{a_n\} = \{1/n\} and \{a_n\} = \{2^n\} from - - -
    - - - - A scatter plot of the first 10 terms in the first sequence for this example. - -

    - The points (n,a_n) are plotted, for n ranging from 1 to 10. - These points follow the graph y=1/x, begining at (1,1), - and descending toward the n axis as n gets large. -

    - -

    - We can see that the value of a_n remains positive, but will continue to decrease as n increases. - This illustrates the fact that the sequence is bounded between 0 and 1. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,...,10}, - ytick={1}, - extra y ticks={.5,.25,.1}, - extra y tick labels={$1/2$,$1/4$,$1/10$}, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {1/x}; - - \draw (axis cs:6,.75) node { $\ds a_n = \frac{1}{n}$}; - - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - The scatter plot for the second sequence in this example, which shows exponential growth. - -

    - This time, the points (n,2^n) are plotted, for values of n between 1 and 8. - These points follow the exponential graph y=2^x; - we can see that the y value will continue to increase without bound, - illustrating the fact that this sequence is not bounded. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=275, - xmin=-.1,xmax=9, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:8,samples=8] {2^x}; - - \draw (axis cs:4,150) node { $\ds a_n = 2^{n}$}; - - \end{axis} - \end{tikzpicture} - - - - -
    -     -
    -     - -     - -

    - The previous example produces some interesting concepts. - First, we can recognize that the sequence \ds\left\{1/n\right\} converges to 0. - This says, informally, that most - of the terms of the sequence are - really close to 0. - This implies that the sequence is bounded, - using the following logic. - First, most terms are near 0, so we could find some sort of bound on these terms - (using , - the bound is \varepsilon). - That leaves a few terms that are not near 0 (, a - finite number of terms). - A finite list of numbers is always bounded. -

    - -

    - This logic implies that if a sequence converges, it must be bounded. - This is indeed true, as stated by the following theorem. -

    - - - Convergent Sequences are Bounded - -

    - Let \ds \left\{a_n\right\} be a convergent sequence. - Then \{a_n\} is bounded. - bounded sequenceconvergence - convergenceof sequence - sequencesconvergent -

    -     -     - - - - - -

    - In - we saw the sequence \ds \{b_n\} = \left\{\left(1+1/n\right)^{n}\right\}, - where it was stated that \lim\limits_{n\to\infty} b_n = e. - (Note that this is simply restating part of . - The limit can also be found using logarithms and L'Hospital's rule.) - Even though it may be difficult to intuitively grasp the behavior of this sequence, - we know immediately that it is bounded. -

    - - - - - -

    - Another interesting concept to come out of - again involves the sequence \{1/n\}. - We stated, without proof, that the terms of the sequence were decreasing. - That is, that a_{n+1} \lt a_n for all n. (This is easy to show. - Clearly n \lt n+1. - Taking reciprocals flips the inequality: - 1/n \gt 1/(n+1). - This is the same as a_n \gt a_{n+1}.) Sequences that either steadily increase or decrease are important, - so we give this property a name. -

    - - - Monotonic Sequences - -

    -

      -
    1. -

      - A sequence \{a_n\} is monotonically increasing - if a_n \leq a_{n+1} for all n, , - - a_1 \leq a_2 \leq a_3 \leq \cdots a_n \leq a_{n+1} \cdots - -

      -
      2. - -
    2. -

      - A sequence \{a_n\} is monotonically decreasing - if a_n \geq a_{n+1} for all n, , - - a_1 \geq a_2 \geq a_3 \geq \cdots a_n \geq a_{n+1} \cdots - -

      -
      3. - -
    3. -

      - A sequence is monotonic - if it is monotonically increasing or monotonically decreasing. - - sequencesmonotonic - monotonic sequence - -

      -
      4. -
    -

    -     -     - - - - - - - Determining monotonicity - -

    - Determine the monotonicity of the following sequences. -

    - -

    -

      -
    1. -

      - \ds \{a_n\} = \left\{\frac{n+1}n\right\} -

      -
      2. - -
    2. -

      - \ds \{a_n\} = \left\{\frac{n^2+1}{n+1}\right\} -

      -
      3. - -
    3. -

      - \ds \{a_n\} = \left\{\frac{n^2-9}{n^2-10n+26}\right\} -

      -
      4. - -
    4. -

      - \ds \{a_n\} = \left\{\frac{n^2}{n!}\right\} -

      -
      5. -
    -

    -     - -

    - In each of the following, we will examine a_{n+1}-a_n. - If a_{n+1}-a_n \geq 0, - we conclude that a_n\leq a_{n+1} and hence the sequence is increasing. - If a_{n+1}-a_n\leq 0, - we conclude that a_n\geq a_{n+1} and the sequence is decreasing. - Of course, a sequence need not be monotonic and perhaps neither of the above will apply. -

    - -

    - We also give a scatter plot of each sequence. - These are useful as they suggest a pattern of monotonicity, - but analytic work should be done to confirm a graphical trend. -

    - -

    -

      -
    1. -

      - - a_{n+1}-a_n \amp = \frac{n+2}{n+1} - \frac{n+1}{n} - \amp = \frac{(n+2)(n)-(n+1)^2}{(n+1)n} - \amp = \frac{-1}{n(n+1)} - \amp \lt 0 \text{ for all \(n\). } - - Since a_{n+1}-a_n\lt 0 for all n, - we conclude that the sequence is decreasing. -

      -
      2. - -
    2. -

      - - a_{n+1}-a_n \amp = \frac{(n+1)^2+1}{n+2} - \frac{n^2+1}{n+1} - \amp = \frac{\big((n+1)^2+1\big)(n+1)- (n^2+1)(n+2)}{(n+1)(n+2)} - \amp = \frac{n^2+3n}{(n+1)(n+2)} - \amp \gt 0 \text{ for all \(n\). } - - Since a_{n+1}-a_n\gt 0 for all n, - we conclude the sequence is increasing. -

      -
      3. - -
    3. -

      - We can clearly see in , - where the sequence is plotted, that it is not monotonic. - However, it does seem that after the first 4 terms it is decreasing. - To understand why, perform the same analysis as done before: - - - a_{n+1}-a_n \amp = \frac{(n+1)^2-9}{(n+1)^2-10(n+1)+26} - \frac{n^2-9}{n^2-10n+26} - \amp = \frac{n^2+2n-8}{n^2-8n+17}-\frac{n^2-9}{n^2-10n+26} - \amp = \frac{(n^2+2n-8)(n^2-10n+26)-(n^2-9)(n^2-8n+17)}{(n^2-8n+17)(n^2-10n+26)} - \amp = \frac{-10n^2+60n-55}{(n^2-8n+17)(n^2-10n+26)} - . -

      - -

      - We want to know when this is greater than, or less than, 0. - The denominator is always positive, - therefore we are only concerned with the numerator. - For small values of n, - the numerator is positive. - As n grows large, - the numerator is dominated by -10n^2, - meaning the entire fraction will be negative; - , for large enough n, a_{n+1}-a_n \lt 0. - Using the quadratic formula we can determine that the numerator is negative for n\geq 5. - In short, the sequence is simply not monotonic, - though it is useful to note that for n\geq 5, - the sequence is monotonically decreasing. -

      -
      4. - -
    4. - -

      - Again, the plot in - shows that the sequence is not monotonic, - but it suggests that it is monotonically decreasing after the first term. - We perform the usual analysis to confirm this. - - a_{n+1}-a_n \amp = \frac{(n+1)^2}{(n+1)!} - \frac{n^2}{n!} - \amp = \frac{(n+1)^2-n^2(n+1)}{(n+1)!} - \amp = \frac{-n^3+2n+1}{(n+1)!} - - When n=1, the above expression is \gt 0; - for n\geq 2, the above expression is \lt 0. - Thus this sequence is not monotonic, - but it is monotonically decreasing after the first term. -

      -
      5. -
    -

    - -
    - Plots of sequences in - - -
    - - - - Plot of the first sequence in this example. It is decreasing and bounded below. - -

    - The scatter plot for a_n = (n+1)/n shows a sequence that is decreasing, - and bounded below by y=1. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ytick={-1,-2,1,2}, - ymin=-.1,ymax=2.5, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {(x+1)/x}; - - \draw (axis cs:6,1.9) node { $\ds a_n = \frac{n+1}{n}$}; - - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - Scatter plot for the second sequence in this example. It is increasing but not bounded. - -

    - The points in the scatter plot for this example get larger as n increases, - illustrating the fact that this is an increasing sequence. - In fact, for large n we can see that a_n \approx n, - and the points in the plot do appear to follow close to the line y=x. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-1,ymax=11, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {(x^2+1)/(x+1)}; - - \draw (axis cs:7.5,1.9) node { $\ds a_n = \frac{n^2+1}{n+1}$}; - - \end{axis} - \end{tikzpicture} - - - - -
    -     - - -
    - - - - Scatter plot for the third sequence in this example. It is not monotonic. - -

    - The plot for a_n = \frac{n^2-9}{n^2-10n+26} shows a sequence that is initially increasing, - but begins to decrease after n=5. -

    - -

    - Although this sequence is not monotonic, it is eventually monotonic, - and bounded below, - which (as we will soon see) is sufficient for the determination of a limit. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-1,ymax=16, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {(x^2-9)/(x^2-10*x+26)}; - - \draw (axis cs:3.4,11) node { $\ds a_n = \frac{n^2-9}{n^2-10n+26}$}; - - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - - - Scatter plot for the last sequence in this example. It is not monotonic. - -

    - The plot for a_n =n^2/n! shows a sequence that is initially increasing, - but begins to decrease after n=2. -

    - -

    - Although this sequence is not monotonic, it is eventually monotonic, - and bounded below, - which (as we will soon see) is sufficient for the determination of a limit. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=2.1, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {(x^2)/(factorial(x))}; - - \draw (axis cs:7,1.25) node { $\ds a_n = \frac{n^2}{n!}$}; - - \end{axis} - - - - \end{tikzpicture} - - - - -
    -     -     -
    -     - -     - -

    - Knowing that a sequence is monotonic can be useful. - Consider, for example, - a sequence that is monotonically decreasing and is bounded below. - We know the sequence is always getting smaller, - but that there is a bound to how small it can become. - This is enough to prove that the sequence will converge, - as stated in the following theorem. -

    - - - Bounded Monotonic Sequences are Convergent - -

    -

      -
    1. -

      - Let \{a_n\} be a monotonically increasing sequence that is bounded above. - Then \{a_n\} converges. -

      -
      2. - -
    2. -

      - Let \{a_n\} be a monotonically decreasing sequence that is bounded below. - Then \{a_n\} converges. - sequencesconvergent - convergenceof monotonic sequences -

      -
      3. -
    -

    -     -     - -

    - Consider once again the sequence \{a_n\} = \{1/n\}. - It is easy to show it is monotonically decreasing and that it is always positive (, bounded below by 0). - Therefore we can conclude by that the sequence converges. - We already knew this by other means, - but in the following section this theorem will become very useful. -

    - -

    - We can replace - with the statement Let \{a_n\} be a bounded, - monotonic sequence. - Then \{a_n\} converges; - , \ds \lim_{n \to\infty}a_n exists. - We leave it to the reader in the exercises to show the theorem and the above statement are equivalent. -

    - - - -

    - Sequences are a great source of mathematical inquiry. - The On-Line Encyclopedia of Integer Sequences () contains thousands of sequences and their formulae. - (As of this writing, there are 328,977 sequences in the database.) - Perusing this database quickly demonstrates that a single sequence can represent several different - real life phenomena. -

    - -

    - Interesting as this is, our interest actually lies elsewhere. - We are more interested in the sum of a sequence. - That is, given a sequence \{a_n\}, - we are very interested in a_1+a_2+a_3+\cdots. - Of course, one might immediately counter with - Doesn't this just add up to infinity? Many times, - yes, but there are many important cases where the answer is no. - This is the topic of series, - which we begin to investigate in . -

    - - - - Terms and Concepts - - - -

    - Use your own words to define a sequence. -

    -     - - - -     - - - - -

    - The domain of a sequence is the numbers. -

    -     - - - - - - - - -     - - - - -

    - Use your own words to describe the - range of a sequence. -

    -     - - - -     - - - - -

    - Describe what it means for a sequence to be bounded. -

    -     - - - -     -     - - - Problems - - - -

    - Give the first five terms of the given sequence. -

    -     - - - - -

    - \ds \{a_n\} = \left\{\frac{4^n}{(n+1)!}\right\} -

    -     - -

    - 2,\frac{8}{3},\frac{8}{3},\frac{32}{15},\frac{64}{45} -

    -     - -     - - - - -

    - \ds \{b_n\} = \left\{\left(-\frac32\right)^n\right\} -

    -     - -

    - -\frac{3}{2},\frac{9}{4},-\frac{27}{8},\frac{81}{16}, - -\frac{243}{32} -

    -     - -     - - - - -

    - \ds \{c_n\} = \left\{-\frac{n^{n+1}}{n+2}\right\} -

    -     - -

    - -\frac{1}{3},-2,-\frac{81}{5},-\frac{512}{3},-\frac{15625}{7} -

    -     - -     - - - - -

    - \ds \{d_n\} = \left\{\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)\right\} -

    -     - -

    - 1, 1, 2, 3, 5 -

    -     - -     - -     - - - -

    - Determine the nth term of the given sequence. -

    -     - - - - -

    - 4, 7, 10, 13, 16, \ldots -

    -     - -

    - a_n = 3n+1 -

    -     - -     - - - - -

    - \ds 3,\, -\frac32,\, \frac34,\, -\frac38,\, \ldots -

    -     - -

    - a_n = (-1)^{n+1}\frac{3}{2^{n-1}} -

    -     - -     - - - - -

    - 10,\, 20,\, 40,\, 80,\, 160,\, \ldots -

    -     - -

    - a_n = 10\cdot 2^{n-1} -

    -     - -     - - - - -

    - \ds 1, 1,\, \frac12,\, \frac16,\, \frac1{24},\, \frac1{120},\, \ldots -

    -     - -

    - a_n = 1/(n-1)! -

    -     - -     - -     - - - -

    - Use the following information to determine the limit of the given sequences. -

    - -

    -

      -
    • -

      - \ds \{a_n\} = \left\{\frac{2^n-20}{2^n}\right\}; - \lim\limits_{n\to\infty} a_n = 1 -

      -
      • - -
    • -

      - \ds \{b_n\} = \left\{\left(1+\frac{2}{n}\right)^n\right\}; - \lim\limits_{n\to\infty} b_n = e^2 -

      -
      • - -
    • -

      - \ds \{c_n\} = \left\{\sin(3/n)\right\}; - \lim\limits_{n\to\infty} c_n = 0 -

      -
      • -
    -

    -     - - - - -

    - \ds\{a_n\} = \left\{ \frac{2^n-20}{7\cdot2^n} \right\} -

    -     - -

    - 1/7 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{ 3b_n-a_n \right\} -

    -     - -

    - 3e^2-1 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{ \sin(3/n)\left(1+\frac2n\right)^n \right\} -

    -     - -

    - 0 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{ \left(1+\frac2n\right)^{2n} \right\} -

    -     - -

    - e^4 -

    -     - -     - -     - - - -

    - Determine whether the sequence converges or diverges. - If convergent, give the limit of the sequence. -

    -     - - - - -

    - \ds\{a_n\} = \left\{(-1)^n\frac{n}{n+1}\right\} -

    -     - -

    - diverges -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\frac{4n^2-n+5}{3n^2+1}\right\} -

    -     - -

    - converges to 4/3 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\frac{4^n}{5^n}\right\} -

    -     - -

    - converges to 0 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\frac{n-1}{n}-\frac{n}{n-1}\right\}, - n\geq 2 -

    -     - -

    - converges to 0 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\ln(n)\right\} -

    -     - -

    - diverges -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\frac{3n}{\sqrt{n^2+1}}\right\} -

    -     - -

    - converges to 3 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\left(1+\frac1n\right)^n\right\} -

    -     - -

    - converges to e -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{5-\frac1n\right\} -

    -     - -

    - converges to 5 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\frac{(-1)^{n+1}}{n}\right\} -

    -     - -

    - converges to 0 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\frac{1.1^n}{n}\right\} -

    -     - -

    - diverges -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\frac{2n}{n+1}\right\} -

    -     - -

    - converges to 2 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{(-1)^n\frac{n^2}{2^n-1}\right\} -

    -     - -

    - converges to 0 -

    -     - -     - -     - - - -

    - Determine whether the sequence is bounded, - bounded above, bounded below, or none of the above. -

    -     - - - - -

    - \ds\{a_n\} = \left\{\sin(n) \right\} -

    -     - -

    - bounded -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\tan(n) \right\} -

    -     - -

    - neither bounded above or below -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{(-1)^n\frac{3n-1}{n}\right\} -

    -     - -

    - bounded -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\frac{3n^2-1}{n}\right\} -

    -     - -

    - bounded below -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{n\cos(n) \right\} -

    -     - -

    - neither bounded above or below -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{2^n-n!\right\} -

    -     - -

    - bounded above -

    -     - -     - -     - - - -

    - Determine whether the sequence is monotonically increasing or decreasing. - If it is not, - determine if there is an m such that it is monotonic for all n\geq m. -

    -     - - - - -

    - \ds\{a_n\} = \left\{\frac{n}{n+2}\right\} -

    -     - -

    - monotonically increasing -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\frac{n^2-6n+9}{n}\right\} -

    -     - -

    - monotonically increasing for n\geq 3 -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{(-1)^n\frac{1}{n^3}\right\} -

    -     - -

    - never monotonic -

    -     - -     - - - - -

    - \ds\{a_n\} = \left\{\frac{n^2}{2^n}\right\} -

    -     - -

    - monotonically decreasing for n\geq 3 -

    -     - -     -     - - - -

    - The following exercises explore further the theory of sequences. -

    -     - - - - -

    - Prove ; that is, - use the definition of the limit of a sequence to show that if \lim\limits_{n\to\infty} \abs{a_n} = 0, then \lim\limits_{n\to\infty} a_n = 0. -

    -     - -

    - Let \{a_n\} be given such that \lim\limits_{n\to\infty} \abs{a_n} = 0. - By the definition of the limit of a sequence, - given any \varepsilon \gt 0, - there is a m such that for all n \gt m, \abs{\abs{a_n} - 0} \lt \varepsilon. - Since \abs{\abs{a_n}-0} = \abs{a_n - 0}, - this directly implies that for all n \gt m, - \abs{a_n - 0} \lt \varepsilon, - meaning that \lim\limits_{n\to\infty} a_n = 0. -

    -     - -     - - - - -

    - Let \{a_n\} and \{b_n\} be sequences such that - \lim\limits_{n\to\infty} a_n = L and \lim\limits_{n\to\infty} b_n = K. -

    -     - - - -

    - Show that if a_n\lt b_n for all n, then L\leq K. -

    -     -     - - - -

    - Give an example where L = K. -

    -     - -

    - a_n = 1/3^n and b_n = 1/2^n -

    -     -     - -     - - - - -

    - Prove the Squeeze Theorem for sequences: Let \{a_n\} and \{b_n\} be such that - \lim\limits_{n\to\infty} a_n = L and \lim\limits_{n\to\infty} b_n = L, - and let \{c_n\} be such that a_n\leq c_n\leq b_n for all n. - Then \lim\limits_{n\to\infty} c_n = L -

    -     - -

    - A sketch of one proof method: -

    - -

    - Let any \epsilon>0 be given. - Since \{a_n\} and \{b_n\} converge, - there exists an N>0 such that for all n\geq N, - both a_n and b_n are within \epsilon/2 of L; - we can conclude that they are at most - \epsilon apart from each other. - Since a_n\leq c_n \leq b_n, - one can show that c_n is within \epsilon of L, - showing that \{c_n\} also converges to L. -

    -     - -     - - - - - -

    - Prove the statement Let \{a_n\} be a bounded, - monotonic sequence. - Then \{a_n\} converges; - , \ds \lim_{n \to\infty}a_n exists. - is equivalent to . - That is, - -

      -
    1. -

      - Show that if is true, - then above statement is true, and -

      -
      2. - -
    2. -

      - Show that if the above statement is true, - then is true. -

      -
      3. -
    -

    -     - -

    - A sketch of one proof method: - -

      -
    1. -

      - Assume that is true, - and let \{a_n\} be bounded and monotonic. - Since \{a_n\} is bounded, - it is bounded both above and below. - If it is increasing, - it is bounded above and we apply ; - if it is decreasing, it is bounded below and we apply the theorem. - Either way, \{a_n\} converges and the statement is true. -

      -
      2. - -
    2. -

      - Assume the statement is true, - and let \{a_n\} be a monotonically increasing sequence that is bounded above. - Since \{a_n\} is monotonically increasing, - a_1\leq a_2\leq \ldots; - that is, a_1 bounds \{a_n\} from below. - Therefore \{a_n\} is bounded and monotonic; - by the statement, \{a_n\} converges. - A similar statement can be made for when \{a_n\} is monotonically decreasing and bounded below. - Therefore is true. -

      -
      3. -
    -

    -     - -     -     -     -     -
    -
    - Infinite Series - - -

    - Given the sequence \{a_n\} = \{1/2^n\} = 1/2,\, 1/4,\, 1/8,\, \ldots, - consider the following sums: - - a_1 \amp= 1/2 \amp = \amp 1/2 - a_1+a_2\amp = 1/2+1/4 \amp = \amp 3/4 - a_1+a_2+a_3 \amp = 1/2+1/4+1/8 \amp =\amp 7/8 - a_1+a_2+a_3+a_4 \amp= 1/2+1/4+1/8+1/16 \amp = \amp 15/16 - -

    - -

    - In general, we can show that - - a_1+a_2+a_3+\cdots +a_n = \frac{2^n-1}{2^n} = 1-\frac{1}{2^n} - . -

    - -

    - Let S_n be the sum of the first n terms of the sequence \{1/2^n\}. - From the above, we see that S_1=1/2, S_2 = 3/4, etc. - Our formula at the end shows that S_n = 1-1/2^n. -

    - -

    - Now consider the following limit: - \lim\limits_{n\to\infty}S_n = \lim_{n\to\infty}\big(1-1/2^n\big) = 1. - This limit can be interpreted as saying something amazing: - the sum of all the terms of the sequence \{1/2^n\} is 1. -

    - -

    - This example illustrates some interesting concepts that we explore in this section. - We begin this exploration with some definitions. -

    -     - - - Convergence of series - - - Infinite Series, <m>n</m>th Partial Sums, Convergence, Divergence - -

    - Let \{a_n\} be a sequence, beginning at some index value n=k. -

    - -

    -

      -
    1. -

      - The sum \ds \sum_{n=k}^\infty a_n is called an - infinite series - (or, simply series). -

      -
      2. - -
    2. -

      - Let S_n denote the sum of the first n terms in the sequence \{a_n\}, - known as the nth partial sum of the sequence. - We can then define the sequence \{S_n\} of partial sums of \{a_n\}. -

      -
      3. - -
    3. -

      - If the sequence \{S_n\} converges to L, - we say the series \ds \sum_{n=k}^\infty a_n - converges to L, - and we write \ds \sum_{n=k}^\infty a_n = L. -

      -
      4. - -
    4. -

      - If the sequence \{S_n\} diverges, - the series \ds \sum_{n=k}^\infty a_n diverges. - seriesdefinition - seriespartial sums - seriesconvergent - seriesdivergent - convergenceof series - divergenceof series -

      -
      5. -
    -

    -     -     - -

    - Using our new terminology, - we can state that the series \ds \infser 1/2^n converges, - and \ds \infser 1/2^n = 1. -

    - -

    - Note that in the definition above, we do not necessarily assume that our sum begins with n=1. - In fact, it is quite common to have a series beginning at n=0, - and in some cases we may need to consider other values as well. - The nth partial sum S_n will always denote the sum of the first n terms: - For example, \infser 1/n has - - S_n = \overbrace{1+\frac12+\cdots + \frac1n}^{n \text{ terms}} - , - while \sum_{n=0}^\infty 3^{-n} has - - S_n = \overbrace{1+\frac13+\cdots + \frac{1}{3^{n-1}}}^{n \text{ terms}} - , - and \sum_{n=3}^\infty \frac{1}{n^2-2n} has - - S_n = \overbrace{\frac{1}{3}+\frac{1}{8}+\cdots + \frac{1}{(n+2)^2-2(n+2)}}^{n \text{ terms}} - . - In general, for the series \ds\sum_{n=k}^\infty a_n, - the nth partial sum will be \ds S_n = \sum_{i=k}^{k+n-1}a_i. -

    - -

    - We will explore a variety of series in this section. - We start with two series that diverge, - showing how we might discern divergence. -

    - - - Showing series diverge - -

    -

      -
    1. -

      - Let \{a_n\} = \{n^2\}. - Show \ds \infser a_n diverges. -

      -
      2. - -
    2. -

      - Let \{b_n\} = \{(-1)^{n+1}\}. - Show \ds \infser b_n diverges. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - Consider S_n, the nth partial sum. - - S_n \amp = a_1+a_2+a_3+\cdots+a_n - \amp = 1^2+2^2+3^2\cdots + n^2. - By , this is - \amp = \frac{n(n+1)(2n+1)}{6} - . - Since \lim\limits_{n\to\infty}S_n = \infty, - we conclude that the series \ds \infser n^2 diverges. - It is instructive to write \ds \infser n^2=\infty for this tells us - how the series diverges: - it grows without bound. - - A scatter plot of the sequences \{a_n\} and \{S_n\} is given in . - The terms of \{a_n\} are growing, - so the terms of the partial sums \{S_n\} are growing even faster, - illustrating that the series diverges. -

      -
      2. - -
    2. -

      - The sequence \{b_n\} starts with 1, -1, 1, -1, - \ldots. - Consider some of the partial sums S_n of \{b_n\}: - - S_1 \amp = 1 - S_2 \amp = 0 - S_3 \amp = 1 - S_4 \amp = 0 - - This pattern repeats; - we find that S_n = \begin{cases} 1 \amp n\, \text{ is odd } \\, 0 \amp n\, \text{ is even } \end{cases}. - As \{S_n\} oscillates, - repeating 1, 0, 1, 0, \ldots, - we conclude that \lim\limits_{n\to\infty}S_n does not exist, - hence \ds\infser (-1)^{n+1} diverges. - - A scatter plot of the sequence \{b_n\} and the partial sums \{S_n\} is given in . - When n is odd, - b_n = S_n so the marks for b_n are drawn oversized to show they coincide. -

      - -
      - Scatter plots relating to - -
      - - - - Scatter plots of the sequence, and corresponding partial sums, for the first part of this example. - -

      - The scatter plot shows the first 10 terms in the sequence a_n=n^2, which follow the graph y=x^2. - On the same plot, the first 10 terms in the sequence of partial sums S_n are shown. - As one might expect, the value of S_n grows much more rapidly than that of a_n. -

      -       - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - ymin=-10,%ymax=2.1, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,secondcolor,mark size={2.4pt},mark=square*,domain=1:10,samples=10] {x*(x+1)*(2*x+1)/6}; - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {x^2}; - - \end{axis} - - - \node[shift={(0,-25pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
      - -
      - - - - Scatter plots of the sequence, and corresponding partial sums, for the second part of this example. - -

      - This scatter plot shows the terms in the sequence b_n, along with the corresponding partial sums. - Since the value of b_n alternates between -1 and 1, - we see one sequence of dots along the line y=1, and another along the line y=-1. -

      - -

      - When n is even, S_n=0, so we also see a sequence of dots along the n axis. - When n is odd, S_n=a_n, so the dots for b_n and S_n overlap. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-1.1,ymax=1.1, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {(-1)^(x+1)}; - \addplot [only marks,secondcolor,mark=square*,mark size={2.5pt}] coordinates {(1,1)(2,0)(3,1)(4,0)(5,1)(6,0)(7,1)(8,0)(9,1)(10,0)}; - - \end{axis} - - - - \node[shift={(0,-15pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $b_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
      -       -
      -
      3. -
    -

    -     - -     - - - -

    - While it is important to recognize when a series diverges, - we are generally more interested in the series that converge. - In this section we will demonstrate a few general techniques for determining convergence; - later sections will delve deeper into this topic. -

    -     - - - Geometric Series -

    - One important type of series is a - geometric series. -

    - - - Geometric Series - -

    - A geometric series is a series of the form - - \infser[0] r^n = 1+r+r^2+r^3+\cdots+r^n+\cdots - -

    - -

    - Note that the index starts at n=0, not n=1. - seriesgeometric - geometric series -

    -     -     - - - -

    - We started this section with a geometric series, - although we dropped the first term of 1. - One reason geometric series are important is that they have nice convergence properties. -

    - - - Geometric Series Test - -

    - Consider the geometric series \ds \infser[0] r^n. -

    - -

    -

      -
    1. - -

      - For r\neq 1, the nth partial sum is: - - S_n = 1+r+r^2+\cdots + r^{n-1} = \frac{1-r^{n}}{1-r} - . - When r=1, S_n = n. -

      -
      2. - -
    2. -

      - The series converges if, and only if, \abs{r} \lt 1. - When \abs{r}\lt 1, - - seriesgeometric - geometric series - convergenceof geometric series - divergenceof geometric series - - \infser[0] r^n = \frac{1}{1-r} - . -

      -
      3. -
    -

    -     - -

    - We begin by proving the formula for the simplied form for the partial sums. - Consider the nth partial sum of the geometric series, - S_n=\sum_{i=0}^n r^i: - - S_n \amp = 1+r+r^2+\cdots+r^{n-2}+r^{n-1} - Multiply both sides by r: - r\cdot S_n \amp = r+r^2+r^3+\dots+r^{n-1}+r^{n} - Now subtract the second line from the first and solve for S_n: - S_n-r\cdot S_n \amp = 1-r^n - S_n(1-r) \amp = 1-r^{n} - S_n \amp = \frac{1-r^{n}}{1-r} - . - We have shown Part - of . -

    - -

    - Now, examining the partial sums, - we consider five cases to determine when S_n converges: - -

      -
    1. -

      - If \abs{r}\lt 1, then - r^n \to 0 as n \to \infty, - so we have \inflim S_n=\frac{1-0}{1-r}=\frac{1}{1-r}, - a convergent sequence of partial sums. -

      -
      2. - -
    2. -

      - If r \gt 1, then r^n \to \infty - as n \to \infty, so - - S_n = \frac{1-r^n}{1-r}=\frac{r^n}{r-1}-\frac{1}{r-1} - - diverges to infinity. (Note that r-1 is a positive constant.) -

      -
      3. - -
    3. -

      - If r \lt -1, then r^n will oscillate between large positive - and large negative values as n increases. - The same will be true of S_n, so \inflim S_n does not exist. -

      -
      4. - -
    4. - If r=1, - then S_n = \frac{1-1^{n+1}}{1-1} is undefined. - However, examining S_n = 1+r+r^2+\dots+r^n for r=1, - we can see that the partial sums simplify to S_n=n, - and this sequence diverges to \infty. -
      5. - -
    5. - If r=-1, then S_n = \frac{1-(-1)^{n}}{2}. - For even values of n, the partial sums are always 0. - For odd values of n, the partial sums are always 1. - So the sequence of partial sums diverges. -
      6. -
    - - Therefore, a geometric series converges if and only if \abs{r} \lt 1. -

    -     - -     - - - -

    - According to , - the series - - \ds\infser[0] \frac{1}{2^n} =\infser[0] \left(\frac 12\right)^2= 1+\frac12+\frac14+\cdots - - converges as r=1/2 \lt 1, - and \ds \infser[0] \frac{1}{2^n} = \frac{1}{1-1/2} = 2. - This concurs with our introductory example; - while there we got a sum of 1, we skipped the first term of 1. -

    - - - Exploring geometric series - -

    - Check the convergence of the following series. - If the series converges, find its sum. -

    - -

    -

      -
    1. -

      - \ds \sum_{n=2}^\infty \left(\frac34\right)^n -

      -
      2. - -
    2. -

      - \ds \infser[0] \left(\frac{-1}{2}\right)^n -

      -
      3. - -
    3. -

      - \ds \infser[0] 3^n -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - Since r=3/4\lt 1, this series converges. - By , we have that - - \infser[0] \left(\frac34\right)^n = \frac{1}{1-3/4} = 4 - . - However, note the subscript of the summation in the given series: - we are to start with n=2. - Therefore we subtract off the first two terms, giving: - - \sum_{n=2}^\infty \left(\frac34\right)^n = 4 - 1 - \frac34 = \frac94 - . - This is illustrated in . -

      -
      - Scatter plots for the series in - - - Scatter plots of the sequence, and corresponding partial sums, for the first part of this example. - -

      - Scatter plots for both a_n = (3/4)^n and S_n are shown together in the same image. - The point for a_2 is not visible, as it is covered up by S_2, which has the same value. -

      - -

      - The points in the plot for a_n illustrate a sequence that decreases toward 0, - while the points in the plot for S_n show that the partial sums are increasing toward a value slightly larger than 2. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={2,4,6,8,10}, - ymin=-.5,ymax=2.5, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt}] coordinates {(2,0.5625)(3,0.9844)(4,1.301)(5,1.538)(6,1.716)(7,1.85)(8,1.95)(9,2.025)(10,2.081)}; - \addplot [only marks,firstcolor,mark size={2.4pt},domain=2:10,samples=9] {(.75)^x}; - - \end{axis} - - - - \node[shift={(0,-10pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
      -
      2. - -
    2. -

      - Since \abs{r} = 1/2 \lt 1, - this series converges, and by , - - \infser[0] \left(\frac{-1}{2}\right)^n = \frac{1}{1-(-1/2)} = \frac23 - . - The partial sums of this series are plotted in . - Note how the partial sums are not purely increasing as some of the terms of the sequence \{(-1/2)^n\} are negative. -

      -
      - Scatter plots for the series in - - - Scatter plots of the sequence, and corresponding partial sums, for the second part of this example. - -

      - Scatter plots for a_n = (-1/2)^n and the corresponding partial sums are shown together in the same image. - The numbers a_n alternate in sign between positive and negative values, but get closer to 0 as n increases. -

      - -

      - The y coordinates for the points in the scatter plot for S_n also oscillate, - but the oscillations get steadily smaller, and appear to settle down toward a common y value. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={2,4,6,8,10}, - ymin=-1.1,ymax=1.1, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,secondcolor,mark=square*,mark size={2.4pt}] coordinates {(0,1.)(1,0.5)(2,0.75)(3,0.625)(4,0.6875)(5,0.6563)(6,0.6719)(7,0.6641)(8,0.668)(9,0.666)(10,0.667)}; - \addplot [only marks,firstcolor,mark size={2.4pt},domain=0:10,samples=11] {(-.5)^(x)}; - - \end{axis} - - - - \node[shift={(0,15pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
      -
      3. - -
    3. -

      - Since r \gt 1, the series diverges. - (This makes common sense; we expect the sum - - 1+3+9+27 + 81+243+\cdots - - to diverge.) - This is illustrated in . -

      -
      - Scatter plots for the series in - - - Scatter plots of the sequence, and corresponding partial sums, for the third part of this example. - -

      - Scatter plots for the sequence a_n=3^n and the corresponding partial sums are shown together in the same image. - Both plots follow curves that appear to be exponential in nature, with the points for S_n slightly above those for a_n. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-50,%ymax=1.1, - xmin=-1,xmax=7, - xlabel={$n$} - ] - - \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt}] coordinates {(0,1)(1,3)(2,12)(3,39)(4,120)(5,363)(6,1092)}; - \addplot [only marks,firstcolor,mark size={2.4pt},domain=0:6,samples=7] {3^(x)}; - - \end{axis} - - - - \node[shift={(0,-20pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
      -
      4. -
    -

    -     - -     -     - - - <m>p</m>-Series -

    - Another important type of series is the p-series. -

    - - - <m>p</m>-Series, General <m>p</m>-Series - -

    -

      -
    1. -

      - A p-series is a series of the form - - \infser \frac{1}{n^p}, \qquad \text{ where \(p \gt 0\). } - -

      -
      2. - -
    2. -

      - A general p-series is a series of the form - - seriesp@p-series - p@p-series - p@p-series - - - \infser \frac{1}{(an+b)^p} - , - where p \gt 0 and a, b are real numbers such that a\neq 0 and an+b\gt 0 for all n\geq 1. -

      -
      3. -
    -

    -     -     - -

    - Like geometric series, - one of the nice things about p-series is that they have easy to determine convergence properties. -

    - - - <m>p</m>-Series Test - -

    - A general p-series - \ds\infser \frac{1}{(an+b)^p} will converge if, - and only if, - p \gt 1. - seriesp@p-series - p@p-series - convergenceof p@of p-series - divergenceof p@of p-series -

    -     -     - - - - - Determining convergence of series - -

    - Determine the convergence of the following series. -

    - -

    -

      -
    1. -

      - \ds\infser \frac{1}{n} -

      -
      2. - -
    2. -

      - \ds\infser \frac{1}{n^2} -

      -
      3. - -
    3. -

      - \ds\infser \frac{1}{\sqrt{n}} -

      -
      4. - -
    4. -

      - \ds\infser \frac{(-1)^n}{n} -

      -
      5. - -
    5. -

      - \ds\sum_{n=11}^\infty \frac{1}{(\frac12n-5)^3} -

      -
      6. - -
    6. -

      - \ds\infser \frac{1}{2^n} -

      -
      7. -
    -

    -     - -

    -

      -
    1. -

      - This is a p-series with p=1. - By , this series diverges. - - This series is a famous series, - called the Harmonic Series, - so named because of its relationship to harmonics - in the study of music and sound. -

      -
      2. - -
    2. -

      - This is a p-series with p=2. - By , it converges. - Note that the theorem does not give a formula by which we can determine - what the series converges to; - we just know it converges. - A famous, unexpected result is that this series converges to \ds{\pi^2}/{6}. -

      -
      3. - -
    3. -

      - This is a p-series with p=1/2; - the theorem states that it diverges. -

      -
      4. - -
    4. -

      - This is not a p-series; - the definition does not allow for alternating signs. - Therefore we cannot apply . - (Another famous result states that this series, - the Alternating Harmonic Series, - converges to \ln(2).) -

      -
      5. - -
    5. -

      - This is a general p-series with p=3, - therefore it converges. -

      -
      6. - -
    6. -

      - This is not a p-series, - but a geometric series with r=1/2. - It converges. -

      -
      7. -
    -

    -     - -     - - - -

    - Later sections will provide tests by which we can determine whether or not a given series converges. - This, in general, is much easier than determining what - a given series converges to. - There are many cases, though, - where the sum can be determined. -

    - - - Telescoping series - -

    - Evaluate the sum \ds \infser \left(\frac1n-\frac1{n+1}\right). - - seriestelescoping - telescoping series -

    -     - -

    - It will help to write down some of the first few partial sums of this series. - - S_1 \amp = \frac11-\frac12 = 1-\frac12 - S_2 \amp = \left(\frac11-\frac12\right) + \left(\frac12-\frac13\right) = 1-\frac13 - S_3 \amp = \left(\frac11-\frac12\right) + \left(\frac12-\frac13\right)+\left(\frac13-\frac14\right) = 1-\frac14 - S_4 \amp = \left(\frac11-\frac12\right) + \left(\frac12-\frac13\right)+\left(\frac13-\frac14\right) +\left(\frac14-\frac15\right) = 1-\frac15 - -

    - -

    - Note how most of the terms in each partial sum are canceled out! - In general, we see that \ds S_n = 1-\frac{1}{n+1}. - The sequence \{S_n\} converges, - as \lim\limits_{n\to\infty}S_n = \lim_{n\to\infty}\left(1-\frac1{n+1}\right) = 1, - and so we conclude that \ds \infser \left(\frac1n-\frac1{n+1}\right) = 1. - Partial sums of the series are plotted in . -

    - -
    - Scatter plots relating to the series of - - - Scatter plots of the sequence, and corresponding partial sums, for this example. - -

    - A scatter plot for the sequence a_n = \frac1n-\frac{1}{n+1}=\frac{1}{n(n+1)} is shown. - These points begin at (1,1/2) and then descend toward the n axis. -

    - -

    - Also shown is the scatter plot for the sequence S_n of partial sums of this series. - These terms are shown to be given by S_n = 1-\frac{1}{n+1}, - and we see that the points on the scatter plot begin at (1,1/2) - and then rise toward the line y=1. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={2,4,6,8,10}, - ymin=-.1,ymax=1.1, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt},domain=1:10,samples=10] {1-1/(x+1)}; - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {1/x-1/(x+1)}; - - \end{axis} - - - - \node[shift={(0,-15pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
    -     - -     - -

    - The series in - is an example of a telescoping series. - Informally, a telescoping series is one in which most terms cancel with preceding or following terms, - reducing the number of terms in each partial sum. - The partial sum S_n did not contain n terms, - but rather just two: 1 and 1/(n+1). - seriestelescoping - telescoping series -

    - -

    - When possible, we seek a way to write an explicit formula for the - nth partial sum S_n. - This makes evaluating the limit - \lim\limits_{n\to\infty} S_n much more approachable. - We do so in the next example. -

    - - - Evaluating series - -

    - Evaluate each of the following infinite series. -

    - -

    -

      -
    1. -

      - \ds \infser \frac{2}{n^2+2n} -

      -
      2. - -
    2. -

      - \ds \infser \ln\left(\frac{n+1}{n}\right) -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - We can decompose the fraction 2/(n^2+2n) as - - \frac2{n^2+2n} = \frac1n-\frac1{n+2} - . - (See , Partial Fraction Decomposition, - to recall how this is done, if necessary.) - - - - Expressing the terms of \{S_n\} is now more instructive: - - S_1 \amp = 1-\frac13 - S_2 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right) - \amp = 1+\frac12-\frac13-\frac14 - S_3 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right) - \amp = 1+\frac12-\frac14-\frac15 - S_4 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\left(\frac14-\frac16\right) - \amp = 1+\frac12-\frac15-\frac16 - S_5 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\left(\frac14-\frac16\right)+\left(\frac15-\frac17\right) - \amp = 1+\frac12-\frac16-\frac17 - - We again have a telescoping series. - In each partial sum, - most of the terms cancel and we obtain the formula \ds S_n = 1+\frac12-\frac1{n+1}-\frac1{n+2}. - Taking limits allows us to determine the convergence of the series: - - \lim_{n\to\infty}S_n = \lim_{n\to\infty} \left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right) = \frac32 - , - so \infser \frac1{n^2+2n} = \frac32. - This is illustrated in . -

      -
      2. - -
    2. -

      - We begin by writing the first few partial sums of the series: - - S_1 \amp = \ln\left(2\right) - S_2 \amp = \ln\left(2\right)+\ln\left(\frac32\right) - S_3 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right) - S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) - - At first, this does not seem helpful, - but recall the logarithmic identity: - \ln(x) +\ln(y) = \ln(xy). - Applying this to S_4 gives: - - S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) - \amp = \ln\left(\frac21\cdot\frac32\cdot\frac43\cdot\frac54\right) = \ln\left(5\right) - . - We can conclude that \{S_n\} = \big\{\ln(n+1)\big\}. - This sequence does not converge, - as \lim\limits_{n\to\infty}S_n=\infty. - Therefore \ds\infser \ln\left(\frac{n+1}{n}\right)=\infty; - the series diverges. - Note in how the sequence of partial sums grows slowly; - after 100 terms, it is not yet over 5. - Graphically we may be fooled into thinking the series converges, - but our analysis above shows that it does not. -

      -
      3. -
    -

    - -
    - Scatter plots relating to the series in - - -
    - - - - Scatter plots of the sequence, and corresponding partial sums, for the first part of this example. - -

    - The scatter plots given appear very similar to the previous example. - Again, points plotted for the terms in the sequence a_n, - which begin at (1,2/3), descend toward the n axis, - while the points for the sequence of partial sums ascend from the same point, - in this case getting closer and closer to the line y=3/2. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={2,4,6,8,10}, - ymin=-.1,ymax=1.6, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt},domain=1:10,samples=10] {1.5-1/(x+1)-1/(x+2)}; - \addplot [only marks,firstcolor,mark size={2.4pt},domain=1:10,samples=10] {2/(x^2+2*x)}; - - \end{axis} - - - - \node[shift={(0,-17pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
    - -
    - - - - Scatter plots of the sequence, and corresponding partial sums, for the second part of this example. - -

    - Scatter plots are shown for both a_n = \ln(1+1/n) and S_n = \ln(n+1), - for n=10,20,\ldots, 100. - The points for the sequence a_n are all very close to the n axis. - The points for the sequence S_n of partial sums follow the graph y=\ln(x+1); - although this may appear to be bounded, we know that the value of the logarithm - will eventually approach infinity, even if it does so very slowly. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=5, - xmin=-10,xmax=110, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=10:100,samples=10] {ln((x+1)/x)}; - \addplot [only marks,secondcolor,mark=square*,mark size={2.4pt},domain=10:100,samples=10] {ln(x+1)}; - - \end{axis} - - - - \node[shift={(0,-15pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
    -     -
    -     - -     - -

    - We are learning about a new mathematical object, the series. - As done before, we apply old - mathematics to this new topic. -

    - - - Properties of Infinite Series - -

    - Let \ds \infser a_n = L,\ds\infser b_n = K, - and let c be a constant. -

    - -

    -

      -
    1. -

      - Constant Multiple Rule: - \ds\infser c\cdot a_n = c\cdot\infser a_n = c\cdot L. - - Constant Multiple Ruleof series -

      -
      2. - -
    2. -

      - Sum/Difference Rule: - \ds\infser \big(a_n\pm b_n\big) = \infser a_n \pm \infser b_n = L \pm K. - - seriesproperties - Sum/Difference Ruleof series -

      -
      3. -
    -

    -     -     - -

    - Before using this theorem, we provide a few famous series. -

    - - - Important Series -

    -

      -
    1. -

      - \ds\infser[0] \frac1{n!} = e. - (Note that the index starts with n=0.) -

      -
      2. - -
    2. -

      - \ds\infser \frac1{n^2} = \frac{\pi^2}{6}. -

      -
      3. - -
    3. -

      - \ds\infser \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}. -

      -
      4. - -
    4. -

      - \ds\infser[0] \frac{(-1)^{n}}{2n+1} = \frac{\pi}{4}. -

      -
      5. - -
    5. -

      - \ds\infser \frac{1}{n} diverges. - (This is called the Harmonic Series.) - - Harmonic Series -

      -
      6. - -
    6. -

      - \ds\infser \frac{(-1)^{n+1}}{n} = \ln(2). - (This is called the Alternating Harmonic Series.) - Alternating Harmonic Series -

      -
      7. -
    -

    -     - - - Evaluating series - -

    - Evaluate the given series. -

    - -

    -

      -
    1. -

      - \ds\infser \frac{(-1)^{n+1}\big(n^2-n\big)}{n^3} -

      -
      2. - -
    2. -

      - \ds\infser \frac{1000}{n!} -

      -
      3. - -
    3. -

      - \ds \frac1{16}+\frac1{25}+\frac1{36}+\frac1{49}+\cdots -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - We start by using algebra to break the series apart: - - \infser \frac{(-1)^{n+1}\big(n^2-n\big)}{n^3} \amp = \infser\left(\frac{(-1)^{n+1}n^2}{n^3}-\frac{(-1)^{n+1}n}{n^3}\right) - \amp = \infser\frac{(-1)^{n+1}}{n}-\infser\frac{(-1)^{n+1}}{n^2} - \amp = \ln(2) - \frac{\pi^2}{12} \approx -0.1293 - . - This is illustrated in . -

      -
      2. - -
    2. -

      - This looks very similar to the series that involves e in . - Note, however, - that the series given in this example starts with n=1 and not n=0. - The first term of the series in the Key Idea is 1/0! = 1, - so we will subtract this from our result below: - - \infser \frac{1000}{n!} \amp = 1000\cdot\infser \frac{1}{n!} - \amp = 1000\cdot (e-1) \approx 1718.28 - . - This is illustrated in . - The graph shows how this particular series converges very rapidly. -

      - -
      - Scatter plots relating to the series in - - -
      - - - - Scatter plots of the sequence, and corresponding partial sums, for the first part of this example. - -

      - The points in the scatter plot for the sequence a_n - appear to alternate betwee positive and negative y values, - and they converge toward the n axis from both sides. -

      - -

      - The points in the scatter plot for the sequence of partial sums S_n - also appear to oscillate, although they all have negative y coordinates. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={10,20,30,40,50}, - ymin=-.3,ymax=.3, - xmin=-1,xmax=55, - xlabel={$n$} - ] - - \addplot [only marks,firstcolor,mark size={2.4pt},domain=5:50,samples=10] {((-1)^(x+1)*(x^2-x))/(x^3)}; - \addplot [only marks,secondcolor,mark=square*,mark size={2.4pt}] coordinates {(5,-0.05528)(10,-0.1723)(15,-0.09917)(20,-0.1525)(25,-0.1105)(30,-0.1452)(35,-0.1156)(40,-0.1414)(45,-0.1186)(50,-0.139)}; - - \end{axis} - - - - \node[shift={(0,0pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
      - -
      - - - - Scatter plots of the sequence, and corresponding partial sums, for the first part of this example. - -

      - Scatter plots are shown for the sequence a_n = 1000/n!, and the corresponding partial sums. - The value of a_n approaches zero quite rapidly, - and as a result, we also see that the points in the plot for S_n quickly converge toward the limiting value. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.3,ymax=2100, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt}] coordinates {(1,1000.)(2,1500.)(3,1667.)(4,1708.)(5,1717.)(6,1718.)(7,1718.)(8,1718.)(9,1718.)(10,1718.)}; - \addplot [only marks,firstcolor,mark size={2.4pt}] coordinates {(1,1000.)(2,500.)(3,166.7)(4,41.67)(5,8.333)(6,1.389)(7,0.1984)(8,0.0248)(9,0.002756)(10,0.0002756)}; - - \end{axis} - - - - \node[shift={(0,-25pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
      -       -
      -
      3. - -
    3. -

      - The denominators in each term are perfect squares; - we are adding \ds \sum_{n=4}^\infty \frac{1}{n^2} - (note we start with n=4, not n=1). - This series will converge. - Using the formula from , - we have the following: - - \infser \frac1{n^2} \amp = \sum_{n=1}^3 \frac1{n^2} +\sum_{n=4}^\infty \frac1{n^2} - \infser \frac1{n^2} - \sum_{n=1}^3 \frac1{n^2} \amp =\sum_{n=4}^\infty \frac1{n^2} - \frac{\pi^2}{6} - \left(\frac11+\frac14+\frac19\right) \amp = \sum_{n=4}^\infty \frac1{n^2} - \frac{\pi^2}{6} - \frac{49}{36} \amp = \sum_{n=4}^\infty \frac1{n^2} - 0.2838\amp \approx \sum_{n=4}^\infty \frac1{n^2} - -

      -
      4. -
    -

    -     - -     - -

    - It may take a while before one is comfortable with this statement, - whose truth lies at the heart of the study of infinite series: - it is possible that the sum of an infinite list of nonzero numbers is finite. - We have seen this repeatedly in this section, - yet it still may take some getting used to. -

    - -

    - As one contemplates the behavior of series, a few facts become clear. -

    - -

    -

      -
    1. -

      - In order to add an infinite list of nonzero numbers and get a finite result, - most of those numbers must be very near 0. -

      -
      2. - -
    2. -

      - If a series diverges, - it means that the sum of an infinite list of numbers is not finite - (it may approach \pm \infty or it may oscillate), - and: -

      - -

      -

        -
      1. -

        - The series will still diverge if the first term is removed. -

        -
        2. - -
      2. -

        - The series will still diverge if the first 10 terms are removed. -

        -
        3. - -
      3. -

        - The series will still diverge if the first 1,000,000 terms are removed. -

        -
        4. - -
      4. -

        - The series will still diverge if any finite number of terms from anywhere in the series are removed. -

        -
        5. -
      -

      -
      3. -
    -

    - -

    - These concepts are very important and lie at the heart of the next two theorems. -

    - - - <m>n</m>th-Term Test for Divergence - -

    - Consider the series \ds\infser a_n. - If \lim\limits_{n\to\infty}a_n \neq 0, - then \ds\infser a_n diverges. - - seriesnth@nth-term test - nth@nth-term test - convergencenth@nth-term test - divergencenth@nth-term test -

    -     -     - -

    - Important! This theorem does not state - that if \ds \lim_{n\to\infty} a_n = 0 then \ds \sum_{n=1}^\infty a_n converges. - The standard example of this is the Harmonic Series, - as given in . - The Harmonic Sequence, \{1/n\}, converges to 0; - the Harmonic Series, \ds \sum_{n=1}^\infty \frac1n, - diverges. -

    - -

    - Looking back, - we can apply this theorem to the series in . - In that example, - the nth terms of both sequences do not converge to 0, therefore we can quickly conclude that each series diverges. -

    - -

    - One can rewrite - to state If a series converges, - then the underlying sequence converges to 0. - While it is important to understand the truth of this statement, - in practice it is rarely used. - It is generally far easier to prove the convergence of a sequence than the convergence of a series. -

    - - - Infinite Nature of Series - -

    - The convergence or divergence of an infinite series remains unchanged by the addition or subtraction of any finite number of terms. - That is: -

    - -

    -

      -
    1. -

      - A divergent series will remain divergent with the addition or subtraction of any finite number of terms. -

      -
      2. - -
    2. -

      - A convergent series will remain convergent with the addition or subtraction of any finite number of terms. - (Of course, the sum will likely change.) -

      -
      3. -
    -

    -     -     - -

    - Consider once more the Harmonic Series \ds\infser \frac1n which diverges; - that is, the sequence of partial sums \{S_n\} grows (very, - very slowly) without bound. - One might think that by removing the large - terms of the sequence that perhaps the series will converge. - This is simply not the case. - For instance, - the sum of the first 10 million terms of the Harmonic Series is about 16.7. - Removing the first 10 million terms from the Harmonic Series changes the nth partial sums, - effectively subtracting 16.7 from the sum. - However, a sequence that is growing without bound will still grow without bound when 16.7 is subtracted from it. -

    - -

    - The equations below illustrate this. - The first line shows the infinite sum of the Harmonic Series split into the sum of the first 10 million terms plus the sum of - everything else. - The next equation shows us subtracting these first 10 million terms from both sides. - The final equation employs a bit of - psuedo-math: - subtracting 16.7 from infinity - still leaves one with infinity. - - \infser \frac1n \amp = \sum_{n=1}^{10,000,000}\frac1n + \ds\sum_{n=10,000,001}^\infty \frac1n - \infser \frac1n - \sum_{n=1}^{10,000,000}\frac1n\amp = \ds\sum_{n=10,000,001}^\infty \frac1n - \infty - 16.7 \amp = \infty - . -

    - -

    - Just for fun, - we can show that the Harmonic Series diverges algebraically - (without the use of ). -

    - - - Divergence of the harmonic series -

    - If you just consider the partial sums - - S_1, S_2, S_3, \dots, S_{1000}, S_{1001}, \dots - , - it is not apparent that the partial sums diverge. - Indeed they do diverge, but very, very slowly. - (If you graph them on a logarithmic scale however, - you can clearly see the divergence of the partial sums.) - Instead, we will consider the partial sums, - indexed by powers of 2. - That is, we will consider S_2,S_4, S_8, S_{16}, \dots. - - S_2=1+\frac12 - S_4=1+\frac12+\frac13+\frac14 - S_8=1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac18 - -

    - -

    - Next, we consider grouping together terms in each partial sum. - We will use these groupings to set up inequalities. - - S_2=1+\frac12 - S_4=1+\frac12+\left(\frac13+\frac14\right) - S_8=1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right) - -

    - -

    - In the partial sum S_4, - we note that since 1/3\gt 1/4, we can say - - S_4=1+\frac12+\left(\frac13+\frac14\right)\gt 1+\frac12+\underbrace{\left(\frac14+\frac14\right)}_{1/2}=1+\frac22 - . - Do the same in S_8 and also note that every term in the group - \left(\frac15+\frac16+\frac17+\frac18\right) is larger than 1/8. - So - - S_8 \amp = 1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right) - \amp \gt 1+\frac12+\underbrace{\left(\frac14+\frac14\right)}_{1/2}+\underbrace{\left(\frac18+\frac18+\frac18+\frac18\right)}_{1/2}=1+\frac32 - -

    - -

    - Generally, we can see that S_{2^n} \gt 1+\frac{n}2. - (In order to really show this, - we should employ proof by induction.) - Since the sequence of partial sums clearly diverges, - so does the series \infser 1/n. -

    -     - -

    - This section introduced us to series and defined a few special types of series whose convergence properties are well known: - we know when a p-series or a geometric series converges or diverges. - Most series that we encounter are not one of these types, - but we are still interested in knowing whether or not they converge. - The next three sections introduce tests that help us determine whether or not a given series converges. -

    -     - - - - Terms and Concepts - - - -

    - Use your own words to describe how sequences and series are related. -

    -     - -

    - Answers will vary. -

    -     - -     - - - - -

    - Use your own words to define a partial sum. -

    -     - - - -     - - - - -

    - Given a series \ds \infser a_n, - describe the two sequences related to the series that are important. -

    -     - - - -

    - One sequence is the sequence of terms \{a_\}. - The other is the sequence of nth partial sums, - \{S_n\} = \{\sum_{i=1}^n a_i\}. -

    -     - -     - - - - -

    - Use your own words to explain what a geometric series is. -

    -     - - - -     - - - - -

    - If \{a_n\} is convergent, - then \ds \infser a_n is also convergent. -

    -     - - -     - - - -

    - If \{a_n\} converges to 0, then \ds \sum_{n=0}^\infty a_n converges. -

    -     - -     -     - - - Problems - - - -

    - A series \ds\infser a_n is given. -

    - -

    -

      -
    1. -

      - Give the first 5 partial sums of the series. -

      -
      2. - -
    2. -

      - Give a graph of the first 5 terms of a_n and S_n on the same axes. -

      -
      3. -
    -

    -     - - - - -

    - \ds \infser \frac{(-1)^n}{n} -

    -     - -

    -

      -
    1. -

      - -1,-\frac{1}{2},-\frac{5}{6},-\frac{7}{12},-\frac{47}{60} -

      -
      2. - -
    2. -

      - Plot omitted -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{1}{n^2} -

    -     - -

    -

      -
    1. -

      - 1,\frac{5}{4},\frac{49}{36},\frac{205}{144},\frac{526 - 9}{3600} -

      -
      2. - -
    2. -

      - Plot omitted -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \cos(\pi n) -

    -     - -

    -

      -
    1. -

      - -1,0,-1,0,-1 -

      -
      2. - -
    2. -

      - Plot omitted -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser n -

    -     - -

    -

      -
    1. -

      - 1,3,6,10,15 -

      -
      2. - -
    2. -

      - Plot omitted -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{1}{n!} -

    -     - -

    -

      -
    1. -

      - 1,\frac{3}{2},\frac{5}{3},\frac{41}{24},\frac{103}{60} -

      -
      2. - -
    2. -

      - Plot omitted -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{1}{3^n} -

    -     - -

    -

      -
    1. -

      - \frac{1}{3},\frac{4}{9},\frac{13}{27},\frac{40}{81},\frac{121}{243} -

      -
      2. - -
    2. -

      - Plot omitted -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \left(-\frac{9}{10}\right)^n -

    -     - -

    -

      -
    1. -

      - -0.9,-0.09,-0.819,-0.1629,-0.75339 -

      -
      2. - -
    2. -

      - Plot omitted -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \left(\frac{1}{10}\right)^n -

    -     - -

    -

      -
    1. -

      - 0.1,0.11,0.111,0.1111,0.11111 -

      -
      2. - -
    2. -

      - Plot omitted -

      -
      3. -
    -

    -     - -     - -     - - - -

    - Use - to show the given series diverges. -

    -     - - - - -

    - \ds \infser \frac{3n^2}{n(n+2)} -

    -     - -

    - \lim\limits_{n\to\infty}a_n = 3; - by the series diverges. -

    -     - -     - - - - -

    - \ds \infser \frac{2^n}{n^2} -

    -     - -

    - \lim\limits_{n\to\infty}a_n = \infty; - by the series diverges. -

    -     - -     - - - - -

    - \ds \infser \frac{n!}{10^n} -

    -     - -

    - \lim\limits_{n\to\infty}a_n = \infty; - by the series diverges. -

    -     - -     - - - - -

    - \ds \infser \frac{5^n-n^5}{5^n+n^5} -

    -     - -

    - \lim\limits_{n\to\infty}a_n = 1; - by the series diverges. -

    -     - -     - - - - -

    - \ds \infser \frac{2^n+1}{2^{n+1}} -

    -     - -

    - \lim\limits_{n\to\infty}a_n = 1/2; - by the series diverges. -

    -     - -     - - - - -

    - \ds \infser \left(1+\frac1n\right)^n -

    -     - -

    - \lim\limits_{n\to\infty}a_n = e; - by the series diverges. -

    -     - -     - -     - - - -

    - State whether the given series converges or diverges. -

    -     - - - - -

    - \ds \infser \frac{1}{n^5} -

    -     - -

    - Converges; p-series with p=5. -

    -     - -     - - - - -

    - \ds \infser[0] \frac{1}{5^n} -

    -     - -

    - Converges; geometric series with r=1/5. -

    -     - -     - - - - -

    - \ds \infser[0] \frac{6^n}{5^n} -

    -     - -

    - Diverges; geometric series with r=6/5. -

    -     - -     - - - - -

    - \ds \infser n^{-4} -

    -     - -

    - Converges; p-series with p=4. -

    -     - -     - - - - -

    - \ds \infser \sqrt{n} -

    -     - -

    - Diverges; fails nth term test. -

    -     - -     - - - - -

    - \ds \infser \frac{10}{n!} -

    -     - -

    - Converges; by - and , - series converges to 10e. -

    -     - -     - - - - -

    - \ds \infser \left(\frac{1}{n!}+\frac1n\right) -

    -     - -

    - Diverges; although the series \ds\infser \frac{1}{n!} converges, \ds \infser \frac1n is the (divergent) harmonic series. We can only use the sum rule for series if both parts converge separately. -

    -     - -     - - - - -

    - \ds \infser \frac{2}{(2n+8)^2} -

    -     - -

    - Converges; general p-series with p=2. -

    -     - -     - - - - -

    - \ds \infser \frac{1}{2n} -

    -     - -

    - Diverges; by - this is half the Harmonic Series, - which diverges by growing without bound. - Half of growing without bound - is still growing without bound. -

    -     - -     - - - - -

    - \ds \infser \frac{1}{2n-1} -

    -     - -

    - Diverges; general p-series with p=1. -

    -     - -     - -     - - - -

    - A series is given. -

    - -

    -

      -
    1. -

      - Find a formula for S_n, - the nth partial sum of the series. -

      -
      2. - -
    2. -

      - Determine whether the series converges or diverges. - If it converges, state what it converges to. -

      -
      3. -
    -

    -     - - - - -

    - \ds \infser[0] \frac{1}{4^n} -

    -     - -

    -

      -
    1. -

      - S_n = \frac{1-(1/4)^{n+1}}{3/4} -

      -
      2. - -
    2. -

      - Converges to 4/3. -

      -
      3. -
    -

    -     - -     - - - -

    - \ds \sum_{n=1}^\infty 2 -

    -     - -

    -

      -
    1. -

      - S_n = 2n -

      -
      2. - -
    2. -

      - Diverges. -

      -
      3. -
    -

    -     -     - - - - - -

    - \ds 1^3+2^3+3^3+4^3+\cdots -

    -     - -

    -

      -
    1. -

      - S_n = \left(\frac{n(n+1)}{2}\right)^2 -

      -
      2. - -
    2. -

      - Diverges -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser (-1)^n n -

    -     - -

    -

      -
    1. -

      - - S_n = \left\{\begin{array}{cc} -\frac{n+1}{2} \amp n \text{ is odd } \\ - \frac{n}{2} \amp n \text{ is even } - \end{array} \right. - -

      -
      2. - -
    2. -

      - Diverges -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser[0] \frac{5}{2^n} -

    -     - -

    -

      -
    1. -

      - S_n = 5\frac{1-1/2^(n+1)}{1/2} -

      -
      2. - -
    2. -

      - Converges to 10. -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser[0] e^{-n} -

    -     - -

    -

      -
    1. -

      - S_n = \frac{1-(1/e)^{n+1}}{1-1/e}. -

      -
      2. - -
    2. -

      - Converges to 1/(1-1/e) = e/(e-1). -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds 1-\frac13+\frac19-\frac{1}{27}+\frac{1}{81}+\cdots -

    -     - -

    -

      -
    1. -

      - S_n = \frac{1-(-1/3)^(n+1)}{4/3} -

      -
      2. - -
    2. -

      - Converges to 3/4. -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{1}{n(n+1)} -

    -     - -

    -

      -
    1. -

      - With partial fractions, a_n = \frac1n-\frac1{n+1}. - Thus S_n = 1-\frac1{n+1}. -

      -
      2. - -
    2. -

      - Converges to 1. -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{3}{n(n+2)} -

    -     - -

    -

      -
    1. -

      - With partial fractions, - a_n = \frac32\left(\frac1n-\frac1{n+2}\right). - Thus S_n = \frac32\left(\frac32-\frac1{n+1}-\frac1{n+2}\right). -

      -
      2. - -
    2. -

      - Converges to 9/4 -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{1}{(2n-1)(2n+1)} -

    -     - -

    -

      -
    1. -

      - With partial fractions, - a_n = \frac12\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right). - Then S_n = \frac12\left(1-\frac1{2n+1}\right) = \frac{n}{2n+1}. -

      -
      2. - -
    2. -

      - Converges to 1/2. -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \ln\left(\frac{n}{n+1}\right) -

    -     - -

    -

      -
    1. -

      - S_n = \ln\big(1/(n+1)\big) -

      -
      2. - -
    2. -

      - Diverges - (to -\infty). -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{2n+1}{n^2(n+1)^2} -

    -     - -

    -

      -
    1. -

      - S_n = 1-\frac{1}{(n+1)^2} -

      -
      2. - -
    2. -

      - Converges to 1. -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \frac{1}{1\cdot 4}+\frac1{2\cdot5}+\frac1{3\cdot6}+\frac1{4\cdot7}+\cdots -

    -     - -

    -

      -
    1. -

      - a_n = \frac1{n(n+3)}; - using partial fractions, - the resulting telescoping sum reduces to S_n = \frac13\left(1+\frac12+\frac13-\frac1{n+1}-\frac1{n+2}-\frac1{n+3}\right) -

      -
      2. - -
    2. -

      - Converges to 11/18. -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds 2+\left(\frac12+\frac13\right) + \left(\frac14+\frac19\right) + \left(\frac18+\frac1{27}\right)+\cdots -

    -     - -

    -

      -
    1. -

      - a_n = 1/2^n+1/3^n for n\geq 0. - Thus S_n = \frac{1-1/2^2}{1/2}+\frac{1-1/3^n}{2/3}. -

      -
      2. - -
    2. -

      - Converges to 2+3/2 = 7/2. -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \sum_{n=2}^\infty \frac{1}{n^2-1} -

    -     - -

    -

      -
    1. -

      - With partial fractions, - a_n = \frac12\left(\frac1{n-1}-\frac1{n+1}\right). - Thus S_n = \frac12\left(3/2-\frac1n-\frac{1}{n+1}\right). -

      -
      2. - -
    2. -

      - Converges to 3/4. -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \infser[0] \big(\sin(1) \big)^n -

    -     - -

    -

      -
    1. -

      - S_n=\frac{1-(\sin(1) )^{n+1}}{1-\sin(1) } -

      -
      2. - -
    2. -

      - Converges to \frac{1}{1-\sin(1) }.. -

      -
      3. -
    -

    -     - -     -     - - - - -

    - Break the Harmonic Series into the sum of the odd and even terms: - - \infser \frac1n = \infser \frac{1}{2n-1}+\infser \frac{1}{2n} - . -

    - -

    - The goal is to show that each of the series on the right diverge. -

    -     - - - -

    - Show why \ds \infser \frac{1}{2n-1} \gt \infser \frac{1}{2n}. -

    - -

    - (Compare each nth partial sum.) -

    -     - -

    - The nth partial sum of the odd series is 1+\frac13+\frac15+\cdots+\frac{1}{2n-1}. - The nth partial sum of the even series is \frac12+\frac14 + \frac16 + \cdots +\frac1{2n}. - Each term of the even series is less than the corresponding term of the odd series, - giving us our result. -

    -     -     - - - -

    - Show why \ds\infser \frac{1}{2n-1}\lt 1+\infser \frac{1}{2n} -

    -     - -

    - The nth partial sum of the odd series is 1+\frac13+\frac15+\cdots+\frac1{2n-1}. - The nth partial sum of 1 plus the even series is 1+\frac12+\frac14+\cdots + \frac{1}{2(n-1)}. - Each term of the even series is now greater than or equal to the corresponding term of the odd series, - with equality only on the first term. - This gives us the result. -

    -     -     - - - -

    - Explain why (a) and (b) demonstrate that the series of odd terms is convergent, if, - and only if, the series of even terms is also convergent. - (That is, show both converge or both diverge.) -

    -     - -

    - If the odd series converges, - the work done in (a) shows the even series converges also. - (The sequence of the nth partial sum of the even series is bounded and monotonically increasing.) - Likewise, (b) shows that if the even series converges, - the odd series will, too. - Thus if either series converges, the other does. -

    - -

    - Similarly, (a) and (b) can be used to show that if either series diverges, - the other does, too. -

    -     -     - - - -

    - Explain why knowing the Harmonic Series is divergent determines that the even and odd series are also divergent. -

    -     - -

    - If both the even and odd series converge, - then their sum would be a convergent series. - This would imply that the Harmonic Series, their sum, is convergent. - It is not. - Hence each series diverges. -

    -     -     - -     - - - - -

    - Show the series \ds \infser \frac{n}{(2n-1)(2n+1)} diverges. -

    -     - -

    - Using partial fractions, - we can show that a_n = \frac14\left(\frac1{2n-1}+\frac{1}{2n+1}\right). - The series is effectively twice the sum of the odd terms of the Harmonic Series which was shown to diverge in . - Thus this series diverges. -

    -     - -     -     -     -
    -
    - Integral and Comparison Tests - -

    - Knowing whether or not a series converges is very important, - especially when we discuss Power Series in . - Theorems - and - give criteria for when Geometric and p-series converge, - and - gives a quick test to determine if a series diverges. - There are many important series whose convergence cannot be determined by these theorems, though, - so we introduce a set of tests that allow us to handle a broad range of series. - We start with the Integral Test. -

    -     - - - Integral Test -

    - We stated in - that a sequence \{a_n\} is a function a(n) whose domain is \mathN, - the set of natural numbers. - If we can extend a(n) to \mathbb{R}, - the real numbers, - and it is both positive and decreasing on [1,\infty), - then the convergence of \ds \infser a_n is the same as \ds\int_1^\infty a(x)\, dx. -

    - - - Integral Test - -

    - Let a sequence \{a_n\} be defined by a_n=a(n), - where a(n) is continuous, - positive and decreasing on [1,\infty). - Then \ds \infser a_n converges, if, - and only if, \ds\int_1^\infty a(x)\, dx converges. - seriesIntegral Test - Integral Test - convergenceIntegral Test - divergenceIntegral Test -

    -     -     - - - - -

    - We can demonstrate the truth of the Integral Test with two simple graphs. - In , - the height of each rectangle is - a(n)=a_n for n=1,2,\ldots, - and clearly the rectangles enclose more area than the area under y=a(x). - Therefore we can conclude that - - \ds \int_1^\infty a(x)\, dx \lt \infser a_n - . -

    - -
    - Illustrating the truth of the Integral Test - - -
    - - - - First of two graphs illustrating why the integral test works. - -

    - The graph of a positive, decreasing function y=a(x) is shown. - Also shown are the first four rectangles in a Riemann sum with partition points at each integer, - and using left endpoints. Since a(x) is decreasing, this is an overestimate, - which is illustrated by the fact that the rectangles all reach above the graph. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,2,3,4,5}, - ytick={1,2}, - ymin=-.1,ymax=2.5, - xmin=-.5,xmax=5.6 - ] - - \addplot+ [domain=-.5:5.6,samples=50] ({x},{2/(x+1)}); - - \draw [thick,secondcolor] (axis cs: 1,0) -- (axis cs:1,1) -- (axis cs: 2,1) -- (axis cs: 2,0) - (axis cs: 2,0) -- (axis cs:2,.667) -- (axis cs: 3,.667) -- (axis cs: 3,0) - (axis cs: 3,0) -- (axis cs:3,.5) -- (axis cs: 4,.5) -- (axis cs: 4,0) - (axis cs: 4,0) -- (axis cs:4,.4) -- (axis cs: 5,.4) -- (axis cs: 5,0); - - \draw (axis cs: .95,1.75) node { $y=a(x)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - Second of two graphs illustrating why the integral test works. - -

    - The graph of a positive, decreasing function y=a(x) is shown. - Also shown are the first four rectangles in a Riemann sum with partition points at each integer, - and using right endpoints. Since a(x) is decreasing, this is an underestimate, - which is illustrated by the fact that the rectangles all lie below the graph. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,2,3,4,5}, - ytick={1,2}, - ymin=-.1,ymax=2.5, - xmin=-.5,xmax=5.6 - ] - - \addplot+ [domain=-.5:5.6,samples=50] ({x},{2/(x+1)}); - - \draw [thick,secondcolor] (axis cs: 1,0) -- (axis cs:1,.667) -- (axis cs: 2,.667) -- (axis cs: 2,0) - (axis cs: 2,0) -- (axis cs:2,.5) -- (axis cs: 3,.5) -- (axis cs: 3,0) - (axis cs: 3,0) -- (axis cs:3,.4) -- (axis cs: 4,.4) -- (axis cs: 4,0) - (axis cs: 4,0) -- (axis cs:4,.333) -- (axis cs: 5,.333) -- (axis cs: 5,0); - - \draw (axis cs: .95,1.75) node { $y=a(x)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - In , - we draw rectangles under y=a(x) with the Right-Hand rule, - starting with n=2. - This time, the area of the rectangles is less than the area under y=a(x), - so \ds\sum_{n=2}^\infty a_n \lt \int_1^\infty a(x)\, dx. - Note how this summation starts with n=2; - adding a_1 to both sides lets us rewrite the summation starting with n=1: - - \infser a_n \lt a_1 +\int_1^\infty a(x)\, dx - . -

    - -

    - Combining Equations and , we have - - \infser a_n\lt a_1 +\int_1^\infty a(x)\, dx \lt a_1 + \infser a_n - . -

    - -

    - From Equation we can make the following two statements: -

    - -

    -

      -
    1. -

      - If \ds \infser a_n diverges, - so does \ds\int_1^\infty a(x)\, dx (because \ds \infser a_n \lt a_1 +\int_1^\infty a(x)\, dx) -

      -
      2. - -
    2. -

      - If \ds \infser a_n converges, - so does \ds\int_1^\infty a(x)\, dx (because \ds \ds \int_1^\infty a(x)\, dx \lt \infser a_n.) -

      -
      3. -
    -

    - -

    - Therefore the series and integral either both converge or both diverge. - - allows us to extend this theorem to series where a(n) is positive and decreasing on [b,\infty) for some b \gt 1. - A formal proof of the is shown below. -

    - - - Proof of the Integral Test -

    - Let a(x)=a_x be a postive, - continuous, decreasing function on [1,\infty). - We will consider how the partial sums of - \infser a_n compare to the integral \int_0^\infty a(x)\, dx . - We first consider the case where \int_1^{\infty}a(x)\, dx diverges. -

    - -

    -

      -
    1. -

      - Suppose that \int_1^{\infty}a(x)\, dx diverges. - Using , - we can say that S_n=\sum_{i=1}^{n}a_i\gt \int_1^{n+1}a(x)\, dx. - If we let n \to \infty in this inequality, - we know that \int_1^{n+1}a(x)\, dx will get arbitrarily large as n \to \infty (since - a(x) \gt 0 and \int_1^{\infty}a(x)\, dx diverges). - Therefore we conclude that - S_n=\sum_{i=1}^{n}a_i will also get arbitrarily large as n \to \infty, - and thus \infser a_n diverges. -

      -
      2. - -
    2. -

      - Now suppose that \int_1^{\infty}a(x)\, dx converges to M, - where M is some positive, finite number. - Using , - we can say that 0 \lt S_n=\sum_{i=1}^{n}a_i \lt \int_1^{\infty} a(x)\, dx=M. - Therefore our sequence of partial sums, S_n is bounded. - Furthermore, - S_n is a monotonically increasing sequence since all of the terms a_n are positive. - Since S_n is both bounded and monotonic, - S_n converges by and by , - the series \infser a_n converges as well. -

      -
      3. -
    -

    -     - - - Using the Integral Test - -

    - Determine the convergence of - \ds\infser \frac{\ln(n) }{n^2}. (The terms of the sequence - \{a_n\} = \{\ln(n) /n^2\} and the nth partial sums are given in .) -

    -     - -

    - - implies that a(n) = (\ln(n) )/n^2 is positive and decreasing on [2,\infty). - We can determine this analytically, too. - We know a(n) is positive as both \ln(n) and n^2 are positive on [2,\infty). - Treating a(n) as a continuous function of n defined on [1, \infty), - consider a'(n) = (1-2\ln(n) )/n^3, - which is negative for n\geq 2. - Since a'(n) is negative, - a(n) is decreasing for n\geq 2. - We can still use the integral test since a finite number of terms will not affect convergence of the series. -

    - -
    - Plotting the sequence and series in - - - Scatter plots of the sequence used in this example, and the corresponding sequence of partial sums. - -

    - A scatter plot of the sequence a_n=\ln(n)/n^2 is shown. - Aside from an initial point (1,0), this is a decreasing sequence, - descending from the point (2,\ln(2)/4) toward the n axis. -

    - -

    - The scatter plot for the sequence S_n of partial sums is also shown; - this is an increasing sequence, which appears to be bounded above by some value less than 1. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={2,4,...,20}, - ymin=-.05,ymax=.85, - xmin=-1,xmax=20.9, - xlabel={$n$} - ] - - \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt}] coordinates {(1.,0)(2.,0.1733)(3.,0.2954)(4.,0.382)(5.,0.4464)(6.,0.4961)(7.,0.5359)(8.,0.5684)(9.,0.5955)(10.,0.6185)(11.,0.6383)(12.,0.6556)(13.,0.6708)(14.,0.6842)(15.,0.6963)(16.,0.7071)(17.,0.7169)(18.,0.7258)(19.,0.734)(20.,0.7415)}; - \addplot [only marks,firstcolor,mark size={2.4pt}] coordinates {(1.,0)(2.,0.1733)(3.,0.1221)(4.,0.08664)(5.,0.06438)(6.,0.04977)(7.,0.03971)(8.,0.03249)(9.,0.02713)(10.,0.02303)(11.,0.01982)(12.,0.01726)(13.,0.01518)(14.,0.01346)(15.,0.01204)(16.,0.01083)(17.,0.009804)(18.,0.008921)(19.,0.008156)(20.,0.007489)}; - - \end{axis} - - \node[shift={(0,-20pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
    - -

    - Applying the Integral Test, - we test the convergence of \ds \int_1^\infty \frac{\ln(x) }{x^2}\, dx. - Integrating this improper integral requires the use of Integration by Parts, - with u = \ln(x) and dv = 1/x^2\, dx. - - \int_1^\infty \frac{\ln(x) }{x^2}\, dx \amp = \lim_{b\to\infty} \int_1^b \frac{\ln(x) }{x^2}\, dx - \amp = \lim_{b\to\infty} -\frac1x\ln(x) \Big|_1^b + \int_1^b\frac1{x^2}\, dx - \amp = \lim_{b\to\infty} -\frac1x\ln(x) -\frac 1x\Big|_1^b - \amp = \lim_{b\to\infty}1-\frac1b-\frac{\ln(b) }{b}. \text{ Apply L'Hospital's Rule: } - \amp = 1 - . -

    - -

    - Since \ds \int_1^\infty \frac{\ln(x) }{x^2}\, dx converges, - so does \ds \infser \frac{\ln(n) }{n^2}. -

    -     - -     - - - -

    - - was given without justification, - stating that the general p-series - \ds \infser \frac 1{(an+b)^p} converges if, and only if, - p \gt 1. - In the following example, - we prove this to be true by applying the Integral Test. -

    - - - Using the Integral Test to establish <xref ref="thm_pseries"/> - -

    - Let a, b be real numbers such that a\neq 0 and an+b\gt 0 for all n\geq 1. - Use the Integral Test to prove that - \ds \infser \frac1{(an+b)^p} converges if, and only if, - p \gt 1. -

    -     - -

    - Consider the integral \ds\int_1^\infty \frac1{(ax+b)^p}\, dx; - assuming p\neq 1 and a\neq 0, - - \int_1^\infty \frac1{(ax+b)^p}\, dx \amp = \lim_{c\to\infty} \int_1^c \frac1{(ax+b)^p}\, dx - \amp = \lim_{c\to\infty} \frac{1}{a(1-p)}(ax+b)^{1-p}\Big|_1^c - \amp = \lim_{c\to\infty} \frac{1}{a(1-p)}\big((ac+b)^{1-p}-(a+b)^{1-p}\big) - . -

    - -

    - This limit converges if, - and only if, p \gt 1 so that 1-p \lt 0. - It is easy to show that the integral also diverges in the case of p=1. - (This result is similar to the work preceding .) -

    - -

    - Therefore \ds \infser \frac 1{(an+b)^p} converges if, - and only if, - p \gt 1. -

    -     - -     - -

    - We consider two more convergence tests in this section, - both comparison tests. - That is, we determine the convergence of one series by comparing it to another series with known convergence. -

    -     - - - Direct Comparison Test - - Direct Comparison Test - -

    - Let \{a_n\} and \{b_n\} be positive sequences where - a_n\leq b_n for all n\geq N, for some N\geq 1. - seriesDirect Comparison Test - Direct Comparison Testfor series - convergenceDirect Comparison Test - divergenceDirect Comparison Test -

    - -

    -

      -
    1. -

      - If \ds \infser b_n converges, - then \ds \infser a_n converges. -

      -
      2. - -
    2. -

      - If \ds \infser a_n diverges, - then \ds \infser b_n diverges. -

      -
      3. -
    -

    -     - -

    - Let 0\lt a_n\leq b_n for all n\geq N \geq 1. - Note that both partial sums for both series are positive and increasing since the terms of both sequences are positive. -

    - -

    -

      -
    1. -

      - Suppose that \ds \infser b_n converges, - so \ds \infser b_n=S, where S is a finite, - positive number. - (S must be positive since b_n \gt 0.) -

      - -

      - Comparing the partial sums, - we must have \ds \sum_{i=N}^n a_i \leq \sum_{i=N}^n b_i since - a_n\leq b_n for all n\geq N. - Furthermore since \ds \infser b_n converges to S, - our partial sums for a_n are bounded - (note that the partial sums started at i=N, - but a finite number of terms will not affect the boundedness of the partial sums). - - 0\lt \sum_{i=N}^n a_i \leq \sum_{i=N}^n b_i \lt S - . - Since the sequence of partial sums, - s_n=\sum_{i=1}^n a_i is both monotonically increasing and bounded, - we can say that s_n converges - (by ), - and therefore so does \infser a_n . -

      -
      2. - -
    2. -

      - Suppose that \ds \infser a_n diverges, - so \ds \sum_{i=1}^n a_n=\infty. (We can say that the series diverges to \infty since the terms of the series are always positive). - Comparing the partial sums, we have - - \ds \sum_{i=N}^n a_i \leq \sum_{i=N}^n b_i - - Then applying limits, we get - - \ds \lim\limits_{n \to \infty}\sum_{i=N}^n a_i \leq \lim\limits_{n \to \infty}\sum_{i=N}^n b_i - . - Since the limit on the left side diverges to \infty, - we can say that \lim\limits_{n \to \infty}\sum_{i=N}^n b_i also diverges to \infty. -

      -
      3. -
    -

    -     - -     - - - - - - - Applying the Direct Comparison Test - -

    - Determine the convergence of \ds\infser \frac1{3^n+n^2}. -

    -     - -

    - This series is neither a geometric or p-series, - but seems related. - We predict it will converge, - so we look for a series with larger terms that converges. - (Note too that the Integral Test seems difficult to apply here.) -

    - -

    - Since 3^n \lt 3^n+n^2, - \ds \frac1{3^n} \gt \frac1{3^n+n^2} for all n\geq1. - The series \ds\infser \frac{1}{3^n} is a convergent geometric series; - by , - \ds \infser \frac1{3^n+n^2} converges. -

    -     - -     - - - Applying the Direct Comparison Test - -

    - Determine the convergence of \ds\infser \frac{1}{n-\ln(n) }. -

    -     - -

    - We know the Harmonic Series \ds\infser \frac1n diverges, - and it seems that the given series is closely related to it, - hence we predict it will diverge. -

    - -

    - Since n\geq n-\ln(n) for all n\geq 1, - \ds \frac1n \leq \frac1{n-\ln(n) } for all n\geq 1. -

    - -

    - The Harmonic Series diverges, - so we conclude that \ds\infser \frac{1}{n-\ln(n) } diverges as well. -

    -     - -     - - - -

    - The concept of direct comparison is powerful and often relatively easy to apply. - Practice helps one develop the necessary intuition to quickly pick a proper series with which to compare. - However, it is easy to construct a series for which it is difficult to apply the Direct Comparison Test. -

    - -

    - Consider \ds\infser \frac1{n+\ln(n) }. - It is very similar to the divergent series given in . - We suspect that it also diverges, - as \ds \frac 1n \approx \frac1{n+\ln(n) } for large n. - However, the inequality that we naturally want to use - goes the wrong way: - since n\leq n+\ln(n) for all n\geq 1, - \ds\frac1n \geq \frac{1}{n+\ln(n) } for all n\geq 1. - The given series has terms less than - the terms of a divergent series, - and we cannot conclude anything from this. -

    - -

    - Fortunately, - we can apply another test to the given series to determine its convergence. -

    - - -     - - - Limit Comparison Test - - Limit Comparison Test - -

    - Let \{a_n\} and \{b_n\} be positive sequences. - seriesLimit Comparison Test - Limit Comparison Testfor series - convergenceLimit Comparison Test - divergenceLimit Comparison Test -

    - -

    -

      -
    1. -

      - If \lim\limits_{n\to\infty} \dfrac{a_n}{b_n} = L, - where L is a positive real number, - then \ds \infser a_n and - \ds \infser b_n either both converge or both diverge. -

      -
      2. - -
    2. -

      - If \lim\limits_{n\to\infty} \dfrac{a_n}{b_n} = 0, - then if \ds \infser b_n converges, - then so does \ds \infser a_n. -

      -
      3. - -
    3. -

      - If \lim\limits_{n\to\infty} \dfrac{a_n}{b_n} = \infty, - then if \ds \infser b_n diverges, - then so does \ds \infser a_n. -

      -
      4. -
    -

    -     -     - -

    - - is most useful when the convergence of the series from \{b_n\} is known and we are trying to determine the convergence of the series from \{a_n\}. -

    - - - -

    - We use the Limit Comparison Test in the next example to examine the series - \ds\infser \frac1{n+\ln(n) } which motivated this new test. -

    - - - Applying the Limit Comparison Test - -

    - Determine the convergence of - \ds\infser \frac1{n+\ln(n) } using the Limit Comparison Test. -

    -     - -

    - We compare the terms of \ds\infser \frac1{n+\ln(n) } to the terms of the Harmonic Sequence \ds\infser \frac1{n}: - - \lim_{n\to\infty}\frac{1/(n+\ln(n) )}{1/n} \amp = \lim_{n\to\infty} \frac{n}{n+\ln(n) } - \amp = 1 \text{ (after applying L'Hospital's Rule) } - . -

    - -

    - Since the Harmonic Series diverges, - we conclude that \ds\infser \frac1{n+\ln(n) } diverges as well. -

    -     - -     - - - Applying the Limit Comparison Test - -

    - Determine the convergence of \ds\infser \frac1{3^n-n^2} -

    -     - -

    - This series is similar to the one in , - but now we are considering 3^n-n^2 - instead of 3^n+n^2. - This difference makes applying the Direct Comparison Test difficult. -

    - -

    - Instead, we use the Limit Comparison Test and compare with the series \ds\infser \frac1{3^n}: - - \lim_{n\to\infty}\frac{1/(3^n-n^2)}{1/3^n} \amp = \lim_{n\to\infty}\frac{3^n}{3^n-n^2} - \amp = 1 \text{ (after applying L'Hospital's Rule twice) } - . -

    - -

    - We know \ds\infser \frac1{3^n} is a convergent geometric series, - hence \ds\infser \frac1{3^n-n^2} converges as well. -

    -     - -     - -

    - As mentioned before, - practice helps one develop the intuition to quickly choose a series with which to compare. - A general rule of thumb is to pick a series based on the dominant term in the expression of \{a_n\}. - It is also helpful to note that factorials dominate exponentials, - which dominate algebraic functions (e.g., polynomials), - which dominate logarithms. - In the previous example, - the dominant term of \ds\frac{1}{3^n-n^2} was 3^n, - so we compared the series to \ds \infser \frac1{3^n}. - It is hard to apply the Limit Comparison Test to series containing factorials, though, - as we have not learned how to apply L'Hospital's Rule to n!. -

    - - - - - Applying the Limit Comparison Test - -

    - Determine the convergence of \ds\infser \frac{\sqrt{n}+3}{n^2-n+1}. -

    -     - -

    - We naïvely attempt to apply the rule of thumb given above and note that the dominant term in the expression of the series is 1/n^2. - Knowing that \ds \infser \frac1{n^2} converges, - we attempt to apply the Limit Comparison Test: - - \lim_{n\to\infty}\frac{(\sqrt{n}+3)/(n^2-n+1)}{1/n^2} \amp = \lim_{n\to\infty}\frac{n^2(\sqrt n+3)}{n^2-n+1} - \amp = \infty \text{ (Apply L'Hospital's Rule) } - . -

    - -

    - - part (3) only applies when \ds\infser b_n diverges; - in our case, it converges. - Ultimately, our test has not revealed anything about the convergence of our series. -

    - -

    - The problem is that we chose a poor series with which to compare. - Since the numerator and denominator of the terms of the series are both algebraic functions, - we should have compared our series to the dominant term of the numerator divided by the dominant term of the denominator. -

    - -

    - The dominant term of the numerator is n^{1/2} and the dominant term of the denominator is n^2. - Thus we should compare the terms of the given series to n^{1/2}/n^2 = 1/n^{3/2}: - - \lim_{n\to\infty}\frac{(\sqrt{n}+3)/(n^2-n+1)}{1/n^{3/2}} \amp = \lim_{n\to \infty} \frac{n^{3/2}(\sqrt n+3)}{n^2-n+1} - \amp = 1 \text{ (Apply L'Hospital's Rule) } - . -

    - -

    - Since the p-series \ds\infser \frac1{n^{3/2}} converges, - we conclude that \ds\infser \frac{\sqrt{n}+3}{n^2-n+1} converges as well. -

    -     - -     - -

    - We mentioned earlier that the Integral Test did not work well with series containing factorial terms. - The next section introduces the Ratio Test, - which does handle such series well. - We also introduce the Root Test, - which is good for series where each term is raised to a power. -

    -     - - - - Terms and Concepts - - - -

    - In order to apply the Integral Test to a sequence \{a_n\}, - the function a(n) = a_n must be - , , - and . -

    -     - - - - - - ["continuous","positive","decreasing"].includes(ans) - - - - - - - ["continuous","positive","decreasing"].includes(ans) && !ans_array.slice(0,1).includes(ans) - - - - - ans_array.slice(0,1).includes(ans) - - You already gave that answer. - - - - - - ["continuous","positive","decreasing"].includes(ans) && !ans_array.slice(0,2).includes(ans) - - - - - ans_array.slice(0,2).includes(ans) - - You already gave that answer. - - - - -     - - - - -

    - The Integral Test can be used to determine the sum of a convergent series. -

    -     - -     - - - - -

    - What test(s) in this section do not work well with factorials? -

    -     - - - -

    - The Integral Test - (we do not have a continuous definition of n! yet) - and the Limit Comparison Test - (same as above, hence we cannot take its derivative). -

    -     - -     - - - - -

    - Suppose \ds \infser[0] a_n is convergent, - and there are sequences \{b_n\} and \{c_n\} such that b_n \leq a_n \leq c_n for all n. - What can be said about the series - \ds \infser[0] b_n and \ds \infser[0] c_n? -

    -     - - - -

    - \ds \infser[0] b_n converges; - we cannot conclude anything about \ds \infser[0] c_n -

    -     - -     -     - - - Problems - - - -

    - Use the Integral Test to determine the convergence of the given series. -

    -     - - - - -

    - \ds \infser \frac{1}{2^n} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds \infser \frac{1}{n^4} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds \infser \frac{n}{n^2+1} -

    -     - -

    - Diverges -

    -     - -     - - - - -

    - \ds \sum_{n=2}^\infty \frac{1}{n\ln(n) } -

    -     - -

    - Diverges -

    -     - -     - - - - -

    - \ds \infser \frac{1}{n^2+1} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds \sum_{n=2}^\infty \frac{1}{n(\ln(n) )^2} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds \infser \frac{n}{2^n} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds \infser \frac{\ln(n) }{n^3} -

    -     - -

    - Converges -

    -     - -     - -     - - - -

    - Use the Direct Comparison Test to determine the convergence of the given series; - state what series is used for comparison. -

    -     - - - - -

    - \ds \infser \frac{1}{n^2+3n-5} -

    -     - -

    - Converges; compare to \ds \infser \frac{1}{n^2}, - as 1/(n^2+3n-5) \leq 1/n^2 for all n \gt 1. -

    -     - -     - - - - -

    - \ds \infser \frac{1}{4^n+n^2-n} -

    -     - -

    - Converges; compare to \ds \infser \frac{1}{4^n}, - as 1/(4^n+n^2-n) \leq 1/4^n for all n\geq 1. -

    -     - -     - - - - -

    - \ds \infser \frac{\ln(n) }{n} -

    -     - -

    - Diverges; compare to \ds \infser \frac{1}{n}, - as 1/n \leq \ln(n) /n for all n\geq 3. -

    -     - -     - - - - -

    - \ds \infser \frac{1}{n!+n} -

    -     - -

    - Converges; compare to \ds \infser \frac{1}{n!}, - as 1/(n!+n) \leq 1/n! for all n\geq 1. -

    -     - -     - - - - -

    - \ds \sum_{n=2}^\infty \frac{1}{\sqrt{n^2-1}} -

    -     - -

    - Diverges; compare to \ds \infser \frac{1}{n}. - Since n=\sqrt{n^2} \gt \sqrt{n^2-1}, - 1/n \leq 1/\sqrt{n^2-1} for all n\geq 2. -

    -     - -     - - - - -

    - \ds \sum_{n=5}^\infty \frac{1}{\sqrt{n}-2} -

    -     - -

    - Diverges; compare to \ds \infser \frac{1}{\sqrt n}, - as 1/\sqrt n \leq 1/(\sqrt{n}-2) for all n\geq 5. -

    -     - -     - - - - -

    - \ds \infser \frac{n^2+n+1}{n^3-5} -

    -     - -

    - Diverges; compare to \ds \infser \frac{1}{n}: - - \frac 1n = \frac{n^2}{n^3} \lt \frac{n^2+n+1}{n^3} \lt \frac{n^2+n+1}{n^3-5} - , - for all n\geq 1. -

    -     - -     - - - - -

    - \ds \infser \frac{2^n}{5^n+10} -

    -     - -

    - Converges; compare to \ds \infser \left(\frac{2}{5}\right)^n, - as 2^n/(5^n+10) \lt 2^n/5^n for all n\geq 1. -

    -     - -     - - - - -

    - \ds \sum_{n=2}^\infty \frac{n}{n^2-1} -

    -     - -

    - Diverges; compare to \ds \infser \frac 1n. - Note that - - \frac{n}{n^2-1} = \frac{n^2}{n^2-1}\cdot\frac1n \gt \frac 1n - , - as \frac{n^2}{n^2-1} \gt 1, for all n\geq 2. -

    -     - -     - - - - -

    - \ds \sum_{n=2}^\infty \frac{1}{n^2\ln(n) } -

    -     - -

    - Converges; compare to \ds \infser \frac 1{n^2}, - as 1/(n^2\ln(n) ) \leq 1/n^2 for all n\geq 3. -

    -     - -     - -     - - - -

    - Use the Limit Comparison Test to determine the convergence of the given series; - state what series is used for comparison. -

    -     - - - - -

    - \ds \infser \frac{1}{n^2 -3n+5} -

    -     - -

    - Converges; compare to \ds \infser \frac 1{n^2}. -

    -     - -     - - - - -

    - \ds \infser \frac{1}{4^n-n^2} -

    -     - -

    - Converges; compare to \ds \infser \frac 1{4^n}. -

    -     - -     - - - - -

    - \ds \sum_{n=4}^\infty \frac{\ln(n) }{n-3} -

    -     - -

    - Diverges; compare to \ds \infser \frac {\ln(n) }{n}. -

    -     - -     - - - - -

    - \ds \infser \frac{1}{\sqrt{n^2+n}} -

    -     - -

    - Diverges; compare to \ds \infser \frac {1}{n}. -

    -     - -     - - - - -

    - \ds \infser \frac{1}{n+\sqrt{n}} -

    -     - -

    - Diverges; compare to \ds \infser \frac {1}{n}. -

    -     - -     - - - - -

    - \ds \infser \frac{n-10}{n^2+10n+10} -

    -     - -

    - Diverges; compare to \ds \infser \frac {1}{n}. -

    -     - -     - - - - -

    - \ds \infser \sin\big(1/n\big) -

    -     - -

    - Diverges; compare to \ds \infser \frac {1}{n}. - Just as \lim\limits_{n\to0}\frac{\sin(n) }{n} = 1, - \lim\limits_{n\to\infty}\frac{\sin(1/n)}{1/n} = 1. -

    -     - -     - - - - -

    - \ds \infser \frac{n+5}{n^3-5} -

    -     - -

    - Converges; compare to \ds \infser \frac {1}{n^2}. -

    -     - -     - - - - -

    - \ds \infser \frac{\sqrt{n}+3}{n^2+17} -

    -     - -

    - Converges; compare to \ds \infser \frac {1}{n^{3/2}}. -

    -     - -     - - - - -

    - \ds \infser \frac{1}{\sqrt{n}+100} -

    -     - -

    - Diverges; compare to \ds \infser \frac {1}{n^{1/2}}. -

    -     - -     - -     - - - -

    - Determine the convergence of the given series. - State the test used; more than one test may be appropriate. -

    -     - - - - -

    - \ds \infser \frac{n^2}{2^n} -

    -     - -

    - Converges; Integral Test -

    -     - -     - - - - -

    - \ds \infser \frac{1}{(2n+5)^3} -

    -     - -

    - Converges; Integral Test, - p-Series Test, Direct & Limit Comparison Tests can all be used. -

    -     - -     - - - - -

    - \ds \infser \frac{n!}{10^n} -

    -     - -

    - Diverges; the nth Term Test and Direct Comparison Test can be used. -

    -     - -     - - - - -

    - \ds \infser \frac{\ln(n) }{n!} -

    -     - -

    - Converges; the Direct Comparison Test can be used with sequence 1/(n-1)!. -

    -     - -     - - - - -

    - \ds \infser \frac{1}{3^n+n} -

    -     - -

    - Converges; the Direct Comparison Test can be used with sequence 1/3^n. -

    -     - -     - - - - -

    - \ds \infser \frac{n-2}{10n+5} -

    -     - -

    - Diverges; the nth Term Test can be used, - along with the Limit Comparison Test - (compare with 1/10). -

    -     - -     - - - - -

    - \ds \infser \frac{3^n}{n^3} -

    -     - -

    - Diverges; the nth Term Test can be used, - along with the Integral Test. -

    -     - -     - - - - -

    - \ds \infser \frac{\cos(1/n)}{\sqrt{n}} -

    -     - -

    - Diverges; the Limit Comparison Test can be used with sequence 1/\sqrt{n}. -

    -     - -     -     - - - -

    - Given that \ds \infser a_n converges, - state which of the following series converges, - may converge, or does not converge. -

    -     - - - -

    - \ds \infser \frac{a_n}n -

    -     - -

    - Converges; use Direct Comparison Test as \frac{a_n}{n}\lt n. -

    -     -     - - - -

    - \ds \infser a_n a_{n+1} -

    -     - -

    - Converges; since original series converges, - we know \lim_{n\to\infty}a_n = 0. - Thus for large n, a_na_{n+1} \lt a_n. -

    -     -     - - - -

    - \ds \infser (a_n)^2 -

    -     - -

    - Converges; similar logic to so (a_n)^2\lt a_n. -

    -     -     - - - -

    - \ds \infser na_n -

    -     - -

    - May converge; - certainly na_n \gt a_n but that does not mean it does not converge. -

    -     -     - - - -

    - \ds \infser \frac{1}{a_n} -

    -     - -

    - Does not converge, using logic from part and n^{th} Term Test. -

    -     -     -     - - - -

    - In this exercise, we explore an approximation method for series to which the - applies. -

    -     - - -

    - Let a(x) be a function to which the - applies, - and for which the series \infser a_n converges. -

    -

    - Let R_n = \sum_{n+1}^\infty a_n denote the remainder; - that is, the difference between \infser a_n and the nth partial sum. - (Note that R_n is the size of the error that results if we approximate the series by the nth partial sum.) - Explain why we must have the following inequality: - - \int_n^\infty a(x)\,dx \leq R_n \leq \int_{n+1}^\infty a(x)\,dx - -

    -     - -

    - Sketch a graph to represent y=a(x). What property must a(x) have? - See if you can interpret R_n in terms of left and right handed Riemann sums. -

    -     - -

    - Since the Integral Test applies, we know that a(x) must be a decreasing function. - The sum R_n = a_{n+1}+a_{n+2}+\cdots can be viewed as a right-endpoint Riemann sum - for the integral \int_n^\infty a(x)\,dx, and this must be an underestimate, - since a(x) is a decreasing function. - But R_n can also be viewed as a left-endpoint Riemann sum for the integral - \int_{n+1}^\infty a(x)\,dx, and this must be an overestimate. -

    -     -     - - -

    - Estimate the error involved in using the first 12 terms to approximate the series \sum_{n=1}^\infty 1/n^4. - What is the approximate value of the series? -

    -     - -

    - For the first 12 terms, we take n=12. - We find: - - \int_{12}^\infty \frac{1}{x^4}\,dx \amp = \frac{1}{5184} \approx 0.000193 - \int_{13}^\infty \frac{1}{x^4}\,dx \amp = \frac{1}{6591} \approx 0.000152 - \sum_{1}^{12}\frac{1}{n^4} \amp \approx 1.082153 - . -

    -     -     - - -

    - How many terms must we take to ensure that the nth partial sum approximation for - \infser 1/n^4 is accurate to 5 decimal places? -

    -     - -

    - We have - - R_n\leq \int_n^\infty \frac{1}{x^4}\,dx = \frac{1}{3n^3} - . - For 5 decimal places of accuracy, we need R_n\leq 0.000005. - Setting 0.000005=1/(3n^3) and solving for n gives - n\approx 40.55, so we should take 41 terms. -

    -     -     -     -     -     -
    -
    - Ratio and Root Tests - -

    - The nth-Term Test of - states that in order for a series \ds \infser a_n to converge, - \lim\limits_{n\to\infty}a_n = 0. - That is, the terms of \{a_n\} must get very small. - Not only must the terms approach 0, they must approach 0 fast enough: - while \lim\limits_{n\to\infty}1/n=0, - the Harmonic Series \ds\infser \frac1n diverges as the terms of \{1/n\} do not approach 0 fast enough. -

    - -

    - The comparison tests of the previous section determine convergence by comparing terms of a series to terms of another series whose convergence is known. - This section introduces the Ratio and Root Tests, - which determine convergence by analyzing the terms of a series to see if they approach 0 fast enough. -

    -     - - - Ratio Test - - Ratio Test - -

    - Let \{a_n\} be a positive sequence and consider \lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}. - seriesRatio Comparison Test - Ratio Comparison Testfor series - convergenceRatio Comparison Test - divergenceRatio Comparison Test -

    - -

    -

      -
    1. -

      - If \lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}\lt 1, - then \ds\infser a_n converges. -

      -
      2. - -
    2. -

      - If \lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n} \gt 1 or \lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=\infty, - then \ds\infser a_n diverges. -

      -
      3. - -
    3. -

      - If \lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}=1, - the Ratio Test is inconclusive. -

      -
      4. -
    -

    -     -     - - - - -

    - The principle of the Ratio Test is this: - if \lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n} = L\lt 1, - then for large n, - each term of \{a_n\} is significantly smaller than its previous term which is enough to ensure convergence. -

    - - - Applying the Ratio Test - -

    - Use the Ratio Test to determine the convergence of the following series: -

    - -

    -

      -
    1. -

      - \ds \infser \frac{2^n}{n!} -

      -
      2. - -
    2. \ds \infser \frac{3^n}{n^3}
      3. - -
    3. \ds \infser \frac{1}{n^2+1}
      4. -
    -

    -     - -

    -

      -
    1. -

      - \ds \infser \frac{2^n}{n!}: - - \inflim{\frac{a_{n+1}}{a_n}} \amp = \lim_{n\to\infty}\frac{2^{n+1}/(n+1)!}{2^n/n!} - \amp = \lim_{n\to\infty} \frac{2^{n+1}n!}{2^n(n+1)!} - \amp = \lim_{n\to\infty} \frac{2}{n+1} - \amp =0 - . - Since the limit is 0\lt 1, - by the Ratio Test \ds\infser \frac{2^n}{n!} converges. - The fact that \inflim{\frac{a_{n+1}}{a_n}}=0 can be interpreted to mean that in the long run, - the term a_{n+1} is roughly 0 times as large as a_n. - In other words, not only is a_n decreasing to 0, - it is decreasing very quickly. - That is, the terms of a_n decrease to 0 sufficiently fast enough to guarantee the convergence of \infser a_n. -

      -
      2. - -
    2. -

      - \ds\infser \frac{3^n}{n^3}: - - \inflim{\frac{a_{n+1}}{a_n}} \amp = \lim_{n\to\infty} \frac{3^{n+1}/(n+1)^3}{3^n/n^3} - \amp = \lim_{n\to\infty}\frac{3^{n+1}n^3}{3^n(n+1)^3} - \amp = \lim_{n\to\infty} \frac{3n^3}{(n+1)^3} - \amp = 3 - . - Since the limit is 3 \gt 1, - by the Ratio Test \ds\infser \frac{3^n}{n^3} diverges. - The fact that \inflim{\frac{a_{n+1}}{a_n}}=3 can be interpreted to mean that in the long run, - the term a_{n+1} is roughly 3 times as large as a_n, - so a_n is increasing - by roughly a factor of 3 in the long run. - We could also use - to determine that this series diverges. - The exponential will dominate the polynomial in the long run, - so \inflim 3^n/n^3=\infty. -

      -
      3. - -
    3. -

      - \infser\frac{1}{n^2+1}: - - \inflim\frac{a_{n+1}}{a_n} \amp = \lim_{n\to\infty} \frac{1/\big((n+1)^2+1\big)}{1/(n^2+1)} - \amp = \lim_{n\to\infty} \frac{n^2+1}{(n+1)^2+1} - \amp = 1 - . - Since the limit is 1, the Ratio Test is inconclusive. - We can easily show this series converges using the . - We can also use or , - with each comparing to the series \ds \infser \frac{1}{n^2}. -

      -
      4. -
    -

    -     - -     - - - -

    - The Ratio Test is not effective when the terms of a series only - contain algebraic functions (e.g., polynomials). - It is most effective when the terms contain some factorials or exponentials. - The previous example also reinforces our developing intuition: - factorials dominate exponentials, - which dominate algebraic functions, - which dominate logarithmic functions. - In Part 1 of the example, - the factorial in the denominator dominated the exponential in the numerator, - causing the series to converge. - In Part 2, the exponential in the numerator dominated the algebraic function in the denominator, - causing the series to diverge. -

    - -

    - While we have used factorials in previous sections, - we have not explored them closely and one is likely to not yet have a strong intuitive sense for how they behave. - The following example gives more practice with factorials. -

    - - - Applying the Ratio Test - -

    - Determine the convergence of \ds\infser \frac{n!n!}{(2n)!}. -

    -     - -

    - Before we begin, - be sure to note the difference between (2n)! and 2n!. - When n=4, - the former is 8!=8\cdot7\cdot\ldots\cdot 2\cdot1=40,320, - whereas the latter is 2(4\cdot3\cdot2\cdot1) = 48. -

    - -

    - Applying the Ratio Test: - - \inflim\frac{a_{n+1}}{a_n} \amp = \lim_{n\to\infty} \frac{(n+1)!(n+1)!/\big(2(n+1)\big)!}{n!n!/(2n)!} - \amp = \lim_{n\to\infty}\frac{(n+1)!(n+1)!(2n)!}{n!n!(2n+2)!} - Noting that (n+1)!=(n+1)\cdot n! and (2n+2)! = (2n+2)\cdot(2n+1)\cdot(2n)!, we have - \amp = \lim_{n\to\infty}\frac{(n+1)(n+1)}{(2n+2)(2n+1)} - \amp = 1/4 - . -

    - -

    - Since the limit is 1/4\lt 1, - by the Ratio Test we conclude \ds \infser \frac{n!n!}{(2n)!} converges. -

    - -

    - To find the limit in the second to last line, - recall that we just need to examine the leading terms of the numerator and denominator, - which are n^2 and 4n^2 respectively. -

    -     - -     -     - - - Root Test -

    - The final test we introduce is the Root Test, - which works particularly well on series where each term is raised to a power, - and does not work well with terms containing factorials. -

    - - - - - Root Test - -

    - Let \{a_n\} be a positive sequence, - and consider \lim\limits_{n\to \infty} (a_n)^{1/n}. - seriesRoot Comparison Test - Root Comparison Testfor series - convergenceRoot Comparison Test - divergenceRoot Comparison Test -

    - -

    -

      -
    1. -

      - If \lim\limits_{n\to \infty} (a_n)^{1/n}\lt 1, - then \ds\infser a_n converges. -

      -
      2. - -
    2. -

      - If \lim\limits_{n\to \infty} (a_n)^{1/n} \gt1 or \lim\limits_{n\to \infty} (a_n)^{1/n}=\infty, - then \ds\infser a_n diverges. -

      -
      3. - -
    3. -

      - If \lim\limits_{n\to \infty} (a_n)^{1/n}=1, - the Root Test is inconclusive. -

      -
      4. -
    -

    -     -     - - - Applying the Root Test - -

    - Determine the convergence of the following series using the Root Test: -

    - -

    -

      -
    1. -

      - \ds \infser \left(\frac{3n+1}{5n-2}\right)^n -

      -
      2. - -
    2. -

      - \ds \infser\frac{n^4}{(\ln(n) )^n} -

      -
      3. - -
    3. -

      - \ds \infser \frac{2^n}{n^2} -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - - \inflim \left(a_n\right)^{1/n} \amp = \inflim \left(\left(\frac{3n+1}{5n-2}\right)^n\right)^{1/n} - \amp =\inflim \frac{3n+1}{5n-2} = \frac 35 - . - Since the limit is less than 1, we conclude the series converges. - Note: it is difficult to apply the Ratio Test to this series. -

      -
      2. - -
    2. - -

      - - \inflim \left(a_n \right)^{1/n} \amp =\inflim \left(\frac{n^4}{(\ln(n))^n}\right)^{1/n} - \amp = \inflim \frac{\big(n^{4/n}\big)}{\ln(n)} - The limit of the numerator must be found using L'Hospital's Rule for indeterminate powers - \inflim \left(n^{4/n}\right) \amp = \inflim e^{\ln\left(n^{4/n}\right)} - \amp = \inflim e^{{4\ln\left(n\right)}/n} - Now apply L'Hospital's to the expression in the exponent: - \amp \stackrel{\, \text{ by LHR } \, }{=} \inflim e^{4/n} - \amp = e^0=1 - . - Since the numerator approaches 1 - (by L'Hospital's Rule) - and the denominator - grows to infinity, we have - - \inflim \frac{\big(n^{4/n}\big)}{\ln(n)} =0 - . - Since the limit is less than 1, we conclude the series converges. -

      -
      3. - -
    3. -

      - \inflim \left(\frac{2^n}{n^2}\right)^{1/n} = \inflim \frac{2}{\big(n^{2/n}\big)} = 2. - - Since this is greater than 1, we conclude the series diverges. (Note: The is easy to apply to this series.) -

      - -

      - (Also note: The limit in the denominator is found in a similar fashion as was illustrated in Part. - In general \inflim (n)^{b/n}=1 for any real number b.) -

      -
      4. -
    -

    -     - -     - - - - - -

    - Each of the tests we have encountered so far has required that we analyze series from - positive sequences. - - relaxes this restriction by considering - alternating series, - where the underlying sequence has terms that alternate between being positive and negative. -

    - - -     - - - - Terms and Concepts - - - -

    - The Ratio Test is not effective when the terms of a sequence only contain functions. -

    -     - - - - polynomial|algebraic|rational - - -

    - polynomial, rational, or more generally, algebraic are all good answers here. -

    -     -     -     - -     - - - - -

    - The Ratio Test is most effective when the terms of a sequence contains - and/or functions. -

    -     - - - - - - ["factorial","exponential"].includes(ans) - - - - - - - ["factorial","exponential"].includes(ans) && !ans_array.slice(0,1).includes(ans) - - - - - ans_array.slice(0,1).includes(ans) - - You already gave that answer. - - - - -     - - - - -

    - What three convergence tests do not work well with terms containing factorials? -

    -     - - - -

    - Test for divergence -

    -     -     - - -

    - Comparison test -

    -     -     - - -

    - Limit comparison test -

    -     -     - - -

    - Integral test -

    -     -     - - -

    - Ratio test -

    -     -     - - -

    - Root test -

    -     -     -     - -

    - Integral Test, Limit Comparison Test, and Root Test -

    -     - -     - - - - -

    - The Root Test works particularly well on series where each term is . -

    -     - - - - a power|a power function|raised to a power - - - - -

    - Your answer includes the correct words but has extra or missing text. -

    -     -     -     -     - - -     -     - - Problems - - - -

    - Determine the convergence of the given series using the Ratio Test. - If the Ratio Test is inconclusive, - state so and determine convergence with another test. -

    -     - - - - -

    - \ds\infser[0] \frac{2n}{n!} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser[0] \frac{5^n-3n}{4^n} -

    -     - -

    - Diverges -

    -     - -     - - - - -

    - \ds\infser[0] \frac{n!10^n}{(2n)!} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser \frac{5^n+n^4}{7^n+n^2} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser \frac{1}{n} -

    -     - -

    - The Ratio Test is inconclusive; - the p-Series Test states it diverges. -

    -     - -     - - - - -

    - \ds\infser \frac{1}{3n^3+7} -

    -     - -

    - The Ratio Test is inconclusive; - the Direct Comparison Test with 1/n^3 shows it converges. -

    -     - -     - - - - -

    - \ds\infser \frac{10\cdot5^n}{7^n-3} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser n\cdot\left(\frac35\right)^n -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser \frac{2\cdot4\cdot6\cdot8\cdots 2n}{3\cdot6\cdot9\cdot12\cdots 3n} -

    -     - -

    - Converges; note the summation can be rewritten as \ds\infser \frac{2^nn!}{3^nn!}, - from which the Ratio Test or Geometric Series Test can be applied. -

    -     - -     - - - - -

    - \ds\infser \frac{n!}{5\cdot10\cdot15\cdots (5n)} -

    -     - -

    - Converges; rewrite the summation as - \ds\infser \frac{n!}{5^nn!} then apply the Ratio Test or Geometric Series Test. -

    -     - -     - -     - - - -

    - Determine the convergence of the given series using the Root Test. - If the Root Test is inconclusive, - state so and determine convergence with another test. -

    -     - - - - -

    - \ds\infser \left(\frac{2n+5}{3n+11}\right)^n -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser \left(\frac{0.9n^2-n-3}{n^2+n+3}\right)^n -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser \frac{2^nn^2}{3^n} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser \frac{1}{n^n} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser \frac{3^n}{n^22^{n+1}} -

    -     - -

    - Diverges -

    -     - -     - - - - -

    - \ds\infser \frac{4^{n+7}}{7^n} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser \left(\frac{n^2-n}{n^2+n}\right)^n -

    -     - -

    - Diverges. - The Root Test is inconclusive, - but the nth-Term Test shows divergence. - (The terms of the sequence approach e^{-2}, - not 0, as n\to\infty.) -

    -     - -     - - - - -

    - \ds\infser \left(\frac1n-\frac{1}{n^2}\right)^n -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser \frac1{\big(\ln(n) \big)^n} -

    -     - -

    - Converges -

    -     - -     - - - - -

    - \ds\infser \frac{n^2}{\big(\ln(n) \big)^n} -

    -     - -

    - Converges -

    -     - -     - -     - - - -

    - Determine the convergence of the given series. - State the test used; more than one test may be appropriate. -

    -     - - - - -

    - \ds\infser \frac{n^2+4n-2}{n^3+4n^2-3n+7} -

    -     - -

    - Diverges; Limit Comparison Test with the harmonic series 1/n. -

    -     - -     - - - - -

    - \ds\infser \frac{n^44^n}{n!} -

    -     - -

    - Converges; Ratio Test -

    -     - -     - - - - -

    - \ds\infser \frac{n^2}{3^n+n} -

    -     - -

    - Converges; Ratio Test or Limit Comparison Test with 1/3^n. -

    -     - -     - - - - -

    - \ds\infser \frac{3^n}{n^n} -

    -     - -

    - Converges; Root Test -

    -     - -     - - - - -

    - \ds\infser \frac{n}{\sqrt{n^2+4n+1}} -

    -     - -

    - Diverges; nth-Term Test or Limit Comparison Test with 1. -

    -     - -     - - - - -

    - \ds\infser \frac{n!n!n!}{(3n)!} -

    -     - -

    - Converges; Ratio Test -

    -     - -     - - - - -

    - \ds\infser[2] \frac{1}{\ln(n) } -

    -     - -

    - Diverges; Direct Comparison Test with 1/n -

    -     - -     - - - - -

    - \ds\infser \left(\frac{n+2}{n+1}\right)^n -

    -     - -

    - Diverges; nth-Term Test (nth term approaches e.) -

    -     - -     - - - - -

    - \ds\infser[2] \frac{n^3}{\big(\ln(n) \big)^n} -

    -     - -

    - Converges; Root Test -

    -     - -     - - - - -

    - \ds\infser \left(\frac1n-\frac1{n+2}\right) -

    -     - -

    - Converges; Limit Comparison Test with 1/n^2 - (get common denominator first). - It is also a Telescoping Series. -

    -     - -     - -     -     -     -
    -
    - Alternating Series and Absolute Convergence -

    - All of the series convergence tests we have used require that the underlying sequence \{a_n\} be a positive sequence. - (We can relax this with - and state that there must be an N \gt 0 such that a_n \gt 0 for all n \gt N; - that is, \{a_n\} is positive for all but a finite number of values of n.) -

    - -

    - In this section we explore series whose summation includes negative terms. - We start with a very specific form of series, - where the terms of the summation alternate between being positive and negative. -

    - - - Alternating Series - -

    - Let \{a_n\} be a positive sequence. - An alternating series - is a series of either the form - seriesalternating - - \infser (-1)^na_n\qquad \text{ or } \qquad \infser (-1)^{n+1}a_n - . -

    -     -     - -

    - Recall the terms of Harmonic Series come from the Harmonic Sequence \{a_n\} = \{1/n\}. - An important alternating series is the - Alternating Harmonic Series: - - \infser (-1)^{n+1}\frac1n = 1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots - -

    - -

    - Geometric Series can also be alternating series when r\lt 0. - For instance, if r=-1/2, the geometric series is - - \infser[0] \left(\frac{-1}{2}\right)^n = 1-\frac12+\frac14-\frac18+\frac1{16}-\frac1{32}+\cdots - -

    - -

    - - states that geometric series converge when \abs{r}\lt 1 and gives the sum: - \ds \infser[0] r^n = \frac1{1-r}. - When r=-1/2 as above, we find - - \infser[0] \left(\frac{-1}{2}\right)^n = \frac1{1-(-1/2)} = \frac 1{3/2} = \frac23 - . -

    - -

    - A powerful convergence theorem exists for other alternating series that meet a few conditions. -

    - - - Alternating Series Test - -

    - Let \{a_n\} be a positive, - decreasing sequence where \lim\limits_{n\to\infty}a_n=0. - Then - seriesAlternating Series Test - Alternating Series Test - convergenceAlternating Series Test - divergenceAlternating Series Test - - \infser (-1)^{n}a_n \qquad \text{ and } \qquad \infser (-1)^{n+1}a_n - - converge. -

    -     -     - - - - - - - - - -

    - The basic idea behind - is illustrated in . - A positive, decreasing sequence \{a_n\} is shown along with the partial sums - - S_n = \sum_{i=1}^n(-1)^{i+1}a_i =a_1-a_2+a_3-a_4+\cdots+(-1)^{n+1}a_n - . -

    - -

    - Because \{a_n\} is decreasing, - the amount by which S_n bounces up/down decreases. - Moreover, the odd terms of S_n form a decreasing, - bounded sequence, - while the even terms of S_n form an increasing, - bounded sequence. - Since bounded, monotonic sequences converge - (see ) - and the terms of \{a_n\} approach 0, one can show the odd and even terms of S_n converge to the same common limit L, - the sum of the series. -

    - -
    - Illustrating convergence with the Alternating Series Test - - - Scatter plots of a positive, decreasing sequence and its alternating sequence of partial sums. - -

    - On one set of coordinate axes, two scatter plots are shown. - The first is the plot of a positive, decreasing sequence a_n; - the second is the plot of the sequence of partial sums for the corresponding alternating sequence (-1)^na_n. -

    - -

    - The scatter plots illustrate why an alternating series converges: - as n increases, the partial sums oscillate back and forth across a horizontal line marked L - (the limiting value). - Since a_n is a decreasing sequence, the oscillations get smaller as n increases, - and the points in the scatter plot for S_n get closer and closer to the line y=L. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.25,ymax=1.25, - xmin=-1,xmax=11, - xlabel={$n$} - ] - - \addplot [only marks,secondcolor,mark=square*,mark size={2.75pt}] coordinates{(1,1)}; - \addplot [only marks,firstcolor,mark size={2.4pt}] coordinates{(1,1)}; - \addplot [only marks,firstcolor,mark size={2.4pt},domain=2:10,samples=9] {2/(x+1)}; - \addplot [only marks,secondcolor,mark=square*,mark size={2.4pt},domain=2:10,samples=9] coordinates {(2., 0.333333)(3., 0.833333)(4., 0.433333)(5.,0.766667)(6., 0.480952)(7., 0.730952)(8., 0.50873)(9.,0.70873)(10., 0.526912)}; - \addplot [dashed,domain=-.1:10] {.614} node [pos=0,left] { $L$}; - - \end{axis} - - - \node[shift={(0,-10pt)},draw] at (myplot.south) - {\begin{tikzpicture} - - \fill [firstcolor] (0,0) circle (2pt) node [right,black] { $a_n$}; - \fill [secondcolor] (1,0) circle (0.1pt) node [right, black] { $S_n$}; - \draw [secondcolor] plot [only marks,mark=square*] coordinates {(1,0)}; - - \end{tikzpicture}}; - - \end{tikzpicture} - - - - -
    - -
    - A visual representation of adding terms of an alternating series. The arrows represent the length and direction of each term of the sequence. - - - An illustration of the partial sums in an alternating series, using line segments to demonstrate convergence. - -

    - On a set of coordinate axes, a sequence of horizontal line segments is shown. - At the top of the image is a line segment indicating in increase from 0 to a_1. - An arrow pointing to the right is at the end of the segment. -

    - -

    - Below this segment is a shorter segment with an arrow pointing to the left. - The right end of this segment aligns with the right end of the first segment, - indicating that we obtain a_1-a_2 by starting at a_1 and then moving to the left by a distance a_2. -

    - -

    - The third segment is below the second. Its left end aligns with the left end of the second segment, - and it points to the right, indicating the act of adding a_3 to the partial sum. - The length of this segment is shorter than that of the second, indicating the fact that the sequence a_n is decreasing. -

    - -

    - As we move down, additional segments are drawn, alternating between pointing left and right, - and getting shorter with each step. Below the segments is a horizontal axis, on which the values - S_1, S_2, \ldots of the partial sums are shown, as well as the limit L. - The illustration overall is intended to convey the idea that the line segments will shrink toward the limiting value L - as n increases. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - extra x ticks={1, 0.3333, 0.83333, 0.43333, 0.76667, 0.4809, 0.7309, 0.612}, - extra x tick labels={ $S_1$, $S_2$, $S_3$, $S_4$, $S_5$, $S_6$, $S_7$, $L$ }, - ytick=\empty, - xlabel=\empty, - ylabel=\empty, - ymin=-.2,ymax=2.25, - xmin=-.2,xmax=1.2, - x post scale=2, - ] - - \addplot[firstcurvestyle,infiniteright,domain=0:1] ({x},{2}) node [pos=1, right] {$a_1$}; - \addplot[firstcurvestyle,infiniteleft,domain=0.333:1] ({x},{1.75}) node [pos=0, left] {$-a_2$}; - \addplot[firstcurvestyle,infiniteright,domain=0.333:.833] ({x},{1.5}) node [pos=1, right] {$a_3$}; - \addplot[firstcurvestyle,infiniteleft,domain=0.433:.8333] ({x},{1.25}) node [pos=0, left] {$-a_4$}; - \addplot[firstcurvestyle,infiniteright,domain=0.4333:.76667] ({x},{1}) node [pos=1, right] {$a_5$}; - \addplot[firstcurvestyle,infiniteleft,domain=0.4809:.76667] ({x},{.75}) node [pos=0, left] {$-a_6$}; - \addplot[firstcurvestyle,infiniteright,domain=0.4809:.7309] ({x},{.5}) node [pos=1, right] {$a_7$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - - - Applying the Alternating Series Test - -

    - Determine if the Alternating Series Test applies to each of the following series. -

    - -

    -

      -
    1. -

      - \ds \infser (-1)^{n+1}\frac1n -

      -
      2. - -
    2. -

      - \ds \infser (-1)^n\frac{\ln(n) }{n} -

      -
      3. - -
    3. -

      - \ds \infser (-1)^{n+1}\frac{\abs{\sin(n) }}{n^2} -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - This is the Alternating Harmonic Series as seen previously. - The underlying sequence is \{a_n\} = \{1/n\}, - which is positive, decreasing, - and approaches 0 as n\to\infty. - Therefore we can apply the Alternating Series Test and conclude this series converges. - - While the test does not state what the series converges to, we will see later that \ds \infser (-1)^{n+1}\frac1n=\ln(2). -

      -
      2. - -
    2. -

      - The underlying sequence is \{a_n\} = \{\ln(n) /n\}. - This is positive and approaches 0 as n\to\infty - (use L'Hospital's Rule). - However, the sequence is not decreasing for all n. - It is straightforward to compute a_1=0, a_2\approx0.347, - a_3\approx 0.366, and a_4\approx 0.347: - the sequence is increasing for at least the first 3 terms. - - We do not immediately conclude that we cannot apply the Alternating Series Test. Rather, consider the long-term behavior of \{a_n\}. Treating a_n=a(n) as a continuous function of n defined on [1,\infty), we can take its derivative: - - a'(n) = \frac{1-\ln(n) }{n^2} - . - The derivative is negative for all n\geq 3 (actually, - for all n \gt e), - meaning a(n)=a_n is decreasing on [3,\infty). - We can apply the Alternating Series Test to the series when we start with n=3 and conclude that \ds \sum_{n=3}^\infty(-1)^n\frac{\ln(n) }{n} converges; - adding the terms with n=1 and n=2 do not change the convergence (, we apply ). - The important lesson here is that as before, - if a series fails to meet the criteria of the Alternating Series Test on only a finite number of terms, - we can still apply the test. -

      -
      3. - -
    3. -

      - The underlying sequence is \{a_n\} = \abs{\sin(n) }/n. - This sequence is positive and approaches 0 as n\to\infty. - However, it is not a decreasing sequence; - the value of \abs{\sin(n) } oscillates between 0 and 1 as n\to\infty. - We cannot remove a finite number of terms to make \{a_n\} decreasing, - therefore we cannot apply the Alternating Series Test. - - Keep in mind that this does not mean we conclude the series diverges; in fact, it does converge. We are just unable to conclude this based on . We will be able to show that this series converges shortly. -

      -
      4. -
    -

    -     - -     - -

    - - gives the sum of some important series. - Two of these are - - \infser \frac1{n^2} =\frac{\pi^2}6 \approx 1.64493 \text{ and } \infser \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}\approx 0.82247 - . -

    - -

    - These two series converge to their sums at different rates. - To be accurate to two places after the decimal, - we need 202 terms of the first series though only 13 of the second. - To get 3 places of accuracy, - we need 1069 terms of the first series though only 33 of the second. - Why is it that the second series converges so much faster than the first? -

    - -

    - While there are many factors involved when studying rates of convergence, - the alternating structure of an alternating series gives us a powerful tool when approximating the sum of a convergent series. -

    - - - The Alternating Series Approximation Theorem - -

    - Let \{a_n\} be a sequence that satisfies the hypotheses of the Alternating Series Test, - and let S_n and L be the - nth partial sums and sum, respectively, - of either \ds \infser (-1)^{n}a_n or \ds \infser (-1)^{n+1}a_n. - Then - - seriesalternating!Approximation Theorem - -

    - -

    -

      -
    1. -

      - E_n=\abs{S_n-L} \lt a_{n+1}, and -

      -
      2. - -
    2. -

      - L is between S_n and S_{n+1}. -

      -
      3. -
    -

    -     -     - - - -

    - Part 1 of - states that the nth partial sum of a convergent alternating series will be within a_{n+1} of its total sum. - You can see this visually in . - Look at the distance between S_6 and L. - Clearly this distance is less than the length of the arrow corresponding to a_7. -

    - -

    - Also consider the alternating series we looked at before the statement of the theorem, - \ds \infser \frac{(-1)^{n+1}}{n^2}. - Since a_{14} = 1/14^2 \approx 0.0051, - we know that S_{13} is within 0.0051 of the total sum. -

    - -

    - Moreover, Part 2 of the theorem states that since - S_{13} \approx 0.8252 and S_{14}\approx 0.8201, - we know the sum L lies between 0.8201 and 0.8252. - One use of this is the knowledge that S_{14} is accurate to two places after the decimal. -

    - -

    - Some alternating series converge slowly. - In - we determined the series \ds\infser (-1)^{n+1}\frac{\ln(n) }{n} converged. - With n=1001, we find \ln(n) /n \approx 0.0069, - meaning that S_{1000} \approx 0.1633 is accurate to one, - maybe two, places after the decimal. - Since S_{1001} \approx 0.1564, - we know the sum L is 0.1564\leq L\leq0.1633. -

    - - - Approximating the sum of convergent alternating series - -

    - Approximate the sum of the following series, - accurate to within 0.001. -

    - -

    -

      -
    1. -

      - \ds \infser (-1)^{n+1}\frac{1}{n^3} -

      -
      2. - -
    2. -

      - \ds \infser (-1)^{n+1}\frac{\ln(n) }{n} -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - Using , - we want to find n where 1/n^3 \leq 0.001. - That is, we want to find the the first time a term in the sequence a_n is smaller than the desired level of error: - - \frac1{n^3} \amp \leq 0.001=\frac{1}{1000} - n^3 \amp \geq 1000 - n \amp \geq \sqrt[3]{1000} - n \amp \geq 10 - . - Let L be the sum of this series. - By Part 1 of the theorem, - \abs{S_9-L}\lt a_{10} = 1/1000. (We found a_{10}=a_{n+1}\lt 0.0001, - so n=9). - We can compute S_9=0.902116, - which our theorem states is within 0.001 of the total sum. - - We can use Part 2 of the theorem to obtain an even more accurate result. As we know the 10th term of the series is (-1)^n/10^3=-1/1000, we can easily compute S_{10} = 0.901116. Part 2 of the theorem states that L is between S_9 and S_{10}, so 0.901116 \lt L\lt 0.902116. -

      -
      2. - -
    2. -

      - We want to find n where \ln(n)/n \lt 0.001. - We start by solving \ln(n)/n = 0.001 for n. - This cannot be solved algebraically, - so we will use Newton's Method to approximate a solution. - (Note: we can also use a Brute Force technique. - That is, we can guess and check numerically until we find a solution.) - - Let f(x) = \ln(x)/x-0.001; we want to know where f(x) = 0. We make a guess that x must be large, so our initial guess will be x_1=1000. Recall how Newton's Method works: given an approximate solution x_n, our next approximation x_{n+1} is given by - - x_{n+1} = x_n - \frac{f(x_n)}{\fp(x_n)} - . - We find \fp(x) = \big(1-\ln(x)\big)/x^2. - This gives - - x_2 \amp = 1000 - \frac{\ln(1000)/1000-0.001}{\big(1-\ln(1000)\big)/1000^2} - \amp = 2000 - . - Using a computer, - we find that Newton's Method seems to converge to a solution x=9118.01 after 8 iterations. - Taking the next integer higher, - we have n=9119, where \ln(9119)/9119 =0.000999903\lt 0.001. - - Again using a computer, we find S_{9118} = -0.160369. Part 1 of the theorem states that this is within 0.001 of the actual sum L. Already knowing the 9{,}119th term, we can compute S_{9119} = -0.159369, meaning -0.159369 \lt L \lt -0.160369. -

      -
      3. -
    -

    - -

    - Notice how the first series converged quite quickly, - where we needed only 10 terms to reach the desired accuracy, - whereas the second series took over 9,000 terms. -

    -     - -     - -

    - One of the famous results of mathematics is that the Harmonic Series, - \ds \infser \frac1n diverges, - yet the Alternating Harmonic Series, - \ds \infser (-1)^{n+1}\frac1n, converges. - The notion that alternating the signs of the terms in a series can make a series converge leads us to the following definitions. - Alternating Harmonic Series -

    - - - Absolute and Conditional Convergence - -

    -

      -
    1. -

      - A series \ds \infser a_n - converges absolutely - if \ds \infser \abs{a_n} converges. - - convergenceabsolute - convergenceconditional - seriesabsolute convergence - seriesconditional convergence -

      -
      2. - -
    2. -

      - A series \ds \infser a_n - converges conditionally - if \ds \infser a_n converges but \ds \infser \abs{a_n} diverges. -

      -
      3. -
    -

    -     -     - - -

    - Thus we say the Alternating Harmonic Series converges conditionally. -

    - - - - - Determining absolute and conditional convergence - -

    - Determine if the following series converge absolutely, - conditionally, or diverge. -

    - -

    -

      -
    1. \ds \infser (-1)^n\frac{n+3}{n^2+2n+5}
      2. - -
    2. - \ds \infser (-1)^n\frac{n^2+2n+5}{2^n} -
      3. - -
    3. - \ds \sum_{n=3}^\infty (-1)^n\frac{3n-3}{5n-10} -
      4. -
    -

    -     - -

    -

      -
    1. -

      - We can show the series - - \ds \infser \abs{(-1)^n\frac{n+3}{n^2+2n+5}}= \infser \frac{n+3}{n^2+2n+5} - - diverges using the Limit Comparison Test, - comparing with 1/n. - - The series \ds \infser (-1)^n\frac{n+3}{n^2+2n+5} converges using the Alternating Series Test; we conclude it converges conditionally. -

      -
      2. - -
    2. -

      - We can show the series - - \ds \infser \abs{(-1)^n\frac{n^2+2n+5}{2^n}}=\infser \frac{n^2+2n+5}{2^n} - - converges using the Ratio Test. - - Therefore we conclude \ds \infser (-1)^n\frac{n^2+2n+5}{2^n} converges absolutely. -

      -
      3. - -
    3. -

      - The series - - \ds \sum_{n=3}^\infty \abs{(-1)^n\frac{3n-3}{5n-10}} = \sum_{n=3}^\infty \frac{3n-3}{5n-10} - - diverges using the nth Term Test, - so it does not converge absolutely. - The series \ds \sum_{n=3}^\infty (-1)^n\frac{3n-3}{5n-10} fails the conditions of the Alternating Series Test as - (3n-3)/(5n-10) does not approach 0 as n\to\infty. - We can state further that this series diverges; - as n\to\infty, - the series effectively adds and subtracts 3/5 over and over. - This causes the sequence of partial sums to oscillate and not converge. - Therefore the series \ds \infser (-1)^n\frac{3n-3}{5n-10} diverges. -

      -
      4. -
    -

    -     - -     - -

    - Knowing that a series converges absolutely allows us to make two important statements, - given in below. - The first is that absolute convergence is stronger - than regular convergence. - That is, just because \infser a_n converges, - we cannot conclude that \infser \abs{a_n} will converge, - but knowing a series converges absolutely tells us that \infser a_n will converge. -

    - -

    - One reason this is important is that our convergence tests all require that the underlying sequence of terms be positive. - By taking the absolute value of the terms of a series where not all terms are positive, - we are often able to apply an appropriate test and determine absolute convergence. - This, in turn, - determines that the series we are given also converges. -

    - -

    - The second statement relates to - rearrangements - - rearrangements of series - seriesrearrangements of series. - When dealing with a finite set of numbers, - the sum of the numbers does not depend on the order which they are added. - (So 1+2+3 = 3+1+2.) - One may be surprised to find out that when dealing with an infinite set of numbers, - the same statement does not always hold true: - some infinite lists of numbers may be rearranged in different orders to achieve different sums. - The theorem states that the terms of an absolutely convergent series can be rearranged in any way without affecting the sum. -

    - - - - Absolute Convergence Theorem - -

    - Let \ds \infser a_n be a series that converges absolutely. - - convergenceabsolute - Absolute Convergence Theorem - seriesAbsolute Convergence Theorem - rearrangements of series - seriesrearrangements - -

    - -

    -

      -
    1. - -

      - \ds \infser a_n converges. -

      -
      2. - -
    2. -

      - Let \{b_n\} be any rearrangement of the sequence \{a_n\}. - Then - - \infser b_n = \infser a_n - . -

      -
      3. -
    -

    -     - -

    - We will provide a proof for Part - of . - Suppose that \infser \abs{a_n} converges. - We start by noting that for any sequence a_n, we have - - -\abs{a_n} \leq a_n \leq \abs{a_n} - If we add \abs{a_n} to all three sides: - 0\leq a_n +\abs{a_n} \leq 2\abs{a_n} - . - We are now in a position to apply the to the series \infser \left(a_n+\abs{a_n}\right). - Since \infser \abs{a_n} converges by our supposition, - so does \infser 2\abs{a_n} - (the scalar multiple of a convergent series also converges by ). - Therefore \infser \left(a_n+\abs{a_n}\right) converges by the . -

    - -

    - Now we turn our attention to \infser a_n. - We can say - - \infser a_n \amp =\infser \left(a_n+\abs{a_n}-\abs{a_n}\right) - \amp =\infser \left(a_n+\abs{a_n}\right)-\infser \abs{a_n} - . - The last line is the difference between two convergent series, - which is also convergent by . - Therefore \infser a_n converges. -

    -     - -     - -

    - In , - we determined the series in Part converges absolutely. - tells us the series converges - (which we could also determine using the Alternating Series Test). -

    - -

    - The theorem states that rearranging the terms of an absolutely convergent series does not affect its sum. - This implies that perhaps the sum of a conditionally convergent series can change based on the arrangement of terms. - Indeed, it can. - The Riemann Rearrangement Theorem - (named after Bernhard Riemann) - states that any conditionally convergent series can have its terms rearranged so that the sum is any desired value, - including \infty! -

    - -

    - As an example, - consider the Alternating Harmonic Series once more. - We have stated that - - \infser (-1)^{n+1}\frac1n = 1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17\cdots = \ln(2) - , -

    - -

    - (see - or ). -

    - -

    - Consider the rearrangement where every positive term is followed by two negative terms: - - 1-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac1{10}-\frac1{12}\cdots - -

    - -

    - (Convince yourself that these are exactly the same numbers as appear in the Alternating Harmonic Series, - just in a different order.) Now group some terms and simplify: - - \left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\frac1{12}+\cdots \amp = - \frac12-\frac14+\frac16-\frac18+\frac1{10}-\frac{1}{12}+\cdots \amp = - \frac12\left(1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots\right) \amp = \frac12\ln(2) - . -

    - -

    - By rearranging the terms of the series, - we have arrived at a different sum! (One could try - to argue that the Alternating Harmonic Series does not actually converge to \ln(2), - because rearranging the terms of the series - shouldn't change the sum. - However, the Alternating Series Test proves this series converges to L, - for some number L, - and if the rearrangement does not change the sum, - then L = L/2, implying L=0. - But the Alternating Series Approximation Theorem quickly shows that L \gt 0. - The only conclusion is that the rearrangement did - change the sum.) This is an incredible result. -

    - -

    - We end here our study of tests to determine convergence. - A table summarizing the tests can be found at the end of the text, - in . -

    - -

    - While series are worthy of study in and of themselves, - our ultimate goal within calculus is the study of Power Series, - which we will consider in the next section. - We will use power series to create functions where the output is the result of an infinite summation. -

    - - - - Terms and Concepts - - - -

    - Why is \ds\infser \sin(n) not an alternating series? -

    -     - - - -

    - The signs of the terms do not alternate; - in the given series, some terms are negative and the others positive, - but they do not necessarily alternate. -

    -     - -     - - - - -

    - A series \ds\infser (-1)^na_n converges when \{a_n\} is , - and \lim\limits_{n\to\infty}a_n =. -

    -     - - - - - ["positive", "decreasing"].includes(ans) - - - - - - - ["positive", "decreasing"].includes(ans) && !ans_array.slice(0,1).includes(ans) - - - - - ans_array.slice(0,1).includes(ans) - - You already gave that answer. - - - - - 0|zero - - - - -     - - - - -

    - Give an example of a series where - \ds \infser[0] a_n converges but \ds \infser[0] \abs{a_n} does not. -

    -     - - - -

    - Many examples exist; one common example is a_n = (-1)^n/n. -

    -     - -     - - - - -

    - The sum of a convergent - series can be changed by rearranging the order of its terms. -

    -     - - - - - - - - -

    - Your answer includes the correct word but has extra text. -

    -     -     -     -     - -     -     - - Problems - - - -

    - An alternating series \ds \sum_{n=i}^\infty a_n is given. -

    - -

    -

      -
    1. -

      - Determine if the series converges or diverges. -

      -
      2. - -
    2. -

      - Determine if \ds \infser[0] \abs{a_n} converges or diverges. -

      -
      3. - -
    3. -

      - If \ds \infser[0] a_n converges, - determine if the convergence is conditional or absolute. -

      -
      4. -
    -

    -     - - - - -

    - \ds \infser \frac{(-1)^{n+1}}{n^2} -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - converges (p-Series) -

      -
      3. - -
    3. -

      - absolute -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{(-1)^{n+1}}{\sqrt{n!}} -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - converges (use Ratio Test) -

      -
      3. - -
    3. -

      - absolute -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser[0] (-1)^{n}\frac{n+5}{3n-5} -

    -     - -

    -

      -
    1. -

      - diverges (limit of terms is not 0) -

      -
      2. - -
    2. -

      - diverges -

      -
      3. - -
    3. -

      - n/a; diverges -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser (-1)^{n}\frac{2^n}{n^2} -

    -     - -

    -

      -
    1. -

      - diverges (limit of terms is not 0) -

      -
      2. - -
    2. -

      - diverges -

      -
      3. - -
    3. -

      - n/a; diverges -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser[0] (-1)^{n+1}\frac{3n+5}{n^2-3n+1} -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - diverges (Limit Comparison Test with 1/n) -

      -
      3. - -
    3. -

      - conditional -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{(-1)^{n}}{\ln(n) +1} -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - diverges (Direct Comparison Test with 1/n) -

      -
      3. - -
    3. -

      - conditional -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \sum_{n=2}^\infty (-1)^n\frac{n}{\ln(n) } -

    -     - -

    -

      -
    1. -

      - diverges (limit of terms is not 0) -

      -
      2. - -
    2. -

      - diverges -

      -
      3. - -
    3. -

      - n/a; diverges -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{(-1)^{n+1}}{1+3+5+\cdots+(2n-1) } -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - converges (the sum in the denominator is n^2) -

      -
      3. - -
    3. -

      - absolute -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser \cos\big(\pi n\big) -

    -     - -

    -

      -
    1. -

      - diverges (terms oscillate between \pm 1) -

      -
      2. - -
    2. -

      - diverges -

      -
      3. - -
    3. -

      - n/a; diverges -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser[2] \frac{\sin\big((n+1/2)\pi\big)}{n\ln(n) } -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - diverges (Integral Test) -

      -
      3. - -
    3. -

      - conditional -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser[0] \left(-\frac23\right)^n -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - converges (Geometric Series with r=2/3) -

      -
      3. - -
    3. -

      - absolute -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser[0] (-e)^{-n} -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - converges (Geometric Series with r=1/e) -

      -
      3. - -
    3. -

      - absolute -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser[0] \frac{(-1)^nn^2}{n!} -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - converges (Ratio Test) -

      -
      3. - -
    3. -

      - absolute -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser[0] (-1)^n2^{-n^2} -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - converges (Ratio Test) -

      -
      3. - -
    3. -

      - absolute -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{(-1)^n}{\sqrt{n}} -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - diverges (p-Series Test with p=1/2) -

      -
      3. - -
    3. -

      - conditional -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \infser \frac{(-1000)^n}{n!} -

    -     - -

    -

      -
    1. -

      - converges -

      -
      2. - -
    2. -

      - converges (Ratio Test) -

      -
      3. - -
    3. -

      - absolute -

      -
      4. -
    -

    -     - -     - -     - - - -

    - Let S_n be the n^{ th } partial sum of a series. - A convergent alternating series is given and a value of n. - Compute S_n and S_{n+1} and use these values to find bounds on the sum of the series. -

    -     - - - - -

    - \ds \infser \frac{(-1)^n}{\ln(n+1)}, n=5 -

    -     - -

    - S_5 = -1.1906; S_{6} = -0.6767; -

    - -

    - \ds -1.1906 \leq \infser \frac{(-1)^n}{\ln(n+1)} \leq -0.6767 -

    -     - -     - - - - -

    - \ds \infser \frac{(-1)^{n+1}}{n^4}, n=4 -

    -     - -

    - S_4 = 0.9459; S_5 = 0.9475; -

    - -

    - \ds 0.9459 \leq \infser \frac{(-1)^n}{n^4} \leq 0.9475 -

    -     - -     - - - - -

    - \ds \infser[0] \frac{(-1)^{n}}{n!}, n=6 -

    -     - -

    - S_6 = 0.3681; S_7 = 0.3679; -

    - -

    - \ds 0.3679 \leq \infser[0] \frac{(-1)^{n}}{n!} \leq 0.3681 -

    -     - -     - - - - -

    - \ds \infser[0] \left(-\frac12\right)^n, n=9 -

    -     - -

    - S_9 = 0.666016; S_{10} = 0.666992; -

    - -

    - \ds 0.666016 \leq \infser[0] \left(-\frac12\right)^n \leq 0.666992 -

    -     - -     - -     - - - -

    - A convergent alternating series is given along with its sum and a value of \varepsilon. - Use - to find n such that the - nth partial sum of the series is within - \varepsilon of the sum of the series. -

    -     - - - - -

    - \ds \infser \frac{(-1)^{n+1}}{n^4} = \frac{7\pi^4}{720}, - \varepsilon = 0.001 -

    -     - -

    - n=5 -

    -     - -     - - - - -

    - \ds \infser[0] \frac{(-1)^{n}}{n!} = \frac1e, - \varepsilon = 0.0001 -

    -     - -

    - n=7 -

    -     - -     - - - - -

    - \ds \infser[0] \frac{(-1)^{n}}{2n+1}=\frac{\pi}4, - \varepsilon = 0.001 -

    -     - -

    - Using the theorem, - we find n=499 guarantees the sum is within 0.001 of \pi/4. - (Convergence is actually faster, - as the sum is within \varepsilon of \pi/24 when n\geq 249.) -

    -     - -     - - - - -

    - \ds \infser[0] \frac{(-1)^{n}}{(2n)!}=\cos(1), - \varepsilon = 10^{-8} -

    -     - -

    - n=5 ((2n)! \gt 10^8 when n\geq 6) -

    -     - -     - -     -     -     -
    -
    - Power Series -

    - So far, our study of series has examined the question of - Is the sum of these infinite terms finite?, - , Does the series converge? - We now approach series from a different perspective: as a function. - Given a value of x, - we evaluate f(x) by finding the sum of a particular series that depends on x - (assuming the series converges). - We start this new approach to series with a definition. -

    - - - - - Power Series - -

    - Let \{a_n\} be a sequence, - let x be a variable, and let c be a real number. - power series - seriespower -

    - -

    -

      -
    1. -

      - The power series in x is the series - - \infser[0] a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+\ldots - -

      -
      2. - -
    2. -

      - The power series in x centered at c is the series - - \infser[0] a_n(x-c)^n = a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\ldots - -

      -
      3. -
    -

    -     -     - - - Examples of power series - -

    - Write out the first five terms of the following power series: -

    - -

    -

      -
    1. \infser[0] x^n
      2. -
    2. \infser (-1)^{n+1}\frac{(x+1)^n}n
      3. -
    3. \infser[0] (-1)^{n+1} \frac{(x-\pi)^{2n}}{(2n)!}
      4. -
    -

    -     - -

    -

      -
    1. -

      - One of the conventions we adopt is that x^0=1 regardless of the value of x. - Therefore - - \infser[0] x^n = 1+x+x^2+x^3+x^4+\ldots - - This is a geometric series in x with r=x. -

      -
      2. - -
    2. -

      - This series is centered at c=-1. - Note how this series starts with n=1. - We could rewrite this series starting at n=0 with the understanding that a_0=0, - and hence the first term is 0. - - \amp \infser (-1)^{n+1}\frac{(x+1)^n}n - \amp =(x+1) - \frac{(x+1)^2}{2} + \frac{(x+1)^3}{3} - \frac{(x+1)^4}{4}+\frac{(x+1)^5}{5}\ldots - -

      -
      3. - -
    3. -

      - This series is centered at c=\pi. - Recall that 0!=1. - - \amp \infser[0] (-1)^{n+1} \frac{(x-\pi)^{2n}}{(2n)!} - \amp = -1+\frac{(x-\pi)^2}{2} - \frac{(x-\pi)^4}{24}+ \frac{(x-\pi)^6}{6!}-\frac{(x-\pi)^8}{8!}\ldots - -

      -
      4. -
    -

    -     - -     - -

    - We introduced power series as a type of function, - where a value of x is given and the sum of a series is returned. - Of course, not every series converges. - For instance, in part 1 of , - we recognized the series \ds \infser[0] x^n as a geometric series in x. - - states that this series converges only when \abs{x}\lt 1. -

    - -

    - This raises the question: - For what values of x will a given power series converge?, - which leads us to a theorem and definition. -

    - - - Convergence of Power Series - -

    - Let a power series \ds \infser[0] a_n(x-c)^n be given. - Then one of the following is true: - convergenceof power series - power seriesconvergence - seriespower - -

    - -

    -

      -
    1. -

      - The series converges only at x=c. -

      -
      2. - -
    2. -

      - There is an R \gt 0 such that the series converges for all x in - - (c-R,c+R) and diverges for all x\lt c-R and x \gt c+R. -

      -
      3. - -
    3. -

      - The series converges for all x. -

      -
      4. -
    -

    -     -     - - -

    - The value of R is important when understanding a power series, - hence it is given a name in the following definition. - Also, note that part 2 of - makes a statement about the interval (c-R,c+R), - but the not the endpoints of that interval. - A series may/may not converge at these endpoints. -

    - - - Radius and Interval of Convergence - -

    -

      -
    1. -

      - The number R given in - is the radius of convergence of a given series. - When a series converges for only x=c, - we say the radius of convergence is 0, , R=0. - When a series converges for all x, - we say the series has an infinite radius of convergence, - , R=\infty. -

      -
      2. - -
    2. -

      - The interval of convergence - is the set of all values of x for which the series converges. - - convergenceradius of - convergenceinterval of - radius of convergence - interval of convergence - seriesradius of convergence - seriesinterval of convergence -

      -
      3. -
    -

    -     -     - -

    - To find the interval of convergence, we start by using the ratio test to find the radius of convergence R. - If 0\lt R\lt\infty, we know the series converges on (c-R,c+R), - and it remains to check for convergence at the endpoints. -

    - -

    - Given \infser[0] a_n(x-c)^n we apply the ratio test to \infser[0]\abs{a_n(x-c)^n} - since the ratio test requires positive terms. - We find - - \lim_{n\to\infty}\frac{\abs{a_{n+1}(x-c)^{n+1}}}{\abs{a_n(x-c)^n}} = L\abs{x-c} - , - where L=\lim_{n\to\infty}\frac{\abs{a_{n+1}}}{\abs{a_n}}. - It follows that the series converges absolutely (and therefore converges) - for any x such that L\abs{x-c}\lt 1; - that is, for x in \left(c-\frac1L,c+\frac1L\right). -

    - -

    - On the other hand, suppose for some x that L\abs{x-c}\gt 1. - Then, for sufficiently large n, \abs{a_{n+1}}\gt \abs{a_n}. - This means that the terms of \infser[0] a_n(x-c)^n are growing in absolute value, - and therefore cannot converge to zero. This means that the series diverges, by . -

    - -

    - From the above observations, it follows that R=\frac{1}{L} must be the radius of convergence. -

    - - - Determining the Radius and Interval of Convergence - -

    - Given the power series \ds \infser[0] a_n(x-c)^n, - apply the ratio test to the series \ds \infser[0]\abs{a_n (x-c)^n}. - The result will be L\abs{x-c}, where \ds L=\lim_{n\to\infty}\frac{\abs{a_{n+1}}}{\abs{a_n}}. -

    - -

    -

      -
    1. -

      - If L=0, then the power series converges for every x - by the ratio test, since L\abs{x-c}=0\lt 1. -

      -
      2. - -
    2. -

      - If L=\infty, then power series converges only when x=c. -

      -
      3. - -
    3. -

      - If 0\lt L\lt \infty, then R=1/L is the radius of convergence: - by the ratio test, the series converges when \abs{x-c}\lt R. -

      - -

      - To determine the interval of convergence, plug the endpoints (x=c-R and x=c+R) - into the power series, and test the resulting series for convergence. - If the series converges, we include the endpoint. If it diverges, we exclude the endpoint. -

      -
      4. -
    -

    -     -     - - - -

    - - allows us to find the radius of convergence R of a series by applying the Ratio Test - (or any applicable test) - to the absolute value of the terms of the series. - We practice this in the following example. -

    - - - Determining the radius and interval of convergence - -

    - Find the radius and interval of convergence for each of the following series: -

    - -

    -

      -
    1. -

      - \ds \infser[0] \frac{x^n}{n!} -

      -
      2. - -
    2. -

      - \ds \infser (-1)^{n+1}\frac{x^n}{n} -

      -
      3. - -
    3. -

      - \ds \infser[0] 2^n(x-3)^n -

      -
      4. - -
    4. -

      - \ds \infser[0] n!x^n -

      -
      5. -
    -

    -     - -

    -

      -
    1. -

      - We apply the Ratio Test to the series \ds \infser \abs{\frac{x^n}{n!}}: - - \lim_{n\to\infty} \frac{\abs{x^{n+1}/(n+1)!}}{\abs{x^n/n!}} \amp = \lim_{n\to\infty} \abs{\frac{x^{n+1}}{x^n}\cdot\frac{n!}{(n+1)!}} - \amp = \lim_{n\to\infty} \abs{\frac x{n+1}} - \amp = 0 \text{ for all } x - . - The Ratio Test shows us that regardless of the choice of x, - the series converges. - Therefore the radius of convergence is R=\infty, - and the interval of convergence is (-\infty,\infty). -

      -
      2. - -
    2. -

      - We apply the Ratio Test to the series \ds \infser \abs{(-1)^{n+1}\frac{x^n}{n}} = \infser \abs{\frac{x^n}{n}}: - - \lim_{n\to\infty} \frac{\abs{x^{n+1}/(n+1)}}{\abs{x^n/n}} \amp = \lim_{n\to\infty} \abs{\frac{x^{n+1}}{x^n}\cdot \frac{n}{n+1}} - \amp = \lim_{n\to\infty} (\frac{n}{n+1})\abs{x} - \amp = \abs{x} - . - The Ratio Test states a series converges if the limit of \abs{a_{n+1}/a_n} = L\lt 1. - We found the limit above to be \abs{x}; - therefore, the power series converges when \abs{x} \lt 1, - or when x is in (-1,1). - Thus the radius of convergence is R=1. - - - To determine the interval of convergence, we need to check the endpoints of (-1,1). When x=-1, - we have the opposite of the Harmonic Series: - - \infser (-1)^{n+1}\frac{(-1)^n}{n} \amp = \infser \frac{-1}{n} - \amp = -\infty - . - The series diverges when x=-1. - - When x=1, we have the series \ds \infser (-1)^{n+1}\frac{(1)^n}{n}, - which is the Alternating Harmonic Series, which converges. Therefore the interval of convergence is (-1,1]. -

      -
      3. - -
    3. -

      - We apply the Ratio Test to the series \ds\infser[0] \abs{2^n(x-3)^n}: - - \lim_{n\to\infty} \frac{\abs{ 2^{n+1}(x-3)^{n+1}}}{\abs{2^n(x-3)^n}} \amp = \lim_{n\to\infty} \abs{\frac{2^{n+1}}{2^n}\cdot\frac{(x-3)^{n+1}}{(x-3)^n}} - \amp =\lim_{n\to\infty} \abs{2(x-3)} - . - According to the Ratio Test, - the series converges when \abs{2(x-3)}\lt 1 \implies \abs{x-3} \lt 1/2. - The series is centered at 3, and x must be within 1/2 of 3 in order for the series to converge. - Therefore the radius of convergence is R=1/2, - and we know that the series converges absolutely for all x in (3-1/2,3+1/2) = (2.5, 3.5). - We check for convergence at the endpoints to find the interval of convergence. - When x=2.5, we have: - - \infser[0] 2^n(2.5-3)^n \amp = \infser[0] 2^n(-1/2)^n - \amp =\infser[0] (-1)^n - , - which diverges. - A similar process shows that the series also diverges at x=3.5. - Therefore the interval of convergence is (2.5, 3.5). -

      -
      4. - -
    4. -

      - We apply the Ratio Test to \ds \infser[0] \abs{n!x^n}: - - \lim_{n\to\infty} \frac{\abs{ (n+1)!x^{n+1}}}{\abs{n!x^n}} \amp = \lim_{n\to\infty} \abs{(n+1)x} - \amp = \infty\, \text{ for all \(x\), except \(x=0\). } - - The Ratio Test shows that the series diverges for all x except x=0. - Therefore the radius of convergence is R=0. -

      -
      5. -
    -

    -     - -     - - - -

    - We can use a power series to define a function: - - f(x) = \infser[0] a_nx^n - - where the domain of f is a subset of the interval of convergence of the power series. - One can apply calculus techniques to such functions; - in particular, we can find derivatives and antiderivatives. -

    - - - Derivatives and Indefinite Integrals of Power Series Functions - -

    - Let \ds f(x) = \infser[0] a_n(x-c)^n be a function defined by a power series, - with radius of convergence R. -

    - -

    -

      -
    1. -

      - f(x) is continuous and differentiable on (c-R,c+R). -

      -
      2. - -
    2. -

      - \ds \fp(x) = \infser a_n\cdot n\cdot (x-c)^{n-1}, - with radius of convergence R. -

      -
      3. - -
    3. -

      - \ds \int f(x)\, dx = C+\infser[0] a_n\frac{(x-c)^{n+1}}{n+1}, - with radius of convergence R. -

      -
      4. -
    - seriespower!derivatives and integrals - integrationof power series - derivativepower series - power seriesderivatives and integrals - -

    -     -     - - -

    - A few notes about : -

    - -

    -

      -
    1. -

      - The theorem states that differentiation and integration do not change the radius of convergence. - It does not state anything about the - interval of convergence. - They are not always the same. -

      -
      2. - -
    2. -

      - Notice how the summation for \fp(x) starts with n=1. - This is because the constant term a_0 of f(x) becomes 0 through differentiation. -

      -
      3. - -
    3. -

      - Differentiation and integration are simply calculated term-by-term using the Power Rules. -

      -
      4. -
    -

    - - - Derivatives and indefinite integrals of power series - -

    - Let \ds f(x) = \infser[0] x^n. - Find \fp(x) and \ds F(x) =\int f(x)\, dx, - along with their respective intervals of convergence. -

    -     - -

    - We find the derivative and indefinite integral of f(x), - following . -

    - -

    -

      -
    1. -

      - - \fp(x) \amp = \infser nx^{n-1} = 1+2x+3x^2+4x^3+\cdots - \amp = \infser[0](n+1)x^n - . - In , - we recognized that \ds \infser[0] x^n is a geometric series in x. - We know that such a geometric series converges when \abs{x}\lt 1; - that is, the interval of convergence is (-1,1). - To determine the interval of convergence of \fp(x), - we consider the endpoints of (-1,1): - - \fp(-1) = 1-2+3-4+\cdots, \text{ which diverges. } - - - \fp(1) = 1+2+3+4+\cdots, \text{ which diverges. } - - Therefore, the interval of convergence of \fp(x) is (-1,1). -

      -
      2. - -
    2. -

      - \ds F(x) = \int f(x)\, dx = C+\infser[0] \frac{x^{n+1}}{n+1} = C+ x+\frac{x^2}{2}+\frac{x^3}3+\cdots - - To find the interval of convergence of F(x), - we again consider the endpoints of (-1,1): - - F(-1) = C-1+1/2-1/3+1/4+\cdots - - The value of C is irrelevant; - notice that the rest of the series is an Alternating Series that whose terms converge to 0. - By the Alternating Series Test, - this series converges. (In fact, - we can recognize that the terms of the series after C are the opposite of the Alternating Harmonic Series. - We can thus say that F(-1) = C-\ln(2).) - - F(1) = C+1+1/2+1/3+1/4+\cdots - - Notice that this summation is C\, + the Harmonic Series, - which diverges. - Since F converges for x=-1 and diverges for x=1, - the interval of convergence of F(x) is [-1,1). -

      -
      3. -
    -

    -     - -     - -

    - The previous example showed how to take the derivative and indefinite integral of a power series without motivation for why we care about such operations. - We may care for the sheer mathematical enjoyment that we can, - which is motivation enough for many. - However, we would be remiss to not recognize that we can learn a great deal from taking derivatives and indefinite integrals. -

    - -

    - Recall that \ds f(x) = \infser[0] x^n in is a geometric series. - According to , - this series converges to 1/(1-x) when \abs{x}\lt 1. - Thus we can say - - f(x) = \infser[0] x^n = \frac 1{1-x}, \text{ on } (-1,1) - . -

    - -

    - Integrating the power series, - (as done in ,) - we find - - F(x) = C_1+\infser[0] \frac{x^{n+1}}{n+1} - , - while integrating the function f(x) = 1/(1-x) gives - - F(x) = -\ln\abs{1-x} + C_2 - . -

    - -

    - Equating Equations and , we have - - F(x) = C_1+\infser[0] \frac{x^{n+1}}{n+1} = -\ln\abs{1-x} + C_2 - . -

    - -

    - Letting x=0, we have F(0) = C_1 = C_2. - This implies that we can drop the constants and conclude - - \infser[0] \frac{x^{n+1}}{n+1} = -\ln\abs{1-x} - . -

    - -

    - We established in - that the series on the left converges at x=-1; - substituting x=-1 on both sides of the above equality gives - - -1+\frac12-\frac13+\frac14-\frac15+\cdots = -\ln(2) - . -

    - -

    - On the left we have the opposite of the Alternating Harmonic Series; - on the right, we have -\ln(2). - We conclude that - - 1-\frac12+\frac13-\frac14+\cdots = \ln(2) - . -

    - -

    - Important: We stated in - (in ) - that the Alternating Harmonic Series converges to \ln(2), - and referred to this fact again in - of . - However, we never gave an argument for why this was the case. - The work above finally shows how we conclude that the Alternating Harmonic Series converges to \ln(2). - - Alternating Harmonic Series -

    - -

    - We use this type of analysis in the next example. -

    - - - Analyzing power series functions - -

    - Let \ds f(x) = \infser[0] \frac{x^n}{n!}. - Find \ds \fp(x) and \ds \int f(x)\, dx, - and use these to analyze the behavior of f(x). -

    -     - -

    - We start by making two notes: - first, in , - we found the interval of convergence of this power series is (-\infty,\infty). - Second, we will find it useful later to have a few terms of the series written out: - - \infser[0] \frac{x^n}{n!} = 1 + x + \frac{x^2}2+\frac{x^3}{6} + \frac{x^4}{24} +\cdots - -

    - -

    - We now find the derivative: - - \fp(x) \amp = \infser n\frac{x^{n-1}}{n!} - \amp =\infser \frac{x^{n-1}}{(n-1)!} = 1+x+\frac{x^2}{2!}+\cdots. - Since the series starts at n=1 and each term refers to (n-1), we can re-index the series starting with n=0: - \amp = \infser[0] \frac{x^{n}}{n!} - \amp = f(x) - . -

    - -

    - We found the derivative of f(x) is f(x). - The only functions for which this is true are of the form y=ce^x for some constant c. - As f(0) = 1 - (see Equation), - c must be 1. - Therefore we conclude that - - f(x) = \infser[0] \frac{x^n}{n!} = e^x - - for all x. -

    - -

    - We can also find \ds \int f(x)\, dx: - - \int f(x)\, dx \amp = C+\infser[0] \frac{x^{n+1}}{n!(n+1)} - \amp = C+ \infser[0] \frac{x^{n+1}}{(n+1)!} - -

    - -

    - We write out a few terms of this last series: - - C+ \infser[0] \frac{x^{n+1}}{(n+1)!} = C+ x+ \frac{x^2}2+\frac{x^3}{6}+\frac{x^4}{24}+\cdots - -

    - -

    - The integral of f(x) differs from f(x) only by a constant, - again indicating that f(x) = e^x. -

    -     - -     - -

    - - and the work following - established relationships between a power series function and regular - functions that we have dealt with in the past. - In general, given a power series function, it is difficult - (if not impossible) - to express the function in terms of elementary functions. - We chose examples where things worked out nicely. -

    - -

    - In this section's last example, - we show how to solve a simple differential equation with a power series. -

    - - - Solving a differential equation with a power series - -

    - Give the first 4 terms of the power series solution to \yp = 2y, - where y(0) = 1. -

    -     - -

    - The differential equation \yp = 2y describes a function y=f(x) where the derivative of y is twice y and y(0)=1. - This is a rather simple differential equation; - with a bit of thought one should realize that if y=Ce^{2x}, - then \yp = 2Ce^{2x}, and hence \yp = 2y. - By letting C=1 we satisfy the initial condition of y(0)=1. -

    - -

    - Let's ignore the fact that we already know the solution and find a power series function that satisfies the equation. - The solution we seek will have the form - - f(x) = \infser[0] a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+\cdots - - for unknown coefficients a_n. - We can find \fp(x) using : - - \fp(x) = \infser a_n\cdot n\cdot x^{n-1} = a_1+2a_2x+3a_3x^2+4a_4x^3\cdots - . -

    - -

    - Since \fp(x) = 2f(x), we have - - a_1+2a_2x+3a_3x^2+4a_4x^3\cdots \amp = 2\big(a_0+a_1x+a_2x^2+a_3x^3+\cdots\big) - \amp =2a_0+2a_1x+2a_2x^2+2a_3x^3+\cdots - -

    - -

    - The coefficients of like powers of x must be equal, - so we find that - - a_1 = 2a_0, 2a_2 = 2a_1, 3a_3 = 2a_2, 4a_4 = 2a_3, \text{ etc. } - -

    - -

    - The initial condition y(0) = f(0) = 1 indicates that a_0 = 1; - with this, we can find the values of the other coefficients: - - a_0 = 1 \text{ and } a_1=2a_0 \amp \Rightarrow a_1 = 2; - a_1 = 2 \text{ and } 2a_2 = 2a_1 \amp \Rightarrow a_2=4/2 =2; - a_2=2 \text{ and } 3a_3 = 2a_2 \amp \Rightarrow a_3=8/(2\cdot3)=4/3; - a_3=4/3 \text{ and } 4a_4 = 2a_3 \amp \Rightarrow a_4 =16/(2\cdot3\cdot4)= 2/3 - . -

    - -

    - Thus the first 5 terms of the power series solution to the differential equation \yp=2y is - - f(x) = 1+ 2x+2x^2 + \frac43x^3+\frac23x^4+\cdots - -

    - -

    - In , - as we study Taylor Series, - we will learn how to recognize this series as describing y=e^{2x}. -

    -     -     - -

    - Our last example illustrates that it can be difficult to recognize an elementary function by its power series expansion. - It is far easier to start with a known function, - expressed in terms of elementary functions, - and represent it as a power series function. - One may wonder why we would bother doing so, - as the latter function probably seems more complicated. - In the next two sections, - we show both how to do this and why - such a process can be beneficial. -

    - - - - Terms and Concepts - - - -

    - We adopt the convention that x^0=, - regardless of the value of x. -

    -     - - - - - - - - -

    - We define x^0=1 (even when x=0). -

    -     - -     - - - - -

    - What is the difference between the radius of convergence and the interval of convergence? -

    -     - - - -

    - The radius of convergence is a value - R such that a power series, centered at x=c, - converges for all values of x in (c-R,c+R). - The interval of convergence is an interval - on which the power series converges; - it may differ from (c-R,c+R) only at the endpoints. -

    -     - -     - - - - -

    - If the radius of convergence of - \ds\infser[0] a_nx^n is 5, what is the radius of convergence of \ds\infser n\cdot a_nx^{n-1}? -

    -     - -

    - It is still 5: the derivative of a power series has the same radius of convergence as the original power series. -

    -     - -     - - - - -

    - If the radius of convergence of - \ds\infser[0] a_nx^n is 5, - then the radius of convergence of \ds\infser[0] (-1)^na_nx^n is - - . -

    -     - - - - - - - - -

    - The radius of convergence will not change, since it is measured using absolute convergence. - However, it's possible that the endpoints of the interval of convergence might change. -

    -     -     -     -     - - -     -     - - Problems - - - -

    - Write out the sum of the first 5 terms of the given power series. -

    -     - - - - -

    - \ds\infser[0] 2^nx^n -

    -     - -

    - 1+2x+4x^2+8x^3+16x^4 -

    -     - -     - - - - -

    - \ds\infser \frac{1}{n^2}x^n -

    -     - -

    - x+\frac{x^2}{4}+\frac{x^3}{9}+\frac{x^4}{16}+\frac{x^5}{25} -

    -     - -     - - - - -

    - \ds\infser[0] \frac{1}{n!}x^n -

    -     - -

    - 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24} -

    -     - -     - - - - -

    - \ds\infser[0] \frac{(-1)^n}{(2n)!}x^{2n} -

    -     - -

    - 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320} -

    -     - -     - -     - - - -

    - A power series is given. -

    - -

    -

      -
    1. -

      - Find the radius of convergence. -

      -
      2. - -
    2. -

      - Find the interval of convergence. -

      -
      3. -
    -

    -     - - - - -

    - \ds\infser[0] \frac{(-1)^{n+1}}{n!}x^{n} -

    -     - -

    -

      -
    1. -

      - R=\infty -

      -
      2. - -
    2. -

      - (-\infty,\infty) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] nx^{n} -

    -     - -

    -

      -
    1. -

      - R=1 -

      -
      2. - -
    2. -

      - (-1,1) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser \frac{(-1)^n(x-3)^{n}}{n} -

    -     - -

    -

      -
    1. -

      - R=1 -

      -
      2. - -
    2. -

      - (2,4] -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] \frac{(x+4)^{n}}{n!} -

    -     - -

    -

      -
    1. -

      - R=\infty -

      -
      2. - -
    2. -

      - (-\infty,\infty) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] \frac{x^{n}}{2^n} -

    -     - -

    -

      -
    1. -

      - R=2 -

      -
      2. - -
    2. -

      - (-2,2) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] \frac{(-1)^n(x-5)^{n}}{10^n} -

    -     - -

    -

      -
    1. -

      - R=10 -

      -
      2. - -
    2. -

      - (-5,15) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] 5^n(x-1)^{n} -

    -     - -

    -

      -
    1. -

      - R=1/5 -

      -
      2. - -
    2. -

      - (4/5,6/5) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] (-2)^nx^{n} -

    -     - -

    -

      -
    1. -

      - R=1/2 -

      -
      2. - -
    2. -

      - (-1/2,1/2) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] \sqrt{n}x^{n} -

    -     - -

    -

      -
    1. -

      - R=1 -

      -
      2. - -
    2. -

      - (-1,1) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] \frac{n}{3^n}x^{n} -

    -     - -

    -

      -
    1. -

      - R=3 -

      -
      2. - -
    2. -

      - (-3,3) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] \frac{3^n}{n!}(x-5)^{n} -

    -     - -

    -

      -
    1. -

      - R=\infty -

      -
      2. - -
    2. -

      - (-\infty,\infty) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] (-1)^nn!(x-10)^{n} -

    -     - -

    -

      -
    1. -

      - R=0 -

      -
      2. - -
    2. -

      - x=10 -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser \frac{x^n}{n^2} -

    -     - -

    -

      -
    1. -

      - R=1 -

      -
      2. - -
    2. -

      - [-1,1] -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser \frac{(x+2)^n}{n^3} -

    -     - -

    -

      -
    1. -

      - R=1 -

      -
      2. - -
    2. -

      - [-3,-1] -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] n!\left(\frac x{10}\right)^n -

    -     - -

    -

      -
    1. -

      - R=0 -

      -
      2. - -
    2. -

      - x=0 -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] n^2\left(\frac{x+4}{4}\right)^n -

    -     - -

    -

      -
    1. -

      - R=4 -

      -
      2. - -
    2. -

      - x=(-8,0) -

      -
      3. -
    -

    -     - -     - -     - - - -

    - A function \ds f(x) = \infser[0] a_nx^n is given. -

    - -

    -

      -
    1. -

      - Give a power series for \fp(x) and its interval of convergence. -

      -
      2. - -
    2. -

      - Give a power series for \int f(x)\, dx and its interval of convergence. -

      -
      3. -
    -

    -     - - - - -

    - \ds\infser[0] nx^n -

    -     - -

    -

      -
    1. -

      - \ds \fp(x) = \infser n^2x^{n-1}; (-1,1) -

      -
      2. - -
    2. -

      - \ds \int f(x)\, dx = C+\infser[0] \frac{n}{n+1}x^{n+1}; - (-1,1) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser \frac{x^n}{n} -

    -     - -

    -

      -
    1. -

      - \ds \fp(x) = \infser x^{n-1}; (-1,1) -

      -
      2. - -
    2. -

      - \ds \int f(x)\, dx = C+\infser \frac{1}{n(n+1)}x^{n+1}; - [-1,1] -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] \left(\frac{x}{2}\right)^n -

    -     - -

    -

      -
    1. -

      - \ds \fp(x) = \infser \frac{n}{2^n}x^{n-1}; (-2,2) -

      -
      2. - -
    2. -

      - \ds \int f(x)\, dx = C+\infser[0] \frac{1}{(n+1)2^n}x^{n+1}; - [-2,2) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] (-3x)^n -

    -     - -

    -

      -
    1. -

      - \ds \fp(x) = \infser n(-3)^nx^{n-1}; (-1/3,1/3) -

      -
      2. - -
    2. -

      - \ds \int f(x)\, dx = C+\infser[0] \frac{(-3)^n}{n+1}x^{n+1}; - (-1/3,1/3] -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] \frac{(-1)^nx^{2n}}{(2n)!} -

    -     - -

    -

      -
    1. -

      - \ds \fp(x) = \infser \frac{(-1)^nx^{2n-1}}{(2n-1)!} =\infser[0] \frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}; - (-\infty,\infty) -

      -
      2. - -
    2. -

      - \ds \int f(x)\, dx = C+\infser[0] \frac{(-1)^nx^{2n+1}}{(2n+1)!}; - (-\infty,\infty) -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds\infser[0] \frac{(-1)^nx^{n}}{n!} -

    -     - -

    -

      -
    1. -

      - \ds \fp(x) = \infser \frac{(-1)^nx^{n-1}}{(n-1)!} =\infser[0] \frac{(-1)^{n+1}x^{n}}{n!}; - (-\infty,\infty) -

      -
      2. - -
    2. -

      - \ds \int f(x)\, dx = C+\infser[0] \frac{(-1)^{n}x^{n+1}}{(n+1)!}; - (-\infty,\infty) -

      -
      3. -
    -

    -     - -     - -     - - - -

    - Give the first 5 terms of the series that is a solution to the given differential equation. -

    -     - - - - -

    - y\,'=3y,y(0)=1 -

    -     - -

    - 1+3x+\frac92x^2+\frac92x^3+\frac{27}{8}x^4 -

    -     - -     - - - - -

    - y\,'=5y,y(0)=5 -

    -     - -

    - 5+25x+\frac{125}2x^2+\frac{625}{6}x^3+\frac{3125}{24}x^4 -

    -     - -     - - - - -

    - y\,'=y^2,y(0)=1 -

    -     - -

    - 1+x+x^2+x^3+x^4 -

    -     - -     - - - - -

    - y\,'=y+1,y(0)=1 -

    -     - -

    - 1+2x+x^2+\frac13{x^3}+\frac1{12}{x^4} -

    -     - -     - - - - -

    - y\,''=-y,y(0)=0, y\,'(0) = 1 -

    -     - -

    - 0+x+0x^2-\frac16x^3+0x^4 -

    -     - -     - - - - -

    - y\,''=2y,y(0)=1, y\,'(0) = 1 -

    -     - -

    - 1+x+x^2+\frac13x^3+\frac16x^4 -

    -     - -     - -     -     -     -
    -
    - Taylor Polynomials - - - -

    - Consider a function y=f(x) and a point \big(c,f(c)\big). - The derivative, \fp(c), - gives the instantaneous rate of change of f at x=c. - Of all lines that pass through the point \big(c,f(c)\big), - the line that best approximates f at this point is the tangent - line; that is, the line whose slope - (rate of change) - is \fp(c). -

    - -

    - In , - we see a function y=f(x) graphed. - The table in - shows that f(0)=2 and \fp(0) = 1; - therefore, the tangent line to f at x=0 is - p_1(x) = 1(x-0)+2 = x+2. - The tangent line is also given in the figure. - Note that near x=0, - p_1(x) \approx f(x); - that is, the tangent line approximates f well. -

    - - - -
    - - A graph of f(x) and its tangent line at 0 - - The graph of a generic function is shown, along with the tangent line to the graph at x=0. - -

    - The image shows the graph of a function f(x) and its tangent line at (0,f(0)). - The precise features of the graph of f(x) are unimportant for this illustration. - What is relevant is that points on the tangent line, which is the graph of a linear function labeled as p_1(x), - are close to the corresponding points on the graph of f(x), near the point (0,f(0)). - In other words, the image illustrates the fact that when x is close to zero, - the value of p_1(x) is close to the value of f(x). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-5.5,ymax=8.5, - xmin=-4.2,xmax=4.4 - ] - - \addplot [firstcurvestyle,infiniteleft,domain=-4:2,samples=60] {exp(x)*sin(deg(x))*cos(deg(x))+2}; - \addplot [firstcurvestyle,domain=2:3] {exp(x)*sin(deg(x))*cos(deg(x))+2}; - \addplot [firstcurvestyle,infiniteright,domain=3:3.4,samples=10] {exp(x)*sin(deg(x))*cos(deg(x))+2} node [pos=.8,left] { $y=f(x)$}; - - \addplot [secondcurvestyle,domain=-4:4] {x+2} node [pos=0,below right] { $y=p_1(x)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - Derivatives of f evaluated at 0 - -

    - - f(0) \amp = 2 \amp \fp''(0) \amp =-1 - \fp(0) \amp = 1 \amp f^{(4)}(0) \amp = -12 - \fpp(0) \amp = 2 \amp f^{(5)}(0) \amp = -19 - -

    -     - -
    -     - -

    - One shortcoming of this approximation is that the tangent line only - matches the slope of f; - it does not, for instance, match the concavity of f. - We can find a polynomial, p_2(x), - that does match the concavity near 0 without much difficulty, though. - The table in - gives the following information: - - f(0) = 2 \qquad \fp(0) = 1\qquad \fp'(0) = 2 - . -

    - -

    - Therefore, we want our polynomial p_2(x) to have these same - properties. That is, we need - - p_2(0) = 2 \qquad p_2'(0) = 1 \qquad p_2''(0) = 2 - . -

    - -

    - This is simply an initial-value problem. - We can solve this using the techniques first described in . - To keep p_2(x) as simple as possible, - we'll assume that not only p_2''(0)=2, - but that p_2''(x)=2. - That is, the second derivative of p_2 is constant, - meaning p_2 is a quadratic function. -

    - -

    - If p_2''(x) = 2, then - p_2'(x) = 2x+C for some constant C. - Since we have determined that p_2'(0) = 1, - we find that C=1 and so p_2'(x) = 2x+1. - Finally, we can compute p_2(x) = x^2+x+C. - Using our initial values, we know p_2(0) = 2 so C=2. - We conclude that p_2(x) = x^2+x+2. - This function is plotted with f in . -

    - - - -

    - We can repeat this approximation process by creating polynomials of - higher degree that match more of the derivatives of f at x=0. - In general, a polynomial of degree n can be created to match the - first n derivatives of f. - - shows p_4(x)= -x^4/2-x^3/6+x^2+x+2, - whose first four derivatives at 0 match those of f. (Using the table in , - start with p_4^{(4)}(x)=-12 and solve the related initial-value problem.) -

    - - - - - -
    - Plotting f, p_2 and p_4 - - - The graph of a function and two polynomials that approximate the function near x=0. - -

    - The graph of a function f(x) is shown. It is the same function as the first image in this section, - but again, the precise details of the graph are unimportant. -

    - -

    - Also shown are the graphs of two functions p_2(x) and p_4(x). - The function p_2(x) is quadratic, and its graph is a parabola that opens upward. - The function p_4(x) is a polynomial of degree 4. -

    - -

    - All three graphs intersect at the point (0,f(0)), - and the values of both p_2(x) and p_4(x) are close to the value of f(x) when x is close to 0. - Two observations are important: first, both of these polynomial graphs appear to lie more closely to the graph of f(x) - than the tangent line in the first image. Second, the graph of p_4(x) is a good approximation to f(x) - over a larger interval than the graph of p_2(x). -

    - -

    - In particular, the point (0,f(0)) appears to be a local minimum, - and there is a corresponding minimum in the graphs of both p_2(x) and p_4(x). - But the graph of f(x) also appears to have a local maximum near x=1. - Near x=1, the graph of p_2(x) separates from that of f(x): - the first continues to increase, while the second begins to decrease. - Near x=1, p_2(x) is no longer a good approximation to f(x). -

    - -

    - However, the graph of p_4(x) also has a maximum near x=1, - and p_4(x) appears to be a good approximation to f(x) at least until x=2. -

    -     - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-5.5,ymax=8.5, - xmin=-4.5,xmax=4.4 - ] - - \addplot+ [infinite,domain=-4:3.2,samples=120] {exp(x)*sin(deg(x))*cos(deg(x))+2} node [pos=0.6,left] { $y=f(x)$}; - \addplot+ [domain=-3:2] {x^2+x+2} node [pos=0.2, below left] { $y=p_2(x)$}; - \addplot+ [domain=-2:2] {-x^4/2-x^3/6+x^2+x+2} node [pos=0.1, below left] { $y=p_4(x)$}; - - \end{axis} - - \end{tikzpicture} - - - -
    - - - - - - - - - - - - - -

    - As we use more and more derivatives, - our polynomial approximation to f gets better and better. - In this example, the interval on which the approximation is - good gets bigger and bigger. - shows p_{13}(x); - we can visually affirm that this polynomial approximates f very - well on [-2,3]. - The polynomial p_{13}(x) is not - particularly nice. It is - - p_{13}(x)\amp=\frac{16901x^{13}}{6227020800}+\frac{13x^{12}}{1209600}-\frac{1321x^{11}}{39916800}-\frac{779x^{10}}{1814400} -\frac{359x^9}{362880} - \amp +\frac{x^8}{240}+\frac{139x^7}{5040}+\frac{11 x^6}{360}-\frac{19x^5}{120}-\frac{x^4}{2}-\frac{x^3}{6}+x^2+x+2 - . -

    - -
    - Plotting f and p_{13} - - - The graph of a function is shown, along with the graph of a degree 13 polynomial that closely approximates the function. - -

    - The graph of a function f(x) is shown. It is the same function as in the previous two images. - Also shown is the graph of a degree 13 polynomial p_{13}(x). - We can see that the values of f(x) and p_{13}(x) are very close over an interval - from -2 to 3. For x \lt -3, the graph of f(x) appears to approach a horizontal asymptote, - while the graph of p_{13}(x) approaches -\infty, so it ceases to be a good approximation to f(x) at this point. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-5.5,ymax=8.5, - xmin=-4.5,xmax=4.4 - ] - - \addplot+ [infinite,domain=-3.5:3.3,samples=120] {exp(x)*sin(deg(x))*cos(deg(x))+2} node [pos=0, above] { $y=f(x)$}; - \addplot+ [infinite,smooth] coordinates {(-3.52,-6.364)(-3.36,-2.334)(-3.2,-0.1706)(-3.04,0.9569)(-2.88,1.528)(-2.72,1.809)(-2.56,1.944)(-2.4,2.009)(-2.24,2.038)(-2.08,2.048)(-1.92,2.046)(-1.76,2.031)(-1.6,2.006)(-1.44,1.969)(-1.28,1.924)(-1.12,1.872)(-0.96,1.82)(-0.8,1.775)(-0.64,1.747)(-0.48,1.747)(-0.32,1.783)(-0.16,1.866)(0,2.)(0.16,2.185)(0.32,2.411)(0.48,2.662)(0.64,2.908)(0.8,3.112)(0.96,3.227)(1.12,3.202)(1.28,2.988)(1.44,2.546)(1.6,1.855)(1.76,0.9264)(1.92,-0.1926)(2.08,-1.406)(2.24,-2.565)(2.4,-3.474)(2.56,-3.891)(2.72,-3.542)(2.88,-2.133)(3.04,0.6461)(3.2,5.139)(3.36,11.79)} node [pos=0.2, right] {$p_{13}(x)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - The polynomials we have created are examples of - Taylor polynomials, - named after the British mathematician Brook Taylor who made important discoveries about such functions. - While we created the above Taylor polynomials by solving initial-value problems, - it can be shown that Taylor polynomials follow a general pattern that make their formation much more direct. - This is described in the following definition. -

    - - - - - Taylor Polynomial, Maclaurin Polynomial - -

    - Let f be a function whose first n derivatives exist at - x=c. - Taylor Polynomialdefinition - Maclaurin Polynomialdefinition - Maclaurin Polynomial|see{Taylor Polynomial} - -

      -
    1. -

      - The Taylor polynomial of degree n of f at - x=c is - - p_n(x) \amp = f(c) + \fp(c)(x-c) + \frac{\fpp(c)}{2!}(x-c)^2 - \amp \quad+\frac{\fp''(c)}{3!}(x-c)^3+\cdots+\frac{f\,^{(n)}(c)}{n!}(x-c)^n - . -

      -
      2. - -
    2. -

      - A special case of the Taylor polynomial is the Maclaurin - polynomial, where c=0. - That is, the Maclaurin polynomial of degree n of - f is - - p_n(x) = f(0) + \fp(0)x + \frac{\fpp(0)}{2!}x^2+\frac{\fp''(0)}{3!}x^3+\cdots+\frac{f\,^{(n)}(0)}{n!}x^n - . -

      -
      3. -
    -

    -     -     - - - - - - - - - -

    - We will practice creating Taylor and Maclaurin polynomials in the - following examples. -

    - - - Finding and using Maclaurin polynomials - -

    -

      -
    1. -

      - Find the nth Maclaurin polynomial for f(x) = e^x. -

      -
      2. - -
    2. -

      - Use p_5(x) to approximate the value of e. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - We start with creating a table of the derivatives of e^x - evaluated at x=0. - In this particular case, this is relatively simple, - as shown in . -

      - -
      - The derivatives of f(x)=e^x evaluated at x=0 - -

      - - f(x) \amp = e^x \amp f(0) \amp = 1 - \fp(x) \amp = e^x \amp \fp(0) \amp =1 - \fp'(x) \amp = e^x \amp \fp'(0) \amp =1 - \amp \vdots \amp \amp \vdots - f^{(n)}(x) \amp = e^x \amp f^{(n)}(0) \amp =1 - -

      -       - -
      - -

      - By the definition of the Maclaurin polynomial, we have - - p_n(x) \amp = f(0) + \fp(0)x + \frac{\fpp(0)}{2!}x^2+\frac{\fp''(0)}{3!}x^3+\cdots+\frac{f\,^{(n)}(0)}{n!}x^n - \amp = 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3 + \frac{1}{24}x^4 + \cdots + \frac{1}{n!}x^n - . -

      -
      2. - -
    2. -

      - Using our answer from part 1, we have - - e^x\approx p_5(x) = 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5 - . - To approximate the value of e, - note that e = e^1 = f(1) \approx p_5(1). - It is very straightforward to evaluate p_5(1): - - p_5(1) = 1+1+\frac12+\frac16+\frac1{24}+\frac1{120} = \frac{163}{60} \approx 2.71667 - . - A plot of f(x)=e^x and p_5(x) is given in - . - To 5 decimal places, the actual value of e is - 2.71828. So this approximation agrees to two decimal places. - -

      -
      3. -
    -

    -
    - A plot of f(x)=e^x and its 5^{th} degree - Maclaurin polynomial p_5(x) - - - The graph of the exponential function and its degree 5 Maclaurin polynomial. - -

    - The graph of f(x)=e^x is shown on the interval [-3,2]. - Also shown is the graph of p_5(x), the degree 5 Maclaurin polynomial of f(x). - For -2\lt x\lt 2, there is very little visible difference between the two graphs. - Near x=3 we can see that the two graphs begin to separate. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-3,ymax=11, - xmin=-3.75,xmax=2.9 - ] - - \addplot+ [infinite,domain=-3.2:2.2,samples=40] {e^x} node [pos=0,above right] { $y=f(x)$}; - \addplot+ [infinite,domain=-3.5:2.4,samples=40] {1+x+(1/2)*x^2+(1/6)*x^3+(1/24)*x^4+(1/120)*x^5} node [pos=0.4,below right] { $y=p_{5}(x)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -     - - - - - Finding and using Taylor polynomials - -

    -

      -
    1. -

      - Find the nth Taylor polynomial of y=\ln(x) at x=1. -

      -
      2. - -
    2. -

      - Use p_6(x) to approximate the value of \ln(1.5). -

      -
      3. - -
    3. -

      - Use p_6(x) to approximate the value of \ln(2). -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - We begin by creating a table of derivatives of \ln(x) - evaluated at x=1. - While this is not as straightforward as it was in the previous - example, a pattern does emerge (for n\ge 1), - as shown in . - - Notice in the table below that each time we take a derivative - (starting at the second derivative), - we apply the power rule and bring down - the exponent to multiply by the previous coefficent. - So the 6 in the 4^{th} derivative is actually - 1\cdot 2\cdot 3=3!. -

      - -
      - Derivatives of \ln(x) evaluated at x=1 - -

      - - f(x) \amp = \ln(x) \amp f(1) \amp = 0 - \fp(x) \amp = \frac1x \amp \fp(1) \amp = 1 - \fp'(x) \amp = -\frac{1}{x^2} \amp \fp'(1) \amp = -1 - \fp''(x) \amp = \frac{2}{x^3} \amp \fp''(1) \amp = 2 - f^{(4)}(x) \amp = -\frac{6}{x^4} \amp f^{(4)}(1) \amp = -6 - \amp \vdots \amp \amp \vdots - f^{(n)}(x) \amp = \amp f^{(n)}(1) \amp = - (-1)^{n+1}\amp\frac{(n-1)!}{x^n} \amp (-1)^{n+1}\amp(n-1)! - -

      -       - -
      - -

      - Notice that the coefficients alternate in sign starting at n=1. - Using , we have - - p_n(x) \amp = f(c) + \fp(c)(x-c) + \frac{\fpp(c)}{2!}(x-c)^2+\dots - \amp \dots\frac{\fp''(c)}{3!}(x-c)^3+\cdots+\frac{f\,^{(n)}(c)}{n!}(x-c)^n - \amp = 0 + \frac{0!}{1!}(x-1) - \frac{1!}{2!}(x-1)^2+\dots - \amp \dots\frac{2!}{3!}(x-1)^3+\cdots+\frac{(-1)^{n+1}\cdot (n-1)!}{n!}(x-1)^n - \amp = (x-1)-\frac12(x-1)^2+\frac13(x-1)^3-\dots - \amp \dots \frac14(x-1)^4+\cdots+\frac{(-1)^{n+1}}{n}(x-1)^n - . - Note how the coefficients of the (x-1) terms turn out to - be nice. -

      -
      2. - -
    2. -

      - We can compute p_6(x) using our work above: - - p_6(x) \amp = (x-1)-\frac12(x-1)^2+\frac13(x-1)^3 - \amp \quad\quad -\frac14(x-1)^4+\frac15(x-1)^5-\frac16(x-1)^6 - . - Since p_6(x) approximates \ln(x) well near x=1, - we approximate \ln(1.5) \approx p_6(1.5): - - p_6(1.5) \amp = (1.5-1)-\frac12(1.5-1)^2+\frac13(1.5-1)^3+\dots - \amp \dots -\frac14(1.5-1)^4 +\frac15(1.5-1)^5-\frac16(1.5-1)^6 - \amp =\frac{259}{640} - \amp \approx 0.404688 - . - This is a good approximation as a calculator shows that - \ln(1.5) \approx 0.4055. - below - plots y=\ln(x) with y=p_6(x). - We can see that \ln(1.5) \approx p_6(1.5). - - -

      -
      3. - -
    3. -

      - We approximate \ln 2 with p_6(2): - - p_6(2) \amp = (2-1)-\frac12(2-1)^2+\frac13(2-1)^3-\frac14(2-1)^4+\cdots - \amp \cdots +\frac15(2-1)^5-\frac16(2-1)^6 - \amp = 1-\frac12+\frac13-\frac14+\frac15-\frac16 - \amp = \frac{37}{60} - \amp \approx 0.616667 - . - This approximation is not terribly impressive: - a hand held calculator shows that \ln(2) \approx 0.693147. - The graph in - shows that p_6(x) provides less accurate approximations - of \ln(x) as x gets close to 0 or 2. - - Surprisingly enough, - even the 20th degree Taylor polynomial fails to - approximate \ln(x) for x\gt 2 very well, - as shown in . - We'll soon discuss why this is. - - - -

      -
      4. -
    -

    - -
    - A plot of y=\ln(x) and its 6th degree Taylor - polynomial at x=1 - - - A graph of the natural logarithm function and its degree 6 Taylor polynomial centered at x=1. - -

    - The graph of y=\ln(x) is shown on the interval (0,3). - Also shown is the graph of p_6(x), the degree 6 Taylor polynomial of \ln(x), - centered at x=1. - The graph of p_6(x) is very close to the graph of \ln(x) on the interval (0.5,1.5), - but does not approximate the logarithm very well outside of this interval. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-4.5,ymax=2.4, - xmin=-.5,xmax=3.2 - ] - - \addplot+ [infinite,domain=0.01:2.8,samples=101] {ln(x)} node [pos=1,above left] { $y=\ln(x)$}; - \addplot+ [infinite,domain=-0.22:2.8,samples=40] {(x-1)-.5*(x-1)^2+(1/3)*(x-1)^3-(1/4)*(x-1)^4+(1/5)*(x-1)^5-(1/6)*(x-1)^6} node [pos=1,left] { $y=p_{6}(x)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - A plot of y=\ln(x) and its 20^{th} degree - Taylor polynomial at x=1 - - - A graph of the natural logarithm function and its degree 20 Taylor polynomial centered at x=1. - -

    - The graph y=\ln(x) is shown again, this time with the degree 20 Taylor polynomial of \ln(x), centered at x=1. - Unlike with the example involving y=e^x, increasing the degree of the Taylor polynomial did not do much to improve - how well it approximates the logarithm. The approximation appears to be accurate over a slightly larger interval - than the degree 6 approximation, but there is not a significant difference between the two images. -

    -     - - \begin{tikzpicture} - \begin{axis}[ymax=3] - \addplot+[infinite,domain=-4.6:1.1,variable=t] ({e^t},{t}) node [pos=1,above left] {$y=\ln(x)$}; - \addplot+[infinite,domain=-0.1:2.3,smooth] {(x-1) - (x-1)^2/2 + (x-1)^3/3 - (x-1)^4/4 + (x-1)^5/5 - (x-1)^6/6 + (x-1)^7/7 - (x-1)^8/8 + (x-1)^9/9 - (x-1)^10/10 + (x-1)^11/11 - (x-1)^12/12 + (x-1)^13/13 - (x-1)^14/14 + (x-1)^15/15 - (x-1)^16/16 + (x-1)^17/17 - (x-1)^18/18 + (x-1)^19/19 - (x-1)^20/20} node [pos=0.8,below right] {$y=p_{20}(x)$}; - \end{axis} - \end{tikzpicture} - - - -
    -     -     - -     - - - -

    - Taylor polynomials are used to approximate functions f(x) in mainly - two situations: -

      -
    1. -

      - When f(x) is known, but perhaps - hard to compute directly. - For instance, we can define - the cosine of an angle as either the ratio of sides of a right - triangle (adjacent over hypotenuse) or using the definition - in terms of the unit circle. - However, neither of these provides a convenient way of computing - \cos(2). - A Taylor polynomial of sufficiently high degree can provide a - reasonable method of computing such values using only operations - usually hard-wired into a computer - (+, -, and \div). -

      -
      2. - -
    2. -

      - When f(x) is not known, - but information about its derivatives is known. - This occurs more often than one might think, - especially in the study of differential equations. -

      -
      3. -
    -

    - -

    - In both situations, a critical piece of information to have is - How good is my approximation? - If we use a Taylor polynomial to compute \cos(2), - how do we know how accurate the approximation is? -

    - - - -

    - We had the same problem when studying Numerical Integration. - - provided bounds on the error when using, - say, Simpson's Rule to approximate a definite integral. - These bounds allowed us to determine that, for instance, - using 10 subintervals provided an approximation within \pm 0.01 of the exact value. - The following theorem gives similar bounds for Taylor - (and hence Maclaurin) - polynomials. -

    - - - - - Taylor's Theorem - -

    -

      -
    1. -

      - Let f be a function whose - (n+1)\text{th } derivative exists on an interval I - and let c be in I. - Then, for each x in I, - there exists z_x between x and c such that - - f(x) = f(c) + \fp(c)(x-c) + \cdots - +\frac{f^{(n)}(c)}{n!}(x-c)^n+R_n(x) - , - where \ds R_n(x) = \frac{f^{(n+1)}(z_x)}{(n+1)!}(x-c)^{(n+1)}. - Taylor PolynomialTaylor's Theorem - Taylor's Theorem -

      -
      2. - -
    2. -

      - \ds \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{(x-c)^{(n+1)}}, - where z is in I. -

      -
      3. -
    -

    -     -     - - - - - - - -

    - The first part of Taylor's Theorem states that f(x) = p_n(x) + R_n(x), - where p_n(x) is the - nth order Taylor polynomial and R_n(x) is the remainder, - or error, in the Taylor approximation. - The second part gives bounds on how big that error can be. - If the (n+1)th derivative is large on I, - the error may be large; - if x is far from c, - the error may also be large. - However, the (n+1)! term in the denominator tends to ensure that - the error gets smaller as n increases. -

    - -

    - The following example computes error estimates for the approximations of - \ln(1.5) and \ln(2) made in . -

    - - - Finding error bounds of a Taylor polynomial - -

    - Use - to find error bounds when approximating - \ln(1.5) and \ln(2) with p_6(x), - the Taylor polynomial of degree 6 of f(x)=\ln(x) at x=1, - as calculated in . -

    -     - -

    -

      -
    1. -

      - We start with the approximation of - \ln(1.5) with p_6(1.5). - The theorem references an open interval I that contains - both x and c. - The smaller the interval we use the better; - it will give us a more accurate - (and smaller!) - approximation of the error. - We let I = (0.9,1.6), - as this interval contains both c=1 and x=1.5. - The theorem references \max\abs{f^{(n+1)}(z)}. - In our situation, this is asking - How big can the 7th derivative of - y=\ln(x) be on the interval (0.9,1.6)? - The seventh derivative is y = -6!/x^7. - The largest absolute value it attains on I is about 1506. - (There are no critical numbers of f^{(7)} in the interval - so we evaluate the endpoints: - f^{(7)}(0.9)\approx 1506 and f^{(7)}(1.6)\approx 27.) - In particular, - we are evaluating at x=1.5, so we let x=1.5. - Thus we can bound the error as: - - \abs{R_6(1.5)} \amp \leq \frac{\max\abs{f^{(7)}(z)}}{7!}\abs{(1.5-1)^7} - \amp \leq \frac{1506}{5040}\cdot\frac1{2^7} - \amp \approx 0.0023 - . - We computed p_6(1.5) = 0.404688; - using a calculator, we find \ln(1.5) \approx 0.405465, - so the actual error is about 0.000778, - which is less than our bound of 0.0023. - This affirms Taylor's Theorem; - the theorem states that our approximation would be within about - 2 thousandths of the actual value, - whereas the approximation was actually closer. - only gives an upper - bound on the error. -

      -
      2. - -
    2. -

      - We again find an interval I that contains both c=1 - and x=2; we choose I = (0.9,2.1). - The maximum value of the seventh derivative of f on this - interval is again about 1506 - (as the largest values come near x=0.9). - Thus - - \abs{ R_6(2)} \amp \leq \frac{\max\abs{f^{(7)}(z)}}{7!}\abs{(2-1)^7} - \amp \leq \frac{1506}{5040}\cdot1^7 - \amp \approx 0.30 - . - This bound is not as nearly as good as before. - Using the degree 6 Taylor polynomial at x =1 will bring - us within 0.3 of the correct answer. - As p_6(2)\approx 0.61667, - our error estimate guarantees that the actual value of - \ln(2) is somewhere between 0.31667 and 0.91667. - These bounds are not particularly useful. - In reality, our approximation was only off by about 0.07. - However, we are approximating ostensibly because we do not know - the real answer. - In order to be assured that we have a good approximation, - we would have to resort to using a polynomial of higher degree. -

      -
      3. -
    -

    -     - -     - -

    - We practice again. - This time, we use Taylor's theorem to find n that guarantees our - approximation is within a certain amount. -

    - - - Finding sufficiently accurate Taylor polynomials - -

    - Find n such that the nth Taylor polynomial of - f(x)=\cos(x) at x=0 approximates \cos(2) to - within 0.001 of the actual answer. - What is p_n(2)? -

    -     - -

    - Following Taylor's theorem, - we need bounds on the size of the derivatives of f(x)=\cos(x). - In the case of this trigonometric function, this is easy. - All derivatives of cosine are - \pm \sin(x) or \pm \cos(x). - In all cases, - these functions are never greater than 1 in absolute value. - We want the error to be less than 0.001. - To find the appropriate n, - consider the following inequalities: - - \frac{\max\abs{f^{(n+1)}(z)}}{(n+1)!}\abs{(2-0)^{(n+1)}} \amp \leq 0.001 - \frac1{(n+1)!}\cdot2^{(n+1)} \amp \leq 0.001 - . -

    - -

    - We find an n that satisfies this last inequality with - trial-and-error. - When n=8, - we have \ds \frac{2^{8+1}}{(8+1)!} \approx 0.0014; - when n=9, - we have \ds \frac{2^{9+1}}{(9+1)!} \approx 0.000282 \lt 0.001. - Thus we want to approximate \cos(2) with p_9(2). -

    - -

    - We now set out to compute p_9(x). - We again need a table of the derivatives of - f(x)=\cos(x) evaluated at x=0. - A table of these values is given in . -

    - - - -
    - A table of the derivatives of f(x)=\cos(x) evaluated at x=0 - -

    - - f(x) \amp = \cos(x) \amp f(0) \amp = 1 - f'(x) \amp = -\sin(x) \amp \fp(0) \amp = 0 - \fp'(x) \amp = -\cos(x) \amp \fp'(0) \amp = -1 - \fp''(x) \amp = \sin(x) \amp \fp''(0) \amp = 0 - f^{(4)}(x) \amp = \cos(x) \amp f^{(4)}(0) \amp =1 - f^{(5)}(x) \amp = -\sin(x) \amp f^{(5)}(0) \amp =0 - f^{(6)}(x) \amp = -\cos(x) \amp f^{(6)}(0) \amp =-1 - f^{(7)}(x) \amp = \sin(x) \amp f^{(7)}(0) \amp =0 - f^{(8)}(x) \amp = \cos(x) \amp f^{(8)}(0) \amp =1 - f^{(9)}(x) \amp = -\sin(x) \amp f^{(9)}(0) \amp =0 - -

    -     - -
    - - - -

    - Notice how the derivatives, evaluated at x=0, follow a certain - pattern. All the odd powers of x in the Taylor polynomial will - disappear as their coefficient is 0. - While our error bounds state that we need p_9(x), - our work shows that this will be the same as p_8(x). -

    - -

    - Since we are forming our polynomial at x=0, - we are creating a Maclaurin polynomial, and: - - p_8(x) \amp = f(0) + \fp(0)x + \frac{\fpp(0)}{2!}x^2 + \frac{\fp''(0)}{3!}x^3 + \cdots +\frac{f^{(8)}(0)}{8!}x^8 - \amp = 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+\frac{1}{8!}x^8 - . -

    - -

    - We finally approximate \cos(2): - - \cos(2) \approx p_8(2) = -\frac{131}{315} \approx -0.41587 - . -

    - -

    - Our error bound guarantee that this approximation is within - 0.001 of the correct answer. - Technology shows us that our approximation is actually within about - 0.0003 of the correct answer. -

    - -

    - - shows a graph of y=p_8(x) and y=\cos(x). - Note how well the two functions agree on about (-\pi,\pi). -

    - -
    - A graph of f(x)= \cos(x) and its degree 8 Maclaurin - polynomial - - - A graph of the cosine function is shown, along with its degree 8 Maclaurin polynomial approximation. - -

    - The graph y=\cos(x) is given on the interval [-5,5], - along with the graph of p_8(x), the degree 8 Maclaurin polynomial approximation of \cos(x). -

    - -

    - The image illustrates how well the Maclaurin polynomials approximate the cosine function. - With a degree 8 polynomial, there is little to no visible difference between the graphs over the interval (-\pi,\pi). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-5,-4,-3,-2,-1,1,2,3,4,5}, - ymin=-1.5,ymax=1.5, - xmin=-5.5,xmax=5.5 - ] - - \addplot+[infinite,domain=-5:5,samples=101] {cos(deg(x))} node[pos=0.6, right] { $y=cos(x)$}; - \addplot+[infinite,smooth] coordinates {(-5.,2.528)(-4.8,1.6)(-4.6,0.8894)(-4.4,0.343)(-4.2,-0.0768)(-4.,-0. - 3968)(-3.8,-0.6355)(-3.6,-0.8052)(-3.4,-0.9146)(-3.2,-0.9695)(-3.,-0. - 9748)(-2.8,-0.9345)(-2.6,-0.8532)(-2.4,-0.7357)(-2.2,-0.5878)(-2.,-0. - 4159)(-1.8,-0.2271)(-1.6,-0.02917)(-1.4,0.17)(-1.2,0.3624)(-1.,0.5403) - (-0.8,0.6967)(-0.6,0.8253)(-0.4,0.9211)(-0.2,0.9801)(0,1.)(0.2,0.9801) - (0.4,0.9211)(0.6,0.8253)(0.8,0.6967)(1.,0.5403)(1.2,0.3624)(1.4,0.17)( - 1.6,-0.02917)(1.8,-0.2271)(2.,-0.4159)(2.2,-0.5878)(2.4,-0.7357)(2.6,- - 0.8532)(2.8,-0.9345)(3.,-0.9748)(3.2,-0.9695)(3.4,-0.9146)(3.6,-0. - 8052)(3.8,-0.6355)(4.,-0.3968)(4.2,-0.0768)(4.4,0.343)(4.6,0.8894)(4. - 8,1.6)(5.,2.528)} node[pos=0.1,right] { $y=p_8(x)$} ; - - \end{axis} - - \end{tikzpicture} - - - - -
    - - -     - - -     - - - - - Finding and using Taylor polynomials - -

    -

      -
    1. -

      - Find the degree 4 Taylor polynomial, - p_4(x), for f(x)=\sqrt{x} at x=4. -

      -
      2. - -
    2. -

      - Use p_4(x) to approximate \sqrt{3}. -

      -
      3. - -
    3. -

      - Find bounds on the error when approximating \sqrt{3} - with p_4(3). -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - We begin by evaluating the derivatives of f at x=4. - This is done in . -

      - -
      - A table of the derivatives of f(x)=\sqrt{x} evaluated at x=4 - -

      - - f(x) \amp = \sqrt{x} \amp f(4) \amp = 2 - \fp(x) \amp = \frac{1}{2\sqrt{x}} \amp \fp(4) \amp = \frac14 - \fp'(x) \amp = \frac{-1}{4x^{3/2}} \amp \fp'(4) \amp = \frac{-1}{32} - \fp''(x) \amp = \frac{3}{8x^{5/2}} \amp \fp''(4) \amp = \frac{3}{256} - f^{(4)}(x) \amp = \frac{-15}{16x^{7/2}} \amp f^{(4)}(4) \amp = \frac{-15}{2048} - -

      -       - -
      - - - -

      - These values allow us to form the Taylor polynomial p_4(x): - - p_4(x) = 2 \amp + \frac14(x-4) +\frac{-1/32}{2!}(x-4)^2+\dots - \amp \dots \frac{3/256}{3!}(x-4)^3+\frac{-15/2048}{4!}(x-4)^4 - . -

      -
      2. - -
    2. -

      - As p_4(x) \approx \sqrt{x} near x=4, - we approximate \sqrt{3} with p_4(3) = 1.73212. -

      -
      3. - -
    3. -

      - To find a bound on the error, - we need an open interval that contains x=3 and x=4. - We set I = (2.9,4.1). - The largest value the fifth derivative of - f(x)=\sqrt{x} takes on this interval is near x=2.9, - at about 0.0273. (We often graph the - (n+1)^{th} derivative to find its extrema. - In this case is f^{(5)}(x)=105/(32x^{9/2}) is always - decreasing, so the maximum occurs at 2.9.) Thus - - \abs{R_4(3)} \leq \frac{0.0273}{5!}\abs{(3-4)^5} \approx 0.00023 - . - This shows our approximation is accurate to at least the first 2 - places after the decimal. - (It turns out that our approximation is actually accurate to 4 - places after the decimal.) - A graph of f(x)=\sqrt x and p_4(x) is given in - . - Note how the two functions are nearly indistinguishable on - (2,7). - - - - -

      -
      4. -
    -

    - -
    - A graph of f(x)=\sqrt{x} and its degree 4 Taylor - polynomial at x=4 - - - The graph of the square root function and its degree 4 Taylor polynomial, centered at x=4. - -

    - The graph y=\sqrt{x} is shown, for x in the interval [0,10]. - Also shown is the degree 4 Taylor polynomial p_4(x), centered at x=4. - The Taylor polynomial appears to approximate \sqrt{x} very well over the interval (2,7). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=3.6, - xmin=-1,xmax=11.2 - ] - - \addplot+ [rightarrow,domain=0:3.16,samples=40] ({x^2},{x}) node [pos=1,above left] { $y=\sqrt{x}$}; - \addplot+ [rightarrow] coordinates {(0,0.5469)(0.2,0.6536)(0.4,0.7551)(0.6,0.8518)(0.8,0.944)(1.,1.032)(1. - 2,1.116)(1.4,1.196)(1.6,1.273)(1.8,1.346)(2.,1.417)(2.2,1.485)(2.4,1. - 55)(2.6,1.613)(2.8,1.673)(3.,1.732)(3.2,1.789)(3.4,1.844)(3.6,1.897)( - 3.8,1.949)(4.,2.)(4.2,2.049)(4.4,2.098)(4.6,2.145)(4.8,2.191)(5.,2. - 236)(5.2,2.28)(5.4,2.324)(5.6,2.366)(5.8,2.408)(6.,2.448)(6.2,2.488)( - 6.4,2.527)(6.6,2.565)(6.8,2.602)(7.,2.637)(7.2,2.672)(7.4,2.705)(7.6, - 2.737)(7.8,2.768)(8.,2.797)(8.2,2.824)(8.4,2.849)(8.6,2.873)(8.8,2. - 894)(9.,2.913)(9.2,2.929)(9.4,2.942)(9.6,2.953)(9.8,2.96)(10.,2.964)} node [pos=0.8,below right] { $y=p_{4}(x)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -     - -

    - Our final example gives a brief introduction to using Taylor polynomials - to solve differential equations. -

    - - - Approximating an unknown function - -

    - A function y=f(x) is unknown save for the following two facts. - -

      -
    1. -

      - y(0) = f(0) = 1, and -

      -
      2. - -
    2. -

      - \yp= y^2 -

      -
      3. -
    -

    - -

    - (This second fact says that amazingly, - the derivative of the function is actually the function squared!) -

    - -

    - Find the degree 3 Maclaurin polynomial p_3(x) of y=f(x). -

    -     - -

    - One might initially think that not enough information is given to - find p_3(x). - However, note how the second fact above actually lets us know what - \yp(0) is: - - \yp = y^2 \Rightarrow \yp(0) = y^2(0) - . -

    - -

    - Since y(0) = 1, we conclude that \yp(0) = 1. -

    - -

    - Now we find information about \yp'. - Starting with \yp=y^2, - take derivatives of both sides, - with respect to x. - That means we must use implicit differentiation. - - \yp \amp = y^2 - \frac{d}{dx}\big(\yp\big) \amp = \frac{d}{dx}\big(y^2\big) - \yp' \amp = 2y\cdot \yp. - Now evaluate both sides at x=0: - \yp'(0) \amp = 2y(0)\cdot \yp(0) - \yp'(0) \amp = 2 - . -

    - -

    - We repeat this once more to find \yp''(0). - We again use implicit differentiation; - this time the Product Rule is also required. - - \frac{d}{dx}\big(\yp'\big) \amp = \frac{d}{dx} \big(2y\yp\big) - \yp'' \amp = 2\yp\cdot \yp + 2y\cdot \yp'. - Now evaluate both sides at x=0: - \yp''(0) \amp = 2\yp(0)^2 + 2y(0)\yp'(0) - \yp''(0) \amp = 2+4=6 - . -

    - -

    - In summary, we have: - - y(0) = 1 \qquad \yp(0) = 1 \qquad \yp'(0) = 2 \qquad \yp''(0) = 6 - . -

    - -

    - We can now form p_3(x): - - p_3(x) \amp = 1 + x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 - \amp = 1+x+x^2+x^3 - . - - - -

    - -

    - It turns out that the differential equation we started with, - \yp=y^2, - where y(0)=1, can be solved without too much difficulty: - y = \frac{1}{1-x}. - - shows this function plotted with p_3(x). - Note how similar they are near x=0. -

    - -
    - A graph of y=-1/(x-1) and y=p_3(x) from - - - - The graph of the exact solution to the differential equation in this example, and a polynomial approximation. - -

    - The graph y=1/(1-x) is shown, for x from -1 to about 0.6. - Also shown is the graph of p_3(x), the Maclaurin polynomial obtained as an approximate solution to the differential equation in this example. - As expected, the polynomial is a good approximation to the exact solution when x is close to 0. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=3.5, - xmin=-1.1,xmax=1.4 - ] - - \addplot+ [infinite,domain=-1:.65,samples=40] {-1/(x-1)} node[pos=1,left]{ $\displaystyle y= \frac{1}{1-x}$}; - \addplot+ [infinite,domain=-1:0.81,samples=40] {1+x+x^2+x^3} node [pos=1, below right]{ $y=p_3(x)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -     - -

    - It is beyond the scope of this text to pursue error analysis when using - Taylor polynomials to approximate solutions to differential equations. - This topic is often broached in introductory Differential Equations - courses and usually covered in depth in Numerical Analysis courses. - Such an analysis is very important; - one needs to know how good their approximation is. - We explored this example simply to demonstrate the usefulness of Taylor - polynomials. -

    - -

    - Most of this chapter has been devoted to the study of infinite series. - This section has taken a step back from this study, - focusing instead on finite summation of terms. - In the next section, we explore Taylor Series, - where we represent a function with an infinite series. -

    - - - - - - Terms and Concepts - - - - -

    - What is the difference between a Taylor polynomial and a - Maclaurin polynomial? -

    - - -     - - - -

    - The Maclaurin polynomial is a special case of Taylor polynomials. - Taylor polynomials are centered at a specific x-value; - when that x-value is 0, it is a Maclauring polynomial. -

    -     - -     - - - - -

    - In general, - p_n(x) approximates f(x) better and better as - n gets larger. - -

    -     - -

    - True: this is essentially the content of Taylor's Theorem. - For most well-behaved functions, the size of the remainder, - which quantifies the error in our approximation, gets smaller. -

    -     - -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $p = Formula("6+3*x-4*x^2"); - - -

    - For some function f(x), - the Maclaurin polynomial of degree 4 is - p_4(x) = 6+3x-4x^2+5x^3-7x^4. - What is p_2(x)? -

    - -

    - -

    -     - -

    - A higher-degree Maclaurin polynomial begins with the same terms - as any lower-degree - Maclaurin polynomial of the same function. Therefore, - p_2(x) = 6+3x-4x^2. -

    -     -     -     - - - - - - -

    - For some function f(x), - the Maclaurin polynomial of degree 4 is - p_4(x) = 6+3x-4x^2+5x^3-7x^4. - What is \fpp'(0)? -

    - -

    - -

    -     - -

    - Since the coefficient of x^3 in a Maclaurin polynomial - is equal to \frac{\fpp'(0)}{3!}, - we have 5=\frac{\fpp'(0)}{6}, so - \fpp'(0)=30. -

    -     -     -     -     - - - Problems - - - -

    - Find the Maclaurin polynomial of degree n for the given function. -

    -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $macpoly = Formula("1-x+(1/2)*x^2-(1/6)*x^3"); - - -

    - Degree n=3, - for f(x) = e^{-x}. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $macpoly = Formula("x-(1/6)*x^3+(1/120)*x^5-(1/5040)*x^7"); - - -

    - Degree n=8, - for f(x) = \sin(x). -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $macpoly = Formula("x+x^2+(1/2)*x^3+(1/6)*x^4+(1/24)*x^5"); - - -

    - Degree n=5, - for f(x) = x\cdot e^x. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $macpoly = Formula("x+(1/3)*x^3+(2/15)*x^5"); - - -

    - Degree n=6, - for f(x) = \tan(x). -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $macpoly = Formula("1+2*x+2*x^2+(4/3)*x^3+(2/3)*x^4"); - - -

    - Degree n=4, - for f(x) = e^{2x}. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $macpoly = Formula("1+x+x^2+x^3+x^4"); - - -

    - Degree n=4, - for \ds f(x) = \frac1{1-x}. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $macpoly = Formula("1-x+x^2-x^3+x^4"); - - -

    - Degree n=4, - for \ds f(x) = \frac1{1+x}. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $macpoly = Formula("1-x+x^2-x^3+x^4-x^5+x^6-x^7"); - - -

    - Degree n=7, - for \ds f(x) = \frac1{1+x}. -

    - -

    - -

    -     -     -     -     - - - -

    - Find the Taylor polynomial of degree n, - at x=c, for the given function. -

    -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $taypoly = Formula("1+(1/2)*(x-1)-(1/8)*(x-1)^2+(1/16)*(x-1)^3-(5/128)*(x-1)^4")->reduce(); - - -

    - f(x) = \sqrt x, - degree n=4, at c=1. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $c = random(2,15,1); - $c1 = $c +1; - $a0 = Compute("ln($c1)"); - - $f = Formula("ln(x+1)")->reduce(); - - $deg = 4; - - $taypoly = Formula("$a0 + (x - $c)/$c1 - (x-$c)^2/(2*$c1^2) + (x - $c)^3/(3*$c1^3) - (x-$c)^4/(4*$c1^4)")->reduce(); - - -

    - f(x) = \ln(x+1), degree , at c=. -

    - -

    - -

    - - - Note: if your answer is not accepted, - try replacing \ln() with its approximation to 5 decimal places. - -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $taypoly = Formula("(1/sqrt(2))-(1/sqrt(2))*(x-pi/4)-(1/(2*sqrt(2)))*(x-pi/4)^2+(1/(6*sqrt(2)))*(x-pi/4)^3+(1/(24*sqrt(2)))*(x-pi/4)^4-(1/(120*sqrt(2)))*(x-pi/4)^5-(1/(720*sqrt(2)))*(x-pi/4)^6")->reduce(); - - -

    - f(x) = \cos(x), degree 6, at c=\pi/4. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $taypoly = Formula("(1/2)+(sqrt(3)/2)*(x-pi/6)-(1/4)*(x-pi/6)^2-(sqrt(3)/12)*(x-pi/6)^3+(1/48)*(x-pi/6)^4+(sqrt(3)/240)*(x-pi/6)^5")->reduce(); - - -

    - f(x) = \sin(x), degree 5, at c=\pi/6. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $c = random(2,5,1); - $taypoly = Formula("(1/$c)-(1/$c^2)*(x-$c)+(1/$c^3)*(x-$c)^2-(1/$c^4)*(x-$c)^3+(1/$c^5)*(x-$c)^4-(1/$c^6)*(x-$c)^5")->reduce(); - - -

    - f(x) = \frac1x, degree 5, at c=. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $taypoly = Formula("1-2*(x-1)+3*(x-1)^2-4*(x-1)^3+5*(x-1)^4-6*(x-1)^5+7*(x-1)^6-8*(x-1)^7+9*(x-1)^8")->reduce(); - - -

    - \ds f(x) = \frac{1}{x^2}, degree 8, at c=1. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $taypoly = Formula("1/2+(1/2)*(x+1)+(1/4)*(x+1)^2")->reduce(); - - -

    - \ds f(x) = \frac{1}{x^2+1}, degree 3, at c=-1. -

    - -

    - -

    -     -     -     - - - - - Context("Fraction")->noreduce('(-x)+y') ; - Context()->flags->set( - reduceConstants => 0, - reduceConstantFunctions => 0, - ); - $taypoly = Formula("-pi^2-2*pi*(x-pi)+((pi^2-2)/2)*(x-pi)^2")->reduce(); - - -

    - f(x) = x^2\cos(x), degree 2, at c=\pi. -

    - -

    - -

    -     -     -     -     - - - -

    - Approximate the function value with the indicated Taylor polynomial - and give approximate bounds on the error. -

    -     - - - - -

    - Approximate \sin(0.1) with the Maclaurin polynomial of - degree 3. -

    -     - -

    - p_3(x) =x-\frac{x^3}{6}; - p_3(0.1) = 0.09983. - Error is bounded by - \pm \frac{1}{4!}\cdot0.1^4 \approx \pm 0.000004167. -

    -     - -     - - - - -

    - Approximate \cos(1) with the Maclaurin polynomial of - degree 4. -

    -     - -

    - p_4(x) =1-\frac{x^2}{2}+\frac{x^4}{24}; - p_4(1) = 13/24\approx 0.54167. - Error is bounded by - \pm \frac{1}{5!}\cdot1^5 \approx \pm 0.00833 -

    -     - -     - - - - -

    - Approximate \sqrt{10} with the Taylor polynomial of - degree 2 centered at x=9. -

    -     - -

    - p_2(x) =3+\frac{1}{6} (-9+x)-\frac{1}{216} (-9+x)^2; - p_2(10) = 3.16204. - The third derivative of f(x) =\sqrt x is bounded on - (8,11) by 0.003. - Error is bounded by \pm \frac{0.003}{3!}\cdot1^3 = \pm 0.0005. -

    -     - -     - - - - -

    - Approximate \ln(1.5) with the Taylor polynomial of - degree 3 centered at x=1. -

    -     - -

    - p_3(x) =-1+x-\frac{1}{2} (-1+x)^2+\frac{1}{3} (-1+x)^3; - p_3(1.5) = 0.41667. - The fourth derivative of f(x) =\ln(x) is bounded on - (0.9,2) by 10. - Error is bounded by \pm \frac{10}{4!}\cdot 0.5^4 = \pm 0.026. -

    -     - -     -     - - - -

    - The following exercises ask for an n to be found such that - p_n(x) approximates f(x) within a certain bound of accuracy. -

    -     - - - - -

    - Find n such that the Maclaurin polynomial of degree n of - f(x)= e^x approximates e within 0.0001 of the actual value. -

    -     - -

    - The nth derivative of - f(x)=e^x is bounded by 3 on intervals containing 0 and 1. - Thus \abs{R_n(1)}\leq \frac{3}{(n+1)!}1^{(n+1)}. - When n=7, this is less than 0.0001. -

    -     - -     - - - - -

    - Find n such that the Taylor polynomial of degree n of - f(x)= \sqrt x, centered at x=4, - approximates \sqrt 3 within 0.0001 of the actual value. -

    -     - -

    - The nth derivative of - f(x)=\sqrt x is bounded by 0.1 on intervals containing 3 and 4. - Thus \abs{R_n(\pi)}\leq \frac{0.1}{(n+1)!}(1)^{(n+1)}. - When n=4, this is less than 0.0001. -

    -     - -     - - - - -

    - Find n such that the Maclaurin polynomial of degree - n of f(x)= \cos(x) approximates - \cos(\pi/3) within 0.0001 of the actual value. -

    -     - -

    - The nth derivative of - f(x)=\cos(x) is bounded by 1 on intervals - containing 0 and \pi/3. - Thus \abs{R_n(\pi/3)}\leq \frac{1}{(n+1)!}(\pi/3)^{(n+1)}. - When n=7, this is less than 0.0001. - Since the Maclaurin polynomial of \cos(x) only uses even - powers, we can actually just use n=6. -

    -     - -     - - - - -

    - Find n such that the Maclaurin polynomial of degree - n of f(x)= \sin(x) approximates - \cos(\pi) within 0.0001 of the actual value. -

    -     - -

    - The nth derivative of - f(x)=\sin(x) is bounded by 1 on intervals - containing 0 and \pi. - Thus \abs{R_n(\pi)}\leq \frac{1}{(n+1)!}(\pi)^{(n+1)}. - When n=12, this is less than 0.0001. - Since the Maclaurin polynomial of \sin(x) only uses odd powers, - we can actually just use n=11. -

    -     - -     -     - - - -

    - Find the nth term of the indicated Taylor polynomial. -

    -     - - - - -

    - Find a formula for the nth term of the Maclaurin - polynomial for f(x)=e^x. -

    -     - -

    - The nth term is \frac{1}{n!}x^n. -

    -     - -     - - - - -

    - Find a formula for the nth term of the Maclaurin - polynomial for f(x)=\cos(x). -

    -     - -

    - The nth term is: - when n is even, \frac{(-1)^{n/2}}{n!}x^n; - when n is odd, 0. -

    -     - -     - - - -

    - Find a formula for the nth term of the Maclaurin polynomial - for f(x)=\sin(x). -

    -     - -

    - The nth term is: - when n even, 0; when n is odd, - \frac{(-1)^{(n-1)/2}}{n!}x^n. -

    -     -     - - - - -

    - Find a formula for the nth term of the Maclaurin - polynomial for \ds f(x)=\frac{1}{1-x}. -

    -     - -

    - The nth term is x^n. -

    -     - -     - - - - -

    - Find a formula for the nth term of the Maclaurin - polynomial for \ds f(x)=\frac{1}{1+x}. -

    -     - -

    - The nth term is (-1)^nx^n. -

    -     - -     - - - - -

    - Find a formula for the nth term of the Taylor - polynomial for \ds f(x)=\ln(x) - centered at x=1. -

    -     - -

    - The nth term is (-1)^n\frac{(x-1)^n}{n}. -

    -     - -     -     - - - -

    - Approximate the solution to the given differential equation with a - degree 4 Maclaurin polynomial. -

    -     - - - - -

    - \yp=y, y(0) = 1 -

    -     - -     - - - - -

    - \yp=5y, y(0) = 3 -

    -     - -     - - - - -

    - \ds \yp=\frac2y, y(0) = 1 -

    -     - -     - -     -     -     -
    -
    - Taylor Series -

    - In , - we showed how certain functions can be represented by a power series function. - In , - we showed how we can approximate functions with polynomials, - given that enough derivative information is available. - In this section we combine these concepts: - if a function f(x) is infinitely differentiable, - we show how to represent it with a power series function. -

    - - - - - Taylor and Maclaurin Series - -

    - Let f(x) have derivatives of all orders at x=c. - Taylor Seriesdefinition - Maclaurin Seriesdefinition - Maclaurin Series|see{Taylor Series} - seriesTaylor - seriesMaclaurin -

    - -

    -

      -
    1. -

      - The Taylor Series of f(x), - centered at c is - - \infser[0] \frac{f^{(n)}(c)}{n!}(x-c)^n - . -

      -
      2. - -
    2. -

      - Setting c=0 gives the Maclaurin Series of f(x): - - \infser[0] \frac{f^{(n)}(0)}{n!}x^n - . -

      -
      3. -
    -

    -     -     - -

    - If p_n(x) is the nth degree Taylor polynomial for f(x) centered at x=c, - we saw how f(x) is approximately equal - to p_n(x) near x=c. - We also saw how increasing the degree of the polynomial generally reduced the error. -

    - -

    - We are now considering series, - where we sum an infinite set of terms. - Our ultimate hope is to see the error vanish and claim a function is - equal to its Taylor series. -

    - -

    - When creating the Taylor polynomial of degree n for a function f(x) at x=c, - we needed to evaluate f, - and the first n derivatives of f, at x=c. - When creating the Taylor series of f, - it helps to find a pattern that describes the - nth derivative of f at x=c. - We demonstrate this in the next two examples. -

    - - - The Maclaurin series of <m>f(x) = \cos(x)</m> - -

    - Find the Maclaurin series of f(x)=\cos(x). -

    -     - -

    - In - we found the 8th degree Maclaurin polynomial of \cos(x). - In doing so, - we created the table shown in . -

    - -
    - Derivatives of f(x)=\cos (x) evaluated at x=0 - -

    - - f(x) \amp = \cos(x) \amp f(0) \amp = 1 - f'(x) \amp = -\sin(x) \amp \fp(0) \amp = 0 - \fp'(x) \amp = -\cos(x) \amp \fp'(0) \amp = -1 - \fp''(x) \amp = \sin(x) \amp \fp''(0) \amp = 0 - f^{(4)}(x) \amp = \cos(x) \amp f^{(4)}(0) \amp =1 - f^{(5)}(x) \amp = -\sin(x) \amp f^{(5)}(0) \amp =0 - f^{(6)}(x) \amp = -\cos(x) \amp f^{(6)}(0) \amp =-1 - f^{(7)}(x) \amp = \sin(x) \amp f^{(7)}(0) \amp =0 - f^{(8)}(x) \amp = \cos(x) \amp f^{(8)}(0) \amp =1 - f^{(9)}(x) \amp = -\sin(x) \amp f^{(9)}(0) \amp =0 - -

    -     - -
    - -

    - Notice how f^{(n)}(0)=0 when n is odd, - f^{(n)}(0)=1 when n is divisible by 4, - and f^{(n)}(0)=-1 when n is even but not divisible by 4. - Thus the Maclaurin series of \cos(x) is - - 1-\frac{x^2}2+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!} - \cdots - -

    - -

    - We can go further and write this as a summation. - The coefficients alternate between positive and negative. - Since we only need the terms where the power of x is even, - we write the power series in terms of x^{2n}: - - \infser[0] (-1)^{n}\frac{x^{2n}}{(2n)!} - . -

    - -

    - This Maclaurin series is a special type of power series. - As such, we should determine its interval of convergence. - Applying the Ratio Test, we have - - \lim_{n\to\infty}\frac{\abs{(-1)^{n+1}\dfrac{x^{2(n+1)}}{\big(2(n+1)\big)!}}}{\abs{(-1)^n\frac{x^{2n}}{(2n)!}}} \amp = \lim_{n\to\infty}\abs{\frac{x^{2n+2}}{x^{2n}}}\frac{(2n)!}{(2n+2)!} - \amp = \lim_{n\to\infty} \frac{\abs{x}^2}{(2n+2)(2n+1)} - . -

    - -

    - For any fixed x, this limit is 0. - Therefore this power series has an infinite radius of convergence, - converging for all x. - It is important to note what we have, and have not, determined: - we have determined the Maclaurin series for \cos(x) along with its interval of convergence. - We have not shown that \cos(x) is - equal to this power series. -

    -     - -     - -

    - - found the Taylor Series representation of \cos(x). - We can easily find the Taylor Series representation of \sin(x) by recognizing that - \int \cos(x)\, dx=\sin(x) and apply . -

    - - - The Taylor series of <m>f(x)=\ln(x)</m> at <m>x=1</m> - -

    - Find the Taylor series of f(x) = \ln(x) centered at x=1. -

    -     - -

    - - shows the nth derivative of \ln(x) evaluated at x=1 for n=0,\ldots,5, - along with an expression for the nth term: - - f^{(n)}(1) = (-1)^{n+1}(n-1)! \text{ for \(n\geq 1\). } - -

    - -

    - Remember that this is what distinguishes Taylor series from Taylor polynomials; - we are very interested in finding a pattern for the nth term, - not just finding a finite set of coefficients for a polynomial. -

    - -
    - Derivatives of \ln(x) evaluated at x=1 - -

    - - f(x) \amp = \ln(x) \amp f(1) \amp = 0 - \fp(x) \amp = \frac1x \amp \fp(1) \amp = 1 - \fp'(x) \amp = -\frac{1}{x^2} \amp \fp'(1) \amp = -1 - \fp''(x) \amp = \frac{2}{x^3} \amp \fp''(1) \amp = 2 - f^{(4)}(x) \amp = -\frac{6}{x^4} \amp f^{(4)}(1) \amp = -6 - f^{(5)}(x) \amp = \frac{24}{x^5} \amp f^{(5)}(1) \amp = 24 - \amp \vdots \amp \amp \vdots - f^{(n)}(x) \amp = \amp f^{(n)}(1) \amp = - (-1)^{n+1}\amp\frac{(n-1)!}{x^n} \amp (-1)^{n+1}\amp (n-1)! - -

    -     - -
    - -

    - Since f(1) = \ln(1) = 0, - we skip the first term and start the summation with n=1, - giving the Taylor series for \ln(x), - centered at x=1, as - - \infser (-1)^{n+1}(n-1)!\frac{1}{n!}(x-1)^n = \infser (-1)^{n+1}\frac{(x-1)^n}{n} - . -

    - -

    - We now determine the interval of convergence, using the Ratio Test. - - \lim_{n\to\infty} \frac{\abs{(-1)^{n+2}\dfrac{(x-1)^{n+1}}{n+1}}}{\abs{(-1)^{n+1}\dfrac{(x-1)^n}{n}}} \amp = \lim_{n\to\infty} \abs{\frac{(x-1)^{n+1}}{(x-1)^n}}\frac{n}{n+1} - \amp = \abs{x-1} - . -

    - -

    - By the Ratio Test, - we have convergence when \abs{x-1} \lt 1: - the radius of convergence is 1, and we have convergence on (0,2). - We now check the endpoints. -

    - -

    - At x=0, the series is - - \sum_{n=1}^\infty (-1)^{n+1}\frac{(-1)^n}{n} = -\sum_{n=1}^\infty \frac1n - , - which diverges (it is the Harmonic Series times (-1).) -

    - -

    - At x=2, the series is - - \sum_{n=1}^\infty (-1)^{n+1}\frac{(1)^n}{n} = \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n} - , - the Alternating Harmonic Series, which converges. -

    - -

    - We have found the Taylor series of \ln x centered at x=1, - and have determined the series converges on (0,2]. - We cannot (yet) say that \ln x is equal to this Taylor series on (0,2]. -

    -     - -     - -

    - It is important to note that - defines a Taylor series given a function f(x), but makes no claim about their equality. - We will find that most of the time they are equal, - but we need to consider the conditions that allow us to conclude this. -

    - -

    - - states that the error between a function f(x) and its - nth-degree Taylor polynomial p_n(x) is R_n(x), where - - \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{(x-c)^{(n+1)}} - . -

    - -

    - If R_n(x) goes to 0 for each x in an interval I as n approaches infinity, - we conclude that the function is equal to its Taylor series expansion. -

    - - - Function and Taylor Series Equality - -

    - Let f(x) have derivatives of all orders at x=c, - let R_n(x) be as stated in , - and let I be an interval on which the Taylor series of f(x) converges. - If \lim\limits_{n\to\infty} R_n(x) = 0 for all x in I, then - Taylor Seriesequality with generating function - - f(x) = \infser[0] \frac{f^{(n)}(c)}{n!}(x-c)^n\, \text{ on \(I\). } - -

    -     -     - - - -

    - We demonstrate the use of this theorem in an example. -

    - - - Establishing equality of a function and its Taylor series - -

    - Show that f(x) = \cos(x) is equal to its Maclaurin series, - as found in , for all x. -

    -     - -

    - Given a value x, - the magnitude of the error term R_n(x) is bounded by - - \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{x^{n+1}} - . -

    - -

    - Since all derivatives of \cos(x) are - \pm \sin(x) or \pm\cos(x), - whose magnitudes are bounded by 1, we can state - - \abs{R_n(x)} \leq \frac{1}{(n+1)!}\abs{x^{n+1}} - - which implies - - -\frac{\abs{x^{n+1}}}{(n+1)!} \leq R_n(x) \leq\frac{\abs{x^{n+1}}}{(n+1)!} - . -

    - -

    - For any x, - \lim\limits_{n\to\infty} \frac{x^{n+1}}{(n+1)!} = 0. - Applying the Squeeze Theorem to Equation, - we conclude that \lim\limits_{n\to\infty} R_n(x) = 0 for all x, - and hence - - \cos(x) = \infser[0] (-1)^{n}\frac{x^{2n}}{(2n)!} \text{ for all \(x\) } - . -

    -     -     - -

    - It is natural to assume that a function is equal to its Taylor series on the series' interval of convergence, - but this is not always the case. - In order to properly establish equality, - one must use . - This is a bit disappointing, - as we developed beautiful techniques for determining the interval of convergence of a power series, - and proving that R_n(x)\to 0 can be difficult. - For instance, - it is not a simple task to show that \ln x equals its Taylor series on (0,2] as found in ; - in the Exercises, - the reader is only asked to show equality on (1,2), - which is simpler. -

    - -

    - There is good news. - A function f(x) that is equal to its Taylor series, - centered at any point the domain of f(x), - is said to be an analytic function, - analytic function - and most, if not all, - functions that we encounter within this course are analytic functions. - Generally speaking, - any function that one creates with elementary functions (polynomials, - exponentials, trigonometric functions, - etc.) that is not piecewise defined is probably analytic. - For most functions, - we assume the function is equal to its Taylor series on the series' interval of convergence and only use - when we suspect something may not work as expected. -

    - -

    - We develop the Taylor series for one more important function, - then give a table of the Taylor series for a number of common functions. - Binomial Series - seriesBinomial -

    - - - The Binomial Series - -

    - Find the Maclaurin series of f(x) = (1+x)^k, k\neq 0. -

    -     - -

    - When k is a positive integer, - the Maclaurin series is finite. - For instance, when k=4, we have - - f(x) = (1+x)^4 = 1+4x+6x^2+4x^3+x^4 - . -

    - -

    - The coefficients of x when k is a positive integer are known as the - binomial coefficients, - giving the series we are developing its name. -

    - -

    - When k=1/2, we have f(x) = \sqrt{1+x}. - Knowing a series representation of this function would give a useful way of approximating \sqrt{1.3}, - for instance. -

    - -

    - To develop the Maclaurin series for - f(x) = (1+x)^k for any value of k\neq0, - we consider the derivatives of f evaluated at x=0: -

    - -

    - - f(x) \amp = (1+x)^k \amp f(0) \amp = 1 - \fp(x) \amp = k(1+x)^{k-1} \amp \fp(0) \amp =k - \fp'(x) \amp = k(k-1)(1+x)^{k-2} \amp \fp'(0) \amp =k(k-1) - \fp''(x) \amp = k(k-1)(k-2)(1+x)^{k-3} \amp \fp''(0) \amp =k(k-1)(k-2) - \amp \vdots \amp \amp \vdots - - For a general n, - - f^{(n)}(x) = k(k-1)\cdots\bigl(k-(n-1)\bigr)(1+x)^{k-n} - , - giving f^{(n)}(0) =k(k-1)\cdots\bigl(k-(n-1)\bigr). -

    -

    - Thus the Maclaurin series for f(x) = (1+x)^k is - - (1+x)^k\amp =1+ kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \ldots - \amp \dots + \frac{k(k-1)\cdots\big(k-(n-1)\big)}{n!}(x-c)^n+\ldots - -

    - -

    - It is important to determine the interval of convergence of this series. - With - - a_n = \frac{k(k-1)\cdots\big(k-(n-1)\big)}{n!}x^n - , - we apply the Ratio Test: - - \lim_{n\to\infty}\frac{\abs{a_{n+1}}}{\abs{a_n}}\amp =\lim_{n\to\infty} \frac{\abs{\dfrac{k(k-1)\cdots(k-(n-1))(k-n)}{(n+1)!}x^{n+1}}}{\abs{\dfrac{k(k-1)\cdots\big(k-(n-1)\big)}{n!}x^n}} - \amp =\lim_{n\to\infty} \abs{\frac{k-n}{n+1}x} - \amp = \abs{x} - . -

    - -

    - The series converges absolutely when the limit of the Ratio Test is less than 1; - therefore, we have absolute convergence when \abs{x}\lt 1. -

    - -

    - While outside the scope of this text, - the interval of convergence depends on the value of k. - When k \gt 0, the interval of convergence is [-1,1]. - When -1\lt k\lt 0, the interval of convergence is [-1,1). - If k\leq -1, the interval of convergence is (-1,1). -

    -     - -     - -

    - We learned that Taylor polynomials offer a way of approximating a - difficult to compute - function with a polynomial. - Taylor series offer a way of exactly representing a function with a series. - One probably can see the use of a good approximation; - is there any use of representing a function exactly as a series? -

    - -

    - While we should not overlook the mathematical beauty of Taylor series - (which is reason enough to study them), - there are practical uses as well. - They provide a valuable tool for solving a variety of problems, - including problems relating to integration and differential equations. -

    - -

    - In - we give a table of the Taylor series of a number of common functions. - We then give a theorem about the - algebra of power series, that is, - how we can combine power series to create power series of new functions. - This allows us to find the Taylor series of functions like - f(x) = e^x\cos(x) by knowing the Taylor series of e^x and \cos(x). -

    - -

    - Before we investigate combining functions, - consider the Taylor series for the arctangent function - (see ). - Knowing that \tan^{-1}(1) = \pi/4, - we can use this series to approximate the value of \pi: - - \frac{\pi}4 \amp = \tan^{-1}(1) = 1-\frac13+\frac15-\frac17+\frac19-\cdots - \pi \amp = 4\left(1-\frac13+\frac15-\frac17+\frac19-\cdots\right) - -

    - -

    - Unfortunately, - this particular expansion of \pi converges very slowly. - The first 100 terms approximate \pi as 3.13159, - which is not particularly good. -

    - -

    - - - Important Taylor Series Expansions - - - Function and Series - First Few Terms - Interval of - - - Convergence - - - \ds e^x = \infser[0] \frac{x^n}{n!} - \ds 1+ x+\frac{x^2}{2!} + \frac{x^3}{3!}+\cdots - (-\infty,\infty) - - - \ds \sin(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!} - \ds x-\frac{x^3}{3!}+\frac{x^5}{5!} - \frac{x^7}{7!}+\cdots - (-\infty,\infty) - - - \ds \cos(x) = \infser[0] (-1)^n\frac{x^{2n}}{(2n)!} - \ds 1-\frac{x^2}{2!}+\frac{x^4}{4!} - \frac{x^6}{6!} +\cdots - (-\infty,\infty) - - - \ds \ln(x) = \infser(-1)^{n+1}\frac{(x-1)^n}{n} - \ds (x-1)- \frac{(x-1)^2}{2} +\frac{(x-1)^3}{3}-\cdots - (0,2] - - - \ds \frac{1}{1-x} = \infser[0] x^n - \ds 1+x+x^2+x^3+\cdots - (-1,1) - - - \ds \tan^{-1}(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{2n+1} - \ds x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots - [-1,1] - - - \ds (1+x)^k=\infser[0] \binom{k}{n}x^n - \ds 1+kx+\frac{k(k-1)}{2!}x^2 + \cdots - (-1,1) - - - -

    - Note that for (1+x)^k, the interval of convergence may contain one or both endpoints, - depending on the value of k, and we are using the generalized binomial coefficients - - \binom{k}{n} = \frac{k(k-1)\cdots (k-(n-1))}{n!} - . - Taylor Seriescommon series -

    - - - - - - Algebra of Power Series - -

    - Let \ds f(x) = \infser[0] a_nx^n and - \ds g(x) = \infser[0] b_nx^n converge absolutely for \abs{x}\lt R, - and let h(x) be a polynomial function. - power seriesalgebra of -

    - -

    -

      -
    1. -

      - \ds f(x)\pm g(x) = \infser[0] (a_n\pm b_n)x^n for \abs{x}\lt R. -

      -
      2. - -
    2. -

      - \ds f(x)g(x) = \left(\infser[0] a_nx^n\right)\left(\infser[0] b_nx^n\right) = \infser[0]\big(a_0b_n+a_1b_{n-1}+\ldots a_nb_0\big)x^n for \abs{x}\lt R. -

      -
      3. - -
    3. -

      - \ds f\big(h(x)\big) = \infser[0] a_n\big(h(x)\big)^n for \abs{h(x)}\lt R. -

      -
      4. -
    -

    -     -     - - - - - Combining Taylor series - -

    - Write out the first 3 terms of the Taylor Series for - f(x) = e^x\cos(x) using - and . -

    -     - -

    - informs us that - - e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \text{ and } \cos(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots - . -

    - -

    - Applying , we find that - - e^x\cos(x) = \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right) - . - Distribute the right hand expression across the left: - - e^x\cos(x)\amp = 1\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)+x\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right) - \amp + \frac{x^2}{2!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)+\frac{x^3}{3!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right) - \amp + \frac{x^4}{4!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)+\cdots - - If we distribute again and collect like terms, we find - - e^x\cos(x) = 1 + x -\frac{x^3}{3}-\frac{x^4}{6} - \frac{x^5}{30}+\frac{x^7}{630}+\cdots - . -

    - -

    - While this process is a bit tedious, - it is much faster than evaluating all the necessary derivatives of - e^x\cos(x) and computing the Taylor series directly. -

    - -

    - Because the series for e^x and \cos(x) both converge on (-\infty,\infty), - so does the series expansion for e^x\cos(x). -

    -     - -     - - - - - - Creating new Taylor series - -

    - Use - to create series for y=\sin(x^2) and y=x^3/(3+x^4). -

    -     - -

    - Given that - - \sin(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!} = x-\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}+\cdots - , - we simply substitute x^2 for x in the series, giving - - \sin(x^2) \amp = \infser[0] (-1)^n\frac{(x^2)^{2n+1}}{(2n+1)!} - \amp \infser[0] (-1)^n\frac{(x^{4n+2}}{(2n+1)!} - \amp =x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!} -\frac{x^{14}}{7!}\cdots - . -

    - -

    - Since the Taylor series for \sin(x) has an infinite radius of convergence, - so does the Taylor series for \sin(x^2). -

    - -

    - For y=x^3/(3+x^4), we begin with the geometric series expansion - - \frac{1}{1-x} = \sum_{n=0}^\infty x^n - . - Note that we can write - - \frac{1}{3+x^4} = \frac13\cdot\frac{1}{1+x^4/3} = \frac13 \cdot \frac{1}{1-(-x^4/3)} - . - Substituting -x^4/3 into the geometric series expansion, we get - - \frac{1}{3+x^4} = \sum_{n=0}^\infty (-x^4/3)^n = \sum_{n=0}^\infty \frac{(-1^n)x^{4n}}{3^n} - . - Finally, we can multiply both sides of the above equation by x^3 to obtain - - \frac{x^3}{3+x^4} = x^3\sum_{n=0}^\infty \frac{(-1^n)x^{4n}}{3^n} = \sum_{n=0}^\infty \frac{(-1)^nx^{4n+3}}{3^n} - . -

    -     - -     - - - - - A (somewhat foolish) combination of Taylor series - -

    - Discuss possible methods for obtaining a Taylor series expansion for f(x)=\ln(\sqrt{x}). -

    -     - -

    - Since f(x) is a composition, our first instict might be to apply - to the problem. However, \sqrt{x} is not a polynomial function, - and neither \ln(x) nor \sqrt{x} have Maclaurin series expansions. -

    - -

    - You might already see a simple way to proceed, but let us first consider the following: - \sqrt{x}=(1+(x-1))^{1/2} can be expanded as a binomial series centered at x=1. - We also know the Taylor series for \ln(x) at x=1, and note that \sqrt{1}=1, - so when x is near 1, so is \sqrt{x}. -

    - -

    - What happens if we take the Taylor series - - \ln(x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{(x-1)^n}{n} = (x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\cdots - - and substitute in - - \sqrt{x} = \sum_{n=0}^\infty \binom{1/2}{n}(x-1)^n = 1+\frac12(x-1)-\frac14(x-1)^2+\cdots - , - where \binom{1/2}{n}=\frac{1/2(1/2-1)\cdots (1/2-(n-1))}{n!} denotes the binomial coefficient? -

    - -

    - Short answer: a mess. We have to replace each occurence of x-1 - in the power series for \ln(x) with \sqrt{x}-1 = \frac12(x-1)-\frac14(x-1)^2+\frac{1}{16}(x-1)^3+\cdots, - and then expand, and collect terms. - If we do this, keeping only terms up to (x-1)^3, we find: - - \ln(\sqrt{x}) \amp = \left(\frac12(x-1)-\frac14(x-1)^2+\frac{1}{16}(x-1)^3+\cdots\right) - \amp \quad -\frac12\left(\frac12(x-1)-\frac14(x-1)^2+\frac{1}{16}(x-1)^3+\cdots\right)^2 - \amp \quad + \frac13\left(\frac12(x-1)-\frac14(x-1)^2+\frac{1}{16}(x-1)^3+\cdots\right)+\cdots - \amp = \frac12(x-1)-\frac14(x-2)^2+\frac16(x-1)^3-\cdots - . -

    - -

    - But of course, there was a better way all along: - - \ln(\sqrt{x}) = \ln(x^{1/2}) = \frac12\ln(x) - - using properties of the logarithm, and indeed, - the result above is the same as the one we would have obtained by simply multiplying the Taylor series for \ln(x) by \frac12. - Power series manipulation is a powerful technique, but one should not apply it blindly. -

    -     -     - - - Using Taylor series to approximate a composition - -

    - Use Taylor series to determine a degree 5 Taylor polynomial approximation to f(x)=e^{\sin(x)}. -

    -     - -

    - Here we want to apply , but h(x)=\sin(x) is not a polynomial. - However, we are interested in approximation, so we replace \sin(x) - by the Maclaurin polynomial - - q(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - . - The Maclaurin series for f(x)=e^x is given by - - e^x = \sum_{n=0}^\infty \frac{x^n}{n!} - . - Next, we substitute q(x) into the series for e^x. - The algebra gets very messy, but we can simplify things: - since we want the degree 5 approximation, there is no need to write down terms involving x^6 or higher powers. - - e^{\sin(x)} \amp \approx 1 + q(x) + \frac{1}{2!}q(x)^2+\frac{1}{3!}q(x)^3+\frac{1}{4!}q(x)^4+\frac{1}{5!}q(x)^5 - \amp = 1 + \left(x-\frac{x^3}{6}+\frac{x^5}{120}\right) + \frac12\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right) - \amp \quad + \frac{1}{6}\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right)^3 + \frac{1}{24}\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right)^4 - \amp \quad\quad + \frac{1}{120}\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right)^5 - \amp = 1+x-\frac{x^3}{6}+\frac{x^5}{120}+\frac12\left(x^2-\frac13x^4+\cdots\right)+\frac16\left(x^3-\frac12x^5+\cdots\right) - \amp \quad + \frac{1}{24}\left(x^4+\cdots\right) + \frac{1}{120}\left(x^5+\cdots\right) - \amp = 1+x+\frac12x^2-\frac18x^4-\frac{1}{15}x^5+\cdots - . -

    -

    - While the algebra is a bit of a mess, it is often less work than computing the Taylor polynomial directly, - as the derivatives of a composite function quickly get complicated. - The function f(x)=e^{\sin(x)} and its approximation are plotted in below. - Note that our polynomial approximation is very good on [-1,1]. -

    -
    - A graph of f(x) and its degree 5 Maclaurin polynomial - - A graph of the function from this example, and its degree 5 Maclaurin polynomial approximation. - -

    - The image shows the graph of f(x)=e^{\sin(x)} on the interval [-2,2], - along with the graph of p_5(x), the degree 5 Maclaurin polynomial approximation of f(x). - As we have come to expect, there is very little noticeable difference between the two graphs near x=0, - and it appears that the polynomial approximation remains good over the interval (-1,1). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-0.5,ymax=3, - xmin=-2.1,xmax=2.2 - ] - - \addplot [firstcurvestyle,domain=-2:2,samples=60] {exp(sin(deg(x)))} node [pos=.8,left] { $y=f(x)$}; - - \addplot [secondcurvestyle,domain=-2:2] {1+x+x^2/2-x^4/8-x^5/15} node [pos=0,below right] { $y=p_5(x)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -     - -

    - In the previous example, the reader might be left wondering why we would bother with all that algebra, - when the computer could have given us the result in seconds. - One reason is simply that it lets us see how these different pieces fit together. - Computing a Taylor polynomial by combining existing results will give the same polynomial as computing derivatives. - Also we see that we can compute an approximation by replacing both parts of a composition with approximations. - In the last couple of examples in this chapter, we see another reason: - often we have to define functions in terms of power series derived through integration, - or the solution of a differential equation, where there is no known function we can simply plug into the computer. -

    - - - Using Taylor series to evaluate definite integrals - -

    - Use the Taylor series of e^{-x^2} to evaluate \ds \int_0^1e^{-x^2}\, dx. -

    -     - -

    - We learned, when studying Numerical Integration, - that e^{-x^2} does not have an antiderivative expressible in terms of elementary functions. - This means any definite integral of this function must have its value approximated, - and not computed exactly. -

    - -

    - We can quickly write out the Taylor series for - e^{-x^2} using the Taylor series of e^x: - - e^x \amp = \infser[0] \frac{x^n}{n!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots - and so - e^{-x^2} \amp = \infser[0] \frac{(-x^2)^n}{n!} - \amp = \infser[0] (-1)^n\frac{x^{2n}}{n!} - \amp = 1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots - . -

    - -

    - We use to integrate: - - \int e^{-x^2}\, dx = C + x - \frac{x^3}{3}+\frac{x^5}{5\cdot2!}-\frac{x^7}{7\cdot3!}+\cdots +(-1)^n\frac{x^{2n+1}}{(2n+1)n!}+\cdots - -

    - -

    - This is the antiderivative of e^{-x^2}; - while we can write it out as a series, - we cannot write it out in terms of elementary functions. - We can evaluate the definite integral - \ds \int_0^1e^{-x^2}\, dx using this antiderivative; - substituting 1 and 0 for x and subtracting gives - - \int_0^1e^{-x^2}\, dx = 1-\frac{1}{3}+\frac{1}{5\cdot 2!}-\frac{1}{7\cdot3!} + \frac{1}{9\cdot4!}\cdots - . -

    - -

    - Summing the 5 terms shown above give the approximation of 0.74749. - Since this is an alternating series, - we can use the Alternating Series Approximation Theorem, - (), - to determine how accurate this approximation is. - The next term of the series is 1/(11\cdot5!) \approx 0.00075758. - Thus we know our approximation is within - 0.00075758 of the actual value of the integral. - This is arguably much less work than using Simpson's Rule to approximate the value of the integral. -

    -     - -     - - - Using Taylor series to solve differential equations - -

    - Solve the differential equation \yp=2y in terms of a power series, - and use the theory of Taylor series to recognize the solution in terms of an elementary function. -

    -     - -

    - We found the first 5 terms of the power series solution to this differential equation in - in . - These are: - - a_0=1, a_1 = 2, a_2 = \frac42=2, a_3=\frac{8}{2\cdot3}=\frac43, a_4=\frac{16}{2\cdot3\cdot4} = \frac23 - . -

    - -

    - We include the unsimplified - expressions for the coefficients found in - as we are looking for a pattern. - It can be shown that a_n = 2^n/n!. - Thus the solution, written as a power series, is - - y = \infser[0] \frac{2^n}{n!}x^n = \infser[0] \frac{(2x)^n}{n!} - . -

    - -

    - Using - and , - we recognize f(x) = e^{2x}: - - e^x = \infser[0] \frac{x^n}{n!} \qquad \Rightarrow \qquad e^{2x} = \infser[0] \frac{(2x)^n}{n!} - . -

    -     - -     - -

    - Finding a pattern in the coefficients that match the series expansion of a known function, - such as those shown in , - can be difficult. - What if the coefficients in the previous example were given in their reduced form; - how could we still recover the function y=e^{2x}? -

    - -

    - Suppose that all we know is that - - a_0=1, a_1=2, a_2=2, a_3=\frac43, a_4=\frac23 - . -

    - -

    - - states that each term of the Taylor expansion of a function includes an n!. - This allows us to say that - - a_2=2=\frac{b_2}{2!}, a_3 = \frac43=\frac{b_3}{3!}, \text{ and } a_4 = \frac23=\frac{b_4}{4!} - - for some values b_2, b_3 and b_4. - Solving for these values, - we see that b_2=4, b_3 = 8 and b_4=16. - That is, we are recovering the pattern we had previously seen, - allowing us to write - - f(x) = \infser[0] a_nx^n \amp = \infser[0] \frac{b_n}{n!}x^n - \amp = 1+2x+ \frac{4}{2!}x^2 + \frac{8}{3!}x^3+\frac{16}{4!}x^4 + \cdots - -

    - -

    - From here it is easier to recognize that the series is describing an exponential function. -

    - -

    - There are simpler, - more direct ways of solving the differential equation \yp = 2y, - as discussed in . - We applied power series techniques to this equation to demonstrate its utility, - and went on to show how sometimes - we are able to recover the solution in terms of elementary functions using the theory of Taylor series. - Most differential equations faced in real scientific and engineering situations are much more complicated than this one, - but power series can offer a valuable tool in finding, - or at least approximating, the solution. -

    - -

    - This chapter introduced sequences, - which are ordered lists of numbers, - followed by series, wherein we add up the terms of a sequence. - We quickly saw that such sums do not always add up to - infinity, but rather converge. - We studied tests for convergence, - then ended the chapter with a formal way of defining functions based on series. - Such series-defined functions - are a valuable tool in solving a number of different problems throughout science and engineering. -

    - -

    - Coming in the next chapters are new ways of defining curves in the plane apart from using functions of the form y=f(x). - Curves created by these new methods can be beautiful, useful, - and important. -

    - - - - Terms and Concepts - - - -

    - What is the difference between a Taylor polynomial and a Taylor series? -

    -     - - - -

    - A Taylor polynomial is a polynomial, - containing a finite number of terms. - A Taylor series is a series, - the summation of an infinite number of terms. -

    -     - -     - - - - -

    - What theorem must we use to show that a function is equal to its Taylor series? -

    -     - - - -

    - , - entitled Function and Taylor Series Equality -

    -     - -     -     - - Problems - - -

    - - gives the nth term of the Taylor series of common functions. - Verify the formula given in the Key Idea by finding - the first few terms of the Taylor series of the given function and identifying a pattern. -

    -     - - - - -

    - f(x) = e^x; c=0 -

    -     - -

    - All derivatives of e^x are e^x which evaluate to 1 at x=0. -

    - -

    - The Taylor series starts 1+x+\frac12x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+\cdots; -

    - -

    - the Taylor series is \ds \infser[0] \frac{x^n}{n!} -

    -     - -     - - - - -

    - f(x) = \sin(x); c=0 -

    -     - -

    - All derivatives of \sin(x) are either - \pm\cos(x) or \pm \sin(x), - which evaluate to \pm 1 or 0 at x=0. - The Taylor series starts 0+x+0x^2-\frac16x^3+0x^4+\frac1{120}x^5; -

    - -

    - the Taylor series is \ds \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!} -

    -     - -     - - - - -

    - f(x) = 1/(1-x); c=0 -

    -     - -

    - The nth derivative of 1/(1-x) is f^{(n)}(x) = (n)!/(1-x)^{n+1}, - which evaluates to n! at x=0. -

    - -

    - The Taylor series starts 1+x+x^2+x^3+\cdots; -

    - -

    - the Taylor series is \ds \infser[0] x^n -

    -     - -     - - - - -

    - f(x) = \tan^{-1}(x); c=0 -

    -     - -

    - The derivative of \tan^{-1}(x) is 1/(1+x^2). - Taking successive derivatives using the Quotient Rule, - the derivatives of \tan^{-1}(x) fall into two categories in terms of their evaluation at x=0. -

    - -

    - When n is even, - \ds f^{(n)}(x) = (-1)^{(n-1)/2}\frac{p(x)}{(1+x^2)^n}, - where p(x) is a polynomial such that p(0) = 0. - Hence f^{(n)}(0) = 0 when n is even. -

    - -

    - When n is odd, - \ds f^{(n)}(x) = (-1)^{(n-1)/2}\frac{p(x)}{(1+x^2)^n}, - where p(x) is a polynomial such that p(0) = (n-1)!. - Hence f^{(n)}(0) = (-1)^{(n-1)/2}(n-1)! when n is odd. (The unusual power of (-1) is such that every other odd term is negative.) -

    - -

    - The Taylor series starts x-\frac13x^3+\frac15x^5+\cdots; - by reindexing to only obtain odd powers of x, we get -

    - -

    - the Taylor series is \ds \infser[0] (-1)^n\frac{x^{2n+1}}{2n+1}. -

    -     - -     - -     - - - -

    - Find a formula for the nth term of the Taylor series of f(x), - centered at c, - by finding the coefficients of the first few powers of x and looking for a pattern. - (The formulas for several of these are found in ; - show work verifying these formula.) -

    -     - - - - -

    - f(x) = \cos(x); c=\pi/2 -

    -     - -

    - The Taylor series starts 0-(x-\pi/2)+0x^2+\frac16(x-\pi/2)^3+0x^4-\frac1{120}(x-\pi/2)^5; -

    - -

    - the Taylor series is \ds \infser[0] (-1)^{n+1}\frac{(x-\pi/2)^{2n+1}}{(2n+1)!} -

    -     - -     - - - - -

    - f(x) = 1/x; c=1 -

    -     - -

    - The Taylor series starts 1-(x-1)+(x-1)^2-(x-1)^3+(x-1)^4-(x-1)^5; -

    - -

    - the Taylor series is \ds \infser[0] (-1)^{n}(x-1)^n -

    -     - -     - - - - -

    - f(x) = e^{-x}; c=0 -

    -     - -

    - f^{(n)}(x) = (-1)^ne^{-x}; - at x=0, f^{(n)}(0)=-1 when n is odd and - f^{(n)}(0)=1 when n is even. -

    - -

    - The Taylor series starts 1-x+\frac12x^2-\frac1{3!}x^3+\cdots; -

    - -

    - the Taylor series is \ds \infser[0] (-1)^n\frac{x^n}{n!}. -

    -     - -     - - - - -

    - f(x) = \ln(1+x); c=0 -

    -     - -

    - f^{(n)}(x) = (-1)^{n+1}\frac{(n-1)!}{(1+x)^n}; - at x=0, f^{(n)}(0)=(-1)^{n+1}(n-1)! -

    - -

    - The Taylor series starts x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots; -

    - -

    - the Taylor series is \ds \infser (-1)^{n+1}\frac{x^n}{n}. -

    -     - -     - - - - -

    - f(x) = x/(x+1); c=1 -

    -     - -

    - f^{(n)}(x) = (-1)^{n+1}\frac{n!}{(x+1)^{n+1}}; - at x=1, f^{(n)}(1)=(-1)^{n+1}\frac{n!}{2^{n+1}} -

    - -

    - The Taylor series starts \frac12+\frac14(x-1)-\frac18(x-1)^2+\frac1{16}(x-1)^3\cdots; -

    - -

    - the Taylor series is \ds \frac12+\infser (-1)^{n+1}\frac{(x-1)^n}{2^{n+1}}. -

    -     - -     - - - - -

    - f(x) = \sin(x); c=\pi/4 -

    -     - -

    - The derivatives of \sin(x) are - \pm \cos(x) and \pm \sin(x); - at x=\pi/4, - these derivatives evaluate to \pm \sqrt{2}/2. -

    - -

    - The Taylor series starts \frac{\sqrt{2}}2+\frac{\sqrt{2}}2(x-\pi/4) - \frac{\sqrt{2}}2\frac{(x-\pi/4)^2}{2}-\frac{\sqrt{2}}2\frac{(x-\pi/4)^3}{3!}+\frac{\sqrt{2}}2\frac{(x-\pi/4)^4}{4!}+\frac{\sqrt{2}}2\frac{(x-\pi/4)^5}{5!}\cdots. - Note how the signs are even, - even, odd, odd, even, even, odd, - odd,\ldots We saw signs like these in - of ; - one way of producing such signs is to raise (-1) to a special quadratic power. - While many possibilities exist, - one such quadratic is (n+3)(n+4)/2. -

    - -

    - Thus the Taylor series is \ds \infser[0] (-1)^{\frac{(n+3)(n+4)}{2}}\frac{\sqrt2}{2}\frac{(x-\pi/4)^n}{n!}. -

    -     - -     - -     - - - -

    - Show that the Taylor series for f(x), - as given in , - is equal to f(x) by applying ; - that is, show \lim\limits_{n\to\infty}R_n(x) =0. -

    -     - - - - -

    - f(x) = e^x -

    -     - -

    - Given a value x, - the magnitude of the error term R_n(x) is bounded by - - \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{x^{(n+1)}} - , - where z is between 0 and x. -

    - -

    - If x \gt 0, - then z\lt x and f^{(n+1)}(z) =e^z\lt e^x. - If x\lt 0, - then x\lt z\lt 0 and f^{(n+1)}(z) =e^z\lt 1. - So given a fixed x value, - let M = \max\{e^x,1\}; - f^{(n)}(z)\lt M. - This allows us to state - - \abs{R_n(x)} \leq \frac{M}{(n+1)!}\abs{x^{(n+1)}} - . -

    - -

    - For any x, - \lim\limits_{n\to\infty} \frac{M}{(n+1)!}\abs{x^{(n+1)}}= 0. - Thus by the Squeeze Theorem, - we conclude that \lim\limits_{n\to\infty} R_n(x) = 0 for all x, - and hence - - e^x = \infser[0] \frac{x^{n}}{n!} \text{ for all \(x\) } - . -

    -     - -     - - - - -

    - f(x) = \sin(x) -

    -     - -

    - The following argument is essentially the same as that given for - f(x) = \cos(x) in . -

    - -

    - Given a value x, - the magnitude of the error term R_n(x) is bounded by - - \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{x^{(n+1)}} - . -

    - -

    - Since all derivatives of \sin(x) are - \pm \cos(x) or \pm\sin(x), - whose magnitudes are bounded by 1, we can state - - \abs{R_n(x)} \leq \frac{1}{(n+1)!}\abs{x^{(n+1)}} - . -

    - -

    - For any x, - \lim\limits_{n\to\infty} \frac{x^{n+1}}{(n+1)!} = 0. - Thus by the Squeeze Theorem, - we conclude that \lim\limits_{n\to\infty} R_n(x) = 0 for all x, - and hence - - \sin(x) = \infser[0] (-1)^{n}\frac{x^{2n+1}}{(2n+1)!} \text{ for all \(x\) } - . -

    -     - -     - - - - -

    - f(x) = \ln(x) (show equality only on (1,2)) -

    -     - -

    - Per the statement of the problem, - we only consider the case 1\lt x\lt 2. -

    - -

    - If 1\lt x\lt 2, then - 1\lt z\lt x and f^{(n+1)}(z) =\frac{n!}{z^{n+1}}\lt n!. - Thus - - \abs{R_n(x)} \leq \frac{n!}{(n+1)!}\abs{(x-1)^{(n+1)}}= \frac{(x-1)^{n+1}}{n+1}\lt \frac1{n+1} - . -

    - -

    - Thus - - \lim_{n\to\infty} \abs{R_n(x)} \lt \lim_{n\to\infty} \frac1{n+1}=0 - , - hence - - \ln(x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{(x-1)^n}n\,\text{ on } \,(1,2) - . -

    -     - -     - - - - -

    - f(x) = 1/(1-x) (show equality only on (-1,0)) -

    -     - -

    - Given a value x, - the magnitude of the error term R_n(x) is bounded by - - \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{x^{(n+1)}} - , - where z is between 0 and x. -

    - -

    - Note that \abs{f^{(n+1)}(x)} = \frac{(n+1)!}{(1-x)^{n+2}}. -

    - -

    - If -1\lt x\lt 0, then - x\lt z\lt 0 and f^{(n+1)}(z) =\frac{(n+1)!}{(1-z)^{n+2}}\lt \frac{(n+1)!}{(1-x)^{n+2}}. - Thus - - \abs{R_n(x)} \leq \frac{(n+1)!}{(1-x)^{n+2}}\frac{1}{(n+1)!}\abs{x^{n+1}}= \frac{(x-1)^{n+1}}{n+1} - . -

    - -

    - For a fixed x, - - \lim_{n\to\infty} \frac{(x-1)^{n+1}}{n+1}=0 - - since \abs{x}\lt 1, hence - - \frac{1}{1-x} = \infser[0] x^n \text{ on } (-1,0) - . -

    -     - -     - -     - - - -

    - Use the Taylor series given in to verify the given identity. -

    -     - - - - -

    - \cos(-x) = \cos(x) -

    -     - -

    - Given \ds \cos(x) = \infser[0] (-1)^n\frac{x^{2n}}{(2n)!}, -

    - -

    - \ds\cos(-x) = \infser[0] (-1)^n\frac{(-x)^{2n}}{(2n)!}=\infser[0] (-1)^n\frac{x^{2n}}{(2n)!}=\cos(x), - as all powers in the series are even. -

    -     - -     - - - - -

    - \sin(-x) = -\sin(x) -

    -     - -

    - Given \ds \sin(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!}, -

    - -

    - \ds\sin(-x) = \infser[0] (-1)^n\frac{(-x)^{2n+1}}{(2n+1)!}=\infser[0] (-1)^n\frac{-x^{2n+1}}{(2n+1)!}=-\sin(x), - as all powers in the series are odd. -

    -     - -     - - - - -

    - \frac{d}{dx}\big(\sin(x) \big) = \cos(x) -

    -     - -

    - Given \ds \sin(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!}, -

    - -

    - \ds\frac{d}{dx}\big(\sin(x) \big) = \frac{d}{dx}\left(\infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!}\right)=\infser[0] (-1)^n\frac{(2n+1)x^{2n}}{(2n+1)!}=\infser[0] (-1)^n\frac{x^{2n}}{(2n)!}=\cos(x). (The summation still starts at n=0 as there was no constant term in the expansion of \sin(x)). -

    -     - -     - - - - -

    - \frac{d}{dx}\big(\cos(x) \big) = -\sin(x) -

    -     - -

    - Given \ds \cos(x) = \infser[0] (-1)^n\frac{x^{2n}}{(2n)!}, -

    - -

    - \ds\frac{d}{dx}\big(\cos(x) \big) = \frac{d}{dx}\left(\infser[0] (-1)^n\frac{x^{2n}}{(2n)!}\right)=\infser (-1)^n\frac{(2n)x^{2n-1}}{(2n)!}=\infser (-1)^n\frac{x^{2n-1}}{(2n-1)!}. - We can re-index this summation to start at n=0 by replacing n with n+1 in the summation: - - \infser (-1)^n\frac{x^{2n-1}}{(2n-1)!} =\infser[0] (-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!} - . -

    - -

    - Note that this series has the opposite sign of the Taylor series for \sin(x); - thus \frac{d}{dx}(\cos(x) ) = -\sin(x). -

    -     - -     - -     - - - -

    - Write out the first 5 terms of the Binomial series with the given k-value. -

    -     - - - - -

    - k=1/2 -

    -     - -

    - \ds 1+\frac x2-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128} -

    -     - -     - - - - -

    - k=-1/2 -

    -     - -

    - \ds 1-\frac x2+\frac{3x^2}{8}-\frac{5x^3}{16}+\frac{35x^4}{128} -

    -     - -     - - - - -

    - k=1/3 -

    -     - -

    - \ds 1+\frac x3-\frac{x^2}{9}+\frac{5x^3}{81}-\frac{10x^4}{243} -

    -     - -     - - - - -

    - k=4 -

    -     - -

    - \ds 1+4x+6x^2+4x^3+x^4 (note the series is finite, - and the formula still applies) -

    -     - -     - -     - - - -

    - Use the Taylor series given in - to create the Taylor series of the given functions. -

    -     - - - - -

    - f(x) = \cos\big(x^2\big) -

    -     - -

    - \ds \infser[0] (-1)^n\frac{(x^2)^{2n}}{(2n)!} = \infser[0] (-1)^n\frac{x^{4n}}{(2n)!}. -

    -     - -     - - - - -

    - f(x) = e^{-x} -

    -     - -

    - \ds \infser[0] \frac{(-x)^n}{n!}. -

    -     - -     - - - - -

    - f(x) = \sin\big(2x+3\big) -

    -     - -

    - \ds \infser[0] (-1)^n\frac{(2x+3)^{2n+1}}{(2n+1)!}. -

    -     - -     - - - - -

    - f(x) = \tan^{-1}\big(x/2\big) -

    -     - -

    - \ds \infser[0] (-1)^n\frac{(x/2)^{2n+1}}{(2n+1)}. -

    -     - -     - - - - -

    - f(x) = e^x\sin(x)(only find the first 4 terms) -

    -     - -

    - \ds x+x^2+\frac{x^3}{3}-\frac{x^5}{30} -

    -     - -     - - - - -

    - f(x) = (1+x)^{1/2}\cos(x)(only find the first 4 terms) -

    -     - -

    - \ds 1+\frac x2-\frac{5x^2}{8}-\frac{3x^3}{16} -

    -     - -     - -     - - - -

    - Approximate the value of the given definite integral by using the first 4 nonzero terms of the integrand's Taylor series. -

    -     - - - - -

    - \ds \int_0^{\sqrt{\pi}} \sin\big(x^2\big)\, dx -

    -     - -

    - \ds \int_0^{\sqrt{\pi}} \sin\big(x^2\big)\, dx \approx \int_0^{\sqrt{\pi}} \left(x^2-\frac{x^6}6+\frac{x^{10}}{120}-\frac{x^{14}}{5040}\right) dx = 0.8877 -

    -     - -     - - - - -

    - \ds \int_0^{\sqrt[3]{\pi}} \cos\left(x^3\right)\, dx -

    -     - -

    - \ds \int_0^{\sqrt[3]{\pi}} \cos\big(x^3\big)\, dx \approx \int_0^{\pi^2/4} \left(1-\frac{x^6}{2!}+\frac{x^{12}}{4!}-\frac{x^{18}}{6!}\right)\, dx = 0.7864. -

    -     - -     - -     -     -     -
    -     - - - Curves in the Plane - -

    - We have explored functions of the form y=f(x) closely throughout this text. - We have explored their limits, - their derivatives and their antiderivatives; - we have learned to identify key features of their graphs, - such as relative maxima and minima, - inflection points and asymptotes; - we have found equations of their tangent lines, - the areas between portions of their graphs and the x-axis, - and the volumes of solids generated by revolving portions of their graphs about a horizontal or vertical axis. -

    - -

    - Despite all this, - the graphs created by functions of the form y=f(x) are limited. - Since each x-value can correspond to only 1 y-value, - common shapes like circles cannot be fully described by a function in this form. - Fittingly, the vertical line test - excludes vertical lines from being functions of x, - even though these lines are important in mathematics. -

    - -

    - In this chapter we'll explore new ways of drawing curves in the plane. - We'll still work within the framework of functions, - as an input will still only correspond to one output. - However, our new techniques of drawing curves will render the vertical line test pointless, - and allow us to create important and beautiful new curves. - Once these curves are defined, - we'll apply the concepts of calculus to them, - continuing to find equations of tangent lines and the areas of enclosed regions. -

    -     - -
    - Conic Sections - -

    - The ancient Greeks recognized that interesting shapes can be formed by intersecting a plane with a - double napped cone (, two identical cones placed tip-to-tip as shown in the following figures). - As these shapes are formed as sections of conics, - they have earned the official name conic sections. -

    - - - -

    - The three most interesting - conic sections are given in the top row of . - They are the parabola, the ellipse - (which includes circles) - and the hyperbola. - In each of these cases, the plane does not intersect the tips of the cones - (usually taken to be the origin). -

    - -
    - Conic Sections - - -
    - Parabola - - A diagonal plane intersecting a double napped cone, forming a parabola - -

    - Two cones stacked tip-to-tip, intersected by a plane. - A diagonal plane intersects the top cone. - The plane does not touch the tip of either cones. - The top edge of the plane intersects the top edge of the cone. - The parts of the plane intersecting the cone are highlighted, which forms a parabola in the plane. -

    -     - -
    - -
    - Ellipse - - A diagonal plane intersecting a double napped cone, forming an ellipse. - -

    - Two cones stacked tip-to-tip, intersected by a plane. - A diagonal plane intersects the top cone. - The plane does not touch the tip of either cones. - The edge of the plane does not touch the top edge of the cone. - The parts of the plane intersecting the cone are highlighted, which forms an ellipse in the plane. -

    -     - -
    - -
    - Circle - - A horizontal plane intersecting a double napped cone, forming a circle. - -

    - Two cones stacked tip-to-tip, intersected by a plane. - A horizontal plane intersects the top cone. - The parts of the plane intersecting the cone are highlighted, which forms a circle in the plane. -

    -     - -
    - -
    - Hyperbola - - A vertical plane intersecting a double napped cone, forming a hyperbola. - -

    - Two cones stacked tip-to-tip, intersected by a vertical plane. - The plane does not pass through the tips of the cones. - The parts of the plane intersecting the cones are highlighted, which forms a hyperbola in the plane. - The hyperbola contains two seperate segments, with one placed vertically above the other with a space between them. - The top segment is in the shape of a rounded v, and the bottom segment is a vertical reflection of the top segment. -

    -     - -
    -     - - -
    - Point - - A horizontal plane intersecting the tips in a double napped cone. - -

    - Two cones stacked tip-to-tip, intersected by a horizontal plane. - The plane passes through the point where the tips of the two cones are touching. - The plane touches the cone at only that single point, forming a point in the plane. -

    -     - -
    - -
    - Line - - A diagonal plane intersecting a double napped cone, forming a line in the plane - -

    - Two cones stacked tip-to-tip, intersected by a diagonal plane. - The plane extends diagonally through the point where the tips of the cones are touching. - The plane only touches the outer edges of the cones. - The parts of the plane touching the cones are highlighted, forming a straight line in the plane. -

    -     - -
    - -
    - Crossed Lines - - A vertical plane intersecting a double napped cone, forming crossed lines in the plane. - -

    - Two cones stacked tip-to-tip, intersected by a vertical plane. - The plane extends vertically, passing through the points at which the tips of the cones touch. - The parts of the plane intersecting the cones are highlighted, forming crossed straight lines in the plane. - The image on the plane is straight lines in an X shape. -

    -     - -
    -     -     -
    - - -

    - When the plane does contain the origin, - three degenerate cones can be formed as shown the bottom row of : - a point, a line, and crossed lines. - We focus here on the nondegenerate cases. - conic sections - conic sectionsdegenerate -

    - -

    - While the above geometric constructs define the conics in an intuitive, - visual way, - these constructs are not very helpful when trying to analyze the shapes algebraically or consider them as the graph of a function. - It can be shown that all conics can be defined by the general second-degree equation - - Ax^2+Bxy+Cy^2+Dx+Ey+F=0 - . -

    - -

    - While this algebraic definition has its uses, - most find another geometric perspective of the conics more beneficial. -

    - -

    - Each nondegenerate conic can be defined as the locus, or set, - of points that satisfy a certain distance property. - These distance properties can be used to generate an algebraic formula, - allowing us to study each conic as the graph of a function. -

    -     - - - Parabolas - - - - - Parabola - -

    - A parabola is the locus of all points equidistant from a point - (called a focus) - and a line - (called the directrix) - that does not contain the focus. - conic sectionsparabola - paraboladefinition - directrix - focus -

    -     -     - -
    - Illustrating the definition of the parabola and establishing an algebraic formula - - - A sketch of a parabola with key components labeled. - -

    - An upward opening parabola with labels for key components. - At the bottom of the parabola a point is labeled as the vertex. - A dashed vertical line, labeled the axis of symmetry, separates the parabola into two halves. - The axis of symmetry crosses the parabola through the vertex. - Below the parabola, a horizontal line is labeled the Directrix. - The distance between the directrix and the vertex is labeled p. - Above the vertex, a point is drawn on the axis of symmetry and is labeled the "focus". - The distance between the vertex and the focus is p, the same distance between the vertex and the directrix. - On the right side of the parabola, a point is labeled (x,y). - The distance between the point and the focus, and the distance between the point and the directrix are equal, labeled d. -

    -     - - - \begin{tikzpicture}[scale=.88] - - \draw [thick,firstcolor] (-3,-1) node [above,black,shift={(12pt,0pt)}] { Directrix} -- (3,-1); - \filldraw [black] (0,1) circle (2.4pt) node [above left] { Focus}; - - \draw [thick,secondcolor] (-3,2.25) parabola bend (0,0) (3,2.25); - \filldraw (0,0) circle (2.4pt) node [below left] { Vertex}; - - \draw (.3,.5) node[] { $\left.\rule{0pt}{12pt}\right\}p$}; - \draw (.3,-.5) node[] { $\left.\rule{0pt}{12pt}\right\}p$}; - - \coordinate (A) at (2.5,1.5625); - - \filldraw [black] (A) circle (2.4pt) node [right] { $(x,y)$}; - \draw [thick,dashed] (0,1) -- (A) node [pos=.5,above] { $d$} -- (2.5,-1) node [pos=.5,right] { $d$}; - \draw [thick,dashed] (0,-1.5) -- node [pos=.85,above,rotate=90] { Axis of} node [pos=.85,below,rotate=90] { Symmetry} (0,3.2); - - \end{tikzpicture} - - - - -
    - -

    - illustrates this definition. - The point halfway between the focus and the directrix is the vertex. - The line through the focus, - perpendicular to the directrix, - is the axis of symmetry, - as the portion of the parabola on one side of this line is the mirror-image of the portion on the opposite side. -

    - -

    - The definition leads us to an algebraic formula for the parabola. - Let P=(x,y) be a point on a parabola whose focus is at F=(0,p) and whose directrix is at y=-p. (We'll assume for now that the focus lies on the y-axis; - by placing the focus p units above the x-axis and the directrix p units below this axis, - the vertex will be at (0,0).) -

    - -

    - We use the Distance Formula to find the distance d_1 between F and P: - - d_1=\sqrt{(x-0)^2+(y-p)^2} - . -

    - -

    - The distance d_2 from P to the directrix is more straightforward: - - d_2=y-(-p) = y+p - . -

    - -

    - These two distances are equal. - Setting d_1=d_2, we can solve for y in terms of x: - - d_1\amp = d_2 - \sqrt{x^2+(y-p)^2} \amp = y+p - Now square both sides. - x^2+(y-p)^2 \amp = (y+p)^2 - x^2+y^2-2yp+p^2 \amp = y^2+2yp+p^2 - x^2 \amp =4yp - y\amp = \frac{1}{4p}x^2 - . -

    - -

    - The geometric definition of the parabola has led us to the familiar quadratic function whose graph is a parabola with vertex at the origin. - When we allow the vertex to not be at (0,0), - we get the following standard form of the parabola. -

    - - - General Equation of a Parabola -

    -

      -
    1. -

      - Vertical Axis of Symmetry: - The equation of the parabola with vertex at (h,k) and directrix y=k-p in standard form is - - y=\frac{1}{4p}(x-h)^2+k - . - The focus is at (h,k+p). -

      -
      2. - -
    2. -

      - Horizontal Axis of Symmetry: - The equation of the parabola with vertex at (h,k) and directrix x=h-p in standard form is - - x=\frac{1}{4p}(y-k)^2+h - . - The focus is at (h+p,k). -

      -
      3. -
    -

    - -

    - Note: p is not necessarily a positive number. - parabolageneral equation -

    -     - - - Finding the equation of a parabola - -

    - Give the equation of the parabola with focus at (1,2) and directrix at y=3. -

    -     - -

    - The vertex is located halfway between the focus and directrix, - so (h,k) = (1,2.5). - This gives p=-0.5. - Using - we have the equation of the parabola as - - y=\frac{1}{4(-0.5)}(x-1)^2+2.5 = -\frac12(x-1)^2+2.5 - . -

    - -
    - The parabola described in - - - A downward opening parabola with a vertex in the first quadrant. - -

    - A downward opening parabola with a vertex in the first quadrant. - The vertex of the parabola is at (1,2.5). - The parabola crosses the x-axis at around x = -1.2 and x = 3.2. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-6.5,ymax=3.0, - xmin=-3.5,xmax=5.5 - ] - - \addplot+ [domain=-3:5,samples=40] {-.5*(x-1)^2+2.5}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - The parabola is sketched in . -

    -     - -     - - - Finding the focus and directrix of a parabola - -

    - Find the focus and directrix of the parabola x=\frac18y^2-y+1. - The point (7,12) lies on the graph of this parabola; - verify that it is equidistant from the focus and directrix. -

    -     - -

    - We need to put the equation of the parabola in its general form. - This requires us to complete the square: - - x \amp = \frac18y^2-y+1 - \amp = \frac18\big(y^2-8y+8\big) - \amp = \frac18\big(y^2-8y+16 -16+8\big) - \amp = \frac18\big((y-4)^2 - 8\big) - \amp = \frac18(y-4)^2 -1 - . -

    - -

    - Hence the vertex is located at (-1,4). - We have \frac18=\frac1{4p}, so p=2. - We conclude that the focus is located at (1,4) and the directrix is x=-3. - The parabola is graphed in , - along with its focus and directrix. -

    - -
    - The parabola described in . The distances from a point on the parabola to the focus and directrix are given. - - - A parabola opening to the right with its directrix and focus drawn. - -

    - A rightwards opening parabola with a vertex at (-1,4). - The focus of the parabola is drawn at the point (1,4). - The directrix is drawn as the vertical line at x=-3. - The point (7,12) is drawn lying on the parabola. - The distance between the point and the directrix is drawn. - The distance between the point and the focus is drawn. - Both distances are labeled as 10 units long. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-5.5,ymax=14.5, - xmin=-11,xmax=14 - ] - - \addplot+ [domain=-5:14,samples=40] ({x^2/8-x+1},x); - \addplot+ [solid,domain=-5:14] (-3,x); - - \draw [dashed,firstcolor!70,thick] (axis cs:-3,12) -- node [above,pos=.5,black] { 10} (axis cs:7,12) -- node [pos=.6,right,black] { 10} (axis cs: 1,4); - - \filldraw [secondcolor] (axis cs: 1,4) circle (2.4pt) (axis cs: 7,12) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - The point (7,12) lies on the graph and is - 7-(-3)=10 units from the directrix. - The distance from (7,12) to the focus is: - - \sqrt{(7-1)^2 + (12-4)^2} = \sqrt{100}=10 - . -

    - -

    - Indeed, the point on the parabola is equidistant from the focus and directrix. -

    -     - -     - - - Reflective Property -

    - One of the fascinating things about the nondegenerate conic sections is their reflective properties. - Parabolas have the following reflective property: -

    - -
    -

    - Any ray emanating from the focus that intersects the parabola reflects off along a line perpendicular to the directrix. -

    -
    - -

    - This is illustrated in . The following theorem states this more rigorously. -

    - -
    - Illustrating the parabola's reflective property - - - A rightward opening parabola demonstrating the reflective property. - -

    - A parabola opening towards the right. - The directrix is drawn as a vertical line to the left of the parabola. - The focus is drawn as a point to the right of the vertex. - Four different rays are emanating from the focus. - The first line extends towards the top right, where it then reflects off the parabola and extends to the right. - The second line extends upwards, where it then reflects off of the parabola and extends towards the right. - The third line extends towards the lower left, where it then reflects off the parabola and extends towards the right. - The fourth line exends twards the lower right, where it once again reflects off the parbola and extends towards the right. - All of the rays extending towards the right are perpendicular to the directrix. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis y line=none, - axis x line=none, - ymin=-6.5,ymax=14.5, - xmin=-4,xmax=12 - ] - - \coordinate (focus) at (axis cs: 1,4); - - \addplot+ [domain=-6:14,samples=40] ({x^2/8-x+1},x); - \addplot+ [solid,domain=-6:14] (-3,x); - - \draw [thick] (axis cs: 12,12) -- (axis cs: 7,12) -- (focus) - (axis cs: 12,8) -- (axis cs: 1,8) -- (focus) - (axis cs: 12,3) -- (axis cs: -.875,3) - (axis cs: -.875,3) -- (focus) - (axis cs: 12,-3) -- (axis cs: 5.125,-3) -- (focus); - - \filldraw [secondcolor] (focus) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    - - - Reflective Property of the Parabola - -

    - Let P be a point on a parabola. - The tangent line to the parabola at P makes equal angles with the following two lines: -

    - -

    -

      -
    1. -

      - The line containing P and the focus F, and -

      -
      2. - -
    2. -

      - The line perpendicular to the directrix through P. - parabolareflective property -

      -
      3. -
    -

    -     -     - -

    - Because of this reflective property, - paraboloids (the 3D analogue of parabolas) make for useful flashlight reflectors as the light from the bulb, - ideally located at the focus, - is reflected along parallel rays. - Satellite dishes also have paraboloid shapes. - Signals coming from satellites effectively approach the dish along parallel rays. - The dish then focuses these rays at the focus, - where the sensor is located. -

    -     -     - - - Ellipses - - - - Ellipse - -

    - An ellipse is the locus of all points whose sum of distances from two fixed points, - each a focus of the ellipse, is constant. - conic sectionsellipse - ellipsedefinition - focus -

    -     -     - -

    - An easy way to visualize this construction of an ellipse is to pin both ends of a string to a board. - The pins become the foci. - Holding a pencil tight against the string places the pencil on the ellipse; - the sum of distances from the pencil to the pins is constant: - the length of the string. - See . -

    - -
    - Illustrating the construction of an ellipse with pins, pencil and string - - - A demonstration of the creation of an ellipse from two foci. - -

    - Two points are drawn, around which an ellipse is sketched. - A pencil is drawn, representing the sketching of the ellipse. - The point of the pencil is touching the ellipse in the top right of the ellipse. - The distances between the point of the pencil and the two foci are drawn. - The distance between the left focus is labeled d_2, and the distance between the right focus is labeled d_1. - It is clear that d_2 is larger than d_1. - Beneath the foci, it is written that d_1 + d_2 = \mathrm{constant}. -

    -     - - - \begin{tikzpicture}[scale=1.32] - - \draw [thick,dashed] (0,0) circle [x radius=2,y radius=1.5]; - - \filldraw (1.3,0) circle (2.4pt) - (-1.3,0) circle (2.4pt); - - \draw (1.3,0) -- node [right,pos=.5] { $d_1$} (1.39,1.07) -- node [above,pos=.5] { $d_2$} (-1.3,0); - \draw (0,-.7) node { $d_1+d_2=$ constant}; - - \begin{scope}[shift={(1.9cm,1.225cm)}] - \begin{scope}[rotate=120] - \begin{scope}[xscale=.25,yscale=.5] - - \draw (0,0) -- (0,-2) -- (1,-2)--(1,0) -- (.5,1) -- cycle; - \draw [fill=black] (.3,.6) -- ( .5,1)--(.7,.6)--cycle; - \draw (0,0) cos (.5,-.1) sin (1,0); - - \end{scope} - \end{scope} - \end{scope} - - \end{tikzpicture} - - - - -
    - -

    - We can again find an algebraic equation for an ellipse using this geometric definition. - Let the foci be located along the x-axis, - c units from the origin. - Let these foci be labeled as - F_1 = (-c,0) and F_2=(c,0). - Let P=(x,y) be a point on the ellipse. - The sum of distances from F_1 to P (d_1) and from F_2 to P (d_2) is a constant d. - That is, d_1+d_2=d. - Using the Distance Formula, we have - - \sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = d - . -

    - -

    - Using a fair amount of algebra can produce the following equation of an ellipse (note that the equation is an implicitly defined function; - it has to be, as an ellipse fails the Vertical Line Test): - - \frac{x^2}{\left(\frac d2\right)^2} + \frac{y^2}{\left(\frac d2\right)^2-c^2} = 1 - . -

    - -

    - This is not particularly illuminating, - but by making the substitution a=d/2 and b=\sqrt{a^2-c^2}, - we can rewrite the above equation as - - \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 - . -

    - -

    - This choice of a and b is not without reason; - as shown in , - the values of a and b have geometric meaning in the graph of the ellipse. -

    - -
    - Labeling the significant features of an ellipse - - - An ellipse with labels for key components. - -

    - An ellipse is drawn with labels for key features. - The ellipse is separated into 4 equal sections by a vertical and a horizontal line, which both pass through the center of the ellipse. - The vertical line dividing the ellipse in half vertically is labeled as the "minor axis." - The horizontal dividing the ellipse in half horizontally is labeled as the "major axis." - The points on the major axis touching the ellipse are labeled as "vertices". - Two points on the major axis are labeled as "foci". The foci both lie a short distance away from the vertices. - The distance between the foci and their respective vertices is equal. - The distance between the foci and the center is also equal. - The distance between the left vertex and the minor axis is labeled a. The length of the major axis is 2a. - The length between the center and the top of the minor axis is labeled b. The length of the minor axis is 2b - The distance between the center and the right focus is labeled c. -

    -     - - - \begin{tikzpicture}[>=latex] - - \draw [thick,firstcolor] (0,0) circle [x radius=2,y radius=1.5]; - \draw [thick,dashed] (-2.4,0) -- (2.4,0) (0,1.8) -- (0,-1.9); - - \filldraw [secondcolor] (1.3,0) circle (2.4pt) (-1.3,0) circle (2.4pt); - - \draw [->] (-2,-2) node [fill=white] { Major axis} -- (-2.2,-.1); - \draw [->] (2,-2) node [fill=white] { Minor axis} -- (.1,-1.8); - \draw [->] (-2,2) -- (-2,.3); - \draw [->] (-2,2) node [fill=white] { Vertices} -- (1.9,.1); - \draw [->] (2,2) -- (-1.2,.1); - \draw [->] (2,2) node [fill=white] { Foci} -- (1.3,.1); - - \filldraw [firstcolor] (2,0) circle (2.4pt) - (-2,0) circle (2.4pt); - - \draw (-1,-.25) node { $\underbrace{\rule{1.8cm}{0pt}}_a$}; - \draw (.65,-.25) node { $\underbrace{\rule{1.1cm}{0pt}}_c$}; - \draw (-.25,.75) node [] { $b\left\{\rule[-.65cm]{0pt}{1.3cm}\right.$}; - - \end{tikzpicture} - - - - -
    - -

    - In general, the two foci of an ellipse lie on the - major axis of the ellipse, - and the midpoint of the segment joining the two foci is the center. - The major axis intersects the ellipse at two points, - each of which is a vertex. - The line segment through the center and perpendicular to the major axis is the minor axis. - The constant sum of distances - that defines the ellipse is the length of the major axis, - , 2a. -

    - -

    - Allowing for the shifting of the ellipse gives the following standard equations. -

    - - - Standard Equation of the Ellipse -

    - The equation of an ellipse centered at (h,k) with major axis of length 2a and minor axis of length 2b in standard form is: -

    - -

    -

      -
    1. -

      - Horizontal major axis: - \ds \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1. -

      -
      2. - -
    2. -

      - Vertical major axis: \ds \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1. -

      -
      3. -
    -

    - -

    - The foci lie along the major axis, - c units from the center, where c^2=a^2-b^2. - ellipsestandard equation -

    -     - - - Finding the equation of an ellipse - -

    - Find the general equation of the ellipse graphed in . -

    - -
    - The ellipse used in - - - An ellipse centered in the second quadrant, lying entirely in the second and third quadrants. - -

    - An ellipse with a center at (-3,1). - The ellipse has vertices at (-3,6) and (-3,4). - The ellipse appears to have a minor axis of length 4. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - ytick={-4,-2,2,4,6}, - ymin=-4.9,ymax=6.9, - xmin=-6.9,xmax=6.9 - ] - - \addplot+ [domain=0:360,samples=60] ({2*cos(x)-3},{5*sin(x)+1}); - \addplot [soliddot,color=black,fill=black] coordinates {(-3,1)}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - The center is located at (-3,1). - The distance from the center to a vertex is 5 units, hence a=5. - The minor axis seems to have length 4, so b=2. - Thus the equation of the ellipse is - - \frac{(x+3)^2}{4}+\frac{(y-1)^2}{25} = 1 - . -

    -     - -     - - - Graphing an ellipse - -

    - Graph the ellipse defined by 4x^2+9y^2-8x-36y=-4. -

    -     - -

    - It is simple to graph an ellipse once it is in standard form. - In order to put the given equation in standard form, - we must complete the square with both the x and y terms. - We first rewrite the equation by regrouping: - - 4x^2+9y^2-8x-36y=-4 \Rightarrow (4x^2-8x) + (9y^2-36y) = -4 - . -

    - -

    - Now we complete the squares. - - (4x^2-8x) + (9y^2-36y) \amp = -4 - 4(x^2-2x) + 9(y^2-4y) \amp = -4 - 4(x^2-2x +1 - 1) + 9(y^2-4y+4-4) \amp = - 4 - 4\big((x-1)^2-1\big) + 9\big((y-2)^2-4\big) \amp = -4 - 4(x-1)^2 -4 + 9(y-2)^2-36 \amp = -4 - 4(x-1)^2 + 9(y-2)^2 \amp = 36 - \frac{(x-1)^2}{9} + \frac{(y-2)^2}{4} \amp = 1 - . -

    - -

    - We see the center of the ellipse is at (1,2). - We have a=3 and b=2; - the major axis is horizontal, - so the vertices are located at (-2,2) and (4,2). - We find c=\sqrt{9-4} = \sqrt{5}\approx 2.24. - The foci are located along the major axis, - approximately 2.24 units from the center, at (1\pm 2.24,2). - This is all graphed in -

    - -
    - Graphing the ellipse in - - - An ellipse centered at (1,2), with vertices at (-2,2) and (4,2). - -

    - An ellipse centered at (1,2). - The vertices of the ellipse are drawn at (-2,2) and (4,2). - Two foci are drawn at (1 - 2.24, 2) and (1 + 2.24, 2). - The minor axis of the ellipse has a length of 4. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-2,-1,1,2,3,4}, - ytick={-1,1,2,3,4}, - ymin=-2,ymax=5, - xmin=-3,xmax=5 - ] - - \addplot+ [domain=0:360,samples=60] ({3*cos(x)+1},{2*sin(x)+2}); - - \filldraw (axis cs:1,2) circle (2.4pt) - (axis cs:3.24,2) circle (1pt) - (axis cs:-1.24,2) circle (1pt); - - \filldraw [firstcolor] (axis cs: 4,2) circle (2.4pt) - (axis cs: -2,2) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -     - - - Eccentricity -

    - When a=b, we have a circle. - The general equation becomes - - \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{a^2} = 1 \Rightarrow (x-h)^2 + (y-k)^2 = a^2 - , - the familiar equation of the circle centered at (h,k) with radius a. - Since a=b, c = \sqrt{a^2-b^2}=0. - The circle has two foci, - but they lie on the same point, the center of the circle. -

    - -

    - Consider , - where several ellipses are graphed with a=1. - In , we have c=0 and the ellipse is a circle. - As c grows, - the resulting ellipses look less and less circular. - A measure of this noncircularness - is eccentricity. -

    - -
    - Understanding the eccentricity of an ellipse - - -
    - - - - An ellipse of eccentricity 0. - -

    - A circle of radius 1 centered at the orgin. - The major and minor axis both have a length of 2. - Below the ellipse it is written that e = 0. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-1,1}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.35,xmax=1.35 - ] - - \addplot+ [domain=0:360,samples=60] ({cos(x)},{sin(x)}); - - \filldraw (axis cs:0,0) circle (2.4pt); - - \draw (axis cs:.8,-.9) node { $e=0$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - An ellipse of eccentricity 0.3. - -

    - An ellipse with foci at around (-0.2, 0) and (0.2,2). - The major axis has a length of 2 and the minor axis has a length of about 1.8. - It is written that e=0.3 -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-1,1}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.35,xmax=1.35 - ] - - \addplot+ [domain=0:360,samples=60] ({cos(x)},{.95*sin(x)}); - - \filldraw (axis cs:.3,0) circle (2.4pt) - (axis cs:-.3,0) circle (2.4pt); - - \draw (axis cs:.8,-.9) node { $e=0.3$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - - -
    - - - - An ellipse of eccentricity 0.8. - -

    - An ellipse with foci drawn at (-0.8,0) and (0.8,0). - The major axis has a length of 2, and the minor axis has a length of around 1. - Below the ellipse it is written that e = 0.8. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-1,1}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.35,xmax=1.35 - ] - - \addplot+ [domain=0:360,samples=60] ({cos(x)},{.6*sin(x)}); - - \filldraw (axis cs:.8,0) circle (2.4pt) - (axis cs:-.8,0) circle (2.4pt); - - \draw (axis cs:.8,-.9) node { $e=0.8$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - An ellipse of eccentricity 0.99. - -

    - An ellipse with eccentricity 0. - The vertices and the foci are at the same points, one at (-1,0) and (1,0). - The major axis has a length of 2, and the minor axis has a length of about 0.2. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-1,1}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.35,xmax=1.35 - ] - - \addplot+ [domain=0:360,samples=60] ({cos(x)},{.141*sin(x)}) ; - - \draw (axis cs:.8,-.9) node { $e=0.99$}; - - \filldraw (axis cs:.99,0) circle (2.4pt) - (axis cs:-.99,0) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -     - -
    - - - Eccentricity of an Ellipse - -

    - The eccentricity e of an ellipse is \ds e=\frac{c}{a}. - ellipseeccentricity - eccentricity -

    -     -     - -

    - The eccentricity of a circle is 0; - that is, a circle has no noncircularness. - As c approaches a, - e approaches 1, giving rise to a very noncircular ellipse, - as seen in . -

    - -

    - It was long assumed that planets had circular orbits. - This is known to be incorrect; - the orbits are elliptical. - Earth has an eccentricity of 0.0167 it has a nearly circular orbit. - Mercury's orbit is the most eccentric, with e=0.2056. - (Pluto's eccentricity is greater, at e=0.248, - the greatest of all the currently known dwarf planets.) - The planet with the most circular orbit is Venus, with e=0.0068. - The Earth's moon has an eccentricity of e=0.0549, - also very circular. -

    -     - - - Reflective Property -

    - The ellipse also possesses an interesting reflective property. - Any ray emanating from one focus of an ellipse reflects off the ellipse along a line through the other focus, - as illustrated in . - This property is given formally in the following theorem. -

    - -
    - Illustrating the reflective property of an ellipse - - - An ellipse with rays emanating from the foci, demonstrating the reflective property of an ellipse. - -

    - An ellipse with the foci drawn. - The left focus is labeled F_1, and the right focus is labeled F_2. - A ray is emanating from F_1 towards the bottom right. - It then reflects off the ellipse to the upper right, where it then passes through F_2. - A ray also emanates from F_2 towards the upper right. - It then reflects off the ellipse towards the bottom left, where it then passes through F_1. -

    -     - - - \begin{tikzpicture}[scale=1.2,>=latex,scale=1.32] - - \draw [thick] (0,0) circle [x radius=2cm,y radius = 1.2cm]; - - \filldraw (1.6,0) circle (2.4pt) node [above left] { $F_2$} - (-1.6,0) circle (2.4pt) node [above] { $F_1$}; - - \draw [->,firstcolor,thick] (1.6,0) -- ({2*cos(30)},{1.2*sin(30)}); - \draw [->,firstcolor,thick] ({2*cos(30)},{1.2*sin(30)}) -- (-1.55,0); - \draw [->,secondcolor,thick] (-1.6,0) -- ({2*cos(250)},{1.2*sin(250)}); - \draw [->,secondcolor,thick] ({2*cos(250)},{1.2*sin(250)}) -- (1.55,0); - - \end{tikzpicture} - - - - -
    - - - Reflective Property of an Ellipse - -

    - Let P be a point on a ellipse with foci F_1 and F_2. - The tangent line to the ellipse at P makes equal angles with the following two lines:ellipsereflective property -

    - -

    -

      -
    1. -

      - The line through F_1 and P, and -

      -
      2. - -
    2. -

      - The line through F_2 and P. -

      -
      3. -
    -

    -     -     - -

    - This reflective property is useful in optics and is the basis of the phenomena experienced in whispering halls. -

    -     -     - - - Hyperbolas - - - -

    - The definition of a hyperbola is very similar to the definition of an ellipse; - we essentially just change the word - sum to difference. -

    - - - Hyperbola - -

    - A hyperbola is the locus of all points where the absolute value of difference of distances from two fixed points, - each a focus of the hyperbola, is constant. - conic sectionshyperbola - hyperboladefinition - focus -

    -     -     - -

    - We do not have a convenient way of visualizing the construction of a hyperbola as we did for the ellipse. - The geometric definition does allow us to find an algebraic expression that describes it. - It will be useful to define some terms first. -

    - -

    - The two foci lie on the transverse axis of the hyperbola; - the midpoint of the line segment joining the foci is the - center of the hyperbola. - The transverse axis intersects the hyperbola at two points, - each a vertex of the hyperbola. - The line through the center and perpendicular to the transverse axis is the - conjugate axis. - This is illustrated in . - It is easy to show that the constant difference of distances used in the definition of the hyperbola is the distance between the vertices, - , 2a. -

    - -
    - Labeling the significant features of a hyperbola - - - A hyperbola with labels for key components - -

    - A hyperbola with labels for key components. - The hyperbola is composed of two seperate segments with a space between them. - The first shape is on the left. The shape begins in the upper left, moving almost linearly towards the lower right. - The curve then begins to bend gently backwards before moving linearly towards the bottom left. - The right curve is a reflection of the left curve. It opens towards the right. - The entire hyperbola is separated into 4 equal segments by two axes. - The axis dividing the hyperola in half vertically is labeled as the "conjugate axis." - The axis dividing th ehyperbola in half horizontally is labeled as the "transverse axis." - The point in which the axes touch is the center of the hyperbola. - The points in which the curves intersect the transverse axis are labeled as foci. - Two foci lie on the transverse axis. - The first focus is to the left of the left curve. - The right focus is to the right of the right curve. - The distance between the left vertex and the center is labeled as a. - The distance between the center and the right focus is labeled as c. -

    -     - - - \begin{tikzpicture}[>=latex] - - \begin{axis}[ - axis y line=none, - axis x line=none, - ymin=-3.2,ymax=3.2, - xmin=-3.2,xmax=3.2 - ] - - \addplot [firstcurvestyle,domain=-70:70,samples=40] ({sec(x)},{tan(x)}); - \addplot [firstcurvestyle,domain=-70:70,samples=40] ({-sec(x)},{tan(x)}); - - \filldraw (axis cs:0,0) circle (2.4pt); - - \draw [thick,dashed] (axis cs:-3.2,0) -- node [above,pos=.88] { Transverse} node [below,pos=.88] { axis} (axis cs:3.2,0) - (axis cs:0,-3.2) -- node [below,rotate=90,pos=.8] { axis }node [above,rotate=90,pos=.8] { Conjugate} (axis cs:0,3.2); - - \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt) - (axis cs:-1,0) circle (2.4pt); - - \filldraw [secondcolor] (axis cs:1.4,0) circle (2.4pt) - (axis cs:-1.4,0) circle (2.4pt); - - \draw [->] (axis cs: 1.25,-2.8) -- (axis cs: -1.35,-.1); - \draw [->] (axis cs: 1.25,-2.8) node [fill=white] { Foci } -- (axis cs: 1.4,-.1); - - \draw [->] (axis cs: -1.25,-2.8) -- (axis cs: -1,-.1); - \draw [->] (axis cs: -1.25,-2.8) node [fill=white] { Vertices} -- (axis cs: .95,-.1); - - \draw (axis cs:-.5,.4) node { $\overbrace{\rule{18pt}{0pt}}^a$}; - \draw (axis cs:.7,.4) node { $\overbrace{\rule{25pt}{0pt}}^c$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - - - Standard Equation of a Hyperbola -

    - The equation of a hyperbola centered at (h,k) in standard form is: - hyperbolastandard equation -

    - -

    -

      -
    1. -

      - Horizontal Transverse Axis: - \ds \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1. -

      -
      2. - -
    2. -

      - Vertical Transverse Axis: - \ds \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2} = 1. -

      -
      3. -
    -

    - -

    - The vertices are located a units from the center and the foci are located c units from the center, - where c^2 = a^2+b^2. -

    -     - - - Graphing Hyperbolas -

    - Consider the hyperbola \frac{x^2}9-\frac{y^2}1 = 1. - Solving for y, we find y=\pm\sqrt{x^2/9-1}. - As x grows large, - the -1 part of the equation for y becomes less significant and y\approx \pm\sqrt{x^2/9} = \pm x/3. - That is, as x gets large, - the graph of the hyperbola looks very much like the lines y=\pm x/3. - These lines are asymptotes of the hyperbola, - as shown in . -

    - -
    - Graphing the hyperbola \frac{x^2}9-\frac{y^2}1 = 1 along with its asymptotes, y=\pm x/3 - - - A graph of a hyperbola and its asymptotes. - -

    - A graph of the hyperbola \frac{x^2}{9} - \frac{y^2}{1} = 1, along with its two asymptotes. - The hyperbola is centered at the origin, with vertices at(-3,0) and (3,0). - Two asymptotes are drawn from the equations y = \frac{x}{3} and y = -\frac{x}{3}. - As the hyperbola extends outwards from the origin, it becomes close to the asymptotes. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-3.2,ymax=3.2, - xmin=-8.2,xmax=8.2 - ] - - \addplot [firstcurvestyle,domain=-70:70,samples=40] ({3*sec(x)},{tan(x)}); - \addplot [firstcurvestyle,domain=-70:70,samples=40] ({-3*sec(x)},{tan(x)}); - - \addplot [secondcurvestyle,domain=-8:8] {x/3}; - \addplot [secondcurvestyle,domain=-8:8] {-x/3}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - This is a valuable tool in sketching. - Given the equation of a hyperbola in general form, - draw a rectangle centered at (h,k) with sides of length 2a parallel to the transverse axis and sides of length 2b parallel to the conjugate axis. - (See - for an example with a horizontal transverse axis.) - The diagonals of the rectangle lie on the asymptotes. -

    - -
    - Using the asymptotes of a hyperbola as a graphing aid - - - An illustration of sketching a hyperbola with a rectangle. - -

    - A hyperbola sketched using a rectangle. - A center of the rectangle is drawn at the point (h,k). - The rectangle has a width of 2a and a height of 2b. - The vertices are drawn at the points (h-a,k) and (h+a,k). - The asymptotes are drawn as the lines passing through the center and the corners of the rectangle. - The hyperbola is then drawn, following the asymptotes, touching the vertices, and then moving away from the center to follow the asymptotes. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick=\empty, - extra x ticks={2,8,5}, - extra x tick labels={$h-a$, $h+a$,$h$}, - ytick=\empty, - extra y ticks={2,4,6}, - extra y tick labels={$k-b$, $k$, $k+b$}, - ymin=-2.2,ymax=10.2, - xmin=-3.2,xmax=13.2 - ] - - \addplot [firstcurvestyle,domain=-70:70,samples=40] ({3*sec(x)+5},{2*tan(x)+4}); - \addplot [firstcurvestyle,domain=-70:70,samples=40] ({-3*sec(x)+5},{2*tan(x)+4}); - - \draw [thick,dashed] (axis cs: 2,6) -- (axis cs:8,6) -- (axis cs:8,2) -- (axis cs: 2,2) -- cycle; - - \addplot [secondcurvestyle,domain=-3:13] {2*(x-5)/3+4}; - \addplot [secondcurvestyle,domain=-3:13] {-2*(x-5)/3+4}; - - \filldraw (axis cs:5,4) circle (2.4pt); - - \filldraw [firstcolor] (axis cs:2,4) circle (2.4pt); - \filldraw [firstcolor] (axis cs:8,4) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - These lines pass through (h,k). - When the transverse axis is horizontal, - the slopes are \pm b/a; - when the transverse axis is vertical, - their slopes are \pm a/b. - This gives equations: -

    - - - - - - -

    Horizontal Transverse Axis

    Vertical Transverse Axis -     - - \ds y=\pm\frac ba(x-h)+k - \ds y=\pm\frac ab(x-h)+k. - -     - - - Graphing a hyperbola - -

    - Sketch the hyperbola given by \ds \frac{(y-2)^2}{25}-\frac{(x-1)^2}{4}=1. -

    -     - -

    - The hyperbola is centered at (1,2); - a=5 and b=2. - In - we draw the prescribed rectangle centered at (1,2) along with the asymptotes defined by its diagonals. - The hyperbola has a vertical transverse axis, - so the vertices are located at (1,7) and (1,-3). - This is enough to make a good sketch. -

    - -
    - Graphing the hyperbola in - - - A graph of the hyperbola in this example - -

    - A graph of the hyperbola given in . - The hyperbola has the equation \frac{(y-2)^2}{25} - \frac{(x-1)^2}{4} = 1. - The center of the hyperbola is drawn at (1,2). - A rectangle is drawn around the center, with a height of 10 and a width of 4. - At the top and bottom of the rectangle, the vertices of the hyperbola are drawn at (1,7) and (1,-3). - The asymptotes of the hyperbola are drawn passing through the corners of the rectangle and the center. - The hyperbola is then drawn above and below the rectangle, opening upwards and downwards respectively. - The foci are also drawn a short distance above and below the vertices. - The top focus is near the point (1, 7.4) and the bottom focus is near the point (1,-3.4). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=4, - minor y tick num=4, - ymin=-7.2,ymax=11.2, - xmin=-5.2,xmax=7.2 - ] - - \addplot [firstcurvestyle,domain=-70:70,samples=40] ({2*tan(x)+1},{5*sec(x)+2}); - \addplot [firstcurvestyle,domain=-70:70,samples=40] ({2*tan(x)+1},{-5*sec(x)+2}); - - \draw [thick,dashed] (axis cs: -1,7) -- (axis cs:3,7) -- (axis cs:3,-3) -- (axis cs: -1,-3) -- cycle; - - \addplot [secondcurvestyle,domain=-3:13] {5*(x-1)/2+2}; - \addplot [secondcurvestyle,domain=-3:13] {-5*(x-1)/2+2}; - - \filldraw (axis cs:1,2) circle (2.4pt); - - \filldraw [firstcolor] (axis cs:1,7) circle (2.4pt); - \filldraw [firstcolor] (axis cs:1,-3) circle (2.4pt); - \filldraw [secondcolor] (axis cs:1,7.4) circle (2.4pt); - \filldraw [secondcolor] (axis cs:1,-3.4) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - We also find the location of the foci: - as c^2= a^2+b^2, we have c=\sqrt{29}\approx 5.4. - Thus the foci are located at - (1,2\pm 5.4) as shown in the figure. -

    -     - -     - - - Graphing a hyperbola - -

    - Sketch the hyperbola given by 9x^2-y^2+2y=10. -

    -     - -

    - We must complete the square to put the equation in general form. - (We recognize this as a hyperbola since it is a general quadratic equation and the x^2 and y^2 terms have opposite signs.) - - 9x^2-y^2+2y \amp =10 - 9x^2- (y^2-2y) \amp = 10 - 9x^2 - (y^2-2y+1-1) \amp = 10 - 9x^2 -\big((y-1)^2-1\big) \amp = 10 - 9x^2 - (y-1)^2 \amp = 9 - x^2 - \frac{(y-1)^2}{9} \amp =1 - -

    - -
    - Graphing the hyperbola in - - - The hyperbola described in this example - -

    - The hyperbola described in . - The hyperbola comes from the equation 9x^2 - y^2 + 2y = 10. - The hyperbola is centered at (0,1), around which a rectangle is drawn. - The rectangle has a height of 6 and a width of 2. - The vertices are drawn on the left and right sides of the rectangle, at the points (-1,1) and (1,1). - The hyperbola is the drawn, opening to the left and right and following the diagonal asymptotes, which are also drawn. - The foci are also drawn at the points (-3.2,1) and (3.2,1). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - minor y tick num=4, - ymin=-10.9,ymax=11.9, - xmin=-4.2,xmax=4.2 - ] - - \addplot [firstcurvestyle,domain=-75:75,samples=40] ({sec(x)},{3*tan(x)+1}); - \addplot [firstcurvestyle,domain=-75:75,samples=40] ({-sec(x)},{3*tan(x)+1}); - - \draw [thick,dashed] (axis cs: -1,4) -- (axis cs:1,4) -- (axis cs:1,-2) -- (axis cs: -1,-2) -- cycle; - - \addplot [secondcurvestyle,domain=-4:4] {3*(x-0)/1+1}; - \addplot [secondcurvestyle,domain=-4:4] {-3*(x-0)/1+1}; - - \filldraw (axis cs:0,1) circle (2.4pt); - - \filldraw [firstcolor] (axis cs:-1,1) circle (2.4pt); - \filldraw [firstcolor] (axis cs:1,1) circle (2.4pt); - \filldraw [secondcolor] (axis cs:3.2,1) circle (2.4pt); - \filldraw [secondcolor] (axis cs:-3.2,1) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - We see the hyperbola is centered at (0,1), - with a horizontal transverse axis, - where a=1 and b=3. - The appropriate rectangle is sketched in - along with the asymptotes of the hyperbola. - The vertices are located at (\pm 1,1). - We have c=\sqrt{10}\approx 3.2, - so the foci are located at - (\pm 3.2,1) as shown in the figure. -

    -     - -     -     - - - Eccentricity -

    - The eccentricity of a hyperbola is defined in the same way as an ellipse. -

    - - Eccentricity of a Hyperbola - -

    - The eccentricity of a hyperbola is \ds e=\frac ca. - hyperbolaeccentricity - eccentricity -

    -     -     - -

    - Note that this is the definition of eccentricity as used for the ellipse. - When c is close in value to a (, e\approx 1), - the hyperbola is very narrow - (looking almost like crossed lines). - - shows hyperbolas centered at the origin with a=1. - The graph in has c=1.05, - giving an eccentricity of e=1.05, which is close to 1. - As c grows larger, - the hyperbola widens and begins to look like parallel lines, - as shown in . -

    - -
    - Understanding the eccentricity of a hyperbola - - -
    - - - - A hyperbola with eccentricity 1.05 - -

    - A hyperbola with eccentricity 1.05. The hyperbola is centered at the origin. - The vertices are drawn to the left and right of the center. - The foci are also drawn, and are shown to be very close to the vertices. - The hyperbola is quite narrow, only slowing spreading out vertically as the curve moves away from the center. - Below the hyperbola it is written that e=1.05. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-10,-5,5,10}, - minor x tick num=4, - ytick={-10,-5,5,10}, - minor y tick num=4,%extra y ticks={-5,-3,...,7}, - ymin=-11.9,ymax=11.9, - xmin=-11.9,xmax=11.9 - ] - - \addplot [firstcurvestyle,domain=-85:85,samples=40] ({sec(x)},{.32*tan(x)}); - \addplot [firstcurvestyle,domain=-85:85,samples=40] ({-sec(x)},{.32*tan(x)}); - - \filldraw (axis cs:0,0) circle (2.4pt); - - \filldraw [firstcolor] (axis cs:-1,0) circle (2.4pt); - \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt); - \filldraw [secondcolor] (axis cs:1.05,0) circle (2.4pt); - \filldraw [secondcolor] (axis cs:-1.05,0) circle (2.4pt); - - \draw (axis cs: 9,-6) node { $e = 1.05$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - A hyperbola with eccentricity 1.5 - -

    - A hyperbola with eccentricity 1.5. - The hyperbola is centered at the origin. - The vertices are drawn to the left and right of the center. - The foci are drawn next to the vertices. - The foci are a short distance away from the vertices. - The hyperbola is of moderate width, increasing horizontally and vertically at a similar rate. - Below the hyperbola it is written that e=1.5 -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-10,-5,5,10}, - minor x tick num=4, - ytick={-10,-5,5,10}, - minor y tick num=4,%extra y ticks={-5,-3,...,7}, - ymin=-11.9,ymax=11.9, - xmin=-11.9,xmax=11.9 - ] - - \addplot [firstcurvestyle,domain=-85:85,samples=40] ({sec(x)},{1.12*tan(x)}); - \addplot [firstcurvestyle,domain=-85:85,samples=40] ({-sec(x)},{1.12*tan(x)}); - - \filldraw (axis cs:0,0) circle (2.4pt); - - \filldraw [firstcolor] (axis cs:-1,0) circle (2.4pt); - \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt); - \filldraw [secondcolor] (axis cs:1.5,0) circle (2.4pt); - \filldraw [secondcolor] (axis cs:-1.5,0) circle (2.4pt); - - \draw (axis cs: 9,-6) node { $e = 1.5$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - - -
    - - - - A hyperbola with eccentricity 3 - -

    - A hyperbola with eccentricity 3, centered at the origin. - The vertices are drawn to the left and right of the center. - The foci are drawn, and are a moderate distance away from the vertices. - The hyperbola is quite wide, quickly increasing vertically as the curve moves further from the origin. - Below the hyperbola it is written that e = 3. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-10,-5,5,10}, - minor x tick num=4, - ytick={-10,-5,5,10}, - minor y tick num=4,%extra y ticks={-5,-3,...,7}, - ymin=-11.9,ymax=11.9, - xmin=-11.9,xmax=11.9 - ] - - \addplot [firstcurvestyle,domain=-85:85,samples=40] ({sec(x)},{2.83*tan(x)}); - \addplot [firstcurvestyle,domain=-85:85,samples=40] ({-sec(x)},{2.83*tan(x)}); - - \filldraw (axis cs:0,0) circle (2.4pt); - - \filldraw [firstcolor] (axis cs:-1,0) circle (2.4pt); - \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt); - \filldraw [secondcolor] (axis cs:3,0) circle (2.4pt); - \filldraw [secondcolor] (axis cs:-3,0) circle (2.4pt); - - \draw (axis cs: 9,-6) node { $e = 3$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - A hyperbola with eccentricity 10 - -

    - A hyperbola of eccentricity 10, centered at the origin. - The vertices are drawn to the left and right of the center. - The foci are drawn a large distance form the vertices. - The hyperbola is very wide, increasing vertically significantly more quickly than horizontally. - Below the hyperbola it is written that e=10. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-10,-5,5,10}, - minor x tick num=4, - ytick={-10,-5,5,10}, - minor y tick num=4,%extra y ticks={-5,-3,...,7}, - ymin=-11.9,ymax=11.9, - xmin=-11.9,xmax=11.9 - ] - - \addplot [firstcurvestyle,domain=-85:85,samples=40] ({sec(x)},{9.95*tan(x)}); - \addplot [firstcurvestyle,domain=-85:85,samples=40] ({-sec(x)},{9.95*tan(x)}); - - \filldraw (axis cs:0,0) circle (2.4pt); - - \filldraw [firstcolor] (axis cs:-1,0) circle (2.4pt); - \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt); - \filldraw [secondcolor] (axis cs:10,0) circle (2.4pt); - \filldraw [secondcolor] (axis cs:-10,0) circle (2.4pt); - - \draw (axis cs: 9,-6) node { $e = 10$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -     -
    -     - - - Reflective Property -

    - Hyperbolas share a similar reflective property with ellipses. - However, in the case of a hyperbola, - a ray emanating from a focus that intersects the hyperbola reflects along a line containing the other focus, - but moving away from that focus. - This is illustrated in - (on the next page). - Hyperbolic mirrors are commonly used in telescopes because of this reflective property. - It is stated formally in the following theorem. -

    - - - Reflective Property of Hyperbolas - -

    - Let P be a point on a hyperbola with foci F_1 and F_2. - The tangent line to the hyperbola at P makes equal angles with the following two lines:hyperbolareflective property -

    - -

    -

      -
    1. -

      - The line through F_1 and P, and -

      -
      2. - -
    2. -

      - The line through F_2 and P. -

      -
      3. -
    -

    -     -     -     - - - Location Determination -

    - Determining the location of a known event has many practical uses - (locating the epicenter of an earthquake, an airplane crash site, - the position of the person speaking in a large room, etc.). -

    - -

    - To determine the location of an earthquake's epicenter, - seismologists use trilateration - (not to be confused with triangulation). - A seismograph allows one to determine how far away the epicenter was; - using three separate readings, - the location of the epicenter can be approximated. -

    - -

    - A key to this method is knowing distances. - What if this information is not available? - Consider three microphones at positions A, - B and C which all record a noise (a person's voice, - an explosion, etc.) created at unknown location D. - The microphone does not know - when the sound was created, - only when the sound was detected. - How can the location be determined in such a situation? -

    - -
    - Illustrating the reflective property of a hyperbola - - - A hyperbola demonstrating the reflective property. - -

    - A hyperbola with rays emanating from the left focus. - The ray moves towards the right, reflecting off a point on the right side of the hyperbola. - The reflected ray moves towards the upper left. - A dashed line is drawn between the right focus and the point where the ray touches the hyperbola. - The reflected ray is moving in the same direction as the dashed line. -

    -     - - - \begin{tikzpicture}[>=latex] - - \begin{axis}[ - axis y line=none, - axis x line=none, - ymin=-11.9,ymax=11.9, - xmin=-5,xmax=5 - ] - - \addplot [firstcurvestyle,domain=-85:85,samples=40] ({sec(x)},{2.83*tan(x)}); - \addplot [firstcurvestyle,domain=-85:85,samples=40] ({-sec(x)},{2.83*tan(x)}); - - \filldraw (axis cs:0,0) circle (2.4pt); - - \filldraw [firstcolor] (axis cs:-1,0) circle (2.4pt); - \filldraw [firstcolor] (axis cs:1,0) circle (2.4pt); - - \draw [secondcolor,thick,->] (axis cs:-3,0) -- (axis cs: 1.4,2.83); - \draw [secondcolor,thick,->] (axis cs:1.4,2.83) -- (axis cs: -1.8,8.49); - - \draw [dashed,secondcolor,thick] (axis cs:1.4,2.83)--(axis cs: 3,0); - - \filldraw (axis cs:1.4,2.83) circle (2.4pt); - - \filldraw (axis cs:3,0) circle (2.4pt) node [below] { $F_2$}; - \filldraw (axis cs:-3,0) circle (2.4pt) node [below] { $F_1$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - If each location has a clock set to the same time, - hyperbolas can be used to determine the location. - Suppose the microphone at position A records the sound at exactly 12:00, - location B records the time exactly 1 second later, - and location C records the noise exactly 2 seconds after that. - We are interested in the difference of times. - Since the speed of sound is approximately 340 m/s, we can conclude quickly that the sound was created 340 meters closer to position A than position B. - If A and B are a known distance apart (as shown in ), - then we can determine a hyperbola on which D must lie. -

    - -

    - The difference of distances is 340; - this is also the distance between vertices of the hyperbola. - So we know 2a= 340. - Positions A and B lie on the foci, so 2c=1000. - From this we can find b\approx 470 and can sketch the hyperbola, - given in . - We only care about the side closest to A. (Why?) -

    - -

    - We can also find the hyperbola defined by positions B and C. - In this case, - 2a = 680 as the sound traveled an extra 2 seconds to get to C. - We still have 2c=1000, - centering this hyperbola at (-500,500). - We find b\approx 367. - This hyperbola is sketched in . - The intersection point of the two graphs is the location of the sound, - at approximately (188,-222.5). -

    - -
    - - - -
    - - - Three points drawn and labeled on a plane - -

    - Three points drawn and labeled on a plane. - A is at the point (500,0). - B is at the point (-500,0). - C is at the point (-500,1000). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1100,ymax=1100, - xmin=-1100,xmax=1100 - ] - - \filldraw (axis cs:500,0) circle (2.4pt) node [above] { $A$}; - \filldraw (axis cs:-500,0) circle (2.4pt) node [above] { $B$}; - \filldraw (axis cs:-500,1000) circle (2.4pt) node [left] { $C$}; - - \end{axis} - - \end{tikzpicture} - - - -
    - - -
    - - - A hyperbola drawn from points A and B to illustrate the location property - -

    - A hyperbola drawn from points A and B in . - A and B are the foci of the hyperbola. - The vertices of the hyperbola are 170 units away from the origin. - The left half of the hyperbola is drawn with a dashed line, while the right half of the parabola is drawn with a solid line. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1100,ymax=1100, - xmin=-1100,xmax=1100 - ] - - \addplot+ [domain=-85:85] ({170*sec(x)},{470*tan(x)}); - \addplot+ [domain=-85:85] ({-170*sec(x)},{470*tan(x)}); - - \filldraw (axis cs:500,0) circle (2.4pt) node [above] { $A$}; - \filldraw (axis cs:-500,0) circle (2.4pt) node [above] { $B$}; - \filldraw (axis cs:-500,1000) circle (2.4pt) node [left] { $C$}; - - \end{axis} - - \end{tikzpicture} - - - -
    - - -
    - - - A fourth point found from hyperbolas given by points in the previous figures - -

    - A hyperbola drawn from points B and C in , intersecting with the hyperbola drawn in . - The hyperbola uses B and C as the foci. The vertices of the hyperbola are 640 units apart. - The bottom hyperbola, with focus B, extends towards the right. - The point at which it intersects with the hyperbola with focus A is labeled as D. -

    -     - - - \begin{tikzpicture} - \begin{axis}[ - axis on top, - ymin=-1100,ymax=1100, - xmin=-1100,xmax=1100 - ] - - \addplot+ [domain=-85:85] ({170*sec(x)},{470*tan(x)}); - \addplot+ [domain=-85:85] ({367*tan(x)-500},{340*sec(x)+500}); - \addplot [secondcurvestyle,domain=-85:85] ({367*tan(x)-500},{-340*sec(x)+500}); - - \filldraw (axis cs:500,0) circle (2.4pt) node [above] { $A$}; - \filldraw (axis cs:-500,0) circle (2.4pt) node [above] { $B$}; - \filldraw (axis cs:-500,1000) circle (2.4pt) node [left] { $C$}; - \filldraw (axis cs:188,-222) circle (2.4pt) node [below left] { $D$}; - - \end{axis} - - \end{tikzpicture} - - - -
    - -     -
    -     - -

    - This chapter explores curves in the plane, - in particular curves that cannot be described by functions of the form y=f(x). - In this section, - we learned of ellipses and hyperbolas that are defined implicitly, - not explicitly. - In the following sections, - we will learn completely new ways of describing curves in the plane, - using parametric equations - and polar coordinates, - then study these curves using calculus techniques. -

    -     - - - - Terms and Concepts - - - - -

    - What is the difference between degenerate and nondegenerate conics? -

    -     - - - -

    - When defining the conics as the intersections of a plane and a double napped cone, - degenerate conics are created when the plane intersects the tips of the cones - (usually taken as the origin). - Nondegenerate conics are formed when this plane does not contain the origin. -

    -     - -     - - - - -

    - Use your own words to explain what the eccentricity of an ellipse measures. -

    -     - - - -     - - - - -

    - What has the largest eccentricity: - an ellipse or a hyperbola? -

    -     - - - -

    - Ellipse -

    -     -     - - -

    - Hyperbola -

    -     -     -     - -     - - - - -

    - Explain why the following is true: - If the coefficient of the x^2 term in the equation of an ellipse in standard form is smaller than the coefficient of the y^2 term, - then the ellipse has a horizontal major axis. -

    -     - - - -

    - With the equation \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1, - the ellipse has a horizontal major axis if a \gt b. - But the coefficient of the x^2 term is 1/a^2 - (not a^2), - so if 1/a^2\lt 1/b^2, then a \gt b and the major axis is horizontal. -

    -     - -     - - - - -

    - Explain how one can quickly look at the equation of a hyperbola in standard form - and determine whether the transverse axis is horizontal or vertical. -

    -     - - - -

    - With a horizontal transverse axis, - the x^2 term has a positive coefficient; - with a vertical transverse axis, - the y^2 term has a positive coefficient. -

    -     - -     - - - -

    - Fill in the blank: It can be said that ellipses and hyperbolas share the - same reflective property: - A ray emanating from one focus will reflect off the conic along a - that contains the other focus. -

    -     - - - - - - - - -

    - Your answer includes the correct word but has extra text. -

    -     -     -     -     -     -     - - - Problems - - - -

    - Find the equation of the parabola defined by the given information. - Sketch the parabola. -

    -     - - - - -

    - Focus: (3,2); directrix: y=1 -

    -     - -

    - y=\frac12(x-3)^2+\frac32 -

    -     - -     - - - - -

    - Focus: (-1,-4); directrix: y=2 -

    -     - -

    - y=\frac{-1}{12}(x+1)^2-1 -

    -     - -     - - - - -

    - Focus: (1,5); directrix: x=3 -

    -     - -

    - x=-\frac14(y-5)^2+2 -

    -     - -     - - - - -

    - Focus: (1/4,0); directrix: x=-1/4 -

    -     - -

    - x=y^2 -

    -     - -     - - - - -

    - Focus: (1,1); vertex: (1,2) -

    -     - -

    - y=-\frac14(x-1)^2+2 -

    -     - -     - - - - -

    - Focus: (-3,0); vertex: (0,0) -

    -     - -

    - x=-\frac1{12}y^2 -

    -     - -     - - - - -

    - Vertex: (0,0); directrix: y=-1/16 -

    -     - -

    - y=4x^2 -

    -     - -     - - - - -

    - Vertex: (2,3); directrix: x=4 -

    -     - -

    - x=-\frac18(y-3)^2+2 -

    -     - -     - -     - - - -

    - The equation of a parabola and a point on its graph are given. - Find the focus and directrix of the parabola, - and verify that the given point is equidistant from the focus and directrix. -

    -     - - - - -

    - y=\frac14x^2, P=(2,1) -

    -     - -

    - focus: (0,1); directrix: y=-1. - The point P is 2 units from each. -

    -     - -     - - - - -

    - x=\frac18(y-2)^2+3, P=(11,10) -

    -     - -

    - focus: (5,2); directrix: x=1. - The point P is 10 units from each. -

    -     - -     - -     - - - -

    - Sketch the ellipse defined by the given equation. - Label the center, foci and vertices. -

    -     - - - - -

    - \ds \frac{(x-1)^2}{3}+\frac{(y-2)^2}{5}=1 -

    -     - - -     - - - - -

    - \ds \frac{1}{25}x^2+\frac{1}{9}(y+3)^2=1 -

    -     - - -     - -     - - - -

    - Find the equation of the ellipse shown in the graph. - Give the location of the foci and the eccentricity of the ellipse. -

    -     - - - - - - - - An ellipse centered in the second quadrant. - -

    - An ellipse centered at the point (1,2). - The major axis has a length of 6. - The minor axis has a length of 4. - The vertices of the graph are at the points (-4,2) and (2,2). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-1.2,ymax=5.2, - xmin=-5.2,xmax=3.2 - ] - - \addplot+ [domain=0:360,samples=60] ({3*cos(x)-1},{2*sin(x)+2}); - - \end{axis} - - \end{tikzpicture} - - - - - -     - -

    - \frac{(x+1)^2}{9}+\frac{(y-2)^2}{4}=1; - foci at (-1\pm\sqrt{5},2); e=\sqrt{5}/3 -

    -     - -     - - - - - - - - A vertical ellipse centered on the x-axis. - -

    - An ellipse centered on the point (1,0). - The vertices of the ellipse are at the points (1,3) and (1,-3). - The major axis appears to have a length of 6. - The minor axis has a length of 1. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-3.5,ymax=3.5, - xmin=-1.2,xmax=2.2 - ] - - \addplot+ [domain=0:360,samples=60] ({.5*cos(x)+1},{3*sin(x))}); - - \end{axis} - - \end{tikzpicture} - - - - - -     - -

    - \frac{(x-1)^2}{1/4}+\frac{y^2}{9}=1; - foci at (1,\pm \sqrt{8.75}); - e=\sqrt{8.75}/3\approx 0.99 -

    -     - -     - -     - - - -

    - Find the equation of the ellipse defined by the given information. - Sketch the elllipse. -

    -     - - - - -

    - Foci: (\pm 2,0); vertices: (\pm 3,0) -

    -     - -

    - \frac{x^2}{9}+\frac{y^2}{5}=1 -

    -     - -     - - - - -

    - Foci: (-1,3) and (5,3); vertices: - (-3,3) and (7,3) -

    -     - -

    - \frac{(x-2)^2}{25}+\frac{(y-3)^2}{16}=1 -

    -     - -     - - - - -

    - Foci: (2,\pm 2); vertices: (2,\pm 7) -

    -     - -

    - \frac{(x-2)^2}{45}+\frac{y^2}{49}=1 -

    -     - -     - - - - -

    - Focus: (-1,5); vertex: - (-1,-4); center: (-1,1) -

    -     - -

    - \frac{(x+1)^2}{9}+\frac{(y-1)^2}{25}=1 -

    -     - -     - -     - - - -

    - Write the equation of the given ellipse in standard form. -

    -     - - - - -

    - x^2-2x+2y^2-8y=-7 -

    -     - -

    - \frac{(x-1)^2}{2}+(y-2)^2=1 -

    -     - -     - - - - -

    - 5x^2+3y^2=15 -

    -     - -

    - \frac{x^2}{3}+\frac{y^2}{5}=1 -

    -     - -     - - - - -

    - 3x^2+2y^2-12y+6=0 -

    -     - -

    - \frac{x^2}{4}+\frac{(y-3)^2}{6}=1 -

    -     - -     - - - - -

    - x^2+y^2-4x-4y+4=0 -

    -     - -

    - \frac{(x-2)^2}{4}+\frac{(y-2)^2}{4}=1 -

    -     - -     - -     - - - -

    - Find the equation of the hyperbola shown in the graph. -

    -     - - - - - - - - A hyperbola centered at the origin with a horizontal transverse axis. - -

    - A hyperbola centered at the origin with a horizontal transverse axis. - The foci of the hyperbola are at the points (-2,0) and (2,0). - The vertices of the hyperbola are at the points (-1,0) and (1,0) -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,1,2,-2}, - ymin=-3.5,ymax=3.5, - xmin=-3.5,xmax=3.5 - ] - - \addplot [firstcurvestyle,domain=-70:70] ({sec(x)},{1.73*tan(x))}); - \addplot [firstcurvestyle,domain=-70:70] ({-sec(x)},{1.73*tan(x))}); - - \filldraw (axis cs: 2,0) circle (2.4pt) - (axis cs: -2,0) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - - -     - -

    - x^2-\frac{y^2}{3}=1 -

    -     - -     - - - - - - - - A hyperbola centered at the origin with a vertical transverse axis. - -

    - A hyperbola centered at the origin with a vertical transverse axis. - The hyperbola has foci at the points (0,5) and (0,-5). - The vertices of the ellipse are at the points (0,1) and (0,-1). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - minor x tick num=4, - minor y tick num=4, - ymin=-6.5,ymax=6.5, - xmin=-8.5,xmax=8.5 - ] - - \addplot [firstcurvestyle,domain=-70:70] ({4.9*tan(x)},{sec(x))}); - \addplot [firstcurvestyle,domain=-70:70] ({4.9*tan(x)},{-sec(x))}); - - \filldraw (axis cs: 0,5) circle (2.4pt) - (axis cs: 0,-5) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - - -     - -

    - y^2-\frac{x^2}{24}=1 -

    -     - -     - - - - - - - - A hyperbola with center (1,3) and a vertical transverse axis. - -

    - A hyperbola with center (1,3) and a vertical transverse axis. - The foci of the hyperbola are not drawn. - The vertices of the hyperbola are at the points (1,5) and (1,1). - A box is drawn around the origin with a width of 6 and a height of 4. - The diagonal asymptotes are drawn as the lines passing through the corners of the box, passing through the center. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - minor x tick num=4, - ymin=-1.5,ymax=7.5, - xmin=-4.5,xmax=6.5 - ] - - \addplot [firstcurvestyle,domain=-70:70] ({3*tan(x)+1},{2*sec(x)+3}); - \addplot [firstcurvestyle,domain=-70:70] ({3*tan(x)+1},{-2*sec(x)+3}); - - \addplot [secondcurvestyle,domain=-8:8] (x,{2/3*(x-1)+3}); - \addplot [secondcurvestyle,domain=-8:8] (x,{-2/3*(x-1)+3}); - - \draw [thick,dashed,gray] (axis cs: -2,5) -- (axis cs:4,5) -- (axis cs:4,1) -- (axis cs: -2,1) --cycle; - - \filldraw (axis cs: 1,3) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - - -     - -

    - \frac{(y-3)^2}{4}-\frac{(x-1)^2}{9}=1 -

    -     - -     - - - - - - - - A hyperbola centered at the point (1,3) with a horizontal transverse axis - -

    - A hyperbola centered at the point (1,3) with a horizontal transverse axis. - The foci of the hyperbola are not drawn. - The vertices of the hyperbola are at the points (-2,3) and (4,3). - A box is drawn around the center with a width of 6 and a height of 4. - The diagonal asymptotes are drawn as the lines passing through the corners of the box, passing through the center. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=4, - ymin=-1.5,ymax=7.5, - xmin=-4.5,xmax=6.5 - ] - - \addplot [firstcurvestyle,domain=-70:70] ({3*sec(x)+1},{2*tan(x)+3}); - \addplot [firstcurvestyle,domain=-70:70] ({-3*sec(x)+1},{2*tan(x)+3}); - - \addplot [secondcurvestyle,domain=-8:8] (x,{2/3*(x-1)+3}); - \addplot [secondcurvestyle,domain=-8:8] (x,{-2/3*(x-1)+3}); - - \draw [thick,dashed,gray] (axis cs: -2,5) -- (axis cs:4,5) -- (axis cs:4,1) -- (axis cs: -2,1) --cycle; - - \filldraw (axis cs: 1,3) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - - -     - -

    - \frac{(x-1)^2}{9}-\frac{(y-3)^2}{4}=1 -

    -     - -     - -     - - - -

    - Sketch the hyperbola defined by the given equation. - Label the center and foci. -

    -     - - - - -

    - \ds \frac{(x-1)^2}{16}-\frac{(y+2)^2}{9}=1 -

    -     - - -     - - - - -

    - \ds (y-4)^2-\frac{(x+1)^2}{25}=1 -

    -     - - -     - -     - - - -

    - Find the equation of the hyperbola defined by the given information. - Sketch the hyperbola. -

    -     - - - - -

    - Foci: (\pm 3,0); vertices: (\pm 2, 0) -

    -     - -

    - \frac{x^2}{4}-\frac{y^2}{5}=1 -

    -     - -     - - - - - - -

    - Foci: (0,\pm 3); vertices: (0,\pm 2) -

    -     - -

    - \frac{y^2}{4}-\frac{x^2}{5}=1 -

    -     -     -     - - - - -

    - Foci: (-2,3) and (8,3); vertices: - (-1,3) and (7,3) -

    -     - -

    - \frac{(x-3)^2}{16}-\frac{(y-3)^2}{9}=1 -

    -     - -     - - - - -

    - Foci: (3,-2) and (3,8); vertices: - (3,0) and (3,6) -

    -     - -

    - \frac{(y-3)^2}{9}-\frac{(x-3)^2}{16}=1 -

    -     - -     - -     - - - -

    - Write the equation of the hyperbola in standard form. -

    -     - - - - -

    - 3x^2-4y^2=12 -

    -     - -

    - \frac{x^2}{4}-\frac{y^2}{3}=1 -

    -     - -     - - - - -

    - 3x^2-y^2+2y=10 -

    -     - -

    - \frac{x^2}{3}-\frac{(y-1)^2}{9}=1 -

    -     - -     - - - - -

    - x^2-10y^2+40y=30 -

    -     - -

    - (y-2)^2-\frac{x^2}{10}=1 -

    -     - -     - - - - -

    - (4y-x)(4y+x)=4 -

    -     - -

    - 4y^2-\frac{x^2}{4}=1 -

    -     - -     -     - - - - -

    - Consider the ellipse given by \ds \frac{(x-1)^2}{4}+\frac{(y-3)^2}{12}=1. -

    -     - - - -

    - Verify that the foci are located at (1,3\pm 2\sqrt{2}). -

    -     - -

    - c=\sqrt{12-4} = 2\sqrt{2}. -

    -     -     - - - -

    - The points P_1 = (2,6) and P_2 = (1+\sqrt{2},3+\sqrt{6}) \approx (2.414,5.449) lie on the ellipse. - Verify that the sum of distances from each point to the foci is the same. -

    -     - -

    - The sum of distances for each point is 2\sqrt{12}\approx 6.9282. -

    -     -     - -     - - - - -

    - Johannes Kepler discovered that the planets of our solar system have elliptical orbits with the Sun at one focus. - The Earth's elliptical orbit is used as a standard unit of distance; - the distance from the center of Earth's elliptical orbit to one vertex is 1 Astronomical Unit, or A.U. -

    - -

    - The following table gives information about the orbits of three planets. -

    - - - - Planet - Distance fromcenter to vertex - Orbiteccentricity - - - Mercury - 0.387 A.U. - 0.2056 - - - Earth - 1 A.U. - 0.0167 - - - Mars - 1.524 A.U. - 0.0934 - - -     - - - -

    - In an ellipse, - knowing c^2=a^2-b^2 and e=c/a allows us to find b in terms of a and e. - Show b=a\sqrt{1-e^2}. -

    -     - -

    - Solve for c in e=c/a: c=ae. - Thus a^2e^2=a^2-b^2, and b^2=a^2-a^2e^2. - The result follows. -

    -     -     - - - -

    - For each planet, - find equations of their elliptical orbit of the form - \ds\frac{x^2}{a^2}+\frac{y^2}{b^2}=1. (This places the center at (0,0), - but the Sun is in a different location for each planet.) -

    -     - -

    - Mercury: x^2/(0.387)^2 + y^2/(0.3787)^2=1 - - Earth: x^2+y^2/(0.99986)^2 = 1 - - Mars: x^2/(1.524)^2+y^2/(1.517)^2=1 -

    -     -     - - - -

    - Shift the equations so that the Sun lies at the origin. - Plot the three elliptical orbits. -

    -     - -

    - Mercury: (x-0.08)^2/(0.387)^2 + y^2/(0.3787)^2=1 - - Earth: (x-0.0167)^2+y^2/(0.99986)^2 = 1 - - Mars: (x-0.1423)^2/(1.524)^2+y^2/(1.517)^2=1 -

    -     -     - -     - - - - -

    - A loud sound is recorded at three stations that lie on a line as shown in the figure below. - Station A recorded the sound 1 second after Station B, - and Station C recorded the sound 3 seconds after B. - Using the speed of sound as 340m/s, determine the location of the sound's origination. -

    - - - A number line with 3 points drawn. - -

    - A number line with 3 seperate points drawn. - A is the left most point. - B is the point in the middle. The distance between A and B is 1000\mathrm{m}. - C is the left most point. The distance between B and C is 2000 \mathrm{m}. -

    -     - - - \begin{tikzpicture}[scale=1.3] - - \draw [fill=black] (0,0) circle (2.4pt) node [below] {\(A\)} -- node [pos=.5,above] {1000m} (1.5,0) circle (2.4pt) node [below] {\(B\)} -- node [pos=.5,above] {2000m} (4.5,0) circle (2.4pt) node [below] {\(C\)}; - - \end{tikzpicture} - - - -     - -

    - The sound originated from a point approximately 31m to the right of B and 1390m above or below it. (Since the three points are collinear, we cannot distinguish whether the sound originated above/below the line containing the points.) -

    -     - -     -     -     -
    -
    - Parametric Equations - - -

    - We are familiar with sketching shapes, - such as parabolas, by following this basic procedure: -

    - -
    - Plotting a graph y=f(x) - - Diagram outlining the steps involved in plotting a function. - -

    - A diagram containing mostly text, that outlines the process for plotting a function. - The steps illustrated are as follows: -

      -
    1. -

      - Choose x -

      -
      2. -
    2. -

      - Use function f to find y (y=f(x)) -

      -
      3. -
    3. -

      - Plot point (x,y) -

      -
      4. -
    -

    - -

    - This is not really indicative of how a person would sketch a graph by hand - (better techniques for doing so were covered in ), - but it is not far off from how a computer might generate a plot, - or how a student who is first learning about functions might use a table of values to create a plot. -

    -     - - - \begin{tikzpicture} - - \draw (0,0) node (A) {Choose \(x\)} - (5.25,0) node (B) {Use a function \(f\) to find \(y\) \(\big(y=f(x)\big)\)} - (11.25,0) node (C) {Plot point \((x,y)\)}; - - \draw [->] (A) -- (B); - \draw [->] (B) -- (C); - - \end{tikzpicture} - - - -
    - -

    - The rectangular equation - y=f(x) works well for some shapes like a parabola with a vertical axis of symmetry, - but in the previous section we encountered several shapes that could not be sketched in this manner. (To plot an ellipse using the above procedure, - we need to plot the top and bottom - separately.)curverectangular equation -

    - -

    - In this section we introduce a new sketching procedure: -

    - -
    - Plotting a curve (x(t),y(t)) - - Diagram illustrating the process for plotting a parametric curve. - -

    - This is another text-based diagram, which illustrates the method for plotting curves that will be covered in this section. - In this method, the coordinates x and y are both defined as functions of a third variable, t. - Four phrases are arranged in a diamond shape, with text left, right, top, and bottom. -

    - -

    - To the left of the diagram is the text, Choose t. - An arrow points up and to the left from this text to the phrase Use a function f to find x (x=f(t)), - which is located at the top of the diagram. - Another arrow points down and to the left from Choose t to the phrase - Use a function g to find y (y=g(t)), which is located at the bottom of the diagram. -

    - -

    - From the phrases at the top and bottom of the diagram there are two more arrows pointing to the right, - to a final phrase, which states, Plot point (x,y). -

    -     - - - \begin{tikzpicture}[every node/.append style={font=\large}] - - \draw (0,0) node (A) {Choose \(t\)} - (3,1) node (B1) {Use a function \(f\) to find \(x\) \(\big(x=f(t)\big)\)} - (3,-1) node (B2) {Use a function \(g\) to find \(y\) \(\big(y=g(t)\big)\)} - (6.25,0) node (C) {Plot point \((x,y)\)}; - - \draw [->] (A) -- (B1); - \draw [->] (A) -- (B2); - \draw [->] (B1) -- (C); - \draw [->] (B2) -- (C); - - \end{tikzpicture} - - - -
    - -

    - Here, x and y are found separately but then plotted together: - for each value of the input t, we plot the output - the point (x(t),y(t)). -

    -     - - - Plotting parametric curves -

    - The procedure outlined in leads us to a definition. -

    - - - Parametric Equations and Curves - -

    - Let f and g be continuous functions on an interval I. - The set of all points \big(x,y\big) = \big(f(t),g(t)\big) in the Cartesian plane, - as t varies over I, - is the graph of the - parametric equations - x=f(t) and y=g(t), - where t is the parameter. - A curve is a graph along with the parametric equations that define it. - curveparametrically defined - parametric equationsdefinition -

    -     -     - -

    - This is a formal definition of the word curve. - When a curve lies in a plane - (such as the Cartesian plane), - it is often referred to as a plane curve. - Examples will help us understand the concepts introduced in the definition. -

    - - - Plotting parametric functions - -

    - Plot the graph of the parametric equations x=t^2, - y=t+1 for t in [-2,2]. -

    -     - -

    - We plot the graphs of parametric equations in much the same manner as we plotted graphs of functions like y=f(x): - we make a table of values, plot points, - then connect these points with a - reasonable looking curve. - shows such a table of values; - note how we have 3 columns. -

    - -

    - The points (x,y) from the table are plotted in . - The points have been connected with a smooth curve. - Each point has been labeled with its corresponding t-value. - These values, along with the two arrows along the curve, - are used to indicate the orientation of the graph. - This information helps us determine the direction in which the graph is moving. -

    - -
    - A table of values of the parametric equations in along with a sketch of their graph - -
    - - - t - x - y - - - - - - - - -2 - 4 - -1 - - - -1 - 1 - 0 - - - 0 - 0 - 1 - - - 1 - 1 - 2 - - - 2 - 4 - 3 - - -
    -
    - - - - Plot of a parabola with its vertex at the point (0,1), opening to the right, with several marked points. - -

    - The parametric curve obtained from the table of values in is shown. - On a set of coordinate axes, the points corresponding to the values t=-2,-1,0,1,2 are plotted. - These are, respectively, (4,-1), (1,0), (0,1), (1,2), (4,3). -

    - -

    - The curve joining these points is plotted; it appears to be a parabola opening to the right, - with its vertex at the point (0,1). - Arrow heads on the curve indicate a direction of travel corresponding to an increasing value of t. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-2.2,ymax=4.2, - xmin=-.9,xmax=5.2 - ] - - \addplot+ [domain=-2:2,samples=60] ({x^2},{x+1}); - - \draw [->,>=latex] (axis cs:3.9204,2.98) -- (axis cs:3.96,2.99); - \draw [->,>=latex] (axis cs:2.25,-.5) -- (axis cs:2.22,-.49); - - \filldraw (axis cs:4,-1) circle (2.4pt) node [below] { $t=-2$}; - \filldraw (axis cs:1,0) circle (2.4pt) node [above] { $t=-1$}; - \filldraw (axis cs:0,1) circle (2.4pt) node [left] { $t=0$}; - \filldraw (axis cs:1,2) circle (2.4pt) node [above] { $t=1$}; - \filldraw (axis cs:4,3) circle (2.4pt) node [above ] { $t=2$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    -     - -     - -

    - We often use the letter t as the parameter as we often regard t as representing time. - Certainly there are many contexts in which the parameter is not time, - but it can be helpful to think in terms of time as one makes sense of parametric plots and their orientation (for instance, - At time t=0 the position is (1,2) and at time t=3 the position is (5,1).). -

    - - - Plotting parametric functions - -

    - Sketch the graph of the parametric equations x=\cos^2(t), - y=\cos(t) +1 for t in [0,\pi]. -

    -     - -

    - We again start by making a table of values in , - then plot the points (x,y) on the Cartesian plane in . -

    - -
    - A table of values of the parametric equations in along with a sketch of their graph - -
    - - - t - x - y - - - - - - - - 0 - 1 - 2 - - - \pi/4 - 1/2 - 1+\sqrt{2}/2 - - - \pi/2 - 0 - 1 - - - 3\pi/4 - 1/2 - 1-\sqrt{2}/2 - - - \pi - 1 - 0 - - -
    -
    - - - - Plot of a parabola with its vertex at the point (0,1), opening to the right, with several marked points. - -

    - The parametric curve obtained from the table of values in is shown. - The curve is very similar in appearance to the one in ; - again it appears to be a parabola opening to the right, with its vertex at the point (0,1). -

    - -

    - The primary difference is that the marked points now correspond to parameter values - t = 0, \pi/4, \pi/2, 3\pi/4, \pi. The direction of travel is opposite to that in . -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=2.1, - xmin=-.1,xmax=1.6 - ] - - \addplot+ [domain=0:180,samples=70] ({(cos(x))^2},{cos(x)+1}); - - \draw [->,>=latex] (axis cs:0.998271, 0.00086485) -- (axis cs:0.999534, 0.000233112); - \draw [->,>=latex] (axis cs:0.770151, 1.87758) -- (axis cs:0.75311, 1.86782); - - \filldraw (axis cs:1,2) circle (2.4pt) node [right] { $t=0$}; - \filldraw (axis cs:.5,1.71) circle (2.4pt) node [below right] { $t=\pi/4$}; - \filldraw (axis cs:0,1) circle (2.4pt) node [right] { $t=\pi/2$}; - \filldraw (axis cs:.5,.292) circle (2.4pt) node [above right] { $t=3\pi/4$}; - \filldraw (axis cs:1,0) circle (2.4pt) node [above right] { $t=\pi$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    - -

    - It is not difficult to show that the curves in Examples - and - are portions of the same parabola. - While the parabola is the same, - the curves are different. - In , - if we let t vary over all real numbers, - we'd obtain the entire parabola. - In this example, - letting t vary over all real numbers would still produce the same graph; - this portion of the parabola would be traced, - and re-traced, infinitely many times. - The orientation shown in - shows the orientation on [0,\pi], - but this orientation is reversed on [\pi,2\pi]. -

    - -

    - These examples begin to illustrate the powerful nature of parametric equations. - Their graphs are far more diverse than the graphs of functions produced by - y=f(x) functions. -

    -     -     - - - - - -

    - Technology Note: Most graphing utilities can graph functions given in parametric form. - Often the word parametric - is abbreviated as PAR - or PARAM in the options. - The user usually needs to determine the graphing window (i.e, the minimum and maximum x- and y-values), - along with the values of t that are to be plotted. - The user is often prompted to give a t minimum, - a t maximum, - and a t-step or \Delta t. - Graphing utilities effectively plot parametric functions just as we've shown here: - they plots lots of points. - A smaller t-step plots more points, - making for a smoother graph - (but may take longer). - In , the t-step is 1; - in , - the t-step is \pi/4. -

    - -

    - One nice feature of parametric equations is that their graphs are easy to shift. - While this is not too difficult in the - y=f(x) context, - the resulting function can look rather messy. - (Plus, to shift to the right by two, - we replace x with x-2, - which is counter-intuitive.) - The following example demonstrates this. -

    - - - Shifting the graph of parametric functions - -

    - Sketch the graph of the parametric equations x=t^2+t, - y=t^2-t. - Find new parametric equations that shift this graph to the right 3 places and down 2. -

    -     - -

    - The graph of the parametric equations is given in . - It is a parabola with a axis of symmetry along the line y=x; - the vertex is at (0,0). -

    - -

    - In order to shift the graph to the right 3 units, - we need to increase the x-value by 3 for every point. - The straightforward way to accomplish this is simply to add 3 to the function defining x: - x = t^2+t+3. - To shift the graph down by 2 units, - we wish to decrease each y-value by 2, so we subtract 2 from the function defining y: - y = t^2-t-2. - Thus our parametric equations for the shifted graph are x=t^2+t+3, - y=t^2-t-2. - This is graphed in . - Notice how the vertex is now at (3,-2). -

    - -
    - Illustrating how to shift graphs in - -
    - - - - Plot of the "unshifted" parametric curve in this example. - -

    - A plot of the curve x=t^2+1, y=t^2-t is shown. - The shape of the curve is similar to that of a parabola, but slightly distorted. -

    - -

    - The curve begins near the point (2,6), and then descends to a y intercept at (0,2). - It briefly enters the second quadrant before passing through the origin into the fourth quadrant. - A small portion of the curve lies in the fourth quarant before it crosses the x axis at (2,0); - after this point, it continues into the first quadrant, exiting the image near the point (10,5). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={2,4,6,8,10}, - ymin=-2.6,ymax=6.5, - xmin=-.6,xmax=10.5 - ] - - \addplot+ [domain=-3:3,samples=60] ({x^2+x},{x^2-x}); - - \draw [->,>=latex] (axis cs:0.96, 4.16) -- (axis cs:0.9381, 4.1181); - \draw [->,>=latex] (axis cs:6,2) -- (axis cs:6.05,2.03); - - \draw (axis cs: 4,5) node [align=left] { $x=t^2+t$\\ $y=t^2-t$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - The shifted version of the curve in this example. - -

    - The curve in this image is identical to the one beside it in ; - the only difference is that it has been shifted 3 units to the right, and 2 units down. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={2,4,6,8,10}, - ymin=-2.6,ymax=6.5, - xmin=-.6,xmax=10.5 - ] - - \addplot+ [domain=-3:3,samples=60] ({x^2+x+3},{x^2-x-2}); - - \draw [->,>=latex] (axis cs:3.96, 2.16) -- (axis cs:3.9381, 2.1181); - \draw [->,>=latex] (axis cs:9,0) -- (axis cs:9.05,0.03); - - \draw (axis cs: 2.5,5) node [align=left] { $x=t^2+t+3$\\ $y=t^2-t-2$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    -     -     - -

    - Because the x- and y-values of a graph are determined independently, - the graphs of parametric functions often possess features not seen on - y=f(x) type graphs. - The next example demonstrates how such graphs can arrive at the same point more than once. -

    - - - Graphs that cross themselves - -

    - Plot the parametric functions x=t^3-5t^2+3t+11 and - y=t^2-2t+3 and determine the t-values where the graph crosses itself. -

    -     - -

    - Using the methods developed in this section, - we again plot points and graph the parametric equations as shown in . - It appears that the graph crosses itself at the point (2,6), - but we'll need to analytically determine this. -

    - -
    - A graph of the parametric equations from - - - Plot of a more complicated parametric curve that has a self-intersection. - -

    - The curve in this image could describe the path of a fly that turns sharply in a loop, - before heading off in another direction. - It begins in the second quadrant, traveling down and to the right, - and crossing the y axis at approximately (0,6.5). -

    - -

    - The curve continues in this direction until it turns sharply near the point (12,2). - The curve then travels to the left and bends upwards, crossing itself at the point (2,6). - After this point of self-intersection, the curve continues in the first quadrant, - traveling up and to the right. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=4, - minor y tick num=4, - ymin=-.9,ymax=16, - xmin=-6,xmax=16 - ] - - \addplot+ [domain=-2:5,samples=70] ({x^3-5*x^2+3*x+11},{x^2-2*x+3}); - - \draw [->,>=latex] (axis cs:-4.62287, 7.5225) -- (axis cs:-4.4041, 7.4756); - \draw [->,>=latex] (axis cs:7,11) -- (axis cs:7.1107, 11.0601); - - \draw (axis cs: 7,14) node [align=left] { $x=t^3-5t^2+3t+11$\\ $y=t^2-2t+3$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - We are looking for two different values, say, - s and t, where - x(s) = x(t) and y(s) = y(t). - That is, the x-values are the same precisely when the y-values are the same. - This gives us a system of 2 equations with 2 unknowns: - - s^3-5s^2+3s+11 \amp = t^3-5t^2+3t+11 - s^2-2s+3 \amp = t^2-2t+3 - -

    - -

    - Solving this system is not trivial but involves only algebra. - Using the quadratic formula, - one can solve for t in the second equation and find that \ds t = 1\pm \sqrt{s^2-2s+1}. - This can be substituted into the first equation, - revealing that the graph crosses itself at t=-1 and t=3. - We confirm our result by computing - x(-1) = x(3)=2 and y(-1) = y(3) = 6. -

    -     -     - - -     - - - Converting between rectangular and parametric equations -

    - It is sometimes useful to rewrite equations in rectangular form (, y=f(x)) into parametric form, - and vice-versa. - Converting from rectangular to parametric can be very simple: - given y=f(x), the parametric equations x=t, - y=f(t) produce the same graph. - As an example, given y=x^2, - the parametric equations x=t, - y=t^2 produce the familiar parabola. - However, other parametrizations can be used. - The following example demonstrates one possible alternative. -

    - - - Converting from rectangular to parametric - -

    - Consider y=x^2. - Find parametric equations x=f(t), - y=g(t) for the parabola where t=\frac{dy}{dx}. - That is, t=a corresponds to the point on the graph whose tangent line has slope a. -

    -     - -

    - We start by computing \frac{dy}{dx}: \yp = 2x. - Thus we set t=2x. - We can solve for x and find x= t/2. - Knowing that y=x^2, we have y= t^2/4. - Thus parametric equations for the parabola y=x^2 are - - x=t/2 y=t^2/4 - . -

    - -

    - To find the point where the tangent line has a slope of -2, - we set t=-2. - This gives the point (-1, 1). - We can verify that the slope of the line tangent to the curve at this point indeed has a slope of -2. -

    -     - -     - -

    - We sometimes choose the parameter to accurately model physical behavior. -

    - - - Converting from rectangular to parametric - -

    - An object is fired from a height of 0 feet and lands 6 seconds later, 192 feet away. - Assuming ideal projectile motion, the height, in feet, - of the object can be described by h(x) = -x^2/64+3x, - where x is the distance in feet from the initial location. (Thus - h(0) = h(192) = 0 feet.) Find parametric equations x=f(t), - y=g(t) for the path of the projectile where x is the horizontal distance the object has traveled at time t - (in seconds) - and y is the height at time t. -

    -     - -

    - Physics tells us that the horizontal motion of the projectile is linear; - that is, the horizontal speed of the projectile is constant. - Since the object travels 192 in 6, - we deduce that the object is moving horizontally at a rate of 32, - giving the equation x=32t. - As y=-x^2/64+3x, we find y= -16t^2+96t. - We can quickly verify that \yp'=-32 , - the acceleration due to gravity, - and that the projectile reaches its maximum at t=3, - halfway along its path. -

    - -

    - These parametric equations make certain determinations about the object's location easy: - 2 seconds into the flight the object is at the point \big(x(2),y(2)\big) = \big(64,128\big). - That is, it has traveled horizontally 64 - and is at a height of 128, - as shown in . -

    - -
    - Graphing projectile motion in - - - A parabolic arc describing the motion of a projectile fired up and to the right from the origin. - -

    - The curve is a parabola that opens downward from a vertex at (96,144). - It illustrates the path of a projectile that is fired from the origin and travels up and to the right - before reaching its maximum height at the vertex of the parabola. - It then travels back down, returning to the x axis at the point (192,0). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=0,ymax=160, - xmin=0,xmax=210 - ] - - \addplot+ [domain=0:6,samples=40] ({32*x},{-16*x^2+96*x}); - - \draw [->,>=latex] (axis cs:32,80) -- (axis cs:35.2, 86.24); - - \filldraw (axis cs:64, 128) circle (2.4pt) node [above left] { $t=2$}; - - \draw (axis cs: 100,50) node [align=left] { $x=32t$\\ $y=-16t^2+96t$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -     - -

    - It is sometimes necessary to convert given parametric equations into rectangular form. - This can be decidedly more difficult, - as some simple looking parametric equations can have very - complicated rectangular equations. - This conversion is often referred to as - eliminating the parameter, - as we are looking for a relationship between x and y that does not involve the parameter t. -

    - - - Eliminating the parameter - -

    - Find a rectangular equation for the curve described by - - x= \frac{1}{t^2+1} \text{ and } y=\frac{t^2}{t^2+1} - . -

    -     - -

    - There is not a set way to eliminate a parameter. - One method is to solve for t in one equation and then substitute that value in the second. - We use that technique here, then show a second, simpler method. -

    - -

    - Starting with x= 1/(t^2+1), solve for t: - t = \pm\sqrt{1/x-1}. - Substitute this value for t in the equation for y: - - y \amp = \frac{t^2}{t^2 +1} - \amp = \frac{1/x-1}{1/x-1+1} - \amp = \frac{1/x - 1}{1/x} - \amp = \left(\frac1x-1\right)\cdot x - \amp = 1-x - . -

    - -
    - Graphing parametric and rectangular equations for a graph in - - - A graph of two overlapping lines illustrating rectangular and parametric descriptions of a curve. - -

    - Two lines are plotted. The first is the line y=1-x, - which is drawn from the point (-1,2) to the point (2,-1). -

    - -

    - A second, thicker line is drawn over the first line, from (0,1) to (1,0). - At (0,1) there is a hollow (white-filled) dot, - indicating that the second line approaches, but does not reach, this point. - At (1,0) there is a solid dot, indicating the fact that the parametric version of the line - reaches this point, but does not go past it. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.1,ymax=2.1, - xmin=-2.1,xmax=2.1 - ] - - \addplot+ [semithick,domain=-2:2] ({x},{1-x}); - \addplot [firstcurvestyle,very thick,domain=0:1] ({x},{1-x}); - - \draw [fill=white] (axis cs:0,1) circle (2.4pt); - \filldraw [fill=black] (axis cs:1,0) circle (2.4pt) node [align=left,shift={(5pt,30pt)}] { $\displaystyle x=\frac{1}{t^2+1}$\\[2pt] $\displaystyle y=\frac{t^2}{t^2+1}$}; - - \draw (axis cs: -1,1.4) node { $y=1-x$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Thus y=1-x. - One may have recognized this earlier by manipulating the equation for y: - - y = \frac{t^2}{t^2+1} = 1-\frac{1}{t^2+1} = 1-x - . -

    - -

    - This is a shortcut that is very specific to this problem; - sometimes shortcuts exist and are worth looking for. -

    - -

    - We should be careful to limit the domain of the function y=1-x. - The parametric equations limit x to values in (0,1], - thus to produce the same graph we should limit the domain of y=1-x to the same. -

    - -

    - The graphs of these functions is given in . - The portion of the graph defined by the parametric equations is given in a thick line; - the graph defined by y=1-x with unrestricted domain is given in a thin line. -

    -     - -     - - - Eliminating the parameter - -

    - Eliminate the parameter in - x=4\cos(t) +3, y= 2\sin(t) +1 -

    -     - -

    - We should not try to solve for t in this situation as the resulting algebra/trig would be messy. - Rather, we solve for \cos(t) and \sin(t) in each equation, - respectively. - This gives - - \cos(t) = \frac{x-3}{4} \text{ and } \sin(t) =\frac{y-1}{2} - . -

    - -

    - The Pythagorean Theorem gives \cos^2(t) +\sin^2(t) =1, so: - - \cos^2(t) +\sin^2(t) \amp =1 - \left(\frac{x-3}{4}\right)^2 +\left(\frac{y-1}{2}\right)^2 \amp =1 - \frac{(x-3)^2}{16}+\frac{(y-1)^2}{4} \amp =1 - -

    - -
    - Graphing the parametric equations x=4\cos(t) +3, y=2\sin(t) +1 in - - - Graph of the ellipse obtained from the parametric equations in this example. - -

    - The plot shows an ellipse. The major axis runs from (-1,1) to (7,1), - and the minor axis runs from (3,-1) to (3,5). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-3.1,ymax=5.1, - xmin=-1.1,xmax=8.5 - ] - - \addplot+ [domain=0:360,samples=70] ({4*cos(x)+3},{2*sin(x)+1}); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - This final equation should look familiar it is the equation of an ellipse! - - plots the parametric equations, - demonstrating that the graph is indeed of an ellipse with a horizontal major axis and center at (3,1). -

    -     - -     - - - - - -

    - The Pythagorean Theorem can also be used to identify parametric equations for hyperbolas. - We give the parametric equations for ellipses and hyperbolas in the following Key Idea. -

    - - - Parametric Equations of Ellipses and Hyperbolas -

    -

      -
    • -

      - The parametric equations - - x=a\cos(t) +h, y=b\sin(t) +k - - define an ellipse with horizontal axis of length 2a and vertical axis of length 2b, - centered at (h,k). - ellipseparametric equations -

      -
      • - -
    • -

      - The parametric equations - - x= a\tan(t) +h, y=\pm b\sec(t) +k - - define a hyperbola with vertical transverse axis centered at (h,k), and - - x=\pm a\sec(t) +h, y=b\tan(t) + k - - defines a hyperbola with horizontal transverse axis. - Each has asymptotes at y=\pm b/a(x-h)+k. - - - hyperbolaparametric equations - - -

      -
      • -
    -

    -     -     - - - Special Curves -

    - - gives a small gallery of interesting and famous - curves along with parametric equations that produce them. - Interested readers can begin learning more about these curves through internet searches. -

    - -

    - One might note a feature shared by two of these graphs: - sharp corners, or cusps. - We have seen graphs with cusps before and determined that such functions are not differentiable at these points. - This leads us to a definition. -

    - -
    - A gallery of interesting planar curves - - -
    - - Astroid where x=\cos^3(t) and y=\sin^3(t) - - Plot of an astroid: a curve like a 4-pointed star. - -

    - A plot of the parametric curve x=\cos^3(t), y=\sin^3(x), - known as an astroid. - The curve has four cusps, at the points (1,0), (0,1), (-1,0), and (0,-1). - The appearance is not unlike what one would get if one held these four points fixed on the unit circle, - and then bent the rest of the circle so that it curves inward (toward the origin) rather than outward. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.1 - ] - - \addplot+ [domain=0:360,samples=80] ({(cos(x))^3},{(sin(x))^3}); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - Rose Curve where x=\cos(t)\sin(4t) and y=\sin(t)\sin(4t) - - Plot of a rose curve with 8 loops. - -

    - The plot of the curve x=\cos(t)\sin(4t), y=\sin(t)\sin(4t) resembles a flower. - It consists of eight loops, equal in size and shape, each passing through the origin. - The shape of each loop resembles a narrow tear-drop. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-1,1}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.1 - ] - - \addplot+ [domain=0:360,samples=300] ({sin(4*x)*cos(x)},{(sin(4*x)*sin(x)}); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - - -
    - - Hypotrochoid where x=2\cos(t)+5\cos(2t/3) and y=2\sin(t)-5\sin(2t/3) - - Plot of a hypotrochoid, which resembles a smooth, five-pointed star. - -

    - The curve x=2\cos(t)+5\cos(2t/3), y=2\sin(t)-5\sin(2t/3), known as a hypotrochoid, is shown. - The curve is symmetric about the x axis, and resembles a five-pointed star, - with one point on the x axis at (7,0). -

    - -

    - The points of the star are rounded, rather than sharp. - The curve passes through itself several times, - in a manner similar to how a pentagram is drawn. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-7.1,ymax=7.1, - xmin=-7.1,xmax=7.5 - ] - - \addplot+[domain=0:1440,samples=120] ({2*cos(x)+5*cos(2*x/3)},{2*sin(x)-5*sin(2*x/3)}); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - Epicycloid where x=4\cos(t)-\cos(4t) and y=4\sin(t)-\sin(4t) - - Plot of an epicycloid, which looks somewhat like a puffy clover leaf. - -

    - The epicycloid x=4\cos(t)-\cos(4t), y=4\sin(t)-\sin(4t) is shown. - The curve travels once around the origin, and consists of three arcs, - with each pair of arcs meeting at a cusp. -

    - -

    - The effect is not unlike the appearance of a three-leaf clover, - if the leaves were much more puffy, and further from the center. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-7.1,ymax=7.1, - xmin=-7.1,xmax=7.5 - ] - - \addplot+ [domain=0:720,samples=240] ({4*cos(x)-cos(4*x)},{4*sin(x)-sin(4*x)}); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -     -
    - - - Smooth - -

    - A curve C defined by x=f(t), - y=g(t) is smooth - on an interval I if \fp and \gp are continuous on I and not simultaneously 0 - (except possibly at the endpoints of I). - A curve is piecewise smooth - on I if I can be partitioned into subintervals where C is smooth on each subinterval. - curvesmooth - smoothcurve - cusp -

    -     -     - -

    - Consider the astroid, given by x=\cos^3(t), y=\sin^3(t). - Taking derivatives, we have: - - x' = -3\cos^2(t) \sin(t) \text{ and } \yp = 3\sin^2(t) \cos(t) - . -

    - -

    - It is clear that each is 0 when t=0,\, \pi/2,\, \pi,\ldots. - Thus the astroid is not smooth at these points, - corresponding to the cusps seen in the figure. -

    - -

    - We demonstrate this once more. -

    - - - Determine where a curve is not smooth - -

    - Let a curve C be defined by the parametric equations - x=t^3-12t+17 and y=t^2-4t+8. - Determine the points, if any, where it is not smooth. -

    -     - -

    - We begin by taking derivatives. - - x' = 3t^2-12, \yp = 2t-4 - . -

    - -

    - We set each equal to 0: - - x' \amp = 0 \Rightarrow 3t^2-12=0 \Rightarrow t=\pm 2 - \yp \amp = 0 \Rightarrow 2t-4 = 0 \Rightarrow t=2 - -

    - -

    - We see at t=2 both x' and \yp are 0; - thus C is not smooth at t=2, - corresponding to the point (1,4). - The curve is graphed in , - illustrating the cusp at (1,4). -

    - -
    - Graphing the curve in ; note it is not smooth at (1,4) - - - Plot of a parametric curve with a sharp cusp. - -

    - A plot of the curve x=t^3-12t+17, y=t^2-4t+8 is shown. - Despite the fact that x and y are both differentiable functions of t, - there is a sharp cusp at the point corresponding to t=2, - which happens to be the point where the derivatives of x and y with respect to t are simultaneously zero. -

    - -

    - The appearance of the curve is not unlike that of a set of tweezers. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=4, - ymin=-1,ymax=9, - xmin=-1,xmax=11 - ] - - \addplot+ [domain=-2:3.5] ({x^3-12*x+17},{x^2-4*x+8}); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -     - -

    - If a curve is not smooth at t=t_0, - it means that x'(t_0)=\yp(t_0)=0 as defined. - This, in turn, means that rate of change of x - (and y) - is 0; - that is, at that instant, neither x nor y is changing. - If the parametric equations describe the path of some object, - this means the object is at rest at t_0. - An object at rest can make a sharp change in direction, - whereas moving objects tend to change direction in a smooth fashion. -

    - -

    - One should be careful to note that a sharp corner - does not have to occur when a curve is not smooth. - For instance, one can verify that x=t^3, - y=t^6 produce the familiar y=x^2 parabola. - However, in this parametrization, the curve is not smooth. - A particle traveling along the parabola according to the given parametric equations comes to rest at t=0, - though no sharp point is created. -

    - -

    - Our previous experience with cusps taught us that a function was not differentiable at a cusp. - This can lead us to wonder about derivatives in the context of parametric equations and the application of other calculus concepts. - Given a curve defined parametrically, - how do we find the slopes of tangent lines? - Can we determine concavity? - We explore these concepts and more in the next section. -

    -     - - - - Terms and Concepts - - - -

    - When sketching the graph of parametric equations, - the x and y values are found separately, - then plotted together. - -

    -     - -     - - - - -

    - The direction in which a graph is moving - is called the of the graph. -

    -     - - - - - - - - -     - - - - -

    - An equation written as y=f(x) is said to be written in - form. -

    -     - - - - - - - - -     - - - - -

    - Create parametric equations x=f(t), - y=g(t) and sketch their graph. - Explain any interesting features of your graph based on the functions f and g. -

    -     - - - -     -     - - - Problems - - - -

    - Sketch the graph of the given parametric equations by hand, - making a table of points to plot. - Be sure to indicate the orientation of the graph. -

    -     - - - - -

    - x=t^2+t,y=1-t^2,-3\leq t\leq 3 -

    -     - - - - Sketch of the parametric curve in this exercise. - -

    - The sketch for this exercise is a curve that lies mostly in the fourth quadrant. - It resembles part of a slingshot orbit for a comet passing around the sun: - the curve passes through the origin from below, turns quickly in the second quadrant, - crossing the y axis at (0,1), and then the x axis at (2,0), - where it returns to the fourth quadrant. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=4, - minor y tick num=4, - ymin=-9,ymax=2, - xmin=-1,xmax=13 - ] - - \addplot+ [domain=-3:3,samples=40] ({x^2+x},{1-x^2}); - - \draw [>=latex,->,thick] (axis cs: 6,-3) -- (axis cs:6.05,-3.04); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=1,y=5\sin(t),-\pi/2\leq t\leq \pi/2 -

    -     - - - - The vertical line x=1. - -

    - The curve for this exercise is the vertical line x=1. - An arrow on the line indicates that the direction of travel is up. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - minor y tick num=4, - ymin=-5.5,ymax=5.5, - xmin=-.5,xmax=1.6 - ] - - \addplot+ [domain=-5:5] ({1},{x}); - - \draw [>=latex,->,thick] (axis cs: 1,1) -- (axis cs:1,1.1); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=t^2,y=2,-2\leq t\leq 2 -

    -     - - - - The horizontal line y=2, marked with two arrows. - -

    - The horizontal line y=2. - On the line there are two arrows pointing in opposite directions. - These indicate that the direction of travel is to the left when t\lt 0, - and to the right when t\gt 0. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=4, - minor y tick num=4, - ymin=-.5,ymax=2.5, - xmin=-.5,xmax=2.5 - ] - - \addplot+ [domain=-2:2] ({x^2},{2}); - - \draw [>=latex,<->,thick] (axis cs: .9,2) -- (axis cs:1.5,2); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=t^3-t+3,y=t^2+1,-2\leq t\leq 2 -

    -     - - - - The solution curve for this exercise. - -

    - The curve begins to the left of the y axis, and crosses near (0,4). - It passes through the first quadrant to the point (3,2); - it then bends downward and makes a teardrop-shaped loop before passing through (3,2) a second time, - and then continuing up and to the right, through the first quadrant. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=4.5, - xmin=-.5,xmax=5.5 - ] - - \addplot+ [domain=-2:2,samples=50] ({x^3-x+3},{x^2+1}); - - \draw [>=latex,->,thick] (axis cs: 3.897,2.69) -- (axis cs:3.938,2.7161); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - -     - - - -

    - Sketch the graph of the given parametric equations; - using a graphing utility is advisable. - Be sure to indicate the orientation of the graph. -

    -     - - - - -

    - x=t^3-2t^2,y=t^2,-2\leq t \leq 3 -

    -     - - - - Computer-generated sketch of the parametric curve in this exercise. - -

    - A curve resembling a check mark, with a cusp at the origin. - Direction of travel is from the second quadrant toward the cusp, - and then up from the cusp to a y intercept at (0,4), - and then into the first quadrant. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=0.1,ymax=9, - xmin=-11,xmax=11 - ] - - \addplot+ [domain=-2:3,samples=40] ({x^3-2*x^2},{x^2}); - - \draw [>=latex,->,thick] (axis cs:-1.183,1.69)-- (axis cs:-1.18411,1.7161); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=1/t,y=\sin(t),0\lt t \leq 10 -

    -     - - - - Computer-generated sketch of the parametric curve in this exercise. - -

    - A curve resembling a sine wave with a period that gets longer for large values of x. - The direction of travel is that of decreasing x value, with the y value oscillating. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.1,ymax=1.1, - xmin=-0.5,xmax=1.5 - ] - - \addplot+ [domain=0.1:10,samples=101] ({1/x},{sin(deg(x))}); - - \draw [>=latex,->,thick] (axis cs:0.25,-.756802)-- (axis cs:0.249377,-.763301); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=3\cos(t),y=5\sin(t),0\leq t \leq 2\pi -

    -     - - - - Computer-generated sketch of the parametric curve in this exercise. - -

    - The curve is an ellipse, centered at the origin, with counter-clockwise direction of travel. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-5.5,ymax=5.5, - xmin=-5.5,xmax=5.5 - ] - - \addplot+ [domain=0:360,samples=60] ({3*cos(x)},{5*sin(x)}); - - \draw [>=latex,->,thick] (axis cs:2.12,3.5355)-- (axis cs:2.1,3.5707); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=3\cos(t) +2,y=5\sin(t) +3,0\leq t \leq 2\pi -

    -     - - - - Computer-generated sketch of the parametric curve in this exercise. - -

    - An ellipse with center (2,3) and counter-clockwise direction of travel. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=4, - minor y tick num=4, - ymin=-5.5,ymax=8.5, - xmin=-5.5,xmax=7.5 - ] - - \addplot+ [domain=0:360,samples=60] ({3*cos(x)+2},{5*sin(x)+3}); - - \draw [>=latex,->,thick] (axis cs:4.12,6.5355) -- (axis cs:4.1,6.5707); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=\cos(t),y=\cos(2t),0\leq t \leq \pi -

    -     - - - - Computer-generated sketch of the parametric curve in this exercise. - -

    - The curve resembles a parabola, with vertex at (0,-1). - The direction of travel is from right to left. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.1 - ] - - \addplot+ [domain=0:180,samples=60] ({cos(x)},{cos(x*2)}); - - \draw [>=latex,->,thick] (axis cs:0.8845,0.5646) -- (axis cs:0.8798,0.548024); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=\cos(t),y=\sin(2t),0\leq t \leq 2\pi -

    -     - - - - Computer-generated sketch of the parametric curve in this exercise. - -

    - A figure-eight curve, centered at the origin. - The orientation is counter-clockwise in the fourth and first quadrants; - once the curve passes through the origin (a point of self-intersection) - this direction becomes clockwise it the second and third quadrants. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.1] - - \addplot+ [domain=0:360,samples=120] ({cos(x)},{sin(x*2)}); - - \draw [>=latex,->,thick] (axis cs:0.8845,0.82534) -- (axis cs:0.8798,0.8365); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=2\sec(t),y=3\tan(t),-\pi/2\lt t \lt \pi/2 -

    -     - - - - Computer-generated sketch of the parametric curve in this exercise. - -

    - The curve resembles one branch of a hyperbola, opening to the right, with a vertex at (2,0). - The direction of travel is that of increasing y value. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=4, - minor y tick num=4, - ymin=-16.,ymax=16, - xmin=-1,xmax=11 - ] - - \addplot+ [domain=-80:80] ({2*sec(x)},{3*tan(x)}); - - \draw [>=latex,->,thick] (axis cs:5.32,7.395) -- (axis cs:5.4549,7.61254); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=\cosh(t),y=\sinh(t),-2\leq t \leq 2 -

    -     - - - Computer-generated sketch of the parametric curve in this exercise. - -

    - The curve resembles one branch of a hyperbola, opening to the right, with a vertex at (1,0). - The direction of travel is that of increasing y value. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - minor x tick num=4, - minor y tick num=4, - ymin=-4.5, - ymax=4.5, - xmin=-.1, - xmax=4.5 - ] - - \addplot+ [domain=-2:2,samples=60] ({cosh(x)},{sinh(x)}); - \draw[>=latex,->,thick] (axis cs:2.35241, -2.12928)-- (axis cs:2.33123, -2.10586); - \end{axis} - - \end{tikzpicture} - - -     -     - - - - - -

    - x=\cos(t) +\frac14\cos(8t),y=\sin(t) +\frac14\sin(8t),0\leq t \leq 2\pi -

    -     - - - - Computer-generated sketch of the parametric curve in this exercise. - -

    - A flower-shaped curve, with 7 petals. - Each petal is an arc that loops around and intersects itself before continuing to the next arc. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.5,ymax=1.5, - xmin=-1.5,xmax=1.5 - ] - - \addplot+ [domain=0:360,samples=280] ({cos(x)+cos(8*x)/4},{sin(x)+sin(8*x)/4}); - - \draw [>=latex,->,thick] (axis cs:.9571,.7071) -- (axis cs:.9534,.7341); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - x=\cos(t) +\frac14\sin(8t),y=\sin(t) +\frac14\cos(8t),0\leq t \leq 2\pi -

    -     - - - - Computer-generated sketch of the parametric curve in this exercise. - -

    - A curve that spirals around the origin, with several self-intersecting loops. - This curve has 9 arcs and 9 loops. It is similar to the curve in the previous exercise, - except that this time, the arcs bend inward toward the origin, rather than outward. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.5,ymax=1.5, - xmin=-1.5,xmax=1.5 - ] - - \addplot+ [domain=0:360,samples=260] ({cos(x)+sin(8*x)/4},{sin(x)+cos(8*x)/4}); - - \draw [>=latex,->,thick] (axis cs:0.3613,.6771) -- (axis cs:.33215,.68275); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - -     - - - -

    - Four sets of parametric equations are given. - Describe how their graphs are similar and different. - Be sure to discuss orientation and ranges. -

    -     - - - - - -

    - x=t y=t^2, -\infty\lt t\lt \infty -

    -     - -

    - Traces the parabola y=x^2, moves from left to right. -

    -     -     - - -

    - x=\sin(t) y=\sin^2(t), - -\infty\lt t\lt \infty -

    -     - -

    - Traces the parabola y=x^2, - but only from -1\leq x\leq 1; - traces this portion back and forth infinitely. -

    -     -     - - -

    - x=e^t y=e^{2t}, - -\infty\lt t\lt \infty -

    -     - -

    - Traces the parabola y=x^2, - but only for 0\lt x. - Moves left to right. -

    -     -     - - -

    - x=-t y=t^2, -\infty\lt t\lt \infty -

    -     - -

    - Traces the parabola y=x^2, moves from right to left. -

    -     -     - -     - - - - - -

    - x=\cos(t) y=\sin(t), 0\leq t\leq 2\pi -

    -     - -

    - Traces a circle of radius 1 counterclockwise once. -

    -     -     - - -

    - x=\cos(t^2) y=\sin(t^2), 0\leq t\leq 2\pi -

    -     - -

    - Traces a circle of radius 1 counterclockwise over 6 times. -

    -     -     - - -

    - x=\cos(1/t) y=\sin(1/t), 0\lt t\lt 1 -

    -     - -

    - Traces a circle of radius 1 clockwise infinite times. -

    -     -     - - -

    - x=\cos(\cos(t) ) y=\sin(\cos(t) ), 0\leq t\leq 2\pi -

    -     - -

    - Traces an arc of a circle of radius 1, from an angle of -1 radians to 1 radian, twice. -

    -     -     - -     - -     - - - -

    - Eliminate the parameter in the given parametric equations. -

    -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("3x+2y=17",limits=>[[-3,3],[6,11]]); - - -

    - x=2t+5, y=-3t+1 -

    - - - Eliminate the parameter t. - -

    - -

    -     -     -     - - - - -

    - x=\sec(t), y=\tan(t) -

    -     - -

    - x^2-y^2=1 -

    -     - -     - - - - -

    - x=4\sin(t) +1, y=3\cos(t) -2 -

    -     - -

    - \frac{(x-1)^2}{16}+\frac{(y+2)^2}{9}=1 -

    -     - -     - - - - -

    - x=t^2, y=t^3 -

    -     - -

    - y=x^{3/2} -

    -     - -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("y-2x=3",limits=>[[-3,3],[0,6]]); - - -

    - x=\frac{1}{t+1}, y=\frac{3t+5}{t+1} -

    - - - Eliminate the parameter t. - -

    - -

    -     -     -     - - - - -

    - \ds x=e^t, \ds y=e^{3t}-3 -

    -     - -

    - y=x^3-3 -

    -     - -     - - - - -

    - \ds x=\ln(t), \ds y=t^2-1 -

    -     - -

    - y=e^{2x}-1 -

    -     - -     - - - - -

    - \ds x=\cot(t), \ds y=\csc(t) -

    -     - -

    - y^2-x^2=1 -

    -     - -     - - - - -

    - \ds x=\cosh(t), \ds y=\sinh(t) -

    -     - -

    - x^2-y^2=1 -

    -     - -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("x=1-2y^2",limits=>[[-1,2],[-2,2]]); - - -

    - x=\cos(2t), y=\sin(t) -

    - - - Eliminate the parameter t. - -

    - -

    -     -     -     - -     - - - -

    - Eliminate the parameter in the given parametric equations. - Describe the curve defined by the parametric equations based on its rectangular form. -

    -     - - - - -

    - \ds x=at+x_0, \ds y=bt+y_0 -

    -     - -

    - y=\frac{b}{a}(x-x_0)+y_0; - line through (x_0,y_0) with slope b/a. -

    -     - -     - - - - -

    - \ds x=r\cos(t), \ds y=r\sin(t) -

    -     - -

    - x^2+y^2=r^2; - circle centered at (0,0) with radius r. -

    -     - -     - - - - -

    - \ds x=a\cos(t) +h, \ds y=b\sin(t) +k -

    -     - -

    - \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1; - ellipse centered at (h,k) with horizontal axis of length 2a and vertical axis of length 2b. -

    -     - -     - - - - -

    - \ds x=a\sec(t) +h, \ds y=b\tan(t) +k -

    -     - -

    - \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1; - hyperbola centered at (h,k) with horizontal transverse axis and asymptotes with slope b/a. - The parametric equations only give half of the hyperbola. - When a>0, the right half; - when a\lt 0, the left half. -

    -     - -     - -     - - - -

    - Find parametric equations for the given rectangular equation using the parameter \ds t=\frac{dy}{dx}. - Verify that at t=1, - the point on the graph has a tangent line with slope of 1. -

    -     - - - - - Context()->variables->are(t=>'Real'); - $x=Formula("(t+11)/6"); - $y=Formula("(t^2-97)/12"); - Context("Point"); - $point=Point("(2,-8)"); - Context("Numeric"); - $yp=Formula("6x-11"); - $yp1=Real("1"); - - -

    - y=3x^2-11x+2 -

    - - - Give the value for x(t): - - -

    - -

    - - - Give the value for y(t): - - -

    - -

    - - - Give the point on the curve when t=1. - - -

    - -

    - - - Determine the formula for \lz{y}{x}: - - -

    - -

    - - - Find the value of \lz{y}{x} when t=1: - - -

    - -

    -     -     -     - - - - - Context()->variables->are(t=>'Real'); - Context()->variables->set(t=>{limits=>[0.1,10]}); - $x=Formula("ln(t)"); - $y=Formula("t"); - Context("Point"); - $point=Point("(0,1)"); - Context("Numeric"); - $yp=Formula("e^x"); - $yp1=Real("1"); - - -

    - y=e^x -

    - - - Give the value for x(t): - - -

    - -

    - - - Give the value for y(t): - - -

    - -

    - - - Give the point on the curve when t=1. - - -

    - -

    - - - Determine the formula for \lz{y}{x}: - - -

    - -

    - - - Find the value of \lz{y}{x} when t=1: - - -

    - -

    -     -     -     - - - - - Context()->variables->are(t=>'Real'); - Context()->variables->set(t=>{limits=>[-0.9,0.9]}); - $x=Formula("cos^-1(t)"); - $y=Formula("sqrt(1-t^2)"); - Context("Point"); - $point=Point("(0,0)"); - Context("Numeric"); - $yp=Formula("cos(x)"); - $yp1=Real("1"); - - -

    - y=\sin(x) -

    - - - Give the value for x(t): - - -

    - -

    - - - Give the value for y(t): - - -

    - -

    - - - Give the point on the curve when t=1. - - -

    - -

    - - - Determine the formula for \lz{y}{x}: - - -

    - -

    - - - Find the value of \lz{y}{x} when t=1: - - -

    - -

    -     -     -     - - - - -

    - y=\sqrt{x} on [0,\infty) -

    -     - -

    - x=1/(4t^2), y=1/(2t). - At t=1, x=1/4, y=1/2. -

    - -

    - \yp=1/(2\sqrt{x}); when x=1/4, \yp=1. -

    -     - -     - -     - - - -

    - Find the values of t where the graph of the parametric equations crosses itself. -

    -     - - - - - - $crossings=List("-1,1"); - Context("Point"); - $points=List("(3,-2)"); - - -

    - x=t^3-t+3, y=t^2-3 -

    - - - Give the value(s) of t where the graph crosses itself. - Enter your answer as a comma-separated list of numbers. - - -

    - -

    - - - Give the point(s) in the plane where the crossings happen. - If there is more than one, use commas to separate them. - - -

    - -

    -     -     -     - - - - -

    - x=t^3-4t^2+t+7,y=t^2-t -

    -     - -

    - t=-1,\, 2 -

    -     - -     - - - - -

    - x=\cos(t),y=\sin(2t) on [0,2\pi] -

    -     - -

    - t=\pi/2, 3\pi/2 -

    -     - -     - - - - -

    - x=\cos(t) \cos(3t),y=\sin(t) \cos(3t) on [0,\pi] -

    -     - -

    - t=\pi/6, \pi/2, 5\pi/6 -

    -     - -     - -     - - - -

    - Find the value(s) of t where the curve defined by the parametric equations is not smooth. -

    -     - - - - -

    - x=t^3+t^2-t,y=t^2+2t+3 -

    -     - -

    - t=-1 -

    -     - -     - - - - - - $cusps=List("2"); - Context("Point"); - $points=List("(-4,-8)"); - - -

    - x=t^2-4t, y=t^3-2t^2-4t -

    - - - Give the values of t where the curve is not smooth. - (If there is more than one, use commas to separate them.) - - -

    - -

    - - - Give the corresponding points in the plane where the curve is not smooth. - (If there is more than one, use commas to separate them.) - - -

    - -

    -     -     -     - - - - -

    - x=\cos(t),y=2\cos(t) -

    -     - -

    - t=n\pi, where n is an integer -

    -     - -     - - - - - $cusps=List("0"); - Context("Point"); - $points=List("(1,0)"); - - -

    - x=2\cos(t)-\cos(2t), - y=2\sin(t)-\sin(2t) -

    - - - Give the values of t where the curve is not smooth. - (If there is more than one, use commas to separate them.) - - -

    - -

    - - - Give the corresponding points in the plane where the curve is not smooth. - (If there is more than one, use commas to separate them.) - - -

    - -

    -     - -

    - t=n\pi, where n is an even integer -

    -     -     -     - -     - - - -

    - Find parametric equations that describe the given situation. -

    -     - - - - -

    - A projectile is fired from a height of 0, - landing 16 away in 4. -

    -     - -

    - x=4t, y=-16t^2 + 64t -

    -     - -     - - - - -

    - A projectile is fired from a height of 0, - landing 200 away in 4. -

    -     - -

    - x=50t, y=-16t^2 + 64t -

    -     - -     - - - - -

    - A projectile is fired from a height of 0, - landing 200 away in 20. -

    -     - -

    - x=10t, y=-16t^2 + 320t -

    -     - -     - - - - - Context()->variables->are(t=>'Real'); - $x=Formula("2cos(t)"); - $y=Formula("-2sin(t)"); - Context()->flags->set(tolType=>'absolute',tolerance=>0.001); - $multians=MultiAnswer($x,$y)->with( - singleResult=>1, - checker=>sub{ - my ($correct,$student,$self)=@_; - my ($xcor,$ycor)=@{$correct}; - my ($xstu,$ystu)=@{$student}; - my $eliminated=Formula("($xstu/2)^2 + ($ystu/2)^2"); - return 0 unless ($eliminated == Formula("1")); - - #normalize answers - $xstu=$xstu/2; - $ystu=-$ystu/2; - $xcor=$xcor/2; - $ycor=-$ycor/2; - - #shift student answers - my $xstu0=$xstu->eval(t=>0); - my $ystu0=$ystu->eval(t=>0); - my $phi=($xstu0 >= 0)?Compute("arcsin($ystu0)"):Compute("pi-arcsin($ystu0)"); - $xstu=$xstu->substitute(t=>Formula("t-$phi")); - $ystu=$ystu->substitute(t=>Formula("t-$phi")); - - return ($xstu == $xcor and $ystu == $ycor); - } - ); - - -

    - Find parametric equations that describe a circle of radius 2, - centered at the origin, - that is traced clockwise once at constant speed on [0,2\pi]. -

    - - Enter the function for x. - -

    - -

    - - Enter the function for y. - -

    - -

    -     -     -     - - - - - Context()->variables->are(t=>'Real'); - $x=Formula("3cos(2 pi t)+1"); - $y=Formula("3sin(2 pi t)+1"); - Context()->flags->set(tolType=>'absolute',tolerance=>0.001); - $multians=MultiAnswer($x,$y)->with( - singleResult=>1, - checker=>sub{ - my ($correct,$student,$self)=@_; - my ($xcor,$ycor)=@{$correct}; - my ($xstu,$ystu)=@{$student}; - my $eliminated=Formula("(($xstu-1)/3)^2 + (($ystu-1)/3)^2"); - return 0 unless ($eliminated == Formula("1")); - - #normalize answers - $xstu=($xstu-1)/3; - $ystu=($ystu-1)/3; - $xcor=($xcor-1)/3; - $ycor=($ycor-1)/3; - - #shift student answers - my $xstu0=$xstu->eval(t=>0); - my $ystu0=$ystu->eval(t=>0); - my $phi=($xstu0 >= 0)?Compute("arcsin($ystu0)"):Compute("pi-arcsin($ystu0)"); - $xstu=$xstu->substitute(t=>Formula("t-$phi/(2pi) - ")); - $ystu=$ystu->substitute(t=>Formula("t-$phi/(2pi)")); - - return ($xstu == $xcor and $ystu == $ycor); - } - ); - - -

    - Find parametric equations that describe a circle of radius 3, - centered at (1,1), - that is traced once counter-clockwise at constant speed on [0,1]. -

    - - Enter the function for x. - -

    - -

    - - Enter the function for y. - -

    - -

    -     -     -     - - - - - Context()->variables->are(t=>'Real'); - $x=Formula("3cos(2 pi t)+1"); - $y=Formula("3sin(2 pi t)+1"); - $multians = MultiAnswer($x, $y)->with( - singleResult => 1, - checker => sub { - my ( $correct, $student, $self ) = @_; - my ( $xstu, $ystu ) = @{$student}; - my $dx = $xstu->D('t'); - my $dy = $ystu->D('t'); - return 0 if ($dx == Formula("0") or $dy == Formula("0")); - my $eliminated = Formula("($xstu-1)^2 + ($ystu/3 - 1)^2"); - return ($eliminated == Formula("1")); - } - ); - - -

    - Find parametric equations that describe an ellipse centered at (1,3), - with vertical major axis of length 6 and minor axis of length 2. -

    - - Enter the function for x. - -

    - -

    - - Enter the function for y. - -

    - -

    -     -     -     - - - - -

    - An ellipse with foci at (\pm 1,0) and vertices at (\pm 5,0). -

    -     - -

    - x=5\cos(t), y=\sqrt{24}\sin(t); - other answers possible -

    -     - -     - - - - -

    - A hyperbola with foci at (5,-3) and (-1,-3), - and with vertices at (1,-3) and (3,-3). -

    -     - -

    - x=\pm\sec(t) +2, y=\sqrt{8}\tan(t) -3; - other answers possible -

    -     - -     - - - - -

    - A hyperbola with vertices at - (0,\pm 6) and asymptotes y=\pm 3x. -

    -     - -

    - x=2\tan(t), y=\pm 6\sec(t); - other answers possible -

    -     - -     - -     -     -     -
    -
    - Calculus and Parametric Equations - -

    - The previous section defined curves based on parametric equations. - In this section we'll employ the techniques of calculus to study these curves. -

    - -

    - We are still interested in lines tangent to points on a curve. - They describe how the y-values are changing with respect to the x-values, - they are useful in making approximations, - and they indicate instantaneous direction of travel. -

    - - - -     - - - Tangent lines to parametric curves - -

    - The slope of the tangent line is still \frac{dy}{dx}, - and the Chain Rule allows us to calculate this in the context of parametric equations. - If x=f(t) and y=g(t), - the Chain Rule states that - - \frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt} - . -

    - -

    - Solving for \frac{dy}{dx}, we get - - \frac{dy}{dx} = \frac{dy}{dt}\Bigg/\frac{dx}{dt} = \frac{\gp(t)}{\fp(t)} - , - provided that \fp(t)\neq 0. - This is important so we label it a Key Idea. -

    - - - Finding <m>\frac{dy}{dx}</m> with Parametric Equations -

    - Let x=f(t) and y=g(t), - where f and g are differentiable on some open interval I and \fp(t)\neq 0 on I. - Then - parametric equationsfinding \frac{dy}{dx} - derivativeparametric equations - - \frac{dy}{dx} = \frac{\gp(t)}{\fp(t)} - . -

    -     - -

    - We use this to define the tangent line. -

    - - - Tangent and Normal Lines - -

    - Let a curve C be parametrized by x=f(t) and y=g(t), - where f and g are differentiable functions on some interval I containing t=t_0. - The tangent line to C at t=t_0 is the line through - \big(f(t_0),g(t_0)\big) with slope m=\gp(t_0)/\fp(t_0), - provided \fp(t_0)\neq 0. -

    - -

    - The normal line to C at t=t_0 is the line through - \big(f(t_0),g(t_0)\big) with slope m=-\fp(t_0)/\gp(t_0), - provided \gp(t_0)\neq 0. - tangent line - normal line - parametric equationstangent line - parametric equationsnormal line -

    -     -     - - - -

    - The definition leaves two special cases to consider. - When the tangent line is horizontal, - the normal line is undefined by the above definition as \gp(t_0)=0. - Likewise, when the normal line is horizontal, - the tangent line is undefined. - It seems reasonable that these lines be defined - (one can draw a line tangent to the right side - of a circle, for instance), - so we add the following to the above definition. -

    - -

    -

      -
    1. -

      - If the tangent line at t=t_0 has a slope of 0, the normal line to C at t=t_0 is the line x=f(t_0). -

      -
      2. - -
    2. -

      - If the normal line at t=t_0 has a slope of 0, the tangent line to C at t=t_0 is the line x=f(t_0). -

      -
      3. -
    -

    - - - Tangent and Normal Lines to Curves - -

    - Let x=5t^2-6t+4 and y=t^2+6t-1, - and let C be the curve defined by these equations. -

    - -

    -

      -
    1. -

      - Find the equations of the tangent and normal lines to C at t=3. -

      -
      2. - -
    2. -

      - Find where C has vertical and horizontal tangent lines. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - We start by computing \fp(t) = 10t-6 and \gp(t) =2t+6. - Thus - - \frac{dy}{dx} = \frac{2t+6}{10t-6} - . - Make note of something that might seem unusual: - \frac{dy}{dx} is a function of t, not x. - Just as points on the curve are found in terms of t, - so are the slopes of the tangent lines. - The point on C at t=3 is (31,26). - The slope of the tangent line is m=1/2 and the slope of the normal line is m=-2. - Thus, - -

        -
      • -

        - the equation of the tangent line is \ds y=\frac12(x-31)+26, and -

        -
        • - -
      • -

        - the equation of the normal line is \ds y=-2(x-31)+26. -

        -
        • -
      -

      - -

      - This is illustrated in . -

      - -
      - Graphing tangent and normal lines in - - - Plot of a parametric curve and its tangent and normal lines at one point. - -

      - The curve x=5t^2-6t+4, y=t^2+6t-1 is shown. - The shape of the curve is that of a distored parabola, opening to the right from a vertex near (2,3). - (The precise location is \left(\frac{11}{5},\frac{74}{25}\right).) - The curve begins below the x axis, traveling left until it reaches the vertex, - after which it continues up and to the right. -

      - -

      - At the point (31,26), the tangent and normal lines to the curve are shown. - The tangent line has a positive slope, while the normal line has a negative slope, - and as expected, the two lines are perpendicular. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-20,ymax=55, - xmin=-5,xmax=90 - ] - - \addplot+ [domain=-3.5:5,samples=40] ({5*x^2-6*x+4},{x^2+6*x-1}); - - \addplot [secondcurvestyle,solid,domain=-5:90] ({x},{.5*(x-31)+26}); - \addplot [secondcurvestyle,solid,domain=-20:55] ({x},{-2*(x-31)+26}); - - \draw [>=latex,->] (axis cs: 12.14,-5.204) -- (axis cs: 12,-5.16); - - \filldraw (axis cs: 67,-10) circle (2.4pt) - (axis cs: 2.2,2.96) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
      - -
      2. - -
    2. -

      - To find where C has a horizontal tangent line, - we set \frac{dy}{dx}=0 and solve for t. - In this case, - this amounts to setting \gp(t)=0 and solving for t (and making sure that \fp(t)\neq 0). - - \gp(t)=0 \Rightarrow 2t+6=0 \Rightarrow t=-3 - . - The point on C corresponding to t=-3 is (67,-10); - the tangent line at that point is horizontal - (hence with equation y=-10). - To find where C has a vertical tangent line, - we find where it has a horizontal normal line, - and set -\frac{\fp(t)}{\gp(t)}=0. - This amounts to setting \fp(t)=0 and solving for t (and making sure that \gp(t)\neq 0). - - \fp(t)=0 \Rightarrow 10t-6=0 \Rightarrow t=0.6 - . - The point on C corresponding to t=0.6 is (2.2,2.96). - The tangent line at that point is x=2.2. - The points where the tangent lines are vertical and horizontal are indicated on the graph in . -

      -
      3. -
    -

    -     - -     - - - Tangent and Normal Lines to a Circle - -

    -

      -
    1. -

      - Find where the unit circle, - defined by x=\cos(t) and y=\sin(t) on [0,2\pi], - has vertical and horizontal tangent lines. -

      -
      2. - -
    2. -

      - Find the equation of the normal line at t=t_0. -

      -
      3. -
    -

    -

    - - -

    -

      -
    1. -

      - We compute the derivative following : - - \frac{dy}{dx} = \frac{\gp(t)}{\fp(t)} = -\frac{\cos(t) }{\sin(t) } - . - The derivative is 0 when \cos(t) = 0; - that is, when t=\pi/2,\, 3\pi/2. - These are the points (0,1) and (0,-1) on the circle. - - The normal line is horizontal (and hence, the tangent line is vertical) when \sin(t) =0; that is, when t= 0,\,\pi,\,2\pi, corresponding to the points (-1,0) and (0,1) on the circle. These results should make intuitive sense. -

      -
      2. - -
    2. -

      - The slope of the normal line at t=t_0 is \ds m=\frac{\sin(t_0) }{\cos(t_0) } = \tan(t_0). - This normal line goes through the point (\cos(t_0) ,\sin(t_0) ), - giving the line - - y \amp =\frac{\sin(t_0) }{\cos(t_0) }(x-\cos(t_0) ) + \sin(t_0) - \amp = (\tan(t_0) )x - , - as long as \cos(t_0) \neq 0. - It is an important fact to recognize that the normal lines to a circle pass through its center, - as illustrated in . - Stated in another way, - any line that passes through the center of a circle intersects the circle at right angles. -

      - -
      - Illustrating how a circle's normal lines pass through its center - - - Sketch of the unit circle and one normal line, which passes through the circle's center. - -

      - A sketch of the unit circle is shown. - At a point on the circle in the first quadrant - (corresponding to an angle that appears to be slightly more than \pi/3), - a normal line is drawn. -

      - -

      - The normal line passes through the center of the circle, illustrating the conclusion of this example. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.3,xmax=1.3 - ] - - \addplot+ [domain=0:360,samples=70] ({cos(x)},{sin(x)}); - \addplot+ [solid,domain=-1:1] ({x},{2*x}); - - \filldraw (axis cs: .447,.894) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      3. -
    -

    - - -     - - - Tangent lines when <m>\frac{dy}{dx}</m> is not defined - -

    - Find the equation of the tangent line to the astroid x=\cos^3(t), - y=\sin^3(t) at t=0, - shown in . -

    -
    - A graph of an astroid - - - Graph of an astroid. - -

    - Graph of the astroid curve x^{2/3}+y^{2/3}=1. - It has cusps at the vertices (1,0),(0,1),(-1,0) and (0,-1). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-1,1}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.1 - ] - - \addplot+ [domain=0:360,samples=70] ({(cos(x))^3},{(sin(x))^3}); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - We start by finding x'(t) and \yp(t): - - x'(t) = -3\sin(t) \cos^2(t) , \qquad \yp(t) = 3\cos(t) \sin^2(t) - . -

    - -

    - Note that both of these are 0 at t=0; - the curve is not smooth at t=0 forming a cusp on the graph. - Evaluating \frac{dy}{dx} at this point returns the indeterminate form of 0/0. -

    - -

    - We can, however, - examine the slopes of tangent lines near t=0, - and take the limit as t\to 0. - - \lim_{t\to0} \frac{\yp(t)}{x'(t)} \amp =\lim_{t\to0} \frac{3\cos(t) \sin^2(t) }{-3\sin(t) \cos^2(t) } \text{ (We can cancel as \(t\neq 0\).) } - \amp = \lim_{t\to0} -\frac{\sin(t) }{\cos(t) } - \amp = 0 - . -

    - -

    - We have accomplished something significant. - When the derivative \frac{dy}{dx} returns an indeterminate form at t=t_0, - we can define its value by setting it to be - \lim\limits_{t\to t_0}\frac{dy}{dx}, if that limit exists. - This allows us to find slopes of tangent lines at cusps, - which can be very beneficial. -

    - -

    - We found the slope of the tangent line at t=0 to be 0; - therefore the tangent line is y=0, the x-axis. -

    -     - -     - - - - -     - - - Concavity -

    - We continue to analyze curves in the plane by considering their concavity; - that is, we are interested in \frac{d^2y}{dx^2}, - the second derivative of y with respect to x. To find this, - we need to find the derivative of - \frac{dy}{dx} with respect to x; that is, - - \frac{d^2y}{dx^2}=\frac{d}{dx}\left[\frac{dy}{dx}\right] - , - but recall that \frac{dy}{dx} is a function of t, - not x, making this computation not straightforward. - concavity - parametric equationsconcavity -

    - -

    - To make the upcoming notation a bit simpler, - let h(t) = \frac{dy}{dx}. - We want \frac{d}{dx}[h(t)]; - that is, we want \frac{dh}{dx}. - We again appeal to the Chain Rule. - Note: - - \frac{dh}{dt} = \frac{dh}{dx}\cdot\frac{dx}{dt} \Rightarrow \frac{dh}{dx} = \frac{dh}{dt}\Bigg/\frac{dx}{dt} - . -

    - -

    - In words, to find \frac{d^2y}{dx^2}, - we first take the derivative of \frac{dy}{dx} - with respect to t, - then divide by x'(t). - We restate this as a Key Idea. -

    - - - Finding <m>\frac{d^2y}{dx^2}</m> with Parametric Equations -

    - Let x=f(t) and y=g(t) be twice differentiable functions on an open interval I, - where \fp(t)\neq 0 on I. - Then - parametric equationsfinding \frac{d^2y}{dx^2} - - \frac{d^2y}{dx^2} = \frac{d}{dt}\left[\frac{dy}{dx}\right]\Bigg/\frac{dx}{dt} = \frac{d}{dt}\left[\frac{dy}{dx}\right]\Bigg/\fp(t) - . -

    -     - -

    - Examples will help us understand this Key Idea. -

    - - - Concavity of Plane Curves - -

    - Let x=5t^2-6t+4 and - y=t^2+6t-1 as in . - Determine the t-intervals on which the graph is concave up/down. -

    -     - -

    - Concavity is determined by the second derivative of y with respect to x, - \frac{d^2y}{dx^2}, - so we compute that here following . -

    - -

    - In , - we found \ds\frac{dy}{dx} = \frac{2t+6}{10t-6} and \fp(t) = 10t-6. - So: - - \frac{d^2y}{dx^2} \amp = \frac{d}{dt}\left[\frac{2t+6}{10t-6}\right]\Bigg/(10t-6) - \amp = -\frac{72}{(10t-6)^2}\Bigg/(10t-6) - \amp = -\frac{72}{(10t-6)^3} - \amp = -\frac{9}{(5t-3)^3} - -

    - -
    - Graphing the parametric equations in to demonstrate concavity - - - Sketch of a parametric curve illustrating concavity. - -

    - The curve shown is the same distorted parabola from , - but without the tangent and normal lines. -

    - -

    - Below the vertex, which corresponds to t=3/5, - the curve is concave up, and there is a label on the curve indicating that for t\lt 3/5, the curve is concave up. - Above the vertex, the curve is concave down, and there is a label on the curve indicating that for t\gt 3/5, the curve is concave down. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-20,ymax=55, - xmin=-5,xmax=90 - ] - - \addplot+ [domain=-3.5:5,samples=50] ({5*x^2-6*x+4},{x^2+6*x-1}); - - \draw [>=latex,->] (axis cs: 12.14,-5.204) -- (axis cs: 12,-5.16); - - \filldraw (axis cs: 2.2,2.96) circle (2.4pt); - - \draw (axis cs: 31,26) node [above,rotate=26.56] { $t>3/5$; concave down}; - \draw (axis cs: 50.25,-9.75) node [below] { $t<3/5$; concave up}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - The graph of the parametric functions is concave up when - \frac{d^2y}{dx^2} \gt 0 and concave down when \frac{d^2y}{dx^2} \lt 0. - We determine the intervals when the second derivative is greater/less than 0 by first finding when it is 0 or undefined. -

    - -

    - As the numerator of \ds -\frac{9}{(5t-3)^3} is never 0, \frac{d^2y}{dx^2} \neq 0 for all t. - It is undefined when 5t-3=0; - that is, when t= 3/5. - Following the work established in , - we look at values of t greater/less than 3/5 on a number line: -

    - - - A number line for the sign of the second derivative in this example. - -

    - A number line is shown, on which the value 3/5 is marked. - To the left of this point there is text indicating that \frac{d^2y}{dx^2}\gt 0, and that the curve is concave up. - To the right of this point there is text indicating that \frac{d^2y}{dx^2}\lt 0, and that the curve is concave down. -

    -     - - - \begin{tikzpicture} - \begin{axis}[numberline, - xmin=-0.9,xmax=0.9, - after end axis/.code={ - \path (axis cs:0,0) - node [anchor=north,yshift=-0.075cm] {\footnotesize $3/5$}; - },] - \addplot[guideline] coordinates {(0,0) (0,2)}; - \addplot[mark=none] coordinates {(-0.5,1)} node {\parbox{8em}{\centering \small $\frac{d^2y}{dx^2}> 0$\\concave up}}; - \addplot[mark=none] coordinates {(0.5,1)} node {\parbox{8em}{\centering \small $\frac{d^2y}{dx^2}<0$\\concave down}}; - \end{axis} - - \end{tikzpicture} - - - - -

    - Reviewing , - we see that when t=3/5=0.6, - the graph of the parametric equations has a vertical tangent line. - This point is also a point of inflection for the graph, - illustrated in . -

    - - - -     - -     - - - Concavity of Plane Curves - -

    - Find the points of inflection of the graph of the parametric equations x=\sqrt{t}, - y=\sin(t), for 0\leq t\leq 16. -

    -     - -

    - We need to compute \frac{dy}{dx} and \frac{d^2y}{dx^2}. - - \frac{dy}{dx} = \frac{\yp(t)}{x'(t)} = \frac{\cos(t) }{1/(2\sqrt{t})} = 2\sqrt{t}\cos(t) - . - - \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{x'(t)} = \frac{\cos(t) /\sqrt{t}-2\sqrt{t}\sin(t) }{1/(2\sqrt{t})}=2\cos(t) -4t\sin(t) - . -

    - -

    - The points of inflection are found by setting \frac{d^2y}{dx^2}=0. - This is not trivial, - as equations that mix polynomials and trigonometric functions generally do not have nice solutions. -

    - -

    - In we see a plot of the second derivative. - It shows that it has zeros at approximately t=0.5,\,3.5,\,6.5,\,9.5,\,12.5 and 16. - These approximations are not very good, - made only by looking at the graph. - Newton's Method provides more accurate approximations. - Accurate to 2 decimal places, we have: - - t=0.65,\,3.29,\,6.36,\,9.48,\,12.61\,\text{ and } \,15.74 - . -

    - -

    - The corresponding points have been plotted on the graph of the parametric equations in . - Note how most occur near the x-axis, - but not exactly on the axis. -

    - -
    - In (a), a graph of \frac{d^2y}{dx^2}, showing where it is approximately 0. In (b), graph of the parametric equations in along with the points of inflection - -
    - - - - Graph of the second derivative is a sinusoid with increasing amplitude. - -

    - The image shows the graph y = 2\cos(t)-4t\sin(t), which is a graph of \frac{d^2y}{dx^2}. - The curve is sinusoidal, but with an amplitude that increases as t increases. - The graph allows us to estimate the values of t where the second derivative is zero. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=4, - minor y tick num=4, - ymin=-59,ymax=59, - xmin=-.5,xmax=16.9, - xlabel={$t$} - ] - - \addplot+ [domain=0:16,samples=120] ({x},{2*cos(deg(x))-4*x*sin(deg(x))}); - - \draw (axis cs: 5,-40) node { $y=2\cos(t) -4t\sin(t) $}; - - \end{axis} - \end{tikzpicture} - - - - -
    - -
    - - - - Graph of the parametric curve in this example, with points of inflection marked. - -

    - The graph shows a curve that appears to be sinusoidal, but with a frequency that increases with x. - On the graph are several marked points, indicating where the inflection points occur. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={3,1,2,4}, - ymin=-1.1,ymax=1.1, - xmin=-.5,xmax=4.5 - ] - - \addplot+ [domain=0:.5,samples=40] ({sqrt(x)},{sin(deg(x))}); - \addplot [firstcurvestyle,domain=.5:3,samples=50] ({sqrt(x)},{sin(deg(x))}); - \addplot [firstcurvestyle,domain=3:18,samples=120] ({sqrt(x)},{sin(deg(x))}); - - \filldraw (axis cs: 0.808252, 0.607787) circle (2.4pt) - (axis cs: 1.81447, -0.150147) circle (2.4pt) - (axis cs: 2.52223, 0.0783547) circle (2.4pt) - (axis cs: 3.07855, -0.0526833) circle (2.4pt) - (axis cs: 3.55049, 0.0396324) circle (2.4pt) - (axis cs:3.96733, -0.0317508) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    -     -     - - -     - - - Arc Length -

    - We continue our study of the features of the graphs of parametric equations by computing their arc length. -

    - - -

    - Recall in - we found the arc length of the graph of a function, - from x=a to x=b, to be - - L = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx - . -

    - -

    - We can use this equation and convert it to the parametric equation context. - Letting x=f(t) and y=g(t), - we know that \frac{dy}{dx} = \gp(t)/\fp(t). - It will also be useful to calculate the differential of x: - - dx = \fp(t)dt \qquad \Rightarrow \qquad dt = \frac{1}{\fp(t)}\cdot dx - . -

    - -

    - Starting with the arc length formula above, consider: - - L \amp = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx - \amp = \int_a^b \sqrt{1+\frac{\gp(t)^2}{\fp(t)^2}}\, dx. - Factor out the \fp(t)^2: - \amp = \int_a^b \sqrt{\fp(t)^2+\gp(t)^2}\cdot\underbrace{\frac1{\fp(t)}\, dx}_{=dt} - \amp = \int_{t_1}^{t_2} \sqrt{\fp(t)^2+\gp(t)^2}\, dt - . -

    - -

    - Note the new bounds - (no longer x bounds, - but t bounds). - They are found by finding t_1 and t_2 such that a= f(t_1) and b=f(t_2). - This formula is important, so we restate it as a theorem. -

    - - - Arc Length of Parametric Curves - -

    - Let x=f(t) and y=g(t) be parametric equations with \fp and \gp continuous on [t_1,t_2], - on which the graph traces itself only once. - The arc length of the graph, from t=t_1 to t=t_2, - is - parametric equationsarc length - arc length - - L = \int_{t_1}^{t_2} \sqrt{\fp(t)^2+\gp(t)^2}\, dt - . -

    -     -     - - -

    - As before, these integrals are often not easy to compute. - We start with a simple example, - then give another where we approximate the solution. -

    - - - Arc Length of a Circle - -

    - Find the arc length of the circle parametrized by x=3\cos(t), - y=3\sin(t) on [0,3\pi/2]. -

    -     - -

    - By direct application of , we have - - L \amp = \int_0^{3\pi/2} \sqrt{(-3\sin(t) )^2 +(3\cos(t) )^2} \, dt. - Apply the Pythagorean identity. - \amp = \int_0^{3\pi/2} 3 \, dt - \amp = 3t\Big|_0^{3\pi/2} = 9\pi/2 - . -

    - -

    - This should make sense; - we know from geometry that the circumference of a circle with radius 3 is 6\pi; - since we are finding the arc length of 3/4 of a circle, - the arc length is 3/4\cdot 6\pi = 9\pi/2. -

    -     - -     - - - Arc Length of a Parametric Curve - -

    - The graph of the parametric equations x=t(t^2-1), - y=t^2-1 crosses itself as shown in , - forming a teardrop. Find the arc length of the teardrop. -

    -
    - A graph of the parametric equations in , where the arc length of the teardrop is calculated - - - Graph of a "teardrop" curve that intersects itself at the origin. - -

    - The curve given by x=t(t^2-1), y=t^2-1 forms a teardrop shape. - The curve crosses itself at the origin, and the teardrop portion of the graph lies below the x axis. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,-1}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.1 - ] - - \addplot+ [domain=-1.5:1.5,samples=60] ({x^3-x},{x^2-1}); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - We can see by the parametrizations of x and y that when t=\pm 1, - x=0 and y=0. - This means we'll integrate from t=-1 to t=1. - Applying , we have - - L \amp = \int_{-1}^1\sqrt{(3t^2-1)^2+(2t)^2}\, dt - \amp = \int_{-1}^1 \sqrt{9t^4-2t^2+1} \, dt - . -

    - -

    - Unfortunately, - the integrand does not have an antiderivative expressible by elementary functions. - We turn to numerical integration to approximate its value. - Using 4 subintervals, Simpson's Rule approximates the value of the integral as 2.65051. - Using a computer, more subintervals are easy to employ, - and n=20 gives a value of 2.71559. - Increasing n shows that this value is stable and a good approximation of the actual value. -

    -     - -     - - - - -     - - - Surface Area of a Solid of Revolution -

    - Related to the formula for finding arc length is the formula for finding surface area. - We can adapt the formula found in - from - in a similar way as done to produce the formula for arc length done before. -

    - - - Surface Area of a Solid of Revolution - -

    - Consider the graph of the parametric equations x=f(t) and y=g(t), - where \fp and \gp are continuous on an open interval I containing t_1 and t_2 on which the graph does not cross itself. - surface areasolid of revolution - integrationsurface area - parametric equationssurface area -

    - -

    -

      -
    1. -

      - The surface area of the solid formed by revolving the graph about the x-axis is (where - g(t)\geq 0 on [t_1,t_2]): - - \text{ Surface Area } = 2\pi\int_{t_1}^{t_2} g(t)\sqrt{\fp(t)^2+\gp(t)^2}\, dt - . -

      -
      2. - -
    2. -

      - The surface area of the solid formed by revolving the graph about the y-axis is (where - f(t)\geq 0 on [t_1,t_2]): - - \text{ Surface Area } = 2\pi\int_{t_1}^{t_2} f(t)\sqrt{\fp(t)^2+\gp(t)^2}\, dt - . -

      -
      3. -
    -

    -     -     - - - Surface Area of a Solid of Revolution - -

    - Consider the teardrop shape formed by the parametric equations x=t(t^2-1), - y=t^2-1 as seen in . - Find the surface area if this shape is rotated about the x-axis, - as shown in . -

    -
    - Rotating a teardrop shape about the x-axis in - - - - The surface obtained by revolving the teardrop shape from the previous example about the x axis. - -

    - The image, which is interactive, shows the surface in three dimensions that is obtained when the teardrop curve - from is revolved about the x axis. - The result is a donut-like shape, but with a pinch-point in the center rather than a hole. -

    -     - - - - - //ASY file for figparcalc8_3D.asy in Chapter 9 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,3,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.,1.); - pair zbounds=(-1.25,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$x$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$y$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //surface t(t^2-1), t^2-1 revolved around x axis //({(x^2-1)*cos(y)},{x^3-x},{(x^2-1)*sin(y)}); - triple f(pair t) { - return ((t.x^2-1)*cos(t.y),(t.x^3-t.x),(t.x^2-1)*sin(t.y)); - } - surface s=surface(f,(-1,0),(1,2*pi),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //curve on the surfaces - triple g(real t) {return (0,t^3-t,-1+t^2);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - - - -
    -     - -

    - The teardrop shape is formed between t=-1 and t=1. - Using , - we see we need for g(t)\geq 0 on [-1,1], - and this is not the case. - To fix this, we simplify replace g(t) with -g(t), - which flips the whole graph about the x-axis - (and does not change the surface area of the resulting solid). - The surface area is: - - \text{ Area } \,S \amp = 2\pi\int_{-1}^1 (1-t^2)\sqrt{(3t^2-1)^2+(2t)^2}\, dt - \amp = 2\pi\int_{-1}^1 (1-t^2)\sqrt{9t^4-2t^2+1} \, dt - . -

    - -

    - Once again we arrive at an integral that we cannot compute in terms of elementary functions. - Using Simpson's Rule with n=20, - we find the area to be S=9.44. - Using larger values of n shows this is accurate to 2 places after the decimal. -

    -     -     - -

    - After defining a new way of creating curves in the plane, - in this section we have applied calculus techniques to the parametric equation defining these curves to study their properties. - In the next section, - we define another way of forming curves in the plane. - To do so, we create a new coordinate system, - called polar coordinates, - that identifies points in the plane in a manner different than from measuring distances from the y- and x- axes. -

    -     - - - - Terms and Concepts - - - -

    - Given parametric equations x=f(t) and y=g(t), - \lz{y}{x} = \fp(t)/\gp(t), - as long as \gp(t) \neq 0. - -

    -     - -     - - - - -

    - Given parametric equations x=f(t) and y=g(t), - the derivative \frac{dy}{dx} as given in is a function of what variable? -

    -     - - - -

    - x -

    -     -     - - -

    - y -

    -     -     - - -

    - t -

    -     -     -     - -     - - - - -

    - Given parametric equations x=f(t) and y=g(t), - to find \lzn{2}{y}{x}, - one simply computes \lzoo{t}{\lz{y}{x}}. - -

    -     - -     - - - - -

    - If \lz{y}{x}=0 at t=t_0, - then the normal line to the curve at t=t_0 is a vertical line. - -

    -     - -     -     - - - Problems - - - -

    - Parametric equations for a curve are given. -

    - -

    -

      -
    1. -

      - Find \ds\frac{dy}{dx}. -

      -
      2. - -
    2. -

      - Find the equations of the tangent and normal line(s) at the point(s) given. -

      -
      3. - -
    3. -

      - Sketch the graph of the parametric functions along with the found tangent and normal lines. -

      -
      4. -
    -

    -     - - - - -

    - x=t, y=t^2;t=1 -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = 2t -

      -
      2. - -
    2. -

      - Tangent line: y= 2(x-1)+1; normal line: - y = -1/2(x-1)+1 -

      -
      3. -
    -

    -     - -     - - - - -

    - x=\sqrt{t}, y=5t+2;t=4 -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = 10\sqrt{t} -

      -
      2. - -
    2. -

      - Tangent line: y= 20(x-2)+22; normal line: - y = -1/20(x-2)+22 -

      -
      3. -
    -

    -     - -     - - - - -

    - x=t^2-t, y=t^2+t;t=1 -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = \frac{2 t+1}{2 t-1} -

      -
      2. - -
    2. -

      - Tangent line: y= 3x+2; - normal line: y = -1/3x+2 -

      -
      3. -
    -

    -     - -     - - - - -

    - x=t^2-1, y=t^3-t;t=0 and t=1 -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = \frac{3t^2}{2 t} -

      -
      2. - -
    2. -

      - t=0: Tangent line: x=-1; - normal line: y = 0 - - t=1: Tangent line: y=x; normal line: y=-x -

      -
      3. -
    -

    -     - -     - - - - -

    - x=\sec(t), - y=\tan(t) on (-\pi/2,\pi/2);t=\pi/4 -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = \csc(t) -

      -
      2. - -
    2. -

      - t=\pi/4: Tangent line: - y=\sqrt{2}(x-\sqrt{2})+1; normal line: - y = -1/\sqrt{2}(x-\sqrt{2})+1 -

      -
      3. -
    -

    -     - -     - - - - -

    - x=\cos(t), - y=\sin(2t) on [0,2\pi];t=\pi/4 -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = -2\cos(2t)\csc(t) -

      -
      2. - -
    2. -

      - t=\pi/4: Tangent line: - y=1; normal line: x=\sqrt{2}/2 -

      -
      3. -
    -

    -     - -     - - - - -

    - x=\cos(t) \sin(2t), - y=\sin(t) \sin(2t) on [0,2\pi]; t=3\pi/4 -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = \frac{\cos(t) \sin(2t)+2\sin(t) \cos(2t)}{-\sin(t) \sin(2t)+2\cos(t) \cos(2t)} -

      -
      2. - -
    2. -

      - Tangent line: y=x-\sqrt{2}; - normal line: y=-x -

      -
      3. -
    -

    -     - -     - - - - -

    - x=e^{t/10}\cos(t), y=e^{t/10}\sin(t); t=\pi/2 -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = \frac{\sin(t)+10 \cos(t)}{\cos(t)-10 \sin(t)} -

      -
      2. - -
    2. -

      - Tangent line: y=-x/10+e^{\pi/20}; normal line: - y=10x+e^{\pi/20} -

      -
      3. -
    -

    -     - -     - -     - - - -

    - Find the t-values where the curve defined by the given parametric equations has a horizontal tangent line. - Note: these are the same equations as in Exercises. -

    -     - - - - -

    - x=t, y=t^2 -

    -     - -

    - t=0 -

    -     - -     - - - - -

    - x=\sqrt{t}, y=5t+2 -

    -     - -

    - t=0 (though this uses a one-sided limit, - as x(t) is not defined for t\lt 0. -

    -     - -     - - - - - - $hort=List("-1/2"); - Context("Point"); - $points=List("(3/4,-1/4)"); - - -

    - x=t^2-t, - y=t^2+t -

    - - - Give the t values where the curve has a horizontal tangent line. - (If there is more than one, use commas to separate them.) - - -

    - -

    - - - Give the corresponding point(s) in the plane where the tangent line is horizontal. - (If there is more than one, use commas to separate them.) - - -

    - -

    -     -     -     - - - - -

    - x=t^2-1, y=t^3-t -

    -     - -

    - t=\pm 1/\sqrt{3} -

    -     - -     - - - - -

    - x=\sec(t), y=\tan(t) on (-\pi/2,\pi/2) -

    -     - -

    - The graph does not have a horizontal tangent line. -

    -     - -     - - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $hort=List(Formula("pi/4"),Formula("3pi/4"),Formula("5pi/4"),Formula("7pi/4")); - Context("Point"); - $points=Compute("(sqrt(2)/2,1), (-sqrt(2)/2,-1), (-sqrt(2)/2,1), (sqrt(2)/2,-1)"); - - -

    - x=\cos(t), y=\sin(2t), on [0,2\pi) -

    - - - Give the t values where the curve has a horizontal tangent line. - (If there is more than one, use commas to separate them.) - - -

    - -

    - - - Give the corresponding point(s) in the plane where the tangent line is horizontal. - (If there is more than one, use commas to separate them.) - - -

    - -

    -     -     -     - - - - -

    - x=\cos(t) \sin(2t), y=\sin(t) \sin(2t) on [0,2\pi] -

    -     - -

    - The solution is non-trivial; - use identities \sin(2t)=2\sin(t) \cos(t) and - \cos(2t)=\cos^2(t) -\sin^2(t) to rewrite \gp(t) = 2\sin(t) (2\cos^2(t) -\sin^2(t) ). - On [0,2\pi], \sin(t) = 0 when t=0,\pi,2\pi, - and 2\cos^2(t) -\sin^2(t) =0 when t=\tan^{-1}(\sqrt{2}),\,\pi \pm \tan^{-1}(\sqrt{2}),\,2\pi-\tan^{-1}(\sqrt{2}). -

    -     - -     - - - - -

    - x=e^{t/10}\cos(t), y=e^{t/10}\sin(t) -

    -     - -

    - t=\tan^{-1}(-10),\,\tan^{-1}(-10)+\pi -

    -     - -     - -     - - - -

    - Find the point t=t_0 where the graph of the given parametric equations is not smooth, - then find \lim\limits_{t\to t_0}\frac{dy}{dx}. -

    -     - - - - - - $t0 = Real("0"); - $L = Real("0"); - - -

    - x=\frac{1}{t^2+1}, - y=t^3 -

    - - - Find the point t_0 where the curve is not smooth. - - -

    - -

    - - - Find \lim\limits_{t\to t_0}\lz{y}{x}. - -

    - -

    -     -     -     - - - - - - $t0 = Real("2"); - $L = Real("1"); - - -

    - x=-t^3+7t^2-16t+13, - y=t^3-5t^2+8t-2 -

    - - - Find the point t_0 where the curve is not smooth. - - -

    - -

    - - - Find \lim\limits_{t\to t_0}\lz{y}{x}. - -

    - -

    -     -     -     - - - - -

    - x=t^3-3t^2+3t-1,y=t^2-2t+1 -

    -     - -

    - t_0=1; \lim_{t\to 1} \frac{dy}{dx} = \infty. -

    -     - -     - - - - -

    - \ds x=\cos^2(t),y=1-\sin^2(t) -

    -     - -

    - t_0=\ldots,-\pi/2,0,\pi/2,\pi,\ldots; - \lim_{t\to 0} \frac{dy}{dx} = 1. -

    -     - -     - -     - - - -

    - For the given parametric equations for a curve, - find \frac{d^2y}{dx^2}, - then determine the intervals on which the graph of the curve is concave up/down. - Note: these are the same equations as in Exercises. -

    -     - - - - -

    - x=t,y=t^2 -

    -     - -

    - \frac{d^2y}{dx^2}=2; always concave up -

    -     - -     - - - - -

    - x=\sqrt{t},y=5t+2 -

    -     - -

    - \frac{d^2y}{dx^2}=10; always concave up -

    -     - -     - - - - - - - Context()->variables->are(t=>'Real'); - $ypp=Formula("-(4/(2t-1)^3)"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up=List("(-inf,1/2]"); - $down=List("[1/2,inf)"); - - -

    - x=t^2-t, - y=t^2+t -

    - - - Find \lzn{2}{y}{x} as a function of t. - - -

    - -

    - - - Give the interval(s) where the curve is concave up. - (If there is more than one interval, - use commas to separate them.) - -

    - -

    - - - Give the interval(s) where the curve is concave down. - (If there is more than one interval, - use commas to separate them.) - - -

    - -

    -     -     -     - - - - -

    - x=t^2-1,y=t^3-t -

    -     - -

    - \frac{d^2y}{dx^2}=\frac{3t^2+1}{4t^3}; - concave down on (-\infty,0); - concave up on (0,\infty). -

    -     - -     - - - - -

    - x=\sec(t),y=\tan(t) on (-\pi/2,\pi/2) -

    -     - -

    - \frac{d^2y}{dx^2}=-\cot^3(t); - concave up on (-\pi/2,0); - concave down on (0,\pi/2). -

    -     - -     - - - - - - - Context()->variables->are(t=>'Real'); - $ypp=Formula("2 (sin(t)*-2sin(2t) - cos(2t)*cos(t)) / sin^3(t)"); - Context()->flags->set(ignoreEndpointTypes => 1); - $up=Compute("[pi/2,pi], [3pi/2,2pi]"); - $down=Compute("[0,pi/2], [pi,3pi/2]"); - - -

    - x=\cos(t), - y=\sin(2t), on - [0,2\pi) -

    - - - Find \lzn{2}{y}{x} as a function of t. - - -

    - -

    - - - Give the interval(s) where the curve is concave up. - (If there is more than one interval, - use commas to separate them.) - -

    - -

    - - - Give the interval(s) where the curve is concave down. - (If there is more than one interval, - use commas to separate them.) - - -

    - -

    -     -     -     - - - - -

    - \ds x=\cos(t) \sin(2t),y=\sin(t) \sin(2t) on [-\pi/2,\pi/2] -

    -     - -

    - \frac{d^2y}{dx^2} = \frac{4(13+3\cos(4t))}{(\cos(t) +3\cos(3t))^3}, - obtained with a computer algebra system; - concave up on \big(-\tan^{-1}(\sqrt{2}/2),\tan^{-1}(\sqrt{2}/2)\big), - concave down on \big(-\pi/2,-\tan^{-1}(\sqrt{2}/2)\big)\cup\big(\tan^{-1}(\sqrt{2}/2),\pi/2\big) -

    -     - -     - - - - -

    - x=e^{t/10}\cos(t),y=e^{t/10}\sin(t) -

    -     - -

    - \frac{d^2y}{dx^2}=\frac{1010}{e^{t/10}\big(\cos(t) -10\sin(t) \big)^3}; - concavity switches at t=\tan^{-1}(1/10)+n\pi, - where n is an integer. -

    -     - -     - -     - - - -

    - Find the arc length of the graph of the parametric equations on the given interval(s). -

    -     - - - - - Context()->flags->set(reduceConstants=>0); - $length=Formula("6pi"); - - -

    - x=-3\sin(2t), - y=3\cos(2t) on [0,\pi] -

    - -

    - -

    -     -     -     - - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $length1=Formula("sqrt(101)(e^(pi/5)-1)"); - $length2=Formula("sqrt(101)(e^(2pi/5)-e^(pi/5))"); - - -

    - x=e^{t/10}\cos(t), - y=e^{t/10}\sin(t) on - [0,2\pi] and [2\pi,4\pi]. -

    - - - Give the arc length on [0,2\pi]: - - -

    - -

    - - - Give the arc length on [2\pi,4\pi]: - - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $length=Formula("2sqrt(34)"); - - -

    - x=5t+2, - y=1-3t on [-1,1] -

    - -

    - -

    -     -     -     - - - - -

    - x=2t^{3/2},y=3t on [0,1] -

    -     - -

    - L=4\sqrt{2}-2 -

    -     - -     - -     - - - -

    - Numerically approximate the given arc length. -

    -     - - - - -

    - Approximate the arc length of one petal of the rose curve - x=\cos(t) \cos(2t),y=\sin(t) \cos(2t) using Simpson's Rule and n=4. -

    -     - -

    - L\approx 2.4416 (actual value: L=2.42211) -

    -     - -     - - - - -

    - Approximate the arc length of the bow tie curve - x=\cos(t),y=\sin(2t) using Simpson's Rule and n=6. -

    -     - -

    - L\approx 9.73004 (actual value: L=9.42943) -

    -     - -     - - - - -

    - Approximate the arc length of the parabola x=t^2-t,y=t^2+t on [-1,1] using Simpson's Rule and n=4. -

    -     - -

    - L\approx 4.19216 (actual value: L=4.18308) -

    -     - -     - - - - -

    - A common approximate of the circumference of an ellipse given by - x=a\cos(t),y=b\sin(t) is \ds C\approx 2\pi\sqrt{\frac{a^2+b^2}2}. - Use this formula to approximate the circumference of x=5\cos(t), - y=3\sin(t) and compare this to the approximation given by Simpson's Rule and n=6. -

    -     - -

    - Formula: C\approx 25.9062; Simpson's Rule: - C\approx 25.4786 (actual value: C=25.527) -

    -     - -     - -     - - - -

    - A solid of revolution is described. - Find or approximate its surface area as specified. -

    -     - - - - -

    - Find the surface area of the sphere formed by rotating the circle - x=2\cos(t),y=2\sin(t) about: -

    -     - - - -

    - The x-axis. -

    -     -     - - - -

    - The y-axis. -

    -     -     - -

    - The answer is 16\pi for both - (of course), - but the integrals are different. -

    -     - -     - - - - -

    - Find the surface area of the torus - (or donut) - formed by rotating the circle - x=\cos(t) +2,y=\sin(t) about the y-axis. -

    -     - -

    - 8\pi^2. -

    -     - -     - - - - -

    - Approximate the surface area of the solid formed by rotating the - upper right half - of the bow tie curve x=\cos(t),y=\sin(2t) on - [0,\pi/2] about the x-axis, - using Simpson's Rule and n=4. -

    -     - -

    - SA\approx 8.50101 (actual value SA = 8.02851 -

    -     - -     - - - - -

    - Approximate the surface area of the solid formed by rotating the one petal of the rose curve - x=\cos(t) \cos(2t),y=\sin(t) \cos(2t) on - [0,\pi/4] about the x-axis, - using Simpson's Rule and n=4. -

    -     - -

    - SA\approx 1.36751 (actual value SA = 1.36707) -

    -     - -     - -     -     -     -
    -
    - Introduction to Polar Coordinates - - - -

    - We are generally introduced to the idea of graphing curves by relating x-values to y-values through a function f. - That is, we set y=f(x), - and plot lots of point pairs (x,y) to get a good notion of how the curve looks. - This method is useful but has limitations, - not least of which is that curves that - fail the vertical line test - cannot be graphed without using multiple functions. -

    - -

    - The previous two sections introduced and studied a new way of plotting points in the x,y-plane. - Using parametric equations, - x and y values are computed independently and then plotted together. - This method allows us to graph an extraordinary range of curves. - This section introduces yet another way to plot points in the plane: - using polar coordinates. -

    -     - - - Polar Coordinates -

    - Start with a point O in the plane called the pole - (we will always identify this point with the origin). - From the pole, draw a ray, called the initial ray - (we will always draw this ray horizontally, - identifying it with the positive x-axis). - A point P in the plane is determined by the distance r that P is from O, - and the angle \theta formed between the initial ray and the segment \overline{OP} - (measured counter-clockwise). - We record the distance and angle as an ordered pair (r,\theta). - To avoid confusion with rectangular coordinates, - we will denote polar coordinates with the letter P, - as in P(r,\theta). - This is illustrated in - polar coordinates - polarcoordinates - coordinatespolar -

    - -
    - Illustrating polar coordinates - - Illustration of polar coordinates relative to a pole and initial ray. - -

    - A description of polar coordinates. - An initial point O is shown, and from O, - a ray pointing to the right, called the initial ray. -

    - -

    - An additional line segment is drawn from O in a direction up and to the right. - An angle is measured from the initial ray to the line segment, and labeled as \theta. - The length of the line segment is labeled r. -

    - -

    - At the other end of the line segment, a point P is labeled, and assigned the coordinates (r,\theta). -

    -     - - - \begin{tikzpicture}[scale=1.24] - - \draw[thick,->] (0,0) node [below] {\(O\)} -- (3,0) node [below] {initial ray}; - - \filldraw (0,0) circle (2.4pt); - \filldraw [rotate=55] (2,0) circle (2.4pt); - - \draw [thick,rotate=55] (0,0)-- node [rotate=55,pos=.5,above] {\(r\)} (2,0) node [above] {\(P=P(r,\theta)\)}; - \draw [->] (.75,0) arc(0:55:.75); - \draw [rotate=27.5] (1,0) node {\(\theta\)}; - - \end{tikzpicture} - - - - -
    - -

    - Practice will make this process more clear. -

    - - - Plotting Polar Coordinates - -

    - Plot the following polar coordinates: - polar coordinatesplotting points - - A = P(1,\pi/4),\, B=P(1.5,\pi),\, C = P(2,-\pi/3),\, D = P(-1,\pi/4) - -

    -     - -

    - To aid in the drawing, - a polar grid is provided below. - To place the point A, go out 1 unit along the initial ray - (putting you on the inner circle shown on the grid), - then rotate counter-clockwise \pi/4 radians - (or 45^\circ). - Alternately, one can consider the rotation first: - think about the ray from O that forms an angle of \pi/4 with the initial ray, - then move out 1 unit along this ray - (again placing you on the inner circle of the grid). -

    - - - Drawing of a polar grid: a set of concentric circles, and rays from their common center. - -

    - An illustration of the polar grid system. - It is intended as a polar adaptation of the Cartesian grid system. - Instead of x and y coordinate axes, - we see an origin O, and an initial ray, pointing horizontally to the right. - On the initial ray, points 1, 2, and 3 are marked. -

    - -

    - Through each of these points passes a circle centered at O, - representing the polar coordinate values r=1, r=2, and r=3. - Also drawn are several dashed lines through the origin. - Viewed as rays from the origin, each of these lines represents a fixed value for the polar coordinate \theta. -

    -     - - - \begin{tikzpicture}[scale=0.75] - - \draw [dashed,gray] (-3.1,0) -- (0,0); - \draw [thick,->,>=stealth] (0,0) node [below] {\(O\)} -- (3.5,0); - - \filldraw (0,0) circle (2.4pt); - - \foreach \x in {1,2,3} - { - \draw (0,0) circle (\x cm); - \draw (\x,0) node [below right] {\x}; - } - - \foreach \x in {30,45,60,90,120,135,150} - { - \draw [rotate=\x,dashed,gray] (-3.1,0) -- (3.1,0); - } - - \end{tikzpicture} - - - - -

    - To plot B, - go out 1.5 units along the initial ray and rotate \pi radians - (180^\circ). -

    - -

    - To plot C, go out 2 units along the initial ray then rotate - clockwise \pi/3 radians, - as the angle given is negative. -

    - -
    - Plotting polar points in - - - A plot of several points on a polar grid. - -

    - The polar grid from the previous image is shown, with the dashed lines removed. - On the circle of radius 1, points A and D are marked. - The segment OA makes an angle of \pi/4 with the initial ray, - while the point D is on the opposite end of the diameter through A. -

    - -

    - The point B lies between the circles of radius 1 and 2, - to the left of the point O. - The point C is on the circle of radius 2 - and lies below the initial ray, so that the segment OC - makes an angle of -\pi/3 with the initial ray. -

    -     - - - \begin{tikzpicture}[scale=.75,>=latex] - \draw[thick,->] (0,0) node [below] {$O$} -- (3.5,0) ; - \filldraw (0,0) circle (2.4pt); - \foreach \x in {1,2,3} - {\draw (0,0) circle (\x cm); - \draw (\x,0) node [below right] {\x}; - } - \filldraw [rotate=45] (1,0) circle (2.4pt) node [above right] {$A$}; - \filldraw [rotate=180] (1.5,0) circle (2.4pt) node [above] {$B$}; - \filldraw [rotate=-60] (2,0) circle (2.4pt) node [below right] {$C$}; - \filldraw [rotate=45] (-1,0) circle (2.4pt) node [below] {$D$}; - - \end{tikzpicture} - - - - -
    - -

    - To plot D, move along the initial ray -1 - units in other words, - back up 1 unit, then rotate counter-clockwise by \pi/4. - The results are given in . -

    -     -     - - - -

    - Consider the following two points: - A = P(1,\pi) and B = P(-1,0). - To locate A, - go out 1 unit on the initial ray then rotate \pi radians; - to locate B, - go out -1 units on the initial ray and don't rotate. - One should see that A and B are located at the same point in the plane. - We can also consider C=P(1,3\pi), or D = P(1,-\pi); - all four of these points share the same location. -

    - -

    - This ability to identify a point in the plane with multiple polar coordinates is both a - blessing and a curse. - We will see that it is beneficial as we can plot beautiful functions that intersect themselves - (much like we saw with parametric functions). - The unfortunate part of this is that it can be difficult to determine when this happens. - We'll explore this more later in this section. -

    -     - - - Polar to Rectangular Conversion -

    - It is useful to recognize both the rectangular (or, Cartesian) coordinates of a point in the plane and its polar coordinates. - - shows a point P in the plane with rectangular coordinates (x,y) and polar coordinates P(r,\theta). - Using trigonometry, - we can make the identities given in the following Key Idea. -

    - -
    - Converting between rectangular and polar coordinates - - - A triangle illustrating the conversion between rectangular and polar coordinates. - -

    - An right-angled triangle is shown, with the right angle in the bottom-right corner. - The vertex on the left is labeled O, and the angle at that vertex is labeled \theta. - The label on the hypotenuse indicates a length of r, - and the opposite end of the hypotenuse is labeled P. -

    - -

    - The bottom of the triangle (the side adjacent to the angle \theta) is labeled with the length x, - and the altitude of the triangle (the side oppsite the angle \theta) is labeled with the length y. -

    -     - - - \begin{tikzpicture}[scale=0.93] - - \draw [thick] (0,0) -- node [below,pos=.5] {$x$} (4,0) -- node [right,pos=.5] {$y$} (4,2) -- node [above,rotate=26.57,pos=.5] {$r$} (0,0); - - \draw [->,thick] (1.5,0) arc (0:26:1.5); - \draw [rotate=13] (1.8,0) node {$\theta$}; - - \filldraw (0,0) circle (2.4pt) node [below] {$O$} - (4,2) circle (2.4pt) node [above] {$P$}; - - \end{tikzpicture} - - - - -
    - - - Converting Between Rectangular and Polar Coordinates -

    - Given the polar point P(r,\theta), - the rectangular coordinates are determined by - - x=r\cos(\theta) \qquad y=r\sin(\theta) - . -

    - -

    - Given the rectangular coordinates (x,y), - the polar coordinates are determined by - - r^2=x^2+y^2\qquad \tan(\theta) = \frac yx - . -

    -     - - - Converting Between Polar and Rectangular Coordinates - -

    -

      -
    1. -

      - Convert the polar coordinates P(2,2\pi/3) and - P(-1,5\pi/4) to rectangular coordinates. -

      -
      2. - -
    2. -

      - Convert the rectangular coordinates (1,2) and (-1,1) to polar coordinates. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      -

        -
      1. -

        - We start with P(2,2\pi/3). - Using , we have - - x= 2\cos(2\pi/3) = -1\qquad y = 2\sin(2\pi/3) = \sqrt{3} - . - So the rectangular coordinates are (-1,\sqrt{3}) \approx (-1,1.732). -

        -
        2. - -
      2. -

        - The polar point P(-1,5\pi/4) is converted to rectangular with: - - x=-1\cos(5\pi/4) = \sqrt{2}/2\qquad y= -1\sin(5\pi/4) = \sqrt{2}/2 - . - So the rectangular coordinates are (\sqrt{2}/2,\sqrt{2}/2) \approx (0.707,0.707). -

        -
        3. -
      -

      - -

      - These points are plotted in . - The rectangular coordinate system is drawn lightly under the polar coordinate system so that the relationship between the two can be seen. -

      - -
      - Plotting rectangular and polar points in - -
      - - - - Overlaid rectangular and polar grid systems, with marked points. - -

      - A polar grid is shown, using a relatively thick line width. - Also shown is a rectangular grid, with a much thinner line width, so that the polar grid is emphasized. -

      - -

      - The point P(-1,5\pi/4) is marked on the circle of radius 1; - it lies in the first quadrant of the rectangular grid, - since 5\pi/4 is in the third quadrant, but r=-1 is negative. -

      - -

      - The point P(2,2\pi/3) is marked on the circle of radius 2 in the second quadrant. -

      -       - - - \begin{tikzpicture} - - \draw [thin,gray] (-2.5,-2.5) grid (2.5,2.5); - \draw [thick,->] (0,0) node [below] {$O$} -- (2.5,0); - - \foreach \x in {1,2} - { - \draw [very thick] (0,0) circle (\x cm); - } - - \filldraw (0,0) circle (2.4pt); - \filldraw [rotate=120] (2,0) circle (2.4pt) node [ left] { $P(2,\frac{2\pi}{3})$}; - \filldraw [rotate=225] (-1,0) circle (2.4pt) node [shift={(0pt,12pt)}] {$P(-1,\frac{5\pi}{4})$}; - - \end{tikzpicture} - - - - -
      - -
      - - - - Overlaid rectangular and polar grid systems, with marked points. - -

      - Rectangular and polar grids are shown; - this time the rectangular grid is emphasized with a thicker line width. -

      - -

      - The points (0,0), (1,2), and (-1,1) are marked on the rectangular grid. - Also indicated are the angles each point makes with the initial ray. -

      -       - - - \begin{tikzpicture} - - \draw [very thick] (-2.5,-2.5) grid (2.5,2.5); - \draw [thin,gray,->] (0,0) node [below left,black] {$(0,0)$} -- (2.5,0); - - \foreach \x in {1,2} - { - \draw [thin,gray] (0,0) circle (\x cm); - } - - \filldraw (0,0) circle (2.4pt); - - \filldraw [] (1,2) circle (2.4pt) node [above left] { $(1,2)$}; - \filldraw [] (-1,1) circle (2.4pt) node [above right] {$(-1,1)$}; - - \draw [gray,dashed] (0,0) -- ($(0,0)!2.2!(-1,1)$); - \draw [gray,->] (.5,0) arc (0:135:.5); - \draw [rotate=35,gray ] (.65,0) node [] {$\frac{3\pi}{4}$}; - - \draw [gray,dashed] (0,0) -- ($(0,0)!1.2!(1,2)$); - \draw [gray,->] (.5,0) arc (0:135:.5); - \draw [gray,->] (-.5,0) arc (180:135:.5); - \draw [rotate=35,gray ] (.65,0) node [] {$\frac{3\pi}{4}$}; - \draw [rotate=-30,gray ] (-.65,0) node [shift={(-3pt,0pt)}] {$\frac{-\pi}{4}$}; - - \draw [gray,->] (1.5,0) arc (0:63.4:1.5); - \draw [rotate=28,gray ] (1.75,0) node [] {$1.11$}; - - \end{tikzpicture} - - - - -
      -       -
      - -
      2. - -
    2. -

      -

        -
      1. -

        - To convert the rectangular point (1,2) to polar coordinates, - we use the Key Idea to form the following two equations: - - 1^2+2^2 = r^2 \qquad \tan(\theta) = \frac{2}{1} - . - The first equation tells us that r=\sqrt{5}. - Using the inverse tangent function, we find - - \tan(\theta) = 2 \Rightarrow \theta = \tan^{-1}(2) \approx 1.11\approx 63.43^\circ - . - Thus polar coordinates of (1,2) are P(\sqrt{5},1.11). -

        -
        2. - -
      2. -

        - To convert (-1,1) to polar coordinates, we form the equations - - (-1)^2+1^2=r^2 \qquad \tan(\theta) = \frac{1}{-1} - . - Thus r=\sqrt{2}. - We need to be careful in computing \theta: - using the inverse tangent function, we have - - \tan(\theta) = -1 \Rightarrow \theta = \tan^{-1}(-1) = -\pi/4 = -45^\circ - . - This is not the angle we desire. - The range of \tan^{-1}(x) is (-\pi/2,\pi/2); - that is, it returns angles that lie in the - 1st and 4th quadrants. - To find locations in the 2nd and 3rd quadrants, - add \pi to the result of \tan^{-1}(x). - So \pi+(-\pi/4) puts the angle at 3\pi/4. - Thus the polar point is P(\sqrt{2},3\pi/4). - - An alternate method is to use the angle \theta given by arctangent, but change the sign of r. Thus we could also refer to (-1,1) as - P(-\sqrt{2},-\pi/4). -

        -
        3. -
      -

      - -

      - These points are plotted in . - The polar system is drawn lightly under the rectangular grid with rays to demonstrate the angles used. -

      -
      3. -
    -

    -     -     - - - - -     - - - Polar Functions and Polar Graphs -

    - Defining a new coordinate system allows us to create a new kind of function, - a polar function. Rectangular coordinates lent themselves well to creating functions that related x and y, - such as y=x^2. - Polar coordinates allow us to create functions that relate r and \theta. - Normally these functions look like r=f(\theta), - although we can create functions of the form \theta = f(r). - The following examples introduce us to this concept. - polarfunctions - polarfunctions!graphing -

    - - - Introduction to Graphing Polar Functions - -

    - Describe the graphs of the following polar functions. -

    - -

    -

      -
    1. -

      - r = 1.5 -

      -
      2. - -
    2. -

      - \theta = \pi/4 -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - The equation r=1.5 describes all points that are 1.5 units from the pole; - as the angle is not specified, - any \theta is allowable. - All points 1.5 units from the pole describes a circle of radius 1.5. - We can consider the rectangular equivalent of this equation; - using r^2=x^2+y^2, we see that 1.5^2=x^2+y^2, - which we recognize as the equation of a circle centered at (0,0) with radius 1.5. - This is sketched in . -

      -
      2. - -
    2. -

      - The equation \theta = \pi/4 describes all points such that the line through them and the pole make an angle of \pi/4 with the initial ray. - As the radius r is not specified, it can be any value - (even negative). - Thus \theta = \pi/4 describes the line through the pole that makes an angle of - \pi/4 = 45^\circ with the initial ray. - We can again consider the rectangular equivalent of this equation. - Combine \tan(\theta) =y/x and \theta =\pi/4: - - \tan(\pi) /4 = y/x \Rightarrow x\tan(\pi) /4 = y \Rightarrow y = x - . - This graph is also plotted in . -

      - -
      - Plotting standard polar plots - - - Polar plot showing curves of constant radius and constant angle - -

      - A polar grid is shown. On the grid, two curves are drawn. - The curve r=1.5 is a circle centered at the origin of radius 1.5. - The curve \theta = \pi/4 is a line through the origin that makes an angle of \pi/4 with the initial ray. -

      -       - - - \begin{tikzpicture} - - \draw [thick,->] (0,0) node [below] {$O$} -- (2.5,0); - - \foreach \x in {1,2} - { - \draw [thin] (0,0) circle (\x cm); - \draw (\x,0) node [shift={(3pt,-4pt)}] {\x}; - } - - \filldraw (0,0) circle (2.4pt); - - \draw [very thick] (0,0) circle (1.5); - \draw[rotate=90] (1.7,0) node { $r=1.5$}; - - \draw [thick] (-2,-2) -- node [pos=1,left] { $\theta = \frac{\pi}4$} (2,2); - - \end{tikzpicture} - - - - -
      -
      3. -
    -

    -     - -     - -

    - The basic rectangular equations of the form x=h and y=k create vertical and horizontal lines, - respectively; - the basic polar equations r= h and - \theta =\alpha create circles and lines through the pole, - respectively. - With this as a foundation, - we can create more complicated polar functions of the form r=f(\theta). - The input is an angle; the output is a length, - how far in the direction of the angle to go out. -

    - -

    - We sketch these functions much like we sketch rectangular and parametric functions: - we plot lots of points and connect the dots with curves. - We demonstrate this in the following example. -

    - - - Sketching Polar Functions - -

    - Sketch the polar function r=1+\cos(\theta) on [0,2\pi] by plotting points. -

    -     - -

    - A common question when sketching curves by plotting points is - Which points should I plot? - With rectangular equations, - we often choose easy values integers, - then add more if needed. - When plotting polar equations, - start with the common angles multiples of \pi/6 and \pi/4. - - gives a table of just a few values of \theta in [0,\pi]. -

    - -

    - Consider the point P(2,0) determined by the first line of the table. - The angle is 0 radians we do not rotate from the initial ray then we go out 2 units from the pole. - When \theta=\pi/6, - r = 1.866 (actually, it is 1+\sqrt{3}/2); - so rotate by \pi/6 radians and go out 1.866 units. -

    - -

    - The graph shown uses more points, - connected with straight lines. - (The points on the graph that correspond to points in the table are signified with larger dots.) - Such a sketch is likely good enough to give one an idea of what the graph looks like. -

    - -
    - Graphing a polar function in by plotting points - -
    - - - \thetar=1+\cos(\theta) - - - 02 - - - \pi/61.86603 - - - \pi/21 - - - 4\pi/30.5 - - - 7 \pi/41.70711 - - -
    - -
    - - - - A rough sketch of a polar function. - -

    - On a polar grid, the points from the table in are plotted. - Each point is joined to an adjacent point by a line segement to form a rough sketch of the polar curve r=1+\cos(\theta). - The curve appears to be roughly heart-shaped, with a cusp at the origin. -

    -     - - - \begin{tikzpicture}[scale=1.06] - - \draw [dashed,gray] (-2.1,0) -- (0,0); - \draw[thick,->,>=stealth] (0,0) node [below] {$O$} -- (2.5,0); - - \filldraw (0,0) circle (2.4pt); - - \foreach \x in {1,2} - { - \draw (0,0) circle (\x cm); - \draw (\x,0) node [shift={(3pt,-4pt)}] {\x}; - } - - \foreach \x in {30,45,60,90,120,135,150} - { - \draw [rotate=\x,dashed,gray] (-2.3,0) -- (2.3,0); - } - - \draw [thick,firstcolor] (2,0)--(1.616,0.933)--(1.207,1.207) -- (0.75,1.299) -- (0,1.) -- (-0.25,0.433) -- (-0.2071,0.2071) -- (-0.116,0.06699) -- (0,0) -- (-0.116,-0.06699) -- (-0.2071,-0.2071) -- (-0.25,-0.433) -- (0,-1) -- (0.75,-1.299) -- (1.207,-1.207) -- (1.616,-0.933) -- (2,0); - - \filldraw (2.,0) circle (2pt) - (1.616,0.933) circle (2pt) - (1.207,1.207) circle (1pt) - (0.75,1.299) circle (1pt) - (0,1.) circle (2pt) - (-0.25,0.433) circle (1pt) - (-0.2071,0.2071) circle (1pt) - (-0.116,0.06699) circle (1pt) - (0,0) circle (1pt) - (-0.116,-0.06699) circle (1pt) - (-0.2071,-0.2071) circle (1pt) - (-0.25,-0.433) circle (2pt) - (0,-1.) circle (1pt) - (0.75,-1.299) circle (1pt) - (1.207,-1.207) circle (2pt) - (1.616,-0.933) circle (1pt); - - \end{tikzpicture} - - - - -
    -     -
    - -     - -     - - -

    - Technology Note: Plotting functions in this way can be tedious, - just as it was with rectangular functions. - To obtain very accurate graphs, - technology is a great aid. - Most graphing calculators can plot polar functions; - in the menu, - set the plotting mode to something like polar or POL, - depending on one's calculator. - As with plotting parametric functions, - the viewing window no longer determines the x-values that are plotted, - so additional information needs to be provided. - Often with the window settings are the settings for the beginning and ending \theta values - (often called \theta_{\text{ min } } and \theta_{\text{ max } }) - as well as the \theta_{\text{ step } } that is, - how far apart the \theta values are spaced. - The smaller the \theta_{\text{ step } } value, - the more accurate the graph - (which also increases plotting time). - Using technology, we graphed the polar function - r=1+\cos(\theta) from - in . -

    - -
    - Using technology to graph a polar function - - - A computer-generated sketch of a polar function. - -

    - On a polar grid, the curve from is plotted, using technology. - The result is a smooth version of the plot in . - The curve appears to be roughly heart-shaped, with a cusp at the origin. -

    -     - - - \begin{tikzpicture}[scale=1.1] - - \draw [dashed,gray] (-2.1,0) -- (0,0); - \draw [thick,->,>=stealth] (0,0) node [below] {$O$} -- (2.5,0); - - \filldraw (0,0) circle (2.4pt); - - \foreach \x in {1,2} - { - \draw (0,0) circle (\x cm); - \draw (\x,0) node [shift={(3pt,-4pt)}] {\x}; - } - - \foreach \x in {30,45,60,90,120,135,150} - { - \draw [rotate=\x,dashed,gray] (-2.3,0) -- (2.3,0); - } - - \draw [firstcolor,thick,domain=0:360,samples=60] plot ({cos(\x)*(1+cos(\x))},{sin(\x)*(1+cos(\x))}); - - \end{tikzpicture} - - - - -
    - - - Sketching Polar Functions - -

    - Sketch the polar function r=\cos(2\theta) on [0,2\pi] by plotting points. -

    -     - -

    - We start by making a table of - \cos(2\theta) evaluated at common angles \theta, - as shown in . - These points are then plotted in . - This particular graph moves - around quite a bit and one can easily forget which points should be connected to each other. - To help us with this, - we numbered each point in the table and on the graph. -

    - -
    - Table of points for plotting a polar curve in - - Pt.\theta\cos(2\theta) - 101 - 2\pi/60.5 - 3\pi/40 - 4\pi/3-0.5 - 5\pi/2-1 - 62\pi/3-0.5 - 73\pi/40 - 85\pi/60.5 - 9\pi1 - 107\pi/60.5 - 115\pi/40 - 124\pi/3-0.5 - 133\pi/2-1 - 145\pi/3-0.5 - 157\pi/40 - 1611\pi/60.5 - 172\pi1 - -
    - -

    - Using more points - (and the aid of technology) - a smoother plot can be made as shown in . - This plot is an example of a rose curve. -

    - -
    - Polar plots from - -
    - - - - An initial rough sketch of a polar curve obtained by connecting points on the curve. - -

    - On a polar grid, the points from the table in are plotted. - The points are labeled consecutively from 1 to 17, using the values from the table. - These points are then joined, in order, using straight lines, in the matter of a - connect-the-dots drawing. -

    - -

    - The result is four diamond-shaped leaves. - Each diamond has one point at the origin. - The diamonds then point left, right, up, and down. -

    -     - - - \begin{tikzpicture}[scale=1.1] - - \draw [dashed,gray] (-2.1,0) -- (0,0); - \draw [thick,->,>=latex] (0,0) node [below] {} -- (2.5,0); - - \filldraw (0,0) circle (2.4pt); - - \foreach \x/\y in {1/{},2/{}} - { - \draw (0,0) circle (\x cm); - \draw (\x,0) node [shift={(3pt,-4pt)}] { $\y$}; - } - - \foreach \x in {30,45,60,90,120,135,150} - { - \draw [rotate=\x,dashed,gray] (-2.3,0) -- (2.3,0); - } - - \draw [thick,firstcolor] plot coordinates {(2.,0)(0.866,0.5)(0,0)(-0.5,-0.866)(0,-2.)(0.5,-0.866)(0,0)(-0.866,0.5)(-2.,0)(-0.866,-0.5)(0,0)(0.5,0.866)(0,2.)(-0.5,0.866)(0,0)(0.866,-0.5)(2,0)}; - - \foreach \x/\y/\z/\w in { - 2./0/1/{above right},0.866/0.5/2/above,0/0/3/above,-0.5/-0.866/4/above, - 0/-2./5/above, 0.5/-0.866/6/above,0/0/7/left,-0.866/0.5/8/above,-2./0/9/above left, - -0.866/-0.5/10/above,0/0/11/right,0.5/0.866/12/above,0/2./13/above,-0.5/0.866/14/above, - 0/0/15/below,0.866/-0.5/16/above,2/0/17/{below right}} - { - \filldraw (\x,\y) circle (2.4pt) node [\w] { \z}; - } - - \end{tikzpicture} - - - - -
    - -
    - - - - A smoothed version of the plot for this example. - -

    - The plot in is replaced by a smoothed version. - The diamonds are replaced by rounded curves, each resembling a leaf or flower petal. - Each loop (or leaf) has a cusp at the origin; however, the curve as a whole is traced smoothly. -

    - -

    - Beginning at the rectangular point (1,0) (on the positive x axis, when \theta=0), - the curve bends up and to the left, before turning downward, and eventually reaching the origin. - The curve then continues down and to the left into the third quadrant, - until it bends to the right, reaching the y axis at the rectangular point (0,-1). - It then loops back up through the fourth quadrant to the origin; - it then continues into the second quadrant, intersecting the x axis at (-1,0), - and passing through the third quadrant to form the left-hand loop as it returns to the origin. -

    - -

    - From there, we get the upper loop in a similar fashion, in the first quadrant and then the second, - and finally, the curve passes through the origin one more time into the fourth quadrant, - where it completes bottom of the right-hand loop. -

    -     - - - \begin{tikzpicture}[scale=1.1] - - \draw [dashed,gray] (-2.1,0) -- (0,0); - \draw [thick,->,>=latex] (0,0) node [below] {$O$} -- (2.5,0); - - \filldraw (0,0) circle (2.4pt); - - \foreach \x/\y in {1/{},2/1} - { - \draw (0,0) circle (\x cm); - \draw (\x,0) node [shift={(3pt,-4pt)}] { $\y$}; - } - - \foreach \x in {30,45,60,90,120,135,150} - { - \draw [rotate=\x,dashed,gray] (-2.3,0) -- (2.3,0); - } - - \draw [thick,firstcolor,domain=0:360,samples=130] plot ({2*cos(\x)*cos(2*\x)},{2*sin(\x)*cos(2*\x)}); - - \foreach \x/\y in {2./0,0.866/0.5,0/0,-0.5/-0.866,0/-2.,0.5/-0.866,0/0, - -0.866/0.5,-2./0,-0.866/-0.5,0/0,0.5/0.866,0/2.,-0.5/0.866,0/0,0.866/-0.5} - { - \filldraw (\x,\y) circle (2.4pt); - } - - \end{tikzpicture} - - - - -
    -     - -
    -     - -     - -

    - It is sometimes desirable to refer to a graph via a polar equation, - and other times by a rectangular equation. - Therefore it is necessary to be able to convert between polar and rectangular functions, - which we practice in the following example. - We will make frequent use of the identities found in . -

    - - - Converting between rectangular and polar equations - - -

    - Convert from rectangular to polar. -

      -
    1. -

      - y=x^2 -

      -
      2. - -
    2. -

      - xy = 1 -

      -
      3. -
    -

    - -

    - Convert from polar to rectangular. -

      -
    1. -

      - \ds r=\frac{2}{\sin(\theta) -\cos(\theta) } -

      -
      2. - -
    2. -

      - r=2\cos(\theta) -

      -
      3. -
    -

    -     -     - -

    -

      -
    1. -

      - Replace y with r\sin(\theta) and replace x with r\cos(\theta), giving: - - y \amp =x^2 - r\sin(\theta) \amp = r^2\cos^2(\theta) - \frac{\sin(\theta) }{\cos^2(\theta) } \amp = r - - We have found that r=\sin(\theta) /\cos^2(\theta) = \tan(\theta) \sec(\theta). - The domain of this polar function is (-\pi/2,\pi/2); - plot a few points to see how the familiar parabola is traced out by the polar equation. -

      -
      2. - -
    2. -

      - We again replace x and y using the standard identities and work to solve for r: - - xy \amp = 1 - r\cos(\theta) \cdot r\sin(\theta) \amp = 1 - r^2 \amp = \frac{1}{\cos(\theta) \sin(\theta) } - r \amp = \frac{1}{\sqrt{\cos(\theta) \sin(\theta) }} - - This function is valid only when the product of \cos(\theta) \sin(\theta) is positive. - This occurs in the first and third quadrants, - meaning the domain of this polar function is (0,\pi/2) \cup (\pi,3\pi/2). - - We can rewrite the original rectangular equation xy=1 as y=1/x. This is graphed in ; note how it only exists in the first and third quadrants. -

      - -
      - Graphing xy=1 from - - - Graph of the hyperbola y=1/x. - -

      - The graph is that of the standard hyperbola y=1/x. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - minor x tick num=4, - minor y tick num=4, - ymin=-5.3,ymax=5.3, - xmin=-5.3,xmax=5.3 - ] - - \addplot [firstcurvestyle,domain=-5:-.1,samples=60] ({x},{1/x}); - \addplot [firstcurvestyle,domain=.1:5,samples=60] ({x},{1/x}); - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      3. - -
    3. -

      - There is no set way to convert from polar to rectangular; - in general, we look to form the products - r\cos(\theta) and r\sin(\theta), - and then replace these with x and y, respectively. - We start in this problem by multiplying both sides by \sin(\theta) -\cos(\theta): - - r \amp = \frac{2}{\sin(\theta) -\cos(\theta) } - r(\sin(\theta) -\cos(\theta) ) \amp = 2 - r\sin(\theta) -r\cos(\theta) \amp = 2. \qquad \text{ Now replace with \(y\) and \(x\): } - y-x \amp = 2 - y \amp = x+2 - . - The original polar equation, - r=2/(\sin(\theta) -\cos(\theta) ) does not easily reveal that its graph is simply a line. - However, our conversion shows that it is. - The upcoming gallery of polar curves gives the general equations of lines in polar form. -

      -
      4. - -
    4. -

      - By multiplying both sides by r, - we obtain both an r^2 term and an r\cos(\theta) term, - which we replace with x^2+y^2 and x, respectively. - - r \amp =2\cos(\theta) - r^2 \amp = 2r\cos(\theta) - x^2+y^2 \amp = 2x. - We recognize this as a circle; by completing the square we can find its radius and center. - x^2-2x+y^2 \amp = 0 - (x-1)^2 + y^2 \amp =1 - . - The circle is centered at (1,0) and has radius 1. - The upcoming gallery of polar curves gives the equations of - some circles in polar form; - circles with arbitrary centers have a complicated polar equation that we do not consider here. -

      -
      5. -
    -

    -     - -     - -

    - Some curves have very simple polar equations but rather complicated rectangular ones. - For instance, the equation - r=1+\cos(\theta) describes a cardioid - (an important shape for modeling the sensitivity of microphones, among other things; - one is graphed in the gallery in the Limaçon section). - It's rectangular form is not nearly as simple; - it is the implicit equation - x^4+y^4+2x^2y^2-2xy^2-2x^3-y^2=0. - The conversion is not hard, but takes several steps, - and is left as a problem in the Exercise section. -

    - - - Gallery of Polar Curves - -

    - There are a number of basic and - classic polar curves, - famous for their beauty and/or applicability to the sciences. - polarfunction!gallery of graphs - This section ends with a small gallery of some of these graphs. - We encourage the reader to understand how these graphs are formed, - and to investigate with technology other types of polar functions. -

    - -
    - Lines in polar coordinates - -
    - Through the origin: \theta = \alpha - - - A line through the origin with positive slope. - -

    - A line through the origin is shown on a set of rectangular axes. - Also shown is an angle \alpha, measured from the positive x axis to the line. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,rotate=50] (-2.1,0) -- (2.1,0); - - \draw [->] (.5,0) arc (0:50:.5); - \draw [rotate=25] (.7,0) node { $\alpha$}; - - \end{tikzpicture} - - - - -
    -
    - Horizontal line: r=a\csc(\theta) - - - A horizontal line. - -

    - The graph is that of a horizontal line, plotted on a set of rectangular axes. - The graph is labeled with the value a, which is the distance from the x axis to the line. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor] (-2,.6) -- (2,.6); - - \draw (-.2,.3) node { $a\left\{\rule[-.23cm]{0pt}{.23cm}\right.$}; - - \end{tikzpicture} - - - - -
    - -
    - Vertical line: r=a\sec(\theta) - - - A vertical line. - -

    - A vertical line is plotted against a set of rectangular axes. - The distance from the line to the y axis is labeled a. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor] (.6,-2) -- (.6,2); - - \draw (.3,.2) node { $\overbrace{\rule{.55cm}{0pt}}$}; - \draw (.3,.4) node { $a$}; - - \end{tikzpicture} - - - - -
    -
    - Not through origin: \ds r=\frac{b}{\sin(\theta) -m\cos(\theta)} - - - A line with positive slope m, and y intercept b. - -

    - A line is plotted against a set of rectangular axes. - The line has a positive slope m, which is indicated next to the line. - Also shown is a distance b from the origin to the y intercept of the line, - indicating that this particular line does not pass through the origin. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,shift={(0,.6)},rotate=50] (-2.1,0) -- (1.7,0) node [pos=.82,rotate=50,black,shift={(0,-5pt)}] { slope $=m$}; - - \draw (.2,.3) node { $\left.\rule[-.23cm]{0pt}{.23cm}\right\}b$}; - - \end{tikzpicture} - - - - -
    -     -
    - -
    - Circles and Spirals - -
    - Centered on x-axis: r=a\cos(\theta) - - - A circle centered on the positive x axis and passing through the origin. - -

    - A circle is shown, with its center on the positive x axis. - The circle passes through the origin, and the diameter of the circle is labeled as a. - This suggests that the circle has center (a/2,0) and radius a/2. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor] (.9,0) circle (.9); - - \draw (.9,.1) node { $\overbrace{\rule{1.7cm}{0cm}}$}; - \draw (.9,.3) node { $a$}; - - \end{tikzpicture} - - - - -
    -
    - Centered on y-axis: r=a\sin(\theta) - - - A circle centered on the positive y axis and passing through the origin. - -

    - A circle is shown, with its center on the positive y axis. - The circle passes through the origin, and the diameter of the circle is labeled as a. - This suggests that the circle has center (0,a/2) and radius a/2. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor] (0,.9) circle (.9); - - \draw (-.2,.9) node { $a\left\{\rule[-.8cm]{0cm}{0.8cm}\right.$}; - - \end{tikzpicture} - - - - -
    -
    - Centered on origin: r=a - - - A circle centered at the origin. - -

    - On a set of rectangular axes, a circle centered at the origin is plotted. - The radius a of the circle is labeled. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor] (0,0) circle (.9); - - \draw (.45,.1) node { $\overbrace{\rule{.8cm}{0cm}}$}; - \draw (.45,.3) node { $a$}; - - \end{tikzpicture} - - - - -
    -
    - Archimedean spiral: r=\theta - - - A counter-clockwise spiral that begins at the origin. - -

    - The curve begins at the origin and spirals outward, moving in a counter-clockwise direction. - The distance from the curve to the origin increases with the angle, - and three full revolutions of the spiral are shown. -

    - -

    - The appearance of the curve is not unlike that of a snail's shell. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:18.85,samples=200] plot ({cos(deg(\x))*(\x/9.5)},{sin(deg(\x))*(\x/9.5)}); - - \end{tikzpicture} - - - - -
    -     -
    - -
    - Limaçons - -
    - With inner loop: \ds \frac ab \lt 1 - - - A limaçon curve with a loop at the origin, and symmetric about the x axis. - -

    - The first of several curves in the limaçcon family. - The curve intersects the x axis three times: at the origin, which it passes through twice, - and at two other points on the positive x axis. -

    - -

    - The curve consists of two loops that are joined at the origin. - The larger, outer loop passes through the x intercept with the larger value of x. - It is somewhat heart-shaped, with a cusp at the origin. -

    - -

    - The smaller curve lies inside the larger curve, and is shaped like a teardrop. - The two curves join together in such a way that, although each individually has a cusp, - the curve as a whole is traced out smoothly, with the two cusps occurring at the origin, - where the curve intersects itself. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:360,samples=90] plot ({cos(\x)*(.6+1.2*cos(\x))},{sin(\x)*(.6+1.2*cos(\x))}); - - \end{tikzpicture} - - - - -
    -
    - Cardioid: \ds \frac ab=1 - - - A heart-shaped curve in the limaçon family, known as a cardioid. - -

    - This is also a limaçon curve, and it is again symmetric about the x axis. - Its appearance is similar to the previous curve, but with the inner loop removed. -

    - -

    - The resulting cusp at the origin gives this curve a heart-like shape; - as a result, the curve is also known as a cardioid. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:360,samples=90] plot ({cos(\x)*(.9+.9*cos(\x))},{sin(\x)*(.9+.9*cos(\x))}); - - \end{tikzpicture} - - - - -
    -
    - Dimpled: \ds 1\lt \frac ab \lt 2 - - - A dimpled limaçon curve. - -

    - A third curve in the limaçon family, also symmetric about the x axis. - This curve does not pass through the origin. - It has two x intercepts, one positive, and one negative. - The positive x intercept has a larger absolute value (it is further from the origin). -

    - -

    - To the right of the y axis, the curve appears to be almost oval-like in shape. - But as it crosses to the left of the y axis, it begins to bend more sharply toward the second x intercept. - On this side, the curve bends inward, forming a dimple, or dent. - The dent is smooth, however, and is not a cusp. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:360,samples=90] plot ({cos(\x)*(1+.8*cos(\x))},{sin(\x)*(1+.8*cos(\x))}); - - \end{tikzpicture} - - - - -
    -
    - Convex: \ds \frac ab \gt 2 - - - A convex limaçon curve. - -

    - A fourth limaçon curve. It is also symmetric about the x axis. - Like the dimpled limaçon, it has two x intercepts and does not pass through the origin. -

    - -

    - Unlike the dimpled limaçon, to the left of the y axis, the curve flattens out, - but does not bend inward. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:360,samples=80] plot ({cos(\x)*(1.3+.6*cos(\x))},{sin(\x)*(1.3+.6*cos(\x))}); - - \end{tikzpicture} - - - - -
    -     -
    - -

    - Symmetric about x-axis: r=a\pm b\cos(\theta) -

    -

    - Symmetric about y-axis: r=a\pm b\sin(\theta); a,b \gt 0 -

    - -
    - Rose curves - -
    - r=a\cos(2\theta) - - - A rose curve with four leaves along the coordinate axes. - -

    - The curve r=\cos(2\theta) is four-leaf rose curve; it appears to be the same as the one in . - Two of the leaves lie along the x axis, and two leaves lie along the y axis. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:360,samples=180] plot ({cos(\x)*(1.9*cos(2*\x))},{sin(\x)*(1.9*cos(2*\x))}); - - \end{tikzpicture} - - - - -
    -
    - r=a\sin(2\theta) - - - A rose curve with four leaves, each in one of the quadrants. - -

    - The curve r=\sin(2\theta) is a rose curve with four leaves. - It is has the same appearance as the rose curve in , - but it is rotated by 45 degrees, so that each leaf lies in one of the four quadrants. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:360,samples=180] plot ({cos(\x)*(1.9*sin(2*\x))},{sin(\x)*(1.9*sin(2*\x))}); - - \end{tikzpicture} - - - - -
    -
    - r=a\cos(3\theta) - - - A rose curve with three leaves, symmetric about the x axis. - -

    - The curve r=\cos(3\theta) is a rose curve with three leaves. - The leaves are narrower in appearance than those of the four-leaf rose curves. -

    - -

    - The curve overall is symmetric about the x axis. - One leaf lies along the positive x axis, - while the other two leaves are in the second and third quadrants. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:360,samples=160] plot ({cos(\x)*(1.9*cos(3*\x))},{sin(\x)*(1.9*cos(3*\x))}); - - \end{tikzpicture} - - - - -
    -
    - r=a\sin(3\theta) - - - A rose curve with three leaves, symmetric about the y axis. - -

    - Another rose curve with three leaves, this time given by r=\sin(3\theta). - The curve is symmetric about the y. - One leaf lies along the negative y axis, - while the other two leaves are in the first and second quadrants. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:360,samples=160] plot ({cos(\x)*(1.9*sin(3*\x))},{sin(\x)*(1.9*sin(3*\x))}); - - \end{tikzpicture} - - - - -
    -     -
    - -

    - Symmetric about x-axis: r=a \cos(n\theta) -

    - -

    - Symmetric about y-axis: r=a\sin(n\theta) -

    - -

    - Curve contains 2n petals when n is even and n petals when n is odd. -

    - -
    - Special Curves - -
    - Rose curve: r=a\sin(\theta/5) - - - A polar curve consisting of three loops. - -

    - The curve r=\sin(\theta/5) is also known as a rose curve, - but it is more similar to a limaçon in appearance, - except that it has more loops. -

    - -

    - The curve is symmetric about the y axis. - There are three loops in total, each of which intersects the y axis twice. - The smallest loop has a teardrop shape. It lies inside the middle loop, - which is heart-shaped. - The middle loop, in turn, lies inside of the largest loop, - which appears nearly circular, except that it has a slight cusp at the bottom, - where it joins with the middle loop. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:900,samples=200] plot ({cos(\x)*(1.9*sin(\x/5))},{sin(\x)*(1.9*sin(\x/5))}); - - \end{tikzpicture} - - - - -
    -
    - Rose curve: r=a\sin(2\theta/5) - - - An elaborate polar curve consisting of many intersecting loops. - -

    - The curve r=\sin(2\theta/5) is also known as a rose curve, - but it is more elaborate than any of the earlier examples. -

    - -

    - The curve is symmetric about both axes, and consists of many loops of varying size. - These loops intesect each other several times. - The overall appearance is similar to a sort of braided knot. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=0:1800,samples=500] plot ({cos(\x)*(1.9*cos(2*\x/5))},{sin(\x)*(1.9*cos(2*\x/5))}); - - \end{tikzpicture} - - - - -
    -
    - Lemniscate: r^2=a^2\cos(2\theta) - - - A lemniscate curve, which has the shape of a figure-eight. - -

    - The lemniscate r^2 = \cos^2(2\theta) is a figure-eight curve. - Since r^2 cannot be positive, - the points on the curve all correspond to angles where \cos(2\theta) is positive. - The curve is symmetric about the x axis, and looks much like the symbol for infinity. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=-45:45,samples=70] plot ({cos(\x)*(1.9*sqrt(cos(2*\x)))},{sin(\x)*(1.9*sqrt(cos(2*\x)))}); - \draw [thick,firstcolor,domain=-45:45,samples=70] plot ({-cos(\x)*(1.9*sqrt(cos(2*\x)))},{-sin(\x)*(1.9*sqrt(cos(2*\x)))}); - - \end{tikzpicture} - - - - -
    -
    - Eight Curve: r^2=a^2\sec^4(\theta) \cos(2\theta) - - - A polar curve in the shape of a figure-eight. - -

    - This final curve in the gallery is also a figure-eight curve. - Like the lemniscate, it is symmetric about the x axis, - but it is flatter at the ends than the lemniscate. -

    -     - - - \begin{tikzpicture} - - \draw [<->] (-2.1,0) -- (2.1,0); - \draw [<->] (0,-2.1) -- (0,2.1); - - \draw [thick,firstcolor,domain=-45:45,samples=80] plot ({cos(\x)*(1.9*sqrt(sec(\x)^4*cos(2*\x)))},{sin(\x)*(1.9*sqrt(sec(\x)^4*cos(2*\x)))}); - \draw [thick,firstcolor,domain=-45:45,samples=80] plot ({-cos(\x)*(1.9*sqrt(sec(\x)^4*cos(2*\x)))},{-sin(\x)*(1.9*sqrt(sec(\x)^4*cos(2*\x)))}); - - \end{tikzpicture} - - - - -
    -     -
    -     - -

    - Earlier we discussed how each point in the plane does not have a unique representation in polar form. - This can be a good thing, - as it allows for the beautiful and interesting curves seen in the preceding gallery. - However, it can also be a bad thing, - as it can be difficult to determine where two curves intersect. -

    - - - Finding points of intersection with polar curves - -

    - Determine where the graphs of the polar equations - r=1+3\cos(\theta) and r=\cos(\theta) intersect. -

    -     - -

    - As technology is generally readily available, - it is usually a good idea to start with a graph. - We have graphed the two functions in ; - to better discern the intersection points, - zooms in around the origin. -

    - -
    - Graphs to help determine the points of intersection of the polar functions given in - -
    - - - - Overlapping plots of a circle and a limaçon, showing their points of intersection. - -

    - The two curves for this example are plotted. - The curve r=1+3\cos(\theta) is a limaçon with an inner loop, - while r=\cos(\theta) is a circle with radius 1/2 and center (1/2,0). -

    - -

    - The circle is significantly smaller than the limaçon: - it intersects the x axis at (0,0) and (1,0), - while the inner loop of the limaçon intersects the x axis at (0,0) and (2,0). -

    - -

    - Near the origin, the inner loop of the limaçon narrows faster than the circle, - and in addition to the origin, two other points of intersection can be seen. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-2.6,ymax=2.6, - xmin=-1.5,xmax=4.5 - ] - - \addplot+ [domain=0:360,samples=120] ({cos(x)*(1+3*cos(x))},{sin(x)*(1+3*cos(x))}); - \addplot+ [solid,domain=0:180,samples=40] ({cos(x)*cos(x)},{sin(x)*cos(x)}); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
    - -
    - - - - Zoomed in view of the intersections between the two polar curves. - -

    - The second image is a zoomed in view of , - in which the three points of intersection can better be seen. - One intersection is at the origin, and the other two points of intersection lie on opposite sides of the x axis. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.6,ymax=.6, - xmin=-.72,xmax=.72 - ] - - \addplot+ [domain=0:360,samples=180] ({cos(x)*(1+3*cos(x))},{sin(x)*(1+3*cos(x))}); - \addplot+ [solid,domain=0:180,samples=80] ({cos(x)*cos(x)},{sin(x)*cos(x)}); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - To start we set the functions equal to each other and solve for \theta: - - 1+3\cos(\theta) \amp = \cos(\theta) - 2\cos(\theta) \amp = -1 - \cos(\theta) \amp = -\frac12 - \theta \amp = \frac{2\pi}{3}, \frac{4\pi}{3} - . -

    - -

    - (There are, of course, - infinitely many solutions to the equation \cos(\theta) =-1/2; - as the limaçon is traced out once on [0,2\pi], - we restrict our solutions to this interval.) -

    - -

    - We need to analyze this solution. - When \theta = 2\pi/3 we obtain the point of intersection that lies in the 4th quadrant. - When \theta = 4\pi/3, - we get the point of intersection that lies in the second quadrant. - There is more to say about this second intersection point, however. - The circle defined by r=\cos(\theta) is traced out once on [0,\pi], - meaning that this point of intersection occurs while tracing out the circle a second time. - It seems strange to pass by the point once and then recognize it as a point of intersection only when arriving there a second time. - The first time the circle arrives at this point is when \theta = \pi/3. - It is key to understand that these two points are the same: - (\cos(\pi/3),\pi/3) and (\cos(4\pi/3),4\pi/3). -

    - -

    - To summarize what we have done so far, - we have found two points of intersection: - when \theta=2\pi/3 and when \theta=4\pi/3. - When referencing the circle r=\cos(\theta), - the latter point is better referenced as when \theta=\pi/3. -

    - -

    - There is yet another point of intersection: the pole - (or, the origin). - We did not recognize this intersection point using our work above as each graph arrives at the pole at a different \theta value. -

    - -

    - A graph intersects the pole when r=0. - Considering the circle r=\cos(\theta), - r=0 when \theta = \pi/2 - (and odd multiples thereof, as the circle is repeatedly traced). - The limaçon intersects the pole when 1+3\cos(\theta) =0; - this occurs when \cos(\theta) = -1/3, - or for \theta = \cos^{-1}(-1/3). - This is a nonstandard angle, - approximately \theta = 1.9106 = 109.47^\circ. - The limaçon intersects the pole twice in [0,2\pi]; - the other angle at which the limaçon is at the pole is the reflection of the first angle across the x-axis. - That is, \theta = 4.3726 = 250.53^\circ. -

    -     - -     - -

    - If all one is concerned with is the (x,y) coordinates at which the graphs intersect, - much of the above work is extraneous. - We know they intersect at (0,0); - we might not care at what \theta value. - Likewise, using \theta =2\pi/3 and - \theta=4\pi/3 can give us the needed rectangular coordinates. - However, in the next section we apply calculus concepts to polar functions. - When computing the area of a region bounded by polar curves, - understanding the nuances of the points of intersection becomes important. -

    -     - - - - Terms and Concepts - - - -

    - In your own words, - describe how to plot the polar point P(r,\theta). -

    -     - - -     - - - - -

    - When plotting a point with polar coordinate P(r,\theta), - r must be positive. - -

    -     - -     - - - - -

    - Every point in the Cartesian plane can be represented by a polar coordinate. - -

    -     - -     - - - - -

    - Every point in the Cartesian plane can be represented uniquely by a polar coordinate. - -

    -     - -     -     - - - Problems - - - -

    - Plot the points with the given polar coordinates. -

    - -

    -

      -
    1. -

      - A=P(2,0) -

      -
      2. - -
    2. -

      - B=P(1,\pi) -

      -
      3. - -
    3. -

      - C=P(-2,\pi/2) -

      -
      4. - -
    4. -

      - D=P(1,\pi/4) -

      -
      5. -
    -

    -     - - - The four points plotted in this exercise. - -

    - On a polar grid, four points are plotted. - The point A is at the intersection of the initial ray and the circle of radius 2. - Points B and D are both on the circle of radius 1. - The point B is on the same line as the initial ray, but in the opposite direction. - The point D lies above the initial ray, making an angle of \pi/4. - Finally, the point C is at the bottom of the circle of radius 2. -

    -     - - - \begin{tikzpicture} - - \foreach \x in {1,2} - { - \draw (0,0) circle (\x); - \draw (\x,0) node [below right] {\(\x\)}; - } - \draw [thick,->,>=stealth] (0,0) node [below] {\(O\)} -- (2.5,0); - - \filldraw (xyz polar cs: angle=0,radius=2) circle (2pt) node [above right] {\(A\)} - (xyz polar cs: angle=180,radius=1)circle (2pt) node [right] {\(B\)} - (xyz polar cs: angle=90,radius=-2)circle (2pt) node [right] {\(C\)} - (xyz polar cs: angle=45,radius=1)circle (2pt) node [right] {\(D\)}; - - \end{tikzpicture} - - - -     - -     - - - - -

    - Plot the points with the given polar coordinates. -

    - -

    -

      -
    1. -

      - A=P(2,3\pi) -

      -
      2. - -
    2. -

      - B=P(1,-\pi) -

      -
      3. - -
    3. -

      - C=P(1,2) -

      -
      4. - -
    4. -

      - D=P(1/2,5\pi/6) -

      -
      5. -
    -

    -     - - - The four points plotted in this exercise. - -

    - On a polar grid, four points are plotted. - Points A and B are both on the same line as the initial ray, - but to the left of the origin O, with A on the circle of radius 2, - and B on the circle of radius 1. - The point C is on the circle of radius 1, - and makes an angle slightly greater than a right angle with the initial ray, - placing it above and just to the left of the origin O. - The point D lies almost immediately below the point C; - the circle of radius 1/2 is not marked as part of the grid. -

    -     - - - \begin{tikzpicture} - - \foreach \x in {1,2} - { - \draw (0,0) circle (\x); - \draw (\x,0) node [below right] {\(\x\)}; - } - \draw [thick,->,>=stealth] (0,0) node [below] {\(O\)} -- (2.5,0); - - \filldraw (xyz polar cs: angle=180,radius=2) circle (2pt) node [above] {\(A\)} - (xyz polar cs: angle=180,radius=1)circle (2pt) node [above] {\(B\)} - (xyz polar cs: angle=114.6,radius=1)circle (2pt) node [right] {\(C\)} - (xyz polar cs: angle=150,radius=.5)circle (2pt) node [right] {\(D\)}; - - \end{tikzpicture} - - - -     - -     - - - - -

    - For each of the given points give two sets of polar coordinates that identify it, - where 0\leq \theta\leq 2\pi. -

    - - - Four points plotted on a polar grid. - -

    - A standard polar grid is given, with circles of radius 1, 2, and 3. - Rays marking the angles \pi/6, \pi/4, \pi/3, and \pi/2 are also shown, - along with rays for the corresponding angles in the other quadrants. -

    - -

    - The point A lies on the ray \theta=\pi/4, - at a distance half way between the circles of radius 2 and 3. -

    - -

    - The point B lies on the ray \theta=-\pi/6, - and on the circle r=1. -

    - -

    - The point C lies on the ray \theta = 4\pi/3, - and on the circle r=3. -

    - -

    - The point D lies on the ray \theta = 2\pi/3, - at a distance half way between the circles of radius 1 and 2. -

    -     - - - \begin{tikzpicture} - - \draw [dashed,gray] (-3.1,0) -- (0,0); - \draw [thick,->,>=stealth] (0,0) node [below] {\(O\)} -- (3.5,0); - - \filldraw (0,0) circle (2.4pt); - - \foreach \x in {1,2,3} - { - \draw (0,0) circle (\x cm); - \draw (\x,0) node [below right] {\x}; - } - - \foreach \x in {30,45,60,90,120,135,150} - { - \draw [rotate=\x,dashed,gray] (-3.1,0) -- (3.1,0); - } - - \filldraw (xyz polar cs: angle=45,radius=2.5) circle (2pt) node [above] {\(A\)} - (xyz polar cs: angle=-30,radius=1)circle (2pt) node [below] {\(B\)} - (xyz polar cs: angle=240,radius=3)circle (2pt) node [right] {\(C\)} - (xyz polar cs: angle=120,radius=1.5)circle (2pt) node [right] {\(D\)}; - - \end{tikzpicture} - - - - -     - -

    - A=P(2.5,\pi/4) and P(-2.5,5\pi/4); -

    - -

    - B=P(-1,5\pi/6) and P(1,11\pi/6); -

    - -

    - C=P(3,4\pi/3) and P(-3,\pi/3); -

    - -

    - D=P(1.5,2\pi/3) and P(-1.5,5\pi/3); -

    -     - -     - - - - - Context()->variables->are(t=>"Real"); - @A=(Compute("(2,pi/6)"),Compute("(-2,-(5*pi)/6)")); - $Alist=List($A[0],$A[1]); - @B=(Compute("(1,-pi/3)"),Compute("(-1,(2*pi)/3)")); - $Blist=List($B[0],$B[1]); - @C=(Compute("(2,(3*pi)/4)"),Compute("(-2,-pi/4)")); - $Clist=List($C[0],$C[1]); - @D=(Compute("(5/2,pi)"),Compute("(5/2,-pi)")); - $Dlist=List($D[0],$D[1]); - - -

    - For each of the given points give two sets of polar coordinates that identify it, - where -\pi\lt \theta\leq \pi. -

    - - - Four points plotted on a polar grid. - -

    - A standard polar grid is given, with circles of radius 1, 2, and 3. - Rays marking the angles \pi/6, \pi/4, \pi/3, and \pi/2 are also shown, - along with rays for the corresponding angles in the other quadrants. -

    - -

    - The point A lies on the ray \theta=\pi/6, - and on the circle r=1. -

    - -

    - The point B lies on the ray \theta=-\pi/3, - and on the circle r=1. -

    - -

    - The point C lies on the ray \theta = 3\pi/4, - and on the circle r=2. -

    - -

    - The point D lies on the ray \theta = \pi, - at a distance half way between the circles of radius 2 and 3. -

    -     - - \begin{tikzpicture} - \draw [dashed,gray] (-3.1,0) -- (0,0); - \draw[thick,->,>=stealth] (0,0) node [below] {\(O\)} -- (3.5,0) ; - \filldraw (0,0) circle (2.4pt); - \foreach \x in {1,2,3} - {\draw (0,0) circle (\x cm); - \draw (\x,0) node [below right] {\x}; - } - \foreach \x in {30,45,60,90,120,135,150} - {\draw [rotate=\x,dashed,gray] (-3.1,0) -- (3.1,0); - } - \filldraw (xyz polar cs: angle=30,radius=2) circle (2pt) node [above] {\(A\)} - (xyz polar cs: angle=-60,radius=1)circle (2pt) node [below] {\(B\)} - (xyz polar cs: angle=135,radius=2)circle (2pt) node [right] {\(C\)} - (xyz polar cs: angle=180,radius=2.5)circle (2pt) node [below] {\(D\)}; - \end{tikzpicture} - - - - - Enter two sets of coordinates for the point A. Separate your answers by a comma. - -

    - -

    - - - Enter two sets of coordinates for the point B. Separate your answers by a comma. - -

    - -

    - - - Enter two sets of coordinates for the point C. Separate your answers by a comma. - -

    - -

    - - - Enter two sets of coordinates for the point D. Separate your answers by a comma. - -

    - -

    -     - -     -     - - - - - Context("Point"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - @xy=(Compute("(sqrt(2),sqrt(2))"), Compute("(sqrt(2),-sqrt(2))")); - @rt=(Compute("(sqrt(5),arctan(-1/2))"), Compute("(sqrt(5),pi+arctan(-1/2))")); - @rtev=map{$_->cmp(showCoordinateHints => 0, checker => sub { - my ($correct,$student,$ansHash) = @_; - my ($sr,$st) = $student->value; - my ($cr,$ct) = $correct->value; - return (($sr*cos($st) == $cr*cos($ct)) and ($sr*sin($st) == $cr*sin($ct))); - })}(@rt); - - -

    - Convert each of the following polar coordinates to rectangular, - and each of the following rectangular coordinates to polar. -

    - -

    -

      -
    1. -

      - A=P(2,\pi/4) -

      - -

      - (x,y)= -

      -
      2. - -
    2. -

      - B=P(2,-\pi/4) -

      - -

      - (x,y)= -

      -
      3. - -
    3. -

      - C=(2,-1) -

      - -

      - P(r,\theta)=P -

      -
      4. - -
    4. -

      - D=(-2,1) -

      - -

      - P(r,\theta)=P -

      -
      5. -
    -

    -     - -

    - A=(\sqrt{2},\sqrt{2}), - B=(\sqrt{2},-\sqrt{2}), - C=P(\sqrt{5},-0.46), - D=P(\sqrt{5},2.68). -

    -     -     -     - - - - - Context("Point"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - @xy=(Compute("(-3,0)"), Compute("(-1/2,sqrt(3)/2)")); - @rt=(Compute("(4,pi/2)"), Compute("(2,-pi/3)")); - @rtev=map{$_->cmp(showCoordinateHints => 0, checker => sub { - my ($correct,$student,$ansHash) = @_; - my ($sr,$st) = $student->value; - my ($cr,$ct) = $correct->value; - return (($sr*cos($st) == $cr*cos($ct)) and ($sr*sin($st) == $cr*sin($ct))); - })}(@rt); - - -

    - Convert each of the following polar coordinates to rectangular, - and each of the followingrectangular coordinates to polar. -

    - -

    -

      -
    1. -

      - A=P(3,\pi) -

      - -

      - (x,y)= -

      -
      2. - -
    2. -

      - B=P(1,2\pi/3) -

      - -

      - (x,y)= -

      -
      3. - -
    3. -

      - C=(0,4) -

      - -

      - P(r,\theta)=P -

      -
      4. - -
    4. -

      - D=(1,-\sqrt{3}) -

      - -

      - P(r,\theta)=P -

      -
      5. -
    -

    -     - -

    - A=(-3,0), - B=(-1/2,\sqrt{3}/2), - C=P(4,\pi/2), - D=P(2,-\pi/3). -

    -     -     -     - - - - -

    - Graph the polar function on the given interval. -

    -     - - - - -

    - r=2,0\leq \theta\leq \pi/2 -

    -     - - - - Portion of the circle of radius 2, centered at the origin, in the first quadrant. - -

    - An arc of the circle r=2 is shown, for 0\leq \theta\leq \pi/2. - This is the quarter of a circle of radius 2, centered at the origin, that lies in the first quadrant. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.5,ymax=2.1, - xmin=-.5,xmax=2.6 - ] - - \addplot+ [domain=0:90] ({2*cos(x)},{2*sin(x)}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - \theta=\pi/6,-1\leq r\leq 2 -

    -     - - - - A line segment through the origin with positive slope. - -

    - A line segment through the origin is shown. - The segment makes an angle of \pi/6 with the positive x axis, - and it extends a distance of 2 units into the first quadrant, - and one unit into the third quadrant. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,2,-1,-2}, - ymin=-2.1,ymax=2.1, - xmin=-2.5,xmax=2.5 - ] - - \addplot+ [domain=-.866:1.73] ({x},{.577*x}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=1-\cos(\theta),[0,2\pi] -

    -     - - - - A cardioid, symmetric about the x axis, with x intercepts at -2 and 0. - -

    - The curve is a cardioid that is symmetric about the x axis. - The cusp is at the origin, and the other x intercept is at (-2,0). - (It is in the opposite direction of the example in the gallery of polar curves.) -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-2.1,ymax=2.1, - xmin=-2.5,xmax=2.5 - ] - - \addplot+ [domain=0:360,samples=80] ({cos(x)*(1-cos(x))},{sin(x)*(1-cos(x))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=2+\sin(\theta),[0,2\pi] -

    -     - - - - A convex limaçon, symmetric about the y axis. - -

    - The curve is a convex limaçon. This is the fourth type of limaçon in the gallery of polar curves. - In this case, the limaçon is symmetric about the y axis, - with the flattened part of the curve at the bottom. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-2.6,ymax=3.6, - xmin=-3.7,xmax=3.7 - ] - - \addplot+ [domain=0:360,samples=80] ({cos(x)*(2+sin(x))},{sin(x)*(2+sin(x))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=2-\sin(\theta),[0,2\pi] -

    -     - - - - A convex limaçon, symmetric about the y axis. - -

    - The curve is a convex limaçon. This is the fourth type of limaçon in the gallery of polar curves. - In this case, the limaçon is symmetric about the y axis, - with the flattened part of the curve at the top. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-3.6,ymax=2.6, - xmin=-3.7,xmax=3.7 - ] - - \addplot+ [domain=0:360,samples=70] ({cos(x)*(2-sin(x))},{sin(x)*(2-sin(x))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=1-2\sin(\theta),[0,2\pi] -

    -     - - - - A limaçon with an inner loop, symmetric about the y axis. - -

    - The curve is a limaçon with an inner loop. - It is symmetric about the y axis. - The inner loop lies below the x axis, with y intercepts at y=0 and y=-1. - The outer loop has its other y intercept at y=-3. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-3.6,ymax=2.6, - xmin=-3.7,xmax=3.7 - ] - - \addplot+ [domain=0:360,samples=90] ({cos(x)*(1-2*sin(x))},{sin(x)*(1-2*sin(x))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=1+2\sin(\theta),[0,2\pi] -

    -     - - - - A limaçon with an inner loop, symmetric about the y axis. - -

    - The curve is a limaçon with an inner loop. - It is symmetric about the y axis. - The inner loop lies above the x axis, with y intercepts at y=0 and y=1. - The outer loop has its other y intercept at y=3. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-2.6,ymax=3.6, - xmin=-3.7,xmax=3.7 - ] - - \addplot+ [domain=0:360,samples=90] ({cos(x)*(1+2*sin(x))},{sin(x)*(1+2*sin(x))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=\cos(2\theta),[0,2\pi] -

    -     - - - - A rose curve with four petals. - -

    - The curve is a rose curve with four loops that all pass through the origin. - The loops are symmetric about the two coordinate axes, - with their second intercepts at (\pm 1,0) and (0,\pm 1). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.3,xmax=1.3 - ] - - \addplot+ [domain=0:360,samples=150] ({cos(x)*(cos(2*x))},{sin(x)*(cos(2*x))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=\sin(3\theta),[0,\pi] -

    -     - - - - A rose curve with three loops, symmetric about the y axis. - -

    - A rose curve with three loops that all pass through the origin. - One loop is along the negative y axis, with a y intercept at (-1,0). - The other two loops lie in the first and second quadrants. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.3,xmax=1.3 - ] - - \addplot+ [domain=0:180,samples=101] ({cos(x)*(sin(3*x))},{sin(x)*(sin(3*x))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=\cos(\theta/3),[0,3\pi] -

    -     - - - - A limaçon with an inner loop. - -

    - The curve is a limaçon with an inner loop, - symmetric about the x axis, - but it is shifted horizontally, to the left, - relative to the example in the gallery of polar curves. -

    - -

    - The point of self-intersection is at (-1/2, 0). - The other end of the inner loop is at the origin, - and the far end of the outer loop is at the point (1,0). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.3,xmax=1.3 - ] - - \addplot+ [domain=0:540,samples=120] ({cos(x)*(cos(x/3))},{sin(x)*(cos(x/3))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=\cos(2\theta/3),[0,6\pi] -

    -     - - - - An elaborate rose curve with many self-intersections and four primary loops. - -

    - This is a more complicated curve. - It passes several times through the origin, and has eight other points of self-intersection. - The largest loops in the curve are similar to cardioids; - there are four of these passing through the origin, with a second intercept at one of the four points (\pm 1, 0), (0,\pm 1). - As these loops intersect each other, they create four other loops of intermediate size, - and four smaller loops in the center. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.3,xmax=1.3 - ] - - \addplot+ [domain=0:1080,samples=300] ({cos(x)*(cos(2*x/3))},{sin(x)*(cos(2*x/3))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=\theta/2,[0,4\pi] -

    -     - - - - A counter-clockwise spiral - -

    - The curve is a counter-clockwise spiral that begins at the origin. - It makes two full revolutions, and ends at the point (2\pi, 0). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-5.9,ymax=4.9, - xmin=-6.5,xmax=6.5 - ] - - \addplot+ [domain=0:720,samples=150] ({cos(x)*(x*3.14/360)},{sin(x)*(x*3.14/360)}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=3\sin(\theta),[0,\pi] -

    -     - - - - A circle passing through the origin with its center on the positive y axis. - -

    - A circle of radius 3/2 with its center at (0,3/2). - It passes through the origin and the point (0,3). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ytick={2,3,1}, - ymin=-.9,ymax=3.9, - xmin=-2.9,xmax=2.9 - ] - - \addplot+ [domain=0:180,samples=60] ({cos(x)*(3*sin(x))},{sin(x)*(3*sin(x))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - -

    - r=2\cos(\theta),[0,\pi/2] -

    -     - - - A semi-circle in the first quadrant. - -

    - The curve is a semi-circle in the first quadrant. - Its endpoints are (0,0) and (2,0), - making the center (1,0) and radius 1. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-2.3,ymax=2.3,% - xmin=-2.5,xmax=2.5% - ] - - \addplot+ [domain=0:90,samples=60] ({cos(x)*(2*cos(x))},{sin(x)*(2*cos(x))}); - - \end{axis} - - \end{tikzpicture} - - -     -     - - - - - -

    - r=\cos(\theta) \sin(\theta),[0,2\pi] -

    -     - - - - A four-leaf rose with one petal in each quadrant. - -

    - The curve is a four-leafed rose that lies within the circle r=1/2. - One leaf lies in each of the four quadrants. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.1,ymax=1.1, - xmin=-1.3,xmax=1.3 - ] - - \addplot+ [domain=0:360,samples=120] ({cos(x)*(cos(x)*sin(x))},{sin(x)*(cos(x)*sin(x))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - r=\theta^2-(\pi/2)^2,[-\pi,\pi] -

    -     - - - - A curve with two loops: a circle inside a larger leaf shape. - -

    - The curve for this exercise is a fairly strange shape. - It is symmetric about the x axis. - There is a large, outer loop that looks like a leaf or a raindrop. - It has a cusp at (-7,0), and also passes through the origin. - There is a smaller, inner loop that looks almost like a circle. - It passes through the origin and the point (-3,0). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-3.75,ymax=3.75, - xmin=-8.1,xmax=.9 - ] - - \addplot+ [domain=-3.14:3.14,samples=120] ({cos(deg(x))*(x^2-2.4674)},{sin(deg(x))*(x^2-2.4674)}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - \ds r=\frac{3}{5\sin(\theta) -\cos(\theta) },[0,2\pi] -

    -     - - - - A straight line with positive slope. - -

    - The curve is a straight line with x intercept at (-3,0) - and y intercept (0,3/5). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-4.9,ymax=4.96, - xmin=-5.9,xmax=5.9 - ] - - \addplot+ [domain=.1:3] ({cos(deg(x))*(3/(5*sin(deg(x))-cos(deg(x)))},{sin(deg(x))*(3/(5*sin(deg(x))-cos(deg(x)))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - \ds r=\frac{-2}{3\cos(\theta) -2\sin(\theta) },[0,2\pi] -

    -     - - - - A straight line with positive slope. - -

    - The curve is a straight line with x intercept at (-2/3,0) - and y intercept (0,1). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-4.9,ymax=4.96, - xmin=-5.9,xmax=5.9 - ] - - \addplot+ [domain=.1:3] ({cos(deg(x))*(-2/(3*cos(deg(x))-2*sin(deg(x))))},{sin(deg(x))*(-2/(3*cos(deg(x))-2*sin(deg(x))))}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - \ds r=3\sec(\theta),(-\pi/2,\pi/2) -

    -     - - - - A vertical line with x=3. - -

    - The curve is the vertical line x=3. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-4.9,ymax=4.96, - xmin=-5.9,xmax=5.9 - ] - - \addplot+ [domain=-5:5] ({3},{x}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - -

    - \ds r=3\csc(\theta),(0,\pi) -

    -     - - - - The horizontal line y=4. - -

    - The curve is the horizontal line y=4. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-4.9,ymax=4.96, - xmin=-5.9,xmax=5.9 - ] - - \addplot+ [domain=-5:5] ({x},{4}); - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - -     - - - -

    - Convert the polar equation to a rectangular equation. -

    -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("(x-3)^2+y^2=9", limits=>[[-1,7],[-4,4]]); - - -

    - Convert the polar equation to a rectangular equation. -

    - -

    - r=6\cos(\theta) -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("x^2+(y+2)^2=4", limits=>[[-3,3],[-5,1]]); - - -

    - Convert the polar equation to a rectangular equation. -

    - -

    - r=-4\sin(\theta) -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("(x-1/2)^2+(y-1/2)^2=1/2", limits=>[[-1,2],[-1,2]]); - - -

    - Convert the polar equation to a rectangular equation. -

    - -

    - r=\cos(\theta) +\sin(\theta) -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("y=2/5x+7/5", limits=>[[-3,3],[-2,4]]); - - -

    - Convert the polar equation to a rectangular equation. -

    - -

    - r=\dfrac{7}{5\sin(\theta)-2\cos(\theta)} -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("x=3", limits=>[[0,6],[-3,3]]); - - -

    - Convert the polar equation to a rectangular equation. -

    - -

    - r=\dfrac{3}{\cos(\theta)} -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("y=4", limits=>[[-3,3],[1,7]]); - - -

    - Convert the polar equation to a rectangular equation. -

    - -

    - r=\dfrac{4}{\sin(\theta)} -

    - -

    - -

    -     -     -     - - - - -

    - \ds r=\tan(\theta) -

    -     - -

    - x^4+x^2y^2x^2-y^2=0 -

    -     - -     - - - -

    - \ds r=\cot\theta -

    -     - -

    - y^4+x^2y^2-x^2=0 -

    -     -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("x^2+y^2=4", limits=>[[-3,3],[-3,3]]); - - -

    - Convert the polar equation to a rectangular equation. -

    - -

    - r=2 -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - $eq=ImplicitEquation("y=x/sqrt(3)", limits=>[[-3,3],[-3,3]]); - - -

    - Convert the polar equation to a rectangular equation. -

    - -

    - \theta=\pi/6 -

    - -

    - -

    -     -     -     - -     - - - -

    - Convert the rectangular equation to a polar equation. -

    -     - - - - - Context("ImplicitEquation"); - Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); - $eq=ImplicitEquation("theta=pi/4", limits=>[[0,3],[0,3]]); - - -

    - Convert the rectangular equation to a polar equation. - Type theta for \theta. -

    - -

    - y=x -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); - $eq=ImplicitEquation("r=7/(sin(theta) -4cos(theta) )", limits=>[[1.5,3],[pi()/2,3*pi()/2]]); - - -

    - Convert the rectangular equation to a polar equation. - Type theta for \theta. -

    - -

    - y=4x+7 -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); - $eq=ImplicitEquation("r=5sec(theta)", limits=>[[4,10],[-pi()/4,pi()/4]]); - - -

    - Convert the rectangular equation to a polar equation. - Type theta for \theta. -

    - -

    - x=5 -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); - $eq=ImplicitEquation("r=5csc(theta)", limits=>[[4,10],[pi()/4,3*pi()/4]]); - - -

    - Convert the rectangular equation to a polar equation. - Type theta for \theta. -

    - -

    - y=5 -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); - $eq=ImplicitEquation("r=cos(theta)/sin^2(theta)", limits=>[[0,4],[-pi()/2,pi()/2]]); - - -

    - Convert the rectangular equation to a polar equation. - Type theta for \theta. -

    - -

    - x=y^2 -

    - -

    - -

    -     -     -     - - - - -

    - \ds x^2y=1 -

    -     - -

    - r=1/\sqrt[3]{\cos^2(\theta) \sin(\theta) } -

    -     - -     - - - - - Context("ImplicitEquation"); - Context()->variables->are(r=>'Real',theta=>['Real',TeX=>'\theta']); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $eq=ImplicitEquation("r=sqrt(7)", limits=>[[0,4],[0,2*pi()]]); - - -

    - Convert the rectangular equation to a polar equation. - Type theta for \theta. -

    - -

    - x^2+y^2=7 -

    - -

    - -

    -     -     -     - - - - -

    - \ds (x+1)^2+y^2=1 -

    -     - -

    - r=-2\cos(\theta) -

    -     - -     - -     - - - -

    - Find the points of intersection of the polar graphs. -

    -     - - - - - Context("Point"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->operators->add('P'=>{precedence=>6,associativity=>'left',type=>'unary',string=>"P",class=>'Parser::UOP::plus'}); - $rts=Compute("P(sqrt(3)/2,pi/6), P(0,pi/2), P(-sqrt(3)/2,5pi/6)"); - $rtsev=$rts->cmp(showCoordinateHints => 0, checker => sub { - my ($correct,$student,$ansHash) = @_; - my ($sr,$st) = $student->value; - my ($cr,$ct) = $correct->value; - return (($sr*cos($st) == $cr*cos($ct)) and ($sr*sin($st) == $cr*sin($ct))); - }); - - -

    - Find the points where r=\sin(2\theta) intersects r=\cos(\theta) on [0,\pi], - expressed in polar coordinates with notation P(r,\theta). -

    - -

    - -

    -     -     -     - - - - -

    - r=\cos(2\theta) and r=\cos(\theta) on [0,\pi] -

    -     - -

    - P(1,0), P(0,\pi/2)=P(0,\pi/4), P(-1/2,2\pi/3) -

    -     - -     - - - - - Context("Point"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->operators->add('P'=>{precedence=>6,associativity=>'left',type=>'unary',string=>"P",class=>'Parser::UOP::plus'}); - $rts=Compute("P(0,0), P(sqrt(2),pi/4)"); - $rtsev=$rts->cmp(showCoordinateHints => 0, checker => sub { - my ($correct,$student,$ansHash) = @_; - my ($sr,$st) = $student->value; - my ($cr,$ct) = $correct->value; - return (($sr*cos($st) == $cr*cos($ct)) and ($sr*sin($st) == $cr*sin($ct))); - }); - - -

    - Find the points where r=2\cos(\theta) intersects r=2\sin(\theta) on [0,\pi], - expressed in polar coordinates with notation P(r,\theta). -

    - -

    - -

    -     -     -     - - - - -

    - r=\sin(\theta) and r=\sqrt{3}+3\sin(\theta) on [0,2\pi] -

    -     - -

    - P(\sqrt{3}/2,\pi/3)=P(-\sqrt{3}/2,4\pi/3), - P(\sqrt{3}/2,2\pi/3)=P(-\sqrt{3}/2,5\pi/3), P(0,\pi/2) -

    -     - -     - - - - -

    - r=\sin(3\theta) and r=\cos(3\theta) on [0,\pi] -

    -     - -

    - P(0,0)=P(0,\pi/6), - P(\sqrt{2}/2,\pi/12), - P(-\sqrt{2}/2,5\pi/12), - P(\sqrt{2}/2,3\pi/4) -

    -     - -     - - - - - Context("Point"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->operators->add('P'=>{precedence=>6,associativity=>'left',type=>'unary',string=>"P",class=>'Parser::UOP::plus'}); - $rts=Compute("P(3/2,pi/3), P(3/2,-pi/3), P(0,pi)"); - $rtsev=$rts->cmp(showCoordinateHints => 0, checker => sub { - my ($correct,$student,$ansHash) = @_; - my ($sr,$st) = $student->value; - my ($cr,$ct) = $correct->value; - return (($sr*cos($st) == $cr*cos($ct)) and ($sr*sin($st) == $cr*sin($ct))); - }); - - -

    - Find the points where r=3\cos(\theta) intersects - r=1+\cos(\theta) on [-\pi,\pi], - expressed in polar coordinates with notation P(r,\theta). -

    - -

    - -

    -     -     -     - - - - -

    - r=1 and r=2\sin(2\theta) on [0,2\pi] -

    -     - -

    - For all points, r=1; - \theta = \pi/12,\,5\pi/12,\,7\pi/12,\,11\pi/12,\,13\pi/12,\,17\pi/12,\,19\pi/12,\,23\pi/12. -

    -     - -     - - - - -

    - r=1-\cos(\theta) and r=1+\sin(\theta) on [0,2\pi] -

    -     - -

    - P(0,0)=P(0,3\pi/2), - P(1+\sqrt{2}/2,3\pi/4), - P(1-\sqrt{2}/2,7\pi/4) -

    -     - -     - -     - - - - -

    - Pick a integer value for n, where n\neq 2,3, - and use technology to plot - \ds r=\sin\left(\frac mn\theta\right) for three different integer values of m. - Sketch these and determine a minimal interval on which the entire graph is shown. -

    -     - -

    - Answers will vary. - If m and n do not have any common factors, - then an interval of 2n\pi is needed to sketch the entire graph. -

    -     - -     - - - - -

    - Create your own polar function, - r=f(\theta) and sketch it. - Describe why the graph looks as it does. -

    -     - -

    - Answers will vary. -

    -     - -     -     -     -
    -
    - Calculus and Polar Functions - - -

    - The previous section defined polar coordinates, - leading to polar functions. - We investigated plotting these functions and solving a fundamental question about their graphs, namely, - where do two polar graphs intersect? -

    - -

    - We now turn our attention to answering other questions, - whose solutions require the use of calculus. - A basis for much of what is done in this section is the ability to turn a polar function - r=f(\theta) into a set of parametric equations. - Using the identities x=r\cos(\theta) and y=r\sin(\theta), - we can create the parametric equations x=f(\theta)\cos(\theta), - y=f(\theta)\sin(\theta) and apply the concepts of . -

    -     - - - Polar Functions and <m>dy/dx</m> -

    - We are interested in the lines tangent to a given graph, - regardless of whether that graph is produced by rectangular, - parametric, or polar equations. - In each of these contexts, - the slope of the tangent line is \frac{dy}{dx}. - Given r=f(\theta), - we are generally not concerned with r\,'=\fp(\theta); - that describes how fast r changes with respect to \theta. - Instead, we will use x=f(\theta)\cos(\theta), - y=f(\theta)\sin(\theta) to compute \frac{dy}{dx}. -

    - -

    - Using we have - - \frac{dy}{dx} = \frac{dy}{d\theta}\Big/\frac{dx}{d\theta} - . -

    - -

    - Each of the two derivatives on the right hand side of the equality requires the use of the Product Rule. - We state the important result as a Key Idea. -

    - - - Finding <m>\frac{dy}{dx}</m> with Polar Functions -

    - Let r=f(\theta) be a polar function. - With x=f(\theta)\cos(\theta) and y=f(\theta)\sin(\theta), - polarfunctions!finding \frac{dy}{dx} - tangent line - - \frac{dy}{dx} = \frac{\fp(\theta)\sin(\theta) +f(\theta)\cos(\theta) }{\fp(\theta)\cos(\theta) -f(\theta)\sin(\theta) } - . -

    -     - - - - - Finding <m>\frac{dy}{dx}</m> with polar functions - -

    - Consider the limaçon r=1+2\sin(\theta) on [0,2\pi]. -

    - -

    -

      -
    1. -

      - Find the equations of the tangent and normal lines to the graph at \theta=\pi/4. -

      -
      2. - -
    2. -

      - Find where the graph has vertical and horizontal tangent lines. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - We start by computing \frac{dy}{dx}. - With \fp(\theta) = 2\cos(\theta), we have - - \frac{dy}{dx} \amp = \frac{2\cos(\theta) \sin(\theta) + \cos(\theta) (1+2\sin(\theta) )}{2\cos^2(\theta) -\sin(\theta) (1+2\sin(\theta) )} - \amp = \frac{\cos(\theta) (4\sin(\theta) +1)}{2(\cos^2(\theta) -\sin^2(\theta) )-\sin(\theta) } - . - When \theta=\pi/4, \frac{dy}{dx}=-2\sqrt{2}-1 - (this requires a bit of simplification). - In rectangular coordinates, - the point on the graph at \theta=\pi/4 is (1+\sqrt{2}/2,1+\sqrt{2}/2). - Thus the rectangular equation of the line tangent to the limaçon at \theta=\pi/4 is - - y=(-2\sqrt{2}-1)\big(x-(1+\sqrt{2}/2)\big)+1+\sqrt{2}/2 \approx -3.83 x+8.24 - . - The limaçon and the tangent line are graphed in . - - The normal line has the opposite-reciprocal slope as the tangent line, so its equation is - - y \approx \frac{1}{3.83}x+1.26 - . -

      - -
      - The limaçon in with its tangent line at \theta=\pi/4 and points of vertical and horizontal tangency - - - A limaçon with an inner loops, symmetric about the y axis, and a tangent line. - -

      - The curve is a limaçon with an inner loop. - It is symmetric about the y axis, with intercepts on the positive y axis. - There are several marked points on the curve, indicating the location of horizontal and vertical tangent lines. -

      - -

      - Points marking horizontal tangent lines are at the top of both loops, - at the points (0,3) (outer loop) and (0,1) (inner loop). - There are also two points marking horizontal tangent lines at the bottom of the outer loop; - these lie below the x axis, on either side of the y axis. -

      - -

      - There are also four points marking vertical tangent lines. - These are on the left and right sides of each loop. -

      - -

      - Just above the point where the vertical tangent on the right side of the outer loop is marked, - a tangent line to the curve is drawn. The tangent line has a relatively large negative slope. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.35,ymax=3.4, - xmin=-2.1,xmax=2.1 - ] - - \addplot+ [domain=0:360,samples=101] ({cos(x)*(1+2*sin(x))},{sin(x)*(1+2*sin(x))}); - \addplot+ [solid,domain=0.1:2] ({x},{-3.82843*x+8.24264}); - - \filldraw (axis cs: 0,3) circle (2.4pt) - (axis cs: 0,1) circle (2.4pt) - (axis cs: -.493,-0.125) circle (2.4pt) - (axis cs: .483,-0.125) circle (2.4pt); - - \filldraw [fill=white,draw=black,thick] - (axis cs: 1.76011, 1.31048) circle (2.4pt) - (axis cs: -1.76011, 1.31048) circle (2.4pt) - (axis cs: 0.368977, 0.572639) circle (2.4pt) - (axis cs: -0.369009, 0.578743) circle (2.4pt); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
      -
      2. - -
    2. -

      - To find the horizontal lines of tangency, - we find where \frac{dy}{dx}=0; - thus we find where the numerator of our equation for \frac{dy}{dx} is 0. - - \cos(\theta) (4\sin(\theta) +1)=0 \Rightarrow \cos(\theta) =0 \text{ or } 4\sin(\theta) +1=0 - . - On [0,2\pi], - \cos(\theta) =0 when \theta=\pi/2,\,3\pi/2. - - Setting 4\sin(\theta) +1=0 gives \theta=\sin^{-1}(-1/4)\approx -0.2527 = -14.48^\circ. - We want the results in [0,2\pi]; - we also recognize there are two solutions, - one in the third quadrant and one in the fourth. - Using reference angles, the two solutions are - \theta =3.39 and 6.03 radians. - The four points we obtained where the limaçon has a horizontal tangent line are given in - with black-filled dots. - - To find the vertical lines of tangency, - we set the denominator of \frac{dy}{dx}=0. - - 2(\cos^2(\theta) -\sin^2(\theta) )-\sin(\theta) \amp = 0 . - Convert the \cos^2(\theta) term to 1-\sin^2(\theta): - 2(1-\sin^2(\theta) -\sin^2(\theta) )-\sin(\theta) \amp = 0 - 4\sin^2(\theta) + \sin(\theta) -2 \amp = 0. - Recognize this as a quadratic in the variable \sin(\theta). Using the quadratic formula, we have - \sin(\theta) \amp = \frac{-1\pm\sqrt{33}}{8} - . - We solve \sin(\theta) = \frac{-1+\sqrt{33}}8 and \sin(\theta) = \frac{-1-\sqrt{33}}8: - - \sin(\theta) \amp =\frac{-1+\sqrt{33}}8 \amp \sin(\theta) \amp = \frac{-1-\sqrt{33}}{8} - \theta \amp = \sin^{-1}\left(\frac{-1+\sqrt{33}}8\right) \amp \theta \amp = \sin^{-1}\left(\frac{-1-\sqrt{33}}8\right) - \theta \amp = 0.6349 \amp \theta \amp = -1.0030 - - In each of the solutions above, - we only get one of the possible two solutions as - \sin^{-1}(x) only returns solutions in [-\pi/2,\pi/2], - the 4th and 1st quadrants. - Again using reference angles, we have: - - \sin(\theta) = \frac{-1+\sqrt{33}}8 \Rightarrow \theta = 0.6349,\,2.5067 \text{ radians } - - and - - \sin(\theta) = \frac{-1-\sqrt{33}}8 \Rightarrow \theta = 4.1446,\,5.2802 \text{ radians. } - - These points are also shown in - with white-filled dots. -

      -
      3. -
    -

    -     - -     - - - -

    - When the graph of the polar function r=f(\theta) intersects the pole, - it means that f(\alpha)=0 for some angle \alpha. - Thus the formula for \frac{dy}{dx} in such instances is very simple, - reducing simply to - - \frac{dy}{dx} = \tan\alpha - . -

    - -

    - This equation makes an interesting point. - It tells us the slope of the tangent line at the pole is \tan\alpha; - some of our previous work (see, for instance, - ) shows us that the line through the pole with slope - \tan\alpha has polar equation \theta=\alpha. - Thus when a polar graph touches the pole at \theta=\alpha, - the equation of the tangent line at the pole is \theta=\alpha. -

    - - - Finding tangent lines at the pole - -

    - Let r=1+2\sin(\theta), a limaçon. - Find the equations of the lines tangent to the graph at the pole. -

    -     - -

    - We need to know when r=0. - - 1+2\sin(\theta) \amp = 0 - \sin(\theta) \amp = -1/2 - \theta \amp = \frac{7\pi}{6},\,\frac{11\pi}6 - . -

    - -

    - Thus the equations of the tangent lines, in polar, - are \theta = 7\pi/6 and \theta = 11\pi/6. - In rectangular form, the tangent lines are - y=\tan(7\pi/6)x and y=\tan(11\pi/6)x. - The full limaçon can be seen in ; - we zoom in on the tangent lines in . -

    - -
    - Graphing the tangent lines at the pole in - - - A zoomed in view of a limaçon near the origin, and two tangent lines at that point. - -

    - The curve is the same limaçon as in , but zoomed in near the origin. - The origin is a point of self-intersection for the curve, and there are two tangent lines: - one for the first time the curve passes through the origin, and one for the second. -

    - -

    - The two lines together make an X shape at the origin. - Due to the symmetry of the curve, one tangent line has positive slope, - and the other has a negative slope of the same absolute value. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=1.37, - xmin=-1.1,xmax=1.1 - ] - - \addplot+ [domain=0:360,samples=120] ({cos(x)*(1+2*sin(x))},{sin(x)*(1+2*sin(x))}); - \addplot [secondcurvestyle,solid,domain=-1:1] ({x},{.577*x}); - \addplot [secondcurvestyle,solid,domain=-1:1] ({x},{-.577*x}); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
    -     - -     -     - - - Area -

    - When using rectangular coordinates, - the equations x=h and y=k defined vertical and horizontal lines, - respectively, and combinations of these lines create rectangles - (hence the name rectangular coordinates). - It is then somewhat natural to use rectangles to approximate area as we did when learning about the definite integral. - polarfunctions!area -

    - -

    - When using polar coordinates, - the equations \theta=\alpha and r=c form lines through the origin and circles centered at the origin, - respectively, - and combinations of these curves form sectors of circles. - It is then somewhat natural to calculate the area of regions defined by polar functions by first approximating with sectors of circles. -

    - -

    - Consider - where a region defined by - r=f(\theta) on [\alpha,\beta] is given. - (Note how the sides of the region are the lines - \theta=\alpha and \theta=\beta, - whereas in rectangular coordinates the sides - of regions were often the vertical lines x=a and x=b.) -

    - - - -

    - Partition the interval [\alpha,\beta] into n equally spaced subintervals as \alpha = \theta_0 \lt \theta_1 \lt \cdots \lt \theta_{n}=\beta. - The length of each subinterval is \Delta\theta = (\beta-\alpha)/n, - representing a small change in angle. - The area of the region defined by the ith subinterval - [\theta_{i-1},\theta_{i}] can be approximated with a sector of a circle with radius f(c_i), - for some c_i in [\theta_{i-1},\theta_{i}]. - The area of this sector is \frac12f(c_i)^2\Delta\theta. - This is shown in , - where [\alpha,\beta] has been divided into 4 subintervals. - We approximate the area of the whole region by summing the areas of all sectors: - - \text{ Area } \approx \sum_{i=1}^n \frac12f(c_i)^2\Delta\theta - . -

    - -

    - This is a Riemann sum. - By taking the limit of the sum as n\to\infty, - we find the exact area of the region in the form of a definite integral. -

    - -
    - Computing the area of a polar region - -
    - - - - Plot of a generic polar function and the area it encloses between two angles. - -

    - The plot of some unknown polar curve r=f(\theta) is shown. - The curve lies entirely in the first quadrant, and has a somewhat wavy shape, - although the particular shape of the curve is unimportant. -

    - -

    - Two rays labeled \theta=\alpha and \theta=\beta are drawn, - and the area bounded by the two rays and the polar curve is shaded. - Several rays corresponding to angles between \alpha and \beta - are also shown as dashed lines. -

    -     - - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.1 - ] - - \addplot+ [areastyle,domain=18:72] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}) -- (axis cs:0,0) -- cycle; - \addplot [firstcurvestyle,domain=0:90,samples=50] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}); - - \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs: 0.905831, 0.294322) node [pos=.7,below,rotate=16,black] { $\theta=\alpha$}; - \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs:0.313792, 0.965751) node [pos=.7,above,rotate=72,black] { $\theta=\beta$}; - - \draw [thick,firstcolor,dashed] (axis cs:0,0) -- (axis cs: 0.862592, 0.528597) - (axis cs:0,0) -- (axis cs: 0.732107, 0.732107) - (axis cs:0,0) -- (axis cs: 0.497095, 0.811186); - - \draw (axis cs:.8,.85) node { $r=f(\theta)$}; - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
    - -
    - - - - Plot of a region bounded by a polar curve, and its approximation by cicular wedges. - -

    - A curve r=f(\theta) is shown; it is the same curve as . - The rays \theta=\alpha and \theta=\beta are also shown, - as well as the shaded region bounded by these rays and the polar curve. -

    - -

    - Overlaid on the shaded region are four circular wedges. - The angle between \theta=\alpha and \theta=\beta is divided into four smaller angles; - these correspond to the rays shown as dashed lines in . - Each wedge is a sector of a circle that spans one of these angles, - with radius corresponding some the value of f(\theta). -

    - -

    - Overall, the image illustrates the fact that the area bounded by the polar curve and the two rays - is approximated by the sum of the areas of the four circular wedges. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=18:72] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}) -- (axis cs:0,0) -- cycle; - - \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs:0.313792, 0.965751) node [pos=.7,above,rotate=72,black] { $\theta=\beta$}; - - \addplot [fill=white,domain=18:31.5] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; - \addplot [fill=white,domain=31.5:45] ({1.05*cos(x)},{1.05*sin(x)}) -- (axis cs:0,0) -- cycle; - \addplot [fill=white,domain=45:58.5] ({1.0*cos(x)},{1*sin(x)}) -- (axis cs:0,0) -- cycle; - \addplot [fill=white,domain=58.5:72] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; - - \addplot [secondcurvestyle,areastyle,domain=18:31.5] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; - \addplot [secondcurvestyle,areastyle,domain=31.5:45] ({1.05*cos(x)},{1.05*sin(x)}) -- (axis cs:0,0) -- cycle; - \addplot [secondcurvestyle,areastyle,domain=45:58.5] ({1.0*cos(x)},{1*sin(x)}) -- (axis cs:0,0) -- cycle; - \addplot [secondcurvestyle,areastyle,domain=58.5:72] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; - - \addplot [secondcurvestyle,solid,domain=18:31.5] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; - \addplot [secondcurvestyle,solid,domain=31.5:45] ({1.05*cos(x)},{1.05*sin(x)}) -- (axis cs:0,0) -- cycle; - \addplot [secondcurvestyle,solid,domain=45:58.5] ({1.0*cos(x)},{1*sin(x)}) -- (axis cs:0,0) -- cycle; - \addplot [secondcurvestyle,solid,domain=58.5:72] ({.96*cos(x)},{.96*sin(x)}) -- (axis cs:0,0) -- cycle; - - \draw (axis cs:.8,.85) node { $r=f(\theta)$}; - - \draw [thick,secondcolor] (axis cs:0,0) -- (axis cs: 0.905831, 0.294322) node [pos=.7,below,rotate=16,black] { $\theta=\alpha$}; - - \addplot [firstcurvestyle,domain=0:90,samples=50] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
    -     - -
    - - - - - Area of a Polar Region - -

    - Let f be continuous and non-negative on [\alpha,\beta], - where 0\leq \beta-\alpha\leq 2\pi. - The area A of the region bounded by the curve r=f(\theta) and the lines - \theta=\alpha and \theta=\beta is - - A \,=\, \frac12\int_\alpha^\beta f(\theta)^2 \, d\theta\, =\, \frac12\int_\alpha^\beta r^{\,2} \, d\theta - -

    -     -     - -

    - The theorem states that 0\leq \beta-\alpha\leq 2\pi. - This ensures that region does not overlap itself, - which would give a result that does not correspond directly to the area. -

    - - - Area of a polar region - -

    - Find the area of the circle defined by r=\cos(\theta). - (Recall this circle has radius 1/2.) -

    -     - -

    - This is a direct application of . - The circle is traced out on [0,\pi], leading to the integral - - \text{ Area } \amp = \frac12\int_0^\pi \cos^2(\theta)\, d\theta - \amp = \frac12\int_0^\pi \frac{1+\cos(2\theta)}{2}\, d\theta - \amp = \frac14\big(\theta +\frac12\sin(2\theta)\big)\Bigg|_0^\pi - \amp = \frac14\pi - . -

    - -

    - Of course, we already knew the area of a circle with radius 1/2. - We did this example to demonstrate that the area formula is correct. -

    -     - -     - - - - - Area of a polar region - -

    - Find the area of the cardioid r=1+\cos(\theta) bound between - \theta=\pi/6 and \theta=\pi/3, - as shown in . -

    - -
    - Finding the area of the shaded region of a cardioid in - - - A cardioid curve, within which is a shaded region bounded by the curve and two rays. - -

    - The polar curve r=1+\cos(\theta) is a cardioid that is symmetric about the x axis, - with a cusp at the origin, and second x intercept at (2,0). -

    - -

    - The rays \theta=\pi/6 and \theta=\pi/3 are drawn from the origin to where they meet the cardioid. - The region bounded by the cardioid and the two rays is shaded. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ytick={1}, - ymin=-.7,ymax=1.5, - xmin=-.5,xmax=2.1 - ] - - \addplot [firstcurvestyle,areastyle,domain=30:60] ({cos(x)*(1+cos(x))},{sin(x)*(1+cos(x))}) -- (axis cs:0,0) -- cycle; - \addplot [firstcurvestyle,domain=0:360,samples=101] ({cos(x)*(1+cos(x))},{sin(x)*(1+cos(x))}); - - \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs: 1.61603, 0.933013) node [pos=.6,below,rotate=30,black] { $\theta=\pi/6$}; - \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs:0.75, 1.29904) node [pos=.6,above,rotate=60,black] { $\theta=\pi/3$}; - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
    -     - -

    - This is again a direct application of . - - \text{ Area } \amp = \frac12\int_{\pi/6}^{\pi/3} (1+\cos(\theta) )^2\, d\theta - \amp = \frac12\int_{\pi/6}^{\pi/3} (1+2\cos(\theta) +\cos^2(\theta) )\, d\theta - \amp = \frac12\left(\theta+2\sin(\theta) +\frac12\theta+\frac14\sin(2\theta)\right)\Bigg|_{\pi/6}^{\pi/3} - \amp = \frac18\big(\pi+4\sqrt{3}-4\big) \approx 0.7587 - . -

    -     - -     - - - Area Between Curves - -

    - Our study of area in the context of rectangular functions led naturally to finding area bounded between curves. - We consider the same in the context of polar functions. - polarfunctions!area between curves -

    - -

    - Consider the shaded region shown in . - We can find the area of this region by computing the area bounded by - r_2=f_2(\theta) and subtracting the area bounded by - r_1=f_1(\theta) on [\alpha,\beta]. - Thus - - \text{ Area } \,= \,\frac12\int_\alpha^\beta r_2^{\,2}\, d\theta - \frac12\int_\alpha^\beta r_1^{\,2}\, d\theta = \frac12\int_\alpha^\beta \big(r_2^{\,2}-r_1^{\,2}\big)\, d\theta - . -

    - -
    - Illustrating area bound between two polar curves - - - Illustration of a region bounded by two polar curves and two rays. - -

    - Polar curves r=f_1(\theta) and r=f_2(\theta) are drawn in the first quadrant. - The functions are not specified, but the curves are intended to appear as generic polar curves: - both begin on the positive x axis and end on the positive y axis, - and both are somewhat wavy, with a value of r that oscillates on the interval [0,\pi/2]. -

    - -

    - The curve r=f_1(\theta) lies closer to the origin than r=f_2(\theta) at all points. - Two rays \theta=\alpha and \theta=\beta are shown. - The region bounded by the two rays and the two polar curves is shaded. - (The region corresponds to \alpha\leq\theta\leq \beta and f_1(\theta)\leq r\leq f_2(\theta).) -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=1.2, - xmin=-.1,xmax=1.2 - ] - - \addplot [firstcurvestyle,areastyle] coordinates { - (0.866,0.5)(0.8626,0.5286)(0.8577,0.557) - (0.8509,0.5848)(0.8417,0.6116)(0.83,0.6369) - (0.8155,0.6604)(0.7983,0.6818)(0.7785,0.7009) - (0.7563,0.7177)(0.7321,0.7321)(0.7063,0.7443) - (0.6795,0.7546)(0.652,0.7634)(0.6244,0.7711) - (0.5971,0.7782)(0.5705,0.7852)(0.5449,0.7928) - (0.5204,0.8013)(0.4971,0.8112)(0.475,0.8227) - (0.35,0.6062)(0.3617,0.5902)(0.3728,0.5741) - (0.3836,0.5582)(0.3942,0.5425)(0.4046,0.5273) - (0.4151,0.5126)(0.4257,0.4984)(0.4366,0.4849) - (0.4479,0.4719)(0.4596,0.4596)(0.4719,0.4479) - (0.4849,0.4366)(0.4984,0.4257)(0.5126,0.4151) - (0.5273,0.4046)(0.5425,0.3942)(0.5582,0.3836) - (0.5741,0.3728)(0.5902,0.3617)(0.6062,0.35)}; - - \addplot [firstcurvestyle,domain=0:90,samples=60] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}); - \addplot [secondcurvestyle,solid,domain=0:90,samples=40] ({cos(x)*(.7+.05*sin(6*x))},{sin(x)*(.7+.05*sin(6*x))}); - - \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs: 0.866025, 0.5) node [pos=.4,below,rotate=30,black] { $\theta=\alpha$}; - \draw [thick,firstcolor] (axis cs:0,0) -- (axis cs:0.475, 0.822724) node [pos=.4,above,rotate=60,black] { $\theta=\beta$}; - - \draw (axis cs:.8,.85) node { $r_2=f_2(\theta)$} - (axis cs:.2,.8) node { $r_1=f_1(\theta)$}; - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
    - - - Area Between Polar Curves -

    - The area A of the region bounded by - r_1=f_1(\theta) and r_2=f_2(\theta), - \theta=\alpha and \theta=\beta, - where f_1(\theta)\leq f_2(\theta) on [\alpha,\beta], is - - A = \frac12\int_\alpha^\beta \big(r_2^{\,2}-r_1^{\,2}\big)\, d\theta - . -

    -     - - - - - Area between polar curves - -

    - Find the area bounded between the curves - r=1+\cos(\theta) and r=3\cos(\theta), - as shown in . -

    - -
    - Finding the area between polar curves in - - - A circle and a cardioid enclose a region that is inside the circle but outside the cardioid. - -

    - Two polar curves are plotted. The first is a circle with center at (3/2,0) and radius 3/2. - (The circle intercepts the x axis at (0,0) and (3,0) and is symmetric about the x axis.) - The second curve is a cardioid; it is symmetric about the x axis, - with its cusp at the origin, and a second x intercept at (2,0). -

    - -

    - The two curves intersect at two points on opposite sides of the x axis. - The cardioid covers up a significant portion of the circle, for x between 0 and 2. - The portion of the circle that is not covered by the cardioid is shaded. - This is the region that lies outside of the cardioid, but inside the circle. - It has a shape similar to that of a crescent moon. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-1.6,ymax=1.6, - xmin=-.6,xmax=3.3 - ] - - \addplot [firstcurvestyle,areastyle] coordinates { - (0.75,-1.299)(0.8719,-1.293)(0.9947,-1.273)(1.117,-1.24) - (1.237,-1.194)(1.353,-1.135)(1.464,-1.063)(1.567,-0.9793) - (1.663,-0.884)(1.748,-0.7783)(1.823,-0.6634)(1.885,-0.5406) - (1.935,-0.4113)(1.971,-0.277)(1.993,-0.1393)(2.,0)(1.993,0.1393) - (1.971,0.277)(1.935,0.4113)(1.885,0.5406)(1.823,0.6634) - (1.748,0.7783)(1.663,0.884)(1.567,0.9793)(1.464,1.063) - (1.353,1.135)(1.237,1.194)(1.117,1.24)(0.9947,1.273) - (0.8719,1.293)(0.75,1.299)(0.75,1.299)(0.9381,1.391) - (1.137,1.455)(1.343,1.492)(1.552,1.499)(1.76,1.477) - (1.964,1.427)(2.158,1.348)(2.339,1.244)(2.504,1.115) - (2.649,0.9642)(2.772,0.7949)(2.87,0.6101)(2.942,0.4135) - (2.985,0.2088)(3.,0)(2.985,-0.2088)(2.942,-0.4135) - (2.87,-0.6101)(2.772,-0.7949)(2.649,-0.9642)(2.504,-1.115) - (2.339,-1.244)(2.158,-1.348)(1.964,-1.427)(1.76,-1.477) - (1.552,-1.499)(1.343,-1.492)(1.137,-1.455)(0.9381,-1.391)(0.75,-1.299)}; - - \addplot [firstcurvestyle,domain=0:180,samples=70] ({cos(x)*(3*cos(x))},{sin(x)*(3*cos(x))}); - \addplot [secondcurvestyle,solid,domain=0:360,samples=80] ({cos(x)*(1+cos(x))},{sin(x)*(1+cos(x))}); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
    -     - -

    - We need to find the points of intersection between these two functions. - Setting them equal to each other, we find: - - 1+\cos(\theta) \amp = 3\cos(\theta) - \cos(\theta) \amp =1/2 - \theta \amp = \pm \pi/3 - -

    - -

    - Thus we integrate \frac12\big((3\cos(\theta) )^2-(1+\cos(\theta) )^2\big) on [-\pi/3,\pi/3]. - - \text{ Area } \amp = \frac12\int_{-\pi/3}^{\pi/3} \big((3\cos(\theta) )^2-(1+\cos(\theta) )^2\big)\, d\theta - \amp = \frac12\int_{-\pi/3}^{\pi/3} \big( 8\cos^2(\theta) -2\cos(\theta) -1\big)\, d\theta - \amp = \frac12\big(2\sin(2\theta) - 2\sin(\theta) +3\theta\big)\Bigg|_{-\pi/3}^{\pi/3} - \amp = \pi - . -

    - -

    - Amazingly enough, - the area between these curves has a nice value. -

    -     - -     - - - Area defined by polar curves - -

    - Find the area bounded between the polar curves r=1 and r=2\cos(2\theta), - as shown in . -

    -
    - The region bounded by the functions in - - A zoomed in view of a region bounded by a circle, a rose curve, and the x axis. - -

    - Two polar curves are plotted: a four-leaf rose curve, and a the unit circle. - The image is zoomed in to give a better view of the region whose area is being computed in this example. - One leaf of the rose curve is along the positive x axis. - It intersects the unit circle at a point in the first quadrant. -

    - -

    - A shaded region is also shown. The region is bounded below by the x axis. - To the left of the point where the rose curve intersects the circle, - the region is bounded above by the rose curve. - To the right of this point, until the circle meets the x axis at (1,0), - the region is bounded between the x axis and the circle. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-1.15,ymax=1.15, - xmin=-.5,xmax=2.2 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(1.,0)(0.9986,0.05234)(0.9945,0.1045)(0.9877,0.1564)(0.9781,0.2079)(0.9659,0.2588)(0.9511,0.309)(0.9336,0.3584)(0.9135,0.4067)(0.891,0.454)(0.866,0.5)(0.7742,0.4744)(0.6822,0.443)(0.5907,0.406)(0.5,0.3633)(0.4107,0.3151)(0.3232,0.2617)(0.2379,0.2032)(0.1554,0.1399)(0.07593,0.07205)(0,0)}; - - \addplot [firstcurvestyle,domain=0:360,samples=180] ({cos(x)*(2*cos(2*x))},{sin(x)*(2*cos(2*x))}); - \addplot [secondcurvestyle,solid,domain=0:360,samples=70] ({cos(x)*(1)},{sin(x)*(1)}); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - -
    -     - -

    - We need to find the point of intersection between the two curves. - Setting the two functions equal to each other, we have - - 2\cos(2\theta) = 1 \Rightarrow \cos(2\theta) = \frac12 \Rightarrow 2\theta = \pi/3 \Rightarrow \theta=\pi/6 - . -

    - -
    - Breaking the region bounded by the functions in into its component parts - - A zoomed in view of a polar region, showing it divided into two parts. - -

    - The region in is shown again, but zoomed in even further. - The point at which the rose curve intersects the circle is found to have polar coordinates (1,\pi/6). - The ray \theta=\pi/6 is shown as a dashed line; it divides the region into two parts. -

    - -

    - The first part corresponds to 0\leq \theta\leq \pi/6; - it is the part of the region that lies below the ray \theta=\pi/6. - It is bounded by the x axis, the unit circle, and the ray \theta=\pi/6. - The second part of the region lies above the ray \theta=\pi/6, - corresponding to angles \pi/6\leq \theta\leq \pi/4. - This part of the region lies between the rose curve and the ray \theta=\pi/6. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1, - xmin=-.1,xmax=1.2 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(0.866,0.5)(0.7742,0.4744)(0.6822,0.443)(0.5907,0.406)(0.5,0.3633)(0.4107,0.3151)(0.3232,0.2617)(0.2379,0.2032)(0.1554,0.1399)(0.07593,0.07205)(0,0)}; - \addplot [secondcurvestyle,areastyle] coordinates {(1.,0)(0.9986,0.05234)(0.9945,0.1045)(0.9877,0.1564)(0.9781,0.2079)(0.9659,0.2588)(0.9511,0.309)(0.9336,0.3584)(0.9135,0.4067)(0.891,0.454)(0.866,0.5)(0,0)}; - - \addplot [firstcurvestyle,domain=0:360,samples=230] ({cos(x)*(2*cos(2*x))},{sin(x)*(2*cos(2*x))}); - \addplot [secondcurvestyle,solid,domain=0:360,samples=80] ({cos(x)*(1)},{sin(x)*(1)}); - - \draw [thick,dashed] (axis cs: 0,0) -- (axis cs:.866,.5); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - -
    - -

    - In , - we zoom in on the region and note that it is not really bounded - between two polar curves, - but rather by two polar curves, - along with \theta=0. - The dashed line breaks the region into its component parts. - Below the dashed line, the region is defined by r=1, - \theta=0 and \theta = \pi/6. - (Note: the dashed line lies on the line \theta=\pi/6.) - Above the dashed line the region is bounded by - r=2\cos(2\theta) and \theta =\pi/6. - Since we have two separate regions, - we find the area using two separate integrals. -

    - -

    - Call the area below the dashed line A_1 and the area above the dashed line A_2. - They are determined by the following integrals: - - A_1 = \frac12\int_0^{\pi/6} (1)^2\, d\theta\qquad A_2 = \frac12\int_{\pi/6}^{\pi/4} \big(2\cos(2\theta)\big)^2\, d\theta - . -

    - -

    - (The upper bound of the integral computing A_2 is \pi/4 as - r=2\cos(2\theta) is at the pole when \theta=\pi/4.) -

    - -

    - We omit the integration details and let the reader verify that - A_1 = \pi/12 and A_2 = \pi/12-\sqrt{3}/8; - the total area is A = \pi/6-\sqrt{3}/8. -

    -     - -     -     -     - - - Arc Length -

    - As we have already considered the arc length of curves defined by rectangular and parametric equations, - we now consider it in the context of polar equations. - Recall that the arc length L of the graph defined by the parametric equations x=f(t), - y=g(t) on [a,b] is - arc length - polarfunction!arc length - - L = \int_a^b \sqrt{\fp(t)^2 + \gp(t)^2}\, dt = \int_a^b \sqrt{x'(t)^2+\yp(t)^2}\, dt - . -

    - -

    - Now consider the polar function r=f(\theta). - We again use the identities x=f(\theta)\cos(\theta) and - y=f(\theta)\sin(\theta) to create parametric equations based on the polar function. - We compute x'(\theta) and - \yp(\theta) as done before when computing \frac{dy}{dx}, - then apply Equation. -

    - -

    - The expression x'(\theta)^2+\yp(\theta)^2 can be simplified a great deal; - we leave this as an exercise and state that - - x'(\theta)^2+\yp(\theta)^2 = \fp(\theta)^2+f(\theta)^2 - . -

    - -

    - This leads us to the arc length formula. -

    - - - Arc Length of Polar Curves - -

    - Let r=f(\theta) be a polar function with \fp continuous on [\alpha,\beta], - on which the graph traces itself only once. - The arc length L of the graph on [\alpha,\beta] is - - L = \int_\alpha^\beta \sqrt{\fp(\theta)^2+f(\theta)^2}\, d\theta = \int_\alpha^\beta\sqrt{(r\,')^2+ r^2}\, d\theta - . -

    -     -     - - - Arc length of a limaçon - -

    - Find the arc length of the limaçon r=1+2\sin(\theta). -

    -     - -

    - With r=1+2\sin(\theta), we have r\,' = 2\cos(\theta). - The limaçon is traced out once on [0,2\pi], - giving us our bounds of integration. - Applying , we have - - L \amp = \int_0^{2\pi} \sqrt{(2\cos(\theta))^2+(1+2\sin(\theta))^2}\, d\theta - \amp = \int_0^{2\pi} \sqrt{4\cos^2\theta+4\sin^2\theta +4\sin(\theta)+1}\, d\theta - \amp = \int_0^{2\pi} \sqrt{4\sin(\theta)+5}\, d\theta - \amp \approx 13.3649 - . -

    - -
    - The limaçon in whose arc length is measured - - - A limaçon with an inner loop that is symmetric about the y axis. - -

    - The limaçon r=1+2\sin(\theta) has an inner loop, - and is symmetric about the y axis. - The point of self-intersection is at the origin; - the inner loop meets the y axis again at (0,1), - and the outer loop meets the y axis again at (0,3). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=3.3, - xmin=-2.22,xmax=2.22 - ] - - \addplot+ [domain=0:360,samples=101] ({cos(x)*(1+2*sin(x))},{sin(x)*(1+2*sin(x))}); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
    - -

    - The final integral cannot be solved in terms of elementary functions, - so we resorted to a numerical approximation. - (Simpson's Rule, with n=4, - approximates the value with 13.0608. - Using n=22 gives the value above, - which is accurate to 4 places after the decimal.) -

    -     - -     -     - - - Surface Area -

    - The formula for arc length leads us to a formula for surface area. - The following Theorem is based on . -

    - - - Surface Area of a Solid of Revolution - -

    - Consider the graph of the polar equation r=f(\theta), - where \fp is continuous on [\alpha,\beta], - on which the graph does not cross itself. - surface areasolid of revolution - polarfunction!surface area - integrationsurface area -

    - -

    -

      -
    1. -

      - The surface area of the solid formed by revolving the graph about the initial ray (\theta=0) is: - - \text{ Surface Area } = 2\pi\int_\alpha^\beta f(\theta)\sin(\theta)\sqrt{\fp(\theta)^2+f(\theta)^2}\, d\theta - . -

      -
      2. - -
    2. -

      - The surface area of the solid formed by revolving the graph about the line \theta=\pi/2 is: - - \text{ Surface Area } = 2\pi\int_\alpha^\beta f(\theta)\cos(\theta)\sqrt{\fp(\theta)^2+f(\theta)^2}\, d\theta - . -

      -
      3. -
    -

    -     -     - - - Surface area determined by a polar curve - -

    - Find the surface area formed by revolving one petal of the rose curve - r=\cos(2\theta) about its central axis, - as shown in . -

    -
    - Finding the surface area of a rose-curve petal that is revolved around its central axis - -
    - - - - A four leaf rose curve, with leaves along the coordinate axes. - -

    - The four leaf rose curve r=\cos(2\theta) is plotted. - Although the entire curve is plotted, the portion of the right-hand leaf - that lies in the first quadrant is highlighted. - This is the part of the curve corresponding to 0\leq \theta\leq \pi/4; - it begins at (1,0) and ends at the origin. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,1}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.3,xmax=1.3 - ] - - \addplot+ [domain=0:45,samples=30] ({cos(x)*(cos(2*x))},{sin(x)*(cos(2*x))}); - \addplot+ [solid,domain=45:360,samples=120] ({cos(x)*(cos(2*x))},{sin(x)*(cos(2*x))}); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - -
    - - - - -
    - - - The surface of revolution obtained by revolving one leaf of a rose curve about the x axis. - -

    - A three-dimensional plot shows the surface of revolution obtained when the portion of the rose curve - from the previous image is revolved about the x axis. - The surface is similar in appearance to a teardrop, - or perhaps a blimp. -

    -     - - - - - //ASY file for figparcalc8_3D.asy in Chapter 9 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,-8,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,1.25); - pair ybounds=(-.25,.25); - pair zbounds=(-.25,.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice));//,Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - //label("$x$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$y$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - label("\ ",(0,0,1.5)); - label("\ ",(0,-1.5,0)); - - //surface r=cos(2 theta) revolved around x axis // ({cos(y)*(cos(2*y))},{(sin(y)*(cos(2*y)))*cos(x)},{(sin(y)*(cos(2*y)))*sin(x)}); - triple f(pair t) { - return (cos(t.y)*cos(2*t.y),sin(t.y)*cos(2*t.y)*cos(t.x),sin(t.y)*cos(2*t.y)*sin(t.x)); - } - surface s=surface(f,(0,0),(2pi,pi/4),8,16,Spline); - //Should use apexmeshpen for p but that was causing a core dump somehow - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //curve on the surfaces //({cos(x)*cos(2*x)},{0},{sin(x)*cos(2*x)}); - triple g(real t) {return (cos(t)*cos(2*t),0,sin(t)*cos(2*t));} - path3 mypath=graph(g,0,pi/4,operator ..);draw(mypath,bluepen+linewidth(2)); - - - -
    - -     -
    -     - -

    - We choose, as implied by the figure, - to revolve the portion of the curve that lies on - [0,\pi/4] about the initial ray. - Using - and the fact that \fp(\theta) = -2\sin(2\theta), we have - - \text{ Surface Area } \amp = 2\pi\int_0^{\pi/4} \cos(2\theta)\sin(\theta)\sqrt{\big(-2\sin(2\theta)\big)^2+\big(\cos(2\theta)\big)^2}\, d\theta - \amp \approx 1.36707 - . -

    - -

    - The integral is another that cannot be evaluated in terms of elementary functions. - Simpson's Rule, with n=4, - approximates the value at 1.36751. -

    -     -     - -

    - This chapter has been about curves in the plane. - While there is great mathematics to be discovered in the two dimensions of a plane, - we live in a three dimensional world and hence we should also look to do mathematics in 3D that is, - in space. - The next chapter begins our exploration into space by introducing the topic of vectors, - which are incredibly useful and powerful mathematical objects. -

    -     - - - - Terms and Concepts - - - -

    - Given polar equation r=f(\theta), - how can one create parametric equations of the same curve? -

    -     - - - -

    - Using x=r\cos(\theta) and y=r\sin(\theta), - we can write x=f(\theta)\cos(\theta), - y=f(\theta)\sin(\theta). -

    -     - -     - - - - -

    - With rectangular coordinates, - it is natural to approximate area with ; - with polar coordinates, it is natural to approximate area with . -

    -     - - - - - - - - - wedges|circular wedges|sectors of circles|sections of circles|circular sectors|circular sections - - - - -     -     - - Problems - - - -

    - Find \lz{y}{x} - (in terms of \theta). - Then find the equations of the tangent and normal lines to the curve at the indicated \theta-value. -

    -     - - - - - - - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $dydx=Formula("-cot(theta)"); - Context("Numeric")->variables->add(y=>'Real'); - parser::Assignment->Allow; - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $tan=Formula("y=-(x-sqrt(2)/2)+sqrt(2)/2"); - $nor=Formula("y=x"); - - -

    - r=1, \theta=\pi/4 -

    - - - Find \lz{y}{x} in terms of \theta. - Type theta for \theta. - -

    - -

    - - - Give the equation of the tangent line. - - -

    - -

    - - - Give the equation of the normal line. - - -

    - -

    -     -     -     - - - - - - - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $dydx=Formula("1/2(tan(theta)-cot(theta))"); - Context("Numeric")->variables->add(y=>'Real'); - parser::Assignment->Allow; - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $tan=Formula("y=1/2"); - $nor=Formula("x=1/2"); - - -

    - r=\cos(\theta), \theta=\pi/4 -

    - - - Find \lz{y}{x} in terms of \theta. - Type theta for \theta. - -

    - -

    - - - Give the equation of the tangent line. - - -

    - -

    - - - Give the equation of the normal line. - - -

    - -

    -     -     -     - - - - -

    - r=1+\sin(\theta), \theta = \pi/6 -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = \frac{\cos(\theta) (1+2\sin(\theta) )}{\cos^2(\theta) -\sin(\theta) (1+\sin(\theta) )} -

      -
      2. - -
    2. -

      - tangent line: x=3\sqrt{3}/4; - normal line: y=3/4 -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds r=1-3\cos(\theta), \theta = 3\pi/4 -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = \frac{3 \sin^2(t)+(1-3 \cos(t)) \cos(t)}{3 \sin(t) \cos(t)-\sin(t) (1-3 \cos(t))} -

      -
      2. - -
    2. -

      - tangent line: - y=\frac{1}{1+3\sqrt{2}}(x+(1/\sqrt{2}+3/2))+1/\sqrt{2}+3/2 \approx y=0.19(x+2.21)+2.21; - normal line: - y=-(1+3\sqrt{2})(x+(1/\sqrt{2}+3/2))+1/\sqrt{2}+3/2 -

      -
      3. -
    -

    -     - -     - - - - - - - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $dydx=Formula("(theta cos(theta) + sin(theta))/(cos(theta)-theta sin(theta))"); - Context("Numeric")->variables->add(y=>'Real'); - parser::Assignment->Allow; - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $tan=Formula("y=-2/pi x + pi/2"); - $nor=Formula("y=pi/2 x + pi/2"); - - -

    - r=\theta, \theta=\pi/2 -

    - - - Find \lz{y}{x} in terms of \theta. - Type theta for \theta. - -

    - -

    - - - Give the equation of the tangent line. - - -

    - -

    - - - Give the equation of the normal line. - - -

    - -

    -     -     -     - - - - - - - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $dydx=Formula("(cos(theta)cos(3theta)-3sin(theta)sin(3theta))/(-cos(3theta)sin(theta)-3cos(theta)sin(3theta))"); - Context("Numeric")->variables->add(y=>'Real'); - parser::Assignment->Allow; - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $tan=Formula("y=x/sqrt(3)"); - $nor=Formula("y=-sqrt(3)x"); - - -

    - r=\cos(3\theta), \theta=\pi/6 -

    - - - Find \lz{y}{x} in terms of \theta. - Type theta for \theta. - -

    - -

    - - - Give the equation of the tangent line. - - -

    - -

    - - - Give the equation of the normal line. - - -

    - -

    -     -     -     - - - - - - - Context()->variables->are(theta=>['Real',TeX=>'\theta']); - $dydx=Formula("(4sin(theta)cos(4theta)+sin(4theta)cos(theta))/(4cos(theta)cos(4theta)-sin(theta)sin(4theta))"); - Context("Numeric")->variables->add(y=>'Real'); - parser::Assignment->Allow; - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $tan=Formula("y=5sqrt(3)(x+sqrt(3)/4)-3/4"); - $nor=Formula("y=-1/(5sqrt(3))(x+sqrt(3)/4)-3/4"); - - -

    - r=\sin(4\theta), \theta=\pi/3 -

    - - - Find \lz{y}{x} in terms of \theta. - Type theta for \theta. - -

    - -

    - - - Give the equation of the tangent line. - - -

    - -

    - - - Give the equation of the normal line. - - -

    - -

    -     -     -     - - - - -

    - \ds r=\frac1{\sin(\theta) -\cos(\theta) };\theta = \pi -

    -     - -

    -

      -
    1. -

      - \frac{dy}{dx} = 1 -

      -
      2. - -
    2. -

      - tangent line: y=x+1; normal line: y=-x-1 -

      -
      3. -
    -

    -     - -     - -     - - - -

    - Find the values of \theta in the given interval where the graph of the polar function has horizontal and vertical tangent lines. -

    -     - - - - -

    - \ds r=3; [0,2\pi] -

    -     - -

    - horizontal: \theta=\pi/2,3\pi/2; -

    - -

    - vertical: \theta = 0,\pi,2\pi -

    -     - -     - - - - -

    - \ds r=2\sin(\theta); [0,\pi] -

    -     - -

    - horizontal: \theta=0,\pi/2,\pi; -

    - -

    - vertical: \theta = \pi/4,3\pi/4 -

    -     - -     - - - - -

    - \ds r=\cos(2\theta); [0,2\pi] -

    -     - -

    - horizontal: \theta=\tan^{-1}(1/\sqrt{5}),\,\pi/2,\,\pi-\tan^{-1}(1/\sqrt{5}),\,\pi+\tan^{-1}(1/\sqrt{5}),\,3\pi/2,\,2\pi-\tan^{-1}(1/\sqrt{5}); -

    - -

    - vertical: \theta = 0,\,\tan^{-1}(\sqrt{5}),\,\pi-\tan^{-1}(\sqrt{5}),\,\pi,\,\pi+\tan^{-1}(\sqrt{5}),\,2\pi-\tan^{-1}(\sqrt{5}) -

    -     - -     - - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); - $hor=List(Formula("pi/3"), Formula("pi"), Formula("5pi/3")); - $ver=List(Formula("0"), Formula("2pi/3"), Formula("4pi/3")); - - -

    - r=1+\cos(\theta); [0, 2\pi) -

    - - - Give the values of \theta where the tangent line is horizontal. - Enter your answer as a comma-separated list. - -

    - -

    - - - Give the values of \theta where the tangent line is vertical. - Enter your answer as a comma-separated list. - -

    - -

    -     - -

    - horizontal: \theta=\pi/3,\,5\pi/3; -

    - -

    - vertical: \theta = 0,\,2\pi/3,\,4\pi/3,\,2\pi -

    - -

    - At \theta=\pi, \frac{dy}{dx} = 0/0; - apply L'Hopital's Rule to find that - \frac{dy}{dx}\rightarrow 0 as \theta\rightarrow \pi. -

    -     -     -     - -     - - - -

    - Find the equation of the lines tangent to the graph at the pole. -

    -     - - - - -

    - \ds r=\sin(\theta);[0,\pi] -

    -     - -

    - In polar: \theta = 0 \,\cong \,\theta = \pi -

    - -

    - In rectangular: y=0 -

    -     - -     - - - - -

    - \ds r=\sin(3\theta);[0,\pi] -

    -     - -

    - In polar: \theta = 0,\,\theta = \pi/3,\,\theta = 2\pi/3. -

    - -

    - In rectangular: y=0, - y=\sqrt{3}x, and y= -\sqrt{3}x. -

    -     - -     - -     - - - -

    - Find the area of the described region. -

    -     - - - - -

    - Enclosed by the circle: \ds r=4\sin(\theta) -

    -     - -

    - area = 4\pi -

    -     - -     - - - - -

    - Enclosed by the circle \ds r=5 -

    -     - -

    - area = 25\pi -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); - $area=Formula("pi/12"); - - -

    - Find the area enclosed by one petal of r=\sin(3\theta). -

    - -

    - -

    -     -     -     - - - -

    - Enclosed by one petal of the rose curve r=\cos (n\,\theta), - where n is a positive integer. -

    -     - -

    - area = \pi/(4n) -

    -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); - $area=Formula("3pi/2"); - - -

    - Find the area enclosed by the cardioid r=1-\sin(\theta). -

    - -

    - -

    -     -     -     - - - - - -

    - Enclosed by the inner loop of the limaçon \ds r=1+2\cos(\theta) -

    -     - -

    - area = \pi-3\sqrt{3}/2 -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); - $area=Formula("2pi+3sqrt(3)/2"); - - - -

    - Find the area enclosed by the outer loop of the limaçon r=1+2\cos(\theta) - (including area enclosed by the inner loop). -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); - $area=Formula("pi+3sqrt(3)"); - - - -

    - Find the area enclosed between the inner and outer loop of the limaçon r=1+2\cos(\theta). -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); - $area=Formula("1"); - - -

    - Find the area enclosed by r=2\cos(\theta), - r=2\sin(\theta), and the x-axis, as shown: -

    - - - Two circles, one along each axis, bound a region in the first quadrant. - -

    - There are two circles of radius 1. - The first circle lies along the x axis, with intercepts at (0,0) and (2,0). - The second circle lies along the y axis, with intercepts at (0,0) and (0,2). - The two circles intersect in the first quadrant, at the point (1,1). -

    - -

    - The shaded region consists of everything that is between the x axis and the first circle, - except for the portion of the first circle that overlaps with the second circle. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.1,ymax=2.1,% - xmin=-1.4,xmax=2.4% - ] - - \addplot [firstcurvestyle,areastyle,domain=0:45,samples=20] ({cos(x)*2*cos(x)},{sin(x)*2*cos(x)}) -- (axis cs:0,0); - \addplot [fill=white, smooth,domain=0:50,samples=20] ({cos(x)*2*sin(x)},{sin(x)*2*sin(x)}); - \addplot [firstcurvestyle,domain=0:180,samples=60] ({cos(x)*2*cos(x)},{sin(x)*2*cos(x)}); - \addplot [secondcurvestyle,domain=0:180,samples=40] ({cos(x)*2*sin(x)},{sin(x)*2*sin(x)}); - - \end{axis} - \end{tikzpicture} - - - -

    - The area is . -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); - $area=Formula("1/32(4pi-3sqrt(3))"); - - -

    - Find the area enclosed by r=\cos(\theta) and r=\sin(2\theta), - as shown: -

    - - - A semi-circle and one leaf of a rose curve, both in the first quadrant, overlapping in a shaded area. - -

    - Two curves are shown. One is a semi-circular arc, from (0,0) to (1,0), in the first quadrant. - The other curve the leaf of the four-leaf rose r=\sin(2\theta) that lies in the first quadrant. -

    - -

    - The two curves overlap in a region that lies above the rose curve, but below the circle. - This region is shaded. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xtick={1}, - ytick={1}, - ymin=-.1,ymax=1.1,% - xmin=-.1,xmax=1.34% - ] - - \addplot [firstcurvestyle,areastyle,domain=90:30,samples=20] ({cos(x)*cos(x)},{sin(x)*cos(x)}) -- (axis cs:0,0); - \addplot [firstcurvestyle,areastyle,domain=0:30,samples=20] ({cos(x)*sin(2*x)},{sin(x)*sin(2*x)}) -- (axis cs:0,0); - \addplot [firstcurvestyle,domain=0:90,samples=60] ({cos(x)*cos(x)},{sin(x)*cos(x)}); - \addplot [secondcurvestyle,domain=0:90,samples=40] ({cos(x)*sin(2*x)},{sin(x)*sin(2*x)}); - - \end{axis} - \end{tikzpicture} - - - -

    - The area is . -

    -     -     -     - - - - -

    - Enclosed by r=\cos(3 \theta) and r=\sin(3\theta), as shown: -

    - - - Overlapping leaves of two different three-leaf rose curves. - -

    - Two curves are shown. The first is the leaf of the three-leaf rose curve r=\sin(3\theta) that lies in the first quadrant. - The second is the leaf of the three-leaf rose curve r=\cos(3\theta) that lies along the positive x axis. - The shaded region is the interior of the leaf of r=\sin(3\theta), - except for the part that overlaps with the leaf of r=\cos(3\theta). -

    -     - - \begin{tikzpicture} - \begin{axis}[ - xtick={1}, - ytick={.5}, - ymin=-.25,ymax=.7,% - xmin=-.08,xmax=1.1% - ] - - \addplot [firstcurvestyle,areastyle,domain=15:60,samples=20] ({cos(x)*sin(3*x)},{sin(x)*sin(3*x)}) -- (axis cs:0,0); - \addplot [white,fill=white, smooth,domain=0:30,samples=20] ({cos(x)*cos(3*x)},{sin(x)*cos(3*x)}); - \addplot [firstcurvestyle, domain=0:60,samples=60] ({cos(x)*sin(3*x)},{sin(x)*sin(3*x)}); - \addplot [secondcurvestyle,domain=-30:30,samples=60] ({cos(x)*cos(3*x)},{sin(x)*cos(3*x)}); - - \end{axis} - \end{tikzpicture} - - -     - -

    - area = \ds \int_{\pi/12}^{\pi/3} \frac12 \sin^2(3\theta)\,d\theta - \int_{\pi/12}^{\pi/6}\frac12\cos^2(3\theta)\,d\theta = \frac1{12}+\frac{\pi}{24} -

    -     - -     - - - - -

    - Enclosed by r=\cos(\theta) and r=1-\cos(\theta), as shown: -

    - - - A cardioid and a circle, and a region of overlap in the first quadrant. - -

    - Two polar curves are plotted. The first is a circle with center (1/2,0) and radius 1/2; - it is symmetric about the x axis and passes through (0,0) and (1,0). -

    - -

    - The second curve is a cardioid. It is larger than the circle, and points in the opposite direction. - That is, it is symmetric about the x axis, with its cusp at the origin, - but its other x intercept lies on the negative x axis, at (-2,0). -

    - -

    - The two curves overlap in a small region in the first quadrant, which is shaded. - The region extends from the origin to the point in the first quadrant where the two curves intersect. - It is bounded above (and to the left) by the circle, and below (and to the right) by the cardioid. -

    -     - - \begin{tikzpicture} - \begin{axis}[ - ymin=-1.4,ymax=1.4,% - xmin=-2.2,xmax=1.2% - ] - - \addplot [firstcurvestyle,areastyle,domain=60:90,samples=60] ({cos(x)*cos(x)},{sin(x)*cos(x)}); - \addplot [firstcurvestyle,areastyle,domain=0:60,samples=40] ({cos(x)*(1-cos(x))},{sin(x)*(1-cos(x))}); - \addplot [firstcurvestyle,domain=0:180,samples=60] ({cos(x)*cos(x)},{sin(x)*cos(x)}); - \addplot [secondcurvestyle,domain=0:360,samples=60] ({cos(x)*(1-cos(x))},{sin(x)*(1-cos(x))}); - - \end{axis} - \end{tikzpicture} - - -     - -

    - area = \ds \int_{0}^{\pi/3} \frac12 (1-\cos(\theta) )^2\,d\theta +\int_{\pi/3}^{\pi/2} \frac12 (\cos(\theta) )^2\,d\theta =\frac{7\pi}{24}-\frac{\sqrt{3}}2\approx 0.0503 -

    -     - -     - -     - - - -

    - Answer the questions involving arc length. -

    -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); - $length=Formula("4pi"); - - -

    - Use the arc length formula to compute the arc length of the circle r=2. -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFormulas=>0); - $length=Formula("4pi"); - - -

    - Use the arc length formula to compute the arc length of the circle r=4\sin(\theta). -

    - -

    - -

    -     -     -     - - - -

    - Use the arc length formula to compute the arc length of r=\cos(\theta)+\sin(\theta). -

    -     - -

    - \sqrt{2}\pi -

    -     -     - - - -

    - Use the arc length formula to compute the arc length of the cardioid r=1+\cos(\theta). - (Hint: apply the formula, simplify, - then use a Power-Reducing Formula to convert 1+\cos(\theta) into a square.) -

    -     - -

    - 8 -

    -     -     - - - - - $length=OneOf("2.2592,2.22748"); - - -

    - Approximate the arc length of one petal of the rose curve - r=\sin(3\theta) with Simpson's Rule and n=4. -

    - -

    - -

    -     -     -     - - - - - - - -

    - Let x(\theta) = f(\theta)\cos(\theta) and y(\theta)=f(\theta)\sin(\theta). - Show, as suggested by the text, that - - x\,'(\theta)^2+y\,'(\theta)^2 = \fp(\theta)^2+f(\theta)^2 - . -

    -     - -

    - x\,'(\theta) = \fp(\theta)\cos(\theta) -f(\theta)\sin(\theta), - y\,'(\theta) = \fp(\theta)\sin(\theta) + f(\theta)\cos(\theta). - Square each and add; - applying the Pythagoran Theorem twice achieves the result. -

    -     - -     - - -     - - - -

    - Answer the questions involving surface area. -

    -     - - - - -

    - Use - to find the surface area of the sphere formed by revolving the circle r=2 about the initial ray. -

    -     - -

    - SA = 16\pi -

    -     - -     - - - - -

    - Use - to find the surface area of the sphere formed by revolving the circle - r=2\cos(\theta) about the initial ray. -

    -     - -

    - SA = 4\pi -

    -     - -     - - - - -

    - Find the surface area of the solid formed by revolving the cardioid - r=1+\cos(\theta) about the initial ray. -

    -     - -

    - SA = 32\pi/5 -

    -     - -     - - - - -

    - Find the surface area of the solid formed by revolving the circle - r=2\cos(\theta) about the line \theta=\pi/2. -

    -     - -

    - SA = 4\pi^2 -

    -     - -     - - - - -

    - Find the surface area of the solid formed by revolving the line r=3\sec(\theta), - -\pi/4\leq\theta\leq\pi/4, - about the line \theta=\pi/2. -

    -     - -

    - SA = 36\pi -

    -     - -     - - - -

    - Find the surface area of the solid formed by revolving the line r=3\sec\theta, - 0\leq\theta\leq\pi/4, about the initial ray. -

    -     - -

    - SA = 9\pi -

    -     -     - -     -     -     -
    -     - - - Vectors - - - -

    - This chapter begins with moving our mathematics out of the plane and into space. That is, - we begin to think mathematically not only in two dimensions, - but in three. - With this foundation, - we can explore vectors both in the plane and in space. -

    -     - -
    - Introduction to Cartesian Coordinates in Space - -

    - Up to this point in this text we have considered mathematics in a 2-dimensional world. - We have plotted graphs on the xy-plane using rectangular and polar coordinates and found the area of regions in the plane. - We have considered properties of solid objects, - such as volume and surface area, - but only by first defining a curve in the plane and then rotating it out of the plane. -

    - -

    - While there is wonderful mathematics to explore in - 2D, we live in a 3D - world and eventually we will want to apply mathematics involving this third dimension. - In this section we introduce Cartesian coordinates in space and explore basic surfaces. - This will lay a foundation for much of what we do in the remainder of the text. -

    - - - -

    - Each point P in space can be represented with an ordered triple, - P=(a,b,c), where a, - b and c represent the relative position of P along the x-, y- and z-axes, - respectively. - Each axis is perpendicular to the other two. -

    - -

    - Visualizing points in space on paper can be problematic, - as we are trying to represent a 3-dimensional concept on a 2-dimensional medium. - We cannot draw three lines representing the three axes in which each line is perpendicular to the other two. - Despite this issue, - standard conventions exist for plotting shapes in space that we will discuss that are more than adequate. -

    - -

    - One convention is that the axes must conform to the - right hand rule. - This rule states that when the index finger of the right hand is extended in the direction of the positive x-axis, - and the middle finger - (bent inward so it is perpendicular to the palm) - points along the positive y-axis, - then the extended thumb will point in the direction of the positive z-axis. (It may take some thought to verify this, - but this system is inherently different from the one created by using the - left hand rule.)right hand ruleof Cartesian coordinates -

    - -

    - As long as the coordinate axes are positioned so that they follow this rule, - it does not matter how the axes are drawn on paper. - There are two popular methods that we briefly discuss. -

    - -
    - Plotting the point P=(2,1,3) in space - - - - Plot of point P = (2, 1, 3) in space. - - -

    - The x and y axes are drawn from -2 to 2 and the z - axis is drawn from -1 to 3. The point P= (2,1,3) is drawn in space. - A dashed cuboid is drawn with one vertex at the origin, three of its edges along the coordinate axes, - and the point P on the corner opposite the origin. -

    -     - - - - - //ASY file for figcartcoord1.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11,5,2.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1,2,3}; - real[] myychoice={-2,-1,1,2}; - real[] myzchoice={-2,-1,1,2}; - defaultpen(0.5mm); - pair xbounds=(-3,3.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$z$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$x$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$y$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // Draw the lines - draw((0,0,1)--(0,2,1)--(3,2,1)--(3,0,1)--(0,0,1), redpen+dashed+linewidth(.5));//top - draw((0,0,0)--(0,2,0)--(3,2,0)--(3,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//bottom - draw((3,0,0)--(3,0,1), redpen+dashed+linewidth(.5));//up1 - draw((0,2,0)--(0,2,1), redpen+dashed+linewidth(.5));//up2 - draw((3,2,0)--(3,2,1), redpen+dashed+linewidth(.5));//up3 - label("$P$",(3,2,1),E); - dotfactor=3; dot((3,2,1),bluepen); - - - - -
    - -

    - In - we see the point P=(2,1,3) plotted on a set of axes. - The basic convention here is that the xy-plane is drawn in its standard way, - with the z-axis down to the left. - The perspective is that the paper represents the xy-plane and the positive z axis is coming up, - off the page. - This method is preferred by many engineers. - Because it can be hard to tell where a single point lies in relation to all the axes, - dashed lines have been added to let one see how far along each axis the point lies. -

    - -

    - One can also consider the xy-plane as being a horizontal plane in, say, - a room, where the positive z-axis is pointing up. - When one steps back and looks at this room, - one might draw the axes as shown in . - The same point P is drawn, again with dashed lines. - This point of view is preferred by most mathematicians, - and is the convention adopted by this text. -

    - -

    - Just as the x- and y-axes divide the plane into four quadrants, - the x-, y-, and z-coordinate planes divide space into eight octants. - The octant in which x, y, - and z are positive is called the first octant. - We do not name the other seven octants in this text. - octantfirst - first octant -

    - -
    - Plotting the point P=(2,1,3) in space with a perspective used in this text - - - - Plot of point P = (2,1,3) in space with perspective used in the text. - - -

    - The x and y axes are drawn from -2 to 2 and the z - axis is drawn from -1 to 3. The point P= (2,1,3) is drawn in space. - A dashed cuboid is drawn with one vertex at the origin, three of its edges along the coordinate axes, - and the point P on the corner opposite the origin. - The z axis is shown as the vertical line. -

    -     - - - - - //ASY file for figcartcoord2.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={-2,3}; - defaultpen(0.5mm); - pair xbounds=(-2,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-1.5,4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // Draw the lines - //draw((2,0,0)--(2,1,0), redpen+dashed+linewidth(.5)); - //draw((2,1,0)--(2,1,3)--(0,0,3), redpen+dashed+linewidth(.5)); - //draw((0,1,0)--(2,1,0), redpen+dashed+linewidth(.5)); - draw((0,0,3)--(0,1,3)--(2,1,3)--(2,0,3)--(0,0,3), redpen+dashed+linewidth(.5));//top - draw((0,0,0)--(0,1,0)--(2,1,0)--(2,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//top - draw((2,0,0)--(2,0,3), redpen+dashed+linewidth(.5));//up1 - draw((0,1,0)--(0,1,3), redpen+dashed+linewidth(.5));//up2 - draw((2,1,0)--(2,1,3), redpen+dashed+linewidth(.5));//up - label("$P$",(2,1,3),E); - dotfactor=3; dot((2,1,3),bluepen); - - - - -
    -     - - - Measuring Distances -

    - It is of critical importance to know how to measure distances between points in space. - The formula for doing so is based on measuring distance in the plane, - and is known - (in both contexts) - as the Euclidean measure of distance. -

    - - - Distance In Space - -

    - Let P=(x_1,y_1,z_1) and Q = (x_2,y_2,z_2) be points in space. - The distance D between P and Q is distancebetween points in space - - D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} - . -

    -     -     - - - -

    - We refer to the line segment that connects points P and Q in space as \overline{PQ}, - and refer to the length of this segment as \norm{\overline{PQ}}. - The above distance formula allows us to compute the length of this segment. -

    - - - - - Length of a line segment - -

    - Let P = (1,4,-1) and let Q = (2,1,1). - Draw the line segment \overline{PQ} and find its length. -

    -     - -

    - The points P and Q are plotted in ; - no special consideration need be made to draw the line segment connecting these two points; - simply connect them with a straight line. - One cannot actually measure this line on the page and deduce anything meaningful; - its true length must be measured analytically. - Applying , we have - - \norm{\overline{PQ}} = \sqrt{(2-1)^2+(1-4)^2+(1-(-1))^2} = \sqrt{14}\approx 3.74 - . -

    - -
    - Plotting points P and Q in - - - - Plotting point P = (1, 4, -1) and point Q = (2, 1, 1). - - -

    - The x axis is drawn from 0 to 2, the z axis is - drawn from -2 to 2 and the y axis is drawn from 0 - to 4. Two points P = (1, 4, -1) and Q = (2, 1, 1) are - drawn in space and are connected by a straight line. -

    -     - - - - - //ASY file for figspace1.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,1); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2,4}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - pair xbounds=(-1,2.5); - pair ybounds=(-1,4.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // Draw the lines for Q=(2,1,1) - draw((2,0,0)--(2,1,0)--(0,1,0), redpen+dashed+linewidth(.5));//top Q - //draw((0,0,1)--(0,1,1)--(2,1,1)--(2,0,1)--(0,0,1), redpen+dashed+linewidth(.5));//top Q - //draw((0,0,0)--(0,1,0)--(2,1,0)--(2,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//bottom Q - //draw((2,0,0)--(2,0,1), redpen+dashed+linewidth(.5));//up1 Q - //draw((0,1,0)--(0,1,1), redpen+dashed+linewidth(.5));//up2 Q - draw((2,1,0)--(2,1,1), redpen+dashed+linewidth(.5));//up3 Q - label("$Q$",(2,1,1),E); - dotfactor=3; dot((2,1,1),bluepen); - - // Draw the lines for P=(1,4,-1) - //draw((0,0,-1)--(0,4,-1)--(1,4,-1)--(1,0,-1)--(0,0,-1), redpen+dashed+linewidth(.5));//top P - //draw((0,0,0)--(0,4,0)--(1,4,0)--(1,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//bottom P - //draw((1,0,0)--(1,0,-1), redpen+dashed+linewidth(.5));//up1 P - //draw((0,4,0)--(0,4,-1), redpen+dashed+linewidth(.5));//up2 P - draw((0,4,0)--(1,4,0)--(1,0,0), redpen+dashed+linewidth(.5));//bottom P - draw((1,4,0)--(1,4,-1), redpen+dashed+linewidth(.5));//up3 P - label("$P$",(1,4,-1),E); - dotfactor=3; dot((1,4,-1),bluepen); - - //line from P to Q - draw((1,4,-1)--(2,1,1), bluepen); - - - - -
    -     - -     -     - - - Spheres -

    - Just as a circle is the set of all points in the plane equidistant from a given point (its center), a sphere is the set of all points in space that are equidistant from a given point. - - allows us to write an equation of the sphere. - sphere -

    - -

    - We start with a point C = (a,b,c) which is to be the center of a sphere with radius r. - If a point P=(x,y,z) lies on the sphere, - then P is r units from C; that is, - - \norm{\overline{PC}} = \sqrt{(x-a)^2+(y-b)^2+(z-c)^2} = r - . -

    - -

    - Squaring both sides, - we get the standard equation of a sphere in space with center at - C=(a,b,c) with radius r, - as given in the following Key Idea. -

    - - - Standard Equation of a Sphere in Space -

    - The standard equation of the sphere with radius r, - centered at C=(a,b,c), is - - (x-a)^2+(y-b)^2+(z-c)^2=r^2 - . -

    -     - - - - - Equation of a sphere - -

    - Find the center and radius of the sphere defined by x^2+2x+y^2-4y+z^2-6z=2. -

    -     - -

    - To determine the center and radius, - we must put the equation in standard form. - This requires us to complete the square - (three times). - - x^2+2x+y^2-4y+z^2-6z\amp =2 - (x^2+2x+1) + (y^2-4y+4)+ (z^2-6z+9) - 14 \amp = 2 - (x+1)^2 + (y-2)^2 + (z-3)^2 \amp = 16 - -

    - -

    - The sphere is centered at (-1,2,3) and has a radius of 4. -

    -     - -     - -

    - The equation of a sphere is an example of an implicit function defining a surface in space. - In the case of a sphere, - the variables x, y and z are all used. - We now consider situations where surfaces are defined where one or two of these variables are absent. -

    -     - - - Introduction to Planes in Space -

    - The coordinate axes naturally define three planes - (shown in ), - the coordinate planes: - the xy-plane, - the yz-plane and the xz-plane. - The xy-plane is characterized as the set of all points in space where the z-value is 0. - planescoordinate plane - planesintroduction - This, in fact, gives us an equation that describes this plane: z=0. - Likewise, the xz-plane is all points where the y-value is 0, characterized by y=0. -

    - - - -
    - The xy-plane in (a), the yz-plane in (b) and the xz-plane in (c) - -
    - - - - - Graph of xy plane in space. - - -

    - The axes are uncalibrated. The plane is drawn along the x and y axes - at z=0, the normal is along the z axis. -

    -     - - - - - //ASY file for figspacexy.pdf in Chapter 10 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the xy-plane - triple f(pair t) { - return (t.x,t.y,0); - } - surface s=surface(f,(-2,-2),(2,2),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - - - -
    - -
    - - - - - Graph of yz plane in space. - - -

    - The axes are uncalibrated. The plane is drawn along the y and z - axes at x=0, the normal is along the x axis. -

    -     - - - - - //ASY file for figspaceyz.pdf in Chapter 10 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the yz-plane - triple f(pair t) { - return (0,t.x,t.y); - } - surface s=surface(f,(-2,-2),(2,2),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - - - -
    - -
    - - - - - Graph of xz plane in space. - - -

    - The axes are uncalibrated. The plane is drawn along the x and z axes at - y =0, the normal is along the y axis. -

    -     - - - - - //ASY file for figspacexz.pdf in Chapter 10 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the xz-plane - triple f(pair t) { - return (t.x,0,t.y); - } - surface s=surface(f,(-2,-2),(2,2),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - - - -
    -     -
    - -

    - The equation x=2 describes all points in space where the x-value is 2. - This is a plane, - parallel to the yz-coordinate plane, - shown in . -

    - -
    - The plane x=2 - - - - Graph of plane x=2 in space. - - -

    - The y and the z axes are uncalibrated, the x axis is - drawn from -2 to 2. The yz plane is drawn at x=-2. -

    -     - - - - - //ASY file for figspace2.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2.5,3.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the x=2 plane - triple f(pair t) { - return (2,t.x,t.y); - } - surface s=surface(f,(-2,-2),(2,2),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - - - -
    - - - Regions defined by planes - -

    - Sketch the region defined by the inequalities -1\leq y\leq 2. -

    -     - -

    - The region is all points between the planes y=-1 and y=2. - These planes are sketched in , - which are parallel to the xz-plane. - Thus the region extends infinitely in the x and z directions, - and is bounded by planes in the y direction. -

    - -
    - Sketching the boundaries of a region in - - - - Graph of two planes that are parallel to the xz plane. - - -

    - The y and z axes are uncalibrated, the x axis is drawn - from -2 and 2. Two planes are drawn parallel to the xz - plane at y=-1 and y =2, with normal along the y axis. -

    -     - - - - - //ASY file for figspace3.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,2.5,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the y=-1 plane - triple f(pair t) { - return (t.x,-1,t.y); - } - surface s=surface(f,(-2,-2),(2,2),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - //Draw the y=2 plane - triple f(pair t) { - return (t.x,2,t.y); - } - surface s=surface(f,(-2,-2),(2,2),8,8,Spline); - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - - - -
    -     - -     -     - - - Cylinders -

    - The equation x=1 obviously lacks the y and z variables, - meaning it defines points where the y and z coordinates can take on any value. - Now consider the equation x^2+y^2=1 - in space. In the plane, - this equation describes a circle of radius 1, centered at the origin. - In space, the z coordinate is not specified, - meaning it can take on any value. - In , - we show part of the graph of the equation - x^2+y^2=1 by sketching 3 circles: - the bottom one has a constant z-value of -1.5, - the middle one has a z-value of 0 and the top circle has a z-value of 1. - By plotting all possible z-values, - we get the surface shown in . - This surface looks like a tube, or a cylinder; - mathematicians call this surface a cylinder - for an entirely different reason. -

    - - - -
    - Sketching x^2+y^2=1 - -
    - - - - - Graph of three circles with radius 1 with centres on the z axis along different levels. - - -

    - The y and z axes are uncalibrated, the x axis is drawn from -2 - to 2. There are three circles with radius of 1 and centres all along the z - axis and are laid parallel to the xy plane. The circle in the middle has its centre on the origin. -

    -     - - - - - //ASY file for figspacecylinder1.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the circle at height z=1 - triple g(real t) {return (cos(t),sin(t),1);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - //Draw the circle at height z=0 - triple g(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - //Draw the circle at height z=-1.5 - triple g(real t) {return (cos(t),sin(t),-1.5);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - - - -
    - -
    - - - - - Graph of a hollow cylinder made with the three circles shown in the previous graph. - - -

    - The y and z axes are uncalibrated, the x axis is drawn from -2 to 2. - There are three circles with radius of 1 and centres all along the z axis and are laid - parallel to the xy plane. These circles form the area of cross-section and a cylinder of equation - x^2+y^2 =1 is drawn that includes all three circles. -

    -     - - - - - //ASY file for figspacecylinder1b.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the circle at height z=1 - triple g(real t) {return (cos(t),sin(t),1);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - //Draw the circle at height z=0 - triple g(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - //Draw the circle at height z=-1.5 - triple g(real t) {return (cos(t),sin(t),-1.5);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - //Draw the cylinder on top - triple f(pair t) { - return (cos(t.x),sin(t.x),t.y); - } - surface s=surface(f,(0,-1.5),(2*pi,1.2),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -
    -     - -
    - - - Cylinder - -

    - Let C be a curve in a plane and let L be a line not parallel to C. - A cylinder is the set of all lines parallel to L that pass through C. - The curve C is the directrix of the cylinder, - and the lines are the rulings. - cylinder - directrix -

    -     -     - -

    - In this text, - we consider curves C that lie in planes parallel to one of the coordinate planes, - and lines L that are perpendicular to these planes, - forming right cylinders. - Thus the directrix can be defined using equations involving 2 variables, - and the rulings will be parallel to the axis of the third variable. -

    - -

    - In the example preceding the definition, - the curve x^2+y^2=1 in the xy-plane is the directrix and the rulings are lines parallel to the z-axis. - (Any circle shown in - can be considered a directrix; - we simply choose the one where z=0.) - Sample rulings can also be viewed in . - More examples will help us understand this definition. -

    - - - Graphing cylinders - -

    - Graph the following cylinders. -

    - -

    -

      -
    1. z=y^2
      2. -
    2. x=\sin(z)
      3. -
    -

    -     - -

    -

      -
    1. -

      - We can view the equation z=y^2 as a parabola in the yz-plane, - as illustrated in . - As x does not appear in the equation, - the rulings are lines through this parabola parallel to the x-axis, - shown in . - These rulings give a general idea as to what the surface looks like, - drawn in . -

      - -
      - Sketching the cylinder defined by z=y^2 - -
      - - - - - Graph of cylinder in space with formula z = y^2. - - -

      - The x, y and z are uncalibrated. The graph shows a parabola z = y^2 - drawn on the zy plane. The parabola has its vertex on the origin. -

      -       - - - - - //ASY file for figspace4a.pdf in Chapter 10 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2.25,2.25); - pair ybounds=(-2.25,2.25); - pair zbounds=(-0.25,4.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the parabola z=y^2 - triple g(real t) {return (0,t,t^2);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,bluepen); - - //Draw the line for y=-1.9 on the parabola z=y^2 - triple g(real t) {return (t,-1.9,(1.9^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); - - //Draw the line for y=1.7 on the parabola z=y^2 - triple g(real t) {return (t,1.7,(1.7^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); - - - - -
      - -
      - - - - - Graph of parabola in space along with a group of lines on the parabola. - - -

      - The x, y and z axis are uncalibrated. The parabola is drawn - on the y and z axis, z being a function of y. There - is a group of parallel lines called rulings equidistant from each other that are - drawn on the parabola parallel to the xz plane, these lines give an idea - of the surface. -

      -       - - - - - //ASY file for figspace4b.pdf in Chapter 10.1 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2.25,2.25); - pair ybounds=(-2.25,2.25); - pair zbounds=(-0.25,4.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the parabola z=y^2 for t from -2 to 2 - triple g(real t) {return (0,t,t^2);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,bluepen); - - //Draw the line for y=-1.9 on the parabola z=y^2 - triple g(real t) {return (t,-1.9,(1.9^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=-1.6 on the parabola z=y^2 - triple g(real t) {return (t,-1.6,(1.6^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=-1.3 on the parabola z=y^2 - triple g(real t) {return (t,-1.3,(1.3^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=-1 on the parabola z=y^2 - triple g(real t) {return (t,-1,(1^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=-0.7 on the parabola z=y^2 - triple g(real t) {return (t,-0.7,(0.7^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=-0.4 on the parabola z=y^2 - triple g(real t) {return (t,-0.4,(0.4^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=-0.1 on the parabola z=y^2 - triple g(real t) {return (t,-0.1,(0.1^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=0.2 on the parabola z=y^2 - triple g(real t) {return (t,0.2,(0.2^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=0.5 on the parabola z=y^2 - triple g(real t) {return (t,0.5,(0.5^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=0.8 on the parabola z=y^2 - triple g(real t) {return (t,0.8,(0.8^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=1.1 on the parabola z=y^2 - triple g(real t) {return (t,1.1,(1.1^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=1.4 on the parabola z=y^2 - triple g(real t) {return (t,1.4,(1.4^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Draw the line for y=1.7 on the parabola z=y^2 - triple g(real t) {return (t,1.7,(1.7^2));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - - - -
      - -
      - - - - - Graph of surface with parabolic area of cross-section. - - -

      - The x, y and z axis are uncalibrated. The parabola is drawn on the y and z axis, - z being a function of y. A surface with a parabolic area of cross-section parallel to the yz - plane is shown. -

      -       - - - - - //ASY file for figspace4c.pdf in Chapter 10.1 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2.25,2.25); - pair ybounds=(-2.25,2.25); - pair zbounds=(-0.25,4.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the parabola z=y^2 for t from -2 to 2 - triple g(real t) {return (0,t,t^2);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,bluepen); - - //Draw the cylinder z=y^2 - triple f(pair t) { - return (t.x,t.y,(t.y)^2); - } - surface s=surface(f,(-2,-2),(2,2),8,32,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -
      -       -
      -
      2. - -
    2. - -

      - We can view the equation x=\sin(z) as a sine curve that exists in the xz-plane, - as shown in . - The rules are parallel to the y axis as the variable y does not appear in the equation x=\sin(z); - some of these are shown in . - The surface is shown in . -

      - -
      - Sketching the cylinder defined by x=\sin(z) - -
      - - - - - Graph of function x= sin (z) drawn in space. - - -

      - The x, y and z axes are uncalibrated. A function x= sin(z) - is drawn on the xz plane where x is a function of z. The z - axis is positioned vertically, and two sine waves are drawn on it one along the positive - z axis and one along the negative z axis. The two waves connect at the origin. -

      -       - - - - - //ASY file for figspace4d.pdf in Chapter 10.1 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,5,13); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2*pi-.5,2*pi+.75); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve x=sin z - triple g(real t) {return (sin(t),0,t);} - path3 mypath=graph(g,-2*pi,2*pi,operator ..); draw(mypath,bluepen); - - //Draw red lines on x=sin(z) for z in{-6.28,-4.71,-1.57,1.57,4.71,6.28} - triple g(real t) {return (sin(2*pi),t,2*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); - triple g(real t) {return (sin(1.5*pi),t,1.5*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); - triple g(real t) {return (sin(0.5*pi),t,0.5*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); - triple g(real t) {return (sin(-0.5*pi),t,-0.5*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); - triple g(real t) {return (sin(-1.5*pi),t,-1.5*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); - triple g(real t) {return (sin(-2*pi),t,-2*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,invisible); - - - - -
      - -
      - - - - - Graph of function x= sin (z) drawn in space along with the rules. - - -

      - The x, y and z axes are uncalibrated. A function x= sin(z) - is drawn on the xz plane where x is a function of z. The z - axis is positioned vertically, and two sine waves are drawn on it one along the positive - z axis and one along the negative z axis. The two waves connect at the origin. - The rules are drawn on the curve and are parallel to the y axis. -

      -       - - - - - //ASY file for figspace4e.pdf in Chapter 10.1 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,5,13); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2*pi-.5,2*pi+.75); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve x=sin z - triple g(real t) {return (sin(t),0,t);} - path3 mypath=graph(g,-2*pi,2*pi,operator ..); draw(mypath,bluepen); - - //Draw red lines on x=sin(z) for z in{-6.28,-4.71,-1.57,1.57,4.71,6.28} - triple g(real t) {return (sin(2*pi),t,2*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (sin(1.5*pi),t,1.5*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (sin(0.5*pi),t,0.5*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (sin(-0.5*pi),t,-0.5*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (sin(-1.5*pi),t,-1.5*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (sin(-2*pi),t,-2*pi);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - - - -
      - -
      - - - - - Graph of function x= sin (z) drawn in space along with the rules. - - -

      - The x, y and z axes are uncalibrated. The sine function - x=\sin(z) described previously is used as the area of cross-section to - form the surface. -

      -       - - - - - //ASY file for figspace4f.pdf in Chapter 10.1 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,5,13); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2*pi-.5,2*pi+.75); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve x=sin z - triple g(real t) {return (sin(t),0,t);} - path3 mypath=graph(g,-2*pi,2*pi,operator ..); draw(mypath,bluepen); - - //Draw the cylinder z=y^2 - triple f(pair t) { - return (sin(t.y),t.x,t.y); - } - surface s=surface(f,(-2,-2*pi),(2,2*pi),16,32,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -
      -       -
      -
      3. -
    -

    -     - -     -     - - - Surfaces of Revolution -

    - One of the applications of integration we learned previously was to find the volume of solids of revolution solids formed by revolving a curve about a horizontal or vertical axis. - We now consider how to find the equation of the surface of such a solid. -

    - - - -

    - Consider the surface formed by revolving - y=\sqrt{x} about the x-axis. - Cross-sections of this surface parallel to the yz-plane are circles, - as shown in . - Each circle has equation of the form - y^2+z^2=r^2 for some radius r. - The radius is a function of x; - in fact, it is r(x) = \sqrt{x}. - Thus the equation of the surface shown in is y^2+z^2=(\sqrt{x})^2. -

    - -
    - Introducing surfaces of revolution - -
    - - - - - Graph showing cross- section for surface of revolution. - - -

    - The y and x axes are drawn from -2 to 2 and the x axis is drawn from 0 to 4. - There are two planes drawn parallel to the yz plane and both of them have circles - outlined inside the plane. The first plane at x=1 has a smaller circle, while the - one at x=4 is bigger, both circles have formula y^2 + z^2 = r^2 for some radius r. - There is also half of a parabola drawn on the xy plane with x being a function of y, - this half parabola passes through both the circles intersecting them. -

    -     - - - - - //ASY file for figsurf_rev_intro.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,5,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={-2,2}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - pair xbounds=(-0.25,4.5); - pair ybounds=(-2.3,2.3); - pair zbounds=(-2.3,2.3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve y=sqrt(x) - triple g(real t) {return (t,sqrt(t),0);} - path3 mypath=graph(g,0,5,operator ..); draw(mypath,bluepen); - - //Draw red circles for x=1,4 - triple g(real t) {return (1,sin(t),cos(t));} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - triple g(real t) {return (4,2*sin(t),2*cos(t));} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Draw the x=1 plane with no spline - triple f(pair t) { - return (1,t.x,t.y); - } - surface s=surface(f,(-1.25,-1.25),(1.25,1.25)); - //pen p=apexmeshpen; - draw(s,surfacepen,nolight); - - //Draw the x=4 plane with no spline - triple f(pair t) { - return (4,t.x,t.y); - } - surface s=surface(f,(-2.25,-2.25),(2.25,2.25)); - //pen p=apexmeshpen; - draw(s,surfacepen,nolight); - - - - -
    - -
    - - - - - Graph showing a surface of revolution. - - -

    - The y and z axes are drawn from -2 to 2 and the - x axis is drawn from 0 to 4. There are two planes drawn - parallel to the yz plane and both of them have circles outlined inside - the plane. The first plane at x =1 has a smaller circle, while the one at - x=4 is bigger. There is also half of a parabola drawn on the xy - plane with x being a function of y, this half parabola passes - through both the circles intersecting the planes. This half parabola when rotated - over the x axis gives a hollow dome that opens along the positive x axis. -

    -     - - - - - //ASY file for figsurf_rev_introb.pdf in Chapter 10 - - size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,5,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={-2,2}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,4.5); - pair ybounds=(-2.3,2.3); - pair zbounds=(-2.3,2.3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve y=sqrt(x) - triple g(real t) {return (t,sqrt(t),0);} - path3 mypath=graph(g,0,5,operator ..); draw(mypath,bluepen); - - //Draw the revolved surface y^2 + z^2 = x - triple f(pair t) { - return (t.x,sqrt(t.x)*cos(t.y),sqrt(t.x)*sin(t.y)); - } - surface s=surface(f,(0,0),(5,2*pi),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -
    -     -
    - -

    - We generalize the above principles to give the equations of surfaces formed by revolving curves about the coordinate axes. -

    - - - Surfaces of Revolution, Part 1 -

    - Let r be a radius function. - surface of revolution -

      -
    1. -

      - The equation of the surface formed by revolving y=r(x) or z=r(x) about the x-axis is y^2+z^2=r(x)^2. -

      -
      2. - -
    2. -

      - The equation of the surface formed by revolving x=r(y) or z=r(y) about the y-axis is x^2+z^2=r(y)^2. -

      -
      3. - -
    3. -

      - The equation of the surface formed by revolving x=r(z) or y=r(z) about the z-axis is x^2+y^2=r(z)^2. -

      -
      4. -
    -

    -     - - - Finding equation of a surface of revolution - -

    - Let y=\sin(z) on [0,\pi]. - Find the equation of the surface of revolution formed by revolving - y=\sin(z) about the z-axis. -

    -     - -

    - Using , - we find the surface has equation x^2+y^2=\sin^2(z). - The curve is sketched in and the surface is drawn in . -

    - -

    - Note how the surface - (and hence the resulting equation) - is the same if we began with the curve x=\sin(z), - which is also drawn in . -

    -
    - Revolving y=\sin(z) about the z-axis in - -
    - - - - - Graph of two functions y = sin(z) and x = sin(z). - - -

    - The x and y axes are drawn from -1 to 1 and the - z axis is drawn from 0 to 3. Two functions x= \sin(z) - and y= \sin(z) are shown. The two curves are perpendicular to each other as - x=\sin(z) is drawn on the xz plane and y= \sin(z) is drawn on - the yz plane. Both curves start at the origin and end at the same point - (0, \pi/2, 0). -

    -     - - - - - //ASY file for figsurfrev1a.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,1.25); - pair zbounds=(-0.25,pi+.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve x=sin(z) in red - triple g(real t) {return (sin(t),0,t);} - path3 mypath=graph(g,0,pi,operator ..); draw(mypath,redpen); - label("$x=\sin(z)$",(.5,0,1.75)); - - //Draw the curve y=sin(z) in blue - triple g(real t) {return (0,sin(t),t);} - path3 mypath=graph(g,0,pi,operator ..); draw(mypath,bluepen); - label("$y=\sin(z)$",(0,.5,1.75)); - - - - -
    - -
    - - - - - Graph showing surface formed by revolving y = sin (z) about the z axis. - - -

    - The x and y axes are drawn from -1 to 1 and the z - axis is drawn from 0 to 3. The function y= \sin(z) is rotated - around the z axis and it forms a sphere with tapering top and bottom. -

    -     - - - - - //ASY file for figsurfrev1b.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,1.25); - pair zbounds=(-0.25,pi+.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve y=sin(z) in red - triple g(real t) {return (0,sin(t),t);} - path3 mypath=graph(g,0,pi,operator ..); draw(mypath,bluepen); - - //Draw the revolved surface x^2 + y^2 = sin^2(z) - triple f(pair t) { - return (sin(t.x)*cos(t.y),sin(t.x)*sin(t.y),t.x); - } - surface s=surface(f,(0,0),(pi,2*pi),16,32,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -
    -     -
    -     - -     - -

    - This particular method of creating surfaces of revolution is limited. - For instance, in - of - we found the volume of the solid formed by revolving - y=\sin(x) about the y-axis. - Our current method of forming surfaces can only rotate - y=\sin(x) about the x-axis. - Trying to rewrite y=\sin(x) as a function of y is not trivial, - as simply writing x=\sin^{-1}(y) only gives part of the region we desire. -

    - -

    - What we desire is a way of writing the surface of revolution formed by rotating y=f(x) about the y-axis. - We start by first recognizing this surface is the same as revolving z=f(x) about the z-axis. - This will give us a more natural way of viewing the surface. -

    - -

    - A value of x is a measurement of distance from the z-axis. - At the distance r, we plot a z-height of f(r). - When rotating f(x) about the z-axis, - we want all points a distance of r from the z-axis in the xy-plane to have a z-height of f(r). - All such points satisfy the equation r^2=x^2+y^2; - hence r=\sqrt{x^2+y^2}. - Replacing r with \sqrt{x^2+y^2} in f(r) gives z=f(\sqrt{x^2+y^2}). - This is the equation of the surface. -

    - - - Surfaces of Revolution, Part 2 -

    - Let z=f(x), x\geq 0, - be a curve in the xz-plane. - The surface formed by revolving this curve about the z-axis has equation z=f\big(\sqrt{x^2+y^2}\big). - surface of revolution -

    -     - - - Finding equation of surface of revolution - -

    - Find the equation of the surface found by revolving - z=\sin(x) about the z-axis. -

    -     - -

    - Using , - the surface has equation z=\sin\big(\sqrt{x^2+y^2}\big). - The curve and surface are graphed in . -

    - -
    - Revolving z=\sin(x) about the z-axis in - -
    - - - - - Graph of sine wave of amplitude 1 in the xz plane. - - -

    - The x and y axes are drawn from -5 to 5 and the z - axis is drawn from -1 to 1. The curve z= \sin(x) is drawn in the - xz plane, it is a wave with amplitude of z=1. The curve starts at the origin, - curves up and reaches a peak close to x=2, then it decreases and crosses the x - axis close to x=4, it decreases till it reaches a depth of z=-1, after which - it increases again to meet the x axis close to x=6. -

    -     - - - - - //ASY file for figsurfrev2a.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-5,5}; - real[] myychoice={-5,5}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-7,7); - pair ybounds=(-7,7); - pair zbounds=(-1.25,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve z=sin(x) in blue - triple g(real t) {return (t,0,sin(t));} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - - - -
    - -
    - - - - - Graph showing surface of revolution of function z= sin(x) around x axis. - - -

    - The surface of equation z= \sin(\sqrt {x^2 +y^2}) is made by revolving - the wave of formula z= \sin(x) about the z axis. -

    -     - - - - - //ASY file for figsurfrev2b.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-5,5}; - real[] myychoice={-5,5}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-7,7); - pair ybounds=(-7,7); - pair zbounds=(-1.25,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve z=sin(x) in blue - triple g(real t) {return (t,0,sin(t));} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen+.7mm); - - //Draw the revolved surface z = sin(sqrt(x^2 + y^2)) - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),sin(sqrt(((t.y*cos(t.x))^2+(t.y*sin(t.x))^2)))); - } - surface s=surface(f,(0,0),(2*pi,2*pi),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -
    -     -
    -     -     -     - - - Quadric Surfaces -

    - Spheres, planes and cylinders are important surfaces to understand. - We now consider one last type of surface, - a quadric surface. - The definition may look intimidating, - but we will show how to analyze these surfaces in an illuminating way. -

    - - - - - Quadric Surface - -

    - A quadric surface is the graph of the general second-degree equation in three variables: - quadric surfacedefinition - - Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0 - . -

    -     -     - -

    - When the coefficients D, - E or F are not zero, - the basic shapes of the quadric surfaces are rotated in space. - We will focus on quadric surfaces where these coefficients are 0; - we will not consider rotations. - There are six basic quadric surfaces: - the elliptic paraboloid, - elliptic cone, ellipsoid, hyperboloid of one sheet, - hyperboloid of two sheets, and the hyperbolic paraboloid. -

    - -
    - The elliptic paraboloid z=x^2/4+y^2 - - - - Graph of elliptic paraboloid z = x^2/4 +y^2. - - -

    - The axes are uncalibrated. There are two parabolas shown one in the plane x=0 - and the other in y=0. There is a circle drawn in the plane z=d. The - elliptical paraboloid z= x^2/4 +y^2 has both the parabolas and the circle included - in the surface. -

    -     - - - - - //ASY file for figquadric_parb.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-0.25,5.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z = x^2/4+y^2 - triple f(pair t) { - return (2*sqrt(t.y)*cos(t.x),sqrt(t.y)*sin(t.x),t.y);// - } - surface s=surface(f,(0,0),(2*pi,4),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the trace for y=0 in red - triple g(real t) {return (t,0,t^2/4);} - path3 mypath=graph(g,-4,4,operator ..); draw(mypath,redpen); - //Draw the trace for x=0 in red - triple g(real t) {return (0,t,t^2);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - //Draw the trace for z=2 in red - triple g(real t) {return (2*sqrt(2)*cos(t),sqrt(2)*sin(t),2);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent \centering In plane\\ $y=0$",(4,0,6),N);//label("$y=0$",(4,0,5.5),N); - draw((4.25,0,5.5)--(4,0,4.25),linewidth(.75),Arrow3); - label("\noindent \centering In plane\\ $x=0$",(0,2,6),N);//label("$x=0$",(0,2,5.5),N); - draw((0,2.25,5.5)--(0,2,4.25),linewidth(.75),Arrow3); - label("\noindent \centering In plane\\ $z=d$",(-1,3,1),N);//label("$z=d$",(-1,2,0.5),E); - draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -
    - -

    - We study each shape by considering traces, - trace - quadric surfacetrace - that is, intersections of each surface with a plane parallel to a coordinate plane. - For instance, - consider the elliptic paraboloid z= x^2/4+y^2, - shown in . - If we intersect this shape with the plane z=d (, replace z with d), - we have the equation: - - d \amp = \frac{x^2}4+y^2. - Divide both sides by d: - 1 \amp = \frac{x^2}{4d} + \frac{y^2}{d} - . -

    - -

    - This describes an ellipse so cross sections parallel to the xy-coordinate plane are ellipses. - This ellipse is drawn in the figure. -

    - -

    - Now consider cross sections parallel to the xz-plane. - For instance, - letting y=0 gives the equation z=x^2/4, clearly a parabola. - Intersecting with the plane x=0 gives a cross section defined by z=y^2, - another parabola. - These parabolas are also sketched in the figure. -

    - -

    - Thus we see where the elliptic paraboloid gets its name: - some cross sections are ellipses, and others are parabolas. -

    - -

    - Such an analysis can be made with each of the quadric surfaces. - We give a sample equation of each, - provide a sketch with representative traces, - and describe these traces. -

    - -

    -

    -
  • - Elliptic Paraboloid -

    - z=\frac{x^2}{a^2}+\frac{y^2}{b^2} -

    -
    • -
    -

    - - - - - - Graph of elliptic paraboloid. - - -

    - The axes are uncalibrated. The elliptical parabola opens along the positive z axis. -

    -     - - - - - //ASY file for figquadric_par.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-0.25,5.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z = x^2/4+y^2 - triple f(pair t) { - return (2*sqrt(t.y)*cos(t.x),sqrt(t.y)*sin(t.x),t.y);// - } - surface s=surface(f,(0,0),(2*pi,4),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - label("\noindent \centering In plane\\ $y=0$",(4,0,6),N,invisible); - label("\noindent \centering In plane\\ $x=0$",(0,2,6),N,invisible); - label("\noindent \centering In plane\\ $z=0$",(-1,3,1),N,invisible); - - - - - - - Plane - - Trace - - - - - - - - x=d - - Parabola - - - y=d - - Parabola - - - z=d - - Ellipse - - - - - - Graph of elliptic paraboloid. - - -

    - The axes are uncalibrated. There are two parabolas shown one in the plane x=0 - and the other in y=0. There is a circle drawn in the plane z=d. The - elliptical paraboloid has both the parabolas and the circle included in the surface. - It opens along the positive z axis. -

    -     - - - - - //ASY file for figquadric_parb.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-0.25,5.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z = x^2/4+y^2 - triple f(pair t) { - return (2*sqrt(t.y)*cos(t.x),sqrt(t.y)*sin(t.x),t.y);// - } - surface s=surface(f,(0,0),(2*pi,4),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the trace for y=0 in red - triple g(real t) {return (t,0,t^2/4);} - path3 mypath=graph(g,-4,4,operator ..); draw(mypath,redpen); - //Draw the trace for x=0 in red - triple g(real t) {return (0,t,t^2);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - //Draw the trace for z=2 in red - triple g(real t) {return (2*sqrt(2)*cos(t),sqrt(2)*sin(t),2);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent \centering In plane\\ $y=0$",(4,0,6),N);//label("$y=0$",(4,0,5.5),N); - draw((4.25,0,5.5)--(4,0,4.25),linewidth(.75),Arrow3); - label("\noindent \centering In plane\\ $x=0$",(0,2,6),N);//label("$x=0$",(0,2,5.5),N); - draw((0,2.25,5.5)--(0,2,4.25),linewidth(.75),Arrow3); - label("\noindent \centering In plane\\ $z=d$",(-1,3,1),N);//label("$z=d$",(-1,2,0.5),E); - draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -     - -

    - One variable in the equation of the elliptic paraboloid will be raised to the first power; - above, this is the z variable. The paraboloid will open in the direction of this variable's axis. - Thus x= y^2/a^2+z^2/b^2 is an elliptic paraboloid that opens along the x-axis. - Multiplying the right hand side by (-1) defines an elliptic paraboloid that opens in the opposite direction. - quadric surfacegallery - quadric surfaceelliptic paraboloid -

    - -

    -

    -
  • - Elliptic Cone -

    - z^2=\frac{x^2}{a^2}+\frac{y^2}{b^2} -

    -
    • -
    -

    - - - - - - - Graph showing elliptic cone opening along the z axis. - - -

    - The axes are uncalibrated. Two hollow elliptic cones are drawn with vertices at the origin, - one opening along the positive z axis and the other along the negative z axis. -

    -     - - - - - //ASY file for figquadric_cone.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,1); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-3,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the top half of the surface z^2 = x^2/(1.5)^2+y^2 - triple f(pair t) { - return (1.5*t.y*cos(t.x),t.y*sin(t.x),t.y);// - } - surface s=surface(f,(0,-2),(2*pi,2),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - label("\noindent\centering In plane\\ $y=d$",(4,0,-1),invisible); - label("\noindent\centering In plane\\ $y=0$",(0,4,1.5),invisible); - label("\noindent\centering In plane\\ $z=d$",(4,0,-1),invisible); - - - - - - - Plane - - Trace - - - - - - - - x=0 - - Crossed Lines - - - y=0 - - Crossed Lines - - - - - - x=d - - Hyperbola - - - y=d - - Hyperbola - - - z=d - - Ellipse - - -     - - - - - - Graph showing elliptic cone opening along the z axis. - - -

    - The axes are uncalibrated. Two hollow elliptic cones are drawn with vertices at the origin, - one opening along the positive z axis and the other along the negative z axis. - The graph shows three traces, on the plane y=0, the trace is a straight line passing through - the vertices of the cones, when the plane is z=d the trace is an ellipse. -

    -     - - - - //ASY file for figquadric_coneb.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,1); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-3,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the top half of the surface z^2 = x^2/(1.5)^2+y^2 - triple f(pair t) { - return (1.5*t.y*cos(t.x),t.y*sin(t.x),t.y);// - } - surface s=surface(f,(0,-2),(2*pi,2),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for y=0 in red - triple g(real t) {return (t,0,t/1.5);} - path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); - triple g(real t) {return (t,0,-t/1.5);} - path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); - - //Draw the trace for z=-2 in red - triple g(real t) {return (1.5*1.5*cos(t),1.5*sin(t),-1.5);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent\centering In plane\\ $y=0$",(0,4,1.5)); - draw((.1,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - draw((-.1,4,1.5)--(-3,0,2),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $z=d$",(4,0,-1)); - draw((4,0,-1)--(1.6,1.1,-1.5),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $y=d$",(4,0,-1),invisible); - - - - - - - - Graph showing elliptic cone opening along the z axis. - - -

    - The axes are uncalibrated. Two hollow elliptic cones are drawn with vertices at the origin, - one opening along the positive z axis and the other along the negative z axis. - A hyperbola is shown in the y=d plane, the two parts of the hyperbola are traced on the elliptic cone. -

    -     - - - - - //ASY file for figquadric_conec.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,1); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-3,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 = x^2/(1.5)^2+y^2 - triple f(pair t) { - return (1.5*t.y*cos(t.x),t.y*sin(t.x),t.y);// - } - surface s=surface(f,(0,-2),(2*pi,2),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for y=d in red - triple g(real t) {return (1.5*.5*tan(t),0.5,0.5/cos(t));} - path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); - triple g(real t) {return (1.5*.5*tan(t),0.5,-0.5/cos(t));} - path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $y=d$",(4,0,-1)); - draw((4,0,-1)--(1.3,.5,-1),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $y=0$",(0,4,1.5),invisible); - label("\noindent\centering In plane\\ $z=d$",(4,0,-1),invisible); - - - - -     -     -

    - One can rewrite the equation as z^2-x^2/a^2-y^2/{b^2} = 0. - The one variable with a positive coefficient corresponds to the axis that the cones open along. - quadric surfaceelliptic cone -

    - -

    -

    -
  • - Ellipsoid -

    - \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 -

    -
    • -
    -

    - - - - - - Graph of the three dimensional ellipsoid. - - -

    - The x, y and z axes are uncalibrated. Graph of an ellipsoid with - centre at origin. -

    -     - - - - - //ASY file for figquadric_ellipsoid.pdf in Chapter 10.1 - //STILL NEED TO FIX LABELS - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - xaxis3("",-3,3,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",-3,3,invisible,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",-3,3,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 = x^2/(1.5)^2+y^2 - triple f(pair t) { - return (1.5*cos(t.y)*cos(t.x),cos(t.y)*sin(t.x),sin(t.y));//({cos(x)*1.5*cos(y)},{sin(x)*cos(y)},{sin(y)}) - } - surface s=surface(f,(-pi,-pi/2),(1*pi,pi/2),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for y=d in red - //triple g(real t) {return (1.5*.5*tan(t),0.5,0.5/cos(t));} - //path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); - //triple g(real t) {return (1.5*.5*tan(t),0.5,-0.5/cos(t));} - //path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent\centering In plane\\ $y=0$",(2,0,-1),invisible); - label("\noindent\centering In plane\\ $x=0$",(0,2,-.5),invisible); - label("\noindent\centering In plane\\ $z=0$",(0,2,1.5),invisible); - - - - - - - Plane - - Trace - - - - - - - - x=d - - Ellipse - - - y=d - - Ellipse - - - z=d - - Ellipse - - - - - - Graph of the three dimensional ellipsoid. - - -

    - The x, y and z axes are uncalibrated. There are three ellipses drawn. - The first one is on the xz plane, y=0. The second is on the yz plane, - with x=0. The third on the xy plane, with z=0. - Filling in the traces gives the ellipsoid. -

    -     - - - - - //ASY file for figquadric_ellipsoidb.pdf in Chapter 10 - //STILL NEED TO FIX LABELS - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - xaxis3("",-3,3,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",-3,3,invisible,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",-3,3,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 = x^2/(1.5)^2+y^2 - triple f(pair t) { - return (1.5*cos(t.y)*cos(t.x),cos(t.y)*sin(t.x),sin(t.y));// - } - surface s=surface(f,(-pi,-pi/2),(1*pi,pi/2),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the trace for x=0 in red - triple g(real t) {return (0,sin(t),cos(t));} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - //Draw the trace for y=0 in red - triple g(real t) {return (1.5*sin(t),0,cos(t));} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - //Draw the trace for z=0 in red - triple g(real t) {return (1.5*sin(t),cos(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent\centering In plane\\ $y=0$",(2,0,-1),S); - draw((2,0,-1)--(1.3,0,-.5),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $x=0$",(0,2,-.5),S); - draw((0,2,-.5)--(0,.86,-.5),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $z=0$",(0,2,1.5),N); - draw((0,2,1.5)--(1.05,.75,0),linewidth(.75),Arrow3); - - - - -     - -

    - If a=b=c\neq0, the ellipsoid is a sphere with radius a; compare to . - quadric surfaceellipsoid - quadric surfacesphere -

    - -

    -

    -
  • - Hyperboloid of One Sheet -

    - \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1 -

    -
    • -
    -

    - - - - - - Graph showing hyperboloid of one sheet. - - -

    - The three axes are uncalibrated. Graph shows a hyperboloid of one sheet. -

    -     - - - - - //ASY file for figquadric_hyp_one_sheet.pdf in Chapter 10.1 - //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? - //STILL NEED TO FIX LABELS - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - xaxis3("",-3,3,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",-3,3,invisible,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",-3,3,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 = x^2/(1.5)^2+y^2 - triple f(pair t) { - return (cos(t.y)/cos(t.x),sin(t.y)/cos(t.x),tan(t.x));//({cos(y)*sec(x)},{sec(x)*sin(y)},{tan(x)}); - } - - bool cond(pair t) {return tan(t.x)<1;} - - surface s=surface(f,(-1,0),(1,2*pi),32,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for y=d in red - //triple g(real t) {return (1.5*.5*tan(t),0.5,0.5/cos(t));} - //path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); - //triple g(real t) {return (1.5*.5*tan(t),0.5,-0.5/cos(t));} - //path3 mypath=graph(g,-1.315,1.315,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent\centering In plane\\ $y=0$",(2,0,-1),S,invisible); - label("\noindent\centering In plane\\ $x=0$",(0,2,-.5),S,invisible); - label("\noindent\centering In plane\\ $z=0$",(-1.6,1.6,.4),N,invisible); - - - - - - - Plane - - Trace - - - - - - - - x=d - - Hyperbola - - - y=d - - Hyperbola - - - z=d - - Ellipse - - - - - - Graph showing hyperboloid of one sheet. - - -

    - The three axes are uncalibrated. Graph shows a hyperboloid of one sheet. The hyperboloid of one sheet - is drawn about the z axis. It appears to be a cylinder with a narrow middle. In the middle of - the sheet there is a circle drawn on the plane z=o. There are two hyperbolas drawn on the plane - x=0 and y=0. -

    -     - - - - - //ASY file for figquadric_hyp_one_sheetb.pdf in Chapter 10.1 - //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? - //STILL NEED TO FIX LABELS - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - xaxis3("",-3,3,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",-3,3,invisible,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",-3,3,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 = x^2/(1.5)^2+y^2 - triple f(pair t) { - return (cos(t.y)/cos(t.x),sin(t.y)/cos(t.x),tan(t.x));//({cos(y)*sec(x)},{sec(x)*sin(y)},{tan(x)}); - } - surface s=surface(f,(-1,0),(1,2*pi),32,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - triple g(real t) {return (0,1/cos(t),tan(t));} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (0,-1/cos(t),tan(t));} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for y=0 in red - triple g(real t) {return (1/cos(t),0,tan(t));} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (-1/cos(t),0,tan(t));} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for z=0 in red - triple g(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent\centering In plane\\ $y=0$",(2,0,-1),S); - draw((2,0,-1)--(1.2,0,-.5),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $x=0$",(0,2,-.5),S); - draw((0,2,-.5)--(0,1.2,-.5),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $z=0$",(-1.6,1.6,.4),N); - draw((-1.6,1.6,.4)--(-.75,.75,0),linewidth(.75),Arrow3); - - - - -     - -

    - The one variable with a negative coefficient corresponds to the axis that the hyperboloid opens along. - quadric surfacehyperboloid of one sheet -

    - -

    -

    -
  • - Hyperboloid of Two Sheets -

    - \frac{z^2}{c^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 -

    -
    • -
    -

    - - - - - - Graph showing hyperboloid of two sheets. - - -

    - The three axes are uncalibrated. Graph shows a hyperboloid of two sheets. -

    -     - - - - - //ASY file for figquadric_hyp_two_sheetb.pdf in Chapter 10.1 - //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? - //STILL NEED TO FIX LABELS - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 - x^2 - y^2=1 - triple f(pair t) { - return (cos(t.y)*tan(t.x),sin(t.y)*tan(t.x),1/cos(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) - } - surface s=surface(f,(-1,0),(1,pi),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - triple f(pair t) { - return (cos(t.y)*tan(t.x),sin(t.y)*tan(t.x),-1/cos(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) - } - - bool cond(pair t) {return 1/cos(t.x) <= 1.6;} - - surface s=surface(f,(-1,0),(1,pi),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - //triple g(real t) {return (0,1/cos(t),tan(t));} - //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - //triple g(real t) {return (0,-1/cos(t),tan(t));} - //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for y=0 in red - //triple g(real t) {return (1/cos(t),0,tan(t));} - //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - //triple g(real t) {return (-1/cos(t),0,tan(t));} - //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for z=0 in red - //triple g(real t) {return (cos(t),sin(t),0);} - //path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent\centering In plane\\ $y=0$",(1.5,0,-1),N,invisible); - label("\noindent\centering In plane\\ $x=0$",(0,1.5,-1),N,invisible); - label("\noindent\centering In plane\\ $z=d$",(1.1,1.1,-2),S,invisible); - - - - - - - Plane - - Trace - - - - - - - - x=d - - Hyperbola - - - y=d - - Hyperbola - - - z=d - - Ellipse - - - - - - Graph showing hyperboloid of two sheets. - - -

    - The three axes are uncalibrated. Graph shows a hyperboloid of two sheets. The hyperboloids of - two sheets are drawn about the z axis. The first sheet opens along the positive z - axis. The second sheet to the bottom opens along the negative z axis. Both plates have two - hyperbolas drawn one in the zy plane and one in the xz plane. In the bottom sheet - there is a circle drawn on the plane z=d. -

    -     - - - - - //ASY file for figquadric_hyp_two_sheetb.pdf in Chapter 10.1 - //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? - //STILL NEED TO FIX LABELS - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 - x^2 - y^2=1 - triple f(pair t) { - return (cos(t.y)*tan(t.x),sin(t.y)*tan(t.x),1/cos(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) - } - surface s=surface(f,(-1,0),(1,pi),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - triple f(pair t) { - return (cos(t.y)*tan(t.x),sin(t.y)*tan(t.x),-1/cos(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) - } - surface s=surface(f,(-1,0),(1,pi),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - triple g(real t) {return (0,tan(t),1/cos(t));} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (0,tan(t),-1/cos(t));} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for y=0 in red - triple g(real t) {return (tan(t),0,1/cos(t));} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (tan(t),0,-1/cos(t));} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for z=-1.7 in red - triple g(real t) {return (1.37*cos(t),1.37*sin(t),-1.7);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent\centering In plane\\ $y=0$",(1.5,0,-1),N); - draw((1.5,0,-1)--(1,0,-1.35),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $x=0$",(0,1.5,-1),N); - draw((0,1.5,-1)--(0,1,-1.35),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $z=d$",(1.1,1.1,-2),S); - draw((1.1,1.1,-2)--(.95,.95,-1.7),linewidth(.75),Arrow3); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -     - -

    - The one variable with a positive coefficient corresponds to the axis that the hyperboloid opens along. - In the case illustrated, when \abs{d}\lt \abs{c}, there is no trace. - quadric surfacehyperboloid of two sheets -

    - -

    -

    -
  • - Hyperbolic Paraboloid -

    - z=\frac{x^2}{a^2}-\frac{y^2}{b^2} -

    -
    • -
    -

    - - - - - - - Graph of the hyperbolic paraboloid. - - -

    - The three axes are uncalibrated. Graph shows a hyperbolic paraboloid. -

    -     - - - - - //ASY file for figquadric_hyp_par.pdf in Chapter 10.1 - //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? - //STILL NEED TO FIX LABELS - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(2,5,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 - x^2 - y^2=1 - triple f(pair t) { - return (t.x,t.y,(t.x)^2-(t.y)^2);//({x},{y},{x^2-y^2}); - } - surface s=surface(f,(-1,-1),(1,1),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - //triple g(real t) {return (0,1/cos(t),tan(t));} - //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - //triple g(real t) {return (0,-1/cos(t),tan(t));} - //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for y=0 in red - //triple g(real t) {return (1/cos(t),0,tan(t));} - //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - //triple g(real t) {return (-1/cos(t),0,tan(t));} - //path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for z=0 in red - //triple g(real t) {return (cos(t),sin(t),0);} - //path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - - - - Plane - - Trace - - - - - - - - x=d - - Parabola - - - y=d - - Parabola - - - z=d - - Hyperbola - - -     - - - - - - Graph of the hyperbolic paraboloid. - - -

    - The three axes are uncalibrated. There are two parabolas drawn, one in plane - y=0 opening up along the positive z axis in the yz plane - and the other in x=0 opening down along the negative z axis in the - xz plane. Both parabolas have vertices at the origin. - Filling the traces gives the hyperbolic paraboloid. -

    -     - - - - - //ASY file for figquadric_hyp_parb.pdf in Chapter 10.1 - //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? - //STILL NEED TO FIX LABELS - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(2,5,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 - x^2 - y^2=1 - triple f(pair t) { - return (t.x,t.y,(t.x)^2-(t.y)^2);//({x},{y},{x^2-y^2}); - } - surface s=surface(f,(-1,-1),(1,1),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - triple g(real t) {return (0,t,-t^2);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for y=0 in red - triple g(real t) {return (t,0,t^2);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent\centering In plane\\ $y=0$",(.5,0,.6),N); - draw((.5,0,.6)--(.15,0,.05),linewidth(.75),Arrow3); - label("\noindent\centering In plane\\ $x=0$",(0,1.5,0.3),E); - draw((0,1.5,.3)--(0,.7,-.4),linewidth(.75),Arrow3); - //draw((-.2,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - - - - - Graph of the hyperbolic paraboloid. - - -

    - The three axes are uncalibrated. Graph shows the hyperbolic paraboloid along with two hyperbolas. There are two hyperbolas drawn, - in plane z=d. For d>0 the two hyperbolas opening up along the positive and negative x axis. For d<0 - the other hyperbola opens along the positive and negative y axis. -

    -     - - - - - //ASY file for figquadric_hyp_parb.pdf in Chapter 10.1 - //LOOKS A LITTLE STRANGE AT THE EDGES. PERHAPS CHANGE t RANGE? - //STILL NEED TO FIX LABELS - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(2,5,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //xaxis3("",-2,2,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); - //yaxis3("",-2,2,invisible,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",-2,2,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 - x^2 - y^2=1 - triple f(pair t) { - return (t.x,t.y,(t.x)^2-(t.y)^2);//({x},{y},{x^2-y^2}); - } - surface s=surface(f,(-1,-1),(1,1),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for z=1/4 in red - triple g(real t) {return (0.5/cos(t),0.5*tan(t),0.25);} - path3 mypath=graph(g,-1.05,1.05,operator ..); draw(mypath,redpen); - triple g(real t) {return (-0.5/cos(t),0.5*tan(t),0.25);} - path3 mypath=graph(g,-1.05,1.05,operator ..); draw(mypath,redpen); - - //Draw the traces for z=-1/4 in red - triple g(real t) {return (0.5*tan(t),0.5/cos(t),-0.25);} - path3 mypath=graph(g,-1.05,1.05,operator ..); draw(mypath,redpen); - triple g(real t) {return (0.5*tan(t),-0.5/cos(t),-0.25);} - path3 mypath=graph(g,-1.05,1.05,operator ..); draw(mypath,redpen); - - //Add labels - label("\noindent\centering Hyperbolas in \\plane $z=d$",(.7,0,1),N); - label("\noindent\centering $(d>0)$",(.5,0,.6),N); - draw((.52,0,.6)--(.5,0,.25),linewidth(.75),Arrow3); - draw((.48,0,.6)--(-.5,0,.25),linewidth(.75),Arrow3); - label("\noindent\centering $(d<0)$",(1,0,-0.5),S); - draw((1,.1,-0.5)--(.8,.9,-.25),linewidth(.75),Arrow3); - draw((1,-.1,-0.5)--(.8,-.9,-.25),linewidth(.75),Arrow3); - - - - -     -     -

    - The parabolic traces will open along the axis of the one variable that is raised to the first power. - quadric surfacegallery - quadric surfacehyperbolic paraboloid -

    - - - Sketching quadric surfaces - -

    - Sketch the quadric surface defined by the given equation. -

    - -

    -

      -
    1. y=\frac{x^2}{4}+\frac{z^2}{16}
      2. -
    2. x^2+\frac{y^2}{9}+\frac{z^2}{4}=1
      3. -
    3. \ds z=y^2-x^2
      4. -
    -

    -     - -

    -

      -
    1. -

      - \ds y=\frac{x^2}{4}+\frac{z^2}{16}: We first identify the quadric by pattern-matching with the equations given previously. - Only two surfaces have equations where one variable is raised to the first power, - the elliptic paraboloid and the hyperbolic paraboloid. - In the latter case, the other variables have different signs, - so we conclude that this describes a hyperbolic paraboloid. - As the variable with the first power is y, - we note the paraboloid opens along the y-axis. - To make a decent sketch by hand, - we need only draw a few traces. - In this case, - the traces x=0 and z=0 form parabolas that outline the shape. -

      -

      - x=0: The trace is the parabola - y=z^2/16 -

      -

      - z=0: The trace is the parabola y=x^2/4. -

      - -

      - Graphing each trace in the respective plane creates a sketch as shown in . - This is enough to give an idea of what the paraboloid looks like. - The surface is filled in in . -

      - -
      - Sketching an elliptic paraboloid - -
      - - - - - Graph of two parabolas that form the elliptic paraboloid. - - -

      - The x and z axes are drawn from -4 to 4 and the y axis - is drawn from 0 to 2. Two parabolas are drawn that are perpendicular to each other, - one in the xy plane and the other in the yz plane. Both have vertices at origin. -

      -       - - - - - //ASY file for figspace5ab.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(15,5,4.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-4,4}; - real[] myychoice={1,2}; - real[] myzchoice={-4,4}; - defaultpen(0.5mm); - pair xbounds=(-4.5,4.5); - pair ybounds=(-0.5,2.5); - pair zbounds=(-4.5,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface y=x^2/4+z^2/16 - //triple f(pair t) { - // return (1.5*cos(t.y)*cos(t.x),cos(t.y)*sin(t.x),sin(t.y));//({cos(x)*1.5*cos(y)},{sin(x)*cos(y)},{sin(y)}) - //} - //surface s=surface(f,(0,-2),(pi,2),32,16); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - triple g(real t) {return (0,t^2/16,t);} - path3 mypath=graph(g,-4,4,operator ..); draw(mypath,redpen); - - //Draw the traces for z=0 in red - triple g(real t) {return (t,t^2/4,0);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -
      - -
      - - - - - Graph of elliptic paraboloid. - - -

      - The three axes are uncalibrated. Two parabolas are drawn that are perpendicular to - each other, one in the xy plane and the other in the yz plane. Both have vertices at origin. - Filling in the trace gives the elliptic paraboloid. -

      -       - - - - - //ASY file for figspace5a.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(15,5,4.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-4.5,4.5); - pair ybounds=(-0.5,2.5); - pair zbounds=(-4.5,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface y=x^2/4+z^2/16 - triple f(pair t) { - return (2*t.y*cos(t.x),(t.y)^2,4*t.y*sin(t.x));//({2*cos(x)*y},{y^2},{4*y*sin(x)}); - } - surface s=surface(f,(0,-1),(pi,1),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - triple g(real t) {return (0,t^2/16,t);} - path3 mypath=graph(g,-4,4,operator ..); draw(mypath,redpen); - - //Draw the traces for z=0 in red - triple g(real t) {return (t,t^2/4,0);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -
      -       - -
      -
      2. - -
    2. -

      - \ds x^2+\frac{y^2}{9}+\frac{z^2}{4}=1: This is an ellipsoid. - We can get a good idea of its shape by drawing the traces in the coordinate planes. -

      - -

      - x=0: The trace is the ellipse \ds\frac{y^2}{9}+\frac{z^2}{4}=1. - The major axis is along the y-axis with length 6 (as b=3, - the length of the axis is 6); - the minor axis is along the z-axis with length 4. -

      - -

      - y=0: The trace is the ellipse \ds x^2+\frac{z^2}{4}=1. - The major axis is along the z-axis, - and the minor axis has length 2 along the x-axis. - z=0: The trace is the ellipse \ds x^2+\frac{y^2}{9}=1, - with major axis along the y-axis. -

      - -

      - Graphing each trace in the respective plane creates a sketch as shown in . Filling in the surface gives . -

      - -
      - Sketching an ellipsoid - -
      - - - - - Graph of trace for the ellipsoid. - - -

      - The x, y and z axes are drawn from -3 to 3. - There are three ellipses drawn. The first one is on the xz plane, y=0 - with equation x^2 + z^2/4 =1. The second is on the yz plane, with x=0 - with the equation y^2/9 + z^2/4 =1. The third on the xy plane, with z=0 - and has an equation x^2 +y^2 /9 =1. -

      -       - - - - - //ASY file for figspace5b.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-3,3}; - real[] myychoice={-3,3}; - real[] myzchoice={-3,3}; - defaultpen(0.5mm); - pair xbounds=(-4.5,4.5); - pair ybounds=(-4.5,4.5); - pair zbounds=(-4.5,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2+y^2/9+z^2/4=1 - //triple f(pair t) { - // return (cos(t.y)*cos(t.x),3*sin(t.y)*cos(t.x),2sin(t.x));// - //} - //surface s=surface(f,(0,0),(2*pi,2*pi),32,16,Spline); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - triple g(real t) {return (0,3*cos(t),2*sin(t));} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Draw the traces for y=0 in red - triple g(real t) {return (cos(t),0,2*sin(t));} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Draw the traces for z=0 in red - triple g(real t) {return (cos(t),3*sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -
      - -
      - - - - - Graph of the three dimensional ellipsoid. - - -

      - The x, y and z axes are drawn from -3 to 3. There are three ellipses drawn. The first one is on the - xz plane, y=0 with equation x^2 + z^2/4 =1. The second is on the yz plane, with x=0 with the equation - y^2/9 + z^2/4 =1. The third on the xy plane, with z=0 and has an equation x^2 +y^2 /9 =1. - Filling in the surface gives the ellipsoid. -

      -       - - - - - //ASY file for figspace5bb.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-4.5,4.5); - pair ybounds=(-4.5,4.5); - pair zbounds=(-4.5,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2+y^2/9+z^2/4=1 - triple f(pair t) { - return (cos(t.y)*cos(t.x),3*sin(t.y)*cos(t.x),2sin(t.x));// - } - surface s=surface(f,(-pi,-pi/2),(pi,pi/2),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - triple g(real t) {return (0,3*cos(t),2*sin(t));} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Draw the traces for y=0 in red - triple g(real t) {return (cos(t),0,2*sin(t));} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Draw the traces for z=0 in red - triple g(real t) {return (cos(t),3*sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,redpen); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -
      -       -
      -
      3. - -
    3. -

      - \ds z=y^2-x^2: This defines a hyperbolic paraboloid, - very similar to the one shown in the gallery of quadric sections. - Consider the traces in the y-z and x-z planes: -

      - -

      - x=0: The trace is z=y^2, - a parabola opening up in the y-z plane. -

      - -

      - y=0: The trace is z=-x^2, - a parabola opening down in the x-z plane. -

      - -

      - Sketching these two parabolas gives a sketch like that in , and filling in the surface gives a sketch like . -

      - -
      - Sketching a hyperbolic paraboloid - -
      - - - - - Graph showing traces for a hyperbolic paraboloid. - - -

      - The x, y and z axes are drawn from -1 to 1. - There are two parabolas drawn, one in plane x=0 with equation z= y^2 - opening up in the yz plane and the other in y=0 with equation z=-x^2 - opening down in the xz plane. Both parabolas have vertices at the origin. -

      -       - - - - - //ASY file for figspace5c.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4.5,3,2.7); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 = y^2 - x^2 - //triple f(pair t) { - // return (t.x,t.y,-(t.x)^2+(t.y)^2);// - //} - //surface s=surface(f,(-1,-1),(1,1),32,16,Spline); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - triple g(real t) {return (0,t,t^2);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for y=0 in red - triple g(real t) {return (t,0,-t^2);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -
      - -
      - - - - - Graph of the hyperbolic paraboloid. - - -

      - The three axes are uncalibrated. There are two parabolas drawn, one in - plane x=0 with equation z= y^2 opening up in the yz - plane and the other in y=0 with equation z=-x^2 opening down - in the xz plane. Both parabolas have vertices at the origin. - Filling the traces gives the hyperbolic paraboloid. -

      -       - - - - - //ASY file for figspace5cb.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4.5,3,2.7); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2 = y^2 - x^2 - triple f(pair t) { - return (t.x,t.y,-(t.x)^2+(t.y)^2);// - } - surface s=surface(f,(-1,-1),(1,1),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces for x=0 in red - triple g(real t) {return (0,t,t^2);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the traces for y=0 in red - triple g(real t) {return (t,0,-t^2);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -
      -       -
      -
      4. -
    -

    -     - -     - - - Identifying quadric surfaces - -

    - Consider the quadric surface shown in . - Which of the following equations best fits this surface? -

    - -

    -

      -
    1. \ds x^2-y^2-\frac{z^2}{9}=0
      2. -
    2. \ds x^2-y^2-z^2=1
      3. -
    3. \ds z^2-x^2-y^2=1
      4. -
    4. 4x^2-y^2-\frac{z^2}9=1
      5. -
    -

    - -
    - A possible equation of this quadric surface is found in - - - - Graph showing hyperboloid of two sheets. - - -

    - The z and y axes are drawn from -3 to 3 and the x axis is - drawn from -1 to 1. The y and z axes are drawn from -3 to - 3. The hyperboloids of two sheets are drawn about the x axis. The first sheet - has a centre at x =0.5 and opens along the positive y axis. The second sheet has - a centre at x=-0.5 and opens along the negative y axis. -

    -     - - - - - //ASY file for figspace6.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,9.5,5.7); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-3,3}; - real[] myzchoice={-3,3}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-4.5,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface 4x^2 - y^2 - z^2/9 = 1 - triple f(pair t) { - return (0.5/cos(t.x),cos(t.y)*tan(t.x),3*sin(t.y)*tan(t.x));// - } - surface s=surface(f,(-pi/3,0),(pi/3,pi),16,24); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - triple f(pair t) { - return (-0.5/cos(t.x),cos(t.y)*tan(t.x),3*sin(t.y)*tan(t.x));// - } - surface s=surface(f,(-pi/3,0),(pi/3,pi),16,24); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -
    -     - -

    - The image clearly displays a hyperboloid of two sheets. - The gallery informs us that the equation will have a form similar to \frac{z^2}{c^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. -

    - -

    - We can immediately eliminate option (a), - as the constant in that equation is not 1. -

    - -

    - The hyperboloid opens along the x-axis, - meaning x must be the only variable with a positive coefficient, - eliminating (c). -

    - -

    - The hyperboloid is wider in the z-direction than in the y-direction, - so we need an equation where c \gt b. - This eliminates (b), leaving us with (d). - We should verify that the equation given in (d), - 4x^2-y^2-\frac{z^2}9=1, fits. -

    - -

    - We already established that this equation describes a hyperboloid of two sheets that opens in the x-direction and is wider in the z-direction than in the y. - Now note the coefficient of the x-term. - Rewriting 4x^2 in standard form, we have: - \ds 4x^2 = \frac{x^2}{(1/2)^2}. - Thus when y=0 and z=0, - x must be 1/2; - , each hyperboloid starts at x=1/2. - This matches our figure. -

    - -

    - We conclude that \ds 4x^2-y^2-\frac{z^2}9=1 best fits the graph. -

    -     -     - - - -

    - This section has introduced points in space and shown how equations can describe surfaces. - The next sections explore vectors, - an important mathematical object that we'll use to explore curves in space. -

    -     - - - - Terms and Concepts - - - - -

    - Axes drawn in space must conform to the rule. -

    -     - - - - - - - - -     - - - - - - - - -

    - In the plane, - the equation x=2 defines a ; - in space, x=2 defines a . -

    -     - - - - - - - - - - - - - -     - - - - -

    - In the plane, the equation y=x^2 defines a ; - in space, y=x^2 defines a . -

    -     - - - - curve|parabola - - - - - surface|cylinder|parabolic cylinder - - - - -     - - - - - -

    - Which quadric surface looks like a Pringleschip? -

    - -     - - - -

    - Elliptic paraboloid -

    -     -     - - -

    - Elliptic cone -

    -     -     - - -

    - Ellipsoid -

    -     -     - - -

    - Hyperboloid of one sheet -

    -     -     - - -

    - Hyperboloid of two sheets -

    -     -     - - -

    - Hyperbolic paraboloid -

    -     -     -     - -     - - - - -

    - Consider the hyperbola x^2-y^2=1 in the plane. - If this hyperbola is rotated about the x-axis, - what quadric surface is formed? -

    -     - - - -

    - A hyperboloid of two sheets -

    -     - -     - - - - -

    - Consider the hyperbola x^2-y^2=1 in the plane. - If this hyperbola is rotated about the y-axis, - what quadric surface is formed? -

    -     - - - -

    - A hyperboloid of one sheet -

    -     - -     -     - - - Problems - - - - parserPopUp.pl - - - Context("Point"); - $A=Point("(1,4,2)"); - $B=Point("(2,6,3)"); - $C=Point("(4,3,1)"); - $AB=$B-$A; - $BC=$C-$B; - $CA=$A-$C; - @ABcoords=$AB->value; - @BCcoords=$BC->value; - @CAcoords=$CA->value; - Context()->flags->set(reduceConstantFunctions=>0); - $magAB=Formula("sqrt(($ABcoords[0])^2+($ABcoords[1])^2+($ABcoords[2])^2)"); - $magBC=Formula("sqrt(($BCcoords[0])^2+($BCcoords[1])^2+($BCcoords[2])^2)"); - $magCA=Formula("sqrt(($CAcoords[0])^2+($CAcoords[1])^2+($CAcoords[2])^2)"); - ($a,$b,$c)=num_sort($magAB,$magBC,$magCA); - $dodonot=(($a)**2+($b)**2==($c)**2)?'do':'do not'; - $yn=DropDown(['do','do not'],$dodonot,showInStatic=>0); - - - -

    - The points A=(1,4,2), - B=(2,6,3) and C=(4,3,1) form a triangle in space. - Find the distances between each pair of points and determine if the triangle is a right triangle. -

    - - Find the distance \left\lVert\overline{AB}\right\rVert - -

    - -

    - - Find the distance \left\lVert\overline{BC}\right\rVert - -

    - -

    - - Find the distance \left\lVert\overline{CA}\right\rVert - -

    - -

    - - Use the drop-down to indicate whether or not the three points form a right triangle. - -

    - -

    -     - -     -     - - - - -

    - The points A=(1,1,3), B=(3,2,7), - C=(2,0,8) and D = (0,-1,4) form a quadrilateral ABCD in space. - Is this a parallelogram? -

    -     - -

    - Yes, as opposite sides have equal length. - \norm{\overline{AB}} = \sqrt{21}=\norm{\overline{CD}}; - \norm{\overline{BC}} = \sqrt{6}=\norm{\overline{AD}}. -

    -     - -     - - - - - Context("Point"); - $center=Point("(4,-1,0)"); - $radius=Real("3"); - - - -

    - Find the center and radius of the sphere defined by - - x^2-8x+y^2+2y+z^2+8=0: - -

    - - Enter the center of the sphere. -

    - -

    - - Enter the radius of the sphere. - -

    - -

    -     -     -     - - - - - Context("Point"); - $center=Point("(-2,1,2)"); - Context()->flags->set(reduceConstantFunctions=>0); - $radius=Formula("sqrt(5)"); - - - -

    - Find the center and radius of the sphere defined by - - x^2+y^2+z^2+4x-2y-4z+4=0: - -

    - - Enter the center of the sphere. -

    - -

    - - Enter the radius of the sphere. - -

    - -

    -     -     -     - - - - -

    - Describe the region in space defined by the inequalities. -

    -     - - - - -

    - x^2+y^2+z^2\lt 1 -

    -     - -

    - Interior of a sphere with radius 1 centered at the origin. -

    -     - -     - - - - - - - -

    - 0\leq x\leq 3 -

    -     - -

    - Region bounded between the planes x=0 - (the y-z coordinate plane) - and x=3. -

    -     -     -     - - - - -

    - x\geq 0,\ y\geq0, \ z\geq0 -

    -     - -

    - The first octant of space; - all points (x,y,z) where each of x, - y and z are positive. - (Analogous to the first quadrant in the plane.) -

    -     - -     - - - - -

    - y\geq 3 -

    -     - -

    - All points in space where the y value is greater than 3; - viewing space as often depicted in this text, - this is the region to the right - of the plane y=3 (which is parallel to the x-z coordinate plane.) -

    -     - -     - -     - - - - -

    - Sketch the cylinder in space. -

    -     - - - - -

    - z=x^3 -

    -     - - - - A surface obtained by translating the cubic curve along the y axis - -

    - Viewed with the y axis pointing out the page, - the surface looks like the graph z=x^3 in the xz plane. - The rest of the surface is generated by sliding this curve back and forth along the y axis. - The result is a sheet that looks something like a lounge chair. -

    -     - - - - - //ASY file for fig10_01_ex_153D.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,1.25); - pair zbounds=(-1.25,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the parabola z=x^3 for t from -2 to 2 - triple g(real t) {return (t,0,t^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,bluepen); - - //Draw the cylinder z=x^3 - triple f(pair t) { - return (t.x,t.y,(t.x)^3); - } - surface s=surface(f,(-1,-1),(1,1),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces (in this case lines) - triple g(real t) {return (-0.2,t,(-0.2)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (-0.4,t,(-0.4)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (-0.6,t,(-0.6)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (-0.8,t,(-0.8)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (-1,t,(-1)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (0.2,t,(0.2)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (0.4,t,(0.4)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (0.6,t,(0.6)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (0.8,t,(0.8)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (1,t,(1)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - triple g(real t) {return (t,0,(t)^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,bluepen); - - - - -     - -     - - - - -

    - y=\cos(z) -

    -     - - - - A surface generated by translating the curve y=cos(z) along the x axis - -

    - In the yz plane, the surface appears to be a cosine curve. - This curve is translated back and forth along the x axis, - so that when viewed from other angles, it appears to be a wavy sheet. -

    -     - - - - - //ASY file for fig10_01_ex_173D.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={-3,3}; - real[] myzchoice={-6,6}; - defaultpen(0.5mm); - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2*pi,2*pi); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the cylinder on top - triple f(pair t) { - return (t.x,cos(t.y),t.y); - } - surface s=surface(f,(-2,-2*pi),(2,2*pi),8,32,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces (in this case vertical lines) - triple g(real t) {return (t,cos(-3*pi/2),-3*pi/2);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (t,cos(-pi/2),-pi/2);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (t,cos(0),0);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (t,cos(3*pi/2),3*pi/2);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (t,cos(pi/2),pi/2);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (0,cos(t),t);} - path3 mypath=graph(g,-2*pi,pi*2,operator ..); draw(mypath,bluepen); - - - - -     - -     - - - - -

    - \ds \frac{x^2}{4}+\frac{y^2}{9}=1 -

    -     - - - - A cylindrical tube with cross sections in the shape of an ellipse - -

    - A classic cylinder, in the shape of a tube. - Cross-sections parallel to the xy plane are ellipses. -

    -     - - - - - //ASY file for fig10_01_ex_173D.pdf in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-3,3}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-4,4); - pair ybounds=(-4,4); - pair zbounds=(-4,4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the cylinder on top - triple f(pair t) { - return (2*cos(t.x),3*sin(t.x),t.y); - } - surface s=surface(f,(0,-2),(2*pi,2),32,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces (in this case vertical lines) - triple g(real t) {return (2*cos(0),3*sin(0),t);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (2*cos(2*pi/10),3*sin(2*pi/10),t);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (2*cos(2*2*pi/10),3*sin(2*2*pi/10),t);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (2*cos(3*2*pi/10),3*sin(3*2*pi/10),t);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (2*cos(4*2*pi/10),3*sin(4*2*pi/10),t);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (2*cos(5*2*pi/10),3*sin(5*2*pi/10),t);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (2*cos(6*2*pi/10),3*sin(6*2*pi/10),t);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (2*cos(7*2*pi/10),3*sin(7*2*pi/10),t);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (2*cos(8*2*pi/10),3*sin(8*2*pi/10),t);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - triple g(real t) {return (2*cos(9*2*pi/10),3*sin(9*2*pi/10),t);} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,redpen); - - triple g(real t) {return (2*cos(t),3*sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - - - -     - -     - - - - -

    - \ds y=\frac1x -

    -     - - - - A pair of sheets that look like a hyperbola when viewed from above - -

    - Viewed along the z axis, the surface appears to be the hyperbola y=1/x. - From other angles, it is a pair of curved sheets, bent in the shape of a hyperbola. -

    -     - - - - - //ASY file for fig10_01_ex_20.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(-8,41,2.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-10,10}; - real[] myychoice={-10,10}; - real[] myzchoice={-5,5}; - defaultpen(0.5mm); - - pair xbounds=(-12,12); - pair ybounds=(-12,12); - pair zbounds=(-6,6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface y=1/x - triple f(pair t) { - return (t.x,1/t.x,t.y);//({y^2},{cos(x)*y},{3*sin(x)*y}); - } - surface s=surface(f,(0.1,-3),(10,3),16,8); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (t.x,1/t.x,t.y);//({y^2},{cos(x)*y},{3*sin(x)*y}); - } - surface s=surface(f,(-10,-3),(-.1,3),16,8); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the traces (in this case vertical lines) - triple g(real t) {return (1,1,t);} - path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); - triple g(real t) {return (-1,-1,t);} - path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); - triple g(real t) {return (5,.2,t);} - path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); - triple g(real t) {return (-5,-.2,t);} - path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); - triple g(real t) {return (10,.1,t);} - path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); - triple g(real t) {return (-10,-.1,t);} - path3 mypath=graph(g,-3,3,operator ..); draw(mypath,redpen); - triple g(real t) {return (t,1/t,0);} - path3 mypath=graph(g,.1,10,operator ..); draw(mypath,bluepen); - triple g(real t) {return (t,1/t,0);} - path3 mypath=graph(g,-10,-.1,operator ..); draw(mypath,bluepen); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -     - -     - -     - - - -

    - Give the equation of the surface of revolution described. -

    -     - - - - - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - $eq=ImplicitEquation("x^2+z^2=(1/(1+y^2))^2"); - - - -

    - Give the equation of the surface formed by revolving - z=\frac{1}{1+y^2} in the yz-plane about the y-axis. -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - $eq=ImplicitEquation("y^2+z^2=x^4"); - - - -

    - Give the equation of the surface formed by revolving y=x^2 in the xy-plane about the x-axis. -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - $eq=ImplicitEquation("x^2+y^2=z"); - - - -

    - Give the equation of the surface formed by revolving z=x^2 in the xz-plane about the z-axis. -

    - -

    - -

    -     -     -     - - - - - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - $eq=ImplicitEquation("x^2+y^2=1/z^2"); - - - -

    - Give the equation of the surface formed by revolving z=1/x in the xz-plane about the z-axis. -

    - -

    - -

    -     -     -     - -     - - - -

    - A quadric surface is sketched. - Determine which of the given equations best fits the graph. -

    -     - - - - - - - - Graph showing a paraboloid opening along the x axis. - - -

    - The z and y axes are drawn from -3 to 3 and the x axis is - drawn from -1 to 1. The y and z axes are drawn from -3 to 3. The elliptic paraboloid is shown with centre at the origin and it opens along the positive x axis. -

    -     - - - - - //ASY file for fig10_01_ex_19.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(3,12,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={-3,3}; - real[] myzchoice={-3,3}; - defaultpen(0.5mm); - - pair xbounds=(-.5,1.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-3.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2 + y^2/9 + z^2/4 = 1 - triple f(pair t) { - return (t.y^2,cos(t.x)*t.y,3*sin(t.x)*t.y);//({y^2},{cos(x)*y},{3*sin(x)*y}); - } - surface s=surface(f,(0,0),(2*pi,1),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -

    - (a) \ds x=y^2+\frac{z^2}{9} (b) \ds x=y^2+\frac{z^2}{3} -

    -     - -

    - (a)\ds x=y^2+\frac{z^2}{9} -

    -     - -     - - - - - - - - Graph showing elliptic cone opening along the y axis. - - -

    - The axes are drawn from -1 to 1. Two hollow elliptic cones are drawn with - vertices at the origin, one opening along the positive y axis and the other along - the negative y axis. -

    -     - - - - - //ASY file for fig10_01_ex_20.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,3,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2 + y^2/9 + z^2/4 = 1 - triple f(pair t) { - return (cos(t.x)*(t.y)^2,(t.y)^2,sin(t.x)*(t.y)^2);//({cos(x)*(y)^2},{y^2},{sin(x)*(y)^2}); - } - surface s=surface(f,(0,0),(2*pi,1),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (cos(t.x)*(t.y)^2,-(t.y)^2,sin(t.x)*(t.y)^2);//({cos(x)*(y)^2},{y^2},{sin(x)*(y)^2}); - } - surface s=surface(f,(0,0),(2*pi,1),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -

    - (a) \ds x^2-y^2-z^2=0 (b) x^2-y^2+z^2=0 -

    -     - -

    - (b) x^2-y^2+z^2=0 -

    -     - -     - - - - - - - - Graph of an ellipsoid. - - -

    - The three axes are uncalibrated. An ellipsoid is shown. -

    -     - - - - - //ASY file for fig10_01_ex_21.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-3,3}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-3.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2 + y^2/9 + z^2/4 = 1 - triple f(pair t) { - return (cos(t.x)*cos(t.y),3*sin(t.x)*cos(t.y),2*sin(t.y));//({cos(x)*cos(y)},{sin(x)*3*cos(y)},{2*sin(y)}); - } - surface s=surface(f,(-pi,-pi/2),(pi,pi/2),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -

    - (a) \ds x^2+\frac{y^2}3+\frac{z^2}2=1 (b) \ds x^2+\frac{y^2}9+\frac{z^2}4=1 -

    -     - -

    - (b) \ds x^2+\frac{y^2}9+\frac{z^2}4=1 -

    -     - -     - - - - - - - - Graph showing hyperboloid with two sheets. - - -

    - The x, y and z axes are drawn from -2 to 2. - The hyperboloid of two sheets is drawn about the y axis. The first sheet - has a centre at y =1 and opens along the positive y axis. The second - sheet has a centre at y=-1 and opens along the negative y axis. -

    -     - - - - - //ASY file for fig10_01_ex_22.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface 4x^2 - y^2 - z^2/9 = 1 - triple f(pair t) { - return (cos(t.y)*tan(t.x),1/cos(t.x),sin(t.y)*tan(t.x));//({cos(y)*tan(x)},{sec(x)},{tan(x)*sin(y)}); - } - surface s=surface(f,(0,0),(1,2*pi),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (cos(t.y)*tan(t.x),-1/cos(t.x),sin(t.y)*tan(t.x));//({cos(y)*tan(x)},{sec(x)},{tan(x)*sin(y)}); - } - surface s=surface(f,(0,0),(1,2*pi),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Add labels - //label("In plane $y=0$",(0,4,1.5),E); - //draw((0,4,1.5)--(3,0,2),linewidth(.75),Arrow3); - //draw((0,5,1.75)--(-3,0,2),linewidth(.75),Arrow3); - //label("In plane $z=d$",(4,0,-1),W); - //draw((-1,3,1.75)--(-2.1,1.1,2),linewidth(.75),Arrow3); - - - - -

    - (a) y^2-x^2-z^2=1 (b) y^2+x^2-z^2=1 -

    -     - -

    - (a) y^2-x^2-z^2=1 -

    -     - -     - -     - - - -

    - Sketch the quadric surface. -

    -     - - - - -

    - \ds z-y^2+x^2=0 -

    -     - - - - A hyperbolic paraboloid (the classic "Pringles chip") - -

    - The surface is a standard hyperbolic paraboloid, opening upward along the y axis, - and downward along the x axis. -

    -     - - - - - //ASY file for fig10_01_ex_283D.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,3,1); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.25,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the hyp par on top - //Draw the surface z^2 - y^2 - x^2=0 - triple f(pair t) { - return (t.x,t.y,-(t.x)^2+(t.y)^2); - } - surface s=surface(f,(-1,-1),(1,1),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -     - -     - - - - -

    - \ds z^2=x^2+\frac{y^2}4 -

    -     - - - - An elliptical cone - -

    - The surface is a cone, opening along the z axis. - Cross-sections parallel to the xy plane are ellipses. -

    -     - - - - - //ASY file for fig10_01_ex_263D.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,5,1); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-4,4}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - pair xbounds=(-5,5); - pair ybounds=(-5,5); - pair zbounds=(-3,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2/9 - y^2 + z^2/25 = 1 - triple f(pair t) { - return (t.y*cos(t.x),2*t.y*sin(t.x),t.y);// - } - surface s=surface(f,(0,-2),(2*pi,2),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -     - -     - - - - -

    - x=-y^2-z^2 -

    -     - - - - A circular paraboloid, opening along the negative x axis - -

    - The surface is a circular paraboloid, opening along the negative x axis. -

    -     - - - - - //ASY file for fig10_01_ex_233D.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.25,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x=-y^2-z^2 - triple f(pair t) { - return (-t.x,sqrt(t.x)*cos(t.y),sqrt(t.x)*sin(t.y)); - } - surface s=surface(f,(0,0),(1,2*pi),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -     - -     - - - - -

    - \ds 16x^2-16y^2-16z^2=1 -

    -     - - - - A hyperboloid of two sheets, opening along the x axis. - -

    - A hyperboloid of two sheets with circular cross sections, opening along the x axis. -

    -     - - - - - //ASY file for fig10_01_ex_263D.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,5,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-.25,.25}; - real[] myychoice={-.5,.5}; - real[] myzchoice={-.5,.5}; - defaultpen(0.5mm); - pair xbounds=(-1,1); - pair ybounds=(-1,1); - pair zbounds=(-1,1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2/9 - y^2 + z^2/25 = 1 - triple f(pair t) { - return (.25/cos(t.x),.25*cos(t.y)*tan(t.x),.25*sin(t.y)*tan(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) - } - surface s=surface(f,(-pi/3,0),(pi/3,pi),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (-.25/cos(t.x),.25*cos(t.y)*tan(t.x),.25*sin(t.y)*tan(t.x));//({cos(y)*tan(x)},{tan(x)*sin(y)},{sec(x)}) - } - surface s=surface(f,(-pi/3,0),(pi/3,pi),32,16); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -     - -     - - - - -

    - \ds \frac{x^2}9-y^2+\frac{z^2}{25}=1 -

    -     - - - - A hyperboloid of one sheet, opening along the y axis - -

    - A hyperboloid of one sheet. It opens along the y axis, - and cross sections parallel to the xz plane are ellipses. -

    -     - - - - - //ASY file for fig10_01_ex_263D.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(13,5,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-3,3}; - real[] myychoice={-1,1}; - real[] myzchoice={-5,5}; - defaultpen(0.5mm); - pair xbounds=(-6,6); - pair ybounds=(-1.5,1.5); - pair zbounds=(-6,6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2/9 - y^2 + z^2/25 = 1 - triple f(pair t) { - return (3*cos(t.y)/cos(t.x),tan(t.x),5*sin(t.y)/cos(t.x)); - } - surface s=surface(f,(-1,0),(1,2*pi),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -     - -     - - - - -

    - \ds 4x^2+2y^2+z^2=4 -

    -     - - - - An ellpsoid, somewhat in the shape of a squashed olive. - -

    - An ellpsoid, somewhat in the shape of a squashed olive. -

    -     - - - - - //ASY file for fig10_01_ex_263D.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,5,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-sqrt(2),sqrt(2)}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2/9 - y^2 + z^2/25 = 1 - triple f(pair t) { - return (cos(t.y)*cos(t.x),sqrt(2)*cos(t.y)*sin(t.x),2*sin(t.y));//({cos(x)*1.5*cos(y)},{sin(x)*cos(y)},{sin(y)}) - } - surface s=surface(f,(-pi,-pi/2),(1*pi,pi/2),32,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -     - -     - -     -     -     -
    -
    - An Introduction to Vectors -

    - Many quantities we think about daily can be described by a single number: - temperature, speed, cost, weight and height. - There are also many other concepts we encounter daily that cannot be described with just one number. - For instance, - a weather forecaster often describes wind with its speed and its direction - (\ldots with winds from the southeast gusting up to 30 mph \ldots). - When applying a force, - we are concerned with both the magnitude and direction of that force. - In both of these examples, direction is important. - Because of this, we study vectors, - mathematical objects that convey both magnitude and direction information. -

    - - - -

    - One bare-bones definition of a vector is based on what we wrote above: - a vector is a mathematical object with magnitude and direction parameters. - This definition leaves much to be desired, - as it gives no indication as to how such an object is to be used. - Several other definitions exist; - we choose here a definition rooted in a geometric visualization of vectors. - It is very simplistic but readily permits further investigation. -

    - - - Vector - -

    - A vector is a directed line segment. -

    - -

    - Given points P and Q - (either in the plane or in space), - we denote with \overrightarrow{PQ} the vector - from P to Q. - The point P is said to be the - initial point of the vector, - and the point Q is the terminal point. -

    - -

    - The magnitude, - length or norm - of \overrightarrow{PQ} is the length of the line segment \overline{PQ}: - \norm{\overrightarrow{PQ}} = \norm{\overline{PQ}}. -

    - -

    - Two vectors are equal - if they have the same magnitude and direction. - vectors - vectorsdefinition - vectorsnorm - vectorsmagnitude - norm - magnitude of vector - terminal point - initial point -

    -     -     - -

    - - shows multiple instances of the same vector. - Each directed line segment has the same direction and length (magnitude), - hence each is the same vector. -

    - -
    - Drawing the same vector with different initial points - - - - Graph shows the same vectors drawn at different initial points. - - -

    - The x and y axes are drawn from -4 to 4. There are four vectors drawn. - In the first quadrant is the first vector drawn from point (0,0) to (3,1). The second - vector is drawn in the second quadrant from point (-4,1) to point (-1,2). In the fourth - quadrant the vector is drawn from (2, -4) to (5, -3). The last vector is drawn from - (-2, -3) to (1, -2) and it crosses the third . -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - ymin=-5.5,ymax=5.5, - xmin=-5.5,xmax=5.5 - ] - - \draw [very thick,->] (axis cs:0,0) -- (axis cs:3,1); - \draw [very thick,->] (axis cs:-2,-3) -- (axis cs:1,-2); - \draw [very thick,->] (axis cs:-4,1) -- (axis cs:-1,2); - \draw [very thick,->] (axis cs:2,-4) -- (axis cs:5,-3); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - We use \mathbb{R}^2 - (pronounced r two) - to represent all the vectors in the plane, - and use \mathbb{R}^3 - (pronounced r three) - to represent all the vectors in space. - r@\mathbb{R} -

    - -
    - Illustrating how equal vectors have the same displacement - - - - Graphs showing equal vectors have the same displacement. - - -

    - The x and y axes are drawn from -4 to 4. There are two vectors - PQ and RS. The vector PQ is drawn in the first quadrant from point - P=(1,0) and Q=(3,1). The second vector is in the second quadrant from point - R=(-3, 1) to S=(-1,2). -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - ymin=-4.5,ymax=4.5, - xmin=-4.5,xmax=4.5 - ] - - \draw [very thick,->] (axis cs:1,0) node [above] { $P$} -- (axis cs:3,1) node [above] { $Q$}; - \draw [very thick,->] (axis cs:-3,1) node [above] { $R$} -- (axis cs:-1,2) node [above] { $S$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Consider the vectors \overrightarrow{PQ} and \overrightarrow{RS} as shown in . - The vectors look to be equal; - that is, they seem to have the same length and direction. - Indeed, they are. - Both vectors move 2 units to the right and 1 unit up from the initial point to reach the terminal point. - One can analyze this movement to measure the magnitude of the vector, - and the movement itself gives direction information - (one could also measure the slope of the line passing through P and Q or R and S). - Since they have the same length and direction, - these two vectors are equal. -

    - -

    - This demonstrates that inherently all we care about is displacement; - that is, how far in the x, - y and possibly z directions the terminal point is from the initial point. - Both the vectors \overrightarrow{PQ} and \overrightarrow{RS} in - have an x-displacement of 2 and a y-displacement of 1. - This suggests a standard way of describing vectors in the plane. - A vector whose x-displacement is a and whose y-displacement is b will have terminal point (a,b) when the initial point is the origin, - (0,0). - This leads us to a definition of a standard and concise way of referring to vectors. -

    - - - Component Form of a Vector - -

    -

      -
    1. -

      - The component form of a vector \vec{v} in \mathbb{R}^2, - whose terminal point is (a,b) when its initial point is (0,0), - is \la a,b\ra. -

      -
      2. - -
    2. -

      - The component form of a vector \vec{v} in \mathbb{R}^3, - whose terminal point is (a,b,c) when its initial point is (0,0,0), - is \la a,b,c\ra. -

      -
      3. -
    -

    - -

    - The numbers a, b - (and c, respectively) - are the components of \vec v. - vectorscomponent form -

    -     -     - -

    - It follows from the definition that the component form of the vector \overrightarrow{PQ}, - where P=(x_1,y_1) and Q=(x_2,y_2) is - - \overrightarrow{PQ} = \la x_2-x_1, y_2-y_1\ra; - - in space, where P=(x_1,y_1,z_1) and Q=(x_2,y_2,z_2), - the component form of \overrightarrow{PQ} is - - \overrightarrow{PQ} = \la x_2-x_1, y_2-y_1,z_2-z_1\ra - . -

    - -

    - We practice using this notation in the following example. -

    - - - Using component form notation for vectors - -

    -

      -
    1. -

      - Sketch the vector \vec v=\la 2,-1\ra starting at P=(3,2) and find its magnitude. -

      -
      2. - -
    2. -

      - Find the component form of the vector \vec w whose initial point is - R=(-3,-2) and whose terminal point is S=(-1,2). -

      -
      3. - -
    3. -

      - Sketch the vector \vec u = \la 2,-1,3\ra starting at the point Q = (1,1,1) and find its - magnitude. -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - Using P as the initial point, - we move 2 units in the positive x-direction and -1 units in the positive y-direction to arrive at the terminal point P\,'=(5,1), - as drawn in . - - The magnitude of \vec v is determined directly from the component form: - - \norm{\vec v} =\sqrt{2^2+(-1)^2} = \sqrt{5} - . -

      -
      - Graphing vectors in - -
      - - - - - Graph showing two vectors RS and PP';. - - -

      - The x and y axes are drawn from -4 to 4. Two vectors are drawn. - The vector PP’ is in the first quadrant it is downward facing, it starts from - P=(3, 2) and ends at P'=(5, 1). The second vector RS from (-3, -2) - to (-1, 2). This vector lies midway between the second and the third quadrant. -

      -       - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - minor x tick num=4, - minor y tick num=4, - ymin=-5.5,ymax=5.5, - xmin=-5.5,xmax=5.9 - ] - - \draw [very thick,->] (axis cs:3,2) node [left] { $P$} -- (axis cs:5,1) node [shift={(5pt,0pt)}] { $P\,'$}; - \draw [very thick,->] (axis cs:-3,-2) node [below left] { $R$} -- (axis cs:-1,2) node [above] { $S$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      - -
      - - - - - Graph showing vector QQ'. - - -

      - The x and y axes are drawn from 0 to 2. The z axis is drawn - from 0 to 4. The points Q = (1, 1, 1) and Q'= (3, 0, 4) along with - the vector are drawn in space. - A dashed cube is drawn with Q at the vertex furthest away from the origin. -

      -       - - - - - //ASY file for figvectintro3b.pdf in Chapter 10.1 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,9,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={-2,4}; - defaultpen(0.5mm); - - pair xbounds=(-1,3.5); - pair ybounds=(-1,2.5); - pair zbounds=(-1,5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // Draw the lines for Q=(1,1,1) - draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1), redpen+dashed+linewidth(.5));//top Q - draw((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//bottom Q - draw((1,0,0)--(1,0,1), redpen+dashed+linewidth(.5));//up1 Q - draw((0,1,0)--(0,1,1), redpen+dashed+linewidth(.5));//up2 Q - draw((1,1,0)--(1,1,1), redpen+dashed+linewidth(.5));//up3 Q - label("$Q$",(1,1,1),N); - //dotfactor=3; dot((2,1,1),bluepen); - - // Draw the lines for Q'=(3,0,4) - draw((0,0,4)--(0,0,4)--(3,0,4)--(3,0,4)--(3,0,4), redpen+dashed+linewidth(.5));//top P - draw((0,0,0)--(3,0,0)--(3,0,0)--(3,0,0)--(0,0,0), redpen+dashed+linewidth(.5));//bottom P - draw((3,0,0)--(3,0,4), redpen+dashed+linewidth(.5));//up1 P - draw((0,0,0)--(0,0,3), redpen+dashed+linewidth(.5));//up2 P - draw((3,0,0)--(3,0,3), redpen+dashed+linewidth(.5));//up3 P - label("$Q'$",(3,0,4),N); - //dotfactor=3; dot((1,4,-1),bluepen); - - //line from P to Q - draw((1,1,1)--(3,0,4), black,Arrow3(size=2mm)); - - - - -
      -       -
      -
      2. - -
    2. -

      - Using the note following , we have - - \overrightarrow{RS} = \la -1-(-3), 2-(-2)\ra = \la 2,4\ra - . - One can readily see from that the x- and y-displacement of \overrightarrow{RS} is 2 and 4, respectively, - as the component form suggests. -

      -
      3. - -
    3. -

      - Using Q as the initial point, - we move 2 units in the positive x-direction, - -1 unit in the positive y-direction, - and 3 units in the positive z-direction to arrive at the terminal point Q' = (3,0,4), - illustrated in . - - The magnitude of \vec u is: - - \norm{\vec u} = \sqrt{2^2+(-1)^2+3^2} = \sqrt{14} - . -

      -
      4. -
    -

    -     - -     - - - -

    - Now that we have defined vectors, - and have created a nice notation by which to describe them, - we start considering how vectors interact with each other. - That is, we define an algebra on vectors. -

    - - - Vector Algebra - -

    -

      -
    1. -

      - Let \vec u = \la u_1,u_2\ra and - \vec v = \la v_1,v_2\ra be vectors in \mathbb{R}^2, - and let c be a scalar. - vectorsalgebra of -

      - -

      -

        -
      1. -

        - The addition, or sum, - of the vectors \vec u and \vec v is the vector - - \vec u+\vec v = \la u_1+v_1, u_2+v_2\ra - . -

        -
        2. - -
      2. -

        - The multiplication of a scalar c and a vector \vec v is the vector - - c\vec v = c\la v_1,v_2\ra = \la cv_1,cv_2\ra - . -

        -
        3. -
      -

      -
      2. - -
    2. -

      - Let \vec u = \la u_1,u_2,u_3\ra and - \vec v = \la v_1,v_2,v_3\ra be vectors in \mathbb{R}^3, - and let c be a scalar. -

      - -

      -

        -
      1. -

        - The addition, or sum, - of the vectors \vec u and \vec v is the vector - - \vec u+\vec v = \la u_1+v_1, u_2+v_2, u_3+v_3\ra - . -

        -
        2. - -
      2. -

        - The multiplication of a scalar c and a vector \vec v is the vector - - c\vec v = c\la v_1,v_2,v_3\ra = \la cv_1,cv_2,cv_3\ra - . -

        -
        3. -
      -

      -
      3. -
    -

    -     -     - -

    - In short, we say addition and scalar multiplication are computed - component-wise. -

    - - - - - Adding vectors - -

    - Sketch the vectors \vec u = \la1,3\ra, - \vec v = \la 2,1\ra and - \vec u+\vec v all with initial point at the origin. -

    -     - -

    - We first compute \vec u +\vec v. - - \vec u+\vec v \amp = \la 1,3\ra + \la 2,1\ra - \amp = \la 3,4\ra - . -

    - -
    - Graphing the sum of vectors in - - - - Graph showing sum of two vectors u and v. - - -

    - The x and y axes are drawn from 0 to 4. Two vectors \vec u and \vec v - are shown along with the vector addition of the two. The \vec u vector is drawn from point - (0,0) to (1,3). The \vec v vector is drawn from origin to (2,1). The vector - \vec u is longer than the \vec v vector. The \vec u + \vec v vector is drawn from - the origin to (3, 4). - The \vec u + \vec v vector is the longest and is in between the two vectors. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=4.1, - xmin=-.5,xmax=4.1 - ] - - \draw [very thick,->] (axis cs:0,0) -- (axis cs:1,3) node [pos=.5,above,rotate=70] { $\vec u$}; - \draw [very thick,->] (axis cs:0,0) -- (axis cs:2,1) node [pos=.5,above,rotate=40] { $\vec v$}; - \draw [very thick,->] (axis cs:0,0) -- (axis cs:3,4) node [pos=.5,above,rotate=50] { $\vec u+\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - These are all sketched in . -

    -     - -     - -

    - As vectors convey magnitude and direction information, - the sum of vectors also convey length and magnitude information. - Adding \vec u+\vec v suggests the following idea: -

    - -
    -

    - Starting at an initial point, go out \vec u, then go out \vec v. -

    -
    - -

    - This idea is sketched in , - where the initial point of \vec v is the terminal point of \vec u. - This is known as the Head to Tail Rule of adding vectors. - Vector addition is very important. - For instance, - if the vectors \vec u and \vec v represent forces acting on a body, - the sum \vec u+\vec v gives the resulting force. - Because of various physical applications of vector addition, - the sum \vec u+\vec v is often referred to as the - resultant vector, - or just the resultant. - vectorsresultant - Parallelogram Law - vectorsParallelogram Law - Head To Tail Rule - vectorsHead To Tail Rule - vectorszero vector -

    - -
    - Illustrating how to add vectors using the Head to Tail Rule and Parallelogram Law - - - - Graph showing sum of two vectors u and v. - - -

    - The x and y axes are drawn from 0 to 4. Two vectors \vec u and - \vec v are shown along with the vector addition of the two. The \vec u vector is drawn - from point (0,0) to (1,3). The \vec v vector is drawn from origin to (2,1). -

    -

    - The \vec v vector is translated to start from the point (1, 3) to (3, 4) and it - forms a triangle with \vec u and u+v. The \vec u vector is translated to start - from point (2, 1) to point (3, 4). The vector \vec u is longer than the \vec v - vector and it forms a triangle with \vec u and u+v. The \vec u + \vec v vector is drawn - from the origin to (3, 4). - The \vec u + \vec v vector is the longest and is in between the two vectors. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=4.1, - xmin=-.5,xmax=4.1 - ] - - \draw [very thick,->] (axis cs:0,0) -- (axis cs:1,3) node [pos=.5,above,rotate=70] { $\vec u$}; - \draw [very thick,->] (axis cs:0,0) -- (axis cs:2,1) node [pos=.5,above,rotate=30] { $\vec v$}; - - \draw [very thick,->,gray] (axis cs:1,3) -- (axis cs:3,4) node [pos=.5,above,rotate=30] { $\vec v$}; - \draw [very thick,->,gray] (axis cs:2,1) -- (axis cs:3,4) node [pos=.4,above,rotate=70] { $\vec u$}; - - \draw [very thick,->] (axis cs:0,0) -- (axis cs:3,4) node [pos=.5,above,rotate=50] { $\vec u+\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Analytically, - it is easy to see that \vec u+\vec v = \vec v+\vec u. - - also gives a graphical representation of this, using gray vectors. - Note that the vectors \vec u and \vec v, - when arranged as in the figure, form a parallelogram. - Because of this, - the Head to Tail Rule is also known as the Parallelogram Law: - the vector \vec u+\vec v is defined by forming the parallelogram defined by the vectors \vec u and \vec v; - the initial point of \vec u+\vec v is the common initial point of parallelogram, - and the terminal point of the sum is the common terminal point of the parallelogram. -

    - -

    - While not illustrated here, - the Head to Tail Rule and Parallelogram Law hold for vectors in \mathbb{R}^3 as well. -

    - -

    - It follows from the properties of the real numbers and that - - \vec u-\vec v = \vec u + (-1)\vec v - . -

    - -

    - The Parallelogram Law gives us a good way to visualize this subtraction. - We demonstrate this in the following example. -

    - - - Vector Subtraction - -

    - Let \vec u = \la 3,1\ra and \vec v=\la 1,2\ra. - Compute and sketch \vec u-\vec v. -

    -     - -

    - The computation of \vec u-\vec v is straightforward, - and we show all steps below. - Usually the formal step of multiplying by (-1) is omitted and we just subtract. - - \vec u-\vec v \amp = \vec u + (-1)\vec v - \amp = \la 3,1\ra + \la -1,-2\ra - \amp = \la 2,-1\ra - . -

    - -
    - Illustrating how to subtract vectors graphically - - - - Graph showing subtraction of two vectors u from v. - - -

    - The x axis is drawn from 0 to 4 and the y axis is drawn from -1 to - 3. The \vec u vector is drawn from origin to point (1, 2). The \vec v vector - is drawn from the origin to (2,1). The vector \vec u - \vec v is drawn from origin to - (2, -1) and it lies in the fourth quadrant. -

    -

    - The \vec u - \vec v vector is translated to start from point (1, 2) and ends at point - (3, 1). The \vec v vector is also translated but in the opposite direction and it starts - from (3, 1) and ends at point (2, -1). -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - ymin=-1.5,ymax=3.1, - xmin=-.5,xmax=4.1 - ] - - \draw [very thick,->] (axis cs:0,0) -- (axis cs:3,1) node [pos=.5,above,rotate=20] { $\vec u$}; - \draw [very thick,->] (axis cs:0,0) -- (axis cs:1,2) node [pos=.5,above,rotate=60] { $\vec v$}; - \draw [very thick,->] (axis cs:0,0) -- (axis cs:2,-1) node [pos=.5,below,rotate=-20] { $\vec u-\vec v$}; - - \draw [very thick,->,gray] (axis cs:3,1) -- (axis cs:2,-1) node [black,pos=.7,below,rotate=60] { $-\vec v$}; - \draw [very thick,->,gray] (axis cs:1,2) -- (axis cs:3,1) node [black,pos=.5,above,rotate=-20] { $\vec u-\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - illustrates, - using the Head to Tail Rule, - how the subtraction can be viewed as the sum \vec u + (-\vec v). - The figure also illustrates how - \vec u-\vec v can be obtained by looking only at the terminal points of \vec u and \vec v - (when their initial points are the same). -

    -     - -     - - - Scaling vectors - -

    -

      -
    1. -

      - Sketch the vectors \vec v = \la 2,1\ra and 2\vec v with initial point at the origin. -

      -
      2. - -
    2. -

      - Compute the magnitudes of \vec v and 2\vec v. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - We compute 2\vec v: - - 2\vec v \amp = 2\la 2,1\ra - \amp = \la 4,2\ra - . -

      - -
      - Graphing vectors \vec v and 2\vec v in - - - - Graph showing two similar vectors of different magnitudes. - - -

      - The x axis is drawn from 0 to 4 and the y axis is drawn from -1 - to 3. The \vec v vector is drawn from origin to point (2,1), another vector - 2\vec v is also drawn from origin to point (4, 2). The two vectors have the same - direction. -

      -       - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=3.1, - xmin=-.5,xmax=4.1 - ] - - \draw [gray,very thick,->] (axis cs:0,0) -- (axis cs:4,2) node [pos=.7,above,rotate=30,black] { $2\vec v$}; - - \draw [very thick,->] (axis cs:0,0) -- (axis cs:2,1) node [pos=.5,above,rotate=30] { $\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      - -

      - Both \vec v and 2\vec v are sketched in . - Make note that 2\vec v does not start at the terminal point of \vec v; - rather, its initial point is also the origin. -

      -
      2. - -
    2. -

      - The figure suggests that 2\vec v is twice as long as \vec v. - We compute their magnitudes to confirm this. - - \norm{\vec v} \amp = \sqrt{2^2+1^2} - \amp = \sqrt{5}. - \norm{2\vec v}\amp =\sqrt{4^2+2^2} - \amp = \sqrt{20} - \amp = \sqrt{4\cdot 5} = 2\sqrt{5} - . - As we suspected, 2\vec v is twice as long as \vec v. -

      -
      3. -
    -

    -     - -     - -

    - The zero vector is the vector whose initial point is also its terminal point. - It is denoted by \vec 0. - Its component form, in \mathbb{R}^2, is \la 0,0\ra; - in \mathbb{R}^3, it is \la 0,0,0\ra. - Usually the context makes is clear whether \vec 0 is referring to a vector in the plane or in space. -

    - - - -

    - Our examples have illustrated key principles in vector algebra: - how to add and subtract vectors and how to multiply vectors by a scalar. - The following theorem states formally the properties of these operations. -

    - - - Properties of Vector Operations - -

    - The following are true for all scalars c and d, - and for all vectors \vec u, - \vec v and \vec w, where \vec u, - \vec v and \vec w are all in - \mathbb{R}^2 or where \vec u, - \vec v and \vec w are all in - \mathbb{R}^3: - vectorsalgebraic properties -

    - -

    -

      -
    1. -

      - \vec u+\vec v = \vec v+\vec u Commutative Property -

      -
      2. - -
    2. -

      - (\vec u+\vec v)+\vec w = \vec u+(\vec v+\vec w) Associative Property -

      -
      3. - -
    3. -

      - \vec v+\vec 0 = \vec v Additive Identity -

      -
      4. - -
    4. -

      - (cd)\vec v= c(d\vec v) -

      -
      5. - -
    5. -

      - c(\vec u+\vec v) = c\vec u+c\vec v Distributive Property -

      -
      6. - -
    6. -

      - (c+d)\vec v = c\vec v+d\vec v Distributive Property -

      -
      7. - -
    7. -

      - 0\vec v = \vec 0 -

      -
      8. - -
    8. - -

      - \norm{c\vec v} = \abs{c}\cdot\norm{\vec v} -

      -
      9. - -
    9. - -

      - \vnorm u = 0 if, and only if, \vecu = \vec 0. -

      -
      10. -
    -

    -     -     - - - -

    - As stated before, - each nonzero vector \vec v conveys magnitude and direction information. - We have a method of extracting the magnitude, - which we write as \norm{\vec v}. - Unit vectors are a way of extracting just the direction information from a vector. -

    - - - Unit Vector - -

    - A unit vector is a vector \vec v with a magnitude of 1; - that is, - vectorsunit vector - unit vector - - \norm{\vec v}=1 - . -

    -     -     - - - -

    - Consider this scenario: - you are given a vector \vec v and are told to create a vector of length 10 in the direction of \vec v. - How does one do that? - If we knew that \vec u was the unit vector in the direction of \vec v, - the answer would be easy: 10\vec u. - So how do we find \vec u? -

    - -

    - Property - of holds the key. - If we divide \vec v by its magnitude, - it becomes a vector of length 1. - Consider: - - \snorm{\frac{1}{\norm{\vec v}}\vec v} \amp = \frac{1}{\norm{\vec v}}\norm{\vec v} \amp \text{ (we can pull out \(\ds \frac{1}{\norm{\vec v}}\) as it is a positive scalar)} - \amp = 1 - . -

    - -

    - So the vector of length 10 in the direction of \vec v is \ds 10\frac{1}{\norm{\vec v}}\vec v. - An example will make this more clear. -

    - - - Using Unit Vectors - -

    - Let \vec v= \la 3,1\ra and let \vec w = \la 1,2,2\ra. -

    - -

    -

      -
    1. -

      - Find the unit vector in the direction of \vec v. -

      -
      2. - -
    2. -

      - Find the unit vector in the direction of \vec w. -

      -
      3. - -
    3. -

      - Find the vector in the direction of \vec v with magnitude 5. -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - We find \norm{\vec v} = \sqrt{10}. - So the unit vector \vec u in the direction of \vec v is - - \vec u = \frac{1}{\sqrt{10}}\vec v = \la \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\ra - . -

      -
      2. - -
    2. -

      - We find \norm{\vec w} = 3, - so the unit vector \vec z in the direction of \vec w is - - \vec u = \frac13\vec w = \la \frac13,\frac23,\frac23\ra - . -

      -
      3. - -
    3. -

      - To create a vector with magnitude 5 in the direction of \vec v, - we multiply the unit vector \vec u by 5. - Thus 5\vec u = \la 15/\sqrt{10},5/\sqrt{10}\ra is the vector we seek. - This is sketched in . -

      - -
      - Graphing vectors in . All vectors shown have their initial point at the origin - - - - Graph showing three similar vectors of different magnitudes. - - -

      - The x axis is drawn from 0 to 5 and the y axis is drawn from 0 - to 3. The \vec u vector is drawn from origin to point (1, 0.4), another vector - \vec v is also drawn from origin to point (3, 1). The third vector 5\vec u is - also drawn from the origin to point (5, 1.5). The three vectors have the same direction. -

      -       - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=3.1, - xmin=-.5,xmax=5.1 - ] - - \draw [gray,very thick,->] (axis cs:0,0) -- (axis cs:4.74,1.58) node [pos=.8,above,rotate=30,black] { $5\vec u$}; - - \draw [very thick,->] (axis cs:0,0) -- (axis cs:3,1) node [pos=.6,above,rotate=30] { $\vec v$}; - - \draw [very thick,->,gray] (axis cs:0,0) -- (axis cs:.949,.316) node [pos=.5,above,rotate=30,black] { $\vec u$}; - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      4. -
    -

    -     - -     - -

    - The basic formation of the unit vector \vec u in the direction of a vector \vec v leads to a interesting equation. - It is: - - \vec v = \norm{\vec v}\frac{1}{\norm{\vec v}}\vec v - . -

    - -

    - We rewrite the equation with parentheses to make a point: - - \vec v = \underbrace{\norm{\vec v}}_{\text{magnitude } }\cdot\underbrace{\left(\frac{1}{\norm{\vec v}}\vec v\right)}_{\text{direction } } - . -

    - -

    - This equation illustrates the fact that a nonzero vector has both magnitude and direction, - where we view a unit vector as supplying - only direction information. - Identifying unit vectors with direction allows us to define - parallel vectors. -

    - - - Parallel Vectors - -

    -

      -
    1. -

      - Unit vectors \vec u_1 and - \vec u_2 are parallel - if \vec u_1 = \pm \vec u_2. - vectorsparallel - parallel vectors -

      -
      2. - -
    2. -

      - Nonzero vectors \vec v_1 and - \vec v_2 are parallel - if their respective unit vectors are parallel. -

      -
      3. -
    -

    -     -     - - - -

    - It is equivalent to say that vectors \vec v_1 and - \vec v_2 are parallel if there is a scalar c\neq 0 such that \vec v_1 = c\vec v_2 - (see marginal note). -

    - -

    - If one graphed all unit vectors in - \mathbb{R}^2 with the initial point at the origin, - then the terminal points would all lie on the unit circle. - Based on what we know from trigonometry, - we can then say that the component form of all unit vectors in \mathbb{R}^2 is - \la \cos(\theta) ,\sin(\theta) \ra for some angle \theta. -

    - -

    - A similar construction in \mathbb{R}^3 shows that the terminal points all lie on the unit sphere. - These vectors also have a particular component form, - but its derivation is not as straightforward as the one for unit vectors in \mathbb{R}^2. - Important concepts about unit vectors are given in the following Key Idea. -

    - - - Unit Vectors -

    -

      -
    1. -

      - The unit vector in the direction of a nonzero vector \vec v is - - \vec u = \frac1{\norm{\vec v}} \vec v - . -

      -
      2. - -
    2. -

      - A vector \vec u in \mathbb{R}^2 is a unit vector if, - and only if, - its component form is \la \cos\theta,\sin\theta\ra for some angle \theta. - unit vectorproperties - vectorsunit vector -

      -
      3. - -
    3. -

      - A vector \vec u in \mathbb{R}^3 is a unit vector if, - and only if, - its component form is \la \sin(\theta) \cos(\varphi) ,\sin(\theta) \sin(\varphi) ,\cos(\theta) \ra for some angles \theta and \varphi. -

      -
      4. -
    -

    -     - - - -

    - These formulas can come in handy in a variety of situations, - especially the formula for unit vectors in the plane. -

    - - - Finding Component Forces - -

    - Consider a weight of 50lb hanging from two chains, - as shown in . - One chain makes an angle of 30^\circ with the vertical, - and the other an angle of 45^\circ. - Find the force applied to each chain. -

    - -
    - A diagram of a weight hanging from 2 chains in - - - - Image shows weight suspended with two chains. - - -

    - Image is of a weight of 50 pounds suspended by two chains. The chain on the left forms an angle - of 30 degrees with the vertical and the chain on the left forms a degree of 45 with the - vertical. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.29] - - \filldraw[thick,black,fill=gray!30] (-.5,0) -- (.5,0) -- (1,-1) -- (-1,-1) -- cycle; - \draw (0,-.5) node {50lb}; - - \draw [thick] (-1.5,1.5) -- (2,1.5); - \clip (-1.5,1.5) rectangle (2,-1.25); - - \draw [thick,rotate=120] (0,0) -- (3,0); - \draw [thick,rotate=45] (0,0) -- (3,0); - \draw [dashed] (0,0) -- (0,2); - \draw [rotate=45,->] (.8,0) arc (0:45:.8); - \draw [rotate=67] (1,0) node { $45^\circ$}; - \draw [rotate=90,->] (1,0) arc (0:30:1); - \draw [rotate=105] (1.2,0) node { $30^\circ$}; - - \end{tikzpicture} - - - - -
    -     - -

    - Knowing that gravity is pulling the 50lb weight straight down, - we can create a vector \vec F to represent this force. - - \vec F = 50\la 0,-1\ra = \la 0,-50\ra - . -

    - -

    - We can view each chain as pulling the weight up, - preventing it from falling. - We can represent the force from each chain with a vector. - Let \vec F_1 represent the force from the chain making an angle of 30^\circ with the vertical, - and let \vec F_2 represent the force form the other chain. - Convert all angles to be measured from the horizontal - (as shown in ), - and apply . - As we do not yet know the magnitudes of these vectors, - (that is the problem at hand), - we use m_1 and m_2 to represent them. - - \vec F_1 = m_1\la \cos(120^\circ) ,\sin(120^\circ) \ra - - - \vec F_2 = m_2\la \cos(45^\circ) ,\sin(45^\circ) \ra - -

    - -

    - As the weight is not moving, we know the sum of the forces is \vec 0. - This gives: - - \vec F + \vec F_1 + \vec F_2 \amp = \vec 0 - \la 0,-50\ra + m_1\la \cos(120^\circ) ,\sin(120^\circ) \ra + m_2\la \cos(45^\circ) ,\sin(45^\circ) \ra \amp =\vec 0 - -

    - -
    - A diagram of the force vectors from - - - - Image showing force vectors for this example. - - -

    - Image shows the force vectors from the exercise. The vector \vec F_2 is at an angle 45 - from the horizontal and the vector \vec F_1 forms an angle of 120 from the horizontal. - A third vector representing the downward pull by gravity marked as \vec F. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.75] - - \draw [thick,rotate=120,->] (0,0) -- (1.5,0) node [left] { $\vec F_1$}; - \draw [thick,rotate=45,->] (0,0) -- (1.75,0) node [right] { $\vec F_2$}; - \draw [thick,rotate=-90,->] (0,0) -- (1,0) node [right] { $\vec F$}; - - \draw [dashed] (0,0) -- (0,1.5); - \draw [dashed] (-1.25,0) -- (1.25,0); - - \filldraw [black] (0,0) circle (2.4pt); - - \draw [dashed,->] (.6,0) arc (0:120:.6); - \draw [rotate=20] (.9,0) node { $120^\circ$}; - \draw [dashed,->] (1.5,0) arc (0:45:1.5); - \draw [rotate=15] (1.8,0) node { $45^\circ$}; - - \end{tikzpicture} - - - - -
    - -

    - The sum of the entries in the first component is 0, and the sum of the entries in the second component is also 0. - This leads us to the following two equations: - - m_1\cos(120^\circ) + m_2\cos(45^\circ) \amp =0 - m_1\sin(120^\circ) + m_2\sin(45^\circ) \amp =50 - -

    - -

    - This is a simple 2-equation, 2-unknown system of linear equations. - We leave it to the reader to verify that the solution is - - m_1=50(\sqrt{3}-1) \approx 36.6;\qquad m_2=\frac{50\sqrt{2}}{1+\sqrt{3}} \approx 25.88 - . -

    - -

    - It might seem odd that the sum of the forces applied to the chains is more than 50lb. - We leave it to a physics class to discuss the full details, - but offer this short explanation. - Our equations were established so that the vertical - components of each force sums to 50lb, - thus supporting the weight. - Since the chains are at an angle, - they also pull against each other, - creating an additional horizontal force while holding the weight in place. -

    -     - -     - -

    - Unit vectors were very important in the previous calculation; - they allowed us to define a vector in the proper direction but with an unknown magnitude. - Our computations were then computed component-wise. - Because such calculations are often necessary, - the standard unit vectors can be useful. -

    - - - Standard Unit Vectors - -

    -

      -
    1. -

      - In \mathbb{R}^2, - the standard unit vectors are - vectorsstandard unit vectorunit vectorstandard unit vector - - \veci = \la 1,0\ra \text{ and } \vecj = \la 0,1\ra - . -

      -
      2. - -
    2. -

      - In \mathbb{R}^3, the standard unit vectors are - - \veci = \la 1,0,0\ra \text{ and } \vecj = \la 0,1,0\ra \text{ and } \veck = \la 0,0,1\ra - . -

      -
      3. -
    -

    -     -     - - - Using standard unit vectors - -

    -

      -
    1. -

      - Rewrite \vec v = \la 2,-3\ra using the standard unit vectors. -

      -
      2. - -
    2. -

      - Rewrite \vec w = 4\veci - 5\vecj +2\veck in component form. -

      -
      3. -
    -

    -

    - - -

    -

      -
    1. -

      - \displaystyle \begin{aligned}\vec v \amp = \la 2,-3\ra \\ - \amp = \la 2,0\ra + \la 0,-3\ra \\ - \amp = 2\la 1,0\ra -3\la 0,1\ra\\ - \amp = 2\veci - 3\vecj - \end{aligned} -

      -
      2. - -
    2. -

      - \displaystyle \begin{aligned}\vec w \amp = 4\veci - 5\vecj +2\veck\\ - \amp = \la 4,0,0\ra +\la 0,-5,0\ra + \la 0,0,2\ra \\ - \amp = \la 4,-5,2\ra - \end{aligned} -

      -
      3. -
    -

    - -

    - These two examples demonstrate that converting between component form and the standard unit vectors is rather straightforward. - Many mathematicians prefer component form, - and it is the preferred notation in this text. - Many engineers prefer using the standard unit vectors, - and many engineering text use that notation. -

    - -     - - - Finding Component Force - -

    - A weight of 25lb is suspended from a chain of length 2ft while a wind pushes the weight to the right with constant force of 5lb as shown in . - What angle will the chain make with the vertical as a result of the wind's pushing? - How much higher will the weight be? -

    - -
    - A figure of a weight being pushed by the wind in - - - - Image showing weight suspended by a chain and wind action on the weight. - - -

    - Image shows a weight of 25 pounds being suspended by a chain. The chain forms an angle of - \theta with the vertical, and an angle \varphi with the horizontal. The wind is pushing - the weight to the right with force \vec F_w. The chain is of length 2 feet. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.29] - - \draw [dashed] (-1.5,-.2) -- (2,-.2); - - \draw (1.75,.65) node { 2ft $\left\}\rule{0pt}{1.2cm}\right.$}; - - \filldraw[thick,black,fill=gray!30] (-.5,0) -- (.5,0) -- (1,-1) -- (-1,-1) -- cycle; - \draw (0,-.5) node {25lb}; - - \draw [thick] (-1.5,1.5) -- (2,1.5); - \clip (-1.5,1.5) rectangle (2,-1.25); - - \draw [thick,rotate=110] (0,0) -- (3,0); - \draw [dashed] (0,0) -- (0,2); - \draw [rotate=90,->] (1,0) arc (0:20:1); - \draw [->] (.6,0) arc (0:109:.6); - \draw [rotate=50] (.75,0) node { $\varphi$}; - \draw [rotate=99] (1.2,0) node { $\theta$}; - \draw [thick,->] (-1.45,-.5) -- (-.8,-.5) node [pos=.18,below] { $\vec F_w$}; - - \end{tikzpicture} - - - - -
    -     - -

    - The force of the wind is represented by the vector \vec F_w = 5\veci. - The force of gravity on the weight is represented by \vec F_g = -25\vecj. - The direction and magnitude of the vector representing the force on the chain are both unknown. - We represent this force with - - \vec F_c = m\la \cos(\varphi) ,\sin(\varphi) \ra = m\cos(\varphi) \, \veci + m\sin(\varphi) \,\vecj - - for some magnitude m and some angle with the horizontal \varphi. - (Note: \theta is the angle the chain makes with the vertical; - \varphi is the angle with the horizontal.) -

    - -

    - As the weight is at equilibrium, - the sum of the forces is \vec0: - - \vec F_c + \vec F_w + \vec F_g \amp = \vec 0 - m\cos(\varphi) \, \veci + m\sin(\varphi) \,\vecj + 5\veci - 25\vecj \amp =\vec 0 - -

    - -

    - Thus the sum of the \veci and \vecj components are 0, leading us to the following system of equations: - - \begin{split} - 5+m\cos\varphi \amp = 0\\ - -25+m\sin\varphi \amp = 0 - \end{split} - -

    - -

    - This is enough to determine \vec F_c already, - as we know m\cos(\varphi) = -5 and m\sin(\varphi) =25. - Thus F_c = \la -5,25\ra. - We can use this to find the magnitude m: - - m = \sqrt{(-5)^2+25^2} = 5\sqrt{26}\approx 25.5\text{ lb } - . -

    - -

    - We can then use either equality from Equation to solve for \varphi. - We choose the first equality as using arccosine will return an angle in the 2nd quadrant: - - 5 + 5\sqrt{26}\cos(\varphi) = 0 \Rightarrow \varphi = \cos^{-1}\left(\frac{-5}{5\sqrt{26}}\right) \approx 1.7682\approx 101.31^\circ - . -

    - -

    - Subtracting 90^\circ from this angle gives us an angle of 11.31^\circ with the vertical. -

    - -

    - We can now use trigonometry to find out how high the weight is lifted. - The diagram shows that a right triangle is formed with the 2ft chain as the hypotenuse with an interior angle of 11.31^\circ. - The length of the adjacent side - (in the diagram, the dashed vertical line) - is 2\cos(11.31^\circ) \approx 1.96ft. - Thus the weight is lifted by about 0.04ft, almost 1/2in. -

    -     -     - -

    - The algebra we have applied to vectors is already demonstrating itself to be very useful. - There are two more fundamental operations we can perform with vectors, - the dot product and the cross product. - The next two sections explore each in turn. -

    - - - - Terms and Concepts - - - - -

    - Name two different things that cannot be described with just one number, - but rather need 2 or more numbers to fully describe them. -

    -     - - - -     - - - - -

    - What is the difference between (1,2) and \la 1,2\ra? -

    -     - - - -

    - (1,2) is a point; - \la 1,2\ra is a vector that describes a displacement of 1 unit in the x-direction and 2 units in the y-direction. -

    -     - -     - - - - -

    - What is a unit vector? -

    -     - - - -

    - A vector with magnitude 1. -

    -     - -     - - - -

    - Unit vectors can be thought of as conveying what type of information? -

    -     - - - -

    - Direction -

    -     -     - - - - - -

    - What does it mean for two vectors to be parallel? -

    -     - - - -

    - Their respective unit vectors are parallel; - unit vectors \vec u_1 and - \vec u_2 are parallel if \vec u_1 = \pm \vec u_2. -

    - -

    - Alternatively, two vectors are parallel if one is a scalar multiple of the other. -

    -     - -     - - - - -

    - What effect does multiplying a vector by -2 have? -

    -     - - - -

    - It stretches the vector by a factor of 2, and points it in the opposite direction. -

    -     - -     - -     - - Problems - - - -

    - Points P and Q are given. - Write the vector \overrightarrow{PQ} in component form and using the standard unit vectors. -

    -     - - - - - - Context("Vector2D"); - Context()->constants->undefine(i,j); - $comp=Vector("<1,6>"); - Context("Vector2D"); - Context()->flags->set(ijk=>1); - Context()->parens->undefine('<','(','{'); - $stan=Vector("i+6j"); - - - -

    - If P=(2,-1) and Q = (3,5), - write the vector \overrightarrow{PQ}: -

    -     - - - -

    - in component form. -

    - -

    - -

    -     -     - - - -

    - using the standard unit vectors. -

    - -

    - -

    -     -     -     -     - - - - - - Context("Vector2D"); - Context()->constants->undefine(i,j); - $comp=Vector("<4,-4>"); - Context("Vector2D"); - Context()->flags->set(ijk=>1); - Context()->parens->undefine('<','(','{'); - $stan=Vector("4i-4j"); - - - -

    - If P=(3,2) and Q = (7,-2), - write the vector \overrightarrow{PQ}: -

    -     - - - -

    - in component form. -

    - -

    - -

    -     -     - - - -

    - using the standard unit vectors. -

    - -

    - -

    -     -     -     -     - - - - - - Context("Vector"); - Context()->constants->undefine(i,j,k); - $comp=Vector("<6,-1,6>"); - Context("Vector"); - Context()->flags->set(ijk=>1); - Context()->parens->undefine('<','(','{'); - $stan=Vector("6i-j+6k"); - - - -

    - If P=(0,3,-1) and Q = (6,2,5), - write the vector \overrightarrow{PQ}: -

    -     - - - -

    - in component form. -

    - -

    - -

    -     -     - - - -

    - using the standard unit vectors. -

    - -

    - -

    -     -     -     -     - - - - - - Context("Vector"); - Context()->constants->undefine(i,j,k); - $comp=Vector("<2,2,0>"); - Context("Vector"); - Context()->flags->set(ijk=>1); - Context()->parens->undefine('<','(','{'); - $stan=Vector("2i+2j"); - - - -

    - If P=(2,1,2) and Q = (4,3,2), - write the vector \overrightarrow{PQ}: -

    -     - - - -

    - in component form. -

    - -

    - -

    -     -     - - - -

    - using the standard unit vectors. -

    - -

    - -

    -     -     -     -     - -     - - - - -

    - Let \vec u = \la 1,-2\ra and \vec v= \la 1,1\ra. -

    -     - - - -

    - Find \vec u+\vec v, - \vec u-\vec v, 2\vec u-3\vec v. -

    -     - -

    - \vec u+\vec v = \la 2,-1\ra; - \vec u -\vec v = \la0,-3\ra; - 2\vec u-3\vec v = \la -1,-7\ra. -

    -     -     - - - -

    - Sketch the above vectors on the same axes, - along with \vec u and \vec v. -

    -     -     - - - -

    - Find \vec x where \vec u+\vec x = 2\vec v-\vec x. -

    -     - -

    - \vec x = \la 1/2,2\ra. -

    -     -     - -     - - - - -

    - Let \vec u = \la 1,1,-1\ra and \vec v= \la 2,1,2\ra. -

    -     - - - -

    - Find \vec u+\vec v, - \vec u-\vec v, \pi\vec u-\sqrt{2}\vec v. -

    -     - -

    - \vec u+\vec v = \la 3,2,1\ra; - \vec u -\vec v = \la-1,0,-3\ra; - \pi\vec u-\sqrt{2}\vec v = \la \pi-2\sqrt{2},\pi-\sqrt{2},-\pi-2\sqrt{2}\ra. -

    -     -     - - - -

    - Sketch the above vectors on the same axes, - along with \vec u and \vec v. -

    -     -     - - - -

    - Find \vec x where \vec u+\vec x = \vec v+2\vec x. -

    -     - -

    - \vec x = \la-1,0,-3\ra. -

    -     -     - -     - - - - -

    - Sketch \vec u, \vec v, - \vec u+\vec v and \vec u-\vec v on the same axes. -

    -     - - - - - - - - Graph shows two vectors in the plane. - - -

    - The x and y axes are uncalibrated. Two vectors \vec u and \vec v - are shown, both starting at the origin and facing away. The vector \vec u is in the - first quadrant, and is bent close to the positive x axis, the \vec v vector is - in the third quadrant and is bent close to the negative y axis. The vector \vec v - appears to be slightly longer than \vec u. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - xtick=\empty, - ytick=\empty, - ymin=-4.5,ymax=4.5, - xmin=-4.5,xmax=4.5 - ] - - \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: 3,1) node [above,black] { $\vec u$}; - \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: -1,-4) node [black, left] { $\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - - - - Solution image for this exercise - -

    - The solution image shows the same two vectors as in the statement of the problem. - The vector \vec{u}-\vec{v} points from the tip of \vec{v} to the tip of \vec{u}. - The vector \vec{u}+\vec{v} points between the two original vectors; - the tail is at the origin, and in this case the tip is in the 4th quadrant. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - xtick=\empty, - ytick=\empty, - ymin=-4.5,ymax=4.5, - xmin=-4.5,xmax=4.5 - ] - - \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: 3,1) node [above,black] { $\vec u$}; - \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: -1,-4)node [black, left] { $\vec v$}; - - \draw [->,thick,gray] (axis cs:0,0) -- (axis cs: 2,-3) node [black,below right] { $\vec u+\vec v$}; - \draw [->,thick,gray] (axis cs:-1,-4) -- (axis cs: 3,1) node [black,below,pos=.3,sloped] { $\vec u-\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - - - - - Graph shows two vectors in the plane. - - -

    - The x and y axes are uncalibrated. Two vectors \vec u and \vec v are shown, - both start at the origin and are facing away from each other. The \vec u vector is in the first - quadrant while the \vec v is in the third quadrant, the \vec v vector appears to be 1 / 4 - th \vec u. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - xtick=\empty, - ytick=\empty, - ymin=-4.5,ymax=4.5, - xmin=-4.5,xmax=4.5 - ] - - \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: 3,3) node [above,black] { $\vec u$}; - \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: -1,-1) node [black, below] { $\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - - - - Solution image for this exercise - -

    - The solution image shows the same two vectors as in the statement of the problem. - The vector \vec{u}+\vec{v} points in the same direction as \vec{u}, but is not as long. - The vector \vec{u}-\vec{v} is drawn below the two original vectors. - It points in the same direction as \vec{u}, but its length is the sum of the lengths of \vec{u} and \vec{v}. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - xtick=\empty, - ytick=\empty, - ymin=-4.5,ymax=4.5, - xmin=-4.5,xmax=4.5 - ] - - \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: 3,3) node [above,black] { $\vec u$}; - \draw [->,thick,firstcolor] (axis cs:0,0) -- (axis cs: -1,-1) node [black, left] { $\vec v$}; - - \draw [->,thick,gray] (axis cs:0,0) -- (axis cs: 2,2) node [black,above left] { $\vec u+\vec v$}; - \draw [->,thick,gray] (axis cs:-1,-2) -- (axis cs: 3,2) node [black,below,pos=.3,sloped] { $\vec u-\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -

    - Sketch of \vec u-\vec v shifted for clarity. -

    -     - -     - - - - - - - - Graph shows two vectors in space. - - -

    - The x, y and z axes are uncalibrated. Two vectors \vec u and - \vec v are shown, both start at the origin and face away from each other. The vector - \vec u is longer and appears to be in the zy plane, and the vector \vec v - is shorter and appears to be in the xz plane. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - axis lines=center, - view={135}{35}, - xtick=\empty, - ytick=\empty, - ztick=\empty, - ymin=-1.5,ymax=1.5, - xmin=-1.5,xmax=1.5, - zmin=-0.5, zmax=1.5, - every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, - xlabel={ $x$}, - every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, - ylabel={ $y$}, - every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, - zlabel={ $z$} - ] - - \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:-1,1,0) node [black,right] { $\vec u$}; - \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:1,0,1) node [black,left] { $\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - - - - Solution image for this exercise - -

    - The solution image shows the same two vectors as in the statement of the problem. - The vector \vec{u}-\vec{v} points from the tip of \vec{v} to the tip of \vec{u}. - The vector \vec{u}+\vec{v} points between the two original vectors, in the plane containing them. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - axis lines=center, - view={135}{35}, - xtick=\empty, - ytick=\empty, - ztick=\empty, - ymin=-1.5,ymax=1.5, - xmin=-1.5,xmax=1.5, - zmin=-0.5, zmax=1.5, - every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, - xlabel={ $x$}, - every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, - ylabel={ $y$}, - every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, - zlabel={ $z$} - ] - - \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:-1,1,0) node [black,right] { $\vec u$}; - \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:1,0,1) node [black,left] { $\vec v$}; - - \draw [thick,->,gray] (axis cs:0,0,0) -- (axis cs:0,1,1) node [black,above] { $\vec u+\vec v$}; - \draw [thick,->,gray] (axis cs:1,0,1) -- (axis cs:-1,1,0) node [black,above,pos=.9] { $\vec u-\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - - - - - - - - - Graph shows two vectors in space. - - -

    - The x, y and z axes are uncalibrated. Two vectors \vec u and - \vec v are shown, both start at the origin. The \vec u vector is along the - positive z axis and the \vec v vector is along the positive y axis. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis lines=center, - view={135}{35}, - xtick=\empty, - ytick=\empty, - ztick=\empty, - ymin=-1.5,ymax=1.5, - xmin=-1.5,xmax=1.5, - zmin=-0.5, zmax=1.5, - every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, - xlabel={ $x$}, - every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, - ylabel={ $y$}, - every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, - zlabel={ $z$} - ] - - \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:0,0,1) node [black,left] { $\vec u$}; - \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:0,1,0) node [black,below] { $\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - - - - Solution image for this exercise - -

    - The solution image shows the same two vectors as in the statement of the problem. - The vector \vec{u}-\vec{v} points from the tip of \vec{v} to the tip of \vec{u}. - The vector \vec{u}+\vec{v} points between the two original vectors, in the plane containing them. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - width=175pt, - axis lines=center, - view={135}{35}, - xtick=\empty, - ytick=\empty, - ztick=\empty, - ymin=-1.5,ymax=1.5, - xmin=-1.5,xmax=1.5, - zmin=-0.5, zmax=1.5, - every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, - xlabel={ $x$}, - every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, - ylabel={ $y$}, - every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, - zlabel={ $z$} - ] - - \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:0,0,1) node [black,left] { $\vec u$}; - \draw [thick,->,firstcolor] (axis cs:0,0,0) -- (axis cs:0,1,0) node [black,below] { $\vec v$}; - - \draw [thick,->,gray] (axis cs:0,0,0) -- (axis cs:0,1,1) node [black,above] { $\vec u+\vec v$}; - \draw [thick,->,gray] (axis cs:0,1,0) -- (axis cs:0,0,1) node [black,right,pos=.2] { $\vec u-\vec v$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -     - -     - - - -

    - Find - \norm{\vec u}, \norm{\vec v}, - \norm{\vec u+\vec v} and \norm{\vec u-\vec v}. -

    -     - - - - - Context()->flags->set(reduceConstantFunctions=>0); - $normu=Formula("sqrt(5)"); - $normv=Formula("sqrt(13)"); - $normupv=Formula("sqrt(26)"); - $normumv=Formula("sqrt(10)"); - - - -

    - \vec u=\la 2,1\ra, \vec v = \la 3,-2\ra. -

    - - - Enter the value of \norm{\vec u}. - - -

    - -

    - - - Enter the value of \norm{\vec v}. - - -

    - -

    - - - Enter the value of \norm{\vec{u} +\vec{v}}. - - -

    - -

    - - - Enter the value of \norm{\vec{u}-\vec{v}}. - - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstantFunctions=>0); - $normu=Formula("sqrt(17)"); - $normv=Formula("sqrt(3)"); - $normupv=Formula("sqrt(14)"); - $normumv=Formula("sqrt(26)"); - - - -

    - \vec u=\la -3,2,2\ra, \vec v = \la 1,-1,1\ra. -

    - - - Enter the value of \norm{\vec u}. - - -

    - -

    - - - Enter the value of \norm{\vec v}. - - -

    - -

    - - - Enter the value of \norm{\vec{u} +\vec{v}}. - - -

    - -

    - - - Enter the value of \norm{\vec{u}-\vec{v}}. - - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstantFunctions=>0); - $normu=Formula("sqrt(5)"); - $normv=Formula("3sqrt(5)"); - $normupv=Formula("2sqrt(5)"); - $normumv=Formula("4sqrt(5)"); - - - -

    - \vec u=\la 1,2\ra, \vec v = \la -3,-6\ra. -

    - - - Enter the value of \norm{\vec u}. - - -

    - -

    - - - Enter the value of \norm{\vec v}. - - -

    - -

    - - - Enter the value of \norm{\vec{u} +\vec{v}}. - - -

    - -

    - - - Enter the value of \norm{\vec{u}-\vec{v}}. - - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstantFunctions=>0); - $normu=Formula("7"); - $normv=Formula("35"); - $normupv=Formula("42"); - $normumv=Formula("28"); - - - -

    - \vec u=\la 2,-3,6\ra, \vec v = \la 10,-15,30\ra. -

    - - - Enter the value of \norm{\vec u}. - - -

    - -

    - - - Enter the value of \norm{\vec v}. - - -

    - -

    - - - Enter the value of \norm{\vec{u} +\vec{v}}. - - -

    - -

    - - - Enter the value of \norm{\vec{u}-\vec{v}}. - - -

    - -

    -     -     -     - -     - - - - -

    - Under what conditions is \norm{\vec u}+\norm{\vec v} = \norm{\vec u+\vec v}? -

    -     - -

    - When \vec u and \vec v have the same direction. - (Note: parallel is not enough.) -

    -     - -     - - - - -

    - Find the unit vector \vec u in the direction of \vec v. -

    -     - - - - - Context("Vector2D"); - Context()->flags->set(reduceConstantFunctions=>0); - $u=Compute("<3/sqrt(58), 7/sqrt(58)>"); - - - -

    - Find the unit vector \vec u in the direction of \vec v = \la 3,7\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector2D"); - Context()->flags->set(reduceConstantFunctions=>0); - $u=Compute("<0.6, 0.8>"); - - - -

    - Find the unit vector \vec u in the direction of \vec v = \la 6,8\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstantFunctions=>0,reduceConstants=>0); - $u=Compute("<1/3,-2/3,2/3>"); - - - -

    - Find the unit vector \vec u in the direction of \vec v = \la 1,-2,2\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstantFunctions=>0,reduceConstants=>0); - $u=Compute("<1/sqrt(3),-1/sqrt(3),1/sqrt(3)>"); - - - -

    - Find the unit vector \vec u in the direction of \vec v = \la 2,-2,2\ra. -

    - -

    - -

    -     -     -     - -     - - - - - Context("Vector2D"); - Context()->flags->set(reduceConstantFunctions=>0,reduceConstants=>0); - $u=Compute("<cos(50*pi/180),sin(50*pi/180)>"); - - - -

    - Find the unit vector in the first quadrant of \mathbb{R}^2 that makes a - 50^{\circ} angle with the x-axis. -

    - -

    - -

    -     -     -     - - - - - Context("Vector2D"); - Context()->flags->set(reduceConstantFunctions=>0,reduceConstants=>0); - $u=Compute("<-1/2,sqrt(3)/2>"); - - - -

    - Find the unit vector in the second quadrant of \mathbb{R}^2 that makes a - 30^{\circ} angle with the y-axis. -

    - -

    - -

    -     -     -     - - - - -

    - Verify, from , - that - - \vec u=\la \sin(\theta) \cos(\varphi) ,\sin(\theta) \sin(\varphi) ,\cos(\theta) \ra - - is a unit vector for all angles \theta and \varphi. -

    -     - -

    - - \norm{\vec u} \amp = \sqrt{\sin^2(\theta) \cos^2(\varphi) +\sin^2(\theta) \sin^2(\varphi) +\cos^2(\theta) } - \amp = \sqrt{\sin^2(\theta) (\cos^2(\varphi) +\sin^2(\varphi) )+\cos^2(\theta) } - \amp = \sqrt{\sin^2(\theta) +\cos^2(\theta) } - \amp =1 - . -

    -     - -     - - - - -

    - A weight of 100lb is suspended from two chains, - making angles with the vertical of \theta and \varphi as shown in the figure below. -

    - - - - Image shows weight suspended with two chains. - - -

    - Image is of a weight of 100 pounds suspended by two chains. The chain on the left forms an - angle of \varphi degrees with the vertical and the chain on the right forms a degree of - \theta with the vertical. -

    -     - - \begin{tikzpicture} - \filldraw[thick,black,fill=gray!30] (-.5,0) -- (.5,0) -- (1,-1) -- (-1,-1)--cycle; - \draw (0,-.5) node {100lb}; - - \draw [thick] (-1.75,1.5) -- (1.75,1.5); - \clip (-1.5,1.5) rectangle (2,-1.25); - \draw [thick,rotate=135] (0,0) -- (3,0);% node[below,sloped,pos=.5] { $\vec F_2$}; - \draw [thick,rotate=45] (0,0) -- (3,0); %node[below,sloped,pos=.5] { $\vec F_1$};; - \draw [dashed] (0,0) -- (0,2); - \draw [rotate=45,->,>=stealth] (.8,0) arc (0:45:.8); - \draw [rotate=67] (1,0) node { $\theta$}; - \draw [rotate=90,->,>=stealth] (1,0) arc (0:45:1); - \draw [rotate=105] (1.2,0) node { $\varphi$}; - \end{tikzpicture} - - - -

    - Angles \theta and \varphi are given. - Find the magnitude of the force applied to each chain. -

    -     - - - - -

    - \theta = 30^\circ,\varphi=30^\circ -

    -     - -

    - The force on each chain is 100/\sqrt{3}\approx 57.735lb. -

    -     - -     - - - - -

    - \theta = 60^\circ,\varphi=60^\circ -

    -     - -

    - The force on each chain is 100lb. -

    -     - -     - - - - -

    - \theta = 20^\circ,\varphi=15^\circ -

    -     - -

    - The force on the chain with angle \theta is approx. - 45.124lb; - the force on the chain with angle \varphi is approx. - 59.629lb. -

    -     - -     - - - - -

    - \theta = 0^\circ,\varphi=0^\circ -

    -     - -

    - The force on each chain is 50lb. -

    -     - -     - -     - - - -

    - A weight of plb is suspended from a chain of length \ell while a constant force of - \vec F_w pushes the weight to the right, - making an angle of \theta with the vertical, - as shown in the figure below. -

    - - - - Image showing weight suspended by a chain and wind action on the weight. - - -

    - Image shows a weight of p pounds being suspended by a chain. The chain forms an angle of - \theta with the vertical. The wind is pushing the weight to the right with force \vec F_w. - The chain is of length \ell feet. -

    -     - - \begin{tikzpicture} - \draw [dashed] (-1.5,-.2) -- (2,-.2); - \draw (1.75,.65) node { $\ell$\ ft $\left.\rule{0pt}{.8cm}\right\}$}; - \filldraw[thick,black,fill=gray!30] (-.5,0) -- (.5,0) -- (1,-1) -- (-1,-1)--cycle; - \draw (0,-.5) node {$p$ lb}; - \draw [thick] (-1.5,1.5) -- (2,1.5); - \clip (-1.5,1.5) rectangle (2,-1.25); - \draw [thick,rotate=110] (0,0) -- (3,0); - \draw [dashed] (0,0) -- (0,2); - \draw [rotate=90,->,>=stealth] (1,0) arc (0:20:1); - \draw [rotate=99] (1.2,0) node { $\theta$}; - \draw [thick,->,>=stealth] (-1.45,-.5) -- (-.8,-.5) node [pos=.15,below] { $\vec F_w$}; - \end{tikzpicture} - - - -

    - A force \vec F_w and length \ell are given. - Find the angle \theta and the height the weight is lifted as it moves to the right. -

    -     - - - - -

    - \vec F_w=1lb, \ell = 1ft, p = 1lb -

    -     - -

    - \theta = 45^\circ; the weight is lifted 0.29 ft - (about 3.5in). -

    -     - -     - - - - -

    - \vec F_w=1lb, \ell = 1ft, p = 10lb -

    -     - -

    - \theta = 5.71^\circ; the weight is lifted 0.005 ft - (about 1/16th of an inch). -

    -     - -     - - - - -

    - \vec F_w=1lb, \ell = 10ft, p = 1lb -

    -     - -

    - \theta = 45^\circ; the weight is lifted 2.93 ft. -

    -     - -     - - - - -

    - \vec F_w=10lb, \ell = 10ft, p = 1lb -

    -     - -

    - \theta = 84.29^\circ; - the weight is lifted 9 ft. -

    -     - -     -     -     -     -
    -
    - The Dot Product -

    - The previous section introduced vectors and described how to add them together and how to multiply them by scalars. - This section introduces a - multiplication on vectors called the dot product. -

    - - - - - Dot Product - -

    -

      -
    1. -

      - Let \vec u = \la u_1,u_2\ra and - \vec v = \la v_1,v_2\ra in \mathbb{R}^2. - The dot product of \vec u and \vec v, - denoted \dotp uv, is - - \dotp uv = u_1v_1+u_2v_2 - . -

      -
      2. - -
    2. -

      - Let \vec u = \la u_1,u_2,u_3\ra and - \vec v = \la v_1,v_2,v_3\ra in \mathbb{R}^3. - The dot product of \vec u and \vec v, - denoted \dotp uv, is - - \dotp uv = u_1v_1+u_2v_2+u_3v_3 - . -

      -
      3. -
    - dot productdefinition - vectorsdot product -

    -     -     - - - -

    - Note how this product of vectors returns a - scalar, not another vector. - We practice evaluating a dot product in the following example, - then we will discuss why this product is useful. -

    - - - Evaluating dot products - -

    -

      -
    1. -

      - Let \vec u=\la 1,2\ra, - \vec v=\la 3,-1\ra in \mathbb{R}^2. - Find \dotp uv. -

      -
      2. - -
    2. -

      - Let \vec x = \la 2,-2,5\ra and - \vec y = \la -1, 0, 3\ra in \mathbb{R}^3. - Find \dotp xy. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - Using , we have - - \dotp uv = 1(3)+2(-1) = 1 - . -

      -
      2. - -
    2. -

      - Using the definition, we have - - \dotp xy = 2(-1) -2(0) + 5(3) = 13 - . -

      -
      3. -
    -

    -     - -     - -

    - The dot product, as shown by the preceding example, - is very simple to evaluate. - It is only the sum of products. - While the definition gives no hint as to why we would care about this operation, - there is an amazing connection between the dot product and angles formed by the vectors. - Before stating this connection, - we give a theorem stating some of the properties of the dot product. -

    - - - Properties of the Dot Product - -

    - Let \vec u, - \vec v and \vec w be vectors in \mathbb{R}^2 or - \mathbb{R}^3 and let c be a scalar. - dot productproperties - vectorsdot product -

    - -

    -

      -
    1. -

      - \dotp uv = \dotp vu (Commutative Property) -

      -
      2. - -
    2. -

      - \vec u\cdot(\vec v+\vec w) = \dotp uv + \dotp uw (Distributive Property) -

      -
      3. - -
    3. -

      - c(\dotp uv) = (c\vec u)\cdot \vec v = \vec u \cdot (c\vec v) -

      -
      4. - -
    4. -

      - \dotp 0v = 0 -

      -
      5. - -
    5. -

      - \dotp vv=\norm{\vec v}^2 -

      -
      6. -
    -

    -     -     - -

    - The last statement of the theorem makes a handy connection between the magnitude of a vector and the dot product with itself. - Our definition and theorem give properties of the dot product, - but we are still likely wondering - What does the dot product mean? - It is helpful to understand that the dot product of a vector with itself is connected to its magnitude. -

    - - - -

    - The next theorem extends this understanding by connecting the dot product to magnitudes and angles. - Given vectors \vec u and \vec v in the plane, - an angle \theta is clearly formed when \vec u and \vec v are drawn with the same initial point as illustrated in . (We always take \theta to be the angle in [0,\pi] as two angles are actually created.) -

    - -
    - Illustrating the angle formed by two vectors with the same initial point - -
    - - - - - The angle formed by two vectors. - - -

    - Image shows the angle \theta formed by two vectors, \vec u - and \vec v. The angle is formed from \vec u to \vec v. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.48] - - \draw [rotate=-15,->] (0,0) -- (2,0) node (A) [right] { $\vec u$}; - \draw [rotate=35,->] (0,0) -- (2.5,0) node (B) [right] { $\vec v$}; - \draw [rotate=-15,->] (.75,0) arc (0:50:.75); - \draw [rotate=10] (.9,0) node { $\theta$}; - - \end{tikzpicture} - - - - -
    - -
    - - - - - - Graph showing angle formed by two vectors on a plane containing both vectors. - - -

    - The three axes are shown and all are uncalibrated. Two vectors \vec u and - \vec v are shown, along with the place that contains both vectors. The two - vectors form an angle and it is labeled as \theta. The pane lies at an angle - in the positive values of all axes. -

    -     - - - - - //ASY file for figdotpangle3D.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-0.5,2.5); - pair ybounds=(-0.5,2.5); - pair zbounds=(-0.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw theplane z=.1*x+y+1; - triple f(pair t) { - return (t.x,t.y,-.1*t.x+.3*t.y+.5); - } - surface s=surface(f,(0,0),(2,2),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - // Draw the vectors - draw((0.5,0.5,.6)--(1.5,0.25,.425), black+linewidth(2),Arrow3(size=3mm));//u - draw((0.5,0.5,.6)--(0.25,1.8,1.015), black+linewidth(2),Arrow3(size=3mm));//v - label("$\vec{u}$",(1.5,0.25,.425),N); - label("$\vec{v}$",(0.25,1.8,1.015),N); - dotfactor=3; dot((0.5,0.5,.6),black); - - //Draw the arc with theta - draw((.99,.38,.51)..(.81,.78,.65)..(.41,.97,.75)); - label("$\theta$",(0.5,0.5,.7),W); - - - - -
    -     -
    - -

    - The same is also true of 2 vectors in space: - given \vec u and \vec v in - \mathbb{R}^3 with the same initial point, - there is a plane that contains both \vec u and \vec v. - (When \vec u and \vec v are co-linear, - there are infinitely many planes that contain both vectors.) - In that plane, - we can again find an angle \theta between them - (and again, 0\leq \theta\leq \pi). - This is illustrated in . -

    - -

    - The following theorem connects this angle \theta to the dot product of \vec u and \vec v. -

    - - - The Dot Product and Angles - -

    - Let \vec u and \vec v be nonzero vectors in - \mathbb{R}^2 or \mathbb{R}^3. - Then - - \dotp uv = \norm{\vec u}\,\norm{\vec v} \cos(\theta) - , - where \theta, 0\leq\theta\leq \pi, - is the angle between \vec u and \vec v. - dot productproperties - vectorsdot product -

    -     -     - - - -

    - Using , - we can rewrite this theorem as - - \frac{\vec u}{\norm{\vec u}}\cdot \frac{\vec v}{\norm{\vec v}} = \cos(\theta) - . -

    - - - -

    - Note how on the left hand side of the equation, - we are computing the dot product of two unit vectors. - Recalling that unit vectors essentially only provide direction information, - we can informally restate - as saying The dot product of two directions gives the cosine of the angle between them. -

    - -

    - When \theta is an acute angle (, 0\leq \theta \lt \pi/2), - \cos(\theta) is positive; - when \theta = \pi/2, \cos(\theta) = 0; - when \theta is an obtuse angle (\pi/2\lt \theta \leq \pi), - \cos(\theta) is negative. - Thus the sign of the dot product gives a general indication of the angle between the vectors, - illustrated in . -

    - -
    - Illustrating the relationship between the angle between vectors and the sign of their dot product - - - - - Image shows the relation between the angle between two vectors and the sign of their dot product. - - -

    - Each image has the two vectors \vec u and \vec v along with \theta shown. - The dot product along with three possible cases of <, > and equal to 0 are shown. - In the first image, the dot product has a positive value, the angle between the two vectors is acute. - In the second image, the dot product is equal to 0, the angle is 90 degrees. - In the third image, the dot product has a negative value and the angle is obtuse. -

    -     - - - \begin{tikzpicture}[scale=1.5,>=stealth] - - \begin{scope} - \draw[->] (0,0) -- (1,0) node [pos=.5,below] { $\vec u\cdot \vec v >0$} node [right] { $\vec u$}; - \draw[->] (0,0) -- (.5,.866) node [above right] { $\vec v$}; - \draw[->] (.3,0) arc (0:60:.3); - \draw[rotate=30] (.45,0) node { $\theta$}; - \end{scope} - - \begin{scope}[shift={(3cm,0)}] - \draw[->] (0,0) -- (1,0) node [pos=.5,below] { $\vec u\cdot \vec v =0$}node [right] { $\vec u$}; - \draw[->] (0,0) -- (0,1) node [above ] { $\vec v$}; - \draw[->] (.3,0) arc (0:90:.3); - \draw[rotate=22.5] (.7,.2) node { $\theta=\pi/2$}; - \end{scope} - - \begin{scope}[shift={(6.5cm,0)}] - \draw[->] (0,0) -- (1,0) node [pos=.5,below] { $\vec u\cdot \vec v <0$}node [right] { $\vec u$}; - \draw[->] (0,0) -- (-.707,.707) node [above left] { $\vec v$}; - \draw[->] (.3,0) arc (0:135:.3); - \draw[rotate=67.5] (.45,0) node { $\theta$}; - \end{scope} - - \end{tikzpicture} - - - - -     -
    - -

    - We can use to compute the dot product, - but generally this theorem is used to find the angle between known vectors - (since the dot product is generally easy to compute). - To this end, we rewrite the theorem's equation as - - \cos(\theta) = \frac{\dotp uv}{\norm{\vec u}\norm{\vec v}} \Leftrightarrow \theta = \cos^{-1}\left(\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right) - . -

    - -

    - We practice using this theorem in the following example. -

    - - - Using the dot product to find angles - -

    - Let \vec u = \la 3,1\ra, - \vec v = \la -2,6\ra and \vec w = \la -4,3\ra, - as shown in . - Find the angles \alpha, - \beta and \theta. -

    -
    - Vectors used in - - - - Graph shows three vectors and the angles between them. - - -

    - The x axis is drawn from -4 to 4 and the y - axis is drawn from 0 to 6. The \vec u, \vec v - and \vec w are shown. -

    -

    - The \vec u starts at the origin and ends at point (3, 1). - The \vec v also starts at the origin and ends at point (-2, 6) - and \vec w is also drawn from the origin and ends at point (-4, 3). -

    -

    - The angle \alpha is drawn between \vec u and \vec v, - angle \beta is drawn between \vec v and \vec w and angle - \theta is drawn between \vec u and \vec w. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - minor x tick num=4, - ymin=-1.8,ymax=6.99, - xmin=-5.5,xmax=5.5, - ] - - \draw [thick,->] (axis cs:0,0) -- (axis cs: 3,1) node [above] {$\vec u$}; - \draw [thick,->] (axis cs:0,0) -- (axis cs: -2,6) node [above] {$\vec v$}; - \draw [thick,->] (axis cs:0,0) -- (axis cs: -4,3) node [above] {$\vec w$}; - - \draw [->] (axis cs: .949,.316) arc (18.4:108.4:12pt); - \draw (axis cs:0.680986, 1.33651) node { $\alpha$}; - - \draw [->] (axis cs: -0.474342, 1.42302) arc (108.4:155.1:12pt); - \draw (axis cs:-1.13929, 1.58257) node { $\beta$}; - - \draw [->] (axis cs: 2.3725, 0.79) arc (18.4:143.1:30pt); - \draw (axis cs:1.5, 2.59808) node { $\theta$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - We start by computing the magnitude of each vector. - - \norm{\vec u} = \sqrt{10}; \norm{\vec v} = 2\sqrt{10}; \norm{\vec w} = 5 - . -

    - -

    - We now apply to find the angles. - - \alpha \amp = \cos^{-1}\left(\frac{\dotp uv}{(\sqrt{10})(2\sqrt{10})}\right) - \amp = \cos^{-1}(0) = \frac{\pi}2 = 90^\circ - . - - \beta \amp = \cos^{-1}\left(\frac{\dotp vw}{(2\sqrt{10})(5)}\right) - \amp = \cos^{-1}\left(\frac{26}{10\sqrt{10}}\right) - \amp \approx 0.6055 \approx 34.7^\circ. - \theta \amp = \cos^{-1}\left(\frac{\dotp uw}{(\sqrt{10})(5)}\right) - \amp = \cos^{-1}\left(\frac{-9}{5\sqrt{10}}\right) - \amp \approx 2.1763 \approx 124.7^\circ - -

    -     - -     - -

    - We see from our computation that \alpha + \beta = \theta, - as indicated by . - While we knew this should be the case, - it is nice to see that this non-intuitive formula indeed returns the results we expected. -

    - -

    - We do a similar example next in the context of vectors in space. -

    - - - Using the dot product to find angles - -

    - Let \vec u = \la 1,1,1\ra, - \vec v = \la -1,3,-2\ra and \vec w = \la -5,1,4\ra, - as illustrated in . - Find the angle between each pair of vectors. -

    - -
    - Vectors used in - - - - Graph showing three vectors drawn in space. - - -

    - The x axis is drawn from -5 to 5, - the y axis is drawn from 0 to 4 and - the z axis is drawn from -2 to 4. - Three vectors \vec u, \vec v and \vec w are - shown all starting at the origin, along with separate hollow dashed - cuboids that represent their three axes values and its three sides. - The \vec u has value, \vec v has value and \vec w has value. -

    -     - - - - - //ASY file for figdotp3.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(23,7.4,3.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-5,5}; - real[] myychoice={2,4}; - real[] myzchoice={-2,2,4}; - defaultpen(0.5mm); - pair xbounds=(-5.5,5.5); - pair ybounds=(-1,4.5); - pair zbounds=(-2.5,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // Draw the lines for vec{u}=<1,1,1> - draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1), dashed+linewidth(.5)+bluepen);//top - draw((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0), dashed+linewidth(.5)+bluepen);//bottom - draw((1,0,0)--(1,0,1), dashed+linewidth(.5)+bluepen);//up1 - draw((0,1,0)--(0,1,1), dashed+linewidth(.5)+bluepen);//up2 - draw((1,1,0)--(1,1,1), dashed+linewidth(.5)+bluepen);//up3 - draw((0,0,0)--(1,1,1), bluepen,Arrow3(size=2mm)); - label("$\vec{u}$",(1,1,1),E); - //dotfactor=3; dot((2,1,1),bluepen); - - // Draw the lines for \vec{v}=<-1,3,-2> - draw((0,0,-2)--(0,3,-2)--(-1,3,-2)--(-1,0,-2)--(0,0,-2), dashed+linewidth(.5)+redpen);//top - draw((0,0,0)--(0,3,0)--(-1,3,0)--(-1,0,0)--(0,0,0), dashed+linewidth(.5)+redpen);//bottom - draw((-1,0,0)--(-1,0,-2), dashed+linewidth(.5)+redpen);//up1 - draw((0,3,0)--(0,3,-2), dashed+linewidth(.5)+redpen);//up2 - draw((-1,3,0)--(-1,3,-2), dashed+linewidth(.5)+redpen);//up3 - draw((0,0,0)--(-1,3,-2),redpen,Arrow3(size=2mm)); - label("$\vec{v}$",(-1,3,-2),E); - - // Draw the lines for \vec{w}=<-5,1,4> - draw((0,0,4)--(0,1,4)--(-5,1,4)--(-5,0,4)--(0,0,4), dashed+linewidth(.5)+orange);//top - draw((0,0,0)--(0,1,0)--(-5,1,0)--(-5,0,0)--(0,0,0), dashed+linewidth(.5)+orange);//bottom - draw((-5,0,0)--(-5,0,4), dashed+linewidth(.5)+orange);//up1 - draw((0,1,0)--(0,1,4), dashed+linewidth(.5)+orange);//up2 - draw((-5,1,0)--(-5,1,4), dashed+linewidth(.5)+orange);//up3 - draw((0,0,0)--(-5,1,4), orange,Arrow3(size=2mm)); - label("$\vec{w}$",(-5,1,4),E); - - - - -
    -     - -

    -

      -
    1. -

      - Between \vec u and \vec v: - - \theta \amp = \cos^{-1}\left(\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right) - \amp = \cos^{-1}\left(\frac{0}{\sqrt{3}\sqrt{14}}\right) - \amp = \frac{\pi}2 - . -

      -
      2. - -
    2. -

      - Between \vec u and \vec w: - - \theta \amp = \cos^{-1}\left(\frac{\dotp uw}{\norm{\vec u}\norm{\vec w}}\right) - \amp = \cos^{-1}\left(\frac{0}{\sqrt{3}\sqrt{42}}\right) - \amp = \frac{\pi}2 - . -

      -
      3. - -
    3. -

      - Between \vec v and \vec w: - - \theta \amp = \cos^{-1}\left(\frac{\dotp vw}{\norm{\vec v}\norm{\vec w}}\right) - \amp = \cos^{-1}\left(\frac{0}{\sqrt{14}\sqrt{42}}\right) - \amp = \frac{\pi}2 - . -

      -
      4. -
    -

    - -

    - While our work shows that each angle is \pi/2, , 90^\circ, - none of these angles looks to be a right angle in . - Such is the case when drawing three-dimensional objects on the page. -

    -     -     - -

    - All three angles between these vectors was \pi/2, - or 90^\circ. - We know from geometry and everyday life that - 90^\circ angles are nice - for a variety of reasons, - so it should seem significant that these angles are all \pi/2. - Notice the common feature in each calculation (and also the calculation of \alpha in ): - the dot products of each pair of angles was 0. - We use this as a basis for a definition of the term orthogonal, - which is essentially synonymous to perpendicular. -

    - - - Orthogonal - -

    - Nonzero vectors \vec u and \vec v are orthogonal - if their dot product is 0. - orthogonal - perpendicular|see{orthogonal} - vectorsorthogonal -

    -     -     - - - - - Finding orthogonal vectors - -

    - Let \vec u = \la 3,5\ra and \vec v = \la 1,2,3\ra. -

    - -

    -

      -
    1. -

      - Find two vectors in \mathbb{R}^2 that are orthogonal to \vec u. -

      -
      2. - -
    2. -

      - Find two non-parallel vectors in - \mathbb{R}^3 that are orthogonal to \vec v. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - Recall that a line perpendicular to a line with slope m has slope -1/m, - the opposite reciprocal slope. - We can think of the slope of \vec u as 5/3, - its rise over run. A vector orthogonal to \vec u will have slope -3/5. - There are many such choices, though all parallel: - - \la -5,3\ra \text{ or } \la 5,-3\ra \text{ or } \la -10,6\ra \text{ or } \la 15,-9\ra,\text{ etc. } - -

      -
      2. - -
    2. -

      - There are infinitely many directions in space orthogonal to any given direction, - so there are an infinite number of non-parallel vectors orthogonal to \vec v. - Since there are so many, we have great leeway in finding some. - One way is to arbitrarily pick values for the first two components, - leaving the third unknown. - For instance, let \vec v_1 = \la 2,7,z\ra. - If \vec v_1 is to be orthogonal to \vec v, - then \vec v_1\cdot\vec v = 0, so - - 2+14+3z=0 \Rightarrow z = \frac{-16}{3} - . - So \vec v_1 = \la 2, 7, -16/3\ra is orthogonal to \vec v. - We can apply a similar technique by leaving the first or second component unknown. - Another method of finding a vector orthogonal to \vec v mirrors what we did in part 1. - Let \vec v_2 = \la-2,1,0\ra. - Here we switched the first two components of \vec v, - changing the sign of one of them - (similar to the opposite reciprocal concept before). - Letting the third component be 0 effectively ignores the third component of \vec v, - and it is easy to see that - - \vec v_2\cdot\vec v = \la -2,1,0\ra\cdot\la 1,2,3\ra = 0 - . - Clearly \vec v_1 and \vec v_2 are not parallel. -

      -
      3. -
    -

    -     - -     - -

    - An important construction is illustrated in , - where vectors \vec u and \vec v are sketched. - In , - a dotted line is drawn from the tip of \vec u to the line containing \vec v, - where the dotted line is orthogonal to \vec v. - In , - the dotted line is replaced with the vector \vec z and \vec w is formed, - parallel to \vec v. - It is clear by the diagram that \vec u = \vec w+\vec z. - What is important about this construction is this: - \vec u is decomposed - as the sum of two vectors, - one of which is parallel to \vec v and one that is perpendicular to \vec v. - It is hard to overstate the importance of this construction - (as we'll see in upcoming examples). -

    - -

    - The vectors \vec w, - \vec z and \vec u as shown in form a right triangle, - where the angle between \vec v and \vec u is labeled \theta. - We can find \vec w in terms of \vec v and \vec u. -

    - -

    - Using trigonometry, we can state that - - \norm{\vec w} = \norm{\vec u}\cos(\theta) - . -

    - -
    - Developing the construction of the orthogonal projection - -
    - - - - - Diagram shows two vectors and the angle between them. - - -

    - Two vectors \vec u and \vec v are shown that start from the - same position, the angle between them is marked \theta, the vector - \vec u is shorter. From the end of \vec u a dashed perpendicular - is drawn on \vec v. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=0.93] - - \draw [thick,->] (0,0) -- (4,2) node [right] { $\vec v$}; - \draw [thick,->] (0,0) -- (1,3) node [above ] { $\vec u$}; - \draw [dotted,thick] (1,3) -- (2,1); - \draw (1.82111, 0.910557) -- ({1.73167, 1.08944}) -- (1.91056, 1.17889); - \draw (.4,.2) arc (26.5:71.6:.45); - \draw [rotate=49] (.6,0) node { $\theta$}; - - \end{tikzpicture} - - - - -
    - -
    - - - - - Diagram shows two vectors and the angle between them. - - -

    - Two vectors \vec u and \vec v are shown that - start from the same position, the angle between them is marked - \theta, the vector \vec u is shorter. From the end - of \vec u a dashed perpendicular is drawn on \vec v. - The vector \vec w and \vec z are also added. From the - initial point the vector \vec w is drawn in the same direction - as \vec v but it ends at the start of the perpendicular. The - perpendicular starts at the tip of \vec w and ends at the tip - of \vec u and is labeled \vec z. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=0.93] - - \draw [thick,->] (2,1) -- (4,2) node [right] { $\vec v$}; - \draw [thick,->] (0,0) -- (1,3) node [above ] { $\vec u$}; - \draw [thick,->,gray] (0,0) -- (2,1) node [below,black] { $\vec w$}; - \draw [<-,thick] (1,3) -- (2,1) node [right,pos=.5] { $\vec z$}; - \draw (1.82111, 0.910557) -- ({1.73167, 1.08944})--(1.91056, 1.17889); - \draw (.4,.2) arc (26.5:71.6:.45); - \draw [rotate=49] (.6,0) node { $\theta$}; - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - We also know that \vec w is parallel to to \vec v; that is, - the direction of \vec w is the direction of \vec v, - described by the unit vector \vec v/\norm{\vec v}. - The vector \vec w is the vector in the direction - \vec v/\norm{\vec v} with magnitude \norm{\vec u}\cos(\theta): - - \vec w \amp = \Big(\norm{\vec u}\cos(\theta) \Big)\frac{1}{\norm{\vec v}}\vec v \amp \amp - \amp = \left(\norm{\vec u}\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right)\frac{1}{\norm{\vec v}} \vec v \amp \quad \amp \text{ (Replacing } \cos(\theta) \text{ using } \text{)} - \amp = \frac{\dotp uv}{\norm{\vec v}^2}\vec v \amp \amp - \amp = \frac{\dotp uv}{\dotp vv}\vec v \amp \amp \text{ (Applying }\text{)} - . -

    - -

    - Since this construction is so important, - it is given a special name. -

    - - - Orthogonal Projection - -

    - Let nonzero vectors \vec u and \vec v be given. - The orthogonal projection of \vec u onto \vec v, - denoted \proj uv, is - orthogonal projection - vectorsorthogonal projection - - \proj uv = \frac{\dotp uv}{\dotp vv}\vec v - . -

    -     -     - - - - - Computing the orthogonal projection - -

    -

      -
    1. -

      - Let \vec u= \la -2,1\ra and \vec v=\la 3,1\ra. - Find \proj uv, - and sketch all three vectors with initial points at the origin. -

      -
      2. - -
    2. -

      - Let \vec w = \la 2,1,3\ra and \vec x = \la 1,1,1\ra. - Find \proj wx, - and sketch all three vectors with initial points at the origin. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - Applying , we have - - \proj uv \amp = \frac{\dotp uv}{\dotp vv}\vec v - \amp = \frac{-5}{10}\la 3,1\ra - \amp = \la -\frac32,-\frac12\ra - . - Vectors \vec u, \vec v and - \proj uv are sketched in . - Note how the projection is parallel to \vec v; - that is, it lies on the same line through the origin as \vec v, - although it points in the opposite direction. - That is because the angle between \vec u and \vec v is obtuse (, greater than 90^\circ). -

      - -
      - Sketching the three vectors in Part of - - - Graph shows two vectors u and v, and the projection of u on v. - - -

      - The x axis is drawn from -2 to 3 and the - y axis is drawn from -2 to 2. Two vectors - \vec u and \vec v are shown, both start at the origin. - The vector u ends in point (-2, 1) and lies in the - second quadrant and \vec v ends in point ( 3, 1) and - lies in the first quadrant. The projection of \vec u on - \vec vis shown and it lies in the third quadrant. -

      -       - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - xtick={-3,-2,-1,1,2,3}, - ytick={1,2,-1,-2}, - ymin=-2.8,ymax=2.8, - xmin=-2.9,xmax=3.9, - ] - - \draw [thick,->] (axis cs:0,0) -- (axis cs: -2,1) node [above] {$\vec u$}; - \draw [thick,->] (axis cs:0,0) -- (axis cs: 3,1) node [above] {$\vec v$}; - \draw [thick,->,gray] (axis cs:0,0) -- (axis cs: -1.5,-.5) node [below,black] {$\text{proj}_{\, \vec v}\, \vec u$}; - - \draw [dotted,thick] (axis cs: -2,1) -- (axis cs:-1.5,-.5); - - \end{axis} - - \end{tikzpicture} - - - -
      -
      2. - -
    2. -

      - Apply the definition: - - \proj wx \amp = \frac{\dotp wx}{\dotp xx}\vec x - \amp = \frac{6}{3}\la 1,1,1\ra - \amp = \la 2,2,2\ra - . - These vectors are sketched in , - and again in from a different perspective. - Because of the nature of graphing these vectors, - the sketch in makes it difficult to recognize that the drawn projection has the geometric properties it should. - The graph shown in illustrates these properties better. -

      - -
      - Sketching the three vectors in Part of - -
      - - - - - Graph shows two vectors w and x in space, and the projection of w on x. - - -

      - The x, y and z axes are drawn from 0 to - 2. Two vectors \vec w and \vec x are drawn, - both start at the origin, \vec w ends at (2, 1, 3) - and \vec x ends at (1, 1, 1). The projection of - \vec w on \vec x is shown, it also starts at the - origin and ends at (2, 2, 2). -

      -       - - - - - //ASY file for figdotp4b.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - - // The text calls for two viewpoints of this picture to show orthogonality - // The two projections below give those two views - // view "b", saved as figdotp4b_3D - currentprojection=orthographic(8,2,5); - - // view "c", saved as figdotp4c_3D - //currentprojection=orthographic(2,8,6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-0.5,2.5); - pair ybounds=(-1,2.5); - pair zbounds=(-0.5,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // Draw the lines for vec{x}=<1,1,1> - //draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1), dashed+linewidth(.5));//top - draw((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0), dashed+linewidth(.5));//bottom - //draw((1,0,0)--(1,0,1), dashed+linewidth(.5));//up1 - //draw((0,1,0)--(0,1,1), dashed+linewidth(.5));//up2 - draw((1,1,0)--(1,1,1), dashed+linewidth(.5));//up3 - draw((0,0,0)--(1,1,1), Arrow3(size=3mm)); - label("$\vec{x}$",(1,1,1),N); - //dotfactor=3; dot((2,1,1),bluepen); - - // Draw the lines for projection of w onto x as <2,2,2> - //draw((0,0,2)--(0,2,2)--(2,2,2)--(2,0,2)--(0,0,2), dashed+linewidth(.5));//top - draw((0,0,0)--(0,2,0)--(2,2,0)--(2,0,0)--(0,0,0), dashed+linewidth(.5));//bottom - //draw((2,0,0)--(2,0,2), dashed+linewidth(.5));//up1 - //draw((0,2,0)--(0,2,2), dashed+linewidth(.5));//up2 - draw((2,2,0)--(2,2,2), dashed+linewidth(.5));//up3 - draw((0,0,0)--(2,2,2), gray,Arrow3(size=3mm)); - label("$\textrm{proj}_{\vec{x}} \vec{w}$",(2,2,2),N); - - // Draw the lines for \vec{w}=<2,1,3> - //draw((0,0,3)--(0,1,3)--(2,1,3)--(2,0,3)--(0,0,3), dashed+linewidth(.5));//top - draw((0,0,0)--(0,1,0)--(2,1,0)--(2,0,0)--(0,0,0), dashed+linewidth(.5));//bottom - //draw((2,0,0)--(2,0,3), dashed+linewidth(.5));//up1 - //draw((0,1,0)--(0,1,3), dashed+linewidth(.5));//up2 - draw((2,1,0)--(2,1,3), dashed+linewidth(.5));//up3 - draw((0,0,0)--(2,1,3), Arrow3(size=3mm)); - label("$\vec{w}$",(2,1,3),N); - - - - -
      - -
      - - - - - - Graph shows two vectors w and x in space, and the projection of w on x. - - -

      - The x, y and z axes are drawn from 0 to - 2. Two vectors \vec w and \vec x are drawn, - both start at the origin, \vec w ends at (2, 1, 3) - and \vec x ends at (1, 1, 1). The projection of - \vec w on \vec x is shown, it also starts at the - origin and ends at (2, 2, 2). -

      -       - - - - - //ASY file for figdotp4b.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - - // The text calls for two viewpoints of this picture to show orthogonality - // The two projections below give those two views - // view "b", saved as figdotp4b_3D - //currentprojection=orthographic(8,2,5); - - // view "c", saved as figdotp4c_3D - currentprojection=orthographic(2,8,6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-0.5,2.5); - pair ybounds=(-1,2.5); - pair zbounds=(-0.5,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // Draw the lines for vec{x}=<1,1,1> - //draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1), dashed+linewidth(.5));//top - draw((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0), dashed+linewidth(.5));//bottom - //draw((1,0,0)--(1,0,1), dashed+linewidth(.5));//up1 - //draw((0,1,0)--(0,1,1), dashed+linewidth(.5));//up2 - draw((1,1,0)--(1,1,1), dashed+linewidth(.5));//up3 - draw((0,0,0)--(1,1,1), Arrow3(size=3mm)); - label("$\vec{x}$",(1,1,1),N); - //dotfactor=3; dot((2,1,1),bluepen); - - // Draw the lines for projection of w onto x as <2,2,2> - //draw((0,0,2)--(0,2,2)--(2,2,2)--(2,0,2)--(0,0,2), dashed+linewidth(.5));//top - draw((0,0,0)--(0,2,0)--(2,2,0)--(2,0,0)--(0,0,0), dashed+linewidth(.5));//bottom - //draw((2,0,0)--(2,0,2), dashed+linewidth(.5));//up1 - //draw((0,2,0)--(0,2,2), dashed+linewidth(.5));//up2 - draw((2,2,0)--(2,2,2), dashed+linewidth(.5));//up3 - draw((0,0,0)--(2,2,2), gray,Arrow3(size=3mm)); - label("$\textrm{proj}_{\vec{x}} \vec{w}$",(2,2,2),N); - - // Draw the lines for \vec{w}=<2,1,3> - //draw((0,0,3)--(0,1,3)--(2,1,3)--(2,0,3)--(0,0,3), dashed+linewidth(.5));//top - draw((0,0,0)--(0,1,0)--(2,1,0)--(2,0,0)--(0,0,0), dashed+linewidth(.5));//bottom - //draw((2,0,0)--(2,0,3), dashed+linewidth(.5));//up1 - //draw((0,1,0)--(0,1,3), dashed+linewidth(.5));//up2 - draw((2,1,0)--(2,1,3), dashed+linewidth(.5));//up3 - draw((0,0,0)--(2,1,3), Arrow3(size=3mm)); - label("$\vec{w}$",(2,1,3),N); - - - - -
      -       -
      -
      3. -
    -

    -     - -     - -

    - We can use the properties of the dot product found in - to rearrange the formula found in : - - \proj uv \amp = \frac{\dotp uv}{\dotp vv}\vec v - \amp = \frac{\dotp uv}{\norm{\vec v}^2}\vec v - \amp = \left(\vec u \cdot \frac{\vec v}{\norm{\vec v}}\right) \frac{\vec v}{\norm{\vec v}} - . -

    - -

    - The above formula shows that the orthogonal projection of \vec u onto \vec v is only concerned with the - direction of \vec v, - as both instances of \vec v in the formula come in the form \vec v/\norm{\vec v}, - the unit vector in the direction of \vec v. -

    - -

    - A special case of orthogonal projection occurs when \vec v is a unit vector. - In this situation, - the formula for the orthogonal projection of a vector \vec u onto \vec v reduces to just \proj uv = (\vec u\cdot\vec v)\vec v, - as \vec v\cdot\vec v = 1. -

    - -

    - This gives us a new understanding of the dot product. - When \vec v is a unit vector, - essentially providing only direction information, - the dot product of \vec u and \vec v gives - how much of \vec u is in the direction of \vec v. - This use of the dot product will be very useful in future sections. -

    - -

    - Now consider - where the concept of the orthogonal projection is again illustrated. - It is clear that - - \vec u = \proj uv + \vec z - . -

    - -
    - Illustrating the orthogonal projection - - - - Diagram shows two vectors and the angle between them. - - -

    - Two vectors \vec u and \vec v are shown that start from the - same position, the angle between them is marked \theta, the vector - u is shorter. From the end of \vec u a dashed perpendicular - is drawn on \vec v. The projection of \vec u and \vec z - are also added. The projection of \vec u is in the same direction as - \vec v but it ends at the start of the perpendicular. The perpendicular - starts at the tip of the projection of \vec u and ends at the tip of - \vec u and is labeled \vec z. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.32] - - \draw [thick,->] (2,1) -- (4,2) node [below] { $\vec v$}; - \draw [thick,->] (0,0) -- (1,3) node [above] { $\vec u$}; - \draw [thick,->,gray] (0,0) -- (2,1) node [pos=.5,below right,black] { $\text{proj}_{\, \vec v}\, \vec u$}; - \draw [<-,thick] (1,3) -- (2,1) node [right,pos=.5] { $\vec z$}; - \draw (1.82111, 0.910557) -- ({1.73167, 1.08944}) -- (1.91056, 1.17889); - - \end{tikzpicture} - - - - -
    - -

    - As we know what \vec u and \proj uv are, - we can solve for \vec z and state that - - \vec z = \vec u - \proj uv - . -

    - -

    - This leads us to rewrite Equation in a seemingly silly way: - - \vec u = \proj uv + (\vec u - \proj uv) - . -

    - -

    - This is not nonsense, as pointed out in the following Key Idea. - (Notation note: the expression \parallel \vec y - means is parallel to \vec y. - We can use this notation to state - \vec x\parallel\vec y which means - \vec x is parallel to \vec y. - The expression \perp \vec y - means is orthogonal to \vec y, and is used similarly.) -

    - - - Orthogonal Decomposition of Vectors -

    - Let nonzero vectors \vec u and \vec v be given. - Then \vec u can be written as the sum of two vectors, - one of which is parallel to \vec v, - and one of which is orthogonal to \vec v: - orthogonal decomposition of vectors - orthogonaldecomposition - vectorsorthogonal decomposition - - \vec u = \underbrace{\proj uv}_{\parallel\ \vec v}\ +\ (\underbrace{\vec u-\proj uv}_{\perp\ \vec v}) - . -

    -     - - - -

    - We illustrate the use of this equality in the following example. -

    - - - Orthogonal decomposition of vectors - -

    -

      -
    1. -

      - Let \vec u = \la -2,1\ra and - \vec v = \la 3,1\ra as in . - Decompose \vec u as the sum of a vector parallel to \vec v and a vector orthogonal to \vec v. -

      -
      2. - -
    2. -

      - Let \vec w =\la 2,1,3\ra and - \vec x =\la 1,1,1\ra as in . - Decompose \vec w as the sum of a vector parallel to \vec x and a vector orthogonal to \vec x. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - In , - we found that \proj uv = \la -1.5,-0.5\ra. - Let - - \vec z = \vec u - \proj uv = \la -2,1\ra - \la -1.5,-0.5\ra = \la-0.5, 1.5\ra - . - Is \vec z orthogonal to \vec v? (, is - \vec z \perp\vec v?) We check for orthogonality with the dot product: - - \dotp zv = \la -0.5,1.5\ra \cdot \la 3,1\ra =0 - . - Since the dot product is 0, we know \vec z \perp \vec v. - Thus: - - \vec u \amp = \proj uv\ +\ (\vec u - \proj uv) - \la -2,1\ra \amp = \underbrace{\la -1.5,-0.5\ra}_{\parallel\ \vec v}\ +\ \underbrace{\la -0.5,1.5\ra}_{\perp \ \vec v} - . -

      -
      2. - -
    2. -

      - We found in - that \proj wx = \la 2,2,2\ra. - Applying the Key Idea, we have: - - \vec z = \vec w - \proj wx = \la 2,1,3\ra - \la 2,2,2\ra = \la 0,-1,1\ra - . - We check to see if \vec z \perp \vec x: - - \dotp zx = \la 0,-1,1\ra \cdot \la 1,1,1\ra = 0 - . - Since the dot product is 0, we know the two vectors are orthogonal. - We now write \vec w as the sum of two vectors, - one parallel and one orthogonal to \vec x: - - \vec w \amp = \proj wx\ +\ (\vec w - \proj wx) - \la 2,1,3\ra \amp = \underbrace{\la 2,2,2\ra}_{\parallel\ \vec x}\ +\ \underbrace{\la 0,-1,1\ra}_{\perp \ \vec x} - -

      -
      3. -
    -

    -     - -     - -

    - We give an example of where this decomposition is useful. -

    - - - Orthogonally decomposing a force vector - -

    - Consider , - showing a box weighing 50lb on a ramp that rises 5ft over a span of 20ft. - Find the components of force, - and their magnitudes, - acting on the box (as sketched in ): -

    - -
    - Sketching the ramp and box in . Note: The vectors are not drawn to scale. - -
    - - - - - Diagram showing a box on a ramp. - - -

    - Image of a box being placed on an inclined plane. The downward - force for gravity is marked as \vec g. The surface of the - ramp is labeled as \vec r base of the ramp is labeled 20 - and the height is marked as 5. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{scope}[scale=.25] - \draw [thick,->] (20,5) -- node [right,pos=.5] { $5$} (20,0) -- node [below,pos=.6] { 20} (0,0) -- (20,5) node [above] { $\vec r$}; - \draw [thick] (10,2.5) -- (9.25,5.5) -- (12.25,6.25) -- (13,3.25); - \draw [thick,->] (11.125,4.375) -- (11.125,-3.625) node [below] { $\vec g$}; - \end{scope} - - \end{tikzpicture} - - - - -
    - -
    - - - - - Diagram showing a box on a ramp. - - -

    - Image of a box being placed on an inclined plane. The downward force for - gravity is marked as \vec g. The surface of the ramp is labeled as - \vec r base of the ramp is labeled 20 and the height is marked - as 5. The projection of \vec g on \vec r is shown. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{scope}[scale=.25] - \draw [thick,->] (20,5) -- node [right,pos=.5] { $5$} (20,0) -- node [below,pos=.6] { 20}(0,0) -- (20,5) node [above] { $\vec r$}; - \draw [thick] (10,2.5) -- (9.25,5.5) -- (12.25,6.25) -- (13,3.25); - \draw [thick,->] (11.125,4.375) -- (11.125,-3.625) node [below] { $\vec g$}; - \draw [gray,thick,->] (11.125,4.375) -- (13,-3.15) node [right,pos=.4,black] { $\vec z$}; - \draw [gray,thick,->] (13,-3.15) -- (11.125,-3.625) node [shift={(12pt,-8pt)} ,black,pos=0] { $\proj gr$}; - \end{scope} - - \end{tikzpicture} - - - - -
    -     -
    - -

    -

      -
    1. -

      - in the direction of the ramp, and -

      -
      2. - -
    2. -

      - orthogonal to the ramp. -

      -
      3. -
    -

    -     - -

    - As the ramp rises 5ft over a horizontal distance of 20ft, - we can represent the direction of the ramp with the vector \vec r= \la 20,5\ra. - Gravity pulls down with a force of 50lb, - which we represent with \vec g = \la 0,-50\ra. -

    - -

    -

      -
    1. -

      - To find the force of gravity in the direction of the ramp, - we compute \proj gr: - - \proj gr \amp = \frac{\dotp gr}{\dotp rr}\vec r - \amp = \frac{-250}{425}\la 20,5\ra - \amp = \la -\frac{200}{17},-\frac{50}{17}\ra \approx \la -11.76,-2.94\ra - . - The magnitude of \proj gr is \norm{\proj gr} = 50/\sqrt{17} \approx 12.13\text{ lb }. - Though the box weighs 50lb, - a force of about 12lb is enough to keep the box from sliding down the ramp. -

      -
      2. - -
    2. -

      - To find the component \vec z of gravity orthogonal to the ramp, - we use . - - \vec z \amp = \vec g - \proj gr - \amp = \la \frac{200}{17},-\frac{800}{17}\ra \approx \la 11.76,-47.06\ra - . - The magnitude of this force is \norm{\vec z} \approx 48.51lb. - In physics and engineering, - knowing this force is important when computing things like static frictional force. - (For instance, - we could easily compute if the static frictional force alone was enough to keep the box from sliding down the ramp.) -

      -
      3. -
    -

    -     - -     - - - Application to Work -

    - In physics, the application of a force F to move an object in a straight line a distance d produces work; - the amount of work W is W=Fd, - (where F is in the direction of travel). - The orthogonal projection allows us to compute work when the force is not in the direction of travel. -

    - -
    - Finding work when the force and direction of travel are given as vectors - - - - Diagram showing the two vectors for force and displacement used to find work. - - -

    - A box is shown that is being pushed to the right by a force. The displacement is - shown as a horizontal vector and is labeled \vec d. Force is drawn from the - middle of the box at an angle and is labeled as \vec F. The projection of - the \vec F is also shown, it is parallel to the displacement vector. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \draw [thick] (0,0) rectangle (2,2); - \draw [thick,->] (0,-.25) -- (5,-.25) node [right] { $\vec d$}; - \draw [thick,->] (1,1) -- (3,2) node [shift={(5pt,2.5pt)}] { $\vec F$}; - \draw [thick,gray,->] (1,1) -- (3,1) node [right,black] { $\proj Fd$}; - - \end{tikzpicture} - - - - -
    - -

    - Consider , - where a force \vec F is being applied to an object moving in the direction of \vec d. - (The distance the object travels is the magnitude of \vec d.) - The work done is the amount of force in the direction of \vec d, - \norm{\proj Fd}, times \vnorm d: - - \norm{\proj Fd}\cdot\vnorm d \amp = \norm{\frac{\dotp Fd}{\dotp dd}\vec d}\cdot \vnorm d - \amp = \abs{\frac{\dotp Fd}{\norm{\vec{d}}^2}}\cdot \vnorm d\cdot\vnorm d - \amp = \frac{\abs{\dotp Fd}}{\norm{\vec{d}}^2}\norm{\vec{d}}^2 - \amp = \abs{\dotp Fd} - . -

    - -

    - The expression \dotp Fd will be positive if the angle between \vec F and \vec d is acute; - when the angle is obtuse - (hence \dotp Fd is negative), - the force is causing motion in the opposite direction of \vec d, - resulting in negative work. - We want to capture this sign, - so we drop the absolute value and find that W = \dotp Fd. -

    - - - Work - -

    - Let \vec F be a constant force that moves an object in a straight line from point P to point Q. - Let \vec d = \overrightarrow{PQ}. - The work W done by \vec F along \vec d is W = \dotp Fd. - work -

    -     -     - - - Computing work - -

    - A man slides a box along a ramp that rises 3ft over a distance of 15ft by applying 50lb of force as shown in . - Compute the work done. -

    -
    - Computing work when sliding a box up a ramp in - - - - Diagram showing a box on a ramp. - - -

    - Image of a box being placed on an inclined plane or ramp. The base of the - ramp is labeled 15 and the height is marked as 3. Force is - drawn from the middle of the box at an angle of 30 from the horizontal - and is labeled as \vec F. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=.35] - - \draw [thick] (0,0) -- node [below,pos=.5] { 15} (15,0) -- node [right, pos=.5] { 3} (15,3) -- cycle; - \draw [thick] (5,1) -- (4.55,3.25) -- (6.8,3.7) -- (7.25,1.45); - - \begin{scope}[shift={(5.9,2.35)}] - \draw [thick,rotate=30,->] (0,0) -- (5,0) node [right] { $\vec F$}; - \draw [thick,dashed] (-2,0) -- (4,0); - \draw (2,0) arc (0:30:2); - \draw [rotate=15] (3,0) node { $30^\circ$}; - \end{scope} - - \end{tikzpicture} - - - - -
    -     - -

    - The figure indicates that the force applied makes a - 30^\circ angle with the horizontal, - so \vec F = 50\la \cos(30^\circ) ,\sin(30^\circ) \ra \approx \la 43.3,25\ra. - The ramp is represented by \vec d = \la 15,3\ra. - The work done is simply - - \dotp Fd = 50\la \cos(30^\circ) ,\sin(30^\circ) \ra \cdot \la 15,3\ra \approx 724.5 \text{ ft--lb } - . -

    - -

    - Note how we did not actually compute the distance the object traveled, - nor the magnitude of the force in the direction of travel; - this is all inherently computed by the dot product! -

    -     -     - -

    - The dot product is a powerful way of evaluating computations that depend on angles without actually using angles. - The next section explores another - product on vectors, - the cross product. Once again, - angles play an important role, - though in a much different way. -

    -     - - - - Terms and Concepts - - - -

    - The dot product of two vectors is a , not a vector. -

    -     - - - - - - - - -     - - - - -

    - How are the concepts of the dot product and vector magnitude related? -

    -     - - - -

    - The magnitude of a vectors is the square root of the dot product of a vector with itself; that is, - \norm{\vec v} = \sqrt{\vec v\cdot \vec v}. -

    -     - -     - - - - -

    - How can one quickly tell if the angle between two vectors is acute or obtuse? -

    -     - - - -

    - By considering the sign of the dot product of the two vectors. - If the dot product is positive, the angle is acute; - if the dot product is negative, the angle is obtuse. -

    -     - -     - - - - -

    - Give a synonym for orthogonal. -

    -     - - - -

    - Perpendicular is one answer, - although we tend to use perpendicular to refer to lines, - and orthogonal to refer to vectors. -

    -

    - Note that the zero vector is orthogonal to any vector, - but it doesn't really make sense to say it is perpendicular. -

    -     - -     -     - - - Problems - - - -

    - Find the dot product of the given vectors. -

    -     - - - - - Context("LimitedNumeric"); - $dot=Real("-22"); - - - -

    - \vec u = \la 2,-4\ra, \vec v = \la 3,7\ra -

    - -

    - -

    -     -     -     - - - - - Context("LimitedNumeric"); - $dot=Real("33"); - - - -

    - \vec u = \la 5,3\ra, \vec v = \la 6,1\ra -

    - -

    - -

    -     -     -     - - - - - Context("LimitedNumeric"); - $dot=Real("3"); - - - -

    - \vec u = \la 1,-1,2\ra, - \vec v = \la 2,5,3\ra -

    - -

    - -

    -     -     -     - - - - - Context("LimitedNumeric"); - $dot=Real("0"); - - - -

    - \vec u = \la 3,5,-1\ra, - \vec v = \la 4,-1,7\ra -

    - -

    - -

    -     -     -     - - - - -

    - \vec u = \la 1,1\ra, \vec v = \la 1,2,3\ra -

    -     - -

    - not defined -

    -     - -     - - - - - Context("LimitedNumeric"); - $dot=Real("0"); - - - -

    - \vec u = \la 1,2,3\ra, - \vec v = \la 0,0,0\ra -

    - -

    - -

    -     -     -     - -     - - - - -

    - Create your own vectors \vec u, - \vec v and \vec w in - \mathbb{R}^2 and show that \vec u\cdot (\vec v+\vec w) = \vec u\cdot \vec v + \vec u\cdot \vec w. -

    -     - -

    - Answers will vary. -

    -     - -     - - - - -

    - Create your own vectors \vec u and \vec v in - \mathbb{R}^3 and scalar c and show that c(\vec u\cdot \vec v) = \vec u\cdot (c\vec v). -

    -     - -

    - Answers will vary. -

    -     - -     - - - - -

    - Find the measure of the angle between the two vectors in radians. -

    -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $theta=Formula("arccos(3/sqrt(10))"); - - - -

    - \vec u = \la 1,1\ra and - \vec v = \la 1,2\ra. -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $theta=Formula("arccos(-1/sqrt(170))"); - - - -

    - \vec u = \la -2,1\ra and - \vec v = \la 3,5\ra. -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $theta=Formula("pi/4"); - - - -

    - \vec u = \la 8,1,-4\ra and - \vec v = \la 2,2,0\ra. -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $theta=Formula("pi/2"); - - - -

    - \vec u = \la 1,7,2\ra and - \vec v = \la 4,-2,5\ra. -

    - -

    - -

    -     -     -     - -     - - - -

    - A vector \vec v is given. - Give two vectors that are orthogonal to \vec v. -

    -     - - - - - Context("Vector"); - $v=Vector("<4,7>"); - $u=Vector("<-7,4>"); - $w=Vector("<14,-8>"); - $multians = MultiAnswer($u,$v)->with( - singleResult => 0, - checker => sub { - my ($correct,$student,$self) = @_; - my ($s1, $s2) = @{$student}; - $ret[0] = (($s1 . $v == 0) and ($s1 != Vector("<0,0>"))); - $ret[1] = (($s2 . $v == 0) and ($s2 != Vector("<0,0>")) and ($s1 != $s2)); - return ~~@ret; - } - ); - - - -

    - Find two nonzero vectors orthogonal to \vec v = \la 4,7\ra. -

    - -

    - -

    - -

    - -

    -     - -

    - Answers will vary; two possible answers are - \la -7,4\ra and \la 14,-8\ra. -

    -     -     -     - - - - - Context("Vector"); - $v=Vector("<-3,5>"); - $u=Vector("<5,3>"); - $w=Vector("<-15,-9>"); - $multians = MultiAnswer($u,$v)->with( - singleResult => 0, - checker => sub { - my ($correct,$student,$self) = @_; - my ($s1, $s2) = @{$student}; - $ret[0] = (($s1 . $v == 0) and ($s1 != Vector("<0,0>"))); - $ret[1] = (($s2 . $v == 0) and ($s2 != Vector("<0,0>")) and ($s1 != $s2)); - return ~~@ret; - } - ); - - - -

    - Find two nonzero vectors orthogonal to \vec v = \la -3,5\ra. -

    - -

    - -

    - -

    - -

    -     - -

    - Answers will vary; two possible answers are - \la 5,3\ra and \la -15,-9\ra. -

    -     -     -     - - - - - Context("Vector"); - $v=Vector("<1,1,1>"); - $u=Vector("<1,0,-1>"); - $w=Vector("<4,5,-9>"); - $multians = MultiAnswer($u,$v)->with( - singleResult => 0, - checker => sub { - my ($correct,$student,$self) = @_; - my ($s1, $s2) = @{$student}; - $ret[0] = (($s1 . $v == 0) and ($s1 != Vector("<0,0>"))); - $ret[1] = (($s2 . $v == 0) and ($s2 != Vector("<0,0>")) and ($s1 != $s2)); - return ~~@ret; - } - ); - - - -

    - Find two nonzero vectors orthogonal to \vec v = \la 1,1,1\ra. -

    - -

    - -

    - -

    - -

    -     - -

    - Answers will vary; two possible answers are - \la 1,0,-1\ra and \la 4,5,-9\ra. -

    -     -     -     - - - - - Context("Vector"); - $v=Vector("<1,-2,3>"); - $u=Vector("<2,1,0>"); - $w=Vector("<1,1,1/3>"); - $multians = MultiAnswer($u,$v)->with( - singleResult => 0, - checker => sub { - my ($correct,$student,$self) = @_; - my ($s1, $s2) = @{$student}; - $ret[0] = (($s1 . $v == 0) and ($s1 != Vector("<0,0>"))); - $ret[1] = (($s2 . $v == 0) and ($s2 != Vector("<0,0>")) and ($s1 != $s2)); - return ~~@ret; - } - ); - - - -

    - Find two nonzero vectors orthogonal to \vec v = \la 1,-2,3\ra. -

    - -

    - -

    - -

    - -

    -     - -

    - Answers will vary; two possible answers are - \la 2,1,0\ra and \la 1,1,1/3\ra. -

    -     -     -     - -     - - - -

    - Vectors \vec u and \vec v are given. - Find \proj uv, - the orthogonal projection of \vec u onto \vec v, - and sketch all three vectors with the same initial point. -

    -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<1,2>"); - $v=Vector("<-1,3>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); - - - -

    - \vec u = \la 1,2\ra and \vec v = \la -1,3\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<5,5>"); - $v=Vector("<1,3>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); - - - -

    - \vec u = \la 5,5\ra and \vec v = \la 1,3\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<-3,2>"); - $v=Vector("<1,1>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); - - - -

    - \vec u = \la -3,2\ra and \vec v = \la 1,1\ra -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<-3,2>"); - $v=Vector("<2,3>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); - - - -

    - \vec u = \la -3,2\ra and \vec v = \la 2,3\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<1,5,1>"); - $v=Vector("<1,2,3>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den,$nums[2]/$den>"); - - - -

    - \vec u = \la 1,5,1\ra and \vec v = \la 1,2,3\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<3,-1,2>"); - $v=Vector("<2,2,1>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den,$nums[2]/$den>"); - - - -

    - \vec u = \la 3,-1,2\ra and \vec v = \la 2,2,1\ra. -

    - -

    - -

    -     -     -     -     - - - -

    - Vectors \vec u and \vec v are given. - Write \vec u as the sum of two vectors, - one of which is parallel to \vec v - (or is zero) - and one of which is orthogonal to \vec v. - Note: these are the same pairs of vectors as found in Exercises. -

    -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<1,2>"); - $v=Vector("<-1,3>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); - @uc=$u->value; - @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1); - $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den>"); - $multians = MultiAnswer($projuv,$comp)->with( - singleResult => 0, - checker => sub { - my ($correct,$student,$self) = @_; - my ($s1, $s2) = @{$student}; - my ($c1, $c2) = @{$correct}; - if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) - {return [1,1];} - elsif ($s1+$s2 != $u) - {$self->setMessage(1,"These vectors do not sum to u."); - $self->setMessage(2,"These vectors do not sum to u."); - return [0,0]; - } - elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is parallel to v."); - $self->setMessage(2,"But this vector is not orthogonal to v."); - return [1,0]; - } - elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is not orthogonal to v."); - $self->setMessage(2,"This vector is parallel to v."); - return [0,1]; - } - elsif ($s1 . $v == 0) - {$self->setMessage(1,"This vector is orthogonal to v."); - $self->setMessage(2,"But this vector is not parallel to v."); - return [1,0]; - } - elsif ($s2 . $v == 0) - {$self->setMessage(1,"This vector is not parallel to v."); - $self->setMessage(2,"This vector is orthogonal to v."); - return [0,1]; - } - else { - return [0,0]; - }; - } - ); - - - -

    - \vec u = \la 1,2\raand \vec v = \la -1,3\ra. -

    - -

    - \vec u={}+{} -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<5,5>"); - $v=Vector("<1,3>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); - @uc=$u->value; - @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1); - $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den>"); - $multians = MultiAnswer($projuv,$comp)->with( - singleResult => 0, - checker => sub { - my ($correct,$student,$self) = @_; - my ($s1, $s2) = @{$student}; - my ($c1, $c2) = @{$correct}; - if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) - {return [1,1];} - elsif ($s1+$s2 != $u) - {$self->setMessage(1,"These vectors do not sum to u."); - $self->setMessage(2,"These vectors do not sum to u."); - return [0,0]; - } - elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is parallel to v."); - $self->setMessage(2,"But this vector is not orthogonal to v."); - return [1,0]; - } - elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is not orthogonal to v."); - $self->setMessage(2,"This vector is parallel to v."); - return [0,1]; - } - elsif ($s1 . $v == 0) - {$self->setMessage(1,"This vector is orthogonal to v."); - $self->setMessage(2,"But this vector is not parallel to v."); - return [1,0]; - } - elsif ($s2 . $v == 0) - {$self->setMessage(1,"This vector is not parallel to v."); - $self->setMessage(2,"This vector is orthogonal to v."); - return [0,1]; - } - else { - return [0,0]; - }; - } - ); - - - -

    - \vec u = \la 5,5\ra and \vec v = \la 1,3\ra. -

    - -

    - \vec u={}+{} -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<-3,2>"); - $v=Vector("<1,1>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); - @uc=$u->value; - @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1); - $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den>"); - $multians = MultiAnswer($projuv,$comp)->with( - singleResult => 0, - checker => sub { - my ($correct,$student,$self) = @_; - my ($s1, $s2) = @{$student}; - my ($c1, $c2) = @{$correct}; - if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) - {return [1,1];} - elsif ($s1+$s2 != $u) - {$self->setMessage(1,"These vectors do not sum to u."); - $self->setMessage(2,"These vectors do not sum to u."); - return [0,0]; - } - elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is parallel to v."); - $self->setMessage(2,"But this vector is not orthogonal to v."); - return [1,0]; - } - elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is not orthogonal to v."); - $self->setMessage(2,"This vector is parallel to v."); - return [0,1]; - } - elsif ($s1 . $v == 0) - {$self->setMessage(1,"This vector is orthogonal to v."); - $self->setMessage(2,"But this vector is not parallel to v."); - return [1,0]; - } - elsif ($s2 . $v == 0) - {$self->setMessage(1,"This vector is not parallel to v."); - $self->setMessage(2,"This vector is orthogonal to v."); - return [0,1]; - } - else { - return [0,0]; - }; - } - ); - - - -

    - \vec u = \la -3,2\ra and \vec v = \la 1,1\ra. -

    - -

    - \vec u={}+{} -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<-3,2>"); - $v=Vector("<2,3>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den>"); - @uc=$u->value; - @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1); - $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den>"); - $multians = MultiAnswer($projuv,$comp)->with( - singleResult => 0, - checker => sub { - my ($correct,$student,$self) = @_; - my ($s1, $s2) = @{$student}; - my ($c1, $c2) = @{$correct}; - if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) - {return [1,1];} - elsif ($s1+$s2 != $u) - {$self->setMessage(1,"These vectors do not sum to u."); - $self->setMessage(2,"These vectors do not sum to u."); - return [0,0]; - } - elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is parallel to v."); - $self->setMessage(2,"But this vector is not orthogonal to v."); - return [1,0]; - } - elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is not orthogonal to v."); - $self->setMessage(2,"This vector is parallel to v."); - return [0,1]; - } - elsif ($s1 . $v == 0) - {$self->setMessage(1,"This vector is orthogonal to v."); - $self->setMessage(2,"But this vector is not parallel to v."); - return [1,0]; - } - elsif ($s2 . $v == 0) - {$self->setMessage(1,"This vector is not parallel to v."); - $self->setMessage(2,"This vector is orthogonal to v."); - return [0,1]; - } - else { - return [0,0]; - }; - } - ); - - - -

    - \vec u = \la -3,2\ra and \vec v = \la 2,3\ra. -

    - -

    - \vec u={}+{} -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<1,5,1>"); - $v=Vector("<1,2,3>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den,$nums[2]/$den>"); - @uc=$u->value; - @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1,2); - $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den,$compnums[2]/$den>"); - $multians = MultiAnswer($projuv,$comp)->with( - singleResult => 0, - checker => sub { - my ($correct,$student,$self) = @_; - my ($s1, $s2) = @{$student}; - my ($c1, $c2) = @{$correct}; - if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) - {return [1,1];} - elsif ($s1+$s2 != $u) - {$self->setMessage(1,"These vectors do not sum to u."); - $self->setMessage(2,"These vectors do not sum to u."); - return [0,0]; - } - elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is parallel to v."); - $self->setMessage(2,"But this vector is not orthogonal to v."); - return [1,0]; - } - elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is not orthogonal to v."); - $self->setMessage(2,"This vector is parallel to v."); - return [0,1]; - } - elsif ($s1 . $v == 0) - {$self->setMessage(1,"This vector is orthogonal to v."); - $self->setMessage(2,"But this vector is not parallel to v."); - return [1,0]; - } - elsif ($s2 . $v == 0) - {$self->setMessage(1,"This vector is not parallel to v."); - $self->setMessage(2,"This vector is orthogonal to v."); - return [0,1]; - } - else { - return [0,0]; - }; - } - ); - - - -

    - \vec u = \la 1,5,1\ra and \vec v = \la 1,2,3\ra. -

    - -

    - \vec u={}+{} -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $u=Vector("<3,-1,2>"); - $v=Vector("<2,2,1>"); - $num=$u . $v; - $den=$v . $v; - @vc=$v->value; - @nums=map{$_*$num}(@vc); - $projuv=Compute("<$nums[0]/$den,$nums[1]/$den,$nums[2]/$den>"); - @uc=$u->value; - @compnums=map{$uc[$_]*$den-$nums[$_]}(0,1,2); - $comp=Compute("<$compnums[0]/$den,$compnums[1]/$den,$compnums[2]/$den>"); - $multians = MultiAnswer($projuv,$comp)->with( - singleResult => 0, - checker => sub { - my ($correct,$student,$self) = @_; - my ($s1, $s2) = @{$student}; - my ($c1, $c2) = @{$correct}; - if ((($s1==$c1) and ($s2==$c2)) or (($s1==$c2) and ($s2==$c1))) - {return [1,1];} - elsif ($s1+$s2 != $u) - {$self->setMessage(1,"These vectors do not sum to u."); - $self->setMessage(2,"These vectors do not sum to u."); - return [0,0]; - } - elsif (abs($s1 . $v) == sqrt($s1 . $s1)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is parallel to v."); - $self->setMessage(2,"But this vector is not orthogonal to v."); - return [1,0]; - } - elsif (abs($s2 . $v) == sqrt($s2 . $s2)*sqrt($v . $v)) - {$self->setMessage(1,"This vector is not orthogonal to v."); - $self->setMessage(2,"This vector is parallel to v."); - return [0,1]; - } - elsif ($s1 . $v == 0) - {$self->setMessage(1,"This vector is orthogonal to v."); - $self->setMessage(2,"But this vector is not parallel to v."); - return [1,0]; - } - elsif ($s2 . $v == 0) - {$self->setMessage(1,"This vector is not parallel to v."); - $self->setMessage(2,"This vector is orthogonal to v."); - return [0,1]; - } - else { - return [0,0]; - }; - } - ); - - - -

    - \vec u = \la 3,-1,2\ra and \vec v = \la 2,2,1\ra. -

    - -

    - \vec u={}+{} -

    -     -     -     - -     - - - - -

    - A 10lb box sits on a ramp that rises 4ft over a distance of 20ft. - How much force is required to keep the box from sliding down the ramp? -

    -     - -

    - 1.96lb -

    -     - -     - - - - -

    - A 10lb box sits on a 15ft ramp that makes a - 30^\circ angle with the horizontal. - How much force is required to keep the box from sliding down the ramp? -

    -     - -

    - 5lb -

    -     - -     - - - - -

    - How much work is performed in moving a box horizontally 10ft with a force of 20lb applied at an angle of 45^\circ to the horizontal? -

    -     - -

    - 141.42ftlb -

    -     - -     - - - - -

    - How much work is performed in moving a box horizontally 10ft with a force of 20lb applied at an angle of 10^\circ to the horizontal? -

    -     - -

    - 196.96ftlb -

    -     - -     - - - - -

    - How much work is performed in moving a box up the length of a ramp that rises 2ft over a distance of 10ft, - with a force of 50lb applied horizontally? -

    -     - -

    - 500ftlb -

    -     - -     - - - - -

    - How much work is performed in moving a box up the length of a ramp that rises 2ft over a distance of 10ft, - with a force of 50lb applied at an angle of 45^\circ to the horizontal? -

    -     - -

    - 424.26ftlb -

    -     - -     - - - - -

    - How much work is performed in moving a box up the length of a 10ft ramp that makes a 5^\circ angle with the horizontal, - with 50lb of force applied in the direction of the ramp? -

    -     - -

    - 500ftlb -

    -     - -     -     -     -
    -
    - The Cross Product - -

    - Orthogonality is immensely important. - A quick scan of your current environment will undoubtedly reveal numerous surfaces and edges that are perpendicular to each other - (including the edges of this page). - The dot product provides a quick test for orthogonality: - vectors \vec u and \vec v are perpendicular if, - and only if, \dotp uv=0. -

    - - - -

    - Given two non-parallel, - nonzero vectors \vec u and \vec v in space, - it is very useful to find a vector \vec w that is perpendicular to both \vec u and \vec v. - There is an operation, called the cross product, - that creates such a vector. - This section defines the cross product, - then explores its properties and applications. -

    - - - Cross Product - -

    - Let \vec u =\la u_1,u_2,u_3\ra and - \vec v = \la v_1,v_2,v_3\ra be vectors in \mathbb{R}^3. - The cross product of \vec u and \vec v, - denoted \crossp uv, - is the vector vectorscross productcross productdefinition - - \crossp uv = \la u_2v_3-u_3v_2,-(u_1v_3-u_3v_1),u_1v_2-u_2v_1\ra - . -

    -     -     - -

    - This definition can be a bit cumbersome to remember. - After an example we will give a convenient method for computing the cross product. - For now, careful examination of the products and differences given in the definition should reveal a pattern that is not too difficult to remember. - (For instance, - in the first component only 2 and 3 appear as subscripts; - in the second component, only 1 and 3 appear as subscripts. - Further study reveals the order in which they appear.) -

    - -

    - Let's practice using this definition by computing a cross product. -

    - - - Computing a cross product - -

    - Let \vec u = \la 2,-1,4\ra and \vec v = \la 3,2,5\ra. - Find \crossp uv, - and verify that it is orthogonal to both \vec u and \vec v. -

    -     - -

    - Using , we have - - \crossp uv = \la (-1)5-(4)2,-\big((2)5-(4)3\big), (2)2-(-1)3\ra = \la -13,2,7\ra - . -

    - -

    - (We encourage the reader to compute this product on their own, - then verify their result.) -

    - -

    - We test whether or not \crossp uv is orthogonal to \vec u and \vec v using the dot product: - - \big(\crossp uv\big) \cdot \vec u = \la -13,2,7\ra \cdot \la 2,-1,4\ra = 0 - , - - \big(\crossp uv\big) \cdot \vec v = \la -13,2,7\ra \cdot \la 3,2,5 \ra = 0 - . -

    - -

    - Since both dot products are zero, - \crossp uv is indeed orthogonal to both \vec u and \vec v. -

    -     - -     - -

    - A convenient method of computing the cross product starts with forming a particular 3\times 3 - matrix, or rectangular array. - The first row comprises the standard unit vectors \veci, - \vecj, and \veck. - The second and third rows are the vectors \vec u and \vec v, - respectively. - Using \vec u and \vec v from , - we begin with: - - \begin{matrix} \veci\amp \vecj\amp \veck \\ 2\amp -1\amp 4\\3\amp 2\amp 5 - \end{matrix} - - -

    - -

    - Now repeat the first two columns after the original three: - - \begin{matrix} \veci\amp \vecj\amp \veck\amp \veci\amp \vecj \\ 2\amp -1\amp 4\amp 2\amp -1\\3\amp 2\amp 5\amp 3\amp 2 - \end{matrix} - - This gives three full upper left to lower right diagonals, - and three full upper right to lower left diagonals, - as shown. - Compute the products along each diagonal, - then add the products on the right and subtract the products on the left: -

    - - - Schematic diagram for computing the cross product by multiplying along diagonal arrows. - -

    - The image illustrates how to use the array given above, with the two repeated columns, - to compute the cross product. - From the top row, diagonal arrows are drawn down and to the right, - starting in the first three columns from \veci, \vecj, and vec k. - Diagonal arrows are also drawn down and to the left, - starting in the last three columns from \veck, \veci, and \vecj. - (Recall that the last two columns are copies of the first two.) -

    - -

    - The cross product is computed by multiplying along each arrow. - The quantities obtained from the right-pointing arrows are added, - while the quantities obtained from the left-pointing arrows are subtracted. - The result of the product along each arrow is indicated at the tip of that arrow. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.32] - - \node at (0,0) {\(\begin{array}{ccccc} \ \veci\ \amp \ \vecj\ \amp \ \veck\ \amp \ \veci\ \amp \ \vecj\ \\ 2\amp -1\amp 4\amp 2\amp -1\\3\amp 2\amp 5\amp 3\amp 2 - \end{array} \)}; - - \draw[->, thin] (-1.3,.4) -- (.5,-.8) node[below] {\(-5\veci\)}; - \draw[->, thin] (-.7,.4) -- (1.3,-.8) node[below] {\(12\vecj\)}; - \draw[->, thin] (0,.4) -- (2,-.8) node[below] {\(4\veck\)}; - \draw[->, thin] (0,.4) -- (-2,-.8) node[below] {\(-3\veck\)}; - \draw[->, thin] (.7,.4) -- (-1.3,-.8) node[below] {\(8\veci\)}; - \draw[->, thin] (1.3,.4) -- (-.5,-.8) node[below] {\(10\vecj\)}; - - \end{tikzpicture} - - - - -

    - - \crossp uv = \big(-5\veci+12\vecj+4\veck\,\big) - \big(-3\veck+8\veci+10\vecj\,\big) = -13\veci+2\vecj+7\veck = \la -13,2,7\ra - . -

    - - - -

    - We practice using this method. -

    - - - Computing a cross product - -

    - Let \vecu=\la 1,3,6\ra and \vec v = \la -1,2,1\ra. - Compute both \crossp uv and \crossp vu. -

    -     - -

    - To compute \crossp uv, - we form the matrix as prescribed above, - complete with repeated first columns: - - \begin{matrix} \ \veci\ \amp \ \vecj\ \amp \ \veck\ \amp \ \veci\ \amp \ \vecj\ \\ 1\amp 3\amp 6\amp 1\amp 3\\-1\amp 2\amp 1\amp -1\amp 2 \end{matrix} - -

    - -

    - We let the reader compute the products of the diagonals; - we give the result: - - \crossp uv = \big(3\veci-6\vecj+2\veck\,\big) - \big(-3\veck + 12\veci+\vecj\,\big) = \la -9,-7,5\ra - . -

    - -

    - To compute \crossp vu, - we switch the second and third rows of the above matrix, - then multiply along diagonals and subtract: - - \begin{matrix} \ \veci\ \amp \ \vecj\ \amp \ \veck\ \amp \ \veci\ \amp \ \vecj\ \\-1\amp 2\amp 1\amp -1\amp 2\\ 1\amp 3\amp 6\amp 1\amp 3 \end{matrix} - -

    - -

    - Note how with the rows being switched, - the products that once appeared on the right now appear on the left, - and vice-versa. - Thus the result is: - - \crossp vu = \big(12\veci+\vecj-3\veck\,\big) - \big(2\veck + 3\veci-6\vecj\,\big) = \la 9,7,-5\ra - , - which is the opposite of \crossp uv. - We leave it to the reader to verify that each of these vectors is orthogonal to \vec u and \vec v. -

    -     - -     -     - - - Properties of the Cross Product -

    - It is not coincidence that - \crossp vu = -(\crossp uv) in the preceding example; - one can show using - that this will always be the case. - The following theorem states several useful properties of the cross product, - each of which can be verified by referring to the definition. -

    - - - Properties of the Cross Product - -

    - Let \vecu, \vecv and \vecw be vectors in - \mathbb{R}^3 and let c be a scalar. - The following identities hold: - vectorscross product - cross productproperties -

    - -

    -

      -
    1. -

      - \crossp uv = -(\crossp vu) (Anticommutative Property) -

      -
      2. - -
    2. -

      -

        -
      1. -

        - (\vec u+\vec v)\times \vecw = \crossp uw+\crossp vw (Distributive Properties) -

        -
        2. - -
      2. -

        - \vec u \times (\vec v+\vec w) = \crossp uv+\crossp uw -

        -
        3. -
      -

      -
      3. - -
    3. -

      - c(\crossp uv) = (c\vecu) \times \vec v = \vecu \times (c\vecv) -

      -
      4. - -
    4. -

      -

        -
      1. -

        - (\crossp uv)\cdot \vecu = 0 (Orthogonality Properties) -

        -
        2. - -
      2. -

        - (\crossp uv)\cdot \vecv = 0 -

        -
        3. -
      -

      -
      5. - -
    5. -

      - \crossp uu = \vec 0 -

      -
      6. - -
    6. -

      - \crossp u0 = \vec 0 -

      -
      7. - -
    7. -

      - \vecu \cdot (\vecv\times\vecw) = (\crossp uv)\cdot \vecw (Triple Scalar Product) -

      -
      8. -
    -

    -     -     - - - -

    - We introduced the cross product as a way to find a vector orthogonal to two given vectors, - but we did not give a proof that the construction given in satisfies this property. - asserts this property holds; - we leave it as a problem in the Exercise section to verify this. -

    - - - -

    - Property 5 from the theorem is also left to the reader to prove in the Exercise section, - but it reveals something more interesting than - the cross product of a vector with itself is \vec 0. - Let \vec u and \vec v be parallel vectors; - that is, let there be a scalar c such that \vecv = c\vecu. - Consider their cross product: - - \crossp uv \amp = \vecu \times (c\vec u)\amp \amp - \amp = c(\crossp uu) \amp \amp\text{ (by Property 3 of } \text{)} - \amp = \vec 0 \amp \amp\text{ (by Property 5 of } \text{)} - . -

    - -

    - We have just shown that the cross product of parallel vectors is \vec 0. - This hints at something deeper. - - related the angle between two vectors and their dot product; - there is a similar relationship relating the cross product of two vectors and the angle between them, - given by the following theorem. -

    - - - The Cross Product and Angles - -

    - Let \vec u and \vec v be nonzero vectors in \mathbb{R}^3. - Then - - \norm{\crossp uv} = \vnorm u\, \vnorm v \sin(\theta) - , - where \theta, 0\leq \theta \leq \pi, - is the angle between \vecu and \vecv. - vectorscross product - cross productproperties -

    -     -     - - - -

    - Note that this theorem makes a statement about the - magnitude of the cross product. - When the angle between \vecu and \vecv is 0 or \pi (, the vectors are parallel), - the magnitude of the cross product is 0. - The only vector with a magnitude of 0 is \vec 0 - (see Property - of ), - hence the cross product of parallel vectors is \vec 0. -

    - - - - - -

    - We demonstrate the truth of this theorem in the following example. -

    - - - The cross product and angles - -

    - Let \vec u = \la 1,3,6\ra and - \vec v = \la -1,2,1\ra as in . - Verify by finding \theta, - the angle between \vecu and \vecv, - and the magnitude of \crossp uv. -

    -     - -

    - We use - to find the angle between \vecu and \vecv. - - \theta \amp = \cos^{-1}\left(\frac{\dotp uv}{\vnorm u\, \vnorm v}\right) - \amp = \cos^{-1}\left(\frac{11}{\sqrt{46}\sqrt{6}}\right) - \amp \approx 0.8471 = 48.54^\circ - . -

    - -

    - Our work in - showed that \crossp uv = \la -9,-7,5\ra, - hence \norm{\crossp uv} = \sqrt{155}. - Is \norm{\crossp uv} = \vnorm u\, \vnorm v\sin(\theta)? - Using numerical approximations, we find: - - \norm{\crossp uv} \amp =\sqrt{155} \amp \vnorm u\,\vnorm v \sin(\theta) \amp = \sqrt{46}\sqrt{6}\sin(0.8471) - \amp \approx 12.45. \amp \amp \approx 12.45 - . -

    - -

    - Numerically, they seem equal. - Using a right triangle, one can show that - - \sin\left(\cos^{-1}\left(\frac{11}{\sqrt{46}\sqrt{6}}\right)\right) = \frac{\sqrt{155}}{\sqrt{46}\sqrt{6}} - , - which allows us to verify the theorem exactly. -

    -     -     - - - Right Hand Rule -

    - The anticommutative property of the cross product demonstrates that \crossp uv and - \crossp vu differ only by a sign these vectors have the same magnitude but point in the opposite direction. - When seeking a vector perpendicular to \vec u and \vec v, - we essentially have two directions to choose from, - one in the direction of \crossp uv and one in the direction of \crossp vu. - Does it matter which we choose? - How can we tell which one we will get without graphing, etc.? -

    - -

    - Another wonderful property of the cross product, as defined, - is that it follows the right hand rule. - Given \vec u and \vec v in - \mathbb{R}^3 with the same initial point, - point the index finger of your right hand in the direction of \vecu and let your middle finger point in the direction of \vecv - (much as we did when establishing the right hand rule for the 3-dimensional coordinate system). - Your thumb will naturally extend in the direction of \crossp uv. - One can practice this using . - If you switch, - and point the index finder in the direction of \vecv and the middle finger in the direction of \vecu, - your thumb will now point in the opposite direction, - allowing you to visualize - the anticommutative property of the cross product. - right hand ruleof the cross product -

    - -
    - Illustrating the Right Hand Rule of the cross product - - - - Three-dimensional image illustrating the right-hand rule for the direction of the cross product. - -

    - On a three-dimensional coordinate system, with x, y, and z axes, - vectors \vec u and \vec v are plotted with their tails at the origin. - The plane passing through the origin that contains these two vectors is also plotted, - along with a normal vector to the plane. -

    - -

    - The normal vector is upward-pointing, and is equal to the cross product \vec{u}\times\vec{v}. - When the plane is viewed from the side on which the normal vector is placed, - the position of the vector \vec v corresponds to a counter-clockwise rotation from the position of \vec u. -

    -     - - - - - //ASY file for figcrossp_rhr.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-0.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw theplane z=(-2*x+5*y)/27; - triple f(pair t) { - return (t.x,t.y,(-2*t.x+5*t.y)/27); - } - surface s=surface(f,(-1.5,-1.5),(1.5,1.5),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - // Draw vec{u}=<-1/5,1,1/5> - draw((0,0,0)--(-1/5,1,1/5), bluepen,Arrow3(size=3mm)); - label("$\vec{u}$",(-1/5,1,1/5),N); - - // Draw \vec{v}=<-6/5,3/5,1/5> - draw((0,0,0)--(-6/5,3/5,1/5), bluepen,Arrow3(size=3mm)); - label("$\vec{v}$",(-6/5,3/5,1/5),N); - - // Draw the cross product of u and v <2,-5,27> but scale it as <2,-5,27>/25 - draw((0,0,0)--(2/25,-5/25,27/25), redpen,Arrow3(size=3mm)); - label("$\vec{u} \times \vec{v}$",(2/25,-5/25,27/25),NW); - - - - -
    -     -     - - - Applications of the Cross Product -

    - There are a number of ways in which the cross product is useful in mathematics, - physics and other areas of science beyond just - finding a vector perpendicular to two others. - We highlight a few here. - cross productapplications -

    - - - Area of a Parallelogram -

    - It is a standard geometry fact that the area of a parallelogram is A = bh, - where b is the length of the base and h is the height of the parallelogram, - as illustrated in . - As shown when defining the Parallelogram Law of vector addition, - two vectors \vecu and \vecv define a parallelogram when drawn from the same initial point, - as illustrated in . - Trigonometry tells us that h = \vnorm u \sin(\theta), - hence the area of the parallelogram is - - A = \vnorm u\,\vnorm v\sin(\theta) = \norm{\crossp uv} - , - where the second equality comes from . -

    - -
    - Using the cross product to find the area of a parallelogram - -
    - - - - A parallelogram with base b and height h labeled. - -

    - A parallelogram with a horizontal base is shown. - The base is labeled with the quantity b. - A dashed line is drawn from the top-left vertex of the parallelogram to the base; - this line is perpendicular to the base, - and its length is equal to the height of the parallelogram, - which is labeled h. -

    -     - - - \begin{tikzpicture}[scale=1.24] - - \draw (0,0) -- node [below,pos=.5] { $b$} (2,0) -- (3,1.5) -- (1,1.5) -- (0,0); - \draw [dashed] (1,1.5) -- (1,0) node [pos=.5,right] {$h$}; - \draw (.8,0) -- (.8,.2) -- (1,.2); - - \end{tikzpicture} - - - - -
    - -
    - - - - Another parallelogram, with two adjacent sides labeled as vectors. - -

    - The parallelogram is the same one as , - but with different labeling. - This time, the two sides adjacent to the bottom-right vertex are labeled with vectors. - The bottom of the parallelogram is labeled with a vector \vec v, - and the left side is labeled with a vector \vec u. - The angle between these vectors is labeled \theta. - The same altitude of the parellelogram is drawn as a dashed line with height h. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.24] - - \draw [->](0,0) -- node [below,pos=1] { $\vec v$} (2,0); - \draw (2,0) -- (3,1.5) -- (1,1.5); - \draw (.3,.175) node { $\theta$}; - \draw [->](0,0) -- (1,1.5) node [above] { $\vec u$}; - \draw [dashed] (1,1.5) -- (1,0) node [pos=.5,right] {$h$}; - \draw (.8,0) -- (.8,.2) -- (1,.2); - - \end{tikzpicture} - - - - -
    -     -
    - -

    - We illustrate using Equation in the following example. - cross productapplications!area of parallelogram -

    - - - Finding the area of a parallelogram - -

    -

      -
    1. -

      - Find the area of the parallelogram defined by the vectors - \vecu = \la 2,1\ra and \vecv = \la 1,3\ra. -

      -
      2. - -
    2. -

      - Verify that the points A = (1,1,1), B = (2,3,2), - C = (4,5,3) and D = (3,3,2) are the vertices of a parallelogram. - Find the area of the parallelogram. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - sketches the parallelogram defined by the vectors \vec u and \vec v. - We have a slight problem in that our vectors exist in - \mathbb{R}^2, not \mathbb{R}^3, - and the cross product is only defined on vectors in \mathbb{R}^3. - We skirt this issue by viewing \vec u and \vecv as vectors in the x-y plane of \mathbb{R}^3, - and rewrite them as \vec u = \la 2,1,0\ra and \vecv =\la 1,3,0\ra. - We can now compute the cross product. - It is easy to show that \crossp uv = \la 0,0,5\ra; therefore the area of the parallelogram is A = \norm{\crossp uv} = 5. -

      - -
      - Sketching the parallelograms in - -
      - - - - A parallelogram in the plane, spanned by vectors u and v. - -

      - The first quadrant in \R^2 is shown. - Two vectors a drawn with their tails at the origin. - The vector \vec u ends at the point (2,1), - while the vector \vec v ends at the point (1,3). -

      - -

      - These vectors make up two of the four sides of the parallelogram. - The side from (1,3) to (3,4) is parallel to \vec u, - and the side from (2,1) to (3,4) is parallel to \vec v. -

      -       - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4,5}, - ymin=-.5,ymax=5.5, - xmin=-.5,xmax=4.5 - ] - - \draw [thick,->] (axis cs:0,0) -- (axis cs:2,1) node [right,] { $\vec u$}; - \draw [thick,->] (axis cs:0,0) -- (axis cs:1,3) node [above] { $\vec v$}; - - \draw [thick] (axis cs:2,1) -- (axis cs:3,4) -- (axis cs: 1,3); - - \end{axis} - - \end{tikzpicture} - - - - -
      - -
      - - - - - A three-dimensional image of a parallelogram in space, with vertices A, B, C, and D. - -

      - A three-dimensional coordinate system is shown, with x, y, and z axes. - A parallelogram is shown in space relative to these axes, - with vertices labeled A, B, C, and D. -

      -       - - - - - //ASY file for figcrossp4a.asy in Chapter 10 - // CAN'T FILL IT IN WITH ''fill'' (only in 2d) - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - //currentprojection=orthographic(4,4,2); - currentprojection=orthographic(7.5,14,4); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-0.5,4.5); - pair ybounds=(-0.5,4.5); - pair zbounds=(-0.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the edges of the parallelogram with corners at (1,1,1),(2,3,2),(4,5,3),(3,3,2) - draw((1,1,1)--(2,3,2)--(4,5,3)--(3,3,2)--cycle); - label("A",(1,1,1),S); - label("B",(2,3,2),N); - label("C",(4,5,3),N); - label("D",(3,3,2),W); - - // Fill in the parallelogram - //I changed the viewpoint to (1,4,1), then added://Hartman comment - triple f(pair t) {return (1+t.x+2t.y,1+2t.x+2t.y,1+t.x+t.y);} - surface s=surface(f,(0,0),(1,1),1,1); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -
      -       -
      -
      2. - -
    2. -

      - To show that the quadrilateral ABCD is a parallelogram (shown in ), - we need to show that the opposite sides are parallel. - We can quickly show that \overrightarrow{AB} =\overrightarrow{DC} = \la 1,2,1\ra and \overrightarrow{BC} = \overrightarrow{AD} = \la 2,2,1\ra. - We find the area by computing the magnitude of the cross product of \overrightarrow{AB} and \overrightarrow{BC}: - - \overrightarrow{AB} \times \overrightarrow{BC} = \la 0,1,-2\ra \Rightarrow \norm{\overrightarrow{AB}\times\overrightarrow{BC}} = \sqrt{5} \approx 2.236 - . -

      -
      3. -
    -

    -     - -     - -

    - This application is perhaps more useful in finding the area of a triangle - (in short, triangles are used more often than parallelograms). - We illustrate this in the following example. -

    - - - Area of a triangle - -

    - Find the area of the triangle with vertices A=(1,2), - B=(2,3) and C=(3,1), - as pictured in . -

    -
    - Finding the area of a triangle in - - - A triangle in the plane, with vertices A(1,2), B(2,3), and C(3,1). - -

    - The triangle for this problem is sketched in the first quadrant of the plane. - The vertices are labeled, with A at (1,2), - B at (2,3), and C at (3,1). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3}, - ymin=-.1,ymax=3.5, - xmin=-.1,xmax=3.9 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(1,2) (2,3) (3,1) (1,2)}; - \addplot [firstcurvestyle,-] coordinates {(1,2) (2,3) (3,1) (1,2)}; - - \draw (axis cs:1,2) node [left] { $A$} (axis cs:2,3) node [above] { $B$} (axis cs:3,1) node [below] { $C$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - We found the area of this triangle in - to be 1.5 using integration. - There we discussed the fact that finding the area of a triangle can be inconvenient using the - \frac12bh - formula as one has to compute the height, - which generally involves finding angles, etc. - Using a cross product is much more direct. -

    - -

    - We can choose any two sides of the triangle to use to form vectors; - we choose \overrightarrow{AB} = \la 1,1\ra and \overrightarrow{AC}=\la 2,-1\ra. - As in the previous example, - we will rewrite these vectors with a third component of 0 so that we can apply the cross product. - The area of the triangle is - - \frac12\norm{\overrightarrow{AB}\times\overrightarrow{AC}} = \frac12\norm{\la 1,1,0\ra \times \la 2,-1,0\ra} = \frac12\norm{\la 0,0,-3\ra} = \frac32 - . -

    - -

    - We arrive at the same answer as before with less work. -

    -     - -     - - -     - - - Volume of a Parallelepiped -

    - The three dimensional analogue to the parallelogram is the - parallelepiped. - Each face is parallel to the opposite face, - as illustrated in . - By crossing \vec v and \vec w, - one gets a vector whose magnitude is the area of the base. - Dotting this vector with \vecu computes the volume - of the parallelepiped! (Up to a sign; - take the absolute value.) -

    - -
    - A parallelepiped is the three dimensional analogue to the parallelogram - - - - A parellelepiped in three dimensions. It is the box-like object spanned by three vectors. - -

    - Three vectors \vec u, \vec v, and \vec w are plotted in three dimensions. - These vectors make up three edges of a three-dimensional solid known as a parallelepiped. - Each side of the parallelepiped is a parallelogram, and opposite sides are equal parallelograms lying in parallel planes. - One pair of opposite sides comes from the parallelogram spanned by \vec u and \vec v, - another from the parallelogram spanned by \vec v and \vec w, - and the last pair from the parallelogram spanned by \vec u and \vec w. -

    -     - - - - - //ASY file for figcrosspparallelpiped.asy in Chapter 10 - // NOT SHADED!! - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - //real[] myxchoice={}; - //real[] myychoice={}; - //real[] myzchoice={}; - defaultpen(0.5mm); - //xaxis3("$x$",-0.5,1,black,OutTicks(myxchoice),Arrow3(size=3mm)); - //yaxis3("$y$",-1,1,black,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("$z$",-0.5,1.5,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //Draw the parallelepiped with u=<1,1,0>, v=<-1,1,0>, w=<0,1,1> - draw((1,1,0)--(0,0,0), Arrow3(size=3mm));// u - label("$\vec{u}$",(0,0,0),W); - draw((1,1,0)--(0,2,0), Arrow3(size=3mm));// v - label("$\vec{v}$",(0,2,0),S); - draw((1,1,0)--(1,2,1), Arrow3(size=3mm));// w - label("$\vec{w}$",(1,2,1),W); - //shifted u to get the other edges of the box - draw((-1,1,0)--(0,2,0),bluepen);// u shifted to v - draw((0,1,1)--(1,2,1),bluepen);// u shifted to w - draw((-1,2,1)--(0,3,1),bluepen);// u shifted to v+w - //shifted v to get the other edges of the box - draw((0,0,0)--(-1,1,0),bluepen);// v shifted to u - draw((0,1,1)--(-1,2,1),bluepen);// v shifted to w - draw((1,2,1)--(0,3,1),bluepen);// v shifted to u+w - //shifted w to get the other edges of the box - draw((0,0,0)--(0,1,1),bluepen);// w shifted to u - draw((-1,1,0)--(-1,2,1),bluepen);// w shifted to v - draw((0,2,0)--(0,3,1),bluepen);// w shifted to u+v - - // MIGHT BE ABLE TO USE THIS TO FILL IN THE 6 SIDES - // Fill in the parallelogram - //I changed the viewpoint to (1,4,1), then added://Hartman comment - //triple f(pair t) {return (1+t.x+2t.y,1+2t.x+2t.y,1+t.x+t.y);} - //surface s=surface(f,(0,0),(1,1),1,1); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - //OR THIS FROM THE WEB???? - //http://tex.stackexchange.com/questions/137153/fill-a-region-between-two-coplanar-paths-in-asymptote - //settings.outformat="pdf"; - //settings.render=0; - //import three; - //size(5cm); - //path3 p = (0,-2,-2)-- (0,2,-2) -- (0,2,2) -- (0,-2,2) -- (0,-2,-2) -- (0,-2,-2); - //path3 q = (0,-.25,-1.1) -- (0,.25,-1.1) -- (0,.25,1.1) -- (0,-.25,1.1) -- (0,-.25,-1.1); - //draw(surface(p -- reverse(q) -- cycle), emissive(yellow)); - //draw(p ^^ q, black); - - //my try - import three; - path3 p = (0,0,0)-- (1,1,0) -- (0,2,0) -- (-1,1,0); //Left - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0,0,0)-- (1,1,0) -- (1,2,1) -- (0,1,1); //bottom - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (-1,1,0)-- (0,2,0) -- (0,3,1) -- (-1,2,1); //right - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0,0,0)-- (-1,1,0) -- (-1,2,1) -- (0,1,1); //back - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (1,1,0)-- (0,2,0) -- (0,3,1) -- (1,2,1); //front - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0,1,1)-- (1,2,1) -- (0,3,1) -- (-1,2,1); //top - draw(surface(p -- cycle), simplesurfacepen); - - - - -
    - -

    - cross productapplications!volume of parallelepiped -

    - -

    - Thus the volume of a parallelepiped defined by vectors \vecu, - \vecv and \vec w is - - V = \abs{\vecu\cdot (\crossp vw)} - . -

    - -

    - Note how this is the Triple Scalar Product, - first seen in . - Applying the identities given in the theorem shows that we can apply the Triple Scalar Product in any order - we choose to find the volume. - That is, - - V = \abs{\vecu\cdot(\crossp vw)} = \abs{\vec u\cdot (\crossp wv)} = \abs{(\crossp uv)\cdot \vecw}, \text{ etc. } - -

    - - - Finding the volume of parallelepiped - -

    - Find the volume of the parallelepiped defined by the vectors \vecu = \la 1,1,0\ra, - \vecv = \la -1,1,0\ra and \vecw = \la 0,1,1\ra. -

    -     - -

    - We apply Equation. - We first find \crossp vw =\la 1,1,-1\ra. - Then - - \abs{\vec u\cdot(\crossp vw)} = \abs{\la 1,1,0\ra \cdot \la1,1,-1\ra} = 2 - . -

    - -

    - So the volume of the parallelepiped is 2 cubic units. -

    - -
    - A parallelepiped in - - - - The parallelepiped whose volume is computed in this example. - -

    - A parallelpiped is plotted against a three-dimensional coordinate system, - with x, y, and z axes. - Vectors \vec{u}, \vec{v}, \vec{w} are plotted with their tails at the origin, - although the vector \vec{v} is not visible in the default view of the parallelepiped. - (It can be seen once the image is rotated.) -

    - -

    - These vectors make up the three edges of the parallelepiped that are adjacent to the origin. -

    -     - - - - - //ASY file for figcrossp6.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={2}; - real[] myzchoice={0.5,1}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(0,3); - pair zbounds=(-0.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the parallelepiped with u=<1,1,0>, v=<-1,1,0>, w=<0,1,1> - draw((0,0,0)--(1,1,0), Arrow3(size=3mm));// u - label("$\vec{u}$",(1,1,0),W); - draw((0,0,0)--(-1,1,0), Arrow3(size=3mm));// v - label("$\vec{v}$",(-1,1,0),N); - draw((0,0,0)--(0,1,1), Arrow3(size=3mm));// w - label("$\vec{w}$",(0,1,1),W); - //shifted u to get the other edges of the box - draw((-1,1,0)--(0,2,0),bluepen);// u shifted to v - draw((0,1,1)--(1,2,1),bluepen);// u shifted to w - draw((-1,2,1)--(0,3,1),bluepen);// u shifted to v+w - //shifted v to get the other edges of the box - draw((1,1,0)--(0,2,0),bluepen);// v shifted to u - draw((0,1,1)--(-1,2,1),bluepen);// v shifted to w - draw((1,2,1)--(0,3,1),bluepen);// v shifted to u+w - //shifted w to get the other edges of the box - draw((1,1,0)--(1,2,1),bluepen);// w shifted to u - draw((-1,1,0)--(-1,2,1),bluepen);// w shifted to v - draw((0,2,0)--(0,3,1),bluepen);// w shifted to u+v - - //my try at shading. - import three; - path3 p = (0,0,0)-- (1,1,0) -- (0,2,0) -- (-1,1,0); //Left - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0,0,0)-- (1,1,0) -- (1,2,1) -- (0,1,1); //bottom - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (-1,1,0)-- (0,2,0) -- (0,3,1) -- (-1,2,1); //right - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0,0,0)-- (-1,1,0) -- (-1,2,1) -- (0,1,1); //back - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (1,1,0)-- (0,2,0) -- (0,3,1) -- (1,2,1); //front - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0,1,1)-- (1,2,1) -- (0,3,1) -- (-1,2,1); //top - draw(surface(p -- cycle), simplesurfacepen); - - - - -
    -     - -     -     - -

    - While this application of the Triple Scalar Product is interesting, - it is not used all that often: - parallelepipeds are not a common shape in physics and engineering. - The last application of the cross product is very applicable in engineering. -

    - - - Torque -

    - Torque is a measure of the turning force applied to an object. - A classic scenario involving torque is the application of a wrench to a bolt. - When a force is applied to the wrench, the bolt turns. - When we represent the force and wrench with vectors \vec F and \vec \ell, - we see that the bolt moves - (because of the threads) - in a direction orthogonal to \vec F and \vec \ell. - Torque is usually represented by the Greek letter \tau, or tau, - and has units of N\cdotm, - a Newtonmeter, or ft\cdotlb, a footpound. - cross productapplications!torque - torque -

    - -

    - While a full understanding of torque is beyond the purposes of this book, - when a force \vec F is applied to a lever arm - \vec \ell, the resulting torque is - - \vec \tau = \crossp \ell F - . -

    - - - Computing torque - -

    - A lever of length 2ft makes an angle with the horizontal of 45^\circ. - Find the resulting torque when a force of 10lb is applied to the end of the level where: -

    - -
    - Showing a force being applied to a lever in - - - Two adjacent images, each showing a pair of vectors, one of which represents a lever, and the other, a force. - -

    - Two images are shown side-by-side. - Both images show a pair of vectors, one of which is labeled \vec l (the lever), - and the other, which is labeled \vec F (the force). -

    - -

    - In each image, the vector \vec l has its tail at a pivot point. - A circular arrow is drawn around this point to show the direction of rotation of the lever about the pivot. - The direction is clockwise in each image. -

    - -

    - The force vector \vec F is drawn with its tip at the tip of the vector \vec l. - In both images, the vector \vec l points up and to the right, - and the vector \vec F points down and to the right. - The difference is that in the first image, the vectors meet at a 90^\circ angle, - while in the second image, the angle is 60^\circ. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.1] - - \draw [thick,->,rotate=45] (-2,0) -- (0,0) node [below,pos=.5] { $\vec\ell$}; - \draw [rotate=45,->] (-1.8,0) arc (0:-270:.2); - \draw [rotate=45] (-.4,0) arc (180:90:.4); - \draw [rotate=0] (-.6,0) node { $90^\circ$}; - - \filldraw [black,rotate=45] (-2,0) circle (2.4pt); - - \begin{scope}[shift={(-.05,.05)}] - \draw [thick,->,rotate=-45] (-1.5,0)node [left] { $\vec F$} -- (0,0); - \end{scope} - - \begin{scope}[shift={(3,0)}] - - \draw [thick,->,rotate=45] (-2,0) -- (0,0) node [below,pos=.5] { $\vec\ell$}; - \draw [rotate=45,->] (-1.8,0) arc (0:-270:.2); - \draw [rotate=45] (-.4,0) arc (180:120:.4); - \draw [rotate=15] (-.6,0) node { $60^\circ$}; - - \filldraw [black,rotate=45] (-2,0) circle (2.4pt); - - \begin{scope}[shift={(-.05,.05)}] - \draw [thick,->,rotate=-15] (-1.5,0) node [above] { $\vec F$} -- (0,0); - \end{scope} - - \end{scope} - - \end{tikzpicture} - - - - -
    - -

    -

      -
    1. -

      - the force is perpendicular to the lever, and -

      -
      2. - -
    2. -

      - the force makes an angle of 60^\circ with the lever, - as shown in . -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - We start by determining vectors for the force and lever arm. - Since the lever arm makes a - 45^\circ angle with the horizontal and is 2ft long, - we can state that \vec \ell = 2\la \cos(45^\circ) ,\sin(45^\circ) \ra = \la \sqrt2,\sqrt2\ra. - - Since the force vector is perpendicular to the lever arm (as seen in the left hand side of ), - we can conclude it is making an angle of -45^\circ with the horizontal. - As it has a magnitude of 10lb, we can state \vec F = 10\la \cos(-45^\circ), \sin(-45^\circ)\ra = \la 5\sqrt2,-5\sqrt2\ra. - - Using Equation to find the torque requires a cross product. We again let the third component of each vector be 0 and compute the cross product: - - \vec\tau \amp = \crossp \ell F - \amp = \la \sqrt2,\sqrt2,0\ra \times \la 5\sqrt2,-5\sqrt2,0\ra - \amp = \la 0,0,-20\ra - - This clearly has a magnitude of 20 ft-lb. - We can view the force and lever arm vectors as lying on the page; - our computation of \vec\tau shows that the torque goes into the page. - This follows the Right Hand Rule of the cross product, - and it also matches well with the example of the wrench turning the bolt. - Turning a bolt clockwise moves it in. -

      -
      2. - -
    2. -

      - Our lever arm can still be represented by \vec \ell = \la \sqrt2,\sqrt2\ra. - As our force vector makes a - 60^\circ angle with \vec \ell, we can see - (referencing the right hand side of the figure) - that \vec F makes a - -15^\circ angle with the horizontal. - Thus - - \vec F = 10\la \cos-15^\circ,\sin-15^\circ\ra \amp = \la \frac{5(1+\sqrt3)}{\sqrt2},\frac{5(-1+\sqrt3)}{\sqrt2}\ra - \amp \approx \la 9.659,-2.588\ra - . - We again make the third component 0 and take the cross product to find the torque: - - \vec\tau \amp = \crossp \ell F - \amp = \la \sqrt2,\sqrt2,0\ra \times \la \frac{5(1+\sqrt3)}{\sqrt2},\frac{5(-1+\sqrt3)}{\sqrt2},0\ra - \amp = \la 0,0,-10\sqrt3\ra - \amp \approx \la 0,0,-17.321\ra - . - As one might expect, when the force and lever arm vectors - are orthogonal, - the magnitude of force is greater than when the vectors - are not orthogonal. -

      -
      3. -
    -

    -     -     -     - -

    - While the cross product has a variety of applications - (as noted in this chapter), - its fundamental use is finding a vector perpendicular to two others. - Knowing a vector is orthogonal to two others is of incredible importance, - as it allows us to find the equations of lines and planes in a variety of contexts. - The importance of the cross product, - in some sense, relies on the importance of lines and planes, - which see widespread use throughout engineering, physics and mathematics. - We study lines and planes in the next two sections. -

    -     - - - - Terms and Concepts - - - - -

    - The cross product of two vectors is a , not a scalar. -

    -     - - - - - - - - -     - - - - -

    - One can visualize the direction of \vec u\times\vec v using the rule. -

    -     - - - - right hand|right-hand - - - - -     - - - - -

    - Give a synonym for orthogonal. -

    -     - - - -

    - Perpendicular is one answer, - although we tend to use perpendicular to refer to lines, - and orthogonal to refer to vectors. -

    -

    - Note that the zero vector is orthogonal to any vector, - but it doesn't really make sense to say it is perpendicular. -

    -     - -     - - - - - -

    - A fundamental principle of the cross product is that - \vec u\times\vec v is orthogonal to \vec u and \vec v. - -

    -     - -     - - - - -

    - is a measure of the turning force applied to an object. -

    -     - - - - - - - - - -     - - - -

    - If \vec u and \vec v are parallel, - then \vec u\times\vec v=\vec 0. -

    -     -     - -     - - Problems - - - -

    - Vectors \vec u and \vec v are given. - Compute \vec u\times\vec v and check that this vector is orthogonal to both \vec u and \vec v. -

    -     - - - - - Context("Vector"); - $cp=Vector("<12, -15, 3>"); - - - -

    - Let \vec u = \la 3,2,-2\ra, - \vec v = \la 0,1,5\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - $cp=Vector("<11, 1, -17>"); - - - -

    - Let \vec u = \la 5, -4, 3\ra, - \vec v = \la 2, -5, 1\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - $cp=Vector("<-5, -31, 27>"); - - - -

    - Let \vec u = \la 4, -5, -5\ra, - \vec v = \la 3, 3, 4\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - $cp=Vector("<47, -36, -44>"); - - - -

    - Let \vec u = \la -4, 7, -10\ra, - \vec v = \la 4, 4, 1\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - $cp=Vector("<0, -2, 0>"); - - - -

    - Let \vec u = \la 1, 0, 1\ra, - \vec v = \la 5, 0, 7\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - $cp=Vector("<0, 0, 0>"); - - - -

    - Let \vec u = \la 1, 5, -4\ra, - \vec v = \la -2, -10, 8\ra. -

    - -

    - -

    -     -     -     - - - -

    - \vec u = \langle a,b,0\rangle, - \vec v = \langle c,d,0\rangle -

    -     - -

    - \vec u\times \vec v = \langle 0,0,ad-bc\rangle -

    -     -     - - - - - - Context("Vector"); - Context()->flags->set(ijk=>1); - $cp=Vector("k"); - - - -

    - \vec u = \hat\imath, - \vec v = \hat\jmath. -

    - -

    - -

    - -

    - Check this is orthogonal to both \vec u and \vec v. -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(ijk=>1); - $cp=Vector("-j"); - - - -

    - \vec u = \hat\imath, \vec v = \hat{k}. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(ijk=>1); - $cp=Vector("i"); - - - -

    - \vec u = \hat\jmath, \vec v = \hat{k}. -

    - -

    - \vec u\times\vec v= -

    -     -     -     - -     - - - - -

    - Pick any vectors \vec u, - \vec v and \vec w in - \mathbb{R}^3 and show that \vec u \times (\vec v+\vec w) = \vec u\times \vec v+\vec u\times \vec w. -

    -     - -

    - Answers will vary. -

    -     - -     - - - - -

    - Pick any vectors \vec u, - \vec v and \vec w in - \mathbb{R}^3 and show that \vec u \cdot (\vec v\times\vec w) = (\vec u\times \vec v)\cdot \vec w. -

    -     - -

    - Answers will vary. -

    -     - -     - - - - -

    - The magnitudes of vectors \vec u and \vec v in \mathbb{R}^3 are given, - along with the angle \theta between them. - Use this information to find the magnitude of \vec u\times\vec v. -

    -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $normucv=Formula("5"); - - - -

    - If \vnorm{u} = 2, \vnorm{v} = 5, - and \theta = 30^{\circ} is the angle between \vec u and \vec v, - then \norm{\vec u\times\vec v}= -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $normucv=Formula("21"); - - - -

    - If \vnorm{u} = 3, \vnorm{v} = 7, - and \theta = \pi/2 is the angle between \vec u and \vec v, - then \norm{\vec u\times\vec v}= -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $normucv=Formula("0"); - - - -

    - If \vnorm{u} = 3, \vnorm{v} = 4, - and \theta = \pi is the angle between \vec u and \vec v, - then \norm{\vec u\times\vec v}= -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $normucv=Formula("5"); - - - -

    - If \vnorm{u} = 2, \vnorm{v} = 5, - and \theta = 5\pi/6 is the angle between \vec u and \vec v, - then \norm{\vec u\times\vec v}= -

    -     -     -     - -     - - - -

    - Find the area of the parallelogram defined by the given vectors. -

    -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $area=Formula("sqrt(14)"); - - - -

    - Find the area of the parallelogram defined by \vec u = \la 1,1,2\ra, - and \vec v = \la 2,0,3\ra. -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $area=Formula("sqrt(230)"); - - - -

    - Find the area of the parallelogram defined by \vec u = \la -2,1,5\ra, - and \vec v = \la -1,3,1\ra. -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $area=Formula("3"); - - - -

    - Find the area of the parallelogram defined by \vec u = \la 1,2\ra, - and \vec v = \la 2,1\ra. -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $area=Formula("6"); - - - -

    - Find the area of the parallelogram defined by \vec u = \la 2,0\ra, - and \vec v = \la 0,3\ra. -

    - -

    - -

    -     -     -     - -     - - - -

    - Find the area of the triangle with the given vertices. -

    -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $area=Formula("5sqrt(2)/2"); - - - -

    - Find the area of the triangle with vertices (0,0,0), - (1,3,-1) and (2,1,1). -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $area=Formula("3sqrt(30)"); - - - -

    - Find the area of the triangle with vertices (5,2,-1), - (3,6,2) and (1,0,4). -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $area=Formula("1"); - - - -

    - Find the area of the triangle with vertices (1,1), - (1,3) and (2,2). -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $area=Formula("5/2"); - - - -

    - Find the area of the triangle with vertices (3,1), - (1,2) and (4,3). -

    - -

    - -

    -     -     -     - -     - - - -

    - Find the area of the quadrilateral with the given vertices. - (Hint: break the quadrilateral into two triangles.) -

    -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $area=Formula("7"); - - - -

    - Find the area of the quadrilateral with vertices (0,0), - (1,2), - (3,0), and (4,3). -

    - -

    - -

    -     - -

    - Break the quadrilateral into two triangles. -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $area=Formula("4sqrt(14)"); - - - -

    - Find the area of the quadrilateral with vertices (0,0,0), - (2,1,1), (-1,2,-8), and (1,-1,5). -

    - -

    - -

    -     - -

    - Break the quadrilateral into two triangles. -

    -     -     -     - -     - - - -

    - Find the volume of the parallelepiped defined by the given vectors. -

    -     - - - - - $volume=Compute("2"); - - - -

    - Find the volume of the parallelepiped defined by \vec u = \la 1,1,1\ra, - \vec v=\la 1,2,3\ra, and \vec w = \la 1,0,1\ra. -

    - -

    - -

    -     -     -     - - - - - $volume=Compute("15"); - - - -

    - Find the volume of the parallelepiped defined by \vec u = \la -1,2,1\ra, - \vec v=\la 2,2,1\ra, and \vec w = \la 3,1,3\ra. -

    - -

    - -

    -     -     -     - -     - - - -

    - Find a unit vector orthogonal to both \vec u and \vec v. -

    -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $w=Compute("<1/sqrt(6),1/sqrt(6),-2/sqrt(6)>"); - $w=OneOf("$w,-$w"); - - - -

    - Find a unit vector orthogonal to both \vec u = \la 1,1,1\ra, - and \vec v=\la 2,0,1\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $w=Compute("<-2/sqrt(21),1/sqrt(21),4/sqrt(21)>"); - $w=OneOf("$w,-$w"); - - - -

    - Find a unit vector orthogonal to both \vec u = \la 1,-2,1\ra, - and \vec v=\la 3,2,1\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $w=Compute("<0,1,0>"); - $w=OneOf("$w,-$w"); - - - -

    - Find a unit vector orthogonal to both \vec u = \la 5,0,2\ra, - and \vec v=\la -3,0,7\ra. -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0, reduceConstantFunctions=>0); - $u=Compute("<1,-2,1>"); - $w=Compute("<2/sqrt(5),1/sqrt(5),0>"); - $ev=$w->cmp(checker => sub { - my ($correct,$student,$self) = @_; - return 1 if (($student . $student == 1) and ($student . $u == 0)); - }); - - - -

    - Find a unit vector orthogonal to both \vec u = \la 1,-2,1\ra, - and \vec v=\la -2,4,-2\ra. -

    - -

    - -

    -     - -

    - Since \vec u and \vec v are parallel, - any unit vector orthogonal to \vec u works - (such as \frac{1}{\sqrt{2}}\la 1,0,-1\ra). -

    -     -     -     - -     - - - - -

    - A bicycle rider applies 150lb of force, straight down, - onto a pedal that extends 7in horizontally from the crankshaft. - Find the magnitude of the torque applied to the crankshaft. -

    -     - -

    - 87.5ftlb -

    -     - -     - - - - -

    - A bicycle rider applies 150lb of force, - straight down, onto a pedal that extends 7in from the crankshaft, - making a 30^\circ angle with the horizontal. - Find the magnitude of the torque applied to the crankshaft. -

    -     - -

    - 43.75\sqrt{3}\approx 75.78ftlb -

    -     - -     - - - - -

    - To turn a stubborn bolt, 80lb of force is applied to a 10in wrench. - What is the maximum amount of torque that can be applied to the bolt? -

    -     - -

    - 200/3\approx 66.67ftlb -

    -     - -     - - - - -

    - To turn a stubborn bolt, 80lb of force is applied to a 10in wrench in a confined space, - where the direction of applied force makes a - 10^\circ angle with the wrench. - How much torque is subsequently applied to the wrench? -

    -     - -

    - 11.58ftlb -

    -     - -     - - - - -

    - Show, using the definition of the Cross Product, - that \vec u\cdot(\vec u\times\vec v)=0; - that is, that \vec u is orthogonal to the cross product of \vec u and \vec v. -

    -     - -

    - With \vec u = \la u_1,u_2,u_3\ra and \vec v = \la v_1,v_2,v_3\ra, we have - - \vec u\cdot(\vec u\times\vec v) \amp = \la u_1,u_2,u_3\ra\cdot (\la u_2v_3-u_3v_2,-(u_1v_3-u_3v_1),u_1v_2-u_2v_1\ra) - \amp = u_1(u_2v_3-u_3v_2) - u2(u_1v_3-u_3v_1)+u3(u_1v_2-u_2v_1) - \amp =0 - . -

    -     - -     - - - - -

    - Show, using the definition of the Cross Product, - that \vec u\times\vec u=\vec 0. -

    -     - -

    - With \vec u = \la u_1,u_2,u_3\ra, we have - - \vec u\times\vec u \amp = \la u_2u_3-u_3u_2,-(u_1u_3-u_3u_1),u_1u_2-u_2u_1\ra) - \amp = \la 0,0,0\ra - \amp =\vec 0 - . -

    -     - -     -     -     -
    -
    - Lines - -

    - lines - To find the equation of a line in the xy-plane, we need two pieces of information: a point and the slope. - The slope conveys direction information. - As vertical lines have an undefined slope, - the following statement is more accurate: -

    - -

    - To define a line, one needs a point on the line and the direction of the line. -

    - -

    - This holds true for lines in space. -

    - - -     - - - Lines in space -

    - Let P be a point in space, - let \vec p be the vector with initial point at the origin and terminal point at P (, \vec p - points to P), - and let \vec d be a vector. - Consider the points on the line through P in the direction of \vec d. -

    - -

    - Clearly one point on the line is P; - we can say that the vector - \vec p lies at this point on the line. - To find another point on the line, - we can start at \vec p and move in a direction parallel to \vec d. - For instance, - starting at \vec p and traveling one length of \vec d places one at another point on the line. - Consider - where certain points along the line are indicated. -

    - -
    - Defining a line in space - - - - In three dimensions, vectors p (position), d (direction), and p+d are plotted. A line passes through the tips of p and p+d. - -

    - In a three-dimensional coordinate system, vectors \vec d and \vec p are plotted with their tails at the origin, - along with their sum, \vec{p}+\vec{d}. - The vector \vec{p} is the position vector for a line, which is also shown. - The line passes through the tips of \vec{p} and \vec{p}+\vec{d}, - since the difference (\vec{p}+\vec{d})-\vec{p}=\vec d is the direction vector of the line. - Also plotted is the vector \vec{p}-2\vec{d}, which points to another point on the line. -

    -     - - - - - //ASY file for figlines_intro.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-1,1); - pair ybounds=(-1,3.5); - pair zbounds=(-1,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //draw vector d=<-1,0,1> - draw((0,0,0)--(-1,0,1),redpen,Arrow3(size=2mm)); - label("$\vec{d}$",(-1,0,1),N); - //draw vector p=<0,2,1> - draw((0,0,0)--(0,2,1),black,Arrow3(size=2mm)); - label("$\vec{p}$",(0,2,1),N); - //draw vector p+d - draw((0,0,0)--(-1,2,2),black,Arrow3(size=2mm)); - label("$\vec{p}+\vec{d}$",(-1,2,2),W); - //draw vector p-2d - draw((0,0,0)--(2,2,-1),black,Arrow3(size=2mm)); - label("$\vec{p}-2\vec{d}$",(2,2,-1),W); - //draw the line ({-x},{2},{1+x}); - draw((2.2,2,-1.2)--(-1.5,2,2.5),bluepen); - - - - -
    - -

    - The figure illustrates how every point on the line can be obtained by starting with \vec p and moving a certain distance in the direction of \vec d. - That is, we can define the line as a function of t: - - \vec\ell(t) = \vec p + t\ \vec d - . -

    - -

    - In many ways, this is not a new concept. - Compare Equation to the familiar - y=mx+b equation of a line: -

    - -
    - Understanding the vector equation of a line - - A text description, comparing the elements the equations of a line in the plane, and in space. - -

    - On the left is the equation y=mx+b for a line in the plane with slope m and y intercept b. - On the right is the equation \vec{\ell}(t) = \vec p + t\vec d for a line in space - through the point \vec p with direction vector \vec d. -

    - -

    - Above the two equations is the text Starting Point. - From this text are two arrows, pointing to the value b in the plane equation, and the value \vec p in the space equation. - Also above the equations is the text Direction. - From this text, two arrows point to the value m in the plane equation, and the value \vec d in the space equation. -

    - -

    - Below the two equations is the text How Far To Go In That Direction. - Arrows point from this text to the value x in the plane equation, and the value t in the space equation. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=0.75] - - \draw (0,0) node (L) {\(y\ =\ b\ +\ m x\)}; - \draw (5,0) node (R) {\(\vec \ell(t)\ =\ \vec p\ +\ t \vec d\)}; - - \node (A) at (1,1.5) {Starting Point}; - \node (B) at (4,1.5) {Direction}; - \node (C) at (2.5,-1.5) {How Far To Go In That Direction}; - - \draw [->,thick] (A) -- (5,.2); - \draw [->,thick] (A) -- (0,.2); - \draw [->,thick] (B) -- (.9,.2); - \draw [->,thick] (B) -- (6.45,.25); - \draw [->,thick] (C) -- (1.3,-.15); - \draw [->,thick] (C) -- (6.1,-.3); - - \end{tikzpicture} - - - -
    - -

    - The equations exhibit the same structure: - they give a starting point, - define a direction, and state how far in that direction to travel. -

    - - - -

    - Equation is an example of a - vector-valued function; - the input of the function is a real number and the output is a vector. - We will cover vector-valued functions extensively in the next chapter. -

    - -

    - There are other ways to represent a line. - Let P = (x_0,y_0,z_0), \vec p = \la x_0,y_0,z_0\ra, and let \vec d = \la a,b,c\ra. - Then the equation of the line through P in the direction of \vec d is: - - \vec\ell(t) \amp = \vec p + t\vec d - \amp = \la x_0,y_0,z_0\ra + t\la a,b,c\ra - \amp = \la x_0 + at, y_0+bt, z_0+ct\ra - . -

    - -

    - The last line states that the x values of the line are given by x=x_0+at, - the y values are given by y = y_0+bt, - and the z values are given by z = z_0 + ct. - These three equations, taken together, - are the parametric equations of the line - through \vec p in the direction of \vec d. -

    - -

    - Finally, each of the equations for x, - y and z above contain the variable t. - We can solve for t in each equation: - - x = x_0+at \amp \Rightarrow t=\frac{x-x_0}{a}, - y=y_0+bt \amp \Rightarrow t = \frac{y-y_0}{b}, - z = z_0+ct \amp \Rightarrow t = \frac{z-z_0}{c} - , - assuming a,b,c\neq 0. - Since t is equal to each expression on the right, - we can set these equal to each other, - forming the symmetric equations of the line - through \vec p in the direction of \vec d: - - \frac{x-x_0}{a} = \frac{y-y_0}{b}=\frac{z-z_0}{c} - . -

    - -

    - Each representation has its own advantages, depending on the context. - We summarize these three forms in the following definition, - then give examples of their use. -

    - - - Equations of Lines in Space - -

    - Let P = (x_0,y_0,z_0) and let \vec p = \la x_0,y_0,z_0\ra. Consider the line in space that passes through - P in the direction of \vec d = \la a,b,c\ra. - linesequations for -

    - -

    -

      -
    1. -

      - The vector equation of the line is - - \vec \ell(t) = \vec p+t\vec d - . -

      -
      2. - -
    2. -

      - The parametric equations of the line are - - x = x_0+at, y=y_0+bt, z = z_0+ct - . -

      -
      3. - -
    3. -

      - The symmetric equations of the line are - - \frac{x-x_0}{a} = \frac{y-y_0}{b}=\frac{z-z_0}{c} - . -

      -
      4. -
    -

    -     -     - - - Finding the equation of a line - -

    - Give all three equations, - as given in , - of the line through P = (2,3,1) in the direction of \vec d = \la -1,1,2\ra. - Does the point Q=(-1,6,6) lie on this line? -

    -     - -

    - We identify the point P=(2,3,1) with the vector \vec p =\la 2,3,1\ra. - Following the definition, we have -

    - -

    -

      -
    • -

      - the vector equation of the line is \vec\ell(t) = \la 2,3,1\ra + t\la -1,1,2\ra; -

      -
      • - -
    • -

      - the parametric equations of the line are - - x = 2-t, y = 3+t, z = 1+2t; \text{ and } - -

      -
      • - -
    • -

      - the symmetric equations of the line are - - \frac{x-2}{-1}=\frac{y-3}{1} = \frac{z-1}{2} - . -

      -
      • -
    -

    - -
    - Graphing a line in - - - - A line in space with direction vector d passes through a point P. - -

    - A set three-dimensional coordinate axes is shown. - A vector \vec d is plotted with its tail at the origin of this coordinate system; - it is the direction vector for a line. The line is also plotted, - and is shown passing through a point P. - An additional point Q is in the image, but it is not on the line. -

    -     - - - - - //ASY file forfiglines1.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={5}; - real[] myzchoice={5}; - defaultpen(0.5mm); - pair xbounds=(-1,2.5); - pair ybounds=(-1,5.5); - pair zbounds=(-1,5.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //draw vector d=<-1,1,2> - draw((0,0,0)--(-1,1,2),black,Arrow3(size=2mm)); - label("$\vec{d}$",(-1,0,1),N); - // dot at P=(2,3,1) - dotfactor=3;dot((2,3,1)); - label("P",(2,3,1),N); - // dot at Q=(-1,6,6) - dotfactor=3;dot((-1,6,6)); - label("Q",(-1,6,6),S); - //draw the line ({2-t},{3+t},{1+2t}); - draw((3,2,-1)--(-1,6,7),bluepen); - - - - -
    - -

    - The first two equations of the line are useful when a t value is given: - one can immediately find the corresponding point on the line. - These forms are good when calculating with a computer; - most software programs easily handle equations in these formats. (For instance, - the graphics program that made - can be given the input (2-t,3+t,1+2*t) - for -1\leq t\leq 3.). -

    - -

    - Does the point Q = (-1,6,6) lie on the line? - The graph in - makes it clear that it does not. - We can answer this question without the graph using any of the three equation forms. - Of the three, - the symmetric equations are probably best suited for this task. - Simply plug in the values of x, - y and z and see if equality is maintained: - - \frac{-1-2}{-1} \stackrel{?}{=} \frac{6-3}{1} \stackrel{?}{=} \frac{6-1}{2} \Rightarrow 3=3\neq2.5 - . -

    - -

    - We see that Q does not lie on the line as it did not satisfy the symmetric equations. -

    -     - -     - - - Finding the equation of a line through two points - -

    - Find the parametric equations of the line through the points - P=(2,-1,2) and Q = (1,3,-1). -

    -     - -

    - Recall the statement made at the beginning of this section: - to find the equation of a line, - we need a point and a direction. - We have two points; - either one will suffice. - The direction of the line can be found by the vector with initial point P and terminal point Q: - \overrightarrow{PQ} = \la -1,4,-3\ra. -

    - -

    - The parametric equations of the line \ell through P in the direction of \overrightarrow{PQ} are: - - \ell: x= 2-t y=-1+4t z=2-3t - . -

    - -
    - A graph of the line in - - - - A line in space passes through points P and Q. The vector PQ between these points is also shown. - -

    - Two points P and Q are plotted in space, against a set of three-dimensional coordinate axes. - The vector \overrightarrow{PQ} is plotted as an arrow pointing from P to Q. - The line passing through P and Q is also plotted; - between P and Q, it overlaps with the vector \overrightarrow{PQ}. -

    -     - - - - - //ASY file for figlines6.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,17,9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={-4,4}; - real[] myzchoice={-4,4}; - defaultpen(0.5mm); - pair xbounds=(-1,3); - pair ybounds=(-4.5,4.5); - pair zbounds=(-4.5,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //draw vector PQ=<-1,4,-3> - draw((2,-1,2)--(1,3,-1),redpen,Arrow3(size=2mm)); - label("$\overrightarrow{PQ}$",(1.5,1,0.5),S); - // dot at P=(2,-1,2) - dotfactor=3;dot((2,-1,2));label("P",(2,-1,2),N); - // dot at Q=(1,3,-1) - dotfactor=3;dot((1,3,-1));label("Q",(1,3,-1),N); - //draw the line ({2-t},{-1+4t},{2-3t}) in two parts, before P and after Q - draw((2.5,-3,3.5)--(2,-1,2),bluepen); - draw((1,3,-1)--(0.5,5,-2.5),bluepen); - - - - -
    - -

    - A graph of the points and line are given in . - Note how in the given parametrization of the line, - t=0 corresponds to the point P, - and t=1 corresponds to the point Q. - This relates to the understanding of the vector equation of a line described in . - The parametric equations start at the point P, - and t determines how far in the direction of \overrightarrow{PQ} to travel. - When t=0, we travel 0 lengths of \overrightarrow{PQ}; - when t=1, we travel one length of \overrightarrow{PQ}, - resulting in the point Q. -

    -     - -     -     - - - Parallel, Intersecting and Skew Lines - -

    - In the plane, two distinct - lines can either be parallel or they will intersect at exactly one point. - In space, given equations of two lines, - it can sometimes be difficult to tell whether the lines are distinct or not (, the same line can be represented in different ways). - Given lines \vec\ell_1(t) = \vec p_1 + t\vec d_1 and \vec \ell_2(t) = \vec p_2+t\vec d_2, - we have four possibilities: - \vec \ell_1 and \vec \ell_2 are - linesskew - linesparallel - linesintersecting -

    - -

    -

    -
  • - the same line -

    they share all points

    -
    • - -
  • - intersecting lines -

    they share only 1 point;

    -
    • - -
  • - parallel lines -

    \vec d_1\parallel \vec d_2, no points in common;

    -
    • - -
  • - skew lines -

    \vec d_1\nparallel \vec d_2, no points in common.

    -
    • -
    -

    - - - -

    - The next two examples investigate these possibilities. -

    - - - Comparing lines - -

    - Consider lines \ell_1 and \ell_2, - given in parametric equation form: - - \ell_1:\, \begin{matrix} - x\amp =\amp 1+3t \\ - y\amp =\amp 2-t\\ - z\amp =\amp t - \end{matrix}\qquad\qquad - \ell_2:\, \begin{matrix} - x\amp =\amp -2+4s\\ - y\amp =\amp 3+s\\ - z\amp =\amp 5+2s - \end{matrix} - . -

    - -

    - Determine whether \ell_1 and \ell_2 are the same line, - intersect, are parallel, or skew. -

    -     - -

    - We start by looking at the directions of each line. - Line \ell_1 has the direction given by - \vec d_1=\la 3,-1,1\ra and line \ell_2 has the direction given by \vec d_2 = \la 4,1,2\ra. - It should be clear that \vec d_1 and \vec d_2 are not parallel, - hence \ell_1 and \ell_2 are not the same line, - nor are they parallel. - verifies this fact - (where the points and directions indicated by the equations of each line are identified). -

    - -

    - We next check to see if they intersect - (if they do not, they are skew lines). - To find if they intersect, - we look for t and s values such that the respective x, - y and z values are the same. - That is, we want s and t such that: - - \begin{matrix} 1+3t \amp =\amp -2+4s\\ 2-t\amp =\amp 3+s\\ t\amp =\amp 5+2s \end{matrix} - . -

    - -
    - Sketching the lines from - - - - A three dimensional plot of two lines in space. The lines are not parallel, and they do not intersect. - -

    - A set of three-dimensional coordinate axes are shown, - with labels x, y, and z, as usual. - Two points P_1 and P_2 are plotted. - The point P_1 is in the xy plane, - while the point P_2 is higher up, near the top of the figure. -

    - -

    - Through each of the two points, a line is drawn. - The line through P_1 has a direction vector \vec{d}_1 - that points in a different direction from that of the direction vector \vec{d}_2 of the line through P_2, - so the lines are not parallel. It also appears from the figure that the lines do not intersect. -

    -     - - - - - //ASY file for figlines2.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-5,5}; - real[] myychoice={-5,5}; - real[] myzchoice={-5,5}; - defaultpen(0.5mm); - pair xbounds=(-5.5,5.5); - pair ybounds=(-5.5,5.5); - pair zbounds=(-1,6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //draw P1 and vector d1 at P1 - dotfactor=3;dot((1,2,0));label("$P_1$",(1,2,0),S); - draw((1,2,0)--(4,1,1),black,Arrow3(size=2mm)); - label("$\vec{d}_1$",(4,1,1),N); - - //draw P2 and vector d2 at P2 - dotfactor=3;dot((-2,3,5));label("$P_2$",(-2,3,5),N); - draw((-2,3,5)--(2,4,7),black,Arrow3(size=2mm)); - label("$\vec{d}_2$",(2,4,7),S); - - //draw the line L1 ({1+3t},{2-t},{t}) in two parts, before P1 and after P1+d1 - draw((-2,3,-1)--(1,2,0),bluepen); - draw((4,1,1)--(7,0,2),bluepen); - label("$\ell_1$",(7,0,2),S); - - //draw the line L2 ({-2+4t},{3+t},{5+2t}) in two parts, before P2 and after P2+d2 - draw((-6,2,3)--(-2,3,5),bluepen); - draw((2,4,7)--(6,5,9),bluepen); - label("$\ell_2$",(-6,2,3),S); - - - - -
    - -

    - This is a relatively simple system of linear equations. - Since the last equation is already solved for t, - substitute that value of t into the equation above it: - - 2-(5+2s) = 3+s \Rightarrow s=-2,\ t=1 - . -

    - -

    - A key to remember is that we have - three equations; - we need to check if s=-2,\ t=1 satisfies the first equation as well: - - 1+3(1) \neq -2+4(-2) - . -

    - -

    - It does not. - Therefore, we conclude that the lines \ell_1 and \ell_2 are skew. -

    -     - -     - - - Comparing lines - -

    - Consider lines \ell_1 and \ell_2, - given in parametric equation form: - - \ell_1:\, \begin{matrix} x\amp =\amp -0.7+1.6t \\ y\amp =\amp 4.2+2.72t\\z\amp =\amp 2.3-3.36t \end{matrix} \qquad\qquad \ell_2:\,\begin{matrix} x\amp =\amp 2.8-2.9s\\y\amp =\amp 10.15-4.93s\\z\amp =\amp -5.05+6.09s. \end{matrix} - -

    - -

    - Determine whether \ell_1 and \ell_2 are the same line, - intersect, are parallel, or skew. -

    -     - -

    - It is obviously very difficult to simply look at these equations and discern anything. - This is done intentionally. - In the real world, most equations that are used do not have nice, - integer coefficients. - Rather, there are lots of digits after the decimal and the equations can look messy. -

    - -

    - We again start by deciding whether or not each line has the same direction. - The direction of \ell_1 is given by - \vec d_1 = \la 1.6,2.72,-3.36\ra and the direction of \ell_2 is given by \vec d_2 = \la -2.9,-4.93,6.09\ra. - When it is not clear through observation whether two vectors are parallel or not, - the standard way of determining this is by comparing their respective unit vectors. - Using a calculator, we find: - - \vec u_1 \amp = \frac{\vec d_1}{\norm{\vec d_1}} = \la 0.3471,0.5901,-0.7289\ra - \vec u_2 \amp = \frac{\vec d_2}{\norm{\vec d_2}} = \la -0.3471,-0.5901,0.7289\ra - . -

    - -

    - The two vectors seem to be parallel - (at least, their components are equal to 4 decimal places). - In most situations, - it would suffice to conclude that the lines are at least parallel, - if not the same. - One way to be sure is to rewrite \vec d_1 and - \vec d_2 in terms of fractions, not decimals. - We have - - \vec d_1 =\la \frac{16}{10},\frac{272}{100},-\frac{336}{100}\ra \qquad \vec d_2 = \la -\frac{29}{10},-\frac{493}{100},\frac{609}{100}\ra - . -

    - -

    - One can then find the magnitudes of each vector in terms of fractions, - then compute the unit vectors likewise. - After a lot of manual arithmetic - (or after briefly using a computer algebra system), - one finds that - - \vec u_1 = \la \sqrt{\frac{10}{83}},\frac{17}{\sqrt{830}},-\frac{21}{\sqrt{830}}\ra \qquad \vec u_2 = \la -\sqrt{\frac{10}{83}},-\frac{17}{\sqrt{830}},\frac{21}{\sqrt{830}}\ra - . -

    - -

    - We can now say without equivocation that these lines are parallel. -

    - -

    - Are they the same line? - The parametric equations for a line describe one point that lies on the line, - so we know that the point P_1 = (-0.7,4.2,2.3) lies on \ell_1. - To determine if this point also lies on \ell_2, - plug in the x, - y and z values of P_1 into the symmetric equations for \ell_2: - - \frac{(-0.7)-2.8}{-2.9} \amp \stackrel{?}{=} \frac{(4.2)-10.15}{-4.93} \stackrel{?}{=} \frac{(2.3)-(-5.05)}{6.09} - 1.2069\amp=1.2069=1.2069 - . -

    - -
    - Graphing the lines in - - - - A plot in space of a single line through two points, with two parallel direction vectors shown next to it. - -

    - Two points P_1 and P_2 are plotted in a three-dimensional coordinate system. - A line is shown passing through these two points. - Next to the line, two vectors are plotted. - The vector \vec{d}_1 points downward, in the direction of the line. - The vector \vec{d}_2 points in the opposite direction, and is twice as long. -

    - -

    - The diagram shows that the line through P_1 in the direction of \vec{d}_1 - is in fact the same as the line through P_2 in the direction of \vec{d}_2. -

    -     - - - - - //ASY file for figlines3.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(10,26,7); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={10}; - real[] myzchoice={-5,5}; - defaultpen(0.5mm); - pair xbounds=(-1,4.5); - pair ybounds=(-1,11); - pair zbounds=(-6,6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //draw P1 and vector d1 at P1 - dotfactor=3;dot((-0.7,4.2,2.3));label("$P_1$",(-0.7,4.2,2.3),E); - //draw((-0.7,4.2,2.3)--(0.9,6.92,-1.06),black,Arrow3(size=2mm)); - label("$\vec{d}_1$",(0.9+.5,6.92,-1.06),W); - - //draw P2 and vector d2 at P2 - //dotfactor=3;dot((5.7,15.08,-11.14));label("$P_2$",(5.7,15.08,-11.14),E); - //draw((5.7,15.08,-11.14)--(2.8,10.15,-5.05),black,Arrow3(size=2mm)); - //label("$\vec{d}_2$",(2.8,10.15,-5.05),E); - - dotfactor=3;dot((2.8,10.15,-5.05));label("$P_2$",(2.8,10.15,-5.05),W); - //draw((2.8,10.15,-5.05)--(-.1,5.22,1.04),black,Arrow3(size=2mm)); - label("$\vec{d}_2$",(-.1-.5,5.22,1.04),E); - - //draw the line L ({-0.7+1.6t},{4.2+2.72t},{2.3-3.36t}) - draw((3.3,11,-6.1)--(-2.3,1.48,5.66),bluepen); - - draw((2.8-.5,10.15,-5.05)--(-.1-.5,5.22,1.04),redpen,Arrow3(size=2mm)); - draw((-0.7+.5,4.2,2.3)--(0.9+.5,6.92,-1.06),redpen,Arrow3(size=2mm)); - - - - -
    - -

    - The point P_1 lies on both lines, - so we conclude they are the same line, - just parametrized differently. - - graphs this line along with the points and vectors described by the parametric equations. - Note how \vec d_1 and \vec d_2 are parallel, - though point in opposite directions - (as indicated by their unit vectors above). -

    -     -     -     - - - Distances -

    - Given a point Q and a line \vec\ell(t) = \vec p+t\vec d in space, - it is often useful to know the distance from the point to the line. - (Here we use the standard definition of distance, - , the length of the shortest line segment from the point to the line.) - Identifying \vec p with the point P, - - will help establish a general method of computing this distance h. -

    - -
    - Establishing the distance from a point to a line - - - A line through a point P, with direction d. Near the line is a point Q, whose distance to the line is shown as the height h of a triangle. - -

    - A generic line is shown, without coordinates. - On the line is a point P and a direction vector \vec{d}. - Near the line is a point Q. - The vector \overrightarrow{PQ} points from the point P on the line to the point Q off the line. - This vector forms the hypotenuse of a right-angled triangle whose base is part of the line, - and whose height, h, is shown as a dashed line segment from Q to the line. - The angle \theta between \vec d and \overrightarrow{PQ} is also shown. - The height h is the perpendicular distance from Q to the line. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.32] - - \draw [thin,firstcolor] (-2,-1) -- (2,1); - \draw [thick,->,secondcolor] (0,0) -- (1,.5) node [below,black] { $\vec d$}; - - \filldraw [black] (0,3) circle (2.4pt) node [above] { $Q$} - (0,0) circle (2.4pt) node [below] { $P$}; - - \draw [thick,dashed] (0,3) -- (1.2,.6) node [right,pos=.5] { $h$}; - \draw [thick,->] (0,0) -- (0,3) node [pos=.5,left] { $\overrightarrow{PQ}$}; - \draw [rotate=58.3] (.4,0) node { $\theta$}; - - \end{tikzpicture} - - - - -
    - -

    - From trigonometry, we know h = \norm{\overrightarrow{PQ}}\sin(\theta). - We have a similar identity involving the cross product: - \norm{\overrightarrow{PQ}\times \vec d} = \norm{\overrightarrow{PQ}}\, \vnorm{d}\sin(\theta). - Divide both sides of this latter equation by \vnorm{d} to obtain h: - - h = \frac{\norm{\overrightarrow{PQ}\times \vec d}}{\vnorm{d}} - . -

    - - - -

    - It is also useful to determine the distance between lines, - which we define as the length of the shortest line segment that connects the two lines - (an argument from geometry shows that this line segments is perpendicular to both lines). - Let lines \vec\ell_1(t) = \vec p_1 + t\vec d_1 and \vec\ell_2(t) = \vec p_2 + t\vec d_2 be given, - as shown in . - To find the direction orthogonal to both \vec d_1 and \vec d_2, - we take the cross product: - \vec c = \vec d_1\times \vec d_2. - The magnitude of the orthogonal projection of - \overrightarrow{P_1P_2} onto \vec c is the distance h we seek: - - h\amp = \norm{\operatorname{proj}_{\vec c}\overrightarrow{P_1P_2}} - \amp = \norm{\frac{\overrightarrow{P_1P_2}\cdot\vec c}{\dotp cc}\vec c} - \amp =\frac{\abs{\overrightarrow{P_1P_2}\cdot \vec c}}{\vnorm c^2}\vnorm c - \amp =\frac{\abs{\overrightarrow{P_1P_2}\cdot \vec c}}{\vnorm c} - . -

    - -
    - Establishing the distance between lines - - - - Two skew lines in space are plotted with respective points and direction vectors. The distance h between them is indicated. - -

    - Two lines in space are shown, without a coordinate system. - The lines pass, respectively, through points P_1 and P_2, - and have respective direction vectors \vec{d}_1 and \vec{d}_2. -

    - -

    - The vector \overrightarrow{P_1P_2} is shown pointing from the first line to the second. - Also shown is the cross product \vec{c}=\vec{d}_1\times\vec{d}_2; - this vector is perpendicular to both lines. - The distance h between the lines is shown as a segment parallel to \vec c. -

    -     - - - - - //ASY file for figlines_dist23D.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((-.14,-9.67,3.98),(0.003,0.01,0.026),(0,0,0),1,(-0.09,-0.2)); - defaultrender.merge=true; - - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-3,3); - pair ybounds=(-3,3); - pair zbounds=(-3,3); - - xaxis3("",xbounds.x,xbounds.y,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,invisible,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - triple p1=(0,0,0); - triple p2=(0,0,2); - triple p12=p2-p1; - triple d1=(1,0,0); - triple d2=(1,1,-1); - - // line 1 - triple f(real t) { - return (p1+t*d1);} - - // line 2 - triple g(real t) { - return (p2+t*d2);} - - //draw P1 (t=0) and vector d1 at P1 (t=1) - dotfactor=3; - dot(p1);label("$P_1$",p1,S); - draw(f(0)--f(.6),redpen+.7mm,Arrow3(size=2mm)); - label("$\vec{d}_1$",f(.6),S); - - //draw P2 (t=-1.5) and vector d2 at P2 (t=-0.5) - dotfactor=3; - dot(p2);label("$P_2$",p2,N); - draw(g(0)--g(.6),redpen+.7mm,Arrow3(size=2mm)); - label("$\vec{d}_2$",g(.6),N); - - //draw vector P1 to P2 - draw(p1--p2,black,Arrow3(size=2mm)); - //label("$\overrightarrow{P_1 P_2}$",(p1+p2)/2,W); - label("$\overrightarrow{P_1 P_2}$",(p1+p2)/2,W); - - //draw the lines - draw(f(-1)--f(0),bluepen);//L1 - draw(f(.6)--f(2),bluepen);//L1 - draw(g(-1)--g(0),bluepen);//L2 - draw(g(.6)--g(2),bluepen);//L2 - - triple c=cross(d1,d2); - c=c/sqrt(dot(c,c)); - - //draw perp line - draw(f(1)--f(1)+c*abs(dot(p12,c)),black); - - // draw perp vector - draw(f(.9)--f(.9)+c*abs(dot(p12,c))*.4,redpen,Arrow3(size=2mm)); - label("$\vec c$",f(.9)+c*abs(dot(p12,c))*.4,W); - - label("$h$",(f(1)+f(1)+c*abs(dot(p12,c)))/2,E); - - // draw squares on perp line - triple sq1a = f(1)+c*abs(dot(p12,c))*.15; - triple sq1b = sq1a+.15*d1; - triple sq1c = sq1b+(f(1)-sq1a); - draw(sq1a -- sq1b -- sq1c); - - triple sq2a = f(1)+c*abs(dot(p12,c))*.85; - triple sq2b = sq2a+.15*d2/length(d2); - triple sq2c = sq2b+(f(1)+c*abs(dot(p12,c))-sq2a); - draw(sq2a -- sq2b -- sq2c); - - - - -
    - -

    - A problem in the Exercise section is to show that this distance is 0 when the lines intersect. - Note the use of the Triple Scalar Product: - \overrightarrow{P_1P_2}\cdot \vec c = \overrightarrow{P_1P_2}\cdot (\vec d_1\times \vec d_2). -

    - - - -

    - The following Key Idea restates these two distance formulas. -

    - - - Distances to Lines -

    -

      -
    1. -

      - Let P be a point on a line \ell that is parallel to \vec d. - The distance h from a point Q to the line \ell is: - distancebetween point and line - distancebetween lines - linesdistances between - - h =\frac{\norm{\overrightarrow{PQ}\times \vec d}}{\vnorm{d}} - . -

      -
      2. - -
    2. -

      - Let P_1 be a point on line \ell_1 that is parallel to \vec d_1, - and let P_2 be a point on line \ell_2 parallel to \vec d_2, - and let \vec c = \vec d_1\times \vec d_2, - where lines \ell_1 and \ell_2 are not parallel. - The distance h between the two lines is: - - h=\frac{\abs{\overrightarrow{P_1P_2}\cdot \vec c}}{\vnorm c} - . -

      -
      3. -
    -

    -     - - - - - Finding the distance from a point to a line - -

    - Find the distance from the point - Q=(1,1,3) to the line \vec\ell(t) = \la 1,-1,1\ra+t\la 2,3,1\ra. -

    -     - -

    - The equation of the line gives us the point - P=(1,-1,1) that lies on the line, - hence \overrightarrow{PQ} = \la 0,2,2\ra. - The equation also gives \vec d= \la 2,3,1\ra. - Following , - we have the distance as - - h \amp = \frac{\norm{\overrightarrow{PQ}\times \vec d}}{\vnorm{d}} - \amp = \frac{\norm{\la -4,4,-4\ra}}{\sqrt{14}} - \amp =\frac{4\sqrt{3}}{\sqrt{14}} \approx 1.852 - . -

    - -

    - The point Q is approximately 1.852 units from the line \vec\ell(t). -

    -     - -     - - - Finding the distance between lines - -

    - Find the distance between the lines - - \ell_1: \begin{matrix} x\amp =\amp 1+3t \\ y\amp =\amp 2-t\\z\amp =\amp t \end{matrix} \qquad\qquad - \ell_2:\begin{matrix} x\amp =\amp -2+4s\\y\amp =\amp 3+s\\z\amp =\amp 5+2s. \end{matrix} - -

    -     - -

    - These are the sames lines as given in , - where we showed them to be skew. - The equations allow us to identify the following points and vectors: - - P_1 = (1,2,0) P_2 = (-2,3,5) \Rightarrow \overrightarrow{P_1P_2} = \la -3,1,5\ra - . - - \vec d_1 = \la 3,-1,1\ra \vec d_2 = \la 4,1,2\ra \Rightarrow \vec c = \vec d_1\times \vec d_2 = \la -3,-2,7\ra - . -

    - -

    - From - we have the distance h between the two lines is - - h \amp = \frac{\abs{\overrightarrow{P_1P_2}\cdot \vec c}}{\vnorm c} - \amp =\frac{42}{\sqrt{62}} \approx 5.334 - . -

    - -

    - The lines are approximately 5.334 units apart. -

    -     - -     - -

    - One of the key points to understand from this section is this: - to describe a line, we need a point and a direction. - Whenever a problem is posed concerning a line, - one needs to take whatever information is offered and glean point and direction information. - Many questions can be asked - (and are asked in the Exercise section) - whose answer immediately follows from this understanding. -

    - -

    - Lines are one of two fundamental objects of study in space. - The other fundamental object is the plane, - which we study in detail in the next section. - Many complex three dimensional objects are studied by approximating their surfaces with lines and planes. -

    -     - - - - Terms and Concepts - - - -

    - To find an equation of a line, - what two pieces of information are needed? -

    -     - - - -

    - A point on the line and the direction of the line. -

    -     - -     - - - - -

    - Two distinct lines in the plane can intersect or be . -

    -     - - - - - - - - -     - - - - -

    - Two distinct lines in space can intersect, - be or be . -

    -     - - - - - - ["parallel","skew"].includes(ans) - - - - - - - ["parallel","skew"].includes(ans) && !ans_array.slice(0,1).includes(ans) - - - - - ans_array.slice(0,1).includes(ans) - - You already gave that answer. - - - - -     - - - - -

    - Use your own words to describe what it means for two lines in space to be skew. -

    -     - - - -     -     - - - Problems - - - -

    - Write the vector, - parametric and symmetric equations of the lines described. -

    -     - - - - -

    - Passes through P=(2,-4,1), - parallel to \vec d=\la 9,2,5\ra. -

    -     - -

    - vector: \ell(t) = \la 2,-4,1\ra + t\la 9,2,5\ra -

    - -

    - parametric: x= 2+9t, y=-4+2t, z = 1+5t -

    - -

    - symmetric: (x-2)/9 =(y+4)/2 = (z-1)/5 -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(t=>"Real"); - $v=Compute("(6,1,7)+t<-3,2,5>"); - $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; - $ds=Formula($st)->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; - $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return ($par and $tch); - }); - Context("Numeric"); - Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); - parser::Assignment->Allow; - $p=List(Formula("x=6-3t"),Formula("y=1+2t"),Formula("z=7+5t")); - $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - }} - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - }} - if(!$aH->{isPreview}){ - push(@err,"You need to provide more parametrizations")if $i<3; - push(@err,"You have given too many parametrizations")if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - Context("Numeric"); - Context()->variables->add(y=>'Real',z=>'Real'); - Context()->operators->redefine('=',using=>',',from=>'Numeric'); - Context()->operators->set('='=>{string=>' = ',TeX=>'='}); - Context()->lists->set(List=>{separator=>" = "}); - $s=List("(x-6)/-3","(y-1)/2","(z-7)/5"); - $sev=$s->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$dv;my$incpt;my%uses; - if($entry->class ne "Formula"){push(@err,"Your $ith quantity $entry is not a formula");next;} - else{$uses{'x'}=$entry->usesOneOf("x");$uses{'y'}=$entry->usesOneOf("y");$uses{'z'}=$entry->usesOneOf("z"); - if($uses{'x'}+$uses{'y'}+$uses{'z'}!=1){push(@err,"Your $ith quantity is not a formula in only x, y, or z");next;}; - $dv=$entry->D('x')->eval(x=>0,y=>0,z=>0)+$entry->D('y')->eval(x=>0,y=>0,z=>0)+$entry->D('z')->eval(x=>0,y=>0,z=>0); - $incpt=$entry->eval(x=>0,y=>0,z=>0)/$dv; - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->class eq "Formula"){ - $uj{'x'}=$st->[$j]->usesOneOf("x"); - $uj{'y'}=$st->[$j]->usesOneOf("y"); - $uj{'z'}=$st->[$j]->usesOneOf("z"); - if(($uses{'x'} and $uj{'x'}) or ($uses{'y'} and $uj{'y'}) or ($uses{'z'} and $uj{'z'})){ - push(@err,"Your $ith quantity uses the same variable as a previous one")unless $aH->{isPreview}; - $used=1;last; - }}} - if(!$used){ - if($uses{'x'}){$sV=$sV+Formula("<t/$dv-$incpt,0,0>")}; - if($uses{'y'}){$sV=$sV+Formula("<0,t/$dv-$incpt,0>")}; - if($uses{'z'}){$sV=$sV+Formula("<0,0,t/$dv-$incpt>")}; - }} - if(!$aH->{isPreview}){ - push(@err,"You need to provide more symmetric quantities") if $i<3; - push(@err,"You have given too many symmetric quantities") if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant) {$ds = Vector("$ds");} else {return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - - - - -

    - \ell is a line that passes through P=(6,1,7), - parallel to \vec d=\la -3,2,5\ra. -

    - - - Give the right-hand side (without \ell(t)=) of a vector equation for \ell. - -

    - -

    - - - Give the parametric equations for \ell - (separated using commas). - -

    - -

    - - - Give the symmetric equations for \ell. - (Enter DNE if they are not defined.) - -

    - -

    -     -     -     - - - - -

    - Passes through P=(2,1,5) and Q = (7,-2,4). -

    -     - -

    - Answers can vary: - vector: \ell(t) = \la 2,1,5\ra + t\la 5,-3,-1\ra -

    - -

    - parametric: x= 2+5t, y=1-3t, z = 5-t -

    - -

    - symmetric: (x-2)/5 =-(y-1)/3 = -(z-5) -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(t=>"Real"); - $v=Compute("(1,-2,3)+t<4,7,2>"); - $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; - $ds=Formula($st)->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; - $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return ($par and $tch); - }); - Context("Numeric"); - Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); - parser::Assignment->Allow; - $p=List(Formula("x=1+4t"),Formula("y=-2+7t"),Formula("z=3+2t")); - $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{ - ($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - } - } - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - } - } - if(!$aH->{isPreview}){ - push(@err,"You need to provide more parametrizations")if $i<3; - push(@err,"You have given too many parametrizations")if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - Context("Numeric"); - Context()->variables->add(y=>'Real',z=>'Real'); - Context()->operators->redefine('=',using=>',',from=>'Numeric'); - Context()->operators->set('='=>{string=>' = ',TeX=>'='}); - Context()->lists->set(List=>{separator=>" = "}); - $s=List("(x-1)/4","(y+2)/7","(z-3)/2"); - $sev=$s->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$dv;my$incpt;my%uses; - if($entry->class ne "Formula"){push(@err,"Your $ith quantity $entry is not a formula");next;} - else { - $uses{'x'}=$entry->usesOneOf("x"); - $uses{'y'}=$entry->usesOneOf("y"); - $uses{'z'}=$entry->usesOneOf("z"); - if($uses{'x'}+$uses{'y'}+$uses{'z'}!=1){push(@err,"Your $ith quantity is not a formula in only x, y, or z");next;}; - $dv=$entry->D('x')->eval(x=>0,y=>0,z=>0)+$entry->D('y')->eval(x=>0,y=>0,z=>0)+$entry->D('z')->eval(x=>0,y=>0,z=>0); - $incpt=$entry->eval(x=>0,y=>0,z=>0)/$dv; - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->class eq "Formula"){ - $uj{'x'}=$st->[$j]->usesOneOf("x"); - $uj{'y'}=$st->[$j]->usesOneOf("y"); - $uj{'z'}=$st->[$j]->usesOneOf("z"); - if(($uses{'x'} and $uj{'x'}) or ($uses{'y'} and $uj{'y'}) or ($uses{'z'} and $uj{'z'})){ - push(@err,"Your $ith quantity uses the same variable as a previous one")unless $aH->{isPreview}; - $used=1;last; - } - } - } - if(!$used){ - if($uses{'x'}){$sV=$sV+Formula("<t/$dv-$incpt,0,0>")}; - if($uses{'y'}){$sV=$sV+Formula("<0,t/$dv-$incpt,0>")}; - if($uses{'z'}){$sV=$sV+Formula("<0,0,t/$dv-$incpt>")}; - } - } - if(!$aH->{isPreview}){ - push(@err,"You need to provide more symmetric quantities") if $i<3; - push(@err,"You have given too many symmetric quantities") if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant) {$ds = Vector("$ds");} else {return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - - - - -

    - \ell is a line that passes through - P=(1,-2,3) and Q = (5,5,5). -

    - - - Give the right-hand side (without \ell(t)=) of a vector equation for \ell. - -

    - -

    - - - Give the parametric equations for \ell - (separated using commas). - -

    - -

    - - - Give the ymmetric equations for \ell. - (Enter DNE if they are not defined.) - -

    - -

    -     -     -     - - - - -

    - Passes through P=(0,1,2) and orthogonal to both -

    - -

    - \vec d_1=\la 2,-1,7\ra and \vec d_2=\la 7,1,3\ra. -

    -     - -

    - Answers can vary; - here the direction is given by \vec d_1\times \vec d_2: - vector: \ell(t) = \la 0,1,2\ra + t\la -10,43,9\ra -

    - -

    - parametric: x= -10t, y=1+43t, z = 2+9t -

    - -

    - symmetric: -x/10 =(y-1)/43 = (z-2)/9 -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(t=>"Real"); - $v=Compute("(5,1,9)+t<0,-1,0>"); - $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; - $ds=Formula($st)->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; - $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return ($par and $tch); - }); - Context("Numeric"); - Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); - parser::Assignment->Allow; - $p=List(Formula("x=5"),Formula("y=1-t"),Formula("z=9")); - $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{ - ($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - } - } - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - } - } - if(!$aH->{isPreview}){ - push(@err,"You need to provide more parametrizations")if $i<3; - push(@err,"You have given too many parametrizations")if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - - - -

    - \ell is a line that passes through - P=(5,1,9) and is orthogonal to both - \vec d_1=\la 1,0,1\ra and \vec d_2=\la 2,0,3\ra. -

    - - - Give the right-hand side (without \ell(t)=) of a vector equation for \ell. - -

    - -

    - - - Give parametric equations for \ell - (separated using commas). - -

    - -

    - - - Give symmetric equations for \ell. - (Enter DNE if they are not defined.) - -

    - -

    -     -     -     - - - - - Context("Vector");Context()->variables->are(t=>"Real"); - $v=Compute("(7,2,-1)+t<1,-1,2>"); - $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; - $ds=Formula($st)->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; - $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return($par and $tch); - }); - Context("Numeric");Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); - parser::Assignment->Allow; - $p=List(Formula("x=7+t"),Formula("y=2-t"),Formula("z=-1+2t")); - $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - }} - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - }} - if(!$aH->{isPreview}){push(@err,"You need to provide more parametrizations")if $i<3;push(@err,"You have given too many parametrizations")if $i>3;} - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return(3*($par and $tch),@err); - }); - Context("Numeric");Context()->variables->add(y=>'Real',z=>'Real'); - Context()->operators->redefine('=',using=>',',from=>'Numeric'); - Context()->operators->set('='=>{string=>' = ',TeX=>'='}); - Context()->lists->set(List=>{separator=>" = "}); - $s=List("x-7","2-y","(z+1)/2"); - $sev=$s->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector");Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$dv;my$incpt;my%uses; - if($entry->class ne "Formula"){push(@err,"Your $ith quantity $entry is not a formula");next;} - else{$uses{'x'}=$entry->usesOneOf("x");$uses{'y'}=$entry->usesOneOf("y");$uses{'z'}=$entry->usesOneOf("z"); - if($uses{'x'}+$uses{'y'}+$uses{'z'}!=1){push(@err,"Your $ith quantity is not a formula in only x, y, or z");next;}; - $dv=$entry->D('x')->eval(x=>0,y=>0,z=>0)+$entry->D('y')->eval(x=>0,y=>0,z=>0)+$entry->D('z')->eval(x=>0,y=>0,z=>0); - $incpt=$entry->eval(x=>0,y=>0,z=>0)/$dv;}; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->class eq "Formula"){ - $uj{'x'}=$st->[$j]->usesOneOf("x");$uj{'y'}=$st->[$j]->usesOneOf("y");$uj{'z'}=$st->[$j]->usesOneOf("z"); - if(($uses{'x'} and $uj{'x'}) or ($uses{'y'} and $uj{'y'}) or ($uses{'z'} and $uj{'z'})){push(@err,"Your $ith quantity uses the same variable as a previous one")unless $aH->{isPreview};$used=1;last;} - }} - if(!$used){ - if($uses{'x'}){$sV=$sV+Formula("<t/$dv-$incpt,0,0>")}; - if($uses{'y'}){$sV=$sV+Formula("<0,t/$dv-$incpt,0>")}; - if($uses{'z'}){$sV=$sV+Formula("<0,0,t/$dv-$incpt>")}; - }} - if(!$aH->{isPreview}){push(@err,"You need to provide more symmetric quantities") if $i<3;push(@err,"You have given too many symmetric quantities") if $i>3;} - $sV=Formula($sV);$ds=$sV->D('t')->reduce; - if($ds->isConstant) {$ds = Vector("$ds");} else {return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0);$tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return(3*($par and $tch),@err); - }); - - - -

    - \ell is a line that passes through the intersection of - \vec\ell_1(t)=\la 2,1,1\ra+t\la 5,1,-2\ra and \vec\ell_2(t)=\la -2,-1,2\ra+t\la 3,1,-1\ra, - and is orthogonal to both lines. -

    - - - Give the right-hand side (without \ell(t)=) of a vector equation for \ell. - -

    - -

    - - - Give parametric equations for \ell - (separated using commas). - -

    - -

    - - - Give symmetric equations for \ell. - (Enter DNE if they are not defined.) - -

    - -

    -     -     -     - - - - - Context("Vector");Context()->variables->are(t=>"Real"); - $v=Compute("(2,2,3)+t<5,-1,-3>"); - $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; - $ds=Formula($st)->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; - $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return($par and $tch); - }); - Context("Numeric");Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); - parser::Assignment->Allow; - $p=List(Formula("x=2+5t"),Formula("y=2-t"),Formula("z=3-3t")); - $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - }} - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - }} - if(!$aH->{isPreview}){push(@err,"You need to provide more parametrizations")if $i<3;push(@err,"You have given too many parametrizations")if $i>3;} - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return(3*($par and $tch),@err); - }); - Context("Numeric");Context()->variables->add(y=>'Real',z=>'Real'); - Context()->operators->redefine('=',using=>',',from=>'Numeric'); - Context()->operators->set('='=>{string=>' = ',TeX=>'='}); - Context()->lists->set(List=>{separator=>" = "}); - $s=List("(x-2)/5","-(y-2)","-(z-3)/3"); - $sev=$s->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector");Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$dv;my$incpt;my%uses; - if($entry->class ne "Formula"){push(@err,"Your $ith quantity $entry is not a formula");next;} - else{$uses{'x'}=$entry->usesOneOf("x");$uses{'y'}=$entry->usesOneOf("y");$uses{'z'}=$entry->usesOneOf("z"); - if($uses{'x'}+$uses{'y'}+$uses{'z'}!=1){push(@err,"Your $ith quantity is not a formula in only x, y, or z");next;}; - $dv=$entry->D('x')->eval(x=>0,y=>0,z=>0)+$entry->D('y')->eval(x=>0,y=>0,z=>0)+$entry->D('z')->eval(x=>0,y=>0,z=>0); - $incpt=$entry->eval(x=>0,y=>0,z=>0)/$dv;}; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->class eq "Formula"){ - $uj{'x'}=$st->[$j]->usesOneOf("x");$uj{'y'}=$st->[$j]->usesOneOf("y");$uj{'z'}=$st->[$j]->usesOneOf("z"); - if(($uses{'x'} and $uj{'x'}) or ($uses{'y'} and $uj{'y'}) or ($uses{'z'} and $uj{'z'})){push(@err,"Your $ith quantity uses the same variable as a previous one")unless $aH->{isPreview};$used=1;last;} - }} - if(!$used){ - if($uses{'x'}){$sV=$sV+Formula("<t/$dv-$incpt,0,0>")}; - if($uses{'y'}){$sV=$sV+Formula("<0,t/$dv-$incpt,0>")}; - if($uses{'z'}){$sV=$sV+Formula("<0,0,t/$dv-$incpt>")}; - }} - if(!$aH->{isPreview}){push(@err,"You need to provide more symmetric quantities") if $i<3;push(@err,"You have given too many symmetric quantities") if $i>3;} - $sV=Formula($sV);$ds=$sV->D('t')->reduce; - if($ds->isConstant) {$ds = Vector("$ds");} else {return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0);$tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return(3*($par and $tch),@err); - }); - - - -

    - \ell is a line that passes through the intersection of - \vec\ell_1(t)=\begin{cases}x\amp=t\\y\amp=-2+2t\\z\amp=1+t\end{cases} and \vec\ell_2(t)=\begin{cases}x\amp=2+t\\y\amp=2-t\\z\amp=3+2t\end{cases}, - and is orthogonal to both lines. -

    - - - Give the right-hand side (without \ell(t)=) of a vector equation for \ell. - -

    - -

    - - - Give parametric equations for \ell - (separated using commas). - -

    - -

    - - - Give symmetric equations for \ell. - (Enter DNE if they are not defined.) - -

    - -

    -     -     -     - - - - -

    - Passes through P=(1,1), - parallel to \vec d = \la 2,3\ra. -

    -     - -

    - vector: \ell(t) = \la 1,1\ra + t\la 2,3\ra -

    - -

    - parametric: x= 1+2t, y=1+3t -

    - -

    - symmetric: (x-1)/2=(y-1)/3 -

    -     - -     - - - - - Context("Vector");Context()->variables->are(t=>"Real"); - $v=Compute("(-2,5)+t<0,1>"); - $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; - $ds=Formula($st)->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; - $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return($par and $tch); - }); - Context("Numeric");Context()->variables->add(y=>"Real",t=>"Real"); - parser::Assignment->Allow; - $p=List(Formula("x=-2"),Formula("y=5+t")); - $pev=$p->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0)+t*<0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y")) or $for->usesOneOf("x","y")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - }} - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0>")} - else{$sV=$sV+Formula("<0,$for>")}; - }} - if(!$aH->{isPreview}){push(@err,"You need to provide more parametrizations")if $i<2;push(@err,"You have given too many parametrizations")if $i>2;} - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return(2*($par and $tch),@err); - }); - - - -

    - \ell is a line that passes through P=(-2,5), - parallel to \vec d = \la 0,1\ra. -

    - - - Give the right-hand side (without \ell(t)=) of a vector equation for \ell. - -

    - -

    - - - Give parametric equations for \ell - (separated using commas) - -

    - -

    - - - Give symmetric equations for \ell. - (Enter DNE if they are not defined.) - -

    - -

    -     -     -     - -     - - - -

    - Determine if the described lines are the same line, - parallel lines, intersecting or skew lines. - If intersecting, give the point of intersection. -

    -     - - - - - Context("Point"); - Context()->strings->add(same=>{},parallel=>{},skew=>{}); - $answer=Compute("parallel"); - - - -

    - \vec\ell_1(t) = \la 1,2,1\ra + t\la 2,-1,1\ra and - \vec\ell_2(t) = \la 3,3,3\ra + t\la -4,2,-2\ra. -

    - - You can answer with same, - parallel, - skew, - or give the point of intersection. - - -

    - -

    -     -     -     - - - - - Context("Point"); - Context()->strings->add(same=>{},parallel=>{},skew=>{}); - $answer=Compute("(12,3,7)"); - - - -

    - \vec\ell_1(t) = \la 2,1,1\ra + t\la 5,1,3\ra and - \vec\ell_2(t) = \la 14,5,9\ra + t\la 1,1,1\ra. -

    - - You can answer with same, - parallel, - skew, - or give the point of intersection. - - -

    - -

    -     -     -     - - - - -

    - \vec\ell_1(t) = \la 3,4,1\ra + t\la 2,-3,4\ra, -

    - -

    - \vec\ell_2(t) = \la -3,3,-3\ra + t\la 3,-2,4\ra. -

    -     - -

    - intersecting; - \vec\ell_1(3) = \vec\ell_2(4) = \la 9,-5,13\ra -

    -     - -     - - - - - Context("Point"); - Context()->strings->add(same=>{},parallel=>{},skew=>{}); - $answer=Compute("same"); - - - -

    - \vec\ell_1(t) = \la 1,1,1\ra + t\la 3,1,3\ra and - \vec\ell_2(t) = \la 7,3,7\ra + t\la 6,2,6\ra. -

    - - You can answer with same, - parallel, - skew, - or give the point of intersection. - - -

    - -

    -     -     -     - - - - - Context("Point"); - Context()->strings->add(same=>{},parallel=>{},skew=>{}); - $answer=Compute("skew"); - - - -

    - \vec\ell_1(t) = \begin{cases}x\amp = 1+2t\\ y\amp = 3-2t\\ z\amp = t\end{cases} and - \vec\ell_2(t) = \begin{cases}x\amp = 3-t\\ y\amp = 3+5t\\ z\amp = 2+7t\end{cases}. -

    - - You can answer with same, - parallel, - skew, - or give the point of intersection. - - -

    - -

    -     -     -     - - - - - Context("Point"); - Context()->strings->add(same=>{},parallel=>{},skew=>{}); - $answer=Compute("parallel"); - - - -

    - \vec\ell_1(t) = \begin{cases}x\amp = 1.1+0.6t\\ y\amp = 3.77+0.9t\\ z\amp = -2.3+1.5t\end{cases} and - \vec\ell_2(t) = \begin{cases}x\amp = 3.11+3.4t\\ y\amp = 2+5.1t\\ z\amp = 2.5+8.5t\end{cases}. -

    - - You can answer with same, - parallel, - skew, - or give the point of intersection. - - -

    - -

    -     -     -     - - - - -

    - \ell_1 = \left\{\begin{aligned}x\amp = 0.2+0.6t\\ y\amp = 1.33-0.45t\\ z\amp = -4.2+1.05t - \end{aligned} \right. and - \ell_2 = \left\{\begin{aligned}x\amp = 0.86+9.2t\\ y\amp = 0.835-6.9t\\ z\amp = -3.045+16.1t - \end{aligned} \right. -

    -     - -

    - same -

    -     - -     - - - - - Context("Point"); - Context()->strings->add(same=>{},parallel=>{},skew=>{}); - $answer=Compute("skew"); - - - -

    - \vec\ell_1(t) = \begin{cases}x\amp = 0.1+1.1t\\ y\amp = 2.9-1.5t\\ z\amp = 3.2+1.6t\end{cases} and - \vec\ell_2(t) = \begin{cases}x\amp = 4-2.1t\\ y\amp = 1.8+7.2t\\ z\amp = 3.1+1.1t\end{cases}. -

    - - You can answer with same, - parallel, - skew, - or give the point of intersection. - - -

    - -

    -     -     -     - -     - - - -

    - Find the distance from the point to the line. -

    -     - - - - -

    - Q=(1,1,1), \vec\ell(t) = \la 2,1,3\ra + t\la 2,1,-2\ra -

    -     - -

    - \sqrt{41}/3 -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $distance=Formula("3sqrt(2)"); - - - -

    - Find the distance from the point - Q=(2,5,6) to the line \vec\ell(t) = \la -1,1,1\ra + t\la 1,0,1\ra. -

    - -

    - -

    -     -     -     - - - - -

    - Q=(0,3), \vec\ell(t) = \la 2,0\ra + t\la 1,1\ra -

    -     - -

    - 5\sqrt{2}/2 -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $distance=Formula("5"); - - - -

    - Find the distance from the point Q=(1,1) to the line \vec\ell(t) = \la 4,5\ra + t\la -4,3\ra. -

    - -

    - -

    -     -     -     - -     - - - -

    - Find the distance between the two lines. -

    -     - - - - -

    - \vec\ell_1(t) = \la 1,2,1\ra + t\la 2,-1,1\ra, -

    - -

    - \vec\ell_2(t) = \la 3,3,3\ra + t\la 4,2,-2\ra. -

    -     - -

    - 3/\sqrt{2} -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $distance=Formula("2"); - - - -

    - Find the distance between the line - \vec\ell_1(t) = \la 0,0,1\ra + t\la 1,0,0\ra and the line \vec\ell_2(t) = \la 0,0,3\ra + t\la 0,1,0\ra. -

    - -

    - -

    -     -     -     - -     - - - -

    - The following exercises explore special cases of the distance formulas found in . -

    -     - - - - -

    - Let Q be a point on the line \vec\ell(t). - Show why the distance formula correctly gives the distance from the point to the line as 0. -

    -     - -

    - Since both P and Q are on the line, - \overrightarrow{PQ} is parallel to \vec d. - Thus \overrightarrow{PQ}\times \vec d = \vec 0, - giving a distance of 0. -

    -     - -     - - - - -

    - Let lines \vec\ell_1(t) and - \vec\ell_2(t) be intersecting lines. - Show why the distance formula correctly gives the distance between these lines as 0. -

    -     - -

    - (Note: this solution is easier once one has studied .) Since the two lines intersect, - we can state P_2= P_1 + a\vec d_1+b\vec d_2 for some scalars a and b. - (Here we abuse notation slightly and add points to vectors.) - Thus \overrightarrow{P_1P_2} = a\vec d_1+b\vec d_2. - Vector \vec c is the cross product of \vec d_1 and \vec d_2, - hence is orthogonal to both, and hence is orthogonal to \overrightarrow{P_1P_2}. - Thus \overrightarrow{P_1P_2}\cdot\vec c = 0, - and the distance between lines is 0. -

    -     - -     - - - - -

    - Let lines \vec\ell_1(t) and \vec\ell_2(t) be parallel. -

    -     - - - -

    - Show why the distance formula for distance between lines cannot be used as stated to find the distance between the lines. -

    -     - -

    - The distance formula cannot be used because since - \vec d_1 and \vec d_2 are parallel, - \vec c is \vec 0 and we cannot divide by \vnorm{0}. -

    -     -     - - - -

    - Show why letting \vec c=(\overrightarrow{P_1P_2}\times\vec d_2)\times\vec d_2 allows one to the use the formula. -

    -     - -

    - Since \vec d_1 and \vec d_2 are parallel, - \overrightarrow{P_1P_2} lies in the plane formed by the two lines. - Thus \overrightarrow{P_1P_2}\times\vec d_2 is orthogonal to this plane, - and \vec c=(\overrightarrow{P_1P_2}\times\vec d_2)\times \vec d_2 is parallel to the plane, - but still orthogonal to both \vec d_1 and \vec d_2. - We desire the length of the projection of \overrightarrow{P_1P_2} onto \vec c, - which is what the formula provides. -

    -     -     - - - -

    - Show how one can use the formula for the distance between a point and a line to find the distance between parallel lines. -

    -     - -

    - Since the lines are parallel, - one can measure the distance between the lines at any location on either line - (just as to find the distance between straight railroad tracks, - one can use a measuring tape anywhere along the track, - not just at one specific place.) - Let P=P_1 and Q=P_2 as given by the equations of the lines, - and apply the formula for distance between a point and a line. -

    -     -     - -     -     -     -     -
    -
    - Planes - -

    - Any flat surface, such as a wall, - table top or stiff piece of cardboard can be thought of as representing part of a plane. - Consider a piece of cardboard with a point P marked on it. - One can take a nail and stick it into the cardboard at P such that the nail is perpendicular to the cardboard; - see . -

    - -
    - Illustrating defining a plane with a sheet of cardboard and a nail - - - - A nail sticks into a flat, rectangular sheet of cardboard. - -

    - A flat, rectangular surface represents a sheet of cardboard. - The image of a nail is drawn sticking into the sheet, at a point P, which is marked. - The nail is standing upright, in a position perpendicular to the sheet of cardboard. -

    -     - - - - - //ASY file for figlines3.asy in Chapter 10 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((-4,27,7),(.012,-0.002,0.015),(0,0,0),.9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2,2); - - xaxis3("",xbounds.x,xbounds.y,invisible,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,invisible,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,invisible,OutTicks(myzchoice),Arrow3(size=3mm)); - - defaultpen(0.5mm); - - real f(pair z) {return 0;} - surface s=surface(f,(-2,-2),(2,2),1,1); - pen p=bluepen+.3mm; - draw(s,surfacepen,meshpen=p,render(merge=true)); - - draw(scale(.1,.1,.2)*rotate(180,Y)*shift(-1Z)*unitcone,surfacepen=white); - draw(scale(.1,.1,.5)*shift(.4Z)*unitcylinder,surfacepen=white); - draw(scale(.15,.15,.05)*shift(14Z)*unitcylinder,surfacepen=white); - - //real f(pair z) {return 0;} - //surface s=surface(f,(-2,-2),(2,2)); - //draw(s,white,meshpen=p,render(merge=true)); - draw(scale3(.15)*shift(4.7Z)*unitdisk,surfacepen=white); - draw(scale3(.15)*shift(5Z)*unitdisk,surfacepen=white); - - draw(scale3(.05)*unitdisk); - label("$P$",(0,0,0),NW); - - //draw((0,0,.9)--(0,0,0),gray+1mm,Arrow3(size=2mm)); - //draw((0,0,.9)--(0,0,1),gray+1.2mm); - - - - -
    - -

    - This nail provides a handle for the cardboard. - Moving the cardboard around moves P to different locations in space. - Tilting the nail - (but keeping P fixed) - tilts the cardboard. - Both moving and tilting the cardboard defines a different plane in space. - In fact, we can define a plane by: 1) the location of P in space, - and 2) the direction of the nail. -

    - - - -

    - The previous section showed that one can define a line given a point on the line and the direction of the line - (usually given by a vector). - One can make a similar statement about planes: - we can define a plane in space given a point on the plane and the direction the plane faces - (using the description above, - the direction of the nail). - Once again, the direction information will be supplied by a vector, - called a normal vector, - that is orthogonal to the plane. - vectorsnormal vector - normal vector - planesnormal vector -

    - -

    - What exactly does orthogonal to the plane mean? - Choose any two points P and Q in the plane, - and consider the vector \overrightarrow{PQ}. - We say a vector \vec n is orthogonal to the plane if \vec n is perpendicular to \overrightarrow{PQ} for all choices of P and Q; - that is, if \vec n\cdot \overrightarrow{PQ}=0 for all P and Q. -

    - -

    - This gives us way of writing an equation describing the plane. - Let P=(x_0,y_0,z_0) be a point in the plane and let - \vec n = \la a,b,c\ra be a normal vector to the plane. - A point Q = (x,y,z) lies in the plane defined by P and \vec n if, - and only if, \overrightarrow{PQ} is orthogonal to \vec n. - Knowing \overrightarrow{PQ} = \la x-x_0,y-y_0,z-z_0\ra, consider: - - \overrightarrow{PQ}\cdot\vec n \amp = 0 - \la x-x_0,y-y_0,z-z_0\ra\cdot \la a,b,c\ra \amp =0 - a(x-x_0)+b(y-y_0)+c(z-z_0) \amp =0 - . - Equation defines an implicit function describing the plane. More algebra produces: - - ax+by+cz = ax_0+by_0+cz_0 - . - The right hand side is just a number, so we replace it with d: - - ax+by+cz = d - . - As long as c\neq 0, we can solve for z: - - z = \frac1c(d-ax-by) - . -

    - -

    - Equation is especially useful as many computer programs can graph functions in this form. - Equations and have specific names, - given next. -

    - - - Equations of a Plane in Standard and General Forms - -

    - The plane passing through the point P=(x_0,y_0,z_0) with normal vector - \vec n=\la a,b,c\ra can be described by an equation with - standard form - - a(x-x_0)+b(y-y_0)+c(z-z_0) =0; - - the equation's general form - is planesequations of - - ax+by+cz = d - . -

    -     -     - - - -

    - A key to remember throughout this section is this: - to find the equation of a plane, - we need a point and a normal vector. - We will give several examples of finding the equation of a plane, - and in each one different types of information are given. - In each case, - we need to use the given information to find a point on the plane and a normal vector. -

    - - - Finding the equation of a plane - -

    - Write the equation of the plane that passes through the points P=(1,1,0), - Q = (1,2,-1) and R = (0,1,2) in standard form. -

    -     - -

    - We need a vector \vec n that is orthogonal to the plane. - Since P, Q and R are in the plane, - so are the vectors \overrightarrow{PQ} and \overrightarrow{PR}; - \overrightarrow{PQ}\times\overrightarrow{PR} is orthogonal to \overrightarrow{PQ} and \overrightarrow{PR} and hence the plane itself. -

    - -

    - It is straightforward to compute \vec n = \overrightarrow{PQ}\times\overrightarrow{PR} = \la 2,1,1\ra. - We can use any point we wish in the plane - (any of P, Q or R will do) - and we arbitrarily choose P. - Following , - the equation of the plane in standard form is - - 2(x-1) + (y-1)+z = 0 - . -

    - -

    - The plane is sketched in . -

    - -
    - Sketching the plane in - - - - Three points P, Q, and R are plotted in space, along with the plane containing them. - -

    - In a three-dimensional coordinate system, three points P, Q, and R are plotted. - Three vectors are drawn with their tails at P: - \overrightarrow{PQ}, \overrightarrow{PR}, and the cross product \overrightarrow{PQ}\times\overrightarrow{PR}. - The plane containing P, Q, and R is also shown. - The vectors \overrightarrow{PQ} and \overrightarrow{PR} are parallel to the plane, - while their cross product is perpendicular to the plane. -

    -     - - - - - //ASY file for figlines_dist2.asy in Chapter 10 - //ASY file for figplanes1.asy in Chapter 10 - - //size(200,200,IgnoreAspect); - size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((1.92,19.3,4),(0,-0.015,0.06),(0,0,0),1); - defaultrender.merge=true; - - - // setup and draw the axes - real[] myxchoice={-4,4}; - real[] myychoice={-4,4}; - real[] myzchoice={-4,4}; - defaultpen(0.5mm); - pair xbounds=(-3.5,3.5); - pair ybounds=(-4.5,4.5); - pair zbounds=(-4.5,6.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0),S); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane - triple f(pair t) { - return (t.x,t.y,-2*t.x-t.y+3); - } - surface s=surface(f,(-1,-2),(2,3),8,8,Spline); - pen p=apexmeshpen+.2mm; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - //Draw points P=(1,1,0), Q=(1,2,-1), R=(0,1,2) - dotfactor=3; - dot((1,1,0));label("$P$",(1,1,0),N); - dot((1,2,-1));label("$Q$",(1,2,-1),W); - dot((0,1,2));label("$R$",(0,1,2),W); - //Draw Vectors - draw((1,1,0)--(1,2,-1),bluepen,Arrow3(size=2mm));//PQ - draw((1,1,0)--(0,1,2),bluepen,Arrow3(size=2mm));//PR - draw((1,1,0)--(3,2,1),bluepen,Arrow3(size=2mm));//P to PQxPR - label("$\overrightarrow{PQ}\times \overrightarrow{PR}$",(3,2,1),N); - - //test i,j,k moved to P - //draw((1,1,0)--(2,1,0),bluepen,Arrow3(size=2mm));//P by i - //draw((1,1,0)--(1,2,0),bluepen,Arrow3(size=2mm));//P by j - //draw((1,1,0)--(1,1,1),bluepen,Arrow3(size=2mm));//P by k - - // Use lines L1 1+3t,2-t,t and L2 -2+4t,3+t,5+2t - //draw P1 (t=0) and vector d1 at P1 (t=1) - //dotfactor=3;dot((1,2,0));label("$P_1$",(1,2,0),N); - //draw((1,2,0)--(4,1,1),redpen,Arrow3(size=2mm)); - //label("$\vec{d}_1$",(4,1,1),N); - //draw P2 (t=-1.5) and vector d2 at P2 (t=-0.5) - //dotfactor=3;dot((-8,1.5,2));label("$P_2$",(-8,1.5,2),N); - //draw((-8,1.5,2)--(-4,2.5,4),redpen,Arrow3(size=2mm)); - //label("$\vec{d}_2$",(-4,2.5,4),N); - - //draw vector P1 to P2 - //draw((1,2,0)--(-8,1.5,2),black,Arrow3(size=2mm)); - //label("$\overrightarrow{P_1 P_2}$",(-3.5,1.75,1),W); - - //draw the lines 1+3t,2-t,t and -2+4t,3+t,5+2t - //draw((-5,4,-2)--(7,0,2),bluepen);//L1 - //draw((-14,0,-1)--(2,4,7),bluepen);//L2 - - - - -
    -     - -     - -

    - We have just demonstrated the fact that any three non-collinear points define a plane. - (This is why a three-legged stool does not rock; - it's three feet always lie in a plane. - A four-legged stool will rock unless all four feet lie in the same plane.) -

    - - - Finding the equation of a plane - -

    - Verify that lines \ell_1 and \ell_2, - whose parametric equations are given below, intersect, - then give the equation of the plane that contains these two lines in general form. - - \ell_1: \begin{matrix} x\amp =\amp -5+2s \\ y\amp =\amp 1+s \\ z\amp =\amp -4+2s \end{matrix} \qquad\qquad \ell_2: \begin{matrix} x \amp =\amp 2+3t\\ y\amp =\amp 1-2t \\ z\amp =\amp 1+t \end{matrix} - -

    -     - -

    - The lines clearly are not parallel. - If they do not intersect, they are skew, - meaning there is not a plane that contains them both. - If they do intersect, there is such a plane. -

    - -

    - To find their point of intersection, we set the x, - y and z equations equal to each other and solve for s and t: - - \begin{matrix} -5+2s \amp =\amp 2+3t \\ 1+s \amp =\amp 1-2t \\ -4+2s \amp =\amp 1+t \end{matrix} \Rightarrow s=2, t=-1 - . -

    - -

    - When s=2 and t=-1, - the lines intersect at the point P= (-1,3,0). -

    - -

    - Let \vec d_1 = \la 2,1,2\ra and - \vec d_2=\la 3,-2,1\ra be the directions of lines \ell_1 and \ell_2, - respectively. - A normal vector to the plane containing these the two lines will also be orthogonal to \vec d_1 and \vec d_2. - Thus we find a normal vector \vec n by computing \vec n = \vec d_1 \times \vec d_2= \la 5,4-7\ra. -

    - -

    - We can pick any point in the plane with which to write our equation; - each line gives us infinite choices of points. - We choose P, the point of intersection. - We follow - to write the plane's equation in general form: - - 5(x+1) +4(y-3) -7z \amp = 0 - 5x + 5 + 4y-12 -7z \amp = 0 - 5x+4y-7z \amp = 7 - . -

    - -

    - The plane's equation in general form is 5x+4y-7z=7; - it is sketched in . -

    - -
    - Sketching the plane in - - - - In three dimensions, two lines are plotted, intersecting at a point P, along with the plane containing both lines. - -

    - Three-dimensional coordinate axes are shown, with the points (5,0,0), (0,5,0), and (0,0,-5) indicated for scale. - A point P is plotted in space. Two lines \ell_1 and \ell_2 are shown intersecting at this point. - The plane containing both lines is also plotted, along with a normal vector, located with its tail at P. -

    -     - - - - - //ASY file for figplanes2.asy in Chapter 10 - - size(282,282,Aspect); - //size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((-16,18.6,6.76),(0.014,-0.015,0077),(0,0,0),.99); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={5}; - real[] myychoice={5}; - real[] myzchoice={-5}; - defaultpen(0.5mm); - pair xbounds=(-4,6); - pair ybounds=(-1,6); - pair zbounds=(-6,5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane - triple f(pair t) { - return (t.x,t.y,(5/7)*t.x+(4/7)*t.y-1); - } - surface s=surface(f,(-5,-6),(5,6),8,8,Spline); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - //Draw points P=(-1,3,0) - dotfactor=3;dot((-1,3,0));label("$P$",(-1,3,0),E); - - //Draw normal vector at P, n=(5,4,-7) - draw((-1,3,0)--(4,7,-7),black,Arrow3(size=2mm));//n at P - label("$\vec{n}$",(4,7,-7),W); - - //draw the lines -5+2t,1+t,-4+2t and 2+3t,1-2t,1+t - draw((5,6,6)--(-5,1,-4),bluepen);label("$\ell_1$",(5,6,6),N);//L1 - draw((5,-1,2)--(-4.9,5.6,-1.3),bluepen);label("$\ell_2$",(5,-1,2),N);//L2 - - - - -
    -     -     - - - Finding the equation of a plane - -

    - Give the equation, in standard form, - of the plane that passes through the point - P=(-1,0,1) and is orthogonal to the line with vector equation \vec \ell(t) = \la -1,0,1\ra + t\la 1,2,2\ra. -

    -     - -

    - As the plane is to be orthogonal to the line, - the plane must be orthogonal to the direction of the line given by \vec d = \la 1,2,2\ra. - We use this as our normal vector. - Thus the plane's equation, in standard form, is - - (x+1) +2y+2(z-1)=0 - . -

    - -

    - The line and plane are sketched in . -

    - -
    - The line and plane in - - - - A plane is plotted in a three-dimensional coordinate system, along with a line passing through it at a point P. - -

    - A plane is plotted against a set of three-dimensional coordinate axes. - A line passes through the plane, intersecting it at a point P. - The line is perpendicular to the plane. -

    -     - - - - - //ASY file for figplanes3.asy in Chapter 10 - - size(282,282,Aspect); - //size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-3,3}; - real[] myychoice={-3,3}; - real[] myzchoice={-3,3}; - defaultpen(0.5mm); - pair xbounds=(-4,4); - pair ybounds=(-4,4); - pair zbounds=(-4,4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane z=-(x+1)/2-y+1 - triple f(pair t) { - return (t.x,t.y,(-1/2)*(t.x+1)-t.y+1); - } - surface s=surface(f,(-3,-3),(3,3),8,8,Spline); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - //Draw points P=(-1,0,1) - dotfactor=3;dot((-1,0,1));label("$P$",(-1,0,1),N); - - //draw the line -1+t,2t,1+2t - draw((1,4,5)--(-3,-4,-3),bluepen);//L t=2 to t=-2 - - //Draw normal vector at P, n=(1,2,2) - //draw((-1,3,0)--(4,7,-7),black,Arrow3(size=2mm));//n at P - //label("$\vec{n}$",(4,7,-7),W); - - - - -
    -     - -     - - - Finding the intersection of two planes - -

    - Give the parametric equations of the line that is the intersection of the planes p_1 and p_2, where: - - p_1: x-(y-2)+(z-1) =0 - p_2: -2(x-2)+(y+1)+(z-3)=0 - -

    -     - -

    - To find an equation of a line, - we need a point on the line and the direction of the line. -

    - -

    - We can find a point on the line by solving each equation of the planes for z: - - p_1: z = -x+y-1 - p_2: z = 2x-y-2 - -

    - -

    - We can now set these two equations equal to each other (, we are finding values of x and y where the planes have the same z value): - - -x+y-1 \amp = 2x-y-2 - 2y \amp = 3x-1 - y \amp = \frac12(3x-1) - -

    - -

    - We can choose any value for x; - we choose x=1. - This determines that y=1. - We can now use the equations of either plane to find z: - when x=1 and y=1, - z=-1 on both planes. - We have found a point P on the line: P= (1,1,-1). -

    - -

    - We now need the direction of the line. - Since the line lies in each plane, - its direction is orthogonal to a normal vector for each plane. - Considering the equations for p_1 and p_2, - we can quickly determine their normal vectors. - For p_1, - \vec n_1 = \la 1,-1,1\ra and for p_2, - \vec n_2 = \la -2,1,1\ra. - A direction orthogonal to both of these directions is their cross product: - \vec d = \vec n_1\times \vec n_2 = \la -2,-3,-1\ra. -

    - -

    - The parametric equations of the line through - P=(1,1,-1) in the direction of d=\la -2,-3,-1\ra is: - - \ell: x= -2t+1 y = -3t+1 z=-t-1 - . -

    - -

    - The planes and line are graphed in . -

    - -
    - Graphing the planes and their line of intersection in - - - - Two planes are shown intersecting along a common line. - -

    - Two planes are plotted in a three-dimensional coordinate system. - The planes intersect along a line, which is also plotted, - along with a point P that lies on the line, as well as on both planes. -

    -     - - - - - //ASY file for figplanes43D.asy in Chapter 10 - - //size(282,282,Aspect); - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8,14.7,14); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={-5}; - defaultpen(0.5mm); - pair xbounds=(-3,3); - pair ybounds=(-3,3); - pair zbounds=(-6,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the planes z=-x+y-1 and z=2x-y-2 - triple f(pair t) { - return (t.x,t.y,-t.x+t.y-1); - } - surface s=surface(f,(-3,-3),(3,3),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p+.2mm,nolight,render(merge=true)); - triple f(pair t) { - return (t.x,t.y,2*t.x-t.y-2); - } - surface s=surface(f,(-3,-3),(3,3),8,8,Spline); - pen pp=apexmeshpen+.2mm; - //draw(s,rgb(1,.4,.7)+opacity(.7),meshpen=pp,nolight,render(merge=true)); // HOT PINK - - pen q=redcurvepen+.1mm; - draw(s,surfacepen2,meshpen=q,nolight,render(merge=true)); // better red - - //Draw point P=(1,1,-1) - dotfactor=3;dot((1,1,-1));label("$P$",(1,1,-1),N); - - //draw the line 1-2t,1-3t,-1-t - draw((4,5.5,0.5)--(-2,-3.5,-2.5),bluepen);//L t=-1.5 to t=1.5 - - - - -
    - -     - -     - - - Finding the intersection of a plane and a line - -

    - Find the point of intersection, if any, - of the line \ell(t) = \la 3,-3,-1\ra +t\la-1,2,1\ra and the plane with equation in general form 2x+y+z=4. -

    -     - -

    - The equation of the plane shows that the vector - \vec n = \la 2,1,1\ra is a normal vector to the plane, - and the equation of the line shows that the line moves parallel to \vec d = \la -1,2,1\ra. - Since these are not orthogonal, - we know there is a point of intersection. - (If there were orthogonal, - it would mean that the plane and line were parallel to each other, - either never intersecting or the line was in the plane itself.) -

    - -

    - To find the point of intersection, - we need to find a t value such that \ell(t) satisfies the equation of the plane. - Rewriting the equation of the line with parametric equations will help: - - \ell(t) = \left\{\begin{aligned}x\amp = 3-t\\ y\amp =-3+2t\\ z\amp = -1+t \end{aligned} \right. - . -

    - -

    - Replacing x, - y and z in the equation of the plane with the expressions containing t found in the equation of the line allows us to determine a t value that indicates the point of intersection: - - 2x+y+z \amp =4 - 2(3-t) + (-3+2t) + (-1+t) \amp = 4 - t\amp =2 - . -

    - -

    - When t=2, - the point on the line satisfies the equation of the plane; - that point is \ell(2) = \la 1,1,1\ra. - Thus the point (1,1,1) is the point of intersection between the plane and the line, - illustrated in . -

    - -
    - Illustrating the intersection of a line and a plane in - - - - A line in three dimensions passes through a plane, intersecting it at a point. - -

    - A plane is plotted against a three-dimensional coordinate system. - A line, labeled \ell(t), is also plotted. - The line appears to be almost, but not quite, parallel to the plane. - It passes through the plane, intersecting it at a point that is plotted, but not labeled. -

    -     - - - - - //ASY file for figplanes53D.asy in Chapter 10 - - //size(282,282,Aspect); - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-3,4); - pair ybounds=(-3,4); - pair zbounds=(-2,7); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane z=4-2x-y - triple f(pair t) { - return (t.x,t.y,4-2*t.x-t.y); - } - surface s=surface(f,(-2,-2),(3,3),8,8,Spline); - pen p=apexmeshpen+.2mm; - draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - //Draw point P=(1,1,1) - dotfactor=3;dot((1,1,1),redpen);//label("$P$",(1,1,1),N); - - //draw the line 3-t,-3+2t,-1+t - draw((3,-3,-1)--(0,3,2),redpen);//L t=0 to t=3 - label("$\ell(t)$",(3,-3,-1),N); - - - - -
    -     - -     -     - - - Distances -

    - Just as it was useful to find distances between points and lines in the previous section, - it is also often necessary to find the distance from a point to a plane. -

    - -

    - Consider , - where a plane with normal vector \vec n is sketched containing a point P and a point Q, - not on the plane, is given. - We measure the distance from Q to the plane by measuring the length of the projection of \overrightarrow{PQ} onto \vec n. - That is, we want: - - \snorm{\text{ proj } _{\,\vec n}\,{\overrightarrow{PQ}}} = \snorm{\frac{\vec n\cdot \overrightarrow{PQ}}{\vnorm n^2}\vec n} = \frac{\abs{\vec n\cdot \overrightarrow{PQ}}}{\vnorm n} - -

    - - - -

    - Equation is important as it does more than just give the distance between a point and a plane. - We will see how it allows us to find several other distances as well: - the distance between parallel planes and the distance from a line and a plane. - Because Equation is important, - we restate it as a Key Idea. -

    - -
    - Illustrating finding the distance from a point to a plane - - - - A plane, on which a point P and normal vector n are marked, along with a point Q not on the plane, and its distance from the plane. - -

    - A generic plane is shown, along with a point P on the plane. - The normal vector \vec n is plotted with its tail at P. - A point Q is shown above the plane, and the vector \overrightarrow{PQ} from P to Q is drawn. - A perpendicular is drawn from the point Q to the plane, and labeled with the distance h. -

    -     - - - - - //ASY file for figplanes_dist3D.asy in Chapter 10 - - size(282,282,Aspect); - //size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4.5,4.5,1.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-4,4}; - real[] myychoice={-4,4}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-5,5); - pair ybounds=(-5,5); - pair zbounds=(-5,5); - - //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane xy-plane - triple f(pair t) { - return (t.x,t.y,0); - } - surface s=surface(f,(-2,-1),(2,1.5),1,1,Spline); - pen p=blackmeshpen; - //pen p=apexmeshpen; - draw(s,rgb(1,1,1)+opacity(0),meshpen=p+thick(),nolight,render(merge=true)); - //draw(s,surfacepen,meshpen=p,nolight,render(merge=true)); - - //Draw origin point and P=(1,1,1) - dotfactor=3;dot((0,0,0),black);dot((0,1,0),black);label("$P$",(0,1,0),E); - - //Draw the normal at P - draw((0,1,0)--(0,1,1),linewidth(.75),Arrow3(size=2mm));label("$\vec{n}$",(0,1,1),E); - - //Draw the dashed line from origin to Q=(0,0,2) with label h - draw((0,0,0)--(0,0,2),dashed+linewidth(.75));dot((0,0,2),black);label("$Q$",(0,0,2),W); - label("$h$",(0,0,1),W); - - //Draw the vector PQ stopping just short of Q - draw((0,1,0)--(0,.05,1.95),linewidth(.75),Arrow3(size=2mm));//label("$\vec{n}$",(0,1,1),E); - - //Draw the perpendicular symbol as two small lines - draw((0,0,0.2)--(0,0.2,0.2),linewidth(.75)); - draw((0,0.2,0)--(0,0.2,0.2),linewidth(.75)); - - - - -
    - - - Distance from a Point to a Plane -

    - Let a plane with normal vector \vec n be given, - and let Q be a point. - The distance h from Q to the plane is - - h = \frac{\abs{\vec n\cdot \overrightarrow{PQ}}}{\vnorm n} - , - where P is any point in the plane. - planesdistance between point and plane - distancebetween point and plane -

    -     - - - - - Distance between a point and a plane - -

    - Find the distance between the point - Q = (2,1,4) and the plane with equation 2x-5y+6z=9. -

    -     - -

    - Using the equation of the plane, - we find the normal vector \vec n = \la 2,-5,6\ra. - To find a point on the plane, - we can let x and y be anything we choose, - then let z be whatever satisfies the equation. - Letting x and y be 0 seems simple; - this makes z = 1.5. - Thus we let P = \la 0,0,1.5\ra, - and \overrightarrow{PQ} = \la 2,1,2.5\ra. -

    - -

    - The distance h from Q to the plane is given by : - - h \amp = \frac{\abs{\vec n\cdot \overrightarrow{PQ}}}{\vnorm n} - \amp = \frac{\abs{\la 2,-5,6\ra \cdot \la 2,1,2.5\ra}}{\norm{\la 2,-5,6\ra}} - \amp = \frac{ \abs{14}}{\sqrt{65}} - \amp \approx 1.74 - . -

    -     - -     - -

    - We can use to find other distances. - Given two parallel planes, - we can find the distance between these planes by letting P be a point on one plane and Q a point on the other. - If \ell is a line parallel to a plane, - we can use the Key Idea to find the distance between them as well: - again, let P be a point in the plane and let Q be any point on the line. - (One can also use .) - The Exercise section contains problems of these types. -

    - -

    - These past two sections have not explored lines and planes in space as an exercise of mathematical curiosity. - However, there are many, many applications of these fundamental concepts. - Complex shapes can be modeled (or, - approximated) using planes. - For instance, - part of the exterior of an aircraft may have a complex, - yet smooth, shape, - and engineers will want to know how air flows across this piece as well as how heat might build up due to air friction. - Many equations that help determine air flow and heat dissipation are difficult to apply to arbitrary surfaces, - but simple to apply to planes. - By approximating a surface with millions of small planes one can more readily model the needed behavior. -

    -     - - - - Terms and Concepts - - - -

    - In order to find the equation of a plane, - what two pieces of information must one have? -

    -     - - - -

    - A point in the plane and a normal vector (, a direction orthogonal to the plane). -

    -     - -     - - - - -

    - What is the relationship between a plane and one of its normal vectors? -

    -     - - - -

    - A normal vector is orthogonal to the plane. -

    -     - -     -     - - Problems - - - -

    - Give any two points in the given plane. -

    -     - - - - -

    - 2x-4y+7z=2 -

    -     - -

    - Answers will vary. -

    -     - -     - - - - - Context("Point"); - $points=List("(-2,9,0),(2,9,3)"); - $pev=$points->cmp(list_checker => sub { - my ($correct,$student,$ansHash,$value) = @_; - my $n = scalar(@$student); - my $score = 0; - my @errors = (); - my $i, $j; - for ($i = 0; $i < $n; $i++) { - my $ith = Value::List->NameForNumber($i+1); - my $p = $student->[$i]; # i-th student answer - if ($p->type ne "Point") { - push(@errors,"Your $ith entry is not a point"); - next; - } - for ($j = 0, $used = 0; $j < $i; $j++) { - if ($student->[$j]->type eq "Point" and $student->[$j] == $p) { - push(@errors,"Your $ith point is the same as a previous one") unless $ansHash->{isPreview}; - $used = 1; last; - } - } - if (!$used) { - my ($a,$b,$c) = $p->value; - if (3*($a+2)+5*($b-9)-4*$c==0) {$score++} else { - push(@errors,"Your $ith point is not correct") unless $ansHash->{isPreview} - } - } - } - if (!$ansHash->{isPreview}) { - push(@errors,"You need to provide more points") if $i < 2; - push(@errors,"You have given too many points") if $score > 2 and $i != $score; - } - return ($score,@errors); - }); - - - -

    - List any two points in the plane with equation 3(x+2)+5(y-9)-4z=0. -

    - -

    - -

    -     -     -     - - - - -

    - x=2 -

    -     - -

    - Answers will vary. -

    -     - -     - - - - - Context("Point"); - $points=List("(0,-2,6),(1,-2,6)"); - $pev=$points->cmp(list_checker => sub { - my ($correct,$student,$ansHash,$value) = @_; - my $n = scalar(@$student); - my $score = 0; - my @errors = (); - my $i, $j; - for ($i = 0; $i < $n; $i++) { - my $ith = Value::List->NameForNumber($i+1); - my $p = $student->[$i]; # i-th student answer - if ($p->type ne "Point") { - push(@errors,"Your $ith entry is not a point"); - next; - } - for ($j = 0, $used = 0; $j < $i; $j++) { - if ($student->[$j]->type eq "Point" and $student->[$j] == $p) { - push(@errors,"Your $ith point is the same as a previous one") unless $ansHash->{isPreview}; - $used = 1; last; - } - } - if (!$used) { - my ($a,$b,$c) = $p->value; - if (4*($b+2)-($c-6)==0) {$score++} else { - push(@errors,"Your $ith point is not correct") unless $ansHash->{isPreview} - } - } - } - if (!$ansHash->{isPreview}) { - push(@errors,"You need to provide more points") if $i < 2; - push(@errors,"You have given too many points") if $score > 2 and $i != $score; - } - return ($score,@errors); - }); - - - -

    - List any two points in the plane with equation 4(y+2)-(z-6)=0. -

    - -

    - -

    -     -     -     - -     - - - -

    - Give the equation of the described plane in standard and general forms. -

    -     - - - - -

    - Passes through (2,3,4) and has normal vector -

    - -

    - \vec n= \la 3,-1,7\ra. -

    -     - -

    - Standard form: 3(x-2)-(y-3)+7(z-4)=0 -

    - -

    - general form: 3x-y+7z=31 -

    -     - -     - - - - - sub grouped { - my $op = shift; - return 1 if ($op->{name} or $op->{isConstant}); - if ($op->{uop}) {return grouped($op->{op});}; - my $context=Context(); - my $lft=$context->Package("Formula")->new($context,$op->{lop}); - my $rgt=$context->Package("Formula")->new($context,$op->{rop}); - return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); - if ($op->{bop} eq '+' or $op->{bop} eq '-') { - return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); - }; - return (grouped($op->{lop}) and grouped($op->{rop})); - }; - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); - $st=ImplicitEquation("2(y-3)+4(z-5)=0"); - $stev=$st->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - #return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); - return $correct == $student; - }); - $ge=ImplicitEquation("2y+4z=26"); - $geev=$ge->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); - Value::Error("Your answer is not in general form") unless $right->isConstant; - return $correct == $student; - }); - - - -

    - A plane passes through (1,3,5) and has normal vector \vec n= \la 0,2,4\ra. -

    - - - An equation for this plane in standard form is: - -

    - -

    - - - An equation for this plane in general form is: - -

    - -

    -     -     -     - - - - -

    - Passes through the points (1,2,3), - (3,-1,4) and (1,0,1). -

    -     - -

    - Answers may vary; -

    - -

    - Standard form: 8(x-1)+4(y-2)-4(z-3)=0 -

    - -

    - general form: 8x+4y-4z=4 -

    -     - -     - - - - - - sub grouped { - my $op = shift; - return 1 if ($op->{name} or $op->{isConstant}); - if ($op->{uop}) {return grouped($op->{op});}; - my $context=Context(); - my $lft=$context->Package("Formula")->new($context,$op->{lop}); - my $rgt=$context->Package("Formula")->new($context,$op->{rop}); - return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); - if ($op->{bop} eq '+' or $op->{bop} eq '-') { - return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); - }; - return (grouped($op->{lop}) and grouped($op->{rop})); - }; - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); - $st=ImplicitEquation("-5(x-5)+3(y-3)+2(z-8)=0"); - $stev=$st->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - #return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); - return $correct == $student; - }); - $ge=ImplicitEquation("-5x+3y+2z=0"); - $geev=$ge->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); - Value::Error("Your answer is not in general form") unless $right->isConstant; - return $correct == $student; - }); - - - -

    - A plane passes through the points (5,3,8), - (6,4,9) and (3,3,3). -

    - - - An equation for this plane in standard form is: - -

    - -

    - - - An equation for this plane in general form is: - -

    - -

    -     -     -     - - - - -

    - Contains the intersecting lines -

    - -

    - \vec\ell_1(t) = \la 2,1,2\ra + t\la 1,2,3\ra and -

    - -

    - \vec\ell_2(t) = \la 2,1,2\ra + t\la 2,5,4\ra. -

    -     - -

    - Answers may vary; -

    - -

    - Standard form: -7(x-2)+2(y-1)+(z-2)=0 -

    - -

    - general form: -7x+2y+z=-10 -

    -     - -     - - - - - - sub grouped { - my $op = shift; - return 1 if ($op->{name} or $op->{isConstant}); - if ($op->{uop}) {return grouped($op->{op});}; - my $context=Context(); - my $lft=$context->Package("Formula")->new($context,$op->{lop}); - my $rgt=$context->Package("Formula")->new($context,$op->{rop}); - return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); - if ($op->{bop} eq '+' or $op->{bop} eq '-') { - return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); - }; - return (grouped($op->{lop}) and grouped($op->{rop})); - }; - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); - $st=ImplicitEquation("3(x-5)+3(z-3)=0"); - $stev=$st->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - #return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); - return $correct == $student; - }); - $ge=ImplicitEquation("3x+3z=24"); - $geev=$ge->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); - Value::Error("Your answer is not in general form") unless $right->isConstant; - return $correct == $student; - }); - - - -

    - A plane contains the intersecting lines - \vec\ell_1(t) = \la 5,0,3\ra + t\la -1,1,1\ra and \vec\ell_2(t) = \la 1,4,7\ra + t\la 3,0,-3\ra. -

    - - - An equation for this plane in standard form is: - -

    - -

    - - - An equation for this plane in general form is: - -

    - -

    -     -     -     - - - - -

    - Contains the parallel lines -

    - -

    - \vec\ell_1(t) = \la 1,1,1\ra + t\la 1,2,3\ra and -

    - -

    - \vec\ell_2(t) = \la 1,1,2\ra + t\la 1,2,3\ra. -

    -     - -

    - Answers may vary; -

    - -

    - Standard form: 2(x-1)-(y-1)=0 -

    - -

    - general form: 2x-y=1 -

    -     - -     - - - - - - sub grouped { - my $op = shift; - return 1 if ($op->{name} or $op->{isConstant}); - if ($op->{uop}) {return grouped($op->{op});}; - my $context=Context(); - my $lft=$context->Package("Formula")->new($context,$op->{lop}); - my $rgt=$context->Package("Formula")->new($context,$op->{rop}); - return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); - if ($op->{bop} eq '+' or $op->{bop} eq '-') { - return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); - }; - return (grouped($op->{lop}) and grouped($op->{rop})); - }; - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); - $st=ImplicitEquation("2(x-1)+(y-1)-3(z-1)=0"); - $stev=$st->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - #return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); - return $correct == $student; - }); - $ge=ImplicitEquation("2x+y-3z=0"); - $geev=$ge->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); - Value::Error("Your answer is not in general form") unless $right->isConstant; - return $correct == $student; - }); - - - -

    - A plane contains the parallel lines - \vec\ell_1(t) = \la 1,1,1\ra + t\la 4,1,3\ra and \vec\ell_2(t) = \la 2,2,2\ra + t\la 4,1,3\ra. -

    - - - An equation for this plane in standard form is: - -

    - -

    - - - An equation for this plane in general form is: - -

    - -

    -     -     -     - - - - -

    - Contains the point (2,-6,1) and the line -

    - -

    - \vec\ell(t) = \left\{\begin{aligned}x\amp =2+5t \\ - y\amp =2+2t \\ - z\amp =-1+2t - \end{aligned} \right. -

    -     - -

    - Answers may vary; -

    - -

    - Standard form: 2(x-2)-(y+6)-4(z-1)=0 -

    - -

    - general form: 2x-y-4z=6 -

    -     - -     - - - - - - sub grouped { - my $op = shift; - return 1 if ($op->{name} or $op->{isConstant}); - if ($op->{uop}) {return grouped($op->{op});}; - my $context=Context(); - my $lft=$context->Package("Formula")->new($context,$op->{lop}); - my $rgt=$context->Package("Formula")->new($context,$op->{rop}); - return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); - if ($op->{bop} eq '+' or $op->{bop} eq '-') { - return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); - }; - return (grouped($op->{lop}) and grouped($op->{rop})); - }; - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); - $st=ImplicitEquation("4(x-5)-2(y-7)-2(z-3)=0"); - $stev=$st->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - #return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); - return $correct == $student; - }); - $ge=ImplicitEquation("4x-2y-2z=0"); - $geev=$ge->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); - Value::Error("Your answer is not in general form") unless $right->isConstant; - return $correct == $student; - }); - - - -

    - A plane contains the point (5,7,3) and the line \vec\ell(t) = \begin{cases}x\amp=t\\y\amp=t\\z\amp=t\end{cases}. -

    - - - An equation for this plane in standard form is: - -

    - -

    - - - An equation for this plane in general form is: - -

    - -

    -     -     -     - - - - - - sub grouped { - my $op = shift; - return 1 if ($op->{name} or $op->{isConstant}); - if ($op->{uop}) {return grouped($op->{op});}; - my $context=Context(); - my $lft=$context->Package("Formula")->new($context,$op->{lop}); - my $rgt=$context->Package("Formula")->new($context,$op->{rop}); - return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); - if ($op->{bop} eq '+' or $op->{bop} eq '-') { - return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); - }; - return (grouped($op->{lop}) and grouped($op->{rop})); - }; - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); - $st=ImplicitEquation("(x-5)+(y-7)+(z-3)=0"); - $stev=$st->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - #return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); - return $correct == $student; - }); - $ge=ImplicitEquation("x+y+z=15"); - $geev=$ge->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); - Value::Error("Your answer is not in general form") unless $right->isConstant; - return $correct == $student; - }); - - - -

    - A plane contains the point (5,7,3) and is orthogonal to the line \vec\ell(t) = \la 4,5,6\ra+ t\la 1,1,1\ra. -

    - - - An equation for this plane in standard form is: - -

    - -

    - - - An equation for this plane in general form is: - -

    - -

    -     -     -     - - - - - - sub grouped { - my $op = shift; - return 1 if ($op->{name} or $op->{isConstant}); - if ($op->{uop}) {return grouped($op->{op});}; - my $context=Context(); - my $lft=$context->Package("Formula")->new($context,$op->{lop}); - my $rgt=$context->Package("Formula")->new($context,$op->{rop}); - return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); - if ($op->{bop} eq '+' or $op->{bop} eq '-') { - return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); - }; - return (grouped($op->{lop}) and grouped($op->{rop})); - }; - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); - $st=ImplicitEquation("4(x-4)+(y-1)+(z-1)=0"); - $stev=$st->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - #return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); - return $correct == $student; - }); - $ge=ImplicitEquation("4x+y+z=18"); - $geev=$ge->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); - Value::Error("Your answer is not in general form") unless $right->isConstant; - return $correct == $student; - }); - - - -

    - A plane contains the point (4,1,1) and is orthogonal to the line \begin{cases}x\amp=4+4t\\y\amp=1+t\\z\amp=1+t\end{cases}. -

    - - - An equation for this plane in standard form is: - -

    - -

    - - - An equation for this plane in general form is: - -

    - -

    -     -     -     - - - - - - sub grouped { - my $op = shift; - return 1 if ($op->{name} or $op->{isConstant}); - if ($op->{uop}) {return grouped($op->{op});}; - my $context=Context(); - my $lft=$context->Package("Formula")->new($context,$op->{lop}); - my $rgt=$context->Package("Formula")->new($context,$op->{rop}); - return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); - if ($op->{bop} eq '+' or $op->{bop} eq '-') { - return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); - }; - return (grouped($op->{lop}) and grouped($op->{rop})); - }; - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); - $st=ImplicitEquation("3(x+4)+8(y-7)-10(z-2)=0"); - $stev=$st->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - #return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); - return $correct == $student; - }); - $ge=ImplicitEquation("3x+8y-10z=24"); - $geev=$ge->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); - Value::Error("Your answer is not in general form") unless $right->isConstant; - return $correct == $student; - }); - - - -

    - A plane contains the point - (-4,7,2) and is parallel to the plane 3(x-2)+8(y+1) -10z=0. -

    - - - An equation for this plane in standard form is: - -

    - -

    - - - An equation for this plane in general form is: - -

    - -

    -     -     -     - - - - - - sub grouped { - my $op = shift; - return 1 if ($op->{name} or $op->{isConstant}); - if ($op->{uop}) {return grouped($op->{op});}; - my $context=Context(); - my $lft=$context->Package("Formula")->new($context,$op->{lop}); - my $rgt=$context->Package("Formula")->new($context,$op->{rop}); - return 0 if (($lft->usesOneOf('x') and $rgt->usesOneOf('x')) or ($lft->usesOneOf('y') and $rgt->usesOneOf('y')) or ($lft->usesOneOf('z') and $rgt->usesOneOf('z'))); - if ($op->{bop} eq '+' or $op->{bop} eq '-') { - return 0 unless (($lft->usesOneOf('x','y','z') and $rgt->usesOneOf('x','y','z')) or ($op->{lop}{name} and $op->{rop}{isConstant}) or ($op->{rop}{name} and $op->{lop}{isConstant})); - }; - return (grouped($op->{lop}) and grouped($op->{rop})); - }; - Context("ImplicitEquation"); - Context()->variables->are(x=>'Real',y=>'Real',z=>'Real'); - Context()->variables->set(x=>{limits=>[-100,100]},y=>{limits=>[-100,100]},z=>{limits=>[-100,100]}); - $st=ImplicitEquation("x-1=0"); - $stev=$st->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - #return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in standard form") unless ($right == 0 and grouped($left->{tree})); - return $correct == $student; - }); - $ge=ImplicitEquation("x=1"); - $geev=$ge->cmp(checker=>sub{ - my ( $correct, $student, $ansHash ) = @_; - return 0 if $ansHash->{isPreview} || $correct != $student; - my $context = Context(); - my $left=$context->Package("Formula")->new($context,$student->{tree}{lop}); - my $right=$context->Package("Formula")->new($context,$student->{tree}{rop}); - Value::Error("Your answer is not in general form") unless ($left->eval(x=>0,y=>0,z=>0)==0); - Value::Error("Your answer is not in general form") unless $right->isConstant; - return $correct == $student; - }); - - - -

    - A plane contains the point (1,2,3) and is parallel to the plane x=5. -

    - - - An equation for this plane in standard form is: - -

    - -

    - - - An equation for this plane in general form is: - -

    - -

    -     -     -     - -     - - - -

    - Give the equation of the line that is the intersection of the given planes. -

    -     - - - - -

    - p1:\ 3 (x - 2) + (y - 1) + 4 z=0, and -

    - -

    - p2:\ 2 (x - 1) - 2 (y + 3) + 6 (z - 1)=0. -

    -     - -

    - Answers may vary: -

    - -

    - \ell = \left\{\begin{aligned}x \amp = 14t\\ - y \amp = -1-10t\\ - z\amp = 2-8t - \end{aligned} \right. -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(t=>"Real"); - $v=Compute("(1,3,3.5)+t<20,2,-26>"); - $vev=$v->cmp(checker=>sub{my($c,$st,$aH)=@_; - $ds=Formula($st)->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");}else{return 0;}; - $dc=$c->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=Formula($st)->eval(t=>0);$cp=$c->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return ($par and $tch); - }); - - - -

    - Give the equation of the line - (in vector form) - that is the intersection of the planes 5 (x - 5) + 2 (y + 2) + 4 (z - 1)=0, - and 3 x - 4 (y - 1) + 2 (z - 1)=0. -

    - -

    - -

    -     -     -     - -     - - - -

    - Find the point of intersection between the line and the plane. -

    -     - - - - -

    -

      -
    • -

      - line: \la 5,1,-1\ra + t\la 2,2,1\ra -

      -
      • -
    • -

      - plane: 5x-y-z=-3 -

      -
      • -
    -

    -     - -

    - (-3,-7,-5) -

    -     - -     - - - - - Context("Point"); - Context()->strings->add('the entire line'=>{}); - $point=Point("(3,1,1)"); - - - -

    -

      -
    • -

      - line: \la 4,1,0\ra + t\la 1,0,-1\ra -

      -
      • -
    • -

      - plane: 3x+y-2z=8 -

      -
      • -
    -

    - - If the plane contains the line, - enter the phrase the entire line. - If the line does not intersect the plane at all, enter none. - - -

    - -

    -     -     -     - - - - -

    -

      -
    • -

      - line: \la 1,2,3\ra + t\la 3,5,-1\ra -

      -
      • -
    • -

      - plane: 3x-2y-z=4 -

      -
      • -
    -

    -     - -

    - No point of intersection; the plane and line are parallel. -

    -     - -     - - - - - Context("Point"); - Context()->strings->add('the entire line'=>{}); - $point=Compute('the entire line'); - - - -

    -

      -
    • -

      - line: \la 1,2,3\ra + t\la 3,5,-1\ra -

      -
      • -
    • -

      - plane: 3x-2y-z=-4 -

      -
      • -
    -

    - - If the plane contains the line, - enter the phrase the entire line. - If the line does not intersect the plane at all, enter none. - - -

    - -

    -     -     -     - -     - - - -

    - Find the indicated distance. -

    -     - - - - -

    - The distance from the point (1,2,3) to the plane -

    - -

    - 3(x-1)+(y-2)+5(z-2)=0. -

    -     - -

    - \sqrt{5/7} -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $distance=Formula("8/sqrt(21)"); - - - -

    - Find the distance from the point (2,6,2) to the plane 2(x-1)-y+4(z+1)=0. -

    - -

    - -

    -     -     -     - - - - -

    - The distance between the parallel planes -

    - -

    - x+y+z=0 and -

    - -

    - (x-2)+(y-3)+(z+4)=0 -

    -     - -

    - 1/\sqrt{3} -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $distance=Formula("3"); - - - -

    - Find the distance between the parallel planes - 2(x-1)+2(y+1)+(z-2)=0 and 2(x-3)+2(y-1)+(z-3)=0. -

    - -

    - -

    -     -     -     - -     - - - - -

    - Show why if the point Q lies in a plane, - then the distance formula correctly gives the distance from the point to the plane as 0. -

    -     - -

    - If P is any point in the plane, - and Q is also in the plane, - then \overrightarrow{PQ} lies parallel to the plane and is orthogonal to \vec n, - the normal vector. - Thus \vec n\cdot \overrightarrow{PQ}=0, giving the distance as 0. -

    -     - -     - - - - -

    - How is - in - easier to answer once we have an understanding of planes? -

    -     - -

    - The intersecting lines define a plane with normal vector \vec n = \vec c = \vec d_1\times \vec d_2. - Since points P_1 and P_2 lie in the plane, - \vec c is orthogonal to \overrightarrow{P_1P_2}, - hence \overrightarrow{P_1P_2}\cdot\vec c = 0, giving a distance of 0. - Knowing the principles of planes, - especially their normal vectors, makes this simpler. -

    -     - -     -     -     -
    -     - - - Vector Valued Functions - -

    - In the previous chapter, - we learned about vectors and were introduced to the power of vectors within mathematics. - In this chapter, - we'll build on this foundation to define functions whose input is a real number and whose output is a vector. - We'll see how to graph these functions and apply calculus techniques to analyze their behavior. - Most importantly, we'll see why - we are interested in doing this: - we'll see beautiful applications to the study of moving objects. -

    -     - -
    - Vector-Valued Functions - -

    - We are very familiar with real valued functions, that is, - functions whose output is a real number. - This section introduces vector-valued functions - functions whose output is a vector. -

    - - - - - Vector-Valued Functions - -

    - A vector-valued function - is a function of the form - - \vec r(t) = \la\, f(t),g(t)\,\ra \text{ or } \vec r(t) = \la \,f(t),g(t),h(t)\,\ra - , - where f, g and h are real valued functions. -

    - -

    - The domain of \vec r is the set of all values of t for which \vec r(t) is defined. - The range of \vec r is the set of all possible output vectors \vec r(t). - vector-valued functiondefinition - functionvector-valued -

    -     -     -     - - - Evaluating and Graphing Vector-Valued Functions -

    - Evaluating a vector-valued function at a specific value of t is straightforward; - simply evaluate each component function at that value of t. - For instance, if \vec r(t) = \la t^2,t^2+t-1\ra, - then \vec r(-2) = \la 4,1\ra. - We can sketch this vector, - as is done in . - Plotting lots of vectors is cumbersome, though, - so generally we do not sketch the whole vector but just the terminal point. - The graph of a vector-valued function is the set of all terminal points of \vec r(t), - where the initial point of each vector is always the origin. - In we sketch the graph of \vec r; we can indicate individual points on the graph with their respective vector, - as shown. - vector-valued functiongraphing -

    - -
    - Sketching the graph of a vector-valued function - -
    - - - - - Graph of the vector \vec r(-2) =\la 4,1\ra. - This vector starts at the origin and ends at the point (4,1). - - Graph of a vector whose terminal point corresponds to a point on the curve. - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,2,3,4,5}, - ytick={-3,-2,-1,1,2,3}, - ymin=-3.5,ymax=3.5, - xmin=-.5,xmax=5.5 - ] - - \draw [->,thick,secondcolor] (axis cs: 0,0) -- (axis cs:4,1) node [black,pos=.5,above] { $\vec r(-2)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - The image contains the plot of the vector valued function \vec r(t) = \la t^2,t^2+t-1\ra. - The image also contains the graph of the vector \vec r(-2) =\la 4,1\ra, which ends at a point on the function \vec r(t). - The function \vec r(t) resembles a parabola which has been rotated about 45 degrees clockwise. - The function is plotted for t approximately between -2.5 and 1.5. - The function begins near the point (5,2) and then slopes down crossing the x-axis at approximately x=2.5. - The slanted parabola then curves upwards around the point (0,-1), from which point it begins increasing in both x and y as t increases. - - Graph of the vector and vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,2,3,4,5}, - ytick={-3,-2,-1,1,2,3}, - ymin=-3.5,ymax=3.5, - xmin=-.5,xmax=5.5 - ] - - \addplot+ [domain=-2.5:2.5,samples=50] ({x^2},{x^2+x-1}); - - \draw [->,thick] (axis cs:1,-1) -- (axis cs: 0.9025,-1.0475); - \draw [->,thick,secondcolor] (axis cs: 0,0) -- (axis cs:4,1) node [black,pos=.5,above] { $\vec r(-2)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - Vector-valued functions are closely related to parametric equations of graphs. - While in both methods we plot points - \big(x(t), - y(t)\big) or \big(x(t),y(t),z(t)\big) to produce a graph, - in the context of vector-valued functions each such point represents a vector. - The implications of this will be more fully realized in the next section as we apply calculus ideas to these functions. -

    - - - Graphing vector-valued functions - -

    - Graph \ds \vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra, - for -2\leq t\leq 2. - Sketch \vec r(-1) and \vec r(2). -

    -     - -

    - We start by making a table of t, - x and y values as shown in . - Plotting these points gives an indication of what the graph looks like. - In , - we indicate these points and sketch the full graph. - We also highlight \vec r(-1) and \vec r(2) on the graph. -

    - -
    - Sketching the vector-valued function of - -
    - - - - t - t^3-t - \ds \frac{1}{t^2+1} - - - - - - - - -2 - -6 - 1/5 - - - -1 - 0 - 1/2 - - - 0 - 0 - 1 - - - 1 - 0 - 1/2 - - - 2 - 6 - 1/5 - - -
    - -
    - - - - - The image contains the plot of the vector valued function \vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra for -2\leq t\leq 2. - The image also contains the vectors \vec r(-1) =\la 0,\frac12 \ra and \vec r(2) =\la 6,\frac15 \ra, which both end at a point on the function \vec r(t). - The function \vec r(t) begins at the point (-6,\frac15) corresponding to the lower bound of t=-2, from which it slowly slopes upwards until crossing the y-axis at y=\frac12. - From here, the function slopes upwards and slightly outwards away from the y-axis until curving back and crossing the y-axis once again at y=1. - The function then slopes downwards and slightly away from the y-axis until curving back and crossing the y-axis again at y=\frac12, completing a loop. - After crossing the y-axis, the function continues slowly sloping downwards until reaching the point (6,\frac15) which corresponds to the upper bound of t=2. - The function is also symmetric about the y-axis. - - Graph of two vectors and vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[% - xtick={-6,-4,-2,2,4,6}, - ymin=-.1,ymax=1.1,% - xmin=-7,xmax=7% - ] - - \addplot+[domain=-2:2] ({x^3-x},{1/(x^2+1)}); - - \filldraw [black] (axis cs: -6,.2) circle (2.4pt); - \filldraw [black] (axis cs: 6,.2) circle (2.4pt); - \filldraw [black] (axis cs: 0,.5) circle (2.4pt); - \filldraw [black] (axis cs: 0.,1) circle (2.4pt); - - \draw [->,thick,secondcolor] (axis cs:0,0)--(axis cs: 0,.5) node [,black,pos=.4,rotate=90,below] { $\vec r(-1)$}; - \draw [->,thick,secondcolor] (axis cs:0,0)--(axis cs: 6,.2) node [black,pos=.4,above] { $\vec r(2)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    -     - -     - - - Graphing vector-valued functions - -

    - Graph \vec r(t) = \la \cos(t) ,\sin(t) ,t\ra for 0\leq t\leq 4\pi. -

    -     - -

    - We can again plot points, - but careful consideration of this function is very revealing. - Momentarily ignoring the third component, - we see the x and y components trace out a circle of radius 1 centered at the origin. - Noticing that the z component is t, - we see that as the graph winds around the z-axis, - it is also increasing at a constant rate in the positive z direction, - forming a spiral. - This is graphed in . - In the graph \vec r(7\pi/4)\approx (0.707,-0.707,5.498) is highlighted to help us understand the graph. -

    - -
    - The graph of \vec r(t) in - - - - - The image contains the plot of the vector valued function \vec r(t) = \la \cos(t) ,\sin(t) ,t\ra for 0\leq t\leq 4\pi. - The image also contains the vector \vec r(7\pi/4)\approx (0.707,-0.707,5.498), which corresponds to a point on the function \vec r(t). - The function \vec r(t) begins at the point (1,0,0) corresponding to the lower bound of t=0. - Ignoring the z coordinate of the vector valued function, the curve is simply a circle centered at the origin in the xy plane. - The addition of the z coordinate given by t then makes this curve linearly increase at t increases. - Accounting for the z coordinate, the function \vec r(t) resembles a linearly increasing circular spiral, which completes two full revolutions. - The spiral ends at the same x and y coordinates as it started, but is 4\pi units above in the z direction, ending at the point (1,0,4\pi). - - Graph of a vector and the three-dimensional vector valued function from the example. - - - - - //ASY file for figvvf23D.asy in Chapter 11 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4.5,4.5,20); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={5,10}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-0.5,15); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the helix <cos t,sin t,t> for t from 0 to 4pi - triple g(real t) {return (cos(t),sin(t),t);} - path3 mypath=graph(g,0,4*pi,operator ..); draw(mypath,bluepen); - - //Draw the vector for r(7pi/4) - draw((0,0,0)--(0.707,-0.707,5.498),redpen+linewidth(2),Arrow3(size=3mm)); - label("$\vec{r}(7 \pi/4)$",(0.707,-0.707,5.498),W); - - - - -
    -     - -     -     - - - Algebra of Vector-Valued Functions - - Operations on Vector-Valued Functions - -

    - Let \vec r_1(t)=\la f_1(t),g_1(t)\ra and - \vec r_2(t)=\la f_2(t),g_2(t)\ra be vector-valued functions in - \mathbb{R}^2 and let c be a scalar. - Then: -

    - -

    -

      -
    1. -

      - \vec r_1(t) \pm \vec r_2(t) = \la\, f_1(t)\pm f_2(t),g_1(t)\pm g_2(t)\,\ra. -

      -
      2. - -
    2. -

      - c\vec r_1(t) = \la\, cf_1(t),cg_1(t)\,\ra. -

      -
      3. -
    -

    - -

    - A similar definition holds for vector-valued functions in \mathbb{R}^3. - vector-valued functionalgebra of -

    -     -     - -

    - This definition states that we add, - subtract and scale vector-valued functions component-wise. - Combining vector-valued functions in this way can be very useful - (as well as create interesting graphs). -

    - - - - - Adding and scaling vector-valued functions - -

    - Let \vec r_1(t) = \la\,0.2t,0.3t\,\ra, - \vec r_2(t) = \la\,\cos(t) ,\sin(t) \,\ra and \vec r(t) = \vec r_1(t)+\vec r_2(t). - Graph \vec r_1(t), \vec r_2(t), - \vec r(t) and 5\vec r(t) on -10\leq t\leq10. -

    -     - -

    - We can graph \vec r_1 and - \vec r_2 easily by plotting points - (or just using technology). - Let's think about each for a moment to better understand how vector-valued functions work. -

    - -

    - We can rewrite \vec r_1(t) = \la\, 0.2t,0.3t\,\ra as \vec r_1(t) = t\la 0.2,0.3\ra. - That is, the function \vec r_1 scales the vector \la 0.2,0.3\ra by t. - This scaling of a vector produces a line in the direction of \la 0.2,0.3\ra. -

    - -

    - We are familiar with \vec r_2(t) = \la\, \cos(t) ,\sin(t) \,\ra; - it traces out a circle, centered at the origin, of radius 1. - graphs - \vec r_1(t) and \vec r_2(t). -

    - -

    - Adding \vec r_1(t) to - \vec r_2(t) produces \vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra, - graphed in . - The linear movement of the line combines with the circle to create loops that move in the direction of - \la 0.2,0.3\ra. (We encourage the reader to experiment by changing - \vec r_1(t) to \la 2t,3t\ra, - etc., and observe the effects on the loops.) -

    - -
    - Graphing the functions in - -
    - - - - - The image contains the plot of the vector valued functions \vec r_1(t) = \la 0.2t,0.3t \ra, \vec r_2(t) = \la \cos(t) ,\sin(t) \ra on -10\leq t\leq10. - The function \vec r_1(t) = \la 0.2t,0.3t \ra is a line which begins at the point (-2,-3) corresponding to t=-10 and ends when t=10 at the point (2,3). - The second function \vec r_2(t) = \la \cos(t) ,\sin(t) \ra looks like a circle of radius 1, but as -10\leq t\leq10 the function completes \frac{20}{2\pi} full circular rotations, which cannot be seen in the graph. - - Graph of the two vector valued functions prior to adding them together. - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-4.1,ymax=4.1, - xmin=-4.9,xmax=4.9 - ] - - \addplot+ [domain=-10:10] ({.2*x},{.3*x}); - \addplot+ [solid,domain=0:360,samples=60] ({cos(x)},{sin(x)}); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - The image contains the plot of the vector valued function \vec r(t) = \vec r_1(t)+\vec r_2(t). - The function is given by \vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra on -10\leq t\leq10. - The function begins in the third quadrant, near the point (-3,-2). - Near this point, the function is concave up, starting with a downward slope and later beginning to slope upwards. - The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate. - The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant. - The function nearly completes a third loop, before ending near the point (3,1.5). - - Graph of the vector valued function coming from adding the circle and line vector valued functions. - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-4.1,ymax=4.1, - xmin=-4.9,xmax=4.9 - ] - - \addplot+ [domain=-10:10,samples=120] ({.2*x+cos(deg(x))},{.3*x+sin(deg(x))}); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - The image contains the plot of the vector valued function 5\vec r(t) = 5\vec r_1(t)+5\vec r_2(t). - The function is now given by \vec r(t) = \la\,5\cos(t) + t,5\sin(t) +1.5t\,\ra on -10\leq t\leq10. - Compared to the unscaled function \vec r(t), the function 5\vec r(t) is exactly 5 times larger than the original function. - The function begins in the third quadrant, near the point (-15,-10). - Like the unscaled function, near this point 5\vec r(t) is concave up, starting with a downwards slope and later beginning to slope upwards. - The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate. - The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant. - The function nearly completes a third loop, before ending near the point (15,7.5). - The entirety of the original function \vec r(t) can be seen to fit between the two loops of the scaled function 5\vec r(t). - - Graph of the vector valued function coming from scaling the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-20,-10,10,20}, - ymin=-20.5,ymax=20.5, - xmin=-24.5,xmax=24.5 - ] - - \addplot+ [domain=-10:10,samples=60] ({.2*x+cos(deg(x))},{.3*x+sin(deg(x))}); - \addplot+ [solid,domain=-10:10,samples=120] ({x+5*cos(deg(x))},{1.5*x+5*sin(deg(x))}); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - Multiplying \vec r(t) by 5 scales the function by 5, producing 5\vec r(t) = \langle 5\cos(t) +t,5\sin(t) +1.5t\rangle, - which is graphed in along with \vec r(t). - The new function is 5 times bigger than \vec r(t). - Note how the graph of 5\vec r(t) in looks identical to the graph of \vec r(t) in . - This is due to the fact that the x and y bounds of the plot in - are exactly 5 times larger than the bounds in . -

    -     - -     - - - Adding and scaling vector-valued functions - -

    - A cycloid is a graph traced by a point p on a rolling circle, - as shown in . - Find an equation describing the cycloid, - where the circle has radius 1. - cycloid -

    - - -
    - Tracing a cycloid - - - The image shows the cycloid which comes from tracking a point p on a rolling circle of radius 1 on a flat surface. - The point p is initially at the top of the circle of radius 1. - Once the circle rolls, the point p is tracked to create a graph of a cycloid. - The graph coming from tracking the point p first begins to decrease, until the point p is at the bottom of the circle, at which point the point p touches the surface the ball is rolling on. - After this point, the graph begins to increase. - The circle continues to roll, with the point p once again becoming the top of the circle, after which the graph begins to decrease. - The circle continues rolling, with the point p becoming the lowest point on the circle two more times, after which the graph stops. - Between the starting point and the point at which p is at the bottom of the circle, the graph resembles a slightly stretched quarter of a circle. - The part of the graph that comes from the remaining two full revolutions of the circle resembles two horizontally stretched semi-circles. - - Image showing the cycloid which comes from tracking a point on a rolling circle. - - - \begin{tikzpicture} - - \draw [thick] (0,1) circle (1); - - \filldraw (1,1) circle (2pt) node [right] {\large $p$}; - - \draw [firstcolor,thick,smooth,domain=-1:14,samples=60] plot ({cos(\x r)+\x},{-sin(\x r)+1}); - - \draw [thick] (-1,0) -- (15,0); - \draw [thick,->,>=stealth] (-1.3,1) arc (180:90:1.3); - \draw [thick,white] (0,2.5) -- (1,2.5); - - \end{tikzpicture} - - - -
    - -     - -

    - This problem is not very difficult if we approach it in a clever way. - We start by letting \vec p(t) describe the position of the point p on the circle, - where the circle is centered at the origin and only rotates clockwise (, it does not roll). - This is relatively simple given our previous experiences with parametric equations; - \vec p(t) = \la \cos(t) , -\sin(t) \ra. -

    - -

    - We now want the circle to roll. - We represent this by letting - \vec c(t) represent the location of the center of the circle. - It should be clear that the y component of \vec c(t) should be 1; - the center of the circle is always going to be 1 if it rolls on a horizontal surface. -

    - -

    - The x component of \vec c(t) is a linear function of t: - f(t) = mt for some scalar m. - When t=0, f(t) = 0 - (the circle starts centered on the y-axis). - When t=2\pi, the circle has made one complete revolution, - traveling a distance equal to its circumference, - which is also 2\pi. - This gives us a point on our line f(t) = mt, - the point (2\pi, 2\pi). - It should be clear that m=1 and f(t) = t. - So \vec c(t) = \la t, 1\ra. -

    - -

    - We now combine \vec p and \vec c together to form the equation of the cycloid: - \vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra, - which is graphed in . -

    -
    - The cycloid in - - - - The graph shows the function \vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra, which is the graph of a cycloid which comes from tracking a point p on a rolling a circle of radius 1 on a flat surface. - The curve begins near the point (0,2), after which it begins to decrease until it reaches its lowest point near the point (2,0). - After this point, the curve begins increasing, until it reaches its highest point when p is at the top of the circle, after which it decreases until reaching another minimum near the point (8,0). - The curve continues in the same fashion, reaching another minimum near the point (14,0), after which it continues for a slight duration in the same fashion as before. - - Graph showing the cycloid which comes from tracking a point on a rolling circle. - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=14, - xmin=-.5,xmax=16 - ] - - \addplot+ [domain=-1:16,samples=80] ({cos(deg(x))+x},{-sin(deg(x))+1}); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -     -     - - - Displacement - -

    - A vector-valued function - \vec r(t) is often used to describe the position of a moving object at time t. - At t=t_0, the object is at \vec r(t_0); - at t=t_1, the object is at \vec r(t_1). - Knowing the locations \vec r(t_0) and - \vec r(t_1) give no indication of the path taken between them, - but often we only care about the difference of the locations, - \vec r(t_1)-\vec r(t_0), - the displacement. - displacement - vector-valued functiondisplacement -

    - - - Displacement - -

    - Let \vec r(t) be a vector-valued function and let - t_0\lt t_1 be values in the domain. - The displacement \vec d of \vec r, - from t=t_0 to t=t_1, is - - \vec d=\vec r(t_1)-\vec r(t_0) - . -

    -     -     - - - -

    - When the displacement vector is drawn with initial point at \vec r(t_0), - its terminal point is \vec r(t_1). - We think of it as the vector which points from a starting position to an ending position. -

    - - - Finding and graphing displacement vectors - -

    - Let \vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra. - Graph \vec r(t) on -1\leq t\leq 1, - and find the displacement of \vec r(t) on this interval. -

    -     - -

    - The function \vec r(t) traces out the unit circle, - though at a different rate than the usual - \la \cos(t) ,\sin(t) \ra parametrization. - At t_0=-1, we have \vec r(t_0) = \la 0,-1\ra; - at t_1=1, we have \vec r(t_1) = \la 0,1\ra. - The displacement of \vec r(t) on [-1,1] is thus \vec d = \la 0,1\ra - \la 0,-1\ra = \la 0,2\ra. -

    - -
    - Graphing the displacement of a position function in - - - - Graph of the function \vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra on -1\leq t\leq 1. - The function \vec r(t) is the right half of the unit circle, beginning at the point (0,-1), crossing the x-axis at the point (1,0), and ending at the point (0,-1). - The graph also contains the displacement vector, which begins at the start of the curve at the point (0,-1) and heads directly upwards until reaching the endpoint of the curve at the point (0,1). - - Graph of the semicircle coming from plotting the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.32,xmax=1.32 - ] - - \addplot+ [domain=-90:90,samples=40] ({cos(x)},{sin(x)}); - - \draw [thick,->,secondcolor,>=stealth] (axis cs: 0,-1) -- (axis cs:0,1) node [left,pos=.7,black]{ $\vec d$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - A graph of \vec r(t) on [-1,1] is given in , - along with the displacement vector \vec d on this interval. -

    -     - -     - -

    - Measuring displacement makes us contemplate related, - yet very different, concepts. - Considering the semi-circular path the object in took, - we can quickly verify that the object ended up a distance of 2 units from its initial location. - That is, we can compute \vnorm{d} = 2. - However, measuring distance from the starting point - is different from measuring distance traveled. - Being a semi-circle, - we can measure the distance traveled by this object as \pi\approx 3.14 units. - Knowing distance from the starting point - allows us to compute average rate of change. -

    - - - Average Rate of Change - -

    - Let \vec r(t) be a vector-valued function, - where each of its component functions is continuous on its domain, - and let t_0\lt t_1. - The average rate of change - of \vec r(t) on [t_0,t_1] is - average rate of change - vector-valued functionaverage rate of change - - \text{ average rate of change } = \frac{\vec r(t_1) - \vec r(t_0)}{t_1-t_0} - . -

    -     -     - - - Average rate of change - -

    - Let \vec r(t) = \la \cos(\frac{\pi}2t),\sin(\frac{\pi}2t)\ra as in . - Find the average rate of change of - \vec r(t) on [-1,1] and on [-1,5]. -

    -     - -

    - We computed in - that the displacement of \vec r(t) on [-1,1] was \vec d = \la 0,2\ra. - Thus the average rate of change of \vec r(t) on [-1,1] is: - - \frac{\vec r(1) -\vec r(-1)}{1-(-1)} = \frac{\la 0,2\ra}{2} = \la 0,1\ra - . -

    - -

    - We interpret this as follows: - the object followed a semi-circular path, - meaning it moved towards the right then moved back to the left, - while climbing slowly, then quickly, then slowly again. - On average, however, - it progressed straight up at a constant rate of \la 0,1\ra per unit of time. -

    - -

    - We can quickly see that the displacement on [-1,5] is the same as on [-1,1], - so \vec d = \la 0,2\ra. - The average rate of change is different, though: - - \frac{\vec r(5)-\vec r(-1)}{5-(-1)} = \frac{\la 0,2\ra}{6} = \la 0,1/3\ra - . -

    - -

    - As it took 3 times as long - to arrive at the same place, - this average rate of change on [-1,5] is 1/3 the average rate of change on [-1,1]. -

    -     - -     - -

    - We considered average rates of change in Sections - and - as we studied limits and derivatives. - The same is true here; - in the following section we apply calculus concepts to vector-valued functions as we find limits, - derivatives, and integrals. - Understanding the average rate of change will give us an understanding of the derivative; - displacement gives us one application of integration. -

    -     - - - - Terms and Concepts - - - -

    - Vector-valued functions are closely related to - of curves. -

    -     - - - - - - - - - - - - - -     - - - - -

    - When sketching vector-valued functions, - technically one isn't graphing points, but rather . -

    -     - - - - - - - - -     - - - - -

    - It can be useful to think of as a vector that points from a starting position to an ending position. -

    -     - - - - - - - - -     - - - -

    - In the context of vector-valued functions, - average rate of change is divided by time. -

    -     - - - - - - - -     -     - - - Problems - - - -

    - Sketch the vector-valued function on the given interval. -

    -     - - - - -

    - \vec r(t) = \la t^2,t^2-1\ra, - for -2\leq t\leq 2. -

    -     - - - - - Graph of the function \vec r(t) = \la t^2,t^2-1\ra for -2\leq t\leq 2. - The graph of the function \vec r(t) is a line which begins at the point (0,-1) and ends at the point (4,3). - The function begins at the point (4,3) when t=-2, linearly decreases until the point (0,-1) and once again follows the same linear path until ending at the point (4,3). - - Graph of the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,2,3,4}, - ytick={-1,1,2,3,4}, - ymin=-1.5,ymax=4.5, - xmin=-.5,xmax=4.5 - ] - - \addplot+ [domain=-2:2] ({x^2},{x^2-1}); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - - - - -

    - \vec r(t) = \la t^2,t^3\ra, - for -2\leq t\leq 2. -

    -     - - - - - Graph of the function \vec r(t) = \la t^2,t^3\ra, for -2\leq t\leq 2. - The graph of the function \vec r(t) begins at the point (4,-8), and upwards and to the left in a concave down fashion until reaching the origin. - After reaching the origin, the function is concave and ends when it reaches the point (4,8). - The function is also symmetric about the x-axis, meaning that the upper and lower half of the curves are mirror images of each other. - - Graph of the vector valued function from the example. - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - xtick={1,2,3,4}, - ymin=-8.9,ymax=8.9, - xmin=-.5,xmax=4.5 - ] - - \addplot+ [domain=-2:2] ({x^2},{x^3}); - - \draw [->,thick,firstcolor] (axis cs: 1,-1) -- (axis cs:0.98,-.97); - \draw [->,thick,firstcolor] (axis cs: 0.98,.97) -- (axis cs: 1,1); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - - - - -

    - \vec r(t) = \la 1/t,1/t^2\ra, - for -2\leq t\leq 2. -

    -     - - - - - Graph of the function \vec r(t) = \la 1/t,1/t^2\ra, for -2\leq t\leq 2. - The graph of the function \vec r(t) begins at the point (-\frac12,\frac14) corresponding to t=2, heads upwards and to the left as t increases, following the path of the left side of the parabola y=x^2. - At t=2 the function starts at the point (\frac12,\frac14) and as t decreases follows the path of the right side of the parabola given by y=x^2. - The function is also mirrored about y-axis, and is equivalent to the parabola y=x^2 outside of the region -\frac12 \leq x \frac12. - - Graph of the vector valued function from the example. - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-.5,ymax=10.5, - xmin=-5.5,xmax=5.5 - ] - - \addplot+ [domain=-2:-.1,samples=40] ({1/x},{1/x^2}); - \addplot+ [solid,domain=.1:2,samples=40] ({1/x},{1/x^2}); - - \draw [->,thick,firstcolor] (axis cs: -1,1) -- (axis cs:-1.01,1.02); - \draw [->,thick,secondcolor] (axis cs: 1.01,1.02) -- (axis cs: 1,1); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - - - - -

    - \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, - for -2\pi\leq t\leq 2\pi. -

    -     - - - - - Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi. - The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi. - As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease. - The function continues to decrease, crossing the xaxis at x=1. - After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin. - The function now increases upwards and to the right until reaching a maximum near point (0.25,1). - After this point, the curve decreases, crossing the x-axis once again at x=1. - The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0). - The function is also symmetric about the x-axis. - - Graph of the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.1,ymax=1.1, - xmin=-.5,xmax=4.5 - ] - - \addplot+ [domain=-6.28:6.28,samples=120] ({.1*x^2},{sin(deg(x))}); - - \draw [->,thick,firstcolor,>=stealth] (axis cs: 0.1, -0.841471) -- (axis cs:0.09801, -0.836026); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - - - - -

    - \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, - for -2\pi\leq t\leq 2\pi. -

    -     - - - - - Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi. - The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi. - As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease. - The function continues to decrease, crossing the xaxis at x=1. - After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin. - The function now increases upwards and to the right until reaching a maximum near point (0.25,1). - After this point, the curve decreases, crossing the x-axis once again at x=1. - The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0). - The function is also symmetric about the x-axis. - - Graph of the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,2,3,4}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-0.1,xmax=4.1 - ] - - \addplot+ [domain=-6.28:6.28,samples=70] ({x*x/10},{sin(deg(x))}); - - \draw [->,thick,firstcolor,>=stealth] (axis cs:3.61201, 0.2698) -- (axis cs: 3.6, 0.279415); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - - - - -

    - \vec r(t) = \la 3\sin(\pi t),2\cos(\pi t)\ra, on [0,2]. -

    -     - - - - - Graph of the function \vec r(t) = \la 3\sin(\pi t),2\cos(\pi t)\ra, on [0,2]. - The graph of the function \vec r(t) is an oval having a horizontal width of 6 and a height of 4 centered at the origin. - The rightmost point of the oval is the point (3,0). - The leftmost point of the oval is the point (-3,0). - The highest point of the oval is the point (0,2), while the lowest is the point (0,-2). - - Graph of the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-3,-2,-1,1,2,3}, - ytick={-1,-2,1,2}, - ymin=-2.5,ymax=2.5, - xmin=-3.5,xmax=3.5 - ] - - \addplot+ [domain=0:2,samples=70] ({3*sin(deg(3.14159*x))},{2*cos(deg(3.14159*x))}); - - \draw [->,thick,firstcolor,>=stealth] (axis cs:1.64607, -1.67205)--(axis cs: 1.62091, -1.68294); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - - - - -

    - \vec r(t) = \la 3\cos(t) ,2\sin(2 t)\ra, on [0,2\pi]. -

    -     - - - - - Graph of the function \vec r(t) = \la 3\cos(t) ,2\sin(2 t)\ra, on [0,2\pi]. - The graph of the function \vec r(t) resembles the symbol \infty. - The curve begins at the point (3,0) which corresponds to t=0. - The curve then begins going upwards and to the left, until it reaches a maximum near the point (2,2). - From here the curve begins going downwards and to the left, passing through the origin, until reaching a minimum near (-2,-2). - The curve then begins going upwards to the left, passing through the point (-3,0), after which it begins going to the right. - The curve reaches a maximum near the point (-2,2), after which it begins decreasing and going to the right, once again crossing the origin. - Finally, the curve reaches another minimum near the point (2,-2), after which it begins going upwards to the right, until reaching its starting point given by (3,0). - The curve is symmetric about both the x and y axes. - - Graph of the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-3,-2,-1,1,2,3}, - ytick={-1,-2,1,2}, - ymin=-2.5,ymax=2.5, - xmin=-3.5,xmax=3.5 - ] - - \addplot+ [domain=0:6.28,samples=120] ({3*cos(deg(x))},{2*sin(deg(2*x))}); - - \draw [->,thick,firstcolor] (axis cs:1.62091, -1.81859) -- (axis cs: 1.64607, -1.83488); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - - - - -

    - \vec r(t) = \la 2\sec(t) ,\tan(t) \ra, on [-\pi,\pi]. -

    -     - - - - - Graph of the function \vec r(t) = \la 2\sec(t) ,\tan(t) \ra, on [-\pi,\pi]. - The graph of the function is symmetric about both the x and y axes. - The curve crosses the x-axis at the point x=2 corresponding to t=\pi. - After this point the upper part of the curve goes upwards and to the right in a mostly linear fashion. - Since the curve is symmetric about the y-axis, the lower part of the curve is the mirrored version of the upper part, meaning that it goes downwards and to the left. - As the curve is symmetric about the x-axis, the left side of the curve is the mirrored version of the right part, meaning that it crosses the x-axis at x=-2. - From this point, the upper part of the left side of the curve goes upwards and to the left, while the lower part goes downwards and to the left. - - Graph of the vector valued function from the example. - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-5.5,ymax=5.5, - xmin=-11,xmax=11 - ] - - \addplot+ [domain=-1.4708:1.4708,samples=40] ({2*sec(deg(x))},{tan(deg(x))}); - \addplot+ [solid,domain=1.6708:4.61239,samples=40] ({2*sec(deg(x))},{tan(deg(x))}); - - \draw [->,thick,firstcolor] (axis cs:2.27899, 0.546302) -- (axis cs: 2.29162, 0.559359); - \draw [->,thick,secondcolor] (axis cs: -2.02022, -0.1425477) -- (axis cs: -2.01744, -0.132358); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - -     - - - -

    - Sketch the vector-valued function on the given interval in \mathbb{R}^3. - Technology may be useful in creating the sketch. -

    -     - - - - -

    - \vec r(t) = \la 2\cos(t) , t, 2\sin(t) \ra, on [0,2\pi]. -

    -     - - - - - Graph of the function \vec r(t) = \la 2\cos(t) , t, 2\sin(t) \ra, on [0,2\pi]. - The graph of the function is a circular spiral centered about the y-axis. - Ignoring the y-axis, the curve is simply a singular circle of radius 2 in the xz-plane. - Incorporating the y coordinate then creates the linearly increasing spiral, which begins at the point (2,0,0), completes precisely one full revolution and ends at the point (2,2\pi,0). - - Graph of the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - axis lines=center, - view={145}{25}, - xtick={1}, - ymin=-.1,ymax=6.5, - xmin=-2.1,xmax=2.1, - zmin=-2.1,zmax=2.1, - every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, - xlabel={ $x$}, - every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, - ylabel={ $y$}, - every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, - zlabel={ $z$} - ] - - \addplot3 [firstcurvestyle,domain=0:6.28,samples y=0] ({2*cos(deg(x))},{x},{2*sin(deg(x))}); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - - - - -

    - \vec r(t) = \la 3\cos(t) , \sin(t) , t/\pi\ra on [0,2\pi]. -

    -     - - - - - Graph of the function \vec r(t) = \la 3\cos(t) , \sin(t) , t/\pi\ra on [0,2\pi]. - The graph of the function is an oval-shaped spiral centered about the z-axis. - Ignoring the z-axis, the curve is simply an oval having a horizontal width of 6 and a height of 2 in the xy-plane. - Incorporating the z coordinate then creates the linearly increasing oval spiral, which begins at the point (3,0,0), completes precisely one full revolution and ends at the point (3,0,2). - - Graph of the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - axis lines=center, - view={145}{25}, - ymin=-1.1,ymax=1.1, - xmin=-3.5,xmax=3.5, - zmin=-0.1, zmax=2.1, - every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, - xlabel={ $x$}, - every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, - ylabel={ $y$}, - every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, - zlabel={ $z$} - ] - - \addplot3 [firstcurvestyle,domain=0:6.28,samples y=0,samples=60] ({3*cos(deg(x))},{sin(deg(x))},{x/3.14159}); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - - - - -

    - \vec r(t) = \la \cos(t) , \sin(t) ,\sin(t) \ra on [0,2\pi]. -

    -     - - - - - Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(t) \ra on [0,2\pi]. - The graph of the function is an oval lying in the plane coming from rotating the xy plane 45 degrees towards the z-axis. - The oval lying in this plane has a horizontal width of \sqrt{2} and a height of 1. - Ignoring the z coordinate, the curve is a unit circle in the xy plane. - Similarly ignoring the y coordinate, the curve is a unit circle in the xz plane. - If we now ignore the x coordinate, the resulting curve is a diagonal line given by z=y in the yz plane. - This line turns back on itself, which can be seen in the image of the oval when considering all three coordinate axes. - - Graph of the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - axis lines=center, - view={145}{25}, - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.1, - zmin=-1.1, zmax=1.1, - every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, - xlabel={ $x$}, - every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, - ylabel={ $y$}, - every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, - zlabel={ $z$} - ] - - \addplot3 [firstcurvestyle,domain=0:6.28,samples y=0,samples=60] ({cos(deg(x))},{sin(deg(x))},{sin(deg(x))}); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - - - - -

    - \vec r(t) = \la \cos(t) , \sin(t) ,\sin(2t)\ra on [0,2\pi]. -

    -     - - - - - Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(2t)\ra on [0,2\pi]. - The graph of the function resembles a saddle centered at the origin whose height is defined by the z-axis. - The two sides of the saddle that taper off fall into negative z and lie in the second and third quadrants in the xy plane. - Ignoring the z coordinate, the curve is a unit circle in the xy plane. - Ignoring the x or y coordinates individually, the curve looks like the \infty symbol in the yz and the xz planes, respectively. - We now describe the z coordinate with respect to travelling along the unit circle in the xy plane. - Starting at t=, the function begins at the point (1,0,0). - As t increases and we travel along the unit circle in the x and y coordinates, z increases until we get to t=\frac{\pi}{2} at which z=1. - Then, continuing along the unit circle, z decreases until it reaches a minimum of z=-1 when t=\frac{3\pi}{4}. - Continuing along the circle, z begins to increase once again, reaching one more maximum of z=1 when t=\frac{5\pi}{4}. - Finally, z begins to decrease, reaching its last minimum of z=1 when t=\frac{7\pi}{4}, after which z increases, and the curve ends where it began, at the point (1,0,0). - - Graph of the vector valued function from the example. - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - axis lines=center, - view={155}{25}, - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.1, - zmin=-1.1, zmax=1.1, - every axis x label/.style={at={(axis cs:\pgfkeysvalueof{/pgfplots/xmax},0,0)},xshift=-3pt,yshift=-3pt}, - xlabel={ $x$}, - every axis y label/.style={at={(axis cs:0,\pgfkeysvalueof{/pgfplots/ymax},0)},xshift=5pt,yshift=-2pt}, - ylabel={ $y$}, - every axis z label/.style={at={(axis cs:0,0,\pgfkeysvalueof{/pgfplots/zmax})},xshift=0pt,yshift=4pt}, - zlabel={ $z$} - ] - - \addplot3 [firstcurvestyle,domain=0:6.28,samples y=0,samples=120] ({cos(deg(x))},{sin(deg(x))},{sin(deg(2*x))}); - - \end{axis} - - \end{tikzpicture} - - - - - - -     - -     - - - -

    - Find \norm{\vec r(t)}. -

    -     - - - - - Context()->variables->are(t=>'Real'); - $norm=Formula("|t| sqrt(1+t^2)"); - - - -

    - If \vec r(t) = \la t,t^2\ra, - then \norm{\vec r(t)}=. -

    -     -     -     - - - - -

    - \vec r(t) = \la 5\cos(t) ,3\sin(t) \ra. -

    -     - -

    - \norm {\vec r(t)} = \sqrt{25\cos^2(t) +9\sin^2(t) }. -

    -     - -     - - - - - Context()->variables->are(t=>'Real'); - $norm=Formula("sqrt(4+t^2)"); - - - -

    - If \vec r(t) = \la 2\cos(t) ,2\sin(t) ,t\ra, - then \norm{\vec r(t)}=. -

    -     -     -     - - - - -

    - \vec r(t) = \la \cos(t) ,t,t^2\ra. -

    -     - -

    - \norm {\vec r(t)} = \sqrt{\cos^2(t) +t^2+t^4}. -

    -     - -     - -     - - - -

    - Create a vector-valued function whose graph matches the given description. -

    -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - $vvf=Formula("<2 cos(t)+1, 2 sin(t) + 2>"); - Context()->flags->set(tolType=>'absolute',tolerance=>0.001); - $vvfev=$vvf->cmp(checker=>sub{ - my ($correct,$student,$self)=@_; - my ($sx,$sy)=$student->value; - my ($cx,$cy)=$correct->value; - my $eliminated=Formula("(($sx-1)/2)^2 + (($sy-2)/2)^2"); - return 0 unless ($eliminated == Formula("1")); - #normalize answers - $sx=($sx-1)/2; - $sy=($sy-2)/2; - $cx=($cx-1)/2; - $cy=($cy-2)/2; - #shift student answers - my $sx0=$sx->eval(t=>0); - my $sy0=$sy->eval(t=>0); - my $phi=($sx0 >= 0)?Compute("arcsin($sy0)"):Compute("pi-arcsin($sy0)"); - $sx=$sx->substitute(t=>Formula("t-$phi")); - $sy=$sy->substitute(t=>Formula("t-$phi")); - return ($sx == $cx and $sy == $cy); - } - ); - - - -

    - A circle of radius 2, - centered at (1,2), - traced counter-clockwise once at constant speed on [0,2\pi). -

    - - - \vec r(t)= - -

    - -

    -     -     -     - - - - -

    - A circle of radius 3, centered at (5,5), - traced clockwise once on [0,2\pi]. -

    -     - -

    - Answers may vary; three solutions are -

    - -

    - \vec r(t) = \la 3\sin(t) +5,3\cos(t) +5\ra, -

    - -

    - \vec r(t) = \la -3\cos(t) +5,3\sin(t) +5\ra and -

    - -

    - \vec r(t) = \la 3\cos(t) +5,-3\sin(t) +5\ra. -

    -     - -     - - - - -

    - An ellipse, centered at (0,0) with vertical major axis of length 10 and minor axis of length 3, traced once counter-clockwise on [0,2\pi]. -

    -     - -

    - Answers may vary, though most direct solution is -

    - -

    - \vec r(t) = \la 1.5\cos(t) ,5\sin(t) \ra. -

    -     - -     - - - - -

    - An ellipse, centered at (3,-2) with horizontal major axis of length 6 and minor axis of length 4, traced once clockwise on [0,2\pi]. -

    -     - -

    - Answers may vary, though most direct solutions are -

    - -

    - \vec r(t) = \la -3\cos(t) +3,2\sin(t) -2\ra, -

    - -

    - \vec r(t) = \la 3\cos(t) +3,-2\sin(t) -2\ra and -

    - -

    - \vec r(t) = \la 3\sin(t) +3,2\cos(t) -2\ra. -

    -     - -     - - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - $vvf=ParametricLine("<t+2, 5t+3>"); - - - -

    - A line through (2,3) with a slope of 5. -

    - - - \vec r(t)= - -

    - -

    -     -     -     - - - - -

    - A line through (1,5) with a slope of -1/2. -

    -     - -

    - Answers may vary, though most direct solutions are -

    - -

    - \vec r(t) = \la t,-1/2(t-1)+5\ra, -

    - -

    - \vec r(t) = \la t+1,-1/2t+5\ra, -

    - -

    - \vec r(t) = \la -2t+1,t+5\ra and -

    - -

    - \vec r(t) = \la 2t+1,-t+5\ra. -

    -     - -     - - - -

    - The line through points (1,2,3) and (4,5,6), where -

    - -

    - \vec r(0) = \la 1,2,3\ra and \vec r(1) = \la 4,5,6\ra. -

    -     - -

    - Specific forms may vary, though most direct solutions are -

    - -

    - \vec r(t) = \la 1,2,3\ra +t\la 3,3,3\ra and -

    - -

    - \vec r(t) = \la 3t+1, 3t+2, 3t+3\ra. -

    -     -     - - - -

    - The line through points (1,2) and (4,4), where -

    - -

    - \vec r(0) = \la 1,2\ra and \vec r(1) = \la 4,4\ra. -

    -     - -

    - Specific forms may vary, though most direct solutions are -

    - -

    - \vec r(t) = \la 1,2\ra +t\la 3,2\ra and -

    - -

    - \vec r(t) = \la 3t+1, 2t+2\ra. -

    -     -     - - - - - Context("Vector"); - Context()->variables->are(t=>'Real'); - $vvf=Formula("<2 cos(t), 2 sin(t), 2t>"); - Context()->flags->set(tolType=>'absolute',tolerance=>0.001); - $vvfev=$vvf->cmp(checker=>sub{ - my ($correct,$student,$self)=@_; - my ($sx,$sy,$sz)=$student->value; - my ($cx,$cy,$cz)=$correct->value; - return 0 unless ($sx == $cx and $sy == $cy); - $szd=$sz->reduce->D('t'); - return 0 unless ($szd == Formula("2")); - my $r = ($sz->eval(t=>0))/(4*pi); - return 0 unless ($r == int($r)); - } - ); - - - -

    - A vertically oriented helix with radius of 2 that starts at (2,0,0) and ends at - (2,0,4\pi) after one revolution on [0,2\pi]. -

    - - - \vec r(t)= - -

    - -

    -     -     -     - - - - -

    - A vertically oriented helix with radius of 3 that starts at (3,0,0) and ends at (3,0,3) after 2 revolutions on [0,1]. -

    -     - -

    - Answers may vary, though most direct solution is -

    - -

    - \vec r(t) = \la 3\cos(4\pi t),3\sin(4\pi t),3t\ra. -

    -     - -     - -     - - - -

    - Find the average rate of change of - \vec r(t) on the given interval. -

    -     - - - - - Context("Vector2D"); - $arc=Vector("<1,0>"); - - - -

    - \vec r(t) = \la t,t^2\ra on [-2,2]. -

    - -

    - -

    -     -     -     - - - - -

    - \vec r(t) = \la t,t+\sin(t) \ra on [0,2\pi]. -

    -     - -

    - \la 1,1\ra -

    -     - -     - - - - - Context("Vector"); - $arc=Vector("<0,0,1>"); - - - -

    - \vec r(t) = \la 3\cos(t) ,2\sin(t) ,t\ra on [0,2\pi]. -

    - -

    - -

    -     -     -     - - - - -

    - \vec r(t) = \la t,t^2,t^3\ra on [-1,3]. -

    -     - -

    - \la 1,2,7\ra -

    -     - -     - -     -     -     -
    -
    - Calculus and Vector-Valued Functions - -

    - The previous section introduced us to a new mathematical object, - the vector-valued function. - We now apply calculus concepts to these functions. - We start with the limit, - then work our way through derivatives to integrals. -

    -     - - - Limits of Vector-Valued Functions -

    - The initial definition of the limit of a vector-valued function is a bit intimidating, - as was the definition of the limit in . - The theorem following the definition shows that in practice, - taking limits of vector-valued functions is no more difficult than taking limits of real-valued functions. -

    - - - - Limits of Vector-Valued Functions - -

    - Let I be an open interval containing c, - and let \vec r(t) be a vector-valued function defined on I, - except possibly at c. - The limit of \vec r(t), - as t approaches c, - is \vec L, expressed as - - \lim_{t\to c} \vec r(t) = \vec L - , - means that given any \varepsilon \gt 0, - there exists a \delta \gt 0 such that for all t\neq c, - if \abs{t-c} \lt \delta, - we have \norm{\vec r(t) - \vec L} \lt \varepsilon. - - vector-valued functionlimits - limitof vector-valued functions -

    -     -     - -

    - Note how the measurement of distance between real numbers is the absolute value of their difference; - the measure of distance between vectors is the vector norm, - or magnitude, of their difference. -

    - -

    - - states that we can compute limits of vector-valued functions component-wise. -

    - - - Limits of Vector-Valued Functions - -

    -

      -
    1. -

      - Let \vec r(t) = \la \,f(t),g(t)\,\ra be a vector-valued function in - \mathbb{R}^2 defined on an open interval I containing c, - except possibly at c. - Then - vector-valued functionlimits - limitof vector-valued functions - - - \lim_{t\to c} \vec r(t) = \la \lim_{t\to c}f(t)\, , \,\lim_{t\to c} g(t)\ra - . -

      -
      2. - -
    2. -

      - Let \vec r(t) = \la \,f(t),g(t),h(t)\,\ra be a vector-valued function in - \mathbb{R}^3 defined on an open interval I containing c, - except possibly at c. - Then - - \lim_{t\to c} \vec r(t) = \la \lim_{t\to c}f(t)\, , \,\lim_{t\to c} g(t)\,, \,\lim_{t\to c} h(t)\ra - -

      -
      3. -
    -

    -     -     - - - - - Finding limits of vector-valued functions - -

    - Let \ds\vec r(t) = \la \frac{\sin(t) }{t},\, t^2-3t+3,\,\cos(t) \ra. - Find \lim\limits_{t\to 0}\vec r(t). -

    -     - -

    - We apply the theorem and compute limits component-wise. - - \lim_{t\to0} \vec r(t) \amp = \la \lim_{t\to 0}\frac{\sin(t) }{t}\, , \, \lim_{t\to 0} t^2-3t+3\, , \, \lim_{t\to 0} \cos(t) \ra - \amp = \la 1,3,1\ra - . -

    -     - -     -     - - - Continuity - - Continuity of Vector-Valued Functions - -

    - Let \vec r(t) be a vector-valued function defined on an open interval I containing c. - vector-valued functioncontinuity - continuous functionvector-valued - -

      -
    1. -

      - \vec r(t) is continuous at c - if \lim\limits_{t\to c} \vec r(t) = r(c). -

      -
      2. - -
    2. -

      - If \vec r(t) is continuous at all c in I, - then \vec r(t) is continuous on I. -

      -
      3. -
    -

    -     -     - - -

    - We again have a theorem that lets us evaluate continuity component-wise. -

    - - - Continuity of Vector-Valued Functions - -

    - Let \vec r(t) be a vector-valued function defined on an open interval I containing c. - Then \vec r(t) is continuous at c if, and only if, - each of its component functions is continuous at c. - vector-valued functioncontinuity - continuous functionvector-valued -

    -     -     - - - - - Evaluating continuity of vector-valued functions - -

    - Let \ds\vec r(t) = \la \frac{\sin(t) }{t},\, t^2-3t+3,\,\cos(t) \ra. - Determine whether \vec r is continuous at t=0 and t=1. -

    -     - -

    - While the second and third components of - \vec r(t) are defined at t=0, - the first component, (\sin(t) )/t, is not. - Since the first component is not even defined at t=0, - \vec r(t) is not defined at t=0, - and hence it is not continuous at t=0. -

    - -

    - At t=1 each of the component functions is continuous. - Therefore \vec r(t) is continuous at t=1. -

    -     -     -     - - - Derivatives -

    - Consider a vector-valued function \vec r defined on an open interval I containing t_0 and t_1. - We can compute the displacement of \vec r on [t_0,t_1], - as shown in . - Recall that dividing the displacement vector by t_1-t_0 gives the average rate of change on [t_0,t_1], - as shown in . -

    - -
    - Illustrating displacement, leading to an understanding of the derivative of vector-valued functions - -
    - - - - - Graph of an arbitrary vector-valued \vec r on the interval which includes [t_0,t_1]. - The function \vec r is a small part of a concave down circular arc. - The graph includes the vectors \vec r (t_0) and \vec r (t_1), which begin from the origin and end at the corresponding point of the function \vec r . - The graph also includes the vector \vec r (t_1) - \vec r (t_0), which begins where \vec r (t_0) ends, and then terminates at the same termination point as \vec r (t_1). - The three vectors \vec r (t_0), \vec r (t_1) and \vec r (t_1) - \vec r (t_0) for a triangle, where following the path of \vec r (t_0) and \vec r (t_1) - \vec r (t_0) takes you to the same point as \vec r (t_1). - The vector \vec r (t_0) terminates on the left side of the circular arc, while \vec r (t_1) terminates further on the right side of the circular arc of given by the function \vec r. - - Illustration of a vector-valued function on a given interval. - - - \begin{tikzpicture}[>=stealth] - - \draw [thick,firstcolor] (2,1) arc [x radius=4,y radius=3,start angle=145,end angle=135] node (A) {}; - \draw [thick,firstcolor] (A.center) arc [x radius=4,y radius=3,start angle=135,end angle=70] node (B) {}; - \draw [thick,firstcolor] (B.center) arc [x radius=4,y radius=3,start angle=70,end angle=60]; - - \draw [thick,->] (0,0) -- (A.center) node [left,pos=.6] { $\vec r(t_0)$}; - \draw [thick,->] (0,0) -- (B.center) node [below,pos=.5] { $\vec r(t_1)$}; - \draw [->,thick,secondcolor] (A.center) -- (B.center) node [above,black,pos=.5,sloped] { $\vec r(t_1)-\vec r(t_0)$}; - - \end{tikzpicture} - - - - -
    - -
    - - - - - Graph of the same arbitrary vector-valued \vec r as well as the vectors \vec r (t_0) and \vec r (t_1) as described in the previous image. - This time the graph includes two additional vectors \vrp (t_0) and \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} . - The vector \vrp (t_0) begins at the termination point of \vec r (t_0), and is tangent to the function \vec r at this point. - The vector \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} also begins at the termination point of \vec r (t_0), and follows the path of the vector \vec r (t_1) - \vec r (t_0) from the previous image. - However, this vector does not end at the termination point of \vec r (t_1) but instead terminates at some point further away in the same direction as the vector \vec r (t_1) - \vec r (t_0). - - Illustration of a vector-valued function on a given interval showcasing a derivative vector. - - - \begin{tikzpicture}[>=stealth] - - \draw [thick,firstcolor] (2,1) arc [x radius=4,y radius=3,start angle=145,end angle=135] node (A) {}; - \draw [thick,firstcolor] (A.center) arc [x radius=4,y radius=3,start angle=135,end angle=70] node (B) {}; - \draw [thick,firstcolor] (B.center) arc [x radius=4,y radius=3,start angle=70,end angle=60]; - - \draw [thick,->] (0,0) -- (A.center) node [left,pos=.6] { $\vec r(t_0)$}; - \draw [thick,->] (0,0) -- (B.center) node [below,pos=.5] { $\vec r(t_1)$}; - \draw [->,thick,secondcolor] (A.center) -- ($(A)!1.2!(B)$) node [above,black,pos=.99] { $\displaystyle\frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0}$}; - - \draw [->,thick,secondcolor] (A.center)--($(A)!.4!29:(B)$) node [above,black] { $\vrp(t_0)$}; - - \end{tikzpicture} - - - - -
    -     -
    - -

    - The derivative of a vector-valued function is a measure of the - instantaneous rate of change, - measured by taking the limit as the length of [t_0,t_1] goes to 0. - Instead of thinking of an interval as [t_0,t_1], - we think of it as [c,c+h] for some value of h - (hence the interval has length h). - The average rate of change is - - \frac{\vec r(c+h)-\vec r(c)}{h} - - for any value of h\neq0. - We take the limit as h\to0 to measure the instantaneous rate of change; - this is the derivative of \vec r. -

    - - - Derivative of a Vector-Valued Function - -

    - Let \vec r(t) be continuous on an open interval I containing c. - vector-valued functionderivatives - derivativevector-valued functions - -

      -
    1. -

      - The derivative of \vec r at t=c is - - \vrp (c) = \lim_{h\to 0} \frac{\vec r(c+h) - \vec r(c)}{h} - . -

      -
      2. - -
    2. -

      - The derivative of \vec r is - - \vrp (t) = \lim_{h\to 0} \frac{\vec r(t+h) - \vec r(t)}{h} - . -

      -
      3. -
    -

    -     -     - - -

    - If a vector-valued function has a derivative for all c in an open interval I, - we say that \vec r(t) is - differentiable on I. -

    - -

    - Once again we might view this definition as intimidating, - but recall that we can evaluate limits component-wise. - The following theorem verifies that this means we can compute derivatives component-wise as well, - making the task not too difficult. -

    - - - - Derivatives of Vector-Valued Functions - -

    -

      -
    1. -

      - Let \vec r(t) = \la \, f(t), g(t)\,\ra. - Then - - \vrp(t) = \la\, \fp(t), \gp(t)\, \ra - . -

      -
      2. - -
    2. -

      - Let \vec r(t) = \la \, f(t), g(t), h(t)\,\ra. - Then - vector-valued functionderivatives - derivativevector-valued functions - - \vrp(t) = \la\, \fp(t), \gp(t), h'(t)\, \ra - . -

      -
      3. -
    -

    -     -     - - - - - Derivatives of vector-valued functions - -

    - Let \vec r(t) = \la t^2,t\ra. -

    - -

    -

      -
    1. -

      - Sketch \vec r(t) and \vrp(t) on the same axes. -

      -
      2. - -
    2. -

      - Compute \vrp(1) and sketch this vector with its initial point at the origin and at \vec r(1). -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - - allows us to compute derivatives component-wise, so - - \vrp(t) = \la 2t, 1\ra - . - \vec r(t) and \vrp(t) are graphed together in . - Note how plotting the two of these together, - in this way, is not very illuminating. - When dealing with real-valued functions, - plotting f(x) with \fp(x) gave us useful information as we were able to compare f and \fp at the same x-values. - When dealing with vector-valued functions, - it is hard to tell which points on the graph of \vrp correspond to which points on the graph of \vec r. -

      -
      2. - -
    2. -

      - We easily compute \vrp(1) = \la 2,1\ra, - which is drawn in - with its initial point at the origin, - as well as at \vec r(1) = \la 1,1\ra. - These are sketched in . -

      - -
      - Graphing the derivative of a vector-valued function in - -
      - - - - - Graph of the vector-valued function \vec r(t) = \la t^2,t\ra. - The graph of the function \vec r(t) is simply the graph of the parabola y=x^2, but instead of opening towards the positive y-axis, the function \vec r(t) opens towards the positive x-axis. - The function \vec r(t) is plotted on the interval [-2,2], where \vec r(-2)= \la 4,-2\ra and \vec r(2)= \la 4,2\ra. - The graph also includes the derivative function \vrp(t) = \la 2t, 1\ra which takes the path of the horizontal line y=1 going from left to right in the standard coordinate axes. - - Graph of the vector-valued function from the example and its derivative. - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-2.2,ymax=2.2, - xmin=-4.54,xmax=4.54 - ] - - \addplot+ [domain=-2:2,samples=40] ({x^2},{x}); - - \draw (axis cs:2,1.7) node { $\vec r(t)$}; - - \draw [thick,secondcolor] (axis cs: -4,1) -- (axis cs:4,1) node [below,pos=.9,black]{ $\vrp(t)$}; - \draw [thick,->,secondcolor] (axis cs: -2,1) -- (axis cs:-1.99,1); - - \draw [thick,->,firstcolor] (axis cs:1,-1) -- (axis cs:0.9801,-.99); - - \end{axis} - - \end{tikzpicture} - - - - -
      - -
      - - - - - Graph of the vector-valued function \vec r(t) = \la t^2,t\ra described in the previous image. - The graph also includes two copies of the vector \vrp(1) = \la 2,1\ra. - The first copy of the vector \vrp(1) = \la 2,1\ra begins at the origin, and ends at the point (2,1), which is also a point on the derivative function \vrp(t) = \la 2t, 1\ra from the previous image. - The second copy of the vector \vrp(1) = \la 2,1\ra begins at the point (1,1), which corresponds to the termination point of \vec r(1) = \la 1,1 \ra. - The second copy of the vector \vrp(1) = \la 2,1\ra is tangent to the function \vec r(t) = \la t^2,t\ra at the point (1,1) corresponding to when t=1 in the function \vec r(t). - - Graph of the vector-valued function from the example and its derivative. - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-2.2,ymax=2.2, - xmin=-4.54,xmax=4.54 - ] - - \addplot+ [domain=-2:2,samples=40] ({x^2},{x}); - - \draw [thick,->,secondcolor] (axis cs: 0,0) -- (axis cs:2,1) node [shift={(15pt,0)},pos=.6,black] { $\vrp(1)$}; - \draw [thick,->,secondcolor] (axis cs: 1,1) -- (axis cs:3,2) node [left,pos=.8,black] { $\vrp(1)$}; - - \draw [thick,->,firstcolor] (axis cs:1,-1) -- (axis cs:0.9801,-.99); - - \end{axis} - - \end{tikzpicture} - - - - -
      -       -
      -
      3. -
    -

    -     - -     - - - Derivatives of vector-valued functions - -

    - Let \vec r(t) = \la \cos(t) , \sin(t) , t\ra. - Compute \vrp(t) and \vrp(\pi/2). - Sketch \vrp(\pi/2) with its initial point at the origin and at \vec r(\pi/2). -

    -     - -

    - We compute \vrp as \vrp(t) = \la -\sin(t) , \cos(t) , 1\ra. - At t= \pi/2, we have \vrp(\pi/2) = \la -1,0,1\ra. - - shows a graph of \vec r(t), - with \vrp(\pi/2) plotted with its initial point at the origin and at \vec r(\pi/2). -

    - -
    - Viewing a vector-valued function and its derivative at one point - - - Graph of the vector-valued function \vec r(t) = \la \cos(t) , \sin(t) , t\ra. - The function is a circular spiral which climbs the z-axis. - Looking at the graph from above the z-axis, the function resembles a unit circle in the xy-plane. - Adding the z coordinate then creates the linearly increasing spiral. - The graph also includes two copies of the vector \vrp(\pi/2) = \la -1,0,1\ra - The first copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the origin, and ends at the point (-1,0,1). - The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the point (0,1,\pi/2), which corresponds to the termination point of \vec r(\pi/2). - The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra is also tangent to the function \vec r(t) at the point (0,1,\pi/2) corresponding to when t=\pi/2 in the function \vec r(t). - - Graph of the vector-valued function from the example and its derivative at a point. - - - - - //ASY file for figvvf23D.asy in Chapter 11 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,5,10); - //currentprojection=orthographic(-5.6,3.8,9.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myzchoice={0,2,4,6}; - real[] myychoice={-1,1}; - - defaultpen(0.5mm); - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,1.25); - pair zbounds=(-0.5,7); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple g(real t) {return (cos(t),sin(t),t);} - path3 mypath=graph(g,0,2pi,operator ..); - - defaultpen(0.75mm); - draw(O--(-1,0,1),redpen,Arrow3(size=4mm)); - - draw(g(pi/2)--g(pi/2)+(-1,0,1),redpen,Arrow3(size=4mm)); - - draw(mypath,bluepen,Arrow3(size=4mm)); - - - - -
    -     -     - -

    - In Examples - and , - sketching a particular derivative with its initial point at the origin did not seem to reveal anything significant. - However, when we sketched the vector with its initial point on the corresponding point on the graph, - we did see something significant: - the vector appeared to be tangent to the graph. - We have not yet defined what tangent - means in terms of curves in space; - in fact, we use the derivative to define this term. -

    - - - Tangent Vector, Tangent Line - -

    - Let \vec r(t) be a differentiable vector-valued function on an open interval I containing c, - where \vrp(c)\neq \vec 0. - tangent line - vector-valued functiontangent line -

    - -

    -

      -
    1. -

      - A vector \vec v is tangent to the graph of \vec r(t) at t=c - if \vec v is parallel to \vrp(c). -

      -
      2. - -
    2. -

      - The tangent line to the graph of - \vec r(t) at t=c is the line through - \vec r(c) with direction parallel to \vrp(c). - An equation of the tangent line is - - \vec \ell(t) = \vec r(c) + t\,\vrp(c) - . -

      -
      3. -
    -

    -     -     - - - - - Finding tangent lines to curves in space - -

    - Let \vec r(t) = \la t,t^2,t^3\ra on [-1.5,1.5]. - Find the vector equation of the line tangent to the graph of \vec r at t=-1. -

    -     - - -

    - To find the equation of a line, - we need a point on the line and the line's direction. - The point is given by \vec r(-1) = \la -1,1,-1\ra. - (To be clear, \la -1,1,-1\ra is a - vector, not a point, - but we use the point pointed to by this vector.) -

    - -

    - The direction comes from \vrp(-1). - We compute, component-wise, - \vrp(t) = \la 1,2t, 3t^2\ra. - Thus \vrp(-1) = \la 1,-2,3\ra. -

    - -
    - Graphing a curve in space with its tangent line - - - - - Graph of the vector-valued function \vec r(t) = \la t,t^2,t^3\ra on [-1.5,1.5]. - The function begins at the point (-1.5,2.25,-3.375), from which it begins to increase linearly in the x and z coordinates, and decrease in the y coordinate. - The function then curves towards the origin. - After passing through the origin, the function begins to increase in all x,y and z coordinates until it reaches the point (1.5,2.25,3.375), at which it ends. - The graph also contains the line \ell(t) = \la -1,1,-1\ra + t\la 1,-2,3\ra, which is the tangent line to the function at \vec r(-1), which is the point (-1,1,-1). - The line can be described as the line which passes through the point (-1,1,-1), moving in the direction of the vector \la 1,-2,3\ra. - Additionally, the line is defined for all t, so it also moves in the opposite direction of the vector \la 1,-2,3\ra, or in other words in the direction of \la -1,2,-3\ra from the point (-1,1,-1). - - Graph of the vector-valued function from the example and the tangent line to a point on the curve. - - - - - //ASY file for figvvfderiv1.asy in Chapter 11 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - pair xbounds=(-2.75,2.75); - pair ybounds=(-2.75,2.75); - pair zbounds=(-2.75,2.75); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve <t,t^2,t^3> for t from -1.5 to 1.5 - triple g(real t) {return (t,t^2,t^3);} - path3 mypath=graph(g,-1.5,1.5,operator ..); draw(mypath,bluepen); - - //Draw the line <-1,1,-1>+t<1,-2,3> for t=-0.75 to t=1 - draw((-1.75,2.5,-3.25)--(0,-1,2),redpen+linewidth(1)); - //label("$\vec{r}\,'(\pi/2)$",(-1,0,1),N); - - - - -
    - -

    - The vector equation of the line is \ell(t) = \la -1,1,-1\ra + t\la 1,-2,3\ra. - This line and \vec r(t) are sketched in . -

    -     -     - - - Finding tangent lines to curves - -

    - Find the equations of the lines tangent to - \vec r(t) = \la t^3,t^2\ra at t=-1 and t=0. -

    -     - -

    - We find that \vrp(t) = \la 3t^2,2t\ra. - At t=-1, we have - - \vec r(-1) = \la -1,1\ra \text{ and } \vrp(-1) = \la 3,-2\ra - , - so the equation of the line tangent to the graph of \vec r(t) at t=-1 is - - \ell(t) = \la -1,1\ra + t\la 3,-2\ra - . -

    - -

    - This line is graphed with \vec r(t) in . -

    - -
    - Graphing \vec r(t) and its tangent line in - - - - Graph of the vector-valued function \vec r(t) = \la t^3,t^2\ra. - The function begins near the point (-3,2), from which it is concave down and sloping downwards towards the origin. - After passing through the origin, the curve is concave down and begins increasing in both x and y coordinates. - The curve is also symmetric about the y-axis. - The graph also contains the line \ell(t) = \la -1,1\ra + t\la 3,-2\ra , which is tangent to the function \vec r(t) = \la t^3,t^2\ra at the point (-1,1) corresponding to when t=1. - - Graph of the vector-valued function from the example and the tangent line to a point on the curve. - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-2.96,ymax=2.96, - xmin=-3.5,xmax=3.5 - ] - - \addplot+ [domain=-1.5:1.5,samples=40] ({x^3},{x^2}); - \addplot+ [solid,domain=-1:1.5] ({3*x-1},{-2*x+1}); - - \draw (axis cs:2,2) node { $\vec r(t)$}; - \draw (axis cs:2,-1.7) node { $\vec \ell(t)$}; - - \draw [thick,firstcolor,->] (axis cs: 1,1) -- (axis cs:1.03,1.02); - \draw [thick,secondcolor,->] (axis cs:2,-1)--(axis cs:2.03,-1.02); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - At t=0, we have \vrp(0) = \la 0,0\ra=\vec 0! - This implies that the tangent line - has no direction. - We cannot apply , - hence cannot find the equation of the tangent line. -

    -     - -     - -

    - We were unable to compute the equation of the tangent line to - \vec r(t)= \la t^3,t^2\ra at t=0 because \vrp(0) = \vec 0. - The graph in - shows that there is a cusp at this point. - This leads us to another definition of smooth, - previously defined by - in . -

    - - - Smooth Vector-Valued Functions - -

    - Let \vec r(t) be a differentiable vector-valued function on an open interval I where \vrp(t) is continuous on I. - \vec r(t) is smooth - on I if \vrp(t)\neq \vec 0 on I. - smooth - vector-valued functionsmooth -

    -     -     - - - -

    - Having established derivatives of vector-valued functions, - we now explore the relationships between the derivative and other vector operations. - The following theorem states how the derivative interacts with vector addition and the various vector products. -

    - - - Properties of Derivatives of Vector-Valued Functions - -

    - Let \vec r and \vec s be differentiable vector-valued functions, - let f be a differentiable real-valued function, - and let c be a real number. - vector-valued functionderivatives - derivativevector-valued functions - dot productand derivatives - cross productand derivatives -

    - -

    -

      -
    1. -

      - \ds \frac{d}{dt}\Big(\vec r(t) \pm \vec s(t)\Big) = \vrp(t) \pm \vec s\,'(t) -

      -
      2. - -
    2. -

      - \ds \frac{d}{dt}\Big(c\vec r(t)\Big) = c\vrp(t) -

      -
      3. - -
    3. -

      - \ds \frac{d}{dt}\Big(f(t)\vec r(t)\Big) = \fp(t)\vec r(t) + f(t)\vrp(t) Product Rule -

      -
      4. - -
    4. -

      - \ds \frac{d}{dt}\Big(\vec r(t)\cdot \vec s(t) \Big) = \vrp(t)\cdot \vec s(t) + \vec r(t)\cdot \vec s\,'(t) Product Rule -

      -
      5. - -
    5. -

      - \ds \frac{d}{dt}\Big(\vec r(t)\times \vec s(t) \Big) = \vrp(t)\times \vec s(t) + \vec r(t)\times \vec s\,'(t) Product Rule -

      -
      6. - -
    6. -

      - \ds \frac{d}{dt}\Big(\vec r\big(f(t)\big)\Big) = \vrp\big(f(t)\big)\fp(t) Chain Rule -

      -
      7. -
    -

    -     -     - - - - - Using derivative properties of vector-valued functions - -

    - Let \vec r(t) = \la t, t^2-1\ra and let - \vec u(t) be the unit vector that points in the direction of \vec r(t). -

    - -

    -

      -
    1. -

      - Graph \vec r(t) and - \vec u(t) on the same axes, on [-2,2]. -

      -
      2. - -
    2. -

      - Find \vec u\,'(t) and sketch \vec u\,'(-2), - \vec u\,'(-1) and \vec u\,'(0). - Sketch each with initial point the corresponding point on the graph of \vec u. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - To form the unit vector that points in the direction of \vec r, - we need to divide \vec r(t) by its magnitude. - - \norm{\vec r(t)} = \sqrt{t^2+(t^2-1)^2} \Rightarrow \vec u(t) = \frac{1}{\sqrt{t^2+(t^2-1)^2}}\la t,t^2-1\ra - . - \vec r(t) and \vec u(t) are graphed in . - Note how the graph of \vec u(t) forms part of a circle; - this must be the case, as the length of - \vec u(t) is 1 for all t. -

      - -
      - Graphing \vec r(t) and \vec u(t) in - - - - Graph of the vector-valued function \vec r(t) = \la t,t^2 -1\ra on [-2,2]. - The function looks like the parabola y=x^2 -1, which takes the path going towards positive x. - The function begins at the point (-2,3), decreases until reaching the point (0,-1), and then increases until ending at the point (2,3). - The graph also contains the function \vec u(t) which is the unit vector that points in the direction of \vec r(t). - The function \vec u(t) looks like the unit circle which is missing a piece of the top. - The circular arc from the graph of \vec u(t) begins at the point where the vector \vec r(-2)=\la -2,3\ra crosses the unit circle. - The circular arc then goes counterclockwise following the path of the unit circle, until ending at the point where the vector\vec r(2)=\la 2,3\ra crosses the unit circle. - - Graph of the vector-valued function with the unit vector function which points in the direction of the function. - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-1.1,ymax=3.1, - xmin=-2.5,xmax=2.5 - ] - - \addplot+ [domain=-2:2,samples=40] ({x},{x^2-1}); - \addplot+ [solid,domain=-2:2,samples=60] ({x/sqrt(x^2+(x^2-1)^2)},{(x^2-1)/sqrt(x^2+(x^2-1)^2)}); - - \draw (axis cs:-2,1.5) node { $\vec r(t)$}; - \draw (axis cs:.7,1.1) node { $\vec u(t)$}; - - \draw [thick,->,secondcolor] (axis cs: -0.768,.64) -- (axis cs:-.7737,.633); - \draw [thick,->,firstcolor] (axis cs:-1.5,1.25) -- (axis cs:-1.49,1.22); - - \end{axis} - - \end{tikzpicture} - - - - -
      -
      2. - -
    2. -

      - To compute \vec u\,'(t), - we use , writing - - \vec u(t) = f(t)\vec r(t), \text{ where } f(t) = \frac{1}{\sqrt{t^2+(t^2-1)^2}}=\big(t^2+(t^2-1)^2\big)^{-1/2} - . - (We could write - - \vec u(t) = \la \frac{t}{\sqrt{t^2+(t^2-1)^2}}, \frac{t^2-1}{\sqrt{t^2+(t^2-1)^2}}\ra - - and then take the derivative. - It is a matter of preference; - this latter method requires two applications of the Quotient Rule where our method uses the Product and Chain Rules.) We find \fp(t) using the Chain Rule: - - \fp(t) \amp = -\frac12\big(t^2+(t^2-1)^2\big)^{-3/2}\big(2t+2(t^2-1)(2t)\big) - \amp = -\frac{2t(2t^2-1)}{2\big(\sqrt{t^2+(t^2-1)^2}\,\big)^3} - - We now find \vec u\,'(t) using part 3 of : - - \vec u\,'(t) \amp = \fp(t)\vec u(t) + f(t)\vec u\,'(t) - \amp = -\frac{2t(2t^2-1)}{2\big(\sqrt{t^2+(t^2-1)^2}\,\big)^3}\la t,t^2-1\ra + \frac{1}{\sqrt{t^2+(t^2-1)^2}}\la 1,2t\ra - . - This is admittedly very messy; - such is usually the case when we deal with unit vectors. - We can use this formula to compute \vec u\,'(-2), - \vec u\,'(-1) and \vec u\,'(0): - - \vec u\,'(-2) \amp = \la-\frac{15}{13 \sqrt{13}},-\frac{10}{13 \sqrt{13}}\ra \approx \la -0.320,-0.213\ra - \vec u\,'(-1) \amp = \la 0,-2\ra - \vec u\,'(0) \amp = \la 1,0\ra - -

      - -
      - Graphing some of the derivatives of \vec u(t) in - - - - Graph of the function \vec u(t) which is the unit vector that points in the direction of \vec r(t). - The graph also contains three unit vectors, \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0). - The vector \vec u\,'(-2)\approx \la -0.320,-0.213\ra begins at the starting point of the function \vec u(t) and points in the direction tangent to the circular arc at t=-2. - The vector \vec u\,'(-1)= \la 0,-2\ra begins at the point (-1,0) corresponding to the termination point of \vec u(-1) - From here, the vector points straight down, as it is tangent to the leftmost point of the circular arc. - The vector \vec u\,'(0)= \la 1,0\ra begins at the point (0,-1) corresponding to the termination point of \vec u(0). - From here the vector points to the right, as it is tangent to the bottom point of the circular arc. - If the vectors \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0) were extended to lines, all three lines would be tangent to a point on the nearly complete unit circle given by \vec u(t). - - Graph of the circular unit vector function along with three derivative vectors of the function. - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-2.1,ymax=1.1, - xmin=-1.92,xmax=1.92 - ] - - \addplot+ [domain=-2:2,samples=80] ({x/sqrt(x^2+(x^2-1)^2)},{(x^2-1)/sqrt(x^2+(x^2-1)^2)}); - - \draw (axis cs:.5,.6) node { $\vec u(t)$}; - \draw [thick,firstcolor,->] (axis cs: -0.98226, 0.187522) -- (axis cs:-0.985435, 0.170055); - \draw [thick,firstcolor,->] (axis cs:-1.5,1.25) -- (axis cs:-1.49,1.22); - \draw [thick,secondcolor,->] (axis cs:-0.5547, 0.83205) -- (axis cs:-0.87472, 0.618704); - \draw [thick,secondcolor,->] (axis cs:-1., 0) -- (axis cs:-1., -2.); - \draw [thick,secondcolor,->] (axis cs:0., -1.) -- (axis cs:1., -1.); - - \end{axis} - - \end{tikzpicture} - - - - -
      - -

      - Each of these is sketched in . - Note how the length of the vector gives an indication of how quickly the circle is being traced at that point. - When t=-2, the circle is being drawn relatively slow; - when t=-1, the circle is being traced much more quickly. -

      -
      3. -
    -

    -     - -     - -

    - It is a basic geometric fact that a line tangent to a circle at a point P is perpendicular to the line passing through the center of the circle and P. - This is illustrated in ; - each tangent vector is perpendicular to the line that passes through its initial point and the center of the circle. - Since the center of the circle is the origin, - we can state this another way: - \vec u\,'(t) is orthogonal to \vec u(t). -

    - -

    - Recall that the dot product serves as a test for orthogonality: - if \vec u\cdot \vec v = 0, - then \vec u is orthogonal to \vec v. - Thus in the above example, \vec u(t)\cdot \vec u\,'(t)=0. -

    - -

    - This is true of any vector-valued function that has a constant length, that is, - that traces out part of a circle. - It has important implications later on, - so we state it as a theorem (and leave its formal proof as an Exercise.) -

    - - - Vector-Valued Functions of Constant Length - -

    - Let \vec r(t) be a vector-valued function of constant length that is differentiable on an open interval I. - That is, \norm{\vec r(t)} = c for all t in I; equivalently, - \vec r(t)\cdot \vec r(t) = c^2 for all t in I. - Then \vec r(t)\cdot\vrp(t) = 0 for all t in I. - vector-valued functionof constant length -

    -     -     -     - - - Integration -

    - Before formally defining integrals of vector-valued functions, - consider the following equation that our calculus experience tells us - should be true: - - \int_a^b \vrp(t)\, dt = \vec r(b) - \vec r(a) - . -

    - -

    - That is, the integral of a rate of change function should give total change. - In the context of vector-valued functions, - this total change is displacement. - The above equation is true; - we now develop the theory to show why. -

    - -

    - We can define antiderivatives and the indefinite integral of vector-valued functions in the same manner we defined indefinite integrals in . - However, we cannot define the definite integral of a vector-valued function as we did in . - That definition was based on the signed area between a function y=f(x) and the x-axis. - An area-based definition will not be useful in the context of vector-valued functions. - Instead, we define the definite integral of a vector-valued function in a manner similar to that of , - utilizing Riemann sums. -

    - - - Antiderivatives, Indefinite and Definite Integrals of Vector-Valued Functions - -

    - Let \vec r(t) be a continuous vector-valued function on [a,b]. - An antiderivative of \vec r(t) is a function - \vec R(t) such that \vec R'(t) = \vec r(t). -

    - -

    - The set of all antiderivatives of \vec r(t) is the - indefinite integral of \vec r(t), denoted by - - \int \vec r(t)\, dt - . -

    - -

    - The definite integral of \vec r(t) on [a,b] is - - \int_a^b \vec r(t)\, dt =\lim_{\norm{\Delta t}\to 0} \sum_{i=1}^n\vec r(c_i)\Delta t_i - , - where \Delta t_i is the length of the - ith subinterval of a partition of [a,b], - \norm{\Delta t} is the length of the largest subinterval in the partition, - and c_i is any value in the - ith subinterval of the partition. - antiderivativeof vector-valued function - definite integralof vector-valued function - indefinite integralof vector-valued function - integrationof vector-valued function -

    -     -     - -

    - It is probably difficult to infer meaning from the definition of the definite integral. - The important thing to realize from the definition is that it is built upon limits, - which we can evaluate component-wise. -

    - -

    - The following theorem simplifies the computation of definite integrals; - the rest of this section and the following section will give meaning and application to these integrals. -

    - - - Indefinite and Definite Integrals of Vector-Valued Functions - -

    - Let \vec r(t) = \la f(t),g(t)\ra be a vector-valued function in - \mathbb{R}^2 that is continuous on [a,b]. -

    - -

    -

      -
    1. -

      - \ds \int \vec r(t)\, dt = \la \int f(t)\, dt, \int g(t)\, dt\ra -

      -
      2. - -
    2. -

      - \ds \int_a^b \vec r(t)\, dt = \la \int_a^b f(t)\, dt, \int_a^b g(t)\, dt\ra -

      -
      3. -
    -

    - -

    - A similar statement holds for vector-valued functions in \mathbb{R}^3. - vector-valued functionintegration - integrationof vector-valued functions -

    -     -     - - - - - Evaluating a definite integral of a vector-valued function - -

    - Let \vec r(t) = \la e^{2t},\sin(t) \ra. - Evaluate \ds \int_0^1 \vec r(t) \,dt. -

    -     - -

    - We follow . - - \int_0^1 \vec r(t) \,dt \amp = \int_0^1 \la e^{2t},\sin(t) \ra \,dt - \amp = \la \int_0^1 e^{2t}\,dt\,, \int_0^1 \sin(t) \,dt \ra - \amp = \la \frac12e^{2t}\Big|_0^1\,, -\cos(t) \Big|_0^1\ra - \amp = \la \frac12(e^2-1)\,, -\cos(1)+1\ra - \amp \approx \la 3.19,0.460\ra - . -

    -     - -     - - - Solving an initial value problem - -

    - Let \vrp'(t) = \la 2, \cos(t) , 12t\ra. - Find \vec r(t) where: -

    - -

    -

      -
    • -

      - \vec r(0) = \la-7,-1,2\ra and -

      -
      • - -
    • -

      - \vrp(0) = \la 5,3,0\ra. -

      -
      • -
    -

    -     - -

    - Knowing \vrp'(t) = \la 2,\cos(t) , 12t\ra, - we find \vrp(t) by evaluating the indefinite integral. - - \int \vrp'(t)\,dt \amp = \la \int 2\,dt\,, \int \cos(t) \,dt\,, \int 12t\,dt\ra - \amp = \la 2t+C_1, \sin(t) + C_2, 6t^2 + C_3\ra - \amp = \la 2t,\sin(t) ,6t^2 \ra + \la C_1,C_2,C_3\ra - \amp = \la 2t,\sin(t) ,6t^2 \ra + \vec C - . -

    - -

    - Note how each indefinite integral creates its own constant which we collect as one constant vector \vec C. - Knowing \vrp(0) = \la 5,3,0\ra allows us to solve for \vec C: - - \vrp(t) \amp = \la 2t,\sin(t) ,6t^2 \ra + \vec C - \vrp(0) \amp = \la 0,0,0 \ra + \vec C - \la 5,3,0\ra \amp = \vec C - . -

    - -

    - So \vrp(t) = \la 2t,\sin(t) ,6t^2\ra + \la 5,3,0\ra = \la 2t+5, \sin(t) + 3, 6t^2\ra. - To find \vec r(t), we integrate once more. - - \int \vrp(t)\,dt \amp = \la \int 2t+5\,dt, \int \sin(t) + 3\,dt, \int 6t^2\,dt \ra - \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \vec C - . -

    - -

    - With \vec r(0) = \la -7,-1,2\ra, - we solve for \vec C: - - \vec r(t) \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \vec C - \vec r(0) \amp = \la 0,-1,0\ra + \vec C - \la -7,-1,2\ra \amp = \la 0,-1,0\ra + \vec C - \la -7,0,2\ra \amp = \vec C - . -

    - -

    - So - - \vec r(t) \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \la -7,0,2\ra - \amp = \la t^2+5t-7,-\cos(t) +3t,2t^3+2\ra - . -

    -     - -     - -

    - What does the integration of a vector-valued function mean? - There are many applications, - but none as direct as the area under the curve - that we used in understanding the integral of a real-valued function. -

    - -

    - A key understanding for us comes from considering the integral of a derivative: - - \int_a^b \vrp(t)\, dt = \vec r(t)\Big|_a^b = \vec r(b)-\vec r(a) - . -

    - -

    - Integrating a rate of change - function gives displacement. - displacement -

    - -

    - Noting that vector-valued functions are closely related to parametric equations, - we can describe the arc length of the graph of a vector-valued function as an integral. - Given parametric equations x=f(t), - y=g(t), the arc length on [a,b] of the graph is - - \text{ Arc Length } = \int_a^b\sqrt{\fp(t)^2+\gp(t)^2}\, dt - , - as stated in - in . - If \vrt = \la f(t), g(t)\ra, - note that \sqrt{\fp(t)^2+\gp(t)^2} = \norm{\vrp(t)}. - Therefore we can express the arc length of the graph of a vector-valued function as an integral of the magnitude of its derivative. -

    - - - Arc Length of a Vector-Valued Function - -

    - Let \vrt be a vector-valued function where \vrp(t) is continuous on [a,b]. - The arc length L of the graph of \vrt is - vector-valued functionarc length - arc length - - L = \int_a^b \norm{\vrp(t)}\, dt - . -

    -     -     - -

    - Note that we are actually integrating a scalar-function here, - not a vector-valued function. -

    - -

    - The next section takes what we have established thus far and applies it to objects in motion. - We will let \vrt describe the path of an object in the plane or in space and will discover the information provided by \vrp(t) and \vrp'(t). -

    -     - - - - Terms and Concepts - - - -

    - Limits, derivatives and integrals of vector-valued functions are all evaluated -wise. -

    -     - - - - - - - - -     - - - - -

    - The definite integral of a velocity function gives . -

    -     - - - - - - - - -     - - - - -

    - Why is it generally not useful to graph both - \vec r(t) and \vrp(t) on the same axes? -

    -     - - - -

    - It is difficult to identify the points on the graphs of - \vec r(t) and \vrp(t) that correspond to each other. -

    -     - -     - - - -

    - contains three product rules. - What are the three different types of products used in these rules? -

    -     - - - -

    - A scalar-vector product, a dot product and a cross product. -

    -     -     -     - - - Problems - - - - -

    - Evaluate the given limit. -

    -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $L=Compute("<11,74,sin(5)>"); - - - -

    - \lim\limits_{t\to 5} \la 2t+1,3t^2-1,\sin(t) \ra -

    -

    - -

    -     -     -     - - - - -

    - \lim\limits_{t\to 3} \la e^t,\frac{t^2-9}{t+3}\ra -

    -     - -

    - \la e^3,0\ra -

    -     - -     - - - - - Context("Vector2D"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $L=Compute("<1,e>"); - - - -

    - \lim\limits_{t\to 0} \la \frac{t}{\sin(t) }, (1+t)^{\frac1t}\ra -

    -

    - -

    -     -     -     - - - - -

    - \lim\limits_{h\to 0} \frac{\vec r(t+h)-\vec r(t)}{h}, where - \vec r(t) = \la t^2,t,1\ra. -

    -     - -

    - \la2t,1,0\ra -

    -     - -     - -     - - - -

    - Identify the interval or union of intervals on which \vec r(t) is continuous. -

    -     - - - - - Context("Interval"); - $dom=Compute("(-inf,0)U(0,inf)"); - - - -

    - \vec r(t) = \la t^2,1/t\ra -

    - -

    - -

    -     -     -     - - - - -

    - \vec r(t) = \la \cos(t) , e^t, \ln(t) \ra -

    -     - -

    - (0,\infty) -

    -     - -     - -     - - - -

    - Find the derivative of the given function. -

    -     - - - - - Context("Vector"); - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $d = Compute("< -sin(t), e^t, 1/t>"); - - - -

    - \vec r(t) = \la \cos(t) , e^t, \ln(t) \ra -

    - -

    - -

    -     -     -     - - - - -

    - \ds \vec r(t) = \la \frac 1t, \frac {2t-1}{3t+1}, \tan(t) \ra -

    -     - -

    - \vrp(t) = \la -1/t^2, 5/(3t+1)^2, \sec^2(t) \ra -

    -     - -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $d = Compute("< 2t sin(t)+t^2cos(t), 6t^2+10t>"); - - - -

    - \vec r(t) = (t^2)\la \sin(t) , 2t+5\ra -

    - -

    - -

    -     -     -     - - - - - Context("Numeric"); - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $d = Compute("(t^2+1)cos(t) +2tsin(t) + 4t+3"); - - - -

    - \vec r(t) = \la t^2+1, t-1\ra\cdot \la \sin(t) , 2t+5\ra -

    - -

    - -

    -     -     -     - - - - - Context("Vector"); - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $d = Compute("<-1,cos(t)-2t,6t^2+10t+2+cos(t)-sin(t)-tcos(t)>"); - - - -

    - \vec r(t) = \la t^2+1, t-1,1\ra\times \la \sin(t) , 2t+5,1\ra -

    - -

    - -

    -     -     -     - - - -

    - \ds \vec r(t) = \la \cosh t, \sinh t\ra -

    -     - -

    - \vrp(t) = \la \sinh t,\cosh t\ra -

    -     -     - -     - - - -

    - First, find \vrp (t). - Then sketch \vec r(t) and \vrp(1), - with the initial point of \vrp(1) at \vec r(1). -

    -     - - - - -

    - \ds \vec r(t) = \la t^2+t, t^2-t\ra -

    -     - - - - - Graph of the function \vec r(t) = \la t^2+t, t^2-t\ra. - The function \vec r(t) resembles a standard parabola which has been rotated 45 degrees clockwise. - The curve passes through the point (2,0), the origin, and the point (0,2) in the given order. - The graph also contains \vrp(1) beggining at the point (2,0) corresponding to the termination point of \vec r(1). - The vector \vrp(1)= \la 3,1 \ra and is tangent to \vec r(t) = \la t^2+t, t^2-t\ra corresponding to the point where t=1. - - Graph of the vector-valued function and its derivative at a point. - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.5,ymax=6.5, - xmin=-1.5,xmax=6.5 - ] - - \addplot+ [domain=-2:2,samples=40] ({x^2+x},{x^2-x}); - - \draw [thick,secondcolor,->,>=stealth] (axis cs: 2,0) -- (axis cs:5,1) node [black,right] { $\vrp(1)$}; - - \end{axis} - - \end{tikzpicture} - - - - - -

    - \vrp(t) = \la 2t+1,2t-1\ra -

    -     - -     - - - - -

    - \ds \vec r(t) = \la t^2-2t+2,t^3-3t^2+2t\ra -

    -     - - - - - Graph of the function \vec r(t) = \la t^2-2t+2,t^3-3t^2+2t\ra. - The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). - After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). - From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. - The graph of the function is also symmetric about the y-axis. - The graph also contains \vrp(1) beginning at the point (1,0) corresponding to the termination point of \vec r(1). - The vector \vrp(1)= \la 0,-1 \ra and is tangent to \vec r(t) corresponding to the point where t=1. - - Graph of the vector-valued function and its derivative at a point. - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.5,ymax=1.5, - xmin=-.5,xmax=3.5 - ] - - \addplot+ [domain=-.5:2.5,samples=60] ({x^2-2*x+2},{x^3-3*x^2+2*x}); - - \draw [thick,secondcolor,->,>=stealth] (axis cs: 1,0) -- (axis cs:1,-1) node [black,right] { $\vrp(1)$}; - - \end{axis} - - \end{tikzpicture} - - - - - -

    - \vrp(t) = \la 2t-2,3t^2-6t+2\ra -

    -     - -     - - - - -

    - \ds \vec r(t) = \la t^2+1,t^3-t\ra -

    -     - - - - - Graph of the function \vec r(t) = \la t^2+1,t^3-t\ra. - The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). - After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). - From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. - The graph of the function is also symmetric about the y-axis. - The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1). - The vector \vrp(1)= \la 2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1. - - Graph of the vector-valued function and its derivative at a point. - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-2.5,ymax=2.5, - xmin=-.5,xmax=4.5 - ] - - \addplot+[domain=-.5:2.5,samples=50] ({x^2-2*x+2},{x^3-3*x^2+2*x}); - - \draw [thick,secondcolor,->,>=stealth] (axis cs: 2,0) -- (axis cs:4,2) node [black,above] { $\vrp(1)$}; - - \end{axis} - - \end{tikzpicture} - - - - - -

    - \vrp(t) = \la 2t,3t^2-1\ra -

    -     - -     - - - - -

    - \ds \vec r(t) = \la t^2-4t+5,t^3-6t^2+11t-6\ra -

    -     - - - - - Graph of the function \vec r(t) = \la t^2-4t+5,t^3-6t^2+11t-6\ra. - The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). - After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). - From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. - The graph of the function is also symmetric about the y-axis. - The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1). - The vector \vrp(1)= \la -2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1. - - Graph of the vector-valued function and its derivative at a point. - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-2.5,ymax=2.5, - xmin=-.5,xmax=4.5 - ] - - \addplot+ [domain=-.5:2.5,samples=40] ({x^2-2*x+2},{x^3-3*x^2+2*x}); - - \draw [thick,secondcolor,->,>=stealth] (axis cs: 2,0) -- (axis cs:0,2) node [black,right] { $\vrp(1)$}; - - \end{axis} - - \end{tikzpicture} - - - - - -

    - \vrp(t) = \la 2t-4,3t^2-12t+11\ra -

    -     - -     - -     - - - -

    - Give the equation of the line tangent to the graph of - \vec r(t) at the given t value. -

    -     - - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - $L=ParametricLine("<2+3t, t>"); - - - -

    - \vec r(t) = \la t^2+t, t^2-t\ra, at t=1 -

    - - - \vec\ell(t)= - -

    - -

    -     -     -     - - - - - - Context("Vector2D"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->are(t=>'Real'); - $L=ParametricLine("(3sqrt(2)/2,sqrt(2)/2) + t<-3sqrt(2)/2,sqrt(2)/2 >"); - - - -

    - \vec r(t) = \la 3\cos(t) ,\sin(t) \ra, at t=\pi/4 -

    - - - \vec\ell(t)= - -

    - -

    -     -     -     - - - - -

    - \ds \vec r(t) = \la 3\cos(t) ,3\sin(t) ,t\ra at t=\pi. -

    -     - -

    - \ell(t) = \la -3,0,\pi\ra + t\la0,-3,1\ra -

    -     - -     - - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->are(t=>'Real'); - $L=ParametricLine("(1,0,0) + t<1,1,1 >"); - - - -

    - \vec r(t) = \la e^t,\tan(t) ,t\ra, at t=0. -

    - - - \vec\ell(t)= - -

    - -

    -     -     -     - -     - - - -

    - Find the value(s) of t for which \vec r(t) is not smooth. -

    -     - - - - -

    - \ds \vec r(t) = \la \cos(t) ,\sin(t) - t\ra -

    -     - -

    - t=2n\pi, where n is an integer; - so t = \ldots-4\pi,-2\pi,0,2\pi,4\pi,\ldots -

    -     - -     - - - - - $notsmooth=List("1"); - - - -

    - \vec r(t) = \la t^2-2t+1,t^3+t^2-5t+3\ra -

    - - - Enter the value(s) of t for which \vec r(t) is not smooth. - -

    - -

    -     -     -     - - - - -

    - \ds \vec r(t) = \la \cos(t) -\sin(t) , \sin(t) - \cos(t) ,\cos(4t)\ra -

    -     - -

    - \vec r(t) is not smooth at t=3\pi/4+n\pi, - where n is an integer -

    -     - -     - - - - - $notsmooth=List("1,-1"); - - - -

    - \vec r(t) = \la t^3-3t+2, -\cos(\pi t),\sin^2(\pi t) \ra -

    - - - Enter the value(s) of t for which \vec r(t) is not smooth. - -

    - -

    -     -     -     - -     - - - -

    - The following exercises ask you to verify parts of . - In each let f(t) = t^3, - \vec r(t) =\la t^2,t-1,1\ra and \vec s(t) = \la \sin(t) , e^t,t\ra. - Compute the various derivatives as indicated. -

    -     - - - - -

    - Simplify f(t)\vec r(t), then find its derivative; - show this is the same as \fp(t)\vec r(t) + f(t)\vrp(t). -

    -     - -

    - Both derivatives return \la 5t^4,4t^3-3t^2,3t^2\ra. -

    -     - -     - - - - -

    - Simplify \vec r(t)\cdot\vec s(t), then find its derivative; - show this is the same as \vrp(t)\cdot\vec s(t) + \vec r(t)\cdot\vec s\,'(t). -

    -     - -

    - Both derivatives return 2\sin(t) +t^2\cos(t) + te^t+1. -

    -     - -     - - - - -

    - Simplify \vec r(t)\times\vec s(t), then find its derivative; - show this is the same as \vrp(t)\times\vec s(t) + \vec r(t)\times\vec s\,'(t). -

    -     - -

    - Both derivatives return \la 2t-e^t-1,\cos(t) -3t^2,(t^2+2t)e^t-(t-1)\cos(t) -\sin(t) \ra. -

    -     - -     - - - -

    - Simplify \ds \vec r\big(f(t)\big), then find its derivative; - show this is the same as \ds \vrp\big(f(t)\big)\fp(t). -

    -     - -

    - Both derivatives return \la 6t^5,3t^2,0\ra -

    -     -     - -     - - - -

    - Evaluate the given definite or indefinite integral. -

    -     - - - - -

    - \ds \int \la t^3,\cos(t) , te^t\ra\,dt -

    -     - -

    - \la \frac14t^4,\sin(t) ,te^t-e^t\ra + \vec C -

    -     - -     - - - - -

    - \ds \int \la \frac{1}{1+t^2},\sec^2(t) \ra\,dt -

    -     - -

    - \la \tan^{-1}(t) ,\tan(t) \ra + \vec C -

    -     - -     - - - - - Context("Vector2D"); - $int=Compute("<-2,0>"); - - - -

    - \displaystyle\int_0^{\pi} \la -\sin(t) ,\cos(t) \ra\,dt=. -

    -     -     -     - - - - -

    - \ds \int_{-2}^{2} \la 2t+1,2t-1\ra\,dt -

    -     - -

    - \la4,-4 \ra -

    -     - -     - -     - - - -

    - Solve the given initial value problems. -

    -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - $r=Compute("<t^2/2+2,-cos(t)+3>"); - - - -

    - Find \vec r(t), given that - \vrp(t) = \la t,\sin(t) \ra and \vec r(0) = \la 2,2\ra. -

    - -

    - \vec r(t)= -

    -     -     -     - - - - -

    - Find \vec r(t), - given that \vrp(t) = \la 1/(t+1),\tan(t) \ra and -

    - -

    - \vec r(0) = \la 1,2\ra. -

    -     - -

    - \vec r(t) = \la \ln\abs{t+1} + 1, -\ln\abs{\cos(t) } + 2\ra -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(t=>'Real'); - $r=Compute("<t^4/12+t+4,t^3/6+2t+5,t^2/2+3t+6>"); - - - -

    - Find \vec r(t), given that \vrp'(t) = \la t^2,t,1\ra, - \vrp(0) = \la 1,2,3\ra and \vec r(0) = \la 4,5,6\ra. -

    - -

    - \vec r(t)= -

    -     -     -     - - - - -

    - Find \vec r(t), - given that \vrp'(t) = \la \cos(t) ,\sin(t) ,e^t\ra, -

    - -

    - \vrp(0) = \la 0,0,0\ra and \vec r(0) = \la 0,0,0\ra. -

    -     - -

    - \vec r(t) = \la-\cos(t) +1,t-\sin(t) ,e^t-t-1 \ra -

    -     - -     - -     - - - -

    - Find the arc length of \vec r(t) on the indicated interval. -

    -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstatnFunctions=>0); - $l=Formula("2sqrt(13)pi"); - - - -

    - \vec r(t) = \la 2\cos(t) , 2\sin(t) , 3t \ra on [0,2\pi]. -

    - -

    - -

    -     -     -     - - - - -

    - \vec r(t) = \la 5\cos(t) , 3\sin(t) , 4\sin(t) \ra on [0,2\pi]. -

    -     - -

    - 10\pi -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstatnFunctions=>0); - $l=Formula("1/54(22^(3/2)-8)"); - - - -

    - \vec r(t) = \la t^3,t^2,t^3 \ra on [0,1]. -

    - -

    - -

    -     -     -     - - - - -

    - \vec r(t) = \la e^{-t}\cos(t) ,e^{-t}\sin(t) \ra on [0,1]. -

    -     - -

    - \sqrt{2}(1-e^{-1}) -

    -     - -     - -     - - - - -

    - Prove ; - that is, show if \vec r(t) has constant length and is differentiable, - then \vec r(t)\cdot \vrp(t)=0. (Hint: - use the Product Rule to compute \frac{d}{dt}\big(\vec r(t)\cdot\vec r(t)\big).) -

    -     - -

    - As \vec r(t) has constant length, - \vec r(t)\cdot\vec r(t)=c^2 for some constant c. - Thus - - \vec r(t)\cdot\vec r(t) \amp = c^2 - \frac{d}{dt}\big(\vec r(t)\cdot\vec r(t)\big) \amp = \frac{d}{dt}\big(c^2\big) - \vrp(t)\cdot \vec r(t)+\vec r(t)\cdot \vrp(t) \amp = 0 - 2\vec r(t)\cdot \vrp(t) \amp =0 - \vec r(t)\cdot \vrp(t) \amp =0 - . -

    -     - -     -     -     -
    -
    - The Calculus of Motion - -

    - A common use of vector-valued functions is to describe the motion of an object in the plane or in space. - A position function - \vec r(t) gives the position of an object at time t. More formally, let O = \vec 0 (either in the plane or in space) and suppose an object is at point P_c at time t=t_c. Then \vec r\big(t_c\big) = \overrightarrow{OP_c}; that is, the vector \vec r\big(t_c\big) points to the location of the object at a given time. - This section explores how derivatives and integrals are used to study the motion described by such a function. -

    - - - - - Velocity, Speed and Acceleration - -

    - Let \vec r(t) be a position function in - \mathbb{R}^2 or \mathbb{R}^3. -

    - -

    -

    -
  • - Velocity -

    - The instantaneous rate of position change, denoted \vec v(t); - that is, \vec v(t) = \vrp(t). -

    -
    • - -
  • - Speed -

    - The magnitude of velocity: \norm{\vec v(t)}. -

    -
    • - -
  • - Acceleration -

    - The instantaneous rate of velocity change, denoted \vec a(t); - that is, \vec a(t) = \vec v\,'(t) = \vrp'(t). - velocity - speed - acceleration - vector-valued functiondescribing motion -

    -
    • -
    -

    -     -     - - - Finding velocity and acceleration - -

    - An object is moving with position function - \vec r(t) = \la t^2-t,t^2+t\ra, -3\leq t\leq 3, - where distances are measured in feet and time is measured in seconds. -

    - -

    -

      -
    1. -

      - Find \vvt and \vat. -

      -
      2. - -
    2. -

      - Sketch \vrt; plot \vec v(-1), - \vec a(-1), \vec v(1) and \vec a(1), - each with their initial point at their corresponding point on the graph of \vrt. -

      -
      3. - -
    3. -

      - When is the object's speed minimized? -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - Taking derivatives, we find - - \vvt = \vrp(t) =\la 2t-1,2t+1\ra \text{ and } \vat = \vrp'(t) = \la 2,2\ra - . - Note that acceleration is constant. -

      -
      2. - -
    2. -

      - \vec v(-1) = \la -3,-1\ra, - \vec a(-1) = \la 2,2\ra; - \vec v(1) = \la 1,3\ra, - \vec a(1) = \la 2,2\ra. - These are plotted with \vrt in . - - We can think of acceleration as pulling - the velocity vector in a certain direction. - At t=-1, the velocity vector points down and to the left; - at t=1, - the velocity vector has been pulled in the \la 2,2\ra - direction and is now pointing up and to the right. - In we plot more velocity/acceleration vectors, - making more clear the effect acceleration has on velocity. -

      - -
      - Graphing the position, velocity and acceleration of an object in - -
      - - - - The curve corresponding the position function for this example, with velocity and acceleration vectors shown at two points. - -

      - The curve given by \vec r(t) = \la t^2-t,t^2+t\ra is plotted, - for -3\leq t\leq 3. - The shape of the curve is parabolic, but rotated, so that it is symmetric about the line y=x. - There are intercepts at (2,0), (0,0), and (0,2) (given in order of increasing t). -

      - -

      - At each of the points (2,0) and (0,2), a pair of vectors is plotted. - One vector in each pair corresponds to velocity, and is tangent to the curve. - The other vector represents acceleration; this vector is always parallel to the line of symmetry for the curve. -

      -       - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-1.9,ymax=13, - xmin=-1.9,xmax=13 - ] - - \addplot+ [domain=-3:3,samples=60] ({x^2-x},{x^2+x}); - - \filldraw (axis cs:2,0) circle (2.4pt); - \filldraw (axis cs:0,2) circle (2.4pt); - - \draw [thick,->,secondcolor] (axis cs:2,0) -- (axis cs:-1,-1); - \draw [thick,->,secondcolor] (axis cs:0,2) -- (axis cs:1,5); - \draw [thick,->] (axis cs:2,0) -- (axis cs:4,2); - \draw [thick,->] (axis cs:0,2) -- (axis cs:2,4); - \draw [thick,->,firstcolor] (axis cs:8.75,3.75) -- (axis cs: 8.69,3.71); - - \end{axis} - - \end{tikzpicture} - - - - -
      - -
      - - - - The same curve as the previous image, with more vectors plotted. - -

      - The curve in is the same as the curve from . - The difference in this image is that additional pairs of velocity and acceleration vectors are plotted. - Since the acceleration is constant, this vector is the same at every point. - The velocity vector is always tangent to the curve, - and is smaller in magnitude the closer the curve is to the origin. -

      -       - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-1.9,ymax=13, - xmin=-1.9,xmax=13 - ] - - \addplot+ [domain=-3:3,samples=60] ({x^2-x},{x^2+x}); - - \filldraw (axis cs:6,2) circle (2.4pt); - \filldraw (axis cs:2,0) circle (2.4pt); - \filldraw (axis cs:0,0) circle (2.4pt); - \filldraw (axis cs:0,2) circle (2.4pt); - \filldraw (axis cs:2,6) circle (2.4pt); - \filldraw (axis cs:12,6) circle (2.4pt); - \filldraw (axis cs:6,12) circle (2.4pt); - - \draw [thick,->,secondcolor] (axis cs:6,2) -- (axis cs:1,-1); - \draw [thick,->,secondcolor] (axis cs:2,0) -- (axis cs:-1,-1); - \draw [thick,->,secondcolor] (axis cs:0,0) -- (axis cs:-1,1); - \draw [thick,->,secondcolor] (axis cs:0,2) -- (axis cs:1,5); - \draw [thick,->,secondcolor] (axis cs:2,6) -- (axis cs:5,11); - - \draw [thick,->] (axis cs:6,2) -- (axis cs:8,4); - \draw [thick,->] (axis cs:2,0) -- (axis cs:4,2); - \draw [thick,->] (axis cs:0,0) -- (axis cs:2,2); - \draw [thick,->] (axis cs:0,2) -- (axis cs:2,4); - \draw [thick,->] (axis cs:2,6) -- (axis cs:4,8); - - \draw [thick,->,firstcolor] (axis cs:8.75,3.75) -- (axis cs: 8.69,3.71); - - \end{axis} - - \end{tikzpicture} - - - - -
      -       -
      - -

      - Since \vat is constant in this example, - as t grows large \vvt becomes almost parallel to \vat. - For instance, when t=10, - \vec v(10) = \la 19,21\ra, - which is nearly parallel to \la 2,2\ra. -

      -
      3. - -
    3. -

      - The object's speed is given by - - \norm{\vvt} = \sqrt{(2t-1)^2+(2t+1)^2} =\sqrt{8t^2+2} - . - To find the minimal speed, we could apply calculus techniques - (such as set the derivative equal to 0 and solve for t, etc.) - but we can find it by inspection. - Inside the square root we have a quadratic which is minimized when t=0. - Thus the speed is minimized at t=0, - with a speed of \sqrt{2} . - - The graph in also implies speed is minimized here. - The filled dots on the graph are located at integer values of t between -3 and 3. - Dots that are far apart imply the object traveled a far distance in 1 second, indicating high speed; - dots that are close together imply the object did not travel far in 1 second, - indicating a low speed. The dots are closest together near t=0, - implying the speed is minimized near that value. -

      -
      4. -
    -

    -     - -     - - - Analyzing Motion - -

    - Two objects follow an identical path at different rates on [-1,1]. - The position function for Object 1 is \vec r_1(t) = \la t, t^2\ra; - the position function for Object 2 is \vec r_2(t) = \la t^3, t^6\ra, - where distances are measured in feet and time is measured in seconds. - Compare the velocity, - speed and acceleration of the two objects on the path. -

    -     - -

    - We begin by computing the velocity and acceleration function for each object: - - \vec v_1(t) \amp = \la 1,2t\ra \amp \vec v_2(t) \amp = \la 3t^2,6t^5\ra - \vec a_1(t) \amp = \la 0,2\ra \amp \vec a_2(t) \amp =\la 6t,30t^4\ra - -

    - -

    - We immediately see that Object 1 has constant acceleration, - whereas Object 2 does not. -

    - -

    - At t=-1, - we have \vec v_1(-1) = \la 1,-2\ra and \vec v_2(-1) = \la 3,-6\ra; - the velocity of Object 2 is three times that of Object 1 and so it follows that the speed of Object 2 is three times that of Object 1 (3\sqrt{5} ft/s compared to \sqrt{5} ft/s.) -

    - -
    - Plotting velocity and acceleration vectors for Object 1 in - - - A parabola with vertex at the origin, opening upward. Several pairs of velocity and acceleration vectors are plotted. - -

    - The plot shows the parabola y=x^2, for -1\leq x\leq 1. - At five points along the curve, velocity and acceleration vectors are plotted. - The acceleration vector is the same at every point, always pointing straight up. - The velocity vector is tangent to the curve at each point, - but always points somewhat to the right, indicating motion along the curve from x=-1 to x=1. - The magnitude of the velocity vector is largest at the two ends of the parabola, and smallest at the origin. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-1.1,ymax=3.1, - xmin=-2.1,xmax=2.1 - ] - - \addplot+ [domain=-1:1,samples=40] ({x},{x^2}); - - \filldraw (axis cs:-1.,1.) circle (2.4pt); - \filldraw (axis cs:-0.5,0.25) circle (2.4pt); - \filldraw (axis cs:0.,0.) circle (2.4pt); - \filldraw (axis cs:0.5,0.25) circle (2.4pt); - \filldraw (axis cs:1.,1.) circle (2.4pt); - - \draw [thick,->,secondcolor] (axis cs:-1.,1.) -- (axis cs:0.,-1.); - \draw [thick,->,secondcolor] (axis cs:-0.5,0.25) -- (axis cs:0.5,-0.75); - \draw [thick,->,secondcolor] (axis cs:0.,0.) -- (axis cs:1.,0.); - \draw [thick,->,secondcolor] (axis cs:0.5,0.25) -- (axis cs:1.5,1.25); - \draw [thick,->,secondcolor] (axis cs:1.,1.) -- (axis cs:2.,3.); - - \draw [thick,->] (axis cs:-1.,1.) -- (axis cs:-1.,3.); - \draw [thick,->] (axis cs:-0.5,0.25) -- (axis cs:-0.5,2.25); - \draw [thick,->] (axis cs:0.,0.) -- (axis cs:0.,2.); - \draw [thick,->] (axis cs:0.5,0.25) -- (axis cs:0.5,2.25); - \draw [thick,->] (axis cs:1.,1.) -- (axis cs:1.,3.); - - \draw [thick,->,firstcolor] (axis cs:-.9,.81) -- (axis cs: -.89,.7921); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - At t=0, the velocity of Object 1 is - \vec v(1) = \la 1,0\ra and the velocity of Object 2 is \vec 0! - This tells us that Object 2 comes to a complete stop at t=0. -

    - -

    - In , - we see the velocity and acceleration vectors for Object 1 plotted for t=-1, -1/2, 0, 1/2 and t=1. - Note again how the constant acceleration vector seems to pull - the velocity vector from pointing down, right to up, right. - We could plot the analogous picture for Object 2, but the velocity and acceleration vectors are rather large (\vec a_2(-1) = \la -6,30\ra!) -

    - -

    - Instead, we simply plot the locations of Object 1 and 2 on intervals of 1/5th of a second, - shown in and . - Note how the x-values of Object 1 increase at a steady rate. - This is because the x-component of \vec a(t) is 0; - there is no acceleration in the x-component. - The dots are not evenly spaced; - the object is moving faster near t=-1 and t=1 than near t=0. -

    - -
    - Comparing the positions of Objects 1 and 2 in - -
    - - - - The parabola y equals x squared, with several points plotted, corresponding to equally-spaced values of t. - -

    - Another plot of the curve y=x^2 is shown, for -1\leq x\leq 1, - this time with points plotted on the curve, - corresponding to equally-spaced values of t, in increments of 0.2. - The points are closer together near the origin, and further apart near the ends of the curve. - This illustrates the fact that the object described by the vector-valued function \vec{r}(t) = \la t,t^2\ra - is moving slowest near the origin: for equal intervals of time, - points will be closer together when the object is moving slowly, - and further apart when the object is moving quickly. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=1.1, - xmin=-1.1,xmax=1.1 - ] - - \addplot+ [domain=-1:1,samples=40] ({x},{x^2}); - - \filldraw (axis cs:-1.,1.) circle (2.4pt); - \filldraw (axis cs:-0.8,0.64) circle (2.4pt); - \filldraw (axis cs:-0.6,0.36) circle (2.4pt); - \filldraw (axis cs:-0.4,0.16) circle (2.4pt); - \filldraw (axis cs:-0.2,0.04) circle (2.4pt); - \filldraw (axis cs:0.,0.) circle (2.4pt); - \filldraw (axis cs:0.2,0.04) circle (2.4pt); - \filldraw (axis cs:0.4,0.16) circle (2.4pt); - \filldraw (axis cs:0.6,0.36) circle (2.4pt); - \filldraw (axis cs:0.8,0.64) circle (2.4pt); - \filldraw (axis cs:1.,1.) circle (2.4pt); - - \draw [thick,->,firstcolor,>=stealth] (axis cs:-.9,.81) -- (axis cs: -.89,.7921); - \draw (axis cs: .5,.6) node { $\vec r_1(t)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - The parabola y equals x squared, with several points plotted, corresponding to equally-spaced values of t. - -

    - The curve is the same as the one plotted in . - However, the distribution of points corresponding to equally-spaced values of t is very different. - Since the motion of this object is given by \vec{r}(t)=\la t^3,t^6\ra, - the points closest to the origin are much closer together. - For example, when t changes from 0 to 0.1, - x changes from 0 to 0.001, which is a very small change. - However, when t changes from 0.9 to 1, - x changes from 0.729 to 1, a difference that is 271 times as large! -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=1.1, - xmin=-1.1,xmax=1.1 - ] - - \addplot+ [domain=-1:1,samples=40] ({x},{x^2}); - - \filldraw (axis cs:-1.,1.) circle (2.4pt); - \filldraw (axis cs:-0.512,0.262144) circle (2.4pt); - \filldraw (axis cs:-0.216,0.046656) circle (2.4pt); - \filldraw (axis cs:-0.064,0.004096) circle (2.4pt); - \filldraw (axis cs:-0.008,0.000064) circle (2.4pt); - \filldraw (axis cs:0.,0.) circle (2.4pt); - \filldraw (axis cs:0.008,0.000064) circle (2.4pt); - \filldraw (axis cs:0.064,0.004096) circle (2.4pt); - \filldraw (axis cs:0.216,0.046656) circle (2.4pt); - \filldraw (axis cs:0.512,0.262144) circle (2.4pt); - \filldraw (axis cs:1.,1.) circle (2.4pt); - - \draw [thick,->,firstcolor,>=stealth] (axis cs:-.9,.81) -- (axis cs: -.89,.7921); - \draw (axis cs: .5,.6) node { $\vec r_2(t)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - In , we see the points plotted for Object 2. - Note the large change in position from t=-1 to t=-0.8; - the object starts moving very quickly. - However, it slows considerably at it approaches the origin, - and comes to a complete stop at t=0. - While it looks like there are 3 points near the origin, - there are in reality 5 points there. -

    - -

    - Since the objects begin and end at the same location, - they have the same displacement. - Since they begin and end at the same time, - with the same displacement, - they have the same average rate of change (, they have the same average velocity). - Since they follow the same path, - they have the same distance traveled. - Even though these three measurements are the same, - the objects obviously travel the path in very different ways. -

    -     -     - - - Analyzing the motion of a whirling ball on a string - -

    - A young boy whirls a ball, - attached to a string, above his head in a counter-clockwise circle. - The ball follows a circular path and makes 2 revolutions per second. - The string has length 2. -

    - -

    -

      -
    1. -

      - Find the position function - \vec r(t) that describes this situation. -

      -
      2. - -
    2. -

      - Find the acceleration of the ball and give a physical interpretation of it. -

      -
      3. - -
    3. -

      - A tree stands 10 in front of the boy. - At what t-values should the boy release the string so that the ball hits the tree? -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - The ball whirls in a circle. - Since the string is 2ft long, - the radius of the circle is 2. - The position function \vrt = \la 2\cos(t) , 2\sin(t) \ra describes a circle with radius 2, - centered at the origin, - but makes a full revolution every 2\pi seconds, - not two revolutions per second. - We modify the period of the trigonometric functions to be 1/2 by multiplying t by 4\pi. - The final position function is thus - - \vrt = \la 2\cos(4\pi t), 2\sin(4\pi t)\ra - . - (Plot this for 0\leq t\leq 1/2 to verify that one revolution is made in 1/2 a second.) -

      -
      2. - -
    2. -

      - To find \vat, we take the derivative of \vrt twice. - - \vvt = \vrp(t) \amp = \la -8\pi \sin(4\pi t), 8\pi \cos(4\pi t)\ra - \vat =\vrp'(t) \amp = \la -32\pi^2 \cos(4\pi t), -32\pi^2 \sin(4\pi t) \ra - \amp = -32\pi^2\la \cos(4\pi t), \sin(4\pi t)\ra - . - Note how \vat is parallel to \vrt, - but has a different magnitude and points in the opposite direction. - Why is this? -

      - -

      - Recall the classic physics equation, - Force = mass acceleration. - A force acting on a mass induces acceleration (, the mass moves); - a mass that is accelerating is being acted upon by a force. - Thus force and acceleration are closely related. - A moving ball wants to travel in a straight line. - Why does the ball in our example move in a circle? - It is attached to the boy's hand by a string. - The string applies a force to the ball, affecting its motion: - the string accelerates the ball. - This is not acceleration in the sense of - it travels faster; rather, - this acceleration is changing the velocity of the ball. - In what direction is this force/acceleration being applied? - In the direction of the string, towards the boy's hand. -

      - -

      - The magnitude of the acceleration is related to the speed at which the ball is traveling. - A ball whirling quickly is rapidly changing direction/velocity. - When velocity is changing rapidly, - the acceleration must be large. -

      -
      3. - -
    3. -

      - When the boy releases the string, - the string no longer applies a force to the ball, - meaning acceleration is \vec 0 and the ball can now move in a straight line in the direction of \vec v(t). -

      - -

      - Let t=t_0 be the time when the boy lets go of the string. - The ball will be at \vec r(t_0), - traveling in the direction of \vec v(t_0). - We want to find t_0 so that this line contains the point (0,10) - (since the tree is 10 directly in front of the boy). -

      - -
      - Modeling the flight of a ball in - - - A diagram showing a coniferous tree, a circle, and a tangent line from the circle to the tree. - -

      - A large tree is at the top of the image; it appears to be pine or spruce. - Below the tree, a circle is shown, with a diameter of 2 feet labeled. - A line is drawn, tangent to the circle, from the circle to a point on the trunk of the tree. - The line is labeled with the vector \la 0, 10\ra - \vrp(t_0). -

      -       - - - \begin{tikzpicture} - - \begin{scope}[scale=.5,shift={(0,10)}] - - \draw[fill=treestump,ultra thick] - (.75,-1) - .. controls (.5,.5) and (.5,3) .. (0.5,4) -- (-0.5,4) - .. controls (-.5,3) and (-.5,.5) .. (-.75,-1) - ; - - \draw[ultra thick,fill=treetop] - (0,10) - .. controls (0,8) and (1,7) .. (1.5,7) - .. controls (1,7) and (1,7) .. (0.5,7.25) - .. controls (1.5,5) and (2.5,4) .. (3,4) - .. controls (2,4) and (1.25,4) .. (1,4.5) - .. controls (2,2) and (3.5,2) .. (4,2) - .. controls (1,1) and (-1,1) .. (-4,2) - .. controls (-3.5,2) and (-2,2) .. (-1,4.5) - .. controls (-1.25,4) and (-2,4) .. (-3,4) - .. controls (-2.5,4) and (-1.5,5) .. (-0.5,7.25) - .. controls (-1,7) and (-1,7) .. (-1.5,7) - .. controls (-1,7) and (0,8) .. (0,10) - ; - - \end{scope} - - \begin{scope}[scale=.5] - - \coordinate (ball) at (215:2cm); - \coordinate (release) at (11.5:2cm); - - \draw [rotate=11.5] (0,0) -- (1.9,0) node [above,sloped,pos=.5] { 2 ft}; - \draw [thick,dashed] (0,0) circle (2cm); - \draw [thick,->] (ball) arc (215:225:2cm); - - \draw (0,10) circle (2pt); - \draw [dotted,thick] (release) -- (0,10) node [sloped, pos=.5, above] { $\langle 0,10\rangle - \vec r(t_0)$}; - - \filldraw [draw=black,fill=white,thick] (release) circle (.1); - - \end{scope} - - \end{tikzpicture} - - - - -
      - -

      - There are many ways to find this time value. - We choose one that is relatively simple computationally. - As shown in , - the vector from the release point to the tree is \la 0,10\ra - \vec r(t_0). - This line segment is tangent to the circle, - which means it is also perpendicular to \vec r(t_0) itself, - so their dot product is 0. - - \amp \vec r(t_0) \cdot \big(\la 0,10\ra - \vec r(t_0)\big) =0 - \amp \la 2\cos(4\pi t_0), 2\sin(4\pi t_0)\ra \cdot \la -2\cos(4\pi t_0),10-2\sin(4\pi t_0)\ra =0 - \amp -4\cos^2(4\pi t_0) + 20\sin(4\pi t_0)-4\sin^2(4\pi t_0) = 0 - \amp 20\sin(4\pi t_0) - 4 =0 - \amp \sin(4\pi t_0) =1/5 - \amp 4\pi t_0 = \sin^{-1}(1/5) - \amp 4\pi t_0 \approx 0.2 + 2\pi n, - where n is an integer. Solving for t_0 we have: - \amp t_0 \approx 0.016 + n/2 - - This is a wonderful formula. - Every 1/2 second after t=0.016\,\text{s} the boy can release the string - (since the ball makes 2 revolutions per second, - he has two chances each second to release the ball). -

      -
      4. -
    -

    -     - -     - - - Analyzing motion in space - -

    - An object moves in a helix with position function \vrt = \la \cos(t) , \sin(t) , t\ra, - where distances are measured in meters and time is in minutes. - Describe the object's speed and acceleration at time t. -

    -     - -

    - With \vrt = \la \cos(t) ,\sin(t) , t\ra, we have: - - \vvt \amp = \la -\sin(t) , \cos(t) , 1\ra \text{ and } - \vat \amp = \la -\cos(t) , -\sin(t) , 0\ra - . -

    - -

    - The speed of the object is \norm{\vvt} = \sqrt{(-\sin(t) )^2+\cos^2(t) +1} = \sqrt{2} - ; - it moves at a constant speed. - Note that the object does not accelerate in the z-direction, - but rather moves up at a constant rate of 1. -

    -     -     - -

    - The objects in Examples - and traveled at a constant speed. - That is, \norm{\vvt} = c for some constant c. - Recall , - which states that if a vector-valued function \vrt has constant length, - then \vrt is perpendicular to its derivative: - \vrt\cdot\vrp(t) = 0. - In these examples, the velocity function has constant length, - therefore we can conclude that the velocity is perpendicular to the acceleration: - \vvt\cdot\vat = 0. - A quick check verifies this. - vector-valued functionof constant length -

    - -

    - There is an intuitive understanding of this. - If acceleration is parallel to velocity, - then it is only affecting the object's speed; - it does not change the direction of travel. - (For example, consider a dropped stone. - Acceleration and velocity are parallel straight down and the direction of velocity never changes, - though speed does increase.) - If acceleration is not perpendicular to velocity, - then there is some acceleration in the direction of travel, - influencing the speed. - If speed is constant, then acceleration must be orthogonal to velocity, - as it then only affects direction, and not speed. -

    - - - Objects With Constant Speed -

    - If an object moves with constant speed, - then its velocity and acceleration vectors are orthogonal. - That is, \vvt\cdot\vat=0. - vector-valued functionof constant length -

    -     -     - - - Projectile Motion -

    - An important application of vector-valued position functions is - projectile motion: - the motion of objects under only the influence of gravity. - We will measure time in seconds, - and distances will either be in meters or feet. - We will show that we can completely describe the path of such an object knowing its initial position and initial velocity (, where it is - and where it is going.) - vector-valued functionprojectile motion - projectile motion -

    - - - -

    - Suppose an object has initial position - \vec r(0) = \la x_0,y_0\ra and initial velocity \vec v(0) = \la v_x,v_y\ra. - It is customary to rewrite - \vec v(0) in terms of its speed v_0 and direction \vec u, - where \vec u is a unit vector. - Recall all unit vectors in - \mathbb{R}^2 can be written as \la \cos(\theta) ,\sin(\theta) \ra, - where \theta is an angle measure counter-clockwise from the x-axis. - (We refer to \theta as the angle of elevation.) - angle of elevation - Thus \vec v(0) = v_0\la \cos(\theta),\sin(\theta)\ra. -

    - -

    - Since the acceleration of the object is known, - namely \vat = \la 0,-g\ra, - where g is the gravitational constant, - we can find \vrt knowing our two initial conditions. - We first find \vvt: -

    - -

    - - \vec v(t) \amp = \int \vat \, dt - \vvt \amp = \int \la 0,-g\ra \, dt - \vvt \amp = \la 0,-gt\ra + \vec C - . -

    - -

    - Knowing \vec v(0) = v_0\la \cos(\theta) ,\sin(\theta) \ra, - we have \vec C = v_0\la \cos(t) ,\sin(t) \ra and so - - \vec v(t) = \la v_0\cos(\theta) , -gt+v_0\sin(\theta) \ra - . -

    - -

    - We integrate once more to find \vrt: - - \vrt \amp = \int \vvt\,dt - \vrt \amp = \int \la v_0\cos(\theta) , -gt+v_0\sin(\theta) \ra\,dt - \vrt \amp = \la \big(v_0\cos(\theta) \big)t, -\frac12gt^2+\big(v_0\sin(\theta) \big)t\ra + \vec C. - Knowing \vec r(0) = \la x_0,y_0\ra, we conclude \vec C = \la x_0,y_0\ra and - \vrt \amp = \la \big(v_0\cos(\theta) \big)t+x_0\,, -\frac12gt^2+\big(v_0\sin(\theta) \big)t+y_0\,\ra - . -

    - - - Projectile Motion -

    - The position function of a projectile propelled from an initial position of \vec r_0=\la x_0,y_0\ra, - with initial speed v_0, - with angle of elevation \theta and neglecting all accelerations but gravity is - vector-valued functionprojectile motion - projectile motion - - \vrt = \la \big(v_0\cos(\theta) \big)t+x_0\,, -\frac12gt^2+\big(v_0\sin(\theta) \big)t+y_0\,\ra - . -

    - -

    - Letting \vec v_0 = v_0\la \cos(\theta) ,\sin(\theta) \ra, - \vrt can be written as - - \vrt = \la 0,-\frac12gt^2\ra + \vec v_0t+\vec r_0 - . -

    -     - -

    - We demonstrate how to use this position function in the next two examples. -

    - - - Projectile Motion - -

    - Sydney shoots her Red Ryderbb gun across level ground from an elevation of 4, - where the barrel of the gun makes a 5^\circ angle with the horizontal. - Find how far the bb travels before landing, - assuming the bb is fired at the advertised rate of 350 and ignoring air resistance. -

    -     - -

    - A direct application of gives - - \vrt \amp = \la (350\cos(5^\circ) )t, -16t^2 + (350\sin(5^\circ) )t + 4\ra - \amp \approx \la 346.67t, -16t^2+30.50t+4\ra - , - where we set her initial position to be \la 0,4\ra. - We need to find when the bb lands, - then we can find where. - We accomplish this by setting the y-component equal to 0 and solving for t: - - -16t^2+30.50t+4 \amp = 0 - t \amp = \frac{-30.50 \pm \sqrt{30.50^2-4(-16)(4)}}{-32} - t \amp \approx 2.03\,\text{s} - . -

    - -

    - (We discarded a negative solution that resulted from our quadratic equation.) -

    - -

    - We have found that the bb lands 2.03 after firing; - with t=2.03, - we find the x-component of our position function is 346.67(2.03) = 703.74\,\text{ft}. - The bb lands about 704 feet away. -

    -     - -     - - - Projectile Motion - -

    - Alex holds his sister's bb gun at a height of 3 - and wants to shoot a target that is 6 above the ground, - 25 away. - At what angle should he hold the gun to hit his target? - (We still assume the muzzle velocity is 350.) -

    -     - -

    - The position function for the path of Alex's bb is - - \vrt = \la (350\cos(\theta) )t, -16t^2+(350\sin(\theta) )t+3\ra - . -

    - -

    - We need to find \theta so that - \vrt =\la 25,6\ra for some value of t. - That is, we want to find \theta and t such that - - (350\cos(\theta) )t = 25 \text{ and } -16t^2+(350\sin(\theta) )t+3 = 6 - . -

    - -

    - This is not trivial - (though not hard). - We start by solving each equation for - \cos(\theta) and \sin(\theta), respectively. - - \cos(\theta) = \frac{25}{350t} \text{ and } \sin(\theta) = \frac{3+16t^2}{350t} - . -

    - -

    - Using the Pythagorean Identity \cos^2(\theta) +\sin^2(\theta) =1, we have - - \left(\frac{25}{350t}\right)^2 + \left(\frac{3+16t^2}{350t}\right)^2 \amp =1 - Multiply both sides by (350t)^2: - 25^2 + (3+16t^2)^2 \amp =350^2t^2 - 256t^4-122,404t^2+634 \amp =0. - This is a quadratic in t^2. That is, we can apply the quadratic formula to find t^2, then solve for t itself. - t^2 \amp = \frac{122,404\pm\sqrt{122,404^2-4(256)(634)}}{512} - t^2 \amp = 0.0052,\,478.135 - t \amp = \pm 0.072,\,\pm 21.866 - -

    - -

    - Clearly the negative t values do not fit our context, - so we have t=0.072 and t=21.866. - Using \cos(\theta) = 25/(350 t), - we can solve for \theta: - - \theta \amp = \cos^{-1}\left(\frac{25}{350\cdot 0.072}\right) \text{ and } \cos^{-1}\left(\frac{25}{350\cdot 21.866}\right) - \theta \amp = 7.03^\circ \text{ and } 89.8^\circ - . -

    - -

    - Alex has two choices of angle. - He can hold the rifle at an angle of about 7^\circ with the horizontal and hit his target - 0.07 after firing, - or he can hold his rifle almost straight up, - with an angle of 89.8^\circ, - where he'll hit his target about 22 later. - The first option is clearly the option he should choose. -

    -     -     -     - - - Distance Traveled -

    - Consider a driver who sets her cruise-control to 60, - and travels at this speed for an hour. - We can ask: -

    - -

    -

      -
    1. -

      - How far did the driver travel? -

      -
      2. - -
    2. -

      - How far from her starting position is the driver? -

      -
      3. -
    -

    - -

    - The first is easy to answer: - she traveled 60 miles. - The second is impossible to answer with the given information. - We do not know if she traveled in a straight line, - on an oval racetrack, or along a slowly-winding highway. -

    - -

    - This highlights an important fact: - to compute distance traveled, - we need only to know the speed, - given by \norm{\vvt}. -

    - - - Distance Traveled - -

    - Let \vvt be a velocity function for a moving object. - The distance traveled by the object on [a,b] is: - distancetraveled - vector-valued functiondistance traveled - integrationdistance traveled - - \text{ distance traveled } = \int_a^b \norm{\vvt}\, dt - . -

    -     -     - -

    - Note that this is just a restatement of : - arc length is the same as distance traveled, - just viewed in a different context. -

    - - - - - Distance Traveled, Displacement, and Average Speed - -

    - A particle moves in space with position function \vrt = \la t,t^2,\sin(\pi t)\ra on [-2,2], - where t is measured in seconds and distances are in meters. - Find: -

    - -

    -

      -
    1. -

      - The distance traveled by the particle on [-2,2]. -

      -
      2. - -
    2. -

      - The displacement of the particle on [-2,2]. -

      -
      3. - -
    3. -

      - The particle's average speed. -

      -
      4. -
    -

    -     - -

    -

      -
    1. -

      - We use to establish the integral: - - \text{ distance traveled } \amp = \int_{-2}^2 \norm{\vvt}\, dt - \amp = \int_{-2}^2 \sqrt{1+(2t)^2+ \pi^2\cos^2(\pi t)}\, dt - . - This cannot be solved in terms of elementary functions so we turn to numerical integration, - finding the distance to be 12.88. -

      -
      2. - -
    2. -

      - The displacement is the vector - - \vec r(2)-\vec r(-2) = \la 2,4,0\ra - \la -2,4,0\ra = \la 4,0,0\ra - . - That is, the particle ends with an x-value increased by 4 and with y- and z-values the same - (see ). -

      -
      3. - -
    3. -

      - We found above that the particle traveled 12.88 over 4 seconds. - We can compute average speed by dividing: 12.88/4 = 3.22\,\text{m/s}. -

      - -
      - The path of the particle in - - - - An illustration of the path followed by the particle in this example. - -

      - This three-dimensional image illustrates the path of the particle in . - In the default view, the path appears to be quite complicated. - But if the image is rotated so that the view is along the y axis, - the curve appears sinusoidal, while it appears parabolic when viewed from above. -

      -       - - - - - //ASY file for figmotion63D.asy in Chapter 11 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5.6,10.7,4); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={2,4}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-3,3); - pair ybounds=(-1,5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve <t,t^2,sin(pi*t)> for t from -2 to 2 - triple g(real t) {return (t,t^2,sin(pi*t));} - path3 mypath=graph(g,-2,2,operator ..); draw(mypath,bluepen); - path3 mypath=graph(g,-2,-1.2,operator ..); draw(mypath,bluepen,Arrow3(size=5mm)); - //path3 mypath=graph(g,-2,-0.4,operator ..); draw(mypath,bluepen,Arrow3(size=5mm)); - //path3 mypath=graph(g,-2,0.8,operator ..); draw(mypath,bluepen,Arrow3(size=5mm)); - //path3 mypath=graph(g,-2,1.6,operator ..); draw(mypath,bluepen,Arrow3(size=5mm)); - - - - -
      - -

      - We should also consider - of , - which says that the average value of a function f on [a,b] is \frac{1}{b-a}\int_a^b f(x)\, dx. - In our context, the average value of the speed is - - \text{average speed}\, = \frac{1}{2-(-2)}\int_{-2}^2 \norm{\vvt}\, dt \approx \frac14 12.88 = 3.22\,\text{m/s} - . - Note how the physical context of a particle traveling gives meaning to a more abstract concept learned earlier. -

      -
      4. -
    -

    -     - -     - -

    - In - of - we defined the average value of a function f(x) on [a,b] to be - - \frac{1}{b-a}\int_a^bf(x)\, dx - . -

    - -

    - Note how in - we computed the average speed as - - \frac{\text{ distance traveled } }{\text{ travel time } } = \frac1{2-(-2)}\int_{-2}^2\norm{\vvt}\, dt - ; - that is, we just found the average value of \norm{\vvt} on [-2,2]. -

    - -

    - Likewise, given position function \vrt, - the average velocity on [a,b] is - - \frac{\text{displacement}}{\text{travel time}}\, = \frac1{b-a}\int_a^b \vec{r}\,'(t)\, dt = \frac{\vec r(b)-\vec r(a)}{b-a} - ; - that is, it is the average value of \vrp(t), - or \vvt, on [a,b]. -

    - - - Average Speed, Average Velocity -

    - Let \vec r(t) be a differentiable position function on [a,b]. -

    - -

    - The average speed is: - - \frac{\text{distance traveled}}{\text{travel time}}\, = \frac{\int_a^b \norm{\vec{r}\,'(t)}\, dt}{b-a} = \frac1{b-a}\int_a^b\norm{\vvt}\, dt - . -

    - -

    - The average velocity is: - - \frac{\text{ displacement}}{\text{travel time}}\, = \frac{\int_a^b \vec{r}\,'(t)\, dt}{b-a} = \frac1{b-a}\int_a^b\vec{r}\,'(t)\, dt - . -

    -     - -

    - The next two sections investigate more properties of the graphs of vector-valued functions and we'll apply these new ideas to what we just learned about motion. -

    -     - - - - Terms and Concepts - - - -

    - How is velocity different from speed? -

    -     - - - -

    - Velocity is a vector, - indicating an objects direction of travel and its rate of distance change (, its speed). - Speed is a scalar. -

    -     - -     - - - - -

    - What is the difference between displacement - and distance traveled? -

    -     - - - -

    - Displacement is a vector, - indicating the difference between the starting and ending positions of an object. - Distance traveled is a scalar, - indicating the arc length of the path followed. -

    -     - -     - - - - -

    - What is the difference between average velocity - and average speed? -

    -     - - - -

    - The average velocity is found by dividing the displacement by the time traveled it is a vector. - The average speed is found by dividing the distance traveled by the time traveled it is a scalar. -

    -     - -     - - - - -

    - Distance traveled is the same as , just viewed in a different context. -

    -     - - - - arc length|arclength - - - - -     - - - - -

    - Describe a scenario where an object's average speed is a large number, - but the magnitude of the average velocity is not a large number. -

    -     - - - -

    - One example is traveling at a constant speed s in a circle, - ending at the starting position. - Since the displacement is \vec 0, - the average velocity is \vec 0, hence \vnorm 0=0. - But traveling at constant speed s means the average speed is also s \gt 0. -

    -     - -     - - - - -

    - Explain why it is not possible to have an average velocity with a large magnitude but a small average speed. -

    -     - - - -

    - Distance traveled is always greater than or equal to the magnitude of displacement, - therefore average speed will always be at least as large as the magnitude of the average velocity. -

    -     - -     -     - - Problems - - - -

    - A position function \vrt is given. - Find \vvt and \vat. -

    -     - - - - -

    - \vrt = \la 2t+1, 5t-2, 7\ra -

    -     - -

    - \vvt = \la 2,5,0\ra, \vat = \la 0,0,0\ra -

    -     - -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - $velocity=Vector("<6t-2,-2t+1>"); - $acceleration=Vector("<6,-2>"); - - - -

    - \vrt = \la 3t^2-2t+1, -t^2+t+14\ra -

    - - - The velocity function \vvt is: - -

    - -

    - - - The acceleration function \vat is: - -

    - -

    -     -     -     - - - - -

    - \vrt = \la \cos(t) ,\sin(t) \ra -

    -     - -

    - \vvt = \la -\sin(t) , \cos(t) \ra, - \vat = \la -\cos(t) , -\sin(t) \ra -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $velocity=Vector("<1/10,sin(t),cos(t)>"); - $acceleration=Vector("<0,cos(t),-sin(t)>"); - - - -

    - \vrt = \la t/10,-\cos(t) ,\sin(t) \ra -

    - - - The velocity function \vvt is: - -

    - -

    - - - The acceleration function \vat is: - -

    - -

    -     -     -     - -     - - - -

    - A position function \vrt is given. - Sketch \vrt on the indicated interval. - Find \vvt and \vat, - then add \vec v(t_0) and \vec a(t_0) to your sketch, - with their initial points at \vec r(t_0), - for the given value of t_0. -

    -     - - - - -

    - \ds \vrt = \la t,\sin(t) \ra on [0,\pi/2]; - t_0= \pi/4 -

    -     - -

    - \vvt = \la 1,\cos(t) \ra, - \vat = \la 0,-\sin(t) \ra -

    - -     - -     - - - - -

    - \ds \vrt = \la t^2,\sin(t^2) \ra on [0,\pi/2]; - t_0=\sqrt{\pi/4} -

    -     - -

    - \vvt = \la 2 t,2 t \cos\left(t^2\right)\ra, - \vat = \la 2,2 \left(\cos\left(t^2\right)-2 t^2 \sin\left(t^2\right)\right)\ra -

    - -     - -     - - - - -

    - \ds \vrt = \la t^2+t,-t^2+2t \ra on [-2,2]; - t_0=1 -

    -     - -

    - \vvt = \la 2t+1,-2t+2\ra, \vat = \la 2,-2\ra -

    - -     - -     - - - - -

    - \ds \vrt = \la \frac{2t+3}{t^2+1},t^2\ra on [-1,1]; - t_0= 0 -

    -     - -

    - \vvt = \la -\frac{2 \left(t^2+3 \ - t-1\right)}{\left(t^2+1\right)^2},2 t\ra, - \vat = \la \frac{2 \left(2 t^3+9 t^2-6 t-3\right)}{\left(t^2+1\right)^3},2\ra -

    - -     - -     - -     - - - -

    - A position function \vrt of an object is given. - Find the speed of the object in terms of t, - and find where the speed is minimized/maximized on the indicated interval. -

    -     - - - - -

    - \ds \vrt = \la t^2,t \ra on [-1,1] -

    -     - -

    - \norm{\vvt} = \sqrt{4t^2+1}. -

    - -

    - Min at t=0; Max at t=\pm 1. -

    -     - -     - - - - - Context()->variables->are(t=>'Real'); - $speed=Formula("|t| sqrt(9t^2-12t+8)"); - $mins=List("0"); - $maxs=List("-1"); - - - -

    - \vrt = \la t^2,t^2-t^3 \ra on [-1,1] -

    - - Find the speed of the object in terms of t. - -

    - -

    - - - Find where the speed is minimized on the given interval. - -

    - -

    - - - Find where the speed is maximized on the given interval. - -

    - -

    -     -     -     - - - - -

    - \ds \vrt = \la 5\cos(t) ,5\sin(t) \ra on [0,2\pi] -

    -     - -

    - \norm{\vvt} = 5. -

    - -

    - Speed is constant, so there is no difference between min/max -

    -     - -     - - - - -

    - \ds \vrt = \la 2\cos(t) ,5\sin(t) \ra on [0,2\pi] -

    -     - -

    - \norm{\vvt} = \sqrt{4\sin^2(t) +25\cos^2(t) }. -

    - -

    - min: t=\pi/2,\,3\pi/2; - max: t=0,\,2\pi -

    -     - -     - - - - - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $speed=Formula("|sec(t)| sqrt(tan^2(t)+sec^2(t))"); - $mins=List("0"); - $maxs=Compute("pi/4"); - - - -

    - \vrt=\la \sec(t) ,\tan(t) \ra on [0,\pi/4]. -

    - - Find the speed of the object in terms of t. - -

    - -

    - - - Find where the speed is minimized on the given interval. - -

    - -

    - - - Find where the speed is maximized on the given interval. - -

    - -

    -     -     -     - - - - -

    - \ds \vrt = \la t+\cos(t) ,1-\sin(t) \ra on [0,2\pi] -

    -     - -

    - \norm{\vvt} = \sqrt{2-2\sin(t) }. -

    - -

    - min: t=\pi/2; max: t=3\pi/2 -

    -     - -     - - - - -

    - \ds \vrt = \la 12t,5\cos(t) ,5\sin(t) \ra on [0,4\pi] -

    -     - -

    - \norm{\vvt} = 13. -

    - -

    - speed is constant, so there is no difference between min/max -

    -     - -     - - - - - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $speed=Formula("sqrt(8t^2+3)"); - $mins=List("0"); - $maxs=Compute("1"); - - - -

    - \vrt=\la t^2-t,t^2+t,t \ra on [0,1]. -

    - - Find the speed of the object in terms of t. - -

    - -

    - - - Find where the speed is minimized on the given interval. - -

    - -

    - - - Find where the speed is maximized on the given interval. - -

    - -

    -     -     -     - - - - -

    - \ds \vrt = \la t,t^2,\sqrt{1-t^2}\ra on [-1,1] -

    -     - -

    - \norm{\vvt} = \sqrt{4t^2+1+t^2/(1-t^2)}. -

    - -

    - min: t=0; max: there is no max; - speed approaches \infty as t\to\pm 1 -

    -     - -     - - - - -

    - Projectile Motion: - \ds \vrt = \la (v_0\cos(\theta) )t,-\frac12gt^2+(v_0\sin(\theta) )t \ra on \ds \left[0,\frac{2v_0\sin(\theta) }g\right] -

    -     - -

    - \norm{\vvt} = \sqrt{g^2t^2-(2gv_0\sin(\theta) )t+v_0^2}. -

    - -

    - min: t=(v_0\sin(\theta) )/g; max: - t=0, t=(2v_0\sin(\theta) )/g -

    -     - -     - -     - - - -

    - Position functions \vec r_1(t) and - \vec r_2(s) for two objects are given that follow the same path on the respective intervals. -

    - -

    -

      -
    1. -

      - Show that the positions are the same at the indicated t_0 and s_0 values; - , show \vec r_1(t_0) = \vec r_2(s_0). -

      -
      2. - -
    2. -

      - Find the velocity, - speed and acceleration of the two objects at t_0 and s_0, - respectively. -

      -
      3. -
    -

    -     - - - - -

    - \vec r_1(t) = \la t,t^2\ra on [0,1]; t_0 = 1 -

    - -

    - \vec r_2(s) = \la s^2,s^4\ra on [0,1]; s_0 = 1 -

    -     - -

    -

      -
    1. -

      - \vec r_1(1) = \la 1,1\ra; - \vec r_2(1) = \la 1,1\ra -

      -
      2. - -
    2. -

      - \vec v_1(1) = \la 1,2\ra; - \norm{\vec v_1(1)} = \sqrt{5}; - \vec a_1(1) = \la 0,2\ra - - \vec v_2(1) = \la 2,4\ra; \norm{\vec v_2(1)} = 2\sqrt{5}; \vec a_2(1) = \la 2,12\ra -

      -
      3. -
    -

    -     - -     - - - - -

    - \vec r_1(t) = \la 3\cos(t) ,3\sin(t) \ra on [0,2\pi]; - t_0 = \pi/2 -

    - -

    - \vec r_2(s) = \la 3\cos(4s),3\sin(4s)\ra on [0,\pi/2]; - s_0 = \pi/8 -

    -     - -

    -

      -
    1. -

      - \vec r_1(\pi/2) = \la0,3\ra; - \vec r_2(\pi/8) = \la 0,3\ra -

      -
      2. - -
    2. -

      - \vec v_1(\pi/2) = \la -3,0\ra; - \norm{\vec v_1(\pi/2)} = 3; - \vec a_1(\pi/2) = \la 0,-3\ra - - \vec v_2(\pi/8) = \la -12,0\ra; \norm{\vec v_2(\pi/8)} = 12; \vec a_2(\pi/8) = \la 0,-48\ra -

      -
      3. -
    -

    -     - -     - - - - -

    - \vec r_1(t) = \la 3t,2t\ra on [0,2]; t_0 = 2 -

    - -

    - \vec r_2(s) = \la 6s-6,4s-4\ra on [1,2]; - s_0 = 2 -

    -     - -

    -

      -
    1. -

      - \vec r_1(2) = \la6,4\ra; - \vec r_2(2) = \la 6,4\ra -

      -
      2. - -
    2. -

      - \vec v_1(2) = \la 3,2\ra; - \norm{\vec v_1(2)} = \sqrt{13}; - \vec a_1(2) = \la 0,0\ra - - \vec v_2(2) = \la 6,4\ra; \norm{\vec v_2(2)} = 2\sqrt{13}; \vec a_2(2) = \la 0,0\ra -

      -
      3. -
    -

    -     - -     - - - - -

    - \vec r_1(t) = \la t,\sqrt{t}\ra on [0,1]; - t_0 = 1 -

    - -

    - \vec r_2(s) = \la \sin(s) ,\sqrt{\sin(s) }\ra on [0,\pi/2]; - s_0 = \pi/2 -

    -     - -

    -

      -
    1. -

      - \vec r_1(1) = \la 1,1\ra; - \vec r_2(\pi/2) = \la 1,1\ra -

      -
      2. - -
    2. -

      - \vec v_1(1) = \la 1,1/2\ra; - \norm{\vec v_1(1)} = \sqrt{5}/2; - \vec a_1(1) = \la 0,-1/4\ra - - \vec v_2(\pi/2) = \la 0,0\ra; \norm{\vec v_2(\pi/2)} = 0; \vec a_2(\pi/2) = \la -1,-1/2\ra -

      -
      3. -
    -

    -     - -     - -     - - - -

    - Find the position function of an object given its acceleration and initial velocity and position. -

    -     - - - - -

    - \vat = \la 2,3\ra;\vec v(0) = \la 1,2\ra,\vec r(0) = \la 5,-2\ra -

    -     - -

    - \vvt = \la 2t+1,3t+2\ra, - \vrt = \la t^2+t+5,3t^2/2+2t-2\ra -

    -     - -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - $pos=Compute("<t^2-t+5,3t^2/2-t-5/2>"); - - - -

    - Given \vat = \la 2,3\ra, - \vec v(1) = \la 1,2\ra, - and \vec r(1) = \la 5,-2\ra, - find the position function \vrt. -

    - -

    - -

    -     -     -     - - - - -

    - \vat = \la \cos(t) ,-\sin(t) \ra;\vec v(0) = \la 0,1\ra,\vec r(0) = \la 0,0\ra -

    -     - -

    - \vvt = \la \sin(t) ,\cos(t) \ra, - \vrt = \la 1-\cos(t) ,\sin(t) \ra -

    -     - -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - $pos=Compute("<10t,-16t^2+50t>"); - - - -

    - Given \vat =\la 0,-32\ra, - \vec v(0) = \la 10,50\ra, - and \vec r(0) = \la 0,0\ra, - find the position function \vrt. -

    - -

    - -

    -     -     -     - -     - - - -

    - Find the displacement, distance traveled, - average velocity and average speed of the described object on the given interval. -

    -     - - - - -

    - An object with position function \vrt = \la 2\cos(t) ,2\sin(t) , 3t\ra, - where distances are measured in feet and time is in seconds, - on [0,2\pi]. -

    -     - -

    - Displacement: \la 0,0,6\pi\ra; distance traveled: - 2\sqrt{13}\pi \approx 22.65ft; average velocity: - \la 0,0,3\ra; average speed: - \sqrt{13} \approx 3.61ft/s -

    -     - -     - - - - - Context("Vector2D"); - $disp=Compute("<-10,0>"); - $dist=Compute("5pi"); - $avvel=Compute("<-10/pi,0>"); - $avspe=Compute("5"); - - - -

    - An object has position function \vrt = \la 5\cos(t) ,-5\sin(t) \ra, - where distances are measured in feet and time is in seconds. - Over [0,\pi]. -

    - - What is the displacement? - -

    - -

    - - What is the distance traveled? - -

    - -

    - - What is the average velocity? - -

    - -

    - - What is the average speed? - -

    - -

    -     -     -     - - - - -

    - An object with velocity function \vvt = \la \cos(t) ,\sin(t) \ra, - where distances are measured in feet and time is in seconds, - on [0,2\pi]. -

    -     - -

    - Displacement: \la 0,0\ra; distance traveled: - 2\pi \approx 6.28ft; average velocity: - \la 0,0\ra; average speed: 1ft/s -

    -     - -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $disp=Compute("<10,20,-10>"); - $dist=Compute("10sqrt(6)"); - $avvel=Compute("<1,2,-1>"); - $avspe=Compute("sqrt(6)"); - - - -

    - An object has velocity function \vvt = \la 1,2,-1 \ra, - where distances are measured in feet and time is in seconds. - Over [0,10]. -

    - - What is the displacement? - -

    - -

    - - What is the distance traveled? - -

    - -

    - - What is the average velocity? - -

    - -

    - - What is the average speed? - -

    - -

    -     -     -     - -     - - - -

    - The following exercises ask you to solve a variety of problems based on the principles of projectile motion. -

    -     - - - - -

    - A boy whirls a ball, attached to a 3 ft string, - above his head in a counter-clockwise circle. - The ball makes 2 revolutions per second. -

    - -

    - At what t-values should the boy release the string so that the ball heads directly for a tree standing 10 ft in front of him? -

    -     - -

    - At t-values of \sin^{-1}(9/30)/(4\pi) + n/2 \approx 0.024 + n/2 seconds, - where n is an integer. -

    -     - -     - - - - -

    - David faces Goliath with only a stone in a 3 ft sling, - which he whirls above his head at 4 revolutions per second. - They stand 20 ft apart. -

    -     - - - -

    - At what t-values must David release the stone in his sling in order to hit Goliath? -

    -     - -

    - t=\sin^{-1}(3/20)/(8\pi) + n/4 \approx 0.006 + n/4, - where n is an integer -

    -     - -

    - The stone, while whirling, - can be modeled by \vrt = \la 3\cos(8\pi t),3\sin(8\pi t)\ra. -

    - -

    - He can releast the stone for t-values t=\sin^{-1}(3/20)/(8\pi) + n/4 \approx 0.006 + n/4, - where n is an integer. -

    -     -     - - - -

    - What is the speed at which the stone is traveling when released? -

    -     - -

    - \norm{\vrp(t)} = 24\pi\approx 51.4 ft/s -

    -     -     - - - -

    - Assume David releases the stone from a height of 6ft and Goliath's forehead is 9 ft above the ground. - What angle of elevation must David apply to the stone to hit Goliath's head? -

    -     - -

    - 0.27 radians, or 15.69^\circ -

    -     - -

    - At t=0.006, - the stone is approximately 19.77 ft from Goliath. - Using the formula for projectile motion, - we want the angle of elevation that lets a projectile starting at - \la 0,6\ra with a initial velocity of 51.4 ft/s arrive at \la 19.77,9\ra. - The desired angle is 0.27 radians, or 15.69^\circ. -

    -     -     - -     - - - - - Context()->flags->set(tolerance=>0.005); - $angle=NumberWithUnits("0.013 radians"); - $lead=NumberWithUnits("11.7 ft"); - - - -

    - A hunter aims at a deer which is 40 yards away. - Her crossbow is at a height of 5 ft, - and she aims for a spot on the deer 4 ft above the ground. - The crossbow fires her arrows at 300 ft/s. -

    -     - - - -

    - At what angle of elevation should she hold the crossbow to hit her target? -

    - -

    - -

    -     -     - - - -

    - If the deer is moving perpendicularly to her line of sight at a rate of 20 mph, - by approximately how much should she lead the deer in order to hit it in the desired location? (How far ahead of the deer should she aim?) -

    - -

    - -

    -     -     -     -     - - - - -

    - A baseball player hits a ball at 100 mph, - with an initial height of 3 ft and an angle of elevation of - 20^\circ, at Boston's Fenway Park. - The ball flies towards the famed Green Monster, - a wall 37 ft high located 310 ft from home plate. -

    -     - - - -

    - Show that as hit, the ball hits the wall. -

    -     - -

    - The position function of the ball is \vrt = \la (146.67\cos(\theta) )t,-16t^2+(146.67\sin(\theta) )t+3\ra, - where \theta is the angle of elevation. -

    - -

    - With \theta=20^\circ, - the ball reaches 310 ft from home plate in 2.25 seconds; - at this time, the height of the ball is 34.9 ft, - not enough to clear the Green Monster. -

    -     -     - - - -

    - Show that if the angle of elevation is 21^\circ, - the ball clears the Green Monster. -

    -     - -

    - With \theta=21^\circ, - the ball reaches 310 ft from home plate in 2.26 s, - with a height of 40 ft, clearing the wall. -

    -     -     - -     - - - - -

    - A Cessna flies at 1000 ft at 150 mph and drops a box of supplies to the professor - (and his wife) - on an island. - Ignoring wind resistance, - how far horizontally will the supplies travel before they land? -

    -     - -

    - The position function is \vrt = \la 220t,-16t^2+1000\ra. - The y-component is 0 when t=7.9; - \vec r(7.9) = \la 1739.25,0\ra, - meaning the box will travel about 1740 ft horizontally before it lands. -

    -     - -     - - - - -

    - A football quarterback throws a pass from a height of 6 ft, - intending to hit his receiver 20 yds away at a height of 5 ft. -

    -     - - - -

    - If the ball is thrown at a rate of 50mph, - what angle of elevation is needed to hit his intended target? -

    -     - -

    - The position function of the ball is \vrt = \la (v_0\cos(\theta) )t,-16t^2+(v_0\sin(\theta) )t+6\ra, - where \theta is the angle of elevation and v_0 is the initial ball speed. -

    - -

    - With v_0 = 73.33 ft/s, there are two angles of elevation possible. - An angle of \theta = 9.47^\circ delivers the ball in 0.83 s, - while an angle of 79.57^\circ delivers the ball in 4.5 s. -

    -     -     - - - -

    - If the ball is thrown at with an angle of elevation of 8^\circ, - what initial ball speed is needed to hit his target? -

    -     - -

    - With \theta=8^\circ, - the initial speed must be 53.8 mph\approx 78.9 ft/s. -

    -     -     - -     -     -     -     -
    -
    - Unit Tangent and Normal Vectors - - Unit Tangent Vector -

    - Given a smooth vector-valued function \vrt, - we defined in - that any vector parallel to \vrp(t_0) is tangent - to the graph of \vrt at t=t_0. - It is often useful to consider just the direction - of \vrp(t) and not its magnitude. - Therefore we are interested in the unit vector in the direction of \vrp(t). - This leads to a definition. -

    - - - Unit Tangent Vector - -

    - Let \vrt be a smooth function on an open interval I. - The unit tangent vector \unittangent(t) is - unit tangent vectordefinition - unit vectorunit tangent vector - - \unittangent(t) = \frac{1}{\norm{\vrp(t)}}\vrp(t) - . -

    -     -     - - - - - Computing the unit tangent vector - -

    - Let \vrt = \la 3\cos(t) , 3\sin(t) , 4t\ra. - Find \unittangent(t) and compute - \unittangent(0) and \unittangent(1). -

    -     - -

    - We apply to find \unittangent(t). - - \unittangent(t) \amp = \frac{1}{\norm{\vrp(t)}}\vrp(t) - \amp =\frac{1}{\sqrt{\big(-3\sin(t) \big)^2+\big(3\cos(t) \big)^2+ 4^2}}\la -3\sin(t) ,3\cos(t) , 4\ra - \amp = \la -\frac35\sin(t) ,\frac35\cos(t) ,\frac45\ra - . -

    - -

    - We can now easily compute \unittangent(0) and \unittangent(1): - - \unittangent(0) = \la 0,\frac35,\frac45\ra\,; \unittangent(1) = \la -\frac35\sin(1) ,\frac35\cos(1) ,\frac45\ra \approx \la -0.505,0.324,0.8\ra - . -

    - -

    - These are plotted in - with their initial points at - \vec r(0) and \vec r(1), respectively. - (They look rather short - since they are only length 1.) -

    - -
    - Plotting unit tangent vectors in - - - - A 3D plot of a portion of a helix, with two unit tangent vectors shown. - -

    - The curve \vec{r}(t) = \la 3\cos(t), 3\sin(t), 4t\ra is a circular helix of radius 3. - One revolution of the helix is shown, representing the interval -\pi\leq t\leq \pi. - Two unit tangent vectors are shown, plotted at the points corresponding to \vec{r}(0) - and \vec{r}(1). -

    - -

    - The length of the unit tangent vectors in the plot are short relative to the length of the curve, - so one has to look closely to tell that the curve does indeed bend away from the tip of the unit tangent vector. -

    -     - - //ASY file for figtannorm13D.asy in Chapter 11 - //ASPECT RATIO IS MAKING THE UNIT NORMAL LOOK LONGER THAN THE UNIT TANGENT - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(10,10,43); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={-10,10}; - defaultpen(0.5mm); - pair xbounds=(-4,4); - pair ybounds=(-4,4); - pair zbounds=(-15,15); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve <3cos t,3sin t,4t> for t from -pi to pi - triple g(real t) {return (3*cos(t),3*sin(t),4*t);} - path3 mypath=graph(g,-pi,pi,operator ..); draw(mypath,bluepen); - //Draw the unit tangent and norma on the curve at t=pi/2 - draw((3,0,0)--(3,.6,.8),redpen+linewidth(2),Arrow3(size=3mm));//T - draw((1.62,2.52,4)--(1.12,2.85,4.8),redpen+linewidth(2),Arrow3(size=3mm));//N - - - - -
    - -

    - The unit tangent vector \unittangent(t) always has a magnitude of 1, though it is sometimes easy to doubt that is true. - We can help solidify this thought in our minds by computing \norm{\unittangent(1)}: - - \norm{\unittangent(1)} \approx \sqrt{(-0.505)^2+0.324^2+0.8^2} = 1.000001 - . -

    - -

    - We have rounded in our computation of \unittangent(1), - so we don't get 1 exactly. - We leave it to the reader to use the exact representation of - \unittangent(1) to verify it has length 1. -

    -     - -     - -

    - In many ways, the previous example was too nice. - It turned out that \vrp(t) was always of length 5. - In the next example the length of \vrp(t) is variable, - leaving us with a formula that is not as clean. -

    - - - Computing the unit tangent vector - -

    - Let \vrt=\la t^2-t,t^2+t\ra. - Find \unittangent(t) and compute - \unittangent(0) and \unittangent(1). -

    -     - -

    - We find \vrp(t) = \la 2t-1,2t+1\ra, and - - \norm{\vrp(t)} = \sqrt{(2t-1)^2+(2t+1)^2} = \sqrt{8t^2+2} - . -

    - -

    - Therefore - - \unittangent(t) = \frac{1}{\sqrt{8t^2+2}}\la 2t-1,2t+1\ra = \la \frac{2t-1}{\sqrt{8t^2+2}},\frac{2t+1}{\sqrt{8t^2+2}}\ra - . -

    - -

    - When t=0, - we have \unittangent(0) = \la -1/\sqrt{2},1/\sqrt{2}\ra; - when t=1, - we have \unittangent(1) = \la 1/\sqrt{10}, 3/\sqrt{10}\ra. - We leave it to the reader to verify each of these is a unit vector. - They are plotted in -

    - -
    - Plotting unit tangent vectors in - - - A rotated parabola, along with two of its unit tangent vectors. - -

    - The curve in this example is the same one that we saw in . - It is a parabola that has been rotated so that it is symmetric about the line y=x, - and it has intercepts at the points (2,0), (0,0), and (0,2). -

    - -

    - There are unit tangent vectors plotted at (0,0) (corresponding to t=0), - and (0,2) (corresponding to t=1). - The tangent vector at the origin points up and to the left, - and the tangent vector at (0,2) points up and to the right. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-1.1,ymax=6.5, - xmin=-2.1,xmax=7 - ] - - \addplot+ [domain=-2:2,samples=40] ({x^2-x},{x^2+x}); - - \draw [thick,->,secondcolor] (axis cs:0,0) -- (axis cs:-.707,.707); - \draw [thick,->,secondcolor] (axis cs:0,2) -- (axis cs:0.32,2.95); - \draw [thick,->,firstcolor] (axis cs:3.79,.77) -- (axis cs:3.75,.75); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -     -     - - - Unit Normal Vector -

    - Just as knowing the direction tangent to a path is important, - knowing a direction orthogonal to a path is important. - When dealing with real-valued functions, - we defined the normal line at a point to the be the line through the point that was perpendicular to the tangent line at that point. - We can do a similar thing with vector-valued functions. - Given \vrt in \mathbb{R}^2, - we have 2 directions perpendicular to the tangent vector, - as shown in . - It is good to wonder Is one of these two directions preferable over the other? -

    - -
    - Given a direction in the plane, there are always two directions orthogonal to it - - - A generic plane curve with a tangent vector, along with two possible normal vectors. - -

    - A portion of a curve is shown in the first quadrant. - The curve is smooth, and forms a large arc that is parabolic in shape. - A tangent vector is shown at one point on the curve. - At that point, two possible normal vectors are plotted. - One points to the inside of the curve, in the direction in which it is turning. - The other points to the outside of the curve, - away from the direction in which the curve is turning. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ytick=\empty, - ymin=-.1,ymax=1.9, - xmin=-.1,xmax=2.3 - ] - - \addplot+ [domain=-1:1,samples=40] ({x^2-x+.6},{x^2+x+.3}); - - \draw[thick,->,black] (axis cs:.51,.41) -- (axis cs:.233,.83); - \draw[thick,->,secondcolor] (axis cs:.51,.41) -- (axis cs:.1,0.13); - \draw[thick,->,secondcolor] (axis cs:.51,.41) -- (axis cs:.93,.69); - \draw[thick,->,firstcolor] (axis cs:0.71, 0.21) -- (axis cs:.698,.218); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Given \vrt in \mathbb{R}^3, - there are infinitely many vectors orthogonal to the tangent vector at a given point. - Again, we might wonder Is one out of this infinite number of choices preferable over the others? - Is one of these the right choice? -

    - -

    - The answer in both \mathbb{R}^2 and \mathbb{R}^3 is Yes, - there is one vector that is not only preferable, - it is the right one to choose. - Recall , - which states that if \vrt has constant length, - then \vrt is orthogonal to \vrp(t) for all t. - We know \unittangent(t), the unit tangent vector, has constant length. - Therefore \unittangent(t) is orthogonal to - \unittangentprime(t). - vector-valued functionof constant length -

    - -

    - We'll see that \unittangentprime(t) is more than just a convenient choice of vector that is orthogonal to \vrp(t); - rather, it is the right choice. - Since all we care about is the direction, - we define this newly found vector to be a unit vector. -

    - - - - Unit Normal Vector - -

    - Let \vrt be a vector-valued function where the unit tangent vector, - \unittangent(t), is smooth on an open interval I. - The unit normal vector - \unitnormal(t) is - unit normal vectordefinition - unit vectorunit normal vector - - \unitnormal(t) = \frac1{\norm{\unittangentprime(t)}}\unittangentprime(t) - . -

    -     -     - - - - - Computing the unit normal vector - -

    - Let \vrt = \la 3\cos(t) , 3\sin(t) , 4t\ra as in . - Sketch both \unittangent(\pi/2) and - \unitnormal(\pi/2) with initial points at \vec r(\pi/2). -

    -     - -

    - In , - we found - - \unittangent(t) = \la (-3/5)\sin(t) ,(3/5)\cos(t) ,4/5\ra - . - Therefore - - \unittangentprime(t) = \la -\frac35\cos(t) ,-\frac35\sin(t) ,0\ra \text{ and } \norm{\unittangentprime(t)} = \frac35 - . -

    - -

    - Thus - - \unitnormal(t) = \frac{\unittangentprime(t)}{3/5} = \la -\cos(t) ,-\sin(t) ,0\ra - . -

    - -

    - We compute \unittangent(\pi/2) = \la -3/5,0,4/5\ra and \unitnormal(\pi/2) = \la 0,-1,0\ra. - These are sketched in . -

    - -
    - Plotting unit tangent and normal vectors in - - - - A three-dimensional helix. At one point, the unit tangent and normal vectors are shown. - -

    - The plot is three-dimensional, and shows a helix revolving about the z axis. - At a point above the y axis, the unit tangent and normal vectors are plotted. - The unit tangent vector points in the direction of travel, - while the unit normal vector points inward, toward the z axis. -

    -     - - //ASY file for figtannorm13D.asy in Chapter 11 - //ASPECT RATIO IS MAKING THE UNIT NORMAL LOOK LONGER THAN THE UNIT TANGENT - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(10,10,34); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-3,3}; - real[] myychoice={-3,3}; - real[] myzchoice={-10,10}; - defaultpen(0.5mm); - //pair xbounds=(-5,5); - //pair ybounds=(-5,5); - //pair zbounds=(-1,9); - - pair xbounds=(-4,4); - pair ybounds=(-4,4); - pair zbounds=(-15,15); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve <3cos t,3sin t,4t> for t from -pi to pi - triple g(real t) {return (3*cos(t),3*sin(t),4*t);} - //path3 mypath=graph(g,0,pi/2+pi/4,operator ..); draw(mypath,bluepen); - path3 mypath=graph(g,-pi,pi,operator ..); draw(mypath,bluepen); - - //Draw the unit tangent and norma on the curve at t=pi/2 - //draw((0,3,2*pi)--(-3/5,3,2*pi+4/5),redpen+linewidth(2),Arrow3(size=3mm));//T - draw((0,3,2*pi)--(-.942,3,2*pi+1.256),redpen+linewidth(2),Arrow3(size=3mm));//T - draw((0,3,2*pi)--(0,2,2*pi),redpen+linewidth(2),Arrow3(size=3mm));//N - - - - -
    -     - -     - - - -

    - The previous example was once again - too nice. In general, - the expression for \unittangent(t) contains fractions of square roots, - hence the expression of \unittangentprime(t) is very messy. - We demonstrate this in the next example. -

    - - - Computing the unit normal vector - -

    - Let \vrt=\la t^2-t,t^2+t\ra as in . - Find \unitnormal(t) and sketch \vrt with the unit tangent and normal vectors at t=-1,0 and 1. -

    -     - -

    - In , we found - - \unittangent(t) = \la \frac{2t-1}{\sqrt{8t^2+2}},\frac{2t+1}{\sqrt{8t^2+2}}\ra - . -

    - -

    - Finding \unittangentprime(t) requires two applications of the Quotient Rule: - - T\,'(t) \amp = \la \frac{\sqrt{8t^2+2}(2)-(2t-1)\left(\frac12(8t^2+2)^{-1/2}(16t)\right)}{8t^2+2},\right. - \amp \left.\frac{\sqrt{8t^2+2}(2)-(2t+1)\left(\frac12(8t^2+2)^{-1/2}(16t)\right)}{8t^2+2} \ra - \amp = \la \frac{4 (2 t+1)}{\left(8 t^2+2\right)^{3/2}},\frac{4 - (1-2 t)}{\left(8 t^2+2\right)^{3/2}}\ra - -

    - -

    - This is not a unit vector; to find \unitnormal(t), - we need to divide \unittangentprime(t) by its magnitude. - - \norm{\unittangentprime(t)} \amp = \sqrt{\frac{16(2t+1)^2}{(8t^2+2)^3}+\frac{16(1-2t)^2}{(8t^2+2)^3}} - \amp = \sqrt{\frac{16(8t^2+2)}{(8t^2+2)^3}} - \amp = \frac{4}{8t^2+2} - . -

    - -

    - Finally, - - \unitnormal(t) \amp = \frac1{4/(8t^2+2)}\la \frac{4 (2 t+1)}{\left(8 t^2+2\right)^{3/2}},\frac{4 - (1-2 t)}{\left(8 t^2+2\right)^{3/2}}\ra - \amp = \la \frac{2t+1}{\sqrt{8t^2+2}},-\frac{2t-1}{\sqrt{8t^2+2}}\ra - . -

    - -

    - Using this formula for \unitnormal(t), - we compute the unit tangent and normal vectors for t=-1,0 and 1 and sketch them in . -

    - -
    - Plotting unit tangent and normal vectors in - - - A rotated parabola, with unit tangent and normal vectors plotted at three points. - -

    - The curve is the same rotated parabola seen previously, - with symmetry about the line y=x, - and intercepts at the points (2,0), (0,0), and (0,2). - At each of the intercepts, the unit tangent and normal vectors are plotted. - The tangent vectors point along the direction of travel, - while the normal vectors point inward, toward the line of symmetry. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-1.1,ymax=6.5, - xmin=-2.1,xmax=7 - ] - - \addplot+ [domain=-2:2,samples=60] ({x^2-x},{x^2+x}); - - \draw [thick,->,secondcolor] (axis cs:0,0) -- (axis cs:-.707,.707); - \draw [thick,->,secondcolor] (axis cs:0,2) -- (axis cs:0.32,2.95); - \draw [thick,->,secondcolor] (axis cs:2,0) -- (axis cs:1.05,-.32); - - \draw [thick,->,secondcolor] (axis cs:0,0) -- (axis cs:.707,.707); - \draw [thick,->,secondcolor] (axis cs:0,2) -- (axis cs:.95,1.68); - \draw [thick,->,secondcolor] (axis cs:2,0) -- (axis cs:1.68,0.95); - - \draw [thick,->,firstcolor] (axis cs:3.79,.77) -- (axis cs:3.75,.75); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -     - -

    - The final result for \unitnormal(t) in - is suspiciously similar to \unittangent(t). - There is a clear reason for this. - If \vec u = \la u_1,u_2\ra is a unit vector in \mathbb{R}^2, - then the only unit vectors orthogonal to \vec u are - \la -u_2,u_1\ra and \la u_2,-u_1\ra. - Given \unittangent(t), we can quickly determine - \unitnormal(t) if we know which term to multiply by (-1). -

    - -

    - Consider again , - where we have plotted some unit tangent and normal vectors. - Note how \unitnormal(t) always points - inside the curve, - or to the concave side of the curve. - This is not a coincidence; this is true in general. - Knowing the direction that \vec r(t) turns - allows us to quickly find \unitnormal(t). -

    - - - Unit Normal Vectors in <m>\mathbb{R}^2</m> - -

    - Let \vec r(t) be a vector-valued function in \mathbb{R}^2 where - \unittangentprime(t) is smooth on an open interval I. - Let t_0 be in I and - \unittangent(t_0) = \la t_1,t_2\ra Then \unitnormal(t_0) is either - - \unitnormal(t_0) = \la -t_2,t_1\ra \text{ or } \unitnormal(t_0) = \la t_2,-t_1\ra - , - whichever is the vector that points to the concave side of the graph of \vec r. - unit tangent vectorin \mathbb{R}^2 - unit normal vectorin \mathbb{R}^2 -

    -     -     -     - - - Application to Acceleration -

    - Let \vrt be a position function. - It is a fact (stated later in ) - that acceleration, \vat, - lies in the plane defined by \unittangent and \unitnormal. - That is, there are scalar functions - a_\text{T}(t) and a_\text{N}(t) such that - - \vat = a_\text{T}(t)\unittangent(t) + a_\text{N}(t)\unitnormal(t) - . -

    - -

    - We generally drop the of t - part of the notation and just write - a_\text{T} and a_\text{N}. -

    - -

    - The scalar a_\text{T} measures how much - acceleration is in the direction of travel, that is, - it measures the component of acceleration that affects the speed. - The scalar a_\text{N} measures how much - acceleration is perpendicular to the direction of travel, that is, - it measures the component of acceleration that affects the direction of travel. - unit tangent vectorand acceleration - unit normal vectorand acceleration -

    - -

    - We can find a_\text{T} using the orthogonal projection of - \vec a(t) onto \unittangent(t) - (review - in if needed). - Recalling that since \unittangent(t) is a unit vector, - \unittangent(t)\cdot\unittangent(t)=1, so we have - - \proj{a(t)}{T(t)} = \frac{\vec a(t)\cdot\unittangent(t)}{\unittangent(t)\cdot\unittangent(t)}\unittangent(t) = \underbrace{\big(\vec a(t)\cdot\unittangent(t)\big)}_{a_\text{T}}\unittangent(t) - . -

    - -

    - Thus the amount of \vat in the direction of - \unittangent(t) is a_\text{T}=\vat\cdot\unittangent(t). - The same logic gives a_\text{N} = \vat\cdot\unitnormal(t). -

    - -

    - While this is a fine way of computing a_\text{T}, - there are simpler ways of finding a_\text{N} - (as finding \unitnormal itself can be complicated). - The following theorem gives alternate formulas for - a_\text{T} and a_\text{N}. -

    - - - - Acceleration in the Plane Defined by <m>\unittangent</m> and <m>\unitnormal</m> - -

    - Let \vrt be a position function with acceleration \vat and unit tangent and normal vectors - \unittangent(t) and \unitnormal(t). - Then \vat lies in the plane defined by - \unittangent(t) and \unitnormal(t); - that is, there exists scalars - a_\text{T} and a_\text{N} such that - - \vat = a_\text{T} \unittangent(t) + a_\text{N} \unitnormal(t) - . -

    - -

    - Moreover, - - a_\text{T} \amp = \vat\cdot\unittangent(t) = \frac{d}{dt}\Big(\norm{\vec v(t)}\Big) - a_\text{N} \amp = \vat\cdot \unitnormal(t) = \sqrt{\norm{\vec a(t)}^2-a_\text{T} ^2} = \frac{\norm{\vat\times\vvt}}{\norm{\vvt}} = \norm{\vvt}\,\norm{\unittangentprime(t)} - -

    -     -     - -

    - unit tangent vectorand acceleration - unit normal vectorand acceleration - an@a_\text{N} - at@a_\text{T} - unit normal vectoran@a_\text{N} - unit tangent vectorat@a_\text{T} -

    - -

    - Note the second formula for a_\text{T}: - \ds \frac{d}{dt}\Big(\norm{\vvt}\Big). - This measures the rate of change of speed, - which again is the amount of acceleration in the direction of travel. -

    - - - - - Computing <m>a_T</m> and <m>a_N</m> - -

    - Let \vrt = \la 3\cos(t) , 3\sin(t) , 4t\ra as in Examples - and . - Find a_\text{T} and a_\text{N}. -

    -     - -

    - The previous examples give \vat = \la -3\cos(t) ,-3\sin(t) ,0\ra and - - \unittangent(t) = \la -\frac35\sin(t) ,\frac35\cos(t) ,\frac45\ra \text{ and } \unitnormal(t) = \la -\cos(t) ,-\sin(t) ,0\ra - . -

    - -

    - We can find a_\text{T} and - a_\text{N} directly with dot products: - - a_\text{T} \amp = \vat\cdot \unittangent(t) = \frac95\cos(t) \sin(t) -\frac95\cos(t) \sin(t) +0 = 0. - a_\text{N} \amp = \vat\cdot \unitnormal(t) = 3\cos^2(t) +3\sin^2(t) + 0 = 3 - . -

    - -

    - Thus \vat = 0\unittangent(t) + 3\unitnormal(t) = 3\unitnormal(t), - which is clearly the case. -

    - -

    - What is the practical interpretation of these numbers? - a_\text{T} =0 means the object is moving at a constant speed, - and hence all acceleration comes in the form of direction change. -

    -     - -     - - - Computing <m>a_T</m> and <m>a_N</m> - -

    - Let \vrt=\la t^2-t,t^2+t\ra as in Examples - and . - Find a_\text{T} and a_\text{N}. -

    -     - -

    - The previous examples give \vat = \la 2,2\ra and - - \unittangent(t) = \la \frac{2t-1}{\sqrt{8t^2+2}},\frac{2t+1}{\sqrt{8t^2+2}}\ra \text{ and } \unitnormal(t) = \la \frac{2t+1}{\sqrt{8t^2+2}},-\frac{2t-1}{\sqrt{8t^2+2}}\ra - . -

    - -

    - While we can compute a_\text{N} using \unitnormal(t), - we instead demonstrate using another formula from . - - a_\text{T} \amp = \vat\cdot\unittangent(t) = \frac{4t-2}{\sqrt{8t^2+2}}+\frac{4t+2}{\sqrt{8t^2+2}} = \frac{8t}{\sqrt{8t^2+2}}. - a_\text{N} \amp = \sqrt{\norm{\vat}^2-a_\text{T} ^2} = \sqrt{8-\left(\frac{8t}{\sqrt{8t^2+2}}\right)^2}=\frac{4}{\sqrt{8t^2+2}} - . -

    - -

    - When t=2, - \ds a_\text{T} = \frac{16}{\sqrt{34}}\approx 2.74 and \ds a_\text{N} = \frac{4}{\sqrt{34}} \approx 0.69. - We interpret this to mean that at t=2, - the particle is accelerating mostly by increasing speed, - not by changing direction. - As the path near t=2 is relatively straight, - this should make intuitive sense. - - gives a graph of the path for reference. -

    - -
    - Graphing \vec r(t) in - - - A rotated parabola, with two points marked, corresponding to parameter values t=0 and t=2. - -

    - The curve is again a rotated parabola that is symmetric about y=x. - There are two marked points: the origin, which is labeled with the parameter value t=0, - and the point (2,6), which is labeled with the parameter value t=2. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.1,ymax=6.5, - xmin=-2.1,xmax=7 - ] - - \addplot+ [domain=-2.1:2.1,samples=60] ({x^2-x},{x^2+x}); - - \filldraw [black] (axis cs: 2,6) circle (2.4pt) node [below right] { $t=2$}; - \filldraw [black] (axis cs: 0,0) circle (2.4pt) node [above right] { $t=0$}; - - \draw (axis cs: 4.2,1.8) node { $\vec r(t)$}; - \draw [thick,->,firstcolor,>=stealth] (axis cs:3.79,.77) -- (axis cs:3.75,.75); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Contrast this with t=0, - where a_\text{T} = 0 and a_\text{N} = 4/\sqrt{2}\approx 2.82. - Here the particle's speed is not changing and all acceleration is in the form of direction change. -

    -     - -     - - - Analyzing projectile motion - -

    - A ball is thrown from a height of 240 - with an initial speed of 64 - and an angle of elevation of 30^\circ. - Find the position function \vrt of the ball and analyze - a_\text{T} and a_\text{N}. - projectile motion -

    -     - -

    - Using - of - we form the position function of the ball: - - \vrt = \la \big(64\cos(30^\circ) \big)t, -16t^2+\big(64\sin(30^\circ) \big)t+240\ra - , - which we plot in . -

    - -
    - Plotting the position of a thrown ball, with 1s increments shown - - - An inverted parabola with vertex in the first quadrant. Points corresponding to several parameter values are marked. - -

    - The path of a ball is shown; the ball travels in a parabolic arc, - beginning, when t=0 at the point (0,240). - It reaches its vertex when t=1, at the point (32\sqrt{3},256), - and then travels down and to the right, - meeting the x axis at the point (160\sqrt{3},0), when t=5. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-10,ymax=260, - xmin=-10,xmax=310 - ] - - \addplot+ [domain=0:5,samples=40] ({64*cos(30)*x},{-16*x^2+64*sin(30)*x+240}); - - \filldraw [black] (axis cs: 0,240) circle (2.4pt) node [below right] { $t=0$}; - \filldraw [black] (axis cs: 55.4,256) circle (2.4pt) node (A)[ ] {}; - \filldraw [black] (axis cs: 111,240) circle (2.4pt) node [above right] { $t=2$}; - \filldraw [black] (axis cs: 166,192) circle (2.4pt) node [above right] { $t=3$}; - \filldraw [black] (axis cs: 222,112) circle (2.4pt) node [above right] { $t=4$}; - \filldraw [black] (axis cs: 277,0) circle (2.4pt) node (B) [] {}; - - \draw [thick,->,firstcolor] (axis cs:149.649, 209.76) -- (axis cs:152,207); - - \end{axis} - - \draw (A) node[ above] {{ $t=1$}}; - \draw (B) node [shift={(-16pt,8pt)}] { $t=5$}; - - \end{tikzpicture} - - - - -
    - -

    - From this we find \vvt = \la 64\cos(30^\circ) , -32t+64\sin(30^\circ) \ra and \vat = \la 0,-32\ra. - Computing \unittangent(t) is not difficult, - and with some simplification we find - - \unittangent(t) = \la \frac{\sqrt{3}}{\sqrt{t^2-2t+4}}, \frac{1-t}{\sqrt{t^2-2t+4}}\ra - . -

    - -

    - With \vat as simple as it is, - finding a_\text{T} is also simple: - - a_\text{T} = \vat\cdot \unittangent(t) = \frac{32t-32}{\sqrt{t^2-2t+4}} - . -

    - -

    - We choose to not find \unitnormal(t) and find - a_\text{N} through the formula a_\text{N} = \sqrt{\norm{\vat}^2-a_\text{T}^2\,}: - - a_\text{N} = \sqrt{32^2-\left(\frac{32t-32}{\sqrt{t^2-2t+4}}\right)^2} = \frac{32\sqrt{3}}{\sqrt{t^2-2t+4}} - . -

    - -

    - - gives a table of values of - a_\text{T} and a_\text{N}. - When t=0, we see the ball's speed is decreasing; - when t=1 the speed of the ball is unchanged. - This corresponds to the fact that at t=1 the ball reaches its highest point. -

    - -

    - After t=1 we see that - a_\text{N} is decreasing in value. - This is because as the ball falls, - its path becomes straighter and most of the acceleration is in the form of speeding up the ball, - and not in changing its direction. -

    - -
    - A table of values of a_T and a_N in - - - t - a_\text{T} - a_\text{N} - - - - - - - - 0 - -16 - 27.7 - - - 1 - 0 - 32 - - - 2 - 16 - 27.7 - - - 3 - 24.2 - 20.9 - - - 4 - 27.7 - 16 - - - 5 - 29.4 - 12.7 - - -
    -     -     - -

    - Our understanding of the unit tangent and normal vectors is aiding our understanding of motion. - The work in - gave quantitative analysis of what we intuitively knew. -

    - -

    - The next section provides two more important steps towards this analysis. - We currently describe position only in terms of time. - In everyday life, - though, we often describe position in terms of distance - (The gas station is about 2 miles ahead, on the left.). - The arc length parameter - allows us to reference position in terms of distance traveled. -

    - -

    - We also intuitively know that some paths are straighter than others and some are curvier than others, - but we lack a measurement of curviness. - The arc length parameter provides a way for us to compute curvature, - a quantitative measurement of how curvy a curve is. -

    -     - - - - Terms and Concepts - - - -

    - If \unittangent(t) is a unit tangent vector, - what is \norm{\unittangent(t)}? -

    - -

    - -

    -     - - - - - - - -

    - What is the length of any unit vector? -

    -     -     -     -     - -     - - - - -

    - If \unitnormal(t) is a unit normal vector, - what is \unitnormal(t)\cdot \vrp(t)? -

    - -

    - -

    -     - - - - - -

    - Any vector of constant magnitude is orthogonal to its derivative. -

    -     -     -     -     - -     - - - - -

    - The acceleration vector \vat lies in the plane defined by what two vectors? -

    -     - - - -

    - \unittangent(t) and \unitnormal(t). -

    -     - -     - - - - -

    - a_\text{T} measures how much the acceleration is affecting the of an object. -

    -     - - - - - - - - -     -     - - - Problems - - - -

    - Given \vrt, - find \unittangent(t) and evaluate it at the indicated value of t. -

    -     - - - - -

    - \vrt = \la 2t^2,t^2-t\ra,t=1 -

    -     - -

    - \unittangent(t) = \la\frac{4 t}{\sqrt{20 t^2-4t+1}},\frac{2 t-1}{\sqrt{20 t^2-4t+1}}\ra; - \unittangent(1) = \la 4/\sqrt{17},1/\sqrt{17}\ra -

    -     - -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $T=Compute("<1/sqrt(1+sin^2(t)),-sin(t)/sqrt(1+sin^2(t))>"); - $Ta=Compute("<sqrt(2/3),-1/sqrt(3)>"); - - - -

    - \vrt = \la t,\cos(t) \ra, t=\pi/4. -

    - - Find \unittangent(t). - -

    - -

    - - Find \unittangent(\pi/4). - -

    - -

    -     -     -     - - - - -

    - \vrt = \la \cos^3(t) ,\sin^3(t) \ra,t=\pi/4 -

    -     - -

    - \unittangent(t) = \frac{\cos(t) \sin(t) }{\sqrt{\cos^2(t) \sin^2(t) }}\la -\cos(t) ,\sin(t) \ra. (Be careful; - this cannot be simplified as just - \la -\cos(t) ,\sin(t) \ra as \sqrt{\cos^2(t) \sin^2(t) }\neq \cos(t) \sin(t), - but rather \abs{\cos(t) \sin(t) }.) - \unittangent(\pi/4) = \la -\sqrt{2}/2,\sqrt{2}/2\ra -

    -     - -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $T=Compute("<-sin(t), cos(t)>"); - $Ta=Compute("<0,-1>"); - - - -

    - \vrt = \la \cos(t) , \sin(t) \ra, t=\pi. -

    - - Find \unittangent(t). - -

    - -

    - - Find \unittangent(\pi). - -

    - -

    -     -     -     - -     - - - -

    - Find the equation of the line tangent to the curve at the indicated t-value using the unit tangent vector. - Note: these are the same problems as in Exercises. -

    -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - $L=Compute("(2,0) + t*<4/sqrt(17),1/sqrt(17)>"); - - - -

    - Find the vector equation of the line tangent to - \vrt = \la 2t^2,t^2-t \ra at t=1 using the unit tangent vector. -

    - - - \ell(t)= - -

    - -

    -     -     -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - $L=Compute("(pi/4,sqrt(2)/2) + t*<sqrt(2/3),-1/sqrt(3)>"); - - - -

    - Find the vector equation of the line tangent to - \vrt = \la t,\cos(t) \ra at t=\pi/4 using the unit tangent vector. -

    - - - \ell(t)= - -

    - -

    -     -     -     - - - - -

    - \vrt = \la \cos^3(t) ,\sin^3(t) \ra,t=\pi/4 -

    -     - -

    - \ell(t) = \la \sqrt{2}/4,\sqrt{2}/4\ra + t\la -\sqrt{2}/2,\sqrt{2}/2\ra; - in parametric form, -

    - -

    - \ell(t) = \left\{\begin{array}{ccc} x\amp =\amp \sqrt{2}/4-\sqrt{2}t/2 \\ y \amp =\amp \sqrt{2}/4+\sqrt{2}t/2 - \end{array} \right. -

    -     - -     - - - - - Context("Vector2D"); - Context()->variables->are(t=>'Real'); - $L=Compute("(-1,0) + t*<0,-1>"); - - - -

    - Find the vector equation of the line tangent to - \vrt = \la \cos(t) ,\sin(t) \ra at t=\pi using the unit tangent vector. -

    - - - \ell(t)= - -

    - -

    -     -     -     - -     - - - -

    - Find \unitnormal(t) using . - Confirm the result using . -

    -     - - - - -

    - \vrt = \la 3\cos(t) , 3\sin(t) \ra -

    -     - -

    - \unittangent(t) = \la -\sin(t) ,\cos(t) \ra; - \unitnormal(t) = \la -\cos(t) ,-\sin(t) \ra -

    -     - -     - - - - -

    - \vrt = \la t, t^2 \ra -

    -     - -

    - \unittangent(t) = \la \frac{1}{\sqrt{1+4t^2}},\frac{2t}{\sqrt{1+4t^2}}\ra; - \unitnormal(t) = \la -\frac{2t}{\sqrt{1+4t^2}},\frac{1}{\sqrt{1+4t^2}}\ra -

    -     - -     - - - - -

    - \vrt = \la \cos(t) ,2\sin(t) \ra -

    -     - -

    - \unittangent(t) = \la -\frac{\sin(t) }{\sqrt{4\cos^2(t) +\sin^2(t) }},\frac{2\cos(t) }{\sqrt{4\cos^2(t) +\sin^2(t) }}\ra; - \unitnormal(t) = \la -\frac{2\cos(t) }{\sqrt{4\cos^2(t) +\sin^2(t) }},-\frac{\sin(t) }{\sqrt{4\cos^2(t) +\sin^2(t) }}\ra -

    -     - -     - - - - -

    - \vrt = \la e^t,e^{-t} \ra -

    -     - -

    - \unittangent(t) = \la \frac{e^t}{\sqrt{e^{2t}+e^{-2t}}},-\frac{e^{-t}}{\sqrt{e^{2t}+e^{-2t}}}\ra; - \unitnormal(t) = \la \frac{e^{-t}}{\sqrt{e^{2t}+e^{-2t}}},\frac{e^{t}}{\sqrt{e^{2t}+e^{-2t}}}\ra -

    -     - -     - -     - - - -

    - A position function \vrt is given along with its unit tangent vector - \unittangent(t) evaluated at t=a, - for some value of a. -

    - -

    -

      -
    1. -

      - Confirm that \unittangent(a) is as stated. -

      -
      2. - -
    2. -

      - Using a graph of \vrt and , - find \unitnormal(a). -

      -
      3. -
    -

    -     - - - - -

    - \vrt = \la 3\cos(t) ,5\sin(t) \ra;\ds \unittangent(\pi/4) = \la -\frac3{\sqrt{34}}, \frac5{\sqrt{34}}\ra. -

    -     - -

    -

      -
    1. -

      - Be sure to show work -

      -
      2. - -
    2. -

      - \unitnormal(\pi/4) = \la -5/\sqrt{34}, -3/\sqrt{34}\ra -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \vrt = \la t,\frac{1}{t^2+1}\ra;\ds \unittangent(1) = \la \frac2{\sqrt{5}}, -\frac1{\sqrt{5}}\ra. -

    -     - -

    -

      -
    1. -

      - Be sure to show work -

      -
      2. - -
    2. -

      - \unitnormal(1) = \la \frac1{\sqrt{5}}, \frac2{\sqrt{5}}\ra -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \vrt = (1+2\sin(t) )\la \cos(t) ,\sin(t) \ra;\ds \unittangent(0) = \la \frac2{\sqrt{5}}, \frac1{\sqrt{5}}\ra. -

    -     - -

    -

      -
    1. -

      - Be sure to show work -

      -
      2. - -
    2. -

      - \unitnormal(0) = \la -\frac1{\sqrt{5}}, \frac2{\sqrt{5}}\ra -

      -
      3. -
    -

    -     - -     - - - - -

    - \ds \vrt = \la \cos^3(t) ,\sin^3(t) \ra;\ds \unittangent(\pi/4) = \la -\frac1{\sqrt{2}}, \frac1{\sqrt{2}}\ra. -

    -     - -

    -

      -
    1. -

      - Be sure to show work -

      -
      2. - -
    2. -

      - \unitnormal(\pi/4) = \la \frac1{\sqrt{2}}, \frac1{\sqrt{2}}\ra -

      -
      3. -
    -

    -     - -     - -     - - - -

    - Find \unitnormal(t). -

    -     - - - - -

    - \vrt = \la 4t,2\sin(t) ,2\cos(t) \ra -

    -     - -

    - \unittangent(t) = \frac{1}{\sqrt{5}}\la 2,\cos(t) ,-\sin(t) \ra; - \unitnormal(t) = \la 0,-\sin(t) ,-\cos(t) \ra -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $N=Compute("<-cos(t), -3/5 sin(t), -4/5sin(t)>"); - - - -

    - If \vrt = \la 5\cos(t) ,3\sin(t) ,4\sin(t) \ra, - find \unitnormal(t). -

    - -

    - -

    -     -     -     - - - - -

    - \vrt = \la a\cos(t) ,a\sin(t) ,b t \ra; a \gt 0 -

    -     - -

    - \unittangent(t) = \frac{1}{\sqrt{a^2+b^2}}\la -a\sin(t) ,a\cos(t) ,b \ra; - \unitnormal(t) = \la -\cos(t) ,-\sin(t) ,0\ra -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(t=>'Real',a=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $N=Compute("<-cos(at), -sin(at), 0>"); - - - -

    - If \vrt = \la \cos(at),\sin(at),t \ra, find \unitnormal(t). -

    - -

    - -

    -     -     -     - -     - - - -

    - Find a_\text{T} and a_\text{N} given \vrt. - Be sure you can sketch \vrt on the indicated interval, - and comment on the relative sizes of a_\text{T} and - a_\text{N} at the indicated t values. -

    -     - - - - -

    - \vrt = \la t,t^2 \ra on [-1,1]; - consider t=0 and t=1. -

    -     - -

    - a_\text{T} = \frac{4t}{\sqrt{1+4t^2}} and a_\text{N} = \sqrt{4-\frac{16t^2}{1+4t^2}} -

    - -

    - At t=0, - a_\text{T} = 0 and a_\text{N} = 2; -

    - -

    - At t=1, - a_\text{T} = 4/\sqrt{5} and a_\text{N} = 2/\sqrt{5}. -

    - -

    - At t=0, - all acceleration comes in the form of changing the direction of velocity and not the speed; - at t=1, - more acceleration comes in changing the speed than in changing direction. -

    -     - -     - - - - - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->set(t=>{limits=>[0.1,4]}); - $aT=Compute("-2/t^5/sqrt(1+1/t^4)"); - $aN=Compute("2/t^3/sqrt(1+1/t^4)"); - $aT1=Formula("-sqrt(2)"); - $aN1=Formula("sqrt(2)"); - $aT2=Formula("-1/(4sqrt(17))"); - $aN2=Formula("1/sqrt(17)"); - - - -

    - \vrt = \la t,1/t \ra on (0,4]; - consider t=1 and t=2. -

    - - Find a_{T}: - -

    - -

    - - Find a_{N}: - -

    - -

    - - Find a_{T}(1): - -

    - -

    - - Find a_{N}(1): - -

    - -

    - - Find a_{T}(2): - -

    - -

    - - Find a_{N}(2): - -

    - -

    -     - -

    - At t=1, - acceleration comes from changing speed and changing direction in - equal measure; at t=2, - acceleration is nearly \vec 0 as it is; - the low value of a_\text{T} shows that the speed is nearly constant and the low value of - a_\text{N} shows the direction is not changing quickly. -

    -     -     -     - - - - -

    - \vrt = \la 2\cos(t) ,2\sin(t) \ra on [0,2\pi]; - consider t=0 and t=\pi/2. -

    -     - -

    - a_\text{T} = 0 and a_\text{N} = 2 -

    - -

    - At t=0, - a_\text{T} = 0 and a_\text{N} = 2; -

    - -

    - At t=\pi/2, - a_\text{T} = 0 and a_\text{N} = 2. -

    - -

    - The object moves at constant speed, - so all acceleration comes from changing direction, - hence a_\text{T}=0. - \vat is always parallel to \unitnormal(t), - but twice as long, hence a_\text{N}=2. -

    -     - -     - - - - - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $aT=Compute("2"); - $aN=Compute("4t^2"); - $aT1=Formula("2"); - $aN1=Formula("2pi"); - $aT2=Formula("2"); - $aN2=Formula("4pi"); - - - -

    - \vrt = \la \cos(t^2),\sin(t^2) \ra on (0,2\pi]; - consider t=\pi/2 and t=\pi. -

    - - Find a_{T}: - -

    - -

    - - Find a_{N}: - -

    - -

    - - Find a_{T}\mathopen{}\left(\sqrt{\pi/2}\right)\mathclose{}: - -

    - -

    - - Find a_{N}\mathopen{}\left(\sqrt{\pi/2}\right)\mathclose{}: - -

    - -

    - - Find a_{T}\mathopen{}\left(\sqrt{\pi}\right)\mathclose{}: - -

    - -

    - - a_{N}\mathopen{}\left(\sqrt{\pi}\right)\mathclose{}: - -

    - -

    -     - -

    - The object moves at increasing speed - (increasing at a constant rate of acceleration), - hence a_\text{T}=2. - Since the object is increasing speed yet always traveling in a circle of radius 1, the direction must change more quickly; - the amount of acceleration that changes direction increases over time. -

    -     -     -     - - - - -

    - \vrt = \la a\cos(t) ,a\sin(t) , bt \ra on [0,2\pi], - where a,b \gt 0; - consider t=0 and t=\pi/2. -

    -     - -

    - a_\text{T} = 0 and a_\text{N} = a -

    - -

    - At t=0, - a_\text{T} = 0 and a_\text{N} = a; -

    - -

    - At t=\pi/2, - a_\text{T} = 0 and a_\text{N} = a. -

    - -

    - The object moves at constant speed, - meaning that a_\text{T} is always 0. - The object rises along the z-axis at a constant rate, - so all acceleration comes in the form of changing direction circling the z-axis. - The greater the radius of this circle the greater the acceleration, - hence a_\text{N}=a. -

    -     - -     - - - - - Context()->variables->are(t=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $aT=Compute("0"); - $aN=Compute("5"); - $aT1=Formula("0"); - $aN1=Formula("5"); - $aT2=Formula("0"); - $aN2=Formula("5"); - - - -

    - \vrt = \la 5\cos(t) ,4\sin(t) , 3\sin(t) \ra on [0,2\pi]; - consider t=0 and t=\pi/2. -

    - - Find a_{T}: - -

    - -

    - - Find a_{N}: - -

    - -

    - - Find a_{T}(0): - -

    - -

    - - Find a_{N}(0): - -

    - -

    - - Find a_{T}(\pi/2): - -

    - -

    - - Find a_{N}(\pi/2): - -

    - -

    -     - -

    - The object moves at constant speed, - meaning that a_\text{T} is always 0. - Acceleration is thus always perpendicular to the direction of travel; - in this particular case, - it is always 5 times the unit vector pointing orthogonal to the direction of travel. -

    -     -     -     - -     -     -     -
    -
    - The Arc Length Parameter and Curvature - - The Arc Length Parametrization -

    - In normal conversation we describe position in terms of both - time and distance. - For instance, imagine driving to visit a friend. - If she calls and asks where you are, - you might answer I am 20 minutes from your house, - or you might say I am 10 miles from your house. - Both answers provide your friend with a general idea of where you are. -

    - - - -

    - Currently, our vector-valued functions have defined points with a parameter t, - which we often take to represent time. - Consider , - where \vrt = \la t^2-t,t^2+t\ra is graphed and the points corresponding to t=0,\ 1 and 2 are shown. - Note how the arc length between t=0 and t=1 is smaller than the arc length between t=1 and t=2; - if the parameter t is time and \vec r is position, - we can say that the particle traveled faster on [1,2] than on [0,1]. -

    - -
    - Introducing the arc length parameter - -
    - - - - A parabola, rotated to be symmetric about the line y=x, with three marked points. - -

    - It is our old friend, the parabolic curve, rotated so that it lies primarily in the first quadrant, - with y=x as its line of symmetry. - There are three marked points: (0,0), (0,2), and (2,6), - corresponding to parameter values t=0, t=1, and t=2, respectively. -

    - -

    - The distance between the first and second points is much less than the distance between the second and third. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.1,ymax=6.5, - xmin=-2.1,xmax=7 - ] - - \addplot+ [domain=-2:2.1,samples=40] ({x^2-x},{x^2+x}); - - \filldraw [secondcolor] (axis cs: 0,0) circle (2.4pt) node [black,above right] { $t=0$}; - \filldraw [secondcolor] (axis cs: 0,2) circle (2.4pt) node [black, right] { $t=1$}; - \filldraw [secondcolor] (axis cs: 2,6) circle (2.4pt) node [black, right] { $t=2$}; - - \draw [thick,->,firstcolor,>=stealth] (axis cs:3.79,.77) -- (axis cs:3.75,.75) node [above ,black] { $\vrt$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - The same parabola as the previous image, this time with six marked points, all equally spaced along the curve. - -

    - The curve generated by \vec{r}(s) is the same as the curve generated by \vec{r}(t) in . - However, as a description of particle motion, the vector-valued function is quite different. - In this image we see six marked points, corresponding to parameter values s=0 through s=6. - The distance between successive points is equal, illustrating the fact that the particle speed is constant in this case. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-1.1,ymax=6.5, - xmin=-2.1,xmax=7 - ] - - \addplot+ [domain=-2:2.1,samples=40] ({x^2-x},{x^2+x}); - - \filldraw [secondcolor] (axis cs: 0,0) circle (2.4pt) node [black,above right] { $s=0$}; - \filldraw [secondcolor] (axis cs: -.24,.94) circle (2.4pt) node [black, right] { $s=1$}; - \filldraw [secondcolor] (axis cs: -.03,1.91) circle (2.4pt) node [black, right] { $s=2$}; - \filldraw [secondcolor] (axis cs: .33,2.84) circle (2.4pt) node [black, right] { $s=3$}; - \filldraw [secondcolor] (axis cs: .75,3.75) circle (2.4pt) node [black, right] { $s=4$}; - \filldraw [secondcolor] (axis cs: 1.21,4.63) circle (2.4pt) node [black, right] { $s=5$}; - \filldraw [secondcolor] (axis cs: 1.71, 5.51) circle (2.4pt) node [black, right] { $s=6$}; - - \draw [thick,->,firstcolor,>=stealth] (axis cs:3.79,.77)--(axis cs:3.75,.75)node [above,black] { $\vec r(s)$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - Now consider , - where the same graph is parametrized by a different variable s. - Points corresponding to s=0 through s=6 are plotted. - The arc length of the graph between each adjacent pair of points is 1. - We can view this parameter s as distance; - that is, the arc length of the graph from s=0 to s=3 is 3, the arc length from s=2 to s=6 is 4, etc. - If one wants to find the point 2.5 units from an initial location (, s=0), - one would compute \vec r(2.5). - This parameter s is very useful, - and is called the arc length parameter. - arc length parameter - arc length -

    - -

    - How do we find the arc length parameter? -

    - -

    - Start with any parametrization of \vec r. - We can compute the arc length of the graph of \vec r on the interval [0,t] with - - \text{arc length}\, = \int_0^t\norm{\vrp(u)}\, du - . -

    - -

    - We can turn this into a function: - as t varies, - we find the arc length s from 0 to t. - This function is - - s(t) = \int_0^t \norm{\vrp(u)}\, du - . -

    - -

    - This establishes a relationship between s and t. - Knowing this relationship explicitly, - we can rewrite \vrt as a function of s: - \vec r(s). - We demonstrate this in an example. -

    - - - Finding the arc length parameter - -

    - Let \vrt = \la 3t-1,4t+2\ra. - Parametrize \vec r with the arc length parameter s. -

    -     - -

    - Using Equation, we write - - s(t) = \int_0^t \norm{\vrp(u)}\, du - . -

    - -

    - We can integrate this, - explicitly finding a relationship between s and t: - - s(t) \amp = \int_0^t \norm{\vrp(u)}\, du - \amp = \int_0^t \sqrt{3^2+4^2}\, du - \amp = \int_0^t 5\, du - \amp = 5t - . -

    - -

    - Since s=5t, - we can write t=s/5 and replace t in \vrt with s/5: - - \vec r(s) = \la 3(s/5)-1, 4(s/5)+2\ra = \la \frac35s-1,\frac45s+2\ra - . -

    - -

    - Clearly, as shown in , - the graph of \vec r is a line, - where t=0 corresponds to the point (-1,2). - What point on the line is 2 units away from this initial point? - We find it with \vec r(2) = \la 1/5, 18/5\ra. -

    - -
    - Graphing \vec r in with parameters t and s - - - A straight line, with positive slope, and several equally-spaced marked points. - -

    - The curve generated by \vec{r}(t) is shown; - it is a straight line with slope 3/4 and y intercept 10/3. - There are six equally-spaced marked points on the line. - The first, at (-1,2), corresponds to both t=0 and s=0. - The last, at (2,6), corresponds to both t=1 and s=5. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,2,-1,-2}, - ymin=-1.1,ymax=6.5, - xmin=-2.1,xmax=2.9 - ] - - \addplot+ [domain=-2.5:2.5] ({3*x-1},{4*x+2}); - - \filldraw (axis cs:-1,2) circle (2.4pt) node [left] { $t=0$} - (axis cs:2,6) circle (2.4pt) node [left] { $t=1$} - (axis cs: -.4,2.8) circle (2.4pt) node [right,secondcolor] { $s=1$} - (axis cs: .2,3.6) circle (2.4pt) node [right,secondcolor] { $s=2$} - (axis cs: .8,4.4) circle (2.4pt) node [right,secondcolor] { $s=3$} - (axis cs: 1.4,5.2) circle (2.4pt) node [right,secondcolor] { $s=4$} - (axis cs: 2,6) circle (2.4pt) node [right,secondcolor] { $s=5$}; - - \draw (axis cs:-1,2) node [secondcolor,shift={(10pt,-5pt)}] { $s=0$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Is the point (1/5,18/5) really 2 units away from (-1,2)? - We use the Distance Formula to check: - - d = \sqrt{\left(\frac15-(-1)\right)^2+ \left(\frac{18}5-2\right)^2} = \sqrt{\frac{36}{25}+\frac{64}{25}} = \sqrt{4}=2 - . -

    - -

    - Yes, \vec r(2) is indeed 2 units away, - in the direction of travel, from the initial point. -

    -     - -     - -

    - Things worked out very nicely in ; - we were able to establish directly that s=5t. - Usually, the arc length parameter is much more difficult to describe in terms of t, - a result of integrating a square root. - There are a number of things that we can learn about the arc length parameter from Equation, though, - that are incredibly useful. -

    - -

    - First, take the derivative of s with respect to t. - The Fundamental Theorem of Calculus - (see ) - states that - - \frac{ds}{dt}=s\primeskip'(t) = \norm{\vrp(t)} - . -

    - -

    - Letting t represent time and \vrt represent position, - we see that the rate of change of s with respect to t is speed; - that is, the rate of change of distance traveled is speed, - which should match our intuition. -

    - -

    - The Chain Rule states that - - \frac{d\vec r}{dt} \amp = \frac{d\vec r}{ds}\cdot\frac{ds}{dt} - \vrp(t) \amp = \vrp(s)\cdot \norm{\vrp(t)} - . -

    - -

    - Solving for \vrp(s), we have - - \vrp(s) = \frac{\vrp(t)}{\norm{\vrp(t)}} = \unittangent(t) - , - where \unittangent(t) is the unit tangent vector. - Equation is often misinterpreted, - as one is tempted to think it states \vrp(t) = \unittangent(t), - but there is a big difference between \vrp(s) and \vrp(t). - The key to take from it is that \vrp(s) is a unit vector. - In fact, the following theorem states that this characterizes the arc length parameter. -

    - - - Arc Length Parameter - -

    - Let \vec r(s) be a vector-valued function. - The parameter s is the arc length parameter if, - and only if, \norm{\vrp(s)} = 1. - arc length parameter -

    -     -     - - -     - - - Curvature -

    - Consider points A and B on the curve graphed in . - One can readily argue that the curve curves more sharply at A than at B. - It is useful to use a number to describe how sharply the curve bends; - that number is the curvature of the curve. -

    - -
    - Establishing the concept of curvature - -
    - - - - Plot of a parametric curve that bends rapidly at one point, but is otherwise relatively flat. - -

    - An L-shaped curve is shown. There are two marked points, labeled A and B. - The point A is at the corner of the curve. - The corner is smooth, not sharp, but represents a significant bend in the curve. - Near the point B, the curve bends much more gently. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=2.9, - xmin=-.1,xmax=3.5 - ] - - \addplot+ [domain=-1:1.5,samples=40] ({x^2-x+1},{x^3+1}); - - \draw [->,firstcolor,thick,>=stealth] (axis cs:2.44, 0.488) -- (axis cs: 2.4141, 0.506961); - - \filldraw (axis cs:.77,1.04) circle (2.4pt) node [below] { $A$} - (axis cs: 1.75, 0.875) circle (2.4pt) node [below] { $B$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - Plot of a parametric curve that bends rapidly at one point, but is otherwise relatively flat, along with some tangent vectors. - -

    - The same L-shaped curve is shown as in , along with the marked points A and B. - Additional points are marked on either side of A and B, - and tangent vectors are plotted at each of these points. - The image illustrates how the directions of the tangent vectors are not that different near B, - but the directions near A are significantly different. -

    -     - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - ymin=-.1,ymax=2.9, - xmin=-.1,xmax=3.5 - ] - - \addplot+ [domain=-1:1.5,samples=40] ({x^2-x+1},{x^3+1}); - - \draw [->,secondcolor,thick] (axis cs:0.8461, 1.53144) -- (axis cs:1.14654, 2.48524); - \draw [->,secondcolor,thick] (axis cs:1.2684, 0.989352) -- (axis cs:0.273445, 1.08968); - \draw [->,secondcolor,thick] (axis cs:2.20203, 0.649597) -- (axis cs:1.35163, 1.17574); - \draw [->,firstcolor,thick] (axis cs:2.44, 0.488) -- (axis cs: 2.4141, 0.506961); - - \filldraw (axis cs:.77,1.04) circle (2.4pt) node [below] { $A$} - (axis cs:0.8461, 1.53144) circle (2.4pt) - (axis cs: 1.2684, 0.989352) circle (2.4pt) - (axis cs: 1.75, 0.875) circle (2.4pt) node [below] { $B$} - (axis cs: 2.20203, 0.649597) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - We derive this number in the following way. - Consider , - where unit tangent vectors are graphed around points A and B. - Notice how the direction of the unit tangent vector changes quite a bit near A, - whereas it does not change as much around B. - This leads to an important concept: - measuring the rate of change of the unit tangent vector with respect to arc length gives us a measurement of curvature. -

    - - - Curvature - -

    - Let \vec r(s) be a vector-valued function where s is the arc length parameter. - The curvature \kappa of the graph of \vec r(s) is - curvature - unit tangent vectorand curvature - - \kappa = \snorm{\frac{d\unittangent}{ds}} = \snorm{\unittangentprime(s)} - . -

    -     -     - - - -

    - If \vec r(s) is parametrized by the arc length parameter, then - - \unittangent(s) = \frac{\vrp(s)}{\norm{\vrp(s)}} \text{ and } \unitnormal(s) = \frac{\unittangentprime(s)}{\norm{\unittangentprime(s)}} - . -

    - -

    - Having defined \norm{\unittangentprime(s)} =\kappa, - we can rewrite the second equation as - - \unittangentprime(s) = \kappa\unitnormal(s) - . -

    - -

    - We already knew that \unittangentprime(s) is in the same direction as \unitnormal(s); - that is, we can think of \unittangent(s) as being pulled - in the direction of \unitnormal(s). - How hard is it being pulled? - By a factor of \kappa. - When the curvature is large, - \unittangent(s) is being pulled hard - and the direction of \unittangent(s) changes rapidly. - When \kappa is small, - T(s) is not being pulled hard and hence its direction is not changing rapidly. -

    - -

    - We use - to find the curvature of the line in . -

    - - - Finding the curvature of a line - -

    - Use - to find the curvature of \vrt = \la 3t-1,4t+2\ra. -

    -     - -

    - In , - we found that the arc length parameter was defined by s=5t, - so \vec r(s) =\la 3s/5-1, 4s/5+2\ra parametrized \vec r with the arc length parameter. - To find \kappa, we need to find \unittangentprime(s). - - \unittangent(s) \amp = \vrp(s) \text{ (recall this is a unit vector) } - \amp = \la 3/5, 4/5\ra. - Therefore - \unittangentprime(s) \amp = \la 0,0\ra - and - \kappa\amp =\snorm{\unittangentprime(s)} = 0 - . -

    - -

    - It probably comes as no surprise that the curvature of a line is 0. (How - curvy is a line? - It is not curvy at all.) -

    -     - -     - -

    - While the definition of curvature is a beautiful mathematical concept, - it is nearly impossible to use most of the time; - writing \vec r in terms of the arc length parameter is generally very hard. - Fortunately, - there are other methods of calculating this value that are much easier. - There is a tradeoff: the definition is easy - to understand though hard to compute, - whereas these other formulas are easy to compute though it may be hard to understand why they work. -

    - - - Formulas for Curvature - -

    - Let C be a smooth curve in the plane or in space. - curvatureequations for -

      -
    1. -

      - If C is defined by y=f(x), then - - \kappa = \frac{\abs{\fp'(x)}}{\Big(1+\big(\fp(x)\big)^2\Big)^{3/2}} - . -

      -
      2. - -
    2. -

      - If C is defined as a vector-valued function in the plane, - \vrt = \la x(t), y(t)\ra, then - - \kappa = \frac{\abs{x'\yp'-x''\yp}}{\big((x')^2+(\yp)^2\big)^{3/2}} - . -

      -
      3. - -
    3. -

      - If C is defined in space by a vector-valued function \vrt, then - - \kappa = \frac{\norm{\unittangentprime(t)}}{\norm{\vrp(t)}} = \frac{\norm{\vrp(t)\times\vrpp(t)}}{\norm{\vrp(t)}^3} = \frac{\vec a(t)\cdot \unitnormal(t)}{\norm{\vec v(t)}^2} - . -

      -
      4. -
    -

    -     -     - -

    - We practice using these formulas. -

    - - - Finding the curvature of a circle - -

    - Find the curvature of a circle with radius r, - defined by \vec c(t) = \la r\cos(t), r\sin(t)\ra. - curvatureof circle -

    -     - -

    - Before we start, - we should expect the curvature of a circle to be constant, - and not dependent on t. (Why?) -

    - -

    - We compute \kappa using the second part of . - - \kappa \amp = \frac{\abs{(-r\sin(t) )(-r\sin(t) ) - (-r\cos(t) )(r\cos(t) )}}{\big( (-r\sin(t) )^2+(r\cos(t) )^2\big)^{3/2}} - \amp = \frac{r^2(\sin^2(t) +\cos^2(t) )}{\big(r^2(\sin^2(t) +\cos^2(t) )\big)^{3/2}} - \amp = \frac{r^2}{r^3} = \frac1r - . -

    - -

    - We have found that a circle with radius r has curvature \kappa = 1/r. -

    -     - -     - -

    - gives a great result. - Before this example, if we were told - The curve has a curvature of 5 at point A, - we would have no idea what this really meant. - Is 5 big does is correspond to a really sharp turn, - or a not-so-sharp turn? - Now we can think of 5 in terms of a circle with radius 1/5. - Knowing the units - (inches vs.miles, for instance) - allows us to determine how sharply the curve is curving. -

    - -

    - Let a point P on a smooth curve C be given, - and let \kappa be the curvature of the curve at P. - A circle that: -

    - -

    -

      -
    • -

      - passes through P, -

      -
      • - -
    • -

      - lies on the concave side of C, -

      -
      • - -
    • -

      - has a common tangent line as C at P and -

      -
      • - -
    • -

      - has radius r=1/\kappa (hence has curvature \kappa) -

      -
      • -
    -

    - -

    - is the osculating circle, - or circle of curvature, - to C at P, and r is the - radius of curvature. - curvatureof circle - osculating circle - circle of curvature - radius of curvature - curvatureradius of - shows the graph of the curve seen earlier in - and its osculating circles at A and B. - A sharp turn corresponds to a circle with a small radius; - a gradual turn corresponds to a circle with a large radius. - Being able to think of curvature in terms of the radius of a circle is very useful. -

    - -
    - Illustrating the osculating circles for the curve seen in - - - A curve with two marked points, corresponding to points of large and small curvature, along with osculating circles at those points. - -

    - The curve shown is the same L-shaped curve as in , - with the same two marked points, labeled A and B. - The osculating circle at A is relatively small, - indicating a large curvature at that point. - At the point B, where the curvature is small, - the osculating circle is large. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=2.9, - xmin=-.1,xmax=3.5 - ] - - \addplot+ [domain=-1:1.5,samples=40] ({x^2-x+1},{x^3+1}); - \addplot [secondcurvestyle,solid,domain=0:360] ({.08*cos(x)+.83},{.08*sin(x)+1.09}); - \addplot [secondcurvestyle,solid,domain=0:360,samples=101] ({2.17*cos(x)+.99},{2.17*sin(x)+-1.15}); - - \draw [->,firstcolor,thick,>=stealth] (axis cs:2.44, 0.488) -- (axis cs: 2.4141, 0.506961); - - \filldraw (axis cs:.77,1.04) circle (2.4pt) node [below] { $A$} - (axis cs: 1.75, 0.875) circle (2.4pt) node [below] { $B$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - - - -

    - (The word osculating comes from a Latin word related to kissing; - an osculating circle kisses - the graph at a particular point. - Many beautiful ideas in mathematics have come from studying the osculating circles to a curve.) -

    - - - Finding curvature - -

    - Find the curvature of the parabola defined by y=x^2 at the vertex and at x=1. -

    -     - -

    - We use the first formula found in . - - \kappa(x) \amp = \frac{\abs{2}}{\big(1+(2x)^2\big)^{3/2}} - \amp = \frac2{\big(1+4x^2\big)^{3/2}} - . -

    - -
    - Examining the curvature of y=x^2 - - - A parabola along with osculating circles at two points. - -

    - The parabola y=x^2 is shown, along with two osculating circles. - At the origin, where the curvature is largest, - the osculating circle is small, and sits completely within the parabola. - At the point (1,1) the curvature is much less, - leading to a large osculating circle that intercepts the parabola twice. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-2.5,ymax=10.1, - xmin=-10.2,xmax=4.9 - ] - - \addplot+ [domain=-3.2:3.2,samples=40] ({x},{x^2}); - \addplot [secondcurvestyle,solid,domain=0:360] ({.5*cos(x)},{.5*sin(x)+.5}); - \addplot [secondcurvestyle,solid,domain=0:360,samples=70] ({5.6*cos(x)-4},{5.6*sin(x)+3.5}); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - At the vertex (x=0), the curvature is \kappa = 2. - At x=1, - the curvature is \kappa = 2/(5)^{3/2} \approx 0.179. - So at x=0, - the curvature of y=x^2 is that of a circle of radius 1/2; - at x=1, - the curvature is that of a circle with radius \approx 1/0.179 \approx 5.59. - This is illustrated in . - At x=3, the curvature is 0.009; - the graph is nearly straight as the curvature is very close to 0. -

    -     -     - - - Finding curvature - -

    - Find where the curvature of \vrt = \la t, t^2, 2t^3\ra is maximized. -

    -     - -

    - We use the third formula in - as \vrt is defined in space. - We leave it to the reader to verify that - - \vrp(t) =\la 1,2t,6t^2\ra, \vrp'(t) = \la 0,2,12t\ra, \text{ and } \vrp(t)\times \vrpp(t) = \la 12t^2,-12t,2\ra - . -

    - -

    - Thus - - \kappa(t) \amp = \frac{\norm{\vrp(t)\times\vrp'(t)}}{\norm{\vrp(t)}^3} - \amp = \frac{\norm{\la 12t^2,-12t,2\ra}}{\norm{\la 1,2t,6t^2\ra}^3} - \amp = \frac{\sqrt{144t^4+144t^2+4}}{\left(\sqrt{1+4t^2+36t^4\ }\right)^3} - -

    - -

    - While this is not a particularly nice formula, - it does explicitly tell us what the curvature is at a given t value. - To maximize \kappa(t), - we should solve \kappa'(t)=0 for t. - This is doable, but very time consuming. - Instead, consider the graph of - \kappa(t) as given in . - We see that \kappa is maximized at two t values; - using a numerical solver, we find these values are t\approx\pm 0.189. - In we graph \vrt and indicate the points where curvature is maximized. -

    - -
    - Understanding the curvature of a curve in space - -
    - The curvature of \vec{r}(t) - - - A plot of the curvature as a function of the parameter t. - -

    - The curve illustrates the value of the curvature \kappa as a function of t. - The graph is symmetric about the y axis, and has the shape of a tall cowboy hat, - with local maxima on either side of the y axis, and a local minimum in between. - The t axis is a horizontal asymptote, - showing that the curve becomes almost straight when t is large in absolute value. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-.1,ymax=2.5, - xmin=-2.1,xmax=2.1, - xlabel = {$t$} - ] - - \addplot+ [smooth] coordinates {(-1.5,0.0122)(-1.45,0.01404)(-1.4,0.01623)(-1.35,0.01887)(-1.3,0.02207)(-1.25,0.02596)(-1.2,0.03074)(-1.15,0.03666)(-1.1,0.04403)(-1.05,0.05331)(-1.,0.06509)(-0.95,0.08019)(-0.9,0.09974)(-0.85,0.1253)(-0.8,0.159)(-0.75,0.204)(-0.7,0.2643)(-0.65,0.3457)(-0.6,0.4557)(-0.55,0.6035)(-0.5,0.7989)(-0.45,1.049)(-0.4,1.352)(-0.35,1.686)(-0.3,2.006)(-0.25,2.246)(-0.2,2.353)(-0.15,2.318)(-0.1,2.191)(-0.05,2.057)(0,2.)(0.05,2.057)(0.1,2.191)(0.15,2.318)(0.2,2.353)(0.25,2.246)(0.3,2.006)(0.35,1.686)(0.4,1.352)(0.45,1.049)(0.5,0.7989)(0.55,0.6035)(0.6,0.4557)(0.65,0.3457)(0.7,0.2643)(0.75,0.204)(0.8,0.159)(0.85,0.1253)(0.9,0.09974)(0.95,0.08019)(1.,0.06509)(1.05,0.05331)(1.1,0.04403)(1.15,0.03666)(1.2,0.03074)(1.25,0.02596)(1.3,0.02207)(1.35,0.01887)(1.4,0.01623)(1.45,0.01404)(1.5,0.0122)}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - A plot of the curve \vec{r}(t)=\la t, t^2, 2t^3\ra - - - - A plot of the vector-valued function in this example, with points of maximum curvature marked. - -

    - A three-dimensional plot of the curve generated by the vector-valued function \vec{r}(t) = \la t, t^2, 3t^3\ra. - In the default viewing angle, the curve sweeps in from the foreground, bending toward the origin. - At the origin, the curve bends sharply and appears to tend upward toward the z axis. -

    - -

    - Rotating the viewpoint reveals that the curve begins below the xy plane, - bends in toward the origin, and then heads upward, above the xy plane. - It actually moves outward, away from the z axis. -

    -     - - - - - //ASY file for figcurvature43D.asy in Chapter 11 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-2,2); - pair ybounds=(-1,2); - pair zbounds=(-2,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the curve <t, t^2, 2*t^3> for t from -1 to 1 - triple g(real t) {return (t, t^2, 2*t^3);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,bluepen); - //path3 mypath=graph(g,-1,-0.9,operator ..); draw(mypath,bluepen,Arrow3(size=4mm)); - path3 mypath=graph(g,-1,-0.7,operator ..); draw(mypath,bluepen,Arrow3(size=4mm)); - //path3 mypath=graph(g,-1,0.7,operator ..); draw(mypath,bluepen,Arrow3(size=4mm)); - //path3 mypath=graph(g,-1,0.9,operator ..); draw(mypath,bluepen,Arrow3(size=4mm)); - - //Draw points of maximum curvature (0.189,(0.189)^2,2*(0.189)^3) - // cooresponding to values t=+/-0.189 - dotfactor=4; - dot(((0.189,(0.189)^2,2*(0.189)^3)),redpen); - dot(((-0.189,(-0.189)^2,2*(-0.189)^3)),redpen); - - - - -
    -     - -
    - -     -     -     - - - Curvature and Motion -

    - Let \vrt be a position function of an object, - with velocity \vvt = \vrp(t) and acceleration \vec a(t)=\vrpp(t). - In - we established that acceleration is in the plane formed by - \unittangent(t) and \unitnormal(t), - and that we can find scalars a_\text{T} and - a_\text{N} such that - curvatureand motion - unit tangent vectorand curvature - unit normal vectorand curvature - at@a_\text{T} - an@a_\text{N} - - \vat = a_\text{T} \unittangent(t) + a_\text{N} \unitnormal(t) - . -

    - -

    - - gives formulas for a_\text{T} and a_\text{N}: - - a_\text{T} = \frac{d}{dt}\Big(\norm{\vvt}\Big) \text{ and } a_\text{N} = \frac{\norm{\vvt\times \vat}}{\norm{\vvt}} - . -

    - -

    - We understood that the amount of acceleration in the direction of \unittangent relates only to how the speed of the object is changing, - and that the amount of acceleration in the direction of \unitnormal relates to how the direction of travel of the object is changing. - (That is, if the object travels at constant speed, a_\text{T} =0; - if the object travels in a constant direction, a_\text{N} =0.) -

    - -

    - In Equation at the beginning of this section, we found - s\primeskip'(t) = \norm{\vvt}. - We can combine this fact with the above formula for a_\text{T} to write - - a_\text{T} = \frac{d}{dt}\Big(\norm{\vvt}\Big) = \frac{d}{dt}\big( s\primeskip'(t)\big) = s\primeskip''(t) - . -

    - -

    - Since s\primeskip'(t) is speed, - s\primeskip''(t) is the rate at which speed is changing with respect to time. - We see once more that the component of acceleration in the direction of travel relates only to speed, - not to a change in direction. -

    - -

    - Now compare the formula for - a_\text{N} above to the formula for curvature in : - - a_\text{N} = \frac{\norm{\vvt\times \vat}}{\norm{\vvt}} \text{ and } \kappa = \frac{\norm{\vrp(t)\times\vrpp(t)}}{\norm{\vrp(t)}^3}=\frac{\norm{\vvt\times \vat}}{\norm{\vvt}^3} - . -

    - -

    - Thus - - a_\text{N} \amp = \kappa \norm{\vvt}^2 - \amp = \kappa\Big(s\primeskip'(t)\Big)^2 - -

    - -

    - This last equation shows that the component of acceleration that changes the object's direction is dependent on two things: - the curvature of the path and the speed of the object. -

    - -

    - Imagine driving a car in a clockwise circle. - You will naturally feel a force pushing you towards the door - (more accurately, - the door is pushing you as the car is turning and you want to travel in a straight line). - If you keep the radius of the circle constant but speed up (, increasing s\primeskip'(t)), - the door pushes harder against you - (a_\text{N} has increased). - If you keep your speed constant but tighten the turn (, increase \kappa), - once again the door will push harder against you. -

    - -

    - Putting our new formulas for - a_\text{T} and a_\text{N} together, we have - - \vat = s\primeskip''(t)\unittangent(t) + \kappa\norm{\vvt}^2\unitnormal(t) - . -

    - -

    - This is not a particularly practical way of finding - a_\text{T} and a_\text{N}, - but it reveals some great concepts about how acceleration interacts with speed and the shape of a curve. -

    - - - Curvature and road design - -

    - The minimum radius of the curve in a highway cloverleaf is determined by the operating speed, - as given in the table in . - For each curve and speed, compute a_\text{N}. -

    -     - - - Operating speed and minimum radius in highway cloverleaf design - - - - OperatingSpeed (mph) - MinimumRadius (ft) - - - 35 - 310 - - - 40 - 430 - - - 45 - 540 - - - -
    - -

    - Using Equation, - we can compute the acceleration normal to the curve in each case. - We start by converting each speed from - miles per hour to feet per second - by multiplying by 5280/3600. - - \text{35 mph, 310 ft} \amp \Rightarrow 51.33\,\text{ft/s},\, \kappa = 1/310 - a_\text{N} \amp = \kappa\, \norm{\vvt}^2 - \amp = \frac1{310}\big(51.33\big)^2 - \amp = 8.50\,\text{ft/s}^2 - . - - \text{40 mph, 430 ft} \amp \Rightarrow 58.67\,\text{ft/s},\, \kappa = 1/430 - a_\text{N} \amp = \frac1{430}\big(58.67\big)^2 - \amp = 8.00\,\text{ft/s}^2 - . - - \text{45 mph, 540 ft} \amp \Rightarrow 66\,\text{ft/s},\, \kappa = 1/540 - a_\text{N} \amp = \frac1{540}\big(66\big)^2 - \amp = 8.07\,\text{ft/s}^2 - . -

    - -

    - Note that each acceleration is similar; this is by design. - Considering the classic Force = mass acceleration formula, - this acceleration must be kept small in order for the tires of a vehicle to keep a - grip on the road. - If one travels on a turn of radius 310 - at a rate of 50, - the acceleration is double, at 17.35. - If the acceleration is too high, - the frictional force created by the tires may not be enough to keep the car from sliding. - Civil engineers routinely compute a - safe design speed, - then subtract 5-10 mph to create the posted speed limit for additional safety. -

    -     -     - -

    - We end this chapter with a reflection on what we've covered. - We started with vector-valued functions, - which may have seemed at the time to be just another way of writing parametric equations. - However, we have seen that the vector perspective has given us great insight into the behavior of functions and the study of motion. - Vector-valued position functions convey displacement, - distance traveled, speed, velocity, - acceleration and curvature information, - each of which has great importance in science and engineering. -

    -     - - - - Terms and Concepts - - - -

    - It is common to describe position in terms of both - and/or . -

    -     - - - - - ["time","distance"].includes(ans) - - - - - - - ["time","distance"].includes(ans) && !ans_array.slice(0,1).includes(ans) - - - - - ans_array.slice(0,1).includes(ans) - - You already gave that answer. - - - - -     - - - - -

    - A measure of the curviness - of a curve is . -

    -     - - - - - - - - -     - - - - -

    - Give two shapes with constant curvature. -

    -     - - - -

    - Answers may include lines, circles, helixes -

    -     - -     - - - - -

    - Describe in your own words what an - osculating circle is. -

    -     - - - -

    - Answer should mention the circle is tangent to the curve and has the same curvature as the curve at that point. -

    -     - -     - - - -

    - Rearrange the blocks to form a valid identity. -

    -     - - \unittangentprime(s) - = - \kappa - \unitnormal(s) - \unittangentprime(t) - \gamma - \tau - \pi - \unitnormal(t) - -     - - - - - - -

    - Given a position function \vrt, - how are a_\text{T} and - a_\text{N} affected by the curvature? -

    -     - - - -

    - a_\text{T} is not affected by curvature; - the greater the curvature, the larger a_\text{N} becomes. -

    -     - -     -     - - Problems - - - -

    - A position function \vrt is given, - where t=0 corresponds to the initial position. - Find the arc length parameter s, - and rewrite \vrt in terms of s; - that is, find \vec r(s). -

    -     - - - - -

    - \vrt = \la 2t, t, -2t\ra -

    -     - -

    - s = 3t, so \vec r(s) = \la 2s/3, s/3, -2s/3\ra -

    -     - -     - - - - - Context("Vector2D"); - Context()->variables->are(s=>'Real',t=>'Real'); - $s=Formula("7t"); - $rs=Compute("<7cos(s/7),7sin(s/7)>"); - - - -

    - \vrt = \la 7\cos(t) ,7\sin(t) \ra. -

    - - Find the arc length paramter s. - -

    - -

    - - - Rewrite \vrt in terms of s. - -

    - -

    -     -     -     - - - - -

    - \vrt = \la 3\cos(t) ,3\sin(t) , 2t\ra -

    -     - -

    - s = \sqrt{13}t, - so \vec r(s) = \la 3\cos(s/\sqrt{13}), 3\sin(s/\sqrt{13}), 2s/\sqrt{13}\ra -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(s=>'Real',t=>'Real'); - $s=Formula("13t"); - $rs=Compute("<5cos(s/13),13sin(s/13),12cos(s/13)>"); - - - -

    - \vrt = \vrt = \la 5\cos(t) ,13\sin(t) , 12\cos(t) \ra. -

    - - Find the arc length paramter s. - -

    - -

    - - - Rewrite \vrt in terms of s. - - -

    - -

    -     -     -     - -     - - - -

    - A curve C is described along with 2 points on C. -

    - -

    -

      -
    1. -

      - Using a sketch, - determine at which of these points the curvature is greater. -

      -
      2. - -
    2. -

      - Find the curvature \kappa of C, - and evaluate \kappa at each of the 2 given points. -

      -
      3. -
    -

    -     - - - - -

    - C is defined by y = x^3-x; - points given at x=0 and x=1/2. -

    -     - -

    - \kappa = \frac{\abs{6x}}{\left(1+(3x^2-1)^2\right)^{3/2}}; -

    - -

    - \kappa(0) = 0, - \kappa(1/2) = \frac{192}{17\sqrt{17}} \approx 2.74. -

    -     - -     - - - - - parserPopUp.pl - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $comp=DropDown(['greater than', 'equal to', 'less than'],0,showInStatic=>0); - $kappa=Compute("|(6x^2-2)/(x^2+1)^3|/(1+4x^2/(x^2+1)^4)^(3/2)"); - $k[0]=Compute("2"); - $k[1]=Formula("2750/641^(3/2)"); - - - -

    - C is defined by y=\frac{1}{x^2+1}; points given at - x=0 and x=2. -

    - -

    - The curvature at x=0 is - the curvature at x=2. -

    - - - Give the curvature as a function of x. - - -

    - -

    - - - Give the value of \kappa(0). - -

    - \kappa(0)= -

    - - - Give the value of \kappa(2). - -

    - -

    -     -     -     - - - - -

    - C is defined by \ds y = \cos(x); - points given at x=0 and x=\pi/2. -

    -     - -

    - \kappa = \frac{\abs{\cos(x) }}{\left(1+\sin^2(x) \right)^{3/2}}; -

    - -

    - \kappa(0) = 1, \kappa(\pi/2) = 0 -

    -     - -     - - - - -

    - C is defined by \ds y = \sqrt{1-x^2} on (-1,1); - points given at x=0 and x=1/2. -

    -     - -

    - \kappa = 1; -

    - -

    - \kappa(0) = 1, \kappa(1/2) = 1 -

    -     - -     - - - - - parserPopUp.pl - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->are(t=>'Real'); - $comp=DropDown(['greater than', 'equal to', 'less than'],2,showInStatic=>0); - $kappa=Compute("|2cos(t)cos(2t)+4sin(t)sin(2t)|/(4cos^2(2t)+sin^2(t))^(3/2)"); - $k[0]=Formula("1/4"); - $k[1]=Formula("8"); - - - -

    - C is defined by \vrt = \la \cos(t) , \sin(2t)\ra; - points given at t=0 and t=\pi/4. -

    - -

    - The curvature at t=0 is - the curvature at t=\pi/4. -

    - - - Give the curvature as a function of t. - - -

    - -

    - - - Give the value of \kappa(0). - -

    - \kappa(0)= -

    - - - Give the value of \kappa(pi/4). - -

    - -

    -     -     -     - - - - -

    - C is defined by \ds \vrt = \la \cos^2(t) , \sin(t) \cos(t) \ra; - points given at t=0 and t=\pi/3. -

    -     - -

    - \kappa = 2; -

    - -

    - \kappa(0) = 2, \kappa(\pi/3) = 2 -

    -     - -     - - - - -

    - C is defined by \ds \vrt = \la t^2-1,t^3-t\ra; - points given at t=0 and t=5. -

    -     - -

    - \kappa = \frac{\abs{6t^2+2}}{\left(4t^2+(3t^2-1)^2\right)^{3/2}}; -

    - -

    - \kappa(0) = 2, - \kappa(5) = \frac{19}{1394\sqrt{1394}}\approx 0.0004 -

    -     - -     - - - - - parserPopUp.pl - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->are(t=>'Real'); - $comp=DropDown(['greater than', 'equal to', 'less than'],0,showInStatic=>0); - $kappa=Compute("|sec^3(t)|/(sec^4(t)+sec^2(t)tan^2(t))^(3/2)"); - $k[0]=Formula("1"); - $k[1]=Formula("3sqrt(3)/(5sqrt(5))"); - - - -

    - C is defined by \vrt = \la \tan(t) ,\sec(t) \ra; - points given at t=0 and t=\pi/6. -

    - -

    - The curvature at t=0 is - the curvature at t=\pi/6. -

    - - - Give the curvature as a function of t. - - -

    - -

    - - - Give the value of \kappa(0). - -

    - \kappa(0)= -

    - - - Give the value of \kappa(\pi/6). - -

    - -

    -     -     -     - - - - -

    - C is defined by \ds \vrt = \la 4t+2,3t-1,2t+5\ra; - points given at t=0 and t=1. -

    -     - -

    - \kappa = 0; -

    - -

    - \kappa(0) = 0, \kappa(1) = 0 -

    -     - -     - - - - - parserPopUp.pl - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->are(t=>'Real'); - $comp=DropDown(['greater than', 'equal to', 'less than'],0,showInStatic=>0); - $kappa=Compute("2sqrt(18t^4+15t^2+1)/(18t^4-2t^2+1)^(3/2)"); - $k[0]=Formula("2"); - $k[1]=Formula("2sqrt(2)/17"); - - - -

    - C is defined by \vrt = \la t^3-t,t^3-4,t^2-1\ra; - points given at t=0 and t=1. -

    - -

    - The curvature at t=0 is - the curvature at t=1. -

    - - - Give the curvature as a function of t. - - -

    - -

    - - - Give the value of \kappa(0). - -

    - \kappa(0)= -

    - - - Give the value of \kappa(1). - -

    - -

    -     -     -     - - - - -

    - C is defined by \ds \vrt = \la 3\cos(t) ,3\sin(t) , 2t\ra; - points given at t=0 and t=\pi/2. -

    -     - -

    - \kappa = \frac{3}{13}; -

    - -

    - \kappa(0) = 3/13, \kappa(\pi/2) = 3/13 -

    -     - -     - - - - - parserPopUp.pl - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - Context()->variables->are(t=>'Real'); - $comp=DropDown(['greater than', 'equal to', 'less than'],1,showInStatic=>0); - $kappa=Formula("1/13"); - $k[0]=Formula("1/13"); - $k[1]=Formula("1/13"); - - - -

    - C is defined by \vrt = \la 5\cos(t) ,13\sin(t) , 12\cos(t) \ra. - Points given at t=0 and t=\pi/2. -

    - -

    - The curvature at t=0 is - the curvature at t=\pi/2. -

    - - - Give the curvature as a function of t. - - -

    - -

    - - - Give the value of \kappa(0). - -

    - \kappa(0)= -

    - - - Give the value of \kappa(\pi/2). - -

    - -

    -     -     - -     - - - -

    - Find the value of x or t where curvature is maximized. -

    -     - - - - - - parser::Root->Enable; - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $max=List(Formula("sqrt(2)/root(4,5)"), Formula("-sqrt(2)/root(4,5)")); - - - -

    - y=\frac{1}{6}x^3 -

    - -

    - -

    -     -     -     - - - - -

    - \ds y=\sin(x) -

    -     - -

    - maximized at x=\ldots -3\pi/2, -\pi/2, \pi/2, \ldots -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $max=Formula("1/4"); - - - -

    - \vrt = \la t^2+2t,3t-t^2\ra -

    - -

    - -

    -     -     -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $max=List(Formula("sqrt(5)"),Formula("-sqrt(5)")); - - - -

    - \vrt = \la t, 4/t, 3/t\ra -

    - -

    - -

    -     -     -     - -     - - - -

    - Find the radius of curvature at the indicated value. -

    -     - - - - -

    - y=\tan(x), at x=\pi/4 -

    -     - -

    - radius of curvature is 5\sqrt{5}/4. -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $radius=Formula("5sqrt(10)"); - - - -

    - y=x^2+x-3 at x=1 -

    - -

    - -

    -     -     -     - - - - -

    - \vrt = \la \cos(t) , \sin(3t)\ra, at t=0 -

    -     - -

    - radius of curvature is 9. -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $radius=Formula("1/45"); - - - -

    - \vrt = \la 5\cos(3 t), t\ra at t=0 -

    - -

    - -

    -     -     -     - -     - - - -

    - Find the equation of the osculating circle to the curve at the indicated t-value. -

    -     - - - - -

    - \vrt = \la t,t^2\ra, at t=0 -

    -     - -

    - x^2+(y-1/2)^2 = 1/4, - or \vec c(t) = \la 1/2\cos(t) , 1/2\sin(t) + 1/2\ra -

    -     - -     - - - - - Context("ImplicitEquation"); - Context()->variables->set(x=>{limits=>[2,4]}); - Context()->variables->set(y=>{limits=>[-1,1]}); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $osc=ImplicitEquation("(x-8/3)^2+y^2 = 1/9"); - - - -

    - \vrt = \la 3\cos(t) , \sin(t) \ra at t=0 -

    - -

    - -

    -     -     -     - - - - -

    - \vrt = \la 3\cos(t) , \sin(t) \ra, at t=\pi/2 -

    -     - -

    - x^2+(y+8)^2 = 81, - or \vec c(t) = \la 9\cos(t) , 9\sin(t) -8\ra -

    -     - -     - - - - - Context("ImplicitEquation"); - Context()->variables->set(x=>{limits=>[-1,2]}); - Context()->variables->set(y=>{limits=>[-1,2]}); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $osc=ImplicitEquation("(x-1/2)^2+(y-1/2)^2 = 1/2"); - - - -

    - \vrt = \la t^2-t,t^2+t\ra at t=0 -

    - -

    - -

    -     -     -     - -     -     -     -
    -     - - - Functions of Several Variables - -

    - A function of the form y=f(x) is a function of a single variable; - given a value of x, we can find a value y. - Even the vector-valued functions of - are single-variable functions; - the input is a single variable though the output is a vector. -

    - -

    - There are many situations where a desired quantity is a function of two or more variables. - For instance, - wind chill is measured by knowing the temperature and wind speed; - the volume of a gas can be computed knowing the pressure and temperature of the gas; - to compute a baseball player's batting average, - one needs to know the number of hits and the number of at-bats. -

    - -

    - This chapter studies multivariable functions, that is, - functions with more than one input. -

    -     - -
    - Introduction to Multivariable Functions - - - Function of Two Variables - -

    - Let D be a subset of \mathbb{R}^2. - A function f of two variables - is a rule that assigns each pair (x,y) in D a value - z=f(x,y) in \mathbb{R}. - D is the domain of f; - the set of all outputs of f is the range. - multivariable function - multivariable functiondomain - multivariable functionrange - functionof two variables -

    -     -     - - - - - - - Understanding a function of two variables - -

    - Let z=f(x,y) = x^2-y. - Evaluate f(1,2), f(2,1), and f(-2,4); - find the domain and range of f. -

    -     - -

    - Using the definition f(x,y) = x^2-y, we have: - - f(1,2) \amp = 1^2-2 = -1 - f(2,1) \amp = 2^2-1 = 3 - f(-2,4) \amp = (-2)^2-4 = 0 - -

    - -

    - The domain is not specified, - so we take it to be all possible pairs in - \mathbb{R}^2 for which f is defined. - In this example, f is defined for - all pairs (x,y), - so the domain D of f is \mathbb{R}^2. -

    - -

    - The output of f can be made as large or small as possible; - any real number r can be the output. (In fact, - given any real number r, - f(0,-r)=r.) So the range R of f is \mathbb{R}. -

    -     -     - - - Understanding a function of two variables - -

    - Let f(x,y) = \sqrt{1-\frac{x^2}9-\frac{y^2}4}. - Find the domain and range of f. -

    -     - -

    - The domain is all pairs (x,y) allowable as input in f. - Because of the square root, - we need (x,y) such that 0\leq1-\frac{x^2}9-\frac{y^2}4: - - 0\amp \leq 1-\frac{x^2}9-\frac{y^2}4 - \frac{x^2}9+\frac{y^2}4 \amp \leq 1 - -

    - -

    - The above equation describes an ellipse and its interior as shown in . - We can represent the domain D graphically with the figure; - in set notation, - we can write D = \{(x,y)|\,\frac{x^2}9+\frac{y^2}4 \leq 1\}. -

    - -
    - Illustrating the domain of f(x,y) in - - - An ellipse, centered at the origin, with shaded interior, illustrating the domain of a function. - -

    - The ellipse \frac{x^2}{9}+\frac{y^2}{4}=1 is plotted in the xy plane. - It is centered at the origin, with intercepts at (\pm 3,0) and (0,\pm 2). - The interior of the ellipse is shaded, to illustrate the domain of - f(x,y) = \sqrt{1-\frac{x^2}{9}-\frac{y^2}{4}}. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-5.9,ymax=5.9, - xmin=-5.9,xmax=5.9 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:360] ({3*cos(x)},{2*sin(x)}); - \addplot [firstcurvestyle,domain=0:360,samples=60] ({3*cos(x)},{2*sin(x)}); - - \draw (axis cs: 3,3) node { $\displaystyle \frac{x^2}9+\frac{y^2}4=1$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - The range is the set of all possible output values. - The square root ensures that all output is \geq 0. - Since the x and y terms are squared, then subtracted, - inside the square root, - the largest output value comes at x=0, - y=0: f(0,0) = 1. - Thus the range R is the interval [0,1]. -

    -     -     -     - - - Graphing Functions of Two Variables -

    - The graph of a function f of two variables is the set of all points - \big(x,y,f(x,y)\big) where (x,y) is in the domain of f. - This creates a surface in space. -

    - -
    - Graphing a function of two variables - -
    - - - - - A collection of points plotted in space, against a set of three-dimensional coordinate axes. - -

    - About two dozen points are plotted in space, - along with a set of three-dimensional coordinate axes. - The location of the points can be better observed by rotating the figure, - but the precise arrangement of the points is unimportant. - The main observation is that these points all lie on the surface plotted in the next image. -

    -     - - - - - //ASY file for figmultigraph_intro3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8,8,1.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={0.5,1}; - defaultpen(0.5mm); - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.25,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw 25 points on the surface z=1/(1+x^2+y^2) - dotfactor=3; - real[] A={-2,-1,0,1,2}; - for (int i=0; i<5; ++i) - { - for (int j=0; j<5; ++j) - { - dot((A[i],A[j],1/(1+(A[i])^2+(A[j])^2)),dotblue); - } - } - - - - -
    - -
    - - - - - A bell-shaped surface in three dimensions. It is symmetric about the z axis and has a peak at the point (0,0,1) - -

    - A three-dimensional plot the surface given by a graph z=f(x,y). - The surface is a bell-shaped hill, with its peak at (0,0,1) on the z axis. - There is rotational symmetry about the z axis, - and as the suface descends, it flattens out and tapers off toward the xy plane. -

    -     - - - - - //ASY file for figmultigraph_introb3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8,8,1.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={0.5,1}; - defaultpen(0.5mm); - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.25,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z=1/(1+x^2+y^2) - triple f(pair t) { - return (t.x,t.y,1/(1+(t.x)^2+(t.y)^2)); - } - surface s=surface(f,(-2,-2),(2,2),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -
    -     - -
    - -

    - One can begin sketching a graph by plotting points, but this has limitations. - Consider where 25 points have been plotted of f(x,y) = \frac1{x^2+y^2+1}. - More points have been plotted than one would reasonably want to do by hand, - yet it is not clear at all what the graph of the function looks like. - Technology allows us to plot lots of points, - connect adjacent points with lines and add shading to create a graph like which does a far better job of illustrating the behavior of f. -

    - -

    - While technology is readily available to help us graph functions of two variables, - there is still a paper-and-pencil approach that is useful to understand and master as it, - combined with high-quality graphics, - gives one great insight into the behavior of a function. - This technique is known as sketching level curves. -

    -     - - - Level Curves -

    - It may be surprising to find that the problem of representing a three dimensional surface on paper is familiar to most people - (they just don't realize it). - multivariable functionlevel curves - level curves - contour lines - Topographical maps, - like the one shown in , - represent the surface of Earth by indicating points with the same elevation with contour lines. - The elevations marked are equally spaced; - in this example, - each thin line indicates an elevation change in 50ft increments and each thick line indicates a change of 200ft. - When lines are drawn close together, elevation changes rapidly - (as one does not have to travel far to rise 50ft). - When lines are far apart, - such as near Aspen Campground, - elevation changes more gradually as one has to walk farther to rise 50ft. -

    - - - -
    - A topographical map displays elevation by drawing contour lines, along with the elevation is constant. - USGS 1:24000-scale Quadrangle for Chrome Mountain, MT 1987. - - - A topographical map of Chrome Mountain in Montana. - -

    - An excerpt from a topographical map of Chrome Mountain in Montana. - The map illustrates the concept of contour lines through its use of lines of elevation. - There is a river that runs through the middle of the map, from top to bottom. - To the right of the river, the lines of elevation are more spread out, - indicating that the land is more gently-sloped in this region, - which includes an area marked as Aspen Campground. -

    - -

    - Near the left and right edges of the map, the lines of elevation are much closer together, - which indicates that these regions consist of steeply-sloped mountainsides. -

    -     - - -
    - -

    - Given a function f(x,y), - we can draw a topographical map - of the graph z=f(x,y) by drawing level curves - (or, contour lines). - A level curve at z=c is a curve in the xy-plane such that for all points (x,y) on the curve, - f(x,y) = c. -

    - -

    - When drawing level curves, - it is important that the c values are spaced equally apart as that gives the best insight to how quickly the - elevation is changing. - Examples will help one understand this concept. -

    - - - Drawing Level Curves - -

    - Let f(x,y) = \sqrt{1-\frac{x^2}9-\frac{y^2}4}. - Find the level curves of f for c=0, - 0.2, 0.4, 0.6, 0.8 and 1. -

    -     - -

    - Consider first c=0. - The level curve for c=0 is the set of all points (x,y) such that 0=\sqrt{1-\frac{x^2}9-\frac{y^2}4}. - Squaring both sides gives us - - \frac{x^2}9+\frac{y^2}4=1 - , - an ellipse centered at (0,0) with horizontal major axis of length 6 and minor axis of length 4. - Thus for any point (x,y) on this curve, f(x,y) = 0. -

    - -

    - Now consider the level curve for c=0.2 - - 0.2 \amp = \sqrt{1-\frac{x^2}9-\frac{y^2}4} - 0.04 \amp = 1-\frac{x^2}9-\frac{y^2}4 - \frac{x^2}9+\frac{y^2}4 \amp =0.96 - \frac{x^2}{8.64}+\frac{y^2}{3.84} \amp =1 - . -

    - -

    - This is also an ellipse, - where a = \sqrt{8.64}\approx 2.94 and b=\sqrt{3.84}\approx 1.96. -

    - -

    - In general, for z=c, the level curve is: - - c \amp = \sqrt{1-\frac{x^2}9-\frac{y^2}4} - c^2 \amp = 1-\frac{x^2}9-\frac{y^2}4 - \frac{x^2}9+\frac{y^2}4 \amp =1-c^2 - \frac{x^2}{9(1-c^2)}+\frac{y^2}{4(1-c^2)} \amp =1 - , - ellipses that are decreasing in size as c increases. - A special case is when c=1; - there the ellipse is just the point (0,0). -

    - -

    - The level curves are shown in . - Note how the level curves for c=0 and c=0.2 are very, - very close together: - this indicates that f is growing rapidly along those curves. -

    - -
    - Graphing the level curves in - -
    - - - - A family of concentric ellipses, centered at the origin. - -

    - This plot, in the xy plane, illustrates several level curves for the function f(x,y) = \sqrt{1-\frac{x^2}{9}-\frac{y^2}{4}}. - The level curve c=1 is a single point: the origin (0,0). - The other level curves form a family of concentric ellipses centered about the origin. - The largest ellipse is the boundary of the domain of f, given by \frac{x^2}{9}+\frac{y^2}{4}=1. - Level curves near the boundary are close together, while those nearer to the origin are further apart. - This illustrates the fact that the surface is relative flat near the top, - while it becomes steeply-sloped along its sides. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-1,1,2,-2,3,-3}, - ytick={-1,1,-2,2,3,-3}, - ymin=-2.5,ymax=2.5, - xmin=-3.2,xmax=3.2 - ] - - \addplot [firstcurvestyle,domain=0:360,samples=80] ({3*(cos(x))},{2*(sin(x))}); - \addplot [firstcurvestyle,domain=0:360,samples=80] ({2.93*(cos(x))},{1.96*(sin(x))}); - \addplot [firstcurvestyle,domain=0:360,samples=80] ({2.75*(cos(x))},{1.83*(sin(x))}); - \addplot [firstcurvestyle,domain=0:360,samples=70] ({2.4*(cos(x))},{1.6*(sin(x))}); - \addplot [firstcurvestyle,domain=0:360,samples=60] ({1.8*(cos(x))},{1.2*(sin(x))}); - - \filldraw [firstcolor] (axis cs:0,0) circle (1pt) node [black, above right] { $c=1$}; - - \draw [->,>=stealth] (axis cs:1.7,1.13) -- (axis cs:2.6,2) node [above] { $c=0.6$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - A surface in the shape of an elliptical dome. - -

    - The surface given by the graph z=\sqrt{1--\frac{x^2}{9}-\frac{y^2}{4}} is plotted. - Its shape is that of an elliptical dome, with its peak on the z axis at the point (0,0,1). - The bottom of the dome lies in the xy plane, and its intersection with this plane is an ellipse. - Near the top, the surface is relatively flat, - but the sides become close to vertical near the xy plane. -

    -     - - - - - //ASY file for figlevelcurve13D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(9,9,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={1,2}; - defaultpen(0.5mm); - pair xbounds=(-3.5,3.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.25,2.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw top half of surface x^2/9+y^2/4+z^2=1 - triple f(pair t) { - //return (cos(t.x)*3*cos(t.y),sin(t.x)*2*cos(t.y),sin(t.y)); - return (3*sin(t.y),sin(t.x)*2*cos(t.y),cos(t.x)*cos(t.y)); - } - surface s=surface(f,(-pi/2,-pi/2),(pi/2,pi/2),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw level curves for surface x^2/9+y^2/4+z^2=1 at c=0,0.2,0.4,0.6,0.8,1 - triple g(real t) {return (3*cos(t)*sqrt(1-(0.8)^2),2*sin(t)*sqrt(1-(0.8)^2),0.8);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - triple g(real t) {return (3*cos(t)*sqrt(1-(0.6)^2),2*sin(t)*sqrt(1-(0.6)^2),0.6);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - triple g(real t) {return (3*cos(t)*sqrt(1-(0.4)^2),2*sin(t)*sqrt(1-(0.4)^2),0.4);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - triple g(real t) {return (3*cos(t)*sqrt(1-(0.2)^2),2*sin(t)*sqrt(1-(0.2)^2),0.2);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - triple g(real t) {return (3*cos(t)*sqrt(1-(0)^2),2*sin(t)*sqrt(1-(0)^2),0);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - dotfactor=3; - dot((0,0,1),bluepen); - - - - -
    -     - -
    - -

    - In , - the curves are drawn on a graph of f in space. - Note how the elevations are evenly spaced. - Near the level curves of c=0 and c=0.2 we can see that f indeed is growing quickly. -

    -     -     - - - Analyzing Level Curves - -

    - Let f(x,y) = \frac{x+y}{x^2+y^2+1}. - Find the level curves for z=c. -

    -     - -

    - We begin by setting f(x,y)=c for an arbitrary c and seeing if algebraic manipulation of the equation reveals anything significant. - - \frac{x+y}{x^2+y^2+1} \amp = c - x+y \amp = c(x^2+y^2+1). - We recognize this as a circle, though the center and radius are not yet clear. By completing the square, we can obtain: - \left(x-\frac{1}{2c}\right)^2+\left(y-\frac1{2c}\right)^2\amp =\frac{1}{2c^2}-1 - , - a circle centered at \big(1/(2c),1/(2c)\big) with radius \sqrt{1/(2c^2)-1}, - where \abs{c}\lt 1/\sqrt{2}. - The level curves for c=\pm 0.2,\,\pm 0.4 and \pm0.6 are sketched in . - To help illustrate elevation, - we use thicker lines for c values near 0, and dashed lines indicate where c\lt 0. -

    - -

    - There is one special level curve, when c=0. - The level curve in this situation is x+y=0, the line y=-x. -

    - -

    - In we see a graph of the surface. - Note how the y-axis is pointing away from the viewer to more closely resemble the orientation of the level curves in . -

    - -
    - Graphing the level curves in - -
    - - - - A line through the origin in the plane, and two sets of nested circles, one on either side of the line. - -

    - This plot in the xy plane shows several level curves for the function f(x,y) = \frac{x+y}{x^2+y^2+1}. - The level curve c=0 is the line y=-x passing through the origin. - Above this line there is a family of nested, but not concentric, circles. - The smallest circles are closest to the origin. These represent level curves for positive values of c. - Across the line y=-x is another set of circles that is the mirror image of the first. - These are the level curves for negative values of c. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-5.1,ymax=5.1, - xmin=-6.1,xmax=6.1 - ] - - \addplot [thin,firstcolor,smooth] coordinates {(1.457,0.8333)(1.443,0.963)(1.403,1.087)(1.338,1.2)(1.251,1.297)(1.145,1.373)(1.026,1.426)(0.8985,1.454)(0.7681,1.454)(0.6406,1.426)(0.5215,1.373)(0.4161,1.297)(0.3288,1.2)(0.2636,1.087)(0.2234,0.963)(0.2097,0.8333)(0.2234,0.7037)(0.2636,0.5797)(0.3288,0.4668)(0.4161,0.3699)(0.5215,0.2933)(0.6406,0.2402)(0.7681,0.2131)(0.8985,0.2131)(1.026,0.2402)(1.145,0.2933)(1.251,0.3699)(1.338,0.4668)(1.403,0.5797)(1.443,0.7037)(1.457,0.8333)}; - \addplot [thick,firstcolor,smooth] coordinates {(2.708,1.25)(2.676,1.553)(2.582,1.843)(2.429,2.107)(2.225,2.333)(1.979,2.512)(1.7,2.636)(1.402,2.7)(1.098,2.7)(0.7995,2.636)(0.5211,2.512)(0.2746,2.333)(0.07067,2.107)(-0.08171,1.843)(-0.1759,1.553)(-0.2077,1.25)(-0.1759,0.9469)(-0.08171,0.6571)(0.07067,0.3932)(0.2746,0.1667)(0.5211,-0.01244)(0.7995,-0.1364)(1.098,-0.1998)(1.402,-0.1998)(1.7,-0.1364)(1.979,-0.01244)(2.225,0.1667)(2.429,0.3932)(2.582,0.6571)(2.676,0.9469)(2.708,1.25)}; - \addplot [very thick,firstcolor,smooth] coordinates {(5.891,2.5)(5.817,3.205)(5.598,3.879)(5.244,4.493)(4.769,5.02)(4.196,5.437)(3.548,5.725)(2.854,5.873)(2.146,5.873)(1.452,5.725)(0.8044,5.437)(0.2309,5.02)(-0.2435,4.493)(-0.598,3.879)(-0.8171,3.205)(-0.8912,2.5)(-0.8171,1.795)(-0.598,1.121)(-0.2435,0.5067)(0.2309,-0.02013)(0.8044,-0.4368)(1.452,-0.7252)(2.146,-0.8726)(2.854,-0.8726)(3.548,-0.7252)(4.196,-0.4368)(4.769,-0.02013)(5.244,0.5067)(5.598,1.121)(5.817,1.795)(5.891,2.5)}; - \addplot [thin,firstcolor,dashed,smooth] coordinates {(-0.2097,-0.8333)(-0.2234,-0.7037)(-0.2636,-0.5797)(-0.3288,-0.4668)(-0.4161,-0.3699)(-0.5215,-0.2933)(-0.6406,-0.2402)(-0.7681,-0.2131)(-0.8985,-0.2131)(-1.026,-0.2402)(-1.145,-0.2933)(-1.251,-0.3699)(-1.338,-0.4668)(-1.403,-0.5797)(-1.443,-0.7037)(-1.457,-0.8333)(-1.443,-0.963)(-1.403,-1.087)(-1.338,-1.2)(-1.251,-1.297)(-1.145,-1.373)(-1.026,-1.426)(-0.8985,-1.454)(-0.7681,-1.454)(-0.6406,-1.426)(-0.5215,-1.373)(-0.4161,-1.297)(-0.3288,-1.2)(-0.2636,-1.087)(-0.2234,-0.963)(-0.2097,-0.8333)}; - \addplot [thick,firstcolor,dashed,smooth] coordinates {(0.2077,-1.25)(0.1759,-0.9469)(0.08171,-0.6571)(-0.07067,-0.3932)(-0.2746,-0.1667)(-0.5211,0.01244)(-0.7995,0.1364)(-1.098,0.1998)(-1.402,0.1998)(-1.7,0.1364)(-1.979,0.01244)(-2.225,-0.1667)(-2.429,-0.3932)(-2.582,-0.6571)(-2.676,-0.9469)(-2.708,-1.25)(-2.676,-1.553)(-2.582,-1.843)(-2.429,-2.107)(-2.225,-2.333)(-1.979,-2.512)(-1.7,-2.636)(-1.402,-2.7)(-1.098,-2.7)(-0.7995,-2.636)(-0.5211,-2.512)(-0.2746,-2.333)(-0.07067,-2.107)(0.08171,-1.843)(0.1759,-1.553)(0.2077,-1.25)}; - \addplot [very thick,firstcolor,dashed,smooth] coordinates {(0.8912,-2.5)(0.8171,-1.795)(0.598,-1.121)(0.2435,-0.5067)(-0.2309,0.02013)(-0.8044,0.4368)(-1.452,0.7252)(-2.146,0.8726)(-2.854,0.8726)(-3.548,0.7252)(-4.196,0.4368)(-4.769,0.02013)(-5.244,-0.5067)(-5.598,-1.121)(-5.817,-1.795)(-5.891,-2.5)(-5.817,-3.205)(-5.598,-3.879)(-5.244,-4.493)(-4.769,-5.02)(-4.196,-5.437)(-3.548,-5.725)(-2.854,-5.873)(-2.146,-5.873)(-1.452,-5.725)(-0.8044,-5.437)(-0.2309,-5.02)(0.2435,-4.493)(0.598,-3.879)(0.8171,-3.205)(0.8912,-2.5)}; - \addplot [very thick,firstcolor] {-x}; - - \draw (axis cs:5,-3.5) node { $c=0$}; - \draw (axis cs:3.8,4.6) node { $c=0.2$}; - \draw (axis cs:1.5,3) node { $c=0.4$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - The surface given by the graph whose level curves were plotted in the previous image. - -

    - The surface given by the graph z=\frac{x+y}{x^2+y^2+1} is plotted in three dimensions. - For points that are far from the z axis, - the surface lies close to the xy plane and is relatively flat. - Near the origin, the surface rises sharply to a peak that lies above the line y=x in the xy plane. - On the opposite side of the origin, the surface drops into a well that is the mirror image of the peak. -

    -     - - - - - //ASY file for figlevelcurve23D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(16,-12,1.4); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-5,5}; - real[] myychoice={-5,5}; - real[] myzchoice={-0.7,0.7}; - defaultpen(0.5mm); - pair xbounds=(-6,6); - pair ybounds=(-6,6); - pair zbounds=(-0.8,.8); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=(x+y)/(1+x^2+y^2) - triple f(pair t) { - return (t.x,t.y,(t.x+t.y)/(1+(t.x)^2+(t.y)^2)); - } - surface s=surface(f,(-6,-6),(6,6),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw level curves for surface x^2/9+y^2/4+z^2=1 at c=0.2,0.4,0.6 - real[] A={0.2,0.4,0.6}; - for (int i=0; i<3; ++i) - { - triple g(real t) { - return (1/(2*A[i])+(cos(t))*sqrt(-1+(1/(2*A[i]^2))),1/(2*A[i])+(sin(t))*sqrt(-1+(1/(2*A[i]^2))),A[i]);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - //Negative values too - triple g(real t) { - return (1/(-2*A[i])+(cos(t))*sqrt(-1+(1/(2*A[i]^2))),1/(-2*A[i])+(sin(t))*sqrt(-1+(1/(2*A[i]^2))),-A[i]);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - } - - draw((-6,6,0)--(6,-6,0),bluepen); - - //triple g(real t) { - //return (1/(2*0.6)+(cos(t))*sqrt(-1+(1/(2*0.6^2))),1/(2*0.6)+(sin(t))*sqrt(-1+(1/(2*0.6^2))),0.6);} - //path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - - - -
    -     - -
    - -

    - Seeing the level curves helps us understand the graph. - For instance, - the graph does not make it clear that one can walk - along the line y=-x without elevation change, - though the level curve does. -

    -     -     -     - - - Functions of Three Variables -

    - We extend our study of multivariable functions to functions of three variables. - (One can make a function of as many variables as one likes; - we limit our study to three variables.) -

    - - - Function of Three Variables - -

    - Let D be a subset of \mathbb{R}^3. - A function f of three variables - is a rule that assigns each triple (x,y,z) in D a value - w=f(x,y,z) in \mathbb{R}. - D is the domain of f; - the set of all outputs of f is the range. - multivariable function - functionof three variables - multivariable functiondomain - multivariable functionrange -

    -     -     - -

    - Note how this definition closely resembles that of . -

    - - - Understanding a function of three variables - -

    - Let f(x,y,z) = \frac{x^2+z+3\sin(y) }{x+2y-z}. - Evaluate f at the point (3,0,2) and find the domain and range of f. -

    -     - -

    - To evaluate the function we simply set x=3, y=0, and z=3 in the definition of f: - - f(3,0,2) = \frac{3^2+2+3\sin(0) }{3+2(0)-2} = 11 - . -

    - -

    - As the domain of f is not specified, - we take it to be the set of all triples (x,y,z) for which f(x,y,z) is defined. - As we cannot divide by 0, - we find the domain D is - - D = \{(x,y,z)\,|\,x+2y-z\neq 0\} - . -

    - -

    - We recognize that the set of all points in - \mathbb{R}^3 that are not - in D form a plane in space that passes through the origin - (with normal vector \la 1,2,-1\ra). -

    - -

    - We determine the range R is \mathbb{R}; - that is, all real numbers are possible outputs of f. - There is no set way of establishing this. - Rather, to get numbers near 0 we can let y=0 and choose z \approx -x^2. - To get numbers of arbitrarily large magnitude, - we can let z\approx x+2y. -

    -     -     -     - - - Level Surfaces -

    - It is very difficult to produce a meaningful graph of a function of three variables. - A function of one variable is a curve - drawn in 2 dimensions; - a function of two variables is a surface - drawn in 3 dimensions; - a function of three variables is a hypersurface - drawn in 4 dimensions. - multivariable functionlevel surface - level surface -

    - -

    - There are a few techniques one can employ to try to picture - a graph of three variables. - One is an analogue of level curves: - level surfaces. - Given w=f(x,y,z), - the level surface at w=c is the surface in space formed by all points (x,y,z) where f(x,y,z)=c. -

    - - - Finding level surfaces - -

    - If a point source S is radiating energy, - the intensity I at a given point P in space is inversely proportional to the square of the distance between S and P. - That is, when S=(0,0,0), - I(x,y,z) = \frac{k}{x^2+y^2+z^2} for some constant k. -

    - -

    - Let k=1; find the level surfaces of I. -

    -     - -

    - We can (mostly) answer this question using - common sense. If energy (say, - in the form of light) is emanating from the origin, - its intensity will be the same at all points equidistant from the origin. - That is, at any point on the surface of a sphere centered at the origin, - the intensity should be the same. - Therefore, the level surfaces are spheres. -

    - -

    - We now find this mathematically. - The level surface at I=c is defined by - - c \amp = \frac{1}{x^2+y^2+z^2}. - A small amount of algebra reveals - x^2+y^2+z^2 \amp = \frac1c - . -

    - -

    - Given an intensity c, - the level surface I=c is a sphere of radius - 1/\sqrt{c}, centered at the origin. -

    - -
    - A table of c values and the corresponding radius r of the spheres of constant value in - - - c - r - - - - - - - 16 - 0.25 - - - 8 - 0.35 - - - 4 - 0.5 - - - 2 - 0.71 - - - 1 - 1 - - - 0.5 - 1.41 - - - 0.25 - 2 - - - 0.125 - 2.83 - - - 0.0625 - 4 - - -
    - -

    - - gives a table of the radii of the spheres for given c values. - Normally one would use equally spaced c values, - but these values have been chosen purposefully. - At a distance of 0.25 from the point source, the intensity is 16; - to move to a point of half that intensity, - one just moves out 0.1 to 0.35 not much at all. - To again halve the intensity, - one moves 0.15, a little more than before. -

    - -

    - Note how each time the intensity if halved, - the distance required to move away grows. - We conclude that the closer one is to the source, - the more rapidly the intensity changes. -

    - -     -     - -

    - In the next section we apply the concepts of limits to functions of two or more variables. -

    -     - - - - Terms and Concepts - - - - -

    - Give two examples - (other than those given in the text) - of real world functions that require more than one input. -

    -     - - -     - - - - -

    - The graph of a function of two variables is a . -

    -     - - - - - - - - -     - - - - -

    - Most people are familiar with the concept of level curves in the context of maps. -

    -     - - - - - - - -

    - Double-check your spelling before you go hunting for the answer. -

    -     -     -     -     - -     - - - - -

    - Along a level curve, the output of a function does not change. -

    -     - -     - - - - -

    - The analogue of a level curve for functions of three variables is a level . -

    -     - - - - - - - - -     - - - - -

    - What does it mean when level curves are close together? - Far apart? -

    -     - - - -

    - When level curves are close together, - it means the function is changing z-values rapidly. - When far apart, it changes z-values slowly. -

    -     - -     -     - - - Problems - - - -

    - Give the domain and range of the multivariable function. -

    -     - - - - -

    - f(x,y) = x^2+y^2+2 -

    -     - -

    - domain: \mathbb{R}^2 -

    - -

    - range: z\geq 2 -

    -     - -     - - - - -

    - f(x,y) = x+2y -

    -     - -

    - domain: \mathbb{R}^2 -

    - -

    - range: \mathbb{R} -

    -     - -     - - - - -

    - f(x,y) = x-2y -

    -     - -

    - domain: \mathbb{R}^2 -

    - -

    - range: \mathbb{R} -

    -     - -     - - - - -

    - \ds f(x,y) = \frac{1}{x+2y} -

    -     - -

    - domain: x\neq 2y; in set notation, - \{(x,y)\,|\, x\neq 2y\} -

    - -

    - range: z\neq 0 -

    -     - -     - - - - -

    - \ds f(x,y) = \frac{1}{x^2+y^2+1} -

    -     - -

    - domain: \mathbb{R}^2 -

    - -

    - range: 0\lt z\leq 1 -

    -     - -     - - - - -

    - \ds f(x,y) = \sin(x) \cos(y) -

    -     - -

    - domain: \mathbb{R}^2 -

    - -

    - range: -1\leq z\leq 1 -

    -     - -     - - - - -

    - \ds f(x,y) = \sqrt{9-x^2-y^2} -

    -     - -

    - domain: \{(x,y)\,|\, x^2+y^2\leq 9\}, - , the domain is the circle and interior of a circle centered at the origin with radius 3. -

    - -

    - range: 0\leq z\leq 3 -

    -     - -     - - - - -

    - \ds f(x,y) = \frac1{\sqrt{x^2+y^2-9}} -

    -     - -

    - domain: \{(x,y)\,|\, x^2+y^2\geq 9\}, - , the domain is the exterior of the circle - (not including the circle itself) - centered at the origin with radius 3. -

    - -

    - range: 0\lt z\lt \infty, or (0,\infty) -

    -     - -     - -     - - - -

    - Describe in words and sketch the level curves for the function and given c values. -

    -     - - - - -

    - \ds f(x,y) = 3x-2y; c = -2,0,2 -

    -     - -

    - Level curves are lines y = (3/2)x-c/2. -

    - -     - -     - - - - -

    - \ds f(x,y) = x^2-y^2; c = -1,0,1 -

    -     - -

    - Level curves are hyperbolas - \frac{x^2}{c}-\frac{y^2}{c}=1, except for c=0, - where the level curve is the pair of lines y=x, - y=-x. -

    - -     - -     - - - - -

    - \ds f(x,y) = x-y^2; c = -2,0,2 -

    -     - -

    - Level curves are parabolas x=y^2+c. -

    - -     - -     - - - - -

    - \ds f(x,y) = \frac{1-x^2-y^2}{2y-2x}; c = -2,0,2 -

    -     - -

    - Level curves are hyperbolas (x-c)^2-(y-c)^2=1, - drawn in graph in different styles to differentiate the curves. -

    - -     - -     - - - - -

    - \ds f(x,y) = \frac{2x-2y}{x^2+y^2+1}; c = -1,0,1 -

    -     - -

    - When c\neq 0, the level curves are circles, - centered at (1/c,-1/c) with radius \sqrt{2/c^2-1}. - When c=0, the level curve is the line y=x. -

    - -     - -     - - - - -

    - \ds f(x,y) = \frac{y-x^3-1}{x}; - c = -3,-1,0,1,3 -

    -     - -

    - Level curves are cubics of the form y=x^3+cx+1. - Note how each curve passes through (0,1) and that the function is not defined at x=0. -

    - -     - -     - - - - -

    - \ds f(x,y) = \sqrt{x^2+4y^2}; c = 1,2,3,4 -

    -     - -

    - Level curves are ellipses of the form \frac{x^2}{c^2}+\frac{y^2}{c^2/4}=1, - , a=c and b=c/2. -

    - -     - -     - - - - -

    - \ds f(x,y) = x^2+4y^2; c = 1,2,3,4 -

    -     - -

    - Level curves are ellipses of the form \frac{x^2}{c}+\frac{y^2}{c/4}=1, - , a=\sqrt{c} and b=\sqrt{c}/2. -

    - -     - -     - -     - - - -

    - Give the domain and range of the functions of three variables. -

    -     - - - - -

    - \ds f(x,y,z) = \frac{x}{x+2y-4z} -

    -     - -

    - domain: x+2y-4z\neq 0; - the set of points in \mathbb{R}^3 NOT in the domain form a plane through the origin. -

    - -

    - range: \mathbb{R} -

    -     - -     - - - - -

    - \ds f(x,y,z) = \frac{1}{1-x^2-y^2-z^2} -

    -     - -

    - domain: x^2+y^2+z^2\neq 1; - the set of points in \mathbb{R}^3 NOT in the domain form a sphere of radius 1. -

    - -

    - range: (-\infty,0)\cup[1,\infty) -

    -     - -     - - - - -

    - \ds f(x,y,z) = \sqrt{z-x^2+y^2} -

    -     - -

    - domain: z\geq x^2-y^2; - the set of points in \mathbb{R}^3 above - (and including) - the hyperbolic paraboloid z=x^2-y^2. -

    - -

    - range: [0,\infty) -

    -     - -     - - - - -

    - \ds f(x,y,z) = z^2\sin(x) \cos(y) -

    -     - -

    - domain: \mathbb{R}^3 -

    - -

    - range: \mathbb{R} -

    -     - -     - -     - - - -

    - Describe the level surfaces of the given functions of three variables. -

    -     - - - - -

    - \ds f(x,y,z) = x^2+y^2+z^2 -

    -     - -

    - The level surfaces are spheres, - centered at the origin, with radius \sqrt{c}. -

    -     - -     - - - - -

    - \ds f(x,y,z) = z-x^2+y^2 -

    -     - -

    - The level surfaces are hyperbolic paraboloids of the form z=x^2-y^2+c; - each is shifted up/down by c. -

    -     - -     - - - - -

    - \ds f(x,y,z) = \frac{x^2+y^2}{z} -

    -     - -

    - The level surfaces are paraboloids of the form z=\frac{x^2}{c}+\frac{y^2}{c}; - the larger c, the wider the paraboloid. -

    -     - -     - - - - -

    - \ds f(x,y,z) = \frac{z}{x-y} -

    -     - -

    - The level surfaces are planes through the origin of the form cx-cy-z=0, that is, - planes through the origin with normal vector \la c,-c,-1\ra. -

    -     - -     - - - - -

    - Compare the level curves of Exercises - and . - How are they similar, and how are they different? - Each surface is a quadric surface; - describe how the level curves are consistent with what we know about each surface. -

    -     - -

    - The level curves for each surface are similar; - for z=\sqrt{x^2+4y^2} the level curves are ellipses of the form \frac{x^2}{c^2}+\frac{y^2}{c^2/4}=1, - , a=c and b=c/2; - whereas for z=x^2+4y^2 the level curves are ellipses of the form \frac{x^2}{c}+\frac{y^2}{c/4}=1, - , a=\sqrt{c} and b=\sqrt{c}/2. - The first set of ellipses are spaced evenly apart, - meaning the function grows at a constant rate; - the second set of ellipses are more closely spaced together as c grows, - meaning the function grows faster and faster as c increases. -

    - -

    - The graph z=\sqrt{x^2+4y^2} can be rewritten as z^2=x^2+4y^2, - an elliptic cone; - the graph z=x^2+4y^2 is a paraboloid, - each matching the description above. -

    -     - -     - -     -     -     -
    -
    - Limits and Continuity of Multivariable Functions - -

    - We continue with the pattern we have established in this text: - after defining a new kind of function, - we apply calculus ideas to it. - The previous section defined functions of two and three variables; - this section investigates what it means for these functions to be continuous. -

    - -

    - We begin with a series of definitions. - We are used to open intervals such as (1,3), - which represents the set of all x such that 1\lt x\lt 3, - and closed intervals such as [1,3], - which represents the set of all x such that 1\leq x\leq 3. - We need analogous definitions for open and closed sets in the xy-plane. -

    - - -     - - - Open and Closed Subsets in Higher Dimensions - - - - Open Disk, Boundary and Interior Points, Open and Closed Sets, Bounded Sets - -

    - An open disk B in \mathbb{R}^2 centered at - (x_0,y_0) with radius r is the set of all points (x,y) such that \ds\sqrt{(x-x_0)^2+(y-y_0)^2} \lt r. -

    - -

    - Let S be a set of points in \mathbb{R}^2. - A point P in \mathbb{R}^2 is a - boundary point - of S if all open disks centered at P contain both points in S and points not in S. -

    - -

    - A point P in S is an - interior point - of S if there is an open disk centered at P that contains only points in S. -

    - -

    - A set S is open - if every point in S is an interior point. -

    - -

    - A set S is closed - if it contains all of its boundary points. -

    - -

    - A set S is bounded - if there is an M \gt 0 such that the open disk, - centered at the origin with radius M, contains S. - A set that is not bounded is unbounded. - open - closed - open disk - closed disk - boundary point - interior point - bounded set - unbounded set -

    -     -     - -

    - - shows several sets in the xy-plane. - In each set, - point P_1 lies on the boundary of the set as all open disks centered there contain both points in, - and not in, the set. - In contrast, - point P_2 is an interior point for there is an open disk centered there that lies entirely within the set. -

    - -
    - Illustrating open and closed sets in the xy-plane - -
    - - - - A region in the first quadrant is shaded, and its boundary is shown as a solid curve. - -

    - A region in the first quadrant of the xy plane is shown. - It is somewhat in the shape of a pond, although the shape is not important. - The interior of the region is shaded, indicating that all the interior points are part of the set. - The boundary of the region is drawn as a solid curve, indicating that all points on the boundary are also part of the set. - Two points are plotted: a point P_1 on the boundary, and a point P_2 in the interior. - Both points are part of the set. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - ytick=\empty, - ymin=-1,ymax=5, - xmin=-1,xmax=7.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; - \addplot [firstcurvestyle,smooth] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; - - \filldraw (axis cs: 3.7,4.062) circle (2.4pt) node [shift={(0,9pt)}] { $P_1$}; - \draw [dashed] (axis cs: 3.7,4.062) circle (5pt); - - \filldraw (axis cs: 2,2) circle (2.4pt) node [shift={(0,9pt)}] { $P_2$}; - \draw [dashed] (axis cs: 2,2) circle (5pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - A region in the first quadrant is shaded, and its boundary is shown as a dashed curve. - -

    - A region in the first quadrant of the xy plane is shown. - It is the same region as the one depicted in . - The difference is that the boundary of the region is drawn as a dashed curve, - indicating that none of the points on the boundary are part of the set. - The same two points are plotted: a point P_1 on the boundary, and a point P_2 in the interior. - This time, P_1 is not part of the set, but P_2 is. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - ytick=\empty, - ymin=-1,ymax=5, - xmin=-1,xmax=7.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; - \addplot [firstcurvestyle,dashed,smooth] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; - - \filldraw (axis cs: 3.7,4.062) circle (2.4pt) node [shift={(0,9pt)}] { $P_1$}; - \draw [dashed] (axis cs: 3.7,4.062) circle (5pt); - - \filldraw (axis cs: 2,2) circle (2.4pt) node [shift={(0,9pt)}] { $P_2$}; - \draw [dashed] (axis cs: 2,2) circle (5pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - A region in the first quadrant is shaded, and its boundary is shown as a partially solid, partially dashed curve. - -

    - A region in the first quadrant of the xy plane is shown. - It is the same region as the one depicted in . - This time, part of the boundary is drawn as a solid curve, and part of the boundary is drawn as a dashed curve. - This indicates that some of the points on the boundary are part of the set, while other points are not. - The same two points are plotted: a point P_1 on the boundary, and a point P_2 in the interior. - This time, P_1 is not part of the set, but P_2 is. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - ytick=\empty, - ymin=-1,ymax=5, - xmin=-1,xmax=7.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; - \addplot [firstcurvestyle,dashed,smooth] coordinates {(1.,3.)(1.045,3.492)(1.167,3.889)(1.346,4.196)(1.56,4.414)(1.79,4.546)(2.015,4.596)(2.226,4.574)(2.425,4.501)(2.615,4.397)(2.799,4.282)(2.98,4.177)(3.16,4.099)(3.34,4.054)(3.52,4.042)(3.7,4.062)(3.88,4.114)(4.06,4.198)(4.242,4.306)(4.432,4.415)(4.636,4.503)(4.859,4.544)(5.107,4.518)(5.385,4.402)(5.681,4.203)(5.979,3.938)(6.261,3.624)(6.508,3.279)(6.705,2.92)(6.84,2.562)(6.917,2.211)(6.943,1.873)(6.922,1.553)(6.861,1.258)(6.766,0.9942)(6.634,0.7662)(6.46,0.5803)(6.238,0.4424)(5.962,0.3583)(5.626,0.3337)(5.229,0.3693)}; - - \addplot [firstcurvestyle,-,smooth] coordinates {(5.229,0.3693)(4.781,0.4503)(4.292,0.5591)(3.774,0.6783)(3.24,0.7904)(2.701,0.8805)(2.182,0.9817)(1.717,1.168)(1.342,1.517)(1.091,2.102)(1.,3.)}; - - \filldraw (axis cs: 3.7,4.062) circle (2.4pt) node [shift={(0,9pt)}] { $P_1$}; - \draw [dashed] (axis cs: 3.7,4.062) circle (5pt); - - \filldraw (axis cs: 2,2) circle (2.4pt) node [shift={(0,9pt)}] { $P_2$}; - \draw [dashed] (axis cs: 2,2) circle (5pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     -
    - -

    - The set depicted in is a closed set as it contains all of its boundary points. - The set in is open, for all of its points are interior points - (or, equivalently, it does not contain any of its boundary points). - The set in is neither open nor closed as it contains some of its boundary points. -

    - - - Determining open/closed, bounded/unbounded - -

    - Determine if the domain of the function f(x,y)=\sqrt{1-x^2/9-y^2/4} is open, closed, - or neither, and if it is bounded. -

    -     - -

    - This domain of this function was found in - to be D = \{(x,y)\,|\,\frac{x^2}9+\frac{y^2}4\leq 1\}, - the region bounded by the ellipse \frac{x^2}9+\frac{y^2}4=1. - Since the region includes the boundary - (indicated by the use of \leq), - the set contains all of its boundary points and hence is closed. - The region is bounded as a disk of radius 4, centered at the origin, - contains D. -

    -     -     - - - Determining open/closed, bounded/unbounded - -

    - Determine if the domain of - f(x,y) = \frac1{x-y} is open, closed, or neither. -

    -     - -

    - As we cannot divide by 0, we find the domain to be D = \{(x,y)\,|\,x-y\neq 0\}. - In other words, the domain is the set of all points (x,y) - not on the line y=x. -

    - -
    - Sketching the domain of the function in - - - An image of the xy plane. It is almost entirely shaded, except for the line y=x. - -

    - The image shows the xy plane with its coordinate axes. - The entire plane is shaded, except for a very thin strip along the line y=x. - This is intended as an illustration of the set of all points (x,y) for which y\neq x. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - ytick=\empty, - ymin=-1,ymax=1, - xmin=-1,xmax=1 - ] - - \filldraw [firstcolor!15,fill=firstcolor!15] (axis cs:-1,-1) rectangle (axis cs: 1,1); - - \addplot [ultra thick,white] coordinates {(-1.,-1.)(1,1)}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - The domain is sketched in . - Note how we can draw an open disk around any point in the domain that lies entirely inside the domain, - and also note how the only boundary points of the domain are the points on the line y=x. - We conclude the domain is an open set. - The set is unbounded. -

    -     -     -     - - - Limits -

    - Recall a pseudo-definition of the limit of a function of one variable: -

    -

    - \lim\limits_{x\to c}f(x) = L -

    -

    - means that if x is really close to c, - then f(x) is really close to L. - A similar pseudo-definition holds for functions of two variables. - We'll say that -

    - -

    - \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y) = L -

    - -

    - means if the point (x,y) is really close to the point (x_0,y_0), - then f(x,y) is really close to L. - The formal definition is given below. -

    - - - - Limit of a Function of Two Variables - -

    - Let S be a set containing - P=(x_0,y_0) where every open disk centered at P contains points in S other than P, - let f be a function of two variables defined on S, - except possibly at P, and let L be a real number. - The limit of f(x,y) as (x,y) approaches - (x_0,y_0) is L, denoted - - \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y) = L - , - means that given any \varepsilon \gt 0, - there exists \delta \gt 0 such that for all (x,y) in S, - where (x,y)\neq (x_0,y_0), - if (x,y) is in the open disk centered at - (x_0,y_0) with radius \delta, - then \abs{f(x,y) - L}\lt \varepsilon. - limitof multivariable function - multivariable functionlimit -

    -     -     - -

    - The concept behind - is sketched in . - Given \varepsilon \gt 0, - find \delta \gt 0 such that if (x,y) is any point in the open disk centered at - (x_0,y_0) in the xy-plane with radius \delta, - then f(x,y) should be within \varepsilon of L. -

    - -
    - Illustrating the definition of a limit. The open disk in the xy-plane has radius \delta. Let (x,y) be any point in this disk; f(x,y) is within \varepsilon of L. - - - - An image that illustrates the concept of the limit of a function of two variables. - -

    - A three-dimensional image, showing the three coordinate axes, - along with a surface, and some additional details. - The surface is plotted in the first octant, and lies above the xy plane. - The shape of the surface is similar to a portion of an umbrella. -

    - -

    - Below the surface, in the xy plane, a small circle with dashed boundary is drawn. - The center of the circle is at a point marked as (x_0,y_0,0). - Above the circle, on the surface, is a circular curve consisting of all the points of the form (x,y,f(x,y)) - on the surface, where the point (x,y,0) lies on the circle in the xy plane. - At the center of this curve is a point marked (x_0,y_0,L). -

    - -

    - Along the z axis there are three points marked, - with labels L-\varepsilon, L, and L+\varepsilon. - This illustrates that for each point (x,y,0) interior to the circle in the xy plane, - the value z=f(x,y) lies between L-\varpsilon and L+\varepsilon. - In other words, the z coordinate of each point on the surface that lies on the - interior of the circular curve must be within \varepsilon of L. -

    -     - - - - - //ASY file for figmultilimit_def3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic((10.3,-3.3,2.4),(-0.004,0.001,0.016),(0,0,0),1,(0,0)); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-0.5,3); - pair ybounds=(-0.5,3); - pair zbounds=(-0.1,2.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=2-x^2/4-y^2/4 - triple f(pair t) { - return (sqrt(4*(2-t.y))*cos(t.x),sqrt(4*(2-t.y))*sin(t.x),t.y); - } - surface s=surface(f,(0,0.75),(pi/2,2),10,10,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //draw the dot on the surface and in the xy-plane - dotfactor=3;dot((1,1,1.5));dot((1,1,0)); - - //draw the dashed circle in the xy-plane of radius 0.5 with label - triple g(real t) {return (1+.24*cos(t),1+.24*sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,linetype(new real[] {4,4})+bluepen+linewidth(1.5)); - label("$(x_0,y_0,0)$",(2,2,0)); - draw((1.6,1.6,0)--(1.1,1.1,0),linewidth(.75),Arrow3(size=3mm)); - - //draw the dashed circle on the surface of radius 0.5 with label - triple g(real t) { - return (1+.24*cos(t),1+.24*sin(t),2-((1+.24*cos(t))^2)/4-((1+.24*sin(t))^2)/4); - } - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,linetype(new real[] {4,4})+bluepen+linewidth(1.5)); - label("$(x_0,y_0,L)$",(2.,2,2.5)); - draw((1.8,1.8,2.2)--(1.1,1.1,1.6),linewidth(0.75),Arrow3(size=3mm)); - - draw((0,-.1,1.5)--(0,0,1.5)); - label("$L$",(0,-.1,1.5),W); - - draw((0,-.1,1.7)--(0,0,1.7)); - label("$L+\varepsilon$",(0,-.1,1.7),W); - draw((0,-.1,1.3)--(0,0,1.3)); - label("$L-\varepsilon$",(0,-.1,1.3),W); - - //draw((1,1,0)--(1+.5*cos(3*pi/2),1+.5*sin(3*pi/2),0),dashed+bluepen+linewidth(1)); - //label("$\delta$",((1,1,0)+(1+.5*cos(3*pi/2),1+.5*sin(3*pi/2),0))/2,N); - - - - -
    - -

    - Computing limits using this definition is rather cumbersome. - The following theorem allows us to evaluate limits much more easily. -

    - - - Basic Limit Properties of Functions of Two Variables - -

    - Let b, x_0, y_0, - L and K be real numbers, - let n be a positive integer, - and let f and g be functions with the following limits: - - \lim_{(x,y)\to (x_0,y_0)}f(x,y) = L \textit{ and } \lim_{(x,y)\to (x_0,y_0)} g(x,y) = K - . -

    - -

    - The following limits hold. - limitof multivariable function - limitproperties - multivariable functionlimit - -

      -
    1. -

      - Constants: \lim\limits_{(x,y)\to (x_0,y_0)} b = b -

      -
      2. - -
    2. -

      - Identity \lim\limits_{(x,y)\to (x_0,y_0)} x = x_0; - \lim\limits_{(x,y)\to (x_0,y_0)} y = y_0 -

      -
      3. - -
    3. -

      - Sums/Differences: - \lim\limits_{(x,y)\to (x_0,y_0)}\big(f(x,y)\pm g(x,y)\big) = L\pm K -

      -
      4. - -
    4. -

      - Scalar Multiples: \lim\limits_{(x,y)\to (x_0,y_0)} b\cdot f(x,y) = bL -

      -
      5. - -
    5. -

      - Products: \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y)\cdot g(x,y) = LK -

      -
      6. - -
    6. -

      - Quotients: \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y)/g(x,y) = L/K, (K\neq 0) -

      -
      7. - -
    7. -

      - Powers: \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y)^n = L^n -

      -
      8. -
    -

    -     -     - -

    - This theorem, - combined with Theorems - and - of , - allows us to evaluate many limits. -

    - - - Evaluating a limit - -

    - Evaluate the following limits: -

      -
    1. \lim_{(x,y)\to (1,\pi)} \left(\frac yx + \cos(xy)\right)
      2. -
    2. \lim_{(x,y)\to (0,0)} \frac{3xy}{x^2+y^2}
      3. -
    -

    -     - -

    -

      -
    1. -

      - The aforementioned theorems allow us to simply evaluate - y/x+\cos(xy) when x=1 and y=\pi. - If an indeterminate form is returned, - we must do more work to evaluate the limit; - otherwise, the result is the limit. - Therefore - - \lim_{(x,y)\to (1,\pi)} \frac yx + \cos(xy) \amp = \frac\pi{1}+\cos(\pi) - \amp = \pi -1 - . -

      -
      2. - -
    2. -

      - We attempt to evaluate the limit by substituting 0 in for x and y, - but the result is the indeterminate form - 0/0. To evaluate this limit, - we must do more work, but we have not yet learned what - kind of work to do. - Therefore we cannot yet evaluate this limit. -

      -
      3. -
    -

    -     -     - -

    - When dealing with functions of a single variable we also considered one-sided limits and stated - - \lim_{x\to c}f(x) = L \text{ if, and only if, } \lim_{x\to c^+}f(x) =L \textbf{ and} \lim_{x\to c^-}f(x) =L - . -

    - -

    - That is, the limit is L if and only if f(x) approaches L when x approaches c from either - direction, the left or the right. -

    - -

    - In the plane, - there are infinitely many directions from which (x,y) might approach (x_0,y_0). - In fact, we do not have to restrict ourselves to approaching - (x_0,y_0) from a particular direction, - but rather we can approach that point along a path that is not a straight line. - It is possible to arrive at different limiting values by approaching - (x_0,y_0) along different paths. - If this happens, - we say that \lim\limits_{(x,y)\to(x_0,y_0) } f(x,y) does not exist - (this is analogous to the left and right hand limits of single variable functions not being equal). -

    - -

    - Our theorems tell us that we can evaluate most limits quite simply, - without worrying about paths. - When indeterminate forms arise, - the limit may or may not exist. - If it does exist, - it can be difficult to prove this as we need to show the same limiting value is obtained regardless of the path chosen. - The case where the limit does not exist is often easier to deal with, - for we can often pick two paths along which the limit is different. -

    - - - Showing limits do not exist - -

    -

      -
    1. -

      - Show \lim\limits_{(x,y)\to (0,0)} \frac{3xy}{x^2+y^2} does not exist by finding the limits along the lines y=mx. -

      -
      2. - -
    2. -

      - Show \lim\limits_{(x,y)\to (0,0)} \frac{\sin(xy)}{x+y} does not exist by finding the limit along the path y=-\sin(x). -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - Evaluating \lim\limits_{(x,y)\to (0,0)} \frac{3xy}{x^2+y^2} along the lines y=mx means replace all y's with mx and evaluating the resulting limit: - - \lim_{(x,mx)\to (0,0)} \frac{3x(mx)}{x^2+(mx)^2} \amp =\lim_{x\to 0} \frac{3mx^2}{x^2(m^2+1)} - \amp = \lim_{x\to 0} \frac{3m}{m^2+1} - \amp = \frac{3m}{m^2+1} - . - While the limit exists for each choice of m, - we get a different limit for each choice of m. - That is, along different lines we get differing limiting values, - meaning the limit does not exist. -

      -
      2. - -
    2. -

      - Let f(x,y) = \frac{\sin(xy)}{x+y}. - We are to show that \lim\limits_{(x,y)\to (0,0)} f(x,y) does not exist by finding the limit along the path y=-\sin(x). - First, however, - consider the limits found along the lines y=mx as done above. - - \lim_{(x,mx)\to (0,0)} \frac{\sin\big(x(mx)\big)}{x+mx} \amp = \lim_{x\to 0} \frac{\sin(mx^2)}{x(m+1)} - \amp = \lim_{x\to 0} \frac{\sin(mx^2)}{x}\cdot\frac1{m+1} - . - By applying L'Hospital's Rule, - we can show this limit is 0 except when m=-1, - that is, along the line y=-x. - This line is not in the domain of f, - so we have found the following fact: - along every line y=mx in the domain of f, - \lim\limits_{(x,y)\to(0,0)} f(x,y)=0. - Now consider the limit along the path y=-\sin(x): - - \lim_{(x,-\sin(x) )\to (0,0)} \frac{\sin\big(-x\sin(x) \big)}{x-\sin(x) } \amp = \lim_{x\to0} \frac{\sin\big(-x\sin(x) \big)}{x-\sin(x) } - - Now apply L'Hospital's Rule twice: - - \amp = \lim_{x\to 0}\frac{\cos\big(-x\sin(x) \big)(-\sin(x) -x\cos(x) )}{1-\cos(x) } \quad \left(0/0\right) - \amp = \lim_{x\to 0}\frac{\begin{matrix}-\sin\big(-x\sin(x) \big)(-\sin(x) -x\cos(x) )^2\phantom{\cos(x)}\\ - \phantom{\cos(x)}+\cos\big(-x\sin(x) \big)(-2\cos(x) +x\sin(x) )\end{matrix}}{\sin(x) } - - . - This last limit is of the form 2/0, - which suggests that the limit does not exist. - Step back and consider what we have just discovered. - Along any line y=mx in the domain of the f(x,y), - the limit is 0. - However, along the path y=-\sin(x), - which lies in the domain of f(x,y) for all x\neq 0, - the limit does not exist. - Since the limit is not the same along every path to (0,0), - we say \lim\limits_{(x,y)\to (0,0)}\frac{\sin(xy)}{x+y} does not exist. -

      -
      3. -
    -

    -     -     - - - Finding a limit - -

    - Let f(x,y) = \frac{5x^2y^2}{x^2+y^2}. - Find \lim\limits_{(x,y)\to (0,0)} f(x,y). -

    -     - -

    - It is relatively easy to show that along any line y=mx, - the limit is 0. - This is not enough to prove that the limit exists, - as demonstrated in the previous example, - but it tells us that if the limit does exist then it must be 0. -

    - -

    - To prove the limit is 0, we apply . - Let \varepsilon \gt 0 be given. - We want to find \delta \gt 0 such that if \sqrt{(x-0)^2+(y-0)^2} \lt \delta, - then \abs{f(x,y)-0} \lt \varepsilon. -

    - -

    - Set \delta \lt \sqrt{\varepsilon/5}. - Note that \abs{\frac{5y^2}{x^2+y^2}} \lt 5 for all (x,y)\neq (0,0), - and that if \sqrt{x^2+y^2} \lt \delta, - then x^2\lt \delta^2. -

    - -

    - Let \sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2}\lt \delta. - Consider \abs{f(x,y)-0}: - - \abs{f(x,y)-0} \amp = \abs{\frac{5x^2y^2}{x^2+y^2}-0} - \amp = \abs{x^2\cdot\frac{5y^2}{x^2+y^2}} - \amp \lt \delta^2\cdot 5 - \amp \lt \frac{\varepsilon}{5}\cdot 5 - \amp = \varepsilon - . -

    - -

    - Thus if \sqrt{(x-0)^2+(y-0)^2}\lt \delta then \abs{f(x,y)-0}\lt \varepsilon, - which is what we wanted to show. - Thus \lim\limits_{(x,y)\to(0,0)} \frac{5x^2y^2}{x^2+y^2} = 0. -

    -     -     -     - - - Continuity -

    - - defines what it means for a function of one variable to be continuous. - In brief, it meant that the graph of the function did not have breaks, holes, - jumps, etc. - We define continuity for functions of two variables in a similar way as we did for functions of one variable. -

    - - - Continuous - -

    - Let a function f(x,y) be defined on a set S containing the point (x_0,y_0). -

    - -

    -

      -
    1. -

      - f is continuous - at (x_0,y_0) if \lim\limits_{(x,y)\to(x_0,y_0)} f(x,y) = f(x_0,y_0). - continuous function - multivariable functioncontinuity -

      -
      2. - -
    2. -

      - f is continuous on S - if f is continuous at all points in S. - If f is continuous at all points in \mathbb{R}^2, - we say that f is continuous everywhere. -

      -
      3. -
    -

    -     -     - - - Continuity of a function of two variables - -

    - Let f(x,y) = \left\{ \begin{array}{rl} \frac{\cos(y) \sin(x) }{x} \amp x\neq 0 \\ - \cos(y) \amp x=0 - \end{array} \right.. - Is f continuous at (0,0)? - Is f continuous everywhere? -

    -     - -

    - To determine if f is continuous at (0,0), - we need to compare \lim\limits_{(x,y)\to (0,0)} f(x,y) to f(0,0). -

    - -

    - Applying the definition of f, - we see that f(0,0) = \cos(0) = 1. -

    - -

    - We now consider the limit \lim\limits_{(x,y)\to (0,0)} f(x,y). - Substituting 0 for x and y in - (\cos(y) \sin(x) )/x returns the indeterminate form 0/0, - so we need to do more work to evaluate this limit. -

    - -

    - Consider two related limits: - \lim\limits_{(x,y)\to (0,0)} \cos(y) and \lim\limits_{(x,y)\to(0,0)} \frac{\sin(x) }x. - The first limit does not contain x, - and since \cos(y) is continuous, - - \lim\limits_{(x,y)\to (0,0)} \cos(y) =\lim_{y\to 0} \cos(y) = \cos(0) = 1 - . -

    - -

    - The second limit does not contain y. - By we can say - - \lim_{(x,y)\to (0,0)} \frac{\sin(x) }{x} = \lim_{x\to 0} \frac{\sin(x) }{x} = 1 - . -

    - -

    - Finally, - of this section states that we can combine these two limits as follows: - - \lim_{(x,y)\to (0,0)} \frac{\cos(y) \sin(x) }{x} \amp = \lim_{(x,y)\to (0,0)} (\cos(y) )\left(\frac{\sin(x) }{x}\right) - \amp =\left(\lim_{(x,y)\to (0,0)} \cos(y) \right)\left(\lim_{(x,y)\to (0,0)} \frac{\sin(x) }{x}\right) - \amp = (1)(1) - \amp =1 - . -

    - -

    - We have found that \lim\limits_{(x,y)\to (0,0)} \frac{\cos(y) \sin(x) }{x} = f(0,0), - so f is continuous at (0,0). -

    - -

    - A similar analysis shows that f is continuous at all points in \mathbb{R}^2. - As long as x\neq0, we can evaluate the limit directly; - when x=0, - a similar analysis shows that the limit is \cos(y). - Thus we can say that f is continuous everywhere. - A graph of f is given in . - Notice how it has no breaks, jumps, etc. -

    - -
    - A graph of f(x,y) in - - - - A graph of the piecewise-defined function in this example, illustrating its continuity. - -

    - A plot of the surface given by the graph z=f(x,y) for the function f in this example. - The surface consists of many peaks and valleys, with the largest peaks lying along the y axis. - Viewing the graph shows us that although the surface is quite bumpy, - there are no jumps or breaks in the surface, illustrating the fact that the function f is continuous. -

    -     - - - - - //ASY file for figmulticont13D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(31,31,2.3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-10,10}; - real[] myychoice={-10,10}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-12,12); - pair ybounds=(-12,12); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=cos(y)sin(x)/x - triple f(pair t) { - return (t.x,t.y,cos(t.y)*sin(t.x)/t.x); - } - surface s=surface(f,(-10,-10),(10,10),15,15,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //triple f(pair t) { - // return (t.x,t.y,cos(t.y)*sin(t.x)/t.x); - //} - //surface s=surface(f,(-11,-11),(0.1,11),16,16,Spline); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - - - -
    -     -     - -

    - The following theorem is very similar to , - giving us ways to combine continuous functions to create other continuous functions. -

    - - - Properties of Continuous Functions - -

    - Let f and g be continuous on a set S, - let c be a real number, - and let n be a positive integer. - The following functions are continuous on S. - continuous functionproperties - multivariable functioncontinuity - -

      -
    1. -

      - Sums/Differences: f\pm g -

      -
      2. - -
    2. -

      - Constant Multiples: c\cdot f -

      -
      3. - -
    3. -

      - Products: f\cdot g -

      -
      4. - -
    4. -

      - Quotients: f/g { (as longs as g\neq 0 on S)} -

      -
      5. - -
    5. -

      - Powers: f^n -

      -
      6. - -
    6. -

      - Roots: \sqrt[n]{f} (if n is even then f\geq 0 on S; - if n is odd, then true for all values of f on S.) -

      -
      7. - -
    7. -

      - Compositions:Adjust the definitions of f and g to: Let f be continuous on S, - where the range of f on S is J, - and let g be a single variable function that is continuous on J. - Then g\circ f, , g(f(x,y)), - is continuous on S. -

      -
      8. -
    -

    -     -     - - - Establishing continuity of a function - -

    - Let f(x,y) = \sin(x^2\cos(y) ). - Show f is continuous everywhere. -

    -     - -

    - We will apply both Theorems - and . - Let f_1(x,y) = x^2. - Since y is not actually used in the function, - and polynomials are continuous - (by ), - we conclude f_1 is continuous everywhere. - A similar statement can be made about f_2(x,y) = \cos(y). - Part 3 of - states that f_3=f_1\cdot f_2 is continuous everywhere, - and Part 7 of the theorem states the composition of sine with f_3 is continuous: - that is, \sin(f_3) = \sin(x^2\cos(y) ) is continuous everywhere. -

    -     -     -     - - - Functions of Three Variables -

    - The definitions and theorems given in this section can be extended in a natural way to definitions and theorems about functions of three - (or more) - variables. - We cover the key concepts here; - some terms from Definitions - and - are not redefined but their analogous meanings should be clear to the reader. -

    - - - Open Balls, Limit, Continuous - -

    -

      -
    1. -

      - An open ball in \mathbb{R}^3 centered at - (x_0,y_0,z_0) with radius r is the set of all points (x,y,z) such that - - \sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2} = r - . - multivariable functionlimit - limitof multivariable function - multivariable functioncontinuity - open ball -

      -
      2. - -
    2. -

      - Let D be a set in \mathbb{R}^3 containing - (x_0,y_0,z_0) where every open ball centered at - (x_0,y_0,z_0) contains points of D other than (x_0,y_0,z_0), - and let f(x,y,z) be a function of three variables defined on D, - except possibly at (x_0,y_0,z_0). - The limit of f(x,y,z) as (x,y,z) approaches (x_0,y_0,z_0) is L, denoted - - \lim_{(x,y,z)\to (x_0,y_0,z_0)} f(x,y,z) = L - , - means that given any \varepsilon \gt 0, - there is a \delta \gt 0 such that for all (x,y,z) in D, - (x,y,z)\neq(x_0,y_0,z_0), - if (x,y,z) is in the open ball centered at - (x_0,y_0,z_0) with radius \delta, - then \abs{f(x,y,z) - L}\lt \varepsilon. -

      -
      3. - -
    3. -

      - Let f(x,y,z) be defined on a set D containing (x_0,y_0,z_0). - We say f is continuous - at (x_0,y_0,z_0) if - - \lim\limits_{(x,y,z)\to (x_0,y_0,z_0)} f(x,y,z) = f(x_0,y_0,z_0) - . - If f is continuous at all points in D, - we say f is continuous on D. -

      -
      4. -
    -

    -     -     - -

    - These definitions can also be extended naturally to apply to functions of four or more variables. - - also applies to function of three or more variables, - allowing us to say that the function - - f(x,y,z) = \frac{e^{x^2+y}\sqrt{y^2+z^2+3}}{\sin(xyz)+5} - - is continuous everywhere. -

    - -

    - When considering single variable functions, - we studied limits, then continuity, then the derivative. - In our current study of multivariable functions, - we have studied limits and continuity. - In the next section we study derivation, - which takes on a slight twist as we are in a multivariable context. -

    -     - - - - Terms and Concepts - - - - -

    - Describe in your own words the difference between the boundary and interior points of a set. -

    -     - - - -     - - - - -

    - Use your own words to describe (informally) what \lim\limits_{(x,y)\to (1,2)} f(x,y) = 17 means. -

    -     - - - -

    - Answers will vary. - One answer is As (x,y) gets close to (1,2), - f(x,y) gets close to 17. -

    -     - -     - - - - -

    - Give an example of a closed, bounded set. -

    -     - - - -

    - Answers will vary. -

    - -

    - One possible answer: \{(x,y) | x^2+y^2\leq 1\} -

    -     - -     - - - - -

    - Give an example of a closed, unbounded set. -

    -     - - - -

    - Answers will vary. -

    - -

    - One possible answer: \{(x,y) | y\geq x^2 \} -

    -     - -     - - - - -

    - Give an example of a open, bounded set. -

    -     - - - -

    - Answers will vary. -

    - -

    - One possible answer: \{(x,y) | x^2+y^2\lt 1 \} -

    -     - -     - - - - -

    - Give an example of a open, unbounded set. -

    -     - - - -

    - Answers will vary. -

    - -

    - One possible answer: \{(x,y) | y \gt x^2 \} -

    -     - -     -     - - Problems - - - -

    - A set S is given. -

    - -

    -

      -
    1. -

      - Give one boundary point and one interior point, - when possible, of S. -

      -
      2. - -
    2. -

      - State whether S is open, closed, or neither. -

      -
      3. - -
    3. -

      - State whether S is bounded or unbounded. -

      -
      4. -
    -

    -     - - - - -

    - \ds S = \left\{(x,y)\,\left| \, \frac{(x-1)^2}{4}+\frac{(y-3)^2}{9}\leq 1\right.\right\} -

    -     - -

    -

      -
    1. -

      - Answers will vary. - - interior point: (1,3) - - boundary point: (3,3) -

      -
      2. - -
    2. -

      - S is a closed set -

      -
      3. - -
    3. -

      - S is bounded -

      -
      4. -
    -

    -     - -     - - - - - -

    - S = \left\{(x,y)\mid y\neq x^2\right\} -

    - - -     - -

    -

      -
    1. -

      - Answers will vary. Interior point: (1,0) (any point with y\neq x^2 will do). - Boundary point: (1,1) (any point with y=x^2 will do). -

      -
      2. -
    2. -

      - S is an open set. -

      -
      3. -
    3. -

      - S is unbounded. -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds S = \left\{(x,y)\,| \, x^2+y^2=1\right\} -

    -     - -

    -

      -
    1. -

      - Answers will vary. - - interior point: none - - boundary point: (0,-1) -

      -
      2. - -
    2. -

      - S is a closed set, consisting only of boundary points -

      -
      3. - -
    3. -

      - S is bounded -

      -
      4. -
    -

    -     - -     - - - - - -

    - S = \left\{(x,y)\mid y \gt \sin(x)\right\}. -

    - - -     - -     - -     - - - -

    - For the given function: -

    - -

    -

      -
    1. -

      - Find the domain D of the function. -

      -
      2. - -
    2. -

      - State whether D is an open or closed set. -

      -
      3. - -
    3. -

      - State whether D is bounded or unbounded. -

      -
      4. -
    -

    -     - - - - -

    - \ds f(x,y) = \sqrt{9-x^2-y^2} -

    -     - -

    -

      -
    1. -

      - D = \left\{(x,y)\, |\, 9-x^2-y^2\geq 0\right\}. -

      -
      2. - -
    2. -

      - D is a closed set. -

      -
      3. - -
    3. -

      - D is bounded. -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds f(x,y) = \sqrt{y-x^2} -

    -     - -

    -

      -
    1. -

      - D = \left\{(x,y)\, |\, y\geq x^2\right\}. -

      -
      2. - -
    2. -

      - D is a closed set. -

      -
      3. - -
    3. -

      - D is unbounded. -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds f(x,y) = \frac{1}{\sqrt{y-x^2}} -

    -     - -

    -

      -
    1. -

      - D = \left\{(x,y)\, |\, y \gt x^2\right\}. -

      -
      2. - -
    2. -

      - D is an open set. -

      -
      3. - -
    3. -

      - D is unbounded. -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds f(x,y) = \frac{x^2-y^2}{x^2+y^2} -

    -     - -

    -

      -
    1. -

      - D = \left\{(x,y)\, |\, (x,y)\neq (0,0) \right\}. -

      -
      2. - -
    2. -

      - D is an open set. -

      -
      3. - -
    3. -

      - D is unbounded. -

      -
      4. -
    -

    -     - -     - -     - - - -

    - A limit is given. - Evaluate the limit along the paths given, - then state why these results show the given limit does not exist. -

    -     - - - - -

    - \lim\limits_{(x,y)\to(0,0)} \frac{x^2-y^2}{x^2+y^2} -

    -     - - - -

    - Along the path y=0. -

    -     -     - - - -

    - Along the path x=0. -

    -     -     - -

    -

      -
    1. -

      - Along y=0, the limit is 1. -

      -
      2. - -
    2. -

      - Along x=0, the limit is -1. -

      -
      3. -
    -

    - -

    - Since the above limits are not equal, the limit does not exist. -

    -     - -     - - - - -

    - \lim\limits_{(x,y)\to(0,0)} \frac{x+y}{x-y} -

    - -

    - Along the path y=mx. -

    -     - -

    -

      -
    1. -

      - Along y=mx, the limit is \frac{m+1}{1-m}. -

      -
      2. -
    -

    - -

    - Since the above limit varies according to what m is used, - each limit is different, meaning the overall limit does not exist. -

    -     - -     - - - - -

    - \lim\limits_{(x,y)\to(0,0)} \frac{xy-y^2}{y^2+x} -

    -     - - - -

    - Along the path y=mx. -

    -     -     - - - -

    - Along the path x=0. -

    -     -     - -

    -

      -
    1. -

      - Along y=mx, the limit is \ds\lim_{x\to 0}\frac{mx(1-m)}{m^2x+1}=0 for all m -

      -
      2. - -
    2. -

      - Along x=0, the limit is -1. -

      -
      3. -
    -

    - -

    - Since the above limits are not equal, the limit does not exist. -

    -     - -     - - - - -

    - \lim\limits_{(x,y)\to(0,0)} \frac{\sin(x^2)}{y} -

    -     - - - -

    - Along the path y=mx. -

    -     -     - - - -

    - Along the path y=x^2. -

    -     -     - -

    -

      -
    1. -

      - Along y=mx, the limit is: - - \lim_{(x,y)\to(0,0)} \frac{\sin(x^2)}{y} \amp = \lim_{x\to 0} \frac{\sin(x^2)}{mx} - apply L'Hospital's Rule - \amp = \lim_{x\to 0} \frac{2x\cos(x^2)}{m} - \amp = 0 - . -

      -
      2. -
    2. -

      - Along y=x^2, the limit is: - - \lim_{(x,y)\to(0,0)} \frac{\sin(x^2)}{y} = \lim_{x\to 0} \frac{\sin(x^2)}{x^2} - . - This can be evaluated with L'Hospital's Rule or from known limits; it is 1. -

      -
      3. -
    -

    - -

    - Since the limits along the lines y=mx are not the same as the limit along y=x^2, - the overall limit does not exist. -

    -     - -     - - - - -

    - \lim\limits_{(x,y)\to(1,2)} \frac{x+y-3}{x^2-1} -

    -     - - - -

    - Along the path y=2. -

    -     -     - - - -

    - Along the path y=x+1. -

    -     -     - -

    -

      -
    1. -

      - Along y=2, the limit is: - - \lim_{(x,y)\to(1,2)} \frac{x+y-3}{x^2-1} \amp = \lim_{x\to 1} \frac{x-1}{x^2-1} - \amp = \lim_{x\to 1} \frac{1}{x+1} - \amp = 1/2 - . -

      -
      2. - -
    2. -

      - Along y=x+1, the limit is: - - \lim_{(x,y)\to(1,2)} \frac{x+y-3}{x^2-1} \amp = \lim_{x\to 1} \frac{2(x-1)}{x^2-1} - \amp = \lim_{x\to 1} \frac{2}{x+1} - \amp = 1 - . -

      -
      3. -
    -

    - -

    - Since the limits along the lines y=2 and y=x+1 differ, - the overall limit does not exist. -

    -     - -     - - - - -

    - \lim\limits_{(x,y)\to(\pi,\pi/2)} \frac{\sin(x) }{\cos(y) } -

    -     - - - -

    - Along the path x=\pi. -

    -     -     - - - -

    - Along the path y=x-\pi/2. -

    -     -     - -

    -

      -
    1. -

      - Along x=\pi, the limit is: - - \lim_{(x,y)\to(\pi,\pi/2)} \frac{\sin(x) }{\cos(y) } \amp = \lim_{y\to \pi/2} \frac{0}{\cos(y) } - \amp = 0 - . -

      -
      2. - -
    2. -

      - Along y=x-\pi/2, the limit is: - - \lim_{(x,y)\to(\pi,\pi/2)} \frac{\sin(x) }{\cos(y) } \amp = \lim_{x\to \pi} \frac{\sin(x) }{\cos(x-\pi/2)} - Apply L'Hospital's Rule: - \amp = \lim_{x\to \pi} \frac{\cos(x) }{\sin(x-\pi/2)} - \amp = 1 - . -

      -
      3. -
    -

    - -

    - Since the limits along the lines x=\pi and y=x-\pi differ, - the overall limit does not exist. -

    -     - -     - -     -     -     -
    -
    - Partial Derivatives - -

    - Let y be a function of x. - We have studied in great detail the derivative of y with respect to x, - that is, \frac{dy}{dx}, - which measures the rate at which y changes with respect to x. - Consider now z=f(x,y). - It makes sense to want to know how z changes with respect to x and/or y. - This section begins our investigation into these rates of change. -

    - - -     - - - First-order partial derivatives -

    - Consider the function f(x,y) = x^2+2y^2, - as graphed in . - By fixing y=2, - we focus our attention to all points on the surface where the y-value is 2, shown in both and . - These points form a curve in the plane y=2: - z = f(x,2) = x^2+8 which defines z as a function of just one variable. - We can take the derivative of z with respect to x along this curve and find equations of tangent lines, etc. -

    - -
    - By fixing y=2, the surface z=f(x,y) = x^2+2y^2 is a curve in space - -
    - - - - - An elliptic paraboloid plotted over a rectangular domain. The trace y=2 is highlighted on the surface. - -

    - The graph z=x^2+2y^2 is an elliptic paraboloid, opening upward, with its vertex at the origin. - The plot uses a rectangular domain, so we see peaks at the corners of the domain. - Mesh curves corresponding to the traces where either x or y are constant are shown, - and one of these curves, the trace y=2, is highlighted. -

    -     - - - - - //ASY file for figpartialintro3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(16,16,36); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-4,-2,2,4}; - real[] myychoice={-4,-2,2,4}; - real[] myzchoice={10,20}; - defaultpen(0.5mm); - pair xbounds=(-5,5); - pair ybounds=(-5,5); - pair zbounds=(-1,22); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=x^2+2*y^2 - triple f(pair t) { - return (t.x,t.y,t.x^2+2*t.y^2); - } - surface s=surface(f,(-3,-3),(3,3),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}, - vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the trace for y=2 - triple g(real t) {return (t,2,t^2+8);} - path3 mypath=graph(g,-3,3,operator ..); draw(mypath,bluepen); - - - - -
    - -
    - - - - - The trace y=2 on the surface of the ellipic paraboloid from the previous image, with the surface removed. - -

    - The curve given by the trace y=2 in the surface z=x^2+2y^2 is plotted in three dimensions. - This is the same curve shown in , - but this time only the curve is plotted. The surface has been removed, to help visualize the curve. -

    - -

    - The curve itself has the shape of a parabola, opening upwards. - It is hanging in space, with its vertex above the mark on the y axis for y=2. - The curve is viewed in perspective but is clearly a parabola. -

    -     - - - - - //ASY file for figpartialintrob3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(16,16,36); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-4,-2,2,4}; - real[] myychoice={-4,-2,2,4}; - real[] myzchoice={10,20}; - defaultpen(0.5mm); - pair xbounds=(-5,5); - pair ybounds=(-5,5); - pair zbounds=(-1,22); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the trace for y=2 - triple g(real t) {return (t,2,t^2+8);} - path3 mypath=graph(g,-3,3,operator ..); draw(mypath,bluepen); - - - - -
    -     -
    - -

    - The key notion to extract from this example is: - by treating y as constant - (it does not vary) - we can consider how z changes with respect to x. - In a similar fashion, - we can hold x constant and consider how z changes with respect to y. - This is the underlying principle of - partial derivatives. - We state the formal, limit-based definition first, - then show how to compute these partial derivatives without directly taking limits. -

    - - - - - - Partial Derivative - -

    - Let z=f(x,y) be a continuous function on a set S in \mathbb{R}^2. -

    - -

    -

      -
    1. -

      - The partial derivative of f with respect to x is: - - f_x(x,y) = \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}h - . -

      -
      2. - -
    2. -

      - The partial derivative of f with respect to y is: - - f_y(x,y) = \lim_{h\to 0} \frac{f(x,y+h) - f(x,y)}h - . -

      -
      3. -
    -

    - -

    - partial derivative - derivativepartial -

    -     -     - - - Computing partial derivatives with the limit definition - -

    - Let f(x,y) = x^2y + 2x+y^3. - Find f_x(x,y) using the limit definition. -

    -     - -

    - Using , we have: - - f_x(x,y) \amp = \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}{h} - \amp = \lim_{h\to 0} \frac{(x+h)^2y+2(x+h)+y^3 - (x^2y+2x+y^3)}{h} - \amp = \lim_{h\to 0} \frac{x^2y+2xhy+h^2y+2x+2h+y^3-(x^2y+2x+y^3)}{h} - \amp = \lim_{h\to 0} \frac{2xhy+h^2y+2h}{h} - \amp =\lim_{h\to 0} 2xy+hy+2 - \amp = 2xy+2 - . -

    - -

    - We have found f_x(x,y) = 2xy+2. -

    -     -     - -

    - - found a partial derivative using the formal, - limit-based definition. - Using limits is not necessary, though, - as we can rely on our previous knowledge of derivatives to compute partial derivatives easily. - When computing f_x(x,y), - we hold y fixed it does not vary. - Therefore we can compute the derivative with respect to x by treating y as a constant or coefficient. -

    - -

    - Just as \frac{d}{dx}\big(5x^2\big) = 10x, - we compute \frac{\partial}{\px}\big(x^2y\big) = 2xy. - Here we are treating y as a coefficient. -

    - -

    - Just as \frac{d}{dx}\big(5^3\big) = 0, - we compute \frac{\partial}{\px}\big(y^3\big) = 0. - Here we are treating y as a constant. - More examples will help make this clear. -

    - - - Finding partial derivatives - -

    - Find f_x(x,y) and f_y(x,y) in each of the following. -

    - -

    -

      -
    1. f(x,y) = x^3y^2+ 5y^2-x+7
      2. -
    2. f(x,y) = \cos(xy^2)+\sin(x)
      3. -
    3. f(x,y) = e^{x^2y^3}\sqrt{x^2+1}
      4. -
    -

    -     - -

    -

      -
    1. -

      - We have f(x,y) = x^3y^2+ 5y^2-x+7. - Begin with f_x(x,y). - Keep y fixed, - treating it as a constant or coefficient, as appropriate: - - f_x(x,y) = 3x^2y^2-1 - . - Note how the 5y^2 and 7 terms go to zero. - To compute f_y(x,y), we hold x fixed: - - f_y(x,y) = 2x^3y+10y - . - Note how the -x and 7 terms go to zero. -

      -
      2. - -
    2. -

      - We have f(x,y) = \cos(xy^2)+\sin(x). - - Begin with f_x(x,y). We need to apply the Chain Rule with the cosine term; y^2 is the coefficient of the x-term inside the cosine function. - - f_x(x,y) = -\sin(xy^2)(y^2)+\cos(x) = -y^2\sin(xy^2)+\cos(x) - . - To find f_y(x,y), - note that x is the coefficient of the y^2 term inside of the cosine term; - also note that since x is fixed, - \sin(x) is also fixed, - and we treat it as a constant. - - f_y(x,y) = -\sin(xy^2)(2xy) = -2xy\sin(xy^2) - . -

      -
      3. - -
    3. -

      - We have f(x,y) = e^{x^2y^3}\sqrt{x^2+1}. - Beginning with f_x(x,y), - note how we need to apply the Product Rule. - - f_x(x,y) \amp = e^{x^2y^3}(2xy^3)\sqrt{x^2+1} + e^{x^2y^3}\frac12\big(x^2+1\big)^{-1/2}(2x) - \amp = 2xy^3e^{x^2y^3}\sqrt{x^2+1}+\frac{xe^{x^2y^3}}{\sqrt{x^2+1}} - . - Note that when finding f_y(x,y) we do not have to apply the Product Rule; - since \sqrt{x^2+1} does not contain y, - we treat it as fixed and hence becomes a coefficient of the e^{x^2y^3} term. - - f_y(x,y) = e^{x^2y^3}(3x^2y^2)\sqrt{x^2+1} = 3x^2y^2e^{x^2y^3}\sqrt{x^2+1} - . -

      -
      4. -
    -

    -     -     - - - -

    - We have shown how to compute a partial derivative, - but it may still not be clear what a partial derivative means. - Given z=f(x,y), - f_x(x,y) measures the rate at which z changes as only x varies: - y is held constant. - partial derivativemeaning -

    - -

    - Imagine standing in a rolling meadow, - then beginning to walk due east. - Depending on your location, you might walk up, - sharply down, or perhaps not change elevation at all. - This is similar to measuring z_x: - you are moving only east - (in the x-direction) - and not north/south at all. - Going back to your original location, - imagine now walking due north - (in the y-direction). - Perhaps walking due north does not change your elevation at all. - This is analogous to z_y=0: - z does not change with respect to y. - We can see that z_x and z_y do not have to be the same, - or even similar, - as it is easy to imagine circumstances where walking east means you walk downhill, - though walking north makes you walk uphill. -

    - - - -

    - The following example helps us visualize this more. -

    - - - Evaluating partial derivatives - -

    - Let z=f(x,y)=-x^2-\frac12y^2+xy+10. - Find f_x(2,1) and f_y(2,1) and interpret their meaning. -

    -     - -

    - We begin by computing f_x(x,y) = -2x+y and f_y(x,y) = -y+x. - Thus - - f_x(2,1) = -3 \text{ and } f_y(2,1) = 1 - . -

    - -

    - It is also useful to note that f(2,1) = 7.5. - What does each of these numbers mean? -

    - -

    - Consider f_x(2,1)=-3, - along with . - If one stands on the surface at the point - (2,1,7.5) and moves parallel to the x-axis (, only the x-value changes, - not the y-value), - then the instantaneous rate of change is -3. - Increasing the x-value will decrease the z-value; - decreasing the x-value will increase the z-value. -

    - -
    - Illustrating the meaning of partial derivatives - -
    - - - - - On a surface in space, a trace curve is highlighted. At one pointon this curve, a tangent line is drawn. - -

    - The surface given by the graph z=f(x,y) is shown, - for f(x,y) = -x^2-\frac12 y^2+xy+10. - It is a portion of an elliptic paraboloid, opening downward. -

    - -

    - Along the surface a curve is drawn, corresponding to the trace y=1. - This is a curve that moves along the surface as x varies, while y is held constant. - At the point (2,1,f(2,1)) on the surface, a line is drawn, tangent to the curve. - This is a line in space, but its slope, relative to x, - is given by the partial derivative f_x(2,1). -

    -     - - - - - //ASY file for figpartial3a3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8,8,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={5,10}; - defaultpen(0.5mm); - pair xbounds=(-1,3); - pair ybounds=(-1,3); - pair zbounds=(-1,12); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=-x^2-(1/2)*y^2+xy+10 - triple f(pair t) { - return (t.x,t.y,-t.x^2-(1/2)*t.y^2+t.x*t.y+10); - } - surface s=surface(f,(0,0),(3,3),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the point (2,1,7.5) trace for y=1 - dotfactor=3;dot((2,1,7.5)); - triple g(real t) {return (t,1,-t^2-(1/2)+t+10);} - path3 mypath=graph(g,0,3,operator ..); draw(mypath,bluepen); - - //Draw the tan line (y=1) L=(2,1,7.5)+t(1,0,-3) for t=-1.5,1 - draw((0.5,1,12)--(3,1,4.5),redpen); - - - - -
    - -
    - - - - - On a surface in space, a trace curve is highlighted. At one pointon this curve, a tangent line is drawn. - -

    - The surface given by the graph z=f(x,y) is shown, - for f(x,y) = -x^2-\frac12 y^2+xy+10. - It is a portion of an elliptic paraboloid, opening downward. -

    - -

    - Along the surface a curve is drawn, corresponding to the trace x=. - This is a curve that moves along the surface as y varies, while x is held constant. - At the point (2,1,f(2,1)) on the surface, a line is drawn, tangent to the trace x=2. - This tangent line is perpendicular to the one drawn in . - It is a line in space, but its slope, relative to y, - is given by the partial derivative f_y(2,1). -

    -     - - - - - //ASY file for figpartial3a3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8,8,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={5,10}; - defaultpen(0.5mm); - pair xbounds=(-1,3); - pair ybounds=(-1,3); - pair zbounds=(-1,12); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=-x^2-(1/2)*y^2+xy+10 - triple f(pair t) { - return (t.x,t.y,-t.x^2-(1/2)*t.y^2+t.x*t.y+10); - } - surface s=surface(f,(0,0),(3,3),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the point (2,1,7.5) trace for x=2 - dotfactor=3;dot((2,1,7.5)); - triple g(real t) {return (2,t,-4-(1/2)*t^2+2*t+10);} - path3 mypath=graph(g,0,3,operator ..); draw(mypath,bluepen); - - //Draw the tan line (x=2) L=(2,1,7.5)+t(0,1,1) for t=-1,2 - draw((2,0,6.5)--(2,3,9.5),redpen); - - - - -
    -     -
    - -

    - Now consider f_y(2,1)=1, - illustrated in . - Moving along the curve drawn on the surface, - , parallel to the y-axis and not changing the x-values, - increases the z-value instantaneously at a rate of 1. - Increasing the y-value by 1 would increase the z-value by approximately 1. -

    - -

    - Since the magnitude of f_x is greater than the magnitude of f_y at (2,1), - it is steeper in the x-direction than in the y-direction. -

    -     -     -     - - - Tangent Planes -

    - Another way to interpret partial derivatives is in terms of the tangent plane. - Consider the graph of a function f(x,y), such as the one in . - Setting x=a, y=b defines a point (a,b,f(a,b)) on the graph. - Through the point (a,b), we have the lines x=a+s, y=b, and x=a, y=b+t, - parallel to the x and y axes, respectively (where s,t are parameters). -

    - -

    - Using the function f(x,y) we define two vector-valued functions: - - \vec{r}_1(s) \amp = \la a+s, b, f(a+s,b)\ra - \vec{r}_2(t) \amp = \la a, b+t, f(a,b+t)\ra - . - Both vector-valued functions define space curves that lie on the surface z=f(x,y), - and these curves intersect at the point (a,b,f(a,b)), when s=t=0. -

    - -

    - Now consider computing \vec{r}_1'(s). - The first two components of this derivative are found in a straightforward manner: - they are 1 and 0, respectively. - To find the third component of the derivative, - notice that in \vec{r}_1(s) we vary the x-component of f while holding the y-component constant. - Using the Chain Rule and , - we find that the third component is f_x(a+s,b). Altogether, we have - - \vec{r}_1'(s) = \la 1,0,f_x(a+s,b)\ra - . - Evaluating this at s=0 gives - - \vec{v}=\vec{r}_1'(0) = \la 1,0,f_x(a,b)\ra - . - We can perform a similar process with \vec{r}_2(t), ultimately leading to - - \vec{w}=\vec{r}_2'(0) = \la 0,1,f_y(a,b)\ra - . - From , - we know that \vec{r}_1'(0) defines a tangent vector to the curve \vec{r}_1(s) when s=0, - and similarly, \vec{r}_2'(0) defines a tangent vector to the curve \vec{r}_2(t) when t=0. -

    - -

    - It seems reasonable that any vector that is tangent to these curves, - which lie on our surface, should also be considered tangent to that surface. - The vectors \vec{v} and \vec{w} are therefore tangent to z=f(x,y) at (a,b,f(a,b)), - and they are definitely not parallel. - From we know that any two non-parallel vectors at a point define a plane through that point. - We also know that taking the cross product of these two vectors gives us a normal vector: - the cross product gives us - - \vec{n}=\vec{v}\times\vec{w}=\la -f_x(a,b), -f_y(a,b), 1\ra - . -

    - -

    - The equation of the plane through (a,b,f(a,b)) with normal vector \vec{n}=\la -f_x(a,b),-f_y(a,b),1\ra is - - -f_x(a,b)(x-a)-f_y(a,b)(y-b)+(z-f(a,b))=0 - . - It is customary to solve for z in this equation and make the following definition. -

    - - - -

    - Let f(x,y) be a function whose first-order partial derivatives exist at (a,b). - The tangent plane to the surface z=f(x,y) at the point (a,b,f(a,b)) is the plane defined by the equation - - z = f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) - . - tangent plane - tangent planeto a graph -

    -     -     - - - Finding a tangent plane equation - -

    - Find the equation of tangent plane to the surface z=x^2+3y^2 at (x,y)=(1,-1). -

    -     - -

    - Our function is f(x,y)=x^2+3y^2, and we have f(1,-1)=4, - so the point on the surface is (1,-1,4). - The partial derivatives are f_x(x,y)=2x and f_y(x,y)=6y, - so f_x(1,-1)=2, f_y(1,-1)=-6. - Using , our plane is given by - - z = 4+2(x-1)-6(y+1) - . -

    -     -     - -

    - Notice the similarity between the tangent plane equation in - and the single variable tangent line equation y = f(c)+f'(c)(x-c). - As with functions of one variable, this suggests a connection between derivatives and linear approximation. - We explore this connection in , - where we'll see that should be strengthed to require that the partial derivatives of f be continuous. -

    -     - - - Second-order partial derivatives -

    - Let z=f(x,y). - We have learned to find the partial derivatives f_x(x,y) and f_y(x,y), - which are each functions of x and y. - Therefore we can take partial derivatives of them, - each with respect to x and y. - We define these second partials along with the notation, - give examples, then discuss their meaning. -

    - - - Second Partial Derivative, Mixed Partial Derivative - -

    - Let z=f(x,y) be continuous on a set S. -

    - -

    -

      -
    1. -

      - The second partial derivative of f with respect to x then x is - - \frac{\partial}{\partial x}\left(\frac{\partial f}{\px}\right) = \frac{\partial^2 f}{\px^2} = \big(\,f_x\,\big)_x = f_{xx} - -

      -
      2. - -
    2. -

      - The second partial derivative of f with respect to x then y is - - \frac{\partial}{\partial y}\left(\frac{\partial f}{\px}\right) = \frac{\partial^2f}{\py\px} = \big(\,f_x\,\big)_y = f_{xy} - -

      -
      3. -
    -

    - -

    - Similar definitions hold for - \frac{\partial^2f}{\py^2} = f_{yy} and \frac{\partial^2f}{\px\py} = f_{yx}. -

    - -

    - The second partial derivatives f_{xy} and f_{yx} are - mixed partial derivatives. - partial derivativemixed - partial derivativesecond derivative - derivativemixed partial -

    -     -     - -

    - The notation of second partial derivatives gives some insight into the notation of the second derivative of a function of a single variable. - If y=f(x), then \fp'(x) = \frac{d^2 y}{dx^2}. - The d^2y portion means - take the derivative of y twice, while dx^2 - means with respect to x both times. - When we only know of functions of a single variable, - this latter phrase seems silly: - there is only one variable to take the derivative with respect to. - Now that we understand functions of multiple variables, - we see the importance of specifying which variables we are referring to. -

    - - - - Second partial derivatives - -

    - For each of the following, find all six first and second partial derivatives. - That is, find - - f_x, f_y, f_{xx}, f_{yy}, f_{xy} \text{ and } f_{yx}\, - . -

    - -

    -

      -
    1. f(x,y) = x^3y^2 + 2xy^3+\cos(x)
      2. - -
    2. f(x,y) = \frac{x^3}{y^2}
      3. - -
    3. f(x,y)=e^{x}\sin(x^2y)
      4. -
    -

    -     - -

    - In each, we give f_x and f_y immediately and then spend time deriving the second partial derivatives. -

    - -

    -

      -
    1. -

      - - f(x,y) \amp = x^3y^2+2xy^3+\cos(x) - f_x(x,y) \amp = 3x^2y^2+2y^3-\sin(x) - f_y(x,y) \amp = 2x^3y+6xy^2 - f_{xx}(x,y) \amp = \frac{\partial}{\px}\big(f_x\big) = \frac{\partial}{\px}\big(3x^2y^2+2y^3-\sin(x) \big) - \amp = 6xy^2-\cos(x) - f_{yy}(x,y) \amp = \frac{\partial}{\py}\big(f_y\big) = \frac{\partial}{\py}\big(2x^3y+6xy^2\big) - \amp = 2x^3+12xy - f_{xy}(x,y) \amp = \frac{\partial}{\py}\big(f_x\big) = \frac{\partial}{\py}\big(3x^2y^2+2y^3-\sin(x) \big) - \amp = 6x^2y+6y^2 - f_{yx}(x,y) \amp = \frac{\partial}{\px}\big(f_x\big) = \frac{\partial}{\px}\big(2x^3y+6xy^2\big) - \amp = 6x^2y+6y^2 - -

      -
      2. - -
    2. -

      - - f(x,y) \amp = \frac{x^3}{y^2} = x^3y^{-2} - f_x(x,y) \amp = \frac{3x^2}{y^2} - f_y(x,y) \amp = -\frac{2x^3}{y^3} - f_{xx}(x,y) \amp = \frac{\partial}{\px}\big(f_x\big) = \frac{\partial}{\px}\big(\frac{3x^2}{y^2}\big) - \amp = \frac{6x}{y^2} - f_{yy}(x,y) \amp = \frac{\partial}{\py}\big(f_y\big) = \frac{\partial}{\py}\big(-\frac{2x^3}{y^3}\big) - \amp = \frac{6x^3}{y^4} - f_{xy}(x,y) \amp = \frac{\partial}{\py}\big(f_x\big) = \frac{\partial}{\py}\big(\frac{3x^2}{y^2}\big) - \amp = -\frac{6x^2}{y^3} - f_{yx}(x,y) \amp = \frac{\partial}{\px}\big(f_x\big) = \frac{\partial}{\px}\big(-\frac{2x^3}{y^3}\big) - \amp = -\frac{6x^2}{y^3} - -

      -
      3. - -
    3. -

      - f(x,y) = e^x\sin(x^2y) Because the following partial derivatives get rather long, - we omit the extra notation and just give the results. - In several cases, - multiple applications of the Product and Chain Rules will be necessary, - followed by some basic combination of like terms. - - f_x(x,y) \amp = e^x\sin(x^2y) + 2xye^x\cos(x^2y) - f_y(x,y) \amp = x^2e^x\cos(x^2y) - f_{xx}(x,y) \amp = e^x\sin(x^2y)+4xye^x\cos(x^2y)+2ye^x\cos(x^2y)-4x^2y^2e^x\sin(x^2y) - f_{yy}(x,y) \amp = -x^4e^x\sin(x^2y) - f_{xy}(x,y) \amp = x^2e^x\cos(x^2y)+2xe^x\cos(x^2y)-2x^3ye^x\sin(x^2y) - f_{yx}(x,y) \amp = x^2e^x\cos(x^2y)+2xe^x\cos(x^2y)-2x^3ye^x\sin(x^2y) - -

      -
      4. -
    -

    -     -     - - - -

    - Notice how in each of the three functions in , - f_{xy} = f_{yx}. - Due to the complexity of the examples, - this likely is not a coincidence. - The following theorem states that it is not. -

    - - - Mixed Partial Derivatives - -

    - Let f be defined such that f_{xy} and f_{yx} are continuous on a set S. - Then for each point (x,y) in S, - f_{xy}(x,y) = f_{yx}(x,y). -

    -     -     - -

    - Finding f_{xy} and f_{yx} independently and comparing the results provides a convenient way of checking our work. -

    -     - - - Understanding Second Partial Derivatives -

    - Now that we know how to find second partials, - we investigate what they tell us. -

    - -

    - Again we refer back to a function y=f(x) of a single variable. - The second derivative of f is - the derivative of the derivative, - or the rate of change of the rate of change. - The second derivative measures how much the derivative is changing. - If \fp'(x)\lt 0, - then the derivative is getting smaller (so the graph of f is concave down); - if \fp'(x) \gt 0, then the derivative is growing, - making the graph of f concave up. -

    - -

    - Now consider z=f(x,y). - Similar statements can be made about f_{xx} and f_{yy} as could be made about \fp'(x) above. - When taking derivatives with respect to x twice, - we measure how much f_x changes with respect to x. - If f_{xx}(x,y)\lt 0, - it means that as x increases, f_x decreases, - and the graph of f will be concave down - in the x-direction. - Using the analogy of standing in the rolling meadow used earlier in this section, - f_{xx} measures whether one's path is concave up/down when walking due east. -

    - -

    - Similarly, f_{yy} measures the concavity in the y-direction. - If f_{yy}(x,y) \gt 0, - then f_y is increasing with respect to y and the graph of f will be concave up in the y-direction. - Appealing to the rolling meadow analogy again, - f_{yy} measures whether one's path is concave up/down when walking due north. -

    - -

    - We now consider the mixed partials f_{xy} and f_{yx}. - The mixed partial f_{xy} measures how much f_x changes with respect to y. - Once again using the rolling meadow analogy, - f_{x} measures the slope if one walks due east. - Looking east, begin walking north - (side-stepping). - Is the path towards the east getting steeper? - If so, f_{xy} \gt 0. - Is the path towards the east not changing in steepness? - If so, then f_{xy}=0. - A similar thing can be said about f_{yx}: - consider the steepness of paths heading north while side-stepping to the east. -

    - -

    - The following example examines these ideas with concrete numbers and graphs. -

    - - - Understanding second partial derivatives - -

    - Let z=x^2-y^2+xy. - Evaluate the 6 first and second partial derivatives at - (-1/2,1/2) and interpret what each of these numbers mean. -

    -     - -

    - We find that: -

    - -

    - f_x(x,y) = 2x+y,f_y(x,y) = -2y+x,f_{xx}(x,y) = 2, - f_{yy}(x,y) = -2 and f_{xy}(x,y) = f_{yx}(x,y) = 1. - Thus at (-1/2,1/2) we have - - f_x(-1/2,1/2) = -1/2,\qquad f_y(-1/2,1/2) = -3/2 - . -

    - -

    - The slope of the tangent line at - (-1/2, 1/2, -1/4) in the direction of x is -1/2: - if one moves from that point parallel to the x-axis, - the instantaneous rate of change will be -1/2. - The slope of the tangent line - at this point in the direction of y is -3/2: - if one moves from this point parallel to the y-axis, - the instantaneous rate of change will be -3/2. - These tangents lines are graphed in and , - respectively, where the tangent lines are drawn in a solid line. -

    - -
    - Understanding the second partial derivatives in - -
    - - - - - A hyperbolic paraboloid, or saddle surface. A trace of constant x value is shown, which has the shape of a downward parabola. - -

    - The surface z=x^2-y^2+xy is a hyperbolic paraboloid, or saddle surface. - It is plotted along with the trace x=-\frac12. - This is a curve lying on the surface that has the shape of a downward-opening parabola. - The fact that this curve is concave down, when viewed along the x axis, - corresponds to the fact that f_{yy} is negative. -

    - -

    - At three points along this trace, tangent lines are drawn. - These lines are tangent to curves on the surface given by traces of constant y, - for three different values of y. Each one has negative slope, relative to x. - For larger values of y, the slope of these lines, relative to x, becomes less negative. - This suggests that the slope, given by f_x is increasing with y, - and this corresponds to the fact that f_{xy} is positive. -

    -     - - - - - //ASY file for figpartial63D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(2.5,8.4,3.4); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-2,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=x^2-y^2+xy - triple f(pair t) { - return (t.x,t.y,t.x^2-t.y^2+t.x*t.y); - } - surface s=surface(f,(-1,-1),(1,1),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the point (-0.5,0.5,-0.25) trace for x=-0.5 - dotfactor=3;dot((-0.5,0.5,-0.25)); - triple g(real t) {return (-0.5,t,0.25-t^2-0.5*t);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the tangent lines in x direction, slope -0.5 - // L=(-0.5,0.5,-0.25)+t(1,0,-0.5) for t=-1,1 - draw((-1.5,0.5,0.25)--(0.5,0.5,-0.75),bluepen); - - //Draw the tangent lines in x direction, slope -0.75 and dot - // L=(-0.5,0.25,0.065)+t(1,0,-0.75) for t=-1,1 - dot((-0.5,0.25,0.065)); - draw((-1.5,0.25,0.815)--(0.5,0.25,-0.685),dashed+bluepen); - - //Draw the tangent lines in x direction, slope -0.25 and dot - // L=(-0.5,0.75,-0.6875)+t(1,0,-0.25) for t=-1,1 - dot((-0.5,0.75,-0.6875)); - draw((-1.5,0.75,-0.4375)--(0.5,0.75,-0.9375),dashed+bluepen); - - - - -
    - -
    - - - - - A hyperbolic paraboloid, or saddle surface. A trace of constant y value is shown, which has the shape of an upward parabola. - -

    - The surface z=x^2-y^2+xy is a hyperbolic paraboloid, or saddle surface. - It is plotted along with the trace y=\frac12. - This is a curve lying on the surface that has the shape of an upward-opening parabola. - The fact that this curve is concave up, when viewed along the y axis, - corresponds to the fact that f_{xx} is positive. -

    - -

    - At three points along this trace, tangent lines are drawn. - These lines are tangent to curves on the surface given by traces of constant x, - for three different values of x. Each one has negative slope, relative to y. - For larger values of x, the slope of these lines, relative to y, becomes less negative. - This suggests that the slope, given by f_y is increasing with x, - and this corresponds to the fact that f_{yx} is positive. -

    -     - - - - - //ASY file for figpartial6b3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(2.5,8.4,3.); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-2,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=x^2-y^2+xy - triple f(pair t) { - return (t.x,t.y,t.x^2-t.y^2+t.x*t.y); - } - surface s=surface(f,(-1,-1),(1,1),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the point (-0.5,0.5,-0.25) and trace for y=0.5 - dotfactor=3;dot((-0.5,0.5,-0.25)); - triple g(real t) {return (t,0.5,t^2-0.25+0.5*t);} - path3 mypath=graph(g,-1,1,operator ..); draw(mypath,redpen); - - //Draw the tangent lines in y direction, slope -1.5 - // L=(-0.5,0.5,-0.25)+t(0,1,-1.5) for t=-1,1 - draw((-0.5,-0.3,.95)--(-0.5,1.3,-1.45),bluepen); - - //Draw the tangent lines in y direction, slope -1.75 and dot - // L=(-0.75,0.5,-0.0625)+t(0,1,-1.75) for t=-1,1 - dot((-0.75,0.5,-0.0625)); - draw((-0.75, -0.3, 1.3375)--(-0.75, 1.3, -1.4625),dashed+bluepen); - - //Draw the tangent lines in y direction, slope -1.25 and dot - // L=(-0.25,0.5,-0.3125)+t(0,1,-1.25) for t=-1,1 - dot((-0.25,0.5,-0.3125)); - draw((-0.25, -0.3, 0.6875)--(-0.25, 1.3, -1.3125),dashed+bluepen); - - - - -
    -     -
    - -

    - Now consider only . - Three directed tangent lines are drawn - (two are dashed), - each in the direction of x; - that is, each has a slope determined by f_x. - Note how as y increases, - the slope of these lines get closer to 0. - Since the slopes are all negative, - getting closer to 0 means the slopes are increasing. - The slopes given by f_x are increasing as y increases, - meaning f_{xy} must be positive. -

    - -

    - Since f_{xy}=f_{yx}, - we also expect f_y to increase as x increases. - Consider where again three directed tangent lines are drawn, - this time each in the direction of y with slopes determined by f_y. - As x increases, the slopes become less steep - (closer to 0). - Since these are negative slopes, - this means the slopes are increasing. -

    - -

    - Thus far we have a visual understanding of f_x, - f_y, and f_{xy}=f_{yx}. - We now interpret f_{xx} and f_{yy}. - In , - we see a curve drawn where x is held constant at x=-1/2: - only y varies. - This curve is clearly concave down, - corresponding to the fact that f_{yy}\lt 0. - In part of the figure, - we see a similar curve where y is constant and only x varies. - This curve is concave up, corresponding to the fact that f_{xx} \gt 0. -

    -     -     -     - - - Partial Derivatives and Functions of Three Variables -

    - The concepts underlying partial derivatives can be easily extend to more than two variables. - We give some definitions and examples in the case of three variables and trust the reader can extend these definitions to more variables if needed. -

    - - - Partial Derivatives with Three Variables - -

    - Let w=f(x,y,z) be a continuous function on a set D in \mathbb{R}^3. -

    - -

    - The partial derivative of f with respect to x is: - - f_x(x,y,z) = \lim_{h\to 0} \frac{f(x+h,y,z)-f(x,y,z)}{h} - . -

    - -

    - Similar definitions hold for - f_y(x,y,z) and f_z(x,y,z). - derivativepartial - partial derivative -

    -     -     - -

    - By taking partial derivatives of partial derivatives, - we can find second partial derivatives of f with respect to z then y, - for instance, just as before. -

    - - - Partial derivatives of functions of three variables - -

    - For each of the following, - find f_x, f_y, f_z, f_{xz}, f_{yz}, - and f_{zz}. -

    - -

    -

      -
    1. f(x,y,z) = x^2y^3z^4+x^2y^2+x^3z^3+y^4z^4
      2. - -
    2. f(x,y,z) = x\sin(yz)
      3. -
    -

    -     - -

    -

      -
    1. -

      - - f_x(x,y,z) \amp = 2xy^3z^4+2xy^2+3x^2z^3 - f_y(x,y,z) \amp = 3x^2y^2z^4+2x^2y+4y^3z^4 - f_z(x,y,z) \amp = 4x^2y^3z^3+3x^3z^2+4y^4z^3 - f_{xz}(x,y,z) \amp = 8xy^3z^3+9x^2z^2 - f_{yz}(x,y,z) \amp = 12x^2y^2z^3+16y^3z^3 - f_{zz}(x,y,z) \amp = 12x^2y^3z^2+6x^3z+12y^4z^2 - -

      -
      2. - -
    2. -

      - f_x = \sin(yz); f_y = xz\cos(yz); f_z = xy\cos(yz), and - - f_{xz}(x,y,z) \amp = y\cos(yz) - f_{yz}(x,y,z) \amp = x\cos(yz) - xyz\sin(yz) - f_{zz}(x,y,z) \amp = -xy^2\sin(yz) - -

      -
      3. -
    -

    -     -     -     - - - Higher Order Partial Derivatives -

    - We can continue taking partial derivatives of partial derivatives of partial derivatives of ; - we do not have to stop with second partial derivatives. - These higher order partial derivatives do not have a tidy graphical interpretation; - nevertheless they are not hard to compute and worthy of some practice. - partial derivativehigh order -

    - -

    - We do not formally define each higher order derivative, - but rather give just a few examples of the notation. - - f_{xyx}(x,y) \amp = \frac{\partial}{\px}\left(\frac{\partial}{\py}\left(\frac{\pf}{\px}\right)\right) \text{ and } - f_{xyz}(x,y,z) \amp =\frac{\partial}{\partial z}\left(\frac{\partial}{\py}\left(\frac{\pf}{\px}\right)\right) - . -

    - - - Higher order partial derivatives - -

    -

      -
    1. -

      - Let f(x,y) = x^2y^2+\sin(xy). - Find f_{xxy} and f_{yxx}. -

      -
      2. - -
    2. -

      - Let f(x,y,z) = x^3e^{xy}+\cos(z). - Find f_{xyz}. -

      -
      3. -
    -

    -

    - - -

    -

      -
    1. -

      - To find f_{xxy}, we first find f_x, - then f_{xx}, then f_{xxy}: - - f_x(x,y) \amp = 2xy^2+y\cos(xy) - f_{xx}(x,y) \amp = 2y^2-y^2\sin(xy) - f_{xxy}(x,y) \amp = 4y-2y\sin(xy) - xy^2\cos(xy) - . - To find f_{yxx}, we first find f_y, - then f_{yx}, then f_{yxx}: - - f_y(x,y) \amp = 2x^2y+x\cos(xy) - f_{yx}(x,y) \amp = 4xy + \cos(xy) - xy\sin(xy) - f_{yxx}(x,y) \amp = 4y-y\sin(xy) - \big(y\sin(xy) + xy^2\cos(xy)\big) - \amp = 4y-2y\sin(xy)-xy^2\cos(xy) - . - Note how f_{xxy} = f_{yxx}. -

      -
      2. - -
    2. -

      - To find f_{xyz}, we find f_x, - then f_{xy}, then f_{xyz}: - - f_x(x,y,z) \amp = 3x^2e^{xy}+ x^3ye^{xy} - f_{xy}(x,y,z) \amp= 3x^3e^{xy}+x^3e^{xy}+x^4ye^{xy} - \amp = 4x^3e^{xy}+x^4ye^{xy} - f_{xyz}(x,y,z) \amp = 0 - . -

      -
      3. -
    -

    - -     - -

    - In the previous example we saw that f_{xxy} = f_{yxx}; - this is not a coincidence. - While we do not state this as a formal theorem, - as long as each partial derivative is continuous, - it does not matter the order in which the partial derivatives are taken. - For instance, f_{xxy} = f_{xyx} = f_{yxx}. -

    - -

    - This can be useful at times. - Had we known this, - the second part of - would have been much simpler to compute. - Instead of computing f_{xyz} in the x, - y then z orders, - we could have applied the z, - then x then y order - (as f_{xyz} = f_{zxy}). - It is easy to see that f_z = -\sin(z); - then f_{zx} and f_{zxy} are clearly 0 as f_z does not contain an x or y. -

    - -

    - A brief review of this section: - partial derivatives measure the instantaneous rate of change of a multivariable function with respect to one variable. - With z=f(x,y), - the partial derivatives f_x and f_y measure the instantaneous rate of change of z when moving parallel to the x- and y-axes, - respectively. - How do we measure the rate of change at a point when we do not move parallel to one of these axes? - What if we move in the direction given by the vector \la 2,1\ra? - Can we measure that rate of change? - The answer is, of course, yes, we can. - This is the topic of . - First, we need to define what it means for a function of two variables to be - differentiable. -

    -     - - - - Terms and Concepts - - - -

    - What is the difference between a constant and a coefficient? -

    -     - - - -

    - A constant is a number that is added or subtracted in an expression; - a coefficient is a number that is being multiplied by a nonconstant function. -

    - -

    - Coefficients are often constant, but they don't have to be. -

    -     - -     - - - - -

    - Given a function f(x,y), - explain in your own words how to compute f_x. -

    -     - - - -

    - Answers will vary; - each should include something about treating y as a constant or a coefficient. -

    -     - -     - - - - - -

    - In the mixed partial fraction f_{xy}, - which is computed first, f_x or f_y? -

    - - -     - - - -

    - f_x -

    -     -     - - -

    - f_y -

    -     -     -     - -     - - - - - - -

    - In the mixed partial fraction \frac{\partial^2f}{\partial x\partial y}, - which is computed first, f_x or f_y? -

    - - -     - - - -

    - f_x -

    -     -     - - -

    - f_y -

    -     -     -     - -     -     - - - Problems - - - -

    - Evaluate f_x(x,y) and - f_y(x,y) at the indicated point. -

    -     - - - - -

    - f(x,y) = x^2y-x+2y+3 at (1,2) -

    -     - -

    - f_x=2xy-1, f_y=x^2+2 -

    - -

    - f_x(1,2) = 3, f_y(1,2) = 3 -

    -     - -     - - - - - $fx=0; - $fy=0; - - - -

    - f(x,y) = x^3-3x+y^2-6y at (-1,3). -

    - - - Find f_x(-1,3). - -

    - -

    - - - Find f_y(-1,3). - -

    - -

    -     -     -     - - - - -

    - f(x,y) = \sin(y) \cos(x) at (\pi/3,\pi/3) -

    -     - -

    - f_x=-\sin(x) \sin(y), f_y=\cos(x) \cos(y) -

    - -

    - f_x(\pi/3,\pi/3) = -3/4, - f_y(\pi/3,\pi/3) = 1/4 -

    -     - -     - - - - - Context("Fraction"); - $fx=Fraction(-1,2); - $fy=Fraction(-1,3); - - - -

    - f(x,y) = \ln(xy) at (-2,-3) - Find: -

    - - - Find f_x(-2,-3). - -

    - -

    - - - Find f_y(-2,-3). - -

    - -

    -     -     -     - -     - - - -

    - Find f_x, f_y, - f_{xx}, f_{yy}, - f_{xy} and f_{yx}. -

    -     - - - - -

    - f(x,y) = x^2y+3x^2+4y-5 -

    -     - -

    - f_x=2xy+6x, f_y=x^2+4 -

    - -

    - f_{xx}=2y+6, f_{yy}=0 -

    - -

    - f_{xy}=2x, f_{yx}=2x -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $fx=Compute("3x^2+6xy+3y^2"); - $fy=Compute("3x^2+6xy+3y^2"); - $fxx=Compute("6x+6y"); - $fxy=Compute("6x+6y"); - $fyx=Compute("6x+6y"); - $fyy=Compute("6x+6y"); - - - -

    - f(x,y) = y^3+3xy^2+3x^2y+x^3 -

    - - - Find f_{x}(x,y) - -

    - -

    - - - Find f_{y}(x,y). - -

    - -

    - - - Find f_{xx}(x,y). - -

    - -

    - - - Find f_{xy}(x,y). - -

    - -

    - - - Find f_{yx}(x,y) - -

    - -

    - - - Find f_{yy}(x,y). - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) = \frac xy -

    -     - -

    - f_x=1/y, f_y=-x/y^2 -

    - -

    - f_{xx}=0, f_{yy}=2x/y^3 -

    - -

    - f_{xy}=-1/y^2, f_{yx}=-1/y^2 -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $fx=Compute("-4/(x^2y)"); - $fy=Compute("-4/(xy^2)"); - $fxx=Compute("8/(x^3y)"); - $fxy=Compute("4/(x^2y^2)"); - $fyx=Compute("4/(x^2y^2)"); - $fyy=Compute("8/(xy^3)"); - - - -

    - f(x,y) = \frac{4}{xy} -

    - - - Find f_{x}(x,y) - -

    - -

    - - - Find f_{y}(x,y). - -

    - -

    - - - Find f_{xx}(x,y). - -

    - -

    - - - Find f_{xy}(x,y). - -

    - -

    - - - Find f_{yx}(x,y) - -

    - -

    - - - Find f_{yy}(x,y). - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) = e^{x^2+y^2} -

    -     - -

    - f_x=2xe^{x^2+y^2}, f_y=2ye^{x^2+y^2} -

    - -

    - f_{xx}=2e^{x^2+y^2}+4x^2e^{x^2+y^2}, - f_{yy}=2e^{x^2+y^2}+4y^2e^{x^2+y^2} -

    - -

    - f_{xy}=4xye^{x^2+y^2}, - f_{yx}=4xye^{x^2+y^2} -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $fx=Compute("e^(x+2y)"); - $fy=Compute("2e^(x+2y)"); - $fxx=Compute("e^(x+2y)"); - $fxy=Compute("2e^(x+2y)"); - $fyx=Compute("2e^(x+2y)"); - $fyy=Compute("4e^(x+2y)"); - - - -

    - f(x,y) = e^{x+2y} -

    - - - Find f_{x}(x,y) - -

    - -

    - - - Find f_{y}(x,y). - -

    - -

    - - - Find f_{xx}(x,y). - -

    - -

    - - - Find f_{xy}(x,y). - -

    - -

    - - - Find f_{yx}(x,y) - -

    - -

    - - - Find f_{yy}(x,y). - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) = \sin(x) \cos(y) -

    -     - -

    - f_x=\cos(x) \cos(y), f_y=-\sin(x) \sin(y) -

    - -

    - f_{xx}=-\sin(x) \cos(y), - f_{yy}=-\sin(x) \cos(y) -

    - -

    - f_{xy}=-\sin(y) \cos(x), - f_{yx}=-\sin(y) \cos(x) -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $fx=Compute("3(x+y)^2"); - $fy=Compute("3(x+y)^2"); - $fxx=Compute("6(x+y)"); - $fxy=Compute("6(x+y)"); - $fyx=Compute("6(x+y)"); - $fyy=Compute("6(x+y)"); - - - -

    - f(x,y) = (x+y)^3 -

    - - - Find f_{x}(x,y) - -

    - -

    - - - Find f_{y}(x,y). - -

    - -

    - - - Find f_{xx}(x,y). - -

    - -

    - - - Find f_{xy}(x,y). - -

    - -

    - - - Find f_{yx}(x,y) - -

    - -

    - - - Find f_{yy}(x,y). - -

    - -

    -     -     -     - - - - -

    - f(x,y) = \cos(5xy^3) -

    -     - -

    - f_x=-5y^3\sin(5xy^3), f_y=-15xy^2\sin(5xy^3) -

    - -

    - f_{xx}=-25y^6\cos(5xy^3), - f_{yy}=-225x^2y^4\cos(5xy^3)-30xy\sin(5xy^3) -

    - -

    - f_{xy}=-75xy^5\cos(5xy^3)-15y^2\sin(5xy^3), - f_{yx}=-75xy^5\cos(5xy^3)-15y^2\sin(5xy^3) -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $fx=Compute("10x cos(5x^2+2y^3)"); - $fy=Compute("6y^2 cos(5x^2+2y^3)"); - $fxx=Compute("10 cos(5x^2+2y^3)-100x^2 sin(5x^2+2y^3)"); - $fxy=Compute("-60xy^2 sin(5x^2+2y^3)"); - $fyx=Compute("-60xy^2 sin(5x^2+2y^3)"); - $fyy=Compute("12y cos(5x^2+2y^3)-36y^4 sin(5x^2+2y^3)"); - - - -

    - f(x,y) = \sin\mathopen{}\left(5x^2+2y^3\right)\mathclose{} -

    - - - Find f_{x}(x,y) - -

    - -

    - - - Find f_{y}(x,y). - -

    - -

    - - - Find f_{xx}(x,y). - -

    - -

    - - - Find f_{xy}(x,y). - -

    - -

    - - - Find f_{yx}(x,y) - -

    - -

    - - - Find f_{yy}(x,y). - -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real'); - $fx=Compute("2y^2/sqrt(4xy^2+1)"); - $fy=Compute("4xy/sqrt(4xy^2+1)"); - $fxx=Compute("-4y^4/sqrt(4xy^2+1)^3"); - $fxy=Compute("-8xy^3/sqrt(4xy^2+1)^3+4y/sqrt(4xy^2+1)"); - $fyx=Compute("-8xy^3/sqrt(4xy^2+1)^3+4y/sqrt(4xy^2+1)"); - $fyy=Compute("-16x^2y^2/sqrt(4xy^2+1)^3+4x/sqrt(4xy^2+1)"); - - - -

    - f(x,y) = \sqrt{4xy^2+1} -

    - - - Find f_{x}(x,y) - -

    - -

    - - - Find f_{y}(x,y). - -

    - -

    - - - Find f_{xx}(x,y). - -

    - -

    - - - Find f_{xy}(x,y). - -

    - -

    - - - Find f_{yx}(x,y) - -

    - -

    - - - Find f_{yy}(x,y). - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) =(2x+5y)\sqrt{y} -

    -     - -

    - f_x=2\sqrt{y}, f_y=5\sqrt{y}+\frac{2x+5y}{2\sqrt{y}} -

    - -

    - f_{xx}=0, - f_{yy}=\frac{5}{\sqrt{y}}-\frac{2x+5y}{4y^{3/2}} -

    - -

    - f_{xy}=\frac{1}{\sqrt{y}}, - f_{yx}=\frac{1}{\sqrt{y}} -

    -     - -     - - - - -

    - \ds f(x,y) =\frac{1}{x^2+y^2+1} -

    -     - -

    - f_x=-\frac{2x}{(x^2+y^2+1)^2}, - f_y=-\frac{2y}{(x^2+y^2+1)^2} -

    - -

    - f_{xx}=\frac{8x^2}{(x^2+y^2+1)^3}-\frac{2}{(x^2+y^2+1)^2}, - f_{yy}=\frac{8y^2}{(x^2+y^2+1)^3}-\frac{2}{(x^2+y^2+1)^2} -

    - -

    - f_{xy}=\frac{8xy}{(x^2+y^2+1)^3}, - f_{yx}=\frac{8xy}{(x^2+y^2+1)^3} -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $fx=Compute("5"); - $fy=Compute("-17"); - $fxx=Compute("0"); - $fxy=Compute("0"); - $fyx=Compute("0"); - $fyy=Compute("0"); - - - -

    - f(x,y) = 5x-17y -

    - - - Find f_{x}(x,y) - -

    - -

    - - - Find f_{y}(x,y). - -

    - -

    - - - Find f_{xx}(x,y). - -

    - -

    - - - Find f_{xy}(x,y). - -

    - -

    - - - Find f_{yx}(x,y) - -

    - -

    - - - Find f_{yy}(x,y). - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) =3x^2+1 -

    -     - -

    - f_x=6x, f_y=0 -

    - -

    - f_{xx}=6, f_{yy}=0 -

    - -

    - f_{xy}=0, f_{yx}=0 -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $fx=Compute("2x/(x^2+y)"); - $fy=Compute("1/(x^2+y)"); - $fxx=Compute("-4x^2/(x^2+y)^2+2/(x^2+y)"); - $fxy=Compute("-2x/(x^2+y)^2"); - $fyx=Compute("-2x/(x^2+y)^2"); - $fyy=Compute("-1/(x^2+y)^2"); - - - -

    - f(x,y) = \ln(x^2+y) -

    - - - Find f_{x}(x,y) - -

    - -

    - - - Find f_{y}(x,y). - -

    - -

    - - - Find f_{xx}(x,y). - -

    - -

    - - - Find f_{xy}(x,y). - -

    - -

    - - - Find f_{yx}(x,y) - -

    - -

    - - - Find f_{yy}(x,y). - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) =\frac{\ln(x) }{4y} -

    -     - -

    - f_x=\frac{1}{4xy}, f_y=-\frac{\ln(x) }{4y^2} -

    - -

    - f_{xx}=-\frac{1}{4x^2y}, - f_{yy}=\frac{\ln(x) }{2y^3} -

    - -

    - f_{xy}=-\frac{1}{4xy^2}, - f_{yx}=-\frac{1}{4xy^2} -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $fx=Compute("5e^x sin(y)"); - $fy=Compute("5e^x cos(y)"); - $fxx=Compute("5e^x sin(y)"); - $fxy=Compute("5e^x cos(y)"); - $fyx=Compute("5e^x cos(y)"); - $fyy=Compute("-5e^x sin(y)"); - - - -

    - f(x,y) = 5e^x\sin(y)+9 -

    - - - Find f_{x}(x,y) - -

    - -

    - - - Find f_{y}(x,y). - -

    - -

    - - - Find f_{xx}(x,y). - -

    - -

    - - - Find f_{xy}(x,y). - -

    - -

    - - - Find f_{yx}(x,y) - -

    - -

    - - - Find f_{yy}(x,y). - -

    - -

    -     -     -     - -     - - - -

    - Form a function f(x,y) such that f_x and f_y match those given. -

    -     - - - - -

    - f_x = \sin(y) +1,f_y = x\cos(y) -

    -     - -

    - f(x,y) = x\sin(y) + x+ C, - where C is any constant. -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $f=Compute("1/2x^2+xy+1/2y^2"); - $fev=$f->cmp(upToConstant=>1); - - - -

    - f_x = x+y and f_y = x+y -

    - -

    - -

    -     -     -     - - - - -

    - f_x = 6xy-4y^2,f_y = 3x^2-8xy+2 -

    -     - -

    - f(x,y) = 3x^2y-4xy^2+2y +C, - where C is any constant. -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $f=Compute("ln(x^2+y^2)"); - $fev=$f->cmp(upToConstant=>1); - - - -

    - f_x = \frac{2x}{x^2+y^2} and f_y = \frac{2y}{x^2+y^2} -

    - -

    - -

    -     -     -     - -     - - - -

    - Find f_x, - f_y, f_z, f_{yz} and f_{zy}. -

    -     - - - - -

    - \ds f(x,y,z) = x^2e^{2y-3z} -

    -     - -

    - f_x = 2xe^{2y-3z}, - f_y = 2x^2e^{2y-3z}, f_z = -3x^2e^{2y-3z} -

    - -

    - f_{yz} = -6x^2e^{2y-3z}, - f_{zy} = -6x^2e^{2y-3z} -

    -     - -     - - - - - Context()->variables->add(y=>'Real',z=>'Real'); - $fx=Compute("3x^2y^2+3x^2z"); - $fy=Compute("2x^3y+2yz"); - $fz=Compute("x^3+y^2"); - $fyz=Compute("2y"); - $fzy=Compute("2y"); - - - -

    - f(x,y,z) = x^3y^2+x^3z+y^2z -

    - - - Find f_{x}(x,y,z). - -

    - -

    - - - Find f_{y}(x,y,z). - -

    - -

    - - - Find f_{z}(x,y,z). - -

    - -

    - - - Find f_{yz}(x,y,z). - -

    - -

    - - - Find f_{zy}(x,y,z). - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y,z) = \frac{3x}{7y^2z} -

    -     - -

    - f_x = \frac{3}{7y^2z}, - f_y = -\frac{6x}{7y^3z}, - f_z = -\frac{3x}{7y^2z^2} -

    - -

    - f_{yz} = \frac{6x}{7y^3z^2}, - f_{zy} = \frac{6x}{7y^3z^2} -

    -     - -     - - - - - Context()->variables->add(y=>'Real',z=>'Real'); - $fx=Compute("1/x"); - $fy=Compute("1/y"); - $fz=Compute("1/z"); - $fyz=Compute("0"); - $fzy=Compute("0"); - - - -

    - f(x,y,z) = \ln(xyz) -

    - - - Find f_{x}(x,y,z). - -

    - -

    - - - Find f_{y}(x,y,z). - -

    - -

    - - - Find f_{z}(x,y,z). - -

    - -

    - - - Find f_{yz}(x,y,z). - -

    - -

    - - - Find f_{zy}(x,y,z). - -

    - -

    -     -     -     - -     -     -     -
    -
    - Differentiability and the Total Differential - -

    - We studied differentials - in , - where - states that if y=f(x) and f is differentiable, - then dy=\fp(x)dx. - One important use of this differential is in Integration by Substitution. - Another important application is approximation. - Let \dx = dx represent a change in x. - When dx is small, dy\approx \dy, - the change in y resulting from the change in x. - Fundamental in this understanding is this: - as dx gets small, - the difference between \dy and dy goes to 0. - Another way of stating this: - as dx goes to 0, the error - in approximating \dy with dy goes to 0. -

    - -

    - We extend this idea to functions of two variables. - Let z=f(x,y), and let - \dx = dx and \dy=dy represent changes in x and y, - respectively. - Let \ddz = f(x+dx,y+dy) - f(x,y) be the change in z over the change in x and y. - Recalling that f_x and f_y give the instantaneous rates of z-change in the x- and y-directions, - respectively, - we can approximate \ddz with dz = f_xdx+f_ydy; - in words, the total change in z is approximately the change caused by changing x plus the change caused by changing y. - In a moment we give an indication of whether or not this approximation is any good. - First we give a name to dz. -

    -     - - - The Total Differential - - - Total Differential - -

    - Let z=f(x,y) be continuous on a set S. - Let dx and dy represent changes in x and y, - respectively. - Where the partial derivatives f_x and f_y exist, - the total differential of z is - total differential - partial derivativetotal differential - - dz = f_x(x,y)\,dx + f_y(x,y)\,dy - . -

    -     -     - - - Finding the total differential - -

    - Let z = x^4e^{3y}. - Find dz. -

    -     - -

    - We compute the partial derivatives: - f_x = 4x^3e^{3y} and f_y = 3x^4e^{3y}. - Following , we have - - dz = 4x^3e^{3y}dx+3x^4e^{3y}dy - . -

    -     -     - - - -

    - We can approximate \ddz with dz, - but as with all approximations, there is error involved. - A good approximation is one in which the error is small. - At a given point (x_0,y_0), - let E_x and E_y be functions of dx and dy such that - E_xdx+E_ydy describes this error. - Then - - \ddz \amp = dz + E_xdx+ E_ydy - \amp = f_x(x_0,y_0)dx+f_y(x_0,y_0)dy + E_xdx+E_ydy - . -

    - -

    - If the approximation of \ddz by dz is good, - then as dx and dy get small, - so does E_xdx+E_ydy. - The approximation of \ddz by dz is even better if, - as dx and dy go to 0, so do E_x and E_y. - This leads us to our definition of differentiability. -

    - - - Multivariable Differentiability - -

    - Let z=f(x,y) be defined on a set S containing (x_0,y_0) where - f_x(x_0,y_0) and f_y(x_0,y_0) exist. - Let dz be the total differential of z at (x_0,y_0), - let \ddz = f(x_0+dx,y_0+dy) - f(x_0,y_0), - and let E_x and E_y be functions of dx and dy such that - - \ddz = dz + E_xdx + E_ydy - . -

    - -

    -

      -
    1. -

      - We say f is differentiable at (x_0,y_0) if, - given \varepsilon \gt 0, - there is a \delta \gt 0 such that if \norm{\la dx,dy\ra} \lt \delta, - then \norm{\la E_x,E_y\ra} \lt \varepsilon. - That is, as dx and dy go to 0, so do E_x and E_y. -

      -
      2. - -
    2. -

      - We say f is differentiable on S - if f is differentiable at every point in S. - If f is differentiable on \mathbb{R}^2, - we say that f is differentiable everywhere. - differentiable - derivativemultivariable differentiability - multivariable functiondifferentiability -

      -
      3. -
    -

    -     -     - - - - - Showing a function is differentiable - -

    - Show f(x,y) = xy+3y^2 is differentiable using . -

    -     - -

    - We begin by finding f(x+dx,y+dy), - \ddz, f_x and f_y. - - f(x+dx,y+dy) \amp = (x+dx)(y+dy) + 3(y+dy)^2 - \amp = xy + xdy+ydx+dxdy + 3y^2+6ydy+3dy^2 - . -

    - -

    - \ddz = f(x+dx,y+dy) - f(x,y), so - - \ddz = xdy + ydx + dxdy + 6ydy+3dy^2 - . -

    - -

    - It is straightforward to compute f_x = y and f_y = x+6y. - Consider once more \ddz: - - \ddz \amp = xdy + ydx + dxdy + 6ydy+3dy^2 \qquad \text{ (now reorder) } - \amp = ydx + xdy+6ydy+ dxdy + 3dy^2 - \amp = \underbrace{(y)}_{f_x}dx + \underbrace{(x+6y)}_{f_y}dy + \underbrace{(dy)}_{E_x}dx+\underbrace{(3dy)}_{E_y}dy - \amp = f_xdx + f_ydy + E_xdx+E_ydy - . -

    - -

    - With E_x = dy and E_y = 3dy, - it is clear that as dx and dy go to 0, E_x and E_y also go to 0. - Since this did not depend on a specific point (x_0,y_0), - we can say that f(x,y) is differentiable for all pairs (x,y) in \mathbb{R}^2, - or, equivalently, that f is differentiable everywhere. -

    -     -     - - - -

    - Our intuitive understanding of differentiability of functions y=f(x) of one variable was that the graph of f was smooth. - A similar intuitive understanding of functions - z=f(x,y) of two variables is that the surface defined by f is also smooth, - not containing cusps, edges, breaks, etc. - The following theorem states that differentiable functions are continuous, - followed by another theorem that provides a - more tangible way of determining whether a great number of functions are differentiable or not. -

    - - - Continuity and Differentiability of Multivariable Functions - -

    - Let z=f(x,y) be defined on a set S containing (x_0,y_0). - If f is differentiable at (x_0,y_0), - then f is continuous at (x_0,y_0). - multivariable functiondifferentiability - multivariable functioncontinuity -

    -     -     - - - Differentiability of Multivariable Functions - -

    - Let z=f(x,y) be defined on a set S. - If f_x and f_y are both continuous on S, - then f is differentiable on S. - multivariable functiondifferentiability -

    -     -     - -

    - The above theorems assure us that essentially all functions that we see in the course of our studies here are differentiable - (and hence continuous) - on their natural domains. - There is a difference between - and , though: - it is possible for a function f to be differentiable yet f_x and/or f_y is - not continuous. - Such strange behavior of functions is a source of delight for many mathematicians, - but in practical situations we want to avoid it, leading to the following definition. -

    - - - Continouously Differentiable Function - - -

    - Let U be an open subset of \R^2. - We say that a function f is continuously differentiable on U - if f_x and f_y are defined and continuous at each point in U. - continuously differentiable -

    - -

    - A similar statement applies for functions of three variables in \R^3. -

    -     -     - -

    - When f_x and f_y exist at a point but are not continuous at that point, - we need to use other methods to determine whether or not f is differentiable at that point. -

    - -

    - For instance, consider the function - - f(x,y) = \begin{cases} - \frac{xy}{x^2+y^2} \amp (x,y)\neq (0,0) \\ - 0 \amp (x,y) = (0,0) - \end{cases} - . -

    - -

    - We can find f_x(0,0) and - f_y(0,0) using : - - f_x(0,0) \amp = \lim_{h\to 0} \frac{f(0+h,0) - f(0,0)}{h} - \amp = \lim_{h\to 0} \frac{0}{h^2} = 0; - f_y(0,0) \amp = \lim_{h\to 0} \frac{f(0,0+h) - f(0,0)}{h} - \amp = \lim_{h\to 0} \frac{0}{h^2} = 0 - . -

    - -

    - Both f_x and f_y exist at (0,0), - but they are not continuous at (0,0), as - - f_x(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)^2} \qquad \text{ and } \qquad f_y(x,y) = \frac{x(x^2-y^2)}{(x^2+y^2)^2} - - are not continuous at (0,0). (Take the limit of f_x as - (x,y)\to(0,0) along the x- and y-axes; - they give different results.) So even though f_x and f_y exist - at every point in the xy-plane, they are not continuous. - Therefore it is possible, - by , - for f to not be differentiable. -

    - -

    - Indeed, it is not. - One can show that f is not continuous at (0,0) - (see ), - and by , - this means f is not differentiable at (0,0). -

    -     - - - Approximating with the Total Differential -

    - By the definition, - when f is differentiable dz is a good approximation for \ddz when dx and dy are small. - We give some simple examples of how this is used here. -

    - - - Approximating with the total differential - -

    - Let z = \sqrt{x}\sin(y). - Approximate f(4.1,0.8). -

    -     - -

    - Recognizing that \pi/4 \approx 0.785\approx 0.8, - we can approximate f(4.1,0.8) using f(4,\pi/4). - We can easily compute f(4,\pi/4) = \sqrt{4}\sin(\pi/4) = 2\left(\frac{\sqrt{2}}2\right) = \sqrt{2}\approx 1.414. - Without calculus, - this is the best approximation we could reasonably come up with. - The total differential gives us a way of adjusting this initial approximation to hopefully get a more accurate answer. -

    - -

    - We let \ddz = f(4.1,0.8) - f(4,\pi/4). - The total differential dz is approximately equal to \ddz, so - - f(4.1,0.8) - f(4,\pi/4) \approx dz \Rightarrow f(4.1,0.8) \approx dz + f(4,\pi/4) - . -

    - -

    - To find dz, we need f_x and f_y. - - f_x(x,y) \amp = \frac{\sin(y) }{2\sqrt{x}} \Rightarrow\amp - f_x(4,\pi/4) \amp = \frac{\sin(\pi) /4}{2\sqrt{4}} - \amp \amp \amp = \frac{\sqrt{2}/2}{4} = \sqrt{2}/8. - f_y(x,y) \amp = \sqrt{x}\cos(y) \Rightarrow\amp - f_y(4,\pi/4) \amp = \sqrt{4}\frac{\sqrt{2}}2 - \amp \amp \amp = \sqrt{2} - . -

    - -

    - Approximating 4.1 with 4 gives dx = 0.1; - approximating 0.8 with \pi/4 gives dy \approx 0.015. - Thus - - dz \amp = f_x(4,\pi/4)(0.1) + f_y(4,\pi/4)(0.015) - \amp = \frac{\sqrt{2}}8(0.1) + \sqrt{2}(0.015) - \amp \approx 0.039 - . -

    - -

    - Returning to Equation, we have - - f(4.1,0.8) \approx 0.039 + 1.414 = 1.4531 - . -

    - -

    - We, of course, - can compute the actual value of f(4.1,0.8) with a calculator; - the actual value, accurate to 5 places after the decimal, - is 1.45254. - Obviously our approximation is quite good. -

    -     -     - -

    - The point of the previous example was not - to develop an approximation method for known functions. - After all, we can very easily compute - f(4.1,0.8) using readily available technology. - Rather, it serves to illustrate how well this method of approximation works, - and to reinforce the following concept: -

    - -

    - New position = old position + amount of change, so -

    - -

    - New position \approx old position + approximate amount of change. -

    - -

    - In the previous example, - we could easily compute f(4,\pi/4) and could approximate the amount of z-change when computing f(4.1,0.8), - letting us approximate the new z-value. -

    - -

    - It may be surprising to learn that it is not uncommon to know the values of f, - f_x and f_y at a particular point without actually knowing the function f. - The total differential gives a good method of approximating f at nearby points. -

    - - - Approximating an unknown function - -

    - Given that f(2,-3) = 6, - f_x(2,-3) = 1.3 and f_y(2,-3) = -0.6, - approximate f(2.1,-3.03). -

    -     - -

    - The total differential approximates how much f changes from the point (2,-3) to the point (2.1,-3.03). - With dx = 0.1 and dy = -0.03, we have - - dz \amp = f_x(2,-3)dx + f_y(2,-3)dy - \amp = 1.3(0.1) + (-0.6)(-0.03) - \amp = 0.148 - . -

    - -

    - The change in z is approximately 0.148, - so we approximate f(2.1,-3.03)\approx 6.148. -

    -     -     -     - - - Tangent Plane Approximation -

    - Recall from that in one variable, - the essence of differentiability is the tangent line approximation. - This idea is emphasized in , - where we first introduced the differential. -

    - -

    - In we saw that the partial derivatives of a function f(x,y) - can be used to define the tangent plane to a graph z=f(x,y). - We will now see that this plane plays the same role for functions of two variables as the tangent line to a graph y=f(x) for a function of one variable. -

    - -

    - Recall from that for a function f(x), - when x is near c we have the linear approximation f(x)\approx \ell(x), - where - - \ell(x) = f(c)+f'(c)(x-c) - - is the linearization of f at c. If we set dx=\dx = x-c, - and evaluate the differential dy = f'(x)\,dx at c, - then we have - - \dy \amp = f(x)-f(c) - dy \amp = \ell(x)-f(c) - . -

    - -

    - Given the graph y=f(x), we know that y=\ell(x) gives the tangent line to the graph at c. - For the graph z=f(x,y) of a function of two variables, we similarly have the tangent plane - - z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) - - defined in , - suggesting that we define the two variable linearization - - \ell(x,y) = f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) - . -

    - -

    - Consider the total differential dz at (a,b): - - dz = f_x(a,b)\,dx + f_y(a,b)\,dy - . - If we assume that (x,y) is close to (a,b), - and set dx = x-a, dy = y-b, - then we have - - dz = f_x(a,b)\,dx + f_y(a,b)\,dy = f_x(a,b)(x-a)+f_y(a,b)(y-b) - . - Since \ell(a,b)=f(a,b), we have - \ell(x,y)-\ell(a,b) = dz, - which agrees with the one-variable situation, - and reinforces the concept of the differential as the linear change in a function. -

    - -

    - If we recast in the language of tangent planes, - we can more easily see the analogy with functions of a single variable. - We can now say that f(x,y) is differentiable at (a,b) if it has a valid tangent plane approximation at (a,b). - Note that f(x,y)-\ell(x,y) is equal to the error term E_x\,dx+E_y\,dy. -

    - -

    - By , we know that the tangent plane at (a,b,f(a,b)) - exists, and gives a good approximation to the graph z=f(x,y), - as long as the partial derivatives of f exist and are continuous at (a,b). -

    -     - - - Error/Sensitivity Analysis -

    - The total differential gives an approximation of the change in z given small changes in x and y. - We can use this to approximate error propagation; - that is, if the input is a little off from what it should be, - how far from correct will the output be? - We demonstrate this in an example. -

    - -

    - sensitivity analysis - total differentialsensitivity analysis -

    - - - Sensitivity analysis - -

    - A cylindrical steel storage tank is to be built that is 10ft tall and 4ft across in diameter. - It is known that the steel will expand/contract with temperature changes; - is the overall volume of the tank more sensitive to changes in the diameter or in the height of the tank? -

    -     - -

    - A cylindrical solid with height h and radius r has volume V = \pi r^2h. - We can view V as a function of two variables, r and h. - We can compute partial derivatives of V: - - \frac{\partial V}{\partial r} = V_r(r,h) = 2\pi rh \qquad \text{ and } \qquad \frac{\partial V}{\partial h} = V_h(r,h) = \pi r^2 - . -

    - -

    - The total differential is dV = (2\pi rh)dr + (\pi r^2)dh. - When h = 10 and r = 2, - we have dV = 40\pi dr + 4\pi dh. - Note that the coefficient of dr is 40\pi\approx 125.7; - the coefficient of dh is a tenth of that, - approximately 12.57. - A small change in radius will be multiplied by 125.7, whereas a small change in height will be multiplied by 12.57. - Thus the volume of the tank is more sensitive to changes in radius than in height. -

    -     -     - -

    - The previous example showed that the volume of a particular tank was more sensitive to changes in radius than in height. - Keep in mind that this analysis only applies to a tank of those dimensions. - A tank with a height of 1ft and radius of 5ft would be more sensitive to changes in height than in radius. -

    - -

    - One could make a chart of small changes in radius and height and find exact changes in volume given specific changes. - While this provides exact numbers, - it does not give as much insight as the error analysis using the total differential. -

    -     - - - Differentiability of Functions of Three Variables -

    - The definition of differentiability for functions of three variables is very similar to that of functions of two variables. - We again start with the total differential. -

    - - - Total Differential - -

    - Let w=f(x,y,z) be continuous on a set D. - Let dx, - dy and dz represent changes in x, - y and z, respectively. - Where the partial derivatives f_x, - f_y and f_z exist, - the total differential of w is - total differential - partial derivativetotal differential - - dw = f_x(x,y,z)\,dx + f_y(x,y,z)\,dy+f_z(x,y,z)\,dz - . -

    -     -     - -

    - This differential can be a good approximation of the change in w when w = f(x,y,z) is - differentiable. -

    - - - Multivariable Differentiability - -

    - Let w=f(x,y,z) be defined on a set D containing - (x_0,y_0,z_0) where f_x(x_0,y_0,z_0), - f_y(x_0,y_0,z_0) and f_z(x_0,y_0,z_0) exist. - Let dw be the total differential of w at (x_0,y_0,z_0), - let \Delta w = f(x_0+dx,y_0+dy,z_0+dz) - f(x_0,y_0,z_0), - and let E_x, - E_y and E_z be functions of dx, - dy and dz such that - differentiable - derivativemultivariable differentiability - multivariable functiondifferentiability - - \Delta w = dw + E_xdx + E_ydy + E_zdz - . -

    - -

    -

      -
    1. -

      - We say f is differentiable at (x_0,y_0,z_0) if, - given \varepsilon \gt 0, - there is a \delta \gt 0 such that if \norm{\la dx,dy,dz\ra} \lt \delta, - then \norm{\la E_x,E_y,E_z\ra} \lt \varepsilon. -

      -
      2. - -
    2. -

      - We say f is differentiable on B - if f is differentiable at every point in B. - If f is differentiable on \mathbb{R}^3, - we say that f is differentiable everywhere. -

      -
      3. -
    -

    -     -     - -

    - Just as before, - this definition gives a rigorous statement about what it means to be differentiable that is not very intuitive. - We follow it with a theorem similar to . -

    - - - Continuity and Differentiability of Functions of Three Variables - -

    - Let w=f(x,y,z) be defined on a set D containing (x_0,y_0,z_0). -

    - -

    -

      -
    1. -

      - If f is differentiable at (x_0,y_0,z_0), - then f is continuous at (x_0,y_0,z_0). -

      -
      2. - -
    2. -

      - If f_x, - f_y and f_z are continuous on D, - then f is differentiable on D. - multivariable functiondifferentiability - multivariable functioncontinuity -

      -
      3. -
    -

    -     -     - - - -

    - This set of definition and theorem extends to functions of any number of variables. - The theorem again gives us a simple way of verifying that most functions that we encounter are differentiable on their natural domains. -

    - -

    - This section has given us a formal definition of what it means for a functions to be differentiable, - along with a theorem that gives a more accessible understanding. - The following sections return to notions prompted by our study of partial derivatives that make use of the fact that most functions we encounter are differentiable. -

    -     - - - - Terms and Concepts - - - - -

    - If f(x,y) is differentiable on S, - the f is continuous on S. -

    -     - -     - - - - -

    - If f_x and f_y are continuous on S, - then f is differentiable on S. -

    -     - -     - - - - -

    - If z=f(x,y) is differentiable, - then the change in z over small changes dx and dy in x and y is approximately dz. -

    -     - -     - - - - -

    - Finish the sentence: - The new z-value is approximately the old z-value plus the approximate . -

    -     - - - - change|amount of change|change in z - - - - -     -     - - - Problems - - - -

    - Find the total differential dz. -

    -     - - - - -

    - z = x\sin(y) + x^2 -

    -     - -

    - dz = (\sin(y) + 2x)dx + (x\cos(y) )dy -

    -     - -     - - - - -

    - z = (2x^2+3y)^2 -

    -     - -

    - dz = 8x(2x^2+3y)dx + 6(2x^2+3y)dy -

    -     - -     - - - - -

    - z = 5x-7y -

    -     - -

    - dz = 5dx -7dy -

    -     - -     - - - - -

    - z = xe^{x+y} -

    -     - -

    - dz = (e^{x+y}+xe^{x+y})dx +xe^{x+y}dy -

    -     - -     - -     - - - -

    - A function f(x,y) is given. - Give the indicated approximation using the total differential. -

    -     - - - - -

    - f(x,y) = \sqrt{x^2+y}. - Approximate f(2.95,7.1) knowing f(3,7) = 4. -

    -     - -

    - dz = \frac{x}{\sqrt{x^2+y}}dx + \frac{1}{2\sqrt{x^2+y}}dy, - with dx = -0.05 and dy = .1. - At (3,7), dz = 3/4(-0.05) + 1/8(.1) = -0.025, - so f(2.95,7.1) \approx -0.025+4 = 3.975. -

    -     - -     - - - - -

    - f(x,y) = \sin(x) \cos(y). - Approximate f(0.1,-0.1) knowing f(0,0) = 0. -

    -     - -

    - dz = (\cos(x) \cos(y) )dx - (\sin(x) \sin(y) )dy, - with dx = 0.1 and dy = -0.1. - At (0,0), dz = 1(.1) - (0)(-0.1) = 0.1, - so f(0.1,-0.1) \approx 0.1+0 = 0.1. -

    -     - -     - - - - -

    - f(x,y) = x^2y-xy^2. - Approximate f(2.04,3.06) knowing f(2,3) = -6. -

    -     - -

    - dz = (2xy-y^2)dx + (x^2-2xy)dy, - with dx = 0.04 and dy = 0.06. - At (2,3), dz = 3(0.04) + (-8)(0.06) = -0.36, - so f(2.04,3.06) \approx -0.36-6 = -6.36. -

    -     - -     - - - - -

    - f(x,y) = \ln(x-y). - Approximate f(5.1,3.98) knowing f(5,4) = 0. -

    -     - -

    - dz = \frac1{x-y}dx -\frac{1}{x-y}dy, - with dx = 0.1 and dy = -0.02. - At (5,4), dz = 1(0.1) + (-1)(-0.02) = 0.12, - so f(5.1,3.98) \approx 0.12+0 = 0.12. -

    -     - -     - -     - - - - -

    - Find the total differential dw. -

    -     - - - - -

    - w= x^2yz^3 -

    -     - -

    - dw = 2xyz^3\,dx + x^2z^3\,dy + 3x^2yz^2\,dz -

    -     - -     - - - - -

    - w= e^x\sin(y) \ln(z) -

    -     - -

    - dw = e^x\sin(y) \ln(z) \,dx + e^x\cos(y) \ln(z) \,dy + e^x\sin(y) \frac1z\,dz -

    -     - -     - -     - - - -

    - Use the information provided and the total differential to make the given approximation. -

    -     - - - - -

    - f(3,1) = 7, - f_x(3,1) = 9, f_y(3,1) = -2. - Approximate f(3.05, 0.9). -

    -     - -

    - dx = 0.05, dy = -0.1. - dz = 9(.05)+(-2)(-0.1) = 0.65. - So f(3.05,0.9) \approx 7+0.65=7.65. -

    -     - -     - - - - -

    - f(-4,2) = 13, - f_x(-4,2) = 2.6, f_y(-4,2) = 5.1. - Approximate f(-4.12, 2.07). -

    -     - -

    - dx = -0.12, dy = 0.07. - dz = 2.6(-.12)+(5.1)(0.07) = 0.045. - So f(-4.12, 2.07) \approx 13+0.045=13.045. -

    -     - -     - - - - -

    - f(2,4,5) = -1, f_x(2,4,5) = 2, - f_y(2,4,5) = -3, f_z(2,4,5) = 3.7. - Approximate f(2.5,4.1,4.8). -

    -     - -

    - dx = 0.5, dy = 0.1, dz = -0.2. -

    - -

    - dw = 2(0.5) + (-3)(0.1) + 3.7(-0.2) = -0.04, - so f(2.5, 4.1, 4.8) \approx -1-0.04 = -1.04. -

    -     - -     - - - - -

    - f(3,3,3) = 5, f_x(3,3,3) = 2, - f_y(3,3,3) = 0, f_z(3,3,3) = -2. - Approximate f(3.1, 3.1,3.1). -

    -     - -

    - dx = 0.1, dy = 0.1, dz = 0.1. -

    - -

    - dw = 2(0.1) + (0)(0.1) + (-2)(.1) = 0, - so f(3.1,3.1,3.1) \approx 5+0=5. -

    -     - -     - -     - - - - -

    - The following exercises ask a variety of questions dealing with approximating error and sensitivity analysis. -

    -     - - - - -

    - A cylindrical storage tank is to be 2ft tall with a radius of 1ft. - Is the volume of the tank more sensitive to changes in the radius or the height? -

    -     - -

    - The total differential of volume is dV = 4\pi dr + \pi dh. - The coefficient of dr is greater than the coefficient of dh, - so the volume is more sensitive to changes in the radius. -

    -     - -     - - - - -

    - Projectile Motion: The x-value of an object moving under the principles of projectile motion is x(\theta,v_0,t)= (v_0\cos(\theta) )t. - A particular projectile is fired with an initial velocity of v_0=250ft/s and an angle of elevation of \theta = 60^\circ. - It travels a distance of 375ft in 3 seconds. -

    - -

    - Is the projectile more sensitive to errors in initial speed or angle of elevation? -

    -     - -

    - Distance of the projectile is a function of two variables (leaving t=3): - D(v_0,\theta) = 3v_0\cos(\theta). - The total differential of D is dD = 3\cos(\theta) dv_0-3v_0\sin(\theta) d\theta. - The coefficient of d\theta has a much greater magnitude than the coefficient of dv_0, - so a small change in the angle of elevation has a much greater effect on distance traveled than a small change in initial velocity. -

    -     - -     - - - - -

    - The length \ell of a long wall is to be approximated. - The angle \theta, as shown in the diagram - (not to scale), - is measured to be 85^\circ, - and the distance x is measured to be 30'. Assume that the triangle formed is a right triangle. -

    - -

    - Is the measurement of the length of \ell more sensitive to errors in the measurement of x or in \theta? -

    - - - A right-angled triangle illustrating a wall of unknown height, a horizontal distance, and an angle. - -

    - The diagram illustrates a simple right-angled triangle. - The base of the triangle is labeled x. The height of the triangle is labeled \ell. - The angle opposite the wall is labeled \theta. -

    -     - - - \begin{tikzpicture} - - \draw [ultra thick] (1,-1) -- node [pos=.5,right] { \(\ell=\)?}(1,1); - \draw [dashed] (1,1) -- (-1,-1) node [xshift=10pt,yshift=5pt] { \(\theta\)} -- node [pos=.5,below] { \(x\)} (1,-1); - - \end{tikzpicture} - - - -     - -

    - Using trigonometry, \ell = x\tan(\theta), - so d\ell = \tan(\theta) dx + x\sec^2(\theta) d\theta. - With \theta = 85^\circ and x=30, - we have d\ell = 11.43dx+3949.38d\theta. - The measured length of the wall is much more sensitive to errors in \theta than in x. - While it can be difficult to compare sensitivities between measuring feet and measuring degrees - (it is somewhat like comparing apples to oranges), - here the coefficients are so different that the result is clear: - a small error in degree has a much greater impact than a small error in distance. -

    -     - -     - - - - -

    - It is common sense that it is far better to measure a long distance with a long measuring tape rather than a short one. - A measured distance D can be viewed as the product of the length \ell of a measuring tape times the number n of times it was used. - For instance, - using a 3' tape 10 times gives a length of 30'. To measure the same distance with a 12' tape, - we would use the tape 2.5 times. (, - 30=12\times 2.5.) Thus D = n\ell. -

    - -

    - Suppose each time a measurement is taken with the tape, - the recorded distance is within 1/16'' of the actual distance. (, d\ell = 1/16'' \approx 0.005ft). - Using differentials, - show why common sense proves correct in that it is better to use a long tape to measure long distances. -

    -     - -

    - With D = n\ell, - the total differential is dD = \ell\, dn+ n\,d\ell. - If one measures with a short tape, - n must be large and hence - n\,d\ell is going to be greater than when a large tape is used - (wherein n will be small). -

    -     - -     -     -     -     -
    -
    - The Multivariable Chain Rule - -

    - Consider driving an off-road vehicle along a dirt road. - As you drive, your elevation likely changes. - What factors determine how quickly your elevation rises and falls? - After some thought, - generally one recognizes that one's velocity - (speed and direction) - and the terrain influence your rise and fall. -

    - -

    - One can represent the terrain as the surface defined by a multivariable function f(x,y); - one can represent the path of the off-road vehicle, as seen from above, - with a vector-valued function \vec r(t) = \langle x(t), - y(t)\rangle; - the velocity of the vehicle is thus \vrp(t) = \langle x'(t),\yp(t)\rangle. -

    - -

    - Consider - in which a surface z=f(x,y) is drawn, - along with a dashed curve in the xy-plane. - Restricting f to just the points on this circle gives the curve shown on the surface (, - the path of the off-road vehicle.) The derivative - \frac{df}{dt} gives the instantaneous rate of change of f with respect to t. - If we consider an object traveling along this path, - \frac{df}{dt}=\frac{dz}{dt} gives the rate at which the object rises/falls (, - the rate of elevation change - of the vehicle.) Conceptually, - the Multivariable Chain Rule combines terrain and velocity information properly to compute this rate of elevation change. -

    - -
    - Understanding the application of the Multivariable Chain Rule - - - - A plot of a surface in space, and a parametric curve on the surface. - -

    - A surface is plotted in space against a set of three-dimensional coordinate axes. - In the xy plane, a dashed circle is shown. - Each point (x,y,0) on this circle determines a point (x,y,f(x,y)) on the surface. - Together, these points form a curve on the surface; this is also illustrated in the image. -

    -     - - - - - //ASY file for figmchain_intro3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-1,5); - pair ybounds=(-1,5); - pair zbounds=(0,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=-0.2*(x-1)^2-0.05*y^2+2 //{-.2*(x-1)^2-.05*y^2+2}; - triple f(pair t) { - return (t.x,t.y,-0.2*(t.x-1)^2-0.05*t.y^2+2); - } - surface s=surface(f,(0,0),(4,4),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the circle in xy-plane - triple g(real t) {return (cos(t)+2,sin(t)+2,0);} - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,dashed+bluepen+linewidth(.75)); - - //Draw the circle on surface - triple g(real t) { - return (cos(t)+2,sin(t)+2,-0.2*(cos(t)+2-1)^2-0.05*(sin(t)+2)^2+2); - } - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - - - -
    - -

    - Abstractly, let z be a function of x and y; - that is, z=f(x,y) for some function f, - and let x and y each be functions of t. - By choosing a t-value, - x- and y-values are determined, - which in turn determine z: - this defines z as a function of t. - The Multivariable Chain Rule gives a method of computing \frac{dz}{dt}. -

    - - -     - - - Multivariable Chain Rule, Part I - - - - Multivariable Chain Rule, Part I - -

    - Let z=f(x,y), x=g(t) and y=h(t), where f, - g and h are differentiable functions. - Then z = f(x,y) = f\big(g(t),h(t)\big) is a function of t, - and - derivativeChain Rule - Chain Rulemultivariable - - \frac{dz}{dt} = \frac{df}{dt} \amp = f_x(x,y)\frac{dx}{dt}+f_y(x,y)\frac{dy}{dt} - \amp = \frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt} - \amp = \langle\, f_x,f_y\rangle \cdot \langle x',\yp\rangle - . -

    -     -     - -

    - The Chain Rule of - states that - - \frac{d}{dx}\Big(f\big(g(x)\big)\Big) = \fp\big(g(x)\big)\gp(x) - . - If t=g(x), we can express the Chain Rule as - - \frac{df}{dx} = \frac{df}{dt}\frac{dt}{dx}; - - recall that the derivative notation is deliberately chosen to reflect their fraction-like properties. - A similar effect is seen in . - In the second line of equations, - one can think of the dx and \partial x as - sort of canceling out, - and likewise with dy and \partial y. -

    - -

    - Notice, too, - the third line of equations in . - The vector \langle\,f_x,f_y\rangle contains information about the surface (terrain); - the vector \langle x',\yp\rangle can represent velocity. - In the context measuring the rate of elevation change of the off-road vehicle, - the Multivariable Chain Rule states it can be found through a product of terrain and velocity information. -

    - -

    - We now practice applying the Multivariable Chain Rule. -

    - - - - - Using the Multivariable Chain Rule - -

    - Let z=x^2y+x, where x=\sin(t) and y=e^{5t}. - Find \frac{dz}{dt} using the Chain Rule. -

    -     - -

    - Following , we find - - f_x(x,y) = 2xy+1,\qquad f_y(x,y) = x^2,\qquad \frac{dx}{dt} = \cos(t) ,\qquad \frac{dy}{dt}= 5e^{5t} - . -

    - -

    - Applying the theorem, we have - - \frac{dz}{dt} = (2xy+1)\cos(t) + 5x^2e^{5t} - . -

    - -

    - This may look odd, as it seems that - \frac{dz}{dt} is a function of x, y and t. - Since x and y are functions of t, - \frac{dz}{dt} is really just a function of t, - and we can replace x with \sin(t) and y with e^{5t}: - - \frac{dz}{dt} = (2xy+1)\cos(t) + 5x^2e^{5t} = (2\sin(t)e^{5t}+1)\cos(t) +5e^{5t}\sin^2(t) - . -

    -     -     - -

    - The previous example can make us wonder: - if we substituted for x and y at the end to show that - \frac{dz}{dt} is really just a function of t, - why not substitute before differentiating, - showing clearly that z is a function of t? -

    - -

    - That is, z = x^2y+x = (\sin(t) )^2e^{5t}+\sin(t) . - Applying the Chain and Product Rules, we have - - \frac{dz}{dt} = 2\sin(t) \cos(t) \, e^{5t}+ 5\sin^2(t) \,e^{5t}+\cos(t) - , - which matches the result from the example. -

    - -

    - This may now make one wonder What's the point? - If we could already find the derivative, - why learn another way of finding it? In some cases, - applying this rule makes deriving simpler, - but this is hardly the power of the Chain Rule. - Rather, in the case where z=f(x,y), - x=g(t) and y=h(t), - the Chain Rule is extremely powerful when - we do not know what f, - g and/or h are. - It may be hard to believe, - but often in the real world - we know rate-of-change information (, information about derivatives) without explicitly knowing the underlying functions. - The Chain Rule allows us to combine several rates of change to find another rate of change. - The Chain Rule also has theoretic use, - giving us insight into the behavior of certain constructions - (as we'll see in the next section). -

    - -

    - We demonstrate this in the next example. -

    - - - Applying the Multivariable Chain Rule - -

    - An object travels along a path on a surface. - The exact path and surface are not known, - but at time t=t_0 it is known that : - - \frac{\partial z}{\partial x} = 5,\qquad \frac{\partial z}{\partial y}=-2,\qquad \frac{dx}{dt}=3\qquad \text{ and } \qquad \frac{dy}{dt}=7 - . -

    - -

    - Find \frac{dz}{dt} at time t_0. -

    -     - -

    - The Multivariable Chain Rule states that - - \frac{dz}{dt} \amp = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} - \amp = 5(3)+(-2)(7) - \amp =1 - . -

    - -

    - By knowing certain rates-of-change information about the surface and about the path of the particle in the xy-plane, - we can determine how quickly the object is rising/falling. -

    -     -     - -

    - We next apply the Chain Rule to solve a max/min problem. -

    - - - Applying the Multivariable Chain Rule - -

    - Consider the surface z=x^2+y^2-xy, a paraboloid, - on which a particle moves with x and y coordinates given by - x=\cos(t) and y=\sin(t). - Find \frac{dz}{dt} when t=0, - and find where the particle reaches its maximum/minimum z-values. -

    -     - -

    - It is straightforward to compute - - f_x(x,y) \amp = 2x-y \amp f_y(x,y) \amp = 2y-x - \frac{dx}{dt} \amp = -\sin(t) \amp \frac{dy}{dt} \amp = \cos(t) - . -

    - -

    - Combining these according to the Chain Rule gives: - - \frac{dz}{dt} = -(2x-y)\sin(t) + (2y-x)\cos(t) - . -

    - -
    - Plotting the path of a particle on a surface in - - - - A circular paraboloid, opening upward, plotted over a rectangular domain. A curve lying on the surface is also plotted. - -

    - The graph z=x^2+y^2-xy is shown; this is a circular paraboloid. - The surface is plotted using a rectangular domain, resulting in peaks at the corners of the domain. - A parametric curve is also plotted; this curve lies on the surface. - When viewed from above, the curve is circular, - but the height of the curve varies as it follows the contours of the surface. -

    -     - - - - - //ASY file for figmchain23D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(6.5,3.4,1.9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={1,2,3}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(0,4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=x^2+y^2-xy - triple f(pair t) { - return (t.x,t.y,t.x^2+t.y^2-t.x*t.y); - } - surface s=surface(f,(-1.1,-1.1),(1.1,1.1),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the circle on surface - triple g(real t) { - return (cos(t),sin(t),(cos(t))^2+(sin(t))^2-cos(t)*sin(t)); - } - path3 mypath=graph(g,0,2*pi,operator ..); draw(mypath,bluepen); - - //plot points for min and max along the curve - dotfactor=3; - dot((cos(pi/4),sin(pi/4),1-cos(pi/4)*sin(pi/4))); - dot((cos(5*pi/4),sin(5*pi/4),1-cos(5*pi/4)*sin(5*pi/4))); - dot((cos(3*pi/4),sin(3*pi/4),1-cos(3*pi/4)*sin(3*pi/4))); - dot((cos(7*pi/4),sin(7*pi/4),1-cos(7*pi/4)*sin(7*pi/4))); - - - - -
    - -

    - When t=0, x=1 and y=0. - Thus \ds\frac{dz}{dt} = -(2)(0)+ (-1)(1) = -1. - When t=0, the particle is moving down, - as shown in . -

    - -

    - To find where z-value is maximized/minimized on the particle's path, - we set \frac{dz}{dt}=0 and solve for t: - - \frac{dz}{dt} =0 \amp = -(2x-y)\sin(t) + (2y-x)\cos(t) - 0\amp = -(2\cos(t) -\sin(t) )\sin(t) +(2\sin(t) -\cos(t) )\cos(t) - 0\amp = \sin^2(t) -\cos^2(t) - \cos^2(t) \amp =\sin^2(t) - t\amp = n\frac{\pi}4 \text{ (for odd \(n\)) } - -

    - -

    - We can use the First Derivative Test to find that on [0,2\pi], - z has reaches its absolute minimum at t=\pi/4 and 5\pi/4; - it reaches its absolute maximum at t=3\pi/4 and 7\pi/4, - as shown in . -

    -     -     - -

    - We can extend the Chain Rule to include the situation where z is a function of more than one variable, - and each of these variables is also a function of more than one variable. - The basic case of this is where z=f(x,y), - and x and y are functions of two variables, - say s and t. -

    - - - Multivariable Chain Rule, Part II - -

    -

      -
    1. -

      - Let z=f(x,y), - x=g(s,t) and y=h(s,t), where f, - g and h are differentiable functions. - Then z is a function of s and t, and -

      - -

      -

        -
      • \frac{\partial z}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial s}
        • - -
      • \frac{\partial z}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}
        • -
      - derivativeChain Rule - Chain Rulemultivariable -

      -
      2. - -
    2. -

      - Let z = f(x_1,x_2,\ldots,x_m) be a differentiable function of m variables, - where each of the x_i is a differentiable function of the variables t_1,t_2,\ldots,t_n. - Then z is a function of the t_i, and - - \frac{\partial z}{\partial t_i} = \frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial t_i} + \frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial t_i} + \cdots + \frac{\partial f}{\partial x_m}\frac{\partial x_m}{\partial t_i} - . -

      -
      3. -
    -

    -     -     - - - Using the Multivariable Chain Rule, Part II - -

    - Let z=x^2y+x, x=s^2+3t and y=2s-t. - Find \frac{\partial z}{\partial s} and \frac{\partial z}{\partial t}, - and evaluate each when s=1 and t=2. -

    -     - -

    - Following , - we compute the following partial derivatives: - - \frac{\partial f}{\partial x} = 2xy+1\qquad\qquad \frac{\partial f}{\partial y} = x^2 - , - - \frac{\partial x}{\partial s} = 2s \qquad\qquad \frac{\partial x}{\partial t} = 3\qquad\qquad \frac{\partial y}{\partial s} = 2 \qquad\qquad \frac{\partial y}{\partial t} = -1 - . -

    - -

    - Thus - - \ds \frac{\partial z}{\partial s} = (2xy+1)(2s) + (x^2)(2) = 4xys+2s + 2x^2, \text{ and } - - - \ds \frac{\partial z}{\partial t} = (2xy+1)(3) + (x^2)(-1) = 6xy-x^2+3 - . -

    - -

    - When s=1 and t=2, - x= 7 and y= 0, so - - \frac{\partial z}{\partial s} = 100\qquad \text{ and } \qquad \frac{\partial z}{\partial t} = -46 - . -

    -     -     - - - - - Using the Multivariable Chain Rule, Part II - -

    - Let w = xy+z^2, where x= t^2e^s, - y= t\cos(s), and z=s\sin(t). - Find \frac{\partial w}{\partial t} when s=0 and t=\pi. -

    -     - -

    - Following , - we compute the following partial derivatives: - - \frac{\partial f}{\partial x} \amp = y \amp \frac{\partial f}{\partial y} \amp = x \amp \frac{\partial f}{\partial z} \amp = 2z - \frac{\partial x}{\partial t} \amp = 2te^s \amp \frac{\partial y}{\partial t} \amp = \cos(s) \amp \frac{\partial z}{\partial t} \amp = s\cos(t) - . -

    - -

    - Thus - - \frac{\partial w}{\partial t} = y(2te^s) + x(\cos(s) ) + 2z(s\cos(t) ) - . -

    - -

    - When s=0 and t=\pi, - we have x=\pi^2, y=\pi and z=0. - Thus - - \frac{\partial w}{\partial t} = \pi(2\pi) + \pi^2 = 3\pi^2 - . -

    -     -     -     - - - Implicit Differentiation -

    - We studied finding \frac{dy}{dx} when y is given as an implicit function of x in detail in . - We find here that the Multivariable Chain Rule gives a simpler method of finding \frac{dy}{dx}. -

    - -

    - For instance, consider the implicit function x^2y-xy^3=3. - We learned to use the following steps to find \frac{dy}{dx}: - - \frac{d}{dx}\Big(x^2y-xy^3\Big) \amp = \frac{d}{dx}\Big(3\Big) - 2xy + x^2\frac{dy}{dx}-y^3-3xy^2\frac{dy}{dx} \amp = 0 - \frac{dy}{dx} = -\frac{2xy-y^3}{x^2-3xy^2} - . -

    - -

    - Instead of using this method, - consider z=x^2y-xy^3. - The implicit function above describes the level curve z=3. - Considering x and y as functions of x, - the Multivariable Chain Rule states that - - \frac{dz}{dx} = \frac{\partial z}{\partial x}\frac{dx}{dx}+\frac{\partial z}{\partial y}\frac{dy}{dx} - . -

    - -

    - Since z is constant - (in our example, z=3), - \frac{dz}{dx} = 0. - We also know \frac{dx}{dx} = 1. - Equation becomes - - 0 \amp = \frac{\partial z}{\partial x}(1) + \frac{\partial z}{\partial y}\frac{dy}{dx} \Rightarrow - \frac{dy}{dx} \amp = -\frac{\partial z}{\partial x}\Big/\frac{\partial z}{\partial y} - \amp = -\frac{\,f_x\,}{f_y} - . -

    - -

    - Note how our solution for \frac{dy}{dx} in Equation is just the partial derivative of z with respect to x, - divided by the partial derivative of z with respect to y, - all multiplied by (-1). -

    - -

    - We state the above as a theorem. -

    - - - Implicit Differentiation - -

    - Let f be a differentiable function of x and y, - where f(x,y)=c defines y as an implicit function of x, - for some constant c. - Then - derivativeimplicit - implicit differentiation - - \frac{dy}{dx} = - \frac{f_x(x,y)}{f_y(x,y)} - . -

    -     -     - -

    - We practice using - by applying it to a problem from . -

    - - - Implicit Differentiation - -

    - Given the implicitly defined function \sin(x^2y^2)+y^3=x+y, - find \yp. - Note: this is the same problem as given in - of , - where the solution took about a full page to find. -

    -     - -

    - Let f(x,y) = \sin(x^2y^2)+y^3-x-y; - the implicitly defined function above is equivalent to f(x,y)=0. - We find \frac{dy}{dx} by applying . - We find - - f_x(x,y) \amp = 2xy^2\cos(x^2y^2)-1 - f_y(x,y) \amp = 2x^2y\cos(x^2y^2)+3y^2-1 - , - so - - \frac{dy}{dx} = -\frac{2xy^2\cos(x^2y^2)-1}{2x^2y\cos(x^2y^2)+3y^2-1} - , - which matches our solution from . -

    -     -     - - - -

    - We can also do implicit differentiation for functions of three variables. - In the same way that a level curve f(x,y)=c is used to implicitly define y as a function of x, - a level surface f(x,y,z)=c can be viewed as implicitly defining z as a function of x and y. -

    - -

    - Suppose the equation f(x,y,z)=c, where c is a constant, - defines the function z=g(x,y). - Then we can use the chain rule to compute the derivatives of f(x,y,z) - with respect to x and y, where we set x=x, y=y, and z=g(x,y). - Since f(x,y,z) is constant, we have - - 0 \amp = \frac{\partial}{\partial x}f(x,y,z) - \amp = f_x(x,y,z)\plz{x}{x}+f_y(x,y,z)\plz{y}{x}+f_z(x,y,z)\plz{z}{x} - \amp = f_x(x,y,z)(1)+f_y(x,y,z)(0)+f_z(x,y,z)\plz{z}{x} - . - Solving for \plz{z}{x} gives us - - \plz{z}{x} = -\frac{f_x(x,y,z)}{f_z(x,y,z)} - , - and similarly, - - \plz{z}{y} = -\frac{f_y(x,y,z)}{f_z(x,y,z)} - . -

    - - - -

    - In we saw that we can use partial - derivatives to determine the equation of the tangent plane to a graph z=f(x,y). - Using implicit differentiation, we can do the same for a level surface f(x,y,z)=c. -

    - - Implicit Differentiation with three variables - -

    - Given that the equation - - x^2yz^3-\sin(x-3z)+4xy^2-3yz=0 - - defines z implicitly as a function of x and y, - compute \plz{z}{x} and \plz{z}{y} using implicit differentiation. - Then, determine the equation of the tangent plane to the surface at the point (3,0,1). -

    -     - -

    - There are two ways to proceed. One is to use implicit differentiation as before, - but using partial derivatives. Whenever we differentiate a function of z, - we multiply by the appropriate partial derivative of z. - The other option is to use the formula derived above. - We will use the first method for the x derivative, and the second for y. -

    -

    - We first take the partial derivative of both sides of Equation with respect to x: - - \frac{\partial}{\partial x}(x^2yz^3-\sin(x-3z)+4xy^2-3yz) \amp = 0 - 2xyz^3 + x^2y(3z^2)\plz{z}{x}-\cos(x-3z)\left(1-3\plz{z}{x}\right)+4y^2-3y\plz{z}{x} \amp = 0 - . - Note that we treated y as a constant, since the derivative is with respect to x. - Next, we collect terms: - - \plz{z}{x}\left(3x^2yz^2+3\cos(x-3z)-3y\right) = -2xyz^3+\cos(x-3z)-4y^2 - . - Lastly, we solve for \plz{z}{x}: - - \plz{z}{x} = \frac{-2xyz^3+\cos(x-3z)-4y^2}{3x^2yz^2+3\cos(x-3z)-3y} - . -

    -

    - For the y derivative, we will use the result given above. - Setting f(x,y,z) = x^2yz^3-\sin(x-3z)+4xy^2-3yz, we have \plz{z}{y} = -\frac{f_y(x,y,z)}{f_z(x,y,z)}. - Therefore, - - \plz{z}{y} = -\frac{x^2z^3+8xy-3z}{3x^2yz^2+3\cos(x-3z)-3y} - . - The second method certainly seems simpler! - The reader is invited to try each part with the other method, and compare answers. -

    -

    - Finally, we consider the problem of the tangent plane. - First, we check that the point (3,0,1) is indeed on the surface: f(3,0,1)=0, - as required. Next we note that z=1 is given to us from this point. - So if f(x,y,z)=c implicitly defines the graph z=g(x,y), - then we must have g(3,0)=1. Next, we have - - g_x(3,0) \amp = \plzoa{z}{x}{(3,0)} = \frac{0+1-0}{0+3-0}=\frac13 - g_y(3,0) \amp = \plzoa{z}{y}{(3,0)} = -\frac{9+0-3}{0+3-0} = -2 - . - The equation of the tangent plane is therefore - - z = g(3,0)+g_x(3,0)(x-3)+g_y(3,0)(y-0) = 1+\frac13(x-3)-2y - . -

    -     -     - -

    - In - we learned how partial derivatives give certain instantaneous rate of change information about a function f(x,y). - In that section, - we measured the rate of change of f by holding one variable constant and letting the other vary - (such as, holding y constant and letting x vary gives f_x). - We can visualize this change by considering the surface defined by f at a point and moving parallel to the x-axis. -

    - -

    - What if we want to move in a direction that is not parallel to a coordinate axis? - Can we still measure instantaneous rates of change? - Yes; we find out how in . - In doing so, - we'll see how the Multivariable Chain Rule informs our understanding of these - directional derivatives. -

    -     - - - - Terms and Concepts - - - -

    - Let a level curve of z=f(x,y) be described by x=g(t), - y = h(t). - Explain why \frac{dz}{dt}=0. -

    -     - - - -

    - Because the parametric equations describe a level curve, - z is constant for all t. - Therefore \frac{dz}{dt}=0. -

    -     - -     - - - - -

    - Fill in the blank: The single variable Chain Rule states - \ds\frac{d}{dx}\Big(f\big(g(x)\big)\Big) = \fp\big(g(x)\big)\cdot. -

    -     - - - - - - - -

    - Type in the correct mathematical expression using plain text. -

    -     -     -     -     - -     - - - - -

    - Put the blocks in order to form a correct chain rule statement. -

    -     - - \frac{df}{dt} - = - \frac{\partial f}{\partial x} - \frac{dx}{dt} - \frac{dt}{dx} - + - \frac{\partial f}{\partial y} - \frac{dy}{dt} - \frac{\partial y}{\partial t} - - -     - - - - - - -

    - If z=f(x,y), where x=g(t) and y=h(t), - we can substitute and write z as an explicit function of t. -

    - -

    - Using the Multivariable Chain Rule to find - \frac{dz}{dt} is sometimes easier than first substituting and then taking the derivative. -

    -     - -     - - - - -

    - The Multivariable Chain Rule is only useful when all the related functions are known explicitly. -

    -     - - -     - - - - - -

    - The Multivariable Chain Rule allows us to compute implicit derivatives easily by just computing two derivatives. -

    -     - - - - - - - - -     -     - - - Problems - - - -

    - Given the functions z=f(x,y), x=g(t) and y=h(t): -

    - -

    -

      -
    1. -

      - Use the Multivariable Chain Rule to compute \lz{z}{t}. -

      -
      2. - -
    2. -

      - Evaluate \lz{z}{t} at the indicated t-value. -

      -
      3. -
    -

    -     - - - - -

    - z=3x+4y, x=t^2, y=2t; t=1 -

    -     - -

    -

      -
    1. -

      - \frac{dz}{dt} = 3(2t)+4(2) = 6t+8. -

      -
      2. - -
    2. -

      - At t=1, \frac{dz}{dt} = 14. -

      -
      3. -
    -

    -     - -     - - - - - Context()->variables->add(y=>'Real',t=>'Real'); - $x=Compute("t"); - $y=Compute("t^2-1"); - $dzdt=Compute("2x-4yt"); - $reduced=$dzdt->substitute(x=>$x,y=>$y); - $dzdt1=$reduced->eval(t=>1,x=>0,y=>0); - $ev=$dzdt->cmp(checker => sub { - my ($correct,$student,$ansHash) = @_; - return 1 if ($correct == $student); - my $sreduced=$student->substitute(x=>$x,y=>$y); - Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($reduced == $sreduced); - return 0; - }); - - - -

    - z=x^2-y^2, x=t, - and y=t^2-1; t=1 -

    - - - Use the Multivariable Chain Rule to compute \lz{z}{t}. - -

    - -

    - - - Evaluate \lz{z}{t} at the indicated t-value. - -

    - -

    -     -     -     - - - - -

    - \ds z=5x+2y, x=2\cos(t) +1, - y=\sin(t) -3; t=\pi/4 -

    -     - -

    -

      -
    1. -

      - \frac{dz}{dt} = 5(-2\sin(t) )+2(\cos(t) ) = -10\sin(t) +2\cos(t) -

      -
      2. - -
    2. -

      - At t=\pi/4, \frac{dz}{dt} = -4\sqrt{2}. -

      -
      3. -
    -

    -     - -     - - - - - Context()->variables->add(y=>'Real',t=>'Real'); - $x=Compute("cos(t)"); - $y=Compute("sin(t)"); - $dzdt=Compute("-sin(t)/(1+y^2)-2xycos(t)/(y^2+1)^2"); - $reduced=$dzdt->substitute(x=>$x,y=>$y); - $ev=$dzdt->cmp(checker => sub { - my ($correct,$student,$ansHash) = @_; - return 1 if ($correct == $student); - my $sreduced=$student->substitute(x=>$x,y=>$y); - Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($reduced == $sreduced); - return 0; - }); - Context("Fraction"); - $dzdt1=Fraction($reduced->eval(t=>pi/2,x=>0,y=>0)); - - - -

    - z=\frac{x}{y^2+1}, x=\cos(t), - and y=\sin(t); t=\pi/2 -

    - - Use the Multivariable Chain Rule to compute \lz{z}{t}. - -

    - -

    - - - Evaluate \lz{z}{t} at the indicated t-value. - -

    - -

    -     -     -     - - - - -

    - \ds z=x^2+2y^2, x=\sin(t), - y=3\sin(t); t=\pi/4 -

    -     - -

    -

      -
    1. -

      - \ds\frac{dz}{dt} = 2x(\cos(t) ) + 4y(3\cos(t) ). -

      -
      2. - -
    2. -

      - At t=\pi/4, x=\sqrt{2}/2, - y=3\sqrt{2}/2, and \frac{dz}{dt} = 19. -

      -
      3. -
    -

    -     - -     - - - - - Context()->variables->add(y=>'Real',t=>'Real'); - $x=Compute("pi t"); - $y=Compute("2pi t + pi/2"); - $dzdt=Compute("-sin(x) sin(y) (pi) + cos(x) cos(y) (2pi)"); - $reduced=$dzdt->substitute(x=>$x,y=>$y); - $ev=$dzdt->cmp(checker => sub { - my ($correct,$student,$ansHash) = @_; - return 1 if ($correct == $student); - my $sreduced=$student->substitute(x=>$x,y=>$y); - Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($reduced == $sreduced); - return 0; - }); - Context("Fraction"); - $dzdt1=Fraction($reduced->eval(t=>3,x=>0,y=>0)); - - - -

    - z=\cos(x) \sin(y), x=\pi t, - and y=2\pi t+\pi/2; t=3 -

    - - - Use the Multivariable Chain Rule to compute \lz{z}{t}. - -

    - -

    - - - Evaluate \lz{z}{t} at the indicated t-value. - -

    - -

    -     -     -     - -     - - - -

    - Functions z=f(x,y), x=g(t) and y=h(t) are given. - Find the values of t where \frac{dz}{dt}=0. - Note: these are the same surfaces/curves as found in Exercises. -

    -     - - - - -

    - \ds z=3x+4y, x=t^2, y=2t -

    -     - -

    - t=-4/3; this corresponds to a minimum -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $zeros=Compute("-sqrt(3/2), 0, sqrt(3/2)"); - - - -

    - Given z=x^2-y^2, x=t, and y=t^2-1, - at what values of t does \lz{z}{t}=0? -

    - -

    - -

    -     -     -     - - - - -

    - \ds z=5x+2y, x=2\cos(t) +1, y=\sin(t) -3 -

    -     - -

    - t=\tan^{-1}(1/5) +n\pi, - where n is an integer -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $zeros=Compute("0, pi"); - - - -

    - Given z=\frac{x}{y^2+1}, - x=\cos(t), and y=\sin(t), - at what values of t in - [0,2\pi) does \lz{z}{t}=0? -

    - -

    - -

    -     - -

    - We find that - - \frac{dz}{dt} = -\frac{2\cos^2(t) \sin(t) }{(1+\sin^2(t) )^2}-\frac{\sin(t) }{1+\sin^2(t) } - . -

    - -

    - Setting this equal to 0, - finding a common denominator and factoring out \sin(t), we get - - \sin(t) \left(\frac{2\cos^2(t) +\sin^2(t) +1}{(1+\sin^2(t) )^2}\right)=0 - . -

    - -

    - We have \sin(t) = 0 when t = \pi n, - where n is an integer. - The expression in the parenthesis above is always positive, - and hence never equal 0. - So all solutions are t=\pi n, n is an integer. - In [0,2\pi), the only solutions are 0 and \pi. -

    -     -     -     - - - - -

    - \ds z=x^2+2y^2, x=\sin(t), y=3\sin(t) -

    -     - -

    - We find that - - \frac{dz}{dt} = 38\cos(t) \sin(t) - . -

    - -

    - Thus \frac{dz}{dt} = 0 when t=\pi n or \pi n+\pi/2, - where n is any integer. -

    -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $zeros=Compute("0, 1/pi*arctan(sqrt(5)), 1- 1/pi*arctan(sqrt(5)), 1, 1+1/pi*arctan(sqrt(5)), 2-1/pi*arctan(sqrt(5))"); - - - -

    - Given z=\cos(x) \sin(y), - x=\pi t, and y=2\pi t+\pi/2, - at what values of t in [0,2) does \lz{z}{t}=0? -

    - -

    - -

    -     - -

    - We find that - - \frac{dz}{dt} = -\pi\sin(\pi t)\sin(2\pi t+\pi/2)+2\pi\cos(\pi t)\cos(2\pi t+\pi/2) - . -

    - -

    - One can easily see that when t is an integer, - \sin(\pi t) =0 and \cos(2\pi t+\pi/2)=0, - hence \frac{dz}{dt}=0 when t is an integer. - There are other places where z has a relative max/min that require more work. - First, verify that \sin(2\pi t+\pi/2) = \cos(2\pi t), - and \cos(2\pi t+\pi/2) = -\sin(2\pi t). - This lets us rewrite \frac{dz}{dt} = 0 as - - -\sin(\pi t)\cos(2\pi t)-2\cos(\pi t)\sin(2\pi t)=0 - . -

    - -

    - By bringing one term to the other side of the equality then dividing, - we can get - - 2\tan(2\pi t) = -\tan(\pi t) - . -

    - -

    - Using the angle sum/difference formula for tangent, we know - - \tan(2\pi t) = \tan(\pi t + \pi t) = \frac{\tan(\pi t)+\tan(\pi t)}{1-\tan^2(\pi t)} - . -

    - -

    - Thus we write - - 2\frac{\tan(\pi t)+\tan(\pi t)}{1-\tan^2(\pi t)} = -\tan(\pi t) - . -

    - -

    - Solving for \tan^2(\pi t), we find - - \tan^2(\pi t) = 5 \implies \tan(\pi t) = \pm\sqrt{5} - , - and so - - \pi t = \tan^{-1}(\pm\sqrt{5}) = \pm\tan^{-1}(\sqrt{5}) - . -

    - -

    - Since the period of tangent is \pi, - we can adjust our answer to be - - \pi t = \pm\tan^{-1}(\sqrt{5})+ n\pi,\text{ where \(n\) is an integer. } - -

    - -

    - Dividing by \pi, we find - - t = \pm\frac1\pi\tan^{-1}(\sqrt{5})+ n,\text{ where \(n\) is an integer. } - -

    - -

    - So the solutions in [0,2) are 0, - \frac{1}{pi}\arctan(sqrt{5}), - 1-\frac{1}{pi}\arctan(sqrt{5}), 1, - 1+\frac{1}{pi}\arctan(sqrt{5}), - and 2-\frac{1}{pi}\arctan(sqrt{5}). -

    -     -     -     - -     - - - -

    - Given the - functions z=f(x,y), - x=g(s,t) and y=h(s,t): -

    - -

    -

      -
    1. -

      - Use the Multivariable Chain Rule to compute - \frac{\partial z}{\partial s} and \frac{\partial z}{\partial t}. -

      -
      2. - -
    2. -

      - Evaluate \frac{\partial z}{\partial s} and - \frac{\partial z}{\partial t} at the indicated s and t values. -

      -
      3. -
    -

    -     - - - - -

    - \ds z=x^2y, x=s-t, y=2s+4t; - s=1, t=0 -

    -     - -

    -

      -
    1. -

      - \frac{\partial z}{\partial s} = 2xy(1) + x^2(2) = 2xy+2x^2; - - \frac{\partial z}{\partial t} = 2xy(-1) + x^2(4) = -2xy+4x^2 -

      -
      2. - -
    2. -

      - With s=1, t=0, x=1 and y=2. - Thus \frac{\partial z}{\partial s} = 6 and \frac{\partial z}{\partial t} = 0 -

      -
      3. -
    -

    -     - -     - - - - - Context()->variables->add(y=>'Real',s=>'Real', t=>'Real'); - $x=Compute("st^2"); - $y=Compute("s^2t"); - $dzds=Compute("-pi sin(pi x + pi y/2)(t^2)-1/2pi sin(pi x + pi y/2)(2st)"); - $dzdt=Compute("-pi sin(pi x + pi y/2)(2st)-1/2pi sin(pi x + pi y/2)(s^2)"); - $dzdsreduced=$dzds->substitute(x=>$x,y=>$y); - $dzdtreduced=$dzdt->substitute(x=>$x,y=>$y); - $dzds1=Compute("2pi"); - $dzdt1=Compute("5pi/2"); - $dzdsev=$dzds->cmp(checker => sub { - my ($correct,$student,$ansHash) = @_; - return 1 if ($correct == $student); - my $sreduced=$student->substitute(x=>$x,y=>$y); - Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdsreduced == $sreduced); - return 0; - }); - $dzdtev=$dzdt->cmp(checker => sub { - my ($correct,$student,$ansHash) = @_; - return 1 if ($correct == $student); - my $sreduced=$student->substitute(x=>$x,y=>$y); - Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdtreduced == $sreduced); - return 0; - }); - - - -

    - z=\cos\mathopen{}\left(\pi x+\frac{\pi}2y\right)\mathclose{}, - x=st^2, and y=s^2t; s=1, t=0 -

    - - - Use the Multivariable Chain Rule to compute \plz{z}{s}. - -

    - -

    - - - Use the Multivariable Chain Rule to compute \plz{z}{t}. - -

    - -

    - - - Evaluate \plz{z}{s} at s=1, t=0. - -

    - -

    - - - Evaluate \plz{z}{t} at s=1, t=0. - -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real',s=>'Real', t=>'Real'); - $x=Compute("st^2"); - $y=Compute("s^2t"); - $dzds=Compute("2x cos(t) + 2y sin(t)"); - $dzdt=Compute("-2xs sin(t) + 2ys cos(t)"); - $dzdsreduced=$dzds->substitute(x=>$x,y=>$y); - $dzdtreduced=$dzdt->substitute(x=>$x,y=>$y); - $dzds1=Compute("4"); - $dzdt1=Compute("0"); - $dzdsev=$dzds->cmp(checker => sub { - my ($correct,$student,$ansHash) = @_; - return 1 if ($correct == $student); - my $sreduced=$student->substitute(x=>$x,y=>$y); - Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdsreduced == $sreduced); - return 0; - }); - $dzdtev=$dzdt->cmp(checker => sub { - my ($correct,$student,$ansHash) = @_; - return 1 if ($correct == $student); - my $sreduced=$student->substitute(x=>$x,y=>$y); - Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdtreduced == $sreduced); - return 0; - }); - - - -

    - z=x^2+y^2, x=s\cos(t), - and y=s\sin(t); s=2, t=\pi/4 -

    - - - Use the Multivariable Chain Rule to compute \plz{z}{s}. - -

    - -

    - - - Use the Multivariable Chain Rule to compute \plz{z}{t}. - -

    - -

    - - - Evaluate \plz{z}{s} at s=2, t=\pi/4. - -

    - -

    - - - Evaluate \plz{z}{t} at s=2, t=\pi/4. - -

    - -

    -     -     -     - - - - - Context()->variables->add(y=>'Real',s=>'Real', t=>'Real'); - $x=Compute("st^2"); - $y=Compute("s^2t"); - $dzds=Compute("-2 y t^2 e^(-(x^2+y^2))"); - $dzdt=Compute("-2 x e^(-(x^2+y^2)) - 4 s t y e^(-(x^2+y^2))"); - $dzdsreduced=$dzds->substitute(x=>$x,y=>$y); - $dzdtreduced=$dzdt->substitute(x=>$x,y=>$y); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $dzds1=Compute("-2/e^2"); - $dzdt1=Compute("-6/e^2"); - $dzdsev=$dzds->cmp(checker => sub { - my ($correct,$student,$ansHash) = @_; - return 1 if ($correct == $student); - my $sreduced=$student->substitute(x=>$x,y=>$y); - Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdsreduced == $sreduced); - return 0; - }); - $dzdtev=$dzdt->cmp(checker => sub { - my ($correct,$student,$ansHash) = @_; - return 1 if ($correct == $student); - my $sreduced=$student->substitute(x=>$x,y=>$y); - Value::Error("Mathematically correct, but give the expression that the Multivariable Chair Rule would give if you did not substitute expressions for x and y") if ($dzdtreduced == $sreduced); - return 0; - }); - - - -

    - z=e^{-(x^2+y^2)}, x=t, - and y=st^2, s=1, t=1 -

    - - - Use the Multivariable Chain Rule to compute \plz{z}{s}. - -

    - -

    - - - Use the Multivariable Chain Rule to compute \plz{z}{t}. - -

    - -

    - - - Evaluate \plz{z}{s} at s=1, t=1. - -

    - -

    - - - Evaluate \plz{z}{t} at s=1, t=1. - -

    - -

    -     -     -     - -     - - - -

    - The given equation defines y implicitly as a function of x. - Find \lz{y}{x} using Implicit Differentiation and . -

    -     - - - - -

    - x^2\tan(y) = 50 -

    -     - -

    - f_x = 2x\tan(y), f_y = x^2\sec^2(y); -

    - -

    - \ds\frac{dy}{dx} = -\frac{2\tan(y) }{x\sec^2(y) } -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $dydx=Compute("-x/y^2"); - - - -

    - \left(3x^2+2y^3\right)^4=2 -

    - - Find \lz{y}{x}: - -

    - -

    -     -     -     - - - - -

    - \ds \frac{x^2+y}{x+y^2}=17 -

    -     - -

    - \ds f_x = \frac{(x+y^2)(2x)-(x^2+y)(1)}{(x+y^2)^2}, - \ds f_y = \frac{(x+y^2)(1)-(x^2+y)(2y)}{(x+y^2)^2}; -

    - -

    - \ds\frac{dy}{dx} = -\frac{2x(x+y^2)-(x^2+y)}{x+y^2-2y(x^2+y)} -

    -     - -     - - - - - Context()->variables->add(y=>'Real'); - $dydx=Compute("-(2x+y)/(2y+x)"); - - - -

    - \ln\mathopen{}\left(x^2+xy+y^2\right)\mathclose{}=1 -

    - - Find \lz{y}{x}: - - -

    - -

    -     -     -     - -     - - - -

    - Find \lz{z}{t}, - or \plz{z}{s} and \plz{z}{t}, - using the supplied information. -

    -     - - - - -

    - \ds\frac{\partial z}{\partial x} = 2,\ds\frac{\partial z}{\partial y} = 1,\ds\frac{dx}{dt} = 4,\ds\frac{dy}{dt} = -5 -

    -     - -

    - \frac{dz}{dt} = 2(4)+1(-5) = 3. -

    -     - -     - - - - - $dzdt=Compute("0"); - - - -

    - \plz{z}{x} = 1, - \plz{z}{y} = -3, \lz{x}{t} = 6, and \lz{y}{t} = 2. -

    - - Find \lz{z}{t}: - -

    - -

    -     -     -     - - - - -

    - \ds\frac{\partial z}{\partial x} = -4,\ds\frac{\partial z}{\partial y} = 9, -

    - -

    - \ds\frac{\partial x}{\partial s} = 5,\ds\frac{\partial x}{\partial t} = 7,\ds \frac{\partial y}{\partial s} = -2,\ds \frac{\partial y}{\partial t} = 6 -

    -     - -

    - \frac{\partial z}{\partial s} = -4(5)+9(-2) = -38, -

    - -

    - \frac{\partial z}{\partial t} = -4(7)+9(6) = 26. -

    -     - -     - - - - - - $dzds=Compute("-2"); - $dzdt=Compute("5"); - - - -

    - \plz{z}{x} = 2, - \plz{z}{y} = 1, \plz{x}{s} = -2, - \plz{x}{t} = 3, \plz{y}{s} = 2 and \plz{y}{t} = -1 -

    - - - Find \plz{z}{s}: - -

    - -

    - - - Find \plz{z}{t}: - -

    - -

    -     -     -     - -     -     -     -
    -
    - Directional Derivatives - -

    - Partial derivatives give us an understanding of how a surface changes when we move in the x and y directions. - We made the comparison to standing in a rolling meadow and heading due east: - the amount of rise/fall in doing so is comparable to f_x. - Likewise, the rise/fall in moving due north is comparable to f_y. - The steeper the slope, the greater in magnitude f_y. -

    - -

    - But what if we didn't move due north or east? - What if we needed to move northeast and wanted to measure the amount of rise/fall? - Partial derivatives alone cannot measure this. - This section investigates directional derivatives, - which do measure this rate of change. -

    - - -     - - - Functions of Two Variables -

    - We begin with a definition. -

    - - - Directional Derivatives - -

    - Let z=f(x,y) be continuous on a set S and let \vec u = \la u_1,u_2\ra be a unit vector. - For all points (x,y), - the directional derivative of f at (x,y) in the direction of \vec u - is derivativedirectional - directional derivative - - D_{\vec u\,}f(x,y) = \lim_{h\to 0} \frac{f(x+hu_1,y+hu_2) - f(x,y)}h - . -

    -     -     - -

    - The partial derivatives f_x and f_y are defined with similar limits, - but only x or y varies with h, not both. - Here both x and y vary with a weighted h, - determined by a particular unit vector \vec u. - This may look a bit intimidating but in reality it is not too difficult to deal with; - it often just requires extra algebra. - However, the following theorem reduces this algebraic load. -

    - - - Directional Derivatives - -

    - Let z=f(x,y) be differentiable on a set S containing (x_0,y_0), - and let \vec u = \la u_1,u_2\ra be a unit vector. - The directional derivative of f at - (x_0,y_0) in the direction of \vec u is derivativedirectional - directional derivative - - D_{\vec u\,}f(x_0,y_0)=f_x(x_0,y_0)u_1 + f_y(x_0,y_0)u_2 - . -

    -     -     - - - Computing directional derivatives - -

    - Let z= 14-x^2-y^2 and let P=(1,2). - Find the directional derivative of f, - at P, in the following directions: -

    - -

    -

      -
    1. -

      - toward the point Q=(3,4), -

      -
      2. - -
    2. -

      - in the direction of \la 2,-1\ra, and -

      -
      3. - -
    3. -

      - toward the origin. -

      -
      4. -
    -

    -     - -

    - The surface is plotted in , - where the point P=(1,2) is indicated in the x,y-plane as well as the point (1,2,9) which lies on the surface of f. - We find that f_x(x,y) = -2x and f_x(1,2) = -2; - f_y(x,y) = -2y and f_y(1,2) = -4. -

    - -
    - Understanding the directional derivative in - - - - A surface is plotted in 3D. Three vectors in the plane are shown corresponding to three tangent vectors on the surface. - -

    - A surface in space is shown, lying above the xy plane in the first octant. - In the xy plane a grid is drawn, along with points P and Q. - At the point P there are also three vectors drawn, parallel to the xy plane. - The vector \vec{u}_1 points toward the point Q. -

    - -

    - A point on the surface is marked, directly above the point P. - At this point, we see three vectors drawn. Each vector is tangent to the surface. - The x and y components of these vectors match the corresponding vectors \vec{u}_1,\vec{u}_2,\vec{u}_3 in the plane, - but they also have a z component given by the directional derivative. - We see that, depending on what direction one moves in the plane, - our path on the surface might climb, or descend, or remain level. -

    -     - - - - - //ASY file for figdirect13D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(16,5,25); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={4}; - real[] myychoice={4}; - real[] myzchoice={10}; - defaultpen(0.5mm); - pair xbounds=(-0.5,4.75); - pair ybounds=(-0.5,4.75); - pair zbounds=(0,15); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=14-x^2-y^2 - triple f(pair t) { - return (t.x,t.y,14-t.x^2-t.y^2); - } - surface s=surface(f,(0,0),(2,2.5),6,5,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point P and on curve and point Q - dotfactor=3;dot((1,2,9)); - dot((1,2,0)); - label("$P$",(1,2,0),N); - dot((3,4,0)); - label("$Q$",(3,4,0),N); - - //draw grid lines - draw((0,1,0)--(4.5,1,0),gray+linewidth(.5)); - draw((0,2,0)--(4.5,2,0),gray+linewidth(.5)); - draw((0,3,0)--(4.5,3,0),gray+linewidth(.5)); - draw((0,4,0)--(4.5,4,0),gray+linewidth(.5)); - draw((1,0,0)--(1,4.5,0),gray+linewidth(.5)); - draw((2,0,0)--(2,4.5,0),gray+linewidth(.5)); - draw((3,0,0)--(3,4.5,0),gray+linewidth(.5)); - draw((4,0,0)--(4,4.5,0),gray+linewidth(.5)); - - //Draw the vectors in the plane from P - draw((1,2,0)--(1.707,2.707,0),Arrow3(size=2mm));//P to u1 - label("$\vec{u}_1$",(1.707,2.707,0),E);//u1 - draw((1,2,0)--(1.894,1.553,0),Arrow3(size=2mm));//P to u2 - label("$\vec{u}_2$",(1.894,1.553,0),S);//u2 - draw((1,2,0)--(0.552,1.106,0),Arrow3(size=2mm));//P to u3 - label("$\vec{u}_3$",(0.552,1.106,0),W);//u3 - - //Draw the vectors on the surface above P - draw((1,2,9)--(1.707,2.707,4.76),Arrow3(size=2mm));//P to u1 - draw((1,2,9)--(1.894,1.553,9),Arrow3(size=2mm));//P to u2 - draw((1,2,9)--(0.552,1.106,13.47),Arrow3(size=2mm));//P to u3 - - - - -
    - -

    -

      -
    1. -

      - Let \vec u_1 be the unit vector that points from the point (1,2) to the point Q=(3,4), - as shown in the figure. - The vector \overrightarrow{PQ} = \la 2,2\ra; - the unit vector in this direction is \vec u_1=\la 1/\sqrt{2}, 1/\sqrt{2}\ra. - Thus the directional derivative of f at (1,2) in the direction of \vec u_1 is - - D_{\vec u_1}f(1,2) = -2(1/\sqrt{2}) +(-4)(1/\sqrt{2}) = -6/\sqrt{2}\approx -4.24 - . - Thus the instantaneous rate of change in moving from the point (1,2,9) on the surface in the direction of \vec u_1 - (which points toward the point Q) - is about -4.24. - Moving in this direction moves one steeply downward. -

      -
      2. - -
    2. -

      - We seek the directional derivative in the direction of \la 2,-1\ra. - The unit vector in this direction is \vec u_2 = \la 2/\sqrt{5},-1/\sqrt{5}\ra. - Thus the directional derivative of f at (1,2) in the direction of \vec u_2 is - - D_{\vec u_2}f(1,2) = -2(2/\sqrt{5})+(-4)(-1/\sqrt{5}) = 0 - . - Starting on the surface of f at (1,2) and moving in the direction of \la 2,-1\ra - (or \vec u_2) - results in no instantaneous change in z-value. - This is analogous to standing on the side of a hill and choosing a direction to walk that does not change the elevation. - One neither walks up nor down, - rather just along the side of the hill. - Finding these directions of no elevation change is important. -

      -
      3. - -
    3. -

      - At P=(1,2), - the direction towards the origin is given by the vector \la -1,-2\ra; - the unit vector in this direction is \vec u_3=\la -1/\sqrt{5},-2/\sqrt{5}\ra. - The directional derivative of f at P in the direction of the origin is - - D_{\vec u_3}f(1,2) = -2(-1/\sqrt{5}) + (-4)(-2/\sqrt{5}) = 10/\sqrt{5} \approx 4.47 - . - Moving towards the origin means - walking uphill quite steeply, - with an initial slope of about 4.47. -

      -
      4. -
    -

    -     -     - - - -

    - As we study directional derivatives, - it will help to make an important connection between the unit vector - \vec u = \la u_1,u_2\ra that describes the direction and the partial derivatives f_x and f_y. - We start with a definition and follow this with a Key Idea. -

    - - - Gradient - -

    - Let z=f(x,y) be differentiable on a set S that contains the point (x_0,y_0). - gradient -

      -
    1. -

      - The gradient of f - is \nabla f(x,y) = \la f_x(x,y),f_y(x,y)\ra. -

      -
      2. - -
    2. -

      - The gradient of f at - (x_0,y_0) is \nabla f(x_0,y_0) = \la f_x(x_0,y_0),f_y(x_0,y_0)\ra. -

      -
      3. -
    -

    -     -     - - -

    - To simplify notation, - we often express the gradient as \nabla f = \la f_x, f_y\ra. - The gradient allows us to compute directional derivatives in terms of a dot product. -

    - - - The Gradient and Directional Derivatives -

    - The directional derivative of - z=f(x,y) in the direction of \vec u is - gradient - directional derivative - derivativedirectional - - D_{\vec u\,}f = \nabla f\cdot \vec u - . -

    -     - -

    - The properties of the dot product previously studied allow us to investigate the properties of the directional derivative. - Given that the directional derivative gives the instantaneous rate of change of z when moving in the direction of \vec u, - three questions naturally arise: -

    - -

    -

      -
    1. -

      - In what direction(s) is the change in z the greatest (, the - steepest uphill)? -

      -
      2. - -
    2. -

      - In what direction(s) is the change in z the least (, the - steepest downhill)? -

      -
      3. - -
    3. -

      - In what direction(s) is there no change in z? -

      -
      4. -
    -

    - -

    - Using the key property of the dot product, we have - - \nabla f\cdot \vec u = \norm{\nabla f}\,\vnorm u \cos(\theta) = \norm{\nabla f}\cos(\theta) - , - where \theta is the angle between the gradient and \vec u. - (Since \vec u is a unit vector, \vnorm{u} = 1.) - This equation allows us to answer the three questions stated previously. -

    - -

    -

      -
    1. -

      - Equation - is maximized when \cos(\theta) =1, - , when the gradient and \vec u have the same direction. - We conclude the gradient points in the direction of greatest z change. -

      -
      2. - -
    2. -

      - Equation - is minimized when \cos(\theta) = -1, - , when the gradient and \vec u have opposite directions. - We conclude the gradient points in the opposite direction of the least z change. -

      -
      3. - -
    3. -

      - Equation - is 0 when \cos(\theta) = 0, - , when the gradient and \vec u are orthogonal to each other. - We conclude the gradient is orthogonal to directions of no z change. -

      -
      4. -
    -

    - -

    - This result is rather amazing. - Once again imagine standing in a rolling meadow and face the direction that leads you steepest uphill. - Then the direction that leads steepest downhill is directly behind you, - and side-stepping either left or right (, moving perpendicularly to the direction you face) does not change your elevation at all. -

    - -

    - Recall that a level curve is defined as a curve in the xy-plane along which the z-values of a function do not change. - Let a surface z=f(x,y) be given, - and let's represent one such level curve as a vector-valued function, - \vrt = \la x(t), y(t)\ra. - As the output of f does not change along this curve, - f\big(x(t),y(t)\big) = c for all t, - for some constant c. -

    - -

    - Since f is constant for all t, \frac{df}{dt} = 0. - By the Multivariable Chain Rule, we also know - - \frac{df}{dt} \amp = f_x(x,y)x'(t) + f_y(x,y)\yp(t) - \amp = \la f_x(x,y),f_y(x,y)\ra \cdot \la x'(t),\yp(t)\ra - \amp = \nabla f \cdot \vrp(t) - \amp =0 - . -

    - -

    - This last equality states \nabla f \cdot \vrp(t) = 0: - the gradient is orthogonal to the derivative of \vec r, - meaning the gradient is orthogonal to the graph of \vec r. - Our conclusion: at any point on a surface, - the gradient at that point is orthogonal to the level curve that passes through that point. -

    - -

    - We restate these ideas in a theorem, - then use them in an example. -

    - - - The Gradient and Directional Derivatives - -

    - Let z=f(x,y) be differentiable on a set S with gradient \nabla f, - let P=(x_0,y_0) be a point in S and let \vec u be a unit vector. -

    - -

    -

      -
    1. -

      - The maximum value of D_{\vec u\,}f(x_0,y_0) is \norm{\nabla f(x_0,y_0)}; - the direction of maximal z increase is \nabla f(x_0,y_0). -

      -
      2. - -
    2. -

      - The minimum value of D_{\vec u\,}f(x_0,y_0) is -\norm{\nabla f(x_0,y_0)}; - the direction of minimal z increase is -\nabla f(x_0,y_0). -

      -
      3. - -
    3. -

      - At P, - \nabla f(x_0,y_0) is orthogonal to the level curve passing through \big(x_0,y_0\big). - gradient - directional derivative - derivativedirectional - level curves - gradientand level curves -

      -
      4. -
    -

    -     -     - - - Finding directions of maximal and minimal increase - -

    - Let f(x,y) = \sin(x) \cos(y) and let P=(\pi/3,\pi/3). - Find the directions of maximal/minimal increase, - and find a direction where the instantaneous rate of z change is 0. -

    -     - -

    - We begin by finding the gradient. - f_x = \cos(x) \cos(y) and f_y = -\sin(x) \sin(y), thus - - \nabla f = \la \cos(x) \cos(y) ,-\sin(x) \sin(y) \ra \text{ and, at \(P\), } \nabla f\left(\frac{\pi}3,\frac{\pi}3\right) = \la\frac14,-\frac34\ra - . -

    - -

    - Thus the direction of maximal increase is \la 1/4, -3/4\ra. - In this direction, - the instantaneous rate of z change is \norm{\la 1/4,-3/4\ra} = \sqrt{10}/4 \approx 0.79. -

    - -

    - - shows the surface plotted from two different perspectives. - In each, the gradient is drawn at P with a dashed line - (because of the nature of this surface, - the gradient points into the surface). - Let \vec u = \la u_1, u_2\ra be the unit vector in the direction of \nabla f at P. - Each graph of the figure also contains the vector \la u_1, u_2, \norm{\nabla f\,}\ra. - This vector has a run of 1 - (because in the xy-plane it moves 1 unit) - and a rise of \norm{\nabla f}, - hence we can think of it as a vector with slope of - \norm{\nabla f} in the direction of \nabla f, - helping us visualize how steep - the surface is in its steepest direction. -

    - -
    - Graphing the surface and important directions in - -
    - - - - - A portion of the graph f(x,y) = sin(x)cos(y), along with a trace curve, and tangent and gradient vectors. - -

    - The surface given by the graph z=\sin(x)\cos(y) resembles rolling hills, - with sinusoidal oscillations in both the x and y directions. - One of the peaks is shown; near the top of the peak a trace of constant z value is drawn. - At one point on this curve, a point is marked, and at this point there are three vectors plotted: -

      -
    • -

      - A vector tangent to the curve given by the trace. -

      -
      • -
    • -

      - A copy of the gradient vector. This vector is dashed and is covered by the surface, - making it somewhat hard to see in the default perspective. - (Note that the actual gradient vector is a vector in the plane; - a copy of it is drawn at ths point on the surface, parallel to the xy plane.) -

      -
      • -
    • -

      - A vector whose x and y components correspond to the gradient, - but whose z component is given by the magnitude of the gradient. - This vector lies tangent to the surface, but is orthogonal to the tangent vector. - It points in a direction moving up along the surface. -

      -
      • -
    -

    -     - - - - - //ASY file for figdirect23D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - - // for the default version, figdirect2 - currentprojection=orthographic(12,13,3); - - // for the second, zoomed in view, figdirect2b - //currentprojection=orthographic((5.6,16.3,2.6),(0,0,1),(0,0,0),1.6,(-0.03,-0.34)); - - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-1,6); - pair ybounds=(-1,5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=sin(x)*cos(y) - triple f(pair t) { - return (t.x,t.y,sin(t.x)*cos(t.y)); - } - surface s=surface(f,(-1.5,-1.5),(4,4),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3;dot((pi/3,pi/3,sqrt(3)/4)); - - //Draw the vectors on the surface - draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+3/sqrt(10),pi/3+1/sqrt(10),sqrt(3)/4),Arrow3(size=2mm));//T at point - draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+1/sqrt(10),pi/3-3/sqrt(10),sqrt(3)/4),linetype(new real[] {2,2}),Arrow3(size=2mm));//N at point - //draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+1/sqrt(10),pi/3-3/sqrt(10),sqrt(3)/4),linetype(new real[] {2,2}),Arrow3(size=2mm));//N at point - draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+1/sqrt(10),pi/3-3/sqrt(10),0.79+sqrt(3)/4),Arrow3(size=2mm));//tangent - - //plot level curve - draw((0.4485,0,0.4336)..(0.4693,0.2857,0.434)..(0.5417,0.5714,0.4337)..(0.7143,0.8481,0.4333)..(0.9331,1.,0.4341)..(1.143,1.075,0.4331)..(1.429,1.118,0.4333)..(1.714,1.118,0.4333)..(2.,1.074,0.4331)..(2.209,1.,0.4341)..(2.429,0.8467,0.4333)..(2.6,0.5714,0.4333)..(2.673,0.2857,0.4333)..(2.694,0,0.4332)..(2.673,-0.2857,0.4333)..(2.615,-0.5276,0.4341)..(2.531,-0.7143,0.4333)..(2.347,-0.9187,0.433)..(2.098,-1.044,0.4341)..(1.821,-1.107,0.4333)..(1.551,-1.122,0.4335)..(1.237,-1.094,0.4338)..(0.9783,-1.022,0.4329)..(0.7185,-0.8529,0.433)..(0.5714,-0.639,0.4341)..(0.4838,-0.3734,0.4331)..(0.4517,-0.1197,0.4334)..(0.4485,-0.01989,0.4335),bluepen+linewidth(2)); - - - - -
    - -
    - - - - - A zoomed-in view of the same surface, curve, and vectors as the previous image. - -

    - This is the same image as . - The only difference is that this image is zoomed in closer to the point at which the three vectors are drawn, - and the perspective is changed slightly. -

    -     - - - - - //ASY file for figdirect23D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - - // for the default version, figdirect2 - //currentprojection=orthographic(12,13,3); - - // for the second, zoomed in view, figdirect2b - currentprojection=orthographic((5.6,16.3,2.6),(0,0,1),(0,0,0),1.6,(-0.03,-0.25)); - - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-1,6); - pair ybounds=(-1,5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=sin(x)*cos(y) - triple f(pair t) { - return (t.x,t.y,sin(t.x)*cos(t.y)); - } - surface s=surface(f,(-1.5,-1.5),(4,4),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3;dot((pi/3,pi/3,sqrt(3)/4)); - - //Draw the vectors on the surface - draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+3/sqrt(10),pi/3+1/sqrt(10),sqrt(3)/4),Arrow3(size=2mm));//T at point - draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+1/sqrt(10),pi/3-3/sqrt(10),sqrt(3)/4),linetype(new real[] {2,2}),Arrow3(size=2mm));//N at point - draw((pi/3,pi/3,sqrt(3)/4)--(pi/3+1/sqrt(10),pi/3-3/sqrt(10),0.79+sqrt(3)/4),Arrow3(size=2mm));//tangent - - //plot level curve - draw((0.4485,0,0.4336)..(0.4693,0.2857,0.434)..(0.5417,0.5714,0.4337)..(0.7143,0.8481,0.4333)..(0.9331,1.,0.4341)..(1.143,1.075,0.4331)..(1.429,1.118,0.4333)..(1.714,1.118,0.4333)..(2.,1.074,0.4331)..(2.209,1.,0.4341)..(2.429,0.8467,0.4333)..(2.6,0.5714,0.4333)..(2.673,0.2857,0.4333)..(2.694,0,0.4332)..(2.673,-0.2857,0.4333)..(2.615,-0.5276,0.4341)..(2.531,-0.7143,0.4333)..(2.347,-0.9187,0.433)..(2.098,-1.044,0.4341)..(1.821,-1.107,0.4333)..(1.551,-1.122,0.4335)..(1.237,-1.094,0.4338)..(0.9783,-1.022,0.4329)..(0.7185,-0.8529,0.433)..(0.5714,-0.639,0.4341)..(0.4838,-0.3734,0.4331)..(0.4517,-0.1197,0.4334)..(0.4485,-0.01989,0.4335),bluepen+linewidth(2)); - - - - -
    -     - -
    - -

    - The direction of minimal increase is \la -1/4,3/4\ra; - in this direction the instantaneous rate of z change is -\sqrt{10}/4 \approx -0.79. -

    - -

    - Any direction orthogonal to - \nabla f is a direction of no z change. - We have two choices: the direction of - \la 3,1\ra and the direction of \la -3,-1\ra. - The unit vector in the direction of - \la 3,1\ra is shown in each graph of the figure as well. - The level curve at z=\sqrt{3}/4 is drawn: - recall that along this curve the z-values do not change. - Since \la 3,1\ra is a direction of no z-change, - this vector is tangent to the level curve at P. -

    -     -     - - - Understanding when <m>\nabla f = \vec 0</m> - -

    - Let f(x,y) = -x^2+2x-y^2+2y+1. - Find the directional derivative of f in any direction at P=(1,1). -

    -     - -

    - We find \nabla f = \la -2x+2, -2y+2\ra. - At P, we have \nabla f(1,1) = \la 0,0\ra. - According to , - this is the direction of maximal increase. - However, \la 0,0\ra is directionless; - it has no displacement. - And regardless of the unit vector \vec u chosen, - D_{\vec u\,}f = 0. -

    - -

    - - helps us understand what this means. - We can see that P lies at the top of a paraboloid. - In all directions, the instantaneous rate of change is 0. -

    - -

    - So what is the direction of maximal increase? - It is fine to give an answer of \vec 0 = \la 0,0\ra, - as this indicates that all directional derivatives are 0. -

    - -
    - At the top of a paraboloid, all directional derivatives are 0 - - - - A downward-opening circular paraboloid, with a point P marked at the vertex. - -

    - The image shows the graph of a circular paraboloid, opening downward. - A point P is marked at the vertex of the paraboloid. - Since this is the highest point, there is no direction in which f can increase. -

    -     - - - - - //ASY file for figdirect93D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(9,3.3,3.2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={1,2}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-0.5,2.5); - pair ybounds=(-0.5,2.5); - pair zbounds=(-0.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=-x^2+2x-y^2+2y+1 -x^2-y^2+2*x+2*y+1 - triple f(pair t) { - return (t.x,t.y,-t.x^2+2*t.x-t.y^2+2*t.y+1); - } - surface s=surface(f,(0,0),(2,2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=4;dot((1,1,3)); - - label("$P$",(1,1,3),N); - - - - -
    -     -     - -

    - The fact that the gradient of a surface always points in the direction of steepest increase/decrease is very useful, - as illustrated in the following example. -

    - - - The flow of water downhill - -

    - Consider the surface given by the graph of f(x,y)= 20-x^2-2y^2. - Water is poured on the surface at (1,1/4). - What path does it take as it flows downhill? -

    -     - -

    - Let \vrt = \la x(t), - y(t)\ra be the vector-valued function describing the path of the water in the xy-plane; - we seek x(t) and y(t). - We know that water will always flow downhill in the steepest direction; - therefore, at any point on its path, - it will be moving in the direction of -\nabla f. - (We ignore the physical effects of momentum on the water.) - Thus \vrp(t) will be parallel to \nabla f, - and there is some constant c such that c\nabla f = \vrp(t) = \la x'(t), - \yp(t)\ra. -

    - -

    - We find \nabla f = \la -2x, -4y\ra and write x'(t) as - \frac{dx}{dt} and \yp(t) as \frac{dy}{dt}. - Then - - c\nabla f \amp = \la x'(t), \yp(t)\ra - \la -2cx, -4cy \ra \amp = \la \frac{dx}{dt}, \frac{dy}{dt}\ra - . -

    - -

    - This implies - - -2cx = \frac{dx}{dt} \text{ and } -4cy =\frac{dy}{dt}, \text{i.e.,} - - - c = -\frac{1}{2x}\frac{dx}{dt} \text{ and } c =-\frac{1}{4y}\frac{dy}{dt} - . -

    - -

    - As c equals both expressions, we have - - \frac{1}{2x}\frac{dx}{dt} =\frac{1}{4y}\frac{dy}{dt} - . -

    - -

    - To find an explicit relationship between x and y, - we can integrate both sides with respect to t. - Recall from our study of differentials that \frac{dx}{dt}dt = dx. - Thus: - - \int \frac{1}{2x}\frac{dx}{dt}\,dt \amp =\int \frac{1}{4y}\frac{dy}{dt}\,dt - \int \frac{1}{2x}\,dx \amp =\int\frac{1}{4y}\,dy - \frac 12\ln\abs{x} \amp = \frac14\ln\abs{y} +C_1 - 2\ln\abs{x} \amp = \ln\abs{y} +C_1 - \ln\abs{x^2} \amp = \ln\abs{y}+C_1 - Now raise both sides as a power of e: - x^2 \amp = e^{\ln\abs{y}+C_1} - x^2 \amp = e^{\ln\abs{y}}e^{C_1}\qquad \text{(Note that \(e^{C_1}\) is just a constant.)} - x^2 \amp = yC_2 - \frac1{C_2}x^2 \amp =y \qquad \text{ (Note that \(1/C_2\) is just a constant.) } - Cx^2 \amp = y - . -

    - -

    - As the water started at the point (1,1/4), - we can solve for C: - - C(1)^2 = \frac14 \Rightarrow C = \frac14 - . -

    - -
    - A sketch of the surface described in along with the path in the xy-plane with the level curves - -
    - - - - - A sector of a downward-opening elliptic paraboloid, along with a path that begins near the top, and travels downward. - -

    - The surface is an elliptic paraboloid, opening downward, with a vertex at (0,0,20). - A point is marked on the surface, a little bit below the vertex. - A curve originating at this point shows the trajectory, down along the surface, - followed by water flowing downhill due to gravity. -

    -     - - - - - //ASY file for figdirect33D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(3.6,18.5,24); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={10,20}; - defaultpen(0.5mm); - pair xbounds=(-1,5); - pair ybounds=(-1,5); - pair zbounds=(-0.5,22); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=20-x^2-2y^2 - triple f(pair t) { - return (t.x,t.y,20-t.x^2-2*t.y^2); - } - - surface s=surface(f,(0,0),(4,4),16,16,Spline); - - triple g(pair t) { - return (sqrt(20)*cos(t.x)*t.y,sqrt(10)*sin(t.x)*t.y,20-(sqrt(20)*cos(t.x)*t.y)^2-2*(sqrt(10)*sin(t.x)*t.y)^2); - } - - surface ss=surface(g,(0,0),(pi/2,1),16,16,Spline); - - pen p=apexmeshpen; - draw(ss,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3;dot((1,1/4,18.875)); - - //Draw a trace on the surface - triple g(real t) {return (t,t^2/4,20-t^2-t^4/8);} - path3 mypath=graph(g,1,3.044,operator ..); draw(mypath,bluepen); - - - - -
    - -
    - - - - A contour plot of the paraboloid in this example, and the corresponding trajectory in the plane. - -

    - The contour plot for the function f(x,y)=20-x^2-2y^2 - consists of a family of concentric ellipses, centered at the origin. - The path followed along the surface by the water projects to a path in the xy plane; - we see that this path appears to be parabolic, and intersects each level curve orthogonally. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - ymin=-3.5,ymax=3.5, - xmin=-4.2,xmax=4.2 - ] - - \addplot [secondcurvestyle,solid,domain=0:360,samples=60] ({2*cos(x)},{sqrt(2)*sin(x)}); - \addplot [secondcurvestyle,solid,domain=0:360,samples=60] ({2*sqrt(2)*cos(x)},{2*sin(x)}); - \addplot [secondcurvestyle,solid,domain=0:360,samples=80] ({2*sqrt(3)*cos(x)},{sqrt(6)*sin(x)}); - \addplot [secondcurvestyle,solid,domain=0:360,samples=80] ({4*cos(x)},{2*sqrt(2)*sin(x)}); - - \addplot [firstcurvestyle,-,domain=1:3.1] ({x},{x^2/4}); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - Thus the water follows the curve y=x^2/4 in the xy-plane. - The surface and the path of the water is graphed in . - In , - the level curves of the surface are plotted in the xy-plane, - along with the curve y=x^2/4. - Notice how the path intersects the level curves at right angles. - As the path follows the gradient downhill, - this reinforces the fact that the gradient is orthogonal to level curves. -

    -     -     - - -     - - - Functions of Three Variables -

    - The concepts of directional derivatives and the gradient are easily extended to three - (and more) - variables. - We combine the concepts behind Definitions - and - and into one set of definitions. -

    - - - Directional Derivatives and Gradient with Three Variables - -

    - Let w=F(x,y,z) be differentiable on a set D and let \vec u be a unit vector in \mathbb{R}^3. - gradient - directional derivative - derivativedirectional - -

      -
    1. -

      - The gradient of F is \nabla F = \la F_x,F_y,F_z\ra. -

      -
      2. - -
    2. -

      - The directional derivative of F in the direction of \vec u is - - D_{\vec u\,}F=\nabla F\cdot \vec u - . -

      -
      3. -
    -

    -     -     - -

    - The same properties of the gradient given in , - when f is a function of two variables, - hold for F, a function of three variables. -

    - - - The Gradient and Directional Derivatives with Three Variables - -

    - Let w=F(x,y,z) be differentiable on a set D, - let \nabla F be the gradient of F, - and let \vec u be a unit vector. - gradient - directional derivative - derivativedirectional - -

      -
    1. -

      - The maximum value of D_{\vec u\,}F is \norm{\nabla F}, - obtained when the angle between - \nabla F and \vec u is 0, , the direction of maximal increase is \nabla F. -

      -
      2. - -
    2. -

      - The minimum value of D_{\vec u\,}F is -\norm{\nabla F}, - obtained when the angle between - \nabla F and \vec u is \pi, - , the direction of minimal increase is -\nabla F. -

      -
      3. - -
    3. -

      - D_{\vec u\,}F = 0 when - \nabla F and \vec u are orthogonal. -

      -
      4. -
    -

    -     -     - -

    - We interpret the third statement of the theorem as - the gradient is orthogonal to level surfaces, - the three-variable analogue to level curves. -

    - - - Finding directional derivatives with functions of three variables - -

    - If a point source S is radiating energy, - the intensity I at a given point P in space is inversely proportional to the square of the distance between S and P. - That is, when S=(0,0,0), - I(x,y,z) = \frac{k}{x^2+y^2+z^2} for some constant k. -

    - -

    - Let k=1, - let \vec u = \la 2/3, 2/3, 1/3\ra be a unit vector, - and let P = (2,5,3). - Measure distances in inches. - Find the directional derivative of I at P in the direction of \vec u, - and find the direction of greatest intensity increase at P. -

    -     - -

    - We need the gradient \nabla I, - meaning we need I_x, I_y and I_z. - Each partial derivative requires a simple application of the Quotient Rule, giving - - \nabla I \amp = \la \frac{-2x}{(x^2+y^2+z^2)^2},\frac{-2y}{(x^2+y^2+z^2)^2},\frac{-2z}{(x^2+y^2+z^2)^2}\ra - \nabla I(2,5,3) \amp = \la \frac{-4}{1444},\frac{-10}{1444},\frac{-6}{1444}\ra \approx \la -0.003,-0.007,-0.004\ra - D_{\vec u\,}I \amp = \nabla I(2,5,3)\cdot \vec u - \amp = -\frac{17}{2166} \approx -0.0078 - . -

    - -

    - The directional derivative tells us that moving in the direction of \vec u from P results in a decrease in intensity of about -0.008 units per inch. - (The intensity is decreasing as \vec u moves one farther from the origin than P.) -

    - -

    - The gradient gives the direction of greatest intensity increase. - Notice that - - \nabla I(2,5,3) \amp = \la \frac{-4}{1444},\frac{-10}{1444},\frac{-6}{1444}\ra - \amp = \frac{2}{1444}\la -2,-5,-3\ra - . -

    - -

    - That is, the gradient at (2,5,3) is pointing in the direction of \la -2,-5,-3\ra, - that is, towards the origin. - That should make intuitive sense: - the greatest increase in intensity is found by moving towards to source of the energy. -

    -     -     - -

    - The directional derivative allows us to find the instantaneous rate of z change in any direction at a point. - We can use these instantaneous rates of change to define lines and planes that are - tangent to a surface at a point, - which is the topic of the next section. -

    -     - - - - Terms and Concepts - - - -

    - What is the difference between a directional derivative and a partial derivative? -

    -     - - - -

    - A partial derivative is essentially a special case of a directional derivative; - it is the directional derivative in the direction of x or y, - , \la 1,0\ra or \la 0,1\ra. -

    -     - -     - - - - - -

    - For f(x,y), - for what \vec u is D_{\vec u}\, f = f_x? -

    - - -     - - - -

    - \veci -

    -     -     - - -

    - \vecj -

    -     -     - - -

    - \veck -

    -     -     -     - -     - - - - - -

    - For f(x,y), - for what \vec u is D_{\vec u}\, f = f_y? -

    - - -     - - - -

    - \veci -

    -     -     - - -

    - \vecj -

    -     -     - - -

    - \veck -

    -     -     -     - -     - - - - - -

    - The gradient is to level curves. -

    -     - - - - - - - perpendicular - -

    - There is a synonym for this that is more appropriate when referring to vectors. -

    -     -     -     -     - -     - - - - -

    - The gradient points in the direction of increase. -

    -     - - - - maximal|maximum|greatest|largest - - - - -     - - - - - -

    - It is generally more informative to view the directional derivative not as the result of a limit, - but rather as the result of a product. -

    -     - - - - - - - - -     -     - - - Problems - - - -

    - A function f(x,y) is given. - Find \nabla f. -

    -     - - - - -

    - f(x,y) = -x^2y+xy^2+xy -

    -     - -

    - \nabla f = \la -2xy+y^2+y, -x^2+2xy+x\ra -

    -     - -     - - - - - Context("Vector2D"); - $gradient=Compute("<cos(x)cos(y),-sin(x)sin(y)>"); - - - -

    - Find \nabla f, where f(x,y) = \sin(x)\cos(y). -

    - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) = \frac{1}{x^2+y^2+1} -

    -     - -

    - \nabla f = \la \frac{-2x}{(x^2+y^2+1)^2}, \frac{-2y}{(x^2+y^2+1)^2}\ra -

    -     - -     - - - - - Context("Vector2D"); - $gradient=Compute("<-4,3>"); - - - -

    - Find \nabla f, where f(x,y) = -4x+3y. -

    - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) = x^2+2y^2-xy-7x -

    -     - -

    - \nabla f = \la 2x-y-7,4y-x\ra -

    -     - -     - - - - - Context("Vector2D"); - $gradient=Compute("<2xy^3-2,3x^2y^2>"); - - - -

    - Find \nabla f, where f(x,y) = x^2y^3-2x. -

    - -

    - -

    -     -     -     - -     - - - -

    - A function f(x,y) and a point P are given. - Find the directional derivative of f in the indicated directions. - Note: these are the same functions as in Exercises. -

    -     - - - - -

    - f(x,y) = -x^2y+xy^2+xy, P= (2,1) -

    -     - - -

    - In the direction of \vec v = \la 3,4\ra -

    -     - -

    - 2/5 -

    -     - -

    - \nabla f = \la -2xy+y^2+y, -x^2+2xy+x\ra; - \nabla f(2,1) = \la -2,2\ra. - The unit vector in the direction of \vec v is \vec u = \la 3/5,4/5\ra. - Therefore, D_{\vec u}f(2,1) = -2(3/5)+2(4/5)=2/5. -

    -     -     - - -

    - In the direction toward the point Q = (1,-1). -

    -     - -

    - -2/\sqrt{5} -

    -     - -

    - \nabla f = \la -2xy+y^2+y, -x^2+2xy+x\ra; - \nabla f(2,1) = \la -2,2\ra. - The vector from P to Q is \vec v = \la 1-2, -1-1\ra = \la -1,-2\ra. - The unit vector in the direction of \vec v is \vec u = \la -1/\sqrt{5},-2/\sqrt{5}\ra. - Therefore, D_{\vec{u}}f(2,1)=-2(-1/\sqrt{5})+2(-2/\sqrt{5})=-2/\sqrt{5}. -

    -     -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $a=Compute("1/4 (1-sqrt(3))"); - $b=Compute("(4sqrt(3)-3)/(10 sqrt(2))"); - - - -

    - Consider f(x,y) = \sin(x)\cos(y), - at P = \left(\frac{\pi}{4},\frac{\pi}{3}\right). -

    -     - - -

    - In the direction of \vec v=\la 1,1\ra. -

    - -

    - -

    -     -     - - - -

    - In the direction toward the point Q = (0,0). -

    - -

    - -

    -     -     -     -     - - - - -

    - \ds f(x,y) = \frac{1}{x^2+y^2+1}, P = (1,1). -

    -     - - -

    - In the direction of \vec v = \la 1,-1\ra. -

    -     - -

    - 0 -

    -     - -

    - \nabla f = \la \frac{-2x}{(x^2+y^2+1)^2}, \frac{-2y}{(x^2+y^2+1)^2}\ra; - \nabla f(1,1) = \la -2/9,-2/9\ra. - The unit vector in the direction of \vec v is \vec u = \la 1/\sqrt{2}, -1/\sqrt{2}\ra. - The directional derivative is D_{\vec u}f(1,1) = -2/9(1\/sqrt{2})-2/9(-1/\sqrt{2})=0. -

    -     -     - - -

    - In the direction toward the point Q = (-2,-2). -

    -     - -

    - 2\sqrt{2}/9 -

    -     - -

    - \nabla f = \la \frac{-2x}{(x^2+y^2+1)^2}, \frac{-2y}{(x^2+y^2+1)^2}\ra; - \nabla f(1,1) = \la -2/9,-2/9\ra. - The vector from P to Q is \vec v = \la -2-1,-2-1\ra = \la -3, -3\ra. - The unit vector in the direction of \vec v is \vec u = \la -1/\sqrt{2}, -1/\sqrt{2}\ra. - The directional derivative is D_{\vec u}f(1,1) = -2/9(-1\/sqrt{2})-2/9(-1/\sqrt{2})=2\sqrt{2}/9. -

    -     -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $a=Compute("-9/sqrt(10)"); - $b=Compute("27/sqrt(34)"); - - - -

    - Consider f(x,y) = -4x+3y, at P = (5,2). -

    -     - - -

    - In the direction of \vec v=\la 3,1\ra. -

    - -

    - -

    -     -     - - -

    - In the direction toward the point Q = (2,7). -

    - -

    - -

    -     -     -     -     - - - - -

    - \ds f(x,y) = x^2+2y^2-xy-7x, P = (4,1) -

    -     - - -

    - In the direction of \vec v = \la -2,5\ra -

    -     - -

    - 0 -

    -     - -

    - \nabla f = \la 2x-y-7,4y-x\ra; - \nabla f(4,1) = \la 0,0\ra. - Since the gradient is zero, so is the directional derivative. -

    -     -     - - -

    - In the direction toward the point Q = (4,0). -

    -     - -

    - 0 -

    -     - -

    - \nabla f = \la 2x-y-7,4y-x\ra; - \nabla f(4,1) = \la 0,0\ra. - Since the gradient is zero, so is the directional derivative. -

    -     -     - -     - - - - - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $a=Compute("3/sqrt(2)"); - $b=Compute("3"); - - - -

    - Consider f(x,y) = x^2y^3-2x, at P = (1,1). -

    -     - - - -

    - Find the directional derivative in the direction of \vec v=\la 3,3\ra. -

    - -

    - -

    -     -     - - - -

    - Find the directional derivative in the direction toward the point Q = (1,2). -

    - -

    - -

    -     -     -     -     - -     - - - -

    - A function f(x,y) and a point P are given. - Investigate the directions of maximal increase and decrease, as indicated. -

    - -

    - Note: these are the same functions and points as in Exercises. -

    -     - - - - -

    - f(x,y) = -x^2y+xy^2+xy, P= (2,1) -

    -     - - -

    - Find the direction of maximal increase of f at P. -

    -     - -

    - \nabla f(2,1) = \la -2,2\ra -

    -     - -

    - The gradient of f is \nabla f(x,y) = \la -2xy+y^2+y, -x^2+2xy+x\ra. - The direction of maximal increase is \nabla f(2,1) = \la -2,2\ra. -

    -     -     - - -

    - What is the maximal value of D_{\vec u}\,f at P? -

    -     - -

    - \sqrt{8} -

    -     - -

    - The maximal value is the norm of the gradient at P, - which is \norm{\nabla f(2,1)} = \norm{\la -2,2\ra} = \sqrt{8}. -

    -     -     - - -

    - Find the direction of maximal decrease in f at P. -

    -     - -

    - \la 2, -2\ra -

    -     - -

    - This is the direction opposite the direction of maximal increase; therefore, \la 2,-2\ra. -

    -     -     - - -

    - Give a direction \vec u such that D_{\vec u}\,f=0 at P. -

    -     - -

    - \vec u = \la 1/\sqrt{2},1/\sqrt{2}\ra -

    -     - -

    - We want \vec u = \la a, b\ra such that \nabla f(2,1)\cdot \vec u = -2a+2b=0. - One such unit vector is \vec u = \la 1/\sqrt{2},1/\sqrt{2}\ra. -

    -     -     - -     - - - - - Context("Vector2D"); - Context()->flags->set(reduceConstants=>0,redcuceConstantFunctions=>0); - $maxdir=Compute("<1/(2sqrt(2)), -1/2 sqrt(3/2)>"); - $maxdirev=$maxdir->cmp(parallel=>1,sameDirection=>1); - $maxval=Compute("1/sqrt(2)"); - $mindir=Compute("<-1/(2sqrt(2)), 1/2 sqrt(3/2)>"); - $mindirev=$mindir->cmp(parallel=>1,sameDirection=>1); - $nochange=Compute("<1/2 sqrt(3/2), 1/(2 sqrt(2))>"); - $nochangeev=$nochange->cmp(parallel=>1); - - - -

    - f(x,y) = \sin(x)\cos(y), - P = \left(\frac{\pi}{4},\frac{\pi}{3}\right): -

    -     - - - -

    - Find the direction of maximal increase of f at P. -

    - -

    - -

    -     -     - - -

    - What is the maximal value of D_{\vec u}\,f at P? -

    - -

    - -

    -     -     - - -

    - Find the direction of maximal decrease in f at P. -

    - -

    - -

    -     -     - - -

    - Give a direction \vec u such that D_{\vec u}\,f=0 at P. -

    - -

    - -

    -     -     -     -     - - - - -

    - \ds f(x,y) = \frac{1}{x^2+y^2+1}, P = (1,1). -

    -     - - - -

    - Find the direction of maximal increase of f at P. -

    -     - -

    - \nabla f(1,1) = \la -2/9,-2/9\ra -

    -     - -

    - The gradient of f is \nabla f = \la \frac{-2x}{(x^2+y^2+1)^2}, \frac{-2y}{(x^2+y^2+1)^2}\ra. - The direction of maximal increase is \nabla f(1,1) = \la -2/9,-2/9\ra. -

    -     -     - - -

    - What is the maximal value of D_{\vec u}\,f at P? -

    -     - -

    - 2\sqrt{2}/9 -

    -     - -

    - The maximal value is the norm of the gradient at P, - which is \norm{\nabla f(1,1)} = \norm{\la -2/9,-2/9\ra}=2\sqrt{2}/9. -

    -     -     - - -

    - Find the direction of maximal decrease in f at P. -

    -     - -

    - \la 2/9,2/9\ra -

    -     - -

    - This is the direction opposite the direction of maximal increase; therefore, \la 2/9,2/9\ra. -

    -     -     - - -

    - Give a direction \vec u such that D_{\vec u}\,f=0 at P. -

    -     - -

    - \vec u = \la 1/\sqrt{2},-1/\sqrt{2}\ra -

    -     - -

    - We want \vec u = \la a, b\ra such that \nabla f(1,1)\cdot \vec u = -\frac29 a-\frac29 b=0. - One such unit vector is \vec u = \la 1/\sqrt{2},-1/\sqrt{2}\ra. -

    -     -     - -     - - - - - Context("Vector2D"); - Context()->flags->set(reduceConstants=>0,redcuceConstantFunctions=>0); - $maxdir=Compute("<-4, 3>"); - $maxdirev=$maxdir->cmp(parallel=>1,sameDirection=>1); - $maxval=Compute("5"); - $mindir=Compute("<4,-3>"); - $mindirev=$mindir->cmp(parallel=>1,sameDirection=>1); - $nochange=Compute("<3,4>"); - $nochangeev=$nochange->cmp(parallel=>1); - - - -

    - f(x,y) = -4x+3y, P = (5,4) -

    -     - - - -

    - Find the direction of maximal increase of f at P. -

    - -

    - -

    -     -     - - -

    - What is the maximal value of D_{\vec u}\,f at P? -

    - -

    - -

    -     -     - - -

    - Find the direction of maximal decrease in f at P. -

    - -

    - -

    -     -     - - -

    - Give a direction \vec u such that D_{\vec u}\,f=0 at P. -

    - -

    - -

    -     -     -     -     - - - - -

    - \ds f(x,y) = x^2+2y^2-xy-7x, P = (4,1) -

    -     - - - -

    - Find the direction of maximal increase of f at P. -

    -     - -

    - No such direction -

    -     - -

    - The gradient of f is \nabla f(x,y) = \la 2x-y-7,4y-x\ra. - At the point P we get \nabla f(4,1) = \la 0,0\ra, - so P is a critical point; there is no such direction. -

    -     -     - - -

    - What is the maximal value of D_{\vec u}\,f at P? -

    -     - -

    - 0 -

    -     - -

    - The maximal value is the norm of the gradient at P, - which is \norm{\nabla f(4,1)} = 0. -

    -     -     - - -

    - Find the direction of maximal decrease in f at P. -

    -     - -

    - No such direction -

    -     - -

    - There is no direction of maximal decrease, since P is a critical point. -

    -     -     - - -

    - Give a direction \vec u such that D_{\vec u}\,f=0 at P. -

    -     - -

    - All directions -

    -     - -

    - Since \nabla f(4,1)=\la 0,0\ra, this is true in all directions. -

    -     -     - -     - - - - - Context("Vector2D"); - Context()->flags->set(reduceConstants=>0,redcuceConstantFunctions=>0); - $maxdir=Compute("<0, 3>"); - $maxdirev=$maxdir->cmp(parallel=>1,sameDirection=>1); - $maxval=Compute("3"); - $mindir=Compute("<0,-3>"); - $mindirev=$mindir->cmp(parallel=>1,sameDirection=>1); - $nochange=Compute("<1,0>"); - $nochangeev=$nochange->cmp(parallel=>1); - - - -

    - Given f(x,y) = x^2y^3-2x, P = (1,1): -

    -     - - - -

    - Find the direction of maximal increase of f at P. -

    - -

    - -

    -     -     - - -

    - What is the maximal value of D_{\vec u}\,f at P? -

    - -

    - -

    -     -     - - -

    - Find the direction of maximal decrease in f at P. -

    - -

    - -

    -     -     - - -

    - Give a direction \vec u such that D_{\vec u}\,f=0 at P. -

    - -

    - -

    -     -     -     -     - -     - - - -

    - A function F(x,y,z), - a vector \vec v and a point P are given. -

    - -

    - Compute the gradient of F, and the derivative of F in the direction of \vec v at P. -

    -     - - - - -

    - \ds F(x,y,z) = 3x^2z^3+4xy-3z^2, - \vec v = \la 1,1,1\ra, P = (3,2,1) -

    -     - - - -

    - Compute the gradient of F. -

    -     - -

    - \nabla F(x,y,z) = \la 6xz^3+4y, 4x, 9x^2z^2-6z\ra -

    -     -     - - -

    - Find the derivative of F at P in the direction of \vec v. -

    -     - -

    - 113/\sqrt{3} -

    -     - -

    - The gradient of F at P is \nabla F(3,2,1) = \la 26, 12, 75\ra. - A unit vector in the direction of \vec v is \vec u = \la 1/\sqrt{3},1/\sqrt{3},1/\sqrt{3}\ra. - The directional derivative is D_{\vec u}f(3,2,1) = \nabla F(3,2,1)\cdot \vec u = 113/\sqrt{3}. -

    -     -     - -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,redcuceConstantFunctions=>0); - $gradient=Compute("<cos(x) cos(y) e^z, -sin(x) sin(y) e^z, sin(x) cos(y) e^z>"); - $directional=Compute("2/3"); - - - -

    - F(x,y,z) = \sin(x)\cos(y)e^z, - \vec v = \la 2,2,1\ra, P = (0,0,0). -

    -     - - - -

    - Find \nabla F(x,y,z). -

    - -

    - -

    -     -     - - - -

    - Find D_{\vec u}\,F at P. -

    - -

    - -

    -     -     -     -     - - - - -

    - \ds F(x,y,z) = x^2y^2-y^2z^2, - \vec v = \la -1,7,3\ra, P = (1,0,-1) -

    -     - - - -

    - Compute the gradient of F. -

    -     - -

    - \nabla F(x,y,z) = \la 2xy^2, 2y(x^2-z^2), -2y^2z\ra -

    -     -     - - -

    - Find the derivative of F at P in the direction of \vec v. -

    -     - -

    - 0 -

    -     - -

    - The gradient of F at P is \nabla F(1,0,-1) = \la 0,0,0\ra. - Therefore, the derivative of F at P in any direction is 0. -

    -     -     - -     - - - - - Context("Vector"); - Context()->flags->set(reduceConstants=>0,redcuceConstantFunctions=>0); - $gradient=Compute("<-4x/(x^2+y^2+z^2)^2, -4y/(x^2+y^2+z^2)^2, -4z/(x^2+y^2+z^2)^2>"); - $directional=Compute("0"); - - - -

    - Given F(x,y,z) = \frac{2}{x^2+y^2+z^2}, - \vec v = \la 1,1,-2\ra, P = (1,1,1): -

    -     - - - -

    - Find \nabla F(x,y,z). -

    - -

    - -

    -     -     - - - -

    - Find D_{\vec u}\,F at P. -

    - -

    - -

    -     -     -     -     - -     -     -     -
    -
    - Tangent Lines, Normal Lines, and Tangent Planes - - Tangent Lines - -

    - Derivatives and tangent lines go hand-in-hand. - Given y=f(x), - the line tangent to the graph of f at x=x_0 is the line through - \big(x_0,f(x_0)\big) with slope \fp(x_0); - that is, the slope of the tangent line is the instantaneous rate of change of f at x_0. -

    - -

    - When dealing with functions of two variables, - the graph is no longer a curve but a surface. - At a given point on the surface, - it seems there are many lines that fit our intuition of being - tangent to the surface. -

    - -

    - In we introduced the concept of the tangent plane, - which could be thought of as consisting of all possible lines tangent to the surface at a given point. - In this section, we explore this idea in more detail. -

    - -
    - Showing various lines tangent to a surface - - - - A surface in three dimensions, including curves on the surface, and tangent lines to those curves. - -

    - A curved surface in space is plotted against a three-dimensional coordinate system. - It appears to be a portion of a downward-opening elliptic paraboloid. - A point is plotted on the surface, and there are several curves on the surface passing through that point. - At this point where the curves meet, a tangent line is drawn to each of the curves. - Since the curves lie in the surface, we can say that these lines are tangent to the surface itself, - and it appears that they all lie in a common plane. -

    -     - - - - - //ASY file for figspace_tangent_intro3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8,8,13); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={10}; - defaultpen(0.5mm); - pair xbounds=(-0.5,3); - pair ybounds=(-0.5,3); - pair zbounds=(-0.5,11); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=-x^2-.5*(y^2)+x*y+10 //(x,y,{-x^2-.5*(y^2)+x*y+10}); - triple f(pair t) { - return (t.x,t.y,-t.x^2-.5*(t.y^2)+t.x*t.y+10); - } - surface s=surface(f,(0,0),(3,3),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3;dot((2,1,7.5)); - - //Draw traces on the surface - triple g(real t) {return (2,t,-4-0.5*t^2+2*t+10);}//-x^2-.5*(y^2)+x*y+10 - path3 mypath=graph(g,0,3,operator ..); draw(mypath,bluepen); - triple g(real t) {return (t,1,-t^2-0.5+t+10);}//-x^2-.5*(y^2)+x*y+10 - path3 mypath=graph(g,0,3,operator ..); draw(mypath,bluepen); - triple g(real t) {return (t,3-t,-t^2-0.5*(3-t)^2+(3-t)*t+10);} - path3 mypath=graph(g,0,3,operator ..); draw(mypath,bluepen); - - //Draw tangent lines on the surface with gradient (-3,1) - //L1 = (2,1,7.5)+t(1,0,-3) for t=-1,1 - draw((1,1,10.5)--(3,1,4.5),redpen); - //L2 = (2,1,7.5)+t(0,1,1) for t=-1,1 - draw((2,0,6.5)--(2,2,8.5),redpen); - //L3 = (2,1,7.5)+t(0.707,-0.707,-2.828) for t=-1,1 - draw((1.293,1.707,10.328)--(2.707,0.293,4.672),redpen); - - - - -
    - -

    - In - we see lines that are tangent to curves in space. - Since each curve lies on a surface, - it makes sense to say that the lines are also tangent to the surface. - The next definition formally defines what it means to be - tangent to a surface. -

    - - - Directional Tangent Line - -

    - Let z=f(x,y) be differentiable on a set S containing - (x_0,y_0) and let \vec u = \la u_1, u_2\ra be a unit vector. - tangent linedirectional -

    - -

    -

      -
    1. -

      - The line \ell_x through - \big(x_0,y_0,f(x_0,y_0)\big) parallel to \la 1,0,f_x(x_0,y_0)\ra is the - tangent line to f in the direction of x at (x_0,y_0). -

      -
      2. - -
    2. -

      - The line \ell_y through - \big(x_0,y_0,f(x_0,y_0)\big) parallel to \la 0,1,f_y(x_0,y_0)\ra is the - tangent line to f in the direction of y at (x_0,y_0). -

      -
      3. - -
    3. -

      - The line \ell_{\vec u} through - \big(x_0,y_0,f(x_0,y_0)\big) parallel to \la u_1,u_2,D_{\vec u\,}f(x_0,y_0)\ra is the - tangent line to f in the direction of \vec u at (x_0,y_0). -

      -
      4. -
    -

    -     -     - -

    - It is instructive to consider each of three directions given in the definition in terms of slope. - The direction of \ell_x is \la 1,0,f_x(x_0,y_0)\ra; - that is, the run is one unit in the x-direction and the rise - is f_x(x_0,y_0) units in the z-direction. - Note how the slope is just the partial derivative with respect to x. - A similar statement can be made for \ell_y. - The direction of \ell_{\vec u} is \la u_1,u_2,D_{\vec u\,}f(x_0,y_0)\ra; - the run is one unit in the \vec u direction - (where \vec u is a unit vector) - and the rise is the directional derivative of z in that direction. -

    - -

    - - leads to the following parametric equations of directional tangent lines: -

    - -

    - - \ell_x(t) \amp = \left\{\begin{array}{l} x=x_0+t \\ y=y_0\\z=z_0+f_x(x_0,y_0)t - \end{array} \right. - \ell_y(t) \amp = \left\{\begin{array}{l} x=x_0 \\ y=y_0+t\\z=z_0+f_y(x_0,y_0)t - \end{array} \right. - \ell_{\vec u}(t) \amp = \left\{\begin{array}{l} x=x_0+u_1t \\ y=y_0+u_2t\\z=z_0+D_{\vec u\,}f(x_0,y_0)t - \end{array} \right. - . -

    - - - Finding directional tangent lines - -

    - Find the lines tangent to the surface z=\sin(x) \cos(y) at - (\pi/2,\pi/2) in the x and y directions and also in the direction of \vec v = \la -1,1\ra. -

    -     - -

    - The partial derivatives with respect to x and y are: - - f_x(x,y) = \cos(x) \cos(y) \amp \Rightarrow f_x(\pi/2,\pi/2) = 0 - f_y(x,y) = -\sin(x) \sin(y) \amp \Rightarrow f_y(\pi/2,\pi/2)=-1 - . -

    - -

    - At (\pi/2,\pi/2), the z-value is 0. -

    - -

    - Thus the parametric equations of the line tangent to f at - (\pi/2,\pi/2) in the directions of x and y are: - - \ell_x(t) = \left\{\begin{array}{l} x=\pi/2 + t\\ y=\pi/2 \\z=0 - \end{array} \right. \text{ and } - \ell_y(t) = \left\{\begin{array}{l} x=\pi/2 \\ y=\pi/2+t \\z=-t - \end{array} \right. - . -

    - -

    - The two lines are shown with the surface in . -

    - -
    - A surface and directional tangent lines in - -
    - - - - - A bumpy surface. At a point on the surface, the tangent lines given by the partial derivatives are shown. - -

    - One of the hills in the bumpy surface given by f(x,y) = \sin(x)\cos(y) is shown, - as well parts of adjacent hills and valleys. - A point is plotted on one face of this hill. - Two lines are drawn through this point, tangent to the surface. - One line appears to be parallel to the x axis; - its slope relative to z is given by the partial derivative with respect to x. - The other line lies in a plane parallel to the yz plane, - and has slope given by the partial derivative with respect to y. -

    -     - - - - - //ASY file for figpartial4a3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-0.5,5); - pair ybounds=(-0.5,4.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=sin(x)cos(y) - triple f(pair t) { - return (t.x,t.y,sin(t.x)*cos(t.y)); - } - surface s=surface(f,(-1,-1),(4,4),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3;dot((pi/2,pi/2,0)); - - //Draw lines on the surface - //L1 = (pi/2,pi/2,0)+t(1,0,0) for t=-1,1 - draw((0.57,1.57,0)--(2.57,1.57,0),redpen); - //L2 = (pi/2,pi/2,0)+t(0,1,-1) for t=-1,1 - draw((1.57,.57,1)--(1.57,2.57,-1),redpen); - - - - -
    - -
    - - - - - A curve lies along a bumpy surface. At one point on the curve, a tangent line to the curve is drawn. - -

    - The same bumpy surface as is shown. - This time, we see a curve that lies in the surface; - the curve lies above the line x+y=\pi in the xy plane, - and passes through the same point as in the previous image. - The tangent line to the curve at this point is plotted; - it is the line tangent to the surface at (\pi/2, \pi/2) in the direction of \langle -1, 1\rangle. -

    -     - - - - - //ASY file for figpartial4b3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-0.5,5); - pair ybounds=(-0.5,4.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=sin(x)cos(y) - triple f(pair t) { - return (t.x,t.y,sin(t.x)*cos(t.y)); - } - surface s=surface(f,(-1,-1),(4,4),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3;dot((pi/2,pi/2,0)); - - //Draw traces on the surface - triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} - path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); - - //Draw lines on the surface - //L1 = (pi/2,pi/2,0)+t(-0.707,0.707,-0.707) for t=-1,1 - draw((2.28,.864,0.707)--(.864,2.28,-0.707),redpen); - - - - -
    -     -
    - -

    - To find the equation of the tangent line in the direction of \vec v, - we first find the unit vector in the direction of \vec v: - \vec u = \la -1/\sqrt{2},1/\sqrt{2}\ra. - The directional derivative at - (\pi/2,\pi,2) in the direction of \vec u is - - D_{\vec u\,}f(\pi/2,\pi,2) = \la 0,-1\ra \cdot \la -1/\sqrt{2},1/\sqrt 2\ra = -1/\sqrt 2 - . -

    - -

    - Thus the directional tangent line is - - \ell_{\vec u}(t) = \left\{\begin{array}{l} x= \pi/2 -t/\sqrt{2}\\ y = \pi/2 + t/\sqrt{2} \\ z= -t/\sqrt{2} - \end{array} \right. - . -

    - -

    - The curve through (\pi/2,\pi/2,0) in the direction of \vec v is shown in along with \ell_{\vec u}(t). -

    -     -     - - - Finding directional tangent lines - -

    - Let f(x,y) = 4xy-x^4-y^4. - Find the equations of all - directional tangent lines to f at (1,1). -

    -     - -

    - First note that f(1,1) = 2. - We need to compute directional derivatives, - so we need \nabla f. - We begin by computing partial derivatives. - - f_x = 4y-4x^3 \Rightarrow f_x(1,1) = 0; f_y = 4x-4y^3\Rightarrow f_y(1,1) = 0 - . -

    - -

    - Thus \nabla f(1,1) = \la 0,0\ra. - Let \vec u = \la u_1,u_2\ra be any unit vector. - The directional derivative of f at (1,1) will be D_{\vec u\,}f(1,1) = \la 0,0\ra\cdot \la u_1,u_2\ra = 0. - It does not matter what direction we choose; - the directional derivative is always 0. - Therefore - - \ell_{\vec u}(t) = \left\{\begin{array}{l} x= 1 +u_1t\\ y = 1+ u_2 t\\ z= 2 - \end{array} \right. - . -

    - -

    - - shows a graph of f and the point (1,1,2). - Note that this point comes at the top of a hill, - and therefore every tangent line through this point will have a slope of 0. -

    - -
    - Graphing f in - - - - A graph of the function in this example; it has the shape of a steep hill. - -

    - The graph z = 4xy-x^4-y^4 is plotted against a set of three-dimensional coordinate axes. - The shape of the surface given by the graph is that of a steep hill: - the degree-four terms in the function result in a surface that descends more rapidly than a paraboloid. -

    - -

    - The peak of the hill appears to be at the point (1,1,2). -

    -     - - - - - //ASY file for figpartial4b3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8,2.6,4.7); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={1,2}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-0.5,2.5); - pair ybounds=(-0.5,2.5); - pair zbounds=(-3.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=\addplot3[domain=-.1:2,y domain=-.1:2,surf,colormap={mp2}{\colormapplaneone}//,shader=faceted,faceted color=black!40,samples y=20,very thin,z buffer=sort,%opacity=.6, - //samples=20,] (x,y,{4*x*y-x^4-y^4}); - - bool cond(pair t) {return 4*t.x*t.y-(t.x)^4-(t.y)^4 >-3;} - - //triple f(pair t) { - // return (t.x,t.y,4*t.x*t.y-(t.x)^4-(t.y)^4); - //} - //surface s=surface(f,(-.1,-.1),(1.5,1.5),32,32,Spline); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (cos(t.x)*t.y+1,sin(t.x)*t.y+1,4*(cos(t.x)*t.y+1)*(sin(t.x)*t.y+1)-(cos(t.x)*t.y+1)^4-(sin(t.x)*t.y+1)^4); - } - surface s=surface(f,(0,0),(2*pi,.75),32,32,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3; - dot((1,1,2)); - - //Draw traces on the surface - //triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} - //path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); - - - - -
    - -

    - That is, consider any curve on the surface that goes through this point. - Each curve will have a relative maximum at this point, - hence its tangent line will have a slope of 0. - The following section investigates the points on surfaces where all tangent lines have a slope of 0. -

    -     -     -     - - - Normal Lines -

    - When dealing with a function y=f(x) of one variable, - we stated that a line through (c,f(c)) was tangent - to f if the line had a slope of \fp(c) and was - normal (or, perpendicular, - orthogonal) to f if it had a slope of -1/\fp(c). - We extend the concept of normal, - or orthogonal, to functions of two variables. -

    - -

    - Let z=f(x,y) be a differentiable function of two variables. - By , - at (x_0,y_0), - \ell_x(t) is a line parallel to the vector \vec d_x=\la 1,0,f_x(x_0,y_0)\ra and - \ell_y(t) is a line parallel to \vec d_y=\la 0,1,f_y(x_0,y_0)\ra. - Since lines in these directions through \big(x_0,y_0,f(x_0,y_0)\big) are - tangent to the surface, - a line through this point and orthogonal to these directions would be orthogonal, - or normal, to the surface. - We can use this direction to create a normal line. -

    - -

    - The direction of the normal line is orthogonal to \vec d_x and \vec d_y, - hence the direction is parallel to \vec d_n = \vec d_x\times \vec d_y. - It turns out this cross product has a very simple form: - - \vec d_x\times \vec d_y = \la 1,0,f_x\ra \times \la 0,1,f_y\ra = \la -f_x,-f_y,1\ra - . -

    - -

    - It is often more convenient to refer to the opposite of this direction, - namely \la f_x,f_y,-1\ra. - This leads to a definition. -

    - - - Normal Line - -

    - Let z=f(x,y) be differentiable on a set S containing (x_0,y_0) where - - a = f_x(x_0,y_0) \text{ and } b=f_y(x_0,y_0) - - are defined. - normal line - orthogonal -

    - -

    -

      -
    1. -

      - A nonzero vector parallel to \vec n=\la a,b,-1\ra is - orthogonal to f at P=\big(x_0,y_0,f(x_0,y_0)\big). -

      -
      2. - -
    2. -

      - The line \ell_n through P with direction parallel to \vec n is the - normal line to f at P. -

      -
      3. -
    -

    -     -     - -

    - Thus the parametric equations of the normal line to a surface z=f(x,y) at \big(x_0,y_0,f(x_0,y_0)\big) is: - - \ell_{n}(t) = \left\{\begin{array}{l} x= x_0+at\\ y = y_0 + bt \\ z = f(x_0,y_0) - t - \end{array} \right. - . -

    - - - Finding a normal line - -

    - Find the equation of the normal line to z=-x^2-y^2+2 at (0,1). -

    -     - -

    - We find z_x(x,y) = -2x and z_y(x,y) = -2y; - at (0,1), we have z_x = 0 and z_y = -2. - We take the direction of the normal line, - following , - to be \vec n=\la 0,-2,-1\ra. - The line with this direction going through the point (0,1,1) is - - \ell_n(t) = \left\{\begin{array}{l} x=0\\y=-2t+1\\z=-t+1 - \end{array} \right. \text{ or } \ell_n(t)=\la 0,-2,-1\ra t+\la 0,1,1\ra - . -

    - -
    - Graphing a surface with a normal line from - - - - A circular paraboloid, opening downward. At one point on the surface, a normal line is drawn. - -

    - The graph z=-x^2-y^2+2 is a circular paraboloid with vertex at (0,0,2), - and opening downward. - At the point (0,1,1) on the surface, a point is plotted, - and through this point, a normal line is drawn. - We can see that the normal line is perpendicular to the surface at the point where they meet. -

    -     - - - - - //ASY file for figpartial4b3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8,4.8,2.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=\addplot3[domain=-.1:2,y domain=-.1:2,surf,colormap={mp2}{\colormapplaneone}//,shader=faceted,faceted color=black!40,samples y=20,very thin,z buffer=sort,%opacity=.6, - //samples=20,] (x,y,{4*x*y-x^4-y^4}); - - bool cond(pair t) {return 2-(t.x)^2-(t.y)^2 >-2;} - - //triple f(pair t) { - // return (t.x,t.y,2-t.x^2-t.y^2); - //} - - triple f(pair t) { - return (cos(t.x)*t.y,sin(t.x)*t.y,2-t.y^2); - } - surface s=surface(f,(0,0),(2*pi,2),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - draw ((0,2.5,1.75)--(0,-1,0),redpen+2bp); - - //triple f(pair t) { - // return (cos(t.x)*t.y,sin(t.x)*t.y,4*cos(t.x)*t.y*sin(t.x)*t.y-(cos(t.x)*t.y)^4-(sin(t.x)//*t.y)^4); - //} - //surface s=surface(f,(-.1,0),(pi/2+.1,1.6),32,32,cond); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3; - dot((0,1,1)); - - //Draw traces on the surface - //triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} - //path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); - - - - -
    - -

    - The surface z=-x^2-y^2+2, - along with the found normal line, - is graphed in . -

    -     -     - -

    - The direction of the normal line has many uses, - one of which is the definition of the - tangent plane which we define shortly. - Another use is in measuring distances from the surface to a point. - Given a point Q in space, - it is a general geometric concept to define the distance from Q to the surface as being the length of the shortest line segment - \overline{PQ} over all points P on the surface. - This, in turn, - implies that \overrightarrow{PQ} will be orthogonal to the surface at P. - Therefore we can measure the distance from Q to the surface z=f(x,y) by finding a point P on the surface such that \overrightarrow{PQ} is parallel to the normal line to f at P. -

    - - - Finding the distance from a point to a surface - -

    - Let f(x,y) = 2-x^2-y^2 and let Q = (2,2,2). - Find the distance from Q to the surface defined by f. -

    -     - -

    - This surface is used in , - so we know that at (x,y), - the direction of the normal line will be \vec d_n = \la -2x,-2y,-1\ra. - A point P on the surface will have coordinates (x,y,2-x^2-y^2), - so \overrightarrow{PQ} = \la 2-x,2-y,x^2+y^2\ra. - To find where \overrightarrow{PQ} is parallel to - \vec d_n, we need to find x, - y and c such that c\overrightarrow{PQ} = \vec d_n. - - c\overrightarrow{PQ} \amp = \vec d_n - c\la 2-x,2-y,x^2+y^2\ra \amp = \la -2x,-2y,-1\ra. - This implies - c(2-x) \amp = -2x - c(2-y) \amp = -2y - c(x^2+y^2) \amp = -1 - -

    - -

    - In each equation, we can solve for c: - - c = \frac{-2x}{2-x} = \frac{-2y}{2-y} = \frac{-1}{x^2+y^2} - . -

    - -

    - The first two fractions imply x=y, - and so the last fraction can be rewritten as c=-1/(2x^2). - Then - - \frac{-2x}{2-x} \amp = \frac{-1}{2x^2} - -2x(2x^2) \amp = -1(2-x) - 4x^3 \amp = 2-x - 4x^3+x-2 \amp =0 - . -

    - -

    - This last equation is a cubic, - which is not difficult to solve with a numeric solver. - We find that x= 0.689, - hence P = (0.689,0.689, 1.051). - We find the distance from Q to the graph of f is - - \norm{\overrightarrow{PQ}} = \sqrt{(2-0.689)^2 +(2-0.689)^2+(2-1.051)^2} = 2.083 - . -

    -     -     - -

    - We can take the concept of measuring the distance from a point to a surface to find a point Q a particular distance from a surface at a given point P on the surface. -

    - - - Finding a point a set distance from a surface - -

    - Let f(x,y) = x-y^2+3. - Let P = \big(2,1,f(2,1)\big) = (2,1,4). - Find points Q in space that are 4 units from the graph of f at P. - That is, find Q such that - \norm{\overrightarrow{PQ}}=4 and \overrightarrow{PQ} is orthogonal to f at P. -

    -     - -

    - We begin by finding partial derivatives: - - f_x(x,y) =1 \qquad \amp \Rightarrow \qquad f_x(2,1) = 1 - f_y(x,y) = -2y \qquad \amp \Rightarrow \qquad f_y(2,1) = -2 - -

    - -

    - The vector \vec n=\la 1,-2,-1\ra is orthogonal to f at P. - For reasons that will become more clear in a moment, - we find the unit vector in the direction of \vec n: - - \vec u = \frac{\vec n}{\vnorm n} = \la 1/\sqrt{6},-2/\sqrt{6},-1/\sqrt{6}\ra \approx \la 0.408,-0.816,-0.408\ra - . -

    - -

    - Thus a the normal line to f at P can be written as - - \ell_n(t) = \la 2,1,4\ra + t\la 0.408,-0.816,-0.408 \ra - . -

    - -

    - An advantage of this parametrization of the line is that letting t=t_0 gives a point on the line that is \abs{t_0} units from P. - (This is because the direction of the line is given in terms of a unit vector.) - There are thus two points in space 4 units from P: - - Q_1 \amp = \ell_n(4) \amp Q_2 \amp = \ell_n(-4) - \amp \approx \la 3.63, -2.27, 2.37\ra \amp \amp \approx\la 0.37, 4.27, 5.63\ra - -

    - -
    - Graphing the surface in along with points 4 units from the surface - - - - A parabolic cylinder, and a normal line to the surface at one point. - -

    - The surface is a parabolic cylinder: - the shape one would obtain by bending a sheet of paper so that its cross-sections are parabolas. - At one point on the surface, a normal line has been drawn. - The normal line extends a distance of four units on either side of the surface; - additional points are plotted at the two ends. -

    -     - - - - - //ASY file for figpartial4b3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(15,10,9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={2,4}; - defaultpen(0.5mm); - pair xbounds=(-1,5.5); - pair ybounds=(-2,4.5); - pair zbounds=(-1,5.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // -y^2+x+3 - - bool cond(pair t) {return 4*t.x*t.y-(t.x)^4-(t.y)^4 >-3;} - - triple f(pair t) { - return (t.x,t.y,3+t.x-t.y^2); - } - surface s=surface(f,(-.5,-2.1),(4,2),8,16,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic},usplinetype=Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3; - dot((2,1,4)); - dot((3.63,-2.27,2.37)); - dot((.37,4.27,5.63)); - - draw ((.37,4.27,5.63)--(3.63,-2.27,2.37),redpen+2bp); - - //Draw traces on the surface - //triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} - //path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); - - - - -
    - -

    - The surface is graphed along with points P, Q_1, - Q_2 and a portion of the normal line to f at P. -

    -     -     -     - - - Tangent Planes -

    - We can use the direction of the normal line to define a plane. - With a=f_x(x_0,y_0), - b=f_y(x_0,y_0) and P = \big(x_0,y_0,f(x_0,y_0)\big), - the vector \vec n=\la a,b,-1\ra is orthogonal to f at P. - (See .) - The plane through P with normal vector \vec n is therefore - tangent to f at P. -

    - - - - - Tangent Plane - -

    - Let z=f(x,y) be differentiable on a set S containing (x_0,y_0), - where a = f_x(x_0,y_0), b=f_y(x_0,y_0), - \vec n= \la a,b,-1\ra and P=\big(x_0,y_0,f(x_0,y_0)\big). -

    - -

    - The plane through P with normal vector \vec n is the - tangent plane to f at P. - The standard form of this plane is - tangent plane - planestangent - - a(x-x_0) + b(y-y_0) - \big(z-f(x_0,y_0)\big) = 0 - . -

    -     -     - - - Finding tangent planes - -

    - Find the equation of the tangent plane to z=-x^2-y^2+2 at (0,1). -

    -     - -

    - Note that this is the same surface and point used in . - There we found \vec n = \la 0,-2,-1\ra and P = (0,1,1). - Therefore the equation of the tangent plane is - - -2(y-1)-(z-1)=0 - . -

    - -

    - The surface z=-x^2-y^2+2 and tangent plane are graphed in . -

    - -
    - Graphing a surface with tangent plane from - - - - A downward-opening circular paraboloid, and a tangent plane to this surface at one point. - -

    - The surface is a circular paraboloid, opening downward. - One point is marked on the surface, and at this point, a tangent plane is drawn. - The tangent plane is illustrated as a rectangle pinned to the surface. -

    - -

    - By rotating the image, we are able to observe how the tangent plane lies close to the surface, - near the point of tangency. -

    -     - - - - - //ASY file for figpartial4b3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(13,10,9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - pair xbounds=(-3,3); - pair ybounds=(-3,3); - pair zbounds=(-3,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // -x^2-y^2+2 - - triple f(pair t) { - return (cos(t.x)*t.y,sin(t.x)*t.y,2-t.y^2); - } - surface s=surface(f,(0,0),(2*pi,2),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3; - dot((0,1,1)); - - // -2*y+3 - triple g(pair t) { - return (t.x,t.y,-2*t.y+3);} - surface ss=surface(g,(-1,.5),(1,1.5),1,1); - pen q=redcurvepen+.1mm; - draw(ss,surfacepen2,meshpen=q,nolight,render(merge=true)); - - //Draw traces on the surface - //triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} - //path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); - - - - -
    -     -     - - - Using the tangent plane to approximate function values - -

    - The point (3,-1,4) lies on the graph of an unknown differentiable function f where - f_x(3,-1) = 2 and f_y(3,-1) = -1/2. - Find the equation of the tangent plane to f at P, - and use this to approximate the value of f(2.9,-0.8). -

    -     - -

    - Knowing the partial derivatives at (3,-1) allows us to form the normal vector to the tangent plane, - \vec n = \la 2,-1/2,-1\ra. - Thus the equation of the tangent line to f at P is: - - 2(x-3)-1/2(y+1) - (z-4) = 0 \Rightarrow z = 2(x-3)-1/2(y+1)+4 - . -

    - -

    - Just as tangent lines provide excellent approximations of curves near their point of intersection, - tangent planes provide excellent approximations of surfaces near their point of intersection. - So f(2.9,-0.8) \approx z(2.9,-0.8) = 3.7. -

    - -

    - This is not a new method of approximation. - Compare the right hand expression for z in Equation to the total differential: - - dz = f_xdx + f_ydy \text{ and } z = \underbrace{\underbrace{2}_{f_x}\underbrace{(x-3)}_{dx}+\underbrace{-1/2}_{f_y}\underbrace{(y+1)}_{dy}}_{dz}+4 - . -

    - -

    - Thus the new z-value - is the sum of the change in z (, dz) and the old z-value (4). - As mentioned when studying the total differential, - it is not uncommon to know partial derivative information about a unknown function, - and tangent planes are used to give accurate approximations of the function. -

    -     -     -     - - - The Gradient and Normal Lines, Tangent Planes -

    - The methods developed in this section so far give a straightforward method of finding equations of normal lines and tangent planes for surfaces with explicit equations of the form z=f(x,y). - However, they do not handle implicit equations well, - such as x^2+y^2+z^2=1. - There is a technique that allows us to find vectors orthogonal to these surfaces based on the gradient. -

    - - - Gradient - -

    - Let w=F(x,y,z) be differentiable on a set D that contains the point (x_0,y_0,z_0). - gradient -

    - -

    -

      -
    1. -

      - The gradient of F - is \nabla F(x,y,z) = \la f_x(x,y,z),f_y(x,y,z),f_z(x,y,z)\ra. -

      -
      2. - -
    2. -

      - The gradient of F at (x_0,y_0,z_0) is - - \nabla F(x_0,y_0,z_0) = \la f_x(x_0,y_0,z_0),f_y(x_0,y_0,z_0),f_z(x_0,y_0,z_0)\ra - . -

      -
      3. -
    -

    -     -     - -

    - Recall that when z=f(x,y), - the gradient \nabla f = \la f_x,f_y\ra is orthogonal to level curves of f. - An analogous statement can be made about the gradient - \nabla F, where w= F(x,y,z). - Given a point (x_0,y_0,z_0), - let c = F(x_0,y_0,z_0). - Then F(x,y,z) = c is a level surface - that contains the point (x_0,y_0,z_0). - The following theorem states that - \nabla F(x_0,y_0,z_0) is orthogonal to this level surface. -

    - - - The Gradient and Level Surfaces - -

    - Let w=F(x,y,z) be differentiable on a set D containing - (x_0,y_0,z_0) with gradient \nabla F, - where F(x_0,y_0,z_0) = c. -

    - -

    - The vector \nabla F(x_0,y_0,z_0) is orthogonal to the level surface - F(x,y,z)=c at (x_0,y_0,z_0). - gradientand level surfaces - level surface -

    -     -     - -

    - The gradient at a point gives a vector orthogonal to the surface at that point. - This direction can be used to find tangent planes and normal lines. -

    - - - Using the gradient to find a tangent plane - -

    - Find the equation of the plane tangent to the ellipsoid - \frac{x^2}{12} +\frac{y^2}{6}+\frac{z^2}{4}=1 at P = (1,2,1). -

    -     - -

    - We consider the equation of the ellipsoid as a level surface of a function F of three variables, - where F(x,y,z) = \frac{x^2}{12} +\frac{y^2}{6}+\frac{z^2}{4}. - The gradient is: - - \nabla F(x,y,z) \amp = \la F_x, F_y,F_z\ra - \amp = \la \frac x6, \frac y3, \frac z2\ra - . -

    - -

    - At P, - the gradient is \nabla F(1,2,1) = \la 1/6, 2/3, 1/2\ra. - Thus the equation of the plane tangent to the ellipsoid at P is - - \frac 16(x-1) + \frac23(y-2) + \frac 12(z-1) = 0 - . -

    - -

    - The ellipsoid and tangent plane are graphed in . -

    - -
    - An ellipsoid and its tangent plane at a point - - - - An ellipsoid centered at the origin in space, and a tangent plane at one point. - -

    - An ellipsoid is plotted against a set of three-dimensional coordinate axes, - with its center at the origin. The ellipsoid is egg-like in shape. -

    - -

    - At one point on the ellipsoid, a point is marked, and a tangent plane is drawn at that point. - The tangent plane is illustrated as a flat rectangle stuck to the surface at the point of tangency. - By rotating the image, we are able to observe how the tangent plane lies close to the surface, - as long as we are not far from the point of tangency. -

    -     - - - - - //ASY file for figpartial4b3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14,11,4.7); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-3.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - // ({cos(y)*(3.46*cos(x))},{cos(y)*(2.45*sin(x))},{2*sin(y)}) - - triple f(pair t) { - return (cos(t.y)*(3.46*cos(t.x)),cos(t.y)*(2.45*sin(t.x)),2*sin(t.y)); - } - surface s=surface(f,(-pi,-pi/2),(pi,pi/2),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=3; - dot((1,2,1)); - - // 1-(x-1)/3-4*(y-2)/3 1. - 0.980581 x, 2. + 0.156893 x + 0.6 y, 1. + 0.11767 x - 0.8 y - //triple g(pair t) { - // return (t.x,t.y,1-(t.x-1)/3-4*(t.y-2)/3);} - //surface ss=surface(g,(.5,1.5),(1.5,2.5),1,1); - triple g(pair t) { - return (1.-0.980581*t.x, 2.+0.156893*t.x+0.6*t.y,1.+0.11767*t.x-0.8*t.y);} - surface ss=surface(g,(-.75,-.75),(.75,.75),1,1); - pen q=redcurvepen+.1mm; - draw(ss,surfacepen2,meshpen=q,nolight,render(merge=true)); - - //Draw traces on the surface - //triple g(real t) {return (t,pi-t,sin(t)*cos(pi-t));} - //path3 mypath=graph(g,0,3.75,operator ..); draw(mypath,bluepen); - - - - -
    -     -     - -

    - To understand why is true, - recall the method of implicit differentiation given in . - A level surface f(x,y,z)=0 can be viewed as defining z=g(x,y) implicitly. - We found that the partial derivatives of z with respect to x and y are then given by - - \plz{z}{x} = g_x(x,y) = -\frac{f_x(x,y,z)}{f_z(x,y,z)}\quad \plz{z}{y} = g_y(x,y) = -\frac{f_y(x,y,z)}{f_z(x,y,z)} - . - If we plug these values into the tangent plane equation - - z = g(a,b)+g_x(a,b)(x-a)+g_y(a,b)(y-b) - - we get, with c=g(a,b), - - z = c - \frac{f_x(a,b,c)}{f_z(a,b,c)}(x-a)-\frac{f_y(a,b,c)}{f_z(a,b,c)}(y-b) - . - If we move everything to the left-hand side of the equation and multiply through by f_z(a,b,c), - we get - - f_x(a,b,c)(x-a)+f_y(a,b,c)(y-b)+f_z(a,b,c)(z-c)=0 - , - which is the equation of a plane with normal vector \nabla f(a,b,c). -

    - - - Finding the tangent plane of a level surface - -

    - Determine the equation of the tangent plane to the level surface - x^2yz^3-\sin(x-3z)+4xy^2-3yz=0 at the point (3,0,1). - (Note that this is the same problem as .) -

    -     - -

    - With f(x,y,z)=x^2yz^3-\sin(x-3z)+4xy^2-3yz we have - - f_x(x,y,z) \amp = 2xyz^3-\cos(x-3z)+4y^2 \amp f_x(3,0,1)\amp = -1 - f_y(x,y,z) \amp = x^2z^3+8xy-3z \amp f_y(3,0,1)\amp = 6 - f_z(x,y,z) \amp = 3x^2yz^2+3\cos(x-3z)-3y \amp f_z(3,0,1) \amp = 3 - . -

    -

    - The equation of the tangent plane is therefore - - -1(x-3)+6y+3(z-1)=0 - . - Note that solving for z gives z=1+\frac13(x-3)-2y, - which is the same result as . -

    -     -     - -

    - Tangent lines and planes to surfaces have many uses, - including the study of instantaneous rates of changes and making approximations. - Normal lines also have many uses. - In this section we focused on using them to measure distances from a surface. - Another interesting application is in computer graphics, - where the effects of light on a surface are determined using normal vectors. -

    - -

    - The next section investigates another use of partial derivatives: - determining relative extrema. - When dealing with functions of the form y=f(x), - we found relative extrema by finding x where \fp(x) = 0. - We can start finding relative extrema of - z=f(x,y) by setting f_x and f_y to 0, but it turns out that there is more to consider. -

    -     - - - - Terms and Concepts - - - - -

    - Explain how the vector \vec v=\la 1,0,3\ra can be thought of as having a slope of 3. -

    -     - - - -

    - Answers will vary. - The displacement of the vector is one unit in the x-direction and 3 units in the z-direction, - with no change in y. - Thus along a line parallel to \vec v, - the change in z is 3 times the change in x , a slope of 3. - Specifically, - the line in the xz-plane parallel to z has a slope of 3. -

    -     - -     - - - - -

    - Explain how the vector \vec v=\la 0.6,0.8, -2\ra can be thought of as having a - slope of -2. -

    -     - - - -

    - Answers will vary. - Let \vec u = \la 0.6,0.8\ra; - this is a unit vector. - The displacement of the vector is one unit in the \vec u-direction and -2 units in the z-direction. - In the plane containing the z-axis and the vector \vec u, - the line parallel to \vec v has slope -2. -

    -     - -     - - - - - -

    - Let z=f(x,y) be differentiable at P. - If \vec n is a normal vector to the tangent plane of f at P, - then \vec n is orthogonal to \ell_x and \ell_y at P. - -

    -     - -     - - - - -

    - Explain in your own words why we do not refer to the - tangent line to a surface at a point, - but rather to directional - tangent lines to a surface at a point. -

    -     - - - -

    - On a surface through a point, - there are many different smooth curves, - each with a tangent line at the point. - Each of these tangent lines is also - tangent to the surface. - There is not just one tangent line, - but many, each in a different direction. - Therefore we refer to directional tangent lines, - not just the tangent line. -

    -     - -     -     - - - Problems - - - -

    - A function f(x,y), - a vector \vec v and a point P are given. - Give the parametric equations of the following directional tangent lines to z=f(x,y) at P: -

    - -

    -

      -
    1. -

      - \ell_x(t) -

      -
      2. - -
    2. -

      - \ell_y(t) -

      -
      3. - -
    3. -

      - \ell_{\vec u\,}(t), - where \vec u is the unit vector in the direction of \vec v. -

      -
      4. -
    -

    -     - - - - -

    - f(x,y) = 2x^2y-4xy^2, \vec v = \la 1,3\ra, - P=(2,3). -

    -     - -

    -

      -
    1. -

      - \ell_x(t) = \left\{\begin{array}{l} x=2+t\\ y=3 \\ z = -48-12t - \end{array} \right. -

      -
      2. - -
    2. -

      - \ell_y(t) = \left\{\begin{array}{l} x=2\\ y=3+t \\ z = -48-40t - \end{array} \right. -

      -
      3. - -
    3. -

      - \ell_{\vec u\,}(t) = \left\{\begin{array}{l} x=2+t/\sqrt{10}\\ y=3+3t/\sqrt{10} \\ z = -48-66\sqrt{2/5}t - \end{array} \right. -

      -
      4. -
    -

    -     - -     - - - - - - Context("Vector"); - Context()->variables->add(t=>"Real"); - $v[0]=Compute("(pi/3,pi/6,3/4)+t<1,0,-3sqrt(3)/4>"); - $v[1]=Compute("(pi/3,pi/6,3/4)+t<0,1,3sqrt(3)/4>"); - $v[2]=Compute("(pi/3,pi/6,3/4)+t<1/sqrt(5),2/sqrt(5),3sqrt(3/5)/4>"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - parser::Assignment->Allow; - $p[0]=List(Formula("x=pi/3+t"),Formula("y=pi/6"),Formula("z=3/4-3sqrt(3)/4t")); - $p[1]=List(Formula("x=pi/3"),Formula("y=pi/6+t"),Formula("z=3/4+3sqrt(3)/4t")); - $p[2]=List(Formula("x=pi/3+t/sqrt(5)"),Formula("y=pi/6+2t/sqrt(5)"),Formula("z=3/4+3sqrt(3/5)/4t")); - for my $k (0..2) { - $pev[$k]=$p[$k]->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - }} - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - }} - if(!$aH->{isPreview}){ - push(@err,"You need to provide more parametrizations")if $i<3; - push(@err,"You have given too many parametrizations")if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v[$k]->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v[$k]->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - }; - - - -

    - f(x,y) = 3\cos(x) \sin(y), \vec v = \la 1,2\ra, P=(\pi/3, \pi/6) -

    - - - Find parametric equations for the directional tangent line \ell_{x} at P. - -

    - -

    - - - Find parametric equations for the directional tangent line \ell_{y} at P. - -

    - -

    - - Find parametric equations for the directional tangent line \ell_{\vec{v}} at P. - -

    - -

    -     -     -     - - - - -

    - f(x,y) = 3x-5y, \vec v = \la 1,1\ra, P=(4,2). -

    -     - -

    -

      -
    1. -

      - \ell_x(t) = \left\{\begin{array}{l} x = 4+t\\ y = 2 \\ z = 2 + 3t - \end{array} \right. -

      -
      2. - -
    2. -

      - \ell_y(t) = \left\{\begin{array}{l} x = 4\\ y = 2+t\\ z = 2-5t - \end{array} \right. -

      -
      3. - -
    3. -

      - \ell_{\vec u\,}(t) = \left\{\begin{array}{l} x = 4+t/\sqrt{2}\\ y = 2+t/\sqrt{2} \\ z = 2 -\sqrt{2}t - \end{array} \right. -

      -
      4. -
    -

    -     - -     - - - - - - Context("Vector"); - Context()->variables->add(t=>"Real"); - $v[0]=Compute("(1,2,3)+t<1,0,0>"); - $v[1]=Compute("(1,2,3)+t<0,1,0>"); - $v[2]=Compute("(1,2,3)+t<1/sqrt(2),1/sqrt(2),0>"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - parser::Assignment->Allow; - $p[0]=List(Formula("x=1+t"),Formula("y=2"),Formula("z=3")); - $p[1]=List(Formula("x=1"),Formula("y=2+t"),Formula("z=3")); - $p[2]=List(Formula("x=1+t/sqrt(2)"),Formula("y=2+t/sqrt(2)"),Formula("z=3")); - for my $k (0..2) { - $pev[$k]=$p[$k]->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - }} - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - }} - if(!$aH->{isPreview}){ - push(@err,"You need to provide more parametrizations")if $i<3; - push(@err,"You have given too many parametrizations")if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v[$k]->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v[$k]->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - }; - - - -

    - f(x,y) = x^2-2x-y^2+4y, \vec v = \la 1,1\ra, P=(1, 2) -

    - - - Find parametric equations for the directional tangent line \ell_{x} at P. - -

    - -

    - - Find parametric equations for the directional tangent line \ell_{y} at P. - -

    - -

    - - Find parametric equations for the directional tangent line \ell_{\vec{v}} at P. - -

    - -

    -     -     -     - -     - - - -

    - A function f(x,y) and a point P are given. - Find the equation of the normal line to z=f(x,y) at P. - Note: these are the same functions as in Exercises. -

    -     - - - - -

    - f(x,y) = 2x^2y-4xy^2, P=(2,3). -

    -     - -

    - \ell_{\vec n}(t) = \left\{\begin{array}{l} x=2-12t\\ y=3-40t \\ z = -48-t - \end{array} \right. -

    -     - -     - - - - - - Context("Vector"); - Context()->variables->are(t=>"Real"); - $v=Compute("(pi/3,pi/6,3/4)+t<-3sqrt(3)/4,3sqrt(3)/4,-1>"); - Context("Numeric"); - Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - parser::Assignment->Allow; - $n=List(Formula("x=pi/3-3sqrt(3)/4t"),Formula("y=pi/6+3sqrt(3)/4t"),Formula("z=3/4-t")); - $nev=$n->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - }} - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - }} - if(!$aH->{isPreview}){ - push(@err,"You need to provide more parametrizations")if $i<3; - push(@err,"You have given too many parametrizations")if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - - - -

    - f(x,y) = 3\cos(x) \sin(y) and P=(\pi/3, \pi/6) -

    - - Find parametric equations for the normal line at P. - - -

    - -

    -     -     -     - - - - -

    - f(x,y) = 3x-5y, P=(4,2). -

    -     - -

    - \ell_{\vec n}(t) = \left\{\begin{array}{l} x = 4+3t\\ y = 2-5t \\ z = 2 -t - \end{array} \right. -

    -     - -     - - - - - - Context("Vector"); - Context()->variables->are(t=>"Real"); - $v=Compute("(1,2,3)+t<0,0,-1>"); - Context("Numeric"); - Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - parser::Assignment->Allow; - $n=List(Formula("x=1"),Formula("y=2"),Formula("z=3-t")); - $nev=$n->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - }} - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - }} - if(!$aH->{isPreview}){ - push(@err,"You need to provide more parametrizations")if $i<3; - push(@err,"You have given too many parametrizations")if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - - - -

    - f(x,y) = x^2-2x-y^2+4y and P=(1,2) -

    - - Find parametric equations for the normal line at P. - -

    - -

    -     -     -     - -     - - - -

    - A function f(x,y) and a point P are given. - Find the two points that are 2 units from the surface z=f(x,y) at P. - Note: these are the same functions as in Exercises. -

    -     - - - - -

    - f(x,y) = 2x^2y-4xy^2, P=(2,3). -

    -     - -

    - (1.425, 1.085, -48.078), - (2.575, 4.915, -47.952) -

    -     - -     - - - - -

    - f(x,y) = 3\cos(x) \sin(y), P=(\pi/3, \pi/6). -

    -     - -

    - (-0.195,1.766,-0.206) and (2.289,-0.719, 1.706) -

    -     - -     - - - - -

    - f(x,y) = 3x-5y, P=(4,2). -

    -     - -

    - (5.014, 0.31, 1.662) and (2.986, 3.690, 2.338) -

    -     - -     - - - - -

    - f(x,y) = x^2-2x-y^2+4y, P=(1,2). -

    -     - -

    - (1,2,1) and (1,2,5) -

    -     - -     - -     - - - -

    - A function f(x,y) and a point P are given. - Find an equation of the tangent plane to z=f(x,y) at P. - Note: these are the same functions as in Exercises. -

    -     - - - - -

    - f(x,y) = 2x^2y-4xy^2, P=(2,3). -

    -     - -

    - -12(x-2)-40(y-3) -(z+48) = 0 -

    -     - -     - - - - - Context("ImplicitPlane"); - $t=Compute("-3sqrt(3)/4(x-pi/3)+3sqrt(3)/4(y-pi/6)-(z-3/4)=0"); - - - -

    - f(x,y) = 3\cos(x) \sin(y), P=(\pi/3, \pi/6). -

    - - - Find an equation of the tangent plane to z=f(x,y) at P. - -

    - -

    -     -     -     - - - - -

    - f(x,y) = 3x-5y, P=(4,2). -

    -     - -

    - 3(x-4)-5(y-2) - (z-2) = 0 (Note that this tangent plane is the same as the original function, - a plane.) -

    -     - -     - - - - - Context("ImplicitPlane"); - $t=Compute("z=3"); - - - -

    - f(x,y) = x^2-2x-y^2+4y, P=(1,2) -

    - - Find an equation of the tangent plane to z=f(x,y) at P. - -

    - -

    -     -     -     - -     - - - -

    - An implicitly defined function of x, - y and z is given, along with a point P that lies on the surface. - Use the gradient \nabla F to: - -

      -
    1. -

      - find the equation of the normal line to the surface at P, and -

      -
      2. - -
    2. -

      - find the equation of the plane tangent to the surface at P. -

      -
      3. -
    -

    -     - - - - -

    - \ds \frac{x^2}{8}+\frac{y^2}4+\frac{z^2}{16}=1, - at P = (1,\sqrt{2},\sqrt{6}) -

    -     - -

    - \nabla F = \la x/4, y/2, z/8\ra; - at P, - \nabla F = \la 1/4, \sqrt{2}/2, \sqrt{6}/8\ra -

    - -

    -

      -
    1. -

      - \ell_{\vec n}(t) = \left\{\begin{array}{l} x= 1+ t/4 \\ y = \sqrt{2}+ \sqrt{2}t/2\\ z = \sqrt{6} + \sqrt{6}t/8 - \end{array} \right. -

      -
      2. - -
    2. -

      - \frac14(x-1) + \frac{\sqrt{2}}{2}(y-\sqrt{2}) + \frac{\sqrt{6}}8(z-\sqrt{6}) = 0. -

      -
      3. -
    -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(t=>"Real"); - $v=Compute("(4,-3,sqrt(5))+t<-2,2/3,2sqrt(5)>"); - Context("Numeric"); - Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - parser::Assignment->Allow; - $n=List(Formula("x=4-2t"),Formula("y=-3+2/3t"),Formula("z=sqrt(5)+2sqrt(5)t")); - $nev=$n->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - }} - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - }} - if(!$aH->{isPreview}){ - push(@err,"You need to provide more parametrizations")if $i<3; - push(@err,"You have given too many parametrizations")if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - Context("ImplicitPlane"); - $t=Compute("-2(x-4) + 2/3(y+3) + 2sqrt(5)(z-sqrt(5)) = 0"); - - - -

    - z^2-\frac{x^2}{4} - \frac{y^2}9=0, - at P = (4,-3,\sqrt{5}) -

    - - Find parametric equations for the normal line. - -

    - -

    - - Find an equation of the tangent plane. - -

    - -

    -     -     -     - - - - -

    - \ds xy^2-xz^2=0, at P = (2,1,-1) -

    -     - -

    - \nabla F = \la y^2-z^2, 2xy, -2xz\ra; - at P, \nabla F = \la 0, 4, 4\ra -

    - -

    -

      -
    1. -

      - \ell_{\vec n}(t) = \left\{\begin{array}{l} x= 2 \\ y = 1+4t\\ z = -1+4t - \end{array} \right. -

      -
      2. - -
    2. -

      - 4(y-1) + 4(z+1) = 0. -

      -
      3. -
    -

    -     - -     - - - - - Context("Vector"); - Context()->variables->are(t=>"Real"); - $v=Compute("(2,pi/12,4)+t<pi/(8sqrt(3)),-sqrt(3),-pi/(8sqrt(3))>"); - Context("Numeric"); - Context()->variables->add(y=>"Real",z=>"Real",t=>"Real"); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - parser::Assignment->Allow; - $n=List(Formula("x=2+pi/(8sqrt(3))t"),Formula("y=pi/12-sqrt(3)t"),Formula("z=4-pi/(8sqrt(3))t")); - $nev=$n->cmp(list_checker=>sub{my($c,$st,$aH,$value)=@_; - my$n=scalar(@$st); - my@err=();my$i,$j; - Context("Vector"); - Context()->variables->add(t=>"Real"); - my$sV=Formula("(0,0,0)+t*<0,0,0>"); - for($i=0;$i<$n;$i++){ - my$ith=Value::List->NameForNumber($i+1); - my$entry=$st->[$i]; - my$var;my$for; - if($entry->type ne "Assignment"){push(@err,"Your $ith entry is not a parametrization");next;} - else{($var,$for)=split('=',"$entry"); - $var=Formula("$var");$for=Formula("$for"); - if(!($var!=Formula("x") or $var!=Formula("y") or $var!=Formula("z")) or $for->usesOneOf("x","y","z")) - {push(@err,"Your $ith entry is not a parametrization")unless $aH->{isPreview};next;} - }; - for($j=0,$used=0;$j<$i;$j++){ - if($st->[$j]->type eq "Assignment"){ - my($vj,$fj)=split('=',$st->[$j]->string); - $vj=Formula("$vj"); - if($vj==$var) {push(@err,"Your $ith entry parametrizes the same variable as a previous one") unless $aH->{isPreview};$used = 1;last;} - }} - if(!$used){ - if($var eq 'x'){$sV=$sV+Formula("<$for,0,0>")} - elsif($var eq 'y'){$sV=$sV+Formula("<0,$for,0>")} - else{$sV=$sV+Formula("<0,0,$for>")}; - }} - if(!$aH->{isPreview}){ - push(@err,"You need to provide more parametrizations")if $i<3; - push(@err,"You have given too many parametrizations")if $i>3; - } - $sV=Formula($sV); - $ds=$sV->D('t')->reduce; - if($ds->isConstant){$ds=Vector("$ds");} else{return 0;}; - $dc=$v->D('t')->reduce;$dc=Vector("$dc");$par=$ds->isParallel($dc); - $sp=$sV->eval(t=>0);$cp=$v->eval(t=>0); - $tch=(($sp==$cp) or $dc->isParallel($sp-$cp)); - return (3*($par and $tch),@err); - }); - Context("ImplicitPlane"); - $t=Compute("pi/(8sqrt(3))(x-2) -sqrt(3)(y-pi/12) -pi/(8sqrt(3))(z-4) = 0"); - - - -

    - \sin(xy)+\cos(yz)=1, - at P = (2, \pi/12, 4) -

    - - - Find parametric equations for the normal line. - -

    - -

    - - Find an equation of the tangent plane. - -

    - -

    -     -     -     - -     -     -     -
    -
    - Extreme Values - - Critical Points of Functions of Two Variables - -

    - Given a function f(x,y), - we are often interested in points where z=f(x,y) takes on the largest or smallest values. - For instance, if f represents a cost function, - we would likely want to know what (x,y) values minimize the cost. - If f represents the ratio of a volume to surface area, - we would likely want to know where f is greatest. - This leads to the following definition. -

    - - - Relative and Absolute Extrema - -

    - Let z=f(x,y) be defined on a set S containing the point P=(x_0,y_0). - maximumrelative/local - minimumrelative/local - extremarelative - extremaabsolute - maximumabsolute - minimumabsolute - -

      -
    1. -

      - If f(x_0,y_0)\geq f(x,y) for all (x,y) in S, - then f has an absolute maximum - at P If f(x_0,y_0)\leq f(x,y) for all (x,y) in S, - then f has an absolute minimum at P. -

      -
      2. - -
    2. -

      - If there is an open disk D containing P such that - f(x_0,y_0) \geq f(x,y) for all points (x,y) that are in both D and S, - then f has a relative maximum at P. - If there is an open disk D containing P such that - f(x_0,y_0) \leq f(x,y) for all points (x,y) that are in both D and S, - then f has a relative minimum at P. -

      -
      3. - -
    3. -

      - If f has an absolute maximum or minimum at P, - then f has an absolute extremum at P. - If f has a relative maximum or minimum at P, - then f has a relative extremum at P. -

      -
      4. -
    -

    -     -     - -

    - If f has a relative or absolute maximum at (x_0,y_0), - it means every curve on the graph of f through (x_0,y_0,f(x_0,y_0)) will also have a relative or absolute maximum at P. - Recalling what we learned in , - the slopes of the tangent lines to these curves at P must be 0 or undefined. - Since directional derivatives are computed using f_x and f_y, - we are led to the following definition and theorem. -

    - - - Critical Point - -

    - Let z = f(x,y) be continuous on a set S. - A critical point - P=(x_0,y_0) of f is a point in S such that, - at P, - critical point -

      -
    • -

      - f_x(x_0,y_0) = 0 and f_y(x_0,y_0) = 0, or -

      -
      • - -
    • -

      - f_x(x_0,y_0) and/or f_y(x_0,y_0) is undefined. -

      -
      • -
    -

    -     -     - - - Critical Points and Relative Extrema - -

    - Let z=f(x,y) be defined on an open set S containing P=(x_0,y_0). - If f has a relative extrema at P, - then P is a critical point of f. - extremarelative - critical point -

    -     -     - -

    - Therefore, to find relative extrema, - we find the critical points of f and determine which correspond to relative maxima, - relative minima, or neither. - The following examples demonstrate this process. -

    - - - - - Finding critical points and relative extrema - -

    - Let f(x,y) = x^2+y^2-xy-x-2. - Find the relative extrema of f. -

    -     - -

    - We start by computing the partial derivatives of f: - - f_x(x,y) = 2x-y-1 \qquad \text{ and } \qquad f_y(x,y) = 2y-x - . -

    - -

    - Each is never undefined. - A critical point occurs when f_x and f_y are simultaneously 0, leading us to solve the following system of linear equations: - - 2x-y-1 = 0\qquad \text{ and } \qquad -x+2y = 0 - . -

    - -

    - This solution to this system is x=2/3, - y=1/3. (Check that at (2/3,1/3), - both f_x and f_y are 0.) -

    - -
    - The surface in with its absolute minimum indicated - - - - A circular paraboloid plotted over a rectangular domain. It is bowl-shaped, with peaks at the corners of the domain. - -

    - The image shows a three-dimensional plot of the surface z = x^2+y^2-xy-x-2. - This is a circular paraboloid, opening upward. - It is bowl-shaped, and because it is plotted over a rectangular domain, - we see peaks at points on the surface over the four corners of the domain. - The bottom of the bowl shape is the location of the absolute minimum. -

    -     - - - - - //ASY file for figmulti_extreme13D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={5}; - defaultpen(0.5mm); - pair xbounds=(-3,3); - pair ybounds=(-3,3); - pair zbounds=(-1,6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=x^2+y^2-xy-x-2 - triple f(pair t) { - return (t.x,t.y,t.x^2+t.y^2-t.x*t.y-t.x-2); - } - surface s=surface(f,(-1,-1),(2,2),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=4;dot((2/3,1/3,-7/3)); - - - - -
    - -

    - The graph in - shows f along with this critical point. - It is clear from the graph that this is a relative minimum; - further consideration of the function shows that this is actually the absolute minimum. -

    -     -     - - - Finding critical points and relative extrema - -

    - Let f(x,y) = -\sqrt{x^2+y^2}+2. - Find the relative extrema of f. -

    -     - -

    - We start by computing the partial derivatives of f: - - f_x(x,y) = \frac{-x}{\sqrt{x^2+y^2}}\qquad \text{ and } \qquad f_y(x,y) = \frac{-y}{\sqrt{x^2+y^2}} - . -

    - -

    - It is clear that f_x=0 when x=0 & y\neq0, - and that f_y=0 when y=0 & x\neq0. - At (0,0), both f_x and f_y are not - 0, but rather undefined. - The point (0,0) is still a critical point, - though, because the partial derivatives are undefined. - This is the only critical point of f. -

    - -
    - The surface in with its absolute maximum indicated - - - - An inverted cone, with a maximum corresponding to a sharp peak on the z axis. - -

    - The surface z = -\sqrt{x^2+y^2}+2 is an inverted circular cone. - The image shows a plot of this graph over a rectangular domain. - At the point (0,0,2) there is a sharp peak; - this is an absolute maximum corresponding to a point where the partial derivatives are undefined. -

    -     - - - - - //ASY file for figmulti_extreme23D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-3,3); - pair ybounds=(-3,3); - pair zbounds=(-1,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=2-sqrt(x^2+y^2) - triple f(pair t) { - return (t.x,t.y,2-sqrt(t.x^2+t.y^2)); - } - surface s=surface(f,(-2,-2),(2,2),32,32); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=4;dot((0,0,2)); - - - - -
    - -

    - The graph of f is plotted in - along with the point (0,0,2). - The graph shows that this point is the absolute maximum of f. -

    -     -     - -

    - In each of the previous two examples, - we found a critical point of f and then determined whether or not it was a relative - (or absolute) - maximum or minimum by graphing. - It would be nice to be able to determine whether a critical point corresponded to a max or a min without a graph. - Before we develop such a test, - we do one more example that sheds more light on the issues our test needs to consider. -

    - - - Finding critical points and relative extrema - -

    - Let f(x,y) = x^3-3x-y^2+4y. - Find the relative extrema of f. -

    -     - -

    - Once again we start by finding the partial derivatives of f: - - f_x(x,y) = 3x^2-3\qquad \text{ and } \qquad f_y(x,y) = -2y+4 - . -

    - -

    - Each is always defined. - Setting each equal to 0 and solving for x and y, we find - - f_x(x,y) = 0 \amp \Rightarrow x=\pm 1 - f_y(x,y) = 0 \amp \Rightarrow y = 2 - . -

    - -

    - We have two critical points: - (-1,2) and (1,2). - To determine if they correspond to a relative maximum or minimum, - we consider the graph of f in . -

    - -
    - The surface in with both critical points marked - - - - The cubic surface studied in this example. - -

    - The surface plotted is the graph of the cubic function f(x,y) = x^3-3x-y^2+4y. - It shape is somewhat wave-like, with some interesting features worth noting. - To the right (in the perspective used for the image) there is a peak; - near this peak, the surface looks similar to a downward-opening elliptic paraboloid. -

    - -

    - To the left of the image, the surface dips down to what initially appears to be a minimum, - if we are thinking of the graph of a cubic function in two dimensions. - However, this is only the case if we view the surface along the y axis. - Rotating the image shows that while the surface curves upward in the x direction, - it curves downward in the y direction. - Near this point, the shape of the surface is more like the saddle surface defined by a hyperbolic paraboloid. -

    -     - - - - - //ASY file for figmulti_extreme33D.asy in Chapter 12 - - size(300,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(3,10,9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={2}; - real[] myzchoice={5}; - defaultpen(0.5mm); - pair xbounds=(-1.5,2); - pair ybounds=(-.5,3.5); - pair zbounds=(-2,6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=x^3-3x-y^2+4y - triple f(pair t) { - return (t.x,t.y,t.x^3-3*t.x-t.y^2+4*t.y); - } - surface s=surface(f,(-2,1),(2,3),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=4;dot((-1,2,6)); - dot((1,2,2)); - - - - -
    - -

    - The critical point (-1,2) clearly corresponds to a relative maximum. - However, the critical point at (1,2) is neither a maximum nor a minimum, - displaying a different, interesting characteristic. -

    - -

    - If one walks parallel to the y-axis towards this critical point, - then this point becomes a relative maximum along this path. - But if one walks towards this point parallel to the x-axis, - this point becomes a relative minimum along this path. - A point that seems to act as both a max and a min is a saddle point. - A formal definition follows. -

    -     -     - - - Saddle Point - -

    - Let P=(x_0,y_0) be in the domain of f where f_x=0 and f_y=0 at P. - We say P is a saddle point of f if, - for every open disk D containing P, - there are points (x_1,y_1) and - (x_2,y_2) in D such that - f(x_0,y_0) \gt f(x_1,y_1) and f(x_0,y_0)\lt f(x_2,y_2). - saddle point - critical point -

    -     -     - -

    - At a saddle point, - the instantaneous rate of change in all directions is 0 and there are points nearby with z-values both less than and greater than the z-value of the saddle point. -

    - -

    - Before - we mentioned the need for a test to differentiate between relative maxima and minima. - We now recognize that our test also needs to account for saddle points. - To do so, we consider the second partial derivatives of f. -

    - -

    - Recall that with single variable functions, - such as y=f(x), if \fp'(c) \gt 0, - then f is concave up at c, and if \fp(c) =0, - then f has a relative minimum at x=c. - (We called this the Second Derivative Test.) - Note that at a saddle point, - it seems the graph is both - concave up and concave down, - depending on which direction you are considering. -

    - -

    - It would be nice if the following were true: -

    - - - - f_{xx} and f_{yy} \gt 0 - \Rightarrow - relative minimum - - - f_{xx} and f_{yy} \lt 0 - \Rightarrow - relative maximum - - - f_{xx} and f_{yy} have opposite signs - \Rightarrow - saddle point. - - - -

    - However, this is not the case. - Functions f exist where f_{xx} and f_{yy} are both positive but a saddle point still exists. - In such a case, - while the concavity in the x-direction is up (, - f_{xx} \gt 0) and the concavity in the y-direction is also up (, f_{yy} \gt 0), - the concavity switches somewhere in between the x- and y-directions. -

    - -

    - To account for this, consider D = f_{xx}f_{yy}-f_{xy}f_{yx}. - Since f_{xy} and f_{yx} are equal when continuous - (refer back to ), - we can rewrite this as D = f_{xx}f_{yy}-f_{xy}^{\,2}. - D can be used to test whether the concavity at a point changes depending on direction. - If D \gt 0, - the concavity does not switch (, at that point, - the graph is concave up or down in all directions). - If D\lt 0, the concavity does switch. - If D=0, - our test fails to determine whether concavity switches or not. - We state the use of D in the following theorem. -

    - - - Second Derivative Test - -

    - Let R be an open set on which a function - z=f(x,y) and all its first and second partial derivatives are defined, - let P = (x_0,y_0) be a critical point of f in R, and let - Second Derivative Test - maximumrelative/local - minimumrelative/local - saddle point - - D = f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^{\,2}(x_0,y_0) - . -

    - -

    -

      -
    1. -

      - If D \gt 0 and f_{xx}(x_0,y_0) \gt 0, - then f has a relative minimum at P. -

      -
      2. - -
    2. -

      - If D \gt 0 and f_{xx}(x_0,y_0)\lt 0, - then f has a relative maximum at P. -

      -
      3. - -
    3. -

      - If D\lt 0, then f has a saddle point at P. -

      -
      4. - -
    4. -

      - If D=0, the test is inconclusive. -

      -
      5. -
    -

    -     -     - - - -

    - We first practice using this test with the function in the previous example, - where we visually determined we had a relative maximum and a saddle point. -

    - - - Using the Second Derivative Test - -

    - Let f(x,y) = x^3-3x-y^2+4y as in . - Determine whether the function has a relative minimum, maximum, - or saddle point at each critical point. -

    -     - -

    - We determined previously that the critical points of f are (-1,2) and (1,2). - To use the Second Derivative Test, - we must find the second partial derivatives of f: - - f_{xx} = 6x;\qquad f_{yy} = -2;\qquad f_{xy} = 0 - . -

    - -

    - Thus D(x,y) = -12x. -

    - -

    - At (-1,2): D(-1,2) = 12 \gt 0, - and f_{xx}(-1,2) = -6. - By the Second Derivative Test, - f has a relative maximum at (-1,2). -

    - -

    - At (1,2): D(1,2) = -12 \lt 0. - The Second Derivative Test states that f has a saddle point at (1,2). -

    - -

    - The Second Derivative Test confirmed what we determined visually. -

    -     -     - - - Using the Second Derivative Test - -

    - Find the relative extrema of f(x,y) = x^2y+y^2+xy. -

    -     - -

    - We start by finding the first and second partial derivatives of f: - - f_x \amp = 2xy+y \amp f_y \amp = x^2+2y+x - f_{xx} \amp = 2y \amp f_{yy} \amp = 2 - f_{xy} \amp = 2x+1 \amp f_{yx} \amp = 2x+1 - . -

    - -

    - We find the critical points by finding where f_x and f_y are simultaneously 0 - (they are both never undefined). - Setting f_x=0, we have: - - f_x=0 \Rightarrow 2xy+y=0 \Rightarrow y(2x+1)=0 - . -

    - -

    - This implies that for f_x=0, - either y=0 or 2x+1=0. -

    - -

    - Assume y=0 then consider f_y=0: - - f_y \amp = 0 - x^2+2y+x \amp = 0, \qquad \text{ and since \(y=0\), we have } - x^2+x \amp = 0 - x(x+1) \amp = 0 - . -

    - -

    - Thus if y=0, we have either x=0 or x=-1, - giving two critical points: - (-1,0) and (0,0). -

    - -

    - Going back to f_x, now assume 2x+1=0, - , that x=-1/2, then consider f_y=0: - - f_y \amp = 0 - x^2+2y+x \amp = 0, \qquad \text{ and since \(x=-1/2\), we have } - 1/4+2y-1/2 \amp = 0 - y\amp = 1/8 - . -

    - -

    - Thus if x=-1/2, - y=1/8 giving the critical point (-1/2,1/8). -

    - -

    - With D = 4y-(2x+1)^2, - we apply the Second Derivative Test to each critical point. -

    - -

    - At (-1,0), D \lt 0, - so (-1,0) is a saddle point. -

    - -

    - At (0,0), D\lt 0, - so (0,0) is also a saddle point. -

    - -

    - At (-1/2,1/8), D \gt 0 and f_{xx} \gt 0, - so (-1/2,1/8) is a relative minimum. -

    - -
    - Graphing f from and its relative extrema - - - - A plot of the graph of the function used in this example. It is a more complicated-looking surface. - -

    - The surface given by the graph of f(x,y)=x^2y+y^2+xy in is rather hard to describe. - It resembles what one would obtain if one started with a rectangular sheet of metal (or some sufficiently flexible material), - bent the two corners on one side upwards, and bent the two corners on the opposite side downwards. - In the middle the surface appears to be relatively flat, - although there are three points plotted there, where the critical points occur. - At the point where the relative minimum occurs, it is very difficult to see that it is indeed a relative minimum. -

    -     - - - - - //ASY file for figmulti_extreme53D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(24,.7,11.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2}; - real[] myychoice={-0.2,0.2,0.4}; - real[] myzchoice={}; - defaultpen(0.5mm); - pair xbounds=(-3,3); - pair ybounds=(-0.3,0.5); - pair zbounds=(-1,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=x^2y+y^2+xy - triple f(pair t) { - return (t.x,t.y,t.x^2*t.y+t.y^2+t.x*t.y); - } - surface s=surface(f,(-2,-0.3),(1.5,0.5),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //plot point on surface - dotfactor=4;dot((0,0,0));dot((-1,0,0));dot((-1/2,1/8,-.016)); - - - - -
    - -

    - - shows a graph of f and the three critical points. - Note how this function does not vary much near the critical points that is, - visually it is difficult to determine whether a point is a saddle point or relative minimum - (or even a critical point at all!). - This is one reason why the Second Derivative Test is so important to have. -

    -     -     - - -     - - - Constrained Optimization -

    - When optimizing functions of one variable such as y=f(x), - we made use of , - the Extreme Value Theorem, that said that over a closed interval I=[a,b], - a continuous function has both a maximum and minimum value. - To find these maximum and minimum values, - we evaluated f at all critical points in the interval, - as well as at the endpoints - (the boundary) - of the interval. - constrained optimization - optimizationconstrained -

    - -

    - A similar theorem and procedure applies to functions of two variables. - A continuous function over a closed set also attains a maximum and minimum value - (see the following theorem). - We can find these values by evaluating the function at the critical values in the set and over the boundary of the set. - After formally stating this extreme value theorem, we give examples. -

    - - - Extreme Value Theorem - -

    - Let z=f(x,y) be a continuous function on a closed, - bounded set S. - Then f has a maximum and minimum value on S. - Extreme Value Theorem -

    -     -     - - - Finding extrema on a closed set - -

    - Let f(x,y) = x^2-y^2+5 and let S be the triangle with vertices (-1,-2), - (0,1) and (2,-2). - Find the maximum and minimum values of f on S. -

    -     - -

    - It can help to see a graph of f along with the set S. - In the triangle defining S is shown in the xy-plane in a dashed line. - Above it is the graph of f; - we are only concerned with the portion of the surface z=f(x,y) enclosed by the - triangle. -

    - -
    - Plotting the graph of f along with the restricted domain S in - -
    - - - - - A saddle surface on which a triangular curve is plotted. - -

    - There are several things going on in this image. - We have the usual three-dimensional coordinate axes; - against these axes, we see the plot of a hyperbolic paraboloid (saddle surface). - Below the surface, in the xy plane, there is a triangle plotted with dashed lines. - Each point on this triangle corresponds to a point on the surface, - and these points together form a triangular curve on the surface, which is also plotted. -

    -     - - - - - //ASY file for figconopt1.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5.6,-7.8,20.3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,2}; - real[] myychoice={-2,1,2}; - real[] myzchoice={5}; - defaultpen(0.5mm); - pair xbounds=(-1.5,3); - pair ybounds=(-2.5,2.5); - pair zbounds=(-1,10); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=x^2 - y^2 + 5 - triple f(pair t) { - return (t.x,t.y,t.x^2-t.y^2+5); - } - surface s=surface(f,(-1.5,-2.25),(2.25,1.75),16,16,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic},usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw triangle in plane - draw((-1,-2,0)--(0,1,0)--(2,-2,0)--cycle,redpen+dashed+linewidth(1)); - - //Draw triangle on surface - triple g(real t) {return (t,-2,t^2+1);} - path3 mypath=graph(g,-1,2,operator ..); draw(mypath,redpen); //side 1 - triple g(real t) {return (t,3*t+1,t^2-(3*t+1)^2+5);} - path3 mypath=graph(g,-1,0,operator ..); draw(mypath,redpen); //side 2 - triple g(real t) {return (t,-3*t/2+1,t^2-(-3*t/2+1)^2+5);} - path3 mypath=graph(g,0,2,operator ..); draw(mypath,redpen); //side 3 - - - - -
    - -
    - - - - A triangle in the plane representing the domain of the function in this example. - -

    - This plot in the xy plane shows the domain used in . - The domain is a triangle, consisting of three lines: -

      -
    1. -

      - The line y=-2, from the point (-1,-2) to the point (2,-2) -

      -
      2. -
    2. -

      - The line y=-\frac32 x+1, from the point (2,-2) to the point (0,1) -

      -
      3. -
    3. -

      - The line y = 3x+1, from the point (-1,-2) to the point (0,1) -

      -
      4. -
    -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-2,2}, - extra x ticks={-1,1}, - ymin=-2.5,ymax=1.5, - xmin=-2.1,xmax=2.1 - ] - - \addplot [firstcurvestyle,domain=-1:0] {3*x+1} node [black,pos=.2,above,sloped] { $y=3x+1$}; - \addplot [firstcurvestyle,domain=0:2] {-(3/2)*x+1} node [black,pos=.6,below,sloped] { $y=-3/2x+1$}; - \addplot [firstcurvestyle,domain=-1:2] {-2} node [black,pos=.5,below] { $y=-2$}; - - \fill [black] (axis cs:-1,-2) circle (2.4pt); - \fill [black] (axis cs:2,-2) circle (2.4pt); - \fill [black] (axis cs:0,1) circle (2.4pt); - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - We begin by finding the critical points of f. - With f_x = 2x and f_y = -2y, - we find only one critical point, at (0,0). -

    - -

    - We now find the maximum and minimum values that f attains along the boundary of S, that is, - along the edges of the triangle. - In we see the triangle sketched in the plane with the equations of the lines forming its edges labeled. -

    - -

    - Start with the bottom edge, along the line y=-2. - If y is -2, - then on the surface, we are considering points f(x,-2); - that is, our function reduces to f(x,-2) = x^2-(-2)^2+5 = x^2+1=f_1(x). - We want to maximize/minimize - f_1(x)=x^2+1 on the interval [-1,2]. - To do so, we evaluate f_1(x) at its critical points and at the endpoints. -

    - -

    - The critical points of f_1 are found by setting its derivative equal to 0: - - \fp_1(x)=0\qquad \Rightarrow x=0 - . -

    - -

    - Evaluating f_1 at this critical point, - and at the endpoints of [-1,2] gives: - - f_1(-1) = 2 \qquad\amp \Rightarrow\qquad f(-1,-2) = 2 - f_1(0) = 1 \qquad\amp \Rightarrow \qquad f(0,-2) = 1 - f_1(2) = 5 \qquad\amp \Rightarrow \qquad f(2,-2) = 5 - . -

    - -

    - Notice how evaluating f_1 at a point is the same as evaluating f at its corresponding point. -

    - -

    - We need to do this process twice more, - for the other two edges of the triangle. -

    - -

    - Along the left edge, along the line y=3x+1, - we substitute 3x+1 in for y in f(x,y): - - f(x,y) = f(x,3x+1) = x^2-(3x+1)^2+5 = -8x^2-6x+4 = f_2(x) - . -

    - -

    - We want the maximum and minimum values of f_2 on the interval [-1,0], - so we evaluate f_2 at its critical points and the endpoints of the interval. - We find the critical points: - - \fp_2(x) = -16x-6=0 \qquad \Rightarrow \qquad x=-3/8 - . -

    - -

    - Evaluate f_2 at its critical point and the endpoints of [-1,0]: - - f_2(-1) = 2 \qquad\amp \Rightarrow\qquad f(-1,-2) = 2 - f_2(-3/8) = 41/8=5.125 \qquad\amp \Rightarrow \qquad f(-3/8,-0.125) = 5.125 - f_2(0) = 4 \qquad\amp \Rightarrow \qquad f(0,1) = 4 - . -

    - -

    - Finally, we evaluate f along the right edge of the triangle, - where y = -3/2x+1. - - f(x,y) = f(x,-3/2x+1) = x^2-(-3/2x+1)^2+5 = -\frac54x^2+3x+4=f_3(x) - . -

    - -

    - The critical points of f_3(x) are: - - \fp_3(x) = 0 \qquad \Rightarrow \qquad x=6/5=1.2 - . -

    - -

    - We evaluate f_3 at this critical point and at the endpoints of the interval [0,2]: - - f_3(0) = 4 \qquad\amp \Rightarrow\qquad f(0,1) = 4 - f_3(1.2) = 5.8 \qquad\amp \Rightarrow \qquad f(1.2,-0.8) = 5.8 - f_3(2) = 5 \qquad\amp \Rightarrow \qquad f(2,-2) = 5 - . -

    - -

    - One last point to test: the critical point of f, - (0,0). - We find f(0,0) = 5. -

    - -
    - The graph of f along with important points along the boundary of S and the interior in - - - Another view of the hyperbolic paraboloid in this example; this time with points of interest indicated. - -

    - This image is another view of the hyperbolic paraboloid in . - We again see the triangular domain in the xy plane, - and the corresponding curve along the surface, outlining the portion of the surface given by the domain. -

    - -

    - There are several points added to the plot this time: one point where the surface meets the z axis; - this corresponds to the critical point in the interior of the domain. - There are also three points corresponding to the three corners of the triangles, - as well as three points on the sides of the triangular curve, - where potential boundary maxima or minima are located. -

    -     - - - - - //ASY file for figconopt1c3D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(2.6,-7.,24); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,2}; - real[] myychoice={-2,1,2}; - real[] myzchoice={5}; - defaultpen(0.5mm); - pair xbounds=(-1.5,3); - pair ybounds=(-2.5,2.5); - pair zbounds=(-1,10); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw surface z=x^2 - y^2 + 5 - triple f(pair t) { - return (t.x,t.y,t.x^2-t.y^2+5); - } - surface s=surface(f,(-1.5,-2.25),(2.25,1.75),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw triangle in plane - draw((-1,-2,0)--(0,1,0)--(2,-2,0)--cycle,redpen+dashed+linewidth(1)); - - //Draw triangle on surface - triple g(real t) {return (t,-2,t^2+1);} - path3 mypath=graph(g,-1,2,operator ..); draw(mypath,redpen); //side 1 - triple g(real t) {return (t,3*t+1,t^2-(3*t+1)^2+5);} - path3 mypath=graph(g,-1,0,operator ..); draw(mypath,redpen); //side 2 - triple g(real t) {return (t,-3*t/2+1,t^2-(-3*t/2+1)^2+5);} - path3 mypath=graph(g,0,2,operator ..); draw(mypath,redpen); //side 3 - - //Dots at 7 points - dotfactor=4; - dot((-1,-2,2));dot((2,-2,5)); dot((0,1,4)); //at corner of triangles - dot((0,0,5)); //maximum - dot((0,-2,1)); //minimum - dot((1.2,-0.8,5.8));dot((-3/8,-1/8,5.125)); //along edges of triangles - - - - -
    - -

    - We have evaluated f at a total of 7 different places, - all shown in . - We checked each vertex of the triangle twice, - as each showed up as the endpoint of an interval twice. - Of all the z-values found, - the maximum is 5.8, found at (1.2,-0.8); - the minimum is 1, found at (0,-2). -

    -     -     - -

    - This portion of the text is entitled - Constrained Optimization - because we want to optimize a function (, find its maximum and/or minimum values) subject to a constraint - some limit to what values the function can attain. - In the previous example, - we constrained ourselves by considering a function only within the boundary of a triangle. - This was largely arbitrary; - the function and the boundary were chosen just as an example, - with no real meaning behind the function or the chosen constraint. -

    - -

    - However, solving constrained optimization problems is a very important topic in applied mathematics. - The techniques developed here are the basis for solving larger problems, - where more than two variables are involved. -

    - - - -

    - We illustrate the technique once more with a classic problem. -

    - - - Constrained Optimization - -

    - The U.S. Postal Service states that the girth+length of Standard Post Package must not exceed 130''. Given a rectangular box, - the length is the longest side, - and the girth is twice the width+height. -

    - -

    - Given a rectangular box where the width and height are equal, - what are the dimensions of the box that give the maximum volume subject to the constraint of the size of a Standard Post Package? -

    -     - -

    - Let w, h and \ell denote the width, - height and length of a rectangular box; - we assume here that w=h. - The girth is then 2(w+h) = 4w. - The volume of the box is V(w,\ell) = wh\ell = w^2\ell. - We wish to maximize this volume subject to the constraint 4w+\ell\leq 130, - or \ell\leq 130-4w. - (Common sense also indicates that \ell \gt 0, w \gt 0.) -

    - -

    - We begin by finding the critical values of V. - We find that V_w = 2w\ell and V_\ell = w^2; - these are simultaneously 0 at points of the form (0,\ell). - These give a volume of 0, so we can ignore these critical points. -

    - -

    - We now consider the volume along the constraint \ell=130-4w. - Along this line, we have: - - V(w,\ell) = V(w,130-4w) = w^2(130-4w) = 130w^2-4w^3 = V_1(w) - . -

    - -

    - The constraint is applicable on the w-interval - [0,32.5] as indicated in the figure. - Thus we want to maximize V_1 on [0,32.5]. -

    - -

    - Finding the critical values of V_1, - we take the derivative and set it equal to 0: - - V\,'_1(w) = 260w-12w^2 = 0 \Rightarrow w(260-12w)= 0 \Rightarrow w=0,\frac{260}{12}\approx 21.67 - . -

    - -

    - We found two critical values: - when w=0 and when w=21.67. - We again ignore the w=0 solution; - the maximum volume, subject to the constraint, - comes at w=h=21.67, \ell = 130-4(21.6) =43.33. - This gives a volume of V(21.67,43.33) \approx 20,343in^3. -

    - -
    - Graphing the volume of a box with girth 4w and length \ell, subject to a size constraint - - - - A three-dimensional plot of the volume in this example, as a function of length and width. - -

    - A set of three-dimensional coordinate axes are given. - The vertical axis is labeled V (in thousands). - The two horizontal axes are labeled w and \ell. -

    - -

    - The surface has the appearance of what one would obtain by taking a rectangular sheet of rubber, - located in the first quadrant of the \ell w plane, - and stretching the corner furthers from the origin upwards. -

    - -

    - In the \ell w plane, a dashed line corresponding to the contraint curve 4w+\ell=130 is shown. - A corresponding curve lies along the surface, - and we see that the height along this curve is zero where it meets the two axes, - but it rises to a maximum value in the middle of the surface. -

    -     - - - - - //ASY file for figconopt23D.asy in Chapter 12 - - size(200,200,IgnoreAspect); - //size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(176,-2075,580); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={20}; - real[] myychoice={50,100}; - real[] myzchoice={20,40,60,80}; - defaultpen(0.5mm); - pair xbounds=(-5,40); - pair ybounds=(-2,150); - //pair zbounds=(0,100000); - pair zbounds=(0,100); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$w$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$\ell$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("\centering $V$\\ (in thousands)",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x)),N); - - //Draw surface z=x^2*y - triple f(pair t) { - return (t.x,t.y,t.x^2*t.y/1000); - } - surface s=surface(f,(0,0),(32.5,130),16,16,Spline); - //s=crop(s,(-1,-1,-1),(33,135,100)); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw path in plane - draw((32.5,0,0)--(0,130,0),redpen+dashed+linewidth(1)); - - //Draw path on surface - triple g(real t) {return (t,130-4*t,t^2*(130-4*t)/1000);} - path3 mypath=graph(g,0,32.5,operator ..); draw(mypath,redpen); //side 1 - - //Dots at max point - dotfactor=4; dot((21.67,43.33,19.4)); - - - - -
    - -

    - The volume function V(w,\ell) is shown in - along with the constraint \ell = 130-4w. - As done previously, - the constraint is drawn dashed in the xy-plane and also along the graph of the function. - The point where the volume is maximized is indicated. -

    -     -     - -

    - It is hard to overemphasize the importance of optimization. - In the real world, we routinely seek to make - something better. - By expressing the something - as a mathematical function, - making something better - means optimize some function. -

    - -

    - The techniques shown here are only the beginning of an incredibly important field. - Many functions that we seek to optimize are incredibly complex, - making the step of find the gradient and set it equal to \vec 0 highly nontrivial. - Mastery of the principles here are key to being able to tackle these more complicated problems. -

    -     - - - - Terms and Concepts - - - - - -

    - - states that if f has a critical point at P, - then f has a relative extrema at P. - -

    -     - -     - - - - - -

    - A point P is a critical point of f if f_x and f_y are both 0 at P. - -

    -     - -     - - - - - -

    - A point P is a critical point of f if f_x or f_y are undefined at P. - -

    -     - -     - - - - -

    - Explain what it means to solve a constrained optimization problem. -

    -     - - - -

    - Answers will vary. - A good answer will state that we are optimizing a function subject to a constraint, - or limit, on the domain of the function. - We are looking to maximize/minimize the function while looking - at only a certain part of the domain. -

    -     - -     -     - - Problems - - - - -

    - Find the critical points of the given function. - Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, - or saddle point. -

    -     - - - - -

    - f(x,y) = \frac12x^2+2y^2-8y+4x -

    -     - -

    - One critical point at (-4,2); - f_{xx} = 1 and D = 4, - so this point corresponds to a relative minimum. -

    -     - -     - - - - - Context("Point"); - $max=List("None"); - $min=List("None"); - $sad=List("(7,-6)"); - $inc=List("None"); - - - -

    - f(x,y) = x^2+4x+y^2-9y+3xy -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test indicates the critical point is a relative maximum. - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test - indicates the critical point is a relative minimum. - - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test - indicates the critical point is a saddle point. - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test is inconclusive. - - -

    - -

    -     -     -     - - - - -

    - f(x,y) = x^2+3y^2-6y+4xy -

    -     - -

    - One critical point at (6,-3); - D = -4, so this point corresponds to a saddle point. -

    -     - -     - - - - - Context("Point"); - $max=List("(0,0)"); - $min=List("None"); - $sad=List("None"); - $inc=List("None"); - - - -

    - f(x,y) = \frac{1}{x^2+y^2+1} -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test indicates the critical point is a relative maximum. - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test - indicates the critical point is a relative minimum. - - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test - indicates the critical point is a saddle point. - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test is inconclusive. - - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) = x^2+y^3-3y+1 -

    -     - -

    - Two critical points: at (0,-1); - f_{xx} = 2 and D = -12, - so this point corresponds to a saddle point; -

    - -

    - at (0,1), f_{xx} = 2 and D = 12, - so this corresponds to a relative minimum. -

    -     - -     - - - - - Context("Point"); - $max=List("(-1,-2)"); - $min=List("(1,2)"); - $sad=List("(1,-2), (-1,2)"); - $inc=List("None"); - - - -

    - f(x,y) = \frac13x^3-x+\frac13y^3-4y -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test indicates the critical point is a relative maximum. - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test - indicates the critical point is a relative minimum. - - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test - indicates the critical point is a saddle point. - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test is inconclusive. - - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) = x^2y^2 -

    -     - -

    - There are infinitly many critical points, - whenever x=0 or y=0. - With D = -12x^2y^2, - at each critical point D = 0 and the test is inconclusive. - (Some elementary thought shows that each is the absolute minimum, - since f(x,y)\geq 0 for all (x,y).) -

    -     - -     - - - - - Context("Point"); - $max=List("(0,-3)"); - $min=List("(-1,3), (1,3)"); - $sad=List("(-1,-3), (0,3), (1,-3)"); - $inc=List("None"); - - - -

    - f(x,y) = x^4-2x^2+y^3-27y-15 -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test indicates the critical point is a relative maximum. - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test - indicates the critical point is a relative minimum. - - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test - indicates the critical point is a saddle point. - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test is inconclusive. - - -

    - -

    -     -     -     - - - - -

    - \ds f(x,y) = \sqrt{16-(x-3)^2-y^2} -

    -     - -

    - One critical point: f_x = 0 when x=3; - f_y = 0 when y = 0, - so one critical point at (3,0), - which is a relative maximum, where - f_{xx} = \frac{y^2-16}{(16-(x-3)^2-y^2)^{3/2}} and D = \frac{16}{(16-(x-3)^2-y^2)^{2}}. -

    - -

    - Both f_x and f_y are undefined along the circle (x-3)^2+y^2=16; - at any point along this curve, - f(x,y)=0, the absolute minimum of the function. -

    -     - -     - - - - - Context("Point"); - $max=List("None"); - $min=List("None"); - $sad=List("None"); - $inc=List("(0,0)"); - - - -

    - f(x,y) = \sqrt{x^2+y^2} -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test indicates the critical point is a relative maximum. - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test - indicates the critical point is a relative minimum. - - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test - indicates the critical point is a saddle point. - -

    - -

    - - List all critical points - (or report that there are none) - of f where the Second Derivative Test is inconclusive. - - -

    - -

    -     -     -     - -     - - - -

    - Find the absolute maximum and minimum of the function subject to the given constraint. -

    -     - - - - - Context("Point"); - $maxout = Compute("3"); - $maxin = Compute("(0,1)"); - $minout = Compute("3/4"); - $minin = Compute("(0,-1/2)"); - - - -

    - Let f(x,y) = x^2+y^2+y+1, - constrained to the triangle with vertices (0,1), - (-1,-1) and (1,-1). -

    - - - The absolute maximum value is: - -

    - -

    - - The absolute maximum happens for input: - -

    - -

    - - - The absolute minimum value is: - -

    - -

    - - - The absolute minimum happens for input: - -

    - -

    -     - -

    - The triangle is bound by the lines y=-1, - y=2x+1 and y=-2x+1. -

    - -

    - Along y=-1, there is a critical point at (0,-1). -

    - -

    - Along y=2x+1, there is a critical point at (-3/5,-1/5). -

    - -

    - Along y=-2x+1, there is a critical point at (3/5,-1/5). -

    - -

    - The function f has one critical point, - irrespective of the constraint, at (0,-1/2). -

    - -

    - Checking the value of f at these four points, - along with the three vertices of the triangle, - we find the absolute maximum is at (0,1,3) and the absolute minimum is at (0,-1/2,3/4). -

    -     -     -     - - - - - Context("Point"); - $maxout = Compute("25/28"); - $maxin = Compute("(5/14,25/196)"); - $minout = Compute("-12"); - $minin = Compute("(-1,1)"); - - - -

    - Let f(x,y) = 5x-7y, - constrained to the region bounded by y=x^2 and y=1. -

    - - - The absolute maximum value is: - -

    - -

    - - The absolute maximum happens for input: - -

    - -

    - - - The absolute minimum value is: - -

    - -

    - - - The absolute minimum happens for input: - -

    - -

    -     - -

    - The region has two corners - at (1,1) and (-1,1). -

    - -

    - Along y=1, there is no critical point. -

    - -

    - Along y=x^2, - there is a critical point at (5/14,25/196)\approx (0.357,0.128). -

    - -

    - The function f itself has no critical points. - Checking the value of f at the corners (1,1), - (-1,1) and the critical point (5/14,25/196), - we find the absolute maximum is at - (5/14,25/196,25/28) \approx (0.357,0.128,0.893) and the absolute minimum is at (-1,1,-12). -

    -     -     -     - - - - -

    - \ds f(x,y) = x^2+2x+y^2+2y, - constrained to the region bounded by the circle x^2+y^2=4. -

    -     - -

    - The region has no corners - or vertices, just a smooth edge. -

    - -

    - To find critical points along the circle x^2+y^2=4, - we solve for y^2: y^2=4-x^2. - We can go further and state y=\pm\sqrt{4-x^2}. -

    - -

    - We can rewrite f as f(x)=x^2+2x + (4-x^2) + 2\sqrt{4-x^2} = 2x+4+2\sqrt{4-x^2}. - (We will return and use -\sqrt{4-x^2} later.) - Solving f\,'(x)=0, we get x=\sqrt{2} \Rightarrow y=\sqrt{2}. - f\,'(x) is also undefined at x=\pm 2, where y=0. -

    - -

    - Using y=-\sqrt{4-x^2}, - we rewrite f(x,y) as f(x) = 2x+4-2\sqrt{4-x^2}. - Solving \fp(x) =0, we get x=-\sqrt{2},\, y=-\sqrt{2}. - Again, \fp(x) is undefined at x=\pm 2. -

    - -

    - The function f(x,y) itself has a critical point at (-1,-1). -

    - -

    - Checking the value of f at (-1,-1), - (\sqrt{2},\sqrt{2}), - (-\sqrt{2},-\sqrt{2}), - (2,0) and (-2,0), - we find the absolute maximum is at (\sqrt2,\sqrt2,4+4\sqrt2) and the absolute minimum is at (-1,-1,-2). -

    -     - -     - - - - -

    - \ds f(x,y) = 3y-2x^2, - constrained to the region bounded by the parabola - y=x^2+x-1 and the line y=x. -

    -     - -

    - The region has two corners - at (-1,-1) and (1,1). -

    - -

    - Along the line y=x, f(x,y) becomes f(x) = 3x-2x^2. - Along this line, we have a critical point at (3/4,3/4). -

    - -

    - Along the curve y=x^2+x-1, - f(x,y) becomes f(x) =x^2+3x-3. - There is a critical point along this curve at (-3/2, -1/4). - Since x=-3/2 lies outside our bounded region, - we ignore this critical point. -

    - -

    - The function f itself has no critical points. -

    - -

    - Checking the value of f at (-1,-1), - (1,1), (3/4,3/4), - we find the absolute maximum is at - (3/4,3/4,9/8) and the absolute minimum is at (-1,-1,-5). -

    -     - -     - -     -     -     -
    - - - -     - - - Multiple Integration - -

    - introduced multivariable functions and we applied concepts of differential calculus to these functions. - We learned how we can view a function of two variables as a surface in space, - and learned how partial derivatives convey information about how the surface is changing in any direction. -

    - -

    - In this chapter we apply techniques of integral calculus to multivariable functions. - In - we learned how the definite integral of a single variable function gave us - area under the curve. - In this chapter we will see that integration applied to a multivariable function gives us - volume under a surface. - And just as we learned applications of integration beyond finding areas, - we will find applications of integration in this chapter beyond finding volume. -

    -     - -
    - Iterated Integrals and Area - -

    - In - we found that it was useful to differentiate functions of several variables with respect to one variable, - while treating all the other variables as constants or coefficients. - We can integrate functions of several variables in a similar way. - For instance, if we are told that f_x(x,y) = 2xy, - we can treat y as staying constant and integrate to obtain f(x,y): - - f(x,y) \amp = \int f_x(x,y)\, dx - \amp = \int 2xy\, dx - \amp = x^2y + C - . -

    - -

    - Make a careful note about the constant of integration, - C. - This constant is something with a derivative of 0 with respect to x, - so it could be any expression that contains only constants and functions of y. - For instance, if f(x,y) = x^2y+ \sin(y) + y^3 + 17, - then f_x(x,y) = 2xy. - To signify that C is actually a function of y, - we write: - integrationof multivariable functions - - f(x,y) = \int f_x(x,y)\, dx = x^2y+C(y) - . -

    - -

    - Using this process we can even evaluate definite integrals. -

    -     - - - Iterated integrals - - - Integrating functions of more than one variable - -

    - Evaluate the integral \ds \int_1^{2y} 2xy\, dx. -

    -     - -

    - We find the indefinite integral as before, - then apply the Fundamental Theorem of Calculus to evaluate the definite integral: - - \int_1^{2y} 2xy\, dx \amp = x^2y\Big|_1^{2y} - \amp = (2y)^2y - (1)^2y - \amp = 4y^3-y - . -

    -     -     - -

    - We can also integrate with respect to y. - In general, - - \int_{h_1(y)}^{h_2(y)} f_x(x,y)\, dx = f(x,y)\Big|_{h_1(y)}^{h_2(y)} = f\big(h_2(y),y\big)-f\big(h_1(y),y\big) - , - and - - \int_{g_1(x)}^{g_2(x)} f_y(x,y)\, dy = f(x,y)\Big|_{g_1(x)}^{g_2(x)} = f\big(x,g_2(x)\big)-f\big(x,g_1(x)\big) - . -

    - -

    - Note that when integrating with respect to x, - the bounds are functions of y (of the form x=h_1(y) and - x=h_2(y)) and the final result is also a function of y. - When integrating with respect to y, - the bounds are functions of x (of the form y=g_1(x) and - y=g_2(x)) and the final result is a function of x. - Another example will help us understand this. -

    - - - Integrating functions of more than one variable - -

    - Evaluate \ds \int_1^x\big(5x^3y^{-3}+6y^2\big)\, dy. -

    -     - -

    - We consider x as staying constant and integrate with respect to y: - - \int_1^x\big(5x^3y^{-3}+6y^2\big)\, dy \amp = \left(\frac{5x^3y^{-2}}{-2}+\frac{6y^3}{3}\right)\Bigg|_1^x - \amp = \left(-\frac52x^3x^{-2}+2x^3\right) - \left(-\frac52x^3+2\right) - \amp = \frac92x^3-\frac52x-2 - . -

    - -

    - Note how the bounds of the integral are from y=1 to y=x and that the final answer is a function of x. -

    -     -     - -

    - In the previous example, - we integrated a function with respect to y and ended up with a function of x. - We can integrate this as well. - This process is known as iterated integration, - or multiple integration. -

    - - - Evaluating an integral - -

    - Evaluate \ds \int_1^2\left(\int_1^x\big(5x^3y^{-3}+6y^2\big)\, dy\right)\, dx. -

    -     - -

    - We follow a standard order of operations - and perform the operations inside parentheses first (which is the integral evaluated in .) - - \int_1^2\left(\int_1^x\big(5x^3y^{-3}+6y^2\big)\, dy\right)\, dx \amp = \int_1^2 \left(\left[\frac{5x^3y^{-2}}{-2}+\frac{6y^3}{3}\right]\Bigg|_1^x\right)\, dx - \amp = \int_1^2 \left(\frac92x^3-\frac52x-2\right)\, dx - \amp = \left(\frac98x^4-\frac54x^2-2x\right)\Bigg|_1^2 - \amp = \frac{89}8 - . -

    - -

    - Note how the bounds of x were x=1 to x=2 and the final result was a number. -

    -     -     - -

    - The previous example showed how we could perform something called an iterated integral; - we do not yet know why - we would be interested in doing so nor what the result, - such as the number 89/8, means. - Before we investigate these questions, - we offer some definitions. -

    - - - Iterated Integration - -

    - Iterated integration - is the process of repeatedly integrating the results of previous integrations. - Evaulating one integral is denoted as follows. -

    - -

    - Let a, b, - c and d be numbers and let g_1(x), - g_2(x), - h_1(y) and h_2(y) be functions of x and y, - respectively. - Then: - integrationiterated - integrationmultiple - iterated integration - multiple integration|see{iterated integration} - integrationnotation -

    - -

    -

      -
    1. -

      - \ds \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy = \int_c^d\left(\int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\right) dy. -

      -
      2. - -
    2. -

      - \ds \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\right) dx. -

      -
      3. -
    -

    -     -     - -

    - Again make note of the bounds of these iterated integrals. -

    - -

    - With \ds \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy, - x varies from h_1(y) to h_2(y), - whereas y varies from c to d. - That is, the bounds of x are curves, - the curves x=h_1(y) and x=h_2(y), - whereas the bounds of y are constants, - y=c and y=d. - It is useful to remember that when setting up and evaluating such iterated integrals, - we integrate from curve to curve, - then from point to point. -

    - -

    - We now begin to investigate why - we are interested in iterated integrals and - what they mean. -

    -     - - - Area of a plane region -

    - Consider the plane region R bounded by - a\leq x\leq b and g_1(x)\leq y\leq g_2(x), - shown in . - We learned in - that the area of R is given by - integrationarea - - \int_a^b \big(g_2(x)-g_1(x)\big)\, dx - . -

    - -
    - Calculating the area of a plane region R with an iterated integral - - - A region in the first quadrant is bounded above and below by graphs, and left and right by vertical lines. - -

    - Two graphs are plotted in the first quadrant of the plane. The graph y=g_1(x) lies entirely below the graph y=g_2(x). - There are also two vertical lines: x=a on the left, and x=b on the right. - Together, the two graphs and two lines bound a region in the plane, labeled R. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - ytick=\empty, - extra x ticks={.1,.9}, - extra x tick labels={$a$,$b$}, - ymin=-.1,ymax=1, - xmin=-.1,xmax=1 - ] - - \addplot [firstcurvestyle,domain=0:1] ({x},{.2*cos(deg(3*x))+.7}) node [pos=.4,above right,black] { $y=g_2(x)$}; - \addplot [firstcurvestyle,domain=0:1] ({x},{.3*x^2+.1}) node [pos=.4,below right,black] { $y=g_1(x)$}; - - \draw [thick,secondcolor] (axis cs: .1,0) -- (axis cs: .1,1); - \draw [thick,secondcolor] (axis cs: .9,0) -- (axis cs: .9,1); - - \draw (axis cs:.5,.5) node { $R$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - We can view the expression \big(g_2(x)-g_1(x)\big) as - - \big(g_2(x)-g_1(x)\big) = \int_{g_1(x)}^{g_2(x)} 1\, dy =\int_{g_1(x)}^{g_2(x)} \, dy - , - meaning we can express the area of R as an iterated integral: - - \text{ area of } R = \int_a^b \big(g_2(x)-g_1(x)\big)\, dx = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} \, dy\right) dx =\int_a^b\int_{g_1(x)}^{g_2(x)} \, dy\, dx - . -

    - -

    - In short: a certain iterated integral can be viewed as giving the area of a plane region. -

    - -

    - A region R could also be defined by - c\leq y\leq d and h_1(y)\leq x\leq h_2(y), - as shown in . - Using a process similar to that above, we have - - \text{ the area of } R = \int_c^d\int_{h_1(y)}^{h_2(y)} \, dx\, dy - . -

    - -
    - Calculating the area of a plane region R with an iterated integral - - - A region in the first quadrant is bounded left and right by graphs, and above and below by horizontal lines. - -

    - Two curves are plotted in the first quadrant of the plane. - Both curves are graphs, but with x as a function of y. - The graph x=h_1(y) lies entirely to the left of the graph x=h_2(y). - There are also two horizontal lines: y=c at the bottom, and y=d at the top. - Together, the two graphs and two lines bound a region in the plane, labeled R. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick=\empty, - ytick=\empty, - extra y ticks={.1,.9}, - extra y tick labels={$c$,$d$}, - ymin=-.1,ymax=1, - xmin=-.1,xmax=1 - ] - - \addplot [firstcurvestyle,domain=0:1] ({.3*x^2+.1},x) node [pos=.7,right,black] { $x=h_1(y)$}; - \addplot [firstcurvestyle,domain=0:1] ({.2*sin(deg(3*x))+.7},{x}) node [pos=.7,left,black] { $x=h_2(y)$}; - - \draw [thick,secondcolor] (axis cs: 0,.1) -- (axis cs: 1,.1); - \draw [thick,secondcolor] (axis cs: 0,.9) -- (axis cs: 1,.9); - - \draw (axis cs:.5,.5) node { $R$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - We state this formally in a theorem. -

    - - - Area of a plane region - -

    -

      -
    1. -

      - Let R be a plane region bounded by - a\leq x\leq b and g_1(x)\leq y\leq g_2(x), - where g_1 and g_2 are continuous functions on [a,b]. - The area A of R is - - A = \int_a^b\int_{g_1(x)}^{g_2(x)} \, dy\, dx - . -

      -
      2. - -
    2. -

      - Let R be a plane region bounded by - c\leq y\leq d and h_1(y)\leq x\leq h_2(y), - where h_1 and h_2 are continuous functions on [c,d]. - The area A of R is - - integrationarea - - - A = \int_c^d\int_{h_1(y)}^{h_2(y)} \, dx\, dy - . -

      -
      3. -
    -

    -     -     - -

    - The following examples should help us understand this theorem. -

    - - - Area of a rectangle - -

    - Find the area A of the rectangle with corners (-1,1) and (3,3), - as shown in . -

    -
    - Calculating the area of a rectangle with an iterated integral in - - - A rectangle in the plane, spanning x values from -1 to 3, and y values from 1 to 3. - -

    - A rectangle, labeled R, is plotted in the xy plane. - It is defined by the inequalities -1\leq x\leq 3 and 1\leq y\leq 3. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,1,2,3}, - ymin=-.1,ymax=3.5, - xmin=-1.5,xmax=3.5 - ] - - \draw [thick,firstcolor] (axis cs: -1,1) -- (axis cs:3,1)--(axis cs:3,3) -- (axis cs:-1,3) -- cycle; - \draw (axis cs:1,2) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - Multiple integration is obviously overkill in this situation, - but we proceed to establish its use. -

    - -

    - The region R is bounded by x=-1, - x=3, y=1 and y=3. - Choosing to integrate with respect to y first, we have - - A = \int_{-1}^3\int_1^3 1\, dy\, dx = \int_{-1}^3 \left(y\, \Big|_1^3\right)\, dx = \int_{-1}^3 2\, dx = 2x\Big|_{-1}^3=8 - . -

    - -

    - We could also integrate with respect to x first, giving: - - A = \int_1^3\int_{-1}^3 1\, dx \, dy =\int_1^3 \left(x\, \Big|_{-1}^3\right)\, dy = \int_1^3 4\, dy = 4y\Big|_1^3 = 8 - . -

    - -

    - Clearly there are simpler ways to find this area, - but it is interesting to note that this method works. -

    -     -     - - - Area of a triangle - -

    - Find the area A of the triangle with vertices at (1,1), - (3,1) and (5,5), - as shown in . -

    - -
    - Calculating the area of a triangle with iterated integrals in - - - An obtuse triangle in the plane. The base is horizontal, and the other sides are lines with positive slope. - -

    - The image shows a triangle in the first quadrant. - It is an obtuse triangle, with a horizontal base. - The base lies along the line y=1, for 1\leq x\leq 3. - The other two sides of triangle are lines with positive slope: - the line y=x, from the point (1,1) to the point (5,5), - and the line y=2x-5, from the point (1,1) to the point (5,5). -

    - -

    - Note that the upper vertex lies to the right of the right-most vertex at the bottom. - This means that an attempt to integrate first with respect to y will require dividing the triangle into two regions. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={4,5,1,2,3}, - ytick={1,2,3,4,5}, - ymin=-.5,ymax=5.9, - xmin=-.5,xmax=5.9 - ] - - \addplot [firstcurvestyle,domain=1:3] {1} node [pos=.5,black,below] { $y=1$}; - \addplot [firstcurvestyle,domain=1:5] {x} node [pos=.5,black,above,sloped] { $y=x$}; - \addplot [firstcurvestyle,domain=3:5] {2*x-5} node [pos=.5,black,below,sloped] { $y=2x-5$}; - - \draw (axis cs:2.5,1.75) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - The triangle is bounded by the lines as shown in the figure. - Choosing to integrate with respect to x first gives that x is bounded by x=y to x = \frac{y+5}2, - while y is bounded by y=1 to y=5. (Recall that since x-values increase from left to right, - the leftmost curve, x=y, - is the lower bound and the rightmost curve, x=(y+5)/2, - is the upper bound.) The area is - - A \amp = \int_1^5\int_{y}^{\frac{y+5}2}\, dx\, dy - \amp = \int_1^5\left(x\, \Big|_y^{\frac{y+5}2}\right)\, dy - \amp = \int_1^5 \left(-\frac12y+\frac52\right)\, dy - \amp = \left(-\frac14y^2+\frac52y\right)\Big|_1^5 - \amp =4 - . -

    - -

    - We can also find the area by integrating with respect to y first. - In this situation, though, - we have two functions that act as the lower bound for the region R, - y=1 and y=2x-5. - This requires us to use two iterated integrals. - Note how the x-bounds are different for each integral: - - A \amp = \int_1^3\int_1^x 1\, dy \, dx \amp +\amp \amp \amp \int_3^5\int_{2x-5}^x1\, dy\, dx - \amp = \int_1^3\big(y\big)\Big|_1^x\, dx \amp + \amp \amp \amp \int_3^5\big(y\big)\Big|_{2x-5}^x\, dx - \amp = \int_1^3\big(x-1\big)\, dx \amp + \amp \amp \amp \int_3^5\big(-x+5\big)\, dx - \amp = 2 \amp + \amp \amp \amp 2 - \amp =4 - . -

    - -

    - As expected, we get the same answer both ways. -

    -     -     - - - Area of a plane region - -

    - Find the area of the region enclosed by y=2x and y=x^2, - as shown in . -

    - -
    - Calculating the area of a plane region with iterated integrals in - - - A region in the first quadrant of the plane, bounded by a line and a parabola. - -

    - A region R in the first quadrant of the plane is shown. -

      -
    • -

      - It is bounded by two curves. -

      -
      • -
    • -

      - The curve that lies above (or alternatively, to the left of) the region is the line y=2x. -

      -
      • -
    • -

      - The curve that lies below (or alternatively, to the right of) the region is the parabola y=x^2. -

      -
      • -
    -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={4,5,1,2,3}, - ytick={1,2,3,4,5}, - ymin=-.5,ymax=4.5, - xmin=-.5,xmax=2.5 - ] - - \addplot [firstcurvestyle,domain=0:2.5] ({x},{2*x}) node [pos=.4,sloped,above,black] { $y=2x$}; - \addplot [firstcurvestyle,domain=0:2.5] ({x},{x^2}) node [pos=.3,sloped,below,black] { $y=x^2$}; - - \draw (axis cs:1,1.5) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - Once again we'll find the area of the region using both orders of integration. -

    - -

    - Using dy\, dx: - - \int_0^2\int_{x^2}^{2x}1\, dy \, dx = \int_0^2(2x-x^2)\, dx = \big(x^2-\frac13x^3\big)\Big|_0^2 = \frac43 - . -

    - -

    - Using dx\, dy: - - \int_0^4\int_{y/2}^{\sqrt{y}} 1\, dx\, dy = \int_0^4 (\sqrt{y}-y/2)\, dy = \left(\frac23y^{3/2} - \frac14y^2\right)\Big|_0^4 = \frac43 - . -

    -     -     -     - - - Changing Order of Integration - -

    - In each of the previous examples, - we have been given a region R and found the bounds needed to find the area of R using both orders of integration. - We integrated using both orders of integration to demonstrate their equality. - iterated integrationchanging order -

    - -

    - We now approach the skill of describing a region using both orders of integration from a different perspective. - Instead of starting with a region and creating iterated integrals, - we will start with an iterated integral and rewrite it in the other integration order. - To do so, we'll need to understand the region over which we are integrating. -

    - -

    - The simplest of all cases is when both integrals are bound by constants. - The region described by these bounds is a rectangle - (see ), - and so: - - \int_a^b\int_c^d 1\, dy\, dx = \int_c^d\int_a^b1\, dx\, dy - . -

    - -

    - When the inner integral's bounds are not constants, - it is generally very useful to sketch the bounds to determine what the region we are integrating over looks like. - From the sketch we can then rewrite the integral with the other order of integration. -

    - -

    - Examples will help us develop this skill. -

    - - - Changing the order of integration - -

    - Rewrite the iterated integral - \ds \int_0^6\int_0^{x/3} 1\, dy\, dx with the order of integration dx\, dy. -

    -     - -

    - We need to use the bounds of integration to determine the region we are integrating over. -

    - -

    - The bounds tell us that y is bounded by 0 and x/3; - x is bounded by 0 and 6. - We plot these four curves: y=0, y=x/3, - x=0 and x=6 to find the region described by the bounds. - shows these curves, - indicating that R is a triangle. -

    - -
    - Sketching the region R described by the iterated integral in - - - A triangular region in the plane, bounded by lines y=0, x=6, and y=x/3. - -

    - The region R in the plane is a right-angled triangle. -

      -
    • -

      - The base of the triangle is the x axis, from the origin to (6,0) -

      -
      • -
    • -

      - The vertical side of the triangle is the line x=6, from (6,0) to (6.2) -

      -
      • -
    • -

      - The hypotenuse of the triangle is the line y=x/3, from (0,0) to (6,2) -

      -
      • -
    -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=2.5, - xmin=-.5,xmax=6.9 - ] - - \draw [very thick,secondcolor] (axis cs:-.5,0) -- (axis cs:6.9,0) - (axis cs:-.5,-.1666) -- node [above,black,pos=.5,sloped] { $y=x/3$} - (axis cs: 6.9, 2.3); - - \draw [very thick,firstcolor] (axis cs:0,-.5) -- (axis cs:0,2.5) - (axis cs:6,-.5) -- (axis cs:6,2.5); - - \draw (axis cs:4,.5) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - To change the order of integration, - we need to consider the curves that bound the x-values. - We see that the lower bound is x=3y and the upper bound is x=6. - The bounds on y are 0 to 2. - Thus we can rewrite the integral as - \ds \int_0^2\int_{3y}^6 1\, dx \, dy. -

    -     -     - - - Changing the order of integration - -

    - Change the order of integration of \ds\int_0^4\int_{y^2/4}^{(y+4)/2}1\, dx\, dy. -

    -     - -

    - We sketch the region described by the bounds to help us change the integration order. - x is bounded below and above (, to the left and right) by x=y^2/4 and x=(y+4)/2 respectively, - and y is bounded between 0 and 4. - Graphing the previous curves, - we find the region R to be that shown in . -

    - -
    - Drawing the region determined by the bounds of integration in - - - A sketch of the region of integration for this example. - -

    - A region R is plotted in the first quadrant of the plane. - It is bounded by several curves. -

      -
    • -

      - The left boundary of the region is the parabola x=y^2/4. - This is a parabola with its vertex at the origin that opens to the right. -

      -
      • -
    • -

      - The right boundary of the region is the line x=(y+4)/2. - This line intersects the parabola at the point (4,4), - and intersects the x axis at the point (2,0). -

      -
      • -
    • -

      - There is also a bottom boundary, given by the portion of the x axis from (0,0) to (2,0). -

      -
      • -
    -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=4.5, - xmin=-.5,xmax=4.5 - ] - - \addplot [firstcurvestyle,domain=-.5:4.5,samples=30] ({x^2/4},{x}) node [pos=.4,sloped,above,black] {$x=y^2/4$}; - \addplot [firstcurvestyle,domain=-.5:4.5] ({((x+4))/2},{x}) node [pos=.4,sloped,below,black] { $x= (y+4)/2$}; - - \draw [very thick,secondcolor] (axis cs:-.5,0) -- (axis cs:4.5,0) - (axis cs:-.5,4) -- (axis cs:4.5,4); - - \draw (axis cs:2,1.75) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - To change the order of integration, - we need to establish curves that bound y. - The figure makes it clear that there are two lower bounds for y: - y=0 on 0\leq x\leq 2, - and y=2x-4 on 2\leq x\leq 4. - Thus we need two double integrals. - The upper bound for each is y=2\sqrt{x}. - Thus we have - - \int_0^4\int_{y^2/4}^{(y+4)/2}1\, dx\, dy = \int_0^2\int_0^{2\sqrt{x}} 1\, dy\, dx + \int_2^4\int_{2x-4}^{2\sqrt{x}}1\, dy\, dx - . -

    -     -     - -

    - This section has introduced a new concept, the iterated integral. - We developed one application for iterated integration: - area between curves. - However, this is not new, - for we already know how to find areas bounded by curves. -

    - -

    - In the next section we apply iterated integration to solve problems we currently do not know how to handle. - The real goal of this section was not to learn a new way of computing area. - Rather, our goal was to learn how to define a region in the plane using the bounds of an iterated integral. - That skill is very important in the following sections. -

    -     - - - - Terms and Concepts - - - - -

    - When integrating f_x(x,y) with respect to x, - the constant of integration C is really which: - C(x) or C(y)? - What does this mean? -

    -     - - - -

    - C(y), meaning that instead of being just a constant, - like the number 5, it is a function of y, - which acts like a constant when taking derivatives with respect to x. -

    -     - -     - - - - -

    - Evaluating a double integral in steps is called - - . -

    -     - - - - - - - - - - - - - -     - - - - -

    - When evaluating an iterated integral, - we integrate from to , - then from to . -

    -     - - - - - - - - - - - - - - - - - - - - - - - -     - - - - -

    - One understanding of an iterated integral is that - \ds \int_a^b\int_{g_1(x)}^{g_2(x)} \, dy\, dx gives the of a plane region. -

    -     - - - - - - - - -     -     - - - Problems - - - -

    - Evaluate the integral and subsequent iterated integral. -

    -     - - - - - -

    - \ds \int_2^{5} \big(6x^2+4xy-3y^2\big)\, dy -

    -     - -

    - 18x^2+42x-117 -

    -     -     - - - -

    - \ds \int_{-3}^{-2} \int_2^{5} \big(6x^2+4xy-3y^2\big)\, dy\, dx -

    -     - -

    - -108 -

    -     -     - -     - - - - - - Context()->variables->are(x=>'Real',y=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $a=Compute("2+pi^2cos(y)"); - $b=Formula("pi^2+pi"); - - - - -

    - \ds \int_0^{\pi} (2x\cos(y) + \sin(x) )\,dx -

    - -

    - -

    -     -     - - - -

    - \ds \int_{0}^{\pi/2} \int_0^{\pi} (2x\cos(y) + \sin(x) )\,dx\,dy -

    -

    - -

    -     -     -     -     - - - - - -

    - \ds \int_1^x \big(x^2y - y+2\big)\, dy -

    -     - -

    - x^4/2-x^2+2x-3/2 -

    -     -     - - - -

    - \ds \int_0^2\int_1^x \big(x^2y - y+2\big)\, dy\, dx -

    -     - -

    - 23/15 -

    -     -     - -     - - - - - - Context()->variables->are(x=>'Real',y=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $a=Compute("y^4/2-y^3+y^2/2"); - $b=Formula("8/15"); - - - - -

    - \ds \int_y^{y^2} (x-y)\,dx -

    -

    - -

    -     -     - - - -

    - \ds \int_{-1}^1\int_y^{y^2} (x-y)\,dx\,dy -

    -

    - -

    -     -     -     -     - - - - - - -

    - \ds \int_0^{y} \big(\cos(x) \sin(y) \big)\, dx -

    -     - -

    - \sin^2(y) -

    -     -     - - - -

    - \ds \int_0^\pi \int_0^{y} \big(\cos(x) \sin(y) \big)\, dx\, dy -

    -     - -

    - \pi/2 -

    -     -     - -     - - - - - - Context()->variables->are(x=>'Real',y=>'Real'); - Context()->flags->set(reduceConstants=>0,reduceConstantFunctions=>0); - $a=Compute("x/(1+x^2)"); - $b=Formula("1/2 ln(5/2)"); - - - - -

    - \ds \int_0^{x} \left(\frac{1}{1+x^2}\right)\,dy -

    -

    - -

    -     -     - - - -

    - \ds \int_1^2 \int_0^{x} \left(\frac{1}{1+x^2}\right)\,dy\,dx -

    -

    - -

    -     -     -     -     - -     - - - -

    - A graph of a planar region R is given. - Give the iterated integrals, - with both orders of integration dy\, dx and dx\, dy, - that give the area of R. - Evaluate one of the iterated integrals to find the area. -

    -     - - - - - - - The region R is a rectangle, with x from 1 to 4, and y from -2 to 1. - -

    - A sketch of the region R for this exercise. - R is a rectangle, with bounds 1\leq x\leq 4 and -2\leq y\le1 1. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={-2,-1,1}, - ymin=-2.5,ymax=1.5, - xmin=-.5,xmax=4.5 - ] - - \draw [very thick,firstcolor] (axis cs:1,-2) -- (axis cs: 4,-2) -- (axis cs: 4,1) -- (axis cs: 1,1) -- cycle; - - \draw (axis cs: 2.5,-1) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \ds \int_1^4\int_{-2}^1\, dy\, dx and \ds\int_{-2}^1\int_1^4\, dx\, dy. -

    - -

    - area of R = 9u^2 -

    -     - -     - - - - - - - The region R is a right triangle with vertices (1,1), (4,1), and (4,3) - -

    - A sketch of the region R for this exercise. - The boundary of R is a right-angle triangle, - with vertices at (1,1), (4,1), and (4,3). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3}, - ymin=-.5,ymax=3.5, - xmin=-.5,xmax=4.5 - ] - - \draw [very thick,firstcolor] (axis cs:1,1) -- (axis cs: 4,1) -- (axis cs: 4,3) -- cycle; - - \draw (axis cs: 3,1.75) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \ds\int_1^4\int_{1}^{\frac23x+\frac13}\, dy\, dx and \ds\int_{1}^3\int_{\frac32y-\frac12}^4\, dx\, dy. -

    - -

    - area of R = 3u^2 -

    -     - -     - - - - - - - A triangular region R. The vertices are (2,5), (2,1), and (4,3). - -

    - The region R for this exercise is a triangle. - The left side of the triangle is vertical, from (2,1) to (2,5). - The other vertex is at (4,3) -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3,4}, - ytick={1,2,3,4,5}, - ymin=-.5,ymax=5.5, - xmin=-.5,xmax=4.5% - ] - - \draw [very thick,firstcolor] (axis cs:2,1) -- (axis cs: 4,3) -- (axis cs: 2,5) -- cycle; - - \draw (axis cs: 2.75,3) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \ds\int_2^4\int_{x-1}^{7-x}\, dy\, dx. - The order dx\, dy needs two iterated integrals as x is bounded above by two different functions. - This gives: - - \int_{1}^3\int_{2}^{y+1}\, dx\, dy+\int_{3}^5\int_{2}^{7-y}\, dx\, dy - . - area of R = 4u^2 -

    -     - -     - - - - - - - The region R is bounded to the left by a parabola opening along the x axis, and the line x=12. - -

    - The region R for this exercise is bounded by a parabola and a line. -

      -
    • -

      - The parabola is x=y^2/3, which has vertex (0,0) and opens to the right. -

      -
      • -
    • -

      - The line is the vertical line x=12. -

      -
      • -
    -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={2,4,...,12}, - ytick={-2,-4,-6,6,4,2}, - ymin=-6.9,ymax=6.9, - xmin=-.5,xmax=12.9 - ] - - \draw (axis cs: 6,2) node {$R$}; - - \addplot+ [domain=-6:6,samples=60] ({x^2/3},{x}) node [pos=.7,sloped,above,black] { $x=y^2/3$} -- cycle; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \ds\int_0^{12}\int_{-\sqrt{3x}}^{\sqrt{3x}}\, dy\, dx and \ds \int_{-6}^6\int_{y^2/3}^{12}\, dx\, dy -

    - -

    - area of R = 96u^2 -

    -     - -     - - - - - - - A region in the first quadrant bounded by two curves that intersect at the points (0,0) and (1,1) - -

    - The region R is bounded by two intersecting curves. -

      -
    • -

      - The curve that lies below (and to the right) of the region is y=x^4. - This curve is increasing and concave up. -

      -
      • -
    • -

      - The curve that lies above (and to the left) of the region is y=\sqrt{x}. - This curve is increasing and concave down. -

      -
      • -
    • -

      - The two curves intersect at the origin, and again at the point (1,1). -

      -
      • -
    -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=1.1, - xmin=-.5,xmax=1.1 - ] - - \addplot [firstcurvestyle,domain=0:1.05] ({x^2},{x}) node [pos=.5,sloped,above,black] { $y=\sqrt{x}$}; - \addplot [firstcurvestyle,domain=-.1:1.05,samples=40] ({x},{x^4}) node [pos=.5,sloped,below,black] { $y=x^4$}; - - \draw (axis cs: .5,.4) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \ds\int_0^{1}\int_{x^4}^{\sqrt{x}}\, dy\, dx and \ds\int_{0}^1\int_{y^2}^{\sqrt[4]{y}}\, dx\, dy -

    - -

    - area of R = 7/15u^2 -

    -     - -     - - - - - - - A region R bounded above by a line, and below by a cubic curve. - -

    - The region R lies in the first quadrant. It is bounded by two curves that intersect twice. -

      -
    • -

      - Above (and to the left of) the region is the line y=4x. -

      -
      • -
    • -

      - Below (and to the right of) the region is the curve y=x^3. -

      -
      • -
    • -

      - The two curves meet at the origin. The curve y=x^3 begins below the line y=4x, - but bends upward to meet the line again at the point (2,8). -

      -
      • -
    -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=8.9, - xmin=-.5,xmax=2.5 - ] - - \addplot [firstcurvestyle,domain=-.1:2.1] ({x},{x^3}) node [pos=.2,sloped,below right,black] { $y=x^3$}; - \addplot [firstcurvestyle,domain=-.1:2.1] {4*x} node [pos=.4,sloped,above,black] { $y=4x$}; - - \draw (axis cs: 1,2.25) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -     - -

    - \ds\int_0^{2}\int_{x^3}^{4{x}}\, dy\, dx and \ds\int_{0}^8\int_{y/4}^{\sqrt[3]{y}}\, dx\, dy -

    - -

    - area of R = 4u^2 -

    -     - -     - -     - - - -

    - Iterated integrals are given that compute the area of a region R in the xy-plane. - Sketch the region R, - and give the iterated integral(s) that give the area of R with the opposite order of integration. -

    -     - - - - -

    - \ds \int_{-2}^2\int_0^{4-x^2} \, dy \, dx -

    -     - - - - A region R bounded by a downward-opening parabola and the x axis. - -

    - The curve y=4-x^2 is a parabola centered on the y axis. - Its vertex is at (0,4), and the parabola opens downward, meeting the x axis at (-2,0) and (2,0). - The parabola forms the upper boundary of a region R; - the lower boundary is the portion of the x axis between the two intercepts at (-2,0) and (2,0). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=4.5, - xmin=-2.5,xmax=2.5 - ] - - \addplot+ [domain=-2:2,samples=60] ({x},{4-x^2}) node [pos=.1,right,black] { $y=4-x^2$} -- cycle; - - \draw (axis cs: 1,1) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -

    - area of R = \ds\int_{0}^{4}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\, dx\, dy -

    -     - -     - - - - -

    - \ds \int_{0}^{1}\int_{5-5x}^{5-5x^2} \, dy \, dx -

    -     - - - - The region R is in the first quadrant, between a parabola and a line. - -

    - The upper boundary of the region R is the parabola y=5-5x^2. - The vertex of this parabola is at (0,5), and the parabola opens downward. -

    - -

    - The lower boundary of R is the line y=5-5x. - This line passes through the vertex at (0,5), - and meets the parabola again at (1,0). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ytick={1,2,3,4,5}, - ymin=-.5,ymax=5.9, - xmin=-.1,xmax=1.1 - ] - - \addplot [firstcurvestyle,domain=0:1] ({x},{5-5*x^2}) node [pos=.05,above right,black] { $y=5-5x^2$}; - \addplot [firstcurvestyle,domain=0:1] {-5*x+5} node [pos=.5,below,sloped,black] { $y=-5x+5$}; - - \draw (axis cs: .5,3.3) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -

    - area of R = \ds \int_0^5\int_{1-y/5}^{\sqrt{1-y/5}} \, dx \, dy -

    -     - -     - - - - -

    - \ds \int_{-2}^2\int_{0}^{2\sqrt{4-y^2}} \, dx \, dy -

    -     - - - - A region bounded by the y axis and the right half of an ellipse centered at the origin. - -

    - The curve x=2\sqrt{4-y^2} is the right half of the ellipse \frac{x^2}{16}+\frac{y^2}{4}=1. - The ellipse intersects the y axis at (0,2) and (0,-2), - and the x axis at (4,0). -

    - -

    - The region R is bounded to the right by the ellipse, and to the left by the portion of the y axis - between the points (0,-2) and (0,2). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-2.5,ymax=2.5, - xmin=-.5,xmax=4.5 - ] - - \draw (axis cs: 1,1) node {$R$}; - - \addplot+ [domain=-90:90,samples=40] ({4*cos(x)},{2*sin(x)}) node [pos=.25,above left,black] { $x^2/16+y^2/4=1$} -- cycle; - - \end{axis} - - \end{tikzpicture} - - - - -

    - area of R = \ds \int_0^4\int_{-\sqrt{4-x^2/4}}^{\sqrt{4-x^2/4}} \, dy \, dx -

    -     - -     - - - - -

    - \ds \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \, dy \, dx -

    -     - - - - A circle of radius 3, centered at the origin. - -

    - The boundary of the region described by the bounds in this integral - is the circle x^2+y^2=9. - Its center is the origin, and its radius is 3. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-3.1,ymax=3.1, - xmin=-3.7,xmax=3.7 - ] - - \draw (axis cs: 1,1) node {$R$}; - - \addplot+ [domain=0:360,samples=80] ({3*cos(x)},{3*sin(x)}) node [pos=.4,below right,black] { $x^2+y^2=9$}; - - \end{axis} - - \end{tikzpicture} - - - - -

    - area of R = \ds \int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \, dx \, dy -

    -     - -     - - - - -

    - \ds \int_{0}^{1}\int_{-\sqrt{y}}^{\sqrt{y}} \, dx \, dy+\int_{1}^{4}\int_{y-2}^{\sqrt{y}} \, dx \, dy -

    -     - - - - The region between a right-opening parabola and a vertical line. - -

    - The region of integration for this exercise is bounded to the left by the curve x=y^2, - which is a parabola opening to the right, with its vertex at the origin. - The region is bounded to the right by the vertical line x=1. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,1,2}, - ytick={1,2,3,4}, - ymin=-.1,ymax=4.5, - xmin=-1.1,xmax=2.1 - ] - - \draw (axis cs: .5,1.5) node {$R$}; - - \addplot [firstcurvestyle,domain=-1:2,] ({x},{x^2}) node [pos=.5,below right,black] { $y=x^2$}; - \addplot [firstcurvestyle,domain=-1:2] {x+2} node [pos=.5,sloped,above,black] { $y=x+2$}; - - \end{axis} - - \end{tikzpicture} - - - - -

    - area of R = \ds \int_{-1}^2\int_{x^2}^{x+2} \, dy \, dx -

    -     - -     - - - - -

    - \ds \int_{-1}^{1}\int_{(x-1)/2}^{(1-x)/2} \, dy \, dx -

    -     - - - - A region bounded by a triangle, with vertices at (-1,1), (-1,-1), and (1,0). - -

    - The boundary of the region of integration for this problem is a triangle. - It is an isosceles triangle, with a vertical side from (-1,-1) to (-1,1), - and remaining vertex on the x axis at (1,0). -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,1}, - ytick={-1,1}, - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.1 - ] - - \draw (axis cs: -.5,.25) node {$R$}; - - \draw [firstcolor,very thick] (axis cs: -1,1) -- (axis cs:1,0) -- (axis cs: -1,-1) -- (axis cs: -1,1); - - \end{axis} - - \end{tikzpicture} - - - - -

    - area of R = \ds \int_{-1}^0\int_{0}^{2y+1} \, dx \, dy + \int_{0}^1\int_{0}^{1-2y} \, dx \, dy -

    -     - -     - -     -     -     -
    -
    - Double Integration and Volume -

    - The definite integral of f over [a,b], \int_a^b f(x)\, dx, - was introduced as the signed area under the curve. - We approximated the value of this area by first subdividing [a,b] into n subintervals, - where the ith subinterval has length \dx_i, - and letting c_i be any value in the ith subinterval. - We formed rectangles that approximated part of the region under the curve with width \dx_i, - height f(c_i), and hence with area f(c_i)\dx_i. - Summing all the rectangle's areas gave an approximation of the definite integral, - and stated that - - \int_a^bf(x)\, dx = \lim_{\norm{\Delta x}\to 0}\sum f(c_i)\dx_i - , - connecting the area under the curve with sums of the areas of rectangles. -

    - -

    - We use a similar approach in this section to find volume under a surface. -

    - - - -

    - Let R be a closed, - bounded region in the xy-plane and let - z=f(x,y) be a continuous function defined on R. - We wish to find the signed volume under the graph of f over R. - (We use the term signed volume - to denote that space above the xy-plane, - under f, will have a positive volume; - space above f and under the xy-plane will have a - negative volume, - similar to the notion of signed area used before.) -

    - -

    - We start by partitioning R into n rectangular subregions as shown in . - For simplicity's sake, - we let all widths be \dx and all heights be \dy. - Note that the sum of the areas of the rectangles is not equal to the area of R, - but rather is a close approximation. - Arbitrarily number the rectangles 1 through n, - and pick a point (x_i,y_i) in the ith subregion. -

    - -
    - Developing a method for finding signed volume under a surface - -
    - - - - An illustration of the partition of a plane region into small rectangles. - -

    - A region in the plane is plotted against x and y coordinate axes. -

      -
    • -

      - The region is symmetric about the x axis, - and extends from the origin to the point (2,0). -

      -
      • -
    • -

      - It is close to a circle in shape, but it is not a true circle: - it is more egg-like, being more sharply curved near the origin. -

      -
      • -
    • -

      - The precise shape is not that important in this context. -

      -
      • -
    -

    - -

    - A grid of small rectangles is drawn over the region, - to illustrate how it can be approximated by subdividing it into rectangles. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.85,ymax=.85, - xmin=-.3,xmax=2.2 - ] - - \addplot+ [domain=-90:90,samples=80] ({cos(x)*(1+cos(2*x))},{sin(x)*(1+cos(2*x))}); - - \draw (axis cs: 0,-.3) -- (axis cs:0,.3) - (axis cs: .2,-.4) -- (axis cs:.2,.4) - (axis cs: 0.4,-.6) -- (axis cs:.4,.6) - (axis cs: 0.6,-.7) -- (axis cs:.6,.7) - (axis cs: .8,-.8) -- (axis cs:.8,.8) - (axis cs: 1,-.8) -- (axis cs:1,.8) - (axis cs: 1.2,-.8) -- (axis cs:1.2,.8) - (axis cs: 1.4,-.8) -- (axis cs:1.4,.8) - (axis cs: 1.6,-.7) -- (axis cs:1.6,.7) - (axis cs: 1.8,-.6) -- (axis cs:1.8,.6) - (axis cs: 2,-.4) -- (axis cs:2,.4); - - \draw (axis cs: .8,.8) -- (axis cs:1.4,.8) - (axis cs: .6,.7) -- (axis cs:1.6,.7) - (axis cs: .4,.6) -- (axis cs:1.8,.6) - (axis cs: .4,.5) -- (axis cs:1.8,.5) - (axis cs: .2,.4) -- (axis cs:2,.4) - (axis cs: 0,.3) -- (axis cs:2,.3) - (axis cs: 0,.2) -- (axis cs:2,.2) - (axis cs: 0,.1) -- (axis cs:2,.1) - (axis cs: .8,-.8) -- (axis cs:1.4,-.8) - (axis cs: .6,-.7) -- (axis cs:1.6,-.7) - (axis cs: .4,-.6) -- (axis cs:1.8,-.6) - (axis cs: .4,-.5) -- (axis cs:1.8,-.5) - (axis cs: .2,-.4) -- (axis cs:2,-.4) - (axis cs: 0,-.3) -- (axis cs:2,-.3) - (axis cs: 0,-.2) -- (axis cs:2,-.2) - (axis cs: 0,-.1) -- (axis cs:2,-.1); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - A surface in three dimensions lies over a partitioned region in the xy plane. A rectangular prism extends from one rectangle to the surface. - -

    - A surface is plotted in space, against three-dimensional coordinate axes. - In the xy plane, there is a copy of the region from , - including a rendering of the rectangular grid representing the partition of the region into small rectangles. -

    - -

    - The surface lies above the xy plane. - On the surface, we can see a curve that corresponds to the boundary of the region in the xy plane. - The grid lines on the surface also appear to correspond to the partition of the region below the surface. -

    - -

    - Over one of the small rectangles in the xy plane, - there is a rectangular box. The base of the box is the rectangle in the plane. - The box extends upward to the surface, - with its height equal to the z coordinate on the surface at one point in the rectangle. -

    -     - - - - - //ASY file for figdouble_intro23D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.5,3.1,3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={-1,1}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-1,2.5); - pair ybounds=(-1.25,1.25); - pair zbounds=(0,2.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface//{z=-.5*(x-1)^2-.5*(y)^2+2}; - triple f(pair t) { - return (t.x,t.y,-.5*(t.x-1)^2-.5*(t.y)^2+2); - } - surface s=surface(f,(-0.221,-1),(2.2,1),12,20,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic},usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - pen pp=linewidth(.25mm); - //draw the grid in the xy-plane - //along fixed x - draw((0,-.3,0) -- (0,.3,0),pp); - draw((.2,-.4,0) -- (.2,.4,0),pp); - draw((0.4,-.6,0) -- (.4,.6,0),pp); - draw((0.6,-.7,0) -- (.6,.7,0),pp); - draw((.8,-.8,0) -- (.8,.8,0),pp); - draw((1,-.8,0) -- (1,.8,0),pp); - draw((1.2,-.8,0) -- (1.2,.8,0),pp); - draw((1.4,-.8,0) -- (1.4,.8,0),pp); - draw((1.6,-.7,0) -- (1.6,.7,0),pp); - draw((1.8,-.6,0) -- (1.8,.6,0),pp); - draw((2,-.4,0) -- (2,.4,0),pp); - //along fixed y - draw((.8,.8,0) -- (1.4,.8,0),pp); - draw((.6,.7,0) -- (1.6,.7,0),pp); - draw((.4,.6,0) -- (1.8,.6,0),pp); - draw((.4,.5,0) -- (1.8,.5,0),pp); - draw((.2,.4,0) -- (2,.4,0),pp); - draw((0,.3,0) -- (2,.3,0),pp); - draw((0,.2,0) -- (2,.2,0),pp); - draw((0,.1,0) -- (2,.1,0),pp); - draw((.8,-.8,0) -- (1.4,-.8,0),pp); - draw((.6,-.7,0) -- (1.6,-.7,0),pp); - draw((.4,-.6,0) -- (1.8,-.6,0),pp); - draw((.4,-.5,0) -- (1.8,-.5,0),pp); - draw((.2,-.4,0) -- (2,-.4,0),pp); - draw((0,-.3,0) -- (2,-.3,0),pp); - draw((0,-.2,0) -- (2,-.2,0),pp); - draw((0,-.1,0) -- (2,-.1,0),pp); - - //Draw curve on top of the grid in xy plane ({cos(x)*(1+cos(2*x))},{sin(x)*(1+cos(2*x))},0); - triple g(real t) {return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),0);} - path3 mypath=graph(g,-pi/2,pi/2,operator ..); - draw(mypath,bluepen); - - // draw curve on surface - triple g(real t) { - return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),-.5*(cos(t)*(1+cos(2*t))-1)^2-.5*(sin(t)*(1+cos(2*t)))^2+2); - } - path3 mypath=graph(g,-pi/2,pi/2,operator ..); - draw(mypath,bluepen); - - //surface s=surface(f,(-pi/2,0),(pi/2,2*pi),8,8,Spline); - //pen p=apexmeshpen; - //draw(s,surfacepen,bluepen+linewidth(2)); - //path3 mypath=graph(g,0,32.5,operator ..); draw(mypath,bluepen); - - //Draw the top of the column on the surface - //draw((1.6,0.3,1.865)--(1.8,.3,1.725),bluepen+linewidth(2)); - //draw((1.8,0.3,1.7)--(1.8,.4,1.7),bluepen+linewidth(2)); - //draw((1.8,0.4,1.7)--(1.6,.4,1.7),bluepen+linewidth(2)); - //draw((1.6,0.4,1.7)--(1.6,.3,1.7),bluepen+linewidth(2)); - draw((1.6,.3,1.775) -- (1.6,.4,1.74) -- (1.8,.4,1.6) -- (1.8,.3,1.635) --cycle,bluepen+.4mm); - - //Draw the rectangular column - //base - draw((1.6,0.3,0)--(1.8,.3,0)--(1.8,.4,0)--(1.6,.4,0)--cycle,redpen+.4mm); - - // //verticals plotted to average of 4 function heights - //draw((1.6,0.3,0)--(1.6,.3,1.7),redpen+.4mm);//base - //draw((1.8,0.3,0)--(1.8,.3,1.7),redpen+.4mm);//base - //draw((1.8,0.4,0)--(1.8,.4,1.7),redpen+.4mm);//base - //draw((1.6,0.4,0)--(1.6,.4,1.7),redpen+.4mm);//base - //top - draw((1.6,0.3,1.7)--(1.8,.3,1.7)--(1.8,.4,1.7)--(1.6,.4,1.7)--cycle,redpen); - - pen q=redcurvepen+.4mm; - // draw(ss,surfacepen2,meshpen=q,nolight,render(merge=true)); - // emissive(redpen+opacity(0.7)) - - //Shade the column now - //import three; - - path3 p = (1.6,0.3,0)--(1.8,.3,0) -- (1.8,0.3,1.7)--(1.6,.3,1.7); //Left - draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); - path3 p = (1.6,0.3,1.7)--(1.8,.3,1.7)--(1.8,.4,1.7)--(1.6,.4,1.7); //top - draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); - //draw(surface(p -- cycle), emissive(redpen+opacity(0.7))); - path3 p = (1.6,0.4,0)--(1.8,.4,0) -- (1.8,0.4,1.7)--(1.6,.4,1.7); //right - draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); - //draw(surface(p -- cycle), emissive(redpen+opacity(0.7))); - path3 p = (1.6,0.3,0)--(1.6,.4,0) -- (1.6,0.4,1.7)--(1.6,.3,1.7); //back - draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); - //draw(surface(p -- cycle), emissive(redpen+opacity(0.7))); - path3 p = (1.8,0.3,0)--(1.8,.4,0) -- (1.8,0.4,1.7)--(1.8,.3,1.7); //front - draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); - //draw(surface(p -- cycle), emissive(redpen+opacity(0.7))); - path3 p = (1.6,0.3,0)--(1.8,.3,0)--(1.8,.4,0)--(1.6,.4,0); //bottom - draw(surface(p -- cycle),surfacepen2,meshpen=q,nolight,render(merge=true)); - //draw(surface(p -- cycle), emissive(redpen+opacity(0.7))); - - - - -
    -     - -
    - -

    - The volume of the rectangular solid whose base is the - ith subregion and whose height is - f(x_i,y_i) is V_i=f(x_i,y_i)\dx\dy. - Such a solid is shown in . - Note how this rectangular solid only approximates the true volume under the surface; - part of the solid is above the surface and part is below. -

    - -

    - For each subregion R_i used to approximate R, - create the rectangular solid with base area \dx\dy and height f(x_i,y_i). - The sum of all rectangular solids is - - \ds \sum_{i=1}^n f(x_i,y_i)\dx\dy - . -

    - -

    - This approximates the signed volume under f over R. - As we have done before, - to get a better approximation we can use more rectangles to approximate the region R. -

    - -

    - In general, each rectangle could have a different width \dx_j and height \dy_k, - giving the ith rectangle an area \Delta A_i = \dx_j\dy_k and the - ith rectangular solid a volume of f(x_i,y_i)\Delta A_i. - Let \norm{\Delta A} denote the length of the longest diagonal of all rectangles in the subdivision of R; - \norm{\Delta A}\to 0 means each rectangle's width and height are both approaching 0. - If f is a continuous function, - as \norm{\Delta A} shrinks - (and hence n\to\infty) - the summation \ds \sum_{i=1}^n f(x_i,y_i)\Delta A_i approximates the signed volume better and better. - This leads to a definition. -

    - - - - - Double Integral, Signed Volume - -

    - Let z=f(x,y) be a continuous function defined over a closed, - bounded region R in the xy-plane. - The signed volume V under f over R is denoted by the - double integral - - V = \iint_R f(x,y)\, dA - . -

    - -

    - Alternate notations for the double integral are - integrationdouble - double integral - iterated integration - signed volume - volume - - \iint_R f(x,y)\, dA=\iint_R f(x,y)\, dx\, dy=\iint_R f(x,y)\, dy\, dx - . -

    -     -     - - - -

    - does not state how to find the signed volume, - though the notation offers a hint. - We need the next two theorems to evaluate double integrals to find volume. -

    - - - Double Integrals and Signed Volume - -

    - Let z=f(x,y) be a continuous function defined over a closed , bounded region R in the xy-plane. - Then the signed volume V under f over R is - - V = \iint_R f(x,y)\, dA = \lim_{\norm{\Delta A}\to 0}\sum_{i=1}^n f(x_i,y_i)\Delta A_i - . -

    -     -     - -

    - This theorem states that we can find the exact signed volume using a limit of sums. - The partition of the region R is not specified, - so any partitioning where the diagonal of each rectangle shrinks to 0 results in the same answer. -

    - -

    - This does not offer a very satisfying way of computing volume, though. - Our experience has shown that evaluating the limits of sums can be tedious. - We seek a more direct method. -

    - -

    - Recall - in . - This stated that if A(x) gives the cross-sectional area of a solid at x, - then \int_a^b A(x)\, dx gave the volume of that solid over [a,b]. -

    - -

    - Consider , - where a surface z=f(x,y) is drawn over a region R. - Fixing a particular x value, - we can consider the area under f over R where x has that fixed value. - That area can be found with a definite integral, namely - - A(x)=\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy - . -

    - -
    - Finding volume under a surface by sweeping out a cross-sectional area - - - - A parabolic surface in space is intersected by a plane of constant x value. - -

    - The plot begins with a set of three-dimensional coordinate axes. - In the xy plane, there is a curve that appears to be a circle of radius 1, centered at (1,0,0). -

    - -

    - The circle is the boundary of a region R. - Above this region, there is a surface; the surface has the shape of a downward-opening circular paraboloid. -

    - -

    - A portion of a vertical plane is also drawn. - The plane corresponds to a fixed value of x, and extends in the y direction across the circle. - The plane is shown extending upward to the surface, which it intersects along a curve. - We can think of this portion of the plane as a slice of the solid that lies below the paraboloid and above the circle. -

    -     - - - - - //ASY file for figdouble_intro23D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5,5.5,1.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={-1,1}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-1,2.5); - pair ybounds=(-1.25,1.25); - pair zbounds=(0,2.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Create a shaded slice - - pen q=redcurvepen+.4mm; - path3 p = (1.6,-0.65,0)--(1.6,0.65,0) -- (1.6,0.65,1.65)..(1.6,0,1.8)..(1.6,-0.65,1.65)--cycle; - draw(surface(p), surfacepen2,meshpen=invisible,nolight,render(merge=true)); - draw(p,q); - - //Draw the surface//{z=-.5*(x-1)^2-.5*(y)^2+2}; - triple f(pair t) { - return (t.x,t.y,-.5*(t.x-1)^2-.5*(t.y)^2+2); - } - surface s=surface(f,(-0.221,-1),(2.2,1),12,20,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic},usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw curve in xy plane ({cos(x)*(1+cos(2*x))},{sin(x)*(1+cos(2*x))},0); - triple g(real t) {return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),0);} - path3 mypath=graph(g,-pi/2,pi/2,operator ..); - draw(mypath,.4mm+bluepen); - label("$R$",(0.5,0.6,0),E); - - //Draw curve on the surface //{z=-.5*(cos(t)*(1+cos(2*t))-1)^2-.5*(sin(t)*(1+cos(2*t)))^2+2}; - triple gg(real t) { - return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),-.5*(cos(t)*(1+cos(2*t))-1)^2-.5*(sin(t)*(1+cos(2*t)))^2+2); - } - path3 mypath=graph(gg,-pi/2,pi/2,operator ..); - draw(mypath,.4mm+bluepen); - - //draw the edge in thick red pen q=redcurvepen+.4mm; - //draw((1.6,-0.65,0)--(1.6,0.65,0)--(1.6,0.65,1.625)..(1.6,0,1.825)..(1.6,-0.65,1.625)--(1.6,-0.65,0),redcurvepen+.4mm); - - - - -
    - -

    - Remember that though the integrand contains x, - we are viewing x as fixed. - Also note that the bounds of integration are functions of x: - the bounds depend on the value of x. -

    - -

    - As A(x) is a cross-sectional area function, - we can find the signed volume V under f by integrating it: - - V = \int_a^b A(x)\, dx = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\right)dx = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx - . -

    - -

    - This gives a concrete method for finding signed volume under a surface. - We could do a similar procedure where we started with y fixed, - resulting in an iterated integral with the order of integration dx\, dy. - The following theorem states that both methods give the same result, - which is the value of the double integral. - It is such an important theorem it has a name associated with it. -

    - - - Fubini's Theorem - -

    - Let R be a closed, - bounded region in the xy-plane and let - z=f(x,y) be a continuous function on R. - double integral - iterated integration - signed volume - volume - Fubini's Theorem -

    - -

    -

      -
    1. -

      - If R is bounded by a\leq x\leq b and g_1(x)\leq y\leq g_2(x), - where g_1 and g_2 are continuous functions on [a,b], then - - \iint_R f(x,y)\, dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx - . -

      -
      2. - -
    2. -

      - If R is bounded by c\leq y\leq d and h_1(y)\leq x\leq h_2(y), - where h_1 and h_2 are continuous functions on [c,d], then - - \iint_R f(x,y)\, dA = \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy - . -

      -
      3. -
    -

    -     -     - -

    - Note that once again the bounds of integration follow the curve to curve, - point to point pattern discussed in the previous section. - In fact, one of the main points of the previous section is developing the skill of describing a region R with the bounds of an iterated integral. - Once this skill is developed, - we can use double integrals to compute many quantities, - not just signed volume under a surface. -

    - - - - - Evaluating a double integral - -

    - Let f(x,y) = xy+e^y. - Find the signed volume under f on the region R, - which is the rectangle with corners (3,1) and (4,2) pictured in , - using Fubini's Theorem and both orders of integration. -

    -
    - Finding the signed volume under a surface in - - - - A ramp-like surface in space, with the portion over a rectangular domain highlighted. - -

    - The surface given by the graph of f(x,y)=xy+e^y has the appearance of a curved, twisted, and very steep ramp. - Along the x axis we have the constant value z=1. - For small, fixed values of x, the traces are exponential curves that increase with y. - For larger values of x, the xy term contributes, and the rate of growth is larger. -

    - -

    - In the xy plane, a rectangular region R is shown. - Dashed lines move upward from the boundary of this region to the surface. - Along the surface, a corresponding rectangle-shaped curve is shown. - The two rectangular curves, and the dashed lines between corresponding corners, - illustrate the region whose volume is computed in this example. -

    -     - - - - - //ASY file for figdouble13D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(10,-8,25); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3,4}; - real[] myychoice={1,2}; - real[] myzchoice={10}; - defaultpen(0.5mm); - - pair xbounds=(0,5); - pair ybounds=(0,2.5); - pair zbounds=(0,15); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface//{z=xy+e^y}; - triple f(pair t) { - return (t.x,t.y,t.x*t.y+exp(t.y)); - } - surface s=surface(f,(0,0),(4.25,2.25),12,20,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw rectangle in plane - draw((3,1,0)--(4,1,0)--(4,2,0)--(3,2,0)--(3,1,0),bluepen+linewidth(2)); - label("$R$",(3.5,1.5,0)); - - //Draw rectangle on surface - draw((3,1,3+exp(1))--(4,1,4+exp(1))..(4,1.5,6+exp(1.5))..(4,2,8+exp(2))--(3,2,6+exp(2))..(3,1.5,4.5+exp(1.5))..(3,1,3+exp(1)),bluepen+linewidth(2)); - - //Draw vertical lines - draw((3,1,0)--(3,1,3+exp(1)),bluepen+dashed+linewidth(0.75)); - draw((3,2,0)--(3,2,6+exp(2)),bluepen+dashed+linewidth(0.75)); - draw((4,1,0)--(4,1,4+exp(1)),bluepen+dashed+linewidth(0.75)); - draw((4,2,0)--(4,2,8+exp(2)),bluepen+dashed+linewidth(0.75)); - - - - -
    -     - -

    - We wish to evaluate \iint_R \big(xy+e^y\big)\, dA. - As R is a rectangle, - the bounds are easily described as - 3\leq x\leq 4 and 1\leq y\leq 2. -

    - -

    - Using the order dy\, dx: - - \iint_R\big(xy+e^y\big) \, dA \amp = \int_3^4\int_1^2\big(xy+e^y\big)\, dy \, dx - \amp = \int_3^4 \left(\left.\left[\frac12xy^2+e^y\right]\right|_1^2\, \right) dx - \amp = \int_3^4\left(\frac 32x + e^2-e\right)dx - \amp = \left.\left(\frac 34x^2 + \big(e^2-e\big)x\right)\right|_3^4 - \amp = \frac {21}4+ e^2-e\approx 9.92 - . -

    - -

    - Now we check the validity of Fubini's Theorem by using the order dx\, dy: - - \iint_R\big(xy+e^y\big) \, dA \amp = \int_1^2\int_3^4\big(xy+e^y\big)\, dx \, dy - \amp = \int_1^2\left(\left.\left[\frac12x^2y+xe^y\right]\right|_3^4\right)dy - \amp = \int_1^2\left(\frac72y+e^y\right)\, dy - \amp = \left.\left(\frac74y^2+e^y\right)\right|_1^2 - \amp =\frac{21}4+e^2-e\approx 9.92 - . -

    - -

    - Both orders of integration return the same result, as expected. -

    -     -     - - - Evaluating a double integral - -

    - Evaluate \iint_R \big(3xy-x^2-y^2+6\big)\, dA, - where R is the triangle bounded by x=0, - y=0 and x/2+y=1, - as shown in . -

    -
    - Finding the signed volume under the surface in - - - - A surface in space, plotted over a triangular domain. - -

    - The graph z = 3xy-x^2-y^2+6 is plotted in the first octant of a three-dimensional coordinate system. - It appears to be a hyperbolic paraboloid, - with slight upward curvature when viewed along the plane y=1-x, - and a larger downward curvature when viewed along the plane y=x. -

    - -

    - In the xy plane, a triangular region is plotted. - This region is bounded by the x and y axes, - and a line segment running from (2,0,0) to (0,1,0). - There are also dashed, vertical lines running from these two points up to the surface. -

    - -

    - On the surface, there are curves corresponding to the three edges of the triangle in the xy plane. - These curves illustrate the portion of the surface under which the solid whose volume we are finding lies. - Each curve has the shape of a downward-opening parabola. -

    -     - - - - - //ASY file for figdouble23D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(1.6,-5.5,12.4); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3,4}; - real[] myychoice={1}; - real[] myzchoice={2,4,6,8}; - defaultpen(0.5mm); - - pair xbounds=(0,2.5); - pair ybounds=(0,1.5); - pair zbounds=(0,9); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface//{z=3xy-x^2-y^2+6}; - triple f(pair t) { - return (t.x,t.y,3*t.x*t.y-t.x^2-t.y^2+6); - } - surface s=surface(f,(-0.1,-0.1),(2.1,1.1),12,20,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw triangle in plane - draw((0,0,0)--(2,0,0)--(0,1,0)--(0,0,0),bluepen+linewidth(2)); - label("$R$",(0,0.5,0),E); - - //Draw triangle on surface - triple g(real t) {return (t,0,6-t^2);} - path3 mypath=graph(g,0,2,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (0,t,6-t^2);} - path3 mypath=graph(g,0,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (2-2*t,t,3*(2-2*t)*(t)-(2-2*t)^2-(t)^2+6);} - path3 mypath=graph(g,0,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //Draw vertical lines - draw((2,0,0)--(2,0,2),bluepen+dashed+linewidth(0.75)); - draw((0,1,0)--(0,1,5),bluepen+dashed+linewidth(0.75)); - - - - -
    -     - -

    - While it is not specified which order we are to use, - we will evaluate the double integral using both orders to help drive home the point that it does not matter which order we use. -

    - -

    - Using the order dy\, dx: - The bounds on y go from curve to curve, - , 0\leq y\leq 1-x/2, - and the bounds on x go from - point to point, , 0\leq x\leq 2. - - \iint_R \big(3xy-x^2-y^2+6\big)\, dA \amp = \int_0^2\int_0^{-\frac x2+1} \big(3xy-x^2-y^2+6\big)\, dy\, dx - \amp = \int_0^2\left.\left(\frac32xy^2-x^2y-\frac13y^3+6y\right)\right|_0^{-\frac x2+1}dx - \amp = \int_0^2 \left(\frac{11}{12}x^3-\frac{11}{4}x^2-x+\frac{17}3\right)dx - \amp = \left.\left(\frac{11}{48}x^4-\frac{11}{12}x^3-\frac12x^2+\frac{17}3x\right)\right|_0^2 - \amp = \frac{17}3=5.\overline{6} - . -

    - -

    - Now lets consider the order dx \, dy. - Here x goes from curve to curve, 0\leq x\leq 2-2y, - and y goes from point to point, 0\leq y\leq 1: - - \iint_R \big(3xy-x^2-y^2+6\big)\, dA \amp = \int_0^1\int_0^{2-2y} \big(3xy-x^2-y^2+6\big)\, dx\, dy - \amp = \int_0^1\left.\left(\frac32x^2y-\frac13x^3-xy^2+6x\right)\right|_0^{2-2y} dy - \amp = \int_0^1\left(\frac{32}3y^3-22y^2+2y+\frac{28}3\right)dy - \amp =\left.\left(\frac83y^4-\frac{22}3y^3+y^2+\frac{28}3y\right)\right|_0^1 - \amp =\frac{17}3=5.\overline{6} - . -

    - -

    - We obtained the same result using both orders of integration. -

    -     -     - - -

    - Note how in these examples that the bounds of integration depend only on R; - the bounds of integration have nothing to do with f(x,y). - This is an important concept, - so we include it as a Key Idea. -

    - - - Double Integration Bounds -

    - When evaluating \iint_R f(x,y)\, dA using an iterated integral, - the bounds of integration depend only on R. - The function f does not determine the bounds of integration. -

    -     - -

    - Before doing another example, - we give some properties of double integrals. - Each should make sense if we view them in the context of finding signed volume under a surface, - over a region. -

    - - - Properties of Double Integrals - -

    - Let f and g be continuous functions over a closed, - bounded plane region R, - and let c be a constant. - double integralproperties - iterated integrationproperties -

    - -

    -

      -
    1. -

      - \ds \iint_Rc\,f(x,y)\, dA = c\iint_Rf(x,y)\, dA. -

      -
      2. - -
    2. -

      - \ds \iint_R \big(f(x,y)\pm g(x,y)\big)\, dA = \iint_R f(x,y)\, dA \pm \iint_R g(x,y)\, dA -

      -
      3. - -
    3. -

      - If f(x,y)\geq 0 on R, - then \ds \iint_R f(x,y)\, dA\geq 0. -

      -
      4. - -
    4. -

      - If f(x,y)\geq g(x,y) on R, - then \ds \iint_R f(x,y)\, dA\geq \iint_R g(x,y)\, dA. -

      -
      5. - -
    5. - -

      - Let R be the union of two nonoverlapping regions, - R = R_1\bigcup R_2 - (see ). - Then - - \iint_R f(x,y)\, dA = \iint_{R_1}f(x,y)\, dA+ \iint_{R_2}f(x,y)\, dA - . -

      - -
      - R is the union of two nonoverlapping regions, R_1 and R_2 - - - Illustration of a pond-like region in the plane that has been divided into two sub-regions. - -

      - The sketch shows a generic region R in the plane. - The precise shape of the region is unimportant, - but its shape is like that of a pond with a rounded but uneven boundary. -

      - -

      - A straight line passes through the middle of the region, - dividing it into two smaller, nonoverlapping regions R_1 and R_2. -

      -       - - - \begin{tikzpicture}[scale=0.88] - - \draw [firstcolor,thick,fill=firstcolor!15,smooth] plot coordinates {(0,2.)(0.06639,2.527)(0.2407,2.839)(0.4857,3.009)(0.7641,3.112)(1.039, - 3.221)(1.294,3.367)(1.532,3.53)(1.759,3.689)(1.98,3.823)(2.2,3.915)(2. - 42,3.967)(2.64,3.992)(2.86,4.)(3.08,4.)(3.305,3.987)(3.543,3.93)(3. - 808,3.797)(4.107,3.558)(4.45,3.185)(4.815,2.703)(5.17,2.152)(5.483,1. - 577)(5.723,1.018)(5.872,0.5022)(5.938,0.02083)(5.93,-0.4363)(5.861,-0. - 88)(5.741,-1.319)(5.58,-1.742)(5.389,-2.128)(5.179,-2.456)(4.96,-2. - 705)(4.737,-2.865)(4.502,-2.953)(4.243,-2.991)(3.95,-3.)(3.614,-2.999) - (3.244,-2.984)(2.86,-2.935)(2.484,-2.829)(2.137,-2.646)(1.832,-2.378)( - 1.558,-2.06)(1.299,-1.734)(1.039,-1.443)(0.7641,-1.222)(0.4857,-0. - 9779)(0.2407,-0.5015)(0.06639,0.4201)(0,2.)}; - - \draw [thick,firstcolor] (3,4) -- (4,-3); - - \draw (2,1) node {$R_1$}; - \draw (5,.5) node {$R_2$}; - \draw (1,-2) node {$R$}; - - \end{tikzpicture} - - - - -
      -
      6. -
    -

    -     -     - - - - - Evaluating a double integral - -

    - Let f(x,y) = \sin(x) \cos(y) and R be the triangle with vertices (-1,0), - (1,0) and (0,1) - (see ). - Evaluate the double integral \iint_Rf(x,y)\, dA. -

    - -
    - Finding the signed volume under a surface in - - - - A wave-like surface on which a triangular curve is drawn, corresponding to a domain in the plane z=0 below. - -

    - The graph z=\sin(x)\cos(y) is plotted against a set of three-dimensional coordinate axes. - It is a wave-like surface, but no crests or troughs are visible, - as we only see the portion of the surface with -1\leq x\leq 1 and 0\leq y\leq 1. -

    - -

    - In the xy plane, the domain of f(x,y) for this integral is shown as a triangle, - with vertices at (-1,0,0), (1,0,0), and (0,1,0). - In the default view of the image, this triangle is only partially visible below the surface. -

    - -

    - On the surface, we can see a triangular curve given by the points on the graph that lie above the boundary of the domain. -

    -     - - - - - //ASY file for figdouble33D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(.92,4,3.7); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={1}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(0,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface//{z=sin(x)cos(y)}; - triple f(pair t) { - return (t.x,t.y,sin(t.x)*cos(t.y)); - } - surface s=surface(f,(-1.25,-0.25),(1.25,1.5),12,20,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw triangle in plane - draw((-1,0,0)--(1,0,0)--(0,1,0)--(-1,0,0),bluepen+dashed+linewidth(0.75)); - label("$R$",(0.8,0.25,0),E); - - //Draw triangle on surface - triple g(real t) {return (t,0,sin(t));} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t-1,t,sin(t-1)*cos(t));} - path3 mypath=graph(g,0,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,1-t,sin(t)*cos(1-t));} - path3 mypath=graph(g,0,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //Draw vertical lines - //draw((-1,0,0)--(-1,0,sin(-1)),bluepen+dashed+linewidth(0.75)); - //draw((1,0,0)--(1,0,sin(1)),bluepen+dashed+linewidth(0.75)); - - - - -
    -     - -

    - If we attempt to integrate using an iterated integral with the order dy\, dx, - note how there are two upper bounds on R meaning we'll need to use two iterated integrals. - We would need to split the triangle into two regions along the y-axis, - then use , - Part. -

    - -

    - Instead, let's use the order dx\, dy. - The curves bounding x are y-1\leq x\leq 1-y; - the bounds on y are 0\leq y\leq 1. - This gives us: - - \iint_R f(x,y)\, dA \amp = \int_0^1\int_{y-1}^{1-y}\sin(x) \cos(y) \, dx\, dy - \amp = \int_0^1\left.\Big( -\cos(x) \cos(y) \Big)\right|_{y-1}^{1-y}\,dy - \amp = \int_0^1 \cos(y) \Big(-\cos(1-y) + \cos(y-1)\Big)dy - . -

    - -

    - Recall that the cosine function is an even function; - that is, \cos(x) = \cos(-x). - Therefore, from the last integral above, - we have \cos(y-1) = \cos(1-y). - Thus the integrand simplifies to 0, and we have - - \iint_R f(x,y)\, dA \amp = \int_0^1 0\, dy - \amp = 0 - . -

    - -

    - It turns out that over R, - there is just as much volume above the xy-plane as below - (look again at ), - giving a final signed volume of 0. -

    -     -     - - - - - Evaluating a double integral - -

    - Evaluate \iint_R (4-y)\, dA, - where R is the region bounded by the parabolas y^2=4x and x^2=4y, - graphed in . -

    - -
    - Finding the volume under the surface in - - - - A rectangular portion of a plane in space, and a petal-like curve on the plane. - -

    - The graph z=4-y is a plane. - It is plotted as a rectangular region in the first octant, - intersecting the xy plane along the line y=4, - and the xz plane along the line z=4. -

    - -

    - The parabolas y^2=4x and x^2=4y are sketched in the xy plane, - from the origin to their intersection at (4,4,0). - On the plane, we see the corresponding curves, - which are also parabolas. The region between them has the shape of a narrow leaf or petal. -

    -     - - - - - //ASY file for figdouble43D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(16,6.2,5.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={2,4}; - defaultpen(0.5mm); - - pair xbounds=(0,5); - pair ybounds=(0,5); - pair zbounds=(0,5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface//{z=4-y}; - triple f(pair t) { - return (t.x,t.y,4-t.y); - } - surface s=surface(f,(-0.25,-0.25),(4.5,4.25),12,20,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw parabolas in plane - triple g(real t) {return (t^2/4,t,0);} - path3 mypath=graph(g,0,4.25,operator ..); - draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,t^2/4,0);} - path3 mypath=graph(g,0,4.25,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //Draw parabolas on surface - triple g(real t) {return (t^2/4,t,4-t);} - path3 mypath=graph(g,0,4.,operator ..); - draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,t^2/4,4-t^2/4);} - path3 mypath=graph(g,0,4.,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //Draw vertical lines - //draw((-1,0,0)--(-1,0,sin(-1)),bluepen+dashed+linewidth(0.75)); - //draw((1,0,0)--(1,0,sin(1)),bluepen+dashed+linewidth(0.75)); - - - - -
    -     - -

    - Graphing each curve can help us find their points of intersection. - Solving analytically, the second equation tells us that y=x^2/4. - Substituting this value in for y in the first equation gives us x^4/16 = 4x. - Solving for x: - - \frac{x^4}{16} \amp = 4x - x^4-64x \amp =0 - x(x^3-64) \amp =0 - x\amp = 0,\, 4 - . -

    - -

    - Thus we've found analytically what was easy to approximate graphically: - the regions intersect at (0,0) and (4,4), - as shown in . -

    - -

    - We now choose an order of integration: - dy\, dx or dx\, dy? - Either order works; since the integrand does not contain x, - choosing dx\, dy might be simpler at least, - the first integral is very simple. -

    - -

    - Thus we have the following curve to curve, - point to point bounds: - - y^2/4\leq x\leq 2\sqrt y, \text{ and } 0\leq y\leq 4 - . - Therefore, - - \iint_R (4-y)\, dA \amp = \int_0^4\int_{y^2/4}^{2\sqrt{y}}(4-y)\, dx\, dy - \amp = \int_0^4 \big(x(4-y)\big)\Big|_{y^2/4}^{2\sqrt{y}} dy - \amp = \int_0^4 \Big(\big(2\sqrt{y}-\frac{y^2}{4}\big)\big(4-y\big)\Big)\, dy - \amp = \int_0^4 \Big( \frac{y^3}{4}-y^2-2y^{3/2}+8y^{1/2}\Big)\, dy - \amp = \left.\left(\frac{y^4}{16}-\frac{y^3}{3}-\frac{4y^{5/2}}5+\frac{16y^{3/2}}3\right)\right|_0^4 - \amp = \frac{176}{15} = 11.7\overline{3} - . -

    - -

    - The signed volume under the surface z=f(x,y) is about 11.7 cubic units. -

    -     -     - - -

    - In the previous section we practiced changing the order of integration of a given iterated integral, - where the region R was not explicitly given. - Changing the bounds of an integral is more than just an test of understanding. - Rather, there are cases where integrating in one order is really hard, - if not impossible, - whereas integrating with the other order is feasible. -

    - - - Changing the order of integration - -

    - Rewrite the iterated integral - \ds \int_0^3\int_y^3 e^{-x^2}\, dx\, dy with the order dy\, dx. - Comment on the feasibility to evaluate each integral. -

    -     - -

    - Once again we make a sketch of the region over which we are integrating to facilitate changing the order. - The bounds on x are from x=y to x=3; - the bounds on y are from y=0 to y=3. - These curves are sketched in , - enclosing the region R. -

    - -
    - Determining the region R determined by the bounds of integration in - - - Several lines are plotted in the plane. Three of these lines form a triangle. - -

    - The x and y coordinate axes are plotted in the plane, - with the origin to the bottom-left of the image, - so that the first quadrant is shown. -

    - -

    - Four lines are plotted: the horizontal lines y=0 (along the x axis) and y=3, - the vertical line x=3, and the diagonal line y=x. - The line y=x divides the square 0\leq x,y\leq 3 into two triangles, - one below the line, and one above. -

    - -

    - The line y=3 illustrates that the upper bound for y is 3, - but this line is otherwise not really needed. - The remaining lines y=0, x=3, and y=x form a triangle with vertices at - (0,0), (3,0), and (3,3). -

    - -

    - A label R placed inside the triangle hepls to illustrate that the region lies below y=x, - and not above. -

    -     - - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.5,ymax=3.5, - xmin=-.5,xmax=3.5 - ] - - \addplot+ [very thick,domain=-.5:3.5] {x} node [pos=.5,sloped,above,black] { $y=x$}; - - \draw [very thick,firstcolor] (axis cs:3,-.5) -- (axis cs:3,3.5); - - \draw [very thick,secondcolor] (axis cs:-.5,0) -- (axis cs:3.5,0) - (axis cs:-.5,3) -- (axis cs:3.5,3); - - \draw (axis cs:2,1) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - To change the bounds, - note that the curves bounding y are y=0 up to y=x; - the triangle is enclosed between x=0 and x=3. - Thus the new bounds of integration are - 0\leq y\leq x and 0\leq x\leq 3, - giving the iterated integral \ds \int_0^3\int_0^x e^{-x^2}\, dy\, dx. -

    - -

    - How easy is it to evaluate each iterated integral? - Consider the order of integrating dx\, dy, - as given in the original problem. - The first indefinite integral we need to evaluate is \int e^{-x^2}\, dx; - we have stated before - (see ) - that this integral cannot be evaluated in terms of elementary functions. - We are stuck. -

    - -

    - Changing the order of integration makes a big difference here. - In the second iterated integral, - we are faced with \int e^{-x^2}\, dy; - integrating with respect to y gives us ye^{-x^2}+C, - and the first definite integral evaluates to - - \int_0^x e^{-x^2}\, dy = xe^{-x^2} - . -

    - -

    - Thus - - \int_0^3\int_0^x e^{-x^2}\, dy\, dx = \int_0^3\Big(xe^{-x^2}\Big)dx - . -

    - -

    - This last integral is easy to evaluate with substitution, - giving a final answer of \frac12(1-e^{-9})\approx 0.5. - - shows the surface over R. -

    - -
    - Showing the surface z=f(x,y) defined in over its region R - - - - A wave-like surface in three dimensions, and a curve on the surface illustrating a triangular domain. - -

    - In two dimensions, the graph y=e^{-x^2} is a bell curve, with a peak at (0,1). - The surface z=e^{-x^2} is a cylinder through this curve. - We are shown a portion of this surface with x\geq 0, - which has the appearance of a wave parallel to the y axis that reaches its crest when x=0. -

    - -

    - The triangular domain is sketched in the xy plane, - with vertices at (0,0,0), (3,0,0), and (3,3,0), - but it is mostly obscured by the surface. -

    - -

    - On the surface itself we see the corresponding triangular curve, - illustrating the region under which we are computing a volume with this integral. -

    -     - - - - - //ASY file for figdouble6b3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(6.3,9.6,2.4); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(0,4); - pair ybounds=(0,3); - pair zbounds=(0,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface//{z=exp(-x^2)}; - triple f(pair t) { - return (t.x,t.y,exp(-t.x^2)); - } - surface s=surface(f,(-0.25,0),(3.5,3.5),12,20,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw lines in plane - draw((0,0,0)--(3,0,0)--(3,3,0)--cycle,bluepen+dashed+linewidth(0.75)); - - //Draw lines on surface - draw((3,0,0)--(3,3,0),bluepen+linewidth(2)); - triple g(real t) {return (t,0,exp(-t^2));} - path3 mypath=graph(g,0,3,operator ..); - draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,t,exp(-t^2));} - path3 mypath=graph(g,0,3,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //Draw vertical lines - //draw((-1,0,0)--(-1,0,sin(-1)),bluepen+dashed+linewidth(0.75)); - //draw((1,0,0)--(1,0,sin(1)),bluepen+dashed+linewidth(0.75)); - - - - -
    - -

    - In short, evaluating one iterated integral is impossible; - the other iterated integral is relatively simple. -

    -     -     - - - -

    - - defines the average value of a single-variable function f(x) on the interval [a,b] as - - \text{ average value of \(f(x)\) on \([a,b]\) } = \frac1{b-a}\int_a^b f(x)\, dx; - - that is, it is the area under f over an interval divided by the length of the interval. - We make an analogous statement here: - the average value of z=f(x,y) over a region R is the volume under f over R divided by the area of R. -

    - - - The Average Value of <m>f</m> on <m>R</m> - -

    - Let z=f(x,y) be a continuous function defined over a closed, - bounded region R in the xy-plane. - The average value of f on R is - average value of a function - - \text{ average value of \(f\) on \(R\) } = \frac{\ds \iint_R f(x,y)\, dA}{\ds\iint_R \, dA} - . -

    -     -     - - - Finding average value of a function over a region <m>R</m> - -

    - Find the average value of f(x,y) = 4-y over the region R, - which is bounded by the parabolas y^2=4x and x^2=4y. - Note: this is the same function and region as used in . -

    -     - -

    - In we found - - \iint_R f(x,y)\, dA = \int_0^4\int_{y^2/4}^{2\sqrt{y}}(4-y)\, dx\, dy = \frac{176}{15} - . -

    - -

    - We find the area of R by computing \iint_R \, dA: - - \iint_R \, dA = \int_0^4\int_{y^2/4}^{2\sqrt{y}} \, dx\, dy = \frac{16}{3} - . -

    - -

    - Dividing the volume under the surface by the area gives the average value: - - \text{ average value of \(f\) on \(R\) } = \frac{176/15}{16/3} = \frac{11}5 = 2.2 - . -

    - -

    - While the surface, as shown in , - covers z-values from z=0 to z=4, - the average z-value on R is 2.2. -

    - -
    - Finding the average value of f in - - - - A rectangular portion of a plane in space, and a petal-like curve on the plane. - -

    - This is the same image as the one used in . - We repeat the description here for convenience. -

    - -

    - The graph z=4-y is a plane. - It is plotted as a rectangular region in the first octant, - intersecting the xy plane along the line y=4, - and the xz plane along the line z=4. -

    - -

    - The parabolas y^2=4x and x^2=4y are sketched in the xy plane, - from the origin to their intersection at (4,4,0). - On the plane, we see the corresponding curves, - which are also parabolas. The region between them has the shape of a narrow leaf or petal. -

    -     - - - - - //ASY file for figdouble43D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(16,6.2,5.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={2,4}; - defaultpen(0.5mm); - - pair xbounds=(0,5); - pair ybounds=(0,5); - pair zbounds=(0,5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface//{z=4-y}; - triple f(pair t) { - return (t.x,t.y,4-t.y); - } - surface s=surface(f,(-0.25,-0.25),(4.5,4.25),12,20,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw parabolas in plane - triple g(real t) {return (t^2/4,t,0);} - path3 mypath=graph(g,0,4.25,operator ..); - draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,t^2/4,0);} - path3 mypath=graph(g,0,4.25,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //Draw parabolas on surface - triple g(real t) {return (t^2/4,t,4-t);} - path3 mypath=graph(g,0,4.,operator ..); - draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,t^2/4,4-t^2/4);} - path3 mypath=graph(g,0,4.,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //Draw vertical lines - //draw((-1,0,0)--(-1,0,sin(-1)),bluepen+dashed+linewidth(0.75)); - //draw((1,0,0)--(1,0,sin(1)),bluepen+dashed+linewidth(0.75)); - - - - -
    -     -     - -

    - The previous section introduced the iterated integral in the context of finding the area of plane regions. - This section has extended our understanding of iterated integrals; - now we see they can be used to find the signed volume under a surface. -

    - -

    - This new understanding allows us to revisit what we did in the previous section. - Given a region R in the plane, - we computed \iint_R 1\, dA; - again, our understanding at the time was that we were finding the area of R. - However, we can now view the graph z=1 as a surface, - a flat surface with constant z-value of 1. - The double integral \iint_R 1\, dA finds the volume, - under z=1, over R, - as shown in . - Basic geometry tells us that if the base of a general right cylinder has area A, - its volume is A\cdot h, - where h is the height. - In our case, the height is 1. - We were actually computing the volume of a solid, - though we interpreted the number as an area. -

    - -
    - Showing how an iterated integral used to find area also finds a certain volume - - - - A circular cylinder, bounded below by the xy plane, and above by the plane z=1. - -

    - A circular cylinder is plotted in three dimensions. - The cylinder appears to lie over a circle with center (1,0,0) and radius 1 in the xy plane. -

    - -

    - The plane z=1 sits at the top of the cylinder. -

    - -

    - The image is used to illustrate that for any region in the plane, - there is a cylinder that lies over the region, between the planes z=0 and z=1. - A double integral of the form \iint_R 1 \, dA can be thought of as either computing the area of the region R, - or the volume of the corresponding cylinder of height 1 over R. -

    -     - - - - - //ASY file for figdouble_intro23D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8.2,4.6,1.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={-1,1}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-1,2.5); - pair ybounds=(-1.25,1.25); - pair zbounds=(-.05,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface//{z=-.5*(x-1)^2-.5*(y)^2+2}; - //triple f(pair t) { - // return (t.x,t.y,-.5*(t.x-1)^2-.5*(t.y)^2+2); - //} - //surface s=surface(f,(-0.221,-1),(2.2,1),12,20,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic},usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - //Draw curve on top of the grid in xy plane ({cos(x)*(1+cos(2*x))},{sin(x)*(1+cos(2*x))},0); - triple g(real t) {return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),0);} - path3 mypath=graph(g,-pi/2,pi/2,operator ..); - draw(mypath,bluepen); - - // draw curve on surface - triple g(real t) { - return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),1); - } - path3 mypath=graph(g,-pi/2,pi/2,operator ..); - draw(mypath,bluepen); - - pen q=redcurvepen; - triple gg(pair t) {return (cos(t.x)*(1+cos(2*t.x)),sin(t.x)*(1+cos(2*t.x)),t.y);} - surface s=surface(gg,(-pi/2,0),(pi/2,1),16,4,Spline); - draw(s,surfacepen2,meshpen=q,nolight,render(merge=true)); - - triple f(pair t) { - return (t.x,t.y,1); - } - surface s=surface(f,(-.1,ybounds.x),(xbounds.y,ybounds.y),8,8); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - - -
    - -

    - The next section extends our abilities to find - volumes under surfaces. Currently, - some integrals are hard to compute because either the region R we are integrating over is hard to define with rectangular curves, - or the integrand itself is hard to deal with. - Some of these problems can be solved by converting everything into polar coordinates. -

    - - - - Terms and Concepts - - - - -

    - An integral can be interpreted as giving the signed area over an interval; - a double integral can be interpreted as giving the signed over a region. -

    -     - - - - - - - - -     - - - - -

    - Explain why the following statement is false: - Fubini's Theorem states that - - \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx = \int_a^b\int_{g_1(y)}^{g_2(y)} f(x,y)\, dx\, dy - . -

    -     - - - -

    - When switching the order of integration, - the bounds integrals must change to reflect the bounds of the region of integration. - You cannot merely change the letters x and y in a few places. -

    -     - -     - - - - -

    - Explain why if f(x,y) \gt 0 over a region R, then -

    - -

    - \iint_Rf(x,y)\, dA \gt 0. -

    -     - - - -

    - The double integral gives the signed volume under the surface. - Since the surface is always positive, - it is always above the xy-plane and hence produces only - positive volume. -

    -     - -     - - - - -

    - If \iint_R f(x,y)\, dA = \iint_R g(x,y)\, dA, - does this imply f(x,y) = g(x,y)? -

    -     - - - -

    - No. - It means that there is the same amount of signed volume under f and g over R, - but the functions could be very different. -

    -     - -     -     - - - Problems - - - -

    - For the given integral, - -

      -
    1. -

      - Evaluate the given iterated integral, and -

      -
      2. - -
    2. -

      - rewrite the integral using the other order of integration. -

      -
      3. -
    -

    -     - - - - -

    - \ds \int_1^2\int_{-1}^1\left(\frac xy+3\right)\, dx\, dy -

    -     - -

    - 6; \ds \int_{-1}^1\int_{1}^2\left(\frac xy+3\right)\, dy\, dx -

    -     - -     - - - - - -

    - \ds \int_{-\pi/2}^{\pi/2}\int_{0}^\pi\left(\sin(x) \cos(y) \right)\,dx\,dy -

    - - - - -     - -     - - - - -

    - \ds \int_{0}^{4}\int_{0}^{-x/2+2}\left(3x^2-y+2\right)\, dy\, dx -

    -     - -

    - 112/3; \ds \int_{0}^{2}\int_{0}^{4-2y}\left(3x^2-y+2\right)\, dx\, dy -

    -     - -     - - - - -

    - \ds \int_{1}^{3}\int_{y}^{3}\left(x^2y-xy^2\right)\,dx\,dy -

    - - - - -     - -     - - - - -

    - \ds \int_{0}^{1}\int_{-\sqrt{1-y}}^{\sqrt{1-y}}\left(x+y+2\right)\, dx\, dy -

    -     - -

    - 16/5; \ds \int_{-1}^{1}\int_{0}^{1-x^2}\left(x+y+2\right)\, dy\, dx -

    -     - -     - - - - -

    - \ds \int_{0}^{9}\int_{y/3}^{\sqrt{y}}\left(xy^2\right)\,dx\,dy -

    - - - - - -     - -     - -     - - - -

    - For the given integral: - -

      -
    1. -

      - Sketch the region R given by the problem. -

      -
      2. - -
    2. -

      - Set up the iterated integrals, in both orders, - that evaluate the given double integral for the described region R. -

      -
      3. - -
    3. -

      - Evaluate one of the iterated integrals to find the signed volume under the surface - z=f(x,y) over the region R. -

      -
      4. -
    -

    -     - - - - -

    - \ds \iint_R x^2y\, dA, - where R is bounded by y=\sqrt{x} and y=x^2. -

    -     - -

    -

      -
    1. - - - - A region in the first quadrant of the plane, bounded by two parabolas. - -

      - The correct region is in the first quadrant of the xy plane. - It is a leaf-shaped region, bounded by the parabolas y=x^2 and x=y^2. - The region is symmetric about the line y=x. - The two parabolas intersect at (0,0) and (1,1). - The parabola y=x^2 lies below the line y=x, - and the parabola y=\sqrt{x} lies above. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1}, - ytick={1}, - ymin=-.1,ymax=1.5, - xmin=-.1,xmax=1.5 - ] - - \addplot [firstcurvestyle,domain=-.1:1.1] ({x^2},{x}) node [pos=.5,sloped,above,black] {$y=\sqrt{x}$}; - \addplot [firstcurvestyle,domain=-.1:1.1] ({x},{x^2}) node [pos=.5,sloped,below,black] {$y=x^2$}; - - \draw (axis cs: .5,.5) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - - -
      2. - -
    2. -

      - \ds \int_0^1\int_{x^2}^{\sqrt{x}}x^2y\, dy\, dx = \int_0^1\int_{y^2}^{\sqrt{y}}x^2y\, dx\, dy. -

      -
      3. - -
    3. -

      - \frac3{56} -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \iint_R x^2y\, dA, - where R is bounded by y=\sqrt[3]{x} and y=x^3. -

    -     - -

    -

      -
    1. - - - - A region in the first quadrant of the plane, bounded by a cubic function and its inverse. - -

      - The region for this integral lies in the first quadrant. -

      - -

      - The functions given by y=x^3 and y=\sqrt[3]{x} are inverses of each other, - so the region is symmetric about the line y=x. -

      - -

      - The curve y=x^3 lies below the line y=x, - and the curve y=\sqrt[3]{x} lies above this line. -

      -       - - - \begin{tikzpicture}[>=stealth] - - \begin{axis}[ - axis on top, - xtick={1}, - ytick={1}, - ymin=-1.5,ymax=1.5, - xmin=-1.5,xmax=1.5 - ] - - \addplot [firstcurvestyle,domain=-1.1:1.1,samples=60] ({x^3},{x}) node [pos=.85,above left,black] {$y=\sqrt[3]{x}$}; - \addplot [firstcurvestyle,domain=-1.1:1.1,samples=60] ({x},{x^3}) node [pos=.75,below right,black] {$y=x^3$};; - - \draw (axis cs: -.5,.5) node (A) {$R$}; - - \draw [->] (A) -- (axis cs: .5,.5); - \draw [->] (A) -- (axis cs: -.5,-.5); - - \end{axis} - - \end{tikzpicture} - - - - - -
      2. - -
    2. -

      - \ds \int_0^1\int_{x^3}^{\sqrt[3]{x}}x^2y\, dy\, dx + \int_{-1}^0\int_{\sqrt[3]{x}}^{x^3}x^2y\, dy\, dx \ds = \int_0^1\int_{y^3}^{\sqrt[3]{y}}x^2y\, dx\, dy +\int_{-1}^0\int_{\sqrt[3]{y}}^{y^3}x^2y\, dx\, dy. -

      -
      3. - -
    3. -

      - 0 -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \iint_R x^2-y^2\, dA, - where R is the rectangle with corners (-1,-1), - (1,-1), - (1,1) and (-1,1). -

    -     - -

    -

      -
    1. - - - - A square of side length 2, with its center at the origin. - -

      - A rectangle in the plane. In fact, it is a square, with side length 2. -

      - -

      - The vertices are at (\pm 1,0) and (0,\pm 1), - so that the square is centered at the origin. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-1,1}, - ytick={1,-1}, - ymin=-1.1,ymax=1.1, - xmin=-1.1,xmax=1.1 - ] - - \draw [very thick,firstcolor] (axis cs:-1,-1) -- (axis cs:1,-1) -- (axis cs:1,1) -- (axis cs:-1,1) -- cycle; - - \draw (axis cs: .5,.5) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - - -
      2. - -
    2. -

      - \ds \int_{-1}^1\int_{-1}^{1}x^2-y^2\, dy\, dx = \int_{-1}^1\int_{-1}^{1}x^2-y^2\, dx\, dy. -

      -
      3. - -
    3. -

      - 0 -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \iint_R ye^x\, dA, - where R is bounded by x=0, - x=y^2 and y=1. -

    -     - -

    -

      -
    1. - - - - A region in the plane that lies above a rightward-opening parabola, and below a horizontal line. - -

      - The parabola x=y^2 has its vertex at the origin, and opens toward the right. -

      - -

      - The region for this integral is in the first quadrant. - It lies above the parabola, and below the line y=1. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1}, - ytick={1}, - ymin=-.5,ymax=1.2, - xmin=-.5,xmax=1.2 - ] - - \addplot [firstcurvestyle,domain=-.25:1.1] ({x},{1}); - \addplot [firstcurvestyle,domain=-.25:1.1] ({x^2},{x}) node [pos=.5,below right,black] { $x=y^2$}; - \addplot [firstcurvestyle,domain=-.25:1.1] ({0},{x}); - - \draw (axis cs: .25,.75) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - - -
      2. - -
    2. -

      - \ds \int_{0}^1\int_{0}^{y^2}ye^x\, dx\, dy = \int_{0}^1\int_{\sqrt{x}}^{1}ye^x\, dy\, dx. -

      -
      3. - -
    3. -

      - e/2-1 -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \iint_R \big(6-3x-2y\big)\, dA, - where R is bounded by x=0, - y=0 and 3x+2y=6. -

    -     - -

    -

      -
    1. - - - - A triangular region in the plane, with vertices at (0,0), (2,0), and (0,3). - -

      - The region for this integral is a triangle in the first quadrant, - bounded by the coordinate axes and the line 3x+3y=6. -

      - -

      - The vertices of the triangle are at (0,0), (2,0), and (0,3). -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2}, - ytick={1,2,3}, - ymin=-.5,ymax=3.5, - xmin=-.5,xmax=2.5 - ] - - \addplot [firstcurvestyle,domain=-.25:2.25] ({x},{0}); - \addplot [firstcurvestyle,domain=-.25:2.25] ({x},{3-3/2*x}) node [pos=.4,sloped,above,black] { $3x+2y=6$}; - \addplot [firstcurvestyle,domain=-.25:3.25] ({0},{x}); - - \draw (axis cs: .75,1) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - - -
      2. - -
    2. -

      - \ds \int_{0}^2\int_{0}^{3-3/2x}\big(6-3x-2y\big)\, dy\, dx = \int_{0}^3\int_{0}^{2-2/3y}\big(6-3x-2y\big)\, dx\, dy. -

      -
      3. - -
    3. -

      - 6 -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \iint_R e^y\, dA, where R is bounded by y=\ln(x) and -

    - -

    - \ds y=\frac{1}{e-1}(x-1). -

    -     - -

    -

      -
    1. - - - - A region in the plane bounded by the graph of the natural logarithm and a straight line. - -

      - The graph y=\ln(x) meets the x axis at (1,0). - It tends toward -\infty as it approaches the y axis, - and grows slowly upward as x increases past x=1. -

      - -

      - The line y=\frac{1}{e-1}(x-1) also meets the x axis at (1,0), - and it intercepts the graph y=\ln(x) a second time when x=e. -

      - -

      - The region for this integral therefore lies below y=\ln(x) - and above y=\frac{1}{e-1}(x-1), for 1\leq x\leq e. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3}, - ytick={1,2,3}, - ymin=-.5,ymax=1.25, - xmin=-.5,xmax=3.25 - ] - - \addplot [firstcurvestyle,domain=.5:3.25] ({x},{ln(x)}) node [pos=.7,sloped,above,black] { $y=\ln(x)$}; - \addplot [firstcurvestyle,domain=-.25:3.25] ({x},{1/e*(x-1)}) node [pos=.6,below right,black] { $y=\frac{1}{1-e}(x-1)$}; - - \draw (axis cs: 1,.5) node (A) {$R$}; - - \draw[->,>=stealth] (A) -- (axis cs:1.8,.5); - - \end{axis} - - \end{tikzpicture} - - - - - -
      2. - -
    2. -

      - \ds \int_{1}^e\int_{\frac{x-1}{e-1}}^{\ln(x) }e^y\, dy\, dx = \int_{0}^1\int_{e^y}^{y(e-1)+1}e^y\, dx\, dy. -

      -
      3. - -
    3. -

      - -\frac12e^2+2e-\frac32 -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \iint_R \big(x^3y-x\big)\, dA, - where R is the half of the circle - x^2+y^2=9 in the first and second quadrants. -

    -     - -

    -

      -
    1. - - - - A region bounded by the x axis and the upper half of a semicircle of radius 3. - -

      - The region for this integral is bounded above by the semicircle y=\sqrt{9-x^2}, - which has center (0,0) and radius 3. - The region is bounded below by the x axis, for -3\leq x\leq 3. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={-3,3}, - ytick={-3,3}, - ymin=-3.1,ymax=3.1, - xmin=-3.7,xmax=3.7 - ] - - \addplot+ [domain=0:180,samples=40] ({3*cos(x)},{3*sin(x)}) --cycle; - - \draw (axis cs: 1,1) node (A) {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - - -
      2. - -
    2. -

      - \ds \int_{-3}^3\int_{0}^{\sqrt{9-x^2}}\big(x^3y-x\big)\, dy\, dx = \int_{0}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\big(x^3y-x\big)\, dx\, dy. -

      -
      3. - -
    3. -

      - 0 -

      -
      4. -
    -

    -     - -     - - - - -

    - \ds \iint_R \big(4-3y\big)\, dA, - where R is bounded by y=0, - y=x/e and y=\ln(x). -

    -     - -

    -

      -
    1. - - - - The region above the graph of the natural logarithm, but below its tangent line at x=e. - -

      - For 0\leq x\leq 1, the region is bounded below by the x axis. - For 1\leq x\leq e, the lower bound of the region is y=\ln(x). -

      - -

      - The upper bound of the region is the line y=x/e. - This line is tangent to y=ln(x) at the point (e,1), - and lies above y=\ln(x). - The region is bounded by the portion of this line from the point of tangency to the origin. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2}, - extra x ticks={2.718}, - extra x tick labels={$e$}, - ytick={1}, - %%extra y ticks={-5,-3,...,7},% - ymin=-.5,ymax=1.2,% - xmin=-.5,xmax=3.5% - ] - - \addplot [firstcurvestyle,domain=.5:3.25] ({x},{ln(x)}); - \addplot [firstcurvestyle,domain=-.5:3.25] ({x},{0}); - \addplot [firstcurvestyle,domain=-.5:3.25] ({x},{x/e}); - - \draw (axis cs: .8,.1) node (A) {$R$}; - - \filldraw (axis cs:2.718,1) circle (2.4pt); - - \draw (axis cs:2.718,1) node [below right] { $(e,1)$}; - - \end{axis} - - \end{tikzpicture} - - - - - -
      2. - -
    2. -

      - \ds \int_{0}^1\int_{ey}^{e^y}\big(4-3y\big)\, dx\, dy = \int_{0}^1\int_{0}^{x/e}\big(4-3y\big)\, dy\, dx+\int_{1}^e\int_{\ln(x) }^{x/e}\big(4-3y\big)\, dy\, dx. -

      -
      3. - -
    3. -

      - 3e-7 -

      -
      4. -
    -

    -     - -     - -     - - - -

    - State why it is difficult/impossible to integrate the iterated integral in the given order of integration. - Change the order of integration and evaluate the new iterated integral. -

    -     - - - - -

    - \ds \int_0^4\int_{y/2}^2 e^{x^2}\, dx\, dy -

    -     - -

    - Integrating e^{x^2} with respect to x is not possible in terms of elementary functions. - \ds \int_0^2\int_0^{2x}e^{x^2}\, dy\, dx = e^4-1. -

    -     - -     - - - - -

    - \ds \int_0^{\sqrt{\pi/2}}\int_{x}^{\sqrt{\pi/2}} \cos\big(y^2\big)\, dy\, dx -

    -     - -

    - Integrating \cos(y^2) with respect to y is not possible in terms of elementary functions. - \ds \int_0^{\sqrt{\pi/2}}\int_0^{y}\cos(y^2)\, dx\, dy = \frac12. -

    -     - -     - - - - -

    - \ds \int_0^{1}\int_y^{1} \frac{2y}{x^2+y^2}\, dx\, dy -

    -     - -

    - Integrating \ds\int_y^{1}\frac{2y}{x^2+y^2}\, dx gives \tan^{-1}(1/y)-\pi/4; - integrating \tan^{-1}(1/y) is hard. -

    - -

    - \ds\int_0^1\int_0^x\frac{2y}{x^2+y^2}\, dy \, dx = \ln(2). -

    -     - -     - - - - -

    - \ds \int_{-1}^{1}\int_1^{2} \frac{x\tan^2(y) }{1+\ln(y) }\, dy\, dx -

    -     - -

    - Integrating in the order shown is hard/impossible. - By changing the order of integration, - we have \ds \int_{1}^{2}\int_{-1}^{1} \frac{x\tan^2(y) }{1+\ln(y) }\, dx\, dy = 0, - since the integrand is an odd function with respect to x. - Thus the iterated integral evaluates to 0. -

    -     - -     - -     - - - -

    - Find the average value of f over the region R. - Notice how these functions and regions are related to the iterated integrals given in - Exercises. -

    -     - - - - -

    - \ds f(x,y) =\frac xy+3;R is the rectangle with opposite corners (-1,1) and (1,2). -

    -     - -

    - average value of f = 6/2 = 3 -

    -     - -     - - - - -

    - \ds f(x,y) = \sin(x) \cos(y); - R is bounded by x=0, x=\pi, - y=-\pi/2 and y=\pi/2. -

    -     - -

    - average value of f= 4/\pi^2 -

    -     - -     - - - - -

    - \ds f(x,y) = 3x^2-y+2;R is bounded by the lines y=0, - y=2-x/2 and x=0. -

    -     - -

    - average value of f= \frac{112/3}{4} =28/3 -

    -     - -     - - - - -

    - \ds f(x,y) =x^2y-xy^2; - R is bounded by y=x, - y=1 and x=3. -

    -     - -

    - average value of f= \frac{76/15}{2} = \frac{38}{15} \approx 2.53 -

    -     - -     - -     -     -     -
    -
    - Double Integration with Polar Coordinates -

    - We have used iterated integrals to evaluate double integrals, - which give the signed volume under a surface, z=f(x,y), - over a region R of the xy-plane. - The integrand is simply f(x,y), - and the bounds of the integrals are determined by the region R. -

    - -

    - Some regions R are easy to describe using rectangular coordinates that is, - with equations of the form y=f(x), x=a, etc. - However, some regions are easier to handle if we represent their boundaries with polar equations of the form r=f(\theta), - \theta = \alpha, etc. -

    - - - -

    - The basic form of the double integral is \iint_R f(x,y)\, dA. - We interpret this integral as follows: - over the region R, - sum up lots of products of heights (given by f(x_i,y_i)) and areas - (given by \Delta A_i). - That is, dA represents a little bit of area. - In rectangular coordinates, - we can describe a small rectangle as having area dx\, dy or dy\, dx the area of a rectangle is simply lengthwidth a small change in x times a small change in y. - Thus we replace dA in the double integral with dx\, dy or dy\, dx. -

    - -
    - Approximating a region R with portions of sectors of circles - -
    - - - - A polar grid in the first quadrant, with one grid sector highlighted. - -

    - A polar grid is shown in the first quadrant of the plane. - Concentric circles are plotted with radii 0.2, 0.4, 0.6, and 0.8. - Rays are drawn at angles \pi/12, \pi/6, \pi/4, \pi/3, and 5\pi/12. -

    - -

    - A polar curve is shown. It is somewhat wavy, - but lies close to the circle r=1, from the x axis to the y axis. - Instead of using the circle r=1 as part of the polar grid, - circular sectors are drawn with radius slightly more or less than 1, - so that they align with this curve. -

    - -

    - One of the sectors in the polar grid is highlighted. - It corresponds to the values 0.6\leq r\leq 0.8, and \pi/6\leq \theta\leq \pi/4. - The sector is approximately rectangular in shape, - but it is not a true rectangle, as two of its sides are circular arcs, - and the other two sides are rays from the origin. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[axis equal image, - ymin=-.1,ymax=1.25, - xmin=-.2,xmax=1.24 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(0.6928,0.4)(0.6857,0.412)(0.6784,0.4239)(0.6709,0.4357)(0.6632,0.4474)(0.6553,0.4589)(0.6472,0.4702)(0.6389,0.4815)(0.6304,0.4925)(0.6217,0.5035)(0.6128,0.5142)(0.6038,0.5248)(0.5945,0.5353)(0.5851,0.5456)(0.5755,0.5557)(0.5657,0.5657)(0.4243,0.4243)(0.4316,0.4168)(0.4388,0.4092)(0.4459,0.4015)(0.4528,0.3936)(0.4596,0.3857)(0.4663,0.3776)(0.4728,0.3694)(0.4792,0.3611)(0.4854,0.3527)(0.4915,0.3441)(0.4974,0.3355)(0.5032,0.3268)(0.5088,0.318)(0.5143,0.309)(0.5196,0.3)(0.6928,0.4)}; - - \addplot [secondcurvestyle,solid,domain=0:15] ({cos(x)*(1.05)},{sin(x)*(1.05)}); - \addplot [secondcurvestyle,solid,domain=15:30] ({cos(x)*(.95)},{sin(x)*(.95)}); - \addplot [secondcurvestyle,solid,domain=30:45] ({cos(x)*(1.05)},{sin(x)*(1.05)}); - \addplot [secondcurvestyle,solid,domain=45:60] ({cos(x)*(.97)},{sin(x)*(.97)}); - \addplot [secondcurvestyle,solid,domain=60:75] ({cos(x)*(1)},{sin(x)*(1)}); - \addplot [secondcurvestyle,solid,domain=75:90] ({cos(x)*(1.05)},{sin(x)*(1.05)}); - \addplot [secondcurvestyle,solid,domain=0:90] ({cos(x)*(.8)},{sin(x)*(.8)}); - \addplot [secondcurvestyle,solid,domain=0:90] ({cos(x)*(.6)},{sin(x)*(.6)}); - \addplot [secondcurvestyle,solid,domain=0:90] ({cos(x)*(.4)},{sin(x)*(.4)}); - \addplot [secondcurvestyle,solid,domain=0:90] ({cos(x)*(.2)},{sin(x)*(.2)}); - \addplot [secondcurvestyle,solid,domain=0:1.05] ({cos(15)*(x)},{sin(15)*(x)}); - \addplot [secondcurvestyle,solid,domain=0:1.05] ({cos(30)*(x)},{sin(30)*(x)}); - \addplot [secondcurvestyle,solid,domain=0:1.05] ({cos(45)*(x)},{sin(45)*(x)}); - \addplot [secondcurvestyle,solid,domain=0:1] ({cos(60)*(x)},{sin(60)*(x)}); - \addplot [secondcurvestyle,solid,domain=0:1.05] ({cos(75)*(x)},{sin(75)*(x)}); - - \addplot [firstcurvestyle,-,domain=0:90,samples=80] ({cos(x)*(1+.05*cos(9*x))},{sin(x)*(1+.05*cos(9*x))}); - - \end{axis} - - \node [right] at (myplot.right of origin) { $0$}; - \node [above] at (myplot.above origin) { $\pi/2$}; - - \end{tikzpicture} - - - - -
    - -
    - - - - A close-up view of one region in a polar grid - -

    - A single sector within a polar grid. - Two rays from the origin are shown; they are separated by an angle marked as \Delta\theta. - The rays form two of the four sides of the sector; - the other two sides are circular arcs. -

    - -

    - The radii of the two circular arcs are also labeled. - The smaller radius is labeled r_1, and the larger r_2. - The region bounded by the rays and the two arcs is shaded. -

    - -

    - One way to think of this region is to imagine the crust at the end of a narrow slice of pizza. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis y line=none, - axis x line=none, - ymin=-.05,ymax=0.65, - xmin=-.1,xmax=.74 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(0.6928,0.4)(0.6857,0.412)(0.6784,0.4239)(0.6709,0.4357)(0.6632,0.4474)(0.6553,0.4589)(0.6472,0.4702)(0.6389,0.4815)(0.6304,0.4925)(0.6217,0.5035)(0.6128,0.5142)(0.6038,0.5248)(0.5945,0.5353)(0.5851,0.5456)(0.5755,0.5557)(0.5657,0.5657)(0.4243,0.4243)(0.4316,0.4168)(0.4388,0.4092)(0.4459,0.4015)(0.4528,0.3936)(0.4596,0.3857)(0.4663,0.3776)(0.4728,0.3694)(0.4792,0.3611)(0.4854,0.3527)(0.4915,0.3441)(0.4974,0.3355)(0.5032,0.3268)(0.5088,0.318)(0.5143,0.309)(0.5196,0.3)(0.6928,0.4)}; - - \addplot [secondcurvestyle,solid,domain=30:45] ({cos(x)*(.8)},{sin(x)*(.8)}); - \addplot [secondcurvestyle,solid,domain=30:45] ({cos(x)*(.8)},{sin(x)*(.8)}); - \addplot [secondcurvestyle,solid,domain=30:45] ({cos(x)*(.6)},{sin(x)*(.6)}); - \addplot [secondcurvestyle,solid,domain=0:.8] ({cos(30)*(x)},{sin(30)*(x)}); - \addplot [secondcurvestyle,solid,domain=0:.8] ({cos(45)*(x)},{sin(45)*(x)}); - - \draw [rotate=30] (axis cs:.31,-.12) node [rotate=30] {$\underbrace{\rule{140pt}{0pt}}_{r_1}$}; - \draw [rotate=45] (axis cs:.41,-.03) node [rotate=45] {$\overbrace{\rule{180pt}{0pt}}^{r_2}$}; - - \draw [thick,->,>=stealth,rotate=24] (axis cs:.45,0) arc (0:13:120pt); - \draw (axis cs: .4,.3) node { $\Delta \theta$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - Now consider representing a region R with polar coordinates. - Consider . - Let R be the region in the first quadrant bounded by the curve. - We can approximate this region using the natural shape of polar coordinates: - portions of sectors of circles. - In the figure, one such region is shaded, - shown again in . -

    - -

    - As the area of a sector of a circle with radius r, - subtended by an angle \theta, - is A = \frac12r^2\theta, - we can find the area of the shaded region. - The whole sector has area \frac12r_2^2\Delta \theta, - whereas the smaller, - unshaded sector has area \frac12r_1^2\Delta \theta. - The area of the shaded region is the difference of these areas: - - \Delta A_i = \frac12r_2^2\Delta\theta-\frac12r_1^2\Delta\theta = \frac12\big(r_2^2-r_1^2\big)\big(\Delta\theta\big) = \frac{r_2+r_1}{2}\big(r_2-r_1\big)\Delta\theta - . -

    - -

    - Note that (r_2+r_1)/2 is just the average of the two radii. -

    - -

    - To approximate the region R, - we use many such subregions; - doing so shrinks the difference r_2-r_1 between radii to 0 and shrinks the change in angle \Delta \theta also to 0. - We represent these infinitesimal changes in radius and angle as dr and d\theta, - respectively. - Finally, as dr is small, - r_2\approx r_1, and so (r_2+r_1)/2\approx r_1. - Thus, when dr and d\theta are small, - - \Delta A_i \approx r_i\, dr\, d\theta - . -

    - -

    - Taking a limit, - where the number of subregions goes to infinity and both r_2-r_1 and \Delta\theta go to 0, we get - - dA = r\, dr\, d\theta - . -

    - -

    - So to evaluate \iint_Rf(x,y)\, dA, - replace dA with r\, dr\, d\theta. - Convert the function f(x,y) to a function with polar coordinates with the substitutions - x=r\cos(\theta), y=r\sin(\theta). - Finally, find bounds g_1(\theta)\leq r\leq g_2(\theta) and - \alpha\leq\theta\leq\beta that describe R. - This is the key principle of this section, - so we restate it here as a Key Idea. -

    - - - Evaluating Double Integrals with Polar Coordinates -

    - Let z=f(x,y) be a continuous function defined over a closed, - bounded region R in the xy-plane, - where R is bounded by the polar equations - \alpha\leq\theta\leq\beta and g_1(\theta)\leq r\leq g_2(\theta). - Then - double integralin polar - - \iint_Rf(x,y)\, dA = \int_\alpha^\beta\int_{g_1(\theta)}^{g_2(\theta)} f\big(r\cos(\theta) ,r\sin(\theta) \big)\, r\, dr\, d\theta - . -

    -     - -

    - Examples will help us understand this Key Idea. -

    - - - Evaluating a double integral with polar coordinates - -

    - Find the signed volume under the plane - z= 4-x-2y over the disk bounded by the circle with equation x^2+y^2=1. -

    -     - -

    - The bounds of the integral are determined solely by the region R over which we are integrating. - In this case, it is a disk with boundary x^2+y^2=1. - We need to find polar bounds for this region. - It may help to review ; - bounds for this disk are 0\leq r\leq 1 and 0\leq \theta\leq 2\pi. -

    - -

    - We replace f(x,y) with f(r\cos(\theta) ,r\sin(\theta) ). - That means we make the following substitutions: - - 4-x-2y \Rightarrow 4-r\cos(\theta) -2r\sin(\theta) - . -

    - -

    - Finally, we replace dA in the double integral with r\, dr\, d\theta. - This gives the final iterated integral, which we evaluate: - - \iint_Rf(x,y)\, dA \amp = \int_0^{2\pi}\int_0^1\big(4-r\cos(\theta) -2r\sin(\theta) \big)r\, dr\, d\theta - \amp = \int_0^{2\pi}\int_0^1\big(4r-r^2(\cos(\theta) -2\sin(\theta) )\big)\, dr\, d\theta - \amp = \int_0^{2\pi}\left.\left(2r^2-\frac13r^3(\cos(\theta) -2\sin(\theta) )\right)\right|_0^1d\theta - \amp = \int_0^{2\pi} \left(2-\frac13\big(\cos(\theta) -2\sin(\theta) \big)\right)\, d\theta - \amp = \left.\left(2\theta -\frac13\big(\sin(\theta) +2\cos(\theta) \big)\right)\right|_0^{2\pi} - \amp = 4\pi \approx 12.566 - . -

    - -
    - Evaluating a double integral with polar coordinates in - - - - An ellipse on a plane in space corresponds to a circle drawn below it in the xy plane. - -

    - Three-dimensional coordinate axes are drawn in space. - In the xy plane, a unit circle is drawn. - The plane z=4-x-2y lies above the xy plane. - On the plane, we see the curve corresponding to the unit circle; - since the plane lies at an angle relative to the xy plane, - the image of the unit circle on the plane is an ellipse. -

    -     - - - - - //ASY file for figdoublepol13D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.2,-1.9,13); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={5}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(0,10); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane //({z=4-x-2y); - triple f(pair t) { - return (t.x,t.y,4-t.x-2*t.y); - } - surface s=surface(f,(-1.5,-1.5),(1.5,1.5),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //draw curve on xy plane - triple g(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //draw curve on plane - triple g(real t) {return (cos(t),sin(t),4-cos(t)-2*sin(t));} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - - - -
    - -

    - The surface and region R are shown in . -

    -     -     - - - - - Evaluating a double integral with polar coordinates - -

    - Find the volume under the paraboloid - z=4-(x-2)^2-y^2 over the region bounded by the circles - (x-1)^2+y^2=1 and (x-2)^2+y^2=4. -

    -     - -

    - At first glance, - this seems like a very hard volume to compute as the region R (shown in ) has a hole in it, - cutting out a strange portion of the surface, - as shown in . - However, by describing R in terms of polar equations, - the volume is not very difficult to compute. -

    - -
    - Showing the region R and surface used in - -
    - - - - A region in the plane between that lies between two circles. - -

    - Two circles are plotted in the plane. The smaller of the two circles lies inside the larger. - The circles meet, and are tangent to each other, at the origin. -

    - -

    - The larger circle is centered at (2,0) and has radius 2. - The smaller circle is centered at (1,0) and has radius 1. - The region that lies between the two circles is shaded. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={2,1,3,4}, - ymin=-2.1,ymax=2.1, - xmin=-.54,xmax=4.5 - ] - - \addplot [firstcurvestyle,areastyle,domain=0:360,samples=40] ({2+2*cos(x)},{2*sin(x)}); - \addplot [firstcurvestyle,domain=0:360,samples=80] ({2+2*cos(x)},{2*sin(x)}); - - \addplot [firstcolor,thick,fill=white,area style,domain=0:360,samples=60] ({1+cos(x)},{sin(x)}); - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - - A region in space that lies outside of a cylinder, and below a circular paraboloid. - -

    - The image shows a region in space between two surfaces. - The first surface is a circular paraboloid; - it has its vertext at the point (2,0,4), - and it meets the xy plane along the larger of the two circles in -

    - -

    - The second surface is a circular cylinder that lies above the smaller of the two circles in . - Overall, the shape is like a circular hill that has had a hole drilled from above on one side of its peak. -

    -     - - - - - //ASY file for figdoublepol2b3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(-2.7,-11,16); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={-2,2}; - real[] myzchoice={5}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,4.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(0,7); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface //({z=4-(x-2)^2-y^2); - //triple f(pair t) { - // return (t.x,t.y,4-(t.x-2)^2-t.y^2); - //} - - //surface s=surface(f,(-0.25,-2.25),(4.5,2.25),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - //,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - //); - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - //draw(s,surfacepen); - - real r(real t) { return sqrt((2*cos(t)+1)^2+(2*sin(t))^2);} - real ang(real tt) {return atan2(2*sin(tt),(2*cos(tt)+1));} - real ra(pair s) {return 1+s.y*(r(s.x)-1);} - - triple g(pair t) { return (cos(ang(t.x))*ra(t)+1,sin(ang(t.x))*ra(t),4-(cos(ang(t.x))*ra(t)+1-2)^2-(sin(ang(t.x))*ra(t))^2);} - - surface s=surface(g,(0,0),(2*pi,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - draw(s,surfacepen,meshpen=p); - - triple g(pair t) { return (cos(ang(t.x))*ra(t)+1,sin(ang(t.x))*ra(t),0);} - - surface s=surface(g,(0,0),(2*pi,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - draw(s,surfacepen,meshpen=p); - - triple ff(pair t) { - return (cos(t.x)+1,sin(t.x),t.y*(4-(cos(t.x)+1-2)^2-(sin(t.x))^2));} - surface ss= surface(ff,(0,0),(2*pi,1),16,4,Spline); - pen q=redcurvepen; - //draw(ss,surfacepen2,meshpen=q,nolight,render(merge=true)); - draw(ss,surfacepen,meshpen=p); - - //draw curves on xy plane - triple g(real t) {return (2*cos(t)+2,2*sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (cos(t)+1,sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+linewidth(1)); - - //draw curve on plane - triple g(real t) {return (cos(t)+1,sin(t),4-(cos(t)-1)^2-sin(t)^2);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+linewidth(1)); - - - - -
    -     - -
    - -

    - It is straightforward to show that the circle - (x-1)^2+y^2=1 has polar equation r=2\cos(\theta), - and that the circle (x-2)^2+y^2=4 has polar equation r=4\cos(\theta). - Each of these circles is traced out on the interval 0\leq\theta\leq\pi. - The bounds on r are 2\cos(\theta) \leq r\leq 4\cos(\theta). -

    - -

    - Replacing x with r\cos(\theta) in the integrand, - along with replacing y with r\sin(\theta), - prepares us to evaluate the double integral \iint_Rf(x,y)\, dA: - - \iint_Rf(x,y)\, dA \amp = \int_0^{\pi}\int_{2\cos(\theta) }^{4\cos(\theta) } \Big(4-\big(r\cos(\theta) -2\big)^2-\big(r\sin(\theta) \big)^2\Big)r\, dr\, d\theta - \amp = \int_0^{\pi}\int_{2\cos(\theta) }^{4\cos(\theta) } \big(-r^3+4r^2\cos(\theta) \big)\, dr\, d\theta - \amp = \int_0^\pi \left.\left(-\frac14r^4+\frac43r^3\cos(\theta) \right)\right|_{2\cos(\theta) }^{4\cos(\theta) }d\theta - \amp =\int_0^\pi \left(\left[-\frac14(256\cos^4(\theta) )+\frac43(64\cos^4(\theta) )\right]-\right. - \amp \quad \left.\left[-\frac14(16\cos^4(\theta) )+\frac43(8\cos^4(\theta) )\right]\right)\, d\theta - \amp =\int_0^\pi\frac{44}3\cos^4(\theta) \, d\theta - - To integrate \cos^4(\theta), - rewrite it as \cos^2(\theta) \cos^2(\theta) and employ the power-reducing formula twice: - - \cos^4(\theta) \amp =\cos^2(\theta) \cos^2(\theta) - \amp = \frac12\big(1+\cos(2\theta)\big)\frac12\big(1+\cos(2\theta)\big) - \amp = \frac14\big(1+2\cos(2\theta)+\cos^2(2\theta)\big) - \amp =\frac14\Big(1+2\cos(2\theta)+\frac12\big(1+\cos(4\theta)\big)\Big) - \amp = \frac38+\frac12\cos(2\theta)+\frac18\cos(4\theta) - . - Picking up from where we left off above, we have - - \iint_R f(x,y)\,dA \amp =\int_0^\pi\frac{44}3\cos^4(\theta) \, d\theta - \amp =\int_0^\pi \frac{44}3\left(\frac38+\frac12\cos(2\theta)+\frac18\cos(4\theta)\right)d\theta - \amp = \left.\frac{44}3\left(\frac{3}8\theta+\frac14\sin(2\theta)+\frac{1}{32}\sin(4\theta)\right)\right|_0^\pi - \amp =\frac{11}2\pi\approx 17.279 - . -

    - -

    - While this example was not trivial, - the double integral would have been much - harder to evaluate had we used rectangular coordinates. -

    -     -     - - - Evaluating a double integral with polar coordinates - -

    - Find the volume under the surface given by the graph of - \ds f(x,y) =\frac1{x^2+y^2+1} over the sector of the circle with radius a centered at the origin in the first quadrant, - as shown in . -

    -
    - The surface and region R used in - - - - A portion of a bell-shaped curve, and a triangular curve on the surface that corresponds to a quarter circle in the xy plane. - -

    - The surface given by the graph z=\frac{1}{1+x^2+y^2} is shaped like a steep mountain, - or a witch's hat, - and has circular symmetry about the z axis. - However, only the portion of the surface in the first octant is shown. -

    - -

    - In the xy plane, a quarter-circle is sketched in the first quadrant. - On the surface, we see a curve illustrating the portion of the surface used in the integral. - The curve appears to be triangular; its sides correspond to the x and y coordinate axes, - and the quarter circle. -

    -     - - - - - //ASY file for figdoublepol53D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8,-13,2.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={0.5,1}; - defaultpen(0.5mm); - - pair xbounds=(0,5.5); - pair ybounds=(0,5.5); - pair zbounds=(0,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface //(z=1/(1+x^2+y^2); - triple f(pair t) { - return (t.x,t.y,1/(1+t.x^2+t.y^2)); - } - surface s=surface(f,(-0.25,-0.25),(5,5),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //draw curves on xy plane - triple g(real t) {return (2.5*cos(t),2.5*sin(t),0);} - path3 mypath=graph(g,0,pi/2,operator ..); - draw(mypath,bluepen+dashed+linewidth(0.75)); - label("$a$",(2.7,-0.2,0),S); - - //draw curves on surface - triple g(real t) {return (2.5*cos(t),2.5*sin(t),1/(1+2.5^2));} - path3 mypath=graph(g,0,pi/2,operator ..); - draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,0,1/(1+t^2));} - path3 mypath=graph(g,0,2.5,operator ..); - draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (0,t,1/(1+t^2));} - path3 mypath=graph(g,0,2.5,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //draw connecting lines - draw((2.5,0,0) -- (2.5,0,1/(1+2.5^2)),bluepen+dashed+linewidth(0.75)); - draw((0,2.5,0) -- (0,2.5,1/(1+2.5^2)),bluepen+dashed+linewidth(0.75)); - - - - -
    -     - -

    - The region R we are integrating over is a circle with radius a, - restricted to the first quadrant. - Thus, in polar, the bounds on R are 0\leq r\leq a, - 0\leq\theta\leq\pi/2. - The integrand is rewritten in polar as - - \frac{1}{x^2+y^2+1} \Rightarrow \frac{1}{r^2\cos^2(\theta) +r^2\sin^2(\theta) +1} = \frac1{r^2+1} - . -

    - -

    - We find the volume as follows: - - \iint_Rf(x,y)\, dA \amp = \int_0^{\pi/2}\int_0^a\frac{r}{r^2+1}\, dr\, d\theta - \amp = \int_0^{\pi/2} \frac12\big(\ln\abs{r^2+1}\big)\Big|_0^a\, d\theta - \amp =\int_0^{\pi/2} \frac12\ln(a^2+1)\, d\theta - \amp = \left.\left(\frac12\ln(a^2+1)\theta\right)\right|_0^{\pi/2} - \amp = \frac{\pi}{4}\ln(a^2+1) - . -

    - - -

    - - shows that f shrinks to near 0 very quickly. - Regardless, as a grows, - so does the volume, without bound. -

    -     -     - - - Finding the volume of a sphere - -

    - Find the volume of a sphere with radius a. -

    -     - -

    - The sphere of radius a, - centered at the origin, has equation x^2+y^2+z^2=a^2; - solving for z, we have z=\sqrt{a^2-x^2-y^2}. - This gives the upper half of a sphere. - We wish to find the volume under this top half, - then double it to find the total volume. -

    - -

    - The region we need to integrate over is the disk of radius a, - centered at the origin. - Polar bounds for this equation are 0\leq r\leq a, - 0\leq\theta\leq2\pi. -

    - -

    - All together, the volume of a sphere with radius a is: - - 2\iint_R\sqrt{a^2-x^2-y^2}\, dA \amp = 2\int_0^{2\pi}\int_0^a\sqrt{a^2-(r\cos(\theta) )^2-(r\sin(\theta) )^2}r\, dr\, d\theta - \amp =2\int_0^{2\pi}\int_0^ar\sqrt{a^2-r^2}\, dr\, d\theta. - We can evaluate this inner integral with substitution. - With u=a^2-r^2, du = -2r\, dr. - The new bounds of integration are u(0) = a^2 to u(a)=0. Thus we have: - - \amp = \int_0^{2\pi}\int_{a^2}^0\big(-u^{1/2}\big)\, du\, d\theta - \amp = \int_0^{2\pi}\left.\left(-\frac23u^{3/2}\right)\right|_{a^2}^0 d\theta - \amp = \int_0^{2\pi}\left(\frac23a^3\right)\, d\theta - \amp = \left.\left(\frac23a^3\theta\right)\right|_0^{2\pi} - \amp = \frac43\pi a^3 - . -

    - -

    - Generally, the formula for the volume of a sphere with radius r is given as 4/3\pi r^3; - we have justified this formula with our calculation. -

    -     -     - - - - - Finding the volume of a solid - -

    - A sculptor wants to make a solid bronze cast of the solid shown in , - where the base of the solid has boundary, - in polar coordinates, r=\cos(3\theta), - and the top is defined by the plane z=1-x+0.1y. - Find the volume of the solid. -

    - -
    - Visualizing the solid used in - - - - A solid whose cross-sections are like a three-leaf clover; it has been sliced at an angle. - -

    - A three-dimensional solid is shown, plotted against a set of three-dimensional coordinate axes. - The base of the solid is a three-leaf rose curve. -

    - -

    - The solid is a cylinder, except that it is cut at an angle, - where it meets the plane z=1-x+0.1y. -

    -     - - - - - //ASY file for figdoublepol43D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2,2); - - //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //plane is z=1-x+0.1y - //Draw the surface - //({cos(3*x)*cos(x)},{cos(3*x)*sin(x)},{(1-cos(3*x)*cos(x)+.1*cos(3*x)*sin(x))*y}); - triple f(pair t) { - return (cos(3*t.x)*cos(t.x),cos(3*t.x)*sin(t.x),(1-cos(3*t.x)*cos(t.x)+.1*cos(3*t.x)*sin(t.x))*t.y); - } - surface s=surface(f,(0,0),(2*pi,1),32,32,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //draw curve on xy plane //({cos(3*x)*cos(x)},{cos(3*x)*sin(x)},{0}) - triple g(real t) {return (cos(3*t)*cos(t),cos(3*t)*sin(t),0);} - path3 mypath=graph(g,0,pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //draw curves on surface - triple g(real t) {return (cos(3*t)*cos(t),cos(3*t)*sin(t),1-(cos(3*t)*cos(t))+0.1*(cos(3*t)*sin(t)));} - path3 mypath=graph(g,0,pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //Shade the bottom - import three; - path3 p = (0,0,0)..(.175,0.064,0)..(.264,.119,0)..(0.683,0.183,0)..(.916,0.121,0)..(1,0,0)..(.916,-0.121,0)..(0.683,-0.183,0)..(.264,-.119,0)..(.175,-0.086,0); - draw(surface(p -- cycle), simplesurfacepen); - - path3 p = (0,0,0)..(-0.1429, 0.1196,0)..(-0.2351, 0.1691,0)..(-0.5000, 0.5000,0)..(-0.5628, 0.7328,0)..(-0.5000, 0.8660,0)..(-0.3532, 0.8538,0)..(-0.1830, 0.6830,0)..(-0.0289, 0.2881,0)..(-0.0130, 0.1946,0); - draw(surface(p -- cycle), simplesurfacepen); - - path3 p = (0,0,0)..(-0.0321,-0.1836,0)..(-0.0289,-0.2881,0)..(-0.1830,-0.6830,0)..(-0.3532,-0.8538,0)..(-0.5000,-0.8660,0)..(-0.5628,-0.7328,0)..(-0.5000,-0.5000,0)..(-0.2351,-0.1691,0)..(-0.1620,-0.1086,0); - draw(surface(p -- cycle), simplesurfacepen); - - //Shade the top - path3 p = (0,0,1.0000)..(-0.1429,0.1196,1.1549)..(-0.2351,0.1691,1.2520)..(-0.5000,0.5000,1.5500)..(-0.5628,0.7328,1.6361)..(-0.5000,0.8660,1.5866)..(-0.3532,0.8538,1.4386)..(-0.1830 ,0.6830 ,1.2513)..(-0.0289,0.2881,1.0578)..(-0.0130,0.1946,1.0325); - - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0 , 0 , 1.0000)..(-0.0321 , -0.1836 , 1.0137)..(-0.0289, -0.2881, 1.0001)..(-0.1830, -0.6830, 1.1147)..(-0.3532 , -0.8538 , 1.2678)..(-0.5000, -0.8660 , 1.4134)..(-0.5628 , -0.7328, 1.4895)..(-0.5000 , -0.5000 , 1.4500)..(-0.2351 , -0.1691 , 1.2181)..(-0.1620 , -0.1086, 1.1511); - draw(surface(p -- cycle), simplesurfacepen); - - path3 p = (0 , 0, 1.0000)..( 0.1750, 0.0640 , 0.8314)..( 0.2640, 0.1190 , 0.7479)..( 0.6830, 0.1830 , 0.3353)..( 0.9160, 0.1210 , 0.0961)..( 1.0000, 0 , 0)..( 0.9160, -0.1210 , 0.0719)..( 0.6830, -0.1830 , 0.2987)..( 0.2640, -0.1190 , 0.7241)..( 0.1750, -0.0860 , 0.8164); - draw(surface(p -- cycle), simplesurfacepen); - - //path3 p = (0,0,0)-- (1,1,0) -- (1,2,1) -- (0,1,1); //bottom - //draw(surface(p -- cycle), simplesurfacepen); - //path3 p = (-1,1,0)-- (0,2,0) -- (0,3,1) -- (-1,2,1); //right - //draw(surface(p -- cycle), simplesurfacepen); - //path3 p = (0,0,0)-- (-1,1,0) -- (-1,2,1) -- (0,1,1); //back - //draw(surface(p -- cycle), simplesurfacepen); - //path3 p = (1,1,0)-- (0,2,0) -- (0,3,1) -- (1,2,1); //front - //draw(surface(p -- cycle), simplesurfacepen); - //path3 p = (0,1,1)-- (1,2,1) -- (0,3,1) -- (-1,2,1); //top - //draw(surface(p -- cycle), simplesurfacepen); - - - - -
    -     - -

    - From the outset, - we should recognize that knowing how to set up - this problem is probably more important than knowing - how to compute the integrals. - The iterated integral to come is not - hard to evaluate, - though it is long, requiring lots of algebra. - Once the proper iterated integral is determined, - one can use readily available technology to help compute the final answer. -

    - -

    - The region R that we are integrating over is bound by 0\leq r\leq \cos(3\theta), - for 0\leq \theta\leq\pi - (note that this rose curve is traced out on the interval [0,\pi], - not [0,2\pi]). - This gives us our bounds of integration. - The integrand is z=1-x+0.1y; - converting to polar, we have that the volume V is: - - V = \iint_R f(x,y)\, dA = \int_0^\pi\int_0^{\cos(3\theta)}\big(1-r\cos(\theta) +0.1r\sin(\theta) \big)r\, dr\, d\theta - . -

    - -

    - Distributing the r, the inner integral is easy to evaluate, - leading to - - \int_0^\pi \left(\frac12\cos^2(3\theta)-\frac13\cos^3(3\theta)\cos(\theta) +\frac{0.1}3\cos^3(3\theta)\sin(\theta) \right)\, d\theta - . -

    - -

    - This integral takes time to compute by hand; - it is rather long and cumbersome. - The powers of cosine need to be reduced, - and products like \cos(3\theta)\cos(\theta) need to be turned to sums using the Product To Sum formulas in the back cover of this text. -

    - -

    - We rewrite \frac12\cos^2(3\theta) as \frac14(1+\cos(6\theta)). - We can also rewrite \frac13\cos^3(3\theta)\cos(\theta) as: - - \frac13\cos^3(3\theta)\cos(\theta) \amp = \frac13\cos^2(3\theta)\cos(3\theta)\cos(\theta) - \amp = \frac13\frac{1+\cos(6\theta)}2\big(\cos(4\theta)+\cos(2\theta)\big) - . -

    - -

    - This last expression still needs simplification, - but eventually all terms can be reduced to the form a\cos(m\theta) or - a\sin(m\theta) for various values of a and m. -

    - -

    - We forgo the algebra and recommend the reader employ technology, - such as WolframAlpha, - to compute the numeric answer. - Such technology gives: - - \int_0^\pi\int_0^{\cos(3\theta)}\big(1-r\cos(\theta) +0.1r\sin(\theta) \big)r\, dr\, d\theta = \frac{\pi}{4} \approx 0.785u^3 - . -

    - -

    - Since the units were not specified, - we leave the result as almost 0.8 cubic units (meters, - feet, etc.) Should the artist want to scale the piece uniformly, - so that each rose petal had a length other than 1, she should keep in mind that scaling by a factor of k scales the volume by a factor of k^3. -

    -     -     - - - -

    - We have used iterated integrals to find areas of plane regions and volumes under surfaces. - Just as a single integral can be used to compute much more than - area under the curve, - iterated integrals can be used to compute much more than we have thus far seen. - The next two sections show two, among many, - applications of iterated integrals. -

    - - - - - Terms and Concepts - - - -

    - Match the correct elements on the left to the corresponding elements on the right, - so that \iint_R f(x,y)\, dA is correctly converted to polar coordinates. -

    -     - - - r\cos(\theta) - x - - - r\sin(\theta) - y - - - r\,dr\,d\theta - dA - - - dr\,d\theta - - - \cos(\theta) - - - \sin(\theta) - - - - -     - - - - -

    - Why would one be interested in evaluating a double integral with polar coordinates? -

    -     - - - -

    - Some regions in the xy-plane are easier to describe using polar coordinates than using rectangular coordinates. - Also, some integrals are easier to evaluate one the polar substitutions have been made. -

    -     - -     -     - - - Problems - - - -

    - A function f(x,y) is given and a region R of the xy-plane is described. - Set up and evaluate \iint_Rf(x,y)\, dA using polar coordinates. -

    -     - - - - - $int=Compute("4pi"); - - - -

    - f(x,y) = 3x-y+4 and R is the region enclosed by the circle x^2+y^2=1. -

    - -

    - -

    -     -     -     - - - - -

    - f(x,y) = 4x+4y; - R is the region enclosed by the circle x^2+y^2=4. -

    -     - -

    - \ds \int_0^{2\pi}\int_0^2 \big(4r\cos(\theta) +4r\sin(\theta) \big)r\, dr\, d\theta = 0 -

    -     - -     - - - - - $int=Compute("16pi"); - - - -

    - f(x,y) = 8-y and R is the region enclosed by the circles with polar equations - r=\cos(\theta) and r=3\cos(\theta). -

    - -

    - -

    -     -     -     - - - - -

    - f(x,y) = 4; - R is the region enclosed by the petal of the rose curve - r=\sin(2\theta) in the first quadrant. -

    -     - -

    - \ds \int_0^{\pi/2}\int_{0}^{\sin(2\theta)} \big(4\big)r\, dr\, d\theta = \pi/2 -

    -     - -     - - - - -

    - f(x,y) = \ln(x^2+y^2); - R is the annulus enclosed by the circles - x^2+y^2=1 and x^2+y^2=4. -

    -     - -

    - \ds \int_0^{2\pi}\int_{1}^{2} \big(\ln(r^2)\big)r\, dr\, d\theta = 2\pi\big(\ln(16) -3/2\big) -

    -     - -     - - - - - $int=Compute("pi/2"); - - - -

    - f(x,y) = 1-x^2-y^2 and R is the region enclosed by the circle x^2+y^2=1. -

    - -

    - -

    -     -     -     - - - - -

    - f(x,y) = x^2-y^2; - R is the region enclosed by the circle - x^2+y^2=36 in the first and fourth quadrants. -

    -     - -

    - \ds \int_{-\pi/2}^{\pi/2}\int_{0}^{6} \big(r^2\cos^2(\theta) -r^2\sin^2(\theta) \big)r\, dr\, d\theta = \int_{-\pi/2}^{\pi/2}\int_{0}^{6} \big(r^2\cos(2\theta)\big)r\, dr\, d\theta= 0 -

    -     - -     - - - - -

    - f(x,y) = (x-y)/(x+y); - R is the region enclosed by the lines y=x, - y=0 and the circle x^2+y^2=1 in the first quadrant. -

    -     - -

    - \ds \int_{0}^{\pi/4}\int_{0}^{1} \left(\frac{\cos(\theta) -\sin(\theta) }{\cos(\theta) +\sin(\theta) }\right)r\, dr\, d\theta = \ln(2) -

    -     - -     - -     - - - -

    - An iterated integral in rectangular coordinates is given. - Rewrite the integral using polar coordinates and evaluate the new double integral. -

    -     - - - - -

    - \ds \int_{0}^{5}\int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}} \sqrt{x^2+y^2}\, dy\, dx -

    -     - -

    - \ds \int_{-\pi/2}^{\pi/2}\int_{0}^{5} \big(r^2\big)\, dr\, d\theta=125\pi/3 -

    -     - -     - - - - - $int=Compute("128/3"); - - - -

    - \ds \int_{-4}^{4}\int_{-\sqrt{16-y^2}}^{0} \big(2y-x\big)\, dx\, dy -

    - -

    - -

    -     -     -     - - - - -

    - \ds \int_{0}^{2}\int_{y}^{\sqrt{8-y^2}} \big(x+y\big)\, dx\, dy -

    -     - -

    - \ds \int_{0}^{\pi/4}\int_{0}^{\sqrt8} \big(r\cos(\theta) +r\sin(\theta) \big)r\, dr\, d\theta=16\sqrt2/3 -

    -     - -     - - - - -

    - \ds \int_{-2}^{-1}\int_{0}^{\sqrt{4-x^2}} \big(x+5\big)\, dy\, dx+\int_{-1}^{1}\int_{\sqrt{1-x^2}}^{\sqrt{4-x^2}} \big(x+5\big)\, dy\, dx +\int_{1}^{2}\int_{0}^{\sqrt{4-x^2}} \big(x+5\big)\, dy\, dx -

    - -

    - Hint: draw the region of each integral carefully and see how they all connect. -

    -     - -

    - \ds \int_{0}^{\pi}\int_{1}^{2} \big(r\cos(\theta) +5\big)r\, dr\, d\theta=15\pi/2 -

    -     - -     - -     - - - - -

    - The next two exercises present special double integrals that are especially well suited for evaluation in polar coordinates. -

    -     - - - - -

    - Consider \ds \iint_R e^{-(x^2+y^2)}\, dA. -

    -     - - - -

    - Why is this integral difficult to evaluate in rectangular coordinates, - regardless of the region R? -

    -     - -

    - This is impossible to integrate with rectangular coordinates as - e^{-(x^2+y^2)} does not have an antiderivative in terms of elementary functions. -

    -     -     - - - -

    - Let R be the region bounded by the circle of radius a centered at the origin. - Evaluate the double integral using polar coordinates. -

    -     - -

    - \ds \int_0^{2\pi}\int_0^a re^{r^2}\, dr\, d\theta = \pi(1-e^{-a^2}). -

    -     -     - - - -

    - Take the limit of your answer from (b), as a\to\infty. - What does this imply about the volume under the surface - z= e^{-(x^2+y^2)} over the entire xy-plane? -

    -     - -

    - \lim\limits_{a\to\infty} \pi(1-e^{-a^2})=\pi. - This implies that there is a finite volume under the surface - z=e^{-(x^2+y^2)} over the entire xy-plane. -

    -     -     - -     - - - - -

    - The surface of a right circular cone with height h and base radius a can be described by the equation \ds f(x,y) = h-h\sqrt{\frac{x^2}{a^2}+\frac{y^2}{a^2}}, - where the tip of the cone lies at (0,0,h) and the circular base lies in the xy-plane, - centered at the origin. -

    - -

    - Confirm that the volume of a right circular cone with height h and base radius a is \ds V = \frac13\pi a^2h by evaluating - \ds \iint_R f(x,y)\, dA in polar coordinates. -

    -     - -

    - - \iint_R f(x,y)\, dA \amp = \int_0^{2\pi}\int_0^a \left(h-h\sqrt{\frac{r^2\cos^2(\theta) }{a^2}+\frac{r^2\sin^2(\theta) }{a^2}}\right)r\, dr\, d\theta - \amp = \int_0^{2\pi}\int_0^a \left(hr-h\frac{r^2}{a}\right)\, dr\, d\theta - \amp = \int_0^{2\pi}\left.\left(\frac12hr^2-\frac{h}{3a}r^3\right)\right|_0^a d\theta - \amp = \int_0^{2\pi} \left(\frac16a^2h\right)\, d\theta - \amp = \frac13\pi a^2h - . -

    -     - -     - -     -     -     -
    -
    - Center of Mass - -

    - We have used iterated integrals to find areas of plane regions and signed volumes under surfaces. - A brief recap of these uses will be useful in this section as we apply iterated integrals to compute the mass - and center of mass of planar regions. -

    - -

    - To find the area of a planar region, - we evaluated the double integral \iint_R\, dA. - That is, summing up the areas of lots of little subregions of R gave us the total area. - Informally, we think of \iint_R\, dA as meaning - sum up lots of little areas over R. -

    - -

    - To find the signed volume under a surface, - we evaluated the double integral \iint_R f(x,y)\, dA. - Recall that the dA - is not just a bookend - at the end of an integral; - rather, it is multiplied by f(x,y). - We regard f(x,y) as giving a height, - and dA still giving an area: - f(x,y)\, dA gives a volume. - Thus, informally, \iint_Rf(x,y)\, dA means - sum up lots of little volumes over R. -

    - -

    - We now extend these ideas to other contexts. -

    -     - - - Mass and Weight -

    - Consider a thin sheet of material with constant thickness and finite area. - Mathematicians - (and physicists and engineers) - call such a sheet a lamina. - So consider a lamina, - as shown in , - with the shape of some planar region R, as shown in . - lamina - mass -

    - -
    - Illustrating the concept of a lamina - -
    - - - - A region in the plane with two vertical sides, a curved top, and a slanted bottom. - -

    - An illustration of a lamina as a region in the plane. - This particular region has a curved top, like a downward-opening parabola, - and two sides given by vertical lines. - The right side is shorter than the left side, - and the bottom is a line segment with positive slope. -

    - -

    - There is no coordinate system or other means of reference in the image. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis y line=none, - axis x line=none, - axis on top, - ymin=-.1,ymax=3.5, - xmin=-.1,xmax=3.5 - ] - - \addplot [name path=A,firstcurvestyle,-,domain=1:3] {-(x-2)^2+3}; - \addplot [name path=B,firstcurvestyle,-,domain=1:3] {x/2}; - \addplot [firstcurvestyle,domain=.5:2] (1,x); - \addplot [firstcurvestyle,domain=1.5:2] (3,x); - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B]; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -
    - - - - A region in the plane, with the same shape as the previous image, this time plotted against coordinate axes. - -

    - The first quadrant in the plane is shown, - with the x axis at the bottom of the image, and the y axis on the left. - The same region as is shown, - but this time with the coordinate axes and labels for reference. -

    - -

    - Now it is specified that the sides of the region are the vertical lines x=1 and x=3. - The line along the bottom is labeled with the equation y=f_1(x), - the curve along the top is labeled y=f_2(x), and the region itself is labeled R. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=3.5, - xmin=-.1,xmax=3.5 - ] - - \addplot [name path=A,firstcurvestyle,domain=-.1:3.5,samples=60] {-(x-2)^2+3} node [pos=.65,above,black] { $y=f_2(x)$}; - \addplot [name path=B,firstcurvestyle,domain=-.1:3.5] {x/2} node [pos=.6,sloped,above,black] { $y=f_1(x)$}; - - \addplot [firstcurvestyle,areastyle] fill between [of=A and B,soft clip={domain=1:3}]; - - \addplot [secondcurvestyle,solid,domain=0:3] (1,x); - \addplot [secondcurvestyle,solid,domain=0:3] (3,x); - - \draw (axis cs:2,2) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -
    - -

    - We can write a simple double integral that represents the mass of the lamina: - \iint_R\, dm, where dm - means a little mass. That is, - the double integral states the total mass of the lamina can be found by - summing up lots of little masses over R. -

    - -

    - To evaluate this double integral, - partition R into n subregions as we have done in the past. - The ith subregion has area \Delta A_i. - A fundamental property of mass is that - mass=densityarea. - If the lamina has a constant density \delta, - then the mass of this ith subregion is \Delta m_i=\delta\Delta A_i. - That is, we can compute a small amount of mass by multiplying a small amount of area by the density. -

    - -

    - If density is variable, with density function \delta= \delta(x,y), - then we can approximate the mass of the - ith subregion of R by multiplying - \Delta A_i by \delta(x_i,y_i), - where (x_i,y_i) is a point in that subregion. - That is, for a small enough subregion of R, - the density across that region is almost constant. -

    - -

    - The total mass M of the lamina is approximately the sum of approximate masses of subregions: - - M \approx \sum_{i=1}^n \Delta m_i = \sum_{i=1}^n \delta(x_i,y_i)\Delta A_i - . -

    - -

    - Taking the limit as the size of the subregions shrinks to 0 gives us the actual mass; - that is, integrating \delta(x,y) over R gives the mass of the lamina. -

    - - - Mass of a Lamina with Variable Density - -

    - Let \delta(x,y) be a continuous density function of a lamina corresponding to a closed, - bounded plane region R. - The mass M of the lamina is - mass - - \text{ mass } M = \iint_R\, dm = \iint_R \delta(x,y)\, dA - . -

    -     -     - - - - - Finding the mass of a lamina with constant density - -

    - Find the mass of a square lamina, - with side length 1, with a density of \delta = 3\,\text{g/cm}^2. -

    -     - -

    - We represent the lamina with a square region in the plane as shown in . - As the density is constant, it does not matter where we place the square. -

    - -
    - A region R representing a lamina in - - - The unit square in the first quadrant, plotted against x and y coordinate axes. - -

    - The x axis is at the bottom of the image, and the y axis along the left, - so that the first quadrant in the plane is shown. -

    - -

    - The unit square, given by 0\leq x\leq 1 and 0\leq y\leq 1 - is drawn and shaded. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.1 - ] - - \filldraw [fill=firstcolor!15,draw=firstcolor,thick] (axis cs:0,0) -- (axis cs:1,0) -- (axis cs: 1,1) -- (axis cs:0,1) -- cycle; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Following , - the mass M of the lamina is - - M = \iint_R 3\, dA = \int_0^1\int_0^1 3\, dx\, dy = 3\int_0^1\int_0^1 \, dx\, dy=3\,\text{g} - . -

    - -

    - This is all very straightforward; - note that all we really did was find the area of the lamina and multiply it by the constant density of - 3. -

    -     -     - - - Finding the mass of a lamina with variable density - -

    - Find the mass of a square lamina, - represented by the unit square with lower lefthand corner at the origin - (see ), - with variable density \delta(x,y) = (x+y+2)\,\text{g/cm}^2. -

    -
    - Graphing the density functions in and - - - - Two planes intersect along a line in three dimensions. - -

    - Two planes are plotted against a set of three-dimensional coordinate axes. - Both planes are drawn sitting about the xy plane, - and they intersect along a line. -

    - -

    - The line of intersection is not important: - each plane is illustrating the graph of a different density function. - One is the constant function \delta(x,y)=3, - while the other is the linear function \delta(x,y)=x+y+2. -

    -     - - - - import labelpath3; - //ASY file for figmass2.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(2.9,-4.6,5.2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={2,4}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,1.25); - pair ybounds=(-0.25,1.25); - pair zbounds=(0,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane z=3 with thick line - triple f(pair t) { - return (t.x,t.y,3); - } - surface s=surface(f,(0,0),(1,1),4,4,Spline); - pen q=redcurvepen; - draw(s,surfacepen2,meshpen=q); - draw((0,0,3)--(1,0,3)--(1,1,3)--(0,1,3)--cycle,redpen+linewidth(1)); - label("$z=3$",(1,0.5,3),S); - path3 gg=(1,0,2.8)--(1,1,2.8); - //draw(labelpath("$z=3$",subpath(gg,.35,.65),angle=-90)); - - //draw plane z=x+y+2 - triple f(pair t) { - return (t.x,t.y,t.x+t.y+2); - } - surface s=surface(f,(0,0),(1,1),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - draw((0,0,2)--(1,0,3)--(1,1,4)--(0,1,3)--cycle,bluepen+linewidth(1)); - label("$z=x+y+2$",(.5,0,2.5),S); - - draw((1,0,3)--(0,1,3),black+dashed+linewidth(1)); - - - - -
    -     - -

    - The variable density \delta, - in this example, is very uniform, - giving a density of 3 in the center of the square and changing linearly. - A graph of \delta(x,y) can be seen in ; - notice how same amount of density is above z=3 as below. - We'll comment on the significance of this momentarily. -

    - -

    - The mass M is found by integrating \delta(x,y) over R. - The order of integration is not important; - we choose dx\, dy arbitrarily. - Thus: - - M = \iint_R(x+y+2)\, dA \amp = \int_0^1\int_0^1 (x+y+2)\, dx\, dy - \amp = \int_0^1\left.\left(\frac 12x^2+x(y+2)\right)\right|_0^1dy - \amp = \int_0^1 \left(\frac52+y\right)\, dy - \amp = \left.\left(\frac52y+\frac12y^2\right)\right|_0^1 - \amp = 3\,\text{g} - . -

    - -

    - It turns out that since the density of the lamina is so uniformly distributed above and below - z=3 that the mass of the lamina is the same as if it had a constant density of 3. - The density functions in - and - are graphed in , - which illustrates this concept. -

    -     -     - - - Finding the weight of a lamina with variable density - -

    - Find the weight of the lamina represented by the disk with radius - 2, - centered at the origin, - with density function \delta(x,y) = (x^2+y^2+1)\,\text{lb/ft}^2. - Compare this to the weight of the lamina with the same shape and density - \delta(x,y) = (2\sqrt{x^2+y^2}+1)\,\text{lb/ft}^2. -

    -     - -

    - A direct application of - states that the weight of the lamina is \iint_R\delta(x,y)\, dA. - Since our lamina is in the shape of a circle, - it makes sense to approach the double integral using polar coordinates. -

    - -

    - The density function \delta(x,y) = x^2+y^2+1 becomes - - \delta(r,\theta) = (r\cos(\theta) )^2+(r\sin(\theta) )^2+1 = r^2+1 - . - The circle is bounded by 0\leq r\leq 2 and 0\leq\theta\leq2\pi. - Thus the weight W is: - - W \amp = \int_0^{2\pi}\int_0^2 (r^2+1)r\, dr\, d\theta - \amp = \int_0^{2\pi} \left.\left(\frac14r^4+\frac12r^2\right)\right|_0^2d\theta - \amp = \int_0^{2\pi} \left(6\right) d\theta - \amp = 12\pi \approx 37.70\text{ lb } - . -

    - -

    - Now compare this with the density function \delta(x,y) = 2\sqrt{x^2+y^2}+1. - Converting this to polar coordinates gives - - \delta(r,\theta) = 2\sqrt{(r\cos(\theta) )^2+(r\sin(\theta) )^2}+1 = 2r+1 - . - Thus the weight W is: - - W \amp = \int_0^{2\pi}\int_0^2 (2r+1)r\, dr\, d\theta - \amp = \int_0^{2\pi} (\frac23r^3+\frac12r^2)\Big|_0^2d\theta - \amp = \int_0^{2\pi} \left(\frac{22}3\right)\, d\theta - \amp = \frac{44}3\pi \approx 46.08\text{ lb } - . -

    - -

    - One would expect different density functions to return different weights, - as we have here. - The density functions were chosen, though, to be similar: - each gives a density of 1 at the origin and a density of 5 at the outside edge of the circle, - as seen in . -

    - -
    - Graphing the density functions in . In (a) is the density function \delta(x,y) = x^2+y^2+1; in (b) is \delta(x,y) = 2\sqrt{x^2+y^2}+1 - -
    - - - - - A circular paraboloid, opening upward, plotted over a circular domain. - -

    - The graph of \delta(x,y)=x^2+y^2+1 is a circular paraboloid, - opening upward, with vertex at (0,0,1). - A circle of radius 2, centered at the origin, is also plotted in the xy plane; - this is the boundary of the domain for the graph. -

    -     - - - - - //ASY file for figmass3a3D.asy in Chapter 13 - - size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={5}; - defaultpen(0.5mm); - - pair xbounds=(-2.75,2.75); - pair ybounds=(-2.75,2.75); - pair zbounds=(0,6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane z=3 with thick line - triple f(pair t) { - return (cos(t.x)*t.y,sin(t.x)*t.y,t.y^2+1); - } - surface s=surface(f,(0,0),(2*pi,2),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //draw circle in plane - triple g(real t) {return (2*cos(t),2*sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - - - -
    - -
    - - - - - A circular cone, opening upward, plotted over a circular domain. - -

    - The graph of \delta(x,y)=2\sqrt{x^2+y^2}+1 is a circular cone, - opening upward, with vertex at (0,0,1). - A circle of radius 2, centered at the origin, is also plotted in the xy plane; - this is the boundary of the domain for the graph. -

    -     - - - - - //ASY file for figmass3b3D.asy in Chapter 13 - - size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={5}; - defaultpen(0.5mm); - - pair xbounds=(-2.75,2.75); - pair ybounds=(-2.75,2.75); - pair zbounds=(0,6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane z=3 with thick line - triple f(pair t) { - return (cos(t.x)*t.y,sin(t.x)*t.y,2*t.y+1); - } - surface s=surface(f,(0,0),(2*pi,2),16,16,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //draw circle in plane - triple g(real t) {return (2*cos(t),2*sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - - - -
    -     -
    - -

    - Notice how x^2+y^2+1 \leq 2\sqrt{x^2+y^2}+1 over the circle; - this results in less weight. -

    -     -     - -

    - Plotting the density functions can be useful as our understanding of mass can be related to our understanding of - volume under a surface. - We interpreted \iint_R f(x,y)\, dA as giving the volume under f over R; - we can understand \iint_R\delta(x,y)\, dA in the same way. - The volume under \delta over R is actually mass; - by compressing the volume - under \delta onto the xy-plane, - we get more mass in some areas than others , areas of greater density. -

    - -

    - Knowing the mass of a lamina is one of several important measures. - Another is the center of mass, which we discuss next. -

    -     - - - Center of Mass -

    - Consider a disk of radius 1 with uniform density. - It is common knowledge that the disk will balance on a point if the point is placed at the center of the disk. - What if the disk does not have a uniform density? - Through trial-and-error, - we should still be able to find a spot on the disk at which the disk will balance on a point. - This balance point is referred to as the - center of mass, - or center of gravity. - It is though all the mass is centered there. - In fact, if the disk has a mass of 3, - the disk will behave physically as though it were a point-mass of - 3 located at its center of mass. - For instance, - the disk will naturally spin with an axis through its center of mass - (which is why it is important to - balance the tires of your car: - if they are out of balance, - their center of mass will be outside of the axle and it will shake terribly). - center of mass - masscenter of -

    - -

    - We find the center of mass based on the principle of a - weighted average. - Consider a college class in which your homework average is 90%, - your test average is 73%, and your final exam grade is an 85%. - Experience tells us that our final grade - is not the average - of these three grades: that is, it is not: - - \frac{0.9+0.73+0.85}{3} \approx 0.837 = 83.7% - . -

    - -

    - That is, you are probably not pulling a B in the course. - Rather, your grades are weighted. - Let's say the homework is worth 10% of the grade, - tests are 60% and the exam is 30%. - Then your final grade is: - - (0.1)(0.9) + (0.6)(0.73)+(0.3)(0.85) = 0.783 = 78.3% - . -

    - -

    - Each grade is multiplied by a weight. -

    - -

    - In general, given values x_1,x_2,\ldots,x_n and weights w_1,w_2,\ldots,w_n, - the weighted average of the n values is - - \sum_{i=1}^n w_ix_i\Bigg/\sum_{i=1}^n w_i - . -

    - -

    - In the grading example above, - the sum of the weights 0.1, 0.6 and 0.3 is 1, so we don't see the division by the sum of weights in that instance. -

    - -

    - How this relates to center of mass is given in the following theorem. -

    - - - Center of Mass of Discrete Linear System - -

    - Let point masses m_1,m_2,\ldots,m_n be distributed along the x-axis at locations x_1,x_2,\ldots,x_n, - respectively. - The center of mass \overline{x} of the system is located at - center of mass - - \overline{x} = \sum_{i=1}^nm_ix_i\Bigg/\sum_{i=1}^n m_i - . -

    -     -     - - - Finding the center of mass of a discrete linear system - -

    -

      -
    1. -

      - Point masses of 2 are located at x=-1, - x=2 and x=3 are connected by a thin rod of negligible weight. - Find the center of mass of the system. -

      -
      2. - -
    2. -

      - Point masses of 10, - 2 and 1 are located at x=-1, - x=2 and x=3, - respectively, are connected by a thin rod of negligible weight. - Find the center of mass of the system. -

      -
      3. -
    -

    -     - -

    -

      -
    1. -

      - Following , - we compute the center of mass as: - - \overline{x}=\frac{2(-1) + 2(2)+2(3)}{2+2+2} = \frac43 = 1.\overline{3} - . - So the system would balance on a point placed at x=4/3, - as illustrated in . -

      - -
      - Illustrating point masses along a thin rod and the center of mass - -
      - - - - A number line with three points marked at -1, 2, and 3, and a marking for the center of mass. - -

      - A number line is marked as the x axis. - There are markings for x values from x=-1 to x=3. - Solid dots are placed at x=-1, x=2, and x=3 to illustrate the presence of point masses. -

      - -

      - A small triangle marks the point 4/3, the location of the center of mass, - which is also labeled as \overline{x}. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - axis y line=none, - xtick={-1,0,1,2,3}, - ymin=-.5,ymax=.5, - xmin=-1.5,xmax=3.5 - ] - - \filldraw (axis cs:-1,0) circle (2pt) - (axis cs: 2,0) circle (2pt) - (axis cs: 3,0) circle (2pt) - (axis cs: 1.3,0) node [above] {$\overline{x}$} -- (axis cs: 1.2,-.05) -- (axis cs: 1.4,-.05) -- cycle; - - \draw [very thick] (axis cs: -1,0) -- (axis cs:3,0); - - \end{axis} - - \end{tikzpicture} - - - - -
      - -
      - - - - A number line with three points marked at -1, 2, and 3, and a marking for the center of mass. - -

      - A number line is marked as the x axis. - There are markings for x values from x=-1 to x=3. - Solid dots are placed at x=-1, x=2, and x=3 to illustrate the presence of point masses. - The dots have different sizes to represent the fact that the correspond to different masses. -

      - -

      - A small triangle just to the left of 0 marks the location of the center of mass, - which is also labeled as \overline{x}. -

      -       - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - axis y line=none, - xtick={-1,0,1,2,3}, - ymin=-.5,ymax=.5, - xmin=-1.5,xmax=3.5 - ] - - \filldraw (axis cs:-1,0) circle (4.5pt) - (axis cs: 2,0) circle (2pt) - (axis cs: 3,0) circle (1.4pt) - (axis cs: -.23,0) node [above] {$\overline{x}$} -- (axis cs: -.33,-.05) -- (axis cs: -.13,-.05) -- cycle; - - \draw [very thick] (axis cs: -1,0) -- (axis cs:3,0); - - \end{axis} - - \end{tikzpicture} - - - - -
      -       - -
      -
      2. - -
    2. -

      - Again following , we find: - - \overline{x} = \frac{10(-1)+2(2)+1(3)}{10+2+1} = \frac{-3}{13} \approx -0.23 - . - Placing a large weight at the left hand side of the system moves the center of mass left, - as shown in . -

      -
      3. -
    -

    -     -     - -

    - In a discrete system (, mass is located at individual points, - not along a continuum) we find the center of mass by dividing the mass into a - moment of the system. - In general, a moment is a weighted measure of distance from a particular point or line. - In the case described by , - we are finding a weighted measure of distances from the y-axis, - so we refer to this as the moment about the y-axis, - represented by M_y. - Letting M be the total mass of the system, - we have \overline{x} = M_y/M. -

    - -

    - We can extend the concept of the center of mass of discrete points along a line to the center of mass of discrete points in the plane rather easily. - To do so, we define some terms then give a theorem. -

    - - - Moments about the <m>x</m> and <m>y</m> Axes - -

    - Let point masses m_1, - m_2,\ldots,m_n be located at points - - (x_1,y_1), (x_2,y_2), \ldots, (x_n,y_n) - , - respectively, in the xy-plane. - moment -

    - -

    -

      -
    1. -

      - The moment about the y-axis, M_y, is - \ds M_y = \sum_{i=1}^n m_ix_i. -

      -
      2. - -
    2. -

      - The moment about the x-axis, M_x, is - \ds M_x = \sum_{i=1}^n m_iy_i. -

      -
      3. -
    -

    -     -     - -

    - One can think that these definitions are backwards - as M_y sums up x distances. - But remember, - x distances are measurements of distance from the y-axis, - hence defining the moment about the y-axis. -

    - -

    - We now define the center of mass of discrete points in the plane. -

    - - - Center of Mass of Discrete Planar System - -

    - Let point masses m_1, - m_2,\ldots,m_n be located at points - - (x_1,y_1), (x_2,y_2),\ldots,(x_n,y_n) - , - respectively, in the xy-plane, - and let \ds M = \sum_{i=1}^n m_i. - center of mass -

    - -

    - The center of mass of the system is at (\overline{x},\overline{y}), where - - \overline{x}= \frac{M_y}{M} \text{ and } \overline{y} = \frac{M_x}{M} - . -

    -     -     - - - Finding the center of mass of a discrete planar system - -

    - Let point masses of 1, - 2 - and 5 be located at points (2,0), - (1,1) and (3,1), respectively, - and are connected by thin rods of negligible weight. - Find the center of mass of the system. -

    -     - -

    - We follow - and to find M, - M_x and M_y: -

    - -

    - M = 1+2+5 = 8\,\text{kg}. -

    - - -

    - - M_x \amp = \sum_{i=1}^n m_iy_i - \amp = 1(0) + 2(1) + 5(1) - \amp = 7 - . -

    -

    - - M_y \amp = \sum_{i=1}^n m_ix_i - \amp = 1(2) + 2(1) + 5(3) - \amp = 19 - . -

    -     - -
    - Illustrating the center of mass of a discrete planar system in - - - A triangle in the plane. The vertices are marked with dots of different sizes, and an interior point marks the center of mass. - -

    - The first quadrant in the plane is shown, with x and y coordinate axes. - A triangle is plotted, with vertices at (1,1), (3,1), and (2,0). - Each vertex is marked with a dot, and the size of each dot corresponds to the mass it represents. -

    - -

    - The center of mass is also marked, as a point in the interior of the triangle, - closest to the vertex (3,1), where the largest of the three masses is located. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - xtick={1,2,3}, - ytick={1,2}, - ymin=-.5,ymax=1.5, - xmin=-.5,xmax=3.5 - ] - - \filldraw (axis cs:2,0) circle (1.4pt) - (axis cs: 1,1) circle (2pt) - (axis cs: 3,1) circle (3.16pt) - (axis cs: 2.375,0.875) circle (1pt) node [below] { $(\overline{x},\overline{y})$}; - - - \draw [very thick] (axis cs: 2,0) -- (axis cs:3,1) -- (axis cs:1,1) -- cycle; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Thus the center of mass is \ds (\overline{x},\overline{y}) = \left(\frac{M_y}{M},\frac{M_x}M\right) = \left(\frac{{19}}8,\frac78\right) =(2.375,0.875), - illustrated in . -

    -     -     - -

    - We finally arrive at our true goal of this section: - finding the center of mass of a lamina with variable density. - While the above measurement of center of mass is interesting, - it does not directly answer more realistic situations where we need to find the center of mass of a contiguous region. - However, understanding the discrete case allows us to approximate the center of mass of a planar lamina; - using calculus, we can refine the approximation to an exact value. -

    - -

    - We begin by representing a planar lamina with a region R in the xy-plane with density function \delta(x,y). - Partition R into n subdivisions, - each with area \Delta A_i. - As done before, we can approximate the mass of the - ith subregion with \delta(x_i,y_i)\Delta A_i, - where (x_i,y_i) is a point inside the ith subregion. - We can approximate the moment of this subregion about the y-axis with x_i\delta(x_i,y_i)\Delta A_i that is, - by multiplying the approximate mass of the region by its approximate distance from the y-axis. - Similarly, we can approximate the moment about the x-axis with y_i\delta(x_i,y_i)\Delta A_i. - By summing over all subregions, we have: - - \text{ mass: } M \amp \approx \sum_{i=1}^n \delta(x_i,y_i)\Delta A_i \text{ (as seen before) } - \text{ moment about the \(x\)-axis: } M_x \amp \approx \sum_{i=1}^n y_i\delta(x_i,y_i)\Delta A_i - \text{ moment about the \(y\)-axis: } M_y \amp \approx \sum_{i=1}^n x_i\delta(x_i,y_i)\Delta A_i - -

    - -

    - By taking limits, - where size of each subregion shrinks to 0 in both the x and y directions, - we arrive at the double integrals given in the following theorem. -

    - - - Center of Mass of a Planar Lamina, Moments - -

    - Let a planar lamina be represented by a closed, - bounded region R in the xy-plane with density function \delta(x,y). - center of mass - moment -

    - -

    -

      -
    1. -

      - \ds \text{ mass: } M = \iint_R\delta(x,y)\, dA -

      -
      2. - -
    2. -

      - \ds \text{ moment about the \(x\)-axis: } M_x = \iint_Ry\delta(x,y)\, dA -

      -
      3. - -
    3. -

      - \ds \text{ moment about the \(y\)-axis: } M_y = \iint_Rx\delta(x,y)\, dA -

      -
      4. - -
    4. -

      - The center of mass of the lamina is - - (\overline{x},\overline{y}) = \left(\frac{M_y}{M},\frac{M_x}M\right) - . -

      -
      5. -
    -

    -     -     - -

    - We start our practice of finding centers of mass by revisiting some of the lamina used previously in this section when finding mass. - We will just set up the integrals needed to compute M, - M_x and M_y and leave the details of the integration to the reader. -

    - - - Finding the center of mass of a lamina - -

    - Find the center mass of a square lamina, - with side length 1, with a density of \delta = 3\,\text{g/cm}^2. - (Note: this is the lamina from .) -

    -     - -

    - We represent the lamina with a square region in the plane as shown in as done previously. -

    - -
    - A region R representing a lamina in - - - The unit square in the first quadrant of the plane. - -

    - The unit square, given by 0\leq x\leq 1 and 0\leq y\leq 1, - plotted in the first quadrant, relative to x and y coordinate axes. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-.1,ymax=1.1, - xmin=-.1,xmax=1.1 - ] - - \filldraw [fill=firstcolor!15,draw=firstcolor,thick] (axis cs:0,0) -- (axis cs:1,0) -- (axis cs: 1,1) -- (axis cs:0,1) -- cycle; - - \end{axis} - - \end{tikzpicture} - - - - -
    - -

    - Following , - we find M, M_x and M_y: - - M \amp = \iint_R 3\, dA = \int_0^1\int_0^1 3\, dx\, dy =3\,\text{g} - M_x \amp = \iint_R 3y\, dA = \int_0^1\int_0^1 3y\, dx\, dy =3/2 = 1.5 - M_y \amp = \iint_R 3x\, dA = \int_0^1\int_0^1 3x\, dx\, dy =3/2 = 1.5 - . -

    - -

    - Thus the center of mass is \ds (\overline{x},\overline{y}) = \left(\frac{M_y}M,\frac{M_x}M\right) = (1.5/3,1.5/3) = (0.5,0.5). - This is what we should have expected: - the center of mass of a square with constant density is the center of the square. -

    -     -     - - - Finding the center of mass of a lamina - -

    - Find the center of mass of a square lamina, - represented by the unit square with lower lefthand corner at the origin - (see ), - with variable density \delta(x,y) = (x+y+2)\,\text{g/cm}^2. - (Note: this is the lamina from .) -

    -     - -

    - We follow , - to find M, M_x and M_y: - - M \amp = \iint_R (x+y+2)\, dA = \int_0^1\int_0^1 (x+y+2)\, dx\, dy =3\,\text{g} - M_x \amp = \iint_R y(x+y+2)\, dA = \int_0^1\int_0^1 y(x+y+2)\, dx\, dy =\frac{19}{12} - M_y \amp = \iint_R x(x+y+2)\, dA = \int_0^1\int_0^1 x(x+y+2)\, dx\, dy =\frac{19}{12} - . -

    - -

    - Thus the center of mass is - - (\overline{x},\overline{y}) = \left(\frac{M_y}M,\frac{M_x}M\right) = \left(\frac{19}{36},\frac{19}{36}\right) \approx (0.528,0.528) - . - While the mass of this lamina is the same as the lamina in the previous example, - the greater density found with greater x and y values pulls the center of mass from the center slightly towards the upper righthand corner. -

    -     -     - - - Finding the center of mass of a lamina - -

    - Find the center of mass of the lamina represented by the circle with radius 2, - centered at the origin, - with density function \delta(x,y) = (x^2+y^2+1)\,\text{lb/ft}^2. - (Note: this is one of the lamina used in .) -

    -     - -

    - As done in , - it is best to describe R using polar coordinates. - Thus when we compute M_y, - we will integrate not x\delta(x,y) = x(x^2+y^2+1), - but rather \big(r\cos(\theta) \big)\delta(r\cos(\theta) ,r\sin(\theta) ) = \big(r\cos(\theta) \big)\big(r^2+1\big). - We compute M, M_x and M_y: - - M \amp = \int_0^{2\pi}\int_0^2 (r^2+1)r\, dr\, d\theta = 12\pi\approx 37.7\,\text{lb} - M_x \amp = \int_0^{2\pi}\int_0^2 (r\sin(\theta) )(r^2+1)r \, dr\, d\theta = 0 - M_y \amp = \int_0^{2\pi}\int_0^2 (r\cos(\theta) )(r^2+1)r \, dr\, d\theta = 0 - . -

    - -

    - Since R and the density of R are both symmetric about the x and y axes, - it should come as no big surprise that the moments about each axis is 0. - Thus the center of mass is (\overline{x},\overline{y})=(0,0). -

    -     -     - - - Finding the center of mass of a lamina - -

    - Find the center of mass of the lamina represented by the region R shown in , - half an annulus with outer radius 6 - and inner radius 5, - with constant density 2. -

    -
    - Illustrating the region R in - - - An annular region, between semi-circles of radii 5 and 6 in the upper half-plane. - -

    - A sketch of the region occupied by the lamina for this example. - It is bounded between two semi-circles and the x axis. - The semi-circles have radii 5 and 6, and lie above the x axis. -

    - -

    - The overall shape is that of a relatively thin arch. - The center of mass is also marked in the diagram. - What is interesting in this example is that the center of mass does not lie within the region of the lamina. - Instead, it is located on the y axis, below the region. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis on top, - ymin=-2,ymax=8.8, - xmin=-6.5,xmax=6.5 - ] - - \addplot [firstcurvestyle,areastyle] coordinates {(6.,0)(5.967,0.6272)(5.869,1.247)(5.706,1.854)(5.481,2.44)(5.196,3.)(4.854,3.527)(4.459,4.015)(4.015,4.459)(3.527,4.854)(3.,5.196)(2.44,5.481)(1.854,5.706)(1.247,5.869)(0.6272,5.967)(0,6.)(-0.6272,5.967)(-1.247,5.869)(-1.854,5.706)(-2.44,5.481)(-3.,5.196)(-3.527,4.854)(-4.015,4.459)(-4.459,4.015)(-4.854,3.527)(-5.196,3.)(-5.481,2.44)(-5.706,1.854)(-5.869,1.247)(-5.967,0.6272)(-6.,0)(-5.,0)(-4.973,0.5226)(-4.891,1.04)(-4.755,1.545)(-4.568,2.034)(-4.33,2.5)(-4.045,2.939)(-3.716,3.346)(-3.346,3.716)(-2.939,4.045)(-2.5,4.33)(-2.034,4.568)(-1.545,4.755)(-1.04,4.891)(-0.5226,4.973)(0,5.)(0.5226,4.973)(1.04,4.891)(1.545,4.755)(2.034,4.568)(2.5,4.33)(2.939,4.045)(3.346,3.716)(3.716,3.346)(4.045,2.939)(4.33,2.5)(4.568,2.034)(4.755,1.545)(4.891,1.04)(4.973,0.5226)(5.,0)(6,0)}; - \addplot [firstcurvestyle] coordinates {(6.,0)(5.967,0.6272)(5.869,1.247)(5.706,1.854)(5.481,2.44)(5.196,3.)(4.854,3.527)(4.459,4.015)(4.015,4.459)(3.527,4.854)(3.,5.196)(2.44,5.481)(1.854,5.706)(1.247,5.869)(0.6272,5.967)(0,6.)(-0.6272,5.967)(-1.247,5.869)(-1.854,5.706)(-2.44,5.481)(-3.,5.196)(-3.527,4.854)(-4.015,4.459)(-4.459,4.015)(-4.854,3.527)(-5.196,3.)(-5.481,2.44)(-5.706,1.854)(-5.869,1.247)(-5.967,0.6272)(-6.,0)(-5.,0)(-4.973,0.5226)(-4.891,1.04)(-4.755,1.545)(-4.568,2.034)(-4.33,2.5)(-4.045,2.939)(-3.716,3.346)(-3.346,3.716)(-2.939,4.045)(-2.5,4.33)(-2.034,4.568)(-1.545,4.755)(-1.04,4.891)(-0.5226,4.973)(0,5.)(0.5226,4.973)(1.04,4.891)(1.545,4.755)(2.034,4.568)(2.5,4.33)(2.939,4.045)(3.346,3.716)(3.716,3.346)(4.045,2.939)(4.33,2.5)(4.568,2.034)(4.755,1.545)(4.891,1.04)(4.973,0.5226)(5.,0)(6,0)}; - - \filldraw (axis cs:0,3.51) circle (2.4pt) node [below right] { $(\overline{x},\overline{y})$}; - - \end{axis} - - \end{tikzpicture} - - - - -
    -     - -

    - Once again it will be useful to represent R in polar coordinates. - Using the description of R and/or the illustration, - we see that R is bounded by - 5\leq r\leq 6 and 0\leq\theta\leq\pi. - As the lamina is symmetric about the y-axis, - we should expect M_y=0. - We compute M, M_x and M_y: - - M \amp = \int_0^{\pi}\int_5^6 (2)r\, dr\, d\theta = 11\pi\,\text{lb} - M_x \amp = \int_0^{\pi}\int_5^6 (r\sin(\theta) )(2)r\, dr\, d\theta = \frac{364}3\approx 121.33 - M_y \amp = \int_0^{\pi}\int_5^6 (r\cos(\theta) )(2)r\, dr\, d\theta = 0 - . -

    - -

    - Thus the center of mass is (\overline{x},\overline{y}) = \left(0,\frac{364}{33\pi}\right) \approx (0,3.51). - The center of mass is indicated in ; - note how it lies outside of R! -

    -     -     - -

    - This section has shown us another use for iterated integrals beyond finding area or signed volume under the curve. - While there are many uses for iterated integrals, - we give one more application in the following section: - computing surface area. -

    -     - - - - Terms and Concepts - - - - -

    - Why is it easy to use mass - and weight interchangeably, - even though they are different measures? -

    -     - - - -

    - Because they are scalar multiples of each other. -

    -     - -     - - - - -

    - Given a point (x,y), - the value of x is a measure of distance from the -axis. -

    -     - - - - - - - - -     - - - - -

    - We can think of \iint_R\, dm as meaning - sum up lots of -

    -     - - - - little masses|small masses - - - - -     - - - - -

    - What is a discrete planar system? -

    -     - - - -

    - A collection of individual masses in the plane. - Each mass is a point mass, - , mass located at a point, not across a region. -

    -     - -     - - - - -

    - Why does M_x use \iint_R y\delta(x,y)\, dA instead of \iint_R x\delta(x,y)\, dA; - that is, why do we use y and not x? -

    -     - - - -

    - M_x measures the moment about the x-axis, - meaning we need to measure distance from the x-axis. - Such measurements are measures in the y-direction. -

    -     - -     - - - - -

    - Describe a situation where the center of mass of a lamina does not lie within the region of the lamina itself. -

    -     - - - -

    - If the lamina is an annulus, - the center of mass will likely be in the middle, outside of the region. - (See .) -

    -     - -     -     - - - Problems - - - -

    - Point masses are given along a line or in the plane. - Find the center of mass \overline{x} or (\overline{x},\overline{y}), - as appropriate. - (All masses are in grams and distances are in cm.) -

    -     - - - - -

    - m_1=4 at x=1; m_2=3 at x=3; - m_3=5 at x=10 -

    -     - -

    - \overline{x}=5.25 -

    -     - -     - - - - -

    - m_1=2 at x=-3; - m_2=2 at x=-1; -

    - -

    - m_3=3 at x=0;m_4=3 at x=7 -

    -     - -

    - \overline{x}=1.3 -

    -     - -     - - - - -

    - m_1=2 at (-2,-2); - m_2=2 at (2,-2); -

    - -

    - m_3=20 at (0,4) -

    -     - -

    - (\overline{x},\overline{y}) = (0,3) -

    -     - -     - - - - -

    - m_1=1 at (-1,-1); - m_2=2 at (-1,1); -

    - -

    - m_3=2 at (1,1);m_4=1 at (1,-1) -

    -     - -

    - (\overline{x},\overline{y}) = (0,1/3) -

    -     - -     - -     - - - -

    - Find the mass/weight of the lamina described by the region R in the plane and its density function \delta(x,y). -

    -     - - - - -

    - R is the rectangle with corners (1,-3), (1,2), - (7,2) and (7,-3); - \delta(x,y) = 5\,\text{g/cm}^2 -

    -     - -

    - M = 150\,text{g}; -

    -     - -     - - - - -

    - R is the rectangle with corners (1,-3), (1,2), - (7,2) and (7,-3); - \delta(x,y) = (x+y^2)\,\text{g/cm}^2 -

    -     - -

    - M = 190\,\text{g} -

    -     - -     - - - - -

    - R is the triangle with corners (-1,0), - (1,0), and (0,1); - \delta(x,y) = 2\,\text{lb/in}^2 -

    -     - -

    - M = 2\,text{lb} -

    -     - -     - - - - -

    - R is the triangle with corners (0,0), - (1,0), and (0,1); - \delta(x,y) = (x^2+y^2+1)\,\text{lb/in}^2 -

    -     - -

    - M = 2/3\,\text{lb} -

    -     - -     - - - - -

    - R is the disk centered at the origin with radius 2; - \delta(x,y) = (x+y+4)\,\text{kg/m}^2 -

    -     - -

    - M = 16\pi\approx 50.27\,\text{kg} -

    -     - -     - - - - -

    - R is the circle sector bounded by - x^2+y^2=25 in the first quadrant; - \delta(x,y) = (\sqrt{x^2+y^2}+1)\,\text{kg/m}^2 -

    -     - -

    - M = 325\pi/12\approx 85\,\text{kg} -

    -     - -     - - - - -

    - R is the annulus in the first and second quadrants bounded by - x^2+y^2=9 and x^2+y^2=36; - \delta(x,y) = 4\,\text{lb/ft}^2 -

    -     - -

    - M = 54\pi\approx 169.65\,\text{lb} -

    -     - -     - - - - -

    - R is the annulus in the first and second quadrants bounded by - x^2+y^2=9 and x^2+y^2=36; - \delta(x,y) = \sqrt{x^2+y^2}\,\text{lb/ft}^2 -

    -     - -

    - M = 63\pi\approx 197.92\,\text{lb} -

    -     - -     - -     - - - -

    - Find the center of mass of the lamina described by the region R in the plane and its density function \delta(x,y). -

    - -

    - Note: these are the same lamina as in - . -

    -     - - - - -

    - R is the rectangle with corners (1,-3), (1,2), - (7,2) and (7,-3); - \delta(x,y) = 5\,\text{g/cm}^2 -

    -     - -

    - M = 150\,\text{g}; M_y=600; M_x=-75; - (\overline{x},\overline{y}) = (4,-0.5) -

    -     - -     - - - - -

    - R is the rectangle with corners (1,-3), (1,2), - (7,2) and (7,-3); - \delta(x,y) = (x+y^2)\,\text{g/cm}^2 -

    -     - -

    - M = 190\,\text{g}; M_y= 850; M_x = -315/2; - (\overline{x},\overline{y}) = (4.47,-0.83) -

    -     - -     - - - - -

    - R is the triangle with corners (-1,0), - (1,0), and (0,1); - \delta(x,y) = 2\,\text{lb/in}^2 -

    -     - -

    - M = 2\,\text{lb}; M_y= 0; M_x = 2/3; - (\overline{x},\overline{y}) = (0,1/3) -

    -     - -     - - - - -

    - R is the triangle with corners (0,0), - (1,0), and (0,1); - \delta(x,y) = (x^2+y^2+1)\,\text{lb/in}^2 -

    -     - -

    - M = 2/3\,\text{lb}; M_y= 7/30; M_x = 7/30; - (\overline{x},\overline{y}) = (0.35,0.35) -

    -     - -     - - - - -

    - R is the disk centered at the origin with radius 2; - \delta(x,y) = (x+y+4)\,\text{kg/m}^2 -

    -     - -

    - M = 16\pi\approx 50.27\,\text{kg}; - M_y= 4\pi; M_x = 4\pi; - (\overline{x},\overline{y}) = (1/4,1/4) -

    -     - -     - - - - -

    - R is the circle sector bounded by - x^2+y^2=25 in the first quadrant; - \delta(x,y) = (\sqrt{x^2+y^2}+1)\,\text{kg/m}^2 -

    -     - -

    - M = 325\pi/12\approx 85\,\text{kg}; M_y= 2375/12; - M_x = 2375/12; - (\overline{x},\overline{y}) = (2.33,2.33) -

    -     - -     - - - - -

    - R is the annulus in the first and second quadrants bounded by - x^2+y^2=9 and x^2+y^2=36; - \delta(x,y) = 4\,\text{lb/ft}^2 -

    -     - -

    - M = 54\pi\approx 169.65\,\text{lb}; - M_y= 0; M_x = 504; - (\overline{x},\overline{y}) = (0,2.97) -

    -     - -     - - - - -

    - R is the annulus in the first and second quadrants bounded by - x^2+y^2=9 and x^2+y^2=36; - \delta(x,y) = \sqrt{x^2+y^2}\,\text{lb/ft}^2 -

    -     - -

    - M = 63\pi\approx 197.92\,\text{lb}; - M_y= 0; M_x = 1215/2; - (\overline{x},\overline{y}) = (0,3.07) -

    -     - -     - -     - - - -

    - The moment of inertia - I is a measure of the tendency of a lamina to resist rotating about an axis or continue to rotate about an axis. - I_x is the moment of inertia about the x-axis, - I_y is the moment of inertia about the y-axis, - and I_O is the moment of inertia about the origin. - These are computed as follows: -

    - -

    -

      -
    • I_x = \iint_R y^2\, dm
      • - -
    • I_y = \iint_R x^2\, dm
      • - -
    • I_O = \iint_R \big(x^2+y^2\big)\, dm
      • -
    -

    - -

    - A lamina corresponding to a planar region R is given with a mass of 16 units. - For each, compute I_x, - I_y and I_O. -

    -     - - - - -

    - R is the 4\times 4 square with corners at (-2,-2) and (2,2) with density \delta(x,y) = 1. -

    -     - -

    - I_x = 64/3; I_y = 64/3; I_O = 128/3 -

    -     - -     - - - - -

    - R is the 8\times 2 rectangle with corners at (-4,-1) and (4,1) with density \delta(x,y) = 1. -

    -     - -

    - I_x = 16/3; I_y = 256/3; I_O = 272/3 -

    -     - -     - - - - -

    - R is the 4\times 2 rectangle with corners at (-2,-1) and (2,1) with density \delta(x,y) = 2. -

    -     - -

    - I_x = 16/3; I_y = 64/3; I_O = 80/3 -

    -     - -     - - - - -

    - R is the disk with radius 2 centered at the origin with density \delta(x,y) = 4/\pi. -

    -     - -

    - I_x = 16; I_y = 16; I_O = 32 -

    -     - -     - -     -     -     -
    -
    - Surface Area -

    - In - we used definite integrals to compute the arc length of plane curves of the form y=f(x). - We later extended these ideas to compute the arc length of plane curves defined by parametric or polar equations. -

    - -

    - The natural extension of the concept of - arc length over an interval - to surfaces is surface area over a region. -

    - -

    - Consider the surface z=f(x,y) over a region R in the xy-plane, - shown in . - Because of the domed shape of the surface, - the surface area will be greater than that of the area of the region R. - We can find this area using the same basic technique we have used over and over: - we'll make an approximation, - then using limits, we'll refine the approximation to the exact value. -

    - -
    - Developing a method of computing surface area - -
    - - - - - A surface is plotted above a circular domain. A partition of the domain corresponds to a grid system on the surface. - -

    - In space, a surface is plotted against a set of three-dimensional coordinate axes. - The surface appears to be a downward-opening elliptic paraboloid. -

    - -

    - A domain is shown in the xy plane; it appears to be bounded by a circle of radius 1, - centered at (1,0,0). - The domain is divided into many small rectangles, illustrating a partition like the ones used to define the double integral. -

    - -

    - The grid of rectangles in the domain corresponds to a grid of curves on the surface. - One of the rectangles in the domain is highlighted, along with the corresponding region on the surface. -

    -     - - - - - //ASY file for figsurfacearea_intro13D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.5,3.1,3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={-1,1}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-1,2.5); - pair ybounds=(-1.25,1.25); - pair zbounds=(0,2.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface//{z=-.5*(x-1)^2-.5*(y)^2+2}; - triple f(pair t) { - return (t.x,t.y,-.5*(t.x-1)^2-.5*(t.y)^2+2); - } - surface s=surface(f,(-0.221,-1),(2.2,1),12,20,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - pen pp=linewidth(.25mm); - //draw the grid in the xy-plane - //along fixed x - draw((0,-.3,0) -- (0,.3,0),pp); - draw((.2,-.4,0) -- (.2,.4,0),pp); - draw((0.4,-.6,0) -- (.4,.6,0),pp); - draw((0.6,-.7,0) -- (.6,.7,0),pp); - draw((.8,-.8,0) -- (.8,.8,0),pp); - draw((1,-.8,0) -- (1,.8,0),pp); - draw((1.2,-.8,0) -- (1.2,.8,0),pp); - draw((1.4,-.8,0) -- (1.4,.8,0),pp); - draw((1.6,-.7,0) -- (1.6,.7,0),pp); - draw((1.8,-.6,0) -- (1.8,.6,0),pp); - draw((2,-.4,0) -- (2,.4,0),pp); - //along fixed y - draw((.8,.8,0) -- (1.4,.8,0),pp); - draw((.6,.7,0) -- (1.6,.7,0),pp); - draw((.4,.6,0) -- (1.8,.6,0),pp); - draw((.4,.5,0) -- (1.8,.5,0),pp); - draw((.2,.4,0) -- (2,.4,0),pp); - draw((0,.3,0) -- (2,.3,0),pp); - draw((0,.2,0) -- (2,.2,0),pp); - draw((0,.1,0) -- (2,.1,0),pp); - draw((.8,-.8,0) -- (1.4,-.8,0),pp); - draw((.6,-.7,0) -- (1.6,-.7,0),pp); - draw((.4,-.6,0) -- (1.8,-.6,0),pp); - draw((.4,-.5,0) -- (1.8,-.5,0),pp); - draw((.2,-.4,0) -- (2,-.4,0),pp); - draw((0,-.3,0) -- (2,-.3,0),pp); - draw((0,-.2,0) -- (2,-.2,0),pp); - draw((0,-.1,0) -- (2,-.1,0),pp); - - //Draw curve on top of the grid in xy plane ({cos(x)*(1+cos(2*x))},{sin(x)*(1+cos(2*x))},0); - triple g(real t) {return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),0);} - path3 mypath=graph(g,-pi/2,pi/2,operator ..); - draw(mypath,bluepen); - - // draw curve on surface - triple g(real t) { - return (cos(t)*(1+cos(2*t)),sin(t)*(1+cos(2*t)),-.5*(cos(t)*(1+cos(2*t))-1)^2-.5*(sin(t)*(1+cos(2*t)))^2+2); - } - path3 mypath=graph(g,-pi/2,pi/2,operator ..); - draw(mypath,bluepen); - - //Draw the box in the plane and on the surface - draw((1.6,.3,1.775) -- (1.6,.4,1.74) -- (1.8,.4,1.6) -- (1.8,.3,1.635) --cycle,bluepen); - draw((1.6,.3,0) -- (1.6,.4,0) -- (1.8,.4,0) -- (1.8,.3,0) --cycle,bluepen); - - - - -
    - -
    - - - - - A zoomed-in view of a portion of the surface from the previous image, and one of the rectangular grid pieces. - -

    - This image is a zoomed-in view of the rectangular region that was highlighted on the surface in . - Below the surface, we see a single rectangle from the partition of the domain. - The sides of this rectangle are marked with lengths \Delta x_i and \Delta y_j. -

    - -

    - The rectangle in the xy plane corresponds to a small, rectangular patch on the surface. - This patch is approximated by a piece of the tangent plane at one point, - spanned by vectors \vec{u} and \vec{v}. - These vectors are labeled, but the point is not. -

    -     - - - - - //ASY file for figsurfacearea_intro23D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(3,3.5,4.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(0,1.25); - pair ybounds=(0,1.2); - pair zbounds=(0,2.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface//{-x^2+2}; - triple f(pair t) { - return (t.x,t.y,-t.x^2+2); - } - surface s=surface(f,(0,0),(1,1.2),12,12,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the box in the plane and on the surface - path3 pp=(.6,.4,0)--(.6,.6,0)--(.8,.6,0)--(.8,.4,0)--cycle; - draw(pp,bluepen); - label("$\Delta y_j$",(.9,.5,0));label("$\Delta x_i$",(.7,.7,0)); - path3 ppp=(.6,.4,2-.6^2)--(.6,.6,2-.6^2)--(.8,.6,2-.8^2)--(.8,.4,2-.8^2)--cycle; - draw(ppp,bluepen); - draw(surface(ppp),darksurfacepen,meshpen=p); - label("$\vec{v}$",(.9,.5,2-.85^2));label("$\vec{u}$",(.7,.7,2-.7^2)); - - - - -
    -     - -
    - -

    - As done to find the volume under a surface or the mass of a lamina, - we subdivide R into n subregions. - Here we subdivide R into rectangles, as shown in the figure. - One such subregion is outlined in the figure, - where the rectangle has dimensions \dx_i and \dy_j, - along with its corresponding region on the surface. -

    - -

    - In part of the figure, we zoom in on this portion of the surface. - When \dx_i and \dy_j are small, - the function is approximated well by the tangent plane at any point (x_i,y_j) in this subregion, - which is graphed in part . - In fact, the tangent plane approximates the function so well that in this figure, - it is virtually indistinguishable from the surface itself! - Therefore we can approximate the surface area S_{ij} of this region of the surface with the area - T_{ij} of the corresponding portion of the tangent plane. -

    - -

    - This portion of the tangent plane is a parallelogram, - defined by sides \vec u and \vec v, as shown. - One of the applications of the cross product from - is that the area of this parallelogram is \norm{\vec u\times \vec v}. - Once we can determine \vec u and \vec v, - we can determine the area. -

    - -

    - The vector \vec u is tangent to the surface in the direction of x, - therefore, from , - \vec u is parallel to \la 1,0,f_x(x_i,y_j)\ra. - The x-displacement of \vec u is \dx_i, - so we know that \vec u = \dx_i\la 1,0,f_x(x_i,y_j)\ra. - Similar logic shows that \vec v = \dy_j\la 0,1,f_y(x_i,y_j)\ra. - Thus: - - \text{surface area \(S_{ij}\)} \amp \approx \,\text{area of \(T_{ij}\)} - \amp = \norm{\vec u\times \vec v} - \amp = \norm{\dx_i\la 1,0,f_x(x_i,y_j)\ra\times\dy_j\la 0,1,f_y(x_i,y_j)\ra} - \amp =\sqrt{1+f_x(x_i,y_j)^2+f_y(x_i,y_j)^2}\dx_i\dy_j - . -

    - -

    - Note that \dx_i\dy_j = \Delta A_{ij}, - the area of the i,jth subregion. -

    - -

    - Summing up all n of the approximations to the surface area gives - - \text{surface area over \(R\)} \approx \sum_{i=1}^m\sum_{j=1}^n \sqrt{1+f_x(x_i,y_j)^2+f_y(x_i,y_j)^2}\Delta A_{ij} - . -

    - -

    - Once again take a limit as all of the \dx_i and \dy_j shrink to 0; - this leads to a double integral. -

    - - - Surface Area - -

    - Let z=f(x,y) where f_x and f_y are continuous over a closed, - bounded region R. - The surface area S over R is - surface area - - S \amp = \iint_R \, dS - \amp =\iint_R \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\, dA - . -

    -     -     - - -

    - We test this definition by using it to compute surface areas of known surfaces. - We start with a triangle. -

    - - - Finding the surface area of a plane over a triangle - -

    - Let f(x,y) = 4-x-2y, - and let R be the region in the plane bounded by x=0, - y=0 and y=2-x/2, - as shown in . - Find the surface area of f over R. -

    - -
    - Finding the area of a triangle in space in - - - - A triangle drawn on a plane in space. - -

    - Three-dimensional coordinate axes are drawn in space. - In the xy plane there is a triangle that illustrates the domain for this problem. - The triangle is bounded by the x and y axes, and the line from (4,0,0) to (0,2,0). -

    - -

    - This line also happens to be where the plane z=4-x-2y intersects the xy plane, - so it makes up one of the three sides of the triangle on the surface for this problem. - The other two sides are the intersections of z=4-x-2y with the xz and yz planes. -

    - -

    - Also shown is a line illustrating the altitude of the triangle, - as measured within the plane z=4-x-2y. -

    -     - - - - - //ASY file for figsurfacearea13D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(10.5,6.1,12); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={2,4}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,4.5); - pair ybounds=(-0.25,4.25); - pair zbounds=(0,5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane//{z=4-x-2y}; - triple f(pair t) { - return (t.x,t.y,4-t.x-2*t.y); - } - surface s=surface(f,(-0.25,-0.25),(4,2),4,4); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //draw the triangle on plane - draw((4,0,0)--(0,2,0)--(0,0,4)--(4,0,0),bluepen); - - //draw the triangle on xy plane - draw((4,0,0)--(0,2,0)--(0,0,0)--(4,0,0),bluepen+dashed+linewidth(2)); - - //draw red line on plane - //draw((0.8,1.6,0)--(0,0,4),redpen+linewidth(1)); - - - - -
    -     - -

    - We follow . - We start by noting that f_x(x,y) = -1 and f_y(x,y) = -2. - To define R, we use bounds - 0\leq y\leq 2-x/2 and 0\leq x\leq 4. - Therefore - - S \amp = \iint_R\, dS - \amp = \int_0^4\int_0^{2-x/2} \sqrt{1+(-1)^2+(-2)^2}\, dy\, dx - \amp = \int_0^4 \sqrt{6}\left(2-\frac x2\right)\, dx - \amp = 4\sqrt{6} - . -

    - -

    - Because the surface is a triangle, - we can figure out the area using geometry. - Considering the base of the triangle to be the side in the xy-plane, - we find the length of the base to be \sqrt{20}. - We can find the height using our knowledge of vectors: - let \vec u be the side in the xz-plane and let \vec v be the side in the xy-plane. - The height is then \norm {\vec u - \proj{u}{v}} = 4\sqrt{6/5}; - this is indicated by the additional line drawn in . - Geometry states that the area is thus - - \frac 12\cdot4\sqrt{6/5}\cdot\sqrt{20} = 4\sqrt{6} - . -

    - -

    - We affirm the validity of our formula. -

    -     -     - -

    - It is common knowledge that the surface area of a sphere of radius r is 4\pi r^2. - We confirm this in the following example, - which involves using our formula with polar coordinates. -

    - - - The surface area of a sphere - -

    - Find the surface area of the sphere with radius a centered at the origin, - whose top hemisphere has equation f(x,y)=\sqrt{a^2-x^2-y^2}. -

    -     - -

    - We start by computing partial derivatives and find - - f_x(x,y) = \frac{-x}{\sqrt{a^2-x^2-y^2}} \text{ and } f_y(x,y) = \frac{-y}{\sqrt{a^2-x^2-y^2}} - . -

    - -

    - As our function f only defines the top upper hemisphere of the sphere, - we double our surface area result to get the total area: - - S \amp = 2\iint_R \sqrt{1+ f_x(x,y)^2+f_y(x,y)^2}\, dA - \amp = 2\iint_R \sqrt{1+ \frac{x^2+y^2}{a^2-x^2-y^2}}\, dA - . -

    - -

    - The region R that we are integrating over is bounded by the circle, - centered at the origin, with radius a: x^2+y^2=a^2. - Because of this region, - we are likely to have greater success with our integration by converting to polar coordinates. - Using the substitutions x=r\cos(\theta), y=r\sin(\theta), - dA = r\, dr\, d\theta and bounds - 0\leq\theta\leq2\pi and 0\leq r\leq a, we have: - - S \amp = 2\int_0^{2\pi}\int_0^a \sqrt{1+\frac{r^2\cos^2(\theta) +r^2\sin^2(\theta) }{a^2-r^2\cos^2(\theta) -r^2\sin^2(\theta) }}\, r\, dr\, d\theta - \amp =2\int_0^{2\pi}\int_0^ar\sqrt{1+\frac{r^2}{a^2-r^2}}\, dr\, d\theta - \amp =2\int_0^{2\pi}\int_0^ar\sqrt{\frac{a^2}{a^2-r^2}}\, dr\, d\theta. - Apply substitution u=a^2-r^2 and integrate the inner integral, giving - \amp = 2\int_0^{2\pi} a^2\, d\theta - \amp = 4\pi a^2 - . -

    - -

    - Our work confirms our previous formula. -

    -     -     - - - - - Finding the surface area of a cone - -

    - The general formula for a right cone with height h and base radius a is - - \ds f(x,y) = h-\frac{h}a\sqrt{x^2+y^2} - , - shown in . - Find the surface area of this cone. -

    - -
    - Finding the surface area of a cone in - - - - A downward-opening circular cone, with vertex on the z axis. - -

    - A circular cone, with vertex on the z axis at a point (0,0,h). - The cone opens downward toward the xy plane, - and ends at the xy plane with a circular base of radius a. -

    -     - - - - - //ASY file for figsurfacearea33D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.25,3.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the cone {x*cos(y)}, {x*sin(y)},{1-x} - triple f(pair t) { - return (t.x*cos(t.y),t.x*sin(t.y),1.5*(2-t.x)); - } - surface s=surface(f,(0,0),(2,2*pi),12,12,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //labels a and h - label("$h$",(.2,-.2,3)); - label("$a$",(2.2,.2,0)); - label("$a$",(.2,2.2,0)); - - - - -
    -     - -

    - We begin by computing partial derivatives. - - f_x(x,y) = -\frac{xh}{a\sqrt{x^2+y^2}} \text{ and } f_y(x,y)=-\frac{yh}{a\sqrt{x^2+y^2}} - . -

    - -

    - Since we are integrating over the disk bounded by x^2+y^2=a^2, - we again use polar coordinates. - Using the standard substitutions, our integrand becomes - - \sqrt{1+ \left(\frac{hr\cos(\theta) }{a\sqrt{r^2}}\right)^2 + \left(\frac{hr\sin(\theta) }{a\sqrt{r^2}}\right)^2} - . -

    - -

    - This may look intimidating at first, - but there are lots of simple simplifications to be done. - It amazingly reduces to just - - \sqrt{1+\frac{h^2}{a^2}} = \frac1a\sqrt{a^2+h^2} - . -

    - -

    - Our polar bounds are 0\leq\theta\leq2\pi and 0\leq r\leq a. - Thus - - S \amp = \int_0^{2\pi}\int_0^ar\frac1a\sqrt{a^2+h^2}\, dr\, d\theta - \amp = \int_0^{2\pi} \left.\left(\frac12r^2\frac1a\sqrt{a^2+h^2}\right)\right|_0^ad\theta - \amp = \int_0^{2\pi} \frac12a\sqrt{a^2+h^2} \, d\theta - \amp = \pi a\sqrt{a^2+h^2} - . -

    - -

    - This matches the formula found in the back of this text. -

    - - -     -     - - - Finding surface area over a region - -

    - Find the area of the surface - f(x,y) = x^2-3y+3 over the region R bounded by - -x\leq y\leq x, 0\leq x\leq 4, - as pictured in . -

    - -
    - Graphing the surface in - - - - A triangular region is shown on a trough-like surface - -

    - The surface z=f(x,y) is a parabolic cylinder; - it curves upward, and has a downward slope relative to the yz plane. -

    - -

    - The surface is plotted against the usual three-dimensional coordinate axes. - In the xy plane we see the triangular outline of the domain for this problem. - The vertices of this triangle are at (4,4,0), (4,-4,0), and (0,0,0). -

    - -

    - On the surface, the curves corresponding to the three sides of this triangle are shown, - illustrating the triangular region whose surface area is computed in this example. -

    -     - - - - - //ASY file for figsurfacearea43D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5.3,17,74); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={-4,-2,2,4}; - real[] myzchoice={20}; - defaultpen(0.5mm); - - pair xbounds=(-.25,5); - pair ybounds=(-5,5); - pair zbounds=(-1,25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2-3y+3 - triple f(pair t) { - return (t.x,t.y,t.x^2-3*t.y+3); - } - surface s=surface(f,(-0.25,-4.5),(4.5,4.5),12,12,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //triangle in xy plane - draw((0,0,0)--(4,4,0)--(4,-4,0)--cycle,bluepen+dashed+linewidth(1)); - - //triangle on surface - triple g(real t) {return (t,t,t^2-3*t+3);} - path3 mypath=graph(g,0,4,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,-t,t^2+3*t+3);} - path3 mypath=graph(g,0,4,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (4,t,16-3*t+3);} - path3 mypath=graph(g,-4,4,operator ..);draw(mypath,bluepen+linewidth(2)); - - - - -
    -     - -

    - It is straightforward to compute - f_x(x,y) = 2x and f_y(x,y) = -3. - Thus the surface area is described by the double integral - - \iint_R \sqrt{1+(2x)^2+(-3)^2}\, dA = \iint_R \sqrt{10+4x^2}\, dA - . -

    - -

    - As with integrals describing arc length, - double integrals describing surface area are in general hard to evaluate directly because of the square-root. - This particular integral can be easily evaluated, - though, with judicious choice of our order of integration. -

    - -

    - Integrating with order dx\, dy requires us to evaluate \int \sqrt{10+4x^2}\, dx. - This can be done, - though it involves Integration By Parts and \sinh^{-1}(x). - Integrating with order dy\, dx has as its first integral \int \sqrt{10+4x^2}\, dy, - which is easy to evaluate: - it is simply y\sqrt{10+4x^2}+C. - So we proceed with the order dy\, dx; - the bounds are already given in the statement of the problem. - - \iint_R\sqrt{10+4x^2}\, dA \amp = \int_0^4\int_{-x}^x\sqrt{10+4x^2}\, dy \, dx - \amp = \int_0^4\left.\big(y\sqrt{10+4x^2}\big)\right|_{-x}^x dx - \amp =\int_0^4\big(2x\sqrt{10+4x^2}\big)\, dx. - Apply substitution with u = 10+4x^2: - \amp = \left.\left(\frac16\big(10+4x^2\big)^{3/2}\right)\right|_0^4 - \amp = \frac13\big(37\sqrt{74}-5\sqrt{10}\big) \approx 100.825\,\text{units}^2 - . -

    - -

    - So while the region R over which we integrate has an area of 16 square units, - the surface has a much greater area as its z-values change dramatically over R. -

    -     -     - -

    - In practice, - technology helps greatly in the evaluation of such integrals. - High powered computer algebra systems can compute integrals that are difficult, - or at least time consuming, by hand, - and can at the least produce very accurate approximations with numerical methods. - In general, just knowing how - to set up the proper integrals brings one very close to being able to compute the needed value. - Most of the work is actually done in just describing the region R in terms of polar or rectangular coordinates. - Once this is done, technology can usually provide a good answer. -

    - -

    - We have learned how to integrate integrals; - that is, we have learned to evaluate double integrals. - In the next section, - we learn how to integrate double integrals that is, - we learn to evaluate triple integrals, - along with learning some uses for this operation. -

    - - - - Terms and Concepts - - - - -

    - Surface area is related to what previously studied concept? -

    -     - - - -     - - - - -

    - To approximate the area of a small portion of a surface, - we computed the area of its plane. -

    -     - - - - - - - - -     - - - - -

    - We interpret \ds \iint_R\, dS as - sum up lots of little . -

    -     - - - - - - - - - surface area - - -

    - Your answer should be plural. -

    -     -     -     - -     - - - - -

    - Why is it important to know how to set up a double integral to compute surface area, - even if the resulting integral is hard to evaluate? -

    -     - - - -

    - Technology makes good approximations accessible, if not exact answers. -

    -     - -     - - - - -

    - Why do z=f(x,y) and z=g(x,y)=f(x,y)+h, - for some real number h, - have the same surface area over a region R? -

    -     - - - -

    - Intuitively, - adding h to f only shifts f up (, parallel to the z-axis) and does not change its shape. - Therefore it will not change the surface area over R. -

    - -

    - Analytically, f_x = g_x and f_y=g_y; - therefore, the surface area of each is computed with identical double integrals. -

    -     - -     - - - - -

    - Let f(x,y) be a function defined over a region R and let g(x,y)=2f(x,y). - Why is the surface area of z=g(x,y) over R not twice the surface area of z=f(x,y) over R? -

    -     - - - -

    - Analytically, g_x = 2f_x and g_y=2f_y. - The double integral to compute the surface area of f over R is \ds \iint_R \sqrt{1+f_x^2+f_y^2}\, dA; - the double integral to compute the surface area of g over R is \ds \iint_R\sqrt{1+4f_x^2+4f_y^2}\, dA, - which is not twice the double integral used to calculate the surface area of f. -

    -     - -     -     - - - Problems - - - -

    - Set up the iterated integral that computes the surface area of the graph of the given function over the region R. -

    -     - - - - -

    - f(x,y) = \sin(x) \cos(y);R is the rectangle with bounds - 0\leq x\leq 2\pi, 0\leq y\leq2\pi. -

    - - - - - A bumpy surface with several peaks and valleys. - -

    - The surface z=\sin(x)\cos(y) is plotted over a rectangular domain. - The domain is large enough that several of the peaks and valleys typical of this surface can be seen. -

    - -

    - Note that the image is somewhat decorative in this case: - the formula for surface area can be applied to the given function without knowing the appearance of the surface. -

    -     - - - - - //ASY file for fig13_05_ex_053D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={5}; - real[] myychoice={5}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,7); - pair ybounds=(-0.5,7); - pair zbounds=(-1.25,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface sin(x)cos(y) - triple f(pair t) { - return (t.x,t.y,sin(t.x)*cos(t.y)); - } - surface s=surface(f,(0,0),(2*pi,2*pi),12,12,Spline); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p); - - //lines in xy plane - draw((0,0,0)--(0,2*pi,0),bluepen+dashed+linewidth(2)); - draw((2*pi,0,0)--(2*pi,2*pi,0),bluepen+dashed+linewidth(2)); - - //lines on surface - triple g(real t) {return (t,0,sin(t));} - path3 mypath=graph(g,0,2*pi,operator ..);draw(mypath,bluepen+dashed+linewidth(2)); - triple g(real t) {return (t,2*pi,sin(t));} - path3 mypath=graph(g,0,2*pi,operator ..);draw(mypath,bluepen+dashed+linewidth(2)); - - - - -     - -

    - \ds SA = \int_0^{2\pi}\int_0^{2\pi} \sqrt{1+ \cos^2(x) \cos^2(y) +\sin^2(x) \sin^2(y) }\, dx\, dy -

    -     - -     - - - - -

    - \ds f(x,y) = \frac{1}{x^2+y^2+1};R is bounded by the circle x^2+y^2=9. -

    - - - - - A steep surface with a single peak, resembling a witch's hat. - -

    - The surface z=\frac{1}{x^2+y^2+1} has rotational symmetry about the z axis. - It resembles a steep mountain with a smooth peak, or perhaps the hat of a witch. -

    - -

    - Again, the image is primarily decorative; it is sufficient to know the function and the region to determine the surface area. -

    -     - - - - - //ASY file for fig13_05_ex_063D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(13,13,1.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-4,-2,2,4}; - real[] myychoice={-4,-2,2,4}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-4.25,4.25); - pair ybounds=(-4.25,4.25); - pair zbounds=(-0.25,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface 1/(1+x^2+y^2) - triple f(pair t) { - return (t.x,t.y,1/(1+t.x^2+t.y^2)); - } - surface s=surface(f,(-4,-4),(4,4),12,12,Spline); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p); - - //circle - triple g(real t) {return (3*cos(t),3*sin(t),1/10);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+dashed+linewidth(1)); - - - - -     - -

    - \ds SA = \int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \sqrt{1+ \frac{4x^2+4y^2}{(1+x^2+y^2)^4}}\, dx\, dy -

    - -

    - Polar offers simpler bounds: -

    - -

    - \ds SA = \int_0^{2\pi}\int_0^3 r\sqrt{1+\frac{4r^2}{(1+r^2)^4}}\, dr\, d\theta -

    -     - -     - - - - -

    - \ds f(x,y) = x^2-y^2;R is the rectangle with opposite corners (-1,-1) and (1,1). -

    - - - - - A saddle surface, with its saddle point at the origin in three dimensions. - -

    - A standard hyperbolic paraboloid, or saddle surface. - The saddle point is at the origin. - The surface curves upward along the x axis, and downward along the y axis. -

    -     - - - - - //ASY file for fig13_05_ex_073D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(2.2,4.1,3.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface x^2-y^2 - triple f(pair t) { - return (t.x,t.y,t.x^2-t.y^2); - } - surface s=surface(f,(-1.2,-1.2),(1.2,1.2),12,12,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p); - - //lines on surface - triple g(real t) {return (t,1,t^2-1);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); - triple g(real t) {return (t,-1,t^2-1);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); - triple g(real t) {return (1,t,1-t^2);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); - triple g(real t) {return (-1,t,1-t^2);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); - - - - -     - -

    - \ds SA = \int_{-1}^{1}\int_{-1}^{1} \sqrt{1+ 4x^2+4y^2}\, dx\, dy -

    -     - -     - - - - -

    - \ds f(x,y) = \frac{1}{e^{x^2}+1};R is the rectangle bounded by -

    - -

    - -5\leq x\leq 5 and 0\leq y\leq 1. -

    - - - - - A mostly flat surface with a ridge along the y axis - -

    - A graph of the function f(x,y) = \frac{1}{e^{x^2}+1} in three dimensions. - The surface is mostly flat, with a ridge along the y axis of height 1/2. -

    -     - - - - - //ASY file for fig13_05_ex_083D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.6,3.9,3.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-5,5}; - real[] myychoice={-1,1}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-6,6); - pair ybounds=(-1,1.5); - pair zbounds=(-0.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface 1/(exp(x^2)+1) - triple f(pair t) { - return (t.x,t.y,1/(exp(t.x^2)+1)); - } - surface s=surface(f,(-6,-0.5),(6,1.5),16,12,Spline); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p); - - //lines on surface - draw((5,0,0)--(5,1,0),bluepen+dashed+linewidth(1.5)); - draw((-5,0,0)--(-5,1,0),bluepen+dashed+linewidth(1.5)); - triple g(real t) {return (t,0,1/(exp(t^2)+1));} - path3 mypath=graph(g,-5,5,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); - triple g(real t) {return (t,1,1/(exp(t^2)+1));} - path3 mypath=graph(g,-5,5,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); - - - - -     - -

    - \ds SA = \int_{-5}^{5}\int_{0}^{1} \sqrt{1+ \frac{4x^2e^{2x^2}}{\big(1+e^{x^2}\big)^4}}\, dy\, dx -

    -     - -     - -     - - - -

    - Find the area of the given surface over the region R. -

    -     - - - - -

    - z = 3x-7y+2; - R is the rectangle with opposite corners (-1,0) and (1,3). -

    -     - -

    - \ds SA = \int_{0}^{3}\int_{-1}^{1} \sqrt{1+ 9+49}\, dx\, dy = 6\sqrt{59} \approx 46.09 -

    -     - -     - - - - -

    - z = 2x+2y+2; - R is the triangle with corners (0,0), - (1,0) and (0,1). -

    -     - -

    - \ds SA = \int_{0}^{1}\int_{0}^{1-x} \sqrt{1+ 4+4}\, dy\, dx = 18 -

    -     - -     - - - - -

    - z = x^2+y^2+10; R is bounded by the circle x^2+y^2=16. -

    -     - -

    - This is easier in polar: - - SA \amp = \int_{0}^{2\pi}\int_{0}^{4} r\sqrt{1+ 4r^2\cos^2(t) +4r^2\sin^2(t) }\, dr\, d\theta - \amp = \int_0^{2\pi}\int_0^4r\sqrt{1+4r^2}\, dr\, d\theta - \amp = \frac{\pi}{6}\big(65\sqrt{65}-1\big) \approx 273.87 - -

    -     - -     - - - - -

    - z = -2x+4y^2+7 over R, - the triangle bounded by y=-x, - y=x, 0\leq y\leq 1. -

    -     - -

    - - SA \amp = \int_{0}^{1}\int_{-y}^{y} \sqrt{1+ 4+64y^2}\, dx\, dy - \amp = \int_0^{1}\big(2y\sqrt{5+64y^2}\big)\, dy - \amp = \frac1{96}\big(69\sqrt{69}-5\sqrt{5}\big)\approx 5.85 - -

    -     - -     - - - - -

    - z = x^2+y over R, - the triangle bounded by y=2x, - y=0 and x=2. -

    -     - -

    - - SA \amp = \int_{0}^{2}\int_{0}^{2x} \sqrt{1+ 1+4x^2}\, dy\, dx - \amp = \int_0^{2}\big(2x\sqrt{2+4x^2}\big)\, dx - \amp = \frac{26}{3}\sqrt{2}\approx 12.26 - -

    -     - -     - - - - -

    - z = \frac23x^{3/2}+2y^{3/2} over R, - the rectangle with opposite corners (0,0) and (1,1). -

    -     - -

    - - SA \amp = \int_{0}^{1}\int_{0}^{1} \sqrt{1+ x+9y}\, dx\, dy - \amp = \int_0^{1}\frac23\Big((9y+2)^{3/2}-(9y+1)^{3/2}\Big)\, dy - \amp = \frac{4}{135}\big(121\sqrt{11}-100\sqrt{10}-4\sqrt{2}+1\big)\approx 2.383 - -

    -     - -     - - - - -

    - z = 10-2\sqrt{x^2+y^2} over the region R - bounded by the circle x^2+y^2=25. - (This is the cone with height 10 and base radius 5; - be sure to compare your result with the known formula.) -

    -     - -

    - This is easier in polar: - - SA \amp = \int_{0}^{2\pi}\int_{0}^{5} r\sqrt{1+ \frac{4r^2\cos^2(t) +4r^2\sin^2(t) }{r^2\sin^2(t) +r^2\cos^2(t) }}\, dr\, d\theta - \amp = \int_0^{2\pi}\int_0^5r\sqrt{5}\, dr\, d\theta - \amp = 25\pi\sqrt{5}\approx 175.62 - -

    -     - -     - - - - -

    - Find the surface area of the sphere with radius 5 by doubling the surface area of f(x,y) = \sqrt{25-x^2-y^2} over the region R - bounded by the circle x^2+y^2=25. - (Be sure to compare your result with the known formula.) -

    -     - -

    - This is easier in polar: - - SA \amp = 2\int_{0}^{2\pi}\int_{0}^{5} r\sqrt{1+ \frac{r^2\cos^2(t) +r^2\sin^2(t) }{25-r^2\sin^2(t) -r^2\cos^2(t) }}\, dr\, d\theta - \amp = 2\int_0^{2\pi}\int_0^5r\sqrt{\frac{1}{25-r^2}}\, dr\, d\theta - \amp = 100\pi\approx 314.16 - -

    -     - -     - - - - -

    - Find the surface area of the ellipse formed by restricting the plane - f(x,y) = cx+dy+h to the region R bounded by - the circle x^2+y^2=1, - where c, d and h are some constants. - Your answer should be given in terms of c and d; - why does the value of h not matter? -

    -     - -

    - Integrating in polar is easiest considering R: - - SA \amp = \int_{0}^{2\pi}\int_{0}^{1} r\sqrt{1+ c^2+d^2}\, dr\, d\theta - \amp = \int_0^{2\pi}\frac12\Big(\sqrt{1+c^2+d^2}\Big)\, dy - \amp = \pi\sqrt{1+c^2+d^2} - . -

    - -

    - The value of h does not matter as it only shifts the plane vertically (, parallel to the z-axis). - Different values of h do not create different ellipses in the plane. -

    -     - -     - -     -     -     -
    -
    - Volume Between Surfaces and Triple Integration - - Volume between surfaces -

    - We learned in - how to compute the signed volume V under a surface - z=f(x,y) over a region R: - V = \iint_R f(x,y)\, dA. - It follows naturally that if f(x,y)\geq g(x,y) on R, - then the volume between f(x,y) and g(x,y) on R is - - V = \iint_R f(x,y)\, dA - \iint_R g(x,y)\, dA = \iint_R \big(f(x,y)-g(x,y)\big)\, dA - . -

    - - - Volume Between Surfaces - -

    - Let f and g be continuous functions on a closed, - bounded region R, - where f(x,y)\geq g(x,y) for all (x,y) in R. - The volume V between f and g over R is - volume - - V =\iint_R \big(f(x,y)-g(x,y)\big)\, dA - . -

    -     -     - - - Finding volume between surfaces - -

    - Find the volume of the space region bounded by the planes z=3x+y-4, - z=8-3x-2y, x=0 and y=0. - - In the planes are drawn; - in , only the defined region is given. -

    - -
    - Finding the volume between the planes given in - -
    - - - - - Two planes in space intersect along a line. - -

    - Two planes are shown in space relative to a set of three-dimensional coordinate axes. - Each plane is plotted over the same rectangular domain in the xy plane, - and drawn in the shape of a parallelogram. -

    - -

    - The planes intersect along a line. - With respect to the parallelogram shapes used in the image, - one pair of opposite corners for each plane lies along the line of intersection. -

    - -

    - When viewed from above, the line of intersection divides the rectangular domain into two triangles. -

    -     - - - - - //ASY file for figtrip13D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5.4,-12.7,20.2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2,4}; - real[] myzchoice={-5,5}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,3); - pair ybounds=(-0.5,5); - pair zbounds=(-5.25,10); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane z=3x+y-4 - triple f(pair t) { - return (t.x,t.y,3*t.x+t.y-4); - } - surface s=surface(f,(0,0),(2,4),1,1,Spline); - pen p=apexmeshpen; - draw(s,simplesurfacepen,meshpen=bluepen+linewidth(1.5)); - - //Draw the plane z=8-3x-2y - triple f(pair t) { - return (t.x,t.y,8-3*t.x-2*t.y); - } - surface s=surface(f,(0,0),(2,4),1,1,Spline); - pen p=apexmeshpen; - draw(s,simplesurfacepen2,meshpen=redpen+linewidth(1.5)); - - //line - draw((0,4,0)--(2,0,2),dashed+linewidth(1.5)); - - //lines on surface - //draw((5,0,0)--(5,1,0),bluepen+dashed+linewidth(1.5)); - //draw((-5,0,0)--(-5,1,0),bluepen+dashed+linewidth(1.5)); - //triple g(real t) {return (t,0,1/(exp(t^2)+1));} - //path3 mypath=graph(g,-5,5,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); - //triple g(real t) {return (t,1,1/(exp(t^2)+1));} - //path3 mypath=graph(g,-5,5,operator ..);draw(mypath,bluepen+dashed+linewidth(1.5)); - - - - -
    - -
    - - - - - Two planes in space plotted over a triangular domain. The planes intersect along one edge of the triangle. - -

    - This is another drawing of the same two planes as , - but this time, the planes are plotted over a triangular domain. - The sides of the triangle correspond to the x and y axes, - and the line of intersection of the two planes. -

    - -

    - Each plane is therefore drawn as a triangle in space. - The two triangles meet along an edge, - and the opposite vertex of each triangle is along the z axis. -

    - -

    - One triangle lies below the other, if we measure height relative to the z axis. - These two triangles make up two of the four faces of a tetrahedron. - The other two faces lie in the xz and yz coordinate planes, respectively. -

    -     - - - - - //ASY file for figtrip1b3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(5.4,-12.7,20.2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2,4}; - real[] myzchoice={-5,5}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,3); - pair ybounds=(-0.5,5); - pair zbounds=(-5.25,10); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //lines of the triangles - draw((0,4,0)--(2,0,2)--(0,0,8)--cycle,redpen+linewidth(1.5)); - draw((0,0,-4)--(0,0,8)--(2,0,2)--cycle,black+linewidth(1.5)); - draw((0,0,-4)--(0,4,0),bluepen+linewidth(1.5)); - - //shade - import three; - path3 p = (0,4,0)--(2,0,2)--(0,0,8); //Left - //draw(surface(p -- cycle), emissive(rgb(1,.4,.7)+opacity(0.7))); - draw(surface(p -- cycle), simplesurfacepen2); - - path3 p = (0,0,-4)--(0,4,0)--(2,0,2); //Left - draw(surface(p -- cycle), simplesurfacepen); - - path3 p = (0,0,-4)--(0,0,8)--(2,0,2); //Left - draw(surface(p -- cycle), rgb(.7,.7,.7)+opacity(.2)); - - path3 p = (0,0,-4)--(0,0,8)--(0,4,0); //Left - draw(surface(p -- cycle), rgb(.7,.7,.7)+opacity(.2)); - - //surfacepen2 - - // to get perspectives right with figtrip1a, draw with invisible pen - - draw((2,0,2)--(2,4,6),invisible); - draw((2,0,2)--(2,4,-6),invisible); - - - - -
    -     - -
    -     - -

    - We need to determine the region R over which we will integrate. - To do so, we need to determine where the planes intersect. - They have common z-values when 3x+y-4=8-3x-2y. - Applying a little algebra, we have: - - 3x+y-4 \amp = 8-3x-2y - 6x+3y \amp =12 - 2x+y \amp =4 - -

    - -

    - The planes intersect along the line 2x+y=4. - Therefore the region R is bounded by x=0, - y=0, and y=4-2x; - we can convert these bounds to integration bounds of - 0\leq x\leq 2, 0\leq y\leq 4-2x. - Thus - - V \amp = \iint_R \big(8-3x-2y-(3x+y-4)\big)\, dA - \amp = \int_0^2\int_0^{4-2x} \big(12-6x-3y\big)\, dy\, dx - \amp = 16\,\text{units}^3 - . -

    - -

    - The volume between the surfaces is 16 cubic units. -

    -     -     - -

    - In the preceding example, we found the volume by evaluating the integral - - \ds \int_0^2\int_0^{4-2x} \big(8-3x-2y-(3x+y-4)\big)\, dy\, dx - . -

    - -

    - Note how we can rewrite the integrand as an integral, - much as we did in : - - 8-3x-2y-(3x+y-4) = \int_{3x+y-4}^{8-3x-2y}\, dz - . -

    - -

    - Thus we can rewrite the double integral that finds volume as - - \int_0^2\int_0^{4-2x} \big(8-3x-2y-(3x+y-4)\big)\, dy\, dx = \int_0^2\int_0^{4-2x}\left(\int_{3x+y-4}^{8-3x-2y}\, dz\right)\, dy\, dx - . -

    - -

    - This no longer looks like a double integral, - but more like a triple integral. - Just as our first introduction to double integrals was in the context of finding the area of a plane region, - our introduction into triple integrals will be in the context of finding the volume of a space region. -

    - -
    - Approximating the volume of a region D in space - -
    - - - - - An ellipsoid in space, centered at the origin. - -

    - An ellipsoid in space, plotted against three-dimensional coordinate axes. - It is egg-shaped, and is longest along the x axis. -

    - -

    - The grid lines on the ellipsoid are like lines of latitude and longitude. -

    -     - - - - - //ASY file for figtripintro3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - - pair xbounds=(-2.25,2.25); - pair ybounds=(-2.25,2.25); - pair zbounds=(-2.25,2.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Ellipsoid - triple f(pair t) { - return (cos(t.x)*2*cos(t.y),sin(t.x)*cos(t.y),sin(t.y)); - } - surface s=surface(f,(-pi,-pi/2),(1*pi,pi/2),16,16,Spline); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p); - - - - -
    - -
    - - - - - A zoomed-in view of a surface, showing grid lines. A small rectangular prism illustrates a small piece of volume. - -

    - A zoomed in view of a portion of the ellipsoid in where it meets the positive y axis. - The grid lines on the surface are horizontal (traces with constant z value) and vertical (traces by planes through the z axis), - like lines of latitude and longitude. -

    - -

    - On one of the rectangles formed by the grid, - a small rectangular box is drawn. - The image illustrates how we can approximate the volume of an object like an ellipsoid by dividing it up - into small cube-like pieces. -

    -     - - - - - //ASY file for figtripintroa3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(0,2.25); - pair ybounds=(0.25,1.75); - pair zbounds=(-0.5,0.5); - - //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Ellipsoid - triple f(pair t) { - return (cos(t.x)*2*cos(t.y),sin(t.x)*cos(t.y),sin(t.y)); - } - surface s=surface(f,(pi/6,-pi/3),(2*pi/3,pi/3),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p); - - //draw red cube - //draw((0., 1.05, -0.105) -- (0,1.05,.105) -- (-.345,1.05,0.105) -- (-0.345, 1.05, -0.105)--cycle,redpen);//front - - draw((0.13, 1.05, 0.26) -- (0.13, 1.05, 0) -- (-0.26, 1.05, 0) -- (-0.25, 1.05, 0.26)--cycle,redpen);//front - - draw((0.13, 0.9, 0.26) -- (0.13, 0.9, 0) -- (-0.26, 0.9, 0) -- (-0.25, 0.9, 0.26)--cycle,redpen);//back - - //draw((-.345,0.979,0.105)--(0,0.979,0.105)--(0,0.979,-0.105)--(-.345,0.979,-0.105)--cycle,redpen);//back - //sides - draw((0.13, 0.9, 0.26)--(0.13, 1.05, 0.26),redpen); - draw((0.13, 0.9, 0)--(0.13, 1.05, 0),redpen); - draw((-0.26, 0.9, 0)--(-0.26, 1.05, 0),redpen); - draw((-0.25, 0.9, 0.26)--(-0.25, 1.05, 0.26),redpen); - - - - -
    -     - -
    - -

    - To formally find the volume of a closed, - bounded region D in space, - such as the one shown in , - we start with an approximation. - Break D into n rectangular solids; - the solids near the boundary of D may possibly not include portions of D and/or include extra space. - In , - we zoom in on a portion of the boundary of D to show a rectangular solid that contains space not in D; - as this is an approximation of the volume, - this is acceptable and this error will be reduced as we shrink the size of our solids. -

    - -

    - The volume \Delta V_i of the - ith solid D_i is \Delta V_i = \dx_i\dy_i\ddz_i, - where \dx_i, - \dy_i and \ddz_i give the dimensions of the rectangular solid in the x, - y and z directions, respectively. - By summing up the volumes of all n solids, - we get an approximation of the volume V of D: - - V \approx \sum_{i=1}^n \Delta V_i = \sum_{i=1}^n \dx_i\dy_i\ddz_i - . -

    - -

    - Let \norm{\Delta D} represent the length of the longest diagonal of rectangular solids in the subdivision of D. - As \norm{\Delta D}\to 0, - the volume of each solid goes to 0, as do each of \dx_i, - \dy_i and \ddz_i, for all i. - Our calculus experience tells us that taking a limit as - \norm{\Delta D}\to 0 turns our approximation of V into an exact calculation of V. - Before we state this result in a theorem, - we use a definition to define some terms. -

    - - - Triple Integrals, Iterated Integration (Part I) - -

    - Let D be a closed, bounded region in space. - Let a and b be real numbers, - let g_1(x) and g_2(x) be continuous functions of x, - and let f_1(x,y) and - f_2(x,y) be continuous functions of x and y. - integrationtriple - triple integral - iterated integration -

    - -

    -

      -
    1. -

      - The volume V of D is denoted by a - triple integral, - - V = \iiint_D dV - . -

      -
      2. - -
    2. -

      - The iterated integral \ds \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \, dz\, dy\, dx is evaluated as - - \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \, dz\, dy\, dx=\int_a^b\int_{g_1(x)}^{g_2(x)}\left(\int_{f_1(x,y)}^{f_2(x,y)} \, dz\right)\, dy\, dx - . - Evaluating the above iterated integral is - triple integration. -

      -
      3. -
    -

    -     -     - -

    - Our informal understanding of the notation \iiint_D\, dV is - sum up lots of little volumes over D, - analogous to our understanding of - \iint_R\, dA and \iint_R\, dm. -

    - - - -

    - We now state the major theorem of this section. -

    - - - Triple Integration (Part I) - -

    - Let D be a closed, - bounded region in space and let - \Delta D be any subdivision of D into n rectangular solids, - where the ith subregion D_i has dimensions - \dx_i\times\dy_i\times\ddz_i and volume \Delta V_i. - integrationtriple - triple integral - iterated integration -

    - -

    -

      -
    1. -

      - The volume V of D is - - V = \iiint_D\, dV = \lim_{\norm{\Delta D}\to0} \sum_{i=1}^n \Delta V_i = \lim_{\norm{\Delta D}\to0} \sum_{i=1}^n \dx_i\dy_i\ddz_i - . -

      -
      2. - -
    2. -

      - If D is defined as the region bounded by the planes x=a and x=b, - the cylinders y=g_1(x) and y=g_2(x), - and the surfaces z=f_1(x,y) and - z=f_2(x,y), where a\lt b, - g_1(x)\leq g_2(x) and f_1(x,y)\leq f_2(x,y) on D, then - - \iiint_D \, dV = \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \, dz\, dy\, dx - . -

      -
      3. - -
    3. -

      - V can be determined using iterated integration with other orders of integration - (there are 6 total), - as long as D is defined by the region enclosed by a pair of planes, - a pair of cylinders, and a pair of surfaces. -

      -
      4. -
    -

    -     -     - -

    - We evaluated the area of a plane region R by iterated integration, - where the bounds were from curve to curve, - then from point to point. - allows us to find the volume of a space region with an iterated integral with bounds - from surface to surface, then from curve to curve, - then from point to point. In the iterated integral - - \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \, dz\, dy\, dx - , - the bounds a\leq x\leq b and - g_1(x)\leq y\leq g_2(x) define a region R in the xy-plane over which the region D exists in space. - However, these bounds are also defining surfaces in space; - x=a is a plane and y=g_1(x) is a cylinder. - The combination of these 6 surfaces enclose, and define, - D. -

    - -

    - Examples will help us understand triple integration, - including integrating with various orders of integration. -

    - - - - - Finding the volume of a space region with triple integration - -

    - Find the volume of the space region in the first octant bounded by the plane z=2-y/3-2x/3, - shown in , - using the order of integration dz\, dy\, dx. - Set up the triple integrals that give the volume in the other 5 orders of integration. -

    -
    - The region D used in - - - - A triangular portion of a plane in space. It lies in the first octant and has vertices on the coordinate axes. - -

    - A plane in space is illustrated, by showing the portion of the plane the lies in the first octant. - This portion of the plane is a triangle, whose vertices lie on the coordinate axes. - Using the image (or the equation z=2-y/3-2x/3), - we can determine that the plane meets the coordinate axes at the points (3,0,0), (0,6,0), and (0,0,2). -

    - -

    - The region in space whose volume is computed in this problem is a tetrahedron, - whose faces are the plane described above, - as well as the triangles formed in the three coordinate planes by the coordinate axes and the edges of the first triangle. -

    -     - - - - - //ASY file for figtrip23D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.2,10.1,6.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={5}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,5); - pair ybounds=(-0.5,7); - pair zbounds=(-0.5,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //triangle in space - draw((3,0,0)--(0,6,0)--(0,0,2)--cycle,bluepen+linewidth(1.5)); - - //shade - import three; - path3 p = (3,0,0)--(0,6,0)--(0,0,2); //Left - draw(surface(p -- cycle), simplesurfacepen); - - - - -
    -     - -

    - Starting with the order of integration dz\, dy\, dx, - we need to first find bounds on z. - The region D is bounded below by the plane z=0 - (because we are restricted to the first octant) - and above by z=2-y/3-2x/3; - 0\leq z\leq 2-y/3-2x/3. -

    - -

    - To find the bounds on y and x, - we collapse the region onto the xy-plane, - giving the triangle shown in . (We know the equation of the line y=6-2x in two ways. - First, by setting z=0, - we have 0 = 2-y/3-2x/3 \Rightarrow y=6-2x. - Secondly, we know this is going to be a straight line between the points (3,0) and (0,6) in the xy-plane.) -

    - -
    - The region found by collapsing D onto the xy-plane - - - - A triangle in the xy plane, plotted in space relative to 3D coordinate axes. - -

    - A set of three-dimensional coordinate axes is shown, with the z axis pointing upward. - In the xy plane a triangle is drawn. - The sides of the triangle are formed by the x axis, the y axis, - and the line from (3,0,0) to (0,6,0), - which is one edge of the triangle in space shown in . - This line is labeled with the equation y-6-2x. -

    - -

    - This triangle represents the shadow of the plane in space on the xy plane, - when viewed directly from above. - It will be used to determine the bounds in the triple integral when we integrate first with respect to z. -

    -     - - - - - //ASY file for figtrip2b3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.2,10.1,6.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={5}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,5); - pair ybounds=(-0.5,7); - pair zbounds=(-0.5,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //triangle in space - draw((3,0,0)--(0,6,0)--(0,0,0)--cycle,bluepen+linewidth(1.5)); - - //shade - import three; - path3 p = (3,0,0)--(0,6,0)--(0,0,0); //Left - draw(surface(p -- cycle), simplesurfacepen); - label("$y=6-2x$",(2.5,3.5,0)); - - - - -
    - -

    - We define that region R, - in the integration order of dy\, dx, - with bounds 0\leq y\leq 6-2x and 0\leq x\leq 3. - Thus the volume V of the region D is: - - V \amp = \iiint_D \, dV - \amp = \int_0^3\int_0^{6-2x}\int_0^{2-\frac 13y-\frac 23x}\, dz\, dy\, dx - \amp = \int_0^3\int_0^{6-2x}\left(\int_0^{2-\frac 13y-\frac 23x}\, dz\right)\, dy\, dx - \amp =\int_0^3\int_0^{6-2x}z\Big|_0^{2-\frac 13y-\frac 23x}\, dy\, dx - \amp = \int_0^3\int_0^{6-2x}\left(2-\frac 13y-\frac 23x\right)\, dy\, dx - . - From this step on, we are evaluating a double integral as done many times before. We skip these steps and give the final volume, V=6. -

    - -

    - The order dz\, dx\, dy: -

    - -

    - Now consider the volume using the order of integration dz\, dx\, dy. - The bounds on z are the same as before, - 0\leq z\leq 2-y/3-2x/3. - Collapsing the space region on the xy-plane as shown in , - we now describe this triangle with the order of integration dx\, dy. - This gives bounds 0\leq x\leq 3-y/2 and 0\leq y\leq 6. - Thus the volume is given by the triple integral - - V = \int_0^6\int_0^{3-\frac12y}\int_0^{2-\frac13y-\frac23x}\, dz\, dx\, dy - . -

    - -

    - The order dx\, dy\, dz: -

    - -

    - Following our surface to surface\ldots strategy, - we need to determine the x-surfaces - that bound our space region. - To do so, approach the region from behind, - in the direction of increasing x. - The first surface we hit as we enter the region is the yz-plane, - defined by x=0. - We come out of the region at the plane z=2-y/3-2x/3; - solving for x, we have x= 3-y/2-3z/2. - Thus the bounds on x are: - 0\leq x\leq 3-y/2-3z/2. -

    - -

    - Now collapse the space region onto the yz-plane, - as shown in . (Again, - we find the equation of the line z=2-y/3 by setting x=0 in the equation - x=3-y/2-3z/2.) We need to find bounds on this region with the order dy\, dz. - The curves that bound y are y=0 and y=6-3z; - the points that bound z are 0 and 2. - Thus the triple integral giving volume is: -

    - - -

    - - 0 \amp \leq x\leq 3-y/2-3z/2 - 0 \amp \leq y\leq 6-3z - 0 \amp \leq z\leq 2 - -

    - -

    - \Rightarrow -

    - -

    - - \int_0^2\int_0^{6-3z}\int_0^{3-y/2-3z/2}\, dx\, dy\, dz - . -

    -     - -

    - The order dx\, dz\, dy: -

    - -
    - The region D in is collapsed onto the yz-plane in (a); in (b), the region is collapsed onto the xz-plane - -
    - - - - - A triangle plotted in the yz plane, relative to a set of 3D coordinate axes. - -

    - A three-dimensional coordinate system is shown, with the z axis pointing upward. - A triangle is plotted in the yz plane. - The edges of the triangle are the y and z coordinate axes, - and the line from (0,6,0) to (0,0,2), - where the plane in space from meets the yz plane. - This line is labeled with the equation z=2-y/3. -

    - -

    - This is the shadow of the region of integration for this problem on the yz plane, - when viewed along the x axis. - It is used to set up the bounds of integration when we integrate first with respect to x. -

    -     - - - - - //ASY file for figtrip2c3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.2,10.1,6.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={5}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,5); - pair ybounds=(-0.5,7); - pair zbounds=(-0.5,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //triangle in space - draw((0,0,2)--(0,6,0)--(0,0,0)--cycle,bluepen+linewidth(1.5)); - - //shade - import three; - path3 p = (0,0,2)--(0,6,0)--(0,0,0); //Left - draw(surface(p -- cycle), simplesurfacepen); - label("$z=2-y/3$",(0,2,1.5),E); - - - - -
    - -
    - - - - - A triangle plotted in the xz plane, relative to a set of 3D coordinate axes. - -

    - A three-dimensional coordinate system is shown, with the z axis pointing upward. - A triangle is plotted in the xz plane. - The edges of the triangle are the x and z coordinate axes, - and the line from (3,0,0) to (0,0,2), - where the plane in space from meets the xz plane. - This line is labeled with the equation z=2-2x/3. -

    - -

    - This is the shadow of the region of integration for this problem on the xz plane, - when viewed along the y axis. - It is used to set up the bounds of integration when we integrate first with respect to y. -

    -     - - - - - //ASY file for figtrip2d3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.2,10.1,6.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={5}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,5); - pair ybounds=(-0.5,7); - pair zbounds=(-0.5,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //triangle in space - draw((0,0,2)--(3,0,0)--(0,0,0)--cycle,bluepen+linewidth(1.5)); - - //shade - import three; - path3 p = (0,0,2)--(3,0,0)--(0,0,0); //Left - draw(surface(p -- cycle), simplesurfacepen); - label("$z=2-2x/3$",(1,0,1.5),W); - - - - -
    -     - -
    - -

    - The x-bounds are the same as the order above. - We now consider the triangle in and describe it with the order dz\, dy: - 0\leq z\leq 2-y/3 and 0\leq y\leq 6. - Thus the volume is given by: -

    - - -

    - - 0 \amp \leq x\leq 3-y/2-3z/2 - 0 \amp \leq z\leq 2-y/3 - 0 \amp \leq y\leq 6 - -

    - -

    - \Rightarrow -

    - -

    - - \int_0^6\int_0^{2-y/3}\int_0^{3-y/2-3z/2}\, dx\, dz\, dy - . -

    -     - -

    - The order dy\, dz\, dx: -

    - -

    - We now need to determine the y-surfaces that determine our region. - Approaching the space region from behind - and moving in the direction of increasing y, - we first enter the region at y=0, - and exit along the plane z= 2-y/3-2x/3. - Solving for y, this plane has equation y = 6-2x-3z. - Thus y has bounds 0\leq y\leq 6-2x-3z. -

    - -

    - Now collapse the region onto the xz-plane, - as shown in . - The curves bounding this triangle are z=0 and z=2-2x/3; - x is bounded by the points x=0 to x=3. - Thus the triple integral giving volume is: -

    - - -

    - - 0 \amp \leq y\leq 6-2x-3z - 0 \amp \leq z\leq 2-2x/3 - 0 \amp \leq x\leq 3 - -

    - -

    - \Rightarrow -

    - -

    - - \int_0^3\int_0^{2-2x/3}\int_0^{6-2x-3z}\, dy\, dz\, dx - . -

    -     - -

    - The order dy\, dx\, dz: -

    - -

    - The y-bounds are the same as in the order above. - We now determine the bounds of the triangle in using the order dy\, dx\, dz. - x is bounded by x=0 and x=3-3z/2; - z is bounded between z=0 and z=2. - This leads to the triple integral: -

    - - -

    - - 0 \amp \leq y\leq 6-2x-3z - 0 \amp \leq x\leq 3-3z/2 - 0 \amp \leq z\leq 2 - -

    - -

    - \Rightarrow -

    - -

    - - \int_0^2\int_0^{3-3z/2}\int_0^{6-2x-3z}\, dy\, dx\, dz - . -

    -     - -

    - This problem was long, but hopefully useful, - demonstrating how to determine bounds with every order of integration to describe the region D. - In practice, - we only need 1, but being able to do them all gives us flexibility to choose the order that suits us best. -

    -     -     - -

    - In the previous example, - we collapsed the surface into the x-y, x-z, - and yz-planes as we determined the curve to curve, - point to point bounds of integration. - Since the surface was a triangular portion of a plane, - this collapsing, or projecting, was simple: - the projection of a straight line in space onto a coordinate plane is a line. -

    - -

    - The following example shows us how to do this when dealing with more complicated surfaces and curves. -

    - - - Finding the projection of a curve in space onto the coordinate planes - -

    - Consider the surfaces z=3-x^2-y^2 and z=2y, - as shown in . - The curve of their intersection is shown, - along with the projection of this curve into the coordinate planes, - shown dashed. - Find the equations of the projections into the coordinate planes. -

    -
    - Finding the projections of the curve of intersection in - -
    - - - - - A surface in space intersects with a plane through the origin, forming a parabolic curve in space. - -

    - Two surfaces are plotted relative to a set of three-dimensional coordinate axes. - One surface is a portion of a downward-opening circular paraboloid; - the other is a plane through the origin. -

    - -

    - The plane cuts the paraboloid along a curve, which is a parabola in space. -

    -     - - - - - //ASY file for fig3d_proj3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(6.7,1.9,11.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={1,2}; - real[] myzchoice={2,4}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,2.5); - pair ybounds=(-0.5,2.25); - pair zbounds=(-0.5,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //plane z=2y - triple f(pair t) { - return (t.x,t.y,2*t.y); - } - surface s=surface(f,(0,0),(2,1.5),1,1,Spline); - pen p=redpen+.1mm; - draw(s,surfacepen2,meshpen=p); - - //z=3-x^2-y^2 - triple f(pair t) { - return (t.x,t.y,3-t.x^2-t.y^2); - } - surface s=surface(f,(-.1,-.1),(2,1.5),8,8,Spline); - //triple f(pair t) { - // return (cos(t.x)*t.y,sin(t.x)*t.y,3-t.y^2-t.y^2); - //} - // - //surface s=surface(f,(-.1,0),(pi/2+.1,2),8,8,Spline); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p); - - //lines and curves on the surfaces - draw((0,0,0)--(0,1,2),bluepen+dashed+linewidth(1)); - triple g(real t) {return (t,0,2*(1-(t/2)^2));} - path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+dashed+linewidth(1)); - triple g(real t) {return (t,1-(t/2)^2,2*(1-(t/2)^2));} - path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,1-(t/2)^2,0);} - path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+dashed+linewidth(1)); - - - - -
    - -
    - - - - - A curve in space, and its projections onto the three coordinate planes. - -

    - The curve of intersection of the two surfaces in - is plotted relative to a set of three dimensional coordinate axes, - with the surfaces removed to highlight the curve. -

    - -

    - The curve is a parabolic arc in space, from the point (2,0,0) on the x axis, - to the point (0,1,2) on the yz plane. -

    - -

    - Three other curves are shown, these are the projections of the original curve onto the coordinate planes. -

      -
    • -

      - In the xy plane, there is a parabolic arc from (2,0,0) on the x axis, to (0,1,0) on the y axis. -

      -
      • -
    • -

      - In the xz plane, there is a parabolic arc from (2,0,0) on the x axis, to (0,0,2) on the z axis. -

      -
      • -
    • -

      - The projection onto the yz plane is not a parabola; - rather, it is a straight line segment, from the origin to the point (0,1,2). -

      -
      • -
    -

    -     - - - - - //ASY file for fig3d_projb3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(6.7,1.9,11.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={1,2}; - real[] myzchoice={2,4}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,2.5); - pair ybounds=(-0.5,2.25); - pair zbounds=(-0.5,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //plane z=2y - triple f(pair t) { - return (t.x,t.y,2*t.y); - } - surface s=surface(f,(0,0),(2,1.5),1,1,Spline); - pen p=apexmeshpen; - //draw(s,rgb(1,.4,.7)+opacity(.7),meshpen=p); - - //z=3-x^2-y^2 - triple f(pair t) { - return (t.x,t.y,3-t.x^2-t.y^2); - } - surface s=surface(f,(0,0),(2,1.5),8,8,Spline); - pen p=apexmeshpen; - draw(s,invisible,meshpen=invisible); - - //lines and curves on the surfaces - draw((0,0,0)--(0,1,2),bluepen+dashed+linewidth(1)); - triple g(real t) {return (t,0,2*(1-(t/2)^2));} - path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+dashed+linewidth(1)); - triple g(real t) {return (t,1-(t/2)^2,2*(1-(t/2)^2));} - path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,1-(t/2)^2,0);} - path3 mypath=graph(g,0,2,operator ..);draw(mypath,bluepen+dashed+linewidth(1)); - - - - -
    -     - -
    -     - -

    - The two surfaces are z=3-x^2-y^2 and z=2y. - To find where they intersect, - it is natural to set them equal to each other: 3-x^2-y^2=2y. - This is an implicit function of x and y that gives all points (x,y) in the xy-plane where the z values of the two surfaces are equal. -

    - -

    - We can rewrite this implicit function by completing the square: - - 3-x^2-y^2=2y \Rightarrow y^2+2y+x^2=3 \Rightarrow (y+1)^2+x^2=4 - . -

    - -

    - Thus in the xy-plane the projection of the intersection is a circle with radius 2, centered at (0,-1). -

    - -

    - To project onto the xz-plane, - we do a similar procedure: - find the x and z values where the y values on the surface are the same. - We start by solving the equation of each surface for y. - In this particular case, it works well to actually solve for y^2: -

    - -

    - z=3-x^2-y^2 \Rightarrow y^2=3-x^2-z -

    - -

    - z=2y \Rightarrow y^2=z^2/4. -

    - -

    - Thus we have (after again completing the square): - - 3-x^2-z = z^2/4 \Rightarrow \frac{(z+2)^2}{16}+\frac{x^2}4=1 - , - and ellipse centered at (0,-2) in the xz-plane with a major axis of length 8 and a minor axis of length 4. -

    - -

    - Finally, to project the curve of intersection into the yz-plane, - we solve equation for x. - Since z=2y is a cylinder that lacks the variable x, - it becomes our equation of the projection in the yz-plane. -

    - -

    - All three projections are shown in . -

    -     -     - - - - - Finding the volume of a space region with triple integration - -

    - Set up the triple integrals that find the volume of the space region D bounded by the surfaces x^2+y^2=1, - z=0 and z=-y, - as shown in , - with the orders of integration dz\, dy\, dx, - dy\, dx\, dz and dx\, dz\, dy. -

    - -
    - The region D in is shown in (a); in (b), it is collapsed onto the xy-plane - -
    - - - - - A cylindrical wedge in space, resembling a wedge cut from a tree. - -

    - An illustration of the region of integration in . - The region is a three-dimensional solid cut from a cylinder. -

    - -

    - The plane z=-y meets the xy plane along the x axis. - The region that lies between these planes, and within the cylinder x^2+y^2=1, - is a semi-circular wedge. - It resembles the shape of a piece cut from a tree that is being cut down. -

    -     - - - - - //ASY file for figtrip33D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.1,1.9,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,-0.5}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,0.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple f(pair t) {return (cos(t.x),sin(t.x),-sin(t.x)*t.y);} - surface s=surface(f,(-pi,0),(0,1),16,2,Spline); - pen p=apexmeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - triple f(pair t) {return (cos(t.x)*t.y,sin(t.x)*t.y,-sin(t.x)*t.y);} - surface s=surface(f,(-pi,0),(0,1),16,2,Spline); - pen p=redpen+.1mm; - draw(s,simplesurfacepen2,meshpen=p); - - //disk x^2+y^2=1 in 3rd and 4th quadrants - triple g(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(g,pi,2*pi,operator ..);draw(mypath,bluepen+linewidth(1.5)); - - //surface above disk x^2+y^2=1 in 3rd and 4th quadrants - triple g(real t) {return (cos(t),sin(t),-sin(t));} - path3 mypath=graph(g,pi,2*pi,operator ..); - draw(mypath,bluepen+linewidth(1.5)); - - //label the surfaces - label("$x^2+y^2=1$",(1.4,-.5,0)); - draw((1.1,-.6,0)--(.7,-.7,.35),Arrow3(size=2mm)); - label("$z=-y$",(0,-.5,1.25)); - draw((0,-.5,1.15)--(0,-.5,.5),Arrow3(size=2mm)); - - - - -
    - -
    - - - - - The projection of the cylindrical wedge in the previous image onto the xy plane. It is a semi-circular region. - -

    - In three dimensions, the projection of the solid in onto the xy plane is shown. - The projection forms a semi-circular region in the plane, - between the x axis and the half of the circle x^2+y^2=1 with y\leq 0. -

    -     - - - - - //ASY file for figtrip3b3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.1,1.9,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,-0.5}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,0.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //disk x^2+y^2=1 in 3rd and 4th quadrants - triple g(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(g,pi,2*pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - //line - draw((-1,0,0)--(1,0,0),bluepen+linewidth(2)); - - int k=24;//number of panels - - //shade bottom - import three; - for (int i=k; i<2*k; ++i) - { - path3 p = (0,0,0)--(cos(i*pi/k),sin(i*pi/k),0)--(cos((i+1)*pi/k),sin((i+1)*pi/k),0); //Left - draw(surface(p -- cycle), simplesurfacepen); - } - - - - -
    -     - -
    -     - -

    - The order dz\, dy\, dx: -

    - -

    - The region D is bounded below by the plane z=0 and above by the plane z=-y. - The cylinder x^2+y^2=1 does not offer any bounds in the z-direction, - as that surface is parallel to the z-axis. - Thus 0\leq z\leq -y. -

    - -

    - Collapsing the region into the xy-plane, - we get part of the circle with equation - x^2+y^2=1 as shown in . - As a function of x, - this half circle has equation y=-\sqrt{1-x^2}. - Thus y is bounded below by - -\sqrt{1-x^2} and above by y=0: - -\sqrt{1-x^2}\leq y\leq 0. - The x bounds of the half circle are -1\leq x\leq 1. - All together, - the bounds of integration and triple integral are as follows: -

    - - -

    - - 0 \amp \leq z\leq -y - -\sqrt{1-x^2}\leq \amp y\leq 0 - -1 \amp \leq x\leq 1 - -

    - -

    - \Rightarrow -

    - -

    - - \int_{-1}^1\int_{-\sqrt{1-x^2}}^{0}\int_0^{-y}\, dz\, dy\, dx - . -

    -     - -

    - We evaluate this triple integral: - - \int_{-1}^1\int_{-\sqrt{1-x^2}}^{0}\int_0^{-y}\, dz\, dy\, dx \amp = \int_{-1}^1\int_{-\sqrt{1-x^2}}^{0}\big(-y\big)\, dy\, dx - \amp =\int_{-1}^1\big(-\frac12y^2\big)\Big|_{-\sqrt{1-x^2}}^{0}\, dx - \amp = \int_{-1}^1 \frac12\big(1-x^2\big)\, dx - \amp = \left.\left(\frac12\left(x-\frac13x^3\right)\right)\right|_{-1}^1 - \amp = \frac23\text{ units } ^3 - . -

    - -

    - With the order dy\, dx\, dz: -

    - -

    - The region is bounded below - in the y-direction by the surface x^2+y^2=1 \Rightarrow y=-\sqrt{1-x^2} and above - by the surface y=-z. - Thus the y bounds are -\sqrt{1-x^2}\leq y\leq -z. -

    - -

    - Collapsing the region onto the xz-plane gives the region shown in ; - this half disk is bounded by z=0 and x^2+z^2=1. - (We find this curve by solving each surface for y^2, - then setting them equal to each other. - We have y^2=1-x^2 and y=-z\Rightarrow y^2=z^2. - Thus x^2+z^2=1.) - It is bounded below by x=-\sqrt{1-z^2} and above by x=\sqrt{1-z^2}, - where z is bounded by 0\leq z\leq 1. - All together, we have: -

    - - -

    - - -\sqrt{1-x^2} \amp \leq y\leq -z - -\sqrt{1-z^2} \amp \leq x\leq \sqrt{1-z^2} - 0 \amp \leq z\leq 1 - -

    - -

    - \Rightarrow -

    - -

    - - \int_{0}^1\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\int_{-\sqrt{1-x^2}}^{-z}\, dy\, dx\, dz - . -

    -     - -
    - The region D in is shown collapsed onto the xz-plane in (a); in (b), it is collapsed onto the yz-plane - -
    - - - - - The projection of the solid in this example onto the xy plane; it is a semi-circular region. - -

    - On a set of three-dimensional coordinate axes, - the projection of the solid in onto the xz plane is shown. - It is a semi-circular region, bounded below by the x axis, - and bounded above by the semi-circle x^2+z^2=1, z\geq 0. -

    -     - - - - - //ASY file for figtrip3c3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.1,1.9,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,-0.5}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,0.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //disk x^2+y^2=1 in 3rd and 4th quadrants - triple g(real t) {return (cos(t),0,-sin(t));} - path3 mypath=graph(g,pi,2*pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - //line - draw((-1,0,0)--(1,0,0),bluepen+linewidth(2)); - - int k=24;//number of panels - - //shade bottom - import three; - for (int i=k; i<2*k; ++i) - { - path3 p = (0,0,0)--(cos(i*pi/k),0,-sin(i*pi/k))--(cos((i+1)*pi/k),0,-sin((i+1)*pi/k)); //Left - draw(surface(p -- cycle), simplesurfacepen); - } - - - - -
    - -
    - - - - - The projection of the solid in this example onto the yz plane; it is a triangle. - -

    - Using the same three-dimensional coordinate system, - the projection of the solid in onto the yz plane is shown. - This projection is a triangle. The base of the triangle lies along the negative y axis, - from the origin to (0,-1,0). - Another side is the vertical line y=-1 in the yz plane, from (0,-1,0) to (0,-1,1). - The remaining side is the line z=-y, from (-1,0,1) to the origin. -

    -     - - - - - //ASY file for figtrip3d3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.1,1.9,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,-0.5}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,0.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //triangle in space - draw((0,0,0)--(0,-1,0)--(0,-1,1)--cycle,bluepen+linewidth(2)); - - //shade triangle - import three; - - path3 p = (0,0,0)--(0,-1,0)--(0,-1,1); //Left - draw(surface(p -- cycle), simplesurfacepen); - - - - -
    -     - -
    - -

    - With the order dx\, dz\, dy: -

    - -

    - D is bounded below by the surface - x=-\sqrt{1-y^2} and above by \sqrt{1-y^2}. - We then collapse the region onto the yz-plane and get the triangle shown in . (The hypotenuse is the line z=-y, - just as the plane.) Thus z is bounded by - 0\leq z\leq -y and y is bounded by -1\leq y\leq 0. - This gives: -

    - - -

    - - -\sqrt{1-y^2} \amp \leq x\leq \sqrt{1-y^2} - 0 \amp \leq z\leq -y - -1 \amp \leq y\leq 0 - -

    - -

    - \Rightarrow -

    - -

    - - \int_{-1}^0\int_{0}^{-y}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\, dx\, dz\, dy - . -

    -     -     -     - -

    - The following theorem states two things that should make - common sense to us. - First, using the triple integral to find volume of a region D should always return a positive number; - we are computing volume here, - not signed volume. - Secondly, to compute the volume of a - complicated region, - we could break it up into subregions and compute the volumes of each subregion separately, - summing them later to find the total volume. -

    - - - Properties of Triple Integrals - -

    - Let D be a closed, bounded region in space, - and let D_1 and D_2 be non-overlapping regions such that D=D_1\bigcup D_2. - triple integralproperties - iterated integrationproperties -

    - -

    -

      -
    1. -

      - \ds \iiint_D \, dV \geq 0 -

      -
      2. - -
    2. -

      - \ds \iiint_D\, dV = \iiint_{D_1}\, dV + \iiint_{D_2}\, dV. -

      -
      3. -
    -

    -     -     - -

    - We use this latter property in the next example. -

    - - - Finding the volume of a space region with triple integration - -

    - Find the volume of the space region D bounded by the coordinate planes, - z=1-x/2 and z=1-y/4, - as shown in . - Set up the triple integrals that find the volume of D in all 6 orders of integration. -

    - -
    - The region D in is shown in (a); in (b), it is collapsed onto the xy-plane - -
    - - - - - A pyramid with a rectangular base in the plane z=0 and its peak on the z axis. - -

    - The region of integration for is a pyramid with a rectangular base. - The base lies in the xy plane, and the peak is on the z axis at (0,0,1). - The other four sides are triangles that lie above the xy plane: -

      -
    • -

      - One side is in the xz plane, with vertices (0,0,0), (2,0,0) and (0,0,1). -

      -
      • -
    • -

      - One side is in the yz plane, with vertices (0,0,0), (0,4,0), and (0,0,1) -

      -
      • -
    • -

      - Another side lies in a plane labeled with the equation z=1-\frac12 x. - The vertices of this triangle are (2,0,0), (2,4,0), and (0,0,1). -

      -
      • -
    • -

      - The last side lies in a plane labeled with the equation z=1-\frac14 y. - The vertices of this triangle are (0,4,0), (2,4,0), and (0,0,1). -

      -
      • -
    -

    -     - - - - - //ASY file for figtrip43D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11.2,8.3,2.7); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.1,4.25); - pair ybounds=(-0.1,4.25); - pair zbounds=(-0.1,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //lines for tetrahedron - draw((0,0,0)--(2,0,0)--(2,4,0)--(0,4,0)--cycle,bluepen+linewidth(2)); - draw((0,0,0)--(0,0,1),bluepen+linewidth(2)); - draw((2,0,0)--(0,0,1),bluepen+linewidth(2)); - draw((0,4,0)--(0,0,1),bluepen+linewidth(2)); - draw((2,4,0)--(0,0,1),bluepen+linewidth(2)); - - //shade rectangle and sides - import three; - path3 p = (0,0,0)--(2,0,0)--(2,4,0)--(0,4,0); - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0,0,0)--(2,0,0)--(0,0,1); - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0,0,0)--(0,4,0)--(0,0,1); - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0,4,0)--(2,4,0)--(0,0,1); - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (2,0,0)--(2,4,0)--(0,0,1); - draw(surface(p -- cycle), simplesurfacepen); - - //labels and arrows - label("$z=1-\frac{1}{2}x$",(3.5,1.65,0)); - draw((3,1.5,0)--(1.8,2,.1),Arrow3(size=2mm)); - label("$z=1-\frac{1}{4}y$",(0,3,1)); - draw((0,3,0.8)--(1,3,0.25),Arrow3(size=2mm)); - - - - -
    - -
    - - - - - The base of the pyramid shown in the previous image. It is a rectangle, divided into two triangles along one diagonal. - -

    - The base of the pyramid in is shown in the xy plane, - in a three-dimensional coordinate system. -

    - -

    - A dashed line in the plane runs from the origin to the opposite corner of the rectangle, at the point (2,4,0). - This line divides the rectangle into two triangles. - One triangle lies between the x axis and the dashed line, and is labeled R_1. - The other triangle lies between the y axis and the dashed line, and is labeled R_2. -

    -     - - - - - //ASY file for figtrip4d3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11.2,8.3,2.7); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={4}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.1,4.25); - pair ybounds=(-0.1,4.25); - pair zbounds=(-0.1,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //rectangle in xy plane w diagonal and labels - draw((0,0,0)--(2,0,0)--(2,4,0)--(0,4,0)--cycle,bluepen+linewidth(2)); - draw((0,0,0)--(2,4,0),bluepen+dashed+linewidth(1)); - label("$R_1$",(1.5,1,.1)); - label("$R_2$",(1,3,.1)); - - //shade rectangle - import three; - path3 p = (0,0,0)--(2,0,0)--(2,4,0)--(0,4,0); //Left - draw(surface(p -- cycle), simplesurfacepen); - - - - -
    -     - -
    -     - -

    - Following the bounds-determining strategy of - surface to surface, curve to curve, - and point to point, we can see that the most difficult orders of integration are the two in which we integrate with respect to z first, - for there are two upper - surfaces that bound D in the z-direction. - So we start by noting that we have - - 0\leq z\leq 1-\frac12x \text{ and } 0\leq z\leq 1-\frac14y - . -

    - -

    - We now collapse the region D onto the xy-axis, - as shown in . - The boundary of D, - the line from (0,0,1) to (2,4,0), - is shown in as a dashed line; - it has equation y=2x. (We can recognize this in two ways: - one, in collapsing the line from (0,0,1) to (2,4,0) onto the xy-plane, - we simply ignore the z-values, - meaning the line now goes from (0,0) to (2,4). - Secondly, the two surfaces meet where z=1-x/2 is equal to z=1-y/4: - thus 1-x/2=1-y/4 \Rightarrow y=2x.) -

    - -

    - We use the second property of to state that - - \iiint_D \, dV = \iiint_{D_1}\, dV + \iiint_{D_2}\, dV - , - where D_1 and D_2 are the space regions above the plane regions R_1 and R_2, - respectively. - Thus we can say - - \iiint_D\, dV = \iint_{R_1}\left(\int_0^{1-x/2}\, dz\right)dA + \iint_{R_2}\left(\int_0^{1-y/4}\, dz\right)dA - . -

    - -

    - All that is left is to determine bounds of R_1 and R_2, - depending on whether we are integrating with order dx\, dy or dy\, dx. - We give the final integrals here, - leaving it to the reader to confirm these results. -

    - -

    - dz\, dy\, dx: -

    - -

    - - \amp\quad 0 \leq z\leq 1-x/2 \amp \amp\quad 0 \leq z\leq 1-y/4 - \amp\quad 0 \leq y\leq 2x \amp \amp\quad 2x \leq y\leq 4 - \amp\quad 0 \leq x\leq 2 \amp \amp\quad 0 \leq x\leq 2 - \amp\amp\amp - \iiint_D\, dV = \amp \int_0^2\int_0^{2x}\int_0^{1-x/2}\, dz\, dy\, dx \amp +\quad \amp \int_0^2\int_{2x}^4\int_0^{1-y/4}\, dz\, dy\, dx - -

    - -

    - dz\, dx\, dy: -

    - -

    - - \amp\quad 0 \leq z\leq 1-x/2\amp \amp\quad 0 \leq z\leq 1-y/4 - \amp\quad y/2 \leq x\leq 2\amp \amp\quad 0 \leq x\leq y/2 - \amp\quad 0 \leq y\leq 4\amp \amp\quad 0 \leq y\leq 4 - \amp\amp\amp - \iiint_D\, dV = \amp \int_0^4\int_{y/2}^{2}\int_0^{1-x/2}\, dz\, dx\, dy \amp +\quad \amp \int_0^4\int_{0}^{y/2}\int_0^{1-y/4}\, dz\, dx\, dy - -

    - -

    - The remaining four orders of integration do not require a sum of triple integrals. - In - we show D collapsed onto the other two coordinate planes. - Using these graphs, we give the final orders of integration here, - again leaving it to the reader to confirm these results. -

    - -
    - The region D in is shown collapsed onto the xz-plane in (a); in (b), it is collapsed onto the yz-plane - -
    - - - - - The projection of the pyramid onto the plane y=0 is a triangle. - -

    - In a three-dimensional coordinate system, - the side of the pyramid in in the xz plane is shown. - This represents the projection of the solid onto the xz plane, - and it is a triangle, with vertices (0,0,0), (2,0,0), and (0,0,1). -

    -     - - - - - //ASY file for figtrip4b3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11.2,8.3,2.7); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.1,4.25); - pair ybounds=(-0.1,4.25); - pair zbounds=(-0.1,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //lines for tetrahedron - draw((0,0,0)--(2,0,0)--(0,0,1)--cycle,bluepen+linewidth(2)); - - //shade rectangle and sides - import three; - path3 p = (0,0,0)--(2,0,0)--(0,0,1); - draw(surface(p -- cycle), simplesurfacepen); - - - - -
    - -
    - - - - - The projection of the pyramid onto the plane x=0 is a triangle. - -

    - In a three-dimensional coordinate system, - the side of the pyramid in in the yz plane is shown. - This represents the projection of the solid onto the yz plane, - and it is a triangle, with vertices (0,0,0), (0,4,0), and (0,0,1). -

    -     - - - - - //ASY file for figtrip4c3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={2,4}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.1,4.25); - pair ybounds=(-0.1,4.25); - pair zbounds=(-0.1,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //lines for tetrahedron - draw((0,0,0)--(0,4,0)--(0,0,1)--cycle,bluepen+linewidth(2)); - - //shade rectangle and sides - import three; - path3 p = (0,0,0)--(0,4,0)--(0,0,1); - draw(surface(p -- cycle), simplesurfacepen); - - - - -
    -     - -
    - -

    - dy\, dx\, dz: -

    - - -

    - - 0 \amp \leq y\leq 4-4z - 0 \amp \leq x\leq 2-2z - 0 \amp \leq z\leq 1 - -

    - -

    - \Rightarrow -

    - -

    - - \int_0^1\int_{0}^{2-2z}\int_0^{4-4z}\, dy\, dx\, dz - . -

    -     - -

    - dy\, dz\, dx: -

    - - -

    - - 0 \amp \leq y\leq 4-4z - 0 \amp \leq z\leq 1-x/2 - 0 \amp \leq x\leq 2 - -

    - -

    - \Rightarrow -

    - -

    - - \int_0^2\int_{0}^{1-x/2}\int_0^{4-4z}\, dy\, dx\, dz - . -

    -     - -

    - dx\, dy\, dz: -

    - - -

    - - 0 \amp \leq x\leq 2-2z - 0 \amp \leq y\leq 4-4z - 0 \amp \leq z\leq 1 - -

    - -

    - \Rightarrow -

    - -

    - - \int_0^1\int_{0}^{4-4z}\int_0^{2-2z}\, dx\, dy\, dz - . -

    -     - -

    - dx\, dz\, dy: -

    - - -

    - - 0 \amp \leq x\leq 2-2z - 0 \amp \leq z\leq 1-y/4 - 0 \amp \leq y\leq 4 - -

    - -

    - \Rightarrow -

    - -

    - - \int_0^4\int_{0}^{1-y/4}\int_0^{2-2z}\, dx\, dz\, dy - . -

    -     -     -     - -

    - We give one more example of finding the volume of a space region. -

    - - - Finding the volume of a space region - -

    - Set up a triple integral that gives the volume of the space region D bounded by - z= 2x^2+2 and z=6-2x^2-y^2. - These surfaces are plotted in and , - respectively; - the region D is shown in . -

    - -
    - The region D is bounded by the surfaces shown in (a) and (b); D is shown in (c) - -
    - - - - - An elliptic paraboloid, opening downward, is plotted over a rectangular domain. - -

    - A three-dimensional sketch of a surface in space. - The surface is the elliptic paraboloid z=6-2x^2-y^2, - which opens downward and has its vertex at (0,0,6). -

    -     - - - - - //ASY file for figtrip5b3D.asy in Chapter 13 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8.4,6.8,12.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={5}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.5,7); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple f(pair t) { - return (t.x,t.y,6-2*t.x^2-t.y^2); - } - surface s=surface(f,(-1,-2),(1,2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p); - - //lines for tetrahedron - //draw((0,0,0)--(0,4,0)--(0,0,1)--cycle,bluepen+linewidth(2)); - - //shade rectangle and sides - //import three; - //path3 p = (0,0,0)--(0,4,0)--(0,0,1); - //draw(surface(p -- cycle), simplesurfacepen); - - - - -
    - -
    - - - - - A parabolic cylinder in space. It opens upward, and is symmetric about the y axis. - -

    - The surface z=2x^2+2 is a parabolic cylinder: - it has the shape of a long trough with parabolic cross-sections. - It opens upward, and is symmetric about the y axis. -

    -     - - - - - //ASY file for figtrip5c3D.asy in Chapter 13 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8.4,6.8,12.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={5}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.5,7); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple f(pair t) { - return (t.x,t.y,2*t.x^2+2); - } - surface s=surface(f,(-1,-2),(1,2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=redpen+.1mm; - draw(s,surfacepen2,meshpen=p); - - //lines for tetrahedron - //draw((0,0,0)--(0,4,0)--(0,0,1)--cycle,bluepen+linewidth(2)); - - //shade rectangle and sides - //import three; - //path3 p = (0,0,0)--(0,4,0)--(0,0,1); - //draw(surface(p -- cycle), simplesurfacepen); - - - - -
    - -
    - - - - - The region in space bounded by the paraboloid and cylinder in the previous two images. - -

    - A solid in space. - It is bounded above by the elliptic paraboloid in , - and below by the parabolic cylinder in . -

    - -

    - The overall shape of the solid is interesting. - The portion of the parabolic cylinder on the bottom looks like a taco shell. - The paraboloid on top is like an elongated dome. - Perhaps the shape is what you get if you completely fill a taco with a scoop of ice cream, - or make a really thick madeline cookie. -

    -     - - - - - //ASY file for figtrip53D.asy in Chapter 13 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8.4,6.8,12.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={5}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.5,7); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //paraboloid 6-2*t.x^2-t.y^2 - triple f(pair t) { - return (cos(t.x)*t.y,2*sin(t.x)*t.y,6-(2*(cos(t.x)*t.y)^2+(2*sin(t.x)*t.y)^2)); - } - surface s=surface(f,(0,0),(2*pi,1),16,8,Spline); - pen p=apexmeshpen+.1mm; - draw(s,surfacepen,meshpen=p); - - //parametric trough - triple f(pair t) { - return (cos(t.x)*t.y,2*sin(t.x)*t.y,(2*(cos(t.x)*t.y)^2+2)); - } - surface s=surface(f,(0,0),(2*pi,1),16,16,Spline); - pen p=redpen+.1mm; - draw(s,surfacepen2,meshpen=p); - - //ellipse in xy plane - triple g(real t) {return (cos(t),2*sin(t),0);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - //ellipse up on the surface intersection - triple g(real t) {return (cos(t),2*sin(t),2*cos(t)^2+2);} - path3 mypath=graph(g,0,2*pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - - - -
    -     - -
    -     - -

    - The main point of this example is this: - integrating with respect to z first is rather straightforward; - integrating with respect to x first is not. -

    - -

    - The order dz\, dy\, dx: -

    - -

    - The bounds on z are clearly 2x^2+2\leq z\leq 6-2x^2-y^2. - Collapsing D onto the xy-plane gives the ellipse shown in . - The equation of this ellipse is found by setting the two surfaces equal to each other: - - 2x^2+2 = 6-2x^2-y^2 \Rightarrow 4x^2+y^2=4 \Rightarrow x^2+\frac{y^2}4=1 - . -

    - -

    - We can describe this ellipse with the bounds - - -\sqrt{4-4x^2} \leq y\leq \sqrt{4-4x^2} \text{ and } -1\leq x\leq 1 - . -

    - -

    - Thus we find volume as -

    - - -

    - - 2x^2+2 \amp \leq z\leq 6-2x^2-y^2 - -\sqrt{4-4x^2} \amp \leq y\leq \sqrt{4-4x^2} - -1 \amp \leq x\leq 1 - -

    - -

    - \Rightarrow -

    - -

    - - \int_{-1}^1\int_{-\sqrt{4-4x^2}}^{\sqrt{4-4x^2}}\int_{2x^2+2}^{6-2x^2-y^2}\, dz\, dy\, dx - . -

    -     - -

    - The order dy\, dz\, dx: -

    - -

    - Integrating with respect to y is not too difficult. - Since the surface z=2x^2+2 is a cylinder whose directrix is the y-axis, - it does not create a border for y. - The paraboloid z=6-2x^2-y^2 does; - solving for y, we get the bounds - - -\sqrt{6-2x^2-z}\leq y\leq \sqrt{6-2x^2-z} - . -

    - -

    - Collapsing D onto the xz-plane gives the region shown in ; - the lower curve is from the cylinder, - with equation z=2x^2+2. - The upper curve is from the paraboloid; - with y=0, the curve is z=6-2x^2. - Thus bounds on z are 2x^2+2\leq z\leq 6-2x^2; - the bounds on x are -1\leq x\leq 1. - Thus we have: -

    - - -

    - - -\sqrt{6-2x^2-z} \amp \leq y\leq \sqrt{6-2x^2-z} - 2x^2+2 \amp \leq z\leq 6-2x^2 - -1 \amp \leq x\leq 1 - -

    - -

    - \Rightarrow -

    - -

    - - \int_{-1}^1\int_{2x^2+2}^{6-2x^2}\int_{-\sqrt{6-2x^2-z}}^{\sqrt{6-2x^2-z}}\, dy\, dz\, dx - . -

    -     - -
    - The region D in is collapsed onto the xz-plane in (a); in (b), it is collapsed onto the yz-plane - -
    - - - - - The shadow of the solid for this example on the plane y=0. - -

    - In a three-dimensional coordinate system, - a sketch is given of the shadow obtained by projecting the solid in - onto the xz plane. -

    - -

    - It is a plane region, drawn in perspective. - The region is bounded above by a parabola opening downward, with its vertex at (0,0,6), - and is bounded below by a parabola opening upward, with its vertex at (0,0,2). -

    -     - - - - - //ASY file for figtrip5e3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8.4,6.8,12.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={2,4,6}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.5,7); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //curves in xy plane - triple g(real t) {return (t,0,2*t^2+2);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,0,6-2*t^2);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - //shade area inbetween - int k=12; - import three; - for (int i=0; i<k; ++i) - { - path3 p = (i/k,0,2*(i/k)^2+2)-- ((i+1)/k,0,2*((i+1)/k)^2+2)--((i+1)/k,0,6-2*((i+1)/k)^2) -- ((i)/k,0,6-2*(i/k)^2); //Left - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (-i/k,0,2*(i/k)^2+2)-- (-(i+1)/k,0,2*((i+1)/k)^2+2)--(-(i+1)/k,0,6-2*((i+1)/k)^2) -- (-(i)/k,0,6-2*(i/k)^2); //Left - draw(surface(p -- cycle), simplesurfacepen); - } - - - - -
    - -
    - - - - - The shadow of the solid for this example on the plane x=0. It is divided into two regions. - -

    - In a three-dimensional coordinate system, - a sketch is given of the shadow obtained by projecting the solid in - onto the yz plane. -

    - -

    - It is a plane region, drawn in perspective. - The region is divided into two parts. - The bottom part is labeled R_1. - It is bounded below by a horizontal line parallel to the y axis, with z=2. - This corresponds to the bottom of the parabolic cylinder. - The upper bound of R_1 is a parabola that opens downward, with its vertex at (0,0,4). - This parabola comes from the curve that is formed by the intersection of the two surfaces; - it is labeled with the equation z=4-y^2/2. -

    - -

    - The upper part is labeled R_2. This region is bounded below by the parabola that forms the upper bound of R_1. - It is bounded above by a second parabola that also opens downward, with its vertex at (0,0,6). - This parabola corresponds to the x=0 trace through the elliptic paraboloid. - It is labeled with the equation z=6-y^2. -

    -     - - - - - //ASY file for figtrip5d3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8.4,6.8,12.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={2,4,6}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.5,7); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //curves in xy plane - triple g(real t) {return (0,t,6-t^2);} - path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (0,t,4-t^2/2);} - path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - draw((0,-2,2)--(0,2,2),bluepen+linewidth(2)); - - //shade area inbetween - int k=12; - import three; - for (int i=-2*k; i<2*k; ++i) - { - path3 p = (0,i/k,6-(i/k)^2)-- (0,(i+1)/k,6-((i+1)/k)^2)--(0,(i+1)/k,4-(0.5)*((i+1)/k)^2) -- (0,(i)/k,4-(0.5)*((i)/k)^2); //top in blue - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0,i/k,2)-- (0,(i+1)/k,2)--(0,(i+1)/k,4-(0.5)*((i+1)/k)^2) -- (0,(i)/k,4-(0.5)*((i)/k)^2); //bottom in pink - draw(surface(p -- cycle), simplesurfacepen2); - } - - //labels - label("$R_1$",(0.1,.5,3)); - label("$R_2$",(0.1,.5,5)); - label("$z=4-y^2/2$",(0,-1.5,5.5)); - label("$z=6-y^2$",(0,2,6)); - draw((0.2,-1.5,5.3)--(0,-1,3.55),Arrow3(size=2mm)); - draw((0.2,2,5.7)--(0,1,5.05),Arrow3(size=2mm)); - - - - -
    -     - -
    - -

    - The order dx\, dz\, dy: -

    - -

    - This order takes more effort as D must be split into two subregions. - The two surfaces create two sets of upper/lower bounds in terms of x; - the cylinder creates bounds - - -\sqrt{z/2-1}\leq x\leq \sqrt{z/2-1} - - for region D_1 and the paraboloid creates bounds - - -\sqrt{3-y^2/2-z^2/2}\leq x\leq \sqrt{3-y^2/2-z^2/2} - - for region D_2. -

    - -

    - Collapsing D onto the yz-axes gives the regions shown in . - We find the equation of the curve - z=4-y^2/2 by noting that the equation of the ellipse seen in has equation - - x^2+y^2/4=1 \Rightarrow x = \sqrt{1-y^2/4} - . -

    - -

    - Substitute this expression for x in either surface equation, - z=6-2x^2-y^2 or z=2x^2+2. - In both cases, we find - - z=4-\frac12y^2 - . -

    - -

    - Region R_1, corresponding to D_1, has bounds - - 2\leq z\leq 4-y^2/2, -2\leq y\leq 2 - - and region R_2, corresponding to D_2, has bounds - - 4-y^2/2\leq z\leq 6-y^2, -2\leq y\leq 2 - . -

    - -

    - Thus the volume of D is given by: - - \int_{-2}^2\int_2^{4-y^2/2}\int_{-\sqrt{z/2-1}}^{\sqrt{z/2-1}}\, dx\, dz\, dy \, +\, \int_{-2}^2\int_{4-y^2/2}^{6-y^2}\int_{-\sqrt{3-y^2/2-z^2/2}}^{\sqrt{3-y^2/2-z^2/2}}\, dx\, dz\, dy - . -

    -     -     - -

    - If all one wanted to do in - was find the volume of the region D, - one would have likely stopped at the first integration setup - (with order dz\, dy\, dx) - and computed the volume from there. - However, we included the other two methods 1) to show that it could be done, - messy or not, - and 2) because sometimes we have - to use a less desirable order of integration in order to actually integrate. -

    -     - - - Triple Integration and Functions of Three Variables -

    - There are uses for triple integration beyond merely finding volume, - just as there are uses for integration beyond - area under the curve. - These uses start with understanding how to integrate functions of three variables, - which is effectively no different than integrating functions of two variables. - This leads us to a definition, followed by an example. -

    - -

    - - - Iterated Integration, (Part II) - -

    - Let D be a closed, - bounded region in space, - over which g_1(x), g_2(x), f_1(x,y), - f_2(x,y) and h(x,y,z) are all continuous, - and let a and b be real numbers. -

    - -

    - The iterated integral - \ds \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\, dz\, dy\, dx is evaluated as - integrationtriple - triple integral - iterated integration - - \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\, dz\, dy\, dx = \int_a^b\int_{g_1(x)}^{g_2(x)}\left(\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\, dz\right) dy\, dx - . -

    - - - - - Evaluating a triple integral of a function of three variables - -

    - Evaluate \ds \int_0^1\int_{x^2}^x\int_{x^2-y}^{2x+3y} \big(xy+2xz\big)\, dz\, dy\, dx. -

    -     - -

    - We evaluate this integral according to . -

    - -

    - - \amp \int_0^1\int_{x^2}^x\int_{x^2-y}^{2x+3y} \big(xy+2xz\big)\, dz\, dy\, dx - \amp = \int_0^1\int_{x^2}^x\left(\int_{x^2-y}^{2x+3y} \big(xy+2xz\big)\, dz\right)\, dy\, dx - \amp = \int_0^1\int_{x^2}^x\left(\big(xyz+ xz^2\big)\Big|_{x^2-y}^{2x+3y}\right)\, dy\, dx - \amp = \int_0^1\int_{x^2}^x\Bigg(xy(2x+3y)+x(2x+3y)^2-\Big(xy(x^2-y)+x(x^2-y)^2\Big)\Bigg)\, dy\, dx - \amp =\int_0^1\int_{x^2}^x\Big(-x^5+x^3y+4x^3+14x^2y+12xy^2\Big)\, dy\, dx - . -

    - -

    - We continue as we have in the past, showing fewer steps. - - \amp = \int_0^1\Bigg(-\frac72x^7-8x^6-\frac72x^5+15x^4\Bigg)\, dx - \amp = \frac{281}{336}\approx 0.836 - . -

    -     -     - -

    - We now know how to evaluate a triple integral of a function of three variables; - we do not yet understand what it means. - We build up this understanding in a way very similar to how we have understood integration and double integration. -

    - -

    - Let h(x,y,z) be a continuous function of three variables, - defined over some space region D. - We can partition D into n rectangular-solid subregions, - each with dimensions \dx_i\times\dy_i\times\ddz_i. - Let (x_i,y_i,z_i) be some point in the ith subregion, - and consider the product h(x_i,y_i,z_i)\dx_i\dy_i\ddz_i. - It is the product of a function value (that's the - h(x_i,y_i,z_i) part) and a small volume \Delta V_i - (that's the \dx_i\dy_i\ddz_i part). - One of the simplest understanding of this type of product is when h describes the density of an object, - for then h\times\text{ volume } =\text{ mass }. -

    - -

    - We can sum up all n products over D. - Again letting \norm{\Delta D} represent the length of the longest diagonal of the n rectangular solids in the partition, - we can take the limit of the sums of products as \norm{\Delta D}\to 0. - That is, we can find - - S = \lim_{\norm{\Delta D}\to 0} \sum_{i=1}^n h(x_i,y_i,z_i)\Delta V_i=\lim_{\norm{\Delta D}\to 0} \sum_{i=1}^n h(x_i,y_i,z_i)\dx_i\dy_i\ddz_i - . -

    - -

    - While this limit has lots of interpretations depending on the function h, - in the case where h describes density, - S is the total mass of the object described by the region D. -

    - -

    - We now use the above limit to define the - triple integral, - give a theorem that relates triple integrals to iterated iteration, - followed by the application of triple integrals to find the centers of mass of solid objects. -

    - - - Triple Integral - -

    - Let w=h(x,y,z) be a continuous function over a closed, - bounded region D in space, - and let \Delta D be any partition of D into n rectangular solids with volume \Delta V_i. - The triple integral of h over D - is - integrationtriple - triple integral - iterated integration - - \iiint_Dh(x,y,z)\, dV = \lim_{\norm{\Delta D}\to 0}\sum_{i=1}^n h(x_i,y_i,z_i)\Delta V_i - . -

    -     -     - -

    - The following theorem assures us that the above limit exists for continuous functions h and gives us a method of evaluating the limit. -

    - - - Triple Integration (Part II) - -

    - Let w=h(x,y,z) be a continuous function over a closed, - bounded region D in space, - and let \Delta D be any partition of D into n rectangular solids with volume V_i. - integrationtriple - triple integral - iterated integration -

      -
    1. -

      - The limit \lim\limits_{\norm{\Delta D}\to 0}\sum_{i=1}^n h(x_i,y_i,z_i)\Delta V_i exists. -

      -
      2. - -
    2. -

      - If D is defined as the region bounded by the planes x=a and x=b, - the cylinders y=g_1(x) and y=g_2(x), - and the surfaces z=f_1(x,y) and - z=f_2(x,y), where a\lt b, - g_1(x)\leq g_2(x) and f_1(x,y)\leq f_2(x,y) on D, then - - \iiint_D h(x,y,z)\, dV = \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\, dz\, dy\, dx - . -

      -
      3. -
    -

    -     -     - - -

    - Note: In an aside in , - we showed how the summation of rectangles over a region R in the plane could be viewed as a double sum, - leading to the double integral. - Likewise, we can view the sum - - \sum_{i=1}^nh(x_i,y_i,z_i)\dx_i\dy_i\ddz_i - - as a triple sum, - - \sum_{k=1}^p\sum_{j=1}^n\sum_{i=1}^mh(x_i,y_j,z_k)\dx_i\dy_j\ddz_k - , - which we evaluate as - - \sum_{k=1}^p\left(\sum_{j=1}^n\left(\sum_{i=1}^mh(x_i,y_j,z_k)\dx_i\right)\dy_j\right)\ddz_k - . -

    - -

    - Here we fix a k value, - which establishes the z-height of the rectangular solids on one level - of all the rectangular solids in the space region D. - The inner double summation adds up all the volumes of the rectangular solids on this level, - while the outer summation adds up the volumes of each level. -

    - -

    - This triple summation understanding leads to the - \iiint_D notation of the triple integral, - as well as the method of evaluation shown in . -

    - -

    - We now apply triple integration to find the centers of mass of solid objects. -

    -     - - - Mass and Center of Mass -

    - One may wish to review - for a reminder of the relevant terms and concepts. -

    - - - Mass, Center of Mass of Solids - -

    - Let a solid be represented by a closed, - bounded region D in space with variable density function \delta(x,y,z). - moment - center of mass - mass -

    - -

    -

      -
    1. -

      - The mass of the object is \ds M= \iiint_D \, dm=\iiint_D \delta(x,y,z)\, dV. -

      -
      2. - -
    2. -

      - The moment about the yz-plane - is \ds M_{yz}=\iiint_D x\delta(x,y,z)\, dV. -

      -
      3. - -
    3. -

      - The moment about the xz-plane - is \ds M_{xz}=\iiint_D y\delta(x,y,z)\, dV. -

      -
      4. - -
    4. -

      - The moment about the xy-plane - is \ds M_{xy}=\iiint_D z\delta(x,y,z)\, dV. -

      -
      5. - -
    5. -

      - The center of mass of the object is - - \big(\overline{x},\overline{y},\overline{z}\big) = \left(\frac{M_{yz}}M,\frac{M_{xz}}M,\frac{M_{xy}}M\right) - . -

      -
      6. -
    -

    -     -     - - - Finding the center of mass of a solid - -

    - Find the mass and center of mass of the solid represented by the space region bounded by the coordinate planes and z=2-y/3-2x/3, - shown in , - with constant density \delta(x,y,z)=3\,\text{g/cm}^3. - (Note: this space region was used in .) -

    - -
    - Finding the center of mass of the solid in - - - - A triangle in space, formed by the portion of a plane that lies in the first octant. - -

    - This is the same surface shown in . - We repeat the description given there. -

    - -

    - A plane in space is illustrated, by showing the portion of the plane the lies in the first octant. - This portion of the plane is a triangle, whose vertices lie on the coordinate axes. - Using the image (or the equation z=2-y/3-2x/3), - we can determine that the plane meets the coordinate axes at the points (3,0,0), (0,6,0), and (0,0,2). -

    - -

    - The region in space whose volume is computed in this problem is a tetrahedron, - whose faces are the plane described above, - as well as the triangles formed in the three coordinate planes by the coordinate axes and the edges of the first triangle. -

    -     - - - - - //ASY file for figtrip23D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.2,10.1,6.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2,4}; - real[] myychoice={5}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,5); - pair ybounds=(-0.5,7); - pair zbounds=(-0.5,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //triangle in space - draw((3,0,0)--(0,6,0)--(0,0,2)--cycle,bluepen+linewidth(1.5)); - - //shade - import three; - path3 p = (3,0,0)--(0,6,0)--(0,0,2); //Left - draw(surface(p -- cycle), simplesurfacepen); - - - - -
    -     - -

    - We apply . - In , - we found bounds for the order of integration - dz\, dy\, dx to be 0\leq z\leq 2-y/3-2x/3, - 0\leq y\leq 6-2x and 0\leq x\leq 3. - We find the mass of the object: - - M \amp = \iiint_D \delta(x,y,z)\, dV - \amp = \int_0^3\int_0^{6-2x}\int_0^{2-y/3-2x/3} \big(3\big)\, dz\, dy\, dx - \amp = 3\int_0^3\int_0^{6-2x}\int_0^{2-y/3-2x/3} \, dz\, dy\, dx - \amp = 3(6) = 18\,\text{g} - . -

    - -

    - The evaluation of the triple integral is done in , - so we skipped those steps above. - Note how the mass of an object with constant density is simply - densityvolume. -

    - -

    - We now find the moments about the planes. - - M_{xy} \amp = \iiint_D 3z\, dV - \amp = \int_0^3\int_0^{6-2x}\int_0^{2-y/3-2x/3} \big(3z\big)\, dz\, dy\, dx - \amp = \int_0^3\int_0^{6-2x} \frac32\big(2-y/3-2x/3\big)^2\, dy\, dx - \amp = \int_0^3 -\frac49\big(x-3\big)^3\, dx - \amp = 9 - . -

    - -

    - We omit the steps of integrating to find the other moments. - - M_{yz} \amp = \iiint_D 3x\, dV - \amp = \frac{27}2. - M_{xz} \amp = \iiint_D 3y\, dV - \amp = 27 - . -

    - -

    - The center of mass is - - \big(\overline{x},\overline{y},\overline{z}\big) = \left(\frac{27/2}{18},\frac{27}{18},\frac{9}{18}\right) = \big(0.75,1.5,0.5\big) - . -

    -     -     - - - Finding the center of mass of a solid - -

    - Find the center of mass of the solid represented by the region bounded by the planes z=0 and z=-y and the cylinder x^2+y^2=1, - shown in , - with density function \delta(x,y,z) = 10+x^2+5y-5z. - (Note: this space region was used in .) -

    - -
    - Finding the center of mass of the solid in - - - - A cylindrical wedge in space, resembling a wedge cut from a tree. - -

    - This is the same solid in space that was depicted in . - We repeat the description that was given there. -

    -

    - An illustration of the region of integration in . - The region is a three-dimensional solid cut from a cylinder. -

    - -

    - The plane z=-y meets the xy plane along the x axis. - The region that lies between these planes, and within the cylinder x^2+y^2=1, - is a semi-circular wedge. - It resembles the shape of a piece cut from a tree that is being cut down. -

    -     - - - - - //ASY file for figtrip33D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.1,1.9,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,-0.5}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,0.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple f(pair t) {return (cos(t.x),sin(t.x),-sin(t.x)*t.y);} - surface s=surface(f,(-pi,0),(0,1),16,2,Spline); - pen p=apexmeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - triple f(pair t) {return (cos(t.x)*t.y,sin(t.x)*t.y,-sin(t.x)*t.y);} - surface s=surface(f,(-pi,0),(0,1),16,2,Spline); - pen p=redpen+.1mm; - draw(s,simplesurfacepen2,meshpen=p); - - //disk x^2+y^2=1 in 3rd and 4th quadrants - triple g(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(g,pi,2*pi,operator ..);draw(mypath,bluepen+linewidth(1.5)); - - //surface above disk x^2+y^2=1 in 3rd and 4th quadrants - triple g(real t) {return (cos(t),sin(t),-sin(t));} - path3 mypath=graph(g,pi,2*pi,operator ..); - draw(mypath,bluepen+linewidth(1.5)); - - //label the surfaces - label("$x^2+y^2=1$",(1.4,-.5,0)); - draw((1.1,-.6,0)--(.7,-.7,.35),Arrow3(size=2mm)); - label("$z=-y$",(0,-.5,1.25)); - draw((0,-.5,1.15)--(0,-.5,.5),Arrow3(size=2mm)); - - - - -
    -     - -

    - As we start, consider the density function. - It is symmetric about the yz-plane, - and the farther one moves from this plane, the denser the object is. - The symmetry indicates that \overline x should be 0. -

    - -

    - As one moves away from the origin in the y or z directions, - the object becomes less dense, - though there is more volume in these regions. -

    - -

    - Though none of the integrals needed to compute the center of mass are particularly hard, - they do require a number of steps. - We emphasize here the importance of knowing how to set up the proper integrals; - in complex situations we can appeal to technology for a good approximation, - if not the exact answer. - We use the order of integration dz\, dy\, dx, - using the bounds found in . - (As these are the same for all four triple integrals, - we explicitly show the bounds only for M.) - - M \amp = \iiint_D \big(10+x^2+5y-5z\big)\, dV - \amp = \int_{-1}^1\int_{-\sqrt{1-x^2}}^0\int_0^{-y} \big(10+x^2+5y-5z\big)\, dV - \amp = \frac{64}5-\frac{15\pi}{16} \approx 3.855. - M_{yz} \amp = \iiint_D x\big(10+x^2+5y-5z\big)\, dV - \amp =0. - M_{xz} \amp = \iiint_D y\big(10+x^2+5y-5z\big)\, dV - \amp = 2-\frac{61\pi}{48}\approx -1.99. - M_{xy} \amp = \iiint_D z\big(10+x^2+5y-5z\big)\, dV - \amp = \frac{61\pi}{96}-\frac{10}9\approx 0.885 - . -

    - -

    - Note how M_{yz}=0, as expected. - The center of mass is - - \big(\overline{x},\overline{y},\overline{z}\big) = \left(0,\frac{-1.99}{3.855},\frac{0.885}{3.855}\right) \approx \big(0,-0.516, 0.230\big) - . -

    -     -     - -

    - As stated before, - there are many uses for triple integration beyond finding volume. - When h(x,y,z) describes a rate of change function over some space region D, - then \ds \iiint_D h(x,y,z)\, dV gives the total change over D. - Our one specific example of this was computing mass; - a density function is simply a rate of mass change per volume function. - Integrating density gives total mass. -

    - -

    - While knowing how to integrate is important, - it is arguably much more important to know - how to set up integrals. - It takes skill to create a formula that describes a desired quantity; - modern technology is very useful in evaluating these formulas quickly and accurately. -

    - -

    - In , we learn about two new coordinate systems - (each related to polar coordinates) - that allow us to integrate over closed regions in space more easily than when using rectangular coordinates. -

    -     - - - - Terms and Concepts - - - -

    - The strategy for establishing bounds for triple integrals is - to , - to and - to . -

    -     - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    - surface to surface, curve to curve and point to point -

    -     - -     - - - - -

    - Give an informal interpretation of what - \ds \iiint_D\, dV means. -

    -     - - - -

    - One possible answer is sum up lots of little volumes over D. -

    -     - -     - - - - -

    - Give two uses of triple integration. -

    -     - - - -

    - Answers can vary. - From this section we used triple integration to find the volume of a solid region, - the mass of a solid, and the center of mass of a solid. -

    -     - -     - - - - -

    - If an object has a constant density \delta and a volume V, - what is its mass? -

    -     - - - -

    - \delta V. -

    -     - -     -     - - - Problems - - - -

    - Two functions f_1(x,y) and - f_2(x,y) and a region R in the x, - y plane are given. - Set up and evaluate the double integral that finds the volume - between the surfaces given by the graphs of these two functions over R. -

    -     - - - - -

    - f_1(x,y) = 8-x^2-y^2, f_2(x,y) = 2x+y; -

    - -

    - R is the square with corners (-1,-1) and (1,1). -

    -     - -

    - V = \int_{-1}^1\int_{-1}^1 \big(8-x^2-y^2-(2x+y)\big)\, dx\, dy = 88/3 -

    -     - -     - - - - - $int=Compute("52"); - - - -

    - z=f_1(x,y) = x^2+y^2 and z=f_2(x,y) = -x^2-y^2; -

    -

    - R is the square with corners (0,0) and (2,3). -

    - -

    - -

    -     -     -     - - - - -

    - f_1(x,y) = \sin(x) \cos(y), - f_2(x,y) = \cos(x) \sin(y) +2; -

    - -

    - R is the triangle with corners (0,0), - (\pi,0) and (\pi,\pi). -

    -     - -

    - V = \int_{0}^{\pi}\int_{0}^x \big(\cos(x) \sin(y) +2-\sin(x) \cos(y) \big)\, dy\, dx = \pi^2-\pi\approx 6.728 -

    -     - -     - - - - - $int=Compute("3pi/2"); - - - -

    - f_1(x,y) = 2x^2+2y^2+3 and f_2(x,y) = 6-x^2-y^2; -

    - -

    - R is the disc x^2+y^2\leq1. -

    - -

    - -

    -     -     -     - -     - - - -

    - A domain D is described by its bounding surfaces, - along with a graph. - Set up the triple integrals that give the volume of D in all 6 orders of integration, - and find the volume of D by evaluating the indicated triple integral. -

    -     - - - - -

    - D is bounded by the coordinate planes and -

    - -

    - z=2-2x/3-2y. -

    - -

    - Evaluate the triple integral with order dz\, dy\, dx. -

    - - - - - A tetrahedron in the first octant, with one vertex at the origin, and the others on the coordinate axes. - -

    - The solid is a tetrahedron, plotted in the first octant relative to the usual three-dimensional coordinate axes. - The vertices of the tetrahedron are at (0,0,0), (3,0,0), (0,1,0), and (0,0,2). - Three of the four faces lie in the coordinate planes; - the remaining face is in the first octant, and is labeled with the equation z=2-\frac23 x-2y. -

    -     - - - - - //ASY file for fig13_06_ex_07.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3}; - real[] myychoice={1,2,3}; - real[] myzchoice={1,2}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,3.5); - pair ybounds=(-0.5,3.5); - pair zbounds=(-0.5,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //edges of tetrahedron - draw((0,0,0)--(3,0,0)--(0,1,0)--cycle,bluepen+linewidth(2)); - draw((3,0,0)--(0,0,2),bluepen+linewidth(2)); - draw((0,0,0)--(0,0,2),bluepen+linewidth(2)); - draw((0,1,0)--(0,0,2),bluepen+linewidth(2)); - - //shade faces - import three; - path3 p=(0,0,0)--(3,0,0)--(0,1,0); - draw(surface(p -- cycle), simplesurfacepen); - path3 p=(0,0,0)--(3,0,0)--(0,0,2); - draw(surface(p -- cycle), simplesurfacepen); - path3 p=(0,0,0)--(0,1,0)--(0,0,2); - draw(surface(p -- cycle), simplesurfacepen); - path3 p=(0,0,0)--(3,0,0)--(0,0,2); - draw(surface(p -- cycle), simplesurfacepen); - - //labels and arrow - label("$z=2-\frac{2}{3}x-2y$",(3,3,0)); - draw((2.5,2.5,.25)--(1.5,0.5,0.7),Arrow3(size=2mm)); - - - - -     - -

    - dz\, dy\, dx: - \ds\int_0^3\int_0^{1-x/3}\int_0^{2-2x/3-2y}\, dz\, dy\, dx -

    - -

    - dz\, dx\, dy: - \ds\int_0^1\int_0^{3-3y}\int_0^{2-2x/3-2y}\, dz\, dx\, dy -

    - -

    - dy\, dz\, dx: - \ds\int_0^3\int_0^{2-2x/3}\int_0^{1-x/3-z/2}\, dy\, dz\, dx -

    - -

    - dy\, dx\, dz: - \ds\int_0^2\int_0^{3-3z/2}\int_0^{1-x/3-z/2}\, dy\, dx\, dz -

    - -

    - dx\, dz\, dy: - \ds\int_0^1\int_0^{2-2y}\int_0^{3-3y-3z/2}\, dx\, dz\, dy -

    - -

    - dx\, dy\, dz: - \ds\int_0^2\int_0^{1-z/2}\int_0^{3-3y-3z/2}\, dx\, dy\, dz -

    - -

    - \ds V = \int_0^3\int_0^{1-x/3}\int_0^{2-2x/3-2y}\, dz\, dy\, dx =1. -

    -     - -     - - - - -

    - D is bounded by the planes y=0, - y=2, x=1, z=0 and -

    - -

    - z=(3-x)/2. -

    - -

    - Evaluate the triple integral with order dx\, dy\, dz. -

    - - - - - A triangular prism. The two triangular sides are parallel to the xz plane. - -

    - The solid is a triangular prism. The base is a square, and it lies in the xy plane. - The corners of the square are at (1,0,0), (3,0,0), (3,2,0), and (1,2,0). -

    - -

    - The two triangular faces are parallel to the xz plane. - One face is in the plane y=0, with vertices (1,0,0), (3,0,0), and (3,0,1). - The other face is in the plane y=2, with vertices (1,2,0), (3,2,0), and (3,2,1). -

    - -

    - The front face (according to how the solid is oriented in the image) - is a rectangle in the plane z=\frac12(3-x). - Two edges are parallel to the y axis, and two edges are parallel to the hypotenuse of the triangular faces. -

    - -

    - The last face is at the back, in the plane x=1. - This is a vertical rectangle, with 0\leq y\leq 2 and 0\leq z\leq 1. -

    -     - - - - - //ASY file for fig13_06_ex_083D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8.8,7.8,3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3}; - real[] myychoice={1,2,3}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,3.5); - pair ybounds=(-0.5,3.5); - pair zbounds=(-0.5,1.75); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //edges of object - draw((1,0,0)--(1,2,0)--(3,2,0)--(3,0,0)--cycle,bluepen+linewidth(2)); - draw((3,0,0)--(1,0,1)--(1,2,1)--(3,2,0),bluepen+linewidth(2)); - draw((3,0,0)--(1,0,1),bluepen+linewidth(2)); - draw((1,0,0)--(1,0,1),bluepen+linewidth(2)); - draw((1,2,0)--(1,2,1),bluepen+linewidth(2)); - - //shade faces - import three; - path3 p=(1,0,0)--(1,2,0)--(3,2,0)--(3,0,0); - draw(surface(p -- cycle), simplesurfacepen); - path3 p=(1,0,1)--(1,2,1)--(3,2,0)--(3,0,0); - draw(surface(p -- cycle), simplesurfacepen); - path3 p=(1,0,0)--(1,2,0)--(1,2,1)--(1,0,1); - draw(surface(p -- cycle), simplesurfacepen); - path3 p=(1,0,0)--(3,0,0)--(1,0,1); - draw(surface(p -- cycle), simplesurfacepen); - path3 p=(1,2,0)--(3,2,0)--(1,2,1); - draw(surface(p -- cycle), simplesurfacepen); - - //labels and arrow - label("$z=\frac{1}{2}(3-x)$",(3,0,1.1)); - draw((3,0.25,1)--(2,1,0.45),Arrow3(size=2mm)); - - - - -     - -

    - dz\, dy\, dx: - \ds\int_1^3\int_0^{2}\int_0^{(3-x)/2}\, dz\, dy\, dx -

    - -

    - dz\, dx\, dy: - \ds\int_0^2\int_1^{3}\int_0^{(3-x)/2}\, dz\, dx\, dy -

    - -

    - dy\, dz\, dx: - \ds\int_1^3\int_0^{(3-x)/2}\int_0^{2}\, dy\, dz\, dx -

    - -

    - dy\, dx\, dz: - \ds\int_0^1\int_1^{3-2z}\int_0^{2}\, dy\, dx\, dz -

    - -

    - dx\, dz\, dy: - \ds\int_0^2\int_0^{1}\int_1^{3-2z}\, dx\, dz\, dy -

    - -

    - dx\, dy\, dz: - \ds\int_0^1\int_0^{2}\int_1^{3-2z}\, dx\, dy\, dz -

    - -

    - \ds V = \int_0^1\int_0^{2}\int_1^{3-2z}\, dx\, dy\, dz =2. -

    -     - -     - - - - -

    - D is bounded by the planes x=0, - x=2, z=-y and by -

    - -

    - z=y^2/2. -

    - -

    - Evaluate the triple integral with the order dy\, dz\, dx. -

    - - - - - A solid bounded above by a plane, and below by a parabolic cylinder. - -

    - On a set of three-dimensional coordinate axes, a solid is given between two surfaces. -

    - -

    - One surface is a plane. It is illustrated as a rectangle. -

      -
    • -

      - The bottom of the rectangle lies along the x axis, for x between 0 and 2. -

      -
      • -
    • -

      - The sides are negatively sloped, relative to the yz: - the equation of the plane is z=-y, - and setting x=0 or x=2 in this plane produces the sides. -

      -
      • -
    • -

      - The top of the rectangle is parallel to the x axis; - it is a line segment with y=-2, z=2, and 0\leq x\leq 2. -

      -
      • -
    -

    - -

    - The other surface is a parabolic cylinder. - Cross-sections are the half of the cylinder z=\frac12 y^2 with y\leq 0. - This surface meets the rectangle in the plane at the top and bottom edges of the rectangle. - It has the appearance of a sheet that has been hung below the rectangle and attached at either end. -

    - -

    - Projecting the solid onto either the xy or xz planes produces a rectangle, - while projection onto the yz plane produces a region that is bounded above by the line z=-y, - and below by the parabola z=\frac12 y^2. -

    -     - - - - - //ASY file for fig13_06_ex_093D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={}; - real[] myzchoice={1,2}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,2.5); - pair ybounds=(-2.5,0.5); - pair zbounds=(-0.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //edges of plane - draw((0,0,0)--(2,0,0)--(2,-2,2)--(0,-2,2)--cycle,bluepen+linewidth(2)); - - //shade plane - import three; - path3 p=(0,0,0)--(2,0,0)--(2,-2,2)--(0,-2,2); - draw(surface(p -- cycle), simplesurfacepen); - - //edges of parabolic cylinder - triple g(real t) {return (2,t,0.5*t^2);} - path3 mypath=graph(g,-2,0,operator ..); - draw(mypath,redpen+linewidth(2)); - triple g(real t) {return (0,t,0.5*t^2);} - path3 mypath=graph(g,-2,0,operator ..); - draw(mypath,redpen+linewidth(2)); - - //Shade the cylinder - int k=10; - for (int i=0; i<2*k; ++i) - { - path3 p=(2,-i/k,0.5*(i/k)^2)--(2,-(i+1)/k,0.5*((i+1)/k)^2)--(0,-(i+1)/k,0.5*((i+1)/k)^2)--(0,-i/k,0.5*(i/k)^2); - draw(surface(p -- cycle), surfacepen2); - } - - //labels and arrow - label("$z=-y$",(.5,0.5,2),E);draw((.5,.75,1.8)--(.75,-1,1),Arrow3(size=2mm)); - label("$z=\frac{1}{2}y^2$",(2.25,-1.5,.75),W); - draw((2.25,-1.5,.7)--(2,-0.85,.32),Arrow3(size=2mm)); - - - - -     - -

    - dz\, dy\, dx: - \ds\int_0^2\int_{-2}^{0}\int_{y^2/2}^{-y}\, dz\, dy\, dx -

    - -

    - dz\, dx\, dy: - \ds\int_{-2}^0\int_0^{2}\int_{y^2/2}^{-y}\, dz\, dx\, dy -

    - -

    - dy\, dz\, dx: - \ds\int_0^2\int_0^{2}\int_{-\sqrt{2z}}^{-z}\, dy\, dz\, dx -

    - -

    - dy\, dx\, dz: - \ds\int_0^2\int_0^{2}\int_{-\sqrt{2z}}^{-z}\, dy\, dx\, dz -

    - -

    - dx\, dz\, dy: - \ds\int_{-2}^0\int_{y^2/2}^{-y}\int_0^{2}\, dx\, dz\, dy -

    - -

    - dx\, dy\, dz: - \ds\int_0^2\int_{-\sqrt{2z}}^{-z}\int_0^{2}\, dx\, dy\, dz \ds V = \int_0^2\int_0^{2}\int_{-\sqrt{2z}}^{-z}\, dy\, dz\, dx =4/3. -

    -     - -     - - - - -

    - D is bounded by the planes z=0, - y=9, x=0 and by -

    - -

    - z=\sqrt{y^2-9x^2}. -

    - -

    - Do not evaluate any triple integral. -

    - - - - - One quarter of an elliptic cone that opens along the y axis. - -

    - The solid for this exercise is bounded by the xy and yz coordinate planes, - and the elliptic cone z=\sqrt{y^2-9x^2}. -

    - -

    - The cone opens along the y axis. - Only a quarter of the cone is shown in the image; - namely, that portion that lies in the first octant. -

    - -

    - The projection of the solid onto the xy plane is a triangle, - with vertices at (0,0,0), (0,9,0), and (3,9,0). -

    - -

    - The solid also meets the yz plane in a triangle, - with vertices at (0,0,0), (0,9,0), and (0,9,9). -

    - -

    - The projection of the solid onto the xz plane is a quarter ellipse. - It is the region bounded by the ellipse 9x^2+z^2=81, - and the positive x and z axes. -

    -     - - - - - //ASY file for fig13_06_ex_103D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14,10,21); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={3}; - real[] myychoice={3,6,9}; - real[] myzchoice={3,6,9}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,5); - pair ybounds=(-0.25,10); - pair zbounds=(-0.25,10); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //edges of object - draw((0,0,0)--(0,9,0)--(3,9,0)--cycle,bluepen+linewidth(2)); - draw((0,9,0)--(0,9,9)--(0,0,0)--cycle,bluepen+linewidth(2)); - triple g(real t) {return (t,9,sqrt(81-9*t^2));} - path3 mypath=graph(g,0,3,operator ..);draw(mypath,bluepen+linewidth(2)); - - triple f(pair t) { - return (t.x/3*t.y,t.x,sqrt(t.x^2-9(t.x/3*t.y)^2)); - } - surface s=surface(f,(0,0),(9,1),8,12,Spline); - pen p=apexmeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - //label and arrow - label("$z=\sqrt{y^2-9x^2}$",(4,4.5,0)); - draw((3.3,4,.2)--(1.25,4,1.4),Arrow3(size=2mm)); - - - - -     - -

    - dz\, dy\, dx: - \ds\int_0^3\int_{3x}^{9}\int_{0}^{\sqrt{y^2-9x^2}}\, dz\, dy\, dx -

    - -

    - dz\, dx\, dy: - \ds\int_{0}^9\int_0^{y/3}\int_{0}^{\sqrt{y^2-9x^2}}\, dz\, dx\, dy -

    - -

    - dy\, dz\, dx: - \ds\int_0^3\int_0^{\sqrt{81-9x^2}}\int_{\sqrt{z^2+9x^2}}^{9}\, dy\, dz\, dx -

    - -

    - dy\, dx\, dz: - \ds\int_0^9\int_0^{\sqrt{9-z^2/9}}\int_{\sqrt{z^2+9x^2}}^{9}\, dy\, dx\, dz -

    - -

    - dx\, dz\, dy: - \ds\int_{0}^9\int_{0}^{y}\int_0^{\frac13\sqrt{y^2-z^2}}\, dx\, dz\, dy -

    - -

    - dx\, dy\, dz: - \ds\int_0^9\int_{z}^{9}\int_0^{\frac13\sqrt{y^2-z^2}}\, dx\, dy\, dz -

    -     - -     - - - - -

    - D is bounded by the planes x=2, - y=1, z=0 and -

    - -

    - z=2x+4y-4. -

    - -

    - Evaluate the triple integral with the order dx\, dy\, dz. -

    - - - - - A tetrahedron in space, plotted with respect to a three-dimensional coordinate system. - -

    - The image contains the usual three-dimensional coordinate axes, and a tetrahedron. - Three of the faces of the tetrahedron are parallel to the coordinate planes. -

      -
    • -

      - The bottom is a triangle in the xy plane with vertices (2,0,0), (2,1,0), and (0,1,0). -

      -
      • -
    • -

      - Another face lies in the plane x=2, forming a triangle with vertices (2,0,0), (2,1,0), and (2,1,4). -

      -
      • -
    • -

      - A third face lies in the plane y=1; it is a triangle with vertices (0,1,0), (2,1,0), and (2,1,4). -

      -
      • -
    • -

      - The last face lies in the plane z=2x+4y-4; it is a triangle with vertices (2,0,0), (0,1,0), and (2,1,4). -

      -
      • -
    -

    -     - - - - - //ASY file for fig13_06_ex_113D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.2,-5.1,8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={1,2}; - real[] myzchoice={2,4}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,2.5); - pair ybounds=(-0.25,2.5); - pair zbounds=(-0.25,5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //edges of object - draw((0,1,0)--(2,1,0)--(2,0,0)--cycle,bluepen+linewidth(2)); - draw((0,1,0)--(2,1,4)--(2,1,0)--cycle,bluepen+linewidth(2)); - draw((0,1,0)--(2,1,4)--(2,0,0)--cycle,bluepen+linewidth(2)); - - //shade plane - import three; - path3 p=(0,1,0)--(2,1,4)--(2,0,0); - draw(surface(p -- cycle), simplesurfacepen); - - //label and arrow - label("$z=2x+4y-4$",(1,0.5,4)); - draw((1,0.5,3.6)--(1.2,.65,1.6),Arrow3(size=2mm)); - - - - -     - -

    - dz\, dy\, dx: - \ds\int_0^2\int_{1-x/2}^{1}\int_{0}^{2x+4y-4}\, dz\, dy\, dx -

    - -

    - dz\, dx\, dy: - \ds\int_{0}^1\int_{2-2y}^{2}\int_{0}^{2x+4y-4}\, dz\, dx\, dy -

    - -

    - dy\, dz\, dx: - \ds\int_0^2\int_0^{2x}\int_{z/4-x/2+1}^{1}\, dy\, dz\, dx -

    - -

    - dy\, dx\, dz: - \ds\int_0^4\int_{z/2}^{2}\int_{z/4-x/2+1}^{1}\, dy\, dx\, dz -

    - -

    - dx\, dz\, dy: - \ds\int_{0}^1\int_{0}^{4y}\int_{z/2-2y+2}^2\, dx\, dz\, dy -

    - -

    - dx\, dy\, dz: - \ds\int_0^4\int_{z/4}^{1}\int_{z/2-2y+2}^2\, dx\, dy\, dz - \ds V = \int_0^4\int_{z/4}^{1}\int_{z/2-2y-2}^2\, dx\, dy\, dz = 4/3. -

    -     - -     - - - - -

    - D is bounded by the planes z=0 and z=2y, and by y=4-x^2. -

    - -

    - Evaluate the triple integral with the order dz\, dy\, dx. -

    - - - - - A cylindrical wedge with its edge along the x axis. - -

    - A cylindrical wedge, between the planes z=0 and z=2y. - These planes intersect along the x axis, forming the sharp edge of the wedge. - The round face of the wedge is the parabolic cylinder y=4-x^2, for y\geq 0. -

    - -

    - The projection of the surface onto the xy plane is the region bounded below by the x axis, - and above by the parabola y=4-x^2. -

    - -

    - The projection of the surface onto the yz plane is a triangle, - with vertices at (0,0,0), (0,4,0), and (0,4,8). -

    - -

    - The projection of the surface onto the xz plane is the region bounded below by the x axis, - and above by the parabola z = 8-2x^2. -

    -     - - - - - //ASY file for fig13_06_ex_123D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12.1,-7.1,16); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,-1,1,2}; - real[] myychoice={1,2,3,4}; - real[] myzchoice={2,4,6,8}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-0.25,5); - pair zbounds=(-0.25,10); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - triple g(real t) {return (t,4-t^2,0);} - path3 mypath=graph(g,-2,2,operator ..);draw(mypath,redpen+linewidth(2)); - triple g(real t) {return (t,4-t^2,2*(4-t^2));} - path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - //shade object - import three; - int k=12; - for (int i=-2*k; i<2*k; ++i) - { - path3 p=(i/k,4-(i/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i)/k,4-((i)/k)^2,2*(4-((i)/k)^2)); - draw(surface(p -- cycle), simplesurfacepen2); - path3 p=(i/k,0,0)--(i/k,4-(i/k)^2,2*(4-(i/k)^2))--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i+1)/k,0,0); - draw(surface(p -- cycle), simplesurfacepen); - } - - //label and arrow - label("$z=2y$",(-2,2,7)); - draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - label("$y=4-x^2$",(2.5,2,0)); - draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - - -     - -

    - dz\, dy\, dx: - \ds\int_{-2}^2\int_{0}^{4-x^2}\int_{0}^{2y}\, dz\, dy\, dx -

    - -

    - dz\, dx\, dy: - \ds\int_{0}^4\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\int_{0}^{2x+4y-4}\, dz\, dx\, dy -

    - -

    - dy\, dz\, dx: - \ds\int_{-2}^2\int_0^{8-2x^2}\int_{z/2}^{4-x^2}\, dy\, dz\, dx -

    - -

    - dy\, dx\, dz: - \ds\int_0^8\int_{-\sqrt{4-z/2}}^{\sqrt{4-z/2}}\int_{z/2}^{4-x^2}\, dy\, dx\, dz -

    - -

    - dx\, dz\, dy: - \ds\int_{0}^4\int_{0}^{2y}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\, dx\, dz\, dy -

    - -

    - dx\, dy\, dz: - \ds\int_0^8\int_{z/2}^{4}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\, dx\, dy\, dz - \ds V = \int_{-2}^2\int_{0}^{4-x^2}\int_{0}^{2y}\, dz\, dy\, dx = 512/15. -

    -     - -     - - - - -

    - D is bounded by the coordinate planes and by -

    - -

    - y=1-x^2 and y=1-z^2. -

    - -

    - Do not evaluate any triple integral. - Which order is easier to evaluate: - dz\, dy\, dx or dy\, dz\, dx? - Explain why. -

    - - - - - A region in the first octant bounded by the planes x=0 and z=0, and two parabolic cylinders. - -

    - This solid is bounded by two planes, and two parabolic cylinders. -

      -
    • -

      - The bottom of the solid is in the xy plane, - in a region bounded by the x and y coordinate axes and the parabola y=1-x^2. -

      -
      • -
    • -

      - Another side of the solid lies in the yz plane, - forming a region bounded by the y and z coordinate axes and the parabola y=1-z^2. -

      -
      • -
    • -

      - The solid also has a face in the xz plane. This face is a square, with x and z between 0 and 1. -

      -
      • -
    • -

      - The solid is bounded above by the parabolic cylinder y=1-z^2, - and the remaining face is the parabolic cylinder y=1-x^2. -

      -
      • -
    -

    - -

    - When setting up integrals where the integration is first with respect to y, - it is important to note that the square in the xz plane needs to be divided into two triangles along the line x=z. - For z\geq x, y is bounded above by the cylinder y=1-z^2. - For z\leq x, y is bounded above by the cylinder y=1-x^2. -

    -     - - - - - //ASY file for fig13_06_ex_133D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,1.5); - pair ybounds=(-0.25,1.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - triple g(real t) {return (t,1-t^2,0);} - path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (0,1-t^2,t);} - path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,1-t^2,t);} - path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - //draw square and sideline - draw((0,0,0)--(1,0,0)--(1,0,1)--(0,0,1)--(0,0,0),bluepen+linewidth(2)); - draw((0,0,0)--(0,1,0),bluepen+linewidth(2)); - - //shade object - import three; - int k=12; - for (int i=0; i<k; ++i) - { - path3 p=(0,1-(i/k)^2,i/k)--(0,1-((i+1)/k)^2,(i+1)/k)--((i+1)/k,1-((i+1)/k)^2,(i+1)/k)--((i)/k,1-((i)/k)^2,(i)/k); - draw(surface(p -- cycle), simplesurfacepen2);//pink - - path3 p=(i/k,1-(i/k)^2,0)--((i+1)/k,1-((i+1)/k)^2,0)--((i+1)/k,1-((i+1)/k)^2,(i+1)/k)--((i)/k,1-((i)/k)^2,(i)/k); - draw(surface(p -- cycle), simplesurfacepen);//blue - } - - //label and arrow - label("$y=1-x^2$",(1,1,0)); - draw((0.9,0.9,.1)--(.65,.6,.4),Arrow3(size=2mm)); - label("$y=1-z^2$",(0,1,.85)); - draw((.1,.9,.75)--(.3,.8,.45),Arrow3(size=2mm)); - - - - -     - -

    - dz\, dy\, dx: - \ds\int_{0}^1\int_{0}^{1-x^2}\int_{0}^{\sqrt{1-y}}\, dz\, dy\, dx -

    - -

    - dz\, dx\, dy: - \ds\int_{0}^1\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\, dz\, dx\, dy -

    - -

    - dy\, dz\, dx: - \ds\int_{0}^1\int_0^{x}\int_{0}^{1-x^2}\, dy\, dz\, dx + \int_{0}^1\int_x^{1}\int_{0}^{1-z^2}\, dy\, dz\, dx -

    - -

    - dy\, dx\, dz: - \ds\int_0^1\int_{0}^{z}\int_{0}^{1-z^2}\, dy\, dx\, dz+\int_0^1\int_{z}^{1}\int_{0}^{1-x^2}\, dy\, dx\, dz -

    - -

    - dx\, dz\, dy: - \ds\int_{0}^1\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\, dx\, dz\, dy -

    - -

    - dx\, dy\, dz: - \ds\int_0^1\int_{0}^{1-z^2}\int_{0}^{\sqrt{1-y}}\, dx\, dy\, dz - Answers will vary. Neither order is particularly hard. The order dz\, dy\, dx requires integrating a square root, so powers can be messy; the order dy\, dz\, dx requires two triple integrals, but each uses only polynomials. -

    -     - -     - - - - -

    - D is bounded by the coordinate planes and by -

    - -

    - z=1-y/3 and z=1-x. -

    - -

    - Evaluate the triple integral with order dx\, dy\, dz. -

    - - - - - A pyramid with a rectangular base and vertex on the z axis. - -

    - The solid for this exercise is a pyramid. The base in the xy plane is a rectangle, - with x between 0 and 1, and y between 0 and 3. -

    - -

    - One side of the pyramid lies in the plane z=1-x; this is the upper bound when integrating first with respect to y. -

    - -

    - Another side of the pyramid lies in the plane z=1-y/3; this is the upper bound when integrating first with respect to x. -

    - -

    - The remaining sides are in the other two coordinate planes: -

      -
    • -

      - In the xz plane, we have the triangle with vertices (0,0,0), (1,0,0), and (0,0,1) -

      -
      • -
    • -

      - In the yz plane, we have the triangle with vertices (0,0,0), (0,2,0), and (0,0,1) -

      -
      • -
    -

    - -

    - When integrating first with respect to z, - note that the rectangle in the xy plane needs to be divided into two triangles along the line y=3x. - For y\geq 3x, the upper bound is the plane z=1-y/3. - For y\leq 3x, the upper bound is the plane z=1-x. -

    -     - - - - - //ASY file for fig13_06_ex_143D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={1,2,3}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,2.5); - pair ybounds=(-0.25,3.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //edges of object - draw((0,0,0)--(1,0,0)--(1,3,0)--(0,3,0)--cycle,bluepen+linewidth(2)); - draw((1,0,0)--(0,0,1),bluepen+linewidth(2)); - draw((0,3,0)--(0,0,1),bluepen+linewidth(2)); - draw((0,0,0)--(0,0,1),bluepen+linewidth(2)); - draw((1,3,0)--(0,0,1),bluepen+linewidth(2)); - - //shade object - import three; - path3 p=(1,0,0)--(1,3,0)--(0,0,1); - draw(surface(p -- cycle), simplesurfacepen); - path3 p=(1,3,0)--(0,3,0)--(0,0,1); - draw(surface(p -- cycle), simplesurfacepen2); - - //label and arrow - label("$z=1-x$",(2,1.5,0)); - draw((1.75,1.5,0.1)--(0.75,1,.35),Arrow3(size=2mm)); - label("$z=1-y/3$",(0,2.5,1)); - draw((0.15,2.5,0.85)--(0.35,2.25,.35),Arrow3(size=2mm)); - - - - -     - -

    - dz\, dy\, dx: - \ds\int_{0}^1\int_{0}^{3x}\int_{0}^{1-x}\, dz\, dy\, dx+\int_{0}^1\int_{3x}^{3}\int_{0}^{1-y/3}\, dz\, dy\, dx -

    - -

    - dz\, dx\, dy: - \ds\int_{0}^3\int_{0}^{y/3}\int_{0}^{1-y/3}\, dz\, dy\, dx+\int_{0}^3\int_{y/3}^{1}\int_{0}^{1-x}\, dz\, dx\, dy -

    - -

    - dy\, dz\, dx: - \ds\int_{0}^1\int_0^{1-x}\int_{0}^{3-3z}\, dy\, dz\, dx -

    - -

    - dy\, dx\, dz: - \ds\int_0^1\int_{0}^{1-z}\int_{0}^{3-3z}\, dy\, dx\, dz -

    - -

    - dx\, dz\, dy: - \ds\int_{0}^3\int_{0}^{1-y/3}\int_{0}^{1-z}\, dx\, dz\, dy -

    - -

    - dx\, dy\, dz: - \ds\int_0^1\int_{0}^{3-3z}\int_{0}^{1-z}\, dx\, dy\, dz - \ds V = \int_0^1\int_{0}^{3-3z}\int_{0}^{1-z}\, dx\, dy\, dz = 1. -

    -     - -     - -     - - - -

    - Evaluate the triple integral. -

    -     - - - - -

    - \ds \int_{-\pi/2}^{\pi/2}\int_0^\pi\int_0^\pi\big(\cos(x) \sin(y) \sin(z) \big)\, dz\, dy\, dx -

    -     - -     - - - - - $int=Compute("7/8"); - - - -

    - Evaluate \ds \int_{0}^{1}\int_0^x\int_0^{x+y}(x+y+z)\,dz\,dy\,dx. -

    - -

    - -

    -     -     -     - - - - -

    - \ds \int_{0}^{\pi}\int_{0}^{1}\int_{0}^{z}\big(\sin(yz)\big)\, dx\, dy\, dz -

    -     - -

    - \pi -

    -     - -     - - - - - $int=Compute("0"); - - - -

    - Evaluate \ds \int_{\pi}^{\pi^2}\int_{x}^{x^3}\int_{-y^2}^{y^2}\left(z\frac{x^2y+y^2x}{e^{x^2+y^2}}\right)\,dz\,dy\,dx. -

    - -

    - -

    -     -     -     - -     - - - -

    - Find the center of mass of the solid represented by the indicated space region D with density function \delta(x,y,z). -

    -     - - - - -

    - D is bounded by the coordinate planes and -

    - -

    - z=2-2x/3-2y; \delta(x,y,z) = 10\,\text{g/cm}^3. -

    - -

    - (Note: this is the same region as used in .) -

    -     - -

    - M = 10, M_{yz} = 15/2, - M_{xz}=5/2, M_{xy}=5; -

    - -

    - (\overline{x},\overline{y},\overline{z}) = (3/4, 1/4, 1/2) -

    -     - -     - - - - -

    - D is bounded by the planes y=0, - y=2, x=1, z=0 and -

    - -

    - z=(3-x)/2; \delta(x,y,z) = 2\,\text{g/cm}^3. -

    - -

    - (Note: this is the same region as used in .) -

    -     - -

    - M = 4, M_{yz} = 20/3, - M_{xz}=4, M_{xy}=4/3; -

    - -

    - (\overline{x},\overline{y},\overline{z}) = (5/3, 1, 1/3) -

    -     - -     - - - - -

    - D is bounded by the planes x=2, - y=1, z=0 and -

    - -

    - z=2x+4y-4;\delta(x,y,z) = x^2lb/in^3. -

    - -

    - (Note: this is the same region as used in .) -

    -     - -

    - M = 16/5, M_{yz} = 16/3, - M_{xz}=104/45, M_{xy}=32/9; -

    - -

    - (\overline{x},\overline{y},\overline{z}) = (5/3,13/18,10/9) \approx (1.67,0.72,1.11) -

    -     - -     - - - - -

    - D is bounded by the plane z=2y and by y=4-x^2. -

    - -

    - \delta(x,y,z) = y^2lb/in^3. -

    - -

    - (Note: this is the same region as used in .) -

    -     - -

    - M = \frac{65,536}{15}\approx 208.05, - M_{yz} = 0, - M_{xz}=\frac{2,097,152}{3465}\approx 605.24, - M_{xy}=\frac{2,097,152}{3465}\approx 605.24; -

    - -

    - (\overline{x},\overline{y},\overline{z}) = (0,32/11,32/11) \approx (0,2.91,2.91) -

    -     - -     - -     -     -     -
    -
    - Triple Integration with Cylindrical and Spherical Coordinates - -

    - Just as polar coordinates gave us a new way of describing curves in the plane, - in this section we will see how - cylindrical and spherical - coordinates give us new ways of desribing surfaces and regions in space. -

    - - - - -     - - - Cylindrical Coordinates -

    - In short, cylindrical coordinates can be thought of as a combination of the polar and rectangular coordinate systems. - One can identify a point (x_0,y_0,z_0), - given in rectangular coordinates, - with the point (r_0,\theta_0,z_0), - given in cylindrical coordinates, - where the z-value in both systems is the same, - and the point (x_0,y_0) in the xy-plane is identified with the polar point P(r_0,\theta_0); - see . - So that each point in space that does not lie on the z-axis is defined uniquely, - we will restrict r\geq 0 and 0\leq \theta\leq 2\pi. - cylindrical coordinates - coordinatescylindrical -

    - -
    - Illustrating the principles behind cylindrical coordinates - - A schematic diagram of the cylindrical coordinate system in three dimensions. - -

    - A diagram illustrating how a point is located in the cylindrical coordinate system: -

      -
    • -

      - The usual three-dimensional x,y,z coordinate axes are drawn in space, - with the z axis pointing up. -

      -
      • -
    • -

      - In the xy plane, a line segment is drawn from the origin to a point in the first quadrant. - The line segment is labeled r; it is the same r used for polar coordinates in two dimensions. -

      -
      • -
    • -

      - An angle is drawn in the xy plane, from the positive x axis to the ray labeled r. - The angle is labeled \theta; it is the same \theta used for polar coordinates in two dimensions. -

      -
      • -
    • -

      - From the end of the ray labeled r, a vertical line segment is drawn, - from the xy plane to a point in the first octant. - The line segment is labeled z, and the point is labeled (r,\theta,z). -

      -
      • -
    -

    -     - - - //import apexstyle; - - - //ASY file for figstokes2_3D.asy in Chapter 13 - - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(24,17,11.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.1,1.1); - pair ybounds=(-.1,1.1); - pair zbounds=(-.1,1.1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - draw((.5,sqrt(3)/2,1)--(.5,sqrt(3)/2,0)--(0,0,0),dashed); - - draw(arc((0,0,0),(.5,0,0),(.25,sqrt(3)/4,0)),Arrow3(size=2mm)); - - dot((.5,sqrt(3)/2,1),dotblue); - - label("$\theta$",(.55*cos(pi/6),.55*sin(pi/6),.1)); - - label("$(r,\theta,z)$",(.5,sqrt(3)/2,1.1)); - - label("$z$",(.5,sqrt(3)/2+.1,.5)); - - label("$r$",(.20,sqrt(3)/4,0.05)); - - //draw(s,emissive(gray+opacity(.3)),meshpen=invisible); - - - //draw(s,simplesurfacepen,meshpen=apexmeshpen); - - - - //label("$\mathcal{S}$",(1,1,5)); - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - - -
    - -

    - We use the identity z=z along with the identities found in - to convert between the rectangular coordinate (x,y,z) and the cylindrical coordinate (r,\theta,z), namely: - - \begin{array}{l} \text{ From rectangular to cylindrical: } r=\sqrt{x^2+y^2}, \tan(\theta) = y/x \text{ and } z=z;\\ - \text{ From cylindrical to rectangular: } x=r\cos(\theta), y=r\sin(\theta) \text{ and } z=z. \end{array} - -

    - - - -

    - These identities, - along with conversions related to spherical coordinates, - are given later in . -

    - - - - - Converting between rectangular and cylindrical coordinates - -

    - Convert the rectangular point - (2,-2,1) to cylindrical coordinates, - and convert the cylindrical point (4,3\pi/4,5) to rectangular. -

    -     - -

    - Following the identities given above (and, - later in ), - we have r = \sqrt{2^2+(-2)^2} = 2\sqrt{2}. - Using \tan(\theta) = y/x, - we find \theta = \tan^{-1}(-2/2) =-\pi/4. - As we restrict \theta to being between 0 and 2\pi, - we set \theta = 7\pi/4. - Finally, z = 1, - giving the cylindrical point (2\sqrt2,7\pi/4,1). -

    - -

    - In converting the cylindrical point (4,3\pi/4,5) to rectangular, - we have x = 4\cos\big(3\pi/4\big) = -2\sqrt{2}, - y = 4\sin\big(3\pi/4\big) = 2\sqrt{2} and z=5, - giving the rectangular point (-2\sqrt{2},2\sqrt{2},5). -

    -     -     - -

    - Setting each of r, - \theta and z equal to a constant defines a surface in space, - as illustrated in the following example. -

    - - - Canonical surfaces in cylindrical coordinates - -

    - Describe the surfaces r=1, - \theta = \pi/3 and z=2, - given in cylindrical coordinates. -

    -     - -

    - The equation r=1 describes all points in space that are 1 unit away from the z-axis. - This surface is a tube or cylinder - of radius 1, centered on the z-axis, - as graphed in - (which describes the cylinder x^2+y^2=1 in space). -

    - -

    - The equation \theta=\pi/3 describes the plane formed by extending the line \theta=\pi/3, - as given by polar coordinates in the xy-plane, - parallel to the z-axis. -

    - -

    - The equation z=2 describes the plane of all points in space that are 2 units above the xy-plane. - This plane is the same as the plane described by z=2 in rectangular coordinates. -

    - -
    - Graphing the canonical surfaces in cylindrical coordinates from - - Three surfaces in space, corresponding to fixed values of each of the three cylindrical coordinates. - -

    - Three surfaces are drawn in space relative to a set of three-dimensional coordinate axes. - Each surface is obtained by setting one of the three cylindrical coordinates equal to a constant. -

      -
    • -

      - The surface r=1 is a circular cylinder centered about the z axis. - Its intersection with the xy plane is the unit circle. -

      -
      • -
    • -

      - The surface \theta=\pi/3 is a half-plane that terminates on the z axis. - It cuts the cylinder r=1 in a vertical line that passes through the point given by \theta=\pi/3 - on the unit circle. -

      -
      • -
    • -

      - The surface z=2 is a horizontal plane. It intersects the cylinder r=1 - in a circle. -

      -
      • -
    -

    -     - - - //import apexstyle; - - - //ASY file for figcylindrical1_3D.asy in Chapter 13 - - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(27,12,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={1,2}; - defaultpen(0.5mm); - - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-.5,2.75); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (t.x,t.y,2);// - } - surface s=surface(f,(-1.5,-1.5),(1.5,1.5),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=bluemeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - draw((-1.5,-1.5,2) -- (-1.5,1.5,2) -- (1.5,1.5,2) -- (1.5,-1.5,2) -- (-1.5,-1.5,2),apexmeshpen+.25mm); - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (t.x,t.x*sqrt(3),t.y);// - } - surface s=surface(f,(0,-.5),(1.25,2.25),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen q=redmeshpen2+.1mm; - draw(s,simplesurfacepen2,meshpen=q); - - draw((0,0,-.5) -- (1.25,1.25*sqrt(3),-.5) -- (1.25,1.25*sqrt(3),2.25) -- (0,0,2.25) -- (0,0,-.5),redcurvepen+.25mm); - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (cos(t.x),sin(t.x),t.y);// - } - surface s=surface(f,(0,-.5),(2*pi,2.25),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen q=greenmeshpen+.1mm; - - draw(s,simplesurfacepen3,meshpen=q); - - //dot((.5,sqrt(3)*.5,2),rgb(.1,.1,.1)); - - draw((.5,sqrt(3)*.5,-.5)--(.5,sqrt(3)*.5,2.25)); - - triple g(real t) {return (cos(t),sin(t),2);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,blackmeshpen); - - triple g(real t) {return (cos(t),sin(t),2.25);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,greencurvepen+.25mm); - - triple g(real t) {return (cos(t),sin(t),-.5);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,greencurvepen+.25mm); - - triple g(real t) {return (t,sqrt(3)*t,2);} - path3 mypath=graph(g,0,.866,operator ..); - - draw(mypath,blackmeshpen); - - //draw(s,emissive(gray+opacity(.3)),meshpen=invisible); - - - //draw(s,simplesurfacepen,meshpen=apexmeshpen); - - - - label("$z=2$",(-1.25,1.25,2.3)); - label("$r=1$",(1.1*cos(pi/6),1.1*sin(pi/6),-.75)); - - label("$\theta=\frac{\pi}{3}$",(1.25,1.25*sqrt(3),1)); - - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - -
    - -

    - All three surfaces are graphed in . - Note how their intersection uniquely defines the point P=(1,\pi/3,2). -

    -     -     - -

    - Cylindrical coordinates are useful when describing certain domains in space, - allowing us to evaluate triple integrals over these domains more easily than if we used rectangular coordinates. -

    - -

    - - shows how to evaluate \iiint_Dh(x,y,z)\, dV using rectangular coordinates. - In that evaluation, we use dV = dz\,dy\,dx - (or one of the other five orders of integration). - Recall how, in this order of integration, - the bounds on y are curve to curve - and the bounds on x are point to point: - these bounds describe a region R in the xy-plane. - We could describe R using polar coordinates as done in . - In that section, we saw how we used - dA = r\,dr\,d\theta instead of dA = dy\,dx. -

    - -

    - Considering the above thoughts, - we have dV = dz\big(r\,dr\,d\theta\big) = r\,dz\,dr\,d\theta. - We set bounds on z as surface to surface - as done in the previous section, - and then use curve to curve - and point to point - bounds on r and \theta, respectively. - Finally, using the identities given above, - we change the integrand h(x,y,z) to h(r,\theta,z). -

    - -

    - This process should sound plausible; - the following theorem states it is truly a way of evaluating a triple integral. -

    - - - Triple Integration in Cylindrical Coordinates - -

    - Let w=h(r,\theta,z) be a continuous function on a closed, - bounded region D in space, - bounded in cylindrical coordinates by \alpha \leq \theta \leq \beta, - g_1(\theta)\leq r \leq g_2(\theta) and f_1(r,\theta) \leq z \leq f_2(r,\theta). - Then integrationwith cylindrical coordinates - - \iiint_D h(r,\theta,z)\, dV = \int_\alpha^\beta\int_{g_1(\theta)}^{g_2(\theta)}\int_{f_1(r,\theta)}^{f_2(r,\theta)}h(r,\theta,z) r\,dz\,dr\,d\theta - . -

    -     -     - - - - - Evaluating a triple integral with cylindrical coordinates - -

    - Find the mass of the solid represented by the region in space bounded by z=0, - z=\sqrt{4-x^2-y^2}+3 and the cylinder x^2+y^2=4 - (as shown in ), - with density function \delta(x,y,z) = x^2+y^2+z+1, - using a triple integral in cylindrical coordinates. - Distances are measured in centimeters and density is measured in grams per cubic centimeter. -

    -
    - Visualizing the solid used in - - A circular cylinder capped by a spherical dome. - -

    - The surface depicted in this image looks like the top of a grain silo, - or like the end of a pill capsule. - It consists of a circular cylinder, centered on the z axis, - extending from the xy plane to the plane z=3. -

    - -

    - When z=3, the cylinder meets a hemisphere of the same radius as the cylinder. - The peak of the hemisphere lies on the z axis, at the point (0,0,5). -

    -     - - - //import apexstyle; - - - //ASY file for figcylindrical2_3D.asy in Chapter 13 - - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(22,22,11); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={1,3,5}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-.5,6); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (2*t.y*cos(t.x),2*t.y*sin(t.x),sqrt(4-4*t.y^2)+3);// - } - surface s=surface(f,(0,0),(2pi,1),16,16,Spline); - pen p=bluemeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - //draw((-1.5,-1.5,2) -- (-1.5,1.5,2) -- (1.5,1.5,2) -- (1.5,-1.5,2) -- (-1.5,-1.5,2),apexmeshpen+.25mm); - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (2*cos(t.x),2*sin(t.x),t.y);// - } - surface s=surface(f,(0,0),(2*pi,3),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=bluemeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - - - //draw(s,emissive(gray+opacity(.3)),meshpen=invisible); - - - //draw(s,simplesurfacepen,meshpen=apexmeshpen); - - - - - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - -
    -     - -

    - We begin by describing this region of space with cylindrical coordinates. - The plane z=0 is left unchanged; - with the identity r=\sqrt{x^2+y^2}, - we convert the hemisphere of radius 2 to the equation z=\sqrt{4-r^2}; - the cylinder x^2+y^2=4 is converted to r^2=4, - or, more simply, r=2. - We also convert the density function: - \delta(r,\theta,z) = r^2+z+1. -

    - -

    - To describe this solid with the bounds of a triple integral, - we bound z with 0\leq z\leq \sqrt{4-r^2}+3; - we bound r with 0 \leq r \leq 2; - we bound \theta with 0 \leq \theta \leq 2\pi. -

    - -

    - Using - and , - we have the mass of the solid is - - M=\iiint_D\delta(x,y,z)\, dV \amp = \int_0^{2\pi}\int_0^2\int_0^{\sqrt{4-r^2}+3}\big(r^2+z+1\big)r\,dz\,dr\,d\theta - \amp = \int_0^{2\pi}\int_0^2\big((r^3+4r)\sqrt{4-r^2}+\frac52r^3+\frac{19}2r\big)\,dr\,d\theta - \amp = \frac{1318\pi}{15} \approx 276.04\,\text{g} - , - where we leave the details of the remaining double integral to the reader. -

    -     -     - - - Finding the center of mass using cylindrical coordinates - -

    - Find the center of mass of the solid with constant density whose base can be described by the polar curve - r=\cos(3\theta) and whose top is defined by the plane z=1-x+0.1y, - where distances are measured in feet, - as seen in . - (The volume of this solid was found in .) -

    -
    - Visualizing the solid used in - - A solid whose cross-sections are like a three-leaf clover; it has been sliced at an angle. - -

    - A three-dimensional solid is shown, plotted against a set of three-dimensional coordinate axes. - The base of the solid is a three-leaf rose curve. -

    - -

    - The solid is a cylinder, except that it is cut at an angle, - where it meets the plane z=1-x+0.1y. - We first saw this solid in . -

    -     - - - - //ASY file for figdoublepol43D.asy in Chapter 13 - - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-2,2); - - //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //plane is z=1-x+0.1y - //Draw the surface - //({cos(3*x)*cos(x)},{cos(3*x)*sin(x)},{(1-cos(3*x)*cos(x)+.1*cos(3*x)*sin(x))*y}); - triple f(pair t) { - return (cos(3*t.x)*cos(t.x),cos(3*t.x)*sin(t.x),(1-cos(3*t.x)*cos(t.x)+.1*cos(3*t.x)*sin(t.x))*t.y); - } - surface s=surface(f,(0,0),(2*pi,1),32,32,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //draw curve on xy plane //({cos(3*x)*cos(x)},{cos(3*x)*sin(x)},{0}) - triple g(real t) {return (cos(3*t)*cos(t),cos(3*t)*sin(t),0);} - path3 mypath=graph(g,0,pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - - //draw curves on surface - triple g(real t) {return (cos(3*t)*cos(t),cos(3*t)*sin(t),1-(cos(3*t)*cos(t))+0.1*(cos(3*t)*sin(t)));} - path3 mypath=graph(g,0,pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - - //Shade the bottom - import three; - path3 p = (0,0,0)..(.175,0.064,0)..(.264,.119,0)..(0.683,0.183,0)..(.916,0.121,0)..(1,0,0)..(.916,-0.121,0)..(0.683,-0.183,0)..(.264,-.119,0)..(.175,-0.086,0); - draw(surface(p -- cycle), simplesurfacepen); - - path3 p = (0,0,0)..(-0.1429, 0.1196,0)..(-0.2351, 0.1691,0)..(-0.5000, 0.5000,0)..(-0.5628, 0.7328,0)..(-0.5000, 0.8660,0)..(-0.3532, 0.8538,0)..(-0.1830, 0.6830,0)..(-0.0289, 0.2881,0)..(-0.0130, 0.1946,0); - draw(surface(p -- cycle), simplesurfacepen); - - path3 p = (0,0,0)..(-0.0321,-0.1836,0)..(-0.0289,-0.2881,0)..(-0.1830,-0.6830,0)..(-0.3532,-0.8538,0)..(-0.5000,-0.8660,0)..(-0.5628,-0.7328,0)..(-0.5000,-0.5000,0)..(-0.2351,-0.1691,0)..(-0.1620,-0.1086,0); - draw(surface(p -- cycle), simplesurfacepen); - - //Shade the top - path3 p = (0,0,1.0000)..(-0.1429,0.1196,1.1549)..(-0.2351,0.1691,1.2520)..(-0.5000,0.5000,1.5500)..(-0.5628,0.7328,1.6361)..(-0.5000,0.8660,1.5866)..(-0.3532,0.8538,1.4386)..(-0.1830 ,0.6830 ,1.2513)..(-0.0289,0.2881,1.0578)..(-0.0130,0.1946,1.0325); - - draw(surface(p -- cycle), simplesurfacepen); - path3 p = (0 , 0 , 1.0000)..(-0.0321 , -0.1836 , 1.0137)..(-0.0289, -0.2881, 1.0001)..(-0.1830, -0.6830, 1.1147)..(-0.3532 , -0.8538 , 1.2678)..(-0.5000, -0.8660 , 1.4134)..(-0.5628 , -0.7328, 1.4895)..(-0.5000 , -0.5000 , 1.4500)..(-0.2351 , -0.1691 , 1.2181)..(-0.1620 , -0.1086, 1.1511); - draw(surface(p -- cycle), simplesurfacepen); - - path3 p = (0 , 0, 1.0000)..( 0.1750, 0.0640 , 0.8314)..( 0.2640, 0.1190 , 0.7479)..( 0.6830, 0.1830 , 0.3353)..( 0.9160, 0.1210 , 0.0961)..( 1.0000, 0 , 0)..( 0.9160, -0.1210 , 0.0719)..( 0.6830, -0.1830 , 0.2987)..( 0.2640, -0.1190 , 0.7241)..( 0.1750, -0.0860 , 0.8164); - draw(surface(p -- cycle), simplesurfacepen); - - -
    -     - -

    - We convert the equation of the plane to use cylindrical coordinates: - z= 1-r\cos(\theta)+0.1r\sin(\theta). - Thus the region is space is bounded by 0 \leq z \leq 1-r\cos(\theta) + 0.1r\sin(\theta), - 0 \leq r \leq \cos(3\theta), - 0 \leq \theta \leq \pi (recall that the rose curve - r=\cos(3\theta) is traced out once on [0,\pi]. -

    - -

    - Since density is constant, - we set \delta = 1 and finding the mass is equivalent to finding the volume of the solid. - We set up the triple integral to compute this but do not evaluate it; - we leave it to the reader to confirm it evaluates to the same result found in . - - M = \iiint_D\delta \, dV = \int_0^{\pi}\int_0^{\cos(3\theta)}\int_0^{1-r\cos(\theta)+0.1r\sin(\theta)} r\,dz\,dr\,d\theta = \frac{\pi}{4} - . -

    - -

    - From - we set up the triple integrals to compute the moments about the three coordinate planes. - The computation of each is left to the reader (using technology is recommended): - - M_{yz} = \iiint_D x\,dV \amp = \int_0^{\pi}\int_0^{\cos(3\theta)}\int_0^{1-r\cos(\theta)+0.1r\sin(\theta)} (r\cos(\theta)) r\,dz\,dr\,d\theta - \amp = \frac{-3\pi}{64} \approx -0.147. - - - M_{xz} = \iiint_D y\,dV \amp = \int_0^{\pi}\int_0^{\cos(3\theta)}\int_0^{1-r\cos(\theta)+0.1r\sin(\theta)} (r\sin(\theta)) r\,dz\,dr\,d\theta - \amp = \frac{3\pi}{640} \approx 0.015. - M_{xy} = \iiint_D z\,dV \amp = \int_0^{\pi}\int_0^{\cos(3\theta)}\int_0^{1-r\cos(\theta)+0.1r\sin(\theta)} (z) r\,dz\,dr\,d\theta - \amp = \frac{1903\pi}{12800} \approx 0.467 - . -

    - -

    - The center of mass in rectangular coordinates, - found by dividing the respective moments by the mass, - is approximately located at (-0.188,0.019,0.595), - which lies outside the bounds of the solid. -

    -     -     -     - - - Spherical Coordinates -

    - spherical coordinates - coordinatesspherical -

    - -

    - In short, spherical coordinates can be thought of as a - double application - of the polar coordinate system. - In spherical coordinates, - a point P is identified with (\rho,\theta,\varphi), - where \rho is the distance from the origin to P, - \theta is the same angle as would be used to describe P in the cylindrical coordinate system, - and \varphi is the angle between the xy-plane and the ray from the origin to P; - see . - So that each point in space that does not lie on the z-axis is defined uniquely, - we will restrict \rho \geq 0, - 0 \leq \theta \leq 2\pi and -\pi/2 \leq \varphi \leq \pi/2. -

    - - - - -

    - Note that most mathematics textbooks define \varphi to be measured - from the positive z-axis, with values in [0,\pi], - rather than from the xy-plane. -

    - -

    - We have chosen our convention with a number of considerations in mind: -

      -
    • -

      - The coordinates (\rho,\theta,\varphi) form a right-handed - coordinate system: one in which the orientation matches that of our usual (x,y,z) - coordinates, where the right-hand rule applies. - If \varphi is measured from the z-axis, - the order (\rho, \varphi, \theta) is needed to get a right-handed system. -

      -
      • -
    • -

      - Points of the form (a,\alpha,0) are the same in both cylindrical and spherical coordinates. -

      -
      • -
    • -

      - Some integration problems become slightly easier: - we will see soon that the volume element in spherical coordinates involves \cos(\varphi), - which integrates to \sin(\varphi). - In the usual convention, the volume element involves \sin(\varphi), - which integrates to -\cos(\varphi) a source of many common sign errors. -

      -
      • -
    -

    - -

    - Students of Physics will encounter yet another convention. - In Physics, the variable r is preferred as the radial coordinate, - and spherical coordinates are given as (r,\theta,\varphi); - however, in Physics, \varphi becomes the angle in the xy-plane, - while \theta is the angle measured from the positive z-axis. -

    - -

    - Note that the angle in the xy-plane (\theta, in our case) - is known as the azimuthal angle. - Our angle \varphi is known as the elevation angle. - The angle used in other conventions that is measured from the positive z-axis - (often identified with the north pole) is known as the polar angle. - For further discussion, the Wikipedia article - is quite useful. -

    -     - -
    - Illustrating the principles behind spherical coordinates - - A schematic diagram illustrating the spherical coordinate system relative to the rectangular coordinate axes. - -

    - A diagram illustrating how a point is located in the spherical coordinate system: -

      -
    • -

      - The usual three-dimensional x,y,z coordinate axes are drawn, with the z axis pointing up. -

      -
      • -
    • -

      - A point in space is plotted somewhere in the first octant; it is labelled with the coordinates (\rho,\varphi,\theta). - A line segment is drawn from the origin to this point; its length is labeled with the spherical coordinate \rho. -

      -
      • -
    • -

      - Another line segement is drawn from the point in space, vertically down to the xy plane: - if the rectangular coordinates of the point in space are (x,y,z), - then the point in the plane has rectangular coordinates (x,y,0). - (Note that this is the same vertical segment used to define the cylindrical coordinate z.) -

      -
      • -
    • -

      - Finally, a third line segment is drawn from the origin to the point in the plane that lies directly below the original point in space. - This line segment lies in the xy plane. The three line segments drawn in this diagram form a right angle triangle, - with the segment of length \rho as hypotenuse. -

      -
      • - -
    • -

      - An angle is drawn from the segment in the xy plane to the segment of length \rho. - This angle is labeled with the spherical coordinate \varphi; - it is the angle at the vertex in the right-angled triangle that is at the origin. -

      -
      • - -
    • -

      - Another angle is drawn from the positive x axis to the side of the triangle in the xy plane, - and labeled with the spherical coordinate \theta. -

      -
      • - -
    • -

      - Note that the angle \theta is the same in polar, cylindrical, and spherical coordinates. -

      -
      • -
    -

    -     - - - //import apexstyle; - - - //ASY file for figsphericalintro.asy in Chapter 13 - - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(24,17,11.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.1,1.1); - pair ybounds=(-.1,1.1); - pair zbounds=(-.1,1.1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - draw((.5,sqrt(3)/2,1)--(.5,sqrt(3)/2,0)--(0,0,0)--cycle,dashed); - - draw(arc((0,0,0),(.5,0,0),(.25,sqrt(3)/4,0)),Arrow3(size=2mm)); - - draw(arc((0,0,0),(0.25,sqrt(3)/4,0),(.5,sqrt(3)/2,1)*.5/1.414),Arrow3(size=2mm)); - - guide3 myarc=arc((0,0,0),(0.25,sqrt(3)/4,0),(0.5,sqrt(3)/2,1)*.5/1.414); - - label("$\varphi$",myarc,NE); - - - dot((.5,sqrt(3)/2,1),dotblue); - - label("$\theta$",(.55*cos(pi/6),.55*sin(pi/6),.1)); - - label("$(\rho,\theta,\varphi)$",(.5,sqrt(3)/2,1.1)); - - //label("$z$",(.5,sqrt(3)/2+.1,.5)); - - label("$\rho$",(.5,sqrt(3)/2,1)*.5+(0.0,0.0,.07)); - - //draw(s,emissive(gray+opacity(.3)),meshpen=invisible); - - - //draw(s,simplesurfacepen,meshpen=apexmeshpen); - - -
    - - - - - -

    - The following Key Idea gives conversions to/from our three spatial coordinate systems. -

    - - - Converting Between Rectangular, Cylindrical and Spherical Coordinates -

    -

      -
    • - Rectangular and Cylindrical -

      - - \amp r^2 = x^2+y^2, \amp \amp \tan(\theta) = y/x,\amp z\amp=z - \amp x=r\cos(\theta), \amp \amp y =r\sin(\theta),\amp z\amp =z - -

      -
      • -
    • - Rectangular and Spherical -

      - - \amp \rho = \sqrt{x^2+y^2+z^2}, \amp \amp \tan(\theta) = y/x, \amp \amp \sin(\varphi) = z/\sqrt{x^2+y^2+z^2} - \amp x=\rho\cos(\varphi)\cos(\theta),\amp \amp y=\rho\cos(\varphi)\sin(\theta),\amp \amp z=\rho\sin(\varphi) - -

      -
      • -
    • - Cylindrical and Spherical -

      - - \amp \rho =\sqrt{r^2+z^2},\amp \amp \theta = \theta,\amp \amp \tan(\varphi) = z/r - \amp r=\rho \cos(\varphi),\amp \amp \theta = \theta,\amp \amp z=\rho\sin(\varphi) - -

      -
      • -
    -

    -     - - - - - Converting between rectangular and spherical coordinates - -

    - Convert the rectangular point - (2,-2,1) to spherical coordinates, - and convert the spherical point - (6,\pi/3,0) to rectangular and cylindrical coordinates. -

    -     - -

    - This rectangular point is the same as used in . - Using , - we find \rho = \sqrt{2^2+(-1)^2+1^2} = 3. - Using the same logic as in , - we find \theta = 7\pi/4. - Finally, \sin(\varphi) = 1/3, - giving \varphi = \sin^{-1}(1/3) \approx 0.34, - or about 19.47^\circ. - Thus the spherical coordinates are approximately (3,7\pi/4,0.34). -

    - -

    - Converting the spherical point (6,\pi/3,0) to rectangular, - we have x = 6\cos(0)\cos(\pi/3) = 3, - y = 6\cos(0)\sin(\pi/3) = 3\sqrt{3} and z = 6\sin(0) = 0. - Thus the rectangular coordinates are (3,3\sqrt{3},0). -

    - -

    - To convert this spherical point to cylindrical, - we have r = 6\cos(0) = 6, - \theta = \pi/3 and z = 6\sin(0) =0, - giving the cylindrical point (6,\pi/3,0). -

    -     -     - - - Canonical surfaces in spherical coordinates - -

    - Describe the surfaces \rho=1, - \theta = \pi/3 and \varphi = \pi/3, - given in spherical coordinates. -

    -     - -

    - The equation \rho = 1 describes all points in space that are 1 unit away from the origin: - this is the sphere of radius 1, centered at the origin. -

    - -

    - The equation \theta = \pi/3 describes the same surface in spherical coordinates as it does in cylindrical coordinates: - beginning with the line \theta = \pi/3 in the xy-plane as given by polar coordinates, - extend the line parallel to the z-axis, forming a plane. -

    - -

    - The equation \varphi=\pi/3 describes all points P - in space where the ray from the origin to P makes an angle of \pi/3 with the xy-plane. - This describes a cone, - with the positive z-axis its axis of symmetry, with point at the origin. -

    - -
    - Graphing the canonical surfaces in spherical coordinates from - - A three-dimensional plot showing three surfaces, each of which is obtained by fixing a value of one of the three spherical coordinates. - -

    - Three surfaces are plotted relative to the usual three-dimensional coordinate axes. -

      -
    • -

      - The surface \rho=1 is the unit sphere: a sphere of radius 1, centered at the origin. -

      -
      • -
    • -

      - The surface \varphi = \pi/3 is the top half of a circular cone that opens along the z axis. - The angle between any line through the origin that lies on the cone and the xy plane is \pi/3. - (Equivalently, the angle between any line through the origin that lies on the cone and the z axis is \pi/6.) -

      -
      • -
    • -

      - The surface \theta=\pi/3 is a half-plane that terminates on the z axis, - the same as it was in cylindrical coordinates. -

      -
      • -
    -

    -     - - - //import apexstyle; - - - //ASY file for figspherical1_3D.asy in Chapter 13 - - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(27,13,9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-2,2); - pair ybounds=(-2,2); - pair zbounds=(-1.5,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),sqrt(3)*t.y);// - } - surface s=surface(f,(0,0),(2*pi,1.5/sqrt(3)),16,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=bluemeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - //draw((-1.5,-1.5,2) -- (-1.5,1.5,2) -- (1.5,1.5,2) -- (1.5,-1.5,2) -- (-1.5,-1.5,2),apexmeshpen+.25mm); - - triple g(real t) {return (1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5);} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,apexmeshpen); - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (t.x,t.x*sqrt(3),t.y);// - } - surface s=surface(f,(0,-1.25),(1.25,1.5),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen q=redmeshpen2+.1mm; - draw(s,simplesurfacepen2,meshpen=q); - - draw((0,0,-1.25) -- (1.25,1.25*sqrt(3),-1.25) -- (1.25,1.25*sqrt(3),1.5) -- (0,0,1.5) -- (0,0,-1.5),redcurvepen+.25mm); - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (sin(t.y)*cos(t.x),sin(t.y)*sin(t.x),cos(t.y));// - } - surface s=surface(f,(0,0),(2*pi,pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen q=greenmeshpen+.1mm; - draw(s,simplesurfacepen3,meshpen=q); - - //dot((.5,sqrt(3)*.5,2),rgb(.1,.1,.1)); - - //draw((.5,sqrt(3)*.5,-.5)--(.5,sqrt(3)*.5,2.25)); - - triple g(real t) {return (1/2*cos(t),1/2*sin(t),1/2*sqrt(3));} - path3 mypath=graph(g,0,2*pi,operator ..); - draw(mypath,blackmeshpen); - - //(1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5) - draw((0,0,0) -- (1.5/sqrt(3)*cos(pi/3),1.5/sqrt(3)*sin(pi/3),1.5)); - - triple g(real t) {return (sin(t)*cos(pi/3),sin(t)*sin(pi/3),cos(t));} - path3 mypath=graph(g,0,pi,operator ..); - draw(mypath,blackmeshpen); - - triple g(real t) {return (cos(t),sin(t),-.5);} - path3 mypath=graph(g,0,2*pi,operator ..); - //draw(mypath,greencurvepen+.25mm); - - triple g(real t) {return (t,sqrt(3)*t,2);} - path3 mypath=graph(g,-.866,.866,operator ..); - //draw(mypath,blackmeshpen); - - //draw(s,emissive(gray+opacity(.3)),meshpen=invisible); - - - //draw(s,simplesurfacepen,meshpen=apexmeshpen); - - - - label("$\varphi=\frac{\pi}{3}$",(1.5/sqrt(3)*cos(2pi/3),1.5/sqrt(3)*sin(2pi/3),1.5),E); - label("$\rho=1$",(1.1*cos(pi/200),1.1*sin(pi/200),-.9)); - label("$\theta=\frac{\pi}{3}$",(1.25,1.25*sqrt(3),1)); - - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - -
    - -

    - All three surfaces are graphed in . - Note how their intersection uniquely defines the point P=(1,\pi/3,\pi/6). -

    -     -     - -

    - Spherical coordinates are useful when describing certain domains in space, - allowing us to evaluate triple integrals over these domains more easily than if we used rectangular coordinates or cylindrical coordinates. - The crux of setting up a triple integral in spherical coordinates is appropriately describing the - small amount of volume, - dV, used in the integral. -

    - -

    - Considering , - we can make a small spherical wedge by varying \rho, - \theta and \varphi each a small amount, \Delta\rho, - \Delta\theta and \Delta\varphi, respectively. - This wedge is approximately a rectangular solid when the change in each coordinate is small, - giving a volume of about - - \Delta V \approx \Delta\rho\ \times\ \rho\Delta\varphi\ \times\ \rho\cos(\varphi)\Delta\theta - . -

    - -
    - Approximating the volume of a standard region in space using spherical coordinates - - A schematic diagram illustrating how the spherical volume element is computed. - -

    - An illustration of the volume obtained from a small spherical wedge. - The diagram is quite complicated: -

      -
    • -

      - Along the z axis, a segment of length \rho is indicated. -

      -
      • -
    • -

      - This segement is rotated in a vertical plane (corresponding to a fixed value of \theta). - It sweeps out a circular arc spanned by an angle \Delta \varphi; - the length of the arc is \rho\Delta\varphi. -

      -
      • -
    • -

      - Two rays from the origin are drawn in the xy plane; - the angle between them is labeled \Delta\theta. -

      -
      • -
    • -

      - Above one of these two rays, a right triangle is drawn. - The hypotenuse of the triangle is a ray from the origin to one end of the arc of length \rho\Delta\varphi. - The base of the triangle is the side adjacent to the angle \varphi, so the length of the base is \rho\cos\varphi. -

      -
      • -
    • -

      - The circular arc between the two rays in the xy plane therefore has length \rho\cos\varphi\Delta\theta. -

      -
      • -
    • -

      - The spherical wedge that is depicted looks like a distorted rectangular box. - The dimensions of the box, based on the observations above, are - - \Delta\rho\times\rho\Delta\varphi\times\rho\cos\varphi\Delta\theta - . -

      -
      • -
    -

    -     - - - //import apexstyle; - - - //ASY file for figspherical1_3D.asy in Chapter 13 - - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14.4,2.6,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.1,.6); - pair ybounds=(-.1,.6); - pair zbounds=(-.1,1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - - real t1=pi/3-.2; - real t2=pi/3+.2; - real p1=pi/6-.2; - real p2=pi/6+.2; - real r1=.75; - real r2=.9; - - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (r1*cos(t.x)*sin(t.y),r1*sin(t.x)*sin(t.y),r1*cos(t.y));// - } - surface s=surface(f,(t1,p1),(t2,p2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - - triple f(pair t) { - return (r2*cos(t.x)*sin(t.y),r2*sin(t.x)*sin(t.y),r2*cos(t.y));// - } - surface s=surface(f,(t1,p1),(t2,p2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - triple f(pair t) { - return (t.y*cos(t.x)*sin(p1),t.y*sin(t.x)*sin(p1),t.y*cos(p1));// - } - surface s=surface(f,(t1,r1),(t2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - triple f(pair t) { - return (t.y*cos(t.x)*sin(p2),t.y*sin(t.x)*sin(p2),t.y*cos(p2));// - } - surface s=surface(f,(t1,r1),(t2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - triple f(pair t) { - return (t.y*cos(t1)*sin(t.x),t.y*sin(t1)*sin(t.x),t.y*cos(t.x));// - } - surface s=surface(f,(p1,r1),(p2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - triple f(pair t) { - return (t.y*cos(t2)*sin(t.x),t.y*sin(t2)*sin(t.x),t.y*cos(t.x));// - } - surface s=surface(f,(p1,r1),(p2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - // - // lines for phi - // - draw((r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1)) -- (0,0,0) -- (r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)),black+.25mm+dashed); - - draw(arc((0,0,0),(r1*cos(t1)*sin(p2)/2.5,r1*sin(t1)*sin(p2)/2.5,r1*cos(p2)/2.5),(r1*cos(t1)*sin(p1)/2.5,r1*sin(t1)*sin(p1)/2.5,r1*cos(p1)/2.5)),black+.25mm,Arrow3(size=1.5mm)); - - label("$\Delta\varphi$",.95*(r1*cos(t1)*sin((p1+p2)/2)/2,r1*sin(t1)*sin((p1+p2)/2)/2,r1*cos((p1+p2)/2)/2)); - - - - // - // lines for theta - // - - draw((0,0,0) -- (r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0) -- (r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)),black+.25mm+dashed); - - draw((0,0,0) -- (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),0) -- (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)),black+.25mm+dashed); - - draw(arc((0,0,0),.5*(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0),.5*(r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),0)),black+.25mm,Arrow3(size=1.5mm)); - - label("$\Delta\theta$",1.2*.5*(r1*cos((t1+t2)/2)*sin(p2),r1*sin((t1+t2)/2)*sin(p2),0)); - - // - // lines for rho - // - - draw(arc((0,0,0),(r1*cos(t1)*sin(0),r1*sin(t1)*sin(0),r1*cos(0)),(r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1))),black+.25mm+dashed); - - draw(arc((0,0,0),(r2*cos(t1)*sin(0),r2*sin(t1)*sin(0),r2*cos(0)),(r2*cos(t1)*sin(p1),r2*sin(t1)*sin(p1),r2*cos(p1))),black+.25mm+dashed); - - - label("$\Delta\rho$",((r1+r2)/2*cos(t1)*sin(.65*(p1)),(r1+r2)/2*sin(t1)*sin(.65*(p1)),(r1+r2)/2*cos(.65*(p1)))); - - real poff = -0.05; - - draw((r1*cos(t1)*sin(p1+poff),r1*sin(t1)*sin(p1+poff),r1*cos(p1+poff)) -- (r2*cos(t1)*sin(p1+poff),r2*sin(t1)*sin(p1+poff),r2*cos(p1+poff)),black+.25mm,Arrows3(size=1.5mm)); - - label("$\rho$",(0,-.04,.5*r1)); - - draw((0,-.02,.01) -- (0,-.02,.99*r1),black+.25mm,Arrows3(size=1.5mm)); - - - // - // lines for theta length - // - - draw((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) -- (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05),black+.25mm,Arrows3(size=1.5mm)); - - label("$\rho\cos(\varphi)\Delta\theta$",1.5*((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05)+(0,0,-.05))/2+(0,0,-.25),S); - - draw(1.5*((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05)+(0,0,-.05))/2 + (0,0,-.25) -- 1.05((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05))/2+(0,0,-.01),black+.25mm+dashed,Arrow3(size=1.5mm)); - - // - // lines for phi length - // - - label("$\rho\Delta\varphi$",.85*(r1*cos(t1)*sin((p1+p2)/2),r1*sin(t1)*sin((p1+p2)/2),r1*cos((p1+p2)/2))); - - draw(arc((0,0,0),.95*(r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1)),.95*(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2))),black+.25mm,Arrows3(size=1.5mm)); - - -
    - -

    - Given a region D in space, - we can approximate the volume of D with many such wedges. - As the size of each of \Delta\rho, - \Delta\theta and \Delta\varphi goes to zero, - the number of wedges increases to infinity and the volume of D is more accurately approximated, giving - - dV = d\rho\ \times\ \rho\, d\varphi\ \times\ \rho\cos(\varphi)d\theta = \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi - . -

    - -

    - Again, this development of dV should sound reasonable, - and the following theorem states it is the appropriate manner by which triple integrals are to be evaluated in spherical coordinates. -

    - - - - Triple Integration in Spherical Coordinates - -

    - Let w=h(\rho,\theta,\varphi) be a continuous function on a closed, - bounded region D in space, - bounded in spherical coordinates by \alpha_1 \leq \varphi \leq \alpha_2, - \beta_1 \leq \theta \leq \beta_2 and f_1(\theta,\varphi) \leq \rho \leq f_2(\theta,\varphi). - Then integrationwith spherical coordinates - - \iiint_D h(\rho,\theta,\varphi)\, dV = \int_{\alpha_1}^{\alpha_2}\int_{\beta_1}^{\beta_2}\int_{f_1(\theta,\varphi)}^{f_2(\theta,\varphi)} h(\rho,\theta,\varphi) \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi - . -

    -     -     - - - Establishing the volume of a sphere - -

    - Let D be the region in space bounded by the sphere, - centered at the origin, of radius r. - Use a triple integral in spherical coordinates to find the volume V of D. -

    -     - -

    - The sphere of radius r, - centered at the origin, has equation \rho = r. - To obtain the full sphere, - the bounds on \theta and \varphi are - 0\leq \theta \leq 2\pi and -\pi/2 \leq \varphi \leq \pi/2. - This leads us to: - - V \amp = \iiint_D\, dV - \amp = \int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\int_0^r\big(\rho^2\cos(\varphi)\big)\, d\rho\, d\theta\, d\varphi - \amp = \int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\left(\frac13\rho^3\cos(\varphi)\Big|_0^r\right)\, d\theta\, d\varphi - \amp = \int_{-\pi/2}^{\pi/2}\int_0^{2\pi} \left(\frac13r^3\cos(\varphi)\right)\, d\theta\, d\varphi - \amp = \int_{-\pi/2}^{\pi/2} \left(\frac{2\pi}3r^3\cos(\varphi)\right)\, d\varphi - \amp = \left.\left(\frac{2\pi}3r^3\sin(\varphi)\right)\right|_{-\pi/2}^{\pi/2} - \amp = \frac{4\pi}3r^3 - , - the familiar formula for the volume of a sphere. - Note how the integration steps were easy, - not using square roots nor integration steps such as Substitution. -

    -     -     - - - - - Finding the center of mass using spherical coordinates - -

    - Find the center of mass of the solid with constant density enclosed above by \rho=4 and below by \varphi = \pi/3, - as illustrated in . -

    -
    - Graphing the solid, and its center of mass, from - - A gem-like solid, bounded below by the top half of a circular cone, and above by a spherical cap. - -

    - The solid depicted in this image is reminiscent of a gemstone. - It is bounded below by a circular cone with its vertex at the origin, that opens upwards along the z axis. - The solid is bounded above by a spherical cap. - (Perhaps the solid also resembles a snow cone.) -

    - -

    - Also depicted (but only barely visible) is a point marked on the z axis to indicate the location of the center of mass. -

    -     - - - //import apexstyle; - - - //ASY file for figspherical1_3D.asy in Chapter 13 - - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12,13,3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={4}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-.1,5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (t.y*cos(t.x)*sin(pi/6),t.y*sin(t.x)*sin(pi/6),t.y*cos(pi/6));// - } - surface s=surface(f,(0,0),(2*pi,4),16,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (4*cos(t.x)*sin(t.y),4*sin(t.x)*sin(t.y),4*cos(t.y));// - } - surface s=surface(f,(0,0),(2*pi,pi/6),16,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - dot((0,0,2.799),blackmeshpen+1.2mm); - - - //draw((-1.5,-1.5,2) -- (-1.5,1.5,2) -- (1.5,1.5,2) -- (1.5,-1.5,2) -- (-1.5,-1.5,2),apexmeshpen+.25mm); - - triple g(real t) {return (1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5);} - path3 mypath=graph(g,0,2*pi,operator ..); - //draw(mypath,apexmeshpen); - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (t.x,t.x*sqrt(3),t.y);// - } - surface s=surface(f,(-1.25,-1.5),(1.25,1.5),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen q=redmeshpen; - //draw(s,simplesurfacepen2,meshpen=q); - - //draw((-1.25,-1.25*sqrt(3),-1.5) -- (1.25,1.25*sqrt(3),-1.5) -- (1.25,1.25*sqrt(3),1.5) -- (-1.25,-1.25*sqrt(3),1.5) -- (-1.25,-1.25*sqrt(3),-1.5),redcurvepen+.25mm); - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (sin(t.y)*cos(t.x),sin(t.y)*sin(t.x),cos(t.y));// - } - surface s=surface(f,(0,0),(2*pi,pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen q=greencurvepen; - //draw(s,simplesurfacepen3,meshpen=q); - - //dot((.5,sqrt(3)*.5,2),rgb(.1,.1,.1)); - - //draw((.5,sqrt(3)*.5,-.5)--(.5,sqrt(3)*.5,2.25)); - - triple g(real t) {return (1/2*cos(t),1/2*sin(t),1/2*sqrt(3));} - path3 mypath=graph(g,0,2*pi,operator ..); - //draw(mypath,blackmeshpen); - - //(1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5) - //draw((1.5/sqrt(3)*cos(4*pi/3),1.5/sqrt(3)*sin(4*pi/3),1.5) -- (0,0,0) -- (1.5/sqrt(3)*cos(pi/3),1.5/sqrt(3)*sin(pi/3),1.5)); - - triple g(real t) {return (sin(t)*cos(pi/3),sin(t)*sin(pi/3),cos(t));} - path3 mypath=graph(g,0,2*pi,operator ..); - //draw(mypath,blackmeshpen); - - -
    -     - -

    - We will set up the four triple integrals needed to find the center of mass (, to compute M, - M_{yz}, - M_{xz} and M_{xy}) and leave it to the reader to evaluate each integral. - Because of symmetry, - we expect the x- and y- coordinates of the center of mass to be 0. -

    - -

    - While the surfaces describing the solid are given in the statement of the problem, - to describe the full solid D, - we use the following bounds: - 0 \leq \rho \leq 4, - 0 \leq \theta \leq 2\pi and 0 \leq \varphi \leq \pi/3. - Since density \delta is constant, - we assume \delta =1. -

    - -

    - The mass of the solid: - - M \amp = \iiint_D\, dm = \iiint_D\, dV - \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4\big(\rho^2\cos(\varphi)\big)\, d\rho\, d\theta\, d\varphi - \amp = \frac{64}3\big(2-\sqrt{3}\big)\pi \approx 17.958 - . -

    - -

    - To compute M_{yz}, the integrand is x; - using , - we have x = \rho\cos(\varphi)\cos(\theta). - This gives: - - M_{yz} \amp = \iiint_D x\, dm - \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big((\rho\cos(\varphi)\cos(\theta))\rho^2\cos(\varphi)\big) \, d\rho\, d\theta\, d\varphi - \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big(\rho^3\cos^2(\varphi)\cos(\theta)\big) \, d\rho\, d\theta\, d\varphi - \amp =0 - , - which we expected as we expect \overline{x} = 0. -

    - -

    - To compute M_{xz}, the integrand is y; - using , - we have y = \rho\cos(\varphi)\sin(\theta). - This gives: - - M_{xz} \amp = \iiint_D y\, dm - \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big((\rho\cos(\varphi)\sin(\theta))\rho^2\cos(\varphi)\big) \, d\rho\, d\theta\, d\varphi - \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big(\rho^3\cos^2(\varphi)\sin(\theta)\big) \, d\rho\, d\theta\, d\varphi - \amp = 0 - , - which we also expected as we expect \overline{y} = 0. -

    - -

    - To compute M_{xy}, the integrand is z; - using , - we have z = \rho\sin(\varphi). - This gives: - - M_{xy} \amp = \iiint_D z\, dm - \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big((\rho\sin(\varphi))\rho^2\cos(\varphi)\big) \, d\rho\, d\theta\, d\varphi - \amp = \int_0^{\pi/3}\int_0^{2\pi}\int_0^4 \big(\rho^3\sin(\varphi)\cos(\varphi)\big) \, d\rho\, d\theta\, d\varphi - \amp =16\pi \approx 50.266 - . -

    - -

    - Thus the center of mass is (0,0,M_{xy}/M) \approx (0,0,2.799), - as indicated in . -

    -     -     - - - - - -

    - This section has provided a brief introduction into two new coordinate systems useful for identifying points in space. - Each can be used to define a variety of surfaces in space beyond the canonical surfaces graphed as each system was introduced. -

    - -

    - However, the usefulness of these coordinate systems does not lie in the variety of surfaces that they can describe nor the regions in space these surfaces may enclose. - Rather, cylindrical coordinates are mostly used to describe cylinders and spherical coordinates are mostly used to describe spheres. - These shapes are of special interest in the sciences, especially in physics, - and computations on/inside these shapes is difficult using rectangular coordinates. - For instance, in the study of electricity and magnetism, - one often studies the effects of an electrical current passing through a wire; - that wire is essentially a cylinder, - described well by cylindrical coordinates. -

    - -

    - This chapter investigated the natural follow-on to partial derivatives: - iterated integration. - We learned how to use the bounds of a double integral to describe a region in the plane using both rectangular and polar coordinates, - then later expanded to use the bounds of a triple integral to describe a region in space. - We used double integrals to find volumes under surfaces, - surface area, - and the center of mass of lamina; - we used triple integrals as an alternate method of finding volumes of space regions and also to find the center of mass of a region in space. -

    - -

    - Integration does not stop here. - We could continue to iterate our integrals, - next investigating quadruple integrals - whose bounds describe a region in 4-dimensional space - (which are very hard to visualize). - We can also look back to regular - integration where we found the area under a curve in the plane. - A natural analogue to this is finding the - area under a curve, - where the curve is in space, not in a plane. - These are just two of many avenues to explore under the heading of integration. -

    -     - - - - Terms and Concepts - - -

    - Explain the difference between the roles r, - in cylindrical coordinates, and \rho, - in spherical coordinates, play in determining the location of a point. -

    -     - - - -

    - In cylindrical, - r determines how far from the origin one goes in the xy-plane before considering the z-component. - Equivalently, - if on projects a point in cylindrical coordinates onto the xy-plane, - r will be the distance of this projection from the origin. -

    - -

    - In spherical, - \rho is the distance from the origin to the point. -

    -     -     - - - -

    - Why are points on the z-axis not determined uniquely when using cylindrical and spherical coordinates? -

    -     - - - -

    - If r=0 or \rho=0, - then the point in each coordinate system lies on the z-axis regardless of the value of \theta. -

    -     -     - - - -

    - What surfaces are naturally defined using cylindrical coordinates? -

    -     - - - -

    - Cylinders (tubes) centered at the origin, - parallel to the z-axis; - planes parallel to the z-axis that intersect the z-axis; - planes parallel to the xy-plane. -

    -     -     - - - -

    - What surfaces are naturally defined using spherical coordinates? -

    -     - - - -

    - Spheres centered at the origin; - planes parallel to the z-axis that intersect the z-axis; - cones centered on the z-axis with point at the origin. -

    -     -     -     - - - Problems - - - -

    - Points are given in either the rectangular, - cylindrical or spherical coordinate systems. - Find the coordinates of the points in the other systems. -

    -     - - - - -

    - Points in rectangular coordinates: - (2,2,1) and (-\sqrt{3},1,0) -

    -     - -

    - Cylindrical: - (2\sqrt 2,\pi/4,1) and (2,5\pi/6,0) -

    - -

    - Spherical: - (3,\pi/4,\cos^{-1}(3)) and (2,5\pi/6,0) -

    -     -     - - -

    - Points in cylindrical coordinates: - (2,\pi/4,2) and (3,3\pi/2,-4) -

    -     - -

    - Rectangular: - (\sqrt 2,\sqrt 2,2) and (0,-3,-4) -

    - -

    - Spherical: - (2\sqrt 2,\pi/4,\pi/4) and (5,3\pi/2,-\tan^{-1}(4/3)) -

    -     -     - - -

    - Points in spherical coordinates: - (2,\pi/4,\pi/4) and (1,0,0) -

    -     - -

    - Rectangular: - (1,1,\sqrt{2}) and (1,0,0) -

    - -

    - Cylindrical: - (\sqrt{2},\pi/4,\sqrt{2}) and (1,0,0) -

    -     -     -     - - - - -

    - Points in rectangular coordinates: - (0,1,1) and (-1,0,1) -

    -     - -

    - Cylindrical: (1,\pi/2,1) and (1,\pi,1) -

    -

    - Spherical: - (\sqrt 2,\pi/2,\pi/4) and (\sqrt{2}, \pi, \pi/4) -

    -     -     - - - -

    - Points in cylindrical coordinates: - (0,\pi,1) and (2,4\pi/3,0) -

    -     - -

    - Rectangular: - (0,0,1) and (-1,-\sqrt 3,0) -

    -

    - Spherical: - (1,\pi,\pi/2) and (2,4\pi/3,0) -

    -     -     - - -

    - Points in spherical coordinates: - (2,\pi/6,0) and (3,\pi,-\pi/2) -

    -     - -

    - Rectangular: - (\sqrt 3,1,0) and (0,0,-3) -

    -

    - Cylindrical: - (2,\pi/6,0) and (0,\pi,-3) -

    -     -     -     -     - - - -

    - Describe the curve, - surface or region in space determined by the given bounds in cylindrical coordinates. -

    -     - - - - -

    - r=1, 0\leq \theta\leq 2\pi, 0\leq z\leq 1 -

    -     - -

    - A cylindrical surface or tube, - centered along the z-axis of radius 1, extending from the xy-plane up to the plane z=1 (, the tube has a length of 1). -

    -     -     - - -

    - 1\leq r\leq 2, 0\leq \theta\leq \pi, 0\leq z\leq 1 -

    -     - -

    - This is a region of space, - being half of a tube with thick - walls of inner radius 1 and outer radius 2, centered along the z-axis with a length of 1, where the half below - the xz-plane is removed. -

    -     -     -     - - - - -

    - 1\leq r\leq 2, \theta= \pi/2, 0\leq z\leq 1 -

    -     - -

    - A square portion of the yz-plane with corners at (0,1,0), - (0,1,1), - (0,2,1) and (0,2,0). -

    -     -     - - -

    - r= 2, 0\leq \theta\leq 2\pi, z=5 -

    -     - -

    - This is a curve, a circle of radius 2, centered at (0,0,5), - lying parallel to the xy-plane (, in the plane z=5). -

    -     -     -     -     - - - -

    - Describe the curve, - surface or region in space determined by the given bounds in spherical coordinates. -

    -     - - - - -

    - \rho=3, 0\leq \theta\leq2\pi, - 0\leq\varphi\leq \pi/2 -

    -     - -

    - This is upper half of the sphere of radius 3 centered at the origin (, the upper hemisphere). -

    -     -     - - -

    - 2\leq\rho\leq3, 0\leq \theta\leq2\pi, - -\pi/2\leq\varphi\leq \pi/2 -

    -     - -

    - This is a region of space, - where the ball of radius 2, centered at the origin, - is removed from the ball of radius 3, centered at the origin. -

    -     -     -     - - - - -

    - 0\leq\rho\leq2, 0\leq \theta\leq\pi, \varphi = \pi/4 -

    -     - -

    - This is a region of space, a half of a solid cone with rounded top, - where the rounded top is a portion of the ball of radius 2 centered at the origin and the sides of the cone make an angle of \pi/4 with the positive z-axis. - The bounds on \theta mean only the portion above - the xz-plane are retained. -

    -     -     - - -

    - \rho=2, - 0\leq \theta\leq2\pi, \varphi = \pi/3 -

    -     -     - - -

    - This is a curve, - a circle of radius 1 centered at (0,0,\sqrt 3), - lying parallel to the xy-plane. -

    -     -     -     -     - - - -

    - Standard regions in space, - as defined by cylindrical and spherical coordinates, are shown. - Set up the triple integral that integrates the given function over the graphed region. -

    -     - - - - -

    - Cylindrical coordinates, integrating h(r,\theta,z): -

    - - - A region in space given by constant bounds in cylindrical coordinates. - -

    - The image depicts a solid region in space given by the following inequalities: - - r_1 \amp \leq r \leq r_2 - \theta_1 \amp \leq \theta\leq \theta_2 - z_1 \amp \leq z\leq z_2 - . -

    - -

    - The shape of the solid resembles that of a circular wall. -

    -     - - //import apexstyle; - //ASY file for figspherical1_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(16.5,1,4.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - pair zbounds=(-.1,1.1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - real r1=.4; - real r2=.6; - real t1=pi/6; - real t2=4pi/3; - real z1=.4; - real z2=.8; - - // (t.x*cos(t.y),t.x*sin(t.y),z) - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (r1*cos(t.y),r1*sin(t.y),t.x);// - } - surface s=surface(f,(z1,t1),(z2,t2),4,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - triple f(pair t) { - return (r2*cos(t.y),r2*sin(t.y),t.x);// - } - surface s=surface(f,(z1,t1),(z2,t2),4,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - triple f(pair t) { - return (t.x*cos(t1),t.x*sin(t1),t.y);// - } - surface s=surface(f,(r1,z1),(r2,z2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - triple f(pair t) { - return (t.x*cos(t2),t.x*sin(t2),t.y);// - } - surface s=surface(f,(r1,z1),(r2,z2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - triple f(pair t) { - return (t.x*cos(t.y),t.x*sin(t.y),z1);// - } - surface s=surface(f,(r1,t1),(r2,t2),4,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - triple f(pair t) { - return (t.x*cos(t.y),t.x*sin(t.y),z2);// - } - surface s=surface(f,(r1,t1),(r2,t2),4,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - draw((0,0,0) -- .8*(cos(t1),sin(t1),0),dashed+.25mm); - draw((0,0,0) -- .8*(cos(t2),sin(t2),0),dashed+.25mm); - - draw(arc((0,0,0),.3*(1,0,0),.3*(cos(t2),sin(t2),0),(0,0,1)),black+.25mm,Arrow3(size=2mm)); - draw(arc((0,0,0),.5*(1,0,0),.3*(cos(t1),sin(t1),0),(0,0,1)),black+.25mm,Arrow3(size=2mm)); - - label("$\theta_1$",.6(cos(t1/2),sin(t1/2),-.05)); - label("$\theta_2$",.5(cos(t2/1.5),sin(t2/1.5),0)); - - draw ((r2*cos(t2),r2*sin(t2),z1) -- (r2*cos(t2),r2*sin(t2),0),dashed+.25mm); - draw ((r2*cos(t1),r2*sin(t1),z1) -- (r2*cos(t1),r2*sin(t1),0),dashed+.25mm); - - draw ((0,0,z2) -- (r2*cos(t2/1.5),r2*sin(t2/1.5),z2),dashed+.25mm); - draw ((0,0,z2) -- (r1*cos(t2/3),r1*sin(t2/3),z2),dashed+.25mm); - - label("$r_2$",(r2*.95*cos(t2/1.5),r2*.95*sin(t2/1.5),z2+.05)); - label("$r_1$",(r1*.8*cos(t2/3),r1*.8*sin(t2/3),z2+.05)); - - draw((0,0,z1) -- (r1*cos(t1),r1*sin(t1),z1),dashed+.25mm); - draw((0,0,z2) -- (r1*cos(t1),r1*sin(t1),z2),dashed+.25mm); - - label("$z_1$",(0,0,z1),W); - label("$z_2$",(0,0,z2),W); - - //dot((0,0,2.799),blackmeshpen+1.2mm); - //draw((-1.5,-1.5,2) -- (-1.5,1.5,2) -- (1.5,1.5,2) -- (1.5,-1.5,2) -- (-1.5,-1.5,2),apexmeshpen+.25mm); - - triple g(real t) {return (1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5);} - path3 mypath=graph(g,0,2*pi,operator ..); - //draw(mypath,apexmeshpen); - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (t.x,t.x*sqrt(3),t.y);// - } - surface s=surface(f,(-1.25,-1.5),(1.25,1.5),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen q=redmeshpen; - //draw(s,simplesurfacepen2,meshpen=q); - - //draw((-1.25,-1.25*sqrt(3),-1.5) -- (1.25,1.25*sqrt(3),-1.5) -- (1.25,1.25*sqrt(3),1.5) -- (-1.25,-1.25*sqrt(3),1.5) -- (-1.25,-1.25*sqrt(3),-1.5),redcurvepen+.25mm); - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (sin(t.y)*cos(t.x),sin(t.y)*sin(t.x),cos(t.y));// - } - surface s=surface(f,(0,0),(2*pi,pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen q=greencurvepen; - //draw(s,simplesurfacepen3,meshpen=q); - - //dot((.5,sqrt(3)*.5,2),rgb(.1,.1,.1)); - - //draw((.5,sqrt(3)*.5,-.5)--(.5,sqrt(3)*.5,2.25)); - - triple g(real t) {return (1/2*cos(t),1/2*sin(t),1/2*sqrt(3));} - path3 mypath=graph(g,0,2*pi,operator ..); - //draw(mypath,blackmeshpen); - - //(1.5/sqrt(3)*cos(t),1.5/sqrt(3)*sin(t),1.5) - //draw((1.5/sqrt(3)*cos(4*pi/3),1.5/sqrt(3)*sin(4*pi/3),1.5) -- (0,0,0) -- (1.5/sqrt(3)*cos(pi/3),1.5/sqrt(3)*sin(pi/3),1.5)); - - triple g(real t) {return (sin(t)*cos(pi/3),sin(t)*sin(pi/3),cos(t));} - path3 mypath=graph(g,0,2*pi,operator ..); - //draw(mypath,blackmeshpen); - - -     - -

    - \ds\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}\int_{z_1}^{z_2}h(r,\theta,z)r\, dz\, dr\, d\theta -

    -     -     - - - -

    - Spherical coordinates, integrating h(\rho,\theta,\varphi): -

    - - A region in space given by constant bounds in spherical coordinates. - -

    - A solid region in space is depicted; it illustrates the following inequalities: - - \rho_1 \amp \leq \rho \leq \rho_2 - \varphi_1 \amp \leq \varphi\leq \varphi_2 - \theta_1 \amp \leq \theta \leq \theta_2 - . -

    - -

    - The shape of the solid looks similar to a thick piece of rind peeled from the top of an orange. -

    -     - - - //import apexstyle; - - - //ASY file for figspherical1_3D.asy in Chapter 13 - - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(16,.9,8.1); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.6,.6); - pair ybounds=(-.6,.6); - pair zbounds=(-.1,1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - - real t1=pi/3; - real t2=7pi/6; - real p1=pi/6-.2; - real p2=pi/6+.2; - real r1=.75; - real r2=.9; - - - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (r1*cos(t.x)*sin(t.y),r1*sin(t.x)*sin(t.y),r1*cos(t.y));// - } - surface s=surface(f,(t1,p1),(t2,p2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - - triple f(pair t) { - return (r2*cos(t.x)*sin(t.y),r2*sin(t.x)*sin(t.y),r2*cos(t.y));// - } - surface s=surface(f,(t1,p1),(t2,p2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - triple f(pair t) { - return (t.y*cos(t.x)*sin(p1),t.y*sin(t.x)*sin(p1),t.y*cos(p1));// - } - surface s=surface(f,(t1,r1),(t2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - triple f(pair t) { - return (t.y*cos(t.x)*sin(p2),t.y*sin(t.x)*sin(p2),t.y*cos(p2));// - } - surface s=surface(f,(t1,r1),(t2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - triple f(pair t) { - return (t.y*cos(t1)*sin(t.x),t.y*sin(t1)*sin(t.x),t.y*cos(t.x));// - } - surface s=surface(f,(p1,r1),(p2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - triple f(pair t) { - return (t.y*cos(t2)*sin(t.x),t.y*sin(t2)*sin(t.x),t.y*cos(t.x));// - } - surface s=surface(f,(p1,r1),(p2,r2),4,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,simplesurfacepen,meshpen=p); - - - // - // lines for phi - // - draw((r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1)) -- (0,0,0) -- (r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)),black+.25mm+dashed); - - //draw(arc((0,0,0),(r1*cos(t1)*sin(p1)/2.5,r1*sin(t1)*sin(p1)/2.5,r1*cos(p1)/2.5),(r1*cos(t1)*sin(p2)/2.5,r1*sin(t1)*sin(p2)/2.5,r1*cos(p2)/2.5)),black+.25mm,Arrow3(size=1.5mm)); - - //label("$\Delta\varphi$",.95*(r1*cos(t1)*sin((p1+p2)/2)/2,r1*sin(t1)*sin((p1+p2)/2)/2,r1*cos((p1+p2)/2)/2)); - - - - // - // lines for theta - // - - draw((0,0,0) -- (r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0),black+.25mm+dashed); - - draw((0,0,0) -- (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),0),black+.25mm+dashed); - - draw(arc((0,0,0),.25*(r1,0,0),.5*(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0)),black+.25mm,Arrow3(size=1.5mm)); - - draw(arc((0,0,0),.4*(r1,0,0),.4*(r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),0),(0,0,1)),black+.25mm,Arrow3(size=1.5mm)); - - label("$\theta_1$",.25*(r1*cos(t1/2)*sin(p2),r1*sin(t1/2)*sin(p2),0)); - label("$\theta_2$",.5*(r1*cos(t2/2)*sin(p2),r1*sin(t2/2)*sin(p2),0)); - - // - // lines for rho - // - - draw(arc((0,0,0),(r1*cos(t1)*sin(0),r1*sin(t1)*sin(0),r1*cos(0)),(r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1))),black+.25mm+dashed); - - draw(arc((0,0,0),(r2*cos(t1)*sin(0),r2*sin(t1)*sin(0),r2*cos(0)),(r2*cos(t1)*sin(p1),r2*sin(t1)*sin(p1),r2*cos(p1))),black+.25mm+dashed); - - - label("$\rho_1$",(0,0,r1),W); - label("$\rho_2$",(0,0,r2),W); - - real poff = -0.05; - - //draw((r1*cos(t1)*sin(p1+poff),r1*sin(t1)*sin(p1+poff),r1*cos(p1+poff)) -- (r2*cos(t1)*sin(p1+poff),r2*sin(t1)*sin(p1+poff),r2*cos(p1+poff)),black+.25mm,Arrows3(size=1.5mm)); - - //label("$\rho$",(0,-.04,.5*r1)); - - //draw((0,-.02,.01) -- (0,-.02,.99*r1),black+.25mm,Arrows3(size=1.5mm)); - - - // - // lines for theta length - // - - //draw((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) -- (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05),black+.25mm,Arrows3(size=1.5mm)); - - //label("$\rho\sin(\varphi)\Delta\theta$",1.5*((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05)+(0,0,-.05))/2+(0,0,-.25),S); - - //draw(1.5*((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05)+(0,0,-.05))/2 + (0,0,-.25) -- 1.05((r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2)-.05) + (r1*cos(t2)*sin(p2),r1*sin(t2)*sin(p2),r1*cos(p2)-.05))/2+(0,0,-.01),black+.25mm+dashed,Arrow3(size=1.5mm)); - - // - // lines for phi length - // - - draw(arc((0,0,0),(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0),.95*(r1*cos(t1)*sin(p1),r1*sin(t1)*sin(p1),r1*cos(p1))),black+.25mm,Arrows3(size=1.5mm)); -label("$\varphi_2$",.8*(r1*cos(t1)*sin(pi/2-p1/2-.5),r1*sin(t1)*sin(pi/2-p1/2-.5),r1*cos(pi/2-p1/2-.5)),S); - -label("$\varphi_1$",.63*(r1*cos(t1)*sin(pi/2-p1-(p2-p1)/2),r1*sin(t1)*sin(pi/2-p1-(p2-p1)/2),r1*cos(pi/2-p1-(p2-p1)/2)),S); -draw(arc((0,0,0),.8*(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),0),.7*(r1*cos(t1)*sin(p2),r1*sin(t1)*sin(p2),r1*cos(p2))),black+.25mm,Arrows3(size=1.5mm)); - - -     - -

    - \ds\int_{\varphi_1}^{\varphi_2}\int_{\theta_1}^{\theta_2}\int_{\rho_1}^{\rho_2}h(\rho,\theta,\varphi)\rho^2\sin(\varphi)\, d\rho\, d\theta\, d\varphi -

    -     -     -     - - - -

    - A triple integral in cylindrical coordinates is given. - Describe the region in space defined by the bounds of the integral. -

    -     - - - - -

    - \ds \int_0^{\pi/2}\int_0^2\int_0^2 r\, dz\, dr\, d\theta -

    -     - -

    - The region in space is bounded between the planes z=0 and z=2, - inside of the cylinder x^2+y^2=4, - and the planes \theta = 0 and \theta = \pi/2: - describes a wedge of a cylinder of height 2 and radius 2; - the angle of the wedge is \pi/2, or 90^\circ. -

    -     -     - - - -

    - \ds \int_0^{2\pi}\int_3^4\int_0^5 r\, dz\, dr\, d\theta -

    -     - -

    - Bounded between the planes z=0 and z=5, - between the cylinders x^2+y^2=9 and x^2+y^2=16: - describes a pipe or tube - of length 5, an inner radius of 3 and outer radius of 4. -

    -     -     - - - -

    - \ds \int_0^{2\pi}\int_0^1\int_0^{1-r} r\, dz\, dr\, d\theta -

    -     - -

    - Bounded between the plane z=0 and the cone z=1-\sqrt{x^2+y^2}: - describes an inverted cone, - with height of 1, point at (0,0,1) and base radius of 1. -

    -     -     - - - -

    - \ds \int_0^{\pi}\int_0^1\int_0^{2-r} r\, dz\, dr\, d\theta -

    -     - -

    - Bounded between y\geq 0, - inside the cylinder x^2+y^2=1, - above the plane z=0 and below the cone z = 2-\sqrt{x^2+y^2}: - describes cylindrical solid of height 1 and radius 2, topped with an inverted cone of height 1 and base radius 1 with point at (0,0,2). -

    -     -     - - - -

    - \ds \int_0^{\pi}\int_0^3\int_0^{\sqrt{9-r^2}} r\, dz\, dr\, d\theta -

    -     - -

    - Describes a quarter of a ball of radius 3, centered at the origin; - the quarter resides above the xy-plane and above the xz-plane. -

    -     -     - - - -

    - \ds \int_0^{2\pi}\int_0^a\int_0^{\sqrt{a^2-r^2}+b} r\, dz\, dr\, d\theta -

    -     - -

    - Bounded between the plane z=0, - inside the cylinder x^2+y^2 = a^2, - and below the upper hemisphere z= \sqrt{a^2-x^2-y^2}+b, - with radius a and centered at (0,0,b): - describes a cylindrical solid of radius a and height b, - topped with the upper hemisphere of radius a. -

    -     -     -     - - - -

    - A triple integral in spherical coordinates is given. - Describe the region in space defined by the bounds of the integral. -

    -     - - - - -

    - \ds \int_0^{\pi/2}\int_0^{\pi/2}\int_0^{1} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi -

    -     - -

    - Describes the portion of the unit ball that resides in the first octant. -

    -     -     - - - -

    - \ds \int_{-\pi/2}^{\pi/2}\int_0^{\pi}\int_1^{1.1} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi -

    -     - -

    - Describes half of a spherical shell (, y\geq 0) with inner radius of 1 and outer radius of 1.1 centered at the origin. -

    -     -     - - - -

    - \ds \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_0^{2} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi -

    -     - -

    - Bounded below by the cone z=\sqrt{x^2+y^2} and above by the sphere x^2+y^2+z^2=4: - describes a shape that is somewhat diamond-like; - some think of it as looking like an ice cream cone - (see ). - It describes a cone, - where the side makes an angle of \pi/4 with the xy-plane, - topped by the portion of the ball of radius 2, centered at the origin. -

    -     -     - - - -

    - \ds \int_{\pi/4}^{\pi/3}\int_0^{2\pi}\int_0^{2} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi -

    -     - -

    - It is the region is space bounded below by - z=\sqrt{x^2+y^2} and above by the sphere x^2+y^2+z^2=4, - with the portion above the cone z=\sqrt3\sqrt{x^2+y^2} removed: - it describes a cone, - where the side makes an angle of \pi/4 with the xy-plane, - topped by the portion of the ball of radius 2, centered at the origin, - with the inner cone with angle \pi/3 removed, - along with corresponding portion of the ball of radius 2. -

    -     -     - - - -

    - \ds \int_{\pi/3}^{\pi/2}\int_0^{2\pi}\int_0^{\csc(\varphi)} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi -

    -     - -

    - The region in space is bounded below by the cone - z=\sqrt{3}\sqrt{x^2+y^2} and above by the plane z=1: - it describes a cone, with point at the origin, - centered along the positive z-axis, - with height of 1 and base radius of \cot(\pi/3) = \tan(\pi/6) = 1/\sqrt{3}. -

    -     -     - - - -

    - \ds \int_{\pi/3}^{\pi/2}\int_0^{2\pi}\int_0^{a\csc(\varphi)} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi -

    -     - -

    - The region in space is bounded below by the cone - z=\sqrt{3}\sqrt{x^2+y^2} and above by the plane z=a: - it describes a cone, with point at the origin, - centered along the positive z-axis, - with height of a and base radius of a\cot(\pi/3). -

    -     -     -     - - - -

    - A solid is described along with its density function. - Find the mass of the solid using cylindrical coordinates. -

    -     - - - -

    - Bounded by the cylinder x^2+y^2=4 and the planes z=0 and z=4 with density function \delta(x,y,z) =\sqrt{x^2+y^2}+1. -

    -     - -

    - In cylindrical coordinates, the density is \delta(r,\theta,z) = r+1. - Thus mass is - - \int_0^{2\pi}\int_0^2\int_0^4 (r+1)r\, dz\, dr\, d\theta = 112\pi/3 - . -

    -     -     - - - -

    - Bounded by the cylinders x^2+y^2=4 and x^2+y^2=9, - between the planes z=0 and z=10 with density function \delta(x,y,z) =z. -

    -     - -

    - In cylindrical coordinates, the density is \delta(r,\theta,z) = z. - Thus mass is - - \int_0^{2\pi}\int_2^3\int_0^{10} zr\, dz\, dr\, d\theta = 250\pi - . -

    -     -     - - - -

    - Bounded by y\geq 0, the cylinder x^2+y^2=1, - and between the planes z=0 and z=4-y with density function \delta(x,y,z) =1. -

    -     - -

    - In cylindrical coordinates, the density is \delta(r,\theta,z) = 1. - Thus mass is - - \int_0^{\pi}\int_0^1\int_0^{4-r\sin(\theta)} r\, dz\, dr\, d\theta = 2\pi-2/3\approx 5.617 - . -

    -     -     - - - -

    - The upper half of the unit ball, - bounded between z= 0 and z=\sqrt{1-x^2-y^2}, - with density function \delta(x,y,z) =1. -

    -     - -

    - In cylindrical coordinates, the density is \delta(r,\theta,z) = 1. - Thus mass is - - \int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} r\, dz\, dr\, d\theta = 4\pi/3 - . -

    -     -     -     - - - -

    - A solid is described along with its density function. - Find the center of mass of the solid using cylindrical coordinates. - (Note: these are the same solids and density functions as found in Exercises.) -

    -     - - - -

    - Bounded by the cylinder x^2+y^2=4 and the planes z=0 and z=4 with density function \delta(x,y,z) =\sqrt{x^2+y^2}+1. -

    -     - -

    - In cylindrical coordinates, the density is \delta(r,\theta,z) = r+1. - Thus mass is - - M=\int_0^{2\pi}\int_0^2\int_0^4 (r+1)r\, dz\, dr\, d\theta = 112\pi/3 - . -

    - -

    - We find M_{yz} = 0, M_{xz} = 0, - and M_{xy} = 224\pi/3, - placing the center of mass at (0,0,2). -

    -     -     - - - -

    - Bounded by the cylinders x^2+y^2=4 and x^2+y^2=9, - between the planes z=0 and z=10 with density function \delta(x,y,z) =z. -

    -     - -

    - In cylindrical coordinates, the density is \delta(r,\theta,z) = z. - Thus mass is - - M=\int_0^{2\pi}\int_2^3\int_0^{10} zr\, dz\, dr\, d\theta = 250\pi - . -

    - -

    - We find M_{yz} = 0, M_{xz} = 0, - and M_{xy} = 5000\pi/3, - placing the center of mass at (0,0,20/3). -

    -     -     - - - -

    - Bounded by y\geq 0, the cylinder x^2+y^2=1, - and between the planes z=0 and z=4-y with density function \delta(x,y,z) =1. -

    -     - -

    - In cylindrical coordinates, the density is \delta(r,\theta,z) = 1. - Thus mass is - - \int_0^{\pi}\int_0^1\int_0^{4-r\sin(\theta)} r\, dz\, dr\, d\theta = 2\pi-2/3\approx 5.617 - . -

    - -

    - We find M_{yz} = 0, - M_{xz} = 8/3-\pi/8, and M_{xy} = 65\pi/16-8/3, - placing the center of mass at \approx (0,0.405,1.80). -

    -     -     - - - -

    - The upper half of the unit ball, - bounded between z= 0 and z=\sqrt{1-x^2-y^2}, - with density function \delta(x,y,z) =1. -

    -     - -

    - In cylindrical coordinates, the density is \delta(r,\theta,z) = 1. - Thus mass is - - \int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} r\, dz\, dr\, d\theta = 2\pi/3 - . -

    - -

    - We find M_{yz} = 0, M_{xz} = 0, - and M_{xy} = \pi/4, placing the center of mass at (0,0,3/8). -

    -     -     -     - - - -

    - A solid is described along with its density function. - Find the mass of the solid using spherical coordinates. -

    -     - - - -

    - The upper half of the unit ball, - bounded between z= 0 and z=\sqrt{1-x^2-y^2}, - with density function \delta(x,y,z) =1. -

    -     - -

    - In spherical coordinates, - the density is \delta(\rho,\theta,\varphi) = 1. - Thus mass is - - \int_0^{\pi/2}\int_0^{2\pi}\int_{0}^{1} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = 2\pi/3 - . -

    -     -     - - - -

    - The spherical shell bounded between x^2+y^2+z^2=16 and - x^2+y^2+z^2=25 with density function \delta(x,y,z) = \sqrt{x^2+y^2+z^2}. -

    -     - -

    - In spherical coordinates, - the density is \delta(\rho,\theta,\varphi) = \rho. - Thus mass is - - \int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\int_{4}^{5} (\rho)\rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = 369\pi - . -

    -     -     - - - -

    - The conical region bounded below by z=\sqrt{x^2+y^2} and above by the sphere - x^2+y^2+z^2=1 with density function \delta(x,y,z) = z. -

    -     - -

    - In spherical coordinates, - the density is \delta(\rho,\theta,\varphi) = \rho\sin(\varphi). - Thus mass is - - \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_{0}^{1} \big(\rho\sin(\varphi)\big)\rho^2\sin(\varphi)\, d\rho\, d\theta\, d\varphi = \pi/8 - . -

    -     -     - - - -

    - The cone that lies above the cone z=\sqrt{x^2+y^2} and below the plane z=1 with density function \delta(x,y,z) = z. -

    -     - -

    - In spherical coordinates, - the density is \delta(\rho,\theta,\varphi) = \rho\sin(\varphi). - Thus mass is - - \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_{0}^{\csc(\varphi)} \big(\rho\sin(\varphi)\big)\rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = \pi/4 - . -

    -     -     -     - - - -

    - A solid is described along with its density function. - Find the center of mass of the solid using spherical coordinates. - (Note: these are the same solids and density functions as found in Exercises.) -

    -     - - - -

    - The upper half of the unit ball, - bounded between z= 0 and z=\sqrt{1-x^2-y^2}, - with density function \delta(x,y,z) =1. -

    -     - -

    - In spherical coordinates, - the density is \delta(\rho,\theta,\varphi) = 1. - Thus mass is - - \int_0^{\pi/2}\int_0^{2\pi}\int_{0}^{1} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = 2\pi/3 - . -

    - -

    - We find M_{yz} = 0, M_{xz} = 0, - and M_{xy} = \pi/4, placing the center of mass at (0,0,3/8). -

    -     -     - - - -

    - The spherical shell bounded between x^2+y^2+z^2=16 and - x^2+y^2+z^2=25 with density function \delta(x,y,z) = \sqrt{x^2+y^2+z^2}. -

    -     - -

    - In spherical coordinates, - the density is \delta(\rho,\theta,\varphi) = \rho. - Thus mass is - - \int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\int_{4}^{5} (\rho)\rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = 369\pi - . -

    - -

    - We find M_{yz} = 0, M_{xz} = 0, - and M_{xy} = 0, placing the center of mass at (0,0,0). -

    -     -     - - - -

    - The conical region bounded above z=\sqrt{x^2+y^2} and below the sphere - x^2+y^2+z^2=1 with density function \delta(x,y,z) = z. -

    -     - -

    - In spherical coordinates, - the density is \delta(\rho,\theta,\varphi) = \rho\sin(\varphi). - Thus mass is - - \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_{0}^{1} \big(\rho\sin(\varphi)\big)\rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = \pi/8 - . -

    - -

    - We find M_{yz} = 0, - M_{xz} = 0, and M_{xy} = (4-\sqrt{2})\pi/30, - placing the center of mass at (0,0,4(4-\sqrt2)/15). -

    -     -     - - - -

    - The cone bounded above z=\sqrt{x^2+y^2} and below the plane z=1 with density function \delta(x,y,z) = z. -

    -     - -

    - In spherical coordinates, - the density is \delta(\rho,\theta,\varphi) = \rho\sin(\varphi). - Thus mass is - - \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_{0}^{\csc(\varphi)} \big(\rho\sin(\varphi)\big)\rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi = \pi/4 - . -

    - -

    - We find M_{yz} = 0, M_{xz} = 0, - and M_{xy} = \pi/5, placing the center of mass at (0,0,4/5). -

    -     -     -     - - - -

    - A region is space is described. - Set up the triple integrals that find the volume of this region using rectangular, - cylindrical and spherical coordinates, - then comment on which of the three appears easiest to evaluate. -

    -     - - - -

    - The region enclosed by the unit sphere, x^2+y^2+z^2=1. -

    -     - -

    - Rectangular: - \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}}\, dz\, dy\, dx -

    - -

    - Cylindrical: - \int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}r\, dz\, dr\, d\theta -

    - -

    - Spherical: \int_{-\pi/2}^{\pi/2}\int_0^{2\pi}\int_0^1 \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi -

    - -

    - Spherical appears simplest, - avoiding the integration of square-roots and using techniques such as Substitution; - all bounds are constants. -

    -     -     - - - -

    - The region enclosed by the cylinder - x^2+y^2=1 and planes z=0 and z=1. -

    -     - -

    - Rectangular: - \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{0}^{1}\, dz\, dy\, dx -

    - -

    - Cylindrical: - \int_0^{2\pi}\int_0^1\int_{0}^{1}r\, dz\, dr\, d\theta -

    - -

    - Spherical: \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_0^{\csc(\varphi)} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi + \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_0^{\csc(\varphi)} \rho^2\sin(\varphi)\, d\rho\, d\theta\, d\varphi -

    - -

    - Cylindrical appears simplest, - avoiding the integration of square-roots and two triple integrals; - all bounds are constants. -

    -     -     - - - -

    - The region enclosed by the cone - z=\sqrt{x^2+y^2} and plane z=1. -

    -     - -

    - Rectangular: - \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{\sqrt{x^2+y^2}}^{1}\, dz\, dy\, dx -

    - -

    - Cylindrical: - \int_0^{2\pi}\int_0^1\int_{r}^{1}r\, dz\, dr\, d\theta -

    - -

    - Spherical: \int_{\pi/4}^{\pi/2}\int_0^{2\pi}\int_0^{\csc(\varphi)} \rho^2\cos(\varphi)\, d\rho\, d\theta\, d\varphi -

    - -

    - Cylindrical appears simplest, - avoiding the integration of square-roots that rectangular uses. - Spherical is not difficult, though it requires Substitution, - an extra step. -

    -     -     - - - -

    - The cube enclosed by the planes x=0, - x=1, y=0, y=1, - z=0 and z=1. - (Hint: in spherical, - use order of integration d\rho\, d\varphi\, d\theta.) -

    -     - -

    - Rectangular: - \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\, dz\, dy\, dx -

    - -

    - Cylindrical: - \int_0^{\pi/4}\int_0^{\sec(\theta)}\int_{0}^{1}r\, dz\, dr\, d\theta + \int_{\pi/4}^{\pi/2}\int_0^{\csc(\theta)}\int_{0}^{1}r\, dz\, dr\, d\theta -

    - -

    - Spherical: - - \amp \int_0^{\pi/4}\int_0^{\tan^{-1}(\sec(\theta))}\int_0^{\csc(\varphi)}\rho^2 \cos(\varphi)\, d\rho\, d\varphi\, d\theta - \amp\quad+\int_0^{\pi/4}\int_{\tan^{-1}(\sec(\theta))}^{\pi/2}\int_0^{\sec(\theta)\sec(\varphi)}\rho^2 \cos(\varphi)\, d\rho\, d\varphi\, d\theta - \amp\quad+\int_{\pi/4}^{\pi/2}\int_0^{\tan^{-1}(\csc(\theta))}\int_0^{\csc(\varphi)}\rho^2 \cos(\varphi)\, d\rho\, d\varphi\, d\theta - \amp\quad+\int_{\pi/4}^{\pi/2}\int_{\tan^{-1}(\csc(\theta))}^{\pi/2}\int_0^{\csc(\theta)\sec(\varphi)}\rho^2 \cos(\varphi)\, d\rho\, d\varphi\, d\theta - . -

    - -

    - Rectangular is clearly the simplest. -

    -     -     -     -     -     -
    - -     - - - Vector Analysis - -

    - This chapter explores completely different relationships between vectors and integration. - These relationships will enable us to compute the work done by a magnetic field in moving an object along a path and find how much air moves through an oddly-shaped screen in space, - among other things. -

    - -

    - Our upcoming work with integration will benefit from a review. - We are not concerned here with techniques of integration, - but rather what an integral does - and how that relates to the notation we use to describe it. -

    - - - Integration review -

    - Recall from - that when R is a region in the xy-plane, - \iint_R dA gives the area of the region R. - The integral symbols are elongated esses meaning sum - and dA represents a small amount of area. Taken together, - \iint_R dA means sum up, over R, - small amounts of area. This sum then gives the total area of R. - We use two integral symbols since R is a two-dimensional region. -

    - -

    - Now let z=f(x,y) represent a surface. - The integral \iint_R f(x,y)\, dA means - sum up, over R, - function values (heights) given by f times small amounts of area. - Since height area = volume, - we are summing small amounts of volume over R, - giving the total signed volume under the surface - z=f(x,y) and above the xy-plane. -

    - -

    - This notation does not directly inform us how - to evaluate the double integrals to find an area or a volume. - With additional work, - we recognize that a small amount of area dA can be measured as the area of a small rectangle, - with one side length a small change in x and the other side length a small change in y. - That is, dA = dx\,dy or dA = dy\,dx. - We could also compute a small amount of area by thinking in terms of polar coordinates, - where dA = r\,dr\,d\theta. - These understandings lead us to the iterated integrals we used in . -

    - -

    - Let us back our review up farther. - Note that \int_1^3\, dx = x\big|_1^3 = 3-1 = 2. - We have simply measured the length of the interval [1,3]. - We could rewrite the above integral using syntax similar to the double integral syntax above: - - \int_1^3\, dx = \int_Idx, \text{ where \(I\) = \([1,3]\) } - . -

    - -

    - We interpret \int_I dx as meaning sum up, - over the interval I, - small changes in x. - A change in x is a length along the x-axis, - so we are adding up along I small lengths, - giving the total length of I. -

    - -

    - We could also write \int_1^3f(x)\, dx as \int_I f(x)\, dx, - interpreted as sum up, over I, - heights given by y = f(x) times small changes in x. - Since heightlength = area, - we are summing up areas and finding the total signed area between - y = f(x) and the x-axis. -

    - -

    - This method of referring to the process of integration can be very powerful. - It is the core of our notion of the Riemann Sum. - When faced with a quantity to compute, - if one can think of a way to approximate its value through a sum, - the one is well on their way to constructing an integral (or, - double or triple integral) that computes the desired quantity. - We will demonstrate this process throughout this chapter, - starting with the next section. -

    -     -     - - -
    - Introduction to Line Integrals - -

    - We first used integration to find - area under a curve. In this section, - we learn to do this (again), but in a different context. -

    -     - - - Line Integrals of Functions - -

    - Consider the surface and curve shown in . - The surface is given by f(x,y)=1-\cos(x)\sin(y). - The dashed curve lies in the xy-plane and is the familiar y=x^2 parabola from -1\leq x\leq 1; - we'll call this curve C. - The curve drawn with a solid line in the graph is the curve in space that lies on our surface with x and y values that lie on C. -

    - -

    - The question we want to answer is this: - what is the area that lies below the curve drawn with the solid line? - In other words, - what is the area of the region above C and under the the surface z=f(x,y)? - This region is shown in . -

    - -

    - We suspect the answer can be found using an integral, - but before trying to figure out what that integral is, - let us first try to approximate its value. -

    - -
    - Finding area under a curve in space - -
    - - - A curve in the x,y plane, and the corresponding curve on a surface lying above it. - -

    - A set of three-dimensional coordinate axes are drawn in space. - A surface lies above the xy plane; it is the graph z=1-\cos(x)\sin(y). - The surface looks a little bit like a camping chair, but the precise shape is unimportant. -

    - -

    - In the xy plane, the parabola y=x^2 is plotted.. - On the surface is the corresponding curve, given by parametric equations - x=t, y=t^2, z = 1-\cos(t)\sin(t^2). - These are the points on the surface that lie directly above the parabola. -

    -     - - - - - //ASY file for fig10_01_ex_233D.asy in Chapter 10 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(1.17,4.69,2.35); - //currentprojection=orthographic(2.5,4.6,1.9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={1}; - real[] myzchoice={1,2}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-.2,1.5); - pair zbounds=(-.1,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z=x^2-y^2+1 - triple f(pair t) { - return (t.x,t.y,-sin(t.y)*cos(t.x)+1); - } - surface s=surface(f,(-1,-.1),(1,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return (t,t^2,-cos(t)*sin(t^2)+1);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,t^2,0);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)+dashed); - - - -
    - -
    - - - A cylindrical surface in space that lies above a parabola in the plane, and below a space curve. - -

    - This image shows the surface created by the portion of the cylinder y=x^2 in space - that lies above the plane z=0 and below the surface z=1-\cos(x)\sin(y). -

    - -

    - The surface is like a wall that is bent in the shape of a parabola. - The top of the wall forms a curve that is sinusoidal in shape. - The wall is highest where it meets the z axis, - and then it slopes gently downward toward the ends of the wall, which are of a lower height. -

    -     - - - - - //ASY file for fig10_01_ex_233D.asy in Chapter 10 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(1.17,4.69,2.35); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={1}; - real[] myzchoice={1,2}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-.2,1.5); - pair zbounds=(-.1,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z=x^2-y^2+1 - triple f(pair t) { - return (t.x,t.x^2,t.y*(-sin(t.x^2)*cos(t.x)+1)); - } - surface s=surface(f,(-1,0),(1,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=apexmeshpen; - draw(s,simplesurfacepen,meshpen=p); - - triple g(real t) {return (t,t^2,-cos(t)*sin(t^2)+1);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,t^2,0);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)+dashed); - - - -
    - -
    - - - Approximating the area under a curve in space using planar rectangles. - -

    - The same curve as in is plotted in space. - This time, the cylindrical surface has been removed, - and replaced by four rectangles used to approximate the area under the curve. -

    - -

    - Each rectangle lies in a vertical plane. - The base of each rectangle lies along a secant line between two points of the parabola y=x^2 in the xy plane. - The height of each rectangle is determined by the height of the curve in space at some point above the base. -

    -     - - - - - //ASY file for fig10_01_ex_233D.asy in Chapter 10 - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(1.17,4.69,2.35); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={1}; - real[] myzchoice={1,2}; - defaultpen(0.5mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-.2,1.5); - pair zbounds=(-.1,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z=x^2-y^2+1 - triple f1(pair t) { - return (t.x,-.75/.5*(t.x+1)+1,t.y*(.664)); - } - surface s1=surface(f1,(-1,0),(-.5,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=redpen+linewidth(1); - draw(s1,simplesurfacepen2,meshpen=p); - - //Draw the surface z=x^2-y^2+1 - triple f2(pair t) { - return (t.x,-.5*(t.x+.5)+.25,t.y*(.89)); - } - surface s2=surface(f2,(-.5,0),(0,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=redpen+linewidth(1); - draw(s2,simplesurfacepen2,meshpen=p); - - //Draw the surface z=x^2-y^2+1 - triple f3(pair t) { - return (t.x,.5*(t.x-.5)+.25,t.y*(.89)); - } - surface s3=surface(f3,(0,0),(.5,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=redpen+linewidth(1); - draw(s3,simplesurfacepen2,meshpen=p); - - //Draw the surface z=x^2-y^2+1 - triple f4(pair t) { - return (t.x,.75/.5*(t.x-1)+1,t.y*(.664)); - } - surface s4=surface(f4,(.5,0),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=redpen+linewidth(1); - draw(s4,simplesurfacepen2,meshpen=p); - - triple g(real t) {return (t,t^2,-cos(t)*sin(t^2)+1);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,t^2,0);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)+dashed); - - - -
    -     - -
    - -

    - In , - four rectangles have been drawn over the curve C. - The bottom corners of each rectangle lie on C, - and each rectangle has a height given by the function f(x,y) for some (x,y) pair along C between the rectangle's bottom corners. -

    - -

    - As we know how to find the area of each rectangle, - we are able to approximate the area above C and under f. - Clearly, our approximation will be - an approximation. - The heights of the rectangles do not match exactly with the surface f, - nor does the base of each rectangle follow perfectly the path of C. -

    - -

    - In typical calculus fashion, - our approximation can be improved by using more rectangles. - The sum of the areas of these rectangles gives an approximate value of the true area above C and under f. - As the area of each rectangle is - height width, we assert that the - - \text{ area above \(C\) } \approx \sum (\text{ heights } \times\text{ widths } ) - . -

    - -

    - When first learning of the integral, - and approximating areas with heights widths, - the width was a small change in x: dx. - That will not suffice in this context. - Rather, each width of a rectangle is actually approximating the arc length of a small portion of C. - In , - we used s to represent the arc-length parameter of a curve. - A small amount of arc length will thus be represented by ds. -

    - -

    - The height of each rectangle will be determined in some way by the surface f. - If we parametrize C by s, - an s-value corresponds to an (x,y) pair that lies on the parabola C. - Since f is a function of x and y, - and x and y are functions of s, - we can say that f is a function of s. - Given a value s, we can compute f(s) and find a height. - Thus - - \text{ area under \(f\) and above \(C\) } \amp \approx \sum (\text{ heights } \times\text{ widths } ); - \text{ area under \(f\) and above \(C\) } \amp =\lim_{\norm{\Delta s}\to0}\sum f(c_i)\Delta s_i - \amp =\int_Cf(s)\, ds - . -

    - -

    - Here we have introduce a new notation, - the integral symbol with a subscript of C. - It is reminiscent of our usage of \iint_R. - Using the train of thought found in the Integration Review preceding this section, - we interpret \int_C f(s)\, ds - as meaning sum up, along a curve C, - function values f(s)\timessmall arc lengths. - It is understood here that s represents the arc-length parameter. -

    - -

    - All this leads us to a definition. - The integral found in Equation - is called a line integral. - We formally define it below, but note that the definition is very abstract. - On one hand, - one is apt to say the definition makes sense, while on the other, - one is equally apt to say but I don't know what I'm supposed to do with this definition. - We'll address that after the definition, - and actually find an answer to the area problem we posed at the beginning of this section. -

    - - - Line Integral Over A Scalar Field - -

    - Let C be a smooth curve parametrized by s, - the arc-length parameter, - and let f be a continuous function of s. - A line integral is an integral of the form - - \int_C f(s)\, ds = \lim_{\norm{\Delta s}\to 0}\sum_{i=1}^n f(c_i)\Delta s_i - , - where s_0\lt s_1\lt \ldots\lt s_n is any partition of the s-interval over which C is defined, - c_i is any value in the ith subinterval, - \Delta s_i is the width of the ith subinterval, - and \norm{\Delta s} is the length of the longest subinterval in the partition. - line integralover scalar field -

    -     -     - - -

    - When C is a closed curve, - , a curve that ends at the same point at which it starts, we use - - \oint_C f(s)\, ds \text{ instead of } \int_C f(s)\, ds - . -

    - -

    - The definition of the line integral does not specify whether C is a curve in the plane or space - (or hyperspace), - as the definition holds regardless. - For now, we'll assume C lies in the xy-plane. -

    - -

    - This definition of the line integral doesn't really say anything new. - If C is a curve and s is the arc-length parameter of C on a\leq s\leq b, then - - \int_Cf(s)\, ds = \int_a^bf(s)\, ds - . -

    - -

    - The real difference with this integral from the standard - \int_a^bf(x)\, dx - we used in the past is that of context. - Our previous integrals naturally summed up values over an interval on the x-axis, - whereas now we are summing up values over a curve. - If we can parametrize the curve with the arc-length parameter, - we can evaluate the line integral just as before. - Unfortunately, - parametrizing a curve in terms of the arc-length parameter is usually very difficult, - so we must develop a method of evaluating line integrals using a different parametrization. -

    - -

    - Given a curve C, find any parametrization of C: - x = g(t) and y=h(t), - for continuous functions g and h, - where a\leq t\leq b. - We can represent this parametrization with a vector-valued function, - \vrt = \langle g(t),h(t)\rangle. -

    - -

    - In , - we defined the arc-length parameter in Equation as - - s(t) = \int_0^t \norm{\vec r\,'(u)}\, du - . -

    - -

    - By the Fundamental Theorem of Calculus, - ds = \norm{\vec r\,'(t)}\, dt. - We can substitute the right hand side of this equation for ds in the line integral definition. -

    - -

    - We can view f as being a function of x and y since it is a function of s. - Thus f(s) =f(x,y) =f\big(g(t),h(t)\big). - This gives us a concrete way to evaluate a line integral: - - \int_C f(s)\, ds = \int_a^bf\big(g(t),h(t)\big)\norm{\vec r\,'(t)}\, dt - . -

    - -

    - We restate this as a theorem, - along with its three-dimensional analogue, - followed by an example where we finally evaluate an integral and find an area. -

    - - - Evaluating a Line Integral Over A Scalar Field - -

    -

      -
    • -

      - Let C be a curve parametrized by \vrt =\langle g(t), - h(t)\rangle, - a\leq t\leq b, - where g and h are continuously differentiable, - and let z=f(x,y), where f is continuous over C. - Thenline integralover scalar field - - \int_C f(s)\, ds = \int_a^b f\big(g(t),h(t)\big)\norm{\vec r\,'(t)}\, dt - . -

      -
      • - -
    • -

      - Let C be a curve parametrized by - \vrt =\langle g(t), - h(t),k(t)\rangle, a\leq t\leq b, - where g, - h and k are continuously differentiable, - and let w=f(x,y,z), where f is continuous over C. - Then - - \int_Cf(s)\, ds = \int_a^bf\big(g(t),h(t),k(t)\big)\norm{\vec r\,'(t)}\, dt - . -

      -
      • -
    -

    -     -     - -

    - To be clear, - the first point of - can be used to find the area under a surface - z=f(x,y) and above a curve C. - We will later give an understanding of the line integral when C is a curve in space. -

    - -

    - Let's do an example where we actually compute an area. -

    - - - Evaluating a line integral: area under a surface over a curve - -

    - Find the area under the surface - f(x,y) =\cos(x)+\sin(y)+2 over the curve C, - which is the segment of the line y=2x+1 on -1\leq x\leq 1, - as shown in . -

    -
    - Finding area under a curve in - -
    - - - A curve lies along a surface in space. The curve corresponds to a line drawn in the x,y plane. - -

    - The surface z=\cos(x)+\sin(y)+2 is plotted in space against three-dimensional coordinate axes. - It has the appearance of a saddle surface in the region that is plotted. -

    - -

    - A curve lies along this surface; it is the set of points on the surface - that lie above the line y=2x+1 in the xy plane. - The line in the plane is also plotted for reference. -

    -     - - - - - // ASY file for figlinescalarfield2_3D.asy in Chapter 10 - // The surface is z=cos(x)+sin(y)+2; the path is the line - // y=2x+1. - // This draws the path, the curve in space, and the surface. - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(10,7.25,3.75); - //(7.63,8.65,5); - //currentprojection=orthographic(2.5,4.6,1.9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,3}; - real[] myychoice={-1,3}; - real[] myzchoice={3}; - defaultpen(0.5mm); - pair xbounds=(-1.2,3.4); - pair ybounds=(-1.2,3.4); - pair zbounds=(-.1,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z=cos(x)+sin(y)=2 - triple f(pair t) { - return (t.x,t.y,-sin(t.y)+cos(t.x)+2); - } - surface s=surface(f,(-1,-1),(pi,pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return (t,2*t+1,-sin(2t+1)+cos(t)+2);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,2*t+1,0);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)+dashed); - - - -
    - -
    - - - A curve in space lies above the line y=2x+1 in the x,y plane. The surface that lies between the line and curve is shown. - -

    - The curve in space from is plotted again with respect to three-dimensional coordinate axes. - This time, the surface on which the curve lies is not shown. - Instead, we again see the line y=2x+1 in the xy plane. -

    - -

    - Between the curve in space and the line in the plane there is a surface, which is plotted. - This surface lies in the plane defined by the equation y=2x+1. - It is bounded below by the xy plane, and above by the curve in space. -

    - -

    - From the perspective of a direction normal to the plane y=2x+1, - this is just the area under a curve. -

    -     - - - - - // ASY file for figlinescalarfield2_3D.asy in Chapter 10 - // The surface is z=cos(x)+sin(y)=2; the path is the line - // y=2x+1. - // This draws the surface which we are finding the area of. - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(10,7.25,3.75); - //currentprojection=orthographic(2.5,4.6,1.9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,3}; - real[] myychoice={-1,3}; - real[] myzchoice={3}; - defaultpen(0.5mm); - pair xbounds=(-1.2,3.4); - pair ybounds=(-1.2,3.4); - pair zbounds=(-.1,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z=cos(x)+sin(y)=2 - triple f(pair t) { - return (t.x,2*t.x+1,t.y*(-sin(t.x*2+1)+cos(t.x)+2)); - } - surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=bluecurvepen; - draw(s,simplesurfacepen,meshpen=p); - - triple g(real t) {return (t,2*t+1,-sin(2t+1)+cos(t)+2);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,2*t+1,0);} - path3 mypath=graph(g,-1,1,operator ..); - draw(mypath,bluepen+linewidth(2)+dashed); - - - -
    -     - -
    -     - -

    - Our first step is to represent C with a vector-valued function. - Since C is a simple line, - and we have a explicit relationship between y and x (namely, - that y is 2x+1), - we can let x = t, y = 2t+1, - and write \vrt = \langle t, 2t+1\rangle for -1\leq t\leq 1. -

    - -

    - We find the values of f over C as f(x,y) = f(t,2t+1) = \cos(t)+\sin(2t+1) + 2. - We also need \norm{\vec r\,'(t)}; - with \vrp(t) = \langle 1,2\rangle, - we have \norm{\vrp(t)} = \sqrt{5}. - Thus ds = \sqrt{5}\, dt. -

    - -

    - The area we seek is - - \int_Cf(s)\, ds \amp = \int_{-1}^1 \big(\cos(t)+\sin(2t+1) + 2\big)\sqrt{5}\, dt - \amp = \left.\sqrt{5}\big(\sin(t) - \frac12\cos(2t+1)+2t\big)\right|_{-1}^1 - \amp \approx 14.418\ \text{ units } ^2 - . -

    -     -     - -

    - We will practice setting up and evaluating a line integral in another example, - then find the area described at the beginning of this section. -

    - - - Evaluating a line integral: area under a surface over a curve - -

    - Find the area over the unit circle in the xy-plane and under the graph of f(x,y) = x^2-y^2+3, - shown in . -

    -
    - Finding area under a curve in - -
    - - - A curve lies on a hyperbolic paraboloid above a circle in the plane. - -

    - The hyperbolic paraboloid z=x^2-y^2+3 is plotted in space, - against a set of three-dimensional coordinate axes. - Below the surface, in the xy plane, the unit circle is plotted. -

    - -

    - The points on the surface that lie above the circle form a curve on the surface. - The curve cuts out a portion of the surface that looks exactly like a Pringles chip. -

    -     - - - - - // ASY file for figlinescalarfield2_3D.asy in Chapter 10 - // The surface is z=x^2-y^2+3; the path is the unit circle. - // - // This draws the path, the curve in space, and the surface. - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.5,9,8); - //(7.63,8.65,5); - //currentprojection=orthographic(2.5,4.6,1.9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={4}; - defaultpen(0.5mm); - pair xbounds=(-1.2,1.2); - pair ybounds=(-1.2,1.2); - pair zbounds=(-.1,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z=x^2-y^2+3 - triple f(pair t) { - return (t.x,t.y,t.x^2-t.y^2+3); - } - surface s=surface(f,(-1.1,-1.1),(1.1,1.1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return (cos(t),sin(t),cos(t)^2-sin(t)^2+3);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(2)+dashed); - - - -
    - -
    - - - A cylindrical surface between the x,y plane and a hyperbolic paraboloid. - -

    - A plot of the portion of the cylinder x^2+y^2=1 - that lies between the xy plane and the hyperbolic paraboloid z=x^2-y^2+3. - The surface looks something like a roller coaster track, along with the scaffolding beneath it. -

    -     - - - - - // ASY file for figlinescalarfield2_3D.asy in Chapter 10 - // The surface is z=x^2-y^2+3; the path is the unit circle. - // - // This draws the surface of the area we seek. - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.5,9,8); - //(7.63,8.65,5); - //currentprojection=orthographic(2.5,4.6,1.9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={4}; - defaultpen(0.5mm); - pair xbounds=(-1.2,1.2); - pair ybounds=(-1.2,1.2); - pair zbounds=(-.1,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z=x^2-y^2+3 - triple f(pair t) { - return (cos(t.x),sin(t.x),t.y*(cos(t.x)^2-sin(t.x)^2+3)); - } - surface s=surface(f,(0,0),(2pi,1),16,8,Spline); - pen p=apexmeshpen; - draw(s,simplesurfacepen,meshpen=p); - - triple g(real t) {return (cos(t),sin(t),cos(t)^2-sin(t)^2+3);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(2)+dashed); - - - -
    -     - -
    -     - -

    - The curve C is the unit circle, - which we will describe with the parametrization - \vrt = \langle \cos(t), \sin(t)\rangle for 0\leq t\leq 2\pi. - We find \norm{\vrp(t)} = 1, so ds = 1 dt. -

    - -

    - We find the values of f over C as f(x,y) = f(\cos(t), \sin(t)) = \cos^2t-\sin^2t+3. - Thus the area we seek is (note the use of the \oint f(s) ds notation): - - \oint_C f(s)\, ds \amp = \int_0^{2\pi}\big(\cos^2t-\sin^2t+3\big)\, dt - \amp = 6\pi - . -

    - -

    - (Note: we may have approximated this answer from the start. - The unit circle has a circumference of 2\pi, - and we may have guessed that due to the apparent symmetry of our surface, - the average height of the surface is 3.) -

    -     -     - -

    - We now consider the example that introduced this section. -

    - - - Evaluating a line integral: area under a surface over a curve - -

    - Find the area under f(x,y) = 1-\cos(x)\sin(y) and over the parabola y = x^2, - from -1\leq x\leq 1. -

    -     - -

    - We parametrize our curve C as - \vrt = \langle t,t^2\rangle for -1\leq t\leq 1; - we find \norm{\vrp(t)} = \sqrt{1+4t^2}, - so ds = \sqrt{1+4t^2}\, dt. -

    - -

    - Replacing x and y with their respective functions of t, - we have f(x,y) = f(t,t^2) = 1-\cos(t)\sin(t^2). - Thus the area under f and over C is found to be - - \int_C f(s)\, ds \amp = \int_{-1}^1 \Big(1-\cos(t)\sin\big(t^2\big)\Big)\sqrt{1+t^2}\, dt. - This integral is impossible to evaluate using the techniques developed in this text. We resort to a numerical approximation; accurate to two places after the decimal, we find the area is - \amp = 2.17 - . -

    -     -     - -

    - We give one more example of finding area. -

    - - - Evaluating a line integral: area under a curve in space - -

    - Find the area above the xy-plane and below the helix parametrized by \vrt = \langle \cos(t),2\sin(t),t/\pi\rangle, - for 0\leq t\leq 2\pi, - as shown in . -

    -
    - Finding area under a curve in - - A helical ramp in space. - -

    - A cylindrical surface that lies between the xy plane and a circular helix. - The surface looks somewhat like a spiral staircase that makes one revolution. -

    -     - - - - - // ASY file for figlinescalarfield2_3D.asy in Chapter 10 - // The path is an elliptical helix ⟨ cos t, 2sin t, t/pi ⟩; no specific surface. - // - // This draws the surface of the area we seek. - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(9.8,10.4,5.5); - //(7.63,8.65,5); - //currentprojection=orthographic(2.5,4.6,1.9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-2.2,2.2); - pair ybounds=(-2.2,2.2); - pair zbounds=(-.1,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z=x^2-y^2+3 - triple f(pair t) { - return (cos(t.x),2sin(t.x),t.y*(t.x/pi)); - } - surface s=surface(f,(0,0),(2pi,1),16,4,Spline); - pen p=apexmeshpen; - draw(s,simplesurfacepen,meshpen=p); - - triple g(real t) {return (cos(t),2sin(t),t/pi);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (cos(t),2sin(t),0);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(2)+dashed); - - - -
    -     - -

    - Note how this is problem is different than the previous examples: - here, the height is not given by a surface, but by the curve itself. -

    - -

    - We use the given vector-valued function \vec r(t) to determine the curve C in the xy-plane by simply using the first two components of \vec r(t): - \vec c(t) = \langle \cos(t),2\sin(t)\rangle. - Thus ds = \norm{\vec c\,'(t)}\,dt = \sqrt{\sin^2t + 4\cos^2t}\,dt. -

    - -

    - The height is not found by evaluating a surface over C, - but rather it is given directly by the third component of \vec r(t): - t/\pi. - Thus - - \oint_C f(s)\, ds = \int_0^{2\pi} \frac{t}{\pi}\sqrt{\sin^2t + 4\cos^2t}\,dt \approx 9.69 - , - where the approximation was obtained using numerical methods. -

    -     -     - - - -

    - Note how in each of the previous examples we are effectively finding - area under a curve, - just as we did when first learning of integration. - We have used the phrase area over - a curve C and under a surface, - but that is because of the important role C plays in the integral. - The figures show how the curve C defines another curve on the surface z=f(x,y), - and we are finding the area under that curve. -

    -     - - - Properties of Line Integrals -

    - Many properties of line integrals can be inferred from general integration properties. - For instance, if k is a scalar, - then \int_C k\,f(s)ds = k\int_Cf(s)ds. -

    - -

    - One property in particular of line integrals is worth noting. - If C is a curve composed of subcurves C_1 and C_2, - where they share only one point in common (see , - then the line integral over C is the sum of the line integrals over C_1 and C_2: - - \int_Cf(s)\, ds = \int_{C_1}f(s)\, ds+\int_{C_2}f(s)\, ds - . -

    - -
    - Illustrating properties of line integrals - -
    - - - Two oriented curves in the plane that are joined at a point. - -

    - Two oriented curves in the plane are drawn without the use of coordinate axes. - The first curve C_1 goes from a point A to a point D. - The second curve C_2 goes from the point D to a point B. -

    - -

    - The two curves have the same tangent line at D, - so that the curve obtained by combining them is smooth. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.35] - - \draw [thick,firstcolor] (0,1) arc [start angle=90,end angle=270,radius=1]; - \draw [thick,firstcolor] (0,-1) parabola (1,1); - - \filldraw (0,1) circle [radius=2pt] node [below] {$A$}; - \filldraw (1,1) circle [radius=2pt] node [right] {$B$}; - \filldraw (0,-1) circle [radius=2pt] node [above] {$D$}; - - \draw (-1,0) node [right] {$C_1$}; - \draw (.5,-.5) node [right] {$C_2$}; - \draw [thick,->,firstcolor] (-1,0) -- (-1,-.01); - \draw [thick,->,firstcolor] (.5,-.5) -- (.51,-.48); - \draw (-1,0) node {\phantom{M}}; - - \end{tikzpicture} - - - -
    - -
    - - - Two oriented curves in the plane that are joined at a point. - -

    - Two oriented curves in the plane are drawn without the use of coordinate axes. - The first curve C_1 goes from a point A to a point D. - The second curve C_2 goes from the point D to a point B. -

    - -

    - The curve C_1 is circular, while C_2 is a line segment. - The two curves have the different tangent lines at D, - so that the curve obtained by combining them is not smooth, but is piecewise smooth. -

    -     - - - \begin{tikzpicture}[>=stealth,scale=1.35] - - \draw [thick,firstcolor] (0,1) arc [start angle=90,end angle=270,radius=1]; - \draw [thick,firstcolor] (0,-1) -- (1,1); - - \filldraw (0,1) circle [radius=2pt] node [below] {$A$}; - \filldraw (1,1) circle [radius=2pt] node [right] {$B$}; - \filldraw (0,-1) circle [radius=2pt] node [above left] {$D$}; - - \draw (-1,0) node [right] {$C_1$}; - \draw (.5,0) node [right] {$C_2$}; - \draw [thick,->,firstcolor] (-1,0) -- (-1,-.01); - \draw [thick,->,firstcolor] (.5,0) -- (.51,.02); - \draw (-1,0) node {\phantom{M}}; - - \end{tikzpicture} - - - -
    -     - -
    - -

    - This property allows us to evaluate line integrals over some curves C that are not smooth. - Note how in the curve is not smooth at D, - so by our definition of the line integral we cannot evaluate \int_C f(s)ds. - However, one can evaluate line integrals over C_1 and C_2 and their sum will be the desired quantity. -

    - -

    - A curve C that is composed of two or more smooth curves is said to be - piecewise smooth. - In this chapter, - any statement that is made about smooth curves also holds for piecewise smooth curves. - smooth curvepiecewise - piecewise smooth curve -

    - -

    - We state these properties as a theorem. -

    - - - Properties of Line Integrals Over Scalar Fields - -

    -

      -
    1. -

      - Let C be a smooth curve parametrized by the arc-length parameter s, - let f and g be continuous functions of s, - and let k_1 and k_2 be scalars. - Then line integralproperties over a scalar field - - \ds \int_C\big(k_1f(s)+k_2g(s)\big)\, ds = k_1\int_Cf(s)\, ds +k_2\int_Cg(s)\, ds - . -

      -
      2. - -
    2. -

      - Let C be piecewise smooth, - composed of smooth components C_1 and C_2. - Then - - \int_Cf(s)\, ds = \int_{C_1}f(s)\, ds + \int_{C_2}f(s)\, ds - . -

      -
      3. -
    -

    -     -     -     - - - Mass and Center of Mass -

    - We first learned integration as a method to find area under a curve, - then later used integration to compute a variety of other quantities, - such as arc length, volume, force, etc. - In this section, - we also introduced line integrals as a method to find area under a curve, - and now we explore one more application. -

    - -

    - Let a curve C - (either in the plane or in space) - represent a thin wire with variable density \delta(s). - We can approximate the mass of the wire by dividing the wire (, the curve) into small segments of length - \Delta s_i and assume the density is constant across these small segments. - The mass of each segment is density of the segment its length; - by summing up the approximate mass of each segment we can approximate the total mass: - - \text{ Total Mass of Wire } = \sum \delta(s_i)\Delta s_i - . -

    - -

    - By taking the limit as the length of the segments approaches 0, we have the definition of the line integral as seen in . - When learning of the line integral, - we let f(s) represent a height; - now we let f(s) = \delta(s) represent a density. -

    - -

    - We can extend this understanding of computing mass to also compute the center of mass of a thin wire. - (As a reminder, - the center of mass can be a useful piece of information as objects rotate about that center.) - We give the relevant formulas in the next definition, followed by an example. - Note the similarities between this definition and , - which gives similar properties of solids in space. -

    - - - Mass, Center of Mass of Thin Wire - -

    - Let a thin wire lie along a smooth curve C with continuous density function \delta(s), - where s is the arc length parameter. - masscenter of - mass - -

      -
    1. -

      - The mass of the thin wire is \ds M = \int_C \delta(s)\, ds. -

      -
      2. - -
    2. -

      - The moment about the yz-plane - is \ds M_{yz} = \int_C x\delta(s)\, ds. -

      -
      3. - -
    3. -

      - The moment about the xz-plane - is \ds M_{xz} = \int_C y\delta(s)\, ds. -

      -
      4. - -
    4. -

      - The moment about the xy-plane - is \ds M_{xy} = \int_C z\delta(s)\, ds. -

      -
      5. - -
    5. -

      - The center of mass of the wire is - - (\overline{x},\overline{y},\overline{z}) = \left(\frac{M_{yz}}M, \frac{M_{xz}}M,\frac{M_{xy}}M\right) - . -

      -
      6. -
    -

    -     -     - - - Evaluating a line integral: calculating mass - -

    - A thin wire follows the path - \vrt = \langle 1+\cos(t),1+\sin(t), 1+ \sin(2t)\rangle, - 0\leq t\leq 2\pi. - The density of the wire is determined by its position in space: - \delta(x,y,z) = y+z gm/cm. - The wire is shown in , - where a light color indicates low density and a dark color represents high density. - Find the mass and center of mass of the wire. -

    - -
    - Finding the mass of a thin wire in - - A curve in space representing a wire, and a point representing its center of mass. - -

    - A curve in space is drawn in the first octant, - relative to a set of three-dimensional coordinate axes. -

    - -

    - The curve looks like it could be the intersection of a hyperbolic paraboloid with a cylinder. - From above, the curve appears to be circular, but from the side, - we can see that the curve oscillates up and down. -

    - -

    - The center of mass is shown as a point in space. - This point is not on the curve, but it is surrounded by the curve. -

    -     - - - - import palette; - - // ASY file for figlinescalarfield2_3D.asy in Chapter 10 - // The path is an elliptical helix ⟨ cos t, 2sin t, t/pi ⟩; no specific surface. - // - // This draws the surface of the area we seek. - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14.4,-6.86,4); - //(7.63,8.65,5); - //currentprojection=orthographic(2.5,4.6,1.9); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={2}; - real[] myychoice={2}; - real[] myzchoice={2}; - defaultpen(0.5mm); - pair xbounds=(-.2,2.5); - pair ybounds=(-.2,2.5); - pair zbounds=(-.1,2.1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - real r(real t) {return 3exp(-0.1*t);} - real x(real t) {return 1+cos(t);} - real y(real t) {return 1+sin(t);} - real z(real t) {return 1+sin(2*t);} - - path3 p=graph(x,y,z,0,2*pi,50,operator ..); - - tube T=tube(p,.05); - surface s=T.s; - s.colors(palette(s.map(ypart)+s.map(zpart),Gradient(palered,heavyblue))); - draw(s,render(merge=true)); - //dot((1,1.25,1.2),black); - draw(shift(1,1.25,1.2)*scale3(.05)*unitsphere); - //draw(T.center,thin()); - - - -
    -     - -

    - We compute the density of the wire as - - \delta(x,y,z) = \delta\big(1+\cos(t),1+\sin(t), 1+\sin(2t)\big) = 2+\sin(t)+\sin(2t) - . -

    - -

    - We compute ds as - - ds = \norm{\vrp(t)}\, dt = \sqrt{\sin^2t+\cos^2t+4\cos^2(2t)}\, dt = \sqrt{1+4\cos^2(2t)}\, dt - . -

    - -

    - Thus the mass is - - M = \oint_C \delta(s)\, ds = \int_0^{2\pi} \big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt \approx 21.08\text{ gm } - . -

    - -

    - We compute the moments about the coordinate planes: - - M_{yz} \amp = \oint_C x\delta(s)\, ds - \amp = \int_0^{2\pi}(1+\cos(t))\big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt - \amp \approx 21.08. - M_{xz} \amp = \oint_C y\delta(s)\, ds - \amp = \int_0^{2\pi}(1+\sin(t))\big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt - \amp \approx 26.35 - M_{xy} \amp = \oint_C z\delta(s)\, ds - \amp = \int_0^{2\pi}\big(1+\sin(2 t)\big)\big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt - \amp \approx 25.40 - -

    - -

    - Thus the center of mass of the wire is located at - - (\overline{x},\overline{y},\overline{z}) = \left(\frac{M_{yz}}M, \frac{M_{xz}}M,\frac{M_{xy}}M\right) \approx (1,1.25,1.20) - , - as indicated by the dot in . - Note how in this example, - the curve C is centered - about the point (1,1,1), - though the variable density of the wire pulls the center of mass out along the y and z axes. -

    -     -     - -

    - We end this section with a callback to the Integration Review that preceded this section. - A line integral looks like: \int_C f(s)\, ds. - As stated before the definition of the line integral, - this means sum up, along a curve C, - function values f(s) small arc lengths. - When f(s) represents a height, - we have height length = area. - When f(s) is a density (and we use - \delta(s) by convention), we have density - (mass per unit length) - length = mass. -

    - -

    - In the next section, we investigate a new mathematical object, - the vector field. - The remaining sections of this chapter are devoted to understanding integration in the context of vector fields. -

    -     - - - - Terms and Concepts - - -

    - Explain how a line integral can be used to find the area under a curve. -

    -     - - - -

    - When C is a curve in the plane and f is a function defined over C, - then \int_C f(s)\, ds describes the area under the spatial curve that lies on f, - over C. -

    -     -     - - - -

    - How does the evaluation of a line integral given as - \int_C f(s)\, ds differ from a line integral given as \oint_C f(s)\, ds? -

    -     - - - -

    - The evaluation is the same. - The \oint notation signifies that the curve C is a closed curve, - though the evaluation is the same. -

    -     -     - - - -

    - Why are most line integrals evaluated using instead of directly - as \int_C f(s)\, ds? -

    -     - - - -

    - The variable s denotes the arc-length parameter, - which is generally difficult to use. - allows one to parametrize a curve using another, - ideally easier-to-use, parameter. -

    -     -     - - - -

    - Sketch a closed, - piecewise smooth curve composed of three subcurves. -

    -     - - -     -     - - - Problems - - -

    - A planar curve C is given along with a function f that is defined over C. - Evaluate the line integral \ds \int_Cf(s)\, ds. -

    -     - - - -

    - C is the line segment joining the points (-2,-1) and (1,2); - the function is f(x,y)=x^2+y^2+2. -

    -     - -

    - 12\sqrt{2} -

    -     -     - - - -

    - C is the segment of y=3x+2 on [1,2]; - the function is f(x,y)=5x+2y. -

    -     - -

    - 41\sqrt{10}/2 -

    -     -     - - - -

    - C is the circle with radius 2 centered at the point (4,2); - the function is f(x,y)=3x-y. -

    -     - -

    - 40\pi -

    -     -     - - - -

    - C is the curve given by \vec r(t) = \langle \cos(t)+t\sin(t), \sin(t)-t\cos(t)\rangle on [0,2\pi]; - the function is f(x,y)=5. -

    -     - -

    - 10\pi^2 -

    -     -     - - - -

    - C is the piecewise curve composed of the line segments that connect (0,1) to (1,1), - then connect (1,1) to (1,0); - the function is f(x,y)=x+y^2. -

    -     - -

    - Over the first subcurve of C, - the line integral has a value of 3/2; - over the second subcurve, the line integral has a value of 4/3. - The total value of the line integral is thus 17/6. -

    -     -     - - - -

    - C is the piecewise curve composed of the line segment joining the points (0,0) and (1,1), - along with the quarter-circle parametrized by \langle \cos(t),-\sin(t)+1\rangle on - [0,\pi/2](which starts at the point (1,1) and ends at (0,0); - the function is f(x,y)=x^2+y^2. -

    -     - -

    - Over the first subcurve of C, - the line integral has a value of 2\sqrt{2}/3; - over the second subcurve, the line integral has a value of \pi-2. - The total value of the line integral is thus \pi+2\sqrt{2}/3-2. -

    -     -     -     - - - -

    - A planar curve C is given along with a function f that is defined over C. - Set up the line integral \ds \int_Cf(s)\, ds, - then approximate its value using technology. -

    -     - - - -

    - C is the portion of the parabola y=2x^2+x+1 on [0,1]; - the function is f(x,y)=x^2+2y. -

    -     - -

    - \int_0^1(5t^2+_2t+2)\sqrt{(4t+1)^2+1}\, dt \approx 17.071 -

    -     -     - - - -

    - C is the portion of the curve y=\sin(x) on [0,\pi]; - the function is f(x,y)=x. -

    -     - -

    - \int_0^\pi t\sqrt{1+\cos^2t}\, dt \approx 6.001 -

    -     -     - - - -

    - C is the ellipse given by \vec r(t) = \langle 2\cos(t),\sin(t)\rangle on [0,2\pi]; - the function is f(x,y)=10-x^2-y^2. -

    -     - -

    - \oint_0^{2\pi} \big(10-4\cos^2t-\sin^2t\big)\sqrt{\cos^2t+4\sin^2t}\, dt \approx 74.986 -

    -     -     - - - -

    - C is the portion of y=x^3 on [-1,1]; - the function is f(x,y)=2x+3y+5. -

    -     - -

    - \int_{-1}^{1} \big(3t^3+2t+5\big)\sqrt{9t^4+1}\, dt \approx 15.479 -

    -     -     -     - - - -

    - A parametrized curve C in space is given. - Find the area above the xy-plane that is under C. -

    -     - - - -

    - C: \vec r(t) = \langle 5t,t,t^2\rangle for 1\leq t\leq 2. -

    -     - -

    - 7\sqrt{26}/3 -

    -     -     - - - -

    - C: \vec r(t) = \langle \cos(t),\sin(t), \sin(2t)+1\rangle for 0\leq t\leq 2\pi. -

    -     - -

    - 2\pi -

    -     -     - - - -

    - C: \vec r(t) = \langle 3\cos(t),3\sin(t), t^2\rangle for 0\leq t\leq 2\pi. -

    -     - -

    - 8\pi^3 -

    -     -     - - - -

    - C: \vec r(t) = \langle 3t,4t, - t\rangle for 0\leq t\leq 1. -

    -     - -

    - 5/2 -

    -     -     -     - - - -

    - A parametrized curve C is given that represents a thin wire with density \delta. - Find the mass and center of mass of the thin wire. -

    -     - - - -

    - C: \vec r(t) = \langle \cos(t),\sin(t), t\rangle for 0\leq t\leq 4\pi; - \delta(x,y,z) = z. -

    -     - -

    - M=8\sqrt{2}\pi^2; center of mass is (0,-1/(2\pi), 8\pi/3). -

    -     -     - - - -

    - C: \vec r(t) = \langle t-t^2,t^2-t^3,t^3-t^4\rangle for 0\leq t\leq 1; - \delta(x,y,z) = x+2y+2z. - Use technology to approximate the value of each integral. -

    -     - -

    - M\approx 0.237; - center of mass is approximately (0.173, 0.099,0.065). -

    -     -     -     -     -     -
    -
    - Vector Fields - -

    - We have studied functions of two and three variables, - where the input of such functions is a point - (either a point in the plane or in space) - and the output is a number. -

    - -

    - We could also create functions where the input is a point (again, - either in the plane or in space), - but the output is a vector. - For instance, we could create the following function: - \vec F(x,y) = \langle x+y, x-y\rangle, - where \vec F(2,3) = \langle 5,-1\rangle. - We are to think of \vec F assigning the vector - \langle 5,-1\rangle to the point (2,3); - in some sense, - the vector \langle 5,-1\rangle lies at the point (2,3). -

    - - - -

    - Such functions are extremely useful in any context where magnitude and direction are important. - For instance, - we could create a function \vec F that represents the electromagnetic force exerted at a point by a electromagnetic field, - or the velocity of air as it moves across an airfoil. -

    - -

    - Because these functions are so important, - we need to formally define them. -

    - - - Vector Field - -

    -

      -
    1. -

      - A vector field in the plane - is a function \vec F(x,y) whose domain is a subset of - \mathbb{R}^2 and whose output is a two-dimensional vector: - vector field - - \vec F(x,y) = \langle M(x,y), N(x,y)\rangle - . -

      -
      2. - -
    2. -

      - A vector field in space - is a function \vec F(x,y,z) whose domain is a subset of - \mathbb{R}^3 and whose output is a three-dimensional vector: - - \vec F(x,y,z) = \langle M(x,y,z), N(x,y,z), P(x,y,z)\rangle - . -

      -
      3. -
    -

    -     -     - -

    - This definition may seem odd at first, - as a special type of function is called a field. However, - as the function determines a field of vectors, - we can say the field is defined by the function, - and thus the field is a function. -

    - -
    - Demonstrating methods of graphing vector fields - -
    - - - Several arrows are drawn in the plane to illustrate the concept of a vector field. - -

    - We tend to think of a vector field as a collection of vectors, - with each vector drawn with its tail at the point where it is defined. -

    - -

    - This image illustrates the fact that this conception does not work well in practice. - Eight different vectors are plotted, each with its tail at the point used to compute the vector. - The vectors end up in a couple of configurations resembling tridents, - and the overall effect is not very informative. -

    -     - - - import graph; - size(282,282); - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - - if(incolor) { - colorone = bluepen+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - } else - { - colorone = black+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - }; - - real f(real x) {return exp(x);} - pair F(real x) {return (x,f(x));} - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - real[] myxchoice={-3,-2,-1,1,2,3}; - real[] myychoice={-3,-2,-1,1,2,3}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-2,-2); - pair b=(2,2); - - int[] array={-1,0,1}; - for(int j : array){ - for(int k : array){ - draw((j,k) -- (j+j+k,k+j-k),arrow=Arrow(DefaultHead,size=4),colorone); - } - } - - //pair vectfunction(pair z) {return (z.y+z.x,z.x-z.y);} - //path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - //add(vectorfield(vector,a,b,5,5,bluepen)); - - - -
    - -
    - - - Several arrows are plotted in the plane to illustrate the concept of a vector field. - -

    - This plot shows the same vectors that were plotted in . - The difference is that this time, the vectors have been shifted so that their midpoint, - not their tail, is at the point used to compute each vector. -

    - -

    - This image gives a better idea of the directions represented by the vector field; - however, it is still somewhat confusing due to the length of the vectors. - To help with visualization, the vectors in a vector field are often drawn much shorter than their true length. -

    -     - - - import graph; - size(282,282); - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - - if(incolor) { - colorone = bluepen+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - } else - { - colorone = black+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - }; - - real f(real x) {return exp(x);} - pair F(real x) {return (x,f(x));} - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - real[] myxchoice={-3,-2,-1,1,2,3}; - real[] myychoice={-3,-2,-1,1,2,3}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-2,-2); - pair b=(2,2); - - int[] array={-1,0,1}; - for(int j : array){ - for(int k : array){ - draw((j-(j+k)/2,k-(j-k)/2) -- (j+(j+k)/2,k+(j-k)/2),arrow=Arrow(DefaultHead,size=4),bluepen); - } - } - - //pair vectfunction(pair z) {return (z.y+z.x,z.x-z.y);} - //path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - //add(vectorfield(vector,a,b,5,5,bluepen)); - - - -
    -     - -
    - -

    - Visualizing vector fields helps cement this connection. - When graphing a vector field in the plane, - the general idea is to draw the vector - \vec F(x,y) at the point (x,y). - For instance, - using \vec F(x,y) = \langle x+y,x-y\rangle as before, - at (1,1) we would draw \langle 2,0\rangle. -

    - -

    - In , - one can see that the vector \langle 2,0\rangle is drawn - starting from the point (1,1). - A total of 8 vectors are drawn, - with the x- and y-values of -1,0,1. - In many ways, the resulting graph is a mess; - it is hard to tell what this field looks like. -

    - -

    - In , - the same field is redrawn with each vector \vec F(x,y) drawn - centered on the point (x,y). - This makes for a better looking image, - though the long vectors can cause confusion: - when one vector intersects another, - the image looks cluttered. -

    - -

    - A common way to address this problem is limit the length of each arrow, - and represent long vectors with thick arrows, - as done in . - Usually we do not use a graph of a vector field to determine exactly the magnitude of a particular vector. - Rather, we are more concerned with the relative magnitudes of vectors: - which are bigger than others? - Thus limiting the length of the vectors is not problematic. -

    - -
    - Demonstrating methods of graphing vector fields - -
    - - - Eight vectors are plotted in the plane to represent a vector field, using relative length to decrease clutter. - -

    - This is another representation of the vector field in . - In this version, the vectors plotted are shorter than their true length. - Instead, relative length is used to indicate varying magnitude. -

    - -

    - This change reduces clutter and makes it easier to understand the vector field. - Eight vectors are plotted at the points (1,1), (1,0), (1,-1), - (0,-1), (-1,-1), (-1,0), (-1,1), and (0,1). -

    - -

    - As we move counter-clockwise about the square, - the vectors rotate clockwise. - The vector at (1,1) points to the right, - and the vector at each adjacent point is rotated by 45 degrees as we move around the square. -

    -     - - - import graph; - size(282,282); - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - - if(incolor) { - colorone = bluepen+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - } else - { - colorone = black+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - }; - - real f(real x) {return exp(x);} - pair F(real x) {return (x,f(x));} - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - real[] myxchoice={-3,-2,-1,1,2,3}; - real[] myychoice={-3,-2,-1,1,2,3}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-1,-1); - pair b=(1,1); - - pair vectfunction(pair z) {return (z.y+z.x,z.x-z.y);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,3,3,bluepen)); - - - -
    - -
    - - - Another plot of the same vector field used in every image so far, but this time with many more vectors. - -

    - We get one more plot of the same vector field use in - and the previous images in this section. - This time, instead of eight vectors being plotted, there are about eighty. -

    - -

    - Using more vectors, we can see that magnitude increases with distance from the origin, - and the directions of the vectors appear to follow hyperbolic curves in the plane. -

    -     - - - import graph; - size(282,282); - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen+ linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - - real f(real x) {return exp(x);} - pair F(real x) {return (x,f(x));} - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - real[] myxchoice={-3,-2,-1,1,2,3}; - real[] myychoice={-3,-2,-1,1,2,3}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-3,-3); - pair b=(3,3); - - pair vectfunction(pair z) {return (z.y+z.x,z.x-z.y);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,8,8,mainvect)); - - - -
    -     - -
    - -

    - Drawing arrows with variable thickness is best done with technology; - search the documentation of your favorite graphing program for terms like vector fields - or slope fields to learn how. - Technology obviously allows us to plot many vectors in a vector field nicely; - in , - we see the same vector field drawn with many vectors, - and finally get a clear picture of how this vector field behaves. - (If this vector field represented the velocity of air moving across a flat surface, - we could see that the air tends to move either to the upper-right or lower-left, - and moves very slowly near the origin.) -

    - -

    - We can similarly plot vector fields in space, - as shown in , - though it is not often done. - The plots get very busy very quickly, - as there are lots of arrows drawn in a small amount of space. - In - the field \vec F = \langle -y,x,z\rangle is graphed. - If one could view the graph from above, - one could see the arrows point in a circle about the z-axis. - One should also note how the arrows far from the origin are larger than those close to the origin. -

    - -

    - It is good practice to try to visualize certain vector fields in one's head. - For instance, - consider a point mass at the origin and the vector field that represents the gravitational force exerted by the mass at any point in the room. - The field would consist of arrows pointing toward the origin, - increasing in size as they near the origin - (as the gravitational pull is strongest near the point mass). -

    - -
    - Graphing a vector field in space - - A very chaotic plot of a three-dimensional vector field. - -

    - This plot shows a vector field in three dimensions, relative to the usual coordinate axes. - It is a very cluttered plot. - With the default perspective, it appears to be a large jumble of arrows, - pointing every which way in space. -

    - -

    - Rotating the image reveals a bit more structure: - from above, the arrows appear to follow a circular trajectory. - In fact, the vectors in this vector field are all tangent to helical curves, - with those above the xy plane pointing upward, and those below pointing downward. -

    -     - - - - - size(200,200,IgnoreAspect); - currentprojection=orthographic(13.5,-13,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={1}; - defaultpen(0.25mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - path3 gradient1(triple z){ - return O--(-z.y,z.x,z.z); - } - - triple A=(-1,-1,-1); - triple B=(1,1,1); - - picture VectorPlot3D(path3 vector(triple t), triple a, triple b, - int nx=nmesh, int ny=nx, int nz=nx,bool truesize=false, - real maxlength=truesize ? 0 : min(abs(b.x-a.x)/nx,abs(b.y-a.y)/ny,abs(b.z-a.z)/nz), - // bool cond(pair z)=null, - pen p=currentpen, - arrowbar3 arrow=Arrow3(6), margin3 margin=PenMargin3, - string name="", render render=defaultrender) - { - picture pic; - real dx=1/nx; - real dy=1/ny; - real dz=1/nz; - real scale; - if(maxlength > 0) { - real size(triple t) { - path3 g=vector(t); - return abs(point(g,size(g)-1)-point(g,0)); - } - real max=size((0,0,0)); - - for(int i=0; i <= nx; ++i) { - real x=interp(a.x,b.x,i*dx); - for(int j=0; j <= ny; ++j) - { - real y=interp(a.y,b.y,j*dy); - for(int k=0; k <= nz; ++k) - max=max(max,size((x,y,interp(a.z,b.z,k*dz)))); - }} - scale=max > 0 ? maxlength/max : 1; - } else scale=1; - bool group=name != "" || render.defaultnames; - if(group) - begingroup3(pic,name == "" ? "vectorfield" : name,render); - for(int i=0; i <= nx; ++i) { - real x=interp(a.x,b.x,i*dx); - for(int j=0; j <= ny; ++j) { - real y=interp(a.y,b.y,j*dy); - for(int k=0; k <= nz; ++k) - { triple z=(x,y,interp(a.z,b.z,k*dz)); - { - path3 g=scale3(scale)*vector(z); - string name="vector"; - if(truesize) { - picture opic; - draw(opic,g,p,arrow,margin,name,render); - add(pic,opic,z); - } else - draw(pic,shift(z)*g,p,arrow,margin,name,render); - } - } - }} - if(group) - endgroup3(pic); - return pic; - - } - add(VectorPlot3D(gradient1,A,B,3,3,3,bluepen)); - - - -
    -     - - - Vector Field Notation and Del Operator -

    - - defines a vector field \vec F using the notation - - \vec F(x,y) = \langle M(x,y), N(x,y)\rangle \text{ and } \vec F(x,y,z) = \langle M(x,y,z), N(x,y,z),P(x,y,z)\rangle - . -

    - -

    - That is, the components of \vec F are each functions of x and y - (and also z in space). - As done in other contexts, we will drop the of x, - y and z portions of the notation and refer to vector fields in the plane and in space as - - \vec F = \langle M, N\rangle \text{ and } \vec F = \langle M,N,P\rangle - , - respectively, as this shorthand is quite convenient. -

    - -

    - Another item of notation will become useful: - the del operator. - del operator - Recall in - how we used the symbol \nabla - (pronounced del) - to represent the gradient of a function of two variables. - That is, if z = f(x,y), - then del f = \nabla f = \langle f_x, f_y\rangle. -

    - - - -

    - We now define \nabla to be the del operator. - It is a vector whose components are partial derivative operations. -

    - -

    - In the plane, - \ds\nabla = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right\rangle; - in space, \ds\nabla = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle. -

    - -

    - With this definition of \nabla, - we can better understand the gradient \nabla f. - As f returns a scalar, - the properties of scalar and vector multiplication gives - - \nabla f = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right\rangle f = \left\langle \frac{\partial}{\partial x}\,f, \frac{\partial}{\partial y}\,f\right\rangle = \langle f_x, f_y\rangle - . -

    - - - -

    - Now apply the del operator \nabla to vector fields. - Let \vec F = \langle x+\sin(y),y^2+z,x^2\rangle. - We can use vector operations and find the dot product of \nabla and \vec F: - - \nabla \cdot \vec F \amp = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle\cdot \langle x+\sin(y),y^2+z,x^2\rangle - \amp = \frac{\partial}{\partial x}(x+\sin(y))+ \frac{\partial}{\partial y}(y^2+z) + \frac{\partial}{\partial z}(x^2) - \amp =1+2y - . -

    - -

    - We can also compute their cross products: - - \nabla\times \vec F \amp = \left\langle \frac{\partial}{\partial y}\big(x^2\big)-\frac{\partial}{\partial z}\big(y^2+z\big),\frac{\partial}{\partial z}\big(x+\sin(y)\big)-\frac{\partial}{\partial x}\big(x^2\big),\right. - \amp \phantom{=\frac{\partial}{\partial y}\big(x^2\big)-\frac{\partial}{\partial z}\big(y^2+z\big)}\left.\frac{\partial}{\partial x}\big(y^2+z\big)-\frac{\partial}{\partial y}\big(x+\sin(y)\big)\right\rangle - \amp =\langle -1,-2x,-\cos(y)\rangle - . -

    - -

    - We do not yet know why we would want to compute the above. - However, as we next learn about properties of vector fields, - we will see how these dot and cross products with the del operator are quite useful. -

    -     - - - Divergence and Curl -

    - Two properties of vector fields will prove themselves to be very important: - divergence and curl. - Each is a special derivative of a vector field; - that is, each measures an instantaneous rate of change of a vector field. -

    - - - -

    - If the vector field represents the velocity of a fluid or gas, - then the divergence - divergence - vector fielddivergence of - of the field is a measure of the - compressibility of the fluid. - If the divergence is negative at a point, - it means that the fluid is compressing: - more fluid is going into the point than is going out. - If the divergence is positive, - it means the fluid is expanding: - more fluid is going out at that point than going in. - A divergence of zero means the same amount of fluid is going in as is going out. - If the divergence is zero at all points, - we say the field is incompressible. - incompressible vector field -

    - -

    - It turns out that the proper measure of divergence is simply \nabla \cdot \vec F, - as stated in the following definition. -

    - - - Divergence of a Vector Field - -

    - The divergence of a vector field \vec F is - divergence - vector fielddivergence of - - \divv \vec F = \nabla \cdot \vec F - . -

      -
    • -

      - In the plane, with \vec F = \langle M,N\rangle, - \divv \vec F = M_x+N_y. -

      -
      • - -
    • -

      - In space, with \vec F = \langle M,N,P\rangle, - \divv \vec F = M_x+N_y+P_z. -

      -
      • -
    -

    -     -     - - - -

    - Curlvector fieldcurl of - curl - is a measure of the spinning action of the field. - Let \vec F represent the flow of water over a flat surface. - If a small round cork were held in place at a point in the water, - would the water cause the cork to spin? - No spin corresponds to zero curl; - counterclockwise spin corresponds to positive curl and clockwise spin corresponds to negative curl. -

    - -

    - In space, things are a bit more complicated. - Again let \vec F represent the flow of water, - and imagine suspending a tennis ball in one location in this flow. - The water may cause the ball to spin along an axis. - If so, the curl of the vector field is a vector - (not a scalar, as before), - parallel to the axis of rotation, - following a right hand rule: - when the thumb of one's right hand points in the direction of the curl, - the ball will spin in the direction of the curling fingers of the hand. -

    - -

    - In space, it turns out the proper measure of curl is \nabla \times \vec F, - as stated in the following definition. - To find the curl of a planar vector field \vec F = \langle M,N\rangle, - embed it into space as \vec F = \langle M, N, 0\rangle and apply the cross product definition. - Since M and N are functions of just x and y - (and not z), - all partial derivatives with respect to z become 0 and the result is simply \langle 0,0,N_x-M_y\rangle. - The third component is the measure of curl of a planar vector field. -

    - - - - - Curl of a Vector Field - -

    -

      -
    • -

      - Let \vec F = \langle M,N\rangle be a vector field in the plane. - The curl of \vec F is \curl \vec F = N_x - M_y. - curl -

      -
      • - -
    • -

      - Let \vec F = \langle M,N,P\rangle be a vector field in space. - The curl of \vec F is \curl \vec F = \nabla \times \vec F = \langle P_y-N_z,M_z-P_x,N_x - M_y\rangle. -

      -
      • -
    -

    -     -     - -

    - We adopt the convention of referring to curl as \nabla \times \vec F, - regardless of whether \vec F is a vector field in two or three dimensions. - (Some people prefer to write (\nabla\times \vec F)\cdot \vec k in two dimensions.) -

    - - - -

    - We now practice computing these quantities. -

    - - - Computing divergence and curl of planar vector fields - -

    - For each of the planar vector fields given below, - view its graph and try to visually determine if its divergence and curl are 0. - Then compute the divergence and curl. - -

      -
    1. -

      - \vec F = \langle y,0\rangle (see ) -

      -
      2. - -
    2. -

      - \vec F = \langle -y,x\rangle (see ) -

      -
      3. - -
    3. -

      - \vec F = \langle x,y\rangle (see ) -

      -
      4. - -
    4. -

      - \vec F = \langle \cos(y), \sin(x)\rangle (see ) -

      -
      5. -
    -

    -

    - - -

    -

      -
    1. -

      - The arrow sizes are constant along any horizontal line, - so if one were to draw a small box anywhere on the graph, - it would seem that the same amount of fluid would enter the box as exit. - Therefore it seems the divergence is zero; it is, as - - \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(0) = 0 - . -

      - -
      - The vector fields in parts 1 and 2 in - -
      - - - A vector field of horizontal arrows: to the right above the x axis, and to the left below. - -

      - A two-dimensional vector field is plotted against x and y coordinate axes. - Several things are notable in this image: -

        -
      • -

        - All of the vectors in the vector field are horizontal -

        -
        • -
      • -

        - Vectors above the x axis point to the right, - and vectors below the x axis point to the left. -

        -
        • -
      • -

        - Near the x axis, the magnitude of the vectors is very small, - and the vector field vanishes completely along the x axis. - The vectors get larger as they get further from the x axis. -

        -
        • -
      -

      -       - - - import graph; - size(282,282); - - // vector field F = ⟨ y, 0 ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen+ linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - - real f(real x) {return exp(x);} - pair F(real x) {return (x,f(x));} - - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-1,-1); - pair b=(1,1); - - pair vectfunction(pair z) {return (z.y,0);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,8,12,mainvect)); - - - -
      - -
      - - - A rotational vector field. The vectors appear to describe counter-clockwise circular motion. - -

      - A two-dimensional vector field is plotted against x and y coordinate axes. - The vectors in this vector field appear to describe circular motion. - Each vector could be tangent to a circle centered at the origin, although no circles are depicted in the image. - The directions of the vectors correspond to counter-clockwise motion. -

      - -

      - The magnitudes of the vectors depend on their distance from the origin. - Vectors near the centre of the image are small, while those near the edges are larger. -

      -       - - - import graph; - size(282,282); - - // vector field F = ⟨ -y, x ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen+ linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - - real f(real x) {return exp(x);} - pair F(real x) {return (x,f(x));} - - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-1,-1); - pair b=(1,1); - - pair vectfunction(pair z) {return (-z.y,z.x);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,mainvect)); - - - -
      -       -
      - -

      - At any point on the x-axis, - arrows above it move to the right and arrows below it move to the left, - indicating that a cork placed on the axis would spin clockwise. - A cork placed anywhere above the x-axis would have water above it moving to the right faster than the water below it, - also creating a clockwise spin. - A clockwise spin also appears to be created at points below the x-axis. - Thus it seems the curl should be negative - (and not zero). - Indeed, it is: - - \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(0) - \frac{\partial}{\partial y}(y) = -1 - . -

      -
      2. - -
    2. -

      - It appears that all vectors that lie on a circle of radius r, - centered at the origin, have the same length - (and indeed this is true). - That implies that the divergence should be zero: - draw any box on the graph, - and any fluid coming in will lie along a circle that takes the same amount of fluid out. - Indeed, the divergence is zero, as - - \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(-y) + \frac{\partial}{\partial y}(x) = 0 - . - Clearly this field moves objects in a circle, - but would it induce a cork to spin? - It appears that yes, it would: - place a cork anywhere in the flow, - and the point of the cork closest to the origin would feel less flow than the point on the cork farthest from the origin, - which would induce a counterclockwise flow. - Indeed, the curl is positive: - - \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(x) - \frac{\partial}{\partial y}(-y) = 1-(-1) = 2 - . - Since the curl is constant, - we conclude the induced spin is the same no matter where one is in this field. -

      -
      3. - -
    3. -

      - At the origin, - there are many arrows pointing out but no arrows pointing in. - We conclude that at the origin, - the divergence must be positive - (and not zero). - If one were to draw a box anywhere in the field, - the edges farther from the origin would have larger arrows passing through them than the edges close to the origin, - indicating that more is going from a point than going in. - This indicates a positive - (and not zero) - divergence. - This is correct: - - \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) = 1+1=2 - . - One may find this curl to be harder to determine visually than previous examples. - One might note that any arrow that induces a clockwise spin on a cork will have an equally sized arrow inducing a counterclockwise spin on the other side, - indicating no spin and no curl. - This is correct, as - - \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(y) - \frac{\partial}{\partial y}(x) = 0 - . -

      - -
      - The vector fields in parts 3 and 4 in - -
      - - - A radial vector field. Each vector points away from the origin, and vectors further from the origin are larger. - -

      - A two-dimensional vector field is plotted against x and y coordinate axes. - The vectors in this vector field could be tangent to lines through the origin: - each one points directly away from the origin, - in the same direction as the line from the origin to the point where the vector is plotted. -

      - -

      - The magnitude of each vector depends on its distance from the origin. - Those close to the origin are small, while those further out are larger. -

      -       - - - import graph; - size(282,282); - - // vector field F = ⟨ x,y ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen+ linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - - real f(real x) {return exp(x);} - pair F(real x) {return (x,f(x));} - - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-1,-1); - pair b=(1,1); - - pair vectfunction(pair z) {return (z.x,z.y);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,mainvect)); - - - -
      - -
      - - - A vector field that appears to describe swirling motion, with several vortices. - -

      - A two-dimensional vector field is plotted against x and y coordinate axes. - This vector field is quite chaotic, as one might expect given the oscillatory nature of the sine and cosine functions. -

      - -

      - There appear to be several vortices, where the vectors circulate around certain points. - Between each vortex is an area where the vectors could be tangent to a hyperbola, - like contour lines from a hyperbolic paraboloid. -

      -       - - - import graph; - size(282,282); - - // vector field F = ⟨ cos y, sin x ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen+ linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - real f(real x) {return exp(x);} - pair F(real x) {return (x,f(x));} - - pair xbounds=(-7,7); - pair ybounds=(-7,7); - real[] myxchoice={-6,-3,3,6}; - real[] myychoice={-6,-3,3,6}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-2*pi,-2*pi); - pair b=(2*pi,2*pi); - - pair vectfunction(pair z) {return (cos(z.y),sin(z.x));} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,mainvect)); - - - -
      -       -
      -
      4. - -
    4. -

      - One might find this divergence hard to determine visually as large arrows appear in close proximity to small arrows, - each pointing in different directions. - Instead of trying to rationalize a guess, - we compute the divergence: - - \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(\cos(y)) + \frac{\partial}{\partial y}(\sin(x)) = 0 - . - Perhaps surprisingly, the divergence is 0. - With all the loops of different directions in the field, - one is apt to reason the curl is variable. - Indeed, it is: - - \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(\sin(x)) - \frac{\partial}{\partial y}(\cos(y)) = \cos(x) + \sin(y) - . - Depending on the values of x and y, - the curl may be positive, negative, or zero. -

      -
      5. -
    -

    - -     - - - Computing divergence and curl of vector fields in space - -

    - Compute the divergence and curl of each of the following vector fields. - -

      -
    1. -

      - \vec F = \langle x^2+y+z, -x-z, x+y\rangle -

      -
      2. - -
    2. -

      - \vec F = \langle e^{xy}, \sin(x+z),x^2+y\rangle -

      -
      3. -
    -

    -     - -

    - We compute the divergence and curl of each field following the definitions. - -

      -
    1. -

      - - \divv \vec F \amp = \nabla \cdot \vec F = M_x+N_y+P_z = 2x+0+0= 2x - \curl\vec F \amp = \nabla \times \vec F = \langle P_y-N_z,M_z-P_x,N_x - M_y\rangle - \amp = \langle 1 - (-1), 1-1,-1-(1)\rangle = \langle 2,0,-2\rangle - . - For this particular field, - no matter the location in space, - a spin is induced with axis parallel to \langle 2,0,-2\rangle. -

      -
      2. - -
    2. -

      - - \divv \vec F \amp = \nabla \cdot \vec F = M_x+N_y+P_z = ye^{xy}+0+0= ye^{xy} - \curl\vec F \amp = \nabla \times \vec F = \langle P_y-N_z,M_z-P_x,N_x - M_y\rangle - \amp = \langle 1-\cos(x+z), -2x, \cos(x+z) - xe^{xy}\rangle - -

      -
      3. -
    -

    -     -     - - - - - Creating a field representing gravitational force - -

    - The force of gravity between two objects is inversely proportional to the square of the distance between the objects. - Locate a point mass at the origin. - Create a vector field \vec F that represents the gravitational pull of the point mass at any point (x,y,z). - Find the divergence and curl of this field. -

    -     - -

    - The point mass pulls toward the origin, so at (x,y,z), - the force will pull in the direction of \langle -x, -y, -z\rangle. - To get the proper magnitude, - it will be useful to find the unit vector in this direction. - Dividing by its magnitude, we have - - \vec u = \left\langle \frac{-x}{\sqrt{x^2+y^2+z^2}}, \frac{-y}{\sqrt{x^2+y^2+z^2}},\frac{-z}{\sqrt{x^2+y^2+z^2}}\right\rangle - . -

    - -

    - The magnitude of the force is inversely proportional to the square of the distance between the two points. - Letting k be the constant of proportionality, - we have the magnitude as \ds\frac{k}{x^2+y^2+z^2}. - Multiplying this magnitude by the unit vector above, - we have the desired vector field: - - \vec F = \left\langle \frac{-kx}{(x^2+y^2+z^2)^{3/2}}, \frac{-ky}{(x^2+y^2+z^2)^{3/2}},\frac{-kz}{(x^2+y^2+z^2)^{3/2}}\right\rangle - . -

    - -

    - We leave it to the reader to confirm that - \divv \vec F = 0 and \curl \vec F = \vec 0. -

    - -
    - A vector field representing a planar gravitational force - - A radial vector field. Each vector points toward the origin, and vectors near the origin are larger. - -

    - A two-dimensional vector field is plotted relative to the x and y axes, - with the origin at the center of the image. - This is a radial vector field, like the one in , - but there are some important differences. -

    - -

    -

      -
    • -

      - There are fewer vectors plotted than the previous example of a radial field -

      -
      • -
    • -

      - The vectors point inward, toward the origin -

      -
      • -
    • -

      - The magnitudes of the vectors close to the origin are larger than those further out. -

      -
      • -
    -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ cos y, sin x ⟩ - - defaultpen(fontsize(8)); - - real f(real x) {return exp(x);} - pair F(real x) {return (x,f(x));} - - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(.3,.3); - pair b=(1,1); - - pair vectfunction(pair z) {if ((z.x==0) && (z.y==0)) {return (0,0);} - return (-z.x/(z.x^2+z.y^2)^(3/2),-z.y/(z.x^2+z.y^2)^(3/2));} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,3,3,bluepen)); - - pair aa=(-1,.3); - pair bb=(-.3,1); - - pair vectfunction(pair z) {if ((z.x==0) && (z.y==0)) {return (0,0);} - return (-z.x/(z.x^2+z.y^2)^(3/2),-z.y/(z.x^2+z.y^2)^(3/2));} - path vectorb(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vectorb,aa,bb,3,3,bluepen)); - - pair aa=(-1,-1); - pair bb=(-.3,-.3); - - pair vectfunction(pair z) {if ((z.x==0) && (z.y==0)) {return (0,0);} - return (-z.x/(z.x^2+z.y^2)^(3/2),-z.y/(z.x^2+z.y^2)^(3/2));} - path vectorb(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vectorb,aa,bb,3,3,bluepen)); - - pair aa=(.3,-1); - pair bb=(1,-.3); - - pair vectfunction(pair z) {if ((z.x==0) && (z.y==0)) {return (0,0);} - return (-z.x/(z.x^2+z.y^2)^(3/2),-z.y/(z.x^2+z.y^2)^(3/2));} - path vectorb(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vectorb,aa,bb,3,3,bluepen)); - - - -
    - -

    - The analogous planar vector field is given in . - Note how all arrows point to the origin, - and the magnitude gets very small when - far from the origin. -

    -     -     - -

    - A function f(x,y) naturally induces a vector field, - \vec F = \nabla f = \langle f_x,f_y\rangle. - Given what we learned of the gradient in , - we know that the vectors of \vec F point in the direction of greatest increase of f. - Because of this, f is said to be the - potential function - potential function - vector fieldpotential function of - of \vec F. - Vector fields that are the gradient of potential functions will play an important role in the next section. -

    - - - A vector field that is the gradient of a potential function - -

    - Let f(x,y) = 3-x^2-2y^2 and let \vec F = \nabla f. - Graph \vec F, and find the divergence and curl of \vec F. -

    -     - -

    - Given f, - we find \vec F = \nabla f = \langle -2x,-4y\rangle. - A graph of \vec F is given in . - In , - the vector field is given along with a graph of the surface itself; - one can see how each vector is pointing in the direction of - steepest uphill, - which, in this case, is not simply just - toward the origin. -

    - -
    - A graph of a function f(x,y) and the vector field \vec F = \nabla f in - -
    - - - The gradient vector field of a potential fuction. Vectors point toward the origin, but appear to be tangent to parabolic paths. - -

    - A vector field is plotted in two dimensions relative to the x and y coordinate axes. - The origin is at the center of the image. -

    - -

    - The vectors in this vector field point toward the origin, but it is not a radial vector field. -

    - -

    - The vectors appear to lie tangent to curved paths. - In particular, vectors above the x axis lie tangent to paths that could be a family of parabolas of the form y=kx^2, - where k\gt 0. - Vectors near the x axis are nearly horizontal, and could be tangent to a very wide parabola (with k very small). - Vectors near the y axis are nearly vertical, and could be tangent to a very steep parabola (with k very large). -

    - -

    - Vectors below the x axis describe similar trajectories, - except that they follow parths of downward-opening parabolas (ones where k\lt 0). -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ cos y, sin x ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen+ linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - real f(real x) {return exp(x);} - pair F(real x) {return (x,f(x));} - - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-.9,-.9); - pair b=(.9,.9); - - pair vectfunction(pair z) {return (-2*z.x,-4*z.y);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,mainvect)); - - - -
    - -
    - - - A three-dimensional image showing the graph of a function of two variables, with its gradient vector field plotted in the x,y plane below. - -

    - The image is three-dimensional, with x, y, and z coordinate axes, and the origin at the center of the image. - The vector field from is plotted in the xy plane. - Above the plane is the graph of the function f(x,y)=3-x^2-2y^2; - the graph is a downward-opening elliptic paraboloid. -

    -     - - - - import palette; - - // ASY file for figlinescalarfield2_3D.asy in Chapter 10 - // The path is an elliptical helix ⟨ cos t, 2sin t, t/pi ⟩; no specific surface. - // - // This draws the surface of the area we seek. - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(10.5,-9,18.3); - //(7.63,8.65,5); - //currentprojection=orthographic(10.5,-9,18.3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={3}; - defaultpen(0.5mm); - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - pair zbounds=(-1,4); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - - //currentlight=(1,-1,0.5); - - real f(pair z) {return -z.x^2-2*z.y^2+3;} - - path3 gradient(pair z) { - static real dx=sqrtEpsilon, dy=dx; - return (0,0,0)--((f(z+dx)-f(z-dx))/2dx,(f(z+I*dy)-f(z-I*dy))/2dy,0); - } - - pair a=(-.9,-.9); - pair b=(.9,.9); - - triple F(pair z) {return (z.x,z.y,0);} - - add(vectorfield(gradient,F,a,b,12,12,bluepen)); - - pen p=apexmeshpen; - draw(surface(f,a,b,Spline),surfacepen,meshpen=p); - //draw(surface(f,a,b,Spline),gray+opacity(0.5)); - - //pen p=apexmeshpen; - //draw(s,surfacepen,meshpen=p); - - - -
    -     - -
    - -

    - We leave it to the reader to confirm that - \divv \vec F = -6 and \curl \vec F = 0. -

    -     -     - - - -

    - There are some important concepts visited in this section that will be revisited in subsequent sections and again at the very end of this chapter. - One is: given a vector field \vec F, - both \divv\vec F and - \curl\vec F are measures of rates of change of \vec F. - The divergence measures how much the field spreads (diverges) at a point, - and the curl measures how much the field twists (curls) at a point. - Another important concept is this: - given z=f(x,y), the gradient - \nabla f is also a measure of a rate of change of f. - We will see how the integrals of these rates of change produce meaningful results. -

    - -

    - This section introduces the concept of a vector field. - The next section applies calculus to vector fields. - A common application is this: - let \vec F be a vector field representing a force - (hence it is called a force field, - though this name has a decidedly comic-book feel) - and let a particle move along a curve C under the influence of this force. - What work is performed by the field on this particle? - The solution lies in correctly applying the concepts of line integrals in the context of vector fields. -

    -     - - - - Terms and Concepts - - -

    - Give two quantities that can be represented by a vector field in the plane or in space. -

    -     - - - -

    - Answers will vary. - Appropriate answers include velocities of moving particles (air, - water, ); - gravitational or electromagnetic forces. -

    -     -     - - - -

    - In your own words, - describe what it means for a vector field to have a negative divergence at a point. -

    -     - - - -

    - Specific answers will vary, - though should relate to the idea that - more of the vector field is moving into that point than out of that point. -

    -     -     - - - -

    - In your own words, - describe what it means for a vector field to have a negative curl at a point. -

    -     - - - -

    - Specific answers will vary, - though should relate to the idea that the vector field is spinning clockwise at that point. -

    -     -     - - - -

    - The divergence of a vector field \vec F at a particular point is 0. - Does this mean that \vec F is incompressible? - Why/why not? -

    -     - - - -

    - No; to be incompressible, - the divergence needs to be 0 everywhere, not just at one point. -

    -     -     -     - - - Problems - - -

    - Sketch the given vector field over the rectangle with opposite corners (-2,-2) and (2,2), - sketching one vector for every point with integer coordinates (, at (0,0), - (1,2), etc.). -

    -     - - - -

    - \vec F = \langle x,0\rangle -

    -     - -

    - Correct answers should look similar to -

    - - - A vector field of horizontal vectors, pointing away from the y axis. - -

    - A two-dimmensional vector field is plotted relative to x and y coordinate axes, - with the origin at the center. All of the vectors are horizontal. -

    - -

    - The vectors all point away from the y axis: - those with x\gt 0 point to the right, while those with x\lt 0 point to the left. - Vectors close to the y axis are small, - and those near the left and right edges of the image are largest. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ x,0 ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - pen fgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen+ linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - fgvect = bluepen+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - fgvect = rgb(.6,.6,.6)+linejoin(0); - }; - pair xbounds=(-2.8,2.8); - pair ybounds=(-2.8,2.8); - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-2,-2); - pair b=(2,2); - - pair vectfunction(pair z) {return (z.x,0);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,5,5,fgvect)); - - -     -     - - - -

    - \vec F = \langle 0,x\rangle -

    -     - -

    - Correct answers should look similar to -

    - - - A vertical vector field. Vectors right of the y axis point up, and those to the left point down. - -

    - A two-dimmensioanl vector field is plotted relative to x and y coordinate axes, - with the origin at the center. -

    - -

    - All of the vectors are vertical. - Those with x\gt 0 point up, while those with x\lt 0 point down. - Vectors close to the y axis are small, - and those near the left and right edges of the image are largest. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ 0,x ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - pen fgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen+ linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - fgvect = bluepen+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - fgvect = rgb(.6,.6,.6)+linejoin(0); - }; - pair xbounds=(-2.8,2.8); - pair ybounds=(-2.8,2.8); - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-2,-2); - pair b=(2,2); - - pair vectfunction(pair z) {return (0,z.x);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,5,5,fgvect)); - - -     -     - - - -

    - \vec F = \langle 1,-1\rangle -

    -     - -

    - Correct answers should look similar to -

    - - - A constant vector field. All vectors point down and to the right. - -

    - A two-dimensional vector field is plotted relative to x and y coordinate axes, - with the origin at the center of the image. -

    - -

    - The vector field in this image is constant: - many vectors are plotted, but they all have the same magnitude and direction. - In particular, each vector points down and to the right. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ 1,-1 ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - pen fgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen+ linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - fgvect = bluepen+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - fgvect = rgb(.6,.6,.6)+linejoin(0); - }; - pair xbounds=(-2.8,2.8); - pair ybounds=(-2.8,2.8); - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-2,-2); - pair b=(2,2); - - pair vectfunction(pair z) {return (1,-1);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - add(vectorfield(vector,a,b,5,5,fgvect)); - - -     -     - - - -

    - \vec F = \langle y^2,1\rangle -

    -     - -

    - Correct answers should look similar to -

    - - - A two dimensional vector field. The vectors point up and to the right. Vectors near the x axis are shorter and steeper. - -

    - A two-dimensional vector field is plotted relative to x and y coordinate axes, - with the origin at the center. -

    - -

    - Along the lines y=2 and y=-2, - vectors point up and to the right, with a relatively small slope. - These vectors have the largest magnitude. -

    - -

    - Along the lines y=1 and y=-1 - vectors point up and to the right, but with a steeper slope. - The magnitudes of these vectors are less than the ones at the top and bottom of the image. -

    - -

    - Along the x axis, the vectors are vertical, and have the smallest magnitude. -

    - -

    - It looks like the vectors could lie tangent to a family of cubic curves of the form y = a\sqrt[3]{x}+b. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ y^2,1 ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - pen fgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen+ linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - fgvect = bluepen+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - fgvect = rgb(.6,.6,.6)+linejoin(0); - }; - pair xbounds=(-2.8,2.8); - pair ybounds=(-2.8,2.8); - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-2,-2); - pair b=(2,2); - - pair vectfunction(pair z) {return (z.y^2,1);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,5,5,fgvect)); - - -     -     -     - - - -

    - Find the divergence and curl of the given vector field. -

    -     - - - -

    - \vec F = \langle x,y^2\rangle -

    -     - -

    - \divv \vec F = 1+2y -

    - -

    - \curl \vec F = 0 -

    -     -     - - - -

    - \vec F = \langle -y^2,x\rangle -

    -     - -

    - \divv \vec F = 0 -

    - -

    - \curl \vec F = 1+2y -

    -     -     - - - -

    - \vec F = \langle \cos(xy), \sin(xy)\rangle -

    -     - -

    - \divv \vec F = x\cos(xy)-y\sin(xy) -

    - -

    - \curl \vec F = y\cos(xy)+x\sin(xy) -

    -     -     - - - -

    - \ds\vec F = \la \frac{-2x}{(x^2+y^2)^2},\frac{-2y}{(x^2+y^2)^2}\ra -

    -     - -

    - \divv \vec F = \frac{4}{(x^2+y^2)^2} -

    - -

    - \curl \vec F = 0 -

    -     -     - - - -

    - \ds\vec F = \la x+y,y+z,x+z\ra -

    -     - -

    - \divv \vec F = 3 -

    - -

    - \curl \vec F = \la -1,-1,-1\ra -

    -     -     - - - -

    - \ds\vec F = \la x^2+z^2,x^2+y^2,y^2+z^2\ra -

    -     - -

    - \divv \vec F = 2x+2y+2z -

    - -

    - \curl \vec F = \la 2y,2z,2x\ra -

    -     -     - - - -

    - \vec F = \nabla f, where f(x,y) = \frac12x^2+\frac13y^3. -

    -     - -

    - \divv \vec F = 1+2y -

    - -

    - \curl\vec F = 0 -

    -     -     - - - -

    - \vec F = \nabla f, where f(x,y) = x^2y. -

    -     - -

    - \divv \vec F = 2y -

    - -

    - \curl\vec F = 0 -

    -     -     - - - -

    - \vec F = \nabla f, where f(x,y,z) = x^2y+\sin(z). -

    -     - -

    - \divv \vec F = 2y-\sin(z) -

    - -

    - \curl\vec F = \vec 0 -

    -     -     - - - -

    - \vec F = \nabla f, where \ds f(x,y,z) = \frac1{x^2+y^2+z^2}. -

    -     - -

    - \divv \vec F = \frac{2}{(x^2+y^2+z^2)^2} -

    - -

    - \curl\vec F = \vec 0 -

    -     -     -     -     -     -
    -
    - Line Integrals over Vector Fields - -

    - Suppose a particle moves along a curve C under the influence of an electromagnetic force described by a vector field \vec F. - Since a force is inducing motion, work is performed. - How can we calculate how much work is performed? -

    - -

    - Recall that when moving in a straight line, - if \vec F represents a constant force and \vec d represents the direction and length of travel, - then work is simply W = \vec F\cdot \vec d. - However, we generally want to be able to calculate work even if \vec F is not constant and C is not a straight line. -

    - -

    - As we have practiced many times before, - we can calculate work by first approximating, - then refining our approximation through a limit that leads to integration. -

    - - - -

    - Assume as we did in - that C can be parametrized by the arc length parameter s. - Over a short piece of the curve with length ds, - the curve is approximately straight and our force is approximately constant. - The straight-line direction of this short length of curve is given by \vec T, - the unit tangent vector; - let \vec d = \vec T\, ds, - which gives the direction and magnitude of a small section of C. - Thus work over this small section of C is \vec F \cdot \vec d = \vec F\cdot \vec T\, ds. -

    - -

    - Summing up all the work over these small segments gives an approximation of the work performed. - By taking the limit as ds goes to zero, - and hence the number of segments approaches infinity, - we can obtain the exact amount of work. - Following the logic presented at the beginning of this chapter in the Integration Review, - we see that - - W = \int_C \vec F\cdot \vec T\, ds - , - a line integral. -

    - -

    - This line integral is beautiful in its simplicity, - yet is not so useful in making actual computations - (largely because the arc length parameter is so difficult to work with). - To compute actual work, - we need to parametrize C with another parameter t via a vector-valued function \vec r(t). - As stated in , - ds = \norm{\vrp(t)}\, dt, - and recall that \vec T = \vrp(t)/\norm{\vrp(t)}. - Thus - - W \amp = \int_C \vec F\cdot\vec T\, ds = \int_C \vec F\cdot \frac{\vrp(t)}{\norm{\vrp(t)}}\norm{\vrp(t)}\, dt - \amp = \int_C\vec F\cdot \vrp(t)\, dt = \int_C \vec F\cdot d\vec r - , - where the final integral uses the differential d\vec r for \vrp(t)\,dt. -

    -     - - - Evaluating Line Integrals over Vector Fields -

    - These integrals are known as line integrals over vector fields. - By contrast, - the line integrals we dealt with in - are sometimes referred to as line integrals over scalar fields. - line integralover scalar field - Just as a vector field is defined by a function that returns a vector, - a scalar field is a function that returns a scalar, - such as z = f(x,y). - We waited until now to introduce this terminology so we could contrast the concept with vector fields. -

    - -

    - We formally define this line integral, - then give examples and applications. -

    - - - Line Integral Over A Vector Field - -

    - Let \vec F be a vector field with continuous components defined on a smooth curve C, - parametrized by \vrt, - and let \vec T be the unit tangent vector of \vrt. - The line integral over \vec F along C - is - line integralover vector field - - \int_C \vec F\cdot d\vec r = \int_C \vec F\cdot\vec T\, ds - . -

    -     -     - -

    - In , - note how the dot product \vec F \cdot \vec T is just a scalar. - Therefore, this new line integral is really just a special kind of line integral found in ; - letting f(s) = \vec F(s)\cdot \vec T(s), - the right-hand side simply becomes \int_C f(s)\, ds, - and we can use the techniques of that section to evaluate the integral. - We combine those techniques, - along with parts of Equation, - to clearly state how to evaluate a line integral over a vector field in the following Key Idea. -

    - - - - - Evaluating a Line Integral Over A Vector Field -

    - Let \vec F be a vector field with continuous components defined on a smooth curve C, - parametrized by \vrt, a\leq t\leq b, - where \vec r is continuously differentiable. - Thenvector fieldover vector field - - \int_C\vec F\cdot\vec T\, ds = \int_C \vec F\cdot d\vec r =\int_a^b \vec F\big(\vec r(t)\big) \cdot \vrp(t)\, dt - . -

    -     - -

    - An important concept implicit in this Key Idea: - we can use any continuously differentiable parametrization \vrt of C that preserves the orientation of C: - there isn't a right one. - In practice, choose one that seems easy to work with. -

    - -

    - Notation note: the above Definition and Key Idea implicitly evaluate \vec F along the curve C, - which is parametrized by \vrt. - For instance, - if \vec F = \langle x+y, x-y\rangle and \vrt = \langle t^2,\cos(t)\rangle, - then evaluating \vec F along C means substituting the x- and y-components of \vrt in for x and y, - respectively, in \vec F. - Therefore, along C, - \vec F = \langle x+y,x-y\rangle = \la t^2+\cos(t), t^2-\cos(t)\ra. - Since we are substituting the output - of \vrt for the input of \vec F, - we write this as \vec F\big(\vrt\big). - This is a slight abuse of notation as technically the input of \vec F is to be a point, - not a vector, but this shorthand is useful. -

    - -

    - We use an example to practice evaluating line integrals over vector fields. -

    - - - Evaluating a line integral over a vector field: computing work - -

    - Two particles move from (0,0) to (1,1) under the influence of the force field \vec F = \langle x, x+y\rangle. - One particle follows C_1, the line y=x; - the other follows C_2, - the curve y=x^4, as shown in . - Force is measured in newtons and distance is measured in meters. - Find the work performed by each particle. -

    -
    - Paths through a vector field in - - A vector field is plotted in two dimensions, along with two different paths that begin and end at the same points. - -

    - A two dimensional vector field is plotted relative to x and y coordinate axes. - The origin is at the bottom-left of the image, with the first quadrant visible. -

    - -

    - The vectors in the vector field all point up and to the right. - Those close to the y axis are nearly vertical; - the vectors close to the x axis are at an angle of approximately 45 degrees, - and the directions of the rest lie somewhere in between. - The magnitude of the vectors shrinks to zero as they get close to the origin. -

    - -

    - Two curves are also plotted in the plane. - Both curves meet at the points (0,0) and (1,1). - The curve y=x^4 lies below the line y=x. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ x, x+y ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - - - pen colorone; - pen colortwo; - - if(incolor) { - colorone = bluepen+1.5pt; - colortwo = redpen+1.5pt; - } else - { - colorone = black+1.5pt; - colortwo = gray+1.5pt; - }; - - real f(real x) {return x;} - pair F(real x) {return (x,f(x));} - - real g(real x) {return x^4;} - pair G(real x) {return (x,g(x));} - - - pair xbounds=(-.1,1.1); - pair ybounds=(-.1,1.1); - real[] myxchoice={1}; - real[] myychoice={1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(0,0); - pair b=(1,1); - - pair vectfunction(pair z) {return (z.x,z.x+z.y);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,bgvect)); - - draw(graph(f,0,1,operator ..),colorone); - draw(graph(g,0,1,operator ..),colortwo); - - label("$y=x$",(.4,.55),filltype=Fill(white)); - label("$y=x^4$",(.55,.25),filltype=Fill(white)); - - - -
    -     - -

    - To compute work, we need to parametrize each path. - We use \vec r_1(t) = \langle t,t\rangle to parametrize y=x, - and let \vec r_2(t) =\langle t,t^4\rangle parametrize y=x^4; - for each, 0\leq t\leq 1. -

    - -

    - Along the straight-line path, - \vec F\big(\vec r_1(t)\big) = \langle x, x+y\rangle = \langle t, t+t\rangle = \langle t,2t\rangle. - We find \vrp_1(t) =\langle 1,1\rangle. - The integral that computes work is: - - \int_{C_1} \vec F\cdot d\vec r \amp = \int_0^1 \langle t,2t\rangle\cdot\la 1,1\ra\, dt - \amp = \int_0^1 3t\, dt - \amp = \frac32t^2\Big|_0^1 = 1.5 \text{ joules } - . -

    - -

    - Along the curve y = x^4, - \vec F\big(\vec r_2(t)\big) = \la x, x+y\ra = \la t, t+t^4\ra. - We find \vrp_2(t) = \la 1, 4t^3\ra. - The work performed along this path is - - \int_{C_2} \vec F\cdot d\vec r \amp = \int_0^1 \la t,t+t^4\ra\cdot\la 1,4t^3\ra\, dt - \amp = \int_0^1 \big(t + 4t^4+ 4t^7\big)\, dt - \amp = \big(\frac12t^2 + \frac45t^5 + \frac12t^8\big)\Big|_0^1 = 1.8 \text{ joules } - . -

    - -

    - Note how differing amounts of work are performed along the different paths. - This should not be too surprising: - the force is variable, one path is longer than the other, etc. -

    -     -     - - - Evaluating a line integral over a vector field: computing work - -

    - Two particles move from (-1,1) to (1,1) under the influence of a force field \vec F = \la y, x\ra. - One moves along the curve C_1, - the parabola defined by y = 2x^2-1. - The other particle moves along the curve C_2, - the bottom half of the circle defined by x^2+(y-1)^2=1, - as shown in . - Force is measured in pounds and distances are measured in feet. - Find the work performed by moving each particle along its path. -

    -
    - Paths through a vector field in - - A two-dimensional vector field, and two curves in the plane. The vectors appear to lie tangent to a family of hyperbolas. - -

    - A two-dimensional vector field is plotted relative to x and y coordinate axes, - with the origin at the center of the image. -

    - -

    - The vectors in the vector field appear to follow hyperbolic trajectories: - if we drew a contour plot for the function f(x,y)=x^2-y^2, - the vector field would lie tangent to the curves in that plot. -

    - -

    - Two curves are plotted against this vector field. -

      -
    • -

      - The curves meet at the points (-1,1) and (1,1). -

      -
      • -
    • -

      - One curve is a parabola with its vertex at (0,-1). - It opens upward, and ends at the points (-1,1) and (1,1). -

      -
      • -
    • -

      - The other curve is the bottom half of a circle of radius 1, - with its center at (1,0). -

      -
      • -
    • -

      - The two curves also meet at the points \left(\pm\frac{\sqrt{3}}{2},\frac12\right), - but these intersections are not relevant to the solution of the problem. -

      -
      • -
    -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ y, x ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - - real f(real x) {return 2*x^2-1;} - pair F(real x) {return (x,f(x));} - - //real g(real x) {return x^4;} - //pair G(real x) {return (x,g(x));} - - pair G(real t) {return (cos(t),sin(t)+1);} - - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-1,-1); - pair b=(1,1); - - pair vectfunction(pair z) {return (z.y,z.x);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,bgvect)); - - draw(graph(F,-1,1,operator ..),colorone); - draw(graph(G,pi,2*pi,operator ..),colortwo); - - label("$y=2x^2-1$",(.7,-.9),filltype=Fill(white+opacity(.7))); - label("$x^2+(y-1)^2=1$",(0,.25),filltype=Fill(white+opacity(.7))); - - - -
    -     - -

    - We start by parametrizing C_1: - the parametrization \vec r_1(t) = \la t, 2t^2-1\ra is straightforward, - giving \vrp_1 = \la 1,4t\ra. - On C_1, - \vec F\big(\vec r_1(t)\big) = \la y,x\ra = \la 2t^2-1,t\ra. -

    - -

    - Computing the work along C_1, we have: - - \int_{C_1} \vec F\cdot d\vec r_1 \amp = \int_{-1}^1 \la 2t^2-1,t\ra\cdot\la 1,4t\ra\, dt - \amp = \int_{-1}^1 \big(2t^2-1+4t^2\big)\, dt = 2 \text{ ft-lbs } - . -

    - -

    - For C_2, - it is probably simplest to parametrize the half circle using sine and cosine. - Recall that \vec r(t) = \la \cos(t), \sin(t)\ra is a parametrization of the unit circle on 0\leq t\leq 2\pi; - we add 1 to the second component to shift the circle up one unit, - then restrict the domain to - \pi\leq t\leq 2\pi to obtain only the lower half, - giving \vec r_2(t) = \la \cos(t), \sin(t)+1\ra, - \pi\leq t\leq 2\pi, - and hence \vrp_2(t) = \la -\sin(t), \cos(t)\ra and \vec F\big(\vec r_2(t)\big) = \la y,x\ra = \la \sin(t)+1,\cos(t)\ra. -

    - -

    - Computing the work along C_2, we have: - - \int_{C_2} \vec F\cdot d\vec r_2 \amp = \int_{\pi}^{2\pi} \la \sin(t)+1,\cos(t)\ra\cdot\la -\sin(t),\cos(t)\ra\, dt - \amp = \int_{\pi}^{2\pi} \big(-\sin^2t-\sin(t)+\cos^2t\big)\, dt = 2 \text{ ft-lbs } - . -

    - -

    - Note how the work along C_1 and C_2 in this example is the same. - We'll address why later in this section when - conservative fields - and path independence are discussed. -

    -     -     -     - - - Properties of Line Integrals Over Vector Fields -

    - Line integrals over vector fields share the same properties as line integrals over scalar fields, - with one important distinction. - The orientation of the curve C matters with line integrals over vector fields, - whereas it did not matter with line integrals over scalar fields. -

    - - -

    - It is relatively easy to see why. - Let C be the unit circle. - The area under a surface over C is the same whether we traverse the circle in a clockwise or counterclockwise fashion, - hence the line integral over a scalar field on C is the same irrespective of orientation. - On the other hand, if we are computing work done by a force field, - direction of travel definitely matters. - Opposite directions create opposite signs when computing dot products, - so traversing the circle in opposite directions will create line integrals that differ by a factor of -1. -

    - - - Properties of Line Integrals Over Vector Fields - -

    -

      -
    1. -

      - Let \vec F and \vec G be vector fields with continuous components defined on a smooth curve C, - parametrized by \vrt, and let k_1 and k_2 be scalars. - Thenline integralproperties over a vector field - - \ds \int_C\big(k_1\vec F+k_2\vec G\big)\cdot d\vec r = k_1\int_C\vec F\cdot d\vec r +k_2\int_C\vec G\cdot d\vec r - . -

      -
      2. - -
    2. -

      - Let C be piecewise smooth, - composed of smooth components C_1 and C_2. - Then - - \int_C\vec F\cdot d\vec r = \int_{C_1}\vec F\cdot d\vec r + \int_{C_2}\vec F\cdot d\vec r - . -

      -
      3. - -
    3. -

      - Let C^* be the curve C with opposite orientation, - parametrized by \vec r\,^*. - Then - - \int_C\vec F\cdot d\vec r = -\int_{C^*}\vec F\cdot d\vec r\,^* - . -

      -
      4. -
    -

    -     -     - - - -

    - We demonstrate using these properties in the following example. -

    - - - Using properties of line integrals over vector fields - -

    - Let \vec F = \la 3(y-1/2),1\ra and let C be the path that starts at (0,0), - goes to (1,1) along the curve y=x^3, - then returns to (0,0) along the line y=x, - as shown in . - Evaluate \oint_C \vec F\cdot d\vec r. -

    - -
    - The vector field and curve in - - A two-dimensional vector field of vectors tangent to a family of parabolas, and a pair of curves in the plane. - -

    - A two dimensional vector field is plotted relative to x and y coordinate axes. - The origin is at the bottom-left of the image, with the first quadrant visible. -

    - -

    - The vectors in the vector field appear to lie tangent to a family of parabolas. - These parabolas are not shown, but if they were, - they would all open to the right, with their vertices along the line y=\frac12. -

    - -

    - Two curves are also plotted in the plane; they are the same curves as the ones in . - Both curves meet at the points (0,0) and (1,1). - The curve y=x^4 lies below the line y=x. -

    - -

    - There are arrows on the curves indicating direction of motion. - Along the y=x^4 curve is an arrow pointing away from the origin in the direction of travel toward (1,1). - The arrow on the line y=x points away from (1,1) and back toward the origin. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ 3(y-.5), 1 ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - - pair xbounds=(-.2,1.2); - pair ybounds=(-.2,1.2); - real[] myxchoice={1}; - real[] myychoice={1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-.1,-.1); - pair b=(1.1,1.1); - - pair vectfunction(pair z) {return (3(z.y-.5),1);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,bgvect)); - - real f(real x) {return 2*x^2-1;} - pair F(real x) {return (x,x^3);} - - //real g(real x) {return x^4;} - //pair G(real x) {return (x,g(x));} - - pair G(real t) {return (t,t);} - - draw(graph(F,0,1,operator ..),colorone,arrow=MidArrow(5)); - draw(graph(G,1,0,operator ..),colortwo,arrow=MidArrow(5)); - - //label("$y=2x^2-1$",(.7,-.9),filltype=Fill(white+opacity(.7))); - //label("$x^2+(y-1)^2=1$",(0,.25),filltype=Fill(white+opacity(.7))); - - - -
    -     - -

    - As C is piecewise smooth, - we break it into two components C_1 and C_2, - where C_1 follows the curve y=x^3 and C_2 follows the curve y=x. -

    - -

    - We parametrize C_1 with - \vec r_1(t) = \la t, t^3\ra on 0\leq t\leq 1, - with \vrp_1(t) = \la 1,3t^2\ra. - We will use \vec F\big(\vec r_1(t)\big) = \la 3(t^3-1/2),1\ra. -

    - -

    - While we always have unlimited ways in which to parametrize a curve, - there are 2 direct methods to choose from when parametrizing C_2. - The parametrization \vec r_2(t)=\la t,t\ra, - 0\leq t\leq 1 traces the correct line segment but with the wrong orientation. - Using Property 3 of , - we can use this parametrization and negate the result. -

    - -

    - Another choice is to use the techniques of - to create the line with the orientation we desire. - We wish to start at ( 1,1) and travel in the - \vec d = \la -1,-1\ra direction for one length of \vec d, - giving equation \vec \ell(t) = \la 1,1\ra + t\la -1,-1\ra = \la 1-t,1-t\ra on 0\leq t\leq 1. -

    - -

    - Either choice is fine; we choose - \vec r_2(t) to practice using line integral properties. - We find \vrp_2(t) = \la 1,1\ra and \vec F\big(\vec r_2(t)\big) = \la 3(t-1/2),1\ra. -

    - -

    - Evaluating the line integral (note how we subtract the integral over C_2 as the orientation of \vec r_2(t) is opposite): - - \oint_C \vec F\cdot d\vec r \amp = \int_{C_1}\vec F\cdot d\vec r_1 - \int_{C_2}\vec F\cdot d\vec r_2 - \amp = \int_0^1 \la 3(t^3-1/2),1\ra\cdot \la 1,3t^2\ra dt- \int_0^1 \la 3(t-1/2),1\ra\cdot\la 1,1\ra dt - \amp = \int_0^1\Big(3t^3+3t^2-3/2\Big)\, dt - \int_0^1 \Big(3t-1/2\Big)\, dt - \amp = \big(1/4 \big) - \big(1\big) - \amp = -3/4 - . -

    - -

    - If we interpret this integral as computing work, - the negative work implies that the motion is mostly against - the direction of the force, - which seems plausible when we look at . -

    -     -     - - - - - Evaluating a line integral over a vector field in space - -

    - Let \vec F = \la -y, x, 1\ra, - and let C be the portion of the helix given by \vrt = \langle \cos(t),\sin(t), t/(2\pi)\rangle on [0,2\pi], - as shown in . - Evaluate \int_C\vec F\cdot d\vec r. -

    - -
    - The graph of \vec r(t) in - - A three-dimensional vector field and a portion of a helix curve. - -

    - A set of three-dimensional coordinate axes is plotted, with the origin at the center. - Relative to these axes there is a vector field and a curve. -

    - -

    - From the default perspective, the vector field appears to be rather chaotic. - It is difficult to tell what is going on; - we can tell that all vectors have an upward slope, but the left and right orientation of the vectors seems random. - When the image is rotated to show the perspective from the positive z axis, - the vectors in the vector field appear to follow circular trajectories. -

    - -

    - Also plotted is the helix given by the vector-valued function \vec{r}(t) = \la \cos(t),\sin(t),t/(2\pi)\ra. - The image shows one revolution of the helix, beginning on the x axis at (1,0,0) - and ending at (1,0,1). -

    - -

    - Collectively, the vectors in the vector field appear as though they could be tangent vectors - to a family of such helix curves. -

    -     - - - - - size(200,200,IgnoreAspect); - currentprojection=orthographic(16.8,9.4,8.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={1}; - defaultpen(0.25mm); - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-1.5,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - path3 gradient1(triple z){ - return O--(-z.y,z.x,1); - } - - triple A=(-1,-1,-1); - triple B=(1,1,1); - - picture VectorPlot3D(path3 vector(triple t), triple a, triple b, - int nx=nmesh, int ny=nx, int nz=nx,bool truesize=false, - real maxlength=truesize ? 0 : min(abs(b.x-a.x)/nx,abs(b.y-a.y)/ny,abs(b.z-a.z)/nz), - // bool cond(pair z)=null, - pen p=currentpen, - arrowbar3 arrow=Arrow3(6), margin3 margin=PenMargin3, - string name="", render render=defaultrender) - { - picture pic; - real dx=1/nx; - real dy=1/ny; - real dz=1/nz; - real scale; - if(maxlength > 0) { - real size(triple t) { - path3 g=vector(t); - return abs(point(g,size(g)-1)-point(g,0)); - } - real max=size((0,0,0)); - - for(int i=0; i <= nx; ++i) { - real x=interp(a.x,b.x,i*dx); - for(int j=0; j <= ny; ++j) - { - real y=interp(a.y,b.y,j*dy); - for(int k=0; k <= nz; ++k) - max=max(max,size((x,y,interp(a.z,b.z,k*dz)))); - }} - scale=max > 0 ? maxlength/max : 1; - } else scale=1; - bool group=name != "" || render.defaultnames; - if(group) - begingroup3(pic,name == "" ? "vectorfield" : name,render); - for(int i=0; i <= nx; ++i) { - real x=interp(a.x,b.x,i*dx); - for(int j=0; j <= ny; ++j) { - real y=interp(a.y,b.y,j*dy); - for(int k=0; k <= nz; ++k) - { triple z=(x,y,interp(a.z,b.z,k*dz)); - { - path3 g=scale3(scale)*vector(z); - string name="vector"; - if(truesize) { - picture opic; - draw(opic,g,p,arrow,margin,name,render); - add(pic,opic,z); - } else - draw(pic,shift(z)*g,p,arrow,margin,name,render); - } - } - }} - if(group) - endgroup3(pic); - return pic; - - } - add(VectorPlot3D(gradient1,A,B,3,3,3,bluepen)); - - triple g(real t) {return(cos(t),sin(t),t/(2*pi));} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,curvepen2,MidArrow3(6)); - - //triple g(real t) {return (t,t,-2*t^4+4*t^2);} - //path3 mypath=graph(g,-1,1,operator ..); - - - -
    -     - -

    - A parametrization is already given for C, - so we just need to find \vec F\big(\vec r(t)\big) and \vec r '(t). -

    - -

    - We have \vec F\big(\vec r(t)\big) = \la -\sin(t), \cos(t), 1\ra and \vrp(t) = \la -\sin(t), \cos(t), 1/(2\pi)\ra. - Thus - - \int_C \vec F\cdot d\vec r \amp = \int_0^{2\pi} \la -\sin(t), \cos(t), 1\ra\cdot \la -\sin(t), \cos(t), 1/(2\pi)\ra dt - \amp = \int_0^{2\pi} \Big(\sin^2t+\cos^2t + \frac1{2\pi}\Big)dt - \amp = 2\pi + 1 \approx 7.28 - -

    -     -     - - -     - - - The Fundamental Theorem of Line Integrals -

    - We are preparing to make important statements about the value of certain line integrals over special vector fields. - Before we can do that, - we need to define some terms that describe the domains over which a vector field is defined. - line integralFundamental Theorem - Fundamental Theorem of Line Integrals -

    - - -

    - A region in the plane is connected - connected - if any two points in the region can be joined by a piecewise smooth curve that lies entirely in the region. - In , - sets R_1 and R_2 are connected; - set R_3 is not connected, - though it is composed of two connected subregions. -

    - -

    - A region is simply connected - simply connected - connectedsimply - if every simple closed curve that lies entirely in the region can be continuously deformed (shrunk) to a single point without leaving the region. (A curve is simple - simple curve - if it does not cross itself.) In , - only set R_1 is simply connected. - Region R_2 is not simply connected as any closed curve that goes around the hole - in R_2 cannot be continuously shrunk to a single point. - As R_3 is not even connected, - it cannot be simply connected, - though again it consists of two simply connected subregions. -

    - -

    - We have applied these terms to regions of the plane, - but they can be extended intuitively to domains in space - (and hyperspace). - In , - the domain bounded by the sphere - (at left) - and the domain with a subsphere removed - (at right) - are both simply connected. - Any simple closed path that lies entirely within these domains can be continuously deformed into a single point. - In , - neither domain is simply connected. - A left, the ball has a hole that extends its length and the pictured closed path cannot be deformed to a point. - At right, two paths are illustrated on the torus that cannot be shrunk to a point. -

    - -

    - We will use the terms connected and simply connected in subsequent definitions and theorems. -

    - -
    - R_1 is simply connected; R_2 is connected, but not simply connected; R_3 is not connected - - Several regions in the plane are used to illustrate the concepts of connected, and simply connected. - -

    - Three different regions in the plane are plotted, without the use of coordinate axes. - The regions are pond-like, with boundaries resembling bumpy, poorly-drawn circles. -

    - -

    - The region R_1 is both connected and simply connected. - It is a single pond-like shape, with its interior completely shaded. -

    - -

    - The region R_2 is connected but not simply connected. - It is like a pond that contains an island: there is a region in its interior that is not shaded, - indicating that this part of the interior is not part of R_2. -

    - -

    - The region R_3 is not connected. - It consists of two pond-like shapes, each separately resembling R_1. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - axis y line=none, - axis x line=none, - ymin=-.75,ymax=3, - xmin=-2,xmax=2, - ] - - \draw (axis cs: -1.6,1.4) node {$R_1$} - (axis cs: 1.6,1.1) node {$R_2$} - (axis cs: 0,.6) node {$R_3$}; - - \addplot [firstcurvestyle,areastyle] coordinates {(-0.45,2.)(-0.4429,2.07)(-0.4788,2.134)(-0.5466,2.18)(-0.6142,2.212)(-0.6552,2.251)(-0.6695,2.31)(-0.6802,2.387)(-0.7106,2.456)(-0.766,2.497)(-0.8346,2.509)(-0.9034,2.506)(-0.9684,2.502)(-1.031,2.496)(-1.091,2.476)(-1.144,2.442)(-1.192,2.407)(-1.246,2.388)(-1.318,2.384)(-1.399,2.374)(-1.464,2.337)(-1.49,2.27)(-1.476,2.189)(-1.447,2.115)(-1.435,2.055)(-1.45,2.)(-1.477,1.94)(-1.491,1.874)(-1.482,1.809)(-1.457,1.749)(-1.426,1.691)(-1.389,1.634)(-1.339,1.59)(-1.273,1.57)(-1.2,1.576)(-1.135,1.586)(-1.081,1.573)(-1.03,1.528)(-0.9669,1.474)(-0.8941,1.445)(-0.8255,1.463)(-0.7738,1.519)(-0.737,1.586)(-0.702,1.64)(-0.6605,1.681)(-0.6167,1.722)(-0.5804,1.769)(-0.5524,1.823)(-0.5227,1.877)(-0.4848,1.935)(-0.45,2.)}; 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    - -
    - The domains in (a) are simply connected, while the domains in (b) are not - -
    - - - Two spherical surfaces in space, plotted without coordinate axes. The second sphere has a smaller sphere in its interior. - -

    - Two spheres are plotted side by side in three dimensions. - There are no coordinate axes used in the image. -

    - -

    - On the left is a single sphere; it represents the domain consisting of all points on and inside the sphere. -

    - -

    - The sphere on the right has another smaller sphere inside it. - It represents the domain consisting of all points outside the smaller sphere, but inside the larger sphere. - Despite this hole in the middle of the domain, it is still simply connected. - The sphere on the inside does not obstruct us from transforming one curve inside the large sphere to another, - as we can stretch the curve around the small sphere. -

    -     - - - - - //ASY file for fig10_01_ex_233D.asy in Chapter 10 - - size(200,200); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(0,-5,3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={1}; - real[] myzchoice={1,2}; - defaultpen(0.5mm); - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - pair zbounds=(-1.1,1.1); - - //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2+x^2+y^2=1, a sphere shifted - triple f(pair t) { - return (cos(t.x)*sin(t.y)-1.1,sin(t.x)*sin(t.y),cos(t.y)); - } - surface s=surface(f,(0,0),(2pi,pi),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=apexmeshpen+.09mm; - draw(s,simplesurfacepen,meshpen=p); - - //Draw the surface z^2+x^2+y^2=1, a sphere shifted - triple f(pair t) { - return (cos(t.x)*sin(t.y)+1.1,sin(t.x)*sin(t.y),cos(t.y)); - } - surface s=surface(f,(0,0),(2pi,pi),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - //pen p=rgb(0,0,.2); - draw(s,simplesurfacepen,meshpen=p); - - //Draw the surface z^2+x^2+y^2=1, a smaller sphere shifted - triple f(pair t) { - return (.3*cos(t.x)*sin(t.y)+1.1,.3*sin(t.x)*sin(t.y),.3*cos(t.y)); - } - surface s=surface(f,(0,0),(2pi,pi),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - //pen p=rgb(0,0,.2); - draw(s,emissive(surfacepen),meshpen=p); - - triple g(real t) {return (t,t^2,-cos(t)*sin(t^2)+1);} - path3 mypath=graph(g,-1,1,operator ..); - //draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,t^2,0);} - path3 mypath=graph(g,-1,1,operator ..); - //draw(mypath,bluepen+linewidth(2)+dashed); - - - -
    - -
    - - - Two surfaces in three dimensions. One is a sphere with a cylinder through its center; the other is a torus. - -

    - Two surfaces are shown side by side in three dimensions, without the use of coordinate axes. - Each represents a solid domain in space that is not simply connected. -

    - -

    - The domain on the left consists of the region on and inside a sphere, - except that a cylindrical hole has been cut along one diameter of the sphere. - This surface is not simply connected: - any curve that wraps around the cylinder on the inside cannot be shrunk to a point. -

    - -

    - The domain on the right is a torus. - A torus is a ring-shaped surface, like a donut or a bagel. - The torus is not simply connected because any curve that travels all the way around the hole in the middle cannot be shrunk to a point. -

    - -

    - An additional fun fact: these surfaces may be geometrically different, but topologically, they are the same. - If we squish the sphere vertically, until the cylinder through the middle is shrunk to a circle, - we end up with the torus. -

    -     - - - - - //ASY file for fig10_01_ex_233D.asy in Chapter 10 - - size(200,200); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(0,-5,3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={1}; - real[] myzchoice={1,2}; - defaultpen(0.5mm); - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - pair zbounds=(-1.1,1.1); - - //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the surface z^2+x^2+y^2=1, a sphere - triple f(pair t) { - return (cos(t.x)*sin(t.y)-1.1,sin(t.x)*sin(t.y),cos(t.y)); - } - surface s=surface(f,(0,1/4),(2pi,pi-1/4),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - pen p=apexmeshpen+.09mm; - draw(s,simplesurfacepen,meshpen=p); - - //Draw the surface z^2+x^2+y^2=1, a sphere - triple f(pair t) { - return (cos(t.x)*sin(.25)-1.1,sin(t.x)*sin(.25),cos(.25)+t.y*(cos(pi-.25)-cos(.25))); - } - surface s=surface(f,(0,0),(2pi,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - //pen p=rgb(0,0,.2); - draw(s,simplesurfacepen,meshpen=p); - - triple g(real t) {return (cos(t)-1.1, sin(t),0);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(1)); - - //Draw the torus - real cc=.75; - real aa=.25; - triple f(pair t) { - return ((cc+aa*cos(t.y))*cos(t.x)+1.1,(cc+aa*cos(t.y))*sin(t.x),aa*sin(t.y)); - } - surface s=surface(f,(0,0),(2pi,2pi),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic} - ); - //pen p=apexmeshpen; - draw(s,simplesurfacepen,meshpen=p); - - triple g(real t) {return (cos(t)+1.1,sin(t),0);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(1)); - - real yy = 5pi/4; - - triple g(real t) {return ((cc+aa*cos(t))*cos(yy)+1.1,(cc+aa*cos(t))*sin(yy),aa*sin(t));} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(1)); - - - -
    -     - -
    - -

    - Recall how in - particles moved from A = (-1,1) to - B = (1,1) along two different paths, - wherein the same amount of work was performed along each path. - It turns out that regardless of the choice of path from A to B, - the amount of work performed under the field \vec F = \la y, x\ra is the same. - Since our expectation is that differing amounts of work are performed along different paths, - we give such special fields a name. -

    - - - Conservative Field, Path Independent - -

    - Let \vec F be a vector field defined on an open, - connected domain D in the plane or in space containing points A and B. - If the line integral \int_C \vec F\cdot d\vec r has the same value for all choices of paths C starting at A and ending at B, then - conservative field - vector fieldconservative - path independent - line integralpath independent - -

      -
    • -

      - \vec F is a conservative field and -

      -
      • - -
    • -

      - The line integral \int_C \vec F\cdot d\vec r is - path independent and can be written as - - \int_C \vec F\cdot d\vec r = \int_A^B \vec F\cdot \, d\vec r - . -

      -
      • -
    -

    -     -     - -

    - When \vec F is a conservative field, - the line integral from points A to B is sometimes written as - \int_A^B\vec F\cdot d\vec r to emphasize the independence of its value from the choice of path; - all that matters are the beginning and ending points of the path. -

    - -

    - How can we tell if a field is conservative? - To show a field \vec F is conservative using the definition, - we need to show that all - line integrals from points A to B have the same value. - It is equivalent to show that all - line integrals over closed paths C are 0. - Each of these tasks are generally nontrivial. -

    - -

    - There is a simpler method. - Consider the surface defined by z = f(x,y) = xy. - We can compute the gradient of this function: - \nabla f = \la f_x, f_y\ra = \la y, x\ra. - Note that this is the field from , - which we have claimed is conservative. - We will soon give a theorem that states that a field \vec F is conservative if, - and only if, - it is the gradient of some scalar function f. - To show \vec F is conservative, - we need to determine whether or not - \vec F = \nabla f for some function f. (We'll later see that there is a yet simpler method). - To recognize the special relationship between \vec F and f in this situation, - f is given a name. -

    - - - Potential Function - -

    - Let f be a differentiable function defined on a domain D in the plane or in space (, - z = f(x,y) or w = f(x,y,z)) and let - \vec F = \nabla f, the gradient of f. - Then f is a potential function of \vec F. - potential function - vector fieldpotential function of -

    -     -     - -

    - We now state the Fundamental Theorem of Line Integrals, - which connects conservative fields and path independence to fields with potential functions. -

    - - - Fundamental Theorem of Line Integrals - -

    - Let \vec F be a vector field whose components are continuous on a connected domain D in the plane or in space, - let A and B be any points in D, - and let C be any path in D starting at A and ending at B. - line integralFundamental Theorem - Fundamental Theorem of Line Integrals - conservative field - vector fieldconservative - path independent - line integralpath independent - -

      -
    1. -

      - \vec F is conservative if and only if there exists a differentiable function f such that \vec F = \nabla f. -

      -
      2. - -
    2. -

      - If \vec F is conservative, then - - \int_C\vec F\cdot d\vec r = \int_A^B \vec F\cdot d\vec r = f(B) - f(A) - . -

      -
      3. -
    -

    -     -     - -

    - Once again considering , - we have A = (-1,1), - B = (1,1) and \vec F = \la y,x\ra. - In that example, - we evaluated two line integrals from A to B and found the value of each was 2. - Note that f(x,y) = xy is a potential function for \vec F. - Following the Fundamental Theorem of Line Integrals, - consider f(B) - f(A): - - f(B) - f(A) = f(1,1) - f(-1,1) = 1 - (-1) = 2 - , - the same value given by the line integrals. -

    - -

    - We practice using this theorem again in the next example. -

    - - - Using the Fundamental Theorem of Line Integrals - -

    - Let \vec F = \la 3x^2y+2x, - x^3+1\ra, A = (0,1) and B = (1,4). - Use the first part of the Fundamental Theorem of Line Integrals to show that \vec F is conservative, - then choose any path from A to B and confirm the second part of the theorem. -

    -     - -

    - To show \vec F is conservative, - we need to find z = f(x,y) such that \vec F = \nabla f = \la f_x, f_y\ra. - That is, we need to find f such that - f_x = 3x^2y+2x and f_y = x^3+1. - As all we know about f are its partial derivatives, - we recover f by integration: - - \int \frac{\partial f}{\partial x}\, dx = f(x,y) + C(y) - . -

    - -

    - Note how the constant of integration is more than - just a constant: - it is anything that acts as a constant when taking a derivative with respect to x. - Any function that is a function of y - (containing no x's) - acts as a constant when deriving with respect to x. -

    - -

    - Integrating f_x in this example gives: - - \int \frac{\partial f}{\partial x}\, dx = \int (3x^2y+2x)\, dx = x^3y+x^2 + C_1(y) - . -

    - -

    - Likewise, integrating f_y with respect to y gives: - - \int \frac{\partial f}{\partial y}\, dy = \int( x^3+1)\, dy = x^3y+ y + C_2(x) - . -

    - -

    - These two results should be equal with appropriate choices of C_1(y) and C_2(x): - - x^3y+x^2 + C_1(y) = x^3y+ y + C_2(x) \Rightarrow C_2(x) = x^2 \text{ and } C_1(y) = y - . -

    - -

    - We find f(x,y) = x^3y+x^2+y, - a potential function of \vec F. (If \vec F were not conservative, - no choice of C_2(x) and C_1(y) would give equality.) -

    - -

    - By the Fundamental Theorem of Line Integrals, - regardless of the path from A to B, - - \int_A^B\vec F\cdot d\vec r \amp = f(B) - f(A) - \amp = f(1,4) - f(0,1) - \amp = 9 - 1 = 8 - . -

    - -

    - To illustrate the validity of the Fundamental Theorem, - we pick a path from A to B. - The line between these two points would be simple to construct; - we choose a slightly more complicated path by choosing the parabola y = x^2+2x+1. - This leads to the parametrization \vrt = \la t, t^2+2t+1\ra, - 0\leq t\leq 1, with \vrp(t) = \la t, 2t+2\ra. - Thus - - \int_C \vec F\cdot d\vec r \amp = \int_C\vec F\big(\vrt\big)\cdot\vrp(t)\, dt - \amp = \int_0^1\la 3(t)(t^2+2t+1)+2t, t^3+1\ra\cdot\la t,2t+2\ra\, dt - \amp = \int_0^1 \big(5t^4+8t^3+3t^2+4t+2\big)\, dt - \amp = \big(t^5+2t^4+t^3+2t^2+2t\big)\Big|_0^1 - \amp = 8 - , - which matches our previous result. -

    -     -     - -

    - The Fundamental Theorem of Line Integrals states that we can determine whether or not \vec F is conservative by determining whether or not \vec F has a potential function. - This can be difficult. - A simpler method exists if the domain of \vec F is simply connected - (not just connected as needed in the Fundamental Theorem of Line Integrals), - which is a reasonable requirement. - We state this simpler method as a theorem. -

    - - - - - Curl of Conservative Fields - -

    - Let \vec F be a vector field whose components have continuous partial - derivatives on a simply connected domain D in the plane or in space. - Then \vec F is conservative if and only if - \curl \vec F = 0 or \vec 0, in 2D or 3D, respectively. - curlof conservative fields - conservative field -

    -     -     - -

    - In , - we showed that \vec F =\langle 3x^2y+2x,x^3+1\rangle is conservative by finding a potential function for \vec F. - Using the above theorem, - we can show that \vec F is conservative much more easily by computing its curl: - - \curl \vec F = N_x - M_y = 3x^2 - 3x^2 = 0 - . -

    - -     - - - - Terms and Concepts - - - -

    - In practice, - the evaluation of line integrals over vector fields involves computing the magnitude of a vector-valued function. -

    -     - -

    - Although magnitude might appear in the definition, it cancels out once we parametrize. - It is true for line integrals over scalar fields, though. -

    -     -     - - - -

    - Let \vec F(x,y) be a vector field in the plane and let - \vec r(t) be a two-dimensional vector-valued function. - Why is \vec F\big(\vec r(t)\big) - an abuse of notation? -

    -     - - - -

    - The input of \vec F should be a point in the plane, - not a two dimensional vector. -

    -     -     - - - -

    - The orientation of a curve C matters when computing a line integral over a vector field. -

    -     - -

    - Much like swapping the bounds in a definite integral, - changing the orientation of a curve introduces a minus sign. -

    -     -     - - - -

    - The orientation of a curve C matters when computing a line integral over a scalar field. -

    -     - -

    - For scalar fields, we multiply by the magnitude of the tangent vector, - and the minus sign introduced by a change in orientation does not change the magnitude. -

    -     -     - - - -

    - Under reasonable conditions, - if \curl \vec F = \vec 0, - what can we conclude about the vector field \vec F? -

    -     - - - -

    - We can conclude that \vec F is conservative. -

    -     -     - - - -

    - Let \vec F be a conservative field and let C be a closed curve. - Why are we able to conclude that \oint _C \vec F\cdot d\vec r = 0? -

    -     - - - -

    - By the Fundamental Theorem of Line Integrals, - since \vec F is conservative, - \oint_C \vec F\cdot d\vec r = f(B) - f(A), - where f is a potential function for \vec F and A and B are the initial and terminal points of C, - respectively. - Since C is a closed curve, A = B, - and hence f(B) - f(A) = 0. -

    -     -     -     - - Problems - - -

    - A vector field \vec F and a curve C are given. - Evaluate \ds\int_C\vec F\cdot d\vec r. -

    -     - - - -

    - \vec F = \langle y, y^2\rangle; - C is the line segment from (0,0) to (3,1). -

    -     - -

    - 11/6. (One parametrization for C is - \vec r(t) = \langle 3t,t\rangle on 0\leq t\leq 1.) -

    -     -     - - - -

    - \vec F = \langle x,x+y\rangle; - C is the portion of the parabola y=x^2 from (0,0) to (1,1). -

    -     - -

    - 5/3. (One parametrization for C is - \vec r(t) = \langle t,t^2\rangle on 0\leq t\leq 1.) -

    -     -     - - - -

    - \vec F = \langle y,x\rangle; - C is the top half of the unit circle, - beginning at (1,0) and ending at (-1,0). -

    -     - -

    - 0. (One parametrization for C is - \vec r(t) = \langle \cos(t),\sin(t)\rangle on 0\leq t\leq \pi.) -

    -     -     - - - -

    - \vec F = \langle xy,x\rangle; - C is the portion of the curve y=x^3 on -1\leq x\leq 1. -

    -     - -

    - 2/5. (One parametrization for C is - \vec r(t) = \langle t,t^3\rangle on -1\leq t\leq 1.) -

    -     -     - - - -

    - \vec F = \langle z,x^2,y\rangle; - C is the line segment from (1,2,3) to (4,3,2). -

    -     - -

    - 12. (One parametrization for C is - \vec r(t) = \langle 1,2,3\rangle+t\langle 3,1,-1\rangle on 0\leq t\leq 1.) -

    -     -     - - - -

    - \vec F = \langle y+z,x+z,x+y\rangle; - C is the helix \vec r(t) = \langle \cos(t),\sin(t),t/(2\pi)\rangle on 0\leq t\leq 2\pi. -

    -     - -

    - 1. -

    -     -     -     - - - -

    - Find the work performed by the force field \vec F moving a particle along the path C. -

    -     - - - -

    - \vec F = \langle y,x^2\rangle N; C is the segment of the line y=x from (0,0) to (1,1), - where distances are measured in meters. -

    -     - -

    - 5/6 joules. (One parametrization for C is - \vec r(t) = \langle t,t\rangle on 0\leq t\leq 1.) -

    -     -     - - - -

    - \vec F = \langle y,x^2\rangle N; C is the portion of - y=\sqrt x from (0,0) to (1,1), - where distances are measured in meters. -

    -     - -

    - 13/15 joules. (One parametrization for C is - \vec r(t) = \langle t,\sqrt t\rangle on 0\leq t\leq 1.) -

    -     -     - - - -

    - \vec F = \langle 2xy,x^2,1\rangle lbs; - C is the path from (0,0,0) to (2,4,8) via - \vec r(t) = \langle t, t^2, t^3\rangle on 0\leq t\leq 2, - where distance are measured in feet. -

    -     - -

    - 24 ft-lbs. -

    -     -     - - - -

    - \vec F = \langle 2xy,x^2,1\rangle lbs; - C is the path from (0,0,0) to (2,4,8) via - \vec r(t) = \langle t,2t, 4t\rangle on 0\leq t\leq 2, - where distance are measured in feet. -

    -     - -

    - 24 ft-lbs. -

    -     -     -     - - - -

    - A conservative vector field \vec F and a curve C are given. - -

      -
    1. -

      - Find a potential function f for \vec F. -

      -
      2. - -
    2. -

      - Compute \curl \vec F. -

      -
      3. - -
    3. -

      - Evaluate \ds\int_C \vec F\cdot d\vec r directly, - , using . -

      -
      4. - -
    4. -

      - Evaluate \ds\int_C \vec F\cdot d\vec r using the Fundamental Theorem of Line Integrals. -

      -
      5. -
    -

    -     - - - -

    - \vec F = \langle y+1,x\rangle, - C is the line segment from (0,1) to (1,0). -

    -     - -

    -

      -
    1. -

      - f(x,y) = xy+x -

      -
      2. - -
    2. -

      - \curl \vec F = 0. -

      -
      3. - -
    3. -

      - 1. (One parametrization for C is - \vec r(t) = \langle t,t-1\rangle on 0\leq t\leq 1.) -

      -
      4. - -
    4. -

      - 1 (with A = (0,1) and - B = (1,0), f(B) - f(A) = 1.) -

      -
      5. -
    -

    -     -     - - - -

    - \vec F = \langle 2x+y, 2y+x\rangle, - C is curve parametrized by - \vec r(t) = \langle t^2-t, t^3-t\rangle on 0\leq t\leq 1. -

    -     - -

    -

      -
    1. -

      - f(x,y) = x^2+xy+y^2 -

      -
      2. - -
    2. -

      - \curl \vec F = 0. -

      -
      3. - -
    3. -

      - 0. -

      -
      4. - -
    4. -

      - 0 (with A = (0,0) and - B = (0,0), f(B) - f(A) = 0.) -

      -
      5. -
    -

    -     -     - - - -

    - \vec F = \langle 2xyz,x^2z,x^2y\rangle, - C is curve parametrized by - \vec r(t) = \langle 2t+1,3t-1,t\rangle on 0\leq t\leq 2. -

    -     - -

    -

      -
    1. -

      - f(x,y) = x^2yz -

      -
      2. - -
    2. -

      - \curl \vec F = \vec 0. -

      -
      3. - -
    3. -

      - 250. -

      -
      4. - -
    4. -

      - 250 (with A = (1,-1,0) and B = (5,5,2), - f(B) - f(A) = 250.) -

      -
      5. -
    -

    -     -     - - - -

    - \vec F = \langle 2x, 2y, 2z\rangle, - C is curve parametrized by - \vec r(t) = \langle \cos(t),\sin(t), \sin (2t)\rangle on 0\leq t\leq 2\pi. -

    -     - -

    -

      -
    1. -

      - f(x,y) = x^2+y^2+z^2 -

      -
      2. - -
    2. -

      - \curl \vec F = \vec 0. -

      -
      3. - -
    3. -

      - 0. -

      -
      4. - -
    4. -

      - 0 (with A = (1,0,0) and - B = (1,0,0), f(B) - f(A) = 0.) -

      -
      5. -
    -

    -     -     -     - - - -

    - Prove part of : - let \vec F =\langle M,N,P\rangle be a conservative vector field. - Show that \curl \vec F = 0. -

    -     - -

    - Since \vec F is conservative, - it is the gradient of some potential function. - That is, \nabla f = \langle f_x,f_y,f_z\rangle = \vec F = \langle M, N, P\rangle. - In particular, M = f_x, - N = f_y and P=f_z. -

    - -

    - Note that \curl \vec F = \langle P_y - N_z, M_z-P_x, N_x-M_y \rangle = \langle f_{zy} - f_{yz}, f_{xz} - f_{zx}, f_{yx} - f_{xy}\rangle, which, - by , - is \langle 0,0,0\rangle. -

    -     -     -     -     -
    -
    - Flow, Flux, Green's Theorem and the Divergence Theorem - - Flow and Flux -

    - Line integrals over vector fields have the natural interpretation of computing work when \vec F represents a force field. - It is also common to use vector fields to represent velocities. - In these cases, the line integral - \int_C \vec F\cdot d\vec r is said to represent flow. - flow - flux - circulation -

    - -
    - Illustrating the principles of flow and flux - - A constant vector field in the plane, pointing to the right, and a triangular path in the first quadrant. - -

    - A two-dimensional vector field is plotted relative to x and y coordinate axes. - The origin is at the bottom-left of the image, so that the first quadrant is displayed. -

    - -

    - The vector field is constant, with all arrows of equal magnitude, and pointing to the right. -

    - -

    - A triangular path is drawn in the first quadrant. -

      -
    • -

      - The bottom of the triangle is a line labeled C_1, traveling from left to right. -

      -
      • -
    • -

      - The left side of the triangle is vertical; this line is labeled C_2, - and oriented to travel from bottom to top. -

      -
      • -
    • -

      - The hypotenuse of the triangle is labeled C_3, - and it is oriented to travel from top-left to bottom-right. -

      -
      • -
    -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ 1,0 ⟩ - - defaultpen(fontsize(8)); - - pen colorone; - pen colortwo; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - }; - - pair xbounds=(-.2,1.2); - pair ybounds=(-.2,1.2); - real[] myxchoice={1}; - real[] myychoice={1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-.1,-.1); - pair b=(1.1,1.1); - - pair vectfunction(pair z) {return (1,0);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,6,6,rgb(.6,.6,.6))); - - pair f(real x) {return (.2+x,1-x);} - pair F(real x) {return (x,.2);} - - //real g(real x) {return x^4;} - //pair G(real x) {return (x,g(x));} - - pair G(real t) {return (.2,t);} - - draw(graph(F,0.2,1,operator ..),colorone,arrow=MidArrow(5)); - draw(graph(G,0.2,1,operator ..),colorone,arrow=MidArrow(5)); - draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); - - label("$C_1$",(.8,.15),filltype=Fill(white+opacity(.7))); - label("$C_2$",(.13,.8),filltype=Fill(white+opacity(.7))); - label("$C_3$",(.6,.5),filltype=Fill(white+opacity(.7))); - - - -
    - -

    - Let the vector field \vec F = \la 1,0\ra represent the velocity of water as it moves across a smooth surface, - depicted in . - A line integral over C will compute - how much water is moving - along the path C. -

    - -

    - In the figure, - all of the water above C_1 is moving along that curve, - whereas none of the water above C_2 is moving along that curve - (the curve and the flow of water are at right angles to each other). - Because C_3 has nonzero horizontal and vertical components, - some of the water above that curve is moving along the curve. -

    - -

    - When C is a closed curve, - we call flow circulation, - represented by \oint_C \vec F\cdot d\vec r. -

    - -

    - The opposite of flow is flux, - a measure of how much water is moving - across the path C. - If a curve represents a filter in flowing water, - flux measures how much water will pass through the filter. - Considering again , - we see that a screen along C_1 will not filter any water as no water passes across that curve. - Because of the nature of this field, - C_2 and C_3 each filter the same amount of water per second. -

    - -

    - The terms flow and flux - are used apart from velocity fields, too. - Flow is measured by \int_C \vec F\cdot d\vec r, - which is the same as \int_C \vec F\cdot\vec T\, ds by . - That is, flow is a summation of the amount of \vec F that is - tangent to the curve C. -

    - -

    - By contrast, - flux is a summation of the amount of \vec F that is orthogonal - to the direction of travel. - To capture this orthogonal amount of \vec F, - we use \int_C \vec F \cdot \vec n\, ds to measure flux, - where \vec n is a unit vector orthogonal to the curve C. - (Later, we'll measure flux across surfaces, too. - For example, - in physics it is useful to measure the amount of a magnetic field that passes through a surface.) -

    - - - -

    - How is \vec n determined? - We'll later see that if C is a closed curve, - we'll want \vec n to point to the outside of the curve - (measuring how much is going out). - We'll also adopt the convention that closed curves should be traversed counterclockwise. -

    - -

    - (If C is a complicated closed curve, - it can be difficult to determine what - counterclockwise means. - Consider . - Seeing the curve as a whole, we know which way - counterclockwise is. - If we zoom in on point A, - one might incorrectly choose to traverse the path in the wrong direction. - So we offer this definition: - a closed curve is being traversed counterclockwise if the outside is to the right of the path and the inside is to the left.) -

    - -
    - Determining counterclockwise is not always simple without a good definition - - A closed curve in the plane, in the shape of a cashew or boomerang. - -

    - A set of two-dimensional coordinate axes are plotted, with the origin at the center of the image. - A closed curve is plotted in the plane. - The curve is symmetric about the y axis, - and its shape is like that of a cashew nut (or perhaps a crescent moon, croissant, or boomerang). -

    - -

    - There are intercepts at (1,0), (0,1), (-1,0), - and a fourth point, labeled A, that lies below the point (0,1) on the positive y axis. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ 1,0 ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - - if(incolor) { - colorone = bluepen+.8pt+linejoin(0); - colortwo = redpen+.8pt+linejoin(0); - } else - { - colorone = black+.8pt+linejoin(0); - colortwo = gray+.8pt+linejoin(0); - }; - - pair xbounds=(-1.2,1.2); - pair ybounds=(-1.2,1.2); - real[] myxchoice={-1,1}; - real[] myychoice={1,-1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-.1,-.1); - pair b=(1.1,1.1); - - pair vectfunction(pair z) {return (1,0);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - //add(vectorfield(vector,a,b,5,5,rgb(.6,.6,.6))); - - pair f(real x) {return (cos(x),sin(x))*(1+.5*cos(10x));} - pair F(real x) { - if(x<pi) {return (cos(x),sin(x));} - return (cos(x),-sin(3(x-pi))/3); - } - //real g(real x) {return x^4;} - //pair G(real x) {return (x,g(x));} - - pair G(real t) {return (.2,t);} - - //draw(graph(F,0.2,1,operator ..),colorone,arrow=MidArrow(5)); - draw(graph(F,0,2pi,operator ..),colorone); - //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); - - filldraw(shift(0,.33)*scale(.03)*unitcircle,black); - label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); - //label("$C_2$",(.13,.8),filltype=Fill(white+opacity(.7))); - //label("$C_3$",(.6,.5),filltype=Fill(white+opacity(.7))); - - - -
    - -

    - When a curve C is traversed counterclockwise by \vrt = \la f(t), - g(t)\ra, - we rotate \vec T clockwise 90^\circ to obtain \vec n: - - \vec T = \frac{\la \fp(t),\gp(t)\ra}{\norm{\vrp(t)}} \Rightarrow \vec n = \frac{\la \gp(t),-\fp(t)\ra}{\norm{\vrp(t)}} - . -

    - -

    - Letting \vec F = \la M, N\ra, we calculate flux as: - - \int_C \vec F\cdot \vec n\, ds \amp = \int_C \vec F\cdot \frac{\la \gp(t),-\fp(t)\ra}{\norm{\vrp(t)}} \norm{\vrp(t)}\, dt - \amp = \int_C \la M, N\ra \cdot \la \gp(t),-\fp(t)\ra\, dt - \amp = \int_C \Big(M\,\gp(t) - N\,\fp(t)\Big)dt - \amp = \int_C M\,\gp(t)\, dt - \int_C N\,\fp(t)\, dt. - As the x and y components of \vrt are f(t) and g(t) respectively, the differentials of x and y are dx = \fp(t)dt and dy=\gp(t)dt. We can then write the above integrals as: - \amp = \int_C M\, dy - \int_C N\, dx. - This is often written as one integral (not incorrectly, though somewhat confusingly, as this one integral has two d 's): - \amp =\int_CM\, dy -N\, dx - . -

    - -

    - We summarize the above in the following definition. -

    - - - Flow, Flux - -

    - Let \vec F=\la M,N\ra be a vector field with continuous components defined on a smooth curve C, - parametrized by \vrt =\la f(t),g(t)\ra, - let \vec T be the unit tangent vector of \vrt, - and let \vec n be the clockwise 90^\circdegree rotation of \vec T. - flow - flux - -

      -
    • -

      - The flow of \vec F along C is - - \int_C \vec F\cdot\vec T\, ds=\int_C \vec F\cdot d\vec r - . -

      -
      • - -
    • -

      - The flux of \vec F across C is - - \int_C \vec F\cdot \vec n\, ds = \int_C M\, dy -N\, dx = \int_C\Big(M\,\gp(t) - N\,\fp(t)\Big)dt - . -

      -
      • -
    -

    -     -     - -

    - This definition of flow also holds for curves in space, - though it does not make sense to measure - flux across a curve in space. -

    - -

    - Measuring flow is essentially the same as finding work performed by a force as done in the previous examples. - Therefore we practice finding only flux in the following example. -

    - - - Finding flux across curves in the plane - -

    - Curves C_1 and C_2 each start at (1,0) and end at (0,1), - where C_1 follows the line y=1-x and C_2 follows the unit circle, - as shown in . - Find the flux across both curves for the vector fields - \vec F_1 = \la y, -x+1\ra and \vec F_2 = \la -x, 2y-x\ra. -

    -
    - Illustrating the curves and vector fields in . In (a) the vector field is \vec F_1, and in (b) the vector field is \vec F_2. - -
    - - - Two paths in the plane are plotted relative to a vector field. - -

    - The first quadrant in a two-dimensional coordinate system is shown, - with x and y axes, and the origin at the bottom-left of the image. -

    - -

    - There are two curves plotted; both are paths from the point (1,0) to the point (0,1). - The curve C_1 is a straight line, while C_2 is a quarter of the unit circle. -

    - -

    - A vector field is also shown in the image. - The vectors in the vector field appear to lie tangent to a family of circles centered at the point (1,0). - The vectors indicate a clockwise direction of travel, and their magnitude is smallest near (1,0). - Near the two curves, the vectors all cross the curves in a direction that is up and to the right. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ y,-x+1 ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - pair xbounds=(-.2,1.2); - pair ybounds=(-.2,1.2); - real[] myxchoice={1}; - real[] myychoice={1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-.1,-.1); - pair b=(1.1,1.1); - - pair vectfunction(pair z) {return (z.y,-z.x+1);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,bgvect)); - - pair f(real x) {return (cos(x),sin(x));} - pair F(real x) {return (x, 1-x);} - //real g(real x) {return x^4;} - //pair G(real x) {return (x,g(x));} - - pair G(real t) {return (.2,t);} - - draw(graph(F,1,0,operator ..),colorone,arrow=MidArrow(5)); - draw(graph(f,0,pi/2,operator ..),colortwo,arrow=MidArrow(5)); - //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); - - //filldraw(shift(0,.33)*scale(.03)*unitcircle,black); - //label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); - label("$C_2$",(.75,.77),filltype=Fill(white+opacity(.7))); - label("$C_1$",(.43,.45),filltype=Fill(white+opacity(.7))); - - - -
    - -
    - - - Two paths in the plane are plotted relative to a vector field. - -

    - The first quadrant in a two-dimensional coordinate system is shown, - with x and y axes, and the origin at the bottom-left of the image. -

    - -

    - There are two curves plotted; both are paths from the point (1,0) to the point (0,1). - The curve C_1 is a straight line, while C_2 is a quarter of the unit circle. - (These are same curves as in .) -

    - -

    - A vector field is also shown in the image. - The vectors in the vector field appear to lie tangent to a family of hyperbolas centered at the origin. - Unlike in , the orientations of the vectors relative to the curves change along the curve. - Near (1,0), the vectors are pointing down and to the left, - while near the point (0,1), the vectors are pointing up and to the left. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ -x, 2y-x ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - - pair xbounds=(-.2,1.2); - pair ybounds=(-.2,1.2); - real[] myxchoice={1}; - real[] myychoice={1}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-.1,-.1); - pair b=(1.1,1.1); - - pair vectfunction(pair z) {return (-z.x,2z.y-z.x);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,bgvect)); - - pair f(real x) {return (cos(x),sin(x));} - pair F(real x) {return (x, 1-x);} - //real g(real x) {return x^4;} - //pair G(real x) {return (x,g(x));} - - pair G(real t) {return (.2,t);} - - draw(graph(F,1,0,operator ..),colorone,arrow=MidArrow(5)); - draw(graph(f,0,pi/2,operator ..),colortwo,arrow=MidArrow(5)); - //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); - - //filldraw(shift(0,.33)*scale(.03)*unitcircle,black); - //label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); - label("$C_2$",(.75,.77),filltype=Fill(white+opacity(.7))); - label("$C_1$",(.43,.45),filltype=Fill(white+opacity(.7))); - - - -
    -     - -
    -     - -

    - We begin by finding parametrizations of C_1 and C_2. - As done in , - parametrize C_1 by creating the line that starts at (1,0) and moves in the \la -1,1\ra direction: - \vec r_1(t) = \la 1,0\ra + t\la -1,1\ra = \la 1-t, t\ra, - for 0\leq t\leq 1. - We parametrize C_2 with the familiar - \vec r_2(t) = \la \cos(t),\sin(t)\ra on 0\leq t\leq \pi/2. - For reference later, we give each function and its derivative below: - - \vec r_1(t) = \la 1-t, t\ra, \vrp_1(t) = \la -1,1\ra - . - - \vec r_2(t) = \la \cos(t), \sin(t)\ra, \vrp_2(t) = \la -\sin(t) ,\cos(t)\ra - . -

    - -

    - When \vec F = \vec F_1 = \la y, -x+1\ra (as shown in ), - over C_1 we have M = y =t and N = -x+1 = -(1-t)+1 = t. - Using , we compute the flux: - - \int_{C_1} \vec F\cdot \vec n\, ds \amp = \int_{C_1} \Big(M\,\gp(t) - N\,\fp(t)\Big)\, dt - \amp = \int_0^1 \Big( t(1) - t(-1)\Big)\, dt - \amp = \int_0^1 2t\, dt - \amp = 1 - . -

    - -

    - Over C_2, - we have M = y = \sin(t) and N = -x+1 = 1-\cos(t). - Thus the flux across C_2 is: - - \int_{C_1} \vec F\cdot \vec n\, ds \amp = \int_{C_1} \Big(M\,\gp(t) - N\,\fp(t)\Big)\, dt - \amp = \int_0^{\pi/2}\Big((\sin(t))(\cos(t)) - (1-\cos(t))(-\sin(t))\Big)\, dt - \amp = \int_0^{\pi/2} \sin(t)\, dt - \amp =1 - . -

    - -

    - Notice how the flux was the same across both curves. - This won't hold true when we change the vector field. -

    - -

    - When \vec F = \vec F_2 = \la -x,2y-x\ra (as shown in ), - over C_1 we have M = -x = t-1 and N = 2y-x = 2t-(1-t) = 3t-1. - Computing the flux across C_1: - - \int_{C_1} \vec F\cdot \vec n\, ds \amp = \int_{C_1} \Big(M\,\gp(t) - N\,\fp(t)\Big)\, dt - \amp = \int_0^1 \Big( (t-1)(1) - (3t-1)(-1)\Big)\, dt - \amp = \int_0^1 (4t-2)\, dt - \amp = 0 - . -

    - -

    - Over C_2, - we have M = -x = -\cos(t) and N = 2y-x = 2\sin(t)-\cos(t). - Thus the flux across C_2 is: - - \int_{C_1} \vec F\cdot \vec n\, ds \amp = \int_{C_1} \Big(M\,\gp(t) - N\,\fp(t)\Big)\, dt - \amp = \int_0^{\pi/2}\Big((-\cos(t))(\cos(t)) - (2\sin(t)-\cos(t))(-\sin(t))\Big)\, dt - \amp = \int_0^{\pi/2} \big(2\sin^2 t-\sin(t)\cos(t)-\cos^2t\big)\, dt - \amp =\pi/4 - 1/2\approx 0.285 - . -

    - -

    - We analyze the results of this example below. -

    -     -     - -

    - In , - we saw that the flux across the two curves was the same when the vector field was \vec F_1 = \la y, -x+1\ra. - This is not a coincidence. - We show why they are equal in . - In short, the reason is this: - the divergence of \vec F_1 is 0, and when \divv \vec F = 0, - the flux across any two paths with common beginning and ending points will be the same. -

    - -

    - We also saw in the example that the flux across C_1 was 0 when the field was \vec F_2 = \la -x, 2y-x\ra. - Flux measures how much of the field crosses the path from left to right - (following the conventions established before). - Positive flux means most of the field is crossing from left to right; - negative flux means most of the field is crossing from right to left; - zero flux means the same amount crosses from each side. - When we consider , - it seems plausible that the same amount of - \vec F_2 was crossing C_1 from left to right as from right to left. -

    -     - - - Green's Theorem -

    - There is an important connection between the circulation around a closed region R and the curl of the vector field inside of R, - as well as a connection between the flux across the boundary of R and the divergence of the field inside R. - These connections are described by Green's Theorem and the Divergence Theorem, - respectively. - We'll explore each in turn. - Green's Theorem -

    - - -

    - Green's Theorem states the counterclockwise circulation around a closed region R is equal to the sum of the curls over R. -

    - - - Green's Theorem - -

    - Let R be a closed, - bounded region of the plane whose boundary C is composed of finitely many smooth curves, - let \vec r(t) be a counterclockwise parametrization of C, - and let \vec F =\la M,N\ra where N_x and M_y are continuous over R. - Then - Green's Theorem - - \oint_C \vec F\cdot d\vec r = \iint_R\curl \vec F\, dA - . -

    -     -     - -

    - We'll explore Green's Theorem through an example. -

    - - - Confirming Green's Theorem - -

    - Let \vec F =\la -y,x^2+1\ra and let R be the region of the plane bounded by the triangle with vertices (-1,0), - (1,0) and (0,2), - shown in . - Verify Green's Theorem; that is, - find the circulation of \vec F around the boundary of R and show that is equal to the double integral of \curl \vec F over R. -

    - -
    - The vector field and planar region used in - - A triangular path in the plane is plotted against a two-dimensional vector field. - -

    - A set of coordinate axes in the plane is shown, with the x axis at the bottom of the image, - and the y axis in the center. -

    - -

    - A two-dimensional vector field is plotted. - Near the bottom of the image, the vectors in the field are nearly vertical. - As we move up the image, the vectors rotate; - while they maintain an upward trajectory, the vectors also point more to the left the higher up they are. -

    - -

    - Over the vector field, a triangular path is plotted. - The path is oriented counter-clockwise, - from the point (-1,0), along the x axis to (1,0), - then up to (0,2), and then back down to (-1,0). -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ -y,x^2+1 ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - pair xbounds=(-1.2,1.2); - pair ybounds=(-.2,2.2); - real[] myxchoice={-1,1}; - real[] myychoice={1,2}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-1.1,-.1); - pair b=(1.1,2.1); - - pair vectfunction(pair z) {return (-z.y,z.x^2+1);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,bgvect)); - - //pair f(real x) {return (cos(x),sin(x));} - //real g(real x) {return x^4;} - //pair G(real x) {return (x,g(x));} - - //pair G(real t) {return (.2,t);} - - pair F(real x) {return (x, 0);} - draw(graph(F,-1,1,operator ..),colorone,arrow=MidArrow(5)); - - pair F(real x) {return (x, 2-2x);} - draw(graph(F,1,0,operator ..),colorone,arrow=MidArrow(5)); - - pair F(real x) {return (x, 2+2x);} - draw(graph(F,0,-1,operator ..),colorone,arrow=MidArrow(5)); - - //draw(graph(f,0,pi/2,operator ..),colortwo,arrow=MidArrow(5)); - //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); - - //filldraw(shift(0,.33)*scale(.03)*unitcircle,black); - //label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); - //label("$C_2$",(.75,.77),filltype=Fill(white+opacity(.7))); - label("$R$",(0,.5),filltype=Fill(white+opacity(.7))); - - - -
    -     - -

    - The curve C that bounds R is composed of 3 lines. - While we need to traverse the boundary of R in a counterclockwise fashion, - we may start anywhere we choose. - We arbitrarily choose to start at (-1,0), - move to (1,0), - etc., with each line parametrized by \vec r_1(t), - \vec r_2(t) and \vec r_3(t), respectively. -

    - -

    - We leave it to the reader to confirm that the following parametrizations of the three lines are accurate: -

    - - - - \vec r_1(t) = \la 2t-1,0\ra, - for 0\leq t\leq 1, - with \vrp_1(t) = \la 2,0\ra, - - - \vec r_2(t) = \la 1-t,2t\ra, - for 0\leq t\leq 1, - with \vrp_2(t) = \la -1,2\ra, and - - - \vec r_3(t) = \la -t,2-2t\ra, - for 0\leq t\leq 1, - with \vrp_3(t) = \la -1,-2\ra. - - - -

    - The circulation around C is found by summing the flow along each of the sides of the triangle. - We again leave it to the reader to confirm the following computations: - - \int_{C_1}\vec F\cdot d\vec r_1 \amp = \int_0^1 \la 0,(2t-1)^2+1\ra\cdot \la 2,0\ra dt = 0, - \int_{C_2}\vec F\cdot d\vec r_2 \amp = \int_0^1 \la -2t,(1-t)^2+1\ra\cdot \la -1,2\ra dt = 11/3, \text{ and } - \int_{C_3}\vec F\cdot d\vec r_3 \amp = \int_0^1 \la 2t-2,t^2+1\ra\cdot \la -1,-2\ra dt = -5/3 - . -

    - -

    - The circulation is the sum of the flows: 2. -

    - -

    - We confirm Green's Theorem by computing \iint_R \curl \vec F\, dA. - We find \curl \vec F = 2x+1. - The region R is bounded by the lines y = 2x+2, - y=-2x+2 and y=0. - Integrating with the order dx\, dy is most straightforward, - leading to - - \int_0^2\int_{y/2-1}^{1-y/2} (2x+1)\, dx\, dy = \int_0^2 (2-y)\, dy = 2 - , - which matches our previous measurement of circulation. -

    -     -     - - - - - Using Green's Theorem - -

    - Let \vec F = \la \sin(x),\cos(y)\ra and let R be the region enclosed by the curve C parametrized by - \vec r(t) = \la 2\cos(t)+ \frac1{10}\cos(10t),2\sin(t)+\frac1{10}\sin(10t)\ra on 0\leq t\leq 2\pi, - as shown in . - Find the circulation around C. -

    - -
    - The vector field and planar region used in - - A curve in the plane, like a bumpy circle or flower, and a vector field. - -

    - A coordinate plane with x and y axes, and the origin at the center of the image. -

    - -

    - The vector field is fairly complicated. -

      -
    • -

      - It appears to have two sources at (0,\pi/2) and (0,-\pi/2), where the vector field vanishes. -

      -
      • -
    • -

      - Along the lines y=\pi/2 and y=-\pi/2, the vector field is horizontal, and points away from the y axis. -

      -
      • -
    • -

      - Below y=-\pi/2, the vectors point down and away from the y axis. -

      -
      • -
    • -

      - Above y=\pi/2, the vecors point down and away from the y axis. -

      -
      • -
    • -

      - Between the lines y=\pm \pi/2, the vectors point up and away from the y axis -

      -
      • -
    -

    - -

    - The curve is a bumpy circle, resembling the outline of a flower. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ sin x, cos y ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - - pair xbounds=(-2.8,2.8); - pair ybounds=(-2.8,2.8); - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-2.5,-2.5); - pair b=(2.5,2.5); - - pair vectfunction(pair z) {return (sin(z.x),cos(z.y));} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,bgvect)); - - //pair f(real x) {return (cos(x),sin(x));} - //real g(real x) {return x^4;} - //pair G(real x) {return (x,g(x));} - - //pair G(real t) {return (.2,t);} - - pair F(real x) {return (2*cos(x)+cos(10x)/10,2*sin(x)+sin(10x)/10);} - draw(graph(F,0+.1,2pi+.1,operator ..),colorone,arrow=MidArrow(5)); - - //draw(graph(f,0,pi/2,operator ..),colortwo,arrow=MidArrow(5)); - //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); - - //filldraw(shift(0,.33)*scale(.03)*unitcircle,black); - //label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); - //label("$C_2$",(.75,.77),filltype=Fill(white+opacity(.7))); - label("$R$",(0,.5),filltype=Fill(white+opacity(.7))); - - - -
    -     - -

    - Computing the circulation directly using the line integral looks difficult, - as the integrand will include terms like - \sin\big(2\cos(t) + \frac1{10}\cos(10t)\big). -

    - -

    - Green's Theorem states that \oint_C\vec F\cdot d\vec r = \iint_R \curl\vec F\, dA; - since \curl \vec F = 0 in this example, - the double integral is simply 0 and hence the circulation is 0. -

    - -

    - Since \curl \vec F = 0, - we can conclude that the circulation is 0 in two ways. - One method is to employ Green's Theorem as done above. - The second way is to recognize that \vec F is a conservative field, - hence there is a function f(x,y) wherein \vec F = \nabla f. - Let A be any point on the curve C; - since C is closed, - we can say that C begins - and ends at A. - By the Fundamental Theorem of Line Integrals, - \oint_C \vec F\, d\vec r = f(A)-f(A) = 0. -

    -     -     - - - - - -

    - One can use Green's Theorem to find the area of an enclosed region by integrating along its boundary. - Let C be a closed curve, - enclosing the region R, - parametrized by \vec r(t) = \la f(t),g(t)\ra. - We know the area of R is computed by the double integral \iint_R \, dA, - where the integrand is 1. - By creating a field \vec F where \curl \vec F =1, - we can employ Green's Theorem to compute the area of R as \oint_C \vec F\cdot d\vec r. -

    - -

    - One is free to choose any field \vec F to use as long as \curl\vec F = 1. - Common choices are \vec F = \la 0,x\ra, - \vec F = \la -y,0\ra and \vec F = \la -y/2,x/2\ra. - We demonstrate this below. -

    - - - Using Green's Theorem to find area - -

    - Let C be the closed curve parametrized by - \vrt = \la t-t^3,t^2\ra on -1\leq t\leq 1, - enclosing the region R, - as shown in . - Find the area of R. -

    - -
    - The region R, whose area is found in - - A teardrop-shaped region in the plane. - -

    - The region R is plotted in two dimensions, - with the x axis at the bottom of the image, - and the y axis in the center. -

    - -

    - The region R is a teardrop shape. - It is symmetric about the y axis, - with a cusp at (0,1), and a fairly flat bottom at the origin. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={-1,1}, - ytick={1}, - ymin=-.5,ymax=1.5, - xmin=-1.1,xmax=1.1, - ] - - \addplot+ [domain=-1:1,samples=40] ({x*(x^2-1)},{x^2}); - - \draw (axis cs:.2,.4) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - -
    -     - -

    - We can choose any field \vec F, - as long as \curl \vec F = 1. - We choose \vec F = \la -y,0\ra. - We also confirm - (left to the reader) - that \vrt traverses the region R in a counterclockwise fashion. - Thus - - \text{ Area of \(R\) } \amp = \iint_R\, dA - \amp = \oint_C \vec F\cdot d\vec r - \amp = \int_{-1}^1 \la -t^2,0\ra\cdot \la 1-3t^2,2t\ra\, dt - \amp = \int_{-1}^1 (-t^2)(1-3t^2)\, dt - \amp = \frac8{15} - . -

    -     -     - - - - - - - - -     - - - The Divergence Theorem -

    - Green's Theorem makes a connection between the circulation around a closed region R and the sum of the curls over R. - The Divergence Theorem makes a somewhat - opposite connection: - the total flux across the boundary of R is equal to the sum of the divergences over R. -

    - - - The Divergence Theorem (in the plane) - -

    - Let R be a closed, - bounded region of the plane whose boundary C is composed of finitely many smooth curves, - let \vec r(t) be a counterclockwise parametrization of C, - and let \vec F =\la M,N\ra where M_x and N_y are continuous over R. - ThenDivergence Theoremin the plane - - \oint_C \vec F\cdot \vec n\, ds = \iint_R\divv \vec F\, dA - . -

    -     -     - - - Confirming the Divergence Theorem - -

    - Let \vec F = \la x-y,x+y\ra, - let C be the circle of radius 2 centered at the origin and define R to be the interior of that circle, - as shown in . - Verify the Divergence Theorem; - that is, find the flux across C and show it is equal to the double integral of \divv \vec F over R. -

    - -
    - The region R used in - - A circle in the plane, plotted against a spiral vector field. - -

    - Two-dimensional coordinate axes are plotted, with the origin at the center. - A circle is shown, with its center at the origin, and radius 2. -

    - -

    - The vector field appears to follow paths that spriral outward from the origin. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ x-y,x+y ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - - pair xbounds=(-2.8,2.8); - pair ybounds=(-2.8,2.8); - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - - xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - //label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - //label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-2.5,-2.5); - pair b=(2.5,2.5); - - pair vectfunction(pair z) {return (z.x-z.y,z.x+z.y);} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,bgvect)); - - -
    -     - -

    - We parametrize the circle in the usual way, with \vrt = -

    - -

    - \la 2\cos(t),2\sin(t)\ra, 0\leq t\leq 2\pi. - The flux across C is - - \oint_C \vec F\cdot \vec n\, ds \amp = \oint_C\big(M\gp(t)-N\fp(t)\big)\, dt - \amp = \int_0^{2\pi} \big((2\cos(t)-2\sin(t))(2\cos(t)) - (2\cos(t)+2\sin(t))(-2\sin(t))\big)\, dt - \amp = \int_0^{2\pi} 4\, dt = 8\pi - . -

    - -

    - We compute the divergence of \vec F as \divv \vec F = M_x+N_y = 2. - Since the divergence is constant, - we can compute the following double integral easily: - - \iint_R \divv \vec F\, dA = \iint_R 2\, dA = 2\iint_R\, dA = 2(\text{ area of \(R\) } ) = 8\pi - , - which matches our previous result. -

    -     -     - - - Flux when <m>\divv \vec F = 0</m> - -

    - Let \vec F be any field where \divv \vec F = 0, - and let C_1 and C_2 be any two nonintersecting paths, - except that each begin at point A and end at point B - (see ). - Show why the flux across C_1 and C_2 is the same. -

    -     - -

    - By referencing , - we see we can make a closed path C that combines C_1 with C_2, - where C_2 is traversed with its opposite orientation. - We label the enclosed region R. - Since \divv \vec F = 0, - the Divergence Theorem states that - - \oint_C \vec F\cdot \vec n\, ds = \iint_R \divv \vec F\, dA = \iint_R 0\, dA = 0 - . -

    - -

    - Using the properties and notation given in , - consider: -

    - -
    - As used in , the vector field has a divergence of 0 and the two paths only intersect at their initial and terminal points. - - Two curves in the plane between points A and B, and a vector field with many vortices. - -

    - Two curves in the plane are plotted without reference to coordinate axes. - Both curves travel from a point A at the bottom right of the image to a point B at the top left. -

    - -

    - The vector field is quite complicated, and appears to circle about many different vortices. -

    -     - - - import graph; - size(282,282); - - // vector field F = ⟨ y+2,-2x+1 ⟩ - - defaultpen(fontsize(8)); - - - pen colorone; - pen colortwo; - pen mainvect; - pen bgvect; - - if(incolor) { - colorone = bluepen+1.5pt+linejoin(0); - colortwo = redpen+1.5pt+linejoin(0); - mainvect = bluepen + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - } else - { - colorone = black+1.5pt+linejoin(0); - colortwo = gray+1.5pt+linejoin(0); - mainvect = black + linejoin(0); - bgvect = rgb(.6,.6,.6)+linejoin(0); - }; - pair xbounds=(-.2,1.2); - pair ybounds=(-.2,1.2); - real[] myxchoice={1}; - real[] myychoice={1}; - - //xaxis("",xbounds.x,xbounds.y,Ticks(myxchoice),arrow=Arrow(DefaultHead,size=4)); - //yaxis("",ybounds.x,ybounds.y,Ticks(myychoice),arrow=Arrow(DefaultHead,size=4)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0)); - ////label("\phantom{$x$}",(xbounds.x-0.05*(xbounds.y-xbounds.x),0)); - //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x))); - ////label("\phantom{$y$}",(0,ybounds.x-0.05*(ybounds.y-ybounds.x))); - - pair a=(-.1,-.1); - pair b=(1.1,1.1); - - pair vectfunction(pair z) {return (cos(pi*3*z.y),sin(pi*3*z.x));} - path vector(pair t) {return -vectfunction(t)/2 -- vectfunction(t)/2;} - - add(vectorfield(vector,a,b,12,12,bgvect)); - - pair f(real x) {return (cos(x)+.05*sin(10x),sin(x)+.05*sin(10x));} - pair F(real x) {return (x, (1-x)^2);} - //real g(real x) {return x^4;} - //pair G(real x) {return (x,g(x));} - - pair G(real t) {return (.2,t);} - - draw(graph(F,1,0,operator ..),colorone,arrow=MidArrow(5)); - draw(graph(f,0,pi/2,operator ..),colortwo,arrow=MidArrow(5)); - //draw(graph(f,0,.8,operator ..),colorone+linetype(new real[] {4,2}),arrow=MidArrow(5)); - - dot((1,0),black); - dot((0,1),black); - label("$A$",(1,-.05),filltype=Fill(white+opacity(.7))); - label("$B$",(-.05,1),filltype=Fill(white+opacity(.7))); - - //filldraw(shift(0,.33)*scale(.03)*unitcircle,black); - //label("$A$",(.1,.4));//filltype=Fill(white+opacity(.7))); - label("$C_1$",(.75,.85),filltype=Fill(white+opacity(.7))); - label("$C_2$",(.43,.45),filltype=Fill(white+opacity(.7))); - - - -
    -

    - - 0 \amp = \oint_C \vec F\cdot \vec n\, ds - \amp = \int_{C_1} \vec F\cdot \vec n\, ds +\int_{C_2^*} \vec F\cdot \vec n\, ds - (where C_2^* is the path C_2 traversed with opposite orientation) - \amp = \int_{C_1} \vec F\cdot \vec n\, ds -\int_{C_2} \vec F\cdot \vec n\, ds. - \int_{C_2} \vec F\cdot \vec n\, ds\amp = \int_{C_1} \vec F\cdot \vec n\, ds - . - Thus the flux across each path is equal. -

    -     -     - -

    - In this section, we have investigated flow and flux, - quantities that measure interactions between a vector field and a planar curve. - We can also measure flow along spatial curves, - though as mentioned before, - it does not make sense to measure flux across spatial curves. -

    - -

    - It does, however, - make sense to measure the amount of a vector field that passes across a surface in space i.e, the flux across a surface. - We will study this, - though in the next section we first learn about a more powerful way to describe surfaces than using functions of the form z=f(x,y). -

    -     - - - - Terms and Concepts - - - -

    - Let \vec F be a vector field and let C be a curve. - Flow is a measure of the amount of \vec F going C; - flux is a measure of the amount of \vec F going C. -

    -     - - - - - - - - - - - - -     - - - -

    - What is circulation? -

    -     - - - -

    - It is the measure of flow around the entirety of a closed curve C. -

    -     -     - - - -

    - Green's Theorem states, informally, - that the circulation around a closed curve that bounds a region R is equal to the sum of the - of \vec{F} across R. -

    -     - - - - - - - -     - - - -

    - The Divergence Theorem states, informally, - that the outward flux across a closed curve that bounds a region R - is equal to the sum of the of \vec{F} across R. -

    -     - - - - - - - - -

    - the divergence of \vec F, or \divv \vec F -

    -     -     - - - -

    - Let \vec F be a vector field and let C_1 and C_2 - be any nonintersecting paths except that each starts at point A and ends at point B. - If the of \vec{F} is 0, - then \int_{C_1} \vec F\cdot \vec T\, ds = \int_{C_2} \vec F\cdot \vec T\, ds. -

    -     - - - - - - - -     - - - -

    - Let \vec F be a vector field and let C_1 and C_2 - be any nonintersecting paths except that each starts at point A and ends at point B. - If the of \vec{F} is 0, - then \int_{C_1} \vec F\cdot \vec n\, ds = \int_{C_2} \vec F\cdot \vec n\, ds. -

    -     - - - - - - - -     -     - - - Problems - - -

    - A vector field \vec F and a curve C are given. - Evaluate \int_C \vec F\cdot\vec n\, ds, - the flux of \vec F over C. -

    -     - - - -

    - \vec F = \langle x+y,x-y\rangle; - C is the curve with initial and terminal points (3,-2) and (3,2), - respectively, - parametrized by \vec r(t) = \langle 3t^2,2t\rangle on -1\leq t\leq 1. -

    -     - -

    - 12 -

    -     -     - - - -

    - \vec F = \langle x+y,x-y\rangle; - C is the curve with initial and terminal points (3,-2) and (3,2), - respectively, - parametrized by \vec r(t) = \langle 3,t\rangle on -2\leq t\leq 2. -

    -     - -

    - 12 -

    -     -     - - - -

    - \vec F = \langle x^2,y+1\rangle; - C is line segment from (0,0) to (2,4). -

    -     - -

    - -2/3 -

    -     -     - - - -

    - \vec F = \langle x^2,y+1\rangle; - C is the portion of the parabola y=x^2 from (0,0) to (2,4). -

    -     - -

    - 10/3 -

    -     -     - - - -

    - \vec F = \langle y,0\rangle; - C is the line segment from (0,0) to (0,1). -

    -     - -

    - 1/2 -

    -     -     - - - -

    - \vec F = \langle y,0\rangle; - C is the line segment from (0,0) to (1,1). -

    -     - -

    - 1/2 -

    -     -     -     - - - -

    - A vector field \vec F and a closed curve C, - enclosing a region R, are given. - Verify Green's Theorem by evaluating \oint_C\vec F\cdot d\vec r and - \iint_R \curl \vec F\, dA, showing they are equal. -

    -     - - - -

    - \vec F = \langle x-y,x+y\rangle; - C is the closed curve composed of the parabola y=x^2 on - 0\leq x\leq 2 followed by the line segment from (2,4) to (0,0). -

    -     - -

    - The line integral \oint_C\vec F\cdot d\vec r, - over the parabola, is 38/3; - over the line, it is -10. - The total line integral is thus 38/3-10 = 8/3. - The double integral of \curl \vec F = 2 over R also has value 8/3. -

    -     -     - - - -

    - \vec F = \langle -y,x\rangle; - C is the unit circle. -

    -     - -

    - Both the line integral and double integral have value of 2\pi. -

    -     -     - - - -

    - \vec F = \langle 0,x^2\rangle; - C the triangle with corners at (0,0), - (2,0) and (1,1). -

    -     - -

    - Three line integrals need to be computed to compute \oint_C \vec F\cdot d\vec r. - It does not matter which corner one starts from first, - but be sure to proceed around the triangle in a counterclockwise fashion. -

    - -

    - From (0,0) to (2,0), - the line integral has a value of 0. - From (2,0) to (1,1) the integral has a value of 7/3. - From (1,1) to (0,0) the line integral has a value of -1/3. - Total value is 2. -

    - -

    - The double integral of \curl\vec F over R also has value 2. -

    -     -     - - - -

    - \vec F = \langle x+y,2x\rangle; - C the curve that starts at (0,1), - follows the parabola y=(x-1)^2 to (3,4), - then follows a line back to (0,1). -

    -     - -

    - Two line integrals need to be computed to compute \oint_C \vec F\cdot d\vec r. - Along the parabola, the line integral has value 25.5. - Along the line, the line integral has value -21. - Together, the total value is 4.5 -

    - -

    - The double integral of \curl\vec F over R also has value 4.5. -

    -     -     -     - - - -

    - A closed curve C enclosing a region R is given. - Find the area of R by computing - \oint_C \vec F\cdot d\vec r for an appropriate choice of vector field \vec F. -

    -     - - - -

    - C is the ellipse parametrized by - \vec r(t) = \langle 4\cos(t),3\sin(t)\rangle on 0\leq t\leq 2\pi. -

    -     - -

    - Any choice of \vec F is appropriate as long as \curl \vec F = 1. - When \vec F = \langle -y/2,x/2\rangle, - the integrand of the line integral is simply 6. - The area of R is 12\pi. -

    -     -     - - - -

    - C is the curve parametrized by - \vec r(t) = \langle \cos(t),\sin (2t)\rangle on -\pi/2\leq t\leq \pi/2. -

    -     - -

    - Any choice of \vec F is appropriate as long as \curl \vec F = 1. - The choices of \vec F = \langle -y,0\rangle and - \langle 0,x\rangle each lead to reasonable integrands. - The area of R is 4/3. -

    -     -     - - - -

    - C is the curve parametrized by - \vec r(t) = \langle -t^3+3t^2-2t,2(t-1)^2\rangle on 0\leq t\leq 2. -

    -     - -

    - Any choice of \vec F is appropriate as long as \curl \vec F = 1. - The choices of \vec F = \langle -y,0\rangle, - \langle 0,x\rangle and - \langle -y/2,x/2\rangle each lead to reasonable integrands. - The area of R is 16/15. -

    -     -     - - - -

    - C is the curve parametrized by - \vec r(t) = \langle 2\cos(t)+\frac1{10}\cos(10t),2\sin(t)+\frac1{10}\sin (10t)\rangle on 0\leq t\leq 2\pi. -

    -     - -

    - Any choice of \vec F is appropriate as long as \curl \vec F = 1. - The choice of \vec F = \langle -y/2,x/2\rangle leads to a reasonable integrand after simplification. - The area of R is 41\pi/10. -

    -     -     -     - - - -

    - A vector field \vec F and a closed curve C, - enclosing a region R, are given. - Verify the Divergence Theorem by evaluating \oint_C\vec F\cdot \vec n\, ds and - \iint_R \divv \vec F\, dA, showing they are equal. -

    -     - - - -

    - \vec F = \langle x-y,x+y\rangle; - C is the closed curve composed of the parabola y=x^2 on - 0\leq x\leq 2 followed by the line segment from (2,4) to (0,0). -

    -     - -

    - The line integral \oint_C\vec F\cdot \vec n\, ds, - over the parabola, is -22/3; - over the line, it is 10. - The total line integral is thus -22/3+10 = 8/3. - The double integral of \divv \vec F = 2 over R also has value 8/3. -

    -     -     - - - -

    - \vec F = \langle -y,x\rangle; - C is the unit circle. -

    -     - -

    - Both the line integral and double integral have value of 0. -

    -     -     - - - -

    - \vec F = \langle 0,y^2\rangle; - C the triangle with corners at (0,0), - (2,0) and (1,1). -

    -     - -

    - Three line integrals need to be computed to compute \oint_C \vec F\cdot \vec n\, ds. - It does not matter which corner one starts from first, - but be sure to proceed around the triangle in a counterclockwise fashion. -

    - -

    - From (0,0) to (2,0), - the line integral has a value of 0. - From (2,0) to (1,1) the integral has a value of 1/3. - From (1,1) to (0,0) the line integral has a value of 1/3. - Total value is 2/3. -

    - -

    - The double integral of \divv\vec F over R also has value 2/3. -

    -     -     - - - -

    - \vec F = \langle x^2/2,y^2/2\rangle; - C the curve that starts at (0,1), - follows the parabola y=(x-1)^2 to (3,4), - then follows a line back to (0,1). -

    -     - -

    - Two line integrals need to be computed to compute \oint_C \vec F\cdot \vec n\, ds. - Along the parabola, the line integral has value 159/20. - Along the line, the line integral has value 6. - Together, the total value is 279/20. -

    - -

    - The double integral of \divv\vec F over R also has value 279/20. -

    -     -     -     -     -     -
    -
    - Parametrized Surfaces and Surface Area - -

    - Thus far we have focused mostly on 2-dimensional vector fields, - measuring flow and flux along/across curves in the plane. - Both Green's Theorem and the Divergence Theorem make connections between planar regions and their boundaries. - We now move our attention to 3-dimensional vector fields, - considering both curves and surfaces in space. -

    -     - - - Parametrizing surfaces - - -

    - We are accustomed to describing surfaces as functions of two variables, - usually written as z=f(x,y). - For our coming needs, - this method of describing surfaces will prove to be insufficient. - Instead, we will parametrize our surfaces, - describing them as the set of terminal points of some vector-valued function \vec r(u,v) =\langle f(u,v),g(u,v),h(u,v)\rangle. - The bulk of this section is spent practicing the skill of describing a surface \surfaceSusing a vector-valued function. - Once this skill is developed, - we'll show how to find the surface area S of a parametrically-defined surface \surfaceS, a skill needed in the remaining sections of this chapter. -

    - - - - Parametrized Surface - -

    - Let \vec r(u,v) = \langle\, f(u,v),g(u,v),h(u,v)\rangle - be a vector-valued function that is continuous and one to one on the interior of its domain R in the u-v plane. - The set of all terminal points of \vec r (, the range - of \vec r ) is the surface - \surfaceS, and \vec r along with its domain R form a - parametrization of \surfaceS. -

    - -

    - This parametrization is smooth - on R if \vec r_u and \vec r_v are continuous and - \vec r_u\times \vec r_v is never \vec 0 on the interior of R. - surface - parametric equationsof a surface - parametrized surface - smoothsurface - surfacesmooth -

    -     -     - -

    - Given a point (u_0,v_0) in the domain of a vector-valued function \vec r, - the vectors \vec r_u(u_0,v_0) and - \vec r_v(u_0,v_0) are tangent to the surface \surfaceS at - \vec r(u_0,v_0) (a proof of this is developed later in this section). - The definition of smoothness dictates that \vec r_u\times \vec r_v \neq \vec 0; - this ensures that neither \vec r_u nor \vec r_v are \vec 0, - nor are they ever parallel. - Therefore smoothness guarantees that \vec r_u and - \vec r_v determine a plane that is tangent to \surfaceS. -

    - - - -

    - A surface \surfaceS is said to be orientable - orientable - if a field of normal vectors can be defined on \surfaceS that vary continuously along \surfaceS. This definition may be hard to understand; - it may help to know that orientable surfaces are often called two sided. - A sphere is an orientable surface, - and one can easily envision an inside - and outside of the sphere. - A paraboloid is orientable, - where again one can generally envision - inside and outside sides - (or top and bottom sides) - to this surface. - Just about every surface that one can imagine is orientable, - and we'll assume all surfaces we deal with in this text are orientable. -

    - -

    - Möbius band - - It is enlightening to examine a classic non-orientable surface: - the Möbius band, shown in . - Vectors normal to the surface are given, - starting at the point indicated in the figure. - These normal vectors vary continuously - as they move along the surface. - Letting each vector indicate the - top side of the band, - we can easily see near any vector which side is the top. -

    - -

    - However, if as we progress along the band, - we recognize that we are labeling - both sides of the band as the top; - in fact, there are not two sides to this band, but one. - The Möbius band is a non-orientable surface. -

    - -
    - A Möbius band, a non-orientable surface - - A depction of a Möbius band, showing how the normal vector changes direction while traveling around the surface. - -

    - A Möbius band is the surface one obtains by taking a strip of paper, - giving it a half-twist, and gluing the ends of the strip together. - It is a famous example of a surface with only one side. -

    - -

    - This is illustrated in the image by plotting a collection of normal vectors along the central curve of the surface. - (This is the curve obtained if you draw a line down the middle of your strip of paper before connecting the two ends.) - The normal vector begins its trip pointing up, - but as it travels around the curve it rotates, and ends up pointing down when we return to the starting point of the curve. -

    - -

    - This illustrates that it is impossible to assign a consistent choice of normal vector across the surface, - so that the Möbius band is non-orientable. -

    -     - - - - - //ASY file for fig3d_proj3D.asy in Chapter 13 - - size(282,282,Aspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(6.7,-2,1.9); - defaultrender.merge=true; - - // setup and draw the axes - //real[] myxchoice={2}; - //real[] myychoice={1,2}; - //real[] myzchoice={2,4}; - //defaultpen(0.5mm); - - //pair xbounds=(-0.5,2.5); - //pair ybounds=(-0.5,2.25); - //pair zbounds=(-0.5,4.5); - - //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //plane z=2y - triple f(pair t) { - return ((1+t.y/2*cos(t.x/2))*cos(t.x),(1+t.y/2*cos(t.x/2))*sin(t.x),t.y/2*sin(t.x/2)); - } - surface s=surface(f,(0,-.3),(2*pi,.3),50,5); - pen p=bluepen+.1mm; - draw(s,simplesurfacepen,meshpen=p); - - triple g(real t) { - return f((t,.3)); - } - - path3 mypath=graph(g,-2pi,2pi,operator ..); - draw(mypath,bluepen+linewidth(1)); - - // - // Attempted various ways of getting a normal vector. Turns out all - // were probably ok, it was just that the aspect ratio was off. - // Used ``Aspect'' in the orthographic line above to fix it. - // - - triple n(real t) { - return (.5*cos(t)*sin(t/2),(1/2)*(sin(t/2)*sin(t)),-.5*cos(t/2)); - } - //(.5*cos(t)*sin(t/2),(1/4)*(cos(t/2)-cos(3*t/2)),-.5*cos(t/2)) - - //for(real i=0; i<=2*pi;i=i+2pi/10) { - //draw(f((i,0.)) -- (f((i,0.))+n(i)),rgb(1-i/(2pi),0,i/(2pi))+.2mm,Arrow3); - //} - - //draw((f((0,0)).x,0,0) -- (1,0,-.5),redpen+.2mm,Arrow3); - //draw((0.309017, 0.951057, 0.) -- (0.399835, 1.23057, -0.404508),rgb(.9,0,.1)+.2mm,Arrow3); - //draw((-0.809017, 0.587785, 0.) -- (-1.19373, 0.867294, -0.154508),rgb(.5,0,.5)+.2mm,Arrow3); - //draw((0.309017, -0.951057, 0.) -- (0.399835, -1.23057, 0.404508),rgb(.1,0,.9)+.2mm,Arrow3); - - // - // Final method of drawing normal vectors. - // - - if(incolor) { - for(real i=0; i<=2pi; i=i+2pi/20) { - triple a = f((i+.1,0))-f((i-.1,0)); - triple b = f((i,.1))-f((i,-.1)); - triple c = .3*(a.y*b.z-a.z*b.y,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x)/sqrt(((a.y*b.z-a.z*b.y)^2+(a.z*b.x-a.x*b.z)^2+(a.x*b.y-a.y*b.x)^2)); - draw(f((i,0)) -- (f((i,0))+c),rgb(1-i/(2pi),0,i/(2pi))+.2mm,Arrow3); - } - } else { - for(real i=0; i<=2pi; i=i+2pi/20) { - triple a = f((i+.1,0))-f((i-.1,0)); - triple b = f((i,.1))-f((i,-.1)); - triple c = .3*(a.y*b.z-a.z*b.y,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x)/sqrt(((a.y*b.z-a.z*b.y)^2+(a.z*b.x-a.x*b.z)^2+(a.x*b.y-a.y*b.x)^2)); - draw(f((i,0)) -- (f((i,0))+c),rgb(.75*i/(2pi),.75*i/(2pi),.75*i/(2pi))+.2mm,Arrow3); - } - } - - triple a = f((0+.1,0))-f((0-.1,0)); - triple b = f((0,.1))-f((0,-.1)); - triple c = .3*(a.y*b.z-a.z*b.y,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x)/sqrt(((a.y*b.z-a.z*b.y)^2+(a.z*b.x-a.x*b.z)^2+(a.x*b.y-a.y*b.x)^2)); - label("end",(f((0,0))+c-(0,0,.1))); - - triple a = f((2pi+.1,0))-f((2pi-.1,0)); - triple b = f((2pi,.1))-f((2pi,-.1)); - triple c = .3*(a.y*b.z-a.z*b.y,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x)/sqrt(((a.y*b.z-a.z*b.y)^2+(a.z*b.x-a.x*b.z)^2+(a.x*b.y-a.y*b.x)^2)); - label("start",(f((2pi,0))+c+(0,0,.1))); - - - -
    - -

    - We now practice parametrizing surfaces. -

    - - - Parametrizing a surface over a rectangle - -

    - Parametrize the surface z=x^2+2y^2 over the rectangular region R defined by - -3\leq x\leq 3, -1\leq y\leq 1. -

    -     - -

    - There is a straightforward way to parametrize a surface of the form - z=f(x,y) over a rectangular domain. - We let x=u and y=v, - and let \vec r(u,v) = \langle u,v, f(u,v)\rangle. - In this instance, - we have \vec r(u,v) = \langle u,v,u^2+2v^2\rangle, - for -3\leq u\leq 3, -1\leq v\leq 1. - This surface is graphed in . -

    - -
    - The surface parametrized in - - A portion of an elliptic paraboloid with a rectangular domain. - -

    - A set of three-dimensional coordinate axes are shown, with the z axis pointing up. - In the xy plane a shaded rectangle is plotted; this represents the parameter domain. - Sitting above the rectangle is a portion of an elliptic paraboloid, opening upward. - The appearance of the surface is similar to that of a hammock or sling. -

    -     - - - - - //ASY file for figparsurf1_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11.9,9.6,17.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-3,3}; - real[] myychoice={-3,3}; - real[] myzchoice={5,10}; - defaultpen(0.5mm); - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-1,12); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the top half of the surface z^2 = x^2+2y^2 - triple f(pair t) { - return (t.x,t.y,t.x^2+2*t.y^2);// - } - surface s=surface(f,(-3,-1),(3,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //triple g(real t) {return(cos(t),sin(t),t/(2*pi));} - //path3 mypath=graph(g,0,2pi,operator ..); - draw(surface((3,1,0)--(3,-1,0)--(-3,-1,0)--(-3,1,0)--cycle),curvepen+opacity(.5)); - draw((3,1,0)--(3,-1,0)--(-3,-1,0)--(-3,1,0)--cycle,curvepen); - - - -
    -     -     - - - Parametrizing a surface over a circular disk - -

    - Parametrize the surface z=x^2+2y^2 over the circular region R enclosed by the circle of radius 2 that is centered at the origin. -

    -     - -

    - We can parametrize the circular boundary of R with the vector-valued function \la 2\cos(u),2\sin(u)\ra, - where 0\leq u\leq 2\pi. - We can obtain the interior of R by scaling this function by a variable amount, - , by multiplying by v: - \la 2v\cos(u),2v\sin(u)\ra, - where 0\leq v\leq 1. -

    - -

    - It is important to understand the role of v in the above function. - When v=1, - we get the boundary of R, a circle of radius 2. - When v=0, we simply get the point (0,0), - the center of R - (which can be thought of as a circle with radius of 0). - When v=1/2, - we get the circle of radius 1 that is centered at the origin, - which is the circle halfway - between the boundary and the center. - As v varies from 0 to 1, we create a series of concentric circles that fill out all of R. -

    - -
    - The surface parametrized in - - An elliptic paraboloid plotted over a circular domain. - -

    - A set of three-dimensional coordinate axes are shown, with the z axis pointing up. - In the xy plane a shaded circle is plotted, with its center at the origin; this represents the parameter domain. - Sitting above the circle is a portion of an elliptic paraboloid, opening upward. - Because the domain is circular, the surface appears more bowl-shaped than in the previous example. -

    -     - - - - - //ASY file for figparsurf1_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11.9,9.6,17.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-3,3}; - real[] myychoice={-3,3}; - real[] myzchoice={5,10}; - defaultpen(0.5mm); - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-1,12); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - - //Draw the top half of the surface z^2 = x^2+2y^2 - triple f(pair t) { - return (2*t.y*cos(t.x),2*t.y*sin(t.x),(2*t.y*cos(t.x))^2+2*(2*t.y*sin(t.x))^2);// - } - surface s=surface(f,(0,0),(2pi,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - triple g(real t) { - return (2*cos(t),2*sin(t),0); - } - - //triple g(real t) {return(cos(t),sin(t),t/(2*pi));} - path3 mypath=graph(g,0,2pi,operator ..); - draw(surface(mypath--cycle),curvepen+opacity(.5)); - draw(mypath,curvepen); - - - - -
    - -

    - Thus far, we have determined the x and y components of our parametrization of the surface: - x=2v\cos(u) and y=2v\sin(u). - We find the z component simply by using z = f(x,y) = x^2+2y^2: - - z = (2v\cos(u))^2+2(2v\sin(u))^2 = 4v^2\cos^2u+8v^2\sin^2u - . -

    - -

    - Thus \vec r(u,v) = \langle 2v\cos(u),2v\sin(u),4v^2\cos^2u+8v^2\sin^2u\rangle, - 0\leq u\leq 2\pi, 0\leq v\leq 1, - which is graphed in . - The way that this graphic was generated highlights how the surface was parametrized. - When viewing from above, one can see lines emanating from the origin; - they represent different values of u as u sweeps from an angle of 0 up to 2\pi. - One can also see concentric circles, - each corresponding to a different value of v. -

    -     -     - -

    - Examples - and - demonstrate an important principle when parametrizing surfaces given in the form - z=f(x,y) over a region R: - if one can determine x and y in terms of u and v, - then z follows directly as z=f(x,y). -

    - -

    - In the following two examples, - we parametrize the same surface over triangular regions. - Each will use v as a scaling factor - as done in . -

    - - - Parametrizing a surface over a triangle - -

    - Parametrize the surface z=x^2+2y^2 over the triangular region R enclosed by the coordinate axes and the line y=2-2x/3, - as shown in . -

    -
    - Part (a) shows a graph of the region R, and part (b) shows the surface over R, as defined in - -
    - - - A triangular domain is plotted in the first quadrant of the plane. - -

    - The first quadrant in the plane is shown, with the x axis at the bottom of the image, and the y axis to the left. - A triangular region is shaded and labeled as R. - It is a right triangle, with its base along the x axis, from (0,0) to (3,0), - and another side along the y axis, from (0,0) to (0,2). -

    - -

    - The hypotenuse, from (0,2) to (3,0), is labeled with the equation y=2-2x/3. - A dashed line is also drawn through the region, from (0,1) to (3,0). - This illustrates the parameter value v=\frac12. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,2,3}, - ytick={1,2,3}, - ymin=-.5,ymax=3.5, - xmin=-.5,xmax=3.5, - ] - - \filldraw [thick,draw=firstcolor,fill=firstcolor!15] (axis cs: 0,0) -- (axis cs: 0,2) -- (axis cs: 3,0) node [pos=.5,sloped,above] { $y = 2-2x/3$} -- cycle; - - \draw [thick,dashed,draw=secondcolor] (axis cs: 3,0) -- (axis cs: 0,1); - \draw (axis cs: .5,.5) node {$R$}; - - \end{axis} - - \end{tikzpicture} - - - -
    - -
    - - - An elliptic paraboloid is plotted in space over a triangular domain. - -

    - A set of three-dimensional coordinate axes are drawn, with the z axis pointing up. - The surface z=x^2+2y^2 is plotted over the triangular domain illustrated in . - This is the same surface as the previous two examples, plotted again with a differently-shaped domain. - The domain, as usual, is rendered in the xy plane. -

    - -

    - The shape of the surface is hard to tell from the default perspective of the image. - Rotating the image, so that the viewpoint is from above, - shows that the surface has a cusp at the origin, which is one corner of the domain, - and the surface rises up from there into the first octant. -

    - -

    - The sides of the domain along the x and y axes lead to parabolic curves in the xz and yz planes. - These curves rise to cusps at the other corners of the domain, - and they are joined by a parabolic curve in space that corresponds to the hypotenuse of the triangle. -

    -     - - - - - //ASY file for figparsurf1_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11.9,9.6,17.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-3,3}; - real[] myychoice={-3,3}; - real[] myzchoice={5,10}; - defaultpen(0.5mm); - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-1,12); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the top half of the surface z^2 = x^2+2y^2 - triple f(pair t) { - return (t.x,t.y*(2-.666*t.x),(t.x)^2+2*(t.y*(2-.666*t.x))^2);// - } - surface s=surface(f,(0,0),(3,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //triple g(real t) {return(cos(t),sin(t),t/(2*pi));} - //path3 mypath=graph(g,0,2pi,operator ..); - draw(surface((0,0,0)--(0,2,0)--(3,0,0)--cycle),curvepen+opacity(.5)); - draw((0,0,0)--(0,2,0)--(3,0,0)--cycle,curvepen); - - - -
    -     -
    -     - -

    - We may begin by letting x=u, - 0\leq u\leq 3, and y = 2-2u/3. - This gives only the line on the - upper side of the triangle. - To get all of the region R, - we can once again scale y by a variable factor, - v. -

    - -

    - Still letting x = u, - 0\leq u\leq 3, we let - y = v(2-2u/3), 0\leq v\leq 1. - When v=0, - all y-values are 0, and we get the portion of the x-axis between x=0 and x=3. - When v=1, we get the upper side of the triangle. - When v=1/2, we get the line y=1/2(2-2u/3) = 1-u/3, - which is the line halfway up the triangle, - shown in the figure with a dashed line. -

    - -

    - Letting z = f(x,y) = x^2+2y^2, - we have \vec r(u,v) = \langle u, v(2-2u/3), - u^2+2\big(v(2-2u/3)\big)^2\rangle, - 0\leq u\leq 3, 0\leq v\leq 1. - This surface is graphed in . - Again, when one looks from above, - we can see the scaling effects of v: - the series of lines that run to the point (3,0) each represent a different value of v. -

    - -

    - Another common way to parametrize the surface is to begin with y=u, - 0\leq u\leq 2. - Solving the equation of the line y=2-2x/3 for x, - we have x = 3-3y/2, - leading to using x=v(3-3u/2), 0\leq v\leq 1. - With z=x^2+2y^2, - we have \vec r(u,v) = \langle v(3-3u/2),u, \big(v(3-3u/2)\big)^2+2v^2\rangle, - 0\leq u\leq 2, 0\leq v\leq 1. -

    -     -     - - - Parametrizing a surface over a triangle - -

    - Parametrize the surface z=x^2+2y^2 over the triangular region R enclosed by the lines y=3-2x/3, - y=1 and x=0 as shown in . -

    -
    - Part (a) shows a graph of the region R, and part (b) shows the surface over R, as defined in - -
    - - - A triangular domain in the plane, with vertices at (0,1), (0,3), and (3,1). - -

    - A set of two-dimensional coordinate axes is shown, with the origin in the bottom-left of the image. - A region R is plotted as a shaded rectangle. - The rectangle has vertices (0,1), (3,1), and (0,3). - The hypotenuse of the triangle is labeled with the equation y=3-2x/3. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick={1,2,3}, - ytick={1,2,3}, - ymin=-.5,ymax=3.5, - xmin=-.5,xmax=3.5, - ] - - \filldraw [thick,draw=firstcolor,fill=firstcolor!15] (axis cs: 0,1) -- (axis cs: 0,3) -- (axis cs: 3,1) node [pos=.5,sloped,above] { $y = 3-2x/3$} -- cycle; - - \draw (axis cs: 1,1.75) node {$R$}; - - \end{axis} - \end{tikzpicture} - - - -
    - -
    - - - An elliptic paraboloid is plotted in space over a triangular domain. - -

    - A set of three-dimensional coordinate axes are drawn, with the z axis pointing up. - The surface z=x^2+2y^2 is plotted over the triangular domain illustrated in . - This is the same surface as the previous three examples, plotted again with a different domain. - The domain, as usual, is rendered in the xy plane. - The image is very similar to the one in , - with the bottom cusp on the surface at the point (0,1,2) instead of the origin. -

    - -

    - In the default view, the surface looks something like a triangular tortilla chip. -

    - -

    - The sides of the domain along the x and y axes lead to parabolic curves in the planes x=0 and y=1. - These curves rise to cusps at the other corners of the domain, - and they are joined by a parabolic curve in space that corresponds to the hypotenuse of the triangle. -

    -     - - - - - //ASY file for figparsurf4_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11.9,9.6,17.8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-3,3}; - real[] myychoice={-3,3}; - real[] myzchoice={5,10}; - defaultpen(0.5mm); - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-1,12); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the top half of the surface z^2 = x^2+2y^2 - triple f(pair t) { - return (t.x,1+t.y*(2-.666*t.x),(t.x)^2+2*(1+t.y*(2-.666*t.x))^2);// - } - surface s=surface(f,(0,0),(3,1),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //triple g(real t) {return(cos(t),sin(t),t/(2*pi));} - //path3 mypath=graph(g,0,2pi,operator ..); - draw(surface((0,1,0)--(0,3,0)--(3,1,0)--cycle),curvepen+opacity(.5)); - draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - - -
    -     - -
    -     - -

    - While the region R in this example is very similar to the region R in the previous example, - and our method of parametrizing the surface is fundamentally the same, - it will feel as though our answer is much different than before. -

    - -

    - We begin with letting x=u, 0\leq u\leq 3. - We may be tempted to let y = v(3-2u/3), - 0\leq v\leq 1, but this is incorrect. - When v = 1, - we obtain the upper line of the triangle as desired. - However, when v=0, - the y-value is 0, which does not lie in the region R. -

    - -

    - We will describe the general method of proceeding following this example. - For now, consider y = 1+v(2-2u/3), 0\leq v\leq 1. - Note that when v=1, - we have y=3-2u/3, the upper line of the boundary of R. - Also, when v=0, we have y=1, - which is the lower boundary of R. - With z=x^2+2y^2, - we determine \vec r(u,v) = \langle u, 1+v(2-2u/3), - u^2+2\big(1+v(2-2u/3)\big)^2\rangle, - 0\leq u\leq 3, 0\leq v\leq 1. -

    - -

    - The surface is graphed in . -

    -     -     - -

    - Given a surface of the form z=f(x,y), - one can often determine a parametrization of the surface over a region R in a manner similar to determining bounds of integration over a region R. - Using the techniques of , - suppose a region R can be described by a\leq x\leq b, - g_1(x) \leq y\leq g_2(x), - , the area of R can be found using the iterated integral - - \int_a^b\int_{g_1(x)}^{g_2(x)}\, dy\, dx - . -

    - -

    - When parametrizing the surface, - we can let x=u, a\leq u\leq b, - and we can let y = g_1(u)+v\big(g_2(u)-g_1(u)\big), - 0\leq v\leq 1. - The parametrization of x is straightforward, - but look closely at how y is determined. - When v=0, y=g_1(u) = g_1(x). - When v=1, y= g_2(u)=g_2(x). -

    - -

    - As a specific example, - consider the triangular region R from , - shown in . - Using the techniques of , - we can find the area of R as - - \int_0^3\int_1^{3-2x/3} dy\, dx - . -

    - -

    - Following the above discussion, - we can set x=u, where 0\leq u\leq 3, - and set y = 1+ v\big(3-2u/3-1\big) = 1+v(2-2u/3), - 0\leq v\leq 1, as used in that example. -

    - -

    - One can do a similar thing if R is bounded by c\leq y\leq d, - h_1(y)\leq x\leq h_2(y), - but for the sake of simplicity we leave it to the reader to flesh out those details. - The principles outlined above are given in the following Key Idea for reference. -

    - - - Parametrizing Surfaces -

    - Let a surface \surfaceS be the graph of a function f(x,y), - where the domain of f is a closed, - bounded region R in the xy-plane. - Let R be bounded by a\leq x\leq b, - g_1(x)\leq y\leq g_2(x), - , the area of R can be found using the iterated integral \int_a^b\int_{g_1(x)}^{g_2(x)}\, dy\, dx, - and let h(u,v) = g_1(u)+v\big(g_2(u)-g_1(u)\big). -

    - -

    - \surfaceS can be parametrized as - - \vec r(u,v) = \la u, h(u,v), f\big(u,h(u,v)\big)\ra, a\leq u\leq b,\ 0\leq v\leq 1 - . -

    -     - - - Parametrizing a cylindrical surface - -

    - Find a parametrization of the cylinder x^2 + z^2/4=1, - where -1\leq y\leq 2, - as shown in . -

    -
    - The cylinder parametrized in - - A cylindrical surface centered along the y axis. - -

    - The surface is an elliptical cylinder. It is plotted relative to a set of three-dimensional coordinate axes, - with the y axis running down the center of the cylinder. -

    -     - - - - - //ASY file for figparsurf5_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(20.4,10,4.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-2.5,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the cylinder x^2+z^2/4=1 - triple f(pair t) { - return (cos(t.x),t.y,2sin(t.x));// - } - surface s=surface(f,(0,-1),(2pi,2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return(cos(t),-1,2sin(t));} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,curvepen); - - triple g(real t) {return(cos(t),2,2sin(t));} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,curvepen); - - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - - -
    -     - -

    - The equation x^2+z^2/4=1 can be envisioned to describe an ellipse in the xz-plane; - as the equation lacks a y-term, - the equation describes a cylinder - (recall ) - that extends without bound parallel to the y-axis. - This ellipse has a vertical major axis of length 4, a horizontal minor axis of length 2, and is centered at the origin. - We can parametrize this ellipse using sines and cosines; - our parametrization can begin with - - \vec r(u,v) = \la \cos(u), \text{ ??? } , 2\sin(u)\ra, 0\leq u\leq 2\pi - , - where we still need to determine the y component. -

    - -

    - While the cylinder x^2+z^2/4=1 is satisfied by any y value, - the problem states that all y values are to be between y=-1 and y=2. - Since the value of y does not depend at all on the values of x or z, - we can use another variable, v, to describe y. - Our final answer is - - \vec r(u,v) = \la \cos(u), v, 2\sin(u)\ra, 0\leq u\leq 2\pi, -1\leq v\leq 2 - . -

    -     -     - - - Parametrizing an elliptic cone - -

    - Find a parametrization of the elliptic cone z^2 = \frac{x^2}{4}+\frac{y^2}{9}, - where -2\leq z\leq 3, - as shown in . -

    -
    - The elliptic cone as described in - - An elliptic cone with its cusp at the origin, including portions above and below the x,y plane. - -

    - An elliptic cone is plotted relative to three-dimensional coordinate axes. - The cone is centered along the z axis, and has its cusp at the origin. - Portions of both halves of the cone (both above and below the xy plane) are included. -

    -     - - - - - //ASY file for figparsurf6a_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(20.4,10,4.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-10,-5,5,10}; - real[] myychoice={-10,-5,5,10}; - real[] myzchoice={-3,3}; - defaultpen(0.5mm); - - pair xbounds=(-10.5,10.5); - pair ybounds=(-10.5,10.5); - pair zbounds=(-3.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the cone z^2=x^2/4+y^2/9 - triple f(pair t) { - return (2*t.y*cos(t.x),3*t.y*sin(t.x),t.y);// - } - surface s=surface(f,(0,-2),(2pi,3),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return(-4cos(t),-6sin(t),-2);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,curvepen); - - triple g(real t) {return(6cos(t),9sin(t),3);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,curvepen); - - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - - -
    -     - -

    - One way to parametrize this cone is to recognize that given a z value, - the cross section of the cone at that z value is an ellipse with equation \frac{x^2}{(2z)^2} + \frac{y^2}{(3z)^2}=1. - We can let z=v, for - -2\leq v\leq 3 and then parametrize the above ellipses using sines, - cosines and v. -

    - -

    - We can parametrize the x component of our surface with - x=2z\cos(u) and the y component with y=3z\sin(u), - where 0\leq u\leq 2\pi. - Putting all components together, we have - - \vec r(u,v) = \la 2v\cos(u), 3v\sin(u), v\ra, 0\leq u\leq 2\pi, -2\leq v\leq 3 - . -

    - -

    - When v takes on negative values, - the radii of the cross-sectional ellipses become negative, - which can lead to some surprising results. - Consider , - where the cone is graphed for 0\leq u\leq \pi. - Because v is negative below the xy-plane, - the radii of the cross-sectional ellipses are negative, - and the opposite side of the cone is sketched below the xy-plane. -

    - -
    - The elliptic cone as described in with restricted domain - - A cone plotted parametrically with restricted domain. - -

    - A plot of part of the cone illustrated in . - This image shows the effect of both the restricted domain, - and the fact that the cross-sectional ellipses have negative radius when z\lt 0. -

    - -

    - With 0\leq u\leq \pi, the result is that above the xy plane, - we get the half of the cone where y\geq 0, - while below the xy plane, we get the half of the cone where y\leq 0. -

    -     - - - - - //ASY file for figparsurf6a_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(20.4,10,4.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-10,-5,5,10}; - real[] myychoice={-10,-5,5,10}; - real[] myzchoice={-3,3}; - defaultpen(0.5mm); - - pair xbounds=(-10.5,10.5); - pair ybounds=(-10.5,10.5); - pair zbounds=(-3.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the cone z^2=x^2/4+y^2/9 - triple f(pair t) { - return (2*t.y*cos(t.x),3*t.y*sin(t.x),t.y);// - } - surface s=surface(f,(0,-2),(pi,3),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return(-4cos(t),-6sin(t),-2);} - path3 mypath=graph(g,0,pi,operator ..); - draw(mypath,curvepen); - - triple g(real t) {return(6cos(t),9sin(t),3);} - path3 mypath=graph(g,0,pi,operator ..); - draw(mypath,curvepen); - - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - - -
    -     -     - - - Parametrizing an ellipsoid - -

    - Find a parametrization of the ellipsoid - \frac{x^2}{25}+y^2+\frac{z^2}{4}=1 as shown in . -

    -
    - An ellipsoid in (a), drawn again in (b) with its domain restricted, as described in - -
    - - - The ellipsoid to be parametrized in this example. - -

    - A set of three-dimensional coordinate axes are shown, with the origin in the center of the image. - The ellipsoid has the shape of a smooth, rounded pebble. - It is narrowest in the y direction, and longest in the x direction. -

    -     - - - - - //ASY file for figparsurf7a_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(20.4,10,4.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-5,5}; - real[] myychoice={-1,1}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - - pair xbounds=(-6,6); - pair ybounds=(-3,3); - pair zbounds=(-3,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the ellipoid x^2/25+y^2+z^2/4=1 - triple f(pair t) { - return (5*cos(t.y)*sin(t.x),sin(t.y)*sin(t.x),2cos(t.x));// - } - surface s=surface(f,(0,0),(pi,2pi),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - - -
    - -
    - - - A portion of an ellipsoid in space corresponding to a restricted parameter domain. - -

    - A cutout of part of the ellipsoid in . - It is a U-shaped surface obtained by cutting off the top and bottom of the ellipsoid, - as well as a good part of the end that meets the positive x axis. -

    -     - - - - - //ASY file for figparsurf7b_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(20.4,10,4.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-5,5}; - real[] myychoice={-1,1}; - real[] myzchoice={-2,2}; - defaultpen(0.5mm); - - pair xbounds=(-6,6); - pair ybounds=(-3,3); - pair zbounds=(-3,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the ellipoid x^2/25+y^2+z^2/4=1, restricted domain - triple f(pair t) { - return (5*cos(t.y)*sin(t.x),sin(t.y)*sin(t.x),2cos(t.x));// - } - surface s=surface(f,(pi/4,pi/4),(2pi/3,3pi/2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - - -
    -     - -
    -     - -

    - Recall - from , - which states that all unit vectors in space have the form - \langle \sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta\rangle for some angles \theta and \varphi. - If we choose our angles appropriately, - this allows us to draw the unit sphere. - To get an ellipsoid, - we need only scale each component of the sphere appropriately. -

    - -

    - The x-radius of the given ellipsoid is 5, the y-radius is 1 and the z-radius is 2. - Substituting u for \theta and v for \varphi, - we have - - \vec r(u,v) = \langle 5\sin(u)\cos(v), \sin(u)\sin(v),2\cos(u)\rangle - , - where we still need to determine the ranges of u and v. -

    - -

    - Note how the x and y components of \vec r have \cos(v) and \sin(v) terms, - respectively. - This hints at the fact that ellipses are drawn parallel to the xy-plane as v varies, - which implies we should have v range from 0 to 2\pi. -

    - -

    - One may be tempted to let 0\leq u\leq 2\pi as well, - but note how the z component is 2\cos(u). - We only need \cos(u) to take on values between -1 and 1 once, - therefore we can restrict u to 0\leq u\leq \pi. -

    - -

    - The final parametrization is thus - - \vec r(u,v) = \langle 5\sin(u)\cos(v), \sin(u)\sin(v),2\cos(u)\rangle, 0\leq u\leq\pi, 0\leq v\leq 2\pi - . -

    - -

    - In , - the ellipsoid is graphed on \frac{\pi}{4}\leq u\leq \frac{2\pi}{3}, - \frac{\pi}4\leq v\leq \frac{3\pi}2 to demonstrate how each variable affects the surface. -

    -     -     - -

    - Parametrization is a powerful way to represent surfaces. - One of the advantages of the methods of parametrization described in this section is that the domain of - \vec r(u,v) is always a rectangle; - that is, the bounds on u and v are constants. - This will make some of our future computations easier to evaluate. -

    - -

    - Just as we could parametrize curves in more than one way, - there will always be multiple ways to parametrize a surface. - Some ways will be more natural than others, - but these other ways are not incorrect. - Because technology is often readily available, - it is often a good idea to check one's work by graphing a parametrization of a surface to check if it indeed represents what it was intended to. -

    -     - - - Surface Area - -

    - It will become important in the following sections to be able to compute the surface area of a surface \surfaceS given a smooth parametrization \vec r(u,v), - a\leq u\leq b, c\leq v\leq d. - Following the principles given in the integration review at the beginning of this chapter, - we can say thatsurface areaof parametrized surface - - \text{ Surface Area of }\surfaceS\, =S = \iint_{\surfaceS}\, dS - , - where dS represents a small amount of surface area. - That is, to compute total surface area S, - add up lots of small amounts of surface area dS across the entire surface \surfaceS. The key to finding surface area is knowing how to compute dS. - We begin by approximating. -

    - -

    - In - we used the area of a plane to approximate the surface area of a small portion of a surface. - We will do the same here. -

    - -

    - Let R be the region of the u-v plane bounded by a\leq u\leq b, - c\leq v\leq d as shown in . - Partition R into rectangles of width - \Delta u = \frac{b-a}n and height \Delta v = \frac{d-c}n, - for some n. - Let p=(u_0,v_0) be the lower left corner of some rectangle in the partition, - and let m and q be neighboring corners as shown. -

    - -

    - The point p maps to a point - P = \vec r(u_0,v_0) on the surface \surfaceS, and the rectangle with corners p, - m and q maps to some region - (probably not rectangular) - on the surface as shown in , - where M = \vec r(m) and Q = \vec r(q). - We wish to approximate the surface area of this mapped region. -

    - -

    - Let \vec u = M-P and \vec v = Q-P. - These two vectors form a parallelogram, - illustrated in , - whose area approximates - the surface area we seek. - In this particular illustration, - we can see that parallelogram does not particularly match well the region we wish to approximate, - but that is acceptable; - by increasing the number of partitions of R, - \Delta u and \Delta v shrink and our approximations will become better. -

    - -
    - Illustrating the process of finding surface area by approximating with planes - -
    - - - A rectangular region in the plane, with a smaller subrectangle highlighted in its interior. - -

    - A rectangle R is drawn and shaded in the first quadrant of the uv plane. - Within the rectangle, points p, q, and m are marked. - These points make up three of the four corners of a smaller subrectangle inside of R. -

    - -

    - Along the u axis (the horizontal axis), there are points marked a and b corresponding to the left and right edges of R. - Also marked is a point u_0, which is the u coordinate of both p and q, - and a point u_0+\Delta u, which is the u coordinate of m. -

    - -

    - Along the v axis (the vertical axis), points c and d are marked to indicate the top and bottom of R. - Another point is marked with the value v_0; this is the v coordinate of both p and m. - Also marked is a point with the value v_0+\Delta v, which is the v coordinate of q. -

    - -

    - A smaller subrectangle is drawn inside of R. - It has its bottom left vertex at p, with coordinates (u_0,v_0), - and sides of length \Delta u and \Delta v. -

    -     - - - \begin{tikzpicture} - - \begin{axis}[ - xtick=\empty, - ytick=\empty, - extra x ticks={.5,2.25,1,1.5}, - extra x tick labels={$a$,$b$,$u_0$,$u_0+\Delta u$}, - extra y ticks={.1,2,.75,1}, - extra y tick labels={$c$,$d$,$v_0$,$v_0+\Delta v$}, - ymin=-.1,ymax=2.5, - xmin=-.1,xmax=2.5, - xlabel={$u$}, - ylabel={$v$} - ] - - \filldraw [thick,draw=firstcolor,fill=firstcolor!15] (axis cs: .5,.1) -- (axis cs: .5,2) -- (axis cs: 2.25,2) -- (axis cs: 2.25,.1) -- cycle; - - \coordinate (P) at (axis cs: 1,.75); - \coordinate (Q) at (axis cs: 1.5,.75); - \coordinate (R) at (axis cs: 1,1); - \coordinate (d) at (axis cs: 1.5,1); - - \draw [thick,draw=secondcolor,fill=secondcolor!15] (P) -- (Q) -- (d) -- (R) -- cycle; - - \filldraw (P) circle (1pt) node [below left] {$p$}; - \filldraw (R) circle (1pt) node [above left] {$q$}; - \filldraw (Q) circle (1pt) node [below right] {$m$}; - - \draw (axis cs: .5,2) node [below right] {$R$}; - - \end{axis} - - - \end{tikzpicture} - - - -
    - -
    - - - A surface in space with parameter domain R, highlighting the portion corresponding to a particular subrectangle. - -

    - A surface is drawn in space, relative to a set of three-dimensional coordinate axes. - The precise shape of the surface is unimportant, - but it appears to be a portion of an elliptic paraboloid, - corresponding to the rectangular parameter domain R in . -

    - -

    - There are points on the surface marked P, Q, and M, - corresponding to the points p, q, and m in the parameter domain. - There is also a shaded patch on the surface, representing the portion that corresponds to the subrectangle - drawn in . -

    - -

    - The sides of this patch are curved, since the surface is, but the patch is approximately rectangular. - To emphasize this point, the image also shows the vectors \overrightarrow{PQ} (from P to Q) - and \overrightarrow{PM} (from P to M). - These vectors are straight line approximations of the corresponding sides of the patch on the surface. -

    -     - - - - - //ASY file for figparsurfareaA_3D.asy in Chapter 14, developing surface area - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(15.5,18.6,18.1,true,true); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-.1,1.5); - pair ybounds=(-.1,1.5); - pair zbounds=(-.1,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the top half of the surface z = x^2+2y^2 - triple ff(pair t) { - return (t.y*(cos(t.x)+.1*sin(3t.x)),t.y*sin(t.x),2-(t.y*(cos(t.x)+.1*sin(3t.x)))^2-(t.y*sin(t.x))^2);// - } - - triple f(pair t) { - return -ff(t)+(1,1,3); - } - - surface s=surface(f,(0,0),(2pi/4,1),4,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - surface s=surface(f,(2pi/16,9/16),(4pi/16,11/16),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen2,meshpen=invisible); - - real u0=2pi/16; - real v0=9/16; - real du=2pi/16; - real dv=2/16; - triple ru=f((u0+du,v0))-f((u0,v0)); - triple rv=f((u0+du,v0))-f((u0,v0+dv)); - - draw(f((u0+du,v0))--f((u0,v0)),curvepen2,Arrow3(5)); - draw(f((u0+du,v0))--f((u0+du,v0+dv)),curvepen2,Arrow3(5)); - - label("$P$",f((u0+du,v0))+(.05,-.05,.1)); - label("$M$",f((u0,v0))+(0,+.1,.1)); - label("$Q$",f((u0+du,v0+dv))+(.05,-.05,.1)); - - dot(f((u0+du,v0)),.7mm+black); - dot(f((u0,v0)),.7mm+black); - dot(f((u0+du,v0+dv)),.7mm+black); - - -
    - -
    - - - A zoomed in view of the parallelogram approximation of a small patch on a surface. - -

    - This is a zoomed-in view of the patch illustrated in . - We see how the vectors \vec{u} = \overrightarrow{PQ} and \vec{v} = \overrightarrow{PM} - form two sides of a parallelogram with one vertex at P. -

    - -

    - This parallelogram overlaps with most (but not all) of the patch on the surface given by - u_0\leq u\leq u_0+\Delta u and v_0\leq v\leq v_0+\Delta v, - illustrating how the area of the parallelogram (which we know how to compute) - is a good approximation of the area of the patch on the surface. -

    -     - - - - - //ASY file for figparsurfareaA_3D.asy in Chapter 14, developing surface area - - size(198,198,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(15.5,18.6,18.1,Z,(0,0,0),1,(0,-.15),true,true); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(.3,.8); - pair ybounds=(.3,.8); - pair zbounds=(.5,1.5); - - //xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - //yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - //zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - //label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - //label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - //label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the top half of the surface z = x^2+2y^2 - triple ff(pair t) { - return (t.y*(cos(t.x)+.1*sin(3t.x)),t.y*sin(t.x),2-(t.y*(cos(t.x)+.1*sin(3t.x)))^2-(t.y*sin(t.x))^2);// - } - - triple f(pair t) { - return -ff(t)+(1,1,3); - } - - surface s=surface(f,(pi/16,8/16),(2pi/4*3/4*10/12,12/16),8,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - surface s=surface(f,(2pi/16,9/16),(4pi/16,11/16),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - //draw(s,rgb(1,.5,.5)+opacity(.7),meshpen=invisible); - - real u0=2pi/16; - real v0=9/16; - real du=2pi/16; - real dv=2/16; - triple ru=f((u0+du,v0))-f((u0,v0)); - triple rv=f((u0+du,v0))-f((u0+du,v0+dv)); - - triple aa=f((u0+du,v0)); - triple bb=aa-ru; - triple cc=bb-rv; - triple dd=cc+ru; - - draw(f((u0+du,v0))--f((u0,v0)),curvepen2,Arrow3(5)); - draw(f((u0+du,v0))--f((u0+du,v0+dv)),curvepen2,Arrow3(5)); - draw(f((u0+du,v0+dv))--f((u0+du,v0+dv))-ru--f((u0+du,v0+dv))-ru+rv,curvepen2); - - label("$P$",f((u0+du,v0))+(.005,-.005,.05)); - label("$M$",f((u0,v0))+(0,+.025,.005)); - label("$Q$",f((u0+du,v0+dv))+(.005,-.005,.05)); - - label("$\vec u$",(f((u0+du,v0))+f((u0,v0)))/2+(.05,.005,0)); - label("$\vec v$",(f((u0+du,v0))+f((u0+du,v0+dv)))/2+(.005,-.005,.05)); - - dot(f((u0+du,v0)),.7mm+black); - dot(f((u0,v0)),.7mm+black); - dot(f((u0+du,v0+dv)),.7mm+black); - - draw(surface(aa--bb--cc--dd--cycle),surfacepen2); - - //draw(surface(f((u0,v0))--f((u0+du,v0))--f((u0+du,v0+dv))--f((u0,v0+dv))--cycle),surfacepen2); - - dot(f((u0+du,v0)),.7mm+black); - - surface s=surface(f,(2pi/16,9/16),(4pi/16,9/16),8,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen2,meshpen=curvepen2); - - surface s=surface(f,(2pi/16,11/16),(4pi/16,11/16),8,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen2,meshpen=curvepen2); - - surface s=surface(f,(2pi/16,9/16),(2pi/16,11/16),1,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen2,meshpen=curvepen2); - - surface s=surface(f,(4pi/16,9/16),(4pi/16,11/16),1,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen2,meshpen=curvepen2); - - -
    -     - -
    - -

    - From - we know the area of this parallelogram is \snorm{\vec u\times \vec v}. - If we repeat this approximation process for each rectangle in the partition of R, - we can sum the areas of all the parallelograms to get an approximation of the surface area S: - - \text{ Surface area of } \surfaceS \, =S \approx \sum_{j=1}^n\sum_{i=1}^n \snorm{\vec u_{i,j}\times \vec v_{i,j}} - , - where \vec u_{i,j} = \vec r(u_i+\Delta u,v_j) - \vec r(u_i,v_j) and \vec v_{i,j} = \vec r(u_i,v_j+\Delta v)-\vec r(u_i,v_j). -

    - -

    - From our previous calculus experience, - we expect that taking a limit as - n\to \infty will result in the exact surface area. - However, the current form of the above double sum makes it difficult to realize what the result of that limit is. - The following rewriting of the double summation will be helpful: - - \amp \sum_{j=1}^n\sum_{i=1}^n \snorm{\vec u_{i,j}\times \vec v_{i,j}}= - \amp \sum_{j=1}^n\sum_{i=1}^n \snorm{\big(\vec r(u_i+\Delta u,v_j) - \vec r(u_i,v_j)\big) \times \big(\vec r(u_i,v_j+\Delta v)-\vec r(u_i,v_j)\big)}= - \amp \sum_{j=1}^n\sum_{i=1}^n \snorm{\frac{\vec r(u_i+\Delta u,v_j) - \vec r(u_i,v_j)}{\Delta u} \times \frac{\vec r(u_i,v_j+\Delta v)-\vec r(u_i,v_j)}{\Delta v}}\Delta u\Delta v - . -

    - -

    - We now take the limit as n\to\infty, - forcing \Delta u and \Delta v to 0. - As \Delta u\to 0, - - \frac{\vec r(u_i+\Delta u,v_j) - \vec r(u_i,v_j)}{\Delta u} \to \vec r_u(u_i,v_j) \text{ and } - - - \frac{\vec r(u_i,v_j+\Delta v)-\vec r(u_i,v_j)}{\Delta v} \to \vec r_v(u_i,v_j) - . -

    - -

    - (This limit process also demonstrates that \vec r_u(u,v) and - \vec r_v(u,v) are tangent to the surface \surfaceS at \vec r(u,v). - We don't need this fact now, but it will be important in the next section.) -

    - -

    - Thus, in the limit, the double sum leads to a double integral: - - \lim_{n\to\infty} \sum_{j=1}^n\sum_{i=1}^n \snorm{\vec u_{i,j}\times \vec v_{i,j}}= \int_c^d\int_a^b \snorm{\vec r_u\times\vec r_v}\, du\, dv - . -

    - - - Surface Area of Parametrically Defined Surfaces - -

    - Let \vec r(u,v) be a smooth parametrization of a surface \surfaceS over a closed, - bounded region R of the u-v plane. - surface areaof parametrized surface -

      -
    • -

      - The surface area differential dS is: - dS = \snorm{\vec r_u\times \vec r_v}\, dA. -

      -
      • - -
    • -

      - The surface area S of \surfaceS is - - S = \iint_\surfaceS\, dS = \iint_R \snorm{\vec r_u\times \vec r_v}\, dA - . -

      -
      • -
    -

    -     -     - - - - - Finding the surface area of a parametrized surface - -

    - Using the parametrization found in , - find the surface area of z=x^2+2y^2 over the circular disk of radius 2, centered at the origin. -

    -     - -

    - In , - we parametrized the surface as \vec r(u,v) = \la 2v\cos(u), 2v\sin(u), 4v^2\cos^2u+8v^2\sin^2u\ra, - for 0\leq u\leq 2\pi, 0\leq v\leq 1. - To find the surface area using , - we need \snorm{\vec r_u\times\vec r_v}. - We find: - - \vec r_u \amp = \la -2v\sin(u), 2v\cos(u), 8v^2\cos(u)\sin(u)\ra - \vec r_v \amp = \la 2\cos(u), 2\sin(v), 8v\cos^2 u+16v\sin^2u\ra - \vec r_u\times\vec r_v \amp = \la 16v^2\cos(u), 32v^2\sin(u), -4v\ra - \snorm{\vec r_u\times\vec r_v} \amp = \sqrt{256v^4\cos^2u+1024v^4\sin^2u+16v^2} - . -

    - -

    - Thus the surface area is - - S = \iint_\surfaceS \, dS \amp = \iint_R\snorm{\vec r_u\times \vec r_v}\, dA - \amp = \int_0^1\int_0^{2\pi} \sqrt{256v^4\cos^2u+1024v^4\sin^2u+16v^2}\, du\, dv - \amp \approx 53.59 - . -

    - -

    - There is a lot of tedious work in the above calculations and the final integral is nontrivial. - The use of a computer-algebra system is highly recommended. -

    -     -     - - - -

    - In , - we recalled the arc length differential ds=\snorm{\vrp(t)}dt. - In subsequent sections, we used that differential, - but in most applications the \snorm{\vrp(t)} - part of the differential canceled out of the integrand - (to our benefit, - as integrating the square roots of functions is generally difficult). - We will find a similar thing happens when we use the surface area differential dS in the following sections. - That is, our main goal is not to be able to compute surface area; - rather, surface area is a tool to obtain other quantities that are more important and useful. - In our applications, we will use dS, - but most of the time the \snorm{\vec r_u\times \vec r_v} - part will cancel out of the integrand, - making the subsequent integration easier to compute. -

    -     - - - - Terms and Concepts - - -

    - In your own words, describe what an orientable surface is. -

    -     - - - -

    - Answers will vary, - though generally should meaningfully include terms like two sided. -

    -     -     - - - -

    - Give an example of a non-orientable surface. -

    -     - - - -

    - Many possible answers exist; - the one given by the book is the Möbius band. -

    -     -     -     - - - Problems - - -

    - Parametrize the surface defined by the function - z=f(x,y) over each of the given regions R of the xy-plane. -

    -     - - - -

    - z = 3x^2y; - -

      -
    1. -

      - R is the rectangle bounded by - -1\leq x\leq 1 and 0\leq y\leq 2. -

      -
      2. - -
    2. -

      - R is the circle of radius 3, centered at (1,2). -

      -
      3. - -
    3. -

      - R is the triangle with vertices (0,0), - (1,0) and (0,2). -

      -
      4. - -
    4. -

      - R is the region bounded by the x-axis and the graph of y = 1-x^2. -

      -
      5. -
    -

    -     - -

    -

      -
    1. -

      - \vec r(u,v) = \langle u, v, 3u^2v\rangle on - -1\leq u\leq 1, 0\leq v\leq 2. -

      -
      2. - -
    2. -

      - \vec r(u,v) = \langle 3v\cos(u)+1, 3v\sin(u)+2, 3(3v\cos(u)+1)^2(3v\sin(u)+2)\rangle, - on 0\leq u\leq 2\pi, 0\leq v\leq 1. -

      -
      3. - -
    3. -

      - \vec r(u,v) = \langle u, v(2-2u), 3u^2v(2-2u)\rangle on 0\leq u, v\leq 1. -

      -
      4. - -
    4. -

      - \vec r(u,v) = \langle u, v(1-u^2), 3u^2v(1-u^2)\rangle on - -1\leq u\leq 1, 0\leq v\leq 1. -

      -
      5. -
    -

    -     -     - - - -

    - z = 4x+2y^2; - -

      -
    1. -

      - R is the rectangle bounded by - 1\leq x\leq 4 and 5\leq y\leq 7. -

      -
      2. - -
    2. -

      - R is the ellipse with major axis of length 8 parallel to the x-axis, - and minor axis of length 6 parallel to the y-axis, - centered at the origin. -

      -
      3. - -
    3. -

      - R is the triangle with vertices (0,0), - (2,2) and (0,4). -

      -
      4. - -
    4. -

      - R is the annulus bounded between the circles, - centered at the origin, with radius 2 and radius 5. -

      -
      5. -
    -

    -     - -

    -

      -
    1. -

      - \vec r(u,v) = \langle u, v, 4u+2u^2\rangle on - 1\leq u\leq 4, 5\leq v\leq 7. -

      -
      2. - -
    2. -

      - \vec r(u,v) = \langle 4v\cos(u), 3v\sin(u), 16v\cos(u)+2(3v\sin(u))^2\rangle, - on 0\leq u\leq 2\pi, 0\leq v\leq 1. -

      -
      3. - -
    3. -

      - \vec r(u,v) = \langle u, u+v(4-2u), 4u+2\big(u+v(4-2u)\big)^2\rangle on - 0\leq u\leq 2, 0\leq v\leq 1. -

      -
      4. - -
    4. -

      - \vec r(u,v) = \langle v\cos(u), v\sin(u), 4v\cos(u) + 2(v\sin(u))^2\rangle on - 0\leq u\leq 2\pi, 2\leq v\leq 5. -

      -
      5. -
    -

    -     -     -     - - - -

    - A surface \surfaceS in space is described that cannot be defined as the graph of a function f(x,y). - Give a parametrization of \surfaceS. -

    -     - - - -

    - \surfaceS is the rectangle in space with corners at (0,0,0), - (0,2,0), - (0,2,1) and (0,0,1). -

    -     - -

    - \vec r(u,v) = \langle 0, u, v\rangle with - 0\leq u\leq 2, 0\leq v\leq 1. -

    -     -     - - - -

    - \surfaceS is the triangle in space with corners at (1,0,0), - (1,0,1) and (0,0,1). -

    -     - -

    - \vec r(u,v) = \langle u, 0, 1-u+vu\rangle with - 0\leq u\leq 1, 0\leq v\leq 1. -

    -     -     - - - -

    - \surfaceS is the ellipsoid \ds\frac{x^2}{9} + \frac{y^2}{4}+\frac{z^2}{16} = 1. -

    -     - -

    - \vec r(u,v) = \langle 3\sin(u)\cos(v), 2\sin(u)\sin(v), 4\cos(u)\rangle with - 0\leq u\leq \pi, 0\leq v\leq 2\pi. -

    -     -     - - - -

    - \surfaceS is the elliptic cone \ds y^2= x^2+\frac{z^2}{16}, - for -1 \leq y \leq 5. -

    -     - -

    - Answers may vary; - one solution is \vec r(u,v) = \langle v\cos(u), v, 4v\sin(u)\rangle with - 0\leq u\leq 2\pi, -1\leq v\leq 5. -

    -     -     -     - - - -

    - A domain D in space is given. - Parametrize each of the bounding surfaces of D. -

    -     - - - -

    - D is the domain bounded by the planes z = \frac12(3-x), - x=1, - y=0, y=2 and z=0. -

    - - - A triangular prism plotted in three dimensions. - -

    - A set of three-dimensional coordinate axes are drawn, with the z axis pointing up. - The domain D for this problem is a triangular prism. - Cross-sections parallel to the xz plane are triangles, - while cross-sections parallel to the other coordinate planes are rectangles. -

    - -

    -

      -
    • -

      - The base of the prism is a square in the xy plane, with 1\leq x\leq 3 and 0\leq y\leq 2. -

      -
      • -
    • -

      - There is a vertical face in the plane x=1 that is also a square, with 0\leq y\leq 2 and 0\leq z\leq 2. -

      -
      • -
    • -

      - Two other vertical faces are triangles. The triangle in the plane y=0 has vertices (1,0,0), - (3,0,0), and (1,0,2). The triangle in the plane y=2 has vertices - (1,2,0), (3,2,0), and (3,2,2). -

      -
      • -
    • -

      - The remaining face corrsponds to the hypotenuse in each triangular cross-section; - it lies in the plane z=\frac12(3-x). -

      -
      • -
    -

    -     - - - - - //ASY file for fig13_06_ex_083D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8.8,7.8,3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3}; - real[] myychoice={1,2,3}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,3.5); - pair ybounds=(-0.5,3.5); - pair zbounds=(-0.5,1.75); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //edges of object - draw((1,0,0)--(1,2,0)--(3,2,0)--(3,0,0)--cycle,bluepen+linewidth(2)); - draw((3,0,0)--(1,0,1)--(1,2,1)--(3,2,0),bluepen+linewidth(2)); - draw((3,0,0)--(1,0,1),bluepen+linewidth(2)); - draw((1,0,0)--(1,0,1),bluepen+linewidth(2)); - draw((1,2,0)--(1,2,1),bluepen+linewidth(2)); - - //shade faces - import three; - path3 p=(1,0,0)--(1,2,0)--(3,2,0)--(3,0,0); - draw(surface(p -- cycle), surfacepen); - path3 p=(1,0,1)--(1,2,1)--(3,2,0)--(3,0,0); - draw(surface(p -- cycle), surfacepen); - path3 p=(1,0,0)--(1,2,0)--(1,2,1)--(1,0,1); - draw(surface(p -- cycle), surfacepen); - path3 p=(1,0,0)--(3,0,0)--(1,0,1); - draw(surface(p -- cycle), surfacepen); - path3 p=(1,2,0)--(3,2,0)--(1,2,1); - draw(surface(p -- cycle), surfacepen); - - //labels and arrow - label("$z=\frac{1}{2}(3-x)$",(3,0,1.1)); - draw((3,0.25,1)--(2,1,0.45),Arrow3(size=2mm)); - - - -     - -

    - Answers may vary. -

    - -

    - For z = \frac12(3-x): - \vec r(u,v) = \langle u, v , \frac12(3-u)\rangle, - with 1\leq u\leq 3 and 0\leq v\leq 2. -

    - -

    - For x=1: \vec r(u,v) = \langle 1,u,v\rangle, - with 0\leq u\leq 2, 0\leq v\leq 1 -

    - -

    - For y=0: \vec r(u,v) = \langle u,0,v/2(3-u)\rangle, - with 1\leq u\leq 3, 0\leq v\leq 1 -

    - -

    - For y=2: \vec r(u,v) = \langle u,2,v/2(3-u)\rangle, - with 1\leq u\leq 3, 0\leq v\leq 1 -

    - -

    - For z=0: \vec r(u,v) = \langle u,v,0\rangle, - with 1\leq u\leq 3, 0\leq v\leq 2 -

    -     -     - - - -

    - D is the domain bounded by the planes z=2x+4y-4, - x=2, y=1 and z=0. -

    - - - A tetrahedron with vertices (2,0,0), (0,1,0), (2,1,0), and (2,1,4). - -

    - A tetrahedron is plotted with respect to three-dimensional coordinate axes. -

      -
    • -

      - One face lies in the xy plane, with vertices (2,0,0), (0,1,0), and (2,1,0). -

      -
      • -
    • -

      - One face lies in the plane x=2, with vertices (2,0,0), (2,1,0), and (2,1,4). -

      -
      • -
    • -

      - Another face lies in the plane y=1, with vertices (0,1,0), (2,1,0), and (2,1,4) -

      -
      • -
    • -

      - The last face lies in the plane z=2x+4y-4. - It meets the xy plane along the line from (2,0,0) to (0,1,0), - and has its remaining vertex at (2,1,4). -

      -
      • -
    -

    -     - - - - - //ASY file for fig14_05_ex_10_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(7.2,-5.1,8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={1,2}; - real[] myzchoice={2,4}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,2.5); - pair ybounds=(-0.25,2.5); - pair zbounds=(-0.25,5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //edges of object - draw((0,1,0)--(2,1,0)--(2,0,0)--cycle,bluepen+linewidth(2)); - draw((0,1,0)--(2,1,4)--(2,1,0)--cycle,bluepen+linewidth(2)); - draw((0,1,0)--(2,1,4)--(2,0,0)--cycle,bluepen+linewidth(2)); - - //shade plane - import three; - path3 p=(0,1,0)--(2,1,4)--(2,0,0); - draw(surface(p -- cycle), surfacepen); - - path3 p=(2,0,0)--(2,1,4)--(2,1,0); - draw(surface(p -- cycle), surfacepen); - - path3 p=(0,1,0)--(2,1,4)--(2,1,0); - draw(surface(p -- cycle), surfacepen); - - path3 p=(0,1,0)--(2,1,0)--(2,0,0); - draw(surface(p -- cycle), surfacepen); - - //label and arrow - label("$z=2x+4y-4$",(1,0.5,4)); - draw((1,0.5,3.6)--(1.2,.65,1.6),Arrow3(size=2mm)); - - - -     - -

    - Answers may vary. -

    - -

    - For z=2x+4y-4: - \vec r(u,v) = \langle u, 1-u/2+uv/2, 2u+4(1-u/2+uv/2)-4\rangle, - with 0\leq u\leq 2, 0\leq v\leq 1. -

    - -

    - For x=2: \vec r(u,v) = \langle 2,u,4uv\rangle, - with 0\leq u\leq 1, 0\leq v\leq 1 -

    - -

    - For y=1: \vec r(u,v) = \langle u,1,2uv\rangle, - with 0\leq u\leq 2, 0\leq v\leq 1 -

    - -

    - For z=0: - \vec r(u,v) = \langle u, 1-u/2+uv/2,0\rangle, - with 0\leq u\leq 2, 0\leq v\leq 1 -

    -     -     - - - -

    - D is the domain bounded by z=2y, - y=4-x^2 and z=0. -

    - - - A cylindrical wedge with its edge along the x axis. - -

    - A cylindrical wedge, between the planes z=0 and z=2y. - These planes intersect along the x axis, forming the sharp edge of the wedge. - The round face of the wedge is the parabolic cylinder y=4-x^2, for y\geq 0. -

    - -

    - The face of the surface in the xy plane is the region bounded below by the x axis, - and above by the parabola y=4-x^2. -

    - -

    - Another face is the portion of the plane z=2y that lies above the face in the xy plane. -

    - -

    - The last face is the parabolic cylinder. It is the portion of the cylinder y=4-x^2 - that lies above the plane z=0 and below the plane z=2y. -

    -     - - - - - //ASY file for fig13_06_ex_123D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12.1,-7.1,16); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,-1,1,2}; - real[] myychoice={1,2,3,4}; - real[] myzchoice={2,4,6,8}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-0.25,5); - pair zbounds=(-0.25,10); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - triple g(real t) {return (t,4-t^2,0);} - path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,4-t^2,2*(4-t^2));} - path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - //shade object - //import three; - //int k=12; - //for (int i=-2*k; i<2*k; ++i) - //{ - //path3 p=(i/k,4-(i/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i)/k,4-((i)/k)^2,2*(4-((i)/k)^2)); - //draw(surface(p -- cycle), simplesurfacepen2); - //path3 p=(i/k,0,0)--(i/k,4-(i/k)^2,2*(4-(i/k)^2))--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i+1//)/k,0,0); - //draw(surface(p -- cycle), simplesurfacepen); - //} - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y*(4-t.x^2),2*t.y*(4-t.x^2));// - } - surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,(4-t.x^2),2*t.y*(4-t.x^2));// - } - surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y*(4-t.x^2),0);// - } - surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - label("$z=2y$",(-2,2,7)); - draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - label("$y=4-x^2$",(2.5,2,0)); - draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - Answers may vary. -

    - -

    - For z=2y: \vec r(u,v) = \langle u, v(4-u^2), 2v(4-u^2)\rangle with - -2\leq u\leq 2 and 0\leq v\leq 1. -

    - -

    - For y=4-x^2: \vec r(u,v) = \langle u, 4-u^2, 2v(4-u^2)\rangle with - -2\leq u\leq 2 and 0\leq v\leq 1. -

    - -

    - For z=0: - \vec r(u,v) = \langle u, v(4-u^2), 0\rangle with - -2\leq u\leq 2 and 0\leq v\leq 1. -

    -     -     - - - -

    - D is the domain bounded by y=1-z^2, - y=1-x^2, x=0, y=0 and z=0. -

    - - - A region in the first octant bounded by the planes x=0 and z=0, and two parabolic cylinders. - -

    - This surface has three faces that lie in planes, and two faces that are parts of parabolic cylinders. -

      -
    • -

      - The bottom face is in the xy plane, - in a region bounded by the x and y coordinate axes and the parabola y=1-x^2. -

      -
      • -
    • -

      - Another face lies in the yz plane, - forming a region bounded by the y and z coordinate axes and the parabola y=1-z^2. -

      -
      • -
    • -

      - There is another face in the xz plane. This face is a square, with x and z between 0 and 1. -

      -
      • -
    • -

      - Another part of the surface is the portion parabolic cylinder y=1-z^2, - and the remaining face is the parabolic cylinder y=1-x^2. - Both of these can be viewed as graphs over the xz plane. - The surface y=1-z^2 lies over the triangle bounded by x=z, x=0, and z=1. - The surface y=1-x^2 lies over the triangle bounded by x=z, x=1, and z=0. -

      -
      • -
    -

    -     - - - - - //ASY file for fig13_06_ex_133D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(4,4,2); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={1}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.25,1.5); - pair ybounds=(-0.25,1.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - triple g(real t) {return (t,1-t^2,0);} - path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (0,1-t^2,t);} - path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,1-t^2,t);} - path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - //draw square and sideline - draw((0,0,0)--(1,0,0)--(1,0,1)--(0,0,1)--(0,0,0),bluepen+linewidth(2)); - draw((0,0,0)--(0,1,0),bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,(1-t.x^2),t.y*t.x);// - } - surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.y*t.x,(1-t.x^2),t.x);// - } - surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,0,t.y);// - } - surface s=surface(f,(0,0),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (0,t.y*(1-t.x^2),t.x);// - } - surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y*(1-t.x^2),0);// - } - surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - label("$y=1-x^2$",(1,1,0)); - draw((0.9,0.9,.1)--(.65,.6,.4),Arrow3(size=2mm)); - label("$y=1-z^2$",(0,1,.85)); - draw((.1,.9,.75)--(.3,.8,.45),Arrow3(size=2mm)); - - //shade object - //import three; - //int k=12; - //for (int i=0; i<k; ++i) - //{ - //path3 p=(0,1-(i/k)^2,i/k)--(0,1-((i+1)/k)^2,(i+1)/k)--((i+1)/k,1-((i+1)/k)^2,(i+1)/k)--((i)/k,1-((i)/k)^2,(i)/k); - //draw(surface(p -- cycle), simplesurfacepen2);//pink - - //path3 p=(i/k,1-(i/k)^2,0)--((i+1)/k,1-((i+1)/k)^2,0)--((i+1)/k,1-((i+1)/k)^2,(i+1)/k)--((i)/k,1-((i)/k)^2,(i)/k); - //draw(surface(p -- cycle), simplesurfacepen);//blue - //} - - - -     - -

    - Answers may vary. -

    - -

    - For y=1-z^2: - \vec r(u,v) = \langle u, v(1-u^2), \sqrt{1-v(1-u^2)}\rangle with - 0\leq u\leq 1 and 0\leq v\leq 1. -

    - -

    - For y=1-x^2: - \vec r(u,v) = \langle u, 1-u^2, uv\rangle with - 0\leq u\leq 1 and 0\leq v\leq 1. -

    - -

    - For x=0: - \vec r(u,v) = \langle 0, v(1-u^2),u\rangle with - 0\leq u\leq 1 and 0\leq v\leq 1. -

    - -

    - For y=0: \vec r(u,v) = \langle u, 0,v\rangle with - 0\leq u\leq 1 and 0\leq v\leq 1. -

    - -

    - For z=0: - \vec r(u,v) = \langle u, v(1-u^2), 0rangle with - 0\leq u\leq 1 and 0\leq v\leq 1. -

    -     -     - - - -

    - D is the domain bounded by the cylinder - x^2+y^2/9=1 and the planes z=1 and z=3. -

    - - - An elliptical cylinder, centered on the z axis, capped by planes z=1 and z=3. - -

    - This surface has three components. - Two are faces in the planes z=1 and z=3. - Each of these is bounded by the ellipse x^2+y^2/9=1. -

    - -

    - The remaining component is the portion of the elliptical cylinder x^2+y^2/9=1 - that lies between the planes z=1 and z=3. -

    -     - - - - - //ASY file for fig14_05_ex_21_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14,14,4); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-3,3}; - real[] myzchoice={1,3}; - defaultpen(0.5mm); - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-0.25,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple g(real t) {return (cos(t),3sin(t),3);} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (cos(t),3sin(t),1);} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (cos(t.x),3*sin(t.x),t.y);// - } - surface s=surface(f,(0,1),(2pi,3),16,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (t.y*cos(t.x),t.y*3*sin(t.x),1);// - } - surface s=surface(f,(0,0),(2pi,1),16,2,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (t.y*cos(t.x),t.y*3*sin(t.x),3);// - } - surface s=surface(f,(0,0),(2pi,1),16,2,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - label("$x^2+y^2/9=1$",(2.5,0,1)); - draw((2.25,0,1.25)--(1,0,2),Arrow3(size=2mm)); - - //triple g(real t) {return (0,t,6-t^2);} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - - -     - -

    - Answers may vary. -

    - -

    - For x^2+y^2/9=1: - \vec r(u,v) = \langle \cos(u), 3\sin(u), v\rangle with - 0\leq u\leq 2\pi and 1\leq v\leq 3. -

    - -

    - For z=1: - \vec r(u,v) = \langle v\cos(u), 3v\sin(u), 1\rangle with - 0\leq u\leq 2\pi and 0\leq v\leq 1. -

    - -

    - For z=3: - \vec r(u,v) = \langle v\cos(u), 3v\sin(u), 3\rangle with - 0\leq u\leq 2\pi and 0\leq v\leq 1. -

    -     -     - - - -

    - D is the domain bounded by the cone - x^2+y^2=(z-1)^2 and the plane z=0. -

    - - - An inverted circular cone, together with a disk in the x,y plane - -

    - This surface has two components. - One is the cone x^2+y^2=(z-1)^2; - this is a circular cone; the vertex is at (0,0,1), - and our surface is the part that opens downward. -

    - -

    - The cone meets the xy plane along the unit circle. - The other part of the surface is the disk in the xy plane bounded by the unit circle. -

    -     - - - - - //ASY file for fig14_05_ex_21_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14,14,4); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - triple g(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),1-t.y);// - } - surface s=surface(f,(0,0),(2pi,1),16,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),0);// - } - surface s=surface(f,(0,0),(2pi,1),16,2,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - label("$x^2+y^2=(z-1)^2$",(1,0,1)); - draw((.9,0,.9)--(.55,0,.5),Arrow3(size=2mm)); - - //triple g(real t) {return (0,t,6-t^2);} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - - -     - -

    - Answers may vary. -

    - -

    - For x^2+y^2=(z-1)^2: - \vec r(u,v) = \langle v\cos(u), v\sin(u), 1-v\rangle with - 0\leq u\leq 2\pi and 0\leq v\leq 1. -

    - -

    - For z=0: - \vec r(u,v) = \langle v\cos(u), v\sin(u), 0\rangle with - 0\leq u\leq 2\pi and 0\leq v\leq 1. -

    -     -     - - - -

    - D is the domain bounded by the cylinder z=1-x^2 and the planes y=-1, - y=2 and z=0. -

    - - - A solid that looks like a barn with a parabolic roof. - -

    - The boundary of this solid consists of four surface components. - Overall, the shape looks something like a barn or greenhouse with a curved roof. -

    - -

    - The base is a rectangle in the xy plane, with -1\leq x\leq 1 and -1\leq y\leq 2. -

    - -

    - The two ends lie in the planes y=-1 and y=2. - They are bounded above by the parabola z=1-x^2, and below by z=0. -

    - -

    - The last part is the parabolic cylinder z=1-x^2, for -1\leq x\leq 1, and -1\leq y\leq 2. -

    -     - - - - - //ASY file for fig14_05_ex_23_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14,14,4); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-2,2}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y,1-t.x^2);// - } - surface s=surface(f,(-1,-1),(1,2),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,-1,t.y*(1-t.x^2));// - } - surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,2,t.y*(1-t.x^2));// - } - surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (t.x,t.y,0);// - } - surface s=surface(f,(-1,-1),(1,2),2,2,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return (t,-1,1-t^2);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,2,1-t^2);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - draw((1,-1,0)--(1,2,0)--(-1,2,0)--(-1,-1,0)--cycle,bluepen+linewidth(2)); - - //label and arrow - label("$z=1-x^2$",(0,1,1.55)); - draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); - - //triple g(real t) {return (0,t,6-t^2);} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - - -     - -

    - Answers may vary. -

    - -

    - For z=1-x^2: \vec r(u,v) = \langle u,v,1-u^2\rangle with - -1\leq u\leq 1 and -1\leq v\leq 2. -

    - -

    - For y=-1: - \vec r(u,v) = \langle u,-1,v(1-u^2)\rangle with - -1\leq u\leq 1 and 0\leq v\leq 1. -

    - -

    - For y=2: - \vec r(u,v) = \langle u,2,v(1-u^2)\rangle with - -1\leq u\leq 1 and 0\leq v\leq 1. -

    - -

    - For z=0: \vec r(u,v) = \langle u,v,0\rangle with - -1\leq u\leq 1 and -1\leq v\leq 2. -

    -     -     - - - -

    - D is the domain bounded by the paraboloid - z=4-x^2-4y^2 and the plane z=0. -

    - - - A region bounded above by an elliptic paraboloid, and below by the x,y plane. - -

    - The solid looks something like half of a seed. - It is bounded above by the elliptic paraboloid z=4-x^2-4y^2, - which opens downward from a vertex at (0,0,4), forming a sort of domed roof. -

    - -

    - This paraboloid forms one component of the bounding surface for the solid. - The other part is the elliptical disk formed when the surface meets the xy plane. - It is the region in the xy plane that lies on and inside the ellipse x^2+4y^2=4. -

    -     - - - - - //ASY file for fig14_05_ex_23_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14,14,8); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={4}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.25,4.75); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (2*t.y*cos(t.x),t.y*sin(t.x),4-(2*t.y*cos(t.x))^2-4(t.y*sin(t.x))^2);// - } - surface s=surface(f,(0,0),(2pi,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (2*t.y*cos(t.x),t.y*sin(t.x),0);// - } - surface s=surface(f,(0,0),(2pi,1),16,2,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return (2cos(t),sin(t),0);} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - //label and arrow - label("$z=4-x^2-4y^2$",(0,1.5,3)); - draw((0,1.45,2.9)--(0,1,1),Arrow3(size=2mm)); - - //triple g(real t) {return (0,t,6-t^2);} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - - -     - -

    - Answers may vary. -

    - -

    - For z=4-x^2-4y^2: - \vec r(u,v) = \langle 2v\cos(u),v\sin(u),4-(2v\cos(u))^2-4(v\sin(u))^2\rangle with - 0\leq u\leq 2\pi and 0\leq v\leq 1. -

    - -

    - For z=0: - \vec r(u,v) = \langle 2v\cos(u),v\sin(u),0\rangle with - 0\leq u\leq 2\pi and 0\leq v\leq 1. -

    -     -     -     - - - -

    - Find the surface area S of the given surface \surfaceS. (The associated integrals are computable without the assistance of technology.) -

    -     - - - - -

    - \surfaceS is the plane z=2x+3y over the rectangle - -1\leq x\leq 1, 2\leq y \leq 3. -

    -     - -

    - S = 2\sqrt{14}. -

    -     -     - - - -

    - \surfaceS is the plane z=x+2y over the triangle with vertices at (0,0), - (1,0) and (0,1). -

    -     - -

    - S = \sqrt{6}/2. -

    -     -     - - - -

    - \surfaceS is the plane z=x+y over the circular disk, - centered at the origin, with radius 2. -

    -     - -

    - S = 4\sqrt{3}\pi. -

    -     -     - - - -

    - \surfaceS is the plane z=x+y over the annulus bounded by the circles, - centered at the origin, with radius 1 and radius 2. -

    -     - -

    - S = 3\sqrt{3}\pi. -

    -     -     -     - - - -

    - Set up the double integral that finds the surface area S of the given surface \surfaceS, then use technology to approximate its value. -

    -     - - - -

    - \surfaceS is the paraboloid - z=x^2+y^2 over the circular disk of radius 3 centered at the origin. -

    -     - -

    - S =\int_0^3\int_0^{2\pi}\sqrt{v^2+4v^4}\, du\, dv= (37\sqrt{37}-1)\pi/6 \approx 117.319. -

    -     -     - - - -

    - \surfaceS is the paraboloid - z=x^2+y^2 over the triangle with vertices at (0,0), - (0,1) and (1,1). -

    -     - -

    - S = \int_0^1\int_0^1\sqrt{v^2+4u^2v^2+4v^4}\, du\, dv \approx 0.931. -

    -     -     - - - -

    - \surfaceS is the plane z=5x-y over the region enclosed by the parabola y=1-x^2 and the x-axis. -

    -     - -

    - S =\int_0^1\int_{-1}^{1}\sqrt{(5u^2-5)^2+2(1-u^2)^2}\, du\, dv = 4\sqrt{3}\approx 6.9282. -

    -     -     - - - -

    - \surfaceS is the hyperbolic paraboloid - z=x^2-y^2 over the circular disk of radius 1 centered at the origin. -

    -     - -

    - S =\int_0^1\int_{0}^{2\pi}\sqrt{v^2+4v^4}\, du\, dv = (5\sqrt{5}-1)\pi/6 \approx 5.330. -

    -     -     -     -     -     -
    -
    - Surface Integrals - -

    - Consider a smooth surface \surfaceS that represents a thin sheet of metal. - How could we find the mass of this metallic object? -

    - -

    - If the density of this object is constant, - then we can find mass via mass= density surface area, - and we could compute the surface area using the techniques of the previous section. -

    - -

    - What if the density were not constant, - but variable, described by a function \delta(x,y,z)? - We can describe the mass using our general integration techniques as - - \text{ mass } = \iint_\surfaceS \, dm - , - where dm represents a little bit of mass. That is, - to find the total mass of the object, - sum up lots of little masses over the surface. -

    - -

    - How do we find the little bit of mass dm? - On a small portion of the surface with surface area \Delta S, - the density is approximately constant, - hence dm \approx \delta(x,y,z)\Delta S. - As we use limits to shrink the size of - \Delta S to 0, we get dm = \delta(x,y,z)dS; - that is, a little bit of mass is equal to a density times a small amount of surface area. - Thus the total mass of the thin sheet is - - \text{ mass } =\iint_\surfaceS \delta(x,y,z)\, dS - . -

    - -

    - To evaluate the above integral, - we would seek \vec r(u,v), - a smooth parametrization of \surfaceS over a region R of the u-v plane. - The density would become a function of u and v, - and we would integrate \iint_R \delta(u,v)\snorm{\vec r_u\times \vec r_v}\, dA. -

    - -

    - The integral in Equation is a specific example of a more general construction defined below. -

    -     - - - Surface integrals of scalar fields - - - - - Surface Integral - -

    - Let G(x,y,z) be a continuous function defined on a surface \surfaceS. The - surface integral of G on \surfaceS is - surface integral - - \iint_\surfaceS G(x,y,z)\, dS - . -

    -     -     - -

    - Surface integrals can be used to measure a variety of quantities beyond mass. - If G(x,y,z) measures the static charge density at a point, - then the surface integral will compute the total static charge of the sheet. - If G measures the amount of fluid passing through a screen - (represented by \surfaceS) - at a point, then the surface integral gives the total amount of fluid going through the screen. -

    - - - Finding the mass of a thin sheet - -

    - Find the mass of a thin sheet modeled by the plane - 2x+y+z=3 over the triangular region of the xy-plane bounded by the coordinate axes and the line y=2-2x, - as shown in , - with density function \delta(x,y,z) = x^2+5y+z, - where all distances are measured in cm and the density is given as gm/cm^2. -

    -
    - The surface whose mass is computed in - - A triangular portion of a plane in space. It is the graph of a linear function over a triangular domain in the plane. - -

    - The positive x, y, and z axes are shown in space, with the z axis pointing up. - In the xy plane, a triangular domain is shaded. - It is the region in the plane z=0 bounded by the x and y axes, - and the line segment from (1,0,0) to (0,2,0). -

    - -

    - Above the xy plane is another triangle. - This triangle lies in the plane 2x+y+z=3, - and can be viewed as the graph z=3-2x-y over the triangular domain in the plane. -

    - -

    - The vertices of this triangular surface in space are - (1,0,1), (0,2,1), and (0,0,3). -

    -     - - - - - //ASY file for figsurfint1_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(16,12.5,14); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={2}; - real[] myzchoice={3}; - defaultpen(0.5mm); - - pair xbounds=(-.1,2.5); - pair ybounds=(-.1,2.5); - pair zbounds=(-.1,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane 2x+y+z=3 over the triangle bounded - // by coordinate planes and y=2-2x - triple f(pair t) { - return (t.x,t.y*(2-2*t.x),3-2*t.x-t.y*(2-2t.x));// - } - surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=invisible); - - draw(surface((1,0,0)--(0,0,0)--(0,2,0)--cycle),curvepen+opacity(.5)); - draw((1,0,0)--(0,0,0)--(0,2,0)--cycle,curvepen); - draw(f((0,0))--f((1,0))--f((0,1))--cycle,curvepen); - - -
    -     - -

    - We begin by parametrizing the planar surface \surfaceS. Using the techniques of the previous section, - we can let x=u and y=v(2-2u), - where 0\leq u\leq 1 and 0\leq v\leq 1. - Solving for z in the equation of the plane, - we have z=3-2x-y, hence z = 3-2u-v(2-2u), - giving the parametrization \vec r(u,v) = \langle u, v(2-2u), 3-2u-v(2-2u)\rangle. -

    - -

    - We need dS=\snorm{\vec r_u\times \vec r_v}dA, - so we need to compute \vec r_u, - \vec r_v and the norm of their cross product. - We leave it to the reader to confirm the following: - - \vec r_u = \langle 1,-2v,2v-2\rangle, \vec r_v = \langle 0,2-2u, 2u-2\rangle - , - - \vec r_u\times \vec r_v = \langle 4-4u,2-2u,2-2u\rangle \text{ and } \snorm{\vec r_u\times \vec r_v} = 2\sqrt{6}\sqrt{(u-1)^2} - . -

    - -

    - We need to be careful to not simplify - \snorm{\vec r_u\times \vec r_v} = 2\sqrt{6}\sqrt{(u-1)^2} as 2\sqrt{6}(u-1); - rather, it is 2\sqrt{6}|u-1|. - In this example, u is bounded by 0\leq u\leq 1, - and on this interval |u-1| = 1-u. - Thus dS = 2\sqrt{6}(1-u)dA. -

    - -

    - The density is given as a function of x, y and z, - for which we'll substitute the corresponding components of \vec r (with the slight abuse of notation that we used in previous sections): - - \delta(x,y,z) \amp = \delta\big(\vec r(u,v)\big) - \amp = u^2 + 5v(2-2u)+3-2u-v(2-2u) - \amp = u^2-8uv-2u+8v+3 - . -

    - -

    - Thus the mass of the sheet is: - - M \amp = \iint_\surfaceS\, dm - \amp = \iint_R \delta\big(\vec r(u,v)\big)\snorm{\vec r_u\times \vec r_v}\, dA - \amp = \int_0^1\int_0^1 \big(u^2-8uv-2u+8v+3\big)\big(2\sqrt{6}(1-u)\big)\, du\, dv - \amp = \frac{31}{\sqrt{6}} \approx 12.66 \text{ gm. } - -

    -     -     -     - - - Flux -

    - Let a surface \surfaceS lie within a vector field \vec F. - One is often interested in measuring the flux - of \vec F across \surfaceS; that is, - measuring how much of the vector field passes across \surfaceS. For instance, - if \vec F represents the velocity field of moving air and \surfaceS represents the shape of an air filter, - the flux will measure how much air is passing through the filter per unit time. - flux -

    - -

    - As flux measures the amount of \vec F passing across \surfaceS, we need to find the - amount of \vec F orthogonal to \surfaceS. - Similar to our measure of flux in the plane, - this is equal to \vec F\cdot \vec n, - where \vec n is a unit vector normal to \surfaceS at a point. - We now consider how to find \vec n. -

    - - - -

    - Given a smooth parametrization - \vec r(u,v) of \surfaceS, the work in the previous section showing the development of our method of computing surface area also shows that \vec r_u(u,v) and - \vec r_v(u,v) are tangent to \surfaceS at \vec r(u,v). - Thus \vec r_u\times \vec r_v is orthogonal to \surfaceS, and we let - - \vec n = \frac{\vec r_u\times \vec r_v}{\snorm{\vec r_u\times \vec r_v}} - , - which is a unit vector normal to \surfaceS at \vec r(u,v). -

    - -

    - The measurement of flux across a surface is a surface integral; - that is, to measure total flux we sum the product of - \vec F\cdot\vec n times a small amount of surface area: - \vec F\cdot \vec n\, dS. -

    - -

    - A nice thing happens with the actual computation of flux: - the \snorm{\vec r_u\times \vec r_v} terms go away. - Consider: - - \text{ Flux } \amp = \iint_\surfaceS \vec F\cdot \vec n\, dS - \amp = \iint_R \vec F\cdot \frac{\vec r_u\times \vec r_v}{\snorm{\vec r_u\times \vec r_v}}\snorm{\vec r_u\times \vec r_v}\, dA - \amp = \iint_R \vec F\cdot (\vec r_u\times \vec r_v)\, dA - . -

    - -

    - The above only makes sense if \surfaceS is orientable; - the normal vectors \vec n must vary continuously across \surfaceS. We assume that \vec n does vary continuously. - (If the parametrization \vec r of \surfaceS is smooth, - then our above definition of \vec n will vary continuously.) -

    - - - Flux over a surface - -

    - Let \vec F be a vector field with continuous components defined on an orientable surface \surfaceS with normal vector \vec n. - The flux of \vec F across \surfaceS is - flux - - \text{ Flux } = \iint_\surfaceS \vec F\cdot \vec n\, dS - . -

    - -

    - If \surfaceS is parametrized by \vec r(u,v), - which is smooth on its domain R, then - - \text{ Flux } = \iint_R \vec F\big(\vec r(u,v)\big)\cdot (\vec r_u\times \vec r_v)\, dA - . -

    -     -     - -

    - Since \surfaceS is orientable, - we adopt the convention of saying one passes from the back - side of \surfaceS to the front - side when moving across the surface parallel to the direction of \vec n. - Also, when \surfaceS is closed, - it is natural to speak of the regions of space - inside and outside - \surfaceS. We also adopt the convention that when \surfaceS is a closed surface, - \vec n should point to the outside of \surfaceS. If - \vec n = \vec r_u\times\vec r_v points inside \surfaceS, use \vec n = \vec r_v\times \vec r_u instead. -

    - -

    - When the computation of flux is positive, - it means that the field is moving from the back side of \surfaceS to the front side; - when flux is negative, - it means the field is moving opposite the direction of \vec n, - and is moving from the front of \surfaceS to the back. - When \surfaceS is not closed, - there is not a right and wrong - direction in which \vec n should point, - but one should be mindful of its direction to make full sense of the flux computation. -

    - -

    - We demonstrate the computation of flux, - and its interpretation, in the following examples. -

    - - - Finding flux across a surface - -

    - Let \surfaceS be the surface given in , - where \surfaceS is parametrized by - \vec r(u,v) = \langle u, v(2-2u),3-2u-v(2-2u)\rangle on 0\leq u\leq 1, - 0\leq v\leq 1, and let \vec F = \langle 1, x,-y\rangle, - as shown in . - Find the flux of \vec F across \surfaceS. -

    - -
    - The surface and vector field used in - - A triangular surface in space, and a vector field flowing across the surface. - -

    - The positive x, y, and z coordinate axes are drawn in space. - A triangular surface lies in the first octant above the xy plane; - it is the same surface that was plotted in -

    - -

    - A vector field in space is also plotted. - From the perspective used in the book, - the vectors appear to point mainly downward, - and are largest in magnitude to the right of the image, - where the value of y is largest. -

    -     - - - - - //ASY file for figsurfint1_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(16,12.5,14); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={2}; - real[] myzchoice={3}; - defaultpen(0.5mm); - - pair xbounds=(-.1,2.5); - pair ybounds=(-.1,2.5); - pair zbounds=(-.1,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane 2x+y+z=3 over the triangle bounded - // by coordinate planes and y=2-2x - triple f(pair t) { - return (t.x,t.y*(2-2*t.x),3-2*t.x-t.y*(2-2t.x));// - } - surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=invisible); - - path3 gradient1(triple z){ - return O--(1,z.x,-z.y); - } - - triple A=(0,0,1); - triple B=(1,2,3); - - picture VectorPlot3D(path3 vector(triple t), triple a, triple b, - int nx=nmesh, int ny=nx, int nz=nx,bool truesize=false, - real maxlength=truesize ? 0 : min(abs(b.x-a.x)/nx,abs(b.y-a.y)/ny,abs(b.z-a.z)/nz), - // bool cond(pair z)=null, - pen p=currentpen, - arrowbar3 arrow=Arrow3(6), margin3 margin=PenMargin3, - string name="", render render=defaultrender) - { - picture pic; - real dx=1/nx; - real dy=1/ny; - real dz=1/nz; - real scale; - if(maxlength > 0) { - real size(triple t) { - path3 g=vector(t); - return abs(point(g,size(g)-1)-point(g,0)); - } - real max=size((0,0,0)); - - for(int i=0; i <= nx; ++i) { - real x=interp(a.x,b.x,i*dx); - for(int j=0; j <= ny; ++j) - { - real y=interp(a.y,b.y,j*dy); - for(int k=0; k <= nz; ++k) - max=max(max,size((x,y,interp(a.z,b.z,k*dz)))); - }} - scale=max > 0 ? maxlength/max : 1; - } else scale=1; - bool group=name != "" || render.defaultnames; - if(group) - begingroup3(pic,name == "" ? "vectorfield" : name,render); - for(int i=0; i <= nx; ++i) { - real x=interp(a.x,b.x,i*dx); - for(int j=0; j <= ny; ++j) { - real y=interp(a.y,b.y,j*dy); - for(int k=0; k <= nz; ++k) - { triple z=(x,y,interp(a.z,b.z,k*dz)); - { - path3 g=scale3(scale)*vector(z); - string name="vector"; - if(truesize) { - picture opic; - draw(opic,g,p,arrow,margin,name,render); - add(pic,opic,z); - } else - draw(pic,shift(z)*g,p,arrow,margin,name,render); - } - } - }} - if(group) - endgroup3(pic); - return pic; - - } - add(VectorPlot3D(gradient1,A,B,3,3,3,bluepen)); - - - -
    -     - -

    - Using our work from the previous example, - we have \vec n = \vec r_u\times\vec r_v = \langle 4-4u,2-2u,2-2u\rangle. - We also need \vec F\big(\vec r(u,v)\big) = \langle 1, u, -v(2-2u)\rangle. -

    - -

    - Thus the flux of \vec F across \surfaceS is: - - \text{ Flux } \amp = \iint_\surfaceS \vec F\cdot \vec n\, dS - \amp = \iint_R \langle 1,u,-v(2-2u)\rangle\cdot\langle 4-4u,2-2u,2-2u\rangle\, dA - \amp = \int_0^1\int_0^1 \big(-4u^2v-2u^2+8uv-2u-4v+4\big)\, du\, dv - \amp = 5/3 - . -

    - -

    - To make full use of this numeric answer, - we need to know the direction in which the field is passing across \surfaceS. The graph in helps, - but we need a method that is not dependent on a graph. -

    - -

    - Pick a point (u,v) in the interior of R and consider \vec n(u,v). - For instance, - choose (1/2,1/2) and look at \vec n(1/2,1/2) = \langle 2,1,1\rangle/\sqrt{6}. - This vector has positive x, - y and z components. - Generally speaking, one has some - idea of what the surface \surfaceS looks like, - as that surface is for some reason important. - In our case, - we know \surfaceS is a plane with z-intercept of z=3. - Knowing \vec n and the flux measurement of positive 5/3, - we know that the field must be passing from behind - \surfaceS, , the side the origin is on, - to the front of \surfaceS. -

    -     -     - - - - - Flux across surfaces with shared boundaries - -

    - Let \surfaceS_1 be the unit disk in the xy-plane, - and let \surfaceS_2 be the paraboloid z=1-x^2-y^2, - for z\geq 0, - as graphed in . - Note how these two surfaces each have the unit circle as a boundary. -

    - -
    - The surfaces used in - - A closed surface consisting of a downward-opening circular paraboloid, and a disc in the x,y plane. - -

    - The surface \surfaceS_2 is a circular paraboloid, opening downward, - with its vertex on the z axis at (0,0,1). - It intersects the xy plane in the unit circle. -

    - -

    - The unit circle and its interior form the disk that is \surfaceS_1. - The two surfaces are therefore joined along the unit circle in the xy plane, - forming a closed surface like a dome or a hill. -

    -     - - - - - //ASY file for figsurfint1_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(22.8,20,5.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={2}; - real[] myzchoice={3}; - defaultpen(0.5mm); - - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - pair zbounds=(-.1,1.1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane unit disk - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),0);// - } - surface s=surface(f,(-pi/2,0),(3pi/2,1),8,8,usplinetype=Spline,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=redmeshpen; - draw(s,surfacepen2,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),1-t.y^2);// - } - surface s=surface(f,(0,0),(2pi,1),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(f,0,2pi,operator ..); - draw(mypath,curvepen); - - - -
    - -

    - Let \vec F_1 = \langle 0,0,1\rangle and \vec F_2 = \langle 0,0,z\rangle. - Using normal vectors for each surface that point upward, - , with a positive z-component, - find the flux of each field across each surface. -

    -     - -

    - We begin by parametrizing each surface. -

    - -

    - The boundary of the unit disk in the xy-plane is the unit circle, - which can be described with - \langle \cos(u),\sin(u),0\rangle, 0\leq u\leq 2\pi. - To obtain the interior of the circle as well, - we can scale by v, giving - - \vec r_1(u,v) = \langle v\cos(u),v\sin(u), 0\rangle, 0\leq u\leq 2\pi 0\leq v\leq 1 - . -

    - -

    - As the boundary of \surfaceS_2 is also the unit circle, - the x and y components of - \vec r_2 will be the same as those of \vec r_1; - we just need a different z component. - With z = 1-x^2-y^2, we have - - \vec r_2(u,v) = \langle v\cos(u),v\sin(u), 1-v^2\cos^2u-v^2\sin^2u\rangle = \langle v\cos(u),v\sin(u), 1-v^2\rangle - , - where 0\leq u\leq 2\pi and 0\leq v\leq 1. -

    - -

    - We now compute the normal vectors \vec n_1 and \vec n_2. -

    - -

    - For \vec n_1: - \vec r_{1u}= \langle -v\sin(u), v\cos(u),0\rangle, - \vec r_{1v} = \langle \cos(u),\sin(u),0\rangle, so - - \vec n_1 = \vec r_{1u}\times \vec r_{1v} = \langle 0,0,-v\rangle - . -

    - -

    - As this vector has a negative z-component, we instead use - - \vec n_1 = \vec r_{1v}\times \vec r_{1u} = \langle 0,0,v\rangle - . -

    - -

    - Similarly, \vec n_2: - \vec r_{2u}= \langle -v\sin(u), v\cos(u),0\rangle, - \vec r_{2v} = \langle \cos(u),\sin(u),-2v\rangle, so - - \vec n_2 = \vec r_{2u}\times \vec r_{2v} = \langle -2v^2\cos(u),-2v^2\sin(u),-v\rangle - . -

    - -

    - Again, this normal vector has a negative z-component so we use - - \vec n_2 = \vec r_{2v}\times \vec r_{2u} = \langle 2v^2\cos(u),2v^2\sin(u),v\rangle - . -

    - -

    - We are now set to compute flux. - Over field \vec F_1=\langle 0,0,1\rangle: - - \text{ Flux across } \surfaceS_1 \amp = \iint_{\surfaceS_1} \vec F_1\cdot \vec n_1\, dS - \amp = \iint_R\langle 0,0,1\rangle\cdot\langle 0,0,v\rangle\, dA - \amp = \int_0^1\int_0^{2\pi} (v)\, du\, dv - \amp = \pi - . - - \text{ Flux across } \surfaceS_2 \amp = \iint_{\surfaceS_2} \vec F_1\cdot \vec n_2\, dS - \amp = \iint_R\langle 0,0,1\rangle\cdot\langle 2v^2\cos(u),2v^2\sin(u),v\rangle\, dA - \amp = \int_0^1\int_0^{2\pi} (v)\, du\, dv - \amp = \pi - . -

    - -

    - These two results are equal and positive. - Each are positive because both normal vectors are pointing in the positive z-directions, - as does \vec F_1. - As the field passes through each surface in the direction of their normal vectors, - the flux is measured as positive. -

    - -

    - We can also intuitively understand why the results are equal. - Consider \vec F_1 to represent the flow of air, - and let each surface represent a filter. - Since \vec F_1 is constant, - and moving straight up, - it makes sense that all air passing through \surfaceS_1 also passes through - \surfaceS_2, and vice-versa. -

    - -

    - If we treated the surfaces as creating one piecewise-smooth surface \surfaceS, we would find the total flux across \surfaceS by finding the flux across each piece, - being sure that each normal vector pointed to the outside of the closed surface. - Above, \vec n_1 does not point outside the surface, - though \vec n_2 does. - We would instead want to use -\vec n_1 in our computation. - We would then find that the flux across \surfaceS_1 is -\pi, - and hence the total flux across \surfaceS is -\pi + \pi = 0. - (As 0 is a special number, - we should wonder if this answer has special significance. - It does, which is briefly discussed following this example and will be more fully developed in the next section.) -

    - -

    - We now compute the flux across each surface with \vec F_2=\langle 0,0,z\rangle: - - \text{ Flux across } \surfaceS_1 \amp = \iint_{\surfaceS_1} \vec F_2\cdot \vec n_1\, dS . - Over \surfaceS_1, \vec F_2 = \vec F_2\big(\vec r_2(u,v)\big) = \langle 0,0,0\rangle. Therefore, - \amp = \iint_R\langle 0,0,0\rangle\cdot\langle 0,0,v\rangle\, dA - \amp = \int_0^1\int_0^{2\pi} (0)\, du\, dv - \amp = 0 - . - - \text{ Flux across } \surfaceS_2 \amp = \iint_{\surfaceS_2} \vec F_2\cdot \vec n_2\, dS . - Over \surfaceS_2, \vec F_2 = \vec F_2\big(\vec r_2(u,v)\big) = \langle 0,0,1-v^2\rangle. Therefore, - \amp = \iint_R\langle 0,0,1-v^2\rangle\cdot\langle 2v^2\cos(u),2v^2\sin(u),v\rangle\, dA - \amp = \int_0^1\int_0^{2\pi} (v^3-v)\, du\, dv - \amp = \pi/2 - . -

    - -

    - This time the measurements of flux differ. - Over \surfaceS_1, - the field \vec F_2 is just \vec 0, hence there is no flux. - Over \surfaceS_2, - the flux is again positive as - \vec F_2 points in the positive z direction over - \surfaceS_2, as does \vec n_2. -

    -     -     - -

    - In the previous example, - the surfaces \surfaceS_1 and - \surfaceS_2 form a closed surface that is piecewise smooth. - That the measurement of flux across each surface was the same for some fields - (and not for others) - is reminiscent of a result from , - where we measured flux across curves. - The quick answer to why the flux was the same when considering - \vec F_1 is that \divv \vec F_1 = 0. - In the next section, - we'll see the second part of the Divergence Theorem, - which will more fully explain this occurrence. - We will also explore Stokes' Theorem, - the spatial analogue to Green's Theorem. -

    - - - - - - -     - - - - Terms and Concepts - - - -

    - In the plane, - flux is a measurement of how much of the vector field passes across a ; - in space, flux is a measurement of how much of the vector field passes across a . -

    -     - - - - - - - - - - - - -     - - - -

    - When computing flux, - what does it mean when the result is a negative number? -

    -     - - - -

    - Answers will vary; in general, - it means that more of the vector field passes through the surface opposite the - direction of the normal vector than in the same direction of the normal vector. -

    -     -     - - - -

    - When \surfaceS is a closed surface, - we choose the normal vector so that it points to the of the surface. -

    -     - - - - outside|exterior - - - -     - - - -

    - If \surfaceS is a plane, - and \vec F is always parallel to \surfaceS, - then the flux of \vec F across \surfaceS will be . -

    -     - - - - 0|zero - - - -     -     - - - Problems - - - -

    - A surface \surfaceS that represents a thin sheet of material with density \delta is given. - Find the mass of each thin sheet. -

    -     - - - - -

    - \surfaceS is the plane - z=x+y on -2\leq x\leq 2, - -3\leq y\leq 3, with \delta(x,y,z) = z+10. -

    -     - -

    - 240\sqrt{3} -

    -     -     - - - -

    - \surfaceS is the unit sphere, - with \delta(x,y,z) = x+y+z+10. -

    -     - -

    - 40\pi -

    -     -     -     - - - -

    - A surface \surfaceS and a vector field \vec F are given. - Compute the flux of \vec F across \surfaceS. - (If \surfaceS is not a closed surface, - choose \vec n so that it has a positive z-component, - unless otherwise indicated.) -

    -     - - - -

    - \surfaceS is the plane z = 3x+y on - 0\leq x\leq 1, 1\leq y\leq 4; - \vec F = \langle x^2,-z,2y\rangle. -

    -     - -

    - 24 -

    -     -     - - - -

    - \surfaceS is the plane - z = 8-x-y over the triangle with vertices at (0,0), - (1,0) and (1,5); - \vec F = \langle 3,1,2\rangle. -

    -     - -

    - 15 -

    -     -     - - - -

    - \surfaceS is the paraboloid z = x^2+y^2 over the unit disk; - \vec F = \langle 1,0,0\rangle. -

    -     - -

    - 0 -

    -     -     - - - -

    - \surfaceS is the unit sphere; - \vec F = \langle y-z,z-x,x-y\rangle. -

    -     - -

    - 0 -

    -     -     - - - -

    - \surfaceS is the square in space with corners at (0,0,0), - (1,0,0), - (1,0,1) and (0,0,1) (choose \vec n such that it has a positive y-component); - \vec F = \langle 0,-z,y\rangle. -

    -     - -

    - -1/2 -

    -     -     - - - -

    - \surfaceS is the disk in the yz-plane with radius 1, centered at (0,1,1) (choose \vec n such that it has a positive x-component); - \vec F = \langle y,z,x\rangle. -

    -     - -

    - \pi -

    -     -     - - - -

    - \surfaceS is the closed surface composed of \surfaceS_1, - whose boundary is the ellipse in the xy-plane described by - \frac{x^2}{25}+\frac{y^2}9 = 1 and \surfaceS_2, - part of the elliptical paraboloid f(x,y) = 1-\frac{x^2}{25}-\frac{y^2}9 (see graph); - \vec F = \langle 5,2,3\rangle. -

    - - - An elliptical paraboloid opens downward, intersecting the x,y plane in an ellipse. - -

    - An elliptical paraboloid is plotted relative to the usual three-dimensional coorindate axes. - The paraboloid has its vertex on the z axis at (0,0,1), and opens downward. -

    - -

    - The paraboloid intersects the xy plane along an ellipse; - the surface \surfaceS_2 consists of the portion of the paraboloid with 0\leq z\leq 1. -

    - -

    - The surface \surfaceS_1 is illustrated as the region on and inside the ellipse where \surfaceS_2 meets the xy plane. -

    -     - - - - - //ASY file for - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(19,17,1); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-5,5}; - real[] myychoice={-3,3}; - real[] myzchoice={1}; - defaultpen(0.5mm); - pair xbounds=(-5.75,5.75); - pair ybounds=(-5.75,5.75); - pair zbounds=(-0.25,1.1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the parabola z=y^2 for t from -2 to 2 - triple g(real t) {return (5cos(t),3sin(t),0);} - path3 mypath=graph(g,0,2pi,operator ..); draw(mypath,bluepen); - - //Draw the cylinder z=y^2 - triple f(pair t) { - return (5t.y*cos(t.x),3t.y*sin(t.x),0); - } - surface s=surface(f,(0,0),(2pi,1),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the cylinder z=y^2 - triple f(pair t) { - return (5t.y*cos(t.x),3t.y*sin(t.x),1-t.y^2); - } - surface s=surface(f,(0,0),(2pi,1),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - -     - -

    - 0; the flux over \surfaceS_1 is -45\pi and the flux over \surfaceS_2 is 45\pi. -

    -     -     - - - -

    - \surfaceS is the closed surface composed of \surfaceS_1, - part of the unit sphere and \surfaceS_2, - part of the plane z=1/2 (see graph); - \vec F = \langle x,-y,z\rangle. -

    - - - Approximately three quarters of the unit sphere. The top has been removed and replaced by a disk. - -

    - The unit sphere is plotted with respect to the usual three-dimensional coordnate axes. - However, it is not the entire sphere, but the portion with -1\leq z\leq \frac12. -

    - -

    - The top portion of the sphere has been removed, - an in its place is a disk in the plane z=\frac12. - The center of the disk is at \left(0,0,\frac12\right), - and the radius is \frac{\sqrt{3}}{2}. -

    - -

    - The surface looks something like what you'd get by slicing the top off of an orange. -

    -     - - - - - //ASY file for - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(17,17,11); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,1.25); - pair zbounds=(-1.25,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the parabola z=y^2 for t from -2 to 2 - triple g(real t) {return (sqrt(3)/2*cos(t),sqrt(3)/2*sin(t),.5);} - path3 mypath=graph(g,0,2pi,operator ..); draw(mypath,bluepen); - - //Draw the cylinder z=y^2 - triple f(pair t) { - return (sin(t.x)*cos(t.y),sin(t.x)*sin(t.y),cos(t.x)); - } - surface s=surface(f,(pi/3,0),(pi,2pi),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw the cylinder z=y^2 - triple f(pair t) { - return (t.y*sqrt(3)/2*cos(t.x),t.y*sqrt(3)/2*sin(t.x),.5); - } - surface s=surface(f,(0,0),(2pi,1),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - - -     - -

    - 9\pi/8; the flux over - \surfaceS_1 is 3\pi/4 (use - \vec r(u,v) = \langle \sin(u)\cos(v),\sin(u)\sin(v),\cos(u)\rangle on \pi/3\leq u\leq \pi, - 0\leq v\leq 2\pi) and the flux over - \surfaceS_2 is 3\pi/8 (use \vec r(u,v) = \langle v\sqrt{3}\cos (u)/2, v\sqrt{3}\sin(u)/2,1/2\rangle for - 0\leq u\leq 2\pi, 0\leq v\leq 1. -

    -     -     -     -     -     -
    -
    - The Divergence Theorem and Stokes' Theorem - - The Divergence Theorem -

    - - gives the Divergence Theorem in the plane, - which states that the flux of a vector field across a closed curve - equals the sum of the divergences over the region enclosed by the curve. - Recall that the flux was measured via a line integral, - and the sum of the divergences was measured through a double integral. -

    - -

    - We now consider the three-dimensional version of the Divergence Theorem. - It states, in words, that the flux across a closed surface - equals the sum of the divergences over the domain enclosed by the surface. - Since we are in space - (versus the plane), - we measure flux via a surface integral, - and the sums of divergences will be measured through a triple integral. -

    - - - - - The Divergence Theorem (in space) - -

    - Let D be a closed domain in space whose boundary is an orientable, - piecewise smooth surface \surfaceS with outer unit normal vector \vec n, - and let \vec F be a vector field whose components are differentiable on D. - ThenDivergence Theoremin space - - \iint_\surfaceS \vec F\cdot\vec n\, dS =\iiint_D \divv \vec F\, dV - . -

    -     -     - - - - - Using the Divergence Theorem in space - -

    - Let D be the domain in space bounded by the planes z=0 and z=2x, - along with the cylinder x=1-y^2, - as graphed in , - let \surfaceS be the boundary of D, - and let \vec F = \langle x+y,y^2, 2z\rangle. -

    - -
    - The surfaces used in - - A parabolic wedge, bounded by two planes and a parabolic cylinder. - -

    - The x, y, and z axes are plotted in three dimensions. - The z axis points up, the x axis points back and to the left, - and the y axis points toward us and to the left. -

    - -

    - The surface consists of three components: -

      -
    • -

      - In the xy plane is the region bounded by the y axis - and the parabola x=1-y^2. -

      - -

      - The parabola has its vertex at (1,0,0) and it meets the - y axis at (0,1,0) and (0,-1,0) -

      -
      • - -
    • -

      - Another parabolic region lies in the plane z=2x. - This plane intersects the y axis along the segment -1\leq y\leq 1, - which is the common bottom of the two parabolic regions; - it forms a sharp edge where the two planes meet. -

      - -

      - The parabolic boundary of this region is the curve given by - \vec{r}(t) = \la 1-t^2, t, 2-2t^2\ra. - This is a parabola in the plane z=2x that lies directly above the parabola x=1-y^2 in the xy plane. -

      -
      • - -
    • -

      - The remaining surface is the partion of the parabolic cylinder x=1-y^2 - that lies between the planes z=0 and z=2x. -

      -
      • -
    -

    - -

    - Overall, the surfaces form a wedge shape similar to a segment that might be cut from a tree being chopped down. -

    -     - - - - - //ASY file for divthm_space2_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(-17,32,12.6); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={-1,1}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-.1,1.2); - pair ybounds=(-1.3,1.3); - pair zbounds=(-.1,2.2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the planar parabola - triple f(pair t) { - return (t.y*(1-t.x^2),t.x,0);// - } - surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=redmeshpen; - draw(s,surfacepen2,meshpen=p); - - //Draw plane over parabola - triple f(pair t) { - return (t.y*(1-t.x^2),t.x,2*t.y*(1-t.x^2));// - } - surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw parabolic cylinder side - triple f(pair t) { - return ((1-t.x^2),t.x,2*t.y*(1-t.x^2));// - } - surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(real t) {return (1-t^2,t,0);} - path3 mypath=graph(f,-1,1,operator ..); - draw(mypath,curvepen); - - triple f(real t) {return (1-t^2,t,2*(1-t^2));} - path3 mypath=graph(f,-1,1,operator ..); - draw(mypath,curvepen); - - - -
    - -

    - Verify the Divergence Theorem by finding the total outward flux of \vec F across \surfaceS, and show this is equal to \iiint_D \divv\vec F\, dV. -

    -     - -

    - The surface \surfaceS is piecewise smooth, - comprising surfaces \surfaceS_1, - which is part of the plane z=2x, - surface \surfaceS_2, which is part of the cylinder x=1-y^2, - and surface \surfaceS_3, - which is part of the plane z=0. - To find the total outward flux across \surfaceS, we need to compute the outward flux across each of these three surfaces. -

    - -

    - We leave it to the reader to confirm that surfaces \surfaceS_1, - \surfaceS_2 and \surfaceS_3 can be parametrized by \vec r_1, - \vec r_2 and \vec r_3 respectively as - - \vec r_1(u,v) \amp = \la v(1-u^2), u, 2v(1-u^2)\ra, - \vec r_2(u,v) \amp = \la (1-u^2), u, 2v(1-u^2)\ra, - \vec r_3(u,v) \amp = \la v(1-u^2), u, 0\ra - , - where -1\leq u\leq 1 and - 0\leq v\leq 1 for all three functions. -

    - -

    - We compute a unit normal vector \vec n for each as \frac{\vec r_u\times\vec r_v}{\snorm{\vec r_u\times\vec r_v}}, - though recall that as we are integrating \vec F\cdot \vec n\, dS, - we actually only use \vec r_u\times\vec r_v. - Finally, in previous flux computations, - it did not matter which direction \vec n pointed as long as we made note of its direction. - When using the Divergence Theorem, - we need \vec n to point to the outside of the closed surface, - so in practice this means we'll either use - \vec r_u\times\vec r_v or \vec r_v\times\vec r_u, - depending on which points outside of the closed surface \surfaceS. -

    - -

    - We leave it to the reader to confirm the following cross products and integrations are correct. -

    - -

    - For \surfaceS_1, - we need to use \vec r_{1v}\times\vec r_{1u} = \langle 2(u^2-1),0,1-u^2\rangle. - (Note the z-component is nonnegative as u\leq 1, - therefore this vector always points up, - meaning to the outside, of \surfaceS.) - The flux across \surfaceS_1 is: - - \amp \text{ Flux across }\surfaceS_1: - \amp \quad = \iint_{\surfaceS_1} \vec F\cdot \vec n_1\, dS - \amp \quad = \int_0^1\int_{-1}^1 \vec F\big(\vec r_1(u,v)\big)\cdot \big(\vec r_{1v}\times\vec r_{1u}\big)\, du\, dv - \amp \quad = \int_0^1\int_{-1}^1 \la v(1-u^2)+u, u^2,4v(1-u^2)\ra \cdot \la 2(u^2-1),0,1-u^2\ra\, du\, dv - \amp \quad = \int_0^1\int_{-1}^1 \big(2u^4v+2u^3-4u^2v-2u+2v\big)\, du\, dv - \amp \quad = \frac{16}{15} - . -

    - -

    - For \surfaceS_2, - we use \vec r_{2u}\times\vec r_{2v} = \langle 2(1-u^2), 4u(1-u^2),0\rangle. - (Note the x-component is always nonnegative, - meaning this vector points outside \surfaceS.) - The flux across \surfaceS_2 is: - - \amp \text{ Flux across } \surfaceS_2: - \amp \quad = \iint_{\surfaceS_2} \vec F\cdot \vec n_2\, dS - \amp \quad = \int_0^1\int_{-1}^1 \vec F\big(\vec r_2(u,v)\big)\cdot \big(\vec r_{2u}\times\vec r_{2v}\big)\, du\, dv - \amp \quad = \int_0^1\int_{-1}^1 \la 1-u^2+u, u^2, 4v(1-u^2)\ra \cdot \la 2(1-u^2), 4u(1-u^2),0\ra\, du\, dv - \amp \quad = \int_0^1\int_{-1}^1 \big(4u^5-2u^4-2u^3+4u^2-2u-2\big)\, du\, dv - \amp \quad = \frac{32}{15} - . -

    - -

    - For \surfaceS_3, - we use \vec r_{3u}\times\vec r_{3v} = \langle 0,0,u^2-1\rangle. - (Note the z-component is never positive, - meaning this vector points down, outside of \surfaceS.) - The flux across \surfaceS_3 is: - - \amp\text{ Flux across } \surfaceS_3: - \amp \quad = \iint_{\surfaceS_3} \vec F\cdot \vec n_3\, dS - \amp \quad = \int_0^1\int_{-1}^1 \vec F\big(\vec r_3(u,v)\big)\cdot \big(\vec r_{3u}\times\vec r_{3v}\big)\, du\, dv - \amp \quad = \int_0^1\int_{-1}^1 \la v(1-u^2)+u,u^2,0\ra \cdot \la 0,0,u^2-1\ra\, du\, dv - \amp \quad = \int_0^1\int_{-1}^1 0\, du\, dv - \amp \quad = 0 - . -

    - -

    - Thus the total outward flux, - measured by surface integrals across all three component surfaces of \surfaceS, is 16/15+32/15+0 = 48/15 = 16/5 = 3.2. - We now find the total outward flux by integrating \divv \vec F over D. -

    - -

    - Following the steps outlined in , - we see the bounds of x, - y and z can be set as (thinking surface to surface, - curve to curve, point to point): - - 0\leq z\leq 2x; 0\leq x\leq 1-y^2; -1\leq y\leq 1 - . -

    - -

    - With \divv \vec F = 1+2y+2 = 2y+3, - we find the total outward flux of \vec F over \surfaceS as: - - \text{ Flux = } \iiint_D\divv \vec F\, dV = \int_{-1}^1\int_0^{1-y^2}\int_0^{2x}\big(2y+3\big)\, dz\, dx\, dy = 16/5 - , - the same result we obtained previously. -

    -     -     - -

    - In - we see that the total outward flux of a vector field across a closed surface can be found two different ways because of the Divergence Theorem. - One computation took far less work to obtain. - In that particular case, - since \surfaceS was comprised of three separate surfaces, - it was far simpler to compute one triple integral than three surface integrals - (each of which required partial derivatives and a cross product). - In practice, if outward flux needs to be measured, - one would choose only one method. - We will use both methods in this section simply to reinforce the truth of the Divergence Theorem. -

    - -

    - We practice again in the following example. -

    - - - Using the Divergence Theorem in space - -

    - Let \surfaceS be the surface formed by the paraboloid z=1-x^2-y^2, - z\geq 0, - and the unit disk centered at the origin in the xy-plane, - graphed in , - and let \vec F = \langle 0,0,z\rangle. - (This surface and vector field were used in .) -

    - -
    - The surfaces used in - - A circular paraboloid, opening downward, meets the x,y plane along the unit circle. The interior of the circle is a disk. - -

    - This is the same image used in . - A circular paraboloid has its vertex at (0,0,1) on the z axis and opens downward, - meeting the xy plane along the unit circle. -

    - -

    - The interior of the unit circle is a disk that serves as a cap at the bottom of the paraboloid. -

    -     - - - - - //ASY file for figsurfint1_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(22.8,20,5.5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1}; - real[] myychoice={2}; - real[] myzchoice={3}; - defaultpen(0.5mm); - - pair xbounds=(-1.1,1.1); - pair ybounds=(-1.1,1.1); - pair zbounds=(-.1,1.1); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane unit disk - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),0);// - } - surface s=surface(f,(-pi/2,0),(3pi/2,1),8,8,usplinetype=Spline,vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=redmeshpen; - draw(s,surfacepen2,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),1-t.y^2);// - } - surface s=surface(f,(0,0),(2pi,1),8,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(real t) {return (cos(t),sin(t),0);} - path3 mypath=graph(f,0,2pi,operator ..); - draw(mypath,curvepen); - - - -
    - -

    - Verify the Divergence Theorem; - find the total outward flux across \surfaceS and evaluate the triple integral of \divv \vec F, - showing that these two quantities are equal. -

    -     - -

    - We find the flux across \surfaceS first. - As \surfaceS is piecewise-smooth, - we decompose it into smooth components \surfaceS_1, - the disk, - and \surfaceS_2, the paraboloid, - and find the flux across each. -

    - -

    - In , - we found the flux across \surfaceS_1 is 0. - We also found that the flux across \surfaceS_2 is \pi/2. - (In that example, - the normal vector had a positive z component hence was an outer normal.) - Thus the total outward flux is 0+\pi/2 = \pi/2. -

    - -

    - We now compute \iiint_D \divv \vec F\, dV. - We can describe D as the domain bounded by (think surface to surface, - curve to curve, point to point): - - 0\leq z\leq 1-x^2-y^2, -\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}, -1\leq x\leq 1 - . -

    - -

    - This description of D is not very easy to integrate. - With polar, we can do better. - Let R represent the unit disk, - which can be described in polar simply as r, - where 0\leq r\leq 1 and 0\leq \theta\leq 2\pi. - With x=r\cos \theta and y=r\sin\theta, - the surface \surfaceS_2 becomes - - z=1-x^2-y^2 \Rightarrow 1-(r\cos\theta)^2-(r\sin\theta)^2 \Rightarrow 1-r^2 - . -

    - -

    - Thus D can be described as the domain bounded by: - - 0\leq z\leq 1-r^2, 0\leq r\leq 1, 0\leq \theta\leq 2\pi - . -

    - -

    - With \divv \vec F = 1, - we can integrate, recalling that dV = r\, dz\, dr\, d\theta: - - \iiint_D\divv \vec F\, dV = \int_0^{2\pi}\int_0^1\int_0^{1-r^2} r\, dz\, dr\, d\theta = \frac{\pi}2 - , - which matches our flux computation above. -

    -     -     - - - A <q>paradox</q> of the Divergence Theorem and Gauss's Law - -

    - The magnitude of many physical quantities - (such as light intensity or electromagnetic and gravitational forces) - follow an inverse square law: - the magnitude of the quantity at a point is inversely proportional to the square of the distance to the source of the quantity. - Gauss's Law -

    - -

    - Let a point light source be placed at the origin and let \vec F be the vector field which describes the intensity and direction of the emanating light. - At a point (x,y,z), - the unit vector describing the direction of the light passing through that point is \langle x,y,z\rangle/\sqrt{x^2+y^2+z^2}. - As the intensity of light follows the inverse square law, - the magnitude of \vec F at (x,y,z) is - k/(x^2+y^2+z^2) for some constant k. - Taken together, - - \vec F(x,y,z) = \frac{k}{(x^2+y^2+z^2)^{3/2}}\langle x,y,z\rangle - . -

    - -

    - Consider the cube, centered at the origin, - with sides of length 2a for some a \gt 0 (hence corners of the cube lie at (a,a,a), - (-a,-a,-a), - etc., as shown in ). - Find the flux across the six faces of the cube and compare this to \iint_D \divv\vec F\, dV. -

    - -
    - The cube used in - - A cube centered at the origin in space. - -

    - A cube is drawn in space, relative to three-dimensional coordinate axes. - It is centered at the origin, and its faces are parallel to the coordinate planes. -

    - -

    - One corner of the cube (the nearest one in the perspective used for the image) - is labeled with the coordinates (a,a,a). - The opposite corner is labeled with coordinates (-a,-a,-a). -

    -     - - - - - //ASY file for divthm_space3_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(30,20,12); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={}; - real[] myychoice={}; - real[] myzchoice={}; - defaultpen(0.5mm); - - pair xbounds=(-1.2,1.2); - pair ybounds=(-1.2,1.2); - pair zbounds=(-1.2,1.2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the faces of the cube - pen p=apexmeshpen; - - triple f(pair t) { - return (t.x,t.y,1);// - } - surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (t.x,t.y,-1);// - } - surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (t.x,1,t.y);// - } - surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (t.x,-1,t.y);// - } - surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (1,t.x,t.y);// - } - surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (-1,t.x,t.y);// - } - surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - draw(s,surfacepen,meshpen=p); - - label("$(a,a,a)$",(1.05,1.05,1.1)); - label("$(-a,-a,-a)$",(-1.05,-1.05,-1.1)); - dot((1,1,1),black+1mm); - - dot((-1,-1,-1),black+1mm); - - - -
    -     - -

    - Let \surfaceS_1 be the top face of the cube, - which can be parametrized by \vec r(u,v) = \langle u,v,a\rangle for - -a\leq u\leq a, -a\leq v\leq a. - We leave it to the reader to confirm that \vec r_u\times \vec r_v = \langle 0,0,1\rangle, - which points outside of the cube. -

    - -

    - The flux across this face is: - - \text{ Flux } \amp = \iint_{\surfaceS_1} \vec F\cdot \vec n\, dS - \amp = \int_{-a}^a\int_{-a}^a \vec F\big(\vec r(u,v)\big)\cdot \big(\vec r_u\times \vec r_v\big)\, du\, dv - \amp = \int_{-a}^a\int_{-a}^a \frac{k\ a}{(u^2+v^2+a^2)^{3/2}}\, du\, dv. - This double integral is not trivial to compute, requiring multiple trigonometric substitutions. This example is not meant to stress integration techniques, so we leave it to the reader to confirm the result is - \amp = \frac{2k\pi}3 - . -

    - -

    - Note how the result is independent of a; - no matter the size of the cube, - the flux through the top surface is always 2k\pi/3. -

    - -

    - An argument of symmetry shows that the flux through each of the six faces is 2k\pi/3, - thus the total flux through the faces of the cube is 6\times2k\pi/3 = 4k\pi. -

    - -

    - It takes a bit of algebra, but we can show that \divv\vec F = 0. - Thus the Divergence Theorem would seem to imply that the total flux through the faces of the cube should be - - \text{ Flux } =\iiint_D \divv \vec F\, dV = \iiint_D 0\, dV = 0 - , - but clearly this does not match the result from above. - What went wrong? -

    - -

    - Revisit the statement of the Divergence Theorem. - One of the conditions is that the components of \vec F must be differentiable on the domain enclosed by the surface. - In our case, \vec F is not - differentiable at the origin it is not even defined! - As \vec F does not satisfy the conditions of the Divergence Theorem, - it does not apply, - and we cannot expect \iint_\surfaceS \vec F\cdot\vec n\, dA = \iiint_D \divv\vec F\, dV. -

    - -

    - Since \vec F is differentiable everywhere except the origin, - the Divergence Theorem does apply over any domain that does not include the origin. - Let \surfaceS_2 be any surface that encloses the cube used before, - and let \hat D be the domain between - the cube and \surfaceS_2; - note how \hat D does not include the origin and so the Divergence Theorem does apply over this domain. - The total outward flux over \hat D is thus \iint_{\hat D}\divv \vec F\, dV = 0, - which means the amount of flux coming out of - \surfaceS_2 is the same as the amount of flux coming out of the cube. - The conclusion: the flux across any - surface enclosing the origin will be 4k\pi. -

    - -

    - This has an important consequence in electrodynamics. - Let q be a point charge at the origin. - The electric field generated by this point charge is - - \vec E = \frac{q}{4\pi \epsilon_0}\frac{\la x,y,z\ra}{(x^2+y^2+z^2)^{3/2}} - , - , it is \vec F with k = q/(4\pi \epsilon_0), - where \epsilon_0 is a physical constant - (the permittivity of free space). - Gauss's Law states that the outward flux of \vec E across any surface enclosing the origin is q/\epsilon_0. -

    -     -     - - - -

    - Our interest in the Divergence Theorem is twofold. - First, its truth alone is interesting: - to study the behavior of a vector field across a closed surface, - one can examine properties of that field within the surface. - Secondly, it offers an alternative way of computing flux. - When there are multiple methods of computing a desired quantity, - one has power to select the easiest computation as illustrated next. -

    - - - - - - Using the Divergence Theorem to compute flux - -

    - Let \surfaceS be the cube bounded by the planes x=\pm 1, - y=\pm 1, z=\pm 1, - and let \vec F = \langle x^2y,2yz,x^2z^3\rangle. - Compute the outward flux of \vec F over \surfaceS. -

    -     - -

    - We compute \divv \vec F = 2xy+2z+3x^2z^2. - By the Divergence Theorem, - the outward flux is the triple integral over the domain D enclosed by \surfaceS: - - \text{ Outward flux: } \int_{-1}^1\int_{-1}^1\int_{-1}^1(2xy+2z+3x^2z^2)\, dz\, dy\, dx = \frac83 - . -

    - -

    - The direct flux computation requires six surface integrals, - one for each face of the cube. - The Divergence Theorem offers a much more simple computation. -

    -     -     -     - - - Stokes' Theorem -

    - Just as the spatial Divergence Theorem of this section is an extension of the planar Divergence Theorem, Stokes' Theorem is the spatial extension of Green's Theorem. - Recall that Green's Theorem states that the circulation of a vector field around a closed curve in the plane is equal to the sum of the curl of the field over the region enclosed by the curve. - Stokes' Theorem effectively makes the same statement: - given a closed curve that lies on a surface \surfaceS, the circulation of a vector field around that curve is the same as the sum of - the curl of the field - across the enclosed surface. - We use quotes around the curl of the field - to signify that this statement is not quite correct, - as we do not sum \curl \vec F, - but \curl \vec F\cdot\vec n, - where \vec n is a unit vector normal to \surfaceS. That is, - we sum the portion of \curl \vec F that is orthogonal to \surfaceS at a point. -

    - - - -

    - Green's Theorem dictated that the curve was to be traversed counterclockwise when measuring circulation. - Stokes' Theorem will follow a right hand rule: - when the thumb of one's right hand points in the direction of \vec n, - the path C will be traversed in the direction of the curling fingers of the hand - (this is equivalent to traversing counterclockwise in the plane). -

    - - - Stokes' Theorem - -

    - Let \surfaceS be a piecewise smooth, - orientable surface whose boundary is a piecewise smooth curve C, - let \vec n be a unit vector normal to \surfaceS, let C be traversed with respect to \vec n according to the right hand rule, - and let the components of \vec F have continuous first partial derivatives over \surfaceS. Then - Stokes' Theorem - - \oint_C \vec F\cdot \, d\vec r = \iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS - . -

    -     -     - -

    - In general, the best approach to evaluating the surface integral in Stokes' Theorem is to parametrize the surface \surfaceS with a function \vec r(u,v). - We can find a unit normal vector \vec n as - - \vec n = \frac{\vec r_u\times\vec r_v}{\snorm{\vec r_u\times\vec r_v}} - . -

    - -

    - Since dS = \snorm{\vec r_u\times\vec r_v}\, dA, - the surface integral in practice is evaluated as - - \iint_\surfaceS (\curl \vec F)\cdot (\vec r_u\times\vec r_v)\, dA - , - where \vec r_u\times\vec r_v may be replaced by - \vec r_v\times\vec r_u to properly match the direction of this vector with the orientation of the parametrization of C. -

    - - - Verifying Stokes' Theorem - -

    - Considering the planar surface f(x,y) = 7-2x-2y, - let C be the curve in space that lies on this surface above the circle of radius 1 and centered at (1,1) in the xy-plane, - let \surfaceS be the planar region enclosed by C, - as illustrated in , - and let \vec F = \langle x+y,2y, y^2\rangle. - Verify Stoke's Theorem by showing \oint_C \vec F\cdot \, d\vec r = \iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS. -

    - -
    - As given in , the surface \surfaceSis the portion of the plane bounded by the curve - - An ellipse drawn on a plane in space. It lies over a circle in the x,y plane. - -

    - The usual three-dimensional coordinate axes are shown, - with the x axis toward us and to the left, - the y axis toward us and to the right, and the z axis pointing up. -

    - -

    - A circle is drawn in the xy plane. It lies in the first quadrant; - its center is at (1,1,0), and its radius is 1. - It intercepts the x axis at (1,0,0), and the y axis at (0,1,0). -

    - -

    - A plane in space is drawn in the first octant. - It is shaped like a rectangle; its highest corner is on the z axis at (0,0,7); - the opposite corner is lowest, at (2,2,-1). -

    - -

    - On the plane, a curve C is drawn. - The curve is an ellipse, consisting of all points in the plane that lie above the circle in the xy plane. - The surface \surfaceS for this problem is the region on and inside this ellipse. -

    -     - - - - - //ASY file for figstokes1_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11,19,27); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={1,2}; - real[] myzchoice={2,4,6,8}; - defaultpen(0.5mm); - - pair xbounds=(-.1,2.5); - pair ybounds=(-.1,2.5); - pair zbounds=(-.1,9); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (t.x,t.y,-2t.x-2t.y+7);// - } - surface s=surface(f,(-.1,-.1),(2.2,2.2),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return (cos(t)+1,sin(t)+1,7-2(cos(t)+1)-2(sin(t)+1));} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (cos(t)+1,sin(t)+1,0);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+dashed+linewidth(1)); - - label("$\mathcal{S}$",(1,1,4)); - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - - -
    -     - -

    - We begin by parametrizing C and then find the circulation. - A unit circle centered at (1,1) can be parametrized with x=\cos(t)+1, - y=\sin(t)+1 on 0\leq t\leq 2\pi; - to put this curve on the surface f, - make the z component equal f(x,y): - z = 7-2(\cos(t)+1)-2(\sin(t)+1) = 3-2\cos(t) - 2\sin(t). - All together, - we parametrize C with \vec r(t) = \la \cos(t)+1, \sin(t)+1, 3-2\cos(t)-2\sin(t)\ra. -

    - -

    - The circulation of \vec F around C is - - \oint_C\vec F\cdot \, d\vec r \amp = \int_0^{2\pi}\vec F\big(\vec r(t)\big)\cdot \vrp(t)\, dt - \amp = \int_0^{2\pi}\big(2\sin^3t-2\cos(t)\sin^2t+3\sin^2t-3\cos(t)\sin(t)\big)\, dt - \amp = 3\pi - . -

    - -

    - We now parametrize \surfaceS. (We reuse the letter r - for our surface as this is our custom.) Based on the parametrization of C above, - we describe \surfaceS with \vec r(u,v) = \la v\cos(u)+1, v\sin(u)+1, 3-2v\cos(u)-2v\sin(u)\ra, - where 0\leq u\leq 2\pi and 0\leq v\leq 1. -

    - -

    - We leave it to the reader to confirm that \vec r_u\times \vec r_v = \langle 2v,2v,v\rangle. - As 0\leq v\leq 1, - this vector always has a non-negative z-component, - which the right-hand rule requires given the orientation of C used above. - We also leave it to the reader to confirm \curl\vec F = \langle 2y,0,-1\rangle. -

    - -

    - The surface integral of Stokes' Theorem is thus - - \iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS \amp = \iint_\surfaceS (\curl\vec F)\cdot (\vec r_u\times \vec r_v)\, dA - \amp = \int_0^1\int_0^{2\pi} \langle 2v\sin(u)+2,0,-1\rangle\cdot\langle 2v,2v,v\rangle\, du\, dv - \amp = 3\pi - , - which matches our previous result. -

    -     -     - - -

    - One of the interesting results of Stokes' Theorem is that if two surfaces \surfaceS_1 and - \surfaceS_2 share the same boundary, - then \iint_{\surfaceS_1} (\curl \vec F)\cdot \vec n\, dS = \iint_{\surfaceS_2} (\curl \vec F)\cdot \vec n\, dS. - That is, the value of these two surface integrals is somehow independent of the interior of the surface. - We demonstrate this principle in the next example. -

    - - - Stokes' Theorem and surfaces that share a boundary - -

    - Let C be the curve given in - and note that it lies on the surface z = 6-x^2-y^2. - Let \surfaceS be the region of this surface bounded by C, - and let \vec F = \langle x+y,2y,y^2\rangle as in the previous example. - Compute \iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS to show it equals the result found in the previous example. -

    - -
    - As given in , the surface \surfaceSis the portion of the plane bounded by the curve - -
    - - - The portion of a circular paraboloid in the first octant, and an elliptical curve that lies on the surface. - -

    - This image has elements in common with . - We have the same three-dimensional coordinate axes, - and the same circle in the xy plane, centered at (1,1,0), of radius 1. -

    - -

    - The portion of the circular paraboloid z=6-x^2-y^2 that lies in the first octant is plotted. - It has its vertex at (0,0,6), and opens downward. - On the surface is an ellipse. - Although the surface is curved, the ellipse is the same ellipse that was drawn in . - The surface \surfaceS is the portion of the paraboloid on and inside the ellipse. -

    -     - - - - - //ASY file for figstokes2_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11,19,27); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={1,2}; - real[] myzchoice={2,4,6,8}; - defaultpen(0.5mm); - - pair xbounds=(-.1,2.5); - pair ybounds=(-.1,2.5); - pair zbounds=(-.1,9); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane z=6-x^2-y^2 - triple f(pair t) { - return (t.x,t.y,6-t.x^2-t.y^2);// - } - surface s=surface(f,(-.1,-.1),(2.2,2.2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return (cos(t)+1,sin(t)+1,6-(cos(t)+1)^2-(sin(t)+1)^2);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (cos(t)+1,sin(t)+1,0);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+dashed+linewidth(1)); - - label("$\mathcal{S}$",(1,1,5)); - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - - -
    - -
    - - - The parabolic surface from the previous image is shown intersecting a plane along a common elliptical curve. - -

    - This image shows both the circular paraboloid from - and the plane from . -

    - -

    - The two surfaces are shown intersecting along the same ellipse. - This illustrates how the curve C is the boundary of both surfaces. -

    -     - - - - - //ASY file for figstokes2_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11,19,27); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2}; - real[] myychoice={1,2}; - real[] myzchoice={2,4,6,8}; - defaultpen(0.5mm); - - pair xbounds=(-.1,2.5); - pair ybounds=(-.1,2.5); - pair zbounds=(-.1,9); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw the plane z=7-2x-2y - triple f(pair t) { - return (t.x,t.y,-2t.x-2t.y+7);// - } - surface s=surface(f,(-.1,-.1),(2.2,2.2),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=redmeshpen; - draw(s,surfacepen2,meshpen=p); - - //Draw the plane z=6-x^2-y^2 - triple f(pair t) { - return (t.x,t.y,6-t.x^2-t.y^2);// - } - surface s=surface(f,(-.1,-.1),(2.2,2.2),16,16,usplinetype=new splinetype[] {notaknot,notaknot,monotonic},vsplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return (cos(t)+1,sin(t)+1,6-(cos(t)+1)^2-(sin(t)+1)^2);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (cos(t)+1,sin(t)+1,0);} - path3 mypath=graph(g,0,2pi,operator ..); - draw(mypath,bluepen+dashed+linewidth(1)); - - //label("$\mathcal{S}$",(1,1,5)); - //draw((0,1,0)--(0,3,0)--(3,1,0)--cycle,curvepen); - - - -
    -     - -
    -     - -

    - We begin by demonstrating that C lies on the surface z=6-x^2-y^2. - We can parametrize the x and y components of C with x=\cos(t)+1, - y=\sin(t)+1 as before. - Lifting these components to the surface z=6-x^2-y^2 gives the z component as z = 6-x^2-y^2 = 6-(\cos(t)+1)^2-(\sin(t)+1)^2 = 3-2\cos(t)-2\sin(t), - which is the same z component as found in . - Thus the curve C lies on the surface z=6-x^2-y^2, - as illustrated in . -

    - -

    - Since C and \vec F are the same as in the previous example, - we already know that \oint_C\vec F\cdot\, d\vec r = 3\pi. - We confirm that this is also the value of \iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS. -

    - -

    - We parametrize \surfaceS with - - \vec r(u,v) = \langle v\cos(u)+1,v\sin(u)+1, 6-(v\cos(u)+1)^2-(v\sin(u)+1)^2\rangle - , - where 0\leq u\leq 2\pi and 0\leq v\leq 1, - and leave it to the reader to confirm that - - \vec r_u\times \vec r_v = \la 2v\big(v\cos(u)+1\big), 2v\big(v\sin(u)+1\big),v\ra - , - which also conforms to the right-hand rule with regard to the orientation of C. - With \curl \vec F = \langle 2y,0,-1\rangle as before, we have - - \amp\iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS - \quad\quad\amp = \int_0^1\int_0^{2\pi} \la 2v\sin(u)+2,0,-1\ra\cdot \la 2v\big(v\cos(u)+1\big), 2v\big(v\sin(u)+1\big),v\ra\, du\, dv - \quad\quad\amp =3\pi - . -

    - -

    - Even though the surfaces used in this example and in are very different, - because they share the same boundary, Stokes' Theorem guarantees they have equal sum of curls - across their respective surfaces. -

    -     -     - - -     - - - A Common Thread of Calculus -

    - We have threefold interest in each of the major theorems of this chapter: - the Fundamental Theorem of Line Integrals, Green's, Stokes' and the Divergence Theorems. - First, we find the beauty of their truth interesting. - Second, each provides two methods of computing a desired quantity, - sometimes offering a simpler method of computation. -

    - -

    - There is yet one more reason of interest in the major theorems of this chapter. - These important theorems also all share an important principle with the Fundamental Theorem of Calculus, - introduced in . -

    - -

    - Revisit this fundamental theorem, - adopting the notation used heavily in this chapter. - Let I be the interval [a,b] and let y=F(x) be differentiable on I, - with F\,'(x) = f(x). - The Fundamental Theorem of Calculus states that - - \int_I f(x)\, dx = F(b) - F(a) - . -

    - -

    - That is, the sum of the rates of change of a function F over an interval I can also be calculated with a certain sum of F itself on the boundary of I - (in this case, at the points x=a and x=b). -

    - -

    - Each of the named theorems above can be expressed in similar terms. - Consider the Fundamental Theorem of Line Integrals: - given a function f(x,y), - the gradient \nabla f is a type of rate of change of f. - Given a curve C with initial and terminal points A and B, - respectively, - this fundamental theorem states that - - \int_C \nabla f\, ds = f(B) - f(A) - , - where again the sum of a rate of change of f along a curve C can also be evaluated by a certain sum of f at the boundary of C (, the points A and B). -

    - -

    - Green's Theorem is essentially a special case of Stokes' Theorem, - so we consider just Stokes' Theorem here. - Recalling that the curl of a vector field \vec F is a measure of a rate of change of \vec F, Stokes' Theorem states that over a surface \surfaceS bounded by a closed curve C, - - \iint_\surfaceS \big(\curl \vec F\big)\cdot \vec n\, dS = \oint_C \vec F\cdot d\vec r - , - , the sum of a rate of change of \vec F can be calculated with a certain sum of \vec F itself over the boundary of \surfaceS. In this case, - the latter sum is also an infinite sum, requiring an integral. -

    - -

    - Finally, the Divergence Theorems state that the sum of divergences of a vector field - (another measure of a rate of change of \vec F) - over a region can also be computed with a certain sum of \vec F over the boundary of that region. - When the region is planar, the latter sum of \vec F is an integral; - when the region is spatial, - the latter sum of \vec F is a double integral. -

    - -

    - The common thread among these theorems: - the sum of a rate of change of a function over a region can be computed as another sum of the function itself on the boundary of the region. - While very general, this is a very powerful and important statement. -

    -     - - - - Terms and Concepts - - -

    - What are the differences between the Divergence Theorems of and this section? -

    -     - - - -

    - Answers will vary; in , - the Divergence Theorem connects outward flux over a closed curve in the plane to the divergence of the vector field, - whereas in this section the Divergence Theorem connects outward flux over a closed surface in space to the divergence of the vector field. -

    -     -     - - - -

    - What property of a vector field does the Divergence Theorem relate to flux? -

    -     - - - -

    - Divergence. -

    -     -     - - - -

    - What property of a vector field does Stokes' Theorem relate to circulation? -

    -     - - - -

    - Curl. -

    -     -     - - - -

    - Stokes' Theorem is the spatial version of what other theorem? -

    -     - - - -

    - Green's Theorem. -

    -     -     -     - - - Problems - - - -

    - A closed surface \surfaceS enclosing a domain D and a vector field \vec F are given. - Verify the Divergence Theorem on \surfaceS; that is, - show \iint_\surfaceS \vec F\cdot \vec n\, dS = \iiint_D\divv \vec F\, dV. -

    -     - - - -

    - \surfaceS is the surface bounding the domain D enclosed by the plane - z=2-x/2-2y/3 and the coordinate planes in the first octant; - \vec F = \langle x^2,y^2,x\rangle. -

    - - - A tetrahedron with vertices (0,0,0), (4,0,0), (0,3,0), and (0,0,2). - -

    - A surface consisting of the four faces of a tetrahedron is plotted relative to three-dimensional coordinate axes. - Each face is a triangle; there is one in each coordinate plane, - and the fourth face lies in the plane z=2-x/2-2y/3. -

    - -

    - The plane z=2-x/2-2y/3 is plotted in the first octant. - It meets the coordinate axes at the points (4,0,0), (0,3,0), and (0,0,2). - These are three of the four vertices of the tetrahedron; the remaining vertex is at the origin. -

    -     - - - - - //ASY file for fig14_07_ex_05_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12,12,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={4}; - real[] myychoice={3}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-.5,4.5); - pair ybounds=(-.5,4.5); - pair zbounds=(-0.25,3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - //triple g(real t) {return (2cos(t),2sin(t),0);} - //path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - //triple g(real t) {return (t,4-t^2,2*(4-t^2));} - //path3 mypath=graph(g,-2,2,operator ..); - draw((4,0,0)--(0,3,0)--(0,0,2)--cycle,bluepen+linewidth(2)); - draw((4,0,0)--(0,0,0),bluepen+linewidth(2)); - draw((0,3,0)--(0,0,0),bluepen+linewidth(2)); - draw((0,0,2)--(0,0,0),bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y*(3-3t.x/4),2-t.x/2-2t.y*(3-3t.x/4)/3);// - } - surface s=surface(f,(0,0),(4,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y*(3-3t.x/4),0);// - } - surface s=surface(f,(0,0),(4,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,0,t.y*(2-t.x/2));// - } - surface s=surface(f,(0,0),(4,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (0,t.x,t.y*(2-2t.x/3));// - } - surface s=surface(f,(0,0),(3,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - //label("$z=2y$",(-2,2,7)); - //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - //label("$y=4-x^2$",(2.5,2,0)); - //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - Outward flux across the plane z=2-x/2-2y/3 is 14; - across the plane z=0 the outward flux is -8; - across the planes x=0 and y=0 the outward flux is 0. -

    - -

    - Total outward flux: 14. -

    - -

    - \iint_D\divv\vec F\, dV = \int_0^{4}\int_0^{3-3x/4}\int_0^{2-x/2-2y/3}(2x+2y)\, dz\, dy\, dx = 14. -

    -     -     - - - -

    - \surfaceS is the surface bounding the domain D enclosed by the cylinder - x^2+y^2=1 and the planes z=-3 and z=3; - \vec F = \langle -x,y,z\rangle. -

    - - - A circular cylinder, centered on the z axis, capped by disks in the planes z=3 and z=-3. - -

    - The surface consists of three parts: -

      -
    1. -

      - The circular cylinder x^2+y^2=1, which is symmetric about the z axis, - from z=-3 to z=3. -

      -
      2. -
    2. -

      - The disk bounded by x^2+y^2=1 in the plane z=3; - this is the top cap of the cylinder. -

      -
      3. -
    3. -

      - The disk bounded by x^2+y^2=1 in the plane z=-3; - this is the bottom cap of the cylinder. -

      -
      4. -
    -

    -     - - - - - //ASY file for fig14_07_ex_06_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12,12,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={-3,3}; - defaultpen(0.5mm); - - pair xbounds=(-3.5,3.5); - pair ybounds=(-3.5,3.5); - pair zbounds=(-3.5,3.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - triple g(real t) {return (cos(t),sin(t),3);} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (cos(t),sin(t),-3);} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (cos(t.x),sin(t.x),t.y);// - } - surface s=surface(f,(0,-3),(2pi,3),16,2,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),3);// - } - surface s=surface(f,(0,0),(2pi,1),8,2,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),-3);// - } - surface s=surface(f,(0,0),(2pi,1),8,2,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - //label("$z=2y$",(-2,2,7)); - //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - //label("$y=4-x^2$",(2.5,2,0)); - //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - Outward flux across the cylinder x^2+y^2=1 is 0; - across the plane z=3 the outward flux is 3\pi; - across the plane z=-3 the outward flux is 3\pi. -

    - -

    - Total outward flux: 6\pi. -

    - -

    - \iint_D\divv\vec F\, dV = \int_0^{2\pi}\int_0^{1}\int_{-3}^{3}r\, dz\, dr\, d\theta = 6\pi. -

    -     -     - - - -

    - \surfaceS is the surface bounding the domain D enclosed by - z=xy(3-x)(3-y) and the plane z=0; - \vec F = \langle 3x,4y,5z+1\rangle. -

    - - - A steep hill with a square base in the x-y plane. - -

    - The graph z=xy(3-x)(3-y) is plotted relative to three-dimensional coordinate axes. - This surface meets the xy plane along a square, - with sides given by the x and y axes, - and the lines x=3 and y=3 in the xy plane. -

    - -

    - This square in the plane is part of the surface, - and the other part is the graph. - The graph is shaped like a steep hill, with a single peak in the center of the square. -

    -     - - - - - //ASY file for fig14_07_ex_07_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12,12,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={3}; - real[] myychoice={3}; - real[] myzchoice={4}; - defaultpen(0.5mm); - - pair xbounds=(-.5,3.5); - pair ybounds=(-.5,3.5); - pair zbounds=(-.5,4.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - //triple g(real t) {return (cos(t),sin(t),3);} - //path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - //triple g(real t) {return (cos(t),sin(t),-3);} - //path3 mypath=graph(g,0,2pi,operator ..); - draw((0,0,0) -- (3,0,0) -- (3,3,0) -- (0,3,0)--cycle,bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y,t.x*t.y*(3-t.x)*(3-t.y));// - } - surface s=surface(f,(0,0),(3,3),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y,0);// - } - surface s=surface(f,(0,0),(3,3),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - //label("$z=2y$",(-2,2,7)); - //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - //label("$y=4-x^2$",(2.5,2,0)); - //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - Outward flux across the surface z=xy(3-x)(3-y) is 252; - across the plane z=0 the outward flux is -9. -

    - -

    - Total outward flux: 243. -

    - -

    - \iint_D\divv\vec F\, dV = \int_0^{3}\int_0^{3}\int_{0}^{xy(3-x)(3-y)}12\, dz\, dy\, dx = 243. -

    -     -     - - - -

    - \surfaceS is the surface composed of \surfaceS_1, - the paraboloid z=4-x^2-y^2 for z\geq 0, - and \surfaceS_2, the disk of radius 2 centered at the origin; - \vec F = \langle x,y,z^2\rangle. -

    - - - A familiar dome-shaped surface between a circular paraboloid and the x,y plane. - -

    - The surface consists of the circular paraboloid z=4-x^2-y^2, - which opens downward from (0,0,4) and meets the xy plane along the circle x^2+y^2=4, - as well as the disk in the xy plane that lies on and inside the circle of intersection - between the plane and the paraboloid. -

    -     - - - - - //ASY file for fig14_07_ex_08_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12,12,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={4}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.25,5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - triple g(real t) {return (2cos(t),2sin(t),0);} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - //triple g(real t) {return (t,4-t^2,2*(4-t^2));} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (2*t.y*cos(t.x),t.y*2*sin(t.x),4-4*t.y^2);// - } - surface s=surface(f,(0,0),(2pi,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (2*t.y*cos(t.x),2*t.y*sin(t.x),0);// - } - surface s=surface(f,(0,0),(2pi,1),8,4,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - //label("$z=2y$",(-2,2,7)); - //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - //label("$y=4-x^2$",(2.5,2,0)); - //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - Outward flux across the paraboloid is 112\pi/3; - across the disk the outward flux is 0. -

    - -

    - Total outward flux: 112\pi/3. -

    - -

    - \iint_D\divv\vec F\, dV = \int_0^{2\pi}\int_0^2\int_0^{4-r^2}(2z+2)r\, dz\, dr\, d\theta = 112\pi/3. -

    -     -     -     - - - -

    - A closed curve C that is the boundary of a surface \surfaceS is given along with a vector field \vec F. - Verify Stokes' Theorem on C; - that is, show \oint_C \vec F\cdot d\vec r = \iint_\surfaceS\big(\curl \vec F\,\big)\cdot\vec n\, dS. -

    -     - - - -

    - C is the curve parametrized by - \vec r(t) = \langle \cos(t), \sin(t), 1\rangle and \surfaceS is the portion of - z=x^2+y^2 enclosed by C; - \vec F = \langle z,-x,y\rangle. -

    - - - A bowl-shaped surface given by a circular paraboloid plotted over the unit disk. - -

    - The surface z=x^2+y^2 is plotted relative to a set of three-dimensional coordinate axes. - The domain used for the graph is the unit disk, so that the surface is shaped like a bowl. - The boundary of the surface is a circle in the plane z=1. -

    -     - - - - - //ASY file for fig14_07_ex_09_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12,12,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-0.25,2); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - triple g(real t) {return (cos(t),sin(t),1);} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - //triple g(real t) {return (t,4-t^2,2*(4-t^2));} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),t.y^2);// - } - surface s=surface(f,(0,0),(2pi,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - //label("$z=2y$",(-2,2,7)); - //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - //label("$y=4-x^2$",(2.5,2,0)); - //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - Circulation on C: \oint_C \vec F\cdot d\vec r = -\pi -

    - -

    - \iint_\surfaceS\big(\curl \vec F\big)\cdot\vec n\, dS = -\pi. -

    -     -     - - - -

    - C is the curve parametrized by - \vec r(t) = \langle \cos(t), \sin(t), e^{-1}\rangle and \surfaceS is the portion of - z=e^{-x^2-y^2} enclosed by C; - \vec F = \langle -y,x,1\rangle. -

    - - - A bell-shaped surface in space, with a circular boundary. - -

    - The graph z=e^{-x^2-y^2} is plotted relative to a set of three-dimensional coordinate axes, - for 0\leq x^2+y^2\leq 1. - The surface is shaped like a bell or a dome, opening downward from a vertext at (0,0,1). -

    - -

    - The boundary of the surface is a circle; - the circle lies in the horizontal plane z=e^{-1}, - has radius 1, and is centered on the z axis. -

    -     - - - - - //ASY file for fig14_07_ex_10_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12,12,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - triple g(real t) {return (cos(t),sin(t),exp(-1));} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - //triple g(real t) {return (t,4-t^2,2*(4-t^2));} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),exp(-t.y^2));// - } - surface s=surface(f,(0,0),(2pi,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - //label("$z=2y$",(-2,2,7)); - //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - //label("$y=4-x^2$",(2.5,2,0)); - //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - Circulation on C: \oint_C \vec F\cdot d\vec r = \pi -

    - -

    - \iint_\surfaceS\big(\curl \vec F\big)\cdot\vec n\, dS = \pi. -

    -     -     - - - -

    - C is the curve that follows the triangle with vertices at (0,0,2), - (4,0,0) and (0,3,0), - traversing the the vertices in that order and returning to (0,0,2), - and \surfaceS is the portion of the plane - z=2-x/2-2y/3 enclosed by C; - \vec F = \langle y,-z,y\rangle. -

    - - - A triangular surface in the first octant, with intercepts at (4,0,0), (0,3,0), and (0,0,2). - -

    - The curve C is plotted in space relative to three-dimensional coordinate axes. - It is a triangle, given by the path from (0,0,2) on the z axis, - to (4,0,0) on the x axis, to (0,3,0) on the y axis, and then back to (0,0,2). -

    - -

    - The surface \surfaceS is the portion of the plane in the first octant that lies on and inside this triangle. -

    -     - - - - - //ASY file for fig14_07_ex_10_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12,12,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={4}; - real[] myychoice={3}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-.5,4.5); - pair ybounds=(-.5,4.5); - pair zbounds=(-0.25,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - //triple g(real t) {return (cos(t),sin(t),exp(-1));} - //path3 mypath=graph(g,0,2pi,operator ..); - draw((0,0,2)--(4,0,0)--(0,3,0)--cycle,bluepen+linewidth(2)); - //triple g(real t) {return (t,4-t^2,2*(4-t^2));} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y*(3-3t.x/4),2-t.x/2-2*(t.y*(3-3t.x/4))/3);// - } - surface s=surface(f,(0,0),(4,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - //label("$z=2y$",(-2,2,7)); - //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - //label("$y=4-x^2$",(2.5,2,0)); - //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - Circulation on C: The flow along the line from (0,0,2) to (4,0,0) is 0; - from (4,0,0) to (0,3,0) it is -6, - and from (0,3,0) to (0,0,2) it is 6. - The total circulation is 0+(-6)+6=0. -

    - -

    - \iint_\surfaceS\big(\curl \vec F\big)\cdot\vec n\, dS = \iint_\surfaceS 0 \, dS = 0. -

    -     -     - - - -

    - C is the curve whose x and y coordinates follow the parabola y=1-x^2 from x=1 to x=-1, - then follow the line from (-1,0) back to (1,0), - where the z coordinates of C are determined by f(x,y) = 2x^2+y^2, - and \surfaceS is the portion of - z=2x^2+y^2 enclosed by C; - \vec F = \langle y^2+z,x,x^2-y\rangle. -

    - - - Graph of an elliptic paraboloid over a parabolic domain. - -

    - The usual three-dimensional coordinate axes are drawn. - The surface is the graph z=2x^2+y^2, which is an elliptic paraboloid. -

    - -

    - The domain for the graph is not illustrated, - but it is the region between the x axis and the parabola y=1-x^2. -

    - -

    - The surface is shaped like a hammock or a sail. - Its boundary consists of two parabolic curves that meet at two cusps. - The cusps are the highest points on the surface; - they lie in the xz plane at (1,0,2) and (-1,0,2). -

    - -

    - One part of the boundary is a parabola in the xz plane, - given by z=2x^2 for -1\leq x\leq 1. - The other part is the portion of the elliptic paraboloid z=2x^2+y^2 - that lies above the curve y=1-x^2 in the xy plane. -

    -     - - - - - //ASY file for fig14_07_ex_10_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12,12,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={1}; - real[] myzchoice={2}; - defaultpen(0.5mm); - - pair xbounds=(-1.5,1.5); - pair ybounds=(-1.5,1.5); - pair zbounds=(-0.25,2.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - triple g(real t) {return (t,1-t^2,2t^2+(1-t^2)^2);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,0,2t^2);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - //triple g(real t) {return (t,4-t^2,2*(4-t^2));} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y*(1-t.x^2),2t.x^2+(t.y*(1-t.x^2))^2);// - } - surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - //label("$z=2y$",(-2,2,7)); - //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - //label("$y=4-x^2$",(2.5,2,0)); - //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - Circulation on C: The flow along the parabola is -32/15; - the flow along the line is 4/3. - The total circulation is 4/3-32/15 = -4/5. -

    - -

    - \iint_\surfaceS\big(\curl \vec F\big)\cdot\vec n\, dS = -4/5. -

    -     -     -     - - - -

    - A closed surface \surfaceS and a vector field \vec F are given. - Find the outward flux of \vec F over \surfaceS either through direct computation or through the Divergence Theorem. -

    -     - - - -

    - \surfaceS is the surface formed by the intersections of z=0 and z=(x^2-1)(y^2-1); - \vec F = \langle x^2+1,yz,xz^2\rangle. -

    - - - A domed hill with a square base in the x-y plane. - -

    - The graph z=(x^2-1)(y^2-1) is plotted relative to three-dimensional coordinate axes. - This surface meets the xy plane along a square, - with sides given by the lines x=\pm 1 and y=\pm 1 in the xy plane. -

    - -

    - The graph is shaped like a hill, with its peak on the z axis at (0,0,1). - The boundary of the graph is the square in the xy plane. -

    - -

    - The surface consists of this graph and the region in the xy - plane bounded by the square -1\leq x,y\leq 1. -

    -     - - - - - //ASY file for fig14_07_ex_13_3D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12,12,5); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={-1,1}; - defaultpen(0.5mm); - - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,1.25); - pair zbounds=(-0.25,1.25); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - //triple g(real t) {return (cos(t),sin(t),3);} - //path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - //triple g(real t) {return (cos(t),sin(t),-3);} - //path3 mypath=graph(g,0,2pi,operator ..); - draw((-1,-1,0) -- (-1,1,0) -- (1,1,0) -- (1,-1,0)--cycle,bluepen+linewidth(2)); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y,(t.x^2-1)*(t.y^2-1));// - } - surface s=surface(f,(-1,-1),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y,0);// - } - surface s=surface(f,(-1,-1),(1,1),1,1,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - //label("$z=2y$",(-2,2,7)); - //draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - //label("$y=4-x^2$",(2.5,2,0)); - //draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - 128/225 -

    -     -     - - - -

    - \surfaceS is the surface formed by the intersections of the planes z=\frac12(3-x), - x=1, - y=0, y=2 and z=0; - \vec F = \langle x,y^2,z\rangle. -

    - - - A triangular prism, with two triangular faces and three rectangular faces. - -

    - The surface \surfaceS consists of all five faces of a triangular prism. - The prism is plotted relative to a set of three-dimensional coordinate axes. -

    - -

    - Two of the faces are triangles. These lie in the planes y=0 and y=2. - The vertices of the triangles are at (1,0,0), (3,0,0), and (1,0,1) for the face in the xz plane, - and at (1,2,0), (3,2,0), and (1,2,1) for the face in the plane y=2. -

    - -

    - One face is in the xy plane; - it is a square given by 1\leq x\leq 3 and 0\leq y\leq 2. -

    - -

    - Another face is in the plane x=1; - it is a rectangle with 0\leq y\leq 2 and 0\leq z\leq 1. -

    - -

    - The final face is in the plane z=\frac12(3-x). - This face lies on the hypotenuse of each triangular face. - It is a rectangle, with vertices (3,0,0), (3,2,0), (1,2,1), and (1,0,1). -

    -     - - - - - //ASY file for fig13_06_ex_083D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(8.8,7.8,3); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={1,2,3}; - real[] myychoice={1,2,3}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-0.5,3.5); - pair ybounds=(-0.5,3.5); - pair zbounds=(-0.5,1.75); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //edges of object - draw((1,0,0)--(1,2,0)--(3,2,0)--(3,0,0)--cycle,bluepen+linewidth(2)); - draw((3,0,0)--(1,0,1)--(1,2,1)--(3,2,0),bluepen+linewidth(2)); - draw((3,0,0)--(1,0,1),bluepen+linewidth(2)); - draw((1,0,0)--(1,0,1),bluepen+linewidth(2)); - draw((1,2,0)--(1,2,1),bluepen+linewidth(2)); - - //shade faces - import three; - path3 p=(1,0,0)--(1,2,0)--(3,2,0)--(3,0,0); - draw(surface(p -- cycle), surfacepen); - path3 p=(1,0,1)--(1,2,1)--(3,2,0)--(3,0,0); - draw(surface(p -- cycle), surfacepen); - path3 p=(1,0,0)--(1,2,0)--(1,2,1)--(1,0,1); - draw(surface(p -- cycle), surfacepen); - path3 p=(1,0,0)--(3,0,0)--(1,0,1); - draw(surface(p -- cycle), surfacepen); - path3 p=(1,2,0)--(3,2,0)--(1,2,1); - draw(surface(p -- cycle), surfacepen); - - //labels and arrow - label("$z=\frac{1}{2}(3-x)$",(3,0,1.1)); - draw((3,0.25,1)--(2,1,0.45),Arrow3(size=2mm)); - - - -     - -

    - 8 -

    -     -     - - - -

    - \surfaceS is the surface formed by the intersections of the planes z=2y, - y=4-x^2 and z=0; - \vec F = \langle xz,0,xz\rangle. -

    - - - A wedge-shaped surface given by two intersecting planes and a parabolic cylinder. - -

    - This is another wedge-shaped surface, similar to one in . -

    - -

    - There are three components to the surface. -

      -
    • -

      - The first is the region in the xy plane bounded by the x axis and the parabola y=4-x^2. - It meets the x axis along the interval -2\leq x\leq 2. -

      -
      • -
    • -

      - The second part lies in the plane z=2y. - This is a planar surface, bounded below by the segment -2\leq x\leq 2 along the x axis, - and above by a parabola that lies in the plane z=2y. -

      - -

      - This part of the surface lies directly above the part in the xy plane, - and meets it along the x axis. -

      -
      • -
    • -

      - The last part is the portion of the parabolic cylinder y=4-x^2 - that lies between the planes z=0 and z=2y. -

      -
      • -
    -

    -     - - - - - //ASY file for fig13_06_ex_123D.asy in Chapter 13 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(12.1,-7.1,16); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,-1,1,2}; - real[] myychoice={1,2,3,4}; - real[] myzchoice={2,4,6,8}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-0.25,5); - pair zbounds=(-0.25,10); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //parabola in plane - triple g(real t) {return (t,4-t^2,0);} - path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - triple g(real t) {return (t,4-t^2,2*(4-t^2));} - path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - //shade object - //import three; - //int k=12; - //for (int i=-2*k; i<2*k; ++i) - //{ - //path3 p=(i/k,4-(i/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,0)--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i)/k,4-((i)/k)^2,2*(4-((i)/k)^2)); - //draw(surface(p -- cycle), simplesurfacepen2); - //path3 p=(i/k,0,0)--(i/k,4-(i/k)^2,2*(4-(i/k)^2))--((i+1)/k,4-((i+1)/k)^2,2*(4-((i+1)/k)^2))--((i+1//)/k,0,0); - //draw(surface(p -- cycle), simplesurfacepen); - //} - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y*(4-t.x^2),2*t.y*(4-t.x^2));// - } - surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,(4-t.x^2),2*t.y*(4-t.x^2));// - } - surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y*(4-t.x^2),0);// - } - surface s=surface(f,(-2,0),(2,1),16,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //label and arrow - label("$z=2y$",(-2,2,7)); - draw((-1.8,2,6.8)--(-1,2,4.1),Arrow3(size=2mm)); - label("$y=4-x^2$",(2.5,2,0)); - draw((2.3,2.5,0.2)--(1.55,2.25,2.5),Arrow3(size=2mm)); - - - -     - -

    - 8192/105\approx 78.019 -

    -     -     - - - -

    - \surfaceS is the surface formed by the intersections of the cylinder z=1-x^2 and the planes y=-2, - y=2 and z=0; - \vec F = \langle 0,y^3,0\rangle. -

    - - - A surface resembling a barn or greenhouse with an arched roof. - -

    - There are four components to this surface. - Overall, the surface looks like the exterior of a barn or greenhouse with an arched roof. -

    - -

    -

      -
    • -

      - The graph z=1-x^2 is a parabolic cylinder, opening downwards. - It is shaped like an arch, with its peak along the line with z=1 directly above the y axis. - The cylinder extends to the xy plane, and is plotted for -2\leq y\leq 2. -

      -
      • - -
    • -

      - The ends of the surface lie in the planes y=2 and y=-2. - They are bounded above by the parabola z=1-x^2, - and below by the xy plane, for -1\leq x\leq 1. -

      -
      • - -
    • -

      - The bottom of the surface is a rectangle in the xy plane, - given by -1\leq x\leq 1 and -2\leq y\leq 2. - This is where the parabolic cylinder, and the two ends, meet the xy plane. -

      -
      • -
    -

    -     - - - - - //ASY file for fig14_05_ex_23_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14,14,4); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-2,2}; - real[] myzchoice={1}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-0.25,1.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y,1-t.x^2);// - } - surface s=surface(f,(-1,-2),(1,2),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,-2,t.y*(1-t.x^2));// - } - surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,2,t.y*(1-t.x^2));// - } - surface s=surface(f,(-1,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple f(pair t) { - return (t.x,t.y,0);// - } - surface s=surface(f,(-1,-2),(1,2),2,2,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - triple g(real t) {return (t,-2,1-t^2);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,2,1-t^2);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - draw((1,-2,0)--(1,2,0)--(-1,2,0)--(-1,-2,0)--cycle,bluepen+linewidth(2)); - - //label and arrow - label("$z=1-x^2$",(0,1,1.55)); - draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); - - //triple g(real t) {return (0,t,6-t^2);} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - - -     - -

    - 64/3 -

    -     -     -     - - - -

    - A closed curve C that is the boundary of a surface \surfaceS is given along with a vector field \vec F. - Find the circulation of \vec F around C either through direct computation or through Stokes' Theorem. -

    -     - - - -

    - C is the curve whose x- and y-values are determined by the three sides of a triangle with vertices at (-1,0), - (1,0) and (0,1), traversed in that order, - and the z-values are determined by the function z=xy; - \vec F = \langle z-y^2,x,z\rangle. -

    - - - A twisted-looking surface given by a portion of a hyperbolic paraboloid. - -

    - The graph z=xy is a hyperbolic paraboloid; - it is plotted here relative to the usual three-dimensional coordinate axes, over a triangular domain. -

    - -

    - The surface has a twisted, wavy shape. - It has one edge that lies along the x axis, for -1\leq x\leq 1. - For 0\leq x\leq 1, the surface bends upward, with its highest points above the line y=x. - It then bends back down to meet the y axis. -

    - -

    - The portion for -1\leq x\leq 0 is like a mirror image of the other side, - but it lies below the xy plane. - The lowest points lie below the line y=-x, - and the surface dips down from the x and y axes to meet these points. -

    - -

    - The other part of the boundary begins on the x axis at (1,0,0). - It curves upward in a parabolic arch, meeting the y axis at (0,1,0). - The boundary then dips downward in another parabolic curve - that rises up to meet the x axis again at (-1,0,0). -

    -     - - - - - //ASY file for fig14_07_ex_23_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(-10,22,1); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={1}; - real[] myzchoice={.25}; - defaultpen(0.5mm); - - pair xbounds=(-1.25,1.25); - pair ybounds=(-.25,1.25); - pair zbounds=(-.25,.3); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x-1+t.y*(2-2*t.x),t.x,(t.x-1+t.y*(2-2*t.x))*t.x);// - } - surface s=surface(f,(0,0),(1,1),8,8,usplinetype=new splinetype[] {notaknot,notaknot,monotonic}); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //triple g(real t) {return (t,-2,1-t^2);} - //path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,0,0);} - path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,1-t,t*(1-t));} - path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (t,t+1,t*(t+1));} - path3 mypath=graph(g,-1,0,operator ..);draw(mypath,bluepen+linewidth(2)); - - //draw((-1,0,0)--(1,0,0)--(0,1,0)--cycle,bluepen+linewidth(2)); - - //label and arrow - //label("$z=1-x^2$",(0,1,1.55)); - //draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); - - //triple g(real t) {return (0,t,6-t^2);} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - - -     - -

    - 5/3 -

    -     -     - - - -

    - C is the curve whose x- and y-values are given by - \vec r(t) = \langle 2\cos(t),2\sin(t)\rangle and the z-values are determined by the function z=x^2+y^3-3y+1; - \vec F = \langle -y,x,z\rangle. -

    - - - A surface that looks something like a ladle, if the handle had melted and drooped downward. - -

    - This might be one of the more complicated surfaces in the book. - It is plotted relative to the usual set of coordinate axes in three dimensions. - There is a local minimum that lies along the y axis. - The surface is bowl-shaped near the minimum. -

    - -

    - The boundary of the surface is wavy. - There is a peak on the boundary at the point (0,2,6). - If we travel along the boundary in either direction from this point, - the surface dips down when y is close to 1, and then climbs back up again. -

    - -

    - Above the y axis, the surface climbs from the minimum to a saddle point when y=-1. - The surface climbs up if we move away from the saddle point in the x direction, - and it dips back down if we move in the y direction. -

    - -

    - If the surface were flipped over, it would look like a saddle with a back rest, - and a hump in the front. -

    -     - - - - - //ASY file for fig14_07_ex_18_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14,14,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={6}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-.25,6.5); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.y*cos(t.x),t.y*sin(t.x),(t.y*cos(t.x))^2+(t.y*sin(t.x))^3-3t.y*sin(t.x)+1);// - } - surface s=surface(f,(0,0),(2pi,2),16,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //triple g(real t) {return (t,-2,1-t^2);} - //path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (2cos(t),2sin(t),4cos(t)^2+8sin(t)^3-6sin(t)+1);} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - //triple g(real t) {return (t,1-t,t*(1-t));} - //path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - //triple g(real t) {return (t,t+1,t*(t+1));} - //path3 mypath=graph(g,-1,0,operator ..);draw(mypath,bluepen+linewidth(2)); - - //draw((-1,0,0)--(1,0,0)--(0,1,0)--cycle,bluepen+linewidth(2)); - - //label and arrow - //label("$z=1-x^2$",(0,1,1.55)); - //draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); - - //triple g(real t) {return (0,t,6-t^2);} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - - -     - -

    - 8\pi -

    -     -     - - - -

    - C is the curve whose x- and y-values are given by - \vec r(t) = \langle \cos(t),3\sin(t)\rangle and the z-values are determined by the function z=5-2x-y; - \vec F = \langle -\frac13y,3x,\frac23y-3x\rangle. -

    - - - A region in a plane in space, bounded by an ellipse in that plane. - -

    - The surface is planar; it is bounded by an ellipse that lies in the plane z=5-2x-y. - The ellipse is centered on the z axis. - The surface consists of all points in the plane that lie on or inside this ellipse. -

    -     - - - - - //ASY file for fig14_07_ex_18_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(14,14,10); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-2,2}; - real[] myychoice={-2,2}; - real[] myzchoice={10}; - defaultpen(0.5mm); - - pair xbounds=(-2.5,2.5); - pair ybounds=(-2.5,2.5); - pair zbounds=(-.25,11); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.y*cos(t.x),3*t.y*sin(t.x),-(t.y*cos(t.x))*2-3*(t.y*sin(t.x))+5);// - } - surface s=surface(f,(0,0),(2pi,1),16,8,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //triple g(real t) {return (t,-2,1-t^2);} - //path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - triple g(real t) {return (cos(t),3sin(t),-2cos(t)-3sin(t)+5);} - path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - //triple g(real t) {return (t,1-t,t*(1-t));} - //path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - //triple g(real t) {return (t,t+1,t*(t+1));} - //path3 mypath=graph(g,-1,0,operator ..);draw(mypath,bluepen+linewidth(2)); - - //draw((-1,0,0)--(1,0,0)--(0,1,0)--cycle,bluepen+linewidth(2)); - - //label and arrow - //label("$z=1-x^2$",(0,1,1.55)); - //draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); - - //triple g(real t) {return (0,t,6-t^2);} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - - -     - -

    - 23\pi -

    -     -     - - - -

    - C is the curve whose x- and y-values are sides of the square with vertices at (1,1), - (-1,1), - (-1,-1) and (1,-1), traversed in that order, - and the z-values are determined by the function z=10-5x-2y; - \vec F = \langle 5y^2,2y^2,y^2\rangle. -

    - - - A rectangle lying within a plane in space. - -

    - The image shows a rectangle in space, viewed in perspective, - relative to a set of three-dimensional coordinate axes. -

    - -

    - The sides of the rectangle lie above the square given by -1\leq x\leq 1 and -1\leq y\leq 1. - The z coordinates of the points on the rectangle are given by the plane equation z=10-5x-2y. -

    -     - - - - - //ASY file for fig14_07_ex_20_3D.asy in Chapter 14 - - size(200,200,IgnoreAspect); - //currentprojection=perspective(7,2,1); - currentprojection=orthographic(11,11,80); - defaultrender.merge=true; - - // setup and draw the axes - real[] myxchoice={-1,1}; - real[] myychoice={-1,1}; - real[] myzchoice={5,10,15}; - defaultpen(0.5mm); - - pair xbounds=(-1.25,1.25); - pair ybounds=(-1.25,1.25); - pair zbounds=(-.25,18); - - xaxis3("",xbounds.x,xbounds.y,black,OutTicks(myxchoice),Arrow3(size=3mm)); - yaxis3("",ybounds.x,ybounds.y,black,OutTicks(myychoice),Arrow3(size=3mm)); - zaxis3("",zbounds.x,zbounds.y,black,OutTicks(myzchoice),Arrow3(size=3mm)); - - label("$x$",(xbounds.y+0.05*(xbounds.y-xbounds.x),0,0)); - label("$y$",(0,ybounds.y+0.05*(ybounds.y-ybounds.x),0)); - label("$z$",(0,0,zbounds.y+0.05*(zbounds.y-zbounds.x))); - - //Draw paraboloid over unit disk - triple f(pair t) { - return (t.x,t.y,10-5t.x-2t.y);// - } - surface s=surface(f,(-1,-1),(1,1),3,3,Spline); - pen p=apexmeshpen; - draw(s,surfacepen,meshpen=p); - - //triple g(real t) {return (t,-2,1-t^2);} - //path3 mypath=graph(g,-1,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - //triple g(real t) {return (cos(t),3sin(t),-2cos(t)-3sin(t)+5);} - //path3 mypath=graph(g,0,2pi,operator ..);draw(mypath,bluepen+linewidth(2)); - - //triple g(real t) {return (t,1-t,t*(1-t));} - //path3 mypath=graph(g,0,1,operator ..);draw(mypath,bluepen+linewidth(2)); - - //triple g(real t) {return (t,t+1,t*(t+1));} - //path3 mypath=graph(g,-1,0,operator ..);draw(mypath,bluepen+linewidth(2)); - - draw((1,1,3)--(-1,1,13)--(-1,-1,17)--(1,-1,7)--cycle,bluepen+linewidth(2)); - - //label and arrow - //label("$z=1-x^2$",(0,1,1.55)); - //draw((0,.9,1.45)--(0,.5,1.05),Arrow3(size=2mm)); - - //triple g(real t) {return (0,t,6-t^2);} - //path3 mypath=graph(g,-2,2,operator ..);draw(mypath,bluepen+linewidth(2)); - - - -     - -

    - 0 -

    -     -     -     - - - -

    - The following exercisesare designed to challenge your understanding and require no computation. -

    -     - - - - -

    - Let \surfaceS be any closed surface enclosing a domain D. - Consider \vec F_1 = \langle x,0,0\rangle and \vec F_2=\langle y,y^2,z-2yz\rangle. -

    - -

    - These fields are clearly very different. - Why is it that the total outward flux of each field across \surfaceS is the same? -

    -     - -

    - Each field has a divergence of 1; - by the Divergence Theorem, - the total outward flux across \surfaceS is \iint_D 1\, dS for each field. -

    -     -     - - - - -

    - Green's Theorem can be used to find the area of a region enclosed by a curve by evaluating a line integral with the appropriate choice of vector field \vec F. - What condition on \vec F makes this possible? -

    -     - -

    - \curl\vec F = 1. -

    -     -     - - - -

    - Likewise, Stokes' Theorem can be used to find the surface area of a region enclosed by a curve in space by evaluating a line integral with the appropriate choice of vector field \vec F. - What condition on \vec F makes this possible? -

    -     - -

    - \curl\vec F\cdot \vec n = 1, - where \vec n is a unit vector normal to \surfaceS. -

    -     -     -     - - - -

    - The Divergence Theorem establishes equality between a particular double integral and a particular triple integral. - What types of circumstances would lead one to choose to evaluate the triple integral over the double integral? -

    -     - -

    - Answers will vary. - Often the closed surface \surfaceS is composed of several smooth surfaces. - To measure total outward flux, - this may require evaluating multiple double integrals. - Each double integral requires the parametrization of a surface and the computation of the cross product of partial derivatives. - One triple integral may require less work, - especially as the divergence of a vector field is generally easy to compute. -

    -     -     - - - -

    - Stokes' Theorem establishes equality between a particular line integral and a particular double integral. - What types of circumstances would lead one to choose to evaluate the double integral over the line integral? -

    -     - -

    - Answers will vary. - Often the closed curve C is composed of several smooth curves. - To measure the total circulation, - one may have to evaluate line integrals along each curve. - Each line integral requires the parametrization of its curve. - It may be less work to evaluate one single double (, surface) integral. -

    -     -     -     -     -     -
    -     - - - - - Answers to Selected Exercises - - - Quick Reference - -
    - Differentiation Formulas - - Derivative Rules -
      -
    1. - \lzo{x}(cx)=c -
      2. - -
    2. - \lzo{x}(u\pm v)=u'\pm v' -
      3. - -
    3. - \lzo{x}(u\cdot v)=uv'+ u'v -
      4. - -
    4. - \lzo{x}(\frac uv)=\frac{vu'-uv'}{v^2} -
      5. - -
    5. - \lzo{x}(u(v))=u'(v)v' -
      6. - -
    6. - \lzo{x}(c)=0 -
      7. - -
    7. - \lzo{x}(x)=1 -
      8. -
    -     - - - Derivatives of Elementary Functions -
      -
    1. - \lzo{x}(x^n)=nx^{n-1} -
      2. - -
    2. - \lzo{x}(e^x)=e^x -
      3. - -
    3. - \lzo{x}(a^x)=\ln a\cdot a^x -
      4. - -
    4. - \lzo{x}(\ln x)=\frac{1}{x} -
      5. - -
    5. - \lzo{x}(\log_a x)=\frac{1}{\ln a}\cdot \frac1x -
      6. - -
    6. - \lzo{x}(\sin(x))=\cos(x) -
      7. - -
    7. - \lzo{x}(\cos(x))=-\sin(x) -
      8. - -
    8. - \lzo{x}(\csc(x))=-\csc(x)\cot(x) -
      9. - -
    9. - \lzo{x}(\sec(x))=\sec(x)\tan(x) -
      10. - -
    10. - \lzo{x}(\tan(x))=\sec^2(x) -
      11. - -
    11. - \lzo{x}(\cot(x))=-\csc^2(x) -
      12. - -
    12. - \lzo{x}(\cosh=(x))=\sinh(x) -
      13. - -
    13. - \lzo{x}(\sinh(x))=\cosh(x) -
      14. - -
    14. - \lzo{x}(\sech(x))=-\sech(x)\tanh(x) -
      15. - -
    15. - \lzo{x}(\tanh(x))=\sech^2(x) -
      16. - -
    16. - \lzo{x}(\csch(x))=-\csch(x)\coth(x) -
      17. - -
    17. - \lzo{x}(\coth(x))=-\csch^2(x) -
      18. -
    -     - - - Derivatives of Inverse Functions -
      -
    1. - \lzo{x}(\sin^{-1}(x))=\frac{1}{\sqrt{1-x^2}} -
      2. - -
    2. - \lzo{x}(\cos^{-1}(x))=\frac{-1}{\sqrt{1-x^2}} -
      3. - -
    3. - \lzo{x}(\csc^{-1}(x))=\frac{-1}{\abs{x}\sqrt{x^2-1}} -
      4. - -
    4. - \lzo{x}(\sec^{-1}(x))=\frac{1}{\abs{x}\sqrt{x^2-1}} -
      5. - -
    5. - \lzo{x}(\tan^{-1}(x))=\frac{1}{1+x^2} -
      6. - -
    6. - \lzo{x}(\cot^{-1}(x))=\frac{-1}{1+x^2} -
      7. - -
    7. - \lzo{x}(\cosh^{-1}(x))=\frac1{\sqrt{x^2-1}} -
      8. - -
    8. - \lzo{x}(\sinh^{-1}(x))=\frac1{\sqrt{x^2+1}} -
      9. - -
    9. - \lzo{x}(\sech^{-1}(x))=\frac{-1}{x\sqrt{1-x^2}} -
      10. - -
    10. - \lzo{x}(\csch^{-1}(x))=\frac{-1}{\abs{x}\sqrt{1+x^2}} -
      11. - -
    11. - \lzo{x}(\tanh^{-1}(x))=\frac1{1-x^2} -
      12. - -
    12. - \lzo{x}(\coth^{-1}(x))=\frac1{1-x^2} -
      13. -
    -     -
    - -
    - Integration Formulas - - - Basic Rules -
      -
    1. - \int c\cdot f(x)\,dx=c\int f(x)\,dx -
      2. - -
    2. - \int \bigl(f(x)\pm g(x)\bigr)\,dx = \int f(x)\, dx \pm \int g(x)\, dx -
      3. - -
    3. - \int 0\,dx = C -
      4. - -
    4. - \int 1\,dx=x+C -
      5. -
    -     - - - Integrals of Elementary (non-Trig) Functions -
      -
    1. - \int e^x\,dx=e^x+C -
      2. - -
    2. - \int \ln(x)\,dx=x\ln(x) -x +C -
      3. - -
    3. - \int a^x\,dx=\frac{1}{\ln(a)}\cdot a^x+C -
      4. - -
    4. - \int \frac{1}{x}\,dx =\ln \abs{x} + C -
      5. - -
    5. - \int x^n\,dx=\frac{1}{n+1}x^{n+1}+C, n\neq -1 -
      6. -
    -     - - - Integrals Involving Trigonometric Functions -
      -
    1. - \int \cos(x)\,dx=\sin(x)+C -
      2. - -
    2. - \int \sin(x)\,dx=-\cos(x)+C -
      3. - -
    3. - \int \tan(x)\,dx=-\ln \abs{\cos(x)}+C -
      4. - -
    4. - \int \sec(x)\,dx=\ln \abs{\sec(x)+\tan(x)}+C -
      5. - -
    5. - \int \csc(x)\,dx=-\ln \abs{\csc(x)+\cot(x)}+C -
      6. - -
    6. - \int \cot(x)\,dx=\ln \abs{\sin(x)}+C -
      7. - -
    7. - \int \sec^2(x)\,dx=\tan(x)+C -
      8. - -
    8. - \int \csc^2(x)\,dx=-\cot(x)+C -
      9. - -
    9. - \int \sec(x)\tan(x)\,dx=\sec(x)+C -
      10. - -
    10. - \int \csc(x)\cot(x)\,dx=-\csc(x)+C -
      11. - -
    11. - \int \cos^2(x)\,dx=\frac12x+\frac14\sin\big(2x\big)+C -
      12. - -
    12. - \int \sin^2(x)\,dx=\frac12x-\frac14\sin\big(2x\big)+C -
      13. - -
    13. - \int \frac{1}{x^2+a^2}\,dx = \frac1a\tan^{-1}\left(\frac xa\right)+C -
      14. - -
    14. - \int \frac{1}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac xa\right)+C -
      15. - -
    15. - \int \frac{1}{x\sqrt{x^2-a^2}} = \frac1a\sec^{-1}\left(\frac{\abs{x}}{a}\right)+C -
      16. -
    -     - - - Integrals Involving Hyperbolic Functions -
      -
    1. - \int \cosh(x)\,dx=\sinh(x)+C -
      2. - -
    2. - \int \sinh(x)\,dx=\cosh(x)+C -
      3. - -
    3. - \int \tanh(x)\,dx=\ln(\cosh(x))+C -
      4. - -
    4. - \int \coth(x)\,dx=\ln \abs{\sinh(x)}+C -
      5. - -
    5. - \int \frac{1}{\sqrt{x^2-a^2}}\, dx =\ln\abs{x+\sqrt{x^2-a^2}}+C -
      6. - -
    6. - \int \frac{1}{\sqrt{x^2+a^2}}\, dx=\ln\abs{x+\sqrt{x^2+a^2}}+C -
      7. - -
    7. - \int \frac{1}{a^2-x^2}\, dx =\frac{1}{2a}\ln\abs{\frac{a+x}{a-x}}+C -
      8. - -
    8. - \int \frac{1}{x\sqrt{a^2-x^2}}\, dx = \frac{1}{a}\ln\left(\frac{x}{a+\sqrt{a^2-x^2}}\right)+C -
      9. - -
    9. - \int \frac{1}{x\sqrt{x^2+a^2}}\, = \frac{1}{a}\ln\abs{\frac{x}{a+\sqrt{x^2+a^2}}}+C -
      10. -
    -     -
    - -
    - Trigonometry Reference - - - - The Unit Circle - - A detailed plot of the unit circle, showing angles in both degrees and radians, and the coordinates of the corresponding points on the circle. - -

    - The unit circle x^2+y^2=1 is plotted, along with the x and y coordinate axes. - There are sixteen marked points on the circle, showing various angles and the coordinates of the corresponding points on the circle. - This can be used to evaluate the trigonometric functions at these special angles: - for a point (x,y) on the circle corresponding to an angle \theta, we have x=\cos(\theta) and y=\sin(\theta). -

    - -

    - The values given in the diagram are as follows: -

      -
    • -

      - 0^\circ, 0 radians, point (1,0) -

      -
      • -
    • -

      - 30^\circ, \pi/6 radians, point \left(\frac{\sqrt{3}}{2},\frac12\right) -

      -
      • -
    • -

      - 45^\circ, \pi/4 radians, point \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) -

      -
      • -
    • -

      - 60^\circ, \pi/3 radians, point \left(\frac12,\frac{\sqrt{3}}{2}\right) -

      -
      • -
    • -

      - 90^\circ, \pi/2 radians, point (0,1) -

      -
      • -
    • -

      - 120^\circ, 2\pi/3 radians, point \left(-\frac12,\frac{\sqrt{3}}{2}\right) -

      -
      • -
    • -

      - 135^\circ, 3\pi/4radians, point \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) -

      -
      • -
    • -

      - 150^\circ, 5\pi/6 radians, point \left(-\frac{\sqrt{3}}{2},\frac12\right) -

      -
      • -
    • -

      - 180^\circ, \pi radians, point (-1,0) -

      -
      • -
    • -

      - 210^\circ, 7\pi/6 radians, point \left(-\frac{\sqrt{3}}{2},-\frac12\right) -

      -
      • -
    • -

      - 225^\circ, 5\pi/4 radians, point \left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right) -

      -
      • -
    • -

      - 240^\circ, 4\pi/3 radians, point \left(-\frac12,-\frac{\sqrt{3}}{2}\right) -

      -
      • -
    • -

      - 270^\circ, 3\pi/2radians, point (0,-1) -

      -
      • -
    • -

      - 300^\circ, 5\pi/3 radians, point \left(\frac{\sqrt{3}}{2},-\frac12\right) -

      -
      • -
    • -

      - 315^\circ, 7\pi/4 radians, point \left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right) -

      -
      • -
    • -

      - 330^\circ, 11\pi/6 radians, point \left(\frac{\sqrt{3}}{2},-\frac12\right) -

      -
      • -
    -

    -     - - - \begin{tikzpicture}[scale=3, every node/.append style={font=\footnotesize}] - - \draw [<->,>=latex] (-1.5,0) -- (1.4,0) node [right] { $x$}; - \draw [<->,>=latex] (0,-1.3) -- (0,1.3) node [above] { $y$}; - - \foreach \x / \y / \z / \w / \v in { - 0/0/{1,0}/right/white, - 30/{\pi/6}/{\frac{\sqrt{3}}2,\frac 12}/above right/none,% - 45/{\pi/4}/{\frac{\sqrt{2}}2,\frac{\sqrt{2}}2}/above right/none, - 60/{\pi/3}/{\frac{1}2,\frac{\sqrt{3}}2}/{above right}/none, - 90/{\pi/2}/{0,1}/above/white,% - 120/{2\pi/3}/{-\frac{1}2,\frac{\sqrt{3}}2}/above left/none, - 135/{3\pi/4}/{-\frac{\sqrt{2}}2,\frac{\sqrt{2}}2}/above left/none, - 150/{5\pi/6}/{-\frac{\sqrt{3}}2,\frac{1}2}/above left/none,% - 180/{\pi}/{-1,0}/left/white, - 210/{7\pi/6}/{-\frac{\sqrt{3}}2,-\frac{1}2}/below left/none, - 225/{5\pi/4}/{-\frac{\sqrt{2}}2,-\frac{\sqrt{2}}2}/below left/none, - 240/{4\pi/3}/{-\frac{1}2,-\frac{\sqrt{3}}2}/below left/none, - 270/{3\pi/2}/{0,-1}/below/white, - 300/{5\pi/3}/{\frac{1}2,-\frac{\sqrt{3}}2}/below right/none, - 315/{7\pi/4}/{\frac{\sqrt{2}}2,-\frac{\sqrt{2}}2}/below right/none, - 330/{11\pi/6}/{\frac{\sqrt{3}}2,-\frac{1}2}/below right/none% - } - { - \draw (\x:.65cm) node [fill=\v] { \x$^\circ$}; - \draw (\x:.85cm) node [fill=\v] { $\y$}; - \draw (\x:1cm) node [\w,fill=\v] { $\left(\z\right)$}; - \draw [fill=black] (\x:1) circle (.5pt); - } - - \draw [thick] (0,0) circle (1); - - \end{tikzpicture} - - - -     -     - - - Definitions of the Trigonometric Functions - - Unit Circle Definition - - - An illustration of the correspondence between an angle and a point on the unit circle. - -

    - The unit circle is drawn over the x and y coordinate axes in the plane. - No scale markings are given. -

    - -

    - A point on the circle in the second quadrant is labeled with coordintes (x,y). - A line is drawn from the origin to this point. - There is an arc from the positive x axis to the line segment from the origin to the point; - the arc is labeled with the angle \theta. -

    - -

    - A dashed line is drawn vertically from the point (x,y) to the x axis, and labeled with the coordinate y. - The portion of the x axis between the origin and the point where the dashed line meets the axis is labeled x. -

    -     - - - \begin{tikzpicture}[>=latex,scale=1.73,thick] - - \draw [<->](-1.3,0)--(1.3,0) node [right] {\(x\)}; - \draw [<->] (0,-1.3) -- (0,1.3) node [above] {\(y\)}; - \draw (0,0) circle (1); - \draw [fill= black] (-.6,.8) circle (1pt); - \draw (0,0) -- (-.6,.8) node [above left] {\((x,y)\)}; - \draw [->] (.5,0) arc (0:127:.5); - \draw [dashed,thin] (-.6,.8) -- (-.6,0) node [pos=.5,left] {\(y\)}; - \draw (-.3,0) node [below] {\(x\)}; - \draw (.45,.45) node {\(\theta\)}; - - \end{tikzpicture} - - - - - - - \sin(\theta) = y - \cos(\theta) = x - - - - - - \ds\csc(\theta) = \frac1y - \ds\sec(\theta) = \frac1x - - - - - - \ds\tan(\theta) = \frac yx - \ds\cot(\theta) = \frac xy - - - -     -     - - - Right Triangle Definition - - - A right angle triangle, with sides labeled "opposite", "adjacent", and "hypotenuse". - -

    - A right angle triangle is drawn without reference to a coordinate system. - The angle at the bottom-left vertex is labeled \theta. - The bottom of the triangle is labeled Adjacent. - The right side of the triangle, which is vertical, is labeled Opposite. - The diagonal from bottom-left to top-right is labeled Hypotenuse. -

    -     - - - \begin{tikzpicture}[thick,scale=1.35] - - \draw (0,0) -- (2.5,0) node [below,pos=.5] {Adjacent} -- (2.5,2) node [pos=.5,rotate=-90,shift={(0pt,7pt)}] {Opposite} -- (0,0) node [pos=.5,above,rotate=38.7] {Hypotenuse} node [shift={(20pt,8pt)}] {\(\theta\)}; - \draw[->,>=latex] (1,0) arc (0:38.7:1); - \draw (2.2,0) -- (2.2,.3) -- (2.5,.3); - - \end{tikzpicture} - - - - - - - \ds\sin(\theta) = \frac{\text{O} }{\text{H} } - \ds\csc(\theta) = \frac{\text{H} }{\text{O} } - - - - - - \ds\cos(\theta) = \frac{\text{A} }{\text{H} } - \ds\sec(\theta) = \frac{\text{H} }{\text{A} } - - - - - - \ds\tan(\theta) = \frac{\text{O} }{\text{A} } - \ds\cot(\theta) = \frac{\text{A} }{\text{O} } - - -     -     -     - - - Common Trigonometric Identities - - - Pythagorean Identities -
      -
    1. \sin^2(x)+\cos^2(x)= 1
      2. -
    2. \tan^2(x)+ 1 = \sec^2(x)
      3. -
    3. 1 + \cot^2(x)=\csc^2(x)
      4. -
    -     - - - Double Angle Formulas -
      -
    1. \sin(2x) = 2\sin(x)\cos(x)
      2. - -
    2. -

      - - \cos(2x) \amp = \cos^2(x) - \sin^2(x) \amp \amp - \amp = 2\cos^2(x)-1 \amp \amp - \amp = 1-2\sin^2(x) \amp \amp - -

      -
      3. - -
    3. \tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}
      4. -
    -     -     - - - - Cofunction Identities -
      -
    1. \sin\left(\frac{\pi}{2}-x\right) = \cos(x)
      2. - -
    2. \cos\left(\frac{\pi}{2}-x\right) = \sin(x)
      3. - -
    3. \tan\left(\frac{\pi}{2}-x\right) = \cot(x)
      4. -
    4. \csc\left(\frac{\pi}{2}-x\right) = \sec(x)
      5. - -
    5. \sec\left(\frac{\pi}{2}-x\right) = \csc(x)
      6. - -
    6. \cot\left(\frac{\pi}{2}-x\right) = \tan(x)
      7. -
    -     - - - Even/Odd Identities -
      -
    1. \sin(-x) = -\sin(x)
      2. - -
    2. \cos (-x) = \cos(x)
      3. - -
    3. \tan (-x) = -\tan(x)
      4. - -
    4. \csc(-x) = -\csc(x)
      5. - -
    5. \sec (-x) = \sec(x)
      6. - -
    6. \cot (-x) = -\cot(x)
      7. -
    -     -     - - - - Power-Reducing Formulas -
      -
    1. \sin^2(x) = \frac{1-\cos(2x)}{2}
      2. - -
    2. \cos^2(x) = \frac{1+\cos(2x)}{2}
      3. - -
    3. \tan^2(x) = \frac{1-\cos(2x)}{1+\cos(2x)}
      4. -
    -     - - - Sum to Product Formulas -
      -
    1. \sin(x)+\sin(y) = 2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)
      2. - -
    2. \sin(x)-\sin(y) = 2\sin\left(\frac{x-y}2\right)\cos\left(\frac{x+y}2\right)
      3. - -
    3. \cos(x)+\cos(y) = 2\cos\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)
      4. - -
    4. \cos(x)-\cos(y) = -2\sin\left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)
      5. -
    -     -     - - - Product to Sum Formulas -
      -
    1. \sin(x)\sin(y) = \frac12 \big(\cos(x-y) - \cos (x+y)\big)
      2. - -
    2. \cos(x)\cos(y) = \frac12\big(\cos (x-y) +\cos (x+y)\big)
      3. - -
    3. \sin(x)\cos(y) = \frac12 \big(\sin(x+y) + \sin (x-y)\big)
      4. -
    -     - - - Angle Sum/Difference Formulas -
      -
    1. \sin (x\pm y) = \sin(x)\cos(y) \pm \cos(x)\sin(y)
      2. - -
    2. \cos (x\pm y) = \cos(x)\cos(y) \mp \sin(x)\sin(y)
      3. - -
    3. \tan (x\pm y) = \frac{\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}
      4. -
    -     -     -
    - - -
    - Areas and Volumes - - -

    - Triangles -

      -
    • -

      - h=a\sin(\theta) -

      -
      • -
    • -

      - Area = \frac12bh -

      -
      • -
    • -

      - Law of Cosines: - c^2=a^2+b^2-2ab\cos(\theta) -

      -
      • -
    -

    - - - A schematic diagram of a triangle, labeling three sides, an angle, and an altitude. - -

    - A triangle is drawn without reference to a coordinate system. - The bottom of the triangle is horizontal, and the other two sides are diagonal, - with the upper vertex positioned about two thirds of the way from left to right. -

    - -

    - The right diagonal is labeled a, - the bottom is labeled b, and the left diagonal is labeled c. - A dashed line is drawn from the upper vertex to the bottom, which it meets at a right angle. - This line is the altitude of the triangle, and it is labeled h, for height. - The angle at the bottom-right vertex of the triangle is labeled \theta. -

    -     - - - \begin{tikzpicture}[x=30pt,y=30pt,thick,scale=0.75] - - \draw (0,0) -- node [below,pos=.5] { \(b\)} (3,0) node [shift={(-15pt,8pt)}] {\(\theta\)} -- node [pos=.5,right] { \(a\)} (2,1.5) -- node [pos=.5,above left] { \(c\)} (0,0); - \draw (2.7,0) arc (180:125:.3); - \draw [dashed] (2,1.5) -- (2,0) node [pos=.5,left] {\(h\)}; - \draw (2,.2) -- (1.8,.2) -- (1.8,0); - - \end{tikzpicture} - - - - -

    - Right Circular Cone -

      -
    • -

      - Volume = \frac 13 \pi r^2h -

      -
      • -
    • -

      - Surface Area = - \pi r\sqrt{r^2+h^2} +\pi r^2 -

      -
      • -
    -

    - - - - A schematic diagram of a right circular cone, showing the height and radius. - -

    - A right circular cone is drawn without reference to a coordinate system. - The apex of the cone is at the top, and the circular base is at the bottom. - A dashed line from the center of the base to the edge is labeled r, for radius. - Another dashed line from the center of the base to the apex is labeled h, for height. -

    -     - - - \begin{tikzpicture}[x=13pt,y=15pt,thick,scale=1.13] - - \begin{scope}[xscale=2] - \draw (-1,0) arc (-180:0:1); - \draw [dashed] (1,0) arc (0:180:1); - \draw (-1,.1) -- (0,3) -- (1,.15); - \draw [dashed] (0,3) -- node [pos=.5,left] { \(h\)} (0,0); - \draw [dashed] (0,0) -- (1,0) node [pos=.5,above] { \(r\)}; - \end{scope} - \draw [fill=black] (0,0) circle (1pt); - - \end{tikzpicture} - - - -     - - -

    - Parallelograms -

      -
    • -

      - Area = bh -

      -
      • -
    -

    - - - A generic parallelogram, with base and height labeled. - -

    - A parallelogram is drawn without reference to a coordinate system. - The top and bottom sides are horizontal, while the left and right sides are diagonal, with positive slope. - The bottom side of the parallelogram is labeled b, for base. -

    - -

    - A dashed line is drawn from the top-left vertex to the base, which it meets at a right angle. - This line is labeled h, for the height of the parallelogram. -

    -     - - - \begin{tikzpicture}[x=30pt,y=25pt,thick,scale=0.75] - - \draw (0,0) -- node [below,pos=.5] { \(b\)} (2,0) -- (3,1.5) -- (1,1.5) -- (0,0); - \draw [dashed] (1,1.5) -- (1,0) node [pos=.5,right] {\(h\)}; - \draw (.8,0) -- (.8,.2) -- (1,.2); - - \end{tikzpicture} - - - - -

    - Right Circular Cylinder -

      -
    • -

      - Volume = \pi r^2h -

      -
      • -
    • -

      - Surface Area = - 2\pi rh +2\pi r^2 -

      -
      • -
    -

    - - - A diagram of a right circular cylinder, labeling the radius and height. - -

    - A generic right circular cylinder is drawn without reference to a coordinate system. - The cylinder is oriented vertically, with a circular base at the bottom. -

    - -

    - One side of the cylinder is labeled h, for the height, - and a line segment is drawn from the center of the circular top to the edge, - and labeled r, for radius. -

    -     - - - \begin{tikzpicture}[x=13pt,y=14pt,thick,scale=1.13] - - \begin{scope}[xscale=2] - \draw (-1,0) arc (-180:0:1); - \draw [dashed] (1,0) arc (0:180:1); - \draw (0,2.5) circle (1); - \draw (-1,0) -- (-1,2.5) (1,0)-- (1,2.5) node [right,pos=.5] {\(h\)}; - \draw (0,2.5) -- (1,2.5) node [above,pos=.5] {\(r\)}; - \end{scope} - - \draw [fill=black] (0,2.5) circle (1pt); - - \end{tikzpicture} - - - -     - - -

    - Trapezoids -

      -
    • -

      - Area = \frac12(a+b)h -

      -
      • -
    -

    - - - A schematic diagram of a generic trapezoid. - -

    - A trapezoid is drawn without reference to a coordinate system. - Its two parallel sides are drawn horizontally. -

    - -

    - The top side is shorter, and labeled with its length, a. - The longer bottom side is labeled with the length b. - The other two sides are slanted. -

    - -

    - A dashed line is drawn from the top-left vertex to the base, - perpendicular to the two parallel sides. - This line is labeled h, for the height of the trapezoid. -

    -     - - - \begin{tikzpicture}[x=30pt,y=25pt,thick,scale=0.75] - - \draw (0,0) -- node [below,pos=.7] { \(b\)} (3,0) -- (2.5,1.5) -- node [above,pos=.5] {\(a\)} (1.5,1.5) -- (0,0); - \draw [dashed] (1.5,1.5) -- (1.5,0) node [pos=.5,right] {\(h\)}; - \draw (1.3,0) -- (1.3,.2) -- (1.5,.2); - - \end{tikzpicture} - - - - -

    - Sphere -

      -
    • -

      - Volume = \frac43\pi r^3 -

      -
      • -
    • -

      - Surface Area =4\pi r^2 -

      -
      • -
    -

    - - - A image of a sphere, showing one circumference and its radius. - -

    - A sphere is drawn without reference to a coordinate system. - A line from the center of the sphere to its right edge is labeled r, for the radius. - A great circle is drawn around the middle of the sphere, - corresponding to what would be the equator on Earth. -

    -     - - - \begin{tikzpicture}[x=13pt,y=13pt,thick,scale=1.12] - - \begin{scope}[xscale=2] - \draw (-1,0) arc (-180:0:1); - \draw [dashed] (1,0) arc (0:180:1); - \end{scope} - - \draw (0,0) circle (2); - \draw [dashed] (0,0) -- (2,0) node [pos=.5,above] {\(r\)}; - \draw [fill=black] (0,0) circle (1pt); - - \end{tikzpicture} - - - -     - - -

    - Circles -

      -
    • -

      - Area = \pi r^2 -

      -
      • -
    • -

      - Circumference = 2\pi r -

      -
      • -
    -

    - - - A generic circle with its radius indicated. - -

    - A simple sketch of a circle, without reference to coordinates. - A line segment from the center of the circle to the circumference indicates the radius, r. -

    -     - - - \begin{tikzpicture}[x=30pt,y=30pt,thick] - - \draw (0,0) circle (1); - \draw [dashed] (0,0) -- (1,0) node [pos=.5,above] {\(r\)}; - \draw [fill=black] (0,0) circle (1pt); - - \end{tikzpicture} - - - - -

    - General Cone -

      -
    • -

      - Area of Base = A -

      -
      • -
    • -

      - Volume = \frac13Ah -

      -
      • -
    -

    - - - A drawing of a general cone, with an arbitrary plane region for its base. - -

    - A sketch of a general cone: the base is a closed curve in a plane, - drawn in perspective, but without reference to a coordinate system. -

    - -

    - The base is labeled with its area, A. - A dashed line is drawn from the apex of the cone to the base, and labeled h, for height. -

    -     - - - \begin{tikzpicture}[x=13pt,y=10pt,thick,scale=1.12] - - \begin{scope} - \clip (0,0) rectangle (4,-2.5); - \draw [smooth] plot coordinates {(0,0) (1,1.5) (2,1.5) (4,0) (3,-1) (2,-1.5) (1,-2) (0,0)}; - \end{scope} - - \begin{scope} - \clip (0,0) rectangle (4,2.5); - \draw [smooth,dashed] plot coordinates {(0,0) (1,1.5) (2,1.5) (4,0) (3,-1) (2,-1.5) (1,-2) (0,0)}; - \end{scope} - - \draw (0,0) -- (2,4)--(4,0); - \draw [dashed] (2,0)--(2,4) node [pos=.5,right] {\(h\)}; - \draw [fill=black](2,0) circle (1pt); - \draw (1.5,-.75) node {\(A\)}; - - \end{tikzpicture} - - - -     - - -

    - Sectors of Circles -

      -
    • -

      - \theta in radians -

      -
      • -
    • -

      - Area = \frac12\theta r^2 -

      -
      • -
    • -

      - s=r\theta -

      -
      • -
    -

    - - - A pie-shaped sector of a circle, labeled with angle, radius, and arc length. - -

    - A drawing of a sector of a circle, without reference to a coordinate system. - The sector is shaped like a slice of pie, corresponding to an acute angle. - The angle is labeled \theta, and the line segment on one side of the angle is labeled r, for the radius. - The circular arc opposite the angle is labeled s, for its arc length. -

    -     - - - \begin{tikzpicture}[x=30pt,y=30pt,thick] - - \draw (2,0) arc (0:50:2) -- (0,0); - \draw [] (0,0) -- (2,0) node [pos=.5,below] {\(r\)}; - \draw [fill=black] (0,0) circle (1pt); - \draw (1.95,1.0) node {\(s\)}; - \draw (0,0) node [shift={(15pt,8pt)}] {\(\theta\)}; - - \end{tikzpicture} - - - - -

    - General Right Cylinder -

      -
    • -

      - Area of Base = A -

      -
      • -
    • -

      - Volume = Ah -

      -
      • -
    -

    - - - A sketch of a right cylinder with an arbitrary base. - -

    - A drawing of a general right cylinder without reference to a coordinate system. - The base of the cylinder is a general plane region bounded by a simple, closed curve. - The base is labeled with its area, A. - The sides of the cylinder are vertical, and labeled with a height, h. - The top of the cylinder is another copy of the base. -

    -     - - - \begin{tikzpicture}[x=13pt,y=10pt,thick,scale=1.12] - - \begin{scope} - \clip (0,0) rectangle (4,-2.5); - \draw [smooth] plot coordinates {(0,0) (1,1.5) (2,1.5) (4,0) (3,-1) (2,-1.5) (1,-2) (0,0)}; - \end{scope} - - \begin{scope} - \clip (0,0) rectangle (4,2.5); - \draw [smooth,dashed] plot coordinates {(0,0) (1,1.5) (2,1.5) (4,0) (3,-1) (2,-1.5) (1,-2) (0,0)}; - \end{scope} - - \begin{scope}[shift={(0,4)}] - \draw [smooth] plot coordinates {(0,0) (1,1.5) (2,1.5) (4,0) (3,-1) (2,-1.5) (1,-2) (0,0)}; - \end{scope} - - \draw (0,0) -- (0,4) (4,0) -- (4,4) node [pos=.5,right] {\(h\)}; - \draw (2,0) node {\(A\)}; - - \end{tikzpicture} - - - -     -     -
    - -
    - Algebra - - - Factors and Zeros of Polynomials - -

    - Let p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 be a polynomial. - If p(a)=0, - then a is a zero of the polynomial and a solution of the equation p(x)=0. - Furthermore, (x-a) is a factor of the polynomial. -

    -     - - - Fundamental Theorem of Algebra - -

    - An nth degree polynomial has n - (not necessarily distinct) - zeros. - Although all of these zeros may be imaginary, - a real polynomial of odd degree must have at least one real zero. -

    -     - - - Quadratic Formula - -

    - If p(x) = ax^2 + bx + c, - and 0 \le b^2 - 4ac, - then the real zeros of p are x=(-b\pm \sqrt{b^2-4ac})/2a -

    -     - - - Special Factors - -

    - - x^2 - a^2 \amp = (x-a)(x+a) - x^3 - a^3 \amp= (x-a)(x^2+ax+a^2) - x^3 + a^3 \amp= (x+a)(x^2-ax+a^2) - x^4 - a^4 \amp= (x^2-a^2)(x^2+a^2) - (x+y)^n \amp=x^n + nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2+\cdots +nxy^{n-1}+y^n - (x-y)^n \amp=x^n - nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2-\cdots \pm nxy^{n-1}\mp y^n - -

    -     - - - Binomial Theorem - -

    - - (x+y)^2 \amp= x^2 + 2xy + y^2 - (x-y)^2 \amp= x^2 -2xy +y^2 - (x+y)^3 \amp= x^3 + 3x^2y + 3xy^2 + y^3 - (x-y)^3 \amp= x^3 -3x^2y + 3xy^2 -y^3 - (x+y)^4 \amp= x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 - (x-y)^4 \amp= x^4 - 4x^3y + 6x^2y^2 - 4xy^3 + y^4 - -

    -     - - - Rational Zero Theorem - -

    - If p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 has integer coefficients, - then every rational zero of p is of the form x=r/s, - where r is a factor of a_0 and s is a factor of a_n. -

    -     - - - Factoring by Grouping - -

    - ac x^3 + adx^2 + bcx + bd = ax^2(cx+d)+b(cx+d)=(ax^2+b)(cx+d) -

    -     - - - Arithmetic Operations - -

    - - ab+ac\amp=a(b+c) \amp \frac{a}{b}+\frac{c}{d} \amp= \frac{ad+bc}{bd} \amp \frac{a+b}{c} \amp = \frac{a}{c} + \frac{b}{c} - \frac{\left(\displaystyle\frac{a}{b}\right)}{\left(\displaystyle\frac{c}{d}\right)}\amp=\left(\frac{a}{b}\right)\left(\frac{d}{c}\right)=\frac{ad}{bc} \amp \frac{\left(\displaystyle\frac{a}{b}\right)}{c} \amp = \frac{a}{bc} \amp \frac{a}{\left(\displaystyle\frac{b}{c}\right)} \amp= \frac{ac}{b} - a\left(\frac{b}{c}\right)\amp= \frac{ab}{c}\amp \frac{a-b}{c-d}\amp=\frac{b-a}{d-c}\amp \frac{ab+ac}{a}\amp=b+c - -

    -     - - - Exponents and Radicals - -

    - - a^0\amp =1, \, a \ne 0 \amp (ab)^x\amp=a^xb^x \amp a^xa^y \amp = a^{x+y} \amp \sqrt{a}\amp=a^{1/2} - \frac{a^x}{a^y}\amp=a^{x-y} \amp \sqrt[n]{a}\amp =a^{1/n} \amp \left(\frac{a}{b}\right)^x\amp=\frac{a^x}{b^x} \amp \sqrt[n]{a^m}\amp=a^{m/n} - a^{-x}\amp=\displaystyle\frac{1}{a^x} \amp \sqrt[n]{ab}\amp=\sqrt[n]{a}\sqrt[n]{b} \amp (a^x)^y\amp=a^{xy} \amp \sqrt[n]{\frac{a}{b}}\amp=\frac{\sqrt[n]{a}}{\sqrt[n]{b}} - -

    -     -
    - -
    - Additional Formulas - - - Summation Formulas: - -

    - - - \sum^n_{i=1}{c} \amp = cn \amp \sum^n_{i=1}{i} \amp= \frac{n(n+1)}{2} - - - \sum^n_{i=1}{i^2} \amp = \frac{n(n+1)(2n+1)}{6} \amp \sum^n_{i=1}{i^3} \amp = \left(\frac{n(n+1)}{2}\right)^2 - - -

    -     - - - Trapezoidal Rule: -

    - \int_a^b{f(x)}\, dx \approx \frac{\Delta x}{2}\big[f(x_0)+2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_{n})\big] -

    - -

    - with Error \ds \leq \frac{(b-a)^3}{12n^2}\big[\max \abs{f\,''(x)}\big] -

    -     - - - Simpson's Rule: -

    - \int_a^b{f(x)}\, dx \approx \frac{\Delta x}{3}\big[f(x_0)+4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_{n})\big] -

    - -

    - with Error \ds \leq \frac{(b-a)^5}{180n^4}\big[ \max\abs{f\,^{(4)}(x)}\big] -

    -     - - - Arc Length: -

    - L = \int_a^b\sqrt{1+ f\,'(x)^2}\,dx -

    -     - - - Surface of Revolution: -

    - 2\pi \int_a^b{f(x) \sqrt{1+ f\,'(x)^2}}dx -

    - -

    - (where f(x)\geq 0) -

    - -

    - S = 2\pi \int_a^b{x \sqrt{1+ f\,'(x)^2}}dx -

    - -

    - (where a,b \geq 0) -

    -     - - - Work Done by a Variable Force: -

    - W = \int_a^b{F(x)}dx -

    -     - - - Force Exerted by a Fluid: -

    - F = \int_a^b{w\,d(y)\,\ell(y)}dy -

    -     - - - Taylor Series Expansion for <m>f(x)</m>: -

    - p_n(x) = f(c) + f\,'(c)(x-c) + \frac{f\,''(c)}{2!}(x-c)^2 + \cdots + \frac{f\,^{(n)}(c)}{n!}(x-c)^n + \cdots -

    -     - - - Maclaurin Series Expansion for <m>f(x)</m>, where <m>c=0</m>: -

    - p_n(x) = f(0) + f\,'(0)x + \frac{f\,''(0)}{2!}x^2 + \frac{f\,'''(0)}{3!}x^3 + \cdots + \frac{f\,^{(n)}(0)}{n!}x^n+\cdots -

    -     -
    - -
    - Summary of Tests for Series - - - <tabular> - <col/> - <col/> - <col width="19%"/> - <col width="19%"/> - <col width="22%"/> - <row bottom="minor"> - <cell>Test</cell> - <cell>Series</cell> - <cell><p>Condition(s) of Convergence</p></cell> - <cell><p>Condition(s) of Divergence</p></cell> - <cell>Comment</cell> - </row> - <row bottom="minor"> - <cell><m>n</m>th-Term</cell> - <cell><m>\displaystyle{\sum^\infty_{n=1}{a_n}}</m></cell> - <cell/> - <cell><p><m>\displaystyle{\lim_{n \to \infty} a_n \neq 0}</m></p></cell> - <cell><p>Cannot be used to show convergence.</p></cell> - </row> - <row bottom="minor"> - <cell>Geometric Series</cell> - <cell><m>\displaystyle{\sum^\infty_{n=0}{r^n}}</m></cell> - <cell><p><m>\abs{r} \lt 1</m></p></cell> - <cell><p><m>\abs{r} \geq 1</m></p></cell> - <cell><p><m>\displaystyle{\text{ Sum } = \frac{1}{1-r}}</m></p></cell> - </row> - <row bottom="minor"> - <cell>Telescoping Series</cell> - <cell><m>\displaystyle{\sum^\infty_{n=1}{(b_n-b_{n+a})}}</m></cell> - <cell><p><m>\displaystyle{\lim_{n \to \infty} b_n = L}</m></p></cell> - <cell/> - <cell><p><m>\displaystyle\text{ Sum } = \left(\sum^a_{n=1}b_n\right) -L</m></p></cell> - </row> - <row bottom="minor"> - <cell><m>p</m>-Series</cell> - <cell><m>\displaystyle{\sum^\infty_{n=1}{\frac{1}{(an+b)^p}}}</m></cell> - <cell><p><m>p \gt 1</m></p></cell> - <cell><p><m>p\leq 1</m></p></cell> - <cell/> - </row> - <row bottom="minor"> - <cell>Integral Test</cell> - <cell><m>\displaystyle{\sum^\infty_{n=0}{a_n}}</m></cell> - <cell><p><m>\displaystyle \int_1^\infty a(n)\, dn</m> converges</p></cell> - <cell><p><m>\displaystyle \int_1^\infty a(n)\, dn</m> diverges</p></cell> - <cell><p><m>a_n = a(n)</m> must be continuous</p></cell> - </row> - <row bottom="minor"> - <cell>Direct Comparison</cell> - <cell><m>\displaystyle{\sum^\infty_{n=0}{a_n}}</m></cell> - <cell><p><m>\displaystyle \sum_{n=0}^\infty b_n</m> converges and <m>0\leq a_n\leq b_n</m></p></cell> - <cell><p><m>\displaystyle \sum_{n=0}^\infty b_n</m> diverges and <m>0\leq b_n\leq a_n</m></p></cell> - <cell/> - </row> - <row bottom="minor"> - <cell>Limit Comparison</cell> - <cell><m>\displaystyle{\sum^\infty_{n=0}{a_n}}</m></cell> - <cell><p><m>\displaystyle \sum_{n=0}^\infty b_n</m> converges and <m>\lim\limits_{n\to\infty}\frac{a_n}{b_n} \geq 0</m></p></cell> - <cell><p><m>\displaystyle \sum_{n=0}^\infty b_n</m> diverges and <m>\lim\limits_{n\to\infty}\frac{a_n}{b_n} \gt 0</m></p></cell> - <cell><p>Also diverges if <m>\lim\limits_{n\to\infty}\frac{a_n}{b_n}=\infty</m></p></cell> - </row> - <row> - <cell>Ratio Test</cell> - <cell><m>\displaystyle{\sum^\infty_{n=0}{a_n}}</m></cell> - <cell><p><m>\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n} \lt 1</m></p></cell> - <cell><p><m>\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n} \gt 1</m></p></cell> - <cell><p><m>\{a_n\}</m> must be positive</p></cell> - </row> - <row bottom="minor"> - <cell/> - <cell/> - <cell/> - <cell><p>Also diverges if</p></cell> - <cell><p><m>\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}=\infty</m></p></cell> - </row> - <row> - <cell>Root Test</cell> - <cell><m>\displaystyle{\sum^\infty_{n=0}{a_n}}</m></cell> - <cell><p><m>\displaystyle \lim_{n\to\infty} \big(a_n\big)^{1/n} \lt 1</m></p></cell> - <cell><p><m>\displaystyle \lim_{n\to\infty} \big(a_n\big)^{1/n} \gt 1</m></p></cell> - <cell><p><m>\{a_n\}</m> must be positive</p></cell> - </row> - <row bottom="minor"> - <cell/> - <cell/> - <cell/> - <cell><p>Also diverges if</p></cell> - <cell><p><m>\lim\limits_{n\to\infty} (a_n)^{1/n}=\infty</m></p></cell> - </row> - </tabular> - </table> - </section> -</appendix> - <index label="terminology-index"> - <title>Index - - - - - \ No newline at end of file diff --git a/apex-validation.txt b/apex-validation.txt deleted file mode 100644 index 3c8d4bab0..000000000 --- a/apex-validation.txt +++ /dev/null @@ -1,30735 +0,0 @@ -Validation Report -================= - -Two tools have examined an assembled version of your source: - - (1) "jing" checked conformance with the RELAX-NG schema at - /home/sean/github/pretext/schema/pretext.rng - (a schema can only describe parent-child relationships, - plus the attributes of each element) - (2) the PreTeXt "validation-plus" stylesheet made checks that - no RELAX-NG schema could ever express, and offers advice - besides - -Locations refer to the ASSEMBLED version of your source: your -modular source files have been knitted together, and any version -support has been applied (as elected by a "version" element -within the "source" element of the publication file you -supplied). In particular, an element excluded from the version -being built cannot raise a message here. The assembled source -has been deposited at - /home/sean/github/APEXCalculusPTX/apex-assembled.xml - -Each message locates its problem four ways: - - file: the source file where the problem lies - path: the location within the assembled source - line: the line number within the assembled source - text: an excerpt of the offending content - -Only "file" points into your own source files. In particular, -"line" is a line number of the deposited assembled source named -above, never of one of your files. - -To read a "path", count elements of each name. For example, - - /pretext[1]/book[1]/chapter[7]/section[2]/p[13]/em[2] - -is the second "em" within the thirteenth "p" (paragraph) of the -second "section" of the seventh "chapter" of the book. 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    - -error: attribute "hskip" not allowed here; expected attribute "component", "label", "landscape" or "xml:lang" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[2]/p[2]/m[3] - line: 73734 - text:
    - -error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[1]/exercise[1]/statement[1]/p[1]/ie[1] - line: 74288 - text: - -error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[1]/exercise[2]/statement[1]/p[1]/xref[1] - line: 74314 - text: - -error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/p[1]/m[1] - line: 74349 - text: - -error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/statement[1]/p[1]/m[1] - line: 74377 - text: - -error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/p[1]/m[1] - line: 74406 - text: - -error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/statement[1]/p[1]/m[1] - line: 74435 - text: - -error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/statement[1]/p[1]/m[1] - line: 74464 - text: - -error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/statement[1]/p[1]/m[1] - line: 74492 - text: - -error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/statement[1]/p[1]/m[1] - line: 74520 - text: - -error: element "response" not allowed anywhere; expected the element end-tag or element "answer", "hint" or "solution" - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/statement[1]/p[1]/m[1] - line: 74555 - text: - -error: attribute "vshift" not allowed here; expected attribute "component", "label", "xml:id" or "xml:lang" - file: sec_lhopitals_rule.ptx - path: /pretext[1]/book[1]/chapter[6]/section[7]/subsection[1]/theorem[1]/statement[1]/p[1]/m[11] - line: 75476 - text: elements, if present, or else against the first row. Results may be unpredictable. - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/paragraphs[2]/example[1]/solution[1]/table[1]/tabular[1] - line: 12541 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/statement[1]/image[1] - line: 12915 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/statement[1]/image[1] - line: 12971 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/webwork[1]/statement[1]/image[1] - line: 13029 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/webwork[1]/statement[1]/image[1] - line: 13086 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/webwork[1]/statement[1]/image[1] - line: 13147 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/webwork[1]/statement[1]/image[1] - line: 13204 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/webwork[1]/statement[1]/image[1] - line: 13266 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/webwork[1]/statement[1]/image[1] - line: 13342 - text: - -PTX:WARNING: 'medium' or 'major' table rule attributes will be handled as 'minor' in the output of a WeBWorK PG table produced by WeBWorK's hardcopy production engine - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[1]/webwork[1]/statement[1]/tabular[1]/row[1] - line: 14080 - text: - -PTX:WARNING: 'medium' or 'major' table rule attributes will be handled as 'minor' in the output of a WeBWorK PG table produced by WeBWorK's hardcopy production engine - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[2]/webwork[1]/statement[1]/tabular[1]/row[1] - line: 14170 - text: - -PTX:WARNING: 'medium' or 'major' table rule attributes will be handled as 'minor' in the output of a WeBWorK PG table produced by WeBWorK's hardcopy production engine - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[3]/webwork[1]/statement[1]/tabular[1]/row[1] - line: 14260 - text: - -PTX:WARNING: 'medium' or 'major' table rule attributes will be handled as 'minor' in the output of a WeBWorK PG table produced by WeBWorK's hardcopy production engine - file: sec_limit_continuity.ptx - path: /pretext[1]/book[1]/chapter[1]/section[5]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[4]/webwork[1]/statement[1]/tabular[1]/row[1] - line: 14350 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/introduction[1]/figure[1]/image[1] - line: 14445 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[1]/example[1]/statement[1]/figure[1]/image[1] - line: 14622 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[1]/example[2]/statement[1]/figure[1]/image[1] - line: 14699 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] - line: 14806 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[2]/figure[1]/image[1] - line: 14882 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 15082 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 15186 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 15217 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 15247 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 15487 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 15519 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/subsection[4]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 15558 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/introduction[1]/image[1] - line: 15866 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/introduction[1]/image[1] - line: 15952 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/webwork[1]/introduction[1]/image[1] - line: 16079 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/webwork[1]/introduction[1]/image[1] - line: 16179 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/webwork[1]/introduction[1]/image[1] - line: 16271 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_limit_infty.ptx - path: /pretext[1]/book[1]/chapter[1]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/webwork[1]/introduction[1]/image[1] - line: 16346 - text: - -PTX:WARNING: A run of text contains Unicode characters for double quotation marks (U+201C, decimal 8220; U+201D, decimal 8221). Likely this was introduced in a conversion of source material authored in a word-processor. A matching pair U+201C, U+201D should be replaced by the "" element enclosing content. In rare cases, U+201C might be replaced by the empty element "". In rare cases, U+201D might be replaced by the empty element "". - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/p[4]/md[1] - line: 17355 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] - line: 17499 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] - line: 17534 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] - line: 17571 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/sbsgroup[1]/sidebyside[2]/figure[2]/image[1] - line: 17608 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 17797 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 17900 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[4]/solution[1]/figure[1]/image[1] - line: 18062 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[8]/statement[1]/figure[1]/image[1] - line: 18355 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[8]/solution[1]/figure[1]/image[1] - line: 18463 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[9]/statement[1]/figure[1]/image[1] - line: 18519 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[1]/example[9]/solution[1]/figure[1]/image[1] - line: 18622 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] - line: 18800 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercise[1]/webwork[1]/introduction[1]/image[1] - line: 19519 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercise[2]/webwork[1]/introduction[1]/image[1] - line: 19609 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/statement[1]/image[1] - line: 19698 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/statement[1]/image[1] - line: 19728 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/statement[1]/image[1] - line: 19757 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/statement[1]/image[1] - line: 19789 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[1]/webwork[1]/statement[1]/image[1] - line: 19878 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_intro.ptx - path: /pretext[1]/book[1]/chapter[2]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[2]/webwork[1]/statement[1]/image[1] - line: 19962 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_interpret.ptx - path: /pretext[1]/book[1]/chapter[2]/section[2]/subsection[4]/example[1]/statement[1]/figure[1]/image[1] - line: 20467 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_interpret.ptx - path: /pretext[1]/book[1]/chapter[2]/section[2]/subsection[4]/example[1]/solution[1]/figure[1]/image[1] - line: 20510 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_interpret.ptx - path: /pretext[1]/book[1]/chapter[2]/section[2]/subsection[4]/example[2]/statement[1]/figure[1]/image[1] - line: 20571 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_interpret.ptx - path: /pretext[1]/book[1]/chapter[2]/section[2]/subsection[4]/example[2]/solution[1]/figure[1]/image[1] - line: 20639 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_interpret.ptx - path: /pretext[1]/book[1]/chapter[2]/section[2]/subsection[4]/example[3]/solution[1]/figure[1]/image[1] - line: 20713 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_interpret.ptx - path: /pretext[1]/book[1]/chapter[2]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/statement[1]/image[1] - line: 21362 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_interpret.ptx - path: /pretext[1]/book[1]/chapter[2]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/statement[1]/image[1] - line: 21431 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_interpret.ptx - path: /pretext[1]/book[1]/chapter[2]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/webwork[1]/statement[1]/image[1] - line: 21486 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_interpret.ptx - path: /pretext[1]/book[1]/chapter[2]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/webwork[1]/statement[1]/image[1] - line: 21549 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_basic_rules.ptx - path: /pretext[1]/book[1]/chapter[2]/section[3]/example[1]/solution[1]/figure[1]/image[1] - line: 21813 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_basic_rules.ptx - path: /pretext[1]/book[1]/chapter[2]/section[3]/example[2]/solution[1]/sidebyside[1]/figure[1]/image[1] - line: 22015 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_basic_rules.ptx - path: /pretext[1]/book[1]/chapter[2]/section[3]/example[2]/solution[1]/sidebyside[1]/figure[2]/image[1] - line: 22052 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_prodquot.ptx - path: /pretext[1]/book[1]/chapter[2]/section[4]/example[1]/solution[1]/figure[1]/image[1] - line: 23696 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_prodquot.ptx - path: /pretext[1]/book[1]/chapter[2]/section[4]/example[6]/solution[1]/figure[1]/image[1] - line: 24050 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_chainrule.ptx - path: /pretext[1]/book[1]/chapter[2]/section[5]/figure[1]/image[1] - line: 25631 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_chainrule.ptx - path: /pretext[1]/book[1]/chapter[2]/section[5]/example[4]/solution[1]/figure[1]/image[1] - line: 26034 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_chainrule.ptx - path: /pretext[1]/book[1]/chapter[2]/section[5]/paragraphs[1]/figure[1]/image[1] - line: 26423 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/introduction[1]/figure[1]/image[1] - line: 27590 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 27784 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[3]/solution[1]/figure[1]/image[1] - line: 27910 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[4]/solution[1]/figure[1]/image[1] - line: 28019 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[4]/solution[1]/figure[2]/image[1] - line: 28070 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[5]/solution[1]/figure[1]/image[1] - line: 28152 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[6]/solution[1]/figure[1]/image[1] - line: 28290 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[1]/example[6]/solution[1]/figure[2]/image[1] - line: 28342 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[3]/figure[1]/image[1] - line: 28454 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] - line: 28539 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/webwork[1]/introduction[1]/image[1] - line: 29398 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/webwork[1]/introduction[1]/image[1] - line: 29458 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/webwork[1]/introduction[1]/image[1] - line: 29517 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/webwork[1]/introduction[1]/image[1] - line: 29589 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[5]/webwork[1]/introduction[1]/image[1] - line: 29657 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_implicit.ptx - path: /pretext[1]/book[1]/chapter[2]/section[6]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[6]/webwork[1]/introduction[1]/image[1] - line: 29721 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_inverse_function.ptx - path: /pretext[1]/book[1]/chapter[2]/section[7]/figure[1]/image[1] - line: 30170 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_inverse_function.ptx - path: /pretext[1]/book[1]/chapter[2]/section[7]/figure[2]/image[1] - line: 30232 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_inverse_function.ptx - path: /pretext[1]/book[1]/chapter[2]/section[7]/example[1]/solution[1]/figure[1]/image[1] - line: 30415 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_inverse_function.ptx - path: /pretext[1]/book[1]/chapter[2]/section[7]/figure[3]/sidebyside[1]/image[1] - line: 30484 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_deriv_inverse_function.ptx - path: /pretext[1]/book[1]/chapter[2]/section[7]/figure[3]/sidebyside[1]/image[2] - line: 30512 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 31769 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 31802 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 31841 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/example[1]/statement[1]/figure[1]/image[1] - line: 31921 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/example[2]/statement[1]/figure[1]/image[1] - line: 32051 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/example[3]/statement[1]/figure[1]/image[1] - line: 32108 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/figure[2]/image[1] - line: 32221 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/example[4]/statement[1]/figure[1]/image[1] - line: 32305 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/example[5]/solution[1]/sidebyside[1]/figure[2]/image[1] - line: 32481 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/example[6]/solution[1]/sidebyside[1]/figure[2]/image[1] - line: 32590 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/example[7]/solution[1]/sidebyside[1]/figure[1]/image[1] - line: 32641 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/statement[1]/image[1] - line: 32871 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/statement[1]/image[1] - line: 32958 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/webwork[1]/statement[1]/image[1] - line: 33044 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/webwork[1]/statement[1]/image[1] - line: 33089 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/webwork[1]/statement[1]/image[1] - line: 33146 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/webwork[1]/statement[1]/image[1] - line: 33199 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[5]/webwork[1]/statement[1]/image[1] - line: 33260 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[6]/webwork[1]/statement[1]/image[1] - line: 33316 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[7]/webwork[1]/statement[1]/image[1] - line: 33370 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[3]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[8]/webwork[1]/statement[1]/image[1] - line: 33420 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_mvt.ptx - path: /pretext[1]/book[1]/chapter[3]/section[2]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 33911 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_mvt.ptx - path: /pretext[1]/book[1]/chapter[3]/section[2]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 33945 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_mvt.ptx - path: /pretext[1]/book[1]/chapter[3]/section[2]/figure[1]/image[1] - line: 34047 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_mvt.ptx - path: /pretext[1]/book[1]/chapter[3]/section[2]/example[2]/solution[1]/figure[1]/image[1] - line: 34229 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_incr_decr.ptx - path: /pretext[1]/book[1]/chapter[3]/section[3]/figure[1]/image[1] - line: 34972 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_incr_decr.ptx - path: /pretext[1]/book[1]/chapter[3]/section[3]/figure[2]/sidebyside[1]/image[1] - line: 35094 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_incr_decr.ptx - path: /pretext[1]/book[1]/chapter[3]/section[3]/example[1]/solution[1]/figure[1]/image[1] - line: 35328 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_incr_decr.ptx - path: /pretext[1]/book[1]/chapter[3]/section[3]/example[1]/solution[1]/figure[2]/image[1] - line: 35408 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_incr_decr.ptx - path: /pretext[1]/book[1]/chapter[3]/section[3]/example[1]/solution[1]/figure[3]/image[1] - line: 35480 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_incr_decr.ptx - path: /pretext[1]/book[1]/chapter[3]/section[3]/remark[1]/figure[1]/image[1] - line: 35631 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_incr_decr.ptx - path: /pretext[1]/book[1]/chapter[3]/section[3]/example[2]/solution[1]/figure[1]/image[1] - line: 35792 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_incr_decr.ptx - path: /pretext[1]/book[1]/chapter[3]/section[3]/example[2]/solution[1]/figure[2]/image[1] - line: 35878 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_incr_decr.ptx - path: /pretext[1]/book[1]/chapter[3]/section[3]/example[3]/solution[1]/figure[1]/image[1] - line: 35966 - text: - -PTX:WARNING: You have an image without a description and do not declare the image to be decorative. Because of this, output may not be accessible. If the image does not add information that is not already present, use @decorative="yes". Otherwise, provide a . - file: sec_graph_incr_decr.ptx - path: /pretext[1]/book[1]/chapter[3]/section[3]/example[3]/solution[1]/figure[2]/image[1] - line: 36111 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 37374 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 37401 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[2]/image[1] - line: 37533 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[3]/image[1] - line: 37600 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[4]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] - line: 37674 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[4]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] - line: 37694 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[4]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] - line: 37716 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[4]/sbsgroup[1]/sidebyside[2]/figure[2]/image[1] - line: 37736 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[5]/image[1] - line: 37783 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 37886 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[1]/solution[1]/figure[2]/image[1] - line: 37921 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 38046 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[2]/solution[1]/figure[2]/image[1] - line: 38123 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[3]/statement[1]/figure[1]/image[1] - line: 38191 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/example[3]/solution[1]/figure[1]/image[1] - line: 38258 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[1]/figure[6]/image[1] - line: 38312 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[2]/figure[1]/image[1] - line: 38384 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_concavity.ptx - path: /pretext[1]/book[1]/chapter[3]/section[4]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] - line: 38453 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[1]/solution[1]/figure[1]/image[1] - line: 41502 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[1]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 41588 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[1]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 41615 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[1]/solution[1]/figure[2]/sidebyside[1]/figure[3]/image[1] - line: 41643 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[2]/solution[1]/figure[1]/image[1] - line: 41782 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 41852 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 41884 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[3]/image[1] - line: 41913 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[3]/solution[1]/figure[1]/image[1] - line: 42053 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[3]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 42152 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[3]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 42183 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/example[3]/solution[1]/figure[2]/sidebyside[1]/figure[3]/image[1] - line: 42215 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 42307 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_graph_sketch.ptx - path: /pretext[1]/book[1]/chapter[3]/section[5]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 42331 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_newton.ptx - path: /pretext[1]/book[1]/chapter[4]/section[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 42998 - text: - -PTX:WARNING: A run of text contains Unicode characters for single quotation marks (U+2018, decimal 8216; U+2019, decimal 8217). Likely this was introduced in a conversion of source material authored in a word-processor. A U+2019 in isolation could be an apostrophe. Replace it with the keyboard version: U+0027. A matching pair U+2018, U+2019 should be replaced by the "" element enclosing content. In rare cases, U+2018 might be replaced by the empty element "". In rare cases, U+2019 might be replaced by the empty element "". - file: sec_newton.ptx - path: /pretext[1]/book[1]/chapter[4]/section[1]/figure[1]/sidebyside[1]/figure[1]/image[1]/description[1]/p[2] - line: 43009 - text:

    - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_newton.ptx - path: /pretext[1]/book[1]/chapter[4]/section[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 43035 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_newton.ptx - path: /pretext[1]/book[1]/chapter[4]/section[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 43064 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_newton.ptx - path: /pretext[1]/book[1]/chapter[4]/section[1]/example[1]/solution[1]/figure[1]/image[1] - line: 43275 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_newton.ptx - path: /pretext[1]/book[1]/chapter[4]/section[1]/example[2]/solution[1]/figure[1]/image[1] - line: 43362 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_newton.ptx - path: /pretext[1]/book[1]/chapter[4]/section[1]/paragraphs[1]/figure[1]/image[1] - line: 43485 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_newton.ptx - path: /pretext[1]/book[1]/chapter[4]/section[1]/paragraphs[1]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 43554 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_newton.ptx - path: /pretext[1]/book[1]/chapter[4]/section[1]/paragraphs[1]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 43594 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_newton.ptx - path: /pretext[1]/book[1]/chapter[4]/section[1]/paragraphs[1]/figure[2]/sidebyside[1]/figure[3]/image[1] - line: 43634 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_related_rates.ptx - path: /pretext[1]/book[1]/chapter[4]/section[2]/example[3]/statement[1]/figure[1]/image[1] - line: 44679 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_related_rates.ptx - path: /pretext[1]/book[1]/chapter[4]/section[2]/example[4]/statement[1]/figure[1]/image[1] - line: 44830 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_related_rates.ptx - path: /pretext[1]/book[1]/chapter[4]/section[2]/exercises[1]/subexercises[2]/exercise[5]/webwork[1]/introduction[1]/image[1] - line: 45267 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_related_rates.ptx - path: /pretext[1]/book[1]/chapter[4]/section[2]/exercises[1]/subexercises[2]/exercise[7]/webwork[1]/introduction[1]/image[1] - line: 45452 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_related_rates.ptx - path: /pretext[1]/book[1]/chapter[4]/section[2]/exercises[1]/subexercises[2]/exercise[8]/webwork[1]/introduction[1]/image[1] - line: 45559 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_related_rates.ptx - path: /pretext[1]/book[1]/chapter[4]/section[2]/exercises[1]/subexercises[2]/exercise[10]/webwork[1]/introduction[1]/image[1] - line: 45743 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_optimization.ptx - path: /pretext[1]/book[1]/chapter[4]/section[3]/example[1]/solution[1]/figure[1]/sidebyside[1]/image[1] - line: 46050 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_optimization.ptx - path: /pretext[1]/book[1]/chapter[4]/section[3]/example[1]/solution[1]/figure[1]/sidebyside[1]/image[2] - line: 46100 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_optimization.ptx - path: /pretext[1]/book[1]/chapter[4]/section[3]/example[2]/solution[1]/p[2]/ol[1]/li[1]/figure[1]/sidebyside[1]/image[1] - line: 46322 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_optimization.ptx - path: /pretext[1]/book[1]/chapter[4]/section[3]/example[2]/solution[1]/p[2]/ol[1]/li[1]/figure[1]/sidebyside[1]/image[2] - line: 46369 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_optimization.ptx - path: /pretext[1]/book[1]/chapter[4]/section[3]/example[3]/statement[1]/figure[1]/image[1] - line: 46512 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_optimization.ptx - path: /pretext[1]/book[1]/chapter[4]/section[3]/example[3]/solution[1]/figure[1]/image[1] - line: 46590 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_optimization.ptx - path: /pretext[1]/book[1]/chapter[4]/section[3]/exercises[1]/subexercises[2]/exercise[11]/webwork[1]/statement[1]/image[1] - line: 47137 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_differentials.ptx - path: /pretext[1]/book[1]/chapter[4]/section[4]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 47422 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_differentials.ptx - path: /pretext[1]/book[1]/chapter[4]/section[4]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 47458 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_differentials.ptx - path: /pretext[1]/book[1]/chapter[4]/section[4]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/webwork[1]/introduction[1]/image[1] - line: 48983 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_differentials.ptx - path: /pretext[1]/book[1]/chapter[4]/section[4]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[2]/webwork[1]/introduction[1]/image[1] - line: 49061 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_differentials.ptx - path: /pretext[1]/book[1]/chapter[4]/section[4]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[3]/webwork[1]/introduction[1]/image[1] - line: 49139 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_differentials.ptx - path: /pretext[1]/book[1]/chapter[4]/section[4]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/webwork[1]/statement[1]/image[1] - line: 49215 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_antider.ptx - path: /pretext[1]/book[1]/chapter[5]/section[1]/figure[1]/image[1] - line: 49494 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/figure[1]/image[1] - line: 51076 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/figure[2]/image[1] - line: 51158 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/example[1]/solution[1]/figure[1]/image[1] - line: 51315 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/example[2]/statement[1]/figure[1]/image[1] - line: 51498 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/example[2]/solution[1]/figure[1]/image[1] - line: 51605 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/example[3]/statement[1]/figure[1]/image[1] - line: 51762 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/example[4]/solution[1]/sidebyside[1]/figure[1]/image[1] - line: 51916 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/example[4]/solution[1]/sidebyside[1]/figure[2]/image[1] - line: 51958 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/example[5]/statement[1]/figure[1]/image[1] - line: 52004 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/figure[3]/image[1] - line: 52089 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/introduction[1]/image[1] - line: 52221 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/introduction[1]/image[1] - line: 52319 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/webwork[1]/introduction[1]/image[1] - line: 52422 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/webwork[1]/introduction[1]/image[1] - line: 52528 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/webwork[1]/introduction[1]/image[1] - line: 52629 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/webwork[1]/introduction[1]/image[1] - line: 52711 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/webwork[1]/introduction[1]/image[1] - line: 52801 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/webwork[1]/introduction[1]/image[1] - line: 52898 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/webwork[1]/introduction[1]/image[1] - line: 52988 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/webwork[1]/introduction[1]/image[1] - line: 53081 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/webwork[1]/introduction[1]/image[1] - line: 53185 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_def_int.ptx - path: /pretext[1]/book[1]/chapter[5]/section[2]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[2]/webwork[1]/introduction[1]/image[1] - line: 53259 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/figure[1]/image[1] - line: 53618 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/figure[2]/image[1] - line: 53660 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/figure[3]/image[1] - line: 53726 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 53878 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 53932 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 53986 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/figure[1]/image[1] - line: 54230 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/image[1] - line: 54283 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] - line: 54448 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/figure[2]/image[1] - line: 54620 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] - line: 54760 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_riemann.ptx - path: /pretext[1]/book[1]/chapter[5]/section[3]/subsection[3]/example[4]/solution[1]/figure[1]/image[1] - line: 54981 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/introduction[1]/figure[1]/image[1] - line: 56694 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/introduction[1]/example[1]/statement[1]/figure[1]/image[1] - line: 56747 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/introduction[1]/example[1]/solution[1]/figure[1]/image[1] - line: 56799 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/introduction[1]/example[1]/solution[1]/figure[2]/image[1] - line: 56877 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[2]/example[1]/solution[1]/sidebyside[1]/figure[1]/image[1] - line: 57299 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[2]/example[1]/solution[1]/sidebyside[1]/figure[2]/image[1] - line: 57349 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[4]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 57548 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[4]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 57598 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[4]/example[1]/solution[1]/figure[1]/image[1] - line: 57696 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[1]/image[1] - line: 57772 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 57839 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 57880 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[2]/sidebyside[1]/figure[3]/image[1] - line: 57921 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/example[1]/solution[1]/figure[1]/image[1] - line: 58034 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[3]/sidebyside[1]/image[1] - line: 58111 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[3]/sidebyside[1]/image[2] - line: 58161 - text: - -PTX:WARNING: A run of text contains Unicode characters for double quotation marks (U+201C, decimal 8220; U+201D, decimal 8221). Likely this was introduced in a conversion of source material authored in a word-processor. A matching pair U+201C, U+201D should be replaced by the "" element enclosing content. In rare cases, U+201C might be replaced by the empty element "". In rare cases, U+201D might be replaced by the empty element "". - file: sec_FTC.ptx - path: /pretext[1]/book[1]/chapter[5]/section[4]/subsection[5]/figure[3]/sidebyside[1]/image[2]/shortdescription[1] - line: 58162 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/introduction[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 59796 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/introduction[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 59832 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/introduction[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 59871 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 59985 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 60025 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[1]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 60227 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[1]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 60274 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[2]/figure[1]/image[1] - line: 60359 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[2]/figure[2]/image[1] - line: 60404 - text: - -PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of

    elements, if present, or else against the first row. Results may be unpredictable. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[3]/example[2]/solution[1]/figure[1]/tabular[1] - line: 60705 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[4]/figure[1]/image[1] - line: 60892 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[4]/figure[2]/image[1] - line: 60970 - text: - -PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of elements, if present, or else against the first row. Results may be unpredictable. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[4]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/tabular[1] - line: 61046 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[4]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 61085 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[4]/example[2]/solution[1]/figure[2]/image[1] - line: 61236 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[5]/example[1]/solution[1]/figure[1]/image[1] - line: 61551 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/subsection[5]/example[1]/solution[1]/figure[2]/image[1] - line: 61635 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/webwork[1]/statement[1]/image[1] - line: 63296 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_numerical_integration.ptx - path: /pretext[1]/book[1]/chapter[5]/section[5]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/webwork[1]/statement[1]/image[1] - line: 63372 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_substitution.ptx - path: /pretext[1]/book[1]/chapter[6]/section[1]/subsection[5]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 64607 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_substitution.ptx - path: /pretext[1]/book[1]/chapter[6]/section[1]/subsection[5]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 64647 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_substitution.ptx - path: /pretext[1]/book[1]/chapter[6]/section[1]/subsection[5]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 64742 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_substitution.ptx - path: /pretext[1]/book[1]/chapter[6]/section[1]/subsection[5]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 64785 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_trigint.ptx - path: /pretext[1]/book[1]/chapter[6]/section[3]/subsection[1]/figure[1]/image[1] - line: 68891 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_trig_sub.ptx - path: /pretext[1]/book[1]/chapter[6]/section[4]/insight[1]/p[1]/ol[1]/li[1]/sidebyside[1]/figure[1]/image[1] - line: 70191 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_trig_sub.ptx - path: /pretext[1]/book[1]/chapter[6]/section[4]/insight[1]/p[1]/ol[1]/li[2]/sidebyside[1]/figure[1]/image[1] - line: 70234 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_trig_sub.ptx - path: /pretext[1]/book[1]/chapter[6]/section[4]/insight[1]/p[1]/ol[1]/li[3]/sidebyside[1]/figure[1]/image[1] - line: 70281 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/introduction[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 73018 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/introduction[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 73070 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 73198 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 73252 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[1]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 73326 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[1]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 73384 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[2]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] - line: 73807 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[2]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] - line: 73858 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[2]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] - line: 73918 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_hyperbolic.ptx - path: /pretext[1]/book[1]/chapter[6]/section[6]/subsection[2]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[2]/image[1] - line: 73979 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_lhopitals_rule.ptx - path: /pretext[1]/book[1]/chapter[6]/section[7]/subsection[3]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] - line: 75900 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/introduction[1]/figure[1]/image[1] - line: 77256 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] - line: 77416 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] - line: 77477 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] - line: 77532 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[4]/figure[1]/image[1] - line: 77594 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 77667 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] - line: 77802 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] - line: 77860 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] - line: 77969 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[3]/example[2]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] - line: 78142 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[3]/example[2]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] - line: 78206 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_improper_integration.ptx - path: /pretext[1]/book[1]/chapter[6]/section[8]/subsection[3]/example[3]/solution[1]/figure[1]/image[1] - line: 78350 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 79767 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 79809 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 79850 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/example[1]/statement[1]/figure[1]/image[1] - line: 79944 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/example[2]/statement[1]/figure[1]/image[1] - line: 80006 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/example[3]/statement[1]/figure[1]/image[1] - line: 80096 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/example[3]/solution[1]/figure[1]/image[1] - line: 80173 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/example[4]/statement[1]/figure[1]/image[1] - line: 80269 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/example[5]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 80374 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/example[5]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 80397 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/webwork[1]/statement[1]/image[1] - line: 80611 - text: Graph of the region enclosed by the functions y=\frac12 x +3 and y=\frac12\co... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/webwork[1]/statement[1]/image[1] - line: 80676 - text: Graph of the region enclosed by the functions y=-3x^3+3x+2 and y=x^2+x-1 ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/webwork[1]/statement[1]/image[1] - line: 80804 - text: Graph of the region enclosed by the functions y=\sin(x)+1 and y=\sin(x) b... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/webwork[1]/statement[1]/image[1] - line: 80871 - text: Graph of the region enclosed by the functions y=\sin(4x) and y=\sec^2(x) ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/webwork[1]/statement[1]/image[1] - line: 80937 - text: Graph of the region enclosed by the functions y=\sin(x) and y=\cos(x) bet... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/webwork[1]/statement[1]/image[1] - line: 81003 - text: Graph of the region enclosed by the functions y=2^x and y=4^x between ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/statement[1]/image[1] - line: 81049 - text: Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1Graph of the region enclosed by the functions y=x^2+1, y=\frac14(x-3)^2+1Graph of the region enclosed by the functions y=\sqrt{x}, y=-2x+3 and Graph of the region enclosed by the functions y=x+2 and y=x^2. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/webwork[1]/statement[1]/image[1] - line: 81579 - text: Graph of the region enclosed by the functions x=-\frac12 y+1 and x=\frac12 y^... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[5]/webwork[1]/statement[1]/image[1] - line: 81651 - text: Graph of the region enclosed by the functions y=x^{1/3}, y=\sqrt{x-1/2} a... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[6]/statement[1]/image[1] - line: 81702 - text: Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1 - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_ABC.ptx - path: /pretext[1]/book[1]/chapter[7]/section[1]/exercises[1]/subexercises[2]/exercise[4]/webwork[1]/statement[1]/image[1] - line: 81950 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/figure[1]/image[1] - line: 82009 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[1]/solution[1]/figure[1]/image[1] - line: 82158 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[1]/solution[1]/figure[2]/image[1] - line: 82266 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 82487 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 82564 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 82715 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 82794 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 82907 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 83005 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/figure[3]/image[1] - line: 83144 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 83303 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 83366 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 83442 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 83563 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 83648 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 83730 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/image[1] - line: 83958 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/statement[1]/image[1] - line: 84010 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/image[1] - line: 84059 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/statement[1]/image[1] - line: 84107 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/statement[1]/image[1] - line: 84171 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/statement[1]/image[1] - line: 84221 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/statement[1]/image[1] - line: 84275 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/statement[1]/image[1] - line: 84323 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/statement[1]/image[1] - line: 84759 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/statement[1]/image[1] - line: 84816 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/statement[1]/image[1] - line: 84871 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_disk.ptx - path: /pretext[1]/book[1]/chapter[7]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/statement[1]/image[1] - line: 84922 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 85012 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 85077 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 85146 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 85300 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 85345 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/example[1]/solution[1]/figure[1]/image[1] - line: 85454 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 85555 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 85607 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 85703 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 85855 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 85904 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 85999 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 86136 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 86183 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 86271 - text: - -PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of elements, if present, or else against the first row. Results may be unpredictable. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/insight[2]/tabular[1] - line: 86404 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/image[1] - line: 86543 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/statement[1]/image[1] - line: 86592 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/image[1] - line: 86640 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/statement[1]/image[1] - line: 86688 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/statement[1]/image[1] - line: 86750 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/statement[1]/image[1] - line: 86800 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/statement[1]/image[1] - line: 86848 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_shell_method.ptx - path: /pretext[1]/book[1]/chapter[7]/section[3]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/statement[1]/image[1] - line: 86895 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 87372 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 87406 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[1]/figure[2]/image[1] - line: 87472 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 87633 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 87696 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 87828 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 87876 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] - line: 88086 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/example[2]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 88212 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/example[2]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 88272 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_arc_length.ptx - path: /pretext[1]/book[1]/chapter[7]/section[4]/subsection[2]/example[3]/statement[1]/figure[1]/image[1] - line: 88390 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_work.ptx - path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] - line: 89462 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_work.ptx - path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] - line: 89657 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_work.ptx - path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[3]/figure[1]/image[1] - line: 89773 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_work.ptx - path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] - line: 89848 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_work.ptx - path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[3]/example[3]/statement[1]/figure[1]/image[1] - line: 89930 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_work.ptx - path: /pretext[1]/book[1]/chapter[7]/section[5]/subsection[3]/example[3]/solution[1]/figure[1]/image[1] - line: 89968 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_work.ptx - path: /pretext[1]/book[1]/chapter[7]/section[5]/exercises[1]/subexercises[2]/exercise[19]/statement[1]/image[1] - line: 90640 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_work.ptx - path: /pretext[1]/book[1]/chapter[7]/section[5]/exercises[1]/subexercises[2]/exercise[21]/statement[1]/image[1] - line: 90737 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_work.ptx - path: /pretext[1]/book[1]/chapter[7]/section[5]/exercises[1]/subexercises[2]/exercise[22]/statement[1]/image[1] - line: 90784 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_work.ptx - path: /pretext[1]/book[1]/chapter[7]/section[5]/exercises[1]/subexercises[2]/exercise[23]/statement[1]/image[1] - line: 90828 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/example[1]/statement[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] - line: 90938 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/example[1]/statement[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] - line: 90995 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/figure[1]/image[1] - line: 91143 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/example[2]/statement[1]/figure[1]/image[1] - line: 91288 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/example[2]/solution[1]/p[2]/ol[1]/li[1]/figure[1]/image[1] - line: 91346 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/example[2]/solution[1]/p[2]/ol[1]/li[2]/figure[1]/image[1] - line: 91428 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/example[3]/solution[1]/figure[1]/image[1] - line: 91547 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/example[4]/statement[1]/figure[1]/image[1] - line: 91607 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/image[1] - line: 91773 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/statement[1]/image[1] - line: 91810 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/image[1] - line: 91849 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/statement[1]/image[1] - line: 91894 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/statement[1]/image[1] - line: 91939 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/statement[1]/image[1] - line: 91982 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/statement[1]/image[1] - line: 92025 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/statement[1]/image[1] - line: 92067 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[9]/statement[1]/image[1] - line: 92109 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[10]/statement[1]/image[1] - line: 92148 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/statement[1]/image[1] - line: 92217 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/statement[1]/image[1] - line: 92257 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/statement[1]/image[1] - line: 92306 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/statement[1]/image[1] - line: 92355 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[5]/statement[1]/image[1] - line: 92399 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_fluid_force.ptx - path: /pretext[1]/book[1]/chapter[7]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[6]/statement[1]/image[1] - line: 92443 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 92734 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/figure[1]/image[1] - line: 92967 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] - line: 93090 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] - line: 93166 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[3]/statement[1]/figure[1]/image[1] - line: 93227 - text: - -PTX:WARNING: A run of text contains Unicode characters for single quotation marks (U+2018, decimal 8216; U+2019, decimal 8217). Likely this was introduced in a conversion of source material authored in a word-processor. A U+2019 in isolation could be an apostrophe. Replace it with the keyboard version: U+0027. A matching pair U+2018, U+2019 should be replaced by the "" element enclosing content. In rare cases, U+2018 might be replaced by the empty element "". In rare cases, U+2019 might be replaced by the empty element "". - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[3]/statement[1]/figure[1]/image[1]/shortdescription[1] - line: 93228 - text: Graph of slope field for the logistic differential equation y’=y(1-y) from the example. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[3]/solution[1]/figure[1]/image[1] - line: 93315 - text: - -PTX:WARNING: A run of text contains Unicode characters for single quotation marks (U+2018, decimal 8216; U+2019, decimal 8217). Likely this was introduced in a conversion of source material authored in a word-processor. A U+2019 in isolation could be an apostrophe. Replace it with the keyboard version: U+0027. A matching pair U+2018, U+2019 should be replaced by the "" element enclosing content. In rare cases, U+2018 might be replaced by the empty element "". In rare cases, U+2019 might be replaced by the empty element "". - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[2]/example[3]/solution[1]/figure[1]/image[1]/shortdescription[1] - line: 93316 - text: Graph of slope field for the logistic differential equation y’=y(1-y) with representative solution c... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] - line: 93530 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] - line: 93620 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/subsection[3]/example[3]/solution[1]/figure[1]/image[1] - line: 93764 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/answer[1]/image[1] - line: 94060 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[2]/answer[1]/image[1] - line: 94103 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[3]/answer[1]/image[1] - line: 94144 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/answer[1]/image[1] - line: 94195 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercise[1]/statement[1]/sbsgroup[1]/sidebyside[1]/image[1] - line: 94237 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercise[1]/statement[1]/sbsgroup[1]/sidebyside[1]/image[2] - line: 94270 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercise[1]/statement[1]/sbsgroup[1]/sidebyside[3]/image[1] - line: 94314 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercise[1]/statement[1]/sbsgroup[1]/sidebyside[3]/image[2] - line: 94346 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/answer[1]/image[1] - line: 94425 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/answer[1]/image[1] - line: 94478 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/answer[1]/image[1] - line: 94533 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Graphical_Numerical.ptx - path: /pretext[1]/book[1]/chapter[8]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/answer[1]/image[1] - line: 94581 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Linear.ptx - path: /pretext[1]/book[1]/chapter[8]/section[3]/subsection[1]/figure[1]/image[1] - line: 96129 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Linear.ptx - path: /pretext[1]/book[1]/chapter[8]/section[3]/exercises[1]/subexercises[1]/exercisegroup[4]/exercise[1]/answer[1]/image[1] - line: 96469 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Linear.ptx - path: /pretext[1]/book[1]/chapter[8]/section[3]/exercises[1]/subexercises[1]/exercisegroup[4]/exercise[2]/answer[1]/image[1] - line: 96516 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Modeling.ptx - path: /pretext[1]/book[1]/chapter[8]/section[4]/subsection[1]/figure[1]/image[1] - line: 96583 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Modeling.ptx - path: /pretext[1]/book[1]/chapter[8]/section[4]/subsection[1]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 96957 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Modeling.ptx - path: /pretext[1]/book[1]/chapter[8]/section[4]/subsection[1]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 96999 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_Modeling.ptx - path: /pretext[1]/book[1]/chapter[8]/section[4]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] - line: 97206 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_sequences.ptx - path: /pretext[1]/book[1]/chapter[9]/section[1]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] - line: 97679 - text: Plot of the first four terms of the sequence in part 1 of this example.A plot of the first four terms of the sequence in part 2 of this example.A plot of the first four terms of the sequence in part 3 of this example.A scatter plot showing a representative sample of points from the first sequence i... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_sequences.ptx - path: /pretext[1]/book[1]/chapter[9]/section[1]/example[3]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] - line: 98198 - text: A scatter plot showing a representative sample of points from the second sequence ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_sequences.ptx - path: /pretext[1]/book[1]/chapter[9]/section[1]/example[3]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] - line: 98249 - text: A scatter plot showing a representative sample of points from the third sequence i... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_sequences.ptx - path: /pretext[1]/book[1]/chapter[9]/section[1]/example[4]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] - line: 98406 - text: Scatter plot illustrating the first 20 points in the sequence from the second part... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_sequences.ptx - path: /pretext[1]/book[1]/chapter[9]/section[1]/example[6]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 98678 - text: A scatter plot of the first 10 terms in the first sequence for this example.The scatter plot for the second sequence in this example, which shows exponential ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_sequences.ptx - path: /pretext[1]/book[1]/chapter[9]/section[1]/example[7]/solution[1]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] - line: 99024 - text: Plot of the first sequence in this example. It is decreasing and bounded below.Scatter plot for the second sequence in this example. It is increasing but not bou... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_sequences.ptx - path: /pretext[1]/book[1]/chapter[9]/section[1]/example[7]/solution[1]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] - line: 99097 - text: Scatter plot for the third sequence in this example. It is not monotonic.Scatter plot for the last sequence in this example. It is not monotonic.Scatter plots of the sequence, and corresponding partial sums, for the first part ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 100406 - text: Scatter plots of the sequence, and corresponding partial sums, for the second part... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] - line: 100680 - text: Scatter plots of the sequence, and corresponding partial sums, for the first part ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] - line: 100742 - text: Scatter plots of the sequence, and corresponding partial sums, for the second part... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] - line: 100804 - text: Scatter plots of the sequence, and corresponding partial sums, for the third part ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] - line: 101080 - text: Scatter plots of the sequence, and corresponding partial sums, for this example.Scatter plots of the sequence, and corresponding partial sums, for the first part ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[3]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 101301 - text: Scatter plots of the sequence, and corresponding partial sums, for the second part... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[3]/example[4]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 101510 - text: Scatter plots of the sequence, and corresponding partial sums, for the first part ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[2]/subsection[3]/example[4]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 101562 - text: Scatter plots of the sequence, and corresponding partial sums, for the first part ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_int_comp_tests.ptx - path: /pretext[1]/book[1]/chapter[9]/section[3]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 103227 - text: First of two graphs illustrating why the integral test works. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_int_comp_tests.ptx - path: /pretext[1]/book[1]/chapter[9]/section[3]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 103269 - text: Second of two graphs illustrating why the integral test works. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_int_comp_tests.ptx - path: /pretext[1]/book[1]/chapter[9]/section[3]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 103430 - text: Scatter plots of the sequence used in this example, and the corresponding sequence... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_alt_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[5]/figure[1]/image[1] - line: 106080 - text: Scatter plots of a positive, decreasing sequence and its alternating sequence of p... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_alt_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[5]/figure[2]/image[1] - line: 106136 - text: An illustration of the partial sums in an alternating series, using line segments ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/sidebyside[1]/figure[1]/image[1] - line: 109495 - text: The graph of a generic function is shown, along with the tangent line to the graph... - -PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/sidebyside[1]/figure[2]/sidebyside[1] - line: 109534 - text:

    - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/figure[1]/image[1] - line: 109625 - text: The graph of a function and two polynomials that approximate the function near x=0... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/figure[2]/image[1] - line: 109713 - text: The graph of a function is shown, along with the graph of a degree 13 polynomial t... - -PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1] - line: 109840 - text:

    - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/example[1]/solution[1]/figure[1]/image[1] - line: 109909 - text: The graph of the exponential function and its degree 5 Maclaurin polynomial. normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/example[2]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1] - line: 109992 - text:

    - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/example[2]/solution[1]/sidebyside[1]/figure[1]/image[1] - line: 110116 - text: A graph of the natural logarithm function and its degree 6 Taylor polynomial cente... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/example[2]/solution[1]/sidebyside[1]/figure[2]/image[1] - line: 110153 - text: A graph of the natural logarithm function and its degree 20 Taylor polynomial cent... - -PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/example[4]/solution[1]/figure[1]/sidebyside[1] - line: 110455 - text:

    - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/example[4]/solution[1]/figure[2]/image[1] - line: 110558 - text: A graph of the cosine function is shown, along with its degree 8 Maclaurin polynom... - -PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/example[5]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1] - line: 110649 - text:

    - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/example[5]/solution[1]/figure[1]/image[1] - line: 110737 - text: The graph of the square root function and its degree 4 Taylor polynomial, centered... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_taylor_poly.ptx - path: /pretext[1]/book[1]/chapter[9]/section[7]/example[6]/solution[1]/figure[1]/image[1] - line: 110888 - text: The graph of the exact solution to the differential equation in this example, and ... - -PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the - file: sec_taylor_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[8]/example[1]/solution[1]/figure[1]/sidebyside[1] - line: 111905 - text:

    - -PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("p") and remove the - file: sec_taylor_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[8]/example[2]/solution[1]/figure[1]/sidebyside[1] - line: 112040 - text:

    - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_taylor_series.ptx - path: /pretext[1]/book[1]/chapter[9]/section[8]/example[8]/solution[1]/figure[1]/image[1] - line: 112721 - text: A graph of the function from this example, and its degree 5 Maclaurin polynomial a... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/introduction[1]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] - line: 113876 - text: A diagonal plane intersecting a double napped cone, forming a parabolaA diagonal plane intersecting a double napped cone, forming an ellipse.A horizontal plane intersecting a double napped cone, forming a circle.A vertical plane intersecting a double napped cone, forming a hyperbola.A horizontal plane intersecting the tips in a double napped cone.A diagonal plane intersecting a double napped cone, forming a line in the planeA vertical plane intersecting a double napped cone, forming crossed lines in the p... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[1]/figure[1]/image[1] - line: 114043 - text: A sketch of a parabola with key components labeled. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 114191 - text: A downward opening parabola with a vertex in the first quadrant.A parabola opening to the right with its directrix and focus drawn.A rightward opening parabola demonstrating the reflective property.A demonstration of the creation of an ellipse from two foci. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/figure[2]/image[1] - line: 114536 - text: An ellipse with labels for key components. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] - line: 114639 - text: An ellipse centered in the second quadrant, lying entirely in the second and third... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] - line: 114728 - text: An ellipse centered at (1,2), with vertices at (-2,2) and (4,2).An ellipse of eccentricity 0. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/paragraphs[1]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] - line: 114839 - text: An ellipse of eccentricity 0.3. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/paragraphs[1]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] - line: 114880 - text: An ellipse of eccentricity 0.8. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/paragraphs[1]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[2]/image[1] - line: 114919 - text: An ellipse of eccentricity 0.99. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[2]/paragraphs[2]/figure[1]/image[1] - line: 115004 - text: An ellipse with rays emanating from the foci, demonstrating the reflective propert... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/figure[1]/image[1] - line: 115115 - text: A hyperbola with labels for key components - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[1]/figure[1]/image[1] - line: 115226 - text: A graph of a hyperbola and its asymptotes. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[1]/figure[2]/image[1] - line: 115272 - text: An illustration of sketching a hyperbola with a rectangle. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[1]/example[1]/solution[1]/figure[1]/image[1] - line: 115364 - text: A graph of the hyperbola in this example - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[1]/example[2]/solution[1]/figure[1]/image[1] - line: 115448 - text: The hyperbola described in this example - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[2]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[1]/image[1] - line: 115548 - text: A hyperbola with eccentricity 1.05 - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[2]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] - line: 115596 - text: A hyperbola with eccentricity 1.5 - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[2]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] - line: 115648 - text: A hyperbola with eccentricity 3 - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[2]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[2]/image[1] - line: 115696 - text: A hyperbola with eccentricity 10 - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[4]/figure[1]/image[1] - line: 115817 - text: A hyperbola demonstrating the reflective property. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[4]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 115908 - text: Three points drawn and labeled on a plane - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/subsection[3]/paragraphs[4]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 115942 - text: A hyperbola drawn from points A and B to illustrate the location property A fourth point found from hyperbolas given by points in the previous figuresAn ellipse centered in the second quadrant. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/statement[1]/image[1] - line: 116473 - text: A vertical ellipse centered on the x-axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/exercises[1]/subexercises[2]/exercisegroup[7]/exercise[1]/statement[1]/image[1] - line: 116687 - text: A hyperbola centered at the origin with a horizontal transverse axis.A hyperbola centered at the origin with a vertical transverse axis.A hyperbola with center (1,3) and a vertical transverse axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_conic_sections.ptx - path: /pretext[1]/book[1]/chapter[10]/section[1]/exercises[1]/subexercises[2]/exercisegroup[7]/exercise[4]/statement[1]/image[1] - line: 116844 - text: A hyperbola centered at the point (1,3) with a horizontal transverse axisA number line with 3 points drawn. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/introduction[1]/figure[1]/image[1] - line: 117277 - text: Diagram outlining the steps involved in plotting a function. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/introduction[1]/figure[2]/image[1] - line: 117340 - text: Diagram illustrating the process for plotting a parametric curve.Plot of a parabola with its vertex at the point (0,1), opening to the right, with ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[1]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 117608 - text: Plot of a parabola with its vertex at the point (0,1), opening to the right, with ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[1]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 117750 - text: Plot of the "unshifted" parametric curve in this example. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[1]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 117794 - text: The shifted version of the curve in this example. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[1]/example[4]/solution[1]/figure[1]/image[1] - line: 117860 - text: Plot of a more complicated parametric curve that has a self-intersection.A parabolic arc describing the motion of a projectile fired up and to the right fr... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[2]/example[3]/solution[1]/figure[1]/image[1] - line: 118102 - text: A graph of two overlapping lines illustrating rectangular and parametric descripti... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[2]/example[4]/solution[1]/figure[1]/image[1] - line: 118203 - text: Graph of the ellipse obtained from the parametric equations in this example.Plot of an astroid: a curve like a 4-pointed star. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[3]/figure[1]/sbsgroup[1]/sidebyside[1]/figure[2]/image[1] - line: 118348 - text: Plot of a rose curve with 8 loops. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/subsection[3]/figure[1]/sbsgroup[1]/sidebyside[2]/figure[1]/image[1] - line: 118384 - text: Plot of a hypotrochoid, which resembles a smooth, five-pointed star.Plot of an epicycloid, which looks somewhat like a puffy clover leaf.Plot of a parametric curve with a sharp cusp. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/answer[1]/image[1] - line: 118720 - text: Sketch of the parametric curve in this exercise. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/answer[1]/image[1] - line: 118768 - text: The vertical line x=1. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/answer[1]/image[1] - line: 118813 - text: The horizontal line y=2, marked with two arrows. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/answer[1]/image[1] - line: 118860 - text: The solution curve for this exercise. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_param_eqs.ptx - path: /pretext[1]/book[1]/chapter[10]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/answer[1]/image[1] - line: 118917 - text: Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Computer-generated sketch of the parametric curve in this exercise.Plot of a parametric curve and its tangent and normal lines at one point.Sketch of the unit circle and one normal line, which passes through the circle's c... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_par_calc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[3]/subsection[1]/example[3]/statement[1]/figure[1]/image[1] - line: 120805 - text: Graph of an astroid. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_par_calc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[3]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] - line: 120967 - text: Sketch of a parametric curve illustrating concavity. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_par_calc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[3]/subsection[2]/example[1]/solution[1]/image[1] - line: 121022 - text: A number line for the sign of the second derivative in this example.Graph of the second derivative is a sinusoid with increasing amplitude.Graph of the parametric curve in this example, with points of inflection marked.Graph of a "teardrop" curve that intersects itself at the origin.The surface obtained by revolving the teardrop shape from the previous example abo... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[1]/figure[1]/image[1] - line: 122747 - text: Illustration of polar coordinates relative to a pole and initial ray.Drawing of a polar grid: a set of concentric circles, and rays from their common c... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 122873 - text: A plot of several points on a polar grid. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[2]/figure[1]/image[1] - line: 122958 - text: A triangle illustrating the conversion between rectangular and polar coordinates.<... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 123071 - text: Overlaid rectangular and polar grid systems, with marked points.Overlaid rectangular and polar grid systems, with marked points.Polar plot showing curves of constant radius and constant angle - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 123425 - text: A rough sketch of a polar function. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/figure[1]/image[1] - line: 123514 - text: A computer-generated sketch of a polar function. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[3]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 123608 - text: An initial rough sketch of a polar curve obtained by connecting points on the curv... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[3]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 123665 - text: A smoothed version of the plot for this example. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[4]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] - line: 123814 - text: Graph of the hyperbola y=1/x. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 123923 - text: A line through the origin with positive slope. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 123952 - text: A horizontal line. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 123981 - text: A vertical line. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[1]/sidebyside[1]/figure[4]/image[1] - line: 124010 - text: A line with positive slope m, and y intercept b. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 124046 - text: A circle centered on the positive x axis and passing through the origin.A circle centered on the positive y axis and passing through the origin.A circle centered at the origin. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[2]/sidebyside[1]/figure[4]/image[1] - line: 124134 - text: A counter-clockwise spiral that begins at the origin. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[3]/sidebyside[1]/figure[1]/image[1] - line: 124171 - text: A limaçon curve with a loop at the origin, and symmetric about the x axis.A heart-shaped curve in the limaçon family, known as a cardioid.A dimpled limaçon curve. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[3]/sidebyside[1]/figure[4]/image[1] - line: 124277 - text: A convex limaçon curve. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[4]/sidebyside[1]/figure[1]/image[1] - line: 124321 - text: A rose curve with four leaves along the coordinate axes. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[4]/sidebyside[1]/figure[2]/image[1] - line: 124347 - text: A rose curve with four leaves, each in one of the quadrants. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[4]/sidebyside[1]/figure[3]/image[1] - line: 124374 - text: A rose curve with three leaves, symmetric about the x axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[4]/sidebyside[1]/figure[4]/image[1] - line: 124406 - text: A rose curve with three leaves, symmetric about the y axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[5]/sidebyside[1]/figure[1]/image[1] - line: 124452 - text: A polar curve consisting of three loops. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[5]/sidebyside[1]/figure[2]/image[1] - line: 124489 - text: An elaborate polar curve consisting of many intersecting loops. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[5]/sidebyside[1]/figure[3]/image[1] - line: 124521 - text: A lemniscate curve, which has the shape of a figure-eight. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/paragraphs[1]/figure[5]/sidebyside[1]/figure[4]/image[1] - line: 124550 - text: A polar curve in the shape of a figure-eight. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 124610 - text: Overlapping plots of a circle and a limaçon, showing their points of intersection.... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/subsection[3]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 124657 - text: Zoomed in view of the intersections between the two polar curves.The four points plotted in this exercise. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercise[2]/answer[1]/image[1] - line: 124951 - text: The four points plotted in this exercise. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercise[3]/statement[1]/image[1] - line: 125000 - text: Four points plotted on a polar grid. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercise[4]/webwork[1]/statement[1]/image[1] - line: 125099 - text: Four points plotted on a polar grid. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/answer[1]/image[1] - line: 125349 - text: Portion of the circle of radius 2, centered at the origin, in the first quadrant.<... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/answer[1]/image[1] - line: 125390 - text: A line segment through the origin with positive slope. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/answer[1]/image[1] - line: 125434 - text: A cardioid, symmetric about the x axis, with x intercepts at -2 and 0.A convex limaçon, symmetric about the y axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/answer[1]/image[1] - line: 125518 - text: A convex limaçon, symmetric about the y axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/answer[1]/image[1] - line: 125560 - text: A limaçon with an inner loop, symmetric about the y axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/answer[1]/image[1] - line: 125603 - text: A limaçon with an inner loop, symmetric about the y axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/answer[1]/image[1] - line: 125646 - text: A rose curve with four petals. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[9]/answer[1]/image[1] - line: 125690 - text: A rose curve with three loops, symmetric about the y axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[10]/answer[1]/image[1] - line: 125733 - text: A limaçon with an inner loop. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[11]/answer[1]/image[1] - line: 125784 - text: An elaborate rose curve with many self-intersections and four primary loops.A counter-clockwise spiral - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[13]/answer[1]/image[1] - line: 125871 - text: A circle passing through the origin with its center on the positive y axis.A semi-circle in the first quadrant. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[15]/answer[1]/image[1] - line: 125947 - text: A four-leaf rose with one petal in each quadrant. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[16]/answer[1]/image[1] - line: 125988 - text: A curve with two loops: a circle inside a larger leaf shape. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[17]/answer[1]/image[1] - line: 126033 - text: A straight line with positive slope. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[18]/answer[1]/image[1] - line: 126074 - text: A straight line with positive slope. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[19]/answer[1]/image[1] - line: 126115 - text: A vertical line with x=3. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polar.ptx - path: /pretext[1]/book[1]/chapter[10]/section[4]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[20]/answer[1]/image[1] - line: 126155 - text: The horizontal line y=4. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polarcalc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[1]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] - line: 126932 - text: A limaçon with an inner loops, symmetric about the y axis, and a tangent line.A zoomed in view of a limaçon near the origin, and two tangent lines at that point... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polarcalc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[2]/aside[1]/image[1] - line: 127184 - text: Illustration of a pie-shaped sector of a circle. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polarcalc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[2]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 127237 - text: Plot of a generic polar function and the area it encloses between two angles.Plot of a region bounded by a polar curve, and its approximation by cicular wedges... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polarcalc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[2]/example[2]/statement[1]/figure[1]/image[1] - line: 127442 - text: A cardioid curve, within which is a shaded region bounded by the curve and two ray... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polarcalc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[2]/paragraphs[1]/figure[1]/image[1] - line: 127520 - text: Illustration of a region bounded by two polar curves and two rays.A circle and a cardioid enclose a region that is inside the circle but outside the... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polarcalc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[2]/paragraphs[1]/example[2]/statement[1]/figure[1]/image[1] - line: 127708 - text: A zoomed in view of a region bounded by a circle, a rose curve, and the x axis.A zoomed in view of a polar region, showing it divided into two parts.A limaçon with an inner loop that is symmetric about the y axis.A four leaf rose curve, with leaves along the coordinate axes. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polarcalc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[5]/subsection[4]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 128070 - text: The surface of revolution obtained by revolving one leaf of a rose curve about the... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polarcalc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[5]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[9]/webwork[1]/statement[1]/image[1] - line: 128875 - text: Two circles, one along each axis, bound a region in the first quadrant.A semi-circle and one leaf of a rose curve, both in the first quadrant, overlappin... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polarcalc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[5]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[11]/statement[1]/image[1] - line: 128974 - text: Overlapping leaves of two different three-leaf rose curves. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_polarcalc.ptx - path: /pretext[1]/book[1]/chapter[10]/section[5]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[12]/statement[1]/image[1] - line: 129020 - text: A cardioid and a circle, and a region of overlap in the first quadrant. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/introduction[1]/figure[2]/image[1] - line: 129490 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 129606 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[3]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 129774 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[3]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 129828 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[3]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 129882 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[3]/figure[2]/image[1] - line: 129945 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] - line: 130015 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 130108 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 130167 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 130302 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 130360 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 130465 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 130540 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 130606 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[4]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 130673 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 130765 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 130841 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 130964 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 131024 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 131145 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[5]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 131199 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/figure[1]/image[1] - line: 131299 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[1]/image[1] - line: 131426 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[1]/image[2] - line: 131506 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[1]/sidebyside[1]/image[1] - line: 131600 - text: - -PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of

    elements, if present, or else against the first row. Results may be unpredictable. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[1]/sidebyside[1]/tabular[1] - line: 131653 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[1]/sidebyside[2]/image[1] - line: 131697 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[1]/sidebyside[2]/image[2] - line: 131766 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[2]/image[1] - line: 131852 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[2]/image[2] - line: 131946 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[3]/image[2] - line: 132138 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sidebyside[4]/image[2] - line: 132347 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[2]/sidebyside[2]/image[1] - line: 132558 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/sbsgroup[2]/sidebyside[2]/image[2] - line: 132630 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 132767 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 132837 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 132939 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 133015 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 133117 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 133188 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/subsection[6]/example[2]/statement[1]/figure[1]/image[1] - line: 133285 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/answer[1]/image[1] - line: 133861 - text: A surface obtained by translating the cubic curve along the y axisA surface generated by translating the curve y=cos(z) along the x axisA cylindrical tube with cross sections in the shape of an ellipseA pair of sheets that look like a hyperbola when viewed from above - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/statement[1]/image[1] - line: 134382 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/statement[1]/image[1] - line: 134465 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/statement[1]/image[1] - line: 134539 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[1]/answer[1]/image[1] - line: 134637 - text: A hyperbolic paraboloid (the classic "Pringles chip") - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[2]/answer[1]/image[1] - line: 134700 - text: An elliptical cone - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[3]/answer[1]/image[1] - line: 134762 - text: A circular paraboloid, opening along the negative x axis - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[4]/answer[1]/image[1] - line: 134823 - text: A hyperboloid of two sheets, opening along the x axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[5]/answer[1]/image[1] - line: 134891 - text: A hyperboloid of one sheet, opening along the y axis - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_space_coord.ptx - path: /pretext[1]/book[1]/chapter[11]/section[1]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[6]/answer[1]/image[1] - line: 134953 - text: An ellpsoid, somewhat in the shape of a squashed olive. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/figure[1]/image[1] - line: 135084 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/figure[2]/image[1] - line: 135134 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 135289 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 135328 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/example[2]/solution[1]/figure[1]/image[1] - line: 135538 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/figure[3]/image[1] - line: 135617 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/example[3]/solution[1]/figure[1]/image[1] - line: 135717 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/example[4]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] - line: 135806 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/example[5]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] - line: 136076 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/example[6]/statement[1]/figure[1]/image[1] - line: 136277 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/example[6]/solution[1]/figure[1]/image[1] - line: 136354 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/example[8]/statement[1]/figure[1]/image[1] - line: 136525 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/statement[1]/image[1] - line: 137077 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/solution[1]/image[1] - line: 137115 - text: Solution image for this exercise - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/statement[1]/image[1] - line: 137161 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/solution[1]/image[1] - line: 137198 - text: Solution image for this exercise - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/statement[1]/image[1] - line: 137247 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/solution[1]/image[1] - line: 137294 - text: Solution image for this exercise - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/statement[1]/image[1] - line: 137350 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/solution[1]/image[1] - line: 137395 - text: Solution image for this exercise - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[5]/introduction[1]/image[1] - line: 137843 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_intro.ptx - path: /pretext[1]/book[1]/chapter[11]/section[2]/exercises[1]/subexercises[2]/exercisegroup[6]/introduction[1]/image[1] - line: 137964 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 138258 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 138288 - text: - -PTX:WARNING: A normally does not have a single panel. If this construct is only for layout control, try moving layout onto the element used as panel ("image") and remove the - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[2]/sidebyside[1] - line: 138422 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[2]/sidebyside[1]/image[1] - line: 138423 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/example[2]/statement[1]/figure[1]/image[1] - line: 138496 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/example[3]/statement[1]/figure[1]/image[1] - line: 138601 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[3]/sidebyside[1]/figure[1]/image[1] - line: 138878 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[3]/sidebyside[1]/figure[2]/image[1] - line: 138911 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/example[5]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] - line: 139031 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/example[5]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 139094 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/example[5]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 139180 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/figure[4]/image[1] - line: 139315 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/example[7]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 139492 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/example[7]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 139524 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/paragraphs[1]/figure[1]/image[1] - line: 139631 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_dot_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[3]/paragraphs[1]/example[1]/statement[1]/figure[1]/image[1] - line: 139705 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cross_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[4]/introduction[1]/image[1] - line: 141118 - text: Schematic diagram for computing the cross product by multiplying along diagonal ar... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cross_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[1]/paragraphs[1]/figure[1]/image[1] - line: 141477 - text: Three-dimensional image illustrating the right-hand rule for the direction of the ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cross_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 141581 - text: A parallelogram with base b and height h labeled. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cross_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 141611 - text: Another parallelogram, with two adjacent sides labeled as vectors.A parallelogram in the plane, spanned by vectors u and v. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cross_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[1]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 141738 - text: A three-dimensional image of a parallelogram in space, with vertices A, B, C, and ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cross_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[1]/example[2]/statement[1]/figure[1]/image[1] - line: 141833 - text: A triangle in the plane, with vertices A(1,2), B(2,3), and C(3,1).A parellelepiped in three dimensions. It is the box-like object spanned by three v... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cross_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[2]/example[1]/solution[1]/figure[1]/image[1] - line: 142063 - text: The parallelepiped whose volume is computed in this example. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cross_product.ptx - path: /pretext[1]/book[1]/chapter[11]/section[4]/subsection[2]/paragraphs[3]/example[1]/statement[1]/figure[1]/image[1] - line: 142195 - text: Two adjacent images, each showing a pair of vectors, one of which represents a lev... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_lines.ptx - path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[1]/figure[1]/image[1] - line: 143364 - text: In three dimensions, vectors p (position), d (direction), and p+d are plotted. A l... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_lines.ptx - path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[1]/figure[2]/image[1] - line: 143440 - text: A text description, comparing the elements the equations of a line in the plane, a... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_lines.ptx - path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 143635 - text: A line in space with direction vector d passes through a point P.A line in space passes through points P and Q. The vector PQ between these points ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_lines.ptx - path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] - line: 143922 - text: A three dimensional plot of two lines in space. The lines are not parallel, and th... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_lines.ptx - path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] - line: 144103 - text: A plot in space of a single line through two points, with two parallel direction v... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_lines.ptx - path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[3]/figure[1]/image[1] - line: 144201 - text: A line through a point P, with direction d. Near the line is a point Q, whose dist... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_lines.ptx - path: /pretext[1]/book[1]/chapter[11]/section[5]/subsection[3]/figure[2]/image[1] - line: 144271 - text: Two skew lines in space are plotted with respective points and direction vectors. ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_planes.ptx - path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/figure[1]/image[1] - line: 145926 - text: A nail sticks into a flat, rectangular sheet of cardboard. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_planes.ptx - path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/example[1]/solution[1]/figure[1]/image[1] - line: 146122 - text: Three points P, Q, and R are plotted in space, along with the plane containing the... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_planes.ptx - path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/example[2]/solution[1]/figure[1]/image[1] - line: 146285 - text: In three dimensions, two lines are plotted, intersecting at a point P, along with ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_planes.ptx - path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/example[3]/solution[1]/figure[1]/image[1] - line: 146377 - text: A plane is plotted in a three-dimensional coordinate system, along with a line pas... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_planes.ptx - path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/example[4]/solution[1]/figure[1]/image[1] - line: 146514 - text: Two planes are shown intersecting along a common line. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_planes.ptx - path: /pretext[1]/book[1]/chapter[11]/section[6]/introduction[1]/example[5]/solution[1]/figure[1]/image[1] - line: 146638 - text: A line in three dimensions passes through a plane, intersecting it at a point.A plane, on which a point P and normal vector n are marked, along with a point Q n... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 148255 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 148286 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 148399 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 148469 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 148616 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 148647 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 148681 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[2]/example[2]/statement[1]/figure[1]/image[1] - line: 148745 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] - line: 148818 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] - line: 148912 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/solution[1]/image[1] - line: 149146 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/solution[1]/image[1] - line: 149189 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/solution[1]/image[1] - line: 149235 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/solution[1]/image[1] - line: 149281 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[5]/solution[1]/image[1] - line: 149330 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[6]/solution[1]/image[1] - line: 149380 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[7]/solution[1]/image[1] - line: 149426 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[8]/solution[1]/image[1] - line: 149476 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/solution[1]/image[1] - line: 149534 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/answer[1]/image[1] - line: 149586 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/answer[1]/image[1] - line: 149637 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf.ptx - path: /pretext[1]/book[1]/chapter[12]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/answer[1]/image[1] - line: 149691 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 150414 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 150446 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 150644 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 150682 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] - line: 150744 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[3]/solution[1]/figure[1]/image[1] - line: 150882 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[4]/solution[1]/figure[1]/image[1] - line: 150969 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[5]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/image[1] - line: 151150 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/subsection[3]/example[5]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/image[1] - line: 151226 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/solution[1]/image[1] - line: 151937 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[2]/solution[1]/image[1] - line: 151985 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/solution[1]/image[1] - line: 152035 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_calc.ptx - path: /pretext[1]/book[1]/chapter[12]/section[2]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[4]/solution[1]/image[1] - line: 152085 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_motion.ptx - path: /pretext[1]/book[1]/chapter[12]/section[3]/introduction[1]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 152799 - text: The curve corresponding the position function for this example, with velocity and ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_motion.ptx - path: /pretext[1]/book[1]/chapter[12]/section[3]/introduction[1]/example[1]/solution[1]/p[1]/ol[1]/li[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 152847 - text: The same curve as the previous image, with more vectors plotted.A parabola with vertex at the origin, opening upward. Several pairs of velocity an... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_motion.ptx - path: /pretext[1]/book[1]/chapter[12]/section[3]/introduction[1]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 153055 - text: The parabola y equals x squared, with several points plotted, corresponding to equ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_motion.ptx - path: /pretext[1]/book[1]/chapter[12]/section[3]/introduction[1]/example[2]/solution[1]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 153107 - text: The parabola y equals x squared, with several points plotted, corresponding to equ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_motion.ptx - path: /pretext[1]/book[1]/chapter[12]/section[3]/introduction[1]/example[3]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] - line: 153295 - text: A diagram showing a coniferous tree, a circle, and a tangent line from the circle ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vvf_motion.ptx - path: /pretext[1]/book[1]/chapter[12]/section[3]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/image[1] - line: 153802 - text: An illustration of the path followed by the particle in this example.A 3D plot of a portion of a helix, with two unit tangent vectors shown.A rotated parabola, along with two of its unit tangent vectors. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_tan_norm.ptx - path: /pretext[1]/book[1]/chapter[12]/section[4]/subsection[2]/figure[1]/image[1] - line: 155479 - text: A generic plane curve with a tangent vector, along with two possible normal vector... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_tan_norm.ptx - path: /pretext[1]/book[1]/chapter[12]/section[4]/subsection[2]/example[1]/solution[1]/figure[1]/image[1] - line: 155609 - text: A three-dimensional helix. At one point, the unit tangent and normal vectors are s... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_tan_norm.ptx - path: /pretext[1]/book[1]/chapter[12]/section[4]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] - line: 155740 - text: A rotated parabola, with unit tangent and normal vectors plotted at three points.<... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_tan_norm.ptx - path: /pretext[1]/book[1]/chapter[12]/section[4]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] - line: 156016 - text: A rotated parabola, with two points marked, corresponding to parameter values t=0 ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_tan_norm.ptx - path: /pretext[1]/book[1]/chapter[12]/section[4]/subsection[3]/example[3]/solution[1]/figure[1]/image[1] - line: 156086 - text: An inverted parabola with vertex in the first quadrant. Points corresponding to se... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_curvature.ptx - path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 157230 - text: A parabola, rotated to be symmetric about the line y=x, with three marked points.<... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_curvature.ptx - path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 157273 - text: The same parabola as the previous image, this time with six marked points, all equ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_curvature.ptx - path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 157407 - text: A straight line, with positive slope, and several equally-spaced marked points.Plot of a parametric curve that bends rapidly at one point, but is otherwise relat... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_curvature.ptx - path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 157580 - text: Plot of a parametric curve that bends rapidly at one point, but is otherwise relat... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_curvature.ptx - path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[2]/figure[2]/image[1] - line: 157871 - text: A curve with two marked points, corresponding to points of large and small curvatu... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_curvature.ptx - path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[2]/example[3]/solution[1]/figure[1]/image[1] - line: 157938 - text: A parabola along with osculating circles at two points. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_curvature.ptx - path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 158031 - text: A plot of the curvature as a function of the parameter t. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_curvature.ptx - path: /pretext[1]/book[1]/chapter[12]/section[5]/subsection[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 158067 - text: A plot of the vector-valued function in this example, with points of maximum curva... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_intro.ptx - path: /pretext[1]/book[1]/chapter[13]/section[1]/introduction[1]/example[2]/solution[1]/figure[1]/image[1] - line: 159382 - text: An ellipse, centered at the origin, with shaded interior, illustrating the domain ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_intro.ptx - path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 159444 - text: A collection of points plotted in space, against a set of three-dimensional coordi... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_intro.ptx - path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 159503 - text: A bell-shaped surface in three dimensions. It is symmetric about the z axis and ha... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_intro.ptx - path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[2]/figure[1]/image[1] - line: 159602 - text: A topographical map of Chrome Mountain in Montana. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_intro.ptx - path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 159701 - text: A family of concentric ellipses, centered at the origin. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_intro.ptx - path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 159748 - text: A surface in the shape of an elliptical dome. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_intro.ptx - path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[2]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 159868 - text: A line through the origin in the plane, and two sets of nested circles, one on eit... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_intro.ptx - path: /pretext[1]/book[1]/chapter[13]/section[1]/subsection[2]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 159915 - text: The surface given by the graph whose level curves were plotted in the previous ima... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_limit.ptx - path: /pretext[1]/book[1]/chapter[13]/section[2]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 160990 - text: A region in the first quadrant is shaded, and its boundary is shown as a solid cur... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_limit.ptx - path: /pretext[1]/book[1]/chapter[13]/section[2]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 161035 - text: A region in the first quadrant is shaded, and its boundary is shown as a dashed cu... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_limit.ptx - path: /pretext[1]/book[1]/chapter[13]/section[2]/subsection[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 161080 - text: A region in the first quadrant is shaded, and its boundary is shown as a partially... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_limit.ptx - path: /pretext[1]/book[1]/chapter[13]/section[2]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 161173 - text: An image of the xy plane. It is almost entirely shaded, except for the line y=x.An image that illustrates the concept of the limit of a function of two variables.... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_limit.ptx - path: /pretext[1]/book[1]/chapter[13]/section[2]/subsection[3]/example[1]/solution[1]/figure[1]/image[1] - line: 161805 - text: A graph of the piecewise-defined function in this example, illustrating its contin... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_partial_derivatives.ptx - path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 162937 - text: An elliptic paraboloid plotted over a rectangular domain. The trace y=2 is highlig... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_partial_derivatives.ptx - path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 162998 - text: The trace y=2 on the surface of the ellipic paraboloid from the previous image, wi... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_partial_derivatives.ptx - path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[1]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 163320 - text: On a surface in space, a trace curve is highlighted. At one pointon this curve, a ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_partial_derivatives.ptx - path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[1]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 163391 - text: On a surface in space, a trace curve is highlighted. At one pointon this curve, a ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_partial_derivatives.ptx - path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[4]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 163877 - text: A hyperbolic paraboloid, or saddle surface. A trace of constant x value is shown, ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_partial_derivatives.ptx - path: /pretext[1]/book[1]/chapter[13]/section[3]/subsection[4]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 163962 - text: A hyperbolic paraboloid, or saddle surface. A trace of constant y value is shown, ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_total_differential.ptx - path: /pretext[1]/book[1]/chapter[13]/section[4]/exercises[1]/subexercises[2]/exercisegroup[5]/exercise[3]/statement[1]/image[1] - line: 166781 - text: A right-angled triangle illustrating a wall of unknown height, a horizontal distan... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_chain.ptx - path: /pretext[1]/book[1]/chapter[13]/section[5]/introduction[1]/figure[1]/image[1] - line: 166895 - text: A plot of a surface in space, and a parametric curve on the surface.A circular paraboloid, opening upward, plotted over a rectangular domain. A curve ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_directional_derivative.ptx - path: /pretext[1]/book[1]/chapter[13]/section[6]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 168818 - text: A surface is plotted in 3D. Three vectors in the plane are shown corresponding to ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_directional_derivative.ptx - path: /pretext[1]/book[1]/chapter[13]/section[6]/subsection[1]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 169227 - text: A portion of the graph f(x,y) = sin(x)cos(y), along with a trace curve, and tangen... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_directional_derivative.ptx - path: /pretext[1]/book[1]/chapter[13]/section[6]/subsection[1]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 169324 - text: A zoomed-in view of the same surface, curve, and vectors as the previous image.A downward-opening circular paraboloid, with a point P marked at the vertex.A sector of a downward-opening elliptic paraboloid, along with a path that begins ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_directional_derivative.ptx - path: /pretext[1]/book[1]/chapter[13]/section[6]/subsection[1]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 169670 - text: A contour plot of the paraboloid in this example, and the corresponding trajectory... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_tangent.ptx - path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[1]/figure[1]/image[1] - line: 171151 - text: A surface in three dimensions, including curves on the surface, and tangent lines ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_tangent.ptx - path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 171344 - text: A bumpy surface. At a point on the surface, the tangent lines given by the partial... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_tangent.ptx - path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[1]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 171413 - text: A curve lies along a bumpy surface. At one point on the curve, a tangent line to t... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_tangent.ptx - path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 171549 - text: A graph of the function in this example; it has the shape of a steep hill.A circular paraboloid, opening downward. At one point on the surface, a normal lin... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_tangent.ptx - path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[2]/example[3]/solution[1]/figure[1]/image[1] - line: 171952 - text: A parabolic cylinder, and a normal line to the surface at one point.A downward-opening circular paraboloid, and a tangent plane to this surface at one... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_tangent.ptx - path: /pretext[1]/book[1]/chapter[13]/section[7]/subsection[4]/example[1]/solution[1]/figure[1]/image[1] - line: 172326 - text: An ellipsoid centered at the origin in space, and a tangent plane at one point.A circular paraboloid plotted over a rectangular domain. It is bowl-shaped, with p... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[13]/section[8]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 173706 - text: An inverted cone, with a maximum corresponding to a sharp peak on the z axis.The cubic surface studied in this example. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[13]/section[8]/subsection[1]/example[5]/solution[1]/figure[1]/image[1] - line: 174166 - text: A plot of the graph of the function used in this example. It is a more complicated... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[13]/section[8]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 174298 - text: A saddle surface on which a triangular curve is plotted. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[13]/section[8]/subsection[2]/example[1]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 174366 - text: A triangle in the plane representing the domain of the function in this example.Another view of the hyperbolic paraboloid in this example; this time with points o... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_multi_extreme_values.ptx - path: /pretext[1]/book[1]/chapter[13]/section[8]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] - line: 174710 - text: A three-dimensional plot of the volume in this example, as a function of length an... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/subsection[2]/figure[1]/image[1] - line: 175800 - text: A region in the first quadrant is bounded above and below by graphs, and left and ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/subsection[2]/figure[2]/image[1] - line: 175868 - text: A region in the first quadrant is bounded left and right by graphs, and above and ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] - line: 175964 - text: A rectangle in the plane, spanning x values from -1 to 3, and y values from 1 to 3... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/subsection[2]/example[2]/statement[1]/figure[1]/image[1] - line: 176036 - text: An obtuse triangle in the plane. The base is horizontal, and the other sides are l... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/subsection[2]/example[3]/statement[1]/figure[1]/image[1] - line: 176130 - text: A region in the first quadrant of the plane, bounded by a line and a parabola.A triangular region in the plane, bounded by lines y=0, x=6, and y=x/3.A sketch of the region of integration for this example. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[1]/statement[1]/image[1] - line: 176773 - text: The region R is a rectangle, with x from 1 to 4, and y from -2 to 1.The region R is a right triangle with vertices (1,1), (4,1), and (4,3)A triangular region R. The vertices are (2,5), (2,1), and (4,3).The region R is bounded to the left by a parabola opening along the x axis, and th... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[5]/statement[1]/image[1] - line: 176988 - text: A region in the first quadrant bounded by two curves that intersect at the points ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[6]/statement[1]/image[1] - line: 177055 - text: A region R bounded above by a line, and below by a cubic curve. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/solution[1]/image[1] - line: 177137 - text: A region R bounded by a downward-opening parabola and the x axis.The region R is in the first quadrant, between a parabola and a line.A region bounded by the y axis and the right half of an ellipse centered at the or... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/solution[1]/image[1] - line: 177294 - text: A circle of radius 3, centered at the origin. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_iterated_integrals.ptx - path: /pretext[1]/book[1]/chapter[14]/section[1]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[5]/solution[1]/image[1] - line: 177342 - text: The region between a right-opening parabola and a vertical line.A region bounded by a triangle, with vertices at (-1,1), (-1,-1), and (1,0).An illustration of the partition of a plane region into small rectangles.A surface in three dimensions lies over a partitioned region in the xy plane. A re... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_volume.ptx - path: /pretext[1]/book[1]/chapter[14]/section[2]/figure[2]/image[1] - line: 177891 - text: A parabolic surface in space is intersected by a plane of constant x value.A ramp-like surface in space, with the portion over a rectangular domain highlight... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_volume.ptx - path: /pretext[1]/book[1]/chapter[14]/section[2]/example[2]/statement[1]/figure[1]/image[1] - line: 178186 - text: A surface in space, plotted over a triangular domain. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_volume.ptx - path: /pretext[1]/book[1]/chapter[14]/section[2]/theorem[3]/statement[1]/p[2]/ol[1]/li[5]/figure[1]/image[1] - line: 178390 - text: Illustration of a pond-like region in the plane that has been divided into two sub... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_volume.ptx - path: /pretext[1]/book[1]/chapter[14]/section[2]/example[3]/statement[1]/figure[1]/image[1] - line: 178453 - text: A wave-like surface on which a triangular curve is drawn, corresponding to a domai... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_volume.ptx - path: /pretext[1]/book[1]/chapter[14]/section[2]/example[4]/statement[1]/figure[1]/image[1] - line: 178592 - text: A rectangular portion of a plane in space, and a petal-like curve on the plane.Several lines are plotted in the plane. Three of these lines form a triangle.A wave-like surface in three dimensions, and a curve on the surface illustrating a... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_volume.ptx - path: /pretext[1]/book[1]/chapter[14]/section[2]/example[6]/solution[1]/figure[1]/image[1] - line: 179014 - text: A rectangular portion of a plane in space, and a petal-like curve on the plane.A circular cylinder, bounded below by the xy plane, and above by the plane z=1.A region in the first quadrant of the plane, bounded by two parabolas.A region in the first quadrant of the plane, bounded by a cubic function and its i... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_volume.ptx - path: /pretext[1]/book[1]/chapter[14]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/solution[1]/p[1]/ol[1]/li[1]/image[1] - line: 179763 - text: A square of side length 2, with its center at the origin. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_volume.ptx - path: /pretext[1]/book[1]/chapter[14]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/solution[1]/p[1]/ol[1]/li[1]/image[1] - line: 179835 - text: A region in the plane that lies above a rightward-opening parabola, and below a ho... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_volume.ptx - path: /pretext[1]/book[1]/chapter[14]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[5]/solution[1]/p[1]/ol[1]/li[1]/image[1] - line: 179909 - text: A triangular region in the plane, with vertices at (0,0), (2,0), and (0,3).A region in the plane bounded by the graph of the natural logarithm and a straight... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_volume.ptx - path: /pretext[1]/book[1]/chapter[14]/section[2]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[7]/solution[1]/p[1]/ol[1]/li[1]/image[1] - line: 180067 - text: A region bounded by the x axis and the upper half of a semicircle of radius 3.The region above the graph of the natural logarithm, but below its tangent line at... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_polar.ptx - path: /pretext[1]/book[1]/chapter[14]/section[3]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 180417 - text: A polar grid in the first quadrant, with one grid sector highlighted.A close-up view of one region in a polar grid - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_polar.ptx - path: /pretext[1]/book[1]/chapter[14]/section[3]/example[1]/solution[1]/figure[1]/image[1] - line: 180665 - text: An ellipse on a plane in space corresponds to a circle drawn below it in the xy pl... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_polar.ptx - path: /pretext[1]/book[1]/chapter[14]/section[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 180762 - text: A region in the plane between that lies between two circles. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_polar.ptx - path: /pretext[1]/book[1]/chapter[14]/section[3]/example[2]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 180805 - text: A region in space that lies outside of a cylinder, and below a circular paraboloid... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_polar.ptx - path: /pretext[1]/book[1]/chapter[14]/section[3]/example[3]/statement[1]/figure[1]/image[1] - line: 180969 - text: A portion of a bell-shaped curve, and a triangular curve on the surface that corre... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_double_int_polar.ptx - path: /pretext[1]/book[1]/chapter[14]/section[3]/example[5]/statement[1]/figure[1]/image[1] - line: 181155 - text: A solid whose cross-sections are like a three-leaf clover; it has been sliced at a... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_center_of_mass.ptx - path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 181803 - text: A region in the plane with two vertical sides, a curved top, and a slanted bottom.... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_center_of_mass.ptx - path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 181849 - text: A region in the plane, with the same shape as the previous image, this time plotte... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_center_of_mass.ptx - path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 181981 - text: The unit square in the first quadrant, plotted against x and y coordinate axes.Two planes intersect along a line in three dimensions. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_center_of_mass.ptx - path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[1]/example[3]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 182218 - text: A circular paraboloid, opening upward, plotted over a circular domain.A circular cone, opening upward, plotted over a circular domain.A number line with three points marked at -1, 2, and 3, and a marking for the cent... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_center_of_mass.ptx - path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 182537 - text: A number line with three points marked at -1, 2, and 3, and a marking for the cent... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_center_of_mass.ptx - path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] - line: 182728 - text: A triangle in the plane. The vertices are marked with dots of different sizes, and... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_center_of_mass.ptx - path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[2]/example[3]/solution[1]/figure[1]/image[1] - line: 182877 - text: The unit square in the first quadrant of the plane. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_center_of_mass.ptx - path: /pretext[1]/book[1]/chapter[14]/section[4]/subsection[2]/example[6]/statement[1]/figure[1]/image[1] - line: 183002 - text: An annular region, between semi-circles of radii 5 and 6 in the upper half-plane.<... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_area.ptx - path: /pretext[1]/book[1]/chapter[14]/section[5]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 183771 - text: A surface is plotted above a circular domain. A partition of the domain correspond... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_area.ptx - path: /pretext[1]/book[1]/chapter[14]/section[5]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 183884 - text: A zoomed-in view of a portion of the surface from the previous image, and one of t... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_area.ptx - path: /pretext[1]/book[1]/chapter[14]/section[5]/example[1]/statement[1]/figure[1]/image[1] - line: 184066 - text: A triangle drawn on a plane in space. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_area.ptx - path: /pretext[1]/book[1]/chapter[14]/section[5]/example[3]/statement[1]/figure[1]/image[1] - line: 184271 - text: A downward-opening circular cone, with vertex on the z axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_area.ptx - path: /pretext[1]/book[1]/chapter[14]/section[5]/example[4]/statement[1]/figure[1]/image[1] - line: 184391 - text: A triangular region is shown on a trough-like surface - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_area.ptx - path: /pretext[1]/book[1]/chapter[14]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/image[1] - line: 184689 - text: A bumpy surface with several peaks and valleys. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_area.ptx - path: /pretext[1]/book[1]/chapter[14]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[2]/statement[1]/image[1] - line: 184772 - text: A steep surface with a single peak, resembling a witch's hat. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_area.ptx - path: /pretext[1]/book[1]/chapter[14]/section[5]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/image[1] - line: 184857 - text: A saddle surface, with its saddle point at the origin in three dimensions.A mostly flat surface with a ridge along the y axis - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 185276 - text: Two planes in space intersect along a line. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 185360 - text: Two planes in space plotted over a triangular domain. The planes intersect along o... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 185515 - text: An ellipsoid in space, centered at the origin. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 185573 - text: A zoomed-in view of a surface, showing grid lines. A small rectangular prism illus... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[2]/statement[1]/figure[1]/image[1] - line: 185838 - text: A triangular portion of a plane in space. It lies in the first octant and has vert... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 185919 - text: A triangle in the xy plane, plotted in space relative to 3D coordinate axes.A triangle plotted in the yz plane, relative to a set of 3D coordinate axes.A triangle plotted in the xz plane, relative to a set of 3D coordinate axes.A surface in space intersects with a plane through the origin, forming a parabolic... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[3]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 186423 - text: A curve in space, and its projections onto the three coordinate planes.A cylindrical wedge in space, resembling a wedge cut from a tree.The projection of the cylindrical wedge in the previous image onto the xy plane. I... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 186845 - text: The projection of the solid in this example onto the xy plane; it is a semi-circul... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 186910 - text: The projection of the solid in this example onto the yz plane; it is a triangle.A pyramid with a rectangular base in the plane z=0 and its peak on the z axis.The base of the pyramid shown in the previous image. It is a rectangle, divided in... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 187314 - text: The projection of the pyramid onto the plane y=0 is a triangle. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[5]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 187370 - text: The projection of the pyramid onto the plane x=0 is a triangle. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[6]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 187545 - text: An elliptic paraboloid, opening downward, is plotted over a rectangular domain.A parabolic cylinder in space. It opens upward, and is symmetric about the y axis.... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[6]/statement[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 187670 - text: The region in space bounded by the paraboloid and cylinder in the previous two ima... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[6]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 187853 - text: The shadow of the solid for this example on the plane y=0. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[1]/example[6]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 187923 - text: The shadow of the solid for this example on the plane x=0. It is divided into two ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[3]/example[1]/statement[1]/figure[1]/image[1] - line: 188363 - text: A triangle in space, formed by the portion of a plane that lies in the first octan... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/subsection[3]/example[2]/statement[1]/figure[1]/image[1] - line: 188493 - text: A cylindrical wedge in space, resembling a wedge cut from a tree.A tetrahedron in the first octant, with one vertex at the origin, and the others o... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[2]/statement[1]/image[1] - line: 189014 - text: A triangular prism. The two triangular sides are parallel to the xz plane.A solid bounded above by a plane, and below by a parabolic cylinder.One quarter of an elliptic cone that opens along the y axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[5]/statement[1]/image[1] - line: 189448 - text: A tetrahedron in space, plotted with respect to a three-dimensional coordinate sys... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[6]/statement[1]/image[1] - line: 189575 - text: A cylindrical wedge with its edge along the x axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[7]/statement[1]/image[1] - line: 189712 - text: A region in the first octant bounded by the planes x=0 and z=0, and two parabolic ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_triple_int.ptx - path: /pretext[1]/book[1]/chapter[14]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[8]/statement[1]/image[1] - line: 189868 - text: A pyramid with a rectangular base and vertex on the z axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cylindrical_spherical.ptx - path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[1]/figure[1]/image[1] - line: 190255 - text: A schematic diagram of the cylindrical coordinate system in three dimensions.Three surfaces in space, corresponding to fixed values of each of the three cylind... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cylindrical_spherical.ptx - path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[1]/example[3]/statement[1]/figure[1]/image[1] - line: 190646 - text: A circular cylinder capped by a spherical dome. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cylindrical_spherical.ptx - path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[1]/example[4]/statement[1]/figure[1]/image[1] - line: 190772 - text: A solid whose cross-sections are like a three-leaf clover; it has been sliced at a... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cylindrical_spherical.ptx - path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[2]/figure[1]/image[1] - line: 190994 - text: A schematic diagram illustrating the spherical coordinate system relative to the r... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cylindrical_spherical.ptx - path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[2]/example[2]/solution[1]/figure[1]/image[1] - line: 191230 - text: A three-dimensional plot showing three surfaces, each of which is obtained by fixi... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cylindrical_spherical.ptx - path: /pretext[1]/book[1]/chapter[14]/section[7]/subsection[2]/figure[2]/image[1] - line: 191390 - text: A schematic diagram illustrating how the spherical volume element is computed.A gem-like solid, bounded below by the top half of a circular cone, and above by a... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_cylindrical_spherical.ptx - path: /pretext[1]/book[1]/chapter[14]/section[7]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/statement[1]/image[1] - line: 192277 - text: A region in space given by constant bounds in cylindrical coordinates.A region in space given by constant bounds in spherical coordinates.A curve in the x,y plane, and the corresponding curve on a surface lying above it.... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_intro.ptx - path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 193562 - text: A cylindrical surface in space that lies above a parabola in the plane, and below ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_intro.ptx - path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[1]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 193629 - text: Approximating the area under a curve in space using planar rectangles.A curve lies along a surface in space. The curve corresponds to a line drawn in th... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_intro.ptx - path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[1]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 194042 - text: A curve in space lies above the line y=2x+1 in the x,y plane. The surface that lie... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_intro.ptx - path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[1]/example[2]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 194165 - text: A curve lies on a hyperbolic paraboloid above a circle in the plane.A cylindrical surface between the x,y plane and a hyperbolic paraboloid.A helical ramp in space. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_intro.ptx - path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[2]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 194497 - text: Two oriented curves in the plane that are joined at a point. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_intro.ptx - path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 194536 - text: Two oriented curves in the plane that are joined at a point. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_intro.ptx - path: /pretext[1]/book[1]/chapter[15]/section[1]/subsection[3]/example[1]/statement[1]/figure[1]/image[1] - line: 194736 - text: A curve in space representing a wire, and a point representing its center of mass.... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/introduction[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 195304 - text: Several arrows are drawn in the plane to illustrate the concept of a vector field.... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/introduction[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 195376 - text: Several arrows are plotted in the plane to illustrate the concept of a vector fiel... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/introduction[1]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 195494 - text: Eight vectors are plotted in the plane to represent a vector field, using relative... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/introduction[1]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 195566 - text: Another plot of the same vector field used in every image so far, but this time wi... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/introduction[1]/figure[3]/image[1] - line: 195674 - text: A very chaotic plot of a three-dimensional vector field. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 196071 - text: A vector field of horizontal arrows: to the right above the x axis, and to the lef... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 196157 - text: A rotational vector field. The vectors appear to describe counter-clockwise circul... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 196304 - text: A radial vector field. Each vector points away from the origin, and vectors furthe... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[1]/solution[1]/p[1]/ol[1]/li[3]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 196377 - text: A vector field that appears to describe swirling motion, with several vortices.A radial vector field. Each vector points toward the origin, and vectors near the ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 196712 - text: The gradient vector field of a potential fuction. Vectors point toward the origin,... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/subsection[2]/example[4]/solution[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 196794 - text: A three-dimensional image showing the graph of a function of two variables, with i... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/answer[1]/image[1] - line: 197001 - text: A vector field of horizontal vectors, pointing away from the y axis.A vertical vector field. Vectors right of the y axis point up, and those to the le... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_vector_fields.ptx - path: /pretext[1]/book[1]/chapter[15]/section[2]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/answer[1]/image[1] - line: 197163 - text: A constant vector field. All vectors point down and to the right.A two dimensional vector field. The vectors point up and to the right. Vectors nea... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_vf.ptx - path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[1]/example[1]/statement[1]/figure[1]/image[1] - line: 197676 - text: A vector field is plotted in two dimensions, along with two different paths that b... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_vf.ptx - path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[1]/example[2]/statement[1]/figure[1]/image[1] - line: 197832 - text: A two-dimensional vector field, and two curves in the plane. The vectors appear to... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_vf.ptx - path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] - line: 198071 - text: A two-dimensional vector field of vectors tangent to a family of parabolas, and a ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_vf.ptx - path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[2]/example[2]/statement[1]/figure[1]/image[1] - line: 198235 - text: A three-dimensional vector field and a portion of a helix curve.Several regions in the plane are used to illustrate the concepts of connected, and... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_vf.ptx - path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[3]/figure[2]/sidebyside[1]/figure[1]/image[1] - line: 198510 - text: Two spherical surfaces in space, plotted without coordinate axes. The second spher... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_line_int_vf.ptx - path: /pretext[1]/book[1]/chapter[15]/section[3]/subsection[3]/figure[2]/sidebyside[1]/figure[2]/image[1] - line: 198602 - text: Two surfaces in three dimensions. One is a sphere with a cylinder through its cent... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_greensthm.ptx - path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[1]/figure[1]/image[1] - line: 199501 - text: A constant vector field in the plane, pointing to the right, and a triangular path... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_greensthm.ptx - path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[1]/figure[2]/image[1] - line: 199679 - text: A closed curve in the plane, in the shape of a cashew or boomerang.Two paths in the plane are plotted relative to a vector field. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_greensthm.ptx - path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[1]/example[1]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 199937 - text: Two paths in the plane are plotted relative to a vector field. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_greensthm.ptx - path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] - line: 200181 - text: A triangular path in the plane is plotted against a two-dimensional vector field.<... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_greensthm.ptx - path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[2]/example[2]/statement[1]/figure[1]/image[1] - line: 200354 - text: A curve in the plane, like a bumpy circle or flower, and a vector field.A teardrop-shaped region in the plane. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_greensthm.ptx - path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[3]/example[1]/statement[1]/figure[1]/image[1] - line: 200635 - text: A circle in the plane, plotted against a spiral vector field. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_greensthm.ptx - path: /pretext[1]/book[1]/chapter[15]/section[4]/subsection[3]/example[2]/solution[1]/figure[1]/image[1] - line: 200757 - text: Two curves in the plane between points A and B, and a vector field with many vorti... - -PTX:WARNING: You have an image without a description and do not declare the image to be decorative. Because of this, output may not be accessible. If the image does not add information that is not already present, use @decorative="yes". Otherwise, provide a . - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/aside[1]/image[1] - line: 201395 - text: A rendering of a script S with serifs. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/figure[1]/image[1] - line: 201513 - text: A depction of a Möbius band, showing how the normal vector changes direction while... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[1]/solution[1]/figure[1]/image[1] - line: 201658 - text: A portion of an elliptic paraboloid with a rectangular domain. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[2]/solution[1]/figure[1]/image[1] - line: 201749 - text: An elliptic paraboloid plotted over a circular domain. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[3]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 201862 - text: A triangular domain is plotted in the first quadrant of the plane.An elliptic paraboloid is plotted in space over a triangular domain.A triangular domain in the plane, with vertices at (0,1), (0,3), and (3,1).An elliptic paraboloid is plotted in space over a triangular domain.A cylindrical surface centered along the y axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[6]/statement[1]/figure[1]/image[1] - line: 202351 - text: An elliptic cone with its cusp at the origin, including portions above and below t... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[6]/solution[1]/figure[1]/image[1] - line: 202443 - text: A cone plotted parametrically with restricted domain. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[7]/statement[1]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 202523 - text: The ellipsoid to be parametrized in this example. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[1]/example[7]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 202577 - text: A portion of an ellipsoid in space corresponding to a restricted parameter domain.... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[2]/figure[1]/sidebyside[1]/figure[1]/image[1] - line: 202762 - text: A rectangular region in the plane, with a smaller subrectangle highlighted in its... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[2]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 202832 - text: A surface in space with parameter domain R, highlighting the portion corresponding... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/subsection[2]/figure[1]/sidebyside[1]/figure[3]/image[1] - line: 202925 - text: A zoomed in view of the parallelogram approximation of a small patch on a surface.... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/statement[1]/image[1] - line: 203454 - text: A triangular prism plotted in three dimensions. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[2]/statement[1]/image[1] - line: 203588 - text: A tetrahedron with vertices (2,0,0), (0,1,0), (2,1,0), and (2,1,4).A cylindrical wedge with its edge along the x axis. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/statement[1]/image[1] - line: 203842 - text: A region in the first octant bounded by the planes x=0 and z=0, and two parabolic ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[5]/statement[1]/image[1] - line: 204022 - text: An elliptical cylinder, centered on the z axis, capped by planes z=1 and z=3.An inverted circular cone, together with a disk in the x,y planeA solid that looks like a barn with a parabolic roof. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_parametric_surfaces.ptx - path: /pretext[1]/book[1]/chapter[15]/section[5]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[8]/statement[1]/image[1] - line: 204368 - text: A region bounded above by an elliptic paraboloid, and below by the x,y plane.A triangular portion of a plane in space. It is the graph of a linear function ove... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_integral.ptx - path: /pretext[1]/book[1]/chapter[15]/section[6]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] - line: 204931 - text: A triangular surface in space, and a vector field flowing across the surface.A closed surface consisting of a downward-opening circular paraboloid, and a disc ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_integral.ptx - path: /pretext[1]/book[1]/chapter[15]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[7]/statement[1]/image[1] - line: 205583 - text: An elliptical paraboloid opens downward, intersecting the x,y plane in an ellipse.... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_surface_integral.ptx - path: /pretext[1]/book[1]/chapter[15]/section[6]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[8]/statement[1]/image[1] - line: 205667 - text: Approximately three quarters of the unit sphere. The top has been removed and repl... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/subsection[1]/example[1]/statement[1]/figure[1]/image[1] - line: 205814 - text: A parabolic wedge, bounded by two planes and a parabolic cylinder.A circular paraboloid, opening downward, meets the x,y plane along the unit circle... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/subsection[1]/example[3]/statement[1]/figure[1]/image[1] - line: 206248 - text: A cube centered at the origin in space. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/subsection[2]/example[1]/statement[1]/figure[1]/image[1] - line: 206539 - text: An ellipse drawn on a plane in space. It lies over a circle in the x,y plane.The portion of a circular paraboloid in the first octant, and an elliptical curve ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/subsection[2]/example[2]/statement[1]/figure[1]/sidebyside[1]/figure[2]/image[1] - line: 206762 - text: The parabolic surface from the previous image is shown intersecting a plane along ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[1]/statement[1]/image[1] - line: 207044 - text: A tetrahedron with vertices (0,0,0), (4,0,0), (0,3,0), and (0,0,2).A circular cylinder, centered on the z axis, capped by disks in the planes z=3 and... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[3]/statement[1]/image[1] - line: 207284 - text: A steep hill with a square base in the x-y plane. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[1]/exercise[4]/statement[1]/image[1] - line: 207386 - text: A familiar dome-shaped surface between a circular paraboloid and the x,y plane.A bowl-shaped surface given by a circular paraboloid plotted over the unit disk.A bell-shaped surface in space, with a circular boundary. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[3]/statement[1]/image[1] - line: 207661 - text: A triangular surface in the first octant, with intercepts at (4,0,0), (0,3,0), and... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[2]/exercise[4]/statement[1]/image[1] - line: 207752 - text: Graph of an elliptic paraboloid over a parabolic domain. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[1]/statement[1]/image[1] - line: 207863 - text: A domed hill with a square base in the x-y plane. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[2]/statement[1]/image[1] - line: 207959 - text: A triangular prism, with two triangular faces and three rectangular faces.A wedge-shaped surface given by two intersecting planes and a parabolic cylinder.<... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[3]/exercise[4]/statement[1]/image[1] - line: 208191 - text: A surface resembling a barn or greenhouse with an arched roof. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[1]/statement[1]/image[1] - line: 208330 - text: A twisted-looking surface given by a portion of a hyperbolic paraboloid.A surface that looks something like a ladle, if the handle had melted and drooped ... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: sec_stokes_divergence.ptx - path: /pretext[1]/book[1]/chapter[15]/section[7]/exercises[1]/subexercises[2]/exercisegroup[4]/exercise[3]/statement[1]/image[1] - line: 208539 - text: A region in a plane in space, bounded by an ellipse in that plane.A rectangle lying within a plane in space. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[3]/introduction[1]/assemblage[1]/image[1] - line: 209144 - text: A detailed plot of the unit circle, showing angles in both degrees and radians, an... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[3]/subsection[1]/assemblage[1]/sidebyside[1]/image[1] - line: 209286 - text: An illustration of the correspondence between an angle and a point on the unit cir... - -PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of elements, if present, or else against the first row. Results may be unpredictable. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[3]/subsection[1]/assemblage[1]/sidebyside[1]/tabular[1] - line: 209325 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[3]/subsection[1]/assemblage[2]/sidebyside[1]/image[1] - line: 209352 - text: A right angle triangle, with sides labeled "opposite", "adjacent", and "hypotenuse... - -PTX:WARNING: The rows of this do not all span the same number of columns (counting each cell as its @colspan, or as one column otherwise). Compare the rows against the number of elements, if present, or else against the first row. Results may be unpredictable. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[3]/subsection[1]/assemblage[2]/sidebyside[1]/tabular[1] - line: 209376 - text: - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[1]/image[1] - line: 209544 - text: A schematic diagram of a triangle, labeling three sides, an angle, and an altitude... - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[1]/image[2] - line: 209593 - text: A schematic diagram of a right circular cone, showing the height and radius.A generic parallelogram, with base and height labeled. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[2]/image[2] - line: 209678 - text: A diagram of a right circular cylinder, labeling the radius and height.A schematic diagram of a generic trapezoid. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[3]/image[2] - line: 209773 - text: A image of a sphere, showing one circumference and its radius. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[4]/image[1] - line: 209819 - text: A generic circle with its radius indicated. - -PTX:ADVICE: You have an image description that is more than 125 characters long. Some screen readers will cut off reading alt text after the 125th character. Rewrite the description to be 125 characters or fewer. - file: appendix_back_reference.ptx - path: /pretext[1]/book[1]/backmatter[1]/appendix[1]/section[4]/sbsgroup[1]/sidebyside[4]/image[2] - line: 209856 - text: A drawing of a general cone, with an arbitrary plane region for its base.A pie-shaped sector of a circle, labeled with angle, radius, and arc length.A sketch of a right cylinder with an arbitrary base. From 2ae83ea0dcb49cb9a4b2b4895709bc9c0d31627a Mon Sep 17 00:00:00 2001 From: sean-fitzpatrick Date: Fri, 10 Jul 2026 14:05:13 -0600 Subject: [PATCH 5/5] More schema fixes --- ptx/sec_ABC.ptx | 185 +++++++++++------ ptx/sec_arc_length.ptx | 137 +++++++------ ptx/sec_def_int.ptx | 22 +- ptx/sec_deriv_basic_rules.ptx | 2 +- ptx/sec_deriv_chainrule.ptx | 12 +- ptx/sec_deriv_implicit.ptx | 52 ++--- ptx/sec_deriv_prodquot.ptx | 2 +- ptx/sec_directional_derivative.ptx | 2 +- ptx/sec_newton.ptx | 2 +- ptx/sec_optimization.ptx | 4 +- ptx/sec_substitution.ptx | 2 +- ptx/sec_taylor_poly.ptx | 10 +- ptx/sec_vvf.ptx | 310 ++++++++++++++++------------- ptx/sec_vvf_calc.ptx | 187 +++++++++-------- ptx/sec_work.ptx | 128 +++++++----- 15 files changed, 620 insertions(+), 437 deletions(-) diff --git a/ptx/sec_ABC.ptx b/ptx/sec_ABC.ptx index 863e864ff..847116588 100644 --- a/ptx/sec_ABC.ptx +++ b/ptx/sec_ABC.ptx @@ -485,8 +485,12 @@ - Graph of the shaded region bounded by y=\sqrt{x}+2, y=-(x-1)^2+3 and y=2 with a red horizontal rectangle slice showing integration with respect to y. - The rectangle starts near the beginning of the graph of y=\sqrt{x}+2 at a y-level of about 2.2 and spans across to meet the graph of y=-(x-1)^2+3. +

    + Graph of the shaded region bounded by y=\sqrt{x}+2, + y=-(x-1)^2+3 and y=2 with a red horizontal rectangle slice showing integration with respect to y. + The rectangle starts near the beginning of the graph of y=\sqrt{x}+2 at a + y level of about 2.2 and spans across to meet the graph of y=-(x-1)^2+3. +

    Graph of the shaded region with boundaries relabeled as functions of y. @@ -968,8 +972,13 @@ - Graph of the region enclosed by the functions y=\frac12 x +3 and y=\frac12\cos(x)+1 between x = 0 and x = 2\pi. - The function y=\frac12 x +3 lies above the function y=\frac12\cos(x)+1 for the entirety of the region between x = 0 and x = 2\pi. + +

    + Graph of the region enclosed by the functions y=\frac12 x +3 + and y=\frac12\cos(x)+1 between x = 0 and x = 2\pi. + The function y=\frac12 x +3 lies above the function y=\frac12\cos(x)+1 + for the entirety of the region between x = 0 and x = 2\pi. +

    Graph of the region between two functions between x=0 and x=2pi. @@ -1033,8 +1042,13 @@ - Graph of the region enclosed by the functions y=-3x^3+3x+2 and y=x^2+x-1 between x = -1 and x = 1. - The function y=-3x^3+3x+2 lies above the function y=x^2+x-1 for the entirety of the region between x = -1 and x = 1. + +

    + Graph of the region enclosed by the functions y=-3x^3+3x+2 + and y=x^2+x-1 between x = -1 and x = 1. + The function y=-3x^3+3x+2 lies above the function y=x^2+x-1 + for the entirety of the region between x = -1 and x = 1. +

    Graph of the region enclosed by the two functions between x=0 and x=2pi. @@ -1096,7 +1110,10 @@ - Graph of the region enclosed by horizontal lines y=0 and y=1 between x = 0 and x = \pi. + +

    + Graph of the region enclosed by horizontal lines y=0 and y=1 between x = 0 and x = \pi. +

    Graph of the rectangular region between y=1, y=2, and x=0, x=pi. @@ -1161,8 +1178,12 @@ - Graph of the region enclosed by the functions y=\sin(x)+1 and y=\sin(x) between x = 0 and x = \pi. - The function y=\sin(x)+1 lies above the function y=\sin(x) for the entirety of the region between x = 0 and x = \pi. + +

    + Graph of the region enclosed by the functions y=\sin(x)+1 and y=\sin(x) between x = 0 and x = \pi. + The function y=\sin(x)+1 lies above the function y=\sin(x) + for the entirety of the region between x = 0 and x = \pi. +

    Graph of the region between the two sine graphs between x=0 and x=pi. @@ -1228,8 +1249,12 @@ - Graph of the region enclosed by the functions y=\sin(4x) and y=\sec^2(x) between x = 0 and x = \pi/4. - The function y=\sec^2(x) lies above the function y=\sin(4x) for the entirety of the region between x = 0 and x = \pi/4. + +

    + Graph of the region enclosed by the functions y=\sin(4x) and y=\sec^2(x) between x = 0 and x = \pi/4. + The function y=\sec^2(x) lies above the function y=\sin(4x) + for the entirety of the region between x = 0 and x = \pi/4. +

    Graph of the region enclosed by the two functions between x=0 and x=pi/4. @@ -1294,9 +1319,13 @@ - Graph of the region enclosed by the functions y=\sin(x) and y=\cos(x) between x = \pi/4 and x = 5\pi/4. - The two functions intersect at x = \pi/4 and x = 5\pi/4. - The function y=\sec^2(x) lies above the function y=\sin(4x) for the entirety of the region between x = \pi/4 and x = 5\pi/4. + +

    + Graph of the region enclosed by the functions y=\sin(x) and y=\cos(x) between x = \pi/4 and x = 5\pi/4. + The two functions intersect at x = \pi/4 and x = 5\pi/4. + The function y=\sec^2(x) lies above the function y=\sin(4x) + for the entirety of the region between x = \pi/4 and x = 5\pi/4. +

    Graph of the region enclosed by the two functions between x=pi/4 and x=5 pi/4. @@ -1360,9 +1389,12 @@ - Graph of the region enclosed by the functions y=2^x and y=4^x between x = 0 and x = 1. - The two functions intersect at x = 0 before y=4^x overtakes y=2^x. - The function y=4^x lies above the function y=2^x for the entirety of the region between 0 and 1. + +

    + Graph of the region enclosed by the functions y=2^x and y=4^x between x = 0 and x = 1. + The two functions intersect at x = 0 before y=4^x overtakes y=2^x. + The function y=4^x lies above the function y=2^x for the entirety of the region between 0 and 1. +

    Graph of the region enclosed by the two functions between x=0 and x=1. @@ -1406,14 +1438,17 @@ and y=1.

    - Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1 and the horizontal line y=1. - The curve given by y=\sqrt{x}+1 is drawn starting at the y-axis and ending at the point (1,2) . - The curve given by y=\sqrt{2-x}+1 is drawn starting from the end of the previous curve, at the point (1,2) . - This curve then falls downwards before intersecting the horizontal line y=1 at the point (2,1) . - Both curves lie entirely above the horizontal line y=1. - The curve y=\sqrt{x}+1 also lies to the left of y=\sqrt{2-x}+1 throughout the enclosed region. - - Graph of the region enclosed by the two functions and the line y=1. + +

    + Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1 and the horizontal line y=1. + The curve given by y=\sqrt{x}+1 is drawn starting at the y-axis and ending at the point (1,2) . + The curve given by y=\sqrt{2-x}+1 is drawn starting from the end of the previous curve, at the point (1,2) . + This curve then falls downwards before intersecting the horizontal line y=1 at the point (2,1) . + Both curves lie entirely above the horizontal line y=1. + The curve y=\sqrt{x}+1 also lies to the left of y=\sqrt{2-x}+1 throughout the enclosed region. +

    +     + Graph of the region enclosed by the two functions and the line y=1. \begin{tikzpicture} @@ -1729,12 +1764,15 @@ - Graph of the region enclosed by the functions y=x^2+1, y=\frac14(x-3)^2+1 and the horizontal line y=1. - The curve given by y=x^2+1 is drawn starting at the y-axis and ending at the point (1,2) . - The curve given by y=\frac14(x-3)^2+1 is drawn starting from the end of the previous curve at the point (1,2) . - This curve then falls downwards before intersecting the horizontal line y=1 at the point (3,1) . - Both curves lie above the line y=1 for the entirety of the enclosed region. - The curve y=\frac14(x-3)^2+1 also lies to the right of the curve y=x^2+1 throughout the enclosed region. + +

    + Graph of the region enclosed by the functions y=x^2+1, y=\frac14(x-3)^2+1 and the horizontal line y=1. + The curve given by y=x^2+1 is drawn starting at the y-axis and ending at the point (1,2) . + The curve given by y=\frac14(x-3)^2+1 is drawn starting from the end of the previous curve at the point (1,2) . + This curve then falls downwards before intersecting the horizontal line y=1 at the point (3,1) . + Both curves lie above the line y=1 for the entirety of the enclosed region. + The curve y=\frac14(x-3)^2+1 also lies to the right of the curve y=x^2+1 throughout the enclosed region. +

    Graph of the region enclosed by the two functions and the line y=1. @@ -1800,13 +1838,16 @@ - Graph of the region enclosed by the functions y=\sqrt{x}, y=-2x+3 and y=-\frac12 x. - The curve given by y=\sqrt{x} is drawn starting at the origin, from which it goes upwards and ends at the point (1,1) . - The curve given by y=-2x+3 is drawn starting from the end of the previous curve at the point (1,1) . - This curve then goes downwards before ending at the point (2,-1) . - The curve given by y=-\frac12 x is drawn starting from the origin, from which it goes downwards until it meets the previous curve at the point (2,-1) . - The curve y=\sqrt{x} graphed between x=0 and x=1 and the line y=-2x+3 graphed between x=1 and x=2 lie entirely above the line y=-\frac12 x. - The curve y=\sqrt{x} between y=0 and y=1 and the line y=-\frac12 x between y=-1 and y=0 lie to the left of the curve y=-2x+3. + +

    + Graph of the region enclosed by the functions y=\sqrt{x}, y=-2x+3 and y=-\frac12 x. + The curve given by y=\sqrt{x} is drawn starting at the origin, from which it goes upwards and ends at the point (1,1) . + The curve given by y=-2x+3 is drawn starting from the end of the previous curve at the point (1,1) . + This curve then goes downwards before ending at the point (2,-1) . + The curve given by y=-\frac12 x is drawn starting from the origin, from which it goes downwards until it meets the previous curve at the point (2,-1) . + The curve y=\sqrt{x} graphed between x=0 and x=1 and the line y=-2x+3 graphed between x=1 and x=2 lie entirely above the line y=-\frac12 x. + The curve y=\sqrt{x} between y=0 and y=1 and the line y=-\frac12 x between y=-1 and y=0 lie to the left of the curve y=-2x+3. +

    Graph of the region enclosed by the three functions. @@ -1865,13 +1906,16 @@ - Graph of the region enclosed by the functions y=x+2 and y=x^2. - The line y=x+2 is drawn starting at the point (-1,1), from which it goes upwards and ends at the point (2,4) . - The parabola given by y=x^2 is also drawn starting from the point (-1,1), from which it slopes downwards until the point (0,0) . - The parabola then goes upwards before meeting the line at the point (2,4) . - The line y=x+2 graphed between x=-1 and x=2 lies entirely above the parabola y=x^2 graphed on the same bounds, only interesecting at the end points. - The line y=\sqrt{x} lies to the left of the parabola y=x^2 between y=1 and y=4. - Between y=0 and y=1, the enclosed region spans from the left side of y=x^2 to the right side of the y=x^2. + +

    + Graph of the region enclosed by the functions y=x+2 and y=x^2. + The line y=x+2 is drawn starting at the point (-1,1), from which it goes upwards and ends at the point (2,4) . + The parabola given by y=x^2 is also drawn starting from the point (-1,1), from which it slopes downwards until the point (0,0) . + The parabola then goes upwards before meeting the line at the point (2,4) . + The line y=x+2 graphed between x=-1 and x=2 lies entirely above the parabola y=x^2 graphed on the same bounds, only interesecting at the end points. + The line y=\sqrt{x} lies to the left of the parabola y=x^2 between y=1 and y=4. + Between y=0 and y=1, the enclosed region spans from the left side of y=x^2 to the right side of the y=x^2. +

    Graph of the region enclosed by the line and parabola. @@ -1936,13 +1980,16 @@ - Graph of the region enclosed by the functions x=-\frac12 y+1 and x=\frac12 y^2. - The line y=x+2 is drawn starting at the point (0.5,1), from which it goes downwards and ends at the point (2,-2) . - The sideways parabola given by x=\frac12 y^2 is also drawn starting from the point (0.5,1), from which it goes backwards until the point (0,0). - The parabola then heads to the right before meeting the line at the point (2,-2) . - The line x=-\frac12 y+1 graphed between x=-0.5 and x=2 lies entirely above the parabola x=\frac12 y^2 between the same bounds, only interesecting at the end points. - However, between x=0 and x=0.5 , both the top and bottom of the enclosed region are made up by the parabola. - The sideways parabola x=\frac12 y^2 lies to the left of the line x=-\frac12 y+1 for the entirety of the region between y=-2 and y=1 . + +

    + Graph of the region enclosed by the functions x=-\frac12 y+1 and x=\frac12 y^2. + The line y=x+2 is drawn starting at the point (0.5,1), from which it goes downwards and ends at the point (2,-2) . + The sideways parabola given by x=\frac12 y^2 is also drawn starting from the point (0.5,1), from which it goes backwards until the point (0,0). + The parabola then heads to the right before meeting the line at the point (2,-2) . + The line x=-\frac12 y+1 graphed between x=-0.5 and x=2 lies entirely above the parabola x=\frac12 y^2 between the same bounds, only interesecting at the end points. + However, between x=0 and x=0.5 , both the top and bottom of the enclosed region are made up by the parabola. + The sideways parabola x=\frac12 y^2 lies to the left of the line x=-\frac12 y+1 for the entirety of the region between y=-2 and y=1 . +

    Graph of the region enclosed by the line and parabola. @@ -2008,13 +2055,16 @@ - Graph of the region enclosed by the functions y=x^{1/3}, y=\sqrt{x-1/2} and the lines y=0, x=1. - The curve y=x^{1/3} is drawn starting at the origin, from which it goes upwards and ends at the point (1,1) . - The curve y=\sqrt{x-1/2} is drawn starting from the point (0.5,0), from which it goes upwards until ending at the point (1,sqrt{\frac12}). - The curve y=x^{1/3} graphed between x=0 and x=0.5 lies above x-axis. - Between x=0.5 and x=1 , the curve y=x^{1/3} lies above the curve y=\sqrt{x-1/2}. - The curve y=x^{1/3} lies to the left of the curve y=\sqrt{x-1/2} for the entirety of the length of the curve y=\sqrt{x-1/2}. - After this point, the curve y=x^{1/3} lies to the left of the boundary line occuring at x=1. + +

    + Graph of the region enclosed by the functions y=x^{1/3}, y=\sqrt{x-1/2} and the lines y=0, x=1. + The curve y=x^{1/3} is drawn starting at the origin, from which it goes upwards and ends at the point (1,1) . + The curve y=\sqrt{x-1/2} is drawn starting from the point (0.5,0), from which it goes upwards until ending at the point (1,sqrt{\frac12}). + The curve y=x^{1/3} graphed between x=0 and x=0.5 lies above x-axis. + Between x=0.5 and x=1 , the curve y=x^{1/3} lies above the curve y=\sqrt{x-1/2}. + The curve y=x^{1/3} lies to the left of the curve y=\sqrt{x-1/2} for the entirety of the length of the curve y=\sqrt{x-1/2}. + After this point, the curve y=x^{1/3} lies to the left of the boundary line occuring at x=1. +

    Graph of the region enclosed by the two functions and y=0 and x=1. @@ -2059,13 +2109,16 @@ and y=1.

    - Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1 and the line y=1. - The curve y=\sqrt{x}+1 is drawn starting at the point (0,1), from which it goes upwards and ends at the point (1,2) . - The curve y=\sqrt{2-x}+1 is drawn starting from the end of the previous curve, at the point (1,2), from which it goes downward until ending at the point (2,1). - Both of these curves lie entirely above the horizontal line y=1. - The curve y=\sqrt{x}+1 lies to the left of the curve y=\sqrt{2-x}+1 for the entirety of the enclosed region. - - Graph of the region enclosed by the two functions and the line y=1. + +

    + Graph of the region enclosed by the functions y=\sqrt{x}+1, y=\sqrt{2-x}+1 and the line y=1. + The curve y=\sqrt{x}+1 is drawn starting at the point (0,1), from which it goes upwards and ends at the point (1,2) . + The curve y=\sqrt{2-x}+1 is drawn starting from the end of the previous curve, at the point (1,2), from which it goes downward until ending at the point (2,1). + Both of these curves lie entirely above the horizontal line y=1. + The curve y=\sqrt{x}+1 lies to the left of the curve y=\sqrt{2-x}+1 for the entirety of the enclosed region. +

    +     + Graph of the region enclosed by the two functions and the line y=1. \begin{tikzpicture} diff --git a/ptx/sec_arc_length.ptx b/ptx/sec_arc_length.ptx index e994cd35e..995ec2f1d 100644 --- a/ptx/sec_arc_length.ptx +++ b/ptx/sec_arc_length.ptx @@ -75,9 +75,12 @@ - Graph of the function y=\sin(x) on [0,\pi]. - The curve y=\sin(x) begins at the point (0,0), from which it slopes upwards until reaching a peak at the point (\frac{\pi}{2},1). - From the point, the curve slopes downwards until reaching the x-axis at the point (\pi,0). +

    + Graph of the function y=\sin(x) on [0,\pi]. + The curve y=\sin(x) begins at the point (0,0), + from which it slopes upwards until reaching a peak at the point (\frac{\pi}{2},1). + From the point, the curve slopes downwards until reaching the x-axis at the point (\pi,0). +

    Graph of the sine function for x between 0 and pi. @@ -109,13 +112,15 @@ - Graph of the function y=\sin(x) on [0,\pi], with four straight lines which will be used to approximate the length of this curve. - The four lines are evenly spaced out in intervals of \frac{\pi}{4} on the x-axis. - Each line begins at a point on the curve y=\sin(x), and ends at a point on the same curve after travelling a distance of \frac{\pi}{4} on the x-axis. - The first line begins at the same point (0,0) as the curve, from which it linearly increases until reaching the point (\frac{\pi}{4},\frac{sqrt2}{2}), which is also a point on the curve. - The second line begins at the same point the first line ends, given by (\frac{\pi}{4},\frac{sqrt2}{2}), from which it linearly increases until reaching the point (\frac{\pi}{2},1), which is the peak of the curve. - The third line begins at the same point the second line ends, given by (\frac{\pi}{2},1), from which it linearly decreases until reaching the point (\frac{3\pi}{4},\frac{sqrt2}{2}), which is also a point on the curve. - The fourth line begins at the same point the third line ends, given by (\frac{3\pi}{4},\frac{sqrt2}{2}), from which it linearly decreases until reaching the point (\pi,0), which is the end of the curve. +

    + Graph of the function y=\sin(x) on [0,\pi], with four straight lines which will be used to approximate the length of this curve. + The four lines are evenly spaced out in intervals of \frac{\pi}{4} on the x-axis. + Each line begins at a point on the curve y=\sin(x), and ends at a point on the same curve after travelling a distance of \frac{\pi}{4} on the x-axis. + The first line begins at the same point (0,0) as the curve, from which it linearly increases until reaching the point (\frac{\pi}{4},\frac{sqrt2}{2}), which is also a point on the curve. + The second line begins at the same point the first line ends, given by (\frac{\pi}{4},\frac{sqrt2}{2}), from which it linearly increases until reaching the point (\frac{\pi}{2},1), which is the peak of the curve. + The third line begins at the same point the second line ends, given by (\frac{\pi}{2},1), from which it linearly decreases until reaching the point (\frac{3\pi}{4},\frac{sqrt2}{2}), which is also a point on the curve. + The fourth line begins at the same point the third line ends, given by (\frac{3\pi}{4},\frac{sqrt2}{2}), from which it linearly decreases until reaching the point (\pi,0), which is the end of the curve. +

    Graph of the sine function for x between 0 and pi and four straight lines approximating this curve. @@ -175,12 +180,14 @@ - Graph of the ith subinterval of the function y=f(x), which is graphed on the interval [x_{i-1},x_{i}]. - The graph contains a line between the start and endpoint of the curve, which will be used to approximate the length of the ith subinterval curve. - The ith subinterval of the curve y=f(x) begins at the point (x_{i-1},y_{i-1}) from which it heads upwards in a concave arc until reaching the point (x_{i},y_{i}). - The straight line then passes through the start and endpoints of the curve, (x_{i-1},y_{i-1}) and (x_{i},y_{i}) respectively. - The line lies below the curve for the entire interval that the curve is plotted on. - The graph also contains the measurements \dx_i and \dy_i, giving the respective length of the change in x and y between the start and endpoint of the subinterval of the curve. +

    + Graph of the ith subinterval of the function y=f(x), which is graphed on the interval [x_{i-1},x_{i}]. + The graph contains a line between the start and endpoint of the curve, which will be used to approximate the length of the ith subinterval curve. + The ith subinterval of the curve y=f(x) begins at the point (x_{i-1},y_{i-1}) from which it heads upwards in a concave arc until reaching the point (x_{i},y_{i}). + The straight line then passes through the start and endpoints of the curve, (x_{i-1},y_{i-1}) and (x_{i},y_{i}) respectively. + The line lies below the curve for the entire interval that the curve is plotted on. + The graph also contains the measurements \dx_i and \dy_i, giving the respective length of the change in x and y between the start and endpoint of the subinterval of the curve. +

    Graph of a portion of a curve with a line between the start and endpoint which is used to approximate the length of the curve. @@ -336,9 +343,11 @@ - Graph of the function f(x) = x^{3/2} on the interval between x=0 and x=4. - The curve f(x) = x^{3/2} begins at the point (0,0) from which it heads upwards in a convex arc until reaching the point (4,8). - A straight line plotted between the start and endpoints of the curve would lie entirely above the curve on the interval between x=0 and x=4 and would showcase the shortest distance between the two points. +

    + Graph of the function f(x) = x^{3/2} on the interval between x=0 and x=4. + The curve f(x) = x^{3/2} begins at the point (0,0) from which it heads upwards in a convex arc until reaching the point (4,8). + A straight line plotted between the start and endpoints of the curve would lie entirely above the curve on the interval between x=0 and x=4 and would showcase the shortest distance between the two points. +

    Graph of the function from the example. @@ -402,10 +411,12 @@ - Graph of the function f(x) =\frac18x^2-\ln(x). - The curve is highlighted on the interval between x=1 and x=2. - The curve f(x) =\frac18x^2-\ln(x) begins near the point (0.4,1) from which it heads downwards in a convex arc until crossing the x-axis at approximately x=1.25. - The curve then continues in the convex arc, until it once again reaches the x-axis at approximately x=3. +

    + Graph of the function f(x) =\frac18x^2-\ln(x). + The curve is highlighted on the interval between x=1 and x=2. + The curve f(x) =\frac18x^2-\ln(x) begins near the point (0.4,1) from which it heads downwards in a convex arc until crossing the x-axis at approximately x=1.25. + The curve then continues in the convex arc, until it once again reaches the x-axis at approximately x=3. +

    Graph of the function from the example. @@ -557,10 +568,12 @@ - Graph of an arbitrary function y=f(x) on the interval [a,b]. - The curve is a concave arc starting at x=a at some arbitrary y value from which it slopes upwards until ending at x=b at some slightly higher y value. - The plot of the graph also contains a subinterval on the x-axis, given by [x_{i-1},x_{i}]. - A line is drawn through the points (x_{i-1},f(x_{i-1})) and (x_{i},f(x_{i})), which approximates the length of the curve y=f(x) on the interval [x_{i-1},x_{i}]. +

    + Graph of an arbitrary function y=f(x) on the interval [a,b]. + The curve is a concave arc starting at x=a at some arbitrary y value from which it slopes upwards until ending at x=b at some slightly higher y value. + The plot of the graph also contains a subinterval on the x-axis, given by [x_{i-1},x_{i}]. + A line is drawn through the points (x_{i-1},f(x_{i-1})) and (x_{i},f(x_{i})), which approximates the length of the curve y=f(x) on the interval [x_{i-1},x_{i}]. +

    Graph of an arbitrary function on the interval from a to b, with a line approximating a small portion of the curve. @@ -605,13 +618,15 @@ - Graph of an arbitrary function y=f(x) on the interval [a,b]. - The line drawn through the points (x_{i-1},f(x_{i-1})) and (x_{i},f(x_{i})) is then rotated about the x-axis. - The resulting shape resembles a part of a cone which is lying horizontally, and can be used to approximate the surface area of the function y=f(x) being rotated about the x-axis on the interval [x_{i-1},x_{i}]. - The plot also contains two vertical measurements. - The first vertical measurement is r, which gives the radius of the cone at x=x_{i-1} and the second is R, which gives the radius of the cone at x=x_{i} - The plot also contains an additional measurement L which gives the length of the line connecting f(x_{i-1}) and f(x_{i}). - The measurement L is also the length of the part of the resulting part of a cone that comes from rotating the line about the x-axis. +

    + Graph of an arbitrary function y=f(x) on the interval [a,b]. + The line drawn through the points (x_{i-1},f(x_{i-1})) and (x_{i},f(x_{i})) is then rotated about the x-axis. + The resulting shape resembles a part of a cone which is lying horizontally, and can be used to approximate the surface area of the function y=f(x) being rotated about the x-axis on the interval [x_{i-1},x_{i}]. + The plot also contains two vertical measurements. + The first vertical measurement is r, which gives the radius of the cone at x=x_{i-1} and the second is R, which gives the radius of the cone at x=x_{i} + The plot also contains an additional measurement L which gives the length of the line connecting f(x_{i-1}) and f(x_{i}). + The measurement L is also the length of the part of the resulting part of a cone that comes from rotating the line about the x-axis. +

    Graph of a function with the line approximating the length of a part of the curve being rotated about the x axis. @@ -826,12 +841,14 @@ - Three dimensional graph of the shape coming from revolving y=\sin(x) on [0,\pi] about the x-axis. - The curve y=\sin(x) is drawn on the interval [0,\pi]. - This concave curve begins at the point (0,0), from which it increases until reaching a maximum at the point (\frac{\pi}{2},1). - The curve then decreases until ending at the x-axis at the point (\pi,0). - The curve is then rotated about the x-axis, which creates the solid of revolution. - This solid has the largest diameter and is symmetric about x=\frac{pi}{2}, from which it shrinks down until closing in on itself at x=0 and x=\pi. +

    + Three dimensional graph of the shape coming from revolving y=\sin(x) on [0,\pi] about the x-axis. + The curve y=\sin(x) is drawn on the interval [0,\pi]. + This concave curve begins at the point (0,0), from which it increases until reaching a maximum at the point (\frac{\pi}{2},1). + The curve then decreases until ending at the x-axis at the point (\pi,0). + The curve is then rotated about the x-axis, which creates the solid of revolution. + This solid has the largest diameter and is symmetric about x=\frac{pi}{2}, from which it shrinks down until closing in on itself at x=0 and x=\pi. +

    Three dimensional graph of the sine curve being rotated about the x axis. @@ -955,12 +972,14 @@     - The image shows the cycloid which comes from tracking a point p on a rolling circle of radius 1 on a flat surface. - The point p is initially at the top of the circle of radius 1. - Once the circle rolls, the point p is tracked to create a graph of a cycloid. - The graph coming from tracking the point p first begins to decrease, until the point p is at the bottom of the circle, at which point the point p touches the surface the ball is rolling on. - After this point, the graph begins to increase. - The circle continues to roll, with the point p once again becoming the top of the circle, after which the graph begins to decrease. - The circle continues rolling, with the point p becoming the lowest point on the circle two more times, after which the graph stops. - Between the starting point and the point at which p is at the bottom of the circle, the graph resembles a slightly stretched quarter of a circle. - The part of the graph that comes from the remaining two full revolutions of the circle resembles two horizontally stretched semi-circles. +

    + The image shows the cycloid which comes from tracking a point p on a rolling circle of radius 1 on a flat surface. + The point p is initially at the top of the circle of radius 1. + Once the circle rolls, the point p is tracked to create a graph of a cycloid. + The graph coming from tracking the point p first begins to decrease, until the point p is at the bottom of the circle, at which point the point p touches the surface the ball is rolling on. + After this point, the graph begins to increase. + The circle continues to roll, with the point p once again becoming the top of the circle, after which the graph begins to decrease. + The circle continues rolling, with the point p becoming the lowest point on the circle two more times, after which the graph stops. + Between the starting point and the point at which p is at the bottom of the circle, the graph resembles a slightly stretched quarter of a circle. + The part of the graph that comes from the remaining two full revolutions of the circle resembles two horizontally stretched semi-circles. +

    Image showing the cycloid which comes from tracking a point on a rolling circle. @@ -637,10 +653,12 @@ - The graph shows the function \vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra, which is the graph of a cycloid which comes from tracking a point p on a rolling a circle of radius 1 on a flat surface. - The curve begins near the point (0,2), after which it begins to decrease until it reaches its lowest point near the point (2,0). - After this point, the curve begins increasing, until it reaches its highest point when p is at the top of the circle, after which it decreases until reaching another minimum near the point (8,0). - The curve continues in the same fashion, reaching another minimum near the point (14,0), after which it continues for a slight duration in the same fashion as before. +

    + The graph shows the function \vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra, which is the graph of a cycloid which comes from tracking a point p on a rolling a circle of radius 1 on a flat surface. + The curve begins near the point (0,2), after which it begins to decrease until it reaches its lowest point near the point (2,0). + After this point, the curve begins increasing, until it reaches its highest point when p is at the top of the circle, after which it decreases until reaching another minimum near the point (8,0). + The curve continues in the same fashion, reaching another minimum near the point (14,0), after which it continues for a slight duration in the same fashion as before. +

    Graph showing the cycloid which comes from tracking a point on a rolling circle. @@ -737,9 +755,11 @@ - Graph of the function \vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra on -1\leq t\leq 1. - The function \vec r(t) is the right half of the unit circle, beginning at the point (0,-1), crossing the x-axis at the point (1,0), and ending at the point (0,-1). - The graph also contains the displacement vector, which begins at the start of the curve at the point (0,-1) and heads directly upwards until reaching the endpoint of the curve at the point (0,1). +

    + Graph of the function \vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra on -1\leq t\leq 1. + The function \vec r(t) is the right half of the unit circle, beginning at the point (0,-1), crossing the x-axis at the point (1,0), and ending at the point (0,-1). + The graph also contains the displacement vector, which begins at the start of the curve at the point (0,-1) and heads directly upwards until reaching the endpoint of the curve at the point (0,1). +

    Graph of the semicircle coming from plotting the vector valued function from the example. @@ -977,9 +997,11 @@ - Graph of the function \vec r(t) = \la t^2,t^2-1\ra for -2\leq t\leq 2. - The graph of the function \vec r(t) is a line which begins at the point (0,-1) and ends at the point (4,3). - The function begins at the point (4,3) when t=-2, linearly decreases until the point (0,-1) and once again follows the same linear path until ending at the point (4,3). +

    + Graph of the function \vec r(t) = \la t^2,t^2-1\ra for -2\leq t\leq 2. + The graph of the function \vec r(t) is a line which begins at the point (0,-1) and ends at the point (4,3). + The function begins at the point (4,3) when t=-2, linearly decreases until the point (0,-1) and once again follows the same linear path until ending at the point (4,3). +

    Graph of the vector valued function from the example. @@ -1020,10 +1042,12 @@ - Graph of the function \vec r(t) = \la t^2,t^3\ra, for -2\leq t\leq 2. - The graph of the function \vec r(t) begins at the point (4,-8), and upwards and to the left in a concave down fashion until reaching the origin. - After reaching the origin, the function is concave and ends when it reaches the point (4,8). - The function is also symmetric about the x-axis, meaning that the upper and lower half of the curves are mirror images of each other. +

    + Graph of the function \vec r(t) = \la t^2,t^3\ra, for -2\leq t\leq 2. + The graph of the function \vec r(t) begins at the point (4,-8), and upwards and to the left in a concave down fashion until reaching the origin. + After reaching the origin, the function is concave and ends when it reaches the point (4,8). + The function is also symmetric about the x-axis, meaning that the upper and lower half of the curves are mirror images of each other. +

    Graph of the vector valued function from the example. @@ -1066,10 +1090,12 @@ - Graph of the function \vec r(t) = \la 1/t,1/t^2\ra, for -2\leq t\leq 2. - The graph of the function \vec r(t) begins at the point (-\frac12,\frac14) corresponding to t=2, heads upwards and to the left as t increases, following the path of the left side of the parabola y=x^2. - At t=2 the function starts at the point (\frac12,\frac14) and as t decreases follows the path of the right side of the parabola given by y=x^2. - The function is also mirrored about y-axis, and is equivalent to the parabola y=x^2 outside of the region -\frac12 \leq x \frac12. +

    + Graph of the function \vec r(t) = \la 1/t,1/t^2\ra, for -2\leq t\leq 2. + The graph of the function \vec r(t) begins at the point (-\frac12,\frac14) corresponding to t=2, heads upwards and to the left as t increases, following the path of the left side of the parabola y=x^2. + At t=2 the function starts at the point (\frac12,\frac14) and as t decreases follows the path of the right side of the parabola given by y=x^2. + The function is also mirrored about y-axis, and is equivalent to the parabola y=x^2 outside of the region -\frac12 \leq x \frac12. +

    Graph of the vector valued function from the example. @@ -1112,15 +1138,17 @@ - Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi. - The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi. - As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease. - The function continues to decrease, crossing the xaxis at x=1. - After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin. - The function now increases upwards and to the right until reaching a maximum near point (0.25,1). - After this point, the curve decreases, crossing the x-axis once again at x=1. - The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0). - The function is also symmetric about the x-axis. +

    + Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi. + The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi. + As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease. + The function continues to decrease, crossing the xaxis at x=1. + After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin. + The function now increases upwards and to the right until reaching a maximum near point (0.25,1). + After this point, the curve decreases, crossing the x-axis once again at x=1. + The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0). + The function is also symmetric about the x-axis. +

    Graph of the vector valued function from the example. @@ -1161,15 +1189,17 @@ - Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi. - The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi. - As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease. - The function continues to decrease, crossing the xaxis at x=1. - After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin. - The function now increases upwards and to the right until reaching a maximum near point (0.25,1). - After this point, the curve decreases, crossing the x-axis once again at x=1. - The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0). - The function is also symmetric about the x-axis. +

    + Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi. + The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi. + As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease. + The function continues to decrease, crossing the xaxis at x=1. + After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin. + The function now increases upwards and to the right until reaching a maximum near point (0.25,1). + After this point, the curve decreases, crossing the x-axis once again at x=1. + The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0). + The function is also symmetric about the x-axis. +

    Graph of the vector valued function from the example. @@ -1211,11 +1241,13 @@ - Graph of the function \vec r(t) = \la 3\sin(\pi t),2\cos(\pi t)\ra, on [0,2]. - The graph of the function \vec r(t) is an oval having a horizontal width of 6 and a height of 4 centered at the origin. - The rightmost point of the oval is the point (3,0). - The leftmost point of the oval is the point (-3,0). - The highest point of the oval is the point (0,2), while the lowest is the point (0,-2). +

    + Graph of the function \vec r(t) = \la 3\sin(\pi t),2\cos(\pi t)\ra, on [0,2]. + The graph of the function \vec r(t) is an oval having a horizontal width of 6 and a height of 4 centered at the origin. + The rightmost point of the oval is the point (3,0). + The leftmost point of the oval is the point (-3,0). + The highest point of the oval is the point (0,2), while the lowest is the point (0,-2). +

    Graph of the vector valued function from the example. @@ -1257,15 +1289,17 @@ - Graph of the function \vec r(t) = \la 3\cos(t) ,2\sin(2 t)\ra, on [0,2\pi]. - The graph of the function \vec r(t) resembles the symbol \infty. - The curve begins at the point (3,0) which corresponds to t=0. - The curve then begins going upwards and to the left, until it reaches a maximum near the point (2,2). - From here the curve begins going downwards and to the left, passing through the origin, until reaching a minimum near (-2,-2). - The curve then begins going upwards to the left, passing through the point (-3,0), after which it begins going to the right. - The curve reaches a maximum near the point (-2,2), after which it begins decreasing and going to the right, once again crossing the origin. - Finally, the curve reaches another minimum near the point (2,-2), after which it begins going upwards to the right, until reaching its starting point given by (3,0). - The curve is symmetric about both the x and y axes. +

    + Graph of the function \vec r(t) = \la 3\cos(t) ,2\sin(2 t)\ra, on [0,2\pi]. + The graph of the function \vec r(t) resembles the symbol \infty. + The curve begins at the point (3,0) which corresponds to t=0. + The curve then begins going upwards and to the left, until it reaches a maximum near the point (2,2). + From here the curve begins going downwards and to the left, passing through the origin, until reaching a minimum near (-2,-2). + The curve then begins going upwards to the left, passing through the point (-3,0), after which it begins going to the right. + The curve reaches a maximum near the point (-2,2), after which it begins decreasing and going to the right, once again crossing the origin. + Finally, the curve reaches another minimum near the point (2,-2), after which it begins going upwards to the right, until reaching its starting point given by (3,0). + The curve is symmetric about both the x and y axes. +

    Graph of the vector valued function from the example. @@ -1307,13 +1341,15 @@ - Graph of the function \vec r(t) = \la 2\sec(t) ,\tan(t) \ra, on [-\pi,\pi]. - The graph of the function is symmetric about both the x and y axes. - The curve crosses the x-axis at the point x=2 corresponding to t=\pi. - After this point the upper part of the curve goes upwards and to the right in a mostly linear fashion. - Since the curve is symmetric about the y-axis, the lower part of the curve is the mirrored version of the upper part, meaning that it goes downwards and to the left. - As the curve is symmetric about the x-axis, the left side of the curve is the mirrored version of the right part, meaning that it crosses the x-axis at x=-2. - From this point, the upper part of the left side of the curve goes upwards and to the left, while the lower part goes downwards and to the left. +

    + Graph of the function \vec r(t) = \la 2\sec(t) ,\tan(t) \ra, on [-\pi,\pi]. + The graph of the function is symmetric about both the x and y axes. + The curve crosses the x-axis at the point x=2 corresponding to t=\pi. + After this point the upper part of the curve goes upwards and to the right in a mostly linear fashion. + Since the curve is symmetric about the y-axis, the lower part of the curve is the mirrored version of the upper part, meaning that it goes downwards and to the left. + As the curve is symmetric about the x-axis, the left side of the curve is the mirrored version of the right part, meaning that it crosses the x-axis at x=-2. + From this point, the upper part of the left side of the curve goes upwards and to the left, while the lower part goes downwards and to the left. +

    Graph of the vector valued function from the example. @@ -1365,10 +1401,12 @@ - Graph of the function \vec r(t) = \la 2\cos(t) , t, 2\sin(t) \ra, on [0,2\pi]. - The graph of the function is a circular spiral centered about the y-axis. - Ignoring the y-axis, the curve is simply a singular circle of radius 2 in the xz-plane. - Incorporating the y coordinate then creates the linearly increasing spiral, which begins at the point (2,0,0), completes precisely one full revolution and ends at the point (2,2\pi,0). +

    + Graph of the function \vec r(t) = \la 2\cos(t) , t, 2\sin(t) \ra, on [0,2\pi]. + The graph of the function is a circular spiral centered about the y-axis. + Ignoring the y-axis, the curve is simply a singular circle of radius 2 in the xz-plane. + Incorporating the y coordinate then creates the linearly increasing spiral, which begins at the point (2,0,0), completes precisely one full revolution and ends at the point (2,2\pi,0). +

    Graph of the vector valued function from the example. @@ -1417,10 +1455,12 @@ - Graph of the function \vec r(t) = \la 3\cos(t) , \sin(t) , t/\pi\ra on [0,2\pi]. - The graph of the function is an oval-shaped spiral centered about the z-axis. - Ignoring the z-axis, the curve is simply an oval having a horizontal width of 6 and a height of 2 in the xy-plane. - Incorporating the z coordinate then creates the linearly increasing oval spiral, which begins at the point (3,0,0), completes precisely one full revolution and ends at the point (3,0,2). +

    + Graph of the function \vec r(t) = \la 3\cos(t) , \sin(t) , t/\pi\ra on [0,2\pi]. + The graph of the function is an oval-shaped spiral centered about the z-axis. + Ignoring the z-axis, the curve is simply an oval having a horizontal width of 6 and a height of 2 in the xy-plane. + Incorporating the z coordinate then creates the linearly increasing oval spiral, which begins at the point (3,0,0), completes precisely one full revolution and ends at the point (3,0,2). +

    Graph of the vector valued function from the example. @@ -1468,13 +1508,15 @@ - Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(t) \ra on [0,2\pi]. - The graph of the function is an oval lying in the plane coming from rotating the xy plane 45 degrees towards the z-axis. - The oval lying in this plane has a horizontal width of \sqrt{2} and a height of 1. - Ignoring the z coordinate, the curve is a unit circle in the xy plane. - Similarly ignoring the y coordinate, the curve is a unit circle in the xz plane. - If we now ignore the x coordinate, the resulting curve is a diagonal line given by z=y in the yz plane. - This line turns back on itself, which can be seen in the image of the oval when considering all three coordinate axes. +

    + Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(t) \ra on [0,2\pi]. + The graph of the function is an oval lying in the plane coming from rotating the xy plane 45 degrees towards the z-axis. + The oval lying in this plane has a horizontal width of \sqrt{2} and a height of 1. + Ignoring the z coordinate, the curve is a unit circle in the xy plane. + Similarly ignoring the y coordinate, the curve is a unit circle in the xz plane. + If we now ignore the x coordinate, the resulting curve is a diagonal line given by z=y in the yz plane. + This line turns back on itself, which can be seen in the image of the oval when considering all three coordinate axes. +

    Graph of the vector valued function from the example. @@ -1522,17 +1564,19 @@ - Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(2t)\ra on [0,2\pi]. - The graph of the function resembles a saddle centered at the origin whose height is defined by the z-axis. - The two sides of the saddle that taper off fall into negative z and lie in the second and third quadrants in the xy plane. - Ignoring the z coordinate, the curve is a unit circle in the xy plane. - Ignoring the x or y coordinates individually, the curve looks like the \infty symbol in the yz and the xz planes, respectively. - We now describe the z coordinate with respect to travelling along the unit circle in the xy plane. - Starting at t=, the function begins at the point (1,0,0). - As t increases and we travel along the unit circle in the x and y coordinates, z increases until we get to t=\frac{\pi}{2} at which z=1. - Then, continuing along the unit circle, z decreases until it reaches a minimum of z=-1 when t=\frac{3\pi}{4}. - Continuing along the circle, z begins to increase once again, reaching one more maximum of z=1 when t=\frac{5\pi}{4}. - Finally, z begins to decrease, reaching its last minimum of z=1 when t=\frac{7\pi}{4}, after which z increases, and the curve ends where it began, at the point (1,0,0). +

    + Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(2t)\ra on [0,2\pi]. + The graph of the function resembles a saddle centered at the origin whose height is defined by the z-axis. + The two sides of the saddle that taper off fall into negative z and lie in the second and third quadrants in the xy plane. + Ignoring the z coordinate, the curve is a unit circle in the xy plane. + Ignoring the x or y coordinates individually, the curve looks like the \infty symbol in the yz and the xz planes, respectively. + We now describe the z coordinate with respect to travelling along the unit circle in the xy plane. + Starting at t=, the function begins at the point (1,0,0). + As t increases and we travel along the unit circle in the x and y coordinates, z increases until we get to t=\frac{\pi}{2} at which z=1. + Then, continuing along the unit circle, z decreases until it reaches a minimum of z=-1 when t=\frac{3\pi}{4}. + Continuing along the circle, z begins to increase once again, reaching one more maximum of z=1 when t=\frac{5\pi}{4}. + Finally, z begins to decrease, reaching its last minimum of z=1 when t=\frac{7\pi}{4}, after which z increases, and the curve ends where it began, at the point (1,0,0). +

    Graph of the vector valued function from the example. diff --git a/ptx/sec_vvf_calc.ptx b/ptx/sec_vvf_calc.ptx index 66bf615b8..e63c127f2 100644 --- a/ptx/sec_vvf_calc.ptx +++ b/ptx/sec_vvf_calc.ptx @@ -226,12 +226,14 @@     - Graph of the vector-valued function \vec r(t) = \la \cos(t) , \sin(t) , t\ra. - The function is a circular spiral which climbs the z-axis. - Looking at the graph from above the z-axis, the function resembles a unit circle in the xy-plane. - Adding the z coordinate then creates the linearly increasing spiral. - The graph also includes two copies of the vector \vrp(\pi/2) = \la -1,0,1\ra - The first copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the origin, and ends at the point (-1,0,1). - The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the point (0,1,\pi/2), which corresponds to the termination point of \vec r(\pi/2). - The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra is also tangent to the function \vec r(t) at the point (0,1,\pi/2) corresponding to when t=\pi/2 in the function \vec r(t). +

    + Graph of the vector-valued function \vec r(t) = \la \cos(t) , \sin(t) , t\ra. + The function is a circular spiral which climbs the z-axis. + Looking at the graph from above the z-axis, the function resembles a unit circle in the xy-plane. + Adding the z coordinate then creates the linearly increasing spiral. + The graph also includes two copies of the vector \vrp(\pi/2) = \la -1,0,1\ra + The first copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the origin, and ends at the point (-1,0,1). + The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the point (0,1,\pi/2), which corresponds to the termination point of \vec r(\pi/2). + The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra is also tangent to the function \vec r(t) at the point (0,1,\pi/2) corresponding to when t=\pi/2 in the function \vec r(t). +

    Graph of the vector-valued function from the example and its derivative at a point. @@ -706,13 +716,15 @@ - Graph of the vector-valued function \vec r(t) = \la t,t^2,t^3\ra on [-1.5,1.5]. - The function begins at the point (-1.5,2.25,-3.375), from which it begins to increase linearly in the x and z coordinates, and decrease in the y coordinate. - The function then curves towards the origin. - After passing through the origin, the function begins to increase in all x,y and z coordinates until it reaches the point (1.5,2.25,3.375), at which it ends. - The graph also contains the line \ell(t) = \la -1,1,-1\ra + t\la 1,-2,3\ra, which is the tangent line to the function at \vec r(-1), which is the point (-1,1,-1). - The line can be described as the line which passes through the point (-1,1,-1), moving in the direction of the vector \la 1,-2,3\ra. - Additionally, the line is defined for all t, so it also moves in the opposite direction of the vector \la 1,-2,3\ra, or in other words in the direction of \la -1,2,-3\ra from the point (-1,1,-1). +

    + Graph of the vector-valued function \vec r(t) = \la t,t^2,t^3\ra on [-1.5,1.5]. + The function begins at the point (-1.5,2.25,-3.375), from which it begins to increase linearly in the x and z coordinates, and decrease in the y coordinate. + The function then curves towards the origin. + After passing through the origin, the function begins to increase in all x,y and z coordinates until it reaches the point (1.5,2.25,3.375), at which it ends. + The graph also contains the line \ell(t) = \la -1,1,-1\ra + t\la 1,-2,3\ra, which is the tangent line to the function at \vec r(-1), which is the point (-1,1,-1). + The line can be described as the line which passes through the point (-1,1,-1), moving in the direction of the vector \la 1,-2,3\ra. + Additionally, the line is defined for all t, so it also moves in the opposite direction of the vector \la 1,-2,3\ra, or in other words in the direction of \la -1,2,-3\ra from the point (-1,1,-1). +

    Graph of the vector-valued function from the example and the tangent line to a point on the curve. @@ -793,11 +805,13 @@ - Graph of the vector-valued function \vec r(t) = \la t^3,t^2\ra. - The function begins near the point (-3,2), from which it is concave down and sloping downwards towards the origin. - After passing through the origin, the curve is concave down and begins increasing in both x and y coordinates. - The curve is also symmetric about the y-axis. - The graph also contains the line \ell(t) = \la -1,1\ra + t\la 3,-2\ra , which is tangent to the function \vec r(t) = \la t^3,t^2\ra at the point (-1,1) corresponding to when t=1. +

    + Graph of the vector-valued function \vec r(t) = \la t^3,t^2\ra. + The function begins near the point (-3,2), from which it is concave down and sloping downwards towards the origin. + After passing through the origin, the curve is concave down and begins increasing in both x and y coordinates. + The curve is also symmetric about the y-axis. + The graph also contains the line \ell(t) = \la -1,1\ra + t\la 3,-2\ra , which is tangent to the function \vec r(t) = \la t^3,t^2\ra at the point (-1,1) corresponding to when t=1. +

    Graph of the vector-valued function from the example and the tangent line to a point on the curve. @@ -983,13 +997,15 @@ - Graph of the vector-valued function \vec r(t) = \la t,t^2 -1\ra on [-2,2]. - The function looks like the parabola y=x^2 -1, which takes the path going towards positive x. - The function begins at the point (-2,3), decreases until reaching the point (0,-1), and then increases until ending at the point (2,3). - The graph also contains the function \vec u(t) which is the unit vector that points in the direction of \vec r(t). - The function \vec u(t) looks like the unit circle which is missing a piece of the top. - The circular arc from the graph of \vec u(t) begins at the point where the vector \vec r(-2)=\la -2,3\ra crosses the unit circle. - The circular arc then goes counterclockwise following the path of the unit circle, until ending at the point where the vector\vec r(2)=\la 2,3\ra crosses the unit circle. +

    + Graph of the vector-valued function \vec r(t) = \la t,t^2 -1\ra on [-2,2]. + The function looks like the parabola y=x^2 -1, which takes the path going towards positive x. + The function begins at the point (-2,3), decreases until reaching the point (0,-1), and then increases until ending at the point (2,3). + The graph also contains the function \vec u(t) which is the unit vector that points in the direction of \vec r(t). + The function \vec u(t) looks like the unit circle which is missing a piece of the top. + The circular arc from the graph of \vec u(t) begins at the point where the vector \vec r(-2)=\la -2,3\ra crosses the unit circle. + The circular arc then goes counterclockwise following the path of the unit circle, until ending at the point where the vector\vec r(2)=\la 2,3\ra crosses the unit circle. +

    Graph of the vector-valued function with the unit vector function which points in the direction of the function. @@ -1059,14 +1075,16 @@ - Graph of the function \vec u(t) which is the unit vector that points in the direction of \vec r(t). - The graph also contains three unit vectors, \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0). - The vector \vec u\,'(-2)\approx \la -0.320,-0.213\ra begins at the starting point of the function \vec u(t) and points in the direction tangent to the circular arc at t=-2. - The vector \vec u\,'(-1)= \la 0,-2\ra begins at the point (-1,0) corresponding to the termination point of \vec u(-1) - From here, the vector points straight down, as it is tangent to the leftmost point of the circular arc. - The vector \vec u\,'(0)= \la 1,0\ra begins at the point (0,-1) corresponding to the termination point of \vec u(0). - From here the vector points to the right, as it is tangent to the bottom point of the circular arc. - If the vectors \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0) were extended to lines, all three lines would be tangent to a point on the nearly complete unit circle given by \vec u(t). +

    + Graph of the function \vec u(t) which is the unit vector that points in the direction of \vec r(t). + The graph also contains three unit vectors, \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0). + The vector \vec u\,'(-2)\approx \la -0.320,-0.213\ra begins at the starting point of the function \vec u(t) and points in the direction tangent to the circular arc at t=-2. + The vector \vec u\,'(-1)= \la 0,-2\ra begins at the point (-1,0) corresponding to the termination point of \vec u(-1) + From here, the vector points straight down, as it is tangent to the leftmost point of the circular arc. + The vector \vec u\,'(0)= \la 1,0\ra begins at the point (0,-1) corresponding to the termination point of \vec u(0). + From here the vector points to the right, as it is tangent to the bottom point of the circular arc. + If the vectors \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0) were extended to lines, all three lines would be tangent to a point on the nearly complete unit circle given by \vec u(t). +

    Graph of the circular unit vector function along with three derivative vectors of the function. @@ -1782,11 +1800,12 @@ - Graph of the function \vec r(t) = \la t^2+t, t^2-t\ra. - The function \vec r(t) resembles a standard parabola which has been rotated 45 degrees clockwise. - The curve passes through the point (2,0), the origin, and the point (0,2) in the given order. - The graph also contains \vrp(1) beggining at the point (2,0) corresponding to the termination point of \vec r(1). - The vector \vrp(1)= \la 3,1 \ra and is tangent to \vec r(t) = \la t^2+t, t^2-t\ra corresponding to the point where t=1. +

    Graph of the function \vec r(t) = \la t^2+t, t^2-t\ra. + The function \vec r(t) resembles a standard parabola which has been rotated 45 degrees clockwise. + The curve passes through the point (2,0), the origin, and the point (0,2) in the given order. + The graph also contains \vrp(1) beggining at the point (2,0) corresponding to the termination point of \vec r(1). + The vector \vrp(1)= \la 3,1 \ra and is tangent to \vec r(t) = \la t^2+t, t^2-t\ra corresponding to the point where t=1. +

    Graph of the vector-valued function and its derivative at a point. @@ -1830,13 +1849,15 @@ - Graph of the function \vec r(t) = \la t^2-2t+2,t^3-3t^2+2t\ra. - The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). - After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). - From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. - The graph of the function is also symmetric about the y-axis. - The graph also contains \vrp(1) beginning at the point (1,0) corresponding to the termination point of \vec r(1). - The vector \vrp(1)= \la 0,-1 \ra and is tangent to \vec r(t) corresponding to the point where t=1. +

    + Graph of the function \vec r(t) = \la t^2-2t+2,t^3-3t^2+2t\ra. + The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). + After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). + From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. + The graph of the function is also symmetric about the y-axis. + The graph also contains \vrp(1) beginning at the point (1,0) corresponding to the termination point of \vec r(1). + The vector \vrp(1)= \la 0,-1 \ra and is tangent to \vec r(t) corresponding to the point where t=1. +

    Graph of the vector-valued function and its derivative at a point. @@ -1880,13 +1901,15 @@ - Graph of the function \vec r(t) = \la t^2+1,t^3-t\ra. - The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). - After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). - From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. - The graph of the function is also symmetric about the y-axis. - The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1). - The vector \vrp(1)= \la 2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1. +

    + Graph of the function \vec r(t) = \la t^2+1,t^3-t\ra. + The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). + After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). + From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. + The graph of the function is also symmetric about the y-axis. + The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1). + The vector \vrp(1)= \la 2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1. +

    Graph of the vector-valued function and its derivative at a point. @@ -1930,13 +1953,15 @@ - Graph of the function \vec r(t) = \la t^2-4t+5,t^3-6t^2+11t-6\ra. - The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). - After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). - From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. - The graph of the function is also symmetric about the y-axis. - The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1). - The vector \vrp(1)= \la -2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1. +

    + Graph of the function \vec r(t) = \la t^2-4t+5,t^3-6t^2+11t-6\ra. + The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0). + After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0). + From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant. + The graph of the function is also symmetric about the y-axis. + The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1). + The vector \vrp(1)= \la -2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1. +

    Graph of the vector-valued function and its derivative at a point. diff --git a/ptx/sec_work.ptx b/ptx/sec_work.ptx index 597785bcb..e92b03bfe 100644 --- a/ptx/sec_work.ptx +++ b/ptx/sec_work.ptx @@ -338,11 +338,13 @@     - Image of a spring showing the spring both before and after streching. - In the first illustration, we see the unstreched spring with a block attached to it on its right side, with a rightward facing force vector labeled F, showing the direction of the force that is applied by the spring. - On the x-axis, the block that the spring is going to move is pictured to be centered at x=1. - In the second illustration, we see the streched spring, with the same block still attached on the right side. - This time, the block is centered at x=6, showing the spring moved the block a total of 5 inches to the right. +

    + Image of a spring showing the spring both before and after streching. + In the first illustration, we see the unstreched spring with a block attached to it on its right side, with a rightward facing force vector labeled F, showing the direction of the force that is applied by the spring. + On the x-axis, the block that the spring is going to move is pictured to be centered at x=1. + In the second illustration, we see the streched spring, with the same block still attached on the right side. + This time, the block is centered at x=6, showing the spring moved the block a total of 5 inches to the right. +

    Image showing the important aspects of strenching a spring which are used in computing work. @@ -533,11 +535,13 @@ - Image of a cylindrical storage tank with a radius of 10 and a height of 30ft. - On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank. - The image also contains the ith subinterval of y, which corresponds to the shaded region between y_{i-1} and y_i on the y-axis. - The height of this subinterval of y given by y_{i}-y_{i-1} is also labeled \Delta y_i. - Additionally, the distance from the top of this subinterval, occuring at y_{i} and the point 5 feet above the tank is given on the y-axis as 35-y_i. +

    + Image of a cylindrical storage tank with a radius of 10 and a height of 30ft. + On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank. + The image also contains the ith subinterval of y, which corresponds to the shaded region between y_{i-1} and y_i on the y-axis. + The height of this subinterval of y given by y_{i}-y_{i-1} is also labeled \Delta y_i. + Additionally, the distance from the top of this subinterval, occuring at y_{i} and the point 5 feet above the tank is given on the y-axis as 35-y_i. +

    Illustration of a cylindrical water tank with measurements used to compute the work required to empty it. @@ -649,11 +653,13 @@ - Image of the same cylindrical storage tank with a radius of 10 and a height of 30. - On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank. - Now instead of containing the ith subinterval of y, the image contains an arbitrarily chosen point y which is contained in the tank. - At this point y, the volume is given as V(y)=100\pi dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y. - Additionally, the distance from the arbitrary point y and the point 5 feet above the tank is given on the y-axis as 35-y. +

    + Image of the same cylindrical storage tank with a radius of 10 and a height of 30. + On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank. + Now instead of containing the ith subinterval of y, the image contains an arbitrarily chosen point y which is contained in the tank. + At this point y, the volume is given as V(y)=100\pi dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y. + Additionally, the distance from the arbitrary point y and the point 5 feet above the tank is given on the y-axis as 35-y. +

    Illustration of a cylindrical water tank with simplified measurements used to compute the work required to empty it. @@ -724,12 +730,14 @@ - Image of a conical storage tank with its tip 10 below ground and the base at ground level with a radius of 2. - This means as we move further below ground level, the area of the circular cross-section of the conical storage tank decreases. - On the left side of the storage tank the y-axis is shown with measurements of y=3 which is 3 feet above the tank, y=0 which is the top of the tank, and y=-10 which is the tip of the cone, 10 feet below ground level. - The image contains an arbitrarily chosen point y between y=-10 and y=0 which is contained in the tank. - At this point y, the volume is given as V(y)=\pi (\frac{y}{5} + 2)^2 dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y. - Additionally, the distance from the arbitrary point y and the point 3 feet above the tank is given on the y-axis as 3-y. +

    + Image of a conical storage tank with its tip 10 below ground and the base at ground level with a radius of 2. + This means as we move further below ground level, the area of the circular cross-section of the conical storage tank decreases. + On the left side of the storage tank the y-axis is shown with measurements of y=3 which is 3 feet above the tank, y=0 which is the top of the tank, and y=-10 which is the tip of the cone, 10 feet below ground level. + The image contains an arbitrarily chosen point y between y=-10 and y=0 which is contained in the tank. + At this point y, the volume is given as V(y)=\pi (\frac{y}{5} + 2)^2 dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y. + Additionally, the distance from the arbitrary point y and the point 3 feet above the tank is given on the y-axis as 3-y. +

    Illustration of a conical water tank with measurements used to compute the work required to empty it. @@ -806,16 +814,18 @@ - Image of the swimming pool described in the example. - The pool is 25 long. - On the leftmost side of the image is the deep end of the pool, which is - 10 long having a depth of 6. - After the 10 mark, - the bottom of the pool slopes up until it rises to a depth of 3 - a distance of 5 away in the x dimension. - The shallow end of the pool has a depth of 3 - and length of 10. - The end of the shallow end marks the end of the pool. +

    + Image of the swimming pool described in the example. + The pool is 25 long. + On the leftmost side of the image is the deep end of the pool, which is + 10 long having a depth of 6. + After the 10 mark, + the bottom of the pool slopes up until it rises to a depth of 3 + a distance of 5 away in the x dimension. + The shallow end of the pool has a depth of 3 + and length of 10. + The end of the shallow end marks the end of the pool. +

    Illustration of the swimming pool from the example. @@ -844,16 +854,18 @@ - Image of the same swimming pool described in the example. - The image now contains the two coordinate axes x and y. - On the y axis are five markings. - The label y=0 marks the bottom of the deep end of the pool, the label y=3 marks the bottom of the shallow end of the pool. - The label y=6 marks the top of the water in the pool, and the label y=8 marks a point 2 above the top of the water. - The image also contains an arbitrary label y lying somewhere below the water level of the pool. - On the x-axis, the label x=0 marks the leftmost edge of the pool, x=10 marks the end of the deep end and after the pool slopes up to a depth of 3, the label x=15 marks the start of the shallow end. - The distance between the leftmost edge and rightmost edge of the pool when y is above y=3, which is the depth of shallow end, is constant, and is given by the length of the pool. - On the other hand, the distance between the same two edges of the pool when y is below y=3 is the length of the deep end, plus the additional distance in the x direction between the end of the deep end, x=10, and the point at which the arbitrary y level meets the sloped portion of the pool. - The sloped portion of the pool is given as the line which passes through the points (10,0) and (15,3), which are the end of the deep end, and the start of the shallow end, respectively. +

    + Image of the same swimming pool described in the example. + The image now contains the two coordinate axes x and y. + On the y axis are five markings. + The label y=0 marks the bottom of the deep end of the pool, the label y=3 marks the bottom of the shallow end of the pool. + The label y=6 marks the top of the water in the pool, and the label y=8 marks a point 2 above the top of the water. + The image also contains an arbitrary label y lying somewhere below the water level of the pool. + On the x-axis, the label x=0 marks the leftmost edge of the pool, x=10 marks the end of the deep end and after the pool slopes up to a depth of 3, the label x=15 marks the start of the shallow end. + The distance between the leftmost edge and rightmost edge of the pool when y is above y=3, which is the depth of shallow end, is constant, and is given by the length of the pool. + On the other hand, the distance between the same two edges of the pool when y is below y=3 is the length of the deep end, plus the additional distance in the x direction between the end of the deep end, x=10, and the point at which the arbitrary y level meets the sloped portion of the pool. + The sloped portion of the pool is given as the line which passes through the points (10,0) and (15,3), which are the end of the deep end, and the start of the shallow end, respectively. +

    Illustration of the swimming pool from the example showing necessary differential elements. @@ -1516,11 +1528,13 @@ - Image of a fuel storage tank. - The storage tank has a base that has a 2 length and 15 width. - The forward, left and back walls of the tank are perpendicular to the base, while the right side of the tank slopes upward further away to the right side. - The top of the tank has a length of 5 due to the slope of the right wall of the fuel tank and the top same 15 width as the base. - The depth of the tank, which is the distance between the top and bottom of the tank is 10. +

    + Image of a fuel storage tank. + The storage tank has a base that has a 2 length and 15 width. + The forward, left and back walls of the tank are perpendicular to the base, while the right side of the tank slopes upward further away to the right side. + The top of the tank has a length of 5 due to the slope of the right wall of the fuel tank and the top same 15 width as the base. + The depth of the tank, which is the distance between the top and bottom of the tank is 10. +

    Illustration of the fuel storage tank from the problem. @@ -1613,10 +1627,12 @@ - Image of a water tank in the shape of a circular truncated cone. - The water tank is laying on its base and is in the shape of a cone, from which a small portion including the top is sliced off paralell to the base. - The water tank has a circular base with a radius of 5. - From the base, the tank has a height of 10, and a circular top with a radius of 2. +

    + Image of a water tank in the shape of a circular truncated cone. + The water tank is laying on its base and is in the shape of a cone, from which a small portion including the top is sliced off paralell to the base. + The water tank has a circular base with a radius of 5. + From the base, the tank has a height of 10, and a circular top with a radius of 2. +

    Illustration of the water tank in the shape of a truncated cone. @@ -1660,10 +1676,12 @@ - Image of the pyramidal water tank. - The water tank is in the shape of a pyramid with a square base which is laying on its tip, meaning that the square base is the top of the tank. - The top of the water tank has a square top with a side length of 2. - From the top, the tank has a depth of 7, which is the shortest distance between the square top of the tank and the peak of the pyramid, which is the bottom of the tank. +

    + Image of the pyramidal water tank. + The water tank is in the shape of a pyramid with a square base which is laying on its tip, meaning that the square base is the top of the tank. + The top of the water tank has a square top with a side length of 2. + From the top, the tank has a depth of 7, which is the shortest distance between the square top of the tank and the peak of the pyramid, which is the bottom of the tank. +

    Illustration of the water tank in the shape of an inverted pyramid. @@ -1704,11 +1722,13 @@ +

    Image of a water tank in the shape of a truncated pyramid. The water tank has a square base with a side length of 2. From the base, the pyramidal water tank expands on all sides until reaching the top of the tank. The top of the water tank is a square with a side length of 5. From the top, the tank has a depth of 9, which is the shortest distance between the bottom and top of the water tank. +

    Illustration of the water tank in the shape of a truncated pyramid.
    - Three dimensional graph of the shape coming from revolving y=x^2 on [0,1] about the x-axis. - The quadratic function y=x^2 is drawn on the interval [0,1]. - This curve begins at the point (0,0), from which it quadratically increases until reaching a maximum at the point (1,1). - The curve is then rotated about the x-axis, creating a solid of revolution. - This shape has a circular vertical cross-section which has a radius of r(x)=x^2 and is hollow on the inside. - The shape also is not closed off on its rightmost boundary at x=1. +

    + Three dimensional graph of the shape coming from revolving y=x^2 on [0,1] about the x-axis. + The quadratic function y=x^2 is drawn on the interval [0,1]. + This curve begins at the point (0,0), from which it quadratically increases until reaching a maximum at the point (1,1). + The curve is then rotated about the x-axis, creating a solid of revolution. + This shape has a circular vertical cross-section which has a radius of r(x)=x^2 and is hollow on the inside. + The shape also is not closed off on its rightmost boundary at x=1. +

    Three dimensional graph of the quadratic function being rotated about the x axis. @@ -1015,11 +1034,13 @@ - Three dimensional graph of the shape coming from revolving y=x^2 on [0,1] about the y-axis. - The quadratic function y=x^2 is drawn on the interval [0,1]. - The curve is then rotated about the y-axis, creating a solid of revolution. - This shape has a circular horizontal cross-section, which has a radius of r(x)=x and is hollow on the inside. - The shape also is not closed off on its top boundary at y=1. +

    + Three dimensional graph of the shape coming from revolving y=x^2 on [0,1] about the y-axis. + The quadratic function y=x^2 is drawn on the interval [0,1]. + The curve is then rotated about the y-axis, creating a solid of revolution. + This shape has a circular horizontal cross-section, which has a radius of r(x)=x and is hollow on the inside. + The shape also is not closed off on its top boundary at y=1. +

    Three dimensional graph of the quadratic function being rotated about the y axis. @@ -1136,12 +1157,14 @@ - Three dimensional graph of the shape coming from revolving y=1/x on [1,\infty) about the x-axis. - The function y=1/x is drawn on the interval [1,\infty). - The curve is then rotated about the x-axis, creating the shape called Gabriel's Horn. - This shape has a circular vertical cross-section, which has a radius of r(x)=1/x and is hollow on the inside. - The shape also is not closed off on its leftmost boundary at x=1. - For calculating the volume, we consider how much space is enclosed by x=1 and Gabriel's Horn itself. +

    + Three dimensional graph of the shape coming from revolving y=1/x on [1,\infty) about the x-axis. + The function y=1/x is drawn on the interval [1,\infty). + The curve is then rotated about the x-axis, creating the shape called Gabriel's Horn. + This shape has a circular vertical cross-section, which has a radius of r(x)=1/x and is hollow on the inside. + The shape also is not closed off on its leftmost boundary at x=1. + For calculating the volume, we consider how much space is enclosed by x=1 and Gabriel's Horn itself. +

    Three dimensional graph of Gabriel's Horn. diff --git a/ptx/sec_def_int.ptx b/ptx/sec_def_int.ptx index b2e5bea50..4d3f02ec3 100644 --- a/ptx/sec_def_int.ptx +++ b/ptx/sec_def_int.ptx @@ -983,8 +983,10 @@ The graph area under the curve is a semicircle with radius 3 on the x axis with centre at origin. - The graph shows the area under the curve that is a semicircle with radius 3 on the x axis - with centre at origin. It lies on the first and the second quadrant. +

    + The graph shows the area under the curve that is a semicircle with radius 3 on the x axis + with centre at origin. It lies on the first and the second quadrant. +

    @@ -1658,16 +1660,16 @@ Graph of function that is a semicircle on the x axis. - - + +

    - The y axis is drawn from 0 to 3 and the x axis is - drawn from 0 to 4. - The graph of function f(x) = \sqrt{4-(x-2)^2} - is a semi circle drawn on the x axis with centre at point - (2,0) and radius of 2. + The y axis is drawn from 0 to 3 and the x axis is + drawn from 0 to 4. + The graph of function f(x) = \sqrt{4-(x-2)^2} + is a semi circle drawn on the x axis with centre at point + (2,0) and radius of 2.

    -     +     \begin{tikzpicture} diff --git a/ptx/sec_deriv_basic_rules.ptx b/ptx/sec_deriv_basic_rules.ptx index 425c41b27..2fa0e35e8 100644 --- a/ptx/sec_deriv_basic_rules.ptx +++ b/ptx/sec_deriv_basic_rules.ptx @@ -812,7 +812,7 @@

    - + diff --git a/ptx/sec_deriv_chainrule.ptx b/ptx/sec_deriv_chainrule.ptx index aeb9fcab1..dbd1aa686 100644 --- a/ptx/sec_deriv_chainrule.ptx +++ b/ptx/sec_deriv_chainrule.ptx @@ -1079,11 +1079,13 @@ 3 gears of various sizes demonstrating the chain rule. - Three gears, connected in the order x,u,y. - x is the largest gear, having 36 teeth. It is rotating counter-clockwise. - u is connected to x, and it has 18 teeth. To the left of the connection is \frac{du}{dx} = 2. - y is connected to u, and it has 6 teeth. Below the connection is \frac{dy}{du}=3. - To the right of the gears is the expression \frac{dy}{dx} = 6. +

    + Three gears, connected in the order x,u,y. + x is the largest gear, having 36 teeth. It is rotating counter-clockwise. + u is connected to x, and it has 18 teeth. To the left of the connection is \frac{du}{dx} = 2. + y is connected to u, and it has 6 teeth. Below the connection is \frac{dy}{du}=3. + To the right of the gears is the expression \frac{dy}{dx} = 6. +

    diff --git a/ptx/sec_deriv_implicit.ptx b/ptx/sec_deriv_implicit.ptx index 25c01e13c..1dbba2e3c 100644 --- a/ptx/sec_deriv_implicit.ptx +++ b/ptx/sec_deriv_implicit.ptx @@ -426,16 +426,18 @@ A curve with two distinct segments and a tangent line with a positive slope - Two curves are drawn in the xy-plane. - The left curve stretches upwards from the left side of the y axis, curving slightly to the left. - As y approaches -2, the curve begins to widen to the left, creating a bump in the curve. - As the curve crosses the x axis, the curve moves towards the right, no longer increasing and becoming more horizontal as x increases. - At the point (0,1), a tangent line is drawn, with a moderate positive slope. - This point corresponds to the corner at which the curve begins to become horizontal. - At this point, the curve passes the vertical line test, but does not at most other points on the graph. - The second curve begins to the right of the y-axis, as a line stretching upwards from the bottom of the y-axis. - As x approaches 1, the curve also begins to become horizontal as x increases. - The entire second curve lies in the fourth quadrant. +

    + Two curves are drawn in the xy-plane. + The left curve stretches upwards from the left side of the y axis, curving slightly to the left. + As y approaches -2, the curve begins to widen to the left, creating a bump in the curve. + As the curve crosses the x axis, the curve moves towards the right, no longer increasing and becoming more horizontal as x increases. + At the point (0,1), a tangent line is drawn, with a moderate positive slope. + This point corresponds to the corner at which the curve begins to become horizontal. + At this point, the curve passes the vertical line test, but does not at most other points on the graph. + The second curve begins to the right of the y-axis, as a line stretching upwards from the bottom of the y-axis. + As x approaches 1, the curve also begins to become horizontal as x increases. + The entire second curve lies in the fourth quadrant. +

    @@ -1995,9 +1997,11 @@ A square with rounded corners and edges with a point in the first quadrant. - A curve that lies in all 4 quadrants. - It has the appearance of a square with rounded sides and corners. - A point is drawn at (\sqrt{0.6},\sqrt{0.8}). +

    + A curve that lies in all 4 quadrants. + It has the appearance of a square with rounded sides and corners. + A point is drawn at (\sqrt{0.6},\sqrt{0.8}). +

    \begin{tikzpicture} @@ -2126,16 +2130,18 @@ An oval with a cusp on the right side. - An oval with a cusp on the right side. - A majority of the curve lies to the left of the y-axis. - From the top of the curve, the curve decreases towards the right. - It enters the first quadrant through the point (0,1). - As the curve nears the x-axis, it bends back toward the y-axis, forming a cusp at the origin. - The curve then bends outwards into the fourth quadrant. - The curve continues downwards and to the left, passing the y-axis through the point (0,-1). - In the third quadrant the curve bends upwards, passing vertically into the second quadrant through the point (-2,0). - The curve bends upwards and to the right, once again meeting the top of the curve. - The point at the top of the curve is drawn at (-\frac{3}{4},\frac{3\sqrt{3}}{4}). +

    + An oval with a cusp on the right side. + A majority of the curve lies to the left of the y-axis. + From the top of the curve, the curve decreases towards the right. + It enters the first quadrant through the point (0,1). + As the curve nears the x-axis, it bends back toward the y-axis, forming a cusp at the origin. + The curve then bends outwards into the fourth quadrant. + The curve continues downwards and to the left, passing the y-axis through the point (0,-1). + In the third quadrant the curve bends upwards, passing vertically into the second quadrant through the point (-2,0). + The curve bends upwards and to the right, once again meeting the top of the curve. + The point at the top of the curve is drawn at (-\frac{3}{4},\frac{3\sqrt{3}}{4}). +

    \begin{tikzpicture} diff --git a/ptx/sec_deriv_prodquot.ptx b/ptx/sec_deriv_prodquot.ptx index 7d310c9a4..330cab68c 100644 --- a/ptx/sec_deriv_prodquot.ptx +++ b/ptx/sec_deriv_prodquot.ptx @@ -918,7 +918,7 @@

    - + diff --git a/ptx/sec_directional_derivative.ptx b/ptx/sec_directional_derivative.ptx index f2aee06b6..2bc8b80ca 100644 --- a/ptx/sec_directional_derivative.ptx +++ b/ptx/sec_directional_derivative.ptx @@ -1304,7 +1304,7 @@ - + perpendicular diff --git a/ptx/sec_newton.ptx b/ptx/sec_newton.ptx index ca241a47e..b94333e79 100644 --- a/ptx/sec_newton.ptx +++ b/ptx/sec_newton.ptx @@ -57,7 +57,7 @@

    A tangent line is drawn from the point (x_0,f(x_0)) to intersect the x axis at x_1, - providing a refined approximation of the function’s root. + providing a refined approximation of the function's root.

    diff --git a/ptx/sec_optimization.ptx b/ptx/sec_optimization.ptx index 8948ed26d..c3eafa040 100644 --- a/ptx/sec_optimization.ptx +++ b/ptx/sec_optimization.ptx @@ -1099,7 +1099,9 @@ A rectangle divided into two parts. - A large rectangle divided into two equal rectangles. +

    + A large rectangle divided into two equal rectangles. +

    \begin{tikzpicture} diff --git a/ptx/sec_substitution.ptx b/ptx/sec_substitution.ptx index 46dc3ba17..e740624be 100644 --- a/ptx/sec_substitution.ptx +++ b/ptx/sec_substitution.ptx @@ -1488,7 +1488,7 @@

    - + diff --git a/ptx/sec_taylor_poly.ptx b/ptx/sec_taylor_poly.ptx index 9915759c6..d9fa63ddf 100644 --- a/ptx/sec_taylor_poly.ptx +++ b/ptx/sec_taylor_poly.ptx @@ -1457,8 +1457,14 @@ 0.0003 of the correct answer.

    -

    - +

    + + shows a graph of y=p_8(x) and y=\cos(x). + Note how well the two functions agree on about (-\pi,\pi). +

    + +

    + shows a graph of y=p_8(x) and y=\cos(x). Note how well the two functions agree on about (-\pi,\pi).

    diff --git a/ptx/sec_vvf.ptx b/ptx/sec_vvf.ptx index 5b7671aaa..50ee1a456 100644 --- a/ptx/sec_vvf.ptx +++ b/ptx/sec_vvf.ptx @@ -62,8 +62,10 @@ - Graph of the vector \vec r(-2) =\la 4,1\ra. - This vector starts at the origin and ends at the point (4,1). +

    + Graph of the vector \vec r(-2) =\la 4,1\ra. + This vector starts at the origin and ends at the point (4,1). +

    Graph of a vector whose terminal point corresponds to a point on the curve. @@ -93,12 +95,14 @@ - The image contains the plot of the vector valued function \vec r(t) = \la t^2,t^2+t-1\ra. - The image also contains the graph of the vector \vec r(-2) =\la 4,1\ra, which ends at a point on the function \vec r(t). - The function \vec r(t) resembles a parabola which has been rotated about 45 degrees clockwise. - The function is plotted for t approximately between -2.5 and 1.5. - The function begins near the point (5,2) and then slopes down crossing the x-axis at approximately x=2.5. - The slanted parabola then curves upwards around the point (0,-1), from which point it begins increasing in both x and y as t increases. +

    + The image contains the plot of the vector valued function \vec r(t) = \la t^2,t^2+t-1\ra. + The image also contains the graph of the vector \vec r(-2) =\la 4,1\ra, which ends at a point on the function \vec r(t). + The function \vec r(t) resembles a parabola which has been rotated about 45 degrees clockwise. + The function is plotted for t approximately between -2.5 and 1.5. + The function begins near the point (5,2) and then slopes down crossing the x-axis at approximately x=2.5. + The slanted parabola then curves upwards around the point (0,-1), from which point it begins increasing in both x and y as t increases. +

    Graph of the vector and vector valued function from the example. @@ -206,13 +210,15 @@ - The image contains the plot of the vector valued function \vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra for -2\leq t\leq 2. - The image also contains the vectors \vec r(-1) =\la 0,\frac12 \ra and \vec r(2) =\la 6,\frac15 \ra, which both end at a point on the function \vec r(t). - The function \vec r(t) begins at the point (-6,\frac15) corresponding to the lower bound of t=-2, from which it slowly slopes upwards until crossing the y-axis at y=\frac12. - From here, the function slopes upwards and slightly outwards away from the y-axis until curving back and crossing the y-axis once again at y=1. - The function then slopes downwards and slightly away from the y-axis until curving back and crossing the y-axis again at y=\frac12, completing a loop. - After crossing the y-axis, the function continues slowly sloping downwards until reaching the point (6,\frac15) which corresponds to the upper bound of t=2. - The function is also symmetric about the y-axis. +

    + The image contains the plot of the vector valued function \vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra for -2\leq t\leq 2. + The image also contains the vectors \vec r(-1) =\la 0,\frac12 \ra and \vec r(2) =\la 6,\frac15 \ra, which both end at a point on the function \vec r(t). + The function \vec r(t) begins at the point (-6,\frac15) corresponding to the lower bound of t=-2, from which it slowly slopes upwards until crossing the y-axis at y=\frac12. + From here, the function slopes upwards and slightly outwards away from the y-axis until curving back and crossing the y-axis once again at y=1. + The function then slopes downwards and slightly away from the y-axis until curving back and crossing the y-axis again at y=\frac12, completing a loop. + After crossing the y-axis, the function continues slowly sloping downwards until reaching the point (6,\frac15) which corresponds to the upper bound of t=2. + The function is also symmetric about the y-axis. +

    Graph of two vectors and vector valued function from the example. @@ -279,13 +285,15 @@ - The image contains the plot of the vector valued function \vec r(t) = \la \cos(t) ,\sin(t) ,t\ra for 0\leq t\leq 4\pi. - The image also contains the vector \vec r(7\pi/4)\approx (0.707,-0.707,5.498), which corresponds to a point on the function \vec r(t). - The function \vec r(t) begins at the point (1,0,0) corresponding to the lower bound of t=0. - Ignoring the z coordinate of the vector valued function, the curve is simply a circle centered at the origin in the xy plane. - The addition of the z coordinate given by t then makes this curve linearly increase at t increases. - Accounting for the z coordinate, the function \vec r(t) resembles a linearly increasing circular spiral, which completes two full revolutions. - The spiral ends at the same x and y coordinates as it started, but is 4\pi units above in the z direction, ending at the point (1,0,4\pi). +

    + The image contains the plot of the vector valued function \vec r(t) = \la \cos(t) ,\sin(t) ,t\ra for 0\leq t\leq 4\pi. + The image also contains the vector \vec r(7\pi/4)\approx (0.707,-0.707,5.498), which corresponds to a point on the function \vec r(t). + The function \vec r(t) begins at the point (1,0,0) corresponding to the lower bound of t=0. + Ignoring the z coordinate of the vector valued function, the curve is simply a circle centered at the origin in the xy plane. + The addition of the z coordinate given by t then makes this curve linearly increase at t increases. + Accounting for the z coordinate, the function \vec r(t) resembles a linearly increasing circular spiral, which completes two full revolutions. + The spiral ends at the same x and y coordinates as it started, but is 4\pi units above in the z direction, ending at the point (1,0,4\pi). +

    Graph of a vector and the three-dimensional vector valued function from the example. @@ -432,9 +440,11 @@ - The image contains the plot of the vector valued functions \vec r_1(t) = \la 0.2t,0.3t \ra, \vec r_2(t) = \la \cos(t) ,\sin(t) \ra on -10\leq t\leq10. - The function \vec r_1(t) = \la 0.2t,0.3t \ra is a line which begins at the point (-2,-3) corresponding to t=-10 and ends when t=10 at the point (2,3). - The second function \vec r_2(t) = \la \cos(t) ,\sin(t) \ra looks like a circle of radius 1, but as -10\leq t\leq10 the function completes \frac{20}{2\pi} full circular rotations, which cannot be seen in the graph. +

    + The image contains the plot of the vector valued functions \vec r_1(t) = \la 0.2t,0.3t \ra, \vec r_2(t) = \la \cos(t) ,\sin(t) \ra on -10\leq t\leq10. + The function \vec r_1(t) = \la 0.2t,0.3t \ra is a line which begins at the point (-2,-3) corresponding to t=-10 and ends when t=10 at the point (2,3). + The second function \vec r_2(t) = \la \cos(t) ,\sin(t) \ra looks like a circle of radius 1, but as -10\leq t\leq10 the function completes \frac{20}{2\pi} full circular rotations, which cannot be seen in the graph. +

    Graph of the two vector valued functions prior to adding them together. @@ -463,13 +473,15 @@ - The image contains the plot of the vector valued function \vec r(t) = \vec r_1(t)+\vec r_2(t). - The function is given by \vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra on -10\leq t\leq10. - The function begins in the third quadrant, near the point (-3,-2). - Near this point, the function is concave up, starting with a downward slope and later beginning to slope upwards. - The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate. - The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant. - The function nearly completes a third loop, before ending near the point (3,1.5). +

    + The image contains the plot of the vector valued function \vec r(t) = \vec r_1(t)+\vec r_2(t). + The function is given by \vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra on -10\leq t\leq10. + The function begins in the third quadrant, near the point (-3,-2). + Near this point, the function is concave up, starting with a downward slope and later beginning to slope upwards. + The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate. + The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant. + The function nearly completes a third loop, before ending near the point (3,1.5). +

    Graph of the vector valued function coming from adding the circle and line vector valued functions. @@ -497,15 +509,17 @@ - The image contains the plot of the vector valued function 5\vec r(t) = 5\vec r_1(t)+5\vec r_2(t). - The function is now given by \vec r(t) = \la\,5\cos(t) + t,5\sin(t) +1.5t\,\ra on -10\leq t\leq10. - Compared to the unscaled function \vec r(t), the function 5\vec r(t) is exactly 5 times larger than the original function. - The function begins in the third quadrant, near the point (-15,-10). - Like the unscaled function, near this point 5\vec r(t) is concave up, starting with a downwards slope and later beginning to slope upwards. - The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate. - The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant. - The function nearly completes a third loop, before ending near the point (15,7.5). - The entirety of the original function \vec r(t) can be seen to fit between the two loops of the scaled function 5\vec r(t). +

    + The image contains the plot of the vector valued function 5\vec r(t) = 5\vec r_1(t)+5\vec r_2(t). + The function is now given by \vec r(t) = \la\,5\cos(t) + t,5\sin(t) +1.5t\,\ra on -10\leq t\leq10. + Compared to the unscaled function \vec r(t), the function 5\vec r(t) is exactly 5 times larger than the original function. + The function begins in the third quadrant, near the point (-15,-10). + Like the unscaled function, near this point 5\vec r(t) is concave up, starting with a downwards slope and later beginning to slope upwards. + The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate. + The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant. + The function nearly completes a third loop, before ending near the point (15,7.5). + The entirety of the original function \vec r(t) can be seen to fit between the two loops of the scaled function 5\vec r(t). +

    Graph of the vector valued function coming from scaling the vector valued function from the example. @@ -564,15 +578,17 @@     Tracing a cycloid - Graph of an arbitrary vector-valued \vec r on the interval which includes [t_0,t_1]. - The function \vec r is a small part of a concave down circular arc. - The graph includes the vectors \vec r (t_0) and \vec r (t_1), which begin from the origin and end at the corresponding point of the function \vec r . - The graph also includes the vector \vec r (t_1) - \vec r (t_0), which begins where \vec r (t_0) ends, and then terminates at the same termination point as \vec r (t_1). - The three vectors \vec r (t_0), \vec r (t_1) and \vec r (t_1) - \vec r (t_0) for a triangle, where following the path of \vec r (t_0) and \vec r (t_1) - \vec r (t_0) takes you to the same point as \vec r (t_1). - The vector \vec r (t_0) terminates on the left side of the circular arc, while \vec r (t_1) terminates further on the right side of the circular arc of given by the function \vec r. +

    + Graph of an arbitrary vector-valued \vec r on the interval which includes [t_0,t_1]. + The function \vec r is a small part of a concave down circular arc. + The graph includes the vectors \vec r (t_0) and \vec r (t_1), which begin from the origin and end at the corresponding point of the function \vec r . + The graph also includes the vector \vec r (t_1) - \vec r (t_0), which begins where \vec r (t_0) ends, and then terminates at the same termination point as \vec r (t_1). + The three vectors \vec r (t_0), \vec r (t_1) and \vec r (t_1) - \vec r (t_0) for a triangle, where following the path of \vec r (t_0) and \vec r (t_1) - \vec r (t_0) takes you to the same point as \vec r (t_1). + The vector \vec r (t_0) terminates on the left side of the circular arc, while \vec r (t_1) terminates further on the right side of the circular arc of given by the function \vec r. +

    Illustration of a vector-valued function on a given interval. @@ -258,11 +260,13 @@     - Graph of the same arbitrary vector-valued \vec r as well as the vectors \vec r (t_0) and \vec r (t_1) as described in the previous image. - This time the graph includes two additional vectors \vrp (t_0) and \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} . - The vector \vrp (t_0) begins at the termination point of \vec r (t_0), and is tangent to the function \vec r at this point. - The vector \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} also begins at the termination point of \vec r (t_0), and follows the path of the vector \vec r (t_1) - \vec r (t_0) from the previous image. - However, this vector does not end at the termination point of \vec r (t_1) but instead terminates at some point further away in the same direction as the vector \vec r (t_1) - \vec r (t_0). +

    + Graph of the same arbitrary vector-valued \vec r as well as the vectors \vec r (t_0) and \vec r (t_1) as described in the previous image. + This time the graph includes two additional vectors \vrp (t_0) and \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} . + The vector \vrp (t_0) begins at the termination point of \vec r (t_0), and is tangent to the function \vec r at this point. + The vector \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} also begins at the termination point of \vec r (t_0), and follows the path of the vector \vec r (t_1) - \vec r (t_0) from the previous image. + However, this vector does not end at the termination point of \vec r (t_1) but instead terminates at some point further away in the same direction as the vector \vec r (t_1) - \vec r (t_0). +

    Illustration of a vector-valued function on a given interval showcasing a derivative vector. @@ -459,10 +463,12 @@ - Graph of the vector-valued function \vec r(t) = \la t^2,t\ra. - The graph of the function \vec r(t) is simply the graph of the parabola y=x^2, but instead of opening towards the positive y-axis, the function \vec r(t) opens towards the positive x-axis. - The function \vec r(t) is plotted on the interval [-2,2], where \vec r(-2)= \la 4,-2\ra and \vec r(2)= \la 4,2\ra. - The graph also includes the derivative function \vrp(t) = \la 2t, 1\ra which takes the path of the horizontal line y=1 going from left to right in the standard coordinate axes. +

    + Graph of the vector-valued function \vec r(t) = \la t^2,t\ra. + The graph of the function \vec r(t) is simply the graph of the parabola y=x^2, but instead of opening towards the positive y-axis, the function \vec r(t) opens towards the positive x-axis. + The function \vec r(t) is plotted on the interval [-2,2], where \vec r(-2)= \la 4,-2\ra and \vec r(2)= \la 4,2\ra. + The graph also includes the derivative function \vrp(t) = \la 2t, 1\ra which takes the path of the horizontal line y=1 going from left to right in the standard coordinate axes. +

    Graph of the vector-valued function from the example and its derivative. @@ -497,11 +503,13 @@ - Graph of the vector-valued function \vec r(t) = \la t^2,t\ra described in the previous image. - The graph also includes two copies of the vector \vrp(1) = \la 2,1\ra. - The first copy of the vector \vrp(1) = \la 2,1\ra begins at the origin, and ends at the point (2,1), which is also a point on the derivative function \vrp(t) = \la 2t, 1\ra from the previous image. - The second copy of the vector \vrp(1) = \la 2,1\ra begins at the point (1,1), which corresponds to the termination point of \vec r(1) = \la 1,1 \ra. - The second copy of the vector \vrp(1) = \la 2,1\ra is tangent to the function \vec r(t) = \la t^2,t\ra at the point (1,1) corresponding to when t=1 in the function \vec r(t). +

    + Graph of the vector-valued function \vec r(t) = \la t^2,t\ra described in the previous image. + The graph also includes two copies of the vector \vrp(1) = \la 2,1\ra. + The first copy of the vector \vrp(1) = \la 2,1\ra begins at the origin, and ends at the point (2,1), which is also a point on the derivative function \vrp(t) = \la 2t, 1\ra from the previous image. + The second copy of the vector \vrp(1) = \la 2,1\ra begins at the point (1,1), which corresponds to the termination point of \vec r(1) = \la 1,1 \ra. + The second copy of the vector \vrp(1) = \la 2,1\ra is tangent to the function \vec r(t) = \la t^2,t\ra at the point (1,1) corresponding to when t=1 in the function \vec r(t). +

    Graph of the vector-valued function from the example and its derivative. @@ -562,14 +570,16 @@     Viewing a vector-valued function and its derivative at one point Illustrating the important aspects of stretching a spring in computing work in